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BEGINNING PARTIAL DIFFERENTIAL EQUATIONS

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BEGINNING PARTIALDIFFERENTIALEQUATIONS

BEGINNING PARTIALDIFFERENTIALEQUATIONS

PETER V. O’NEILUniversity of Alabama at Birmingham

A Wiley-Interscience Publication

JOHN WILEY & SONS, INC.

New York • Chichester • Weinheim • Brisbane • Singapore • Toronto

This book is printed on acid-free paper. V

Copyright q 1999 by John Wiley & Sons, Inc. All rights reserved.

Published simultaneously in Canada.

No part of this publication may be reproduced, stored in a retrievalsystem or transmitted in any form or by any means, electronic,mechanical, photocopying, recording, scanning or otherwise, except aspermitted under Section 107 or 108 of the 1976 United StatesCopyright Act, without either the prior written permission of thePublisher, or authorization through payment of the appropriate per-copyfee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers,MA 01923, (978) 750-8400, fax (978) 750-4744. Requests to thePublisher for permission should be addressed to the PermissionsDepartment, John Wiley & Sons, Inc., 605 Third Avenue, New York,NY 10158-0012, (212) 850-6011, fax (212) 850-6008, E-Mail:[email protected].

Library of Congress Cataloging-in-Publication Data:

O’Neil, Peter V.Beginning partial differential equations / Peter V. O’Neil.

p. cm.‘‘A Wiley-Interscience publication.’’Includes index.ISBN 0-471-23887-2 (cloth : alk. paper)1. Differential equations. Partial. I. Title.

QA377.054 1999 98-446215159.353—dc21

Printed in the United States of America

10 9 8 7 6 5 4 3 2 1

v

CONTENTS

Preface ix

1 First Order Partial Differential Equations 1

1.1 Preliminary Notation and Concepts, 11.2 The Linear Equation, 51.3 The Significance of Characteristics, 141.4 The Quasi-Linear Equation and the Method of Characteristics, 20

2 Linear Second Order Partial Differential Equations 29

2.1 Classification, 292.2 Canonical Form of the Hyperbolic Equation, 312.3 Canonical Form of the Parabolic Equation, 352.4 Canonical Form of the Elliptic Equation, 392.5 Canonical Forms and Equations of Mathematical Physics, 45

2.5.1 The Wave Equation, 452.5.2 The Heat Equation, 492.5.3 Laplace’s Equation, or the Potential Equation, 53

2.6 The Second Order Cauchy Problem, 542.7 Characteristics and the Cauchy Problem, 582.8 Characteristics as Carriers of Discontinuities, 682.9 The Cauchy-Kowalevski Theorem, 70

2.9.1 The General Cauchy Problem, 702.9.2 The Cauchy-Kowalevski Theorem, 722.9.3 An Example, 73

vi CONTENTS

2.9.4 Proof of the Cauchy-Kowalevski Theorem for LinearSystems, 76

2.9.5 Hormander’s Theorem, 86

3 Elements of Fourier Analysis 88

3.1 Why Fourier Series?, 883.2 The Fourier Series of a Function, 923.3 Convergence of Fourier Series, 96

3.3.1 Periodic Functions, 963.3.2 Dirichlet’s Formula, 983.3.3 The Riemann-Lebesgue Lemma, 1003.3.4 Convergence of the Fourier Series, 1023.3.5 Change of Scale, 108

3.4 Sine and Cosine Expansions, 1153.4.1 Fourier Sine Series, 1163.4.2 Fourier Cosine Series, 119

3.5 Bessel’s Inequality and Uniform Convergence, 1273.6 The Fourier Integral, 131

3.6.1 Fourier Cosine and Sine Integrals, 1343.7 The Complex Fourier Integral and the Fourier Transform, 1383.8 Convolution, 1483.9 Fourier Sine and Cosine Transforms, 155

4 The Wave Equation 158

4.1 The Cauchy Problem and d’Alembert’s Solution, 1584.1.1 An Alternate Derivation of d’Alembert’s Formula, 165

4.2 d’Alembert’s Solution as a Sum of Forward and BackwardWaves, 168

4.3 Domain of Dependence and the Characteristic Triangle, 1784.4 The Wave Equation on a Half-Line, 1834.5 A Nonhomogeneous Initial-Boundary Value Problem on a Half-

Line, 1884.6 A Nonhomogeneous Wave Equation, 1914.7 The Wave Equation on a Bounded Interval, 1954.8 A General Problem on a Bounded Interval, 2004.9 Fourier Series Solution on a Bounded Interval, 208

4.9.1 Verification of the Solution, 2134.10 A Nonhomogeneous Problem on a Bounded Interval, 2324.11 Fourier Integral Solution of the Cauchy Problem, 245

4.11.1 Solution by Fourier Transform, 2484.12 A Wave Equation in Two Space Dimensions, 2514.13 Kirchhoff/Poisson Solution in 3-Space and Huygens’ Principle, 2564.14 Hadamard’s Method of Descent, 263

CONTENTS vii

5 The Heat Equation 267

5.1 The Cauchy Problem for the Heat Equation Is Ill Posed, 2675.2 Initial and Boundary Conditions, 2685.3 The Weak Maximum Principle, 2715.4 Solutions on Bounded Intervals, 276

5.4.1 Ends of the Bar at Temperature Zero, 2765.4.2 Temperature in a Bar With Insulated Ends, 2825.4.3 Ends of the Bar at Different Temperatures, 2875.4.4 Diffusion of Charges in a Transistor, 293

5.5 Solutions on Unbounded Domains, 3025.5.1 Heat Conduction in an Infinite Bar, 3025.5.2 Heat Conduction in a Semi-Infinite Bar, 3125.5.3 A Problem Involving a Nonhomogeneous Boundary

