NUMERICAL METHODS FOR EXTERIOR PROBLEMS

Peking University liffiajll Series in Mathematics — Vol. 2 NUMERICAL METHODS FOR EXTERIOR PROBLEMS Ying Lung-An NUME...

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Peking University liffiajll

Series in Mathematics — Vol. 2

NUMERICAL METHODS FOR EXTERIOR PROBLEMS Ying Lung-An

NUMERICAL METHODS FOR EXTERIOR PROBLEMS

PEKING UNIVERSITY SERIES IN MATHEMATICS Series Editor: Kung-Ching Chang (Peking University, China)

Vol. 1: An Introduction to Finsler Geometry by Xiaohuan Mo (Peking University, China) Vol. 2: Numerical Methods for Exterior Problems by Ying Lung-An (Peking University & Xiamen University, China)

Peking University fjqSSskjg?,ies T^y

in

Mathematics — Vol. 2

UMERICAL METHODS Ying Lung-An Peking University & Xiamen University, China

P World Scientific Hr

W H S

NEW JERSEY • LONDON • SINGAPORE • BEIJING • SHANGHAI

- HONG KONG • TAIPEI • CHENNAI

Published by World Scientific Publishing Co. Pte. Ltd. 5 Toh Tuck Link, Singapore 596224 USA office: 27 Warren Street, Suite 401-402, Hackensack, NJ 07601 UK office: 57 Shelton Street, Covent Garden, London WC2H 9HE

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NUMERICAL METHODS FOR EXTERIOR PROBLEMS Peking University Series in Mathematics — Vol. 2 Copyright © 2006 by World Scientific Publishing Co. Pte. Ltd. All rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording or any information storage and retrieval system now known or to be invented, without written permission from the Publisher.

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ISBN 981-270-218-0 ISBN 981-270-526-0 (pbk)

Printed in Singapore by Mainland Press

Preface

Partial differential equations play an important role in mathematical physics. Usually some initial conditions or boundary conditions are imposed on the equations, and we are dealing with initial value problems, boundary value problems, or initial-boundary value problems. There are some different points of view to classify the problems: Linear or nonlinear, first order equations or higher order equations, steady problems or evolution problems, interior problems or exterior problems. The first three are according to the properties of the equations, and the last one is according to the properties of the underlying physical domains. For example, the flow problem in a chamber is an interior problem, and the flow problem around an aircraft is an exterior problem. All partial differential equations possess infinite number of degrees of freedom. To solve a problem numerically, one should approximate the problem so that only a finite number of degrees of freedom are solved in real computation. As a result a truncation error appears in every numerical scheme. This fact is the same for both interior and exterior problems. However when one intends to solve an exterior problem, one faces another kind of difficulty that the domain is infinitely large. Most approaches for the interior problems can not be applied to the exterior problems directly. The importance of exterior problems is obvious, and the challenge is serious. There are two kinds of approaches to deal with the exterior problems: to truncate the domain, or to solve the problem directly on the infinite domain. The former ones are: the introducing of artificial boundary conditions, and the introducing of perfectly matched layers. The later ones are: the infinite element method, and the spectral method. While in the boundary element method both approaches are applied, depending on the degree of complexity of the domains.

V

vi

Numerical Methods for Exterior

Problems

The aim of this monograph is to provide a comprehensive study of different approaches. At the beginning we investigate the mathematical theory of the exterior problems of some typical partial differential equations, which provides the foundation of all numerical methods. Restricted by the space we only state the basic results. The readers can refer to the literature for the details. The main body of the book is on some traditional and new methods. Each of them is effective, and certainly restricted to some particular classes of problems. One omission of the book is that no numerical example is included, because there are too many in the literature, and it is too difficult to choose a few, representing most of the problems and most of the methods. It is also due to the restriction of space. However I am not sure if it is appropriate to do so. This is an area growing rapidly. Limited by the author's knowledge, some important approaches must have been ignored. There is a proverb in China:" Cast away a brick and attract a jadestone". I will be pleased to listen to all comments. My friends, Professors Zhang Guan Quan, Guo Ben Yu, Yu De Hao, and Shen Jie kindly provided me their works on this subject. Without their help I could not finish this work. My students, Fang Nengsheng and Liao Caixiu, helped me to proof-read and illustrate the manuscript. I express my sincere thanks to them. Dr. Sim, the editor, helped me a lot for the style file, and Ms. Zhang Ji, the editor, helped me to prepare the final manuscript. I am grateful to them. Ying Lung-an

Contents

Preface

v

1.

1

Exterior Problems of Partial Differential Equations 1.1 1.2 1.3 1.4 1.5 1.6 1.7

1.8 1.9 1.10 1.11 1.12 2.

Harmonic equation-potential theory Poisson equations Poisson equations-variational formulation Helmholtz equations Linear elasticity Bi-harmonic equations Steady Navier-Stokes equations -linearized problems . . . 1.7.1 Navier-Stokes equations 1.7.2 Stokes equations 1.7.3 Behavior of solutions at the infinity 1.7.4 Stokes paradox 1.7.5 Oseen flow Steady Navier-Stokes equations Heat equation Wave equation Maxwell equations Darwin model

2 12 13 17 25 29 35 35 36 39 41 41 44 49 53 56 61

Boundary Element Method

71

2.1

71 71 75 77

Some typical domains 2.1.1 Harmonic equation 2.1.2 Bi-harmonic equation 2.1.3 Stokes equation vii

viii

Numerical

2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.

Problems

2.1.4 Plane elasticity 2.1.5 Helmholtz equation General domains Subdivision of the domain Dirichlet to Neumann operator Finite part of divergent integrals Numerical approximation Error estimates Domain decomposition Boundary perturbation

80 82 85 93 96 98 103 108 113 114

Infinite Element Method

117

3.1

117 117 120 127 131 133 134 136 138 140 142 147 152 152 155 158 161 162

3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9

3.10 4.

Methods for Exterior

Harmonic equation-two dimensional problems 3.1.1 Infinite element formulation 3.1.2 Tranfer matrix 3.1.3 Further discussion for the transfer matrix 3.1.4 Combined stiffness matrix General elements Harmonic equation-three dimensional problems Inhomogeneous equations Plane elasticity Bi-harmonic equations Stokes equation Darwin model Elliptic equations with variable coefficients 3.9.1 A homogeneous equation 3.9.2 An inhomogeneous equation 3.9.3 General multiply connected domains 3.9.4 Transfer matrices Convergence

Artificial Boundary Conditions

167

4.1 4.2 4.3 4.4 4.5 4.6

167 172 175 176 178 178

Absorbing boundary conditions Some approximations Bayliss-Turkel radiation boundary conditions A lower order absorbing boundary condition Liao extrapolation in space and time Maxwell equations

Contents

4.7 4.8

5.

6.

