Elliptic Differential Equations: Theory and Numerical Treatment (Springer Series in Computational Mathematics, 18)

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Springer Series in

Computational Mathematics

Elliptic Differential Equations

W. Hackbusch

Theory and Numerical Treatment

4

Springer

Wolfgang Hackbusch

Elliptic

Differential Equations

Theory and Numerical Treatment

Translated from the German

by Regine Fadiman and Patrick D.F. Ion

With 40 Figures

4

Springer

Wolfgang Hackbusch

MPI für Mathematik in den Naturwissenschaften lnselstr. 22-26 04103 Leipzig, Germany e-mail: [email protected]

Cataloging-in-Publication Data applied for A catalog record for this book is available from the Library of Congress.

Bibliographic information published by Die Deutsche Bibliothek Die Deutsche Bibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data is available in the Internet at httpd/dnb.ddb.de

Second Printing 2003

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

ISSN 0179-3632 ISBN 3-540-54822-X Springer-Verlag Berlin Heidelberg New York o B.G. Teubner, Stuttgart 1987: W. Hackbusch, Throne und Numenik elliptischer Differentialgleichungen. Mit Genehmigung des Verlagea B.G. Teubner, Stuttgart, veranstaltete, allein autorisierte engliache Obersetzung der deutschen Originalausgabe.

This work Is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilm or in any other way, and storage in data banks. Duplication of this publication or parts thereof is permitted only under the provisions of the German Copyright Law of September 9, 1965, in its current version, and permission for use must always be obtained from Springer-Verlag. Violations are liable for prosecution under the German Copyright Law. Springer-Verlag Berlin Heidelberg New York a member of BertelsmannSpringer Science + Business Media GmbH http-J/www.springer.de

o Springer-Verlag Berlin Heidelberg 1992 Printed in Germany The use of general descriptive names, registered names, trademarks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. Cover design: design&production. Heidelberg

Printed on acid-free paper

46/3142oa-54321

Foreword

This book has developed from lectures that the author gave for mathematics students at the Ruhr-Universität Bochum and the Christian-Albrechts-UniKiel. This edition is the result of the translation and correction of the Gennan edition entitled Theorie und NtLmerik Differentialgieichungen.

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

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

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

iv

Foreword

numerical treatment is the description of the discretisation procedures (B), which give finite-dimensional equations for approximations to the solutions. The subsequent second part of the numerical treatment is numerical

analysis (C) of the procedure in question. In particular it is necessary to find 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 simpie problem, since from to 106 unknowns can oceur. The discussion of this third area of numerical problems is skipped (one may find it, e.g., in Hackbusch [5] and [9)).

The descriptions of discretisation procedures and their analyses are closely connected with corresponding chapters of the theory of elliptic equations. In addition, it is not possible to undertake a well-founded numerical analysis without a basic knowledge of elliptic differential equations. Since the latter cannot, in general, be assumed of a reader, it seems to me necessary to present the numerical study along with the theory of elliptic equations. The book is conceived in the first place as an introduction to the treatment of elliptic boundary-value problems. It should, however, serve to lead the reader to further literature on special topics and and applications. It is intentional 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 a discussion of the Navier-Stokes equations, which are important for fluid mechanics, i8 excluded; however, one can approach these matters via the Stokes equation, 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 further discretisation methods (collocation methods, volume-element methods, spectral methods) and Integral-equation methods (boundary-element methods). The Exercises that are presented, which may be considered as remarks without proofs, are an integral part of the exposition. If this book is used as the text for a course they can be used as student problems. But the reader too should test his understanding of the subject on the exercises. The author wishes to thank his collaborators G. Hofmann, G. Wittwn and J. Burmeister for the help in reading and correcting the manuscript of this book. He thanks Thubner Verlag for their cordial collaboration in producing the first German edition. Kiel, December 1985 W. Hackbusch

This translation contains, in addition to the full text of the original edition, a short Section on the integral-equation method. The bibliography has 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 friendly cooperation. Kiel, March 1992

W. Hackbusch

Table of Contents

Foreword Thble of Contents

v

Notation 1

Partial Differential Equations and Their Classification Into Types 1.1

1.2 1.3 1.4

2

2.2 2.3 2.4

.

I .

4

.

.

.

6

.

.

.

7

Posing the Problem Singularity Function The Mean Value Property and Maximum Principle Continuous Dependence on the Boundary Data . .

12 12 14 17 .

.

The Poisson Equation 3.1

3.2 3.3 3.4 3.5 4

1

Examples Classification of Second-Order Equations into Types Type Classification for Systems of First Order Characteristic Properties of the Different Types

The Potential Equation 2.1

3

xi

Posing the Problem Representation of the Solution by the Green Function The Green Function for the Ball The Neumann Boundary Value Problem The integral Equation Method

23

27 27 .

28

34

35

36

Difference Methods for the Poisson Equation

38

Introduction: The One-Dimensional Case The Five-Point Formula M-matrices, Matrix Norms, Positive Definite Matrices Properties of the Matrix Lh Convergence

38

4.1

4.2 4.3

4.4 4.5

40 44 53 59

Vi

Table of Contents

4.6 4.7

62

Discretisations of Higher Order The Discretisation of the Neumann Boundary Value Problem 4.7.1 One-sided Difference for 9u/ôn 4.7.2 Symmetric Difference for Ou/On 4.7.3 Symmetric Difference for Ou/ôn

65 65 70 71

onanOffsetGrid Proof of the Stability Theorem 7 Discretisation in an Arbitrary Domain 4.8.1 Shortley-Weller Approximation 4.8.2 Interpolation at Points near the Boundary

72

4.7.4 4.8

5

78

78 .

General Boundary Value Problems 5.1

5.2

5.3

Dirichlet Boundary Value Problems for Linear 85 Differential Equations 85 5.1.1 Posmg the Problem 86 5.1.2 Maximum Principle 5.1.3 Uniqueness of the Solution and Continuous 87 Dependence 5.1.4 Difference Methods for the General Differential 90 Equation of Second Order 95 5.1.5 Green's Function 95 General Boundary Conditions 95 5.2.1 Fbrmulating the Boundary Value Problem . . . Difference Methods for General Boundary 5.2.2 98 Conditions 103 Boundary Problems of Higher Order 103 5.3.1 The Biharmonic Differential Equation 5.3.2 General Linear Differential Equations of Order 2m 104 5.3.3 Discretisation of the Bihaimonic Differential . 105 Equation

Ibols from Functional Analysis 6.1

6.2

83

85

.

6

.

.

.

110

Banach Spaces and Hubert Spaces 6.1.1 Normed Spaces 6.1.2 Operators 6.1.3 Banach Spaces 6.1.4 Hilbert Spaces Sobolev Spaces

110

6.2.1 L2(f1) 6.2.2 Hc(O) and

115

6.2.3 Fburier

110 111

112 114

115 116

and Hc(IRI))

119

vii

Table of Contents 6.2.4

123

Dual Spaces 6.3.1 Dual Space of a Normed Space

130

6.3.2 Adjoint Operators 6.3.3 Scales of Hilbert Spaces

132

6.4

Compact Operators

135

6.5

Bilinear Forms

137

Variational Ermulation

144

6.3

7.1

7.2 7.3 7.4 8

122

and Extension Theorems

6.2.5

7

H8(fl) for Real a 0

130 133

144

HIstorical Remarks Equations with Homogeneous Dirichiet Boundary Conditions Inhomogeneous Dirichiet Boundary Conditions Natural Boundary Conditions

145 150 152

The Method of Finite Elements 8.1

8.2 8.3

8.4

161

The Ritz-Galerkin Method Error Estimates FinIte Elements 8.3.1 Introduction: Linear Elements for 1? = (a,b) . 8.3.2 Linear Elements for a c JR2 8.3.3 Bilinear Elements for a C JR2 8.3.4 Quadratic Elements for (1 C It2 8.3.5 Elements for a C Ut3 8.3.6 Handling of Side Conditions Error Estimates for Finite Element Methods 8.4.1 H1-Estimates for Linear Elements 8.4.2 L2 and H Estimates for Linear Elements . Generalisations 8.5.1 Error Estimates for Other Elements 8.5.2 Finite Elements for Equations of Higher Order 8.5.2.1 Introduction: The One-Dimensional Btharmornc Equation 8.5.2.2 The Two-Dimensional Case 8,5.2.3 Estimating Errors Finite Elements for Non-Polygonal Regions Additional Remarks 8.7.1 Non-Conformal Elements .

8.5

8.6 8.7

161

167 171 .

171

174

178 180 182 182 185

185 .

190

193 193 .

194

194 195

196 196

199 199

viii

Thble of Contents 200

8.7.2 The Trefftz Method 8.7.3 Finite-Element Methods for Singular Solutions 8.7.4 Adaptive Thangulation 8.7.5 Hierarchical Bases

201

.

201

202 202

8.7.6 Superconvergence 8.8 9

Properties of the

203

Matrix

208

RegularIty 9.1

Solutions of the Boundary Value Problem

inH'(37),a>m 9.1.1 The Regularity Problem 9.1.2 Regularity Theorems for 9.1.3 Regularity Theorems for = 9.1.4 Regularity Theorems for General 1? C 9.1.5 Regularity for Convex Domains and Domains with Corners 9.1.6 Regularity in the Interior . Regularity Properties of Difference Equations 9.2.1 Discrete H'-Regularity 9.2.2 Consistency 9.2.3 Optimal Error Estimates 9.2.4 Hz-Regularity .

9.2

10

210 215 .

10.2

.

Differential Equations with Discontinuous Coefficients 10.1.1 Fbnnulation 10.1.2 Discretisation A Singular Perturbation Problem 10.2.1 The Convection-Diffusion Equation 10.2.2 Stable Difference Schemes 10.2.3 Finite Elements

11.2

11.3

Formulation of Eigenvalue Problems . Finite Element Discretisation 11.2.1 Discretisation 11.2.2 Qualitative Convergence Results 11.2.3 Quantitative Convergence Results 11.2.4 Complementary Problems Discretisation by Difference Methods .

219

.

226

226

232 238

240 244

Elgenvalue Problems 11.1

.

223 226

SpecIal Differential Equations 10.1

11

.

208 208

244 244

246

247 247 249 251

253 253

.

254 254 256

.

.

.

260

264 .

.

267

Table of Contents 12

ix

Stokes Equations 12.1

12.2

275

275 SyBtems of Elliptic Differential Equations 278 Variational Formulation . 278 . 12.2.1 Weak Formulation of the Stokes Equations 279 12.2.2 Saddllepo4nt Problems 12.2.3 Existence and Uniqueness of the Solution of a 282 Saddlepoint Problem 285 12.2.4 Solvability and Regularity of the Stokes Problem 12.2.5 A Vo-elliptic Variational Formulation of the Stokes 289 Problem Mixed Finite-Element Method for the Stokes Problem . 290 12.3.1 Finite-Element Discretisation of a Saddlepoint 290 Problem 291 12.3.2 Stability Conditions 12.3.3 Stable Finite-Element Spaces for the Stokes Problem 293 12.3.3.1 Stability Criterion 293 12.3.3.2 Finite-Element Discretisations with the Bubble Function 294 12.3.3.3 Stable Discretisations with Linear Elements in Vh 296 12.3.3.4 Error Estimates 297 .

12.3

Bibliography

300

Index

307

Notation

Formula Numbers. Equations in Section X.Y are numbered (X.Y.1), (X.Y.2), etc. The equation (3.2.1) is referred to within Section 3.2 8Irnply as (1). In other Sections of Chapter 3 it is called (2.1).

Theorem Numbering. All Theorems, Definitions, Lemmata etc. are numbered together. In Section 3.2 Lemma 3.2.1 is referred to as Lemma 1.

Special Symbols. The following quantities have fixed meanings:

A, B,... B, B, C

C'(D), Ck(D), C°°(D)

Gr(a) d(u, VN)

dl', diag{d1,d2,.. .}

f h

H"(i?), H'(S?), KR(y)

I L

I,

matrices boundary differential operators (cf. (5.2.la,b), (5.3.6)) the complex numbers Holder- and Lipschitz-continuousiy differentiable functions (cf. Definition 3.2.8) k-fold and infinitely continuously differentiable functions infinitely differentiable functions with compact support.a (cf. (6.2.3)) distance of the function u from the subapace VN (ci. Theorem 8.2.1) surface differentials in surface integrals diagonal matrix with the diagonal eLements d1,d2,... a function; often the right-hand side of a differential equation Green's function (cf. Section 3.2) step size (cf. Sections 4.1 and 4.2) Sobolev spaces (cf. Sections 6.2.2 and 6.2.4) open ball about y with radius R (ci. (2.2.7), Section 6.1.1) identity or inclusion (cf. Sections 6.1.2, 6.1.3) a differential operator (cf. (1.2.6)) or the operator associated with a bilinear form (ci. (7.2.9')) the stiffnees matrix (ef. Section 8.1) matrix of a discrete system of equations (cf. (4.I.9a))

Notation

linear space of bounded operators from X to Y

L(X,Y) L°°(S?) L2(J2) IN

n

n(x)

0 O(.)

P qh

Rh,Rh

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

VN,Vh

x,(x,y),(x,y, z)

x=(xl,...,xn)

z I,

p(A) 1?