Condition, 3215.6 The Nonhomogeneous Heat Equation, 3235.7 The Heat Equation in Several Space Variables, 332

6 Dirichlet and Neumann Problems 337

6.1 The Setting of the Problems, 3376.2 The Cauchy Problem for Laplace’s Equation Is Ill Posed, 3466.3 Some Harmonic Functions, 347

6.3.1 Harmonic Functions in the Plane, 3486.3.2 Harmonic Functions in R3, 351

6.4 Representation Theorems, 3536.4.1 A Representation Theorem in R3, 3536.4.2 A Representation Theorem in R2, 359

6.5 The Mean Value Property, 3616.6 The Maximum Principle, 3636.7 Uniqueness and Continuous Dependence of Solutions on Boundary

Data, 3686.8 Dirichlet Problem for a Rectangle, 3706.9 Dirichlet Problem for a Disk, 373

6.10 Poisson’s Integral Representation for a Disk, 3766.11 Dirichlet Problem for the Upper Half Plane, 384

6.11.1 Solution by Fourier Integral, 3846.11.2 Solution by Fourier Transform, 386

6.12 Dirichlet Problem for the Right Quarter Plane, 3876.13 Dirichlet Problem for a Cube, 3916.14 Green’s Function for a Dirichlet Problem, 393

6.14.1 Green’s Function for a Sphere, 3966.14.2 The Method of Electrostatic Images, 400

6.15 Conformal Mapping Techniques, 4026.15.1 Review of Conformal Mappings, 4036.15.2 Bilinear Transformations, 407

viii CONTENTS

6.15.3 Construction of Conformal Mappings BetweenDomains, 412

6.15.4 An Integral Solution of the Dirichlet Problem for aDisk, 420

6.15.5 Solution of Dirichlet Problems by Conformal Mapping, 4226.16 Existence of a Solution of the Dirichlet Problem, 427

6.16.1 An Example in 3-Space, 4276.16.2 An Example in the Plane, 4306.16.3 The Idea Underlying an Existence Theorem for the Dirichlet

Problem, 4326.16.4 Barrier Functions and an Existence Theorem, 4346.16.5 Dirichlet’s Principle, 435

6.17 The Neumann Problem, 4376.18 Neumann Problem for a Rectangle, 4416.19 Neumann Problem for a Disk, 4446.20 Neumann Problem for the Upper Half Plane, 448

7 Conclusion 451

7.1 Historical Perspective, 4517.2 References for Further Reading, 4547.3 Glossary, 4557.4 Answers to Selected Exercises, 456

Index 497

ix

PREFACE

This book is intended as a first course in partial differential equations. Topicsinclude characteristics, canonical forms, well-posed problems, and propertiesof solutions, as well as techniques for writing expressions for solutions.

Our objective is to provide an introduction to this important field of math-ematics, as well as an entry point, for those who wish it, to the modern, moreabstract elements of partial differential equations.

Because this is an introductory treatment, we attempt a balance betweentheory and technique. Computational facility and theoretical understanding re-inforce each other, and both are important for later work in related areas, suchas mathematical physics, differential geometry, or analytic number theory. Al-though the emphasis is on the mathematics, we also point out physical inter-pretations, for those circumstances in which an initial-boundary value problemmodels a physical setting.

The exercises have two purposes. Some are computational, ranging fromroutine to challenging. Answers for many of these are included at the end ofthe book. Other exercises provide additional information about partial differ-ential equations, or extensions of the material of the text. Many of these comewith hints for the development of a proof or a derivation.

Partial differential equations invite graphical representation and experimen-tation. Sometimes we can visualize a solution as a surface. Further, some partialdifferential equations model physical phenomena, and it is interesting and in-structive to couple mathematics with physical intuition by observing the evo-lution with time of a solution that represents heat conduction or wave motion.We can also experiment with the influence of different parameters on the phys-ical phenomenon represented by the partial differential equation, such as theeffect of the specific heat of a metal bar on the way it conducts heat energy,

x PREFACE

or the way the tension and density of a guitar string influence the way itvibrates. Parts of some exercises pursue such issues, and require the use of acomputational package, such as MAPLE or Mathematica. Readers who do nothave access to suitable hardware and software can skip over these exercises.

Preliminary versions of this book have been tested at the University of Al-abama at Birmingham over the past five years.

PREREQUISITES

The reader should be familiar with standard properties of real-valued functionsof n real variables, vector calculus (theorems of Green and Gauss), topics fromthe standard post-calculus course in elementary ordinary differential equations,and convergence of series and improper integrals. Occasional mention is madeof uniform convergence. Topics from Fourier analysis are not assumed as back-ground and are included. Several discussions (the use of conformal mappingsto solve the Dirichlet problem, contour integration to evaluate certain integrals,and Euler’s formula) use complex numbers and complex function theory.

Access to a computational program, such as MAPLE, is useful both in per-forming calculations and in studying properties of solutions, convergence ofFourier series, and other issues of interest in partial differential equations. Suchaccess is not a prerequisite to reading this book, but some exercises invitingcomputer study and experimentation have been included for those who haveit.

ACKNOWLEDGMENTS

I would like to thank my colleagues at UAB for helpful conversations andsuggestions, several classes of students for tolerating the use of preliminaryversions of this book in the form of course notes, and the editorial staff of JohnWiley & Sons for their professionalism in bringing the project to completion.

1

1FIRST ORDER PARTIALDIFFERENTIAL EQUATIONS

1.1 PRELIMINARY NOTATION AND CONCEPTS

A partial differential equation is an equation that contains at least one partialderivative. For example,

u u 22 x = xuy ,x y

4 2 3 3 w w w w w2 3 1 1 cos(xu) 2 xyzu = 04 3x zu u xyz x

and

2 2 2 h h h1 1 = f(x, y, z)2 2 2x y z

are partial differential equations. Such equations are of interest in mathematics(for example, in the study of curvature and surfaces) and in modeling phenom-ena in the sciences, engineering, economics, ecology, and other areas. For themoment we will be concerned with developing the vocabulary and notationwith which to engage in a discussion of partial differential equations.