Finite difference schemes Stationary Navier-Stokes equations 4.8.1 Homogeneous boundary condition at the infinity . 4.8.2 Inhomogeneous boundary conditions at the infinity 4.8.3 A linear boundary condition

ix

182 183 183 186 187

Perfectly Matched Layer Method

191

5.1 5.2 5.3 5.4 5.5 5.6

191 197 201 208 210 212

Wave equations Berenger's perfectly matched layers Stability analysis Uniaxial perfectly matched layers Maxwell equations Helmholtz equations

Spectral Method

217

6.1 6.2 6.3

217 225 230 230

6.4 6.5 6.6

Introduction Orthogonal systems of polynomials Laguerre spectral methods 6.3.1 Mixed Laguerre-Fourier spectral method 6.3.2 Spherical harmonic-generalized Laguerre spectral method 6.3.3 Generalized Laguerre pseudo-spectral method . . 6.3.4 Nonlinear equations Jacobi spectral methods Rational and irrational spectral methods Error estimates

235 237 239 241 243 245

Bibliography

251

Index

265

Chapter 1

Exterior Problems of Partial Differential Equations

This chapter is devoted to the theory on well posedness of the exterior problems of some important partial differential equations in mathematical physics. The emphasis is the boundary conditions at the infinity, which play an essential role in the mathematical theory, and differ from different equations and dimensions. An appropriate numerical method must reflect these boundary conditions correctly. In this respect this chapter is the foundation of the other chapters. On the other hand this chapter provides some tools for numerical methods. For example, the potential theory is the basic tool in the boundary element method. Another example is the variational formulation, which is the starting point of the Galerkin method, and the infinite element method and the spectral method belong to this catalogue. We investigate some typical equations only. No attempt has been made to provide a comprehensive theory on general equations. For example in many cases we consider equations with constant coefficients instead of that with variable coefficients. The reason is that we intend to show the most important facts by means of some simple models. Secondly it is obvious that the complete theory for some general equations may deserve an entire book, which is not our objective. The theory of the Navier-Stokes equations is the subject of a number of outstanding works. We will briefly present some basic results for the exterior problems of the Navier-Stokes equations. We will consider a bounded domain D.c in two dimensional space M.2 or in three dimensional space K 3 . Let dQ, be its boundary. For simplicity we assume that dfl is a connected curve or surface, which is Lipschitz continuous and piecewise smooth. Let 17 be the exterior part of the closed domain fic, the complement, then Q is an unbounded domain. The dimension of the domain f2 is denoted by dim(fi).

1

Numerical Methods for Exterior

Problems

•x

x

Fig. 1 If the space dimension is one, the domain Q.c is reduced to an interval (a, b), and the domain Q is divided into two independent parts: (—00, a) and (b, +00). We consider the latter one, and it has no harm in assuming b = 0, so we will consider the interval (0, 00). In the sequel it will be assumed that functions and some related spaces will always be real-valued, unless we explicitly state that some functions are complex-valued in some individual sections. We will always denoted by C a generic constant that may not be the same at different occasions.

1.1

Harmonic equation—potential theory

First of all we investigate a typical exterior boundary value problem: the exterior problem of the Laplace equation: Ai

where for two dimensional case, A Q2

(1.1)

0, • + •£=?, and a; = {xl,x2) O2

G

and

O2

for three dimensional case, A = -g^ + -g^ + j ^ , and x = (x\, X2, £3) G . We consider the Dirichlet boundary condition: u = g,

x £ d£l,

(1.2)

Exterior Problems of Partial Differential

Equations

3

where g is a continuous function denned on dQ,. We also consider the Neumann boundary condition:

P = g,

xeffl,

(1.3)

ov where g is also a continuous function defined on dfl, and v is the unit outward normal vector. Let us consider classical solutions of the problems. A function u is a solution to the Dirichlet problem, if it is continuous on Q, twice continuously differentiable on f2, and it satisfies the equation (1.1) and the boundary condition (1.2). A function u is a solution to the Neumann problem, if it and its first order derivatives are continuous on fi, twice continuously differentiable on Q, and it satisfies the equation (1.1) and the boundary condition (1.3). However we are interested in well-posed problems, then some boundary conditions at the infinity are required. J u is a bounded function on fl, dim(fi) — 2, . . \ lim^i^oo u = 0, dim(fi) = 3. We study uniqueness first. The strong maximum principle is applied here. Theorem 1. Let G be a bounded domain, u satisfies the equation (1.1) in G, and u is continuous on the closed domain G, then if u assumes its maximum (minimum) value at an interior point x £ G, then u is a constant on G. Moreover, if the first order derivatives ofu are continuous on G, u is not a constant, and u assumes its maximum (minimum) value at a boundary point x £ dG, then | ^ > 0(< 0) at this point. Theorem 2. Ifu\,U2 are two solutions satisfying (1.1),(1.2) and (1-4), or if dim(Cl) = 3 andui,U2 are two solutions satisfying (1.1),(1.3) and (1-4), then u\ = u 0, then V\QQ > 0. v is also a harmonic function. For an arbitrary x £ Q, let R\ > Ro be large enough, such that x £ B(0, Ri) and v\r=n1 > 0, then by the maximum principle v > 0 on 0,f]B(O,Ri).

4

Numerical Methods for Exterior

Problems

Therefore u(x) > —e\n(r/Ro). Being the same u(x) < eln(r/Ro). But e is arbitrary, so u{x) = 0, and x is arbitrary, so u = 0 on fl. 2. Three dimensional Dirichlet problem The proof is analogous. We take an auxiliary function v = u + £, then following the same lines we can prove v > 0, then get u = 0 on CI. 3. Three dimensional Neumann problem Being the same we take an auxiliary function v = u + e. Then we consider a ball B(0, R\) D Clc. Let i?i be large enough, such that v\r=R1 > 0. If v is a constant on Q. f] B(0, R\), then v > 0 on this domain. If v is not a constant and if v assumes its minimum value at a point x e dCl, then by the strong maximum principle f^ < 0, but ^ = 0 on the boundary dfl, so it is impossible. Therefore v assumes its minimum value at r = Ri and also v > 0 on nf]B(0,Ri). Then in either case u(x) > -e on Qf]B(0,Ri). Being the same u(x) < e. But e is arbitrary, so u = 0 for |a;| < R\. R\ can be arbitrary large, so u = 0 on fl. • Two dimensional Neumann problem is more complicated. We have the following result: Theorem 3. / / dim(Cl) = 2 and (1.1),(1.3) and (1-4) is solvable, then L„ g(x) dsx = 0, and ifui, u^ are two solutions to this problem, then u\—U2 is a constant on Ct. Proof.

Let u be a solution to the problem. Let B(0, Ro) C fic as the R2

previous. We define v(x) = u(r^x), and denote by fii and dfli the image of fi and dfl under the mapping x —* j^px, then v is also a harmonic function on Cii and satisfies the Neumann boundary condition (1.3) on dfli. We notice that O is a boundary point of fii, and we will show that it can be removed, that is, if we set Q.? = fli IJ{^} a n ( ^ define a suitable value of v(0), then v is a harmonic function on 1^2• We consider a small disk B{0, R2) C f^, and solve a Dirichlet boundary value problem on it with the boundary value t i | r = ^ 2 . Let the solution be iv, which can be expressed in terms of the Poisson integral: (see Section 2.1, for example) V ( m - ^ 2 - P 2 [2v W W W j ~ 2TT J0 R* + p*-2R2Pcoa{6-0')

M

'

where (p, 9) are polar coordinates. Using the auxiliary functions v — w ± eln(r/i?2) and the same argument as the previous theorem, we can verify that U = I » O D B(Q,R2) \ {O}. Therefore suffice it to set v(0) = w(0).

Exterior Problems of Partial Differential

5

Equations

Integrating the equation (1.1) on CI2 we get JQa | ^ dsx = 0. Returning to the domain f2, it can be verified that Jdn g(x) dsx = 0. Let u be a solution to the corresponding homogeneous problem. We claim that u is a constant. Because v is the solution to (1.1) on CI2 with homogeneous Neumann boundary condition, then if v were not a constant then it should assume its maximum value at one point of the boundary dQ,\ and | ^ > 0 at this point, which is impossible. Therefore v is a constant and so is u. • We notice that the boundary conditions at the infinity (1.4) are different for different dimensions. This fact is essential. For example let fi = {x; \x\ > 1} and we consider the Dirichlet boundary condition u \dQ = 1- We have one solution u = 1 in two dimensional case. Since bounded solution is unique, there is no solution satisfying lim^i^oo u = 0. On the other hand bounded solutions are not unique for this example in three dimensional case, because we have two solutions u = 1 and u = 1/r, where r = \x\. We apply potential theory to study existence. Consider two dimensional case first. For simplicity we assume that dfl is a smooth curve and we will relax this restriction for the variational formulation later on. It is known that the fundamental solution is u = J- In -, which is the solution to the equation -Au

= 6,

(1.5)

where S is the Dirac ^-function. By this fundamental solution, the solution to the Neumann boundary value problem is expressed in terms of a single layer potential:

U{x) =

hjdnu{y)ln\x^\ds^

(L6)

where sy is the differential on the curve, and the solution to the Dirichlet boundary value problem is expressed in terms of a double layer potential: u x

( ) = 7T /
(L7)

The functions u satisfy the equation (1.1) on R 2 \ <9Q, so we need to determine the densities w and a to make the boundary conditions (1.2)-(1.4) satisfied.

