Hackbusch - Elliptic Differential Equations

Springer Series inComputationalMathematics

EllipticDifferentialEquationsTheory andNumericalTreatment

W. Hackbusch

4 Springer

Wolfgang Hackbusch

Elliptic

Differential

Equations

Theory andNumericalTreatment

Translated from the Germanby Regine Fadiman and Patrick D.F. Ion

With 40 Figures

4 Springer

Wolfgang Hackbusch

MPI für Mathematikin den Naturwissenschaftenlnselstr. 22-2604103 Leipzig, Germany

e-mail: [email protected]

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Second Printing 2003

Mathematics Subject Classification (2000): 35J20, 35J25, 35J30, 35J35,35J50, 35J55, 65N06, 65N12, 65N15, 65N25, 65N30

ISSN 0179-3632ISBN 3-540-54822-X Springer-Verlag Berlin Heidelberg New York

o B.G. Teubner, Stuttgart 1987: W. Hackbusch, Throne und Numenik elliptischer Differen-tialgleichungen. Mit Genehmigung des Verlagea B.G. Teubner, Stuttgart, veranstaltete, alleinautorisierte engliache Obersetzung der deutschen Originalausgabe.

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Foreword

This book has developed from lectures that the author gave for mathematicsstudents at the Ruhr-Universität Bochum and the Christian-Albrechts-Uni-

Kiel. This edition is the result of the translation and correction ofthe Gennan edition entitled Theorie und NtLmerik Differential-gieichungen.

The present work is restricted to the theory of partial differential equa-tions of elliptic type, which otherwise tends to be given a treatment whichis either too superficial or too extensive. The following sketch shows what theproblems are for elliptic differential equations.

The theory of elliptic differential equations (A) is concerned with ques-tions of existence, uniqueness, and properties of so&utions. The first problem of

B: Discretisatlon:Difference Methods,finite elements, etc.

iv Foreword

numerical treatment is the description of the discretisation procedures(B), which give finite-dimensional equations for approximations to the solu-tions. The subsequent second part of the numerical treatment is numericalanalysis (C) of the procedure in question. In particular it is necessary tofind out if, and how fast, the approximation converges to the exact solution.The solution of the finite-dimensional equations (D, E) is in general no sim-pie problem, since from to 106 unknowns can oceur. The discussion ofthis third area of numerical problems is skipped (one may find it, e.g., inHackbusch [5] and [9)).

The descriptions of discretisation procedures and their analyses areclosely connected with corresponding chapters of the theory of elliptic equa-tions. In addition, it is not possible to undertake a well-founded numericalanalysis without a basic knowledge of elliptic differential equations. Since thelatter cannot, in general, be assumed of a reader, it seems to me necessary topresent the numerical study along with the theory of elliptic equations.

The book is conceived in the first place as an introduction to the treat-ment of elliptic boundary-value problems. It should, however, serve to leadthe reader to further literature on special topics and and applications. It is in-tentional that certain topics, which are often handled rather summarily, (e.g.,elgenvalue problems) are treated here in greater detail.

The exposition is strictly limited to linear elliptic equations. Thus adiscussion of the Navier-Stokes equations, which are important for fluid me-chanics, i8 excluded; however, one can approach these matters via the Stokesequation, which is thoroughly treated as an example of an elliptic system.

In order not to exceed the limits of this book, we have not considered fur-ther discretisation methods (collocation methods, volume-element methods,spectral methods) and Integral-equation methods (boundary-element meth-ods).

The Exercises that are presented, which may be considered as remarkswithout proofs, are an integral part of the exposition. If this book is used asthe text for a course they can be used as student problems. But the readertoo should test his understanding of the subject on the exercises.

The author wishes to thank his collaborators G. Hofmann, G. Wittwnand J. Burmeister for the help in reading and correcting the manuscript of thisbook. He thanks Thubner Verlag for their cordial collaboration in producingthe first German edition.

Kiel, December 1985 W. Hackbusch

This translation contains, in addition to the full text of the original edi-tion, a short Section on the integral-equation method. The bibliographyhas also been expanded.

The author wishes to thank the translators, R. Fadiman and P. D. F.Ion, for their pleasant collaboration, and Springer-Verlag for their friendlycooperation.

Kiel, March 1992 W. Hackbusch

Table of Contents

Foreword

Thble of Contents v

Notation xi

1 Partial Differential Equations andTheir Classification Into Types 1

1.1 Examples I

1.2 Classification of Second-Order Equations into Types . 4

1.3 Type Classification for Systems of First Order . . . . 6

1.4 Characteristic Properties of the Different Types . . . 7

2 The Potential Equation 12

2.1 Posing the Problem 12

2.2 Singularity Function 14

2.3 The Mean Value Property and Maximum Principle 17

2.4 Continuous Dependence on the Boundary Data . . . . 23

3 The Poisson Equation 27


3.2 Representation of the Solution by the Green Function . 28

3.3 The Green Function for the Ball 34

3.4 The Neumann Boundary Value Problem 353.5 The integral Equation Method 36

4 Difference Methods for the Poisson Equation 38

4.1 Introduction: The One-Dimensional Case 384.2 The Five-Point Formula 404.3 M-matrices, Matrix Norms, Positive Definite

Matrices 444.4 Properties of the Matrix Lh 534.5 Convergence 59

Vi Table of Contents

4.6 Discretisations of Higher Order 62

4.7 The Discretisation of the NeumannBoundary Value Problem 65

4.7.1 One-sided Difference for 9u/ôn 65

4.7.2 Symmetric Difference for Ou/On 70

4.7.3 Symmetric Difference for Ou/ônonanOffsetGrid 71

4.7.4 Proof of the Stability Theorem 7 72

4.8 Discretisation in an Arbitrary Domain 78

4.8.1 Shortley-Weller Approximation 78

4.8.2 Interpolation at Points near the Boundary . . . 83

5 General Boundary Value Problems 85

5.1 Dirichlet Boundary Value Problems for LinearDifferential Equations 85

5.1.1 Posmg the Problem 85

5.1.2 Maximum Principle 86

5.1.3 Uniqueness of the Solution and ContinuousDependence 87

5.1.4 Difference Methods for the General DifferentialEquation of Second Order 90

5.1.5 Green's Function 95

5.2 General Boundary Conditions 95

5.2.1 Fbrmulating the Boundary Value Problem . . . 95

5.2.2 Difference Methods for General BoundaryConditions 98

5.3 Boundary Problems of Higher Order 103

5.3.1 The Biharmonic Differential Equation 103

5.3.2 General Linear Differential Equations of Order 2m 104

5.3.3 Discretisation of the Bihaimonic DifferentialEquation . . . 105

6 Ibols from Functional Analysis 110

6.1 Banach Spaces and Hubert Spaces 110

6.1.1 Normed Spaces 110

6.1.2 Operators 111

6.1.3 Banach Spaces 112

6.1.4 Hilbert Spaces 114

6.2 Sobolev Spaces 115

6.2.1 L2(f1) 115

6.2.2 Hc(O) and 116

6.2.3 Fburier and Hc(IRI)) 119

Table of Contents vii

6.2.4 H8(fl) for Real a � 0 122

6.2.5 and Extension Theorems 123

6.3 Dual Spaces 130

6.3.1 Dual Space of a Normed Space 130

6.3.2 Adjoint Operators 132

6.3.3 Scales of Hilbert Spaces 133

6.4 Compact Operators 135

6.5 Bilinear Forms 137

7 Variational Ermulation 144

7.1 HIstorical Remarks 144

7.2 Equations with Homogeneous DirichietBoundary Conditions 145

7.3 Inhomogeneous Dirichiet Boundary Conditions 150

7.4 Natural Boundary Conditions 152

8 The Method of Finite Elements 161

8.1 The Ritz-Galerkin Method 161

8.2 Error Estimates 167

8.3 FinIte Elements 171

8.3.1 Introduction: Linear Elements for 1? = (a,b) . . 171

8.3.2 Linear Elements for a c JR2 174

8.3.3 Bilinear Elements for a C JR2 178

8.3.4 Quadratic Elements for (1 C It2 180

8.3.5 Elements for a C Ut3 182

8.3.6 Handling of Side Conditions 182

8.4 Error Estimates for Finite Element Methods 185

8.4.1 H1-Estimates for Linear Elements 185

8.4.2 L2 and H Estimates for Linear Elements . . . 190

8.5 Generalisations 193

8.5.1 Error Estimates for Other Elements 193

8.5.2 Finite Elements for Equations of Higher Order . 194

8.5.2.1 Introduction: The One-DimensionalBtharmornc Equation 194

8.5.2.2 The Two-Dimensional Case 195

8,5.2.3 Estimating Errors 196

8.6 Finite Elements for Non-Polygonal Regions 196

8.7 Additional Remarks 199

8.7.1 Non-Conformal Elements 199

viii Thble of Contents

8.7.2 The Trefftz Method 200

8.7.3 Finite-Element Methods for Singular Solutions . 201

8.7.4 Adaptive Thangulation 201

8.7.5 Hierarchical Bases 202

8.7.6 Superconvergence 202

8.8 Properties of the Matrix 203

9 RegularIty 208

9.1 Solutions of the Boundary Value ProbleminH'(37),a>m 208

9.1.1 The Regularity Problem 208

9.1.2 Regularity Theorems for 210

9.1.3 Regularity Theorems for = 215

9.1.4 Regularity Theorems for General 1? C . . . 219

9.1.5 Regularity for Convex Domains and Domainswith Corners 223

9.1.6 Regularity in the Interior 226

9.2 Regularity Properties of Difference Equations . . . . 226

9.2.1 Discrete H'-Regularity 226

9.2.2 Consistency 232

9.2.3 Optimal Error Estimates 238

9.2.4 Hz-Regularity 240

10 SpecIal Differential Equations 244

10.1 Differential Equations with Discontinuous Coefficients 244

10.1.1 Fbnnulation 244

10.1.2 Discretisation 246

10.2 A Singular Perturbation Problem 247

10.2.1 The Convection-Diffusion Equation 247

10.2.2 Stable Difference Schemes 249

10.2.3 Finite Elements 251

11 Elgenvalue Problems 253

11.1 Formulation of Eigenvalue Problems . . 253

11.2 Finite Element Discretisation 254

11.2.1 Discretisation 254

11.2.2 Qualitative Convergence Results . 256

11.2.3 Quantitative Convergence Results . . 260

11.2.4 Complementary Problems 264

11.3 Discretisation by Difference Methods . . . 267

Table of Contents ix

12 Stokes Equations 275

12.1 SyBtems of Elliptic Differential Equations 275

12.2 Variational Formulation 278

12.2.1 Weak Formulation of the Stokes Equations . . . 278

12.2.2 Saddllepo4nt Problems 279

12.2.3 Existence and Uniqueness of the Solution of aSaddlepoint Problem 282

12.2.4 Solvability and Regularity of the Stokes Problem 285

12.2.5 A Vo-elliptic Variational Formulation of the StokesProblem 289

12.3 Mixed Finite-Element Method for the Stokes Problem . 290

12.3.1 Finite-Element Discretisation of a SaddlepointProblem 290

12.3.2 Stability Conditions 291

12.3.3 Stable Finite-Element Spaces for the Stokes Problem 29312.3.3.1 Stability Criterion 293

12.3.3.2 Finite-Element Discretisations with theBubble Function 294

12.3.3.3 Stable Discretisations withLinear Elements in Vh 296

12.3.3.4 Error Estimates 297

Bibliography 300

Index 307

Notation

Formula Numbers.Equations in Section X.Y are numbered (X.Y.1), (X.Y.2), etc. The equa-

tion (3.2.1) is referred to within Section 3.2 8Irnply as (1). In other Sectionsof Chapter 3 it is called (2.1).

Theorem Numbering.All Theorems, Definitions, Lemmata etc. are numbered together. In Sec-

tion 3.2 Lemma 3.2.1 is referred to as Lemma 1.

Special Symbols. The following quantities have fixed meanings:

A, B,... matricesB, B, boundary differential operators

(cf. (5.2.la,b), (5.3.6))C the complex numbersC'(D), Holder- and Lipschitz-continuousiy differentiable

functions (cf. Definition 3.2.8)Ck(D), C°°(D) k-fold and infinitely continuously differentiable

functionsGr(a) infinitely differentiable

functions with compact support.a (cf. (6.2.3))d(u, VN) distance of the function u from the subapace VN

(ci. Theorem 8.2.1)dl', surface differentials in surface integralsdiag{d1,d2,.. .} diagonal matrix with the diagonal eLements d1,d2,...f a function; often the right-hand side of a differential

equationGreen's function (cf. Section 3.2)

h step size (cf. Sections 4.1 and 4.2)H"(i?), H'(S?), Sobolev spaces (cf. Sections 6.2.2 and 6.2.4)KR(y) open ball about y with radius R

(ci. (2.2.7), Section 6.1.1)I identity or inclusion (cf. Sections 6.1.2, 6.1.3)L a differential operator (cf. (1.2.6)) or the operator as-

sociated with a bilinear form (ci. (7.2.9'))I, the stiffnees matrix (ef. Section 8.1)

matrix of a discrete system of equations (cf. (4.I.9a))

L(X,Y)

L°°(S?)L2(J2)IN

n n(x)0O(.)Pqh

Notation

Rh,Rh

supp(f)u = u(x) = u(xj,Uh

VN,Vhx,(x,y),(x,y, z)x=(xl,...,xn)zI,

p(A)1?

V

8;8/On

)xXx'(., .)(., .)O, (, I lo,

(•, (•, )a I - 1k, I

II112

• j

I1•1100

I

linear space of bounded operators from X to Y(cf. Section 6.1.2)

space of essentially bounded functions (cf. 6.1.3)space of square-integrable functions (cf. Section 6.2.1)the natural numbers {1,2,3, . .

normals (ci. (2.2.3a))the zero matrixLandau symbol: 1(x) = O(g(x)) if � constlg(x)Icf. (8.1.6)a grid function, nght-hand side of the discrete

equation (4.1 .9a)the real numbers, the positive real numbersrestrictions (cf. (4.5.2) and (4.5.5b))the singularity (unction (cf. Section 2.2)the support of the function f (cf. Lemma 6.2.2)

a function, e.g., a solution of a differential equationa grid function (discrete solution; cf Section 4)finite-element spaces (ci. (8.1.3) and Section 8.4.1)independent variablesa vector of independent variablesthe integersthe boundary of 1?the boundary points of a grid (cf. (4.2.lb), (4.8.4))fundamental solution functionspectral radius of the matrix Aan open set In or a domain (cf. DefinitIon 2.1.1)a grid (cf. (4.1.6a), (4.2.la), (4.8.2))the Laplace operator (cf. (2.1.la))the five-point difference operator (ci. (4.2.3))gradient (cf. (2.2.3a))differences (cf. (4.1.2a-c), (4.2.3))differences in the normal direction (cf. (4.7.4))normal derivative (cf. (2.2.3a))acalar product (cf. (2.2.3c), (4.3.14a))duality form (ci. Section 6.1.3)acalar product (cf. Section 6.1.4)

Iscalar product and norms on L2(Q)

scalar products and norms onthe Eudidean norm (cf. (2.2.2))the Eucidean norm and the spectral norm

(cf. Section 4.3)norms equivalent to J

J , (cf. (6.2.15), (6.2.16b))maximum norm (ci. (4.3.3)), row sum norm (4.3.11),

or supremum norm (2.4.1), ci. also Section 6.1.1the vector (1, 1,...) (ci. Section 4.3)

1 Partial Differential Equations andTheir Classification Into Types

1.1 Examples

An ordinary differential equation describes a function which depends on onlyone variable. Unfortunately, for many problems it is not possible to restrictattention to a single variable. Almost all physical quantities depend on thespatial variables z, y , and z and on time t. The time dependence might beomitted for stationary processes, and one might perhaps save one spatial di-mension by special geometric assumptions, but even then there would stillremain at least two independent variables. Equations that contain the firstpartial derivatives

where (1 i � or even higher partial derivatives etc., are calledpartial differential equations.

Unlike ordinary differential equations, partial differential equations can-not be analysed all together. Rather, one distinguishes between three typesof equations which have different properties and also require different numer-ical methods.

Before the characteristics for the types are defined, let us introduce someexamples of partial differential equations.

All of the following examples will contain only two independent variablesx,y. The first two examples are partial differential equations of first order,since only first partial derivatives occur.

Example 1.1.1. Find a solution u(x, y) of

(1.1.1)

It is obvious that u(x, y) must be independent of y, i.e., the solution has theform u(x, y) = Thus u(x, y) = w(x) for some arbitrary is a solution of(1).

Equation (1) is a special case of

Example 1.1.2. Find a solution u(x, y) of

(cconstant). (1.1.2)

2 1 Partial Differential Equations and Their Types

Let u be a solution. Introduce new coordinates = x + cy, = y and define

v(e, := ii), I?)) with the aid of = — = Since

V,3 = + (chain rule) and X,3 —c, y,7 = 1, it follows from (2) that= 0. This equation is analogous to (1), and Example 1 shows that

v(e, = co(e). If one now replaces ,g by y one obtains

u(x,y) = (1.1.3)

Conversely, through (3) one obviously obtains a solution of Equation (2) aslong as is continuously differentiable.

In order to determine uniquely the solution of an ordinary differentialequation t&' — f(u) = 0 one needs an initial value u(x0) u0. The partialdifferential equation (2) can be augmented by the initial-value function

for x€Ut (1.1.4)

on the line with aconstant. The comparison of Equations (3) and(4) shows that = uo(x). Thus is determined by =The unique solution of the initial value problem (2) and (4) reads

u(x, y) = tio(x — c(yo — y)). (1.1.5)

The following three examples involve differential equations of second order.

Example 1.1.3. (Potential equation or Laplace equation) Let 0 bean open subset of Find a solution of

infi. (1.1.6)

If one identifies (z, y) E It2 with the complex number z = x + iy C,the solutions can be given immediately. The real and imaginary parts of anyfunction f(z) holomorphic in flare solutions of Equation (6). Three examplesare Rez0 = 1, Rez2 = x2—y2 and Relog(z—zo)if $1. To determine the solution uniquely one needs the boundary values

for all (x,y) on theboundaryf=Ofloffl.

Example 1.1.4. (Wave equation) All solutions of

(1.1.7)

are given byu(x,y) = w(x + y) + — y) (1.1.8)

where and are arbitrary twice continuously differentiable functions. Suit-able initial values are, for example,

u(x,O) = uO(x), u1(z), (x E It) (1.1.9)

where and u1 are given functions. If one inserts (8) into (9), one finds= p + tq = — , where is the derivative of and infers that

1.1 Examples 3

w'= (u1 +t4,)/2, (t4 —tii)/2.

From this one can determine and up to constants of integration. Oneconstant can be chosen arbitrarily, for example, by s°(O) = 0, and the other isdetermined by ti(O, 0) = = ço(0) +

Exercise 1.1.5. Prove that every solution of the wave equation (7) has the

Example 1.1.6. (Heat equation) Find the solution of

(1.1.10)

(physical interpretation: u = temperature, y = time). The separation of vari-ables u(z, y) = v(x)w(y) gives that for every c E lit

u(x, y) sin(cx) . exp(—c2y). (1.1 .lla)

Another solution of (10) for y > 0 is

u(x y) =L

exp((z — (l.l.Ilb)

where is an arbitrary continuous and bounded function. The initial con-dition matching Equation (10), in contrast to (9). contains only one function:

u(x,0) = uO(x). (1.1.12)

The solution (lib), which initially is defined only for y > 0, can be extendedcontinuously to y =0 and there satisfies the initial value requirement (12).

ExercIse 1.1.7. Let uo be bounded in lit and continuous at x. Then provethat the right side of Equation (lib) converges to uo(z) for y —'0. Hint: Firstshow that tz(x, y) = 110(z) + — 110(z)) exp( —

and then decompose the integral into subintegrals over [z — 6,z +5) and(—oo,z—S)U(z+6,oo).

As with ordinary differential equations, equations of higher order can be

described by systems of first-order equations. In the following we give someexamples.

Example 1.1.8. Let the pair (u, v) be the solution of the system

(1.1.13)

If u and v are twice differentiable, the differentiation of (13) yields the equa-tions + =0, + 0, which together imply that - =0.Thus u is a solution of the wave equation (7). The same be shown for v.


Example 1.1.9. (Cauchy-Riemann differential equations) If u and v satisfy

the systemv2—u5=O in $?cIt2, (1.1.14)

then the same consideration as in Example 8 yIelds that both u and v satisfythe potential equation (6).

Example 1.1.10. If u and v satisfy the system

v1+u=0, (1.1.15)

then v solves the heat equation (10).

A second-order system of interest in fluid mechanics can be found in

Example 1.1.11. (Stokes equations) In the system

(1.1.16a)

+ — = 0, (1.1.16b)

=0 (1.1.16c)

u and v denote the flow velocities in x- and y-directions, while w denotes thepressure.

1.2 Classification of Second-Order Equations into Types

The general linear differential equation of second order in two variables reads

1 2 i+d(x,y)u +e(x,y)uy 1-f(x,y)u-f-g(z,y) = Q

( . . )

DefinitIon 1.2.1. (a) Equation (I) is said to be elliptic at (x,y) if

a(x,y)c(z,y) — b(x,y)2 > 0. (1.2.2a)

(b) Equation (1) is said to be hyperbolic at (x,y), if

a(x,y)c(z, y) — b(x,y)2 <0. (1.2.2b)

(c) Equation (1) is said to be parabolic at (x,y) if

ac—b2=0 and rank[ b =2 at(z,y). (1.2.2c)

(d) Equation (1) is said to be elliptic (hyperbolic, parabolic) in ii C 1R2 if itis elliptic (hyperbolic1 parabolic) at all (x, y) E fl

1.2 ClassificatIon of Second-Order Equations into Types 5

Occasionally the parabolic type is defined only by ac — b2 = 0. But onewould not want to call the equation y) + y) = 0 parabolic, norindeed the purely algebraic equation u(z, y) =0.

Example 1.2.2. The potential equation (1.6) is elliptic, the wave equation(1.7) is of hyperbolic type, while the heat equation (1.10) is parabolic.

The definition of types can easily be generalised to the case where morethan two independent variables occur, say . , . The general linear dif-ferential equation of second order in n variables x = (x1,.. . , reads

+ + a(x)u = 1(x). (1.2.3)

Since = holds for twice continuously differentiable functions, onecan assume that, without loss of generality,

aj3(x) = a,1(x). (1.2.4a)

Thus, the coefficients define a symmetric n x n matrix

A(x) = (1.2.4b)

which therefore has only real eigenvalues.

Definition 1.2.3. (a) Equation (3) is said to be elliptic at x if all nelgenvalues of the matrix A(x) have the same sign (±1) (i.e. if A(x) is positiveor negative definite).(b) EquatIon (3) is said to be hyperbolic at x if n — 1 eigenvalues of A(x)have the same sign (±1) and one eigenvalue has the opposite sign.(c) Equation (3) is said to be parabolic at x if one eigenvalue vanishes, theremaining n —2 eigenvaluen have the same sign, and rank(A(x), a(x)) = nwhere a(x) = (aj(x), .. . ,

(d) Equation (3) is said to be elliptic in Si c if it is elliptic at all x 0.

Definition 3 makes it clear that the three types mentioned by no meanscover all cases. An unclassified equation occurs, for example, if A(x) has twopositive and two negative eigenvalues.

In place of (3) one can also write

Lu=f, (1.2.5)

where

(1.2.6)

is a linear differential operator of second order. The operator


which contains only the highest derivatives of L, is called the pri nd psi partof L.

Remark 1.2.4. The ellipticity or hyperbolicity of Eq. (3) depends only onthe principal part of the differential operator.

ExercIse 1.2.5. (Invariance of the type under coordinate transformations)Let Eq. (3) be defined for x a. The transformation i' fl C IR" SI C 1W'is assumed to have a nonsmgular Jacobian matrix S = E C1(Q) at x.Prove that Eq. (3) does not change its type at x if it is written in the newcoordinates = Hint: the matrix A = (ad) becomes SAST after thetransformation. Use Remark 4 and Sylvester's inertia theorem (cf. Gantmacher11, Chapter

1.3 Type Classification for Systems of First Order

The examples 1.1.8-10 are special cases of the general linear system of firstorder in two variables.

u.,(x,y)— A(x,v)uv(x,v)÷B(x,y)u(x,y) = f(x,y). (1.3.1)

Here, u=(uj,...,um)Tisavectorfunction,andA,Baremxrnmatrjces.Incontrast to Section 1.2, A need not be symmetric and can have complex elgen-values. If the elgenvalues A1,... , are real, and if there exists a decomposi-tion A = with D = diag{A1,. .. ,A,,,}, A is called real-diagonalisable.

DefinitIon 1.3.1. (a) System (1) is said to be hyperbolic at (x,y) ifA(:,y) is real-diagooalisable. (b) System (1) is said to be elliptic at (x,y)if all the elgenvalues of A(x, y) are not real.

If A is real or possesses m distinct real eigenvalues the system is hyper-bolic since these conditions are sufficient for real diagonalisability. A singlereal equation, in particular, is always hyperbolic.

Examples 1.1.1 and 1.1.2 are hyperbolic according to the preceding re-mark. System (1.13) from Example 1.1.8 has the form (1) with

A={01

It is hyperbolic since A is real-diagonalisable:

A— [1 ii [—1 0111 1

1

1.4 CharacteristIc of the Different 7

The Cauchy-Riemann system (1.14), which is dosely connected with the po-tential equation (1.6), is elliptic since it has the form (1) with

A={?

and A has the eigenvalues ±i. The system (1.15) corresponding to the(parabolic) heat equation can be described as system (1) with

The eigenvalues (A1 = A2 =0) may be real but A is not diagonalisable. Hence,system (1.15) is neither hyperbolic nor elliptic.

A more generaL system than (1) is

+ + 13u = f. (1.3.2)

If A1 is invertible then multiplication by Aj' gives the form (1) with A—A' A2. Otherwise one has to investigate the generalised eigenvalue problemdet(XA1 + A2) = 0. However, system (2) with singular A1 cannot be elliptic,as can be seen from the following (cf. (4) with = 1, = 0).

A generalisation of (2) to n independent variables is exhibited in thesystem

(1.3.3)

with m x m matrices . and B = B(x). As aspecial case of a later definition (cf. Sect. 12.1) we obtain the

Definition 1.3.2. The system (3) is said to be elliptic at x if for any vectoro . the following holds:

0. (1.3.4)

1.4 Characteristic Properties of the Different Types

The distinguishing of different types of partial differential equations would bepointless if each type did not have fundamentally different properties. Whendiscussing the examples in Sect. 1.1 we already mentioned that the solutionis uniquely determined if initial values and boundary values are prescribed.


In Example 1.1.2 the hyperbolic differential equation (1.2) is augmentedby the specification (1.4) of u on the line y = oonst (see Fig. la). In the caseof the hyperbolic wave equation (1.7), it.,, must also be prescribed (cf. (1.9))since the equation is of second order.

(a) u,u,,, (b) u,u,,,

FIgure 1.4.1. (a) Initial conditions and (b) initial boundary conditions for hyper-bolic problems

it is also sufficient to give the values it and it,1 on a finite interval [xj, ifu is additionally prescribed on the lateral boundaries of the domain I? of Fig.lb. This prescription of initial boundary values occurs, for example, inthe following physical problem. A vibrating string is described by the lateraldeflection u(x,t) at the point x (xl,z21 at time t. The function it satisfiesthe wave equation (1.7) with the coordinate y corresponding to time t. At theinitial instant in time, t = to, the deflection u(x,O) and velocity aregiven for x1 <x <x2. Under the assumption that the string is firmly dampedat the boundary points xl and x2, one obtains the additional boundary datau(xi,t) =t4x2,t) =0 for alit.

For parabolic equations of second order one can also formulate initialvalue and initial boundary value problems (cf. Fig. 2). However, as initialvalue only the function = tlO(x) may be prescribed. An additionalspecification of is not possible, sinceis already determined by the differential equation (1.10) and by it0.

(a) Initial conditions and (b) initial boundary conditions for parabolicproblems

The heat equation (1.10) with the initial and boundary values

t4x,to)=uo(x) in [xi,X21,

u(xj,t) ti(x21t) = fort >

1.4 Characteristic Properties of the Different Types 9

(cf. Figure 2b) describes the temperature u(x, t) of a wire whose ends at x =and x = x2 have the temperatures and The initial temperaturedistribution at time to is given by uO(x).

Aside from the different number of initial data functions in Figure 1and Figure 2, there also is the following difference between hyperbolic andparabolic equations.

Remark 1.4.1. The shaded area 11 in Figure 1 corresponds to t > to, andin Figure 2 to y > For hyperbolic equations one can solve in the same wayinitial-value and initial-boundary-value problems in the domain t � to, whilstparabolic problems in t <to generally do not have a solution.

If one changes the parabolic equation Ut — = 0 to + = 0, theorientation is reversed: solutions exist in general only for t � to.

For the solution of an elliptic equation, boundary values are pre-scribed (cf. Example 1.1.3, Figure 3). A specification such as that in Figure2b would not uniquely determine the solution of an elliptic problem, while thesolution of a parabolic problem would be overdetermined by the boundaryvalues of Figure 3a..

U

U.(b)

FIgure 1.4.3. Boundary value conditions for an elliptic problem

An elliptic problem with specifications such as in Figure lb in general hasno solution. Let us, for example, impose the conditions u(x, 0) = u(0, y)u(1,y) = 0 and = on the solution of the potential equation(1.6), where u1 is not infinitely often differentiable. If a continuous solutionu existed in 17 = [0,1j x [0,11 one could develop u(z,1) into a sine seriesand the following exercise shows that uj would have to be oftendifferentiable, in contradiction to the aseumption.

Exercise 1.4.2. Let E C°[0, 1) have the Fourier expansion

00

=

Show that: (a) the solution of the potential equation (1.6) in the square(1 = (0,1) x (0,1) with boundary values u(O,y) = u(x,0) = u(1,y) = 0and u(z, 1) = is given by

(a) u


00a,,

u(z,y) = . sm(virx)sinh(viry).

(b) For 0 � � 1 and 0 � y < 1, ti(z, y) is differentiable infinitely often. Hint:f(x) = E13,,sin(ziwz) E C°°[0, 1), if lim/3,t) = 0 for all k E IN.

Conversely it does not make sense to put boundary value constraintsas in Fig. 3a on a hyperbolic problem. Consider as an example the waveequation (1.7) in = [0, 1) x [0, 1/711 with the boundary values u(x,O) =ii(0, y) = u(1, y) = 0 and u(x, 1/jr) = sin(virx) for v IN. The solutionreads u(x, y) = sin(v7rx) sin(viry)/ sin ii. Although the boundary values, forall p IN are bounded by 1, the solution in I? may become arbitrarily largesince sup{1/sinv: v IN} = 00. Such a boundary value problem is called "notwell posed" (cf. Definition 2.4.1).

Exercise 1.4.3. Prove that the set {sinv:v IN) is dense in [—1, 1].

Another distinguishing characteristic is the regularity (smoothness) ofthe solution. Let u be the solution of the potential equation (1.6) in Q C 1R2.As stated in Example 1.1.3, ti is the real part of a function holomorphic inSi. Since holomorphic functions are infinitely differentiable, this property alsoholds for ti.

In the case of the parabolic heat equation (1.10) with boundary valuesu(z,0) = no the solution z* is given by (1.llb). For y >0, ti is infinitely differ-entiable. The smoothness of is of no concern here, nor is the smoothnessof the boundary values in the case of the potential equation.

One finds a completely different result for the hyperbolic wave equation(1.7). The solution reads u(z,y) = — y), where and resultdirectly from the initial data (1.9). Check that u is k-times differentiable if u0is k-times and u1 is (k — 1)-times differentiable.

As already mentioned in this section, in the hyperbolic and parabolicequations (1.1), (1.2), (1.7) and (1.10) the variable y often plays the roleof time. Therefore one calls processes described by hyperbolic and parabolicequations nonstationary. Elliptic equations which only contain space coor-dinates as variables are called stationary. More clearly than in Definitions1.2.lb,c the Definitions 1.2.3b,c distinguish the role of a single variable (time)corresponding to the elgenvalue A = 0 in parabolic equations, and to theelgenvalue with opposite sign in hyperbolic equations.

The connection between the different types becomes more comprehensibleif one relates the elliptic equations in the variables to the parabolicand hyperbolic equations in the variables x1, ,x,,,t.

Remark 1.4.4. Let L be a differential operator (2.6) in the variables x =(x1, . . . , and let it be of elliptic type. Let L be scaled such that the matrixA(x) in (2.4b) has only negative eigenvalues. Then

1.4 CharacterIstic Properties of the Different Types 11

ut+Lu=O (1.4.2)

is a parabolic equation for u(x,t) = In contrast

utt+Lu=O (1.4.3)

is of hyperbolic type.

Conversely, the nonstationary problems (2) or (3) lead to the ellipticequation Lu = 0 if one seeks solutions of (2) or (3) that are independent oftime t. One also obtains elliptic equations if one looks for solutions of (2) or (3)with the aid of a separation of variables u(x, t) ço(t)v(x). Theare

u(x,t) = in case (2),(1.4.4)

incase(3),

where v(x) is the solution of the elliptic eigen value problem

Lv=Av. (1.4.5)

2 The Potential Equation

2.1 Posing the Problem

The potential equation from Example 1.1.3 reads

in.QCUt', (2.1.la)

where 02/Ox? + + is the Laplace operator. In physics,Equation (la) describes the potentials; for example the electric potential whena contains no electric charges, the magnetic potential for vanishing currentdensity, the velocity potential, etc. Equation (is) is also called Laplace'sequation since it was described by P. S. Laplace in his five-volume work"Mécanique Céleste" (1799-1825). However, It was L. Euler who first men-tioned the potential equation in 1752.

The connection between the potential equation for n = 2 and functiontheory has already been pointed out in Example 1.1.3. Not only is the Laplaceoperator an example of an elliptic differential operator, but it actually is theprototype (the so-called normal form). By using a transformation of variables,any elliptic differential operator of second order can be transformed so thatits principal part becomes the Laplace operator (cf. Hellwig Li,

in the following a will alwayB be a domain,

Definition 2.1.1. The region a C lit" is called a domain if a is open andconnected.1

The existence of a second derivative of u is required only in 0, not onthe boundary

r=aaof 0. For a prescription of boundary values

u=cp onE' (2.l.lb)

to be meaningful, one has to aseume continuity of u on = 0 U P. Theseconsiderations lead to

1 1? Is called connected If for any x,y E (1 there is a continuous curve within0 that connects x and y, thai is, a -y: $ (0,11 '7(s) E 0 continuously with'y(O) = x,"y(l) = y.


Definition 2.1.2. The function u is said to be harmonic in .17 if u belongsto C2($?) fl and satisfies the potential equation (la).

Here C°(D) denotes the set of continuous fk-fold con-tinuously differentiable, infinitely often differentiablej functions on D.

In general one should not expect that the solution of (la,b) lies inas we show in the following example.

Example 2.1.3. Let $7 = (0, 1) x (0,1) (cf. Figure 1.4.3a). Let the boundaryvalues be given by (x, y) E I'. A solution of the boundaryvalue problem exists but does not belong to C2(.Q).

PROOF. The existence of a solution u will be discussed in Theorem 7.3.7.If u E C2(Th, then it follows that = = 2 for x 10,1); inparticular 0) = 2. From the analogous result 0) = 0) = 0one may conclude Llu(0, 0) = 2 in contradiction to = 0 in .17. U

In the case under discussion one can also show u That thisstatement is generally false, is shown by

Example 2.1.4. In the domain .17 = (—i, x (—i, \ [0, x [0, (cf.Figure 1) introduce the polar coordinates

y=rsinço.

1'l

ft

ft

FIgure 2.1.1. An 1,-shaped region

(2.1.2)

The function u(r, = is the solution of the potentialequation (la) and has smooth boundary values on F (in particular u = 0

ft

14 2 The Potential Equation

This follows from the fact that, along with and u.,,, the radial derivative

= also has to be bounded, However, u,. = O(r1/3) holds

for r -. 0. In order to check that 41u =0, use the following.

ExercIse 2.1.5. Show that (a) In terms of the polar coordinates (2) in lit2,the Laplace operator takes the form

82 18 182(2.1.3)

(b) In terms of the three-diniensional polar coordinates

y=rsinçosint,b,

the Laplace operator is given by

82 28 1 1 82 8 82(2.1.4)

Note. In the n-dimensional case the transformation to polar coordinates leadsto

82 n—18 1(2.1.5)

where the Beltrami operator B contains only derivatives with respect tothe angle variables.

2.2 Singularity Function

The singularity function is defined by

forn=2,s(x,y) := — f 2

(2.2.la)I.. (n—2)w,,

Orn>

for x,y lit", where

= /F(n/2), = 2w, = 4w, (2.2.Ib)

with I' the Gamma function, is the surface of the n-dimensional unit sphere.The Euclidean norm of x in IR" is denoted by

(xl =1/2

(2.2.2)

2.2.1. RrJlxed y E Ut" the potential equation in 1R"\{y} is 8olvedby s(x,y).

2.2 Singularity Function 15

The proof can be carried out directly. It is simplest to introduce polarcoordinates with y as origin and to use (1.5), since s(x,y) depends only onr=lx—yI.

For the next theorem we need to introduce the normal derivative 8/On.Let be a domain with smooth boundary 1'. Let n(x) E denote the outernormal direction at x F, i.e. n is a unit vector perpendicular to thetangential hyperplane at x and points outwards. The normal derivativeof u at x E F is defined as

= (n(x), Vu(x)), (2.2.3a)

where(2.2.3c)

is the gradient of u and

(x,y) = (2.2.3c)

is the scalar product in Ut". In the case of the sphere KR(y) (cf. (7)) thenormal direction is radial, and Ou/On becomes Ou/Or with respect to r =

— y11 if one uses polar coordinates with the origin at y. It follows fromOs(x, y)/Or = —Ix — that

Os(x,y)/On = for x E (2.2.4)

The first Green formula reads (cf. Green [1J)

L dx =—

j (Vu, Vv) + j

u V E if the domain 1? satisfies suitable condi-tions. Here frn... dl' denotes the surface integral.

Domains for which Equation (5a) holds are called normal domains. Tosee sufficient conditions for this refer to Kellogg (1, Chapter LVI and Hellwig11, 1-1.21.

Functionsu,v E C2(Thinanormaldomain.Qsatisfythe second Greenformula

/ u4v = / viju + / — dl'. (2.2.5b)

Theorem 2.2.2. Let (2 be a normal domain, and Let u E be harmonicthere. Then

u(y) = dl', (2.2.6)

for all y (2. Here 0/On, and dl' refer to the variable x.


PROOF. ByKr(y) := (x E x — <r} (2.2.7)

we denote the ball with centre y and radius r. Since the singularity functions(., y) is not differentiable on x = y, Green's formula (5b) is not directlyapplicable. Let

____

:=

with e so small that C £7. Since £7 is again a normal domain, it followsfrom = Lts = 0 (ci. Lemma 1) and (5b) with v a that

J — = 0. (2.2.8a)

We have = 8flU However, at x E the normal directionsof and of 8K,, (y) differ in their signs. The same holds for the normalderivatives, so that the integral in (8a) can be decomposed into ... =

—The assertion of the theorem would be proved if we

could show that —. —u(y) for e 0. The normal derivative 8u/On

is bounded on and fSK(y) s(x, y) di converges like 0(eI log or

0(e) towards zero, as can be seen from (1) and f8K,(y) dl" = Thus,

we obtain8

--.0 (e —-.0). (2.2.8b)

From fSK dl = and (4) one infers

1 8 y) di', = —u(y). (2.2.8c)J8K,(y) V74X

The continuity of u in y yields

J (u(x) — max — —.0 as e —e 0.xEOK1(y)

(2.2.8d)

Equations 8b,c,d show that fSK(y) — dl' —' —u(y) (e —. 0), so

that (8a) proves the theorem.

Any function of the form

lr(x, y) = 8(X, y) + y) (2.2.9)

is called a fundamental solution, of the potential equation, in .17 if for fixedy E .1'? the function is harmonic in .0 and belongs to C2(Th).

Corollary 2.2.3. Under the conditions of Theorem 2 for fundamentalsolution in.0 the following holds:

2.3 The Mean Value Property and Maximum Principle 17

u(y) =j dPi. (2.2.10)

PROOF. (5b) implies — u&P/OnJ df =0. U

For a weakening ofthecondition = -y—sE C2(Th) for E

C'(FJ) fl C2(.Q) refer to Heliwig (1, t—1.4J.

Exercise 2.2.4. Let 1? .= KR(y). Define

=[Ix —

—

for n> 2(2.2.lla)

for

with E Q, = y + — — y) and show:(a) -y is a fundamental solution in fl,(b) y(x, = x),(c) on the surface 1' = OKR(y) the following holds:

= =— Ix — y12)Ix E I').

(2.2.llb)

2.3 The Mean Value Property and Maximum Principle

DefinitIon 2.3.1. A function u has the mean value property in S? ifU E C°(fl) and if for all x E Q and all R >0 with KR(X) C a the followingequation holds

1u(x)

= J (2.3.1)8KR(x)

Since dl.' = the right side in (1) is the mean value of utaken over the surface of the sphere. An equivalent diaracterisation results ifone averagea over the sphere KR(x).

Exercise 2.3.2. u C0(IJ) has the second mean value property in 1? ifu(x) = for all x a, R> 0 with KR(x) C a. Show thatthis mean value property is equivalent to the mean value property (1). Hint:

IKR(x)= fR

fSK(x) dr.

Functions with the mean value property satisfy a maximum principle, as isknown from the function theory for holomorphic functions:


Theorem 2.3.3. (Maximum-minimum principle) Let (1 be a domainand let u E C°(Th be a nonconstant function which has the mean value prop-erty. Then u takes on neither a maximum nor a minimum in a.

PROOF. (a) It suffices to investigate the case of a maximum since a minimumof u is a maximum of —u, and —u also has the mean value property.(b) For an indirect proof we assume that there exists a maximum in y E a:

u(y)=M�u(x) fora.llxEa.

In (c) we will show u(y') = M for arbitrary y' E I?, i.e. u M in contrast tothe assumption u const.(c) Proof of u(y') = M. Let y' E (1. Since S? is connected, there exists apath connecting y and y' running through a, i.e. there exists a continuous

10,11 (2 with y, = y'. We set

I := {s E 10,11: M for all 0 � t � s}.

I contains at least 0, and is closed since u and ço are continuous. Thus thereexists s' = max{s I}, and the definition of I shows that I = [0, In (4)it is proved that s' =lso that y= E land hence u(y') =M follows.(d) Proof of? = 1. The opposite assumption 8' <lean be made and shown tobe contradicted by proving that ii(x) = M in a neighbourhood of x' :=Since x' (2, there exists R> 0 with C £1. Evidently, it follows thatu = M in KR(x)' if it is shown that u = M on ôKr(X)' for all 0 < r < R.(e) Proof of u = Mon OKr(x'). Equation (1) in x' reads

M = u(x') = IJSKr(X')

In general we have u(C) � M . If one had u(C') <M fore' OKr(X') and thusalso u < M in a neighbourhood of one would have on the right side a meanvalue smaller than M. Therefore, u = M on ÔK4x') has been proved. •

Simple deductions from Theorem 3 are contained in

Corollary 2.3.4. Let (2 be bounded. (a) A function with the mean valueproperty takes its maximum and its minimum on 8(2.(b) If two fsinctioni with the mean value property coincide on the boundary8, they are identicaL

PROOF. (a) The extrema are assumed on the compact set = QU Ofl.According to Theorem 3, the extremum cannot be in flif u is not constanton a connected component of fl. But in this case the assertion is also obvious.(b) If u and v with u = v satisfy the mean value property on 8$? then thelatter is also satisfied for w := u - v. Since w = 0 on 0(2, part (a) indicates

2.3 The Mean Value Property and Maximum Prmciple 19

Lemma 2.3.5. Harmonic functions have the mean value property.

PROOF. Let u be harmonic in Qand y KR(y) C (2. We apply the repre-sentation (2.6) for (2 = KR(y). The value s(x,y) is coostant on 8KR(y):it be denoted by a(R). Because of (2.4), Equation (2.6) becomes

f Ott 1 1u(y) = cr(R) j dl' + udf'.

J8KR(y)

The equation agrees with (1) if the first integral vanishes. The latter followsfrom

Lemma 2.3.6. Let u E C2(Th) be harmonic in a normal domain 0. Thenwe have

(2.3.2)

PROOF. In Green's formula (2.5a) substitute 1 and u for uand v respectively.S

Lemma 5, Theorem 3, and Corollary 4 together imply Theorems 7 and 8:

Theorem 2.3.7. (The maximum-minimum principle for harmonicfunctions) Let u be harmonic in the domain $1 and nonconstant. Thereexists no maximum and no minimum in 11.

Theorem 2.3.8. (Uniqueness). Let $2 be bounded. A function harmonic inI? assumes its maximum and its minimum on 8$? and is uniquely determinedby its values on 811.

The representation (1) of u(y) by the values on OKR(y) is a special caseof the following formula which will be proved at the end of this section andwhich provides Equation (1) for x = y.

Theorem 2.3.9. (Poi8son's integral formula) Assume we have cc EC°(OKR(y)) and n 2. The solution of the boundary value problem

4u = 0 in KR(y), u = cc on OKR(y) (2.3.3)

is given by the function

R2_Ix_. 2u(x) = Yl f for x KR(y), (2.3.4)

8KR(y) IX —

which belongs to C°°(KR(y)) Ii C°(KR(y)).


The mean value property only assumes u E C°(Th, while harmonic func-tions belong to C2(.O) (1C°(fl). This makes the following assertion surprising:

Theorem 2.3.10. A function is harmonic infl if and only if it has the meanvalue property there.

PROOF. Because of Lemma 5 it remains to be shown that a function v withthe mean value property is harmonic. Let x E KR(X) C (1 be given arbitrarily.According to Theorem 9 there exists a function u harmonic in KR(X) with

inKn(x), u=v on.ÔKR(x).

According to Lemma 5, u has the mean value property, as does v, and Corol-lary 4b proves that u = v in i.e. v is harmonic in KR(X). SinceKR(X) C 11 is arbitrary, v is harmonic in (1. U

An important application of Theorem 10 is

Theorem 2.3.11. (Harnack) Let be a sequence of functions har-monic in 1? and converging uniformly in Then u = Uk is harmonicmi?.

PROOF. The limit function is continuous: u The limit process,applied to Uk(X) = yields Equation (1) for u; i.e.u has the mean value property. According to Theorem 10, u is also harmonicintl. N

Theorems 3 and 7 on the maximum-minimum principle pertain to globalextrems. The proof of Theorem 3 does not yet exclude local extrema in theinterior. It merely shows that u is then always constant in a circle KR(y) C 1?.As is known from function theory one can derive from this u(x) = M in

if u is analytic, 1. e., there is a convergent power series expansion in aneighbourhood of any X E I?. Indeed, the following theorem holds whoseproof can be found, for example, in Hellwig (1,11111.5]:

Theorem 2.3.12. A function harmonic in tl is analytic there.

The proof of the Poisson formula still needs to be carried out. (a) First wemust show that u in (4) is a function harmonic in KR(y), i. e., it satisfies =0. Since the integrand is twice continuously differentiable and the domain ofintegration 1' is compact, the Laplace operator commutes with theintegral sign:

= (&i)1 j — Ix — . Ix — for x

(2.3.5)

tim forzEf'.

By Equation (6), putting u 1 in Corollary 2.2.3 gives the identity

R2—(x—y$2 f foraJlxEKR(y).JrLet z e I' be arbitrary. Due to Equation (8) one can then write:

R2—Ix—y12Ju(x) —

= —

We define 1'O Vfl (see Figureexpression (9a) into ti(x) — ço(z) = lo + where

R2—Ix—Y12 I i=40—

2.3 The Mean Value Property and Maximum PrInciple 21

According to Exercise 2.2.4 there exists a fundamental solution i'(x, suchthat

8 8Ix —

E f',x E KR(y).

(2.3.6)From = =0 and (5), one infers that =0.(b) The expression (4) defines u(x) at first only for x KR(y). It still needsto be shown_that u has a continuous extension on KR(y) = KR(y) U I' (i.e.,u E C°(KR(y))) and that the continuously extended values agree with theboundary values w:

(2.3.7)

(2.3.8)

(2.3.9a)

Figure 2.3.1. ConstructIon of


Since

I fro—

froI

— f,. x —

it follows from Equation (8) that

Jo 5 max — (2.3.9b)F.)

Because of the continuity of one can choose p >0 auth that for given e >0

<€/2. (2.3.9c)

Set := and choose x E KR(y) sufficiently close to z suchthat €(p\fl 1

Ix — zf � 6(€)

and x — z( � p/2. The last inequality implies

(seeFigurel).

Together with R2—lx—y12 = � �2RIz — xj � 2R5(€) one obtains

R2—Ix—y12 < 2C., fihI = Jr. Ix —

dJr,

1 di.

f,, dl' df = and the definition of o(€) follows

lId (2.3.9d)

Thus for every 0 there exists a 6(€) > 0 such that Ix — zi � 6(€) impliesthe estimate Iu(x) — = + Iii � lIol + jIg, � (cf. (9c and d)). Hence,(7) has been proved, and the continuous extension of u to KR(y) leads to

N

Exercise 2.3.13. Prove that the function u defined by the Poisson integralformula (4) belongs to and solves 2tu =0 in KR(y) even if ismerely assumed to be a function integrable on F = For every pointof continuity z E F of we have u(x) —' (x — z, XE KR(y)).

Exercise 2.3.14. Let £1 be bounded, and let u1 and be harmonic in (1with boundary values and on F = 0(1. Prove that:(a) � on F implies u1 � in (1.(b) If, furthermore, 11 is connected and if W1(x) holds for at least onepoint x E I' then it follows that u1 <u2 everywhere in (1.

2.4 Continuous Dependence on the Boundary Data 23

2.4 Continuous Dependence on the Boundary Data

Definition 2.4.1. An abstract problem of the form

A(x)=y, zEX, yEY,

is said to be well-posed if for all p E Y it has a unique solution x E X andif the latter depends continuously on p.

It Is important to recognise whether a mathematical problem is well-posedsince otherwise essential may occur in its numerical solution. In thecase of the boundary value problem (1.la,b) Xc is the spaceof functions harmonic in S? and Y C°(f) is the set of continuous boundarydata on F = oa.

The topologies of X and Y are given by the supremum norm8:

:= sup and lIc°IIoo :=r sup (2.4.1)x€1? XEI'

The question of the existence of a solution of (1.la,b) will have to be post-poned (see Section 7). The uniqueness, however, has been confirmed alreadyin Theorem 2.3.8, if a is bounded. That the boundedness of a cannot bedropped without further ado is shown in

Example 2.4.2. The functions

u(x1,x2) = x1 in .17 (O,oo) x Ut (2.4.2a)u(x1, x2) = + in .17 1R2\K1(O) (2.4.2b)u(x1,x2) = sin(x1)sinh(x2) in .17= (O,w) x (O,oo) (2.4.2c)

and also the trivial u =0, are solutions of the boundary value problem =0in a, = o r = aa.

For bounded a the harmonic functions (solutions of (I. la,b)) depend notonly continuously but also Lipschltz-continuously on the boundary data:

Theorem 2.4.3. Let a o€ bounded. If ut and solutions of

= =0 in .0, = and ti" on = 8.0

then

Ru' � fly,' — (2.4.3)

PROOF. v := — U11 is a solution of v = — on 1'.According to Theorem 2.3.8, v takes its maximum and its minimum on F:

—Ru" — � v(x) Ru" — for all x€ Th.


The definition (1) of fl 1100 implies (3).

If —, 0 holds for a sequence of boundary values then Theorem3 shows that the associated solutions satisfy — —.0. The followingtheorem states that the existence of a solution of itu = 0 in $7, ti = on Fneed not be assumed.

Theorem 24.4. Let $7 be bouruleL Let E C°(F) be a sequence of bound-ary data which converge uniforvnly toW: let — 0. Let tin be thesolution of 0 in (1, = S0n on F. Then the functions converge urn-formly in Th to u E C2(17) fl C°(Th), and ti solves the boundary value problem

= o a, u = on I'.

PROOF. Since according to Theorem 3 Nun - UmIIoo � - thesequence is Cauchy convergent. Since is complete, convergesuniformly to a ti E C°(Th. According to Theorem 2.3.11, ti is harmonic (i.e.I

Another problem, just as important for numerical mathematics, is rarelydiscussed in the literature: does the solution also depend continuously on theform of the boundary I'? Figure 1 shows domains if and 11" which approxi-mate $7. A polygonal domain, as shown, for in Figure ic, occurs inthe method of finite elements (see Section 8.6).

Figure 2.4.1. Approximation of $7 by (2' and (2"

F) 0. Here, F) := sup{dist(x, F) : x E f',,}, dist(x, F)inf{Ix y(: y E F}. one must specify when E convergesuniformly to C°(f). We make the definition:ip, E C°(f',,) converges uniformly to E C°(f') if, for each e > 0, thereexist numbers N(c) and 6(c) > 0 such that the following implication holds:

n N(c), x€ F, y€ I',,, � 6(e) e. (2.4.4a)

The sequence un E C°(IZ,) converges uniformly tou E C°(.Q) if

2.4 Continuous Dependence on the Boundary Data 25

lim — u(x)I: x fl I?) = 0. (2.4.4b)

Remark 2.4.5. (a) Let K be a set which is compact (i.e. complete and

bounded) with r c K, and C K for all n. Let E C°(K) converge

uniformly on K to If on and on F then (4a) is satisfied.

(b) Let C 1? hold for all n and let be the following (not continuous)continuation of onto Th = u on Then (4b) is

3quivalent to uniform convergence ii on (1 in the usual sense.

Theorem 2.4.6. Let $1,., C (1, with Si and let 1 -, P. Let the

functions which are harmonic in 5Z, be solutions of

on F,. (2.4.5a)

Let E conve!ye uniformly in the sense of(4a) to Then

the following as8ertwfls hold: —

(a) ifthereezistsasolutionuEC2(J?)flC°(J?) of

u=cp on!' (2.4.5b)

then —, ti holds in the sense of (4b).(b) if conversely —+ ti C°(a) is satisfied in the sense of (4b), then ti isthe solution of (5b).

PROOF. (a) Let the continuation II,, be defined as in Remark 5b. Since u isuniformly continuous on there exists >0 for all e >0 such that

lu(x) — � if lx — � (2.4.6a)

Set 6(e) := with 6 from (4a). Because —+ I' thereexists so that � for n � For n � :=max{Np(e), N(e/2)} (N from (4a)) we want to show that — u(x)lfor x For x E S?\S1,, the estimate is trivial because = u(x). Forall x c a, however, there holds

—u(x)I = — u(x)l � —ul (2.4.6b)

(cf. Theorem 2.3.8), because tin - u is harmonic in It remains to estimateforxE Forx€ withn � N(e)thereexistsy €1' with

— � 6(e/2). Thus we obtain

— u(x)j = — u(x)l S — + — u(x)l �

from (4a) and (6a). Since x i',, is arbitrary it follows that ltin —til S onand (6b) proves the uniform convergence ii,, —' ti on Hence, from Remark5b it follows that (4b) is satisfied.


(b) Let K C 1? be a compact set. Since 17,.. -+ 1' there exists a N(K) suchthat K C for n � N(K). Thus, the sequence {u1,: n N(K)} convergesuniformly in the usual sense on K to u so that one can apply Theorem 2.3.11:

consequently u is harmonic in K. Since K C £1 may be chosen arbitrarily, itfollows that u E C2(S?). By assumption, we already have U E C°(Th). That the

•In Theorem 4 one was able to derive the existence of a solution t& of (5b)

just from ço, -. This inference is not possible for the case of a asthe following example shows.

Let := C a := K1(O)\{O} C 1R2. The boundaries are= 0K1(O) U and V = 0K1(O) U {O}, and satisfy —p The

boundary values

= = 0 on 8K1(O), = 1 on 0) = 1

satisfy the condition —+ (cf. (4a) and Remark 5a). The solutions of(5a) can be given explicitly:

u,..(x) = tog(IxI)/log(1/n).

Obviously, u,(x) u(x) := 0 holds pointwise, but u = 0 satisfies neither (4b)nor the boundary value problem (5b). Conversely, one infers from Theorem6a the following result:

Remark 2.4.7. In a = K1(O)\{0} c 11t2 the potential equation has nosolution u C2(S?) fl C°(J7) which assumes the boundary values u(x) =0 on8K1(O)andu(x)=linx=0.

3 The Poisson Equation

3.1 Posing the Problem

The Poisson equation reads

au—f in!? (3.1.la)

with given f C°(Q). In the physical interpretation f is the source termifor example, the charge density in the case of an electrical potential uJ. Todetermine the solution uniquely one needs a boundary value specification, forexample, the Dirichiet condition

(3.l.lb)

DefinitIon 3.1.1. The function u is called the classical solution of theboundary value problem (la,b) if u E C2(O) fl C°(Th) satisfies the equations(la,b) pointwise.

Until we introduce weak solutions in Section 7, "solution" will alwaysmean "classical solution".

The solution of the boundary value problem (la,b) will in general nolonger satisfy the mean value property and the maximum principle. But theseproperties still hold for the differences of two solutions u, and u2 of theequation, since — u2) = f — f =0. Thus the uniqueness of the solution ofproblem (la,b) immediately follows and Theorem 2.4.3 can be brought over:

Theorem 3.1.2. Let 0 be bounded. (a) The solution of (la,b) is uniquelydeterinineii (b) ffu' and u11 are solutions of the Poisson equationfor boundaryvalues w' and then we have

IIti' — u"II,., < — (3.1.2)

PROOF. (b) The proof of Theorem 2.4.3 can be repeated verbatim here.(a) If u1 and u11 are two solutions of (la,b) then (2) shows that Ilu' S

R

Theorems 2.4.4 and 2.4.6 can be transferred Likewise.

28 3 The Equation

3.2 Representation of the Solution by the GreenFunction

Lemma 3.2.1. Let the solution of (l.la,b) belong to C2(Th, where 1? is anormal domain. Then u may be as

u(x)= —! + J — x)]

(3.2.1)for every fundamental solution in (2.2.9).

The proof is the same as in Theorem 2.2.2 or in Corollary 2.2.3. The termwith A = becomes Since the singularity of

is integrable at = x, -yf converges to as e —.0.

ExercIse 3.2.2. (a) Let Si C ft' be bounded, xo $1, / andIf(x)I � — for 8 <n. Show that f(x)dx exists as an improperintegral.(b) Let ii c be bounded and let E 5? depend continuously on D,with D compact. Let f(x,e) be continuous in E x D with xand let � s <n. Show that

F E C°(D).

In the boundary integral in (1) one may replace by (ci. (1.lb)).The function Ou/On on F, however, is unknown and cannot be specified arbi-trarily either, since the boundary values (1.lb) already determine the solutionuniquely (cf. Theorem 3.1.2). To make vanish one must selectthe fundamental solution so that x) =0 for e F, x €5?.

DefinItion 3.2.3. A fundamental solution g In (2.2.9) is called a Greenfunction (of the first kind) if = 0 for all €F, XE 0.

The existence of a Green function is closely related to the solvability ofthe boundary value problem for the potential equation:

Remark 3.2.4. The Green function exists if and only if for all x E $1 theboundary value problem =0 in 5? and 4i = —s(-,x) on F has a solution

The above consideration results in

Theorem 3.2.5. Let .0 be a normal domain. Let the boundary value problem(1.la,b) have a solutionu C2(Th. Assume the existence of a Green functionof the first kind. Then one osn express u explicitly by

3.2 Representation of the Solution by the Green Function 29

u(x) =— L x) (3.2.2)

In the following we reverse the implication. Let the existence of the Greenfunction be assumed. Then, does function u defined by Equation (2) representthe classical solution of the boundary value problem (1.la,b)? Here it must beproved, in particular, that u E C2(i7) and = f. Firstly, it is not even clearyet whether the function u(x) defined by Equation (2) depends continuouslyon x since the definition of a fundamental solution x) does not requirecontinuity with respect to the second argument x. Despite that, the Greenfunction is x) with respect to x in as the following resultshows (cf. Leis [1, p. 671).

ExercIse 3.2.6. Let a be a normal domain. Let the Green function existfor a, and for fixed y E $1 let g(., y) (weaker conditions arepossible !). Now prove that

g(x, y) g(y, x) for x, y E .0. (3.2.3)

Hint: Apply the Green formula (2.2.5b) with = Ux',x" €0, u(x) := g(x,x'), v(x) :=g(x,x"), and use (2.2.10).

If one tries to reverse the assertion of Theorem 5, one encounters thesurprising difficulty of having to set precise conditions on the source term 1.The natural requirement f E is necessary for u C2(fl), but it is notsufficient, as the following theorem, whose proof will be appended at the endof this section, shows.

Theorem 3.2.7. Even if the boundary F and the boundary values p aresufficienLly smooth and if the Green function exists, there are functions I EC°(72) to which no solutions u

Theorem 7 shows that Equation (2) need not represent a classical solu-tion for / However, a sufficient condition for / to do so is Holdercontinuity.

Definition 3.2.8. 1 E issaid to be HOlder continuous in withthe exponent A E (0,1) if there exists a constant C C(f) such that

ff(x) — � — y E (3.2.4a)

We write I E and define the norm IIfllcA(Th) as the smallest con-stant C which satisfies (4a):

:= sup{

(3.2.4b)

30 3 The Poisson Equation

The function I E Ck(1)) is said to be k-fold Holder continuouslydifferentiable in (with the exponent A), if 171€ C"(fl) for all H � k.Here

with v,,�0, I"l=&'I+'"+"n (3.2.5a)

is a multi-index and

17 = (3.2.5b)1 1)

a IvI-fold partial derivative operator, The k-fold HOlder continuously differon-tiable functions form the linear space Cb+A(Th) with the norm

UI := k}. (3.2.4c)

If 8 = k + A one also writes C'(Th) for The k-fold Lipschitzcontinuously differentiable functions I are the result ofthe choice A 1 in (4a,b). For reasons of completenesa let us add that

ill k} (3.2.4d)

is the norm in for integer k � 0.

Exercise 3.2.9. (a) f Is said to be locally Holder continuous in a if foreach x a exists a neighborhood K((x) such that f E C"(K,(x) ri Q).Prove that if is compact then I E CA(fl) follows from the local HOldercontinuity in Th. Formulate and prove corresponding statements forand Ck.1(Th).

(b) Let 8 > 0. Show that lxi' E C'(Kft(O)), if a otherwise lxi' EHint: 1 —t' � (1—i)' for O< t � 1, a � 0.

The function u from Equation (2) can be decomposed into u1 +u2 whereUI = — f0 91 and ua = — f,, çoôg/On dl'; u is the solution of the boundaryvalue problem (1.la,b) if we are able to show that u1 and u2 are solutions of

iiuj=finf1,

Theorem 3.2.10. LI the Ciw, function exists and satisfies suitable condi-tions then

u(x) =— j x) (3.2.6)

is a classical solution of 4u =0 inS?, withu= on!'.

The proof goes in principle just as for Theorem 2.3.9 (cL Leis [1, p.69J).

32 Representation of the Solution by the Green Function 31

Theorem 3.2.11. Suppose the Green Junctiong(',x) C2(Th\{x})forx ES?exists, and let it be f E C'(Th). Then

u(x) =— J

f in with u = I) on I'.

PROOF. The boundary condition u(x) = 0 for x E I' follows easily from0 and (3). The property u C'(Th) and the representation

result from the

Exercise 3.2.12. Let (2 C be bounded and A := xx}. For the derivatives of / with respect to x aseume E C°(A)

and S — with 3 <n for any k. Prove that thenF(x) := x) and IY'F(x) x) H � k.

To prove u E C2(?)) this step cannot be repeated since x)has a singularity which is not integrable. We write the derivative

in the form

= — — (3.2.8a)

Let 8,F(x) be the difference quotient (F(xE) — F(x))/6, with = x3 + e and= for i j. The product rule = G(x)8,F(x) +

applied to Equation (8a) gives

(x) =—

—

f(x) x) — is integrahie, the limit e —' 0

results in the formula

(x) =— f — x) x)

(3.2.8b)Equation (8b) implies

= Jtf(E) — —

so that it remains only to show that = -1. Choose KR(Z) so thatx E KR(z) C (2. The Green function has the form (2.2.9): g 8 + The firsttwo terms in

f x)= J x)

+ f x)+ J x)

1? O\KR(5)

are harmonic in KR(z), so that fxR(Z) s(e, x) = —1 is what has to beproved.

32 3 The Foieson Equation

Let be defined by s(e, x) = — xJ) (cf. (2.2.1)). For fixed r > 0set

1 'v(x) / (3.2.8c)

(L)flr

For all x OKR(z) (i.e., Ix—zi r) v is harmonic, since is nonsingularOfl äKr(Z) and satisfies = 0. Since s(.,x) is harmonic in Kr(Z) forr < xC, the mean-value property (2.3.1) holds, which can now be written

v(x) = s(z, x) = — for — > r. (3.2.&1)

Using Exercise 3.2.2b we see that v(x) is continuous in so that we alsohave

v(x) = o'(r) for — r. (3.2.8e)

Thus v is harmonic in Kr(X) with the constant boundary values (Se). Theunique solution is therefore

v(x) = for z — xl � r. (3.2.8f)

The equations (8c,d,f) yield

J x) = lz —8K,.(s)

and then, since 0 < lz — xl <R,

f 0 8K..(a)

rn I$xln

R2 lz—x12= — R'o-(R) + 2n - 2n

From this we see that, independently of R, n, and z, there results

U

From Theorems 10 and 11 follows

Theorem 3.2.13. Under the same assumptions as for Theorem, 10 and 11Equation (2) gives a representation for the classiest solution of the boundaryvalue problem (1.Ia,b).

3.2 Representation of the Solution by the Green Function 33

Finally we put forward two inequalities for the Green function as exer-cises:

Exercise 3.2.14. In (1, and Qj C Q2, respectively, let the Green functions9, gj and 92 exist. Show

(a) 0 � g(x,y) � s(x,y) for x,y E tiC n 3 (3.2.9)

What the inequality for n=2?(b) � g2(x,y) for x,y E (lj C (12 (3.2.10)

Hint: Exercise 2.3.14.

Supplement: Proof of Theorem 7. If we make use of a later theorem (Theorem6.1.13) then Theorem 7 follows from

Theorem 3.2.15. The solution u does not depend continuously on f, if thesupremum norm (2.4.1) is used as the norm in and that from (4d) isused as the norm in C12(Th).

PROOF. Let .11 K1(O) c und p =0. The disk Ills a normal region forwhich the Green function is known (cf. Theorem 3.3.1). By Theorem 11 thereexist solutions E of = in (1, 0 on I' for the functions

____

I= 2

p,%(r), r := IxI, p,(r) := mm r, log

which belong to 00(17) and are uniformly bounded: = 1/log 2. ByTheorem 11 we have u"(x) = Since � itfollows from Exercise 12 that

=

where g = a. The first integral is bounded since E C2(17). The derivativeof the singularity function is 0) = — The special choice of

gives for x =0

— L + —

The surface integral K := — 12jEJ5de > 0 does not depend onr (0,11, so that the second integral takes on the form := K r'p0(r)dr.Since dr diverges as e —i 0, we deduce —' oo as n —. oo.Since � it follows that the map I '—' u is not bounded,and thus not continuous.

