ADVANCED ENGINEERING MATHEMATICSdl.booktolearn.com/ebooks2/science/mathematics/...ADVANCED...

ADVANCEDENGINEERINGMATHEMATICSwith Modeling Applications

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CRC Press is an imprint of theTaylor & Francis Group, an informa business

Boca Raton London New York

ADVANCEDENGINEERINGMATHEMATICS

S. Graham Kelly

with Modeling Applications

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CRC PressTaylor & Francis Group6000 Broken Sound Parkway NW, Suite 300Boca Raton, FL 33487-2742

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Dedication

In memory of my parents

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ContentsPreface ......................................................................................................................xiAuthor ................................................................................................................... xiii

Chapter 1 Foundations of mathematical modeling .....................................11.1 Engineering analysis ....................................................................................11.2 Conservation laws and mathematical modeling .....................................61.3 Problem formulation .................................................................................. 101.4 Nondimensionalization ............................................................................. 271.5 Scaling ..........................................................................................................45Problems .................................................................................................................56

Chapter 2 Linear algebra .................................................................................652.1 Introduction.................................................................................................652.2 Three-dimensional space ..........................................................................662.3 Vector spaces ............................................................................................... 732.4 Linear independence .................................................................................772.5 Basis and dimension ..................................................................................802.6 Inner products.............................................................................................862.7 Norms ........................................................................................................... 932.8 Gram-Schmidt orthonormalization .........................................................982.9 Orthogonal expansions ........................................................................... 1052.10 Linear operators ........................................................................................ 1092.11 Adjoint operators ...................................................................................... 1212.12 Positive defi nite operators ....................................................................... 1292.13 Energy inner products ............................................................................. 132Problems ............................................................................................................... 139

Chapter 3 Ordinary differential equations ...............................................1433.1 Linear differential equations .................................................................. 1433.2 General theory for second-order differential equations ..................... 1463.3 Differential equations with constant coeffi cients ................................ 1493.4 Differential equations with variable coeffi cients ................................. 1673.5 Singular points of second-order equations .......................................... 173

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viii Contents

3.6 Bessel functions ........................................................................................ 1803.7 Differential equations whose solutions are expressible

in terms of Bessel functions .................................................................... 1933.8 Legendre functions .................................................................................. 198Problems ............................................................................................................... 201

Chapter 4 Variational methods ....................................................................2094.1 Introduction............................................................................................... 2094.2 The general variational problem ............................................................ 2104.3 Variational solutions of operator equations.......................................... 224

4.3.1 Method of least squares ............................................................... 2244.3.2 Rayleigh-Ritz method ..................................................................226

4.4 Finite-element method ............................................................................. 2424.5 Galerkin’s method .................................................................................... 270Problems ............................................................................................................... 273

Chapter 5 Eigenvalue problems ...................................................................2775.1 Eigenvalue and eigenvector problems ...................................................2775.2 Eigenvalues of adjoint operators ............................................................ 2785.3 Eigenvalues of positive defi nite operators ............................................ 2795.4 Eigenvalue problems for operators in fi nite-dimensional

vector spaces ..............................................................................................2805.5 Second-order differential operators ....................................................... 2875.6 Eigenvector expansions ...........................................................................303

5.6.1 Completeness theorem .................................................................3035.6.2 Trigonometric fourier series ........................................................3095.6.3 Completeness of eigenvectors for self-adjoint and

non-self-adjoint operators ............................................................ 3125.6.4 Solution of nonhomogeneous equations using

eigenvector expansions ................................................................ 3145.7 Fourth-order differential operators ....................................................... 3185.8 Differential operators with eigenvalues in

boundary conditions ................................................................................ 3295.9 Eigenvalue problems involving Bessel functions ................................3365.10 Eigenvalue problems in other infi nite-dimensional

vector spaces ..............................................................................................3495.11 Solvability conditions ..............................................................................3545.12 Asymptotic approximations to solutions of

eigenvalue problems ................................................................................ 3585.13 Rayleigh’s quotient ................................................................................... 3705.14 Rayleigh-Ritz method .............................................................................. 3765.15 Green’s functions ...................................................................................... 394Problems ...............................................................................................................408

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Contents ix

Chapter 6 Partial differential equations ....................................................4196.1 Homogeneous partial differential equations ....................................... 4196.2 Second-order steady-state problems, Laplace’s equation ................... 4216.3 Time-dependent problems: Initial value problems ............................. 4416.4 Nonhomogeneous partial differential equations ................................4556.5 Problems in cylindrical coordinates ...................................................... 4716.6 Problems in spherical coordinates ......................................................... 491Problems ...............................................................................................................508

Index ..................................................................................................................... 517

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PrefaceThis book springs from class notes, I developed for a course called Engineer-ing Analysis which I have taught every other fall semester since 1983 at the University of Akron. The course is targeted to students who are beginning graduate study in engineering. The students enrolled are fi rst- and second-year graduate students in the Department of Mechanical Engineering, although I have taught students from other engineering disciplines.

At the beginning of the fi rst class, I tell my students that I have two objec-tives in teaching the class. The fi rst is to prepare them for subsequent gradu-ate courses in engineering by teaching mathematical methods that are used in major graduate classes. The second objective is to prepare students to read engineering literature, such as journals and monographs. Students engaged in thesis and dissertation research need a foundation in mathematical termi-nology and methods to understand previous work in their research area. The title of the course, Engineering Analysis, is vague, but it essentially means “Advanced Engineering Mathematics with Applications.” The course content and the content of this book is exactly that. Contrary to many engineering mathematics courses, the applications are emphasized as well as the physics behind the applications.

The applications are directed toward problems encountered in graduate engineering classes as well as in emerging areas of practice. An understanding of the modeling methods used to derive the mathematical equations as well as the underlying physics of the problem is usually essential to developing a method to solve the mathematical equations. For this reason, issues of modeling and scaling are discussed along with the analysis methods.

The motivation for the course and this book is to provide students and readers an experience of the marriage of engineering and applied mathemat-ics. As in a marriage, both are equal and complement one another.

A unique feature of this book is the foundation laid for study. Books and courses on real analysis and functional analysis concentrate on theory rather than applications. Graduate engineering courses in areas, such as vibration, stress analysis, fl uid mechanics, and heat transfer, concentrate on applications and use mathematical methods such as eigenvalue analysis and separation of variables as tools without much explanation of the mathematical theory underlying why they can be applied. An engineering student should under-stand the underlying mathematics that explains why these methods work,

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xii Preface

when they can be applied, and what are their limitations, but does not need to understand how to prove every theorem. This book takes such a view.

An underlying foundation is developed using the language of vector spaces and linear algebra. Basic results are derived, such as the existence of energy inner products for self-adjoint operators, the eigenvector expansion theorem, and the Fredholm alternative, and applied to problems for discrete and continuous systems. Yes, the differences between fi nite and infi nite dimensional spaces are addressed, especially the issues of convergence. For this study, a discussion of completeness of eigenvectors of self-adjoint opera-tors and heuristic proofs of convergence with respect to energy norms is suffi cient, and only a limited discussion of pointwise convergence of the trig-onometric Fourier and Fourier-Bessel series is presented. Because the focus is on applied mathematics for beginning graduate students, topics such as continuous spectra of eigenvalues are omitted.

This book is different from other advanced engineering mathematics books in many ways, some of which are listed below:

Applications are presented to provide motivation for the mathematics, whereas most existing books illustrate applications after developing the mathematics.Linear algebra is used to provide a foundation for analysis of discrete and continuous systems.Rigor is used in development of concepts and is used in proving theo-rems for which the proofs themselves are instructive.The view is taken that a general understanding of theory is necessary to develop applications.Applications from emerging technologies are presented.

Theorems and proofs are presented, but without the detail found in many mathematics books. It is not intended to have the development of the theory obscure its application to engineering problems. On the other hand, a full understanding of the solution is not possible unless the theory from which it is developed is understood.

Acknowledgement is due to students who have taken Engineering Analy-sis over the years. Their questions and suggestions helped refi ne the book. I gratefully acknowledge my wife, Seala Fletcher-Kelly, not just for her support, but for signifi cant help in preparing the fi gures. I am grateful to B.J. Clark, for-merly of Taylor & Francis, for his efforts in developing the project, as well as to Michael Slaughter and Jonathan Plant at CRC Press for their continued support during the project. I would also like to express appreciation for the work of Amber Donley, project coordinator, at CRC and Glenon Butler, project editor at Taylor & Francis and others at CRC and Taylor & Francis who aided in the publication of this work.

S. Graham Kelly

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•

•

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AuthorS. Graham Kelly received a BS in Engineering Science and Mechanics from Virginia Tech and a MS and PhD in Engineering Mechanics from Virginia Tech in 1977 and 1979, respectively. He served on the faculty at the University of Notre Dame from 1979 to 1982 and at The University of Akron since, in addi-tion to his academic work, Dr. Kelly served in several administrative positions, including associate provost and dean of Engineering. Dr. Kelly is the author of three previous texts, Fundamentals of Mechanical Vibrations, System Dynamics and Response, and Advanced Vibration Analysis as well as a Schaum’s Outline of Mechanical Vibration. He is a member of the American Society of Mechani-cal Engineers, the American Physical Society, the Society for Industrial and Applied Mathematics, the Society for Engineering Science, and the American Institute for Astronautics and Aeronautics. In addition to writing projects, Dr. Kelly currently enjoys teaching, research, working with students, and writing at The University of Akron.

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1

Chapter 1

Foundations of mathematical modeling

1.1 Engineering analysisEngineering analysis, mathematical modeling, engineering mathematics, and applied mathematics are all related terms. Let us start with engineering analysis, which can simply be defi ned as the analysis used by engineers dur-ing design and in applications. The adjective “engineering” implies that a practical reason exists for the analysis and that physical sciences are used in problem development. However the term “analysis” is vague as to the method of analysis used. Engineers use a variety of analytical techniques, including empirical methods, statistical methods, graphical methods, numerical and computational methods, and mathematical methods. “Engineering analy-sis” can refer to the application of any of these methods.

Mathematical modeling is a procedure in which a system of mathematical equations is developed to simulate the behavior of a physical system. The procedure usually includes a method to solve the resulting equations. Engi-neering mathematics refers to the mathematical tools that engineers use to solve the equations derived during modeling. Engineering mathematics is a subset of applied mathematics.

This study focuses on a subset of “engineering analysis,” engineering mathematics and its interrelationship with mathematical modeling. Indeed, the two are intertwined as the knowledge of the available mathematical tools may drive the modeling process, and the engineering knowledge may drive the solution process. In this context, engineering analysis begins with mathematical modeling of an engineering system. The modeling leads to the formulation of a mathematical problem. The mathematical problem is non-dimensionalized (Section 1.4) to understand the scaling of physical parame-ters and to suggest solution techniques. A mathematical method is applied to obtain the solution of the mathematical problem. The mathematical solution is then used to solve the engineering problem.

The above is a simplifi ed overview of engineering analysis. Perhaps an alternate defi nition is the rational application of basic laws of nature and con-stitutive equations or equations of state to derive a mathematical model for an engineering system and then to apply the methods of applied mathematics.

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2 Advanced engineering mathematics with modeling applications

Both the mathematical modeling and the resulting solution pose challenges. Often these are not independent.

Knowledge of the principles and techniques of applied mathematics can aid in the development of a mathematical model of a physical problem, and often the mathematical solution is guided by the physics. Physics and mathe-matics are married. It is no coincidence that the discoverers of the basic principles of physics, such as Sir Isaac Newton, are also the developers of modern mathematical analysis.

One cannot formulate an exact statement of how to develop a mathematical model. While generalities can be examined, the exact nature of mathe matical modeling is discipline-specifi c. The fi rst step is to identify the problem. Prob-lem identifi cation often requires abstraction of the system to be modeled from larger systems. Problem identifi cation also requires specifi cation of a goal. Consider, for example, the heating of the sphere shown in Figure 1.1. One goal of the modeling may be simply to analyze the unsteady state tem-perature at the center of the sphere or to determine the time required for this temperature to reach a certain value. Another possible goal is to determine a material such that the temperature at the center reaches a certain value in a specifi ed time. Still another goal is to determine how the sphere can be heated such that the minimum energy is used for the center of the sphere to reach the specifi ed temperature. Thus possible objectives of mathematical modeling include analysis, design, and optimization.

The abstraction of the system also leads to identifi cation of system inputs (what is feeding the system, usually assumed to be specifi ed) and system outputs (what is to be determined). Independent variables, the variables on which the output depends, are identifi ed. Time and spatial coordinates referred to a fi xed frame are common examples of independent variables. Dependent variables are the variables which specify the output.

R

k,ρ,c

h, T 8

Figure 1.1 An objective of the mathematical modeling of heat transfer processes leading to the determination of the temperature distribution in the sphere may be analysis, design, or synthesis.

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Chapter 1: Foundations of mathematical modeling 3

A necessary step in problem identifi cation is to specify assumptions or identify possible assumptions. It may not be possible to list all reasonable assumptions until a preliminary model is completed. Assumptions can be modifi ed during the modeling process. Scaling issues may lead to modifi ca-tion of assumptions.

Implicit assumptions are those that are taken for granted and rarely stated. Examples include that the earth is a Newtonian reference frame, that the acceleration of gravity on the surface of the earth is 9.81 m/s2, and that nuclear reactions are not occurring within the system. It should be explicitly stated if a particular implicit assumption is not appropriate.

Implicit assumptions are often questioned in extreme cases. A common implicit assumption is that a continuum model can be used. This assumption implies that properties of matter are continuous. However, as engineering problems emerge at the nanoscale, the validity of the continuum assumption at this scale has been questioned. The radius of a carbon atom is approximately 0.34 nm. Can the continuum assumption be used to model nanowires and nanotubes? Answers lie in empirical results and scaling. That is, the continuum assumption is apparently valid if its application leads to results that compare favorably with empirical results. In addition, the continuum model is valid if length scales are greater than those associated with the vibration of molecules.

The alternative to the continuum assumption is to assume that each atom is a discrete particle and to apply conservation laws to each particle. A limitation to such an approach is that it leads to large numbers of simultaneous equations whose solution requires long computational times. In addition, the motion of individual atoms is not as deterministic as in continuous systems. Often Monte Carlo methods are used for molecular dynamics simulations.

Explicit assumptions are those that are specifi c to a particular model-ing problem. Explicit assumptions are made to simplify the modeling or the subsequent mathematical solution of the resulting model. Many explicit assumptions are made for convenience. For example, when aerodynamic drag and other forms of damping are assumed to be small, modeling of the system shown in Figure 1.2 leads to a linear mathematical model. However, the resulting model is not physically practical because it leads to prediction of perpetual motion for any initial energy input.

Assumptions affect the identifi cation of independent and dependent variables. Four independent variables can be identifi ed for the modeling of the temperature distribution of the sphere shown in Figure 1.1. The sphere is heated until the temperature of the center reaches a certain value. The temperature clearly changes with time. The temperature of a particle in the sphere depends on its location in the sphere, which is described by three inde-pendent spatial coordinates: a radial coordinate r, a circumferential angular coordinate θ, and an azimuthal angular coordinate φ. Thus T = T(r,θ,φ,t). If an assumption is made that the sphere is uniformly heated over its surface and that the sphere is at uniform temperature before heating, then the tempera-ture is independent of θ and φ, and T = T(r,t).

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Five dependent variables identifi ed for modeling of a fl uid fl ow problem are the density ρ, the pressure p, and three components of the velocity vector v. When the fl ow is assumed to be incompressible, the density is constant, reducing the number of independent variables to four. If the system involves fl ow in a circular pipe, the velocity vector is often assumed to be directed along the axis of the pipe, reducing the number of dependent variables by two. In addition, if the pipe is of uniform diameter, then the axial component of velocity depends only on the radial coordinate and time.

Assumptions are often made regarding material behavior. Solids are usu-ally assumed to be linearly elastic, air is assumed to be an ideal gas, ther-mal properties are assumed constant with temperature, and liquids are assumed incompressible. Specifi cations of material behavior may be in the form of a constitutive equation such as Hooke’s law or Newton’s viscosity law, an equation of state such as the ideal gas law, or an empirical law such as Newton’s law of cooling. These laws introduce parameters into the model. Common examples of parameters include spring stiffness, viscous damping coeffi cient, thermal conductivity, and heat transfer coeffi cient. Less common parameters include Poisson’s ratio and coeffi cient of thermal expansion.

Additional parameters are introduced from geometry. These include lengths and angular measures. Other parameters are introduced through sys-tem inputs such as velocity of a free-stream fl ow or frequency and amplitude of a harmonic input.

Once the problem has been identifi ed, with independent variables, depen-dent variables, and parameters specifi ed, the basic conservation laws which govern the system are identifi ed. Conservation laws can be applied macro-scopically to the entire system or a fi nite part of the system, or microscopically to an infi nitesimal volume in the system. Their application usually requires an illustrative diagram. The result of the application of basic conservation laws and perhaps some algebraic manipulation leads to the mathematical model for the system.

x

k

m

Figure 1.2 An assumption of no friction, including sliding friction and aerody-namic drag, is often used to linearize the system. While the resulting model does provide valuable information regarding the system, it leads to predicting physically impossible perpetual motion.

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The mathematics used to solve the mathematical model is the focus of the remaining chapters. The purpose of the mathematics is to provide a repre-sentation of the dependent variables in terms of the independent variables and the parameters which can be used to meet the goals of the modeling.

Mathematics and physics are interrelated. Physical explanations exist for mathematical paradoxes, and vice versa. Some examples of interaction between physics and mathematics are listed below:

The partial differential equation governing the solution of the tempera-ture distribution of the sphere shown in Figure 1.1 is second-order in the radial coordinate r. A second-order differential equation usually requires application of two boundary conditions to determine a unique solution; however, the only boundary condition formulated is r = R, the outer radius of the sphere. It is shown in Chapter 5 that a second boundary condition is not necessary due to the form of the differential equation, but the physics requires that the solution be fi nite everywhere in the sphere.Rayleigh-Ritz and fi nite-element approximations minimize the dif-ference in energies between the exact solution and the approximate solution.The trigonometric Fourier series which is used to represent periodic functions in many applications has a theoretical development which is a result of the solution of an eigenvalue problem.Green’s functions are derived as the response of a system to a singular input using variation of parameters.Solvability conditions for the existence of solutions of nonhomoge-neous problems when a nontrivial solution to the corresponding homo-geneous problem exists have physical interpretations.Eigenvalues equal to zero correspond to rigid-body motion in dynamic systems.

The focus of this study is on the interaction of the modeling process and the mathematical solution of the resulting equations. Applications are used from all fi elds of engineering, but mainly from mechanical engineering. Two applications are emphasized.

The temperature distribution in a solid body is modeled through appli-cation of conservation of energy to a differential volume within the body. The resulting mathematical problem depends on the assumptions used in the modeling. If the temperature distribution changes with time, a parabolic partial differential equation, the diffusion equation, is obtained. If the tem-perature is independent of time, an elliptical partial differential equation, Laplace’s equation, is obtained. Assumptions may also reduce the number of spatial variables on which the temperature depends. A steady-state one-dimensional problem is governed by an ordinary differential equation. This differential equation has constant coeffi cients if the area over which conduc-tion heat transfer occurs is constant, and has variable coeffi cients if the area

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•

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varies over the range of the dependent variable. Heat transfer problems can be formulated for bars, cylinders, spheres, and other common volumes. In all problems, the differential equations are second-order in spatial variables and fi rst-order in time. Such problems and their variations are considered throughout the study.

Vibrations of a beam are governed by a partial differential equation which is fourth-order in a spatial coordinate and second-order in time. A normal-mode solution is assumed, to determine the free response leading to a fourth-order ordinary differential equation. A nontrivial solution of the equation occurs only for certain values of the normal-mode parameter, the natural frequency. The problem is equivalent to an eigenvalue problem in which the natural frequencies are the square roots of the eigenvalues and the mode shapes are the corresponding eigenvectors. The mode-shape vec-tors satisfy an orthogonality condition. Variations of the basic beam problem studied include stretched beams, beams with transverse loading, beams on an elastic foundation, and sets of elastically connected beams.

Heat transfer problems and vibration problems describe different physi-cal phenomena, but the underlying mathematics used in solving the prob-lems is similar. The normal-mode solution used in the vibration problem is a form of separation of variables, which is the method used to solve the heat transfer problem. Product solutions in the heat transfer problem satisfy orthogonality conditions similar to those satisfi ed by the mode shapes in the vibration problem. In both cases, nonhomogeneous problems are solved using expansions in terms of homogeneous solutions. Such expansions are called eigenvector expansions or Fourier series expansions.

The remainder of this chapter focuses on mathematical modeling, includ-ing nondimensionalization and scaling. The theory of vector spaces and lin-ear operators, presented in Chapter 2, provides a mathematical foundation on which solution methods can be based. The general theory of ordinary dif-ferential equations and methods of solution are presented in Chapter 3. Vari-ational methods, approximate solution techniques based on energy methods, including the fundamentals of the fi nite-element method, are developed in Chapter 4. A comprehensive theory of eigenvalue problems for discrete and continuous systems is presented in Chapter 5. The study concludes in Chapter 6 with a study of partial differential equations.

1.2 Conservation laws and mathematical modelingA fi nite number of basic conservation laws are used in the mathematical modeling of physical problems. Listed by name only, they include:

Conservation of massConservation of linear momentumConservation of angular momentumConservation of energy

••••

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The second law of thermodynamicsMaxwell’s equations

Each law has a specifi c meaning, but its manifestation may be different for different problems depending on the assumptions, the viewpoint, the mate-rials used, and whether it is applied macroscopically or microscopically.

There are two viewpoints used in mathematical modeling of dynamic systems. For both viewpoints, imagine yourself as an observer viewing the system. Using the Lagrangian viewpoint, you fi x yourself to a particle and observe the motion of the particle and other properties as you travel with the particle. The representation of the motion of the particle is in terms of time and the initial position of the particle. The Lagrangian viewpoint is best used for systems involving motion of discrete particles or for systems in which a mass can be identifi ed and tracked. The motion of a single particle or rigid body is modeled using the Lagrangian formulation. As illustrated in Figure 1.3, the position vector for a particle as the particle moves within a Cartesian reference frame is tracked as a function of time. The Lagrangian viewpoint is also used to model the time-dependent motion of the simply supported beam shown in Figure 1.4. Let x be the distance from the fi xed end to a parti-cle along the beam’s neutral axis when the beam is undefl ected. The resulting displacement of the particle is a function of time. The defl ection of the beam is described by w(x,t), where x represents the initial position of a particle.

••

x

y

z

r = xi + yj + zk

Figure 1.3 Mathematical modeling using the Lagrangian viewpoint is based on tracking the position vector for a particle as it changes with time. The independent variables used in a Lagrangian formulation are the initial position of the particle and time.

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Using the Eulerian viewpoint, you fi x yourself to a point in space and observe the time-dependent properties of the material surrounding you as particles of mass move past you. The Eulerian or fi eld viewpoint is used in modeling systems in which a transport process occurs, such as the fl ow of a fl uid. As you observe from a fi xed point in space, particles of matter move by you with a velocity v. You observe changes in properties such as temperature caused by changes occurring locally in the fi eld, as well as changes occurring because of changes in temperature of the fl uid particles that pass by you. The total rate of change of temperature is the sum of the local rate of change, ∂T/∂t, and the convective rate of change, v ⋅ ∇T. This form of the rate of change is called a material derivative or a substantial derivative and is written as:

DTDt

Tt

T=∂∂

+ ⋅∇v (1.1)

A system is a macroscopic version of the Lagrangian viewpoint in which conservation laws are applied to a fi xed set of matter. A control volume is the macroscopic version of the Eulerian viewpoint in which conservation laws are applied to a defi ned region in space.

There are several techniques used for mathematical modeling. One method is to begin with a set of equations which have been formulated from the basic conservation laws and, using appropriate assumptions, to apply the conservation laws to the given system. This is often the modeling technique used in transport problems, especially the transport of momentum. The Navier-Stokes equations are a set of partial differential equations governing the fl ow of a Newtonian fl uid and are formulated using the Eulerian view-point. Written in vector form, they are

∂∂

+ ⋅ + ⋅ =ρt

∇ρ ρ∇v v 0 (1.2)

ρ ρ ∇ ∇ μ∇∂∂

+ ⋅( ) = − +v

tv v F vb p 2 (1.3)

x

w(x,t)

Figure 1.4 The Lagrangian viewpoint is used for modeling the vibrations of a beam. The vibrations are described using time and x, the initial distance of a particle along the neutral axis of the beam from its left end, as independent variables.

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Equation 1.2 is called the continuity equation and is a statement of conserva-tion of mass. Equation 1.3 is the vector form of conservation of linear momen-tum. The dependent variables in Equation 1.2 and Equation 1.3 are ρ, the mass density of the fl uid, v, the three-dimensional velocity vector, and p, the pressure. The body forces are represented by Fb, and μ is the dynamic coef-fi cient of viscosity. There are fi ve dependent variables, but Equation 1.2 and Equation 1.3 provide only four scalar equations. If the fl ow is incompressible, then conservation of mass implies that ∇ ⋅ v = 0, and Equation 1.2 implies that the density is constant. In other cases, an equation of state relating pressure and density is used along with Equation 1.2 and Equation 1.3. Equations of state often involve temperature as a variable. If temperature varies, then an energy equation is used in conjunction with Equation 1.2 and Equation 1.3.

A second method of modeling is to apply basic forms of the conservation laws directly to the system. If time is an independent variable in the problem formulation, the conservation laws are applied at an arbitrary instant. In all cases, the conservation laws are applied for arbitrary values of the depen-dent variables. Two of the basic laws are conservation of momentum and conservation of energy.

Conservation of momentum is often applied to a free-body diagram of an infi nitesimal particle drawn at an arbitrary instant. Conservation of momen-tum, Newton’s second law as applied to a particle, is formulated as:

F a∑ = m (1.4)

where ∑ F represents the resultant force acting on a free-body diagram of the particle drawn at an arbitrary instant and a is the acceleration vector. Conser-vation of angular momentum is an appropriate form of the moment equation.

Conservation of energy as applied to a control volume takes the form that the rate of energy accumulation within the control volume is equal to the rate at which energy is transferred into the control volume through its boundaries.

A complete mathematical formulation often requires that the basic conser-vation laws be supplemented by laws specifi c to the system being modeled. For example, conservation of energy can be supplemented by Newton’s law of cooling and Fourier’s conduction law. The Navier-Stokes equations are written for a viscous fl uid which satisfi es Newton’s viscosity law. Constitutive equations are used to represent stress-strain relations for many solids.

The method of application of mathematical modeling is different for differ-ent types of systems, yet the concepts are the same for all systems. The prob-lem is fi rst identifi ed. Basic questions are answered. Why is a mathematical model necessary? What level of abstraction is required? What assumptions are necessary for successful modeling? What are the independent and depen-dent variables? What parameters are used in the modeling? Are numerical values of these parameters necessary? Based on answers to these questions, the problem is refi ned. Basic laws of physics, in some form, are applied to the

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abstracted system. Empirical laws specifi c to the system are applied as well as geometrical constraints.

Finally, a set of equations is developed. Before a solution method is applied, an assessment is necessary. Will solving the equations enable the objectives of the modeling to be met? Can additional physical insights be gained from examining the equations? Can these physical insights be used to aid in the solution of the equations? Answers to the previous two questions require that the mathematical equations be written in a nondimensional form.

A solution method is chosen and applied, resulting in mathematical equations defi ning the state of the system in terms of the independent vari-ables. Another assessment is then required. Can these equations be used to meet the objectives of the modeling? Does the solution contradict any of the assumptions? If all appears to be well with the solution, it is then used to satisfy the objectives of the modeling.

1.3 Problem formulationThis section provides examples of mathematical modeling of physical sys-tems. The differential equations are derived, but the methods of solution are found in later chapters.

Example 1.1 A fi n or extended surface, as illustrated in Figure 1.5, is attached to a body to enhance the rate of heat transfer to or from the body. The base of the fi n is maintained at the surface temperature, while its tip is insulated. The surface of the fi n is surrounded by an ambient medium which is at a constant temperature T∞. The convective heat transfer coeffi cient between the fi n and the ambient is h. The fi n is made of a material of thermal con-ductivity k. The thickness of the fi n is small enough that conduction can be assumed negligible across the thickness of the fi n and a one-dimensional model can be used.

A differential approach is used to derive a differential equation governing the steady-state temperature distribution in the fi n. Apply the model to (a) a straight fi n of constant thickness; (b) an annular fi n of constant thickness; and (c) a straight fi n of triangular profi le. In each case, show how to deter-mine the total heat transfer at the base of the fi n.

Solution The approach used is to apply conservation of energy to a control volume of a differential element drawn at an arbitrary instant. The same basic model can be used for a straight fi n and an annular fi n. First, consider a straight fi n. Let x represent the perpendicular distance from the base of the fi n to a point on the fi n. For a straight fi n, assume that heat conduction occurs only in the x-direction. Let b(x) represent the thickness of the fi n, and let w be the width of the fi n at its base.

Consider a differential slice of the fi n as illustrated in Figure 1.6. Heat is transferred into the slice at a distance x from the wall by conduction at a rate of qA, where q is the rate of conductive heat transfer per unit area and A is the area of the face of the slice. Heat is transferred from the slice by conduction at

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a distance x + dx at a rate of (qA)x + dx = (qA)x + [d(qA)x/dx] dx. Heat is transferred from the slice by convection at its surface. Let P be the perimeter of the surface and q the rate of convective heat transfer per unit area. The total rate of heat transfer from the surface of the element by convection is therefore qPdx.

b

w

Figure 1.5 The fi n of Example 1.1 is attached to a body to enhance the rate of heat transfer to or from the body. A mathematical model is formulated for the tempera-ture distribution in the fi n and the rate of heat transfer from its base.

dx

b

w

(a) qc

(qA)x+dx(qA)

x

(b)

Figure 1.6 (a) Conservation of energy is applied to a differential slice of the fi n. The temperature is assumed to vary only along the axis of the fi n, so the area of the element is constant over the length of the fi n. (b) Graphical illustration of the energy balance shows heat transfer by conduction into and out of the element through its internal faces and heat transfer by convection out of the element over its perimeter.

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The steady-state energy balance for the slice is based on the concept that the rate at which heat is transferred into the slice is equal to the rate at which heat is transferred out of the slice. Applying this concept to the slice shown in Figure 1.6 leads to

qA qAd

dxqA dx qPdxx x x( ) = +⎡

⎣⎢⎢

⎤⎦⎥⎥+( ) ( ) ˆ (a)

Equation a can be simplifi ed to

d

dxqA qP( ) ˆ+ = 0 (b)

The formulation is completed through application of two empirical laws which are assumed to apply to this system. The rate of heat transfer by con-duction is given by Fourier’s conduction law, q = − k (dT / dx), while the rate of heat transfer by convection is give by Newton’s law of cooling, q = h(T − T∞). Substituting these laws into Equation b leads to:

d

dxkA

dTdx

h T T( )− −( )=P ∞ 0 (c)

Equation c governs the steady-state temperature distribution over an extended surface.

The differential equation for the temperature distribution in an annular fi n is derived using the coordinate r, which is measured from the centerline of the cylindrical surface to which the fi n is attached to a point along the axis of the fi n. An energy balance is then applied to the annular ring shown in Figure 1.7. The inner surface of the ring is a distance r from the centerline of the cylinder, while its outer surface is a distance r + dr from the centerline of the cylinder. The rate of heat transfer by conduction into the ring at r is (qA)r , where A is the area of the inner surface of the ring. The rate of heat transfer by conduction out of the ring is (qA)r + d/dr (qA)r dr. The rate at which heat is transferred from the ring by convection is qPdr. Application of an energy balance to the ring leads to Equation b, with x replaced by r. Subsequent use of Fourier’s conduction law and Newton’s law of cooling leads to Equation c, with x replaced by r.

Equation c can be used to model one-dimensional steady-state heat transfer over an extended surface when A is interpreted as the area through which heat conduction occurs and P is the perimeter over which heat transfer occurs.

(a) The conductive area of a straight fi n of constant thickness b is A = bw. The perimeter over which convection occurs is P = 2w + 2b. However, the thickness of the fi n is assumed to be small compared to its width, such that P can be approximated by P = 2w. Substitution into Equation c leads to

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ddx

kbwdTdx

h T T

d Tdx

hkb

T T

( )− −( )=

− −( )

( )2 0

2

w ∞

∞

2

2 (d)

The appropriate boundary conditions are T(0) = T0 and dT/dx (L) = 0, where L is the length of the fi n. The total heat transfer from the base of the fi n is

r

r + dr

(a)

qc

(b)

(qA)r

(qA)r + dr

Figure 1.7 (a) The differential element used in the modeling of the temperature distribution in an annular fi n is an annulus of inner radius r and thickness dr. The height of the annulus is the constant thickness of the fi n. (b) Heat is transferred by conduction through the area of the annulus and out of the annulus though convection over its top and bottom perimeter.

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Q kwbdTdx

b = − ( )0 (e)

(b) The conductive area of an annular fi n of constant thickness b is A = 2πrb. The perimeter over which convection occurs is P = 2(2πr). Substitution into Equation c, replacing x by r, leads to

ddr

kbrdTdr

hr T T

ddr

rdTdr

hkb

r T

2 4

2

π π ∞( )− −( )

( )− −TT∞( )= 0

(f)

The appropriate boundary conditions are T(R1) = T0 and (dT/dr) (R2) = 0, where R1 and R2 are the inner and outer radii of the fi n respectively. The total heat transfer from the base of the fi n is:

Q k R bdTdr

Rb 1 1= − ( )( )2π (g)

(c) For the analysis of the triangular fi n shown in Figure 1.8, it is convenient to locate the origin at the tip of the fi n. The fi n’s profi le is described by b(x) = (b0/L)x, where b0 is the thickness of the fi n at its base (x = L). The conductive

w

L

b0

x

Figure 1.8 The area for heat transfer at the tip of the triangular fi n is zero; thus it is convenient to measure x from the fi n’s tip.

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area is A = (b0w/L)x, while the convection perimeter is P = 2w + 2b(x). Assum-ing b0 � w, the perimeter can be approximated by P = 2w. Substitution into Equation c leads to:

ddx

kbL

xdTdx

wh T T

ddx

xdTdx

hLkb

0

0

2 0

2

( )− −( )=

( )−

∞

TT T−( )=∞ 0 (h)

The area at the tip of the fi n is zero. Thus there is no heat transfer from the tip as long as the temperature is fi nite. The only boundary condition is at the base, T(L) = T0. The total heat transfer from the base of the fi n is

Q kb wdTdx

Lb = − ( )0 (i)

Example 1.2 Derive the governing equation for the heat transfer in the annular fi n of Example 1.1 if the temperature at the base of the fi n varies over its circumference and changes with time.

Solution Defi ne θ as a circumferential coordinate measured counterclockwise from a reference line, 0 ≤ θ ≤ 2π, as illustrated in Figure 1.9. The temperature at

rf(θ, t) θ

Figure 1.9 The annular fi n of Example 1.2 is attached to a cylindrical surface. The temperature varies over the surface and with time.

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the base of the extended surface is a function of θ as well as time, f(θ,t). Let r be a coordinate measured radially from the center of the cylinder, R1 ≤ r ≤ R2. Assum-ing that the extended surface is thin and that conduction can be neglected in the direction transverse to the extended surface, then the temperature of a particle in the extended surface is dependent on its location described by r and θ as well as on time, T = T(r,θ,t).

Consider a differential segment of the cylinder shown in Figure 1.10. The segment is a portion of the annular ring of Example 1.1, but is defi ned by an angle dθ. Heat is transferred by conduction into the differential segment on the surfaces defi ned by constant values of r and θ, while heat is transferred out of the segment on surfaces defi ned by constant values of r + dr and θ + dθ. A concern may be that since T varies with θ, the rate of heat transfer qr may vary across the surface defi ned by r. This is true, but consider the total heat transfer over the surface at constant r from θ to θ + dθ,

Q q r brdr

d

r =

+

∫ ( , )θ θθ

θ θ

(a)

b

(qr A)

r + dr

dθ

(qrA)

r

qcA

c

dr

(qθA)θ+dθ

(qθA)θ

Figure 1.10 Because the temperature varies with both the radial and circumfer-ential coordinates, the differential element of the annular fi n in Example 1.2 is an infi nitesimal segment of the annular ring in Figure 1.7.

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The mean value theorem implies that there exists θ*, θ ≤ θ* ≤ θ + dθ such that

Q q r brdr r= ( , )*θ θ (b)

Noting that the area of the surface is brdθ, a Taylor series expansion of qr(r, θ*) about (r,θ) shows that a correction to account for the variation of Qr with θ over the surface is of order dθ and will be smaller than all other terms in the result-ing equation. Thus the rate of heat transfer can reasonably be assumed to be constant over each surface.

Heat is transferred from the element by convection from its upper and lower surfaces. Internal energy is stored in the differential element. The total internal energy is U = ρudV, where u is the specifi c internal energy and the differential volume is dV = 2πbrdrdθ. The appropriate energy balance is that the rate at which energy is stored in the element is equal to the rate at which heat is transferred into the element minus the rate at which heat is trans-ferred out of the element. Application to the differential element shown in Figure 1.10 leads to

∂ρ θ θ∂

( )= + − +∂∂

( )⎡

⎣⎢⎢

⎤

⎦⎥

tudV q A q A q A

rq A drr r r r r r ⎥⎥

− +∂∂

( )⎡

⎣⎢⎢

⎤

⎦⎥⎥−q A q A d qPdrθ θ θ θθ

θ ˆ (c)

Equation c can be simplifi ed to

∂∂

( ) = −∂∂

( ) −∂∂

( ) −ut

brdrdr

rq bd dr q bdrdρ θ θθ

θθr qqdr rd2 θ( ) (d)

Equation d then reduces to

ρ∂

θθr

ut r

rqq q r

b∂∂

= −∂

( )−∂∂

−r

ˆ2 (e)

The rate of convective heat transfer per unit area is given by Newton’s law of cooling, q = h(T–T∞ ). Fourier’s conduction law is used to determine the rate of heat transfer per unit area on a surface: qn = − k∇T ⋅ n, where qn is the heat transfer per unit area into the surface and n is a unit vector normal to the surface. Noting that in polar coordinates, ∇T = (∂T/∂r) er + 1/r (∂T/∂θ) eθ, then qr = − k(∂T/∂r) and qθ = − (k/r) (∂T/∂θ).

The specifi c internal energy is related to temperature by u = cT, where c is the specifi c heat of the body. Substitution into Equation e, assuming that k and c are constants, results in

ρ∂θ

crTt

kr

rTr

kr

T hrb

∂∂

=∂∂

∂∂

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+

∂−

1 22

2TT T−( )∞ (f)

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Dividing by kr equation f can be rewritten as

ρ

∂ θck

Tt r r

rTr r

T∂∂

=∂∂

∂⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+

∂∂

⎡

⎣⎢⎢1 1

2

2

2

⎤⎤

⎦⎥⎥ − −( )2h

bkT T∞ (g)

The Laplacian operator in polar coordinates is ∇ 2 T = 1/r [∂/∂r (r ⋅ ∂T/∂r)] + 1/r2 (∂2T/∂ θ2). Thus Equation g becomes

ρ

∞ck

Tt

Th

bkT T

∂∂

= ∇ − −( )2 2 (h)

Example 1.3 A rigid mass m is attached to the end of an elastic bar is illus-trated in Figure 1.11. A time-dependent force is applied to the mass. If m is much greater than the mass of the bar, a single-degree-of-freedom model is used in which the inertial effects of the bar are neglected and the bar is mod-eled as a linear spring of stiffness EA/L. This lumped-parameter model is illustrated in Figure 1.12. However, if the mass of the bar is comparable to m, then the distributed-parameter model in Figure 1.13 is used.

(a) Derive the differential equation governing the motion for the lumped-parameter model; (b) Derive the differential equation governing the motion for the distributed-parameter model.

Solution (a) Let y(t) represent the displacement of the mass. The spring is assumed to be linear, and all forms of friction are neglected. Application of Newton’s second law to a free-body diagram of the block, drawn at an arbi-trary instant and illustrated in Figure 1.14, leads to:

F(t)m

Figure 1.11 The longitudinal motion of a discrete particle of mass m attached to a fl exible bar is modeled in Example 1.3.

AE

L

m

x

F(t)

Figure 1.12 A lumped-parameter model is used if the mass of the particle is much greater than the mass of the bar.

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F ma∑ =

F tEAL

y md ydt

( )− =2

2

md ydt

EAL

y F t2

2+ = ( ) (a)

Equation a is supplemented by initial conditions specifying the initial dis-placement and the initial velocity of the block.

(b) Let x, 0 ≤ x ≤ L, be a coordinate along the axis of the bar. Note that the displacement of the rigid mass is y(t) = u(L,t). Consider a differential segment of the bar of length dx. The area of the segment is A, the cross-sectional area of the bar. A free-body diagram of the segment, drawn at an arbitrary instant, is illustrated in Figure 1.15. The force acting on each face of the segment is the resultant of the normal stress distribution σ(x,t), which is assumed uniform over each face. If the force acting normal to the element at x is f(x), then the force acting normal to the element at x + dx is f x dx t f x t f x dx( , ) ( , ) ( ).+ = +∂ ∂/

F(t)mx

u(x, t)

ρ, E

L

u(L, t)

Figure 1.13 A distributed-parameter model is used if the mass of the bar is compa-rable to or greater than the mass of the particle.

N

F(t)

mg

EA

Ly

Figure 1.14 Newton’s second law is applied to the free-body diagram of the particle of Example 1.3, drawn at an arbitrary instant, when the lumped-parameter model is applied.

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The mass of the element is dm = ρAdx, where ρ is the mass density of the mate-rial from which the bar is made. The acceleration of the element a u t= ∂ ∂2 2/ . Application of Newton’s second law to the free-body diagram shown in Figure 1.15 leads to

F dm a∑ =( )

− + +∂∂

=∂∂

f x t f x tfx

dx Adxu

x( , ) ( , ) ρ

2

2

∂∂

=∂∂

fx

Au

tρ

2

2 (b)

If the normal stress is less than the material’s yield stress, then the system behaves elastically, and the normal stress is related to the normal strain ε through Hooke’s law, σ = Eε, where E is the modulus of elasticity or Young’s modulus. By defi nition, the normal strain is the change in displacement per change in length, ε = ∂ ∂u x/ . Thus the resultant force acting normal to the differential element at x is f EA u x= ∂ ∂( / ). This force can be substituted into Equation b, leading to

∂∂

∂∂( )=

∂∂x

EAux

Au

tρ

2

2 (c)

In the case of a uniform bar, Equation c reduces to

∂∂

=∂∂

2

2

2

2

ux E

ut

ρ (d)

The form of Equation d is that of the wave equation.The bar is fi xed at x = 0, thus

u t( , )0 0= (e)

dx

f (x + dx, t )f(x, t )

Figure 1.15 A free-body diagram of a differential element of the bar is drawn at an arbitrary instant when the distributed-parameter model is used in Example 1.3.

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A free-body diagram of a rigid block, abstracted from the bar at an arbitrary instant, is illustrated in Figure 1.16. Application of Newton’s second law to this free-body diagram leads to

−∂∂

+ =∂∂

EAux

L t F t mu

tL t( , ) ( ) ( , )

2

2 (f)

Equation d, Equation e, and Equation f are supplemented by two initial conditions.

Equation 1.2 and Equation 1.3, in general, represent four scalar equations in fi ve dependent variables. For an incompressible fl ow, the rate of change of volume of a constant mass of fl uid is zero, leading to the incompressibility condition

∇⋅ =v 0 (1.5)

The continuity equation then leads to the conclusion that the density is con-stant. The incompressible form of the momentum equation becomes

∂∂

+ ⋅∇( ) = − + ∇v

tv v F vb

1 2

ρ∇ρ

νp

(1.6)

where ν = μ ρ/ is the fl uid’s kinematic viscosity. Equation 1.5 and Equation 1.6 are used to develop models for incompressible fl ow problems.

The basic conservation laws are used to derive principles which are use-ful in problem formulation. For example, integration of Newton’s second law along a particle trajectory leads to an energy principle.

N

F(t )

mg

EA (L,t)x

u

Figure 1.16 The boundary condition at x = L for the distributed-parameter model of Example 1.3 is obtained through application of Newton’s second law to a free-body diagram of the particle drawn at an arbitrary instant.

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Energy methods provide an alternative to the application of basic conserva-tion laws to a differential element. Most energy methods are based on variational calculus. Whereas a differential is an infi nitesimal change in an independent variable, a variation is a change in a dependent variable. For example, if x(t) is a displacement of a particle which travels between x1 and x2 between times t1 and t2, there are infi nitely many paths that the particle could take, but only one which satisfi es conservation laws. A variation in u, δu is a varied path along which the particle may travel between t1 and t2 subject to δu(t1) = 0 and δu(t2) = 0, as illustrated in Figure 1.17.

Variational principles require that the exact path be one which minimizes energy. Hamilton’s Principle states that for a conservative system,

δ Ldtt

t

1

2

∫ = 0 (1.7)

where L, the lagrangian, is

L T V= − (1.8)

in which T is the system’s kinetic energy and V is the system’s potential energy.

Defi ning x1, x2,…,xn as a system of generalized coordinates for an n-degree-of-freedom discrete system, the potential energy is a function of these coordinates, V = V(x1, x2,…,xn), and the kinetic energy is a function of the gener-alized coordinates and their time derivatives, T T x x x x x xn n= ( )1 2 1 2, , , , , ,...,… � � � .

(x1, t1)

(x2,t2)x(t)

x(t) + δ(x)

t

x

Figure 1.17 The true path taken by a particle from t1 to t2 is x(t). The variable path x t x( )+ δ is an alternate path between t1 and t2.

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Application of Hamilton’s Principle to a conservative discrete system leads to Lagrange’s equations,

ddt

Lx

Lx

i ni i

∂∂

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟−

∂= =

�∂

0 1 2, ,..., (1.9)

Lagrange’s equations can be applied to derive the differential equations gov-erning the motion of a discrete system.

Example 1.4 Use Lagrange’s equations to derive the differential equations governing the motion of the system shown in Figure 1.18.

Solution The kinetic energy of the system at an arbitrary instant is

T m x m x m x= + +1

2

1

2

1

21 1

22 2

23 3

2� � � (a)

The potential energy in the system at the same instant is

V k x k x x k x x= + −( ) + −( )1

2

1

2

1

21 1

22 2 1

23 3 2

2 (b)

The Lagrangian for the system is

L T V m x m x m x

k x

= − = + +

− +

1

2

1

2

1

2

1

2

1

2

1 12

2 22

3 32

1 12

� � �

kk x x k x x2 2 1 3 3 2

1

2( ) ( )− + −

⎡⎣⎢⎢

⎤⎦⎥⎥

2 2 (c)

Application of Lagrange’s equations to Equation c gives

ddt

Lx

Lx

∂∂

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟−

∂∂

=�1 1

0

ddt

m x k x k x x1 1 1 1 2 2 1 1 0�( )− − − −( ) −[ ]=( )

m x k k x k x1 1 2 1 1 2 2 0�� + +( ) − = (d)

m2 m3

x1

m1

x2 x3k1 k2 k3

Figure 1.18 Lagrange’s equations are used to derive the differential equations gov-erning the motion of the mechanical system.

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ddt

Lx

Lx

ddt

m x k x

∂∂

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟−

∂∂

=

( )− −

�

�

2 2

2 2 2

0

22 1 3 3 2

2 2 1 1 2 3

1 0−( )− −( ) −[ ]=

− + +(

x k x x

m x k x k k

( )

�� )) − =x k x2 3 3 0 (e)

ddt

Lx

Lx

ddt

m x k x

∂∂

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟−

∂∂

=

( )− −

�

�

3 3

3 3 3

0

33 2

3 3 3 2 3 3

0

0

−( )[ ]=

− + =

x

m x k x k x�� (f)

Equation d, Equation e, and Equation f are the differential equations govern-ing the dynamic response of the system shown in Figure 1.18.

Example 1.5 Use Hamilton’s Principle to derive the differential equation for the longitudinal oscillations of a particle of mass m attached to the end of an elastic bar, as illustrated in Figure 1.11.

Solution Consider a differential element of thickness dx and area dA as illustrated in Figure 1.19. The strain energy stored in this element is

dVE

dAdx=σ2

2 (a)

where σ is the normal stress. Assuming elastic behavior, σ = Eε, and for small displacements, the normal strain is ε = ∂ ∂u x/ , where u(x,t) is the displacement

dx

dA

σ(x + dx,t)

σ(x, t)

Figure 1.19 The normal stress in the elastic bar of Figure 1.11 is uniform across the face of the differential element, but varies across the length of the element.

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of a particle along the longitudinal axis of the bar. The total potential energy is obtained by integrating Equation a over the volume of the bar:

VE

Eux

dAdxA

L

=∂∂( )∫∫ 1

2

2

0

=∂∂( )∫ EA u

xdx

L

2

2

0

(b)

The kinetic energy of the element is

dT Adxut

= ( )∂( )1

2

2

ρ∂

(c)

The total kinetic energy is the kinetic energy of the bar plus the kinetic energy of the particle:

TA u

tdx m

ut

L

L

=∂∂( ) +

∂∂

⎡⎣⎢⎢

⎤⎦⎥⎥∫ ρ

2

1

2

2

0

2

( , )t (d)

Substitution of Equation c and Equation d into Hamilton’s Principle leads to

δρA u

tdx m

ut

L tEA

L

2

1

2 2

2

0

2∂∂( ) +

∂∂

⎡⎣⎢⎢

⎤⎦⎥⎥

−∂∫ ( , )

uux

dx dt

L

t

t

∂( )⎧⎨⎪⎪⎪

⎩⎪⎪⎪

⎫⎬⎪⎪⎪

⎭⎪⎪⎪

=∫∫2

01

2

0 (e)

The calculus of variations shows that it is allowable to interchange the order of the variation and the integration, that is, δ δ∂ ∂( ) = ∂ ∂ ∂ ∂( )[ ]u t u t u t/ / /

22( ) ,

and δ δ∂ ∂( )= ∂ ∂ ( )u t t u/ / . Thus, the fi rst integral in Equation e becomes

δρ

δA u

tdxdt A

ut t

u dxd

L

t

t L

2

2

0 0

2

∂∂( ) =

∂∂

∂∂

( )∫∫ ∫1

ρ ttt

t

1

2

∫ (f)

Interchanging the order of integration and applying integration by parts, ∫ = −∫fdg fg gdf with f = ∂u/∂t and dg = ∂/∂t(δu)dt, leading to

δρ

ρ δA u

tdxdt A

ut

u

L

t

t

t t

t t

20

2

1

2

1

2∂∂( ) =

∂∂

−∂∫∫ =

=2 uut

udtt

t

∂

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟

⎧⎨⎪⎪⎪

⎩⎪⎪⎪

⎫⎬⎪⎪⎪

⎭⎪⎪⎪

∫ 2

1

2

δ ddx

L

0

∫ (g)

9533_C001.indd 259533_C001.indd 25 10/29/08 1:29:54 PM10/29/08 1:29:54 PM


Remembering that δu t u t( ) ( )1 0 0= =and 2δ and again interchanging the order of integration, Equation g becomes

δρ

ρ δA u

tdxdt A

ut

udxdt

L

t

t L

t2

0

2

2

01

2

∂∂( ) = −

∂∂∫∫ ∫

2

11

2t

∫ (h)

Next, consider the contribution of the kinetic energy of the discrete par-ticle. Using the properties mentioned earlier, from the calculus of variations and integration by parts, this term becomes:

δ1

21

22

2m

ut

L t dt mu

tL t

t

t∂∂

⎡⎣⎢⎢

⎤⎦⎥⎥

=∂∂∫ ( , ) ( , )

2

δuu L t dtt

t

( , )

1

2

∫ (i)

The potential-energy term can be written as

δ δEA u

xdxdt EA

ux x

u dxd

L

t

tL

2

2

00

1

2

∂∂( ) =

∂∂

∂∂∫∫ ∫ ( ) tt

t

t

1

2

∫ (j)

Application of integration by parts to the inner integral of the right-hand side of Equation j with f = ∂u/∂x and dg = (∂/∂x)(δu)dx leads to [∂(δu)/∂x]dx

δEA u

xdxdt EA

ux

L t

L

t

t

2

2

01

2

∂∂

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟ =

∂∂∫∫ ( , )) ( , ) ( , ) ( , )δ δu L t EA

ux

t u t

EA

t

t

−∂∂

⎧⎨⎪⎪⎩⎪⎪

−

∫ 0 0

1

2

∂∂∂

⎫⎬⎪⎪⎭⎪⎪∫

2

2

ux

udx dtδ0

L

(k)

Combining Equation e, Equation h, Equation i, and Equation k leads to

−∂∂

+∂∂

⎡

⎣⎢⎢

⎤

⎦⎥⎥

⎧⎨⎪

EAux

L t mu

tL t u L t( , ) ( , ) ( , )

2

2δ⎪⎪

⎩⎪⎪+

∂∂

+∂∂

∫t

t

EAux

t u L t

EAu

x

1

2

0

2

( , ) ( , )δ

22

2

2

0

0−∂∂

⎡

⎣⎢⎢

⎤

⎦⎥⎥

⎫⎬⎪⎪⎭⎪⎪

=∫ ρ δAu

tu dt

L

(l)

Equation l must be true for any varied path (for all possible δu). This implies that

Eu

xA

ut

A∂∂

−∂∂

=2

2

2

20ρ (m)

9533_C001.indd 269533_C001.indd 26 10/29/08 1:29:55 PM10/29/08 1:29:55 PM


Equation l also implies that either EA u t t( )( , )∂ ∂ =/ 0 0 or δu t( , )0 0= and that either EA u x L t m u t L t( )( , ) ( )( , )∂ ∂ + ∂ ∂ =/ /2 2 0 or δu L t( , ) = 0. These choices spec-ify possible boundary conditions. For a bar fi xed at x = 0 and with the discrete mass attached at x = L, the appropriate choices are:

u t( , )0 0= (n)

EAux

L t mu

tL t

∂∂

+∂∂

=( , ) ( , )2

20 (o)

1.4 NondimensionalizationThe motion of the one-degree-of-freedom mass-spring and viscous-damper system shown in Figure 1.20 is governed by the differential equation

mx cx k F t�� + + =x 0 sin( )ω (1.10)

The general solution of Equation 1.10 is the sum of a homogeneous xh(t), the solution of Equation 1.10 obtained if F0 = 0, and a particular solution, xp(t), the solution corresponding to the specifi c function on the right-hand side of Equation 1.10. The general solution is:

x t e C t C tntn n( ) cos sin= −( )+ −( )⎡

⎣⎢−ζ ω ζ ω ζω

12

221 1 ⎤⎤

⎦⎥+ +X tsin( )ω φ (1.11)

k c

x

m

F sin(ωt)

Figure 1.20 The steady-state amplitude and the steady-state phase angle of the mass-spring-viscous-damper system are each a function of fi ve dimensional parameters.

9533_C001.indd 279533_C001.indd 27 10/29/08 1:29:55 PM10/29/08 1:29:55 PM


where C1 and C2 are constants of integration to be determined through appli-cation of initial conditions,

ωnkm

= (1.12)

is the system’s natural frequency, and

ζ =c

m n2 ω (1.13)

is the system’s damping ratio. Equation 1.11 is derived assuming 0 < ζ < 1. As t grows large, the particular solution becomes exponentially small and negligible compared with the particular solution. The steady-state response is defi ned as

x t x t

X t

sst

( ) lim ( )

sin( )

=

= +

→∞

ω φ (1.14)

where the steady-state amplitude is determined as

XF

m n n

=−( ) +

0

2 2 2 22ω ω ζωω( ) (1.15)

and the steady-state phase is

φ = −−

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟

−tan 12 2

2ζωωω ω

n

n (1.16)

The steady-state amplitude is a function of fi ve parameters,

X X F m n= ( )0 , , , ,ω ω ζ (1.17)

while the steady-state phase is a function of three parameters,

φ φ ω ω ζ= ( , , )n (1.18)

Note that in Equation 1.17 and Equation 1.18, the parameters ωn and ζ could be replaced by k and c.

Design and synthesis applications often require an understanding of the behavior of system response as the system parameters vary. The term frequency response refers to the variation of steady-state amplitude and steady-state phase with ω, the frequency of the excitation. It is usually easier to understand these

9533_C001.indd 289533_C001.indd 28 10/29/08 1:29:57 PM10/29/08 1:29:57 PM


variations when the relations are written in a nondimensional form. To this end, the frequency ratio is defi ned as

r =ωωn

(1.19)

Use of Equation 1.19 in the form of ω = r nω in Equation 1.15 and Equation 1.16 leads to:

m X

F r r

nω2

0 2 2 2

1

1 2=

−( ) +( )ζ (1.20)

φ = −−( )−tan 1

2

2

1

ζrr

(1.21)

The left-hand side of Equation 1.20 is a nondimensional parameter, often called the magnifi cation factor, and is a function of the nondimensional parameters r and ζ:

m X

FM rnω

ζ2

0

= ( , )

=( ) +( )

1

1 22 2 2−r rζ (1.22)

Noting that for a given system, the steady-state amplitude is proportional to the magnifi cation factor, Equation 1.22 can be used to understand the fre-quency response. Equation 1.22 can be used algebraically to determine the limits of the steady-state amplitude for small and large frequency ratios, to determine the maximum value of X for a specifi c value of ζ, and to develop the plot shown in Figure 1.21, which summarizes the frequency response using a single set of curves. Using Figure 1.21 and Equation 1.22, the follow-ing can be determined regarding the frequency response:

M(0, ζ) = 1M(r, ζ) is asymptotic to 1/r2 large rM(r, ζ) has a maximum value of 1 2 1 2/ ζ ζ−( ) at r = ζ1 2 2− if ζ < 1 2/M(r, ζ) has no extrema if ζ > 1 2/

The variation of the steady-state phase with frequency ratio is plotted in Figure 1.22. The following can be determined from Equation 1.21 and Figure 1.22:

tan φ < 0 and −π < φ < −π/2 for r < 1φ = −π/2 for r = 1

••••

••

9533_C001.indd 299533_C001.indd 29 10/29/08 1:29:58 PM10/29/08 1:29:58 PM


tan φ > 0 and −π/2 < φ < 0 for r > 1φ → 0 for large r

The frequency response of a mechanical system is an example where it is convenient to nondimensionalize a derived relationship between an output variable and the system parameters. In the above example, the steady-state amplitude and steady-state phase are the output parameters. It is assumed that the dimensional parameters (natural frequency, input frequency, excita-tion amplitude, mass, and damping ratio) vary independently of one another. An objective of nondimensionalization is to determine a set of nondimen-sional parameters such that a nondimensional parameter involving the output variable is represented as a function of independent dimensionless parameters. This is easy to do for the frequency response of a mechanical system when a dimensional relation between variables has been derived.

Dimensional analysis is used to determine empirically the relationship between a dependent parameter and the independent parameters on which it depends. A familiar example is the determination of the variation of a drag

••

0 0.5 1 1.5 2 2.5 3 3.5 40

5

10

15

20

25

30

35

40

r

M(r,

ζ)

Figure 1.21 The magnifi cation factor M is a nondimensional parameter propor-tional to the steady-state amplitude of a mass-spring-viscous-damper system. The magnifi cation factor is a function of two dimensionless parameters, the frequency ratio r and the damping ratio ζ.

9533_C001.indd 309533_C001.indd 30 10/29/08 1:29:59 PM10/29/08 1:29:59 PM


force on a bluff body, as shown in Figure 1.23, with parameters of the fl uid fl ow and geometric parameters of the body. These parameters may include the velocity of the fl ow, v, the mass density of the fl uid, ρ, the dynamic viscosity of the fl uid, μ, the speed of sound in the fl uid, c, and the geometric quantities

0 0.5 1 1.5 2 2.5 3 3.5 4−π

−π/2

r

φ(r,ζ

)

0

Figure 1.22 The steady-state phase angle is itself nondimensional and is a function of the nondimensional frequency ratio and the nondimensional damping ratio.

Figure 1.23 The drag force acting on a bluff body placed in a uniform free stream is dependent on the free-stream velocity, the density and dynamic viscosity of the fl uid, the speed of sound in the fl uid, and a length scale. The relationship can be reformulated nondimensionally as the drag coeffi cient as a function of Mach number and Reynolds number.

9533_C001.indd 319533_C001.indd 31 10/29/08 1:30:01 PM10/29/08 1:30:01 PM


which are represented by a characteristic length, L. Noting that the drag force D is the dependent parameter, it is desired to obtain empirically

D D v c L= ρ( , , , , )μ (1.23)

The drag force can be measured on a model in a wind tunnel. Development of a full-scale prototype is often impractical, and each wind tunnel test is expensive. Thus it is desired to develop a test on a scale model and apply the results to the prototype.

When a model or prototype is tested in a wind tunnel, viscous forces, inertial forces, and drag forces act on the body, and the vector sum of these forces is zero. This application of Newton’s law can be represented as a closed force polygon, as illustrated in Figure 1.24. The force polygon for the model is similar to the force polygon for the prototype when the ratio of any two corresponding sides is the same for both. For example using the drag force D and the inertial force I,

DI

DI( ) =( )

prototype model

(1.24)

Equation 1.24 can be rearranged to yield

D IDI

prototype prototypemodel

= ( ) (1.25)

However, Equation 1.25 is true only when the ratio of the viscous force V to the inertial force is the same for the model and the prototype:

VI

VI( ) =( )

prototype model

(1.26)

Equation 1.24, Equation 1.25, and Equation 1.26 imply that if the force polygon for the model is similar to the force polygon for the prototype, then a mea-surement of the drag force on the model can be used to determine the drag force on the prototype, using Equation 1.25. When the force polygons of the

Inertia

Drag Friction

Figure 1.24 Vector addition of the drag force, the friction force, and the inertia force acting on a bluff body forms a closed polygon.

9533_C001.indd 329533_C001.indd 32 10/29/08 1:30:02 PM10/29/08 1:30:02 PM


model and the prototype are similar, the model and prototype are said to exhibit dynamic similitude.

When running an experiment on a model that is used to predict the per-formance of the prototype, kinematic similitude must also be satisfi ed; mean-ing that the ratio of all time scales for the model and for the prototype must be the same. Kinematic similitude is usually enforced by requiring velocity ratios to be equivalent:

vc

vc( ) =( )

model prototype

(1.27)

Equation 1.24, Equation 1.25, Equation 1.26, and Equation 1.27 lead to the conclusion that

DV

fVI

vc

= ( ), (1.28)

A nondimensional form of Equation 1.28 is

C f MD = (Re, ) (1.29)

where CD, the drag coeffi cient, is the ratio of the drag force to the inertial force,

CD

v LD =

1

22 2ρ

(1.30)

Re, the Reynolds number, is the ratio of the viscous forces to the inertial forces,

Re =ρvLμ

(1.31)

and M, the Mach number, is

Mvc

= (1.32)

The previous discussions illustrate the value of nondimensionalization. A derived mathematical relation between the steady-state amplitude and sys-tem parameters was nondimensionalized to provide a better understanding of the steady-state behavior. The fl uid fl ow problem was nondimensionalized so that empirical results on a model could be used to predict the performance of a prototype. The functional relation between the nondimensional depen-dent variables and the nondimensional independent variables is determined from empirical results.

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The process of mathematical modeling is used to develop a mathemati-cal problem relating output variables to independent variables in terms of parameters. Such problems can be nondimensionalized through introduc-tion of nondimensional independent and dependent variables. The process leads to determination of a set of nondimensional parameters. The physical meanings of the parameters are obtained from the nondimensional problem. The choice of nondimensional variables and nondimensional parameters is not unique. The process of nondimensionalizing a mathematical problem is illustrated in the following examples.

Example 1.6 The differential equation governing the motion of a one-degree-of-freedom mass-spring and viscous damper system is Equation 1.10, repeated following:

mx cx kx F t�� + + = ( )0 sin ω (a)

Nondimensionalize Equation a using the following nondimensional variables:

xx

F k*

/=

0

(b)

t tkm

tn* = =ω (c)

Solution Note that x*and t* are nondimensional. The choices of x*and t* are not unique. For example, an alternative choice to nondimensionalize time is t t* = ω .

Derivatives with respect to dimensional independent variables are con-verted to derivatives with respect to nondimensional independent variables through the chain rule:

ddt

ddt

dtdt

ddtt n= =

*

*ω (d)

ddt

ddt

ddt

ddt

ddt

ddt

n n n

2

22= ( )= ( )=ω ω ω

* * *

2

2 (e)

Substituting Equation b, Equation c, Equation d, and Equation e into Equation a leads to

md

dtFk

x cd

dtFk

x kFk

x Fn nω ω22

2

0 0 0

**

** *( )+ ( )+ ( )= 00 sin

*

ωωt

n

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟ (f)

Simplifi cation of Equation f, noting that from Equation (c), ωn k m= / leads to

d xdt

ck

dxdt

x tn

n

2

2

*

*

*

*

* *sin+⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟ + =

⎛ω ωω⎝⎝

⎜⎜⎜⎞

⎠⎟⎟⎟⎟ (g)

9533_C001.indd 349533_C001.indd 34 10/29/08 1:30:03 PM10/29/08 1:30:03 PM


Note that

c

kckm

nωζ= = 2 (h)

and that r n= ω/ ω , where ζ and r are the nondimensional frequency and damping ratios. Thus Equation f can be rewritten as:

d xdt

dxdt

x rt2

22+ + =ζ sin( ) (i)

It is conventional to drop the superscript * from dimensionless quantities and understand that all variables and parameters are henceforth dimensionless.

When solved, Equation i gives a mathematical representation of the non-dimensional displacement in terms of two nondimensional parameters, the damping ratio and the frequency ratio. The steady-state amplitude is obtained by taking the limit t →∞. This leads to X f r* ( , )= ζ . However, from Equation b, X kX F m F Mn

* = = =/ /02

0ω . Hence the relation among the magnifi cation fac-tor, the damping ratio, and the frequency ratio can be obtained using a priori nondimensionalization of the governing differential equation.

The choices for defi ning nondimensional variables are not unique. For example, time could be nondimensionalized using the input frequency rather than the natural frequency, t t* = ω , in which case the nondimensional differential equation becomes:

�� xc

mx

km

xk

mt+ + = ( )

ω ω ω2 2sin (j)

where all variables are nondimensional. Equation j suggests the defi nition of two nondimensional parameters, π ω π ω1

22= =k m c m/ and / . It should be

noted that π121= /r and π ζ2 2= /r. The two sets of dimensionless param-

eters are related. Equation j is obtained by dividing the differential equation obtained after changing from dimensional to nondimensional variables by the coeffi cient of the inertial term. This suggests that π2 is the ratio of the damping forces to the inertial forces. It is, but on a different time scale than ζ.

Example 1.7 A sphere of radius R is initially at a uniform temperature T0 when it is plunged into a bath of temperature T1, as illustrated in Figure 1.25. The sphere is made of a material of specifi c heat c, mass density ρ, and ther-mal conductivity k. The heat transfer coeffi cient between the sphere and the bath is h. The temperature distribution is assumed to be axisymmetric and thus to depend only on the distance from the center of the sphere, r, and on time, t; T(r,t). The partial differential equation governing the unsteady state heat conduction in the sphere is:

ρcTt

kr r

rT∂

∂=

∂∂ ∂( )2

2 ∂r

(a)

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The initial condition is:

T r T( , )0 0= (b)

The boundary condition at the outer radius is obtained, using an energy balance on the surface of the sphere, as:

kTr

R t h T R t T∂∂

= − −( , ) [ ( , ) ]1 (c)

The radial coordinate varies over the range 0 ≤ ≤r R. The temperature is required to remain fi nite at r = 0.

Introduce nondimensional variables to rewrite Equation a, Equation b, and Equation c in nondimensional forms. Identify relevant nondimensional parameters.

Solution The radius of the sphere is a characteristic length, and thus a non-dimensional radial coordinate can be defi ned as

rrR

* = (d)

There is no parameter with the time dimension, and it is not clear what an appropriate nondimensionalization of the time variable would be. Let A be a combination of parameters with the time dimension, and defi ne

ttA

* = (e)

The form of A will be chosen for convenience.

Rr

T(r, t)

T1

Figure 1.25 The sphere is at a uniform temperature when immersed in a bath at a different temperature. Heat transfer by convection through the surface of the sphere leads to a unsteady state temperature distribution throughout the sphere. The nondimensional temperature is dependent upon the nondimensional Biot number, which is a ratio of the rate of heat transfer by convection to the rate of heat transfer by conduction.

9533_C001.indd 369533_C001.indd 36 10/29/08 1:30:06 PM10/29/08 1:30:06 PM


Temperature is a thermodynamic potential. Heat transfer occurs because of differences in temperature. Therefore, in nondimensionalizing the depen-dent variable of temperature, it is convenient to use a ratio of temperature differences. To this end, a nondimensional temperature, Θ( , )* *r t , can be defi ned as

Θ =−−

T TT T

1

0 1

(f)

The chain rule is used to obtain

∂∂

=∂

∂=

∂∂r r

drdr R r*

*

*

1 (g)

∂∂

=∂

∂=

∂∂t t

dtdt A t*

*

*

1 (h)

Substitution of Equation d, Equation e, Equation f, Equation g, and Equation h into Equation a leads to

ρcA t

T T Tk

R r R rR r

R r∂

∂−( ) +[ ]=

∂∂

∂∂* * *

*0 1 1 2 2

2 21 1Θ

**T T T0 1 1−( ) +[ ]{ }Θ (i)

Simplifi cation of Equation i results in

∂∂

=∂

∂∂∂( )Θ Θ

tkAcR r r

rr* * *

**ρ 2 2

21 (j)

For convenience, choose A such that

kAcRρ 2

1=

AcRk

=ρ 2

(k)

Substituting Equation k into Equation j and dropping the *s from nondimen-sional variables leads to

∂∂

=∂∂

∂∂( )Θ Θ

t r rr

r12

2 (l)

Substitution of Equation f into initial condition b leads to

Θ( , )r 0 1= (m)

9533_C001.indd 379533_C001.indd 37 10/29/08 1:30:07 PM10/29/08 1:30:07 PM


Noting that r = R corresponds to r* = 1, substitution of Equation d, Equation e, Equation f, Equation g, and Equation h into boundary condition c gives

T T k

R rt h t T T0 1

0 11 1−( ) ∂

∂= − −

ΘΘ( , ) ( , )( )

∂∂

= −Θ

Θr

t Bi t( , ) ( , )1 1 (n)

where the Biot number is defi ned as

BihRk

= (o)

and is the ratio of the rate of heat transfer by convection to the rate of heat transfer by conduction on the surface of the sphere.

The nondimensional formulation for the unsteady state heat transfer in the sphere is the partial differential equation, Equation l, the initial condi-tion, Equation m, and the boundary condition, Equation o. In each of these equations, it is understood that the variables are dimensionless.

Example 1.8 One continuum model of multi-walled carbon nanotubes is that of concentric Euler-Bernoulli beams connected by elastic layers which model the van der Waals forces between the layers of atoms. Manufacturing often requires carbon nanotubes to be subject to a tensile force and sur-rounded by a polymer gel. Consider a double-walled nanotube modeled by concentric Euler-Bernoulli beams. The tubes are each of length L. Defi ne x such that 0 ≤ ≤x L is a coordinate along the centerline of the tubes, as illustrated in Figure 1.26. Using the subscript “1” to refer to properties of the inner tube and the subscript “2” to refer to properties of the outer tube, the partial differential equations governing the transverse displacements of the tubes, w1(x,t) and w2(x,t) are

E Iwx

Pwx

k w w Awt

1 1

41

4

21

2 1 1 2 1 1

21

2

∂∂

−∂∂

+ −( )+∂∂

ρ (a)

E Iwx

Pwx

k w w k w Aw

2 2

42

4

22

2 1 2 1 2 2 2 2

2∂∂

−∂∂

+ − + +∂

( ) ρ 22

2∂t (b)

where E is the elastic modulus of the tube (usually 1 TPa), ρ is the mass density of the nanotube (approximately 1.3 g/cm3), A is the cross-sectional area (the radius of a carbon atom is 0.34 nm), I is the cross-sectional moment of inertia, P is the tensile force, k1 is the stiffness per unit length used to model the van der Waals forces, and k2 is the stiffness per unit length of the surrounding polymer gel.

9533_C001.indd 389533_C001.indd 38 10/29/08 1:30:08 PM10/29/08 1:30:08 PM


If the tubes are each fi xed at x = 0 and free at x = L, the appropriate bound-ary conditions are:

w t1 0 0( , ) = (c1)

∂∂

=wx

t1 0 0( , ) (c2)

∂∂

=2

1

20

wx

L t( , ) (c3)

E Iwx

L t Pwx

L t1 1

31

3

1 0∂∂

+∂∂

=( , ) ( , ) (c4)

and

w t2 0 0( , ) = (d1)

∂∂

=wx

t2 0 0( , ) (d2)

∂∂

=2

2

20

wx

L t( , ) (d3)

E Iwx

L t Pwx

L t2 2

32

3

2 0∂∂

+∂∂

=( , ) ( , ) (d4)

Nondimensionalize Equation a, Equation b, Equation c, and Equation d.

x

L

Figure 1.26 A continuum model for the displacement of carbon nanotubes uses concentric Euler-Bernoulli beams connected by elastic layers which model the inter-atomic van der Waals forces.

9533_C001.indd 399533_C001.indd 39 10/29/08 1:30:09 PM10/29/08 1:30:09 PM


Solution The spatial variable and the transverse displacements can be non-dimensionalized using the length of the beam,

xxL

* = (e)

wwL1

1* = (f1)

wwL

22* = (f2)

Let T be a combination of parameters with the dimension of time; a nondi-mensional time can be defi ned according to

ttT

* = (g)

Substitution of Equation e, Equation f, and Equation g into Equation a and Equation b gives

E IL x

LwPL x

Lw k L w1 14

4

4 1 2

2

2 1 1 1

∂∂

−∂

∂+ −

*

*

*

* *( ) ( ) ( wwA

T tLw2

1 1

2

2

2 1*

*

*) ( )+∂

∂=

ρ0 (h)

and

E IL x

LwPL x

Lw k L w2 24

4

4 2 2

2

2 2 1 2

∂∂

−∂

∂+ −

*

*

*

* *( ) ( ) ( ww

k LwA

T tLw

1

2 22 2

2

2

2 2

*

*

*

*

)

( )+ +∂

∂=

ρ0 (i)

Rearranging Equation h leads to

∂∂

−∂∂

+ −4

1

4

2

1 1

21

2

14

1 11 2

wx

PLE I

wx

k LE I

w w*

*

*

** *( ))

*

*+

∂∂

=ρ1 1

4

1 12

21

2

A LE I T

wt

0 (j)

For convenience, choose

ρ1 1

4

1 12

1A L

E I T=

T LA

E I= 2 1 1

1 1

ρ (k)

9533_C001.indd 409533_C001.indd 40 10/29/08 1:30:10 PM10/29/08 1:30:10 PM


and defi ne

ε =PLE I

2

1 1

(l)

η1 =k LE I

14

1 1

(m)

Using Equation k, Equation l, and Equation m, Equation h becomes

∂∂

−∂∂

+ − +∂∂

=4

1

4

21

2 1 1 2

21

2

wx

wx

w wwt

ε η ( ) 0 (n)

where the * has been dropped from nondimensional variables. Substitution of Equation k, Equation l, and Equation m into Equation i leads to

E IL x

LwE IL x

LwE2 2

4

4

4 21 14

2

2 2 11∂

∂−

∂∂

+*

*

*

*( ) ( )ε ηII

Lw w k Lw

A E I

L A t

13 2 1 2 2

2 2 1 1

41 1

2

2

( )

(

* * *

*

− +

+∂

∂

ρ

ρLLw2

* ) = 0 (o)

Multiplication of Equation o by L E I31 1/ and dropping the * from nondimen-

sional variables leads to

μ ε η η β∂∂

−∂∂

( )+ − + +∂∂

42

4

2

2 2 1 2 1 2 2

22w

x xw w w w

wt

( )22

= 0 (p)

where

μ =E IE I

2 1

1 1

(q)

β =ρρ

2AA

2

1 1

(r)

η22

4

1 1

=k LE I

(s)

Substitution of Equation e, Equation f, and Equation l into Equation c4 leads to

E IL

wx

t Pwx

t

wx

1 12

313

1

31

3

1 1 0∂∂

+∂∂

=

∂∂

*

*

*

*( , ) ( , )

(( , ) ( , )1 1 01twx

t+∂∂

=ε (t)

9533_C001.indd 419533_C001.indd 41 10/29/08 1:30:11 PM10/29/08 1:30:11 PM


where the * has been dropped from nondimensional variables. Substitution of Equation e, Equation f, Equation l, and Equation q into Equation d4 leads to

μ ε∂∂

+∂∂

=3

2

3

21 1 0wx

twx

t( , ) ( , ) (u)

The nondimensional boundary conditions can be summarized by

w t1 0 0( , ) = (v1)

∂∂

=wx

t1 0 0( , ) (v2)

∂∂

=2

1

21 0

wx

t( , ) (v3)

∂∂

+∂∂

=3wx

twx

t1

3

11 1 0( , ) ( , )ε (v4)

w t2 0 0( , ) = (w1)

∂∂

=wx

t2 0 0( , ) (w2)

∂∂

=2

2

21 0

wx

t( , ) (w3)

μ ε∂∂

+∂∂

=3wx

twx

t2

3

21 1 0( , ) ( , ) (w4)

The nondimensional problem is comprised of the partial differential equa-tions, Equation n and Equation p, and the eight boundary conditions given by Equation v and Equation w.

One value of nondimensionalization is to determine the order of magni-tude of physical effects. The fourth-order spatial derivatives arise from normal bending stresses, the second-order spatial derivatives occur because of the normal stress due to the axial load, the second-order time derivatives are the inertial terms, and the zeroth-order derivatives occur because of the elastic layer between the tubes and surrounding. Since the parameter ε is obtained as the coeffi cient of the second-order spatial derivative terms after dividing by the coeffi cient of the fourth-order spatial derivative, it represents the ratio of the normal stresses due to axial load to the normal stresses due to bending. If ε � 1, the system is “lightly stretched.” If ε � 1, the beams are highly stretched.

9533_C001.indd 429533_C001.indd 42 10/29/08 1:30:12 PM10/29/08 1:30:12 PM


Similarly, if η � 1, the beams are lightly coupled. Asymptotic solutions as expansions in terms of powers of these independent dimensionless parameters can be used to approximate the natural frequencies and mode shapes.

Analysis of the highly stretched case is best performed with a different nondimensionalization, as illustrated in the Example 1.9.

Example 1.9 The differential equation for the transverse displacement of a nonuniform beam of length L subject to an axial load P, as illustrated in Figure 1.27, is

∂∂

∂∂

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟−

∂∂

+∂2

2

2

2

2

2

2

xEI x

wx

Pw

xA x( ) ( )ρ

wwt∂

=2

0 (a)

(a) Defi ne nondimensional variables such that the nondimensional equation for the transverse displacement is

ε α β∂∂

∂∂

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟−

∂∂

+∂∂

2

2

2

2

2

2

2

xx

wx

wx

xw

( ) ( )tt2

0= (b)

(b) A normal-mode solution of Equation b is of the form

w x t u x e i t( , ) ( )= ω (c)

where ω represents a natural frequency of the system and u(x) is its corre-sponding mode shape. Substitute Equation c into Equation b to obtain an ordinary differential equation for u(x) with ω as a parameter.

x P

L

ρ(x), A(x), E(x), I(x)

Figure 1.27 The nonuniform beam is subject to a constant axial load. When the governing equation is nondimensionalized, the highest-order spatial derivative is multiplied by a parameter defi ned as the ratio of the normal stresses due to axial load to the normal stresses due to bending.

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Solution Defi ne the nondimensional variables,

wwL

* = (d)

xxL

* = (e)

ttT

* = (f)

where T is a characteristic time whose form is to be determined. Using the subscript “0” to refer to the value of a function at x = 0, the following nondi-mensional functions can be defi ned:

α( )( ) ( )*x

E x I xE I

=0 0

(g)

βρ

ρ( )

( ) ( )*xx A x

A=

0 0

(h)

Substitution of Equation d, Equation e, Equation f, Equation g, and Equation h into Equation a leads to

1 1 12

2

2 0 0 2

2

2L xE I x

L xLw P

∂∂

∂∂

⎡

⎣⎢⎢

⎤

⎦⎥⎥−

**

**( ) ( )α

LL xLw

AT

xt

Lw2

2

2

0 0

2

2

20

∂∂

( )+∂

∂=

** *

**( ) ( )

ρβ (i)

Multiplying by L P2/ , Equation i can be rearranged to give

E IPL x

xwx

wx

0 0

2

2

2

2

2

2

2

∂∂

∂∂

⎡

⎣⎢⎢

⎤

⎦⎥⎥−

∂∂*

**

*

*

*( )α ++

∂∂

=ρ

β0 02

2

2

20

A LPT

xwt

( )**

* (j)

Choosing

T LAP

=ρ0 0

(k)

defi ning

ε =E IPL

0 0

2 (l)

and dropping the *’s from nondimensional variables, Equation i is found to be identical to Equation b.

(b) Substitution of Equation c into Equation b gives

ε β ωd

dxd udx

d udx

x u2

2

2

2

2

22 0α

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟− − =( ) (m)

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1.5 Scaling“Small” and “large” are relative terms. The microscale is large compared to the nanoscale, but small compared to the macroscale. When modeling physi-cal systems, one often cannot determine whether an effect is small and poten-tially negligible until it is compared to other effects in the system. Even then it is not possible simply to compare dimensional quantities, and therefore the equation is nondimensionalized and nondimensional parameters introduced. It is often possible, by comparing magnitudes of nondimensional param-eters, to neglect specifi c effects when modeling a physical system. Even this approach does not always work, because it is possible that a seemingly small parameter will be multiplied by large derivatives such that the product is of the same order of magnitude as other terms.

The above discussion suggests the following approach: model, nondimen-sionalize, and compare magnitudes of dimensionless parameters. Such a procedure can also be used, not only to refi ne a model, but also to determine the best way to solve mathematically the equations derived during the math-ematical modeling exercise.

The differential equation governing the unsteady state temperature distri-bution over the extended surface shown in Figure 1.28a is derived by apply-ing an energy balance to the differential volume shown in Figure 1.28b. The result is

kT

xT

yc

Tt

∂∂

+∂∂

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟=

∂∂

2

2

2

2ρ (1.33)

The boundary conditions when the surface at x = 0 is maintained at a constant temperature, the surface at x = L is insulated, and heat transfer with the ambi-ent medium occurs over the top and bottom of the fi n are represented by y = 0 and y = w. The mathematical formulations of the boundary conditions are

T y t T0 0, ,( )= (1.34)

∂∂

( )=Tx

L y t, , 0 (1.35)

kTy

x t h T x t T∂∂

( )− ( )−[ ]=∞, , , ,0 0 0 (1.36)

kTy

x w t h T x w t T∂∂

( )− ( )−[ ]=, , , , ∞ 0 (1.37)

The following nondimensional variables are introduced:

xxL

yyw

tkcL

tT TT T

* * * ,= = = =−−ρ

θ ∞

∞2

0

9533_C001.indd 459533_C001.indd 45 10/29/08 1:30:15 PM10/29/08 1:30:15 PM


which, when substituted into the partial differential equation and boundary conditions, lead to

∂∂

+( ) ∂∂

=∂∂

2

2

2 2

2

θ θ θx

Lw y t

(1.38)

θ( , , )0 0y t = (1.39)

qx(x + dx, y)

dx

qc =hP(T(x, y)–T∞)

qx(x, y)w

qc

(c)

(a)

qx(x + dx, y)

dx

qy(x, y + dy)

qx(x, y)dy

qy(x, y)

(b)

L

To x

y

T 8

w

Figure 1.28 (a) A mathematical model for the temperature distribution in the extended surface is developed using several different sets of assumptions. (b) The temperature distribution is two-dimensional, a function of x and y, when L/w is O(1). (c) The temperature distribution is assumed to be a function of x only. The effect of convective heat transfer from the surface is lumped.

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∂∂

( )=θx

y t1 0, , (1.40)

∂∂

( )−( ) =θ

θy

x thwk

x t, , ( , , )0 0 0 (1.41)

∂∂

( )+( ) =θ

θy

x thwk

x t, , ( , , )1 1 0 (1.42)

If L � w, then L/w is a large quantity. Thus, for all terms in Equation 1.38 to be of the same order of magnitude, ∂ ∂2 2θ/ y must be small. If this term is taken to be zero, solutions of Equation 1.38 are of the form θ( , , ) ( , ) ( , )x y t A x t B x t y= + . Application of the boundary conditions at y = 0 and y = w leads to B = 0. Then the temperature is a function only of x and t. However, simplifying the par-tial differential equation to ∂ ∂ = ∂ ∂2 2θ θ/ /x t implies that the temperature is independent of the heat transfer coeffi cient h. This does not seem reasonable. Instead, the derivation of the governing equation can be modifi ed to account for this assumption. An energy balance is applied to the differential segment of Figure 1.28c which includes heat transferred by convection through the upper and lower faces of the surface. The resulting model is

kwT

xh T T cw

Tt

∂∂

− −( )=∂∂

∞

2

22 ρ (1.43)

Substitution of the nondimensional variables into Equation 1.43 leads to

∂∂

− =∂∂

2

2

θθ

θx

Bit

(1.44)

where Bi = 2hL2/kw is the Biot number.Equation 1.44 subject to the boundary conditions given in Equation 1.39

and Equation 1.40 is not diffi cult to solve and is solved in Chapter 6. Suppose, though, that the surface is of nonuniform cross section, w = w(x). The govern-ing equation then becomes

∂∂

∂∂( )− =

∂∂x

xx

Bi xwt

α θ α( ) ( )θ

(1.45)

where α(x) is a nondimensional function representing the nonuniform cross section. Equation 1.45 may have exact solutions for some forms of α(x), but even when it exists, the exact solution is diffi cult to determine. If the Biot number is small, the rate of heat transfer by convection is small compared with the rate of heat transfer by conduction, and then Equation 1.45 can per-haps be approximated by

∂∂

∂∂( )=

∂∂x

xx

xwt

α α( ) ( )θ

(1.46)

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If Equation 1.46 is used, it provides a fi rst-order approximation of the tem-perature distribution. It is the fi rst term, θ0(x), in the asymptotic expansion of θ(x) in terms of the Biot number,

θ θ θ( ) ( ) ( ) ...x x Bi x= + +0 1 (1.47)

The decision to assume that the temperature is independent of y is based on the scaling of the variables. The decision to expand the nondimensional temperature distribution into an asymptotic expansion in terms of the Biot number can be made only when the problem is written in nondimensional variables. Asymptotic expansions can be made only by using an indepen-dent nondimensional variable as the expansion parameter.

Consider the system shown in Figure 1.11, in which a particle of mass m is attached to the end of an elastic bar of length L and cross-sectional area A. The bar is made from a material of elastic modulus E and mass density ρ. The particle is subjected to a time-dependent force, F t F t( ) sin( )= 0 ω . The system is to be modeled to determine its vibrational properties and its frequency response. Assume that all forms of friction are neglected and that normal stresses developed in the bar remain less than the material’s yield stress.

An important decision in modeling the system of Example 1.3 as illustrated in Figure 1.3 is whether to use a discrete model or a distributed-parameter model for the bar. A discrete model is appropriate if the inertial effects of the bar are small compared with the inertial effects of the particle. However, this is an ambiguous statement which is not easily quantifi ed. If inertial effects are neglected, the bar can be modeled by a massless linear spring of stiffness k EA L= / . The resulting mathematical model is Equation a of Example 1.3. Many vibration texts suggest that if the mass of the bar is not negligible, but not too large, the inertial effects of the bar can be approximated by imagin-ing that a particle whose mass is one-third of that of the bar is added to the particle, giving a better model

ˆ sin( )md ydt

AEL

y F t2

2 0+ = ω (1.48)

where

m m AL= +1

3ρ (1.49)

If the inertial effects of the bar are not neglected, the appropriate model is the partial differential equation given by Equation d of Example 1.3, subject to the boundary conditions of Equation e and Equation f.

Most readers are able to solve the ordinary differential equation derived for the discrete model, but are not able to solve the partial differential equation of the distributed parameter model, although all should be able to do so after completing the study of this book. For this reason, it is instructive to determine

9533_C001.indd 489533_C001.indd 48 10/29/08 1:30:18 PM10/29/08 1:30:18 PM


under what conditions the discrete model yields accurate results. This knowl-edge provides guidance in developing subsequent models.

The following nondimensional variables can be introduced into Equation d, Equation e, and Equation f of Example 1.3:

xxL

* = (1.50a)

uuL

* = (1.50b)

tL

Et* =

1

ρ (1.50c)

which leads to

∂∂

=∂∂

2

2

2

2

ux

ut

(1.51)

u t( , )0 0= (1.52)

−∂∂

+ =∂∂

ux

tF LEA

f tmAL

ut

t( , ) ( ) ( , )1 102 2 2

ρ (1.53)

The current discussion intends to determine the conditions under which the inertia of the bar can be neglected or approximated by an equivalent mass given by Equation b and the use of a one-degree-of-freedom assump-tion in modeling the system. One fundamental difference between the two models is that the one-degree-of-freedom model leads to the prediction of only one natural frequency and only one mode of free vibration, whereas the continuous model predicts an infi nite number of natural frequencies and corresponding mode shapes. Thus, since the one-degree-of-freedom model is limited in this regard and can be used only when the lowest natural frequency is required, one means of comparison is to determine the lowest natural frequency obtained using both models.

The natural frequency obtained using the one-degree-of-freedom model is

ω1 1

3

=+( )EA

L m ALρ

=+

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

EA

ALmAL

ρρ

2 1

3

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=+

1 3

3 1LEρ β

(1.54)

where

βρ

=mAL

(1.55)

is the ratio of the mass of the particle to the mass of the bar.The natural frequency obtained from solution of Equation 1.51, Equation

1.52 and Equation 1.53 is a nondimensional frequency, ω ω ρ* ( )= 1/ /L E . Then the comparison can be made to determine the values of β such that the lowest nondimensional frequency obtained from solving Equation 1.51 is close to 3 3 1/( )β+ .

Solution of Equation 1.51 subject to Equation 1.52 and Equation 1.53 using the methods developed in Chapters 5 and 6 reveals that the lowest natural frequency for a given value of β is the smallest positive solution of the tran-scendental equation,

tan ωβω

=1

(1.56)

Figure 1.29 shows a comparison between the values of ω calculated from Equation k and the values of 3 3 1/( )β+ , whereas Figure 1.30 illustrates the percent error in using a one-degree-of-freedom model to determine the low-est natural frequency.

The above example illustrates the appropriate use of approximations. Development of a mathematical solution to a problem obtained through mathematical modeling often requires an analysis of each term in the equa-tions to determine whether or not its effect is as large as other effects consid-ered in the model. In comparing two terms, it is often convenient simply to look at the magnitudes of the parameters used to multiply each term. If two parameters, say η and v, are comparable in magnitude, then η ν= ( )O , which is read, “η is big ‘O’ of v.” The formal defi nition of the big “O” symbol is that f O g( )ν ν= ( )( ) if lim ( )/ ( )

νν ν

→= =

0f g A such that 0 < <A ∞.

This analysis is used to identify the largest or dominant terms in the model. Obtaining a mathematical solution using only the dominant terms is viewed as a fi rst approximation to the solution. This approximation may be improved by using an asymptotic expansion in terms of a small dimensionless parameter. The problem involving only the dominant terms is often signifi cantly simpler than most problems.

Consider again the system of Example 1.8 in which two stretched beams are connected through an elastic layer. The differential equations governing the beam displacements, Equation n and Equation p of Example 1.8, are repeated

∂∂

−∂∂

+ − +∂∂

=4

1

4

21

2 1 1 2

21

2

wx

wx

w wwt

ε η ( ) 0 (1.57)

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μ ε η η β∂∂

−∂∂

+ − + +∂∂

=4

24

22

2 1 2 1 2 2

22

2

wx

wx

w w wwt

( ) 0 (1.58)

The nondimensional parameters can be recognized as follows:

ε is the ratio of the normal stress due to axial load to the normal stress due to bending.η1 is the ratio of the stiffness of the elastic layer connecting the beams to the stiffness of the fi rst beam.η2 is the ratio of the stiffness of the elastic layer between the second beam and the fi xed support to the stiffness of the fi rst beam.μ is the ratio of the stiffness of the second beam to the stiffness of the fi rst beam.β is the ratio of the inertia of the second beam to the inertia of the fi rst beam.

Equation n and Equation p are coupled through the elastic layer. If η1 � 0(1), then the beams are lightly coupled. A fi rst approximation, obtained by

•

•

•

•

•

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

0.8

1

1.2

1.4

1.6

1.8

2

Discrete model

Continuousmodel

ω

β

Figure 1.29 The agreement between the lowest natural frequency calculated for the system shown in Figure 1.11 using a discrete model and using a continuous model increases as the mass ratio increases.

9533_C001.indd 519533_C001.indd 51 10/29/08 1:30:21 PM10/29/08 1:30:21 PM


neglecting the coupling terms, leads to a set of uncoupled equations which is much easier to solve. The coupling is then a second-order effect, and its effects are included through an asymptotic expansion in terms of powers of η1.

If ε is small, then the beams are lightly stretched. A fi rst approximation, obtained by neglecting the stretching terms, is that of coupled Euler-Bernoulli beams. It is much easier to solve for the defl ections of coupled Euler-Bernoulli beams than for defl ections of coupled stretched beams.

A problem which clearly illustrates the necessity of understanding the effects of scaling is that of a uniform fl ow impinging on a fl at plate, as shown in Figure 1.31. Assuming steady-state two-dimensional fl ow of an incom-pressible fl uid, the Navier-Stokes equations are:

∂∂

+∂∂

=ux

vy

0 (1.59)

ρ μuux

vuy

px

ux

uy

∂∂

+∂∂

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟= −

∂∂

+∂∂

+∂∂

2

2

2

22

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟ (1.60)

0 0.2 0.4 0.6 0.8β

1 1.2 1.4 1.6 1.8 20

1

2

3

4

5

6

7

8

9

10

Perc

ent e

rror

Figure 1.30 The percent error in calculation of the lowest natural frequency of the system shown in Figure 1.11 when using a discrete model is only 10 percent for a small mass ratio and less than 1 percent for a mass ratio of 0.6.

9533_C001.indd 529533_C001.indd 52 10/29/08 1:30:23 PM10/29/08 1:30:23 PM


ρ μuvx

vvy

py

vx

vy

∂∂

+∂∂

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟= −

∂∂

+∂∂

+∂∂

2

2

2

22

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟ (1.61)

Equation 1.59 is a statement of conservation of mass, while Equation 1.60 and Equation 1.61 are components of the conservation-of-momentum equa-tion. The terms on the left-hand side of the latter equations are convective acceleration or inertia terms. The terms on the right-hand sides represent pressure and viscous forces.

Equation 1.59, Equation 1.60, and Equation 1.61 must be supplemented by boundary conditions. The no-slip condition requires that u = 0 on the surface of the plate, while the no-penetration condition requires that v = 0 on the surface of the plate. Far away from the plate, u approaches the free-stream velocity U, while v approaches zero.

Equation 1.59, Equation 1.60, and Equation 1.61 can be nondimensional-ized using x x L y y L u u U v v U p p U* * * * *, , ,= = = = =/ / / / and /ρ 2 , where L is any characteristic length. Use of nondimensional variables leads to:

∂∂

+∂∂

=ux

vy

0 (1.62)

uux

vuy

px

ux

uy

∂∂

+∂∂

= −∂∂

+∂∂

+∂∂

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟

1 2

2

2

2Re ⎟⎟⎟ (1.63)

uvx

vvy

py

vx

vy

∂∂

+∂∂

= −∂∂

+∂∂

+∂∂

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟

1 2

2

2

2Re ⎟⎟⎟ (1.64)

δ = O(Re–1/2)

(a)

(b)

U

Figure 1.31 (a) A uniform fl ow impinging on a fl at plate is analyzed using the Navier-Stokes equation. (b) When the governing equations are nondimensionalized, the highest-order derivatives are multiplied by a small parameter. This leads to the hypothesis of a boundary layer which grows along the plate. Further scaling shows that the boundary-layer thickness is inversely proportional to the square root of the Reynolds number.

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where all variables in Equation 1.62, Equation 1.63, and Equation 1.64 are nondimensional and the Reynolds number is defi ned Re = ρ μUL/ .

Methods for solving Equation 1.62, Equation 1.63, and Equation 1.64 subject to boundary conditions depend on the value of Re. For small Re, 1/Re is large, suggesting that the viscous terms dominate the pressure and inertial terms. An asymptotic solution can be developed using expansions of the form:

p p p O= + + ( )+0 12Re Re ... (1.65)

u u u O= + + ( )+0 12Re Re ... (1.66)

v v v O= + + ( )+0 12Re Re ... (1.67)

For large Re, 1/Re is small, suggesting that the inertial and pressure forces dominate the viscous forces. A fi rst attempt to approximate the solutions might be to try asymptotic expansions of the form:

p p p O= + + ( )+0 1 2

1 1

Re Re... (1.68)

u u u O= + + ( )+0 1 2

1 1

Re Re... (1.69)

v v v O= + + ( )+0 1 2

1 1

Re Re... (1.70)

The hierarchical equations obtained at the lowest order do not include vis-cous forces. However, such an attempt changes the nature of the differential equations at the lowest order. Equations containing the viscous forces are second-order in x and y, whereas equations resulting from neglecting the viscous forces are fi rst-order in both x and y. Mathematically, the no-slip condition cannot be satisfi ed by this approximation.

Physically, the shear stresses are largest near the surface of the plate. Even though 1/Re is small, terms such as ∂ ∂2 2u y/ are as large as Re near the surface of the plate and must be included in solving the equations near the surface. However these terms are small far from the plate and can then be neglected.

The above analysis leads to the hypothesis of a boundary layer, a region near the surface of the plate in which the viscous forces are as large as the pressure and inertial forces. This suggests rescaling the y-coordinate within the boundary layer as

yy**

*

=δ

(1.71)

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where δ is the boundary-layer thickness. Substituting Equation 1.71 into Equation 1.62, Equation 1.63, and Equation 1.64 leads to:

∂∂

+∂∂

=ux

vy

10

δ (1.72)

uux

v uy

px

ux

uy

∂∂

+∂∂

= −∂∂

+∂∂

+∂∂

⎛

⎝⎜⎜⎜δ δ

1 12

2 2

2

2Re

⎞⎞

⎠⎟⎟⎟⎟ (1.73)

uvx

v vy

py

vx

vy

∂∂

+∂∂

= −∂∂

+∂∂

+∂∂

⎛

⎝⎜

δ δ δ1 1 12

2 2

2

2Re⎜⎜⎜

⎞

⎠⎟⎟⎟⎟ (1.74)

where the *’s have again been dropped from nondimensional variablesThe viscous terms should be as large as the inertial and pressure terms

within the boundary layer. However, conservation of mass must also be sat-isfi ed. Because δ is a small dimensionless quantity, the two terms in Equation 1.72 are of the same order of magnitude only if v = O(δ). Thus the y- component of velocity can be rescaled by introducing v** = v/δ such that v** = O(1). Substi-tution into Equation 1.72, Equation 1.73, and Equation 1.74 leads to:

∂∂

+∂∂

=ux

vy

0 (1.75)

uux

vuy

px

ux

uy

∂∂

+∂∂

= −∂∂

+∂∂

+∂∂

⎛

⎝⎜⎜⎜

⎞1 12

2 2

2

2Re δ ⎠⎠⎟⎟⎟⎟ (1.76)

uvx

vvy

py

vx

vy

δ δδ

δδ

∂∂

+∂∂

= −∂∂

+∂∂

+∂∂

⎛

⎝1 1 12

2

2

2Re⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟ (1.77)

where the *’s have again been dropped. That is, the variables in Equation 1.75, Equation 1.76, and Equation 1.77 are boundary-layer variables.

The order of magnitude of the boundary-layer thickness is determined by requiring that the viscous forces balance with the inertial and pressure forces in Equation 1.76. Invoking this requirement leads to:

δ = ( )O1

Re (1.78)

Selection of the boundary-layer thickness according to Equation 1.78 makes the largest term in the y-momentum equation, Equation 1.77, equal to ∂p/∂y. Thus the y-component of the momentum equation is satisfi ed to the lowest order in the boundary layer by requiring that the pressure be constant across the boundary layer.

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A similar analysis can be used for the problem of Example 1.9, a fi xed free beam with an axial load. The differential equation governing the displace-ment of the beam is Equation m of Example 1.9 which is repeated below:

ε α β ωd

dxd udx

d udx

x u2

2

2

2

2

22 0

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟− − =( ) (1.79)

If ε is small, and if an asymptotic expansion were attempted for the dis-placement, it would fail because the hierarchical equation obtained at the lowest order is a second-order differential equation which cannot satisfy all four required boundary conditions. Thus there must be a region where

d d u dx dy2 2 2 2[ ( )]α / / is large enough that the bending term balances with at least one other term. This region is also a boundary layer, or due to the phys-ics of the problem, a bending layer. Actually, there are two bending layers, one at each end of the beam. The parameter ω is a natural frequency and has an infi nite but countable number of values. Consider fi rst the case when ω2 1= O( ). Let δ be the boundary-layer thickness, and defi ne the bound-ary-layer variable as x x**= /δ . Equation m can be rewritten in terms of the boundary-layer variable as:

ε

αδ δ

β ω4

2

2

2

2 2

2

221d

dxd udx

d udx

x⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟− − ( ) uu = 0 (1.80)

If δ ε= ( )O 1 2/ , then both the bending and stretching terms are O ε−( )1 , while the inertia term is O(1). Thus the bending and stretching terms interact in the boundary layers.

Suppose ω ε κ/ε2 3 3= =−O( ) . Then Equation 1.80 written in boundary-layer variables becomes

εδ

αδ

βε4

2

2

2

2 2

2

2

1ddx

d udx

d udx

x⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟− − ( )

κ33

0u = (1.81)

If δ ε= ( )O , then both the bending and inertial terms are O ε−( )3 , while the stretching term is O ε−( )2 . Thus the bending and inertial terms interact in the boundary layer for larger frequencies.

The previous two problems are said to have “singularities” such that straightforward asymptotic expansions of the form of Equation 1.68, Equation 1.69, and Equation 1.70 are unsuccessful in obtaining a solution. The nondi-mensionalization and the physics of the problem lead to the discovery of boundary layers.

Problems 1.1. An elastic bar has a static displacement due to an applied axial load

F. The force is suddenly removed, resulting in free oscillations of the bar. Derive the expression governing the displacement u(x, t) of a particle along the axis of the bar.

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a. Specify the governing differential equation as well as all bound-ary and initial conditions necessary to solve for u(x, t).

b. Nondimensionalize the problem by introducing x x L* = / and t t T* = / , where T is to be chosen for convenience.

1.2. The elastic bar of Figure P1.2 is fi xed at x = 0 and has a particle of mass m attached at x = L. The particle is attached to a spring in par-allel with a viscous damper. The differential equation governing the displacement of a particle along the axis of the bar is:

EAu

xA

ut

∂∂

=∂∂

2

2

2

2ρ (a)

a. Specify the boundary conditions at x = 0 and x = L. b. The particle of mass m is displaced a distance δ to the right, held

in this position, and released. Specify the initial conditions for the resulting vibrations.

c. Introduce nondimensional variables, x* = x/L, u* = u/L and t* = t/T. Substitute the nondimensional variables into Equation a, the boundary conditions, and the initial conditions. Choose T such that the resulting partial differential equation is:

∂∂

=∂∂

2

2

2

2

ux

ut

(b)

where all variables in Equation b are nondimensional.

m

ρ,A,E

L

u(x, t)

x

c

k

(a)

m + ρAL

EA

L

y

c

k

(b)

13

Figure P1.2 (a) System for problem 1.2 (b) Discrete system model for problem 1.2(e).

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d. Identify and physically defi ne all nondimensional parameters which appear in the nondimensional formulation of the bound-ary conditions.

e. If the inertia of the elastic bar is small compared to the inertia of the particle, the system shown in Figure P1.2a can be modeled by the discrete system shown in Figure P1.2b. The inertia effects of the bar can be approximated by adding a particle of mass 1 3/ ρAL to the existing particle. Derive the differential equation govern-ing the displacement y(t) of the particle. Express the appropriate initial conditions when the particle is displaced as in part (b).

f. Nondimensionalize the differential equation obtained in part (e). g. Discuss a strategy to determine the accuracy of the approxima-

tion using the mass ratio m AL/ρ . 1.3. The shaft shown in Figure P1.3a is nonuniform such that its polar

moment of inertia varies with x, J = J(x). Let θ( , )x t represent the angular displacement of the plane a distance x from the left support. A differential element of thickness dx is illustrated in Figure P1.3b.

a. Use the free-body diagram to derive the partial differential equation governing the torsional oscillations of the shaft. Specify appropriate boundary conditions.

b. Introduce nondimensional variables x x L t t T* *= =/ and / and a nondimensional function α( ) ( )x J x J= / 0 , where J0 is the polar moment of inertia of the shaft at x = 0. Substitute these variables into the differential equation and choose T such that the govern-ing differential equation contains no parameters.

c. The shaft is circular with a radius which varies according to r x r x( ) = +( )0 1 μ , where r0 is the shaft radius at x = 0 and μ is the rate of increase (or decrease if μ is negative) of the radius and has dimensions of (length)−1. Write the nondimensional differential equation obtained in part (b) for this case.

x

L

ρ, G, J(x)

θ(x, t)

(a) (b)

M(x + dx)M(x)

dx

Figure P1.3 (a) System of problems 1.3 and 1.4. (b) Free-body diagram of differential element at an arbitrary instant.

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1.4. Consider the shaft shown in Figure P1.3a when it is subject to a harmonic torque M t M t( ) sin= ( )0 ω applied at its free end. The radius of the shaft varies linearly as described in problem 1.3c. Let Θ represent the steady-state amplitude of the end of the shaft. The amplitude is a function of the parameters ρ,G,L,r0,μ,M0 and ω. That is, Θ Θ= ( , , , , , )ρ μ ωG L r M0 0 , . Determine a set of nondimensional parameters π1,π2,π3 and π4 for a non-dimensional formulation of the relationship between the parameters, Θ Θ= ( )π π π π1 2 3, , , 4 .

1.5. The steady-state amplitude of the mass attached to the bar shown in Figure P1.5 is:

XF L

A L E LE

m L LE

= ⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟−

⎛

⎝⎜

0

2ω ρ ωρ

ω ωρ

cos sin⎜⎜⎜⎞

⎠⎟⎟⎟⎟

(a)

Equation a shows X X F A L m E= ( )0 , , , , , ,ω ρ .

a. Determine a set of nondimensional parameters which can be used to formulate Equation a in a nondimensional form.

b. Rewrite Equation a in terms of these nondimensional parameters.

1.6. The temperature distribution over the extended surface of Figure P1.6 is:

T x T T Th

kwx

hkw

L( ) ( ) cosh tanh sinh= + − ( )− ( )∞ ∞12 2 2hh

kwx( )⎡

⎣⎢⎢

⎤⎦⎥⎥ (a)

The rate of heat transfer from the base of the extended surface is

Q kwdTdx

= − � ( )0 (b)

Develop a nondimensional relationship between the heat transfer at the base and appropriate nondimensional parameters. Speculate on the physical meanings of the nondimensional parameters.

Fosin(ωt)m

ρ, A, E

L x(t)

Figure P1.5 System of problem 1.5.

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1.7. The differential equation governing the transverse defl ection of a pipe with an inviscid fl uid fl owing with a velocity U is:

EIw

xMU

wx

MUw

x tM m g

wx

M∂∂

+∂∂

+∂∂ ∂

+ +∂∂

+4

42

2

2

2

2 ( ) ( ++∂∂

=mwt

)2

20 (a)

where E is the elastic modulus of the pipe, I is the area moment of inertia of the pipe’s cross section, M is the mass of the fl uid in the pipe per unit length, and m is the mass of the pipe per unit length. Defi ne the following nondimensional variables:

xxL

wwL

tEI

M mtL

* * *= = =+

⎡⎣⎢⎢

⎤⎦⎥⎥

1

2

2 (b)

Nondimensionalize Equation a through introduction of the nondi-mensional variables of. Write the resulting nondimensional equa-tion in terms of the nondimensional parameters, β = +M M m/( )

=[ ]u M EI ULand / /1 2 . 1.8. The general form of the momentum equation for the fl ow of a viscous

fl uid near a free surface is:

ρ μ ρDDt

gv

v p j= ∇ −∇ +2 (a)

x

L

w

T 8,h

Ik


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where v i j k= + +u v w is the velocity vector, ρ is the mass density of the fl uid, p is the fl uid pressure, μ is the dynamic viscosity, and g is the acceleration due to gravity. Let L be a characteristic length of fl ow, V be a characteristic velocity, and p∞ be a characteristic pressure, and defi ne nondimensional variables as:

vv* =V

(b)

t tVL

* = (c)

pp p

V* =

− ∞

ρ 2 (d)

xxL

yyL

zzL

* * *, ,= = = (e)

a. Substitute Equation b, Equation c, Equation d, and Equation e into Equation a to obtain

DDt

pFr

vv j= −∇ + ∇ +

1 12

Re (f)

where Re is the Reynolds number, Fr is the Froude number, and all variables are taken as nondimensional.

b. Defi ne the Froude number mathematically and physically. c. Write the component equations represented by Equation f.

1.9. The momentum equation for the fl ow of a fl uid with free convection is:

ρ ρDDt

T T gv

v= ∇ − −μ β ∞2 ( ) j (a)

where, in addition to the variables defi ned in problem 1.8, β is the coeffi cient of thermal expansion and T is temperature. The non-dimensional variables defi ned in problem 1.8 are used as well Θ = − −( )/( )T T T T∞ ∞1 where T1 and T∞ are reference temperatures.

a. Nondimensionalize Equation a. b. The nondimensional formulation of Equation a contains two non-

dimensional parameters, Re and Gr, the Grashopf number. Defi ne the Grashopf number both mathematically and physically.

1.10. Use Lagrange’s equations to derive the differential equations gov-erning the motion of the system shown in Figure P1.10.

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1.11. A non-uniform bar of length L is in a medium of ambient temperature T∞. One end of the bar is insulated, while the other end is maintained at a temperature of T0. The bar is made of a smart material, which produces an internal heat generation proportional to the gradient of the temperature. Assuming one-dimensional heat generation along the length of the bar and defi ning x as a coordinate along the length of the bar, the differential equation governing the temperature dis-tribution in the bar is

k

ddx

A xdTdx

dTdx

hP x T T( ) ( )( )( )+ + − =∞η 0

(a)

where k is the thermal conductivity of the bar, h is the heat transfer coeffi cient between the bar and the ambient, and η is a parameter related to the internal heat generation. The appropriate boundary conditions are

T(0) = T0 (b)

dTdx

L( ) = 0

(c)

(a) Non-dimensionalize Equation a, Equation b, and Equation c through introduction of appropriate non-dimensional variables.

(b) Identify mathematically and physically all parameters appearing in the non-dimensional formulation.

(c) Write a non-dimensional formulation of the problem when

A x A

xL

( ) = −⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟0

2

1 μ

(d)

and

P x w b

xL

( ) = + −⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟2 2 10 μ

(e)

x1 x2

k 2k

x3

2k

k km 2m

m


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(d) Suppose the ratio of the rate of conduction to the rate of con-vection is assumed to be small, 1/Bi = O(ε), where Bi is the Biot number and ε is a small non-dimensional parameter. Speculate whether a boundary layer exists. If so, determine the thickness of the boundary layer and which terms physically interact within the boundary layer.

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65

Chapter 2

Linear algebra

2.1 IntroductionLinear algebra is the algebra used for analysis of linear systems. It provides a foundation on which solutions to mathematical problems can be developed. Success in obtaining a solution to a mathematical problem requires fi nding the specifi c solution among a possible set of solutions, the solution space. An understanding of the solution space and the properties of elements of the solution space leads to the development of solution techniques. It is in this spirit that this review of linear algebra is presented.

An exact solution of a problem is a solution for the dependent variables which satisfi es without error the mathematical problem for all possible values of the independent variables. An exact solution, while desirable, is not always possible. Approximate solutions are sought when an exact solution is not available. Approximate solutions are of two types. Variational methods are used to determine continuous functions of the independent variables which provide in some sense the “best approximation,” chosen from a specifi ed set, to the exact solution. Numerical solutions provide an approximation to the exact solution only at discrete values of independent variables. Linear algebra provides a framework in which these approximate solutions can be devel-oped and in which the error between the exact solution and an approximate solution can be estimated.

Linear algebra provides a framework for developing solutions to linear prob-lems. Modeling of engineering systems often leads to nonlinear mathematical problems. Exact solutions exist for only a few nonlinear problems. Often assump-tions are made such that the nonlinear problem can be approximated by a linear problem. Even if the assumptions that linearize the problem are not valid, some understanding of the solution can be obtained by studying the linearized prob-lem. Indeed, knowledge of the behavior of the linearized system is necessary to understand the effect of nonlinearities on the system behavior. Numerical methods developed to approximate solutions of nonlinear problems are based upon numerical methods used to develop approximate solutions for linear problems. Approximate solutions of nonlinear problems are often assumed to be perturbations of the linear solution. Thus knowledge of linear solutions is necessary to develop approximate solutions for nonlinear problems.

This chapter provides a review of linear algebra and linear operator the-ory. Theorems are presented, with proofs when these proofs are themselves instructive.

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2.2 Three-dimensional spaceEvery point in three-dimensional space has a unique set of x-y-z coordinates defi ned in a fi xed Cartesian reference frame, as shown in Figure 2.1. The location of a particle at any instant of time is defi ned by the Cartesian coor-dinates of the point it occupies in space. A position vector, r, defi ning the location of the particle is a line segment drawn from the origin (0,0,0) of the Cartesian system to the particle. The position vector has a unique direction with respect to the Cartesian frame and a calculable length denoted by |r|.

Let r1 and r2 represent two vectors defi ned in a Cartesian system. Vector addition in three-dimensional space is defi ned geometrically, as illustrated in Figure 2.2a. A vector parallel to v is drawn such that its tail coincides with the head of the second vector u. If w = r1 + r2, then w is the vector drawn from the origin to the head of the vector parallel to r2. Figure 2.2b shows the process repeated, but with a vector parallel to r1 drawn with its tail coincid-ing with the head of v. The resulting sum is the same as that obtained in Figure 2.2a. Thus vector addition is commutative,

r r r r1 2 2 1+ = + (2.1)

Figure 2.2 illustrates the triangle rule for vector addition. The vectors being added and the resultant (sum) are depicted as sides of a triangle. Since the

x

y

z

r

(x,y,z)

Figure 2.1 A point in a Cartesian coordinate system is referenced by three coordi-nates (x,y,z). A position vector is drawn from the origin to the point.

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Chapter 2: Linear algebra 67

length of any side of a triangle must be less than the sum of the lengths of the other two sides,

r r r r1 2 1 2+ ≤ + (2.2)

Equation 2.2 is called the triangle inequality.The concept of multiplication of a vector by a scalar is illustrated in

Figure 2.3. Let α be any real value. The vector αr is the vector parallel to r whose length is α times the length of r. If α is positive, then the vector α r lies in the same direction as r. If α is negative, the vector α r lies in the direction opposite that of r. If α equals zero, then αr is a vector whose length is zero and is called the zero vector, 0.