6

Numerical Methods for Exterior

Problems

The kernel ^ In , * , of the single layer potential is weakly singular, so (1.6) defines a continuous function on R 2 . Besides, one useful property of the single layer potential is the following: Lemma 1. For the single layer potential (1.6), the limit Mva^^^ u{x) = 0 if and only if Jda d>(y) dsv = 0, otherwise the function u tends to infinity as \x\ —> oo. Proof.

The single layer potential can be rewritten as

u x

w ( ) — 7^ (y) In ~, r dsv -\ In -. / K v ' 2* Jen KJ \x-y\ 2TT \x-y0\Jdn

uj(y) dsv, yy > yi

where yo S flc is a fixed point. Since lim|;c|_>00 In -f^zij = 0 uniformly with respect to y £ dQ and limi^i^oo In -n^z—r = —oo, the conclusion follows immediately. • The kernel of the double layer potential is not weakly singular, so it does not define a continuous function on M.2. However one property of the double layer potential is the following: Lemma 2. For the double layer potential (1.7) if a is a continuous function on dQ,, then the following limits exist: lim u{x) = -a(x0) 2 _ > Xo x Gfl,xo £ dCl

+ — / a(y)—\n2?r JdQ dvy

x

lim u(x) = --o-(x0) 2 x->x0 x € f2c, x0 S d£l Proof.

+— 2?r

J™

-dsy, \x0 - y\

a{y)-—ln-dsy. dv y \x° ~ y\

We consider the case of cr = 1 first. For x ^ y it holds that d

1

duy

\x — y\

_ (x - y) • uy _ COS((:E - y),

\x — y\2

vy)

\x — y\

For a fixed point i £ I 2 , we construct polar coordinates (p, 0) with the origin x. Then for y G d£l dB =

_coB{{x-y)uy) \x-y\

Exterior Problems of Partial Differential

7

Equations

Therefore -2TT,

In-

dl/ / .dn y

dsv

-IT,

\x-y\

x G Qc, x G dQ.,

0,

(1.8)

iea

For general cases let XQ G dQ, be a fixed point. We set v x

()

d = 7T2n

Jan

(
dvy

In-

1

\x - y\

ds„,

Let us show t h a t the function v is continuous at the point XQ , t h e n applying (1.8) we get the desired result. Letting e > 0, we consider a local neighborhood U oi XQ, then take an arc s C dQ, XQ G S, and s is small enough so that

^(a(y)-,(*0)) A h _ ! _ * , ,

< e

for all x G U. Then we take x close enough to xo, such t h a t — / (o~(y) — o-(x0)) < —— In 1*Jan\a \dvy \x-y\

: — - — In -. r > ds% < e. y duy \x0 - y\ j

Then

M

TJJ^-'^W*-*l n

i

_ J L In dv„

ds, FO

<3e,

- y\

D

which implies the continuity of v.

We t u r n now to consider the normal derivatives of the single layer potential along the b o u n d a r y dQ. Since we consider the exterior a n d interior domains simultaneously, for defmiteness, let v always be the outward normal vector with respect to Qc, a n d j^qr and -g^= be referred t o t h e derivatives of functions in Q and Qc respectively. T h e discontinuity of the derivatives of single layer potentials along dQ can be characterized as the following: L e m m a 3 . For the single layer potential on dQ, then

dv+ du(x0)

^(x0) 2 = -u(x0) 2

(1.6) ifu> is a continuous

+ — / w(2/)—In 27r xo Jon + — 2TT Jan

uj(y)-—\ndvx

•day,

function

(1.9)

-y\

-dsy. \x0 - y\

(1.10)

8

Numerical Methods for Exterior

Problems

Proof. For each point x £ dQ we consider the straight line perpendicular to the curve, then we take derivatives along this line. By this way we extend the normal derivatives -Jj^- to a neighborhood of dCl. Let v x

( ) = 7T I 2TT

W

(V) \ ^—

Jon

\dvx

ln

1 \x-y\

f + -5— l n 1 dvy

f?

ds

y

\x - y\ J

For xo S dCl the kernel _d_ 1 d 1 _ cos((y - x), vx) + cos((x - y), vy) dvx \x-y\ dvy \x-y\ \^-y\ is bounded near XQ. Then following the same argument as the previous lemma we can show that v is continuous at XQ, then applying the results of the previous lemma we get the jump results. • Having expressed the solutions of the Dirichlet and Neumann boundary value problems in terms of double layer potentials and single layer potentials, we apply the previous lemmas and get the integral equations for the densities a and u>: 1. Dirichlet exterior problem: -a(x)+ *

a(y)K(x,y)dsy=g(x),

(1.11)

Jan

where K(x, y) = £.£- In ^ . 2. Dirichlet interior problem: -l
[ a{y)K{x,y)dsy=g{x). Jan 3. Neumann exterior problem:

(1.12)

1

-\u(x) ^

+ [ w{y)K(y, x) dsy = -g(x). Jan

(1.13)

4. Neumann interior problem: \u{x)+ *

[ Lj(y)K(y,x)dsy=g(x). Jan

(1.14)

K(x, y) is a continuous function on dtt x dCl, so it defines a compact operator on the space C(dfl). The equations (1.11)-(1.14) are the Fredholm integral equations of the second type. The Fredholm alternative theorem is applied here:

Exterior Problems of Partial Differential

9

Equations

Theorem 4. Let B be a Banach space, and K : B —> B be a compact operator. Then the following alternative holds. Either, 1. There is a unique solution u £ B satisfying the equation u + Ku = f, for all f e B. Or 2. The corresponding homogeneous equation u + Ku = 0 admits k > 0 linearly independent solutions, its dual equation v + K*v = 0, v G B*, also admits k linearly independent solutions, i.e., the dimensions of null spaces are equal, and the equation u + Ku = f is solvable if and only if vf = 0,VveB*,v + K*v = 0. Theorem 5. The integral equations (1.12) and (1.13) admit unique solutions. Proof. Let w be a solution to the homogeneous equation associated with (1.13). Then by the equation - - / u>(x)dsx+ / / cj(y)K(y,x)dsydsx ^ Jan Jan Jdn

= 0.

By exchanging the order of integrals and noting (1.8) we have OJ(X)

dsx = 0.

JdS lan

With this density u> we consider a single layer potential u defined by (1.6). We notice that u> is a solution to the homogeneous equation. Then by Lemma 3 we get ^ j r = 0, . By the uniqueness of Neumann exterior problems u is a constant on CI, then by the uniqueness of Dirichlet interior problems u is equal to the same constant on f2c, so -^r = 0. We apply Lemma 3 again and get w = 0. Then by the Fredholm theorem the equations admit unique solutions. • Theorem 6. The homogeneous integral equation associated with(l.ll)or (1.14) admit one nonzero solution, and any two solutions are linearly dependent. Proof. The homogeneous equation associated with (1.11) has a nonzero solution a = 1. By the Fredholm theorem the homogeneous equation associated with (1.14) also admits nonzero solutions. Let ui be one of them, then we claim that / u>(x) dsx ± 0. Jan

10

Numerical Methods for Exterior

Problems

Otherwise with this density w we consider s a single layer potential u defined by (1.6), then by Lemma 3 we get -^ = 0 . By the uniqueness of Neumann interior problems u is a constant on f2c, then by Lemma 1 and the uniqueness of Dirichlet exterior problems u is equal to the same constant on fi, so -g^f = 0. We apply Lemma 3 again and get u = 0, which is not a nonzero solution. Let u>i,cJ2 be two solutions to the homogeneous equation associated with (1.14). We consider a linear combination c\U\ + C2W2, cf + c\ ^ 0, such that Jgn(ciu>i(x) + C2U>2(x)) dsx — 0. Then by the above argument c\W\ + C2LO2 = 0) so they are linearly dependent. • Theorem 7. The problem (1.1),(1.2),(1.4)

admits a unique solution.