There is some economy in using subscripts to denote partial derivatives. Inthis notion, ux = u/x, uxx = 2u/x2, uxy = 2u/yx, and so on. The partialdifferential equations listed above can be written, respectively,

2 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS

2u 2 xu = xuy ,x y

w 2 3w 1 w 1 cos(xu)w 2 xyzuw = 0xxxx uz uuu zyx x

and

h 1 h 1 h = f(x, y, z).xx yy zz

A solution of a partial differential equation is any function that satisfies theequation. Often we seek a solution satisfying certain conditions, and for theindependent variables confined to a specified set of values.

As an example of a solution, the equation

4u 1 3u 1 u = 0 (1.1)x y

has solution

2x/4u(x, y) = e f(3x 2 4y)

in which f can be any differentiable function of a single variable. We will verifythis by substituting u(x, y) into the partial differential equation. Chain ruledifferentiations yield

1 d d(3x 2 4y)2x/4 2x/4u = 2 e f(3x 2 4y) 1 e [ f(3x 2 4y)]x 4 d(3x 2 4y) dx

1 2x/4 2x/4= 2 e f(3x 2 4y) 1 3e f9(3x 2 4y)4

and

2x/4u = 24e f9(3x 2 4y).y

Upon substitution into equation 1.1 we obtain

2x/4 2x/44u 1 3u 1 u = 2e f(3x 2 4y) 1 12e f9(3x 2 4y)x y

2x/4 2x/42 12e f9(3x 2 4y) 1 e f(3x 2 4y) = 0

for all (x, y).As another example, consider the equation

u 2 9u = 0. (1.2)xx yy

PRELIMINARY NOTATION AND CONCEPTS 3

It is routine to check that

u(x, y) = f(3x 1 y) 1 g(3x 2 y)

is a solution for any twice differentiable functions f and g of a single variable.For example, we could choose f(t) = sin(t) and g(t) = e2t 1 t2 2 cos(t) toobtain the solution

23x1y 2u(x, y) = sin(3x 1 y) 1 e 1 (3x 2 y) 2 cos(3x 2 y).

In view of the latitude in choosing f and g, equation 1.2 has infinitely manysolutions (as does equation 1.1).

The order of a partial differential equation is the order of the highest partialderivative occurring in the equation. Equation 1.1 is of order one and equation1.2 is of order two.

A partial differential equation is linear if it is linear in the unknown functionand its partial derivatives. An equation that is not linear is nonlinear. Forexample,

2x u 2 yu = uxx xy

is a linear partial differential equation, while

2 2x u 2 yu = uxx xy

is nonlinear because of the u2 term, and

1/2(u ) 2 4u = xuxx yy

is nonlinear because of the (uxx)1/2 term.

A partial differential equation is quasi-linear if it is linear in its highest orderderivative terms. For example, the second order equation

3u 1 4yu 2 (u ) 1 u u = cos(u)xx yy x x y

is quasi-linear, being linear in its second derivative terms uxx and uyy. Thisequation is nonlinear because of the cos(u), uxuy and (ux)

3 terms. Of course,any linear equation is also quasi-linear.

We are now ready to begin studying first order partial differential equations.

EXERCISE 1 Show that

1u(x, y, z) =

2 2 2Ïx 1 y 1 z

is a solution of uxx 1 uyy 1 uzz = 0 for (x, y, z) ≠ (0, 0, 0).

4 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS

EXERCISE 2 Show that u(x, t) = f(x 1 ct) 1 g(x 2 ct) is a solution of

2u = c utt xx

for any twice differentiable functions f and g of one variable; c is a positiveconstant.

EXERCISE 3 Show that

x1ct1 1

u(x, t) = (w(x 1 ct) 1 w(x 2 ct)) 1 c(s) dsE2 2c x2ct

is a solution of utt = c2uxx for any w that is twice differentiable and c that isdifferentiable for all real x. c is a positive constant. Show that this solutionsatisfies the conditions

u(x, 0) = w(x), u (x, 0) = c(x)t

for all real x.

EXERCISE 4 Show that, if p is a continuously differentiable function of onevariable, then the first order partial differential equation

u = p(u)ut x

has solution implicitly defined by

u(x, t) = w(x 1 p(u)t),

in which w can be any continuously differentiable function of one variable.Use this idea to determine (perhaps implicitly) a solution of each of the

following equations:

1. ut = kux, with k a nonzero constant2. ut = uux

3. ut = cos(u)ux

4. ut = euux

5. ut = u sin(u)ux

EXERCISE 5 Show that

2 2u(x, y) = ln((x 2 x ) 1 ( y 2 y ) )0 0

satisfies Laplace’s equation uxx 1 uyy = 0 for all pairs (x, y) of real numbersexcept (x0, y0).

THE LINEAR EQUATION 5

EXERCISE 6 Let v and w be solutions of

a(x, y)u 1 b(x, y)u 1 c(x, y)u 1 d(x, y)u 1 e(x, y)u 1 g(x, y)u = 0.xx xy yy x y

Show that aw 1 bv is also a solution for any numbers a and b.

EXERCISE 7 In each of the following, classify the equation as linear, non-linear but quasi-linear, or not quasi-linear.