34 3 The Poiseon Equation

3.3 The Green Function for the Ball

Theorem 3.3.1. The Green function for the bafl KR(y) is gwen by thefunction in (2.2.lia). For f E and 'p C2+A(r) with 0 < A < 1, therepresentation formula (2.2) defines a solution u E of the boundarywJue problem inS?, u=ç' oni'.

The proof of the theorem follows from a result of Schauder, which is citedin Theorem 9.1.20.

in the case n = 2 the plane can be identified with C by the correapondence (x,y) #4 = x + iy. The following considerations are based on

Exerclae3.3.2.be holomorphic. Show

= i" = ÷ (3.3.1)

foiuEC2(S?').

Equation (1) shows, in particular, that a holomorphic transformationof coordinates maps harmonic functions into harmonic functions. An ar-bitrary simply connected region with at least two boundary points can,by the Riemann mapping theorem, be mapped by a conformal mapping

z S? 0(z) K1(0) onto the unit disk such that = 0 for anygiven zo E S?. Let g(C,C') be the Green function for K1(0). One may checkthat G(z, zo) is again a fundamental solution. Now z ôi? im-plies 0(z) 8K1(0), i.e., G(z,zo) = 0. Thus G(z,zo) is the Green functionin 0. This proves

Theorem 3.3.3. Let Si c It2 be simply connected with at least two boundarypoints. Then there exists a Green function of the first kind for (1.

The explicit forms of various Green functions can be found, for example,in the book by Wloka (1, Exercises 21.1—21.81. Of numerical interest mightbe the fact that with conformal mapping one may remove corners which aredisturbing (e.g., reentrant corners) (cf. Gladwell-Wait (1, p.701).

Example 3.3.4. Let 5? be the L-shaped region in Example 2.1.4. Choose= z213: $2 0'. Then # is conformal in 5?. The sides of the angle fo C 01?

(cf. Fig. 2.1.1) are mapped into a single line segment, so that if has no morecomersstiddngln.ThePoiasonequation4u=finflcorrespondetotheequation = in if.

3.4 The Neumann Boundary Value Problem 35

3.4 The Neumann Boundary Value Problem

In (1.lb) and in (2.1.lb) the boundary values u = were given on F. Theseso-called Dirichiet conditions or "Boundary Conditions of the First Kind" arenot the only possibility. An alternative is the Neumann Condition

= on F. (3.4.1)

In physics this second boundary condition, as it is also called, occurs morefrequently than the Dirichiet condition. For example, if u is the velocity po-

tential of a gas, then 0 means that the gas can only move tangentially atthe boundary F. Except in some unusual cases, the boundary value problem

Lu = f in 1? and 8u/On = 'p on F has a unique solution. An exceptional casedoes however occur for L =

Theorem 3.4.1. Let .f? be a normal The Poisson equation = fwith the Neumann boundary condition (1) is only solvable if

L dI't= J f(x) dx. (3.4.2)

If a 8OhLtlOfl t& does exist, then u + c, with c any constant, is also a solution.

PROOF. (1) One may repeat the proof of Lemma 2.3.6 for = I.(2) Obviously n + c satisfies the same equation.

Later, in Example 7.4.8, we wifl show that the Neumann boundary valueproblem for the Poisson equation has a solution if and only if (2) is satisfied,and that two solutions can differ only by a constant.

F,andthenormalderivative Ou/ôn occur. The Green function of the first kind was chosen insuch a way that = 0 for E F. In the case of the second boundaryconditions (1) one makes the assumption that &y(e, = c (c: constant),i.e.

= c —

for

L:=j&'.

Since must be harmonic (i.e., f = 0), from EquatIon (2) we see thatcL +1 =0 is a necessary condition for the existence of Thus the conditionon the Green Function of the Second Kind for the potential equation

36 3 The Equation

Thus the term fru&y/Ondf in (2.1) becomes const• Since u is onlydetermined up to a constant (cf. Theorem 1), one can fix this constant withthe additional condition f1. u dl 0. This gives the following result, if wewrite g for

u(x) — / +

The Green function of the second kind for the ball KR(O) c JR3 can be foundin Leis p. 79).

3.5 The Integral Equation Method

In the representation (2.1) of the Poisson solution the singularity function acan, in particular, be chosen to be -y. If in addition one imposes the givenNeumann data (4.1), one obtains

u(x)= / k(x, dIe + g(x) for x a, (3.5.1)

with the kernel function k(x,.) and the functions

g(x) := g1(x) +

:=

92(x) :=

The right-hand side in Equation (1) with the unknown boundary valueE I', can be used as an ansatz solution:

= j k(x, + g(x). (3.5.2)

The first suminand on the right of(2) is called the double-layer potential(dipole potential); Oi is the single-layer potential, while 92 is avolume potential.

For each u C°(E') the in (2) is a solution of the equation(I.la) in 0. However, is also defined for an argument x in the exteriordomain A closer look at the kernel function shows that itis in fact only weakly singular for the case of smooth boundaries F. Thus #is also defined for x E F. The function which is now defined on allis not continuous at points of the boundary I'. At XO E F there exists bothan interior limit &(XO) for x xo,x E a and an exterior limit forX —' Xii, X E JR"\?. In addition we have the third function value #(Xo) of(2). Their connection is given by the following jump discontinuity relation (cf.Hackbusch [7, Satz 8.2.81):

3.5 The Integral Equation Method 37

— &(x0) = 2u(xo), (3.5.3a)

+ &(xo) = for xo E F. (3.5.3b)

In order that the ansatz (2) does indeed give the solution u in (1), the boundaryvalue &, continued from the interior, must agree with the function u whichis put in the integral: & = u. Now one can solve Equation (3a,b) for &:

= — i4xo). The equation & = u thus leads to

u(xo) = Irk +g(xo) for F. (3.5.4)

Equation (4) is called a Fredhoim integral equation of the secondkind forthe unknown function u C°(fl. The original Neumann boundary-value problem (l.la), (4.1) and the integral eqation (4) are equivalent in thefollowing sense: (a) If u is the solution of the Neumann boundary-value prob-lem, then the boundary values, F, satisfy the integral equation (4).(b) If n E C°(F) is a solution of the integral equation (4), then the expreesion(1) gives a solution of (1.la), (4.1) in the entire domain il.

The transformaUon of a boundary-value problem into an integral equa-tion, and the subsequent solution of the integral equation is referred to asthe integral equation method. It allows, for example, a new approachto existence statements, in that one shows the solvability of (4). The integralequation (4) can also be attacked numerically. If methods similar to the finite-element method described in Section 8 are used, then the result is called theboundary-element method (BEM).

One can find references to the integral equation method in, e.g., Hack-busch 17, and

4 Difference Methods for the PoissonEquation

4.1 Introduction: The One-Dimensional Case

Before developing difference methods for the partial differential Poisson equa-tion, let us first recall the discretisation of ordinary differential equations.The equation a(x)1s"(x) + b(x)u'(x) ÷ c(x)u(x) = f(x) can be supplementedwith initial conditions u(x1) u1,u'(x1) = *4 or with boundary conditionsu(x1) = u(x2) = u2. The ordinary initial value problems correspond to thehyperbolic and parabolic initial value problems, while an ordinary boundaryvalue problem may be viewed as an elliptic boundary value problem in onevariable. In particular one can view

—iil'(x)=f(x) for z€(O,1), (4.1.la)

(4.l.lb)

as the one-dimensional Poisson equation = f in the domain = (0,1)with Dirioblet conditions on the boundary I' = {O, 1).

Difference methods are characterised by the fact that derivatives are re-placed by difference quotients (divided differences), in the following called,for short, "difference? . The first derivative u'(z) can be approximated byseveral (so-called "first") differences, for example, by the forward or rightdifference:

:= (u(z + h) — i4z)J/h, (4.1.2a)

the backward or left difference(8u)(x) := [u(x) — u(x — h)J/h, (4.1.2b)

or the symmetric difference(8°IL)(x) Iu(x + h) — u(x — h)1/(2h), (4.l.2c)

where h > 0 is called the step size. An obvious second difference for u"(x)is

:= [u(x + h) — 2u(x) + u(x — h)J/h2. (4.1.3)

One also calls 8, and index/ D7/emphdifference operators.The product may be viewed as 8 or as o8, i.e.. )u(x)

4.1 Introduction: The One-Dimensional Case 39

Lemma 4.1.1. Let {x—h,x-l-h)CTh. Then

= u'(x) i-hR with < if U E C2(Th) (4.1.4a)

8°u(x) = u'(z) + h2R with IRI if u E (4.i.4b)

= u"(x) + h2R with if u E C4(Th).(4.1.4c)

PROOF. We give the proof only for (4c). If one applies Taylor's formula

E h) u(s) ± hu'(x) + h2u"(x)/2 ± h3u"(z)/6 + h4R4, (4.1.5a)

zjhI?.4 = h4j u"(x±t9h)/4!, (4.l.5b)

with i9 E (0, 1), to Equation (3), the result is (4c) because R =u""(x — i92h)jf24. U

'7'' Slh Figura4.l.1

Gridforh=1/8

grids

= {h,2h, .. .,(n — 1)h = 1— h}, (4.1.6a)

(4.1.6b)

of step size h = 1/n. For x E STlh, only contains the values of u atx,x ± h E Under the assumption that the solution n of Equations (la,b)belongs to C4(Th), (4c) yields the equations

f(s) + 0(h2), z E ha. (4.1.7)

If one neglects the remainder term 0(h2) in Equation (7), one obtains

for

(4.1.8a)

These are n—i equations in n-fl unknowns {u,,(x), z The two mimingequations are supplied by boundary conditions (ib):

uh(O)=p0, (4.1.8b)

Uh is a grid function defined on Its restriction to yields the vector

= — h))T.

40 4 Difference Methods for the Equation

If in (8a) one eliminates the components and tIh(l) with the aid ofEquation (8b), one gets the system of equations

(4.1.9a)

with

2 —1

—1 2 —1

—1 2 —1Lh=h2 (4.1.Db)

—1 2 —1

—1 2

'lb = (1(h) + f(2h), f(3h),.. . ,f(1 — 2h), 1(1 —4) + (4.1 .9c)

4.2 The Five-Point Formula

First we select the unit square

Q={(x,y):0<z<l, O<zy<l}as the fundamental domain. More general domains will be discussed in Section4.8. In the discretiastion process is replaced by the grid

:= {(x,g) (1:x/h, p/hE 1) (4.2.la)

for step size 4= (n E N). The discrete boundary points form the set

F5 := {(x,y) E F:x/h,y/h E Z}. (4.2.lb)

As in (1.lb) we set

Th5 := U = {(x, y) fl5: z/h, p/h Z}. (4.2.lc)

V 0 0 0 0 0 0

• • • • • o •: Points of

• • • • • 0 0: PorntsofFh

• • • . • 0

• • • • • 0

Figure4.2.1 • , • • •A two-dimensional grid

4.2 The Five-Point Formula 41

In the Poisson equation

— = f in 11, (4.2.2a)

(4.2.2b)

the second derivatives and can each be replaced by the respectivedifferences (1.4c) in the x and y directions:

= h21*L(z — h,y) +u(x + h,y)+ u(x,y — h) + u(x,y + h) — 4u(x,y)).

(4.2.3)

Since on the right side of (3) the function is evaluated at five points,is also called the Five- Point Formula. The discretisation of the boundaryvalue problem (2a,b) using leads to the difference equations

= 1(x) for x E (4.2.4a)

tLh(x)=çp(x) forxEl'4. (4.2.4b)

Through (4a7b) one obtains one equation per grid point x = (x, y) Earid hence one equation per component of the grid function uh =Except in the one-dimensional case, there exists no natural arrangement ofgrid points, thus one cannot immediately obtain a matrix representation asin (1.9b). The only natural indexing of is that through x 11h or the pair(i,j) Z2 with x (x,y) = (ih,jh). Let the matrix elements be given by

(4h2—h20 otherwise.

(4.2.5)For x = L, is a diagonal element; in the second case, L4 = —h2,we say that x and are neighbours. If one eliminates the components

x E with the aid of Equation (4b), then Equation (4a) assumes thefollowing form:

= qh(x) for x E ah, (4.2.6a)

where

:= ía + fa(x) := 1(x), Wh(X) := — (4.2.6b)

For the proof split the sum

=Oh

into and .... The second partial sum is and is movedto the right side of the equation.

42 4 Dlfferenee Methods for the Poisson Equation

Remark 4.2.1. fh is the restriction of f to the grid For all points farfrom the boundary we have ph(x) =0; here x E Qh is said to be far fromthe boundary if all neighbours x±(0,h), x±(h,O) belong to flh. In thecase of homogeneous boundary values = 0 we have =

The system of equations (6a) can be expressed in the form (l.9a):

Lhuh =

where the matrixLh =

and the grid functions Uh = (ua(x))xEnh and = are describedby their components (cf. (5), (6b)). Strictly speaking Lh is not a matrix in theusual sense but a linear mapping since the indices x E (1h are not ordered.

Exercise 4.22. (a) An N x N matrix P is called a permutation matrixif w := Pv, for all v E IRN, has the coefficients 1 � i � N,where iv is a permutation of indices {1,. ,N}. Show that P is unitary, i.e.,P-i = PT.(b) Let I be an index set with N elements. Let the "matrix" coefficients

/3 I, be given. To the arrangement al, of indices corresponds theNxN matrix A = with = A be the matrix whichbelongs to a second arrangement &1,. aN of I. Prove that there exists apermutation matrix P such that A = PAPT.(c) Let = ApQ for all a,/3 €1. For each arrangement of the indices thecorresponding matrix is symmetric.

A possible linear enumeration of indices x is lexicographicalordering

(h, h), (2h, h), (3h, h), •. ., (1 — 4, h),(4,24), (24,24), (3h,2h), •.., (I — 4,24),

(4, 1 — 4), (2h, 1 — h), (3h, 1 — 4), •.., (1 — 4,1 — 4). (4.2.7)

Generally, the point x = . , precedes the point y = (yj,.. ,yd) inlexicographical order, if for a j (1,.. , d} the conditions 5d = (i > i) and

<y, hold. Each line in (7) corresponds to a so-called s-row in the grid 1?,,.A vector uh whose (n - 1)2 components are enumerated in the series (7) thusseparates into n —1 blocks (so-called s-blocks). The block decomposition ofthe vectors generates a block decomposition of the matrix which is givenin(8).

Exercise 4.2.3. (a) With the lexicographical numhering of grid points thematrix has the form

4.2 The Five-Point Formula 43

T —I 4 —1

—I T —1 —1 4 —1

Lh=h2 ..• ... .. .•.

—I T —1 —1 4 —1

—I T —1 4(4.2.8)

where T is an (n — 1) x (n — 1) matrix and Lh contains (n — 1)2 blocks. I isthe (n — 1) x (n — 1) identity matrix.(b) Let (1 be the rectangle

(1=(0,a)x(0,b)={(x,y):0<x<a,0<y<b}.Let the step size h satisfy the conditions a = nh and b = mh. Show that thediscretisation (4a,b) in the corresponding grid leads to a matrix which alsohas the form (8). But here Lh contains (m — 1)2 blocks of the size (n — 1) x(n—i).

Another frequently used arrangement is the chequer-board ordering(or red-black ordering). To this end one chessboard pattern divides into"red" and "black" fields:

.Q, :={(x,y)Eflh:(x+y)/hodd},429Qb := {(x,y)E Qh:(z+y)/heven}.

First one numbers the red squares (z,y) E lexicographically, and thenthose of (Ihb . The partition (9) induces a partition of vectors into 2 blocks

and a partition of the matrix Lh into 2.2=4 blocks.

Exercise 4.2.4. WIth respect to the chequer-board ordering, the matrix Lhassumes the form

A

= II2 (4.2.10)

AT

4

where, in general, A is a rectangular block matrix because for n even,and contain a different number of points.

The complete (n — 1)2 x (n — 1)2 matrix in (8) or (10) is needed nei-ther for the theoretical investigation of the system of equations LhUh = norfor its numerical solution. All properties of Lh considered in the followuig areinvariant with respect to re-numbering of the grid points. Even though nuiner-ical methods for the solution of LhU,, = implicitly use an arrangement ofgrid points (with the exception of special algorithms for parallel computers),they never employ the complete (n — 1)2 x (ii — 1)2 matrix Every usable

44 4 DIfference Methods for the Poieson Equation

algorithm must take into account that Lh is sparse, i.e., it has substantiallymore zero than nonzero elements.

In the following we again return to indexing by x Oh. Neverthelese wewill continue to refer to the Lh defined by (5) as a matrix.

The difference operator is also described by the star

1

= h2 1 —4 1 . (4.2.11)1

The general definition of a difference Star (with variable coefficients) reads

co.1(z,y) ci,i(z,y)c_l,o(x,y) y) cj,o(x, y)

c_1,_1(z,y) co,_i(x,y) c1,_1(x,y)

= y)uh(x + ih, y + jh), (4.2.12)

in which the zero coefficients have not been written out.

Attention. The star (11) does not represent a submatrix of Lh! The coeffi-cients of the star appear in each row of Lh.

Remark 4.2.5. Note that the difference operator cannot be equatedwith the matrix Lb sance 4h does not contain information on the type orplace of the boundary conditions. We shall say the matrix Lb belongs toadifference star (12) if the system of equations Lhuh = results from thedifference equations in x E ak after elimination of the Dirichiet boundaryvalues uh(x) = x "h. Even if the vector Ub in LhUh = qh containsonly components Uh(X), x E one occasionally equates ub with the gridfunction on Thh which assumes the prescribed boundary values (4b) on Ph.

4.3 M-matrices, Matrix Norms, Positive DefiniteMatrices

The elements of the matrix A are denoted by E I. Mere A and theindex set I assume the places of Lb and Slh. We writs

A�B foralla,/3€I,

and define analogously A B,A > B,A <B. The zero matrix is denotedby 0.

4.3 M-matrices, MatriX Norms, Po6itive Definite Matrices 45

Definition 4.3.1. A is called an M-matrix if

OaQ>0 forallaEI, Oap�O (4.3.la)

A nonsingular and A' � 0. (4.3.lb)

The inequalities (is) can immediately be proved for Lh (cf. (2.5)). How-ever we still need criteria and auxiliary results to prove (ib).

The index a I is said to be directly connected with I if0. We say that a i is connected with /3 I if there exists a

"connection" (chain of direct connections)

a =ao,al,a2,",ak =/3 with (1 � i � k). (4.3.2)

The index set I together with the direct connections form the graph of A(cf. Fig. 1). A has a symmetrical structure, i.e., 0 holds ifand only if 0. In this case a is (directly) connected with (3 if and onlyif is (directly) connected with a.

Definition 4.3.2. A matrix A is said to be irreducible if every a €1 isconnected with every (3 E I.

1 1 o ,— ® @ Indices

A=011 C)) connections101

FIgure 4.3.1. Graph of the irreducible matrix A

In the case of the matrix A = Lh, two indices x,y are connected ifand only if y = x or if y is a neighbour of x. Arbitrary x, y E can evidentlybe connected by a chain x = . ,x(k) = y of neighbouring points.Thus L5 Is irreducible.

Exezciee 4.3.3. Prove that A is irreducible if and only if there is no orderingof the indices such that the resulting matrix has the form

where A,1 and A22 are respectively square n1 x n1 and n2 x n2 matrices(iii � 1,n2 � 1), and A12 is an n1 x submatrix.

46 4 Difference Methods for the Poisson Equation

The important question as to whether A = Lh is nonsingular can betreated as a special case of the following statement.

Criterion 4.3.4.__(Cerahgorin) Let Kr(z) denote the open disk {C E C: lz<r}, and let Kr(Z) := {C E C: fz — Cl � r) denote the dosed disk.

(a) AU eigenvalues of A lie in

U with r0 = >2

(b) if A is irreducible, the elgenvalues even lie in

UaEI aEf

PROOF. (a) Let A be an eigenvalue of A and u a corresponding eigenvectorwhich, without loss of generality, satisfies =1, where

max{luaI: a E I} (4.3.3)

is the maximum norm. There exists (at least) one I with 1u71 = 1.

Assertion 1. = 1 implies

IA — � >2 � >2 = r7. (*)

From (*) follows A and hence the statement. To prove theassertion use the equation from Au = Au associated to the index -y:

thatis'El

From lts.,I= 1 follows IA—a.rI � By takingthe modulus into the sum and by using lu,l � = 1, (*) follows.(b) Let A be irreducible and let A be an arbitrary elgenvalue of A withassociated eigenvector u which in turn is again normalised by hulk0 =1. The

A E UQEI immediately iee.ds to the statement. Therefore letA be assumed.

Assertion 2. Let 0, i.e., is directly connected with /3; then = 1

and IA — a.rvl = implies = 1 and IA — = r0.

Part (a) proves the existence of a 'y E I with = 1 and IA — � r7.According to the assumption, IA — = mtwt hold so that 2 is

4.3 M-matrlces, Matrix Norms, Positive Definite Matrices 47

applicable to Since A is irreducible, for an arbitrary /3 E I there exists aconnection (2) of with 8: °o = /3, 0. Assertion 2shows

= 1 and A = r0 for all i =0,•in particular, A E for /3 = Since was chosen arbitrarily, itfollows that A E and the statement is proved.Proof of Assertion 2. Besides the inequality chain (*) there also holds IA —

= so that all the inequalities in (*) become equations. In particular

L1a7p1 IupI=

must hold. Since � hulk0 = 1, the identity lupI = must besatisfied for each suminand. Hence 0 implies = 1. The applicationof Assertion 1 to /3 yields A Kr0(aaa)proves IA — = U

Exercise 4.3.5. Let I : a connection (2) exists between a and /3).Show that the eigenvalues of A lie in

U flaEI fiEIa

Definition 4.3.6. (a) A is said to be diagonally dominant if

> <IaQOCI

for all a I. (b) A is said to be irreducibly diagonally dominant if Ais irreducible, the inequality (4a) holds for at least one index a I and

foralla€I. (4.3.4b)

Note that while an irreducible and diagonally dominant matrix is irre-ducibly diagonally dominant, the reverse need not hold.

The matrix Lh from Section 4.2, while not diagonally dominant, is irre-ducibly diagonally dominant, for Lh is irreducible and satisfies (4b). At allpoints near the boundary—i.e., those x E ah, which have a boundary pointy E rh as a neighbour—however, (4a) holds: � 3h2 <4h2

The spectral radius p(A) of a matrix A is given by the eigenvaluethat is largest in modulus:

p(A) := ma.x{IAI: A elgenvalue of A}. (4.3.5)

48 4 Difference Methods for the Ponson Equation

In the following we split A into

A = D — B, D diag {a00:a €1), (4.3.6a)

where D is the diagonal part of A:

= a00, 0 for a /3. (4.3.6b)

B := D - A is the off-diagonal part:

= 0, = for a /3. (4.3.6c)

Criterion 4.3.7. Let (6a-c) hold. Sufficient conditions for

p(D'B) < 1 (4.3.7)

are the diagonal dominance or the irreducible diagonal dominance of A.

PROOF. (a) The coefficients of C := D1B read

C0p =—a0j,/a00 c00=0.

From the diagonal dominance (4a) follows that r0 := Icosi < 1

for all a I. By the Gershgorin Criterion 4a all elgenvalues A of C liein Kr,,(Cao) U0€1 so that IAI < maxr0 < 1 and hencep(C) = p(D'B) < 1 also follows.(b) If A is irreducibly diagonally dominant then rp � 1 for all /3 I andr0 < 1 for at least one a. According to Criterion 4b all eigenvalues of C liein . This set lies in K1(0) UflPE,OKr,(0)CK1(O). At first let us assume that all rp agree: r for all /3. Since r0 < 1

for one a E I, it follows that r < 1 and flfiôKr5(0) = OKr(O) C K1(O). But ifall are not equal then aKr0(o) is empty. Thus in both cases A Eholds and (7) is proved.

Exercise 4.3.8. (a) Weaken irreducible diagonal dominance as fol-lows: Let A satisfy the inequalities (4b) and for all /3 I let the connection(2) exist for an index a E I for which the strict inequality (4a) holds. Provethat even under this assumption p(D'B) < 1 holds.Hint. Use Exercise 5.(b) Show that the geometric series S = converges if and only ifp(C) < 1. Then the following holds: S = (I — C)'. Hint: Represent C inthe form QRQT (Q a unitary, and R an upper triangular matrix) and showIICiIoo K[p(C)Jt'.(c) Let u be a vector. We define as the vector (f) with the entries :=

For two vectors one writes v w if v0 w0 (a E I). Show that:(1)AB�O,ifA?O,B�O;AB>O,ifA>O,B>O;(2) AD >0 if A >0, and D � 0 is a nonsmgular diagonal matrix;(3) A's' Aw ifA �Oandv �w; f(vtL1, ifO�v�w;

4.3 M-rnatrices, Matrix Norms, Positive Definite Matrices 49

(4) Au � lAut Alut if A � 0.

The importance of inequality (7) results from

Lemma 4.3.9. Let A (Ia). Let D and B be defined by (6a-c). A isan M-matrix if and only if p(D'B) < 1.

PROOF. (a) Let C := D'B satisfy p(C) < 1. Then the geometric seriesS := CTM converges (cf. Exercise 8b). From D1 � 0 and B � 0 oneinfers C � 0, C" � 0, and S � 0. Since I = 5(1—C) = SD'D—B(SD)'A, A has the inverse A' = SD'. D1 � 0 and S � 0 result inA' � 0. From this (ib) also results, i.e., A is an M-matrix.(b) Let A be an M-matrix. For an eigenvalue A of D'B select an eigenvectoru 0. According to Exercise 8c we have

tAt liii = tAut = tD'BuI <

Because A-'D � 0 (cf. (la,b)) one obtains IuIso that

= — = A'D(I — D'B)tu) � A'D)u) —

= (1 — IAI)A1DIuI

follows. For � 1 we would get the inequality, tul � 0, i.e., u = 0, incontradiction to the assumption u 0. From this follows IA) < 1 for everyelgenvalue of C = D1B, thus p(D'B) <1. U

Criterion 7 and Lemma 9 imply

Criterion 4.3.10. If a matrix A with the property (la) is diagonally domi-nant or diagonally dominant, then A is an M-matrsx.

Theorem 4.3.11. An irreducible M-matrix A has an element-wise positiveinverse: A' > 0.

PROOF. Let a,$ E I be selected arbitrarily. There exists a connection (2):= ..,c4 =$. Set C :=r D'B. Since >0, it follows that

= . .� Coo Ca cc > 0.

71,,7k—jEI

According to Lemma 9, p(C) <1 holds, so that S := C" converges.Since � > 0 and a, /3 E I are arbitrary, S > 0 is proved. Theassertion results from A' = SD' >0 (cf. proof of Lemma 9). U

In the following we derive norm estimates for A'.


Definition 4.3.12. Let V be a linear space (vector space) over the field ofreal numbers (K := IR) or complex numbers (K := C). The functional . II

called a norm in V if

hull =0 only for u 0, (4.3.8a)

hiu+ vii � hull + Dvii for all u,v E V, (4.3.8b)

iiAuli hAl 111111 for all A K,u V. (4.3&)

Example. Let V = where #1 := is the number of elements of the indexset I. The maximum norm defined in (3) satisfies the norm axioms (8a-c).

If one views the elements u E V as vectors, is called a vector norm.But the matrices also form a linear space. In the Latter case one calls U a

matrix norm. A special class of matrix norms is contained in

Definition 4.3.13. Let V be the vector space with vector norm . fl. Thenone calls

(hAul := sup{flAuhh/hluhl: 0 u V} (4.3.9)

the matrix norm associated with the vector norm fi. fi

ExercIse 4.3.14. Let ifi be defined by (9). Show that: (a) is a norm;(b) the following holds:

IHABIII � luAu DiBhil, (4.3.lOa)

1111111 = 1 (I: unit matrix), (4.3.lOb)

flAull � (fl.A(fl hull, (4.3.lOc)

hAul � p(A). (4.3.lOa)

Example. The matrix norm associated with the maximum norm fi. (cf.(3)) is called the row sum norm and is also denoted by Ii It has theexplicit representation

hhAik = (4.3.11)

ExercIse 4.3.15. (a) Prove (11). (b) For matrices 0 < B � C there holds

In the next theorem we denote by I the vector having only ones as com-ponents:

L=1For the notation v w see Exercise 8c.

Theorem 4.3.16. Let A be an M-matriz and let a vector w exist withAw � 1. Then flA1fl00 <

4.3 M-matrices, Matrix Norms, Positive Definite Matrices 51

PROOF. As in the proof of Lemma 9, let be the vector with the compo-nents u we have lul � � Since A-1 � 0, weobtain

LA'ut � � UuU00A'Aw =

(cf. Exercise 8c) and < Definition 13 implies that

�How to estimate with the aid of a majorising matrix is shown in

Theorem 4.3.17. Let A and A' be M-mainces with A' � A. Then thefollowing holds

0 � A'' and � (4.3.12)

PROOF. A'-' < A1 follows from A' — A'' = A'(A' — A)A'' and

A' 0, A' — A � 0, A'' � 0. The remainder follows from Exercise 15b.IExercise 4.3.18. Prove (12) under the following weaker assumptions: A isan M-matrix, A' satisfies (is) and A' � A. Hint: Repeat the considerationsfrom the first part of the proof of Lemma 9 with the matrices D' and B'associated to A'.

Exercise 4.3.19. Let B a principal submatrix of A, that is, there existsa subset I' C I such that B is given by the entries bap = a, (3 E I'. Provethat if A is an M-matrix, then so is B and 0 � S holds forall a, /3 I'. Hint: Apply Exercise 18 to the following matrix A': a'00 =(a,j9 I'), = (a l\I'), = 0 otherwise.

Another well-known vector norm is the Euclidean norm

11u112 := (4.3.13)

with fixed scaling constant c> 0 (for example, the choice c = h2 in connectionwith the grid functions from Section 4.2 results in the fact that rep-resents an approximation to the integration The matrix norm associatedto •))2 is independent of the factor c. It is called the spectral norm and isalso denoted by )f The name derives from the following characterisation.

Exercise 4.3.20. Prove: (a) for symmetric there holds (IA))2 =p(A) (cf. (5)). (b) For each real matrix holds:

1(A))2 = (p(ATA)Ih/2 = (maximal eigenvalue of ATAI"2.

(c) For each matrix holds � UAII2IIAT 112 Hint: (b) and (lOd).


For the proof in the exercise use the scalar product

(u,v) : CLUaVQ (4.3.14a)ciCI

(c as in (13)) and its properties

(u,u) = (Au,v) = (u1ATv), I(u,v)I � IluII2tIvU2. (4.3.14b)

Here the case K = Ut is always used as basis., i.e., all matrices and vectorsare real.

Definition 4.3.21. A matrix A is said to be positive definite if it issymmetric and

(Au,u) >0 for all (4.3.15)

Exercise 4.3.22. Prove that (a) a symmetric matrix is positive definite ifand only if all eigenvalues are positive.(b) All principal submatrices of a positive definite matrix are positive definite(cf. Exercise 19).(c) The diagonal elements a positive definite matrix are positive.(d) A is called positive semi-definite if the inequality (15) holds with "�"instead of">". A positive semidefinite matrix A has a unique positive semi-definite square root B = A"2, which has the property B2 = A. If A ispositive definite, then so is A1t3.

A corollary to Exercise 22a is

Lemma 4.3.23. A positive definite matrix A u nonsingular and has a pos-itive definite inverse.

The property "A1 is positive definite" is neither necessary nor sufficientto ensure the property "A-' � 0" of an M-matrix. In both cases, however,(irreducible) diagonal dominance is a sufficient criterion (cf. Criterion 10).

Criterion 4.3.24. If a symmetric matrix with positive diagonal entries isdiagonally dominant or diagonally dominant then it is positive def-inite.

PROOF. Since resp. the Gershgorin circles which occurin Criterion 4 do not intersect the semi-axis (—oo, 0), so that all the elgenvaluesmust be positive. By Exercise 22a then A is positive definite,

Lemma 4.3.25. )tmin and be the smallest and largesteigenvalues of a positive definite matrix A. Then there holds

UA(12 = Amex, tIA'1j2 = 1/Arnie. (4.3.16)

4.4 Properties of the Matrix Lh 53

PROOF. Exercise 20a shows that hAil2 p(A) and 11A'hh2 = p(A').From (5) then result p(A) = and = since .Xmin > 0. U

4.4 Properties of the Matrix Lh

Theorem 4.4.1. The matr%x Lh (five-point formuLa) defined in (2.5) has theproperties:

Lh is an M-matrix, (4.4.la)

Lh is positive definite, (4.4.lb)

� 8h2, � 1/8, (4.4.lc)

� 8h2 cos2(irh/2) <8h2, (4.4.ld)

= + 0(h2) � (4.4.le)

PROOF. (a) In Section 4.3 we already noticed that Lh is irreducibly diag-onally dominant and satisfies the inequality (3. la). By Criterion 4.3.10 thenLh is an M-matrix.(b) Since Lh is symmetric and irreducibly diagonally dominant, (ib) followsfrom Criterion 4.3.24.(c) That can be read from (2.5) and (3.11). To estimateone uses Theorem 4.3.16 with w(x,y) = x(1 — z)/2. Then we have Lhw � I(even that (Lhw)(x,y) = 1 unless y = h and y 1 — h) and thwhloo �w(1/2, y) = 1/8.(d) The inequalities (ld,e) result from Lemma 4.3.25 and

Lemma 4.4.2. The (n — 1)2 eigenvectors of L,, are (1 n — 1):

= ((x,y) E Qh). (4.4.2a)

The corresponding eigenvalues are

A,,. — + sin2 (&irh/2)), 1 � v, ,z � n — 1. (4.4.2b)

PROOF. Let Q1D be the one-dimensional grid (1.6a) and let :=sin(,.'irx). Fbr each x E fl1D there holds

= h9sin(z'ir(x — h)) + + h)) — 2sin(virz)J

= 2h2sin(vwx)Icos(virh) — 11

since siri(z/x(x ± h)) = cas(ziirh) ± cos(virx) sin(virh). The identity1 — cost = then implies

54 4 Difference Methods for the Pois&on Equation

= x E

Let LLD be the matrix (1.9b). Note that also involves the bound-ary value u(O) which does not; similarly (88+ti)(1 — h) dependson u(1). However, since u(O) = sin(O) = 0 and ii(1) = 0 we have

= and (2c) can be brought over:

= 4h2s1n2(virh/2)u", 1 v n — 1. (4.4.2c')

The two-dimensional grid function in (2a) can be written as the (tensor)product Now we have that (Lhu")(x, y) is equaL to the sum

+ so that (2b) follows from (2c'). U

In the sequel we want to 8how the analogies between the properties of thePoisson equation (2.2) and the discrete five-point fonnula (2.4a,b).

The analogue of the mean-value property (2.3.1) is the equation

= — h,y)+uh(x + h,y) +uh(x,y — h)+ Uh(X, y + h)]. (4.4.3)

From (2.3) and (2.4a) with I = 0 we obtain the

Remark 4.4.3. The solution Uk of the discrete potential equation (2.4a)with f = 0 satisfies Equation (3) at all grid points (x,y) E .0h.

As in the continuous case the mean-value property (3) implies themaximum-minimum principle.

Remark 4.4.4. Let Uk be a non-constant solution of the discrete poten-tial equation (2.4a) with f 0. The extrema max{uh(x): x E flh} and

are assumed not on ATlh but on

PROOF. If uk were maximal in (x, v) E then because of Equation (3),all neighbouring points (x±h,y) and (x,y±h) would have to carry the samevalues. Since every pair of points can be linked by a chain of neighbouringpoints, it follows that Uk = const, in contradiction to the assumption. U

The last proof indirectly uses the fact that L,1, is irreducible. The irre-ducibility of L,, corresponds to the assumption in Theorem 2.3.7 that 12 is adomain, i.e., connected.

The result of carrying over Theorems 2.4.3 and 3.1.2 reads as follows:

Theorem 4.4.5. Let Uj and be two solutions of (2.4a): = f fordifferent boundary t4 = (i = 1,2). Then the following holds:

— � max — w2(x)l, (4.4.4a)XE

in 12h, if on l's. (4.4.4b)


PROOF. Let wh := —t41. (1) In the case that one has vib � 0 onRemark4provesthatwh=const �Oorwh>O.

(2) Let M be the right side of (4a). —M � wh � M on f'h implies theinequallties —M � � M on .17h. S

The discrete analogue of the Grcen function is Letbe the scaled unit vector

— {h2 X (x, E [lh). (4.4.5a)

The column of the matrix with index E is given by

(eEQh). (4.4.5b)

For ah fixed, gh(•,E) is a grid function defined on The domain ofdefinition is extended to x Thh:

l'p,E (4.4.5()

The are entries of h2L': = The symmetry ofLh implies

Remark 4.4.6. = for all E (ci (3.2.3)).

The representation (3.2.7) is recalled by:

Remark 4.4.7. The solution uh of the system of equations (2.4a) withboundary values 'p =0 reads

uh(X)=h2 (4.4.6)

Equation (6) is the component-wise representation of the equation Uh =The factor h2 compensates for h-2 in (Sa). It was introduced so that

the summation h2 E in (6) approximates the integral f0. Since in Section 3.2we consider the Poisson equation = f, but here the equation = f,the right-hand sides in (3.2.7) and (6) differ in their signs.

The discrete Green is positive also (cf. (3.2.9)):

Remark 4.4.8. 0< h2/8 for x,e üh.

PROOF. The upper bound follows from

gh(X,e) � � IIL;'11001j6d100,


= h2, and (ic). 9h > 0 can be inferred from > 0 (cf. Theorem4.3.11). IThe bound <h2/8 is too pessimistic and can be improved consid-erably.

Remark 4.4.8'.

0< <— � I': (4.4.7)

This estimate via 0(1 log hj) reflects the logarithmic singularity of thesingularity function = —

ExercIse 4.4.9. For the proof of inequality (7) define

:= 1/4 — log(Ix — + 2h2)/logl6, Uh(X) sh(x,O)

and carry out the following steps:(a) sh(x, �0 for all XE Thh,(b) htih(O) = h2,(c) —Lthuh(x) � 0 for all x 1R2 (longer calculation !),(d) � 0 for fixed fIn,(e) � sh(x,e).

Let be defined as in (2.6b). The solution of the discrete potentialequation

htth =0 in flu, lLh = on Ph

is given by Uh := (if one continues the grid function, at first onlydefined on through on Ph; cf. Remark 4.2.5). The representation withthe aid of reads

uh(x)=h2 XEQh.

Since Sph(x) vanishes at all points x far from the boundary, it suffices to extendthe sum over the points near the boundary. If one sums over the neighbouringboundary points of

not acorner point or (1,1)}

instead of over the points near the boundary, then the definition of resultsin the representation

uh(x) = —h XE S?h, (4.4.8)

where 8; is the backward difference with respect to the direction of the normal

4.4 Properties of the Matrix Lh 57

= — hn,x)J

(note that = 0 for E C f's,). The variable — hn ranges over allpoints next to the boundary.

Remark 4.4.10. Equation (8) corresponds to the representation in Theorem3.2.10. The summation approximates the integral f'...

Finally we want to take a closer look at the estimate for the solutionthrough � � IIqhik0/8 (cf. (ic)). According to

(2.6b), fh + contains the right-hand side fh(x) = 1(x) of the discretePoisson equation and the boundary values x E 1h hidden in ThefoLlowing theorem gives a bound in which these components are separated.

Theorem 4.4.11. According to (2.6b), let = fh + be constructed fromf and The discrete solution = of the Poisson boundary valueproblem can be bounded by

< max + max <!max If(x)I + max8 XEOh xEf.,

(4.4.9)

PROOF. Set := and := The estimate for the firstsummand in Uh = U'h + results in the first surnmand in (9). To bounduse the inequality (4a) with uZ, ço1 = and =0, = 0.

The corresponding inequality

� + (4.4.10)

for the solution of the boundary value problem (2.2a,b) has not been men-tioned until now, but will be proved in a more general context in Section 5.1.3.The maximum norm fl. in (9) can be replaced by the Euclidean norms

Jh2 /h lco(x)P.

V x€QAV

Here, h2 and fe,, h and 5,, correspond to each other.

Theorem 4.4.12. Under the assumptions of Theorem 11 there holds

—' 1 1 1� tIL' + S +

(4.4.11)

PROOF. (1) It suffices to consider the case of the potential equation (i.e., f0). Let the restriction of on f'h result in the grid function


for x E Let the mapping = Ltcoh be given by the rectangularmatrix A: uh = Açoh. According to Equation (8) the entries of A read

= = — hn,x) = — hn), XE E fl.

Since A � 0 (cf. Remark 8), one obtains the row sum norm IIAIk0as for the choice of ph(x) = 1 in all x fl',. The

solution vh = then reads = 1 (why?), so it follows that

= IIAwhIko = 111U00 1.

(2) The column sums of A are s(e) := = — hn)

for The grid function Vh h2L'l at the points — hn near theboundary has the values

8(e) — hn), E n normal direction,

as is implied by Remark 7. Let e = E be a point of the left or rightboundary (i.e., = 0 or 1). As mentioned in the proof of (ic), Lhwh � 1

holds for wh(z,y) := x(l — z)/2. Since � 0, then so is wh � andhence vh < h2wh. In particular we have the following estimate

= — hn) — hn) = h2h(1 — h)/2 �

at the point near the boundary eor from the lower boundary one obtains the same estimate if

one uses wh(x, y) := y(l — y)/2. Since the column sums are the row sumsof AT, we have proved

= E FL) h—t/2.

(3) We have IIATAH2 = p(ATA) � � IIAT(IOOIIA,I < (cf.Exercise 4.3.20a, (3.lOd), (3.lOa)) so that the solution Uh = =satisfies the following estimate

= h2 u2(x) = h2

XEOk xEflA

= h2

<h2IIATAII2

¶' 2

Since the assertion follows. S

4.5 Convergence 59

4.5 Convergence

Let U,, be the vector space of the grid functions on The discrete solutionu C°(S1) cannot be compared directly

because of the different domains of definition. In difference methods it is cus-tomary to compare both functions on the grid Th,,. To this end one must mapthe solution u by means of a "restriction"

u U,, (u : continuous solution) (4.5.1)

to U,,. In the following we choose R,, as restriction on

(R,,u)(x) = u(x) for all x (4.5.2)

The limit h Ois made precise as follows. Let H C be a subset withaccumulation point zero: 0 11. For example, the step sizes considered so farform the set H {1/n:n For each h H let (4 be equipped with thenorm 11

DefinItion 4.5.1. (Convergence) The discrete solutionsE converge(with respect to the of norms ft. h E H) to u if

— R,,uII,, —' 0. (4.5.3a)

We have convergence of order k if

— R,,uI(i, = O(hk). (4.5.3b)

The proof of convergence is usually carried out with the aid of the con-cepts of "stability" and "consistency". The discretisation {Lh: h H} is saidlobe stable if

sup <00. (4.5.4)hEll

For the discretisatjon defined in Section 4.2, the stability has been proved in(4.lc) with respect to the row awn norm, and in (4.ld) with respect to thespectral norm.

The grid function in = is the restriction

fh = R,,f (4.5.5a)

of f, where in (2.6b) k,, was chosen as the restriction to 12k,:

(!4f)(x) = 1(x) for all x i?h. (4.5.5b)

The formulation of consistency becomes simpler if instead of the matrix L,,one considers the difference operator —La,, on which it is based (cf. Remark4.2.5). Let I),, be a general difference operator, and let —Dhu,, = .1,, discretise

= I. The discretisation described by the triple (1),,, R,,, !4,) is said to be

80 4 Difference Methods for the Poimon Equation

consistent of order k with respect to fi (consistent with the Laplaceoperato.), if -

— � K? II (4.5.6)

for all u E Here K is independent of h and u.

Remark 4.5.2. Let R,, and be given by (2) and (Sb). The five-pointformula isconBistent of order 2: estimate (6) holds with k =2 and K = 1/6.

PROOF. The expansion (1.4c) can be applied In the x and y directions andyields

= + + 14) with IR,I �(4.5.7)I

Theorem 4.5.3. Let the discretisation (Dh, be consistent of onier k.Let the matrix Lh associated to the difference opelntor D, (cf. Remark 4.2.5)be stable wsth respect to Jj Then the method is of onierk qU E

PROOF. wh := - satisfies the difference equations

—Dhwh = —Dhuh + DhRAU = fh ÷

iflah.Because of the boundary condition wh =0 on we have that

— Ra4*i) (cf. Remark 4.2.1). (4) and (6) imply (3b). IIf one substitutes the concrete values � 1/8 and K = 1/6 one

obtains

Corollary 4.5.4. Let the continuous solution of the boundary value problem(2.2a,b) belong to C4(Th) Let u4 be the discrete solution defined in (2.4a,b).Then the convergence of ub to u is of second order

— (4.5.8)

The u can be weakened to

Corollary 4.5.5. Under the condition u we also have

— � (4.5.8')

PROOF. The remamder term R4 in (1.5b) may also be written as

4.5 Convergence 61

= h' — ,i"(x)J(x ± h —

The Lip6chitz estimate — u"(x)I � — implies R4 �so that in (7), (6), and (8) the norm

by 1

If, however, one further weakens u E C3.1(Th) to u C'(Th), 2 < s <4,one obtains a weaker order of convergence.

Corollary 4.5.6. Under the condition u 2 < a 4, convergesof order a —2:

IIt4h — � (4.5.8")

where K, := 1/[2s(s — 1)1 for 2 a � 3 and K, := 1/(2s(s — 1)(a — 2)J for3<8<4.

The proof results from

Exercise 4.5.7. Show — � 8Kah'211ti11c.(m), 2< 5 � 4.

Even though the proofs of convergence are simple, the results remainunsatisfactory. As can be seen from Example 2.1.3, the continuous solutionof the boundary value problem (2.2a,b) generally does not even satisfy u EC2(Th), although one needs at least u E C'(i7) with a > 2 in Corollary 6 forconvergence. Stronger results can be obtained an analysis which will bediscussed in Section 9.2. That errors of the order of magnitude of 0(h2) occureven under weaker is shown in

Example 4.5.8. If one solves the difference equation (2.4a, b) for -4u = 1

the values ua(4, 4), which are shown in the first column of Table 1. The exactsolution u(4, 4) = 0.0736713... results from a representation that one can findin Example 8.1.11. The quotients of the errors = u(4, 4) — u,(4, 4)

C2(Th.has furthermore the asymptotic expansion ua(4,4) u(4,4)+

h2e(4,4) + 0(h4). The error term e(4, 4) independent of h, is eliminated byusing the Richardson extrapolation

fll\ if ull\ Ill:=

The extrapolated values are already very accurate for h = 1/16 (cf. the lastcolumn of Table 1).

62 4 Difference Methods for the Pouson Equation

Thble 4.5.1 Difference solutions for Example 8

h Quo-tient

Quo-tient

1/81/161/321/64

0.07278260.07344576

0.0736147373

0.0736571855

0.0736668

0.07367106

0.07367133

8.89x0.250

5.66x1050.249

4.5x10606

4.6 Diacretisations of Higher Order

The five-point formula (2.11) is of second order. Even if the solution u belongsto C(17) wIth 8>4, no better bound for - than 0(h2) wouldresult. An obvious method for constructing difference methods of higher orderis the following. As an anaatz for the discretisation of the one-dimensionalequation u" = / choose

(DhU,l)(x) = + vh).

The Thylor expansion provides

(DhRhu)(x) = Ea,.h"_2Q'(x) + ap =

The2kj-Iequataonsao=ai=as=a4=•=a2a=Oanda2=Iformalinear syateni for the 2k + 1 unknown coefficients c,. For k = 1 one obtains theusual difference formula for k =2 a difference of fourth order results:

=

If one applies this for u" to the x andy coordinates, one obtainsfor -4 the difference star

1

h2 —16

— 1 —16 60 —16 1 (4.6.1)12

—16

I

(cf. (2.12)). The difference scheme (1) is of fourth order, but presents dif-ficulties at points near the boundary. To set up the difference formula at

4.6 Discretisations of Higher Order 63

(h, h) E ap,, for example, one needs the values uh(—h, h) and uh(h, —h) out-

side (cf. Figure 1). One poesibility would be to use scheme (1) only at pointsfar from the boundary and to use the five-point formula (2.11) at points nearthe boundary.

The above complications do not occur if one limits oneself to compactnine-point formulae; by thin one means difference methods (2.12) which

are characterised by "cap 0 only for —1 � afi 1" (cf. Fig. 1). Anansatz with the rune free parameters —1 a, 1, leads, however, toa negative result: there is no compact nine-point formula with (Dhu)(x)

+ 0(h3). In this sense the five-point formula is indeed optimal.Neverthelesa. nine-point procedures of fourth order can be obtained if one

also selects the right side f,, of the system of equations (2.6a,b) in a suitablemanner.

N

Difference Scheme (1)•: Compact nine-point formula

* . N.,.N

FIgure 4.0.1. Scheme (1) and the compact nine-point formula

If one applies the compact nine-point scheme

h2 —1 —4 —1

:= —'--i-- —4 20 —4 (4.6.2)—1-4-1

to u E C8(Th, the Thylor expansion results in

—Dhu = (4.6.3)

For the special choice ci the restriction via

- 11/2

fh Rhf := 1/2 4 1/2 f, (4.6.4)1/2

i.e.,

fh(X,y) =(khf)(x,y)

[f(z — h, y) + f(x + h, y)+

f(z, y — h) + f(x, y + h) + 8f(x, (4.6.4')


one obtains the expansion

fb(X, y) = f(x, y) +

f = —4ti, agrees with (3) up to 0(h4).The matrix Lb of the system of equations which results after the elimi-

nation of the boundary values uh(x) = z E has the entries

(20/6

if x — = (Eh, 0) or (0, ±h),otherwise.

The right side of the system of equations Lau4 = Qh

'lb := fh := R,,f according to (4'), := — (4.6.7)

The dlscretisation (Dh, Rh, with D, from (2), Rh from (5.2), from(4') is called the mehrstellen method (cf. Collatz lii).

Exercise 4.6.1. Let D, and Lb be defined respectively by (2) and (6). Provethat(a) Lb Is an M-matrix;(b) 1/8 (stability), � 20h2/3;(c) � ifuE (consistency).

Theorem 4.8.2. Let Sib be defined as in Section 4.2. Let Ub be the solutionpivvided by the mehratelien method Lhuh = with Lb from (6), and from(7). Let the solution is of the boundary value problem (2.2a,b) belong to C6(fl)or C6"(37). Then the following error estimates hold in the respective

11h4 11h4huh — < — �

(4.6.8a)In the aise of the potential equation, i.e. / =0, we even have

Huh — � Kh611u11(,.(Th, huh — RhuhIoo � (4.6.8b)

s/u E reap. U

PROOF. As in Theorem 4.5.3, (8a) results from Exercise (Ib,c). In the casethatf=OtheO(h4)-termalsovanishesin(3). U

4.7 The of the Neumann l3otmdaiy Value Problem 65

4.7 The Discretisation of the NeumannBoundary Value Problem

The Dinchlet boundary conditkrn u(x) = p(x) was directly in the differ-ence method; a diacretisation was not necessary. A different situation arisesfor the Neumann boundary value problem

in s'l=(O,l)x(O,l), Ou/ôn=ço on 1. (4.7.1)

The normal derivative, which reads explicitly

ths for x=(x,O)€f, forx==(x,1)Ef',

t9ufor x=(0,y)Ef, forxr(1,V)Ef',

Like the Laplace operator, must be replaced by a difference. We will investigatethree different discretisations.

4.7.1 One-sided Difference for au/en

The Poisson equation leads to the (n — 1)2 = (1/h — 1)2 equations

= 1(x) for all x E Os, (4.7.3)

which require the values of for all x E where

:= U := I'h\{(O,O), (0,1), (1,0), (1, 1)}.

To obtain a further 4(n—1) equations for {uh(x):XE we replace, at allx the normal derivative eu/en = by the backward difference

Ouh(X) := — tLh(X — lin)J = for x E (4.7.4)

L. I • Discretisation of the Neumann condition

. Discretisation of the Poseson equation

FIgure 4.7.1.

If one inserts the corresponding normal directions n for the four sides ofthe square one obtains

66 4 DIfference Methods for the Pofeson Equation

h hJ —

—u,(h,y)I = 1 for — h 2h 1 — hJ

(4.7.4')Equations (3) and (4) yield (n + 1)2 equations for as many unknowns.

Exercise 4.7.1. After a rescaling of equations (4) to h18;uh(x) =x these equations, together with (3), form a system Lhuh Showthat Lh is symmetric and satisfies (3.la).

As in the Dirichiet problem the variables uh(X), x E can be eliminatedwith the aid of (4) in (3). At the point (h, y) near the boundary, for example,Equation (3) becomes

h2f3tLh(h, v) — tih(h, v — h) — tih(h, y + h) — uh(2h, i')J = f(h, v) -t-h v).

—1

The star h2 —1 4 —1 thus becomes

h2 3 h2 —1 3 —1

—1 0

near, respectively, the left, right, upper and lowe boundaries. At the cornerpoints one even has to replace two boundary values, so that, for example, at

—1

(h, h) Q,, the star reads W2 0 2 —1 . Except for the special case0

h = 1/2, one obtains for the (n — 1)2 values u,(x), x the system ofequations

= (4.7.5a)(4 lfxES?hisfarfromtheboundary,

with = h2 2 if x = (h, h), (h, I — h), (1 — h, h), (1 — h, 1 —

3 otherwise(4.7.5b)

= f —h2 If Thh are neighbours,

and = fh + 4°h, fh(x) = (f4,f)(x) := 1(x), (4.7.5c)(Ph(X) =

4.7 The D tisatlon of the Neumann Boundary Value Problem 67

Remark 4.7.2. (a) L,. is symmetric and satisfies the sign condition (3.Ia).(b) With lexicographical arrangement of the gridpoints of L,, has theform T-I 1

-I T-ILh=h2 ..

—I T -I-I T—I

3 —1

—1 4 —1

T=—1 4 —1

—1 3

The matrix Lh is singular because the system L1ti, = like the contin-uous boundary value problem (1), is, in general, not solvable. The analogueof Theorem 3.4.1 reads as follows.

Theorem 4.7.3. The system of equations (5a) is aauable if and only if

—h2 f(x) = h (4.7.6)XEQh

Any two solutions of(5a) con only differ by a constant:tjL—4 = ci, CE

PROOF. Evidently, L,1 = 0 holds, i.e., 1 E kernel(Lh). Furthermore, Theo-rem 4 will then imply dim(kernel (Lh)) =1. This proves

kernel(Lk) = {cl: C E It) (4.7.7)

and thus the second part of the aseertiori. (ãa) is solvable if and only if thescaler product vanishes for all v E kernel (LI) = range Becauseof LI = Lh and (7)

(1, = 0 or qh(x) =0 (4.7.6')XEOk

is sufficient and According to Definition (5c), (6) and (6') agree. •

Let conditIon (6) be satisfied. System (5a) can be solved as follows. Selectan arbitrary xo E and normalise the solution Uk (determined except forone constant) by

(4.7.8)

Let ilk be the vector without the component Let L,, be the sub-matrix of L,, in which the row and column with index x0 have been left out.Let be constructed likewise. Then

68 4 Methods for the Poisson Equation

(4.7.9)

is a system with (n — 1)2 — 1 equations and unknowns.

Theorem 4.7.4. The system of equations (9) is solvable; in particular, LAis a symmetric M-matrix. Under condition (6), L;14a, supplemented by(8), yields the solution of system (5a).

PROOF. (a.) As a principal submatrix of is symmetric. In flh\{XO}any two grid points can be connected by a disin of neighbouring points sothat LA is irreducible. For all x E üh\{XO} there holds (3.4b); at neighbouringpoints of x0 we even have (3.4a), so that LA is irreducibly diagonally dominant.According to Criterion 4.3.10, LA is an M-matrix, thus nonsingular.(b) If is the solution of (5a), one can assume (8) without loss of generalityso that restricted to also solves Equation (9) and has to agreewith the unique solution

As a corollary of Theorem 4 one obtains that rank(LA) � =(n — j)2

— 1, i.e., dim(kernel (LA)) = 1 and henos (7) holds.Another possibility for the solution of Equation (5a) is to pass to an

extended system of equations

= (4.7.lOa)

11 — {tihl — fQhlwith LII,= j, (4.7.1Db)

where a can be prescribed arbitrarily.

Theorem 4.7.5. Equation (iDa) is always solvable. If for the last corn..ponent of the solution we have A = 0, then condition (6) is satis-fied and uA represents the solution of system (5a) which is normahsed by

= Uh(X) a. However, sf A 0 hoLd, one am interpret Uh as

solution of L,uh = stI&ere = — Al belongs to 1(x) =1(x)— A andfand satisfy condition (6).

PROOF. I is linearly independent of the columns of LA so that rank ILA, 11 =rank (LA) ÷ 1 = (n — 1)2. Likewise, (11,0) is linearly independent of the rowsof ILA, 1) so that rank (Li) = (n — + 1, i.e., is nonsingular. The otherstatements can be read from (lob).

Attention. One should either use Equation (lOa,b) or Equation (9), afterfirst replacing by — (ITqh/1Tl)I. As a justification of this recom-mendation note that the condition for solvability of the continuous problem is

ity condition (6). For einooth functions I and Equation (6) can be shownto hold up to a remainder of order 0(h). Thus it is generally unavoidable to

4.7 The Discretisation of the Neumann Bound&y Value Problem 69

replace f,, and qh by fh —Al and q,, = In the case of Equa-tion (lOa,b) this correction is carried out implicitly. If, however, (9) is usedwithout any correction, the resulting solution can be interpreted as a solutionof Lhuh = with = qh(x) for x Xo and qh(xo) — qh(x).That is, here too, an implicit correction of is carried out, with the differ-ence that the correction is not distributed over all components as before, butis concentrated on If Equation (6) is satisfied up to order 0(h), thenqh(xo) and qh(Xo) differ by 0(h1). Therefore the solution of Equation (9)contains a singularity at the point Xo.

Theorem 4.7.6. (Convergence) Let u E be the solution of the

Neumann problem (1). Let=

be the solution of Equation (lOa).

Then there exists a c E IR and cvnstants C, C' independent of u and h suchthat

S huh — — � Cfhf +(4.7.11)

PROOF. (a) We have

A = + — 1)_2.

Because fa I dx + f1-. di = 0, the first factor consists only of the quadratureerror

(b) According to Theorem 5, is the solution of Lhuh = — Al. Thiscorresponds to the difference equations

lflI?h, 8ntIh=W

The difference wh := uh — satisfies

—LlhWh = + LI hR = LIhRhU + fh — Al- (4.7.12a)

Ifl(ih,

= on (4.7.12b)

The errors of consistency are

Ilchfloo � (of. Remark 4.5.2),

� + (cf. Lemma 4.1.1).

Since the solution exists, condition (6) is satisfied with ch and


a,.

Then wh Wh ci with c = lTwh/1TI is the solution of (12s,b) normalisedby = 0. The application of the following Theorem 4.7.7 to Equation(12a,b) provides the inequality (11). U

Theorem 4.7.7. (Stability) Let condition (6) be soLüfie4. Let the solutionof(3), (4) be norinatised by iTuh = 0. Then there exist constants C1,C2

independent of u and 4 such that

Dt&hlloo C1 max If(x)I + C2 max (4.7.13)xEOj, xE!,,

The proof of this theorem, which corresponds to Theorem 4.4.11, will besupplied in Sect. 4.7.4.

47.2 Symmetric Difference for 8u/On

As can be seen from Theorem 6, the one-sided difference 8; causes the errorterm 0(4). It seema obvious to replace 8; by a symmetric difference. Th thisend the five-point diacretisation is set up at afi points x 47h U Ph (cf.(2.lc)):

= 1k Rhf in Thh, (4.7.14a)

where Rh is the restriction to Thh. The symmetric difference is defined by

+ hn) — uh(x — lrn)J = in x E 1). (4.7.14b)

Here, we assign two normal directions to the corner points, so that for each ofthe corner points two equations of the form (14b) can be set up. In the cornerx = (0,0) one has, for example, the normals

(o\ (—1

The corresponding equations (14b) also contain different (!) values 0) =and ço(0,0+) = For x E F4 the difference for-

mula (14a) needs the values in points x + hii outside of These can beeliminated with the aid of (14b) so that a system of equations Lhuh =remains for the (n + 1)2 components uh(X), x E Thh.

ExercIse 4.7.8. (a) If the grid points of i74 are arranged in lexicographicalorder, L,, has the form

T-21 4-2—I T —I —1 4 —1

L4=h2 .-• .. , T= ..—I T —I —1 4 —1

-21T -24

4.7 The Discretl8ation of the Neumann Boundary Value Problem 71

(b) Lh is not symmetric, but DhLh with Dh = diag {d(x)d(y)}, d(O) d(1) =1/2, d(.) 1 otherwise, is symmetric.

The analogue of Theorem 3 reads:

Theorem 4.7.9. Equation (14a,b) is solvable sf and only if = 0for D,, in Exercise 8b. Any two solutions may differ by only a constant. Theformulation of = 0 with the aid off and reads:

—h2 > d(x)d(y)f(x, y) = 2h >2 d(z)d(y)p(z, y), (4.7.15)(x,v)Erh

where the sumrnand for the corner points occurs twice in the second sum, andboth the different bmits for are taken into account.

Remark 4.7.10. The sums in (15) represent summed trapezoidal formulae.Thus, from / 4od.l' 0 follows Equation (15), except for a remainder

Theorems 4 and 5 can be transferred without difficulty. The convergencetheorem, Theorem 8, becomes

Theorem 4.7.11. LetuE be a solution of(1). Let li,, be

the solution of =With

= }.We have conver-

gence of second

)A) � huh — — (4.7.16)

The proof is essentially the same as for Theorem 6. An additional techni-cal difficulty is the fact that the consetency error must also beCletermined in x E although uls only defined in of treating thedifference equation and the boundary diacretisation separately as in (12a,b)one should directly analyse the equations Lhuh = q,., from which the valuesuh(x + hit) outside of fl have already been eliminated.

4.7.3 SymmetrIc Difference for ôu/On on en Offset Grid

If we offset the above grid by h/2 in the x and v directions, we obtain the grid

in Figure 2. The points near the boundary of are at a distance h/2 fromV. We set

72 4 Difference Methods for the Poleson Equation

1,:= €Zor

(ci. Figure 2). To each point near the boundary x—hn/2 (x "h) correspondsan outlying neighbour x + hn/2.

0000r * — S— —0. —.

• S0: Points inOS S S

°!' ' .:PointsmFh

Figure 4.7.2. Offaet grid

The discretisation of the Poleson equation (1) is

= 1(x) in x (lh, (4.7.17a)

h1[tzh(x + hn/2) — uh(x — hn/2)j = in x E "h• (4.7.1Th)

The difference (iTh) is symmetric with respect to the boundary point x andneverthekes agrees with the backward difference 8; at the grid point x-i-hn/2.

Remark 4.7.12. After eliminating the values u4(x + hn/2), x E Ph, weobtain a system of equations L,uh = where Remark 2 is also valid for thismatrix L,,. In contrast to Section 4.7.1, Lh is of size n2 x n2.

The Theorems 3,4,5, and 7 hold analogously. Theorem 6 hoLds with theinequality (16) instead of (11).

4.7.4 Proof of the Stability Theorem 7

In the case of Dirieblet boundary values the stability statement of Theorem4.4.11 foLlows immediately from the maximum principle and the bound for

The corresponding statement of Theorem 7 for Neumann boundary val-nsa, however, cannot be proved that easily. In the literature one can only findweaker estimates which on the right-hand side of Equation (13) contain anadditional factor I However, this factor is not to be avoided if one usesEquation (9), = without condition (6) being satisfied.

Let the discrete Green function gh(x,e), XE QhUPh, E bedefined by

= (4.7.18a)

forx=el fh'/(4_4h) ifxisneartheboundazy1

%.O otherwlsei t.O if x is far from the boundaryj

4.7 The of the Neumann Boundary Value Problem 73

for XE 1) (4.7.18b)

where and 8; act on x. exists EXEOk = 0 proves thesolvability condition (6').

Lemmi 4.7.13. For arbitrary qp, with = 0 (cf. (6')),

*Lh(X) h2 (4.7.19)

represents a solution of Lhuh =

PROOF. At points far from the boundary x E S?h, (Lhua)(x) = =qh(x). At points near the boundary x E ilh, from 8uh = 0 and (6'), one hasthe Identity

= =— 4—4h

= qh(x). U

Equation (18a,b) determine 9h up to a constant.

Theorem 4.7.14. The Green function om be so selected that

C(i + lIog(Ix -t- for E (4.7.20)

This inequality corresponds to the bound (4.7) in the Dirichiet case. Be-foreTheoreml4isproved,wewanttoshowthatTheorem7followsfromit.

logously K1 and � K2we obtain

h2xEaa

From (19), (20), and = fh + (cL (5c)) thus follows the estimate

� + � Ki max $f(x)( + 2K2 maxXEflA

If Üh := — is the solution of = normalised by=0, one can see that

� 2inf{Ijua — E Jt}

so that the inequality (11) and Theorem 7 are proved.

It remains to prove Theorem 14.

74 4 DIfference Methods for the Equ&tIon

LemmA 4.7.15. Let e be one oJ the unit vectors (1,0) or (0,1). Let 11h

be such that + he E also. For all these a and let there exist a function

Gh(X) = with the properties

(1—1 (.4.7.21a)

otherwise )8;C,1=0 (4.7.21b)

< hC'/(Ix — + h), (4.7.21c)

where C' does not depend oneand Then Theorem 14 holds, andthus alsoTheorem 7.

PROOF. Fbr E let be defined as a solution of

( 1 'I

on f's, —1

L 0 otherwise)

For = E+he, agrees with Gh m Lemma 15. For one

or (0,1). Srnce = (21c) implies the estimate

�If one considers first the case x1 = = x2 � and one applies

1/k � coost (1 + log(k2h) — log(k1h)), one obtains

� C"[1 + — EJ + h)) + I — + h)fl.

In the general case, this bound also comes out, but one must choose theconnection such that Ix — min{lx — x — Since we know that

for SllXEflb,

4 —4h

satisfies the inequality (20). Because 1= (4-4h)/h the equations (18a,b)

are also satisfied, i.e., is the Green function.

We shall construct the function needed In Lemma 15 explicitly. Thebasic building block is the discrete singularity function on thegrid in 1R2

{(x, v) E x/h, y/h 7}.

LAmmi 4.7.16. The singularity function defined by sh(x,e) := Oh(x -and

4.7 The Diecretisatlon of the Neumann Boundary Value Problem 75

1 — 1ah(x) := m;:i I I +

dr1i

— 1 (' (W + x2,12)/(2h))d d—

+

for all E Qh has the property = h-2 and =0 otherwise.

PROOF. Let := exp(i(xith + Note that

— 1) = 4h2e(x, ,,)tsm2øii/2) +

andfT fT I4w2 forx=O

0

For multi-indices a = (a1,a2) E Z2 with � 0, 02 � 0 one defines thepartial difference operators of order Ial = aj +02 by

= (4.7.22)

Starting with the representation in Lemma 16, Thomée (11 proves

Lemma 4.7.17. I(ô°ah)(x)I � + for all x E Qh.