V

8; 8/On

)xXx' (., .) (., .)O, (, (•, (•, )a

II112 •

j

I1•1100

I

I

I

- 1k, I

lo,

I

(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 matrix constlg(x)I Landau symbol: 1(x) = O(g(x)) if cf. (8.1.6) a grid function, nght-hand side of the discrete equation (4.1 .9a) the real numbers, the positive real numbers restrictions (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 equation a grid function (discrete solution; cf Section 4) finite-element spaces (ci. (8.1.3) and Section 8.4.1) independent variables a vector of independent variables the integers the boundary of 1? the boundary points of a grid (cf. (4.2.lb), (4.8.4)) fundamental solution function spectral radius of the matrix A or a domain (cf. DefinitIon 2.1.1) an open set In 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) scalar product and norms on L2(Q) scalar products and norms on the Eudidean norm (cf. (2.2.2)) the Eucidean norm and the spectral norm (cf. Section 4.3) norms equivalent to 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 6.1.1 the vector (1, 1,...) (ci. Section 4.3) J

1 Partial Differential Equations and Their Classification Into Types

1.1 Examples An ordinary differential equation describes a function which depends on only one variable. Unfortunately, for many problems it is not possible to restrict attention to a single variable. Almost all physical quantities depend on the spatial variables z, y , and z and on time t. The time dependence might be omitted for stationary processes, and one might perhaps save one spatial dimension by special geometric assumptions, but even then there would still remain at least two independent variables. Equations that contain the first partial derivatives

where (1

i

or even higher partial derivatives

etc., are called

partial differential equations. Unlike ordinary differential equations, partial differential equations cannot be analysed all together. Rather, one distinguishes between three types of equations which have different properties and also require different numerical methods. Before the characteristics for the types are defined, let us introduce some examples of partial differential equations. All of the following examples will contain only two independent variables

x,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 the form 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 I?)) with the aid of =— = Since v(e, := ii), (chain rule) and X,3 —c, y,7 = 1, it follows from (2) that V,3 = + = 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) as long as is continuously differentiable.

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

for x€Ut

(1.1.4)

with aconstant. The comparison of Equations (3) and on the line (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 be an 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 any function f(z) holomorphic in flare solutions of Equation (6). Three examples are 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 by

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

—

y)

(1.1.8)

where

and are arbitrary twice continuously differentiable functions. Suitable initial values are, for example,

u(x,O) = uO(x), where

=

p+

u1(z),

(x E It)

(1.1.9)

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

1.1 Examples

3

(t4 —tii)/2.

w'= (u1 +t4,)/2,

From this one can determine and up to constants of integration. One constant can be chosen arbitrarily, for example, by s°(O) = 0, and the other is

=

determined 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 variables u(z, y) = v(x)w(y) gives that for every c E lit (1.1 .lla)

sin(cx) . exp(—c2y).

u(x, y)

Another solution of (10) for y > 0 is u(x

y) =

L

(l.l.Ilb)

exp((z —

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

where

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

(1.1.12)

The solution (lib), which initially is defined only for y > 0, can be extended continuously 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 prove that the right side of Equation (lib) converges to uo(z) for y —'0. Hint: First show 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

equations, equations of higher order can be by systems of first-order equations. In the following we give some

with ordinary differential

described

examples.

Example 1.1.8. Let the pair (u, v) be the solution of the system (1.1.13)

If u and v tions

are

+

twice

=0,

differentiable,

+

the

differentiation of

(13) yields

0, which together imply

Thus u is a solution of the wave equation

(7). The same

that

the equa-

-

=0.

be shown for v.

4

1

Partial Differential Equations and Their Types

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

v2—u5=O in $?cIt2,

(1.1.14)

then the same consideration as in Example 8 yIelds that both u and v satisfy the 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 the pressure.

1.2 Classification of Second-Order Equations into Types The general linear differential equation of second order in two variables reads

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

( 1.

Q

2. i )

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 2

(n—2)w,,

(2.2.la)

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 1/2

(xl

by

=

(2.2.2)

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

2.2 Singularity Function

15

The proof can be carried out directly. It is simplest to introduce polar coordinates with y as origin and to use (1.5), since s(x,y) depends only on

r=lx—yI. For the next theorem we need to introduce the normal derivative 8/On.

denote the outer F, i.e. n is a unit vector perpendicular to the tangential hyperplane at x and points outwards. The normal derivative of u at x E F is defined as be a domain with smooth boundary 1'. Let n(x) E

Let

normal direction at x

= (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)) the normal 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 from Os(x, y)/Or = —Ix — that Os(x,y)/On =

for x E

(2.2.4)

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

L u

dx = —

V

E

j (Vu, Vv) + j 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. To see sufficient conditions for this refer to Kellogg (1, Chapter LVI and Hellwig 11, 1-1.21.

Functionsu,v E C2(Thinanormaldomain.Qsatisfythe second Green

formula

/ u4v =

/ viju + /

—

dl'.

(2.2.5b)

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

u(y) = for all y

(2.

Here 0/On, and dl' refer to the variable x.

be harmonic

dl',

(2.2.6)

____ 16

2 The Potential Equation

PROOF. By Kr(y) := (x E

x—

2

(2.2.lla)

for 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:

with

=

=

—

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? if C°(fl) and if for all x E Q and all R >0 with KR(X) C a the following equation holds UE

u(x)

1

J

=

(2.3.1)

8KR(x)

dl.' = the right side in (1) is the mean value of u taken over the surface of the sphere. An equivalent diaracterisation results if Since

one averagea over the sphere KR(x).

Exercise 2.3.2. u C0(IJ) has the second mean value property in 1? if u(x) = for all x a, R> 0 with KR(x) C a. Show that this 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 is known from the function theory for holomorphic functions:

18

2 The Potential Equation

Theorem 2.3.3. (Maximum-minimum principle) Let (1

be a domain and let u E C°(Th be a nonconstant function which has the mean value property. 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 minimum of 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)=Mu(x)

fora.llxEa.

In (c) we will show u(y') = M for arbitrary y' E I?, i.e. u M in contrast to the assumption u const. (c) Proof of u(y') = M. Let y' E (1. Since S? is connected, there exists a path connecting y and y' running through a, i.e. there exists a continuous (2 with y, 10,11 = y'. We set

I := {s

M for all 0 t s}.

E 10,11:

I contains at least 0, and is closed since u and ço are continuous. Thus there In (4) exists s' = max{s I}, and the definition of I shows that I = [0, it is proved that s' =lso that y= E land hence u(y') =M follows. (d) Proof of? = 1. The opposite assumption 8' 0 with C £1. Evidently, it follows that u = 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') =

I

JSKr(X')

In general we have u(C) M . If one had u(C') 0 auth that for given e >0 0, there exist numbers N(c) and 6(c) > 0 such that the following implication holds: ip, E

n

N(c),

x€ F, y€ I',,,

The sequence un

E

6(e)

e. (2.4.4a)

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

2.4 Continuous Dependence on the Boundary Data

Remark 2.4.5.

(a)

Let K be a set which is compact (i.e. complete and E C°(K) converge C K for all n. Let

bounded) with r c K, and uniformly on K to If

on F then (4a) is satisfied. be the following (not continuous) Then (4b) is u on ii on (1 in the usual sense.

and

on

C 1? hold for all n and let (b) Let = onto Th continuation of 3quivalent to uniform convergence

and let 1 which are harmonic in 5Z, be solutions of

Theorem 2.4.6. functions

(2.4.4b)

fl I?) = 0.

— u(x)I: x

lim

25

Let $1,., C (1, with Si

-,

P. Let the

(2.4.5a)

on F,.

Then

conve!ye uniformly in the sense of(4a) to — the following as8ertwfls hold: Let

(a)

E

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

on!'

u=cp

(2.4.5b)

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

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

if lx

lu(x) —

Set 6(e) := exists

(2.4.6a)

—

with 6 from (4a). Because

max{Np(e), N(e/2)} (N from (4a)) we want to show that For x E S?\S1,, the estimate is trivial because for x all x c a, however, there holds —u(x)I =

— u(x)l

—ul

I'

there := — u(x)l = u(x). For —+

For n

for n

so that

(2.4.6b)

It remains to estimate withn N(e)thereexistsy €1' with

(cf. Theorem 2.3.8), because tin - u is harmonic in

forxE —

Forx€

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 on and (6b) proves the uniform convergence ii,, —' ti on Hence, from Remark 5b it follows that (4b) is satisfied.

26

2 The Potential Equation

(b) Let K C 1? be a compact set. Since 17,.. -+ 1' there exists a N(K) such N(K)} converges for n N(K). Thus, the sequence {u1,: n that K C uniformly 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, it follows 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) a as This inference is not possible for the case of just from ço, -. the following example shows. C a := K1(O)\{O} C 1R2. The boundaries are

:= Let = 0K1(O) U

and V = 0K1(O) U {O}, and satisfy

The

—p

boundary values

=

= 0 on 8K1(O),

satisfy the condition —+ (5a) can be given explicitly:

=

1

on

0) =

1

(cf. (4a) and Remark 5a). The solutions

of

u,..(x) = tog(IxI)/log(1/n). u(x) := 0 holds pointwise, but u = 0 satisfies neither (4b) Obviously, u,(x) nor the boundary value problem (5b). Conversely, one infers from Theorem 6a the following result:

solution u

a

= K1(O)\{0} c 11t2 the potential equation has In C2(S?) fl C°(J7) which assumes the boundary values u(x) =0 on

Remark 2.4.7.

8K1(O)andu(x)=linx=0.

no

3 The Poisson Equation

3.1 Posing the Problem The Poisson equation reads

au—f in!?

(3.1.la)

C°(Q). In the physical interpretation f is the source term ifor example, the charge density in the case of an electrical potential uJ. To determine the solution uniquely one needs a boundary value specification, for

with given f

example, the Dirichiet condition (3.l.lb)

DefinitIon 3.1.1. The function u is called the classical solution of the boundary 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 always mean "classical solution".

The solution of the boundary value problem (la,b) will in general no longer satisfy the mean value property and the maximum principle. But these properties still hold for the differences of two solutions u, and u2 of the equation, since =0. Thus the uniqueness of the solution of — u2) = — problem (la,b) immediately follows and Theorem 2.4.3 can be brought over:

f f

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

—

u"II,.,
0.

71,,7k—jEI

to Lemma 9, p(C) 0 for all

(4.3.15)

Exercise 4.3.22. Prove that (a) a symmetric matrix is positive definite if and 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 semidefinite square root B = A"2, which has the property B2 = A. If A is positive 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 sufficient

to 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 is diagonally dominant or inite.

diagonally dominant then it is positive def-

PROOF. Since

resp. the Gershgorin circles which occur in Criterion 4 do not intersect the semi-axis (—oo, 0), so that all the elgenvalues

must be positive. By Exercise 22a then A is positive definite,

Lemma 4.3.25.

be )tmin and the smallest and largest eigenvalues 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

4.4

=

and

From (5) then result p(A) =

= p(A').

p(A) and 11A'hh2

PROOF. Exercise 20a shows that hAil2

since

53

.Xmin

> 0.

U

Properties of the Matrix Lh

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

(4.4.la) (4.4.lb) (4.4.lc) (4.4.ld)

Lh is an M-matrix,

Lh is positive definite,

8h2,

1/8,

8h2 cos2(irh/2) O. M on f'h implies the

(2) Let M be the right side of (4a). —M wh inequallties —M

55

S

M on .17h.

The discrete analogue of the Grcen function

is

Let

be the scaled unit vector

h2

X

(x,

{

—

The column of the matrix

with index

E [lh).

E

(4.4.5a)

is given by

(eEQh).

(4.4.5b)

ah fixed, gh(•,E) is a grid function defined on definition is extended to x Thh:

The domain of

For

The Lh implies

are

entries of h2L':

Remark 4.4.6.

=

= for all

(4.4.5()

E

l'p,

The symmetry of

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) with boundary 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 3.2 we 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
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 be uniformly elliptic in 1? if Q} > 0 (c(x) from (3s')).

Inf{c(x): x

(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, nor is 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. = sin(x)sin(y) are solutions of the boundary Then both u = 0 and value problem. The second solution assumes its maximum at the interior point

(ir/2,ir/2) EQ. In the above example the coefficient 0(x) = sign. As soon asaO, we have

2

(ci. (ib)) has the wrong

Theorem 5.1.3. (Maximum-minimum principle) Let a(x)

0 in the

domain 1?. Assume the coefficients of the elliptic opemtor (ib) are continuous

mO. LetuEC2(Q) .atisfyLu=fandbenonconstant. Then we have: in 4?, no negative minimum of u exists in 1?; (a) if f (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, for example, in Heliwig 11,111-1.11. Here we give a shorter proof, which however, the stronger condition 0(X) 0 in contradiction to the assumption I 0. Part (b) is proved analogously.

ExercIse 5.1.4. The trace is defined by tr(A) := (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.

Prove that:

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 JffOin (JJ and

sf there exists a negative minimum [positive maxzmurnj of u in on the boundary 8fl.

it must lie

Remark 5.1.6. The continuity of the coefficients and a of L in Theorem 3 and COrOllary 5 can be replaced by the assumption a 1) (cf. Bramble-Hubbard 11)). Layton—Morley 111 point in general out 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 E as in (10). The following corollary shows that —Lh is also an M-matrix when is sufficiently small.

Corollary 5.1.16. In addition following hold:

the assumptions of Theorem 14 let the

to

h

(s=1,2).

(5.1.15)

fvvm (10) together with (13') leads to a Then the discretisation of seven-point difference method of second order of consistency such that -L,, is an M-matriz.

PROOF. -Lh satisfies (4.3.la) and is irredudbly diagonally-dominant. ExercIse 5.1.17. The condition su 1a121 + 8ufflcient. Construct a counterexample with a11 = variable, and lad =6 so that L,, is singular.

•

instead of (15) is not 1, 0, h = 1/3,

=

Considerably weaker conditions for the nonsangulanty of L,, than in Theorem 14 and Corollary 16 are needed in Section 9.2 (cf. Exercise 9.2.6, Corollary 11.3.5).

is not a symmetric matrix. Symmetry of L,, is to be exin general, pected only if L is also symmetric: L = Ii. Here the formally adjoint differential operator L' which is associated to L in (2), is defined by

L=

1111

[aj

82

+

0

8

+

82

1

1

_E [(li

8

+

8

+a. (5.1.16)

It is easy to see that a 8ymmetric operator can always be written in the form

8 vxj

(5.1.17)

A difference method for this is given for the case n=2 and a12 =0 by the

five-point star

5.1 Dirichlet Boundary Value Problems for Linear Differential Equations

93

0

o —

a22(x,y

—an(x,y+

—

0

o

o

a(x,y)

0 0

o

0

0

+0

o

(5.1.18)

.

Theorem 5.1.18.