1. u2uxx 1 uy = cos(u)

2. x2ux 1 y2uy 1 uxy = 2xy

3. (x 2 y) 1 uxy = 12ux

4. (x 2 y) 1 2uy = 4y2ux

5. x2uyy 2 yuxx = tan(u)

6. ux 1 2 uxx = 42uy

7. ux 2 uxuy 2 uy = 0

8. uux 1 uxy = u2

9. uxy 2 1 2 sin(ux) = 02 2u ux y

10. uy /ux = x2

EXERCISE 8 Let k be a positive constant. Let

`

1 22(x2j) /4ktu(x, t) = e f(j) dj,E2 pktÏ 2`

in which f is continuous on the real line. Show that ut = kuxx for 2` < x < `,t > 0.

Determine u(x, t) when f(x) [ 1. Hint: Use a change of variables and thestandard result that

`

22we dw = p.ÏE2`

1.2 THE LINEAR EQUATION

Consider the linear first order partial differential equation in two independentvariables:

a(x, y)u 1 b(x, y)u 1 c(x, y)u = f(x, y). (1.3)x y

We assume that a, b, c, and f are continuous in some region of the plane, andthat a(x, y) and b(x, y) are not both zero for the same (x, y).

6 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS

We will show how to solve equation 1.3. The key is to determine a changeof variables

j = w(x, y), h = c(x, y)

which transforms equation 1.3 to the simpler linear equation

w 1 h(j, h)w = F(j, h) (1.4)j

where w(j, h) = u(x(j, h), y(j, h)). We will define this transformation in sucha way that it is one-to-one, at least for all (x, y) in some set $ of points in thex, y-plane. On $, then, we can, at least in theory, solve for x and y as functionsof j and h. To insure this we will require that the Jacobian of the transformationdoes not vanish in $:

w wx yJ = = w c 2 w c ≠ 0x y y xU U

c cx y

for (x, y) in $.Begin the search for a suitable transformation by computing chain rule

derivatives:

u = w j 1 w hx j x h x

and

u = w j 1 w h .y j y h y

Substitute these into equation 1.3 to obtain:

a(w j 1 w h ) 1 b(w j 1 w h ) 1 cw = f.j x h x j y h y

Write this equation as

(aj 1 bj )w 1 (ah 1 bh )w 1 cw = f. (1.5)x y x yj h

This is nearly in the form of equation 1.4 if we choose h = c(x, y) so that

ah 1 bh = 0x y

for (x, y) in $. If hy ≠ 0 this requires that

h bx = 2 .h ay

THE LINEAR EQUATION 7

Suppose for the moment that there is such an h. Putting h(x, y) = c, with can arbitrary constant, then

dh = h dx 1 h dy = 0x y

implies that

dy h bx= 2 = .dx h ay

This means that h = c(x, y) is an integral of the ordinary differential equation

dy b(x, y)= . (1.6)

dx a(x, y)

Equation 1.6 is called the characteristic equation of the linear equation 1.3.The equation h(x, y) = c defines a family of curves in the plane called char-acteristic curves, or characteristics, of equation 1.3. We will say more aboutthese in the next section.

Thus far we have found that we can make the coefficient of wh in thetransformed equation 1.5 vanish if we choose h = c(x, y), with c(x, y) = c anequation defining the general solution of the characteristic equation 1.6. Withthis step alone equation 1.5 comes very close to the transformed equation 1.4we want to achieve. We can now choose j to suit our convenience and thecondition that J ≠ 0. One simple choice is

j = w(x, y) = x.

With this choice,

1 0J = = hyU U

h hx y

and this is nonzero in $ by previous assumption.Now from equation 1.5, the change of variables

j = x, h = c(x, y)

transforms equation 1.3 to

a(x, y)w 1 c(x, y)w = f(x, y).j

8 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS

To complete the transformation to the form of equation 1.4, first write a(x, y),c(x, y), and f(x, y) in terms of j and h to obtain

A(j, h)w 1 C(j, h)w = p(j, h).j

Finally, restricting the variables to a set in which A(j, h) ≠ 0, we have

C pw 1 w =j A A

and this is in the form of equation 1.4 with

C(j, h) p(j, h)h(j, h) = and F(j, h) = .

A(j, h) A(j, h)

Example 1 Consider the linear equation

2x u 1 yu 1 xyu = 1.x y

This is equation 1.3 with a(x, y) = x2, b(x, y) = y, c(x, y) = xy and f(x, y) = 1.We will transform this equation to the simpler equation 1.4.

The characteristic equation is

dy b y= = .2dx a x

Write

1 1dy = dx,2y x

integrate and rearrange terms to obtain

1ln(y) 1 = c

x

for y > 0 and x ≠ 0. This is an integral of the characteristic equation and wechoose

1h = c(x, y) = ln(y) 1 .

x

Graphs of ln(y) 1 1/x = c are the characteristics of this partial differentialequation.

THE LINEAR EQUATION 9

Upon choosing j = x we have the Jacobian

1J = h = ≠ 0y

y

as required.Since j = x,

1h = ln(y) 1 ,

j

so

1ln(y) = h 2

j

and

h21/jy = e .

Now apply the transformation

1j = x, h = ln(y) 1 ,

x

with

w(j, h) = u(x, y).