For the construction of the function Gh In Lemma 15 we select, withoutlcesofgenerality, e=(1,0)andkeepeEQh withe+heEQh.The function

:= hOuh(x — = — — 0h(X — — he) (4.7.23)

satisfies (21a) but not the boundary condition (21b). Let the symmetrisationoperators and 8, be defined by

y) := y)+Uh(h—x, S,uh(z,y) := h—y)j

for (x,y) E Qh. The function

cill......AC'

Is symmetric with respect to the axes y = h/2 and z = h/2. Thus we have(21b) on the left and lower boundary:

= = 0 (4.7.25)

Furthermore, satisfies conditIon (2 La) just sa does. For each 13we define the operator by

76 4 DIfference Methods for the Equation

(Ppuh)(x,y) =

g32L)J,

where L = 2— 2h, and set

— (4.7.26)

Lemma 4.7.18. For /3= (/3i,132) wsth � 2 there holds

for all 2, x E flh. (4.7.27)

Here, Küind ndeni of the choice of the heETh,.

PROOF. According to the definition we have

0) = 0. (4.7.28a)

The operator preserves the symmetry, i.e., Uh = implies

Thus we have that = = and consequently

= = 0. (4.7.28b)

Is the linear combination of for different withx E According to Lemma 17,

hCK'91313 for alt = 2,x E (4.7.28c)

so that (27) follows for 2. Let Ial= 1. DaGh(x) can be written in theform

+ —

with x0 = x, = x, flxk - = h. Each summand has the form

� 2hCK"I/313 for = 1, x E (4.7.28d)

Likewise one infere from (28a) and (28d) the inequality (27) for Iciij =0, whichproves Lemma 18. U

Since E,E12\{o}

Gh(X) (4.7.29)pEZ'

exists. Since = 0 in for all (0,0), Gh aho satisfies Equation(21a). As already mentioned in the proof of (28b), = = S,G5

4.7 The DiscreUsatlon of the Neumann Boundary Value Problem 77

holds so that G,, also satisfies Equation (25): 8Gh =0 at the left and lowerboundary. The proof of 8;Gh =0 at the lower boundaries requires

Lemma 4.7.19. Gh is L-periodic: Ch(x,y) = Gh(x+ L,y) L).

PROOF. Let be abbreviated toDefinition (29) says that

= lim (4.7.30a)k-.oo

Now can be written as a sum over the differences (x-f-131L,

from ph = 0 to ph = x, and over + vh,/32L) from wh = 0

to vh = y. For � 2 the distance between the argumentshe is always � - 1)L. Each summand is thus, by Lemma 17, boundedby O(h2/((piI — 1)2) = O(h2/13?). As a sum of such terms then can beestimated by � for � 2. (4.7.30b)

We want now to show that the following equation (30c) also holds:

= Inn E = urn (4.7.30c)1—k<p1<1+k

The sums (30a) and (30c) differ by Dh := Thevaluesvalue of the sum Dh is then bounded, from (30b), by(2k + 1)O(h/k2) = O(h/k), so that the limits of the doubLe sums in (30a) and(30c) are the same. Changing the variable th to th -1 then transforms (30c)Into

-k�$i(4.7.30d)

The first part of the identity

132L) — GZ(/31L + L,

— +

= ÷ kL, thL) — (Yh'(kL + L, (4.7.30e)

is elementary; the second results from the symmetry = As above,the summands of the last sum can be bounded by O(h/k2) so that (30e)vanishes for k oo. Together with (30d) one obtains

78 4 DIfference Methods for the Poleson Equation

Gb(X, y) = lim + ,91L + L,y + —

= lim 'yØ(x+L,v)=Gk(x+L,y).k.-.oo

The proof of Gh(X, v) = Gh(x, y + L) is analogous. SSince L =2- 2h, the symmetry = S,Gh and the periodicity yield

i.e., 8,1 G,(1,y) = 0 on the upper boundary (l,y) E Likewise we showthat 1) =0 on the right boundary. Thus (21b) is proved.It remains to show (21c): S +h) in The for

llflhIcio <1 arelinearcombinationsofh8;orh(x—E) forE 0h,whith are generated by S,, and P,,. Fbr all E there holds � k-El1so that from Lemma 17 foLlows

� hC'/(Ix — + h) for UJ3IIOO � 1, XE Qh.

On the other hand, Lemma 18 shows that

Ih K/l$13 = hK' — + h).

IIPlk.�2

The assumptions of Lemma 15 are hence satisfied: with (21a-c) exists.Thus theorems 14 and 7 are proved. U

4.8 Discretisation in an Arbitrary Domain

4.8.1 Shortley-Weller Approximation

Let the boundary value problem

—4u=f in 47, cuP (4.8.1)

be given for an arbitrary domain 47. If one places a square grid of step size hover 47 one obtains

:= {(x,y) E 47: x/hE Z and y/h E Z} (4.8.2)

as the set of grid points. By contrast, the set of boundary points must bedefined differently from the case of the square. The left neighbour point of(x, y) 47h reads (x — h, y). If the connecting segment {(x — y): t9 (0, l]}does not lie completely in 47, there exists a boundary point

(x — P with (x,y) E E (0, 1J,(x —t9h,y) €47 for all t9 E [0,s),(4.8.3a)

4.8 Dlseretisation in an Arbitrary DomaIn 79

which now, instead of (x — h, y), is called the left neighbour point of (x, y)(cf. Figure 1). Likewise the right, and upper neighbour points can beboundary points of the following form:

(x+8h,y)E I'(4.8.3b)

(z,y—sh)Ef' with(x,y)Eflh,sE(O,1),(x,y—6h)EQforall

E for all '9€ jO,s).(4.8.3d)

We set

:= {boundary points which satisfy (3a), (3b), (3c), or (3d)}. (4.8.4)

A grid point (x,y) poseeseing a neighbour from 1'h, is said to benear the boundary. All other points of are said to be far from theboundary. As can be seen in Figure 1, a point may be near the boundaryalthough (x ± h, y) and (z, y ± Ii) belong to Oh (namely if not all connectingsegments lie completely in .0).

. E near the boundary• U •EQhfarfromboundary

ciE boundary

U. •a

FIgure 4.8.1. auid "h

ExercIse 4.8.1. For a convex domain .0 show that

Fh={(x,y)EF: x/heZory/hEZ}.

If one wishes to approximate the second derivative u"(x) with the aid ofthe values of u at x' <x <x", one can use Newton's divided differences:

I I IUu (x) = 2[

— —, JJ(x" — x') + Rem. (4.8.5)

ExercIse 4.8.2. Show that (a) the Taylor expansion

IRemI�1 (x" — x)2 + (x —

� max{z" — z,x—

3

(4.8.6)

br the remainder in Equation (5)11 u C3((z',x"J).

80 4 Difference Methods for the Poleson Equation

(b) In Equation (6) one can replace the norm of C3((x', x"I) by that ofC'2"Ux',x"D.(c)Ifx" = x-t-h andx' = x—h thedifferencein(5) agrees wlththe u8ualsecond difference 884n(x).

To set up the difference equation for —4u = / at (z,y) Qh we use thefour neighbouring points

(x — (x + Srh,V), (x,y — (x,y + s0h) U 1',, 0 < � 1,

as defined above. The subscripts to the a are suggest respectively left, right,under and over. Fbr points far from the boundary = 1 holds; for points nearthe boundary (x,y) at least one neighbour lies on 1h and the correspondingdistance a.hmaybesmalierthanh.Equation(&)withx'=x-a,handx"=x + a,.h provides an approximation for Analogously one can replaceby a divided difference. One obtains the difference scheme of Shortleyand Weller (lj:

—+— i4z,y)— ,

2 2— ,

+ Sj) 301.50 +

)*4(x)Y —

(4.8.7)

Remark 4.8.3.

The discrete boundary value problem aesumes the form

= := on (4.8.8a)

with (14f)(x) = 1(x) for x ah,tih=fP OflFh. (4.8.8b)

Theflvecoelflcientson thirightaideofEquation(7)deflnethematrixele-ments for = x and for the four neighbours of x. Otherwise we set

=0. The right side of the system of equations

= qh, (4.8.9a)

= fli +401,, (ph(X) :=— E (4.8.9b)

Again 40h(X) =0 holds for points x Qa far froin the boundary.

4.8 D etisation in an Arbltr&y Domsan 81

Theorem 4.8.4. Let 17 be bounded and contained ma strip (xo,xo+d)xlR or

lRx (im, yo + d) of width d. For the matrix belonging to the Shortley.. Weller

discretisation the following holds:(a) is generally not symmetric;(b) is M-matviz with

� d2/8. (4.8.10)

PROOF. (a) Let x = (x, y) E be near the boundary, but its neigh-bour x' = (x + h,y) be far from the boundary. Then it holds that

= —2h—2/(1 + a,) —h2 if < 1. Other than in Exercise4.7.1, in general no scaling can be found so that DhLh diagonal) becomessymmetric.(b) need not neceesarily be irreducible and hence irreducIbly diagonallydominant. But the weaker condition in Exercise 4.3.8 is satisfied and proventhe M-matrix property.(c) For the proof of (10) we use Theorem 4.3.16. If the domain 17 lies in thestrip (x0, zo + d) x R we select w := (x — z0)(x0 + d —x)/2.The reinwnder in Equation (5) contains only third derivatives that vanish forw. Thus Dhwh agrees with = -1:

The corresponding system of equations reads = := ía -I- wa withIa = I and qj � 0. Thus we have � 1 and Theorem 4.3.16 proves

� flw,i$co � (d/2)2/2 = d2/8.

Exercise 4.8.5. Prove the analogue of the derivative (4.4.9):

HuaNco � � If(x)I

With (10) stabIlity haS been proved. The order of comistency however isonly 1, for at points near the boundary

:= DhRhu —

is of the order of magnitude 0(h'). Here, R4 is the restriction to Uhas been defined in (8a). We want to show that nevertheless there is

convergence of order 2. The difference := - between the discretesolution uh = and the solution ii of (1) satisfies

=— in 17k, (4.8.lIa)

= 0 on (4.8.llb)


is the error of of the x difference (rasp. y difference), in turn,is split into

:= '},Analogously one defines

c'1 + := + :=

The errors are described by (5.7):

� < (4.8.12a)

With K := define

v,=K1 vh=O OflPh, ë,,:=Lhvh.

For XE far from the boundary èh(x) = 0; for XE near the boundary;however, we have

(x E near boundary).

for example, the case x = (z, y) 11h near the boundary with= (x - sjh,y) Ph. According to Exercise the x difference has the

error

� =

The analogous estimate for gives IcL(x)I � Because 4(x) ==0 at the x E far from the boundary one has � Since

� 0 it follows that —vh wI i.e.,

� K = (4.8.12b)

Equations (12a,b) together with + = = - R,,u prove

Theorem 4.8.6. Let (1 the asswnptwn of Theorem 4. Thefor the Shortkij- Weller method is second onier

huh — � +

1(4.8.13)� +

if ti E

4.8 Dlseretlsatlon In an Arbitrary Domaãn 83

Exercise 4.8.7'. Show that if one uses the Shortley-Weller discretisation forall grid points near the boundary, but the mehrstellen method from Section4.6 for all points far from the boundary, we obtain a method with convergenceof third order: — 0(h3).

4.8.2 InterpolatIon at Points near the Boundary

Instead of discretising the Poiseon equation at grid points x E near theboundary, one could also try to determine uh(x) by interpolation from theneighbouring points. If, for example, x = (x, y) E is near the boundary,(z — s1h,y) fh and (x + flh U I),, then linear interpolation yields

tLh(X,y) = Iszuh(X + 8rh,y) + srUh(X — + Si).

Thus, at the point x we set up the equation

+ Sr)Uh(X,V) — siuh(x + srh,y) — SrUh(X — s1h,y) = 0. (4.8.14a)

Since = (x — sgh, y) should be a boundary point, one can replace byIf, however, (x,y + s0h) or (z,y — is a boundary point, we choose

interpolation in the y direction:

÷ So)Ua(X, y) — y + s0h) — Souh(z, y — =0. (4.8. 14b)

At any x 0h far from the boundary the five-point formula (2.3) is used:

= fa(x) = 1(x) (x far from boundary). (4.8.14c)

Theorem 4.8.8. Let (2 satisfy the assumptions of Theorem 4. Let the di.-cretisatlon be given by (14a-c) with the choies between (14a) and (14b) beingmade in such a way that always (at least) one boundary point is used for theinteipolation. Let the system of expiation. resulting after the elimination of theboundary values = E f's, be Lhu,, = Lh is an (in general un-syinmefric) M-matrix tuhsch satisfies the estimate (10). The discrete solutionstih with the order 2, if u E

flua — � + (4.8.15)

PROOF. The M-matrix property and (10) are proved as in Theorem 4.2. Letx E be a point near the boundary at which (ida) is used. The error ofinterpolation is

y) := (sz + y) — 8lRhU(x + y) — — sth, y),

1 2Ich(x,V)I j5r31(8i + Sr)h


With this cL and K := h2flu11C1,I(Th), one can essentially just repeat the proof

of Theorem 6. U

By rescaling and adding the equations (14a,b), one obtains

h2 ((ii + 8,÷

So +Uh(x,V) — !Uh(Z — sjh,y)

SgSr / Ii1 1

— —Uh(X + s,.h, y) — —uh(x, y + s0h)8,. So

— = 0. (4.8.16)

Using this device one can obtain a symmetric matrix Lh, even at arbitrary .12.

ExercIse 4.8.9. Show that the diacretisatiori (14c), (16) leads to asymmetricM-matrix. The estimate of convergence reads

— Rg,t&floo +

In (14a,b) linear interpolation was chonen because the values at the neigh-bouring points were sufficient for it. Constant interpolation by

u(x,y) = u(z — a1h,y) = a1h,y), if(x,y) Ila,(x — s1h,y) E f's,

is less desirable since it only provides first-order convergence: =0(h). By contrast, interpolation of higher order is very applicable indeed (cf.Pereyra-Proekurowakl-Widlund (1)). However, it is desaibed by an equationwblch also contains points at an distance of 2h. Higher boundary appraiu-mations are in particular then necessary If one wants to apply extrapolationmethods (cf. Marthuk-Shaidurov [1, p. 162 ff.J).

5 General Boundary Value Problems

5.1 Dirichlet Boundary Value Problems for LinearDifferential Equations

5.1.1 Posing the Problem

In Section 1.2 we have already formulated the general linear differential equa-tion of second order:

Lu=f in I? (5.1.la)

with

L = + + a(x). (5.1.lb)

We mentioned that, without lose of generality, one can assume

o,,(x) (5.l.lc)

so that the matrixA(x) := •,, (5.1.ld)

is symmetric. A differential operator apparently more general than (ib) is

I , 82 8 82 liii" I 8 8

L =>2 N 8x0x + i— +j+>2 +

j+0.

(5.1.2)But since, for example, = + the operator(2) can be described in the form (ib), provided the coefficients are sufficientlyoften differentiable. According to Definition 1.2.3, Equation (1) is elliptic inS? if all elgenvalues of A(x) have the same sign. One can assume without loseof generality that all elgenvalues are positive so that A(x) is positive definite(cf. Exercise 4.3.22a).

is elliptic m (1 if

forallxEO, (5.1.3a)$,3=1

86 5 Ceneral Boundary Value

For any x E 1? there exists c(x) := : = 1) and it must bepositive (c(x) is the smallest eigenvalue of A(x)!). Hence one can also write

in the form (3a'):

E � c(x) > 0 for all x E 1?, E (5.1.3a')1

Definition 5.1.1. The equation (is), or the operator L, is defined to beuniformly elliptic in 1? if

Inf{c(x): x Q} > 0 (c(x) from (3s')). (5.i.3b)

On P =81? we impose the following Dirichiet boundary value condition:

onl'. (5.1.4)

5.1.2 Maxhnum Principle

In general, the maximum principle does not hold for the equation Lu =1, noris the solution of the boundary value problem (is), (4) uniquely determined.

Example 5.1.2. Let 1? = (O,w) x (O,w), = 0, f = 0, Lu = + 2u.Then both u = 0 and = sin(x)sin(y) are solutions of the boundaryvalue problem. The second solution assumes its maximum at the interior point(ir/2,ir/2) EQ.

In the above example the coefficient 0(x) = 2 (ci. (ib)) has the wrongsign. As soon asa�O, we have

Theorem 5.1.3. (Maximum-minimum principle) Let a(x) 0 in thedomain 1?. Assume the coefficients of the elliptic opemtor (ib) are continuousmO. LetuEC2(Q) .atisfyLu=fandbenonconstant. Then we have:(a) if f in 4?, no negative minimum of u exists in 1?;(b) if f � 0 in 1?, no positive maximum exists in 0.

PROOF. The proof is based on Hopf's lemma, which can be studied, forexample, in Heliwig 11,111-1.11. Here we give a shorter proof, whichhowever, the stronger condition 0(X) <0 instead of a(x) 0. We assumethat u has a minimum at x' E 1? although f � 0. Consequently we have

= 0, i. e.,

f(x) = (Lu)(x) = (5.1.5a)

"4

5.1 Dirichlet Boundary Value Problems for Linear Differential Equations

and the Hesse matrix B := must be positive semidefinite(here, a matrix B is 8aad to be positive semide finite if it is synunetricand satisfies � 0 for all E From Exercise 4 follows

tr(A(x)B) = � 0. (5.1.5b)ij= 1

If the minimum u(x') were negative, one would obtain from (5a,b) the resultf(x') � a(x)ti(x') > 0 in contradiction to the assumption I � 0. Part (b) isproved analogously.

ExercIse 5.1.4. The trace is defined by tr(A) := Prove that:(a) � 0 and tr(A) � 0 if A is positive semideflnite;(b) tr(AB) = tr(BA) =(c) tr(A.B) � 0, if A and B are positive semidefinite.Hint for (c): B112AB112 is positive semidefinite; Exercise 4.3.22d.

As in the case of the potential equation, from Theorem 3 follows the

Corollary 5.1.5. Let I? be bounded (not necessarily a domain); furthennore,as* that the conditions of Theorem3hold Jff<Oin (1 [f>Oin (JJ andsf there exists a negative minimum [positive maxzmurnj of u in it must lieon the boundary 8fl.

Remark 5.1.6. The continuity of the coefficients and a of L inTheorem 3 and COrOllary 5 can be replaced by the assumption a <0 in 1'? orby the stipulation: In every compact set K C (1 let aand be uniformly elliptic.

5.1.3 UnIqueness of the Solution and Continuous Dependence

Lemma 5.1.7. Let I? be bounded, let the coefficient, of L be continuous andlet(3a) be valid anda<Oin 11. Let ul,u2EC2(fl)fl 0(Th) be solutions ofthe boundary value problems

ma, oni' (i=1,2). (5.1.6)

IIft�f2rnaandwl_<4O2onr,thenalaoui<u2inr1.

PROOF. v := — satisfies the equations Lv = 12—11 0 in a and�Oon I'. FromCorollary5and the fact thatv>Oonfone

infers that no negative minimum of v exists. Thus v 0, i. e., � u1. U

88 5 General Boundary Value Problems

Theorem 5.1.8. (Uniqueness) Under the conditions of Lemma 7 the so-lution of the boundary value problem Lu =f infi andu—ço on!' is uniquelydeterminetL

PROOF. Let u1, u2 be two solutions. Lemma 7 with fi 12=1, = =Thus, Ul=u2. U

The next theorem states that the solution depends

L be unsformly elliptic in Si. Under the conditions ofLemma 7 the foil owing holds:

IIui — � — 'P2IIoo + Mflf1 — f21 100 (5.1.7)

for andu2 of(S). In this inequality the vnimberM depends only onK := sup{Iaij(x)I, a(x)I :x E I1}, on the dhpticity constant definedby m := inf{c(x):x E Si) >0 (cf. (3b)) and on the diameter of the domain0.

PROOF.aelect a � 0 so that

ma2—K(a-f-1)� 1

and we define

w(x) := — 1100 + — jf' — /2(100.

on F. From the thoice of KR(z) we have

w(x)�v(x) forxEf'.

Furthermore, the selection made of a ensures:

(Lw)(x) =

—

<0ai(x) a+ — f2H00

�1—{ma2—K(a+I)}((fi

� —1111 — 121100 � fi(x) — /2(x) = (Lv)(x).

Ifoneuses =v, =w, theresultisv�win 0,1 .e.,

— u2(x) v(x) � w(X) � I(wlioo � — 402 lIoo + Mflfj f2Iloo,

5.1 DirichIet Boundary Value Problems for Linear Differential EquatIons 89

where M := Analogously one proves —w v so that (7) follows. U

Exercise 5.1.10. (a) Let Si be bounded; let the coefficients of L be continu-ous in Now show that L is uniformly elliptic in Si if and only if L is elliptic

(b) Theorem 9 holds for the special case =/2 without the

a strip in It's be described by

where x E Ut" is a boundary point of E Ut" with l'il =1 is a unit vector,and S is the strip width. Show that inequality (7) holds with M := ifSi C S and a are as in the proof of Theorem 9.(d) Under the conditions of Theorem 9 show that

� + (5.1.8)

Exercise 5.1.11. Let the coordinate transformation (' x E Th E

and the inverse be continuously differentiable. To the operatorL (in the x coordinates) let Ii correspond in the coordinates. Show that ifL satisfies the conditions of Lemma 7, or of Theorem 9, then so does L'.

The right side f and the boundary values ço are not the only parameterson which the solution u depends. We next investigate how the solution dependson the coefficients and a of the differential operator L.

Theorem 5.1.12. Let the coefficients of V and L" be resp.

Let and be solutions of

LV=L"u"=f in Si, u1=u11 =p on

Let L' saiofpthe conditions of Theorem9andtet U" betongtoC2(Th). WithM from (7) we then have

lu' — � M{( 114 — lItLllc2(Th

+ (5.1.9)

+ ha' —

I/al, = the conditson u11 e C1(i7)flC2(fl) u sufficient; i/also al =then just u" E C°(Th) C2(ii) will do.


PROOF. Set f' := L1ts11. Then = can be boundedby the right side of (9) without the factor M. Theorem 9 applied to = 1'and Vu' = f implies His' — u'%, — filoo.

5.1.4 Difference Methods for the General Differential Equation ofSecond Order

For notational reasons we limit ourselves to the two-dimensional case n =2.General domams 17 E E2 require special discretisations at the boundary asexplained in Section 4.8. Here we only want to discuse the difference formulasat points far from the boundary. Therefore it will suffice to base our comments

on the unit square17=(0,1) x(O,1).

LetLbegnby(Ib).Ifonewihestoobnadifferencemethodofconsistency order 2, the following choice suggests itself, which will later bemodified for reasons of stability.

a11(z, y)08; + 2a12(z, + afl(x,

+ + + a(x,y) (5.1.10)

a12(x,y)/2= h2 alj(x,y) —2[aii(x,y) + all(x,y)

—012(x,y)/2

10 a2(x,v) 0 0 0 0

—al(x,y) 0 a,(x,y) + 0 a(x,y) 00 —02(x,y) 0 0 0 0

Remark 5.1.13. The difference method (10) is a nine-point scheme of con-sistency order 2.

Let Lh be the mMth of the system of difference equations In 17,, associ-ated to (10) (cf. Remark 4.2.5):

Lg,u,,=q,,. (5.1.11)

The solvability of Equation (11), for arbitrary q,,, Is equivalent to uniqueness.In the continuous case, uniqueness was essentially given by the condition a � 0arid the ellipticity condition a,ian >42 (cf. Theorem 8). These conditionsare in general not sufficient to guarantee the solvability of Equation (11). Wethus replace (10) with another discretisation.

Theorem 5.1.14. For the coefficients of L let (3a,b) hold, and asswne that

1a12(x, y)j min{aii (x, y), p)} (5.1.12)

5.1 Dizichlet Boundary Value Problems for Linear Differential Equations 91

and a(x, y) 0 in a. Then there exists for L a seven-point difference method

of consistency oirder 1 such that the associated mattiz Lh is an M-matrixexcept for the sign. In the resulting Equation (11) is solvable.

Note that Condition (12) foLlows from (3a) only ii = >0.

PROOF. At the grid point (x, y) E sia (ef. (4.2.la)) we abbreviate the on-efficients and a(x,v) to Oj, and a. The principal part

(x1 = z,x2 = y) is discretised by the following differencestars:

82._21 82..2 1

1 —2 1 , 0 —2 0(JXj 0 1

a —i i—1 2 —1 , if

a1

1

0 1

012 � 0, we thus obtain

282 1 0

E 8z:h2 —0j3 2(012 —011 —ar) 011—012

1 0

If one introduces := max{a12,0} end := min{a12,O} then both for� 0 and for 012 <0 we get the seven-point star

—aj2 — 101214.2

011 — 10121 2(10121 — 011 — afl) — 10121 . (5.1.13')

4 — 10121

ExercIse 5.1.15. Let Lh be the matrix belonging to the difference formula(13') (note that the coefficients in (13') depend on the spatial coordinates).Assume (3) and (12) hold. Prove that —Lh satisfies the conditions of Exercise4.3.8a.

The first derivative terms are replaced by the forward and back-ward differences, if 0:

28 03 0

:h_1 —aj a 00

element is negative, the others nonnegative. If one adds these terms to (13'),


the resulting matrix —Lh satisfies the conditions of Exercise 4.3.8a. Therefore—Lh is an M-matrix.

it is easy to see that the difference formula (13') is of second order ofconsistency; (14), however, contains the one-sided differences so that the totaldiscretisation is only of first order.

The condition (12) can be avoided if one allows larger difference stars,which also contain the values u(x1 ± vh, £2 ± ph) for fixed v, p E Z (butin general > 1) (cf. Bramble-Hubbard 11)). Layton—Morley 111 pointout that with weaker conditions than (12) one may still obtain a matrix Lh,which, though not an M-matrix, does have a positive inverse.

To obtain a method of consistency order 2, one must discretise Eas in (10). The following corollary shows that —Lh is also an M-matrix when

is sufficiently small.

Corollary 5.1.16. In addition to the assumptions of Theorem 14 let thefollowing hold:

h (s=1,2). (5.1.15)

Then the discretisation of fvvm (10) together with (13') leads to aseven-point difference method of second order of consistency such that -L,, isan M-matriz.

PROOF. -Lh satisfies (4.3.la) and is irredudbly diagonally-dominant. •ExercIse 5.1.17. The condition su � 1a121 + instead of (15) is not8ufflcient. Construct a counterexample with a11 = = 1, 0, h = 1/3,

variable, and lad =6 so that L,, is singular.

Considerably weaker conditions for the nonsangulanty of L,, than in Theo-rem 14 and Corollary 16 are needed in Section 9.2 (cf. Exercise 9.2.6, Corollary11.3.5).

in general, is not a symmetric matrix. Symmetry of L,, is to be ex-pected only if L is also symmetric: L = Ii. Here the formally adjointdifferential operator L' which is associated to L in (2), is defined by

1111 82 0 8 82 1 1 8 8L = [aj + + _E [(li + +a.

(5.1.16)It is easy to see that a 8ymmetric operator can always be written in theform

8(5.1.17)

vxj

A difference method for this is given for the case n=2 and a12 =0 by thefive-point star

5.1 Dirichlet Boundary Value Problems for Linear Differential Equations 93

o 0

— a22(x,y — —an(x,y+o 0

o 0 0

+ 0 a(x,y) 0 . (5.1.18)o o 0

Theorem 5.1.18. Let a22 E The difference method (18) isconsistent of order The associated matrix L,, is symmetric. If a.,, > 0(ellipticity) and a <0, then —L,, is a positive definite M-rnatrix.

PROOF. (a) For the proof of consistency, expand

v(x + h/2) := a11(x + h/2, y)Lt4x + h, y) — u(x, y)j (y fixed)

around x + h/2, and then expand v(x + h/2) — v(x — h/2) around x.(b) The symmetry results from = for x, E (cf. Exercise 4 .2.2c).(c) L,, is irreducibly diagonally dominant so that Criteria 4.3.10 and 4.3.24are applicable. U

The mixed term + can be discretised by

— a12(xB) a12(xC) + a12(XB) 02a12(XA) —21a12(xA) + a12(xC)) 2a12(Xc)

20 a12(XA) + 012(XD) —a12(xc) — a12(xD)

(5.1.19)with xA = (x — XB = (x — + h), 'cc = (x + and x0 =(x+ — h).

Another way of writing (19) is:

(5.1.19')where b := a12.

Exercise 5.1.19. (a) For the coefficients of L in (17) let the following hold:a11 > 0, > 0, = a21 0, a11 + a12 > 0, and also + h/2) +a12(x — h/2, y + h) + a12(x -f h/2, y) > 0, and a 0. Show that the differencescheme which is described by (18) and (19) has consistency order 2 and thatthe ansociated matrix —L,, is a symmetric, irreducibly diagonally-dominantand positive definite M-matrix.(b) What is the suitable discretisatton for the case a12 � 0?

Exercise 5.1.20. The difference formula from Theorem 14 for the operatorL reads + a when

94 5 General Boundary Value Problem8

012 0, � 0, � 0. Let the associated matrix be Lh. Then prove that thetransposed matrix LZ describes a difference method for the adjoint operatorV and also possesses consistency order 1.

In general it is possible to show for regular difference methods that is

a discretisation of L'. The role of regularity is demonstrated in

Example 5.1.21. Let Lu := u" + au' in 11 = (—1, 1) with a(x) � 0 forz 0 and a(z) � 0 for x > 0. According to (14) au' is discretised forz 0 by and for x > 0 by a(x)8u(x). Let the associated matrixbe Lh = Lh,2 + Lh,1, where Lh,2 and Lh,1 correspond to the terms u" andau' respectively. According to the above, should be a discretisation of—(av)'. But the differences at x = 0 and x = h are

h'ta(—h)vh(—h) — a(O)vh(O) — .-(av)' — h10(O)Vh(0) + 0(h),

h'[a(O)vh(O) + a(h)Vh(h) — o(2h)vh(2h)1 = —(av)' + h10(0)Vh(0) + 0(h);

thus they are not consistent. Nevertheless, is a possible discretisation ofL'v = v" — (av)', for it can be shown that the error — has order ofmagnitude 0(h).

To prove stability one has to show � const. Obviously it is suffi-cient to prove this inequality for sufficiently small h. In the proof of Theorem9 we used the fact that

—Lw�l mO, on!'

for w(x) := exp(2Ra) — — + R)). Let Dhuh(x) be the differenceequations from which Lhuh results after elimination of the boundary values.We set we,, := 2Rhw; i. e., wh(x) 2w(x) for x The followrng holds:

-Dhwh = - RhL)W - 2RhLW � 2 - - RhL)w.

Each consistent difference method satisfies fl(DhRh - -' 0. For

sufficiently small ho we thus have

—Dawh(x)�l (XE 0h, h<ho).

This inequality agrees for x £7,, far from the boundary with —Lhw,,(X) �1. For x Qh close to the boundary, —Dhwh(x) also contains the sum

which is not contained in —L,,wh(x). For the discretisa-tions from Theorem 14, Corollary 16, Theorem 18, and Exercise 19, however,

—Lhw,,>—D,,w,,�1

holds for all grid points. Theorem 4.3.16


Theorem 5.1.22. The ducretssations from Theorem 14, Coroflary 16, The-

orem 18, and Exercise 19 are stable under the conditions posed there, i.e.,

const for alt h H = {1/n:n E IN}. According to Theorem 4.5.3

the methods converge. The order of convergence agrees with the correspondingorder of consistency.

5.1.5 Green's Function

The idea of representing the solution by the Green function can be repeatedfor the general differential equation (Ia,b). The Green function (of the firstkind) x) is singular at = x and satisfies

= 0, =0 foe x

forxEIoreEf'.Here, L' is the adjoint differential operator (16). If L L', then g is no longersymmetric: g(x, x). Under suitable conditions the solution of (la-c),(4) can be represented as

u(x) =— L x)

where B = Bt = is a boundary differential operator (n,are the components of the normal vector n = e F). Only when theprincipal part of L agrees with is B the normal derivative.

In the discrete case the inverse L' again corresponds to the Green func-tion g(., .).

5.2 General Boundary Conditions

62.1 FormulatIng the Boundary Value Problem

Let the differential equation be given by (1.la,b). The Dirichlet boundarycondition (1.4) can be written in the form

Bu on I' (5.2.la)

where B is the identity (to be precise: the trace on F). In more general settingsB can be an operator — a so-called boundary differential operator —of order I:

B Ebs(x)o/ax. +bo(x), XE F. (5.2.lb)

If one introduces the vector b(x) = (bj(x), .. By can be written inthe form


Bti = (b(x), Vu(x)) + or B = bTV + b0. (5.2.1W)

Example 5.2.1. (a) From b = 0, b0(x) 0 there results what is known asthe Dirichiet condition u = on F, also known as the boundarycondition of the first kind.(b) The choice b = n, ho = 0 characterises the Neumann condition, alsocalled the boundary condition of the second kind.(c) Equation (Ia) with (b,n) 0, ho 0, is known as the mixed, or theboundary condition of the third kind. Occasionally one just means bya mixed boundary condition: B = unTV ÷ ho = o'O/On + ho with or(x) 0.

Remark 5.2.2. The case (b,n) = 0 is excluded in general. For (b,n) = 0,bTV is a tangential derivative. The boundary condition Bu = w is then verysimilar to a Dirichiet condition. The condition "u = on F" implies "Bu =

:= Bço on F" (Why is Bço defined?).

(a)

FIgure 5.2.1.(a) Boundary value problem with changing boundary condition type(b) Dirichiet problem in a disk with a cut

The normal derivative B = 8/8n (i.e. b = n) is important in connectionwith L = —4. For the general operator L in (l.lb) the so-called conormalderivative B with

b = An (A = (°ij)jJ1,...,n, aq as in (I.lb))

is of greater importance, as we will see in Section 7.4.Statements about existence and uniqueness of the solution always depend

on L and B. We have seen already that for L = —4, B = I (Dirichletcondition) uniqueness is guaranteed (cf. Theorem 3.1.2), while the problemassociated to L = —4, B nTV = 8/t9n is, in general, not solvable (cf.Theorem 3.4.1).

—x)/2

5.2 GeneraL Boundary Conditions 97

The coefficients of B depend on position. Of course, B(x) 0, i.e., b(x) =o and bo(x) = 0, must not occur for any x E I'. But it is poesible thatb(x) = 0 (and bo 0) in C I' and b(x) 0 in t'\'y. Then there is aDirichiet condition u on the piece -y and a boundary condition of fIrstorder on the remaining boundary piece f'\-y. At the points of contact between

and I'\'y the solution generally is not smooth (it has singularities in thederivatives).

Example 5.2.3. Let .0 be the upper semicircle around x = y 0 withradius 1. Let the differential equation and boundary conditions be given asin Figure la. The boundary condition changes its order at x = y = 0. Thesolution in polar coordinates reads: u = r1/2 (cf. (2.1.3)). Check that

= O(r112) and =

The same singularity as in Example 3 occurs in the problem describedin Figure ib; the solution is also r112 This Dirichlet problem andExample 3 are closely connected with each other.

Figure 5.2.2. A domain symmetric with respect to

Example 5.2.4. Let 5? = U Q2 U be as in Figure 2: let the reflectionof S?j in -y result in S?2. If one seeks a solution of Lu = f in (1, Bu onOS?, and if together with u the function II reflected in 'y is also a solution, oneexpects u = This solution then satisfies Lu = f in Ru = onOu/On=Oon'y.

While the boundary condition Bti = on OS? may be of physical origin,Example 4 shows that a Neumann condition may also have a geometric basis.

Another geometrically justified boundary condition is the following. Let5? be given as in Figure 3: and 'Y2 are parts of F =85? with


= {(x,,y):yi � y � Y2}, i = 1,2.

Then, in addition to Bu = on f'\(i'i U 72), we can require the periodicboundary condition

u(x1,y) t4z2,y), uz(Zj,y) = Uz(X2,y) for Yl � y � Y2, (5.2.2)

on and The solution is periodically continuable in the x-direction (withperiod x2 — x1). The origin of periodic boundary conditions is discussed inExample 5,

Example 5.2.5. (a) Let if be an annutus which is described by the p0-lar coordinates r E (ri, r2), w E [0, 2ir). fransformation of the differen-tial equation to polar coordinates gives as the image domain the rectangle11 = (i-i, r2) x (0, 27r). The original boundary conditions on if become bound-ary conditions at the upper and lower boundaries while Equation (2) describesthe periodicity of the angular variable x E (0, 2ir).(b) Instead of on (1' C Ui?, one can also define a boundary value problemon a part of the 2-dimensional surface of a 3-dimensional body. If if lies, forexample, on the surface of the cylinder E + <r2, e IR}, theunfolding of !?' results in a domain I? as in Figure 3. Here too, Equation (2)is justifIed by the fact that x plays the role of an angle variable.(c) If one seeks solutions in the unbounded strip if = JR x one caninstead look for periodic solutions in .0 = (0, 2ir) x (y1y2) with the boundarycondition (2), since these (after periodic continuation) are also solutions of theoriginal problem. Solutions with periodic boundary conditions in the x and ydirections can even be continued onto.0' = 1R2.

5.2.2 Difference Methods for General Boundary Conditions

The boundary conditions posing the least difficulties are the periodicboundary conditions. Define as the set of grid points in .0 and on

(but not on ci. Figure 3).

Z2Z

Figure 5.2.3. Discretisation for periodic boundary conditions


The difference equation at a grid point (x1, y) E has a left neighbour(x1 — h,y) outside Replace uh(zj — h,y) in the difference equation byuh(x2 — h, y). By doing so we have transferred the periodicity onto the differ-ence solution without explicitly discretising Equation (2). Of course, the stepsize must be chosen so that X2 — x1 is a multiple of h.

Exercise 5.2.6. For periodic boundary conditions the structure of the matrixLb changes. Let L = in the square (1 = (0, 1) x (0, 1). Assume Dirichletconditions for the upper and lower boundaries, and for the lateral boundariesthe periodicity condition (2). In analogy to (4.2.8) exhibit the form of thematrix Lb for a lexicographical arrangement of the grid points.

In Section 4.7 we described the discretisation of the boundary conditionB nTV = a/on for the case that .11 is a rectangle and I coincides with thegrid. In the same situation one can discretise the general boundary condition(la,b) as follows.

outer innerLine

(z,y)s

Figure 5.2.4. Discretisation of Bu =

Let E 1h = fl 1' be a grid point on I. The point of intersection+ h, drawn in Figure 4 is given by + Note that

b1 0, since bTV is not to be a tangential derivative. Fbr u E thedirectional derivative bTVu = (b, Vu) in is approximated by

— + + — + 0(h). (5.2.3a)

On the other hand, one obtains + h, under the same conditions byinterpolation up to order 0(h2):

h— (5.2.3b)

100 5 Boundary Value Problems

The combination of (3a) and (3b) leads to the difference formula

— — h)}:=

I/h2 ++ uh(X,y) (5.2.3c)

=

with =

Lemma 5.2.7. Let be the restriction to the grid poinis For a functionU E C2(TJ) one has

— — O(h11uflc3(m)),

i.e., the discretüation of the boundary condition has consistencij order 1.

Equation (3c) has the form

= — — h,V+ h). (5.2.3c')

We assume that the difference equations fh(x) (beforethe elimination of the values E satisfy the inequalities

(5.2.4a)

Here > 0 and S 0 correspond to the sign condition (4.3.la). Thesecond inequality in (4a) agrees with (4.3.4b). The corresponding inequalitiesfor (3c') read:

ct?O, C2�0, (5.2.4b)

Let the difference equations in x E and the boundary equations (3c')given for be combined into Ahuh = (4a,b) implies 0,

� ETherefore, Ah is an M-matrix if A,, is irreducible and (4.3.4a) holds for

at least one x EThe above siderations explain why the boundary discretisation should

satisfy the conditions (4b). (4b) holds for the case b0 =0 if and only if b2/b1 E(0, 11. If b2/b1 [—1,0J, (4b) can be satisfied if one interpolates between and

— h (instead of and + h). If, however, > i.e., if the tangentialcomponent is greater, one could interpolate between andkh), where 1b2/bd � k. Fbr the case b2/b1 � 1, however, it is more practicalto interpolate at the point + h) between + h) and + h,V + h).Generally one should choose as interpolation point the point of intersection ofthe line with the dotted straight line in Figure 4.

In the preceding discussion we started with the case fl = (0, 1) x (0, 1).Iii? is a general region, two discretisation techniques offer themselves (cf.Sections 4.8.1 and 4.8.2).


Line.--—

•: Points in

U: Points in Th

FIgure 5.2.5. FIrst boundary discretisation

First option for discretisation:Let and lj, be chosen as in Section 4.8.1. At near-boundary points thedifferential equation is approximated by a difference scheme which in the caseof L = —LI corresponds to the Shortley-Weller method. For this one needsthe values of Uh at the boundary points E As in Figure 5 let

— sjh, E 1h be a boundary point. Again we can set up an equationanalogous to (3a—c). In Figure 5 has the same position as + h, inFigure 4. In general one has to use the point of intersection of the line whichis also presented by + tb (t IR), and the dotted straight line in Figure5.

Second option for discretisation:Let be the grid aa used above. Now let consist of all points ofthat are far from the boundary. For all (x, y)Qh difference equations (withequidistant step size) are declared which involve For eachpoint 1h := a boundary discretisation must be found (cf.Figure 6). Equation (3a) can be set up with and + instead of

and + h, The arguments of the coefficients b1 and b2 areThis point results implicitly from

—1) = — €1'. (5.2.5)

Remark 6.2.8. (a) At the grid points adjacent to the boundary, the firstdiscretisation requires a difference scheme for Lu = f with nonequidistantstepsizes. The second discretisation requires an approximation of the nonlinearproblem (5). From a programming viewpoint both procedures are undesirablebecause of the case distinctions. (b) If the vector b from B approaches the tan-gential direction, both methods fail since the straight line no longer intersectsthe dotted straight line from Figures 5 and 6. Here, the second discretisationfails earlier than the first one.

F

S— h)


Line

(1+ h,

• Points in

I Points in

FIgure 5.2.6. Second boundary dlecretisation

Another option for avoiding the difficulties described above consists inusing variational difference equations, at least near the boundary (cf. Remark8.6.1).

Occasionally it is also poesible to simplify the boundary conditions byusing coordinate transformations. Let Bu = be prescribed on C P. Ifone finds a transformation (x, y) ,) audi that the equations 0and = are satisfied on one obtains for the transformed problemBu oOu/On + bou = on a vertical boundary piece (8/On = Thediscretisation may be earned out as described in Section 4.7.

The same reasoning as in the preceding sections proves stability andconvergence:

Remark 5.2.9. Let the difference method = ía satisfy the conditionsof Theorem 5.1.22 so that � 1 holds for := K1 + (K>o is constant; wh is as in the proof of Theorem 5.1.22). Let the boundarydiscretisation (3c') satisfy

For sufficiently large K, one then obtains (Bhti,h)(x) � 1 for x E Hence itfollows that � 1, where is the M-matrix defined immediately afterEquation (4b). Since � conat (cf. Theorem 4.3.16), stability has beenproved. The order of convergence is the minimum of 1 (order of consistencyof ; cf. Lemma 7) and the order of consistency of = Ia.

In general it is not poesible to approximate bTV by symmetric differences.To construct a discretisation of Bu = w with consistency order 2 despite thisfact, set:

: — = E

TI


where are three points in If, in order to remove the first twoterms in the Taylor expansion, one uses the differential equation = f in

and the tangential derivative of Bu = one obtains a discretisation oforder 2. For the special case L = 4, B = 0/On + Bramble-Hubbard [2)

proved that the three points (xe, may be chosen such that the inequalities

4 � 0, c0 + C2 ÷ hold and guarantee stability. However, in general, the(xe, do not lie in the direct neighbourhood of f'h. Nonetheless, theconstruction of the discretisation appears too complicated to be recommended

for practical purposes.

5.3 Boundary Problems of Higher Order

5.3.1 The Biharnionic Differential Equation

In elastomechanics the free vibration of rods Leads to (ordinary) differentialequations of second order if longitudinal vibrations (= compression waves)or torsional vibrations are involved. By contrast, transversal vibrations (bending waves) result in an equation of fourth order. Correspondingly, thebending vibration of a plate leads to a partial differential equation of fourthorder. This is the biharmonic equation (plate equation)

42u=f infl (5.3.1)

where 42 = 04/8x4 + 284/8z20y2 + 04/8y4. Here u describes the deflectionof the plate perpendicular to the surface. If the plate is firmly clamped at theedge, one obtains the boundary conditions

u = and Ou/On = on F (5.3.2)

with = 0. A btharmonic probLem (1), (2) also results from thetransformation of the Stokes equations in 11 C in.2 (cf. Remark 12.1.5).

The differential equation (1) may be combined with other boundary val-ues than (2). An example is:

is = and 4u = on F (5.3.3)

(simply supported plate). Other examples can be found in (7.4.12b,e).

Exercise 5.3.1. Show that if one solves the Poisson equations 4v = f in £7,

solution of the boundary value problem (1), (3). Why can Problem (1), (2)not be handled likewise?

Remark 5.3.2. The solutions of L12u = 0 do not satisfy a maximum-minimum pnnciple (counterexamnple: u x2 + y2 in £7 = KR(0)).


5.3.2 General Linear Differential Equations of Order 2m

The partial derivative D° (a multi-index) of order IaI = + +is defined in (3.2.5). A differential operator of order 2m has the form

L = (x Q) (5.3.4a)

and defines the differential equation of order 2m:

Ltt f in (2. (5.3.4b)

Ellipticity has been explained thus far oely for equations of second order (cf.Definition 1.2.3).

Definition 5.3.3. The differential operator L (with real-valued coefficientsaa)iasaidtobe elliptic (oforder2rn) atxE Qif

for (5.3.5a)

Here, is an abbreviation for the polynomial of degreeWe set Evidently (5a) is equivalent to P(x,e)o for all E IR' with = 1. For reasons of continuity either

P(x, <0 must bold. Without loss of generality, we may assume thatP(x,e) > 0; otherwise we scale with the factor —1 (changing from Lu(x)f(x) to —Lu(x) —1(x)). Since the set E = 1) is compact it followsthat c(x) min{P(x, = 1} > 0 and this justifies the formulation of(Sa) as

� c(x) >0. (5.3.5b)frzl.2m

Definition 5.3.4. The differential operator L is said to be uniformlyelliptic in I? if inf{c(x):x E (1} >0 for c(x) from (5b).

ExercIse 5.3.5. (a) L from (1.lb) into the notation of (4a). Whataxe the coefficients Ga for L 112?

(b) Prove that the btharmonic operator 112 is uniformly elliptic.(c) Let aa be real-valued. Why are there no elliptic operators L of odd order?(d) If the coefficients Ga are sufficiently smooth one can write L from (4a) inthe form L = (cf. (1.lb) and (1.2)).

For in = 1 (equation of order 2) we have used one boundary condition; forthe biharmonic equation (in = 2) two boundary conditions occur. In generalone needs m boundary conditions

5.3 Boundary of Higher Order 105

B,u b,0DaIL= (j = 1,2,. ..,rn) on P (5.3.6)

lcxl<m,

with boundary differential operators B1 of order 0 � m3 <2m.

Remark 5.3.6. The boundary operators B2 cannot be arbitrarily,but must be independent of each other (they must form a so-called "normalsystem"; cf. Lions-Magenes [1, p. 113] and Wioka Definition 14.1]). Inparticular the orders in1 must differ pairwise.

We have a Dirichiet boundary condition if B, = =(O/On)-'1 forj 1,•• (ci. (2)).

The representation of the solution using the Green function will notbe discussed at this point. However, it is remarkable that the Green func-tion (and in particular the singularity function) for L is continuous when-ever 2m > n. For L = for example, the singularity function is

Ix — yj (cf. Wloka [1, Exercise 21.91).

5.3.3 Dlscretisation of the Biharmonic Differential Equation

The simplest difference formula for L is the 13-point differencemethod

1

2 —8 2h4 1 —8 20 —8 1 , (5.3.7)

2 —8 2I

which can be represented as the square of the five-point star from (4.2.3).Let the grid C £7 and the boundaries 1h c r, c be defined asin Figure 1. The differenee equation

Dhuh=fh inQh (5.3.8a)

requires the values of uh in £7,, U P,, U 7h• These are determined by theboundary conditions (cf. (2))

Uh(X)=401(X) for x€f, (5.3.8b):= ju,,(x + hn) — Uh(X — hn)J/(2h) = cP2(x)

for (5.3.8c)


0000000Qh:.000000000on.... •00-yb: a

000on.... Sf0

a a • • • • so a = Ui U0000000000000000

Figure 8.3.1. Grid 8r the biharmonic equation In 11= (0,1) x (0,1)

As in Equation (4.7.14b) we assign to the corner points two normal di-rections each. Correspondingly, to each corner point belong two equations(8c). Note that two (m =2) boundary layers occur, fh and 'yb, and that twoboundary conditions are given.

Remark 5.3.7. If with the aid of (8b,c) one eliminates in Equation (8a) theunknowns (uh(e):C e rh U'yh}, one obtains a system of equations Lhuh =for E Li, is not an M-matrix since the sign condition (4.3.la)is violated. But also is in general not � 0 (cf. (4.3.lb)).

Thus the methods of the proof in Section 4 cannot be used here. Insteadwe will prove the analogue of Theorem 4.4.12.

Theorem 5.3.8. Let ui, be the discrete solution of the biharmonic equationin the square a = (0,1) x (0,1) defined by (8a—c). The matrix Lh described mRemai* 7 is symmetric and positive definite. It satisfies

� 64h4, 1/256. (5.3.9a)

If = 0 on "i, (cf (8b)), then be estimated by

1

______

— + (5.3.9b)

where the norms are defined as follows:

Iui,(x)t21"2, := Ih (5.3.9c)

xEOk XEA

Let the sum in contain the corner points twice, where, for ex-ample, the value at x = (0,0) should be intempreted once asand once as (of. comment on (8c)).


PROOF. We limit ourselves to the most important part, the inequality (9b).In the infinite grid Qa {(x,y) E 1R2:x/h,y/h Z} we set

Uh(X) Uh(X) for x E 0h, tih(x) =0 for x Qh\Qh.

After partial summation one obtains

= > =xEQh

(First consider the identities = — E(8v)(x)(8u)(x)in the one-dimensional case.) From the definition of iih it

follows that 0 on U f'h). On 17h one has uh = = 0, and

tih(Xhfl)=Uh(x—hD), Uh(X+hfl)0,uh(X + hn) = uh(x — h(n) +

for x E f'i,. If one sets 0uh(x) := [Uh(X) — Uh(X — hn)1/h = —uh(x— hn)/hat x E then

= —h'8uh(x), = —2h'8uh(x) +

for x E Ph, implies that

(up1, Laup1)Qh (us, 1- —2h 1(a;uh,

where (., and (., are the scalar products belonging to the norms (9c).According to

ab a2 + b2/4 (a, bE JR) (5.3.10)

one estimates

IOn :S ÷ Ico2lr'h/4.

Thus one obtains

+ � (un, fh)f',, + 2h1(a;up1, W2)I',, (5.3.11)

� + +Theorem 4.4.12 showsthat IIUhlfQh � II4huaIInb/16. Thus for = 0

(11) implies � � so that

Iltiafla,. � 16211fhHph. (5.3.12a)

For 0, (11) implies the inequality therefore

IIup1IIs).,, � j IILlhUhIIflh � (5.3.12b)

In the general case we write Up, as t4 +uj,t, where =0, =0, = fp, andf/' = 0. + together with (12a,b) implies (9b). U


Theorem 5.3.9. (Convergence) Let the solution of the bihannonic bound-ary value problem (1), (2) in 17 = (0,1) x (0,1) satisfyu E Then thediscrete solution of equations (8a—c) is convergent (at least) of order

huh — RhulIab � + (5.3.13)

where Rh is the restriction to U 1h•

PROOF. The Thylor expansion shows that

4uhRi,u = + + + h4R, IRI � flh.(5.3.14)

Outside 17h one can specify Rhu arbitrarily since these values do not appearin (13). In l'h we set

Rhu(x-fhn) :=h4 2h5 8

u(x — hn) + + — + x E Ph.

The choice is made so that (14) also holds for x E Th. Applying toyields

= DhR,1u = + in 17h

(kh Is the restriction to flh). The difference Wh := Uk — R,,u (defined in Thh)satisfies

Dhwh = Dhuh — DhRhu

= Rh(f — + =: 9h, (5.3.15a)

wh(x) = uh(x) — (Rhu)(x)= — u(x) = 0 for x E (5.3.15b)

= —

= — +=: in X E 1h. (5.3.15c)

If one sets = and 'I'h = instead of fh andin Theorem 8 (note that çoj = 0!) then (9b) implies assertion (13). IBy using more complicated methods one can replace the factor h1/2 in

(13) by (e > 0 arbitrary) so that the order of convergence isi.e., almost 2.

Remark 5.3.10. Inequality (9a) shows conda(Lh) =O(h4). In general a difference method for a differential equation of order2m leads to the condition

cond(La) = (5.3.16)


This indicates greater sensitivity to errors at higher orders 2m(cf. §4.4 in Stoerili and Bulirsch—Stoer[1J).

Difference methods for boundary value problems of fourth order with vail-able coefficients and general domains fl are discus9ed in an paper by Zlámal[1). There too convergence of oider 0(h3/2) is shown. By contrast, 0(h2)-convergence was proved by Bramble (1J for the 13-point star (7) in a generaldomain if the boundary conditions are suitably discretised.

6 Tools from Functional Analysis

To enable readers without previous knowledge of functional analysis to followthe next chapters we summarise here all definitions and results that will beneeded later on.

6.1 Banach Spaces and Hubert Spaces

6.1.1 Normed Spaces

Let X be a linear space (alternative term: vector space) over K where K = IRor K = C. In the following the normal case K = IR is always intended. K = C

occurs only in connection with Fourier transforms.The notion of a norm . fl: X —. oo) is explained in Definition 4.3.12.

The linear space X equipped with a norm is called a nor med space and isdenoted by the pair (X, fl. Whenever it is clear which norm belongs to X,this norm is called Ii lix and one writes X instead of (X, fl. lix).

Example 6.1.1. (a) The Eudlidean norm H from (4.3.13) and the maximumnorm (4.3.3) are norms on lit".(b) The continuous functions form the (infinite-dimensional) space If1? is bounded, all ii E C°(fl) are bounded so that the supremum norm (2.4.1)is defined and satisfies the norm axioms. If ii is unbounded, the bounded, con-tinuous functions form a proper subset 11C°(& of C°(fl). Insteadof jJ

. we use the traditional notation fl.(c) The Holder-continuous functions introduced in Definition 3.2.8 form thenormed space (C'(J7), ii

The norm defines a topology on X: A C X is open if for all x E A thereexists an >Osothatthe "ball"in A. We write

x or x = urn if — 0.

Example 6.1.2. Let I,,, I The limit procees f, —' f (with respectto denotes the unifonn convergence known from analysis.

6.1 Banach Spaces and Hilbert Spaces 111

Exercise 6.1.3. The norm fl. X —* [0, oo) is continuous; in particular thereversed triangle inequality holds

— IlvIlI � fix — till for x,y E X. (6.1.1)

As Example la shows, several norms can be defined on X. Two norms

fi and ifi .on X are said to be equivalent if there exists a number

o < C <co such that

(jzfi � HlxlD <Cfixfi for all x E X. (6.1.2)

Exercise 6.1.4. Equivalent norms lead to the same topology on X.

6.1.2 Operators

Let X and Y be normed spaces with the norms fi. fix rasp. ff . fly. A linearmapping T: X —. Y is called an operator. If the operator norm

:= sup{IlTxffy/(IzlIx: 0 xE X) (6.1.3)

is finite, T is said to be bounded (ci. (4.3.9)).

Exercise 6.1.5. (a) T is bounded if and only if it is continuous.(b) If is a norm.

With the addition (Tj + T2)x = T1x + T2x and the multiplication byscalars, the bounded operators form a linear space which is denoted byL(X, Y). (L(X, Y), II 11Y4-x) is a normed space. With the additional mul-tiplication (T1T2)x = T1 (T2x), L(X, X) even forms an algebra with unit I,since I always denotes the identity: Ix = z.

Exercise 6.1.6. Prove that

fiTzffy � HTHyi_xIlxflx for all x X, T L(X, Y). (6.1.4a)

If T1 E L(Y,Z) and T2 L(X,Y), then T1T2 E L(X,Z) and

� fiT1 (6.1.4b)

We write T for E L(X,Y) if 0 (convergence inthe operator norm).

112 6 Tools from Functional Analysis

6.1.3 Banach Spaces

A sequence E X:n � 1} is said to be Cauchy convergent, or is calleda Cauchy sequence, if SUp{IIXn — � k} —'0 for k —' 00. A spaceX is said to be complete if any Cauchy-convergent sequence converges toan x E X. A Banach space is a complete norrned space.

Example 6.1.7. (a) is complete, but is not rational numbers).(b) Let D c (C°(D), ((. is a Banach space.(c) II lIc'(D)) for k IN and the Holder spaces (C'(D), It Hc.(D))8 > 0 as well as Ii Ilck.1(D)) are complete, thus Banach spaces.

PRooF of (b). If {u,1} is Cauchy convergent then there exists a limit u(x) :=for all xE D. Since converges uniformly to u, t& must be

continuous, i.e., u C°(D). U

Denote by L°°(D) the set of functions that are bounded and locally iifle-grable on D. Here we do not distinguish between functions which agree almosteverywhere. In this case the supremuin norm is defined by

IHILOQ(D) inf{sup{Iti(x)1 X E D\A} A set of measure zero}}

(L°°(D), H IIL'.'(D)) is a Banach space.

Exercise 6.1.8. Let X be a normed space and Y a Banach space. Show thatL(X,Y) is a Banach space.

Exercise 6.1.9. Let X be a Banach space and Z C X a closed subepace.The quotient space X/Z has as elements the clasass I = {x + z: z Z}. Showthat X/Z with the norm 1(111 Inf{IIx + zIIx: z Z} is a Banach space.

A set A is said to be dense in (X, II fix) if A C X and A = X, i.e.,every x X is the limit of a sequence E A. If (X, fi lix) is a normed,but not complete, space, one calls (X, iI• 112) the completion of X, if Xis dense in X and iIxII.t = llxlIx for all x E X. The completion is uniquelydetermined up to isomorphism, and can be constructed in the same way asone constructs as the completion of

Lemma 6.1.10. Let be a nonned Linear space and subspace of aBanach space (Y, fi. lly) with IlxlIy CtlxiIx for all z X C Y. Let eachCauchy sequence in X with -. 0 aLso satisfy -.0. Thenthere ezütsacompletionX ofXinY:XCXcY.

Frequently it is not easy to describe the image Tx of an operator T EL(X, Y) for all elements x of the Banach space X. The following theorempermits a considerable simplification: It suffices to investigate T on a denseset X0 C X.

6.1 Banach Spaces and Hilbert Spaces 113

Theorem 6.1.11. Let Xo be a dense subspace (or just a dense subset) of the

normed space X. Let Y be a Banach space.

(a) An operator To E L(X0,Y) defined on the subspace Xo withsup{IIToxIIy/flxllx:O x E Xo} has a unique continuation T L(X,Y),i.e., Tx =Toz for all x E Xo.(b) For x,, x E X0, x X) holds Tx =(c) = IIToIIr4_xo.

PROOF. For x E Xo, T is defined by Tx = Tox, while for x X\Xo thereexists a sequence —, X, E X0 and Tx = =It remains to show that urn exists and is independent of the choice ofthe sequence E X0. Since — Tozm(Jy � XmJIX, thesequence T0x1,, is Cauchy convergent. Because of the completeness of Y thereexists y Y with y. Similarly, for a second sequence Xo with

x, there exists for the same reason a y' E Y such that y'.Since y' — y = lim(T04 — and — Toxn(Iy � —

— —, 0, Tx is well defined by (b). Part (c) follows fromIITxHy/IIxIIx = for E XO.

Exercise 6.1.12. Let X0 be dense in (X, fi. (() Let Il( be a second normon Xo, that is equivalent with fl . II on Xo. Show that the completion of X0with respect to results in (X,((f.((() and and (((•ffl are alsoequivalenton X.

A corollary of the "open mapping theorem" (cf. Yoeida ti, §H.5J) is the fol-lowing, important result:

Theorem 6.1.13. Let X, Y be Banach spaces. Let T L(X, Y) beand surjective. Then E L(Y, X) also holds.

Exercise 6.1.14. Let X, Y be Banach spaces, and T E L(X, Y) be injective.Show that:(a) Y0 := range(T) := {Tx:x E X} with as norm is a Banach subspaceof Y.(b) T1 exists on Y0 and is bounded: E L(Yo,X).

Let X and Y be Banach spaces with X C Y. Obviously the inclusiondenoted by I : z X x E Y is a linear mapping. If it is bounded, i.e.,

I E L(X,Y) that is, ((xIly S CIIxI)x for all x E X, (6.1.5)

then X is said to be continuously embedded in Y. If furthermore Xis dense in (Y,($ then X is said to be densely and continuouslyembedded in Y.


Exercise 6.1.15. Let X c V C Z be Banach spaces. Show that if X is[densely audi continuously embedded in Y and Y is (densely audi continuouslyembedded in Z, then X is (densely andj continuously embedded in Z.

6.1.4 Hilbert Spaces

A mapping (., .): Xx X K (K = IR or K = C) is called a scalar producton X if

(z,z)>O (6.1.6a)

t7..x+y,z) = A(x,z)+(y,z) for all A€ K,x,y,z X, (6.1.6b)

forallz,yEX, (6.I.6c)

where denotes the complex conjugate of A.

Exercise 6.1.16. Show that (a) lizil (x,x)1/2 is a norm on X. For theproof use(b) the Schwarz inequality:

flxfl Ifvll for sit z,y X. (6.1.7)

Hint: consider Uz — Ayfl � 0 for A = flxfl2/(x,y), if (x,y) 0.

(c)(.,.):X xX-9Kiscontinuous.

A Banach space X is called a Hubert space if there exists a scalar product')x on X such that

x X X are said to be orthogonal (x I v) if (x,v)x = 0.

If A C X is a subset of the Hubert space X then the orthogonal spaceA' :={XEX:(x,a)x =Oforall aEA}definesaclosedsubspaceofX.

Lemma 6.1.17. Let U be a dosed subspace of a filbert space X. ThenX mn be decomposed into the direct sum X = U U', se.,

a x u u v E U'.= flull3c +

Exercise 6.1.18. Let A be a subset of the Hubert space X. Prove theequivalence of the following statements: (a) A is dense in X; (b) A' = {O};

there exists an a A with (a,z)X 0.


6.2 Sobolev Spaces

in the following .12 is always an open subset of

6.2.1 L2([2)

L2(12) consists of all Lebesgue-measurable functions whose squares on 12 areLebesgue-integrable. Two functions u, V E L2(fl) are considered to be equal(u = v) if u(s) = v(x) for almost all z E 0, i.e., for all x E S?\A, where theexceptional set A has Lebesgue measure = 0.

Theorem 6.2.1. L2(12) forms a Hubert space with the scalar product

:= (u,V)L2(f1) := (6.2.1)

and the norm

___________

luk := 11U11L2(fl) :=v'L Iu(x)I2dx. (6.2.2)

Lemma 6.2.2. The spaces Coo(S?)flL2(12) and are dense in L2(Q).Here

:= {u E C°°(i2): supp(u) compact, supp(u) CC 12}. (6.2.3)

supp(u) := {x E Q:u(x) 0) denotes the support of u. The double zndusionif cc $2 indicates that 777 lies in the interior of Si (i.e., 12' C Si and flOS? = 0).

Let D° be the partial derivative operator (3.2.5b). In the following weneed so-called weak derivatives which are defined in a nonclassical way.

DefinItion 6.2.3. u E L2(0) has a (weak) derivative v := D°u L2(S?) iffor the latter v L2(fl) holds:

(w,v)o for all WE (6.2.4)

Exercise 6.2.4. Show the following: (a) Let u have a weak derivative D"u EL2(Q). If the classical derivative Dcv. exists in (1' C $2, it coincides there(almost everywhere) with the weak one.(b) If u has the weak derivative Va = LPu E L2(S?), and if has the weakderivative = E L2(S?) then is also the weak D(r+&derivativeof u; i.e., =(c) Let f? c be bounded and 0 E 12. u(x) := (a E IR) has weak fIrstderivatives in L2(12) if a =0 or 2a + ii >2.


(d) Fbr C°°(Q) let —4 u E L2(Q) and v L2(f1) in theL2(fl) norm. Then v = D0u.

For further applications we consider

Example 6.2.5. Let 37 = 37k, where the bounded subdomainsare disjoint and have piecewise smooth boundaries. Let Ic IN. A functionu E Ck_1(37) with restrictions u (1 � i � N) has a (weak) kthderivative v0 = D°u L2(I?), Ic, which coincides with the classical one

PROOF. The (Ic — 1)st derivatives exist as classical derivatives so that theassertion need only be shown for k = 1. Let D0 = 0/Ox,, w E Cr(s?). Theassertion is obtained via integration by parts:

—(D0w,u)o = = _Jwx,udX

= - J ={L. —

di]

= J was, dx > J wv0 dx= f dx = (w, va)o,

since every x E also belongs to for some k I with oppositenormal direction =

The inequality I(u, v)oI � IuIoIvIo (cf. (17)) reads explicitly

� L (u(x)I2dxj (6.2.5a)

For v = 1 the result is

u(x) dxj � Iu(x)12 (j&(fl): measure of .0). (6.2.5b)

For a L°°(fl),u,v E L2(D) one now has

I f a(x)u(x)v(x) dxl � IIUIIL2(fl)11V11L2(O). (6.2.5c)

6.2.2 H"(Jl) and

Let Ic E IN U {O}. Let Hk(Q) C L2(Q) be the set of all functions having weakderivatives U-'u L2(J?) for

:= {u E L2(J7): EY'u L2(S2) for IaI � k}.


The Sobolev space denoted here by Hlc(a) is denoted by in

some other places. Usually, there Hk(i1) is a space which only coincides forsufficiently smoothly bounded S? with

Theorem 6.2.6. Hlc(a) forms a Hilbert space with the sailor product

(u,v)k (u,v)Jfh(D) := (6.2.6)

and the (Sobolev) norm

lulk

fl lies densely in HIc(rl).

Lemma 7, whose proof can be found, for example, in Wioka Theorem3.5J permits a second definition of the Sobolev space

Remark 6.2.8. Let Xo := {u C°°((1): jujk <oo}. The completion of Xoin L2(a) with respect to norm (7) results in Hk(i7).

Definition 6.2.9. The completion of in L2(S?) with respect to thenorm (7) is denoted by

Theorem 6.2.10. The Hilbert space is a subspace of Hk(ü) with thesame sailer product (6) and some norm (7). is dense in Fork = 0 there holds

(6.2.8)

PROOF. (a) Cr(Q) C Xo (cf. Remark 8) implies C Hk(S2) C L2(fl).(b) According to the definition CT(.Q) is dense in (c) For k 0 thenorms (2) and (7) coincide. Lemma 2 proves = L2(.Q). Because of

C Hk(a) C (k arbitrary) (8) follows.

6.2.11. For £7 bounded, and

1/2ftLlk,o (6.2.9)

lat=k

are equivalent norms in

PROOF. (a) Evidently,I Ik,O < IIHM(n).