The difference method (18) is Let a22 E The associated matrix L,, is symmetric. If a.,, > 0 consistent of order (ellipticity) and a 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 difference scheme which is described by (18) and (19) has consistency order 2 and that the ansociated matrix —L,, is a symmetric, irreducibly diagonally-dominant and positive definite M-matrix. (b) What is the suitable discretisatton for the case a12 0?

a11

>

0,

Exercise 5.1.20. L reads

The difference formula from Theorem 14 for the operator

+

a when

5 General Boundary Value Problem8

94

0, 0. Let the associated matrix be Lh. Then prove that the 0, transposed matrix LZ describes a difference method for the adjoint operator V and also possesses consistency order 1.

012

In general it is possible to show for regular difference methods that a discretisation of L'. The role of regularity is demonstrated in

is

Example 5.1.21. Let Lu := u" + au' in 11 = (—1, 1) with a(x) 0 for 0 for x > 0. According to (14) au' is discretised for 0 and a(z) z and for x > 0 by a(x)8u(x). Let the associated matrix z 0 by + Lh,1, where Lh,2 and Lh,1 correspond to the terms u" and be Lh = Lh,2 should be a discretisation of au' respectively. According to the above, —(av)'. But the differences at x = 0 and x = h are h'ta(—h)vh(—h) — a(O)vh(O)

h'[a(O)vh(O) + a(h)Vh(h)

—

.-(av)' — h10(O)Vh(0) + 0(h),

—

o(2h)vh(2h)1 = —(av)'

+ h10(0)Vh(0) + 0(h);

is a possible discretisation of thus they are not consistent. Nevertheless, has order of L'v = v" — (av)', for it can be shown that the error — magnitude 0(h). To prove stability one has to show const. Obviously it is sufficient to prove this inequality for sufficiently small h. In the proof of Theorem 9 we used the fact that

—Lwl mO,

on!'

for w(x) := exp(2Ra) — — + R)). Let Dhuh(x) be the difference equations 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:

- RhL)W - 2RhLW 2 -

-Dhwh =

Each consistent difference method satisfies fl(DhRh small ho we thus have

- RhL)w. -

-' 0.

For

sufficiently

—Dawh(x)l (XE 0h, h—D,,w,,1 holds for all grid points. Theorem 4.3.16

5.2 General Boundary Conditions

95

Theorem 5.1.22. The ducretssations from Theorem 14, Coroflary 16, Theorem 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 order of consistency.

agrees

with the corresponding

5.1.5 Green's Function The idea of representing the solution by the Green function can be repeated for the general differential equation (Ia,b). The Green function (of the first x) is singular at = x and satisfies kind)

=

=0 foe

0,

x

forxEIoreEf'. Here, L' is the adjoint differential operator (16). If L L', then g is no longer x). Under suitable conditions the solution of (la-c), symmetric: g(x, (4) can be represented as

u(x) = —

x)

L

where B = Bt =

is a boundary differential operator (n, are the components of the normal vector n = F). Only when the e principal part of L agrees with is B the normal derivative. In the discrete case the inverse L' again corresponds to the Green function 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 boundary condition (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 settings B can be an operator — a so-called boundary differential operator — of order I: B

Ebs(x)o/ax. +bo(x),

If one introduces the vector b(x) = (bj(x), .. the form

XE F.

(5.2.lb)

By can be written in

5 General Boundary Value Problems

96

or B = bTV + b0.

Bti = (b(x), Vu(x)) +

(5.2.1W)

0 there results what is known as on F, also known as the boundary

Example 5.2.1. (a) From b = 0, b0(x)

the Dirichiet condition u = condition of the first kind.

(b) The choice b = n, ho = 0 characterises the Neumann condition, also

called the boundary condition of the second kind. (c) Equation (Ia) with (b,n) 0, ho 0, is known as the mixed, or the boundary condition of the third kind. Occasionally one just means by a 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 very similar to a Dirichiet condition. The condition "u = on F" implies "Bu = := Bço on F" (Why is Bço defined?).

—x)/2

(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 connection with L = —4. For the general operator L in (l.lb) the so-called conormal derivative 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 (Dirichlet condition) uniqueness is guaranteed (cf. Theorem 3.1.2), while the problem associated to L = —4, B nTV = 8/t9n is, in general, not solvable (cf. Theorem 3.4.1).

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 that b(x) = 0 (and bo 0) in C I' and b(x) 0 in t'\'y. Then there is a on the piece -y and a boundary condition of fIrst Dirichiet condition u

order 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 the derivatives).

0 with y radius 1. Let the differential equation and boundary conditions be given as in Figure la. The boundary condition changes its order at x = y = 0. The (cf. (2.1.3)). Check that solution in polar coordinates reads: u = r 1/2 = O(r112) and =

Example 5.2.3. Let .0 be the upper semicircle around x =

The same singularity as in Example 3 occurs in the problem described This Dirichlet problem and in Figure ib; the solution is also r112 Example 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 reflection of S?j in -y result in S?2. If one seeks a solution of Lu = in (1, Bu on

f

OS?, and if together with u the function II reflected in 'y is also a solution, one expects u = This solution then satisfies Lu = in Ru = on

Ou/On=Oon'y.

f

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. Let 5? be given as in Figure 3: and 'Y2 are parts of F =85? with

98

5 General Boundary Value Problems

= {(x,,y):yi y Y2},

Then, in addition to Bu =

on

boundary condition u(x1,y)

t4z2,y),

f'\(i'i

i =

U 72), we

uz(Zj,y) = Uz(X2,y)

1,2.

can require the periodic for

Yl y Y2,

(5.2.2)

The solution is periodically continuable in the x-direction (with and period x2 — x1). The origin of periodic boundary conditions is discussed in Example 5, on

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 differential equation to polar coordinates gives as the image domain the rectangle 11 = (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) describes the periodicity of the angular variable x E (0, 2ir).

(b) Instead of on (1' C Ui?, one can also define a boundary value problem on a part of the 2-dimensional surface of a 3-dimensional body. If if lies, for example, on the surface of the cylinder E e IR}, the + 0 and second inequality in (4a) agrees with (4.3.4b). The corresponding inequalities for (3c') read: Here

ct?O, C20,

(5.2.4b)

and the boundary equations (3c') 0, (4a,b) implies be combined into Ahuh =

Let the difference equations in x E given for

E

Therefore, Ah is an M-matrix if A,, is irreducible and (4.3.4a) holds for at least one x E siderations explain why the boundary discretisation should The above satisfy the conditions (4b). (4b) holds for the case b0 =0 if and only if b2/b1 E [—1,0J, (4b) can be satisfied if one interpolates between and (0, 11. If b2/b1 i.e., if the tangential and + h). If, however, — h (instead of > and component is greater, one could interpolate between k. Fbr the case b2/b1 1, however, it is more practical kh), where 1b2/bd + h) between to interpolate at the point + h) and + h,V + h). Generally one should choose as interpolation point the point of intersection of the 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).

5.2 General Boundary Conditions

101

F

.--

Line

—

S — h)

•: Points in U: Points in Th

FIgure 5.2.5. FIrst boundary discretisation

First option for discretisation: and lj, be chosen as in Section 4.8.1. At near-boundary points the Let differential equation is approximated by a difference scheme which in the case

of L =

corresponds to the Shortley-Weller method. For this one needs As in Figure 5 let the values of Uh at the boundary points E a boundary point. Again we can set up an equation — sjh, be E 1h has the same position as + h, in analogous to (3a—c). In Figure 5 Figure 4. In general one has to use the point of intersection of the line which is also presented by + tb (t IR), and the dotted straight line in Figure —LI

5.

Second option for discretisation: Let consist of all points of be the grid aa used above. Now let that are far from the boundary. For all (x, y) Qh difference equations (with equidistant step size) are declared which involve For each point a boundary discretisation must be found (cf. 1h := Figure 6). Equation (3a) can be set up with and instead of + and The arguments of the coefficients b1 and b2 are + h, This point results implicitly from

—1) =

—

€1'.

(5.2.5)

Remark 6.2.8. (a) At the grid points adjacent to the boundary, the first discretisation requires a difference scheme for Lu = f with nonequidistant

stepsizes. The second discretisation requires an approximation of the nonlinear problem (5). From a programming viewpoint both procedures are undesirable because of the case distinctions. (b) If the vector b from B approaches the tangential direction, both methods fail since the straight line no longer intersects the dotted straight line from Figures 5 and 6. Here, the second discretisation fails earlier than the first one.

102

5

General Boundary Value Problems TI

Line

(1+ h,

• Points in

I Points in FIgure 5.2.6.

Second

boundary dlecretisation

Another option for avoiding the difficulties described above consists in using variational difference equations, at least near the boundary (cf. Remark 8.6.1).

Occasionally it is also poesible to simplify the boundary conditions by using coordinate transformations. Let Bu = be prescribed on C P. If one finds a transformation (x, y) 0 ,) audi that the equations = and are satisfied on one obtains for the transformed problem Bu oOu/On + bou = on a vertical boundary piece (8/On = The discretisation may be earned out as described in Section 4.7. The same reasoning as in the preceding sections proves stability and

convergence: Remark 5.2.9. Let the difference method

= ía satisfy the conditions := K1 + of Theorem 5.1.22 so that 1 holds for (K> o is constant; wh is as in the proof of Theorem 5.1.22). Let the boundary discretisation (3c') satisfy

For sufficiently large K, one then obtains (Bhti,h)(x) 1 for x E Hence it follows that 1, where is the M-matrix defined immediately after Equation (4b). Since conat (cf. Theorem 4.3.16), stability has been proved. The order of convergence is the minimum of 1 (order of consistency of ; 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 this fact, set: :

—

=

E

5.3 Boundary Problems of Higher Order

103

If, in order to remove the first two are three points in = in terms in the Taylor expansion, one uses the differential equation and the tangential derivative of Bu = one obtains a discretisation of Bramble-Hubbard [2) order 2. For the special case L = 4, B = 0/On + may be chosen such that the inequalities proved that the three points (xe, hold and guarantee stability. However, in general, the + C2 ÷ 4 0, c0 where

f

f'h. Nonetheless, the do not lie in the direct neighbourhood of construction of the discretisation appears too complicated to be recommended for practical purposes. (xe,

5.3 Boundary Problems of Higher Order 5.3.1 The Biharnionic Differential Equation In elastomechanics the free vibration of rods Leads to (ordinary) differential equations 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, the bending vibration of a plate leads to a partial differential equation of fourth

order. 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 deflection of the plate perpendicular to the surface. If the plate is firmly clamped at the edge, one obtains the boundary conditions

u=

and Ou/On =

on F

(5.3.2)

with = 0. A btharmonic probLem (1), (2) also results from the transformation of the Stokes equations in 11 C in.2 (cf. Remark 12.1.5). The differential equation (1) may be combined with other boundary values 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 =

minimum pnnciple (counterexamnple: u

0

do not satisfy a maximum-

x2 + y2 in £7 = KR(0)).

5 General Boundary Value Problems

104

5.3.2 General Linear Differential Equations of Order 2m multi-index) of order IaI = The partial derivative D° (a + is defined in (3.2.5). A differential operator of order 2m has the form

(x

L=

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 coefficients

aa)iasaidtobe elliptic (oforder2rn) atxE Qif (5.3.5a)

for

Here,

We set o for all

of degree is an abbreviation for the polynomial Evidently (5a) is equivalent to P(x,e) = 1. For reasons of continuity either E IR' with

P(x, 0; otherwise we scale with the factor —1 (changing from Lu(x) = 1) is compact it follows f(x) to —Lu(x) —1(x)). Since the set E min{P(x, = 1} > 0 and this justifies the formulation of that c(x) (Sa) as

c(x) >0.

(5.3.5b)

frzl.2m

Definition 5.3.4. The differential operator L is said to be uniformly elliptic 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). What axe 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) in the form L = (cf. (1.lb) and (1.2)).

For in = 1 (equation of order 2) we have used one boundary condition; for the biharmonic equation (in = 2) two boundary conditions occur. In general one needs m boundary conditions

of Higher Order

5.3 Boundary

b,0DaIL=

B,u

(j = 1,2,. ..,rn) on P

105

(5.3.6)

lcxl 0 arbitrary) so that the order of convergence is

(13) by i.e., almost 2.

Remark 5.3.10. Inequality (9a) shows conda(Lh) = O(h4). In general a difference method for a differential equation of order 2m leads to the condition

cond(La) =

(5.3.16)

5.3 Boundary Problems of Higher Order

This indicates greater sensitivity to

109

errors at higher orders 2m

(cf. §4.4 in Stoerili and Bulirsch—Stoer[1J). Difference methods for boundary value problems of fourth order with vailable 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 general domain if the boundary conditions are suitably discretised.

6 Tools from Functional Analysis

To enable readers without previous knowledge of functional analysis to follow

the next chapters we summarise here all definitions and results that will be needed 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 = IR or 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 is denoted 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 maximum norm (4.3.3) are norms on lit". (b) The continuous functions form the (infinite-dimensional) space If 1? 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, continuous functions form a proper subset 11C°(& of C°(fl). Instead of jJ . we use the traditional notation fl. (c) The Holder-continuous functions introduced in Definition 3.2.8 form the normed space (C'(J7), ii The norm defines a topology on X: A C X is open if for all x E A there exists an >Osothatthe "ball" in A. We write

x or x = urn

Example 6.1.2. Let I,,, I to

if

—

0.

f

The limit procees f, —' (with respect denotes the unifonn convergence known from analysis.

6.1 Banach Spaces and Hilbert Spaces

Exercise 6.1.3. The norm fl. reversed triangle inequality holds —

IlvIlI

X —* [0, oo)

fix — till

111

is continuous; in particular the

for

x,y E X.

(6.1.1)

As Example la shows, several norms can be defined on X. Two norms on X fi and ifi o < C 0

124

6 Tools from Functionai Analysi8 1 one obtains, for example, For larger s one uses higher interpolation p. 101]). U (ef. Exercise 9.1.13, cf. Wloka

(i.e., by reflection on F). For s = i.e.,

.. , —xc)

formulae for

In the following the restriction is defined only on

is

written in the form -yu. At first, 'y

(-yu)(x) := u(x)

C

for all (x) E

1'.

(6.2.19)

x' =

We write x = H4(P) =

(x1, .

f' is identified with

. .