The following properties follow from the defi nitions of vector addition and multiplication of a vector by a scalar:

i. Associative law of addition: (r1 + r2) + r3 = r1 + (r2 + r3) ii. Commutative law of addition: r1 + r2 = r2 + r1

iii. Addition of zero vector: 0 + r1 = r1

iv. Multiplication by one: (1)r1 = r1

v. Negative vector: −r1 = (−1)r1 r1 + (−r1) = 0 vi. Distributive law of scalar multiplication: ( )α β α β+ = +r r r1 1 1

vii. Associative law of scalar multiplication: ( ) ( )αβ α βr r1 1= viii. Distributive law of addition: α α α( )r r r r1 2 1 2+ = +

A unit vector is a vector whose length is one. Let i , j, and k be a set of unit vectors parallel to the x, y, and z coordinate axes respectively. Using the defi nitions of vector addition and multiplication of a vector by a sca-lar, the position vector for the particle can be written in terms of the unit vectors as:

r i j k= + +x y z (2.3)

r1

(a) (b)

r1 + r2

r1 + r2 r2

r2

r1

Figure 2.2 (a) Vector addition is illustrated using the triangle rule. The resultant is drawn from the tip of r1to the head of r2. (b) Vector addition is commutative.

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Any vector in three-dimensional space may be written as a linear combina-tion of the trio of unit vectors. The vectors i, j, and k form a basis for three-dimensional space.

The Pythagorean theorem is used to determine the length of the position vector as

r = + +x y z2 2 2 (2.4)

A scalar function of two vectors, called the dot product, is defi ned by

r r r r1 2 1 2⋅ = cosθ (2.5)

where θ is the angle made between r1 and r2. The dot product has the geomet-ric interpretation that it is equal to the length of the projection of the vector r1 onto r2. This leads to the calculation of the dot product as:

r r1 2 1 2 1 2 1 2⋅ = + +x x y y z z (2.6)

It should be noted that when computed using Equation 2.6, the dot product has the following properties:

Commutative property: r1 . r2 = r2 . r1

Multiplication by scalar: ( ) ( )α αr r r r1 2 1 2⋅ = ⋅ Distributive property: (r1 + r2) . r3 = r1 . r3 + r2 . r3

Non-negative property: r1 . r1 ≥ 0 and r1 . r1 = 0 if and only if r1 = 0

r

αr, α > 1

(a)

r

αr, α < 0

(b)

Figure 2.3 (a) Multiplication of a vector by a scalar leads to a vector lying in the same direction as the vector, but whose length is multiplied by the scalar. (b) If the scalar is negative, the product vector opposes the original vector.

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It should also be noted that from Equation 2.5 and Equation 2.6,

r r r⋅ = 2 (2.7)

Three-dimensional space, the defi nition of a vector, and the operations of vector addition and multiplication by a scalar are generalized in section 2.3 into the concept of vector spaces. The properties of vectors in a general vector space are defi ned in the same way as the properties satisfi ed by vectors in three-dimensional space. The scalar function of the dot product is generalized in section 2.5 into the concept of inner products.

Example 2.1 A force, F, acting on a particle is represented as a vector in three-dimensional space. The force has a magnitude |F| and lies in the direc-tion of a unit vector e. The resultant force acting on a particle is the vector sum of all forces acting on the particle. A free-body diagram of a particle at a given instant is shown in Figure 2.4. Determine the resultant force acting on the particle at this instant. Specify its magnitude and a unit vector in the direction of the resultant.

Solution The forces are represented as:

F j1 300= − N (a)

F e2 2600= N (b)

F e3 3500= N (c)

z

y

x

300N

(0, 1, 0)

600N

600N

(4, 2, 2)

(–2, 4, 0)

500N

Figure 2.4 Forces on a free-body diagram are drawn in the direction of their line of action with magnitude shown. The forces are thus modeled as vectors.

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where the unit vectors are

ei j k

i j k

22 2 2

4 2 2

4 2 2

6

3

6

6

6

6

=+ +

+ +

= + +

( ) ( ) ( ) (d)

ei j

i j

32 2

2 4

2 4

5

5

2 5

5

=− +

− +

= − +

( ) ( ) (e)

The resultant force acting on the particle is

F F= + +

= −( ) + − + +( ) +

1 2 3

200 6 100 5 300 100 6 200 5 1

F F

i j 000 6

266 3 392 2 244 9

k

i j k

⎡⎣

⎤⎦

= + +[ ]

N

N. . .

(f)

The magnitude of the resultant force is

F = ( ) + +

=

266 3 392 2 244 9

533 6

2 2 2. ( . ) ( . )

.

N

N

(g)

The resultant force is in the direction of the unit vector,

e

i j k

i j

=+ +

= + +

266 3 392 2 244 9

533 6

0 499 0 735 0

. . .

.

. . ..459k

(h)

Example 2.2 The motion of a particle is tracked by its position vector r, a vector drawn from the origin of a Cartesian coordinate system to the position of the particle. Since the particle moves in space, the particle’s position vector is a function of time as well as of the initial position of the particle. The veloc-ity is defi ned as the time rate of change of the position vector,

vr

=ddt

(a)

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By defi nition, the velocity vector is always tangent to the spatial path tra-versed by the particle. The acceleration is defi ned as the time rate of change of the velocity vector,

av r

= =ddt

ddt

2

2 (b)

Equation a and Equation b represent the velocity and acceleration of the par-ticle as seen by an observer attached to the particle. Modeling of a system using the views of observers attached to particles is called a Lagrangian formulation.

The position vector of a particle moving in space, referenced to a Carte-sian frame, is determined to be

r i j k= ( ) + +[ ]cos sin( ) .2 2 0 5t t t m (c)

where t is in seconds.Determine a unit vector tangent to the path of motion at any instant of

time; (b) Determine the component of the acceleration normal to the path of motion.

Solution (a) The velocity vector is determined by applying Equation a to the position vector of Equation c, leading to

v i j k= − + +[ ]2 2 2 2 0 5sin( ) cos( ) .t tm

s (d)

Since the velocity vector is always tangent to the path of motion, it can be expressed as

v v et= (e)

where et is a unit vector tangent to the path. To this end,

v = −[ ] +[ ] +

=

2 2 2 2 0 52 2 2sin( ) cos( ) ( . )t tm

s

2.87m

s

(f)

Using Equation d and Equation f, Equation e can be rearranged to give

e

v

v

i j k

t =

= − + +0 693 2 0 693 2 0 174. sin( ) . cos( ) .t t

(g)

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The acceleration vector can be calculated from Equation b and Equation d as

a i j= − −[ ]4 2 4 2cos( ) sin( )t tm

s2 (h)

The tangential component of the acceleration is

at t= ⋅a e (i)

which using Equation g and Equation h, becomes

a t t tt = − −[ ] − +4 2 4 2 0 693 2 0cos( ) sin( ) . sin( ) .i j ii 6693 2 0 174

0

cos( ) .t j k+[ ]

= (j)

Thus, since the tangential component of the acceleration is zero, the accelera-tion is in a direction normal to the surface.

Example 2.3 A Eulerian formulation is usually used in modeling the fl ow of a fl uid. In a Eulerian formulation, the properties used are those observed at a fi xed location in the fl ow fi eld as different fl uid particles pass through the point. In a Eulerian formulation, the properties are functions of spatial coordinates and time. If v(x,y,z,t) is the Eulerian velocity vector at a point (x,y,z) in the fl ow fi eld, then the acceleration is comprised of two terms. The local acceleration is the rate of change of the velocity and is simply calculated a vl /= ∂ ∂t. The convective acceleration is the rate of change of the velocity at the point (x,y,z) due to the rate at which fl uid particles pass through the point. The convective acceleration is calculated as ac = v . ∇v, where

v v i⋅∇ =∂∂

+∂∂

+∂∂

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟ +v

vx

vvy

vvz

vxx

yx

zx

x∂∂∂

+∂∂

+∂∂

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

+∂∂

+

vx

vvy

vvz

vvx

yy

yz

y

xz

j

vvvy

vvz

yz

zz∂

∂+

∂∂

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟k

The velocity vector for a fl ow fi eld is

v i j= +⎡⎣ ⎤⎦− −xye y et t2 2 2 m

s (a)

(a) Determine the acceleration vector for the fl ow(b) The fl ow rate Q through a surface area is calculated as

Q dAA

= ⋅∫ v n (b)

where n is a unit vector normal to the surface. Determine the fl ow rate through a square whose vertices are (0,2,0),(0,2,− 2), (2,2,− 2), and (2,2,0).

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Solution (a) Noting that the total acceleration is

av

i=∂∂

+∂∂

+∂∂

+⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟ +

∂∂t

vvx

vvy

vvx

xx

yx

xy ++

∂∂

+⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

= − −⎡⎣− −

vvy

xye y e

yy

t t

j

i j2 22 2 2 ⎤⎤⎦ + +⎡⎣ ⎤⎦ + +⎡⎣− −e xy y y x e xy y yt t4 2 4 20 2( ) ( ) ( ) ( )i ⎤⎤⎦{ }

= − +( ) + − +− − −

j

i

m

s2

2 2 2 22 2 4 2 2 3xye xy e y e yt t t ee t−( )⎡⎣ ⎤⎦4 j

m

s2

(c)

(b) The unit normal to the square is n = j, and everywhere on the square, y = 2. Thus

Q e dxdz

es

t

t

=

=

−

−

−

∫∫ 2

16

2 2

0

2

2

0

2 m3

(d)

2.3 Vector spacesVectors in three-dimensional space are defi ned along with a defi nition of vector addition and multiplication by a scalar. These operations yield another vector in three-dimensional space. The defi nitions of these operations, along with the implicit defi nitions of addition and multiplication of real numbers, lead to eight properties that apply to all vectors in three-dimensional space.

Three-dimensional space is illustrated geometrically. However, this is only an example of the more general and abstract concept of a vector space.

Defi nition 2.1 A vector space is a collection of objects called vectors, together with defi ned operations of vector addition and scalar multiplica-tion that satisfy a set of 10 axioms. Let the collection of objects be collectively called V, and let u, v, and w be arbitrary elements of V. Let α and β be arbi-trary elements of an associated scalar fi eld. Then if V is a vector space,

(i) V is closed under addition, that is, u + v is an element of V, (ii) V is closed under scalar multiplication, that is, αu is an element of V, (iii) The associative law of addition, (u + v) = w = u + (v + w), holds, (iv) The commutative law of addition, (u + v) = (v + u), holds, (v) There exists a vector in V, called the zero vector 0, such that u + 0 = u, (vi) There exists a vector in V, −u, such that −u + u = 0, (vii) For the scalar 1, (1)u = u, (viii) The distributive law of scalar multiplication, (α + β)u = αu + βu, holds, (ix) The associative law of scalar multiplication, (αβ)u = α(βu), holds, and (x) The distributive law of addition, α(u + v) = αu + αv, holds.

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If the associated scalar fi eld is the set of all real numbers, then the vector space is said to be a real vector space. If the associated scalar fi eld is the set of complex numbers, then the vector space is called a complex vector space.

Example 2.4 Defi ne Rn as the set of all ordered n-tuples of real numbers.

Vectors u and v in Rn are represented as

u v=

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

=

⎡

⎣

⎢⎢⎢

u

u

u

v

v

vn n

1

2 2

� �

1

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

(a)

The operations of vector addition and scalar multiplication are defi ned such that

u v u+ =

++

+

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

=

u v

u v

u vn n

1 1

2 2

�α

αuu

u

un

1

2α

α�

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

(b)

Show that Rn is a vector space under these defi nitions of vector addition and multiplication by a scalar.

Solution Certainly Rn is closed under vector addition and scalar multipli-

cation. The zero vector is the vector whose components are all zero, and −u is defi ned as (−1)u. It is easy to show that the other axioms defi ning a vector space hold for R

n under these defi nitions of vector addition and scalar multi-

plication. For example, consider the distributive law of scalar multiplication,

( )

( )

( )

( )

α β

α βα β

α β

+ =

++

+

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥u

u

u

un

1

2

� ⎥⎥⎥

(c)

However, the distributive law holds for scalar multiplication of real num-bers, that is, (α + β)ui = αui + βui. Thus

( )α β

α βα β

α β

+ =

++

+

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥u

u u

u u

u un n

1 1

2 2

� ⎥⎥⎥

(d)

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which using the defi nition of vector addition can be written as

( )α β

αα

α

ββ

β

+ =

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

+u

u

u

u

u

u

un

1

2

1

2

� �

nn

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

= +α βu u

(e)

Since the ten axioms of defi nition 2.1 hold, Rn is a vector space.

The vector space Rn is a generalization of the three-dimensional space R

3

discussed in section 2.2. However, vectors in Rn lack the geometric represen-

tation of vectors in R3. Equation a of Example 2.4 provides an alternate way

to Equation 2.3 to write a vector in R3.

Example 2.5 Let Cn[a,b] represent the set of functions of a real variable,

call it x, that are n times differentiable on the real number line within the interval, a ≤ x ≤ b. If f(x) is an element of C

n[a,b], then for any x, a ≤ x ≤ b, f(x) is

a real number. If f(x) and g(x) belong to Cn[a,b] and α is a real scalar, then for

any x, a ≤ x ≤ b, f(x) + g(x) is the real number that is the scalar sum of f(x) + g(x), and αf(x) is the real number defi ned by the scalar multiplication of α and f(x). Show that C

n[a,b] is a vector space under the defi ned operations of vector

addition and scalar multiplication.

Solution Clearly, if f(x) and g(x) are n times differentiable, then f(x) + g(x) and αf(x) are also n times differentiable. The zero function is defi ned as f(x) = 0 for all x, a ≤ x ≤ b and –f(x) is defi ned as (–1)f(x). It is easy to show that the remaining axioms of defi nition 2.1 are satisfi ed and that C

n[a,b]

is a vector space under the given defi nitions of vector addition and scalar multiplication.

Rn and C

n[a,b] are the two vector spaces used most extensively in this

study. Other vector spaces used in applications in this book include the following.

Pn[a,b] is the set of all polynomials of degree n or less defi ned on the inter-

val a ≤ x ≤ b. The defi nitions of vector addition and scalar multiplication for vectors in P

n[a,b] are the same as used in the defi nition of C

n[a,b].

Cn(ℜ) is the set of all functions which are n times continuously differen-

tiable in the spatial volume defi ned by ℜ. Vectors in Cn(ℜ) are functions of three spatial variables, perhaps of the form f(x,y,z), or if cylindrical coordi-nates are used, f(r,θ,z). Since C

n(ℜ)is really a generalization of C

n[a,b], their

defi nitions of vector addition and scalar multiplication are similar. Addition of two vectors in C

n(ℜ) is performed at each point in the spatial volume

defi ned by ℜ.

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S = RkXC

n[a,b] is the set of elements of the form

f x

f x

f xk

1

2

( )

( )

( )

�

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

where f1(x),f2(x),…,fk(x) are each in Cn[a,b]. Defi nitions of vector addition and

scalar multiplication are given by:

f g+ =

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

+

f x

f x

f x

g x

g

k

1

2

1( )

( )

( )

( )

�22

1 1

2( )

( )

( ) ( )

(x

g x

f x g x

f x

k

�

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

=

+)) ( )

( ) ( )

+

+

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

g x

f x g xk k

2

�αff =

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

αα

α

f x

f x

f xk

1

2

( )

( )

( )

�

Defi nition 2.2 Let S be a set of vectors contained in a vector space V. Then S is a subspace of V if S is a vector space in its own right.

To prove that S is a subspace of V, it must be shown that the 10 axioms of defi nition 2.1 are satisfi ed by the vectors in S. However, because all vectors in S are also in V and V is a vector space, then by hypothesis, axioms (iii), (iv), (vii), (viii), (ix), and (x) are satisfi ed. Hence it is only necessary to show that S is closed under vector addition, S is closed under scalar multiplication, the zero vector is in S, and for every u in s, –u is also in S.

Example 2.6 Determine whether each set of vectors is a subspace of R3.

(a) Let S be the set of vectors in R3 whose components sum to zero. That is,

if u is in S, then u1 + u2 + u3 = 0.(b) Let S be the set of vectors in R

3 whose components sum to 1, u1 + u2 +

u3 = 1.

Solution (a) If u and v are in S, then

u v+ =

+++

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

u v

u v

u v

1 1

2 2

3 3

(a)

and the sum of the components of u + v is

( ) ( ) ( ) ( ) (u v u v u v u u u v v v1 1 2 2 3 3 1 2 3 1 2+ + + + + = + + + + + 33

0 0 0

)

= + = (b)

Thus S is closed under vector addition.

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The sum of the components of α u is (α u1) + (α u2) + (α u3) = α (u1 + u2 + u3) = (α)0 = 0. Thus S is closed under scalar multiplication. The zero vector is in S, as is −u. Thus since V is a vector space and axioms (i), (ii), (v), and (vi) of defi nition 2.1 are satisfi ed, then S is a subspace of V.

(b) S is not a subspace of V. S is not closed under either vector addition or under scalar multiplication. If u and v are in S, then u1 + u2 + u3 = 1and v1 + v2 + v3 = 1. Then if w = u + v, w1 + w2 + w3 = (u1 + v1) + (u2 + v2) + (u3 + v3) = 2. Thus, by defi nition of S, w is not in S.

Example 2.7 The boundary-value problem for the nondimensional tem-perature distribution in one dimension, Θ(x), in a wall with both sides fi xed at the same temperature and with a nonuniform internal heat generation, is of the form

ddx

u x2

2

0 0

1 0

Θ

Θ

Θ

=

=

=

( )

( )

( )

The solution must be twice differentiable and satisfy both boundary condi-tions. Let S be the set of functions in C

2[0,1] that satisfy the conditions f(0) = 0

and f(1) = 0. Is S, the set of possible solutions to the boundary value problem, a subspace of C

2[0,1]?

Solution If f(x) and g(x) are in S, then f(0) = 0, f(1) = 0, g(0) = 0, and g(1) = 0. Hence f(0) + g(0) = 0 and f(1) + g(1) = 0. Thus S is closed under addition. More-over, S is closed under scalar multiplication, because αf(0) = 0. The function f(x) = 0 satisfi es f(0) = 0 and f(1) = 0, and hence the zero vector is in S. Also, –f(0) = 0 and –f(1) = 0, thus –f(x) is also in S. Hence since axioms (i), (ii), (v), and (vi) of defi nition 2.1 are satisfi ed, then S is a subspace of C

2[0,1].

Example 2.8 Let V = P4[0,1] ∩S, where S is the subspace of C2[0,1] defi ned in Example 2.7. Show that V is a subspace of S.

Solution An element of V is of the form p(x), a polynomial of degree four or less which satisfi es p(0) = 0 and p(1) = 0. A polynomial is by defi nition in C2[0,1]. Clearly the sum of two such polynomials is also a polynomial of degree four or less, and using an argument similar to that of Example 2.7, must also be zero at x = 0 and x = 1.Thus V is closed under addition. It is also clear that V is closed under scalar multiplication. Furthermore, p(x) = 0 is in V, and –p(x) + p(x) = 0. Hence V is a subspace of S (as well as of P4[0,1]).

2.4 Linear independenceThe mathematical solutions to problems involving discrete dynamic systems reside in a subspace of the vector space Rn. The mathematical solutions to

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problems involving continuous- or distributed-parameter systems reside in a subspace of the vector space Cn[a,b]. To facilitate solutions to these problems, it is imperative to understand properties of vector spaces and of sets of vec-tors within these vector spaces. Approximate solutions are constructed from vectors in a subspace of a vector space in which the exact solution resides. To be able to develop such solutions, it is imperative to understand properties of vector spaces and subspaces. The defi nitions and theorems in this section provide the foundation for these solutions.

Defi nition 2.3 A set of vectors is said to be linearly independent if when a linear combination of the vectors is set equal to zero, all coeffi cients in the linear combination must be zero. If the vectors are not linearly independent, they are said to be linearly dependent.

Defi nition 2.3 states that if a set of vectors, u1,u2,…,uk is linearly indepen-dent, then

ci

i

k

=∑ =

1

0ui (2.8)

implies that c1 = c2 = … = ck = 0. This implies that no vector in the set can be written as a linear combination of the other vectors.

Application of Equation 2.8 for a set of vectors in Rn leads to a set of n

equations summarized by the matrix system

Uc = 0 (2.9)

where U is a matrix with n rows and k columns whose jth column is the vector uj, and c is a column vector of k rows whose ith component is ci. The vectors are linearly independent if and only if a nontrivial solution to the set of linear equations exists. This leads to the following theorem, presented without proof.

Theorem 2.1 Let u1,u2,…,uk be a set of vectors in Rn. Let U be the n × k

matrix whose ith column is ui . If k > n, a nontrivial solution of the system exists, and the set of vectors is linearly dependent. If k = n, the vectors are linearly dependent if and only if U is singular. If k < n, the vectors are linearly dependent only if every square submatrix of U is singular.

Application of Equation 2.8 for a set of functions in C[a,b], f1(x),f2(x),…,fk(x), leads to

c f x c f x c f xk k1 1 2 2 0( ) ( ) ( )+ + =… (2.10)

For a function to be identically zero on the interval a ≤ x ≤ b, it must be zero at every value of x. Differentiating Equation 2.10 k-1 times with respect to x leads to

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c f x c f x c f x

c f x c f

k k1 1 2 2

1 1 2

0′ + ′ + + ′ =

′′ +

( ) ( ) ( )

( )

…

22 0′′ + + ′′ =( ) ( )x c f xk k…

c1

�

�f x c f x c fk kk k

k1

12 2

1 1( ) ( ) (( ) ( )− − −+ + + ))( )x = 0

(2.11)

Equation 2.10 and Equation 2.11 can be summarized in matrix form as

Wc = 0 (2.12)

where W is a k × k matrix called the Wronskian, defi ned by

W =

′ ′ ′f x f x f x f x

f x f x f x f

k1 2 3

1 2 3

( ) ( ) ( ) ( )

( ) ( ) ( )

�

� kk

k

x

f x f x f x f x

f

′

′′ ′′ ′′ ′′( )

( ) ( ) ( ) ( )

(

1 2 3

1

�

� � � � �kk k k

kkx f x f x f x− − − −

⎡

⎣

⎢⎢

12

13

1 1) ( ) ( ) ( )( ) ( ) ( ) ( )�

⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥

(2.13)

and c is a column vector of k rows whose ith component is ci. Equation 2.12 has a nontrivial solution if and only if W is singular for every value of x, a ≤ x ≤ b. If there exists at least one value of x such that W(x) is nonsingular, then the set of vectors is linearly independent. This leads to the following theorem, which is presented without formal proof.

Theorem 2.2 A set of functions in C[a,b], f x f x f xk1 2( ), ( ), , ( )… is linearly dependent if and only if the Wronskian for the set of functions, defi ned by Equation 2.13 is singular for all x, a ≤ x ≤ b.

Example 2.9 Show that the set of functions,

f x f x x f x x1 2 321( ) ( ) ( )= = =

is linearly independent in C[0,1].

Solution The Wronskian of the set can be determined to be

W =

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

1

0 1 2

0 0 2

2x x

x

The determinant of the Wronskian is

W = 2

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Since the Wronskian is nonsingular for at least one x (actually it is nonsingu-lar for all x), the functions are linearly independent.

Example 2.10 The general form of a polynomial of order n, an element of the vector space Pn[a,b], is

p x a x a x a x an nn n( ) = + +…+ +−

+1 21

1 (a)

A polynomial of order n is uniquely defi ned by its coeffi cients, a1,a2,…,an + 1, which defi ne a vector in Rn + 1 of the form

a =

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥

+

a

a

a

an

n

1

2

1

� (b)

Use theorem 2.1 to determine whether or not the polenomials p1(x) = x2 − 1, p2(x) = 2x2 − 4x + 2 and p3(x) = x2 − 4x − 3, are linearly independent.

Solution These quadratic polynomials can be represented as vectors in R3 as

p p1 2

1

0

1

2

4

2

=−

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

= −

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

p3

1

4

3

= −−

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥ (c)

The matrix U as defi ned in theorem 2.1 is

U = − −− −

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

1 2 1

0 4 4

1 2 3

(d)

The determinant of U is calculated as 24. Thus the matrix is nonsingular, and the vectors are linearly independent.

2.5 Basis and dimensionA linear combination of a set of linearly independent vectors in a vector space V, because of closure, is an element of V. A change in any of the coeffi cients in the linear combination leads to a different element of V.

Defi nition 2.4 A set of vectors in a vector space V is said to span V if every vector in V can be written as a linear combination of the vectors in the set.

Theorem 2.3 The span of a set of vectors in a vector space V is a subspace of V.

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Proof Let U = { , , , }u u u1 2 � k be a set of vectors in V. A vector in S = span{U} is written as

a ai

i

k

==

∑ ui

1

Since S is in V, and V is vector space axioms (iii), (iv), (vii), (viii), (ix), and (x) of defi nition 2.1 are true. Consider the sum of two vectors in the span of S:

a b u u u+ = + = += = =

∑ ∑ ∑a b a bi i

i

n

i

i

n

i i i i

i

n

1 1 1

( )

Hence a + b is in S because it can be written as a linear combination of the elements of U. Thus U is closed under vector addition. Closure under scalar multiplication can be similarly shown. The zero vector is obtained by setting each ai = 0, i = 1,2,…,n. The vector −a is obtained as

− = −=

∑a u( )ai i

i

n

1

Hence all ten axioms of defi nition 2.1 hold, and S is a subspace of V.

Example 2.10 P2[0,1] is the space of all polynomials of degree two or less

defi ned on the interval 0 ≤ x ≤ 1. It is easy to show that P2[0,1] is a vector space

and is a subspace of Cn[0,1]. Does the set

P p x p x p x p x p x x= = ={ ( ), ( ), ( )} ( ) , ( )1 2 3 1 21 , p x x32( ) = (a)

span P2[0,1]?

Solution A linear combination of the set of vectors is

p x a bx cx( ) = + + 2 (b)

Every polynomial of degree two or less can be obtained by varying the co effi cients in the linear combination. Hence the set spans P

2[0,1].

The span of a vector space is not unique. Indeed, the number of vectors in a span is also not unique. If the set P of Example 2.10 is augmented to p4(x) = x2 + 2x + 3 then the augmented set also spans P

2[0,1]. It is shown in

Example 2.10 that the vectors in P are linearly independent. However, since p4(x) = 3p1(x) + 2p2(x) + p3(x), the augmented set is not a set of linearly inde-pendent vectors.

Defi nition 2.5 A set of linearly independent vectors that spans a vector space V is called a basis for V.

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The choice of basis vectors is not unique. However, the number of vectors in the basis is unique. The unit vectors i, j, and k defi ned as being parallel to the coordinate axes in three-dimensional space form a basis for R3. An alternate choice for a basis is a set of vectors parallel to the coordinate axes obtained by rotating the x-y axes about the z axis through a counterclockwise angle θ, as illustrated in Figure 2.5. These vectors are related to i, j, and k by i i j j i j k k= ′ − ′ = ′ + ′ ′ =sin cos , cos sin ,θ θ θ θ and . The vectors ′ ′ ′i j k, , and are also a basis for R3.

By defi nition, the set P, defi ned in Example 2.10, spans P2[0,1]. Indeed,

any set of three linearly independent polynomials of degree two or less constitutes a basis for P

2[0,1]. Not all vectors in P

2[0,1] can be written as

a linear combination of only two linearly independent polynomials. A set of four polynomials of degree two or less is not linearly independent. Thus the number of elements in the basis for P

2[0,1] is exactly three.

Defi nition 2.6 The number of vectors in the basis of a vector space V is called the dimension of V.

The vector space Rn has dimension n. The space P

n[a, b] has dimension

n + 1. These are examples of fi nite-dimensional vector spaces.It is not possible to express every element in C

2[0,1] by a linear combina-

tion of a fi nite number of elements of C2[0,1]. A basis for C

2[0,1] contains an

infi nite, number of elements. This is an example of an infi nite-dimensional vector space. Any set of vectors from which any element of an infi nite-dimensional vector space can be written is said to be complete in that space.

z

y

xθ

θ

j'

j

i

i'

k,k'

Figure 2.5 The choice of a basis is not unique. The set of vectors i',j',k' is a basis for R3, as is the set of vectors i,j,k.

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The subspace of Example 2.7 is also an infi nite-dimensional vector space. It is important to determine a complete set of vectors for such a space.

The expression of a vector in an infi nite-dimensional vector space in terms of members of a complete set (or basis) is an infi nite series. Thus questions of convergence arise. Such questions include whether the series converges at every point within the interval [a,b], and if so, then to what value does it con-verge. While these questions are interesting, they are not within the scope of this study. Theorems regarding convergence of series will be presented without proof and subsequently used. Occasionally questions of convergence will not be considered. This does not mean that they are not important, only that their consideration does not improve the understanding of the material being presented. Such a case arises in the proof of theorem 2.4. If the number of vectors in U is not fi nite, then questions about series convergence arise when considering the closure of S under vector addition and scalar multipli-cation. However, it is assumed that the series for a and b converge, and hence their sum converges.Example 2.11 The determination of the natural frequencies and mode shapes for a nonuniform beam requires the solution of a variable-coeffi cient differential equation of the form

ddx

EI xd wdx

A x w2

2

2

2

2 0( ) ( )⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟+ =ρ ω

(a)

where x, the independent variable, is a coordinate along the neutral axis of the beam, measured from its left support, w(x), the dependent variable, is the transverse defl ection of the beam, E is the elastic modulus of the material from which the beam is made, ρ is the mass density of the material, I(x) is the area moment of inertia of the nonuniform cross-section, A(x) is the area of the cross-section, and ω is the natural frequency which is to be determined. After nondimensionalization, Equation a becomes

d

dxx

d wdx

x w2

2

2

2

2 0α β ω( ) ( )⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟+ = (b)

The nondimensional boundary conditions for a fi xed-fi xed beam are

wdwdx

wdwdx

( ) ( )

( ) ( )

0 0 0 0

1 0 1 0

= =

= =

(c)

Since the beam is nonuniform, determination of an exact solution for the mode shapes and natural frequencies is diffi cult. Thus an approximate solution is

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sought. The exact solution resides in a subspace of C4[0,1], call it S, such that if

f(x) is in S, then

fdfd

fdf

( ) ( )

( ) ( )

0 0 0 0

1 0 1 0

= =

= =d

x

x

(d)

An approximate solution is sought from P6[0,1], the space of polynomials of

degree six or less. To satisfy the boundary conditions, the approximate solu-tion must also lie in S. Thus it is desired to seek an approximate solution from the subspace of S and P

6[0,1] defi ned as Q S P= ∩ 6 0 1[ , ] .

Determine the dimension of Q and determine a basis for Q.

Solution Let f(x) be an element of Q. Since f(x) is in P6[0,1], then it is of the

form

f x a x a x a x a x a x a x a( ) = + + + + + +66

55

44

33

22

1 0 (e)

Since f(x) is also in S, it must satisfy the boundary conditions. To this end,

f a( )0 0 0= = (f)

dfdx

a( )0 0 1= = (g)

f a a a a a a a( )1 0 6 5 4 3 2 1 0= = + + + + + + (h)

dfdx

a a a a a a( )1 0 6 5 4 3 26 5 4 3 2 1= = + + + + + (i)

Equation f and Equation g obviously imply that a0 = 0 and a1 = 0. Equation h and Equation i can be manipulated to give

a a a a2 4 5 62 3= + + (j)

a a a a3 4 5 62 3 4= − − − (k)

Thus an arbitrary element of Q is of the form

f x a x a x a x a a a x a a( ) ( ) (= + + − + + + +66

55

44

4 5 63

4 52 3 4 2 ++ 3 62a x) (l)

f x a x x x a x x x a x x( ) ( ) ( ) (= − + + − + + −66 3 2

55 3 2

444 3 3 2 2 33 2+ x ) (m)

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Since f(x) is a linear combination of three linearly independent functions, the dimension of Q is three, and a basis for Q is

f x x x x16 3 24 3( ) = − + (n)

f x x x x25 3 23 2( ) = − + (o)

f x x x x34 3 22( ) = − + (p)

Example 2.12 The nondimensional partial differential equation governing the steady-state temperature distribution Θ( , )x y in a thin square slab is

∂∂

+∂∂

=2

2

2

2

Θ Θx y

u x y( , ) (a)

where u(x,y) is a function describing nondimensional internal heat genera-tion. Consider a slab in which the boundary conditions are

Θ( , )0 0y = (b1)

∂∂

=Θx

y( , )1 0 (b2)

Θ( , )x 0 0= (b3)

∂∂

+ =Θ

Θy

( , ) ( , )x x1 2 1 0 (b4)

where the Biot number, Bi = 2. Equation b1 and Equation b3 represent geo-metric boundary conditions, while Equation b2 and Equation b4 represent natural boundary conditions. Defi ne C2[ℜ] as the vector space of twice dif-ferentiable functions defi ned everywhere in the region ℜ of the x-y plane defi ned such that if (x,y) is in ℜ, then 0 ≤ x ≤ 1 and 0 ≤ x ≤ 1. An element of C2[ℜ] is of the form f(x,y).

Let Q be the subspace of C2[ℜ] spanned by the functions

q x y q x y x q x y y q x y xy q1 2 3 41( , ) , ( , ) , ( , ) , ( , ) ,= = = = 55

26

2

72

8

( , ) , ( , ) ,

( , ) , ( , )

x y x q x y y

q x y xy q x y x

= =

= = 229

310

3y q x y x q x y y, ( , ) ( , )and= =

(a) Defi ne G as a subspace of C2[ℜ] such that if f(x,y) is in G, then f(0,y) = 0 and f(x,0) = 0. That is, if f(x,y) is in N, then f satisfi es the geometric boundary conditions. Determine a basis for Q ∩ G.

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(b) Defi ne N as a subspace of C2[ℜ] such that if f(x,y) is in N, then f (0, y) = 0, ∂f/∂x (1, y) = 0, f (x, 0) = 0, and ∂f/ ∂y (x, 1) + 2f (x, 1) = 0. That is, if f(x,y) is in N, then f satisfi es all boundary conditions. Determine a basis for Q ∩ N.

Solution A vector in Q is of the form

q C C x C y C xy C x C y xy C x y C x= + + + + + + + +1 2 3 4 52

62 2

82

9C733

103+ C y (c)

If q is also in G, then

q y C C y C y C y( , )0 0 1 3 62

103= = + + + (d)

q x C C x C x C x( , )0 0 1 2 52

93= = + + + (e)

Since powers of y are linearly independent, Equation d is satisfi ed for all y only if

C C C C1 3 6 10 0= = = = (f)

Similarly, since powers of x are linearly independent, Equation e is satisfi ed only if

C C C2 5 9 0= = = (g)

Equation c can then be reduced to

q x y C xy C xy C x y( , ) = + +4 72

82 (h)

Thus, one choice of a basis for Q ∩ G is the set {xy,xy2,x2y}.(c) An element of G is also in N. Thus an element in Q ∩ G is of the form of Equation h, but must also satisfy

dqdx

y C C y C y( , )1 0 24 72

8= = + + (i)

Since powers of y are linearly independent, Equation i is satisfi ed only if C4 = C7 = C8 = 0. Hence the intersection of Q and N is the null set.

2.6 Inner productsA function of a vector is an operation that can be performed on all vectors in a vector space. The operation may involve more than one vector from the vector space. If the result of the operation is a scalar, the function is called a scalar function. An example of a scalar function is the dot product defi ned for two vectors in R

3. Properties satisfi ed by the dot product are summarized

in section 2.2. The concept of dot product is generalized into the following defi nition.

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Defi nition 2.7 Let u and v be arbitrary vectors in a vector space V. An inner product of u and v, written (u,v), is a scalar function which satisfi es the fol-lowing properties:

i. (u,v) = (v,u): (Commutative property) ii. (u + v,w) = (u,w) + (v,w): (Distributive property) iii. (α u,v) = α (u,v): (Associative property of scalar multiplication) iv. (u,u– ) ≥ 0 and (u,u– ) = 0 if and only if u = 0: (Non-negativity property)

where α is an arbitrary scalar and an overbar denotes the complex conjugate.

Example 2.13 Which of the following constitute a valid defi nition of an inner product for vectors in R

3?

(a) (u,v) = u1v1 + u2v2 + u3v3

(b) (u,v) = u1v1 + 2u2v2 + u3v3

(c) (u,v) = u1v1–u2v2 + u3v3

(d) (u,v) = u1v1 + u2v2 + u3v2

(e) (u,v) = u1v1 + u2v2 + u3 2 v3

2

Solution To determine whether each of these expressions constitutes a valid inner product on R

3, it must be determined whether or not all the prop-

erties of defi nition 2.7 are satisfi ed.(a) Equation a is the same as the defi nition of the dot product of two vec-

tors, from which the defi nition of inner product is a generalization, and therefore Equation a defi nes a valid inner product for R

3.

(b) Equation b defi nes a valid inner product for R3. Its satisfaction of the

properties of a valid inner product is shown below:

i. ( , ) ( , )u v v u= + + = + + =u v u v u v v u v u v u1 1 2 2 3 3 1 1 2 2 3 32 2

ii. ( , ) ( ) ( ) ( )u v w+ = + + + + +

=

u v w u v w u v w

u w

1 1 1 2 2 2 3 3 3

1 1

2

++ + + + +

= + +

v w u w v w u w v w

u w u w u

1 1 2 2 2 2 3 3 3 3

1 1 2 2 3

2 2

2( ww v w v w v w3 1 1 2 2 3 32) ( )

( , ) ( , )

+ + +

= +u w v w

iii.

( , ) ( ) ( ) ( )

(

α α α α

α

u v = + +

= +

u v u v u v

u v u v

1 1 2 2 3 3

1 1 2

2

2 22 3 3+

=

u v )

( , )α u v

iv. ( , )u u = + +u u u12

22

322 . Since this is the sum of non-negative terms, it is

non-negative for any u. In addition, it is clear that (u,u) ≥ 0 if and only if u = 0.

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(c) Equation c is not a defi nition of a valid inner product for R3. Consider,

for example, the vector

u =

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

1

1

0

Using the proposed defi nition,

(u,u) = (1)(1)–(1)(1) + (0)(0) = 0

Thus there exists a u ≠ 0 such that (u,u) = 0, and property iv is violated.(d) Equation d does not represent a valid inner product for R

3. Consider

the vector

u =

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

0

1

0

Using the proposed defi nition of inner product,

(u,u) = (0)(0) + (1)(0) + (0)(1) = 0

Thus there exists a u ≠ 0 such that (u,u) = 0, and property (iv) is violated.(e) Equation e does not defi ne a valid inner product for R

3. Consider prop-

erty (ii):

( , ) ( ) ( ) ( )u v w+ = + + + + +u v w u v w u v w1 1 1 2 2 2 3 32

32

It is easy to fi nd a set of vectors for which this is not equal to (u,w) + (v,w).The defi nition of an inner product is not unique. There are many valid

defi nitions of an inner product for any vector space. The inner product for R

n defi ned by

( , )u v ==

∑u vi i

i

n

1

(2.14)

is called the standard inner product for Rn. The standard inner product on

C[a,b] is defi ned by

( , ) ( ) ( )f g f x g x dxa

b

= ∫ (2.15)

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Example 2.14 Let S be a subspace of C2[0,1] such that if f(x) is an element of S, then f(0) = 0 and f(1) = 0. Show that the scalar product defi ned as

f gd fdx

gdxL,( ) = ∫2

2

0

1

(a)

is a valid inner product on S. The subscript on the defi nition of the scalar product is used to distinguish its defi nition from that of the standard inner product defi ned for all of C2[0,1].

Solution The scalar product defi ned in Equation a is a valid inner product on S if properties (i)–(iv) are true when u is replaced by f(x) and v is replaced by g(x) for all f(x) and g(x) in S. Considering property (i),

( , ) ( )f gd fdx

g x dxL = ∫2

2

0

1

(b)

Integration by parts ( )∫ = − ∫udv uv vdu is applied to the right-hand side of Equation b with u = g(x) and dv d f dx dx= ( )2 2/ . These choices lead to du = (dg/dx)dx and v = df/dx. Then use of integration by parts for Equation b leads to

f g gdfdx

gdfdx

dfdx

dgdx

dxL, ( ) ( ) ( ) ( )( ) = − −1 1 0 0

0

1

∫∫ (c)

Application of integration by parts to Equation c with u = dg/dx and dv = (df/dx)dx leads to

f g gdfdx

gdfdx

dgdx

fL, ( ) ( ) ( ) ( ) ( )( ) = − − +1 1 0 0 1(1)ddgdx

fd gdx

fdx( ) ( )0 02

2

0

1

+∫ (d)

Since f and g are in S, f(0) = 0, f(1) = 0, g(0) = 0, and g(1) = 0, and Equation d reduces to

( , )

( , )

f gd gdx

fdx

g f

L

L

=

=

∫2

2

0

1

(e)

The conclusion drawn in the last line of Equation e derives from the defi ni-tion of the scalar product in Equation a.

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Considering property (ii) of defi nition 2.7,

( , ) ( )f g hd fdx

d gdx

h x dxL+ = +⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟

=

∫2

2

2

2

0

1

dd fdx

h x dxd gdx

h x dx

f h g hL

2

2

0

12

2

0

1

( ) ( )

, ,

∫ ∫+

=( ) +(( )L

(f)

Considering property (iii) of defi nition 2.7,

α α

α

f gd fdx

g x dx

d fdx

L, ( )( ) =⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟

=

∫2

2

0

1

2

2gg x dx

f g

( )

,

0

1

∫= ( )α

L

(g)

Considering property (iv) of defi nition 2.7,

f fd fdx

f x dxL, ( )( ) = ∫2

2

0

1

(h)

Integration by parts is applied to Equation h, leading to Equation c, but with g(x) replaced by f(x):

f f fdfdx

fdfdx

dfdx

dfdx

dxL, ( ) ( ) ( ) ( )( ) = − −1 1 0 0

0

1

∫∫ (i)

Since f is in S, f(0) = 0 and f(1) = 0, and Equation i becomes

f fdfdx

dxL,( ) = ( )∫2

0

1

(j)

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Equation j shows that ( f,f )L ≥ 0 and (0,0) = 0. Assume that g(x) is an element of S with (g,g) = 0. For this inner product to be zero, Equation j requires that dg/dx = 0 for all x, 0 ≤ x ≤ 1. Thus g(x) = C, a constant. However, since g is in S, g(0) = 0 which implies that C = 0, and thus g(x) = 0. Hence property (iv) of defi nition 2.7 is satisfi ed.

Since the scalar product of Equation a satisfi es the four properties of defi -nition 2.7, Equation a defi nes a valid inner product on S. However, the scalar product has been shown to be a valid inner product on S only. It does not satisfy all properties of an inner product for all of C2[0,1].

Theorem 2.4 (Cauchy-Schwartz Inequality). Let (u,v) represent a valid inner product defi ned for a real vector space V. Then

( , ) ( , )( , )u v u u v v2 ≤ (2.16)

Proof The proof of the Cauchy-Schwartz inequality involves the use of properties of inner products and the algebra of inner products and is thus instructive in its own right. From property (iv) of defi nition 2.7, for any real α and β, it should be noted that

( , )α β α βu v u v− − ≥ 0 (2.17)

Use of property (iii) of defi nition 2.7 leads to

( , ) ( , ) ( , ) ( , )α α α β β α β βu u u v v u v v+ − + − + − − ≥ 0 (2.18)

Since V is a real vector space, property (i) of defi nition 2.7 becomes the com-mutative property. Using properties (i) and (iii) Equation (2.18) is rewritten as

α αβ β2 22 0( , ) ( , ) ( , )u u u v v v− + ≥ (2.19)

Specifi cally defi ne α = (v,v)1/2 and β = (u,u)1/2; substitution into Equation 2.19 leads to

( , )( , ) ( , ) ( , ) ( , ) ( , )( , )v v u u u u v v u v u u v v− + ≥2

1

2

1

2 00

1

2

1

2( , ) ( , ) ( , )u u v v u v≥

(2.20)

which upon squaring becomes Equation 2.16.

Defi nition 2.8 Two vectors u and v in a vector space V are said to be orthog-onal with respect to a defi ned inner product if (u,v) = 0.

The use of the term orthogonality is a generalization of the concept from R

3 in which two vectors are geometrically perpendicular or orthogonal when

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their dot product is zero. However note that, in general, the determination of orthogonality of two vectors is dependent upon the choice of inner product. Orthogonality of two vectors with respect to one inner product defi ned on a vector space V does not guarantee orthogonality with respect to any other valid inner product.

Example 2.15 Consider two vectors in R3 defi ned as

u v=−

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

=−⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

1

3

2

2

1

2

Determine whether these vectors are orthogonal with respect to the inner products defi ned in parts (a) and (b) of Example 2.13.

Solution (a) Checking orthogonality with respect to the inner product of (a) leads to

( , )

( )( ) ( )( ) ( )( )

u v 1= + +

= − + + −

u v u v u v1 2 2 3 3

1 2 3 1 2 2

== −1

Thus the vectors are not orthogonal with respect to this inner product.(b) Checking orthogonality with respect to the inner product of (b) leads to

( , )

( )( ) ( )( ) ( )(

u v = + +

= − + + −

u v u v u v1 1 2 2 3 32

1 2 2 3 1 2 22

0

)

=

Thus the vectors are orthogonal with respect to this inner product.

Example 2.16 Two polynomials in P2[0,1] are defi ned by p1(x) = a1x2 + a2x + a3

and p2(x) = b1x2 + b2x + b3. Recalling that P2[0,1] is a subspace of C2[0,1], Equa-

tion 2.14 is also a valid inner product on P2[0,1]. Example 2.10 shows that a polynomial in P2[0,1] can be represented by a vector of its coeffi cients, an element of R3,whose standard inner product is given by Equation 2.12. Does orthogonality of p1(x) and p2(x) with respect to the standard inner product on C2[0,1] imply orthogonality with respect to the standard inner product on R3 and vice versa?

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Solution The standard inner product on C2[0,1] of p1(x) and p2(x) is

p p p x p x dx

a x a x a b x

C1 2 1 2

0

1

12

2 3 12

2, ( ) ( )( ) =

= + +( )

∫

++ +( )

= + +( ) +

∫ b x b dx

a b x a b a b x a b

2 3

0

1

1 14

1 2 2 13

1 3[ ( ++ +

+ + +

=

∫ a b a b x

a b a b x a b dx

a

2 2 3 12

0

1

2 3 3 2 3 3

11

5

)

( ) ]

bb a b a b a b a b a b

a b

1 1 2 2 1 1 3 2 2 3 1

2

1

4

1

3

1

2

+ +( )+ + +

+

( )

( 33 3 2 3 3+ +a b a b)

(a)

The standard inner product on R3 of these vectors is

p p a b a b a bR1 2 1 1 2 2 3 33,( ) = + + (b)

It can be shown by counterexample, using Equation a and Equation b, that orthogonality with respect to one inner product does not imply orthogonal-ity with respect to another inner product. Suppose p1(x) = 1 and p2(x) = x. Then ( , )p p

R1 2 33 0= , but ( , )p pC1 2 2 1 2= / . Then suppose p1(x) = 1 and p2(x) = 2x–1. In

this case, ( , )p pC1 2 2 0= and (p1, p2) R33 = –2.

2.7 NormsIt is often important to have a measure of the “length” of a vector. This is especially important when estimating the error in approximating one vector by another vector, such as when approximating the solution of a differential equation. A measure of the error might be the “length” of the error vector, which is defi ned as the difference between the exact solution and the approx-imate solution.

The calculation to determine the geometric length of a vector in R3 is an

operation performed on the vector which results in a scalar. Thus the length is a scalar function of the vectors that satisfi es certain properties. This concept

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can be extended to develop a defi nition for the length of a vector in a general vector space.

Defi nition 2.9 A norm of a vector u in a vector space V, written u , is a func-tion of the vector whose result is a real scalar and that satisfi es the following properties:

i. u ≥ 0 and u = 0 if and only if u = 0. ii. α αu u= for any scalar α in the associated scalar fi eld of V. iii. u v u v+ ≤ + , the triangle inequality.

The norm is a generalization of the concept of length of a vector. The fi rst prop-erty requires the length of the vector to be non-negative and that only the zero vector may have a length of zero. The second property is a scaling property: if the vector is multiplied by a scalar, then the length of the resulting product is proportional to the length of the original vector, with the constant of propor-tionality being the absolute value of the scalar. The third property is a gener-alization of the triangle rule for vector addition in three-dimensional space, in which a geometric representation of the addition of two vectors is obtained by placing the head of one vector at the tail of the other, with the resultant vector being drawn from the tail of the fi rst vector to the head of the second, the two vectors and their resultant forming a triangle. The triangle inequality is then a statement that the length of any side of a triangle must be less than the sum of the lengths of the other two sides.

Occasionally a function satisfi es all requirements to be called a norm except that vectors other than the zero vector will have a norm of zero. This case is covered in the following defi nition.

Defi nition 2.10 A semi-norm of a vector u in a vector space V, s(u), is a function of the vector whose result is a real scalar and that satisfi es the fol-lowing properties:

i. s(u) ≥ 0 and s(u) = 0. ii. s s( ) ( )α αu u= any scalar α in the associated scalar fi eld of V. iii. s(u + v) ≤ s(u) + s(v), the triangle inequality.

Example 2.17 Show that the following is a valid defi nition of a norm on C[a,b]:

f f xa x b∞ ≤ ≤

= max ( ) (a)

Solution The proof that Equation a is a valid defi nition of a norm requires that properties(i)–(iii) be shown to hold for all f(x) in C[a,b]. Clearly, from its defi ni-tion, f ∞≥ 0 for all f(x) and 0 0∞= Since f(x) is a continuous function, if f(x) ≠ 0 for some x0, a ≤ x0 ≤ b, then there is a fi nite region around x

0 where f(x) ≠ 0. Thus

it is easy to argue that any f(x) ≠ 0 must have a maximum greater than zero.

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Considering property (ii),

α α

α

α

f f x

f x

f

a x b

a x b

a x b

∞ ≤ ≤

≤ ≤

≤ ≤

=

=

=

max ( )

max ( )

max (xx

f

)

= ∞α

(b)

Finally, considering the triangle inequality,

f g f x g x

f x g x

a x b

a x b

+ = +

≤ +( )

=

∞ ≤ ≤

≤ ≤

max ( ) ( )

max ( ) ( )

mmax ( ) max ( )a x b a x b

f x g x

f g

≤ ≤ ≤ ≤

∞ ∞

+

= +

(c)

Hence, since the three properties of defi nition 2.9 are satisfi ed, f∞

consti-tutes a valid norm on C[0,1].

The defi nition of a norm for a vector space is not unique. One possible defi nition of a norm for C[a,b] is presented in Equation a of Example 2.10. Another scalar function defi ned for vectors in C[a,b] which satisfi es the prop-erties of defi nition 2.9 is

f f x dxa

b

1= ∫ ( )

When comparing the closeness of two vectors, the use of f∞

determines whether the error between the two functions is small at every value of x, whereas the use of the f

1 norm determines whether the functions are close

in an average sense.

Example 2.18 Show that the function

s u u u u u( ) = − +⎡⎣ ⎤⎦12

1 2 22

1

22 (a)

defi ned for a vector u in R2 is a semi-norm for R

2, but is not a valid norm.

Solution Note that

s u u u u u( ) ( )= −⎡⎣ ⎤⎦ = −1 22

1

21 2

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and that if

u =⎡

⎣⎢⎢

⎤

⎦⎥⎥

1

1

then s(u) = (1 - 1) = 0. Thus since s(u) = 0 for u ≠ 0, then if s(u) is a semi-norm, it is not a norm. Clearly s(u) ≥ 0 and s(0) = 0. Then consider

s u u u u( ) ( ) ( )( ) )α α α α αu = − +( )⎡

⎣⎢⎤⎦⎥⎤⎦⎥

⎡⎣ 1

11 2 2

22⎢⎢

=

1

2

α s( )u

and

s u v u v

u v v

u

( ) ( ) ( )

) ( )

(

u v+ = + − +

= − + −

≤ −

1 1 2 2

2 1 2

1

(u1

uu v v

s s

2 1 2) ( )

( ) ( )

+ −

= +u v

Hence s(u) satisfi es all properties of a semi-norm.

Theorem 2.5 Let (u,v) be a valid inner product defi ned on a vector space V. Then u u u=( ), /1 2 is a valid norm on V. Such a norm is called an inner-product-generated norm.

Proof Since (u,v) represents a valid inner product on V, it satisfi es the prop-erties of defi nition 2.7. To prove that u represents a valid norm, it must be shown that the properties of defi nition 2.9 are satisfi ed.

i. u u u=( ) ≥, /1 2 0 u = 0 if and only if u = 0 is true because of property (iv) of defi nition 2.7.

ii. α α α αα α αu u u u u u u u=( ) = ( )[ ] = =, , ( , )/ / /1 2 1 2 1 2

iii. Use of the Cauchy-Schwartz inequality, Equation 2.14, leads to

u v u u u u v v v v+ ≤ ( )+ +⎡

⎣⎢⎢

⎤

⎦⎥⎥

=

, ( , ) ( , ) ( , )

(

21

2

1

2

1

2

uu u v v

u v

, ) ,1

21

2

21

2

+( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

= +

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The f∞

norm and the f1 norm have already been defi ned for C[a,b].

Theorem 2.4 shows that another valid norm for C[a,b] is the inner-product-generated norm,

f f x dxa

b

= [ ]⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥∫ ( )

2

1

2

(2.21)

This norm will be used without subscript and is referred to as the standard inner-product-generated-norm for C[a,b].

Similarly, the standard inner-product-generated norm for Rn is

u =⎡

⎣⎢⎢⎢

⎤

⎦⎥⎥⎥=

∑ui

i

n2

1

1

2

(2.22)

Example 2.19 Let S be the subspace of C2[0,1] spanned by f1(x) = 1, f2(x) = sin(πx), and f3(x) = cos(πx). Determine a set of unit vectors, with respect to the standard inner-product-generated norm for C2[0,1], which span S.

Solution A unit vector is a vector with a magnitude of one with respect to a defi ned norm. A set of unit vectors which span S are f x f x f x f x1 1 2 2( )/ ( ) ( )/ ( )⋅ and

f x f x3 3( ) ( )/ To this end,

f dx1

2

0

11

2

1

1

=⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

=

∫ ( )

,

(a)

f x dx

x

22

0

11

2

1

21 2

= [ ]⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

= − (

∫ sin( )

cos

π

π ))[ ]⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

=

∫ dx0

11

2

1

2,

(b)

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f x dx

x

32

0

11

2

1

21 2

= [ ]⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

= + (

∫ cos( )

cos

π

π ))[ ]⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

=

∫ dx0

11

2

1

2,

(c)

Thus a set of unit vectors which span S are 1 2, sin( )πx , and 2 cos( )πx .

2.8 Gram-Schmidt orthonormalizationThe dot product defi ned in three-dimensional space has a geometric inter-pretation as the length of the projection of one vector onto another multiplied by the length of the second vector. If the two vectors are mutually orthogo-nal, the projection is zero, and the dot product of the vectors is zero. It is convenient to use a set of mutually orthogonal basis vectors to represent any vector in R

3 as a linear combination of these basis vectors. A common choice

for an orthogonal set of basis vectors in R3 is

i j k=

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

=

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

1

0

0

0

1

0

==

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

0

0

1

(2.23)

Any vector u in R3 can be written as

u i j k= + +u u u1 2 3 (2.24)

An orthogonal basis is convenient for the representation of a vector space. It is often easy to obtain a linearly independent basis for a vector space. For exam-ple, it is clear that the vectors p1(x) = 1, p2(x) = x, and p3(x) = x2 form a basis for P

2[a,b]. However, this basis is not an orthogonal basis for P

2[a,b] with respect to

the standard inner product. Thus it is useful to develop a scheme to determine an orthogonal basis which spans the same space as a set of basis vectors.

Defi nition 2.11 A set of vectors is said to be normalized with respect to a norm if the norm of every vector in the set is one.

Defi nition 2.12 A set of vectors is said to be orthonormal with respect to an inner product if the vectors in the set are mutually orthogonal with respect to the inner product and the set is normalized with respect to the inner-product-generated norm.

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If a set of vectors u1,u2,…,un is orthonormal with respect to an inner product, then

( , )u ui j ij= =≠=

⎧⎨⎪⎪⎩⎪⎪

δ0 i j

1 i j (2.25)

for all i,j = 1,2,…,n.A fi nite-dimensional vector space V of dimension n has n basis vectors. The

following theorem provides a procedure by which another basis for V, ortho-normal with respect to any valid inner product on V, may be constructed. If a basis can be found for an infi nite-dimensional vector space, the Gram-Schmidt process, outlined in theorem 2.6, can be used to generate an orthonormal basis for the vector space.

Theorem 2.6 (Gram-Schmidt Orthonormalization) Let u1,u2,… be a fi nite or countably infi nite set of linearly independent vectors in a vector space V with a defi ned inner product (u,v). Let S be the subspace of V spanned by the vectors. There exists an orthonormal set of vectors v

1,v

2,v

3,… whose span is

also S. The members of the orthonormal basis can be calculated sequentially according to

w u vw

w

w u

1 1 11

1

2

= =

= 22 2 1 1 22

2

− =( , )u v v vw

w

�

n n

�

w u= − (uu v v vw

wn i i n

n

n

, )i

n

=

−

∑ =1

1

� �

(2.26)

Proof First consider the following lemma:

Lemma A set of vectors which are elements of a vector space V and which are mutually orthogonal with respect to any valid inner product defi ned on V are linearly independent.

Proof of Lemma Let v1,v2,…,vk be elements of a vector space V which are mutually orthogonal with respect to a valid inner product on V, (u,v), Then (vi,vj) = 0 for i = 1,2,…,k and for j = 1,2,…,k, but j ≠ k. Consider a linear combi-nation of the set of vectors set equal to the zero vector,

0 1 1 2 2= + +…+C C Ck kv v v (a)

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Taking the inner product of both sides of Equation a with vj for an arbitrary j between 1 and k, and using properties of a valid inner product, leads to

0 1 1 2 2 1 1, , , ,v v v v v v vj j j jC C C( )= ( )+ ( )+…+ ( ) (b)

Note that (0,vj) = 0 and that using the mutual orthogonality of the vectors in Equation b leads to Cj = 0. Since j is arbitrary, Cj = 0 for all j, j = 1,2,…,k. Thus the vectors are linearly independent, and the lemma is proved.

Since an orthogonal set of vectors is linearly independent, then if Equa-tions 2.24 generate an orthonormal set, then v1,v2,…,vn,… are linearly inde-pendent. Equations 2.24 also show that each of the vectors in the proposed orthonormal set is a linear combination of the vectors in the basis for S, and thus they are also in S. If S is a space of dimension n, then the proposed set of vectors is composed of n linearly independent vectors, and thus they span S. If S is an infi nite-dimensional space, then it can be shown that the proposed set is complete in S, but that is beyond the scope of this text.

First note that

vw

w wwi

i

i ii= = =

11

Hence the set is normalized. It only remains to show that w1, w2,… form an orthogonal set of vectors. This is done by induction. First, it is shown that w2 is orthogonal to v1

( , ) ( ( , ), )

( , ) ( , )( ,

w v u u v v

u v u v v

2 1 2 2 1 1

2 1 2 1 1

= −

= − vv1)

However, v v v1 1 11 21= = ( , ) / . Thus

( , ) ( , ) ( , )w v u v u v2 1 2 1 2 1 0= − =

Now assume

( , )v vi j ij= = −δ for i,j 1,2, ,k 1…

and consider

( , ) ( ( , ) , )

( , ) (

w v u u v v v

u v u

k i k k j j i

j

k

k i

= −

= −

=

−

∑1

1

kk j j i

j

k

, )( , )v v v=

−

∑1

1

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Since (vi, vj) = 0 for i ≠ j and for i,j < k, the only nonzero term in the sum occurs when j = i. Thus

( , ) ( , ) ( , )( , )

( , ) ( ,

w v u v u v v v

u v u

k j k i k i i i

k i k

= −

= − vv i)

= 0

The theorem is thus proved by induction.If a set of vectors u1, u2,…, uk–1, uk, uk + 1,… is a complete set of linearly inde-

pendent vectors in an infi nite-dimensional vector space V with a defi ned inner product, then the Gram-Schmidt process can be used to determine a set of orthonormal vectors that is complete in V.

Example 2.20 A basis for S, the intersection of P6[0,1] with a subspace of

C4[0,1], defi ned as those functions that satisfy the boundary conditions for

the differential equation governing the vibrations of a fi xed-fi xed beam, is developed in Example 2.15. One basis for S is determined as

f x x x x16 3 24 3( ) = − +

f x x x x25 3 23 2( ) = − + (a)

f x x x x34 3 22( ) = − +

Use the Gram-Schmidt procedure to determine a basis for S that is orthonor-mal with respect to the standard inner product for C

4[0,1].

Solution Normalizing f1,

f x x x dx16 3 2 2

0

11

2

4 3 0 171= − +( )⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

=∫ . (b)

Then

v x

f xf x

v x x x x

11

1

16 3 25 84 23 4 17 5

( )( )

( )

( ) . . .

=

= − +

(c)

Calculating w2

,

w x f x f x v x v x2 2 2 1 1( ) ( ) ( ( ), ( )) ( )= − (d)

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where

( ( ), ( )) ( . .f x v x x x x x x2 15 3 2

0

1

6 33 2 5 84 23 4= − +( ) −∫ ++ =17 5 0 099682. ) .x dx (e)

Thus Equation d becomes

w x x x x x x2

5 3 2 6 33 2 0 009968 5 84 23 4 17( ) . ( . . .= − + − − + 55

0 5823 0 6708 0 253

2

6 5 3 2

x

x x x x

)

. . .= − + − + (f)

Normalizing,

w x x x x x d26 5 3 2 2

0

1

0 5823 0 6708 0 253( ) . . .= − + − +( )∫ xx⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

=

1

2

0 00459. (g)

Thus

v x x x x x2

6 5 3 21

0 004590 5823 0 6708 0 253( )

.. . .= − + − +(( )

= − + − +127 74 219 4 147 2 55 56 5 3 2. . . .x x x x (h)

Calculating w3(x), using Equation (2.26)

w x f x f x v x v x f x v x3 3 3 1 1 3 2( ) ( ) ( ( ), ( )) ( ) ( ( ), ( )= − − )) ( )v x2 (i)

where

( ( ), ( )) ( )( . .f x v x x x x x x3 14 3 2 6 3

0

1

2 5 84 23 4= − + − +∫ 117 5 0 039622. ) .x dx = (j)

( ( ), ( )) ( )( . .f x v x x x x x x3 24 3 2 6 52 127 74 219 4= − + − + −−

+

=

∫ 147 2

55 5

0 00419

3

0

1

2

.

. )

.

x

x dx (k)

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Substituting Equations j and Equation k into Equation i leads to

w x x x x x x x34 3 2 6 32 0 03962 5 84 23 4 17 5( ) . ( . . .= − + − − + 22

6 5 30 00419 127 74 219 4 147 2 55 5

)

. ( . . . .− − + − +x x x x22

6 5 4 3 20 3035 0 9186 0 4580 0 0732

)

. . . .= − + − +x x x x x

(l)

Finally,

w x x x x x x36 5 4 30 3035 0 9186 0 4580 0 0732( ) . . . .= − + − + 22 2

0

11

2

0 0003444

( )⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

=

∫ dx

.

(m)

which when substituted into Equation l, leads to

v x x x x x36 5 4 3881 2 2667 3 2903 6 1329 9 21( ) . . . .= − + − + 22 4 2. x (n)

Example 2.21 Use the Gram-Schmidt process to derive an orthonormal basis for P2[0,1] with respect to the standard inner product on R7. Use the basis f1(x), f2(x), and f3(x) from Example 2.16.

Solution Polynomials in P6[0,1] can be represented as vectors in R7, as illus-trated in Example 2.8. To this end, the polynomials of Example 2.20 can be represented as

p1

1

0

0

4

3

0

0

= −

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

p2

0

1

0

3

2

0

0

= −

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

= −

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥p3

0

0

1

2

1

0

0

⎥⎥⎥⎥⎥⎥⎥⎥

(a)

The standard inner-product-generated norm for p1 is

p12 2 21 4 3 26= + − + =( ) ( ) ( ) (b)

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Thus p1 can be normalized as

vpp

11

1

1

26= = −

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥

1

0

0

4

3

0

0

⎥⎥⎥⎥⎥⎥⎥

(c)

Equation 2.26 is used to calculate a vector orthogonal to p1 as

w p p v v2 2 2 1 1

0

1

0

3

2

0

0

= −( )

= −

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

,

⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

− − − +[ ] −1

263 4 2 3

1

0

0

( )( ) ( )( ) 44

3

0

0

1

1

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

=33

9

13

0

3

1

0

0

−

−−

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

(c)

The norm of w2 is calculated as 260 13/ , and thus

v21

260

9

13

0

3

1

0

0

=

−

−−

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

(d)

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Equation 2.26 is used to construct

w3

0

0

1

2

1

0

0

4= −

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

−113

1

0

0

3

2

0

0

1−

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

−552

9

13

0

3

1

0

0

−

−−

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

=

−−

−

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥1

52

7

13

52

53

21

0

0

⎥⎥⎥⎥⎥⎥⎥⎥

(e)

The norm of w3 is calculated as 6172/52, and thus

v31

6172

7

13

42

53

21

0

0

=

−−

−

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

(f)

Converting the vectors to polynomial form, an orthonormal basis for S with respect to the standard inner product for R7 is

v x x x x

v x

16 3 2

2

0 1961 0 7845 0 5883

0 55

( ) . . .

( ) .

= − +

= − 882 0 8062 0 1861 0 06206 5 3 2x x x x+ − −. . .

v x x x x36 5 40 0819 0 1655 0 6619 0( ) . . .= − − + − .. .6746 0 26733 2x x+

(g)

2.9 Orthogonal expansionsThe Gram-Schmidt Theorem, theorem 2.6, shows that an orthonormal basis for any vector space can be obtained with respect to a valid inner product (u,v). Let S be a fi nite-dimensional vector space of dimension n, and let v1,v2,…,vn be a set of orthonormal vectors that span S with respect to a valid inner product on S. Then any u in S can be written as a linear combination of the vectors in the orthonormal basis

u vi==

∑α i

i

n

1

(2.27)

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Taking the inner product of both sides of Equation 2.27 with vj for an arbi-trary j = 1,2,…,n leads to

( , ) ( , )u v v vj i i

i

n

j==

∑α1

(2.28)

which, using properties (ii) and (iii) of defi nition 2.7, becomes

( , ) ( , )u v v vj i

i

n

i j==

∑α1

(2.29)

Since the basis is an orthonormal basis, the only nonzero term in the summa-tion on the right-hand side of Equation 2.29 corresponds to i = j. Simplifi cation thus leads to

α j j= ( , )u v (2.30)

Substitution of Equation 2.30 into Equation 2.29 leads to

u u v v==

∑( , )i i

i

n

1

(2.31)

Equation 2.31 provides an expansion for u in terms of the vectors in an ortho-normal basis. The coeffi cient α i given by Equation 2.30 is the component of u for the vector vi.

Example 2.22 The function

f x x x x x x( ) = − + − +2 2 9 66 5 4 3 2

is a member of Q, the vector space defi ned in Examples 2.11 and 2.20. Expand f(x) in terms of the orthonormal basis for Q determined in Example 2.20.

Solution Application of Equation 2.31 with n = 3 leads to

f x f x v x v x f x v x v x f( ) ( ( ), ( )) ( ) ( ( ), ( )) ( ) (= + +1 1 2 2 (( ), ( )) ( )x v x v x3 3 (a)

where

( ( ), ( )) ( )( .f x v x x x x x x x1

0

1

6 5 4 3 2 62 2 9 6 5 84= − + − +∫ −− +

=

23 4 17 5

0 322

3 2. . )

.

x x dx

(b)

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( ( ), ( )) ( )( .f x v x x x x x x x26 5 4 3 2 62 2 9 6 127 74= − + − + − ++

− + =

∫ 219 4

147 2 55 5 0 00382

5

0

1

3 2

.

. . .

x

x x dx)

(c)

( ( ), ( )) ( )( .f x v x x x x x x x36 5 4 3 2

0

1

2 2 9 6 881 2= − + − +∫ 66 5 4

3 2

2667 3 2903 6

1329 9 212 4

0

− +

− +

=

. .

. . )

.

x x

x x dx

00006888

(d)

Thus

2 2 9 6 0 322 5 84 23 4 17 56 5 4 3 2 6 3x x x x x x x− + − + = − +. ( . . . xx

x x x

2

6 5 30 00382 127 74 219 4 147 2 55 5

)

. ( . . . .+ − + − + xx

x x x

2

6 5 40 0006888 881 2 2667 3 2903 6 13

)

. ( . . .+ − + − 229 9 212 43 2. . )x x+

(e)

The development of the expansion of a vector in an infi nite-dimensional vector space is complicated by questions of completeness and convergence. A detailed discussion of these topics is important, but outside the scope of this study. However, several defi nitions and concepts are presented and theo-rems are presented without proof.

Defi nition 2.13 Let V be an infi nite-dimensional vector space with a defi ned norm. A sequence of vectors u1, u2, u3,… is said to converge to a vector u if

limn

n→∞

− =u u 0 (2.32)

Defi nition 2.14 Let V be an infi nite-dimensional vector space with a defi ned norm, and let u1, u2, u3,… be a sequence of vectors in V. The sequence is said to be a Cauchy sequence if for each ε > 0, there exists an N such that if m,n ≥ N, u um n− < ε .

A convergent sequence is also a Cauchy sequence, but the converse is not always true. Consider, for example, the sequence of real numbers defi ned by x nn = 1/ . This sequence of real numbers is a Cauchy sequence, but it does not converge to a real number.

Defi nition 2.15 Let V be an infi nite-dimensional vector space with a defi ned norm. The vector space is said to be complete if every Cauchy sequence in the space converges to an element of the space.

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A vector space can be complete with respect to one norm, but not with respect to another. For example, the space C[a,b] is complete with respect to the norm defi ned f f x dxa

b2

1 22= ∫[ [ ( )] ] / , but C[a,b] is not complete with

respect to the norm f f xa x b∞ ≤ ≤

= max ( ) .

Defi nition 2.16 An infi nite-dimensional space which has an inner product defi ned on it and which is complete with respect to the inner-product-gener-ated norm is called a Hilbert space.

Defi nition 2.17 An orthonormal set of vectors, S, is said to be complete in a Hilbert space if there is no other set of vectors of which S is a subset.

The Gram-Schmidt theorem implies that there is an orthonormal basis for every Hilbert space which is complete in the space. However, not every orthonormal set is complete. The following theorem deals with complete-ness of the orthonormal set.

Theorem 2.7 (Parseval’s identity) Let u1, u2, u3,… be an orthonormal set in a Hilbert space. Then if

v v u==

∞

∑( , )i

i

2

1

(2.33)

for every v in the Hilbert space, then the set is complete.Finally, the following theorem shows how to establish an expansion of a

vector in terms of a complete orthonormal set in a Hilbert space.

Theorem 2.8 Let u u u1 2 3, , ,� be a complete orthonormal set in a Hilbert space V, and let v be an arbitrary element of V. Then

v v u u==

∞

∑( , )i i

i 1

(2.34)

Example 2.23 It is shown in chapter 5 that the set of functions defi ned by

f x i xi( ) sin= ( )2 π (a)

is a complete orthonormal set on the Hilbert space V, which is defi ned as the set of all functions in C[0,1] such that if y(x) is in V, then y(0) = 0 and y(1) = 0 with the standard inner product for C[0,1]. Expand f(x) = x2–x, which is a member of V, in terms of this orthonormal set.

Solution The appropriate expansion is

x x i xi

i

2

1

2− ==

∞

∑α πsin( ) (b)

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where

α π

π

i

i

x x i x dx

i

= −

= −( ) −⎡⎣ ⎤⎦

∫ ( ) sin( )2

0

1

3 3

2

2 21 1

(c)

Thus

x xi

i xi

i23 3

1

4 11 1− = − −⎡⎣ ⎤⎦

=

∞

∑ππ( ) sin( ) (d)

The series on the right-hand side of Equation d converges to x2 – x with respect to the standard inner product generated norm defined for C[0, 1] in the sense of definition 2.13.

2.10 Linear operatorsMathematical modeling of physical systems leads to the formulation of a mathematical problem whose solution provides required information about the physical system. It is convenient to examine these equations using a con-sistent formulation. Let u represent a vector of dependent variables which is an element of a vector space D, and let f represent a vector, obtained through the modeling process, which is an element of a vector space R. The relation-ship between u and f can be written as

Lu f= (2.35)

where L is an operator determined from the modeling process. A formal defi nition of an operator follows:

Defi nition 2.18 An operator L is a function by which an element of a vector space D, called the domain of L, is mapped into an element of a vector space R, called the range of L.

Equation 2.35 is the general form of an operator equation. Given a vector f and the defi nition of L, it is desired to fi nd the vector or vectors for which Equation 2.35 is satisfi ed. The vectors u which satisfy Equation 2.35 are said to be solutions of the equation. Before considering how the solutions to Equation 2.35 are obtained, two basic questions are considered. (1) Does a solution exist? (2) If so, how many solutions exist?. The fi rst question can be phrased as, “For a specifi c f in R, is there at least one u in D which solves Equation 2.35?” An alternate form of the second question is, “If a solution exists, is it unique?”. If for each f in R, a unique solution u, a vector in D, exists, then there is a one to one correspondence between the elements of D and R. In this case, an inverse operator L−1 exists whose domain is R and whose range is D, such that

u L f= −1 (2.36)

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Defi nition 2.19 If L is an operator defi ned such that there is a one-to-one correspondence between the elements of its domain D and the elements of its range R, then the inverse of L, denoted by L−1, exists, with domain R and range D, and is defi ned such that if Lu = f, then u = L−1f.

One method of fi nding the solution of Equation 2.35 is to determine the inverse of L and apply Equation 2.36. Note that substitution for f from Equa-tion 2.35 into Equation 2.36 leads to

L Lu u− ( )=1 (2.37)

If R = D, then u is an element of the domain of L−1, and then Equation 2.35 and Equation 2.36 lead to

L L u u−( )=1 (2.38)

Defi nition 2.20 An operator L is said to be a linear operator if for each u and v in D and for all scalars α and β,

L u v Lu Lvα β α β+( )= + (2.39)

An equation of the form of Equation 2.35 in which L is linear is said to be a linear equation. If L is not a linear operator, then Equation 2.35 is a nonlinear equation. The focus of this study is on linear equations.

An n × m matrix is a linear operator whose domain is Rm and whose range is a subspace of Rn. Consider a set of linear equations to solve for a set of m variables x1,x2,x3,…,xm, as illustrated below:

a x a x a x a x y

a x a

m m1 1 1 1 2 2 1 3 3 1 1

2 1 2 2 2

, , , ,

, ,

+ + + + =

+

…

xx a x a x ym m2 2 3 3 2 2+ + + =, ,…

� � � � �

…a x a x a x a x yn n n n m m n, , , ,1 1 2 2 3 3+ + + + =

(2.40)

The matrix formulation of this set of equations is

a a a a

a a a a

a

m

m

n

1 1 1 2 1 3 1

2 1 2 2 2 3 2

1

, , , ,

, , , ,

,

…

…

� � � � �

aa a a

x

x

x

xn n n m

m, , ,2 3

1

2

3

…�

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥

=

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

y

y

yn

1

2

� ⎥⎥⎥

(2.41)

Equation 2.41 can be summarized as

Ax y= (2.42)

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where A is the matrix operator which represents the coeffi cients of the equa-tions of Equation 2.40 arranged in n rows and m columns. The element in the ith row and jth column is identifi ed as ai,j. The solution x is a vector in Rm, while the input vector y is a vector in Rn.

When n > m, the number of linear equations represented by Equation 2.41 is greater than the dimension of the solution vector. If more than m equations are independent, then a solution does not exist for all y. That is, the range of A is not all of Rn When n < m, the number of equations is less than the dimen-sion of the solution vector. In this case, a solution exists, but is not unique; the range is all of Rn, but for each y in Rn, there is more than one x in Rm that solves Equation 2.42.

When n = m and the number of equations is equal to the dimension of the solution vector, then the domain of A is Rn and the range of A is a subspace of Rn. It can be shown that the range is all of Rn when the matrix A is nonsin-gular. That is, its determinant is not equal to zero. When the matrix is non-singular, then A−1 exists and has the property

A Ax x− ( )=1 (2.43)

An associative property can be used on Equation 2.43, giving (A−1A)x = x. The operator in parentheses defi nes the n × n identity matrix (a matrix with ones along the diagonal and zeros for all off-diagonal elements). A formal procedure exists to determine the inverse of a nonsingular square matrix.

When the determinant of a square matrix is zero, the matrix is said to be singular. In this case, the range of A is only a subset of Rn. That is, a solution does not exist for all y in Rn. When a solution does exist for a system with a singular matrix, the solution is not unique.

Example 2.24 The equation of a plane in three-dimensional space is of the form

ax by cz d+ + = (a)

where a, b, c, and d are constants. If n i j k= + +n n nx y z is a unit vector normal to the plane, Equation a can be rewritten as

n x n y n z dx y z+ + = ˆ (b)

where d d a b c= + +/ 2 2 2 . Defi ning r = xi + yj + zk as the position vector from the origin to the point (x,y,z), Equation b can be written in the form

r n⋅ = d (c)

Equation c can be interpreted as a statement that the length of the projection of a position vector from the origin to any point on the plane is the constant d .

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The equations of three separate planes may be written as

a x b y c z d1 1 1 1+ + = (d)

a x b y c z d2 2 2 2+ + = (e)

a x b y c z d3 3 3 3+ + = (f)

Values of x, y, and z which satisfy Equation d, Equation e, and Equation f are the coordinates of a point that lies on all three planes, a point on the intersec-tion of the planes. Equation d, Equation e, and Equation f may be summarized in matrix form as

a b c

a b c

a b c

x

y

z

1 1 1

2 2 2

3 3 3

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥=

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

d

d

d

1

2

3

(g)

Use the matrix formulation given in Equation g to discuss the existence and uniqueness of a point of intersection of three planes.

Solution Three planes may (a) intersect at a single point, (b) intersect along a line, or (c) have no common point of intersection. If the planes intersect at a single point, then Equation g has a unique solution. In this case, the determi-nant of the matrix on the left hand side of Equation g is nonzero. Noting the geometric interpretation of Equation c, this implies that the normal vectors to the planes are linearly independent.

If the vectors are linearly dependent, then without loss of generality it can be assumed that there exist constants α and β such that

n n n1 2 3= +α β (h)

where n1, n2 and n3 are unit vectors normal to the planes, and that a position vector, if it exists, satisfi es equations of the form of Equation c for each plane. These equations are used in the following derivation:

r n

r n n

r n r n

⋅ =

⋅ +( )=

⋅ + ⋅ =

+

1 1

2 3 1

2 3 1

2

ˆ

ˆ

ˆ

ˆ

d

d

d

d

α β

α β

α ββˆ ˆd d1 3=

(i)

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Equation i must be satisfi ed for a solution of Equation g to exist when the normal vectors to the planes are linearly dependent. A condition of the form of Equation i is called a solvability condition. When the solvability condition is satisfi ed, the three planes intersect in a line, and the solution is not unique.

If the solvability condition of Equation i is not satisfi ed, then there is no intersection between the three planes.

Example 2.25 The state of stress at a point in a material is uniquely defi ned by the stress tensor at the point,

σ =

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟σ τ ττ σ ττ τ σ

xx xy xz

yx yy yz

zx zy zz

⎟⎟⎟⎟⎟⎟ (a)

where σxx is the stress acting normal to a plane whose normal is in the + x direction, τ xy and τ xz are the components of the shear stress acting in this plane with τ xy the component in the + y direction, and τ xz is the component in the + z direction. The angular momentum equation can be used to show that the stress tensor is symmetric, τ τ τ τ τ τyx xy xz zx yz zy= = =, and , in the absence of unusual phenomena such as body moments or intrinsic angular momentum.

Let n = nxi + nyj + nzk be a unit vector normal to a plane through the point for which the stress tensor is known. The stress vector acting on the plane is calculated by

σσσ

σ τ τσ τ

τ ττ

x

y

z

xx xy xz

yx yy yz

zx zy

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥=

σσzz

n

n

n

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

1

2

3

(b)

The stress tensor is a matrix operator. In general, its domain and range are R3. However, in application, its domain is restricted to the set of all unit vec-tors in R3. The resulting range is thus also restricted. In operator notation,

An = σ (c)

The stress tensor at a point in a solid is determined to be

σ =−

− −− −

⎛

⎝

⎜⎜⎜⎜⎜

900 600 300

600 1200 600

300 600 1500⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟Pa (d)

Determine (a) the stress vector, (b) the normal stress, and (c) the shear stress acting on a plane whose normal is n i j k= + −2 3 1 3 2 3/ / / .

9533_C002.indd 1139533_C002.indd 113 10/29/08 3:57:12 PM10/29/08 3:57:12 PM


Solution The stress vector acting on this plane can be calculated using Equation c:

σσσ

x

y

z

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥=

−− −900 600 300

600 1200 600

3000 600 1500

2

3

1

3

2

3

− −

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥−

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

=

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥

200

400

1000⎥⎥Pa

(e)

The normal stress is the projection of the stress vector onto the unit vector normal to the plane, or

σ σn = ⋅ n (f)

Substitution of Equation c into Equation f leads to

σn =( )An n, (g)

where the inner product of Equation g is the standard inner product for R3. Thus

σn =( )

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+( )

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+200

2

3400

1

310000

2

3

400

( ) −⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟

= − Pa

(h)

(c) The shear stress is the component of the stress vector perpendicular to the unit normal. Thus if τ is the shear stress vector, then τ ⋅ n = 0. Also τ = σ − σn

n. Hence, for this problem,

τ =

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥− −( )

−

⎡

⎣

⎢⎢⎢200

400

1000

400

2

3

1

3

2

3

⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥

=

⎡

⎣

⎢⎢1

3

1400

1600

2400⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

Pa

(i)

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The magnitude of the shear stress is calculated as

τ τ τ= ⋅ = ×1 2 31 07 10/

. Pa (j)

Example 2.26 Consider an n-degree-of-freedom linear mechanical system as illustrated in Figure 2.6. The displacements of the particles as functions of time are defi ned as x1, x2,…, xn and are referred to as generalized coordi-nates. The potential energy of the system, V, at any instant is a function of the generalized coordinates, V = V(x1, x2,…, xn). The stiffness matrix for a linear system is the matrix K whose element in the ith row and jth column, ki,j, is calculated by

kV

x xi j

i j, =

∂∂ ∂

2

(a)

Since the order of differentiation is interchangeable, ki,j = kj,i , and therefore the stiffness matrix is symmetric.

The potential-energy functional for a linear system can be written in a quadratic form as

V k x xij i j

j

n

i

n

===

∑∑1

211

(b)

(a) Determine the quadratic form of potential energy for the system shown in Figure 2.7 and use it to determine the stiffness matrix for the system.

x1 x2

k1 k2

xn

k3 knm1 m2 mn

Figure 2.6 The stiffness matrix for the system of Example 2.26 is obtained using the potential-energy functional written at an arbitrary instant.

x1 x3x2

k k 3km m m

Figure 2.7 The stiffness matrix for this three-degree-of-freedom mechanical system is a 3 × 3 matrix.

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(b) Show that, for this system, if x is the vector of generalized coordinates at an arbitrary instant,

Kx x,( )= 2V (c)

where the inner product is the standard inner product for Rn.

Solution (a) Note that the potential energy developed in a spring of stiff-ness k when the spring is subject to a force which leads to a change in length of x, measured from the spring’s unstretched length, is V = 1/2(kx2). The potential energy for the system of Figure 2.7 at an arbitrary instant is

V kx k x x k x x

kx

= + −( ) + −( )

= −

1

2

1

2

1

23

2

12

2 12

3 22

121

222 4 6 31 2 2

22 3 3

2kx x kx kx x kx+ − +⎡⎣ ⎤⎦

(d)

The stiffness matrix is obtained through comparison of Equation d with the general quadratic form of Equation c as

K =−

− −−

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

2 0

4 3

0 3 3

k k

k k k

k k

(e)

To obtain Equation d, the symmetry of the stiffness matrix is used, so that two terms within the double summation in Equation b are combined to yield k x x k x x k x xi j i j j i j i i j i j, , ,+ = 2 .

Equation e is used to calculate

Kx =−

− −−

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

⎡

⎣

⎢⎢

2 0

4 3

0 3 3

1

2

3

k k

k k k

k k

x

x

x⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

=−

− + −− +

⎡ 2

4 3

3 3

1 2

1 2 3

2 3

kx kx

kx kx kx

kx kx⎣⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

(f)

The required inner product is evaluated as

Kx x,( )= −( ) + − + −( ) + −2 4 3 31 2 1 1 2 3 2kx kx x kx kx kx x kx22 3 3

12

1 2 22

2 3 32

3

2 2 4 6 3

+( )

= − + − +

kx x

kx kx x kx kx x kx

== 2V

(g)

9533_C002.indd 1169533_C002.indd 116 10/29/08 3:57:14 PM10/29/08 3:57:14 PM


There are many examples of linear operators defi ned for infi nite-dimen-sional vector spaces. The proof of existence and uniqueness of solutions of equations of the form of Equation 2.35 when the domain of L is an infi nite-dimensional vector space is beyond the scope of this study. However, the problems considered are formulated from the viewpoint of the mathematical modeling of a physical system. The physics of the system often dictate that a solution must exist and it must be unique. For example, the temperature distribution in a solid must be continuous and single-valued, requiring a unique solution. If the mathematical problem correctly models the physics, then unique solutions should exist.

Example 2.27 The nondimensional differential equation for the steady-state temperature distribution in a thin rod subject to an internal heat generation is

ddx

Bi x2

2

ΘΘ− = +α β πsin( ) (a)

where α and β are nondimensional constants and Bi = hp/kA is the Biot num-ber. The left end of the rod is maintained at a constant temperature, while the right end is insulated. The temperature distribution satisfi es the boundary conditions

Θ( )0 0= (b)

ddxΘ

( )1 0= (c)

Write this problem in the operator form of Equation 2.35 and defi ne D and R.Show, by obtaining the solution of the differential equation, the existence

and uniqueness of the solution.

Solution (a) Equation a can be written in the form LΘ = f, where LΘ = d 2Θ/ and dx2 − BiΘ and f (x) = α + βsin(πx) The domain D is the subspace of C2[0,1] of all functions g(x) such that g(0) = 0 and dg/dx(1) = 0. The range R is the set of all elements of PC[0,1], the space of all piecewise continuous func-tions defi ned on [0,1].The homogeneous solution of Equation a is of the form

Θh x C Bix C Bix( ) cosh sinh= ( )+ ( )1 2 (d)

where C1 and C2 are arbitrary constants of integration. The particular solu-tion of Equation a is

Θp xBi Bi

x( ) sin= − −+

( )α β

ππ

2 (e)

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The general solution of Equation a is

Θp x C Bix C BixBi Bi

( ) cosh sinh sin= ( )+ ( )− −+

1 2 2

α βπ

ππx( ) (f)

Application of the boundary condition of Equation b to Equation f leads to

0 1 1= − ⇒ =CBi

CBi

α α (g)

Application of the boundary condition of Equation c to Equation f used Equation g leads to

0 2 2

2

= ( )+ ( )++

= −

α βππ

π

BiBi C Bi Bi

Bi

CBi

cosh sinh

βπ22 +( ) ( )

− ( )Bi Bi Bi

Bisinh

cothα

(h)

Substitution of Equation g and Equation h into Equation f leads to

Θ( ) coshsinh

coxBi

BixBi Bi Bi Bi

= ( )++( ) ( )

−α βπ α

π2tth

sinh sin

Bi

BixBi

x

( )⎡

⎣⎢⎢⎢

⎤

⎦⎥⎥⎥

( )− −+

( )α β

π β2π

(i)

Equation i satisfi es the differential equation, Equation a, as well as the bound-ary conditions of Equation b and Equation c. Thus a solution exists. If a sec-ond solution ˆ ( )Θ x exists such that ˆ ( ) ( )Θ Θx x≠ for some values of x, then the temperature is multi-valued at those values, a physical impossibility. There-fore, if Equation a, Equation b, and Equation c are a true mathematical model of the physical system, the solution must be unique.

Example 2.28 The nondimensional differential equations for the trans-verse displacements w x w x1 2( ) ( )and of two elastically coupled, statically loaded beams with an axial load P are

d wdx

d wdx

w w4

1

4

21

2 1 2 0− + − =ε η( ) (a)

μ ε ηd wdx

d wdx

w w f x4

14

21

2 2 1− + − =( ) ( )Λ (b)

9533_C002.indd 1189533_C002.indd 118 10/29/08 3:57:16 PM10/29/08 3:57:16 PM


where ε η μ= = =PL E I kL E I E I E I21 1

41 1 2 2 1 1/ / and /, . The load per unit length,

F(x), is nondimensionalized according to f x F x F( ) ( ) max= / , and then Λ = F L E Imax

31 1/ . The boundary conditions for pinned-pinned beams are:

w w

d wdx

1

21

2

0 0 1 0

0 0

( ) ( )

( )

= =

=

1

d wdx

21

21 0( ) =

(c)

w w

d wdx

2 2

22

2

0 0 1 0

0 0

( ) ( )

( )

= =

=d wdx

22

21 0( ) =

(d)

(a) Write this problem in the operator form of Equation 2.33 and defi ne D and R.

(b) Show, by obtaining the solution of the differential equation, the exis-tence and uniqueness of the solution. Use ε = 2, η = 1, μ = 1 and Λ = 1 with f(x) = sin(πx).

Solution (a) The problem defi ned by Equation a, Equation b, Equation c, and Equation d can be written in the form of Equation 2.35 as Lw = f with w = [w1(x) w2(x)]T,

Lw =− + −

− − +

⎡

⎣

⎢⎢⎢⎢⎢

ddx

ddx

ddx

ddx

4

4

2

2

4

4

2

2

ε η η

η μ ε η⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

⎡

⎣⎢⎢

⎤

⎦⎥⎥

w x

w x1

2

( )

( ) (e)

and

f =⎡

⎣⎢⎢

⎤

⎦⎥⎥

0

Λf x( ) (f)

Let Q be the vector space defi ned as Q = C4 [0,1] × R2. An element of Q is a two-dimensional vector of functions with each element a member of C4 [0,1].

That is, if g is in Q, then g =⎡

⎣⎢⎢

⎤

⎦⎥⎥

g x

g x1

2

( )

( ), where g x g x1 2( ) ( )and each an element

of C4 [0,1]. Defi ne S as the subspace of Q defi ned such that if g is in Q, then

gd gdx

gd gdx

1

21

2 1

21

20 0 0 0 1 0 1( ) , ( ) , ( ) , ( )

= = =

, ( ) , ( ) , ( )= = = =0 0 0 0 0 1 02

22

2 2

2

gd gdx

gd

andgg

dx2

21 0( ) =

Then D = S. The range of L is R = ×C4[ , ]0 1 2R .

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(b) Substitution of the given parameters into Equation a and Equation b leads to

d wdx

d wdx

w w4

1

4

21

2 1 22 0− + − =( ) (g)

d wdx

d wdx

w w x4

24

22

2 2 12− + − =( ) sin( )π (h)

The solutions of Equation g and Equation h are of the form

w w w= +h p (i)

where wh is the homogeneous solution, the solution obtained if f = 0, and wp is the particular solution, the solution particular to the specifi c form of f.

The homogeneous solution is assumed to be of the form

whxa

be=

⎡

⎣⎢⎢

⎤

⎦⎥⎥

α (j)

Substitution of Equation j into Equation g and Equation h with f = 0 leads to

α α4 22 1 0− +( ) − =a b (k)

− + − +( ) =a bα α4 22 1 0 (l)

A nontrivial solution to Equation k and Equation i exists only if

α α

α α

4 2

4 2

2 1 1

1 2 10

− + −− − +

= (m)

Evaluation of the determinant in Equation k leads to the following eighth-order polynomial equation:

α α α α8 6 4 24 6 4 0− + − = (n)

The solutions of Equation l are

α = − ± − ±0 0 1 414 1 414 1 0987 0 4551 1 0987 0, , . , . , . . , .i ..4551i (o)

Equation k implies that for an appropriate value of ∝, α, b = − +( )α α4 22 1 . The homogeneous solution is a linear combination of all possible solutions of the form of Equation j, where the appropriate values of α are given in Equation o.

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Arbitrarily choosing a = 1 in each case and using trigonometric functions to replace complex exponentials leads to a homogeneous solution of

w t

w tC C

1

21 2

1

1

1

1

( )

( )

⎡

⎣⎢⎢

⎤

⎦⎥⎥=

⎡

⎣⎢⎢

⎤

⎦⎥⎥+

⎡

⎣⎢⎢

⎤

⎦⎥⎥⎥

+⎡

⎣⎢⎢

⎤

⎦⎥⎥

+⎡

⎣⎢⎢

⎤

⎦⎥⎥

−t C e C et3

1 4144

1 4141

1

1

1. . tt

C t cosh( . )+−⎡

⎣⎢⎢

⎤

⎦⎥⎥5

1

11 0987 ccos( . )

co

0 4551

1

16

t

C+−⎡

⎣⎢⎢

⎤

⎦⎥⎥

ssh( . )sin( . )

1 0987 0 4551

17

t t

C+−11

1 0987 0 4551⎡

⎣⎢⎢

⎤

⎦⎥⎥sinh( . )cos( . )

t t

sinh( . )sin( .+−⎡

⎣⎢⎢

⎤

⎦⎥⎥

C t8

1

11 0987 0 45511t)

(p)

The particular solution is obtained using the method of undetermined coef-fi cients, discussed in Chapter 3. A particular solution is assumed as

w x

w x

W

Wx1

2

1

2

( )

( )sin( )

⎡

⎣⎢⎢

⎤

⎦⎥⎥ =

⎡

⎣⎢⎢

⎤

⎦⎥⎥ π

(q)

Substitution of Equation q into Equation g and Equation h, using each to develop a relation between W1 and W2

and solving simultaneously leads to

w x

w x1

2

2 4

2 2

2 4

1

1 1

1

1

( )

( )

⎡

⎣⎢⎢

⎤

⎦⎥⎥ =

+( ) +

+( )+( )

π

π

π ++

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥1

sin( )πx

(r)

The general solution is the sum of the homogeneous solution, Equation q and the particular solution Equation r. Application of the boundary conditions, Equation c and Equation d leads to a set of eight simultaneous equations to solve for the constants of integration for which a unique solution is attained.

2.11 Adjoint operatorsEach linear operator has a related operator, called its adjoint, which has important properties.

Defi nition 2.21 Let V be a vector space with an inner product (u, v) defi ned for all u and v in V. Let L be a linear operator whose domain is D, a subspace

9533_C002.indd 1219533_C002.indd 121 10/29/08 3:57:17 PM10/29/08 3:57:17 PM


of V, and whose range is R, also a subspace of V. The adjoint of L with respect to the defi ned inner product, written L*, is an operator whose domain is R and whose range is D, such that

Lu v u L v, ( , )*( )= (2.44)

Defi nition 2.22 If D = R and L* = L, then L is said to be self-adjoint, that is,

Lu v u Lv, ,( )=( ) (2.45)

for all u and v in D.

Example 2.29 Let A be a n × n matrix of the form

A =

a a a a

a a a a

a a

n

n

1 1 1 2 1 3 1

2 1 2 2 2 3 2

3 1 3

, , , ,

, , , ,

, ,

…

…

22 3 3 3

1 2 3

a a

a a a a

n

n n n n n

, ,

, , , ,

…

� � � � �

…

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢

⎤⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥

(a)

(a) Determine A* with respect to the standard inner product on Rn.(b) Under what conditions is A self-adjoint with respect to the standard inner product on Rn?(c) Determine A* with respect to an inner product on Rn defi ned by

( , )u v = + + + +2 1 1 2 2 3 3u v u v u v u vn n… (b)

Solution (a) Let u and v be arbitrary vectors in Rn. Then by defi nition of the standard inner product for Rn,

Au v Au v,

,

( )= ( )

=

=

==

∑

∑∑

i i

i

n

i j j i

j

n

i

n

a u v

1

11

(c)

Since A* is an operator whose domain and range are Rn, it has a matrix rep-resentation of the form

A*

,*

,*

,*

,*

,*

,*

,*

,

=

a a a a

a a a an1 1 1 2 1 3 1

2 1 2 2 2 3 2

…

… nn

n

n n n

a a a a

a a a

*

,*

,*

,*

,*

,*

,*

,

3 1 3 2 3 3 3

1 2 3

…

� � � � �**

,*… an n

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥

(d)

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Then

( , )* *

*,

u A v A v= ( )

=

=

==

∑

∑∑

u

u a v

i ii

n

i j

j

n

i

n

i j

1

11

(e)

Interchanging the names of the indices in Equation e leads to

( , )* *,u A v =

==∑∑ a u vj i j j

i

n

j

n

11

(f)

For the expression in Equation c to be equal to the expression in Equation f for all possible u and v, it is required that

a ai n

j ni j j i,*

,

, , ,

, , ,=

==

⎧⎨⎪⎪⎩⎪⎪

1 2

1 2

…

… (g)

Thus

A*

, , , ,

, , , ,

,=

a a a a

a a a a

a a

n

n

1 1 2 1 3 1 1

1 2 2 2 3 3 2

11 3

…

…

22 3 3 3 3

1 2 3

, , ,

,, , , ,, ,

a a

a a a a

n

n n n n n

…

� � � � �

…

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥

(h)

Thus the adjoint of A with respect to the standard inner product on Rn is the matrix obtained by interchanging the rows and columns of A. Such a matrix is called the transpose matrix, AT. Thus A* = AT.

(b) From Equation g, it is clear that A is self-adjoint with respect to the standard inner product if

a ai n

j ni j j i, ,

, , ,

, , ,=

==

⎧⎨⎪⎪⎩⎪⎪

1 2

1 2

…

… (i)

Such a matrix, whose columns can be interchanged with its rows without changing the matrix, is called a symmetric matrix. Thus an n × n matrix A is self-adjoint with respect to the standard inner product for Rn if and only if A is a symmetric matrix.

9533_C002.indd 1239533_C002.indd 123 10/29/08 3:57:18 PM10/29/08 3:57:18 PM


Following along the same lines as part (a), and using the inner product defi ned in Equation b, leads to

Au v Au Au vi,

,

( )= ( ) + ( )

= +

=

=

∑2

2

1 1

2

1 1

1

v

a u v

ii

n

j j

j

n

∑∑ ∑∑

∑

==

=

= +

a u v

a u v a u

i j j i

j

n

i

n

j j

j

n

i

,

, ,

12

1 1

1

12 11

2 22

v a u vi

i

n

i j j i

j

n

i

n

= ==∑ ∑∑+ ,

(j)

Assuming A* in the form of Equation d leads to

( , )* * *

,*

u A v A v A v= ( ) + ( )

= +

=∑2

2

1 1

2

1 1

u u

u a v

i ii

n

j j

jj

n

i j

j

n

i

n

j j

j

n

u a v

a u v u

i j

= ==

=

∑ ∑∑

∑= +

1 12

1 1

1

2

,

*

,*

ii i

i

n

i j

j

n

i

n

a v u a vi j,* *

,1 1

2 22= ==∑ ∑∑+

(k)

Renaming the indices in Equation k and rearranging leads to

u A v, *

,*

,*( )= + +

= =∑ ∑2 1 1

1

1 1

2

a v u u a v u ai i

i

n

j j

j

n

i j ,,

*i vj

j

n

i

n

==∑∑

12

(l)

Equation j and Equation l are identical for all possible u and v if and only if

a a

a a j n

a a

j j

j

1 1 1 1

1 1

1 1

1

2

2

,*

,

,*

,

,*

=

= =

=

2,3, ,…

,,

,*

, ,

j

j i i j

j n

a a i j

2,3,...,

2,3,...,

=

= = nn

(l)

9533_C002.indd 1249533_C002.indd 124 10/29/08 3:57:19 PM10/29/08 3:57:19 PM


For example, a pair of adjoints with respect to the inner product of Equation b is

A A=

−−−

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

4 2 1 3

2 4 2 4

4 2 1 0

2 3 1 5

**

4 1 2 1

4=

− −

−

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

4 2 3

2 2 1 1

6 4 0 5

(m)

Part (a) of Example 2.29 illustrates that a symmetric n × n matrix is self-adjoint with respect to the standard inner product for Rn. Thus the matrices developed from the stress tensor in Example 2.25 and the stiffness matrix of Example 2.26 are self-adjoint with respect to the standard inner product.Example 2.30 The nondimensional differential equation governing the defl ection, w(x), of a non-uniform beam due to a distributed load per unit length, f(x), is

d

dxx

d wdx

f x2

2

2

2α( ) ( )

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟= (a)

where α(x) is a function describing the variation of material and geomet-ric parameters across the span of the beam. The boundary conditions are dependent on the end constraints. For a beam fi xed at x = 0 and free at x = 1, the appropriate boundary conditions are

w( )0 0= (b1)

dwdx

( )0 0= (b2)

d wdx

2

21 0( ) = (b3)

d

dxx

d wdx

α( ) ( )2

21 0

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟ = (b4)

Equation a may be written in the form of Equation 2.35 with Lw = d2/dx2 (α(x) d2w/dx2). Defi ne V as C4[0,1]. The domain of L is the subspace of V such that if g(x) is in D, then g(x) satisfi es the boundary conditions of Equation b. Consider the range of V to be the same as its domain. The standard inner product on V is

( ( ), ( )) ( ) ( )f x g x f x g x dx= ∫0

1

(c)

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Show that L is self-adjoint on D with respect to the inner product of Equa-tion C.

Solution If L is self-adjoint on D, then (Lf, g) = (g, Lf) for all f(x) and g(x) in D. That is,

d

dxx

d fdx

g x dx f xd

d

2

2

2

2

0

12

α( ) ( ) ( )⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟ =∫ xx

xd gdx

dx2

2

2

0

1

α( )⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟∫ (d)

Recall the integration by parts formula, ∫ = − ∫udv uv vdu . Using integration by parts on the integral on the left-hand side of Equation d with u = g(x) and dv d d f dx dx dx= ⎡

⎣⎢⎤⎦⎥

2 2 2 2( / )/α gives

d

dxx

d fdx

g x dx g xd

d

2

2

2

2

0

1

α( ) ( ) ( )⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟ =∫ xx

d fdx

ddx

d fdxx

x

α α2

20

1 2

2

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟ −

⎛

⎝⎜⎜⎜

=

=

⎜⎜⎞

⎠⎟⎟⎟⎟∫ dg

dxdx

0

1

(e)

Using integration by parts on the remaining integral of Equation e with u = dg/dx and dv d d f dx dx dx= ⎡

⎣⎢⎤⎦⎥( / )/α 2 2 2 leads to

ddx

xd fdx

g x dx g xd

dx

2

2

2

2

0

1

α( ) ( ) ( )⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟ =∫ αα α

d fdx

dgdx

xd fdxx

x2

20

1 2

2

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟ −

⎡

⎣=

=

( )⎢⎢⎢

⎤

⎦⎥⎥

+

=

=

=

∫

x

x

xd fdx

d gdx

dx

gd

d

0

1

2

2

2

2

0

1

1

α( )

( )xx

d fdx

gd

dxd fdx

α α2

2

2

21 0

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟ −

⎛⎝⎜⎜⎜( ) ( )

⎞⎞⎠⎟⎟⎟⎟

− −

( )

( ) ( ) ( ) ( ) ( )

0

1 1 1 0 02

2

dgdx

d fdx

dgdx

α αdd fdx

xd fdx

d gdx

dx

2

2

2

2

2

2

0

1

0( )

( )+∫ α

(f)

Since f(x) and g(x) are both in D, d/dx (α(d2 f/dx2))(1) = 0 g(0) = 0, d2 f/dx2 (1) = 0, and dg/dx (0) = 0. Thus Equation f reduces to

9533_C002.indd 1269533_C002.indd 126 10/29/08 3:57:20 PM10/29/08 3:57:20 PM


d

dxx

d fdx

g x dx xd f2

2

2

2

0

12

α α( ) ( ) ( )⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟ =∫ ddx

d gdx

dx2

2

2

0

1

∫ (g)

Using similar steps, it can be shown that for all f(x) and g(x) in D,

f xd

dxx

d gdx

dx xd fdx

( ) ( ) ( )2

2

2

2

2

2α α

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟ =

dd gdx

dx2

2

0

1

0

1

∫∫ (h)

Thus Equation a is proved, and L is self-adjoint on D with respect to the standard inner product.

Example 2.31 The nondimensional partial differential equation for the steady-state temperature distribution, Θ (r) in a bounded three-dimensional region V is

∇ 2 Θ = – f(r) (a)

where r is a position vector from the origin of the coordinate system to a point in the body and ∇ 2 is the Laplacian operator where Bi is the Biot number. The surface of the region is described by S(r) = 0. The surface of the body is open, and heat transfer occurs though convection, leading to a boundary condition of the form

∇Θ.n = –BiΘ on S (b)

Let Q be the space of all functions defi ned in V. The Laplacian is a linear oper-ator. Defi ne D as the subspace of V consisting of all functions g(r) that satisfy the boundary condition of Equation b. The standard inner product on V is

( ( ), ( )) ( ) ( )f g f g dr r r r= ∫ V

V

(c)

Show that L = ∇2 is a self-adjoint operator on D with respect to the standard inner product for Q.

Solution If L is self-adjoint, then for any vectors f(r) and g(r) both in D,

∇( ) = ∇( )∫ ∫2 2f gd f g dV V

V V

(d)

Recall the vector identity,

∇ ∇( )= ∇ ∇ + ∇. .g f f g g f2 (e)

9533_C002.indd 1279533_C002.indd 127 10/29/08 3:57:20 PM10/29/08 3:57:20 PM


Using the identity of Equation e in the integral on the left-hand side of Equa-tion d leads to

∇( ) = −∇ ∇ + ∇ ∇[ ]∫ ∫2 f gd f g g f dV

V V

V

. .( ) (f)

The divergence theorem implies

∇ ∇ = ∇( )∫ ∫. .( )g f d g f ndV S

V S

(g)

where n is a unit r(a) vector normal to S. Equation g leads to

∇( ) = − ∇ ∇ − ∇∫ ∫ ∫2 f gd f gd g f ndV V V S

V S

. . (h)

Since both f and g are in D, ∇fg . n = − Bif on S and ∇ g · n = −Big on S. Thus Equation h can be written as

∇( ) = − ∇ ∇ −∫ ∫ ∫2 f gd f gd gfdV V Bi S

V V S

. (i)

In a similar fashion it can be shown that

f g d f gd gfd∇( ) = − ∇ ∇ −∫ ∫ ∫2 V V Bi S

V V S

. (j)

The equality of the right-hand sides of Equations i and Equation j proves that L is self-adjoint.

Example 2.32 A Fredholm integral equation is of the form

f x K x y dy g x

x

( ) ( , ) ( )=∫0

(a)

Equation a can be formulated as Lf = g, where Lf f x K x y dy g xx

= ∫ =0

( ) ( , ) ( ). The domain and range of L are C[0,a]. Assuming that the adjoint operator

is of the from L* *( ) ( , )f f x K x y dyx

= ∫0

, determine the form of K*(x,y) using the standard inner product

( , ) ( ) ( )f g f x g x dx

a

= ∫0

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Solution If L* is the adjoint of L with respect to the standard inner product, then

f x K x y dy g x dx f

xa

( ) ( , ) ( )

00

∫∫⎛

⎝

⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟= (( ) ( , ) ( )*x K x y g y dy dx

xa

00

∫∫⎛

⎝

⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟ (b)

Working with the left-hand side of Equation b by interchanging the order of integration leads to

f y K x y dy g x dx f

xa

( ) ( , ) ( )

00

∫∫⎛

⎝

⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟= (( ) ( , ) ( )x K x y g y dxdy

ya

00

∫∫ (c)

Renaming the variables of integration, replacing x by λ and y by τ in Equa-tion c, leads to

( , ) ( ) ( ) ( , )Lf g f g K d d

a

= ∫∫ λ τ λ τ λ τ

τ

00

(d)

Performing similar steps on the right-hand side of Equation a leads to

( , ) ( ) ( ) ( , )

( ) ( )

* *f L g f y g x K x y dydx

f g K

xa

=

=

∫∫00

λ τ **( , )τ λ λ τ

τλ

d d00

∫∫ (e)

It is clear that Equation d and Equation e are equivalent for all f and g if and only if K x y K y x*( , ) ( , )= for all x and y such that 0 ≤ x ≤ a and 0 ≤ y ≤ a.

2.12 Positive defi nite operatorsPositive defi niteness is a property of many linear operators that, if proven for the operator, has many ramifi cations regarding approximate and exact solu-tions of an equation involving the operator.

Defi nition 2.23 Let L be a linear operator whose domain D is a subspace of a vector space V, on which an inner product (u,v) is defi ned. L is positive defi nite with respect to the inner product if (Lu, u) ≥ 0 for all u in D, and (Lu, u) = 0 if and only if u = 0.

Some operators studied will not satisfy the “only if” clause which is required for defi nition 2.23 to apply. For such cases, defi nition 2.23 can be modifi ed as:

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Defi nition 2.24 Let L be a linear operator whose domain D is a subspace of a vector space V, on which an inner product (u,v) is defi ned. L is positive semi-defi nite with respect to the inner product if (Lu, u) ≥ 0 for all u in D, and (Lu, u) = 0 when u = 0, but there are vectors, u ≠ 0, such that (Lu, u) = 0.

Example 2.33 The stiffness matrix for a two-degree-of-freedom system is

K =−

−⎡

⎣⎢⎢

⎤

⎦⎥⎥

4 2

2 3 (a)

Determine whether or not K is positive defi nite with respect to the standard inner product for R2.

Solution Let u be an arbitrary vector in R2. Then

Ku =−

−⎡

⎣⎢⎢

⎤

⎦⎥⎥⎡

⎣⎢⎢

⎤

⎦⎥⎥=

−− +

4 2

2 3

4 2

2 3

1

2

1 2

1

u

u

u u

u uu2

⎡

⎣⎢⎢

⎤

⎦⎥⎥ (b)

and

( , ) ( )Ku u = −( ) + − +

= − +

4 2 2 3

4 4

1 2 1 1 2 2

12

1 2

u u u u u u

u u u 33

2 2 2

2 2

22

12

12

1 2 22

22

12

1

u

u u u u u u

u u u

= + − +( )+

= + −( 222

22) + u

(c)

Thus from Equation c, since (Ku, u) is the sum of three non-negative terms, (Ku, u) ≥ 0 for all u. Clearly the right-hand side of Equation c is zero if u = 0, and if either component of u is not zero, then (Ku, u) is greater than zero. Thus K is positive defi nite according to defi nition 2.23.

Example 2.34 Consider the operator defi ned in Example 2.30 which is used to solve for the defl ection of a beam. Show that the operator is positive defi -nite on the domain specifi ed in Example 2.30 with respect to the standard inner product for C4[0,1]. Note that α(x) > 0 for all x, 0 ≤ x ≤ 1.

Solution Algebraic manipulation of (Lf,g) for arbitrary f and g both in D led to Equation g of Example 2.30, which is repeated following:

( , ) ( ) ( )Lf gd

dxx

d fdx

g x dx=⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟ =∫

2

2

2

2

0

1

α αα( )xd fdx

d gdx

dx2

2

2

2

0

1

∫ (a)

Using Equation a with g = f leads to

( , ) ( )Lf f xd fdx

dx=⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟∫ α

2

2

2

0

1

(b)

9533_C002.indd 1309533_C002.indd 130 10/29/08 3:57:22 PM10/29/08 3:57:22 PM


The integrand of the integral on the right-hand side of Equation b is non-negative for all x within the range of integration. Thus (Lf, f) ≥ 0 for all f. If f(x) = 0, d f dx2 2 0/ = , and the defi nite integral in Equation b is zero. The integral is zero for any f such that d f dx2 2 0/ = for all x, 0 ≤ x ≤ 1. The only continuous function that has a second derivative identically equal to zero is a linear function of the form

f x c c x( ) = +1 2 (c)

However, f must be in D and thus satisfy the boundary conditions of the fi xed-free beam given in Equation b of Example 2.30. Applying f(0) = 0 leads to c1 = 0. Then application of df dx/ ( )0 0= requires c2 = 0. Thus the only linear function in D is f(x) = 0. Thus the only f(x) in D such that (Lf,f) = 0 is f(x) = 0.

From the above and defi nition 2.23, it is clear that L is positive defi nite.

Example 2.35 Reconsider the heat transfer problem of Example 2.31. Defi neLΘ Θ= −∇2 . Determine whether or not L is positive defi nite with respect to the standard inner product for C2 (V) on the domain D defi ned such that if f(r) is in D, then (a) f(r) = 0 everywhere on S, the surface of V(this condi-tion corresponds to the surface at a prescribed temperature), (b) ∇ =f .n 0 everywhere on S (this corresponds to a completely insulated surface), and (c) ∇ = −f Bi f.n ( ) everywhere on S (this corresponds to a condition of heat transfer by convection from the ambient to the surface).

Solution Using the same algebraic manipulations as in Example 2.31, an equation similar to that of Equation i of Example 2.31 can be obtained as

( , )Lf g f gd f gd g f nd= −∇( ) = ∇ ∇ − ∇∫ ∫ ∫2 V V S

V V S

. . (a)

Applying Equation a with g = f leads to

( , )Lf f f fd f f nd

f d f f nd

= ∇ ∇ − ∇

= ∇ − ∇

∫ ∫

∫

. .

.

V S

V S

V S

V S

2 ∫∫ (b)

The integrand of the volume integral of Equation a is non-negative every-where in V, thus the integral is clearly non-negative. It is also equal to zero when f = 0. However, it could also be zero when ∇f is zero everywhere in V. This may occur if f(r) = C, a nonzero constant.

(a) If f(r) = 0 everywhere on S, then the integrand of the surface integral of Equation a is zero everywhere on S and the integral is zero. Since f is pre-scribed to be zero on S, if f(r) = C, then C must be zero. Thus the only f in D such that (L,f,f) = 0 is f = 0. Hence L is positive defi nite on D.

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If ∇ =f .n 0 everywhere on S, then the surface integral of Equation a is identically zero. However, the surface condition is satisfi ed by f(r) = C for any value of C. Thus there exists f(r) ≠ 0 such that (L,f,f) = 0. Hence L is posi-tive semi-defi nite on D.

If ∇ = −f n Bi f. ( ) everywhere on S, then Equation a can be rewritten as

( , )Lf f f d Bi f d= ∇ +∫ ∫2 2V S

V S

(b)

Clearly the right-hand side of Equation b is non-negative. From the bound-ary condition, if ∇f = 0 everywhere in V, which implies that f(r) = C, then f = 0 everywhere on S, which implies C = 0, and thus the right-hand side of Equa-tion b is zero only when f = 0 and L is positive defi nite on D.

2.13 Energy inner productsTheorem 2.10 If L is a positive defi nite and self-adjoint linear operator with respect to a defi ned inner product on a domain D, then an inner product, called an energy inner product, can be defi ned by

( , ) ( , )u v Lu vL = (2.46)

Proof It is required to show that properties (i)–(iv) of defi nition 2.7 are valid for all u, v, and w in D when Equation 2.46 is used as an inner-product defi nition. To this end,

Property (i) (u, v)L = (Lu, v) Defi nition of proposed inner product

= (u, Lv) By hypothesis, L is self-adjoint

= (v, Lu) By hypothesis, (u,v) is a valid inner product,

thus property (i) of defi -nition 2.7 applies

= (v, u)L Defi nition of proposed inner product

Property (ii) (u + v, w)L = (L(u + v), w) Defi nition of proposed inner product

= (Lu + Lv, w) Linearity of L = (Lu, w) + (Lv, w) Property (ii) of Defi nition

2.7 = (u, w)L + (v, w)L Defi nition of proposed

inner product

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Property (iii) ( , ) ( ( ), )α αu v L u vL = Defi nition of proposed inner product

= ( , )αLu v Linearity of L = α( , )Lu v Property (iii) of defi nition

2.7 = α( , )u v Defi nition of proposed

inner product

Property (iv) ( , ) ( , )u u Lu uL = Defi nition of proposed inner product

≥

= =

⎧ 0

0

Dfor all u in

if and only if u 0⎨⎨⎪⎪⎩⎪⎪

L is positive defi nite.