Proof. The uniqueness has been given. We prove existence here. We take a constant c such that Jdn(g(x) + c)ui(x) dsx = 0, where w is a nonzero solution to the homogeneous equation associated with (1.14). Then we solve the equation (1.11) with a right hand side g + c. Owing to the Predholm alternative theorem it is solvable. The desired solution to (1.1),(1.2),(1.4) is u — c, where u is the double layer potential (1.7). • Theorem 8. The sufficient and necessary condition for (1.1),(1.3),(1.4) admitting a solution is Jdn g(x) dsx = 0, and the solutions are unique up to adding an arbitrary constant. Proof. We solve the equation (1.13), then construct a single layer potential u. If Jdng(x)dsx — 0, then one finds that Jgiloj(x)dsx = 0 by integrating both sides of (1.13). Then by Lemma 1 u is bounded, so it satisfies the boundary condition (1.4). • Next we investigate three dimensional problems. Using single layer potential and double layer potential we also get integral equations (1.11)(1.14), where K(x, y ) = X _ | _ _ i _ . Parallel to two dimensional case, we have the following theorems. We will omit most of the proof and only indicate the difference in the argument. Theorem 9. The integral equations (1.12) and (1.13) admit unique solutions. Theorem 10. The homogeneous integral equation associated with (1.11) or (1.14) admit one nonzero solution, and any two solutions are linearly dependent. Proof.

Being the same they admit nonzero solutions.

Exterior Problems of Partial Differential

Equations

11

Let wj,W2 be two solutions to the homogeneous equation associated with (1.14). Let ui,U2 be two single layer potentials with densities wi,W2They are constants in flc. We consider a linear combination c\W\ + C2W2, Cj + c | 7^ 0, such that u — c\Ui + C2U2 = 0 in fic. Since u = 0 on dil, by the uniqueness of the Dirichlet problems, u = 0 on Cl. Therefore ciu>\ + C20J2 = 0. They are linearly dependent. • Theorem 11. The problem (1.1),(1.2),(1.4)

admits a unique solution.

Proof. The uniqueness has been given. We prove existence here. We take a constant c such that fan(g(x) +c)w(x) dsx — 0, where w is a nonzero solution to the homogeneous equation associated with (1-14). Then we solve the equation (1.11) with a right hand side g + c. We may assume that the solution ui constructed in the proof of Theorem 10 satisfies u = l in Q,c. Then the desired solution is u — cui, where u is the double layer potential (1.7). • Theorem 12. The problem (1.1),(1.3),(1.4)

admits a unique solution.

Proof. This is because the corresponding integral equation admits a unique solution. • The results in this section are collected in the following tables: Table 1. Boundary value problems condition for solutions to expressions existence homogeneous problems 2-D Dirichlet none 0 double layer — c 2-D Neumann Jgds = 0 constant single layer +C 3-D Dirichlet none 0 double layer — cu\ single layer 3-D Neumann none 0 Table 2. Integral equations

2-D Dirichlet 2-D Neumann 3-D Dirichlet 3-D Neumann

condition for existence / gu: ds = 0 none J gu> ds = 0

none

solutions to homogeneous equations constant 0 constant 0

12

1.2

Numerical Methods for Exterior

Problems

Poisson equations

We investigate the Poisson equation, -Au

= f,

x£tt,

(1.15)

in this section. For simplicity we assume that / £ Co(fi), that is, / is continuous with a bounded support. The uniqueness of the problem (1.15),(1.2),(1.4) or the problem (1.15),(1.3),(1.4) comes from the results in the previous section directly, so we study existence here. For two dimensional case the Newton potential is 1

«(*) = T" )w / f(v)ln rk -^ v\ dv>

( L16 )

where / is extended to the entire space. If / is extended by zero, the integral is in fact taken on fi. It can be verified that u satisfies the equation (1.15). Analogous to Lemma 1, we have Lemma 4. For the Newton potential (1.16), the limit lim^i-nx, u(x) = 0 if and only if JR2 f(y) dy = 0, otherwise the function u tends to infinity. We extend the function / to Q,c so that JK2 f(y) dy = 0. Let UQ be the corresponding Newton potential, then u — UQ satisfies a boundary value problem of the Laplace equation, so the results in the previous section can be applied here. The Dirichlet problem admits a unique solution. As for the Neumann problem we notice that / f(x)dx Jn

=- / -—^dsx. Jan dv

Thus the sufficient and necessary condition of existing a solution is / f(x) dx+ g(x) dsx = 0, Ja Jan and the solutions are unique up to adding an arbitrary constant. For three dimensional case the Newton potential is

<X) = T- I f(y)r^—\dy-

(L17)

4?r 7 R 3 \x - y\ We can define / = 0 on ftc then verify that both the Dirichlet problem and the Neumann problem are uniquely solvable.

Exterior Problems of Partial Differential

1.3

13

Equations

Poisson equations—variational formulation

To study the variational formulation and weak solutions the (complex) LaxMilgram theorem is applied: Theorem 13. Let H be a Hilbert space, M,a > 0, and a(-, •) be a sesquilinear form on 7i with 1. \a(u,v)\< M\\u\\\\v\\, \/u,v£H, 2. |a(u,u)| > a||u|| 2 , Vu e U. Let f be a bounded linear functional on "H. Then the equation a(u,v) = f(v),

VveTi

admits a unique solution u G 7i, and the mapping f —> u is bounded. We list some notations which are commonly used: Wm'p{Q) and Hm(Q), m £ M.,1 < p < oo, are Sobolev spaces, and || • || m , p , || • || m are norms in these spaces. The semi-norms are denoted by | • | m p, \ • \m respectively. The L 2 inner product is denoted by (•, •). W™'P(Q) and H™^) are the closures of Cg°(Q.) in Wm'P(Q) and Hm(Q) respectively. Hm(Cl) is a Hilbert space which is chosen as the basic space in many variational formulations of partial differential equations. However we have seen two particular solutions in Section 1.1, namely, u = 1 for two dimensional exterior problems and u = 1/r for three dimensional exterior problems. They are certainly not in the space L2(tt). So some different spaces have to be introduced. The following lemmas are useful in introducing some new spaces. Lemma 5. We assume that dim (£1) = 3 and u € Co°(fi), then f T^^

dx
Jn\x-y\

[ \Vu(x)\2 dx,

(1.18)

Jn

where i / 6 K 3 . Proof.

We have 3

»/E*gW>gf£* fc=i 3

dx -Kk J £-? fc=1

,- ,, F-2/1

J,

\*-y?

dx.

14

Numerical Methods for Exterior

Problems

Using the Cauchy inequality to get

' f u2{x) J-, {xk - Vkf

u2(x)

/ 1 Ma dx ^ 2 J \x-y\z

\j

^ u t Y?k=i ^glffi

\x~y\'

\x-y\*£^

=

\Wu(x)\2dx

dx

*' w n i c h g i y e s the result.

a

Lemma 6. We assume that dim (Q) = 2, 5 ( 0 , 1 ) C Qc and u £ Co°(fi), then

u2{x)

r

f Nu(x)

dx<4

L N 2 In2 \x\ Proof.

• dx.

(1.19)

We have

2

2

r

/„£

jM.u{x)^L^2dx dxk

'\x\ In \x

2

JU

f

d

k=idu

2

{x) dxk

Xk 2

Id In

dx=

A ~L

u2(x) dx. n la;!2 In2 |a;|

Then following the same lines we obtain the result.