(b) Let C000(.Q). For an with = k —1 set v := D"u E Thereexists an Rwith Qc KR(O). Foreach XE thus

2

-= f + H)]-R

�2RJ—R

(cf. (Sb)). integration over x I? yields 4RiuI o,is k-fold derivative. Summation over all o with = k—I yields

Now follows likewise for all I � j < k, and

thus � ... Since = 5 for

all u E the statement follows from Exercise 6.1.12. U

Exercise 6.2.12. Show that (a) for bounded (2 and k � 1, Hk((2) andare different. Hint: Consider the constant function u(x) = 1 and use

Lemma 11.(b) Lemma 11 holds even if (2 is bounded in one direction, i.e., if (2 lies on astrip {x E IR": xk1 < R} (k E {1,. .

Theorem 6.2.13. Letm � 1. There exist constants C = C(m) and 17(e) =i)(C, in) snch that

for aU 0 Ic 5 m, u E (6.2.lOa)

+ 77(€))uIo for all e > 0,0 � k < m,u E (6.2.lOb)

PROOF. (a) Partial integration for = I shows that

= = � �Since also it follows that � (n + If one replacesu by with 1, one obtains in the same way

IuI? <CIulg+11u11_i (1 <1 < m; u E (6.2.lOc)

(b) Let 0 be fixed. Set := log IuII,wj := IlVm..+ (in —l)vo)/nz and z1— wj. (bc) leads to 2z1 — — z:+j := log C, where zo =

z = (z1,. , one obtains Az all. Here A is the M-matrix (4.1.9b) forh.= 1.A'implies = exp(vj) � + Wi), i.e., (lOs) with C :=(c) Elementary calculation shows that for each e > 0, 0 0 � Oo < 1 thereexists an Oo) such that

a9b'° � ca + 17(e)b for all a, b � 0. (6.2.lOd)

Formula (lOb) is a corollary of (lOa,d).


By similar means, together with (5.3.10), one proves

+> 0, � m, � k � m, u (6.2.lOe)

�u E (6.2.lOf)

Remark 6.2.14. The set E C°°(47):supp(u) compact, IuIk <oo} is densein Hk(.Q).

PROOF. Let u E Hk(fl), e >0. According to Lemma 7 there exists a functionE C°°(Q) with Iu — u4k <e/2. There exists a E with a(x) = 1

for lxi � 1, a(x) = 0 for lxi � 2. For sufficiently large R, one also has<e/2. Thus there exists v(x) = a(x/R)ue(x) E

withlu—vlk�e. S

Since "supp(u) compact" already implies "supp(u) cc we obtain

Corollary 6.2.15. = for all k � 0.

The Leibniz rule for derivatives of products proves

Theorem 6.2.16. IlatLHHk(a) � for all a EuEHk(fl).

Theorem 16 together with the substitution rule for volume integrals shows

Theorem 6.2.1T. (Transformation theorem). Let T: (1—. if be a one-to-one mappsng onto if with T and ldetdT/dxl � 6>0 inf?. We write v = u o T for v(x) = n(1'(x)). Then u Hk(if) [ualso impises u o T Hk(1l) [€ Hok(S2)j and

flu o TIIHb(a) � (6.2.11)

6.2.3 Fourier and

For tt E one defines the Fourier-transformed function €i by

dx. (6.2.12)

Note that ü is described by a proper integral since the support of u isbounded.


Lemma 6.2.18. ForR—'oo

IR(u; y) := J [J dx]R'

uniformly to u(y) on supp(u).

PROOF. It suffices to discuss the case n = 1 (by Fubim's theorem). Integra-tion with respect to results in

IR(u; y) = (1/7r) J(x — sin(R(x — y))u(x) dx.

Then IR(1;y) = = 1 for all R >0. Since u C°°(LR"),then also w(x, y) := [u(x)—n(y))/(x—y) E The estimates w(x, y) =

and Wz(x,y) = O(1/x2) hold uniformly for y E supp(u). Partialintegration yields

— u(y);y) = IRU — y)w(.,y);y) = sin(R(x — y))w(x,y)dx

= .-(1/ir) J cos(R(x — y))Wx(X,y) dx/R O(1/R).

The statement follows froni

IR(u;y) = u(v)IR(1;y) + IR(u() — u(y);y) = u(y) +O(1/R). •Lemma 6.2.19. (s E L2(JR') and = lulo for all u

PROOF. Lemma 18 shows

J[J [J

= J IR(u;y)u(v)dy—. J Iu(y)I2dy. U

Lemma 6.2.20. The inverse Fourier transformation = u is definedby(13):

Jlim J (6.2.13)


PROOF. We have from Lemma 18

J = J J dyIL"

= u(x) + O(1/R).

Theorem 6.2.21. E with == 1, i.e., 7 is an isometric mapping of L2(IR') onto

itself The scalar prodtict satisfies (u,v)o = (ü,ii)o for all u,v E

PROOF. Since Cr(1R') is dense in L2(JR") (cf. Lemma 2), 7 can be contin-ued to 7: —+ (cf. Theorem 6.1.11). The norm estimate followsfrom Lemma 19. The roles of 7 and are interchangeable (cf. (12) and(13)); thus E L(L2(JRYI), L2(IR")) also holds.

The second statement results from

= — — = + — — = U

Exercise 6.2.22. Prove that: (a) With = .. . there holds

for u E (6.2.14)

(b) There exists C = C(k) such that �E IR".

Lemma 6.2.23. (a) Juk = I%JEcrI<k holds for all u E

(b) A norm on equivalent to 1k is

1u1 := 1(1 + (6.2.15)

PROOF. (a) It suffices to show the statement for is E (cf. Corollary15, Theorem 6.1.11):

- - ==

laI�kV

(b) The statement follows from Exercise 22b. U

Lemma 6.2.24. Let8a,, be the difference opelntorOhju(x) :=where e, is jth unit vector. If u and l49hjUjk C for

all h > 0, 1 j � n, then is E Hk+1(lRn) holds. Conversely, I8hjUIk � Nk+1holds for all u E H'+1 (lit"), h > 0.


PROOF. (a) From +Se,))(e) = follows =Hence, since 4h2s1n2((,h/2) � for h � follows

I

= for � 1/h. Summation over j = 1,.. , n andintegration over then gives

(1u1+i)2 = J(1 +

� J ...+

+ J (1 +

� J . ..+ +

<1J11�1/h

The integral over � 1/h vanishes for h 0 so that the statement follows.

(b) For the converse use (%ju)(x) = (x + the,) cit. S

6.2.4 H'(a) for Real a 0

Let 8 � 0. Fur a = It's one can define the following scalar product (16a) andthe Sobolev norm (16b) for all u E Cr(JR.Th):

(u,v) := J(l + (6.2.16a)

Iut := 11(1 + (6.2.16b)

The completion in also defines the Sobolev space H(IR') for non-integer order s. On the basis of Lemma 23 and Exercise 6.1.12 the newlydefined 11(r) for a U {0} agree with the Sobolev spaces used until thispoht.Letl7cR'.Thenumbers�Ocanbedeconipoeedaas=k+Awith

1. Wedeflne

:={fIaI�k

+ Jf— —

dxdY]7 (6.2.17a)

IuI. := IIUIIH.(sl) := (a = k + A, 0< 6 < 1). (6.2.17b)

The norm is called the Sobolev-SlobodeddT norm. One can define the

The properties of these spaces are summarised in the following theorem (cf.Adams 111, Wloka [11):


Theorem 6.2.25. Let a � 0. (a) For i? = the norms (16b) and (17b)are equivaient, i.e., both norms define the same space(b) {u C°°(.0): supp(o) compact, <oo} is dense in(c) is dense in(d) aD°(bu) E if � a, u H(.0), a b E C(i)),wheret=3EINU{O} ort>s.(e) H'(.O) C Ht(S?), C for a t.(f) In (lOa,b) k and m be reaL

(g) In (11) replace by 1ITUc.(m), with t > k if k t4.

Exercise 6.2.26. Check, the norm (16b), as well as (17b), that thecharacteristic function u(x) = I in u(x) = 0 for (xj > I belongs toIP(IR) if and only ifO � 8<1/2.

6.2.5 Trace and Extension Theorems

The nature of boundary value problems requires that one can form boundaryvalues (restriction of u to I' = 8$?, or trace of u on F = 8$?) in ameaningful way. As can be seen easily, a Holder-continuous function u Ehas a restriction "I r E if only F is sufficiently smooth. But, fromo H(1') does not follow E H(fl. Since the equality u = von only means that u(x) = v(x) almont everywhere in .0, and F is a setof measure zero, u(x) u(x) may hold everywhere on F. Also, the boundaryvalue 0(X) (x F) cannot be defined by a onntinuous extension either since,for example, o H'(.O) need not be continuous (cf. Exercise 4c).

The inverse problem for the definition of is extension: does thereexist, for a given boundary value on 1', a function 0 E If'($?) such thatço ulr? If the answer is negative there exists no solution u H(.0) to theDirichiet boundary value problem.

First we study these problems on the half-space

(6.2.18)Functions u will first be continued to Il E , and then zirestricted to F.

Theorem 6.2.27. Let a 0. There exists an extension operator Esuch that for aU u the extension

andu coincide on

PROOF. For s = 0, set = u on IR!. and = 0 otherwise. Since= I the mapping defined by := 11 is bounded:

= 1. For $ 1 define = by

11 = u on Ti(xi, . —xe) u(x1,.. . for x,, > 0

124 6 Tools from Functionai Analysi8

(i.e., by reflection on F). For s = 1 one obtains, for example,i.e., For larger s one uses higher interpolationformulae for .. , —xc) (ef. Exercise 9.1.13, cf. Wloka p. 101]). U

In the following the restriction is written in the form -yu. At first, 'yis defined only on

C (-yu)(x) := u(x) for all (x) E 1'.

(6.2.19)We write x = x' = (x1, . . . f' is identified with

H4(P) =

Theorem 6.2.28. Let a> 1/2. Then 'y from (19) can be continued toThus we have in particular l'vuIa_i,2 �

for u In the mae n = 1, i.e., -yu = u(0), we have I'yul � C,, lul..

PROOF. It suffices to show h'u1_i,2 � for tz (cf. Theorem6.1.11 and Theorem 25a). Let the Fourier transforms of u E andtu := -yu be = and i1 =Fourier transform). can be written as product where actson x' E and on x,. Therefore has theproperties = and tui = W(.,0). According to Lemma 20,

= W(e',O) = has the representation

= for all.e' (6.2.20)

Since u E := (1 + + lies in L2(IR") (cf.Lemma 23b). Inequality (5a) yields

= JThe first factor may be computed as I(.(1 + with K4 = +

dx < oo, since 8 > 1/2. The second is := IIU(e',because IIUIIL2RØ_I = (Fubin.i's theorem). Together

we have:

(1 + � for E

Integration over E results in

J(i +I � =

= i-


Thus jwI,_j/2 � is proved with = (cf. (16b)). In the

case n = 1, already represents = u(0), and the integration over is

not required. U

Theorem 28 describes the restriction 14(., 0) = -yu to = 0. Evidently,

similarly we have � for any other E 1R, with thesame constant C8. The mapping u(., is continuous tresp. Holder-continuous] in the following sense.

Theorem 6.2.29. For s> 1/2 the following statements hold:

— = 0 for all IR,u E

(6.2.21a)

— < — (6.2.21b)

for all u E 0 A < 1, A � s — 1/2 (and for A = 1 ifs > 3/2).

PROOF. (a) Let u E H8(IR") andset := The function ço,, is continuous in JR andconverges uniformly to since — u(,x)I,_112

— t4,, for all x JR. Thus (21a) follows.(b) + e) — has the Fourier transform =

= — so that = Asin the proof for Theorem 28 set := uui W(.,0). Thefirst integral in the eethnate of now reads

f(i + + e/2) = (1 + l/2—* J(i + t2)' sin2(i7t)

with ,, = + The decomposition of the last integral into subinte-grals over � and fri � 1/u showsThe remainder of the argument follows the same lines as in the proof of The-orem 28. U

Up to this point we have obtained H8(Wt") by completion ofin The next theorem shows that for suffithentiy large $ one can alsocomplete in fl L2(IR") so that contains only claseical func-tions (i.e. continuous, HOlder-continuous, [HOlder-J continuously differentiablefunctions).

Theorem 6.2.30. (Sobolev's lemma) c holds fork EIN Li {0}, s > k + n/2 and H'(IR") C for 0 < t IN, a t + n/2.

PROOF. (a) Let a t + n/2, 0 <t < 1, u E H'(lR"). For given x,yIRTh we want to show iu(x) — u(y)I � C independent of x,y.The coordinates of the 1R" can be rotated so that x = (x1, 0,. . . , 0), y =(yr, 0,.. , 0). An (n— 1)-fold application of Theorem 28 to tzQ, u(., 0), u(., 0,0)


etc., results in u(.,O,0, ,0) Theorem 29 provides thedesired estnnate with C = . . . � thusu and fl . 11c(R") � Cfl . Then u C°(IR') follows from

C(b) Let a t + n/2, 1 < t < 2, u E H(IR"). Part (a) is applicable to

(cf. Theorem 25d,e) with � 1: Dau E for1. Thus u etc.

We return to the statements of Theorems 27 and 28. For all u Ethe restriction = u(.,0) agrees with Completion in ylekis

�This proves the following corollary:

Corollary 6.2.31. Let a > 1/2. For the restriction -iu := u(.,0) we have'y E

With the restriction to = 0 one evidently loses half an order ofdifferentiability. Conversely one gains half an order if one continues w E

suitably in

Theorem 6.2.32. Let a> 1/2, w There eztsts a functionu (t& such that 1t4, � and = w, i.e.,w =

PROOF. Let = and zZ' = be the Fourier transforms. = w isequivalent to (20). For

= := ++ WI2 +

K. := J(i ÷

one checks that (20) and 1u1 = hold. Restriction of uto IR!L proves the parenthetical addition. U

If one replaces with ageneral domain 1? C then x{0} = becomes F 8!?, and the necessity arises of defining the Sobolevspace JP(fl. We begin with the

Definition 6.2.33. Let 0< t IRU{oo} Iresp. k e t4U{0}J. We write (1 Ct(Ii Ck.1j if for every x F := OS? there exists a neighbourhood U Csuch that there exists a bijective mapping —, K1(O) = E < 1}with


çoE Ct(TJ), ço1 E C'(K,(O))E Ck.t(U), (6.2.22a)

fl F) = K,(O): = 0), (6.2.22b)

ço(U n Si) O}, (6.2.22c)

çp(U fl (IRTh\Q)) = K,(0): <0}. (6.2.22d)

Here K,(0) is a circle (ball) if II

is the Euclidean norm. For the maximumnorm Ki(O) is a square (cube). Likewise, K,(0) can be replaced byanother circle KR(z) or any rectangle x x

Figure 6.2.1. Covering neighbourhoods of I' and I?

Example 6.2.34. Let 11 be the circle K,(0) C A neighbourhood off =(1,0) is U' from Fig. 1. The mapping XE U' i-s := E (—1,1) x (—1, 1)with (1 — Z2 = (1 — is bijective andsatisfies (22a-d) with t = oo. The same holds for any x E I'. Thus (1 C°°.

ExercIse 6.2.35. (a) The rectangle Si = (xi, x',') x and the L-domainfrom Example 2.1.4 are domains Si E C°".(b) The cut circle in Figure 5.2.lb does not belong to C°".

Lemma 6.2.36. Let 1? E Ct (5? e Ck.1) be a bounded domain. Then thereexsstsNElN, (1 �i<N) with

NUt open, bounded (0 � i � N), (J U° CC Si, (6.2.23a)

(1 i SN), (JUi =F (6.2.23b)

C bijectivcfori = 1,.. ,N, (6.2.23c)

E Ct(aj(Ut flU3)) (resp. orj' E flU,))I. (6.2.23d)

On W (1 i N) are defined mappings with the properties (22a-d).


PROOF. For every x E F there exist U = U(x) and cc = according toDefinition 33. Let be the restriction to U(x)flf'. Let (i €N) bethe open sets {x (1: dint (x, F) >1/i) Cc fl Let V' be anopen covering of the compact set Th. Therefore there exists a finite coveringthrough U' (1 i N) and at most one V' which is denoted byU0. If one sets U1 := U' n F, = the statements follow from(22a-d).

A set of pairs {(U,,aj): 1 � i N} fulfilling the conditions (23b.d) 18

called a (resp. coordinate system for F.In Example 34 one has N = 4. The map is given by =

E U1; and likewise = cos((3 + ej)ir/2),sin((3 + el)7r/2) e U4, where in each case —1 < < 1. On none obtains = + 1.

Lemma 6.2.37. (Partition of unity) Let � I � N} satisfy (23a).There exist ftinctions E

C U', = 1 for alt x E (6.2.24)

The general construction of the can be found, for example, in Wioka [1,§1.2). In the special case of Figure 1 one may proceed as follows. Let o-(t) :=Ofor ti � 1 and := exp(1/(12 — 1)) for t E (—.1,1). Then E andsupp(a) = [—1,11. In U' from Figure 1 one defines, for example,

:= u(2r — for x = etc.

These cj(x) satisfy (24).A function u on F can be written in the form Each summand

is parametrisable over C :a4(Ug) CC —. IR.

This in turn makes poasible

Definition 6.2.38. Let .0 E fE Assume and satisfy(23b-d) and (24). Let a< t E IN [a � k-fl) or 8< t N, t > 1. TheSobolev space H(F) is defined as the set of all functions U: F IR such that(u,u) E (1 � i � N).

Theorem 6.2.39. (a) H3(f') is a Hubert space with the sci'4ar product

(u, := (u,v)ff.(f) := o

(b) If 1 � � N} is another C'- coordinate system off and{&,) another partition of unity, then the space 11(F) defined &y this is equalto H(F) as a set. The norms of H'(F) and are equivalent.


PROOF. For (b): 'fransformation Theorem 17 Eresp. 25g) ; for I? C°", cf.

Wloka [1, Lemma 4.5).

The trace and extension theorems (Corollary 31 and Theorem 32) can be

extended to any domain with a sufficiently smooth boundary. ')' now denotes

the restriction to P 811: 7U = U Ii'.

Theorem6.2.40. LetQECwithl/2<8<_tEINOfl/2<8<tEreSP.1/2<s=k+IEINI.

(a) The restrtction -yu of u E belongs to y L(H8(Q),H8 — t/2( F)).(b) For each vi E there extsts an extension u e H'(11) with vi =

(c) For each w H'(Jl) there exists a continuation Ew E H8(IR') with

E E L(H8(J7),

PROOF. The proofs follow the same pattern. Let and czj be as inLemma 36. The suminand u E [resp. w8 = has itssupport in [resp. 114 and can be mapped via 40j (reap. onto lresp.

x {O} = There Corollary 31 and Theorem 32 hold. Therestriction to continuation to or can be mapped backagain. The first statement of the theorem is proved in detail as follows.Let := and ü4 := o On the grounds of Theorems 17, 25gthe function üj belongs to Thus the restriction 'y÷üj := Lies

in H8 /2(IR"') (cf. Corollary 31) and has c4(U1) as support. Set :=on =0 on f'\U1. According to Definition 38, vi :=

belongs to H8_h/2(fl. Since represents the restriction on (i one findsfor all u E (Ct(i?) is dense in H'(Q)):

w(x) =

E P. Since all the partial mappings are bounded, one finds that

I.7u1,l/2 = 1w15_i,2 � C3JuJ8. IRemark 6.2.41. (a) Under the conditions of Theorem 40 and the additionalcondition s > IaI + 1/2 there exists a restriction -1D°u E ofthe derivative of u E H8(Q).(b) For each uE with s<l+ l/2one haSyDau=0 ifla(�1.

Theorem 6.2.42. For .0 E C°" holds HJ($1) = {u H1(11): ufr = 0}.

PROOF. (a) u is the limit of v,, all of which satisfy0. Since uf1, with respect to H'/2(fl, it follows that u E

H

130 6 Tools from Analysis

(b) Conversely, let u H'(a) and = 0. The proof that u canbe divided up as follows:(ba) By using the partition of unity over {Ut: 0 � i � N} (cf. Lemma 37), thestatement reduces to the case I? =(bb) Without lose of generality one can amuine n = 1: Si =(be) According to Remark 14, for each > 0 there exists a Ewith finite support such that iii — ü41 q/2. For each 0 there exists

forz�,7,

� 3/ti, 0 � � 1.

The function := — (1 — satisfies E C00(11t÷), = 0,

supp(u€) finite, and lu — Thus X := {v 0,

supp(v) finite} is dense in {u E 111(ffLF):u(O) = O}.(bd) The statement would be proved if were also dense in X withrespect to Let v E X. Evidently, := E holds for all

> 0. Set 0(17) := ilv'I1L3(O,,7) and note that 0(77) 0 for , —' 0. Sincev(z) = for x � it remains to estimate the variables liv — VnhiLa(Oe)

and liv' — t4iIL2(o,,,). Because of v, — Vt = + (p,, — 1)v', one obtains

— v'lf � Ii llL0..(O,n)ilt)11L2(O,,7) +

� + 0(77).

xlSmce v(x) = v (5b) nnphes the estimate iv(x)i for all0 � x � and hence � Then the statement follows fromIv — v,,l? � liv — + liv — � C0(i1) 0. U

When Si C', the normal direction n exists at all boundary points.Analogously to Theorem 42 one proves

Corollary 6.2.43. For (1 E C' and k fI holds

H0k(a)={uE =0foro < <k — 1).

6.3 Dual Spaces

6.3.1 Dual Space of a Normed Space

Let X be a normed, linear space over JR. As a dual space, X' denotes thespace of all bounded, linear mappings of X onto IR:

L(X,IR).

6.3 Duai Spaces 131

According to Exercise 6.1.8, X' is a Banach space with the norm (dual norm)

IIx'IIx' : = zE X}. (6.3.1)

The elements x' X' are called linear functiouals on X. Instead of z'(x)(application of x' to x) one also writes (x, X')XXX', or (x', X)X'xX, and calls

(•, the dual form on X x X':

(X,X')XXX' = (x',x)x'xx x'(x).

Lemma 6.3.1. Let the Bariach spaceX be embedded densely and continuouslyin the Banach space Y. Then Y' is continuously embedded in X'.

PROOF. (a) Let y' E Y'. Because X C Y, y' is defined on X.(b) Since X is a dense subapace of Y, according to Theorem 6.1.11 and (1.5),for each y' E Y' holds:

Ity'IIy' = sup Iy'(x)I/(jxfly � sup Iy'(x)I/IIzfIx

i.e., Y' is embedded continuously in X'.

To the transposed matrix in the finite-dimensional case corresponds thedual mapping (or the dual operator).

Lemma 6.3.2. Let X and V be normed and let T E L(X,Y). For eachE Y'

(X,X')XXXI for all z E X (6.3.2)

defines a unique x' X'. The linear mapping y' x' deft nes the dual operatorT': Y' X' with T'y' = z'. The following holds: T' E L(Y', X') and

= flTfly'...x. (6.3.3)

PROOF. If one writes (2) in the form y'(Tx) = x'(x), one can see that= y' o T. (3) follows from the definitions of the norms:

= sup sup

= sup

IExample 6.3.3. Let SI = (0,1), X = (c°(.??),fl

. and x SI. Themapping E '-U' u(x) E IR is a functional: E (the so-called delta function). The Laplace operator LI to L(C2(Th),


The dual mapping LV is applicable to is char-acterised by = u E C2(fl).

ExercIse 6.3.4. Let S L(X, Y), T L(Y, Z). Show that (TS)' = S'T'.

Exercise 6.3.5. Show that if T L(X, Y) is surjective, then T' is injective.

6.3.2 Adjoint Operators

Let X be a Hubert space (over IR). Every YEX defined by

(x, Y)x

is a linear functional X' wIth IIxi ilylix. The converse also holds(cf. Yosida

Theorem 6.3.6. (Riesz representation theorem) Let X be a Hubertspace andfEX'. There existsauniqne yjEXsuchthat

f(s) = (x,yj)x for ails E X and ilfIix' liv,lix.

Corollary 6.3.7'. Let X be a Hilbert space.(a) There exists a one-to-one correspondence (the Riesz isomorphism)

E L(X,X') with JXJJ = = that preserves the norm, i.e.,liJxilx'-x IlJx 11x4-x' 1.

(b) X" is a Hilbert space with the scahir product (x',y')x' =, ,1/2The dual non's Iix lix' from (1) agrees with the norm induced by (x ,x

(c) One always identifies X with X" because x(x') := x'(x). Prom this follows= Jx = T" = TforT€ L(X,Y) ifY= Y" is also aHiibert

space.(d) One can identify X and X': X = X', .Jx = I.

Let X, Y be Hilbert spaces and T L(X, Y). The mapping defined byE L(Y, X) is called the operator adjoint toT and satisfies

= (x,Ty)x for allxE X, yE Y; = IITUxi-y.(6.3.4)

The adjoint and the dual operator only coincide (i.e., T' = T') if X' is iden-tified with X and Y' with Y.T€ L(X,X) issaidtobe selfadjoint (or symmetric) ifT=T.T E L(X,X) is called a projection if T2 = T. It is an orthogonalprojection if furthermore T is selfadjoint.

Remark 6.3.8. Let X0 be a closed subepace of the Hubert space X. Anorthogonal projection is given by Tx := y X0 with

6.3 Dual Spaces 133

liz — vilx = inf{iiz — Xo}. (6.3.5).

If conversely T L(X, X) is an orthogonal projection with the range Xo :={Tx: z E X} one has (5) for y Tx. An orthogonal projection always has thenorm IIT1IT_x � 1.

PROOF. (a) x can be decomposed uniquely into x = y+z (yE X0, z Xd)(cf. Lemma 6.1.17). y is the unique solution of (5). x X0 implies ythus T2 = T. The analogous decomposition x' y' + z' shows (x,T*XI) =

(Tx,x') = (y,x') = (y,y'+z') = (y,y') = (y+z,y') = (x,Tx'), henceT= T'.(b) Let T be an orthogonal projection with range Xo. Let x = y + z be splitas above. T2 T shows Ty = y. For each y' X0 holds (Tz, y') = (z, =(z,Ty') = (z,y') = 0, thus Tz E Xi-. Together with Tz E Xo follows Tz 0sothatTx=Ty+Tz—y.(c) Tx y and ilxti3c = + � iivlft prove IlTflx4_x � 1. U

6.3.3 Scales of Hilbert Spaces

We assume:

V C U are Hubert spaces with a continuous and dense embedding.(6.3.6)

Lemma 6.3.9. Under assumption (6) U' is embedded continuously anddensely in V'.

PROOF. The continuity of the embedding U' C V' was established in Lemma1. For proof that U' is dense in V', we use Exercise 6.1.18d (A := U',X := V').Let 0 v' V' be arbitrary and u := E V C U. According to thedefinition, u' := JUU U' C V' is characterised by u'(x) (x, U)U for allx€U. follows

(v',u')V' = = = u'(u) >0. U

According to Corollary 7d , U and U' can be identified. By this oneobtains the Gelfand triple

V c U C V' (V c U continuously and densely embedded). (6.3.7)

Corollary 6.3.10. In a Gelfand triple (7) V and U are also continuouslyarid densely embedded in V'.

PROOF. For U C V' see Lemma 9, for V C V' see Exercise 6.2.25.


Attention. Likewise one could identify V with V' and one would obtainU' c V' = V c U. But it is not possible to identify U with W and V with V'simultaneously. In the first case one interprets x(y) = (y, x)uxrj' for x, y E Uas x)u (in particular for x, y E V U), in the second case as (y, z)v.

Exercise 6.3.11. Let (7) be true. Set W := {Jç'u:u U} and define(x, y)w := (Jvx, Jvy)u as the scalar product on W. Show that(a) W is a Hubert space;(b) W C V is a continuous and dense embedding;(c) (v, w)v = (v, Jvvi)u for all v E V, w W;

(d) l(z,y)vI � HxIIuIlyIIw for all x,y E W.

Because U = U' the scalar product (z, y)u can also be written in the formy(x) = (x,y)uxu'. If z E V, then y(x) = (x,y)Vxv' also holds. That meansthat (x,y)u = (X,y)vxv' for all x V,71 E U C V'. Likewise one obtains(x, Y)u = (X,y)v'xv for all x U and y E V. The dense and continuousembedding U C V' proves

Remark 6.3.12. Let V C U C V' be a Gelfand triple. The continuousextension of the scalar product (., .)u to V x V' (V' x VJ results in the dualform (., •)v v' XVI. Therefore the following notation is practical

for x€V,y€V',(x,y)v'xv = (z,y)u for x E V',y E V.

In connection with Sobolev spaces one always chooses U := L2((1) sothat the embeddings read as follows:

H0($2) c L2(Q) c (a � 0), (6.3.8a)

H8($2) c L2(a) c (118(fl))' (a � 0). (6.3.8b)

ExercIse 6.3.13. Show that (8a) and (8b) are Gelfand triples.

The dual space of is also denoted by or

:= H3(J?) := (a � 0).

The norm of H(.Q) according to (1) reads:

:= sup{I(u,v)L2(fl)I/1v18:0 v for a � 0

where (u,v)L2(a) is the dual form on x H'(Q) (cf. Remark 12).

Remark 6.3.14. (a) Let (1 = The norm dual to

:= v €

6.4 Compact Operators 135

i8 equivalent to and has the representation (2.16b) with —s instead of s.(b) The Fourier transform shows E for all 8 IR.

(c) au E H8(Q), if u H8(fl), a C'(i7), where = Isi E IN U {O} or t > (s(.

6.4 Compact Operators

Definition 6.4.1. A subset K of a Banach space is said to be precompact[compact] if each sequence E K (i E IN) contains a convergent subse-quence fand E Kl.

Another definition of compactness reads: Each open covering of K al-ready contains a finite covering of K. Both definitions are equivalent in metricspaces (cf. Dieudooné (1, (3.16.1)]). For the terms "relatively compact" and"precompact" see also Dieudonné (1, (3.17.5)].

Remark 6.4.2. (a) K C is precompact [compactJ if and only if K isbounded (and complete].(b) Let X be a Banach space. The unit sphere {x X: flx(f 1} is compactif and only if dim(X) <co.

Definition 6.4.3. Let X and Y be Banach spaces. The mapping T L(X, Y)is said to be compact if {Tx: x X, Hxllx � 1}, the image of the unit spherein X, is precompact in Y.

Exercise 6.4.4. When is the identity I L(X, X) compact?

Lemma 6.4.5. Let X, Y, and Z be Banach spaces. (a) Let one of the map-pings T1 L(X, Y), T2 L(Y, Z) be compact. Then T2T1 L(X, Z) is alsocompact.(b) T e L(X, Y) is compact if and only if T' e L(Y', X') is compact.

PROOF. (a) Let K1 := {x X: jIz(Ix � 1}. If T1 is compact, i.e., T1(K1) isprecompact, then T2(T1(K1)) is also precompact and thus T2T1 is compact.If, however, T2 is compact, one proves the assertion as follows. Since scalingdoes not change compactness, � 1 can be assumed without lossof generality. Hence T1 (K1) is a subset of the unit sphere in Y and thereforeT2(T1(K1)) is precornpact. (b) cf. Yosida §XJ. I

A special type of compact mapping is a compact embedding:

Definition 6.4.6. Let X C Y be a continuous embedding. X is said tobe compactly embedded in Y if the inclusion I L(X,Y), Ix = x iscompact.


Together with Definitions I and 3 one obtains: X C Y is compactlyembedded if every sequence E X with � 1 contains a subsequenceconvergent in Y.

Example 6.4.7. Let I? be bounded. C is a compact embeddingfor s > 0.

PROOF. Functions E with <1 are equi-continuous anduniformly bounded. The assertion is based on the theorem of Arzelà-Ascoli(cf. Yosida [1, §111.3)).

Analogous results can be obtained for Sobolev spaces (ci Adams (1, p.144), Wloka 11, §7)):

Theorem 6.4.8. Let 5? C IRTh be open and bo'ande.i (a) The embeddingsC (s, t E s > t) are compact.

(b) Fhrther, leti? C°'1. The embeddings Hk(S?) C (k,l E NU TO),k> 1) are compact.(c) Let 0 <t < s and Si E C' (r > t,r> 1) or 5? E (k + 1 > t). Thenthe embedding 11(fl) C ie compact.

Remark 6.4.9. In Theorem 8b one can replace .0 E C°' by the "uniformcone property" (cf. Wioka [1, §2.1)). To ensure Si E it is su.fficient thatthe boundary OS? is piecewise smooth and the inside angles of possible cornersare smaller than

Indented corners (cf. Figure 2.1.1) are thus permitted while a cut domain(cf. Figure 5.2.lb) is excluded.

In Section 6.5 the following situation wdl arise:

V C U C V' Gelfand triple, T E L(V', V). (6.4.1)

Because of the continuous embeddings, T also belongs to L(V', V'), L(U, U),L(V, V), and L(U, V).

Theorem 6.4.10. Let (I) hold. Let V C U be a compact embedding. ThenT L(V',V'), T L(U,U), T L(V,V), T e L(V',U), andT L(U,V)ore compact.

PROOF. As an example, let us do T L(U, V). Since the inclusion IL(V, U) is compact we see I L(U, V') is also compact (cf. Lemma 5b).T L(U, V), as the product of the compact mapping I L(U, V') withT E L(V', V), is compact (cf. Lemma 5a). U

Exercise 6.4.11. Let dirnX <cc or dimY <cc. Show that T L(X,Y) iscompact.


The significance of compact operators T L(X, X) lies in the fact thatthe equation Tx — Ax = y (with z,y E X, y given, x sought) has propertiesanalogous to the finite-dimensional case.

Theorem 6.4.12. (Riesz-Schauder theory) Let X be a Banach space;let T E L(X,X) be compact.(a) For each A (C\{O} one of the following alternatives holds:

(i) (T — AI)' L(X,X) or (ii) A is an eiger&value.

In case (i) the equation Tx — Ax y has a unique solution x E X for ally E X. In case (ii) there exists a finite-dimensional eigenspace T) :=kernel (T — Al) {O}. All x E(A,T)\{O} solve the eigenvalue problemTx = Ax, x 0.(b) The spectrum o'(T) of T consists by definition of all and, if not

e L(X, X), A = 0. There exist at most countably many eigenvalues whichcan only accumulate at zero. A E a(T) if and only if A E o(T').dim(E(A,T)) = dim(E(5,T')) <oo.(c) For A a(T)\{0}, Tx — Ax y has at least one solution x E X if andonly if (y,Z')Xxx' = 0/or all x' E

In Lemma 6.5.18 we need

Lemma 6.4.13. Let X C Y C Z be continuously embedded Banach spacesand let X C V be compactly embedded. Then for every e > 0 there exists asuch that

flxIIy dUxIIx + C€IIxIIz for all z X. (6.4.2)

PROOF. Let e > 0 be fixed. The negation of (2) reads: There exists Xwith (llxiJIy — eIlx4Ix)/IJxdIz 00. For := X we thushave — ')/IIv4Iz —. oo. From this one infers UydIz 0 and flysfIy > 1for sufficiently large i. Since � 1/c and X C Y is a compact embedding,a subsequence ye,, converges to y' V. Now, J)y1JIy > 1 implies Jjy*Jjy � 1,i.e., y 0. On the other hand, also converges in Z to y since Y C Z iscontinuously embedded. —+ 0 gives the contradiction sought: y = 0.

6.5 Bilinear Forms

In the following let us assume that V is a Hubert space. The mappinga(•, .): V x V —* Ut is called a bilinear form if

a(x, y + Az) = a(x, y) + Aa(x, z), a(x + Ay, z) = a(x, z) + Aa(y, z)


for (x, y, z V, A E JR) (in the complex case one speaks of sesquilinear forms:a(x, Ày) = a(.,.) is said to be continuous (or bounded) if thereexists a C3 such that

ja(x,y)I CsIIxJIvI(yIIv for all x,y E V. (6.5.1)

Lemma 6.5.1. (a) To a continuous bilinear form one can assign a uniqueoperator A L(V, V') such that

a(x,y) = for all x,y E V, � C5. (6.5.2)

(b) Let V, and V2 be dense in V. Let a(',.) be defined on V1 x V2 and satisfy(1) with "x E Vj,y E V2" instead of "x,y V'. Then can be extendeduniquely to V x V 80 that (1) holds with the same Cg for all x,y E V.

PROOF. (a) Let x E V be fixed. := a(x,y) defines a functional c°x E V'with IIcPxIIv' � Cs(Ixflv. One sets Ax := çox for x V. hAzily' � Cshlxilvproves l(AiIv.._y � Cs. The definitions show (AX,y)VIXV = (coz,y)v'xv =

= a(x,y). Conversely, for each A L(V,V'), a(x,y) :=(Ax, y)v'xy � hIAxIlvslIyhiy �

(b) According to Theorem 6.1.11, A is also uniquely determined if a(., is onlygiven on V1 x V2. Then (Ax, y)v' v represents the extension.

The proof shows

ilAilv'_v =sup{la(x,y)i:x,yE V,Uxilv = iiyhiv = 1}. (6.5.3)

A is called the operator that is associated to a(.,.). Thebilinear form a(.,.)adjoint to a(.,.) is given by a*(x, y) := a(y,x) (x, y V). The bilinear formis said to be symmetric if a(.,•) = a*(., .).

Exercise 6.5.2. Show that (a) If A belongs to a(.,.), then A' belongs toa(.,.). (b) If a(.,.) is symmetric, then A = A'.

Lemma 6.5.3. Let A E L(V, V') be the operator associated to a continuousbilinear form a(.,.). Then the following statements (i), (ii), (ui) are equivalent:(I) A' E L(V',V)(ii) e, e' > 0 exist such that

inf{sup{ia(x,y)l:yE = 1}:xE V,hlxiiv = 1}=e >0, (6.5.4a)

E = l}:y liyilv 1} = e' >0; (6.5.4b)

(iii) the inequalities (4a) and (4c) hold:

sup{la(x,y)i:xe V,lJxhlv = 1} >0. (6.5.4c)

If one of the statements (i)—(iii) holds, then


C 1/11A'IIv.._vi (e,e' from (4a,b)). (6.5.4d)

Prom (4a) follows

inf{sup{Ia(x,y)I:y E V,flyIlv = 1}:x l',IIxIIv = 1} e >0. (6.5.4e)

Conversely, (4a) follows from (4e) wsth a possibly larger e > 0. (4e) and (4c)

are also called the conditions. (4e) is equivalent to

sup{Ia(x, y)I: y V, lIyItv = 1) � cHxIIv for all x V, (6.5.4e')

because (4e) is equal to (4e') for all x V, IIxIIv = 1. The scaling condition

tIxIIv = 1 can emdently be dropped. The left-hand side in (4e') agrees with thedefinition of the dual norm of Ax so that (4e) and (4e') ore also equivalent to(4e"):

hAzily' � for all x V. (6.5.4e")

PROOF. (a) "(i) (ii)": Let L(V',V) exist. Then (4a) follows from

mf{ } = inf Ia(x,y)I_.

i(Ax,y)laEV Uxllviiyiiv iixilvhiyhlv

— . I(AA'x',y)i — c II A1 lu—i i(x',y)I— mf sup — iv' sup

p€V IIA'ziivliviIv i,€v hIyHv

= = lIE sup 11A'x'ilv/hlx'flv'lCV

= C.

In the same way one shows (4b) with e' = Because A'' =(A')', (3.3), and V" V, it follows that e =(b) "(ii) (iii)": (4c) is a weakening of (4b).(c) "(iii) (i)": e > 0 in (4a) proves that A is injective. We wish to showthat the image W := {Ax: x V} C V' is dosed. For a sequence withI w — w the

W there exists zI, E V with = w1,. From (4a) one infersvia (4e) and (4e") (with x := — that — � flwg, — WpiIv'/f.Since is Cauchy convergent, this property carries over to Thereexists an x' E V with —. f in V. The continuity of A E L(V, V') proves

= Ax,, —+ Ax' so that w' = Ax' E W. According to Lemma 6.1.17 onecan decompose V' into Wt. If A were not surjective (thus W V'), therewould exist a w W1 with w 0. Then y Jv'w = E V would satisfyy 0 (cf. Theorem 6.3.6, Corollary 6.3.7). Since a(z,y) = (Ax,y)v')<v =(Ax, w)v' = 0 for all x V, a contradiction to (4c) would result. Therefore,A is also surjective, and Theorem 6.1.13 proves A' L(V',V).(d) Statement (4d) has already resulted from Part a of the proof. R


It will be shown that for interesting cases conditions (4a) and (4b) areequivalent (cf. Lemma 17). A particularly simple case is in

Exercise 6.5.4. Show that if dim V < 00, then (4a) implies the statement(4b) with e' = e and conversely.

Definition 6.5.5. A bilinear form is said to be V-elliptic if it is continuouson V x V and there is a constant CE such that

a(x,x) � for all x V. (6.5.5)

Exercise 6.5.6. Show that: (a) If W C V is a Hubert subspace with the normequal (or equivalent) to V, then a V-elliptic bilinear form is also W-elliptic.(b) Let a(., .): V x V IR be continuous. If � holds for allx E V0 where Vo is dense in V, then (5) follows with the same CE.(c) Let a(•, .): V x V IR be continuous, symmetric, nonnegative (a(v, v) � 0)and satisfy (4a,c). Then a(.,.) is V-elliptic with � e2/Cg (e from (4a,b),Cs from (1)). Hint: First prove that Ia(u,v)1 � (a(u,u)a(v,v)11/2 (cf. Exercise6.1.16).

Lemma 6.5.7. V-ellipticity (5) implies (I) and (4a,b) with e = e' CE.

PROOF. Lets E V, IIxIIv = 1. V, IIyIIv = 1} � Ia(x,x)I �CE proves (4a) with e CE. (4b) follows analogously. U

The combination of Lemmata 1, 3, 7 together with IIA''tIv,_viIIA1Hv_v' (cf. Lemma 6.3.2) proves

Theorem 6.5.8. Let the bilinear form be V-elliptic [or satisj*j (1), (4a,c)J.Then the corresponding operator A 8atIsfies the conditions

A L(V, V')1 IIA'Iiv"_v = IIAIIv'÷_v � Cs,A1 L(V', V), IIA''llv._v' = 11A'IIv._vi � C'

(6.5.6)

with from (1) and C' = 1/CE fre.sp. C' = 1/c = 1/c'].

With the help of the bilinear form a(.,.) and a functional I V' one canformulate the following problem:

Find x E V with a(x,y) = 1(y) for all y V. (6.5.7)

According to Lemma 1 one can write (7) in the form (Ax—f, y)V'xv = 0

if and only if A' E L(V', V). Hence one obtains

Theorem 6.5.9. Let the bilinear form be continuous (cf. (1)) and satisfy thestability condition (4a,c) [it is sufficient that a(.,.) is V-ellipti4 Then Problem


(7) has exactly one solution x A'f. This satisfies � with

C= 1/e= ife' [resp. C= 1/CEI.

Corollary 6.5.10. Under the assumptions of Theorem 9 the analogous state-ment holds with the same estimate for the adjoint problem

Find xEV with a'(x,y)=f(y) broil yEV. (6.5.8)

PROOF. (cf. Exercise 2a, Theorem 8). U

Exercise 6.5.11. Let a(., .): V xV Lit be continuous. Let V0 be dense in V.Show that the solution x E V of problem (7) is already uniquely determinedby "a(x, y) = f(y) for all y E V0". The same holds for (8).

Problem (7) may be equivalent to a variational problem:

Theorem 6.5.12. Let a(.,.) be V-elliptic and symmetric; furthennore, letf E V'. Then

J(x) := a(x, x) — 2f(x) (x V) (6.5.9)

assumes its unique minimum for the solution x of Equation (7).

PROOF. Let x be the solution of (7). For arbitrary z E V set y := z — x.From

J(z) = J(z + y) = a(x + y, x + y) —2f(x + y)a(x, x) + a(x, y) + a(y, x) + a(y, y) — 2/(x) — 2f(y)

= J(x) + a(y, y) + 2ja(x, y) — 1(y))

J(x) + a(y, y) � .1(x) + = .1(x) + CEIIZ

one can read J(z) > J(x) for all z x. U

The term "V-elliptic" seems to indicate that to elliptic boundary valueproblems correspond V-elliptic bilinear forms. In general this is not the case.Rather, V-coercive forms will be assigned to the elliptic boundary value prob-lems. Their definition necessitates the introduction of a Gelfand triple (cf.(6.3.7)):

V C U C V' (U = U', V C U continuous and densely embedded).

Definition 6.5.13. Let V C U C V' be a Gelfand triple. A bilinear formis said to be V-coercive if it is continuous and if there exists oK lR

and CE > 0 such that

a(x,x) � for all xE V with Ck >0. (6.5.10)


Exercise 6.5.14. Set a(x,y) a(x,y) +CK(x,y)u with CK from (10). LetI: V V' be the inclusion. Show that (a) the coercivity condition (10) isequivalent to the V-ellipticity of a. -(b) If A L(V, V') is associated to a(.,.), then so is A A + CkI to a(.,.).Why does A L(V, V') hold?

The results of Riesz-Schauder theory (Theorem 6.4.12) transfer to A as soonas the embedding V C U is not only continuous but also compact.

Theorem 6.5.15. Let V C U C V' be a Gel/and tr*ple with compact em-bedding V C U. Let the bilinear form a(.,.) be V-coercive with correspond:ngoperator A. Let I: V V' be the inclusion.(a) For each A E C one of the foil owing alternatives holds:

(i) (A—Xf)1 €L(V',V) and (A'—Xiy' EL(V',V),(ii) A is an eigenvo.lue.

in case (i) Ax — Ax = f and A'x' — = f are uniquely solvable for allI E V' (i.e., a(x, y) — A(x, Y)u = 1(y) and a(x',y) v)u = f(y) for ally V). in case (ii) the,e exist finite-dimensional eigenspaces {0} E(A) :=kernel (A — Al) and {0} E'(A) kernel (A' — such that

Ax = Ax for x E(A), i.e., a(z,y) = A(x,y)u for all y V,

(6.5. lie.)

for x*EEI(A), i.e., a*(x*,y)=X(x*,y)u forall y€V.(6.5.1 ib)

(b) The spectrnm o'(A) of A consists of at most countably many eigenvalueswhich cannot accumulate in C. A a(A) if and only if A o'(A'). FhrthermoredimE(A) = <00.(c) For A E r(A), Ax — Ax = f E V' has at least one solution x E V if andonly f I E'(A), i.e., (f,x)v = 0/or all x' E E'(A).

PROOF. With V C U, V C V' is also a compact embedding, i.e., theinclusion I: V —+ V' is compact. A + CKI with from (10) satisfiesA + CKI E L(V, V'), (A + CKI)1 e L(V', V) (cf. Exercise 14 a). Lemma6.5.4 shows that K := (A + V —+ V is compact. Hence the Riesz-Schauder theory is applicable to K — jsl. Since

K — id = —j41 — = + CxIY' {A + —

= —Al)

with A = 1//h — CK, the statements of Theorem 6.4.12 transfer via K —I


Remark 6.5.18. The spectrum has measure zero so that under theconditions of Theorem 15 the solvability of Ax — Ax = f is guaranteed foralmost all A. ProbLem (7) is solvable if not "accidentally" 0 E c(A).

Lemma 6.5.17. Under the conditions of Theorem 15 the inequalities (4a,b)

are equivalent.

PROOF. (4a) proves that A is injective, i.e., 0 (1(A). Theorem iSa showsA' L(V', V) so that (4b) follows from Lemma 3. I

Evidently a(.,.) remains V-coercive if one adds a multiple of (., Gen-

erally, there holds

Lemma 6.5.18. Let a(.,.) be V-coercive where V C U C V'. Then a(.,.) +b(.,.) is also V-coercive if the bilinearformb(.,.) satisfies one of the followingconditions: (a) for every e > 0 exists C( such that

jb(x, x)f for all x E V. (6.5.12a)

(b) Let the embeddings V c X and V C Y be continuous, with at Least one ofthem compact. Let the following hold:

Ib(x,x)I for all x E V. (6.5.12b)

(c) Let the embeddings V C X, V C Y be continuous. Let (12b) hold. Foror Uv assume that for every e >0 there exists a such that

iS ÷ or � eIlxIIv + for x V.

(6.5. 12c)

PROOF. (a) Select e = CE/2 with CE from (10). Then a(•, .)+b(.,.) satisfiesthe V-coercivity condition with CE/2 > 0 and CK + instead of CE andCK.(b) Lemma 6.4.13 proves (12c).(c) Let the first inequality from (12c) hold, for example. Since the em-bedding V C Y is continuous, Cy exists with � CyflxlIv. ChooseC' = in (12c):

5 � +

K Since � I

7 Variational Formulation

'7.1 Historical Remarks

In the preceding chapters it was not possible to establish even for the Dirichietproblem of the potential equation (2.1.la,b) whether, or under what condi-tions, a classical solution u C2(Q) n C°(fl) exists. Green took the viewthat his Green's function, described in 1828, always exists and that It providesthe solution explicitly. This is not the case. Lebesgue proved in 1913 that forcertain domains the Green function does not exist.

Thomson (1847), Kelvin (1847), and Dirichiet offered a different line ofreasoning. The Dirichiet integral

J = J (7.1.1)

describes the energy in physics. With boundary values u = on I' given, oneseeks to minImise 1(u). This variational problem is equivalent to

I(u,v) := =0 for all v with v = 0 on (7.1.2)

The proof of the equivalence results from 1(u+v) = I(u)+21(u,v)+I(v)and 1(v) � 0 for all v (cf. Theorem 6.5.12). Green's formula (2.2.5*) provides

I(u, v) = f0 v/lu dx =0 for all v with v =0 on F such that /lu =0 follows.Thus, like (2), the variational problem 1(u) = miii is equivalent to the Dirichietproblem =0 in a, u = on I'.

The so-called Dirichiet principle states that 1(u), since it is boundedfrom below by 1(u) � 0, must take a minimum for sonie u. According tothe above considerations this would ensure the existence of a solution of the

Dirithlet problem.In 1870, WeierstraB argued against this line of reasoning, stating that

while there may exist an inflinum of 1(u) over {u E fl =on r} it need not necessarily be in this set. Fur example, the Integral

J(u) := fu2(x)dx in {u C°U0,11):u(0) = 0,u(1) = 1} never takes thevalue inf J(u) = 0.

Further, the following example due to Hadainard shows that no finiteinfirnum of the Dirichiet integral need exist. Let r and be the polar coordi-nates In the circle Q K1(0). The function =

in .0 but the integral 1(u) does not exist.

7.2 Equations with Homogeneous Dirichiet Boundary Conditions 145

The above difficulties disappear if one seeks the solutions in the moresuitable Sobolev spaces instead of in C2(Il) fl C°(17).

7.2 Equations with Homogeneous Dirichiet BoundaryConditions

In the following we investigate the elliptic equation

Lt.i = g in Si, (7.2.la)

L= (7.2.lb)PI�m

of order 2m (ef. Section 5.3; Exercise 5.3.5d). The principal part of L is

= (7.2.2)

According to Definition 5.3.4, L is uniformly elliptic in Si if there exists 0

such that

� for all x E Si, E IR". (7.2.3)

Attention. In the case that only is assumed, one needs toreplace "for all XE Si" by "for almost all x Si".

We assume the homogeneous Dirichiet boundary conditions

(7.2.4)

which are only meaningful if F = is sufficiently smooth. Note that in thestandard case m = 1 (an equation of second order) condition (4) becomesU =0.

Since with u = 0 on F the tangential derivatives also vanish, not onlythe kth normal derivatives (k � m — 1) but also all the derivatives of order$ m —1 are equal to zero:

inxEf forlal<m—1. (7.2.4')

Condition (4') no longer requires the existence of a normal direction.According to Corollary 6.2.43, (4') can also be fonnulated as

(7.2.4")

Let it E be a classical solution of(la) and (4). To derivethe variational formulation we take an arbitrary v and consider

146 7 VariatIonal fbrmulation

(Lu, v)0 =

E C8°(.O), the integrand vanishes in the proximity of I' so thatone can integrate by parts:

Dau)dx

without boundary terms occurring. Thus we have found the variational for-mulation

f dx= J g(x)v(x) dx (7.2.5)

IaI,I$I�m a

for all v

since Lu = g. If conversely a with boundary conditions (4)satisfies condition (5), then the partial integration can be reversed andf0(g — Lu)vdx = 0 for all v E Cr(a) proves Lu = g. This means that aclassical solution of the variational problem (5) with boundary condition (4)is also a solution of the original boundary value problem. Hence the differ-ential equation (la,b) and the variational formulation (5) are eqwvalent withrespect to classical solutions.

We introduce the bilinear form

a(u, v) := J (7.2.6)

and the functional

/(v) J g(x)v(x)dx. (7.2.7)a

As remarked above, the boundary condition (4) for classical solutions umeans that a E Thus the "variational formulation"

or "weak formulation" of the boundary value problem (1), (4) reads as follows:

find a with a(u,v) 1(v) for all v E (7.2.8)

A solution of problem (8) which, according to the definition, lies inbut not necessarily in is called a weak solution.

ExercIse 7.2.1. (a) Let 0 be bounded. Show that any claasical solutiona E fl is also a weak solution.(b) With the aid of Example 2.4.2 show that this statement becomes false forunbounded domains.

Theorem 7.2.2. Let E The bilinear form defined by (6) sebOunded on x

7.2 Equations with Homogeneous Dirichiet Boundary Conditions 147

PROOF. Let u,v E The inequality (6.2.Sc) yields

� � const

Since is dense in (ci. Theorem 6.2.10), a(.,.) has an extensionto Hr(.O) x and is bounded by the same constant (cf. Lemma 6.5.lb).

U

The function f(v) is also defined and bounded for v E if, forexample, 9 E L2 ((1). According to Exercise 6.5.11, the variational formulation(8) is equivalent to the following one:

find u with a(u,v) = f(v) for all v E (7.2.9)

One can regain the form Lu = / by applying Lemma 6.5.1. Let L EH—m(a)) and f E Hm(12) = be defined by a(u,v) =

and 1(v) (f,v)ff-.,.(a)XH.,,(a) for all v EEquation (9) states that

Lu—f. (7.2.9')

While (la) represents an equation Lu g in C°(I2) (i.e., for a classical solu-tion), (9') is an equation in Hm(fl).

Theorem 6.5.9 guarantees unique solvability of Equation (9) if a(.,)is We first investigate the standard case m = 1 (equations oforder 2tn = 2).

Theorem 7.2.3. Let 12 be bounded, m 1, aafi L°°(12). Let L (3)(uniform elhpticity) and be equal to the principal part L0, i.e., aap = 0 for

+ )i31 � 1. Then the form a(.,.)

a(u, u) � e' > 0. (7.2.10)

PROOF. Since = = 1 one can identify a and /3 according to == 8/Ox, with indices i,j {1, . .• ,n}. Fbr fixed x 12 use (3)

with =

= E � =

Integration over $2 yields a(u, u) � 1Vu12 dx. Since dx �(cf. Lemma 6.2.11), (10) follows with e' =cC0. U

Corollary 7.2.4. The condi*ion "12 bounded" may be dropped s/for a =

Example 7.2.5. The Helmholtz equation + u = f in 1? leads tothe bilinear form

148 7 Variational Fbrmulatlon

a(u, v) liE (x) + u(x)v(x) j dx = ff(Vu, Vv) + uvj dx.

a(ts,v) is the scalar product tn Ha(S2) (and H'(S?)). The fact that a(u, u) =proves the (S?)-ellipticity.

Exercise 7.2.6. Let the assumptions of Theorem 3 or Corollary 4 be satisfied,except for the fact that the coefficients aao and ao$ = = 1) of thefirst derivatives are arbitrary constants. Show that inequality (10) holdsunchanged.

Theorem 3 cannot easily be extended to the case m> 1.

Theorem 7.2.7. Let the coefficients of the principal part be constant8: 0afl =constforlaj= m. assume that =Oforo < �2m —1, aoo � 0 for a = f3 =0. Let L be uniformly elliptic (cf. (3)). Furtherlet either £2 be bounded or aoo � >0. Then a(.,.) is

PROOF. We continue u through u =0 onto Theorem 6.2.21,Exercise 6.2.22, and inequality (3) show that

a(u,u)-Jaoou2dx= >2 JaafiD0'uD6udxa (aI,(PI=m

= >2 E

= >2

= 11 >2

� efLet aoo � > 0. There exists an e' > 0, so that � e' E1cz1<m fri2for all E From this follows � — and� (cf. Lemma 6.2.23). IfS? is bounded, use Lemma 6.2.11. I

Having shown the of the form a(., •), we are now ableto apply the general proofs after Theorem 6.5.9.

Theorem 7.2.8. (Existence and uniqueness of weak solution8)Ifa(.,.) is Hr(fl)-eflipgic then there exists a solution u E of Problem(9) which satisfies

iUim � (CR (6.5.5)). (7.2.11)

7.2 EquatIons with Homogeneous Dirichiet Boundary Conditions 149

Since (11) holds for all I Hm(.0) and u = (cf (9')), inequaLity (11)is equivalent to

<C := 1/CE. (7.2.11')

The term "variational problem" for (9) goes back to the following state-ment, inferred from Theorem 6.5.12:

Theorem 7.2.9. Let a(.,.) be an and symmetric bilinearform. Then (9) is equivalent to the variational problem

find u E such that J(u) � J(v) for all v E(7.2.12a)

where

J(v) := —1(v). (7.2.12b)

Attention. If a(.,.) is either not or not symmetric, Problem(9) remains meaningful although the does not minimize the functionalJ(u).

Example 7.2.10. The Poisson equation —i.lu = f in S7, u 0 onF, leads to a(u,v) = f0(Vu(x), Vv(x))dx. For a bounded domain .0, a(.,.)is HJ(Q)-elliptic (cf. Theorem 3), so that for any / E there existsexactly one (weak) solution u E H01(Q) of the Poisson equation. This is alsothe solution of the variational problem f0 fVul2dx — f(u) = miii.

A weaker condition than is thea(u,u) � —

Theorem 7.2.11. Letm = 1, and let the coefficients satisfycondition (3) of uniform ellipticity. Then a(.,.) is

PROOF. We write L as L L1 + L11, with Lj satisfying the conditions ofTheorem 3, resp. Corollary 4 ifS? is not bounded, and L,1 containing onlyderivatives of order 1. Then we can apply the following lemma. U

Lemma 7.2.12. Let a(,.) a'(., + a"(.,.) be decompose4 such that a'(.,.)is or perhaps only

with aap E contains only derivatives of order 2m — 1. Then a(.,.)is also

150 7 Variational Formulation

PROOF. (6.5.12c) follows from (6.2.lOb) with X = Y =V = and U = L2(f1), so that Lemma 6.5.18c proves the aseertion. •

The generalisation of Theorem 11 to arbitrary m � 1 requires strongerconditions on the coefficients of the principal part.

Theorem 7.2.13. (Carding's theorem) Let L be uniformly elliptic (cf.(3)) and assume E L°°(a). let the coefficients a,,p withtat = m be uniformly continuous in fl. Then a(.,.) isIf conversely aa$ E C(s?) holds for tat = 1131 = m and E L°°(12) otherwise,then from the follows uniform etlipticity (3).

Details of the proof can be found in Wioka 11, Theorem 19.21. The proofgiven there also holds for unbounded S?, since the coefficients are uniformlycontinuous. For the first part of the theorem one uses a partition of unity.

The significance of coercivity lies in the following statement.

Theorem 7.2.14. Let (1 be bounded and a(.,.) be Thenone of the following alternatives holds:(i) Problem (9) has exactly one (weak) solution u E(ii) The kernels E = kernel(L) and E = kemel(L') are k-dimensional for ak IN, i.e.,

a(e,v)=G, a(v,e*)=O

Further, the eigenualue problem

a(e,v) = A(e,v)L2(O) for all v E (7.2.13)

has countably many eigenvalties which do not accumulate in C.

PROOF. Since for bounded I? the embedding V := C U := L2(Q) iscompact (cf. Theorem 6.4.8b), Theorem 6.5.15 is applicable. U

7.3 Inhomogeneous Dirichiet Boundary Conditions

Next, we consider the boundary value problem

Lu = g in fl, u = w on 1, (7.3.1)

where L is a differential operator of second order (i.e., m = 1). The variationalformulation of the boundary value problem reads:

find u E H'(fl) with u on f' such that (7.3.2a)

a(u,v) = f(v) for all v E (7.3.2b)

7.3 Inhomogeneous Dirichiet Boundary Conditions 151

But according to Section 6.2.5 the restriction uk of is IP (i1) on I' iswell defined as a function in H1/2(F). Thus "is = coon F" must be understoodas the equality = in H1/2(I'). In contrast to the preceding section oneuses a(.,.) in (2b) as a bilinear form on H1(fl) x Hd(S?). It is easy to see thata(.,.) is well-defined and bounded on this product.

Remark 7.3.1. For the solvability of Problem (2a,b) it is necessary that:

there exists a iso H'(i7) with isoll- (7.3.3)

If a function iso with Property (3) is known, a second characterisation of theweak solution results:

Let iso satisfy (3); find w that (7.3.4a)

a(iu,v) = f'(v) := 1(v) — a(uo,v) for all v (7.3.4b)

Remark 73.2. The variational problems (2a,b) and (4a,b) are equivalent. Ifiso and w are the solutions of (4a,b), then is = iso + w is a solution of Problem(2a,b). is is a solution of (2a,b), then, for example, iso = u and w = 0 satisfyProblem (4a,b).

Exercise 7.3.3. Show that f' H'(fl) for f' from (4b) and

Ill—i � Ill—i + (7.3.5)

with C5 from ja(u,v)l � CsIuljIv(t (cf. (6.5.1)).

Remark 7.3.4. The problem (1) and the variational fonnulation (2a,b) havethe same classical solutions if such exist.

PROOF. It suffices to assume v in (2b). Integration by parts canbe carried out as in Section 7.2 and proves the assertion. U

Theorem 7.3.5. (Existence and uniqueness). Let problem (2.9) (withhomogeneous boundary values) be uniquely solvable for all f H1(Q). ThenCondition (3) is sufficient, and necessary, for the unique solvability of Problem

PROOF. If there exists a solution is E H'(.Q) of (2a,b) then (3) is satisfied.However, if (3) holds, one obtains via (4a,b) a unique solution since (4b) agreeswith (2.9). U

Remark 7.3.6. Under the condition 0 C°", (3) is equivalent to

E II 112(f). (7.3.6)

152 7 Variational

PROOF. If vo satisfies Condition (3) then Theorem 6.2.40 shows that E

H'/2(F). If conversely 'p H'/2(f), then the same theorem guarantees anextension E H1(Q) to a with uoIj' = 'p and

luoli 5 (7.3.7)

U

Let inequality (2.11') hold in the case of homogeneous boundary values.Equation (5) shows Iuui S lt'oli + Iwli S Itioli + (Ill—i + C'IuoIj)/€ for thesolution of Problem (4a,b). The estimate (7) proves

Theorem 7.3.7. Let a C°". Let the bilinear form be restricted to H'(a)xand let it satisfy (2.11'). Then for every f H1(Q) and 'p H'12(f)

there exists exactly one solution u H'(a) of Problem (2a,b) with

< + (7.3.8)

ExercIse 7.3.8. Let a(.,.) be symmetric and HJ(Q)-elliptic. Show thatProblem (2a,b) with / = 0 is equivalent to the variational problem: Findti H'(S?) with tip = 'p such that a(is,u) becomes minimal (cf. (1.1)).

7.4 Natural Boundary Conditions

The bilinear form a(.,•) defined in (2.6) is also well-defined on Hm(S?) xHm(Q). In analogy to Theorem 7.2.2 there holds

Theorem 7.4.1. Assume E The bilinear form defined by (2.6)is bounded on Hm(Q) x Hm(Q): ia(u,v)I S Eap 113c*$Ik.o(O)ItImIVIm 101 allu,v E Hm(O).

Now let / be a functional from (Hm(.O))'. Equation (2.7) with gfor example, describes such a functional; but (2.7) is only a special case of thefunctional f subsequently defined in (la), which we want to use as a foundationin the following.

Exercise 7.4.2. Let r be sufficiently smooth and let g L2(a), L2(r)hold. Show that

1(v) := J g(x)v(x) dx + j 'p(x)v(x) dl', v lf'(Q), (7.4.la)

defines a functional in (H'(Q))' with ill Ii(H'(ci))' S L2(D) + UcoIiL2(r)).This implies f E (Hm(a))' for all m> 1. More precisely, the following alsoholds:


� (II))' + IkoItH—1/2(Ii). (7.4. ib)

Frequently, variational problems with a physical background have the form:

find u€Hm(A2), such that (7.4.2a)

a(u,v) = 1(v) for all v Hm(fl). (7.4.2b)

In contrast to the condition u E from Section 7.2, u Hm(47)contains no boundary condition. Nevertheless, Problem (2a,b) has a uniquesolution if a(•,•) is Hm(.O)-elliptic. This condition is easy to satisfy.

Theorem 7.4.3. Under the conditions of Theorem 7.2.3 or 7.2.4a(-,) is H'(fl)-elliptsc: a(u,u) for all u E H1(i2). Problem (2a,b)(with m = 1) has exactly one solution which satisfies the estimate (3):

It&li � (7.4.3)

PROOF. The same as for Theorem 7.2.3 or Corollary 7.2.4, and Theorem7.2.8. U

Corollary 7.4.4. (a) A unique solution which satisfies the estimate (3), alsoexists if instead of H1(12)-ellspticity one assumes: a(.,.) is H'(i2)-coercive,O C°" is bounded, A = 0 is not an eigenvaltie (i.e., a(u, v) 0 for allyE H'(Q) imphesu=O).(b) The combination of inequalities (3) and (ib) results in

luli � + IIWIIff—1/2(r)J (7.4.3')

for the solution of (2a,b), if f is defined by (is) with g (H'(Q))',H'/2(f).

PROOF. (a). According to Theorem 6.4.8b, H'(O) is compactly embeddedin L2(37) so that the statement of Theorem 7.2.14 can be transferred. If A = 0is not an eigenvalue then L1 L((H1(a))', H'(Q)) holds (cf. Theorem6.4.12). U

To find out which classical boundary value problem corresponds to thevariational formuiation (2a,b), we assume that (2a,b) has a classical solutionu E Hm(Q) 11 Further, v E can be assumed also (cf. Lemma6.5.ib). For reasons of simplicity we limit ourselves to the case m 1. Underthe assumption E C1 (1 the followinggeneral Green formula is applicable:

154 7 VariatIonal Formulation

a(u, v)= J {

+ + + aoouvlij=1 4=1 4=1

= [— ÷ + aoou} v

+ f + (17.4.4)

Here, the are the components of the normal direction n = n(z), x E F. Wedefine the boundary differential operator

B := (7.4.5)

when L is described by (2.lb). Equation (4) becomes a(u,v) = f0vLtidx +vBu di. By the formulation of the problem, a(u. v) agrees with 1(v) from

(is). If we first choose v E C H'(f?) the boundary integrals drop outand we obtain Lu = g as in Section 7.2. By this the identity a(u,v) = f(v)reduces to = f1pvdi for all v E H'(i7). According to Theorem6.2.40b, runs over the set H'/2(I') if v runs over H'(O), so that we havefr %b(Bu — dl' 0 for all H'/2(fl; thus Bu = This proves

Theorem 7.4.5. Let F be sufficiently smooth. A classiosJ solution of theproblem (2a,b) with / frvm (is) is also the claaaieal 8olution of the boundaryvalue problem

Lu=g mu, Bu=,o on 1', (7.4.6)

and conversely.

The condition Bit = is called the natural boundary condition.This results from the fact that in (2b) (as distinct from (2.9)) the function vmay assume arbitrary boundary values. Note that the bilinear form determinesL as well as B.