Thus since Properties (i)–(iv) of defi nition 2.7 are satisfi ed by the proposed defi -nition of Equation 2.46, the energy inner product is a valid inner product on D.

Corollary If L is a positive-defi nite and self-adjoint linear operator with respect to an inner product (u,v) defi ned for its domain D, then an energy norm defi ned by

u u uL L= ( )[ ],1

2 (2.47)

is a valid norm on D

Proof Since L is positive defi nite and self adjoint with respect to (u,v), then by theorem 2.10, the energy inner product is a valid inner product on D. Then by theorem 2.5, any inner product can generate a norm of the form of Equation 2.47.

The term “energy” inner product is applied to inner products of the form of Equation 2.46, because positive defi nite and self-adjoint operators that arise from the mathematical modeling of physical systems can often be derived using energy methods. In such cases, (Lu,u) is related to some form of energy.

Example 2.36 The stiffness matrix of Example 2.33 is the stiffness matrix obtained in the modeling of the two-degree-of-freedom system shown Figure 2.8. Since the matrix is symmetric, it is self-adjoint with respect to the standard inner product on R2. It is shown in Example 2.18 that the matrix is positive defi nite. (a) Determine the form of the energy inner product and con-fi rm the commutativity property of the energy inner product. (b) If u1 and u2 represent the displacements of the two blocks, measured from equilibrium, show how the energy norm relates to the potential energy of the system.

Solution (a) Using the stiffness matrix of Example 2.21,

( , ) ,

( ) ( )

u v Ku vK =( )

= − + − +

=

4 2 2 3

4

1 2 1 1 2 2

1

u u v u u v

u v11 2 1 1 2 2 22 2 3− − +u v u v u v

(a)

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The commutativity of the energy inner product is confi rmed by

( , ) ( , )

( ) ( )

v u u KvK =

= − + − +

=

u v v u v v

u v

1 1 2 2 1 2

1

4 2 2 3

4 11 1 2 2 1 2 22 2 3− +− +

=

u v u v u v

( , )u v K

(b)

(b) The potential energy of the system when the blocks have displacements u1 and u2, respectively is

V u u u u

u u u

= + −( ) +

= + −

1

22

1

22

1

2

1

22 2

12

2 12

22

12

2 12( ) ++⎡⎣ ⎤⎦u 2

2

(c)

Comparing Equation c with Equation C of Example 2.33 leads to

V = ( )

=

=

1

2

1

2

1

2

2

Ku u

u u

u

,

( , )K

K

(d)

Example 2.37 The differential equation for the transverse displacement of a beam due to transverse loading is

EId wdx

F x4

4= ( ) (a)

The end conditions for a beam fi xed at x = 0 and with a spring of stiffness k attached at x = L are

w( )0 0= (b1)

dwdx

( )0 0= (b2)

d wdx

L2

20( ) = (b3)

EId wdx

L kw L3

30( ) ( )− = (b4)

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Nondimensionalization of Equation a and Equations b as in Example 2.28 leads to

d wdx

f x4

4= Λ ( ) (c)

w( )0 0= (d1)

dwdx

( )0 0= (d2)

d wdx

2

21 0( ) = (d3)

d wdx

w3

31 1 0( ) ( )− =η (d4)

where η = kL EI3/ .

Defi ne the operator L by Lw d w dx= 4 4/ . The domain of L is S, the subspace of C4[0, 1] defi ned such that if g(x) is in S, then g dg dx d g dx( ) , ( ) , ( )0 0 0 0 1 02 2= = =/ / , and d g dx g3 3 1 1 0/ ( ) ( )− =η . It is easily shown that L is self-adjoint with respect to the standard inner product for C4[0,1].

(a) Show that L is positive defi nite on S with respect to the standard inner product for C4[0,1] and that thus an energy inner product is defi ned on S ( f, g,)E = (Lf, g).

(b) Consider the cross-section shown in Figure 2.8. The normal stress along a line at distance z from the neutral axis (centroidal axis) is

σ =MzI

(e)

where the bending moment M in the cross-section is

M EId wdx

=2

2 (f)

σ(x)A σ(x+dx)Az

Figure 2.8 The normal stress in an Euler-Bernoulli beam varies across the span of the beam.

9533_C002.indd 1359533_C002.indd 135 10/29/08 3:57:26 PM10/29/08 3:57:26 PM


and I z dA= ∫ 2 is the area moment of inertia about the neutral axis.The strain energy per unit volume at a point in the beam, assuming elastic

behavior throughout, is

eE

=σ2

2 (g)

If w(x) is the transverse displacement of the beam, show how (w, w)L is related to the total potential energy of the system (total strain energy plus potential energy developed in the spring).

(c) Discuss the physical meaning of (w, f).

Solution (a) For any f(x) in S,

Lf fd fdx

fdx,( )= ∫4

4

0

1

(h)

Application of integration by parts to the right-hand side of Equation e with u f dv d f dx dx= =and /( )4 4 leads to

Lf f f xd fdx

dfdx

d fdx

dxx

x

, ( )( )= −=

=

∫3

30

1 3

3

0

1

(i)

Application of integration by parts to the right-hand side of Equation i with u = df/dx and dv = d3f/dx3dx gives

Lf f f xd fdx

dfdx

d fdx

d

x

x

x

x

, ( )( )= − +=

=

=

=3

30

1 2

21

1 2 ffdx

dx

fd fdx

2

2

0

1

3

1

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟

=

∫

( )33

3

3

2

21 0 0 1 1( ) ( ) ( ) ( ) ( )

− −fd fdx

dfdx

d fdx

( ) ( )+ +⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟

dfdx

d fdx

d fdx

0 02

2

2

2 ⎟⎟∫2

0

1

dx

(j)

Since f(x) is in S, f df dx d f dx( ) , ( ) ( )0 0 0 0 1 02 2= = =/ and / . Thus Equation j simplifi es to

( , ) ( ) ( )Lf f fd fdx

d fdx

dx= +⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟1 1

3

3

2

2

2

0

1

∫∫ (k)

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Also, since f(x) is in S, d f dx f3 3 1 1/ ( ) ( )= η which when substituted into Equa-tion k gives

Lf f fd fdx

dx, ( )( )= [ ] +⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟∫η 1

22

2

2

0

1

(l)

Equation l clearly shows that since η > 0, (Lf, f) ≥ 0 for all f in S and that (L0,0) = 0. Furthermore, (Lf, f) = 0 implies that d f dx2 2 0/ = for all x, 0 ≤ x ≤ 1. Thus f(x) = ax + b. However, if f is in S, then f(0) = 0, implying b = 0 and df/dx(1) = 0, implying a = 0, and L is positive defi nite on S with respect to the standard inner product for C4[0,1].(a) Since L is self-adjoint and positive defi nite on S with respect to the standard inner product for C4[0,1], an energy inner product of the form ( , ) ,f g Lf gL =( ) can be defi ned on S.(b) Let w(x) be the dimensional transverse displacement of the beam. The potential energy developed in the spring is

V k w Ls = [ ]1

2

2( ) (m)

Combining Equation e, Equation f, and Equation g, the strain energy per volume is

e Ezd wdx

=⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟

1

22

2

2

2

(n)

The total strain energy in the beam is

V edV

Ezd wdx

dA

b =

=⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟

⎡

⎣⎢⎢⎢

⎤

⎦⎥⎥⎥

∫1

22

2

2

2

ddx

E d wdx

z dA

A

L

A

∫∫

∫=⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥

0

2

2

2

2

2⎥⎥⎥∫

0

L

(o)

Recalling that I z dAA= ∫ 2 , where I is the area moment of inertia of the cross sectional area about its centroidal axis Equation o becomes

VEI d w

dxdxb

L

=⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟∫2

2

2

2

0

(p)

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The total potential energy is

V V V k w L EId wdx

dxs b

L

= + = [ ] +⎡

⎣⎢⎢

⎤

⎦⎥⎥∫1

2

1

2

22

2

2

0

( ) (q)

Introduction of nondimensional variables x x L w w L* */ , /= =( ) into Equa-tion q leads to

V kL w EId wdx

L= ⎡⎣ ⎤⎦ +⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟

1

21

1

22 2

2

2

2

**

*( ) ddx*

0

1

∫ (r)

Dividing Equation r by EIL and dropping the *s from the nondimensional variables leads to

V

EILw

d wdx

dx= [ ] +⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟

⎧⎨⎪⎪ ∫1

21 2

2

2

2

0

1

η ( )⎪⎪

⎩⎪⎪⎪

⎫⎬⎪⎪⎪

⎭⎪⎪⎪

(s)

Thus, from Equation l,

V

EILw w= ( )

1

2, L (t)

The energy inner product of the transverse displacement with itself is pro-portional to the total potential energy of the system. The quantity V/(EIL) is a nondimensional potential energy; the energy inner product of the transverse displacement with itself is twice the nondimensional potential energy. This inner product is often called the potential-energy inner product.(c) Consider a differential element of the beam of length dx. Let w be the dimensional displacement of the beam and F(x) the transverse load per length. The work done during application of the load is

dU wF x dx=1

2( ) (u)

The total work done by the external load over the length of the beam is

U dU w x F x dx

Lw x F f x Ldx

L

= =

=

∫ ∫ 1

2

1

2

0

( ) ( )

( ) ( )* *max

* **

( ) ( )

0

1

2

0

1

2

∫

∫=L F

w x f x dxmas

(v)

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By defi ning a nondimensional work variable as U U EIL* /( )= , Equation v can be rewritten as

U EILL F

w x f x dx

U w x f x dx

*

max

*

( ) ( )

( ) ( )

2

0

1

1

2

1

2

=

=

∫

Λ00

1

1

2

∫

= Λ( , )w f

(w)

Problems 2.1. The triangle inequality for all vectors in R3 states that if A = B + C,

then A B C≤ + . Prove that the triangle inequality holds for all vectors A, B, and C in R3.

2.2. Prove that the triangle inequality, as stated in problem 2.1, holds for all vectors in Rn for any fi nite value of n where the norm is the stan-dard inner-product-generated norm.

2.3. Let A = 2i + 3j–k and B = 3j + 2k. Let V be the set of vectors in the span of A and B.

a. Show that V is a subspace of R3. b. Determine an orthonormal basis for V with respect to the stan-

dard inner product for R3.

2.4. Let V be the set of vectors in R3 such that if u = [u1 u2 u3] is in V, then u1 + 2u2 − u3 = 0 Prove or disprove that V is a subspace of R3. If V is a subspace of R3, what is its dimension?

2.5. The boundary conditions satisfi ed by a nondimensional temperature distribution, θ(x), over an extended surface subject to a constant tem-perature at x = 0 and insulated at x = 1 are θ(0) = 0 and dθ/dx(1) = 0. Defi ne S as the set of all functions in C2[0,1] which satisfy these boundary conditions. Show that S is a subspace of C2[0,1].

2.6. Let V be the space of functions which are in both P4[0,1], the space of all polynomials of degree four or less, and the space S as defi ned in problem 2.5. Determine a basis for V.

2.7. Determine an orthonormal basis with respect to the standard inner product on C2[0,1] for the vector space V defi ned in problem 2.6.

2.8. Expand f (x) = x2 (1 − x)2 in terms of (a) the basis vectors obtained in the solution of problem 2.6 and (b) the orthonormal basis obtained in problem 2.7.

2.9. Let p (x) = anxn + an−1xn−1 + … + a1x + a0 and q (x) = bnxn + bn−1xn−1 + … + b1x + b0 be elements of Pn[0,1]. Show that ( , )p q a bi

ni i= =Σ 0 is a valid

inner product defi ned for Pn[0,1].

9533_C002.indd 1399533_C002.indd 139 10/29/08 3:57:29 PM10/29/08 3:57:29 PM


2.10. Determine an orthonormal basis with respect to the inner product as defi ned in problem 2.9 for the vector space V as defi ned in problem 2.6.

2.11. Expand f (x) = x2 (1 − x)2 in terms of the basis vectors obtained in prob-lem 2.10.

2.12. Determine an orthonormal basis for the vector space V as defi ned in problem 2.6 with respect to the inner product ( , )p q a b a bi

ni i= + =2 0 0 1Σ

defi ned on Pn[0,1]. 2.13. Let V be the space of all continuously differentiable functions of two

independent variables x and y over the region 0 ≤ x ≤ 1and 0 ≤ y ≤ 1. Show that ( , ) ( , ) ( , )f g f x y g x y dydx= ∫ ∫

0

1

0

1 is a valid inner product

on V. 2.14. Let S be the subspace of the vector space V of problem 2.13 spanned

by the functions 1 2 2 3 2 2 3, , , , , , , , ,x y x xy y x x y xy y Determine an ortho-normal basis for V with respect to the inner product as defi ned in problem 2.11.

2.15. The defl ection w(x, y) of a square membrane fi xed along its edge is subject to the boundary conditions w y w y( , ) , ( , ) ,0 0 1 0= =w x w x( , ) , ( , )0 0 1 0= = . Let Q be the subspace of the vector space S defi ned in problem 2.14 such that if f(x, y) is in Q, then f y f y f x f x( , ) , ( , ) , ( , ) , ( , )0 0 1 0 0 0 1 0= = = = . Determine a basis for Q.

2.16. Determine an orthonormal basis with respect to the inner product defi ned in problem 2.13 for the vector space Q defi ned in Problem 2.15.

2.17. The boundary conditions for the defl ection, w(x), of a fi xed-free Euler-Bernoulli beam with an axial load are w dw dx d w dx( ) , ( ) ,0 0 0 0 2 2= =/ /

d w dx( ) , ( )1 0 12 3 3= +/ εddwdx( )1 0= . Let Q be the subspace of P6[0,1] such that if f(x) is in Q, f df dx d f dx d f dx( ) , ( ) , ( ) , ( )0 0 0 0 1 0 12 2 3 3= = =/ / /

2+ ddf dx/ ( )1 0= . Determine a basis for Q. 2.18. Determine an orthonormal basis for the vector space Q as defi ned in

Problem 2.15 with respect to the standard inner product on C4[0,1] 2.19. For what value(s) of c3 are the vectors a, b, and c linearly independent?

a b= −−

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

= −

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥2

2

1

4

1

3

⎥⎥⎥⎥

=−−

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

c

2

1

3c

2.20. For what value(s) of c3 are the vectors b and c of problem 2.19 orthog-onal with respect to the standard inner product for R3?

2.21. For what value(s) of c3 are the vectors b and c of problem 2.19 orthogonal with respect to the inner product defi ned by (a,b) = a1b1 + 3a2b2 + 2a3b3?

2.22. What is the adjoint of the matrix

1 3 2

2 2 1

4 2 1

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥ with respect to (a) the

standard inner product for R3 and (b) the inner (x,y) = 2x1y1 + 2x2y2 + 3x3y3?

9533_C002.indd 1409533_C002.indd 140 10/29/08 3:57:30 PM10/29/08 3:57:30 PM


2.23. Determine suffi cient conditions for a 2 × 2 matrix to be self adjoint with respect to the standard inner product on R2.

2.24. Determine suffi cient conditions for a 2 × 2 matrix to be self adjoint with respect to the inner (x,y) = 2x1y1 + x2y2.

2.25. For what values of ε is the operator Lf = –d2f/dx2 positive defi nite with respect to the standard inner product for C2[0,1] if the domain of L is specifi ed such that f(0) = 0 and df/dx(1) + ε f(1) = 0?

2.26. Show that the biharmonic operator ∇ 4φ = ∇ 2(∇ 2 φ) is self adjoint and positive defi nite with respect to the inner product (f(r), g(r)) = ∫R f(x,y) g(x,y)dV when φ = 0 and ∂φ/∂n = 0 everywhere on the boundary of R.

For problems 2.27–2.31, consider the boundary value problem defi ned by

d ydx

dydx

y f x2

22 10+ − = ( ) (a)

y( )0 == 0 (b)

dydx

y( ) ( )1 2 1 0+ = (c)

2.27. Show that Equation a can be rewritten as

−

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+ =

110

2

2

ed

dxe

dydx

y f xx

x ( ) (d)

2.28. Defi ne the operator L whose domain is the subspace of C2[0,1] in

which all elements satisfy the boundary conditions of Equation b

and Equation c by Lye

ddx

edydx

yx

x= −⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+

110

2

2 . Show that L is self-

adjoint with respect to the inner product defi ned by

f g f x g x e dxx, ( ) ( )( )= ∫ 2

0

1

(e)

2.29. Show that L, as defi ned in Problem 2.28 is positive defi nite with respect to the inner product defi ned by Equation e.

2.30. Since L is self-adjoint and positive defi nite with respect to the inner product of Equation e, L may be used to defi ne an energy inner product ( f,g)L. Calculate

x x x x2 33

4

3

5− −

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟,

L (f)

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2.31. Defi ne the vector space Q as the intersection of P 4[0,1] with the domain of L. Determine a basis of L, which is orthonormal with respect to the energy inner product of Problem 2.30.

2.32. Consider the operator L defi ned as

Lf ( )x f y x y dy= ( ) ( )∫ ( )sin sinπ π0

1

(g)

(a) Determine the conditions under which L is self-adjoint with respect to the standard inner product on C[0,1].

(b) Determine the condition under which L is positive defi nite with respect to the standard inner product on C[0,1].

2.33. Demonstrate the defi nition of an energy inner product ( f,g)L derived

from the operator of Problem 2.32 by calculating (x,x2)L.

9533_C002.indd 1429533_C002.indd 142 10/29/08 3:57:32 PM10/29/08 3:57:32 PM

143

Chapter 3

Ordinary differential equations

3.1 Linear differential equationsA differential equation is an operator equation of the form

Lu = f (3.1)

where u = u(x) is an element of SD, a subspace of Cn[a,b], and f = f(x) is an element of SR, a subspace of PC[a,b], the space of piecewise continuous functions on [a,b]. L is a differential operator with domain SD and range SR. The order of the differential equation is the order of the highest-order derivative in L. The differ-ential equation is linear if L is a linear operator; otherwise, it is nonlinear. The differential equation has constant coeffi cients if for arbitrary values of x1 and x2 such that a ≤ x1, x2 ≤ b, the operator defi nition of L is the same for x1 and x2.

The general form of an nth-order linear differential equation is

a xd udx

a xd udx

a xdudx

an

n

n n

n

n( ) ( ) ( )+ + + +−

−

−1

1

1 1� 00( ) ( )x u f x= (3.2)

The differential equation has constant coeffi cients if an, an−1,…, an, a0 are all independent of x. The differential equation is called homogeneous if f(x) = 0; otherwise, it is nonhomogeneous.

A homogeneous solution of Equation 3.2 is a solution of Lu = 0. It should be noted that if u1(x) and u2(x) are homogeneous solutions, then because of the linearity of L,

L L L( ( ) ( )) ( ) ( )α β βu x u x u x u x1 2 1 2+ = +α (3.3)

As a result of the hypothesis that u1(x) and u2(x) are homogeneous solutions of Equation 3.2, the right-hand side of Equation 3.3 is zero. This shows that the homogeneous solution is not unique and that any linear combination of homogeneous solutions is also a homogeneous solution. The homogeneous solution space for a linear differential operator L is the space of functions which satisfy Lu = 0. This space is independent of SD and SR. The dimension of the homogeneous solution space for an nth-order differential operator is n. This is proved in Section 3.2 for n = 2. The proof can be generalized for any n.

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Thus, for an nth-order linear differential equation, the operator’s homoge-neous solution space has a basis of n functions, say u1(x),u2(x),…,un(x), such that the general homogeneous solution is a linear combination of the basis elements and thus is of the form

u x C u x C u x C u xh n n( ) ( ) ( ) ( )= + +1 1 2 2 … (3.4)

where C1, C2, …, Cn are arbitrary constants, often called constants of integration.Let up(x) be a particular solution of Equation 3.4, a solution specifi c to the

form of f(x). A particular solution satisfi es Lup = f. Defi ne

u x u x u xh p( ) ( ) ( )= + (3.5)

Then Lu = L(uh + up) = Luh + Lup = 0 + f = f. Thus a function of the form of Equa-tion 3.5 satisfi es Equation 3.2. This solution is called the general solution. The particular solution is linearly independent of the homogenous solution, because otherwise it could be expressed as a linear combination of the basis vectors of the homogeneous solution space.

The general solution of an nth-order linear ordinary differential equation is of the form

u x C u x C u x C u x u xn n p( ) ( ) ( ) ( ) ( )= + + + +1 1 2 2 � (3.6)

where u1(x),u2(x),…,un(x) is a basis for the homogeneous solution space, C1, C2, …,Cn are arbitrary constants, and up(x) is the particular solution. The constants of integration are determined through application of constraints specifi ed by the physical problem. The constraints are conditions which must be satisfi ed at specifi c values of x. A unique solution can be attained only if n independent constraints are specifi ed. If n constraints are specifi ed, they are applied to the solution of Equation 3.6, leading to a set of n simultaneous linear algebraic equation to solve for the constants of integration. Unless the determinant of the resulting coeffi cient matrix is zero, a unique solution exists for the constants, and a unique solution exists for the differential problem.

An initial value problem is one in which the constraints are developed at a single value of the independent variable x, perhaps x = 0. The range over which the solution is defi ned is 0 ≤ x < ∞. The initial conditions are of the form u u du dx u d u dxn n( ) , / ( ) ,..., / ( )( )0 0 00 0

1 1= = ′ =− − uu n0

1( ).− Application of initial conditions to Equation 3.6 leads to

u u u u

u u un1 2 3

1 2 3

0 0 0 0

0 0 0

( ) ( ) ( ) ... ( )

( ) ( ) ( )′ ′ ′ ′… uu

u u u u

u

n

n

( )

( ) ( ) ( ) ... ( )

0

0 0 0 01 2 3′′ ′′ ′′ ′′

� � � � �

111

21

31 10 0 0 0( ) ( ) ( ) ( )( ) ( ) ( ) ... ( )n n n

nnu u u− − − −

⎡⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥C

C

C

Cn

1

2

3

�

⎥⎥⎥⎥⎥⎥⎥⎥

=

−′ − ′′′− ′′

u u

u u

u u

u

p

p

p

0

0

0

0

0

0

0

( )

( )

( )

(

�nn

pnu− −−

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥1 1 0) ( )( )

(3.7)

9533_C003.indd 1449533_C003.indd 144 10/29/08 3:58:51 PM10/29/08 3:58:51 PM

Chapter 3: Ordinary differential equations 145

The determinant of the coeffi cient matrix in Equation 3.7 is the Wronskian evaluated at x = 0 of the basis functions for the homogeneous solution space. It can be shown that W(0) ≠ 0 and a unique solution of Equation 3.7 exists.

Boundary-value problems are those for which the constraints are speci-fi ed at two values of x, say x = a and x = b. Boundary conditions may specify that a linear combination of the solution and some of its derivatives have spe-cifi c values at a or b. The boundary condition is homogeneous if that specifi c value is zero. The differential operators for most boundary-value problems are of even order. Half the n boundary conditions are applied at x = a, and half are applied at x = b. If all boundary conditions are homogeneous, then SD is defi ned as the subspace of Cn[a,b] consisting of all functions satisfying all boundary conditions. The boundary conditions are applied to Equation 3.6, leading to a set of simultaneous equations to solve for the constants of integration. If the determinant of the coeffi cient matrix is zero, then f(x) must satisfy a solvability condition (Section 5.11) for a solution to exist.

If any boundary condition is nonhomogeneous, then SD is not a vector space in its own right, because the nonhomogeneous boundary condition prevents closure under addition, closure under scalar multiplication, and the existence of the zero vector in SD. In this case, the nonhomogeneity can be transferred to the differential equation through a redefi nition of dependent variables.

The nondimensional differential equation governing the temperature distribution over an extended surface is d2θ/dx2 − Biθ = 0. The end at x = 0 is subject to a fi xed temperature, while the end at x = 1 is subject to a heat fl ux for which the appropriate boundary conditions are θ(0) = 1 and dθ/dx (1) = q. The boundary conditions are nonhomogeneous. A new dependent vari-able is defi ned as φ (x) = θ(x) − 1 − qx. The differential equation governing

φ φ( )x d dx Bi Bi qxis 2 2 1φ / − = +( ) . The boundary conditions become φ(0) = 0 and dφ/dx (1) = 0. Thus the solution for φ(x) is in SD, a subspace of C2[0,1] defi ned such that if f(x) is in SD, then f(0) = 0 and df/dx (1) = 0.

Methods for determining particular solutions are presented in Section 3.3. Superposition of particular solutions can be used for linear problems. Con-sider a linear differential equation of the form

Lu = +f x f x1 2( ) ( ) (3.8)

Let up,1(x) be the particular solution for Lu = f1(x) and let up,2(x) be the particu-lar solution for Lu = f2(x). The linearity of L implies that

L L Lu u u u

f x f x

p p p p, , , ,

( ) ( )

1 2 1 2

1 2

+( )= +

= + (3.9)

Equation 3.9 implies that up,1 + up,2 is the particular solution of Equation 3.8. The general statement of the principle of linear superposition is that if the nonhomogeneous term in a linear ordinary differential equation is the sum of a fi nite number of terms, then the particular solution is the sum of the particular solutions corresponding to each term in the sum.

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3.2 General theory for second-order differential equations

In this section, a general theory for second-order differential equations is developed and then generalized to higher-order equations. The general form of a linear second-order ordinary differential equation is

d ydx

a xdydx

a x y f x2

2 1 0+ + =( ) ( ) ( ) (3.10)

Equation 3.10 is supplemented by two initial conditions, y(0) = y0 and

dy dx y/ ( )0 0= ′ , or by two boundary conditions expressed as linear combina-tions of y and dy/dx. If the interval over which the differential equation is to be solved is a ≤ x ≤ b, one boundary condition is developed at x = a and one at x = b. The following discussion is directed toward initial value problems. However the results can be extended to boundary-value problems, which are the main focus of this book.

The following theorem is fundamental, but is presented without proof because the proof is lengthy and noninstructive toward the goals of this work.

Theorem 3.1 Existence and uniqueness of solutions. If a1 (x), a0 (x) and f (x) are all continuous for all x, then, there exists a unique solution for the dif-ferential equation, Equation 3.1, subject to the initial conditions y (0) = y0 anddy dx y/ ( )0 0= ′.

The homogeneous solution space is the subspace V of C2[0, ∞] consisting of functions which satisfy

d ydx

a xdydx

a x y2

2 1 0 0+ + =( ) ( ) (3.11)

as well as two initial conditions of the form specifi ed in theorem 3.1. The space of functions is not dependent on the values of the initial conditions, but only on the existence of such conditions. It is desired to determine the dimension of V and fi nd a basis for V.

Theorem 3.2 Abel’s formula. Let y1 (x) and y2 (x) be two solutions of Equa-tion 3.11. The Wronskian of these solutions defi ned by

W xy x y x

dydx

dydx

y xdydx

y xd

( )

( ) ( )

( ) ( )= = −1 2

1 2 12

2

yydx

1

(a)

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satisfi es

W x Cea x dx

( )( )

= ∫− 1 (b)

Proof Since y1 (x) and y2 (x) solve Equation 3.11,

d ydx

a xdydx

a x y2

1

2 11

0 1 0+ + =( ) ( ) (c)

d ydx

a xdydx

a x y2

2

2 12

0 2 0+ + =( ) ( ) (d)

Multiplying Equation c by y2 (x) and Equation d by y1 (x) and then subtracting the resulting equations leads to

yd ydx

yd ydx

a x ydydx

ydydx

2

21

2 1

22

2 1 21

12− + −( )=( ) 00 (e)

Noting that dW dx y d y dx y d y dx/ / /= ( )− ( )22

12

12

22

, Equation e becomes

dWdx

a x W+ =1 0( ) (f)

Equation b is a solution of Equation f.The Wronskian of two linearly independent functions is nonzero for some

value of x, while the Wronskian of two linearly dependent functions is zero for all values of x. Let y1 (x) and y2 (x) be two linearly independent functions which solve Equation 3.11. A linear combination of two solutions is also a solution of the differential equation.

Suppose that the dimension of V is 1, and let y1 (x) be a solution of Equation 3.11. The general form of a solution is y (x) = cy1 (x). However, all elements in V must satisfy two initial conditions, which is impossible if the dimension of V is one. Suppose that the dimension of V is two. Let y1 (x) and y2 (x) be two linearly independent solutions of Equation 3.11. Then a linear combination is of the form y (x) = C1y1 (x) + C2y2 (x). Satisfaction of the initial conditions requires that C1y1 (0) + C2y2 (0) = y0 and C dy dx C dy dx y1 1 2 2 00 0/ ( ) / ( )( ) + ( ) = ′. A unique solu-tion of these equations exists as long as W (0) ≠ 0. Therefore, the dimension of V is at least 2. If there are only two linearly independent solutions of Equation 3.11, then the dimension of V is two.

Theorem 3.3 Let V be the vector space of solutions of Equation 3.11 which satisfy two arbitrary initial conditions. The dimension of V is two.

Proof In view of the previous discussion, it is necessary only to show that Equation 3.11 has only two linearly independent solutions, say y1 (x) and y2 (x).

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Suppose that y3 (x) is another solution of Equation 3.11. Then from Abel’s formula, the Wronskian for the solution combination y1 (x) and y3 (x) can be obtained using Abel’s formula as

y xdydx

y xdydx

C ea x dx

13

31

1 31

( ) ( ) ,

( )− = ∫−

(a)

and the Wronskian for the solution combination y2 (x) and y3 (x) is

y xdydx

y xdydx

C ea x dx

23

32

2 31

( ) ( ) ,

( )− = ∫−

(b)

Equation a and Equation b can be rearranged to yield

y xdydx

y xdydx

dydx

y x

11

22

3

3

( )

( ) (

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥ ))

,

,

( )⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥=

⎡

⎣⎢⎢

⎤

⎦⎥⎥

∫−C

Ce

a x dx1 3

2 3

1 (c)

Equation c can be solved, perhaps using Cramer’s rule, for y3 (x), leading to

y xC y x C y x e

y xd

a x dx

32 3 1 1 3 2

1

1

( )( ) ( )

( )

, ,

( )

=−[ ] ∫−

yydx

y xdydx

22

1− ( ) (d)

Note that the denominator of Equation d is the Wronskian of the solutions y1 (x)

and y2 (x), which can be replaced using Abel’s formula by C ea x dx

1 21

,

( ).

−∫ Thus Equation d reduces to

y xCC

y xCC

y x32 3

1 21

1 3

1 22( ) ( ) ( ),

.

,

.

= − (e)

Equation e implies that y3 (x) is not linearly independent of y1 (x) and y2 (x). Thus Equation 3.11 has two linearly independent homogeneous solutions, from which it follows that the dimension of V is 2.

Since the dimension of V is 2, any set of two linearly independent solu-tions of Equation 3.11 is a basis for V.

Abel’s formula provides a method for fi nding a second linearly indepen-dent solution if one is known. Suppose that y1(x) is known, and it is desired to determine a second linearly independent solution. Abel’s formula implies that the second solution, y2 (x), satisfi es

y xdydx

y xdydx

ea x dx

12

21 1

( ) ( )( )

− = ∫− (3.12)

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where the constant C is chosen to be 1. If a different value of C is chosen, say 43.1, the solution for y2 (x) is 43.1 times the solution for C = 1. Dividing Equation 3.12 by y1 (x) leads to

dydx

dydxy

yy

ea x dx2

1

12

1

1 1− = ∫− ( ) (3.13)

Equation 3.13 is a fi rst-order linear differential equation whose solution is y2 (x). Equation 3.13 can be solved using the integrating factor technique where the integrating factor is 1/y1 (x). Multiplying by the integrating factor, Equation 3.13 can be written as

d

dxyy y

ea x dx2

1 12

1 1⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟= ∫− ( )

(3.14)

Integration of Equation 3.14 leads to

y x y xe

ydx Cy x

a x

2 112 1

1

( ) ( ) ( )

( )

=∫

+−

∫ (3.15)

Since y1 (x) has already been identifi ed as a homogeneous solution, the con-stant of integration can be taken to be zero, leading to

y x y xe

ydx

a x

2 112

1

( ) ( )

( )

=∫−

∫ (3.16)

The general solution of Equation 3.10 is a linear combination of the basis functions of V plus a particular solution.

3.3 Differential equations with constant coeffi cientsThe general form of an nth-order linear differential operator with constant coeffi cients is

Ly = + + +−

−

−ad ydx

ad ydx

adydx

a yn

n

n n

n

n1

1

1 1 0… (3.17)

where an, an−1, …, a1 and a0 are called the coeffi cients of the differential operator.A homogeneous solution for a linear differential equation with constant

coeffi cients is assumed to be

y x Cehx( ) = α (3.18)

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where C is an arbitrary constant and α is a constant to be determined. Sub-stitution of Equation 3.18 into Equation 3.17 leads to

a a a a Cenn

nn xα + + +( ) =−−

11

1 0α α α… 0 (3.19)

Since eαx ≠ 0 for all x for any value of ∝, and since C = 0 implies yh (x) = 0 for all x, Equation 3.19 leads to a nontrivial solution if and only if

a a a ann

nnα α α+ + + =−−

11

1 0… 0 (3.20)

Equation 3.20 is an nth-order polynomial equation whose roots are the values of α such that Equation 3.18 is a nontrivial solution of Equation 3.1 when L is defi ned by Equation 3.5. The polynomial in Equation 3.20 has n roots α1, α2, … ,αn, each leading to a solution of the form of Equation 3.18. If all values of α are distinct, then n linearly independent solutions have been found, and thus a basis for the homogeneous solution space is e e ex x xnα α α1 2, , ,… and the general homogeneous solution is

y x C eh ix

i

ni( ) =

=∑ α

1

(3.21)

The coeffi cients of the polynomial of Equation 3.20 are all real, and there-fore complex roots occur in conjugate pairs. Let σ ± iτ be complex conju-gate solutions of Equation 3.20. Their contribution to the homogeneous solution of Equation 3.21 through application of Euler’s identity (eiθ = cos θ + i sin θ) is

D e D e e D e D ei x i x x i x i x1 2 1 2

σ τ σ τ σ τ τ+( ) −( ) −+ = +( )

= eσσ τ τ τ τx D x i x D x i x

e Dx

1 2cos sin cos sin+( )+ −( )[ ]

= σ11 2 1 2

1 2

+( ) + −( )[ ]

= +

D x i D D x

e C x Cx

cos sin

cos s

τ τ

τσ iin τx( )

(3.22)

Let ∝ be a repeated root of Equation 3.20. The second-order differential equa-tion satisfi ed by this repeated root is d2y/dx2 + 2α dy/dx + y = 0. One solution is y1 (x) = eαx. The second solution, which is linearly independent with y1 (x) = eαx, can be obtained using Equation 3.16 as

y x e

e

edx e dx xex

dx

x

x x2

2

2( ) =

∫

( )= =

−

∫ ∫αα

α

α α (3.23)

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The results of Equation 3.23 can be generalized as follows. If α is a root of Equation 3.20 of multiplicity m, the linearly independent solutions correspond-ing to this root are e xe x ex x m xα α α, , ,… −1 . The result can be summarized as

Theorem 3.4 A basis for the homogeneous solution space for an nth-order lin-ear differential equation with constant coeffi cients whose operator is of the form of Equation 3.17 is the set of functions e e ex x xnα α α1 2, , ,… , where αi, i = 1, 2, … , n are the roots of Equation 3.20. If Equation 3.20 has a root of multiplicity m, then m linearly independent basis vectors are e xe x e x ex x m x m xα α α α, , , ,… − −2 1 .

Example 3.1 The free response of the single-degree-of-freedom system shown in Figure 3.1 is governed by the differential equation

md xdt

cdxdt

kx2

20+ + = (a)

The block has a mass of 1 kg, and the spring has a stiffness of 100 N/m. Determine the free response of the block when it is displaced 1 mm from equilibrium and released if (a) c = 4 N · s/m, (b) c = 30 N · s/m, and (c) c = 20 N · s/m.

Solution The initial conditions for the problem are x (0) = 0.001 m and dx/dt (0) = 0 m/s. Substitution of a solution of the form of Equation 3.18 into Equation a leads to the polynomial equation,

α α2 100 0+ + =c (b)

(a) For c = 4 N · s/m, the roots of Equation b are α1 2 2 99 2 9 950, .= − ± − = − ± i.The roots are complex conjugates, and therefore the general solution is obtained using Equation 3.22 as:

m

k c

m = 1kgk = 100 N/m

x

Figure 3.1 The free response of the mass-spring-viscous-damper system of Example 3.1 is governed by a second-order differential equation.

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x t e C t C tt( ) cos . sin( . )= ( )+[ ]−21 29 950 9 950 (c)

Setting x(0) = 0.001 leads to C1 = 0.001, and setting dx/dt (0) = 0 leads to C2 = 2.01 × 10−4. The time-dependent response is

x t e t tt( ) cos . . sin( . )= ( )+[ ]−2 9 950 0 201 9 950 mmm (d)

(b) For c = 30 N · s/m, the roots of Equation b are α1 2 15 125, = − ± or α1 = −26.18 and α2 = −3.819. The general solution of Equation a is

x t C e C et t( ) . .= +− −1

3 8192

26 18 (e)

Setting x(0) = 0.001 leads to C1 + C2 = 0.001, while setting dx/dt (0) = 0 leads to −3.819C1 − 26.18C2 = 0. Simultaneous solution of these equations gives C1 = 1.117 × 10−3 and C2 = −1.708 × 10−4, which when substituted into Equation e give

x t e C et t( ) . .. .= −− −1 117 0 1713 8192

26 18 mm (f)

(c) For c = 20 N · s/m, Equation b has a double root of α = −10. The appropri-ate form of the general solution is

x t C e C tet t( ) = +− −1

102

10 (g)

Setting x(0) = 0.001 in Equation g gives C1 = 0.001, while setting dx/dt (0) = 0 leads to C2 = 0.01. The system response for c = 20 N · s/m is

x t e tet t( ) = +− −10 1010 mm (h)

The system responses represented by Equation d, Equation f, and Equation h are plotted in Figure 3.2. For c = 4 N · s/m, the response is oscillatory with exponentially decaying amplitude. Such a response is said to be underdamped in that the damping force is not large enough to dissipate all the system’s energy over one cycle of oscillation. For c = 20 N · s/m and c = 30 N · s/m, responses exhibit exponential decay, but the response is slower for the larger value of c. The damping force exerted when c = 20 N · s/m is just large enough to dissipate all the system’s energy over one cycle, in which case the response is said to be critically damped. When c = 30 N · s/m, the damping force is larger than required to dissipate all the system’s energy over one cycle, in which case the response is said to be overdamped.

Example 3.2 The differential equation governing the temperature distribu-tion in the straight rectangular fi n shown in Figure 3.3 is

ddx

m2

2

2 0θ

θ− = (a)

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where θ = T − T∞ and m hP hA= / with P = 2(w + t) and A = wt. The fi n is maintained at a constant temperature at its base and is insulated at its tip, leading to the boundary conditions

θ θ( )0 0= (b)

ddx

Lθ

( ) = 0 (c)

Underdamped

Crtically damped

Overdamped

Response of system of example 3.11

0.8

0.6

0.4

0.2

–0.2

–0.4

–0.60 0.1 0.2 0.3 0.4 0.5 0.6

t (sec)

x (m

m)

0.7 0.8 0.9 1

0

Figure 3.2 The underdamped response of the system of Example 3.1 is oscillatory with exponentially decaying amplitude. The critically damped and overdamped responses are not oscillatory and decay exponentially.

x

L

t

T ,h8

wk

θ0

Figure 3.3 The temperature distribution in the extended surface of Example 3.2 is obtained from the solution of a second-order boundary value problem.

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Determine the rate of heat transfer from the base of the fi n.

Solution Substitution of θ(x) = eαx into Equation a leads to the polynomial equa-tion α2 − m2 = 0 whose solutions are α = ± m, leading to a general solution of

θ( )x C e C emx mx= + −1 2 (d)

It should be noted that the hyperbolic trigonometric functions cosh(mx) = 1/2(emx + e−mx) and sinh(mx) = 1/2(emx − e−mx) are linear combinations of the expo-nential functions used in Equation d. Thus sinh(mx) and cosh(mx) also form a basis for the homogeneous solution space, and an alternate formulation of Equation d is

θ( ) cosh( ) sinh( )x D mx D mx= +1 2 (e)

Equation e is more convenient for application of Equation b, the boundary condition at x = 0. Since cosh(0) = 1 and sinh(0) = 0, its application leads to D1 = θ0. Subsequent application of Equation c gives D2 = −θ0 tanh(mL), which when substituted into Equation e, leads to

θ θ( ) cosh( ) tanh( )sinh( )x mx mL mx= −[ ]0 (f)

The rate of heat transfer from the base of the fi n is q = −kA (dθ/dx) (0), which, using Equation f, becomes

q kAm mL

hPkAhPkA

L

tanh( )

tanh

=

=⎛

⎝⎜⎜⎜

⎞

⎠

θ

θ

0

0⎟⎟⎟⎟⎟

(g)

Several methods exist to determine the particular solution for a nonho-mogeneous differential equation with constant coeffi cients. The method of undetermined coeffi cients may be applied when the nonhomogeneous terms have a fi nite number of linearly independent nonzero derivatives. In this case, a particular solution is assumed as a linear combination of all nonzero derivatives with the coeffi cients unspecifi ed. The coeffi cients are determined by substituting the assumed particular solution into the differential equa-tion and equating coeffi cients of like terms. Consider a linear ordinary dif-ferential equation with constant coeffi cients of the form Ly = f, where f has a fi nite number of linearly independent derivatives ′ ′′f x f x f xk( ), ( ),..., ( )( ) . A trial solution for the particular solution of differential equation is assumed to be

y x A f x A f x A f xp kk( ) ( ) ( ) ( )( )= + ′ + +0 1 � (3.24)

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where A0, A1, …, Ak are coeffi cients which are as yet undetermined. Equation 3.24 is then substituted into the differential equation. The resulting equation is rearranged such that each side expresses a linear combination of linearly independent functions. Coeffi cients of functions on the left-hand side are equated with coeffi cients of like functions on the right-hand side. This leads to a set of k + 1 simultaneous equations to solve for the coeffi cients.

The method of undetermined coeffi cients is applied to equations in which the nonhomogeneous terms have a fi nite number of linearly independent derivatives. These include polynomials, exponential functions, trigonomet-ric functions, and products of these functions. Table 3.1 lists functions for which the method of undetermined coeffi cients leads to successful solutions, as well as the form of the trial solutions.

A special case occurs when a nonhomogeneous term is a solution of the corresponding homogeneous equation. In this case, a trial solution of the form of Equation 3.24 leads to Lyp = 0, indicating that the trial solution does not lead to a particular solution. Noting that d xf x dx xd f dx k d f dxk k k k k k[ ( )]/ / /= + − −1 1 , substitution into the operation defi ned by Equation 3.17 leads to

L xf x a xd f

dxn

d f

dxan

n

n

n

n n( )( ) = +⎡

⎣⎢⎢

⎤

⎦⎥⎥ +

−

− −

1

1 1 xxd f

dxn

d f

dx

n

n

n

n

−

−

−

−+ −

⎡

⎣⎢⎢

⎤

⎦⎥⎥

+ +

1

1

2

21( )

� aa xdfdx

f a f

x ad f

dxa

d f

dn

n

n n

n

1 0

1

1

+⎡

⎣⎢⎢

⎤

⎦⎥⎥+

= + −

−

xxa

dfdx

a

nad f

d

n

n

n

−

−

+ + +⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

+

2 1 0

1

�

xx

n ad f

dxa f

x f x f

n n

n

n− −

−

−+ − + +

= +

1 1

2

1 11( )

( )

�

L K

(3.25)

Table 3.1 Trial Solutions for Use in Method of Undetermined Coeffi cients

Nonhomogeneous term Trial solution

xn where n is an integer Cnxn + Cn−1xn−1 + ⋅⋅⋅ + C1x +C0

e ax Ce ax

sin(ax) or cos(ax) C1 sin(ax) + C2 cos(ax)

xn e ax (Cnxn + Cn−1xn−1 + ⋅⋅⋅ + C1x +C0)e ax

xn sin(ax) or xn cos(ax) (Cnxn + Cn−1xn−1 + ⋅⋅⋅ + C1x +C0) sin(ax) + (Dnxn + Dn−1xn−1 + ⋅⋅⋅ + D1x +D0) cos(ax)

ebx sin(ax) or ebx cos(ax) [C1 sin(ax) + C2 cos(ax)]ebx

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Since f(x) is a homogeneous solution, Lf = 0, and Equation 3.25 reduces to

L Kxf x f( ) ( )( )= (3.26)

where Kf is a linear combination of f and its derivatives. This suggests that if f(x) is a homogeneous solution, then an appropriate trial solution is of the form

y x x A f x A f x A f xp kk( ) ( ) ( ) ... ( )( )= + ′ + +⎡⎣ ⎤⎦0 1 (3.27)

If xf(x) is also a homogeneous solution, then the trial solution can be again multiplied by x to obtain the solution using the method of undetermined coeffi cients.

Example 3.3 An extended surface is subject to resistive heating which is modeled as an internal heat source within the surface. The differential equa-tion governing the temperature distribution over the surface is

ddx

m u x2

22θθ− = −Λ ( ) (a)

where Λ = umax L2/k and m is as defi ned in Example 3.2. The surface at x = 0 is maintained at a constant temperature, while the surface at x = L is insulated. Determine the heat transfer across the surface of the fi n if u = x − x2/L.

Solution The general homogeneous solution of Equation a as determined in Example 3.2 is

θh x D mx D mx( ) cosh( ) sinh( )= +1 2 (b)

The assumed form of the particular solution is

θp x A Bx Cx( ) = + + 2 (c)

Substitution of Equation c into Equation a leads to

2 2 2

2

C m A Bx Cx xxL

− + + = −⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟( ) Λ (d)

Collecting coeffi cients of like powers of x leads to

− = −m CL

2 Λ (e)

− =m B2 Λ (f)

2 02C m A− = (g)

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Solving Equation e, Equation f, and Equation g for A, B, and C and substitut-ing into Equation c gives

θp xm L m

xm L

x( ) = − +2

4 2 22Λ Λ Λ

(h)

The general solution is

θ θ θ= +

= + + −

h px x

D mx D mxm L

( ) ( )

cosh( ) sinh( )1 2 4

2Λ Λmm

xm L

x2 2

2+Λ (i)

Application of θ(0) = θ0 to Equation i leads to

θ0 1 4

2= +D

m LΛ

(j)

Application of dθ/dx (L) = 0 to Equation i, using Equation k, gives

02 2

0 4 2 2 2= −( ) + − +m

m LmL D m mL

m m

D

θΛ Λ Λ

sinh( ) cosh( )

22 0 4 3

2= − −( ) −θ

Λ Λm L

mLm mL

tanh( )cosh( )

(k)

The solution of Equation a subject to the boundary conditions of Equation b and Equation c is

θ θ( ) cosh( ) tanh( )sinh( )xm L

mx mL mx= −( ) −[ ]

−

0 4

2Λ

ΛΛ Λm mL

mxm L

m Lx m x3 4

2 2 22cosh( )

sinh( ) ( )+ − +

(l)

The total rate of heat transfer across the surface of the surface is

q h x dx

hm m L

mx mL

L

=

= −( ) −

∫ θ

θ Λ

( )

sinh( ) tanh( )

0

0 4

2ccosh( )

cosh( )cosh( )

mL

hm mL

mx

xx L

xx

[ ]

−

==

==

0

4 0

Λ LL

x

x Lhm L

xm Lx m x

+ − +⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟

=

=Λ4

2 2 2 3

0

22 3

== + − + −⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟

⎡

⎣⎢h

m m mLh mL

Lm

θ Λ0 3

2

31

2sec ( )

⎢⎢⎤

⎦⎥⎥

(m)

9533_C003.indd 1579533_C003.indd 157 10/29/08 3:59:01 PM10/29/08 3:59:01 PM


Example 3.4 The nondimensional differential equation for the transverse defl ection of an axially loaded beam with a transverse load per unit length is

d wdx

d wdx

f x4

4

2

2− =ε Λ ( ) (a)

where ε =PL EI2/ and Λ = F L EI03/ . The boundary conditions for a fi xed-pinned

beam are

w( )0 0= (b)

dwdx

( )0 0= (c)

w( )1 0= (d)

d wdx

2

21 0( ) = (e)

Let Λ = 1 and

f x x( ) sin= ( )π (f)

(a) Determine and plot the transverse defl ection for ε = −4,0 and 4; (b) Deter-mine and plot the transverse defl ection when ε = −π2.

Solution The homogeneous solution is obtained by assuming w (x) = eαx, which when substituted into Equation a, leads to α4 − εα2, whose solutions are α = 0,0, ± √ε. If ε < 0, the load is compressive, and the general form of the homogeneous solution is

w x C x C x C C xh( ) cos sin= −( )+ −( )+ +1 2 3 4ε ε (g)

If ε = 0, the beam is not subject to an axial load, and the general form of the homogeneous solution is

w x C C x C x C xh( ) = + + +1 2 32

43 (h)

If ε > 0, the load is tensile, and the general form of the homogeneous solution is

w x C x C x C C xh( ) cosh sinh= ( )+ ( )+ +1 2 3 4ε ε (i)

Using the method of undetermined coeffi cients, the particular solution of Equation a when the beam is subject to the load per unit length given by Equation f can be assumed to be

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w x A x B xp( ) sin cos= ( )+ ( )π π (j)

Substitution of Equation l into Equation a leads to

π π π π επ π επ π4 4 2 2A x B x A x B xsin( ) cos sin cos+ ( )+ ( )+ (( )= ( )Λsin πx (k)

Equating coeffi cients of sin (πx) on both sides of Equation k gives A = Λ/π2(π2 + ε), while equating coeffi cients of cos(πx) on both sides of Equa-tion k leads to B = 0. It should be noted that ε = −π2 is a special case. The particular solution of Equation a for ε < 0, ε ≠ − π2 is

w x xp( ) sin=+( )

( )Λ

π π επ

2 2 (l)

The general solution for ε < 0 is

w x C x C x C C x( ) cos sin= −( )+ −( )+ + ++( )1 2 3 4 2 2

ε επ π ε

Λssin πx( ) (m)

Application of Equation b, Equation c, Equation d, and Equation e to Equation m leads to

w C C( )0 0 01 3= ⇒ + = (n)

dwdx

C C( )0 0 2 4 2= ⇒ − + = −

−( )ε

π π εΛ

(o)

w C C C C( ) cos sin1 0 01 2 3 4= ⇒ −( ) + −( ) + + =ε ε (p)

d wdx

C C2

2 1 21 0 0( ) cos sin= ⇒ −( ) + −( ) =ε ε ε ε (q)

Equation n, Equation o, Equation p, and Equation q can be solved simultane-ously, leading to

C1 2= −

−( )+( ) − −( )− −( )⎡

⎣⎤⎦

Λsin

cos sin

ε

π π ε ε ε ε (r)

C2 2=

−( )+( ) − −( )− −( )⎡

⎣⎤⎦

Λcos

cos sin

ε

π π ε ε ε ε (s)

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C3 2=

−( )+( ) − −( )− −( )⎡

⎣⎤⎦

Λsin

cos sin

ε

π π ε ε ε ε (t)

C4 2= −

−( )+( ) − −( )− −( )⎡

⎣⎤⎦

Λsin

cos sin

ε

π π ε ε ε ε (u)

The resulting general solution for ε < 0 is

w x( )sin

( ) cos sin

co

=−( )

+ − −( )− −( )⎡⎣

⎤⎦

Λ ε

π π ε ε ε ε2

sscos

sinsin−( )−

−( )−( )

−( )− +⎡

⎣⎢⎢⎢

⎤

⎦⎥⎥ε

ε

εεx x x1

⎥⎥+

+Λ

π π επ

2 2( )sin( )x

(v)

The general solution for ε = 0 is

w x C C x C x C x x( ) sin= + + + + ( )1 2 32

43

4

Λπ

π (w)

Application of the boundary conditions, Equation b, Equation c, Equation d, and Equation e, to Equation w results in

w x x x x x( ) sin= ( )− + −⎡⎣⎢⎢

⎤⎦⎥⎥

Λπ

π ππ π

42 33

2 2 (x)

The general solution for ε > 0 is

w x C x C x C C xh( ) cosh sinh= ( )+ ( )+ + ++(1 2 3 4 2 2

ε επ π ε

Λ))

( )sin πx (y)

Application of Equation b, Equation c, Equation d, and Equation e to Equation y leads to

w x( )sinh

( ) cosh sinh

cos

=( )

+ ( )− ( )⎡⎣

⎤⎦

Λ ε

π π ε ε ε ε2

hhcosh

sinhsinhε

ε

εεx x x( )−

( )( ) ( )−

⎡

⎣⎢⎢⎢

⎤

⎦⎥⎥⎥+

Λππ π ε

π2 2 +( )

( )sin x

(z)

Equation u, Equation x, and Equation z are plotted as functions of x for ε = −4, ε = 0, and ε = 4, respectively in Figure 3.4.

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If ε = −π2, the nonhomogeneous term is also a homogeneous solution. The appropriate form of the particular solution for this special case is

w x Ax x Bx xp( ) sin cos= ( )+ ( )π π (aa)

Substitution of Equation aa into Equation a for ε = −π2 leads to

− ( )+ + +4 43 4 3 4A x Ax x B x Bxπ π π π π π πcos sin( ) sin( ) coss( )

cos sin( ) sin( )

π

π π π

x

A x Ax x B x+ ( )− − −2 22 2π π π π BBx x

x

A x B

π π

π

π π π

2

3 32 2

cos( )

sin( )

cos s

⎡⎣ ⎤⎦

=

− ( )+

Λ

iin( ) sin( )π πx x= Λ

(bb)

Equating coeffi cients of sin(πx) on both sides of Equation bb leads to B = Λ/2π3, while equating coeffi cients of cos(πx) on both sides of Equation bb leads to A = 0, leading to the general solution,

w x C x C x C C x x x( ) cos sin( ) cos= ( )+ + + + (1 2 3 4 32π π

ππ

Λ)) (cc)

8

9 × 10–3

7

6

5

4

W

3

2

1

00 0.1 0.2 0.3 0.4 0.5

x

ε = 0

ε = –4

ε = 4

0.6 0.7 0.8 0.9 1

Figure 3.4 The form of the homogeneous solution for the defl ection of an axially loaded beam of Example 3.4 is dependent on the value of ε. The load is compressive if ε < 1. tensile for ε > 1, and does not exist for ε = 0.

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Application of the boundary conditions, Equation b, Equation c, Equation d, and Equation e, to Equation cc leads to

w C C( )0 0 01 3= ⇒ + = (dd)

dwdx

C C( )0 02

2 4 3= ⇒ + = −π

Λπ

(ee)

w C C C( )1 02

1 3 4 3= ⇒ − + + =

Λπ

(ff)

d wdx

C2

22

11 02

( ) = ⇒ − = −πΛπ

(gg)

Simultaneous solution of Equation dd, Equation ee, Equation ff, and Equation gg leads to

w x x x x x x( ) cos( ) sin cos( )= − ( )− + +⎡⎣⎢

Λ2

41 3

3ππ

ππ π

⎢⎢⎤⎦⎥⎥ (hh)

Figure 3.5 illustrates the displacement for ε = −π2 as calculated from Equa-tion hh.

0 0.1 0.2 0.3 0.4 0.5

Displacement for ε = –π20

–0.002

–0.004

–0.006

–0.008

–0.01

–0.012

–0.014

–0.0160.6

x

0.7 0.8 0.9 1

W

Figure 3.5 When ε = −π2, a special case occurs for the beam of Example 3.4; the nonhomogeneous term is a homogeneous solution.

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The method of variation of parameters is another technique used to derive a particular solution of a nonhomogeneous differential equation. The method can be applied to linear ordinary differential equations with constant or vari-able coeffi cients. As an illustration, consider a linear second-order differen-tial equation of the form

a xd ydx

a xdydx

a x y f x2

2

2 1 0( ) ( ) ( ) ( )+ + = (3.28)

Let y1(x) and y2(x) be linearly independent solutions of the corresponding homogeneous equation. A solution of Equation 3.28 is assumed in the form of

y x C x y x C x y x( ) ( ) ( ) ( ) ( )= +1 1 2 2 (3.29)

where C1(x) and C2(x) are functions to be determined by substitution into Equation 3.28. Differentiation of Equation 3.29 with respect to x leads to

dydx

C x y x C x y x C x y x C= ′ + ′ + ′ +1 1 1 1 2 2 2( ) ( ) ( ) ( ) ( ) ( ) (( ) ( )x y x′2 (3.30)

If the coeffi cients are constants, then dy dx C y x C y x/ ( ) ( )= ′ + ′1 1 2 2 . The same form of dy/dx is achieved if

′ + ′ =C x y x C x y x1 1 2 2 0( ) ( ) ( ) ( ) (3.31)

Differentiation of Equation 3.30 when Equation 3.31 is enforced leads to

d ydx

C x y x C x y x C x y2

2 1 1 1 1 2 2= ′ ′ + ′′ + ′ ′( ) ( ) ( ) ( ) ( ) (xx C x y x) ( ) ( )+ ′′2 2 (3.32)

Substitution of Equation 3.29, Equation 3.30, Equation 3.31, and Equation 3.32 into Equation 3.28 leads to

a x C x y x C x y x C x y x2 1 1 1 1 2 2( ) ( ) ( ) ( ) ( ) ( ) ( )′ ′ + ′′ + ′ ′ ++ ′′[ ]

+

C x y x

a

2 2( ) ( )

11 1 1 2 2( ) ( ) ( ) ( ) ( )

x C x y x C x y x′ + ′[ ]

( ) ( ) ( ) ( ) ( )+ +[ ]=a x C x y x C x y x0 1 1 2 2 ff x( ) (3.33)


C x a x y x a x y x a X y x1 2 1 1 1 0 1( ) ( ) ( ) ( ) ( ) ( ) ( )′′ + ′ +[ ]

+CC x a x y x a x y x a x y x2 2 2 1 2 0 21( ) ( ) ( ) ( ) ( ) ( ) ( )′′ + ′ +[ ]]

+ ′ ′ + ′ ′[ ] =a x C x y x C x y x f x2 1 1 2 2( ) ( ) ( ) ( ) ( ) ( )

(3.34)

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Recalling that y1(x) and y2(x) are homogeneous solutions, Equation 3.34 sim-plifi es to

′ ′ + ′ ′ =y C y Cf x

a x1 1 2 2

2

( )

( ) (3.35)

Equation 3.35 can be summarized in matrix form as

y x y x

y x y xC x

C x1 2

1 2

1

2

( ) ( )

( ) ( )

( )

( )′ ′⎡

⎣⎢⎢

⎤

⎦⎥⎥

′′

⎡

⎣⎣⎢⎢

⎤

⎦⎥⎥=

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

0

2

f xa x

( )

( )

(3.36)

Simultaneous solution of Equation 3.36 leads to

′ = −C xW x

f x y xa x

12

2

1( )

( )

( ) ( )

( ) (3.37)

′ =C xW x

f x y xa x

21

2

1( )

( )

( ) ( )

( ) (3.38)

where W(x) is the Wronskian of the homogeneous solutions,

W x y x y x y x y x( ) ( ) ( ) ( ) ( )= ′ − ′1 2 2 1 (3.39)

Equation 3.37 and Equation 3.38 are then integrated, leading to C1(x) and C2(x), which are used in Equation 3.29.

The method of variation of parameters is easily generalized to determine the particular solution of higher-order differential equations. Let y1, y2, …, yn be n linearly independent homogeneous solutions for an nth-order linear dif-ferential equation. A particular solution can be formed as

y x C x y x C x y x C x y xn n( ) ( ) ( ) ( ) ( ) ( ) ( )= + + +1 1 2 2 � (3.40)

A system of n simultaneous equations is then developed to solve for ′ =C x k nk( ) 1, 2, .., by requiring that d y dx C y x C y xj j j j/ ( ) ( )( ) ( )= + + +1 1 2 2 �

C y xn nj ( )( ) for j = 1, 2, …, n−1. Successive enforcement of these conditions

leads to ′ + ′ + + ′− −C x y x C x y x C x yj jn1 1

12 2

1( ) ( ) ( ) ( ) ( )( ) ( ) � nnj x( )( )−1 for j = 1, 2, … ,n−1.

The nth equation results from eventual substitution into the differential equation which gives ′ + ′ + + ′− − −C y x C y x C y xn n

n nn

1 11

2 21 1( ) ( ) ( )( ) ( ) ( )� == f x a xn( )/ ( ).

The resulting simultaneous equations can be written in matrix form as

y y y

y y y

y y y

n

n

n nnn

1 2

1 2

11

21 1

…

…

� � � �

…

′ ′ ′

− − −( ) ( ) ( )

⎡⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

′′′′

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

C

C

C

C

1

2

3

4

⎥⎥⎥

=

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

0

0

�

f x a xn( )/ ( )

(3.41)

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Equation 3.41 can be solved simultaneously, perhaps using Cramer’s rule. The determinant of the n × n matrix on the right-hand side of Equation 3.41 is the Wronskian of the homogeneous solutions.

Example 3.5 Use the method of variation of parameters to derive a solution for the transverse defl ection of the beam of Example 3.4 when ε = −1 for an arbitrary distributed load per length f(x).

Solution When ε = −1, the homogeneous solutions of Equation a of Example 3.4 are w1(x) = cos(x), w2(x) = sin(x), w3(x) = 1, and w4(x) = x. The general solu-tion is assumed to be of the form

w x C x x C x x C x C x x( ) ( )cos( ) ( )sin( ) ( ) ( )= + + +1 2 3 4 (a)

The derivatives of the coeffi cients are obtained by solving a system of the form WC f′ = , where W is the matrix from which the Wronskian of the homoge-neous solutions is formed, ′C and the vector of derivatives of the coeffi cients,is f =[ ]0 0 0 ( ) .f x T

This system can be expressed as

cos( ) sin( )

sin( ) cos( )

cos( ) sin( )

x x x

x x

x x

1

0 1−

− − 00 0

0 0

1

sin( ) cos( )x x

C

C

−

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥

′

′22

3

4

0

0

0′

′

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥

=

⎡

⎣

⎢⎢⎢⎢⎢C

C f x( )

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥

(b)

The Wronskian is computed as

W x

x x x

x x

x( )

cos( ) sin( )

sin( ) cos( )

cos( ) s=

−− −

1

0 1

iin( )

sin( ) cos( )

x

x x

0 0

0 0

1

−

= (c)

Cramer’s rule is used to solve Equation (b) as

′ =−−

C

x x

x

x

f x x

1

0 1

0 0 1

0 0 0

sin( )

cos( )

sin( )

( ) cos( ) 00 0

= f x x( )sin( ) (d)

′ =−−

C

x x

x

x

x f x

2

0 1

0 0 1

0 0 0

cos( )

sin( )

cos( )

sin( ) ( ) 00 0

= − f x x( )cos( ) (e)

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′ =−− −

C

x x x

x x

x3

0

0 1

cos( ) sin( )

sin( ) cos( )

cos( ) sinn( )

sin( ) cos( ) ( )

( )x

x x f x

f x x0 0

0−

= − (f)

′ =−− −

C

x x

x x

x4

1 0

0 0

cos( ) sin( )

sin( ) cos( )

cos( ) sinn( )

sin( ) cos( ) ( )

( )x

x x f x

f x0 0

0−

= (g)

The variation-of-parameters solution is

w x x f d x f

x

( ) cos( ) ( )sin sin( ) ( )cos= ( ) − ( )∫ ξ ξ ξ ξξ0

dd

f d x f d c

x

x x

ξ

ξ

0

0 0

∫

∫ ∫− + + ( ) ( )ξ ξ ξ ξ 11 2 3 4cos( ) sin( )x c x c c x+ + + (h)

Equation h can be rearranged as

w x x x f d c x c

x

( ) sin( ) cos( ) sin= − + −[ ] ( ) + +∫ ξ ξ ξ ξ0

1 2 (( )x c c x+ +3 4 (i)

Application of the boundary conditions requires calculations of the fi rst and second derivatives of w(x). Leibnitz’s rule for differentiation of a defi nite integral whose limits depend on the variable of differentiation is

d

dxg x d

gx

da x

b x

a x

( , )

( )

( )

( )

ξ ξ ξ∫⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥=

∂∂

bb x

g x a xdadx

g x b xdbdx

( )

( , ( )) ( , ( ))∫ + − (j)

Application of Leibnitz’s rule to Equation i leads to

dwdx

x f d

x

= − −( )[ ] ( )∫ 1

0

cos ξ ξ ξ (k)

d wdx

x f d

x2

2

0

= − − ( )∫ sin( )ξ ξ ξ (l)

Application of the boundary conditions to Equation j, using Equation k and Equation l, leads to

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w c c0 0 01 3( )= ⇒ + = (m)

dwdx

c c0 0 02 4( )= ⇒ + = (n)

w f d c

c

( ) sin( ) cos( )

s

1 0 1 1 1

0

1

1

2

= ⇒ − + −[ ] ( ) +

+

∫ ξ ξ ξ ξ

iin( )1 03 4+ + =c c (o)

d wdx

f d c c2

2

0

1

1 20 1 1= ⇒ − − − −∫ sin( ) ( ) cos( ) sin(ξ ξ ξ 11 0) = (p)

Equation m, Equation n, Equation o, and Equation p are solved simultane-ously for the constants of integration. Subsequent substitution into Equation i leads to the solution of the system for any f(x). The integrals can be evalu-ated for any f(x).

If a solution is desired for only one form of f(x), say f(x) = sin (πx) as in Example 3.5, then substitution into Equation i leads to

w x x x x dx x x

x

( ) cos( ) sin( )sin( ) sin( ) sin( )= −∫ π π0

ccos( ) sin( )

sin

x dx x x dx

x x d

x x

0 0

∫ ∫−

+ ( )

π

π xx C x C x C C x

x

0

1 2 3 4∫ + + + +cos( ) sin( )

(q)

Evaluation of the integrals in Equation q results in

w x x x x dx C x

x

( )( )

sin sin cos( )=+

( ) ( ) +∫1

12 2

0

1π π

π π ++ + +C x C C x2 3 4sin( )

(r)

Equation r is identical to Equation m of Example 3.5 with ε = −1. Boundary conditions are applied to Equation r to obtain Equation v of Example 3.5.

3.4 Differential equations with variable coeffi cientsLet b(x) be a function of an independent variable x Let x0 be a specifi c value of x. The function b(x) is said to be analytic at x0 if there exists a power series expansion of b(x), written as

b x c x xnn

n

( ) = −( )=

∞

∑ 0

0

(3.42)

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such that the expansion converges pointwise to b(x) within an interval x0 − R < x < x0 + R. The value R is called the radius of convergence. The con-stants cn n = 0, 1, 2, … are called the coeffi cients of the expansion.

The function (1 − x)−1 is analytic at x = 0 with a radius of convergence of 1 because it can be expressed using a binomial expansion as

( )1 11 2 1 1− = + + + + + + +− − +x x x x x xk k k� � (3.43)

The series expressed by the right-hand side of Equation 3.43 is a geometric series of ratio x. The series converges if |x| < 1 and diverges if |x| ≥ 1.

Taylor’s theorem provides conditions under which a function has a con-vergent expansion about point x0, a method for calculating the coeffi cients, and an estimate of the remainder after n terms. A form of the theorem pre-sented below is found in most elementary calculus books.

Theorem 3.5 Taylor’s Theorem. Let f(x) be a function defi ned such that f(x) and its fi rst n + 1 derivatives are continuous within a fi nite interval of a real value x0. For an x in this interval,

f x f x x xdfdx

xx x d f

dxx( ) ( ) ( ) ( )

!(= + − +

−( )0 0 0

02 2

2200

03 3

3 03

)!

( )

...

+−( )

+

+ +

x x d fdx

x

x−−( )+

xn

d fdx

x R x xn n

n n0

0 0!

( ) ( , )

(3.44)

where the remainder is

R x xx x

nd fdx

n

n n

n( , )( )

( )!0

01 1

11=

−+

( )+ +

+ ξ (3.45)

for some ξ between x and x0.Taylor’s theorem implies that a function is analytic at x0 if the function and

all its higher-order derivatives exist at x0.The general form of a linear nth-order differential equation with variable

coeffi cients can be written as

d y

dxa x

d y

dxa x

dydx

a x yn

n n

n

n+ + + +−

−

−1

1

1 1 0( ) ( ) ( )� == f x( ) (3.46)

The point x = x0 is said to be an ordinary point of the differential equation if a x a x a x a x f xn n− −1 2 1 0( ), ( ), , ( ), ( ) ( )… and are all analytic at x0. The following theorem, presented without proof, shows that if x0 is an ordinary point of the differential equation, then y(x) is analytic at x0.

Theorem 3.6 Let a x a x a x a x f xn n− −1 2 1 0( ), ( ), , ( ), ( ) ( )… and be coeffi cients in a differential equation of the form of Equation 3.46. If all coeffi cients are

9533_C003.indd 1689533_C003.indd 168 10/29/08 3:59:08 PM10/29/08 3:59:08 PM


analytic at x0, then x = x0 is an ordinary point of the differential equation and all solutions are analytic at x0. Furthermore, if Rm is the minimum of the radii of convergence of the coeffi cients of Equation 3.46, then the radius of convergence of y(x) is at least Rm.

Theorem 3.6 implies that if x0 is an ordinary point of the differential equation, then y(x) has a Taylor series expansion about x0. This suggests that a Taylor series expansion may be assumed for a solution and the method of undetermined coeffi cients applied to determine the coeffi cients in the expansion. Such a method is called a power series method.

The power series method is usually applied to determine homogeneous solutions. If the differential equation is nonhomogenous, once the homoge-neous solutions are determined, the method of variation of parameters may be used to determine the particular solution.

Example 3.6 The differential equation governing the temperature distribution over a nonuniform extended surface is

d

dxx

ddx

mα θ( )θ⎡

⎣⎢⎢

⎤⎦⎥⎥− =2 0 (a)

Use a series solution to determine the homogeneous solutions to Equation a if α(x) = 1 where μ is a constant.

Solution It is noted that x0 = 0 is an ordinary point of Equation a as long as α(0) ≠ 0. (a) A series solution is assumed to be of the form

θ( )x c xnn

n

==

∞

∑0

(b)

Theorem 3.6 implies that Equation b converges pointwise over a fi nite inter-val about x = 0. Thus it can be differentiated term by term, leading to

ddx

nc xnn

n

θ= −

=

∞

∑ 1

1

(c)

ddx

n n c xnn

n

2

2

2

2

1θ

= − −

=

∞

∑ ( ) (d)

Substitution of Equation b, Equation c, and Equation d into Equation a with α(x) = 1 leads to

n n c x m c xnn

n

nn

n

( )− − =−

=

∞

=

∞

∑ ∑1 02

2

2

0

(e)

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The name of an index of summation is arbitrary and does not affect the value of the sum. Thus

n n c x k k c xnn

n

kk

k

( ) ( )( )− = + +−

=

∞

+

=

∞

∑ ∑1 2 12

2

2

0

= + + +

=

( )( )n n c xnn

n

2 1 2

00

∞

∑ (f)

Substituting Equation f into Equation e leads to

( )( )

( )(

n n c x m c x

n n

nn

n

nn

n

+ + − =

+ +

+

=

∞

=

∞

∑ ∑2 1 0

2

2

0

2

0

11 022

0

)c m c xn nn

n

+

=

∞

−⎡⎣⎢

⎤⎦⎥ =∑

(g)

Powers of x are linearly independent, and therefore for Equation g to hold for all x, coeffi cients of all powers of x in the summation in Equation g must be zero:

( )( )n n c m c nn n+ + − = =+2 1 022 0, 1, 2, ... (h)

Equation h is a recurrence relation which is satisfi ed by the coeffi cients. It is convenient to consider even and odd values of n separately. Writing Equation h for the fi rst four even values of n gives

2 02

22

0 2

2

0c m c cm

c− = ⇒ = (i)

( )( )( )( ) ( )( )( )

3 4 03 4 2 3 4

42

2 4

2

2

4

0c m c cm

cm

c− = ⇒ = = ==m

c4

04!

(j)

( )( )( )( ) ( )( )( )(

5 6 05 6 2 3 4 5

62

4 6

2

4

4

c m c cm

cm

− = ⇒ = =))( ) !6 6

6

0=m

c (k)

Mathematical induction with Equation i, Equation j, and Equation k can be used to prove that

cmn

c nn

n

= =!

, , , ...0 2 4 6 (l)

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In a similar fashion, Equation h can be used to show that

cm

nc nn

n

= =−1

1 3 5 7!

, , , ... (m)

The coeffi cients c0 and c1 are arbitrary. They cannot be specifi ed from the differential equation. The solution can be written in terms of these two arbi-trary constants as

y x cmxn

cm

mxn

n

n

n

n

( )! !

, , , , , ,

=( )

+( )

=

∞

=∑0

0 2 4

1

1 3 5

∞∞

∑ (n)

Equation n can then be rewritten as

y x cmx

ncm

mxn

n

n

n

n

( )! ( )!

=( )( )

+( )

+=

∞ +

∑0

2

0

12 1

2 2 1==

∞

∑1

(o)

The functions y x mx nn

n1 0

2 2( ) / != ( ) ( )=∞Σ and

y x mx nn

n2 1

2 1 2 1( ) ( )!= ( ) +=∞ +Σ / are

linearly independent and form a basis for the homogeneous solution space for the differential equation, Equation a.

It should be noted that y1(x) is the Taylor series expansion for cosh(mx) about x = 0 and that y2(x) is the Taylor series expansion for sinh(mx) about x = 0.

Example 3.7 The natural frequencies and mode shapes for the longitudinal motion of a bar of constant area, but of varying density, are obtained from the solution of the non-dimensional differential equation

d udx

x u2

22 0+ =β ω( )

(a)

Determine power series representations of the basis functions for the homo-geneous solution space of Equation a if β μ( )x x= +1 .

Solution: A power series solution of Equation a is assumed as

u x c xkk

k

( ) ==

∞

∑0

(b)

Substitution of Equation b into Equation a leads to

c k k x c x xkk

k

kk k

k

( )( )− + +( )=−

=

∞+

=

∞

∑ ∑1 02

2

2 1

0

ω μ (c)

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Redefi nition of indices in the summations of Equation c leads to

2 1 222

0 22 2

1

1

c c k k c c ck k k

k

+ + + + + +⎡⎣ ⎤⎦+ −

=

ω ω μω( )( )

∞∞

∑ = 0 (d)

Setting coeffi cients of powers of x to zero independently leads to

2 022

0c c+ =ω (e)

( )( )k k c c ck k k+ + + + =+ −1 2 22 2

1ω μω k 1,2,... (f)

Using Equation e and Equation f to solve for c2, c3, c4, c5 and c6 in terms of c0

and c1 leads to

c c2

2

02

= −ω (g)

c c c3

2

1

2

02 3 2 3

= − −ω μω

( )( ) ( )( ) (h)

c c c

c

4

2

2

2

1

4

0

3 4 3 4

2 3 4

= − −

= −

ω μω

ω μω

( )( ) ( )( )

( )( )( )

22

13 4( )( )

c (i)

c c c5

2

3

2

2

4

4 5 4 5

2 3 4 5

= − −

=

ω μω

ω

( )( ) ( )( )

( )( )( )( )) ( )( )( )( ) ( )( )( )c c c1

4

0

4

0

4

2 3 3 4 2 4 5+ +

=

μω μω

ω(( )( )( )( )

( )( )

( )( )( )( )(2 3 4 5

5 3 3

2 3 3 4 51

4

c ++[ ]μω

))c0

(j)

c c c6

2

4

2

3

6

5 6 5 6

2 3 4

= − −

= −

ω μω

ω

( )( ) ( )( )

( )( )( )(55 6 3 4 5 6 2 3 5 60

4

1

4

)( ) ( )( )( )( ) ( )( )( )( )c c c+ +

μω μω11

2 4

02 3 5 6

+μ ω

( )( )( )( )c (k)

It is clear from Equation g, Equation h, Equation i, Equation j and Equation k that

c c ck k k= +α β0 1 (l)

9533_C003.indd 1729533_C003.indd 172 10/29/08 3:59:11 PM10/29/08 3:59:11 PM


where the values of αk and βk are obtained recursively from Equation f. The general solution of Equation a is

u x c x c xkk

k

kk

k

( ) = +=

∞

=

∞

∑ ∑0

0

1

1

α β (m)

3.5 Singular points of second-order equationsA singular point x0 of a second-order differential equation is called a regular singular point if the differential equation can be written as

( ) ( ) ( ) ( )x xd ydx

x x q xdydx

r x y− + − + =02

2

2 0 0 (3.47)

where q(x) and r(x) are analytic at x0.An example of a second-order differential equation with a regular singular

point is the Cauchy-Euler differential equation whose general form is

a xd y

dxa x

d y

dxa x

dydx

a ynn

n

n nn

n

n+ + + +−

−−

−11

1

1 1 0� == 0 (3.48)

Consider a change of independent variables specifi ed by

u x= ln( ) (3.49)

The chain rule is applied to change derivatives with respect to x into deriva-tives with respect to u. The results are

dydx

dydu

dudx

edydu

d ydx

ddu

edydu

e

u

u u

= =

= ( )

−

− −2

2== −

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟

=

−

−

ed ydu

dydu

d ydx

e L y

u

n

nnu

n

22

2

�

(3.50)

where Ln is a linear differential operator of derivatives with respect to u with constant coeffi cients. Noting that xn = enu, substitution of Equation 3.49 into Equation 3.48 leads to

a L y a L y a L y a yn n n n+ + + + =− −1 1 1 1 0 0� (3.51)

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Equation 3.51 is a linear differential equation with constant coeffi cients, and therefore its homogeneous solutions are of the form yh(u) = eαu. Using Equation 3.50, the homogeneous solutions written in terms of x are of the form

y x xh( ) = α (3.52)

If α is a root of multiplicity m of the resulting polynomial, the resulting linearly independent solutions are

x x x x x x x kα α α α, ln( ), [ln( )] ,..., ln( )2 1[ ] − (3.53)

Example 3.8 Solve Example 3.6 if α(x) = (1 + μx)2 and the boundary condi-tions are θ(0) = 1 and dθ/dx (1) = 0.

Solution Equation a of Example 3.6 becomes

d

dxx

ddx

m1 02 2+( )⎡⎣⎢⎢

⎤⎦⎥⎥− =μ

θθ (a)

Defi ne a new independent variable by

z x= +1 μ (b)

Equation a can be rewritten, with z as the independent variable, as

d

dxz

ddz

m2

2

0θ

μθ( )−

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟ = (c)

Expansion of the derivative in Equation c leads to

zddz

zddz

m22

2

2

2 0θ θ

θ+ −⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟ =

μ (d)

Equation d is of the form of a Cauchy-Euler differential equation. A solution is assumed of the form θ = zα which when substituted in Equation d, leads to

α αμ

2

2

0+ −⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟ =

m (e)

Application of the quadratic formula to Equation e leads to

αμ

1 2

21

21 1 4, = − ± +

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

m (f)

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The general form solution to Equation a is

θ α α( )z C z C z= +1 21 2 (g)

which can be written in terms of x as

θ μ μα α( )x C x C x= +( ) + +( )1 21 11 2 (h)

The constants of integration are determined through application of the boundary conditions, leading to

θ μ

μ

μ( )x

m

mx=

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟

− +⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟

+2

1 1 4

1

2

2(( ) − + +

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

1

21 1 4

2mμ

−− +

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟ +

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟1 1 4 2

1

2 2m mμ μ

−− +⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟

+( ) − − +⎛

⎝⎜⎜⎜⎜

⎞

⎠

1 4

12

1

21 1 4

mx

m

μ

μ μ⎟⎟⎟⎟⎟

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

2

(i)

The general form of a second-order Cauchy-Euler equation is

xd ydx

axdydx

by22

20+ + = (3.54)

The general solution of Equation 3.54 is obtained by assuming y = xv, which leads to

ν ν ν( )− + + =1 0a b (3.55)

The form of the solution depends on the values of a and b. If (a − 1)2 > 4b, Equation 3.55 has two real solutions, v1 and v2, such that

y x x y x x1 21 2( ) ( )= =ν ν (3.56)

If in addition a > 1, then at least one root is negative, and the corresponding solution is not analytic at x = 0.

If (a − 1)2 = 4b then Equation 3.55 has a single repeated root ν, and the lin-early independent solutions of Equation 3.54 are

y x x y x x x1 2( ) ( ) ln= =ν ν (3.57)

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If (a − 1)2 < 4b, then Equation 3.55 has a pair of complex conjugate roots, v = vr ± ivi and linearly independent solutions of Equation 3.54 are

y x x x x y x x x xr i r i1 2( ) cos ln ( ) sin ln= ( ) = ( )ν ν ν ν

(3.58)

Equation 3.59 illustrates that solutions of a second-order equation expanded about a regular singular point are not of the form of Equation b of Example 3.7. Indeed, such solutions need not be analytic. Solutions of the Cauchy-Euler equation suggest that solutions about a regular singular point involve noninteger powers of x and perhaps logarithmic terms.

The technique used to obtain series solutions of second-order equations about a regular singular point is called the method of Frobenius and is described by the following theorem, presented without proof.

Theorem 3.7 (Method of Frobenius). Let x0 be a regular singular point of a second-order differential equation, Equation 3.47. A solution to this equation can be expressed as

y x x x c x xkk

k

( ) = −( ) −( )=

∞

∑0 0

0

ν (3.59)

where v is the largest solution of the indicial equation,

ν ν ν( ) ( ) ( )− + + =1 00 0q x r x (3.60)

and c0 ≠ 0. Let μ be the second solution of Equation 3.60. A second linearly independent solution can be obtained dependent on the value of μ:

Case (1) If μ − v is not an integer, then the second solution is

y x x x d x xkk

k

2 0 0

0

( ) = −( ) −( )=

∞

∑μ (3.61)

Case (2) If μ = v, then the second solution is

y x x x d x x y x x xkk

k

2 0 0

0

1 0( ) ( )ln= −( ) −( ) + −=

∞

∑μ (3.62)

Case (3) If μ − v is an integer, then

y x x x d x x cy x x xkk

k

2 0 0

0

1 0( ) ( )ln= −( ) −( ) + −=

∞

∑μ (3.63)

where c is a constant.

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Example 3.9 Apply the method of Frobenius to obtain solutions of Bessel’s equation of order n,

xd ydx

xdydx

x n y22

22 2 0+ + −( ) = (a)

Solution Note that x = 0 is a regular singular point of Bessel’s equation with q(x) = 1 and r(x) = x2 − n2. A solution to Equation a is assumed to be

y x x c xkk

k

( ) ==

∞

∑ν

0

(b)

Noting that

dydx

k c xkk

k

= +( ) + −

=

∞

∑ ν ν 1

0

(c)

d ydx

k k c xkk

k

2

22

0

1= +( ) + −( ) + −

=

∞

∑ ν ν ν (d)

substitution of Equation b into Equation a leads to

k k k n c x c xkk

k

kk+( ) + −( )+ +( )−⎡⎣ ⎤⎦ ++

=

∞+∑ ν ν ν ν ν1 2

0

++

=

∞

∑ =2

0

0k

(e)

Noting that

c x c x c xkk

k

mm

m

kk

k

+ +

=

∞

−+

=

∞

−+

=

∞

∑ ∑ ∑= =ν ν ν2

0

2

2

2

2

(f)

Equation e can be rewritten as

k n c c x c n ck kk

k

+( ) −⎡⎣ ⎤⎦ +{ } + −( )+−

=

∞

∑ ν ν2 22

2

02 2

1 1++( ) −⎡⎣ ⎤⎦ =ν 2 2 0n x (g)

Setting coeffi cients of each power of x to zero independently leads to

ν2 20 0−( ) =n c (h)

1 02 21+( ) −⎡⎣ ⎤⎦ =ν n c (i)

k n c c kk k+( ) −⎡⎣⎢

⎤⎦⎥

+ =−ν2 2

2 2 3, , ... (j)

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Equation h is the indicial equation. Either c0 = 0, or v = ± n. However, the method of Frobenius requires that c0 ≠ 0. The largest solution is v = n.

If c0 ≠ 0, then Equation h and Equation i are satisfi ed by c1 ≠ 0 unless v = −1/2. However, since it is assumed that v > μ, this can never occur, although this phenomenon does provide an alert that n = 1/2 is a special case in that case (3) of theorem 3.7 applies to determine the second solution. The fi rst solution is determined with v = n, c0 ≠ 0, c1 = 0.

The recurrence relation, Equation j, can be rearranged as

cc

k n kk

k= −+

−2

2( ) (k)

Repeated application of Equation k shows that because c1 = 0, ck = 0 for all odd k. Application of Equation k for k = 2,4, and 6 leads to

cc

n2

0

2 2 2= −

+( ) (l)

ccn

cn n

42 0

4 2 4 2 4 2 2 2 4= −

+=

+ +( ) ( )( )( )( ) (m)

ccn

cn n n

64 0

6 2 6 2 4 6 2 2 2 4 2 6= −

+= −

+ + +( ) ( )( )( )( )( )( )) (n)

Use of Equation l, Equation m, and Equation n and mathematical induction lead to

cc

k n nk

k

201

2 4 2 2 2 2 4 2=

−( )[ ] + +( )( )...( ) ( )( )...( nn k

ck n n n k

k

k

+[ ]

=−

+ + +

2

1

2 1 20

2

)

( )

! ( )( )...( ))[ ]

(o)

Substitution of Equation o into Equation a leads to

y x cx

k n n n k

k k n

kk

( )( )

! ( )( )...( )=

−+ + +[ ]

+

0

2

2

1

2 1 2==

∞

∑0

(p)

Equation p is valid for any value of n, integer or noninteger. The form of the second linearly independent solution depends on the value of n. If n is neither an integer nor the midpoint between two integers, then the second solution is obtained from case (1) of theorem 3.7:

9533_C003.indd 1789533_C003.indd 178 10/29/08 3:59:14 PM10/29/08 3:59:14 PM


y x x d xnk

k

k

2

1

( ) = −

=

∞

∑ (q)

A similar procedure is followed with the recurrence relation reduced to

dd

k k nk

k=−

−2

2( ) (r)

Mathematical induction is used to obtain

dd

k n n k nk

k

k20

2

1

2 1 2=

−− − −[ ]

( )

! ( )( )...( ) (s)

and the second solution becomes

y x dx

k n n k n

k k n

k2 0

2

2

1

2 1 2( )

( )

! ( )( )...( )=

−− − −[ ]

−

kk=

∞

∑0

(t)

It is noted that y2 (x) could be obtained by replacing n by −n in the expression for y1 (x). It is clear from Equation t that such a solution cannot be obtained when n is an integer because the denominator will be zero for k = n.

The roots of the indicial equation are equal when n = 0 (ν = μ = 0). The sec-ond solution is obtained using case (2) of theorem 3.7:

y x d x y x xkk

k

2

0

1( ) ( )ln= +=

∞

∑ (u)

Note that d/dx(y1 ln|x|) = ln|x|(dy1/dx) + (1/x)y1 and d2/dx2(y1 ln|x|) = ln|x|(d2y1/dx2) + 2/x (dy1/dx) − (1/x2)y1. Thus,

xd

dxy x x

ddx

y x x y x x xd y2

2

2 1 12

12

2

ln ln ln ln( )+ ( )+ = 11

2

1 21

dxx

dydx

x y+ +⎡

⎣⎢⎢

⎤

⎦⎥⎥

+ =2 21 1xdydx

xdydx

(v)

Thus, substitution of Equation u into Equation a with n = 0 leads to

k k d x kd x d x kckk

k

kk

k

kk

k

k( )− + + +=

∞

=

∞+

=

∞

∑ ∑ ∑1 22 1

2

0

xxk

k=

∞

∑ =1

0 (w)

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where y1(x) is used in the form of Equation b. Equation w can be rearranged to yield

k d kc d x d c xk k kk

k

22

2

1 12 2 0+ +⎡⎣ ⎤⎦ + +( ) =−

=

∞

∑ (x)

Recall that c1 is determined to be zero; thus d1 = 0. The recurrence relation obtained from Equation x is

ddk k

ckk

k= − −−2

2

2 (y)

Without loss of generality, c0 = 1 and d0 = 1. Noting that for n = 0, ck = 0 for odd k and ck = 1/2k[(k/2)!]2 for even k. Mathematical induction and Equation y are used to determine an explicit expression for dk

dd

k

c i

i kk

k in=

−+

− ( )[ ](

+( 1)

k

20

2

2

2

21 2

2( !)

( ) !

!))=

−

∑ 2

0

2

c

k

(z)

3.6 Bessel functionsThe differential equation whose solutions were obtained in Example 3.8 is called Bessel’s equation:

xd ydx

xdydx

x n y22

22 2 0+ + − =( ) (3.64)

The fi rst solution of Equation 3.64 valid for any n is given by Equation p of Example 3.8 as

y xx

k n n k n

k k n

kk

1

2

2

1

2 1 2( )

( )

! ( )( )...( )=

−+ + +[ ]

+

=11

∞

∑ (3.65)

If n is an integer, an alternate representation of Equation 3.65 is

y x n

x

k n kn

kk n

k

1

2

0

21

2( ) !( )

!( )!=

− ( )+

+

=

∞

∑ (3.66)

It should be noted that 2nn! is a constant and can be incorporated into a con-stant of integration.

The Bessel function of the fi rst kind of order n is defi ned for integer n as

J x

x

k n kn

kk n

k

( )( )

!( )!=

− ( )+

+

=

∞

∑1

2

2

0

(3.67)

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Jn(x) is the fi rst solution of Bessel’s equation for integer n.The Bessel function of the fi rst kind for a noninteger value of n is defi ned

in terms of the gamma function,

Γ( )p e t dtt p= − −

∞

∫ 1

0

(3.68)

Integration by parts is applied to derive an important property of the gamma function,

Γ Γ( ) ( )p p p+ =1 (3.69)

Noting that

Γ( )1 1

0

= =−

∞

∫ e dtt (3.70)

Equation 3.69 shows that Γ Γ Γ( ) , ( ) , ( ) ,...2 1 3 2 4 6= = = . Induction can be used to show that for integer n,

Γ( ) ( )!n n= −1 (3.71)

Thus, the gamma function can be thought of as a generalization of the facto-rial to noninteger values.

The properties of the gamma function can be used to rewrite Equation 3.67 as

y x n

x

k n kn

kk n

k

1

2

2 11

21 1

( ) ( )( )

( ) ( )= +

− ( )+ + +

+

=

ΓΓ Γ

00

∞

∑ (3.72)

Equation 3.72 leads to the defi nition of the Bessel function of the fi rst kind of order n as

J x

x

k n kn

kk n

k

( )( )

( ) ( )=

− ( )+ + +

+

=

∞

∑1

21 1

2

0Γ Γ

(3.73)

When n is not an integer or a half integer, the second solution of Bessel’s equation is obtained by replacing n with −n in the fi rst solution. In this case, the Bessel function of the second kind is defi ned by

Y x J x

x

k k nn n

kk n

k

( ) ( )( )

( ) ( )= =

− ( )+ + −

−

−

=

12

1 1

2

0Γ Γ

∞∞

∑ (3.74)

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An alternate and commonly used defi nition for the Bessel function of the second kind is a linear combination of Equation 3.73 and Equation 3.74, defi ned by

Y xn J x J x

nn

n n( )cos( ) ( ) ( )

sin( )=

− −ππ

(3.75)

For an integer value of n, Equation 3.75 can be replaced by

Y xJ x J x

nn

( )cos( ) ( ) ( )

sin( )=

−→

−

ν

ν ννν

limπ

π (3.76)

The series form of Equation 3.76 is

Y x J xx n k

kx

n n

k n

k

( ) ( ) ln( )

( )= +( )−

−+ ( )

−2

2 1 2

2

πγ

ΓΓ

==

−

+

∑

−−( )

+ + + ( )⎡

⎣⎢⎢

⎤

⎦

0

1

21 1

1 1 2

n

k k n

k n kx

π Γ Γ( ) ( )⎥⎥⎥

+⎡

⎣⎢⎢⎢

⎤

⎦⎥⎥⎥

⎧⎨⎪⎪

⎩⎪⎪

⎫⎬⎪⎪

⎭⎪=

+

=∑∑ 1 1

11� �

��

n kk

⎪⎪=

∞

∑k 0

(3.77)

where Euler’s constant is defi ned as

γ = − ( )⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

=→∞

=∑m

m

mlim ln . ..1

0 57721

��

.. (3.78)

The general solution of Bessel’s equation of order n is

y x C J x C Y xn n( ) ( ) ( )= +1 2 (3.79)

Bessel functions of the fi rst and second kind of integer orders are plotted in Figure 3.6 and Figure 3.7, respectively. Bessel functions of the fi rst order are fi nite for all values of x, whereas Bessel functions of the second kind are infi nite at x = 0 and fi nite everywhere else. Both Bessel functions have an infi nite, but countable, number of zeroes. However, they are not periodic.

Consider the equation

xd ydx

xdydx

b x n y22

22 2 2 0+ + − =( ) (3.80)

Let z = bx. Rewriting Equation 3.80 with z as the independent variable leads to

zd ydz

zdydz

z b y22

22 2 0+ + − =( ) (3.81)

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Bessel functions of the first kind

x

–0.5

0.5

1J0(x)

J1(x)J2(x)

J3(x)

0

0 1 2 3 4 5 6 7 8 9 10

Figure 3.6 Bessel functions of the fi rst kind are solutions of Bessel’s equation which are fi nite for all values of x.

Bessel functions of the second kind

Y0(x)Y1(x) Y2(x)

–0.5

–1.5

–1

–2

0.5

1

0

x

1 2 3 4 5 6 7 8 9 10 11

Figure 3.7 Bessel functions of the second kind do not exist at x = 0.

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Equation 3.81 is Bessel’s equation of order n with z as the independent variable. Thus the general solution of Equation 3.80 is

y x C J bx C Y bxn n( ) ( ) ( )= +1 2 (3.82)

A modifi ed form of Bessel’s equation is

xd ydx

xdydx

b x n y22

22 2 2 0+ − + =( ) (3.83)

Let z = ix. Rewriting Equation 3.83 with z as the independent variable leads to

zd ydz

zdydz

b z n y22

22 2 2 0+ + − =( ) (3.84)

The solution of Equation 3.84 is obtained using Equation 3.64. The solution of Equation 3.83 is thus

y z C J ibx C Y ibxn n( ) ( ) ( )= +1 2 (3.85)

Note that for any n,

J ibx

ibx

k n kn

kk n

k

( )( )

( ) ( )=

− ( )+ + +

+

=

∞

∑1

21 1

2

0Γ Γ

=( )

+ + +

+

=

∞

i

bx

k n kn

k n

k

21 1

2

0Γ Γ( ) ( )∑∑

(3.86)

The modifi ed Bessel function of the fi rst kind of order n is defi ned as

I x i J ix

x

k n

nn

n

k n

( ) ( )

( ) (

=

=( )

+ +

−

+

21

2

Γ Γ kkk

+=

∞

∑ 10

) (3.87)

and is one solution of Equation 3.83. The second solution is the modifi ed Bes-sel function of the second kind defi ned by

K xI x I x

nn

n n( )( ) ( )

sin( )= −

− −ππ2

(3.88)

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A limiting process similar to that used to defi ne Yn(x) when n is an integer, Equation 3.88, is used to defi ne Kn(x) for n an integer.

Modifi ed Bessel functions of the fi rst and second kind are plotted in Figure 3.8 and Figure 3.9, respectively. The modifi ed Bessel functions of the fi rst kind are fi nite for all values of their argument, while the modifi ed Bessel functions of the second kind are infi nite when evaluated at zero, but fi nite for all other values of their argument. Contrary to the regular Bessel functions, the modifi ed Bessel functions do not have zeroes.

Example 3.10 The nondimensional differential equation governing the tem-perature distribution in the straight triangular fi n shown in Figure 3.10 is

d

dxx

ddx

mθ

θ( )− =2 0 (a)

where x is nondimensionalized with respect to L, θ = − −∞ ∞( )/( )T T T T0 , and m hL kb= 2 2/ . The boundary condition is

θ( )1 1= (b)

The temperature must be fi nite everywhere over the interval, 0 ≤ x ≤ L. Determine the rate of heat transfer at the base of the fi n.

Modified Bessel functions of the first kind

I1(x)

00

0.5

1.5

2.5

3.5

1

2

3

0.5 1 1.5x

2.52

I0(x)

I2(x)

Figure 3.8 Modifi ed Bessel functions of the fi rst kind are fi nite at all fi nite x.

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Solution Let z = x1/2. Derivatives with respect to x are converted to deriva-tives with respect to z using the chain rule:

ddx

ddz

dzdx

zddz

θ θ

θ

=

=1

2

(c)

x

b

θ = 1

x

h,Tk

8

LI

Figure 3.10 The temperature distribution in a straight triangular fi n is governed by a second-order differential equation with variable coeffi cients whose solution is in terms of modifi ed Bessel functions.

Modified Bessel functions of the second kind

K1(x)

K0(x)

10

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

1.5 2 32.5x

3.5

K2(x)

Figure 3.9 Modifi ed Bessel functions of the second kind do not exist at x = 0.

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ddx

xddx z

ddz

z ddz

θ θ( )= ( )=

1

2 2

11

4

1

4

2

2

ddz z

ddz

θ θ+

(d)

Substitution of Equation d into Equation a leads to

1

4

1

40

2

22d

dz zddz

mθ

+ − =θ

θ (e)

Multiplying Equation e by 4z2 gives

zddz

zddz

m z22

22 24 0

θ+ − =

θθ (f)

Equation f is the same as the modifi ed Bessel’s equation, Equation 3.83, with n = 0 and b = 2m. Thus the general solution of Equation f is

θ( )z C I mz C K mz= ( )+ ( )1 0 2 02 2 (g)

Substituting for z in term of x leads to

θ( )x C I mx C K mx=⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟+

⎛

⎝⎜⎜⎜⎜

⎞1 0

1

22 0

1

22 2⎠⎠⎟⎟⎟⎟

(h)

It should be noted I0(0) = 0, but lim ( )x K x→ = ∞0 0 . Thus, the temperature is fi nite at x = 0 if and only if C2 = 0. Application of Equation b to Equation h with C2 = 0 leads to

CI m1

0

1

2=

( )and

θ( )xI mx

I m=

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

( )

0

1

2

0

2

2 (i)

The rate of heat transfer from the base is

qkb T T

Lddx

=−( )∞� 0

1θ

( ) (j)

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where

ddx

m x

I mI mx

θ=

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

( )′⎛

⎝

−( )2

1

2

22

1

2

00

1

2⎜⎜⎜⎜⎜⎞

⎠⎟⎟⎟⎟

(k)

Note that in Equation k, the prime represents differentiation with respect to the argument of the function. Substitution of Equation k into Equation j gives

qkb m

I mI m=

( )′�

00

22( ) (l)

Term-by-term differentiation of Equation 3.87 for n = 0 leads to

ddx

I xd

dx

x

k

k

k

0

2

2

0

2( )!

[ ]=( )( )

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥=

∞

∑⎥⎥

=( )( )

−

=

∞

∑k

x

k

k

k

2

2 1

2

1!

=+ ( )

+( )[ ]

+ −

=

∞

∑( )

!

( )

px

p

p

p

121

2 1 1

2

0

=( )

+ +

+

=

∞

∑x

p p

p

p

21 2

2 1

1( )!( )!

= I x1( )

(m)

The rate of heat transfer from the base of the fi n is

qkb m

I mI m

L hkb

I Lh

kb

=( )

= ⎛

⎝⎜⎜⎜

⎞

�

�

01

0

22

2

22

( )

⎠⎠⎟⎟⎟⎟

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟I L

hkb

1 22

(n)

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It is shown in Equation m of Example 3.9 that ′ =I x I x0 1( ) ( ). This is an exam-ple of a identity involved Bessel functions which can be derived using the series defi nitions of Bessel functions. These identities are useful in applica-tions where differentiation and integration of Bessel functions are required. Some useful identities are stated below.

1. Derivatives of Bessel functions of order zero

′ = −J x J x0 1( ) ( ) (3.89)

′ = −Y x Y x0 1( ) ( ) (3.90)

′ =I x I x0 1( ) ( ) (3.91)

′ = −K x K x0 1( ) ( ) (3.92)

2. Derivatives of Bessel functions

d

dxx J x x J xn

nn

n( ) ( )⎡⎣ ⎤⎦ = −1 (3.93)

d

dxx Y x x Y xn

nn

n( ) ( )⎡⎣ ⎤⎦ = −1 (3.94)

d

dxx I x x I xn

nn

n( ) ( )⎡⎣ ⎤⎦ = −1 (3.95)

d

dxx K x x K xn

nn

n( ) ( )⎡⎣ ⎤⎦ = − −1 (3.96)

d

dxx J x x J xn

nn

n− −

+⎡⎣ ⎤⎦ = −( ) ( )1 (3.97)

ddx

x Y x x Y xnn

nn

− −+⎡⎣ ⎤⎦ = −( ) ( )1

(3.98)

d

dxx I x x I xn

nn

n− −

+⎡⎣ ⎤⎦ =( ) ( )1 (3.99)

d

dxx K x x K xn

nn

n− −

+⎡⎣ ⎤⎦ = −( ) ( )1 (3.100)

3. Recurrence Relations

J x J xnx

J xn n n+ −+ =1 12

( ) ( ) ( ) (3.101)

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Y x Y xnx

Y xn n n+ −+ =1 1

2( ) ( ) ( ) (3.102)

′ = −−J x J xnx

J xn n n( ) ( ) ( )1 (3.103)

′ = − +J xnx

J x J xn n n( ) ( ) 1( ) (3.104)

Example 3.11 An integral that is useful in applications is x J x dxn0

1 2∫ [ ]( )α .

During the evaluation of the integral, it is convenient to use the identity

d

dxx J x J x x J x

ddx

J xn n n n2

1 122− +⎡⎣ ⎤⎦ = [ ]( ) ( ) ( ) ( ) (a)

(a) Use the properties of Bessel functions, Equations 3.93−3.104, to derive Equation a, (b) use Equation a in the evaluation of ∫ [ ]0

1 2x J x dxn( )α , (c) simplify the result if Jn(α) = 0, and (d) simplify the result if ′ =Jn( )α 0.

Solution (a) The product rule for differentiation is used to expand the derivative on the left-hand side of Equation a as

d

dxx J x J x xJ x J x xn n n n

21 1 1 1

22+ − + −⎡⎣ ⎤⎦ = +( ) ( ) ( ) ( ) ′′ + ′+ − + −J x J x x J x J xn n n n1 12

1 1( ) ( ) ( ) ( ) (b)

Using Equation 3.103 to substitute for ′+J xn 1( ) and Equation 3.104 to substitute for ′−J xn 1( ), Equation b can be rewritten as

ddx

x J x J x xJ x J x xn n n n2

1 1 1 122+ − + −⎡⎣ ⎤⎦ = +( ) ( ) ( ) ( ) JJ x J x

nx

J x

x J

n n n

n

− +

+

−+⎡

⎣⎢⎢

⎤⎦⎥⎥

+

1 1

21

1( ) ( ) ( )

(xxn

xJ x J xn n) ( ) ( )

−−⎡

⎣⎢⎢

⎤⎦⎥⎥−

11

= − + + − + −[ ( ) ( )] ( ) (2 1 1 1 1x x n x n J x Jn n xx

x J x J x J xn n n

)

( ) ( ) ( )+ −[ ]− +2

1 1

= −[ ]− +x J x J x J xn n n2

1 1( ) ( ) ( )

(c)

Addition of Equation 3.103 and Equation 3.104 leads to

′ = −[ ]− +J x J x J xn n n( ) ( ) ( )1

21 1 (d)

Substitution of Equation d on the right-hand side of Equation c leads to Equation a.

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(b) First consider the indefi nite integral

I x J x dxn= [ ]∫ ( )α 2 (e)

The variable of integration is changed from x to z using z = αx, leading to

I z J z dzn= [ ]∫12

2

α( ) (f)

Integration by parts is used to evaluate the integral in Equation f by defi ning

u = [Jn(z)]2 and dv = zdz such that du J x J z dzn n= ′2 ( ) ( ) and v = z2/2. Equation f becomes

Iz

J zz

J z J z dzn n n= [ ] − ′[ ]⎧⎨⎪⎪⎩⎪ ∫1

2 22

2

22

2

α( ) ( ) ( )

⎪⎪

⎫⎬⎪⎪⎭⎪⎪

(g)

Equation a is used to simplify the integral on the right-hand side of Equation g:

Iz

J zddz

z J z J z dn n n= [ ] − ⎡⎣ ⎤⎦+ −1

2

1

22

22 2

1 1α

( ) ( ) ( ) zz

zJ z J zn n

∫⎧⎨⎪⎪⎩⎪⎪

⎫⎬⎪⎪⎭⎪⎪

= [ ] − +

2

2

21

2α( ) ( )JJ zn−[ ]{ }1( )

(h)

Recalling that z = αx, Equation h becomes

Ix

J x J x J xn n n= [ ] −[ ]{ }+ −

22

1 12

( ) ( ) ( )α α α (i)

Noting that Jn(α) = 0, the defi nite integral becomes

x J x dxx

J x J x Jn n n n( ) ( ) ( ) (α α α α[ ] = [ ] −∫ + −2

0

12

21 1

2xx

x

x

)[ ]{ }⎡

⎣⎢⎢

⎤

⎦⎥⎥

=

=

0

1

= [ ] −[ ]{ }+ −1

2

21 1J J Jn n n( ) ( ) ( )α α α

(j)

(c) If Jn(α) = 0, Equation j becomes

x J x dx J Jn n n( ) ( ) ( )α α α[ ] = −∫ + −2

0

1

1 11

2 (k)

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and from Equation 3.103 and Equation 3.104, J Jn n+ = − ′1( ) ( )α α and Jn−1 (α) =′Jn( )α . Thus Equation k becomes

x J x dx Jn n( )α α[ ] = ′ ( )[ ]∫ 2

0

1

1

2

2

1

2

2= ( )[ ]+Jn 1 α (l)

Equation l is the form of the integral used in applications when Jn(α) = 0.(d) If ′ =Jn( )α 0, Equation 3.103 and Equation 3.104 lead to Jn+1(α) = (n/α) Jn(α) and Jn−1(α) = (n/α) Jn(α). In this case, Equation j becomes

x J x dxn

Jn n( ) ( )αα

α[ ] = −⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟[ ]∫ 2

0

12

2

21

21 (m)

Example 3.12 Evaluate ∫ x J x dx52( ) .

Solution Integrals of the form of ∫ x J x dxmn( ) when m and n are integers can

be evaluated using integration by parts in conjunction with Equation 3.93 and Equation 3.94. If m + n is odd, repeated integration by parts will lead to a closed-form evaluation of the integral. If m + n is even, repeated application of integration by parts eventually leads to ∫ J x dx0( ) , which does not have a closed-form evaluation.

The initial strategy in evaluating these integrals is, when using integration by parts, to choose dv such that v can be obtained using either Equation 3.93 or Equation 3.94. For the integral at hand, if dv = x3J2(x), then, from Equation 3.93, v = x3J3(x). For this choice of v, u = x2 and du = 2xdx. Thus application of integration by parts leads to

x J x dx x x J x x J x xdx52

2 33

33 2( ) ( ) ( )∫ ∫= ⎡⎣ ⎤⎦ − ⎡⎣ ⎤⎦

= − ∫x J x x J x dx53

432( ) ( )

(a)

The remaining integral can be evaluated directly from Equation 3.93, leading to

x J x dx x J x x J x52

53

442( ) ( ) ( )∫ = − (b)

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3.7 Differential equations whose solutions are expressible in terms of Bessel functions

Example 3.9 illustrates that some differential equations may, through a change of independent variable, be transformed into Bessel’s equation. There are sev-eral transformations which enable differential equations of certain forms to be rewritten as Bessel’s equation.

Consider the differential equation,

xd ydx

axdydx

b x y22

22 2 0+ + = (3.105)

Equation 3.105 is similar to Bessel’s equation of order zero which has a coef-fi cient of 1 multiplying the x (dy/dx) term. A change of dependent variable y = x (1−a)/2z leads to

xd zdx

xdzdx

b xa

z22

22 2

21

20+ + −

−( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

= (3.106)

Equation 3.106 is Bessel’s equation of order (1− a)/2, and thus the general solution of Equation 3.106 is

y x C x J bx C x Y bxa

a

a

a( ) = ( )+ ( )−

−

−

−1

1

21

2

1

1

21

2

(3.107)

If b2 < 0, then modifi ed Bessel functions are used in Equation 3.106 with argument b x.

Next consider

d

dxx

dydx

b x yr s( )+ =2 0 (3.108)

where r and s are nonnegative. A change of independent variable x u s r= − +2 2/( ) leads to the differential equation

ud ydu

rs r

udydu

bs r

22

221

2 1

2

2+ +

−( )− +

⎡⎣⎢⎢

⎤⎦⎥⎥

+− + 22

02

( ) =y (3.109)

Equation 3.109 is of the form of Equation 3.106 with a r s r= + −( ) − +1 2 1 2/ . Noting that ( )/ ( )/1 2 1 2− = − − +a r s r , the general solution of Equation 3.109 is

y u C u Jb

s ru C u

rs r

rs r

rs( ) =

− +( )+−

− +−

− +

−

1

1

21

2

2

12

2−− +

−− + − +( )r

rs r

Yb

s ru2

1

2

2

2 (3.110)

9533_C003.indd 1939533_C003.indd 193 10/29/08 3:59:26 PM10/29/08 3:59:26 PM


The solution of Equation 3.110 in terms of the original independent variable x is

y x C x Jb

s rx

r

rs r

s r

( ) =− +

⎛

⎝⎜⎜⎜⎜

⎞

⎠

−

−− +

− +

1

1

21

2

2

22

2⎟⎟⎟⎟⎟

+− +

⎛

⎝⎜⎜⎜⎜

⎞−

−− +

− +

C x Jb

s rx

r

rs r

s r

2

1

21

2

2

22

2 ⎠⎠⎟⎟⎟⎟

(3.111)

The differential equation

d

dxx

dydx

b x cx yr s r( )+ +( ) =−2 2 0 (3.112)

has a general solution of

y x C x Jb

s rx C

r s r

( ) =− +

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟+

− − +

1

1

2

2

22

2ν 22

1

2

2

22

2x J

bs r

xr s r− − +

− +

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟ν (3.113)

where

ν =− −− +

( )1 4

2

2r cs r

(3.114)

Example 3.13 The differential equation governing the temperature distri-bution in an annular fi n of hyperbolic profi le is

ddr

Rr

dTdr

hk

r T T4 4

0

0

1

2

3

2π π

∞

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥− −( )= (a)

The boundary conditions are

T R T0 0( )= (b)

dTdr

R1 0( )= (c)

Nondimensional variables are introduced as r* = r/R0 and θ ∞ ∞= − −( )/( )T T T T0 . The nondimensional formulation of the problem is

ddr

rddr

Bi r3

2 0θ

θ⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟−( ) = (d)

θ( )1 1= (e)

ddr

RR

θ 1

0

0⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟= (f)

9533_C003.indd 1949533_C003.indd 194 10/29/08 3:59:27 PM10/29/08 3:59:27 PM


where the Biot number is Bi = hR0/K. Determine the temperature distribution in the fi n.

Solution Equation d is in the form of Equation 3.108 with r = (3/2), s = 1, and b2 = −Bi. Thus the solution is in the form of Equation 3.111, except that the Bessel functions J and Y are replaced by the modifi ed Bessel functions I and K and b is replaced by Bi. The general solution is

θ( )r C r I Bir C r K=⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟+

−

−

−

−1

1

41

3

3

42

1

44

311

3

3

44

3Bir

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

(g)

Application of the boundary condition, Equation f, leads to

C

BiRR

I BiRR

2

1

0

3

4

1

3

1

0

4

3= −

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟ ′ ⎛

⎝⎜⎜⎜

⎞⎠⎟

−⎟⎟⎟

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥−

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟

⎡

−

3

4

1

3

1

0

3

41

4

4

3I Bi

RR

⎣⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟ ′ ⎛

⎝−Bi

RR

K BiRR

1

0

3

4

1

3

1

0

4

3⎜⎜⎜⎜

⎞⎠⎟⎟⎟

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥−

⎛⎝⎜⎜⎜

⎞⎠−

3

4

1

3

1

0

1

4

4

3K Bi

RR

⎟⎟⎟⎟

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

=

3

4

1C

C1μ

(h)

Application of Equation e to Equation g, using Equation h, gives

CI Bi K Bi

1

1

3

1

3

14

3

4

3

=( )− ( )− −

μ (i)

The total heat transfer at the base of the fi n is

Q k RdTdr

R= −⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

( )4 0

3

20π (j)

Equation j can be rewritten in terms of nondimensional variables as

Q Rddr

= −4 10

1

2πθ

( ) (k)

Substitution of Equation g, Equation h, and Equation i into Equation k leads to

Q kR C I Bi BiI Bi K= − − ( )+ ′ ( )−− −

41

4

4

3

4

3 40

1

21 1

3

1

3

πμ

−− ( )⎡

⎣⎢⎢ 1

3

4

3Bi

−

+ ′ ( )⎤⎦⎥⎥1

3

4

3BiK Biμ (l)

9533_C003.indd 1959533_C003.indd 195 10/29/08 3:59:28 PM10/29/08 3:59:28 PM


A differential equation that often occurs in applications is

xd ydx

xdydx

x m m y22

222 1 0+ + − + =[ ( )] (3.115)

where m is an integer. Equation 3.115 is in the form of Equation 3.112 with r = 2, s = 2, b = 1, and c = −m(m + 1). The general solution of Equation 3.115 is obtained using Equation 3.113 as

y x C x J x C x Ym m( ) ( )( / )( / )

( / )( / )= +−

+−

+11 2

1 2 21 2

1 2 (( )x (3.116)

The Bessel functions used in Equation 3.116 are of fractional order which is an odd integer times one half.

Spherical Bessel functions of the fi rst kind and second kind are defi ned as

j xx

J xm m( ) ( )( / )

( / )=⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟ +

π2

1 2

1 2 (3.117)

y xx

Y xm m( ) ( )( / )

( / )=⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟ +

π2

1 2

1 2 (3.118)

Thus, the general solution of Equation 3.115 is written using spherical Bessel functions as

y x C j x C y xm m( ) ( ) ( )= +1 2 (3.119)

Graphs of spherical Bessel functions are presented in Figure 3.11 and Figure 3.12.

It is noted that j0 0 1( ) = while jm( )0 0= for m > 0 and lim ( )x my x→ = −0 ∞. All spherical Bessel functions have an infi nite number of zeroes.Spherical Bessel functions satisfy recurrence relations of the form

j xm

xj x j xm m m+ −−

++ =1 1

2 10( ) ( ) ( ) (3.120)

y xm

xy x y xm m m+ −−

++ =1 1

2 10( ) ( ) ( ) (3.121)

9533_C003.indd 1969533_C003.indd 196 10/29/08 3:59:29 PM10/29/08 3:59:29 PM


Figure 3.11 Spherical Bessel functions of the fi rst kind.