•

Let us define a set HQ'*(Q), the closure of CQ°(£1) with respect to the norm | • |i. With this norm it is a Hilbert space. By Lemma 5 and Lemma 6 it is easy to see that Lemma 7. / / dim (fi) = 2 and H0'*(Q) is equivalent to

B{0,RQ)

C

Q.C,

then the norm \ • \\ in

2\ 2

i,*

•1? + |x|ln|z/,Ro|

Proof. We apply Lemma 6 with a mapping x —> X/RQ and get the result directly. • Lemma 8. / / dim (Q) = 3 and O € fic, then the norm | • |i in H0'*(Cl) is equivalent to 2\ 2

|i,» —

l? +

Exterior Problems of Partial Differential

Equations

15

For simplicity we will always assume that \x\ > Ro in Q later on. We consider the variational formulation of (1.15),(1.2) first. Let g G Hll2{d£l). By the inverse trace theorem of Sobolev spaces there is a function wo G HX{D,), such that uo|an = ,

ffo'*(fi)>

Vv e

(1-20)

where a(u,v) = /

Jn

Vu-Vvdx.

By the Lax-Milgram theorem we get easily: Theorem 14. The problem (1.20) admits a unique solution. The dual product < -, • > can be realized by Lebesgue integrals. Owing to (1.19) and (1.18), if |x| In | i / i ? o | / £ L2(Q) for two dimensional problems, we can set < / , v >= f^fvdx, and \x\f G L2(fl) is suitable for three dimensional problems. To define the variational formulation of the Neumann problems some other spaces are needed. Letting B(0,R\) D f2c and R2 > Ri, we consider a cut-off function £ G C°° such that ((x) = 1 for |x| < i?i, £(x) = 0 for | i | > R2, and 0 < C(z) < 1 for all x. We define H^iSl)

= {u; {u G H^Q f]B(0,

R2)), (1 - Qu G

H^(Sl)}.

Equipped with the norm || • ||iiH, the set iJ x '*(fi) is a Hilbert space. The definition of H1'*(Q.) is independent of the chosen of the cut-off function, because if we take another cut-off function £i, certainly (£ — Ci)u G H1, then the set i71'*(J7) keeps the same. Lemma 9. For two dimensional domains, 1 G i71'*(Q). Proof. Let £ be the function defined as above. Let Ci(x) = ((x/a). is a constant. Then f |VCi(x)|2da:=4 /

\V((x/a)\2dx

= f

|VC(*)| 2 dx.

a>0

16

Numerical Methods for Exterior

Problems

Let a —> oo. £1 is uniformly bounded in iJ 1, *(fi). We extract a subsequence convergent weakly. On the other hand £%o [ | |2! 21, , p ,(ClW - I ) ' ^ = 0. a

^°°Jn

\x\2\n

\x/R0\

Therefore the limit is 1, which is in the space JH"1'*(S7). D Letting g G H~1^2(dQ,), the variational formulation of the problem (1.15)(1.3) is: Find u G Hl'*{n), such that a(u, v) = < / , v >a + < g, v >da, to G H1'*^). (1.21) Certainly the problem (1.21) is not uniquely solvable for two dimensional problems. Because if u is a solution, u + C is also a solution, where C is an arbitrary constant. We modify the problem as the following: Define V = tueHl'*(n);

/ udx = 0\. /fin s(o.iii) J Then consider the problem: Find u g V , such that a(u,«) = < / , v > a + < ff, u >da,

Vw G V.

(1.22)

Lemma 10. The problem (1.22) admits a unique solution. Proof. We show that | • |i is an equivalent norm in V, then by the LaxMilgram theorem the conclusion follows. If \u\i — 0, then u is a constant on ft. Since fnn B,Q R > u dx = 0, u = 0 on il. By the closed graph theorem it is an equivalent norm to || • ||i *. Therefore there is a constant C, such that IMI2,* < C\u\\ = Ca(u, u),

Vw G V.

Returning to the problem (1.21), we have the following: Theorem 15. The sufficient and necessary condition for (1.21) admitting a solution is: < / , 1 >fi + < 5,1 > a a = 0.

(1.23)

Moreover, ifu\ andu^ are two solutions to (1.21), thenu\ — ui is a constant in Cl. Proof. If u is a solution, we take v = 1 in (1.21), then a(u, 1) = 0, which shows (1.23) is necessary. To show it is also sufficient, we solve the problem (1.22) first and get a solution u GV c ff1'*(ft). Letting v be an arbitrary

Exterior Problems of Partial Differential

Equations

17

in i? 1 '*(fi), we take a constant C, such that v + C G V. Then we take v + C as the test function v in (1.22) to obtain a(u, v) = a(u, v+C)-a(u,

C) = a(u, v+C) =< / , v+C >n + < 9, v+C >da

Noting (1.23), (1.21) follows immediately. Let u\, U2 be two solutions to (1.21), then a{u\ — U2,v) — 0. Let v = u\ — U2, then we get \uy — «2|i = 0, which yields that u\ — u-i is a constant in fl. • The situation for three dimensional problems is different: Theorem 16. If dim, (17) = 3 ; the problem (1.21) admits a unique solution. Proof. We need to verify that | • |i is an equivalent norm. Let |u|i = 0 , then u is a constant C in Q,. By the definition of || • Hi,*,

°2 ,

/ •j—pr d,X Ja

< OO,

which gives u = C = 0. Thus | • |i is a norm. By the closed graph theorem, it is equivalent to || • ||i *. •

1.4

Helmholtz equations

We consider the wave equation, d2u

w - A"-

where the variables are normalized to make the wave speed to be 1. Using separation of variables, we set u(x, t) = X(x)T(t), then X and T satisfy T"

AX

The general solution T is T = c\ cos wt + ci sin uit,

X = u>2.

Related to u> the solution u is u = Xl(x)cosLjt

+ X2{x)siviwt

where X(x) = X\{x) + ?X2(a;).

= Re

(X(x)e~iwt),

18

Numerical Methods for Exterior

Problems

To solve X(x) we work with complex-valued functions in this section. In general we investigate the exterior problems of the Helmholtz equations, - ( A + A)u = / ,

AeC,

(1.25)

with the boundary condition, u = 0,

xedfl.

(1.26)

With respect to the operator —A we define two sets: the spectrum, cr(-A) = {A € C;Im A = 0,A > 0}, and the resolvent set, p ( - A ) = C\tr(-A). Theorem 17. Let A £ p(-A), f € L2{Q), then the problem (1.25), (1.26) admits a unique solution u € HQ(Q). Proof. The assertion is proved by applying the Lax-Milgram theorem. Clearly only the coercive condition of the sesqui-linear form, a(u, v) =

(Vu • Vw — Xuv) dx,

Jo.

is to be verified. Let A = Ai + i\2- If A2 = 0, then Ai < 0, then |a(u,u)| > min(l,—A)||u||i. Then existence and uniqueness follows. If A2 =£ 0, we take a constant e € (0,1) and make the following estimation: v2|a(u,u)| >|Rea(u,u)| + |Ima(u,u)| > eKea(u,u) + |Ima(tt,u)|

^ ( H f - A i H ^ + IAal-Hg. Let e be small enough, so that a = \\-z\- e\\ > 0, then the right hand side has a lower bound min(a,e)||u||i, which yields the conclusion. D We turn now to consider the case of A = w2, w > 0. The particular case of A = 0 is the subject of the previous sections, where we know that HQ{Q) is not a suitable working space. To study this problem, we derive two particular solutions to the homogeneous equation first. Let dim (Q) = 3. If u = f(r), r = \x\, then / satisfies the Bessel equation: dr2 r dr There are two particular solutions: ux = - i - e i w r , 4irr

u2 = -?-e-iuJr. 4irr

(1.27)

Exterior Problems of Partial Differential

19

Equations

The first one is known as an "outgoing wave", and the second one is an "incoming wave", because when we return to the wave equation (1.24) and consider the solutions of separation of variables, they are W! = J - e - M « - r )

U2 =

A-KT

i e -M^). VKT

The first one represents a sphere wave spreading with a speed of one, and the second one represents a sphere wave converging to the origin, also with a speed of one. It is observed that for these two particular solutions both w and Vw are not in L2(fl). To choose a suitable space, we consider a weighted norm, ||/»Vu||o + ||pw||o, p = jhp, then the particular solutions are bounded with respect to the norm. However if we want to achieve a result of uniqueness, one of wi or W2 should be excluded in the definition. Relatively the outgoing wave is applied more often, because the scattering problem in scientific computing is: To study the reflecting of an obstacle under a known incoming wave. The outgoing wave satisfies the famous Sommerfield radiation condition: __luJU

0(-),

=

which leads to the inner products, (u,v)0=

/ u-vdx, Jn

(w,u)0,p =

(w, v)v = (u, v)0tP + (Vw, Vv)0tP + I (

(pu,pv)0,

iuju J , (

iuv

and the Hilbert space, V = {u £ #/ o c (Q);w| 9 n = 0, ||w|| v = (u,u)\f2 < oo}. It is easy to see that for the case of UJ = 0, V = H0'*(Q). Let a(x) = 1 + |a;| and Zv(fi) = {/;
VweC 0 °°(fi).