Theorem 6.2.28. Let a> 1/2. Then 'y from (19) can be continued to Thus we have in particular l'vuIa_i,2 In the mae n = 1, i.e., -yu = u(0), we have I'yul C,, lul..

for u

PROOF. It suffices to show h'u1_i,2 for tz (cf. Theorem 6.1.11 and Theorem 25a). Let the Fourier transforms of u E and tu := -yu = and i1 = be Fourier transform). can be written as product where acts on x,. Therefore and on x' E has the properties = and tui = W(.,0). According to Lemma 20, = W(e',O) = has the representation

=

for all.e'

:= (1 + Since u E Lemma 23b). Inequality (5a) yields

(6.2.20) lies in L2(IR") (cf.

+

=

J

The first factor may be computed as I(.(1 + with K4 = + dx < oo, since 8 > 1/2. The second is := IIU(e', (Fubin.i's theorem). Together because IIUIIL2RØ_I = we have:

(1 +

for

Integration over

J(i

+

E

results in =

I

=

i-

E

6.2 Sobolev Spaces

125

(cf. (16b)). In the = is proved with and the integration over is = u(0), represents already U

Thus jwI,_j/2 case n = 1, not required.

= 0. Evidently, Theorem 28 describes the restriction 14(., 0) = -yu to E 1R, with the for any other similarly we have is continuous tresp. Holderu(., same constant C8. The mapping continuous] in the following sense.

Theorem 6.2.29. For s> 1/2 the following statements hold: IR,u E

= 0 for all

—

(6.2.21a)


3/2).

PROOF. (a) Let

u E H8(IR") and The function ço,, is continuous in JR and set := since converges uniformly to — u(,x)I,_112 — t4,, for all x JR. Thus (21a) follows. = + e) — has the Fourier transform (b)

so that = = — := in the proof for Theorem 28 set first integral in the eethnate of now reads

f(i +

+

e/2)

=

(1

+

As uui

W(.,0). The

l/2—* J(i + t2)' sin2(i7t)

The decomposition of the last integral into subinte+ with ,, = grals over 1/u shows and fri The remainder of the argument follows the same lines as in the proof of Theorem 28. U Up to this point we have obtained H8(Wt") by completion of in

The next theorem shows that for suffithentiy large $ one can also complete in fl L2(IR") so that contains only claseical functions (i.e. continuous, HOlder-continuous, [HOlder-J continuously differentiable functions).

Theorem 6.2.30. (Sobolev's lemma) IN

Li {0}, s > k + n/2 and H'(IR") C

PROOF. (a) Let a

c for 0 < t

holds fork E IN, a

t + n/2.

t + n/2, 0 IaI + 1/2 there exists a restriction -1D°u E of the derivative of u E H8(Q). (b) For each uE with

s0.

U

and U' can be identified. By this one

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 continuously arid densely embedded in V'.

PROOF. For U C V' see Lemma 9, for V C V' see Exercise 6.2.25.

134

6 Tools from Functional Analysis

Attention. Likewise one could identify V with V' and one would obtain U' 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 U as 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 form

y(x) = (x,y)uxu'. If z E V, then y(x) = (x,y)Vxv' also holds. That means that (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 continuous embedding U C V' proves

Remark 6.3.12. Let V C U C V' be a Gelfand triple. The continuous extension of the scalar product (., .)u to V x V' (V' x VJ results in the dual XVI. Therefore the following notation is practical form (., •)v v'

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) so that the embeddings read as follows: H0($2) c L2(Q)

c

H8($2) c L2(a) c (118(fl))'

(a 0),

(6.3.8a)

(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

where (u,v)L2(a) is the dual form on

Remark 6.3.14. (a) Let (1 = :=

for a 0

v

x H'(Q) (cf. Remark 12).

The norm dual to v

€

6.4 Compact Operators

135

equivalent to and has the representation (2.16b) with —s instead of s. (b) The Fourier transform shows for all 8 IR. E (c) au E H8(Q), if u H8(fl), a C'(i7), where = Isi E IN U {O} or t > (s(. i8

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 subsequence

fand

E

Kl.

Another definition of compactness reads: Each open covering of K already contains a finite covering of K. Both definitions are equivalent in metric spaces (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 is bounded (and complete]. (b) Let X be a Banach space. The unit sphere {x X: flx(f 1} is compact

if and only if dim(X) t) are compact. (s, t E C°'1. The embeddings Hk(S?) C

(b) Fhrther, leti?

(k,l E NU TO),

k> 1) are compact.

(c) Let 0

0

exist such that

inf{sup{ia(x,y)l:yE E

= 1}:xE V,hlxiiv = 1}=e >0,

(6.5.4a)

= l}:y

(6.5.4b)

liyilv

1}

=

e'

>0;

(iii) the inequalities (4a) and (4c) hold:

sup{la(x,y)i:xe V,lJxhlv = If one of the statements (i)—(iii) holds, then

1}

>0.

(6.5.4c)

139

6.5 Bilinear Forms

(6.5.4d)

(e,e' from (4a,b)).

1/11A'IIv.._vi

C

Prom (4a) follows

l',IIxIIv =

inf{sup{Ia(x,y)I:y E V,flyIlv = 1}:x

1}

e >0.

(6.5.4e)

Conversely, (4a) follows from (4e) wsth a possibly larger e > 0. (4e) and (4c) conditions. (4e) is equivalent to are also called the

for all x

V, lIyItv = 1) cHxIIv

sup{Ia(x, y)I: y

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 the definition of the dual norm of Ax so that (4e) and (4e') ore also equivalent to (4e"): (6.5.4e") for all x V. hAzily'

PROOF. (a) "(i) mf{

}

Ia(x,y)I

= inf aEV

Uxllviiyiiv

.

— mf sup —

L(V',V) exist. Then (4a) follows from

(ii)": Let

i(Ax,y)l

_.

iixilvhiyhlv

I(AA'x',y)i

p€V IIA'ziivliviIv

=

— —

c

II A1 lu—i sup i(x',y)I iv' i,€v

= lIE sup 11A'x'ilv/hlx'flv'l

CV

=

C.

Because A'' =

In the same way one shows (4b) with e' =

(A')', (3.3), and V" (b) "(ii)

hIyHv

V, it follows that e = (iii)": (4c) is a weakening of (4b).

(i)": e > 0 in (4a) proves that A is injective. We wish to show (c) "(iii) with that the image W := {Ax: x V} C V' is dosed. For a sequence the From (4a) one infers W there exists zI, E V with = w1,. that via (4e) and (4e") (with x := — flwg, — WpiIv'/f. — There is Cauchy convergent, this property carries over to Since —. in V. The continuity of A E L(V, V') proves exists an x' E V with = Ax,, —+ Ax' so that w' = Ax' E W. According to Lemma 6.1.17 one Wt. If A were not surjective (thus W V'), there can decompose V' into would exist a w W1 with w 0. Then y Jv'w = E V would satisfy 0 (cf. Theorem 6.3.6, Corollary 6.3.7). Since a(z,y) = (Ax,y)v') J(x) for all z

—

2f(y)

.1(x) + CEIIZ

U

x.

The term "V-elliptic" seems to indicate that to elliptic boundary value problems correspond V-elliptic bilinear forms. In general this is not the case. Rather, V-coercive forms will be assigned to the elliptic boundary value problems. 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 form is said to be V-coercive if it is continuous and if there exists oK and CE > 0 such that

a(x,x)

for

all xE V with Ck >0.

lR

(6.5.10)

142

6 Tools from Functional Analysis

a(x,y) +CK(x,y)u with CK from (10). Let

Exercise 6.5.14. Set a(x,y)

V' be the inclusion. Show that (a) the coercivity condition (10) is I: V equivalent to the V-ellipticity of a.

A + CkI to a(.,.).

L(V, V') is associated to a(.,.), then so is A Why does A L(V, V') hold?

(b) If A

The results of Riesz-Schauder theory (Theorem 6.4.12) transfer to A as soon

as 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 embedding V C U. Let the bilinear form a(.,.) be V-coercive with correspond:ng V' be the inclusion. operator A. Let I: V (a) For each A E C one of the foil owing alternatives holds: (i)

(A—Xf)1 €L(V',V)

(ii)

A is an eigenvo.lue.

and (A'—Xiy' EL(V',V),

f

f

in case (i) Ax — Ax = and A'x' — = are uniquely solvable for all I E V' (i.e., a(x, y) — A(x, Y)u = 1(y) and a(x',y) v)u = f(y) for all E(A) := y V). in case (ii) the,e exist finite-dimensional eigenspaces {0} such that E'(A) kernel (A' — kernel (A — Al) and {0} Ax = Ax

for x

E(A),

a(z,y) = A(x,y)u

i.e.,

for all y

V,

(6.5. lie.)

for x*EEI(A),

a*(x*,y)=X(x*,y)u forall y€V.

i.e.,

(6.5.1 ib) (b) The spectrnm o'(A) of A consists of at most countably many eigenvalues

a(A) if and only if A

which cannot accumulate in C. A

0 exists C( such that

for all x E V.

jb(x, x)f

(6.5.12a)

(b) Let the embeddings V c X and V C Y be continuous, with at Least one of them compact. Let the following hold: for all x E V.

Ib(x,x)I

(6.5.12b)

(c) Let the embeddings V C X, V C Y be continuous. Let (12b) hold. For such that or Uv assume that for every e >0 there exists a or

÷

iS

for x

eIlxIIv +

V.

(6.5. 12c)

PROOF. (a) Select e = CE/2 with CE from (10). Then a(•, .)+b(.,.) satisfies the V-coercivity condition with CE/2 > 0 and CK + instead of CE and CK. (b) Lemma 6.4.13 proves (12c).

(c) Let the first inequality from (12c) hold, for example. Since the embedding V C Y is continuous, Cy exists with CyflxlIv. Choose C' = 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 Dirichiet problem of the potential equation (2.1.la,b) whether, or under what conditions, a classical solution u C2(Q) n C°(fl) exists. Green took the view that his Green's function, described in 1828, always exists and that It provides the solution explicitly. This is not the case. Lebesgue proved in 1913 that for certain domains the Green function does not exist. Thomson (1847), Kelvin (1847), and Dirichiet offered a different line of

reasoning. The Dirichiet integral

J

(7.1.1)

=J

describes the energy in physics. With boundary values u = on I' given, one seeks 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 Dirichiet =0 in a, u = on I'. problem The so-called Dirichiet principle states that 1(u), since it is bounded from below by 1(u) 0, must take a minimum for sonie u. According to the 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 the

value inf J(u) = 0. Further, the following example due to Hadainard shows that no finite infirnum of the Dirichiet integral need exist. Let r and be the polar coordinates 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 more suitable Sobolev spaces instead of in C2(Il) fl C°(17).

7.2 Equations with Homogeneous Dirichiet Boundary Conditions In the following we investigate the elliptic equation (7.2.la)

in Si,

Lt.i = g

L=

(7.2.lb) PIm

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,

Attention. In the case that only

E IR".

(7.2.3)

is assumed, one needs to

replace "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 the

standard case m = 1 (an equation of second order) condition (4) becomes U =0. Since with u = 0 on F the tangential derivatives also vanish, not only the kth normal derivatives (k m — 1) but also all the derivatives of order $ m —1 are equal to zero:

inxEf forlal2

E

>2

11

>2

ef Let aoo > 0. There exists an e' > 0, so that for all E From this follows

e' E1cz1 1. More precisely, the following also holds:

7.4 Natural Boundary Conditions (II))' + IkoItH—1/2(Ii).

153

(7.4. ib)

Frequently, variational problems with a physical background have the form:

such that

find u€Hm(A2),

a(u,v) = 1(v) for all v

(7.4.2a)

Hm(fl).

(7.4.2b)

from Section 7.2, u Hm(47) In contrast to the condition u E contains no boundary condition. Nevertheless, Problem (2a,b) has a unique solution if a(•,•) is Hm(.O)-elliptic. This condition is easy to satisfy. 7.2.4 Theorem 7.4.3. Under the conditions of Theorem 7.2.3 or a(-,) 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): (7.4.3)

It&li

PROOF. The

same

as for Theorem 7.2.3 or Corollary 7.2.4, and Theorem

7.2.8.

U

Corollary 7.4.4. (a) A unique solution which satisfies the estimate (3), also exists 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 all

yE H'(Q) imphesu=O). (b) The combination of inequalities (3) and (ib) results in luli

+

for the solution of (2a,b), if f is defined by (is) with

H'/2(f).

(7.4.3')

IIWIIff—1/2(r)J

g

(H'(Q))',

PROOF. (a). According to Theorem 6.4.8b, H'(O) is compactly embedded in L2(37) so that the statement of Theorem 7.2.14 can be transferred. If A = 0

is not an eigenvalue then L1 6.4.12).

L((H1(a))', H'(Q)) holds (cf. Theorem U

To find out which classical boundary value problem corresponds to the variational formuiation (2a,b), we assume that (2a,b) has a classical solution u E Hm(Q) 11 Further, v E can be assumed also (cf. Lemma 6.5.ib). For reasons of simplicity we limit ourselves to the case m 1. Under the assumption E C1 (1 the following general Green formula is applicable:

154

7

a(u, v)

VariatIonal Formulation

=

+

+

J

{

=

ij=1

+ aoouvl

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. We define 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 out and we obtain Lu = g as in Section 7.2. By this the identity a(u,v) = f(v) = f1pvdi for all v E H'(i7). According to Theorem reduces to runs over the set H'/2(I') if v runs over H'(O), so that we have 6.2.40b,

dl'

fr %b(Bu —

0 for all

H'/2(fl; thus Bu =

This proves

Theorem 7.4.5. Let F be sufficiently smooth. A classiosJ solution of the problem (2a,b) with / frvm (is) is also the claaaieal 8olution of the boundary value 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 v may assume arbitrary boundary values. Note that the bilinear form determines L 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.

00,1 be bounded and a(.,.) be Let Then the statements of Theorem 7.2.14 holds with Hm(fl) instead of

Theorem 7.4.7.

Example 7.4.8. Let (1 a(u,v) =

0°" be a bounded domain. The bilinear form 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 only H'(f?)-coercive. As is known from Theorem 3.4.1, the Neumann boundary

7.4 Natural Boundary Conditions

155

value problem for the Poisson equation (i.e., for c = 0) is not uniquely solvable. 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, the Neumann boundary value problem a(u, v) = 1(v) (v II'(fl)) is solvable if and 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 reads f(1) = g dx + f1. çodl' = 0 (this is Equation (3.4.2), in which I should be

replaced by —g).