0 1 2 3 4 5 6 7 8 9 10

0.8

Spherical bessel functions of the first kind

0.6

0.4

0.2

–0.2

–0.4

0

1j0(x)

j1(x)j2(x)

j3(x)

Figure 3.12 Spherical Bessel functions of the second kind.

y0(x) y1(x) y2(x) y3(x)

2 3 4 5 6 7 8 9 10 11 12

0.4Spherical bessel functions of the second kind

0.2

–0.2

–0.4

–0.6

–0.8

–1

–1.2

–1.4

–1.6

0

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Several of the initial spherical Bessel functions are related to elementary trigonometric functions by

j xx

x0( )

sin( )= (3.122)

j xx

xx

x1 2( )

sin( ) cos( )= − (3.123)

y xx

x0( )

cos( )= − (3.124)

y xx

xx

x1 2( )

cos( ) cos( )= − − (3.125)

3.8 Legendre functionsLegendre’s equation is

d

dxx

dydx

m m y( ) ( )1 1 02−⎡⎣⎢⎢

⎤⎦⎥⎥+ + = (3.126)

The point x = 0 is an ordinary point of Legendre’s equation, and a series solu-tion is assumed to be of the form

y x a xkk

k

( ) ==

∞

∑0

(3.127)

The product rule for differentiation can be used to expand the derivative in Equation 3.126, and Equation 3.127 can be substituted for y(x), leading to

1 1 22 2

0

1

1

−( ) − − + +−

=

∞−

=

∞

∑ ∑x k k a x x ka x m mkk

k

kk

k

( ) ( 11 00

) a xkk

k=

∞

∑ = (3.128)


m m k k a x k k a xkk

k

kk

k

( ) ( ) ( )+ − +[ ] + − ==

∞−

=

∞

∑ ∑1 1 10

2

2

00 (3.129)

Reindexing the second summation in Equation 3.129 leads to

m m k k a k k a xk k

k

k( ) ( ) ( )( )+ − +[ ] + + +{ } =+

=

∞

∑ 1 1 1 2 02

0

(3.130)

9533_C003.indd 1989533_C003.indd 198 10/29/08 3:59:30 PM10/29/08 3:59:30 PM


Setting coeffi cients of powers of x to zero gives

ak k m m

k kak k+ =

+ − ++ +

21 1

1 2

( ) ( )

( )( ) (3.131)

Equation 3.120 is a recursion relation between the coeffi cients. Its application for even values of k leads to defi nition of coeffi cients with even subscripts in terms a0, while its application for odd values of k leads to defi nition of coeffi cients with odd subscripts in terms a1. Leaving these initial coeffi cients arbitrary, the general solution of Equation 3.131 is of the form

y x a y x a y x( ) ( ) ( )= +0 0 1 1 (3.132)

where the expansion for y0(x) contains only even powers of x and the expan-sion for y1(x) contains only odd powers of x.

First consider integer values of m. Equation 3.131 shows that am+2 = 0. If m is even, then y0 (x) is a polynomial of order m, while y1 (x) has an infi nite series expansion. If m is odd, then y1(x) is a polynomial of order m, while y0 (x) has an infi nite series expansion. It thus becomes convenient to rename the func-tions such that for an integer m,

y x c z x c z xm m m( ) ( ) ( ), ,= +1 1 2 2 (3.133)

where z1,m (x) is a polynomial of order n and z2,m (x) has an infi nite series expansion.

The Legendre polynomial of the fi rst kind of order m is defi ned by

P xz xz

m( )( )

( )= 1

1 1 (3.134)

Equation 3.133 and Equation 3.134 are used to evaluate the fi rst four Legendre polynomials as

P x0 1( ) = (3.135a)

P x x1( ) = (3.135b)

P x x221

33 1( ) = −( ) (3.135c)

P x x x331

25 3( ) = −( ) (3.135d)

9533_C003.indd 1999533_C003.indd 199 10/29/08 3:59:30 PM10/29/08 3:59:30 PM


The Legendre function of the second kind of order m is defi ned by

Q xz z x

z z xm( )( ) ( )

( ) ( )=

− ==

1 2

1 2

1

1

m 0, 2, 4,

m 1, 3, 5

…

,,…⎧⎨⎪⎪⎩⎪⎪

(3.136)

It can be shown that the infi nite series representations of the Legendre func-tions of the second kind for an integer order converge within the interval −1 < x < 1 to known functions. For example,

Q x x01( ) tanh ( )= − (3.137a)

Q xx x

x1

2

1

11( ) ln=

+−

− (3.137b)

Legendre’s equation, Equation 3.126, can be used to derive recurrence rela-tions between the Legendre polynomials and Legendre functions as

mP x m xP x m P xm m m( ) ( ) ( ) ( ) ( )− − − −− −2 1 11 2 (3.138a)

mQ x m xQ x m Q xm m m( ) ( ) ( ) ( ) ( )− − − −− −2 1 11 2 (3.138b)

If m is not an integer, a change of variables u = 1 − x is used in Equation 3.126. The resulting equation has a regular singular point at u = 0 (x = 1), with one root of its indicial equation as zero. The second solution is obtained using the method of Frobenius and has a logarithmic singularity at u = 0. The solution which remains fi nite at x = 1, but is undefi ned at x = −1, is called the Legendre function of the fi rst kind of order m, Pm (x). The solution which is undefi ned at x = ± 1 is called the Legendre function of the second kind of order m, Qm(x).

The associated Legendre’s equation is

d

dxx

dydx

m mn

x1 1

12

2

2−( )⎡

⎣⎢⎢

⎤⎦⎥⎥+ + −

−

⎡

⎣⎢⎢

⎤

⎦⎥⎥

( ) yy = 0 (3.139)

The associated Legendre functions of the fi rst kind, P xmn( ), and of the second

kind, Q xmn ( ), of order m and index n satisfy Equation 3.139 and are defi ned as

P x xd P x

dxmn

n nm

n( )( )

= −( )1 2 2 (3.140)

Q x xd Q x

dxmn

n nm

n( )( )

= −( )1 2 2 (3.141)

9533_C003.indd 2009533_C003.indd 200 10/29/08 3:59:31 PM10/29/08 3:59:31 PM


Note that by defi nition, if n and m are integers and n > m, then Pmn x( )= 0 .

The associated Legendre function of the fi rst kind of order m is fi nite at x = ± 1 only if m is an integer.

Problems 3.1. Prove that the homogeneous solution space of a linear differential

operator L is a vector space. 3.2. Consider the set of solutions of Equation (3.10) which satisfy the

boundary conditions specifi ed in theorem 3.1 for arbitrary values of y0 and �y0. That is, the functions in the set satisfy boundary conditions of this form, but not for specifi c values of y0 and �y0. Prove that this set of functions is a vector space.

3.3.−3.13. Solve the following differential equations subject to the given boundary conditions:

3.3. ( / )d y dx y2 2 4 0− = with y (0) = 2 and y (1) = 0.

3.4. ( / )d y dx y2 2 16 0+ = with y (0) = 2 and dy/dx (1) + 3 y (1) = 0.

3.5. ( / ) ( / )d y dx dy dx y2 2 4 4 0+ + = with y (0) = 0 and dy/dx (1) = 1.

3.6. ( / ) ( / )d y dx dy dx y2 2 5 14 0+ − = with dy/dx (0) = 0 and y (1) = 1.

3.7. ( / ) ( / )d y dx dy dx y2 2 5 14 0+ + = with dy/dx (0) = 0 and y (1) = 1.

3.8. ( / ) ( / )d y dx dy dx y2 2 2 10 0+ + = with dy/dx (0) = 0 and (dy/dx) (1) = 1.

3.9. ( / )d y dx y e x2 2 23− = − with y (0) = 0 and y (1) = 0.

3.10. ( / )d y dx y e x2 2 33 4 5+ = +− with y (0) = 0 and dy/dx (1) = 0.

3.11. ( / ) . sin( )d y dx y x2 2 9 0 1 4+ = with y (0) = 1 and y (1) = 0.

3.12. ( / ) sin( )d y dx y e xx2 2 33 2 2− = − with y (0) = 1 and y (1) = 0.

3.13. ( / ) ( / )d y dx dy dx x e x2 2 33 3 2+ = + − with y (0) = 0 and dy/dx (1) = 0. 3.14. The nondimensional equation for the temperature distribution over

the extended surface shown in Figure P3.14 is

ddx

Bi2

20

θ− =θ (a)

x

L

b

TT0

x

Lx* =

88

80

T–T

T –Tθ =

I

Figure P3.14 System of problems 3.14–3.16.

9533_C003.indd 2019533_C003.indd 201 10/29/08 3:59:32 PM10/29/08 3:59:32 PM


Where Bi hL b k b= +( )2 2 � �/ . The face at x = 0 is maintained at a constant nondimensional temperature of 1, θ( )0 1= and is insu-lated at its right end, d dxθ/ ( )1 0= .

a. Determine θ( )x . b. Determine the rate at which heat is transferred from the

base of the surface,

q

k b T TL

ddx

00 0=

− − ∞� ( )( ).

θ

c. Determine the rate at which heat is transferred over the surface by convection,

q h b T T x dxc = + −( )∞∫ 2 0

0

2

( ) ( )� θ .

d. Compare the answers to (b) and (c) and explain. e. The effi ciency of the extended surface is defi ned as the ratio

of total heat transfer from the surface of the fi n as calcu-lated in part (c) to the heat transfer from the surface if were held at the base temperature. Determine the effi ciency for the extended surface.

3.15. Rework problem 3.14 if the surface is subject to an internal heat generation per unit volume, u(x), such that the nondimensional equation for the temperature distribution is

ddx

Bi u x2

2

θ− = −θ Λ ( ),

Where Λ =u L k bmax /2 � . Let Bi = 0.2 and Λ = 0 5. . Solve parts (a)−(c) of problem 3.14 if (i) u(x) = x(1−x), (ii) u(x) = sin(πx), and (iii) u(x) = e −x.

3.16. Rework problem 3.14 if the tip of the fi n is not perfectly insu-lated such that the boundary condition at the right end is

ddx

hLk

θθ( ) ( )1 1 0+ = .

For computational purposes, assume that hl k/ .= 0 5 and Bi = 2.

3.17. An extended surface is composed of two materials as shown in Figure P3.17. Let θ1 (x) be the temperature distribution from

9533_C003.indd 2029533_C003.indd 202 10/29/08 3:59:33 PM10/29/08 3:59:33 PM


the base to x = λ where the material changes, and let θ2 (x) be the temperature distribution in the fi n from x = λ to x = 1, the tip of the extended surface. The governing differential equations are

ddx

Bi2

12 1 1 0

θθ λ− = ≤ ≤0 x (a)

ddx

Bi2

22 2 2 0 1

θθ λ− = ≤ ≤x (b)

where Bi hL b k b12

12= +( )� �/ and Bi hL b k b= +( )2 22� �/ . The appro-

priate boundary conditions are θ1 0 1( ) = and ( / )( )d dxθ2 1 0= . The temperature must be continuous across the surface, requiring that

θ θ λ1 2( ) ( )λ = . The rate of heat transfer must also be continuous

across the surface, requiring that d dx k k d dxθ λ θ λ1 2 1 2/ / ( / )( )= ( ). Rework problem 3.14 for this extended surface. For the computa-tions, assume k k2 1 1 2/ .= and λ = 0 42. .

3.18. The transverse defl ection of the uniform beam shown in Figure P3.18 is governed by the differential equation

EId wdx

f x4

4= ( ) (a)

where f(x) is the distributed load per unit length. Determine the transverse defl ection of a pinned-pinned beam which is subject to the boundary conditions w d w dx( ) ,( / )( )0 0 0 02 2= = , w(L) = 0 and (d2w/dx2) (L) = 0 for (a) f (x) = F0x(L − x) and (b) f (x) =Fm sin(3πx/L).

L

b

h,TT

kk1

20

8

I

λL


9533_C003.indd 2039533_C003.indd 203 10/29/08 3:59:34 PM10/29/08 3:59:34 PM


3.19. Repeat problem 3.18 for a fi xed-pinned beam which has bound-ary a w dw dx w L d w dx L( ) , ( / )( ) , ( ) / )( )0 0 0 0 0 2 2= = = and ( == 0.

3.20. Repeat problem 3.19 for a fi xed-pinned beam with a concentrated load of magnitude F0 applied at the midspan. One approach to solve this problem is to defi ne w1(x) as the transverse defl ection between x = 0 and x = L/2 and w2 (x) as the defl ection between x = L/2 and x = L. The boundary conditions of problem 3.18 apply appropri-ately to w1(0) and w2(L). Matching conditions at x = L/2 are that the defl ection is continuous, w1(L/2) = w2(L/2), the slope of the elastic curve is continuous, (dw1/dx)(L/2) = (dw2/dx) (L/2), the bending moment is continuous, (d2w1/dx2)(L/2) = (d2w2/dx2) (L/2), and the shear force is discontinuous at x = L/2 due to the presence of the concentrated load, EI (d3w1/dx3) (L/2) − EI (d3w2/dx3) (L/2) = F0.

3.21. Repeat problem 3.18 for a beam pinned at x = 0, but with a linear spring of stiffness k attached at x = L. The appropriate bound-ary conditions are w(0) = 0, (d2w/dx3) (0) = 0, (d2w/dx2) (L) = 0, and EI (d3w/dx3) (L) + kw(L) = 0.

3.22. The transverse defl ection of a beam on an elastic foundation (a Winkler foundation), as shown in Figure P3.22, is governed by the nondimensional differential equation

d wdx

w f x4

4+ =η Λ ( ) (a)

f (x)

L

E, I

Figure P3.18 System for problems 3.18–3.21.

f (x)

k

Figure P3.22 System for problems 3.22 and 3.23.

9533_C003.indd 2049533_C003.indd 204 10/29/08 3:59:36 PM10/29/08 3:59:36 PM


Determine w(x) for a fi xed-free beam when (a) f (x) = 1 − x and (b) f (x) = sin(πx). The appropriate boundary conditions are w(0) = x, (dw/dx) (0) = 0, (d2w/dx2) (1) = 0, and (d3w/dx3) (1) = 0. Use η = 4 and Λ = 2.

3.23. Determine the static defl ection of a fi xed-free beam on a Winkler foundation when a force is applied at the free end of the beam such that its defl ection is δ. The problem is (d4w/dx4) + ηw = 0 subject to w(0) = x, (dw/dx) (0) = 0, (d2w/dx2) (1) = 0, and w(1) = δ. Use η = 4. What is the value of the required force which is equal to (d3w/dx3) (1)?

3.24. Solve

ddr

rdwdr

w2 1

20( )+ =

subject to w (1) = 4.

3.25. Solve

ddr

rdwdr

w2 1

20( ) − =

subject to w (0) = 2 and w(1) = 4.

3.26. Solve

ddr

rdwdr

w2 0( )+ =


3.27. Solve

ddr

rdwdr

w2 3 0( )+ =

subject to w (1) = 2 and w (2) = −2.

3.28. Solve

rd wdr

rdwdr

w22

26 2 0− + =


9533_C003.indd 2059533_C003.indd 205 10/29/08 3:59:37 PM10/29/08 3:59:37 PM


3.29. Solve

rd wdr

rdwdr

w22

26 9 0− + =

subject to w (1) = −1 and w (2) = 0.

3.30. Solve

rd wdr

rdwdr

w22

26 10 0− + =

subject to w(0) = 3 and w(1) = 1.

3.31. Solve

rd wdr

rdwdr

w r22

26 10 2 3− + = +

subject to w(1) = 3 and

w(2) = 1. 3.32. Use the power series method to determine two linearly inde-

pendent solutions to Airy’s equation, ( / )d y dx xy2 2 0+ = . 3.33. Use the power series method to determine two linearly inde-

pendent solutions to ( / )d y dx y2 2 4 0+ = . 3.34. Chebyshev functions are solutions of the differential equation

( )1 022

22− − + =x

d ydx

dydx

n y (a)

a. When n is an integer, one solution of Equation a is a poly-nomial of order n. These are called Chebyshev polynomi-als. Determine the polynomial form of these solutions and show that the fi rst four Chebyshev polynomials can be defi ned by

T x T x x T x x T x x0 1 22

31 2 1 4( ) ( ) ( ) ( )= = = − = 33 3− x (b)

b. Show that the Chebyshev polynomials can also be repre-sented as

T x n xn( ) cos cos= ( )−1 (c)

c. Determine the second solution of Equation a for n = 0. d. Determine the second solution of Equation a for n = 3.

9533_C003.indd 2069533_C003.indd 206 10/29/08 3:59:37 PM10/29/08 3:59:37 PM


3.35. Hermite’s equation is

d ydx

xdydx

ny2

22 2 0− + = (a)

a. When n is an integer, one solution of Equation a is a polyno-mial of order n. These are called the Hermite polynomial, Hn (x). Determine the fi rst four Hermite polynomials.

b. Determine the second solution of Equation a for n = 0. c. Determine the second solution of Equation a for n = 3. 3.36. Use a series solution to obtain two linearly independent solu-

tions d2y/dx2 − 4y = 0. 3.37. Use the method of Frobenius to determine solutions of

xd ydx

dydx

xy2

23 0+ + =

3.38. Use the method of Frobenius to determine solutions of

xd ydx

xdydx

x y22

222 0− + =

3.39.–3.41. Determine the solution of each of the following in terms of Bes-sel functions:

3.39. d

dxx

dydx

x y3 24 0( )+ =

3.40. d

dxx

dydx

xy( )− =16 0

3.41. d ydx

x y2

222 0+ =

3.42. Evaluate the integral,

J r rdrn6

2

0

1

λ( )⎡⎣

⎤⎦∫

3.43. Evaluate the integral,

rJ ar dr0

0

1

( )∫

9533_C003.indd 2079533_C003.indd 207 10/29/08 3:59:38 PM10/29/08 3:59:38 PM

209

Chapter 4

Variational methods

4.1 IntroductionThe general form of an equation involving a linear operator L whose domain is D and whose range is R is

Lu = f (4.1)

where f is a vector in R and the solution u is in D. As noted in Chapter 3, one method of solving Equation 4.1 for an operator where D = R and there is a one-to-one correspondence between the elements of the domain and the elements of the range is to obtain the inverse of L and solve u = L−1f. Unfor-tunately, the inverse is not often readily attainable. Other methods of obtain-ing an exact solution to Equation 4.1 are often not successful because of the complexity of L. For example, it is diffi cult to obtain an exact solution of a linear differential equation with variable coeffi cients. Even when an exact solution is available in such cases, it is usually in terms of special functions with which subsequent numerical computations are diffi cult.

Because of this, approximate solutions of Equation 4.1 are often desired. There are three basic categories of approximation methods:

The operator L is approximated by an operator L whose domain and range are the same as L, and an exact solution u is obtained for the approximate equation

ˆ ˆLu f= (4.2)

For example, the differential equation for the vibrations of a beam with a slowly varying cross-section might be approximated by the differen-tial equation for a beam with a uniform cross-section. Obviously when the operator is approximated, u not the exact solution. Improvements to the approximation are often obtained using asymptotic expansions. If, when nondimensionalized, the problem formulation involves a small nondimensional parameter, often denoted by ε, the solution may be expanded into an asymptotic expansion in terms of linearly inde-pendent functions of ε.

A numerical solution is obtained when L is approximated by an opera-tor Ln, whose domain is Rn, and the solution is approximated at discrete

•

•

9533_C004.indd 2099533_C004.indd 209 10/29/08 4:35:35 PM10/29/08 4:35:35 PM


values of independent variables. If L is a differential operator with D = C[a,b], a numerical method is used to provide an approximation to u at a fi nite number of points, a ≤ x1 < x2 < … <xn−1 <xn ≤ b. The range of Ln is Rn.Methods which use the operator L but provide approximations for u at every value of x are called variational methods. Such methods are based on minimizing the distance between the exact solution and the approximate solution.

Variational methods are the topics of this chapter. The general theory behind variational methods is that of calculus of variations. A varied path is any alternate path that the solution can take from the exact solution while sat-isfying the same beginning and ending conditions. The true path is obtained by minimizing the energy between the true path and all alternate paths.

The variational methods discussed in this chapter are approximate meth-ods in which the choice of alternate paths is limited to those that are in the span of a fi nite-dimensional subspace of the vector space in which the true path resides. The best approximation for this subspace is the one in which the distance between the true solution and the approximate solution is min-imized, the distance being measured by an inner-product-generated norm.

The method of least squares uses the standard inner product defi ned for the vector space. If the operator of Equation 4.1 is self-adjoint and positive defi nite with respect to the standard inner product, then an energy inner product can be defi ned. The Rayleigh-Ritz method uses an energy-inner-product-generated norm to minimize the distance between the true solution and other approximate solutions.

It can be shown that when using the energy inner product, it is possible to ease the requirements on the vector space where the possible approximate solutions reside. Only geometric boundary conditions need to be directly satisfi ed by all elements in the vector space of approximate solutions. The continuity requirements on approximate solutions can be eased. These changes lead to approximate solutions being drawn from vector spaces of piecewise continuous functions and the development of a method called the fi nite-element method. This method is well known and extensively used in modeling engineering problems. In keeping with the objectives of this text, only the basic variational theory behind the fi nite-element method is pre-sented, along with several examples.

4.2 The general variational problemLet L be a linear operator defi ned on a domain D which is a subspace of a vector space V. Let (u,v) represent a valid inner product on D. Let Q be an n-dimensional subspace of D that is spanned by a set of vectors v1, v2, … , vn. The Gram-Schmidt process can always be used to generate an orthonormal

•

9533_C004.indd 2109533_C004.indd 210 10/29/08 4:35:36 PM10/29/08 4:35:36 PM

Chapter 4: Variational methods 211

basis for Q using v1, v2, … , vn. Thus, without loss of generality, it can be assumed that the basis is an orthonormal basis for Q.

It is desired to fi nd an approximation to u, the solution of Equation 4.1, from the vectors in Q. This concept is illustrated in Figure 4.1. The quality of the approximation must be assessed by the application of some standard. One possibility is to measure the length of the difference between the exact solution u and the approximate solution. An appropriate norm to use for measuring the length of this difference is an inner-product-generated norm. When the norm of the difference is smaller, the approximation is better. The best approximation to u from the vectors in Q is the vector q in Q such that ||u – q|| is a minimum. Defi ning

λ( )q = −u q (4.3)

the best approximation is the vector that minimizes λ.Since q is in Q, it can be represented by a linear combination of its basis

elements,

q vi==

∑α i

i

n

1

(4.4)

Substituting Equation 4.4 into Equation 4.3 and using the properties of the inner product assuming it only has real value

λ

α α

2 2

1 1

= − = − −

= − −= =

∑ ∑

u q u q u q

u v u vi i

( , )

( i

i

n

i

i

n

, ))

( , ) ( , ) ( , )= − +==

∑∑∑u u u v v vi i j211

α α αi i j

j

n

i

n

(4.5)

Vu

uaQ

u = exact solutionu = best approximation to u from Q

a

Figure 4.1 Variational methods determine the best approximation to a vector u, an element of a vector space V, from Q, a fi nite-dimensional subspace of V.

9533_C004.indd 2119533_C004.indd 211 10/29/08 4:35:37 PM10/29/08 4:35:37 PM


Since the basis is assumed to be orthonormal, (vi,vj) = δij, and the inner sum of the double sum in Equation 4.5 collapses to a single sum, yielding

λ α α2

1

2

1

= − += =

∑ ∑( , ) ( , )u u 2 u vi i

i

n

i

i

n

(4.6)

The right-hand side of Equation 4.6 is a function of the coeffi cients α1, α2, … ,α n. Thus, λ (α1, α2, … ,α n) is stationary when dλ = 0 or

∂∂

=∂∂

= =∂∂

=λα

λα

λα

2

12

2

22

2

20…

n (4.7)

Applying Equation 4.7 to Equation 4.6 leads to

∂∂

= +

=

λα

α

α

2

2 2k

k

k

( , )

( , )

u v

u vk

k (4.8)

Thus, the following theorem is proved.

Theorem 4.1 Fourier Best-Approximation Theorem: Let u be an element of a vector space V with a defi ned inner product (u,v). Let v1, v2, … , vn be an orthonormal basis for Q, a subspace of V. The best approximation to u from among all elements of Q, measured with respect to the inner-product- generated norm, is the vector

q u v vi i==

∑( , )i

n

(4.10)

The Fourier Best-Approximation Theorem provides an algorithm to deter-mine the best approximation to a vector u from a subspace, Q, of the vector space in which the vector resides. The term “best” is measured with respect to an inner-product-generated norm and is relative to the subspace where the approximation resides. Theorem 4.1 does not necessarily provide the “best” approximation available for u, only the best approximation to u from all the vectors in Q. Obviously the absolute “best” approximation to u is u itself. If u is in Q, then application of Theorem 4.1 will lead to resolution of u into its components in terms of the orthonormal basis.

The Fourier Best-Approximation Theorem has many uses. Generically, the application of the Fourier Best-Approximation Theorem is called the method of least squares, which is illustrated in the following examples.

Example 4.1 Find the best approximation to f (x) = x (1–e1–x) on the interval [0,1] with respect to the inner product on C[0,1] from the following subspaces, S, of C[0,1], (a) S is spanned by u1(x) = 1, u2(x) = sin(2πx), u3(x) = cos(2πx). Improve

9533_C004.indd 2129533_C004.indd 212 10/29/08 4:35:38 PM10/29/08 4:35:38 PM


the approximation by including additional basis functions of sin(2πix) and cos(2πix) for i = 2,3,…,50. Develop an expression for the error the approxima-tion. (b) S is spanned by ui(x) = sin(iπx) for i = 1,2,…, n. Develop the expressions for the expansion coeffi cients, (c) S is spanned by u1(x) = 1, u2(x) = x, u3(x) = x2, u4(x) = x3. Use an orthonormal basis for S to develop the approximation.

Solution MATHCAD is used to perform the calculations and develop the comparison graphs.

(a) The basis functions are orthogonal with respect to the standard inner product on C[0,1]. The normalized set of basis functions is v1(x) = 1, v2(x) = 2sin(2πx), v3(x) = 2cos(2πx). The coeffi cients in the approximation can be cal-culated using Equation 4.8 as:

α

α

11

0

1

21

1 0 218

1 2

= − = −

= −

−

−

∫ x e dx

x e

x

x

( ) .

( ) sinn( ) .

( ) cos( )

2 0 024

1 2 2

0

1

31

0

πx dx

x e x dxx

∫ = −

= − −α π

11

0 092∫ = .

(a)

The least-squares approximation for f(x) using this orthonormal basis is

q x x x( ) . . sin( ) . cos( )= − − +0 218 0 024 2 2 0 092 2 2π π (b)

A comparison of f(x) and q(x) is illustrated in Figure 4.2 The norm of the error

of the approximation is e f x q x dx= ∫ − =01 2 0 028( ( ) ( )) . . Much of the error

occurs near the ends of the interval. Only one basis function satisfi es the end conditions of f (x), f (0) = 0 and f (1) = 0.

(b) The basis functions are mutually orthogonal with respect to this inner product. The orthonormal basis functions are vi(x) = 2sin(iπx). The least-squares approximation is of the form of Equation 4.4 with

α

π

i i

x

f x v x dx

x e i x dx

=

= −

=

∫

∫ −

( ) ( )

( ) sin( )

0

1

1

0

1

1 2

πππ

π πi 2

13 1 2

2 2 22 2 2 2

( )( )( ) ( )

ii e ii

++ − − +⎡⎣ ⎤⎦ (c)

9533_C004.indd 2139533_C004.indd 213 10/29/08 4:35:39 PM10/29/08 4:35:39 PM


The least-squares approximation for f (x) using n basis functions is

q x i xn i

i

n

( ) sin( )==

∑α π21

(d)

A comparison of the least-squares approximations using n = 3, 5, and 10 is

shown below. The norm of the error is calculated has e f x q x dxn n= ∫ −0

1 2( ( ) ( )) .

Example 4.1

Part (a)

f(x) : = x ⋅ (1−e1−x)u1(x) : = 1u2(x) : = sin(2 ⋅ πx) Defi nition of basis functionsu3(x) : = cos(2 ⋅ π ⋅ x)

u x) u x dx

u x u x dx

u x u

1 2 0

1 3 0

2

0

1

0

1

( ( )

( ) ( )

( )

⋅ =

⋅ =

⋅

∫

∫33 0

0

1

( )x dx =∫

Demonstrating orthogonality with respect to standard inner product

0 0.2 0.4 0.6 0.80.4

0.3

0.2

0.1

0Least squares approximation for f(x)

f(x)

q(x)

x

Figure 4.2 The least-squares approximation to f(x) = x(1–e1–x) from a subspace spanned by u1(x) = 1, u2(x) = sin(2πx), u3(x) = cos(2πx) on [0,1] is illustrated. The plot was generated using MATHCAD.

9533_C004.indd 2149533_C004.indd 214 10/29/08 4:35:39 PM10/29/08 4:35:39 PM


c u x u x dx

c u x u x d

1 1 1

2 2 2

0

1

: ( ) ( )

: ( ) ( )

= ⋅ =

= ⋅

∫ c1 1

xx

c u x u x dx

0

1

0

1

3 3 3

∫

∫

=

= ⋅ =

: ( ) ( )

c2 0.707

c3 0..707

Calculating inner product generated norms

Developing the orthonormal basis

v (x)x

v2(x)x

11

1

2

2:

( ) :

( ) = =

u

c

u

c :

( )v3(x)

x=

u

c

3

3

Inner product evaluation

α α

α

1 1 1 0 218

2 2

0

1

: .

:

= ( )⋅ ( ) = −

= ( )⋅ ( )

∫ f x v x dx

f x v x dxx

f x v x dx

0

1

0

1

0 24

3 3 0 092

∫

∫

= −

= ( )⋅ ( ) =

α

α α

2

3

.

: .

Least squares Approximation

q(x) : = α1 ⋅ v1(x) + α2 ⋅ v2(x) + α3 ⋅ v3(x)

x : = 0,.02 .. 1

0 0.2 0.4 0.6 0.8 1–0.4

–0.3

–0.2

–0.1

0Least squares approximation for f(x)

f(x)

q(x)

x

9533_C004.indd 2159533_C004.indd 215 10/29/08 4:35:40 PM10/29/08 4:35:40 PM


Error computation

Error Error: ( ) ( )) ( ( ) ( )) = − ⋅ −∫ (f x q x f x q x dx0

1

0.028

v4(x) v5(x)

=

=⋅ ⋅

=⋅ ⋅

:sin( )

:cos(4

2

4π πx

c

x))

: ( ) ( ) .

: (

c

f x v x dx

f

3

4 4 3 127 10

5

0

13α α

α

= ⋅ = − ×

=

∫ −4

xx v x dx) ( ) .⋅ =∫ 5 0 0240

1

α5

g5(x) : = q(x) + α4 ⋅ v4(x) + α5 ⋅ v5(x)

Error Error: ( ( ) ( )) ( ( ) ( ))= − ⋅ −∫ f x g x f x g x dx5 50

1

== 0 014.

f(x)

q(x)

0 0.2 0.4 0.6 0.8 1x

–0.4

–0.3

–0.2

–0.1

0Comparison of 3 and 5 term approximation

g5(x)

v6(x)x

v7(x):sin( )

:c

=⋅ ⋅

=6

2

πc

oos( )

: ( ) ( )

6

2

6 60

1

⋅ ⋅

= ⋅∫

π

α

x

x x x

c

f v d αα

α

6

7 x x x

= − ×

= ⋅

−

∫

9 322 10

7

4

0

1

.

: ( ) ( ) f v d .α6 0 011=

g7(x) : = g5(x) + α6 ⋅ v6(x) + α7 ⋅ v7(x)

Error: = ( ( ) ( )) ( ( ) ( ))f x g x f x g x dx− ⋅ −∫ 7 70

1

Error = 8.411 × 10−3

9533_C004.indd 2169533_C004.indd 216 10/29/08 4:35:41 PM10/29/08 4:35:41 PM


0 0.2 0.4 0.6 0.8 1x

–0.4

–0.3

–0.2

–0.1

0Comparison with 3, 5, and 7 terms

f(x)

q(x)

g5(x)

g7(x)

gh(x) : x x x x= ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅f n d n( ) sin( ) sin( )2 2 2 2π π00

1

1

50

∫∑⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

+ ⋅

=n

f ( )x 22 2 2 20

1

⋅ ⋅ ⋅ ⋅( ) ⋅ ⋅ ⋅ ⋅⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟∫ cos cos( )n dπ πx x n x ⎟⎟⎟⎟+

=∑n 1

50

1α

201 term approximation

–0.4

–0.3

–0.2

–0.1

0

f(x)

gh(x)

0 0.2 0.4 0.6 0.8 1x

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Part (c): Polynomial approximation

u1(x) : = 1u2(x) : = x Basis functionsu3(x) : = x2

u4(x) : = x3

Gram-Schmidt Orthonormalization Process

c1 : u x u x dx c1 1 Norm^2 of u1= ⋅ =∫ 1 10

1

( ) ( )

v1(x) :u x

c=

1

1

( ) v1

b1 : u x v x dx b1 0.5= ⋅ =∫ 2 10

1

( ) ( ) (u2, v1)

w2(x) : = u2(x) − b1 ⋅ v1(x) w2 = u2−(u2, v1)v1

c2 : w x w x dx c2 0.083 Norm^2 of w2= ⋅ =∫ 2 20

1

( ) ( )

v2(x) :w x

c=

2

2

( ) v2

b2 : u x v x dx b2 0.333= ⋅ =∫ 3 10

1

( ) ( ) (u3, v1)

b3 : b3 0.289= ⋅ =∫ u x v x dx3 20

1

( ) ( ) (u3, v2)

w3(x) : = u3(x)−b2 ⋅ v1(x)−b3 ⋅ v2(x) w3 = u3−(u3, v1)v1−(u3, v2)v2

c3 : c3 5.556 10 3= ⋅ = ×∫ −w x w x dx3 30

1

( ) ( ) Norm^2 of w3

v3(x) : =w x

c

3

3

( )

b4 : b4 0.25= ⋅ =∫ u x v x dx4 10

1

( ) ( ) (u4, v1)

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b5 : b5 0.026= ⋅ =∫ u x v x dx4 20

1

( ) ( ) (u4, v2)

b6 : b6 0.112= ⋅ =∫ u x v x dx4 30

1

( ) ( ) (u4, u3)

w4(x) : = u4(x)−b4 ⋅ v1(x)−b5 ⋅ v2(x)−b6 ⋅ v3(x) w4 = u4−(u4, v1)v1 −(u4, u2)v3−(u4, u3)v3

c4 : = ⋅∫ w x w x dx4 40

1

( ) ( ) Norm^2 of w4

v4(x): =w x

c

4

4

( ) v4

v1, v2, v3 and v4 form an orthonormal basis.

Check orthogonality

v x v x dx4 1 3 089 10 15

0

1

( ) ( ) .⋅ = × −∫ Small non-zero value is round-off error

v x v x dx3 2 00

1

( ) ( )⋅ =∫

Orthogonal Polynomials

v1(x)

v2(x)

v3(x)

v4(x)

–4

–2

0

2

4

0 0.2 0.4 0.6 0.8 1x

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Least squares approximation

f (x) : = ⋅ − −x e x( )1 1 Given function

Linear Approximation

β β1 1 1 0 2180

1

: ( ) ( ) .= ⋅ = −∫ f x v x dx (f, v1)

β β2 2 2 0 020

1

: ( ) ( ) .= ⋅ =∫ f x v x dx (f, v2)

gp2(x): = ⋅ + ⋅β β1 1 2 2v x v x( ) ( ) Linear least squares approximation

Error : Erro= −( )⋅ −( )∫ f x gp x f x gp x dx( ) ( ) ( ) ( )2 20

1

rr 0.097= Norm of error of lin-ear approxi-mation

Quadratic Approximation

β β3 3 3 0 0960

1

: ( ) ( ) .= ⋅ =∫ f x v x dx (f, v3)

gp3(x) : = gp2(x) β3 3⋅ ( )v x Quadratic least squares approximation


1

rr 0.013= Norm of

error of quadratic approxi-mation

Cubic Approximation

β β4 4 4 0 0130

1

: ( ) ( ) .= = −∫ f x v x dx (f, v4)

gp4(x) : = gp3(x) β4 4⋅ v x( ) Cubic least squares approximation


1

rr 0.013= Norm

of error of cubic approxima-tion

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Least squares approximation with polynomial basis

f(x)

gp2(x)

gp3(x)

gp4(x)

–0.4

–0.3

–0.2

–0.1

0.1

0

0 0.2 0.4 0.6 0.8 1x

(c) The basis functions in part (b) are not orthogonal, and therefore the Gram-Schmidt procedure is used fi rst to determine an orthogonal basis. The MATHCAD fi le shown above is set up so that the effect of adding each additional term is evident. The linear approximation is very poor whereas the cubic approximation is excellent.

Example 4.2 An engineer runs a test on a centrifugal pump to determine the head developed at a constant speed, but for varying fl ow rate. The data in Table 4.1 are obtained at 2000 rpm.

The engineer wants to use these data to generate a performance curve by fi tting the “best” parabola through the data. Show how the least-squares method can be used to do this.

Solution: It is desired to fi t the best parabola through eight data points. In terms of the previous discussion, it is desired to fi nd the best approximation to an assumed continuous function, h(Q), on C[0,2800] from P2[0,2800].

Table 4.1 Head of a Centrifugal Pump Measured for Varying Flow Rate

Q (gpm) h (ft) Q (gpm) h (ft)

0 115 1600 112

400 122 2000 105

800 122 2400 96

1200 118 2800 87

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Consider the following basis for P2[0,2800]: u1(x) = 1, u2 (x) = x, u3(x) = x2. The standard inner product for C[0,2800] is

( , ) ( ) ( )f g f x g x dx= ∫0

2800

(a)

However, h(Q) is not explicitly known for all values of Q in [0,2800]. It is known only at eight discrete values of Q. A possible alternative to the inner product defi ned by Equation a is

( , ) ( ) ( ) ( ) ( ) ( ) ( ) (f g f g f g f g f= + + +0 0 400 400 800 800 11200 1200

1600 1600

) ( )

( ) ( ) (

g

f g f + + 22000 2000

2400 2400

) ( )

( ) ( ) (

g

f g f + + 22800 2800) ( )g (b)

The inner product suggested by Equation b is not a true inner product for C[0,2800]. Consider the function

f x x x x x x( ) ( )( )( )( )= − − − −400 800 1200 1600

( )( )( )x x x− − −2000 2400 2800 (c)

The function defi ned by Equation c is in C[0,2800]. Using the inner-product defi nition of Equation b, ( f,f ) = 0. Thus, there exists f ≠ 0 such that ( f,f ) = 0. The inner product proposed by Equation b violates property (iv) of the inner-product defi nition. Interestingly, the inner product of Equation b is a valid inner-product defi nition for P2[0,2800], because the only quadratic polyno-mial which can be zero for x = 0, 400, 1200, … , 2800 is the zero polynomial. Since h(Q) is not known exactly, it can be assumed that h(Q) is actually a member of a space of functions that cannot have eight zeros on [0,2800] with-out the function itself being the zero function.

An alternate interpretation is that Equation b could be obtained from Equation a through application of some quadrature formula. In this case, it would be possible to think of h(Q) as a member of R8 and Equation b as the standard inner product on R8.

The fi rst step is to apply the Gram-Schmidt procedure to determine an orthonormal basis for P2[0,2800] with respect to the inner product defi ned in Equation b. To this end,

u

v xuu

1

1

2

11

1

1 1 8

1

8

= =

= =

( , )

( ) (d)

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w x u x u v v

x x

x

2 2 2 1 1

1

2

1

28 8

1

80

( ) ( ) ( , )

( , )

(

= −

= −

= −

− −

))( ) ( ) ( ( ) ( )( ) ( ( )

0 400 1 800 1 1200 1 1600 1+ + + +[

+(( )( ) ( )( ) ( )( ) ( )2000 1 2400 1 2800 1 1

1400

+ + ]

= −x (e)

w x x x2

1

21400 1400

1400 1400 10

( ) ( , )

( )( ) (

= − −

= − − + − 000 1000 600 600

200 200 200

)( ) ( )( )

( )( ) (

−[ + − −

+ − − + ))( ) ( )( )

( )( ) ( )(

200 600 600

1000 1000 1400 1400

+

+ + ))

.

]

=

1

2

2592 3 (f)

v xw xw x

xx2

2

2

41400

2592 33 86 10 5( )

( )

( ) .. .= =

−= × −− 441 10 1× − (g)

w x u x u v v u v v

x x

3 3 3 1 1 3 2 2

2 2 3 86 10

( ) ( )( , ) ( , )

( , .

=

= × −− −

− −

− ×

× − ×

4 1

4

5 41 10

3 86 10 5 41 10

x

x

. )

( . . 11 2

33

3

7 2

1

8

1

8

4 82 10 1

) ( , )

( )( )

( ).

−

= = × −−

x

v xw xw x

x .. .35 10 0 543× +− x (h)

The least-squares approximation is

q x h v v h v v h v v( ) ( , ) ( , ) ( , )

.

= + +

= − × −

1 1 2 2 3 3

67 55 10 xx x2 2 21 03 10 1 16 10+ × + ×−. . (i)

Equation i is illustrated in Figure 4.3 along with the data. The norm of the error is calculated as

e x q h q h q h( ) ( ) ( ) ( ) ( ) ( )= −[ ] + −[ ] + −0 0 400 400 8002 2

(( )

( ) ( ) ( ) (

800

1200 1200 1600 160

2

2

[ ]{

+ −[ ] + −q h q h 00 2000 2000

2400 2400

2 2) ( ) ( )

( ) ( )

[ ] + −[ ]

+ −[ ]

q h

q h 22 21

22800 2800

4 91

+ −[ ] }=

q h( ) ( )

.

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4.3 Variational solutions of operator equationsThe Fourier Best-Approximation Theorem was applied in the previous sec-tion to approximate known vectors. These approximations required evalua-tion of inner products of the exact vector using members of a predetermined basis. However, the approximation of the solution u of an equation of the form Lu = f, where L is a linear operator, is more diffi cult because u is not known and the inner products of the exact solution with the basis vectors cannot be easily determined.

Two methods of approximating solutions of Lu = f are presented in this section. The fi rst method, the method of least squares, provides an approxi-mation to f, while the second, the Rayleigh-Ritz method, is applied to approx-imate solutions to equations in which the operator is positive defi nite and self-adjoint.

4.3.1 Method of least squares

As usual, let L be a linear operator whose domain is a vector space D and whose range is R. An inner product (u,v) is defi ned for all vectors in R. Let Q be a fi nite-dimensional subspace of D with a (not necessarily orthonormal) basis q1, q2, …, qn. An arbitrary element of Q is of the form

0 500 1000 1500 2000 2500 300085

90

95

100

105

110

115

120

125

Q (gpm)

h (ft

)

Centrifugal pump performance curve

test dataleast squares regression

Figure 4.3 A least-squares regression on the data in Table 4.1 is performed using quadratic polynomials.

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q qi i

i

n

==

∑β1

(4.11)

The least-squares approximation provides the approximation to q such that Lq best approximates f. Thus, it is necessary to minimize μ(q) = ||Lq − f||2. Using Equation 4.10 and the properties of inner products,

μ( ) ( , )

( , ) ( , ) ( , ) ( ,

q Lq f Lq f

Lq Lq f Lq Lq f f f

= − −

= − − + ))

,=⎡

⎣⎢⎢⎢

⎤

⎦⎥⎥⎥

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥

= =∑ ∑L q L qi i

i

n

j j

j

n

β β1 1 ⎥⎥

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟−

⎡

⎣⎢⎢⎢

⎤

⎦⎥⎥⎥

⎛

⎝ =∑2

1

L q fi i

i

n

β ,⎜⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟+ ( , )f f (4.12)

Using the linearity of L and further properties of inner products, Equation 4.12 reduces to

μ β β β( ) ( , ) ( , ) ( ,q Lq Lq Lq f fi j i j

j

n

i

n

i i= − +==

∑∑11

2 ffi

n

)=

∑1

(4.13)

Minimization of μ(q) is achieved by requiring that ∂ ∂ = =μ β/ , , , .k k n0 1 2 … To this end,

0 211

=∂

∂

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥−

∂∂

==∑∑β

β βk

i j i j

j

n

i

n

Lq Lq( , )ββ

βk

i i

i

n

i j

j

n

i

Lq f

Lq Lq

( , )

( , )

=

=

∑

∑

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

=

1

1== =∑ ∑∂

∂−

∂∂

=

1 1

2

n

ki j i

j

ni

k

i

Lq f

Lq Lq

ββ β

ββ

( ) ( , )

( , jj

j

n

i

n

j ik i jk i ik

i

n

Lq f)( ) ( , )== =

∑∑ ∑+ −11 1

2β δ β δ δ

== + −= =

∑( , ) ( , ) ( , )Lq Lq Lq Lq Lq fj k j

j

n

i k i k

i

n

β β1 1

2∑∑

∑= −=

2 21

( , ) ( , )Lq Lq Lq fi k i

i

n

kβ (4.14)


( , ) ( , )Lq Lq Lq fi k i

i

n

kβ=

∑ =1

(4.15)

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Equation 4.15 represents a set of simultaneous linear algebraic equations whose solutions are the coeffi cients in the linear combination. The least-squares method fi nds the best approximation to u, interpreted as the vector in Q which minimizes the norm of the difference between Lq and f.

4.3.2 Rayleigh-Ritz method

Recall that if L is a positive defi nite and self-adjoint operator with a defi ned inner product (u,v), then an energy inner product and a corresponding energy-inner-product-generated norm are defi ned by

( , ) ( , )u v Lu vL = (4.16)

u u uL L= ( , ) /1 2 (4.17)

The Rayleigh-Ritz method fi nds the best approximation to u, using the energy-inner-product-generated norm to measure the distance between two vectors. That is, v(q) = ||u – q||L2 is the minimization functional. It should be noted that:

ν( ) ( , )

( ( ), )

( , )

(

q u q u q

L u q u q

Lu Lq u q

L

L= − −

= − −

= − −

= uu u Lu q Lq u Lq q

Lu u Lu q

, ) ( , ) ( , ) ( , )

( , ) ( , ) (

− − +

= − +2 LLq q, )

(4.18)

Noting that for the exact u, Lu = f, substituting for q from Equation 4.11, and applying properties of the inner product leads to

ν β β β( ) ( , ) ( , ) ( , )q f u f q Lq qi i

i

n

i j i j

j

n

= − += =

∑ ∑21 1ii

n

i i

i

n

i uj i j L

j

f u f q q q

=

=

∑

∑= − +

1

1

2( ), ( , ) ( , )β β β===

∑∑11

n

i

n

(4.19)

Minimization of ν(q) requires that ∂ν/∂βk = 0 for all k = 1, 2, … ,n. The minimi-zation procedure is similar to that performed in Equation 4.14 and leads to a set of simultaneous linear algebraic equations to solve for the coeffi cients in the linear combination that leads to the best Rayleigh-Ritz approximation for u. The resulting equations can be summarized by

( , ) ( , ) , , ,q q f q k ni k L

i

n

i k

=∑ = =

1

1 2β … (4.20)

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If the basis vectors are chosen to be an orthonormal set with respect to the energy inner product, then (qi,qk)L = δik, and Equation 4.20 simplifi es to

βk kf q k n= =( , ) , , ,1 2 … (4.21)

Example 4.3 The differential equation for the transverse displacement, w(x), of a uniform beam on an elastic foundation with vertical loading is

EId wdx

kw f x4

4+ = ( ) (a)

where E is the elastic modulus of the material from which the beam is made, I is the moment of inertia of the cross-section about the beam’s neutral axis, k is the stiffness per unit length of the elastic foundation, and f(x) is the load per unit length. Equation a is nondimensionalized through introduction of x* = x/L, w* = w/L and defi nition of f * (x*) = f(Lx)/Fmax, where Fmax is the maxi-mum value of the vertical load. The resulting nondimensional equation is

d wdx

w f x4

4+ =φ λ ( ) (b)

where, as is customary, the *s have been dropped from nondimensional vari-ables and it is understood that all variables and functions in Equation a are nondimensional. The nondimensional parameters are defi ned as

φ =kLEI

3

(c)

λ =F L

EImax

3

(d)

Consider a beam that is fi xed at both ends and is loaded such that

f x x x( ) = − 2 (e)

Equation b is written in the form of Equation 4.1 with Lw = d4w/dx4 + φw. The domain of L, D, is the subspace of C4[0,1] such that all elements of D satisfy the boundary conditions of a fi xed-fi xed beam. It can be shown that L is self-adjoint and positive defi nite on D with respect to the standard inner product for C4[0,1].

Let P 6[0,1] be the space of all polynomials of degree six or less defi ned on [0,1], and let Q be the intersection of D with P 6[0,1]. A basis for Q was deter-mined in Example 3.7 as

u x x x x16 3 24 3( ) = − + (f)

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u x x x x25 3 23 2( ) = − + (g)

u x x x x34 3 22( ) = − + (h)

a. Determine the exact solution for w(x) for φ = 0 and φ = 1. Assume that λ = 1.

b. Determine the best least-squares approximation from Q for w(x) for φ = 1. Use λ = 1 and compare with the exact solutions.

c. Determine the best Rayleigh-Ritz approximation for w(x) from Q, using the basis defi ned in Equation f, Equation g, and Equation h, for φ = 1. Use λ = 1 and compare with the exact solutions.

d. Determine the best Rayleigh-Ritz approximation from Q for w(x) using an orthonormal basis for Q.

Solution: (a) For φ = 0, Equation b reduces to

d wdx

x x4

4

2= − (i)

Equation i is easily integrated four times with respect to x, leading to

w x C C x C x C xx x

( ) = + + + + −1 2 32

43

5 6

120 360 (j)

The constants of integration are obtained by application of the boundary conditions

w C( )0 0 01= ⇒ = (k)

dwdx

C( )0 0 02= ⇒ = (l)

w C C( )1 01

603 4= ⇒ + = − (m)

dwdx

C C( )1 0 2 31

403 4= ⇒ + = − (n)

Equation m and Equation n are solved simultaneously, leading to C3 = –1/40 and C4 = 1/120. Thus,

w xx x x x

( ) = − + +2 3 5 6

40 120 120 360 (o)

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For φ = 1, Equation b becomes

d wdx

w x x4

4

2+ = − (p)

The homogeneous solution of Equation p is obtained by assuming a solu-tion of the form wh(x) = eαx, which leads to α4 + 1 = 0, whose solutions are

2 2 2 2/ /± ( )i and − ±2 2 2 2/ / ( )i . The particular solution is readily obtained as wp(x) = x − x2. Hence the general solution of Equation p is

w x C x x C x x( ) cos cosh cos sinh

= +1 2μ μ μ μ

++ + + −C x x C x x x x3 42sin cosh sin sinhμ μ μ μ (q)

where μ = 2 2/ . Application of the initial conditions leads to

w C( )0 0 01= ⇒ = (r)

dwdx

C C( )0 01

2 3= ⇒ + = −μ

(s)

w C C C( ) cos sinh sin cosh sin sinh1 0 2 3 4= ⇒ + + =μ μ μ μ μ μ 00 (t)

dwdx

C( ) cos cosh sin sinh

1 0 2= ⇒ −( )μ μ μ μ

sin sinh cos cosh+ +( )C3 μ μ μ μ

++ +( )=C41

sin cosh cos sinhμ μ μ μμ

(u)

Equation s, Equation t, and Equation u are solved simultaneously yielding C2 = −0.648, C3 = −0.766, and C4 = 2.017.

MATHCAD is used to provide solutions for parts (b)–(d). Since the exact solution of Equation b is an element of P6[0,1] and D for φ = 0, it is an element of Q. Thus, the best approximation to u from the elements of Q is u itself. Both the least-squares approximation and the Rayleigh-Ritz approximation are identical to u for φ = 0.

The results for φ = 1 are presented in the MATHCAD fi le for Example 4.3. These results indicate excellent agreement among the least-squares approximation, the Rayleigh-Ritz approximation, and the exact solution for φ = 1.

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Example 4.3 Least squares and Rayleigh-Ritz approximations for the dis-placement of a fi xed-fi xed beam on an elastic foundation

Beam properties

f(x) : = x − x2 Non-dimensional load per length

φ : = 1 Ratio of elastic stress to bending stress

λ : = 1 Ratio of stress due to applied load to bending stress

Basis functions

u1(x) : = x6 − 4x3 + 3x2

u2(x) : = x5 − 3x3 + 2x2 Basis functionsu3(x) : = x4 − 2x3 + x2

(a) Least squares method

Lu1(x) :

Lu2(x) :

= + ⋅

=

d

dxu x u x

d

dxu x

4

4

4

4

1 1

2

( ) ( )

(

φ

)) ( )

( ) ( )

+ ⋅

= + ⋅

φ

φ

u x

d

dxu x u x

2

3 34

4Lu3(x) :

Operator acting on basis functions

Evaluation of inner products to calculate coeffi cient matrix

A(i,j) = (Lui,Luj)

A Lu x Lu x dx A Lu x Lu x1 1

0

1

1 21 1 1 2, ,: ( ) ( ) ( ) ( )= ⋅ = ⋅∫ : ddx

A Lu x Lu x dx A Lu x

0

1

1 3

0

1

2 21 3 2

∫

∫= ⋅ = ⋅, ,: ( ) ( ) : ( ) LLu x dx

A Lu x Lu x dx A L

2

2 3

0

1

2 3

0

1

3 3

( )

: ( ) ( ) :, ,

∫

∫= ⋅ = uu x Lu x dx3 30

1

( ) ( )⋅∫Symmetry is used to calculate remaining elements

A A A A A A2 1 1 2 3 1 1 3 3 2 2 3, , , , , ,: : := = =

Coeffi cient matrix of Eq. (4.15)

A =× × ×

× ×2 592 10 1 08 10 2 88 10

1 08 10 4 8 10 1

4 4 3

4 3

. . .

. . ..

. .

44 10

2 88 10 1 44 10 576

3

3 3

×× ×

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

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Calculation of right-hand side vector of Eq. (4.15)

b(i) = (f,Lui)

b f x Lu x dx b f x Lu x dx b1

0

1

2

0

1

31 2: ( ) ( ) : ( ) ( ) := ⋅ = ⋅∫ ∫ == ⋅∫ Lu x f x dx30

1

( ) ( )

b =

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜

18

10

4⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

Solution of simultaneous equations to determine coeffi cients of least squares approximation

Solution is simply obtained by multiplying the inverse of A by b leading to

β

β

:

.

.

.

= ⋅

=− ×

×− ×

−

−

−

−

A b1

3

3

2 778 10

8 333 10

1 341 10 110

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟

The resulting least squares approximation is

w x u x u x u x( ) : ( ) ( ) ( )= ⋅ + ⋅ + ⋅

=

β β β1 2 31 2 3

x : 0,.02 .. 1

0 0.2 0.4 0.6 0.8 1

6 × 10–4

4 × 10–4

2 × 10–4

0

Least squares approximation

w(x)

x

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(b) Rayleigh-Ritz method

Energy inner product evaluations

C(i,j) = (ui,uj)L

C u x Lu x dx C Lu x u x dx1 1

0

1

1 21 1 1 2, ,: ( ) ( ) : ( ) ( )= ⋅ = ⋅∫00

1

1 3

0

1

2 21 3 2 2

∫

∫= ⋅ = ⋅C Lu x u x dx C Lu x u, ,: ( ) ( ) : ( ) (xx dx

C Lu x u x dx C Lu x

)

: ( ) ( ) : ( ), ,

0

1

2 3

0

1

3 32 3 3

∫

∫= ⋅ = ⋅⋅

= ⋅ =

∫

∫

u x dx

C Lu x u x dx C Lu

3

2 1

0

1

2 1

0

1

3 1

( )

: ( ) ( ) :, , 33 1

3 2

0

1

3 2

0

1

( ) ( )

: ( ) ( ),

x u x dx

C Lu x u x dx

⋅

= ⋅

∫

∫

C =

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟

16 9 3 429

9 5 143 2

3 429 2 0 8

.

.

. . ⎟⎟⎟ Symmetry of C is due to self-adjointness of L

Evaluation of right-hand side vector of Eq. (4.20)

d(i) = (f,ui)

d1

0

1

2

0

1

31 2: ( ) ( ) : ( ) ( ) := ⋅ = ⋅ =∫ ∫f x u x dx d f x u x dx d f(( ) ( )

.

.

.

x u x dx⋅

=×

⎛

⎝

⎜⎜⎜⎜

∫

−

3

0 031

0 018

7 143 10

0

1

3

d⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟

Solution for Rayleigh-Ritz coeffi cients

α α:

.

.

.

= ⋅ =− ×

××

⎛−

−

−

−

C d1

3

3

10

2 778 10

8 333 10

1 491 10⎝⎝

⎜⎜⎜⎜⎜⎜⎜⎜

Rayleigh-Ritz approximation

wrr(x) : ( ) ( ) ( )= ⋅ + ⋅ + ⋅α α α1 2 31 2 3u x u x u x

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6 × 10–4

4 × 10–4

2 × 10–4

0

wrr(x)

0 0.2 0.4 0.6 0.8 1

x


(c) Rayleigh-Ritz method using on orthonormal basis with respect to energy inner productThe Gram Schmidt process is used to develop an orthonormal basis with respect to the energy inner product which spans the same space as u1, u2, and u3.

w1(x) := u1(x)

w n Lu x u x dx1 1 10

1

: ( ) ( )= ⋅∫ Energy norm of w1

v1(x) : =

w x

w n

1

1

( ) v1

ip1 : = ⋅∫ Lu x v x dx2 10

1

( ) ( ) (u2,v1)L

w x u x ip v x2 2 1 1( ) : ( ) ( )= − ⋅ w2(x)=u2-*u2,v1)L*v1

Lw2(x) := + ⋅d

dxw x w x

4

42 2( ( )) ( )φ L(w2)

w2n(x) : = ⋅∫ Lw x w x dx2 20

1

( ) ( ) Energy norm of w2

v2(x) : =w x

w n

2

2

( ) v2

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ip2 := ⋅∫ Lu x v x dx3 10

1

( ) ( ) (u3,v1)L

ip3 := ⋅∫ Lu x v x dx3 20

1

( ) ( ) (u3,v2)L

w x u x ip v x ip v x w u u v3 3 2 1 3 2 3 3 3( ) : ( ) ( ) ( ) ( ,= − ⋅ − ⋅ = − 11 1 3 2 2) * ( , ) *L v u u L v−

Lw3(x) := + ⋅d

dxw x w x

4

43 3( ) ( )φ Lw3

w3n := ⋅∫ Lw x w x dx3 30

1

( ) ( ) Energy norm of w3

v3(x) :=w x

w n

3

3

( ) v3

Inner product evaluation for coeffi cients

δ1 10

1

: ( ) ( )= ⋅∫ f x v x dx (f,v1)

δ2 20

1

: ( ) ( )= ⋅∫ f x v x dx (f,v2)

δ3 30

1

: ( ) ( )= ⋅∫ f x v x dx (f,v3)


wgs(x) := ⋅ + ⋅ + ⋅δ δ δ1 1 2 2 3 3v x v x v x( ) ( ) ( )

6 × 10–4

4 × 10–4

2 × 10–4

00 0.2 0.4 0.6 0.8 1

x

Comparison of Rayleigh-Ritz methods

wgs(x)

wrr(x)

9533_C004.indd 2349533_C004.indd 234 10/29/08 4:35:52 PM10/29/08 4:35:52 PM


Exact Solution φ =( )1

μ μ

μ μ

: .= =

= =

=

2

20 707

ct : cos( ) st : sin( )

ch : coosh( ) sh : sinh ( )μ μ=

Defi nitions made for convenience

E := ⋅ ⋅ ⋅⋅ − ⋅ ⋅ + ⋅

1 1 0

ct sh st ch st sh

ct ch st sh st sh ct ch sst ch ct sh⋅ + ⋅

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟

=E

1 1 0

0 584 0. .8819 0 499

0 46 1 457 1 402

.

. . .

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟

e ::=−⎛

⎝

⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⋅

1

0

1

1μ

CEX : CEX = ⋅ =

−−

⎛

⎝

⎜⎜⎜⎜⎜⎜−E e1

0 648

0 766

2 017

.

.

.⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟

w1(x) := CEX1 · cos(μ · x) · sinh(μ · x) + CEX2 · sin(μ · x) · cosh(μ · x)

wexact( ): ( ) sin( ) sinh( )x w x CEX x x x= + ⋅ ⋅ ⋅ ⋅ + −1 3 μ μ xx2

6 × 10–4

4 × 10–4

2 × 10–4

–2 × 10–4

0

0 0.2 0.4 0.6 0.8 1x

wexact(x)

wrr(x)

Exact solution and RR Approx for phi=1

9533_C004.indd 2359533_C004.indd 235 10/29/08 4:35:52 PM10/29/08 4:35:52 PM


The least-squares approximation and the Rayleigh-Ritz approximation are often close to one another, as illustrated in Example 4.3, when the approxi-mations are drawn from the same vector space. However, as shown below, the Rayleigh-Ritz method is more fl exible regarding the vector spaces which can be used to generate an approximation.

Consider the second-order differential operator defi ned by Lf = –d2f/dx2, where f is an arbitrary element of D, the domain of the operator. Consider the standard inner product on D defi ned by ( , ) ( ) ( )f g f x g x dx= ∫ 0

1. Assum-

ing that D is defi ned such that L f L is self-adjoint and positive defi nite with respect to the standard inner product, an energy inner product is defi ned by( , ) ( , ) ( )f g Lf g d f dx gdxL = = ∫ −0

1 2 2/ . Integration by parts on the energy inner product leads to

( , ) ( ) ( ) ( ) ( )Lf g gdfdx

gdfdx

dfdx

dgdx

dxL = − + +1 1 0 0

00

1

∫ (4.22)

Let q1,q2, … ,qn be a set of basis functions chosen for a Rayleigh-Ritz approximation for the solution Lu = f. The approximation requires evalua-tion of energy inner products of the form (qi, qj)L , and since L is self-adjoint, (qi, qj)L = (qi, qj)L. To this end, using Equation 4.22,

( ) ( ) ( ) ( ) ( ),q q qdqdx

qdq

dxdqdxi j L j

ij

j i= − + +1 1 0 0ddq

dxdxj

0

1

∫ (4.23)

( ) ( ) ( ) ( ) ( ),q q qdq

dxq

dq

dx

dq

dxj i L ij

ij j= − + +1 1 0 0

ddqdx

dxi

0

1

∫ (4.24)

The energy inner products are so named because of their relationship to some form of energy in a physical system. For the second-order operator being considered, if u represents the displacement of an elastic bar, then (u,u)L = 2V, where V is the stored strain energy. If u represents the temperature in a conductive heat transfer problem, then (u,u)L represents an internal energy. The stored strain energy in an elastic bar is

Vdudx

dx= ( )∫1

2

2

0

1

(4.25)

Setting g = f = u in Equation 4.22 leads to

( , ) ( ) ( ) ( ) ( )u u ududx

ududx

dudxL = − + +

⎛⎝⎜⎜⎜

⎞⎠

1 1 0 0 ⎟⎟⎟⎟⎟∫2

0

1

dx (4.26)

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Comparison of Equation 4.25 and Equation 4.26 shows that

( , ) ( ) ( ) ( ) ( )u u V u

dudx

ududx

L = − + =2 1 1 0 0 0if

Consider a bar with a discrete spring of nondimensional stiffness k attached at x = 1. When the end of the bar is displaced, a potential energy develops in the spring. In this case, the total potential energy of the system is the sum of the strain energy and the potential energy in the spring,

Vdudx

dx ku=⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟ +∫1

2

1

21

2

2

0

1

( ) (4.27)

Noting that the boundary condition for the displacement of the bar with a discrete spring at its end is du/dx (1) = –ku(1) Equation 4.26 becomes

( , ) ( ) ( ) ( )u u ududx

kududx

dL = + +⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟0 0 1 2

2

xx0

1

∫ (4.28)

and comparison of Equation 4.27 and Equation 4.28 leads to u(0) or du/dx (0) = 0.The boundary condition applied to the end of the bar at x = 1, whether the

bar is free or has an attached spring, is a natural boundary condition. The boundary condition applied to the fi xed end of the bar is a geometric bound-ary condition.

The above discussion leads to the consideration of several points regard-ing the application of the Rayleigh-Ritz method. The Rayleigh-Ritz method may be interpreted as a method which minimizes the difference between the system’s energy for the exact solution and the energy for the approximation. Interpreted in this light, using Equations 4.22–4.28, it appears that the restric-tions on the choices of the basis functions can be relaxed. Evaluation of energy inner products using Equation 4.22 only requires that the basis functions be differentiable, not twice differentiable as required by the exact solution of the differential equation. Furthermore, realizing that, for the cases discussed, the potential energy of the system is independent of the value of du/dx(1), it is not necessary that the basis functions satisfy the boundary condition at x = 1 as long as the potential energy is still represented by Equation 4.27. Under these conditions, an alternate representation of the energy inner product is

( , ) ( ) ( )f g kf gdfdx

dgdx

dxL = +∫1 1

0

1

(4.29)

The above suggests that the requirements for the basis functions used in a Rayleigh-Ritz approximation can be relaxed from requiring that all basis functions be in D, the domain of L. A basis chosen from D is called a set of

9533_C004.indd 2379533_C004.indd 237 10/29/08 4:35:54 PM10/29/08 4:35:54 PM


comparison functions. The functions used in Example 4.3 are a set of com-parison functions. Use of Equation 4.29 for the energy inner product requires only that the basis functions be differentiable, without having to satisfy the natural boundary condition at x = 1. The chosen basis functions must still satisfy the geometric boundary condition. Functions that satisfy such condi-tions are called admissible functions.

The general requirements for admissible functions are that (1) admissible functions satisfy all geometric boundary conditions, and (2) admissible func-tions must have a level of differentiability required by the energy formu-lation. Admissible functions need not satisfy natural boundary conditions, but when using admissible functions that do not satisfy natural boundary conditions, the energy inner product must include, if appropriate, terms to account for nonzero energy from the natural boundary conditions.

If A is the space of admissible functions, then Equation 4.29 is a valid defi nition of an inner product for elements of A. That is, the inner product defi ned in Equation 4.29 satisfi es the four properties required for inner prod-ucts specifi ed in Defi nition 3.7. The inner product of Equation 4.29 is then used for determination of the elements of the coeffi cient matrix, (qi,qj)L, used in the Rayleigh-Ritz approximation.

Example 4.4 The differential equation governing the displacement, w(x), of a beam of variable cross-sectional moment of inertia I(x) due to a uniform distributed load per unit length, f(x), is

d

dxEI

d wdx

f x2

2

2

2

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟= ( ) (a)

Consider a beam fi xed at x = 0 with a spring of stiffness k attached at x = L. (a) Determine the appropriate formulation of the energy inner product to use for a Rayleigh-Ritz approximation. (b) Find the best approximation to w(x) from the space of polynomials of degree four or less that satisfy only the geometric boundary conditions. Use

f x F x L x L( ) ( )max= −⎡⎣ ⎤⎦/ / 2 ,

I x I x L F I( ) . ( ) ,max= +( ) = = × −0

201 0 1 1000 1 10/ with N 55

9 6200 10 1 10

m

/ N/m a

4

2

,

,E k= × = ×N m nnd 2mL =

Solution The boundary conditions at x = 0 are geometric boundary conditions:

w( )0 0= (b)

dwdx

( )0 0= (c)

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The boundary conditions at x = L are natural boundary conditions given by

EI Ld wdx

L( ) ( )2

20= (d)

d

dxEI

d wdx

L kw L2

2

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟ =( ) ( ) (e)

The energy inner product is defi ned by

( , )f gd

dxEI

d fdx

gdxL

L

=⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟∫

2

2

2

2

0

(f)

Using integration by parts twice on Equation f leads to

( , ) ( ) ( ) ( )f g g Ld

dxEI

d f

dxL g

dL =

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

−2

20

ddxEI

d f

dx

dg

2

20

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

−

( )

ddx

L EI Ld wdx

Ldgdx

EId wdx

( ) ( ) ( ) ( ) ( ) ( )

2

2

2

20 0 0+

+ ∫ EId f

dx

d g

dxdx

2

2

2

2

0

1

(g)

Requiring f(x) and g(x) to satisfy the geometric boundary conditions reduces Equation g to

( , ) ( ) ( ) (f g g Ld

dxEI

d f

dxL

dgdxL =

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

−2

2LL EI L

d f

dxL

EId f

dx

d

) ( ) ( )

2

2

2

2

2

+gg

dxdx

2

0

1

∫ (h)

Satisfaction of Equation e for all f and g in D leads to

( , ) ( ) ( ) ( ) ( ) ( )f g kf L g Ldgdx

L EI Ld fdx

L EId

L = − +2

2

2 ffdx

d gdx

dx2

2

2

0

1

∫ (i)

9533_C004.indd 2399533_C004.indd 239 10/29/08 4:35:55 PM10/29/08 4:35:55 PM


If only comparison functions are used in the Rayleigh-Ritz approximation, then Equation i reduces to

( , ) ( ) ( )f g kf L g L EId fdx

d gdx

dxL = +∫2

2

2

2

0

1

(j)

Equation j is the form of the inner product that should be used when admis-sible functions are used. With admissible functions that do not satisfy the natural boundary conditions, Equation d and Equation e, are used, then Equation j is used for inner-product evaluation, because it is clear that this defi nition is a valid defi nition of the inner product on the space of admissible functions.

(b) The subspace of the space of admissible functions from which a Rayleigh-Ritz approximation is sought is the intersection of P 4[0,L] with the space of functions that satisfy the geometric boundary conditions f (0) = 0 and df/dx (0) = 0. It is not hard to show that u1(x) = x2, u2(x) = x3 and u3(x) = x4 form a basis for this space. Hence the form of the Rayleigh-Ritz approximation is

q x x x x( ) = + +α α α12

23

34 (g)

A MATHCAD fi le showing the details of the solution is presented below.

Example 4.4Parameters

E := ⋅200 109

L := 2

I0 := ⋅ −1 10 5

k := ⋅1 107

Fmax := 100

Known functions

f(x) : Fmax

I(x) : I0

= ⋅ −⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟

= ⋅ +

x

L

x

L

2

2

1 0..12

2⋅

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟

x

L

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Basis functions

u1(x) := x2

u2(x) := x3

u3(x) := x4

u(x) :=

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟

u x

u x

u x

1

2

3

( )

( )

( )

Energy inner products

i := 1 .. 3

j := 1 .. 3

Q k u L u L E I xd

dxu x

di j i j

L

i, : ( ) ( )= ⋅ ⋅ + ⋅ ( )⋅ ( ) ⋅∫0

2

2

2

ddxu x dxj2

( )

Q =× × ××

1 765 10 3 704 10 7 757 10

3 704 10 8 435

8 8 8

8

. . .

. . ×× ×× × ×

⎛

⎝

10 1 894 10

7 757 10 1 894 10 4 535 10

8 9

8 9 9

.

. . .

⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟

f :i = ∫ f x u x dxi

L

( ) ( )0

f =

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟

40

53 333

76 19

.

.

Solution

α

α

:

.

.

.

= ⋅

=×

− ××

⎛

⎝

⎜

−

−

−

−

Q f1

6

6

7

2 983 10

2 24 10

4 422 10

⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟

Plot of solution

w(x) := ⋅( )=

∑ α i i

i

u x( )1

3

x := 0,.01 .. 1

9533_C004.indd 2419533_C004.indd 241 10/29/08 4:35:57 PM10/29/08 4:35:57 PM


0 0.2 0.4 0.6 0.8 10

1.5 × 10–6

1 × 10–6

5 × 10–7

w(x)

x

4.4 Finite-element methodThe possibility of using admissible functions as basis functions for the application of the Rayleigh-Ritz method is suggested by Equation 4.29, which results from application of integration by parts to the energy inner product, ( f, g)L. This formulation illustrates that (w, w)L is proportional to the total potential energy if w is the displacement function, and that use of the energy inner product in this form requires a lower order of differentiability than the differential operator. Thus, admissible basis functions are required to satisfy only geometric boundary conditions and have a level of differen-tiability specifi ed by the energy inner product when written in the form of Equation 4.29.

The relaxation of the requirement that the basis functions satisfy the level of differentiability specifi ed by the differential operator enables piecewise-defi ned basis functions with a lower level of differentiability to be chosen. Such basis functions are called interpolating splines. A Rayleigh-Ritz method using splines chosen such that all geometric boundary conditions are satis-fi ed and an appropriate level of differentiability is satisfi ed is called a fi nite-element method.

A set of piecewise linear splines which can be used as basis functions for the Rayleigh-Ritz method applied to second-order differential operators is illustrated in Figure 4.4. The interval 0 ≤ x ≤ 1 is divided into n subintervals (a fi nite number), each of length �. Thus, n = 1/�. The mathematical forms of the n + 1 basis functions are

9533_C004.indd 2429533_C004.indd 242 10/29/08 4:35:57 PM10/29/08 4:35:57 PM


φ

φ

0

1

1 1( ) ( )

( ) ( ) ( )

xx

u x

xx

u x u x

= −( ) − −[ ]

=( ) − −

��

��[[ ]+ −( ) − − −[ ]

= −

=

2 2

2 1

xu x u x

x x

k

��

�

�

( ) ( )

( ) ( )φ φ

φ φφ

φ

1

1 1

( )

( ) ( ) ( )

x k

xx

n u x nn

−

= + −⎡⎣⎢⎢

⎤⎦⎥⎥

− −[ ]

�

�

��

(4.30)

where u(z) is the unit step function defi ned such that

u zz

z( ) =

<>

⎧⎨⎪⎪⎩⎪⎪

0 0

1 0 (4.31)

The Rayleigh-Ritz formulation is

u c xi i

i

n

==

∑ φ ( )0

(4.32)

The geometric boundary condition u(0) = u0 is applied by choosing C0 = u0, and the boundary condition u(1) = u1 is applied by requiring Cn = u1.

1ϕ

n–1(x)ϕn–2(x)ϕ3(x)ϕ2(x)ϕ1(x)ϕ0(x) ϕ

n(x)

1(n–1)l(n–2)l2l 3l 4ll

Figure 4.4 Piecewise-defi ned basis elements for use in the fi nite-element method to approximate the solution of a second-order differential equation.

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Example 4.5 The differential equation governing the vertical displacement of the non-uniform bar of Figure 4.5 under its own weight is

d

dxEA

dudx

gA d

x⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟= ∫ ρ ξ ξ( )

0

(a)

The bar is fi xed at x = 0 and attached to a linear spring of stiffness k at x = L. The resulting boundary conditions are

u( )0 0= (b)

EA Ldudx

L ku L( ) ( ) ( )+ = 0 (c)

Nondimensional variables are introduced by

xxL

∗ = (d)

wwL

∗ = (e)

α(x∗) =AA0

(f)

L

x

k

A(x),E,ρ

Figure 4.5 System of Example 4.5.

9533_C004.indd 2449533_C004.indd 244 10/29/08 4:35:59 PM10/29/08 4:35:59 PM


Substitution of Equation d, Equation e, and Equation f into Equation a, assuming that E is constant, rearranging, and dropping the*’s leads to

d

dxx

dudx

d

x

α μ α ξ ξ( ) ( )⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟= ∫

0

(g)

where

μρ

=gLE

2

(h)

The nondimensional boundary conditions are of the form

u( )0 0= (i)

dudx

u( ) ( )1 1+ =ν 0 (j)

where

ν =kL

EA L( ) (k)

Develop a fi nite-element approximation for the vertical displacement of a bar of length 30.5 m, made from a material of elastic modulus 200 × 109 N/m2 and mass density 7000 kg/m3, if (a) the bar is uniform with area 2 × 10–4 m2, and (b) the area varies linearly according to A(x) = 2 × 10–4 (1 + .02x). Divide the interval from 0 to 1 into fi ve segments of equal length.

Solution (a) If A is constant, then α = 1. For the numerical values given, the nondimensional parameters are μ = 3.194 × 10–4 and v = 0.153. The right-hand side of Equation g is simply μx. The piecewise-defi ned basis functions are illustrated in the MATHCAD fi le which follows the Rayleigh-Ritz approxi-mation is

u x a xi i

i

( ) ( )==

∑ φ0

5

(l)

The geometric boundary condition at x = 0 is imposed by setting a0 = 0. The natural boundary condition at x = 1 is not satisfi ed by the basis functions. Instead, the potential energy of the spring is included in the formulation of the energy inner product. The piecewise-defi ned set of functions can be used as a basis when the energy inner product is written in the form

( , ) ( ) ( ) ( )f g xdfdx

dgdx

dx g g= +∫ α ν0

1

1 1 (m)

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A MATHCAD fi le is developed to obtain the solution. The fi le is written in general so that only the description of the variation of the area with x needs to be changed. The MATHCAD fi le set up for part (b) is illustrated below.

Example 4.5Finite element approximation for displacement of hanging bar with attached spring. The interval from x = 0 to x = 1 is divided into fi ve elements of equal length leading to defi nition of the basis functions as

φ

φ

0 1 5 0 2

1 5

( ) : ( )( ( ) ( . ))

( ) : ( (

x x x x

x x

= − ⋅ − −

= ⋅ ⋅

Φ Φ

Φ xx x x x

x x

) ( . )) ( ) ( ( . ))

( ) : (

− − + − ⋅ ⋅ −

=

Φ Φ0 2 2 5 0 4

2 1φ φ −−

= −

= −

0 2

3 1 0 4

4 1 0 6

5

. )

( ) : ( . )

( ) : ( . )

( )

φ φ

φ φ

x x

x x

xΦ :: ( ) ( . )= ⋅ − ⋅ −5 4 0 8x xΦ

where Φ(x) is the Heaviside function defi ned as 0 for x < 0 and 1 for x > 0. it is also called the unit step function u(x)

x := 0,.01 .. 1

0 0.2 0.4 0.6 0.8 10

0.2

1Basis functions

φ0(x) 0.8

φ1(x)

φ2(x) 0.6

φ3(x)

φ4(x) 0.4

φ5(x)

x

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Dimensional Parameters

ρ := 7000 Mass density in kg/m^3

g := 9.81 Acceleration due to gravity in m/s^2

L := 30.5 Length of bar in m

k := 200000 Stiffness of spring in N/m

E := 200 ⋅ 109 Elastic modulus in N/m^2

A0 := 2 ⋅ 10–4 Area at x = 0 in m^2

Variation of areaArea(xd) : = A0 ⋅ (1 + 0.00xd) Variation of area over length of bar, xd is the

dimensional coordinate along the length of the bar

AL := Area(L) Calculated area at x = L in m^2AL = 2 × 10–4

Nondimensional Parameters

μρ

μ

ν ν

: .

: .

=⋅ ⋅

= ×

=⋅

⋅=

−g L

E

k L

E AL

243 194 10

0 153

Nondimensional functions

α( ) :( )

xArea x L

A=

⋅0

Nondimensional area in terms of nondimensional variable x

f x y dy

x

( ) : ( )= ⋅∫μ α0

Nondimensional loading due to gravity

Inner p roduct evaluations

Ad

dxx

d

dxx x dx1 1 1 1 1 1, : ( ) ( ) ( ) ( )= ( )⋅( )⋅ + ⋅ ⋅φ φ α ν φ φφ1 1 10

0

1

1 1( ) ,∫ =A

Ad

dxx

d

dxx x dx1 2 1 2 1 1, : ( ) ( ) ( ) ( )= ( )⋅( )⋅ + ⋅ ⋅φ φ α ν φ φφ2 1

0

1

( )∫

Ad

dxx

d

dxx x dx1 3 1 3 1 1, : ( ) ( ) ( ) ( )= ( )⋅( )⋅ + ⋅ ⋅φ φ α ν φ φφ3 1

0

1

( )∫

Ad

dxx

d

dxx x dx1 4 1 4 1 1, : ( ) ( ) ( ) ( )= ( )⋅( )⋅ + ⋅ ⋅φ φ α ν φ φφ4 1

0

1

( )∫

Ad

dxx

d

dxx x dx1 5 1 5 1 1, : ( ) ( ) ( ) ( )= ( )⋅( )⋅ + ⋅ ⋅φ φ α ν φ φφ5 1

0

1

( )∫

9533_C004.indd 2479533_C004.indd 247 10/29/08 4:36:02 PM10/29/08 4:36:02 PM


Ad

dxx

d

dxx x dx2 2 2 2 2 1, : ( ) ( ) ( ) ( )= ( )⋅( )⋅ + ⋅ ⋅φ φ α ν φ φφ2 1

0

1

( )∫

Ad

dxx

d

dxx x dx2 3 2 3 3 1, : ( ) ( ) ( ) ( )= ( )⋅( )⋅ + ⋅ ⋅φ φ α ν φ φφ2 1

0

1

( )∫

Ad

dxx

d

dxx x dx2 4 2 4 4 1, : ( ) ( ) ( ) ( )= ( )⋅( )⋅ + ⋅ ⋅φ φ α ν φ φφ2 1

0

1

( )∫

Ad

dxx

d

dxx x dx2 5 2 5 5 1, : ( ) ( ) ( ) ( )= ( )⋅( )⋅ + ⋅ ⋅φ φ α ν φ φφ2 1

0

1

( )∫

Ad

dxx

d

dxx x dx3 3 3 3 3 1, : ( ) ( ) ( ) ( )= ( )⋅( )⋅ + ⋅ ⋅φ φ α ν φ φφ3 1

0

1

( )∫

Ad

dxx

d

dxx x dx3 4 3 4 3 1, : ( ) ( ) ( ) ( )= ( )⋅( )⋅ + ⋅ ⋅φ φ α ν φ φφ4 1

0

1

( )∫

Ad

dxx

d

dxx x dx3 5 3 5 3 1, : ( ) ( ) ( ) ( )= ( )⋅( )⋅ + ⋅ ⋅φ φ α ν φ φφ5 1

0

1

( )∫

Ad

dxx

d

dxx x dx4 4 4 4 4 1, : ( ) ( ) ( ) ( )= ( )⋅( )⋅ + ⋅ ⋅φ φ α ν φ φφ4 1

0

1

( )∫

Ad

dxx

d

dxx x dx4 5 4 5 4 1, : ( ) ( ) ( ) ( )= ( )⋅( )⋅ + ⋅ ⋅φ φ α ν φ φφ5 1

0

1

( )∫

Ad

dxx

d

dxx x dx5 5 5 5 5 1, : ( ) ( ) ( ) ( )= ( )⋅( )⋅ + ⋅ ⋅φ φ α ν φ φφ5 1

0

1

( )∫ A A A A2 1 1 2 3 1 1 3, , , ,: := =

A A A A4 1 1 4 5 1 1 5, , , ,: := =

A A A A3 2 2 3 4 2 2 4, , , ,: := =

A A A A5 2 2 5 4 3 3 4, , , ,: := =

A A A A5 3 3 5 5 4 4 5, , , ,: := =

A =

−− −

− −− −

−

10 5 0 0 0

5 10 5 0 0

0 5 10 5 0

0 0 5 10 4 987

0 0 0 4

.

.. .987 5 159

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

9533_C004.indd 2489533_C004.indd 248 10/29/08 4:36:03 PM10/29/08 4:36:03 PM


b f x x dx b f x x dx

b f

1

0

1

2

0

1

3

1 2: ( ) ( ) : ( ) ( )

:

= ⋅ = ⋅

=

∫ ∫φ φ

(( ) ( ) : ( ) ( )

: ( )

x x dx b f x x dx

b f x

⋅ = ⋅

= ⋅

∫ ∫φ φ3 40

1

4

0

1

5 φφ50

1

( )x dx∫

b =

××××

−

−

−

−

1 276 10

2 556 10

3 836 10

5 102 10

2

5

5

5

5

.

.

.

.

..986 10 5×

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟−

⎟⎟⎟⎟

C

C

:

.

.

.

.

= ⋅

=

×××

−

−

−

−

A b1

5

5

5

2 818 10

5 381 10

7 433 10

8 7117 10

9 005 10

5

5

××

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟

−

−.

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

w x x x x x( ) : ( ) ( ) ( ) ( )= ⋅ + ⋅ + ⋅ + ⋅( )C C C C1 2 3 41 2 3 4φ φ φ φ ++ ⋅C5 5φ ( )x

z := 0,.01 .. 1

0 0.2 0.4 0.6 0.8 10

1 ×10–4

8 ×10–5

6 ×10–5

4 ×10–5

2 ×10–5

w(z)

z

Displacement function of uniform bar

9533_C004.indd 2499533_C004.indd 249 10/29/08 4:36:03 PM10/29/08 4:36:03 PM


(b) For the nonuniform bar described,

α( )

( . )( . )x

x=

× +[ ]×

=

−

−

2 10 1 0 02 30 5

2 10

1

4

4

++ 0 61. x (n)

The right-hand side of Equation g is evaluated as

μ α ξ ξ μ ξ ξ

μ

( ) = +( )

= +( )

∫ ∫d d

x x

x x

0 0

2

1 0 61

0 305

.

. (o)

The values of the nondimensional parameters are μ = 3.194 × 10–4 and v = 0.095. The solution is obtained by modifying the MATHCAD fi le such that xd = 0.02. The resulting fi nite-element solution is illustrated.

A set of basis functions for a Rayleigh-Ritz approximation for the solu-tion of a boundary-value problem involving a fourth-order operator must be twice differentiable, implying that the fi rst derivative of the interpolating splines must be continuous.

Consider an interval 0 ≤ ξ ≤ 4 �. An interpolating spline ψ(ξ) is determined such that ψ(0) = 0, ψ ′(0) = 0, ψ(4 �) = 0, ψ ′(4 �) = 0 and ψ(2 �) = 1. Assuming that ψ(ξ) = C0 + C1ξ + C2ξ2 + C3ξ3 + C4ξ4, an appropriate spline can be determined as

ψ ξξ ξ ξ

( ) = − +2

2

3

3

4

42 16� � � (4.33)

0 0.2 0.4 0.6 0.8 10

1×10−4

8×10−5

6×10−5

4×10−5

2×10−5

Displacement function of variable area bar

w (z)

z

9533_C004.indd 2509533_C004.indd 250 10/29/08 4:36:04 PM10/29/08 4:36:04 PM


The function given in Equation 4.33 is called the generating spline because all basis functions can be defi ned from it.

The nondimensional interval 0 ≤ x ≤ 1 is divided into n intervals, each of length � . A set of basis functions is defi ned by

φ ψi x x i u x i u x i i( ) ( ( ) ) ( ( ) ) ( ( )= − − − − − − +[ ] =3 3 1� � � 00 1 2, ,…n+

Example 4.6 Reconsider Example 4.3 in which a uniform Euler-Bernoulli beam rests on an elastic foundation and is subject to a transverse load. The nondimensional formulation of the governing differential equation is

d wdx

w f x4

4+ =φ λ ( ) (a)

The nondimensional parameters are defi ned as

φ =kLEI

3

(b)

λ =F L

EImax

3

(c)

The beam is loaded so that

f x x x( ) = − 2 (d)

Dividing the interval from 0 to 1 into six segments of equal length, use the fi nite-element method to approximate the transverse defl ection of a beam with (a) both ends free, (b) both ends fi xed, and (c) the end at x = 0 fi xed and the end at x = 1 pinned. In each case, use φ = 1 and λ = 1.

Solution The basis elements are those shown in the MATHCAD fi le used to solve the problem. The appropriate energy-inner-product formulation is

f gd fdx

d gdx

dx f x g x dxL

, ( ) ( )( ) = +∫ ∫2

2

2

2

0

1

0

1

φ (e)

The general form of the Rayleigh-Ritz approximation is

w x a xi i

i

( ) ( )==

∑ φ0

8

(f)

(a) All boundary conditions for a beam which is free at both ends are natu-ral (bending moment and shear force are both zero). Thus, the set of eight

9533_C004.indd 2519533_C004.indd 251 10/29/08 4:36:05 PM10/29/08 4:36:05 PM


interpolating splines serves as basis functions. The results are shown in the attached MATHCAD fi le.(b) Since the beam is fi xed at both ends, all boundary conditions are geo-metric and must be satisfi ed. This could be accomplished at x = 0 by setting a0 = 0, a1 = 0, and a2 = 0. Satisfaction of geometric boundary conditions at x = 1 could be accomplished by setting a6 = 0, a7 = 0, and a8 = 0. Thus, the appropri-ate Rayleigh-Ritz approximation is

w x a x a x a x( ) ( ) ( ) ( )= + +3 3 4 4 5 5φ φ φ (g)

However, when only three basis functions are used, the geometric quanti-ties of defl ection and slope are not independent over the length of the beam. Upon inspection, it can be noted that w ′ (2q) = –w ′ (4q) independently of f(x). This is an artifi cial condition enforced by the choice of basis functions. An expanded set of basis functions can be chosen from the interpolating splines by requiring that

w a

a a a

i i

i

( ) ( )

( ) ( ) ( )

0 0 0

0 0 0

0

8

0 0 1 1 2 2

= =

= + +

=∑ φ

φ φ φ

== + +

= + +

a a a

a a a

0 1 2

0 1 2

3 2

9

16

9

16

ψ ψ ψ( ) ( ) ( )� � �

(h)

′ = = ′

= ′ + ′

=

=∑w a

a a

a

i i

i

( ) ( )

( ) ( )

0 0 0

3

3

4

0

8

0 2

0

φ

ψ ψ� �

−−3

42a (i)

Equation h and Equation i are used to determine a a a0 2 18 9= = − / . Thus, a basis function satisfying the geometric boundary conditions at x = 0 is

u x x x x( ) ( ) ( ) ( )= − + −8

9

8

90 1 2φ φ φ (j)

A similar application of the boundary conditions at x = 1 leads to a basis function of

v x x x x( ) ( ) ( ) ( )= − + −8

9

8

96 7 8φ φ φ (k)

9533_C004.indd 2529533_C004.indd 252 10/29/08 4:36:05 PM10/29/08 4:36:05 PM


An appropriate set of basis functions for a Rayleigh-Ritz approximation is {u(x), φ3(x), φ4(x), φ5(x), v(x)}. MATHCAD results can be seen as below.(c) The boundary conditions at x = 0 for a fi xed-pinned beam are the same as those in part (b), leading to the choice of u(x) in Equation j as a basis function. The geometric boundary condition at x = 1 is satisfi ed by requiring that

w a a a( ) ( ) ( )1 0 3 26 7 8= = ( )+ +

=

ψ ψ ψ� � �

99

16

9

166 7 8a a a+ +

(l)

Equation l is the only condition which must be satisfi ed by the interpolating splines at x = 1. This leads to a7 = −9/16 (a6 + a8) and a choice of a set of six basis functions, namely u x x x x x x( ), ( ), ( ), ( ), ( ) ( / ) ( ),φ φ φ φ φ φ3 4 5 6 7 89 16− −−{ }( / ) ( ) .9 16 7φ x The MATHCAD fi le for the fi nite-element approximation of the displacement of a fi xed-pinned beam on an elastic foundation is shown in Figure 4.11.

Example 4.6 Finite element modeling of beam on an elastic foundation

n := 6 Number of intervals

q :=1n Length of an interval

ψ ξξ ξ ξ

( ) := −⋅

+⋅

2

2

3

3

4

42 16q q q Generating spline

Interpolating splines

φ ψ0 3( ) : ( ) ( ( ) ( ))x x q x x q= + ⋅ ⋅ − −Φ Φ

φ ψ1 2 2( ) : ( ) ( ( ) ( ))x x q x x q= + ⋅ ⋅ − −Φ Φ

φ ψ2 3( ) : ( ) ( ( ) ( ))x x q x x q= + ⋅ − − ⋅Φ Φ

φ ψ3 4( ) : ( ) ( ( ) ( ))x x x x q= ⋅ − − ⋅Φ Φ

φ ψ4 5( ) : ( ) ( ( ) ( ))x x q x q x q= − ⋅ − − − ⋅Φ Φ

φ ψ5 2 2 6( ) : ( ) ( ( ) ( ))x x q x q x q= − ⋅ ⋅ − ⋅ − − ⋅Φ Φ

φ ψ6 3 3 7( ) : ( ) ( ( ) ( ))x x q x q x q= − ⋅ ⋅ − ⋅ − − ⋅Φ Φ

φ ψ7 4 4 8( ) : ( ) ( ( ) ( ))x x q x q x q= − ⋅ ⋅ − ⋅ − − ⋅Φ Φ

φ ψ8 5 5 9( ) : ( ) ( ( ) ( ))x x q x q x q= − ⋅ ⋅ − ⋅ − − ⋅Φ Φ

Plot interpolating splines

x := 0,.01 .. 1

9533_C004.indd 2539533_C004.indd 253 10/29/08 4:36:06 PM10/29/08 4:36:06 PM


0 0.2 0.4 0.6 0.8

0.8

0.6

0.4

0.2

0

φ0(x)

φ1(x)

φ2(x)

φ3(x)

φ4(x)

φ5(x)

φ6(x)

φ7(x)

φ8(x)

x

Interpolating splines for 4th-order problems

Nonhomogeneous term

f(x) : = x − x2

φ := 1

a. Free free beam

Basis functions

A(x) :=

φφφφφφφ

0

1

2

3

4

5

6

( )

( )

( )

( )

( )

( )

( )

x

x

x

x

x

x

x

φφφ

7

8

( )

( )

x

x

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

i := 1 .. 9

j := 1 .. 9

9533_C004.indd 2549533_C004.indd 254 10/29/08 4:36:07 PM10/29/08 4:36:07 PM


Matrix of energy inner products

Qd

dxA x

d

dxA x dxi j i j, : ( ) ( )= ⋅

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟∫

2

20

1 2

2 ⎟⎟+ ⋅ ⋅∫φ A x A x dxi j( ) ( )0

1

Vector of inner products

b f x A x dxi i: ( ) ( )= ⋅∫0

1

Q =

− −−213 313 90 416 151 173 31 053 0 0 0 0 0

90 416 3

. . . .

. 445 735 21 751 302 346 31 053 0 0 0 0

151 173 21 7

. . . .

. .

−− 551 478 158 68 664 302 345 31 053 0 0 0

31 053 30

. . . .

.

− −− 22 346 68 664 691 471 68 664 302 345 31 053 0 0. . . . . .− − −

00 31 053 302 345 68 664 691 471 68 664 302 3. . . . . .− − − − 445 31 053 0

0 0 31 053 302 345 68 664 691 471 68

.

. . . .− − − .. . .

. . .

664 302 346 31 053

0 0 0 31 053 302 345 68 664

−− − 4478 158 21 751 151 173

0 0 0 0 31 053 302 346 21

. . .

. . .

−− 7751 345 735 90 416

0 0 0 0 0 31 053 151 173 90 41

. .

. . .

−− − 66 213 313.

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

b =

× −1 559 10

0 016

0 046

0 073

0 083

0 073

0 046

0

3.

.

.

.

.

.

.

..

.

016

1 559 10 3×

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜−

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

Solution for coeffi cients

W := Q–1·b

W =

××××

−

−

−

−

5 053 10

5 068 10

5 143 10

5 135 10

5

3

3

3

3

.

.

.

.

..

.

.

.

.

183 10

5 135 10

5 143 10

5 068 10

5 0

3

3

3

3

××××

−

−

−

−

553 10 3×

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟

−

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

9533_C004.indd 2559533_C004.indd 255 10/29/08 4:36:07 PM10/29/08 4:36:07 PM


Rayleigh Ritz Approximation

wfree(x): (x)= ⋅=

∑( )W A i

i

1

1

9

x := 0,.01 .. 1

0 0.2 0.4 0.6 0.8x

0.0108

0.0109

0.011

wfree(x)

Deflection of free-free beam

Fixed fi xed beam

Case 1 Use of three basis functions

B(x) :

x

x

x

=

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟

φφφ

3

4

5

( )

( )

( )

i := 1 .. 3 j := 1 .. 3

Rd

dxB x

d

dxB x dxi j i j, : ( ) ( )= ⋅

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟∫

2

2

2

20

1

⎟⎟+ ⋅ ⋅∫φ B x B x dxi j( ) ( )0

1

r f x B x dxi i: ( ) ( )= ⋅∫0

1

R =691 471 68 664 302 345

68 664 691 471 68 6

. . .

. . .

− −− − 664

302 345 68 664 691 471− −

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟. . . ⎟⎟

r =

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟

0 073

0 083

0 073

.

.

.

9533_C004.indd 2569533_C004.indd 256 10/29/08 4:36:08 PM10/29/08 4:36:08 PM


Wfi xed := R–1·r

Wfixed =×××

⎛

⎝

⎜⎜⎜⎜

−

−

−

2 174 10

1 636 10

2 174 10

4

4

4

.

.

.⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟

wfixed(x) := ⋅=

∑( ( ) )Wfixed B xi i

i 1

3

1.10–4

2.10–4

3.10–4

4.10–4

5.10–4

wfixed(x)

Fixed-fixed beam with three basis functions

0 0.2 0.4 0.6 0.8x

Fixed fi xed beam with fi ve basis functions

u x x q x q x q x( ) :=−

⋅ + ⋅( )+ + ⋅( )− ⋅ +( )( )⋅ (8

93 2

8

9ψ ψ ψ Φ ))− −( )( )

= + + ⋅( )− ⋅ +( )( )⋅

Φ x q

u x u x x q x q( ) : ( ) ψ ψ28

9ΦΦ Φ

Φ

x q x q

u x u x x q x

−( )− − ⋅( )( )

= − ⋅ +( )⋅ −

2

8

92( ) : ( ) ψ ⋅⋅( )− − ⋅( )( )q x qΦ 3

v x x q x q x q( ) :=−

⋅ − ⋅( )+ − ⋅( )− ⋅ − ⋅( )( )⋅8

93 4

8

95ψ ψ ψ Φ xx q

v x v x x q x q

− ⋅( )( )

= + − ⋅( )− ⋅ − ⋅( )( )

5

48

93( ) : ( ) ψ ψ ⋅⋅ −( )− − ⋅( )( )

= − ⋅ −( )⋅

Φ Φ

Φ

x q x q

v x v x x q

4 5

8

93( ) : ( ) ψ xx q x q− ⋅( )− − ⋅( )( )3 4Φ

9533_C004.indd 2579533_C004.indd 257 10/29/08 4:36:09 PM10/29/08 4:36:09 PM


–0.8

–0.6

–0.4

–0.2

0Additional basis functions for fixed-fixed beam

u(x)

v(x)

0 0.2 0.4 0.6 0.8x

C(x) :

u x

x

x

x

v x

=

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

( )

( )

( )

( )

( )

φφφ

3

4

5⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

i := 1 .. 5

j := 1 .. 5

S C Ci j i j, : ( ) ( )= ⋅⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟∫ d

dxx

d

dxx dx

2

20

1 2

2 ⎟⎟+ ⋅ ⋅ ( )

= ⋅

∫

∫

φ C C

s f C

i j

i i

( )

: ( ) ( )

x x dx

x x dx

0

1

0

1

s =

− − −−775 284 268 931 299 756 27 603 73 91

268 9

. . . . .

. 331 691 471 68 664 302 345 38 772

299 756 68

. . . .

. .

− − −− 6664 691 471 68 664 255 623

27 603 302 345 68

. . .

. .

−− − − .. . .

. . .

664 691 471 277 889

73 91 38 772 255 623 27

−− − − 77 889 988 055. .

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

s =

−

−

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

0 026

0 073

0 083

0 073

0 026

.

.

.

.

.⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

9533_C004.indd 2589533_C004.indd 258 10/29/08 4:36:09 PM10/29/08 4:36:09 PM


Wfi xeda := S–1·s

Wfixeda =

− ×××

−

−

−

2 24 10

2 063 10

1 736 10

2 103

5

4

4

.

.

.

. ××− ×

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟

−

−

10

5 839 10

4

6.

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

wfixeda(x) : Wfixeda (x)= ⋅=

∑( )i i

i

C1

5

Fixed-fixed beam with five basis functions5.10–4

4.10–4

3.10–4

2.10–4

1.10–4

00 0.2 0.4 0.6 0.8

x

wfixeda(x)

q1(x) : wfixeda(x)=

d

dx

q1(2 q) q1(4 q)⋅ × ⋅ − ×− −= =6 917 7 58. .10 104 4

Fixed Pinned beam

u x x q x q x q( ) : ( ) ( ) ( ) ( (=

−⋅ + ⋅ + + ⋅ − ⋅ +( )⋅

8

93 2

8

9Ψ Ψ Ψ Φ xx x q) ( ))− −Φ

u x u x( ) : ( ) ( ) ( ) ( ( ) (= + + ⋅ − ⋅ +( )⋅ − −Ψ Φ Φx q x q x q x2

8

9Ψ −− ⋅2 q))

u x u x x q x q x q( ) : ( ) ( ) ( ( ) ( ))= − ⋅ + ⋅ − ⋅ − − ⋅

8

92 3Ψ Φ Φ

9533_C004.indd 2599533_C004.indd 259 10/29/08 4:36:10 PM10/29/08 4:36:10 PM


Basis functions

T x

u x

x

x

x

x x

( ) :

( )

( )

( )

( )

( ) ( )

(

=

− ⋅

φ

φ

φ

φ φ

φ

3

4

5

69

167

8 xx x) ( )− ⋅

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜ 9

167φ⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

x := 0,.01 .. 1

0 0.2 0.4 0.6 0.8 1−1

−0.5

0

0.5

1Basis functions for fixed-pinned beam

u (x)

φ3 (x)

φ4 (x)

φ5 (x)

φ6 (x)9

16⋅φ7 (x)−

φ8 (x)9

16⋅φ7 (x)−

x

i := 1 .. 6

h f x T x dxi i: ( ) ( )= ⋅∫

0

1

j

Hd

dxT x

d

dxT x dxi j i j

k

: ..

: ( ) ( ),

( . )

=

=− ⋅

1 6

2

2

2

299 qq

k q

k

( . )− ⋅

=∫∑

01

1

6

H H T x T x dxi j i j i j, ,: ( ) ( )= + ⋅ ⋅∫φ

0

1

9533_C004.indd 2609533_C004.indd 260 10/29/08 4:36:11 PM10/29/08 4:36:11 PM


H=

− −−

753 208 262 878 294 095 29 258 0 0

262 878 66

. . . .

. 99 588 68 616 294 04 33 133 0

294 095 68 616 66

. . . .

. .

− −− 99 588 68 616 312 677 18 637

29 258 294 04

. . . .

. .

− − −− − −668 616 669 588 96 782 198 53

0 33 133 312 677 96

. . . .

. .− .. . .

. . . .

782 546 972 2 493

0 0 18 637 198 53 2 493 411

−− − 667

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

h =

−

− ×

0 026

0 073

0 083

0 073

0 037

7 308

.

.

.

.

.

. 110 3−

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

Wpinned H h

Wpinned

:

.

.

= ⋅

=

− ××

−

−

1

55 663 10

2 119 10−−

−

−

−

×××

−

4

4

4

4

2 693 10

2 394 10

1 653 10

.

.

.

11 2 10 4. ×

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

−

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

wpinned x Wpinned T xi i

i

( ) : ( ( ) )= ⋅=

∑1

6

0 0.2 0.4 0.6 0.8 1

6×10–4

4×10–4

2×10–4

–2×10–4

0

Fixed-pinned beam with six basis functions

wpinned (x)

x

9533_C004.indd 2619533_C004.indd 261 10/29/08 4:36:11 PM10/29/08 4:36:11 PM


Example 4.7 The differential equation governing the nondimensional tem-perature in a variable-area fi n is

d

dxx

ddx

u x13

42−( )⎡

⎣⎢⎢

⎤⎦⎥⎥− =μ

θφ θ Λ ( ) (a)

where φ2 2=hL kb/ and Λ = − ∞u L kb T Tmax / ( )202 � . The surface has a fi xed tem-

perature at x = 0 and is insulated at x = 1, leading to the boundary conditions

θ( )0 1= (b)

ddx

θ( )1 0= (c)

a. Obtain the exact solution for Equation a, Equation b, and Equation c with Λ = 0.

b. Use the method of variation of parameters to determine an exact solu-tion of Equation a, Equation b, and Equation c with u(x) = 1 – x.

c. Use the Rayleigh-Ritz method using fourth-order polynomials which satisfy only the geometric boundary conditions as basis functions to approximate the temperature distribution in the surface for (i) Λ = 0 and (ii) u(x) = 1 – x. Use φ = 1.5 and μ = 0.5.

d. Use a fi nite-element method with fi ve elements to approximate the tem-perature distribution over the surface for (i) Λ = 0 and (ii) u(x) = 1 − x. Use φ = 1.5 and μ = 0.5.

Solution (a) Let z = 1 – μ x. Changing the independent variable in Equation a from x to z leads to

ddz

zddz

u3

4

2

2

θ φμ

θμ

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟−

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟ =

Λ(zz) (d)

The solution of Equation d with Λ = 0 can be obtained through comparison with Equation (3.110) with μ = 3/4 and v = 0. The general solution is

θφμ

( ) //

/ //z C z I z C z K=

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟+1

1 81 5

5 82

1 81

8

555

5 88

5

φμ

z /⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟ (e)

where I z K z1 5 1 5/ ( ) ( )/and are the modifi ed Bessel functions of the fi rst and second kind of order 1/5 and argument z.

9533_C004.indd 2629533_C004.indd 262 10/29/08 4:36:12 PM10/29/08 4:36:12 PM


The boundary condition, Equation, c, can be written in terms of z as dθ/dz (1 – μ) = 0, which when applied to Equation e using Equation f, leads to C2 = − βC1, where

β

φμ

μ=

−( )⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟+ ′ −I I1 5

5 81 5

8

51

8 8

51/

//

φμ

φμ

μμ( )⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟

−( )⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟

5 8

1 55 88

51

/

//K

φμ

μ ++ ′ −( )⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟

8 8

511 5

5 8φμ

φμ

μK //

(f)

The boundary condition, Equation b, can be written in terms of z as θ(z = 1) = 1, which when applied to Equation e and using Equation f, leads to

θ

φμ

βφ

( )

//

/ //

zz I z z K

=

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟−1 8

1 55 8 1 8

1 58

5

8

55

8

5

8

5 8

1 5 1 5

μφμ

βφ

z

I K

/

/ /

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟−

55μ⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟

(g)

Returning to the original independent variable,

θμ

φμ

μ β( )

( ) //

/

xx I x

=− −( )

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟−1

8

511 8

1 55 8 11

8

51

8

1 81 5

5 8

1 5

−( ) −( )⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟μ

φμ

μ

φ

x K x

I

//

/

/55

8

51 5

μβ

φμ

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟−

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟K /

(h)

(b) The particular solution of Equation a is assumed to be of the form

θ μφμ

μ( ) ( )( ) //

/x A x x I x= − −( )⎛⎝⎜⎜⎜

⎞⎠⎟1

8

511 8

1 55 8⎟⎟⎟+ − −( )

⎛⎝⎜⎜⎜

⎞⎠⎟⎟B x x K x( )( ) /

//1

8

511 8

1 55 8μ

φμ

μ ⎟⎟ (i)

Application of the method of variation of parameters leads to

dAdx

u xW x

x K x= − −( )⎛⎝⎜⎜

( )

( )( ) /

//1

8

511 8

1 55 8μ

φμ

μ⎜⎜⎞⎠⎟⎟⎟ (j)

dBdx

u xW x

x I x= − −( )⎛⎝⎜⎜

( )

( )( ) /

//1

8

511 8

1 55 8μ

φμ

μ⎜⎜⎞⎠⎟⎟⎟ (k)

where W(x) is the Wronskian of the homogeneous solutions. Application of Abel’s formula leads to W x e x( )

/

= − −( )1 3 4μ . Solving Equation i and Equation j for A(x) and B(x) and using the solutions in Equation i leads to the general solu-tion of Equation a as

9533_C004.indd 2639533_C004.indd 263 10/29/08 4:36:13 PM10/29/08 4:36:13 PM


θ μφμ

μ( ) ( ) //

/x C x I x= − −( )⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟1

1 81 5

5 81

8

51 ++ − −( )

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟

+

C x K x21 8

1 55 8

18

51( )

(

//

/μφμ

μ

118

511 8

1 55 8 1− −( )

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟

−μφμ

μ ξx I x u e) ( )//

/ μμξ μξφμ

μξ( ) − −( )⎛⎝⎜⎜⎜

⎞⎠⎟⎟

3 4

18

511 8

1 55 8/

( ) //

/K ⎟⎟

+ − −( )⎛⎝⎜⎜⎜

⎞⎠⎟

∫ d

x K x

x

ξ

μφμ

μ

0

1 81 5

5 81

8

51 ( ) /

// ⎟⎟⎟ − −( )−( )u e I( ) ( )

/ //

/ξ μξφμ

μξμξ1 1 81 5

5 83 4

18

51

⎛⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟∫ d

x

ξ0

(l)

Application of Equation b to Equation l leads to

I C K C1 5 1 1 5 28

5

8

51/ /

φμ

φμ

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟ +

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟ = (m)

Application of Equation c to Equation l leads to

1

81

8

51 17 8

1 55 8−( ) −( )

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟+ −−μ

φμ

μφμ μ/

//I (( ) ′ −( )

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

−1 41 5

5 88

51/

//I C

φμ

μ 11

7 81 5

5 81

81

8

51 /

//+ −( ) −( )

⎛⎝⎜⎜⎜

⎞⎠⎟−μ

φμ

μK ⎟⎟⎟+ −( ) ′ −( )⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟

⎡ −φμ μ

φμ

μ18

511 4

1 55 8/

//K

⎣⎣⎢⎢

⎤

⎦⎥⎥

= − −( ) −( )⎛⎝

−

C

I

2

7 81 5

5 81

81

8

51 /

//μ

φμ

μ⎜⎜⎜⎜⎞⎠⎟⎟⎟+ −( ) ′ −( )

⎛⎝⎜⎜⎜

−φμ μ

φμ

μ18

511 4

1 55 8/

//I

⎞⎞⎠⎟⎟⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

−−( ) ( ) ( )/ /u eξ μξμξ1 1 83 4

1 KK d1 55 8

0

1

8

51

1

81

//

φμ

μξ ξ−( )⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟

− −

∫

μμφμ

μφμ μ( ) −( )

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟+ −( )− −7 8

1 55 88

51 1/

//K 11 4

1 55 88

51/

//

′ −( )⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

Kφμ

μ

( ) ( )/ /

//u e Iξ μξ

φμ

μξμξ1 1 81 5

5 83 4

18

51−( ) − −( )

⎛⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟

− −( )

∫−

d

u e

ξ

μμ

0

1

1 12 1 13 4

( ) ( ) // 441 5

5 81 5

58

51

8

51K I/

//

/φμ

μφμ

μ−( )⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟ −( ) 88⎛

⎝⎜⎜⎜

⎞⎠⎟⎟⎟ (n)

Simultaneous solution of Equation m and Equation n leads to C1 and C2.(c) There are two methods to develop a Rayleigh-Ritz approximation to

a homogeneous differential equation with a nonhomogeneous boundary

9533_C004.indd 2649533_C004.indd 264 10/29/08 4:36:14 PM10/29/08 4:36:14 PM


condition. The nonhomogeneity can be transferred to the differential equa-tion through a change of dependent variable. Defi ning Ψ(x) = θ(x)−1, Equation a, Equation b, and Equation c can be rewritten as

d

dxx

ddx

u x1 3 4 2 2−( )⎡⎣⎢⎢

⎤⎦⎥⎥− = −μ φ φ/ ( )

ΨΨ Λ (o)

Ψ( )0 0= (p)

ddxΨ

( )1 0= (q)

A set of basis functions in P4[0,1] which all satisfy the geometric boundary condition of Equation q is φ φ φ φ1 2

23

34

4( ) , ( ) , ( ) ( )x x x x x x x x= = = =and . The Rayleigh-Ritz solution using these basis functions for Λ = 0 and Λ = 1 with u(x) = 1 − x is given in the MATHCAD output as shown below.

Rayleigh Ritz approximations for Example 4.7

Basis functions

ϕ

ϕ

ϕ

ϕ

ϕ

ϕ

1

2

3

4

1

2

3

4

( ) :

( ) :

( ) :

( ) :

( ) :

x x

x x

x x

x x

=

=

=

=

=x

(( )

( )

( )

( )

x

x

x

x

ϕϕϕ

2

3

4

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

Parameters

Λ := 0

η : .= 1 5

μ : .= 0 5

α μ( ) :x x= − ⋅( )13

4 Nondimensional cross sectional area

u x x( ) := −1

f x u x( ) : ( )= ⋅ −Λ η2

9533_C004.indd 2659533_C004.indd 265 10/29/08 4:36:14 PM10/29/08 4:36:14 PM


Matrix and vector calculations

i := 1,2 .. 4

g := ⋅∫ f x x dxi( ) ( )ϕ0

1

j := 1,2 .. 4

K xd

dxx

d

dxx dxi j i j, : ( ) ( ) ( )= ⋅( )⋅( )⎡

⎣⎢⎢

⎤

⎦∫ α ϕ ϕ0

1

⎥⎥⎥

K =

0 803 0 736 0 701 0 681

0 736 0 935 1 021 1 067

0

. . . .

. . . .

.. . . .

. . . .

701 0 021 1 2 1 313

0 681 1 067 1 313 1 483

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

g =

−−

−−

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

1 125

0 75

0 563

0 45

.

.

.

.

⎟⎟⎟⎟⎟⎟⎟⎟

Solution of equations

W K g

W

:

.

.

.

.

= ⋅

=

−⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜

−1

2 254

0 744

0 047

0 154⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

Rayleigh Ritz approximation

Ψ( ) : ( )x W xi i

i

= ⋅( )=

∑ ϕ1

4

x := 0,.01 .. 1

9533_C004.indd 2669533_C004.indd 266 10/29/08 4:36:15 PM10/29/08 4:36:15 PM


0 0.2 0.4 0.6 0.8 1–1.5

–1

–0.5

0

Ψ(x)

x

An alternative to a change of dependent variable is to assume a Rayleigh-Ritz approximation of

θ( )x x x x x= + + + +1 1 22

33

44α α α α (r)

Use of Equation r leads to the same result, but requires a reformulation of the sets of algebraic equations.

(d) The fi nite-element response is obtained using the MATHCAD fi le shown in below.

Finite-element approximations for Example 4.7

Basis functionsThe interval from x = 0 to x = 1 is divided into fi ve elements of equal length leading to defi nition of the basis functions as

ϕ

ϕ

0 1 5 0 2

1 5

( ) : ( )( ( ) ( . ))

( ) :

x x x x

x x x

= − ⋅ − −

= ⋅ ⋅ (

Φ Φ

Φ ))− −( )( )+ − ⋅( )⋅ −( )− −( )( )Φ Φ Φx x x x0 2 2 5 0 2 0 4

2

. . .

ϕ (( ) : .

( ) : .

( ) :

x x

x x

x x

= −( )

= −( )

= −

ϕ

ϕ ϕ

ϕ ϕ

1 0 2

3 1 0 4

4 1 0..

( ) : .

6

5 5 4 0 8

( )

= ⋅ −( )⋅ −( )ϕ x x xΦ

9533_C004.indd 2679533_C004.indd 267 10/29/08 4:36:15 PM10/29/08 4:36:15 PM


where Φ( )x is the Heavyside function defi ned as 0 for x < 0 and 1 for x > 0. it is also called the unit step function u(x)

x := 0,.01 .. 1

0 0.2 0.4 0.6 0.8 1x

0

0.2

0.4

0.6

0.8

1Basis functions

ϕ0(x)

ϕ1(x)

ϕ2(x)

ϕ3(x)

ϕ4(x)

ϕ5(x)

ϕ

ϕϕϕϕϕ

( ) :

( )

( )

( )

( )

( )

x

x

x

x

x

x

=

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

1

2

3

4

5

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

Parameters

Λ := 0

η := 1.5

μ : .= 1 5

α μx x( ) = −( ): 13

4 Nondimensional cross sectional area

u(x) := 1 − x

f(x) : u x= ⋅ −Λ ( ) η2

Matrix and vector calculations

i := 1,2 .. 5

9533_C004.indd 2689533_C004.indd 268 10/29/08 4:36:16 PM10/29/08 4:36:16 PM


g := ⋅ ⋅∫ f x x dxi( ) ( )ϕ0

1

j := 1.2 .. 5

K xd

dxx dxi j j, := ( )⋅ ( )( )⎡

⎣⎢⎢

⎤

⎦⎥⎥∫ α ϕ

0

1

K =

−− −

−

9 237 4 426 0 0 0

4 426 8 455 4 029 0 0

0 4 029 7

. .