(1.28)

To prove well-posedness of the problem, we start by proving Rellich's estimate: L e m m a 1 1 . Let u £ C2 be a solution to the equation Aw + Aw = 0 for \x\ > Ri, A > 0. If u is not identity to zero, then there exist constants

20

Numerical Methods for Exterior

Problems

p > 0 and R2 > Ri, such that

L

\u(x)\2dx

>pR,

VR>R2.

Ri.<\x\
Proof. We consider sphere coordinates and the sphere harmonic operator on a unit sphere, Ao, then A - —2 dr Let Snj,

-— r dr

—A r2

n £ N, j = 0,1, • • • , n be the eigenfunctions to Ao, then -A0Snij

= n(n +

l)Snj.

Setting v = ru, we see v is not identity to zero either, and satisfies / d2

1

\

There exist a r* > RQ, such that v | r = r , is not identity to zero, then there is a sphere harmonic function Snij to make fl(r*) — / Js*

v(r*x)Sntj(x)dx^0.

By this we define a function g{r). We have

o=

L{^+^A°+A)

=g"(r) + \g(r) - " ^

v{rx)Sn j{x) dx

'

1}

J^ v(rx)SnJ(x)

dx

^ ( r ) + ( A _ 2 f e + l)), ( r ) . Therefore g is a solution to the Bessel equation, g(r) = ^{ClJu{V\r) + c2Yu(V\r)), where v = n + | , J„, Yv are Bessel functions, and c\ and ci are two constants. By the fact that g(r*) ^= 0, we have |ci| + | c a| > 0. The asymptotic behavior of the Bessel functions as r —> 00 is: \/rJv{r)

- J-cos(r

- an) + O I - J ,

Exterior Problems of Partial Differential

Equations

21

y/rYv{r) = J - sin(r - an) + O ( - ) , where an — 7r(n + l ) / 2 . Using this relation and / 2 \u(rx)\2r2 dx = 2 \v(rx)\2 dx > \g(r)\2, Js Js we get \u(x)\2dx>

/ JR!<\X\
I

\g{r)\2dr

JRi

[ (|ci|2cos2(V^(r-a„)) + |c2|2sin2(V^(r-a„)))dr * JRX 2 /"^ H— / (cici + c 2 cl) c o s ( \ / A ( r - a „ ) ) sin(\/A(r - a„))-

(| Cl | 2 + | c 2 | 2 ) d r + - /

= - /

(|Cl|2-|c2|2)cos(2yA(r-a„))dr

2 /"^ + - / (C1C2 + c2cT) cos(v/A(r - a n )) sin(\/A(r - an))dr + 0(logr). ^ ./iii

We notice that the second term and the third term are bounded. Thus the conclusion follows. • Lemma 12. If u is a solution to the homogeneous problem (1.28), then u = 0. Proof. The case of A = 0 has been investigated in Section 1.3. It remains to study the case of A > 0. Let B{0,Ri) D flc. We take a cut-off function £ € C£°(R 3 ) such that £ = 1 on B{0,R\). By the properties of solutions to elliptic differential equations, u is an analytic function, then we have a(u, £u) = 0. Let fti = fi P) B(0, R\). We notice that Au + Au = 0 and apply the Green's formula to get a(u, £u) = / (Vu • V(£u) — Au£u) dx Jn = / (Vu • V(Cu) - Au
(Vu • V(C«) - Au(>) da;

— / (Vu • V(C«) — Au£u) dx + /

V u • uuds.

JUi

J\x\ = Ri

22

Numerical Methods for Exterior

Problems

The first integral is real, thus 0 = Im a(u, (u) = Im / Vi J\x\=R1

Ids.

Then it holds that

/ J\x\ = Ri

du dr

J

ds

IU1U

du dr

J\x\=Rr

+ u2\u\2

ds,

which yields u>

J

\2dx<

du dr

IIJOU

< oo.

Then the Rellich's estimate yields u = 0 for r > R\, but u is an analytic function, so u = 0 on ft,. D To study existence we consider the Newton potential, u x

() = /

(1.29)

f{vW{x-y)dy,

JM3

first, where K(x - y) = - L - I ^ e ^ l * - ^

A G c

.

Lemma 13. If |A| < A, u € C 2 (M 3 ) ; \\ufv = \\u\\lp + ||V«||§,p + | | ^ t\/Au||o < oo, / S L 2 (R 3 ), / is compactly supported, and —(A + A)M = / , then u can be expressed in terms of (1.29), and the estimate, (1.30)

ll«llv f f holds, with a constant C, depending on A. Proof. We take a cut-off function C, e C°°(R 3 ) with supp C C and C(x) = 1 for |a;| < 1. For a given x € M3 let a > \x\. Then C{x/a)u(x) = / ((y/a)u{y)8{x

B{0,2),

- y) dy

= - J C(y/a)u(y)(A + X)K(x - y) dy = - J{A(C(y/a)u(y))K(x

- y) + \C(y/a)u(y)K(x

- y)} dy, (1.31)

A{C{y/a)u{y))

= ((y/a)Au(y)

+ u(y)&C(y/a) + 2V<(y/a) • Vu(y)).

Exterior Problems of Partial Differential

23

Equations

We see that f(u(y)A<:(y/a)

+ 2V((y/a)

= j VC(y/a) • ((Vu(y))K(x

• Vu(y))K(x

- y) dy

- y) - u(y)WK(x

= J VC(y/a) • ((Vu(y) - iV\u(y))K(x

- y)) dy

- y) - «(y)(V - iV\)K(x

- y)) dy.

Since u £ V, the right hand side tends to zero as a —> oo. Then letting a - » o o i n (1-31), the expression (1.29) follows. Next let us verify the estimate (1.30), which is a direct consequence of the Holder inequality. For example „du l^-iu-ullo

wl3 d

= J C

1'f/'Mr-V^" P 4' 47r J

-J U

\x — y\*

V

2

dx

or

\x-y^

V

dy

) [J \x-y\^{lV+\y\)M1'2) **'

where f

dy

J \x-y\^{l

f l/2

+ \y\?\y\

dy 5/2

C 5/2

~ J \* ~y\ \y\

~

W

We get

\\--luu\\Q
j

lx_yl3/w

dxdy

+ \y\fdy = C\\f\\la.

The other two terms can be estimated in a similar way.

•

Lemma 14. If f is compactly supported, |A| < A, then the solution u to the problem (1.28) satisfies the estimate, o,nn B(o,R) + ||/||O,CT), where R > R2 > R\, B(0,Ri) and R.