Compute

1 1u = w j 1 w h = w 1 w 2 = w 2 wx x xj h j h S D j h2 2x j

and

1 1u = w j 1 w h = w = w .y y yj h h h h21/jy e

The partial differential equation transforms to

1 12 h21/j h21/jj w 2 w 1 e w 1 je w = 1,S j hD h2 h21/jj e

10 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS

or

2 h21/jj w 1 je w = 1.j

Then

1 1h21/jw 1 e w = ,j 2j j

and this has the form of equation 1.4, in any region of the j, h-plane with j≠ 0. m

The point to transforming equation 1.3 to the form of equation 1.4 is thatwe can solve this transformed equation. Think of

w 1 h(j, h)w = F(j, h)j

as a linear first order ordinary differential equation in j, with h carried alongas a parameter. Following the method for ordinary differential equations, mul-tiply the differential equation by

*h(j,h)dje

to obtain

*h(j,h)dj *h(j,h)dj *h(j,h)dje w 1 h(j, h)e w = F(j, h)e .j

Recognize this as

*h(j,h)dj *h(j,h)dj(e w) = F(j, h)e .j

Integrate with respect to j. Since h is being carried through this process as aparameter, the constant of integration may depend on h. We obtain

*h(j,h)dj *h(j,h)dje w = F(j, h)e dj 1 g(h),Ein which g is any differentiable function of one variable. Then

2*h(j,h)dj *h(j,h)dj 2*h(j,h)djw(j, h) = e F(j, h)e dj 1 g(h)e . (1.7)EThis is the general solution of the transformed equation (by general solution,

THE LINEAR EQUATION 11

we mean one that contains an arbitrary function). Now obtain the general so-lution of the original equation 1.3 by substituting j = j(x, y), h = h(x, y). Thisgeneral solution will have the form

a(x,y)u(x, y) = e [M(x, y) 1 g(c(x, y))]. (1.8)

in which g is any differentiable function of one variable.

Example 2 Consider the constant coefficient equation

au 1 bu 1 cu = 0x y

in which a, b, and c are numbers. Assume that a ≠ 0. The characteristic equationis

dy b=

dx a

with general solution defined by the equation

bx 2 ay = c.

Put

j = x, h = bx 2 ay.

The characteristics of this differential equation are the straight line graphs ofbx 2 ay = c.

With this transformation, we find by a routine calculation that the partialdifferential equation transforms to

aw 1 cw = 0j

or

cw 1 w = 0.j a

Multiply this equation by , or to get*(c/a)dj cj/ae e ,

ccj/a cj/ae w 1 we = 0,j a

hence

cj/a(e w) = 0.j

12 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS

Integrate with respect to j to get

cj/ae w = g(h)

in which g can be any differentiable function of one variable. Then

2cj/aw(j, h) = e g(h).

Finally, transform this solution back in terms of x and y:

2cx/au(x, y) = e g(bx 2 ay).

This solution is readily verified by substitution into the partial differentialequation. m

Observe that the solution in Example 2 has the form specified by equation1.8.

Example 3 Consider

u 1 cos(x)u 1 u = xy.x y

The characteristic equation is

dy= cos(x)

dx

with general solution defined by

y 2 sin(x) = c.

Put

j = x, h = y 2 sin(x).

Graphs of y 2 sin(x) = c are the characteristics of this partial differentialequation.

Now we have

y = h 1 sin(x) = h 1 sin(j)

and the partial differential equation transforms to

w 1 w = j[h 1 sin(j)].j

THE LINEAR EQUATION 13

Multiply this equation by which is e j, to obtain*dje ,

j j j je w 1 we = hje 1 je sin(j).j

Write this equation as

j j j(we ) = hje 1 je sin(j).j

Integrate to obtain

j j jwe = hje dj 1 je sin(j) djE E1 1j j j= he (j 2 1) 1 je (sin(j) 2 cos(j)) 1 e cos(j) 1 g(h).2 2

Then

1 1 2jw(j, h) = h(j 2 1) 1 j(sin(j) 2 cos(j)) 1 cos(j) 1 e g(h).2 2

Finally

1u(x, y) = (y 2 sin(x))(x 2 1) 1 x (sin(x) 2 cos(x))

2

1 2x1 cos(x) 1 e g(y 2 sin(x)),2

in which g is any differentiable function of a single variable. m

Contrast the idea of the general solution for the linear first order ordinarydifferential equation with that for the linear first order partial differential equa-tion. In the former case, the general solution of

y9 1 d(x)y = p(x)

contains an arbitrary constant. Graphs of the solutions obtained by makingchoices of the constant are curves in the (x, y) plane. If we require that y(x0)= y0, then we pick out the unique solution corresponding to the curve passingthrough (x0, y0).

By contrast, if u is the general solution of the linear first order partial dif-ferential equation 1.3, then z = u(x, y) defines a family of surfaces in 3-space,each surface corresponding to a choice of the arbitrary function g in equation

14 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS

1.8. In the next section we will investigate the kind of information that shouldbe given in order to pick out one of these surfaces and determine a uniquesolution.

EXERCISE 9 For each of the following partial differential equations, (a)solve the characteristic equation and sketch graphs of some of the character-istics, (b) define a transformation of the partial differential equation to the formof equation 1.4 and obtain the transformed equation, (c) find the general so-lution of the transformed equation, (d) find the general solution of the givenequation, and (e) verify the solution by substituting it into the partial differentialequation.

1. 3ux 1 5uy 2 xyu = 0

2. ux 2 uy 1 yu = 0

3. ux 1 4uy 2 xu = x

4. 22ux 1 uy 2 yu = 0

5. xux 2 yuy 1 u = x

6. x2ux 2 2uy 2 xu = x2

7. ux 2 xuy = 4

8. x2ux 1 xyuy 1 xu = x 2 y

9. ux 1 uy 2 u = y

10. ux 2 y2uy 2 yu = 0

11. ux 1 yuy 1 xu = 0

12. xux 1 yuy 1 2 = 0

EXERCISE 10 Find the general solution of

1u 1 a(y 2 1)u = bf(x)(y 2 1)ux y 2

in which a and b are real numbers and f is continuous on the real line. Usethe general solution to find the solution satisfying

nu(0, y) = y ,

in which n is a nonnegative integer.

1.3 THE SIGNIFICANCE OF CHARACTERISTICS

In the preceding section we mentioned characteristics but did not actually doanything with them. Now we will investigate their significance, beginning withan example that will be instructive for the point we want to make.