. . .

. .. .

. . .

. .

648 3 619 0

0 0 3 619 6 811 3 184

0 0 0 3 184 3 1

−− −

− 997

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

g =

−−−−

−

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

0 45

0 45

0 45

0 45

0 225

.

.

.

.

.⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

Solution of equations

W K g:

.

.

.

.

.

= ⋅

=

−−−−−

⎛

⎝

−1

0 415

0 765

1 037

1 216

1 282

W

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

Finite element approximation

Ψ x W xi i

i

( ) = ⋅ ( )( )=

∑: ϕ1

5

x := 0,.01 .. 1

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–1.5

–1

–0.5

0

Ψ (x)

x

0 0.2 0.4 0.6 0.8 1

x

4.5 Galerkin’s methodThe least-squares method provides a best approximation in the variational sense for the solution of Lu = f when L is a linear operator by minimizing the distance between Lq and f, where q is a vector from the span of the basis functions. The Rayleigh-Ritz method provides a best approximation in the variational sense for the solution when L is positive defi nite and self-adjoint with respect to a defi ned inner product. The Rayleigh-Ritz approximation minimizes the distance between u and q with respect to the energy norm.

Galerkin’s method is an alternate method for formulating an approximation to the solution of Lu = f, which may be used when L does not satisfy the con-straints of the Rayleigh-Ritz method or perhaps the least-squares method.

Let SD = {φ 1, φ 2, … ,φn} be a set of linearly independent vectors in the domain of L, and let SR = {ψ 1, ψ 2, … ,ψn} be a set of linearly independent vectors in the range of L. Let ( f,g) represent a valid inner product defi ned in R. The Galerkin approximation is of the form

u ==

∑α φi i

i

n

1

(4.36)

The error in the approximation is

e Lu f

L f

= −

=⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟−

=∑

ˆ

α φi i

i

n

1

(4.37)

9533_C004.indd 2709533_C004.indd 270 10/29/08 4:36:17 PM10/29/08 4:36:17 PM


The Galerkin method requires that e be orthogonal to every member of the set SR. That is,

e, , ,...,ψk n( )= =0 1 2k (4.38)

Equation 4.38 represents a set of simultaneous algebraic equations to solve for the coeffi cients in the Galerkin expansion, Equation 4.38. Enforcement of Equation 4.38 implies that the error vector has no projection onto any of the vectors in the set SR.

If L is a linear operator, then Equation 4.38 can be rearranged using the fol-lowing steps allowed by the linearity of L and properties of inner products:

01

= −⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

= (

=∑α φ ψ

α φ ψ

i i

i

n

k

i i k

L f

L

,

, ))−( )=

∑i

n

k

1

f,ψ (4.39)

Equation 4.39 can be rearranged into the following set of simultaneous linear algebraic equations:

L f kφ ψ α ψi k i

i

n

k n, , , ,...,( ) =( ) ==

∑1

1 2 (4.40)

If the sets of vectors SR and SD are identical, then Equation 4.40 becomes

L fφ φ φi k i

i

n

k k n, , , ,...,( ) =( ) ==

∑ α1

1 2 (4.41)

Equation 4.41 is the same set of equations as that developed from the Rayleigh-Ritz formulation when L is a positive defi nite and self-adjoint operator.

Example 4.9 A fi rst-order model of the nondimensional defl ection of a microscale cantilever beam due to electrostatic actuation is

d udx u

4

4 21= −

+φ

( ) (a)

The cantilever beam is subject to the boundary conditions u du dx(0) 1, /=d u dx d u dx(0) 0, / (1) 0 and / (12 2 3 3= = )) 0= . Two linearly independent functions

which satisfy these conditions are

v x x x x12 4 53

4

1

5( ) = − + (b)

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v x x x x23 4 53

2

3

10( ) = − + (c)

Determine a Galerkin approximation for the solution of Equation a with φ = 1, using Equation b and Equation c as basis functions for both SD and SR.

Solution A Galerkin approximation is assumed in the form of

ˆ( ) ( ) ( )u x v x v x= +α α1 1 2 2 (d)

Equation a can be rewritten as

( )1 124

4+ = −u

d udx

(e)

Equation e is nonlinear, and therefore the formulation of Equation 4.41 can-not be used. The error in using Equation d as an approximation is

e x u

d udx

v v x

( ) ( ˆ )ˆ

= + +

= + +[ ] − +

1 1

1 12 24

24

4

1 1 2 22α α (( ) + − +( )[ ]+α α1 236 36 1x (f)

Requiring e(x) to be orthogonal to v1(x) with respect to the standard inner product on C4[0,1] leads to

e x v x dx

v v x

( ) ( )1

0

1

1 1 2 22

1

0

1 12 24

∫ =

+ +[ ] − +( )[{ α α α00

1

2 136 36 1 0∫ + − +( ) ]+ } =x v dxα (g)

Integration and simplifi cation of Equation g lead to

1 23 2 275 0 987 1 292 0 0871 2 1

21 2 2

2. . . . .α α α α α α− + − −

+ 00 202 0 177 0 011 3 606 1013

12

2 1 22 3. . . .α α α α α− + − × − α22

3 0 217` .= − (h)

Requiring e(x) to be orthogonal to v2(x) with respect to the standard inner product on C4[0,1] leads to

9533_C004.indd 2729533_C004.indd 272 10/29/08 4:36:18 PM10/29/08 4:36:18 PM


e x v x dx

v v x

( ) ( )2

0

1

1 1 2 22

1

0

1 12 24

∫ =

+ +[ ] − +( )[{ α α α00

1

2 236 36 1 0∫ + − +( ) ]+ } =x v dxα (i)

Integration and simplifi cation of Equation i lead to

− − − − −0 685 0 254 0 036 0 148 0 0261 2 1 2 1

22. . . . .α α α α α α22

12

23

1 23

230 029 1 11 10 0 037 0+ − − =−. . .α α α α αx

(j)

The solutions of the nonlinear algebraic equations Equation h and Equation j are not unique. One solution is α1 = 0.032 and α2 = 0.080.

Problems 4.1. a. Find the least squares approximation to f(x) = xe − 2x over the

interval −1 ≤ x ≤ 1 from the span of 1, x, x2 using the stan-dard inner product on C[−1,1].

b. Calculate the error in the approximation using the inner product generated norm.

c. Determine the Taylor series expansion for f(x) through the quadratic term. Calculate the error in this approxi-mation and compare to the error in the least squares approximation.

4.2. Repeat Problem 4.1 using the Legendre polynomials P0 (x), P1 (x), P2 (x) as basis functions.

4.3. Find the least squares approximation for f(x) = cos 2 (2x) over the interval 0 ≤ x ≤ 2π using 1, x2, x4 as basis functions and using the standard inner product on C[0,2π].

4.4. Repeat Problem 4.3 using a set of orthonormal functions which have the same span as 1, x2, x4 as basis functions.

4.5. Find the least squares approximation to J0 (x) using 1, x2, x4 as basis functions and using the inner product defi ned by ( , ) ( ) ( ) .f g f x g x xdx= ∫ 0

1

4.6. Find the least squares approximation to J1 (x) using x, x3, x5 as basis functions and using the inner product defi ned by ( , ) ( ) ( ) .f g f x g x xdx= ∫ 0

1

4.6. Determine the least squares quadratic fi t for y as a function of x for the data

X 0 0.25 0.5 0.75 1.0 1.25 1.5 1.75 2.0

Y 3.0 2.2 2.0 1.9 2.5 3.4 5.1 7.5 10.0

9533_C004.indd 2739533_C004.indd 273 10/29/08 4:36:19 PM10/29/08 4:36:19 PM


4.7. The following table presents the data for lattice sums of an atom above a carbon

X (nm) 1.0 1.5 2.0 2.5 3.0 3.5 4.0

S 2.40 0.50 0.17 0.076 0.040 0.025 0.015

By observation it is determined that the data may fi r a curve of the form S = ax b. Taking the natural logarithm of the proposed regression equation leads to ln(s) = ln(a) + b ln(x). Determine the linear least squares approximation for ln(s) as a function of ln(x). Use the regression to approximate S for x = 2.8 nm and extrapo-late S for x = 5.0 nm.

4.8. The fi rst ten zeroes of the Bessel function of the fi rst kind of order zero J0 (x) are 2.405, 5.530, 8.654, 11.792, 14.931, 18.071, 21.212, 24.352, 27.493 and 30.653. Use a quadratic least squares regression to predict the zeroes as a function of the number of the zero.

Problems 4.9–4.17 refer to the system of Figure P4.9. The differ-ential equation for the non-dimensional steady-state tempera-ture in the bar which is subject to an external heat generation over its length is

ddx

xddx

x q xαθ

β θ( ) ( ) ( )⎡⎣⎢⎢

⎤⎦⎥⎥− = (a)

4.9. Consider a uniform bar such that α(x) = 1, β(x) = 0.5 with q(x) = x(1 − x). The bar is subject to a constant temperature at x = 0 and is insulated at x = 1 such that θ(0) = 0, dθ/dx (1) = 0. Let S be the vector space which is the intersection of P3[0,1] with the subspace of C[0,1] of all functions satisfying the homogeneous boundary conditions. Find the best least squares approximation for the temperature distribution in the bar from S.

x

o

1

α(x)=

q(x)β(x)=

A(x)A

hP(x)LkP

o

Figure P4.9

9533_C004.indd 2749533_C004.indd 274 10/29/08 4:36:19 PM10/29/08 4:36:19 PM


4.10. Consider a uniform bar such that α(x) = 1,β(x) = 0.5 with q(x) = x(1−x). The bar is subject to a constant temperature at x = 0 and is insulated at x = 1 such that θ(0) = 0, dθ/dx (1) = 0. Let S be the vector space which is the intersection of P3 [0,1] with the subspace of C[0,1] of all functions satisfying the homogeneous boundary conditions. Fin the best Rayleigh-Ritz approximation for the temperature distribution from S.

4.11. Repeat Problem 4.10 but using an orthonormal basis for S with respect to the energy inner product.

4.12. Consider a uniform bar such that α(x) = 1,β(x) = 0.5 with q(x) = x(1−x). The bar is subject to a constant temperature at x = 0 and is open for heat transfer at x = 1 such that θ(0) = 0, dθ/dx (1) + 0.5θ (1) = 0. Let S be the vector space which is the intersec-tion of P3 [0,1] with the subspace of C[0,1] of all functions sat-isfying the homogeneous boundary conditions. Find the best Rayleigh-Ritz approximation for the temperature distribution in the bar from S.

4.13. Use the Rayleigh-Ritz method to approximate the temperature dis-tribution in a surface with α(x) = (1−0.1x)3/2, and q(x) = x(1 − x). The end x = 0 is maintained at a constant temperature, θ(0) = 0 while its other end is insulated dθ/dx (1) = 0. Use a basis for the intersection of P4 [0,1] with the subspace of C[0,1] of all functions which satisfy the boundary conditions. Compare with the exact solution.

4.14. Use the Rayleigh-Ritz method to approximate the temperature distribution in a surface with α(x) = (1 − 0.1x)3, β(x) = (1 − 0.1x)3/2 and q(x) = x(1 − x). The end x = 0 is maintained at a constant tem-perature, θ(0) = 0 while its other end is insulated dθ/dx (1) = 0. Use a basis for the intersection of P 4 [0,1] with the subspace of C[0,1] of all functions which satisfy the boundary conditions. Compare with the exact solution.

4.15. Use the Rayleigh-Ritz method to approximate the tempera-ture distribution in a surface with α(x) = (1 − 0.1x), β(x) = 1 and q(x) = x(1 − x). The end x = 0 is maintained at a constant tempera-ture, θ(0) = 0

while its other end is insulated dθ/dx (1) = 0. Use a

basis for the intersection of P 4 [0,1] with the subspace of C[0,1] of all functions which satisfy the geometric boundary condition only. Use the energy formulation of the inner products. Com-pare with the exact solution.

4.16. Use the fi nite-element method to solve Problem 4.10. Use fi ve equally spaced elements between x = 0 and x = 1.

4.17. Use the fi nite-element method to solve Problem 4.15. Use fi ve equally spaced elements between x = 0 and x = 1.

Problems 4.18–4.19 refer to the approximate solution of the non-dimensional differential equation for the static defl ection of a

9533_C004.indd 2759533_C004.indd 275 10/29/08 4:36:20 PM10/29/08 4:36:20 PM


stretched beam on an elastic foundation and subject to trans-verse loading, as illustrated in Figure P4.18. In all problems the subspace S refers to the intersection of P6 [0,1] with a subspace of C4 [0,1] which consists of all functions which satisfy the homo-geneous boundary conditions,

ddx

xd wdx

d wdx

w f x2

2

2

2

2

2α ε η( ) ( )⎡

⎣⎢⎢

⎤

⎦⎥⎥− + =

(a)

Use f(x) = 2x2 (1 − x)2 for all problems.

4.18. Use the least squares method to approximate the defl ection of a uniform, α(x) = 1 fi xed-pinned beam w(0) = 0, dw/dx (0) = 0, w(1) = 0, d2w/dx2 (1) = 0 with ε = 0 and η = 2. Use basis functions for S as trial functions.

4.19. Use the Rayleigh-Ritz method to approximate the defl ection of a uniform, α(x) = 1 fi xed-pinned beam w(0) = 0, dw/dx (0) = 0, w(1) = 0, d2w/dx2 (1) = 0 with ε = 0 and η = 2. Use basis functions for S as trial functions.

4.20. Use the Rayleigh-Ritz method to approximate the defl ection of a uniform, α(x) = 1 fi xed-free beam w(0) = 0, dw/dx (0) = 0, d2w/dx2 (1) = 0, d3w/dx3(1) = 0 with ε = 1 and η = 2. Use basis functions for S as trial functions.

4.21. Use the Rayleigh-Ritz method to approximate the defl ection of a uniform, α(x) = 1 fi xed-free beam w(0) = 0, dw/dx (0) = 0, d2w/dx2 (1) = 0, d3w/dx3 (1) = 0 with ε = 1 and η = 2. Use basis functions from P5 [0,1] which satisfy only the geometric boundary condi-tions as basis functions.

4.22. Repeat Problem 4.20, but with α(x) = 1 + 0.2x. 4.23. Solve Problem 4.19 using the fi nite element method. Use fi ve

equally spaced elements. 4.24. Use the fi nite-element method to solve Problem 4.20, but with

α(x) = (1 + 0.25x) 2, Use fi ve equally spaced elements.

k

xP

1

Figure P4.18

9533_C004.indd 2769533_C004.indd 276 10/29/08 4:36:21 PM10/29/08 4:36:21 PM

277

Chapter 5

Eigenvalue problems

5.1 Eigenvalue and eigenvector problemsAn eigenvalue of an operator L is a scalar λ for which a nontrivial solution (u ≠ 0) of

Lu = λu (5.1)

exists. The resulting nontrivial solution is called an eigenvector of L corre-sponding to the eigenvalue λ.

An eigenvector corresponding to an eigenvalue of a linear operator is not unique. Let λ be an eigenvalue of a linear operator L, and let u ≠ 0 be a corresponding eigenvector satisfying Equation 5.1. Defi ne v = cu for an arbitrary c ≠ 0. Consider

Lv L u

Lu

u

v

=

=

=

=

( )c

c

cλ

λ

(5.2)

Equation 5.2 illustrates that v is also an eigenvector of L corresponding to the eigenvalue λ.

Thus, an eigenvector of a linear operator is unique to at best a multiplica-tive constant.

If all the eigenvectors of an operator corresponding to an eigenvalue are multiples of one another, the eigenvalue is nondegenerate. If two or more linearly independent vectors exist for an eigenvalue, that eigenvalue is said to be degenerate.

Let λ1 and λ2 be distinct eigenvalues of a linear operator L with correspond-ing eigenvectors u1 and u2, respectively. If the eigenvectors are linearly depen-dent, then there is some c ≠ 0 such that u2 = cu1. Then

Lu L u2 = ( )c 1 (5.3)

Using the linearity of L and the defi nition of the eigenvalue-eigenvector problem, Equation 5.3 becomes

9533_C005.indd 2779533_C005.indd 277 10/29/08 2:29:05 PM10/29/08 2:29:05 PM


λ λ

λ

2 2 1 1

2

u u

u

=

=

c

1

(5.4)

Equation 5.5 implies that λ1 = λ2, which contradicts the assumption that the eigenvalues are distinct. Thus, the assumption that the vectors are linearly dependent is false. Since the pair of eigenvectors chosen is arbitrary, it is eas-ily shown that if all pairs of vectors in a set are linearly independent, then all vectors in the set are linearly independent. Since the pair of eigenvectors chosen is arbitrary, all pairs of eigenvectors corresponding to distinct eigen-values are linearly independent. Thus the set of eigenvectors corresponding to distinct eigenvalues of a linear operator are linearly independent.

5.2 Eigenvalues of adjoint operatorsLet L be a linear operator whose domain is the vector space SD. Let (u, v) con-stitute a valid inner product on SD. Let L* be the adjoint of L, whose domain is SR , the range of L, with respect to the inner product (u, v). Thus

( , ) ( , )*Lu v u L v= (5.5)

for all u in SD and all v in SR. The eigenvalue-eigenvector problem for L is of the form of Equation 5.1, and the eigenvalue-eigenvector problem for L* is

L v v* = μ (5.6)

where μ is an eigenvalue of L* and v is an eigenvector of L* corresponding to μ.

Taking the inner product of Equation 5.1 with v and using properties of the inner product leads to

Lu v u v

u v

, ,

,

( )=( )

= ( )

λ

λ (5.7)

Using the defi nition of the adjoint on the left-hand side of Equation 5.7 gives

u L v u v, ,*( )= ( )λ (5.8)

However, since v is the eigenvector of L* corresponding to the eigenvalue μ, Equation 5.8 can be rewritten as

u v u v, ,μ λ( )= ( ) (5.9)

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Chapter 5: Eigenvalue problems 279

Noting that the properties of a valid inner product imply that u v u v, ,μ μ( )= ( ), Equation 5.9 can be rearranged to yield

μ λ−( )( )=u v, 0 (5.10)

Equation 5.10 implies that either (u, v) = 0 or λ μ= . This conclusion leads to the following:

Theorem 5.1 (Biorthogonality) If λ is an eigenvalue of a linear operator L with a corresponding eigenvector u, and if μ λ≠ is an eigenvalue of L* with corresponding eigenvector v, then (u, v) = 0.

Proof The theorem follows directly form Equation 5.11.

Theorem 5.2 If L* is the adjoint of a linear operator L with respect to a valid inner product, then the eigenvalues of L* are the complex conjugates of the eigenvalues of L.

Proof Equation 5.11 implies that unless μ λ= , the eigenvectors of L and L* are mutually orthogonal. Suppose that λ is an eigenvalue of L, but that λ is not an eigenvalue of L*. Then the eigenvector u is orthogonal to all eigenvec-tors of L*. It is mentioned in Section 5.6 that the set of eigenvectors of L* is complete on SR. The statement that u is orthogonal to all eigenvectors of L* is thus contradictory, and the theorem is proved by contradiction.

Theorem 5.3 Eigenvalues of a self-adjoint operator are real.

Proof From Theorem 5.2, every eigenvalue of an adjoint operator L* is the complex conjugate of an eigenvalue of the operator L. If λ is an eigenvalue of a self-adjoint operator L, then λ λ= .

Theorem 5.4 Eigenvectors corresponding to distinct eigenvalues of a self-adjoint operator are mutually orthogonal with respect to the inner product for which the operator is self-adjoint.

Proof If the operator is self-adjoint, then u and v are eigenvectors correspon-ding to eigenvalues of L. The theorem follows directly from Equation 5.10.

Theorem 5.3 implies that if L is self-adjoint with respect to any inner prod-uct, then its eigenvalues are real. Theorem 5.4 then implies that if L is self-adjoint with respect to an inner product, then the eigenvectors are mutually orthogonal with respect to that inner product.

5.3 Eigenvalues of positive defi nite operatorsSuppose that L is positive defi nite with respect to a defi ned inner product (u, v). That is,

Lu u,( )≥ 0 (5.11)

9533_C005.indd 2799533_C005.indd 279 10/29/08 2:29:07 PM10/29/08 2:29:07 PM


for all u in SD, and (Lu, u) = 0 if and only if u = 0. Let λ be an eigenvalue of a positive defi nite operator with an eigenvector u, Lu = λu. Taking the inner product of this equation with u leads to (Lu, u) = λ(u, u) or

λ =( )Lu u

u

,2 (5.12)

where the norm is the inner-product-generated norm. Since L is positive def-inite, the right-hand side of Equation 5.12 is that of two positive quantities. This proves

Theorem 5.5 All eigenvalues of a positive defi nite operator are positive.Consider an operator L which has an eigenvalue λ = 0. Then there exists

u ≠ 0 such that Lu = 0. For this u, (Lu,u) = (0,u) = 0. Hence the operator is, at best, non-negative defi nite. If an operator is non-negative defi nite, then Equa-tion 5.13 shows that all its eigenvalues are non-negative. To show that λ = 0 must be an eigenvalue of a non-negative defi nite operator, it is necessary to show that (Lu, u) = 0 implies that u = 0 or Lu = 0. It is shown in Section 5.6 that this is true for self-adjoint operators.

Theorem 5.6 If L is self-adjoint and non-negative defi nite with respect to an inner product (u, v), then λ = 0 is an eigenvalue of L.

5.4 Eigenvalue problems for operators in fi nite-dimensional vector spaces

Let A be a real n × n matrix. The eigenvalue-eigenvector problem for A is of the form

Au u= λ (5.13)

where u is a vector in Rn. Let I be the n × n identity operator, Iu = u. Equation 5.13 can be rewritten as

A I u 0−( ) =λ (5.14)

Equation 5.14 can be expanded into

a a a a

a a a a

a a

n

n

1 1 1 2 1 3 1

2 1 2 2 2 3 2

3 1

, , , ,

, , , ,

,

−−

λλ

…

…

33 2 3 3 3

1 2 3

, , ,

, , , .

a a

a a a a

n

n n n n n

−

⎡

⎣

⎢⎢⎢⎢ λ …

� � � � �

…

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

u

u

u

un

1

2

3

�⎥⎥⎥⎥⎥⎥

=

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥

0

0

0

0

� (5.15)

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Equation 5.15 represents a system of n simultaneous linear equations to solve for the components of the eigenvector. A nontrivial solution exists if and only if the determinant of the coeffi cient matrix is zero, that is,

a a a a

a a a a

a a

n

n

1 1 1 2 1 3 1

2 1 2 2 2 3 2

3 1

, , , ,

, , , ,

,

−−

λλ

…

…

33 2 3 3 3

1 2 3

0, , ,

, , , .

a a

a a a a

n

n n n n n

− =λ …

� � � � �

…

(5.16)

The determinant of a square matrix is a combination of permutations of ele-ments of the matrix. Each permutation is a product of one element from each column. One permutation is the product of the diagonal elements. Thus it is clear that evaluation of the right-hand side of Equation 5.16 yields a polyno-mial in powers of λ with highest power n. The eigenvalues are the roots of this polynomial, called the characteristic polynomial.

Since the eigenvalues are the roots of a polynomial of order n, an n × n matrix has n eigenvalues. All eigenvalues may not be distinct; some may be repeated. If A is a real matrix, then all coeffi cients of its characteristic poly-nomial are real. This implies that if an eigenvalue is complex, that it has a nonzero imaginary part, then its complex conjugate is also an eigenvalue. The eigenvectors are the nontrivial solutions of Equation 5.15 corresponding to each eigenvalue.

It has been shown in chapter 2 that the adjoint of a real n × n matrix A with respect to the standard inner product on Rn is the transpose of the matrix, A* = AT. A matrix is self-adjoint with respect to the standard inner product if it is symmetric. Thus, from Theorems 5.3 and 5.4, all eigenvalues of a sym-metric matrix are real, and the corresponding eigenvectors are mutually orthogonal with respect to the standard inner product on Rn.

Example 5.1 The stress tensor σ≈

defi nes the state of stress at a point in a continuum:

σσ σ σσ σ σσ σ σ

11 12 13

21 22 23

31 32 33

≈=

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟ (a)

The stress tensor represents the stress vector acting on three mutually perpen-dicular planes passing through the point. The stress vector acting on a plane can be resolved into a component normal to the plane (the normal stress) and a component that lies in the plane (the shearing stress), as illustrated in Figure 5.1. The diagonal elements are the normal stresses, the component of stress normal to the planes whose normals are in the x-, y-, and z-directions. The off-diagonal elements are the shear stresses; σij is the stress acting in

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the j-direction on the plane whose normal is in the i-direction. The angular momentum equation can be used to show that the stress tensor is symmetric (σij = σji) in the absence of body coupling or intrinsic angular momentum.

The stress vector acting on a plane can be written in the form

σ=≈σn (b)

where n is a unit vector normal to the plane. The stress vector can be resolved into two components: a normal stress perpendicular to the plane and a shear stress in the plane:

σ= +σ τn m (c)

where σ is the normal stress, τ is the shear stress, and m is a unit vector in the plane.

A principal stress is defi ned as the normal stress on a plane on which the shear stress is zero:

σ= σn (d)

Equating Equation b and Equation d shows that the stress vector acting on a plane of principal stress is

σ σ≈

=n n (e)

τ

σ

Figure 5.1 The stress vector acting on a plane is resolved into a component normal to the plane (the normal stress) and a component in the plane (the shear stress).

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Equation e can be written in matrix form as

σ σ σσ σ σσ σ σ

11 12 13

21 22 23

31 32 33

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟=

⎛

⎝

⎜⎜⎜⎜

n

n

n

n

n

n

1

2

3

1

2

3

σ⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟ (f)

Thus, the principal stresses are the eigenvalues of the stress tensor, and their corresponding eigenvectors are vectors normal to the principal planes.

The principal stresses are eigenvalues of a 3 × 3 symmetric matrix. Thus there are three principal stresses. Since the stress tensor is symmetric, Theorems 5.3 and 5.4 imply that all principal stresses are real and that the planes on which the principal stresses act are mutually orthogonal. Note that the stress tensor is not necessarily positive defi nite, and that hence the principal stresses can be of either sign.

The eigenvectors are unique only to a multiplicative constant. Unit vectors normal to the principal planes are obtained by normalizing the eigenvectors, requiring that their standard norm be one:

n n n12

22

32 1+ + = (f)

Example 5.2 Natural frequencies of discrete systems: The differential equa-tions governing the free vibration of an undamped n-degree-of-freedom mechanical linear system such as that shown in Figure 5.2 are of the form

Mx Kx 0�� + = (a)

where x is the n-dimensional displacement vector, the vector of displacements of designated particles, M is the n × n mass matrix and K is the n × n stiff-ness matrix. As an example, the differential equations of the system shown in Figure 5.2 are

m

m

m

x

x

x

0 0

0 0

0 0 2

1

2

3

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥��

��

��

⎥⎥⎥⎥+

−− −

−

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

3 2 0

2 4 2

0 2 2

1

2

3

k k

k k k

k k

x

x

x

⎡⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥=

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

0

0

0

(b)

m m 2m

k 2k2k

x1 x2 x3

Figure 5.2 The natural frequencies of a multi-degree-of-freedom mechanical sys-tem are the square roots of the inverse of the system’s mass matrix multiplied by its stiffness matrix.

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When an energy formulation is used to derive the differential equations, both the mass and stiffness matrices are symmetric.

A normal-mode solution of Equation a can be assumed as

x X= ei tω (c)

where ω is a natural frequency of the system and X is its mode shape vector. Substitution of Equation c into Equation a leads to

KX MX= ω2 (d)

The mass matrix cannot be singular, hence its inverse M−1 exists, and one can premultiply Equation d by M−1 to obtain

M KX X− =1 2ω (e)

It can be concluded from Equation e that the natural frequencies of an n-degree-of-freedom system are the square roots of the eigenvalues of M−1K and that the mode shapes are their corresponding eigenvectors.

While it is guaranteed that M and K are symmetric matrices, M−1K is not necessarily symmetric. Thus M−1K is not necessarily self-adjoint with respect to the standard inner product on Rn. If v is the velocity vector for an n-degree-of-freedom system at any instant, then it can be shown that (Mv,v) = 2T, where T is the kinetic energy of the system at that instant. How-ever, the system’s kinetic energy is always positive unless its velocity vector is zero. Thus, since v is defi ned arbitrarily, the mass matrix is positive defi -nite with respect to the standard inner product, since it is also symmetric a kinetic-energy inner product is thus defi ned as

u v Mu v, ,( ) =( )M (f)

If u is the displacement vector for an n-degree-of-freedom linear system at any instant, it can be shown that (Ku,u) = 2V, where V is the potential energy of the system at that instant. The potential energy function for a sta-ble system is always positive. Furthermore, if the system is restrained from rigid-body motion, then only a state of zero displacement leads to a potential energy of zero. An unrestrained system such as that shown in Figure 5.3 has a displacement vector u = [1 1]T ≠ 0 such that V = 0. Thus the stiffness matrix is positive defi nite for a stable restrained system and non-negative defi nite for a stable unrestrained system. A potential-energy inner product may be defi ned for a stable restrained system as

u v Ku v, ,( ) =( )K (g)

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Consider the following derivation in which the defi nitions of energy inner products, the self-adjointness of K and M with respect to the standard inner product, and the properties of inner products are used:

M Ku v MM Ku v

Ku v

u Kv

u MM K

− −

−

( ) = ( )= ( )

= ( )

=

1 1

1

, ,

,

,

,

M

vv

Mu M Kv

u M Kv

( )

= ( )

= ( )

−

−

,

,

1

1

M (h)

It can be concluded from Equation h that M−1K is self-adjoint with respect to the kinetic-energy inner product. A similar procedure is used to show that M−1K is self-adjoint with respect to the potential-energy inner product. The positive-defi niteness of M−1K with respect to the kinetic-energy inner product is shown by

M Ku u MM Ku u

Ku u

− −( ) =( )

=( )

1 1, ,

,

M (i)

Equation i implies that if K is positive defi nite with respect to the standard inner product, then M−1K is positive defi nite with respect to the kinetic-energy inner product.

Since M−1K is self-adjoint with respect to the kinetic-energy inner prod-uct, Theorem 5.3 guarantees that all its eigenvalues are real. Theorem 5.5 and the positive-defi niteness of M−1K guarantee that all its eigenvalues are positive. Theorem 5.4 and the self-adjointness of M−1K lead to the conclusion that eigenvectors corresponding to distinct eigenvalues of M−1K are mutually orthogonal with respect to the kinetic-energy inner product.

m m

k

Figure 5.3 The stiffness matrix of an unrestrained system is singular. The lowest natural frequency of the system is zero, which corresponds to a rigid-body mode of vibration.

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Let 0 < λ < λ2 < … < λn be the eigenvalues of M−1K. The natural frequencies are ωk k k n= =λ 1 2, ,..., . Let Xk be the eigenvector (mode shape) correspon-ding to the natural frequency ω k . Mode-shape orthogonality implies that

0 =( ) ≠X Xi i j, j M (j)

The mode shapes are normalized by requiring that

X Xi , i( ) =M 1 (k)

If the mode shapes are normalized according to Equation k, then

X X KX X

MX X

X X

i i i i

i i

i i

, ,

,

,

( ) =( )

=( )

= ( )

=

K

M

ω

ω

ω

i

i

2

2

ii2

(l)

Example 5.3 Determine the natural frequencies and normalized mode-shape vectors for the system shown in Figure 5.2 and demonstrate orthogonality of the mode shapes.

Solution Using the results of Example 5.2, the natural frequencies are the square roots of the eigenvalues of M−1K. Using the mass and stiffness matrices from Equation b of Example 5.2,

M K− =−

− −−

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

1

3 2 0

2 4 2

0 1 1

φ

(a)

where φ = k/m. The eigenvalues of M−1K are determined from

0

3 2 0

2 4 2

0

1= −

=− −

− − −− −

−M K Iλ

φ λ φφ φ λ φ

φ φ λ

(b)

The characteristic equation obtained by expanding the determinant of Equa-tion b is

− + − + =λ φλ φ λ φ3 2 2 38 13 2 0 (c)

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The roots of Equation c are 0.1716 φ, 2 φ, and 5.8284 φ. The natural frequencies, the square roots of the eigenvalues, are

ω ω ω1 2 30 4142 1 4142 2 4142= = =. . .km

km

kmm

(d)

The eigenvector corresponding to the middle eigenvalue is obtained from

3 2 0

2 4 2

0

2

2

2

1

2

φ λ φφ φ λ φ

φ φ

− −− − −

− −

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥λ

X

X

XX3

0

0

0

2 0

2

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥=

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

−−

φ φφ 22 2

0

01

2

3

φ φφ φ

−− −

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥=

X

X

X

00

0

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

(e)

The fi rst line of the matrix system of Equation e leads to X1 = 2X2, while the third line leads to X3 = −X2. Letting X2 be an arbitrary constant C, an eigen-vector corresponding to λ2 is X2 2 1 1= −[ ]C T

. The eigenvector is normal-ized by requiring that X X2 2 M,( ) = 1. To this end,

C m2 2 1 1

1 0 0

0 1 0

0 0 2

2

1

1

−[ ]⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥ −

⎡

⎣

⎢⎢⎢

⎢⎢

⎤

⎦

⎥⎥⎥⎥= 1 (f)

Evaluation of the inner product in Equation f and solving for C gives C m= 0 3780. / , leading to a normalized eigenvector, X2 1 0 7559 0 3980 0 3980= −( )[ . . . ]/ m . The normalized eigenvectors cor-responding to natural frequencies of ω1 and ω3 are obtained using simi-lar procedures. The results are X1

T1/ 0.3366 0.4760 0.5745= m [ ] are

X3 1 0 5615 0 7941 0 1645= −( )[ . . . ]/ m T .

5.5 Second-order differential operatorsThe general form of a linear second-order differential operator is

Ly a xd ydx

a xdydx

a x y= + +2

2

2 1 0( ) ( ) ( ) (5.17)

The right-hand side of Equation 5.17 can be manipulated to yield an alternate representation,

Lyr x

ddx

p xdydx

q x y=⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟+

⎡

⎣⎢⎢

⎤

⎦⎥⎥

1

( )( ) ( ) (5.18)

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The functions in Equation 5.18 are defi ned such that r(x) > 0. Application of the product rule for differentiation to the right-hand side of Equation 5.18 leads to

Lyr x

p xd ydx

dpdx

dydx

q x y= + +⎡

⎣⎢⎢

⎤

⎦⎥⎥

1 2

2( )( ) ( ) (5.19)

The operator in Equation 5.19 is equivalent to the operator in Equation 5.17 if

a xp xr x

2( )( )

( )= (5.20)

a xr x

dpdx

11

( )( )

= (5.21)

aq xr x

0 =( )

( ) (5.22)

Dividing Equation 5.21 by Equation 5.20 leads to

a xa x p x

dpdx

1

2

1( )

( ) ( )= (5.23)

which when integrated gives

ln ˆpaa

dx C= +∫ 1

2

(5.24)

or

p x Ceaa

dx( ) =

∫ 1

2 (5.25)

Without loss of generality, C = 1.Equation 5.20 implies that

r xp xa x

( )( )

( )=

2

(5.26)

and from Equation 5.22

q x a x r x( ) ( ) ( )= 0 (5.27)

If Equation 5.26 leads to r(x) < 0, the negative sign is shifted to p(x) to ensure that r(x) is always positive.

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The operator L is defi ned for all y(x) in SD, a subspace of C2[a, b]. Defi ne an inner product on C2[a, b] as

( , ) ( ) ( ) ( )f g f x g x r x dxra

b

= ∫ (5.28)

Consider

( , )( )

( ) ( )Lf gr x

ddx

p xdtdx

q x fr

a

b

=⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟+∫ 1

(( ) ( ) ( )

( )

x g x r x dx

ddx

p xdtdx

a

b

⎡

⎣⎢⎢

⎤

⎦⎥⎥

=⎡

⎣⎢⎢

⎤

⎦∫ ⎥⎥⎥

+∫g x dx q x f x g x dxa

b

( ) ( ) ( ) ( ) (5.29)

Using integration by parts on the fi rst integral of Equation 5.29 and choosing u = g(x) and dv = (d/dx)[p(x)(df/dx)]dx such that du = (dg/dx)dx and v = p(x)(df/dx) leads to

( , ) ( ) ( ) ( ) (Lf g p x g xdfdx

p xdfdx

dgdx

dx qra

b

a

b

= − +∫ xx f x g x dxa

b

) ( ) ( )∫ (5.30)

Using integration by parts on the fi rst integral of Equation 5.30 and choosing u = p(x)(dy/dx) and dv = (df/dx)dx such that du = d/dx[p(x)(dg/dx)] dx and v = f(x) leads to

( , ) ( ) ( ) ( ) ( )

Lf g p x g xdfdx

p xdgdx

f x

ddx

ra

b

a

b

= −

+ pp xdgdx

f x q x f x g xa

( ) ( ) ( ) ( ) ( )⎡⎣⎢⎢

⎤⎦⎥⎥

+⎡⎣⎢⎢

⎤⎦⎥⎥

bb

a

b

a

b

dx

p x g xdfdx

p xdgdx

f x

r x

∫

= −

+

( ) ( ) ( ) ( )

( )

1

aa

b

ddx

p xdgdx

q x g x f x r x dx∫ ( )+⎡⎣⎢

⎤⎦⎥

=

( ) ( ) ( ) ( ) ( )

pp x g xdfdx

p xdgdx

f x f Lga

b

a

b

r( ) ( ) ( ) ( ) ( , )− +

(5.31)

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It is clear from Equation 5.31 that L is self-adjoint with respect to the inner product defi ned in Equation 5.28 if

p x g xdfdx

p xdgdx

f xa

b

a

b

( ) ( ) ( ) ( )− = 0 (5.32)

for all f(x), g(x) in SD. Note that

p x g xdfdx

p xdgdx

f x

p b g bdfdx

a

b

a

b

( ) ( ) ( ) ( )

( ) ( ) (

−

= bb p a g adfdx

a p bdgdx

b f b p adgd

) ( ) ( ) ( ) ( ) ( ) ( ) ( )− − +xx

a f a

p b g bdfdx

bdgdx

b f b

( ) ( )

( ) ( ) ( ) ( ) ( )= −⎡⎣⎢⎢

⎤⎦⎥⎥⎥− −⎡

⎣⎢⎢

⎤⎦⎥⎥

p a g adfdx

adgdx

a f a( ) ( ) ( ) ( ) ( )

(5.33)

Clearly the right-hand side of Equation 5.33 is zero if p(b) = p(a) = 0.The general form of the boundary conditions at x = a and x = b is

y a y a( ) ( )+ ′ =α 0 (5.34)

y b y b( ) ( )+ ′ =β 0 (5.35)

Since f(x) and g(x) satisfy these conditions, substitution of Equation 5.34 and Equation 5.35 into Equation 5.33 leads to

p b g bdfdx

bdgdx

b f b p a g a( ) ( ) ( ) ( ) ( ) ( ) (−⎡⎣⎢⎢

⎤⎦⎥⎥− )) ( ) ( ) ( )

( ) ( )

dfdx

adgdx

a f a

p b g b f

−⎡⎣⎢⎢

⎤⎦⎥⎥

= − ′ ′β (( ) ( )( ( ))

( ) ( ) ( )

b g b f b

p a g a f a

− ′ − ′[ ]

− − ′ ′ − ′

β

α gg a f a( )( ( ))− ′[ ]=α 0

(5.36)

If α = 0 in Equation 5.33, then the terms in Equation 5.33 evaluated at x = a are identically zero. If α = ∞, such that Equation 5.34 is y′(a) = 0 then the terms in Equation 5.33 evaluated at x = a are identically zero.

Consider next the case where p(a) = p(b) and where SD is defi ned to be com-posed of all functions y(x) such that y(a) = y(b) y′(a) = y′(b) In this case, the right-hand side of Equation 5.32 becomes

p b g b f b g b f b p a g a f a g( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) (′ ′ ′ ′−[ ]− − aa f a

p b g b f b g b f b p a g a

) ( )

( ) ( ) ( ) ( ) ( ) ( ) (

[ ]

− −[ ]+′ ′ )) ( ) ( ) ( )f a g a f a′ ′−[ ]= 0 (5.37)

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The following theorem has been proved:

Theorem 5.7 Any linear second-order differential operator can be written as Ly r x d dx p x dy dx q x y= ( )+[ ]1/ / /( ) ( )( ) ( ) , which is self-adjoint with respect to the inner product ( , ) ( ) ( ) ( )f g f x g x r x dxr a

b= ∫ , provided that

(i) either p(a) = 0 or SD is defi ned such that y a y a( ) ( )+ =α ′ 0 for all y in SD

and either p(b) = 0 or SD is defi ned such that y b y b( ) ( )+ =β ′ 0 for all y in SD

or(ii) p(a) = p(b) and SD is defi ned by

y(a) = y(b)

and y′(a) = y′(b)

Now consider the positive defi niteness of L. Using Equation 5.30, after one application of integration by parts,

( , ) ( ) ( ) ( ) (Lf f p x f xdfdx

p xdfdx

dfdx

dx qra

b

a

b

= − +∫ xx f x f x dxa

b

) ( ) ( )∫ (5.38)

If p(x) ≤ 0, then −∫ [ ] ≥ab p x df dx dx( ) /

20. Now if p(a) = p(b) = 0, then the bound-

ary terms are zero. Also if p(a) = p(b) and if f(a) = f(b), and f ′(a) = f ′(b), then the boundary terms are zero.

Now consider boundary conditions of the form of Equation 5.34 and Equa-tion 5.35. Then from Equation 5.37, the boundary terms simplify to

( ) ( )( ( )) ( ) ( )( ( ))p b y b y b p a y a y a′ β ′ ′ α ′

β

− − −

= − pp b y b p a y a( ) ( ) ( ) ( )′ α ′2 2+ (5.39)

If p(a) < 0 and p(b) < 0, then the boundary terms are non-negative if β ≥ 0 and α ≤ 0. If a boundary condition is simply y(a) = 0 or y(b) = 0, then the corre-sponding boundary term is zero.

From the above and from Equation 5.39, it is clear that if p(x) ≤ 0 and q(x) ≥ 0 for all x, a ≤ x ≤ b, and if boundary conditions are of the form of Equation 5.35 and Equation 5.36 with β ≥ 0 and α ≤ 0, then L is at least non-negative defi nite. Furthermore, Equation 5.39 shows that if q(x) > 0 for some x, then L is positive defi nite. Thus it only remains to determine the conditions under which, if q(x) = 0, there exists an f(x) ≠ 0 such that (Lf,f)r = 0. From Equation 5.39 and Equation 5.40, this is possible only if df/dx = 0 for all x, a ≤ x ≤ b, which requires that f (x) = c0 + c1x. If the boundary conditions are of the form of Equation 5.35 and Equation 5.36, then application of Equation 5.35 leads to c1(1 + α a) + α c0 = 0 and c1(1 + βb) + βc0 = 0. The only solution of these equations is c1 = c0 = 0.

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However, if the boundary conditions are such y′(a) = 0 and y′(b) = 0, then f(x) = c0 satisfi es both conditions and is in the domain of L. In this latter case, L is non-negative defi nite.

If the domain of L is specifi ed by boundary conditions of the form y(a) = y(b) and y′(a) = y′(b), then f(x) = c0 is in the domain of L, and L is non-negative defi nite.

The above results can be summarized in the following theorem:

Theorem 5.8 A second-order differential operator defi ned by Ly = 1/r(x)d[p(x)dy/dx]/dx + [q(x)y] is positive defi nite with respect to the inner product ( , ) ( ) ( ) ( )f y f x g x r x dxa

b= ∫ if p(x) ≤ 0 and q(x) > 0 for all x, a ≤ x ≤ b and SD is defi ned by the conditions of Theorem 5.7 with α ≤ 0 and β ≥ 0. If q(x) = 0 and SD is defi ned by the boundary conditions of Equation 5.35, then L is positive defi nite only if either α or β has a fi nite value. Also, if q(x) = 0, p(a) = p(b) = 0, and if SD is defi ned such that y(a) = y(b) and y′(a) = y′(b). then L is non- negative defi nite.

A second-order differential operator with boundary conditions such that it is self-adjoint and non-negative defi nite with respect to the inner product ( , ) ( ) ( ) ( )f g f x g x r x dxr a

b= ∫ constitutes a Sturm-Liouville system. The eigen-value problem for a Sturm-Liouville system, called a Sturm-Liouville problem, is of the form

1

r xd

dxp x

dydx

q x y y x( )

( ) ( ) ( )( )+⎡⎣⎢⎢

⎤⎦⎥⎥= λ (5.40)

with boundary conditions as specifi ed in Theorems 5.7 and 5.8.The domain of a Sturm-Liouville operator L is specifi ed by boundary

conditions derived from the geometry and physics of a physical problem. Note that if p(a) = 0, then the only condition required at x = a is that y(a) and y′(a) remain fi nite. Often Sturm-Liouville operators obtained from modeling physical systems using cylindrical coordinates have p(r) = r, hence p(0) = 0. There is no boundary in a circular cylinder at r = 0; however, it is expected that the dependent variable remains fi nite at the center of the cylinder.

The conditions y(a) = y(b) and y′(a) = y′(b) are often called periodicity condi-tions in that they require the dependent variable and its spatial derivative to have the same values at both ends of the interval. These conditions are often enforced in the circumferential direction in cylindrical coordinates using the angular coordinate θ, 0 ≤ θ ≤ 2π. For the solution to be single-valued at x = 0, it is required y(0) = y(2π) and y′(0) = y′(2π).

The following theorem is a summary of theorems 5.3, 5.4, 5.7, and 5.8:

Theorem 5.9 The eigenvalues of a Strum-Liouville problem are real and non-negative, 0 1 2 3 1 1≤ ≤ ≤ ≤ ≤ ≤ ≤ ≤− +λ λ λ λ λ λ� �k k k . If conditions, as described in Theorem 5.8, are such that L is positive defi nite on its domain, then the smallest eigenvalue is positive. Furthermore, eigenvectors, w x w xi j( ) ( )and

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corresponding to distinct eigenvalues λ i and λ j of a Sturm-Liouville problem are orthogonal such that ( , )w wi j r = 0.

Example 5.4 A MEMS actuator is modeled as a simply supported uniform column of length L, pinned at both ends, made from a material of elastic modulus E, and having a uniform cross-section of area A and area moment of inertia I. During operation, the column is subject to resistive heating which changes the temperature of the column from its installed temperature T0. The increase in temperature induces thermal stresses in the column leading to a uniform compressive axial load,

P AE T T= −( )α 0 (a)

where α is the coeffi cient of thermal expansion for the column. Let w(x) rep-resent the transverse defl ection of the column. A free-body diagram of a segment of the column of length x is illustrated in Figure 5.4. The internal bending moment is related to the defl ection by

M EId wdx

= −2

2 (b)

Summation of moments about the left support leads to

M Pw+ = 0 (c)

which, when substituted into Equation b, leads to

− =EId wdx

Pw2

2 (d)

P P

(a)

P

R

P

(b)

w(x) M

x

Figure 5.4 (a) The column is subject to a uniform axial load which develops due to change in temperature from an initial temperature. (b) If the column has a transverse defl ection, a free-body diagram of a segment of the column reveals an internal bend-ing moment and axial load.

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Since the column is restrained from displacement at each end,

w( )0 0= (e)

w L( ) = 0 (f)

An obvious solution of Equation d, Equation e, and Equation f is w(x) = 0. The critical buckling load is the smallest value of P for which the transverse displacement of the column is nonzero. Determine the critical buckling tem-perature, the smallest temperature at which the column buckles.

Solution It is clear from Equation d that the critical buckling load is the smallest eigenvalue of the operator

Lw EId wdx

= −2

2 (g)

with a domain defi ned by the boundary conditions of Equation e and Equa-tion f. Comparison of the form of L defi ned in Equation g with the self-adjoint form of Equation 5.18 reveals that p(x) = −EI, r(x) = 1 and q(x) = 0. The opera-tor of Equation d is in the form of a Sturm-Liouville operator; thus, from Theorem 5.7, the operator is self-adjoint with respect to the standard inner product on C2[0, L]. The form of the operator and the boundary conditions are such that the conditions of Theorem 5.8 are satisfi ed and the operator is positive defi nite.

Since the operator is self-adjoint, all eigenvalues must be real, which implies that a critical buckling load exists. The positive defi niteness of the operator confi rms the obvious, that the critical buckling load is positive.

Assuming a solution of the form w(x) = e βx and substituting into the Sturm-Liouville problems leads to β = ±i P EI/ . Thus the general solution of Equation d is

w x CPEI

x CPEI

x( ) cos sin=⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟+

⎛

⎝⎜⎜⎜

⎞

⎠⎟

1 2 ⎟⎟⎟⎟ (h)

Application of Equation e leads to C1 = 0. Then application of Equation f shows that

CPEI

L2 0sin⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟= (i)

A nontrivial solution exists only if

sinPEI

L⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟= 0 (j)

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The critical buckling load, the smallest value of P which satisfi es Equation j, is

PEI

Lcr =

π2

2 (k)

The critical temperature, the temperature at which buckling occurs, can be obtained from Equation a and Equation k as

T TPAE

TI

AL

crcr= +

= +

0

0

2

2

α

πα

(l)

The resulting buckled shape of the column is

w x Cx

L( ) sin= ( )π

(m)

Example 5.5 Consider the boundary-value problem

d ydx

dydx

y2

23 0+ + =λ (a)

y( )0 0= (b)

′ + =y y( ) ( )1 3 1 0 (c)

a. Determine the values of λ for which a nontrivial solution of Equation a, Equation b, and Equation c exists.

b. For each eigenvalue, determine its normalized mode shape, and c. Demonstrate orthogonality of the eigenvectors corresponding to distinct

eigenvalues.

Solution The eigenvalue problem of Equation a, Equation b, and Equation c is in the form of a Sturm-Liouville problem with

Lyd ydx

dydx

= − −2

23 (d)

The operator of Equation e is of the form of Equation 5.18 with a x2 1( ) ,= −a x1 3( ) = − , and a x0 0( ) = . The operator can be rewritten in the form of Equa-tion 5.18 using Equation 5.25, Equation 5.26, and Equation 5.27:

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p x e

e

e

a xa x

dx

dx

x

( )

( )

( )

( )

( )

=∫

=∫

=

−−

1

2

3

1

3

(e)

r x

p xa x

( )( )

( )=

= −

2

e3x

(f)

The function r(x) as determined in Equation f is negative; therefore p(x) and r(x) can be redefi ned as

p x e x( ) = − 3 (g)

r x e x( ) = 3 (h)

Since a0(x) = 0, Equation 5.28 leads to q(x) = 0. The operator can be written in the form of Equation 5.19 as

Le

ddx

edydxx

x= −( )13

3 (i)

The boundary conditions are of the form of Equation 5.34 and Equation 5.35 with α = 0 and β = 3. Thus the conditions of Theorem 5.8 are satisfi ed, and L is a positive defi nite operator.

A solution of Equation a is assumed to be of the form

y x e x( ) = α (j)

Substitution of Equation j into Equation a leads to

α α λ2 3 0+ + = (k)

Application of the quadratic formula to Equation k gives

α λ= − ± −1

23 9 4( ) (l)

Since L is self-adjoint and positive defi nite with respect to an inner product, the values of λ are all real and positive. The mathematical form of the solution depends on the values of α; if the values of α are real and distinct, (λ < 9/4)

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exponential solutions are obtained, whereas if the values of α are complex conjugates, (λ > 9/4) trigonometric solutions are obtained. The case when the values α are real but not distinct, (λ = 9/4), is a special case.

First, suppose that (λ < 9/4), thus defi ning α λ1 1 2 3 9 4= − − −( )/ and α λ2 1 2 3 9 4= − + −( )/ ; the general solution of Equation a is then

y x C e C ex x( ) = +1 21 2α α (m)

where C1 and C2 are constants of integration. Requiring the solution to satisfy the boundary conditions, Equation b and Equation c, leads to

C C1 2 0+ = (n)

α α1 1 2 23 3 0+( ) + +( ) =C C (o)

Equation n and Equation o have only the trivial solution unless the determi-nant of the coeffi cient matrix obtained when formulating the equations in a matrix form is singular. This would require α1 −α2 = 0, which is contrary to the hypothesis of λ < 9/4. Thus there are no eigenvalues such that λ < 9/4.

Assume that λ > 9/4. The general solution of Equation a is

y x C e x C e xx x

( ) cos sin= −( )+ −(− −1

3

22

3

21

24 9

1

24 9λ λ )) (p)

Application of Equation b to Equation p gives C1 = 0. Application of Equation c to the resulting form of Equation p leads to

C23

24 9 4 9 4 9 0sin cosλ λ λ−( )+ − −( )⎡

⎣⎢⎢

⎤⎦⎥⎥= (q)

A nontrivial solution exists only if C2 = 0. Thus, Equation q implies that the eigenvalues are solutions of

− − = −( )2

34 9 4 9λ λtan (r)

Defi ning q = −4 9λ , Equation r is rewritten as

− =2

3q qtan (s)

Equation s is a transcendental equation whose solutions are obtained numer-ically. The solutions correspond to the points of intersection of the graphs of −(2/3)q and tan(q) as illustrated in Figure 5.5. The fi ve smallest values of q which satisfy Equation s and their corresponding values of λ are listed in Table 5.1.

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There are an infi nite, but countable, number of eigenvalues which can be indexed as

0 1 2 1 1< < < < < < <− +λ λ λ λ λ… …k k k (t)

The corresponding eigenvectors are

y x C e xk kx

k( ) sin= −( )−

3

2 4 9λ (u)

0 2 4 6 8 10 12 14 16

–10

–8

–6

–4

–2

0

2

4

6

8

q

f(q)

Figure 5.5 The transcendental equation tan q = −(2/3)q has an infi nite, but count-able, number of positive solutions. As q gets large, the solutions approach odd mul-tiples of π/2.

Table 5.1 Eigenvalues for Example 5.5

i 1 2 3 4 5

q 2.175 5.004 8.038 11.130 14.242

λ 3.433 8.510 18.403 33.219 52.959

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Eigenvectors corresponding to distinct eigenvalues satisfy the orthogonality conditions ( )y yi j r, = 0, which leads to

0 4 9 4 93

2

3

2= −( )⎡

⎣⎢⎢

⎤

⎦⎥⎥ −

− −C e x C ei

xi j

xjsin sinλ λ xx e dx

C C x

x

i j i j

( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

= −( ) −

∫ 3

0

1

4 9 4sin sinλ λ 99

0

1

x dx( )∫ (v)

The eigenvectors are normalized by requiring

1

4 93

2

2

3

=

= −( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

−

( , )

sin

y y

C e x e dx

k k r

kx

kxλ

00

1

2 2

0

1

2

4 9

1

21 2 4

∫

∫= −( )⎡⎣

⎤⎦

= −

C x dx

C

k k

k

sin

cos

λ

λkk

k

kk

x dx

Cx x

−( )⎡⎣

⎤⎦

= +−

−( )⎡

∫ 9

2

1

2 4 92 4 1

0

1

2

λλsin

⎣⎣⎢⎢

⎤

⎦⎥⎥

= +−

−( )⎡

⎣⎢⎢

=

=

x

x

k

kk

C

0

1

2

21

1

2 4 92 4 1

λλsin

⎤⎤

⎦⎥⎥

(w)

Substitution of the double-angle formula for sines and Equation r into Equa-tion w leads to

12

12 4 9 4 9

2 4 9

2

= +−( ) −( )

−

⎡

⎣⎢⎢⎢

⎤

⎦⎥Ck k k

k

sin cosλ λ

λ ⎥⎥⎥

= − −( )⎡⎣⎢⎢

⎤⎦⎥⎥

Ckk

22

21

2

34 9cos λ

(x)

Equation x can be rearranged to yield

Ck =− ( )

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

2

12

342

1

2

cos λ (y)

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The Sturm-Liouville problem of Example 5.5 has an infi nite, but countable, number of eigenvalues and eigenvectors. It can be shown, but the proof is lengthy and noninstructive, that these observations hold true for all Sturm-Liouville problems.

Theorem 5.10 There are an infi nite, but countable, number of eigenvalues of a Sturm-Liouville problem 0 1 2 1 1,< < < < < < <− +λ λ λ λ λ… …k k k .

An analysis of the results of example 5.5 illustrates that the conditions of Theorem 5.8 are suffi cient, but not necessary, conditions for positive defi nite-ness. Consider the substitution, μ = 4λ−9. The results of Example 5.5 show that all values of μ are positive. Using this substitution in the eigenvalue problem Ly = λy leads to 4Ly − 9y = μy. Thus the values of μ are the eigenvalues of the operator 4L − 9. When put into the form of Equation 5.19, q x e x( ) /= − −9 3 2 . Thus, q(x) < 0 and the conditions of Theorem 5.8 are not met, yet all of the eigenvalues of the operator are positive.

Example 5.6 An eigenvalue problem often encountered during the solution of problems using a polar, cylindrical, or spherical coordinate system is

d wd

w2

20

θλ+ = (a)

w w( )0 2= ( )π (b)

dwdx

dwdx

( ) ( )0 2= π (c)

The independent variable θ is a circumferential coordinate whose range is 0 ≤ θ ≤ 2π. Equation b and Equation c enforce the conditions that the depen-dent variable w is single-valued at θ = 0 and θ = 2π. These are called periodicity conditions because their application leads to a solution that is periodic in θ and of period 2π. Determine all eigenvalues and their corresponding nor-malized eigenvectors.

Solution Equation a can be rewritten in the form of Equation 5.18 with

Lwd wdx

= −2

2 (d)

Equation d is in the form of a Sturm-Liouville operator with p(x) = −1, r(x) = 1, and q(x) = 0. The boundary conditions are the periodicity conditions of case (iii) of Theorem 5.7. Thus all eigenvalues are real, and eigenvectors corre-sponding to distinct eigenvalues are self-adjoint with respect to the standard inner product for C2[0,2π].

Application of Theorem 5.8 shows that the operator is non-negative defi -nite on the subspace of C2[0,2π] defi ned by Equation b and Equation c. Thus,

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all eigenvalues are non-negative, but the system may have an eigenvalue of zero.

Assuming a solution to Equation a of the form y(x) = eβ x leads to β λ= ± − . Since the operator is non-negative defi nite, only non-negative values of λ need to be considered. First consider λ = 0, for which the general solution of Equation a is

y x C C x( ) = +1 2 (e)

Application of Equation b gives C1 = C1 + 2πC2, which implies that C2 = 0. Then using y(x) = C1, Equation c is satisfi ed for all values of C1. Thus λ = 0 is an eigenvalue of L with a corresponding eigenvector of y0(x) = C1, a constant. Normalization of the eigenvector leads to

1

2

02

0

2

0

2

12

= [ ]

=

=

∫

∫

y x dx

dx

C

( )

π

π

C12

π

(f)

Equation f can be solved to give C1 1 2= / π , leading to y x0 1 2( ) = / π .If λ > 0, the general solution of Equation a is

y x C x C x( ) cos sin= ( )+ ( )1 2λ λ (g)

Application of Equation b to Equation g leads to

C C C1 1 22 2= ( )+ ( )cos sinπ πλ λ (h)

Application of Equation c to Equation g gives

λ λ π λ λ π λC C C2 1 22 2= − ( )+ ( )sin cos (i)

Equation h and Equation i can be summarized in matrix form as

cos sin

sin cos

2 1 2

2 2 1

π λ π λ

π λ π λ

( )− ( )− ( ) ( )−

⎡

⎣

⎢⎢⎢

⎤

⎦⎦

⎥⎥⎥

⎡

⎣⎢⎢

⎤

⎦⎥⎥=

⎡

⎣⎢⎢

⎤

⎦⎥⎥

C

C1

2

0

0 (j)

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A nontrivial solution of Equation j exists if and only if the determinant of the coeffi cient matrix is zero. To this end,

cos sin

cos

2 1 2 0

2

2 2π λ π λ

π λ

( )−⎡⎣

⎤⎦ + ( )⎡

⎣⎤⎦ =

( )⎡⎣

⎤⎦22 2

2 2 1 2 0

2 1 2

− ( )+ + ( )⎡⎣

⎤⎦ =

− ( )⎡⎣

cos sin

cos

π λ π λ

π λ ⎤⎤⎦ = 0

(k)

The values of λ which solve Equation k are

λ = =n n2 1 2, ,… (l)

Substitution of Equation l into Equation j gives

0 0

0 0

0

0

1

2

⎡

⎣⎢⎢

⎤

⎦⎥⎥⎡

⎣⎢⎢

⎤

⎦⎥⎥=

⎡

⎣⎢⎢

⎤

⎦⎥⎥

C

C (m)

Since Equation m is identically satisfi ed for all values of C1 and C2, the con-stants are independent, and the general form of the eigenvector correspond-ing to λ = n2 is

y x C nx C nxn n n( ) cos sin, ,= ( )+ ( )1 2 (n)

Algebraic calculation reveals (cos(nx), sin(nx) = 0). Also

cos( ) cosnx dx nx dx

x

[ ] = + ( )[ ]

= +

∫ ∫2

0

2

0

2

1

21 2

1

2

1

π π

222

0

2

nnx

x

x

sin( )⎡⎣⎢⎢

⎤⎦⎥⎥

=

=

= π

π

(o)

Equation o leads to a normalization constant of C n1 1, = / π . A similar calcu-lation leads to C n2 1, = / π .

The eigenvalues and normalized eigenvectors for the Sturm-Liouville problem of Equation a, Equation b, and Equation c are

λn n n= =2 0 1 2, , ,... (p)

y x01

2( ) =

π (q)

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y nx

y nxn

n

n

,

,

cos( )

sin( )

1

2

1

1

=

=

⎫

⎬

⎪⎪⎪⎪

⎭⎪⎪⎪⎪

π

π

== 1 2, ,... (r)

The eigenvalue problem described by Equation a, Equation b and Equa-tion c is a positive semi-defi nite Sturm-Liouville problem. The smallest eigenvalue for the system is zero. The eigenvector corresponding to the zero eigenvalue is given by Equation q. As with all Sturm-Liouville problems, the system has an infi nite, but countable, number of eigenvalues. However, all non-zero eigenvalues are degenerate. That is, there is more than one linearly independent eigenvector corresponding to each eigenvalue. The two nor-malized linearly independent eigenvectors corresponding to an eigenvalue of λ = n2, for n = 1,2,… are given in Equation r. For each integer n > 0, the most general solution to Equation a subject to Equation b and Equation c is

y x A nx B nxn n n( ) cos( )/ sin( )/ .= +π π

5.6 Eigenvector expansions5.6.1 Completeness theorem

It is shown in Section 2.9 that if φ φ φ φ φ1 2 1 1, , , , , ,… …k k k− + is a complete ortho-normal set with respect to a valid inner product defi ned on a vector space V, then f, an arbitrary element of V, has a representation of the form

f f i i

i

=∑( , )φ φ (5.41)

The eigenvectors of a self-adjoint operator defi ned on a vector space of fi nite dimension n are orthogonal with respect to a defi ned inner product and span Rn. The eigenvectors form a basis for Rn and are complete in Rn. Thus any vector in Rn can be represented by an eigenvector expansion of the form of Equation 5.41.

Completeness of a set of vectors in an infi nite-dimensional vector space is more diffi cult to prove. A comprehensive general theory of the subject is out-side the objectives of this book. However, some basic results are listed and proved, followed by a general discussion which will guide the development and use of eigenvector expansions.

Theorem 5.10 Let u1,u2,… be a set of orthonormal vectors with respect to an inner product (u,v) defi ned on an infi nite-dimensional vector space V. Then for some u in V, if u u= ∑ =

∞i i1 α i is a convergent series, then α i = ( , )u ui .

Proof Assume that u u= ∑ =∞i i1 α i is a convergent series. Taking the

inner product of both sides of this equation with uk for an arbitrary k

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leads to( , ) ( , )u u u uk i k= ∑ =∞i i1 α . Since the series is convergent, the order of

summation and of taking the inner product may be interchanged, leading to( , ) ( , )u u u uk i k= ∑ =

∞i i1 α . Using the orthonormality of the set of vectors leads

to the desired result.

Theorem 5.11 (Bessel’s inequality) Let u1,u2,… be a set of orthonormal vectors with respect to an inner product (u,v) defi ned on an infi nite- dimensional vector space V. Then for any u in V,

( , )u u ui

i=

∞

∑ ≤1

2 (5.42)

where the norm is the inner-product-generated norm.

Proof The square of any real number is non-negative, thus

u u u u− ≥=

∑( , )i i

i

n

1

2

0 (a)

The defi nition of the inner-product-generated norm and the properties of inner products are used on Equation a, leading to

01 1

≤ − −⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟

= =∑ ∑u u u u u u u u( , ) , ( , )i i

i

n

j j

j

n⎟⎟⎟⎟⎟⎟

=( )− ( )⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟+

=∑u u u u u u, , ,2

1

i i

i

n

uu u u u u u, , ,i i

i

n

j j

j

n

( ) ( )⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟= =∑ ∑

1 1

== − ( )( )+ ( )( )( )=

∑u u u u u u u u u u u2

1

2 , , , , ,i i

i

n

i j i j

jj

n

i

n

==∑∑

11

(b)

The set u1,u2,… is assumed to be orthonormal; thus, u ui j i j, ,( )= δ , which when used in Equation b, leads to

0 22 2

1

2

1

≤ − ( ) + ( )= =

∑ ∑u u u u ui i, ,i

n

i

n

(c)

Rearrangement of Equation c leads to

u u u2, i

i

n

( ) ≤=

∑ 2

1

(d)

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The sequence of partial sums sn in

i= ∑ ( )=12u u, is a bounded sequence of

positive numbers which is not decreasing. Thus, lim n ns→∞ ≤ u 2 , and the theorem is proved.

Theorem 5.12 (Parseval’s Identity) Let u1,u2,… be a set of orthonormal vec-tors with respect to an inner product (u,v) defi ned on an infi nite- dimensional vector space V. Then u1,u2,… is complete on V if and only if

u u u2 2

1

= ( )=

∞

∑ , i

i

(5.43)

for all u in V.

Proof If u1,u2,… is complete in V, then every u in V can be represented by u u= ∑ =

∞i i i1 α , where from Theorem 5.10, α i i=( )u u, . Thus

u u u u u u u2

1 1

= ( ) ( )⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟=

∞

=

∞

∑ ∑, , ,i i

i

j j

j⎟⎟⎟

= ( )( )( )=

∞

=

∞

∑∑ u u u u u u, , ,i j i j

ji 11

(a)

Since u1,u2,… are an orthonormal set, Equation a reduces to

u u u2 2

1

= ( )=

∞

∑ , i

i

(b)

The series on the right-hand side of Equation b is bounded by Bessel’s inequality. Thus Parseval’s identity is established.

If Parseval’s identity is satisfi ed for all u in V, then u has an expansion asu u= ∑ =

∞i i i1 α , and thus the set u1,u2,… is complete in V.

Theorem 5.13 Let u1,u2,… be a set of orthonormal vectors with respect to an inner product (u,v) defi ned on an infi nite-dimensional vector space V. The set is complete in V if and only if v = 0 is the only vector orthogonal to every vector in the set.

Proof First assume the set is complete in V, and let w ≠ 0 be a vector in V orthogonal to all vectors in the set. Then from Parseval’s identity,

w w u2

1

= ( )=

∞

∑ , i

i

(a)

However, since w is orthogonal to all vectors in the set, (w, ui) = 0, and Equa-tion a leads to w = 0. However, the only vector which has a norm of zero is the zero vector. The assumption that w ≠ 0 is thus contradicted.

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The converse can be proved by contradiction. Assume that v = 0 is orthogo-nal to all members of the set, but that the set is not complete in V. Then there exists a vector y in V such that Parseval’s identity is not satisfi ed, and Bessel’s inequality is actually an inequality. The completion is left as an exercise.

Parseval’s identity provides a criterion to determine whether a set of orthonormal vectors is complete in a vector space. It, along with Theorem 5.13, can be used to prove the following:

Theorem 5.14 Let L be a proper Sturm-Liouville operator with eigenvalues 0 1 2 1 1≤ ≤ ≤ ≤ ≤ ≤ ≤− +λ λ λ λ λ… …

k k k and corresponding normalized eigen-vector φ φ φ φ φ1 2 1 1, , , , , ,… …k k k− + . The set of eigenvectors is complete in SD, the domain of L.

Proof The full proof of this theorem is beyond the objective of this study. Using Theorem 5.13, it suffi ces to show that v = 0 is the only vector orthogonal to all elements in the set φ φ φ φ φ1 2 1 1, , , , , ,… …k k k− + . Assume that w ≠ 0 is orthog-onal to each vector in the set. Then for each k, w,φk( ) ≠0. Noting that Lφk = λkφk, then for λk ≠ 0 w w L, ,φk k k( )= ( )1/λ φ . The self-adjointness of L implies that w L Lw, ,φ φk k( )=( ). Thus Lw w−( )=λ φk k, 0. It then remains to show that this

condition is satisfi ed for all k only if w = 0.Theorem 5.14 implies that any vector f in SD has an eigenvector expan-

sion of the form of Equation 5.41. An eigenvector expansion for an infi nite- dimensional vector space is called a Fourier series. Note that the convergence of the series is convergence with respect to the defi nition of the inner- product-generated norm. For a Sturm-Liouville problem, convergence is with respect to the norm f g f x g x r x dxa

b− = ∫ −[ ]⎡⎣⎢

⎤⎦⎥( ) ( ) ( )

2 1 2/. This is a mean convergence.

Pointwise convergence can be considered for each specifi c case.

Example 5.7 (a) Expand f x x e e x( ) ( / )= − +( )− −1 4 3 3 into a series of eigenvec-tors as in Example 5.5. Note that f(x) is in SD. (b) Apply Parseval’s identity to the expansion.

Solution The eigenvectors of Example 5.5 are

y x C e q xk kx

k( ) sin= ( )−

3

2 (a)

where qk is the kth positive solution of

− =2

3q qk ktan( ) (b)

The eigenvectors are normalized by requiring that

( , )

sin( )

y y

C e q x e dx

k k r

kx

kx

=

−

=

⎡

⎣⎢⎢

⎤

⎦⎥⎥∫

1

3

2

2

3

0

1

== 1

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[ ] =

−( )

∫ 1

1

21 2

2 2

0

1

2

C q x dx

C q x d

k k

k k

sin( )

cos( ) xx

Cx

qq x

C

k

kk

x

x

0

1

2

0

1

1

2

1

22 1

∫ =

−⎡

⎣⎢⎢

⎤

⎦⎥⎥

==

=

sin( )

kk

kk

k

kk

qq

Cq

q

2

2

21

1

22 1

21

1

−⎡

⎣⎢⎢

⎤

⎦⎥⎥=

−

sin( )

sin( ))cos( )qk

⎡

⎣⎢⎢

⎤

⎦⎥⎥= 1 (c)

Equation b is used to simplify Equation c, leading to

Cq

k

k

=+ [ ]

2

12

3

2cos( )

(d)

The eigenvector expansion for f(x) is

f x f C e q x C ekx

k

r

kx

( ) , sin( ) s=⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

− −3

2

3

2 iin( )q xk

k=

∞

∑1

(e)

The inner products are evaluated as

f C e q x x e ekx

k

r

x, sin( )− − −

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

= − +3

2 3 31

4(( ) ( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

=

−

∫

∫

C e q x e dx

C xe

kx

kx

k

3

2 3

0

1

0

13

sin

22 3 3

3

2 2

1

4

171

8

x xk

k

k

e e q x dx

Ce q

− +( ) ( )

=−( )

− −

−

sin

ssin( ) cos( )q q q q qk k k k k− −( )⎡⎣⎢⎢

⎤⎦⎥⎥−

71

43 9

49

4

2

++( )qk2

2

(f)

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(b) Application of Parseval’s identity leads to

f k

x e e e dx Ck

e

k

x x

2

1

2 3 33

2

0

1

3

2

2

1

4

2

= ∑

− +∫ =

=

∞

− −

−

( )

α

1171

8

71

43

2 2− − −( ) ( )⎡

⎣⎢⎢

⎤qk qk qk qk qksin( ) cos( )

⎦⎦⎥⎥

( )

⎧

⎨

⎪⎪⎪⎪⎪

⎩

⎪⎪⎪⎪⎪

⎫

⎬

⎪⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

−

+

9

49

4

22

qk

qk ⎪⎪

( )

=

∞

−

∑

=

− −

2

0 709

171

8

7

1

3

2 2

k

Ck

e qk qk qk .

sin( )11

43 9

49

4

2

2

2

− −

+

( )⎡⎣⎢⎢

⎤⎦⎥⎥

( )

⎧

⎨

⎪⎪ qk qk qk

qk

cos( )⎪⎪⎪⎪

⎩

⎪⎪⎪⎪⎪

⎫

⎬

⎪⎪⎪⎪⎪

⎭

⎪⎪⎪⎪⎪

=

∞∑

2

1k

(g)

Example 5.8 Expand the function f x x x( ) ( ) ( )= + −1 4 3 1 32 2 4 4/ /π π in terms of the eigenvectors of Example 5.4.

Solution The eigenvalues and normalized eigenvectors from Example 5.4 are

λn n n= =2 0 1 2, , ,... (a)

y x01

2( ) =

π (b)

y nx

y nxn

n

n

,

,

cos( )

sin( )

1

2

1

1

=

=

⎫

⎬

⎪⎪⎪⎪

⎭⎪⎪⎪⎪

π

π

== 1 2, ,... (c)

The eigenvector expansion is of the form

f x f y y f y y f y yn n n n

n

( ) , , ( , ), , , ,=( ) + ( ) +[ ]=

0 0 1 1 2 2

11

∞

∑ (d)

where

f y x x dx, 0 22

44

0

2

14

3

1

3

1

2

128 15

15

( )= + −( )

=+

∫ π π π

π

π

22π

(e)

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f y x x nx dxn, cos( ),1 22

44

0

2

14

3

1

3

1

6

( )= + −( )

=

∫ π π π

π

−−16

3

2 3

5 4

nn

ππ π

(f)

and

f y x x nx dxn, sin( ),2 22

44

0

2

14

3

1

3

1( )= + −( )

=

∫ π π π

π

56

3−

ππ2 3n

(g)

Substitution of Equation e, Equation f, and Equation g into Equation d leads to

f xnn

( ) =+

+−

( ) −128 15

30

6 16

3

2 3

5 5 3

ππ

ππ π

cos nx56

3 nnnx

n3

1

sin( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

=

∞

∑ (h)

5.6.2 Trigonometric fourier series

The eigenvectors of the Sturm-Liouville problem introduced in Example 5.5 are the basis functions used in a trigonometric Fourier series expansion. If f(x) is a function of period 2π, its Fourier series expansion generated from the eigenvectors is

f x f y y f y y f y yn n n n

n

( ) , , ( , ), , , ,=( ) + ( ) +[ ]=

0 0 1 1 2 2

11

∞

∑ (5.44)

where

f y f x dx, ( )0

0

2

1

2( )= ∫π

π

(5.45)

f y f x nx dxn, ( )cos( ),1

0

2

1( )= ∫π

π

(5.46)

f y f x nx dxn, ( )sin( ),2

0

2

1( )= ∫π

π

(5.47)

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Substitution of Equation 5.45, Equation 5.46, and Equation 5.47 into Equation 5.44 leads to

f xa

a nx b nxn n

n

( ) cos( ) sin( )= + +[ ]=

∞

∑0

12

(5.48)

where the Fourier coeffi cients are defi ned by

a f x nx dx nn = =∫10 1 2

0

2

π

π

( )cos( ) , , ,... (5.49)

b f x nx dx nn = =∫11 2

0

2

π

π

( )sin( ) , ,... (5.50)

Theorems 5.10–5.14 show that the trigonometric Fourier series for f(x) con-verges in the mean to f(x). A general theorem regarding pointwise conver-gence of the Fourier series can be proved.

Theorem 5.15 The trigonometric Fourier series representation of a piece-wise continuous function f(x) defi ned for 0 ≤ x ≤ 2π is that of Equation 5.48, where the Fourier coeffi cients are as calculated in Equation 5.49 and Equation 5.50. The trigonometric Fourier series representation converges to a function which is periodic with period 2π and converges pointwise to f(x) at every x where f(x) is continuous. If f(x) has a jump discontinuity at x0, the Fourier series converges to 1 2 0 0/ ( ) ( )_f x f x+⎡⎣ ⎤⎦

+ at x = x0.Theorem 5.15 implies that the trigonometric Fourier series representation

for f (x) converges to the periodic extension for f (x) outside the interval 0 ≤ x ≤ 2π. The periodic extension fp(x) of f (x) is defi ned such that fp(x ± 2nπ) = f (x) for any integer n = 0,1,2,… and for all x, 0 ≤ x ≤ 2π.

The Fourier series reduces to simpler forms if f(x) is an even function or an odd function.

Theorem 5.16 If f(x) is an even function such that fp(−x) = fp(x) for all x, 0 ≤ x ≤ 2π, then its trigonometric Fourier series representation is given by Equation 5.48 with bn = 0, n = 1,2,…

Theorem 5.17 If f(x)is an odd function such that fp(−x) = −fp(x) for all x, 0 ≤ x ≤ 2π, then its trigonometric Fourier series representation is given by Equation 5.48 with an = 0, n = 0,1,2,…

Example 5.9 (a) Develop the trigonometric Fourier series representation for the function shown in Figure 5.6. (b) Apply Parseval’s identity to the result. (c) Draw the function to which the Fourier series representation converges for −6π ≤ x ≤ 6π.

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Solution (a) The function shown in Figure 5.6 is an odd function. There-fore, according to Theorem 5.17, an = 0, n = 0,1,2,… The coeffi cients of the sine terms can be calculated as

b nx dx nx dx

n

n = + − ( )

= − −

∫ ∫( )sin( ) ( )sin

(

1 1

21

0

2π

π

π

11)n⎡⎣ ⎤⎦

(a)

The trigonometric Fourier series representation of the function shown in Figure 5.6 is

f tn

nx

nnx

n

n

n

( )( )

sin( )

sin( ), ,

=− −

=

=

∞

=

∑21 1

1

1

1 3

455

∞

∑ (b)

(b) Application of Parseval’s identity leads to

f x dxn

n

n

n

( ), ,

, ,

[ ] = ( )

= ( )

∫ ∑=

∞

=

2

0

22

1 3 5

2

1 3 5

4

24

π

π∞∞

=

∞

∑

∑=π8

12

1 3 5n

n , ,

(c)

(c) The trigonometric Fourier series for f(x) converges to the function illus-trated in Figure 5.7. The Fourier series converges to the periodic extension of f(x), fp(x), and converges pointwise to fp(x) at all x except x = ± nπ, at which it converges to zero.

4π3π2ππ

F(t)

t

Figure 5.6 The periodic function of the system of Example 5.9 is illustrated over two periods.

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5.6.3 Completeness of eigenvectors for self-adjoint and non-self-adjoint operators

Let L be an n × n matrix operator. If L is symmetric, it is self-adjoint with respect to the standard inner product for Rn, and all its eigenvalues and eigen-vectors are real. If L is not symmetric, it may still have real eigenvalues and eigenvectors. For example, L could be self-adjoint with respect to an inner product other than the standard inner product. However, not all matrices have real eigenvalues. Such matrices are not self-adjoint with respect to any valid inner product. The associated scalar fi eld for a matrix with complex eigenvalues is the set of complex numbers.

The eigenvectors of a non-self-adjoint matrix are linearly independent and therefore form a basis for Cn, the space of n-dimensional vectors whose elements are complex numbers. The space Rn is a subspace of Cn. Any vector in Rn can be written as a linear combination of eigenvectors of the matrix. However, the coeffi cients in the expansion can be complex.

Recall that the eigenvalues of the adjoint are complex conjugates of the eigenvalues of the matrix and that the eigenvectors satisfy biorthogonality. If u is an eigenvector of a matrix A corresponding to an eigenvector λ and v is an eigenvector of A* corresponding to an eigenvalue μ λ≠ , then (u,v) = 0. Let λ1, λ2,…, λn be the eigenvalues of a matrix A with corresponding eigenvectors u1, u2, …, un. The eigenvalues of A* are λ λ λ1 2, ,..., n which have correspond-ing eigenvectors v1, v2, …, vn. The eigenvectors satisfy the biorthonormality condition,

u vi j i j, ,( )= δ (5.51)

Let u be an arbitrary vector in Rn. Assume that u has an expansion in terms of the eigenvectors of A,

u u==

∑α i i

i

n

1

(5.52)

−6π 6π−5π 5π−4π 4π−3π 3ππ 2π−2π −π

1

−1

Figure 5.7 The Fourier series for the function of Example 5.9 converges to a periodic function of period 2π, which converges pointwise to the periodic extension of F(t) at all t where it is continuous, and to the average value at values of t where it has a jump discontinuity.

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Taking the inner product of both sides of Equation 5.52 with vj leads to

u v u v, ,j i i j

i

n

( )= ( )=

∑α1

(5.53)

Due to the biorthogonality condition, Equation 5.51, the only nonzero term in the summation on the right-hand side of Equation 5.53 corresponds to i = j, leading to

α j j=( )u v, (5.54)

Thus, if the operator is not self-adjoint, the coeffi cients in the eigenvector expansion are the inner products with the vector and the eigenvectors of the adjoint operator.

Theorems have been presented showing that the eigenvectors of a Sturm-Liouville operator are complete in the vector space defi ned as the domain of the operator. Because of this completeness, any vector in the domain of the operator has a convergent eigenvector expansion. The convergence is with respect to the inner-product-generated norm. Without further proof, it is assumed that eigenvectors of all self-adjoint operators whose eigenvalues are countable are complete in the domain of the operator and that a convergent eigenvector expansion exists for any vector in the domain of the operator. This is true for fourth-order differential operators, integral operators, par-tial differential operators, operators defi ned for a set of coupled differential equations, and operators defi ned for a differential equation coupled with an algebraic equation.

There are self-adjoint operators whose eigenvalues are not countable. In such cases, the eigenvectors corresponding to discrete eigenvalues are not complete. An example is an operator of the form Ly d dx p x dy dx= − ( )/ /( )

q x y+ ( ) with boundary conditions at x = a and x = b, but p(x) = 0 for some x, a ≤ x ≤ b. Such a problem is called a turning-point problem and may not have countable eigenvalues.

The proof of Theorem 5.6 from Section 5.3 can now be completed. It is left to show that if (Lu, u) = 0 for some u ≠ 0, then λ = 0 is an eigenvalue of L if L is self-adjoint and all eigenvalues are non-negative.

Since L is self-adjoint, any vector in the domain of L has an eigenvector expansion u u= ∑i i iα . Assume that for some u, (Lu, u) = 0; then

0 =( )

=⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

⎛

⎝

⎜⎜⎜⎜⎜ ∑ ∑

Lu u

L u u

,

,α αi i

i

i j

j

⎞⎞

⎠

⎟⎟⎟⎟⎟⎟ (5.55)

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Using the linearity of L, the properties of inner products, and the orthonor-mality of the eigenvectors, Equation 5.55 reduces to

α λi i

i

2 0∑ = (5.56)

Since the operator is non-negative defi nite, all nonzero eigenvalues must be positive. Thus, Equation 5.56 is true only if one of the eigenvalues is zero and all the coeffi cients in the eigenvector expansion are zero except for the one multiplying the eigenvector corresponding to λ = 0.

5.6.4 Solution of nonhomogeneous equations using eigenvector expansions

Let L be a linear operator defi ned on a domain SD. Assume L is self-adjoint and positive defi nite with respect to a valid inner product ( f, g)r on SD. Let 0 1 2 1 1< ≤ ≤…≤ ≤ ≤ ≤…− +λ λ λ λ λk k k be the eigenvalues of L with corres-ponding normalized eigenvectors φ φ φ φ φ1 2 1 1, , , , ,… k k k− + , … . The eigenvectors are complete in SD.

Consider the nonhomogeneous problem

Lu f= (5.57)

where the domain of L is SD and f is an element of the range of L. Since the eigenvectors of L are complete in SD and u must be in SD, it has an eigenvector expansion of the form

u =∑α φi i

i

(5.58)


L fα φi i

i∑

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

= (5.59)

Since L is a linear operator and by defi nition Lφi = λ i φ i, Equation 5.59 can be rewritten as

α λ φi i i

i∑ = f (5.60)

Taking the inner product of Equation 5.60 with φj for an arbitrary j leads to

α λ φ φ φ

α λ φ φ

i i i

i

j

r

j r

i i i j

∑⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

=, ( , )

( ,

f

)) ( , )r

i

r∑ = f φj

(5.61)

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The orthonormality condition for a set of normalized eigenvectors of a self-adjoint operator is (φi , φj)r = δi, j . Its application in Equation 5.61 leads to

αλ

φjj

j r=1

( , )f (5.62)

The solution of Equation 5.57 can be written as

u f=∑ 1

λφ φ

ii r i

i

( , ) (5.63)

Equation 5.63 provides the description of the inverse operator of a self-adjoint operator,

L f f− =∑1 1

λii r i

i

( , )φ φ (5.64)

Note that the inverse operator does not exist if λ = 0 is an eigenvalue of L.

Example 5.10 The fl exibility matrix A for a discrete mechanical system is the inverse of the stiffness matrix. The stiffness matrix of the system is

K =−

− −−

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

4 2 0

2 3 1

0 1 2

(a)

Use Equation 5.63 to determine A.

Solution The stiffness matrix is symmetric, and therefore it is self-adjoint with respect to the standard inner product on R3. The eigenvalues and nor-malized eigenvectors of K can be determined as

λ1 10 8549

0 4320

0 6793

0 5932

= =

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥.

.

.

.

u ⎥⎥⎥⎥ (b)

λ2 22 4760

0 4913

0 3744

0 7864

= =−

⎡

⎣

⎢⎢.

.

.

.

u ⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥ (c)

λ3 35 6691

0 7563

0 6312

0 1720

= = −

⎡

⎣

⎢.

.

.

.

u ⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥ (d)

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Defi ning f = [ f1 f2 f3 ]T, application of Equation 5.64 leads to

Af = [ ]

⎡

⎣

⎢1

0 85490 4320

1

2

3

.. 0.6973 0.5932

f

f

f

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

⎧

⎨⎪⎪⎪⎪

⎩⎪⎪⎪⎪

⎫

⎬⎪⎪⎪⎪

⎭⎪⎪⎪⎪

0 4320

0 679

.

. 33

0 5932

1

2 46700 4913

.

..

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

+ 0.3744 0.7864−[ ]

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

⎧

⎨⎪⎪⎪⎪

⎩⎪⎪⎪⎪

⎫

⎬

f

f

f

1

2

3

⎪⎪⎪⎪⎪

⎭⎪⎪⎪⎪ −

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥

0.4913

0.3744

0.7864⎥⎥⎥

+ −[ ]

⎡1

5 66910 7563

1

2

3

.. 0.6312 0.1720

f

f

f⎣⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

⎧

⎨⎪⎪⎪⎪

⎩⎪⎪⎪⎪

⎫

⎬⎪⎪⎪⎪

⎭⎪⎪⎪⎪

− .0 7563

00 6312

0 1720

0 4167 0 3333 0

.

.

. . .

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

=11667

0 3333 0 6667 0 3333

0 1667 0 3333 0 6667

. . .

. . .

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

f

f

f

1

2

3

(e)

Example 5.11 The differential equations governing the forced response of a multi-degree-of-freedom discrete system are of the form

Mx Kx F�� + = (a)

where M is the mass matrix, K is the stiffness matrix, x is the displacement vector, and F is the force vector. Recall from Example 5.2 that the natural frequencies are the square roots of the eigenvalues of M−1K and the mode-shape vectors are the corresponding eigenvectors. Also recall that M−1K is self-adjoint with respect to the kinetic- and potential-energy inner prod-ucts and that the eigenvectors are normalized with respect to the kinetic-energy inner product. Let ω1 ≤ ω2 ≤ … ≤ ωn be the natural frequencies of an n-degree-of-freedom system, and let X1, X2, …, Xn be the corresponding normalized mode-shape vectors. The forced response, at any time, is in Rn. Therefore, it has an eigenvector expansion of the form

x Xi==

∑c ti

i

n

( )1

(b)

Show that substitution of Equation b into Equation a leads to a decoupling of the differential equations and a process called modal analysis.

Solution Substitution of Equation b into Equation a and the linearity of the mass and stiffness matrices lead to

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��c ci

i

n

i

i

n

MX KX Fi i

= =∑ ∑+ =

1 1

(c)

Taking the standard inner product of both sides of Equation a with Xj for arbitrary j and using the defi nitions of energy inner products and properties of inner products

��

��

c c

c

i

i

n

i

i

n

MX X KX X F Xi j i j j, , ,( )+ ( )=( )= =

∑ ∑1 1

ii M i Ki

n

i

n

cX X X X F Xi j i j j, , ,( ) + ( ) =( )==

∑∑11

(d)

Orthonormality of the mode shapes implies ( , ) ,X Xi j M i j=δ and ( , ) ,X Xi j K i j i=δ ω2. Using these in Equation d results in

��cj j+ =( )ω2 F Xj, (e)

Equation e represents a set of uncoupled differential equations to solve for the values of each ci.

The procedure in which the forced response of a dynamic system is deter-mined as an eigenvector expansion is called modal analysis. The eigenvalues are the squares of the natural frequencies, and the eigenvectors are their cor-responding mode shapes. The forced response is written as an eigenvector expansion with the coeffi cients as functions of time. Orthogonality proper-ties of the mode shapes are used to derive a set of uncoupled differential equations for the time-dependent coeffi cients. Each differential equation may be solved independently and the forced response obtained from the eigenvector expansion.

The time-dependent coordinates are called principal coordinates, a set of coordinates defi ned so that, when they are used as dependent variables in the differential equations, the equations are uncoupled. They are a set of coordinates because they can be obtained by a linear transformation from the original set of coordinates (called generalized coordinates) represented by the vector x. Defi ne P, the n × n modal matrix as the matrix whose col-umns are the normalized mode-shape vectors. Then

x Pc= (5.65)

where c is the vector of principal coordinates. Since its columns are the normalized mode-shape vectors, which must be linearly independent, P is invertible, and therefore the transformation expressed in Equation 5.65 is one-to-one.

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5.7 Fourth-order differential operatorsConsider the following form of a fourth-order differential operator:

Lyr x

ddx

s xd ydx

ddx

p xdydx

=⎡

⎣⎢⎢

⎤

⎦⎥⎥+

1 2

2

2

2( )( ) ( )

⎡⎡⎣⎢⎢

⎤⎦⎥⎥+

⎧⎨⎪⎪⎩⎪⎪

⎫⎬⎪⎪⎭⎪⎪

q x y( ) (5.66)

Let f(x) and g(x) be arbitrary elements of SD, which is defi ned as all functions in C4[a,b] which satisfy all boundary conditions. Consider the inner product

Lf g f x g x r x dx

ddx

s xd fdx

r

a

b

, ( ) ( ) ( )

( )

( ) =

=⎡

∫2

2

2

2⎣⎣⎢⎢

⎤

⎦⎥⎥

+ ⎡⎣⎢⎢

⎤⎦⎥⎥+g x dx

ddx

p xdfdx

g x dxa

b

( ) ( ) ( )∫∫∫

∫+

a

b

a

b

f x g x dx( ) ( ) (5.67)

Applying integration by parts twice to the fi rst integral in Equation 5.67 and once to the second integral in the equation leads to

Lf g g xd

dxs x

d fdxr, ( ) ( )( ) =

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟

⎡

⎣⎢⎢

⎤

⎦⎥

2

2 ⎥⎥−

⎡

⎣⎢⎢

⎤

⎦⎥⎥

+ ⎡

a

b

a

b

s xdgdx

d fdx

p x g xdfdx

( ) ( ) ( )2

2 ⎣⎣⎢⎢

⎤⎦⎥⎥

+ −

a

b

s xd fdx

d gdx

( )2

2

2

2pp x

dfdx

dgdx

q x f x g x dxa

b

( ) ( ) ( ) ( )+⎡

⎣⎢⎢

⎤

⎦⎥⎥∫ (5.68)

It can be similarly shown that

f g f xd

dxs x

d gdxr, ( ) ( )L( ) =

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟

⎡

⎣⎢⎢

⎤

⎦⎥

2

2 ⎥⎥−

⎡

⎣⎢⎢

⎤

⎦⎥⎥

+ ⎡

a

b

a

b

s xdfdx

d gdx

p x f xdgdx

( ) ( ) ( )2

2 ⎣⎣⎢⎢

⎤⎦⎥⎥

+

a

b

s xd fdx

d gdx

( )2

2

2

2−− +

⎡

⎣⎢⎢

⎤

⎦⎥⎥∫ p x

dfdx

dgdx

q x f x g x dxa

b

( ) ( ) ( ) ( ) (5.69)

The operator L is self-adjoint if the right-hand side of Equation 5.68 is equal to the right-hand side of Equation 5.69. The requirement for self-adjointness becomes

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( ) ( ) (g xd

dxs x

d fdx

sa

b2

2

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

− xxdgdx

d fdx

p x g xdfdxa

b

) ( ) ( )2

2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

+ ⎡⎣⎢⎢

⎤⎦⎥⎥⎥

=⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

a

b

a

b

f xd

dxs x

d gdx

( ) ( )2

2−−

⎡

⎣⎢⎢

⎤

⎦⎥⎥

+ ⎡⎣⎢⎢

s xdfdx

d gdx

p x f xdgdxa

b

( ) ( ) ( )2

2

⎤⎤⎦⎥⎥ a

b

(5.70)

If the boundary conditions are such that Equation 5.70 is satisfi ed for all f and g in SD, then L is self-adjoint with respect to the inner product of Equation 5.67.

Boundary conditions such that the operator is self-adjoint with respect to the inner product of Equation 5.67 are of the form

a y a bd

dxs

d ydx

a y b bx a

1 1

2

2 20( ) ( )−⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟ = −

=22

2

2

1 1

0d

dxs

d ydx

dydx

a s a

x b

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟ =

( )−

=

α β ( )) ( ) ( ) ( )

( )

d ydx

a p a y a

dydx

b s bd y

2

2 1

2 2

2

0+ =

( )−

γ

α βddx

b p b y b2 2 0( ) ( ) ( )+ =γ (5.71)

The positive-defi niteness of L can be examined by substituting g = f in Equa-tion 5.68, leading to

Lf f f xd

dxs x

d fdxr, ( ) ( )( ) =

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟

⎡

⎣⎢⎢

⎤

⎦⎥

2

2 ⎥⎥−

⎡

⎣⎢⎢

⎤

⎦⎥⎥

+ ⎡

a

b

a

b

s xdfdx

d fdx

p x f xdfdx

( ) ( ) ( )2

2 ⎣⎣⎢⎢

⎤⎦⎥⎥

+ −

a

b

s xd fdx

d fdx

( )2

2

2

2pp x

dfdx

dfdx

q x f x g x dxa

b

( ) ( ) ( ) ( )+⎡

⎣⎢⎢

⎤

⎦⎥⎥∫ (5.72)

Equation 5.72 shows that if s x p x( ) , ( ) ,≥ ≤0 0 and q(x) ≥ 0, then if all boundary terms are zero or positive, the operator is non-negative defi nite. L is positive defi nite if q(x) > 0 as long as the boundary terms are zero or positive, or if q(x) = 0 and the boundary terms are positive. If q(x) = 0 and the boundary terms are zero, then the operator is non-negative defi nite if the boundary conditions allow f x c c x( ) = +1 2 to be an element of SD.

The general form of a fourth-order differential operator is

Ly a xd ydx

a xd ydx

a xd ydx

a x= + + +4

4

4 3

2

2 2

2

2 1( ) ( ) ( ) ( )) ( )dydx

a x y+ 0 (5.73)

Expansion of derivatives in Equation 5.73 leads to

Lyr x

s xd ydx

dsdx

d ydx

d sdx

p x= + + +⎛1

24

4

3

3

2

2( )( ) ( )

⎝⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟ + +

⎡

⎣⎢⎢

⎤

⎦⎥⎥

d ydx

dpdx

dydx

q x y2

2( ) (5.74)

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Comparison of Equation 5.74 with Equation 5.73 shows that the fourth-order differential operator may be written in a self-adjoint form if functions s(x), p(x), and q(x) can be found such that

a xs xr x

4( )( )

( )= (5.75)

a3( )( )

( )x

s xr x

=′′

2 (5.76)

a xs x p x

r x2( )

( ) ( )

( )=

′′ + (5.77)

a xp xr x

1( )( )

( )=

′ (5.78)

a xq xr x

0( )( )

( )= (5.79)

Since there are fi ve coeffi cients in the differential equation of Equation 5.73 and only three functions to be determined for the self-adjoint form of Equa-tion 5.74, not all fourth-order differential operators can be rewritten in the self-adjoint form. However, this form is only a suffi cient, not a necessary, con-dition for self-adjointness. It may be possible to determine an inner product for which the operator is self-adjoint without being able to rewrite it in the form of Equation 5.66.

As discussed subsequently, the inner product of Equation 5.67 is an energy inner product. Since L is self-adjoint with respect to this inner product, when it is also positive defi nite, an energy inner product exists of the form

f g f g x r x dxL, ( ) ( )( ) = ( )∫ L

0

1

(5.80)

Theorem 5.18 A fourth-order differential operator of the form of Equation 5.66 is self-adjoint with respect to the inner product defi ned in Equation 5.67 if the domain of the operator S is defi ned such that all f(x) in S satisfy condi-tions of the form of Equation 5.72. Furthermore, the operator is positive defi -nite for appropriate choices of the constants a b a b1 1 2 2 1 1 1 2 2 2, , , , , , , ,α β γ α β γand (see Problem 5.48).

When the operator is self-adjoint, Theorems 5.2–5.6 can be applied, lead-ing to Theorem 5.19.

Theorem 5.19 If the domain of L is defi ned such that the operator of Equa-tion 5.66 is self-adjoint with respect to the inner product defi ned in Equation 5.67, then all eigenvalues of L are real, and eigenvectors corresponding to

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distinct eigenvectors are orthogonal with respect to the inner product of Equation 5.67. The eigenvalues are infi nitely countable and are distinct. Fur-thermore, if the domain is defi ned such that L is positive defi nite, then all eigenvalues of L are positive.

Theorems 5.13 (Bessel’s inequality) and 5.14 (Parseval’s identity) apply when the operator is self-adjoint. The self-adjointness is then used to show completeness of the eigenvectors and to develop an expansion theorem.

Theorem 5.20 Suppose that S, the domain of L, is defi ned such that the operator of Equation 5.66 is self-adjoint with respect to the inner product defi ned in Equation 5.67. Let λ λ λ λ λ1 2 1 1< <…< < < <…− +k k k be the eigen-values of L, with φk(x), the normalized eigenvector (normalized with respect to the inner product of Equation 5.67), corresponding to λk. The set of eigen-vectors is complete on S, and if f(x) is in S, then the expansion ∑ =

∞i i r if1( , )φ φ

converges to f(x).The nondimensional partial differential equation governing the

free vibrations of motion of a nonuniform stretched beam on an elastic foundation is

∂∂

∂∂

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟−

∂∂

∂∂( )+

2

2

2

2xx

wx x

xwx

α ε η( ) ( ) (xx w xwt

) ( )+∂∂

=β2

20 (5.81)

where α(x) is a nondimensional function of the nonuniform bending stiff-ness, ε(x) is a nondimensional axial load allowing for the load to vary over the span of the beam, η(x) is the nondimensional stiffness per length of the elastic foundation, allowing it to vary over the span of the beam, and β(x) is the non-dimensional inertia of the beam. A normal-mode solution for Equation 5.81 can be assumed to be

w x t W x ei t( , ) ( )= ω (5.82)

where ω is a natural frequency and W(x) is its corresponding mode shape. Substitution of Equation 5.82 into Equation 5.81 leads to

d

dxx

d Wdx

xd Wdx

x W2

2

2

2

2

2α η( ) ( ) ( )

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟− +ε == ω β2 ( )x W (5.83)

Dividing Equation 5.83 by β(x) leads to an eigenvalue problem of the form of Equation 5.1 with ω2 as the eigenvalue, W(x) as the eigenvector, and

LWx

ddx

xd Wdx

ddx

xd

=⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟−

1 2

2

2

2βα ε

( )( ) ( )

WWdx

x W( )+⎡

⎣⎢⎢

⎤

⎦⎥⎥

η( ) (5.84)

Equation 5.84 is in the self-adjoint form of Equation 5.66. The problem is self-adjoint if the boundary conditions are defi ned such that Equation 5.70

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is satisfi ed by s x x p x x q x x r x x( ) ( ), ( ) ( ), ( ) ( ) ( ) ( )= = = =α ε η βand . Assuming that the problem is nondimensionalized, a = 0 and b = 1. Consider, for con-venience, a uniform beam in which all properties are constant. Using these simplifi cations, the boundary terms at x = 1 in Equation 5.70 become

α α εgd fdx

dgdx

d fdx

gdfdx

( ) ( ) ( ) ( ) ( ) (1 1 1 1 13

3

2

2− + 11 1 1

1 1 1

3

3

3

3

) ( ) ( )

( ) ( ) ( )

=

− +

α

α ε

fd gdx

dfdx

d gdx

fdggdx

( )1 (5.85)

If the end at x = 1 is fi xed, then if f and g are in SD, then f(1) = g(1) = 0 and df/dx(1) = dg/dx(1) = 0. In this case, both sides of Equation 5.85 are identically zero, and the equation is identically satisfi ed. Such is also the case if the end is pinned. If the end at x = 0 is free, the end moment is zero, and thus d2f/dx2 = d2g/dx2 = 0. The applied load is assumed to remain horizontal. The end of the beam has a slope equal to dw/dx(1). The component of the axial force in the shear direction, the direction perpendicular to the neutral axis, is (nondimensionally) ε(dw/dx(1)). Thus the appropriate boundary condition for a free end at x = 1 is αw d w dx( ) ( )1 03 3+ =ε / . It is then clear that Equation 5.85 is identically satisfi ed when the end is free.

Thus, the fourth-order differential operator is self-adjoint, for a stretched-beam operator is self-adjoint on SD with respect to the inner product

Lf g f x g x x dx, ( ) ( ) ( )( ) = ∫β β0

1

(5.86)

Noting that β(x) is the nondimensional inertia of the beam, the inner prod-uct of Equation 5.86 is a kinetic-energy inner product in that (L(∂w/∂t), ∂w/∂t)β = 2T.

Example 5.12 Determine the natural frequencies and normalized mode shapes of a uniform pinned-fi xed beam subject to a constant axial load, ε = 2, and on a uniform elastic foundation with η = 1. Demonstrate mode-shape orthogonality.

Solution The eigenvalue problem is

d Wdx

d Wdx

W W4

4

2

22− + = λ (a)

W( )0 0= (b)

d Wdx

2

20 0( ) = (c)

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W( )1 0= (d)

dWdx

( )1 0= (e)

A solution of Equation a can be assumed to be the form W x e x( ) = γ , which when substituted into Equation a, leads to

γ γ λ4 22 1 0− + − = (f)

The solution of Equation f is

γ λ= ± ±⎡⎣

⎤⎦1

1

2 (g)

If λ < 1, the values of λ obtained from Equation g are all real, and a nontrivial solution cannot be obtained. The same is true for λ = 0. For λ > 1, Equation g has two real roots, ± = ± +u 1 λ , and two complex roots, ± = ± −iv i λ 1.The general solution of Equation a can be written as

W x C ux C ux C vx C( ) cosh( ) sinh( ) cos( ) sin(= + + +1 2 3 4 vvx) (h)

Application of boundary conditions Equation b, Equation c, Equation d, and Equation e gives

W C C( )0 0 1 3= = + (i)

d Wdx

u C v C2

22

12

30 0( ) = = − (j)

W C u C u C v C v( ) cosh( ) sinh( ) cos( ) sin(1 0 1 2 3 4= = + + + )) (k)

ddx

uC u uC u vC v vW

( ) sinh( ) cosh( ) sin( )1 0 1 2 3= = + − + CC v4 cos( ) (l)

Equation i and Equation j are satisfi ed only by C1 = C3 = 0. Equation k and Equation l can then be written in matrix form as

sinh( ) sin( )

cosh( ) cos( )

u v

u u v v

C

C⎡

⎣⎢⎢

⎤

⎦⎥⎥⎡

⎣⎢ 2

4⎢⎢⎤

⎦⎥⎥=

⎡

⎣⎢⎢

⎤

⎦⎥⎥

0

0 (m)

A nontrivial solution of Equation m exists only if

v v u u u vcos( )sinh( ) cosh( )sin( )− = 0 (n)

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Equation n can be rearranged to yield

v u u vtanh( ) tan( )= (o)

Equation o is a transcendental equation with an infi nite, but countable, number of solutions: 0 1 2 1 1< < <…< < < <…− +λ λ λ λ λk k k . As k gets large, uk/vk approaches 1 and tanh(uk) approaches 1. Thus the larger eigenvalues are such that vk approaches (4k −3)π/4 with λk kv= −( )2 2

1 . The fi ve lowest eigen-values and nondimensional natural frequencies are given in Table 5.2.

The eigenvectors can be determined from Equation h and Equation m as

W x C u xu

vv xk k k

k

kk( ) sinh( )

sinh( )

sin( )sin( )= −

⎡

⎣⎢⎢⎢

⎤

⎦⎥⎥ (p)

The mode shapes are normalized by requiring

W W

C u xu

k k

k kk

,

sinh( )c

( )=

−

1

2 oosh( )

cos( )sin( )

uv v

v x dxk

k kk

⎡

⎣⎢⎢

⎤

⎦⎥⎥

=∫2

0

1

1

cosh

cos

siC

u uv v

kk k

k k=

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟ −2 2

2nn( )

2vv

k

k

⎡

⎣⎢⎢

⎤

⎦⎥⎥

⎧⎨⎪⎪⎩⎪⎪

cosh

( ) coscosh sin c−

+−

82 2

u uu v v v

u vk k

k k k kk k oos sinh

v uk k( )

− +2sinh(22u

uk

k

)}−

1

2

Mode-shape orthogonality is demonstrated by

Table 5.2 Eigenvalues and Natural Frequencies for Example 5.12

k 1 2 3 4 5

λk 2.62 ×102 2.583 ×103 1.106 ×104 3.211×104 7.451×104

uk 4.15 ×100 7.20 ×100 1.03 ×101 1.34 ×101 1.66 ×101

νk 3.90 ×100 7.06 ×100 1.02 ×101 1.32 ×101 1.65 ×101

ωk 1.62 ×101 5.08 ×101 1.05 ×102 1.79 ×102 2.73 ×102

Ck 3.29×10−2 1.54 ×10−3 6.85 ×10−5 3.09×10−6 1.31×10−7

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W W C u xu

vv xk m k k

k

kk, sinh( )

sinh( )

sin( )sin( )( )= −

⎡⎡

⎣⎢⎢

⎤

⎦⎥⎥

−

∫ C

u xu

m

mm

0

1

sinh( )sinh( )

ssin( )sin( )

sinh sin

vv x dx

C C u x

mm

k m k

⎡

⎣⎢⎢

⎤

⎦⎥⎥

= ( ) hhsinh( )

sin( )sin sinhu x

uv

v x u xmk

kk m( )− ( ) ( )

⎡

⎣⎢⎢⎢

−

∫0

1

sinh( )

sin( )sinh

uv

um

mkxx v x

uv

m

k

k

( ) ( )

+

sin

sinh( )

sin( ))

sinh( )

sin( )sin sin

uv

v x v x dxm

mk m( ) ( )

⎤

⎦⎥⎥

The normalized mode shapes are illustrated in Figure 5.8.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1–1.5

–1

–0.5

0

0.5

1

1.5

2

x

w

Figure 5.8 Mode shapes for Example 5.12 are orthonormal with respect to the standard inner product for C4[0,1].

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Example 5.13 A uniform column free at x = 0, but fi xed at x = L, rotates at a constant angular speed Ω about its central axis, as illustrated in Figure 5.9. If the column is stable, it remains vertical, but if it is unstable, it will deform as shown in Figure 5.10.

Let s be a coordinate along the axis of the column, measured from its free end. If the column is stable, then s = x. Defi ne θ as the slope of the elastic curve; then cosθ = dx/ds and sinθ = dr/ds, where r(x) is the distance from the vertical position to the center of the column. Figure 5.11 shows a sectional cut of the column, illustrating the internal shear force which is V = EI(d2θ/ds2). This section of the column also has a centripetal force due to its rotation, F r s gAdss= −∫ 0 2( )Ω ρ . Application of Newton’s second law to this section of the column leads to

x

ΩL

Figure 5.9 The column of Example 5.13 rotates about its neutral axis at a constant angular speed Ω.

Ω

Figure 5.10 The column buckles when the angular speed is greater than a critical speed Ωc r .

VsV

Figure 5.11 The free-body diagram when the column reaches a critical speed shows only an internal shear force. A centripetal inertial force leads to buckling.

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EIdds

gA r s dss

2

22

0

θ= − ∫ρ Ω ( ) (a)

Defi ne u s r s dss

( ) ( )= ∫ 0 . Then du/ds = r(s). Since the goal is to obtain the speed at which the column buckles, it can be reasonably be assumed that at the onset of buckling, θ is small, and therefore sinθ ≈ θ. Combining the above leads to θ = d2u/ds2. Substitution into Equation a gives

EId uds

gA u4

42= ρ Ω (b)

The appropriate boundary conditions specifi ed for a column free at x = 0 and fi xed at x = L are

u( )0 0= (c)

d uds

3

30 0( ) = (d)

duds

L( ) = 0 (e)

d uds

L2

20( ) = (f)

The critical speed is the speed at which instability occurs. It is calculated from

ΩcEIgA

=⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

λρ

1

2

(g)

where λ is the smallest eigenvalue of Lu = d4u/ds4 defi ned on SD whose mem-bers are elements of C4[0,L] which satisfy the boundary conditions of Equa-tion c, Equation d, Equation e, and Equation f. (a) Determine whether or not L is self-adjoint. If not, what is the adjoint problem? (b) Determine the critical buckling speed.

Solution (a) The operator is self-adjoint on SD if Equation 5.69 is true with p(x) = 0 and s(x) = 1. Requiring f(x) and g(x) each to satisfy Equation c, Equa-tion d, Equation e, and Equation f leads to

g Ld fdx

Ldgdx

d fdx

f Ld gdx

L( ) ( ) ( ) ( ) ( ) (3

3

2

2

3

30 0− = )) ( ) ( )−

dfdx

d gdx

0 02

2 (h)

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However, Equation h is not satisfi ed for all f(x) and g(x) in SD, and therefore L is not self-adjoint.

Let f(x) be an element of SD, and let g(x) be an element of SD* , the domain of

L*, the adjoint of L. Since only the boundary terms did not vanish when test-ing for self-adjointness, it is suspected that L* has the same differential form as L, but has a different domain. The appropriate boundary conditions for L* are chosen so that Equation h is satisfi ed for all f(x) in SD and all g(x) in SD

* . This is true only if

dgdx

( )0 0= (i)

d gdx

2

20 0( ) = (j)

g L( ) = 0 (k)

d gdx

L3

30( ) = (l)

(b) Defi ning λ = q4, the general solution of d4u/ds4 = q4u is

u s C qs C qs C qs C( ) cos( ) sin( ) cosh( ) sinh(= + + +1 2 3 4 qqs) (m)

Application of Equation c, Equation d, Equation e, and Equation f to Equation m leads to

C C1 3 0+ = (n)

− + =C C2 4 0 (o)

− + + + =sin( ) cos( ) sinh( ) cosh( )qL C qL C qL C qL C1 2 3 4 00 (p)

− − + + =cos( ) sin( ) cosh( ) sinh( )qL C qL C qL C qL C1 2 3 4 00 (q)

Equation n, Equation o, Equation p, and Equation q have a nontrivial solution if and only if

cos( )cosh( )qL qL = −1 (r)

The smallest value of q which satisfi es Equation r is 1.875/L. Thus λ = q4 = 12.360L4. The critical speed is calculated from Equation g as

ΩcEI

gAL=

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟3 516

4

1

2

.ρ

(s)

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5.8 Differential operators with eigenvalues in boundary conditions

A general formulation of an eigenvalue problem for a second-order differential operator is

Ly y= λ (5.87)

L y a c y aa a( ) ( )= λ (5.88)

L y b c y bb b( ) ( )= λ (5.89)

where L is in the self-adjoint form of Equation 5.18,

Ly = 1/r(x)[d(p(x)dy/dx)/dx] + [q(x)y]

The boundary operators are of the form

L y a y a y aa a a( ) ( ) ( )= ′ +α β (5.90)

L y b y b y bb b b( ) ( ) ( )= ′ +α β (5.91)

and ca and cb are non-negative constants. Consider the inner product defi ned by

( , ) ( ) ( ) ( ) ( ) ( ) ( )f g Lf x g x r s dx L f a g a L f bL a b= +[ ] +[ ]]∫ g ba

b

( ) (5.92)

It is possible to demonstrate that under the conditions required in Section 5.5 for L to be self-adjoint and positive defi nite with respect to the inner product of Equation 5.92, and with αaβa ≤ 0 and αbβb ≥ 0, Equation 5.92 represents a valid inner product. Let λ i and λ j be distinct eigenvalues of the system defi ned by Equation 5.87, Equation 5.88, and Equation 5.89, with corresponding eigenvec-tors yi and yj. Thus

Ly yi i i= λ (5.93)

L y a c y aa i i a i( ) ( )= λ (5.94)

L y b c y bb i i b i( ) ( )= λ (5.95)

Using the inner-product defi nition of Equation 5.92 and Equation 5.93, Equa-tion 5.94, and Equation 5.95.

( , ) ( ) ( ) ( ) ( ) ( )y y Ly y x r x dx L y a y ai j L i j

a

b

a i j= + [ ]∫ ++[ ]

= +

L y b y b

y x y x r x dx c y

b i j

i i j i a i

( ) ( )

( ) ( ) ( ) (λ λ aa y a c y b y bj i b i

a

b

j) ( ) ( ) ( )+∫ λ (5.96)

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In a similar fashion, it can be determined that

( , ) ( ) ( ) ( ) ( ) ( )y y y x y x r x dx c y a y aj i L j i j j a i j= + +λ λ λλ j b i

a

b

jc y b y b( ) ( )∫ (5.97)

Subtracting Equation 5.97 from Equation 5.96 leads to

( ) ( , )

( ) ( ) ( ) ( )

y y y y

y x y x r x dx c

i j L j i L

i j i j a

−

= − +λ λ yy a y a c y b y bi j b i

a

b

j( ) ( ) ( ) ( )+⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥∫ (5.98)

Since the inner product is commutative, Equation 5.98 implies that

( ) ( ) ( ) ( ) ( ) ( ) (λ λi j i j a i j b iy x y x r x dx c y a y a c y b− + + )) ( )

a

b

jy b∫⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥= 0 (5.99)

Since λ i and λ j are distinct,

y x y x r x dx c y a y a c y b yi j a i j b i

a

b

j( ) ( ) ( ) ( ) ( ) ( ) (+ +∫ bb) = 0 (5.100)

Equation 5.100 shows that the eigenvectors corresponding to distinct eigen-values of an eigenvalue problem of the form of Equation 5.87, Equation 5.88, and Equation 5.89 are orthogonal with respect to the inner product of Equation 5.92.

The development of the orthogonality condition leads to the defi nition of the kinetic-energy inner product defi ned by

( , ) ( ) ( ) ( ) ( ) ( ) ( ) ( )f g f x g x r x dx c f a g a c f b g ba bM = + +aa

b

∫ (5.101)

Eigenvectors corresponding to distinct eigenvalues are also orthogonal with respect to the inner product of Equation 5.101.