(1.32)

D QC, and the constant C depends on A

Proof. We take a cut-off function £ € C°°(R 3 ) with supp ( C B(0,R), 0 < C < 1, and £ = 1 for \x\ < R2. The estimate of (u is the standard

24

Numerical Methods for Exterior

Problems

H1-noira estimate for elliptic equations.

ic«i?=(v(c«),v(c«)) =(Vu, V(C2u)) + ((VCK V(C«)) - (V«, (VC)C«) = ( / , V(C 2 u)) + X(u, C2«) + ((VCK V(C«)) - (V(C«), (VC)«) + II(VC)«||§, which gives l u li,nns(o,fl2) - IC u li,«ns(o,fl) < C(llMllo,nriB(0:-R) + ll/llo,nnB(o,fl))(1.33) To estimate the exterior part we notice that the Lemma 13 also holds for all u £ V, which is obtained by approaching u by a sequence of smooth functions. We take another function Ci £ C°°(E 3 ), 0 < Ci < 1, Ci = 1 for |a;| > R<2 and Ci = 0 for |a;| < R\. Extend u by zero to fic, then u is defined on R 3 . (in satisfies - ( A + A)(Ciu) = Ci/ - (ACi)u - 2(VCi) • Vu. By Lemma 13 it holds that Kiu\\v

< C ( | | w | | o , n n B(OtR) + \\f\\o,a

+ || Vu|| 0 ; . B (O,.R 2 )\B(O,.R 1 ))-

Then by (1.33) the inequality (1.32) follows.

D

L e m m a 15. If f is compactly supported, |A| < A, then the solution u to the problem (1.28) satisfies the estimate, \\u\\v < C\\f\\Q,a,

(1-34)

where the constant C depends on A and the domain Q. Proof. If it were not the case, then there would be a sequence {K, fn,un}, such that | | « n | | v = 1,

l|/n||o,<7 < ! / « • •

We take a sub-sequence, still denoted by { A n , / „ , u n } , such that An —> A, un —*• u in V. Once more by Rellich's selection theorem there exists a sub-sequence of this sub-sequence, such that \\un — u||o,nn B(0,R) —> 0- By (1.32) un —> u strongly in V, and u satisfies a(u,v)=0,

Vi;eC 0 °°(fl).

Exterior Problems of Partial Differential

Equations

25

By uniqueness u = 0. On the other hand passing to the limit we get ||u||v = 1- This is a contradiction. • Finally we prove the existence theorem. Theorem 18. The problem (1.28) admits a unique solution. Proof. We prove existence. First we assume that / is compactly supported. We take a constant e > 0 and set Ae = A + ie. By Theorem 17 the problem (1.25),(1.26) admits a unique solution ue £ HQ,{SI) with respect to A£. Let £„ —» +0, then y / A £n —> u>. By Lemmal5 there is a sub-sequence {uEn}, converging weakly to u in V. Taking the limit in (1.28), we verify that u is a solution. For general / we take a sequence {/„} with compact support and converging to / in L^(fi). Then by Lemma 15 the corresponding un is a Cauchy sequence in V. Taking the limit in (1.28), we get the desired solution. • 1.5

Linear elasticity

Let the displacement in an isotropic body be u = [u\, 112,«3)T', where by T we denote transpose of a vector or a matrix. Let the strains be /

/

N

1 / Qui

N

duj

e i i (u)=e J - i (u) = - ( _ + _ ) ,

\
Let (j,j(u), 1 < i,j < 3, be stresses, then the Hooke's law reads Gij{u) = o-ji(u) = A ^ e f c f c ( u ) 1 Sij + 2n€ij(u), where A > 0, fi > 0 are the Lame coefficients and 6^ is the Kronecker's symbol. Let the Young's modulus and the Poisson ratio of the material be E and i>, E > 0, 0 < P < ^, then the relations between them and the Lame coefficients are the following: _ M3A + 2/i) A + /i '

A 2(A + /i)'

Let the external force be / = (/1, /2, / a ) T , then the equilibrium equation is 3

„

- E s ^ » = /i, l < i < 3 . 1=1

J

26

Numerical Methods for Exterior

Problems

The above formulas lead to the Lame equation for u: —p, A u — (A + //)grad divw = / ,

xG

fi.

(1.35)

We consider the Dirichlet boundary condition, u = 0,

x£dfl,

(1.36)

first, which is associated with a fixed boundary. We will work in the space , and the corresponding bilinear form is

f

3

a(u,v) = / y~] (Jij{u)eij{v) dx = / < Adivudivw + 2\i \_, eij{u)eij(v)

f dx.

i,j = l

Let V be the dual space of V. We assume that / € V, for example, / G (Ljoc(Q))3 and JQ r2\f\2 dx < oo. The weak formulation of the problem (1.35),(1.36) is: Find u£V, such that a(u,v) =< f,v >,

VveV.

(1.37)

To study the existence and uniqueness, we need Korn's inequality: Theorem 19. It holds that

t/0(£)'*s'Ei>>*.

<-)

for all u £ V. Proof.

It would be sufficient to consider u G (C^°(Cl)) only. Let Qii •

•^

=eij(u)

+ rij,

(1.39)

where Uj

=

1 dui duj 2{dx~ ~ ~dx~>'

then we have dui dxj

duj dxi

= 4-- r

2

-

(1.40)

Exterior Problems of Partial Differential

Equations

27

Applying the Green's formula and noting that u\ga = 0, we get

f ^i.^idx= Jn dxj

dxi

[ ^i.^idx.

(i.4i)

Jn dxi dxj

By (1.40) and (1.41) we get 3

Y

„

3

I rl{u)dx=

j=iJn

Y

f(divu)2dx<

[^Mdx-

i,j=iJn

<£ f e%(u dx.

Ja

i,j=ijn

(1.42) By (1.39) we also have

(g)!<2(4W+>i). The above two inequalities imply (1.38).

•

Applying the Lax-Milgram theorem, Theorem 19, and Lemma 5 we obtain the following immediately: Theorem 20. The problem (1.37) admits a unique solution. Next let us consider the Neumann boundary condition: 3

2_\aij(u)uj

— 9ii

x

€ 9Q, 1 < i < 3,

(1'43)

which is associated with a distributed force g applied on the boundary d£l. Let V = (H1'*^))3. We assume that / e V and g € (L2(dQ))3, then the problem is: Find u £V, such that a(u, v) = < / , v >n +(g, v)dn.

(1.44)

For the well-posedness of this problem we need the following Korn's inequality: Theorem 21. We assume that Qo is a bounded domain, then there exists a constant C, depending on fio, such that \ x

3

u\h,u0
Y

1/2

I 4 H ^ + Hullo,no |

- n°

.

(1-45)

28

Numerical Methods for Exterior

Problems

This inequality is not trivial. The proof of it can be found in [ Leis, R. (1986)]. Applying Theorem 21 we have the following: T h e o r e m 22. The problem (1-44) admits a unique solution. Proof.

Owing to the Lax-Milgram theorem we need to prove a(u,u) > a||u||i „,

Vw £ V,

a > 0.

It is proved by contradiction. If there exists a sequence {«„} C V, such that ||Mn||i,* = 1 and a(un,un) < 1/n, let B(0,R) be a ball with large R, such that dfl C B(0,R), then {un} is bounded in (H1^))3, fii = D.DB(0, R). Then there is a subsequence, still denoted by {un}, such that un convergence strongly in (L 2 (f2i)) and weakly in (i7 1 (Qi)) . Let the limit be u. Since R is arbitrary, u is denned on Q. Because a(un,un) —* 0 a s n - t oo, €ij(un) converge to zero in L2{Q), 1 < i,j < 3, so e,j(u) = 0, which implies u is a rigid body motion, that is, a linear function. Since Hunlli,*,!^ < 1, we have ||u||i,*,ni < 1> a nd thus ||u||i,* < 1. But u is linear, hence u = 0. Owing to the Korn's inequality (1.45) un converges strongly in (i7 1 (fii)) . We define a cut-off function C, G C£°(f2) such that 0 < £ < 1 and C = 1 if \x\ is large enough. Then a{C,un,C,un) —* 0. Owing to the Korn's inequality (1.38) C,un converges with respect to the norm | • |i, which means (u„ converges in H0'*(£l). We already know that un converges strongly in (i7 1 (fii)) for any large R. Therefore un converges strongly in H1 <*{&). Then ||u||i,» = 1, which contradicts u = 0. • We turn now to two dimensional problems. There are two cases: 1. Plane stress. <7n,(T22)cri2 depend on a;1,a;2 only, and a\s = o~23 = 0-33 = 0.