Exercise 7.4.6. Show that the bilinear form from Example 7.2.10 for =ghas as the natural boundary condition the Neumann condition =0.

Theorem 7.4.7. Let 00,1 be bounded and a(.,.) beThen the statements of Theorem 7.2.14 holds with Hm(fl) instead of

Example 7.4.8. Let (1 0°" be a bounded domain. The bilinear forma(u,v) = associated to the Helmholtz equation

—%iu+cu=f in!? (c>0), 9u/On=O onf,is H'(fl)-elliptic since a(u, u) � inin(1, C)IUI1. For c =0, however, a(.,.) is onlyH'(f?)-coercive. As is known from Theorem 3.4.1, the Neumann boundary


value problem for the Poisson equation (i.e., for c = 0) is not uniquelysolvable. According to alternative (ii) in Theorem 7, there exists a nontrivial

elgenspace E = kernel(L). u E E satisfies a(u, u) = 0, thus Vu = 0. Since (1 is

connected, it follows that u(x) = coust and therefore dimE = 1. Since a(.,.)is symmetric, E :== kernel(L') coincides with E. According to Theorem 7, theNeumann boundary value problem a(u, v) = 1(v) (v II'(fl)) is solvable ifand only if f I E, i.e., f(1) = 0. If f(v) := f0g(x)v(x)dx, 1(1) = 0 reads as

gdx = 0. If, however, / is given by (Ia), the integrability condition readsf(1) = g dx + f1. çodl' = 0 (this is Equation (3.4.2), in which I should bereplaced by —g).

Remark 7.4.9. While the classical formulation of a boundary condition suchas ôu/ôn = 0 requires conditions on the boundary 1', the problem (2a,b) canbe formulated for arbitrary measurable 0.

In the following, we proceed in the opposite direction: does there exist, fora classically formulated boundary value problem Lu = g in .0, Bu = on f',with given L and B, a bilinear form a(., .) such that (2a, b) is the correspondingvariational formulation? This would mean that the freely prescribed boundaryoperator B represents the natural boundary condition.

For m = 1 the general form of the boundary operator reads

B = + bo(x) (x 1). (7.4.7)

With bT = (b1, .. . , one can also write B = (ci. (&2.lb,b')). Here

bTV is not allowed to be a tangential derivative (cf. Remark 5.2.2):

(b(x), n(x)) 0 for all x E 1. (7.4.8)

Remark 7.4.10. Let m = I. Let (8) hold. Let A(x) be the matrix A (ac)(cf. (5.1.ld)). By passing from Bu p to the equivalent scaled equationoBu = with o(x) = (n(x), A(x)n(x))/(n(x), b(x)), one can ensure that(n, ob) = (n, An). Thus in the following it is always assumed that b alreadysatisfies (n, b) = (n, An).

PROOF. Because of (8) a is well defined. For Bu = and oBu =a 0 is required. This is guaranteed by the uniform ellipticity:

(n, An) eInl2 =

Theorem 7.4.11. (Construction of the bilinear form). Let m = 1.

Let L and B be given by (2.lb) and (7), with b satisfljing condition (8). Thenthere exists a bilinear form a(.,.) on H1(s?) x H1(0) such that to the vari-ational problem (2a.,b) corresponds the classical formulation Lu g in 0,Bu=p onf.

156 7 VariatIonal Fonnulation

PROOF. The bilinear form we seek is not uniquely determined. We will givetwo possibilities for its construction. First we discuss the absolute term in (7).On the basis of Remark 10 we assume (n,b) = (n,An).(a) Let the vector function 13(x) := (13i(x),. . E be arbitraryThe differential operator

:= —

++

maps every u C'(S2) into zero: Lju = 0. Thus the operator L can bereplaced by L + L1 without changing the boundary value problem. Let a(.,.)be constructed according to (2.6) from the coefficients of L + Lt. Equation(5) shows that the boundary operator associated with o(.,•) has the absoluteterm

(7.4.9)

If = 0, the choice of = is successful. Otherwise, two other optionsare available.(aa) Select such that on I' the following holds: /3j(x) =Since ml = 1, the term (9) then agrees with bo(x). The practical difficulty inthis method consists in the need to construct a smooth continuation on £1 ofthe boundary values x E F.(ab) Set /3, = -ao' and add a suitable boundary integral:

a(ti, v) := f — + [000 dx

+ (7.4.10)

The integration by parts described above shows that the boundary operatorassociated to (10) reads

b = + b0. (7.4.11)

(b) The operator (11) can be written in the form B = bTV+b0 with b An.Since (ii, b) = (ii, An) = (n, b) has been assumed already, d := b — b isorthogonal to n. To change b to b there are again two options.(ba) Define the n x n-matrix A' on F by A' dziT — i.e., =

— njdj. This A'(x) is skew-symmetric: A'T = —A'. Continue A'(x),which is at first only defined on F, to a skew-symmetric matrix A' E C'(i7).[Here we have the same practical difficulty as in step (aa).J The entries of A'define

8 8 8L2 := E


Again, L,2u = 0 holds for all u E C2(Q), since

— (a, — + + = — = 0.

Thus L can be replaced by L ÷ L2 without changing the boundary valueproblem. The coefficients belonging to L + L2 result in a boundary operatorB whose derivative terms read + = t(A + A')n]TV. By

the construction of A' we have A'n = (dnT — ndT)n d = b — b , since

(n, n) = 1 and (d, n) 0. Since we also have An = b, the derivative termin B gives bTV as desired. The transition L L + L2 does not change theabsolute term in B, so that from (11) follows B = bTV + b0. -(bb) Let B = bTV + b0 be given (cf. (11)) such that d = b b is orthogo-nal to n. From this the boundary operator T := dTV is the derivative in atangential direction if n = 2 [resp. in the tangent hyperplane if n � 3J. if I'is sufficiently smooth then the restriction v ofH'/2(f'). By Remark 6.3.14a, one can show that T E L(H"2(r'), H-'/2(f')).Since T(ulp) Hh1'2(f), is well-defined for H'/2(f), inparticular, for = with v H (fl). Thus

b(u,ts) := j(vtr)T(ulr)dI'

is a bilinear form bounded on H'(fl) x H1(a). We add b(.,.) to a(.,.) in(10). Integration by parts yields the boundary operator B + T = bTV + bo+

U

Remark 7.4.12. Let the bilinear form (2.6) be H'(.Q)-coercive (cf. Theorem3). Let its coefficients, as well as the boundary 1', be sufficiently smooth(€ C'). Then the constructions of the preceding proof again result in anB7jH1(Q)-eoercive form.

PROOF. We go through the steps.(a) In step (aa) only terms of lower order are added so that Lemma 7.2.12 isapplicable. As for step (ab), see step (bb).(ba) In step (ba) one adds b(u,v) f0 + HereLemma 7.2.12 is also applicable, since the skew symmetry of A' results in6(u,u)(bb) In the construction in step (bb), Lemma 7.2.12 is applicable analo-gously. This is easiest to understand in the case n = 2. Let I' be describedby {(z,(s),x2(8)): 0 < s 1}. If Xl,X2 E C'flo, 1J) and =and if we have d E C'(f') for the d in T = dTV, then b(.,.) has therepresentation b(u,v) = where =u(s) = v(x,(s), x2(8)), r E C1((0, 1)). Thanks to periodicity, integrationby parts yields in b(u, v) = — 10' (rV)'lZ ds without boundary terms, so that

b(u,u) ds — = This implies

158 7 Vsriational Fbrmulation

Ib(u, u)I � � CIITIICI(Io,l])

for all s C (1/2,1) (cf. Theorem 6.2.40a). Since � EIUII — one canapply Lemma 6.5.lSc. U

The case m � 2 has been excluded in this section (except for Theorem 1).Boundary value problens of order 2m require m boundary conditions =

on 1' (j = 1,... ,m) (cf. Section 5.3). For m � 2 the proof of Hm(fl)-coercivity becomes more complicated.

In order to carry over Theorem 7.2.13, one needs in addition the so-calledcondition of Agmon (cf. Wloka 1, Theorem 19.3J, Lions-Magenes [1, p. 2101).

The resulting complications can be seen with the aid of the biharmon Icequation.

Example 7.4.13. (a) To the variational problem: find C ff2(fl) with

a(u, v) := J dx = 1(v)

(7.4.12a)

for all v E H2(Q), corresponds the formulation

= g in Q, = and = on F. (7.4.12b)

But the bilinear form a(.,.) Is not H2(.Q)-coercive.(b) To the variational problem: find u C H2(i1) fl HJ(Q) with

for alive(7.4.12c)

(a(.,.) as in (12a)) corresponds the claseical formulation

in!?, u=O (7.4.12d)

The bilinear form is fl(c) The boundary conditions in

= g in S?, u =0 and = p on F (7.4.12e)

are admis8ible. this boundary value problem cannot be writtenin the present form as a variational problem. A variational formulation for(12e) reads:

Find uEH2(fl)flH01(f1) suththat

forallv€H2(!?)with

(7,4.12f)


(a(.,.) as in (12a)). But this does not agree with the present concept since uand v belong to different spaces.

PROOF. The equivalence of the variational and the classical formulation canbe shown via integration by parts:

2 f 01)a(u, v)

= j u dx + j - dl'.

The noncoercivity in part (a) results as follows. Let (1 C IRTh be bounded.For all a E IR., ... = sin(axj)exp(ax2) lies in H2(i7) and satisfies

= 0, hence also a(ua,ua) = 0. If a(.,.) were coercive, there would exista C with 0 a(ua,uo) � for all a, i.e., �The contradiction results from � 102ua/OxiIo = O2IUaIO for sufficientlylarge a.

Natural and Dirichiet conditions can occur together. In Example 13b, u =0 is a Dirichiet condition and = ço a natural boundary condition.Even in the case m = 1 both sorts of boundary conditions can occur.

Example 7.4.14. Let 'y be a nonempty, proper subset of I'. The boundaryvalue problem

= g in 0, = 0 on 'y, Ou/On = çø on (7.4.13)

in the variational formulation reads as follows: find u 114(0) such that

a(u,v) := = 1(v) :=

for ally E where H4 (.0) :={uE H1(l)): ur=Oon-y}. Equation (13)is occasionally termed a Robin problem.

Exercise 7.4.15. Let a(u,v) := f0[(Vu, Vv) +cuvldx with c >0 on V x Vwith V := {u E H'(0):u constant on f'} be defined. Show that(a) a(.,.) is V-elliptic.(b) The weak formulation: u V1 a(u,v) = for all v Vcorresponds to the problem

—4ui- cu = g in (1, u constant on 1', f fcodr. (7.4.14)

which is also called an Adler problem.

Finally we want to point out the difficulty of classically interpreting aweak solution. In the variational formulation (2a,b) the right-hand sides gand ço of the differential equation and the boundary condition are combinedin the functional 1. In the variational formulation the components g and ço are

160 7 VarIational Fbrniulatlon

indistinguishable! u E H'(a) has fist derivatives in L2((J), whose restrictionsto V do not have to make sense. That is why Bu cannot be defined in general;Ru = cannot be viewed as an equality in the space H'/2(f) although

H'/2(F) (cf. Corollary 4b).But even if there Is a classical solution, the following paradox arises. Let u

be a classical solution of Lu =0 in 1?, Bu = on F. to (Ia) defineE (H1(fl))' by f,(v) := One may also view u as a solution

of Lu = in 0, Bu = 0 on I. These equations may even be interpretedclassically in the following way: there exist E C°°(0) with 4 in(H1(i7))'. Let be the claseical solution of = = 0. Thenconverges in H'(O) to the above-mentioned classical solution u.

Incorporating the boundary values Ru = in the differential equationLu = 4 corresponds to a modification of the discretised problem as usedin Section 4. The difference equations Dhuh = fh in and the boundaryconditions uh on F,, resulted in the system of equations L,,u, =fh + (cf. (4.2.6b)). If one defines by = in 0h, uh = 0 onthen satisfies the equations D,,Ti,, = in iZ,, = 0 on Just as thefunctional f cannot be uniquely separated into g and and cannot bereconstructed from In contrast to the discrete case, the separation off intog and is possible, however, provided stronger conditions than g (H'(.Q))'are imposed on g (for example, g L2(fl)).

8 The Method of Finite Elements

In Chapter 7 the variational formulation was introduced only for the purposeof proving the existence of a (weak) solution. It will now turn out that thevariational formulation is the foundation of a new method of discretisation.

8.1 The Ritz-Galerkin Method

Suppose we have a boundary value problem in its variational formulation:

Find u€V, so that a(u,v)=f(v) foiallv€V, (8.1.1)

where we are thiniciug, in particular, of V = and V = H'($?) (cf.

Section 7.2, Section 7.4). Of course, it is assumed that a(.,.) is a boundedbilinear form defined on V x V, and that / V':

Io(u,v)I � CsfItLUvIIVftv for E '€ (8.1.2)

Difference methods arise through discretising the differential operators. Nowwe wish to leave the differential operator hidden in a(.,.) unchanged. The Ritz-Galerkin discretisation consists in replacing the infinite-dimensional space Vwith a finite-dimensional space VN:

VNCV, dhnVN=N<OO. (8.1.3)

VN equipped with the norm fl . is still a Banach space. Since VN C V, botha(u,v) and f(v) are defined for u,v E VN. Thus we may pose the problem (4):

Find uN E VN, so that a(uN,v) = 1(v) for all v E VN. (8.1.4)

The solution of (4), if it exists, is called the Ritz-Galerkin solution (be-longing to VN) of the boundary value problem (1).

To calculate a solution one needs a basis of VN. Let (b1, . . . , bN} be sucha basis, i.e.,

VN= span{bI,...,bN}. (8.1.5)

For each coefficient vector v = {vl,...,vN}T we define

(8.1.6)

162 8 The Method of Finite Elements

Remark 8.1.1. P is an isomorphism between 1RN and VN. The inverseP* VN is thus well-defined on VN.

Lemma 8.1.2. Assuming (5) the problem (4) is equivalent to

Find tLNEVN, sothat foralli=1,...,N. (8.1.7)

PROOF. Putting v = in (4) gives (7). On the other hand, suppose thatv = E VN is arbitrary. Then (7) and the linearity of and fgive (4):

—1(v) = —1(b1)) = 0.

We now seek u so that uN = Pu. The following theorem transformsthe problem (4) [reap. (7)) into a system of linear equations.

Theorem 8.1.3. Assume (5). The NxN-matrixL (L11) and the N-vectorf = (fi,.. . , IN)T are defined by

:= a(b,,b1) (i,j = 1,.. .,N), (8.1.8a)

:= f(b1) (i = 1,..., N), (8.1.8b)

Then the probLems (4) andLu=f (8.1.9)

are equivalent. If u is a solution of (9), then uN : Pu solve8 the problem(4). In the opposite direction, if is a solution of(4), u 1,a solution of(9) (cf. Remark 1).

PROOF. (4) is equivalent to (7). In (7) put = Pu = Eu,bj; thena(uN, b1) = uja(bj, = = = and thus Lu = f. I

In engineering applications where the boundary value problems arise fromcontinuum mechanics (cf. the first paragraph of Section 5.3.1), one calls L thestiffness matrix.

The connections between L and a(.,.), on the one hand, and between fand f(.), on the other, are clear from

Remark 8.1.4. If(u, v) := u1v, is the usual scalar product, then a(u,v) =and f(v) = (f,v) with u = Pu,v = Pv.

A trivial consequence of Theorem 3 is:

Corollary 8.1.5. The Ritz-Galerkin discretisation (4) has a unique solutionfor each f E V' exactly when the matnx L in (8a) is nonsingular.

8.1 The Rits-Galerkin Method 163

The Gelfand triple V C U C VI corresponds to the finite-dimensional

situation VN C UN C where the spaces are the same as sets:

VN = UN VN.

However, the three spaces have different norms:

VN := (VN, fly), UN (VN, U IIu), V, := (VN, II.

VN has fi and UN has fi flu as a norm, while the dual norm

sup{I(v,u)uJ/IJulJv:O u E VN} for v E V'

(in particular for v VN) is defined on V', but is only a norm onsince = 0 for each v orthogonal to VN.

By Lemma 6.5.1 there is an operator associated to a(., .): VN x VN JR

LN:VN

so thata(u,v) = (LNU,v)u for all u,v E Vpj. (8.1.lOa)

Here we are writing (., for (., •)VJXVN (cf. Remark 6.3.12).

The map P is defined by

(Pu,v) for uE VI,vEJRN. (8.1.IOb)

ExercIse 8.1.6. (a) Show: The kernel of P is C U (orthogonal spacerelative to fi flu). The map P: Vk JRN is an isomorphism.(b) Since P V,, JRN exists, we can define

Cp := := U E VN},

where fi .fl Is the Euclidean norm. Show: The matrix P'P: JRN • has aninverse with spectral norm

IRP'PY'II �(c) Show:

QN := P(P*P)tP*:U -, U

is the orthogonal projection (relative to fl• flu) onto UN = VN. In additionwe have QN E L(V', V).

Lemma 8.1.7. Let LN E L(VN,Vk) be the operator conespondin9 tox VN IR, and L L(V,V') that fora(.,.):V x V JR. The

foUowirig relationships hold between the operators L, and the stiffnessmat,ix L:


L = PLP = PLPLN = P'1LP1:VN vii,LN the restriction of QNL and QNLQN to VN.

PROOF. For all u, v E iPI' we have

(Lu,v) = a(Pu,Pv) = (LNPU,PV)U = (P*LNPU,V)=

By Remark 6.3.12 we can write (f, v)u for 1(v). If v = P'v, it followsfrom (lOb) that 1(v) = (f, = (P'f, v). Usmg Remark 4 we then have

f = Pf (8.1.lOc)

for the f in the right side of Eq. (9).In the case of a continuous variational problem the V-ellipticity guaran-

tees its unique The same condition is also suffictent in the discretecase:

Theorem 8.1.8. Auume (3). Suppose the bilinear form is V-elliptic:a(u,u) � CEIIuIft, for ail u E V with CE > 0. Then the mat'*r L in (9)is nonsmgt4ar and the Ritz-Goier*in solution E VN satisfies

IIU?ullv � � (8.1.lla)

PROOF. Lisnonsingularsince andthus

(Lu, u) = a(Pu, Pu) � CEIIPUU?, >0

and so, in particular, Lu 0. By &ercise 6.5.6a a(.,.) is also VN-eIIiptiC withthe same constant CE. From Theorem 6.5.8 there holds �1/CE, i.e., HU"flv � (ha) then results from

ExercIse 8.1.9. Show � II! fly' for any f V'.

Exercise 8.1.10. Show: (a) If a(.,.) is symmetric then so is L.(b) If a(.,.) is symmetric and V-elliptic then L is positive definite. Underthe same assumptions the Ritz-Galerkin solution ut1 solves the followingvariational problem (cf. Theorem 6.5.12):

J(uM) � J(u) :=a(u,u)—2f(u) for all u VN.

Example 8.1.11. (Dirichiet Problem) The boundary value problem is

8.1 The Rltz-Galerkin Method 165

.—zlu(x,y)=1 infl=(0,1)x(O,1), u=O

The weak formulation is given by (1) with V =

a(tt,v) := J (Vu,Vv)dxdy = f1(v) JvdxdY.

The functions

bj (x, y) = sin(irx) sln(wy), b2(x, y) =

bs(x, y) = sin(wx) sm(3iry), b4(x, y) = sin(3wx) sin(3,ry),

fulifi the boundary conditions and so belong to V = Hd(i7). They form a basisof V4 := span {b1,. ..,b4}. The matrix elements = can be workedout to be

L11 = = = 5w2/2 =

In addition the chosen basis is o(., .)-orthogonal:

L is diagonal. Furthermore one may calculate= = dxdy, getting

fi =4/ir2, 12

Hence u = L'f has the components

Ui = u2 = = u4 = 8/(81ir4),

and the Ritz-Galerkin solution is then

=-.[sinwzsiniry + +sinirzsin3iry)

+ sin sin 3iry].

The Ritz-Galerkin solution and the exact solution for x = v = 1/2 are

= = 0.07219140...,

= /[(i + 2v)(1 + 2p)((1 + 2p)2 + (1 +

= 0.0736713.

Example 8.1.12. (Natural boundary conditions) Let the boundaryvalue problem be

166 8 The Method of Finite Elementa

= ir2cosirx in = (0,1) x (0,1), Ou/ôn=Oon F.

The solution is given by u = cos wx + const. The weak fonnulation is in termsof (1) with V = H'(s?),

a(u,v) := +

1(v) :=ir2Jv(x,y)cosirxdxdv.

The boundary value problem has a unique solution in

W:={v€ V:Jvdxdv=0}.

The basis functions

bi(x,y) = z — 1/2, b2(x,y) = (x —

are in W. The stiffness matrix L and the vector f are then

Ii 1/41 1 —2 1L— 9/80]'

so that—L''f—1

— [—20+240/7r2

The solution is t/'(x, y) = (3 — 60/w2)(x — 1/2) — (20 — 240/ir2)(z —

The Ritz-Galerkin solution satisfies the boundary condition öu/8n = 0 andthe differential equation only approximately:

y)/On = 12 — 120/ir2 —0.16.

For x = 1/4 the approximation is v."(1/4,y) = —7/16 + 45/(4ir2) =0.70236..., whereas u(1/4, y) = oosir/4 = 0.7071 ... is the exact value.

In the following we shall consider the case in which o(.,.) is no longerV-elliptic, though it is V-coercive. That a(.,.) is V-coercive guarantees thateither problem (1) is solvable or A = 0 is an elgenvalue. Even if one assumesV-coercivity and the solvability of the problem (1) one can not deduce thesolvability of the discrete problem (4).

Example 8.1.13. a(u,v) := f(u'v' — lOuv)dx is 1)-coercive anda(u,v) = 1(v) := f01 gvdx (v E J.1o1(0, 1)) has a unique solution. Let VNspanned by bj(x) = x(1 — z) E V = Hd(0, 1) (i.e., N = 1). Then the discreteproblem (4) is not solvable since L = 0.

If one replaces the space V in Lemma 6.5.3 with VN, then there followsfrom Exercise 6.5.4


Theorem 8.1.14. The problem (4) is solvable for all f E V' and has aunique solution, Ut", which satisfies the estimate

� S (8.1.llb)EN

if and only if

VN,UVIIV = 1}:u VN,(IUIIV = 1} = CN >0. (8.1.12)

Since (4) is equivalent to the system of equations (9), one has the

Corollary 8.1.15. The matriz L is nonsingtdar if and only if (12) holds.

Exercise 8.1.16. The requirement (12) is equivalent to

flullv � VN,HVUV 1} forallu€ VN. (8.1.12')EN

and

= 1/EN (LN as in (lOa)). (8.1.12")

Warning: The condition (12) for VN does not follow from the analogouscondition (6.4.5a) for V. See, however, Theorem 8.2.8.

The requirement (12") guarantees the existence of L', but it does notsay anything about its condition, cond(L) = which is the decidingfactor for the sensitivity of the system of equations Lu = f. For example, ifone chooses for a(u,v) := u'v'dx the basis (i = 1g.. .,N) then oneobtains the very badly conditioned matrix = iii (i+j —1). The conditioningof L is optimal if one chooses the basis to be a(•, )-orthogonal: a(bj,b,) =as is the case in Example 11, up to a scaling factor.

8.2 Error Estimates

For difference methods the solution u and the grid function uh are defined ondifferent sets. The Ritz-Galerkin solution, is, on the other hand, directlycomparable with is. One can measure the error due to discretisationWith IIu — u?JUv or with — uNItu.

Let is be the solution of (1.1): a(u,v) f(v) for v E V. Suppose — bychance or because of a clever choice of V,1 — that u also belongs to VN; then

is also satisfies (1.4). That means: The discretisation error is zero ifis E We shall now show: The "closer" that is is to VN the smaller is thediscretisation error.


Theorem 8.2.1. (Ceo) Assume (1.2), (1.3), and (1.12) hold. Letu E V be asolution of the problem (1.1), and let uN E VN be the Ritz-Oatev*in solutionof (1.4). Then the following estimate holds:

III' — u"IIv (1+ Cs/CN) iflf — wIly (8.2.1)WEVN

with Cs from (1.2) and from (1.12). Note iflfwEyN IIu—wlIv is the distanceof the function u from VN; it will be abbreviated in the following to

d(u,VN) := ml flu—wIly. (8.2.2)WEVN

PROOF. If u satisfies a(u,v) = /(v) for all v V then it does in particularfor all V E VN. Since we also have a(uN, v) = f(v) for v E Vpj, it follow8 that

forall VEVN. (8.2.3)

For arbitrary v,w E VN with lIvIlv = 1 we can therefore conclude

— w, v) = a((t/' — tij + [u — w], v) = a(u — w, v)

and

— w,v)l 5 — wIIvIfvIlv = CsIlu — why.

From (1.12') we then obtain

—wily � E VN,hIvhIv = 1} � (Cs/CN)flu—wllv.

The triangle inequality then gives

flu � flu — wily + — uNhiv � (1+ Cs/EN)hIu —

SincewEVNisarbitrarywededucetbeaeeertion(I). U

In Theorem I the unique solvability of the problem (1.1) was not assumed,but only the existence of at least one solution.

If the discretisation error is supposed to converge to zero one makes useof a sequence of subspaces V that converge to V In the followingsense:

Theorem 8.2.2. :=VN4 c V(iEt4) be a sequence of subspaces with

lim = 0 for all u V. (8.2.4a)

Asumethat(1.12)holdswitheN4 in addition assume(1.2) (continuity of a(.,.)). Then there exists a unique solutionu of the problem(1.1) and the Ritz-Galer*in solution ud := to

iIu—u'hIv--.O for i—.oo.


Sufficient to ensure (4a) is

UvidensemnV. (8.24b)

PROOF. (a) Assume first the existence of a solution u.(aa) The estimate (1) implies the convergence:

ilu — u'IIv � (1 + d(u, —.0.

(ab) We now wish to show that (4a) follows from (4b). The inclusion ¾-i Cimplies d(u, V1) d(u, Now d(u, will be a null sequence if for eache > 0 there exists an i such that d(u, e. From the assumption (4b) there

is, for each u E V and e > 0, a w E U, with — wily e. Therefore wehave w€ for an i N. That d(u,V3) � — why � e then proves (4a).The convergence u hnplies the uniqueness of the solution u.(b) The next thing to show is that the image W := (Lv: v V} C V' of theoperator L:V —. V1 associated to a(-, -) is closed. Foreachf E Wthereisau EV with Lu = f, so that part (a) of this proof suffices to show the convergence

--. u. Since Itu'IIv � it follows that hiutiv = urn � 11f110'/€.Let E W be a sequence with —. f in V' and f,, = Lug. The Cauchyconvergence Ii!" — shows � ti/v —0 so that

the limit = E V exists. The continuity of L L(V, V') shows thatf = limf,. = lirnLu,, = Lu, and thus that f E W. Hence W is closed.(c) In order to demonstrate the existence of a solution to the problem (1.1)we have to show the surjectivity of L: V —. V'. If L were not surjective (Le.,W V'), there would be an I E W-1- with hIlly' = 1. Let Jv: V —. V' be theRieaz isornorphism (ci. Corollary 6.3.7). Set v := E V. It follows that

f(v)= —(f,f)v' = 1, 0

for all u E V. Let E be the Ritz-Galerkin solutions. They must alsosatisfy v) = 0, i.e.,

a(u',v) — 1(v) = 1.

We split up v into + where v' Vi. By (4a), with v in place of u, onemay guarantee that —.0. This shows

I = /(vi)+o(ui,w)_f(wi)w1) — f(wi)

and

I = v) — f(v)l � [Cshtu'ljv + hhfik"j llw'(Iv.

Since Ilu'IIv � IIfhIv'R is uniformly bounded and Hw'hIv —. (I this amountsto a contradiction. Therefore £ must be surjective, so that for each f E V'there exists a solution u to the equation Lu = I, i.e., to the problem (1.1). •


Corollary 8.2.3. The requirement (1.12) with CN,. � >o from Theorem 2is satisfied with CR if a(.,.) is V-elliptic: a(u,u) �

Exercise 8.2.4. Show that (4a) implies that is dense in V.

Let QN be the orthogonal projection onto VN (cf. Exercise 8.1.6c). Thefactors L:V V', QN:V' —' V

V C V. (8.2.5a)

Exercise 8.2.5. Show there is also the representation

SN = PL'P'L. (8.2.5&')

Lemma 8.2.6. SN is the projection onto VN and is c*zIled the Ritz projec-tion. It sends the solution u of the problem (1.1) to the Ritz-Galerkin solutionuN VN: uN = SNU. Assuming (1.2) and (1.12) we have

IISNIIv+-v � CS/CN. (8.2.5b)

A definition of 5N equivalent to that in (5a) is

SNU E VN and O(SNU,V) = a(u,v) for all v VN, u V. (8.2.5c)

PROOF. (a) Since

a(u,v) = = = (Lu,QNV)U = (QNLU,v)u

for all v VN, it follows that SNtI in (5c) is the Ritz-Galerkin solution for theright-hand side f := QNLU E V', i.e., SNU = QNLu. Conversely one mayargue similarly, and so show the equivalence of the definitions (5a) and (5c).

projection. Inequality (5b) follows from IISNuIIv � (cf. theproof of (1.11)) and = � IILuII'v � CsfIuIfv with C8from (1.2) (cf. Exercise 8.1.9).

Remark 8.2.7. Let a(.,.) be V-elliptic and symmetric. IIIvIlIv := a(v,v)'/2is a norm equivalent to f$. liv. The Ritz projection SN is, with respect to

lily, an orthogonal projection onto Vp,. Thus, in particular, we have

IlISp,iiiv..-v �1. (8.2.5d)

PROOF. The scalar product associated to is a(., .), so that it is to beshown that a(SNv, w) = a(v, SNW). (5c) Implies a(SNv, Sqw) = a(v, SNW),since SNW E VN. The symmetry of a(.,.) and exchanging v and w give

8.3 Finite Elements 171

a(SNv,w) = a(w, SNV) = a(SNW, Spjv) = a(SNV, SNW), so that a(SNv, w) =a(v, SNW). (5d) results from Remark 6.3.8.

Remark 7 shows once again that the Ritz-Oalerkin solution = SNUis the best approximation to u in VN in the sense of the norm fly. This

is equivalent to the variational formulation J(uN) � J(v) for allv VN (cf. Exercise 8,1.lOb).

The condition (1.12), with e,, � e > 0, is difficult to prove, except forV-elliptic bilmear forms. However, the following theorem shows that this con-dition does hold for subepaces approximating well enough.

Theorem 8.2.8. Let the bilinear for,n a(.,.) be V-coercive, where V C U CV' is a continuous, dense, and compact embedding. Let Problem (1.1) be solv-able for all f V'. Assume that (4a) holds for the subspaces C V. Forlarge enough i the stability condition (1.12) is then satisfied with � > 0.

The proof of this will be postponed to a supplement to Lemma 11.2.7.

8.3 Finite Elements

8.3.1 Introduction: Linear Elements for 11 = (o,b)

As soon as the dimension N = dim V?, becomes larger the eseential disadvan-tage of the general Ritz-Calerkin method becomes apparent. The matrix L isin general full, i.e., L11 0 for all i,j = 1, . . ., N. Therefore one needs N2integrations to obtain the values of = o(b,, = fe,. .., whether exactly orapproximately. The final solution of the system of equations Lu = f requires0(N3) operations. As soon as N is no longer small the general Ritz-Galerkinmethod therefore turns out to be unusable.

A glance at the difference method shows that the matrices Lh which occurthere are sparse. Thus it is natural to wonder if it is poesible to choose thebasis {b1, . . . ,bp,r} so that the stiffnees matrix = is also sparse.The best situation would be that the were orthogonal with respect to a(.,.):a(b,,bi) for i j. However, such a basis can be found only for special modelproblems such as the one in Example 8.1.11. Instead we shall base our furtherconsiderations on

Remark 8.3.1. Let the bilinear form a(.,.) be given by (7.2.6). Let Bbe the interior of the support of the basis function i.e., :=

A sufficient condition that ensures = a(bj, b1) = 0 isri B, =

PROOF. The integration = f0... can be restricted to B1 fl B,. •In order to be able to apply Remark 1 the basis functions should have

as small supports as poesible. In constructing them in general one goes about


it from the desired goal: one defines partitions of Q into small pieces, the so-called finite elements, from which the supports of b4 are pieced together.

As an introduction let us investigate the one-dimensional boundary value

problem

—u"(x)=g(x) fora<x<b, u(a)=u(b)=O.

Assume we have a partition of the interval [a, bI given by a = x0 <x1 <...<XN÷1 = b. Denote the interval pieces by 14= (x4_i,x4) (1 � i � N + 1). Forthe subspace VN c let us choose the piecewise linear functions:

VN ={u C°([a,bI):the restriction of u to 13(1 i < N+ 1) (832)islinear;n(a)=u(b)=O}

See the figure below.

FIgure 8.3.1. (a) a piecewise linear function,, (b) a basis function

The continuity "u C°([a, bl)" is equivalent to continuity at the nodesx4 (1 � i � N) : u(x4 + 0) = u(x4 — 0).

Remark 8.3.2. u E VN is uniquely determined by its node values u(x1)

u(x) = —x)+U(x4+t)(x—s4)j/(x4+1 —x4 for zE (8.3.3a)

where u(xO) u(ZN+1) = 0. From Theorem 6.2.42 we condude that VN C(a, b). The (weak) derivative u' L2(a, b) is piecewise constant:

u'(x) = (u(x4+1) — u(xi)j/[z4÷i — x4 for a' 14+1. (8.3.3b)

The basis functions can be defined (see Figure ib) by

I (a' — — x4_1) for <a' <b4(x) = — — a'4) for <a' 1 � I � N (8.3.4)

1.0 otherwise.

(8.3.1)

(a)

a=zo a'i Z2 a'3 a'4

a'


Remark 2 then has 88 a consequence

Remark 8.3.3. One has the representation u = The supportsof the functions b1 are = 1x1_i,xi+i1. The B1 from Remark 1 is nowB1 (xl_l,xl+1).

The weak forirnilation of the boundary value problem (1) is a(u, v) f(v)with

b

a(u,v) j u'v'dx, 1(v) fgvdx for u,v E H01(a,b).

By Remark 1 we have, for the matrix elements, L1, = 0 as soon as �2, since then B1 fl B1 = For Ii —21 � 1 one obtains

1 1= a(b1_i,b1) = / dx = —1/(xj —xs_1),Xs_1 Zj —

L11 = a(b1, b1)= J

(x1 — dx + J — dx (8.3.5a)

= 1/(xg —X4_1) + t/(x1+i —x1),

= —1/(x1÷i —x1).

The right-hand side f = (ft,.. . ,IN)T is given by (5b):

fZ4+jgbgdx

i1 1

= j g(x)(x—xg_i)dx+ Ix1 —x1_1 xi+1 —x,(8.3.5b)

Remark 8.3.4. The system of equations Lu = f, given in (5a,b), is tridiag-onal, thus, in particular, sparse. Fbr an equidistant partitioning x1 a + ihwith h := (b — a)/(N + 1) the coefficients are = —1/h, L1,1 = 2/h,hi = hf0' +th)+g(xg —th)1(1 —t)dt. If g€ C°(a,b), then by the inter-mediate value theorem f1 = hg(x1+&1h), with <1. Therefore the systemof equations Lu = f is, up to a scaling, identical with the difference equationLhuh = fh in Section 4.1, if one defines by := = g(x1 + &1h)instead of by fh(xl) = g(xj): then L = hLh,f = hi,,.

This example shows that it is possible to find basis functions with smallsupports so that L is relatively easy to calculate. In addition, the similarity isapparent between the resulting discretisation method and difference methods.Finally consider the numerical evaluation of the integral (5b),

Exercise 8.3.5. Show that the quadrature formula for gb1 dx

with b1(x) as weight function and one support point is:


(8.3.5c)

What is the formula for a partition of equal intervals?

If one replaces the Dirichiet boundary conditions in (1) by the Neumanncondition then V = H'(a, b). The subapace VN C H'(a, b) results from (2)after removal of the condition u(a) = u(b) 0. In order that dim VN = Nthe numbering has to be changed: The partition of (a, b) is given by a = <X2<,..4(VNb.

Exercise 8.3.6. Let Ou(a)/&z = ga,Du(b)/On = gb, with a(.,.) as before.Suppose the partition of (a,b) Is equidistant: x4 = a+(b — a)(i — 1)/(N — 1).

What is the form of the equation Lu f? Show that Lu = f has at least onebsolution

8.3.2 LInear Elements for £7 C It2

We shall assume:Q c is a polygon. (8.3.6)

As shown in Fig. 2 we divide Si into

t number of N := number of Innersodas (number of all sodas)

Definition 8.3.7. r := {Tj,. . . is called an admimible triangulation of£1 if the following conditions are fulfilled:

(1 � i t) are open triangles ("finit, elements") (8.3.7a)

(8.3.7b)

U = (8.3.7c)

for the iseither


1) empty, or

ii) a common side of the elements and 1',, or

in) a common edge of the elements and T3. (8.3.7d)

Remark 8.3.8. (a) The conditions (7a) and (7c) imply the polygonal shapeof S?, i.e., the assumption (6). (b) Figure 2 shows only admissible triangula-tions. An example of an inadmissible triangulation would be, e.g., a squarepartitioned as follows:

Let r be an admissible triangulation. The point x is called a node (of T)if x is a vertex of one of the T1 E 1-. One distinguishes interior and boundarynodes, according to whether x £2 or x OS?. Let N be the total number ofinterior nodes. We define VN as the subapace of the piecewise linear functions:

with a linear function, i.e., u(x,y) = an + + on T1}.(8.3.8)

Remark 2 can be applied to the two-dimensional case under consideration:

Remark 8.3.9. (a) it is true that VN C (5?). (b) Each function is uniquelydetermined by its node values u(Xj) at the interior nodes xi, 1 � i $ N.

PROOF. (a) Example 6.2.5 shows that VN C H'(fl). Since u = 0 on 8.0,Theorem 6.2.42 proves that VN c HJ(S?).(b) Let x, x', x" be the three vertices of E r. The linear function u(x, y)

is uniquely determined on T1 by the values u(x), u(x'), u(x").a

The converse to Remark 9b is as follows:

Remark 8.3.10. Let x' (1 <i <N) be the interior nodes of r. For anarbitrary (1 � I � N) there exists exactly one u VN with u(xi) =

u = where the basis functions are characterisedby

(8.3.9a)

If T E r is a triangle with the vertices x' = (xi, Vi) as in (9a)) andx' = (x', y'), x" = (s", then

I — '\i'" '\ / 1'tI I I—

— (Xj —x')(y" — 7/) — (vi _7/)(xII — x')on T. (8.3.Qb)

On all T E r that do not have an as a vertex, = 0.

PROOF. Obviously, using (9b) we get a linear function defined on all T E r,which satisfies (9a.). In addition, (9a) forces continuity at all nodes. If the


= (x,,y,) andxk =(xk,yk) are directly connected by the side of atriangle, then (9b) provides the representation + s(xk —xJ)) = sbj(xk) +(1 — with a (0,11 for both triangles that have this side in common.Thus is also continuous on the common edges, so that C°(Q). Applyingthis consideration to two boundary nodes x',x" with = = 0 weIRemark 8.3.11. (a) The dimension of the subepace determined by (8) isthe number of interior nodes of T.(b) The support of the basis function is

= T E r has x' as a wrner}.

(c) Let B, be the interior of There holds fl B, = 0 if and only ifthe nodes and are directly connected by a side.

Corollary 8.3.12. a(u, v) = f0(Vu, Vv) dx is the bilinear form associatedto the Poisson equation. The interala = a(bj, b4) = >k (yb,7 dxare to be taken over the following Tk:(i)aZZTa withx' ifi=j;(ii) all Tk withx4 andx1 ifi1&j.(iii) We have Lgj=O ifx' andx1 are not directly connected by the side ofatriangle.

FIgure 8.3.3. Bask for the node (z07y0)

We get an especially regular triangulation when we first divide intosquares with sides of length h, and then divide these into two triangles (0).The first and second triangulations in Figure 2 are of this sort. We call them"square grid triangulations". The corresponding basis function is depicted inFigure 3. One therefore expects that the matrix L corresponds to a 7-pointformula. For the Laplaes operator, however, one finds the well-known 5-pointformula (4.2.11) from Section 4.2:


Exercise 8.3.13. Let r be a square grid triangulation. Further let a(u, v) =Vv) dx. The basis functions are described by Figure 3. Show that one

has for the entries of the stiffness matrix L

= 4, = —1 if x' x3 = (0, ±h) or (±h,0),

L agrees with from (4.2.8).

Although L h2Lh holds in the case of the Poisson equation =the finite-element discretisation and the difference method still do not coincide,since h2fh has h2g(x') as components and so differs from f dx.

V

FIgure 8.3.4. R.eferenoe triangle T

The integration ft... dx over the triangle T1 E r seems at first difficult.However, for each i one can express fT. ... dx as an integral over the referencetriangle T in Figure 4. The details are in

ExercIse 8.3.14. Let x' =(x',y') (i = 1,2,3) be the vertices oITE r, andlet T be the unit triangle in Figure 4. Show:(a) ,j) x1 + — x') + — x1) maps T onto T.(b) (x2 — x1)(? _yL) — (y2 —y1)(x3 — xl) for all IR.

(c) The substitution rule gives

Jv(x,Y)dxdY=(8.3.10)

In general one evaluates the integral dq over the unit trianglenumerically. Examples of integration formulae can be found in Schwarz (1,§2.4.3) and Ciarlet [1, §4.1). The necessary ordering of the quadrature formulaeis discussed in the same place by Ciarlet (14) and by Wztsth [1).

'7

(0,1)x3

(0,0) (1,0)


In contrast to the difference methods finite-element discretisation offersone the possibility of changing the size of the triangles locally. The thirdtriangulation in Figure 2 contains triangles that become smaller as they arenearer to the intruding corner. This flexibility of the finite-element methodis an essential advantage. On the other hand, one does obtain systems ofequations Lu = f with more complex structures, since (a) u can no longer bestored in a two-dimensional array, (b) L cannot be characterised by a star asin (4.2.12).

Remark 8.3.15. If one replaces the Dirichiet condition u = 0 on 817 bynatural boundary conditions, then the following changes take place:(a) N = dim VN s the number of all nodes (inner and boundary nodes).(b) VN c H'(Q) is given by (8) without the restriction "u = 0 on 811".(c) In Remarks 9—li it should read "nodes" instead of "inner nodes".(d) The matrix elements calculated in Exercise 13 are valid only for innernodes For a Neumann boundary value problem in the square 17= (0,1) x(0,1), L coincides with h2DhLh from Exercise 4.7.8b.(e) In calculating = one has to take account of possible boundaryintegrals over 811 fl if 811 fl 0.

8.3.3 Bilinear Elements for 11 c 1R2

The difference procedures In a square grid are near to partitioning 1? intosquares of side h (cf. Figure 5a). If, more generally, one replaces the squares byparallelograms one obtains partitions like thoee in Figure 5a,b. An admissiblepartition by parallelograms is described by the conditions (7a—d), if in (7a)the expression "triangle" is replaced by "parallelogram".

(a) (b)

FIgure 8.3.5. PartitIon of 0 Into parallelograms

If one were to define the subspace VN by the condition that u must bea linear function in each parallelogram, then there would be only three ofthe four vertex values that could be arbitrarily assigned. In the case of thepartition in Figure 5b one can see that the only piecewise linear function uwith u = 0 on 81118 the null function. Thus in each parallelogram u must


be a function that involves four free parameters. We next consider the case of

a rectangle parallel to the axes P = (z1,x2) x (yj,y4 arid define a bilinearfunction on P by

u(x,y) = + a2z)(aS + a4y). (8.3.lla)

Then u is linear in each direction parallel to the axes — thus, in particular,along the sides of the rectangle. For an arbitrary parallelogram P such as inFigure 6a, the restriction of the function (1 la) to the side of a parallelogramis in general a quadratic function. Therefore one generalises the definition asfollows. Let

i-+ x1 + — x1) + — x1) (8.3.llb)

be the mapping taking the unit square (0,1) x (0,1) onto the parallelogramP (cf. Figure 6a,b). A bilinear function is defined on P by

u(z, y) := y)), v(e, (a + + 5i). (8.3.1 ic)

it is not to calculate v(4r'(x1 y)) explicitly, since all the integrationscan be carried out over (0, 1) x (0, 1) as the reference parallelogram (cf. Exercise14).

y '1

(0,1) (1,1)

x0.

(0,0) (i,0)

FIgure 8.3.6. (a) Parallelogram (b) Unit square as reference parallelogram

If ir = is an partition into parallelograms onedefines VN C H'(a) [rasp. VN C by

VN := {u E C°(Th): on all P u coincides with

a bilinear ftinction} (8.3.12a)reap. VN := C°(Th: ii =0 on 0S?; on all P E u coincides

with a bilinear function} (8.3,12b)

Here N = dixnVN is the number of nodes [rasp. inner nodasj.

£1

180 8 The Method of FInite Elements

AbilinearfünctiononPEwislinearalongeac*lsldeofP. Continuityinthe node points thus already implies continuity in $1.

Remark 8.3.16. The Remarks Sa, 9, 10, and 11 hold with appropriatethanges.

Exercise 8.3.17. Let the bilinear form aesociated to be a(u,v) =f0(Vu, Vv) dx. Aesume (1 = (0,1) x (0,1) is divided into squares of sideh as in Figure 5a. What are the basis functions characterised by (9a)? Show

—1 —1 —1

that the matrix L coincides with the difference star -1 8 -1—1 —1 —1

FIgure 8.3.7. A coinbinatlon of triangles and parallelograms

Remark 8.3.18. Thangle and parallelogram divisions can also be combined.A polygonal domain a can be divided up into both triangles and parallelo-grains (ci. Figure 7). In this case VN is defined as

{n E u linear on the triangles, bilinear on the parallelograma}.

8.3.4 Quadratic Elesnents for fl c

Let 'r be an adinindble triangulation of a polygonal domain I?. We wish toincrease the dimension of the finite-element sulispace by allowing, instead oflinear functions, quadratics

u(x, y) = afl + + + ajtx2 + + on E r

VN={UEC°(fl):u=0000a; u coincides

so that

(8.3.13)

with a quadratic function). (8.3.14)

8.3 FInite Elements 181

(b)

Figure 8.3.8. Nodes for(a) a quadratic ansatz on a triangle,(b) a quadratic a.nsatz of the serendipity clam on a parallelogram

Lemma 8.3.19. (a) Let x1,x2,x3 be the vertices of a triangle T E r,while x4, x5, x6 are the midpoints of the sides (cf. Figure 8a). Each functionquadratic ont is detennined by the values {u(x3):j=1,...,6}.(b) The restriction of the function (13) to a side of T E r gives a one-dimensional qt&adratic function, which is uniquely deterinine4 by three of thenodes lying on this side (e.g., u(x'),u(x4),u(x2) in Figure 8a).(c) flu is quadratic on each E r and continuous at all nodes (i.e., verticesof triangles and mid-points of sides), then u is continuous on 11.

PROOF. (a) The quadratic function can be obtained as the uniquely definedinterpolating polynomial of the form (13). Part (b) is also elementary.(c) Let T and T be neighbouring triangles in Figure 8a. Since UIT andcoincide on x1, x4, x2, by part (b) they represent the same quadratic functiononthecocninonsideofTandT.

We call all [inner] vertices of triangles and mid-poiiits of sides tinner]nodes. By Lemma 18a we can find (or each inner node a basis functIon

By Lemma 18c belongs to VN. This proves

Remark 8.3.20. The number of inner nodes is the dimension of Vjy Lfl14. Each u E VN admits the expression u = where the basisfunction belonging to the node is tharacterised by (9a).

In the sequel we wish to assume that, as in Figure 7, both trianglesand parallelograms are used in the partition. On the triangles the functionu E VN are quadratic. The functions on the parallelograms P must satisfy thefollowing conditions:(a) u(z,y)Ip must be uniquely determined by the values at the 4 vertices and4 mid-points of the sides (cf. Figure 8b).(/3) The restriction to a side of P gIves a (one-dimensional) quadratic function(of the arc length).

By condition (a) the ansatz must contain exactly 8 coefficients. Thequadratic (13) has only 6 while the biquadratic

(a)


has one parameter too many. If one omits the term in the biquadraticansatz one obtains for the unit square the function

= + + as?? + + + aaq2 + are2,i + (8.3.15)

the so-called quadratic ansatz of the serendipity class. The restric-tions to the sides = 0, 1 [resp. = 0, 1] give a quadratic function inn Iresp. The mapping in (1 Ib) (cf. Figure 6a,b) yields the functiontt(x,y) = defined on the parallelogram P, which still satisfiesthe conditions (a) and (f3).

If one were to use the full biquadratic ansatz one would one additionalnode which one could choose as the barycentre of the parallelogram. Cubicansatzes can be carried out in the same way in triangles and parallelograms(cf. Schwarz f 1]).

8.3.5 Elements for Si C It3

In the three-dimensional case assume

isapolyhedron. (8.3.16)

The triangulation from Section 8.3.2 corresponds now to a division of .0 intotetrahedra. it is called admissible, if (7a—d) hold in the appropriate sense: (7a)becomes "Z(1 5 iS t) areopen tetrahedra"; in(7d)itnowshouldread "ForI j, Tg ii T, is either empty, or a common vertex, side, or face of and T,".

Each linear function al + + asy + a4z is uniquely determined by itsvalues at the 4 vertices of the tetrahedron. As a basis for the space

VN={tIEC°(Th): u=O on OS?; u islinearon each tetrahedron Tj (1 I 5 t)}

one chooses with the property (9a): = 1, = 0 (j 1). Thesupport of consists in all tetrahedra that share x' as a vertex. The dimensionN = dim VN is again the number of inner nodes (i.e., vertices of tetrahedra).

As in the two-dimensional case the linear ansatz may be replaced bya quadratic one. Instead of a tetrahedron one can use a parallelipiped or atriangular prism with corresponding ansatses for the functions (cf. Schwarz111).

8.3.6 Handling of Side Conditions

The space V which lies at the foundation of the whole matter may be asubspace of a simply discretisable space W V. A given function w Wbelongs to V if certain side conditions are satisfied. Before we describe thissituation in full two examples will be provided as


Example 8.3.21. If one wishes to make the Neumann boundary-value prob-lem = gin (1 and 8u/8n on 1' uniquely solvable by the addition ofthe side condition u(x) dx =0, one can choose the space V to be

V = {t& H'(a): J u(x) dx = 0). (8.3.17a)

For a bounded domain (1 the form a(u, v) := f0(Vtt, Vii) dx is V-elliptic. Theweak formulation (1.1) with 1(v) :=equation = g, Oufôn = if9 and satisfy the integrability condition/(1) = 0 (cf. (3.4.2)). But even when f(1) 0 there exists a weak solution ofthe corrected equation —4u = := g(x) — [fag dx + dx.

Example 8.3.22. The Adler problem of Exercise 7.4.15 uses

V = {u H1(a): u constant on r}. (8.3.17b)

In both cases W = H1(.Q) is a proper superset of V. Let r be a trian-boundarynodes.Let WhcWbe

the space of linear triangular elements (cf. Remark 15):

Nh := dim Wn = + 4d. (8.3.18a)

The basis functions (bi: 1 i C will be described as usual by= öjj (cf. (9a)).

Astheflnita-elementsubspaceofV we define Vh := Whfl V, which isofsmaller dimension:

Vh:_—W4flV; Mh:—Nh—dirnVh. (8.3.18b)

In the Examples 21 and 22 we have Mh = 1, reap. MA = - 1. Thedifficulty in the numerical solution of the discrete problem (19),

Find i1 E V, with a(,i',v) = 1(v) for all v Vh, (8.3.19)

the for example, in Example 21 no belongs toV,andthuaalsononetoVh,amcefobidx>o.

In principle, it is poesible in the case of (17a) to construct as new basisfunctions linear combinations := V, which have again localised(but a bit larger) supports. In the case of (17b) it is still relatively emiple tofind a practical basis for Va: The basis functions b4 which belong toInner nodes are also in since &j H01(Q) C V. As further elements oneuses bo(x) := E'bj(x), where E' denotes the sum over all boundary nodes.In spite of this, even in this case, it would simply be easier if one could workwith the standard basis of WA.

In order to treat the problem (19) with the aid of E we reintroducethe notation = Pw (w a coefficient vector, of. (1.6)).


Eachv€Vh,CVh} = P'Vh. Thus problem (19) is equivalent to:

Find u E V with a(Pu, Pv) = f(Pv) for all v E V. (8.3.19')

From (1 8a,b) we have dim V = dim Vh = Nh - Mh = dim - Mh. Thespaces V are described by Mh linear conditions

N,,

= 0 (1 � I � Mh); (8.3.20)j=l

V = Ker C = {w E = O}, (8.3.21)

where C = (CU) is an Mh x Nh matrix. In the case of Example 21 we have

Mh=1, ci,=Jb,(x)dx

For Example 22 let x0,. ..,xMh, with Mh = 1, be the boundary nodes.Then C can be defined as follows:

= -1, CU =0 otherwise.

The variation of v over V in (19') can be by the variation of w overIRNh if one couples the conditions (20) by using Lagrange multipliers

(1� which can be put together to form a vector A =(A1,... ,AM,,). The resulting formulation of the probleni is:

find u E and A with Cu =0 and (8.3.22a)

a(Pu, Pw) + (A, Cw) = f(Pw) for all w (8.3.22b)

where (A, = E is the acalar product in

Theorem 8.3.23. The problems (19) and (22a,b) are equivalent in the fol-lowing sense. IfuA is a solution pairfor(22a,b) then u is a solution of(19')and Pu is a solution of (19). Conversely, if uh = Pu isa solution of (19)then there exists precss ely one A E u and A solve

Before we prove this theorem, we give a matrix formulation which isequivalent to (22a,b)

Remark 8.3.24. Let L and f be defined as in (1.8a,b). Further,that B := CT with C from (21). Then (22a,b) is equivalent to the system ofequations

FL Biful Ff1IBT oJ [Aj [oJ (8.3.22')


PROOF. Lu + BA = f is equivalent to Pw) + (A, Cw) = (Lu, w) +(BA,w) = (f,w) = f(Pw) for all w E JR. Note BTU = 0 is the same as

Cu=O.

PROOF (of Theorem p3).

(a) Assume a solution of (22a,b) is given in terns of u, A. Then Cu = 0 implies

U E V and := Pu E Vh. Equation (22b) holds in particular for all w V,so that (19') follows since Cw 0.(b) (19') implies I — Lu E V1. Since V1 = (KerC)1 = (KerBT)'- = Im(B),there exists a A E IRMh with f — Lu = BA, so that (fl'), and thus (22a,b) are

satisfied. For the Nh x Mh matrix 13 with rank we have KerB = {0}, so

that the solution is unique. U

Note that in the case of Example 21 the system of equations (22') isessentially identical to (4.7.lOa,b).

8.4 Error Estimates for Finite Element Methods

8.4.1 for Linear Elements

In this section we shall restrict ourselves to the consideration of the linearelements from Section 8.3.2 and therefore assume:

r is an admissible triangulation of 0 C (8.4.la)

VN is defined by (3.8), if V = (8.4.lb)

VN is as in Remark 8.3.15b, if V = H'(i7). (8.4.lc)

By Theorem 8.2.1 one has to determine d(u, VN) = WE VN}. Todo this one begins by looking at the reference triangle from Figure 8.3.4.

Lemma 8.4.1. Let T = � 0, + , <1 }. For all u H2(T) therehoLds

� lu(0,0)12 + Iu(0, i)J2 + Iu(1,0)12 + J. (8.4.2)

PROOF. (a) First it has to be shown that the bilinear form

a(u, v) := u(0, O)v(O, 0) + u(0, 1)v(0, 1) + u(1,0)v(1, 0) + D°v)L3(T)

is continuous on H2(T) x H2(T) and is H2(T)-coercive. The continuity followsfrom the continuous embedding H2(T) C C°(T), which implies tu(x)l �for all x T (cf. Theorem 6.2.30). Thus we have � (1 + 3O)IuI2IvI,.

186 8 The Method of Finite EUements

Since T E C°", one may apply Theorem 6.4.8b: 112(T) is oompactly embeddedIn H'(T). By Lemma 6.4.13, there exists a constant C,12 such that

� + Cii2IuIo) + for all H2(T).

Since = — the estimate

a(u,u) � — � —

shows that a(.,.) is H2(T)-coercive.(b) Since the embedding 112(T) C L2(T) is also compact one may applyTheorem 6.5.15. The operator A E 112(T)') which is associatedwith a(.,.) either has an inverse A' L(112(T)', 112(T)) or has )'. = 0 as aneigenvalue with an eigenfunction 0 e 112(T). In the latter case e must, inparticular, satisfy the equation a(e,e) =0. From this follows that =0for all lal =2, so that e must be linear: e(x, y) = a + /3x + Furthermore,from a(e,e) = 0 one may conclude that e(O,O) = e(1,0) = e(0,1) = 0, sothat e =0 in contradiction to what was just assumed. From Lemma 6.5.3 andExercfse 6.5.6c one may show the a(u,u) � withCE := + 3O)J > 0 and C from part (a). Thus wehave assertion (2) with C := 1/CE.

Lemmi 8.4.2. Let := hT := {(x,y): xv � 0, x + y � h}. Fbr eachuEH2(Th)therehoidswithlf3l�2

[u2(0, 0) + u2(O, h) + u2(h, O)J (8.4.3)

+leI=2

with C independent of and h.

PROOF. Let := 6 JP(T). The derivatives with respectto (x,y) and to = (x,y)/h are related by = each

lID= JJ = jf= �

+ v2(1,0) + v2(O, 1) +

Since v(0,0) = u(0,0), v(1,0) = u(h,0), v(0, 1) = t40,h), and =one may conclude (3).


Lemma 8.4.3. Let T be an arbitvury truzngle with

Side lengths Interior angles � ao >0. (8.4.4)

For each u and � 2, there holds

� C(cio) +> I(D°u11t3(f)

otT(8.4.5)

where C(ao) depends only on a.o and not on u, or h.

PROOF. Let h � be one of the lengths of the sides of T. In a waysimilar to that in Exercise 8.3.14a, let Th T be the linear transformationthat maps Th onto T. Then := belongs to H2(Th). Underthe conditon (4) we know that the determinant Eis bounded both above and below. From

�Ip,'=I$l

(3), and� C2(oo) 111

IaI=2there then follows (5). U

From Lemma 3 follows the important result:

Theorem 8.4.4. Assume that conditions (Ia-c) hold for r, VN, and V. Letczo be thea whiiehüthe maximum lengthof the sides of all T1€r. Then

IIU—vflHk(a) � (f (ao)h2'11u11u2(a) fork = 0,1 and all u H2(a)nV.

(8.4.6)

PROOF. For a u u E H2(t?) fl if V =one chooses v := VN, i.e., v VN with v(x') = u(x1) at the

implies the estimate

T4Er

=JaI=2


Summation over � k now proves the inequality (6). S

Let be the longest side of T e r, then h := rnax{hp:T E r} is

the longest side-length which occurs in -r. The parameter h is the essentialone and will be used as an index in the sequel: r = rh. Denote by Pr theradius of the inscribed circle in T E r. The ratio hp/p1' tends to infinityexactly when the smallest interior angle tends to zero. If one has the boundmax{hT/p.r: T E T} � const for a family of triangulations, then one calls thefamily quasi-uniform. In this condition the constant should be the samefor all -r in the family {-rh). A family is called uniform if it fulfils thestronger requirement h/ T E r} � const. A family {rh} which is bothunform and for which h —.0 is called regular by Ciarlet 11, §3.1J.

In the construction of triangulations it should be noted that with increas-ing refinement —.0) the interior angle may not also be reduced. Strategiesfor the systematic construction of quasi-uniform sequences V4 can be found,for example, in Hackbusch (1, §3.8.2J.

The case k = 1 of Theorem 4 may now be written as follows:

Theorem 8.4.5. Assume conditions (la—c) hold for a sequence of quasi-unsform triangulations ri,. Then there ezuts a constant C, such that for allh = and Vh = there holds the following estimate:

inf lu — vii � ChIui2 for all u E H2(1l) fl V. (8.4.6')yE V,,

The combination of this theorem with Theorem 8.2.1 gives

Theorem 8.4.6. Assume conditions (la—c) hold for a sequence of quasi-triangulation, ri,. Let the bilinear form fulfil conditions (1.2) and

(1.12). Suppose the constants CN := > 0 from (1.12) are bounded belowby q,, � > 0 (cf. Corollary 8.2.3). Let Problem (1.1) have the solutionu E fl V. Let E V,, be the finite-element solution. Then there ezzstsa constant C, which does not depend on u, h = or v, such that

— uhil � ChIuI2. (8.4.7)

The combination of the Theorems 5 and 8.2.2 can be written:

Theorem 8.4.7. Assume conditions (la-c) hold for a sequence of quasi-untform triangulation. with -' 0. Let the bilinear form fulfil (1.2) and(1.12) with � > 0. Then the problem (1.1) has a unique solution u E V,and the finite-element solution E conveives to U:

IlL — 0 (v —+ oo). (8.4.8)

PROOF. Let u V and e >0 be arbitrary. Since 112(Q) n V is dense in V,there exists U( H2(1?) V with Iu — u4i � e/2. From (6) and h, —. 0 we

8.4 Error for Finite Element Methods 189

see there are i' and with — <e/2; thus Its — � e. Thisproves (2.4a).

Theorem 7 proves convergence without restrictive conditions on u. Onthe other hand the order of convergence 0(h) in the estimate (7) requires theassumption that the solution u V lies in H2(S1). As will later become clear,this assumption is hardly to be taken as fulfilled in every situation. A weakerassumption is u V fl H8(Q) with s E (1,2). The corresponding result is asfollows.

Theorem 8.4.8. Assume that in the assumptions of Theorem 6 u E V fl112(0)ssreplacedbyu€VflH(17)withl LeU? be sufficientlysmooth. Then there holds

— u"ti � (8.4.9)

The proof uses a generalisation of Theorem 4:

Lemma 8.4.9. Under the assumptions of Theorem 4 and suitable conditionson 0, there holds

inf fu—vfi �C"(ao)h''1u15 forsELl,2J,uEH9(.Q)flV. (8.4.10)

PROOF. The proof is based on an interpolation argument that will not befurther explained here. Equation (10) holds for a = 2 (cf. (7)) and for a = 1,

since infju — vii S — Ok = luli. From this follows (10) for a E (1,2)with the norm of the interpolating space [H1(0) fl V, 112(0) fl (cf.Lions-Magenes [1)), which under suitable assumptions on 0 coincides withH(0)nV. U

For a = 1 the right-hand side of (10) becomes ccest.Ni. In fact theestimate inf{Its—vIj:v E V,j is the best poesible. To prove this ,chooeeI (orthogonal with respect On the other aides = 2 is the maximal

value for which the estimates (9) and (10) can hold. Even u E C°°(0) permitsno better order of approximation than 0(h)!

The Ritz projections SN : V —. VN introduced in (2.5a) will now bewritten Sh : V —, Vh. The inequality (9) becomes ju — Shuil/iul. �and this proves

Corollary 8.4.10. Assume conditions (la-c) fori-, Vp, and V. Let the bilinearform fulfil (1.2) and (1.12). The Ritz pmjeetion S,, satisfies the estimate

UI — Ch'. (8.4.11)

Under the asumptions of Lemma 9 there holds

— Sp,i1Jj1(n),...jj.(0)flv for all 1 $ � 2. (8.4.11')

190 8TheMethodolFlniteElementa

8.4.2 L2 and H' Estimates for Linear Elements

According to Theorem 6 0(h), is the optimal order of convergence. This re-sult seems to contradict the 0(h2) convergence of the five-point formula (cf.Section 4.5), since the finite-element method with a particular triangulationis almcot identical to the five-point formula (cf. Exercise 8.3.13). The reasonfor this is that the error u — tie' is measured in the I norm. The estimate

in Theorem 4 suggests the conjecture that the

I bo norm of the error is of the order 0(h2): lu — <Ch2IuI2. However,this statement is false without further assumptions.

Up to this point only the existence of a weak solution u E V (e.g., V =or V = H1(S?)) has been guaranteed. But in Theorem 6 we needed

the stronger assumption u E H2(Q) ii V. A similar regularity condition willalso be imposed on the adjoint problem to (1.1)

Find u E V with o(u,v) = 1(v) for all v E V, (8.4.12)

which usas the adjoint bilinear form a'(u,v) := a(v,u). For f L2(fl) C V',the value 1(v) becomes The regularity condition is:

ForeathfEL2(O) theproblem(12)8413

has a solution is E H2(1l) fl V with � C,4f10.

In Section 9 we shall see that this statement holds for sufficiently smoothdomains fl.

Theorem 8.4.11. (Aubin—Nitsche) Assume (13), (1.2), and (1.12) with =� > 0, and

It' — vfj <CohIuI2 for all is H2(i1) V. (8.4.14)

Let problem (1.1) have the solution is E V. Let E Vh C V be the finite-element solution. Then) with a constant C1 independent of is and h,

— CihIuli. (8.4.15a)

If the solution is also belongs to 1P(.t?) fl V, then these is a constant C2,independent of is and h, such that

— (8.4.15b)

&om Theorem 4, it is sufficent to ensure (14) thai Vh be the space of finiteelements of an admissible and quasi-uniform Inangulation.

PROOF (cf. Nitache [1J). For each e := is — E L2((l) define w H2(S?) fl Vas the solution of (12) for I =

a(v,w) = (e,v)L2(O) for all v E V. (8.4.16a)


Corresponding to w there is, by (14), a w' E V4 with

— w"li � CohlwI2 � COCRhIeI0. (8.4.16b)

Equation (2.3) is a(e, v) = 0 for all v V4; therefore, in particular, we have

a(e,w?&) = 0. (8.4.16c)

From (16a—c) we obtain

= (e, = a(e, w) = a(e, w — w4)

— w"Ii � CsJefiCoCahlelo,

and thusfeb � CSCJOCRh feb1. (8.4.l&J)

From (2.1) one deduces

febi — '11v S (1 + Cg/e4) inf fu — vlv � (1 +vEVh

so that (15a) follows with C1 := CSCOCR(1 + If u E H2(fl), one mayuse the inequality (7), Icli � Chlul2, and deduce (lSb) with C2 := CSCOCRC.ICorollary 8.4.12. The ineqtwlitics (ISa) and (lSb) aie to theproperties (17a) and (17b) of the Ritz projection S4:

JJI — <C1h, (8.4.17a)

flI — � C2h2. (8.4.17b)

The estimates (17a) and (17b) may also be proved directly. The definitionof S4 and the connection between S4 and S4 may be found in

Exercise 8.4.13. lb a(.,.) let the operator L: V —, V', the Ritz projectionS4, and the matrix L be associated. Show:(a) To the adjoint bilinear form a(.,.) belong the stiffness matrix V and theRitz projection

P(LT)_1P*L. (8.4.18a)

(b) There hoklsS4 = (8.4.14b)

PROOF. Second proof of Theorem 11: Let

(u E H2(a): u is a solution of (12) for an I L2(J1)) C V

be the range of E L(L2(Q),H2($1)). Equip the space H2(f1) with thenorm Since L2(fl)' = L2(Q) we have that L—1 E L(L2(f1), is


equivalent to L' E L(H?(fl)',L2(Q)) (ci. Lemma 6.3.2), so that there is aCa with

� Cci. (8.4.19a)

The assumptions (1.2) and (1.12) of Theorem 11 can be brought over withoutchange of the constants to the adjoint problem, so that the statement ofTheorem 8.2.1 can be written in the form flI — ShIIv,..Jp(a) � Forthe adjoint operator we have the estimate

— &) UH2(0'4—V' � (8.4.19b)

The basic assumption (1.2) may be written

IILIIv'.—v � C3. (8.4.19c)

From Equation (18b) one may read the expression

I — = I — = L'(I — = — Sh)'L.