THE SIGNIFICANCE OF CHARACTERISTICS 15

Consider

2u 1 3u 1 8u = 0.x y

The characteristic equation is

dy 3=

dx 2

and the characteristics are the straight line graphs of 3x 2 2y = c.Using the method of Section 1.2, we find that this partial differential equa-

tion has general solution

24xu(x, y) = e g(3x 2 2y)

in which g can be any differentiable function defined over the real line.Notice that simply specifying that the solution is to have a given value at a

particular point does not uniquely determine g, and hence does not determinea unique solution, as occurs with ordinary differential equations.

Now suppose we specify values of u(x, y) along a curve G in the plane. Tobe specific for this example, suppose we choose G as the x-axis and give valuesof u(x, y) at points on G, say

u(x, 0) = sin(x).

We need

24xu(x, 0) = e g(3x) = sin(x)

so

4xg(3x) = e sin(x).

Putting t = 3x,

4t/3g(t) = e sin(t/3).

This determines g and the solution satisfying the condition u(x, 0) = sin(x) onG is

124x 24x 4(3x22y)/3u(x, y) = e g(3x 2 2y) = e e sin (3x 2 2y)S D3

228y/3= e sin x 2 y .S D3

16 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS

In this example, specifying values of u along the x-axis uniquely determinedthe arbitrary function in the general solution, and hence determined the uniquesolution of the partial differential equation having these given values.

Next seek a solution having given values along the line y = x, say

4u(x, x) = x .

From the general solution, this requires that

24x 4u(x, x) = e g(x) = x

and we can choose

4 4xg(x) = x e

to obtain the unique solution

24x 8(x2y) 4u(x, y) = e g(3x 2 2y) = e (3x 2 2y)

satisfying u(x, x) = x4.Despite these two successes, not every curve in the plane can be used to

determine g. Suppose we choose G to be the line 3x 2 2y = 1, and prescribevalues u(x, y) is to have along G, say

1 2u x, (3x 2 1) = x .S D2

Now we must choose g so that

124x 2e g 3x 2 2 (3x 2 1) = xS D2

and this requires that

4x 2g(1) = e x .

This is impossible, hence there is no solution taking the value x2 at points(x, y) on this line.

Why did some choices of G give a solution, and another choice no solution?The difference was that the x-axis and the line y = x are not characteristics ofthe partial differential equation, while the line 3x 2 2y = 1 is a characteristic.

To understand the significance of characteristics in the context of existence

THE SIGNIFICANCE OF CHARACTERISTICS 17

and uniqueness of solutions, go back to the general solution 1.8 of the linearfirst order partial differential equation 1.3. This general solution is

a(x,y)u(x, y) = e [M(x, y) 1 g(c(x, y))].

Suppose we prescribe u(x, y) = q(x) along a characteristic. Now a characteristicis specified by c(x, y) = k. If y = y(x) along this characteristic, then

a(x,y(x))q(x) = e [M(x, y(x)) 1 g(k)]

or

a(x,y(x))q(x) = e [M(x, y(x)) 1 C], (1.9)

in which C is constant. The functions M(x, y) and a(x, y) are determined bythe partial differential equation, and are not under our control, so equation 1.9places a constraint on the given data function q(x). If q(x) is not of this formfor any constant C, then there is no solution taking on these prescribed valueson G. On the other hand, if q(x) is of this form for some C, then there areinfinitely many such solutions, because we can choose for g any differentiablefunction such that g(k) = C.

Example 4 Consider

2 2 xxu 1 2x u 2 u = x e . (1.10)x y

First we will find the general solution. The characteristic equation is

dy= 2x

dx

and this has general solution defined by y 2 x2 = k. The characteristics areparabolas. Let

2j = x, h = y 2 x

to obtain

2 jjw 2 w = j e ,j

which we write as

1 jw 2 w = je .j j

18 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS

Multiply this equation by which is 1/j, to obtain*(21/j)dje ,

1 1 jw 2 w = e ,j 2j j

or

1 jw = e .S Dj j

Integrate with respect to j to get

1 jw = e 1 g(h)j

so

jw = je 1 jg(h).

The general solution of equation 1.10 is

x 2u(x, y) = xe 1 xg(y 2 x ).

We will now attempt to find solutions satisfying given conditions alongvarious curves.

Suppose first we seek a solution such that u(x, y) = sin(x) on the curve y =x2 1 4. Notice that information is being specified along a characteristic. Wewill need

2 xu(x, x 1 4) = xe 1 xg(4) = sin(x).

We must be able to find a constant C such that

xxe 1 Cx = sin(x)

for all x, and this is impossible. There is no solution satisfying the requestedcondition.

Next suppose we want a solution such that u(x, y) = xex 2 x on the parabolay = x2 1 4. Now we need

2 x xu(x, x 1 4) = xe 1 xg(4) = xe 2 x.

This equation requires that g(4) = 21. This problem has infinitely many so-lutions because we can choose g to be any differentiable function of one var-

THE SIGNIFICANCE OF CHARACTERISTICS 19

iable such that g(4) = 21. Even though data is specified on a characteristic,the form of the data allows infinitely many solutions.

Finally, suppose we want a solution such that u(x, y) = cos(x) along the(noncharacteristic) parabola y = x2 1 4x. Now we need

2 xu(x, x 1 4x) = xe 1 xg(4x) = cos(x).

This requires that

xcos(x) 2 xeg(4x) = .

x

Choose

t t/4cos(t/4) 2 e4

g(t) = 4t

for, say, t > 0. The solution of the problem (for x > 0) is

x 2u(x, y) = xe 1 xg(y 2 x )2y 2 x 1 22 ( y2x )/4cos 2 (y 2 x )eS D4 4

x= xe 1 4x S D . m2y 2 x

The problem of finding a solution of equation 1.3 taking on prescribedvalues on a given curve is called a Cauchy problem (for the linear equation),and the given information on the curve is called Cauchy data. Our examplessuggest that we can expect a unique solution of a Cauchy problem if the curveis not characteristic, and no solution or infinitely many solutions if the curveis characteristic.