If the inner product of Equation 5.101 is used to normalize eigenvectors such that

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1

2 2

=( )

= [ ] + [ ] +∫

y y

y x r x dx c f a c f

i i M

i

a

b

a b

,

( ) ( ) ( ) (bb)[ ]2 (5.102)

then

y yi i L i,( ) = λ (5.103)

Example 5.14 The differential equation governing the torsional oscillations of the nonuniform circular shaft shown in Figure 5.12 is

∂∂

∂∂( )=

∂∂x

JGx

Jt

θρ

θ2

2 (a)

where G is the shaft’s shear modulus, ρ is its mass density, and J is the cross-sectional area polar moment of inertia. Since the shaft is fi xed at x = 0,

θ( , )0 0t = (b)

Application of the moment equation to the thin rigid disk at x = L leads to

−∂∂

=∂∂

J L Gx

L t It

L t( ) ( , ) ( , )θ θ2

2 (c)

where I is the mass moment of inertia of the disk about its centroidal axis.

L

x

IJ(x), ρ, G

θ(x, t)

Figure 5.12 The shaft of Example 5.14 has a thin disk attached at its end. Upon appli-cation of the normal-mode solution, an eigenvalue problem is obtained in which the eigenvalue appears in the boundary condition as well as the differential equation.

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Nondimensional variables are introduced according to

xxL

ttT

xJG

J G

xJJ

*

*

*( )

( )

=

=

=

=

α

βρ

ρ

0 0

0 0

(d)

where the subscript “0” refers to a reference value of the quantity. Substitution of Equation d into Equation a, Equation b, and Equation c leads to the nondi-mensional problem formulation,

∂∂

∂∂( )=

∂∂x x t

α βθ θ2

2 (e)

θ 0 0,t( )= (f)

α μ( ) ( , ) ( , )1 1 12

2

∂∂

= −∂∂

θ θx

tt

t (g)

where

μρ

=IJ L0 0

(h)

and t is nondimensionalized by

T LG

=ρ0

0

(i)

Substitution of the normal-mode solution, θ(x, t) = w(x)e i ω t, into Equation e, Equation f, and Equation g leads to the following problem for w(x):

− ( )=d

dxx

dwdx

x wα β ω( ) ( ) 2 (j)

w( )0 0= (k)

α μ( ) ( )1 1 12dwdx

w( )= ω (l)

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The eigenvalue problem of Equation j, Equation k, and Equation l is of the form of Equation 5.87, Equation 5.88, and Equation 5.89 with

Lwx

ddx

xdwdx

= − ( ) =1 2

βα λ ω

( )( ) (m)

L w a w a ca a( ) ( )= = 0 (n)

L w bdwdx

L cb b( ) ( ) ( )= =α μ1 (o)

The inner product of Equation 5.101 becomes

( , ) ( ) ( ) ( ) ( ) ( )f g f x g x x dx f a g aM

L

= +∫ β μ0

(p)

The total kinetic energy of the system is the kinetic energy of the shaft plus the kinetic energy of the disk,

Tt

Jdx It

L t

L

=∂∂( ) +

∂∂( )∫1

2

1

2

2

0

2θρ

θ( , ) (q)

Assuming that θ(x, t) = w(x) Re[e i ω t] = w(x) cos(ωt) and transforming to nondi-mensional variables using Equation d, Equation h, and Equation i leads to

T = ( )( ) [ ] + [ ]1

212 2 0 0 2ω ω β μcos ( ) ( ) ( )t

G JL

w x x dx w 22

0

1

2 2 0 01

2

∫⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

= ( )( )ω ωcos ( , )tG J

Lw w M (r)

Equation r shows that the kinetic energy is proportional to the energy inner product of the system response with itself. Determine the nondimensional natural frequencies and mode shapes of a uniform shaft with a thin disk at the end such that μ = 0.75.

Solution The eigenvalue problem for a uniform shaft is

d wdx

w2

20+ =λ (s)

w( )0 0= (t)

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dwdx

w( ) ( )1 1= μλ (u)

The general solution of Equation s is

w x C x C x( ) cos( ) sin( )= +1 2λ λ (v)

Application of Equation t to Equation v leads to C1 = 0. Application of Equa-tion u to the resulting form of Equation v leads to

λ λ μλ λ

μ λλ

C C2 2

1

cos( ) sin( )

tan( )

=

= (w)

Equation w is a transcendental equation whose solutions for λ are the eigenval-ues of the system. The solutions correspond to the values of the points of inter-section of the curves shown in Figure 5.13. As expected, there are an infi nite, but countable, number of eigenvalues, 0 1 2 1 1< < < < < < <− +λ λ λ λ λ… …k k k .

0 10 20 30λ

40 50 60

–15

–10

–5

0

5

10

15

20

f (λ)

Figure 5.13 The points of intersection are the eigenvalues of Example 5.14.

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As the eigenvalues grow large, the points of intersection of the curves approach integer multiples of π such that lim [( ) ]k k k→∞ = −λ π1 2 . The fi ve low-est values of the eigenvalues and the nondimensional natural frequencies for several values of μ are given in Table 5.3.

The mode shape corresponding to the eigenvalue λk is

w x C xk k k( ) sin( )= λ (x)

Mode-shape orthogonality is demonstrated by

( , ) ( ) ( ) ( ) ( )

si

w w w x w x dx w w

C C

i j M i j i j

i j

= +

=

∫0

1

1 1μ

nn( )sin( ) sin( )sin( )λ λ μ λ λi j i jx x dx0

1

∫ +⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

=−⎡⎣ ⎤⎦−

−+

C Cx

i ji j

i j

i jsin ( )

( )

sin ( )λ λλ λ

λ λ2

xx

C C

i j x

x

i j

⎡⎣ ⎤⎦+

⎧⎨⎪⎪

⎩⎪⎪

⎫⎬⎪⎪

⎭⎪⎪

+

=

=

20

1

( )

si

λ λ

μ nn( )sin( )

sin( )cos( ) cos( )sin

λ λ

λ λ λ

i j

i ji j iC C=

− (( )

( )

sin( )cos( ) cos( )

λλ λ

λ λ λ

j

i j

i j i

2 −

⎧⎨⎪⎪⎩⎪⎪

−+ ssin( )

( )

sin( )sin( )

λλ λ

μ λ λ

j

i j

i j

2 +

+⎫⎬⎪⎪⎭⎪⎪

(y)

Table 5.3 Natural Frequencies of Example 5.14 for Various Values of μ

μ ω1 ω2 ω3 ω4 ω5

0.1 1.429 4.306 7.182 10.200 13.241

0.25 1.265 3.935 6.814 9.812 12.869

0.5 1.077 3.649 6.578 9.630 12.721

0.75 0.951 3.505 6.486 9.563 12.671

1.0 0.860 3.426 6.437 9.529 12.640

2.0 0.653 3.292 6.362 9.477 12.606

10.0 0.311 3.173 6.292 9.435 12.574

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Obtaining a common denominator of λ λi j− and using Equation w to substitute for sin( )λi and sin( )λ j in the last term of Equation y leads to (wi , wj)M = 0.

The eigenvectors are normalized by requiring

1

2 2

0

1

=

= ⎡⎣ ⎤⎦ + ⎡⎣∫

( , )

sin( ) sin( )

w w

C x dx

i i

i i i

M

λ μ λ ⎤⎤⎦

⎧⎨⎪⎪⎪

⎩⎪⎪⎪

⎫⎬⎪⎪⎪

⎭⎪⎪⎪

= −[ ]+ ⎡⎣

2

2 1

2C xi iμ λsin( )⎤⎤⎦{ }

= −⎡

⎣⎢⎢

⎤

⎦⎥⎥+

=

=2

0

1

2 1

21

1

22

x

x

ii

iCλ

λ μsin( ) siin( )

sin( )cos( )

λ

λλ λ

i

ii

i iC

⎡⎣ ⎤⎦{ }= −

⎡

⎣⎢⎢

2

2 1

21

1 ⎤⎤

⎦⎥⎥+ ⎡⎣ ⎤⎦{ }μ λsin( )i

2 (z)

Noting from Equation x that cos( ) sin( )λ μ λ λi i i= , Equation z can be used to determine

Ci

i=

+2

1 2μ λ[sin ] (aa)

The normalized mode shapes corresponding to the fi ve lowest natural fre-quencies for several values of μ are given in Figure 5.14.

5.9 Eigenvalue problems involving Bessel functionsMany applications lead to eigenvalue problems involving Bessel functions. Consider the eigenvalue problem Ly = λy, where L is Bessel’s operator of order n,

Ly = −⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+

1 2

2xd

dxx

dydx

nyx

(5.104)

Equation 5.104 is of the form of the general Sturm-Liouville operator, Equa-tion 5.18, with p(x) = −x, q(x) = n2/x and r(x) = x. If the eigenvalue problem is to be solved over a domain a ≤ x ≤ b, then p(x) < 0 and q(x) > 0 are such that the operator is positive defi nite and self-adjoint with respect to the inner product,

( , ) ( ) ( )f g f x g x xdxa

b

= ∫ (5.105)

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If a = 0, then since p(0) = 0, it is required only at x = 0 that the solution be fi nite.

Consider the eigenvalue problem defi ned by the operator of Equation 5.104 defi ned for 0 ≤ x ≤ 1 with the condition y(1) = 0. The general solution of the eigenvalue problem is

y x C J x C Y xn n( ) ( ) ( )= +1 2λ λ (5.106)

Recalling that Jn(0) is fi nite, but Yn(0) does not exist, it is required that C2 = 0 for the solution to remain fi nite at x = 0. Application of the boundary condi-tion at x = 1 leads to the equation defi ning the eigenvalues:

Jn( )λ = 0 (5.107)

Since the operator is self-adjoint with respect to the inner product of Equation 5.105, it can be concluded that there are an infi nite, but countable,

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1–2–1.5

–1–0.5

00.5

11.5

2

x0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x

–2

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1x

–2

–1.5–1

–0.50

0.51

1.5

μ = 0.25(a)

(c)

(b)

μ = 10.0

μ = 0.75

–1.5–1

–0.50

0.51

1.52

W(x

)

W(x

)

W(x

)

Figure 5.14 Five lowest-mode shapes for the system of Example 5.14 with (a) μ = 0.25, (b) μ = 0.75, and (c) two lowest-mode shapes for μ = 10.0.

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number of solutions of Equation 5.107, λ λ λ λ λ1 2 1 1< < < < < <− +… …k k k . The eigenvectors are indexed with respect to k as

y x C J xk k n k( ) ( )= λ (5.108)

The eigenvectors are normalized by choosing

CJ x J x

J x xdx

k

n k n k

n k

=

=⎧

⎨∫

1

1

1

2

0

1

2

( ( ), ( ))

[ ( )]

λ λ

λ⎪⎪⎪⎪

⎩⎪⎪⎪

⎫

⎬⎪⎪⎪

⎭⎪⎪⎪

=+

1

2

1

2

Jn k( )λ (5.109)

The integral in Equation 5.109 is evaluated using the formulas derived in Example 3.10.

The Bessel functions satisfy the orthonormality relation,

δ λ

λ λ

k m k n k m n m

k m n k n m

C J x C J x

C C J x J

, ( ( ), ( ))

( ) (

=

=

λ

xx dx)

0

1

∫ (5.110)

Noting that the eigenvalues are the square roots of the zeroes of the Bessel functions, and defi ning αk,n as the kth positive zero of Jn(x), the normalized eigenvectors can be written as

yJ xJ

k nn k n

n k n,

,

,

( )

( )=

+2

1

αα

(5.111)

The fi rst ten zeroes for the fi rst fi ve Bessel functions are given in Table 5.4.The Bessel functions J xn k n( ),α for k = 1, 2, … are complete in the subspace V

of C2[0,1], defi ned such that if f(x) is in V, then f(0) = 1 and f(x) is fi nite at x = 0. Thus any function in V has an expansion of the form

f x AJ xJ

kn k n

n k nk

( )( )

( ),

,

=+=

∞

∑ 210

αα

(5.112)

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where

AJ

J x f x xdxkn k n

n k n=+ ∫2

10

1

( )( ) ( )

,,

αα (5.113)

A series of the form of Equation 5.112 is called a Fourier-Bessel expansion. As with the trigonometric Fourier series, it converges pointwise to f(x) for all x at which f(x) is continuous.

Example 5.15 Expand f(x) = 1 − x in a Fourier-Bessel series. Use n = 0, 1, 2, 3, 4, 5 and compare results.

Solution The coeffi cients in the Fourier-Bessel series for f(x) are

A f xJ xJ

J

kn k n

n k n x

=⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

=

+( ),

( )

( ),

,

2

2

1

αα

nn k nn k nJ x x xdx

+−∫

10

1

1( )

( )( ),

,α

α

(a)

The fi rst six Fourier coeffi cients for n = 0, 1, 2, 3, 4 are listed in Table 5.5. For example, the Fourier-Bessel expansion for n = 2 is

f xJ xJ

J( ) .

( . )

( . ).

( .= −2 0 2302

5 131

5 1310 1888

82

3

2 4417

8 4170 0793

11 62

11 623

2

3

xJ

J xJ

)

( . ).

( . )

( . )+

⎡

⎣⎢⎢⎢

− +0 055814 796

14 7960 0431

172

3

2.( . )

( . ).

(J xJ

J .. )

( . ).

( . )

( . )

96

17 960 0339

21 117

21 1173

2

3

xJ

J xJ

−⎤⎤

⎦⎥⎥+ ...

(b)

Table 5.4 Roots of Bessel Functions of Integer Order. αn, m is the mth Positive Zero of Jn(x)

n 0 1 2 3 4

αn,1 2.405 3.832 5.131 6.380 7.558

αn,2 5.520 7.016 8.417 9.761 11.065

αn,3 8.6754 10.173 11.620 13.015 14.373

αn,4 11.792 13.324 14.796 16.223 17.617

αn,5 14.931 16.471 17.940 19.409 20.827

αn,6 18.071 19.616 21.117 22.583 24.019

αn,7 21.212 22.760 24.270 25.748 27.199

αn,8 24.352 25.904 27.421 28.908 30.371

αn,9 27.493 29.047 30.569 32.065 33.537

αn,10 30.653 32.190 33.217 35.219 36.699

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The Fourier-Bessel approximations for f(x) are illustrated in Figure 5.15. The approximation for n = 0 is best near x = 0 because f(0) = 1, whereas Jn(0) = 0 for n > 1. The Fourier series converges pointwise to f(x) for all x except at x = 0 for n > 0.

There are many applications which lead to eigenvalue problems involv-ing Bessel’s equation or equations whose solution involves Bessel functions. The examples below include a buckling problem and a problem to determine natural frequencies and mode shapes for a nonuniform distributed-parameter system.

Table 5.5 Coeffi cients in Fourier-Bessel Series for f(x) = 1 − x. Akn the kth

Coeffi cient in the Expansion Using Jn(αk,n x) as Basis Functions

n 0 1 2 3 4

An1 0.288 0.261 0.230 0.206 0.181

An2 −0.0165 −0.0918 −0.119 −0.129 −0.131

An3 0.0102 0.0560 0.0793 0.0934 0.0985

An4 −0.00285 −0.00359 −0.0558 −0.0683 −0.0762

An5 0.00248 0.0269 0.0431 0.0541 0.0619

An6 −0.00138 −0.0202 −0.0339 −0.0438 −0.0511

1.2

0.8

0.6

0.4

f(x)

0.2

–0.20 0.1 0.2 0.3

Fourier-Bessel representations for f(x) = 1–x

1 – x

n = 0n = 1

n = 2 n = 3

n = 4

0.4 0.5x

0.6 0.7 0.8 0.9 1

0

1

Figure 5.15 The Fourier-Bessel series representation for f(x) = 1 − x is better for smaller n.

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Example 5.16 The differential equation for the slope of the elastic curve of a vertical column under its own weight

d

dxEI

ddx

f xθ

θ⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+ =( ) 0 (a)

where

f x g A d

x

( ) = ( )∫ρ ξ ξ0

(b)

Introducing nondimensional variables,

xxL

wwL

xEI

E I

x

g A d

gA L

L

*

*

( )

( )

=

=

=

=

( )∫

α

β

ρ ξ ξ

ρ

0 0

0

0 0

(c)

into Equation b leads to

d

dxx

ddx

f xαθ

λ θ( ) ( )⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+ = 0 (d)

where

λρ

= 0 03

0 0

gA LE I

(e)

The critical buckling length, Lcr, is the length at which the column buckles under its own weight. Determine the critical buckling length of a fi xed-free column when the column is uniform.

Solution The appropriate boundary conditions for a fi xed-free column are

ddx

θ( )0 0= (f)

θ( )1 0= (g)

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For a uniform column, α(x) = 1, and therefore Equation d simplifi es to

ddx

x2

20

θλ θ+ = (h)

Equation h is of the form of Equation 3.110 with r = 0 and s = 1. The solution of Equation h is of the form of Equation 3.111. The resulting general solution of Equation h is

θ λ λ( ) //

/ //

/x C x J x C x Y x= ( )+ (11 2

1 33 2

21 2

1 33 22

3

2

3 )) (i)

Since the order of the Bessel functions is not an integer or a half-integer an alternate representation of Equation i is

θ λ λ( ) //

/ //

/x C x J x C x J x= ( )+ −11 2

1 33 2

21 2

1 33 22

3

2

3(( ) (j)

The attempted application of Equation g meets with diffi culty. Whereas J−1 3 0/ ( ) is undefi ned, it is not clear what the value of d dx x J x/ ( ( ))/

/1 2

1 3− is at x = 0. It is convenient to use the series expansions of the Bessel functions to apply the boundary condition at x = 0. To this end,

x J x xk k

k1 2

1 33 2 1 22

3

1

14

3

1

3/

// / ( )

( )λ λ( )=

−

+ +( )Γ Γxx

k k

k

k

kk

3 22 1 3

0

2 1 3

11

3

1

//

/

( )

( )

=−( ) ( )

+

+

=

∞

+

∑

λ

Γ Γ ++( )=

∞+∑ 4

30

3 1

k

kx (k)

Then

ddx

x J xkk

1 21 3

3 22

3

1 3 11

3//

/( )

λλ

( )⎡⎣⎢⎢

⎤⎦⎥⎥=

−( ) + (( )+ +( )

+

=

∞

∑2 1 3

0

3

14

3

k

k

k

k kx

/

( )Γ Γ (l)

Equation l is used to determine

ddx

x J xx

1 21 3

3 2

0

1 3

2

3

1

34

3

//

/

/

λλ

( )⎡⎣⎢⎢

⎤⎦⎥⎥

=( )

= Γ(( ) (m)

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Similarly,

x J x xk k

k1 2

1 33 2 1 22

3

1

12

3

1

3/

// / ( )

( )− ( )=

−

+ +( )λ

Γ Γλλ

λ

x

k

k

k

kk

3 22 1 3

0

2 1 3

11

3

1

//

/

( )

( )

=−( ) ( )

+

−

=

∞

−

∑

Γ Γ kkx

k

k

+( )=

∞

∑ 2

30

3

(n)

ddx

x J xkk

1 21 3

3 22

3

1 31

3//

/( )

− ( )⎡⎣⎢⎢

⎤⎦⎥⎥=

−( ) (λ

λ))+ +( )

−

=

∞−∑

2 1 3

0

3 1

12

3

k

k

k

k kx

/

( )Γ Γ (o)

ddx

x J xx

1 21 3

3 2

0

2

30/

//

−=

( )⎡⎣⎢⎢

⎤⎦⎥⎥

=λ (p)

Thus, in view of Equation k and Equation p, application of Equation g to Equation j leads to C1 = 0 and

θ λ( ) //

/x C x J x= ( )−21 2

1 33 22

3 (q)

Application of Equation h to Equation q leads to the transcendental equation

J− ( )=1 32

30/ λ (r)

The smallest value of λ which satisfi es Equation r is λ = 7.833, and therefore, from Equation e, the critical buckling length of a uniform fi xed-free column is

LEIGA

cr =⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟1 986

0

1 3

./

ρ (s)

Example 5.17 Consider the nonuniform shaft with a thin disk and torsional spring attached at one end (x = 1) and fi xed at its other end (x = 0). The non-dimensional problem governing the natural frequencies and mode shapes of the shaft is

d

dxx

ddx

xα ω α( ) ( )Θ

Θ⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+ =2 0 (a)

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Θ( )0 0= (b)

ddxΘ

Θ Θ( ) ( ) ( )1 1 12+ =κ μω (c)

Determine the fi rst fi ve natural frequencies and mode shapes when κ = 1.5 and μ = 1 for α( ) ( . )x x= −1 0 2 2.

Solution Substitution of Equation d into Equation a leads to

d

dxx

ddx

x( . ) ( . )1 0 2 1 0 2 02 2 2−⎡⎣⎢⎢

⎤⎦⎥⎥+ − =

ΘΘω (d)

Defi ne

z x= −1 0 2. (e)

Equation d is rewritten using z as the independent variable as

ddz

zddz

z2 2 0Θ

Θ⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+ =λ (f)

where

λ ω= 25 2 (g)

Using z as the independent variable, the boundary conditions, Equation b and Equation c, become

Θ( )z = =1 0 (h)

ddzΘ

Θ Θ( . ) ( . ) ( . )0 8 5 0 85

0 8+ =κμ

λ (i)

Equation f is of the form of Equation 3.108 with r = 2 and s = 2. The solution to Equation f is obtained by comparison with Equation 3.108 as

Θ( ) ( ) ( )/

//

/z C z J z C z Y z= +−−

−−1

1 21 2 2

1 21 2λ λ (j)

Application of Equation h to Equation j gives

C

JY

C21 2

1 21= − −

−

/

/

( )

( )

λλ

(k)

which when used in Equation j leads to

Θ( ) ( )

( )

( )(/

//

//z z C J z

JY

Y= −−−

−−

1 21 1 2

1 2

1 21 2λ

λλ

λλz)⎡

⎣⎢⎢

⎤

⎦⎥⎥ (l)

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Application of Equation i to Equation l results in

( . . ) ( . )( )

( )/

/

/

0 2 0 6940 0 81 21 2

1 2

λ λλλ

− −−−

−−J

JY

Y 11 2

1 21

0 8

0 8922 0 8

/

//

( . )

. ( . )

λ

λ λ

⎡

⎣⎢⎢

⎤

⎦⎥⎥

= ′ −−−J

J 22

1 21 2 0 8

( )

( )( . )

//

λλ

λY

Y−

−′⎡

⎣⎢⎢

⎤

⎦⎥⎥

(m)

The kinetic-energy inner product for this system is

( , ) ( ) ( )( . ) ( ) ( )f g f x g x x dx f gK = − +∫ 1 0 2 1 12

0

1

(n)

The mode shapes satisfy the orthogonality relation, ( ( ), ( ))Θ Θi jx x i jM for= ≠0 . The mode-shape vectors are normalized by requiring that ( ( ), ( ))Θ Θi ix x M=1.

An alternate representation of the solution of Equation f is in terms of spherical Bessel functions. Equation f is in the form of Equation 3.15 with n = 0, whose solution in terms of spherical Bessel functions is given in terms of Equation 3.116. Thus, the solution of Equation f can be expressed as

Θ( )z C j z C y z= ( )+ ( )1 0 2 0λ λ (o)

Application of boundary conditions, Equation b and Equation c, to Equation o leads to the transcendental equation for the natural frequencies as

− ( ) + + −ω ω ω ω ω ω ω ωj y j y j1 0 0 12

04 5 5 5 1 5 4( ) ( ) ( ) ( . ) ( )yy j y0 0 05 5 4 0( ) ( ) ( )ω ω ω−[ ]= (p)

The differentiation relations ′ =j x j x0 1( ) ( ) and yj x y x0 1( ) ( )= are used in the development of Equation p. The mode shapes are of the form

Θk k k

k

kkx C j x

jy

y( ) ( . )= −( )−( )( )

00

005 1 0 2

5

55ω

ωω

ω (( . )1 0 2−( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

x (q)

The mode shapes are normalized by choosing

C j xjy

yk kk

kk= −( )−

( )( )

−00

005 1 0 2

5

55 1 0ω

ωω

ω( . ) ( .22 1 0 2

4

2

2

0

1

x x dx) ( . )( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

−⎧⎨⎪⎪⎪

⎩⎪⎪⎪

+

∫

j0 ωkkk

kk

jy

y( )−( )( )

( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

⎫⎬⎪⎪

⎭⎪⎪

0

00

25

54

ωω

ω )

−−1

2

(r)

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The following MATHCAD fi le is used to determine the fi ve lowest natural frequencies for the system using Equation p and to plot the corresponding mode shapes.

fl( ) : js(0,5 )ω ω= ⋅ The function js(n,x) is the spherical Bessel function of the fi rst kind of order n and argument x

f2( ) : js(0,4 )ω ω= ⋅

f3( ) : js(1,4 )ω ω= ⋅

gl( ) : ys(0,5 )ω ω= ⋅ The function ys(n,x) is the spherical Bessel function of the second kind of order n and argument x

g2( ) : ys(0,4 )ω ω= ⋅

g3( ) : ys(1,4 )ω ω= ⋅

Transcendental function used to determine natural frequencies

h( ) :ω ω ω ω ω ω ω ω= − ⋅ ⋅ + ⋅ ⋅ + −f g f g3 1 1 3 1 5 2( ) ( ) ( ) ( ) ( . )⋅⋅ ⋅ − ⋅( ( ) ( ) ( ) ( ))f g f g2 1 1 2ω ω ω ω

x: 0.4, 0.41.. 1C=

0 2 4 6 8 10−0.06

−0.04

−0.02

0

0.02

0.04

0.06

h (x)

x

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A graph of h over a small region helps to locate range of roots and to make good initial guesses for iterations

ω : 2= Guess for natural frequency obtained from graph

a : root(h( ), )= ω ω Internal root fi nding program given function and initial guess

a : 2.726= Value of natural frequency

Function defi ning mode shape

f x js a xjs a

ys a( ) [ , ( . )]

( , )

( ,:= −0 5 1 0 2

0 5

0 5⋅ ⋅ ⋅ −

⋅⋅ ))

[ , ( . )]⋅ ⋅ ⋅ − ⋅ys a x0 5 1 0 2

Plotting mode shape

x : 0,.01..1=

0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3Mode shape of Example 5.17 for ω = 0.616

f (x)

x

0 0.2 0.4 0.6 0.8 1−0.2

−0.15

−0.1

–0.05

0Mode shape of Example 5.17 for ω=2.72616

f (x)

x

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0 0.2 0.4 0.6 0.8 1–0.1

–0.05

0

0.05


f (x)

x

0 0.2 0.4 0.6 0.8 1–0.04

–0.02

0

0.02


f (x)

x

0 0.2 0.4 0.6 0.8 1–0.02

–0.01

0

0.01

0.02


f (x)

x

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5.10 Eigenvalue problems in other infi nite-dimensional vector spaces

A model for vibrations of double-walled nanotubes is presented in Example 1.4. The model is that of concentric beams connected by an elastic layer. A generalization of the model is that of a series of beams, not necessarily con-centric, connected by elastic layers. The outermost beams are connected to rigid foundations through elastic layers. The nondimensional differential equations for such a system can be summarized as

μ ε η ηii i

ii i i i i i

wx

wx

w w w w∂∂

−∂∂

+ −( )+ −− − +

4

4

2

2 1 1 1(( )+∂∂

=βiiw

ti n

2

21,2,... (5.114)

The fi rst and nth equations of the form of Equation 5.114 should be modifi ed to take into account that w0 and wn + 1 are zero. A normal-mode solution is assumed to be

w x t u x ei ii t( , ) ( )= ω (5.115)

Use of Equation 5.115 in the equations summarized by Equation 5.114 leads to

μ ε η η η ηii i

i i i i i id udx

d udx

u u u4

4

2

2 1 1 1− − + +( ) −− − − ii i iu i n+ = =12β ω 1,2,..., (5.116)

The differential equations of Equation 5.116 can be summarized in matrix form as

K K u Mu+ =c ω2 (5.117)

where K is the structural-stiffness operator matrix which can be written as

K K K= +b a (5.118)

in which Kb is a diagonal operator matrix representing the bending stiffness with

( ) ,kx

b i i i=∂∂

μ4

4 (5.119)

Ka is a diagonal operator matrix representing the axial stiffness as

( ) ,kx

a i i = −∂∂

ε2

2 (5.120)

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Kc is a tridiagonal matrix of coupling stiffnesses due to the elastic layers with

( ) ,k i nc i i i+ = =1 −η 2,3,..., (5.121a)

( ) , ,...,k i nc i i i= + =−η η1 1 2 (5.121b)

( ) , ,...,,k i nc i i i− −= = −1 1 1 2 1−η (5.121c)

and M is a diagonal mass matrix with

( ) ,m i ni i i= =β 1,2,..., (5.122)

Multiplying by M−1, Equation 5.117 becomes

M K K u u− + =1c( ) ω2 (5.123)

It is clear from Equation 5.123 that the natural frequencies are the square roots of the eigenvalues of the linear operator defi ned by L = M−1(K + Kc). The domain of L is Q, the space of vectors of the form f x f x f xn

T1 2( ) ( ) ( )…[ ] ,

where each fi(x) is an element of S, the subspace of C4[0,1] of functions that satisfy all boundary conditions for the individual beams. The standard inner product on Q is defi ned by

( , )u v v u

0

1

= ∫ T dx (5.124)

The kinetic-energy inner product on Q is defi ned as

u v v Mu

0

1

,( ) = ∫MT dx (5.125)

A potential-energy inner product is defi ned by

u v v Lu,( ) = ( )∫LT dx

0

1

(5.126)

It is easy to show that since Kc and M are symmetric matrices and the stretched-beam operator defi ned by Lbu = d4u/dx4 − ε(d2u/dx2) is self-adjoint with respect to the standard inner product on S, then L is self-adjoint with respect to the kinetic-energy inner product of Equation 5.125 and the potential-energy inner product of Equation 5.126. Furthermore, if either Kc or Lb is positive defi nite, then L is positive defi nite. In this case, Theorems 5.18 and 5.19 imply that all eigenvalues of L are real and positive and that

9533_C005.indd 3509533_C005.indd 350 10/29/08 2:30:19 PM10/29/08 2:30:19 PM


eigenvectors corresponding to distinct eigenvalues are orthogonal with respect to the kinetic-energy inner product.

It is diffi cult to obtain numerical values for the natural frequencies for the general stretched problem. A Rayleigh-Ritz method is presented in Section 5.15. However, it is easy to obtain the natural frequencies and mode shapes for unstretched beams. Thus, for the remainder of this section, set ε = 0, in which case Lb = M−1(Kb + Kc).

Let δk, for k = 1, 2, … be the eigenvalues of the problem d4φ/dx4 = δ4φ subject to the boundary conditions satisfi ed by each of the beams. Let φk(x) be the normalized eigenvectors corresponding to δk. A solution of Equation 5.117 is assumed to be in the form of

u ak k k x= φ ( ) (5.127)

where ak is a vector of constants. Substitution of Equation 5.127 into Equation 5.117 leads to

M K K a a− + =1 2( ˆ )k k kc ω (5.128)

where Kk is a diagonal matrix with

ˆ,

ki i

k i( ) = δ μ4 (5.129)

Equation 5.129 is in the form of an eigenvalue problem for an n × n matrix. Since M K Kand ˆ

k c+ are symmetric, the eigenvalues are all real, and if K Kk + c is positive defi nite, all eigenvalues are positive.

For each k, there are n natural frequencies obtained from the solution of Equation 5.129. Thus the natural frequencies are represented by ωk, j for k = 1, 2, … and j = 1, 2, …, n. Each natural frequency has a corresponding mode shape of uk, j = ak, jφk(x), where ak, j is the eigenvector of Equation 5.128 corre-sponding to the natural frequency ωk, j.

Example 5.18 Consider three unstretched, elastically connected beams with the following mass, bending-stiffness operator, and stiffness-coupling matrices:

M K=

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

=

∂∂

1 0 0

0 1 5 0

0 0 2

0 0

0 2

4

4

. b

x∂∂∂

∂∂

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

=

4

4

4

4

0

0 0 3

1

x

x

Kc

00000 10000 0

10000 30000 20000

0 20000 30000

−− −

−

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥ (a)

9533_C005.indd 3519533_C005.indd 351 10/29/08 2:30:20 PM10/29/08 2:30:20 PM


The normalized mode shape corresponding to a natural frequency ωk of a uniform fi xed-fi xed Euler-Bernoulli beam is

φ ω ω α ωk k k k k kx C x x x( ) {cosh( ) cos( ) [sinh( ) si= − − − nn( )]}ωk x (b)

where

αω ωω ω

kk k

k k=

−−

cosh( ) cos( )

sinh( ) sin( ) (c)

and

C x x x xk k k k k k= − − −⎡cosh( ) cos( ) sinh( ) sin( )ω ω α ω ω⎣⎣ ⎤⎦{ }⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥∫

−

2

0

1 1 2

dx

/

(d)

The lowest natural frequencies and values of δ for a fi xed-fi xed beam are

ω δ

ω δ

1

2

22 37 4 724

61 66

= =

=

. .

.

1

22

3 10.99

=

= =

=

7 852

120 93

4

.

.ω δ

ω 1199 9

2985

.

.

14.144δ

ω

=

= 99 17 295δ = .

(e)

The matrices Kk for k = 1, 2, …, 5 are calculated as

ˆ.

.

.

K1 = ××

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥

500 41 0 0

0 1 00 10 0

0 0 1 50 10

3

3 ⎥⎥

=×

××

⎡

⎣

⎢⎢⎢⎢⎢

ˆ.

.

.

K2

3

3

4

3 80 10 0 0

0 7 60 10 0

0 0 1 14 10

⎤⎤

⎦

⎥⎥⎥⎥⎥

=×

××

⎡ˆ

.

.

.

K3

4

4

4

1 46 10 0 0

0 2 92 10 0

0 0 4 38 10⎣⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

=×

×ˆ.

.

.

K4

4

4

3 99 10 0 0

0 7 99 10 0

0 0 1 200 10

8 92 10 0 0

0 1 78 10

5

3

4

5

×

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

=×

×ˆ.

.K 00

0 0 2 67 105. ×

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

(f)

9533_C005.indd 3529533_C005.indd 352 10/29/08 2:30:21 PM10/29/08 2:30:21 PM


The natural frequencies of the fi rst set of modes for the elastically connected beams are the square roots of the eigenvalues of M K K− +1

c( ˆ )1 . The mode shapes are of the form a1, j φ1(x) for j = 1, 2, 3, where a1, j are the eigenvectors of M K K− +( )1

1ˆ

c normalized with respect to the kinetic-energy inner product (u, v)M = (Mu, v). The fi rst set of natural frequencies and normalized mode shapes is

ω1 1 49 17

0 6069

0 4905

0 3479

, ,.

.

.

.

= =

⎡

⎣

⎢⎢w1 1 ⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

= = −

φ

ω

1

1 2 2111 8

0 7377

( )

.

.

, ,

x

w1 00 1487

0 4597

178 8

1

1 3

.

.

( )

.,

−

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

=

φ

ω

x

w1,

.

.

.

3

0 2958

0 6356

0 3915

= −

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥φ1( )x

(g)

The set ω1,1, ω1,2, and ω1,3 is a set of intramodal frequencies. The fi rst fi ve sets of intramodal frequencies are listed in Table 5.6, while the corresponding normalized mode shapes are given in Table 5.7.

Table 5.6 Sets of Intramodal Frequencies for Example 5.18

k = 1 k = 2 k = 3 k = 4 k = 5

ωk,1 49.17 80.73 139.01 216.52 312.16

ωk,2 111.88 128.79 174.44 254.42 364.15

ωk,3 178.84 190.97 226.45 294.22 395.89

Table 5.7 Normalized Mode-Shape Vectors ak, j, for Example 5.18

k/j 1 2 3 4 5

1 0 6069

0 4905

0 3679

.

.

.

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

0 6550

0 4779

0 3369

.

.

.

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

0 7902

0 4184

0 2377

.

.

.

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

0 9259

0 2852

0 1016

.

.

.

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

0 9776

0 1678

0 0804

.

.

.

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

2 0 7377

0 1487

0 4597

.

.

.

−

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

0 7012

0 1954

0 4749

.

.

.

−−

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

0 5572

0 3297

0 5076

.

.

.

−−

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

0 3438

0 5078

0 4975

.

.

.

−−

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

0 1947

0 6512

0 4037

.

.

.

−−

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

3 0 2958

0 6356

0 3915

.

.

.

−

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

0 2791

0 6325

0 4012

.

.

.

−

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

0 2321

0 6188

0 4311

.

.

.

−

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

0 1563

0 5722

0 4921

.

.

.

−

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

0 0804

0 4630

0 5796

.

.

.

−

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

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5.11 Solvability conditionsLet L be a linear operator defi ned on SD such that λ = 0 is an eigenvalue of L. Then Lu = 0 has a nontrivial (and nonunique) solution. Now consider

Lu f= (5.130)

for a given f. Let u1 be a nontrivial solution to Lu = 0. Then if u is a solution of Equation 5.130,

L u u Lu L u

Lu Lu

f

f

( ) ( )+ = +

= +

= +

=

c c

c

1 1

1

0

(5.131)

and thus, u + cu1 is also a solution of Equation 5.130 for any scalar c.The above shows that if a solution to Equation 5.130 exists, then it is not

unique. However, it is necessary to consider under what circumstances a solu-tion of Equation 5.130 exists. Equation (5.64) presents a representation of the inverse operator for a linear self-adjoint operator constructed from its eigen-values and eigenvectors. Equation (5.64) shows that if the operator has a zero eigenvalue, then its inverse operator does not exist. In such a case, there is not a one-to-one correspondence between the domain of L and the range of L. Equation 5.131 shows that more than one element of the domain is mapped into an element of the range. Conversely, not every element of the range has a corresponding element in the domain. That is, a solution of Equation 5.130 does not exist for all f seemingly in the range of L.

Consider the differential equation

d udx

n u x2

2

2 2 2+ =π (5.132)

where n is an integer, and subject to the boundary conditions

dudx

( )0 0= (5.133)

dudx

( )1 0= (5.134)

The homogeneous solution of Equation 5.132 is

u x C n x C n xh( ) cos( ) sin( )= +1 2π π (5.135)

9533_C005.indd 3549533_C005.indd 354 10/29/08 2:30:24 PM10/29/08 2:30:24 PM


Note that if the boundary conditions of Equation 5.133 and Equation 5.134 are applied to Equation 5.135, C2 = 0, but C1 remains arbitrary, resulting in a nontrivial homogeneous solution of C1 cos(nπx). Thus λ = 0 is an eigenvalue of the operator d2u/dx2 + n2π2u = λu subject to du/dx(0) = 0 and du/dx(1) = 0.

The particular solution of Equation 5.132 is up(x) = (2/n2π2)x, resulting in a general solution of

u x C n x C n xn

x( ) cos( ) sin( )= + +1 2 2 2

2π π

π (5.136)

Application of Equation 5.133 to Equation 5.136 leads to C2 = −(2/n3π3). Subse-quent application of Equation 5.134 to Equation 5.136 then leads to C2 = (−1)n + 1

(2/n3π3). The results for C2 are consistent for even n, but are contradictory for odd n. This implies that a nonunique solution of

u x C n xn

n xn

x( ) cos( ) sin( )= − +1 3 3 2 2

2 2π

ππ

π (5.137)

exists for even values of n, but that no solution exists for odd values of n.The above example illustrates that if λ = 0 is an eigenvalue of an operator

L, then a solution of the nonhomogenous equation of the form of Equation 5.130 exists under certain conditions.

Consider the case when L is self-adjoint with respect to an inner product, (u, v). Assume that u1 is a nontrivial solution of Lu = 0. Taking the inner product of both sides of Equation 5.130 with u1 gives

( ) ( )Lu,u f,u1 1= (5.138)

However, since L is self-adjoint and Lu1 = 0, (Lu, u1) = (u, Lu1) = (u, 0) = 0. Thus

(f,u )1 0= (5.139)

If for a specifi c f, Equation 5.139 is not true, then a contradiction has been obtained, and the assumption that u1 exists is false. Hence, if L is self-adjoint with respect to an inner product, the existence of a solution of Lu = f requires that f be orthogonal to all nontrivial solutions of Lu = 0. This is a solvability condition, a condition which must be satisfi ed for a solution of Equation 5.130 to exist.

Suppose L is not self-adjoint, let L* be the adjoint of L, and let v be a non-trivial solution of

L*v = 0 (5.140)

9533_C005.indd 3559533_C005.indd 355 10/29/08 2:30:25 PM10/29/08 2:30:25 PM


Taking the inner product of both sides of Equation 5.130 with v, using the defi nition of the adjoint operator, and then using Equation 5.140 results in

(

( , )

Lu,v f,v

u,Lv f,v

u, f,v

f v

*

) ( )

( ) ( )

( ) ( )

=

=

=

=

0

0 (5.141)

Equation 5.141 is the appropriate solvability condition when the operator is not self-adjoint. The nonhomogeneous term must be orthogonal to all non-trivial homogeneous solutions of the adjoint equation.

Theorem 5.20 (Fredholm Alternative) If Lu = 0 has a nontrivial solution, call it u1 (λ = 0 is an eigenvalue of L), then Lu = f has a solution if and only if (f, v) = 0, where v is the nontrivial solution of L*v = 0. When a solution of Lu = f exists, it is not unique, because u + cu1 is also a solution for any scalar c.

Let A be an n × n matrix. The system of equations Au = 0 has a nontrivial solution if and only if A is singular A = 0. Thus the solution of Au = f exists and is unique if A is nonsingular. However, if A is singular, a solution of Au = f exists if and only if f is orthogonal to the nontrivial solutions of A*v = 0 or, since the adjoint of a real matrix with respect to the standard inner prod-uct is its transpose, f must be orthogonal to nontrivial solutions of ATu = 0.

Example 5.19 Consider the nonhomogeneous system of algebraic equa-tions defi ned by

1 2

3 6

1

2

1

2

⎡

⎣⎢⎢

⎤

⎦⎥⎥⎡

⎣⎢⎢

⎤

⎦⎥⎥=

⎡

⎣⎢⎢

⎤

⎦⎥⎥

u

u

f

f (a)

The coeffi cient matrix in this system is singular, and therefore one of the eigen-values of matrix is zero. The adjoint of a real n × n matrix with respect to the standard inner product on R

n is its transpose. Thus, the homogeneous adjoint

problem is

1 3

2 6

0

0

1

2

⎡

⎣⎢⎢

⎤

⎦⎥⎥⎡

⎣⎢⎢

⎤

⎦⎥⎥=

⎡

⎣⎢⎢

⎤

⎦⎥⎥

v

v (b)

which has a nontrivial solution, v =−⎡

⎣⎢⎢

⎤

⎦⎥⎥

3

1. The solvability condition that

must be satisfi ed for a solution of Au = f to exist is ( ),3 1 01 2−[ ] [ ] =T Tf f , or 3f1 − f2 = 0.

A vector for which the solvability condition is satisfi ed is f =⎡

⎣⎢⎢

⎤

⎦⎥⎥

1

3, for

which Equation a leads to

u u1 22 1+ = (c)

9533_C005.indd 3569533_C005.indd 356 10/29/08 2:30:26 PM10/29/08 2:30:26 PM


3 6 31 2u u+ = (d)

Equation c can be rearranged to give u1 = 1 − 2u2. Equation d is simply three times Equation c. Thus u2 is arbitrary, call it c, and u1 = 1 − 2c. The solution can be summarized in vector form as

u

uc

1

2

1

0

2

1

⎡

⎣⎢⎢

⎤

⎦⎥⎥=

⎡

⎣⎢⎢

⎤

⎦⎥⎥+

−⎡

⎣⎢⎢

⎤

⎦⎥⎥ (e)

Example 5.20 Example 2.21 illustrates the use of a system of equations to determine the point of intersection of three planes. Consider three planes whose equations are

x y z d+ − = 1 (a)

2 2 2x y z d− + = (b)

− + − =3 3 5 3x y z d (c)

The matrix formulation of Equation a, Equation b, and Equation c is

1 1 1

2 1 2

3 3 5

1−−

− −

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥=

x

y

z

d

dd

d2

3

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥ (d)

The matrix in Equation d is singular. Determine the conditions under which the three planes intersect.

Solution Since the matrix is singular, it has an eigenvalue of zero. The adjoint of the matrix is its transpose. The nontrivial solution of the homoge-neous adjoint problem can be obtained from

1 2 3

1 1 3

1 2 5

0

0

−−

− −

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥=

x

y

z 00

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥ (e)

A solution of Equation e is

v = −−

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

1

2

1

(f)

9533_C005.indd 3579533_C005.indd 357 10/29/08 2:30:27 PM10/29/08 2:30:27 PM


The solvability condition is

d v,( )=

− −[ ]⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥=

− −

0

1 2 1 0

2

1

2

3

1 2 3

d

d

d

d d d == 0 (g)

The general solution of Equation e for this choice of d1, d2 and d3is

x

y

z

c

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥=

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥+ −

−

⎡

⎣

⎢⎢⎢

0

5

3

1

4

3⎢⎢

⎤

⎦

⎥⎥⎥⎥ (h)

for any value of c. Equation h is the vector representation of a line in three-dimensional space represented by the parametric equations

y x

z x

= −

= −

5 4

3 3 (i)

For example, the planes have a line of intersection when d1 = 2, d2 = 1 and d3 = 0.

5.12 Asymptotic approximations to solutions of eigenvalue problems

Consider a problem in which it is desired to determine the eigenvalues of a linear operator of the form L L L= + ε 1, where ε is a small nondimensional parameter. Let λ be an eigenvalue of L with corresponding eigenvector y, Ly = λy. The eigenvalue problem for the perturbed operator is

ˆ ˆ ˆ ˆLy y= λ (5.142)

Approximate solutions to equations containing small parameters are often attempted using asymptotic methods. The solution to an equation containing a small dimensionless parameter is attempted by expanding the unknown in an expansion in terms of functions, often powers, of the small parameter. The expansion is not assumed to be convergent, but asymptotic in which each term is smaller in magnitude than its preceding term. Only a few of the terms in the expansion are determined and the truncation error noted as a function of the dimensionless parameter. The resulting expansion must be uniform for all values of the independent variables and all parameters. Only problems that lead to uniform expansions are considered in this study.

9533_C005.indd 3589533_C005.indd 358 10/29/08 2:30:28 PM10/29/08 2:30:28 PM


Asymptotic expansions for the perturbed eigenvalue and eigenvector are assumed to be

λ λ ελ ε λ= + + +12

2 � (5.143)

and

y y y y= + + +ε ε12

2 � (5.144)

Substituting Equation 5.144 and Equation 5.143 into Equation 5.142 leads to

L L y y y y+( ) + +( )+ = + +( ) + +( )+ε ε λ ελ ε1 1 1 1� � � � � (5.145)

The operators are linear, thus

Ly Ly L y y y y1 1 1 1+ + + = + + +ε ε ε λ ελ ελ εO O( ) ( )2 2 (5.146)

where O(ε2) represents all terms of order ε2 and higher. Collecting coeffi cients of like powers of ε gives

Ly y Ly L y y y− + + − −( )+ =λ ε λ λ ε1 1 1 12( ) 0O (5.147)

Since ε is a small independent dimensionless parameter, powers of ε are linearly independent, and therefore coeffi cients of powers of ε must vanish independently. This leads to a set of hierarchal equations:

O(1) Ly y= λ (5.148)

O( )ε λ λLy y L y y1 1 1 1= − + (5.149)

Let λk,0 and yk,0 constitute an eigenvalue-eigenvector pair which satisfi es Equation 5.148. Defi ne λk,1 as the perturbation of the eigenvalue and y1,k the perturbation in the corresponding eigenvector. The eigenvalue perturbation is the value of λk,1 such that a solution, y1,k, exists of

Ly y L y yk 1 k 1 1 k 0 k 0, , , , , ,= − +λ λk k0 1 (5.150)

Let L* be the adjoint of L with respect to an inner product (u,v). Let Wk,0 be the eigenvector of L* corresponding to its eigenvalue λk ,0 . Application of the Fredholm alternative to Equation 5.150 leads to

− +( )=L y y w1 k 0 k 0 k 0, , , ,,λk 1 0 (5.151)

9533_C005.indd 3599533_C005.indd 359 10/29/08 2:30:28 PM10/29/08 2:30:28 PM


which in turn leads to

λk ,

, ,

, ,

,

,1 =

( )( )L y w

y w1 k 0 k 0

k 0 k 0

(5.152)

If the eigenvectors are normalized, then Equation 5.152 reduces to λk , , ,,1 =( )L y w1 k 0 k 0 .

The perturbation in the eigenvector is obtained using the expansion theo-rem. If the boundary conditions satisfi ed by yk,1 are the same as those satis-fi ed by yk,0 then

y yk 1 j 0, , ,==≠

∑α k j

jj k

1

(5.153)


L yα λ αk j j

jj k

k, , ,0

1

0

=≠

∑⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟− kk j j

jj k

k, , , , ,y y L yk 0 1 k 00

1

1

=≠

∑ = −λ (5.154)

Taking the inner product of both sides of Equation 5.154 with W�,0, with � ≠ jleads to

α λ αk j

jj k

k k j, , , , , , ,, ,Ly w y wj 0 0 j 0� �( )− ( )=≠

∑1

0 0

jjj k

k

k

j

=≠

∑ = −( )

−( )

1

1

0 0

λ

λ λ

, , , ,

, ,

,y L y wk 0 1 k 0 0�

�

==≠

∑ ( )= −1

0 1

j k

k j kα λ, , , , , , ,, ,y w y L y wj 0 k 0 1 k 0 0� �(( )

(5.155)

The principle of bi-orthonormality between the eigenvectors of L and the eigenvectors of L* requires y wj 0, , ,, � �0( )= δ j . Equation 5.155 simplifi es to

αλ λ

kk

,, ,

, ,

,�

�

�= −

( )−

L y w1 k 0 0

0 0

(5.156)

Example 5.21 Consider the eigenvalue problem

− + +⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟=

d ydx

dydx

y y2

22ε λ (a)

9533_C005.indd 3609533_C005.indd 360 10/29/08 2:30:29 PM10/29/08 2:30:29 PM


y(0) 0= (b)

y(1) 0= (c)

Noting that the eigenvalues and normalized eigenvectors of the problem −d2y/dx2 = λy subject to Equation b and Equation c are

λ πn n, ( )02= (d)

y n xn, sin( )01

2= π (e)

determine corrections to the eigenvalues and eigenvectors for small ε.

Solution Equation a can be written as Ly + εL1y = λy, where L1y = dy/dx + 2y. Asymptotic expansions for the eigenvalues and eigenvectors are assumed to be

λ λ ελn n n= + +, ,0 1 � (f)

y yn n n= + +, ,0 1ελ � (g)

The fi rst-order correction for the eigenvalue is obtained using Equation 5.151. Since the O(1) problem is self-adjoint, y = w = yn,0, and since the eigenvectors have been normalized, (yn,0, yn,0) = 1. Thus

λn n n

nn n

y y

dydx

y y dx

, , ,

,, ,

( , )1 0 0

00 0

0

1

2

=

= +( )∫

L1

== +[ ]

=

∫1

22

0

1

n n x n x n x dxπ πcos( ) sin( ) sin( )π π

1

2 (h)

An approximation for the eigenvalues is

λ ηπε

εn = + +( ) ( )2 2

2O (i)

The perturbation in the eigenvectors is obtained using Equation 5.152:

y yn n k k

k n

, , ,1 0=≠

∑α (j)

9533_C005.indd 3619533_C005.indd 361 10/29/08 2:30:30 PM10/29/08 2:30:30 PM


where from Equation 5.155,

απ π

π

n kn ky y

k n

n n x

,, ,( , )

( ) ( )

cos( )

=−

=

+

L1 0 0

2 2

1

2π 22

0

0

1

2 2 2

sin( ) sin( )

n x k x dx

k n

n

π π

( )π

[ ]

−

=

∫

eveen and even or odd and odd

2

k n k

nkk n( )2 2− ππ

n k n keven and odd or odd and even

⎧

⎨⎪⎪⎪⎪

⎩⎪⎪⎪

(k)

Thus, for even values of n,

yn k

k nk xn

k

,

, ,( )

sin( )1 2 2 2

1 3 5

2=

−=∑π

π (l)

and for odd values of n,

yn k

k nk xn

k

,

, ,( )

sin( )1 2 2 2

2 4 6

2=

−=∑π

π (m)

Occasionally the boundary conditions for the O(ε) problem are different than those for the O(1) problem. This difference is usually in the form of a nonhomogeneous boundary condition developed at O(ε). A transformation of the dependent variable can be performed to remove the nonhomogeneity from the boundary condition to the differential equation. An alternative is to take the inner product of both sides of Equation 5.149 with the homogeneous solution of the adjoint problem, as illustrated above. However, the evaluation on the left-hand side leads to a boundary term that does not vanish, which becomes part of the solvability condition. This method is illustrated in the following example.

Example 5.22 The nondimensional natural frequencies of a longitudinal motion of the nonuniform bar shown in Figure 5.16 are obtained from solv-ing the differential equation,

d

dxx

dudx

x uα α ω( ) ( )⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+ =2 0 (a)

subject to the boundary conditions

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u( )0 0= (b)

α η(1)dudx

u( ) ( )1 1 0+ = (c)

Consider a circular bar with a slight linear taper such that

α ε( ) ( )x x= −1 2 (d)

where ε is a small dimensionless parameter.The natural frequencies are the square roots of the eigenvalues of the

operator Lu x d x du dx dx= −1/ / /α α( )[ ( ( ) ) ] . Use an asymptotic method to deter-mine a two-term expansion for the natural frequencies.

Solution Equation a can be written as

− − +⎡⎣⎢⎢

⎤⎦⎥⎥= − +

ddx

x xdudx

x x u( )1 2 1 22 2 2 2ε ε λ( ε ε ) (e)

Asymptotic expansions for the eigenvalues are obtained by assuming

u u u u= + + +0 12

2ε ε � (f)

λ λ ελ ε λ= + + +0 12

2 � (g)

Substitution of Equation f and Equation g into Equation e leads to

− − + + + +⎡

⎣⎢⎢

⎤⎦⎥⎥

=

ddx

x xd

dxu u u( ) ( )1 2 2 2

0 12

2ε ε ε ε �

(( )( )( )λ ελ ε λ ε ε ε ε0 12

22 2

0 12

21 2+ + + − + + + +� �x x u u u (h)

x

α(x)

y(x, t)

Figure 5.16 The natural frequencies of the bar of Example 5.22 are approximated using an asymptotic expansion in terms of the taper rate of the bar.

9533_C005.indd 3639533_C005.indd 363 10/29/08 2:30:32 PM10/29/08 2:30:32 PM


Equation h can be rearranged to give

− + − −⎛⎝⎜⎜⎜

⎞⎠⎟⎟

d udx

d udx

xd udx

dudx

20

2

21

2

20

2

02 2ε ⎟⎟⎟⎡

⎣⎢⎢

+ − − +ε22

2

2

21

2

1 22

02 2d udx

xd udx

dudx

xd udx22

0

0 0 1 0 1

2

2

+⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟⎤

⎦⎥⎥

= + −

xdudx

u u xuλ ε λ λ( 00 0 1

20 2 0 1 0

20 1 1 1 02 2

+

+ − + + − +

λ

ε λ λ λ λ λ

u

u xu x u u xu

)

( λλ2 0u )+� (i)

Setting coeffi cients of like powers of ε to zero leads to the hierarchical equations

Od udx

u( )12

02 0 0− = λ (j)

Od udx

u xd udx

dudx

u x( )ε2

12 0 1

20

20

1 0 02 2 2+ = + − +λ λ λ uu0 (k)

Equation f and Equation g are used in the boundary conditions, Equation b and Equation c, leading to

u0 0 0( ) = (l)

u1 0 0( ) = (m)

u u0 01 1 0′ η( ) ( )+ = (n)

u u u1 1 01 1 2 1′ ′η( ) ( ) ( )+ = (o)

The solution for u0(x) is obtained as

u x c xk k k, ( ) sin0 = λ (p)

where λ βk k= and βk solves

η β βtan = − (q)

Equation p is normalized so that

c x dxk k

k

k k

=⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

=+

∫−

sin ( )

sin

2

0

11

2

2

2

λ

λλ λ

(r)

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For a solution of the O(ε) problem to exist, a solvability condition must be satisfi ed. Taking the inner product of both sides of Equation k with u0(x) leads to

ud udx

u u f x0

21

2 0 1 0, ( , ( ))+⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟=λ (s)

where f x x d u dx du dx u xu( ) ( ) ( )= + − +2 2 220

20 1 0 0 0/ / λ λ . Focusing on the left-

hand side of Equation s leads to

ud udx

u ud udx

u0

21

2 0 1 0

21

2 0 1, +⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟= +

⎛⎝

λ λ⎜⎜⎜⎜⎞⎠⎟⎟⎟⎟ = +∫ ∫ ∫

0

1

0

21

2

0

1

0 0

0

1

1dx ud udx

dx u u dxλ

= −ududx

ududx

01

0

1

10

00

1

1

20

2

0

1

0 0

0

1

1

011

1

+ +

=

∫ ∫ud udx

dx u u dx

udu

d

λ

( )( )

xxu u u u u u u− − + +0 1 1 0 1 0 1

0

1

0 0 1 1 0 0( ) ( ) ( ) ( ) ( ) ( )′ ′ ′ ∫∫ +⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟

d udx

u dx2

0

2 0 0λ (t)

Using boundary conditions for u0 and u1 at x = 0 and x = 1 and noting that the integrand of the integral is identically zero allows Equation t to be simplifi ed to

ud udx

u udu

dxu0

21

2 0 1 011

1, ( )

( )+

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟= +λ η 11

0 0

02

1

1 2 1

2 1

2

( )

( )( ( ))

( ( ))

( si

( )= ′

= −

= −

u u

u

ck

η

η nn )λk2 (u)

The solvability condition is obtained from Equation s and Equation u as

− = + −2 2 22 22

0

2

0η λ λ λc c x xd udx

dudxk k k ksin sin( ), k111 0 02u xuk+

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟λ (v)

Equation v is solved for λ1, leading to

λ η λ λ λ λ12 2 2

0

1

2 2= + ∫c c x x dk k k k k ksin sin( )[ cos( )] xx

ck k

k k

k k

= +

=++

2 2

2

2

1 2

2 1 2

( )sin

( )sin

sin

η λ

λ η λλ λ

(w)

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The fi rst four eigenvalues corresponding to η = 1.5 are calculated from Equa-tion q and Equation w as

λ ε

λ ε

λ

1

2

3

2 175 4 74

5 004 7 079

8 038 7 618

= +

= +

= +

. .

. .

. . εε

λ ε4 11 13 7 795= +. . (x)

Example 5.23 The stability of an elastic column has been considered in Example 5.4, where the critical buckling length is determined for a column subject to a compressive axial load, in Example 5.13 which dealt with the critical speed of a rotating column, and in Example 5.16 which dealt with the critical buckling length of a column under its own weight. Consider a column rotating with an angular speed Ω, subject to an axial load P, and to gravity. The nondimensional differential equation for a uniform column with these combined effects is

EId udx

P gALx Ld udx

A u4

42

2

22+ + =( )ρ ρ Ω (a)

The boundary conditions for the column are

u( )0 0= (b)

d udx

3

30 0( ) = (c)

dudx

( )1 0= (d)

d udx

2

21 0( ) = (e)

It is desired to construct a stability surface, a surface in the parameter space P, Ω, L, which is a boundary between stability and instability. As illustrated in Figure 5.17, the intersection of this surface with the P-axis is at the critical buckling load, its intersection along the L-axis gives the critical buckling length, and its intersection along the Ω-axis gives the critical speed. The sur-face intersects the P-L plane, defi ning a stability curve, if Ω = 0.

The critical values of the parameters are

PEILcr =

π2

24 (f)

LEIgAcr =

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟1 9866

1

3

.ρ

(g)

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Ωcr =⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟3 516

4

1

2

.EI

gALρ (h)

To defi ne the surface better, the following nondimensional parameters are introduced:

L

LL

* =cr

(i)

PP

P LL

* ==

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟cr

cr

2

(j)

ΩΩ

Ω

* ==

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟cr

crLL2

(k)

Substitution of Equation i, Equation j, and Equation k into Equation a leads to

d udx

PL xLd udx

4

4

2 32

29 86 7 833 201 69+ +⎡⎣⎢

⎤⎦⎥ =. . . Ω22 4L u (l)

P

Lcr

L

Ω

Ωcr

Pcr

Figure 5.17 The buckling of a long rotating column subject to an axial load is a func-tion of three parameters. The stability surface is the boundary between stable (inside the surface) and unstable (exterior to the surface) responses in the parameter space.

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Consider the case where gravity is small, but not negligible. To this end, assume L3 = ε << 1. Then Equation m becomes

d udx

P xd udx

4

4

2

32

29 86 7 833 201+ +⎡

⎣⎢⎢⎢

⎤

⎦⎥⎥⎥

=. .ε ε ..69 24

3Ω ε u (m)

P and L are independent, but assume that P = O(ε) = αε where α = O(1). Sub-stitution into Equation m gives

d udx

xd udx

u4

4

2

2

24

39 86 7 833 201 69+ +[ ] =. . .α ε εΩ (n)

Asymptotic expansions are assumed for Ω and u(x) of the form

Ω Ω Ω= + +−

0

2

31

1

3ε ε � (o)

u x u x u x( ) ( ) ( )= + +0 1ε � (p)

Determine Ω1.

Solution Substitution of Equation o and Equation p into Equation n, collect-ing coeffi cients of like powers of ε, and setting them to zero independently leads to

d udx

u4

04 0

20201 69= . Ω (q)

d udx

u u4

1

4 02

1 0 1 0201 69 403 38 9 86 7 8− = − +. . ( . .Ω Ω Ω α 3332

0

2x

d udx

) (r)

Both u0(x) and u1(x) satisfy boundary conditions of the form of Equation b, Equation c, Equation d, and Equation e.

Using the results of Example 5.13,

Ω0 1

2

3 516

201 69

0 2475= =.

( . )

. (s)

and

u x x0 003 768

3 768 3 768( ) cos( . )

sin( . ) sinh( .= +

+Ω

Ω ΩΩΩ Ω

Ω

0

0 0

0

3 768 3 768

3 768

)

cos( . ) cosh( . )

sin( .

+

× x)) cosh( . )

sin( . ) sinh( .

−

++

3 768

3 768 3 768

0

0 0

Ω

Ω Ω

x

))

cos( . ) cosh( . )sinh( . )

3 768 3 7683 768

0 00

Ω ΩΩ

+x (t)

9533_C005.indd 3689533_C005.indd 368 10/29/08 2:30:36 PM10/29/08 2:30:36 PM


The system of Equation s subject to Equation b, Equation c, Equation d, and Equation e constitutes a nonhomogeneous problem for which a nontrivial homogeneous solution exists. Thus a solution exists if and only if a solvability condition is satisfi ed. It is shown in Example 5.16 that the system of Equa-tion q subject to Equation b, Equation c, Equation d, and Equation e is not self-adjoint. Thus the appropriate solvability condition is that the nonhomo-geneous terms of Equation s are orthogonal to the nontrivial solution of the adjoint problem.

The boundary conditions for the adjoint problem are du/dx(0) = 0, d2u/dx2(0) = 0, u(L) = 0, and d3u/dx3(L) = 0. The solution of Equation q subject to these boundary conditions is

u x x0 003 768

3 768 3 76∗ ( ) cos( . )sin( . ) sinh( .

= ++

ΩΩ 88

3 768 3 768

3 768

0

0 0

0

ΩΩ Ω

Ω

)

cos( . ) cosh( . )

sin( .

+

× xx x) cosh( . )

sin( . ) sinh( .

+

−+

3 768

3 768 3 768

0

0

Ω

Ω Ω00

0 00

3 768 3 7683 768

)

cos( . ) cosh( . )sinh( . )

Ω ΩΩ

+x (u)

Requiring the nonhomogeneous term of Equation r to be orthogonal to u0*(x)

leads to

403.38 (9.86 7.833 )0Ω Ω −⎛⎝⎜⎜⎜ 0 1

20

2 0u xd udx

uα + ∗,⎞⎞⎠⎟⎟⎟⎟= 0 (v)

Equation w can be rearranged as

Ω1

20

2 09 86 7 833

403=

+⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟( . . ) ,

.

α ∗xd udx

u

338

9 86 7 833

0 0 0

20

2 0

0

1

Ω ( , )

( . . )

u u

xd udx

u dx

∗

∗α

=

+∫

4403 38

0 302 0 117

0 0 0

0

1

. ( ) ( )

. .

Ω u x u x dx∗

α

∫= − − (w)

Thus,

Ω = − +( )−

0 2475 0 302 0 1172

3

1

3. . .ε ε α (x)

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5.13 Rayleigh’s quotientA functional is a scalar function of a function of a vector. An energy inner product of a vector with itself is a functional. Suppose that L is a linear opera-tor defi ned on a domain SD and that L is a self-adjoint operator with respect to an inner product (f,g). Let y be an arbitrary element of SD. Rayleigh’s quotient is the functional defi ned by

R( ),

,y

Ly y

y y=

( )( )

(5.156)

For each y in SD, R(y) takes on a scalar value.Let u be an eigenvector of L corresponding to an eigenvalue λ. Then

R( )( , )

( , )

( , )

( , )

u =

=

=

Lu u

u u

u u

u u

λ

λ (5.157)

Equation 5.157 shows that the value of Rayleigh’s quotient for an operator L evaluated for an eigenvector of L is the eigenvalue corresponding to the eigenvector.

If L is self-adjoint and positive defi nite, then the expansion theorems imply that its eigenvectors u1, u2, …, uk−1, uk, uk + 1, … are complete in SD. Thus any vector in SD has an expansion of the form

y u=∑α i i

i

(5.158)


Ri

i

j

j( )

,

y

L u ui j

=

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟∑ ∑α α ⎟⎟⎟⎟⎟⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟∑ ∑α αi

i

j

i

u uj j,

(5.159)

Using the linearity of L and properties of inner products, Equation 5.159 becomes

9533_C005.indd 3709533_C005.indd 370 10/29/08 2:30:37 PM10/29/08 2:30:37 PM


R

i j

ji

i j

ji

( )

( , )

( , )y

Lu u

u u

i j

i j

=∑∑∑∑

α α

α α (5.160)

Noting that ui is an eigenvector of L corresponding to the eigenvalue λ i and using the orthonormality of the mode shapes, (u i , uj) = δi,j, Equation 5.160 reduces to

Ri i

i

i

i

( )y =∑∑

λ α

α

2

2 (5.161)

The vector y is dependent on the values of the coeffi cients in the eigenvec-tor expansion thus the value of Rayleigh’s quotient can be viewed as a func-tion of the coeffi cients,

R R k k k( ) ( , , , , , , )y = − +α α α α α1 2 1 1… … (5.162)

Rayleigh’s quotient is stationary when dR = 0 or when

01

12

2=∂∂

+∂∂

+ +∂∂

+R

dR

dR

dk

kα α αα α α� � (5.163)

Since α1, α2, … are all independent parameters, Equation 5.163 shows that R is stationary if

01 2 1 1

=∂∂

=∂∂

= =∂

∂=

∂∂

=∂

∂=

− +

R R R R R

k k kα α α α α� � (5.164)

Setting ∂R/∂αk = 0 in Equation 5.161 gives

λα

αii

ki

j∂∂

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟∑ ∑2

2

αj

⎟⎟−

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

∂∂

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟∑ ∑λ α

αα

i i

i

j

k

22

⎟⎟

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟

=

∑α j

j

2

2 0 (5.165)

9533_C005.indd 3719533_C005.indd 371 10/29/08 2:30:38 PM10/29/08 2:30:38 PM


The left-hand side of Equation 5.165 is zero if its numerator is zero. Simplify-ing leads to

2 2λ α δ αi i i k

i

j

j

,∑ ∑⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟−

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟∑ ∑λ α α δi i

i

j j k

j

2 2 , ⎟⎟⎟⎟⎟=

− =

− =

∑ ∑

∑

0

02 2

2

λ α α α λ α

α λ λ α

k k j

j

k j j

j

k k j j

j

( ) 00 (5.166)

An equation of the form of Equation 5.166 may be written for each k. There are as many independent solutions of the resulting set of equations as there are distinct eigenvalues. The kth solution is of the form αj = δk,j, j = 1, 2, … .

The uniqueness of these solutions is proven by assuming that the kth solution has more than just one αk different from zero. Without loss of generality, assume α α αi, and k� are all nonzero. Then the ith, �th and kth equations lead to

( ) ( )

( ) ( )

λ λ α λ λ α

λ α λ λ α

i i k k

i i k k

− + − =

− + −

� �

� �

2 2

2

0

λ 22

2 2

0

0

=

− + − =( ) ( )λ λ α λ λ αk i i k � � (5.167)

The determinant of the coeffi cient matrix for the system of equations in Equation 5.167 is 2( )( )( )λ λ λ λ λ λi k k i− − −� � . Thus, the only solution of Equation 5.167 is α α αi k= = =� 0. This argument may be repeated with the assumption of any number of nonzero coeffi cients.

The above shows that the only vectors in SD that render R(y) stationary are eigenvectors of L. Equation 5.157 shows that the value of Rayleigh’s quotient of an eigenvector is the eigenvalue corresponding to the eigenvector. Thus the minimum value of Rayleigh’s quotient for a self-adjoint operator is the operator’s smallest eigenvalue.

Rayleigh’s quotient can be used to determine an upper bound for the smallest eigenvalue of a self-adjoint operator. A set of trial vectors, each belonging to SD, is developed. The smallest eigenvalue must be less than the smallest value of Rayleigh’s quotient for all vectors in the set.

Now consider a subspace of SD, call it SD1, in which all vectors are orthogonal to u1, the eigenvector corresponding to the smallest eigenvalue of L. All vectors in SD1 can be represented by Equation 5.158, but with α1 = 0. The process described in Equation 5.159, Equation 5.160, Equation 5.161, Equation 5.162, Equation 5.163, Equation 5.164, Equation 5.165, Equation 5.166,

9533_C005.indd 3729533_C005.indd 372 10/29/08 2:30:39 PM10/29/08 2:30:39 PM


and Equation 5.167 are repeated for all vectors in SD1 with the result that the smallest value of R over all vectors in SD1 is λ2. Similarly, it can be shown that the smallest value of R for all vectors in SD2, the subspace of SD1 in which all elements are orthogonal to u1 and u2, is λ3.

Example 5.24 The differential equation for the transverse defl ection of a uniform column under an axial load has been shown to be

EId ydx

Py2

20+ = (a)

Consider a long vertical column where the axial load is generated due to gravity. The axial load varies over the length of the column as

P gAx= ρ (b)

where ρ is the mass density of the column and A is its cross-sectional area. Thus, Equation a becomes

EId ydx

gAxy2

20+ =ρ (c)

The boundary conditions for a pinned-pinned column are

y( )0 0= (d)

y L( ) = 0 (e)

Nondimensional variables are introduced as

xxL

∗ = (f)

yyL

∗ = (g)

Substitution of Equation e and Equation f into Equation b, Equation c, and Equation d leads to

EId ydx

xy2

20+ =λ (h)

y(0) 0= (i)

y( )1 0= (j)

where

λρ

=gALEI

3

(k)

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The critical buckling length, Lcr, is the length at which the column will buckle under its own weight. Equation h and Equation i constitute a Sturm-Liouville problem with λ as the eigenvalue of the operator,

Ly = −1 2

2xd ydx

(l)

The lowest eigenvalue of the operator defi ned by Equation l acting on the domain defi ned by Equation i and Equation j leads to the critical buckling length from Equation k.

Use Rayleigh’s quotient to determine an upper bound on the critical buck-ling length. Use the trial functions sinπx, x(1–x) and x(1–x2).

Solution The inner product for which L is self-adjoint is

( , ) ( ) ( )f g f x g x xdxr = ∫0

1

(m)

Rayleigh’s quotient is expressed as

Ry y

y y

xd ydx

y x

r

r( )

( , )

( , )

( )

yL

=

=−

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟

1 2

2xxdx

y x y x xdx

y xd ydx

dx

y x

0

1

0

1

2

20

1

∫∫

∫=

−

( ) ( )

( )

[ ( )]]2

0

1

xdx∫ (n)

Application of Rayleigh’s quotient to the suggested trial functions leads to

R xx x dx

x(sin( ))

sin( )[ sin( )]

[sin(

ππ π π

π=

− −∫ 2

0

1

))]2

0

1

xdx∫= 19.739 (o)

R x xx x dx

x x dx( ( ))

( )( )

( )

1

1 2

1

4

0

1

2

0

1− =− − −

−

=

∫∫

(p)

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R x xx x x dx

x x dx( ( ))

( )( )

( )

1

1 6

1

2

2

0

1

2 2

0

1− =− − −

−

=

∫∫

66 (q)

Thus, an upper bound on the critical buckling length can be determined such that

LEIgA

EIgAcr <

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟ =

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟4 1 587

1

3

ρ ρ. ⎟⎟⎟

1

3

(r)

Example 5.25 The differential equations governing the motion of the fi ve-degree-of-freedom system shown in Figure 5.18 are

200 0 0 0 0

0 100 0 0 0

0 0 300 0 0

0 0 0 100 0

0 0 0 0 100

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢

��

��

��

��

��

x

x

x

x

x

1

2

3

4

5

⎤⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥

+

−− −

−

500 100 0 0 0

100 300 200 0 0

0 200 400 −−− −

−

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥

200 0

0 0 200 300 100

0 0 0 100 200

⎥⎥⎥⎥⎥⎥⎥

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥

=

x

x

x

x

x

1

2

3

4

5

0

0

0

0

0

⎡⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥

(a)

The mode shape corresponding to the lowest mode of such a discrete system has no nodes. That is, there is no point in the system except at a fi xed support whose displacement is zero. This implies that all components of the mode-shape vector are of the same sign. Use Rayleigh’s quotient with three trial vectors for the lowest mode shape to determine an upper bound on the lowest natural frequency.

300 kg400

x1 x2 x3 x4 x5

200 kg 100 kg 100 kg 100 kgmN 100 m

N 200 mN 200 m

N 100 mN 100 m

N

Figure 5.18 A Rayleigh-Ritz method is used to approximate the lowest natural frequency and mode shape for the fi ve-degree-of-freedom system.

9533_C005.indd 3759533_C005.indd 375 10/29/08 2:30:41 PM10/29/08 2:30:41 PM


Solution It is shown in Section 5.4 that the natural frequencies are the square roots of the eigenvalues of M−1K, which is self-adjoint with respect to the kinetic-energy inner product (u, v)M = (Mu, v). Thus the appropriate form of Rayleigh’s quotient is

R( )( , )

( , )

( , )

( , )

yM Ky y

y y

Ky y

My y

=

=

−1M

M

(b)

Trial vectors with only positive components are chosen. Since the dis-crete system is constrained at both ends, the displacements of the masses closest to the support should be the smallest. The following trial vectors are used:

y y1 2=

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥

=

⎡

⎣

⎢⎢⎢

1

2

3

2

1

1

4

3

2

1

⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥

=

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥y3

2

3

4

2

1

⎥⎥⎥⎥⎥⎥

(c)

Substitution into Equation b leads to R(y1) = 0.2895, R(y2) = 0.3800 and R(y3) = 0.4143. Thus an upper bound for the lowest natural frequency is

ω1 0 2895 0 5380< =. .rad

s (d)

The exact value of the lowest natural frequency is 0.4633 rad/s.

5.14 Rayleigh-Ritz methodThe Rayleigh-Ritz method used to approximate the solution of Lu = f for a self-adjoint and positive defi nite operator L involves minimizing the energy norm of the difference between the exact solution and the Rayleigh-Ritz approximation. However, this approach does not work for eigenvalue prob-lems because eigenvectors are unique to at most a multiplicative constant.

Let L be a linear operator, defi ned on a domain SD, that is self-adjoint and positive defi nite with respect to an inner product (u, v). Rayleigh’s quotient is defi ned in Equation 5.156 as R(y) = (Ly, y)/(y, y). It was shown in Section 5.13 that R(y) is stationary when y is an eigenvector of L and that the value of Rayleigh’s quotient for an eigenvector is the eigenvalue corresponding to the eigenvector. Let ui i = 1, 2, …, n be a basis for a fi nite-dimensional subspace of SD. A Rayleigh-Ritz approximation for an eigenvector is of the form

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y u==

∑α i i

i

n

1

(5.168)

The Rayleigh-Ritz approximation for the eigenvector is determined as the vec-tor in SD for which the difference between the exact eigenvalue and Rayleigh’s quotient for the vector, R(w) − R(y), is minimized. This is equivalent to fi nd-ing the vector in SD for which Rayleigh’s quotient is a minimum.

The equation defi ning Rayleigh’s quotient can be rewritten as

( , ) ( )( , )Ly y y y y− =R 0 (5.169)

Substitution of Equation 5.168 into Equation 5.169, using the linearity of L and the properties of inner products, leads to

α α α αi j

j

n

i

n

i j i j i j

j

n

R== =

∑∑ ∑−11 1

( , ) ( ) ( , )Lu u y u uii

n

=∑ =

1

0 (5.170)

Differentiating Equation 5.170 with respect to αk for an arbitrary k, k = 1, 2, …, n leads to

∂∂

−∂∂

==∑∑ α

α α α αk

i j i j

j

n

i

n

ki j i j

R( )( , ) (Lu u u u

11α

))

( ) ( )( , )

j

n

i

n

ki j i j

j

n

i

n

R

==

==

∑∑

∑−∂

∂

11

11

y u uα

α α∑∑ = 0 (5.171)

Minimization of R(y) requires ∂R/∂αk = 0. Noting that ∂/∂αk(αiαj) = αiδj,k + αjδi,k , using the self-adjointness of L and the properties of inner products, Equation 5.171 reduces to

α αi i k

i

n

i i k

i

n

R k( , ) ( ) ( , )Lu u y u u= =

∑ ∑= =1 1

1, 2, ...,, n (5.172)

Equation 5.172 represents a system of n algebraic equations for the n coeffi -cients in the Rayleigh-Ritz expansion and can be written in matrix form as

A y Bα α= R( ) (5.173)

where α is the vector of coeffi cients,

( ) ( , ),A Lu ui j i j= (5.174)

9533_C005.indd 3779533_C005.indd 377 10/29/08 2:30:43 PM10/29/08 2:30:43 PM


and

( ) ( , ),B u ui j i j= (5.175)

Equation 5.173 is a homogeneous system of equations with R(y) as a param-eter. When this equation is rewritten as

B A y1− =α αR( ) (5.176)

It is clear that the appropriate values of R(y) are the eigenvalues of the matrix B−1A and the coeffi cients in the Rayleigh-Ritz expansion are the components of the corresponding eigenvectors.

Defi ne μ1, μ2, …, μn as the eigenvalues of B−1A. An eigenvector of B−1A cor-responding to μk is φk with components αi,k, i = 1, 2, …, n. The matrix B is clearly symmetric due to the commutativity of the inner product. The matrix A is symmetric due to the self-adjointness of L. Thus the eigenvalues are the eigen-values of a matrix which is the product of two symmetric matrices. Whereas the product is not necessarily symmetric, it is self-adjoint with respect to an energy inner product defi ned as (u, v)B = (Bu, v) (Section 5.4). Then the eigenvectors are orthogonal with respect to this energy inner product, φ φk B

k, � �( ) = ≠0 . The eigenvectors can be normalized such that (φk , φk)B = 1. The corresponding approximation to an eigenvector of L is

y uk ==

∑α i k i

i

n

,

1

(5.177)

The Rayleigh-Ritz approximation for eigenvalues and eigenvectors is based on minimization of Rayleigh’s quotient over all vectors in the span of the basis. Rayleigh’s quotient is a minimum when the trial vector is the eigenvec-tor corresponding to the lowest eigenvalue. Thus the approximate eigenvector of L obtained from the eigenvector corresponding to the smallest eigenvalue of B−1A approximates the eigenvector corresponding to the smallest eigen-vector of L. It is shown in Section 5.13 that if a trial vector is chosen orthogo-nal to the eigenvector corresponding to the lowest eigenvalue, then an upper bound is obtained for the next lowest eigenvalue. Thus it is reasonable to assume that if the approximate mode-shape vectors are mutually orthogo-nal, then they approximate eigenvectors for the larger eigenvalues, and the eigenvalues of B−1A are approximations of higher eigenvalues of L. To this end, consider

( , ) ,, ,y y uk � �=⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟= =

∑ ∑α αi k i

i

n

j

j

n

1 1

uj⎟⎟⎟⎟

===

∑∑ α αi k j

j

n

i

n

, , ( , )� u ui j

11

(5.178)

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Recall that the elements of the matrix B are the inner products (ui , uj). Then Equation 5.178 becomes

( , ) ( , )

,

y yk � �

�

=

=

φ φ

δ

k

k

B

(5.179)

Equation 5.179 shows that the eigenvector approximations are orthogonal with respect to the inner product of which L is self-adjoint.

Theorem 5.21 Let L be a linear operator whose domain is D such that L is self-adjoint and positive defi nite with respect to an inner product (u,v). Let u1, u2,…, un be a basis for a fi nite-dimensional subspace of D. The Rayleigh-Ritz approximation for the lowest eigenvectors of L, calculated as w uk i

ni k i= ∑ =1 α ,

for k = 1,2,…,n, are mutually orthogonal such that w wi j,( )= 0 for i j n, , ,...,= 1 2 and i ≠ j.

Example 5.26 Consider again the fi ve-degree-of-freedom system of Example 5.25. Use the Rayleigh-Ritz method to approximate the two lowest natural frequencies and mode-shape vectors from the subspace of R5 spanned by u1 1 2 3 2 1=[ ]T and u2

T= − −[ ]1 3 2 1 2 .

Solution As explained in Example 5.25, L = M−1K is self-adjoint with respect to the kinetic-energy inner product. Thus the inner products used in Ray-leigh’s quotient are (u, v) = (Mu, v) and (Lu, v) = (Ku, v). Then A is a 2 × 2 matrix whose elements are ai,j = (Ku i , uj), and B is a 2 × 2 matrix whose elements are bi,j = (Mu i , uj). Calculations lead to

A =× ×

× ×

⎡

⎣⎢⎢

⎤

⎦⎥⎥

1 1 10 9 10

9 10 3 3 10

3 2

2 3

.

. (a)

B =× ×× ×

⎡

⎣⎢⎢

⎤

⎦⎥⎥

3.8 103 2 2 10

2 2 10 2 8 10

3

3 3

.

. . (b)

B A− =−⎡

⎣⎢⎢

⎤

⎦⎥⎥

10 19 0 817

0 172 1 821

. .

. . (c)

The eigenvalues and normalized eigenvectors of B−1A can be calculated as

μ φ1 1

2

30 281

1 7 10

1 937 10= =

×− ×

⎡

⎣⎢⎢

⎤

⎦⎥⎥

−

−.

.

. (d)

μ φ2 2

2

21 729

1 4 10

2 6 10= =

− ××

⎡

⎣⎢⎢

⎤

⎦⎥⎥

−

−.

.

. (e)

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The Rayleigh-Ritz approximations for the lowest natural frequencies are ω μ ω μ1 1 2 20 53 1 315= = = =. .rad/s and rad/s. Approximations for the corresponding normalized mode-shape vectors are

y u u1

2 31 7 10 1 937 10

0 015 0 029 0 048 0

= × − ×

=

− −. .

. . .

1 2

.. .037 0 021[ ]T (f)

y u u2 1 2= − × + ×

= −

− −1 4 10 2 6 10

0 012 0 049 0 010 0

2 2. .

. . . .. .053 0 065−[ ]T (g)

The inner products developed for eigenvalue problems involving self-adjoint operators whose domains are Cn[a,b] are energy inner products. For example, the Sturm-Liouville operator is self-adjoint with respect to an inner product ( , ) ( ) ( ) ( )f g f x g x r x dxr a

b= ∫ , where r(x) is determined from the opera-tor. This is usually a kinetic-energy inner product. Since L is self-adjoint with respect to this inner product, when L is positive defi nite, the inner product ( f, g)L = (Lf, g)r is an energy product, usually a potential-energy inner product. Thus Rayleigh’s quotient,

R r

r

K

M

( )( )

( )

( )

( )

yLy, y

y, y

y, y

y, y

=

=

(5.180)

is the ratio of the potential-energy inner product to the kinetic-energy inner product.

The formulation of Equation 5.180 can be applied to energy inner products developed directly from the kinetic and potential energies of the system as illustrated in the following example.

Example 5.27 The differential equation and boundary conditions used to determine the natural frequencies and boundary conditions for the torsional oscillations of an elastic shaft fi xed at one end and with a thin disk and tor-sional spring attached to its other end are

∂∂

∂∂

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟=

∂∂x

GJx

Jt

θ θρ

2

2 (a)

θ(0, ) 0t = (b)

−∂∂

− =∂∂

GJ Lx

L t k L t It

L tt( ) ( , ) ( , ) ( , )θ

θθ2

2 (c)

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Nondimensional variables are introduced as

xxL

∗ = (d)

t tG

L∗ = 0

02ρ

(e)

α ∗( )( )

( )x

GJ xG J L

=0

(f)

βρρ

∗( )( )

( )x

J xJ L

=0

(g)

The nondimensional formulation of Equation a, Equation b, and Equation c is

∂∂

∂∂

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟=

∂∂x x t

α βθ θ2

2 (h)

θ( , )0 0t = (i)

∂∂

+ = −∂∂

θκθ μ

xt t

t( , ) ( , )1 1

2

2

θ (j)

where κ = ktL/J(L)G and μ = ID/ρJ(L)L. Use of the normal-mode solution, θ(x, t) = Θ(x)e i ω t, in Equation h, Equation i, and Equation j leads to

d

dxddx

α βωΘ

Θ⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+ =2 0 (k)

Θ(0) 0= (l)

ddxΘ

Θ Θ( )1 1 12+ =κ μω( ) ( ) (m)

Equation k is in the form of a Sturm-Liouville operator with p(x) = −α(x), q(x) = 0, and r(x) = β(x). Consider the energy inner product,

( , ) [ ( )] ( )

( ( ) )

f g f x g x dx

ddx

xdfdx

dgdx

d

K =

= −

∫ L

0

1

α xx0

1

∫ (n)

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Application of integration by parts to Equation n leads to

( , ) ( ) ( ) ( ) ( )f g xdfdx

dgdx

dfdx

gK = −∫ α α0

1

1 1 1 (o)

By defi nition α(1) = 1. Substitution of Equation m into Equation o leads to

( , ) ( ) ( ) ( ) ( ) (f g xdfdx

dgdx

f g f gK = + −∫ α μ0

1

21 1 1 1κ ω )) (p)

Using these inner products, Rayleigh’s quotient becomes

R f

xdfdx

dx f f( )

( ) ( ) ( )

(

=( ) + [ ] − [ ]∫ α μω

β

2

0

12 2 2

1 1κ

xx f x dx) ( )[ ]∫ 2

0

1 (q)

Equation q can be rearranged to give

α κ μω( ) ( ) ( )xdfdx

dx f f⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟ + [ ] − [∫

2

0

1

2 21 1 ]] − [ ] =∫2 2

0

1

0R f x f x dx( ) ( ) ( )β (r)

Note that if f is an eigenvector, R( f) = ω2, and Equation r can be written as

α κ β( ) ( ) ( ) ( )xdfdx

dx f R f x⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟ + [ ] −∫

2

0

1

21 ff x dx f( ) ( )[ ] + [ ]

⎧⎨⎪⎪⎪

⎩⎪⎪⎪

⎫⎬⎪⎪⎪

⎭⎪⎪⎪

=∫ 2

0

1

21 0μ (s)

Equation s is minimized as in Section 5.13. The defi nition of the potential-energy inner product used in Equation 5.180 is

( , ) ( ) ( ) ( )f g xdfdx

dgdx

dx f gK = +∫ α κ0

1

1 1 (t)

while the kinetic-energy inner product is defi ned as

( , ) ( ) ( ) ( ) ( ) ( )f g x f x g x dx f gM = +∫ β μ0

1

1 1 (u)

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Consider a shaft made from a uniform material with varying cross- section such that α(x) = β(x) = (1 − δx)2. Consider a shaft with δ = 0.2, κ = 1.5, and u = 1.0. (a) Use a Rayleigh-Ritz method to determine an approximation for the fi ve lowest natural frequencies and mode shapes using the fi ve lowest mode shapes for a uniform shaft as basis functions. (b) Use the fi nite-element method to approximate the fi ve lowest natural frequencies and mode shapes.

Solution (a) The fi ve lowest natural frequencies and mode shapes of the uniform shaft are obtained by solving the problem,

ddx

2

2

2 0Θ

Θ+ =ω (v)

Θ( )0 0= (w)

ddx

xΘ

Θ Θ( ) . ( ) ( )1 1 5 1 2+ = ω (x)

The solution of Equation v subject to Equation w is

Θ( ) sin( )x C x= ω (y)

Application of Equation x to Equation y leads to

tan( ).

ωω

ω=

−2 1 5 (z)

There are an infi nite, but countable, number of natural frequencies which satisfy Equation z; the fi rst fi ve are ω1 = 1.345, ω2 = 3.461, ω3 = 6.443, ω4 = 9.531, ω5 = 12.646. The eigenvectors of the form of Equation j are normalized by requiring

C x dx Ci i i isin( ) sin( )ω ω ω[ ] + [ ] =∫ 2

0

1

2 21 (aa)

The fi rst fi ve normalized mode shapes are

Θ

Θ

1

2

0 885 1 345

1 342 3 461

( ) . sin( . )

( ) . sin( .

x x

x x

=

= ))

( ) . sin( . )

( ) . sin( .

Θ

Θ

3

4

1 396 6 443

1 406 9 53

x x

x

=

= 11

1 410 12 6465

x

x x

)

( ) . sin( . )Θ = (bb)

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The MATHCAD fi le shown below provides the solution to part (a).

Rayleigh Ritz approximations for Example 5.27 (a)

Basis functions

ϕ1(x) := 0.885 ⋅ sin(1.345 ⋅ x) ϕ2(x) := 1.342 ⋅ sin(3.461 ⋅ x) ϕ3(x) := 1.396 ⋅ sin(6.443 ⋅ x) ϕ4(x) := 1.406 ⋅ sin(9.531 ⋅ x) ϕ5(x) := 1.410 ⋅ sin(12.646 ⋅ x)

ϕ

ϕϕϕϕϕ

( ) :

( )

( )

( )

( )

( )

x

x

x

x

x

x

=

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

1

2

3

4

5

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

Parameters

μ := 1 Mass ratio κ := 1.5 Stiffness ratio δ := .2 Taper ratio α(x) := (1 − δ ⋅ x)2 Nondimensional cross section area

Mass and stiffness matrix calculations

i := 1, 2 .. 5

j := 1, 2 .. 5

M x x x xi j i j i j, : ( ) ( ) ( ) ( ) ( )= ⋅ + ⋅ ⋅∫ ϕ ϕ α μ ϕ ϕd 1 10

1

M =

− −−0 985 0 07 0 072 0 044 0 035

0 07 0 858 0 035

. . . . .

. . . 00 024 0 012

0 072 0 035 0 826 0 059 0 011

0 044

. .

. . . . .

.

−

− 00 024 0 059 0 819 0 066

0 035 0 012 0 011 0 066 0

. . . .

. . . .− ..817

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

K xx

xx

x xi j i j i, : ( ) ( ) ( ) ( )= ⋅( )⋅( ) + ⋅ ⋅α ϕ ϕ κ ϕd

dd

dd 1 ϕϕ( )1

0

1

j∫

K =

1 832 0 179 0 197 0 02 0 097

0 179 9 665 1 768 0

. . . . .

. . . .1129 0 434

0 197 1 768 33 627 4 691 0 203

0 02 0 12

.

. . . . .

. . 99 4 691 73 718 9 02

0 097 0 434 0 203 9 02 129 973

. . .

. . . . .

⎛⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

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Natural frequency calculations

D := M−1 ⋅ K

D =

− −1 885 0 832 2 651 3 851 5 436

0 358 11 288 0 1

. . . . .

. . . 117 1 993 2 33

0 05 1 607 40 747 1 088 1 314

0 1

−− −

. .

. . . . .

. 008 0 303 2 664 90 076 2 113

0 033 0 659 0 391 3

− −−

. . . .

. . . .. .582 159 55

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

λ := eigenvals(D)

λ =

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜

159 453

90 141

1 847

11 299

40 807

.

.

.

.

.

⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

Eigenvalues of D

ω := λ0.5

ω =

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

12 627

9 494

1 359

3 361

6 388

.

.

.

.

.⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

Natural frequencies

W := eigenvecs(D)

W =

− −− −0 035 0 047 0 999 0 108 0 074

0 016 0 027 0

. . . . .

. . .0038 0 993 7 047 10

0 011 0 021 2 566 10 0

3

4

− − ××

−

−

. .

. . . .0054 0 996

0 031 0 997 1 366 10 5 376 103 3

−− − × − ×− −

.

. . . . 00 053

0 999 0 051 1 766 10 4 709 10 4 875 3

.

. . . . .− − × × −− − ××

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟−10 3

Q := WT ⋅ M ⋅ W

Q =

×

×

−

−

0 81 0 2 028 10 0 0

0 0 804 0 0 0

2 04 10 0 0 9

15

15

. .

.

. . 991 7 399 10 0

0 0 7 398 10 0 84 0

0 0 0 0 0 81

15

15

.

. .

.

××

⎛

⎝

−

−

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

9533_C005.indd 3859533_C005.indd 385 10/29/08 2:30:50 PM10/29/08 2:30:50 PM


Normalized eigenvectors

w1(x) := 1

1 10 5 1

1

5

( )( ( ) )

,. ,

QW xj j

j

⋅ ⋅=

∑ ϕ

w2(x) := 1

2 20 5 2

1

5

( )( ( ) )

,. ,

QW xj j

j

⋅ ⋅=

∑ ϕ

w3(x) := 1

3 30 5 3

1

5

( )( ( ) )

,. ,

QW xj j

j

⋅ ⋅=

∑ ϕ

w4(x) := 1

4 40 5 4

1

5

( )( ( ) )

,. ,

QW xj j

j

⋅ ⋅=

∑ ϕ

w5(x) := 1

5 50 5 5

1

5

( )( ( ) )

,. ,

QW xj j

j

⋅ ⋅=

∑ ϕ

Mode shape plots

x := 0, .01 .. 1

0 0.2 0.4 0.6 0.8 1–2

–1

0

1

2

w1(x)

w2(x)

w3(x)

w4(x)

w5(x)

x

9533_C005.indd 3869533_C005.indd 386 10/29/08 2:30:51 PM10/29/08 2:30:51 PM


(b) Piecewise linear functions which satisfy the geometric boundary con-dition at x = 0 and are continuous are defi ned as basis elements. The fi nite- element method is applied in the same fashion as the Rayleigh-Ritz method to approximate natural frequencies and mode shapes for this vibration prob-lem. A mass matrix is defi ned from the kinetic-energy inner products of the basis elements. A stiffness matrix is defi ned from the potential-energy inner products of the basis elements. The square roots of the eigenvalues of M−1K are the approximations. The mode-shape approximations are obtained from the eigenvectors, as illustrated in the MATHCAD fi le shown below. This fi le also demonstrates the orthogonality of the fi nite-element approximations for the mode shapes.

Example 5.28 (b) Finite Element approximation for Natural Frequencies and Mode Shapes for Torsional Oscillations of Non uniform shaft Five ele-ments of Equal Length from x = 0 to x = 1

Basis functions

φ0(x) := (1 − 5 ⋅ x)(Φ(x) − Φ(x − 0.2)) φ1(x) := 5 ⋅ x ⋅ (Φ(x) − Φ(x − 0.2)) + (2 − 5 ⋅ x) ⋅ (Φ(x − 0.2) − Φ(x − 0.4)) φ2(x) := φ1(x − 0.2) φ3(x) := φ1(x − 0.4) φ4(x) := φ1(x − 0.6) φ5(x) := (5 ⋅ x − 4) ⋅ Φ(x − 0.8)

x := 0, .01 .. 1

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1Basis functions

φ0(x)

φ1(x)

φ2(x)

φ3(x)

φ4(x)

φ5(x)

x

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u( ) :

( )

( )

( )

( )

( )

x

x

x

x

x

x

=

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

φφφφφ

1

2

3

4

5

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

Column vector of basis functions to facilitate calculation of inner products

Non-uniform area

α(x) := (1 − 0.2 ⋅ x)2 Non-uniform area of shaft μ := 1 Ratio of moment of inertia of disk to moment of

inertia of shaft κ := 1.5 Ratio of torsional stiffness of discrete spring to

torsional stiffness of shaft

Energy inner products

i := 1 .. 5

j := 1 .. 5

B x u x u x dx u ui j i j i j, : ( ) ( ) ( ) ( ) ( )= ⋅ ⋅ + ⋅ ⋅∫ α μ0

1

1 1

A xd

dxu x

d

dxu x dx ui j i j i. : ( ) ( ) ( ) ( )= ⋅( )⋅( ) + ⋅ ⋅α κ 1 uu j( )1

0

1

∫

A =

−− −

−

9 221 4 419 0 0 0

4 419 8 469 4 051 0 0

0 4 051 7

. .

. . .

. .. .

. . .

. .

749 3 699 0

0 0 3 699 7 061 3 353

0 0 0 3 353 4 8

−− −

− 667

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

Potential energy inner products

B =

0 123 0 029 0 0 0

0 029 0 113 0 027 0 0

0 0 027 0 103

. .

. . .

. . 00 025 0

0 0 0 025 0 094 0 022

0 0 0 0 022 1 044

.

. . .

. .

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

Kinetic energy inner products

D := B−1 ⋅ A Natural frequency approximations are square roots of the eigenvalues of D

9533_C005.indd 3889533_C005.indd 388 10/29/08 2:30:53 PM10/29/08 2:30:53 PM


λ := eigenvals(D)

λ =

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜

228 788

117 581

46 351

11 665

1 864

.

.

.

.

.

⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

Eigenvalues of D

wfe : .λ0 5

wfe =

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

15 126

10 843

6 808

3 415

1 365

.

.

.

.

.

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

Q := eigenvals(D)

Q =

− −− −

0 344 0 558 0 566 0 377 0 146

0 583 0 37 0 3

. . . . .

. . . 445 0 617 0 293

0 614 0 363 0 407 0 599 0 431

0 4

. .

. . . . .

.

− −005 0 647 0 626 0 308 0 549

0 015 0 033 0 063 0

. . . .

. . .− − − − .. .152 0 638

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

Eigenvectors of D will provide approximation to mode shapes

Ax := eigenvec(D, λ5)

Ax =

0 146

0 293

0 431

0 549

0 638

.

.

.

.

.

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

Noting that lowest natural frequency corresponds to λ5, the mode shape is determined using its corresponding eigenvector

z1 := 1

Ax B AxT ⋅ ⋅ Normalization with respect to the kinetic

energy inner product

z1 = 1.386

A1 =

0 202

0 406

0 597

0 76

0 884

.

.

.

.

.

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

9533_C005.indd 3899533_C005.indd 389 10/29/08 2:30:53 PM10/29/08 2:30:53 PM


w1(x) ( ( ) )A u xi i

i

11

5

⋅=

∑ Approximation for lowest mode shape

Repeating the above procedure to determine approximations to mode shape of higher modesA × 2 := eigenvec(D, λ4)

A × 2 =

−−−−

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

0 377

0 617

0 599

0 308

0 152

.

.

.

.

.⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

z1 := 1

2 2A B AT× ⋅ ⋅ × z1 = 2.416

A2 := z1 ⋅ A × 2

w2(x) := ( ( ) )A u xi i

i

21

5

⋅=

∑A × 3 := eigenvec(D, λ3)

A × 3 =

−−

−

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

0 566

0 345

0 407

0 626

0 063

.

.

.

.

.⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

z1 := 1

3 3A B AT× ⋅ ⋅ × z1 = 2.822

A3 := z1 ⋅ A × 3

wB(x) := ( ( ) )A u xi i

i

31

5

⋅=

∑A × 4 := eigenvec(D, λ2)

A × 4 =

−

−

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞0 558

0 37

0 363

0 647

0 033

.

.

.

.

. ⎠⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

9533_C005.indd 3909533_C005.indd 390 10/29/08 2:30:54 PM10/29/08 2:30:54 PM


z1 := 1

4 4A B AT× ⋅ ⋅ × z1 = 3.325

A4 := z1 ⋅ A × 4

w4(x) := ( ( ) )A u xi i

i

41

5

⋅=

∑

A × 5: = eigenvec (D, λ1)

A × 5 =

0 344

0 583

0 614

0 405

0 015

.

.

.

.

.

−

−

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

z1 := 1

5 5A B AT× ⋅ ⋅ × z1 = 3.957

A5 := z1 ⋅ A × 5

w5(x) := ( ( ) )A u xi i

i

51

5

⋅=

∑

x := 0, .01 .. 1

0 0.2 0.4 0.6 0.8 1–3

–2

–1

0

1

2

3FE approximations of mode shapes for non-uniform shaft

w2(x)

w1(x)

w3(x)

w4(x)

w5(x)

x

9533_C005.indd 3919533_C005.indd 391 10/29/08 2:30:55 PM10/29/08 2:30:55 PM


w(x) :=

w x

w x

w x

w x

w x

1

2

3

4

5

( )

( )

( )

( )

( )

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

Orthogonality check

i := 1 .. 5

j := 1 .. 5

V x w x w x dx w wi j i j i j, : ( ) ( ) ( ) ( ) ( )= ⋅ ⋅ + ⋅ ⋅∫ α μ 1 10

1

V =

1 5 027 10 1 125 10 1 119 10 5 613 106 6 6. . . .× − × − × − ×− − − −66

6 6 75 027 10 1 1 788 10 5 653 10 2 948 10. . . .× − × × ×− − − −66

6 6 61 125 10 1 788 10 1 8 474 10 7 281− × − × − × − ×− − −. . . . 110

1 119 10 5 653 10 8 474 10 1 9 96

6

6 7 6

−

− − −− × × − × ×. . . . 110

5 613 10 2 948 10 7 281 10 9 96 1

6

6 6 6

−

− − −− × × − × ×. . . . 00 16−

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

K xd

dxw x

d

dxw x dx wi j i j i, : ( ) ( ) ( ) ( )= ⋅( )⋅( ) + ⋅ ⋅α κ 1 ww j( )1

0

1

∫

K =

1 848 2 036 10 0 012 0 016 0 012

2 036 10 11

3

3

. . . . .

.

× −×

−

− .. . . .

.

67 6 945 10 7 857 10 5 753 10

0 012 6

3 3 3− × × ×− −

− − −

.. . . .

.

945 10 46 357 5 289 10 3 526 10

0 016

3 3 3× − × − ×− − −

77 857 10 5 289 10 117 585 2 793 10

0 01

3 3 3. . . .

.

× − × ×− − −

22 5 753 10 3 526 10 2 793 10 228 793 3 3. . . .× − × ×

⎛

⎝

⎜⎜

− − −

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

Example 5.29 Consider again the system of Example 1.3 and Section 5.10 of n stretched beams connected by elastic layers. Numerical computation of the natural frequencies and mode shapes for the general case is diffi cult. While an exact solution for the discrete differential equations is assumed to be of the form aeαx, a polynomial equation of order 4n is obtained to solve for the pos-sible values of α. To make matters worse, the equation’s coeffi cients are func-tions of ω. Thus, a numerical procedure must be applied to fi nd the values of ω for which the boundary conditions are satisfi ed. Numerical computation is diffi cult, partly because it involves evaluation of ex for large values of x.

Develop a Rayleigh-Ritz method as an alternative to the described numer-ical procedure which can be used to approximate the natural frequencies and mode shapes of a series of stretched elastically connected beams. Apply

9533_C005.indd 3929533_C005.indd 392 10/29/08 2:30:56 PM10/29/08 2:30:56 PM


the procedure to the model of a triple-walled nanotube and plot the natural frequencies as a function of ε.

Solution Natural-frequency and mode-shape calculations for elastically connected unstretched Euler-Bernoulli beams are illustrated in Example 5.17. The mode shapes satisfy the same boundary conditions as the mode shapes for the stretched beams, except for the case when one end is free. Thus an appropriate choice of basis functions for a Rayleigh-Ritz approximation of the frequencies of a set of elastically connected stretched beams are the mode shapes for the corresponding unstretched beams.

Using the notation of Section 5.10, the fi rst m sets of natural frequencies of the unstretched beams are defi ned as ωk,j for k = 1, 2, …, m and j = 1, 2, …, n. The corresponding normalized mode shapes are ak,j φk(x). The energy inner products are defi ned by

( , )u v v MuMT= ∫ dx

0

1

(a)

u v v K K K u,( ) = + +( )∫VT

a b c dx0

1

(b)

An advantage of using mode shapes of the unstretched beams as basis func-tions for a Rayleigh-Ritz approximations for the response of the stretched beam is that the mass matrix for both problems is the same. Therefore the basis functions satisfy appropriate orthonormality conditions with respect to the kinetic-energy inner product,

( , ), , ,a ak jφ φ δ δk q j p k qp,q M = (c)

In addition, the mode shapes for the unstretched beams are mutually orthog-onal with respect to the potential-energy inner product for the unstretched beams,

( , ) [( ) ], , , ,a a a K K ack j k p q q L p q

Tp b k j k dxφ φ = +φ φ

0

1

∫∫= ω δ δk j j p k q, , ,

2 (d)

Thus,

( , ) [ (, , , , , ,a a a K ak j k p q q v k j j p k q p qT

pφ φ ω δ δ φ= +2a kk j k dx, )]φ

0

1

∫ (e)

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Reconsider the set of three fi xed-fi xed Euler Bernoulli beams of Example 5.17. The fi rst fi ve sets of intramodal frequencies are given in Table 5.6, while the corresponding normalized mode shapes are given in Table 5.7. These fi f-teen modes will be used as basis functions in the Rayleigh-Ritz approxima-tions for the natural frequencies and mode shapes of a set of three elastically connected stretched fi xed-fi xed beams:

u u u

u u

1 1 1 1

1

1 1 2 2 1 3 3 1

4 2 2 5 2 2

= = =

= =

a a a

a a

, , ,

, ,

φ φ φ

φ φφ φ

φ φ φ

2 6 2 3 2

7 3 3 8 3 2 3 9 3 3 3

10

u

u u u

u

1

=

= = =

=

a

a a a

,

, , ,

aa a a

a a

4 4 11 4 2 4 12 4 3 4

13 5 5 14 5

, , ,

,

1

1

u u

u u

φ φ φ

φ

= =

= = ,, ,2 5 15 5 3 5φ φu = a (f)

The Rayleigh-Ritz approximation is of the form

w u( )x i i

i

==

∑α1

15

(g)

In view of the orthonormality condition of Equation c which is satisfi ed by the basis functions, the matrix B of Equation 5.173 and Equation 5.175 is the 15 × 15 identity matrix. The matrix A is the matrix of potential-energy inner products which are calculated using Equation d. Note that

a K a a ak jp qT

p k j k p qT

pdx xd

, , , ,[ ( )] ( )φ φa

0

12

∫ = −ε φφφk

dxdx

2

0

1

∫ (h)

The A matrix for this example is listed in Table 5.8. The natural-frequency approximations are the square roots of the eigenvalues of the matrix A. The natural frequencies for ε = 50 are listed in Table 5.9.The Rayleigh-Ritz approximations for the mode shapes can be calculated by

w ui i j i

j

p x==

∑ , ( )1

15

(i)

The mode-shape approximations for four modes are illustrated in Figure 5.19.

5.15 Green’s functionsLet L be a linear second-order differential operator defi ned on a domain SD, a subspace of C2[a, b]. The operator is in the self-adjoint form of Equation 5.18.

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The Sturm-Liouville problem corresponding to the operator is Ly = λy. Let λi i = 1, 2, … be the eigenvalues of the Sturm-Liouville problem with corre-sponding normalized eigenvectors yi(x). The eigenvectors are orthonormal with respect to the inner product ( , ) ( ) ( ) ( )f g f x g x r x dxr z

b= ∫ .Consider the nonhomogeneous problem,

Lw f= (5.181)

where w is in SD. The expansion theorem implies that there exist coeffi cients ci i = 1, 2, … . such that

w x c y xi i

i

( ) ( )==

∞

∑1

(5.182)

It is assumed that f(x) is of a form such that it has a convergent eigenvector expansion of the form

Mode shapes corresponding to w_1,1 for ε = 50(a) (b)

(c)

(e)

(d)

Mode shapes corresponding to w_2,2 for ε = 50

Mode shapes corresponding to w_3,1 for ε = 50 Mode shapes corresponding to w_4,3 for ε = 50

Mode shapes corresponding to w_5,3 for ε = 50

0 0.2 0.4 0.6 0.80

0.5

1

w(x)1

w(x)2

w(x)3

w(x)1

w(x)2

w(x)3

w(x)1

w(x)2

w(x)3

w(x)1

w(x)2

w(x)3

w(x)1

w(x)2

w(x)3

x0 0.2 0.4 0.6 0.8

x

0 0.2 0.4 0.6 0.8x

0 0.2 0.4 0.6 0.8x

0 0.2 0.4 0.6 0.8x

–2

–1

0

1

2

–2

–1

0

1

2

–1

–0.5

0

0.5

1

–1

–0.5

0

0.5

1

Figure 5.19 Mode-shape approximations obtained using the Rayleigh-Ritz method for a set of three elastically connected beams, (a) w1,1, (b) w2,2, (c) w3,1, (d) w4,3, and (e) w5,3.

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396 A

dvanced engineering mathem

atics with m

odeling applications

Table 5.8 The A matrix for Rayleigh-Ritz Approximation of Natural Frequencies for a Series of Three Elastically Connected Euler-Bernoulli Beams

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

1 376×103 26.492 7.253 0.07 0.014 28×104 75.602 2.057 1.974 −0.193 0.058 41×103 61.436 33.098 13.291

2 26.492 299×104 81.654 0.024 0.071 0.011 00.116 40.849 31.643 −0.155 −0.145 32×103 58.932 62.122 52.621

3 7.253 81.654 238×104 168×103 0.014 0.06 29.604 86.871 06.777 −0.035 −0.06 −0.157 74.213 19.275 207.38

4 0.07 0.024 168×103 296×108 75.837 36.859 −0.856 −0.047 22×103 65.019 57.892 4.374 −0.765 0.334 −0.028

5 0.014 0.071 0.014 75.837 833×104 96.453 −0.385 −0.754 −0.085 66.075 92.826 10.508 −0.666 −0.476 −0.134

6 28.104 0.011 0.06 36.859 96.453 794×104 −0.055 −0.175 −0.674 101.54 85.948 15.401 −0.188 −0.318 −0.573

7 75.602 00.116 29.604 −0.856 −0.385 −0.055 856×104 37.597 33.381 −2.996 0.202 67×103 033×108 60.883 −9.221

8 2.057 40.849 86.871 −0.047 −0.754 −0.175 37.597 383×104 77.789 −1.3 −2.106 −0.094 86.043 44.461 16.661

9 1.974 −31.643 −306.777 −1.922×10−3 −0.085 −0.674 133.381 577.789 5.436×104 −0.282 −0.615 −2.063 −166.487 −333.284 −674.831

10 −0.193 −0.155 −0.035 −665.019 −466.075 −101.54 −2.996 −1.3 −0.282 5502×104 1.055×103 270.482 1.527 −0.074 2.049×10−3

11 0.058 −0.145 −0.06 157.892 −492.826 −185.948 0.202 −2.106 −0.615 1.055×103 17.008×104 853.548 0.374 0.956 −0.041

12 1.241×10−3 6.732×10−3 −0.157 4.374 10.508 −515.401 3.67×10−3 −0.094 −2.063 270.482 853.548 9.166×104 0.116 0.327 0.899

13 61.436 58.932 74.213 −0.765 −0.666 −0.188 033.103 86.043 66.487 1.527 0.374 0.116 104×105 93.326 61.116

14 33.098 62.122 19.275 0.334 −0.476 −0.318 60.883 44.461 33.284 −0.074 0.956 0.327 93.326 409×105 098×108

15 13.291 52.621 207.38 −0.028 0.134 −0.573 −9.221 16.661 74.831 149×103 −0.041 0.899 61.116 098×108 641×105

9533_C005.indd 3969533_C005.indd 396

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f x f y y xi r i

i

( ) ( , ) ( )==

∞

∑1

(5.183)

Substituting Equation 5.183 and Equation 5.182 into Equation 5.181 leads to

L c y f y yi i

i

i r i

i=

∞

=

∞

∑ ∑⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

=1 1

( , ) (5.184)

Using the linearity properties of L and noting that Lyi = λ i y i , Equation 5.184 can be rearranged to give

[ ( , ) ]λi i i r i

i

c f y y−=

∞

∑1

(5.185)

The eigenvectors are linearly independent. Therefore,

c f yii

i r=1

λ( , ) (5.186)

Table 5.9 Rayleigh-Ritz Approximations for Natural Frequencies of Three Elastically Connected Stretched Beams with ε = 50

k,j ωk, j

1,1 53.54

1,2 113.95

1,3 179.92

2,1 90.90

2,2 135.40

2,3 194.78

3,1 153.20

3,2 184.12

3,3 233.18

4,1 234.44

4,2 264.81

4,3 302.83

5,1 332.28

5,2 375.28

5,3 405.14

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Substituting Equation 5.186 into Equation 5.182, the nonhomogeneous solu-tion can be expressed as

w x f y y

f y y x r

ii r i

i

ii i

( ) ( , )

( ) ( ) ( ) (

=

=

=

∞

∑ 1

1

1λ

λξ ξ ξ))d

a

b

i

ξ∫∑=

∞

1

(5.187)

The series in Equation 5.187 is convergent. Therefore, the order of summation and integration may be changed, leading to

w x K x f r da

b

( ) ( , ) ( ) ( )= ∫ ξ ξ ξ ξ (5.188)

where the Green’s function is

K x y x yi

i i

i

( , ) ( ) ( )ξ ξ==

∞

∑ 1

1λ

(5.190)

The above shows the existence of a Green’s function for a self-adjoint operator. Given the Green’s function, Equation 5.188 is used to determine w(x). In this sense, the integral operator,

Gf K x f r da

b

= ∫ ( , ) ( ) ( )ξ ξ ξ ξ (5.190)

is the inverse operator of L, because L(Gf) = Lw = f and G(Lw) = Gf = w.The general defi nition of a Green’s function for a linear differential operator

L is a function of ξ and x such that if Lw = f, then

w K x f r da

b

= ∫ ( , ) ( ) ( )ξ ξ ξ ξ (5.191)

Equation 5.189 shows that the Green’s function does not exist if the operator is positive semi-defi nite, in which case the smallest eigenvalue is zero. This is consistent with the concept that the Green’s function represents an inverse operator for L. If 0 is an eigenvalue of L, then there exists a nontrivial w such that Lw = 0. If an inverse L−1 exists, then L−1(0) = w. However, if L−1 is linear, then L−1(0) = 0, which contradicts the assumption of a nontrivial w. Also recall

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that if 0 is an eigenvalue of L, then f must satisfy a solvability condition for a solution of Lw = f to exist.

Suppose that the Green’s function exists and is defi ned by Equation 5.190. Consider the conditions under which G is self-adjoint. To this end, consider

( , ) ( ) ( ) ( )

( , ) ( ) ( )

G Gf g f g x r x dx

K x f r d

r

a

b

a

=

=

∫

ξ ξ ξ ξ

bb

a

b

g x r x dx

K x f r

∫∫⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

=

( ) ( )

( , ) ( ) ( )ξ ξ ξaa

b

a

b

g x r x d dx∫∫ ( ) ( ) ξ (5.192)

Interchanging the order of integration in Equation 5.193 leads to

( , ) ( , ) ( ) ( ) ( ) (Gf g K x g x r x dx f rr

a

b

=⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥∫ ξ ξ ξξ ξ)d

a

b

∫ (5.193)

If K(x,ξ) = K(ξ,x), Equation 5.193 becomes

( , ) ( , ) ( ) ( ) ( ) (Gf g K x g x r x dx f rr

a

b

=⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥∫ ξ ξ ξξ ξ

ξ ξ ξ

)

( ) ( ) ( )

d

g f r d

a

b

a

b

∫

∫= G

= ( , )f g rG (5.194)

Equation 5.194 shows that the operator G is self-adjoint with respect to the inner product ( f, g)r if the Green’s function is symmetric, K(x, ξ) = K(ξ, x).