From (19a—c) one has

— � —

(8.4.19d)

Therefore we have proved (17a) with C1 = CaCpC5. Since S5 is a projection,sois I—S5, so that

111 — = II(' —

� Ill — III —

From (17a) and (11) we deduce (ITh). U

The regularity condition (13) is weakened in the following theorem:

Theorem 8.4.14. Assume (1.2), and (1.12) with CN = > 0, and in-(10) for all I � a � 2. In the place of(13) we assume

for some t E 1). If V = H1(a) then assume Es/V = then assume fl V). Then thefinite-element solution satisfies the inequality

flu — u5J(, � htIIuIJ, (0 � t � 1 � 2). (8.4.20)

The Ritz projection satisfies

— � (8.4.21)

PROOF. (For the case V = H'(Q)) In (19a-c) replace L2(I1) by Ht(S?), andby with := {u is the solution to!

Ht(I?)'}. Because of (11'), (19b) becomes s replaced by 2—ti

8.5 Generalisations 193

IRI — HI ShIIv._H.2—.(n) �

As in (19d) one shows that — ShIIH.(O)+.V � Combining this with� Ch'1 (cf. (11')) there follows (21), and thus (20). I

For nonlinear boundary value problems, estimates of the errors in termsof other norms (e.g., L°°(a), are ofinterest. For this we refer to CiarletEl. and Schatz [li.

8.5 Generalisations

8.5.1 Error Estimates for Other Elements

The estimates for the errors for linear functions on triangular elements provedin Section 8.4 are also true for bilinear functions on parallelograms and forcombinations of both sorts of elements (cf. Section 8.3.7). The proofs are alonganalogous lines. Even for tetrahedral elements in a three-dimensional regionS? C 1R3 the same results can be carried over. For the proof one notices thatu J12(Q) has well-defined nodal values even for a C since 2> n/2(n =3, dimension of fl; cf. Theorem 6.2.30).

For quadratic elements (cf. Section 8.3.4) one expects a correspondinglyimproved order of convergence. In general one can show the following: If theansatz function is, in each E r, a polynomial of degree k � 1 (i.e., u(x) =

on T,), then

d(u,Vh)—in({Iu-.vIl:vE focallu€ (8.5.1)

(cf. Ciarlet El, Theorem 3.2.lJ). If one uses parallelograms as a basis anduses ansatz functions that are polynomials of degree at least k, then (I) alsoholds. For example, biquadratic elements and quadratic ansatz functions ofthe serendipity clase fulfil this requirement for k =2. The result correspondingto Theorem 8.4.6 follows from Theorem 8.2.1:

Theorem 8.5.1. Let Vh fulfil (1), and let the bilinear form satisfy (1.2)and (1.12) with =: > 0. Assume problem (1.1) has a solutiona E V n H16+l(fl). Then the finite-element solution a4 E satisfies theinequality

Ia — u411 Chklalk+l. (8.5.2)

The Ritz projection satisfies — � Chk.

The estimate (4.9) now holds for a [l,kJ if a H'(S7)nV. With suitableregularity conditions, there are, as in Theorem 8.4.11, the error estimates

Ia — u"lo � (8.5.3a)II*L — U3iIflb-L(fl), � (8.5.3b)


For (3b) one needs, for instance, '-regularity: For each f E (.0) theadjoint problem (4.12) has a solution u E

Remark 8.5.2. Under suitable assumptions the inequality (4.20) holds for1 — k <t < I <s <k + 1. For negative t the norm should be understoodas the dual norm of Ht(i2).

8.5.2 F'inlte Elements for Equations of Higher Order

8.5.2.1 Introduction: The One-Dimensional Biharmonic EquationAll the spaces of finite elements, Vp,, constructed so far are useless for

equations of order 2m> 2, since Vp, Hm(Q). According to Example 6.2.5,in order to have V,, C H2(.0) it is necessary that not only the function ubut also its derivatives change continuously between elements. The ansatzfunctions must therefore be piecewise smooth and globally in C'(.0).

As a model problem introduce the one-dimensional biharmonic equa-tion

u""(x)=g(x) forO <x <1, u(0)=u'(O)=u(l)=u'(l)=O

which becomes in its weak formulation

a(n,v)=f(v) where

(8.5.4)a(u, v) = / u"v" dx, 1(v) = j gv dx.

Jo Jo

Divide the interval .0 = (0,1) into equal subintervals of length h. Piecewiselinear functions (cf. Figure 8.3.1) can be viewed as linear spline functions, sothat it isnatural todefine Vp, aathespaceof cubic splines (withti=u'=Ofor x = 0 and x = 1) (cf. Stoer 11, §2.4) and Stoer-Bulirsob [1, §2.4J). We cantake as basis functions B-splines, whose supports in general consist of foursubintervals (ci Figure 1). Since cubic spline functions belong to C2(0, 1),they are not just in 1), but even in 1) n 1).

• $ U U U • S

FIgure 8.5.1. B-spline

Simpler yet than working with spline functions is to use cubic Hermiteinterpolation:

Vp, := {u C'(O, 1):u cubic on each subinterval,

u(0) = til(0) = u(1) = u'(1) = 0). (8.5.5)

8.5 Ceneraliaation8 195

To each of the inner nodes x3 = jh there are two basis functions and

with = = = = 1,bM(Z,) = = 0 for

k = 1,2 (j i) (ci. Figure 2).

. I S S U & U

(a) (b)

FIgure 8.5.2. Hermite basis functions

The support consist of only two subintervals. The expressionB are

b14(x) (h — Ix — — x,j + h)/h3 for Zj4 � X

for z

Exercise 8.5.3. Let and u21 (0 < I < 1/h) be the coefficients of the

expression u(x) = + E Show that the coefficientsof the finite-element solution of the problem (4) are given by the equations

+ 24ti1g — + = (856)— 6tLl,s+jJ + + +

where = f2ü = = ih.

The system of equations (6) differs completely from the difference equa-tions (cf. Sectioh 5.3.3), since in Equation (6) there appear values of the

functions at the nodes together with the values of the derivatives at thenodes.

8.5.2.2 The Two-Dimensional CaseThe ansatz (5) can be carried over to (1 C JR2 ifone starts with a partition

into rectangular elements. The ansatz function is bicubic:

vj'=o

on each rectangle of the partition.).

At each inner node one may prescribe the four values u, Conse-quently there are four unknowns and four basis functions y), . . . y)belonging to each node. The latter are products (j, k = 1,2) ofthe one-dimensional basis functions described in Section 8.5.2.1 (cf. Meis-Marcowitz Figure 15.221, Schwarz (1, §2.6.1)).

If one starts from a triangulation r, a fifth-order ansatz gives what iswanted: u(x) = on Er. The number of degrees of freedom is


21 (the number of v Z2 with v 0, lvi � 5). For the nodes one thepoints x1, . . . ,x6 in Figure 8.3.8a. At the vertices xt,x2,x3 one prescribes 6values {D"u(x2): 2}. The 3 remaining degrees of freedom result from giv-ing the normal derivatives at the midpoints of the sides x4, x5, x6.Notice that if two neighbouring triangles (T and T in Figure 8.3.8*) havethe same vertex values � 2, j = 1, 2} and the same normalderivative at x4, then both u and Vu are continuous on the shared side, i.e.,Vh C According to whether V = V = H2(Q) n orV = H2(ul) one has to set u, or the first derivatives, to be zero at the bound-ary nodes

Remark 8.5.4. The finite-element space Vh C H2(Q) described here can ofcourse be used for differential equations of the order 2m =2.

8.5.2.3 Estimating ErrorsInstead of (5.1) one obtains

inf{Iu — E Chk_m+liuIk÷l for all V, (8.5.7)

where k � m depends on the order of the polynomial ansatz (e.g., k = 3 forcubic splines, cubic (resp. bicubicj Hermite interpolation). Here m = 2 for thebiharmonic equation. As in Theorem 8.5.1, there follows from (7) the errorestimate

— t4hlm (8.5.8*)

for the finite-element solution E Vh. Under suitable regularity assumptionsone gets

lu — (2m — k — 1 <t < m <a <k + 1) (8.5.8b)

(cf. Remark 8.5.2). The maximal order of convergence 2(k — m) + 2 results for

solution of the adjoint problem (12) with I must belong toHk+l(a).

8.6 Finite Elements for Non-Polygonal Regions

Since the union of triangles and parallelograine only generates polygonal re-gions a polygonal shape was assumed in (3.6). The finite-element method is,however, in no way restricted to just such regions. On the contrary finiteelements can readily be adapted to curved boundaries.

8.6 Finite Elements for Non-Polygonal Regions 197

Figure 8.6.1. Curvilinear boundaries

Let V = H'(Q) and let .0 be arbitrary. The triangulation 'r can be sochosen that (3.7a,b,d) hold and the "outer" triangles have two vertices on1' = OS? as in Figure 1. In the convex case of Figure la one extends thelinear functions defined on T onto T := TU B. In the case that the boundaryis concave (see Figure ib) one should replace T by T := T\B. One copescorrespondingly in the situation of Figure ic. The nodes and the expressionsfor the basis functions remain undisturbed by these changes. is the union ofthe closures of all the inner triangles E r and all modified triangles T whichlie on the boundary. All the properties and results of Sections 8.3—8.4 extendto the new situation. The only difficulty is of a practical sort: To calculate Land f one has to work out integrals over the triangle T which involves arca.

Assume now V = H0' (0). The previous construction will not be a suc-cess, since the extensions of the linear functions to T will not vanish on theboundary piece r'flOT. Thus Vb C will not be satisfied. As long asthe region is convex, only the situation shown in Figure la occurs, andmay be extended to B C T by uh = 0. in the case of Figure Ib, one mustset the values at the inner nodes to zero, so that uh =0 on T = T\B, and inparticular = 0 on f'flO(T\B). All in all, what results is that the supportof any uh V,, is in a polygonal region inscribed in 0. One interpretationis the following: In the boundary value problem, .0 should be replaced by anapproximating region 0h C 5? (cf. Theorem 2.4.6). The finite-element solu-tion described agrees with that which would result from the smaller region.However, the error estimate in Theorem 8.4.4 only holds for the smaller re-gion IIu — ChIIuIIjJ9(fl). Since v =0 on J?\S?h for anyv V,, one should also estimate IltSIJgl(Q\ah). Now .Q\flh is contained in a

= {x Oa:dist(x,P) � 6} of width 5= max{dist(x,f'):x i2\S?h}.One can estimate as follows:

— = IIUIIHI(O\flh) � II1LIIHI(s1)

If Q\0, consists only of arc segments B as in Figure la (for instance,in the case of a convex region), and ill? E C", then S = 0(h2). From thisfollows the estimate iflfvEVh ((U — VHHI(n) � Ch2lfr4IH2(fl), as for a polygonal

(a) (b) (c)


region. If, however, as in Figure lb the whole triangle is part of Q\Qh, then6 becomes 0(h) and the approximation worsens to infVEvh Vu — VIIHI(Q) �Ch1/2fluHJp(l7) (cf. Strang-Fix p. 192)).

In order to adapt the trianglar or parallelogram elements better to acurved boundary one can use the technique of isoparametric finite ele-

ments. Fiom Figure 8.3.4 we see that the reference triangle T may be mappedinto an arbitrary triangle T E r by an alfine transformation '-+

x' —x') . The linear fresp. in the case of §8.3.4, quadratic)function u(x) on T can be expressed as the image, = of a lin-

ear [resp. quadratic) function v on T. We now replace the affine transformationof triangles by a more general quadratic mapping

( Oj + + + + ÷ T+ + açfl7 + + + 012,72) C

The image T is a curvilinear triangle. The coefficients . . . ,012 are uniquelydetermined by the nodes x11... ,x6 of T which are the images of the verticesand midpoints of the sides of the reference triangle T (cf. Figure 2). Thetriangulation r used so far can be replaced by a "triangulation" by triangleswith curved sides, if neighbouring elements have the same arca as commonboundaries and the midpoints also coincide. The ansatz functions on T Ef have the form u(x) = where is linear Iresp quadratic]in and,7. The resulting finite-element space is the space of isoparametriclinear (reap. quadratic] elements (cf. Strang-Fix [1), Ciarlet [1), Sdiwarz 11],Zienkiewicz[11).

x6

FIgure 8.6.2. MappIng of the reference tangle T to the curvilinear triangle T

In general, there is no reason to use curvilinear triangles in the interiorof .0. As in Figure 3, one chooses ordinary triangles in the interior (i.e., thequadratic transformation is again taken to be linear). A boundary triangle,like T in Figure 3, is, on the other hand, defined as follows: x' and x2 arethe vertices of T which lie on V. One chooses another boundary point x4between x1 and x2 and requires that the boundary of the triangle cuts theboundary I' of the region at x1,x4, and x2.


XX)X4

FIgure 8.8.3. Isoparainetrlc triangulai diseection

8.6.1. Section 5.2.2 shows that, in the case of other than Dinch-let boundary conditions, the construction of difference schemes is increasinglycomplicated. Instead one may restrict the usual difference methods to theinner grid points, and near the boundary discretise by using (e.g., isopara-metric) finite elements.

87 Additional Remarka

it is not poasible to consider here many of the details about, and modificationsof, finite-element methods. In the following, individual cases are just treatedsummarily.

8.'T.l Non-Conformal Elements

Condition (1.3), VA c V, characterises conformal finite-element methods.Discretisations for which VA V are called non-conformal. An exampleof a non-conformal element is the "Wilson rectangle". For this method onestarts with a diseection into rectangles x and assumesthe function to be quadratic on Rj:

VA := {u quadratic on u continuous at vertices of the rectangle R4.

Notice that u VA is required to be continuous only at the nodes, and notin the interior of 11. A function which is not continuous along a side ofcannot belong to H'(fl): VA V = H1(a). A possible basis for Vh is thefollowing. For each node x3, let bj be the basis function belonging to thebilinear elements (cf. Section 8.3.3). The quadratic ansatz chosen here has two


additional degrees of freedom. Because of this one chooses for each rectangle= (Xis, X25) x two further basis functions

y) := (x — — x), y) := (y — —

which vanish at the vertices of so that continuity is ensured there if

and are extended by zero outsideThe first problem which arises for non-conformal elements is the definition

of the stiffness matrix L. A definition in terms of = as in (1.8a),is not possible because bj, V! Therefore, in addition to discretising V toVh, one must replace a(.,.) by a bilinear form ah(., .): Vh x Vh —i ff1.. apartition of £? into rectangles one defines ah(•,•) by

ah(u, v) >2f >2 dx,Rs IaI,t$I�m

if a(.,.) is given by (7.2.6). Notice that a(u,v) = ah(u,v) for u,v V. Con-formal finite-element methods always give consistent discretisations. For non-conformal methods one does not always have consistency. For this problem, inparticular for the so-called "patch test" see Strang-Fix [1, p. 174], Ciarlet

§4.2], Sturnmel (2), Gladwell-Wait [1, p. 83—92], and Thoinasset [11.That an ansatz which seems completely reasonable can possibly yield

completely false discretisations is shown by

Exercise 8.7.1. Let r be an admissible triangulation. Assume that Vh{u constant on each T E r}. Show that

(b) inf{Iu — yb: v E V,1} � ChIuIj for all u E V =(c) Let a(u, v) (Vu, Vv) dx and ah(u, v) := J,(Vu, Vv) dx. Whatis the result for =

Non-conformal elements are, in particular, of importance for equations ofhigher order, since they are lees complicated than conformal elements.

8.7.2 The ¶frefftz Method

In Theorem 6.5.12 it was shown that for symmetric and V-elliptic bilinearforms the weak formulation (1.1) of the boundary value problem is equivalentto the variational problem

Find u E V with J(u) = min{J(v):v V} (8.7.1)

(cf. also Theorem 7.2.9, Exercise 7.3.8). Another variational problem

Find w E W with K(w) = E W} (8.7.2)

is called the dual or complementary variational problem to (1), ifboth have the same solutions u = w V fl W.


For example, the Poisson problem, —Llu f E L2((1) in .0, u =0 on F,Leads to J(v) = lVvI2dx for v E V := Hd(fl). A special dual variationalproblem is due to Trefftz (1926):

K(w) := fVwI2dx for WE W := {w H'(fl):-Ltw f}

(cf. Velte [1, p. 911). Notice that v E V satisfies the boundary condition v =0on F, while for vi E W no boundary condition is required though it mustsatisfy the differential equation Lw = I.

8.7.3 Finite-Element Methods for Singular Solutions

The error estimates provided so far, such as for instance (4.7), depend on theassumption u E H2((2) fl V, winch will be discussed more carefully in Section9.1. That this assumption does not always hold is shown by

Example 8.7.2. (a) The Laplace equation = 0 in the L-shaped regionof Example 2.1.4 has the solution u = r213 — This function only

(b) The solution u = in Example 5.2.3 only belongs to IP(.Q)for a <3/2.

If one replaces, for example in the L-shaped region, the Laplace equationby the more general Poisson equation, = I in .0, u = 0 on F, thenone may still show that for f E L2(a) the solution u may be split up asu = u1 + — ir)/3) with u1 E H2(fl) fl and a E Jt(cf. Strang-Fix [1, p. 275ff.]). In this equation x(r) C°°(fl) is a specifiablefunction with = 1 in 0 r 1/4, x(r) = (I for r � 1/2, such thatx(r)r2"3sin(...) is ensured. An ordinary discretisation by lineartriangular elements can only attain Iu — = 0(h'), a < 2/3. However,if one includes the function sin(...) or similar functions in Vh, thenthe approximation of u in V1, can be improved: d(u, Vh) = 0(h), so that

— uhli = 0(h). For such matters refer to [1] andBlum-Dobrowolski [1). See also Gladwell-Wait [1, p. 119] and Hackbusch L41.

A completely different approach using finite elements which does not needexplicit knowledge of the singularity Is the following. Let r be a triangulationwhose triangles become finer near where the singularity is expected, for exam-pie, at a re-entrant corner (cf. Figure 2.1.1, 5.2.1). For analysis of the errorssee Schatz-Wahlbin

8.7.4 Adaptive Thangulatlon

The sort of triangulation mentioned in the last paragraph can be most easilyobtained by adaptive methods. So far r = rh has been given depending on agiven "step size" h. In an adaptive procedure one starts with a triangulation


r0 and uses a-pceteriori error estimates. Those triangles of r0 that lead tothe largest errors are then divided further so that a new triangulation r isformed. One repeats this procedure until the errors are satisfactory. For theliterature on this we refer to, for example, Rhernboldt's work [11 and the paperof and Rhemboldt referred to therein, as well as to Eriksson—JohnsonEl'.

8.7.5 Hierarchical Bases

Let us give the fundamental idea in the one-dimensional case. Let Vh C1) be, as in (3.2), the space of piecewise linear functions for an equidis-

tant division of the interval f2=(O, 1) intopartsoflengthh. Forh= 1/2 wehave dim(V112) = 1, and b1 as in Figure la is the only basis function. ClearlyV114 C V1,2. For V114 one takes over b1 E V112 as a basis function, and addsthe functions b2, b3 as in Figure ib; {b1, b2, b3} is a basis for Vj,4. SimilarlyV1,8 C Thus b1, b2, b3 supplemented by b4,.. . , (cf. Figure ic) givesa basis — the so-called hierarchical basis — of V118. In contrast to the basisfunctions that have as yet been used, the supports of the are not necessarilyof length 0(h); e.g., supp(bi) = [0,1]. From this it follows that the stiffnessmatrix has more entries than is otherwise usual. In spite of this, the abovechoice of basis does have important advantages:(i) A matrix L = L(Th) already calculated for is a submatrix of the stiff-ness matrix L = for Vh, so that (a global or local) refinement of thetriangulation requires only a minimum of additional effort.(ii) The condition cond(L) = IILII 11L'II of the stiffness matrix isbetter than in the standard case (for example, cond(L) = 0(logh) instead of0(h2); cf. Section 8.8).(iii) There are suitable methods available for solving the system of equationsLu = f. Further details on this can be found in Yserentant [1], Bank-Dupont-Yserentant [1,, and Hackbus& (9].

b1

(a)h= 1/2 (b)h= 1/4 (c)h=1/8FIgure 8.7.1. A hierarchical basis

8.7.6 Superconvergence

In difference methods the error is given by a mesh function Uh — RhU (uh: thedifference solution, u: the exact solution), while for finite-element methods the

8.8 Properties of the Stiffness Matrix 203

solution is defined on the whole region 12, and therefore u can be usedas the error.

A grid function u.1 could in principle be extended to a function ii" definedon all 17, for example, by diooeing the values (x': a grid point of 17h)as the nodal values of the piecewise linear or bilinear finite elements, and thusinterpolating with u5 := Pu5 (for P see (1.6)). In addition to the errors

at the gnd points, there are now the interpolation errors u — PRhuas well: u5 —u = — = P(uh — R5u) — (u — PR5u). As soon as theinterpolation errors are of the same order of magnitude as — Rats, one hasfor Ptih — u the same convergence assertions, otherwise Pu5 — u displays worseconvergence than — Rhu.

Conversely, for finite-element methods, one can ask if one has the sameconvergence properties (e.g., huh — R5uD 0(h2), R5: restriction to thenodes) for the values at the nodes u5 {u"(x4): nodes } or for suitable av-eraged values. If one has indeed better convergence than for u5 —u, one speaksof superconvergence. This notion can also be applied to difference quotients ofnode values, if these are more precise than � (ha(r= 1).For this see Lesaint-Zlárnal [1], Bramble-Schatz [1], Thomée [2], and Louis [1).Greater precision can be achieved with extrapolation techniques which, un-der suitable conditions, can be applied to finite-element problems (cf. Blurn-Rannacher 12]).

8.8 Properties of the Stiffness Matrix

The properties of the matrix L5 have been carefully studied for differenceprocedures since the solvability of the difference equations and the analysis ofconvergence properties depend on them. In the case of finite-element discreti-sation we obtain the corresponding assertions in another way. The factors thatcontrol the solvability and convergence are the subapace V5 and the operator

Vh —. V51,

which was introduced as LN in (1.lOa).The stiffness matrix L, for a given Vp,, depends on the basis {b1,.

chosen. Let

N5 :=dim(V5).

The isomorphism P : V5 defined in (1.6) maps the coefficient vectorv to Pv = E Fbr finite elements, in general, the coefficientscoincide with the nodal values: = (1 � i S Np,). The connectionbetween the matrix L and the operator L5 : Vh —4 is, according to Lemma8.1.7,

L = P'LhP, Lh = P'1LP1. (8.8.1)

The definition of in (1.lOb), (P*u, v) = (u, also depends on thechoice of the scalar product (.,.) in Let us choose here the usual


Nh

(u, v)=

(8.8.2a)

In Exercise 8.1.10 we have already established the following propertiesof L: If a(.,.) is symmetric [and V-ellipticl then L is symmetric (and positivedefinitel. Notice that the difference methods do not always have these proper-ties. For the Poisson problem there is a symmetric form, but nonetheless thematrix Lh in Section 4.8.1 is not symmetric (cf. Theorem 4.8.4).

The condition cond(L) plays an important role in the solution of theequation Lu = f. We would like to show that, under standard conditions, wehave cond(L) = (cf. Remark 5.3.10). First we have to decide on thefundamental norm of ". Up to a scaler factor

N,, 1/2

:= (n: dimension of 17 C (8.8.2b)

is the Eucidean norm coming from the scaler product (2a). The correspondingmatrix norm

sup{IfLvIIh/IIvIIh:O V E (8.8.2c)

is the usual spectral norm of L. As an alternative to let us introducethe norm

:= tIPuIILa(rn for u E (8.8.3)

In some important cases and are equivalent (uniformly in h). Asan consider the linear elements.

Theorem 8.8.1. Let h —40 be a sequence of quasiunifonn triangtdations.Let Vh be the space of linear elements defined in (3.8) with the usual basi8 (see(3.9a)). Then there is a sonstant Cp, which does not depend on h, such that

� � CplIuIIp. (8.8.4)

The basis for the proof of this is

Lemma 8.8.2. Let T = 'i): � 0, + < 1) be the unit ti'iangle (seeIf u is linear on T, then

+u2(1,0)-t-u2(0,1)J sJJ

1

T (8.8.5)

� ..[ti2(O,O) + u2(1,0) + u2(0, 1)].

PROOF. Put ui 1= u(0,0), is9 := u(1,0), u(0, 1). u(e,,i) =uj + (u9 — + (us — it follows that

8.8 Properties of the Stlthiees Matrix 205

12 1 1 fui

U2 us) i 2

The eigenvalues of the symmetrIc 3 x 3 matrix are = A2 -= I and A5 = 4.

Therefore the right-hand side is � and � 4E U

To prove Theorem 1 write as dx = ET.EI.,, IT. (Pu)2 dx.

Let : T —, be the linear maps from T to as in Exercise 8.3.14a.Let u', u", u" be the values of u = Pu at the vertices of T,. The inequality0 < C1h2 < Idet4<I C2h2 and Lemma 2 imply

+ uhI2 + uIhI2) � j IPuI2 dx = det çôJ Jj ,1))12

� (8.8.6)

By definition of a quasiuniforni triangulation all the angles of the triangles are0. Let M > 2w/go. Bath node belongs to at least one triangle, and to

at most M triangles. Since u', U", U" are the components of the vector u, thesummation in Equation (6) over all T1 E Th gives

� = J (Pu)2dx �

so that inequality (4) holds with Cp max(MC2/6, 24/C1)1/2. U

Exercise 8.8.3. Let Vh C be derived from the square grid triangula-tion (cf. Exercise 8.3.13). Show that � Hulla �

The matrixM := P'P, i.e., := J dx (8.8.7)

0is called the mass matrix when it occurs in engineering applications. Con-crete bounds in the inequality (4) have been given by Wathen [1J for variousspecific elements.

Remark 8.8.4. (a) Inequality (4) is equivalent to

IIMII � 11M111 �

(b) We have fluflp � Iih IIuIIh �PROOF. (a) The calculation = f0(Pu)2 dx = (Pu, Pu)o = (P Pu, u)= (Mu,u) S UMII(u,u) = proves the first inequality in(b). The next follows from = u) = �

= since Mis positive definite.


(b) From � (.3 and � (.3, using (b), we deduce theinequality (4). Conversely (Mu,u)112 = IuPjp � CpIIuIIa =implies that C3, since M j8 positive definite (cf. Lemma 4.3.25).Similarly, � follows from II"IIh � U

Since Vh is finite-dimensional Ch := sup{IuIi/IuIo:0 ti is finite.One says that Vh satisfies the inverse estimate if C, = O(h'), that is,

Hi�CihT'Ho for a1lu€Vh (8.8.8)

where C1 does not depend on it as h —, 0.

Theorem 8.8.5. Let rh be a sequence of qwzsiunifonn triangulations. Thenthe space of finite elements inirvdueed in (3.8) sati.tfies the inverse estimate.

PROOF. As in Theorem 1, the proof follows by transforming the integralsfT dx for and � 1 into integrals over T. Here we havein addition to transform D = into (ci. the proof of Lemma 8.4.2). U

Theorem 8.8.6. Suppose 0 C Assume that (1.2), (8), and IIUIIp �CPIIUHh hold. Then we have

IILII � (8.8.9)

PROOF. Multiply the inequality

(u, Lv) a(Pu,Pv) � CgIPuliJPvIi � CgC7h2IPuloIPvIo

by and set u := Lv:

= hTh (Lv, Lv) �

and thus liLvila � that is (9). U

Theorem 8.8.7. Assume (1 C Ft'. Let (1.12) hold with e > 0 andfl.flh<CpII.IIp Thenthere holds

0L'II (8.8.10)

PROOF. We know = and � 1/c (ci.Section 8.1), so that

= =� �


Since

= sup = sup I(P1v,f)I/IvIih

Vj,

= sup I(v, f)I/IPvlj = sup

<Cp sup =

we have that 11L'fUh � for all f; thus we conclude (10). 1The combination of Theorems 6 and 7 leads to

Theorem 8.8.8. Assume (1.2), (1.12) with Ch e > 0, (4), m = 1, and (8).Then we have

oond(L) (8.8.11)

The ideas involved in the proof can, in principle, be carried over to thecase 2m > 2, i.e., to boundary value problems of higher order. The inverseestimate becomes

� for all U E (8.8.12)

In order that inequality (4) hold one must be careful in defining the normV . If all the components u are nodal values (as for instanceis the case for the spline a atz for VhC then ji. llaombedefinedas in (2b). However as soon as the components u involve derivatives(D'Pu)(x'), the in (2b) must be replaced by For example, whenthe Hermits functions of (5.5) are used then u has the components and

(cf. Exercise 8.5.3), where = (Pu)(x4) and = Theappropriate definition of fl. Ha is + with n = I,since (0, 1) C

Exercise 8.8.9. Let V c U C 1" be a Celfand triple and assume V5, C V.Show that the inverse estimate

llullv S CIhmIlulIu for all u (8.8.13a)

implies

Hulk, � CihmIIujlvi and holly � Cih_2mlluhlv, for all u E (8.8.13b)

9 Regularity

9.1 Solutions of the Boundary Value Problem inH8(O), 8> m

9.1.1 The Regularity Problem

The weak formulation of a boundary vaLue problem

Lu—g lull, Bu=p on!' (9.1.1)

asuEV, a(u,v)=f(v) forallvEV (9.1.2)

was, in Section 7, the basis upon which we were able to answer the questionsof existence and uniqueness of the solution. Here, by existence of a solutionwe understand the existence of a weak solution u V.

The error estimates in Section 8.4 made it dear that the statement u E Vis not enough. Under this assumption we can only show — uhllv 0. Themore interesting quantitative estimate Iu — uhIi = 0(h) as in, for example,Theorem 8.4.6 for V = HJ(a) or V = H1(i7), requires the assumption u112(Q) n V. The assertion u E 112(Q) or, more generally u E H'(s?), is astatement of regularity, i.e., a statement about the smoothness of the solution,which will be examined in greater detail in this section.

The regularity proofs in the following sections are very technical. To makethe proof ideas clearer, let us sketch the proof of inequality (4) below for theHelmholtz equation = fin Sic u =OonI'.Step 1: 5? = Since the bilinear form for this situation a(u,v) =

Vv) dx+f0 uvdx is H'(1R2)-elliptic, (4) holds for s = m = 1. Weshallprove (4) by induction for 8 = To this end we take the derivative ofthe differential equation with respect to x, + = If Ithen H3(fl) and the equation + v = by the induction as-sumption, has a unique solution v with IvL_i

If one sets up this inequality for v = and likewise for v =the result is IuIa � IuIe_i + IuxIe_i + Iui,Ia_i � 3C,—11f(8_2. Thus, (4) hasbeen shown for S.

Step 2: 5? = = lRx(O, oo). As above, we can obtain the estimatesince alsosatisfies = and the boundary condition

9.1 Solutions of the Boundary Value Problem in H'(Ii), s > m 209

= 0 on 1. This is not the case for u1. But implies E

E

show This property, however, results from the differentialequation, u — f — E

Step 3: Let 11 be arbitrary, but sufficiently smooth. As in Section 6.2.1, Illsdecomposed into (overlapping) pieces il, which can be mapped into 1R2 or

Correspondingly one splits the solution u into Exsu is a partitionof unity). Then the arguments from steps I and 2 prove inequalities forwhich together result in (4).

Note that only a sketch of the proof was given. Some of the steps of theproof are incomplete. For example, might not the equation + =have a solution in 1R2, E L2(1R2), which does not belong to H' (1R2) andhence does not coincide with the solution v H'(1R2) of + v =

In the following, always let s m. The boundary value problem (2) withV = is sald to be H3-regular if each solution u of problem(2) with f H3_2m(Sl) belongs to and satisfies the estimate

IuI. + (9.1.3)

If L is the operator associated with a(., then it is also said that L isH'-regular.

Remark 9.1.1. (a) H's-regularity always holds.(b) Let the variational problem (2) have a unique solution u with

IUIm < C01fI-rn for all / E H-m(Il). If the boundary value problem is H3-regular, then the weak solution of (2) with f satisfies the inequality

luIs � Cjf13_2m. (9.1.4)

(c) Let L be the operator associated with a(.,.). (4) is equivalent toH8(Q)) and (4'):

—1 I /IlL <C3. (9.1.4)

PROOF. (a) Set Cm = 1. (b) In (3) estimate IUIm by C0 I/I-rn and use the factthat because of s � in the embedding H3_2m(S?) C Hm(Q) is continuoussuch that � C'IfI._2m. Thus we obtain (4) with = C,(1+C0C'). U

The following remark shows that a perturbation of a(.,.) by a smoothterm of order <2m does not change the H3-regularity.

Remark9.1.2. LetLbeH'-regularforalltan operator of order � 2m —1: 5L for all

The assumption on ÔL holds in particular if 5L belongs to the bilinear forma"(.,.) in Lemma 7.2.12 and its coefficients 0afl are sufficiently smooth: forexample, €

210 9 Regularity

PROOF. (By induction) For 8 = m the statement follows from Remark Ia.Assume the assertion for 8 — 1. Let u be the weak solution of (L + ÔL)u =I hence also the solution of Lu = I := I — öLu. Under theinduction assumption the following already holds:

It&L_i � +JUIrn]

such that

IIIfla—2m � IfL—2m + � + ItLimi.

Because of the H-regularity of L, u belongs to H'(fl) and satisfies lu). �1 I.-2m + IUIm). Together these inequalities result in statement (4). U

9.1.2 Regularity Theorems for =

The domain (1 is distinguished by the fact that it has no boundary,and hence no boundary arnditions either.

Theorem 9.1.3. Let m IN, (2= lIt". Let the bilinear form

a(ti,v) := f (9.1.5)°

be Hm(IR")-coercive. For some k IN let the following hold:

E L°°((2) for all a, with H � max(O, k + — in). (9.1.6)

Then everij weak solution u E Hm(IR") of the problem

a(u,v) = (f,v)o for all v (9.1.7)

with / belongs to and satisfies the estimate

ItLIm+k <CkIIfIm+k + lUlmi. (9.1.8)

PROOF. (a) First we must investigate the start of the induction, k = 1. Let8= be the difference operator

:= i.th unit vector ,1 � i � n.

Let u be the solution of (7). v E H"'(lR") set

d(u,v) := + ap(Oh,gu,v), (9.1.9a)

where ap(u,v) := fm,.

9.1 Soluttose of the Boundary Value Problem in H(fl), B > in 211

Let := Prom ouôvdx = fm,. =_a+ u+]v dx = —h av Du dx

we see that

d(u,v) = fP.

— J — + —

IaI�.n P.181="

(9.1.9b)

Since � IvI,÷i (ef. Lemma6.2.24), one bounds the first sum in (9b) withCIUlmI8VIm_i < CIUI ml Vim. One obtains the same bound for the second sum:

ld(u,v)i � ClliimlVlm for v (9.1.9c)

However, if a(.,.) is ap(.,.) is also (cf. Lemma 7.2.12). ForCE >0 and CK from (6.5.10) we thus obtain

op(8u,Ou) + = —a(u,a2u) + d(u,Ou) +(9.1.lOa)

The first summand can be transformed according to (7): a(u, 02u)o = (f, 02u)oand bounded by The inequality JOvl. lvl.+i for 8= rn—iand v = Ou yields

8u)l � ifI_m+lIOh,it4im. (9.1.lOb)

By (9c), the bound for the second suxnmand in (lOa) reads:

� (9.i.iOc)

Finally we have

S lOh,iUlm_liOh,iUIm 5 (9.i.lOd)

(lOa.d) yields

S LifI-m+1 + (Cd + CK)lUlml/CE for all h >0, 1 5 i 5

Lemma 6.2.24 shows that u e Hm+l(JRn) and inequality (8) for k = 1.(b) Now let k = 2. In (9a) substitute EYT8v with ii'l = 1 for and integrateby parts:

212 9 Regularity

d(u, u) a(u, D78v) — v)

= E J dxit',

si

— E J+ J (Dau(x — 4ei))

+ (Oaap)

+ ++ + — (D°v) dx.

As in the first part of the proof, from Id(u,v)I � CdIUIm+lIvIm and

D78v)o( � If12_mID7OVIm_2 � 1112-miVim

follows

� [If I-m+2 + (Cd +

for all hi — 1, h > 0, 1 � I n. Thus, the differences are uni-formly bounded so that Lemma 6.2.24 proves u and IUim+2 �

+ IUIm+jJ. If one uses the estimate lUlrn+1 � CEIfi_m+i + tUImI �CEll + proved in (a), (8) follows for k — 2.

(c) An induction argument is carried out analogously for k � 2.

Corollary 9.1.4. The aoercivity in Theorem 3 ean be replaced (a) by thesufficient conditions/rem Theorem 7.2.11; (b) by the assumption that for someA the bilinear form ap(u,v) + A(u,v)o satisfies the assumption (6.5.4a,b).

Corollary 9.1.5. If, in a(.,.) is Htm(It")-elliptic or if a(.,•) satis-fies condition (6.5.4a,b), then in Theorem 3 the estimate (8) should be replacedby

� CklfI_m+k. (9.1.11)

PROOF. ByRemark2. U

Corollary 9.1.6. Let a(.,.) be Hm(IR')-coercwc. Let the conditions (6) andj E be satisfied fork f4 with k > s + n12 � n/2. Then the weaksolution u of (7) belongs to C(11t"). Hence for s 2m, the weak solution is(1180 a classsa11 solution.

PROOF. The statement results from Sobolev's lemma (Theorem 6.2.30). U

9.1 Solutiom of the Boundary Value Problem in IP(fl), 8> m 213

Corollary 9.1.7. Let a(.,.) be Let the conditions (6) and

/ E be satssfied for all k Ut Then the weak solution of problem

(7) belongs to The conditions sotssfled in particular if f belongs

to and the coefficients constant.

The generalisation of u with k to u e (1R0) with

real s > 0 reads as follows:

Theorem 9.1.8. Let a(.,.) in (5) be Let

s=k+0, 0<0<6<1, t:=k+i9.

For the coefficients (Q = let

aap E Ct Hm(S[j) for k + � m, E L°°(fl) otherwise. (9.1.12)

Then each weak solution of (7) with f E belongs to

and satssfle8 the estimate

IUIm+a <C.( If I—vn+. + IUtmI. (9.1.13)

PROOF. The beginning of the induction is given byk = 0, i.e., 0< s <t < 1.Replace the difference quotient 8 by an approximation to its power 88:

Ru(x) := Raju(x) :=

where is the unit vector. Here =1, and

() = -.s(1 - a)(2 - 8). . . (p -1-

is the binomial coefficient.

ExercIse 9.1.9. Let 0 < 8 < 1. Show that (a) The operator adjoint tois

R'u(x) WhJu(x) = —ph.,).

(c) Fbr the Fburier transform we have

= ((1 —

= [(1 —

Hint: + =

214 9 Regularity

(d) IR, h > 0, 1 � j � n,usimilarly � fUk+.. Hint: Make use of the norms I k and I. (cf.

(6.2.15)), and show � (1 +

By analogy with (9) one obtains

d(u, v) a(u, — ap(RLu, v)

=

+p=1

/ — — —

since the first summand in — = — iihe,) +— — — phe3) belongs to ap(Ru, v). Since we have— — oap(x)I � it follows that

id(u, v)I � CIUImI VIm [1 + ht' � C'iUimIVIm,

for it is true that = and e_M = O(ha_t). Thesame consideration as in Theorem 3 yields

� C[(P � C'EIfi_m+.+itLimj for all h >0, 1 � j � n

(cf. Exercise 9d). To obtain this estimate for jUIm+, we write

(i ++ f (i +

The second integral converges to zero for h -s 0. The first one can be estimatedby plus the sum

� c'J (1 +j=1 3

cuf (1 +3

= C" +

� Of (1 +

9.1 SolnUons of the Boundary Value Problem in H(Q), a> m 215

As in Lemma 6.2.24, u E and (13) follow.(2) As in the proof of Theorem 3, one carries out induction over k and inves-

tigates d(u, v) := a(tL, RD1'v) — v), = k. IExercise 9.1.10. Generalise Corollary 6 with the aid of Theorem 8.

9.1.3 Regularity Theorems for $1

The halfspace in (6.2.18) is characterised by x,, >0. As in Section 7, welimit ourselves to the following two cases: either a Dirichiet problem is givenfor arbitrary m 1, or the natural boundary condition is posed for m =1.

Theorem 9.1.11. (Homogeneous Dirichiet problem). An analogue toTheorem 3 holds for the Dirichlet problem:

U E a(u,v) = (f,v)0 for all V E (9.1.14)

Theorem 8 can also be carried over f one eXdUdC8 the values 8=1/2,3/2,...,m — 1/2.

For the proof we need the following highly technical lemma.

Lemma9.1.12. The nonnl•I.is equivalent to

:= + (9.1.15)

loI=m

PROOF. The relatively elementary case s � m is left to the reader. For0<8< m too, the proof would be considerably simpler if in (15) one were toreplace the dual norm) . of by that ofStep 1. First we prove the statement for .0 = IR" instead of $1 = Accord-ing to Theorem 6.2.25a, for .1? IR" the norm is equivalent to

:= ÷ m)2)h/2

Since = + 1e12)(' + and

0 < C0(1 ÷ < 1 + ( + <C1(l +

Iare also equivalent.

Step 2. For the transition to .0 = the following extensionH'(lR") must be investigated, where x = It":

216 9 Regularity

(4,u)(x):—u(x) for

(4,u)(x', := + u(x', for <0.

We list the properties of 4, as

Exercise 9.1.13. Let the coefficients of 4' be selected as the solutionthe system of equations a,..(vk + = (_l)k (0 k � L — 1). Shothat(a) U E yields 4,u E(b) 4,E for k=0,1,...,L.(c) The operator adjoint to 4, reads

= u(x', xn)+> _VXn)] for x',,>

(d) (8/ôx,2)k(4,*u)(xI,0) = 0 for k = 0,1,..., L —2 and U E(e) E for k = 0,1, . . . , L — 1. Hint: cf. Corollar6.2.43.(f) 4,E for a = 1 —L,2—L,...,0,l,...,L.

Step 3, � L results from D° E (provab.via continuation arguments from Remark 6.3.14b) so that � remairto be shown.Step 4. luL � I4,uI. � 1114'uIIL is true according to Step 1 of the proof. Tbinequality llJ4,uJjL � Cfr., which would finish the proof, reduces to

� for IcwI = m, U E

Let = m. For

fS Li. [(—v) "u(x , + (—i) u(x', —xe/v)] otherwise,

one verifies = As in Exercise 131 one shows that E

over to real a [1 + m — L, L — m] except for the cases — $ E IN (i.ea = —1/2, —3/2,...) (cf. Liona-Magenes [1, p. 54 ffJ). Let L � 2m + 1 anV E ffm—'(JRTh) Since one infers from(Dau, the estimate

I(Da4,u, v)oI = I It?Im_s,

for all V E and therefore (16) with

C := =

9.1 Solutions of the Boundary Value Problem in F1'(ø), 8> m 217

U

PROOF. (of Theorem 11) (a) First let k s = 1. The proof of Theorem3 can be repeated for the differences U = ,n — 1) and impliesthe existence of the derivatives Ou/Ox, E j n. Thus one has

for all = m except for a = (0,... ,0,m).(b)Weset .

m,

where is the coefficient from the bilinear form (5). The remainder of theproof runs as follows. In part (c) we will show that

$Fa(v)I S CIvIm_i for m, V (9.1.17)

Since = (w,D°v)o = (_l)rn(Daw,v)o, (17) means that Eand IDaWIl_m � C for (at = in. According to Lemma 12 it

follows that w E The coercivity of a(.,.) implies uniform ellipticityof L = i.e., � (cf. Theorem7.2.13). For = (0,... ,O, 1) one obtains aoa(x) � Hence it follows fromw E and E J'yJ = 1, that Dàu E Accord-ing to part (a) all other derivatives ((a( m,a a) belong toanyway so that u E has been proved.(c) Proof of (17). For each a a there exists a with ('y( = 1, = 0,

o < y � a (component-wise inequalities). Integration by parts yields

Fa(V) =— J + a&&(D dx

for v E and thus

<C0IVIm_i with Ca : CEIUIm + for all v

Here we used the fact that DTu E aceording to part (a). Sinceis dense in (17) follows for a a.

There remains to investigate a a. We write

F6(v) = a(u, v) — a(u, v) with &(u, v) = J dx,

where E' represents the summation over all pairs (o,f3) (&, a). For eachthere exists with

h'I=l,

As above, one integrates each summand with = in by parts:

218 9 Regularity

dx

=— J dx

— J dx

for v E C = C(u). Alto-

gether one obtains Iâ(u,v)I � CIVIm_l. Together with Ia(u,v)I = �IfI_rn+iHrn—i, (17) also follows for a = &(d) By induction for (k 2,...) one proves in the same way IFa(V)I �CIvlm_k, and from this u E For real a >0, 8 1/2,..., rn— 1/2,one proves correspondingly 1F0(t')I CIVIm_a and hence u

The generalisation of Theorem 11 to inhomogeneous boundary valuesreads as follows.

Theorem 9.1.14. Let the biliriearfortn a(.,.) from (5) beFor an 8 > 0, a {1/2,. . . , m — 1/2} either let (6) hold if a = k E IN, or(12), ifs N. Let u E be the weak solution of the inhomogeneousDirichiet problem

a(u,v) = (f,v)o for all v E (9.1.18a)

8'u/8n1=çoj forl=0,1,...,m—1, (9.1.18b)

where

f (0 � I � m — 1). (9.1.19)

Then it belongs to and satisfies the inequality

IUIm+. � C. [if i...m+a + ÷ IUtm]. (9.1.20)

PROOF. For m = 1 Theorem 6.2.32 guarantees the existence of itowhich satisfies the boundary conditions (18b) (for m> 1 ci.

Wioka [1, Theorem 8.8J). w it — ito is the solution of the homogeneousproblem a(w,v) = F(v) (f,v)o — a(uO,v) (cf. Remark 7.3.2). Theorem 11

may also be carried over to the right-hand side F(v) under discusaion here(instead of (1' and yields w E Hm+a(lRfl) U

By similar means one proves

Theorem 9.1.15. (Natural boundary conditions) Let the bilinearform a(.,•) from (5) be Fbr 8 > 0 either let (6) hold if8 = k N, or (12) ifs N. Let it be the weak solution of theproblem

a(u,v)= 1(v) =1 9(x)v(x)dx+jp(x)v(x)df' for ally €

9.1 SolutIons of the Boundary Value Problem in H'(S?), s> m 219

where( f >

H'—1I-I or — H' 1/2(f)fors<1J'

Then u belongs to and satisfies the estimate

It4i+a � C.ELt9IIH.—1 + 1p1...t/2 + (9.1.21)

If a boundary value problem is in the form (1): Lu = g, Bu = with B =bTV+bo, e > 0 on I = then according to Theorem 7.4.11 one

can find an associated variational formulation and apply Theorem 15.

9.1.4 Regularity Theorems for General 11 C JR"

The following theorems show that the above regularity statements also holdfor (1 C IR" if Q is bounded sufficiently smoothly.

Theorem 9.1.16. Let a for some t � 0. Let the bilinear form (5)be Let 8 � 0 satisjlj

ift€14;0�s<t,For the coefficients let the following hold:

E for all a,13,'y with I-il � max(O,t + (III — m), if t E IN,

(9.1.22)

e for > m — 1, otherwise, if t IN.

Then each weak solution u E of the problem

a(u, v) f f(x)v(x) dx for all v

with f Hm+'(a) belongs to and satisfies the estimate

� C. (If I—m+. + (9.1.23)

For inhomogeneous boundary conditions

= with •,m —1)

instead of u the statement a Hm(a) implies the statement aand the estimate (20).

PROOF. (a) Let {U':i = 0,1,...,N} with U' CS? be a covering of S? asin Lemma 6.2.36. Let with E C U', bethe associated partition of unity from Lemma 6.2.37. There exist derivativesa' which map U' (i � 1) into such that a'(OU' n I) cBy contrast, U° lies in the interior of 5? so that In 81)° =0. The solution u

220 Regularity

can be written as In part (b) of the proof we will treat xou, and inpart (c) xøL for i � 1.(b) We set

d1(u,v) :=a(Xgu,v)—a(u,Xsv) (i=O,l,...,N)

and wish to show the estimate

Ido(u, v)l � CdIUImIVlm_a (u E V E 8 1) (9.1.24)

for 8 = 1. Each suminand of do(u,v) has the form

— J dx.

Since = XoDatj+ lower denvatives of u, one has

— =

with 161 � 161 � 2m—1. One integrates with161 = m by parts, and obteins a bound CIUImIVIm_i, whence (24) follows.

The coefficients a way that the corre-sponding condition (22) Ia satisfied on Denote the resulting bilinear formby v). Since Xo E C°°(.Q) has a support supp(xo) C U°, the extension ofXou through Xou(x) =0 for x poses no probleres. We may formallydefine do(u,v) for v since only the restriction of v to U° is ofany consequence. By (24) do(u, v) can be written, for a fixed u E inthe form

do(u,v) = With d0 Idols_rn � CdlUlm.

Xou is the weak solution of

v) = a(xou, v) = a(tt, xov) + do(u, v) = (f, xov)o + (d0, v)o

= for ally Hm(R").

Theorem 3 (resp. 8J proves xou (thus too and

IXoulrn÷. � CEIXOII_m+. + l4I-m+. + iXOt&Irnl

<C'LIfl_rn+a + CdlUIm + � + lUlmI, (9.1.25a)

wheresisstillrestrictedto8� 1.

(c) The same reasoning as for (i= 1,...,N)shows

= (xJ+di,v)o for V E 1d11_rn+s S CdIUIm.

By assumption the maps a1 : U1 —' and their inverses belongto [reap. Put u(i) tI(x) for i = a1(x), that isü=uo(a')'. Inasimilar way define In

9.1 Solutions of the Boundary Value Problem in H'(I1), 8 > m 221

= >Jone can replace the derivatives with derivatives with respect to thenew coordinates, since t � 0, and thus obtain the form

=

which again is and whose new coefficients &ap satisfythe conditions corresponding to (22). As in (b), Ôap can be continued to

such that the resulting bilinear form isOne can apply Theorem 11 to

= for all ii E

Which yields E and

� Oa[IJI_m+e + +

Transforming back (ci. Theorems 6.2.17, yields

� C, + I-m+. + � C, Ill I—m+. + CdIuIm + CIUIm

and thus

1 i s N, $ � 1. (9.1.25b)

(d) (25a,b) hold for $ 1. Since Hm+. = � es-timate (23) has been proved for a � 1. If the conditions of the theorem al-low an s E (1,2], one proves (23) as follows. Since U E has beenproved already, one can estimate the forms after further integrationby parts via fr4(u,v)I CdItiIm+lIVIm...,. Accordingly, = withd1 Hm+'((?) and � CdJUIm+l. If one inserts the above estimate(23) for s = 1, one obtains (25a,b) and hence also (23) for 1 <s � 2. FUrtherinduction yields (23) for admissible a E (k, k + 1].(e) The case of inhomogeneous boundary values is treated as in Theorem 14.

U

Analogously one may prove

Theorem 9.1.17. (Natural boundary conditions) Let 1? with t � 0.

Exercise 9.1.18. Transfer Corollaries 5,6, and 7 to the situation of Theorems16 and 17. What conditions on a must be added?

222 9 Regularity

Corollary 9.1.19. Let 12 and the coefficients of a(.,.) satisftj the conditionsin Theorem 16, rvsp. 17. An e genft*nction, i.e., a solulsonu E V (V = Hr((i)in the cose of Theorem 16, V = H'(Il) in the of Theorem 17) of

a(u,v)=0 forallvEV, (9.1.26)

PROOF. Use Theorem 16(17) forf =O(andçp=O).

According to Theorem 6.2.30 (Sobolev's lemma) one obtains sufficientconditions via D for u to be a classical solution from

The minimal conditions for U E result from a differenttheoretical approach, which essentially goes back to Schauder. The followingtheorem, for example, can be found in Miranda [1, It shows theregularity of the operator L in (5.1.1).

Theorem 9.1,20. Let k � 2, 0 < A < 1. Let .0 E be a boundeddomain. Let the operutor L = E a beuniformly elliptic in (2 (i.e., (5.1.3a) holds). Let E andf E Then the boundanj value problem Lu = fin (2, u = 'p on F either has a unique (classical) solution uthere exists a finite-dimensionaL eigenspace {O} E C such that forall e E the following holds: Le 0 in .0, e = 0 on F. If a � 0, the firstalternative always holds.

The condition (2€ in Theorem 16 is stronger than necessary. Forthe Dirichiet problem the Lipechitz continuity of F is already sufficient toobtain the following result.

Theorem 9.1.21. (Ne&s 111) Let Si C°'1 be a bounded domain. Let thebilinear form (5) be Let the following hold:

1/2 � t > a > 0.

The coefficients E must belong to if t$I = m. Then theweak solution u of the problem

a(u, v)=

/ f(x)v(x) dx for dlv E

with I E belongs to and satisfies the estimate (23).

The condition of can be replaced by that of uniformellipticity (7.2.3) (cf. Theorems 7.2.11, 7.2.13). The statement of Theorem21 cannot be extended to 8 1/2 since then u Hr'(Si) would containanother boundary condition.

9.1 Solutions of the Boundary Value Problem in H'($2), 8> m 223

The proof of Theorem 21 uses an isomorphism R = R related to (cf.proof of Theorem 8) between and and also between Hr(fl)and such that the form b(u,v) := a(Ru,Rv) isIt is necessary to prove that b(u, v) := a(u, R2v) is alsoWe know f H_rn+8(.11) implies / R2f E Eath solution ofa(u, v) = v)o is also a solution of a(u, = b(u, (f, = (1,so that u E follows.

9.1.5 Regularity for Convex Domains and Domains with Corners

A domain .0 is convex if with x', x" E.0, x' + t(x" — x') belongs to I? for all0 � t � 1 . Convex domains in particular belong to C°", but permit strongerregularity statements than Theorem 21.

Theorem 9.1.22. (Kadlec [1]) Let I? be bounded and convex. Let the bilin-ear form (5) be Let the coefficients of the principal part beLipschitz-continuous:

for all = = 1;

for the remaining ones let the following hol&

E for all a,f3,-y withy � IaI+I/31 <1.

Then every weak solution u E of the problem

a(u, v)= j f(x)v(x) clx for all v E

with f L2(Q) belongs to fl HJ(Q) and satisfies the estimate

C1[IfIo + luli). (9.1.27)

The constant C1 depends only on the diameter of.0.

In Section 8.4.2 IP-regularity was required. A generalisation of (27) in theform of for the biharinonic differential equation with m 2 isknown for convex polygons (cf. Blum—Rannather [1)). For the Poisson equationthe inequality (27) can be reformulated as follows.

Corollary 9.1.23. For the solution of the Poisson equation —Llu = f EL2(Q) in a convex domain.0 with u =0 on F, the following holds:

/� Iflo. (9.1.28)

tal=2

As in Lemma 8.4.1 one shows that the left side of (28) is a norm of112(Q) fl which is equivalent to f 12• Thus, (27) follows. As a model

224 9 Regularity

for the proof of Theorem 22 we carry out the proof for Corollary 23 for thecase (2 C

PROOF. (a) First let us aesume that the convex domain is smooth: (2E C°°.According to Theorem 16 = / L2(.11) has a solution u E H3(i?) nH0'(Q). We want to show

:= > � for all u E (9.1.20)IaI=2

it suffices to prove (29) that for all u in the dense subset {u E C°°(D):u=Oon F) Integration by parts yields

JIz'u12&rdv =

=

=+4k,

where n = (nm, is the normal vector. The tangent direction is given by t =(—n11, n1). Then is the negative tangential derivativeNow and can be expreseed in terms of and Since both is and

2j(uxzn*, - =

= +

Integration by parts of the second suinmand yields

-j -=/ -

The bracketed expresauxi in the last display is the curvature in x I', whichfor a convex domain is always � 0. (28) has thus been proved.

(b) Every convex domain (2can be approximated monotonically by convexWeinterpret V,, :={uE

supp(u) C Th,} as Ritz-Galerkin space c HJ(a). Each U E Cr(.f?) liesin for sufficiently large gi. With C000(Q), is therefore also a densesubset of Fbr every v, the RAtz-Galerkin problem provides the solution

E of = f in = 0 on 13OV. In may becontinued by = 0. Theorem 8.2.2, which is also applicable in the casedim = oo, proves - -. 0, where is is the solution of

9.1 Solutions of the Boundary Value Problem In H(Q), > m fl5

= f in Q. Theorem 9.1.26 will show that for every the restriction of u

on to For each yE C V,,, v we have

if = =

lima,,

this one infers El*.2 � �Ill � il/ iiL2(fl) (ef. (29)), and obtains (28).

What role the H2-regulazity plays for the on asubspaceVhC

Exercise 9.1.24. Let the subepace Vh C satisfy

inf{Iu — vlj:v Vh} <Cohiui2 for all u fl

(cf. (8.4.14)). Let the Poisson problem be 112-regular (aaxrding to Theorem 21convexity of 11 is sufficient). Let Qv: —, Vh C be the orthogonalprojection on Vp, with respect to Show that there exists a Ci such that

— � CihluIi for all u

Hints: (1) With the Poiseon problem the boundary value problem = fin Q and u =0 on F are also H2-regular (cf. Remark 2). The correspondingbilinear form a(.,.) is the scalar product in(2) Qv agrees with the Ritz projection Sh for a(.,.).(3) Use Corollary 8.4.12.

In connection with finite elements one often considers polygonal do-mains a. Since polygons belong to C°", the Dirichiet problem, according toTheorem 21, is with 0 8<1/2. If the polygon is convex (i.e.,if the inner angles are then as in Theorem 22 one has H2-regularity(m = 1). One obtains results between H3/2 and H2 if the maximal inner an-gle of the polygon lies between and 2ir (cf. Schatz-Wahlbin [1)). If a hasa reentrant corner the boundary value problem can no longer be H2-regular(cf. Example 2.1.4; with an inner angle a the solution belongs to for8< 7r/a). Stronger regularity properties may be obtained, however, if specialcompatibility conditions are satisfied in the corners (cf. Kondrat'ev [1J).

Example 9.1.25. Let u be the solution of the Poiseon equation = fin the rectangle a = (0,1) x (0,1) with u = 0 on F. Only for 8 <3 doesf H—2(a) lead to u E H'(fl). Under the additional compatibility conditionthat f vanish at all corners, f(0,0) = 1(0,1) = /(1,0) = 1(1,1) = 0, however,one can also conclude, for I with 3 < 8 <4, that u E H'(a).

226 9 Regulanty

9.1.6 Regularity In the Interior

Up to now all regularity statements have referred to the entire domain (1. In-stead one can also investigate the regularity of the solution u in ho Cc h Thefollowing theorem shows that the answer depends neither on the smoothnessof the boundary nor on the type of boundary condition.

Theorem 9.1.26. Let ho CC hi C (1 and 8 0. Let the bilinear form (5) beFor the coefficients condition (22) with £7 be replaced

byhj and witht�sEt4 ort>s. LetuEVCHm(Q) beaweaksolutionof the problem a(u,v) = for all v E V, where the restrictionbelongs to Then the restriction of u to belongs toand satisfies

<C(s, J?i, IH—"'+(fl1) + I fr'I

PROOF. A special covering of .0 is given by U° = U1 = Thus weobtain the assertion from Part b of the proof of Theorem 16. U

9.2 Regularity Properties of Difference Equations

The convergence estimates for difference equations in Section 4.4 read, forexample, Vu — uhlko � under the condition that u oru This regularity assumption is frequently not satisfied (cf. Ex-arnples 2.1.3-4). A comparison with the error estimate lu — S Ch21u12for the finite-element method suggests that similar estimates also exist fordifference solutions. To obtain the latter, one needs to replace the stability es-timate C (or � C), whith corresponds toL2(fl)), by stronger estimates whith correspond to E L(H1(fl),or L(L2(h), H2(h)).

9.2.1 Discrete H'-REgularlty

For £7 C Ut's the following grids are defined:

Qh := {x v,h, Z}, oh := i7flQh. (9.2.1)

A grid function vh defined on is extended toQ,, by vh =0:

Vh(X) = 0 for x Qh\Qh. (9.2.2)

The Eudidean norm is now called the Li-norm:

rIvalo := := [hm ItJh(X)12J (9.2.3a)

xEQk

9.2 Regularity Properties of Difference Equations 227

It comes from the scalar product

vh(x)wh(x). (9.2.3b)

XEQh

The discrete analogue of H0'(fl) is with the norm

n 1/2

:= IIVhUH1 := + (vh = 0 on Qh\a4) (9.2.3c)

where is the forward difference in the direction. The dual norm reads

Iv,d_i := := sup{I(vh,wh)oI/IwhIj:wh = 0 in QhV?h}. (9.2.3d)

The associated matrix norms = ILhI1.—_1, =etc., are defined by

:= sup{ILhvhL/Ivhlj:0 satisfies (2)) for i,j {—1,0, 1}.

(9.2.3d')

Exercise 9.2.1. Show that (a) is the spectral norm of Lh (cf. §4.3).(b) The following inverse estimates hold:

for (9.2.3e)

The matrix Lh yields the bilinear form

ah(uh,vh) := (9.2.4a)

is said to be if a CE >0 exists such that

uh) � CHIUhI? for all uh and all h >0. (9.2.4b)

Correspondingly, 04(.,.) Is said to be Hg-coercive if there exist >0 andCK E IR with

ah(uh,uh) � — for all and all h >0. (9.2.4c)

As defined in Sect. 4.5, L,, (reap. ah(•,.)) Is said to be Li-stable if

<C0 for all h >0. (9.2.4d)

We call Lh if

� C1 for all h > 0. (9.2.4e)

Exercise 9.2.2. (a) implies Li-stability.(b) = for i,j E {—1,0, 1).(c) If is then so is

228 9 Regularity

(d) If Lh and are stable with respect to i.e., IlLalLo � �C1, then (4d) follows with Co :=(e) implies

The following statement resembles the alternative in Theorem 6.5.15.

Theorem 9.2.3. If is and If is then Lh isalso

PROOF. (a) Let = fh such that ah(uh,uh) = (uh,fh)o.

Coercivity provides

� Eah(uh, uh) + = I(uh, fh)o + C1uh109/CE

� + CIuhIol(uhIo.

On the basis of the stability estimate IUhIO � we obtain �From thisone infers � C and hence �

C (cf. Exercise 2b).(b) Now let Lhuh = fh. Aecording to Part a, One has � CelfhI_1.By estimating = I(fh,uh)oI � lfhI_lIUhIl through +

one obtams the coercivity

� �and hence � C1 with := + U

Instead of the Li-stability in Theorem 3 one can also aseume the solv-ability of the continuous problem and a consistency condition (cf. Corollary11.3.5).

By analogy with Lemma 7.2.12 the following may be proved.

Exercise 9.2.4. If is and if öLhwith IaoI, (cii � conat, contains at most first differences, then Lh + öLhis also Aecording to Exercise 4 it is sufficient to investigate theprincipal part of a difference operator as to its coercivity.

Example 0.2.5. Let i7 C JR2 be bounded. On 0h let Lh be given by thedifference method (5.1.18) (with a = 0):

(Lhuh)(x, y) = ô;afl + y) + y +

(9.2.5)

for (x,y) .17h, where u,4(x,y) = 0 for (x,y) Qh\Qh. Here, let —a11,e >0 in Th. Then ah(.,.) is and is


PROOF. For arbitrary vh, VJh defined on Qh the following rules of summationby parts hold;

= = (9.2.6)

Hence it follows that

ah(tLh,uh) = (Lhuh,uh)O

= — (a11 (.+ •_ (an

� + 101.

As in Lemma 6.2.11 one proves that for bounded $2 the norms Iand

+ are equivalent (uniformly with respect to h)so that (4b) follows. The results from Exercise 2e.

Exercise 9.2.6. Let 1? C be bounded. Let the equation

+ + + = Ibe given on by the difference stars (5.1.18/19), where Uk = 0 in Letthe differential equation be uniformly elliptic in <0, allan — a12a21 �—e(aij + > 0. 1\irther, let C°(TJ) hold. Show that for sufficientlysmall h the associated matrix Lh is for all 4 > 0, isHint: For sufficiently small 4 the following holds:

( li ( h\2—all

for all di, d2 E JR.

The difference methods constructed so far remainif differences of lower order are added (ci. Exercise 4) or if the principal term

+ ... is replaced by + ... with Oh E C1(Th). The abovedifference methods are described by the same difference operator regardlesswhether the grid points are close to or far from the boundary. The homoge-neous Dirichiet boundary condition is discretised by (2): =0 onIf one wants to approximate the boundary condition more aceurately, oneneeds to select special discretisations in the points near the boundary of Qh(cf. Section 4.8.1, 4.8.2). One thus obtains an irregularity which makes a proofof as in Example 5, difficult. We begin with the one-dimensionalcase.

Lemma 9.2.7. Let L,, be the matrix of the one-dimensional Shortley- Welterdiacretisation.s of —u" = f on .0, u =0 on OS?:

230 9 RegularIty

h2 .(..!flh(X) — 2Uh(X+ 8rh)

— 2u,,(x — Srh)1 = 1(x)

I. SiSr Sr(Sr + 3j) + )(9.2.7)

forz (Ih, where 0< < 1 (of. (4.8.7)), u(e) = 811. Forarbitrury 5? C JR there holds � if 0 onThus for bounded 5?, Lh is

PROOF. (a) First, let be aasuined to be connected. Let the grid points of5?hbexi:=xo+lhE5?forl=O,".,k>O.Lettheboundaryporntsof5?be

— s1,0h and Zk ÷ with 81,0, 8r,k E (0, 1). The other factors of Equation(7) in x = x1 are = 1. If one takes into account vh 0 on IR\I? oneobtains the following identity:

(vh,Lhvh)O :=

= + h1 {vh(xo) [(! _2) vh(xO) + (i—

2

)Vh(X1)]

+ vh(xk) [(i- —2) Vh(xk) + (i— 1 i..) Vh(Xk_1)]}i

where := and s,. := Since vh(x0) = and vh(z1) =+ because of vh(x_1) = 0, the first sununand inside

the bracse can be written as

— 2) Vh(ZO) + (i 2

)Vh(Z1)]

I ÷ SI(9.2.8b)

l—8j [8j+2 + 2 +=h 8 Vh(x_1) —c vh(x_1)D vh(zO)1+81 1

For a := (3 := use the inequality —a/3 � —

with = (Si + 2)/SI and note that the function s(1 — .)/L4(1 + 8)(2+ is

bounded by 0.018 (maximum at 8= 1)/2). The second summand insidethe braces is treated analogously and yleith

(vh, Lhvh)O � — + (9.2.8c)

(b) In Part (a) it was aesumed that k > 0. For k =0 one obtains

(vh, LhVh)0 = �SjS,. BISr

so that (8c) also holds.

(c) Let 5? be arbitrary. Let the components of connection be = (a1,

(i Z) with b1 Let be the discretisatiori matrix aseociated withwith support in S1h = QhflQcan be written where

9.2 RegularIty Properties of Difference Equations 231

= 0 outside Let and be the first and last grid points of

:= 11h fl The statement follows from

:S

� = 2(vh, Lhvh)O.