EXERCISE 11 For each of the following partial differential equations, solvethe characteristic equation and sketch graphs of some of the characteristics,find the general solution of the partial differential equation, and attempt to findsolutions satisfying the Cauchy data on the given curves.

1. 3yux 2 2xuy = 0

(a) Find a solution satisfying u(x, y) = x2 on the line y = x.

(b) Find a solution satisfying u(x, y) = 1 2 x2 on the line y = 2x.

(c) Find a solution satisfying u(x, y) = 2x on the ellipse 3y2 1 2x2 = 4.

20 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS

2. ux 2 6uy = y

(a) Find a solution satisfying u(x, y) = ex on the line y = 26x 1 2.

(b) Find a solution satisfying u(x, y) = 1 on the parabola y = 2x2.

(c) Find a solution satisfying u(x, y) = 24x on the line y = 26x.

3. 4ux 1 8uy 2 u = 1

(a) Find a solution satisfying u(x, y) = cos(x) on the line y = 3x.

(b) Find a solution satisfying u(x, y) = x on the line y = 2x.

(c) Find a solution satisfying u(x, y) = 1 2 x on the curve y = x2.

4. 24yux 1 uy 2 yu = 0

(a) Find a solution satisfying u(x, y) = x3 on the line x 1 2y = 3.

(b) Find a solution satisfying u(x, y) = 2y on y2 = x.

(c) Find a solution satisfying u(x, y) = 2 on x 1 2y2 = 1.

5. yux 1 x2uy = xy

(a) Find a solution satisfying u(x, y) = 4x on the curve y = (1/3)x3/2.

(b) Find a solution satisfying u(x, y) = x3 on the curve 3y2 = 2x3.

(c) Find a solution satisfying u(x, y) = sin(x) on the line y = 0.

6. y2ux 1 x2uy = y2

(a) Find a solution satisfying u(x, y) = x on y = 4x.

(b) Find a solution satisfying u(x, y) = 22y on y3 = x3 2 2.

(c) Find a solution satisfying u(x, y) = y2 on y = 2x.

1.4 THE QUASI-LINEAR EQUATION AND THE METHODOF CHARACTERISTICS

Consider the first order quasi-linear partial differential equation

f(x, y, u)u 1 g(x, y, u)u = h(x, y, u) (1.11)x y

in which we seek a solution u as a function of the independent variables x andy. This is linear in the highest partial derivatives (which in the first order caseare of order one), but may be nonlinear in u.

In the linear case we defined characteristics to be certain curves in the x,y-plane. Given a solution u(x, y) of a linear equation and a characteristic G, wecan define a curve on the surface u = u(x, y) whose projection in the x, y-planeis G (Figure 1.1). In this context the curve in x, y, u-space is also called acharacteristic, and its projection into the x, y-plane is called a characteristictrace.

THE QUASI-LINEAR EQUATION 21

FIGURE 1.1 A characteristic C on a solution surface, projecting onto a characteristictrace.

For the quasi-linear equation 1.11, characteristics are curves in x, y, u-spacedefined by

dx dy du= f(x, y, u), = g(x, y, u), = h(x, y, u). (1.12)

dt dt dt

We will show that a solution u(x, y) of a quasi-linear equation may beinterpreted as a surface made up of characteristics. This observation can beused to obtain solutions containing given (noncharacteristic) curves, and henceprovides a way of solving the Cauchy problem when the partial differentialequation is quasi-linear.

We need two observations.

Observation One Suppose u = w(x, y) is a solution of equation 1.11 de-fining a surface (, and that P0 : (x0, y0, u0) is a point on (, so u0 = w(x0, y0).Then the characteristic passing through P0 lies entirely on (.

To see why this is true, suppose the characteristic has parametric equations

x = x(t), y = y(t), u = u(t).

Then for some t0,

x(t ) = x , y(t ) = y , u(t ) = u .0 0 0 0 0 0

22 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS

Because this curve is characteristic, we can use equations 1.12 to compute

dt dx dyw(x(t), y(t)) = w 1 wx ydt dt dt

du= w f(x, y, u) 1 w g(x, y, u) = h(x, y, u) = .x y

dt

Therefore

u(t) = w(x(t), y(t)) 1 k

for some constant k. But P0 is on the surface, so

w(x(t ), y(t )) = u = u(t )0 0 0 0

implies that k = 0. Therefore

u(t) = w(x(t), y(t))

and the characteristic lies on (.

Observation Two If we begin with an arbitrary (but noncharacteristic)curve G and construct the family of characteristics passing through points ofG, as in Figure 1.2, then the resulting surface ( is the graph of a solution ofthe partial differential equation.

To see why this is true, assume that ( is the graph of u = w(x, y). We wantto show that w is a solution of equation 1.11.

Suppose G is parameterized by

x = x(s), y = y(s), z = z(s).

At any (x, y, u) on (,

dx dy du= f(x, y, u), = g(x, y, u), = h(x, y, u)

ds ds ds

because the surface is made up of characteristics. Then

du dx dy= h(x, y, u) = w 1 w = fw 1 gwx y x yds ds ds

so w is a solution.

THE QUASI-LINEAR EQUATION 23

FIGURE 1.2 A solution surface formed by constructing characteristics through pointsof G.

These observations suggest the method of characteristics for solving theCauchy problem when the partial differential equation is quasi-linear. Supposewe want the solution of equation 1.11 assuming prescribed values on a givencurve G that is not characteristic. Construct the characteristic through each pointof G. This defines a surface in three space, and this surface is the graph of thesolution of the Cauchy problem.