Remark 7.4.9. While the classical formulation of a boundary condition such as ôu/ôn = 0 requires conditions on the boundary 1', the problem (2a,b) can be formulated for arbitrary measurable 0. In the following, we proceed in the opposite direction: does there exist, for a 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 corresponding variational formulation? This would mean that the freely prescribed boundary operator 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, .. . , (ci. (&2.lb,b')). Here one can also write B = 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)

(ac) p to the equivalent scaled equation (cf. (5.1.ld)). By passing from Bu oBu = 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 already satisfies (n, b) = (n, An).

Remark 7.4.10. Let m = I. Let (8) hold. Let A(x) be the matrix A

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). Then there exists a bilinear form a(.,.) on H1(s?) x H1(0) such that to the variational problem (2a.,b) corresponds the classical formulation Lu g in 0,

Bu=p onf.

7 VariatIonal

156

Fonnulation

PROOF. The bilinear form we seek is not uniquely determined. We will give two 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),. . be arbitrary E The differential operator

:=

+

—

+

maps every u C'(S2) into zero: Lju = 0. Thus the operator L can be replaced 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 absolute term (7.4.9)

If = 0, the choice of = is successful. Otherwise, two other options are 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 in this method consists in the need to construct a smooth continuation on £1 of the 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 operator associated 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 is

orthogonal 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

L2 :=

E

8

8

8

7.4 Natural Boundary Conditions

157

Again, L,2u = 0 holds for all u E C2(Q), since —

(a,

+

—

=

+

—

= 0.

L2 without changing the boundary value problem. The coefficients belonging to L + L2 result in a boundary operator B whose derivative terms read + = t(A + A')n]TV. By (dnT — ndT)n d = b — b , since the construction of A' we have A'n = (n, n) = 1 and (d, n) 0. Since we also have An = b, the derivative term L + L2 does not change the in B gives bTV as desired. The transition L absolute 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 orthogonal to n. From this the boundary operator T := dTV is the derivative in a tangential direction if n = 2 [resp. in the tangent hyperplane if n 3J. if I' of v is sufficiently smooth then the restriction H'/2(f'). By Remark 6.3.14a, one can show that T E L(H"2(r'), H-'/2(f')). is well-defined for H'/2(f), in Since T(ulp) Hh1'2(f), particular, for = with v H (fl). Thus

Thus L can be replaced by L ÷

b(u,ts) := is

j(vtr)T(ulr)dI'

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. Theorem 3). Let its coefficients, as well as the boundary 1', be sufficiently smooth

(€ C'). Then the constructions of the preceding proof again result in an B7jH1(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 is applicable. As for step (ab), see step (bb). (ba) In step (ba) one adds b(u,v) Here + f0 Lemma 7.2.12 is also applicable, since the skew symmetry of A' results in 6(u,u) (bb) In the construction in step (bb), Lemma 7.2.12 is applicable analogously. This is easiest to understand in the case n = 2. Let I' be described by {(z,(s),x2(8)): 0 < s 1}. If Xl,X2 E C'flo, 1J) and =

if we have d E C'(f') for the d in T = dTV, then b(.,.) has the representation b(u,v) = where = u(s) = v(x,(s), x2(8)), r E C1((0, 1)). Thanks to periodicity, integration by parts yields in b(u, v) = — 10' (rV)'lZ ds without boundary terms, so that and

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 apply Lemma 6.5.lSc.

EIUII —

one can

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-called condition 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 Ic

equation. Example 7.4.13. (a) To the variational problem: find a(u, v) :=

C

ff2(fl) with

dx = 1(v)

J

(7.4.12a)

for all v E H2(Q), corresponds the

=

g

in Q,

=

formulation

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 written in 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)

7.4 Natural Boundary Conditions

159

(a(.,.) as in (12a)). But this does not agree with the present concept since u and v belong to different spaces.

PROOF. The equivalence of the variational and the classical formulation can be shown via integration by parts: a(u, v)

=

j

2

u dx + j

f

01)

-

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 exist a C with 0 a(ua,uo) for all a, i.e., The contradiction results from 102ua/OxiIo = O2IUaIO for sufficiently large a. Natural and Dirichiet conditions can occur together. In Example 13b, u = is a Dirichiet condition and = ço a natural boundary condition. Even in the case m = 1 both sorts of boundary conditions can occur. 0

Example 7.4.14. Let 'y be a nonempty, proper subset of I'. The boundary value problem

=

g

in 0,

= 0 on 'y,

Ou/On =

çø

in the variational formulation reads as follows: find u

a(u,v) :=

on

(7.4.13)

114(0) such that

= 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 V with 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 V corresponds 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 a weak solution. In the variational formulation (2a,b) the right-hand sides g and ço of the differential equation and the boundary condition are combined in the functional 1. In the variational formulation the components g and ço are

7

160

VarIational Fbrniulatlon

indistinguishable! u E H'(a) has fist derivatives in L2((J), whose restrictions to V do not have to make sense. That is why Bu cannot be defined in general;

cannot be viewed as an equality in the space H'/2(f) although

Ru =

H'/2(F)

(cf. Corollary 4b). But even if there Is a classical solution, the following paradox arises. Let u to (Ia) define be a classical solution of Lu =0 in 1?, Bu = on F.

One may also view u as a solution E (H1(fl))' by f,(v) := in 0, Bu = 0 on I. These equations may even be interpreted of Lu = classically in the following way: there exist E C°°(0) with 4 in = 0. Then (H1(i7))'. Let be the claseical solution of = converges in H'(O) to the above-mentioned classical solution u. Incorporating the boundary values Ru = in the differential equation corresponds to a modification of the discretised problem as used Lu = in Section 4. The difference equations Dhuh = fh in and the boundary on F,, resulted in the system of equations L,,u, = conditions uh (cf. (4.2.6b)). If one defines by in 0h, uh = 0 on = fh + then satisfies the equations D,,Ti,, = in iZ,, = 0 on Just as the functional f cannot be uniquely separated into g and and cannot be reconstructed from In contrast to the discrete case, the separation off into g and is possible, however, provided stronger conditions than g (H'(.Q))' are imposed on g (for example, g L2(fl)).

4

8 The Method of Finite Elements

In Chapter 7 the variational formulation was introduced only for the purpose of proving the existence of a (weak) solution. It will now turn out that the variational 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)

and V = H'($?) (cf. where we are thiniciug, in particular, of V = Section 7.2, Section 7.4). Of course, it is assumed that a(.,.) is a bounded bilinear 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. Now we wish to leave the differential operator hidden in a(.,.) unchanged. The RitzGalerkin discretisation consists in replacing the infinite-dimensional space V with a finite-dimensional space VN:

VNCV,

dhnVN=N 0. Then the mat'*r L in (9) is nonsmgt4ar and the Ritz-Goier*in solution

E VN satisfies

(8.1.lla)

IIU?ullv

PROOF. Lisnonsingularsince

and

thus

(Lu, u) = a(Pu, Pu) CEIIPUU?, >0 and so, in particular, Lu 0. By &ercise 6.5.6a a(.,.) is also VN-eIIiptiC with the same constant CE. From Theorem 6.5.8 there holds (ha) then results from 1/CE, i.e., HU"flv

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. Under the same assumptions the Ritz-Galerkin solution ut1 solves the following variational 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

infl=(0,1)x(O,1),

.—zlu(x,y)=1

165

u=O

The weak formulation is given by (1) with V =

a(tt,v) :=

J (Vu,Vv)dxdy =

f

JvdxdY.

1(v)

The functions bj (x, y) = sin(irx) sln(wy), bs(x, y) = sin(wx) sm(3iry),

b2(x, y) = b4(x, y) = sin(3wx) sin(3,ry),

fulifi the boundary conditions and so belong to V = Hd(i7). They form a basis of V4 := span {b1,. ..,b4}. The matrix elements = can be worked

out 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, Hence

u=

12

L'f has the components

Ui

=

u2 =

=

u4 = 8/(81ir4),

and the Ritz-Galerkin solution is then

=-.[sinwzsiniry + +

sin

+sinirzsin3iry)

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 boundary value problem be

8 The Method of Finite Elementa

166

Ou/ôn=Oon F.

= (0,1) x (0,1),

= ir2cosirx in

The solution is given by u = cos wx + const. The weak fonnulation is in terms of (1) with V = H'(s?),

a(u,v) :=

+

:=ir2Jv(x,y)cosirxdxdv.

1(v)

The boundary value problem has a unique solution in

W:={v€ V:Jvdxdv=0}. The basis functions

bi(x,y) = z are

—

b2(x,y) = (x —

1/2,

in W. The stiffness matrix L and the vector f are then L—

Ii

1/41

—2

1

9/80]'

1

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 = the differential equation only approximately:

y)/On =

For x =

1/4

12

—

120/ir2

0

and

—0.16.

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 longer V-elliptic, though it is V-coercive. That a(.,.) is V-coercive guarantees that either problem (1) is solvable or A = 0 is an elgenvalue. Even if one assumes V-coercivity and the solvability of the problem (1) one can not deduce the solvability of the discrete problem (4).

Example 8.1.13. a(u,v) := f(u'v' — lOuv)dx is 1)-coercive and a(u,v) = 1(v) := f01 gvdx (v E J.1o1(0, 1)) has a unique solution. Let VN spanned by bj(x) = x(1 — z) E V = Hd(0, 1) (i.e., N = problem (4) is not solvable since L = 0.

1).

Then the discrete

If one replaces the space V in Lemma 6.5.3 with VN, then there follows from Exercise 6.5.4

8.2 Error Estimates

Theorem 8.1.14.

167

The problem (4) is solvable for all f E V' and has a

unique 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

EN

1}

forallu€ VN.

(8.1.12')

and

= 1/EN

(LN as in (lOa)).

(8.1.12")

Warning: The condition (12) for VN does not follow from the analogous condition (6.4.5a) for V. See, however, Theorem 8.2.8.

The requirement (12") guarantees the existence of L', but it does not which is the deciding say anything about its condition, cond(L) = factor for the sensitivity of the system of equations Lu = f. For example, if (i = 1g.. .,N) then one one chooses for a(u,v) := u'v'dx the basis = iii (i+j —1). The conditioning obtains the very badly conditioned matrix of 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 on is, on the other hand, directly different sets. The Ritz-Galerkin solution,

comparable with is. One can measure the error due to discretisation With IIu — u?JUv or with

—

uNItu.

Let is be the solution of (1.1): a(u,v)

f(v) for v E V. Suppose — by

chance 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 if We shall now show: The "closer" that is is to VN the smaller is the discretisation error. is E

8 The Method of Finite Elements

168

Theorem 8.2.1. (Ceo) Assume (1.2), (1.3), and (1.12) hold. Letu E V be a solution of the problem (1.1), and let uN E VN be the Ritz-Oatev*in solution of (1.4). Then the following estimate holds: III'

— u"IIv

(1+ Cs/CN) iflf

(8.2.1)

— wIly

WEVN

with Cs from (1.2) and from (1.12). Note iflfwEyN IIu—wlIv is the distance of 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 particular

for all V E VN. Since we also have a(uN, v) = f(v) for v E Vpj, it follow8 that

forall VEVN. For arbitrary v,w E VN with lIvIlv = —

= a((t/'

w, v)

1

(8.2.3)

we can therefore conclude

— tij + [u — w], v)

= a(u — w, v)

and

w,v)l 5 From (1.12') we then obtain —

— wIIvIfvIlv = CsIlu

—wily

—

why.

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 use of a sequence of subspaces V that converge to V In the following sense:

Theorem 8.2.2.

:=VN4 c V(iEt4) be a sequence of subspaces with lim

= 0 for all u

V.

Asumethat(1.12)holdswitheN4

(8.2.4a)

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.

169

8.2 Error Estimates

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: d(u,

ilu — u'IIv (1 +

—.0.

(ab) We now wish to show that (4a) follows from (4b). The inclusion ¾-i C Now d(u, will be a null sequence if for each implies d(u, V1) d(u,

e > 0 there exists an i such that d(u, is, for each u E V and e > 0, a w E U,

e. From the assumption (4b) there with e. Therefore we — wily have w€ for an i N. That d(u,V3) e then proves (4a). — why 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 the operator L:V —. V1 associated to a(-, -) is closed. Foreachf E Wthereisau E V 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 in V' and f,, = Lug. The Cauchy E W be a sequence with —. convergence Ii!" — shows —0 so that ti/v the limit E V exists. The continuity of L L(V, V') shows that = 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 the Rieaz isornorphism (ci. Corollary 6.3.7). Set v := E V. It follows that

f

f=

f(v)=

—(f,f)v' =

for all u E V. Let E satisfy v) = 0, i.e.,

0

1,

be the Ritz-Galerkin solutions. They must also a(u',v) — 1(v) =

We split up v into may

+

guarantee that

I

=

1.

where v' Vi. By (4a), —.0. This shows

with

v in

place

of u, one

/(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 amounts to 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). •

170

8 The Method of Finite Elements

>o from

Corollary 8.2.3. The requirement (1.12) with CN,. CR if a(.,.) is V-elliptic: a(u,u) is satisfied with

Theorem 2

is dense in V.

Exercise 8.2.4. Show that (4a) implies that

Let QN be the orthogonal projection onto VN (cf. Exercise 8.1.6c). The V V', QN:V' —' factors L:V V

(8.2.5a)

C V.

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 projection. It sends the solution u of the problem (1.1) to the Ritz-Galerkin solution uN VN: uN = SNU. Assuming (1.2) and (1.12) we have (8.2.5b)

IISNIIv+-v CS/CN.

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

(8.2.5c)

V.

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 the QNLu. Conversely one may right-hand side f := QNLU E V', i.e., SNU = argue similarly, and so show the equivalence of the definitions (5a) and (5c).

projection. Inequality (5b) follows from IISNuIIv proof of (1.11)) and IILuII'v CsfIuIfv = from (1.2) (cf. Exercise 8.1.9).