The above discussion provides some insight into the Green’s function, but does not provide a method for determining the Green’s function. This follows from a physical understanding of the Green’s function. To this end, consider the defl ection of a beam pinned at both ends, as illustrated in Figure 5.20. The differential equation governing the defl ection of the beam due to a distrib-uted load f(x) is

EId wdx

f x4

4= ( ) (5.195)

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The boundary conditions accompanying Equation 5.195 are

w( )0 0= (5.196a)

d wdx

2

20 0( ) = (5.196b)

w L( ) = 0 (5.196c)

d wdx

L2

20( ) = (5.196d)

Suppose that the load is simply a unit concentrated load applied at a distance ξ from the left support. The concentrated load can be mathematically described by f(x) = δ(x − ξ), where δ(x) is the Dirac delta function, also called the unit impulse function. The mathematical problem governing the defl ection is

EId wdx

x4

4= −δ ξ( ) (5.197)

subject to the boundary conditions of Equations 5.196. The solution can be obtained by direct integration of Equation 5.197 or by a number of other methods, including the moment-area method. Using direct integration to solve Equation 5.197, it should be noted that d3w/dx3 is discontinuous at x = ξ. Physically, the shear force developed in the cross-section of the beam is dis-continuous across the concentrated load (Figure 5.21), and the shear force is proportional to d3w/dx3. Mathematically, the integral of the Dirac delta func-tion is ∫ − = −0

xd u xδ τ ξ τ ξ( ) ( ), where u(x − ξ) is the unit step function which

is 0 for x < ξ and 1 for x > ξ. Repeated integration leads to a solution of Equation 5.197 as

q xEI

x u x Cx

Cx

C x C( , ) ( ) ( )ξ ξ ξ= − − + + + +1 1

6 6 23

1

3

2

2

34

⎡⎡

⎣⎢⎢

⎤

⎦⎥⎥ (5.198)

1

x

ξ

Figure 5.20 The kernel for the Green’s function which is used to determine the defl ection of the beam due to a distributed load is determined as the defl ection of the beam at a distance x from its left support due to a concentrated unit load a distance ξ from its left support.

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Application of the boundary conditions, Equation 5.196

q xEI

x u xL

x LL

( , ) ( ) ( )ξ ξ ξξ ξ ξ

= − − + −( ) + −(1 1

61

6 613

3

)) −( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

2ξL

x (5.199)

The value of q(a, b) represents the defl ection at x = a due to a concen-trated unit load at x = b. Conversely q(b, a) represents the defl ection at x = b due to a unit concentrated load applied at x = a. The defl ection obtained if a unit concentrated load is applied at x = a and then a unit concentrated load applied at x = b is the same as if the order of loading were reversed. The total work done by application of the loads in the former case is U q a a q a b q b ba b→ = + +1 2 1 2/ [ ( , )] ( , ) / [ ( , )] , while the total work done by applica-tion of the load fi rst at b and then at a is U q b b q b a q a ab a→ = + +1 2 1 2/ [ ( , )] ( , ) / [ ( , )] . Since the work is the same independent of the order of loading, q(a, b) = q(b, a). Since a and b are arbitrary, a generalization of this result is

q x q x( , ) ( , )ξ ξ= (5.200)

The function q(x, ξ) is called an infl uence function.The defl ection of w(x), a beam due to a superposition of loads f1(x) + f2(x) is

w(x) = w1(x) + w2(x), where w1(x) is the defl ection due to the loading described only by f1(x), and w2(x) is the defl ection due to the loading described only by f2(x). Consider a beam subject to a distributed load per length of f(x). The beam is divided into n segments of equal length, Δx = L/n. Defi ne fi(x) as the load applied over the interval iΔx ≤ x (i + 1)Δx. The distributed load over this interval is equivalent to a concentrated load f f x dxi i x

i xˆ = ∫ +Δ

Δ( ) ( )1 applied at ξi i x

i xi x x i x f x dx= + ∫ −+Δ ΔΔΔ( ) ( ) ( )1 . Using the concept of infl uence functions,

the defl ection along the length of the beam due to the distributed load applied over this interval is w x f q xi

ni i( ) ( , )= ˆ ξ . Using the superposition principle the

defl ection of the beam due to the entire distributed loading is

w x f q xni i

i

n

( ) ( , )==

∑ ˆ ξ1

(5.201)

A distributed load may be replaced by a concentrated load equal to its resul-tant at the centroid of the load distribution for the purposes of calculating internal forces in beams and reactions. However, to determine the defl ection,

(a)

V = RR

V

V = R – 1

(b) 1

R

V

Figure 5.21 The internal shear force in the beam is discontinuous at the location where the concentrated load is applied.

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this replacement must be done continuously rather than discretely. There-fore, Equation 5.201 is not exact. It can be rendered exact by taking the limit as the number of segments grows large (or equivalently, as Δx→0):

w x f q x

f q x d

x i i

i

n

L

( ) lim ( , )

( ) ( , )

=

=

→=

∑

∫

Δ 01

0

ˆ ξ

ξ ξ ξ

(5.202)

Equation 5.202 is an integral representation of the solution in the form of a Green’s function operator. The infl uence function q(x, ξ) is the Green’s func-tion for the problem described by Equation 5.195 and Equation 5.196. The Green’s function for this problem satisfi es the following properties:

q(x, ξ) satisfi es all boundary conditions.q(x, ξ) is a solution of the differential equation Lw = δ(x − ξ) for all x, 0 ≤ x ≤ ξ, and all ξ, ξ < x < L.q(x, ξ), ∂q/∂x(x, ξ), and ∂2q/∂x2(x, ξ) are all continuous for all x, 0 ≤ x ≤ L.∂3q/∂x3 has a jump discontinuity at x = ξ. The value of the discontinuity is such that

EIq

xEI

qx

∂∂

−∂∂

= −+ −3

3

3

31( , ) ( , )ξ ξ ξ ξ (5.203)

Example 5.30 The thin variable-area rod shown in Figure 5.22 is subject to an internal heat generation which varies across its length. The rod is of length L, has a fi xed temperature imposed at one end, and is insulated at the other end. The heat transfer coeffi cient between the rod and the ambient medium is small. Defi ning nondimensional variables by x* = x/L, θ = T − T0/T0, α(x*) = A(x)/A0, and f(x*) = u(x)/umax, where u(x) is the rate of internal heat

••

••

L

x q(x)T0

Figure 5.22 A Green’s function is used to determine the temperature distribution in the variable-area rod.

9533_C005.indd 4029533_C005.indd 402 10/29/08 2:31:06 PM10/29/08 2:31:06 PM


generation across the entire cross-section and umax its maximum value, the formulation of the problem governing the temperature distribution in the bar is

−⎡

⎣⎢⎢

⎤

⎦⎥⎥=β α

θddx

xddx

f x( ) ( ) (a)

θ( )0 0= (b)

ddx

θ( )1 0= (c)

with β = kA0T0/L2umax. Determine the Green’s function for the system with α(x) = 1 + εx and use the Green’s function to determine θ(x) if f(x) = x(1 − x).

Solution The Green’s function is the solution of

− +⎡⎣⎢⎢

⎤⎦⎥⎥= −β ε δ ξ

ddx

xdqdx

x( ) ( )1 (d)

subject to Equation b and Equation c. integration of Equation d gives

β ε ξ( ) ( )1 1+ = − − +xdqdx

u x C (e)

dqdx

u xx

Cx

= −−

++

+( )

( ) ( )

ξβ ε β ε1 1

1 (f)

Application of Equation c to Equation f gives C1 = 1, leading to

dqdx x

u xx

=+

−−

+1

1 1β εξ

β ε( )

( )

( ) (g)

Integration of Equation g results in

q x x u xx

( , ) ln( ) ( )lnξβε

ε ξεεξ

= + − −++

⎛⎝⎜⎜⎜

⎞⎠⎟1

11

1⎟⎟⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥+ C2 (h)

Application of Equation b to Equation h leads to C2 = 0 and the Green’s func-tion of

q x x u xx

( , ) ln( ) ( )lnξβε

ε ξεεξ

= + − −++

⎛⎝⎜⎜⎜

⎞⎠⎟1

11

1⎟⎟⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥ (i)

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Evaluation of the unit step functions of Equation g and Equation i leads to

dqdx

xx

x= +

<

>

⎧

⎨⎪⎪⎪

⎩⎪⎪⎪

1

1

0

β εξ

ξ( ) (j)

q x

x x

x x( , )

ln( )

ln( )

ξβε

ε ξ

βεε ξ

=

+ <

+ >

⎧

⎨

⎪⎪⎪⎪⎪

⎩

11

11

⎪⎪⎪⎪⎪⎪

(k)

Equation j and Equation k show that since the end where x = 1 is insulated and a unit heat source is applied at x = ξ, the rate of heat transfer is zero for x > ξ, leading to a constant temperature in this region. The rate of heat transfer is discontinuous at x = ξ, while the temperature is continuous at this point.

The solution of Equation a for any f(x) is of the form

θ ξ ξ ξ( ) ( ) ( , )x f q x d= ∫0

1

(l)

Substitution of the given f(x) into Equation l leads to

θβε

ξ ξ ε ξεεξ

( ) ( ) ln( ) ( )lnx x u xx

= − + − −++

⎛⎝⎜⎜

11 1

1

1⎜⎜⎞⎠⎟⎟⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

=+

−

∫ d

xd

ξ

εβε

ξ ξ ξ

0

1

0

1

11

ln( )( )∫∫ ∫+ − +

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

ξ ξ εξ ξ( )ln( )1 1

1

dx

(m)

=+

+ + + +( ){ +ln( )

ln(1

3612 6 5 18 15 8 1

4

3 3 2εβε

ε ε ε ε εx

))

( ) (− + + + − +( ) + +( ) +13 66 89 40 6 2 1 3 1 13 2 3 2ε ε ε ε εx x ++ +⎡

⎣⎢⎤⎦⎥

+ + + + + +

ε ε

ε ε ε

)( )

( ) ( ) ( )(

1

4 1 9 1 13 2

x

x x 11+ }εx)

The Green’s function of Example 5.29 satisfi es the following conditions:

q(x, ξ) satisfi es all the boundary conditions.q(x, ξ) is a solution of the differential equation Lw = δ(x–ξ) for all x, 0 ≤ x ≤ ξ and all ξ, ξ < x < 1.

••

9533_C005.indd 4049533_C005.indd 404 10/29/08 2:31:08 PM10/29/08 2:31:08 PM


q(x, ξ) is continuous for all x, 0 ≤ x ≤ 1.∂q/∂x has a jump discontinuity at x = ξ. The value of the discontinuity is such ∂ ∂ −∂ ∂ = − ++ −q x q x/ ( , ) / ( , ) / ( )ξ ξ ξ ξ β εξ1 1 .

The Green’s function for a problem Lw = f subject to appropriate bound-ary conditions is the function q(x, ξ), which is the solution of Lq = δ(x–ξ) that satisfi es all boundary conditions.

The above examples specify the conditions which a Green’s function must satisfy and provide a rubric for the construction of the Green’s function. Con-sider a linear differential operator Lw a x d w dx a x d w dxn

n nn

n n= + −− −( )( ) ( )( )/ /1

1 1 a+ + 1� (( )dw dx a w/ + 0 . The Green’s function satisfi es the following conditions:

q(x, ξ) satisfi es Lq = δ(x − ξ) everywhere except at x = ξ.q(x, ξ) satisfi es all boundary conditions.q(x, ξ), ∂q/∂x, …, ∂n−2q/∂xn−2 are continuous for all x.∂n−1q/∂xn−1 is discontinuous at x = ξ, at which it has a jump discontinuity

∂∂

−∂∂

= −−

−+

−

−−

n

n

n

nn

qx

qx a

1

1

1

1

1( , ) ( , )

( )ξ ξ ξ ξ

ξ (5.204)

The Green’s function can be constructed for any linear differential opera-tor which does not have an eigenvalue of zero, in which case Lw = 0 has a non-trivial solution. If L is self-adjoint, the Green’s function is symmetric, and if the Green’s function is symmetric, the Green’s function operator is self-adjoint.

Example 5.31 Set up the Green’s function solution for Example 5.29, assum-ing the heat transfer coeffi cient between the surface and the ambient is a fi nite value h.

Solution In terms of the nondimensional variables defi ned in Example 5.29, the nondimensional problem governing the temperature distribution in the surface is

− +⎡⎣⎢⎢

⎤⎦⎥⎥+ =β ε

θμθ

ddx

xddx

f x( ) ( )1 (a)

θ( )0 0= (b)

ddx

θ( )1 0= (c)

where μ = hPumaxT0.Since the Green’s function satisfi es the differential equation Lw = δ(x −

ξ) for all x except x = ξ, it is a homogeneous solution for 0 ≤ x ≤ ξ and for ξ < x ≤ 1. Defi ning z = 1 + εx, the homogeneous equation is

ddz

zddz

θν θ

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟− =2 0 (d)

••

••••

9533_C005.indd 4059533_C005.indd 405 10/29/08 2:31:08 PM10/29/08 2:31:08 PM


with ν μ βε2 2= / . The general solution of Equation d is

θ ν ν( ) / /z C I z C K z= ( )+ ( )1 0

1 22 0

1 22 2 (e)

Thus, the Green’s function is of the form

q xC I x C K x

( , )( ) ( )/ /

ξν ε ν ε

=+( )+ +( )1 0

1 22 0

1 22 1 2 1 0 ≤≤ <+( )+ +( ) <

x

C I x C K x x

ξν ε ν ε ξ3 0

1 24 0

1 22 1 2 1( ) ( )/ / ≤≤

⎧⎨⎪⎪

⎩⎪⎪ 1 (f)

Requiring the Green’s function to satisfy the boundary conditions of Equa-tion b and Equation c leads to

q x

C I xIK

K

( , )

( )( )

( )(/

ξ

ν ενν

ν

=

+( )−1 01 2 0

002 1

2

22 1++( )

⎡

⎣⎢⎢

⎤

⎦⎥⎥

≤ <

+

ε ξ

ν

x x

C I

)

(

/1 2

3 0

0

2 1 εεν εν ε

νxIK

K)( )

( )(/

/

/1 2 1

1 2

11 2 0

2 1

2 12( )+

+( )+( )

11 11 2+( )⎡

⎣⎢⎢

⎤

⎦⎥⎥ < ≤

⎧

⎨

⎪⎪⎪⎪⎪⎪

⎩

⎪⎪⎪⎪⎪⎪ε ξx x) /

(f)

The solution is continuous at x = ξ. Thus,

C C H CI

IK

1 3 3

01 2 1

1 2

1

2 12 1

2= =

+( )++( )

ν εξν εν

( )( )/

/

(( )( )

(

//

12 1

2 1

1 2 01 2

0

+( )+( )

⎡

⎣⎢⎢

⎤

⎦⎥⎥

+

εν εξ

ν εξ

K

I )) ( )/ /1 2 0

00

1 22

22 1( )−

( )( )

+( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

IK

Kνν

ν εξ (g)

The form of C3 is obtained by requiring ∂ ∂ −∂ ∂ = ++ −q x q/ ( , ) / ( , ) / ( )ξ ξ ξ ξ ξ β εξ1 1 . Subsequent algebra leads to

CI H K

3

1 2

11 2

1

1

2 1 1 2 1=

++( ) − + +

( )

( ) ( ) (

/

/

εξβν ν εξ ν εξξ) ( )/1 2( ) +⎡⎣ ⎤⎦Q HR

(h)

where Q I K= + +11 2

11 22 1 2 1( ( ) )/ ( ( ) )/ /ν ε ν ε and R I K= 0 02 2( ) ( )ν ν/ . The solution

of the differential equation is

9533_C005.indd 4069533_C005.indd 406 10/29/08 2:31:10 PM10/29/08 2:31:10 PM


θ ξ ξ ξ( ) ( , ) ( )x q x f d= ∫0

1

(i)

Green’s functions can be used to convert differential eigenvalue problems to eigenvalue problems for integral equations. Consider again the transverse defl ections of the simply supported beam shown in Figure 5.20. Suppose that the beam is vibrating with a transverse defl ection w(x, t). The transverse load is due to the inertia of the beam, f(x) = ρA(∂2w/∂t2). Then, from Equation 5.202,

w x t Awt

t q x d

L

( , ) ( , ) ( , )=∂∂∫ ρ ξ ξ ξ

2

2

0

(5.205)

where the Green’s function q(x, ξ) is given in Equation 5.199. Substitution of a normal-mode solution w(x, t) = W(x)ei ω t into Equation 5.205 leads to

W x AW q x d

L

( ) ( ) ( , )= − ∫ω ρ ξ ξ ξ2

0

(5.206)

Noting that the kinetic-energy inner product for the beam is defi ned as( , ) ( ) ( )f g Af x g x dxM

L= ∫ 0 ρ , Equation 5.206 can be rewritten as

W x W q x M( ) ( ( ), ( , ))= −ω ξ ξ2 (5.207)

Dividing Equation 5.207 by ω2 leads to

GW W=1

2ω (5.208)

Equation 5.208 implies that the natural frequencies are the reciprocals of the square roots of the Green’s function operator and that the mode shapes W(x) are the corresponding eigenvectors. Since the Green’s function is sym-metric, the operator is self-adjoint with respect to the kinetic-energy inner product. This implies that the mode shapes corresponding to distinct natural frequencies are mutually orthogonal with respect to the kinetic-energy inner product.

The integral-equation formulation of the eigenvalue problem for the natural frequencies is not easier to solve than the differential equation formulation.

9533_C005.indd 4079533_C005.indd 407 10/29/08 2:31:11 PM10/29/08 2:31:11 PM


Problems 5.1. The differential equations governing the free vibrations of the sys-

tem shown in Figure P5.1 are

m

m

m

x

x

x

0 0

0 2 0

0 0

1

2

3

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥��

��

��

⎥⎥⎥⎥+

−− −

−

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

4 3 0

3 5 2

0 2 2

1

2

3

k k

k k k

k k

x

x

x

⎡⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥=

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

0

0

0

(a)

Recall that the natural frequencies of a discrete system are the square roots of the eigenvalues of M−1K and that the mode shapes are the corresponding eigenvectors (Example 5.2). (a) Determine the natu-ral frequencies for the system; (b) determine the normalized mode shapes for the system.

5.2. The initial conditions for the system of Figure P5.1 and Problem 5.1 are

x

x x

x x

1

2

3

0 0

0

( )

( )

( )

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥=

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥δ

and

�

�

�

x

x x

x x

1

2

3

0 0

0

0

( )

( )

( )

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥=

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥ (b)

Let ω1, ω2 and ω3 be the natural frequencies of the system of Problem 5.1, and let X1, X2, and X3 be their corresponding normalized mode shapes. Let x represent the solution of Equation a of Problem 5.1 subject to the initial conditions of Equation b. At any time, x can be expanded in terms of the normalized mode shapes:

x X X X1 2 3( ) ( ) ( ) ( )t t t t= + +α α α1 2 3 (c)

a. Substitute Equation c into Equation a of Problem 5.1 and take the standard inner product of both sides with respect to Xj for an arbi-trary j. Use the properties of mode-shape orthogonality to derive uncoupled differential equations for αi(t).

b. Substitute Equation c into the initial conditions of Equation c to derive initial conditions for αi(t).

c. Solve the differential equations to obtain αi(t). d. Determine x(t) using Equation c.

m 2m m

k 2k3k

x1 x2 x3

Figure P5.1 Three-degree-of-freedom system of Problems 5.1 and 5.2.

9533_C005.indd 4089533_C005.indd 408 10/29/08 2:31:12 PM10/29/08 2:31:12 PM


5.3. The differential equations governing the motion of the system shown in Figure P5.3 are

m

mx

x

k k

k k

0

0 2

1

2

⎡

⎣⎢⎢

⎤

⎦⎥⎥⎡

⎣⎢⎢

⎤

⎦⎥⎥+

−−⎡

⎣⎢⎢

⎤

⎦⎥

��

�� ⎥⎥⎡

⎣⎢⎢

⎤

⎦⎥⎥=

⎡

⎣⎢⎢

⎤

⎦⎥⎥

x

x1

2

0

0

Determine the natural frequencies and normalized mode shapes for the system.

5.4. The stress tensor at a point in a solid is

S = −

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟200 150 100

150 200 300

100 300 100

⎟⎟⎟⎟⎟⎟⎟kPa

Determine the principal stresses and unit vectors normal to the places on which they act.

5.5. Recall that the stress vector acting on a plane whose unit normal is n can be calculated as Sn. The normal stress acting on a plane, σ, is the component of the stress vector normal to the plane,

σ =( )Sn n, (a)

where the inner product is the standard inner product (the dot product) on R3. Prove that the maximum normal stress at a point is the largest principal stress and that the minimum normal stress is the smallest principal stress by following these steps. The expansion theorem implies that the unit normal can be written as a linear com-bination of the unit normals to the principal planes.

a. Assume a linear combination for the unit normal to a plane, n = α1n1 + α2n2 + α3n3, noting that since the vector is a unit vector, the sum of the squares of the coeffi cients in the linear combina-tion is one. Substitute the expansion into Equation a.

b. Use the properties of inner products and the orthonormality of the unit vectors normal to the principal planes to reduce the resulting equation.

c. The resulting expression is a function of the expansion coeffi -cients. The normal stress is stationary if

m 2m

k

x1 x2

Figure P5.3 Two-degree-of-freedom unrestrained system of Problem 5.3.

9533_C005.indd 4099533_C005.indd 409 10/29/08 2:31:13 PM10/29/08 2:31:13 PM


d d d dσσ

αα

σα

ασ

αα= =

∂∂

+∂

∂+

∂∂

01

12

23

3 (b)

Since the coeffi cients are independent, each of the partial deriva-tives in Equation b must be zero. Calculate the partial derivatives and set them to zero.

d. Determine the solutions of the equations obtained in part (c) and use them to determine the extreme values of the normal stress.

5.6. Using the notation of Problem 5.5, the shear stress vector acting on a plane whose unit normal is n is calculated as

τ σ= −Sn n (c)

The magnitude of the shear stress is

τ τ τ σ σ= = − −( )( , ) ,1

21

2Sn n Sn n (d)

Expand the unit normal in terms of the unit vectors to the principal planes.

a. Use the expansion in Equation d to prove that the maximum normal stress acts on a plane which makes an angle of π/4 to the plane of the largest normal stress.

b. Calculate the maximum shear stress and a unit vector normal to the plane on which it acts to obtain the state of stress described by the stress tensor of Equation a of Problem 5.4.

5.7. The stress vector defi ning the state of stress at a point is

S =− −

− −−

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

100 50 100

50 100 200

100 200 200

⎟⎟⎟⎟⎟⎟⎟⎟⎟kPa (a)

The principal stresses and unit vectors normal to the principal planes are

σ1 1350

0 4082

0 4082

0 8165

= =−⎡

⎣

⎢⎢⎢kPa n

.

.

.⎢⎢

⎤

⎦

⎥⎥⎥⎥

= =σ2 50

0 9129

0 1826

036

kPa n2

.

.

551

200

0

0 8944

04472

2 3

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

= − =−

kPaσ n .

⎡⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

(b)

9533_C005.indd 4109533_C005.indd 410 10/29/08 2:31:14 PM10/29/08 2:31:14 PM


A small change in loading leads to a stress vector of

S =+ − + −

− + − +−

100 10 50 2 100

50 2 100 200

100 200 20

ε εε ε

00

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟kPa (c)

where ε is a small dimensionless parameter. The principal stresses for the new state of stress and the normal vectors can be expanded as

σ σ εμi i i= + (d)

n n mi i i= + ε (e)

Determine the perturbations in principal stresses, μi, i = 1, 2, 3. 5.8. The differential equations governing the motion of the system shown

in Figure P5.8 are

m

m

m

m

x

x

0 0 0

0 2 0 0

0 0 0

0 0 0

1

2

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

��

��

��xx

x

k k

k k k

k k3

4

3 0 0

2 0

0 3

��

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

+

−− −

− −22

0 0 2 2

1

2

3

4

k

k k

x

x

x

x−

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦⎦

⎥⎥⎥⎥⎥⎥

=

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

0

0

0

0

(a)

The lowest natural frequency for the system is 0 4344. /k m , which has a corresponding normalized eigenvector of X = 1/m. [0.1397 0.4005 0.5541 0.5938].

a. Determine the perturbation in the lowest natural frequency if the particle of mass 2m is adjusted to 2m + εm, where ε is a small independent dimensionless parameter.

b. Determine the perturbation in the lowest natural frequency if the stiffness of the spring connecting the leftmost particle of mass m with the particle of mass 2m changes to a stiffness of k −εk, where ε is a small dimensionless parameter.

5.9. a. Determine all eigenvalues and eigenvectors for the matrix

A =−

− −−

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

3 1 0

1 2 1

0 1 2

m 2m m

k

x1 x2 x3 x42k2k k

m

Figure P5.8 Four-degree-of-freedom of Problem 5.8.

9533_C005.indd 4119533_C005.indd 411 10/29/08 2:31:15 PM10/29/08 2:31:15 PM


b. Normalize the eigenvectors and then demonstrate orthonor mality of eigenvectors.

5.10. a. Determine all eigenvalues and eigenvectors of the matrix

A =⎡

⎣⎢⎢

⎤

⎦⎥⎥

1 6

3 4.

b. Determine all eigenvalues and eigenvectors of AT. c. Demonstrate the principle of biorthogonality of the eigenvectors

of A and AT. 5.11. A valid inner product defi ned for R2 is (x, y) = x1 y1 + 2x2 y2. (a) Show

that the matrix A in Problem 5.10 is self-adjoint with respect to this inner product. (b) Demonstrate orthogonality of the eigenvectors of A. (c) Since A is self-adjoint with respect to a valid inner product, its eigenvalues are all real. Use this idea to deduce suffi cient conditions for ensuring that a 2 × 2 matrix has all real eigenvalues. Hint: Is there

any inner product for which 1 6

3 4−⎡

⎣⎢⎢

⎤

⎦⎥⎥ self-adjoint?

5.12. A valid inner product defi ned for R3 is (x, y) = x1 y1 + x2 y2 + 3x3 y3. (a) Give an example of a 3 × 3 matrix A which is self-adjoint with respect to this inner product. (b) Determine the eigenvalues and eigenvectors of this matrix and demonstrate orthogonality of the eigenvectors. (c) Determine the eigenvectors of AT. Note that AT = A* with respect to the standard inner product for R3. Demonstrate the biorthogonal-ity principle. (d) Using this problem as a guide, determine suffi cient conditions for a 3 × 3 matrix to have all real eigenvalues.

Problems 5.13–5.16. For the differential system in each problem, (a) specify an inner product for which the system is self-adjoint; (b) deter-mine all eigenvalues and eigenvectors for the system; (c) normalize the eigenvectors; and (d) demonstrate orthogonality of the eigenvectors.

5.13. d ydx

y2

20+ =λ

dydx

y y

( )

( ) ( )

0 0

1 2 1 0

=

+ = .

5.14. d ydx

dydx

y2

26 0+ + =λ

y

dydx

( )

( )

0 0

1 0

=

= .

9533_C005.indd 4129533_C005.indd 412 10/29/08 2:31:16 PM10/29/08 2:31:16 PM


5.15. xd ydx

xdydx

y22

24 0+ + =λ

y

dydx

( )

( )

1 0

2 0

=

= .

5.16. d ydx

y2

20+ =λ

dydx

dydx

( )

( )

0 0

1 0

=

=

5.17. xd ydx

xdydx

y32

224 0+ + =λ .

Problems 5.18–5.23. refer to the following problem: The differential equation for determination of the natural frequencies and mode shapes for the longitudinal motion of a nonuniform bar is

d

dxx

dydx

x yα ω α( ) ( )⎡

⎣⎢⎢

⎤

⎦⎥⎥+ =2 0 (a)

5.18. Find the fi rst fi ve natural frequencies and normalized mode shapes for a uniform bar, α(x) = 1, which is fi xed at both ends, y(0) = 0 and y(1) = 0.

5.19. Find the fi rst fi ve natural frequencies and normalized mode shapes for a uniform bar, α(x) = 1, which is fi xed at x = 0 and has a linear spring attached at x = 1 such that the boundary conditions are y(0) = 0 and dy/dx(1) + 0.25y(1) = 0.

5.20. Find the fi rst fi ve natural frequencies and normalized mode shapes for a uniform bar, α(x) = 1, which is fi xed at x = 0 and has a discrete particle and a linear spring attached at x = 1 such that the boundary conditions are y(0) = 0 and dy/dx(1) + 0.25y(1) = 0.5ω2y(1).

5.21. Find the fi rst fi ve natural frequencies and normalized mode shapes for a nonuniform bar with α(x) = 1 + 0.5x which is fi xed at both ends, y(0) = 0 and y(1) = 0.

5.22. Find the fi rst fi ve natural frequencies and normalized mode shapes for a nonuniform bar with α(x) = (1–0.5x)2 which is fi xed at x = 0 and free at x = 1 such that y(0) = 0 and dy/dx(1) = 0.

9533_C005.indd 4139533_C005.indd 413 10/29/08 2:31:17 PM10/29/08 2:31:17 PM


5.23. Find the fi rst fi ve natural frequencies and normalized mode shapes for a nonuniform bar with α(x) = (1–0.5x)2 which is fi xed at x = 0 and has a discrete particle and a linear spring attached at x = 1 such that the boundary conditions are y(0) = 0 and α(1)dy/dx(1) + 0.25y(1) = 0.5ω 2y(1).

Problems 5.24–5.29 refer to the following general problem: the natu-ral frequencies ω and mode shapes w(x) of a uniform stretched beam on an elastic foundation are governed by the differential equation

d wdx

d wdx

w w4

4

2

2

2 0− + − =ε η ω (a)

5.24. Determine the fi rst fi ve natural frequencies and mode shapes for a beam with ε = 0 and η = 1 which is pinned at x = 0 and x = 1 such that

y

d ydx

yd ydx

( ) , ( ) , ( ) , ( ) .0 0 0 0 1 0 1 02

2

2

2= = = =

5.25. Determine the fi rst fi ve natural frequencies and mode shapes for a beam with ε = 2 and η = 0 which is pinned at x = 0 and x = 1 such that

y

d ydx

yd ydx

( ) , ( ) , ( ) , ( ) .0 0 0 0 1 0 1 02

2

2

2= = = =

5.26. Determine the fi rst fi ve natural frequencies and mode shapes for a beam with ε = 2 and η = 1 which is pinned at x = 0 and fi xed at x = 1 such that

y

d ydx

ydydx

( ) , ( ) , ( ) , ( ) .0 0 0 0 1 0 1 02

2= = = =

5.27. Determine the fi rst fi ve natural frequencies and mode shapes for a beam with ε = 0 and η = 0 which is fi xed at x = 0 and has a linear spring attached at x = 1 such that

ydydx

d ydx

d ydx

y( ) , ( ) , ( ) , ( ) .0 0 0 0 1 0 1 0 52

2

3

3= = = − (( ) .1 0=

5.28. Determine the fi rst fi ve natural frequencies and mode shapes for a beam with ε = 0.5 and η = 1 which is fi xed at x = 0 and free at x = 1 such that

y

dydx

d ydx

d ydx

d( ) , ( ) , ( ) , ( ) .0 0 0 0 1 0 1 0 5

2

2

3

3= = = +

yydx

( ) .1 0=

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5.29. Determine the fi rst fi ve natural frequencies and mode shapes for a beam with ε = 0 and η = 1 which is free at x = 0 and has an attached mass at x = 1 such that

y

dydx

d ydx

d ydx

( ) , ( ) , ( ) , ( ) .0 0 0 0 1 0 1 0 52

2

3

3= = = = ω22 1 0y( ) .=

5.30. One end of the nonuniform bar shown in Figure P5.30 is fi xed, while its other end is attached to a spring whose other end is attached to a particle. The differential equations governing the displacement of the bar, u(x,t), and the displacement of the particle, y(t), are

∂∂

∂∂

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+ =

xEA

ux

Au xρ ( ) 0 (a)

u t( , )0 0= (b)

EA Lux

L t k y t u L t( ) ( , ) ( ) ( , )∂∂

− −[ ]= 0 (c)

md ydt

k y t u L t2

20+ −[ ]=( ) ( , ) (d)

a. Nondimensionalize the problem through introduction of appro-priate nondimensional variables.

b. Introduce the normal-mode solution,

u x t U x e i t( , ) ( )= ω (e)

y t Ye i t( ) = ω (f)

into the nondimensional equation and determine an eigenvalue-eigenvector problem whose solution leads to the natural frequencies and mode shapes.

L

ρ,E

y(t)

u(x, t)k

m

Figure P5.30 System for Problem 5.30.

9533_C005.indd 4159533_C005.indd 415 10/29/08 2:31:19 PM10/29/08 2:31:19 PM


c. Defi ne the vector space V as the set of vectors of the form gf x

a=

⎡

⎣⎢⎢

⎤

⎦⎥⎥

( ),

where f(x)

is in the subspace of C2[0,1] of all vectors f(x) such that

f(0) = 0 and du/dx(1) + ηu(1) = 0, where η is an appropriate nondi-mensional parameter. Defi ne an operator L whose domain is V by

Lf x

ax

ddx

dfdx

f

( )( )

( )

⎡

⎣⎢⎢

⎤

⎦⎥⎥=

− ( )−

⎡

⎣

⎢⎢

1

1

αα

μα ν⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥ (g)

where μ and ν are appropriate nondimensional constants. Show that the eigenvalue problem is of the form

Lg = ω2g (h)

d. Determine an appropriate inner product for which L is self-adjoint. e. Show that L is positive defi nite with respect to this inner product. f. Determine the fi rst fi ve natural frequencies and mode shapes for a

uniform bar, assuming that all nondimensional constants are one. g. Repeat part (f) if α(x) = (1–0.1x)3.

5.31. Consider the boundary-value problem

d ydx

n y x2

2

2 2+ = (a)

dydx

( )0 0= (b)

dydx

( )1 0= (c)

where n is an integer.

a. Use a solvability condition to determine for what values of n a solu-tion of Equation a subject to Equation b and Equation c exists.

b. Solve Equation a subject to Equation b and Equation c to confi rm the results of part (a).

c. Determine y(x) when a solution exists. 5.32. Show the algebraic steps necessary to obtain Equation 5.56 from

Equation 5.55. 5.33. Use the modal-analysis method derived in Example 5.9 to determine

the response of a two-degree-of-freedom mechanical system whose motion is described by the differential equation

m

m

x

x

k k

k k

0

0 2

21

2

⎡

⎣⎢⎢

⎤

⎦⎥⎥⎡

⎣⎢⎢

⎤

⎦⎥⎥ +

−−

⎡

⎣⎢⎢

⎤

⎦

��

��⎥⎥⎥⎡

⎣⎢⎢

⎤

⎦⎥⎥ =

⎡

⎣⎢⎢

⎤

⎦⎥⎥

x

x F t1

2 0

0

sin( )ω

and with all initial conditions equal to zero.

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5.34. Determine the constraints on the constants a1, b1, a2, b2,α1, β1, γ1, α2, β2 and γ2 as defi ned in Equation 5.71 such that the differential operator of Equation 5.66 is positive defi nite with respect to the inner product defi ned in Equation 5.67.

Problems 5.35–5.44 refer to the problem of fi nding the eigenvalues and eigenvectors of the system

−⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟=

ddx

xdydx

xy3 λ (a)

y( )0 0= (b)

y y( ) ( )1 2 1 0+ ′ = (c)

5.35. Using the exact solution, determine the fi ve lowest eigenvalues and their corresponding eigenvectors.

5.36. Choose two polynomials of order three or less which satisfy the boundary conditions. Use Rayleigh’s quotient to approximate the lowest eigenvalue.

5.37. Choose the elements of a basis for the subspace of C2[0,1] defi ned by the intersection of P4[0,1] with the set of all functions satisfying boundary conditions (b) and (c). Use these functions as a basis for a Rayleigh-Ritz approximation of the lowest eigenvalues of the system.

5.38. Choose the elements of a basis for the subspace of C2[0,1] defi ned by the intersection of P4[0,1] with the set of all functions satisfying boundary condition (b) (the geometric boundary condition). Use these functions as a basis for a Rayleigh-Ritz approximation of the lowest eigenvalues of the system.

5.39. Use a fi nite-element method to approximate the lowest eigenvalues and eigenvectors for the system. Use fi ve equally spaced elements.

Problems 5.40–5.57 refer to the system of Problems 5.35–5.44, but with boundary condition (c) replaced by

y y y( ) ( ) . ( )1 2 1 0 5 1+ ′ = λ (d)

5.40. Develop the appropriate form of the inner product for orthogonality of the eigenvectors.

5.41. Use the exact solution to determine the fi ve lowest eigenvectors. 5.42. Choose two polynomials of order three or less which satisfy the

boundary conditions. Use Rayleigh’s quotient to approximate the lowest eigenvalue.

5.43. Choose the elements of a basis for the subspace of C2[0,1] defi ned by the intersection of P 4[0,1] with the set of all functions satisfying

9533_C005.indd 4179533_C005.indd 417 10/29/08 2:31:21 PM10/29/08 2:31:21 PM


boundary condition (b) (the geometric boundary condition). Use these functions as a basis for a Rayleigh-Ritz approximation of the lowest eigenvalues of the system.

5.44. Use the fi nite-element method to approximate the natural frequencies and mode shapes. Use fi ve equally spaced elements.

5.45. Consider the eigenvalue problem,

( )1 2 301

22

2 2− − + =

−x

d ydx

xdydx

yx

yλ

which is to be solved on the interval −1 ≤ x ≤ 1. (a) Determine the self-adjoint form of the differential operator. (b) What boundary con-ditions must be specifi ed at x = −1 and x = 1? (c) Determine the inner product for eigenvector orthogonality. (d) Determine the eigenval-ues and eigenvectors for this problem (Hint: 30 = (5)(6)).

5.46. Prove that, if subject to appropriate boundary conditions, a sixth-order differential operator of the form

Lyd

dxf x

d y

dxd

dxg x

d y=

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟+

3

3

3

3

2

2

2

( ) ( )ddx

ddx

h xdydx

q x2

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟+

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟+( ) ( ))y

is self-adjoint with respect to the standard inner product on C6[0,1]. 5.47. Under what conditions on f(x), g(x), h(x), and q(x) is the sixth-order

operator of Problem 5.46 positive defi nite with respect to the stan-dard inner product on C6[0,1]?

5.48. Determine the exact value of the critical buckling length for the col-umn of Example 5.22.

5.49. Use the Rayleigh-Ritz method to approximate the critical buckling length of a fi xed-free column. Use a basis of polynomials of order six or less, which satisfy the boundary conditions.

5.50. Use the Rayleigh-Ritz method to approximate the critical rotational speed of a fi xed-free column. Use a basis of polynomials of order six or less, which satisfy the boundary conditions.

5.51. Use the fi nite-element method to approximate the critical buckling length of a fi xed-free column. Use fi ve equally spaced elements.

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419

Chapter 6

Partial differential equations

6.1 Homogeneous partial differential equationsA partial differential equation is a differential equation with more than one independent variable. Although the independent variables can represent a variety of coordinates and scales, this study focuses on partial differential equations in which the independent variables are spatial coordinates and time.

Consider a problem in which it is desired to determine the variation of a dependent variable, say φ, in a region of space R which is bounded by a sur-face S. Let r be a position vector from the origin of the coordinate system to a particle or point in R. Most generally, r is a function of three independent spatial coordinates. The dependent variable may also be a function of time and designated as φ(r,t). The surface S is described by g(r,t) = 0. The problem is an interior problem if g is bounded and the solution is to be obtained for position vectors defi ned within the interior of g, as illustrated in Figure 6.1a. The problem is an exterior problem if the solution is to be obtained for posi-tion vectors defi ned outside of g, as illustrated in Figure 6.1b.

The general form of a linear partial differential equation is

L M G rφ φ φ+ + = f t( , ) (6.1)

where L is a linear operator involving derivatives with respect to spatial vari-ables, M is a linear operator involving derivatives with respect to t, and G is a linear operator involving mixed partial derivatives involving time and spatial variables.

Determination of a solution of Equation 6.1 requires application of appro-priate initial conditions and boundary conditions. The number of initial conditions required depends on the order of M. If M is fi rst-order, then it is necessary to specify φ(r,0). If M is second-order it, is also necessary to specify ∂φ/∂t(r,0). The number and type of boundary conditions depend on the order and form of L. The boundary conditions are prescribed on S. For interior problems, it is often necessary to specify that the depen-dent variable is bounded and single-valued at all points within g(r,t). For exterior problems, it is also required that the dependent variable be single- valued, but conditions as ||r||→ ∞ must be specifi ed. Such condition, called radiation conditions, place restrictions on φ far away from g(r,t). In some

9533_C006.indd 4199533_C006.indd 419 10/29/08 1:57:21 PM10/29/08 1:57:21 PM


cases, the radiation condition is simply that φ approaches zero as ||r||→ ∞. Solutions of the wave equation in exterior regions where refl ection cannot occur must satisfy a condition that all waves are propagating away from the body.

Second-order linear partial differential equations are classifi ed based on their form. The classifi cation of a differential equation with variable coeffi -cients may change with time or spatial position. The equations considered in this study maintain their classifi cation for all r and t. Homogeneous second-order problems considered in this study are of the form of Equation 6.1 with G = 0 and Lφ = − ∇2φ.

(a)

r

g(r,t)

(b)r

g(r,t)

Figure 6.1 (a) An interior problem is one in which the solution is determined for every position vector r everywhere within the volume bounded by g(r,t). (b) An exte-rior problem is one in which the solution is determined for position vectors r outside the closed surface defi ned by g(r,t).

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Chapter 6: Partial differential equations 421

The partial differential equation is elliptic if M = 0 and φ = φ(r) and is independent of t. The resulting equation,

∇ =2 0φ (6.2)

is called Laplace’s equation. Laplace’s equation governs problems such as steady-state heat transfer and potential fl ow of incompressible and inviscid fl uids.

The partial differential equation is parabolic if Mφ = p(r)∂φ/∂t. The resulting equation,

−∇ =∂∂

2φφ

pt

( )r (6.3)

is called the diffusion equation. The diffusion equation governs unsteady state conductive and convective heat and mass transfer problems.

The partial differential equation is hyperbolic if Mφ = p(r)∂2φ/∂t2. The resulting equation,

∇ =∂∂

22

2φ p

t( )r

φ (6.4)

is called the wave equation and governs propagation of waves in solids and fl uids.

A unsteady state problem is homogeneous if f(r,t) = 0 and all boundary con-ditions are homogeneous. A steady-state problem is homogeneous if f(r) = 0 and if nonhomogeneous boundary conditions occur for only one independent variable. A method called separation of variables is applied in subsequent sec-tions to solve homogeneous partial differential equations in bounded domains. Solutions of nonhomogeneous partial differential equations are approached using superposition of solutions and eigenvector expansion methods. Solu-tions are considered in Cartesian, cylindrical, and spherical coordinates.

Solutions in unbounded domains are usually more diffi cult to obtain than in interior regions. A problem in an exterior region is considered at the end of this chapter.

6.2 Second-order steady-state problems, Laplace’s equation

Consider a partial differential equation derived from a mathematical model of a steady-state system; the response of the system is independent of time. The dependent variables are spatial coordinates defi ned by the position

•

•

•

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vector r, measured from the origin of a coordinate system. The dependent variable is φ(r). Assume that the partial differential equation governing φ(r) is of the form

Lφ = 0 (6.5)

where L is a linear operator. It is desired to determine the solution of Equa-tion 6.5 in a region in space R which is bounded by a closed surface described by g(r) = 0.

Determination of a solution of Equation 6.5 requires application of appro-priate boundary conditions. Without loss of generality, assume r = xi + yj + zk such that the solution of Equation 6.5 is expressed using Cartesian coordi-nates. The boundary of the region is described by 0 ≤ x ≤ a, 0 ≤ y ≤ b and 0 ≤ z ≤ c, as illustrated in Figure 6.2. It is desired to determine φ(x,y,z) every-where in the parallelepiped. Nonhomogeneous boundary conditions are pre-scribed on boundary planes in one variable, say z. The number of boundary conditions required depends on the order of the operator L. Nonhomoge-neous boundary conditions applied to other surfaces lead to a nonhomoge-neous problem which is solved using a superposition method as described in Section 6.4. Solution of Equation 6.5 when L is a second-order differential operator requires one boundary condition defi ned on each plane that is part of the boundary of R. The boundary conditions on each plane are a linear

z

x

y

a

b

c

Figure 6.2 Laplace’s equation is solved inside the cubic region.

9533_C006.indd 4229533_C006.indd 422 10/29/08 1:57:23 PM10/29/08 1:57:23 PM


combination of φ and ∂φ/∂n, where n is the unit normal to the boundary. In general,

∂∂

= ∇( )⋅

=∇( )⋅ ∇( )

∇

φn

gg

φ

φ

n

(6.6)

The nonhomogeneous boundary conditions for the problem under discus-sion are of the form

φ αφ

( , , ) , , ( , )x yz

x y f x yz0 0−∂∂

( )= (6.7)

φ βφ

( , , ) , , ( , )x y cz

x y c g x yz+∂∂

( )= (6.8)

Homogeneous boundary conditions are of the form

φ α∂φ

( , , ) , ,0 0 0y zx

y zx−∂

( )= (6.9)

φ βφ

( , , ) , ,a y zx

a y zx+∂∂

( )= 0 (6.10)

φ αφ

( , , ) , ,x zy

x zy0 0 0−∂∂

( )= (6.11)

φ βφ

( , , ) , ,x b zy

x b zy+∂∂

( )= 0 (6.12)

The method of separation of variables assumes the solution can be expressed as a product of the form

φ Φ( , , ) ( , ) ( )x y z x y Z z= (6.13)

The operator L is said to be separable if

L LΦ Φ Φ( , ) ( ) ( ) ( ) ( , ) ( , ) (x y Z z Z z r z x y x y q x[ ]= [ ]+xy ,, ) ( )y Z zLz [ ] (6.14)

where Lxy is a partial differential operator whose domain is defi ned by the boundary conditions specifi ed below, and Lz is an ordinary differential operator defi ned for 0 ≤ z ≤ c. If L is separable, then substitution of Equation 6.13 into Equation 6.5 leads to

9533_C006.indd 4239533_C006.indd 423 10/29/08 1:57:24 PM10/29/08 1:57:24 PM


1 1

ΦΦ

( , ) ( , )( , )

( ) ( )( )

x y r x yx y

Z z q zZ zL Lxy z[ ]= − [ ] (6.15)

The left-hand side of Equation 6.15 is a function of x and y, but not of z. The right-hand side of Equation 6.15 is a function of z only. However, x, y, and z are independent variables. Both sides of Equation 6.15 can be equal for all values of x, y, and z if and only if each side is equal to the same constant; call it λ. Thus Equation 6.15 separates to

Lxy Φ λ Φ( , ) ( , ) ( , )x y r x y x y[ ]= (6.16)

Lz Z z q z Z z( ) ( ) ( )[ ]= −λ (6.17)

The argument used to separate Equation 6.15 into Equation 6.16 and Equa-tion 6.17 is called the separation argument.

Substitution of Equation 6.13 into the boundary conditions, Equation 6.9, gives

Φ αΦ

Φ αΦ

( , ) ( ) , ( )

( , ) ,

0 0 0

0 0

y Z zx

y Z z

yx

x

x

−∂∂

( ) =

−∂∂

yy Z z( )⎡⎣⎢⎢

⎤⎦⎥⎥

=( ) 0

(6.18)

Equation 6.18 must be satisfi ed for all z; thus

Φ αΦ

( , ) ,0 0 0yx

yx−∂∂

( )= (6.19)

Substitution of Equation 6.13 into the remaining boundary conditions, Equa-tion 6.10, Equation 6.11, and Equation 6.12, leads to

Φ βΦ

( , ) ,a yx

a yx+∂∂

( )= 0 (6.20)

Φ αΦ

( , ) ,xy

xy0 0 0−∂∂

( )= (6.21)

Φ βΦ

( , ) ,x by

x by+∂∂

( )= 0 (6.22)

Equation 6.16, Equation 6.19, Equation 6.20, Equation 6.21, and Equation 6.22 constitute a boundary-value problem to solve for Φ(x,y). The system is homo-geneous, and therefore a nontrivial solution exists only for certain values of λ. Thus Equation 6.16 and Equation 6.19, Equation 6.20, Equation 6.21, and Equa-tion 6.22 constitute an eigenvalue problem.

9533_C006.indd 4249533_C006.indd 424 10/29/08 1:57:25 PM10/29/08 1:57:25 PM


Consider fi rst the case in which Lxy is dependent on only one independent variable, say x, in which case Lx is an ordinary differential operator. If it is second-order, it can be written in a self-adjoint form. Let Sx be the domain of Lx specifi ed by the boundary conditions, Equation 6.19 and Equation 6.20. Then Lx is self-adjoint with respect to the inner product ( , ) ( ) ( ) ( )f g f x g x r x dxr

a= ∫0

on Sx, where r(x) is determined by writing the operator in its self-adjoint form. The system has an infi nite, but countable, number of eigenvalues, λ1 ≤ λ2 ≤ λ3 ≤ … ≤ λk–1 ≤ λk ≤ λk+1 ≤ …, with an eigenvector corresponding to λk of Φk (x) for each k. The eigenvectors satisfy an orthogonality condition, (Φi (x), Φj (x))r = 0 for i ≠ j. The eigenvectors can be normalized by requiring that (Φi (x), Φi (x))r = 1.

Let Zk,1(z) and Zk,2 be two linearly independent solutions of Equation 6.16, which is assumed to be of second order, corresponding to λ = λk. Defi ne

φ Φk k k k k kx z A Z z B Z z x( , ) ( ( ) ( )) ( ), ,= +1 2 (6.23)

Equation 6.23 is a solution of Equation 6.5 subject to the boundary condi-tions of Equation 6.9 and Equation 6.10. However, φk(x,y) does not satisfy the boundary conditions of Equation 6.7 and Equation 6.8. The most general solution of a linear homogenous system is a linear combination of all pos-sible solutions. Thus the general solution of Equation 6.5, Equation 6.9, and Equation 6.10 is

φ φ( , ) ( ( ) ( )) ( ), ,x z A Z z B Z z xk k k k k

k

= +=

∞

∑ 1 2

1

(6.24)

Equation 6.24 can also be viewed as an expansion for φ(x,z) in terms of the normalized eigenvectors of Equation 6.16, Equation 6.19, and Equation 6.20. Such an expansion is appropriate because for any z, 0 ≤ z ≤ c, φ(x,z) is in Sx, and its eigenvector expansion is of the form φ( , ) ( ) ( )x z q z xk k k= ∑ =

∞1 φ . How-

ever, Equation 6.16 suggests that qk (z) = Ak Zk,1(z) + Bk Zk,2(z).If the mathematical modeling of the problem is accurate, then f(x) and g(x)

must satisfy boundary conditions 6.9 and 6.10. Thus f(x) and g(x) should be in Sx, in which case they each have an eigenvector expansion which converges pointwise to the functions. Thus

f x u xk k

k

( ) ( )==

∞

∑ Φ1

(6.25)

g x v xk k

k

( ) ( )==

∞

∑ Φ1

(6.26)

where

u f x xk k r=( )( ), ( )Φ (6.27)

9533_C006.indd 4259533_C006.indd 425 10/29/08 1:57:26 PM10/29/08 1:57:26 PM


v f x xk k r=( )( ), ( )Φ (6.28)

Substitution of Equation 6.24 into Equation 6.5, using the eigenvector expan-sion of Equation 6.25, leads to

A Z B Z x A Zk k k k k

k

z k k, , ,( ) ( ) ( ) (1 2

1

10 0 0+( ) − ′=

∞

∑ Φ α )) ( ) ( )

( )

,

,

+ ′( )

=

=

∞

=

∞

∑

∑

B Z x

u x

A Z

k k k

k

k k

k

k k

2

1

1

0 Φ

Φ

11 1 2 20 0 0 0( ) ( ) ( ) ( ), , ,− ′[ ]+ − ′[ ]−α αz k k k z kZ B Z Z ukk k

k

x{ } ==

∞

∑ Φ ( )1

0

(6.29)

The eigenvectors are linearly independent, and therefore each coeffi cient in the linear combination of Equation 6.29 is identically zero, leading to

A Z Z B Z Zk k z k k k z k, , , ,( ) ( ) ( ) ( )1 1 2 20 0 0 0− ′[ ]+ − ′α α[[ ]= uk (6.30)

Similar substitution of Equation 6.24 into Equation 6.5 using Equation 6.28 leads to

A Z c Z c B Z c Z ck k z k k k z k, , , ,( ) ( ) ( ) ( )1 1 2 2+ ′[ ]+ + ′β β[[ ]= vk (6.31)

Equation 6.30 and Equation 6.31 can be solved simultaneously for Ak and Bk.The procedure described above is called separation of variables. The

method is directly applicable when the partial differential operator is sepa-rable and when nonhomogeneous boundary conditions exist only on a por-tion of the boundary described by coordinate curves of one variable.

The following problem illustrates the application of the method of sepa-ration of variables to a two-dimensional problem governed by Laplace’s equation.

Example 6.1 The thin rectangular slab, illustrated in Figure 6.3 is con-strained to have a constant temperature T1 on three sides and a different constant temperature T2 on its fourth side. Determine the steady-state tem-perature distribution in the slab.

Solution The nondimensional partial differential equation governing the temperature distribution in the slab is

∂∂

+∂∂

=2

2

2

20

Θ Θx y

(a)

where

xxa

* = (b)

9533_C006.indd 4269533_C006.indd 426 10/29/08 1:57:27 PM10/29/08 1:57:27 PM


y* =ya

(c)

Θ( , )x yT TT T

=−−

1

2 1

(d)

Note that *’s have been dropped from nondimensional variables in Equation a. The nondimensional boundary conditions are

Θ( , )0 0y = (e)

Θ( , )1 0y = (f)

Θ( ,0) 0x = (g)

Θ xba

,( )= 1 (h)

A product solution of Equation a is assumed to be of the form

Θ( , ) ( ) ( )x y X x Y y= (i)

Substitution of Equation i into Equation a, using the separation argument, leads to

− = =1 12

2

2

2X xd Xdx Y y

d Ydy( ) ( )

λ (j)

Note that the boundary conditions for constant values of x are homoge-neous, while the boundary condition at y = b/a is nonhomogeneous. Thus

b

y

xT

T

a

1 2T

T1

1

Figure 6.3 The temperature of three sides of the thin rectangular slab of Example 6.1 is maintained at T1 while the constant temperature on the fourth side is T2. A steady-state temperature distribution is assumed.

9533_C006.indd 4279533_C006.indd 427 10/29/08 1:57:28 PM10/29/08 1:57:28 PM


the Sturm-Liouville problem is generated in the x-direction. That is, the operator Lx is chosen to be self-adjoint and positive defi nite on an appropri-ate Sx. Such is the case if LxX = − d2X/dx2. Equation j generates two ordinary differential equations:

d Xdx

X2

20+ =λ (k)

d Ydy

Y2

20− =λ (l)

Substituting Equation i into the boundary conditions, Equation e and Equa-tion f, leads to

X( )0 0= (m)

X( )1 0= (n)

The eigenvalues and normalized eigenvectors obtained from solution of the eigenvalue problem of Equation k, Equation m, and Equation n are λk = (kπ )2 and Xk(x) = 2sin(kπx). The eigenvectors are orthogonal with respect to, and are normalized with respect to, the standard inner product on C[0,1].

The general solution of Equation l corresponding to λk = (kπ)2 is

Y x A k y B k yk k k( ) cosh sinh= ( )+ ( )π π (o)

Thus the general solution to Equation a which satisfi es boundary conditions Equation e and Equation f is

Θ( , ) cosh sinh sinx y A k y B k y k xk k

k

= ( )+ ( )[ ] ( )=

π π π211

∞

∑ (p)

Application of Equation g to Equation p results in

Θ π( , ) sinx A k xk

k

0 0 21

= = ( )=

∞

∑ (q)

Equation q is satisfi ed if and only if Ak = 0, which leads to

Θ π π( , ) sinh sinx y B k y k xk

k

= ( ) ( )=

∞

∑ 21

(r)

9533_C006.indd 4289533_C006.indd 428 10/29/08 1:57:29 PM10/29/08 1:57:29 PM


The remaining boundary condition, Equation h, is applied by expanding g(x) = 1 in an eigenvector expansion. However, g(x) = 1 is not in Sx, which would require that g(0) = 0 and g(1) = 0. A convergent eigenvector expansion is available, but it does not converge to g(x) at x = 0 and x = 1. This situation is not a result of the mathematics, but rather of the physics of the model-ing. The boundary conditions specify two values for Θ at the points (0, b/a) and (1, b/a). The boundary conditions at x = 0 and x = 1 specify Θ = 0, while the boundary condition at y = b/a requires Θ = 1. The modeling is inaccurate, and therefore convergence to g(x) is not anticipated at these points. Indeed, the right-hand side of Equation p converges to one value only when evalu-ated at these points.

The constant temperature on each boundary is an assumption made to simplify the problem. In actuality, conduction heat transfer causes the tem-perature on each side near the corners to vary for some distance away from the corners, as illustrated in Figure 6.4. The actual variation of temperature over the boundary is diffi cult to determine; it depends on factors not known from the problem statement, such as how the temperatures are maintained. If the actual temperatures are used as boundary conditions, then the bound-ary conditions on three faces are nonhomogeneous and the method of sepa-ration of variables cannot be directly applied.

Since the boundary conditions violate the laws of physics, the use of an eigenvector expansion for g(x) which does not converge to g(x) at x = 0 and x = 1 provides a reasonable approximation to the actual problem to be solved. To this end,

g x v k xk

k

( ) sin= = ( )=

∞

∑1 21

π (s)

T1

T1 T1

y

x

T2

Figure 6.4 The mathematical model used in Example 6.1 violates the laws of physics because it suggests that the temperature is multivalued at the corners of the region.

9533_C006.indd 4299533_C006.indd 429 10/29/08 1:57:30 PM10/29/08 1:57:30 PM


where

v k x dx

k

k

k

= ( )

= − −( )⎡⎣⎢

⎤⎦⎥

=

∫ ( ) sin1 2

21 1

0

0

1

π

π

k

kk

=

=

⎧

⎨⎪⎪⎪⎪

⎩⎪⎪⎪⎪

2 4 6

2 21 3 5

, , ,...

, , ...π

(t)

Application of Equation h to Equation r using Equation t leads to

v k x B kba

k xk

k

k

k

2 21 1

sin sinh sinπ π π( )= ( ) ( )=

∞

=

∞

∑ ∑ (u)

Equating the coeffi cients of sin(kπx) from both sides of Equation u leads to

Bv

kba

kk=

( )sinh π (v)

Substitution of Equation v into Equation r gives

Θ( , )sinh

sinhsin

, ,

x yk y

k kba

k xk

=( )

( )( )

=

4

1 3π

π

ππ

55

∞

∑ (w)

Now consider a three-dimensional steady-state problem. Let Sxy be the domain of Lxy specifi ed by the boundary conditions of Equation 6.20, Equa-tion 6.21, and Equation 6.22. If Lxy is self-adjoint with respect to an inner product, (F(x,y), G(x,y))xy, defi ned on Sxy, then the system has an infi nite, but countable, number of eigenvalues, λ1 ≤ λ2 ≤ λ3 ≤ … ≤ λk – 1 ≤ λk ≤ λk + 1 ≤ …, with an eigenvector corresponding to λk of Φk(x,y) for each k. The eigenvectors sat-isfy an orthogonality condition (Φi(x,y), Φj(x,y))xy = 0 for i ≠ j.

The eigenvalue problem to determine Φk(x,y) is a partial differential equa-tion, Equation 6.16, subject to the boundary conditions of Equation 6.19, Equa-tion 6.20, Equation 6.21, and Equation 6.22. A product solution for Φk(x,y) is assumed to be of the form

Φ( , ) ( ) ( )x y X x Y y= (6.32)

9533_C006.indd 4309533_C006.indd 430 10/29/08 1:57:31 PM10/29/08 1:57:31 PM


The operator Lxy is separable if L L Lxy x yu x v y g y v y u x f x u x v y( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]= + for all u(x) and v(x) in Sxy. If Lxy is separable, substitution of Equation 6.32 into Equation 6.16 leads to

1 1

X x f xX x

Y x g yY y

( )( )

( ) ( )( )

( )[ ]− = [ ]L Lx yλ (6.33)

Use of the separation argument implies that both sides of Equation 6.33 are equal to the same constant; call it μ. Thus, Equation 6.33 separates into

LxX x X x( ) ( )− =λ μ (6.34)

LyY x Y y( ) ( )= μ (6.35)

Substitution of Equation 6.32 into Equation 6.19, Equation 6.20, Equation 6.21, and Equation 6.22 leads to

X x( )0 0 0− ( )=αdXdx

(6.36)

X adXdx

a( )+ ( )=βx 0 (6.37)

YdYdy

( )0 0 0− ( )=αy (6.38)

Y bddy

b( )+ ( )=βΦ

y 0 (6.39)

Equation 6.35, Equation 6.38, and Equation 6.39 constitute an eigenvalue problem for Y(y) on a domain Sy. Its solution leads to an infi nite number of eigenvalues μ� for � = 1,2,… such that μ1 ≤ μ2 ≤ … ≤ μ� – 1 ≤ μ� ≤μ� + 1 ≤ … Let Y�(y) be the eigenvector corresponding to μ� normalized with respect to an inner product ( f(y),g(y))y defi ned on Sy.

For each �, Equation 6.34, Equation 6.36, and Equation 6.37 constitute an eigenvalue problem. There are an infi nite, but countable, number of eigenvalues, ��,m m = 1,2,…, for each problem, such that λ�,1 ≤ λ�,2 ≤ … ≤ λ�,m−1 ≤ λ�,m ≤ λ�,m+1 ≤ … Let X �, m be the eigenvector corresponding to λ�, m which has been normalized with respect to an inner product (p(x),s(x))x defi ned on Sx.

The eigenvalues of Lx are also the eigenvalues of Lxy. Thus Lxy has a doubly infi nite number of eigenvalues. The eigenvectors of Lxy are

Φ� � �, ,( , ) ( ) ( )m mx y X x Y y= (6.40)

9533_C006.indd 4319533_C006.indd 431 10/29/08 1:57:32 PM10/29/08 1:57:32 PM


Equation 6.17 can be written for each λ� , m � = 1,2,… and m = 1,2,…:

LzZ z Z zm m m� � �, , ,( ) ( )= −λ (6.41)

Equation 6.41 is a linear second-order ordinary differential equation whose general solution is written as

Z z A Z z B Z zm m m m m� � � � �, , , , , , ,( ) ( ) ( )= +1 2 (6.42)

such that a solution of Equation 6.5 subject to Equation 6.7, Equation 6.8, Equation 6.9, and Equation 6.10 is

φ� � � � � �, , , , , , ,( , , ) ( ) ( )m m m m mx y z A Z z B Z z X= +[ ]1 2 ,, ( ) ( )m x Y y� (6.43)

The most general solution of a linear homogeneous problem is a linear com-bination of all possible solutions,

φ( , , ) ( ) ( ) (, , , , , , ,x y z A Z z B Z z Xm m m m m= +[ ]� � � � �1 2 xx Y ym

) ( )�

� =

∞

=

∞

∑∑11

(6.44)

Recall that Lxy is self-adjoint with respect to an inner product on Sxy. Thus any element in Sxy has an expansion in terms of eigenvectors of Lxy. Assume that f(x,y) is in Sxy. Its eigenvector expansion is

f x y u X x Y ym m

m

( , ) ( ) ( ), ,==

∞

=

∞

∑∑ � � �

� 11

(6.45)

where the expansion coeffi cients are

u f x y X x Y yk m xy� � �, ,( , ), ( ) ( )=( ) (6.46)

The function g(x,y) in Equation 6.8 has an eigenvector expansion of the form

g x y v X x Y ym m

m

( , ) ( ) ( ), ,==

∞

=

∞

∑∑ � � �

� 11

(6.47)

Equation 6.44 can be substituted into the boundary condition, Equation 6.7, with f(x) replaced by Equation 6.45, leading to

( ) ( ) ( ) (, , , , , , ,A Z B Z X x Ym m m m m� � � � � �1 20 0+[ ] yy

A Z B Z

m

z m m m m

)

( ), , , , ,

=

∞

=

∞

∑∑

− ′ + ′

11

1 0

�

� � � �α ,, ,

, ,

( ) ( ) ( )

( )

2

11

0[ ]

=

=

∞

=

∞

∑∑ X x Y y

u X x Y

m

m

m m

� �

�

� � ��

�

( )ym=

∞

=

∞

∑∑11

(6.48)

9533_C006.indd 4329533_C006.indd 432 10/29/08 1:57:33 PM10/29/08 1:57:33 PM


Combining the summations in Equation 6.48, and noting the linear indepen-dence of the eigenvectors of Lxy, leads to

Z Z A Z Zm z m m m z� � � �, , , , , , ,( ) ( ) ( )1 1 20 0 0− ′[ ] + − ′α α �� , , , ,( )m m mB u2 0[ ] = (6.49)

A similar procedure used with Equation 6.8 gives

Z c Z c A Z c Zm z m m m z� � � �, , , , , , ,( ) ( ) ( )1 1 2+ ′[ ] + + ′β β �� , , , ,( )m m mc B v2[ ] = (6.50)

Equation 6.49 and Equation 6.50 can then be solved simultaneously for A� , m and B�,m.

Example 6.2 The solid rectangular block shown in Figure 6.5 is in a medium of temperature T∞. The heat transfer coeffi cient between the block and the medium is h. Defi ning

xxa

ya

zza

T TT T

* * *= = = =−−

∞

∞y and Θ

1

(a)

the problem governing the steady nondimensional temperature distribution in the block is

∂∂

+∂∂

+∂∂

=2

2

2

2

2

20

Θ Θ Θx y z

(b)

a

bc

y

z

x

Θ=1

–Bi Θ=0

Θh y

–Bi Θ=0+Bi Θ=0

+Bi Θ=0

Θx

Θx

Θz

Θ–BiΘ=0 y

Figure 6.5 One face of the cube of Example 6.2 is maintained at a constant tempera-ture while the remainder of the surface is subject to heat transfer by convection with the ambient.

9533_C006.indd 4339533_C006.indd 433 10/29/08 1:57:34 PM10/29/08 1:57:34 PM


∂∂

− ( )=Θ

Θx

y z Bi y z( , , ) , ,0 0 0 (c)

∂∂

+ ( )=Θ

Θx

y z Bi y z( , , ) , ,1 1 0 (d)

∂∂

− ( )=Θ

Θy

x z Bi x z( , , ) , ,0 0 0 (e)

∂∂ ( )+ ( )=Θ

Θy

xba

z Bi xba

z, , , , 0 (f)

∂∂

( )− ( )=Θ

Θz

x y Bi x y, , , ,0 0 0 (g)

Θ x yca

, ,( )= 1 (h)

where Bi = ha/k is the Biot number. Determine the steady-state temperature distribution in the block.

Solution The nonhomogeneous boundary condition is on a face of con-stant z, and therefore the appropriate form of the initial product solu-tion is Θ(x,y,z) = Φ(x,y)Z(z), which when substituted into Equation b, leads to

−∂∂

+∂∂

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟=

1 12

2

2

2

2

ΦΦ Φ

( , ) ( )x y x y Z zd Zdz22

(i)

The separation argument applied to Equation i and the introduction of a separation constant λ leads to

−∂∂

+∂∂

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟=

2

2

2

2

Φ ΦλΦ

x y (j)

d Zdz

Z2

2= λ (k)

Use of the product solution in the boundary conditions, Equation c, Equation d, Equation e, and Equation f, gives

∂∂

− ( )=Φx

y Bi y( , ) ,0 0 0Φ (l)

9533_C006.indd 4349533_C006.indd 434 10/29/08 1:57:35 PM10/29/08 1:57:35 PM


∂∂

+ ( )=Φ

Φx

y Bi y( , ) ,1 1 0 (m)

∂∂

− =Φ

Φy

x Bi x( , ) ( , )0 0 0

(n)

∂∂ ( )+ ( )=

ΦΦ

yx

ba

Bi xba

, , 0 (o)

Equation j, Equation l, Equation m, Equation n, and Equation o constitute an eigenvalue problem. The operator defi ned by LxyΦ = − (∂2Φ/∂x2 + ∂2Φ/∂y2) is positive defi nite and self-adjoint on Sxy as defi ned by Equation l, Equation m, Equation n, and Equation o with respect to the inner product

f x y g x y f x y g x y dydx

ba

( , ), ( , ) ( , ) ( , )( ) = ∫ ∫xy

0

1

0

(p)

Thus, 5 all values of λ are real and positive, and eigenvectors correspond-ing to distinct eigenvalues are orthogonal with respect to the inner product defi ned in Equation p.

A product solution for Φ(x,y) is assumed to be of the form Φ(x,y) = X(x)Y(y), which when substituted into Equation j, leads to

1 12

2

2

2X xd Xdx Y

d Ydy( )

+ = −λ (q)

Application of the separation argument to Equation q and introduction of a separation constant μ leads to

d Xdx

X2

20+ −( ) =λ μ (r)

d Ydx

Y2

20+ =μ (s)

Substitution of the product solution into the boundary conditions, Equation l, Equation m, Equation n, and Equation o, gives

dXdx

BiX( )0 0 0− ( )= (t)

dXdx

BiX( )1 1 0+ ( )= (u)

9533_C006.indd 4359533_C006.indd 435 10/29/08 1:57:36 PM10/29/08 1:57:36 PM


dYdy

BiY( )0 0 0− ( )= (v)

dYdy

ba

BiYba( )+ ( )= 0 (w)

The operator defi ned by LyY = − d2Y/dy2 is positive defi nite and self-adjoint on Sy as defi ned by the boundary conditions of Equation v and Equation w with respect to

f y g y f y g y dy

ba

( ), ( ) ( ) ( )( ) = ∫y

0

(x)

which is the standard inner product on C [0, b/a]. Thus all values of μ are real and positive, and eigenvectors corresponding to distinct eigenvalues of Ly are orthogonal with respect to the inner product in Equation w.

The solution of Equation s is

Y y C y C y( ) cos sin= ( )+ ( )1 2μ μ (y)

Application of Equation v to Equation y leads to

CBi

C2 1=μ

(z)

Subsequent application of Equation t results in

tan μμ

ba

BiBi( )=

( ) −2

2 (aa)

Equation aa has an infi nite, but countable, number of solutions. The fi rst fi ve solutions for b/a = 2 and Bi = 4 are given in Table 6.1. The eigenvector corre-sponding to an eigenvalue μ� is

Table 6.1 Eigenvalues for Example 6.2

m 1 2 3 4 5

μm 0.054 3.439 14.88 37.248 60.289

βm 0.22 34.521 86.716 156.487 245.647

λ1,m 0.27 34.575 86.770 156.541 245.702

λ2,m 3.659 37.960 90.155 159.926 249.186

λ3,m 15.10 49.401 101.696 171.367 260.54

λ4,m 37.468 71.769 123.964 193.735 282.895

λ5,m 61.109 94.810 147.005 216.716 306.136

9533_C006.indd 4369533_C006.indd 436 10/29/08 1:57:37 PM10/29/08 1:57:37 PM


Y y C yBi

y� � ��

�( ) cos sin= ( )+ ( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

μμ

μ (bb)

Normalization of the eigenvector with respect to the inner product in Equa-tion x leads to

CBi Bi Bi

ba

��

��

� �

=+ + + −( ) ( )−

4

2 2 11

2 22

μ

μμ

μ μ( ) sin BBiba

cos 2

1

2

μ�( )

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

(cc)

Equation r has nontrivial solutions for each μ� corresponding to eigenvalues λ� , m m = 1,2,… which are obtained from Equation r, Equation t, and Equation u. Note that the solution of this system is obtained by comparison with the system of Equation s, Equation u, and Equation v. The systems are the same if in Equation s, Equation u, and Equation v, y is replaced by x, b/a is replaced by 1, and μ is replaced by λ − μ. Thus the eigenvalues are obtained by solving

tan( )

λ μλ μ

−( )=( ) − −

��

22

BiBi

(dd)

There are an infi nite, but countable, number of solutions of Equation dd for each �. Note that Equation dd may be written as

tan ββ

( )=( ) −

22

BiBi

(ee)

where β = λ − μ�. The solutions of Equation ee, βm m = 1,2,… are independent of μ�.

The corresponding eigenvectors are of the form

X x D xBi

xm m mm

m( ) cos sin= ( )+ ( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

ββ

β (ff)

Equation ff can be normalized by choosing

DBi Bi Bi Bi

mm

m

m

m m

=+ + + −( ) ( )−

4

2 2 11

2 22

β

ββ

β β( ) sin ccos 2

1

2

βm( )

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

(gg)

The eigenvalues of Lxy are given by

λ μ β� �,m = + m (hh)

9533_C006.indd 4379533_C006.indd 437 10/29/08 1:57:38 PM10/29/08 1:57:38 PM


where the μ� are the solutions of Equation aa and the βm are the solutions of Equation ee. The normalized eigenvectors are

Φ

ββ

β

� �

�

, ( ) ( )

cos sin

m m

m mm

m

X x Y y

C D xBi

x

=

= ( )+ ( )⎡

⎣⎢⎢⎢

⎤

⎦⎥⎥ ( )+ ( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

cos sinμμ

μ��

�yBi

y (ii)

For each λ�, m, the general solution of Equation k is

Z z A z B zm m m m m� � � � �, , , , ,( ) cosh sinh= ( )+ ( )λ λ (jj)

The general solution of Equation b is

Θ ββ

β( , , ) cos sinx y z C D xBi

xm mm

m= ( )+ ( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

⎧�⎨⎨

⎪⎪⎩⎪⎪

× (

=

∞

=

∞

∑∑m

y

11�

� cos μ ))+ ( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

×

Biy

A

μμ

��sin

�� , , , ,cosh sinhm m m mz B zλ λ( )+ ( )⎡⎣

⎤⎦}

(kk)

Application of Equation g to Equation kk gives

0 0 0=∂∂

−

=

ΘΘ

λ β

zx y Bi x y

C D B xm m m m

( , , ) ( , , )

cos, ,� � � (( )+ ( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

×

=

∞

=

∞

∑∑ Bix

y

mm

mβ

β

μ

sin

cos

11�

�(( )+ ( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

−

Biy

Bi C D A xm m m

μμ

β

��

� �

sin

cos, (( )+ ( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

×

=

∞

=

∞

∑∑ Bix

y

mm

mβ

β

μ

sin

cos

11�

�(( )+ ( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

= −

Biy

C D B BiAm m m

μμ

λ

��

� � � �

sin

, , ,mm mm

m

m

xBi

x( ) ( )+ ( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

=

∞

=

∞

∑∑ cos sinββ

β11�

cos sin× ( )+ ( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

μμ

μ��

�yBi

y

(ll)

9533_C006.indd 4389533_C006.indd 438 10/29/08 1:57:39 PM10/29/08 1:57:39 PM


Equation ll is satisfi ed if and only if

ABi

Bmm

m��

�,,

,=λ

(mm)

Application of Equation h to Equation kk using Equation mm leads to

Θ x yca

C D xBi

xm mm

m, , cos sin( )= = ( )+ ( )⎡

⎣⎢⎢

⎤

⎦1 � β

ββ ⎥⎥

⎥⎧⎨⎪⎪⎩⎪⎪=

∞

=

∞

∑∑m 11�

cos sin

× ( )+ ( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

μμ

μ��

�yBi

y

cosh sinh,,

,× ( )+BBi

ca

mm

m��

�λ

λ λλ� ,mca( )⎡

⎣⎢⎢

⎤

⎦⎥⎥⎫⎬⎪⎪⎭⎪⎪

(nn)

Equation nn is satisfi ed by expanding f(x,y) in an expansion in terms of the eigenvectors of Lxy,

1 = ( )+ ( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

⎧⎨v C D x

Bixk m m

mm� �, cos sinβ

ββ⎪⎪⎪

⎩⎪⎪

× ( )+ ( )⎡

⎣⎢⎢

⎤

=

∞

=

∞

∑∑m

yBi

y

11�

��

�cos sinμμ

μ⎦⎦⎥⎥⎫⎬⎪⎪⎭⎪⎪

(oo)

where

v C D xBi

xm m mm

m

b

� �, ( ) cos sin= ( )+ ( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

1

0

ββ

βaa

yBi

y dydx

∫∫

× ( )+ ( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

0

1

cos sinμμ

μ��

�

= ( )− ( )+⎡

⎣⎢⎢

⎤

⎦

C D Bi Bim

m

mm

mm

�

�μ ββ

ββ

βsin cos ⎥⎥

⎥

×⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟−

⎛⎝⎜⎜⎜

⎞⎠⎟⎟sin cosμ

μμ�

��

ba

Bi ba⎟⎟⎟+

⎡

⎣⎢⎢

⎤

⎦⎥⎥

Biμ�

(pp)

Equation nn is satisfi ed if and only if

Bv

Bica

mm

mm

��

��

,,

,,cosh sinh

=⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+

λλ λ ,,m

ca

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟

(qq)

9533_C006.indd 4399533_C006.indd 439 10/29/08 1:57:40 PM10/29/08 1:57:40 PM


The solution of Equation b, Equation c, Equation d, Equation e, Equation f, Equation g, and Equation h is

Θ( , , ) cos sin,x y z B C D xBi

xm m mm

m= ( )+ ( )⎡

⎣⎢⎢

⎤� � β

ββ

⎦⎦⎥⎥

⎧⎨⎪⎪⎩⎪⎪

× ( )+ ( )

=

∞

=

∞

∑∑m

yBi

y

11�

��

�cos sinμμ

μ⎡⎡

⎣⎢⎢

⎤

⎦⎥⎥

× (λλ�

�,

,coshmmBi

z))+ ( )⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

⎫⎬⎪⎪

⎭⎪⎪sinh ,λ� m z

(rr)

The eigenvectors of Lxy are products of the eigenvectors of two eigenvalue problems. Each eigenvalue problem is self-adjoint, and therefore each set of eigenvectors satisfi es its own orthonormality conditions,

Y Yp q y p q, ,( ) = δ (6.51)

X Xr s x r s� �, , ,,( ) = δ (6.52)

Mode-shape orthonormality of eigenvectors of Lxy requires that

Y X Y Xp p r q q s xy p q r s, , , ,,( ) = δ δ (6.53)

The inner product used in Equation 6.53 is of the form

f x y g x y f x y g x y h x y dxdyxy

ba

( , ), ( , ) ( , ) ( , ) ( , )( ) =0

∫∫∫0

ba

(6.54)

When Lxy is separable, h(x,y) = hx(x)hy(y). Then

Y X Y X Y y X x Y y X x hp p r q q s xy p p r q q s, , , ,, ( ) ( ) ( ) ( )( ) = xx yx h y dydx

ba

( ) ( )

00

1

∫∫

=⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥∫ X x X x h x dx Yp r q s x p, ,( ) ( ) ( ) (

0

1

yy Y y h y dyq y

ba

) ( ) ( )

0

∫⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

=( ) ( )X X Y Yp r q s x p q y, ,, ,

(6.55)

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The y-subscripted inner product in Equation 6.55 is zero unless p = q. If p = q, the x-subscripted inner product is zero unless r = s. Thus Equation 6.55 agrees with Equation 6.53. In addition,

Y X Y X X X Y Yp p r p p r xy p r p r x p p y, , , ,, , ,( ) = ( ) ( ) (6.56)

Equation 6.56 shows that if the x- and y-eigenvectors are normalized with respect to their individually subscripted inner products, then they are nor-malized with respect to the xy-subscripted inner product.

6.3 Time-dependent problems: Initial value problemsConsider the case when the operator of Equation 6.1 involves derivatives with respect to time, M ≠ 0. The diffusion equation and the wave equation are examples of such partial differential equations. Initial-value problems for partial differential equations occur when a system has an initial state other then its equilibrium state and when the response varies with spatial coordi-nates. The response of the system is then time-dependent.

It is assumed in this section that the partial differential equations, bound-ary conditions, and initial conditions are all nondimensional and that all boundary conditions are homogeneous. The operator L is assumed to be separable in time such that

L r L r r r Lf g t s t g t f h f g t( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]= [ ]+r t [[ ] (6.57)

where Lr is an operator involving only spatial derivatives, Lt is an operator involving only derivatives with respect to time, and s(t) and f(r) are functions determined from the separation procedure.

Consider a problem of the form

L rφ( , )t = 0 (6.58)

where L is an operator separable in time. The region in which φ is defi ned is bounded by a surface described by f(r) = 0. The boundary conditions are of the form

φ αφ

( , ) ( )r rn

t −∂∂

= 0 (6.59)

everywhere on S, the surface of the region. The problem has an appropriate number of initial conditions, at least one of which is nonhomogeneous. A problem whose canonical form is the diffusion equation has one initial con-dition, while a problem whose canonical form is the wave equation requires two initial conditions.

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Since L is separable in time, the method of separation of variables is used, assuming a product solution of

φ Φ( , ) ( ) ( )r rt T t= (6.60)

Substitution of Equation 6.60 into Equation 6.58, while applying the condi-tion for separation, Equation 6.57, leads to

s t T t h T t( ) ( ) ( ) ( ) ( ) ( )L r r r Lr tΦ Φ[ ]+ [ ]= 0 (6.61)


1 1

h s t T tT t

( ) ( )( )

( ) ( )( )

r rL r L

ΦΦr t[ ]= − [ ] (6.62)

The usual separation argument is applied with separation parameter λ, leading to

L r r rr Φ Φ( ) ( ) ( )[ ]= λh (6.63)

Lt T t s t T t( ) ( ) ( )[ ]= −λ (6.64)

Use of the product solution in the boundary condition, Equation 6.59, leads to

Φ αΦ

( , ) ( )r rn

t −∂∂

= 0 (6.65)

on S.Equation 6.63 and Equation 6.64 form an eigenvalue-eigenvector problem.

If Lr is self-adjoint and positive defi nite with respect to the inner product

f g f g h dVV

V

( ), ( ) ( ) ( ) ( )r r r r r( ) = ∫ (6.66)

then all eigenvalues are real and positive, and the eigenvectors correspond-ing to distinct eigenvalues are orthogonal with respect to the inner product defi ned in Equation 6.66.

The solution proceeds in a manner similar to that for steady-state prob-lems. Once all of the eigenvalues and eigenvectors of Lr have been deter-mined, a solution of Equation 6.64 is obtained for each eigenvalue. Each resulting product solution of the form of Equation 6.60 satisfi es the partial differential equation and all boundary conditions. The most general solution

9533_C006.indd 4429533_C006.indd 442 10/29/08 1:57:42 PM10/29/08 1:57:42 PM


is a linear combination of all product solutions. Eigenvector expansions of nonhomogeneous terms in the initial conditions are developed and used to impose the initial conditions.

Example 6.3 A bar of thickness L is at a uniform temperature T0 when it is suddenly submerged in a bath of temperature T∞. Determine the unsteady state temperature distribution in the bar when: (a) the heat transfer coeffi -cient is small and convection heat transfer is neglected; (b) the heat transfer coeffi cient is of an intermediate value and heat convection is of the same order of magnitude as conduction at the ends of the bar, but the cross-sec-tional area is large compared to the perimeter times the length of the bar, such that convection over the length of the bar can be neglected; (c) the heat transfer coeffi cient is large enough such that convection must be included over the length of the bar, but a one-dimensional assumption is used; and (d) the thickness is comparable to the length, meaning that a two dimensional model is required.

Solution Defi ne x* = x/L, t* = ρcpL2/k, and Θ = (T − T∞)/(T0 − T∞). (a) If the heat transfer coeffi cient is small, the temperature at the ends is the same as the temperature of the bath. The problem governing the nondimensional temperature distribution is

∂∂

=∂∂

Θ Θt x

2

2 (a)

Θ( , )0 0t = (b)

Θ( , )1 0t = (c)

Θ( , )x 0 1= (d)

A product solution of Equation a is assumed to be of the form

Θ( , ) ( ) ( )x t X x T t= (e)

which, when substituted into Equation a, leads to

− = −1 12

2Xd Xdx T

dTdt

(f)

The separation argument is applied to Equation f. Defi ning λ as the separa-tion constant, the separated differential equations are

d Xdx

X2

2+ λ (g)

9533_C006.indd 4439533_C006.indd 443 10/29/08 1:57:43 PM10/29/08 1:57:43 PM


dTdt

T+ =λ 0 (h)

Substitution of Equation e into the boundary conditions, Equation b and Equation c, leads to

X( )0 0= (i)

X( )1 0= (j)

The eigenvalues and normalized eigenvectors for the system comprised of Equation g subject to Equation i and Equation j are

λ π� � �=( ) =2

1 2, ,... (k)

X x x� �( ) sin= ( )2 π (l)

The corresponding solutions of Equation h are

T t A e t� �

�( ) = −( )π 2

(m)

The general solution becomes

Θ( , ) sinx t A e xt= ( )−( )

=

∞

∑ 22

1

��

�

�π π (n)

The eigenvector expansion of the nonhomogeneous initial condition is

1 21

= ( )=

∞

∑v x�

�

�sin π (o)

where

v x

x dx

k = ( )( )

= ( )

= − −( )⎡

∫

1 2

2

21 1

0

1

, sin

sin

�

�

��

π

π

π ⎣⎣⎢⎤⎦⎥

===

⎧⎨

2 2 0 2 4 6

1 1 3 5��

�π, , ,...

, , ,...

⎪⎪⎪⎩⎪⎪

(p)

9533_C006.indd 4449533_C006.indd 444 10/29/08 1:57:44 PM10/29/08 1:57:44 PM


Application of the initial condition, Equation d, to Equation n using Equation o and Equation p leads to A� = v�. The time-dependent temperature distribu-tion in the plate is

Θ( ) sint e xt= ( )=

∞−( )∑4 1

1

2

πππ

��

�

� (q)

The expansion of Equation q converges to the nondimensional temper-ature distribution everywhere except at x = 0 and x = 1. This, of course, is because f(x) = 1 is not in the domain of the operator defi ned in Equation h when subject to the boundary conditions of Equation i and Equation j. This analysis is consistent with the mathematical modeling, but it violates physi-cal principles. The modeling suggests that the temperature is discontinuous at the ends of the bar at t = 0. Before submergence in the bath, the bar has a uniform temperature T0, but the modeling suggests that immediately after submergence, the temperature at each end is T∞. The more precise physical model is that used in part (b), in which heat transfer occurs through convec-tion at the ends of the bar and a transient temperature response occurs at the ends of the bar.

(b) The appropriate nondimensional boundary conditions for the model as stated are

∂∂

− =Θ

Θx

t Bi t( , ) ( , )0 0 0 (r)

∂∂

+ =Θ

Θx

t Bi t( , ) ( , )1 1 0 (s)

The problem governed by Equation g, Equation r, and Equation s is the same as that obtained in the solution of Example 6.2. The eigenvalues are the posi-tive solutions of

tan λλ

( )=( ) −

22

Bi

Bi (t)

The normalized eigenvector corresponding to an eigenvalue λ� is

X x D xBi

x� � ��

�( ) cos sin= ( )+ ( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

λλ

λ (u)

where

DBi Bi Bi Bi

��

�

�

� �

=+ + + −( ) ( )−

4

2 2 11

2 22

λ

λλ

λ λ( ) sin ccos 2

1

2

λ�( )

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥

(v)

9533_C006.indd 4459533_C006.indd 445 10/29/08 1:57:45 PM10/29/08 1:57:45 PM


The solution of Equation h corresponding to an eigenvalue λ� is

T t A e t� �

�( ) = −λ (w)

The general solution is

Θ( , ) cos sinx t A e D xBi

xt= ( )+ ( )⎡

⎣⎢⎢

⎤

⎦−

� � ��

��λ λ

λλ ⎥⎥

⎥=

∞

∑� 1

(x)

The eigenvector expansion for the nonhomogeneous function in the bound-ary condition of Equation d is

11

= ( )+ ( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

=

∞

∑v D xBi

x� � ��

�

�

cos sinλλ

λ (y)

where

v D xBi

x dx

D

� � ��

�

�

= ( )+ ( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

=

∫ cos sinλλ

λ

λ

0

1

��

��

��

sin cosλλ

λβ( )− ( )+

⎡

⎣⎢⎢

⎤

⎦⎥⎥

Bi Bi (z)

Application of the initial condition, Equation d, to Equation x using Equa-tion y and Equation z, leads to A� = v�. The solution for the temperature dis-tribution is

Θ( , ) cos sinx t v e D xBi

xt= ( )+ ( )⎡

⎣⎢⎢

⎤

⎦−

� � ��

��λ λ

λλ ⎥⎥

⎥=

∞

∑� 1

(aa)

Evaluation of Equation aa at x = 0 leads to

Θ( , )01

t v e Dt= −

=

∞

∑ � �

�

�λ (bb)

The function f(x) = 1 is still not in the domain of the Sturm-Liouville prob-lem, but the solution of Equation aa leads to a time-dependent temperature at each end.

(c) Let w be the width of the plate and p its thickness. The area over which heat conduction occurs is A = wp, while the perimeter for convec-tion is P = 2(w + p). A second Biot number is defi ned as Bi2 = hPL/kA. The

9533_C006.indd 4469533_C006.indd 446 10/29/08 1:57:45 PM10/29/08 1:57:45 PM


nondimensional governing partial differential equation when convective heat transfer across the surface is included is

∂∂

=∂∂

−Θ Θ

Θt x

Bi2

2 2 (cc)

The boundary conditions are those of Equation r and Equation s and the ini-tial condition is that of Equation d. A product-solution assumption, Θ(x,t) = X(x)T(t), when substituted into Equation cc, leads to

− = − +⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟

1 12

2 2X xd Xdt T

dTdt

Bi( )

(dd)

Application of the usual separation argument to Equation dd, using λ as the separation parameter, leads to

d Xdx

X2

20+ =λ (ee)

dTdt

Bi T+ +( ) =2 0λ (ff)

The problem for X(x) is the same as that of part (b) and thus has the same solution. The only difference in this problem is the solution for Equation ff, which for a specifi c λ� is

T t A e Bi t� �

�( ) = − +( )λ 2 (gg)

The temperature distribution is

Θ( , ) cos sinx t e v e D xBi

xBi t t= ( )+ ( )− −2� � �

��

�λ λλ

λ⎡⎡

⎣⎢⎢

⎤

⎦⎥⎥

=

∞

∑� 1

(hh)

where all terms are as specifi ed in the solution to part (b).(d) The temperature distribution is two-dimensional, Θ(x,y,t). The non-

dimensional governing equation is

∂∂

=∂∂

+∂∂

Θ Θ Θt x y

2

2

2

2 (ii)

The boundary conditions for conduction heat transfer over the entire sur-face are

∂∂

− =Θ

Θx

y t Bi y t( , , ) ( , , )0 1 0 (jj)

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∂∂

+ =Θ

Θx

y t Bi y t( , , ) ( , , )1 1 0 (kk)

∂∂

− =Θ

Θy

x t Bi x t( , , ) ( , , )0 0 0 (ll)

∂∂

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟+

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟

ΘΘ

yx

pL

y t Bi xpL

y t, , , , ⎟⎟ = 0 (mm)

The initial condition is that of Equation d, Θ(x,y,0) = 1.A product solution of Equation ii is assumed to be of the form

Θ Φ( , , ) ( , ) ( )x y t x y T t= (nn)

which when substituted into Equation ii, leads to

−∂∂

+∂∂

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

= −∂∂

1 12

2

2

2ΦΦ Φ

x y TTt

(oo)

Application of the usual separation argument to Equation oo and defi ning a separation parameter λ leads to

∂∂

+∂∂

= −2

2

2

2

Φ ΦΦ

x yλ (pp)

dTdt

T+ =λ 0 (qq)

Use of the product solution in the boundary conditions, Equation jj, Equation kk, Equation ll, and Equation mm, gives

∂∂

− =Φ

Φx

y Bi y( , ) ( , )0 1 0 (rr)

∂∂

+ =Φ

Φx

y Bi y( , ) ( , )1 1 0 (ss)

∂∂

− =Φ

Φy

x Bi x( , ) ( , )0 0 0 (tt)

∂∂ ( )+ ( )=

ΦΦ

yx

pL

y Bi xpL

y, , 0 (uu)

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A product solution is assumed for Equation pp subject to Equation rr, Equa-tion ss, Equation tt, and Equation uu as follows:

Φ( , ) ( ) ( )x y X x Y y= (vv)

Substitution of Equation vv into Equation pp, Equation rr, Equation ss, Equa-tion tt, and Equation uu leads to problems similar to those solved in Example 6.2. The solution, Equation h of Example 6.2, is repeated below:

Φ� �

�

, ( ) ( )

cos sin

m m

m mm

m

X x Y y

C D xBi

x

=

= ( )+ ( )⎡

⎣β

ββ⎢⎢

⎢⎤

⎦⎥⎥ ( )+ ( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

cos sinμμ

μ��

�yBi

y

(ww)

Equations defi ning the parameters in Equation ww are given as Equation z, Equation aa, Equation bb, Equation cc, Equation dd, Equation ee, Equation ff, and Equation gg of Example 6.2.

The details of the remainder of the solution are similar to those in part (a). The solution of Equation qq is Equation w for a specifi c eigenvalue, λ�,m. The general solution of the partial differential equation is

Θ( , , ) cos sin,x y t C D A e xBi

mt

m

mm

m= ( )+−

=

∞

∑ � ��λ β

β1

ββ

μμ

μ

m x

yBi

( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

× ( )+

=

∞

∑�

��

1

cos sin �� y( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

(xx)

Application of the initial condition leads to A� = v�, where, from Equation oo of Example 6.2,

vC D Bi Bi

mm

m

mm

mm

��

�

, sin cos= ( )− ( )+⎡

⎣⎢⎢

⎤

⎦μ ββ

ββ

β⎥⎥⎥

×⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟−

⎛⎝⎜⎜⎜

⎞⎠⎟⎟sin cosμ

μμ�

��

pL

Bi pL⎟⎟⎟+

⎡

⎣⎢⎢

⎤

⎦⎥⎥

Biμ�

(yy)

Example 6.4 The nondimensional problem governing the displacement, u(x,t), of the elastic bar shown in Figure 6.6 is

∂∂

∂∂

⎡

⎣⎢⎢

⎤

⎦⎥⎥=

∂∂x

xux

xu

xα α( ) ( )

2

2 (a)

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The bar is fi xed at x = 0 and free at x = 1, leading to the boundary conditions,

u t( , )0 0= (b)

∂∂

=ux

t( , )1 0 (c)

Before t = 0, the right end of the bar is subject to a force P which leads to a static displacement in the bar, given by

u xP

A Ed

x

( , )( )

01

0 00

= ∫ α ξξ (d)

The force is removed at t = 0, resulting in subsequent motion of the bar. Since the bar is at rest at t = 0,

∂∂

=ut

x( , )0 0 (e)

Determine the response if (a) α(x) = 1 and (b) α(x) = (1 − 0.1x)2.

Solution (a) For α = 1, the partial differential equation and the initial condi-tion, Equation d, become

∂∂

=∂∂

2

2

2

2

ux

ut

(f)

1

x

u(x,t)

P

α(x)

Figure 6.6 The force P is removed from the end of the elastic bar at t = 0. The result-ing motion is obtained through solution of the wave equation with a nonhomogene-ous initial condition.

9533_C006.indd 4509533_C006.indd 450 10/29/08 1:57:48 PM10/29/08 1:57:48 PM


u xPx

A E( , )0

0 0

= (g)

A product solution of Equation a can be assumed to be of the form u(x,t) = X(x)T(t), which when substituted into Equation f, leads to

− = −1 12

2

2

2X xd Xdx T t

d Tdt( ) ( )

(h)

Applying the usual separation argument to Equation h and choosing λ as the separation constant leads to

d Xdx

X2

20+ =λ (i)

d Tdt

T2

20+ =λ (j)

Application of the product solution to the boundary conditions, Equation b and Equation c, results in

X( )0 0= (k)

dXdx

( )1 0= (l)

The eigenvalues and normalized eigenvectors of the system of Equation i, Equation k, and Equation l are

λπ

�

�=

−( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

2 1

2

2

(m)

X x x�

�( ) sin=

−( )⎡

⎣⎢⎢

⎤

⎦⎥⎥2

2 1

2

π (n)

For a specifi c λ�, the general solution of Equation j is

T t A t B t� � �

� �( ) cos sin=

−( )⎡

⎣⎢⎢

⎤

⎦⎥⎥ +

−( )2 1

2

2 1

2

π π⎡⎡

⎣⎢⎢

⎤

⎦⎥⎥ (o)

A solution which satisfi es Equation a, Equation b, and Equation c is

u x t A t B( , ) cos sin=−( )⎡

⎣⎢⎢

⎤

⎦⎥⎥ +

−( )� �

� �2 1

2

2 1

2

π πtt x

⎡

⎣⎢⎢

⎤

⎦⎥⎥

⎧⎨⎪⎪⎩⎪⎪

⎫⎬⎪⎪⎭⎪⎪

−( )⎡

⎣⎢⎢

⎤

⎦2

2 1

2sin

� π ⎥⎥⎥

=

∞

∑� 1

(p)

9533_C006.indd 4519533_C006.indd 451 10/29/08 1:57:49 PM10/29/08 1:57:49 PM


Application of the initial condition, Equation e, to Equation p requires B� = 0.

The initial condition, Equation g, is imposed through an eigenvector expansion,

Px

A Ev x

0 0 1

22 1

2=

−⎡

⎣⎢⎢

⎤

⎦⎥⎥

=

∞

∑ �

�

�sin

( )π (q)

where

vPx

A Ex�

�=

−( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

0 0

22 1

2, sin

π

==⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

−( )⎡

⎣⎢⎢

⎤

⎦⎥⎥∫ Px

A Ex

0 00

1

22 1

2sin

� πddx

P

A E=

−( )−( )[ ]

+

24 1

2 1

1

2

0 0

�

� π

(r)

Application of Equation g to Equation p leads to A� = v�. Then, substituting Equation r into Equation p,

u x tP

A Ex( , ) sin=

−( )−( )

−( )⎡

⎣

+8 1

2 1

2 1

220 0

1

2π

π�

�

�⎢⎢⎢

⎤

⎦⎥⎥

−( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

=

∞

∑ cos2 1

21

�

�

πt (s)

(b) Using α(x) = (1 − 0.1x)2, Equation a and Equation d become

∂∂

−( ) ∂∂

⎡

⎣⎢⎢

⎤

⎦⎥⎥= −( ) ∂

∂xx

ux

xu

t1 0 1 1 0 1

2 22

2. . (t)

u xPx

A E x( , )

( . )0

1 0 10 0

=−

(u)

Substitution of the product solution u(x,t) = X(x)T(t) into Equation t leads to

−−

−⎡

⎣⎢⎢

⎤

⎦⎥⎥= −

1

1 0 11 0 1

12

2

( . ) ( )( . )

x X xd

dxx

dXdx T(( )t

d Tdt

2

2 (v)

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Applying the usual separation argument to Equation v and defi ning the sep-aration parameter as λ results in

−−

−( )⎡

⎣⎢⎢

⎤

⎦⎥⎥=

1

1 0 11 0 1

2

2

( . ).

xd

dxx

dXdx

Xλ (w)

d Tdx

T2

20+ =λ (x)

Equation w with boundary conditions Equation k and Equation l constitutes a Sturm-Liouville problem with p(x) = − (1 − 0.1x)2, q(x) = 0 and r(x) = (1 − 0.1x)2. The results of Section 5.4 prove that Lx is self-adjoint and positive defi nite with respect to the inner product,

f x g x f x g x x dxr

( ), ( ) ( ) ( ) .( ) = −( )∫ 1 0 12

0

1

(y)

The solution of Equation w is aided by a change in independent variable; let

z x= −1 0 1. (z)

Substitution of Equation z into Equation w leads to

ddz

zdXdz

z X2 2100 0⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+ =λ (aa)

Comparison with Equation 3.108 shows that Equation aa has a Bessel-function solution of the form of Equation 3.110 with r = 2, s = 2, and b = 10√λ. The resulting general solution of Equation aa is

X z C z J z C z Y z( ) = ( )+ ( )−

−

−

−1

1

21

2

2

1

21

2

10 10λ λ (bb)

Equation bb may also be written in terms of spherical Bessel functions as

X z C j z C y z( ) = ( )+ ( )1 0 2 010 10λ λ (cc)

Substitution of Equation z into Equation cc leads to

X x C j x C y x( ) . ( . )= −( )( )+ −( )1 0 2 010 1 0 1 10 1 0 1λ λ (dd)

9533_C006.indd 4539533_C006.indd 453 10/29/08 1:57:51 PM10/29/08 1:57:51 PM


Application of Equation k to Equation dd gives

Cj

yC2

0

0

1

10

10= −

( )( )

λ

λ (ee)

Substitution of Equation l into Equation dd using Equation ee leads to a tran-scendental equation for λ,

′ ( ) ( )− ′ ( ) ( )=j y y j0 0 0 09 10 9 10 0λ λ λ λ (ff)

Equation ff has an infi nite number of solutions which are indexed by λ� ,� = 1,2,… . The eigenvector corresponding to λ� is

X x C j xj

yy� � �

�

�

( ) ( . )= −( )−( )( )0

0

0

010 1 0 110

10λ

λ

λ110 1 0 1λ� ( . )−( )

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

x (gg)

The eigenvectors can be normalized by requiring (X�,X�) = 1 or

C j xj

yy� �

�

��

20

0

0

010 1 0 110

1010λ

λ

λλ( . ) (−( )−

( )( )

11 0 1 1 0 1 1

2

0

1

2−( )⎡

⎣⎢⎢⎢

⎤

⎦⎥⎥⎥

−( ) =∫ . ) .x x dx

(hh)

Closed-form solutions of the integrals in Equation hh are not known. For a specifi c λ�, the solution of Equation x is

T t A t B t� � � � �( ) cos sin= ( )+ ( )λ λ (ii)

The remainder of the solution is similar to that for part (a). The general solution is a linear combination of all product solutions. Application of Equation e leads to B� = 0. Application of Equation u leads to A� = v�, where

vPx

A E xX

PxA E

C j

r� �

�

=−

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟

=

0 0

0 00

1 0 1

1

( . ),

00 1 0 110

1010 1 0 1

0

0

0λλ

λλ�

�

��( . ) ( .−( )−

( )( )

−xj

yy xx

x dx

)

.

( )⎡

⎣⎢⎢⎢

⎤

⎦⎥⎥⎥

× −( )

∫0

1

1 0 1

(jj)

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6.4 Nonhomogeneous partial differential equationsDirect application of separation of variables requires that the partial differ-ential operator be separable and that a suffi cient number of boundary con-ditions be homogeneous to generate eigenvalue problems. If the problem is unsteady state, eigenvalue problems must be generated in all spatial direc-tions. If the problem is steady, eigenvalue problems must be generated in all but one of the spatial directions.

Nonhomogeneous problems occur when the governing partial differential equation is nonhomogeneous or when there are too many nonhomogeneous boundary conditions to apply separation of variables directly. Two methods are presented in this section to handle nonhomogeneous terms. A superposi-tion method can often be used for linear problems in which the solution is a superposition of solutions, each of which is obtained through application of separation of variables or through solution of an ordinary differential equa-tion. Nonhomogeneous problems which do not lend themselves to superposi-tion can be solved using an eigenvector expansion method in which a solution is assumed to be an eigenvector expansion of solutions of the corresponding homogeneous problem obtained by removing the nonhomogeneous terms. These approaches are illustrated through the following examples.

Example 6.5 The steady nondimensional temperature distribution in the slab shown in Figure 6.7 is governed by the two-dimensional Laplace’s equation,

∂∂

+∂∂

=2

2

2

20

θ θx y

(a)

The boundary conditions corresponding to Figure 6.7 are

θ( , )0 0y = (b)

θ δ( , )1 1y = (c)

θ( , )x 0 0= (d)

θ α δ( , )x = 2 (e)

Separation of variables cannot be directly applied to solve Equation a sub-ject to the boundary conditions of Equation b, Equation c, Equation d, and Equation e because the system has nonhomogeneous boundary conditions for constant values of both x and y. Consider the superposition,

θ θ θ( , ) ( , ) ( , )x y x y x y= +1 2 (f)

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where θ1(x,y) satisfi es

∂∂

+∂∂

=2

12

21

20

θ θx y

(g)

θ1 0 0( , )y = (h)

θ δ1 11( , )y = (i)

θ1 0 0( , )x = (j)

θ α1 0( , )x = (k)

and θ2(x,y) satisfi es

∂∂

+∂∂

=2

22

222

0θ θx y

(l)

θ2 0 0( , )y = (m)

θ2 1 0( , )y = (n)

θ2 0 0( , )x = (o)

θ α δ2 2( , )x = (p)

α

y

x0

0

1

δ1

δ2

Figure 6.7 The slab has nonhomogeneous boundary conditions on its sides, descri-bed by constant values of two independent variables. Separation of variables is not directly applicable to Laplace’s equation subject to these boundary conditions. There-fore, a superposition method is used.

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It is easy to verify by substitution that θ(x,y) given by Equation f satisfi es Equation a, Equation b, Equation c, Equation d, and Equation e.

The method of separation of variables is used to solve for θ1(x,y) as

θδπ

πα

πα

114

( , )

sin sinh

x y

ny

nx

=

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟

⎛⎝⎜⎜⎜

⎞⎞⎠⎟⎟⎟⎟

( )=

∞

∑ n nn

sinh, ,

π1 3 5

(q)

The solution for θ2(x,y) is obtained by interchanging the roles of x and y, using δ2 in place of δ1, replacing α by 1, and noting that sinh(nπ) should be replaced by sinh(nπa). Thus,

θδπ

π ππα2

2

1

4( , )

sin sinh

sinh,

x yn x n yn n

n

=( ) ( )

( )= 33 5,

∞

∑ (r)

The superposition formula, Equation f of Example 6.5, is obvious, and the choice of problems defi ning the superposition components is obvious.

Example 6.6 The slab shown in Figure 6.8 is subject to an internal heat generation which is a function of x only. The partial differential equation governing the nondimensional temperature distribution is

∂∂

+∂∂

=2

2

2

2

θ θx y

cf x( ) (a)

α

y

x

q(x)

0

1

0

0

1

Figure 6.8 Internal heat generation leads to a nonhomogeneous partial differential equation. Since the nonhomogeneous term is a function of x only, a superposition of a function of x only and a separate function of x and y is developed.

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The boundary conditions corresponding to the slab shown in Figure 6.7 are

θ( , )0 1y = (b)

θ( , )1 0y = (c)

θ( , )x 0 0= (d)

θ α( , )x = 0 (e)

Determine a superposition formula to solve Equation a subject to Equation b, Equation c, Equation d, and Equation e. The problems defi ning each ele-ment of the superposition formula must be directly solvable by separation of variables or be represented by ordinary differential equations.

Solution Since the internal heat generation is a function of x only, it is rea-sonable to assume that it leads to a term dependent only on x, suggesting a superposition formula of the form

θ φ( , ) ( , ) ( )x y x y x= +Φ (f)

Substitution of Equation f into Equation a leads to

∂∂

+∂∂

+ =2

2

2

2

2

2

Φ Φx y

ddx

cf xφ

( ) (g)

It is desired to obtain a homogeneous problem for Φ(x,y). Choosing

∂∂

+∂∂

=2

2

2

20

Φ Φx y

(h)

leads to

ddx

cf x2

2

φ= ( ) (i)

Substitution of Equation f into Equation b, Equation c, Equation d, and Equa-tion e gives

θ φ( , ) ( , ) ( )0 1 0 0y y= = +Φ (j)

θ φ( , ) ( , ) ( )1 0 1 1y y= = +Φ (k)

θ φ( , ) ( , ) ( )x x x0 0 0= = +Φ (l)

θ α α φ( , ) ( , ) ( )x x x= = +0 Φ (m)

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Equation l and Equation m lead to nonhomogeneous boundary conditions at constant values of y for Φ(x,y):

Φ( , ) ( )x x0 = −φ (n)

Φ( , ) ( )x xα φ= − (o)

The problem for Φ(x,y) can be solved directly by separation of variables only if the boundary conditions defi ning its problem at constant values of x are homogeneous,

Φ( , )0 0y = (p)

Φ( , )1 0y = (q)

The choices of Equation p and Equation q, when used in Equation k and Equation l, lead to

φ( )0 1= (r)

φ( )1 0= (s)

The problem governing φ(x) is the ordinary differential equation of Equa-tion i subject to the boundary conditions of Equation r and Equation s. The differential equation is easily solved for a specifi c f(x) by integrating both sides with respect to x twice and then solving for the constants of integration through application of Equation r and Equation s. The problem governing Φ(x,y) is the partial differential equation of Equation h subject to the bound-ary conditions of Equation n, Equation o, Equation p, and Equation q. The problem is directly solvable by separation of variables with the eigenvalue problem in the x-direction. Boundary conditions p and q are applied using eigenvector expansions of φ(x).

Example 6.7 The steady nondimensional temperature distribution of the slab shown in Figure 6.9 is governed by the problem,

∂∂

+∂∂

=2

2

2

20

θ θs s

x y (a)

θs( , )0 0y = (b)

∂∂

+ =θ

θs

xy Bi ys( , ) ( , )1 1 0 (c)

θs( , )x 0 0= (d)

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∂∂

=θ

αsfy

x c q x( , ) ( )1 (e)

where qf(x) is an imposed heat fl ux. The problem described by Equation a, Equation b, Equation c, Equation d, and Equation e can be solved directly by separation of variables. The eigenvalue problem is in the x-direction, and the boundary condition at x = α is imposed using an eigenfunction expansion for qf(x).

At t = 0, an internal heat source begins to function. The heat added is con-stant with x, but a function of y. The initiation of the heat generation leads to a transient temperature distribution, Θ(x,y,t), in the slab. The unsteady state problem is governed by

∂∂

+∂∂

=∂∂

−2

2

2

2 2

Θ Θ Θx y t

c u y( ) (f)

Θ( , , ) ( , )x y x ys0 = θ (g)

Θ( , , )0 0y t = (h)

∂∂

+ =Θ

Θx

y t Bi y t( , , ) ( , , )1 1 0 (i)

Θ( , , )x t0 0= (j)

∂∂

=Θy

x t c q xf( , , ) ( )α 1 (k)

u(y)

y

x

qf(x)

0 +Biθ=0

0

θx

Figure 6.9 The slab has a steady-state temperature distribution when the inter-nal heat generation is initiated. Since the internal heat generation is independent of time, a different steady state will be achieved. Therefore, an initial attempt at a superposition involves a transient temperature distribution and a steady-state temperature distribution. Nonhomogeneous boundary conditions make several additional superpositions necessary.

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Determine a superposition formula which can be used to solve the problem governed by Equation f, Equation g, Equation h, Equation i, Equation j, and Equation k, such that the problem governing each term in the superposition can be solved directly by separation of variables or is an ordinary differen-tial equation. Specify the differential equations and all initial and boundary conditions governing each problem.

Solution The internal heat generation is constant with time. A new steady state will eventually be achieved. The initial transient solution will die out, leaving this new steady state. A reasonable superposition is thus

Θ Φ( , , ) ( , , ) ( , )x y t x y t S x y= + (l)

where Φ(x,y,t) is the transient response and S(x,y) is the eventual steady-state response.

Substitution of Equation l into Equation f leads to

∂∂

+∂∂

+∂∂

+∂∂

=∂∂

−2

2

2

2

2

2

2

2 2

Φ Φ Φx y

Sx

Sy t

c u y( ) (m)

The transient response is a function of three independent variables, whereas the steady-state response is a function of two independent variables. For sim-plicity, it is therefore desirable to choose a homogeneous problem to given in the transient response. The separable partial differential equation governing Φ(x,y) is chosen as

∂∂

+∂∂

=∂∂

2

2

2

2

Φ Φ Φx y t

(n)

Subtract. Equation n from Equation n leads to

∂∂

+∂∂

= −2

2

2

2 2

Sx

Sy

c u y( ) (o)

Substitution of the superposition assumption, Equation l, into the initial con-dition, Equation g, leads to

θs x y x y S x y( , ) ( , , ) ( , )= +Φ 0 (p)

Equation p is satisfi ed by choosing

Φ( , , ) ( , ) ( , )x y x y S x ys0 = −θ (q)

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Substitution of Equation l into the boundary conditions, Equation h, Equa-tion i, Equation j, and Equation k, results in

Θ Φ( , , ) ( , , ) ( , )0 0 0 0y t y t S y= = + (r)

∂∂

+ = =∂∂

+Θ

ΘΦ

Φx

y t Bi y tx

y t Bi( , , ) ( , , ) ( , , ) ( ,1 1 0 1 1 yy tSx

y BiS y, ) ( , ) ( , )+∂∂

+1 1 (s)

Θ Φ( , , ) ( , , ) ( , )x t x t S x0 0 0 0= = + (t)

∂∂

= = +Θ

Φy

x t c q x x t S xf( , , ) ( ) ( , , ) ( , )α α α1 (u)

The problem for Φ(x,y,t) can be solved directly by separation of variables only if all boundary conditions which it satisfi es are homogeneous. To this end, boundary conditions are chosen from Equation r, Equation s, Equation t, and Equation u such that

Φ( , , )0 0y t = (v)

∂∂

+ =Φ

Φx

y t Bi y t( , , ) ( , , )1 1 0 (w)

Φ( , , )x t0 0= (x)

∂∂

=Φx

x t( , , )α 0 (y)

The boundary conditions which must be satisfi ed by S(x,y) are therefore

S y( , )0 0= (z)

∂∂

+ =Sx

y BiS y( , ) ( , )1 1 0 (aa)

S x( , )0 0= (bb)

∂∂

=Sx

x c q xf( , ) ( )α 1 (cc)

The problem for Φ(x,y,t) defi ned by Equation n, Equation q, Equation v, Equation w, Equation x, and Equation y can be directly solved using separation of variables once S(x,y) is known. The problem defi ning S(x,y), Equation o, Equation z, Equation aa, Equation bb, and Equation cc, is nonhomogeneous. The governing partial differential equation has a nonhomogeneous term which is a function of y only. This suggests a superposition formula of the form

S x y A x y B y( , ) ( , ) ( )= + (dd)

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Substitution of Equation dd into Equation o and requiring a homogeneous partial differential equation to be satisfi ed by A(x,y) leads to

∂∂

+∂∂

=2

2

2

20

Ax

Ay

(ee)

d Bdy

c u y2

2 2= − ( ) (ff)

The boundary conditions for S(x,y), Equation z, Equation aa, Equation bb, and Equation cc, are satisfi ed when the superposition of Equation dd is used by choosing

A y B y( , ) ( )0 = − (gg)

∂∂

+ = −Ax

y BiA y BiB y( , ) ( , ) ( )1 1 (hh)

A x( , )0 0= (ii)

∂∂

=A

yx cq xf( , ) ( )α (jj)

B( )0 0= (kk)

dBdy

( )α = 0 (ll)

B(y) is easily obtained by solving the ordinary differential equation, Equation ff, subject to the boundary conditions, Equation kk and Equation ll. However, the problem governing A(x,y), Equation ee, Equation gg, Equation hh, Equation ii, Equation jj, Equation kk, and Equation ll, has nonhomogene-ous boundary conditions at constant values of x and a constant value of y and is itself nonhomogeneous. This fi nal nonhomogeneity is relieved by a simple superposition similar to that used in Example 16.1,

A x y C x y D x y( , ) ( , ) ( , )= + (mm)

where

∂∂

+∂∂

=2

2

2

20

Cx

Cy

(nn)

C y B y( , ) ( )0 = − (oo)

∂∂

+ = −Cx

y BiC y BiB y( , ) ( , ) ( )1 1 (pp)

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C x( , )0 0= (qq)

∂∂

=Cy

x( , )α 0 (rr)

and

∂∂

+∂∂

=2

2

2

20

Dx

Dy

(ss)

D y( , )0 0= (tt)

∂∂

+ =Dx

y BiD y( , ) ( , )1 1 0 (uu)

D x( , )0 0= (vv)

∂∂

=Dy

x c q xf( , ) ( )α 2 (ww)

The fi nal superposition formula is

Θ Φ( , , ) ( , , ) ( ) ( , ) ( , )x y t x y t B y C x y D x y= + + + (xx)

The technique used in the previous example, using a superposition of a transient solution and a steady-state solution, is applicable in problems where an initial steady state is disturbed due to a change in conditions and a new steady state is expected to develop after the transient response decays. The thin slab shown in Figure 6.10 is at a uniform temperature when it is heated from below, leading to a time-dependent temperature in the slab. The resulting problem is nonhomogeneous due to a nonhomogeneous boundary condition. However, if all other sides are insulated, there is no surface for heat transfer with the ambient medium, and instead of reaching a steady state, the temperature in the slab will increase without bound. In this case, a superposition assuming that the temperature is the sum of a transient

q(x )

x

Figure 6.10 The bar has a uniform temperature when the heat fl ux is applied. All other sides are insulated, and therefore a new steady state is not attained. A super-position formula must include a function of time only.