Theorem 9.2.8. Let (1 C 1R2 be boundeL Let Lh be the matrix associatedwith the Shortley- Weller discretisation of the Poisson equation (cf. Section4.8.1). Then Lh is

PROOF. Let be the portion of the z differences Ii' differencceJ suchthat Lh = + The restriction of to a "grid row" {(zih,y) i?:vZ} (y is fixed) corresponds to the matrix Lh in Lemma 7. Thus one ob-tains (vh, � With the analogous inequality (vh, �

one obtains 2(vh,LhVh)O �I

I li——i If

Theorem 9.2.9. Let fl C JR2 be bounded. Let the Poisson equation be dix-cretised with the aid of the five-point formula (4.8.14e) at points far from theboundary and by inteipolation (4.8.16) at points near the boundary. The as-sociated matrix is

PROOF. The corresponding, one-dimensional formulae, except for the scalingfactor ÷ 2, agree with (7) so that the proof of Theorem 8 can easilybe carried over.

Exercise 9.2.10. Show that for a convex domain 4? C JR2 the matrix Lh ofthe Shortley-Weller discretisation satisfies the inequality

(vh,Lhvh)o � +

Theorems 8 and 9 can be strengthened in the following way:

Corollary 9.2.11. Let Lh be as in Theorem 8 or9. Let Dh = diag {d(x):x Ebe a diagonal matrix with

d(x) := 1} (x E 4?h),

where s,., sc,, come from (4.8.7), respectively (4.8.16). For points far fromthe boundary, x E S?h, evidently d(x) = 1 hoLds. The matrix

LI . h h,

232 9 Regularity

which belongs to the re-sa4led system of equations = := Dhfh, is also� C.

PROOF. In the case of the Shortloy-Weller method one needs to combine thecorrection terms (8b) for the x and y directions. The result is that they remain? sothat � �

e > 0. One obtains an analogous estimate for the difference method inSection 4.8.2. I

9.2.2 Consistency

In the followrng we carry over the estimate lUll � ChIul2 C'hlfk,which holds for finite-element solutions, to difference methods. To this endone needs to prove the consistency condition

— := flL11R,1 — � CKh (9.2.9)

for suitable restrictions

R11: H2(S?) fl L2(Q)

To construct the restrictions we first continue u in E2u EAccording to Theorem 6.2.40c one assumes

E2u = on (1, 11E2u11H2(n.2) � (9.2.lOa)

for all U E An analogous continuation E0L2(a) L2(1R2)

with

f := Eof = f on (1, I E L2(fl)(9.2.1Db)

is given, for example, 7 = 0 on 1R2\Q. Let the averaging operatorsC0(Wt ) be defined by

nV2= h1 = h1 I

J—h/2- (9.2.11)

The restrictions R,,, are chosen as follows:

:= i.e., y) = h2 I I U(z + y + dir7

J—h/2 J—h/2

with = F,2u for (x, y) E (9.2.12a)

i.e., (&11f)(x, y)

h4/1112 /11/2 /11/2 fh/2

—11/2 —h/2 —11/2 J—h/2

with 7 = E(Jf for (x,y) (9.2.12b)


The characteristic properties of the convolutions 4 are the subject of

Exercise 9.2.12. Let & be the symmetric difference operator (&ti)(z, y)It4x + h/2, y) — u(x — h/2, y)]/h;, is defined analogously. Show that

C, II4IIHk(1t2)+—Hk(I(2) � C (in particular fork=O,±l,2).(c) := < 11v11L2(R2) for all v L2(1R2).

(d) For a E C°"(1R2) holds — �Here = and =(e) — � for ii, E IN.

(f) flu — � juj IHk+1(Ita) for u E

Consider the differential operator

L = + a4(x)O/8x4 + a(x). (9.2.13a)1,3=1 4=1

First we asswne that S? = JR'S and discretise L with the regular differenceoperator

Lh = +a(x) (9.2.13b)

with an arbitrary combination of the ±-signs.

Lemma 9.2.13. Let £2 = IR". Let Let Land Lh begiven by (13a,b). Then the consistency estimate (9) holds.

PROOF. To simplify the notation let us assume that n =2. Let

= = —81

with � (cf. Exercise 12e,b). Let the term Cor-responding to in (13b) be, for example, all(x; Exercise 12ashows that

== — h, y) = y) —

with &2(z, y) := — h, y) — u(x, y)J = — 4, y) —ux(X,y)] and � � (cf. Exercise12c,b). Since = the error term 52 does not appear if one alsoapproximates by Finally one obtains

= + —

= + — uZju = —

234 9 Regularity

with = < — � (cf.

Exercise 12c,f). Putting this altogether one obtains

— = Si + + 63).

For the first error term the following holds:

1161(1ff_i � C'hJuI2. (9.2.14a)

For arbitrary Vh we have

(vh, (63 + = [alIvhi, 63+ (9.2. 14b)

From a C°"(Jt2) it follows that � +IIOIJCO(R3)(IVPIIIH1 � 80 that

+ Ss)II,çi = suni Kvs, +VkEHL (9.2.14c)

� Cfl63 + C'hIuI2.

From (14a.,c) we have

— � Ch. (9.2.14<1)

Analogously, one shows

110 — 14a Ch. (9.2.14e)

For - similar reasoning result8 in an 0(h)-estimate for thenorms . and Both are upper bounds for thelarger norm 1IH1,_H2(R2) such that

� Ch. (9.2.14f)

Likewise

— RhaIIj,_1I..ff2(Ra) Ch. (9.2.14g)

Statement (9) follows from (14e,f,g). U

When generalising the consistency estimate to more general domains 11 Cthe following difficulty arises. The entries of the matrix Lh according to

(4.8.7) or (14.8.16) are not bounded by Ch-2. Rather, at points near theboundary the inverse of the distance to the boundary point enters, and thisdi8tance may be arbitrarily &nall. One way around this, would be to formulatethe discretisation so that the distances between boundary points and pointsnear near the boundary remain, for example, � h/2. A second possibilitywould be a suitable definition of Ra so that the product appearing in

9.2 Regularity Properties of Difference EquatIons 235

(9) can be estimated (cf. Hackbusch [2)). Here we choose a third option: Lh isreplaced by the re-scaled matrix = DhLh from Corollary 11.

Theorem 9.2.14. Let .0 E C2 (or convex) and bOtind&L For the discreti,a-tion ofLu = f forL = —4 on.0 withu = 0 on F use the discretisation Lhuhaccording to (4.8.7) or (4.8.16). Let = DhLh be defined as in Corollar!J 11.Then the consistency estimate

— ç Ch (9.2.15)

holds.

Here, the matrix L4 from (4.8.7/16) is only taken as an example. Theproof will show that the estimate (9), respectively (15), also holds for otherLh if x near the boundary, represents a second difference. First,two lenunas are needed.

Let 'yh C be the set of points near the boundary. If vh isa grid functiondefined on then we denote by the function

(vhf,,.j(x) vh(x) for x E l'h, (vhi.Th)(x) = 0 for XE

Lenuna 9.2.15. Let .0 E Ce" be bound&L Then there exists a C = C(.0)independent of 4, such that

IVhlmIO �ChIvhII. (9.2.16)

PROOF. .0 E C°'1 follows: there exist numbers K E IN and h0 >0 suchthat for all x with h h0, not all grid points {x+ (vh, ash): —K � i',K}lieinQ.Foreachx€ .0aselectapair(&'o,p0)EZ2with—Kand x + poh) At this grid point x the functions (—K � v, pK) are defined by := vh(x), := 0 for (v,p)Therefore VhI.r,, is a decomposition with = Vh(X) or

=0.Without Ices of generality let us assume that v > 0 and p >0. For x E Qh

we define the chain

X0 = X, X1 = x +(h,0), ... X

= x+ (vh,h), , =According to the definition, either = 0 holds or .0, i.e.

0 (ci. (2)). In both cases we have

<

= 4 — + — + ..

� + + ... ++ + . .. +

236 9 Regulanty

and thus � hflvj + Ii') � Summationover yields estunate (16) for =

In most cases x E already has a direct neighbour in of x 'Yh

such that (16) follows with C = This holds in particular for convexdomains.

Lemma 9.2.16. Let be defined by (12a). Let E2 satisfy (lOs). Then we

have

for all f',u H01((I) fl 112(11),

(9.2.17)where := + x€ 1R2x E (—k, x

PROOF. (a) Let Q = (—n) x (—i, Fbr v 112(Q) one shows

)v(O)— J v(x)dxI C i/j(vz2x + +

since the left-hand side vanishes for linear functions u(x, y) = + fix +Theproofissiniilartotbeonefor(8.4.2).TransformingfromQtoQh

x gives

Ih2v(O)— f v(x) � + + dxdy. (9.2.18)

(b) Let F. Statement (17) follows from (18) with

v(x) := + hx) = ÷ hx), since =0. U

PROOF. (of Theorem 14) (a) Inequality (15) is proved if

j(vh, — � Ch (9.2.19a)

for all vh E and u with IVhIl = = 1. lb this endis split into

In part (b) we show

EL'hRh — � Cih, (9.2.lOb)

The other steps of the proof, (c) and (d), yield

— � C2h, (9.2.19c)

such that (19a) with C = C1 + C2 follows.(b) Lemma 15 shows

Ivklo � = C3h.

9.2 Regularity of Difference Equations 237

For one obtains � + Iv;111 = 1 + fr4j1. The inverse estimate (3e)yields Ivji � Ch11vj0 � CC3, thus

� C4. (9.2.19e)

Let Lh be the (regular) difference operator on the infinite grid

= ph): ii, p Z}.

Since the support of is Qh\'yh, (vi', L'5Wh)O = Lhwh)o holds for all wh.Furthermore, Dhwh)o = (vi', wh)o. This proves the first equality in

—DhRhLJU)oJ = —RhLJii)oI

S � C5hC4C6 := C1h,

where := E2u is the continuation of u to 1R2. Further inequalities resultfrom Lemma 13, (lOe), and

fltIjIH2(R3) � C6jJnflJp(fl) = C6 (9.2.19f)

(cf. (lOa)). Theorem 6.2.40c guarantees the existence of an extension U with(19f) if 11 C2. Another sufficient condition for (19f) is the convexity of (2.(c) The left side of (19c) splits into (va, and Thefirst part is estimated in part (d). Exercise 12c and (19d) yield the secondterm

� � CahIu(2 = CGh.

(9.2.19g)(d) We set wh := Since the support of is contained in 'Yh, wehave = (vi, w,j0. contains differences with respect to the xand y directions. Accordingly we write wh = + i4. In the following welunit ourselves (i) to the Shortley-Weller diacretisation, (ii) to the termand (ii) to the case that

Ii,, (i.e., 8r= 1), x'

The other cases should be treated analogously. We set

:= E H2(E2).

The Shortley-Weller difference in the x direction reads

= a,)

where d,,, is the diagonal element of Since in the equation L'huh fh thevariables F (for example, = x'), have already been eliminated,

has the form

2h

238 9 Regularity

The factor + due to the definition of Dh, remams boundedby 4h2. From Lemma 16 and Kh/2(x') c one infers

— (9.2.19h)

The second divided difference in x = (x, y) can be by

LIx) ath3r

with(2(t—h)/(h2+h2sj),

g(t) = —2(sjh + t)/(h2si(1 + sj)), for —sjh � t � 0,otherwise.

From this one infers

I 11/21 11/2/ /

J(9 2

2h'{jh

+ �(cf. (6.2.5a)). From (19h,i) and the corresponding estimate for one obtainslwh(x)I � such that

= 1s2 IWh(X)12 � C? �xE7h

� (3C7C6)2

From this follows

C34C8 C9h. (9.2.19j)

(19g) and (19j) yield the required inequality (19c). U

Remark 9.2.17'. The proof steps for Theorem 14 can be carried out in thesame manner for more general difference equations (for example, with variablecoefficients, as in Lemma 13).

9.2.3 Optimal Error Estimates

In the following we compare the discrete solution uh = with thestriction := Rhti of the exact solution ii = From the representation

= (fh9220= — Rh!) — —

one immediately obtains


Theorem 9.2.18. Let u E Jf2(Q) hold for the solution of Lu f. Let thenght-hand side fh of the discrete equation = fh be chosen so that

Ifh — Rail_i � C,h. (9.2.21)

If, furthermore, Lh is and if the consistency condition (9) holds,then satisfies the ermr estimate

IUh — ufl1 C1(C1 + CxIuI2)h. (9.2.22)

PROOF. l"h � —Rail_i +ILaRa — U

Corollary 9.2.19. (a) !neqsiality (21) is satisfied in particular if one choosesfa := R,1f.(b) The choice fa(x) := f(x) forx E (ci (4.2.6b)) leads to (21) if f E

or I E In these eases the following even holds:

— Rail0 � — �reap. — Raflo � Ch2Jf 12. (9.2.23)

(c) In Theorem 18 one ean replace the H)-regularity of by that of L',,DhLh (cf Corollary 11), (9) by (15), and (21) by

IDa/h — DaR,,fI_i � C1h. (9.2.21')

The proof of (23) is based on the inequality (18). IError eatimate8 of order 0(h2) can be derived in the same way if one has

consistency conditions of second order. These are, for example, (24a) or (24b):

ILaRa — <Ch2, (9.2.24a)

lLhRh — R,1L1_i4-s � Ch2. (9.2.24b)

Remark 9.2.20. If = Qa (i.e., I? = or 1? = (z',z") x (y',y"), theinequalities (24a,b) can be shown in a way similar to Lemma 13.

Example 9.2.21. The difference method in Example 4.5.8 shows quadraticconvergence. This case can be analysed as follows. Using Remark 20 one shows(24a). In the following section we prove the Re-regularity IL;1 124-0 � C whichfor the symmetric matrix under discussion, Lh, is equivalent with IL;'104-_2 �C (cf. (6.3.3)). Expression (20) leads to

— uj0 � Ch21u12. (9.2.25a)

The corresponding estimate

240 9 Regularity

— � (9.2.25b)

which is based on (24b), fails due to the fact that the solution in Example4.5.8 does not belong to H3(O) (cf. Example 9.1.25).

The verification of (24b) m the presence of irregular discretisationa at theboundary becomes more complicated.

9.2.4

Under suitable conditions on (1, Lu = f L2(f1) has a solution u H2(.O) n

Hd(S?): 1u12 � lo (cf., for example, Theorem 9.1.22). For the discrete so-lution of Lhuh = fh the corresponding question arises: does � CIfhJohold? First one has to define the norm of

If Q = or if the boundary V coincides with grid lines, one can define

1/2

iuhl2 + h2 +XEQh

in general, however, one has to use irregular differences at the boundary.At points far from the boundary x (Ih\-yh, let =At a grid point near the boundary x = (x,y) E for example, withx' := (x — F, xr := (x + h,y) E (hI It appears practical to use thedivided difference (4.8.5). In general, however, this violate the inverseestimate 1uh12 � may become arbitrarily small. Thereforeone defines with the aid of the grid points (x — sjh,y), (z + h,y),(x+2h,y):

2 fUh(X+2h,y)—uh(X+h,V)(2+ &j)h h

— uh(x+h,y) —245(X —sih,y)1s1h

where u,(x — y) =0.If, in addition, F passes between (x + h, y) and (x + 2h, y), one needs

toreplacex+2hbyx+h+srh.If(x+h,y)aieonolongerbelongstoQ,set := h2ua(x). The second differenos is defined accordingly.The mixed difference in the Interior reads := Near the boundaryseveral choices for Dm1, are possible. The definition of reads

(+ + +

xEfl,,

ExercIse 9.2.22. Prove the inverse estimate 12 � Ii.

We call Lh He-regular if S C. An equivalent formulation is11sh12 <CffhIo for uh = E


Exercise 9.2.23. Let Lh be let Lh + 6Lh be and

<C for all h. Show that + 6Lh is also Hz-regular. Hint: Remark9.1.2.

For the proof of one can —in analogy to the proof of Corol-lary 9.1.23— partially sum the scalar product (LhUh, Lhtth)o in order to showthe equivalence of tfhIg ILhuhPo and This technique, however, canonly be applied to a rectangle Si. Here we use a simpler proof which is alsoapplicable to general domains.

Let Ph: L2(a) be the following piecewise constant interpolation:

(Phuh)(X) :={

PhUh(X) X (9.2.26a)

Phuh(x,y) := uh(x',y'), if (x',y') E and (9.2.26b)

x'—h/2<x�x'+h/2,where uh(x) 0 for x E Qh\Qh (cf. (2)).

Lemma 9.2.24. Let Si E Then we have

— I1i#.0 � Ch, (9.2.27a)

� C, (9.2.27b)

� C. (9.2.27c)

PROOF. (a) The estimate (27a) is equivalent to — � Ch (cf.(6.3.3)). Thus we must show IWhIO � ChIuhll for wh := — I)uh. Fromthe following Exercise 25 we obtain the expression

tuh(X) = — for wh = — I)uh, 'C

Exercise 12c shows 111h10 1WIL2(J(2) for w := — I)P,,uh. For everyE (0,1) x (0,1) define the grid function with := w(x

for x E Qh. One may check that y) is a weighted sum of first differencesuh(x,y) — uh(x',y/) where (z',y') E {(x± h,y), (x,y ± h), (z± h,y ± h)}.Thus we have f � for all from which one infers

= 11(2 w(x)12 h2 Jw(x +

= J �(0,1)2

thus ItIhIo � 11010 � h(tsh$1.

= = =

242 9 Regularity

with = {x E Qh: fl V $ one infers

Iwh —tiJhIO � �(cf. part b of the proof and Lemma 15, in which -Ih niay be replaced byTogether with � hIuhIl one obtains Iwhlo �(C' + l)hIuhIl.(b) (27b) with C = 1 follows from lPhuhlo =(c) Second differences of = have already been estimated in (19i). (191)implies � C1u12, i.e. (27c).

ExercIse 9.2.25. Show that the adjoint operators for Ph and Rh areL2(1R2) —i and R: L2(1R2) with

= for x E Qh, =

Furthermore, the following holds: = uh(x) for all x E Qh.The identity

= RhL —'Ph — [(LhRh — RhL)L'Ph + (RhPh 1)1

yields the estimate

� + x

IILhRh — RhLIL4_21L'124-oIPhIo,_o + lRhPh — uI—14—OL

The inverse estimate from Exercise 22 yields 11124—I � Ch1 for the identityI: -s Thgether with the Inequalities (27a—c) follows

Theorem 9.2.26. Let Lh be Hz-regular and satisfy the consistency condition(9). Let Ii be convex or/rem C2 and let L be (i.e., � C).Then Lh is also He-regular.

a E would be a sufficient condition upon 1? guarantee (27a-c). Onthe other hand, the condition given agrees with the conditions for Theorem14 and with the 112-regularity of L.

Corollary 9.2.27. In Theorem 26 one con replace (9) and theof L,, by (15) and the of L'h. The restdt is the ofbothL4

PROOF. For the proof of the of L'h define Phuh(X) := 0 forx and note IExercise 9.2.28. If Si E lit" (n < 3) is bounded and Lh isthen Lh is stable with respect to the row-sum norm: C. Hint: UseC'iUhlO S �

9.2 Regularity of Difference Equations 243

For general Lipechitz domains (1, such as, for example, the L..shapeddomain in Example 2.1.4, L cannot be 112-regular Theorem9.2.21, however, guarantees for s [0, 1/2). For Lh onecan define and prove analogously (cf. HaCkbUSCh [lJ).

10 Special Differential Equations

If the boundary value problems have special properties, one often uses specialdiscretisations for them. We give two examples of this in Sections 10.1 and10.2. For the first example the variational formulation is of decisive irnpor-tance.

10.1 Differential Equations with DiscontinuousCoefficients

10.1.1 Formulation

The seif-adjoint differential equation

+ a(x)n(x) = 1(x) in Si (10.1.la)

(cf. (5.1.17)) occurs often in physics. It can also be written as

with A(x) so that for a = 0 and f = 0 we have theconservation law

In the applications In physics the coeffidents are, in general, constantsof the material. The functions can be varying, if the constitution of thematerial depends on poertion. As soon as we have several materials in contactwith oath other, the coefficients may be discontinuous on the boundaryof contact (see Figure 1).

The equation (Ia) can be understood in a classical sense only if EC1(ii). For discontinuous therefore one uses the variational formulation.If one supplements Equation (is) with the Dirichiet condition

u=0 onl' (1O.l.lb)

then the weak formulation is written

u with a(u,v) = v E (1O.l.2a)

10.1 Differential Equations with Discontinuous Coefficlent8 245

where

a(u,v) — (1O.1.2b)

Note that Equations (2a,b) are defined for arbitrary E

FIgure 10.1.1. An interior boundary = fl

In the sequel we shall assume the situation to be as in Figure 1: i? isby the b mdar into two subregions and Let the co-

efficients be piecewise smooth: E for k = 1,2. Along 'y thecoefficients may be discontinuous so that the one-sided boundary values

lim := lim (x E i')

may be different. In addition let us assume that the solution u is contin-uous, u E C°(Th), but only that it Is piecewise smooth, u E C1(.01) andu C'($?2). The one-sided boundary values of the derivatives are denoted

and (x E With these assumptions iraegration by parts of

— dx gives the result dx — f1, dl'. Sinceii = (ni, n2,...) in Figure 1 is the outgoing normal to but the ingo-ing normal with respect to in £12 there results - dx =f0 dx + J dl'. Putting this together one has

a(ts, v)=j {au

+v / dl'.

Thus from the variational equation (2a) there follows in addition to thedifferential equation (la) also the transition equation

= on (10.1.3)iJ=1

Thus we have shown the following lemma.

246 10 Special Differential Equations

Lemma 10.1.1. Assume C'((4), k = 1,2. If the weak solution u of(2a) in a is continuous and piecewise dsfferentiable in and a2, then it isthe dassicol solution of the differential equation (la) in U (12 = Inaddition to fulfilling the boundary condition (ib) the solution also satisfies thetransition condition (3) on

Corollary 10.1.2. If the coefficients are not continuous then the solution ofthe equation (2a) does not, in general, belong to C2(fl), but will have discon-tinuous derivatives along However, the tangential derivative along 'y maybe continuous.

Example 10.1.3. Suppose the coefficient of a(u, v) := f a(x)u'v1 dx aregiven by a(x) = 1 on the interval and a(x) = 2 on 1). this one-dimensional example the point plays the role of the curve y. The solution ofthe equation (2a) with /(x) = 1 Is

u(x) = — x2](CE)

on 1

1 . —0) = 2 . + 0).

As mentioned at the beginning, the differential equation can be writteninthefonndiv4,=fon(11US?2, whereq4:=A(x)graduon (1andu=OonF. The transition condItion (3) means that (4,, a) is continuous on Since(4,, t) is also continuous for any tangential direction t, 4, is continuous on 'yand thus throughout 0.

The regularity proofs of Section 9.1 can be carried over to the present sit-uation. The smoothnees upon the coefficients of a(.,.) are in eachcase to be required piecewise on i and in addition the dividing line (orhypersurface) must be sufficiently smooth. Then there results the piecewiseregularityu and u E instead of u Bring-ing in the transition conditIon (3) one obtaIns 4' = A(x)graduon the whole region.

10.1.2 Dlscretisatlon

When one discretises using finite elements there are the following difficulties:

Remark 10.1.4. Linear or bilinear elements give a finite-element solutionwith the error estimate Iu — = 0(h1/2). The L2(i?) error bound is ingeneral no better than (ti — u"(O = 0(h).

PROOF. (1,:={tEr: tfl'y}consistainallflniteelementsthat&ejncontact with 'y. Since u has discontinuous derivatives on i', one can have no


better estimate on fl.1 than Vu — Vv = 0(1) (v E Vh). The 8Urface area of £2,is of the order 0(h), so that there results lu — 0(h'/2). lbr the boundon lu ii'(o

Example 10.1.5. In Example 3 choose e (1 + h)/2 and discretise usingpiecewise linear elements of length h. Then there results an error at x = 1/2of u(x) —1z"(z) = ah+0(h2) with a 0.00463. Since u—ui' behaves linearlyin (0,1/2), one obtains the error 0(h) not only for the V norm — u1'jo butalso for the maximum norm IlL — and for the L1 norm Ilu — ti'IItl(ol).

The usual bounds on the errors fu — uhIi = 0(h) and fu — = 0(h2)can however be attained. To do this one must adjust the geometry of thetriangulation to the curve If the dividing line -y is piecewise linear, onemust choose the triangulation such that -y concides with sides of (interior)triangles. If 'y is curvilinear then it may be approximated by isoparametricelements (cf. §8.6).

Similar aseertions hold for the difference scheme (5.1.18).

Example 10.1.6. Let the discontinuity position e in Example 3 be a gridpoint, i.e., E N Then the difference scheme (5.1.18), which here takes theform

h2 —us) = 1

for 1 � ii � h-1 —1, is suited to the equation from Example 3. In general theerror is of the order 0(h2). Since the solution of the differential equation ispiecewise quadratic here, it is in fact exactly given by the difference solution.On the other hand, is not a grid point then the error is of the order 0(h).

In the two-dimensional case one obtains 0(h2) difference solutions if 'ycoincides with lines of the grid. Otherwise the error worsens to 0(h). Anotherpoesible form of difference approximation consists in approximating the dif-ferential equations separately in the regions and .172, and then, to handlethe unknown values on to discretise the transition condition (3).

10.2 A Singular Perturbation Problem

10.2.1 The Convection-Diffusion Equation

In the following we shall consider the boundary value problem

-e4u + = f in .17, u = 0 on 1 (10.2.1)

fore >0. One calls the diffusion term and = (c,Vu) theconvection term. For small e the convection term dominates.

248 10 Differential Equations

Exercise 10.2.1. Let the coefficients be constants. Equation (1) can betransformed by v(x) into the symmetric Hd(fl)-elliptic equation

+ = I in a, v = 0 on 1. (10.2.1')

Equation (1) is elliptic for all > 0. That there is a unique solutionfollows from Exercise 1 (cf. also Theorem 5.1.8). Denote the solution by

Remark 10.2.2. ug(x) and cannot converge uniformly on Th for—40.

PROOF. If and 8u,/8x4 were continuous in one would be able to takethe limit —+0 in Equation (1) and one would obtain the first-order differentialequation

intl (10.2.2)

for := This equation is of hyperbolic type, but not com-patible with the boundary condition u =0 on I' (ci. Section 1.4), i.e., Equation(2) has, in general, no solution with =0 on 1'. U

Equation (2) is called the "reduced equation". Equations (1) and (2)differ in the "perturbation term" -€ilu. Since the equations, (1) and (2), areof different types one speaks of a "singular" perturbation.

The following example will show that *10 = limc..oue exists and satisfiesEquation (2), but that the boundary condition mio =0 is only satisfied on apart Vr, of the boundary r.

Example 10.2.3. (a) The solution of the ordinary boundary value problem

—at" + u' = 1 in (0,1), ti(0) = u(1) = 0 (10.2.3a)

= x_(ea/( 1). On [0,1), converges to *10(z) =x.This function satisfies the reduced equation (2), u' = 1, and the left boundarycondition =0, but not uo(1) 0.(b) The solution of

—at" — u' = 0 in (0, 1), u(0) = 0, *41) = 1 (l0.2.3b)

is Ue(X) = (1 — e_x/e)/(1 — In this case, u0(x) := = 1

satisfies Equation (2), —u' = 0, and uo(1) = 1, but not the boundary conditionat z = 0.

Which boundary condition is fulfilled depends for Equations (3a,b) on thesign of the convection term ±tt'. In the many-dimensional case the decisivefactor is the direction of the vector c = (Cl,...


1-.tAO

0- I-

0 1

FIgure 10.2.1. (a) Solution of Example 3a; (b) Solution of Example 3b

In Figure la,b the solutions of Example 3 are sketched. In the interior Ueis close to only in the neighbourhood of x = 1 (Figure la) [resp. x = 0

(Figure lb)i does differ from in order to satisfy the second boundarycondition. These neighbourhoods, in which the derivatives of attain theorder O(1/€), are called boundary layers. For Example 3 the thidmees ofthe boundary layer is of order

Exercise 10.2.4. The interval [1 — ij, in which the function—1) exceeds the value <1, has the thi&ness =

An analysis of smgularly perturbed problems is to be found, e.g., inKevorkian-Cole [1J and Coering et a!. [1J.

10.2.2 Stable Difference Schemes

The special case of Equation (1) for c = (1,0) is

follows

—E41L+u=f ma, (10.2.4)

(10.2,5)

1

0 1x

The symmetric difference formula (5.1.10) for Equation (4) is

—€

= h2 —€ — h/2 + h/2

For a fixed L, is consistent of order 2. From Corollary 5.1.16 there

Remark 10.2.5. As soon as h the difference scheme (5) leads to anM-matrix.

The property of being an M-matrix was used in Section 5.1.4 to demon-strate the solvability of the discrete equation. It turns out that the difference


equations are also solvable for h � However, with increasing h/€ theredevelops an instability which is made clear in Table 1.

Thble 10.2.1

z 0 1/32 2/32 3/32 4/32 ... 30/32 31/32 1

0.50.10.050.02

1 0.93 0.87 0.81 0.75 ... 0.04 0.02 0

1 0.97 0.95 0.92 0.89 ... 0.22 0.13 01 0.99 0.97 0.96 0.94 ... 0.47 0.31 01 0.99 0.99 0.98 0.98 ... 0.82 0.73 0

i04

1 1.00 0.99 0.99 0.99 ... 0.87 1.11 0

1 1.03 0.99 1.04 0.99 ... 0.23 1.89 0

1 4.95 0.95 5.00 0.90 ... 0.08 5.88 0

1 48.6 0.94 48.7 0.88 ... 0.06 49.6 0

1 4859.5 0.94 4859.6 0.88 ... 0.06 4860.4 0

Table 1 shows the values Ujt,(x, 1/2) of the difference solution ofEquation (4) in the unit square ii = (0,1) x (0,1) with f = 0 and w(x,y)(1 —x)sin(iry). The grid step is h = 1/32. Fbr h/2 = 1/64 = 0.015625, Lhis an (irreducibly diagonally dominant) M-matrix. Because of the maximumprinaple (cf. Remark 4.4.4) the values lie between =0 and maxw =1.The solution of the reduced equation (2) is v) = y) = sinAccordingly the values 1/2) from the first part of Table 1 approach thelimiting value sin ir/2 =1.

In the second part of the table we have h/2. Thus Lh is not anM-matrix. Notice that 1/2) > 1, i.e., even the maximum principledoes not hold. In addition one can see that Ue,h(X, 1/2) converges to 1 — x forthe even multiples x = iih, and thus not to uo(x,y) = sin icy. For the inter-vening odd multiples x = vh an oscillation of amplitude O(h2/2€) develops.

The resulting difficulty is similar to that for initial value problems forstiff (ordinary) differential equations: If one keeps constant and leta it go tozero, the convergence assertions of Section 5.1.4 hold. However, if is verysmall, the condition it < 2€, without which one does not obtain a reasonablesolution, cannot in practice be satisfied.

One way out of this difficulty has already been described in Section 5.1.4.One must apprsdmate the convection term by suitable one-sided differences(5.1.14)

-t += h2 —t— h4 4€ + hIcjI + hIc2I — + hcj (10.2.6)

-€-h4

wherect :=max{0,4}, c


Remark 10.2.6. If one discretises Equation (1) using (6) then one obtains,for all e > 0 and h > 0, an M-matnx Lh. For fixed e, the scheme hasconsistency order 1.

Exercise 10.2.7. The discretisation of Equation (3b) corresponding to (6) is

= 2e +h — hi and this gives the discrete solution uh(x) =— (1 + — (1 +

If one applies the difference operator (6) to a smooth function one obtainsthe Taylor series

= Lu — { + Ic2luyp}h + 0(h2). (10.2.7)

The 0(h) term hIcjIu + hIc2Iu..fl, is called the numerical viscosity (orthe numerical ellipticity) since it amplilles the principal part.

A second remedy consists in replacing the parameter e by a discretisationusing Ch � If one chooses

:= hIcjI/2, hlc2I/2} or Ch e + max (10.2.8)

then the symmetric difference method

-11

02

= —1 4 —1 + 0 cj (10.2.9)C2

leads to an M-matrix. It is true that the convection term has been discretisedto a second order of consistency, but the error of the diffusion term is O(eh —c),which, for the case h > which is relevant in practice, amounts to0(h). Instead of (7) one has

Lu — — c)zlu + 0(h2). (10.2. 10)

The difference — is called the artificial viscosity.In the one.dimensional case the methods using numerical and artificial

viscosities do not differ:

Remark 10.2.8. If, in the one-dimensional case, on chooses according tothe second alternative in (8), then the difference formulae (6) and (9) coincide.

10.2.3 Finite Elements

The difficulties described in the previous section are not restricted to differencemethods.

Exercise 10.2.9. Show that


(a) Linear finite elements on a square grid triangulation applied in the case ofEquation (4) give a discretisation that is identical with the difference method

—1 0 —1/6 1/6L,, = —1 4 —1 + h —1/3 0 1/3 (1O.2.lla)

—1 —1/6 1/6 0

(cf. Exercise 8.3.13).(b) For bilinear elements (cf. Exercise 8.3.17) one obtains

1 —1 —1 h 1 0 +1= —1 8 —1 + — —4 0 +4 . (10.2.llb)

—1 —1 —112

—1 a +1

(c) One-dimensional linear elements for —eu" + u' = f lead to the centraldifference formula

= th' [—1 2 —11+1—1 0 1). (10.2.llc)

The Exercises 9b,c show that finite-element methods correspond to cen-tral difference formulas, and thus can equally well lead to instability.

The method of artificial viscosity corresponds to the finite-element solu-tion of the equation (c, grad *4 = f for appropriate As in Exercise9b one may show

Remark 10.2.10. If one sets := IciIh, Ic2Ih} and uses bilinearelements then the discretisation of Equation (1) leads to an M-matrix.

On the other hand the matrix (ha) has different signs in the sub-diagonaland in the super-diagonal, so that it is not possible to have an M-matrix forany value of €h.

The analogues of one-sided differences are more difficult to construct. Oneapproach is to combine a finite-element method for the diffusion term witha (one-sided) difference method for the convection term (cf. Thomasset [1,§2.4]).

A second possibility is the generalisation of the Galerkin method to thePetrov-Galerkin method, in which the discrete solution of the generalequation (8.1.1) is defined by problem (12):

Find u E V,1, so that a(u,v) = 1(v) for ally e W,, (10.2.12)

(cf. Fletcher [1, §7.2), Thomassetll, §2.2]). Here we have

dimVh = dimWh (but in general Vh Wh).

11 Eigenvalue Problems

11.1 Formulation of Elgenvalue Problems

The classical formula*;ion of an eigenva)ue problem reads

Le Xe in .17, B,e = 0 on F (j 1,•• ,m). (11.1.1)

Here L is an elliptic differential operator of order 2rn, and B, are boundaryoperators. A solution e of (1) is called an eigenfunction if e 0. In thiscase, A is the elgenvalue associated with e.

As in Section 7, one can replace the classical representation (1) by avariational formulation, with a suitable bilinear form a(., .): V xV —+ IR takingthe place of {L,B,}:

Find e V with a(e,v) = A(e, v)0 for all v V. (11.1.2a)

(ti, = dx is the L2($?)-scalar product. Strictly speaking one oughtto replace (., .)o by (., where U is the Hubert space of the Gelfand tripleV c U c V1 (cf. Section 6.3.3). But here we limit ourselves to the standardcase U = L2(fl).

The adjoint eigenvalue problem is formulated as

Find e E V with a(v, e) = for all V E V. (11.1.2b)

Definition 11.1.1. Let A E C. By E(A) one denotes the subepace of alle V which Satisfy Equation (1)[ resp. E(A) is called the eigenspacefor A. With E(A) one denotes the corresponding eigenspace of Equation(2b). A is called an eigenvalue if dim E(A) � 1.

Theorems 6.5.15 and 7.2.14 already contain the following statements:

Theorem 11.1.2. Let V C L2(.17) be continuously, densely and compactlyembedded (for example, let V = with bounded.1?). Let a(.,.) be V-coercive. Then the problems (2a,b) have countably many eigenvalues A E Cwhich may only have an accumulation point at oo. For all A C we have

dimE(A) = dimE(A) <oo. (11.1.3)

254 11 Elgenvalue Problems

Exercise 11.1.3. In addition to the assumptions of Theorem 2 let a(.,.) besymmetric. Show that all eigenvalues are real and problems (2a) and (2b) areidentical so that E(A) =

For ordinary differential equations of second order (i.e., (1 C 1R1, m 1)

it is known that all eigenva.lues are simple: dim E(A) = 1. This statement isincorrect for partial differential equations as the following example shows.

Example 11.1.4. The Poisson equation = Ac in the rectangle (0, a) x(0, b), with Dirichiet boundary values e = 0 on F, has the eigenvalues A =(vir/a)2 + (or/b)2 with 1.', E I'J. The associated eigenfunction reads e(x, y)e"4(x, y) := a = bone obtainsfor &' eigenvalues A = A11"4 = A"1', which have multiplicity at least2 since e1"4 and e14'1' are linearly independent eigenfunctions for the sameelgenvalue. A triple eigenvalue, for example, exists for a = b, A = 50ir2/a2:E(A) = span{e1'7,e7'1,e55}.

The eigenhunctions e E E(A), by definition, belong to V. The regularityinvestigations of Section 9.1 immediately result in a stronger regularity.

Theorem 11.1.5. Let V = Hr(Q) with m � 1, or V = withm = 1. Under the aastsmptions of Theorems 9.1.16 reap. 9.1.17] we haveE(A) c Hm+s(j7).

PROOF. Along with a(.,.), := a(u,v) — A(u,v)o also satisfies theasswnptions. Since aA(e, v) = 0 for e E(A), V V, the statement followsfrom Corollary 9.1.19. I

Besides the standard form (2a) there are generalised elgenvalue problems.An example is the Steklov problem

—& =0 in .1?, tIe/an = Ac on 1',

whose variational formulation reads e E H'(Il), =(v H'(Q)). One can show that all eigenvalues are real and that the state-ments of Theorem 2 hold.

11.2 Finite Element Discretisation

11.2.1 Discretisation

Let Vh C V be a (finite-element) subapace. The Ritz-Galerkin (resp. finite-element) discretisations of the eigenvalue problems (1 .2a,b) read

Vh, = Ah(e",v)O for all v E Vh, (11.2.la)


v Vh. (1l.2.lb)

The discrete elgenspaces are spanned by the solutions of theproblems (in) Iresp. (ib)]. As in Theorem 11.1.2, dirnEh(Ah) = dim

holds. If a(.,.) is symmetric, then Eh(Ah) =As in Section 8.1, the formulation (la,b) can be transcribed into matrix

notation.

Remark 11.2.1. Let e and be the coefficient vectors for = Pe andPe (cf. (8.1.6)). The elgenvalue problems (la,b) are equivalent to

Le = LTe* = 3Me, (11.2.1')

where the stiffness matrix L is defined as in Theorem 8.1.3 and M by (8.8.7).Since in general M I, (1') involves generalised eigenvalue problems.

Exercise 11.2.2. Show that (a) M is positive definite and possesses a de-composition M = ATA (for example, A M"2 or the Cholesky factor).(b) The first problem in (1') is equivalent to the ordinary eigenvalue problem

= AhÔ withL := (AT)1LA', ë = Ae. The second problem in (1')corresponds to LTÔ* = with ë = Ae'.

When investigating convergence, one must watch out for the followingdifficulties:(i) A uniform approximation of all the eigenva.lues and eigenfunctions by dis-crete eigenvalues and eigenfunctions is impossible since the infinitely manyeigenvalues of (1 .2a) are set against the only finitely many of (Ia). It is onlypossible to characterise a fixed elgenvalue A of (1.2a) as an accumulation pointof discrete eigenvalues {Ah: h <O}, and to set up estimates for

A and are the continuous, rasp. discrete, eigenvalues then dim E(A) =dim Eh(Ah) need not hold. It is preferable to limit oneself to the case of sim-ple eigenvalues where dimE(A) = dimEh(Ah) = 1. If diznE(A) = k> 1, itmay well be that the multiple eigenvalue A is approximated by several discreteeigenvalues I = 1,. . , k: dimE(A) = dim The error esti-mates of jA — are then generally worse than for simple aigenvalues. Onlyfor the mean value A := Ar/k does one obtain the usual estimates (ef.

p. 3381).

Exercise 11.2.3. Let the eigenvalue problem be as in Example 11.1.4 with

a = b and A = Let V,, consist of linear elements over a squaregrid triangulation. Show that to the given triple eigenvalue A correspondsa double eigenvalue and a single eigenvalue distinct from it, with!lmn_.o = A (i = 1,2). Hint: the nodal values of the discrete eigenfunctionsagree with the continuous eigenfunctions e1'7, and


11.2.2 Qualitative Convergence Results

This section concerns the question as to whether A and eh e forh 0. The rate of convergence will be discussed in Section 11.2.3. The basicassumptions are the following:

Let a(.,.): V x V lit be V-coercive, (l1.2.2a)

Let V C L2(fl) be continuously, densely, and compactly embedded,(11.2.2b)

so that the theory is applicable (Theorem 11.1.2). Further-more, let a sequence of subapaces -, 0) be given which increasinglyapproximate V (cf. (8.2.4a)):

lirninf{IIu — u E V. (11.2.2c)

We define

aA(.,.):V xV —' aA(u,v) :=a(u,v)— (11.2.3a)

w(A) := mi sup laA(ti,v)I, (11.2.3b)uEV vEV

flvllv=1 ]IvIIvl

wh(A) := inf sup (11.2.3c)vEVg,

ExercIse 11.2.4. Let L and Lh be the operators associated with a(., .): V xV JR and a(.,.):Va x Vh —, lit (cf. (8.1.lOa)). Show that(1) If A is not an eigenvalue we have

= = lI(Lh —At)'1111, (11.2.4)

(cf. Lemma 6.5.3 and ExercIse 8.1.16).(ii) w(A) and are continuous in A E(iii) If in (3b,c) one replaces OA(u, v) by aA(v, u) one obtains the variablesw(A) and which correspond to the a4joint problem. The following holds:

= w(A) and = wh(A). Hint: Use Lemma 6.5.17.

With the aid of (4) and Theorem 6.5.15 one proves the following connec-tion between wa(A), and the eigenvalue problems.

Remark 11.2.5. Let (2a,b) hold. A is an elgenvalue of (1.2a) if and only if0, and Ah is elgenvalue of(la) if and only if wh(Ah)= 0.

Lemm* 11.2.6. Let a(.,.) be V-coercive (ef. (2a)). Then there exists a itsuch that is V-elliptic. In addition, then � 1/CR and �1/CE hold with CE >0 from Definition 6.5.13.


Lemma 11.2.7. Let A C be osmpact. Let (2a-c) hokL Then there exutnumbers C > 0 and ,7(h) > 0, independent of A e A with = 0such that

wh(A) � Cw(A) —i1(h), w(A) � Cc,Jh(A)—rI(h) for aU A E A. (11.2.5)

PROOF. Let the operators Z = Z(A): V —* V and 4 = 4(A): V —* Vh

defined as follows:

z := Z(A)u is the solution of for all v e V,

(11.2.6a)

Zh(A)u is the solution of for all v Vh,

(1L2.6b)

where is chosen acooi-ding to Lemma 6. It shows

for all A A. (11.2.6c)

From aA(u,v) = — z,v) and the definition of w(A) one infers

w(A)IlujIv S sup IaA(u,v)l = 8UP — z,v)l � CsIIu — zIIvvEV vEV

jvIIv=1 IIvIIv=1(11.2.6d)

with Cs : For an arbitrary u E infers from— zh, v), Lemma 6, and (6d) that

sup laA(u,v)i = sup — � —z"llvVEVi. vEVk

IIvIIv=1 IIvIIv=1

� — zIlv — liz — � — HZ — Uultv,

thus Wh (A) � (CECS)'w(A) — flZ — ZhIIVIV/CE. From this follows the firstpart of (5) with C = (CECS)' >0 and ,i(h) = HZ — if

urn sup fl Z(A) — = 0. (11.2.6e)h—so )iEA

The proof of (6e) is earned out indirectly. Its negation reads: there exist e >0,E A, —*0 with H — � >0. Then there exist E V

with

= 1, — � e/2 > 0. (1L2.6f)

Due to (2b) and the of A there exists a subsequence A, e A,Uj V with lixnAj = A5, u' E V' in V'. (2c) and Theorem 8.2.2show H(Z(A) — Z,,(A)u1$v —50. Together with (6c) obtain


— Zh1(A,)ju,IIv

� II[Z(A,) — Z(A*)!uJIIv + II[Zhj(Aj) — Zh,(A))uj$Iv+ —uJHv + IIZh — u*jflv + fl[Z(A) —

<2C1A3 — 1- 2Czflu, — uiIv' + lI[Z(A) — Zh,(A')Ju'jIv 0

in contradiction to (6f).For the proof of the second part of (5) one (6d) and the following

estimate by

sup forall U"EVh,yEV,,

SUp supvEV vEV

IIvIIv=t 11V11v1

� — IIZ(A) — Zh(A)IIv,_v111u40v for all nh e Vh.

Let u E V with IIuflv = I be selected such that

sup inf sup IaA(u,v)l=w(A).oEV tLEV vEV

IvHv=1 IIuIIv=1 flvIIv=1

Since sup IaA(u — u's, � CsIIu — it follows for arbitrary E Vh that

� — IIZ(A) — Zh(A)IJV,...V — CgJJu —

From (2c) and (6e) follows the second part of (5).

A corollary to Lemma 7 is Theorem 8.2.8.11 Problem (8.1.1) for all! V'is solvable, then A =0 cannot be an eigenvalue, i.e., w(0) > 0. Thus it followsthat CN, � WN,(O) � c >0 for sufficiently large i.

A second corollary concerns the convergence of the eigen values.

Theorem 11.2.8. Let (2a-c) hold. If (i —* 00, —s 0) are discreteeigenvdues of(la) with A,11 Ao then ),o is an of(1.2a).

PROOF. If A0 were not an eigenvalue then w(A) � > 0 would be in thec-neighbourhood of since w(A) is continuous (cf. Exercise 4(u)). Therewould exist h0 >0 such that ,1(h) � for all 4 � h0 (C and from(5)). For all E K1Po) with � h0 the contradiction follows from (5):

0= � — � — = >0.

Lemma 11.2.9. (Minimum Principle) Let hold. The functions w(A)and w,1(A) in the interior of Ac C have no pmper positive minimum.

11.2 Finite Element Dlscretisation 259

PROOF. Let L be the operator with o(., .). Let with w(A)>0, be an arbitrary point in the interior of A. For sufficiently small e > 0 we

have C A and > 0 in Thus (L — Al)' is defined inand holomorphic. Caucby'a integral formula says

(L — )J)' = — A)'(L — CI)'dC for all A E Ke(A).2ws

From this one infers

—ES A')

C)IIvv

i.e., w(A) � min{w(():C E (ci. Exercise 4(i)). Thus, w(A) cannotaseume a proper minimum in For wh(A) the conclusion is the same.

The converse of Theorem 8 is contained in

Theorem 11.2.10. Let (2a-c) hold. Let A0 be the eigenvalue of(1.2a). Thenthere exist discrete eigenvalnes of(la) (for all h) stich that Ah = A0.

PROOF. Let e > 0 be arbitrary. Aceording to Theorem 11.1.2, Ao is anisolated eigenvalue: >0 for 0 < IA — Aol � e (e sufficiently small).

Since is continuous and is compact, we have thatIA — Aol = is positive. Because of (5) and w(Ao) = 0 one obtains

for sufficiently small h

Wh(A) � Cw(A) — � Cp, — > v1(h)/C �for all A E Thus must have a proper minimum iii Ke(A0).

By Lemma 9 the minimal value is zero. Thus there exists a Ah Ke(A0)which is a discrete eigenvalue, Wh(Ah) 0. I

The convergence of the eigenfunctions is obtained from

Theorem 11.2.11. Let (2a—c) hold. Let e1' be discrete eigenfunc-Lions with = 1 and A0. Then there exists a subsequencee1" which converges in V to an eigenfunction eE(A0):

e E(Ao), — dlv 0 (i oo), lIellv 1.

PROOF. The functions are uniformly bounded in V. Since V C L2(fl) iscompactly embedded (ci. (2b)), there exists a subeequence whichin L2(Q) to an e L2(Q):

lie — —. 0 (i oo). (11.2.7a)


We define z = Z(Ao)e, = according to (6a,b). According toTheorem 8.2.2 there exists an h1(e) > 0 suth that

liz — <e/2 (11.2.Th)

for <hi(t). The function is a solution of

= (A,4 for all v E (l1.2.7c)

A combination of = (i.e., (6b) for A = Ao) and (7c) yields

— et", v) = := (A0 — — v)o — (A,4 — Ao)(e",

v V,4. Since —' 0 because Ao and (7*), there exists an>0 such that � e/(2CEJ (CE from Lemma 6) and

— <e/2 (11.2.7e)

for � h2(e). (Th) and (7e) show that liz — e for � min{h1(e),thus = z in V. Therefore lime" = z in L2(.Q) C V also

holds. (7*) proves z = e E V such that e = z = becomes a(e, v) =Ao(e,v)o. Therefore e = lime"' Is an elgenfunction of (1.2*). In particular,Idly = limlle"Ilv = 1.

ExercIse 11.2.12. Let (2*-c) hold. Let and e be as in Theorem11. Show that(a) If dim E(Ao) =1 then also dim =1.(b) Let dim E(Ao) = 1. Then we have e in V for := ifI(e",e)vi � 1/2 and := otherwise.

11.2.3 QuantitatIve Convergence Results

The geometric and algebraic multiplicities of an eigenvalue Ao of (1.2*) agreeif

dim Ker(L — Aol) dim Ker(L — Aol)2. (11.2.8)

11.2.13. Let (2a1b) and dim E(Ao) = I hoüL Then (8) is equivalentto (e,e')o 0 forO e E E(Ao), 0 e' E(Ao).

PROOF. We have Ker(L — Aol) = E(Ao) = span {e}. dim Ker(L — Aol)2 > 1holds if and only if there exists a solution v V for (L—AoI)v = e. Accordingto Theorem 6.5.15c this equation has a solution If and only if (e,e)o 0. •

Let E(Ao) = span {e}, E'(Ao) = span {e'}. Under the aesumptlon (8) eand e can be normalised so that

(e,e')o = 1. (11.2.8')

11.2 FnIte Element Dscretiaation 261

We define V := {v E V:(v,e')o = 0}, = {v' e V':(v',e')o O}. Let

be the dual norm for = 11 liv. Fbr Problem (9):

For f V', find u aA(u,v) (f,v)o for all V E (11.2.9)

one defines the variable corresponding to (3b)

:= ml sup laA(u,v)i. (11.2.10)uEc'

Lt"IIv=l 11v11v1

Lemma 11.2.14. Let (2a,b), (8), and dim E(Ao) = 1 hold. Then there existsan e > 0 such that � C > 0 for all IA — e. Problem (9) has exactlyone solution u E V with

ilulIv � Ill Ilv'/2z(A), if > 0.

PROOF. Let L: 1' be the operator aesociated with a(., .): x C.

For 0 < — Aol �e (e sufficiently small) (L — AI)u = f has a unique solutionu E V. From f V' follows

0 = = ([L — A.1]u,e')o = (u, [L — M)'e')o = — X)(u, c')0,

i.e., u E Thus there exists — AI)': —+ as the restriction of(L — )J)-' to V' C V'. For A = Ac Problem (9) has a unique solutioncording to Theorem 6.5.15c. Then there exists (L — Al)-1 for all AAccording to Remark 5 (with V instead of V) , c2i(A) must be positive inKd(Ao). The continuity of proves � C > 0. In analogy to (4) onehas llullv = � The bound by � results from

Exercise 11.2.15. Show that ILemrnk 11.2.16. Let (2a-c), dim E(Ao) = 1, and condition (8) hold. Let A,,be discrete eigenvalues with urn,,....0 A,, = According to Exercise 12b thereexists an e1' E,,(A,,) and e'1' with e" —' e E

E'(7.o), (e,e')o = 1. This enables one to construct the space := {v" EV,,: (v", = 0} and the variable

inf sup IaA(u,v)l.

UuIIv=1

Then there exists a C > 0 independent of h and A E ( such that Wh(A) �Cw,,(A). For sufficiently smallc >0 and h, Wh(A) � irj> Ofor � c.

PROOF. (a) First statement: there exists h0 >0 and C such that

min{iiv+ae"llv:aEC}�IIvIlv/C


The proof is carried out indirectly. The negation reads: there exists a se-quence C, hj —4 0, E with = 1 and + 0.

Thus there exist subsequences with —. a', —* ? in L2(a). Evi-dently, := v, + must have the limit w' = liznwj = v' + a'e'in L2(a). Since limIIw4IL2(n) � = 0, it follows that w = 0,

and thus v' = —a'e'. From 0 = lim(vj,e*h)o = (v',e')o =one infers a' = 0. Thus the contradiction follows from 1 = tim 11v41v �

+ = IIwiIIv =(b) Second statement: L21,,(A)� Ccüh(A) with C from (a). Because aj(u,v)

foruEVh we have

= inf sup IaA(u,v)I/(IIuIIvIIvIIv)

� C inf sup max(aA(u,v+ae")I/[(ItLIIvIjv+ae"flv]aEC

=C inf sup IaA(u,w)I/[IIuhIvIIwUv]O#ioEV,.

�C in! sup

(c) Let e > 0 be chosen such that Ao is the only elgenvalue in Forsufficiently small h, Ah is the only discrete eigenvalue in Ke(A0). In the proof of

From Part b follows that � i := > 0 for A E Accordingto Exercise 12a, aA(u,v) = (f,v)o (v E is solvable for each 1€ and allA E Ke(A0) such that = 0 is excluded. Lemma 9 shows that �IExercise 11.2.17. Let (2a,b) hold. The functions u, v V (u, v)o 0.Let d(eA, Vh) be defined as in (8.2.2). Show that if d(u, Vh) is sufficiently smallthen there exists a u1' Vh with

IIu_uhllv �2d(u,Vh),

Lemmk 11.2.18. Let (2a—c) hOZCL Let Ao be an eigenvalue with (8) anddimE(Ao) =1. For sufficientty h there exists Eh(Ah) with

— � C[IAo — AhI + d(e, Vh)].

PROOF. Let zh := Zh(AO)e be the solution of (6b). Since e = Z(Ao)e, onehas

n h11e—z IIv�CId(e,Vh). (1l.2.lIa)

For all v Vh we have

11.2 Finite Element Dlscretisation 263

v) = (Ao — .&)(e, v)0 — (Ah — v)0

= (.\o — )th)(e,v)o + (Ah — — ehv)0 + (Ah — —

so that= — +(Ah —.p)(e —z',v)o for ally Vh.

(11.2.llb)

According to Lemma 16 we have f(e, � > 0 for sufficiently small h.eh can be scaled so that — = 0.

(lib) corresponds to Problem (9). Lemma 14 proves

— � Wh(Ah) 'Cu)10 — + lie — z"Iivi � C'[ — + d(e, Vh)J.

Together with (1 la) one obtains the statement. U

Theorem 11.2.19. Let (2a—c) hold. Let Ao be an esgenvalue with (8) anddimE(Ao) = 1. Let e E(A0), e E'(10), itchy = 1, (e,e)o = 1. Thenthere exist discrete eigenvalues Xh with

lAo — )'hl � Cd(e, Vh)d(e', Vh). (11.2.12)

PROOF. Choose according to &ercise 17 such that 1Ie — uhhlv �2d(e,Vh), (e — u",e)o = 0. Discrete elgenvalues —i exist by Theo-rem 10. From

0 = a,10(eh,es) = aAh(e",e) —(Ao _Ah)(e!l,e*)o

= —u") —()1.o —

_tiI)_(Ao_Ah)(eh,e*)o= — e, e — u") — (Ao — e, e)o]

follows — Ahi � — eilv[11e — + — AhiJ. By Lemma 18 thereexists an Eh(Ah) such that

P10 — � C'C"[(Ao — Ahl + d(e, Vh)J[I)1o — Ahi + 2d(e*, Vh)J.

From this one obtains (12) with C = 3C'C" for sufficiently small h, sinced(e,Vh)—40. U

Theorem 11.2.20. Under the asswnptions of Theorem 19 there exiet for

e E(Ao), e' E E()10) discrete elgenfunctions E Ea(Ah),with

lie — Cd(e, Vh), lie* — e"lIv � Cd(e, Vh). (11.2.13)

PROOF. Insert (12) with d(e, Vh) � const into Lemma 18. The secondestimate in (13) follows analogously. U

264 11 Eigenvalue Problems

In the following, let V C H'(fl). Theorem 11.1.5 proves

C C H14'(37). (11.2.14a)

Also, let (14b) hold (cf. (8.4.10)):

d(u, Vh) � Ch'IIUIIHl+.(n) for all u E U (l1.2.14b)

Corollary 11.2.21. Let (2a), (2c), (14a,b) hold. Let be the cigenvaluewith(8) and dimE(Ao) = 1. Then there exists Ah,e" Eh(Ah), esh E

such that

IAo — AhJ � Ch2', lie — � Ch', lie5 — e*hIlv � CM. (11.2.15)

Occasionally eigenfunctions may have better regularity than is provenfor ordinary boundary value problems. For example, let —& = Ac be in therectangle (1 = (0,1) x (0, 1) with e= 0 on V. First, Theorem 11.1.5 impliese E H2(.Q) ii thus e C°(17) (cf. Theorem 6.2.30). Thus e = 0 holdsin the corners of .0. AccordIng to Example 9.1.25 it follows that e (j7)for s <4.

As in Section 8.4.2 one obtains better error estimates for e — in theL2-norm. The proof is postponed until after Corollary 29.

Theorem 11.2.22. Let (2a-c), (8), dim E(Ao) = 1, and (14a,b) with $ = 1

hold. Let a(.,.) and a(.,.) be H2 -regular, i.e., fan E L2(Q) , a,h(u,v) =(f,v)o and = (v,f)o (v E Vh, 4 from Lemma 6) have solution8u,u5 E H2(i7). Let e E E(Ao) and e5 E(Ao). Then there exist Ah,e" EEh(Ah), and e*h E with

lie — e"11L2(a) C'h2, lies — e"liV(D) � C'h2.

If (14a,b) also hold with some s> 1, one must replace C'h2 by

11.2.4 Complementary Problems

In Problem (9) we have already encountered a singular equation which never-thelees was solvable. In the following let A0 be the only eigenvalue in the discKr(A0). The equation

aA(u,v)=(f,v)o forallv€V (11.2.16a)

8 singular for A = Ao. For A A0 Equation (16a) is ill-conditioned. In thefollowing we are going to show that Equation (16a) is well-defined and well-conditioned if the right-hand side f lies in the orthogonal complement ofE(Ao):


f ,i.e., (f,e')a =0 for all e' E (11.2.16b)

In the case of A = A0 , with it, it + e (e E E(Ao)) is also the solution. Theuniqueness of the solution is obtained under the conditions (8) and (16c):

E*(A0) (11.2.16c)

Remark 11.2.23. Let (2a,b) and (8) hold. Let A0 be the only eigenvalue inKr(A0). Then (16a,b) has exactly one solution it for all IA — A0I r whichsatisfies (16c). There exists a C independent of f and A such that IIuIIvCII!liv'

PROOF. This follows from Lemma 14 in which the assumption E(Ao) = 1 isnot necessary. U

The finite element discretisation of Equation (16a) reads:

Find U" V,, with = (f,v)o for all v E Vh. (11.2.17)

In general, Equation (17) need not be well-defined, even assuming (16b).For the sake of simplicity we limit ourselves in the following to simple

eigenvalues: d.imE(A0) = 1. Equation (17) is replaced by (18a):

Find E with aA(u",v) = for all v V,, (11.2.18a)

with f(s) .1. (11.2.18b)

(11.2.18c)

Exercise 11.2.24. Show that is equivalent to: Find it" withSA(U1', v) = (f(h), v)o for all v V,, with = fl as in Lemma 16.

Lemma 16 proves the

Remark 11.2.25. Let (2a-c), (8), dim E(A0) = 1 hold. Let A0 be the onlyeigenvalue in Kr(Ao). Then there exists an h0 >0 auth that) for all h � ho andall A Kr(A0), the Problem (18a,b) has a unique solution it" u"(A) whichsatisfies the additional conditions (18c). Further there exists a C independentof h, A, and 1(h) such that � CII!

If E'(A0), f from (16b) need not satisfy condition (18b). IfEh(Ah) and E with (e",e"')o = 1 are known, one can define

f (11.2.19)

satisfies (18b) since Qh represents the projection on (As)'.

266 11 Ezgenvalue Problems

Exercise 11.2.26. Letu 1. = = 1, = 1,= 1. Show that

d(u, E v" J..

� + llullvinf{11e8 —

Theorem 11.2.27. Let (2a-c), dim E(Ao) = dim = i hold. LetAo be the only eigenvatue in Let h be sufficiently small such that(following Remav* 25) the Problem (18a-c) is solvable. For the solutions uand of (i6a-c) and (iSa—c) the ermr estimate

llu—u'iiv � Cfd(u, Vh)+llfllvI inf(lle* Iv': e E —1 lIv'i(il.2.20)

holds, with C independent off, f(h), and h.

PROOF. Repeat the proof of Theorem 8.2.1 for aA(-,-) instead of a(-, .). Hereone must choose w E Vh with w 1. Furthermore, (8.2.3) becomes

— u,v) = — f,v)o for all v E Vh.

CN agrees with � > 0 (A E (cf. Lemma 16). flu — Why isestimated with the aid of Exercise 26, with llullv � il/ liv' being added. ICorollary 11.2.28. If 1(h) is defined by (19), then inequality (20) becomes

flu — � C'[d(u, Vh) + hilly' inf{hle* — e*hIlv: e' E E'(Ao)}J. (i2.2.2ia)

PROOF. —lily' � Cj(f,e")ol = Cl(f,e� Cflf liv' hie* —

Corollary 11.2.29. If additionally the assumptions u E (14a), aridd(u,Vh) S Chitihi÷1 hold forts E then (21a) yields the estimate

flu — � Ch'JIulIffl+.(a). (11.2.21b)

It remains to add the

PROOF. (Theorem 22) For e E(.Ao) there exists e1' Eh(Ah) with f :=e — 1. E(Ao) and Ifhi = Il/ liv � Ch = Ch. According to Remark 25 theproblem OA0(v,w) = (v,f)o has a solution w I E(A0) for all v E V. Theassumption of regularity yields w E H2(I?), 1w12 � Cl/k such that E Vhexists with I. 1w — � Chlwi2 � C'hIflo. The value

11.3 Discretisation by Difference Methods 267

=0—aA0(e5,w")

= (Ao — Aa)(e",w")o— = —

can be bounded by � (cf. (15)). From

= (f, f)o a.x0(f,w) = a.x0(f,w — w") + (Ao —

<C[C'hlflilf to + h2ffIoJ

and Ifti � Ch one infers tf to � C'h2. The same method is then applied to— eshI

ExercIse 11.2.30. Formulate the conditions for and the proof of the errorestimate � from Corollary 28). Hint: Decompose

in fi +12 with ft with and

11.3 Discretisation by Difference Methods

In the following we Limit ourselves to the case of a difference operator of theorder 2m = 2. The differential equation Lu = f with homogeneous Dirichtetboundary condition is replaced, as in Sections 4 arid 5, by the difference equa-tion L5u5 = fh. The eigenvalue equations Le = Ac, Le = Ae are discretisedby

Lheh=Ahes, (11.3.1)

is the transposed and complex-conjugate matrix to L5. The general as-swnptions of the following analysis are:

V = I? C°" bounded, (11.3.2a)

a(u,v) = (Lu,v)o is (11.3.2b)

Consistency condition ILaRa — � Ch. (11.3.2c)

Condition (2c) has been discussed in Section 9.2.2. Assume furthermore thatis For suitable A E IR

:r= L5 — (I : identity matrix)

is thus(LU,hvh,vh)o � (11.3.2d)

Furthermore, letL,.1 := L — p1 H2(Q)-regular, (11.3.2e)

i.e., <C. The boundedness of L and L,, reads

ILI_14_i � C, 1L51_14_i � C. (11.3.2f)


In (9.2.26a,b) we have introduced prolongations Ph: L2(1R2) and—+ L2(S?). Now we need a mapping -.

I Phtlh(x) if Kh12(x) C .0,I. 0 otherwise,

Phtzh(x) := (x E.0),

where Kh,2(x) = {y E lix — <h/2}, Ph ate defined according to(9.2.26b) and according to (9.2.11). Ched that PhUh E and

� C. (11.3.2g)

Let Rh and R,, be defIned as in (9.2.12a,b). They Satisfy

IRaio._o � C, IR,111._1 � C, � C. (11.3.2h)

ExercIse 11.3.1. Show that:

IRh—Rhio.--1 �Ch, (11.3.2i)

— � Ch, 11— RhPhl 0._i � Ch, if — � Ch.(11 .3.2j)

The first inequality in (2j) is equivalent to

l(Phuh,Phvh)o — (uh,vh)oJ � ChjuhIolvhIi. (11.3.2j)

Hint: Exercise 9.2.12 and Lemma 9.2.15.

LemmA 11.3.2. Let (2a,c,g,h,i) hold. (a) it is true that

foraliueH01(f1),

(11.3.2k)

!im — Rh)t40 = 0 for all u L2(.Q), (11.3.21)

— RaLjui_i = 0 for allu E E C. (11.3.2m)

(b) If u and lilnh,oIRauk 0 then u = 0.

PROOF. (a) The proof of (2k,l,m) follows the same pattern, which will be

demonstrated for the case of(21). Fbr >0 one has to show I(Rh — Rh)uI �for h � Since is dense in there exists a ü withlu—ui0 such that i(Rh—Rh)(u—ii)k � By (2i) therefollows (Rh— � � e/2 for h � h(e) := Altogetherthis gives l(Rh —

(b) From (2k) one infers 0 = hmh_.o(Rhu,P,u)o = hmh,o(PhRhu,u)o =(u,u)O thus u = 0. IExercise 11.3.3. LetAh Showthat

11.3 Discretlsation by Difference Methods 269

= (Au,u)o, = (Ahuh,uh)o,

= (v, LAU)O for v = A1LAU E

I= (vh,LA,huh)o for Vh =

hm I(AhRh — k,,A)uI_i = 0 for all u E

The following analysis is tailored to the properties of the difference meth-ods under discussion. We have tried to avoid a more abstract theory thatis applicable to finite elements as well as difference methods. This kind ofapproach can be found in Stummel [1) and Chatelin [1).