(cf. the with C8

Remark 8.2.7. Let a(.,.) be V-elliptic and symmetric. IIIvIlIv := a(v,v)'/2 is 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 be shown 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 = SNU is 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 all

v

VN (cf. Exercise 8,1.lOb).

The condition (1.12), with e,, e > 0, is difficult to prove, except for V-elliptic bilmear forms. However, the following theorem shows that this condition does hold for subepaces approximating well enough.

Theorem 8.2.8. Let the bilinear for,n a(.,.) be V-coercive, where V C U C V' is a continuous, dense, and compact embedding. Let Problem (1.1) be solvable for all f V'. Assume that (4a) holds for the subspaces C V. For large 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 disadvantage of the general Ritz-Calerkin method becomes apparent. The matrix L is in general full, i.e., L11 0 for all i,j = 1, . ., N. Therefore one needs N2 integrations to obtain the values of = o(b,, = fe,. .., whether exactly or approximately. The final solution of the system of equations Lu = f requires 0(N3) operations. As soon as N is no longer small the general Ritz-Galerkin method therefore turns out to be unusable. A glance at the difference method shows that the matrices Lh which occur there are sparse. Thus it is natural to wonder if it is poesible to choose the basis {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 model problems such as the one in Example 8.1.11. Instead we shall base our further considerations on .

.

Remark 8.3.1. Let the bilinear form a(.,.) be given by (7.2.6). Let B be the interior of the support ri B, =

of the basis function i.e., := A sufficient condition that ensures = a(bj, b1) = 0 is

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

172

8 The Method of Finite Elements

it from the desired goal: one defines partitions of Q into small pieces, the socalled finite elements, from which the supports of b4 are pieced together. As an introduction let us investigate the one-dimensional boundary value problem

u(a)=u(b)=O.

—u"(x)=g(x) fora 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 u change continuously between elements. The ansatz but also its derivatives functions must therefore be piecewise smooth and globally in C'(.0). introduce the one-dimensional biharmonic equaAs a model problem tion

u""(x)=g(x) forO m 9.1.1 The Regularity Problem The weak formulation of a boundary vaLue problem

Lu—g lull, Bu=p on!'

(9.1.1)

uEV, a(u,v)=f(v) forallvEV

(9.1.2)

as

was, in Section 7, the basis upon which we were able to answer the questions of existence and uniqueness of the solution. Here, by existence of a solution we understand the existence of a weak solution u V. The error estimates in Section 8.4 made it dear that the statement u E V is not enough. Under this assumption we can only show 0. The — uhllv more 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 u

112(Q) n V. The assertion u E 112(Q) or, more generally u E H'(s?), is a statement 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 make the proof ideas clearer, let us sketch the proof of inequality (4) below for the

Helmholtz 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. Weshall prove (4) by induction for 8 = To this end we take the derivative of the differential equation with respect to x, + = If I then H3(fl) and the equation by the induction as+v= 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) has been shown for S. Step 2: 5? = = lRx(O, oo). As above, we can obtain the estimate since alsosatisfies = and the boundary condition

9.1 Solutions of the Boundary Value Problem in H'(Ii), s > m

=

0

implies

on 1. This is not the case for u1. But

209 E

E

This property, however, results from the differential u— — E equation, Step 3: Let 11 be arbitrary, but sufficiently smooth. As in Section 6.2.1, Ills decomposed into (overlapping) pieces il, which can be mapped into 1R2 or

show

f

is a partition Correspondingly one splits the solution u into Exsu of unity). Then the arguments from steps I and 2 prove inequalities for which together result in (4). Note that only a sketch of the proof was given. Some of the steps of the + proof are incomplete. For example, might not the equation = have a solution in 1R2, E L2(1R2), which does not belong to H' (1R2) and +v= hence does not coincide with the solution v H'(1R2) of In the following, always let s m. The boundary value problem (2) with of problem is sald to be H3-regular if each solution u V= and satisfies the estimate (2) with f H3_2m(Sl) belongs to

(9.1.3)

+

IuI.

If L is the operator associated with a(.,

then it is also said that L is

H'-regular. Remark 9.1.1. (a) H's-regularity always holds. with (b) Let the variational problem (2) have a unique solution u I-rn for all / E H-m(Il). If the boundary value problem is H3< C01f IUIm satisfies the inequality regular, then the weak solution of (2) with f luIs

(9.1.4)

Cjf13_2m.

(c) Let L be the operator associated with a(.,.). (4) is equivalent to H8(Q)) and (4'): IlL

—1

s + n12 n/2. Then the weak jE solution 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

Let the conditions (6) and Ut Then the weak solution of problem be satssfied for all k E sotssfled in particular if f belongs The conditions (7) belongs to constant. and the coefficients to

Corollary 9.1.7. Let a(.,.) be

/

Let

Theorem 9.1.8. Let a(.,.) in (5) be

00. As in Section 7, we The halfspace limit ourselves to the following two cases: either a Dirichiet problem is given for arbitrary m 1, or the natural boundary condition is posed for m =1.

Theorem 9.1.11. (Homogeneous Dirichiet problem). An analogue to Theorem 3 holds for the Dirichlet problem:

a(u,v) = (f,v)0 for all V E

UE

(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.

The nonnl•I.

Lemma9.1.12. is equivalent to

:=

(9.1.15)

+ loI=m

PROOF. The relatively elementary case s m is left to the reader. For 0

0.

if t$I = m. Then the

must belong to of the problem

/ f(x)v(x) dx

belongs to

for dlv

E

and satisfies the estimate (23).

The condition of can be replaced by that of uniform ellipticity (7.2.3) (cf. Theorems 7.2.11, 7.2.13). The statement of Theorem 21 cannot be extended to 8 1/2 since then u Hr'(Si) would contain another boundary condition.

9.1 Solutions of the Boundary Value Problem in H'($2), 8> m

223

(cf. The proof of Theorem 21 uses an isomorphism R = R related to and also between Hr(fl) and proof of Theorem 8) between such that the form b(u,v) := a(Ru,Rv) is and It is necessary to prove that b(u, v) := a(u, R2v) is also

/

We know f H_rn+8(.11) implies R2f E a(u, v) = v)o is also a solution of a(u, = b(u, so that u E follows.

Eath solution of = (1,

(f,

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 all 0 t 1 Convex domains in particular belong to C°", but permit stronger regularity statements than Theorem 21. .

Theorem 9.1.22. (Kadlec [1]) Let I? be bounded and convex. Let the bilinear form (5) be Lipschitz-continuous:

Let the coefficients of the principal part be

for all

=

=

1;

for the remaining ones let the following hol&

for all a,f3,-y withy

E

Then every weak solution u E a(u, v)

of the problem

clx for all v E = j f(x)v(x)

L2(Q) belongs to

with f

IaI+I/31 0 and h > 0, an M-matnx Lh. For fixed e, the scheme has consistency 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 obtains the Taylor series

= Lu — { The

0(h) term hIcjIu

+ Ic2luyp}h + 0(h2).

(10.2.7)

+ hIc2Iu..fl, is called the numerical viscosity (or

the numerical ellipticity) since it amplilles the principal part. A second remedy consists in replacing the parameter e by a discretisation If one chooses using Ch :=

hIcjI/2, hlc2I/2} or Ch

e

+

(10.2.8)

max

then the symmetric difference method

-1 =

—1

—1 +

4

1

02

0

cj

(10.2.9)

C2

leads

to an M-matrix. It is true that the convection term has been discretised

to a second order of consistency, but the error of the diffusion term is O(eh —c),

which is relevant in practice, amounts to

which, for the case h > 0(h). Instead of (7) one has Lu

The difference

—

—

—

c)zlu + 0(h2).

(10.2. 10)

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 to the 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 difference methods.

Exercise 10.2.9. Show that

10 Special Differential Equations

252

(a) Linear finite elements on a square grid triangulation applied in the case of Equation (4) give a discretisation that is identical with the difference method —1

L,, =

4

—1

+h

—1

—1

0 —1/3 —1/6

—1/6 0

1/6 1/3

1/6

0

(1O.2.lla)

(cf. Exercise 8.3.13). (b) For bilinear elements (cf. Exercise 8.3.17) one obtains

=

1

—1

—1

—1

8

—1

—1

—1

—1

h

+— 12

0 0

+1

—4 —1

a

+1

1

+4

(c) One-dimensional linear elements for —eu" + u' = difference formula

=

th' [—1

2

—11+1—1

0

.

(10.2.llb)

f lead to the central (10.2.llc)

1).

The Exercises 9b,c show that finite-element methods correspond to central difference formulas, and thus can equally well lead to instability. The method of artificial viscosity corresponds to the finite-element solution of the equation (c, grad *4 = for appropriate As in Exercise 9b one may show

f

Remark 10.2.10. If one sets

:= IciIh, Ic2Ih} and uses bilinear elements then the discretisation of Equation (1) leads to an M-matrix.

On the other hand the matrix (ha) has different signs in the sub-diagonal and in the super-diagonal, so that it is not possible to have an M-matrix for any value of €h. The analogues of one-sided differences are more difficult to construct. One

approach is to combine a finite-element method for the diffusion term with a (one-sided) difference method for the convection term (cf. Thomasset [1, §2.4]).

A second possibility is the generalisation of the Galerkin method to the

Petrov-Galerkin method, in which the discrete solution of the general equation (8.1.1) is defined by problem (12): Find u E V,1, so that a(u,v) = 1(v) for ally e W,,

(cf. Fletcher [1, §7.2), Thomassetll, §2.2]). Here we have

dimVh = dimWh (but in general Vh

Wh).

(10.2.12)

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 boundary operators. A solution e of (1) is called an eigenfunction if e 0. In this case, A is the elgenvalue associated with e. As in Section 7, one can replace the classical representation (1) by a variational formulation, with a suitable bilinear form a(., .): V xV —+ IR taking the place of {L,B,}: Find e

V with a(e,v) = A(e, v)0

for all v

V.

(11.1.2a)

dx is the L2($?)-scalar product. Strictly speaking one ought = where U is the Hubert space of the Gelfand triple to replace (., .)o by (., V c U c V1 (cf. Section 6.3.3). But here we limit ourselves to the standard case U = L2(fl). The adjoint eigenvalue problem is formulated as (ti,

Find e E V with a(v, e) =

for all V E V.

(11.1.2b)

Definition 11.1.1. Let A E e

C. By E(A) one denotes the subepace of all E(A) is called the eigenspace V which Satisfy Equation (1)[ resp.

for 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 compactly embedded (for example, let V = with bounded.1?). Let a(.,.) be Vcoercive. Then the problems (2a,b) have countably many eigenvalues A E C which may only have an accumulation point at oo. For all A C we have dimE(A) = dimE(A) 0 in C A and have and holomorphic. Caucby'a integral formula says (L —

)J)' = 2ws

—

A)'(L — CI)'dC

for all A E Ke(A).

From this one infers —

ES

C)IIvv

A')

(ci. Exercise 4(i)). Thus, w(A) cannot

i.e., w(A) min{w(():C E

For wh(A) the conclusion is the same.

aseume a proper minimum in

The converse of Theorem 8 is contained in

Theorem 11.2.10. Let (2a-c) hold. Let A0 be the eigenvalue of(1.2a). Then Ah = A0. of(la) (for all h) stich that there exist discrete eigenvalnes

PROOF. Let e > 0 be arbitrary. Aceording to Theorem 11.1.2, Ao is an isolated eigenvalue: >0 for 0 < IA — Aol e (e sufficiently small). is compact, we have that is continuous and Since positive. Because of (5) and w(Ao) = 0 one obtains is IA — Aol = for sufficiently small h Wh(A)

Cw(A)

—

Cp, —

> v1(h)/C

must have a proper minimum iii Ke(A0). Thus By Lemma 9 the minimal value is zero. Thus there exists a Ah Ke(A0) which is a discrete eigenvalue, Wh(Ah) 0. for all A E

I

The convergence of the eigenfunctions is obtained from Theorem 11.2.11. Let (2a—c) hold. Let e1'

be discrete eigenfunc= 1 and A0. Then there exists a subsequence e1" which converges in V to an eigenfunction eE(A0): Lions with

e

E(Ao),

— dlv

0

(i

oo),

lIellv

1.

PROOF. The functions

are uniformly bounded in V. Since V C L2(fl) is compactly embedded (ci. (2b)), there exists a subeequence which in L2(Q) to an e L2(Q): lie —

—. 0

(i

oo).

(11.2.7a)

11 Elgenvalue Problems

260

We define z = Z(Ao)e, = Theorem 8.2.2 there exists an h1(e) > 0

1 holds if and only if there exists a solution v V for (L—AoI)v = e. According to 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) e and e can be normalised so that

(e,e')o = 1.

(11.2.8')

261

11.2 FnIte Element Dscretiaation

= {v' e V':(v',e')o We define V := {v E V:(v,e')o = 0}, be the dual norm for = 11 liv. Fbr Problem (9):

For f

V', find u

(f,v)o for all V

aA(u,v)

O}. Let

(11.2.9)

E

one defines the variable corresponding to (3b)

ml

:=

sup

(11.2.10)

laA(u,v)i.

uEc'

Lt"IIv=l 11v11v1

Lemma 11.2.14. Let (2a,b), (8), and dim E(Ao) = 1 hold. Then there exists e. Problem (9) has exactly C > 0 for all IA — an e > 0 such that one solution u E V with ilulIv Ill Ilv'/2z(A), if

> 0.

C. x PROOF. Let L: 1' be the operator aesociated with a(., .): e (e sufficiently small) (L — AI)u = f has a unique solution For 0 < — Aol u E V. From f V' follows

0=

= ([L — A.1]u,e')o = (u, [L — M)'e')o =

—

X)(u, c')0,

— AI)': —+ Thus there exists as the restriction of For A = Ac Problem (9) has a unique solution cording to Theorem 6.5.15c. Then there exists (L — Al)-1 for all A

i.e., u E (L —

)J)-' to V' C V'.

According to Remark 5 (with V instead of V) , c2i(A) must be positive in Kd(Ao). The continuity of proves C > 0. In analogy to (4) one results from The bound by has llullv =

I

Exercise 11.2.15. Show that

Lemrnk 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 there exists 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" E V,,: (v", = 0} and the variable

inf

sup IaA(u,v)l.