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response and an eventual steady state will not work. If one were attempted, it would become apparent that the nonhomogeneous boundary condition cannot be satisfi ed.

The diffi culty in attaining a new steady state can also be explained by noting that the eigenvalue problem in the spatial direction obtained when solving for the transient response is only non-negative defi nite. The system has a zero eigenvalue due to the imposition of derivative boundary condi-tions at each boundary. Therefore, when the general solution is formed from a superposition of eigenvectors, the transient term corresponding to the zero eigenvalue does not approach zero for large t.

A successful superposition formula for the system shown in Figure 6.10 includes a function of time which is allowed to grow without bound,

θ φ σ( , ) ( , ) ( ) ( )x t x t x t= + +Φ (6.67)

Superposition methods are not successful when the nonhomogeneous term is a function of time or a function of all spatial variables. In such cases, the nonhomogeneity can be addressed using eigenvector expansions. The following two examples illustrate this concept.

Example 6.8 The thin slab shown in Figure 6.11 is subject to an internal heat generation which is a function of x and y. The problem is formulated as

∂∂

+∂∂

=2

2

2

2

θ θx y

f x y( , ) (a)

θ( , )0 0y = (b)

α

y

x

f(x,y)

0

0

0

0

1

Figure 6.11 Internal heat generation leads to a nonhomogeneous differential equa-tion. The nonhomogeneity is a function of both spatial variables, and therefore a superposition in which the temperature is expanded in a series of eigenvectors of the corresponding homogeneous problem is used. Orthogonality of the eigenvectors is used to uncouple the differential equations for the coeffi cients in the expansion.

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θ( , )1 0y = (c)

θ( , )x 0 0= (d)

∂∂

=θ

αy

x( , ) 0 (e)

Determine the solution of Equation a, Equation b, Equation c, Equation d, and Equation e.

Solution Consider the eigenvalue problem, LX = − d2X/dx2 = λX, with X(0) = 0 and X(1) = 1. This is the eigenvalue problem obtained using a separa-tion of variables method a neutral if Equation a is homogeneous and at least one of Equation d and Equation e is nonho mogeneous. The eigenvalues and normalized eigenvectors are λn = n2 π2 and Xn(x) = 2sin(nπx). The eigenvec-tors are orthonormal with respect to the standard inner product on C[0,1].

For any y, θ(x,y) is in the domain of L. Thus it has an eigenvector expansion of the form

θ π( , ) ( ) sin( )x y A y n xn

n

==

∞

∑ 21

(f)

Substitution of Equation f into Equation a leads to

A y n n xd Ady

n xn

n

n( ) sin( ) sin( )−( ) +=

∞

∑ 2 2

1

2

22 2π π π

nn

nn

f x y

d Ady

n A

=

∞

∑ =

−⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

1

2

2

2 2 2

( , )

sinπ (( ) ( , )n x f x yn

π=

∞

∑ =1

(g)

Taking the standard inner product of both sides of Equation g with Xm(x) = 2sin(mπx) for an arbitrary m leads to

X Xd Ady

n A f An mn

n

n

m, ,( ) −⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟=

=

∞

∑2

2

2 2

1

π (( ) (h)

The eigenvectors are orthonormal, (Xn,Xm) = δn,m, and the sum collapses to a single term, resulting in

d Ady

m A g y mmm m

2

2

2 2 1 2− = =π ( ) , ,... (i)

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where

g y f X

f x y m x dx

m m( ) ,

( , ) sin( )

=( )

= ∫ 2

0

1

π (j)

Use of the eigenvector expansion leads to a set of uncoupled differential equations whose solutions are the expansion coeffi cients.

Substitution of Equation f into Equation d leads to

θ π( , ) ( ) sin( )x A n yn

n

0 0 0 21

= ==

∞

∑ (k)

The eigenvectors are linearly independent, and therefore a linear combina-tion can be set equal to zero only if each coeffi cient is zero. Thus, Equation k is valid only if

A nn( ) , ,0 0 1 2= = … (l)

Using a similar argument, Equation e is satisfi ed only if

dAdy

nn ( ) , ,...α = =0 1 2 (m)

The coeffi cients in the eigenvector expansion are obtained by solving Equa-tion g subject to Equation l and Equation m. The homogeneous solution of Equation g is

A C m y C m ym h, cosh( ) sinh( )= +1 2π π (n)

The particular solution is obtained using variation of parameters as

A y m y m g dm p m

y

, ( ) cosh( ) sinh( ) ( )

sin

= −

+

∫π πτ τ τ0

hh( ) cosh( ) ( )

( ) (

m y m g d

m y g

m

y

m

π πτ τ

π τ

τ0

∫

= −[ ]sinh ττ)d

y

τ0

∫

(o)

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The solution for Am(y) obtained by adding Equation n and Equation o and applying the initial conditions is

Am ym

m g dm m= −( )( )

−( )[ ]sinh

coshcosh ( )

ππα

π α τ τ τ0

αα

π τ τ τ

∫

∫+ −[ ] sinh ( ) ( )m y g dm

y

0

(p)

The solution for θ(x,y) is obtained by substitution of Equation j into Equation p and the result into Equation f:

θ πππα

( , ) sin( )sinh

coshx y n x

m ym

n

= −( )( )

⎧⎨⎪⎪

⎩⎪⎪2

==

∞

∑

∫∫× −( )[ ]

1

0

1

0

cosh sin(m nπ α τ

α

ππβ β τ β τ) ,

sinh

f d d( )

+ nn y n f d d

y

π τ πβ β τ β τ( ) sin ( , )−[ ] ( )⎫⎬⎪⎪⎪

⎭⎪⎪⎪

∫∫0

1

0

(q)

Example 6.9 Figure 6.12 illustrates a uniform circular shaft, fi xed at one end, with a thin disk attached at its other end. The system is at rest in equi-librium when a time-dependent torque is applied to the thin disk. The non-dimensional problem governing the resulting torsional oscillations of the shaft and disk is

∂∂

=∂∂

2

2

2

2

θ θx t

(a)

θ( , )0 0t = (b)

L

I

T(t)

ρ,G,J

Figure 6.12 The applied torque leads to a time-dependent term in the boundary condition specifi ed at the end of the shaft. An eigenvector expansion is used to deter-mine the time-dependent angular displacement of the shaft.

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∂∂

+∂∂

=θ

μθ

xt

tf t( , ) ( )1

2

2Λ (c)

θ( , )x 0 0= (d)

∂∂

=θt

( , )x 0 0 (e)

where β = I/ρJL and Λ = T0L/JG, and where T0 is a characteristic value of the applied torque. The nondimensionalization of the problem without the applied torque is developed in Example 5.10.

The problem specifi ed by Equation a, Equation b, Equation c, Equation d, and Equation e cannot be solved by separation of variables because the non-homogeneous term is in a boundary condition rather than an initial condition. Develop a modal analysis solution in which the forced response is obtained through an eigenvector expansion of mode shapes for the free response.

Solution The free response of the system is due to a nonzero initial condi-tion with no subsequent energy input to the system. The free response can be obtained using separation of variables assuming θ(x,t) = X(x)T(t). The result-ing eigenvalue problem defi ning X(x) is

d Xdx

X2

2= −λ (f)

X( )0 0= (g)

dXdx

X( ) ( )1 1= −μλ (h)

The solution of the eigenvalue problem specifi ed in Equation f, Equation g, and Equation h is obtained in Example 5.13. The eigenvalue λk is the kth posi-tive solution of

tan λμ λ

( )=1

(i)

Its corresponding eigenvector is

X x C xk k k( ) sin= ( )λ (j)

It is shown in Example 5.13 that the eigenvectors are orthogonal with respect to the inner product defi ned by

( , ) ( ) ( ) ( ) ( )f g f x g x dx f g= +∫0

1

1 1μ (k)

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Normalizing with respect to the inner product of Equation k leads to a set of orthonormal eigenvectors defi ned by

X x x kk

k

k( )sin

sin , ,..=+ ( )⎡

⎣⎢⎤⎦⎥

( ) =2

11 2

2

μ λλ .. (l)

The solution of Equation a, Equation b, Equation c, Equation d, and Equa-tion e, at every instant of time, satisfi es the same boundary conditions as the eigenvectors obtained from Equation f, Equation g, and Equation h. Since θ(x,t) is in the domain of the operator L as defi ned in Equation f, Equation g, and Equation h, application of the expansion theorem implies that there exist time-dependent coeffi cients, Ak(t), such that

θ( , ) ( ) ( )x t A t X xk k

k

==

∞

∑1

(m)

Substitution of Equation m into Equation a gives

Ad Xdx

d Adt

Xkk

k

kk

k

2

2

1

2

2

1=

∞

=

∞

∑ ∑= (n)

Substituting Equation f into Equation n and rearranging results yields

d Adt

A Xkk k k

k

2

2

1

0+⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟ =

=

∞

∑ λ (o)

Substitution of Equation m into the nonhomogeneous boundary condition, Equation c, leads to

A tdXdx

d Adt

X x f tkk

k

kk

k

( ) ( ) ( ) ( )11

2

2

1=

∞

=

∞

∑ ∑+ =μ Λ (p)

Substitution of Equation h into Equation p leads to

μ λd Adt

A X f tkk k k

k

2

2

1

1+⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟ =

=

∞

∑ ( ) ( )Λ (q)

Multiplying Equation o by Xm(x) and integrating from 0 to 1 leads to

d Adt

A X x X x dxkk k k m

k

2

2

0

1

1

+⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟∫

=

∞

λ ( ) ( )∑∑ = 0 (r)

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Mode-shape orthonormality with respect to the inner product of Equation i implies

X x X x dx X Xk m k m k m( ) ( ) ( ) ( ),

0

1

1 1∫ = −δ μ (s)

which when substituted into Equation r, leads to

d Adt

Ad Adt

A Xmm m

kk k k

2

2

2

21+ = +

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟λ μ λ ( )XXm

k

( )11=

∞

∑ (t)

Substitution of Equation q into Equation t gives

d Adt

A X f tmm m m

2

21+ =λ ( ) ( )Λ (u)

where Xm (1) is evaluated from Equation l, leading to

d Adt

A f tmm m m

m

2

2 2

2

1+ = ( )

+ ( )⎡⎣⎢

⎤⎦⎥

λ λμ λ

Λsinsin

( ) (v)

Initial conditions for Equation v are obtained from substitution of Equation m into Equation d and Equation e, leading to

Am( )0 0= (w)

dAdt

m ( )0 0= (x)

Suppose f(t) = sin(ωt). The solution of Equation v subject to the initial condi-tions of Equation w and Equation x is

6.5 Problems in cylindrical coordinatesCylindrical coordinates are illustrated in Figure 6.13. The coordinates r, θ and z are used to defi ne a point in the cylinder, or even exterior to the cylinder. The coordinate z is measured from a defi ned point along the axis of the cylinder. The coordinate r is the distance from the axis of the cylinder. The coordinate θ is measured circumferentially counter-clockwise from a defi ned plane which includes the z-axis. To describe all points on the surface and interior of a cylinder of length L and radius R0 0 0 2≤ ≤ ≤ ≤ ≤ ≤z L r R, and θ π.

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A differential volume within the cylinder is illustrated in Figure 6.14. The region is formed by fi rst taking a slice of the cylinder of thickness dz, with the slice cut perpendicular to the axis of the cylinder with faces parallel. Then an annular ring is cut from the slice. The inner radius of the ring is r and its outer radius r + dr. Finally, a sector of the ring is cut, defi ning a plug of volume dV. The angle of the sector is dθ with its right face along the radius defi ned by the angle θ. The area of the face of the sector is dA = rdrdθ, while the volume of the plug is dV = rdrdθdz.

z

r

(a)

y

z

x

rsinθr

(b)

rcosθ

θ

θ

Figure 6.13 (a) The cylindrical coordinate system is described by an axial coor-dinate, z, a radial coordinate, r, and a circumferential coordinate, θ. (b) Cylindrical coordinates can be converted to Cartesian coordinates and vice versa.

dz

dr

dθ

dr

dz

Figure 6.14 A differential volume of a cylinder is formed by taking a slice perpen-dicular to the axis of the cylinder, taking an annular ring from the slice, and then cutting the ring along an arc.

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Cylindrical coordinates are related to Cartesian coordinates through

x r= cosθ (6.68)

y r= sin θ (6.69)

z z= (6.70)

Differential equations written in cylindrical coordinates can be derived either by applying appropriate conservation laws to differential volumes, as in Example 1.4, or in the case of systems in which governing equations are written in vector operators, simply by using the following identities to convert the operators to cylindrical coordinates. The gradient in cylindrical coordinates is

∇ =∂∂

+∂∂

+∂∂

φφ φ

θr r z

e e er1 φ

θz (6.71)

The divergence operator written in cylindrical coordinates is

∇⋅ =∂∂

( )+∂∂

+∂∂

v1 1

r rrv

rv v

zrzθ

θ (6.72)

The Laplacian operator in cylindrical coordinates is

∇ =∂∂

∂∂

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+

∂∂

+∂∂

2

2

2

2

2

2

1 1φ

φ φθ

φr r

rr r z

(6.73)

Laplace’s equation, the diffusion equation, and the wave equation are all sep-arable in cylindrical coordinates. Nonhomogeneous equations can be solved using the methods of Section 6.4.

R

L

h,T 8Figure 6.15 The cylinder maintains a steady-state temperature distribution f(r,θ,z) when it is immersed in a medium with an ambient temperature T∞. The resulting unsteady state temperature distribution is obtained by solving the diffusion equa-tion in cylindrical coordinates.

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The operator − ∇2 is shown in Section 2.11 as being self-adjoint with respect to the inner product,

f r z g r z f r z g r z dV

f

V

( , , ), ( , , ) ( , , ), ( , , )θ θ θ θ( )=

=

∫

(( , , ), ( , , )r z g r z rdrd dzθ θ θ∫∫∫

(6.74)

The inner product for a full cylinder is

f r z g r z f r z g r z rdrd d( , , ), ( , , ) ( , , ), ( , , )θ θ θ θ θ( )= zz

RL

00

2

0

∫∫∫π

(6.75)

If the region over which the problem to be solved is a partial cylinder, then the limits of integration of Equation 6.75 must be adjusted accordingly.

It is also shown in Section 2.11 that − ∇2φ is positive defi nite with respect to the inner product of Equation 6.75 if φ or a linear combination of φ and ∇φ ⋅ n are specifi ed over a portion of the surface. If ∇φ ⋅ n is specifi ed everywhere on the surface, the operator is non-negative defi nite.

Example 6.10 The cylinder of Figure 6.15 has a temperature distribution T0f(r,θ,z) when it is immersed in a bath of temperature T∞. The ends of the cylinder are insulated, while its heat is transfered through its circumference with a heat transfer coeffi cient h. The following nondimensional variables are introduced:

rrR

* = (a)

zzR

* = (b)

tct

kR* =

ρ2 (c)

Λ( , , , )( , , , )* * *r z t

T r z t TT T

θθ

=−

−∞

∞0

(d)

where ρ is the mass density of the cylinder, c is its specifi c heat and k is its thermal conductivity.

It is not necessary to nondimensionalize the circumferential coordinate θ because it is measured in radians which are inherently nondimensional. Using the same length scale to nondimensionalize both r and z enables the nondimensional form of the Laplacian to be written as in Equation 6.73, understanding that r and z are nondimensional. Either L or R can be used to

9533_C006.indd 4749533_C006.indd 474 10/29/08 1:58:14 PM10/29/08 1:58:14 PM


defi ne the nondimensional variables. The use of R may be preferable because the ends at constant values of z are insulated, whereas the outer circumfer-ence of the cylinder is subject to convection heat transfer from the bath. In this case, the Biot number is a more accurate measure of the ratio of convec-tive heat transfer to the rate of conductive heat transfer.

The mathematical form of the problem in terms of nondimensional vari-ables with the *’s dropped is the partial differential equation,

∇ =∂∂

2ΛΛt

(e)

subject to the boundary conditions

∂∂

+ =Λ

Λr

z t Bi z t( , , , ) ( , , , )1 1 0θ θ (f)

∂∂

=Λz

r t( , , , )θ 0 0 (g)

∂∂

( )=Λz

r t, , ,θ α 0 (h)

where α = L/R. The initial condition is

Λ r z f r z, , , ( , , )θ θ0( )= (i)

In addition to satisfying Equation f, Equation g, Equation h, and Equation i, the temperature must be bounded and single-valued everywhere in the cylinder. Determine Λ(r,θ,z,t).

Solution The governing partial differential equation is homogeneous and separable. The boundary conditions are all homogeneous. Thus, separation of variables is applicable. To this end, assume a product solution of the form

Λ( , , , ) ( , , ) ( )r z t U r z T tθ θ= (j)

Substitution of Equation j into Equation e gives

T t U UdTdt

UU

TdTdt

( )∇ =

∇ =

2

21 1

(k)

Applying the usual separation argument to Equation k and defi ning the sep-aration constant as − λ leads to

9533_C006.indd 4759533_C006.indd 475 10/29/08 1:58:15 PM10/29/08 1:58:15 PM


dTdt

T+ =λ 0 (l)

and

∇ = −

∂∂

∂∂

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+

2

1

U U

r rr

Ur

λ

112

2

2

2

2rU U

zU

∂∂

+∂∂

= −θ

λ

(m)

Substitution of Equation l into the boundary conditions, Equation f, Equation g, and Equation h, results in

∂∂

( )+ ( )=Ur

z BiU z1 1 0, , , ,θ θ (n)

∂∂

( )=Uz

r, ,θ 0 0 (o)

∂∂

( )=Uz

r, ,θ α 0 (p)

Equation m is a separable partial differential equation and is subject to homogeneous boundary conditions. Assume a product solution of Equation m of the form

U r z R r Z z( , , ) ( ) ( )θ θ= ( )Θ (q)

Substitution of Equation q into Equation m gives

Θ Θ

ΘZr

ddr

rdRdr

RZr

dd

Rd Zdz

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+ + =

2

2

2

2

2θ−−λR ZΘ (r)

Equation r can be multiplied by r2/RΘZ and rearranged, resulting in

rR

ddr

rdRdr

rZ

d Zdz

rdd

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+ + = −

2 2

2

221

λΘ

Θθθ2 (s)

The usual separation argument is applied to Equation s with a separation constant of μ. The resulting equations are

dd

2

20

ΘΘ

θμ+ = (t)

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and

rR

ddr

rdRdr

rZ

d Zdz

r⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+ + −( )=

2 2

2

2 0λ μ (u)

Equation u can be divided by r2 and rearranged to yield

1 1

2

2

2rRddr

rdRdr r Z

d Zdz( )− = − −

μλ (v)

The usual separation argument is used for Equation v with a separation con-stant of − κ, resulting in

d Zdz

Z2

20+ −( ) =λ κ (w)

and

rddr

rdRdr

r R⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+ −( ) =κ μ2 0 (x)

Application of Equation q to Equation n, Equation o, and Equation p leads to

dRdr

BiR( ) ( )1 1 0+ = (y)

dZdz

( )0 0= (z)

dZdz

α( )= 0 (aa).

Separation of variables has been successfully applied. The values of the separation constant λ are the eigenvalues of − ∇2U subject to the boundary conditions, Equation n, Equation o, and Equation p, and the conditions that U remain fi nite and single-valued at every point in the cylinder. The eigen-value problem for − ∇2U can itself be solved using separation of variables. The values of the separation constant μ are the eigenvalues of − d2Θ/dθ2 subject only to the single-valuedness condition. This is a non-negative defi nite Sturm-Liouville problem, which was solved in Example 5.6 Thus there are an infi nite, but countable, number of values of μ: μ μ μ μ μ μ0 1 2 1 1, , , , , , ,...… − +n n n There are then an infi nite number of equations of the form of Equation x to solve for R(r) which is indexed using n. The values of κ are the eigenvalues of the operator

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− [ ]+1 2/ )/ /r d rdR dr r Rn n n( μ subject to Equation y and the requirement that R(r) be fi nite for all r. For each n, this is a Sturm-Liouville problem leading to an infi nite number of eigenvalues, κ κ κ κ κn n n m n m n m, , , , ,, , , , , ,1 2 1 1… …− +

with cor-responding eigenvectors. The values of β = λ − κ are the eigenvalues of − d2Z/dz2, subject to the boundary conditions, Equation z and Equation aa. Since this is also a Sturm-Liouville problem, there are an infi nite number of eigenvalues β β β β β β0 1 2 1 1, , , , , , ,… …− +p p p with corresponding eigenvectors. Thus the eigen-values of − ∇2U are of the form

λ β κn m p p n m, , ,= + (bb)

for n,p = 0,1,2,… and m = 1,2,… The details of the solutions follow.Let U r z R Zn m p n m n p, , ,( , , )θ = Θ be an eigenvector of − ∇2 corresponding to an

eigenvalue λn,m,p. The eigenvector is normalized using the inner product of Equation 6.75. Thus the normalization condition is

R r Z z rdrd dzn m n p, ( ) ( )Θ θ θ

πα

( )⎡⎣ ⎤⎦ =∫∫∫ 2

0

1

0

2

0

1 (cc)

Equation cc can be rewritten as

R r rdr dn m n, ( ) ( )[ ]⎧⎨⎪⎪⎪

⎩⎪⎪⎪

⎫⎬⎪⎪⎪

⎭⎪⎪⎪

[ ]∫ 2

0

1

2Θ θ θθ

π α

0

2

2

0

∫ ∫⎧⎨⎪⎪⎪

⎩⎪⎪⎪

⎫⎬⎪⎪⎪

⎭⎪⎪⎪

⎡⎣ ⎤⎦

⎧⎨⎪⎪

Z z dzp( )⎪⎪

⎩⎪⎪⎪

⎫⎬⎪⎪⎪

⎭⎪⎪⎪

= 1 (dd)

Equation dd shows that if each term in the product is normalized individu-ally, then the eigenvector U r z R Zn m p n m n p, , ,( , , )θ = Θ is normalized. This is the approach taken below.

The solution of Equation t is

Θ( ) cos sinθ μθ μθ= ( )+ ( )C C1 2 (ee)

The two sets of coordinates, (r,θ,z) and (r,θ + 2π,z), represent the same point. Single-valuedness of the temperature requires Θ(θ) = Θ(θ + 2π). Single-valuedness of the rate of heat transfer, q T e r Tθ θ θ= ∇ ⋅ = ∂ ∂( )1/ / , requires d d d dΘ Θ/ /θ θ θ θ π( ) ( )= + 2 . Thus the eigenvector and its derivative must be periodic of period 2π. These are the conditions considered in Example 5.6. The eigenvalues for the periodic response are μn = n2 n = 0,1,2,… The normal-ized eigenvector corresponding to μ = 0 is

Θ0

1

2θ

π( )= (ff)

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An eigenvalue μn = n2 n = 1,2,… has two linearly independent eigenvectors. These normalized eigenvectors are

Θ1

1n n= ( )

πθcos (gg)

Θ2

1n n= ( )

πθsin (hh)

For μn = n2, Equation x becomes

rddr

rdRdr

r n Rnn

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+ −( ) =κ 2 2 0 (ii)

Equation ii can be recognized as Bessel’s equation of order n with a general solution of

R r D J r D Y rn n n n n( ) = ( )+ ( )1 2κ κ (jj)

Rn(0) is fi nite only at n=0 if D2n = 0, leading to Rn(r) = D1nJn( κr). Application of Equation y leads to the transcendental equation whose solutions defi ne the eigenvalues by

κ κ κ′ ( )+ ( )=J BiJn n 0 (kk)

For each n, Equation kk has an infi nite number of solutions indexed as κ κ κ κ κn n n m n m n m, , , , ,1 2 1 1< <…< < < <…− + The corresponding normalized eigenvectors are

R r D J rn m n m n n m, , ,( ) = ( )κ (ll)

where

D

J r rdr

n m

n n m

,

,

=

( )⎡⎣

⎤⎦

⎧⎨⎪⎪⎩⎪⎪

⎫⎬⎪⎪⎭⎪⎪

=

∫

1

2

0

11

2

κ

1

2

2

2+⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

⎡

⎣⎢⎢⎢

⎤

⎦⎥⎥⎥

( )Bi

Jn m

n n mκ

κ,

,

⎧⎧⎨⎪⎪

⎩⎪⎪

⎫⎬⎪⎪

⎭⎪⎪

−1

2

(mm)

9533_C006.indd 4799533_C006.indd 479 10/29/08 1:58:20 PM10/29/08 1:58:20 PM


The eigenvalues of the problems defi ned by Equation w, Equation z, and Equation aa are

βπ

αpp

p=⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟ =

2

0 1 2, , ,... (nn)

The corresponding normalized eigenvectors are

Z z0

1( ) =

α (oo)

Z zp

zp( ) cos=⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟

2

απ

α (pp)

Note that the problem defi ning Z(z) is non-negative defi nite because its low-est eigenvalue is zero.

Summarizing, the eigenvalues for U are

λπ

ακn m p n m

p, , ,=

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟ +

2

(qq)

where κn.m is the mth positive solution of Equation kk for each n. The ranges of the indices in Equation qq are n, p = 0,1,2,… and m = 1,2,… The correspond-ing eigenvectors are

U D J r mm m m0 0 0 0 0

1

21 2, , , , , ,...= ( ) =

πακ (rr1)

U D J rp

z mm p m m0 0 0 0

1, , , , cos ,= ( ) ⎛

⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟πα

κπ

αpp = 1 2, ,... (rr2)

U D J r n n mn m n m n n m, , , , , cos , , ,...0 1

11 2= ( ) ( ) =

πακ θ (rr3)

U D J r n n mn m n m n n m, , , , , sin , , ,...0 2

11 2= ( ) ( ) =

πακ θ (rr3)

U D J r np

z nn m p n m n n m, , , , , cos cos ,12

= ( ) ( ) ( )πακ θ

πα

mm p, , ,...= 1 2 (rr5)

U D J r np

z nn m p n m n n m, , , , , sin cos ,22

= ( ) ( ) ( )πακ θ

πα

mm p, , ,...= 1 2 (rr6)

9533_C006.indd 4809533_C006.indd 480 10/29/08 1:58:21 PM10/29/08 1:58:21 PM


The solution of Equation l leads to

T t A en m p n m ptn m p

, , , ,( ) , ,= −λ (ss)

The general solution is a linear combination of all homogeneous solutions

Λ( , , , ) ( , , ), , , , ,, ,r z t A e U r z Am

tm

mθ θλ= +−0 0 0 0 0

0 0mm p

tm p

pm

e U r zm p, , ,

, , ( , , )−

=

∞

=∑

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

λ θ00

111

00

∞

−

∑

+ e A Un m tn m n m

λ , ,, , , ,, , , , , , ,( , , ) ( , , )0 1 0 0 2

1

r z B U r zn m n m

m

θ θ+[ ]⎧⎨

=

∞

∑⎪⎪⎪

⎩⎪⎪

+

=

∞

=

∞−

∑

∑

n

m

e n m

1

1

λ , ,pptn m p n m p n m p n m pA U r z B U r, , , , , , , , , ,( , , ) ( ,1 2θ θ+ ,, )z

p

[ ]⎫⎬⎪⎪⎪

⎭⎪⎪⎪=

∞

∑1

(tt)

Application of the initial condition, Equation i, to Equation tt leads to

f r z A U r z A U rm m m p m p( , , ) ( , , ) (, , , , , , , ,θ θ= +0 0 0 0 0 0 ,, , )θ zpm =

∞

=

∞

∑∑⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

+

11

AA U r z B U r zn m n m n m n m, , , , , , , , , ,( , , ) ( , ,0 0 1 0 0 2θ θ+ ))[ ]⎧⎨⎪⎪

⎩⎪⎪

+

=

∞

=

∞

=

∞

∑∑

∑

mn

m

11

1

AA U r z B U r zn m p n m p n m p n m p, , , , , , , , , ,( , , ) ( , ,1 2θ θ+ ))[ ]⎫⎬⎪⎪⎪

⎭⎪⎪⎪=

∞

∑p 1

(uu)

The coeffi cients in the summations on the right-hand side of Equation uu are the coeffi cients in the eigenvector expansion for f (r,θ,z). For example,

A D J r np

n m p n m n m, , , , , cos( )cos1 0

2= ( ) ⎛

⎝⎜⎜⎜πα

κ θπ

α⎞⎞⎠⎟⎟⎟⎟∫∫∫ f r z rdrd dz( , , )θ θ

πα

0

1

0

2

0

(vv)

Example 6.11 A sector of the membrane shown in Figure 6.16 has been made from fi bers of a nanomaterial. The natural frequencies of the mem-brane when it is clamped over its entire perimeter are to be measured and the data used to predict the properties of the material. The differential equa-tion governing the vibrations of the membrane is

T wwt

∇ =∂∂

22

2ρ (a)

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where w(r,θ,t) is the displacement of the membrane, T is the tension in the membrane, and ρ is its mass density. The boundary conditions are

w R t( , , )θ = 0 (b)

w r t( , , )0 0= (c)

w r t, ,π6

0⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟= (d)

Nondimensional variables defi ned by r r R w w R t t T R* * *, / ,= = =/ and /ρ 2 can be substituted into Equation a, Equation b, Equation c, and Equation d, leading to

∇ =∂∂

22

2w

wt

(e)

w t( , , )1 0θ = (f)

π6

Figure 6.16 The natural frequencies of the membrane of Example 6.11 are obtained using separation of variables.

9533_C006.indd 4829533_C006.indd 482 10/29/08 1:58:23 PM10/29/08 1:58:23 PM


where *’s have been dropped from nondimensional variables. The boundary conditions represented by Equation c and Equation d have the same equa-tions in nondimensional form. Determine the natural frequencies and nor-malized mode shapes of the membrane.

Solution A product solution of Equation e is assumed to be of the form

w r t Q t U r( , , ) ( ) ( , )θ θ= (g)

Substitution of Equation g into Equation e, rearranging, and using the usual separation argument leads to

d Qdt

Q2

20+ =λ (h)

−∇ =2U Uλ (i)

where λ is a separation constant. Noting that the general solution of Equa-tion h is Q C t C t= +1 2cos( ) sin( )λ λ it is clear that the natural frequencies are the square roots of the values of the separation constants, which are the eigenvalues of − ∇2Q subject to Q Q r Q r( , ) , ( , ) , .1 0 0 0 0θ π/6= = ( )=and

To solve Equation i, a product solution is assumed of the form

Q r R r( , ) ( ) ( )θ θ= Θ (j)

Substitution of Equation j into Equation i leads to

1 1

2

2

2rddr

rdRdr r

R tdd

R⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟ + +Θ Θ( ) ( )θ

Θλ

θ== 0 (k)

Equation k can be rearranged to yield

rR

ddr

rdRdr

rR

ddr

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+ = −

λ 2 2

2

1

ΘΘ

(l)

The usual separation argument is used on Equation l with a separation con-stant μ. The resulting differential equations are

dd

2

20

ΘΘ

θ+ =μ (m)

rddr

rdRdr

r R⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+ −( ) =λ μ2 0 (n)

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Use of the product solution in the boundary conditions leads to

R( )1 0= (o)

Θ(0) 0= (p)

Θπ6

0⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟= (q)

The general solution for Equation m is

Θ( ) cos sinθ = ( )+ ( )C C1 2μθ μθ (r)

Application of Equation p to Equation q leads to C1 = 0. Application of Equa-tion q to Equation r with C1 = 0 leads to

sin μπ6

0⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟= (s)

Equation r is satisfi ed for

μn n n= =36 2 1,2,... (t)

The eigenvectors can be normalized by requiring that ∫ [ ] =0

6

22 2

6 1π/

sin( )C n dθ θ which leads C2 12= /π . The normalized eigenvectors are

Θ( ) sin( )θπ

θ=12

6n (u)

Substitution of Equation t into Equation n leads to

rddr

rdRdr

r n Rnn

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+ −( ) =λ 2 236 0 (v)

The solution of Equation v is

R r D J r D Y rn n n n n( ) = ( )+ ( )1 6 2 6λ λ (w)

The solution is fi nite at r = 0 only if D n2 0= . Subsequent application of Equa-tion o to Equation w leads to

J n6 0λ( )= (x)

9533_C006.indd 4849533_C006.indd 484 10/29/08 1:58:24 PM10/29/08 1:58:24 PM


There are an infi nite, but countable, number of solutions of Equation x for each n. These are indexed as λn,m, which is the mth solution of Equation x for a specifi c n. The eigenvectors are normalized by choosing

D J r rdrn n n m1 6

2

0

11

2

, ,= ( )⎡⎣⎢

⎤⎦⎥

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

∫−

λ

==( )⎡

⎣⎢⎤⎦⎥ − ( ) ( )

⎧

− +

2

6

2

6 1 6 1J J Jn n m n n m n n mλ λ λ, , ,

⎨⎨⎪⎪⎪⎪

⎩⎪⎪⎪⎪

⎫

⎬⎪⎪⎪⎪

⎭⎪⎪⎪⎪

1

2

(y)

It is clear from Equation h that the natural frequencies are the square roots of the eigenvalues of U, ω λm n m n, ,= . The fi rst three natural frequencies for the fi rst fi ve values of n are given in Table 6.2.

Example 6.12 The cylinder shown in Figure 6.17 is at a uniform tem-perature T0 when a uniform heat fl ux is suddenly applied to the right end. The circumference of the cylinder is maintained at T0, while the left end of the cylinder is insulated. Determine the resulting unsteady state tempera-ture distribution in the cylinder.

T0

R

k,ρ,c

q

L

Figure 6.17 The cylinder of Example 6.12 is at a uniform temperature when a heat fl ux begins at its left end. The resulting problem for the unsteady-state temperature distribution is nonhomogeneous, and therefore a superposition solution is used.

Table 6.2 Natural Frequencies of membrane of Example 6.11

n 1 2 3

ωn,1 9.936 16.698 23.257

ωn,2 17.004 20.79 27.698

ωn,3 20.321 24.495 31.65

ωn,4 23.586 24.495 35.375

ωn,5 26.82 31.46 38.965

9533_C006.indd 4859533_C006.indd 485 10/29/08 1:58:25 PM10/29/08 1:58:25 PM


Solution Since the initial temperature is uniform and the applied heat fl ux is uniform over the end of the cylinder, the temperature is constant circumferentially for every z and every r and thus independent of θ. The mathematical formulation for the nondimensional temperature distribu-tion, Θ( , , ) ( ( , , ) )/r z t T r z t T T= − 0 0, in terms of the nondimensional varibles r r L z z L t kt cL* * */ , / /= = =and ρ 2 is

∂∂

+∂∂

+∂∂

=∂∂

2

2

2

2

1Θ Θ Θ Θr r r z t

(a)

Θ( , , )r z 0 0= (b)

ΘRL

z t, ,⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟= 0 (c)

∂∂

=Θz

r t( , , )0 0 (d)

∂∂

=Θz

r t( , , )1 α (e)

where α = qL kT/ 0

The boundary condition at z = 1, Equation e, is nonhomogeneous, and therefore the system of Equation a, Equation b, Equation c, Equation d, and Equation e is nonhomogeneous. Heat can escape from the cylinder through its circumference, and therefore the temperature should approach a steady-state distribution. This suggests that the temperature is a super-position of an unsteady state or transient solution and a steady-state solu-tion of the form

Θ Θ Θ( , , ) ( , ) ( , , )r z t r z r z ts t= + (f)

Substitution of Equation f into Equation a, Equation b, Equation c, Equation d, and Equation e leads to

∂∂

+∂∂

+∂∂

+∂∂

+∂∂

+∂2

2

2

2

2

2

21 1Θ Θ Θ Θ Θ Θs s s t t

r r r z r r rtt t

z t∂=

∂∂2

Θ (g)

Θ Θs tr z r z( , ) ( , , )+ =0 0 (h)

Θ Θs sRL

zRL

z t, , ,⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟= 0 (i)

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∂∂

( )+∂∂

( )=Θ Θs t

zr

zr t, , ,0 0 0 (j)

∂∂

( )+∂∂

( )=Θ Θs t

zr

zr t, , ,1 1 α (k)

The goal of the superposition is to choose a problem for Θt (r,z,t) which can be solved directly using separation of variables. The problem must consist of a homogeneous partial differential equation, a nonhomogeneous initial condition, and homogeneous boundary conditions. The following choice of problem for Θt (r,z,t) meets these requirements:

∂∂

+∂∂

+∂∂

=∂∂

2

2

2

2

1Θ Θ Θ Θt t t t

r r r z t (l)

Θ Θt sr z r z, , ,0( )= − ( ) (m)

ΘtRL

z t, ,⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟= 0 (n)

∂∂

( )=Θt

zr t, ,0 0 (o)

∂∂

( )=Θt

zr t, ,1 0 (p)

Substitution of Equation l, Equation m, Equation n, Equation o, and Equa-tion p into Equation g, Equation h, Equation i, Equation j, Equation k, and Equation l leads to formulating the problem to be solved for the steady-state response as

∂∂

+∂∂

+∂∂

=2

2

2

2

10

Θ Θ Θs s s

r r r z (q)

ΘsRL

z t, ,⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟= 0 (r)

∂∂

( )=Θs

zr t, ,0 0 (s)

∂∂

( )=Θs

zr t, ,1 α (t)

9533_C006.indd 4879533_C006.indd 487 10/29/08 1:58:27 PM10/29/08 1:58:27 PM


The steady-state solution can be obtained by solve Equation q subject to Equation r, Equation s, and Equation t, direct application of separation of variables.

A product solution of Equation q is assumed as

Θs r z R r Z z( , ) ( ) ( )= (u)

Substituting Equation u into Equation q, rearranging, and using the usual separation argument, noting that since the nonhomogeneous boundary con-dition is applied at a constant value of z, the Sturm-Liouville problem must be in the r-coordinate, leads to

d Zdz

Z2

20− =λ (v)

rd Rdr

rdRdr

r R22

2

2 0+ + =λ (w)

RRL

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟= 0 (x)

The nontrivial solutions of Equation w and Equation x subject to the con-straint that the temperature remains fi nite at r = 0 are

R r A J rn n n( ) = ( )0 λ (y)

where the eigenvalue λn is the nth solution of

JRL0 0λ

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟= (z)

and

A J r rdrn n

RL

= ( )⎡⎣

⎤⎦

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

=

∫

−

0

2

0

1

2

λ

λλ

λ

n L

RJRL

n1

1

2

( )⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

(aa)

9533_C006.indd 4889533_C006.indd 488 10/29/08 1:58:28 PM10/29/08 1:58:28 PM


The solution of Equation v which satisfi es the boundary condition at z = 0 is

Z z C zn n n( ) cosh= ( )λ (bb)

The boundary condition at z = 1 is satisfi ed using an eigenvector expansion for f (r) = α, leading to

C R r

A J

n

n n

n r

n n

n n

=( )

( )

=( )

1

10

λ λα

λ λα λ

sinh, ( )

sinhrr rdr

RA JRL

L

RL

n n

n n

( )

=

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟

∫0

1α λ

λ λsinh(( )

(cc)

The steady-state solution is

Θs n n n n

n

r z C z A J r( , ) cosh= ( ) ( )=

∞

∑ λ λ0

1

(dd)

The transient response is obtained through solution of Equation l, Equation m, Equation n, Equation o, and Equation p using separation of variables and using Equation dd to impose the initial condition. To this end, a product solution of Equation l is assumed to be of the form

Θt r z t R r Z z T t( , , ) ˆ ( ) ˆ ( ) ˆ( )= (ee)

Substituting Equation ee into Equation l, rearranging, and using the usual separation arguments leads to

dTdt

Tˆ

ˆ+ =μ 0 (ff)

rd Rdr

rdRdr

r R22

2

2 0ˆ ˆ

ˆ+ + =κ (gg)

d Zdz

Z2

20

ˆˆ+ −( ) =μ κ (hh)

9533_C006.indd 4899533_C006.indd 489 10/29/08 1:58:29 PM10/29/08 1:58:29 PM


It should be noted that the problem defi ning ˆ ( )R r is the same as that defi n-ing R(r), and therefore ˆ ( ) ( )R r R rn n= , which is stated in Equation y, Equa-tion z, and Equation aa. The corresponding normalized solution for ˆ ( )Z zm which satisfi es Equation hh and dZ dz dZ dzˆ/ ( ) ˆ/ ( )0 1 0= = is

ˆ ( )Z z0 1= (ii)

ˆ ( ) cosZ z m z mm = ( ) =

1

2π 1,2,3,... (jj)

The eigenvalues are of the form

μ λ πn m n m, = + 2 2 (kk)

The solutions of Equation ff are of the form

ˆ ( ), ,,T t F en m n m

tn m= −μ (ll)

The general form of the transient solution is the sum of all possible solutions, which, using Equation aa, Equation jj, and Equation ll, is

Θt n n

n

nt

n

r z t A J r

F e Fn

, ,

, ,

( )= ( )⎡

⎣⎢⎢⎢

× +

=

∞

−

∑ 0

1

0

λ

λmm

t

m

m z e n m1

21

cos ,π μ( )⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

⎤

⎦⎥⎥⎥

−

=

∞

∑ (mm)

Application of the initial condition leads to

− ( ) ( )= ( )⎡⎣

=

∞

=

∞

∑ ∑C z A J r A J ri i i i

i

n n

n

cosh λ λ λ0

1

0

1

×× + ( )⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

⎤

⎦⎥⎥

=

∞

∑F F m zn n m

m

, , cos0

1

1

2π

⎥⎥ (nn)

The left-hand side of Equation nn has an eigenvector expansion in terms of the eigenvectors obtained for the transient solution, which leads to

F C z A J r A J rn i i i i n n

i

, cosh ,0 0 0

1

= − ( ) ( ) ( )⎛

⎝ =

∞

∑ λ λ λ⎜⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

= − ( ) ( ) (A A C z J r J ri n i i i ncosh λ λ λ0 0 ))∫∫∑=

∞

rdrdz

RL

i 00

1

1

(oo)

9533_C006.indd 4909533_C006.indd 490 10/29/08 1:58:30 PM10/29/08 1:58:30 PM


Application of orthonormality of eigenvectors in the r-direction,

∫ =0 0 0

R Li n i n i nA A J r J r rdr

/

,( ) ( )λ λ δ , reduces Equation oo to

F C z dz

C

n n n

n

n

n

, sinh

cosh

0

0

1

1

= − ( )

= −⎡⎣⎢

⎤⎦⎥

∫ λ

λλ (pp)

Using a similar analysis,

F C z A J rA

J r m zn m i i i in

n, cosh , cos= − ( ) ( ) ( )λ λ λ π0 02

(( )⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟

= − ( )

=

∞

∑j

i n ii

A A Cz

1

2cosh cλ oos m z J r J r rdrdz

C

i n

RL

i

π λ λ( ) ( ) ( )

= −

∫∫∑=

∞

0 0

00

1

1

nnn

mn n

n

z m z dx

C

2

1

0

1

1

cosh cos( )

( ) sinh

λ π

λλ

( )

=−

∫+

++ m2 2π

(qq)

6.6 Problems in spherical coordinatesThe spherical coordinate system is illustrated in Figure 6.18. The coordinates r, θ and φ are used to locate a point uniquely in the sphere. The radial coor-dinate r is the distance from the center of the sphere to a point such that 0 ≤ r ≤ R. The circumferential coordinate θ is the angle measured counter-clockwise within a plane passing through the center of the sphere from a reference line to the projection of the radius to the point on the plane such that 0 ≤ θ ≤ 2π. The azimuthal coordinate φ is the angle made by the pro-jection to the radius. The azimuthal coordinate may be defi ned by either − π/2 ≤ φ ≤ π/2 or 0 ≤ φ ≤ π.

Spherical coordinates can be converted to Cartesian coordinates by

x r= cos cosφ θ (6.76)

y r= cos sinφ θ (6.77)

z r= sinφ (6.78)

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A differential volume within the sphere is illustrated in Figure 6.19. The vol-ume of this element is

dV r drd d= 2 sinφ θ φ (6.79)

y

x

z

(r,θ,φ)

φ

θ

Figure 6.18 The coordinates of a point in a spherical coordinate system are rep-resented in terms of a radial coordinate, r, a circumferential coordinate, θ, and an azimuthal coordinate, φ.

dθ

dφ

rdφ

rsinφdθ

dr

Figure 6.19 The volume of a differential element in a spherical coordinate system is dV = r2 sinφdrdθdφ.

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Vector operators written in spherical coordinates are

∇ =∂∂

+∂∂

+∂∂

ΛΛ Λ Λr r r

e e er

1 1

sinφ φθ φθ (6.80)

∇⋅ =∂∂

( )+∂∂

+∂

∂G

1 1 12

2

r rr G

rG

rGr

sin sinsin

φ φφθ

θ φ φφ( ) (6.81)

∇ =∂∂

∂∂( )+

∂∂

+∂

∂2

22

2 2

2

2 2

1 1 1Λ

Λ Λr r

rr r rsin sinφ θ φ φφ

sinφφ

∂∂

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟

Λ (6.82)

Laplace’s equation, the diffusion equation, and the wave equation are all separable in spherical coordinates.

The operator − ∇

2 was shown in Example 2.31 to be self-adjoint with respect to the inner product of Equation a of that example. When the bound-ary of the region over which − ∇

2 is bounded by r = 1 with 0 ≤ θ ≤ 2π and 0 ≤ φ ≤ π, the inner product is defi ned as

f r g r f r g r r, , , , , , , , , sinθ φ θ φ θ φ θ φ( ) ( )( )= ( ) ( ) 2

0

1

∫∫∫∫0

2

0

ππ

φ θ φdrd d (6.83)

Example 6.13 The surface of the sphere shown in Figure 6.20 is maintained at a constant temperature distribution, f(θ,φ). Determine the steady-state non-dimensional temperature, Λ(r,θ,φ). Assume that the radial coordinate is non-dimensional with r* = r/R.

Solution Laplace’s equation in spherical coordinates reduces to

∂∂

∂∂

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+

∂∂

+∂

∂rr

r2

2

2

2

1 1Λ Λsin sin

sφ θ φ φ

iinφφ

∂∂

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟ =

Λ0 (a)

T=f(θ,φ)

Figure 6.20 The surface of the sphere is maintained at a constant temperature distribution.

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The nonhomogeneous boundary condition is at r = 1. Therefore, a product solution can be assumed to be of the form

Λ( , , ) ( ) ,r R r Uθ φ θ φ= ( ) (b)

Substitution of Equation b into Equation a leads to

Uddr

rdRdr

R U R2

2

2

2

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+

∂∂

+∂

∂sin sinφ θ φ φssinφ

φ∂∂

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟=

U0 (c)

Dividing Equation c by RU and rearranging leads to

1 1 12

2

2

2Rddr

rdRdr U

UU

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟= −

∂∂

−sin sinφ θ φφ φ

φφ

∂∂

∂∂

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟sin

U (d)

Applying the usual separation argument to Equation d and defi ning the separation constant as λ leads to

rd Rdr

rdRdr

R22

20+ − =λ (e)

−∂∂

−∂

∂∂∂

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟=

1 12

2

2sin sinsin

φ θ φ φφ

φλ

U UUU (f)

It is noted that Equation e is a Cauchy-Euler equation (Section 3.5).A product solution for U is assumed to be of the form

U( , ) ( ) ( )θ φ φ= Θ Φθ (g)

Substitution of Equation g into Equation f and rearranging gives

− =⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟−

1 2

2

2

ΘΘ

ΦΦd

dd

dddθ φ φ

sinsin sin

φφ λ φφ (h)

The usual separation argument is used on Equation h, with μ defi ned as the separation constant, resulting in

1

sinsin

φ φφ

φλ−

μφ

dd

ddΦ⎛

⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+

⎛

⎝⎜⎜⎜

⎞

⎠sin2⎟⎟⎟⎟⎟ =Φ 0 (i)

dd

2

20

ΘΘ

θ+ =μ (j)

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Solutions of Equation e, Equation i, and Equation j are required to be fi nite and single-valued at all points in the sphere.

The coordinates (r,θ + 2nπ,φ) for any integer value of n refer to the same point as the coordinates (r,θ,φ) Thus, for the temperature and rate of heat transfer to be single-valued, Θ(θ) and dΘ/dθ must be periodic of period 2π. Equation j with the periodicity conditions is the eigenvalue problem which defi nes the trigonometric Fourier series. Its eigenvalues, obtained in Section 5.5 and used in Section 6.2, are μn n= 2 for n = 0,1,2,…. The normalized eigen-vector corresponding to n = 0 is

Θ0

1

2=

π (k)

Each eigenvalue for n > 0 has two linearly independent normalized eigen-vectors given by

Θn n, ( ) cos1

1θ

πθ= ( ) (l1)

Θn n, ( ) sin2

1θ

πθ= ( ) (l2)

Equation i can be written as an eigenvalue problem for each value of n,

−⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+ =

1 2

sinsin

φ φφ

φ φd

ddd

nnn

ΦΦ

sin2λ (m)

Equation n is of the form of a Sturm-Liouville problem with p(φ) = − sinφ, q(φ) = n2/sin2φ, and r(φ) = sinφ. For each n, there are an infi nite, but count-able, number of eigenvalues, λn,m for m = 1,2,…. The eigenvectors Φn,m and Φn,p, respectively corresponding to λn,m and λn,p, satisfy the orthogonality condi-tion (Φn,m,Φn,p) = 0, where

Φ Φ Φ Φn m n p n m n p d, , , ,, sin( ) = ( ) ( )∫φ

π

φφ φ φ0

2

(n)

Since 0 ≤ φ ≤ π and p(0) = p(π) = 0, then according to theorem 5.7 the eigen-vectors must only satisfy the conditions that they are fi nite at φ = 0 and φ = π.

Using the change of independent variable, s = cosφ, Equation n can be rewritten as

dds

sdds

ns

n11

22

2−( )⎡

⎣⎢⎢

⎤

⎦⎥⎥+ −

−

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

Φμ ΦΦn = 0 (o)

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Equation o is the associated Legendre’s equation which is solved in Section 3.7. Unless μ = m(m + 1), its solution is not fi nite at s = ±1, which correspond to φ = 0 and φ = π. For μ = m(m + 1), the general solution of Equation o is

Φn m mn

mnC P s C Q s, ( ) ( )= +1 1 ,

where P smn( ) and Q sm

n ( ) are the associated Legendre functions of the fi rst and second kinds of order m and index n. Noting that Q sm

n ( ) is not fi nite at s = ±1, C2 = 0. Thus

Φn m n m mnC P, , cosφ φ φ( )= ( ) ( )sinn

(p)

The eigenvectors can be normalized by choosing

C P dn m mn n

, (cos ) sin= ⎡⎣⎢

⎤⎦⎥

⎧⎨⎪⎪⎪

⎩⎪⎪⎪

+∫ φ φ2 2 1

0

φ

π ⎫⎫⎬⎪⎪⎪

⎭⎪⎪⎪

−1

2

(q)

The differential equation for R(r) now becomes

rd Rdr

rdRdr

m m Rm mm

22

21 0+ − + =( ) (r)

As discussed in Section 3.5, solution of Equation r is assumed as Rm = rα, which when substituted into Equation r, leads to a quadratic equation whose roots are the values of α. The general solution is then obtained as

R r A r B rm mm

mm( ) ( )= + − +1 (s)

The solution must remain fi nite at the center of the sphere, r = 0, requiring Bm = 0.

Using Equation b, Equation g, Equation k, Equation l, Equation m, Equa-tion p, and Equation s, the general solution of Equation a is

Λ r AC

P rmm

m

mm, , cos,

,θ φ φ( )=⎧⎨⎪⎪⎩⎪⎪

( )=

∞

∑ 00

0 2π

+ ( ) ( )CP r An m

mn n m

n,

,cos sinπ

φ φ mm n m

n

Bcos sin,θ θ+( )=

∞

∑1

(t)

The boundary condition Λ(1,θ,φ) = f(θ,φ) is applied using an eigenvector expansion for f (θ,φ). The constants in Equation t can be obtained as

A fC

PC

fmm

mm

00 0

2 2,

, ,, cos (= ( )⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟ =

πφ

πθ,, ) (cos )sinφ φ φP d dm φ θ

ππ

00

2

∫∫ (u)

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A fC

Pn mn m

mn n

,,, cos sin cos= ( ) ( )

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟π

φ φ θ

== ( ) +Cf P d dn m

mn n n, ( , ) (cos )sin cos sin

πθ φ φ φ θ φ1 φ θθ

ππ

00

2

∫∫ (v)

B fC

Pn mn m

mn n

,,, cos sin sin= ( ) ( )

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟π

φ φ θ

== +∫∫C

f P d dn mm

n, ( , ) (cos )sin sinπ

θ φ φ θ φ φ1

00

2

θ

ππ

(w)

Example 6.14 The sphere of Example 6.13 is at a steady state with a tem-perature distribution given by Equation t, Equation u, Equation v, and Equa-tion w when it is suddenly placed in a medium of temperature T∞. The heat transfer coeffi cient between the sphere and the ambient is h. The nondimen-sional problem formulation for the resulting nondimensional temperature, Λ(r,θ,φ,t), is

∇ =∂∂

2ΛΛt

(a)

∂∂

( )+ ( )ΛΛ

rt Bi t1 1, , , , , ,θ φ θ φ (b)

Λ r g r, , , , ,θ φ θ φ0( )= ( ) (c)

where g (r,θ,φ) is Λ (1,θ,θ) of Example 6.13 adjusted for a change in the defi nition in nondimensional temperature. Determine Λ(r,θ,φ,t).

Solution A product solution of Equation a can be assumed to be of the form

Λ( , , , ) ( ) ( , , )r t T t U rθ φ θ φ= (d)

Substitution of Equation d into Equation a and Equation b, rearrangement of the resulting equation, and application of the usual separation argument with the separation parameter defi ned as λ leads to

dTdt

T+ =λ 0 (e)

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−∇ =2U Uλ (f)

∂∂

+ =Ur

BiU( , , ) ( , , )1 1 0θ φ θ φ (g)

The values of the separation constant are the values of the eigenvalues of − ∇2U subject to Equation g. The eigenvectors must be fi nite and single-valued at all points in the sphere. Eigenvectors corresponding to distinct eigenvectors are orthogonal with respect to the inner product,

f r g r f r g r( , , ), ( , , ) ( , , ) ( , , )θ φ θ φ θ φ θ φ( )= ∫∫0

1

0

2π

00

2

π

∫ r drd dsinφ θ φ (h)

A product solution of Equation f is assumed to be of the form

U r R r( , , ) ( ) ( ) ( )θ φ θ φ= Θ Φ (i)

Substituting Equation i into Equation f with ∇2Λ obtained from Equation 6.82 gives

ΘΦ Φ

φΘθ

Θr

ddr

rdRdr

Rr

dd

R2

2

2 2

2

2

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+ +

sin rrd

ddd

R2 sin

sinφ φ

φΦφ

λ Φ⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟= − Θ (j)

Multiplying Equation j by r2/RΘΦ and rearranging leads to

−⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟− = +

1 1 12 2

2

2

2Rddr

rdRdr

rdd

λφΘ

Θθsin ssin

sinφΦ φ

φΦφ

dd

dd

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟ (k)

Applying the usual separation argument to Equation k and denoting the separation constant as μ results in

ddr

rdRdr

r R2 2⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+ −( )λ μ (l)

1 12

2

2sin sinsin

φ θ φΦ φφ

ΦφΘ

Θdd

dd

dd

+⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟= −−μ (m)

Multiplying Equation m by sin2 φ and rearranging leads to

− =⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+

1 2

2

2

ΘΘθ

φΦ φ

φΦφ

μdd

dd

dd

sinsin sin φφ (n)

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Applying the usual separation argument to Equation n and denoting the separation constant as κ results in

dd

2

20

Θθ

κΘ+ = (o)

1

2sinsin

sinφ φφ

Φφ

μκ

φd

ddd

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+ −

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟ =Φ 0 (p)

The values of the separation constant κ are the eigenvalues of Equation o subject to periodicity conditions. The eigenvalues are κn = n2 and the corre-sponding normalized eigenvectors are those of Equation k. Equation l, and Equation m of Example 6.13.

Equation p is the same as that encountered in the steady-state problem of Example 6.13. The eigenvalues are of the form μm = m (m + 1), and the normal-ized eigenvector is expressed in terms of the associated Legendre function of the fi rst kind, as in Equation p of Example 6.13.

The self-adjoint form of equation l is

−⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+

+=

1 12

2

2rddr

rdRdr

m mr

R Rmm m

( )λ (q)

Equation q is in proper Sturm-Liouville form with p(r) = –r2, q(r) = μ/r2, and r (r) = r2. For each m, its eigenvalues are infi nite, but countable, λm,p for p = 1,2,…. The orthogonality condition for Rm,p and Rm,q corresponding to distinct eigen-values λm,p and λm,q is

0

2

0

1

=( )

=

R r R r

R r R r r dr

m p m q r

m p m q

, ,

, ,

( ), ( )

( ) ( )∫∫

(r)

Equation q is the spherical Bessel’s equation discussed in Section 3.6. Its gen-eral solution is

R r D j r E y rm m m m m( ) = ( )+ ( )λ λ (s)

where jm(r) and ym(r) are the spherical Bessel functions of the fi rst and second kinds of order m and argument r. Noting that ym(0) is not fi nite at r = 0 it is required that Em = 0 for the temperature to be fi nite at r = 0. Application of

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the product solution to Equation g leads to dR/dr(1) + BiR(1) = 0, which when applied to Equation s, results in

λ λ λ′ ( )+( ) ( )=j Bi jm m 0 (t)

The eigenvalue λm,p is the pth positive solution of Equation t. The resulting normalized eigenvectors are

R r D j rm p m p m m p, . ,( ) = ( )λ (u)

where

D j r r drm p m m p, ,= ( )⎡⎣⎢

⎤⎦⎥

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

∫−

λ2

2

0

11

2

(v)

The closed-form evaluation of Equation v is in terms of confl uent hypergeo-metric functions and is beyond the scope of this study.

The normalized eigenvectors of Equation f are

UD C

P j rm pm p n m

m m m p02

, ,, ,

,cos= ( ) ( )π

φ λ (w)

UD C

P j rm n pm p n m

mn n

m m p, , ,, ,

,cos sin1 = ( ) ( ) ( )π

φ φ λ ccosθ (x)

UD C

P j rm n pm p n m

mn n

m m p, , ,, ,

,cos sin2 = ( ) ( ) ( )π

φ φ λ ssin θ (y)

Noting that the general solution of Equation e is Tn,m,p(t) = An,m,pe–λm,p t, the

solution of Equation a is

Λ r t e A Um pt

p

m p m p, , , (,, , , ,θ φ λ( )=

⎧⎨⎪⎪

⎩⎪⎪

−

=

∞

∑1

0 0 rr

A U r B U

m

n m p m n p n m p m

, , )

( , , ), , , , , , ,

θ φ

θ φ

=

∞

∑

+ +

1

1 ,, , , ( , , )n p

n

r2

1

θ φ[ ]⎫⎬⎪⎪

⎭⎪⎪=

∞

∑ (z)

Example 6.15 The surface of the sphere shown in Figure 6.20 is pulsating such that the distance from the center of the sphere to a point on the surface of the sphere is

r w t= +1 ( , , )θ φ (a)

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The pulsations of the sphere lead to the initiation and propagation of acous-tic waves in the surrounding inviscid fl uid. If Φ(r,θ,φ,t) is the velocity poten-tial in the fl uid which is related to the velocity vector by

v = ∇Φ (b)

then the waves are governed by the wave equation,

∇ =∂∂

22

2Φ

Φt

(c)

A boundary condition is that the normal component of velocity of the fl uid at the surface of the sphere is equal to the velocity of the particle on the sphere at that point,

∂∂

=∂∂

Φr

twt

t( , , , ) ( , , )1 θ φ θ φ (d)

Since there is no refl ection, the waves are only outgoing.

(a) Determine Φ( , )r t if w t A tp p p p( ) sin( )= ∑ +=∞

0 ω γ ; w(t) is a periodic func-tion with a trigonometric Fourier series representation.

(b) Determine Φ( , , , )r tθ φ if w t f t( , , ) ( , )sin( )θ φ θ φ ω= , where f is a continu-ous function over the surface of the sphere.

Solution (a) The boundary condition on the surface of the sphere is

∂∂

= +=

∞

∑Φr

t A tp p p p

p

( , ) cos( )11

ω ω ν (e)

The pulsations of the body are such that it retains the shape of a sphere, but with changing radius. The pulsations are independent of θ and φ, and there-fore the waves are uniformly propagated from the surface of the sphere and are independent of θ and φ. Equation c becomes

12

22

2r tr

r t∂∂

∂∂

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟=

∂∂

Φ Φ (f)

Then the boundary condition, Equation e, suggests a solution of the form

Φ( , ) sin cosr t g t g tp p p p

p

= ( )+ ( )⎡⎣ ⎤⎦=

∞

∑ 1 2

1

ω ω (g)

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Substitution of Equation g into Equation f leads to

10

22 1 2

1

22

1

2

rddr

rdgdr

g

rd gdr

pp p

p

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟+ =ω

++ + =2 01 2 2

1rdgdr

r gpp pω

(h)

and

rd gdr

rdgdr

r gp pp p

22

2

2

2 2 222 0+ + =ω (i)

The solutions of Equation h and Equation i are written using spherical Bessel functions as

g r C j r C y rp p p p p1 1 0 2 0( ) ( ) ( )= +ω ω (j)

g r C j r C y rp p p p p1 3 0 4 0( ) ( ) ( )= +ω ω (k)

Asymptotic relations for the spherical Bessel functions are

j rr

r Or

r0 2

1 1( ) sinω

ωω ∞ ≈ ( )+ ( ) →as (l)

y rr

r Or

r0 2

1 1( ) cosω

ωω ∞ ≈ − ( )+ ( ) →as (m)

Combining Equation g, Equation i, Equation l, and Equation m, the behavior of the waves for large r is

Φ( , ) sin cos( ) sin(r t

rC r C r t

pp p p p p≈ ( )−[ ]1

1 2ω

ω ω ω ))

sin cos(

{

+ ( )−

=

∞

∑p

p p p pC r C

1

3 4ω ω rr tp r) cos( )[ ] }+ ( )ω 0 12

(n)

Use of trigonometric identities in Equation n leads to

Φ( , ) [ cos( ( )) cos( ( )r tr

C t r C t rp

p p p p≈ − − +1

21 1

ωω ω ))

sin( ( )) sin( ( ))]

[

{

− + − −

+

=

∞

∑p

p p p pC t r C t r

1

2 2ω ω

CC t r C t r

C

p p p p

p p

3 3

4

sin( ( )) sin( ( ))

cos( (

ω ω

ω

+ + −

− tt r C t tp p r− − + }+ ( ))) cos( ( ))]4

10 2ω (o)

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Waves are continuously generated from the surface of the pulsating sphere. A wave is generated at the surface at t = 0. The wave propagates in the radial direction. A wave of the form cos( ( )) sin( ( ))ω ωt r t r− −or has a positive velocity and propagates in the positive radial direction (an outgoing wave), whereas a wave of the form cos( ( )) sin( ( ))ω ωt r t r+ +or has a negative velocity and propagates in the negative radial direction (an incoming wave). However, the region is unbounded, and outgoing waves are not refl ected. Since only outgoing waves are permitted in the solution,

C Cp p3 2 0− = (p)

C Cp p1 4 0+ = (q)

The solution becomes

Φ( , ) [ ( ) ( )]sin( )r t C j r C y r tp p p p p

p

= +{=

∞

1 0 2 0

1

ω ω ω∑∑+ − }[ ( ) ( )]cos( )C j r C y r tp p p p p2 0 1 0ω ω ω (r)

Satisfaction of Equation e requires that

ω ω ωp p p p p p

p

C j C y t

C

[ ( ) ( )]sin( )

[

1 0 2 0

1

2

′ + ′{

+

=

∞

∑ ω

pp p p p p

p p p

j C y t

A t

′ − ′ }

= +

0 1 0( ) ( )]cos( )

cos(

ω ω ω

ω ω ννp

p

)=

∞

∑1

(s)

The trigonometric terms on the right-hand side of Equation s can be expanded using a trigonometric identity; then coeffi cients of like terms are equated for each p, leading to

C j C y Ap p p p p p p p1 0 2 0′ ( )+ ′⎡⎣ ⎤⎦ = −ω ω ω ω( ) sin ν (t)

C j C y Ap p p p p p p p2 0 1 0′ ( )− ′⎡⎣ ⎤⎦ =ω ω ω ω( ) cosν (u)

Equation t and Equation u are solved simultaneously, leading to

C Aj y

jp p

p p p p

p

10 0

0

= −′ ( ) + ′ ( )

′ ( )⎡⎣ ⎤⎦

ω ω

ω

sin cosν ν22

0

2+ ′ ( )⎡⎣ ⎤⎦y pω

(v)

C Aj y

jp p

p p p p

p

20 0

0

2=

′ ( ) − ′ ( )′ ( )⎡⎣ ⎤⎦

ω ω

ω

cos sinν ν

++ ′ ( )⎡⎣ ⎤⎦y p0

2ω

(w)

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(b) The boundary condition on the surface of the sphere is

∂∂

=Φr

t f t( , , , ) ( , )cos( )1 θ φ ω θ φ ω (x)

The form of Equation x and the knowledge that waves must be outgoing only suggest a superposition solution of the form

Φ( , , , ) ( , , )sin( ) ( , , )cos( )r t A r t B r tθ φ θ φ ω θ φ ω= + (y)

Substitution into Equation c and choosing appropriate problems leads to

∇ + =2 2 0A Aω (z)

∇ + =2B Bω2 0 (aa)

A product solution of Equation z is assumed to be of the form

A r R r( , , ) ( ) ( ) ( )θ Φ θ φ= Θ Ψ (bb)

Substitution of Equation bb into Equation z leads to

Θ Θ ΘΨ Ψ

φ θrddr

rdRdr

Rr

dd

R2

2

2 2

2

2

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+ +

sin rrd

ddd

R2

2 0sin

sinφ φ

φΨφ

ω ΘΨ⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+ = (cc)

Multiplying Equation cc by r R2/ ΘΨ and rearranging leads to

−⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+ = +

1 12 2 2

2

2

2Rddr

rdRdr

rdd

ωφ

Θθsin Θ

11

sinsin

φ φφ

ΨφΦ

dd

dd

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟ (dd)

Applying the usual separation argument to Equation dd and denoting the separation constant as μ results in

ddr

rdRdr

r R2 2 2⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+ −( )ω μ (ee)

1 12

2

2sin sinsin

φ θ φ φΨφΘ

ΘΦ

dd

dd

dd

+⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟=φ −−μ (ff)

Multiplying Equation ff by sin2 φ and rearranging leads to

− =⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+

1 2

2

2

Θ θφ

Φ φφ

Ψφ

μdd

dd

dd

Θ sinsin sin φφ (gg)

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Applying the usual separation argument to Equation gg and denoting the separation constant as κ results in

dd

2

20

Θθ

κΘ+ = (hh)

1

2sinsin

sinφ φφ

Φφ

μκ

φd

ddd

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+ −

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟ =Ψ 0 (ii)

The conditions satisfi ed by the solution of Equation hh are that the velocity potential and the velocity be single-valued everywhere within the acoustic fi eld, which lead to the velocity potential and the θ-component of the velocity being periodic in θ with period 2π. The eigenvalues of Equation hh subject to the periodicity conditions are of the form κn = n2

for n = 0,1,2,… .The normalized eigenvector corresponding to n = 0 is

Θ0

1

2=

π (jj)

Each eigenvalue for n > 0 has two linearly independent normalized eigen-vectors given by

Θn n, ( ) cos1

1θ

πθ= ( ) (kk)

Θn n, ( ) sin2

1θ

πθ= ( ) (ll)

The conditions satisfi ed by Equation ii are that the velocity must remain fi nite at every point in the fi eld. Following the discussion of the solution of Equation n of Example 6.13, the eigenvalues of Equation ii subject to these conditions are of the form μm = m (m+1) and their eigenvectors are

Ψn m n m mn nC P, , cos sinφ φ φ( )= ( ) ( ) (mm)

where Cn m, is chosen to normalize the eigenvector. The general solution of Equation ee is then

R r D j r D y rm m m m m( ) = ( )+ ( )1 2ω ω (nn)

The general solution of Equation z is

A rC

D j r D y rmm

m m m m00

1 0 2 0,,

, , , ,, ,θ φπ

ω ω( )= ( )+ ( )[ ]] ( )Pmn(cos )sinφ φ (oo)

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A r

CD j r D y rn m

n mm m m m,

,, , , ,, ,θ φ

πω ω( )= ( )+ ( )

21 1 2 1[[ ]{

+ ( )+ ( )[ ]

cos( )

sin(, , , ,

n

D j r D y rm m m m

θ

ω ω1 2 2 2 nn Pmn nθ φ φ) (cos )sin} ( )

(pp)

The solution of Equation aa is similar to that of Equation oo and Equation pp, except with different arbitrary constants; call them E. Application of the superposition formula, Equation y, leads to

Φ( , , , ) ( ) ( ),, , , ,r t

CD j r D y rm

m m m mθ φπ

ω ω= +[ ]01 0 2 0 PP

CD j r D

m

m

n mm m m

(cos )

( ),, , , ,

φ

πω

⎡⎣⎢⎢

+ +

=

∞

∑0

1 1 22

11

1

1 2 2 2

y r n

D j r D

m

n

m m m

( ) cos( )

( ), , , ,

ω θ

ω

[ ]{

+ +

=

∞

∑

yy r n Pm mn n( ) sin( ) (cos )sin ( )ω θ φ φ ω[ ] } ⎤

⎦⎥⎥sin tt( )

+ +[ ]CE j r E y r Pm

m m m m m0

1 0 2 0,

, , , ,( ) ( ) (cosπ

ω ω φφ

πω ω

)

( ) (,, , , ,

⎡⎣⎢⎢

+ +

=

∞

∑m

n mm m m m

CE j r E y r

0

1 1 2 12

)) cos( )

( ) ( ), , , ,

[ ]{

+ +

=

∞

∑ n

E j r E y r

n

m m m m

θ

ω ω

1

1 2 2 2[[ ] } ( )⎤⎦⎥⎥

( )sin( ) (cos )sinn Pmn nθ φ φ ωcos t

(qq)

Asymptotic relations for the spherical Bessel functions are

j rr

r m Orm( ) sinω

ωω π≈ −

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+

⎛⎝⎜⎜⎜

⎞⎠

1 1

2

12⎟⎟⎟⎟⎟ → ∞as r (rr)

y rr

r m Or0 2

1 1

2

1( ) cosω

ωω π≈ − −

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟⎟+

⎛⎝⎜⎜⎜

⎞⎠⎠⎟⎟⎟⎟ → ∞as r (ss)

The solution includes only outgoing waves if (see part a)

D Em m1 0 2 0 0, , , ,+ = (tt)

E Dm m1 0 2 0 0, , , ,− = (uu)

D Em m1 1 2 1 0, , , ,+ = (vv)

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E Dm m1 1 2 1 0, , , ,− = (ww)

D Em m1 2 2 2 0, , , ,+ = (xx)

E Dm m1 2 2 2 0, , , ,− = (yy)

Application of the boundary condition on the surface of the sphere leads to

CD j D y Pm

m m m m m0

1 0 2 0,

, , , ,( ) ( ) (cos )π

ω ω φ′ + ′[ ]⎡⎣⎢⎢⎢

+ ′ + ′[ ]

=

∞

∑m

n mm m m m

CD j D y

0

1 1 2 12

,, , , ,( ) ( ) co

πω ω ss( )

( ) ( ) sin, , , ,

n

D j D y

n

m m m m

θ

ω ω

{

+ ′ + ′[ ]

=

∞

∑1

1 2 2 2 (( ) (cos )sin ( )

,

n P

CE

mn n

m

θ φ φ ω

π

} ⎤⎦⎥⎥

( )

+

sin t

01,, , , ,( ) ( ) (cos )m m m m m

m

j E y P0 2 0

0

′ + ′[ ]⎡⎣⎢⎢

=

∞

∑ ω ω φ

++ ′ + ′[ ]{CE j E y nn m

m m m m,

, , , ,( ) ( ) cos( )2

1 1 2 1π

ω ω θnn

m m m mE j E y n P

=

∞

∑

+ ′ + ′[ ] }

1

1 2 2 2, , , ,( ) ( ) sin( )ω ω θ mmn n

f t

(cos )sin

( , )cos( )

φ φ ω

θ φ ω

( )⎤⎦⎥⎥

( )

=

cos t

(zz)

Eigenvector expansions are used to obtain

f r F P G n G nm m m n m n( , ) (cos ) cos( ) sin(, , , ,θ φ θ= + +1 2 θθ φ φ) (cos )sin ( )[ ]⎧⎨⎪⎪

⎩⎪⎪

⎫⎬⎪⎪

⎭⎪⎪=

∞

=∑ Pm

n n

nm 10

∞∞

∑ (aaa)

where

F

f P d d

Pm

m

m

=( ) ( )

[ ]

∫∫ θ φ φ φ φ θ

φ

ππ

, cos sin

(cos )

00

2

2 ssinφ φ θππ

d d00

2

∫∫ (bbb)

G

f n P d dm n

nm

n

, ,

, cos( ) cos sin

1

1

0=( ) ( ) +∫ θ φ θ φ φ φ θ

π

00

2

2

0

π

π

θ φ φ φ φ θ

∫∫ ( )⎡⎣ ⎤⎦cos( ) (cos )sin sinn P d dm

n

00

2π

∫ (ccc)

9533_C006.indd 5079533_C006.indd 507 10/29/08 1:58:48 PM10/29/08 1:58:48 PM


G

f n P d dm n

nm

n

, ,

, sin( ) cos sin

2

1

0=( ) ( ) +∫ θ φ θ φ φ φ θ

π

00

2

2

0

π

π

θ φ φ φ φ θ

∫∫ ( )⎡⎣ ⎤⎦sin( ) (cos )sin sinn P d dm

n

00

2π

∫ (ddd)

Equation zz and Equation aaa are used to obtain

C

D j D y Fmm m m m m

01 0 2 0

,, , , ,

πω ω′ ( )+ ′ ( )[ ]= (eee)

C

E j E ymm m m m

01 0 2 0 0,

, , , ,π

ω ω′ ( )+ ′ ( )[ ]= (fff)

C

D j D y Gn mm m m m m n

,, , , , , ,

21 1 2 1 1

πω ω′ ( )+ ′ ( )[ ]= (ggg)

C

E j E ymm m m m

01 0 2 0 0,

, , , ,π

ω ω′ ( )+ ′ ( )[ ]= (hhh)

C

D j D y Gn mm m m m m n

,, , , , , ,

21 2 2 2 2

πω ω′ ( )+ ′ ( )[ ]= (iii)

C

E j E yn mm m m m

,, , , ,

201 2 2 2

πω ω′ ( )+ ′ ( )[ ]= (jjj)

Equation tt, Equation uu, Equation vv, Equation ww, Equation xx, and Equa-tion yy and Equation eee, Equation fff, Equation ggg, and Equation hhh are solved simultaneously to obtain the constants.

Problems 6.1. The nondimensional natural frequencies ω and mode shapes φ(x,y)

of a rectangular membrane can be determined from

∂∂

+∂∂

+ =2

2

2

2

2 0φ φ

ω φx y

(a)

Determine the natural frequencies and normalized mode shapes for a membrane which is clamped along its entire edge. The appropriate boundary conditions are φ(0,y) = 0, φ(1,y) = 0, φ(x,0) = 0 and φ(x,α) = 0 where, is the ratio of the lengths of the sides of the membrane.

9533_C006.indd 5089533_C006.indd 508 10/29/08 1:58:49 PM10/29/08 1:58:49 PM


6.2. Solve Problem 6.1 if the membrane is free along the edge y = 0 such that its boundary condition ∂ ∂ =φ/ ( , )x x 0 0

6.3. Determine the steady-state temperature distribution, Θ( , )x y in the thin square slab shown in Figure P6.3. The problem is governed by Laplace’s equation subject to the boundary conditions Θ( , ) ,0 1y =

∂ ∂ =Θ/ ( , ) ,x y1 0 Θ( , )x 0 0= and ∂ ∂ =Θ/ ( , )y x 1 0.

6.4. Solve Problem 6.3 if the boundary conditions are Θ( , ) ( ),0 1y y y= −∂ ∂Θ/ x ( , ) ,1 0y = Θ( , )x 0 0= and ∂ ∂Θ/ y ( , )x 1 0= .

6.5. Solve Problem 6.3 if the boundary conditions are Θ( , ) ( ),0 1y y y= −∂ ∂ = =Θ Θ/ ( , ) , ( , )x y x1 0 0 0 and ∂ ∂Θ/ ( , )y x 1 + =1 2 1 0. ( , )Θ y .

6.6. The unsteady state temperature distribution in an extended surface with a large heat transfer coeffi cient is governed by

∂∂

−( ) =∂∂

2

2

ΘΘ

Θx

Bit

(a)

The surface has an initial temperature distribution when the ambient temperature suddenly changes to Θ1. The left end of the surface is maintained at the new temperature, while the right end of the surface is insulated. The boundary conditions are Θ Θ( , ) / ( , )0 0 1 0t x t= ∂ ∂ =and . Assume that the initial condition Θ Θ( , ) cos( )x x0 0= π .

6.7. The thin slab shown in Figure P6.7 initially is at a uniform tempera-ture when a heat fl ux is applied to one side. Determine the result-ing unsteady state temperature distribution, which is governed by the partial differential equation ∂ ∂ + ∂ ∂ = ∂ ∂2 2 2 2Θ Θ Θ/ / /x y t subject to the boundary conditions Θ Θ( , , ) , / ( , , ) sin( / ),0 0 1x t x y t y= ∂ ∂ = π αΘ( ,0,x t)) 0 and= Θ( , , )x tα = 0 and the initial condition Θ( , , )x y 0 0= .

y

x

1

1

Figure P6.3 System of Problems 6.3–6.5.

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6.8. A thin slab is initially at a uniform temperature when the tem-perature of one side is suddenly changed. Determine the resulting unsteady state temperature distribution, which is governed by the partial differential equation ∂ ∂ +∂ ∂ = ∂ ∂2 2 2 2Θ Θ Θ/ / /x y t subject to the boundary conditions Θ Θ Θ( , , ) , / ( , , ) ,0 1 1 0x t x y t x t= ∂ ∂ = =( ,0, ) 0and Θ( , , )x tα = 0 and the initial Θ( , , )x y 0 0= .

6.9. A thin slab is initially at a temperature f(x,y) when the temperature of two parallel sides is changed to the same temperature while the other two sides remain insulated. Determine the resulting unsteady state temper-ature distribution, which is governed by the partial differential equa-tion ∂ ∂ +∂ ∂ = ∂ ∂2 2 2 2Θ Θ Θ/ / /x y t subject to the boundary conditions

Θ Θ Θ( , , ) , ( , , ) , /0 0 1 0x t y t y x t= = ∂ ∂ =( ,0, ) 0 and∂ ∂ =Θ/ ( , , )y x tα 0 and the initial condition Θ( , , ) ( , )x y f x y0 = .

6.10. The thin shaft shown in Figure P6.10 is initially subject to a torque at its end which is suddenly removed. Determine the resulting response of the shaft, which is governed by the nondimensional partial differential equation ∂2Θ/∂x2 = ∂2Θ/∂t2 subject to the initial conditions Θ(0,t) = 0, ∂Θ/∂x(1,t) = 0, and Θ(x,0) = Cx, where C is a constant and ∂Θ/∂t(x,0) = 0.

6.11. The disk at the end of the thin shaft shown in Figure P6.11 is initially subject to a torque which is suddenly removed. Determine

α

y

x0

0

0

1

sin(πy/α)

Figure P6.7 System of Problem 6.7.

Mθ(x,t)


9533_C006.indd 5109533_C006.indd 510 10/29/08 1:58:52 PM10/29/08 1:58:52 PM


the resulting response of the shaft, which is governed by the non-dimensional partial differential equation ∂2Θ/∂x2 = ∂2Θ/∂t2 subject to the initial conditions Θ(0,t) = 0, ∂Θ/∂x(1,t) = −μ∂2Θ/∂t2(1,t) and Θ(x,0) = Cx and ∂Θ/∂t(x,0) = 0, where μ and C are constants.

Problems 6.12–6.16 refer to the system shown in Figure P6.12. The beam is subject to a concentrated load at its end, which leads to a static defl ection of the form

w xx x

0

2 3

2 6( ) = −

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟α (a)

When the load is removed, the resulting vibrations are governed by

∂∂

+∂∂

=4

4

2

20

wx

wt

(b)

The boundary conditions are

w t( , )0 0= (c)

∂∂

=wx

t( , )0 0 (d)

∂∂

+ +∂3

3 1

2

21 1 1

wx

t w twt

t( , ) ( , ) ( , )η μα (e)

∂∂

+∂

∂ ∂

2wx

tw

x t2 2

3

21( , ) μ (f)

The initial conditions are

w xx x

( , )02 6

2 3

= −⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟α (g)

∂∂

( )=wt

x,0 0 (h)

M

θ(x,t)


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6.12. Determine w(x,t) if η = μ1 = μ2 = 0. 6.13. Determine w(x,t) if η = 2 and μ1 = μ2 = 0. 6.14. Determine w(x,t) if η = μ1 = 0 and μ2 = 1. 6.15. Determine w(x,t) if μ1 = 0.5 and η = μ2 = 0. 6.16. Determine w(x,t) if μ1 = 0.5, μ1 = 0.5 and η = 2.

Problems 6.17–6.24 refer to the nondimensional temperature dis-tribution in a cylinder which, in general, is governed by

∂∂

+∂∂

+∂∂

+∂∂

=∂∂

2

2 2

2

2 2

2

2

1 1 1Φ Φ Φ Φ Φr r r r r z tθ

(a)

Nondimensional variables are introduced such that 0 ≤ z ≤ 1, 0 ≤ r ≤ R/L = α, and 0 ≤ θ ≤ 2π.

6.17. Determine the steady-state temperature distribution when the temper-ature at each end is determined by different constants and the circum-ference is insulated. The temperature is independent of θ. The boundary conditions for Φ(r,z) are ∂Φ/∂r(α,z) = 0, Φ(r,0) = 0, and Φ(r,1) = 1.

6.18. Determine the steady-state temperature distribution in the cylinder when the temperature at each end is maintained at the same value and a heat fl ux q(θ,z) is applied to the circumference of the cylinder. The boundary conditions for Φ(r,θ,z) are ∂Φ/∂r(α,θ,z) = q(θ,z), Φ(r,θ, 0) = 0, and Φ(r,θ,1) = 0.

6.19. The cylinder is at a uniform temperature when the temperature of the surrounding medium is suddenly changed. Both ends of the cyl-inder are insulated, but heat is transferred by convection to the sur-rounding medium from the circumferential surface of the cylinder. The unsteady state temperature is independent of θ. The boundary conditions for Φ(r,z,t) are ∂Φ/∂r(α,z,t) + BiΦ(α,z,t) = 0, ∂Φ/∂z(r,0,t) = 0, and ∂Φ/∂z(r,1,t) = 0. The initial condition for the system is Φ(r,z,0) = 1.

6.20. The cylinder has a steady-state temperature distribution of f(r,θ,z) when the temperature of the surrounding medium is suddenly

m,I

k

x

w(x,t)

Figure P6.12 System of Problems 6.12–6.16.

9533_C006.indd 5129533_C006.indd 512 10/29/08 1:58:55 PM10/29/08 1:58:55 PM


changed. Both ends of the cylinder are insulated, but heat is trans-ferred by convection to the surrounding medium from the circumfer-ential surface of the cylinder. The boundary conditions for Φ(r,θ,1,t) are ∂Φ/∂r(α,θ,z,t) + BiΦ(α,θ,z,t) = 0, ∂Φ/∂z(r,θ,0,t) = 0, and ∂Φ/∂z(r,θ,1,t) = 0. The initial condition for the system is Φ(r,θ,z,0) = f(r,θ,z).

6.21. The cylinder has a steady-state temperature distribution of f(r,θ,z) when the left end is suddenly subject to a uniform heat fl ux. The right end of the cylinder is insulated, but heat is transferred by convection to the surrounding medium from the circumferential surface of the cylinder. The boundary conditions for is Φ(r,θ,z,t) are ∂Φ/∂r(α,θ,z,t) + BiΦ(α,θ,z,t) = 0, ∂Φ/∂z(r,θ,0,t) = q, and ∂Φ/∂z(r,θ,1,t) = 0. The initial condition for the system is Φ(r,θ,z,0) = f(r,θ,z).

6.22. The cylinder has a uniform temperature when the circumferen-tial surface is suddenly subject to a heat fl ux q(z). Both ends of the cylinder are insulated. The boundary conditions for Φ(r,z,t) are ∂Φ/∂r(α,z,t) = q(z),∂Φ/∂z(r,0,t) = 0, and ∂Φ/∂z(r,1,t) = 0. The initial con-dition for the system is Φ(r,z,0) = 1.

6.23. The cylinder has a steady-state temperature distribution f(r,θ,z) when an internal heat generation u(z) begins in the cylinder. The right end of the cylinder is insulated, but heat is transferred by convection to the surrounding medium from the circumferential surface of the cylinder. The governing partial differential equation is modifi ed to ∂ ∂ + ∂ ∂ + ∂ ∂ + ∂2 2 2 2 2 2 21 1 1Φ Φ Φ Φ/ ( / ) / ( / ) / ( / ) /r r r r rθ ∂∂ =∂ ∂ +z t u z2 Φ/ ( ). The boundary conditions for Φ(r,θ,z,0) are ∂Φ/∂r(α,θ,z,t) + BiΦ(α,θ,z,t) = 0, ∂Φ/∂z(r,θ,0,t) = 0, and ∂Φ/∂z(r,θ,1,t) = 0. The initial condition for the system is Φ(r,θ,z,0) = f(r,θ,z).

6.24. The cylinder has the steady-state temperature distribution obtained in the solution of Problem 6.18 when an internal heat generation begins in the cylinder. The boundary conditions remain the same as in Problem 6.18, but the governing partial differential equation is modifi ed to ∂ ∂ + ∂ ∂ + ∂ ∂ +2 2 2 2 21 1Φ Φ Φ/ ( / ) / ( / ) /r r r r θ ( / ) /1 2 2 2r z∂ ∂ =Φ∂ ∂ +Φ/ t u z( ). Solve the problem to determine Φ(r,θ,z,t).

6.25. Determine the time-dependent response of the system of Example 6.4 when α(x) = (1 − 0.1x)3.

6.26. The disk at the end of the thin shaft shown in Figure P6.11 is at rest in its equilibrium position when a time-dependent uniformly distributed torque is applied across the length of the shaft. Determine the result-ing response of the shaft, which is governed by the nondimensional partial differential equation ∂ ∂ = ∂ ∂ + −2 2 2 2 1Θ Θ/ / ( )sinx t x x tφ ω subject to the boundary conditions Θ(0,t) = 0 and ∂Θ/∂x(1,t) = − μ[∂2Θ/∂t2(1,t)] and the initial conditions Θ(x,0) = 0 and ∂Θ/∂t(x,0) = 0, where μ and φ are constants.

6.27. Determine the steady-state temperature distribution in a sphere whose surface has a temperature distribution f(θ) which is indepen-dent of φ.

9533_C006.indd 5139533_C006.indd 513 10/29/08 1:58:57 PM10/29/08 1:58:57 PM


6.28. The curved surface of a hemisphere, as illustrated in Figure P6.28, is subject to a heat fl ux, q(θ,φ). The fl at surface is maintained at a constant temperature. Determine the steady-state temperature distribution in the sphere, Φ(r,θ,φ), using the boundary conditions ∂Φ/∂r(1,θ,φ) = q(θ,φ) and Φ(r,θ,π/2) = 0.

6.29. Solve Problem 6.28 using the boundary conditions ∂Φ/∂r(1,θ,φ) = q(θ,φ) and Φ(r,π,φ) = 0.

6.30. Determine the steady-state temperature distribution in the spheri-cal shell shown in Figure 6.30 if the inner surface is insulated and the outer surface is maintained at a constant temperature, f(θ,φ).

q(r,θ)

Figure P6.28 System of Problems 6.28 and 6.29.

R2R1

f(θ,φ)


9533_C006.indd 5149533_C006.indd 514 10/29/08 1:58:58 PM10/29/08 1:58:58 PM


The nondimensional boundary conditions are ∂Φ/∂r(1,θ,φ) = 0 and Φ(R2/R1) = f(θ,φ).

6.31. The sphere shown in Figure P6.31 is maintained at a uniform tem-perature T1 when the exterior is subject to a uniform heat fl ux q. Determine the resulting unsteady state temperature distribution in the sphere when (a) q is constant over the surface of the sphere and (b) q = q (θ, φ)

6.32. The sphere shown in Figure P6.31 is maintained at a uniform tem-perature T1 when an internal heat generation u is initiated in the sphere. Determine the resulting unsteady state temperature distri-bution in the sphere when (a) u is a constant throughout the sphere all (b) u = u (r, θ) and (c) u = u (r, θ, φ).

6.33 Rework Example 6.8 by fi rst considering the eigenvalue problem

− =

=

′( )=

d Ydy

Y

Y

Y

2

2

0 0

0

μ

α

( )

Then expand f(x,y) for each x in terms of the eigenvectors for this problem. Show that Equation q of Example 6.8 is attained.

References 1. Abramowitz, M., and I. Stegun. 1965. Handbook of mathematical functions. Mine-

ola, NY: Dover. 2. Arpaci, V. 1966. Conduction heat transfer. Reading, MA: Addison Wesley. 3. Courant, R., and Hilbert, D. 1962. Methods of mathematical physics. New York, NY:

Wiley-Interscience.

R

k,ρ,c

h,T 8

Figure P6.31 System of Problems 6.31 and 6.32.

9533_C006.indd 5159533_C006.indd 515 10/29/08 1:58:59 PM10/29/08 1:58:59 PM


4. Franklin, J. N. 1968. Matrix theory. Englewood Cliffs, NJ: Prentice Hall. 5. Greenberg, M. 1998. Advanced engineering mathematics. 2nd ed. Englewood

Cliffs, NJ: Prentice Hall. 6. Haberman, R. 2003. Applied partial differential equations. 4th ed. Englewood

Cliffs, NJ: Prentice Hall. 7. Haug, E., and K. Choi. 1993. Methods of engineering mathematics. Englewood

Cliffs, NJ: Prentice Hall. 8. Hildebrand, F. 1976. Advanced calculus for applications. 2nd ed. Englewood Cliffs,

NJ: Prentice Hall. 9. Kelly, S. G. 2007. Advanced vibration analysis. Boca Raton, FL: CRC Press. 10. Kreyszig, E. 2005. Advanced engineering mathematics. 9th ed. New York, NY:

Wiley. 11. Langhaar, H. L. 1962. Energy methods in applied mechanics. New York, NY:

Wiley. 12. Meirovitch, L. 1997. Principles and techniques of vibrations. 1997. Upper Saddle

River, NJ: Prentice Hall. 13. Nayfeh, A. 1993. Introduction to perturbation techniques. New York, NY: Wiley. 14. Ozisik, M. 1993. Heat conduction. 2nd ed. New York, NY: Wiley. 15. Pelesko, J., and D. H. Bernstein. 2003. Modeling MEMS and NEMS. Boca Raton,

FL: CRC Press. 16. Prenter, P. 1975. Splines and variational methods. New York, NE: Wiley-

Interscience. 17. Reddy, J. N. 1984. An introduction to the fi nite element method. New York:

McGraw-Hill. 18. Strang, G., and G. J. Fix. 1973. An analysis of the fi nite element method. Englewood

Cliffs, NJ: Prentice Hall. 19. Strang, G. 2005. Linear algebra and its applications. 4th ed. Belmont, CA: Brooks

Cole. 20. Weinstock, R. 1952. Calculus of variations with applications to physics and engineer-

ing. New York, NY: McGraw-Hill. 21. White, F. 2005. Viscous fl uid fl ow. 3rd ed. New York, NY: McGraw-Hill. 22. Wylie, C. R. 1975. Advanced engineering mathematics. New York, NY:

McGraw-Hill. 23. Zhang, Z. 2007. Nano/Microscale heat transfer. New York, NY: McGraw-Hill. 24. Zienkiewicz, O. C., and R. I. Taylor. 1991. The fi nite element method. 4th ed. New

York, NY: McGraw-Hill.

9533_C006.indd 5169533_C006.indd 516 10/29/08 1:59:01 PM10/29/08 1:59:01 PM

517

A

Abel’s formula, 146, 148

Added mass, 48

Adjoint operators

defi nition, 121–122

eigenvalues of, 278

Acceleration due to gravity, 3

Acceleration vector, 71

Admissible functions, 238

Ambient temperature, 10

Analytic function, 167–168

Angle between vectors, 68

Approximate solutions, 65, 84, 212

Associated Legendre’s equation, 200,

499, 505

Associated Legendre functions, 200–201,

500, 505

Associative law of addition, 67, 73

Associative law of scalar multiplication, 67, 73

Assumptions, 3

Associative laws, 73

Asymptotic expansion, 54, 359, 502, 506

Asymptotic solution, 54

Axial coordinate, 472

Axial stiffness, 349

Azimuthal angular coordinate, 2

B

Basis, 81–82

Beam defl ection

axially-loaded beam, 158–162, 165–167

elastic foundation, on,

fi nite-element solution, 251–261

Rayleigh-Ritz solution, 227–235

Green’s function, 400

microscale cantilever, 271–273

non-unifrom beam, 238–242

positive-defi nite operator, 130

self-adjoint operator, 125

strain energy, 135

Bending layer, 56

Bending moment, 135

Bending stiffness, 349

Bessel’s equation

from cylindrical coordinates, 477, 483

series solution, 177–180

Bessel function applications,

acoustic wave propagation from

pulsating sphere

one-dimensional, 502–503

three-dimensional, 504–508

buckling of column due to gravity,

341–343

Fourier-Bessel series, 339–341

Green’s function for temperature

distribution in variable-area slab,

405–406

initial value problem for vibrations of

non-uniform bar, 454–454

heat transfer in annular hyperbolic fi n,

194–195

heat transfer in triangular fi n, 185–188

oscillations of non-uniform shaft,

343–348

steady heat transfer in axi-symmetric

cylinder, 488–491

temperature distribution in

variable-area fi n, 262

unsteady heat transfer in cylinder, 479

unsteady heat transfer in sphere,

499–500

vibrations of a circular sector membrane,

484–485

Bessel functions, 180

derivatives of, 189

eigenvalue problems, 336–348

fi rst kind, of the, 181

integrals of, 190–192

Index

9533_C007.indd 5179533_C007.indd 517 10/29/08 2:25:20 PM10/29/08 2:25:20 PM

518 Index

orthogonality of, 327

recurrence relations, 189–190

second kind, of the, 181

solutions in terms of, 193–194, 263–264

Bessel’s inequality, 304

Best approximation, 211–212

Binomial expansion, 168

Biorthogonality, 279, 312–313, 360

Biot number, 47, 434

Bluff body, 31

Body forces, 9

Boundary conditions

defi ning subspaces of Cn [a, b], 83

for partial differential equations, 419

used to determine constants of

integration, 145

Boundary layer, 54

Boundary layer thickness, 55

Boundary value problems, 145

Buckling, 326

C

C2[ℜ], 85

Cn [a, b]

defi nition, 75

dimension, 82

inner products on, 89

linearly independent vectors, 79

subspaces of, 77

C4[0,1]XR2, 119

Carbon atom, 3

Cartesian coordinates, 422

Carbon nanotubes, 38

Cauchy-Euler equation, 173–175, 496

Cauchy sequence, 107

Cauchy-Schwartz inequality, 91, 96

Centrifugal pump, 221

Centripetal force, 326

Characteristic length, 32, 53

Chebyshev functions, 206

Circumferential angular coordinate, 3, 472

Clamped membrane, 508

Closure properties, 73

Column buckling

due to axial loading, 293–295

due to combined effects, 366–369

due to gravity, 341, 373–375

Communicative law of addition, 67, 73

Comparison functions, 236

Completeness, 107, 305

Complex vector space, 74

Conservation laws, 6

Conservation of angular momentum, 6

Conservation of energy, 6, 10

Conservation of mass, 6, 8, 55

Conservation of momentum, 6, 9

Constitutive equations, 3, 9

Continuum assumption, 3

Convective acceleration, 72

Convergence, 83, 107, 303, 313

Critical buckling load, 295, 366

Critical buckling length, 341, 366, 375

Critical buckling speed, 326, 328, 367

Critically damped, 152–153

Cylindrical coordinates, 471–472

D

Damping coeffi cient, 27

Damping ratio, 28

Degenerate eigenvalues, 303

Determinant, 516

Differential element

annular, 13

cylindrical, 16, 472

slice, 11, 19–20

spherical, 491–492

Differential equation

constant coeffi cients, 149–167

defi nition, 143

homogeneous solution, 149–154

particular solution, 154–167

variable coeffi cients, 167–173

Differential operator, 143

Diffusion equation, 421, 441

Dimension, 82

Dirac delta function, 400

Discrete model, 48

Distributive law of addition, 67, 73

Distributive law of scalar multiplication,

67, 73

Distributed parameter model, 18, 48

Divergence

cylindrical coordinates, 473

Divergence theorem, 128

Domain, 109

Dominant terms, 50

Dot product, 68

Double-walled nanotube, 349

Drag coeffi cient, 33

Dynamic similitude, 32

Dynamic viscosity, 9

E

Eigenvalue

defi nition, 277

9533_C007.indd 5189533_C007.indd 518 10/29/08 2:25:21 PM10/29/08 2:25:21 PM

Index 519

in boundary conditions, 329–331

matrix operator, 280–281

self-adjoint operators, 528

positive defi nite operators, 280

Sturm-Liouville problem, 292–303

Eigenvector

defi nition, 277

Eigenvector expansion, 306

Elastic foundation, 321

Elastic layer, 51

Elastic modulus, 18, 83

Elastically coupled beams, 50, 118, 349, 392

Electrostatic actuation, 271

Elliptic partial differential equations, 421

Empirical laws, 12

Energy balance, 12, 36, 45

Energy inner product

defi nition, 132

fourth-order problems, 251

use in Rayleigh-Ritz, 236

work and energy, 235–236

Energy norm, 133

Equation of plane, 111

Error, 213

Euler-Bernoulli beam, 135

Eulerian viewpoint, 7, 72

Euler’s constant, 378

Euler’s identity, 150

Even function, 310

Exact solution, 65

Existence of negative vector, 67, 73

Existence of solutions, 146

Explicit assumptions, 3

Extended surface, see fi n

External load, 138

F

Factorial, 181

Fin, 10

annular, 12–14

straight, 152–143

triangular, 14–15

Finite-dimensional vector spaces, 82

Finite-element method

basis functions for fourth-order

operator, 250–251

basis functions for second-order

operator, 243

beam on elastic foundation, 251–261

formulation, 242

non-uniform bar with attached spring, 250

temperature distribution in a

variable-area fi n, 267–270

torsional oscillations of non-uniform

shaft with attached disk and

spring, 387–392

uniform bar with attached spring, 246–249

First-order approximation, 51

Fixed-fi xed beam, 83

Fixed-free beam, 59

Fixed-pinned beam, 158

Flexibility matrix, 315

Flow rate, 72

Force, 69

Force polygon, 32

Fourier-Bessel series, 339

Fourier best approximation theorem, 212

Fourier series

convergence, 306

general, 306

trigonometric, 309–312

Fourier coeffi cients, 310

Fourier’s conduction law, 12, 17

Fourth-order operators, 318–328

Fredholm alternative, 356

Fredholm integral equation, 128

Free-body diagram, 9, 19, 69

Free-stream velocity, 53

Frequency ratio, 29

Frequency response, 28

Froude number, 61

Functional, 370

G

Galerkin’s method, 270–273

Gamma function, 181

Geometric boundary condition, 236

Geometric series, 168

Gradient, 473

Gram-Schmidt orthonormalization, 99–105,

218, 222

Grashopf number, 61

Green’s functions, 398

H

Hamilton’s principle, 22

Head, 221

Heat transfer coeffi cient, 10, 47

Heaviside function, 246

Hemisphere, 514

Hierarchal equations, 54, 359

Highly stretched beams, 42

Hilbert space, 108

Homogeneous solution

constant coeffi cients, 139

defi nition, 144

9533_C007.indd 5199533_C007.indd 519 10/29/08 2:25:22 PM10/29/08 2:25:22 PM

520 Index

Hooke’s law, 4

Hyperbolic partial differential equations, 421

Hyperbolic trigonometric functions, 154

I

Implicit assumptions, 3

Inaccurate modeling, 429

Incoming waves, 503

Incompressible fl ows, 21

Inconsistent boundary conditions, 429

Independent variables, 2

Infi nite-dimensional vector spaces, 83

Infl uence function, 400

Initial conditions, 419

Initial value problem, 144, 441

Inner products

Cn [a, b], 89

properties, 87

Rn, 88

Inner product generated norm, 96

Insulated surface, 45, 153

Integral equation, 407

Integrating factor, 149

Integration by parts, 25, 89, 236

Internal energy, 17, 236

Internal heat generation,

straight fi n, 156–157

thin rod, 117–118

Interpolating splines, 242

Intersection of planes, 112, 357

Intersection of vector spaces, 77

Intramodal frequencies, 353

Inverse operator, 110, 315, 398

J

Jump discontinuity, 310

K

Kinematic similitude, 33

Kinetic energy, 333

Kinetic energy inner product

discrete vibrations, 284

elastically connected axially loaded

beams, 350

non-unifrom shaft with attached disk, 382

shaft with attached disk, 333

L

Lagrange’s equations, 23

Lagrangian, 22

Lagrangian viewpoint, 7

Laplace’s equation, 421, 455

Laplacian

cylindrical coorinates, 473

positive-defi niteness, 131

self-adjointness, 127

Least squares

approximation of functions, 212

regression, 222–224

variational method, 224–226, 230–231

Legendre’s equation, 198, 495

Legendre function of the second kind, 200

Legendre polynomial, 199, 495–496

Leibnitz’s rule, 166

Length of vector

R3, 68

general vector spaces, 93–94

measure of error, 211

Lightly coupled beams, 51

Lightly stretched beams, 42

Linear operator, 110

Linearly independent vectors, 78

Linearly dependent vectors, 78

Local acceleration, 73

Longitudinal displacement of bar, 244–250

Lumped parameter model, 18, 46

M

Mach number, 33

Macroscale, 45

Magnifi cation factor, 29

Material derivative, 8

Mass density, 35, 83

Mass matrix, 283

Mass ratio, 50

Mass-spring-viscous damper system, 27,

34, 151

MATHCAD programs

beam on elastic foundation, fi nite-

element method, 253–261

beam on elastic foundation, least squares

and Rayleigh-Ritz, 230–235

least squares approximation for

functions, 214–221


variable-area fi n

fi nite-element method, 267–270

Rayleigh-Ritz method, 265–267

non-uniform beam using

Rayleigh-Ritz with admissible

functions, 240–242

torsional oscillations of non-uniform

shaft

9533_C007.indd 5209533_C007.indd 520 10/29/08 2:25:22 PM10/29/08 2:25:22 PM

Index 521

exact solution,

Rayleight-Ritz method, 386–392

vertical displacement of non-uniform

bar using fi nite-element method,

246–250

Mathematical induction, 170

Matrix, 110

Maxwell’s equations. 7

Mean value theorem, 17

Membrane defl ection, 140

MEMS actuator, 293

Method of Frobenius, 176

Microscale, 45

Method of undetermined coeffi cients, 155–162

Modal analysis, 316

Modal matrix, 316

Mode shape vector

defi nition, 284

orthogonality, 285

normalization, 286

Modifi ed Bessel functions, 184

Molecular dynamics, 3

Moment equation, 9

Moment of inertia, 38, 83

Monte Carlo methods, 3

Multiplication by one, 67, 73

Multi-walled nanotubes

Euler-Bernoulli beam model, 38

non-dimensional equations 42

small parameter analysis, 51–52

Mutually orthogonal vectors, 98

N

Nanoscale, 3, 45

Natural boundary condition, 236

Natural frequencies,

bar with varying density, 171

circular sector membrane, 485


beams, 392–394

elastically connected beams, 353

fi xed-fi xed beam, 352

Green’s functions, using, 407

multi-degree-of-freedom system, 284,

317, 375, 379

non-uniform bar, 366

one-degree-of-freedom system, 49

stretched beam on an elastic foundation,

321–325

torsional oscillations of shaft with

attached spring and disk,

343–348, 380

transverse vibrations of beams, 43

Navier-Stokes equations, 8, 21, 52

Neutral axis, 83

Newtonian reference frame, 3

Newton’s law of cooling, 12, 17

Newton’s second law, 19, 326

Newton’s viscosity law, 9

Nonhomogeneous boundary conditions,

145, 422

Nonhomogeneous problems, 314

Non-negativity property, 68, 87

Norm

||||∞, 94, 108

||||1, 95

defi nition, 94

inner-product-generated, 96

Norm of error, 223

Normal mode solution, 43, 84, 332, 407

Normal strain, 24

Normal stress, 24, 113, 281

Normal to plane, 111

Normalization, 286

Normalized, 98

No-penetration condition, 53

No-slip condition, 53

n-tuples, 74

Null set, 86

Numerical methods, 65

O

O(ε), 50, 359

Odd function, 310

One-dimensional heat conduction, 10

One-degree-of-freedom model, 49

One third order Bessel functions, 342

Ordinary point, 168

Operator, 109

Orthogonal vectors, 91

Orthogonality, 91–93

Orthogonality of eigenvectors

Bessel functions, 338

eigenvalues in boundary conditions, 320

Orthogonal expansions, 105–106

Orthonormal vectors, 98

Outgoing waves, 503

Overdamped, 152–153

P

Pn [a, b]

basis, 81

defi nition, 77

orthogonality of vectors, 93

orthonormal basis, 101

subspaces of, 84

9533_C007.indd 5219533_C007.indd 521 10/29/08 2:25:23 PM10/29/08 2:25:23 PM

522 Index

Parabolic fi t, 221

Parabolic partial differential equations, 421

Parseval’s identity, 108, 305, 308

Particular solution, 144

Perimeter, 11

Periodic boundary conditions, 291, 300

Periodic functions, 309

Periodic extension, 311

Perturbed problem, 358

Phase angle, 29

Piecewise linear splines, 242–243

Pinned-pinned beam, 399

Pipes, 60

Pointwise convergence, 169

Polar coordinate, 17

Polar moment of inertia, 58

Polymer gel, 38

Position vector, 66

Positive-defi nite operator

defi nition, 129

eigenvalues of, 280

Laplacian, 131

matrix, 130

Potential energy

multi-degree-of-freedom system, 115

Potential energy inner product

beam on elastic foundation, 251

beam with attached spring, 134


beams, 330

non-uniform bar with attached spring, 245

stiffness matrix, 133

Power series method, 169–173

Pressure, 53

Principal coordinates, 317

Principal stress, 282

Principal plane, 282

Problem identifi cation, 2

Product solution, 422, 442

Projection, 68, 98

Prototype, 32

Pulsating sphere, 500–508

Pythagorean theorem, 68

Q

Quadratic form, 115

Quadratic polynomial, 222

Quadrature formula, 222

R

Rn

complete basis, 303

defi nition, 74

inner products on, 87

linearly independent vectors, 78

orthogonality of vectors, 92

subspaces of, 76

Radial coordinate, 3, 472

Radiation condition, 419–420

Radius of convergence, 168

Range, 109

Rayleigh’s quotient, 370–376

Rayleigh-Ritz method

beam on elastic foundation, 232–234

eigenvalue problems, 376

formulation, 226

natural frequencies of fi ve-degree-of-

freedom system, 379–380

natural frequencies of non-uniform shaft

with attached disk and spring,

380–387

natural frequencies of elastically

coupled beams, 392–394


variable-area fi n, 265–267

using admissible functions, 238

using piecewise linear functions, 243

variable-area beam, 240–241

Real vector space, 74

Recurrence formula, 170

Regression, 222–224

Regular singular point, 173

Repeated root, 150

Resistive heating, 156

Resultant, 66

Reynolds number, 33, 54

Rotating column, 326–328, 366–369

S

Scalar function, 68, 86

Scalar multiplication

Cartesian coordinate, 66

Scaling, 45

Second law of thermodynamics, 7

Second-order differential operator

self-adjoint form, 291

Second-order effect, 51

Self adjoint operator

defi nition, 122

eigenvalues of, 278–279

fourth-order differential operators, 320

Fredholm integral operator, 128

Green’s functions, 399

Laplacian, 127

matrix operators, 125

second-order differential operators, 288

9533_C007.indd 5229533_C007.indd 522 10/29/08 2:25:24 PM10/29/08 2:25:24 PM

Index 523

Semi-norm, 94

Separable operator, 422

Separation argument, 423

Separation constant, 423, 442

Separation of variables, 422

diffusion equation in cylindrical

coordinates

axisymmetric, 488491

three-dimensional, 474–481

diffusion equation in one dimension,

443–448

diffusion equation in spherical

coordinates, 497–500

diffusion equation in two dimensions,

448–449

Laplace’s equation in spherical

coordinate, 493–497

Laplace’s equation in three dimensions

433–440

Laplace’s equation in two dimensions,

426–429

wave equation in one dimension, 449–454

wave equation in polar coordinates,

481–485

wave equation in spherical coordinates,

504–508

Shear force, 326

Shear stress, 54, 113, 281

Single valued solution, 478

Singular matrix, 78

Singularities, 56

Slope of elastic curve, 326

Solution space

basis, 148

constant coeffi cients, 151

defi nition, 143

dimension, 147

Solutions, 109

Solvability conditions, 355

Span, 80

Specifi c heat, 17, 35

Specifi c internal energy, 17

Speed of sound, 33

Spherical Bessel functions, 196–198,

345–348, 453, 499, 502

Spherical coordinates, 3, 491–492

Spherical shell, 514

Spherical waves, 501–508

Stability surface, 366

Standard inner product

Rn, 88

Cn [a, b], 88–89

Standard inner product generated norms, 97

Steady-state amplitude, 59

Steady-state solution, 461, 464, 486

Stiffness matrix

inverse, 315

positive-defi niteness, 130

potential energy formulation, 115

potential energy inner product, 133, 284

Strain energy, 236

Stress tensor, 113, 281

Stress vector, 113, 282

Structural stiffness operator, 349

Sturm-Liouville problem

complete eigenvectors, 306

defi nition, 292

Subspace

defi ned by boundary conditions, 77

defi nition, 76

intersection of two vector spaces, 84, 86

of Cn [a, b], 77

Superposition, 145

Superposition formula for partial

differential equations

temperature distribution in circular

cylinder with heat fl ux, 485–491


one-dimensional bar, 464–465


two-dimensional slab

applied heat fl ux, 460–464

internal heat generation, 458–459

non-homogeneous boundary

conditions, 455–457

Symmetric matrix, 123, 281

T

Tangential component of acceleration, 72

Taper rate, 363

Taylor series expansion, 17

Taylor’s theorem, 168

Temperature distribution in a cylinder

Nonhomogeneous, 485–491

unsteady, 474–481

Temperature distribution in a fi n

annular, of hyperbolic profi le, 194–195

internal heat source, with, 156–157

straight rectangular

homogeneous solution, 152–154

power series solution, 169–171

triangular, 185–188

variable area straight, 174–175

with internal heat generation, 262–270


Temperature distribution in a rod, 117


9533_C007.indd 5239533_C007.indd 523 10/29/08 2:25:24 PM10/29/08 2:25:24 PM

524 Index

Temperature distribution in a slab

steady three dimensional, 433

steady two-dimensional, 426

Temperature distribution in a sphere

non-dimensional formulation, 35–38

steady three-dimensional, 493–497

unsteady three-dimensional, 497–500

Temperature distribution in a

two-dimensional slab

internal heat generation, 457–459

non-dimensional formulation, 45–47

non-homogeneous, 455–457

two dimensional internal heat

generation, 465–468

unsteady with heat fl ux, 459–464

Temperature distribution in three

dimensions, 127, 433

Thermal conductivity, 10, 35

Three-dimensional space, 66

Transcendental equation, 297–298, 324, 334

Transient solution, 461, 464, 486

Transport problems, 8

Transpose matrix, 123, 281

Trial solutions, 155

Triangle inequality


property of norms, 94

U

Underdamped, 152–153

Undetermined coeffi cients, see Method of

undetermined coeffi cients

Uniform expansion, 358

Unit step function, 243

Unit vector, 67, 97

Uniqueness, 146

Unrestrained system, 284

Unsteady heat transfer

annular fi n, 15–18

uniform plate, 447–449

uniform bar, 443–447

V

Van der Waals forces, 38

Variable coeffi cient differential

equation, 83

Variation, 22

Variation of parameters, 163–167

Variational principles, 22

Vector addition

Cartesian coordinates, 66, 69

Vector length


Vector multiplication, 67

Vector space

Cn [a, b], 75

C2[ℜ], 85

Defi nition, 73

Pn [a, b], 77, 84

Rn, 74

Velocity vector, 70

Velocity potential, 501

Vibrations of continuous systems

longitudinal vibrations of bar

initial value problem, 450–454

non-dimensional formulation, 43

non-uniform bar, 363–366

one-degree-of-freedom

approximation, 49

varying density, 171

sector of membrane, 481–485

stretched beam on elastic foundation,

321–325

torsional oscillations of non-uniform shaft

exact solution, 343–348

fi nite-element approximations,

387–392

Rayleigh-Ritz approximation,

380–387

time-dependent forcing, 469–471

torsional oscillations of uniform shaft,

331–336

use of Hamilton’s principle, 24

Vibrations of discrete systems

forced response, 315–316

natural frequencies of, 284

potential energy, 115

Rayleigh-Ritz method, 379

Rayleigh’s quotient approximations,

375–376

use of Lagrange’s equations, 23

Vibrations of one-degree-of-freedom system

damped free response, 151–152

frequency response, 28

non-dimensional equation, 34–35

steady-state amplitude, 28

Viscous forces, 53

W

Wave equation, 20, 421, 441

Wave propagation, 501–508

Wronskian, 79, 145, 164

Z

Zero eigenvalue, 354

Zero vector, 73

Zeroes of Bessel functions, 339

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