This strategy also suggests why we do not want to specify data along acharacteristic C. If we did so, then the characteristic through each point of Cwould be just C itself, not a surface.

Here are two illustrations of the method of characteristics.

Example 5 We want the solution of

uyu 2 xu = ex y

that passes through the curve G given by y = sin(x), u = 0. This means that werequire

u(x, sin(x)) = 0.

The characteristics of this partial differential equation are specified by

dx dy du u= y, = 2x, = e .dt dt dt

24 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS

From the first two of these equations we can write

dy x= 2 ,

dx y

or

ydy 1 xdx = 0,

with general solution (in terms of t)

x = a cos(t) 1 b sin(t), y = b cos(t) 2 a sin(t),

with a and b constant. From du/dt = eu we obtain

2u2e = t 1 c.

The characteristics therefore have parametric representation

2ux = a cos(t) 1 b sin(t), y = b cos(t) 2 a sin(t), e = c 2 t.

Parameterize G as

x = s, y = sin(s), u = 0.

We use s as parameter on G to distinguish between points on G and points oncharacteristics.

We want to construct a characteristic through each point of G (as in Figure1.2). Let P : (s, sin(s), 0) be a point of G and suppose a characteristic passesthrough P when t = 0 (this is just a scaling of the parameter). Then at t = 0,

0x = a = s, y = b = sin(s), and e = 1 = 0 1 c,

giving us a = s, b = sin(s), and c = 1 at this point of intersection of G with P.Therefore, the characteristic intersecting G at P has parametric equations

2ux = s cos(t) 1 sin(s)sin(t), y = sin(s)cos(t) 2 s sin(t), e = 1 2 t.

Now eliminate t and s from these equations. From the first two equations,

s = x cos(t) 2 y sin(t)

and

sin(s) = y cos(t) 1 x sin(t).

THE QUASI-LINEAR EQUATION 25

FIGURE 1.3(a) Solution of yux 2 xuy = eu containing the curve y = sin(x), u = 0(Example 5).

Therefore

sin(x cos(t) 2 y sin(t)) = y cos(t) 1 x sin(t). (1.13)

But e2u = 1 2 t implies that t = 1 2 e2u. Substitute this into equation (1.13)to get

2u 2u 2u 2usin(x cos(1 2 e ) 2 y sin(1 2 e )) = y cos(1 2 e ) 1 x sin(1 2 e )).

This equation implicitly defines the solution of the Cauchy problem. It is easyto check that y = sin(x), u = 0 satisfies this equation.

Figure 1.3(a) shows a graph of part of the surface (solution), and Figure1.3(b) shows the same surface from a different perspective. m

Example 6 Consider the quasi-linear equation

xu 1 yu = sec(u).x y

We would like the solution passing through

2G :x = s , y = sin(s), u = 0.

The characteristics satisfy

dx dy du= x, = y, = sec(u),

dt dt dt

which have solutions defined by

t tx = Ae , y = Be , sin(u) = t 1 c.

As in Example 5, we want to construct the characteristic through each point

26 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS

FIGURE 1.3(b) Another perspective of the solution surface shown in Figure 1.3(a)(Example 5).

of G. Suppose a characteristic passes through G at P : (s2, sin(s), 0) at t = 0.Then

2x = A = s , y = B = sin(s), sin(0) = 0 = 0 1 c,

so

2A = s , B = sin(s) and c = 0

at this point. Then

2 t tx = s e , y = sin(s)e and sin(u) = t.

We want to eliminate s and t from these equations. Since t = sin(u), then

sin(u)y = sin(s)e

so

2sin(u)sin(s) = ye

and

2sin(u)s = arcsin(ye ).

THE QUASI-LINEAR EQUATION 27

FIGURE 1.4(a) Graph of the curve x = t2, y = sin(t), u = 0 (Example 6).

Then

sin(u) 2sin(u) 2x = e [arcsin(ye )] .

This equation implicitly defines u(x, y) such that u(s2, sin(s)) = 0—that is, thesolution surface contains the data curve G. Figure 1.4(a) shows a graph of G,and Figure 1.4(b) part of the solution surface. m

EXERCISE 12 For each of the following, use the method of characteristicsto find a solution of the partial differential equation that passes through thegiven curve G. The solution may be implicitly defined. Graph part of the so-lution surface.

1. xux 1 yuy = sec(u); G is the curve defined by y = x3, u = 0.

2. ux 2 xuy = 4; G is given by y = 4x, u = 0.

3. ux 2 y2uy = 1; G is given by y = x2 1 2; u = 0.

4. ux 2 y3uy = sec(u); G is given by y = x2, u = 0.

5. ux 1 yuy = u; G is given by y = 1 2 x, u = 1.

6. ux 1 y2uy = cos(u); G is given by x = y2, u = 0.

7. ux 2 uy = u2; G is given by y = 2x 2 1, u = 4.

8. x3ux 2 yuy = u; G is given by x = y2 2 1, u = 0.

9. ux 2 y2uy = u; G is given by y = 1 2 x2, u = 2.

10. xux 1 uy = eu; G is given by y = x 2 1, u = 0.

28 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS

FIGURE 1.4(b) Solution of xux 1 yuy = sec(u) containing the curve shown in Figure1.4(a).

EXERCISE 13 Use the method of characteristics to show that the solutionof the problem

uu 1 u = 0x y

u(x, 0) = f(x)

is defined implicitly by the equation

u(x, y) = f(x 2 u(x, y)y).

Hint: Think of G as the curve x = s, y = s, u = f(s).Next use the idea of Exercise 4 to solve this problem. Compare these

solutions.

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