UuIIv=1

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. Then

PROOF. (a) First statement: there exists h0 >0 and C such that

min{iiv+ae"llv:aEC}IIvIlv/C

262

11 Elgenvalue Problems

The proof is carried out indirectly. The negation reads: there exists a se0. = 1 and with C, hj —4 0, + quence E —* ? in L2(a). Evi—. a', Thus there exist subsequences with must have the limit w' = liznwj = v' + a'e' dently, := v, + = 0, it follows that w = 0, in L2(a). Since limIIw4IL2(n) lim(vj,e*h)o and thus v' = —a'e'. From 0 = = (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 For sufficiently small h, Ah is the only discrete eigenvalue in Ke(A0). In the proof of

From Part b follows that > 0 for A E According i := to Exercise 12a, aA(u,v) = (f,v)o (v E is solvable for each 1€ and all A E Ke(A0) such that = 0 is excluded. Lemma 9 shows that

I

Exercise 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 small then 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) and dimE(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, one has n11e—zh

For all v

Vh we have

IIvCId(e,Vh).

(1l.2.lIa)

11.2 Finite Element Dlscretisation

v) =

= so

(.\o

—

(Ao

.&)(e, v)0

—

— (Ah

263

v)0

—

—

)th)(e,v)o + (Ah —

ehv)0 + (Ah —

—

that

=

+(Ah —.p)(e —z',v)o for ally

—

Vh.

(11.2.llb)

According to Lemma 16 we have f(e, — eh can be scaled so that = 0. (lib) corresponds to Problem (9). Lemma —

+ lie

Wh(Ah) 'Cu)10 —

Together with (1 la)

one

Theorem 11.2.19. =

1.

Let

14 proves

z"Iivi C'[

+ d(e, Vh)J.

—

obtains the statement.

Let

(2a—c) hold. Let

e

lAo —

U

Ao be

an esgenvalue with

E'(10), itchy = 1, (e,e)o =

E(A0), e there exist discrete eigenvalues Xh

dimE(Ao)

—

> 0 for sufficiently small h.

(8) and 1.

Then

with

)'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 exist by Theo—i

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 = follows

—

exists an

— e,

e — u")

—

— eilv[11e

Ahi

(Ao

e,

—

+

—

—

AhiJ.

e)o]

By Lemma 18 there

Eh(Ah) such that

P10 —

C'C"[(Ao

From this one obtains (12)

—

Ahl + d(e, Vh)J[I)1o — Ahi

with

C = 3C'C"

for

+ 2d(e*, Vh)J.

sufficiently small h, since

U

d(e,Vh)—40.

Theorem 11.2.20.

asswnptions of Theorem 19 there exiet for

e

elgenfunctions

E(Ao), e' E

Under the E()10) discrete

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 second estimate in (13) follows analogously. U

11 Eigenvalue Problems

264

In the following, let V C H'(fl). Theorem 11.1.5 proves C H14'(37).

C

(11.2.14a)

Also, let (14b) hold (cf. (8.4.10)): for all u

d(u, Vh) Ch'IIUIIHl+.(n)

E

U

(l1.2.14b)

be the cigenvalue Corollary 11.2.21. Let (2a), (2c), (14a,b) hold. Let with(8) and dimE(Ao) = 1. Then there exists Ah,e" Eh(Ah), esh E

such

that

IAo

— AhJ Ch2',

Ch',

lie —

lie5 — e*hIlv CM.

(11.2.15)

Occasionally eigenfunctions may have better regularity than is proven for ordinary boundary value problems. For example, let —& = Ac be in the rectangle (1 = (0,1) x (0, 1) with e= 0 on V. First, Theorem 11.1.5 implies thus e C°(17) (cf. Theorem 6.2.30). Thus e = 0 holds e E H2(.Q) ii in the corners of .0. AccordIng to Example 9.1.25 it follows that e (j7)

for s 1, one must replace C'h2 by

11.2.4

Complementary Problems

In Problem (9) we have already encountered a singular equation which neverthelees was solvable. In the following let A0 be the only eigenvalue in the disc Kr(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 the following 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 of E(Ao):

11.2 Finite Element Discretisation

f

,i.e.,

(f,e')a =0

for all e' E

265

(11.2.16b)

In the case of A = A0

, with it, it + e (e E E(Ao)) is also the solution. The uniqueness 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 in Kr(A0). Then (16a,b) has exactly one solution it for all IA — A0I r which satisfies (16c). There exists a C independent of f and A such that IIuIIv CII! liv' PROOF. This follows from Lemma 14 in which the assumption E(Ao) = 1 is not necessary. U The finite element discretisation of Equation (16a) reads:

Find U"

V,,

= (f,v)o for all v E Vh.

with

(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,,

with f(s) .1.

(11.2.18a)

(11.2.18b) (11.2.18c)

Exercise 11.2.24. Show that is equivalent to: Find it" v) = (f(h), v)o for all v V,, with fl

SA(U1',

=

with 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 only eigenvalue in Kr(Ao). Then there exists an h0 >0 auth that) for all h ho and all A Kr(A0), the Problem (18a,b) has a unique solution it" u"(A) which satisfies the additional conditions (18c). Further there exists a C independent of h, A, and 1(h) such that CII! If E'(A0), f from (16b) need not satisfy condition (18b). If Eh(Ah) and E with (e",e"')o = 1 are known, one can define

f satisfies (18b) since Qh represents the projection on

(11.2.19)

(As)'.

11 Ezgenvalue Problems

266

=

=

Exercise 11.2.26. Letu 1.

1,

= 1,

= 1. Show that

d(u,

E

v" J..

+ llullvinf{11e8 —

= i hold. Let dim E(Ao) = dim Let h be sufficiently small such that Ao be the only eigenvatue in (following Remav* 25) the Problem (18a-c) is solvable. For the solutions u and of (i6a-c) and (iSa—c) the ermr estimate

Theorem 11.2.27. Let (2a-c),

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(-, .). Here one 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 is estimated with the aid of Exercise 26, with llullv il/ liv' being added.

I

Corollary 11.2.28. If 1(h) is defined by (19), then inequality (20) becomes flu —

PROOF.

C'[d(u, Vh) + hilly' inf{hle* — e*hIlv: e' E E'(Ao)}J. (i2.2.2ia)

—lily' Cj(f,e")ol = Cl(f,e Cflf liv' hie* —

Corollary 11.2.29. If additionally the assumptions u E (14a), arid then (21a) yields the estimate d(u,Vh) S Chitihi÷1 hold forts E 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 the problem OA0(v,w) = (v,f)o has a solution w I E(A0) for all v E V. The assumption of regularity yields w E H2(I?), 1w12 Cl/k such that E Vh exists with I. 1w — Chlwi2 C'hIflo. The value

11.3 Discretisation by Difference Methods

267

=0—aA0(e5,w")

=

(Ao

—

Aa)(e",w")o

—

=

(cf. (15)). From

be bounded by

can

= (f, f)o

—

a.x0(f,w) = a.x0(f,w — w") + (Ao

—

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 formulation

in the next section. For reasons of completeneas we point out the following transformation.

Remark 12.1.5. Let n = 2 and thus u = exists a so-called stream function

(01,02). Because

with uj =

divu = 0 there

02 =

12 Stokes Equations

278

= Insertion in Equation (lat,2) results in the biharmonic equation = 0 on F. This 8f2/Ox1 — Ofj/0x2. The boundary condition (3) means =0 and 845/& =0 on F where ô/8t is the tangential is equivalent to = 0 implies = const on P. Since the constant may be derivative. =0 on P. chosen arbitrarily, one sets =

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

(n-fold product).

x

A corresponding definition holds for H'(S?), H2(Q), etc. The norm associin the following. will again be denoted by ated with According to Remark 12.1.1 the pressure component p of the Stokes problem is not uniquely determined. In order to determine uniquely the constant in p = fi+const, we standardise p by the requirement p dx =0. That is the reason 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 that u and p are classical solutions of the Stokes problem (1.2a,b). Multiplication of the ith equation in (1.2a) with Vj E Cr(s?) and subsequent integration implies that

j =

f

d(x) =

+

{(Vui(x), Vvj(x)) —

dx

dx for

E

(12.2.la)

1 0. Let u with be chosen according to Lemma 13. We continue ti by u = 0 onto Ut . For the restriction on z = 0 we have according to Theorem 6.2.28

CJtzj)1

)IILZ(R)

Let u1(O,

= 1 for 0 < y < and x(v) and Ixto = we have

= Let

.f?(1

0 otherwise. Since ui(0,y) =

lui(O,.)IoIxIo

= {(x,y) E > 0} {(x,y):x = 0,0< y < Because w =1 in and because divtz w we have

=

288

Equations

12

Iw(x)12 dx

1/2

vi div u dx

=J =

=

J

dlv u dx

n) dE' = j (u,

_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 regularity 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 and p be the (weak) solution of the Stokes problem (22) with I E Hk(fl), 9 E Hk+l(a) fl for k E t4 U {O}. Then we have u E Hk÷2(.Q) fl and there exists a C depending only on £2 such that

fl

pE

PROOF. Cf.

(12.2.23b)

Ia + IgIa+i].

+ LPIk+1

El, Chap.llI, §5]

U

In analogy to Theorem 9.1.22 it is sufficient to £2 in order to obtain u E H2(a) and p H'(s?) from f

the convexity of 112(fl).

Theorem 12.2.19. (Cf. Kellogg-Oaborn [1]) Let £2 C JR2 be bounded and convex. 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

CIfIo.

1u12 +

For the more general problem (22) with g the solution satisfies 1u12

1ff

+ lpli

L2(a) and g E

(12.2.23c)

0 in a convex polygonal domain

+

If

(12.2.23d)

IIJ(a). Here, HJ(fl) is the sub8pace of H'(a)

with the following (stronger) norm:

:=

+

with

IaI=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 0). Vo is a closed As kernel for the mapping B = In the i.e., again a Hubert space for the same norm subepace of following we investigate the problem

has been defined in (15) by V0

V0 c

{u

—div

.

Find u

V0

with a(tz,v) = f(v)

(12.2.24)

for all v E V0,

f0(Vu(x), Vv(x)) dx. Problem (24) has the same form as where a(u, v) . , n), only equations the weak formulation of the = (i = has been replaced by Vo. here

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 in depends only on the diameter of 4'?. In particular, problem a(u, u) for 1€ V01. (24) has a unique solution u E V0 with

PROOF. The

of a(.,.) (ci. Lemma 12) carries over to Vo C (cf. Exercise 6.5.6a). This implies the other statements (cf. Theorem

I

6.5.9)

Theorem 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 component 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 that belongs to Vu is H1(.O). The part of

gj(vj)

results in E VI with

a(u,vj) for all vj. E V1 =

E

(V0)1.

Theorem 14 proves the condition (2!'), which results in the bijectivity of B V (cf. Lemma 13).p B'(fj. —gj), by definition, satisfies

= f1(v1)

b(p,vj) =

=

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)

For arbitrary v E V1 E V1,

=0

for all V0 E V0.

to be split into v =

+ V1 with

E V0 and

one obtains

b(p, v) = b(p, vo) + b(p, vI) =

fj(vi)

—

a(u,vj.)

f(vj.) — a(u,v) + a(u, v0)

12 Stokes Equations

290

= f(vj)—a(u, v)+f(vo) .= f(v)—a(u,v), because u E 0 for all to E satisfy the variational formulation (3a—c) of the Stokes problem.

by (24). Since also b(w,u) =

V0,

u and p

Note that Problem (24) is solvable for all bounded domains although Problem (3a—c) depends more sensitively on Al (cf. Counterexample 16). Theorem 21 shows that only the component p has a domain-dependent bound ClfoH Gill_i holds for all .0 C KR(0). Ii'io

function 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 on (cf. Exercise 8.3.14) OT and outside. The map -. T to a general T 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 there dzdy. exists a C >0 independent of h such that dx dy

295

12.3 Mixed Finite-Element Method for the Stokes Problem

We set

linear combinations of the linear elements E H0t (fl)

and the bubble functions for f'

Wa: linear elements E

Vh :=

For the side condition have

(12.3.14)

E

see Section 8.3.6. Since ü..p e H0t(fl), we

C

C

be the quasiunifonn tnangulatson on a bounded and Wh be given by (14). Then the stability conpolygonal domain £7. Let and that the dition (8) is satisfieaL Under the further conditions that Si E Pois8on problem be H2(Si)-regt4ar the Brezzi condition (6b,c) also holds.

Theorem 12.3.11. Let rh

PROOF. (a)

On every T E Ta, Vp is constant:

an arbitrary p We set

Vp = v := >

E

so that

ii := v/lyle

(ü.jr. bubble function (13b) on

= 1. Exercise 10 yields

b(p, v)

J (Vp, v) dx = TEm

+

= 0

f

T

Since lVplo and lpI' are equivalent norris on the subapace H'(Q) fl

it follows that lb(p,v)l

and

C'lplilVpk/jvk. In a

similar way one shows that lvlo

C"(VpIo and obtains 43 := C'/C" independent of h. The left-hand side in (8) is

with so

that

(8) follows.

(b) (9a) is satisfied for Vh (cf. Theorem 8.4.5). The same holds for the inverse estimate (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 in Brezzi-Pitkäranta [11.

12 Stokes Equations

296

12.3.3.3 Stable Discretisations with Linear Elements in Vh If 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 two different triangulations and Th. By decomposing each T rh as in Figure 1, through halving the sides, into four similar triangles, one obtains Th12. We define:

: linear elements for triangulation Th/2,

V,, c

Wh c

H'(fl) fl

(12 3 15)

: linear elements for triangulation

or

V,, c

:

quadratic elements for triangulation Th,

W,, C H'(Q) fl

linear elements for triangulation r11.

FIgure 12.3.1. Theorem 12.3.12.

Th

'12

16

and Th/2

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 of the 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 be the 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 its value by pu,. The mid-pomt of is a node of rh/2. We define the piecewise linear 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.