The variable is now defined by

:= inf sup I(LA,huh,vh)oI = inf ILA,huhI_i. (11.3.3)IuhIl=l Ivhh=L IlahIl=l

As in Exercise 11.2.4 we have

wh(A)=0, ifAisaneigenvalueof Lh;= otherwise. (11.3.3')

The analogue of Lemma 11.2.7 reads

Lemma 11.3.4. Let A C C be compact. Let (2a-m) hold. Then there existvariables C >0 independent ofAEA, such that

wh(A) � Cw(A) — w(A) � Cwh(A) — for all A A, h > 0.(11.3.4)

PROOF. (a) Since A is compact, w(A) is continuous, and wh(A) is equi-continuous in A, it is sufficient to show that Illflh_.owh(A) � Cw(A), and w(A) �Climh.o,Jh(A)forallAEAwithC>0.(b) Define for A E A and uh with IuhIl = 1 and ILA,huhj_1 = wh((A)

u:=Phuh, z,,

with p from (2e). Without loss of generality it can be assumed that p A.We have

— zli = — Zh) + P,,zh — zIl � — ZhIl + IPhzh — zIi,

IPhzh — zli = IPhLZh — Rp,zJ — — PhRJ&]zI1

� — — + —

(ci. (2g,j,m)) so that

luh—zhIl�CxIu—zIl—o(1), C1 >0. (11.3.5a)

270 11 Eigenvalue Problems

As in (2.6d) one obtains

lu — zli � C24,(A)Iuli, C2 > 0. (11.3.5b)

From

= ILA,htLhl_1 = + (p —

� + — uh)o � — Al + uh)o

� + = + C5

with CA := max{lp — Al: A E A) > 0 follows

� —wh(A)j.

Either � from which the statement results directly, or wh(A)CE/2 WhiCh yields

luhlo � Co = ColuhIl with Co := (11.3.5c)

We want to show that there exists ho > 0 and C,, = C,(Co) such that

luali <CplPhuhll for all uh with IuhIo � Coluhil and h � h0. (11.3.5d)

The negation of (5d) reads: there exists with lUhll = 1, lualo � C0 and

lPhuhtl 0 (h —* 0). From IR,M*ualo � � 0 and

IUh — RhP,uhlo � 1 —RhPhIo.lIuhll � Ch —*0 follows lUhlO —*0 in contrasttO lUhlO � C0. Thus we have (5(1)

Together with L,,,,1(uh — zh) = LAhuh, (2d) and (5a,b,d) yield the firstof the inequalities (4):

= = — Zh)I_1 � —

� — zIj — o(l) � C5C1C2w(A)IuIj — o(1)

� Cw(A) — o(1) with C := C5C1C2/C,, >0.

(c) Let > 0 be arbitrary; u with luli = 1 can be selected such thatw(A) � ILAuI_I — Set uh := RhU. According to Exercise 3 it holds that

(v, LAU)O for v = A'LAU E A = I —4= (Vh, for vh :=

From

Rhv — Vh = — + — —10

— L,1,hUh = [ALA — —, 0,

(v, LAu)o — (Rhv, RhLAU)O = (LI — LAU)O —' 0,

for h 0 (ci. (2m), (2k)), one infers —' ILAUI_i and

w(A) � ILAUI_1 — o(1) � Uh(A) — — o(1) (h —*0)


for eache>0. has been proved. ICorollary 11.3.5. Under the assumptions of Lemma 4 the following holds:If there exists for all A A then there exists an ho > (1 such that LA,h

for CI1AE A andh�ho is

PROOF.Wehaveamedw(A)>OinAsomax{w(A):AEA}z:v,>O.Choose h0 according to Lemma 4 such that ws(A) � Cw(A) — � for

h < ho. Then � 2/ (Ct7) for all A E A,h � ho. IThe proof of Theorem 11.2.8 can be earned over without change and

results in

Theorem 11.3.6. Assume (2a—m). If —. 0) are discrete eigenvaiuesof ProbLem (1) with —. then A0 is an eigenvaiue of(1.2a).

Lemma 11.2.9 and Theorem 11.2.10 can also be carried over withoutchanges.

Theorem 11.3.7. Let (2a—m) hold. Let A0 be an cigenvalue of(l.2a). Thenthere exist discrete eigenvalues Ah of(l) [toraH hJ such that = A0.

Theorem 11.3.8. Let (2a—m) hold. Let eh be discrete eigenJunctions with1 for Ah, where Ao (h —. 0). Then there exists a subsequence

such that converges in to an eigenft*nction 0 0 e E E(Ao).we have - —.0.

PROOF. Because C (cf. (2g) the functions := Pheh are uni-formly bounded. is compactly embedded in L2(fl) (cf. (2a) and Theo-rem 6.4.8a) such that a subsequence e5' converges in L2(a) to an e E L2(J2):

— do — 0. We have in particular

Estimate (2c) ytelds

fR,1z — � IRaz — zsIi � ChIeIo 0

for z = (A0 — 1s)L'e, := (A0 —

From — = (Ao — — eh) + (A5 — A0)ea —. 0 in follows— —. 0 such that 1R5(e — 0 (h —. 0) results. Lemma

2b shows that e = z E HJ(a), and 0 = e E E(A0) is excluded because of= IlimeaIi = 1. 1

Theorem 11.3.9. Let (2a—m) hold. Let be the solution of the discretecigenvatuc problem = with = I, A5 = Ao. Then there

272 11 Elgenvalue Pmbleins

exists a subsequence such that in convei9es to on eigenfiznc-tion 0 e' E E*(Ao). Furthermore, — R,4e'Ii —'0.

Proof Sketch. The proof is not analogous to that of Theorem 8 since theconsistency condition (2a,m) does not necessarily imply the correspondingstatements for the adjoint operators. One has to carry out the following steps:(a) := Phek_—, converges in L2(f1)_for a subsequence h = —*0.

(b) For z = (A0 — and Zh := (Ao — the following holds:

z — = — — + —

For each v E L2(fl) we obtain

(v,z — = (Ao — + ([P, —

— —*0

for h = —+ 0 (cf. (2i), (2c)).(c) Izh — —+0 may be inferred from Lh(zh — = (Ao — —.0. Inparticular, (v, — —' 0 for each v E L2(S?).

(d) — = — —'0 for each V E L2(Q) (cf. (21)).

(e) follows (v,z — e')o = 0, thus z = e' E'(A0), where0.

ExercIse 11.3.10. Carry over Exercise 11.2.12 to the present situation.

In Lemma 11.2.16 we defined Now set

:= {uh: (uh,eja = 0} = = —-.

:= inf sup I(LA,auh,vh)oI/(Iu,,I1 Ivali). (11.3.6)O$UkEch

A basic condition for the following is

dim Eh(Ah) = 1, Eh(Ah) = span

E = )huh}, E =uh}.Exercise 10 shows that (7) is valid for h � h0 if dim E(Ao) 1.

Lemma 11.3.11. Let (2a-m), dimE(Ao) = 1, and (2.8) hol4L Then thereexi8t ho>OandaC>Othat is independent ofh<ho andA€Csuch that� forh < h0. For sufficiently sinallc >0 andh, >0for all

PROOF. (a) There exists a C >0 such that


forallv,1E1?'h,aEC,h>O.

Indirect Proof: Assume that there exists a sequence vh with —+0, = 1,E €, := vp4 + 0. For a subsequence of

mL2((I),

P,4wh, —+ := v + = 0

converge. From 0 = = (v,JI — + —+

(v,e)o one infers = 0, & = (w*,e*)o/(e*,e*)o = 0, v = 0. Thecontradiction results from 1 = limIvh4Il = limlwh,I1 = 0.(b) The rest of the proof runs as in Lemma 11.2.16.

Lemma 11.3.12. Let (2a—m), dim E(Ao) = 1, and (2.8) hold. One maychoose0 eE and eh Eh(.Xh) so that IR,,e—esli �

PROOF. We have that e = Assume also that()io — The inequality (9.2.22) implies jRhe — zhJl � ChIeI2.

For sufficiently small h we have (e,,, 0 so that one can scale such that(eh — zh,ejo = 0. From

— zh) = (Ah — + — — Rhe)

= (Ah — + (Ah — — + (R,1 —

one infers that Ieh — ZhIl � C(IAh — + h) (cf. Lemma 11) since I(Rh —= 0(h). The statement follows from this.

11.3.13. Under the a8sumptsons of Lemma 12 we have the inequality

PROOF. Choose such that (R5e — eh, = (eh, — ej0 = 0. Forthe Rayleigh quotient Ah := Rhe*)o/(Rhe, R,1e)o we then have

IAh � — — � eh(h + I.Xo — AhjJ

with Ch := CIRhe —

From (2c, j) one infers that

(LhRp,e,R5e')o — (Le,e*)o

= — RhLJe, + (Le, — = 0(h)

and P'h — = 0(h) such that Rh — Ch + ehlAh — For sufficientlysmall h we have Ch (cf. Theorem 9, Exercise 10) thus Rh —Aol <2Ch. U

Leminata 12 and 13 give


Theorem 11.3.14. Let (2a—m), = span {e}, and (2.8) hohL There

exists eh Eh(Ah) with — � Ch.

Since IR,1e' — = o(1) or even IRhe — 0(h), toTheorem 11.2.19 one expects that Pio — AhI = o(h) [resp. 0(h2)J. In general,this estimate is false as the following counterexample shows.

Example 11.3.15. The eigenvalue problem —*/' + u' = Au in (0, 1) withu(0) = u(1) = 0 has the solution u(x) = exp(Ax)sin(Rx) with A = 1/2. Theeigcnvalue is ).o = + 1/4. One calculates that the discretisation +

= Au has the eigenvalue

= h212 — + — —

+ h_11c08(,rh)eA'hI — 1 +

with A':=log(1—h)/(2h)

such that IA0 — AhI turns out no better than 0(h).

12 Stokes Equations

12.1 Systems of Elliptic Differential Equations

In Example 1.1 11 we have already stated the Stokes equations for 47 C 1R2:

+ Op/Oxi fi, (12.1.lai)

— LIU2 + 12, (12.1.182)

— 8u1/Oxt — = 0. (12.Llb)

In the case of 42 C another equation ÷ Op/0x3 = f3 needs to beadded, and the left-hand side of (Ib) must be supplemented by —ôusJäz3.A representation independent of the dimension can be obtained if one takestogether (u1, Ireep. (Ui, U2, u3)J as a vector u satisfying, in 42, the equations

— 4u+ Vp = I (12.1.2a)

—divu=0. (12.l.2b)

Here, dlv is the divergence operator

divu =

and n is both the dimension of 47 C JR" and the number of components ofu(x) Only n � 3 is of physical interest here. In fluid mechanics theStokes equation describes the flow of an incompreesible medium (neglectingthe inertial terms) and describes the velocity field. With x is thevelocity of the medium in the direction; the function p denotes the

Up to now we have not formulated any boundary conditions. In the fol.lowing we limit ourselves to Dirichlet boundary values:

u=0 onE'. (12.1.3)

This implies that the flow vanishes at the boundary. No boundary conditionis given for p. Since both the pairs (u,p) and (U,p + const) satisfy the Stokesequations (2a,b), (3), one obtains:

Remark 12.1.1. p is determined only up to a constant by the Stokes equation(2a1b) and the boundary conditions (3).

276 12 Stokes Equations

The Stokes equations have been chosen as an example of a system ofdifferential equations. It remains to investigate whether Equation (2a,b) iselliptic in a sense yet to be defined. Even though the functions U4, for givenp, are solutions of the elliptic equations = — indetermining p, (2a,b) do not provide an elliptic equation, in any current senseof elliptic.

A general system of q differential equations for q functions Ui, ... canbe written in the form

(12.1.4a)

with the differential operators

= (1 � i,j � q). (12.1.4b)

The equations (4a) are summarised as Lu f where L is the matrix (Li,) ofdifferential operators and u = (Ui,.. •,uq)7, I The order ofthe operator is at most ks,. Let the numbers ml,••-,mq,m'i,• .•,m'q bechosen so that

k11�mj+m (1<i,j_<q). (12.1.5)

As the principal part of one defines

=0.Thecharacteristic polynomial with reads:

:= E

and forms the matrix function

Definition 12.1.2. (Agmon-Douglis-Nirenberg (1J) Let (5) hold forThe differential operator L = is said to be elliptic in x E a if

0 for all 0 E (12.1.6a)

£ is said to be uniformly elliptic in a if there exists an E >0 such that

� for all XE (12.1.6b)

with 2m := To be more precise one should call £ elliptic withindices 714, since the definition does depend on In connection with

12.1 Systems of Elliptic Thfferential Equations 277

this problem, as well as for an additional condition for x E F, q = 2, see theoriginal paper of Agmon—Douglas—Nirenberg [lJ. Further information on thissubject can be found in Cosner Lii.

Exercise 12.1.3. (a) Show that the numbers m,, are not unique. If mjand satisfy the inequality (5), then so do m +k. The definitionof is independent of k.(b) Show that for q = 1, i.e., for the case of a single equation one recoversfrom (6a) the Definitions 1.2.3a Iresp. (5.1.3a)j; (fib) corresponds to (5.i.3a').(c) For a first-order system (i.e., = 1, mj = 1, m = 0), (6a) coincides withDefinition 1.3.2.

In order to describe the Stokes equations in the form (4a,b) we set

q:=n+i,for

The orders are = 2, = = 1 (i � n), and = 0 otherwise. Thenumbers

for mq=m'q=O

satisfy (5). coincides with L,, and is independent of x:

= = = for i � n, = 0 otherwise.

From this we see

0

IdetL"(e)I = 0

—iei2= with 2m = 2n,

... —en °

so that (6b) is satisfied, with =1.

Exercise 12.1.4. In elasticity theory the so-called displacement function.0 C JR3 JR3 is described by the Lamé system:

in!?, on!'(&, A > 0). Show that this system of three equations is uniformly elliptic:IdetL"(e)I =

For the treatment of Stokes equations we will use a variational formulationin the next section. For reasons of completeneas we point out the followingtransformation.

Remark 12.1.5. Let n = 2 and thus u = (01,02). Because divu = 0 thereexists a so-called stream function with uj = 02 =


Insertion in Equation (lat,2) results in the biharmonic equation =

8f2/Ox1 — Ofj/0x2. The boundary condition (3) means = 0 on F. Thisis equivalent to =0 and 845/& =0 on F where ô/8t is the tangentialderivative. = 0 implies = const on P. Since the constant may bechosen arbitrarily, one sets = =0 on P.

12.2 Variational Formulation

12.2.1 Weak Formulation of the Stokes Equations

Since u = (u1,. •. , u,1) is a vector-valued function, we introduce

Ho'(a) x x x (n-fold product).

A corresponding definition holds for H'(S?), H2(Q), etc. The norm associ-ated with will again be denoted by in the following.

According to Remark 12.1.1 the pressure component p of the Stokes prob-lem is not uniquely determined. In order to determine uniquely the constant inp = fi+const, we standardise p by the requirement p dx =0. That is the rea-son why in the following p will always belong to the subspace C

:= {p E L2(.O): J p(x) dx = O}.

To derive the weak formulation we proceed as in Section 7.1 and assume thatu and p are classical solutions of the Stokes problem (1.2a,b). Multiplicationof the ith equation in (1.2a) with Vj E Cr(s?) and subsequent integrationimplies that

j d(x) = + dx (12.2.la)

=f {(Vui(x), Vvj(x)) — dx for E 1 <i �

Summation over i now gives

J{(Vu(x), Vv(x)) — p(x) div, v(x)} dx= L (1(x), v(x)) dx, (12.2. la')

where the abbreviation

(Vu,Vv) =i=1 *3=1

is used. Equation (1 .2b) is then multiplied with some q and inte-grated, giving

12.2 Va atlonal Formulation 279

_fq(x)divti(x)dx = 0 for all q E (12.2.lb)

With the bilinear forms

a(u,v) J(Vu(x)7Vv(x))dx for u,v E (12.2.2a)

b(p, v)— J p(x)div v(x) dx for p v E (12.2.2b)

we obtain the lVweak formulation of the Stokes problem as (3a-c):

Find u and p E such that (12.2.3a)

a(u,v) + b(p,v) 1(v) := J(f,v)dx for all v E (12.2.3b)

b(q,u) 0 for all q (12.2.3c)

In (3b) we first replace "v E by "v Since both sides of(3b) depend continuously on v E and since Cr(I1) is dense in RW?),(3b) follows for all v E

Remark 12.2.1. A classical solution u E p Eof the Stokes problem (1.2a,b), (1.3) is also a weak solution, i.e., a solution of(3a-c). If conversely (3a-c) has a solution with u C2(Th, p C1 (1?), thenit is also the classical solution of the boundary value problem (i.2a,b), (1.3).

PROOF. (a) The above considerations prove the first part.(b) Equation (3c) implies divu = 0. Let i {1,.. . ,n}. In Equation (3b) onecan choose v with vj =0 for j i. Integration by parts recovers(la) and hence the ith equation in (1.2a). I12.2.2 Saddlepoint Problems

The situation in (3a—'c) is a special case of the following problem. We replacethe spaces and in (3a—'c) by two Hilbert spaces V and W. Let

o(., .): V x V —' IR a continuous bilinear form on V x V, (12.2.4a)

b(., .): W x V JR a continuous bilinear form on W x V, (12.2.4b)

fi E V', 12 E W'. (12.2.4e)

In generalisation of (6.5.1) we call b(., .): W xV JR continuous (or bounded),if there exists a Cb JR such that

CbIjwIIwIIvIIv for all w E W,v E V.

The objective of this chapter is to solve the problem (5a-c):


Find v V and w W with (12.2.5a)

a(v,x) + b(w,x) = fi(x) for all x E V, (12.2.5b)

b(y,v) = f2(Y) for all y W. (12.2.5c)

Formally, (5a—c) can be transformed to the form

Find u E X with c(u,z) = f(z) for all z E X (12.2.6a)

if one sets:

X =V x W, c(u,z) :=a(v,x)+b(to,x)+b(y,v) and

1(z) =fl(x)+f2(y) for u= (x).(12.2.6b)

Exercise 12.2.2. Show that (a) c(., .):X x X —+ IRis a continuous bilinearform.(b) Problems (5a—c) and (6a,b) are equivalent.

That the variational problems (5) and (6) must be handled differentlythan in Section 7 is made clear by the following remark.

Remark 12.2.3. The bilinear form c(•,.) in (6b) cannot be X-elliptic.

PROOF. We have c(u,u)=0 for all tL= (0).

In analogy to (6.5.9) we set

J(v,w) := a(v,v) + 2b(w,v) — 2fi(v) — 212(w),

and therefore J(v, w) = c(u, u) — 21(u) for u = (t').i'or v E V, w E W

we know J(v, w) is neither bounded below nor above. Therefore the solutionv,w' of (5a—c) does not give a minimum of J; however, under suitable con-ditions, v,w' may be a saddle point.

Theorem 12.2.4. Let (4a-c) hold. Let a(.,.) be symmetrsc and V-elliptic.The pair E V, E W is a solution of the problem (5a-c) if and only if

J(v,w) J(v,w') � J(v,w) for all v E V,w W. (12.2.7)

Another equivalent characterisation is

J(v*,w*) = mini(v,w) = max mmJ(v,w). (12.2.8)VEV WEWvEV

PROOF. (aa) Let v',w solve (5). The expreselon in brackets in

12.2 VariatIonal Formulation 281

J(v, w) — w) = a(v — v, — v)

_v)+b(w*,v* —v) —fi(v —v)J

vanishes because of (5b). Since a(v' — v,v — v) >0 for all v v E V, thesecond inequality in (7) follows. One also proves the converse as for Theorem6.5.12: If J(v,w) is minimal for v = then (5b) holds.(ab) If v is a solution of (Sc), then

J(v',w) — J(v', to) = 21b(w* — to,?) — f2(W' — to))

vanishes for all to, which proves the first part of (7) in the stronger formJ(v,w) = J(v,w). If, however, v is not a solution of (Sc), there exists aWE W such that 6 := — J(v,w) 0. Since J(v',w') — J(v',ii,) =—5 for tis := 2w — to, the first inequality cannot be valid. For the conversedefine = to. The first part of (7) implies

0 � J(ve,w) _J(v*,w±) = f2(W)J

for both signs, and so b(w,v') = 12(w). Since to W is arbitrary, one obtains(Sc).

(ba) We set j(w) := mInVEv J(v,w). According to Theorem 6.5.12, j(w) =to), where E V is the solution of (5b). If and are the solutions

for to and to', it follows that

—v,,x) = F(s) := b(w—w',x) for all x V.

Since IIFIlv' — w'Ijy and IIvw — S C'IIFIIv' one obtains

IIvw — vwsflv 5 — w'IIv for all w,w" E W. (12.2.9a)

(bb) By using the definition of in (5b) we can write:

J(v*,w*)_J(vv,w)__a(v*,v*)+2b(w*,v*)_211(v)_212(w*)

— + 2b(w, — — 2f2(w)J

= — v*) — b(w, — v') — v'))+ — .—?) + 2jb(w — to,?) — f2(w — to)]

_w,v*)_12(w* w)j.

(12.2.9b)

(be) Let v', wa be a solution of (5a-c). Because of (Sc) the expression inbrackets in (9b) vanishes and we have

J(v*,w*) = + — —?) � = j(w).

(5b) gives = and so

J(v,w) =j(w') = maxj(w); (12.2.9c)wEW

282 12 Stokss Equations

i.e., (8) holds.(bd) Now let v', w be a solution of (8). If in (9b) one sets w = w', =oneobtainsfromJ(v',w)=j(w')thatv =vw..Henoea(vw_v*,vw_vs)=

— — depends quadratically on — WIIV (ef. (9a)). Thevariation over w := w' — Ày (A IR, y W arbitrary) givee

0 = — = 21b(y,v) — 1(y)],

and so (Sc) is proved. (Sb) has already been established with v = v,L,..

12.2.3 Existence and Uniqueness of the Solution of a SaddlepointProblem

To make the saddlepoint problem (5a-c) somewhat more transparent we in-troduce the operators associated with the bilinear forms:

A L(V,V') with a(v,x) = for v,x V, (12.2.IOa)

BEL(W,V'), B*EL(V,Wl)

with b(w,x) = = (w,Bz)wxw', (12.2.lOb)

C E L(X, X') with c(u, z) = (Cu, for u, z X. (12.2.1(k)

Thus problem (6a,b) now has the form Cu = while (5a-c) can be writtenas

Ày + Bw = fi (12.2.lla)

B4v = 12. (12.2.llb)

If one assumes the existence of A1 L(V', V)1 one can solve (ha) for v:

v = A'(f1 — Bw) (12.2.12a)

and substitute in (lib):

= BA'f1 — 12. (12.2.12b)

The invortibility of A is in no way necessary for the solvability of the saddle-point problem. However, it does simplify the analysis, and does hold true inthe case of the Stokes problem.

Remark 12.2.5. (a) Under the assumptions

E L(V',V), E L(W',W) (12.2.13)

the saddlepoint problem (5a—c) jresp. Equations (1 la,b)] are uniquely solvable.(b) A necessary condition for the existence of (BA'B)' is

B E V') is injective. (12.2.14)


PROOF. (a) Under the assumption of (BA'B)' E L(W',W), (12b) isuniquely solvable for w, and then (12a) yields v. (b) The mjectivity of BA 1Bimplies (14). IAttention. In general, B: W —+ V' is not bijective so that the representationof(BA'B)' as is not possible. The example of the 3x 3 matrix

= A=

and B = shows that a system

of the form (1 la,b) can be solvable even with a singular matrix A. Thereforethe assumption A—1 E L(V', V) is not A closer look reveals thesubspace

V0 := ker B = {v V: B'v = O} = {v V:b(y,v) = 0 for all yE W} C V,(12.2.15)

which, as we noted before, in general is not trivial. The kernel of a continuousmapping is closed so that V, according to Lemma 6.1.17, can be representedas the sum of orthogonal spaces:

V = V0 V1 with := (V0)1. (12.2.16a)

ExercIse 12.2.6. Let (16a) hold. Show that (a) V' can be represented as

where (12.2.16b)

:= {t/ V':v'(v) = 0 for all V E Vj.},

(12.2.16c):= {v' E V':v'(v) = 0 for all v E Vo}.

AsanormonV01andVf oneusesll.IIv'.(b)The Riesz isomorphismJv:V—' V' maps V0onto V0' and V1 ontoVf

(d) The following holds:

= E E Vf (12.2.16d)

The decompositions (16a,b) of V and V' define a block decomposition ofthe operator A:

Aoj1I A A I)L.tilo It11J

A00 L(Vo,V01), A01 L(Vj,Vo'), A10 L(Vo,VI), A11 L(V1,Vfl.

Here, for example, A00 is defined as follows:

forvoEVo

The corresponding decomposition of B into (Ba, B1) is written as(0, B), since = 0 to the definition of V0. Conversely, we haverange (B) C so that B = System (lla,b) thus becomes


Aoovo + A01v1 = ho (12.2.17a)

A10v0 + Ajjvj + Bw = (12.2.1Th)B*VJ. = /2 (12.2.17c)

where v = +vJ.1 E V0, vj E Vj, = ho + fu, hoE E

Theorem 12.2.7. Let (4a—c) hold. Let V0 be defined by (15). A necessaryand sufficient condition for the unique solvability of the saddkpoint problem(5a-c) for all Ii V' is the aristence of the inverses

E E L(Vf,W). (12.2.18)

PROOF. (a) (17a—c) represents a staggered system of equations. (18) impliesB*_l = (B—') L(W',V1) so that one can solve (17c) for vj = B''f2.v0 E V0 is obtained from (17a): vo = — Ao.jvj). Finally, w resultsfrom (17b).

(b) In order to show that (18) is neceesary, we take ho V01 arbitrary, /1.1. = 0and 12 = 0. By hypothesis here we have a solution (vo, vl, w) K Vo x V1 x W.

= 0 implies Vj K Vo, so that = 0 because V0 n V1 = {0}. ThusAoovo = ho has a unique solution vo E Vo for each 110 V01. Since A00: Vo

is bijective and bounded, Theorem 6.1.13 shows that A&J' L(VO', Vo). Ifone takes hi E arbitrary and ito = 0, 12 = 0, one infers vj = 0 and

= 0, so that Bw = Iii has a unique solution w K W. As we did for A00,one also infers that B1 E L(Vf, W).

The formulation of conditions (18) in terms of the bilinear forms resultsin the conditions:

inf{sup{Ia(vo,xo)I:zo E Vo, IIxoIIv = 1}:vo Vo,flvoIIv = 1) ? a >0,(12.2.19a)

sup{Ia(xo,vo)I:xo K Vo, IIxoIIv = 1} >0 for all 0 vO E V0, (12.2.lOb)

V,HxIIv = 1}:w€ W,flwIIw 1}�fl>0. (12.2.19c)

ExercIse 12.2.8. Show that (19a) [resp. (19c)J are equivalent to (19a') [reap.(19c')j:

sup{ia(vo,zo)I:xo K Vo,IIxoIIv = 1} � ajIvoIIv for all v0 V0,

(12.2. 19a')

sup{Ib(w,x)I:z E V, Dxliv 1} � I3Dwllw for all w€ W. (12.2.19c')

LemmR 12.2.9. Let (4a,b) hold. Let V0 be defined by (15). Then condi-tions (18) and (19a—c) ore equivalent. Here we have � 1/a,

1/13.


PROOF. Because of (16d) and = 0 for z E Vo one can write (19c) in

the form

inf{sup{Ib(w, x)j: x E V1. HxItv = l}:w W, llwIIw = 1} � > 0.(12.2. 19d)

For 0 x E V1 one has that x Vo, therefore according to (15):

E = 1) >0 for all 0 xE V1. (12.2.19e)

As in the proof of Lemma 6.5.3, we obtain the equivalence of (19a,b) with

E and of (19d,e) with B' e U

Corollary 12.2.10. (a) Condition (19b) becomes superfluous if a(.,.) issymmetric on Vo x Vo or tf Lemma 6.5.17 applies.(b) Each of the following conditions is sufficient for (19a,b) and hence aLsofor E L(V01, V0):

a(., .): Vo x Vo JR is V0-eliiptic:

a(vo,vo) � for all vO E Vo, (12.2.20a)

a(., .): V x V JR is V-elliptic. (12.2.20b)

PROOF. (a) As in Lemma 6.5.17. (b) (20b) implies (20a); (20a) yields(19a,b). U

ExercIse 12.2.11. Show that under the (4a—c), (18) is alsoequivalent to the existence of C1 L(X',X) (cf. (lOc)). Find a bound forflC'flx._x' in terms of flAoo'flv0*_v0', IIAUv'4_v', and

12.2.4 Solvability and Regularity of the Stokes Problem

Conditions (19a,b) (i.e., E L(V0", Vo)) are easy to 8atisfy for the Stokesproblem:

Lemma 12.2.12. Let.0 be bounded. Then the forms (2a,b) describing theStokes problem the conditions (4a,b) and (19a,b).

PROOF. (4a,b) is self-evident. According to Example 7.2.10, Vv) dxis (.0)-elliptic. From this follows the H,?j(Q)-ellipticity of Corollarylobproves(19a,b). U

It remains to prove condition (19c), which for the Stokes problem assumesthe form

sup{I f w(x)divu(x) dxl: n E luli = l} � I3lwk for all w E

(12.2.21)


or IIVWIIH-1(n) � for all w E (12.2.21')

Lemma 12.2.13. Sufficient and necessary for (21) is that for eachw E

there exists u E such that

w =divu, luli � fl'Iwk. (12.2.21")

PROOF. (a) For w E select u such that (21") holds and set iiThe left-hand side in (21) is � faw(x)divü(x)dx = �(b) If (21) holds one infers as in Section 12.2.3 the bijectivity of B: Vj —, Wwith � 1/ft. Therefore u B'1w satisfies condition (21").

Necäs [2j proves

Theorem 12.2.14. Condition (21) is satisfied If 11 E C°" is bounded. Underthis assumption, the Stokes problem

—4u + Vp = f, — divu = g in $1, u =0 on F (12.2.22)

has a unique solution (u,p) forf E H'(Q) andg E

luli + Info � I/I—i + Igk]. (12.2.23)

Remark 12.2.15. Under the conditions that n = 2 and that 1? E C2 is abounded domain1 the existence proof can be carried out as follows.

PROOF. We need to prove (21"). Fbr w solve = win .0, go =0on F. Theorem 9.1.16 shows that go E Since Vgo H1(f1) and n(x)C2(F) it follows that g := ôu/8n H'/2(f') (cf. Theorem 6.2.40a). From(3.4.2) one infers that = fawdx = 0 since w Integration ofg over F yields C with OG/Ot = g, where is the tangentialderivative. There exists a function %b E with = C and = 0 onF and � CIGI3,2 � � � C"Iwk. We set

Clearly, ul,u2 E H'(.0). Let the nonnal direction at x F be n(x) =(nI(x),n2(x))T. The tangent direction is thus t(x) (n9(x), _ni(x))T. Foru = (u1,u2)T one obtains

(u, n) = —cP5fli = = —g-i-ÔC/8t =0,

(u,t) = —goxn2 — + g011n1 — = —&p/8t — = 0,

since go = 0 on F implies Ogo/àt = 0. (u, a) = (u, t) = 0 yields u = 0 on 1'such that u = (ui, u2) E is proved. One verifies that

12.2 VariationaL Formulation 287

divlh = + = — - + = W

with IUIi � k012 + .The above proof uses the H2-regularity of the Poisson problem and re-

quires corresponding assumptions on (1. Theorem 14 also assumes Q C°'1Since the Poisson equation = I is solvable for any domwn .0 which iscontained in the disk KR(0), or at least in a strip {x E < R}, re-sulting in the inequality CRlf one might conjecture that a similarresult holds for the Stokes problem.

Counterexample 12.2.16. For E (0, 1) let = {(x,y): —1 <z < 1,0 < y < + (1 — €)lxI} (ci. Figure 1). All are located in therectangle (—1, 1) x (0, 1). there exists no 8> 0 such that (21")holds for all 0, w e

1

FIgure 12.2.1. JZ

PROOF. We select w E such that w(s, y) 1 for x > 0, w(x, y) = —1

for x � 0. Let the inequality (21") hold for with > 0. Let uwith � be chosen according to Lemma 13. We continue ti by u = 0onto Ut . For the restriction on z = 0 we have according to Theorem 6.2.28

)IILZ(R) � � CJtzj)1 � �Let = 1 for 0 < y < and x(v) 0 otherwise. Since ui(0,y) =u1(O, and Ixto = we have

= � lui(O,.)IoIxIo �

Let .f?(1 = {(x,y) E > 0} {(x,y):x = 0,0< y < =Because w =1 in and because divtz w we have

-1 0

288 12 Equations

1/2 � Iw(x)12 dx= J vi div u dx

= J dlv u dx= j (u, n) dE'

= _juidf= _Jui(O,v)dv.

The last two inequalities result in 1/2 � from which we infer that>0 cannot be independent of e.

ExercIse 12.2.17. Construct a domain 1? located on the strip JR. x (0, 1)in which the Stokes equations are not solvable. Hint: Join the domains(v€N) from Figure 1.

As for the case of scalar differential equations one obtains stronger regu-larity of the Stokes solution if one more than I E H-'(Q).

Theorem 12.2.18. Let £2 be bounded and sufficiently smooth. Let u andp be the (weak) solution of the Stokes problem (22) with I E Hk(fl), 9 EHk+l(a) fl for k E t4 U {O}. Then we have u E Hk÷2(.Q) flp E fl and there exists a C depending only on £2 such that

+ LPIk+1 � Ia + IgIa+i]. (12.2.23b)

PROOF. Cf. El, Chap.llI, §5] U

In analogy to Theorem 9.1.22 it is sufficient to the convexity of£2 in order to obtain u E H2(a) and p H'(s?) from f 112(fl).

Theorem 12.2.19. (Cf. Kellogg-Oaborn [1]) Let £2 C JR2 be bounded andconvex. If 1€ L2(S?), then the Stokes equation (1.2a,b) has a unique solution

H2(a) n p E H'(a) fl which satisfy the estimate

1u12 + � CIfIo. (12.2.23c)

For the more general problem (22) with g 0 in a convex polygonal domainthe solution satisfies

1u12 + lpli � If + (12.2.23d)

1ff L2(a) and g E IIJ(a). Here, HJ(fl) is the sub8pace of H'(a)with the following (stronger) norm:

:= + withIaI=1

6(x):=min{Ix—eI:e€F corners of thepolygona}.

12.2 Varia*onal Formulation 289

12.2.5 A V0-elliptic Variational Formulation of the Stokes Problem

V0 c has been defined in (15) by V0 {u 0).As kernel for the mapping B = —div Vo is a closedsubepace of i.e., again a Hubert space for the same norm . In thefollowing we investigate the problem

Find u V0 with a(tz,v) = f(v) for all v E V0, (12.2.24)

where a(u, v) f0(Vu(x), Vv(x)) dx. Problem (24) has the same form asthe weak formulation of the equations = (i = . , n), onlyhere has been replaced by Vo.

Lemma 12.2.20. Let be a bounded domain (or bounded in one direction;cf Exercise 6.2.12b). The form a(.,.) is Vo-elliptic. The constant C2 > 0 ina(u, u) � depends only on the diameter of 4'?. In particular, problem(24) has a unique solution u E V0 with for 1€ V01.

PROOF. The of a(.,.) (ci. Lemma 12) carries over to Vo C(cf. Exercise 6.5.6a). This implies the other statements (cf. Theorem

6.5.9) ITheorem 12.2.21. LetS? E C°" be a bounded domain. Assume f E H'(Q).Then the solution u Vo of problem (24) coincides with the solution compo-nent u of the mixed formulation (3a--c).

PROOF. According to Exercise 6 one can split / E in such a way that

f—fo+fi, fo€V0', LEV', fo(v)=Oforv€Vj, fj(v)=Ofor vEV0.

In (24) one can replace f(v) by fo(v). u V0 c results in E

H1(.O). The part of that belongs to Vu is E VI with

gj(vj) a(u,vj) for all vj. E V1 = (V0)1.

Theorem 14 proves the condition (2!'), which results in the bijectivity ofB V (cf. Lemma 13).p B'(fj. —gj), by definition,

satisfies

b(p,vj) = = f1(v1) = fj(vi) —a(u,v1)

for all v1 E V1. Furthermore, since div Vi,) = 0 for v0 E V0, it follows that

b(p, vij) =0 for all V0 E V0.

For arbitrary v E to be split into v = + V1 with E V0 andV1 E V1, one obtains

b(p, v) = b(p, vo) + b(p, vI) = fj(vi) — a(u,vj.) f(vj.) — a(u,v) + a(u, v0)


= f(vj)—a(u, v)+f(vo) .= f(v)—a(u,v),

by (24). Since also b(w,u) = 0 for all to E because u E V0, u and psatisfy the variational formulation (3a—c) of the Stokes problem.

Note that Problem (24) is solvable for all bounded domains althoughProblem (3a—c) depends more sensitively on Al (cf. Counterexample 16). The-orem 21 shows that only the component p has a domain-dependent bound

Ii'io <C01f i—i while Hi ClfoH � Gill_i holds for all .0 C KR(0).Strictly speaking, the variational problem (24) is not equivalent to the

Stokes problem since, for example, for A? from Exercise 17 the Stokes equationsare not solvable, whereas Problem (24) definitely has a solution.

The original formulation (3a—c) may be interpreted as Equation (24), intowhich one has introduced the "side condition" div u = 0 using the Lagrangefunction p (cf. Section 8.3.6).

12.3 Mixed Finite-Element Method for the StokesProblem

12.3.1 FInite-Element Discretisation of a Saddlepoint Problem

One would have an ordinary Ritz-Oalerkin discretisation if in the variationalformulation (2.24) one were to replace the space V0 by a finite-dimensionalsubapace Vp, c V0. But this is not as easy as it sounds.

ExercIse 12.3.1. Let the square (1 = (0,1) x (0,1) be triangulated regularlyas in Figure 8.3.2, 5a or decomposed into grid squares. Let c Hd(i7)be the space of the finite triangular elements (cf. (8.3.8)) [resp. of bilinearelements (ef. (8.3.12b))J. Define the corresponding subepace for the Stokesproblem as Vp, := {u = (ul,u2):uj,u2 E and divu =0} C Vo. Show thatVp, contains only the null function.

The remaining procedure is thus oriented toward the weak formulation(2.3a—c). The space X = V x W is replaced with X,, = Vh x Wh. The discreteproblem

Find ii' V,, and E Wp, with (12.3.la)

+ = fj(z) for all xE Vp,, (12.3.lb)

=f2(y) forallyEWp,, (12.3.lc)

is called a mixed Ritz-Galerkin problem [reap. a mixed finite-element problemif Vp, and Wp, are formed from finite elementsj. To the formulation (2.6a,b)corresponds the equivalent way of writing (la—c)

Find xh Xh with c(x",z)= f(z) for all z E Xp,. (12.3.1')

12.3 Mixed Finite-Element Method for the Stokes Problem 291

In the case of the Stokes equations, (2.3a—c) provides the desired solutionwhich satisfies the side condition div u = 0. However, the finite-element solu-tion of (la-c) generally does not satisfy the condition dlv =0. One can view

from (la—c) as the solution of a nonconformal finite-element discretisationof (2.24), as is shown in the following exercise.

Exercise 12.3.2. Let /2 0 in (ic); set := {z E Vh: b(y,z) = 0for all y E Show that each solution v1' in (la-c) is also a solution of

Find with = fi(x) for all z (12.3.2)

Since in general VO,h V0 (cf. (2.15)), (2) is a nonoonformal discretisation of(2.24).

Let and OfVh and Wh,where

Nv,h dim := dim Wh. (12.3k)Let the coefficients of v E Vh and w E W,, be v and w:

NV.h NW.h

v = Pvv := w = Pww ;= (12.3.3b)

As in Section 8.1, we prove

Theorem 12.3.3. The varw.tionoj problem (la—c) is equivalent to the systemof equations

fAa BhJ IV] In]0 j {wJ =

(12.3.4a)

where the matrices and vectors are given by

(1 � i,j � 1 � k � Nw,h)(12.3.4b)

:= f1(bfl, f2,h := (1 � I � Nv,h; 1 � k � Nw,h). (12.3.4c)

The connection between (la—c) and (4a-c) is given by v" = Pvv, wh Pww.For u (i), f := one obtains the system of equations

a-' •. - a a'-)hUL, I' (123.4a)La J

which corresponds to the fonnutation (1').

12.3.2 Stability Conditions

When selecting the subepaces and one has to be careful, for evenseemingly reasonable spaces lead to singular or unstable systems of equations(la-c). The former occurs in the following example.


Example 12.3.4. Let the Stokes problem be posed for the L—shaped domainin Figure 8.3.2. For all three components uI,u2, and p we use piecewise lineartriangular elements over the triangulation r which is given by the second orthird picture of Figure 8.3.2. Here, let uj = u2 =0 on F and let p Wh satisfythe side condition p i.e., f0pdx = 0. Then the discrete variationalproblem (la—c) is not solvable.

PROOF. The second ithird) triangulation in Figure 8.3.2 has 5 jlO] innernodes x', with each of them carrying values ui(x') and Thus dim Vp, =2 - 5 = 10 frlim Vp, 20]. Because of the side condition p dic — 0 thedimension of Wp, is smaller by one than the number of inner and boundarynodes: dim Wp, = 21 -1 = 20 fdimWh = 21]. In both cases the statementresults from

Lemma 12.3.5. It is necessorij for the solvability of (la—c) that NV,,, � NW,,,,i.e., dixnVh �dlmWh.

PROOF. According to Theorem 3 the solvability of (la-c) is equivalent tothe nonsingularity of the matrix C,, in (4a'). Elementary considerations showthat rank (C,,) � Nv,,, + rank (B,,). Since B,, is an Nv,,, x Nw,,,-matrix itfollows from Nv,,, <Nw,,, that rank (C,,) <Nv,,, + Ny,ç,, and hence C,, issingular. I

Thus an increase in the dimension of W,, does not always lead to anbetter approximation of w W. The choices of Vp, and W,, must be mutuallyadjusted. The inequality dim Vp, � dimW,, corresponds to the requirementthat B in (2.lOb) needs to be injective, but not necessarily to be surjective.

In order to formulate the necessary stability conditions, we define

V0,,, := {v Vp,: b(y, v) = 0 for all y E W,,} (12.3.5)

(cf. Exercise 2). VOJ, is the discrete analogue of the space Vo in (2.15). Theconditions, which can be traced back to Brezzi read:

inf{sup{ja(vo, xo)I: x0 E llxollv 1}: vo E Vo,h, Iivellv = 1} � >0,(12.3.6a)

inf{sup{Ib(w,x)i:x Vp,, = 1}:w W,,, = 1} � >0.(12.3.6b)

Theorem 12.3.6. Let (2.4a-'c) hold and dim Vh <oo. The Brezzi conditions(6a—b) are necessanj and sufficient for the solvability of the discrete problem(la—c). The solution uh = (vhwh) E V,, x Wh satisfies

lIu"llx := + ]112 Chill lix' := ChI IjfilIv' + 1112112W,

(12.3.7)where Ch depends on ap,, Ph and on the bounds Ca and Gb in la(v, x)l �CaiIvtivilxllv and Ib(w,x)I Cbi(wi(w1ixiiv. If


a >01 øh � f3> 0 (12.3.6e)

holds for all parameters h of a sequence of discretisations, then the di8creti-

sation is said to be stable and C,1 remains botrnded: C,1 � C for all h.

PROOF. Theorem 12.2.7 and Lemma 12.2.9 are applicable with V,1, Wh

instead of V0, V, W. (2.19a,c) then are the same as (6a,b), while (2.19b) follows

from (2.19a) because dim <oo (cf. Exercise 6.5.4). ICondition (6a) is trivial for the Stokes problem:

Exercise 12.3.7. Show that Condition (6a) is always satisfied with a constantczh independent of h, if a(., .): V x V JR is V-elliptic.

12.3.3 Stable Finite-Element Spaces for the Stokes Problem

12.3.3.1 Stability Criterion

For the Stokes problem C C must hold. In abounded domain a(•,•) is so that (6a) is satisfied withCE > 0 . It is somewhat more difficult to prove the conditions (6b,c), whichfor the Stokes problem the form

sup{Jb(p,u)I:uE = 1}�J3IpIo forailpE W,1 (12.3.6d)

wherein $> 0 must be independent of p and h. it is simpler to prove themodified condition

sup{Ib(p, u E V,1, Iuk = 1} � /3IpIi for slip E W,1 (12.3.8)

> 0 independent oip and h). Since in (8) the norm occurs, the latterrequires Wh C H1(i7). This excludes, for example, piecewise constant finiteelements.

Together with the usual Condition (8) is sufficient for sta-bility.

Theorem 12.3.8. Let 11 E C°'1 be Let the Poisson problem beregular (it is sufficient that 1? be convex; ef. Theorem 9.1.22). Let V,1 satisfythe approxsmation property

inf{ju — u E 112(11) fl (12.3.9a)

and the inverse estimate

for all E V,1. (12.3.9b)

Then Condition (8) is sufficient for the Brezzi condition (6b,c).


PROOF. (a) For given p there exists u with = I andb(p,u) � I31PIO (cf. Theorem 12.2.14 and (2.21)). According to Exercise 9.1.24,the orthogonal on Vh satisfies the conditions

�IuIi=l,From b(p,uh) b(p,u) — b(p,e) � 13(plo — b(p,e) � — Iplilelo � $Ii'lo —

Chtpli and � 1 one infers

E Vh, Iv"li = 1} � f3lpJo — ChlpIi. (12.3.10)

(b) Because of (8) and (9b) there exists E V,, with 1 and

Ib(p,ti')I � = �From this follows

Vh, = 1} � with := (12.3.11)

If one multiplies (10) with and (11) with the sum reads:

E = 1) � := //(C + (12.3.12)

Since 13 is independent of p and h, (6d) (6b,c)j has been proved.

12.3.3.2 Finite-Element Diacretisatione with the Bubble Function

In the following let fl be a polygonal domain and an admissible trian-gulation. Example 4 shows that linear triangular elements make no sense forand p. We increase the dimension of Vh by adding to it the so-called "bubblefunctions" or "bulb functions".

On the reference triangle T = > < 1} the bubblefunction is defined by

:= — — i) in T, u 0 otherwise. (12.3.13a)

The name derives from the fact that u is positive only in T and vanishes onOT and outside. The map -. T to a general T (cf. Exercise 8.3.14)results in the expression

ü.1(x,y) := (12.3.13b)

for the bubble function on

Exercise 12.3.10. Let be a quasiuniform triangulation. Show that thereexists a C >0 independent of h such that dx dy � dzdy.


We set

linear combinations of the linear elements E H0t (fl)

and the bubble functions for f' E (12.3.14)

Vh := Wa: linear elements E

For the side condition C see Section 8.3.6. Since ü..p e H0t(fl), wehave C

Theorem 12.3.11. Let rh be the quasiunifonn tnangulatson on a boundedpolygonal domain £7. Let and Wh be given by (14). Then the stability con-dition (8) is satisfieaL Under the further conditions that Si E and that thePois8on problem be H2(Si)-regt4ar the Brezzi condition (6b,c) also holds.

PROOF. (a) an arbitrary p On every T E Ta, Vp is constant:Vp = We set

v := > E ii := v/lyle (ü.jr. bubble function (13b) on

so that = 1. Exercise 10 yields

b(p, v)= J (Vp, v) dx = + f0 TEm

T

Since lVplo and lpI' are equivalent norris on the subapace H'(Q) flit follows that lb(p,v)l � and � C'lplilVpk/jvk. In asimilar way one shows that lvlo � C"(VpIo and obtains � with43 := C'/C" independent of h. The left-hand side in (8) is � so that(8) follows.(b) (9a) is satisfied for Vh (cf. Theorem 8.4.5). The same holds for the inverseestimate (9b) (cf. Theorem 8.8.5). Therefore (6b,c) follows from Theorem 8.

U

A general result on the stabilisation by bubble functions can be found inBrezzi-Pitkäranta [11.


12.3.3.3 Stable Discretisations with Linear Elements in VhIf one wants to avoid bubble functions, one must increase the dimension

of V,, in some other way. In this section we shall consider for Vh and Wh twodifferent triangulations and Th. By decomposing each T rh as in Figure1, through halving the sides, into four similar triangles, one obtains Th12. Wedefine:

V,, c : linear elements for triangulation Th/2,(12 3 15)

Wh c H'(fl) fl : linear elements for triangulation

or

V,, c : quadratic elements for triangulation Th,'12 16

W,, C H'(Q) fl linear elements for triangulation r11.

FIgure 12.3.1. Th and Th/2

Theorem 12.3.12. Theorem 11 holds analogowsig for V,, x Wh in (15) or(16).

PROOF. (a) Let Vh x Wh be given by (15). Fbr each inner triangular side ofthe triangulation m there exist two triangles T17,T21, E with =(cf. Figure 2). Let 8/Ot be the derivative in the direction of 'y, let 8/On bethe directional derivative perpendicular to it. There exists and with

+ = 1, 8/Ox = a78/On + b78/Ot, O/Oy = b.1O/On — a70/Ot.

In contrast to Op/On, Op/Otis constant on T17 UT27 U We denote itsvalue by pu,. The mid-pomt of is a node of rh/2. We define the piecewiselinear function over by its values at the nodes

u.1(x7) = w,(x') = 0 at the remaining nodes (12.3.17a)

and set

v := E ( E Vh, v/lv$o. (12.3 ITh)

The sum E7 extends over all interior sides of Tj,. In U T2-, we have

so that

12.3 Mixed Finite—Element Method for the Stokes Problem 297

(Vp,

� Ch2>1pt1.y12.

are the sides of T Th (r5 is quasiunifonn!), then dx �From this one infers that b(p, v) � i IVpIo, as in the proof of Theorem

11, and finishes the proof analogously.(b) In the case of quadratic elements given by (16) one has the same nodesas in (a) (cf. Figure 8.3.8a and Figure 1). Use (17a,b) to define the quadraticfunction V E Vh and carry out the proof as in (a). U

12.3.3.4 Error Estimates

FIgure 12.3.2. Ti.,, Ta.-,

In the following, the condition (8.1.12) should be replaced by the stabilitycondition (6a.-c). In the place of the approximation property (8.4.6') we nowhave the inequalities

inf{Iv — Vh} � Chfi42 for all V E V =

inf{Ip — E W,j � for all p W =

(12.3.18a)

(12.3.18b)

Condition (18a) is equivalent to condition (9a) in Theorem 8.The following theorem applies to general saddlepoint problems, and can

be reduced to Theorem 8.2.1.

Theorem 12.3.13. Let u = X = V x W be the solution of (2.5a—c)[resp. (6a)1. Let the discrete problem (la-c) with Xh = x C X satiify

J (Vp, ( dx = IT17UT2.1


the Brezzi condition (6a.-c) and have the solution = (v's, w"). Then thereexssts a varsable C independent of h such that

— uhjlx � Cinf{IIu — Xh}. (12.3.19)

PROOF. The Brezzi condition (cf. Theorem 6) yields S lix' forall right1-hand sides f X' in (2.5a--c), in particular for all f E Xh =The above inequality means � t' for the operator Lh:Xh —,

which belongs to c(., .): Xh x Xh -. IR. According to Exercise 8.1.16is equivalent to Condition (8.1.12) for c(., .): X x X IR

[instead of a(., .): V x V IRj with 1/n. Theorem 8.2.1 yields the state-ment (19).

For the Stokes problem with (), instead of u = (), =inequality (19) is now rewritten as follows:

— uhi2 + [p C2inf{lu — + _phlo2:uh Vh,p" E Wh}or

lu — + — uhfi + [p Wh}.(12.3.20)

Theorem 12.3.14. Let the Stokes equation (1.2a,b) have a solution u Ep (ci. Theorem 12.2.19). For the suhspaces

C and Wh C l€t the Brezzi condition (6a—c) and the ap-proximation conditions (18a,b) be Then the discrete solutionsatisfies the estimate

lu — + — P"lo � C'hflul9 + Ipli]. (12.3.21)

PROOF. Combine inequalities (20) and (18a,b). N

Using the same reasoning as in the second proof for Theorem 8.4.11 withc(..) instead of a(.,.) one proves

Theorem 12.3.15. For each f L2(a), g E fl H'(fl) let the Stokesproblem ÷ Vp f, —divu = g have a solution u Ep H1(Q) fl with 1u12 + � C[1f10 + 1gb]. Under the asstImptlon8of Theorem 14 we then have the estimates

— + [p — p'i—i C'h[Ju(i + Iplol, (12.3.22a)— ÷ [p — p41.i � C"h2[Iu12 + [p1'] (12.3.22b)

for the finite element solutions. Here lpf—t is the dual norm

Corollary 12.3.16. Combining (22b) and (21) one obtains

12.3 MIxed Finite-Element Method for the Stokes Problem 299

— + —p"Io � + (12.3.22c)

A new exposition of the mixed finite-element discretisatiori can be foundin Brezzi-Fbrtin [1).

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Index

Note: The page numbers given in bold-face refer to occurrences of the termprinted emphatically on the page indicated.

Adjoint problem, 141, 190, 196Adler problem, 159, 183Agmon, condition of, 158AnalytIc, 20

condItion, 139BabuMca-Brezzi condition, 284Banach space, 110, 111, 112, 113,

131, 135, 137, 161Basis, 161, 165, 166, 171, 182, 183,

204, 291hIerarchical, 202

Basis condition, 171, 172, 173, 175,176, 177, 180, 181, 183, 200

Beltrami operator, 14Biharmonic equation. Ses Differen-

tial equationBilinear ftrin, 137, 154, 155, 227,

279adjoint, 138, 190, 191continuous or bounded, 138, 146,

157, 161, 185, 279, 280corresponding to a pde of order

2m, 146, 210corresponding to the Helmholtz

equation, 147, 154, 208corresponding to the Poisson

equation, 149, 154, 165, 176,180, 183

extension of a, 138operator associated to a. See Op-

erator, 138, 140, 142, 163, 209

symmetrIc, 138, 141, 149, 152,164, 170, 200, 204, 208, 280,285

V-elliptic (H"-elliptic, etc.), 140,141, 147, 148, 149, 152, 153,154, 159, 164, 170, 183, 200,204, 212, 228, 280

V-coercive etc.),141, 142, 143, 149, 150, 153,154, 157, 158, 166, 171, 185,210, 211, 212, 213, 218, 219,222, 223, 226, 227, 228, 229,256, 267

Block, block matrix, block partition-log, 283

Boundary condition, 158, 226, 275Dirichlet, 27, 35, 38, 65, 85, 86,

96, 145, 150, 159, 164, 174,178, 215, 222, 229

discretisation of the, 100first, 35,96mixed, 96natural, 152, 154, 155, 159, 168,

178, 218, 221Neumann, 36, 65, 96, 154, 174,

183periodic, 98second, 96third, 96

Boundary element method (BEM),37

Boundary layer, 249

308 Index

Boundary value, 7, 9 10, 12, 26Boundary value problem, 13, 19, 27,

28,85

Brezzi condition, 292, 293, 295, 298Bubble function, 294, 295, 296Cauchy-Riemann equations, 4, 7Chequer-board. See OrderingCoefficient vector, 161, 255Compact (cf. embedding, nine-point

formula, operator), 135, 201,257, 269

Complementary problem, 264

Complete, 24, 112Completion, 112, 117, 122, 125CondItion, 108, 204Cone condition, 136Conservation form, 244Convex. See DomainConsistent, consistency, 59, 64, 69,

93, 94, 200, 228, 232, 233, 235,239, 242, 267

Convection-diffusion equation, 247Convergence, 168

of the difference method, 59, 60,61, 69, 83, 84, 95, 102, 108,

239

of the eigenfunction, 255, 259,260, 271, 272

of the 255, 258, 260super, 202, 203

Coordinate transformation. See Do-main

Corner, re-entrant (cf.domaIn), 34, 223, 225

Covering, 127Dense, 10, 112, 113, 114, 115, 117,

119, 123, 141, 169, 170, 224

Derivative, 104conormal, 96, 201normal, 15, 65, 95, 96, 145

tangential, 96, 99, 103, 145, 224,246, 278

weak, 115, 116, 172

Diagonalisable, real, 6, 7Diagonally dominant, 47, 48, 49, 52

Irreducibly, 47, 48, 49, 52, 53, 92,93, 250

Difference (divided), 38, 49, 62, 79,240

backward or left-sided, 38, 56, 65,72, 91, 250

forward or right-sIded, 38, 91,227, 250

second, 38, 80, 238, 240

symmetric, 38, 70, 71, 102, 233,249, 251, 252

Difference method, 38, 59, 65, 90,

91, 93, 94, 102, 108, 173, 177,178, 202, 204, 247, 249, 267,269

13-point, 108, 109of higher order, 62

Difference operator, 10, 44, 121, 210,213, 228, 233

Difference star (stencil), 44, 62, 91Differential equation(s)

biharmonic, 74, 95, 103, 104,

105, 106, 108, 158, 194, 218,223, 278

of first order, 1, 3, 4, 6, 277

of order 2m, 104, 108, 145, 158of second order, 2, 5, 8, 12, 85,

90,150

ordinary, 1, 2, 3, 38, 248, 254partial, 1, 2, 7system of, 3, 4, 6, 7, 105, 275,

276Differential operator, 5, 59, 60, 75,

85, 89, 104, 253, 276formally ad.joint, 92, 95boundary, 95, 154, 155, 156

Dirichiet integral, 144Dirichiet principle, 144Divergence (operator), 275Domain, 12

convex, 79, 197, 223, 224, 225,231, 235, 236, 242, 288, 293

exterior, 36general, 78, 100, 109, 196, 234,

241, 243L-shaped, 13, 34, 127, 201, 243normal, 15, 16, 19, 28, 33transformation of the, 6, 89unbounded, 23, 146, 147

Double-layer potential. See PotentialDual form, 131, 134

Dual space, 130Eigenfunctlon, 222, 253, 254, 259,

263, 271Elgenvalue, 5, 6, 7, 10, 47, 51, 52,

53, 137, 142, 150, 258, 254,255, 256, 258, 259, 260, 261,262, 264, 265, 271, 274

Eigenvalue problem, 7, 11, 137, 150,253

adjoint, 253, 256elliptic, 253, 254, 255, 274

Ethptic, 4, 5, 6, 7, 9, 10, 11, 12, 85,

104, 276

Index 309

uniformly, 86, 87, 88, 89, 104,145, 147, 148, 149, 150, 155,222, 229, 276, 277

V-. See Bilinear form, 248, 256,280, 285, 289, 293

Embeddingcompact, 185, 136, 142, 143, 150,

153, 171, 186, 253, 256continuous, 118, 131, 133, 134,

136, 137, 141, 143, 171, 185,253, 256

dense, 131, 133, 134, 137, 141,171, 253, 256

Error estimatesfor difference methods, 190, 238,

239for eigenvalue problems, 260, 263,

264for finite element methods, 185,

190, 193, 196, 246, 266, 297,298

for Ritz-Galerkln methods, 167Estimate. See Error estimate, in-

verse estimateExistence. See SolutionExtension. See Bilinear form, opera-

torof a function, 123, 129, 215, 237

Extrapolation (method), 61, 84Finite elements, 171, 172, 174, 251,

254constant(bi)cubic, 182, 194, 195, 196, 296(bi)Ilnear, 172, 174, 175, 178, 179,

182, 183, 185, 190, 194, 199,204, 246, 247, 252, 295, 296

isoparainetrlc, 198,247mixed, 290, 299nonconforjnal, 199, 200, 291of the serendipity class, 181, 182,

193quadratIc, 181, 182, 193

Finite element method, 24Five-point formula, 40, 41, 53, 60,

62, 92, 176, 190, 231Form. See Bilinear form, dual form,

8esquilinear formFourier expansion, 9Fourier transform, 110, 119, 120,

124, 125, 126, 135, 213FunctIonal, 132, 140, 146, 152Fundamental solution, 16, 17, 28Galerkin method. See RJtz-Galerkln,

Petrov-GalerklnG6rding, Theorem of, 150

Gei'fand triple, 41, 133, 134, 136,141, 163, 207, 253

(Ierschgorin, Theorem of, 46, 48Gradient, 15Green's formula, 15, 29, 144, 153Green's function of the first kind,

28, 29, 30, 31, 33, 34, 95, 105,144

discrete, 55, 72, 95Green's function of the second kind,

35Grid, 39, 40, 226Grid function, 39, 44Grid points

close to the boundary, 63, 79, 83far from the boundary, 42,63, 79neighbouring, 41, 42, 80

Grid size (width, step sIze), 39Harmonic, 13, 15, 16, 19, 20, 22,

23, 24, 25, 144Harnack, Theorem of, 20Heat equation, 3, 4, 5, 7, 8, 10Hehnholtz equation, 147, 154, 208hubert space, 110, 114, 115, 117,

122, 132, 134, 279, 289Hökler continuous, 29, 30, 110, 123,

125Hyperbolic, 4, 5, 6, 7, 8, 9, 10, 11,

248Improperly posed. See Well-posedInclusion, 135, 136, 142Initial-boundary value problem, 8,

9, 207Initial value problem, 2, 3, 7, 8, 9,

38Instationaiy problem, 10, 11Integral equation method, 36, 37Interpolation, 84, 124

Hermite, 194, 196, 207spline, 194, 196, 207

Inverse estimate, 206, 221, 240, 293Irreducible, 45, 54Irreducibly diagonally dominant. See

Diagonally dominantLagranglan (6.ctor), 184, 290Lamé differential equations, 277Laplace equation (of potential equa-

tions), 2, 12Laplace operator, 12, 14, 131, 176Lexicographical ordering. See Order-

ingLipschitz continuity, 23, 30Map, mapping. See OperatorMass matrix, 205

310 Index

Maximum (-minimum) prmciple, 17,18, 19, 20, 27, 86, 103, 250,258

Mean-value property, 17, 18, 19, 20,27,54

Mehrstallen method, 64, 83M-matnx, 45, 49, 50, 51, 52, 53,

64, 68, 81, 83, 84, 91, 92, 93,100, 106, 249, 250, 251, 252

Multi-index, 30, 104Neighbour. See Grid pointsNeumann condition. See Boundary

conditionNine-point formula, 90, 180

compact, 63, 63Nodal points, 172, 175, 178, 181,

203Nodal values, 172, 195Norm, 29, 30, 50

dual, 131, 132, 134, 163, 194, 215,227, 298

Euclldean, 14, 51, 57, 110, 163,204, 226

equIvalent, 111, 117, 123, 204,295

matrix, 50assocIated, 50, 204, 227maximum, 46, 110operator, 111row sum, 50, 59, 170, 242Sobolev, 117, 122Sobolev-Slobodeckil, 122, 247spectral, 51, 59, 204supremum, 23, 110, 112vector, 50

Normal (dIrection), 15, 116, 145,224

Normal system, 105Normed space, 110Operator, ill

adjoint, 132, 163, 213, 216, 242associated. See Bilinear form, 209,

256, 282compact, 135, 136, 137difference. See Difference operatordifferential. See Differential opera-

tordual, 131selfadjoint, 132

Ordering (of the grid poInts), 42,43, 45

lexicographical, 42, 67, 70chequer-board, 43

Orthogonal (ef. projection), 114Orthogonal space, 114, 163, 283

Parabolic, 4,5, 5,7,8, 9, 10, 11Partition of unity, 128, 130, 150,

209, 219Petrov-Galerkin method, 252Plate equation (cf. blharmonic), 85,

103Poisson equation, 27, 34, 35, 38, 41,

57, 103, 149, 155, 177, 201,225, 231, 254, 276, 287, 289,293, 295

Poisson's integral formula, 19, 20,22

Polar coordinates, 13, 14, 15, 97,98, 144

Positive definite, 5, 52, 53, 93, 106,164, 204, 255

Positive semidefinite, 52, 87Potential, double-layer (dipole), 36

sIngle-layer, 36volume, 36

Potential equation (of. LaplaceequatIon), 2, 4, 5, 7, 9, 12, 13,14, 16, 28, 35

Precompact, 135Principal part, 6, 12, 95, 145, 147,

148, 150, 223, 228, 251, 276Projection, 132, 170

orthogonal, 132, 133, 163, 170,225, 294

Ritz, 170, 189, 191, 192, 193, 225Rayleigh-Ritz. See RitzReduced equation, 248, 250Reference triangle (reference ele-

ment), 10, 177, 179, 185, 198,204, 294

Regularity, 190, 192, 196, 208, 209,215, 219, 222, 223, 225, 232,246, 254, 264, 267, 285, 287,288, 293, 295

discrete, 227, 228, 229, 230, 231,239, 240, 241, 242, 243, 271

Riesz reprssent.atlon theorem, 132Riesz isomorphism, 132, 283Rlesz-Schauder theory, 1ST, 142Ritz-Galerkln method (cf. solution),

161, 171, 290Robin problem, 159Saddle-point (problem), 279, 280,

282, 284, 290, 297Scaler product, 52, 114, 115, 117,

121, 132, 148, 162, 170, 227Separation (of variables), 3, 11Serendipity class. See Finite ele-

mentsSesquilinear form, 138

Index 311

Seven-point formula, 91, 92, 176Shortley-Weiler discretisation, 78,

80, 81, 83, 229, 231, 237Side condition, 182, 183, 290, 291Single-layer potential. See PotentialSingular perturbation, 247, 248Singularity function, 14, 16, 56

discreteSobolev, lemma of, 125, 212, 222Sobolev space (cf. norm), 114, 117,

122, 128, 133, 134, 136, 145Solution

classical, 27, 30, 32, 145, 146,147, 154, 212, 222, 246, 279

existence of a, 9,13,23,29, 32,35, 37, 67, 137, 142, 144, 151,155, 161, 286, 292

existence and uniqueness of a,137, 141, 142, 147, 148, 149,150, 151, 152, 153, 162, 164,166, 167, 168, 171, 183, 188,222, 248, 282, 284, 285, 286,288, 289, 292

Ritz-Galerkin, 164, 165, 166, 168,170

uniqueness of the, 18, 19, 26, 27,35, 86, 87, 88, 148, 149

weak, 27, 146, 150, 159, 190, 208,209, 210, 212, 213, 218, 219.222, 223, 226, 246, 279

Sparse matrices, 44, 171, 173Spectral radIus, 47Spectrum, 137, 142Stable, stabilIty, 59, 64, 70, 90, 94,

95, 102, 103, 227, 228, 242,249, 252, 291, 293

Star (stencil), 178, 180Stationary problem, 10Steklov problem, 254Step size. See Grid sizeStiffness matrIx, 102, 163, 165, 171,

177, 191, 200, 202, 203, 255Stokes equations, 4, 103, 275, 278,

279, 282, 285, 286, 288, 290,291, 292, 293, 298

Support, 115, 171, 173, 176, 182,183, 194, 195, 202

Time, 8, 9, 10, 11Trace (of a function), 3, 123, 129Trace (of a matrix), 8?Transformation. See Domain,

FourierTransformation theorem, 119, 129Transition condition, 245, 246, 247Trefftz'method, 200

Triangulation, 174adaptive, 201admissible, 174, 175, 179, 185,

190quasi-unifurm, 188, 190, 204, 208,

294, 295regular, 188uniform, 188

Type (of a pde), 1,4,5,6,7,10Type invariance, 6Uniqueness. See SolutionVariational formulation (weak for-

mulation), 144, 146, 149, 150,158, 161, 165, 166, 173, 183,194, 208, 244, 253, 254, 278,288

Variational problem, 141, 152, 159,184, 171

dual/complementary, 200Viscosity, artificial, 251, 252

numerIcal, 251Wave equation, 2, 3, 5, 8, 10Weak formulation. See Variational

formulationWell-posed (problems), 23Wilson's rectangle, 199

This book offers a simultaneous treatment

of the theory and the numerical treatment

of elliptic problems. The subjects dealt

with include the classical theory (Green' s

function, maximum principle, etc.) as well

as the variational formulation. The author

describes and analyses finite difference and

finite element methods. Specific chapters

are devoted to the eigenvalue problem and

the Stokes problem.

ISSN 0179-3632

ISBN 3-540-54822-X

1111119 783540 548225

http:llwww.springer.de

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