The sum E7 extends over all interior sides of Tj,. In

(12.3 ITh)

U T2-, we have

12.3 Mixed Finite—Element Method for the Stokes Problem

297

(Vp,

so that

J (Vp, (

I

dx =

T17UT2.1

Ch2>1pt1.y12. are the sides of T

Th (r5 is quasiunifonn!), then

From this one infers that b(p, v) 11, and finishes the proof analogously.

dx

i IVpIo, as in the proof of Theorem

(b) In the case of quadratic elements given by (16) one has the same nodes as in (a) (cf. Figure 8.3.8a and Figure 1). Use (17a,b) to define the quadratic function V E Vh and carry out the proof as in (a). U

FIgure 12.3.2.

Ti.,, Ta.-,

12.3.3.4 Error Estimates In the following, the condition (8.1.12) should be replaced by the stability condition (6a.-c). In the place of the approximation property (8.4.6') we now have the inequalities inf{Iv —

inf{Ip —

Vh}

E W,j

Chfi42 for all V E V =

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

12 Stokes Equations

298

the Brezzi condition (6a.-c) and have the solution

= (v's, w"). Then there

exssts a varsable C independent of h such that —

uhjlx Cinf{IIu —

(12.3.19)

Xh}.

PROOF. The Brezzi condition (cf. Theorem 6) yields S lix' for all 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 8.1.16 is 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 statement (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 E (ci. Theorem 12.2.19). For the suhspaces

p

and Wh C l€t the Brezzi condition (6a—c) and the approximation conditions (18a,b) be Then the discrete solution satisfies the estimate C

lu —

+

—

P"lo

C'hflul9 + Ipli].

PROOF. Combine inequalities (20) and (18a,b).

(12.3.21)

N

Using the same reasoning as in the second proof for Theorem 8.4.11 with

c(..) instead of a(.,.) one proves

Theorem 12.3.15. For each f

L2(a), g E

fl H'(fl) let the Stokes

problem ÷ Vp f, —divu = g have a solution u E p H1(Q) fl with 1u12 + C[1f10 + 1gb]. Under the asstImptlon8 of Theorem 14 we then have the estimates —

+ [p — p'i—i

C'h[Ju(i + Iplol,

(12.3.22a)

—

÷ [p

C"h2[Iu12 + [p1']

(12.3.22b)

—

p41.i

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 —

+

—p"Io

+

299

(12.3.22c)

A new exposition of the mixed finite-element discretisatiori can be found in Brezzi-Fbrtin [1).

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Index

Note: The page numbers given in bold-face refer to occurrences of the term

printed emphatically on the page indicated. Adjoint problem, 141, 190, 196 Adler problem, 159, 183 Agmon, condition of, 158 AnalytIc, 20 condItion, 139 BabuMca-Brezzi condition, 284 Banach space, 110, 111, 112, 113, 131, 135, 137, 161 Basis, 161, 165, 166, 171, 182, 183, 204, 291 hIerarchical, 202 Basis condition, 171, 172, 173, 175, 176, 177, 180, 181, 183, 200

Beltrami operator, 14 Biharmonic equation. Ses Differential equation Bilinear ftrin, 137, 154, 155, 227, 279

adjoint, 138, 190, 191 continuous or bounded, 138, 146, 157, 161, 185, 279, 280 corresponding to a pde of order 2m, 146, 210 corresponding to the Helmholtz equation, 147, 154, 208 corresponding to the Poisson equation, 149, 154, 165, 176, 180, 183

extension of a, 138 operator associated to a. See Operator, 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 partitionlog, 283 Boundary condition, 158, 226, 275 Dirichlet, 27, 35, 38, 65, 85, 86, 96, 145, 150, 159, 164, 174, 178, 215, 222, 229 discretisation of the, 100

first, 35,96 mixed, 96 natural, 152, 154, 155, 159, 168, 178, 218, 221 Neumann, 36, 65, 96, 154, 174, 183

periodic, 98 second, 96 third, 96 Boundary element method (BEM),

37 Boundary layer, 249

Index

308

Boundary value, 7, 9 10, 12, 26 Boundary value problem, 13, 19,

27,

28,85

Brezzi condition, 292, 293, 295, 298 Bubble function, 294, 295, 296 Cauchy-Riemann equations, 4, 7 Chequer-board. See Ordering Coefficient vector, 161, 255 Compact (cf. embedding, nine-point formula, operator), 135, 201, 257, 269 Complementary problem, 264 Complete, 24, 112 Completion, 112, 117, 122, 125 CondItion, 108, 204 Cone condition, 136 Conservation form, 244 Convex. See Domain Consistent, consistency, 59, 64, 69, 93, 94, 200, 228, 232, 233, 235, 239, 242, 267 Convection-diffusion equation, 247 Convergence, 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, 260 super, 202, 203 Coordinate transformation. See Domain Corner, re-entrant (cf. domaIn), 34, 223, 225 Covering, 127

10, 112, 113, 114, 115, 117, 119, 123, 141, 169, 170, 224

Dense,

Derivative, 104 conormal, 96, 201 normal, 15, 65, 95, tangential,

96,

96, 145

99, 103, 145,

224,

246, 278 weak,

115,

116, 172

Diagonalisable, real, 6, 7 Diagonally 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,

269

108, 109 of higher order, 62 Difference operator, 10, 44, 121, 210, 13-point,

213,

228, 233

Difference

star (stencil), 44, 62, 91

Differential equation(s) biharmonic, 74, 95, 103, 104, 105, 106, 108, 158, 194, 218, 278

223,

order, 1, 3, 4, 6, 277 of order 2m, 104, 108, 145, 158 of second order, 2, 5, 8, 12, 85, of

first

90,150

ordinary, 1, 2, 3, 38, 248, 254 partial, 1, 2, 7 system of, 3, 4, 6, 7, 105, 275, 276

Differential operator, 5, 59, 60, 75, 85, 89, 104, 253, 276 formally ad.joint, 92, 95 boundary, 95, 154, 155, 156 Dirichiet integral, 144 Dirichiet principle, 144 Divergence (operator), 275 12 convex, 79, 197, 223, 224, 225, 231, 235, 236, 242, 288, 293 exterior, 36 general, 78, 100, 109, 196, 234, 241, 243 L-shaped, 13, 34, 127, 201, 243 normal, 15, 16, 19, 28, 33 transformation of the, 6, 89 unbounded, 23, 146, 147 Double-layer potential. See Potential Dual form, 131, 134 Dual space, 130 Eigenfunctlon, 222, 253, 254, 259, Domain,

263, 271 5, 6, 7, 10, 47, 51, 52, 53, 137, 142, 150, 258, 254,

Elgenvalue,

255, 256, 258, 259, 260, 261, 262,

264, 265, 271,

Eigenvalue problem, 7, 56, 65,

72, 91, 250

forward or right-sIded, 38, 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,

91,

274 137, 150,

11,

253

adjoint, 253, 256 elliptic, 253, 254, 255, 274 Ethptic,

4,

104,

5, 6, 7, 9, 10, 11, 12, 85,

276

Index

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 Embedding compact, 185, 136, 142, 143, 150, 153, 171, 186, 253, 256 continuous, 118, 131, 133, 134, 136, 137, 141, 143, 171, 185, 253, 256 dense, 131, 133, 134, 137, 141, 171, 253, 256

Error estimates for difference methods, 190, 238, 239

for eigenvalue problems, 260, 263, 264

for finite element methods, 185, 190, 193, 196, 246, 266, 297, 298

for Ritz-Galerkln methods, 167 Estimate. See Error estimate, inverse estimate Existence. See Solution Extension. See Bilinear form, operator of a function, 123, 129, 215, 237 Extrapolation (method), 61, 84 Finite elements, 171, 172, 174, 251, 254

constant (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, 198,247 mixed, 290, 299 nonconforjnal, 199, 200, 291 of the serendipity class, 181, 182, 193

quadratIc, 181, 182, 193 Finite element method, 24 Five-point formula, 40, 41, 53, 60, 62, 92, 176, 190, 231 Form. See Bilinear form, dual form, 8esquilinear form Fourier expansion, 9 Fourier transform, 110, 119, 120, 124, 125, 126, 135, 213 FunctIonal, 132, 140, 146, 152 Fundamental solution, 16, 17, 28 Galerkin method. See RJtz-Galerkln, Petrov-Galerkln G6rding, Theorem of, 150

309

Gei'fand triple, 41, 133, 134, 136, 141, 163, 207, 253 (Ierschgorin, Theorem of, 46, 48 Gradient, 15 Green's formula, 15, 29, 144, 153 Green's function of the first kind, 28, 29, 30, 31, 33, 34, 95, 105, 144

discrete, 55, 72, 95 Green's function of the second kind, 35 Grid, 39, 40, 226 Grid function, 39, 44 Grid points close to the boundary, 63, 79, 83 far from the boundary, 42,63, 79 80 neighbouring, 41, 42, Grid size (width, step sIze), 39 Harmonic, 13, 15, 16, 19, 20, 22, 23, 24, 25, 144 Harnack, Theorem of, 20 Heat equation, 3, 4, 5, 7, 8, 10 Hehnholtz equation, 147, 154, 208 hubert space, 110, 114, 115, 117, 122, 132, 134, 279, 289 Hökler continuous, 29, 30, 110, 123, 125

Hyperbolic, 4, 5, 6, 7, 8, 9, 10, 11, 248

Improperly posed. See Well-posed Inclusion, 135, 136, 142 Initial-boundary value problem, 8, 9, 207 Initial value problem, 2, 3, 7, 8, 9, 38

Instationaiy problem, 10, 11 Integral equation method, 36, 37 Interpolation, 84, 124 Hermite, 194, 196, 207 spline, 194, 196, 207 Inverse estimate, 206, 221, 240, 293 Irreducible, 45, 54 Irreducibly diagonally dominant. See Diagonally dominant Lagranglan (6.ctor), 184, 290 Lamé differential equations, 277 Laplace equation (of potential equations), 2, 12 Laplace operator, 12, 14, 131, 176 Lexicographical ordering. See Ordering

Lipschitz continuity, 23, 30 Map, mapping. See Operator Mass matrix, 205

Index

310

Maximum (-minimum) prmciple, 17, 18, 19, 20, 27, 86, 103, 250, 258

Mean-value property, 17, 18, 19, 20,

27,54 Mehrstallen method, 64, 83 M-matnx, 45, 49, 50, 51, 52, 53, 64, 68, 81, 83, 84, 91, 92, 93, 100, 106, 249, 250, 251, 252 Multi-index, 30, 104 Neighbour. See Grid points Neumann condition. See Boundary condition Nine-point formula, 90, 180 compact, 63, 63 Nodal points, 172, 175, 178, 181, 203

Nodal values, 172, 195 Norm, 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, 50 assocIated, 50, 204, 227 maximum, 46, 110 operator, 111 row sum, 50, 59, 170, 242 Sobolev, 117, 122 Sobolev-Slobodeckil, 122, 247

spectral, 51, 59, 204 supremum, 23, 110, 112 vector, 50 Normal (dIrection), 15, 116, 145, 224

Normal system, 105 Normed space, 110

Operator, ill adjoint, 132, 163, 213, 216, 242 associated. See Bilinear form, 209, 256, 282 compact, 135, 136, 137 difference. See Difference operator differential. See Differential operator

dual, 131 selfadjoint, 132 Ordering (of the grid poInts), 42, 43, 45 lexicographical, 42, 67, 70 chequer-board, 43 Orthogonal (ef. projection), 114 Orthogonal space, 114, 163, 283

Parabolic, 4,5, 5,7,8, 9, 10, 11 Partition of unity, 128, 130, 150, 209, 219

Petrov-Galerkin method, 252 Plate equation (cf. blharmonic), 85, 103 Poisson 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, 87 Potential, double-layer (dipole), 36 sIngle-layer, 36 volume, 36 Potential equation (of. Laplace equatIon), 2, 4, 5, 7, 9, 12, 13, 14, 16, 28, 35 Precompact, 135 Principal part, 6, 12, 95, 145, 147, 148, 150, 223, 228, 251, 276 Projection, 132, 170 orthogonal, 132, 133, 163, 170, 225, 294

Ritz, 170, 189, 191, 192, 193, 225 Rayleigh-Ritz. See Ritz Reduced equation, 248, 250 Reference triangle (reference element), 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, 132 Riesz isomorphism, 132, 283 Rlesz-Schauder theory, 1ST, 142 Ritz-Galerkln method (cf. solution), 161, 171, 290 Robin problem, 159 Saddle-point (problem), 279, 280, 282, 284, 290, 297 Scaler product, 52, 114, 115, 117, 121, 132, 148, 162, 170, 227 Separation (of variables), 3, 11 Serendipity class. See Finite elements Sesquilinear form, 138

Index

Seven-point formula, 91, 92, 176 Shortley-Weiler discretisation, 78, 80, 81, 83, 229, 231, 237 Side condition, 182, 183, 290, 291 Single-layer potential. See Potential Singular perturbation, 247, 248 Singularity function, 14, 16, 56 discrete Sobolev, lemma of, 125, 212, 222 Sobolev space (cf. norm), 114, 117, 122, 128, 133, 134, 136, 145 Solution 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, 173 Spectral radIus, 47 Spectrum, 137, 142 Stable, stabilIty, 59, 64, 70, 90, 94, 95, 102, 103, 227, 228, 242, 249, 252, 291, 293 Star (stencil), 178, 180 Stationary problem, 10 Steklov problem, 254 Step size. See Grid size Stiffness matrIx, 102, 163, 165, 171, 177, 191, 200, 202, 203, 255 Stokes 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, 11 Trace (of a function), 3, 123, 129 Trace (of a matrix), 8? Transformation. See Domain, Fourier Transformation theorem, 119, 129 Transition condition, 245, 246, 247 Trefftz'method, 200

311

Triangulation, 174 adaptive, 201 admissible, 174, 175, 179, 185, 190

quasi-unifurm, 188, 190, 204, 208, 294, 295

regular, 188 uniform, 188

Type (of a pde), 1,4,5,6,7,10 Type invariance, 6 Uniqueness. See Solution Variational formulation (weak formulation), 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, 200 Viscosity, artificial, 251, 252 numerIcal, 251 Wave equation, 2, 3, 5, 8, 10 Weak formulation. See Variational formulation Well-posed (problems), 23 Wilson'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

111111

9 783540 548225 http:llwww.springer.de