We eliminate ei3, €23, £33 in the Hooke's law, then expresses the stresses as

o-i:j(u) = o-ji(u) = 1_f>2

( ^2€kk(u)

j Si:i + Y ^ - r e y ( u ) -

Exterior Problems of Partial Differential

Equations

29

for i, j = 1,2. Therefore if we denote -

PE

1"="^'

E ^~ 2(1 + 9)'

the equation for u keeps the same as (1.35). Only the parameters are different. 2. Plane strain. £11,622,612 depend on xi,x% only, and ei3 = £23 = £33 = 0. The Hooke's law is

<Jij(u) = crji(u) = A I ^efefe(u) J Sij + 2fj,eij(u),

for i, j = 1,2. The parameters in the equation keeps the same as the three dimensional case. The argument for the existence and uniqueness of the Dirichlet and Neumann problems is analogous to that in Section 1.3. We only give the results here. The Dirichlet problem admits a unique solution. The sufficient and necessary condition for the Neumann problem admitting a solution is: < f> 1 >fi + < Q-i 1 >dn= 0) a n d the solutions are not unique. An arbitrary constant vector is a solution to the homogeneous Neumann problem, and if u\ and «2 are two solutions, then ui — u
1.6

Bi-harmonic equations

We investigate two dimensional bi-harmonic equations, A2u = / ,

(1.46)

in this section. One background of this equation is the bending problem of thin plates, where u is the deflection, the displacement perpendicular to the plates. Another background is plane elasticity. If the stresses are expressed by CTU = f^f, <*\i = gf^, 022 = f^r, where A is the Airy stress function, then A satisfies (1.46). First of all let us derive the corresponding bilinear form. Suppose that u e C4(f2) and we take v G CQ°(£1). Integrating by parts we get

30

Numerical Methods for Exterior

/

Problems

A2U-vdx

JO.

r dAu

,

r „A

f dAu

,

f _„A

= / —-—-yds— I Jan ov Jn

„

,

vAu-Vvdx

„

,

= / —-— • vds — / vVAu • Vv dx Jan dv Jn fn j

n

_. /<93u <9v \9a; 3 3a;i

<93u dv dx\dx2 0x2

d3w <9v dx\dx\dx\

d 3 u dv . dx\dx2}

f dAu , /" _A 3», ; / —— • vds — / z/Aw • — as Van oz/ JdQ dv Jdi I an + Jnf

f dAu

/ ( 9 2 u <9v \dx\ dx\

<92u d v dx\dx2 dx2

<92u dv dx\dx2 dx\

[n_-)(&u&v_ JQ \dx\dx\

,

f _A

d2u dv dx\ dx2

\ )

d2ud2v\ dx\dx\)

_cPv__&v_ dx\dx2dx\dx2

dv ,

= / —— • vds — / vAu • dv— as Jan dv Jan f Jan f /,

-\ (d2u \9xj

2

* 2

2

-s /" d u

Jn

<92u dxidx2 d "

JQ

*

d2u 2 \ dv dx\V<1) dv

2

/ 2

d2u

2N

\dx\dx\

A dv ,

dx\dx2dx\dx2

dx\dx\)

f dAu , /• _A <% = / —-— • vds — / vAu • 7— as Jan o " ./en <^ /" „

-x f°2u

2

dQ

~s d f ds \

d2u dx2

d2u dx\dx2

7n

\<9a:2(9:r2

Jn

„ ^2«

°2u . ,

2\

o.

dv

j <92u dx\

dx\0x2dx\dx2

1

dx\dx\)

where v = (vi,v2) is the unit outward normal vector, and s is the unit clockwise tangential vector.

Exterior Problems of Partial Differential

31

Equations

Fig. 2 For plane elasticity problems, v is the Poisson ratio 0 < v < ^, and for the Stokes problem in fluid dynamics, v = 1. We define

a(u,v) = / Jil

vAu-Avdx _ fd2ud2v \dx\dx\

[ Jn

d2u d2v dx\dxi dx\dxi

d2u d2v\ dx\dx\) (1.47)

the lateral shearing force Tu--

dAu dv

(1_-}9_(&u

dx$VlV2 (1.48)

dx\dx-i ("* ~ "*) ~

ds \dx\

and the bending moment load, Mu = vAu

+(1-D)

'd2u dxl"1

,

„

d2U

32U 1 2

^ *dx\dx
r, 2

' '

(1.49)

then a(u, v) — I

Jan

Tu • vds — I

Jen

Mu • v— ds = / fv dx.

"

Jn

(1.50)

For plate problems there are several kind of boundary conditions, namely, 1. Dirichlet boundary condition, u = gi, | ^ = ff2! 2. Neumann boundary condition, Tu = gi, Mu = g^; 3. simply supported boundary condition, u — gi, Mu = g%.

32

Numerical Methods for Exterior

Problems

To study these boundary value problems we define some spaces with respect to the exterior domain f2, and at first we prove a lemma. Lemma 16. We assume that B(0,1) C fic and u G „2^ A Ar |Vu(z)| 2 U{X)

f

L i d 4 In4 Id Proof.

dx< -[^

-VJn\x\ 2 In2 \x

CQ°(Q),

then

dx.

(1.51)

We have 2

2r

/£

f

du{x) u(x)dxk

d« 2 (:r) dxk

2

Xk

dx

I4 In 3 Xk

• da;

4

I In 3 I

= f u2(x)( — + 7n V ; I |x| 4 In3 N ' H 4 In4 W

dx

2 >3 / -» 4 M 4 Ju \x\| In |x| Using the Cauchy inequality to get

u2(x) '/

• dx

x\4 In 4 |z|

But £ 2

(xkV

fc=i 1W

I f

-

u2(x)

[Xk?

\J N 4 l n 4 N ^ ,-

dx

J Id 2 In

dx D

1, which gives the result.

Let us define a space H0'*(£l), the closure of C^°(Cl) with respect to the norm [ - 12 - Equipped with this norm it is a Hilbert space. By Lemma 6 and Lemma 16 it is easy to see that Lemma 17. If to |2,*

=

C £lc, then the norm |-|2 in H0'*(Q) is equivalent

B(0,RQ)

\l +

|a;|ln|x/i?o|

+

|x| 2 ln"|x/.Ro|

(1.52)

Letting B(0, R\) D fic and R2 > Ri, w e consider a cut-off function £ G C°° such that C(x) = 1 for | i | < Ri, ((x) = 0 for \x\ > R2, and 0 < £(x) < 1 for all x. We define H2'*(tl) = {u;C« e

tf2(fif|i?(0,£2)),(l 2

- 0 « G #o'*(tt)}-

Equipped with the norm || • ||2,* the set H '*(Cl) is a Hilbert space. Being analogous to the space H1'*(fl), the definition of H2'*{0) is independent of the chosen of the cut-off function £.

Exterior Problems of Partial Differential

33

Equations

Lemma 18. Let p be a polynomial, the order of which is not greater than one, denoted by p G Pi{Q), then p G H2'*(Q). Proof. Let £ be the cut-off function defined as above. Let £i(x) = C,{x/a). a > 0 is a constant. Then d2(tiP) __ d2Ci +^i^P_ ^_^P_ + By Lemma 9 the semi-norm | • |i of the second and the third terms are uniformly bounded with respect to a. For the first term we have

L{^fit)dx^L{^k{x/a)p{!c)}/x = _

2 /

\ «

(x)p(ax) \

dx,

which is also uniformly bounded. Let a —> oo, then Q\p is uniformly bounded in H2'*(£l). We extract a subsequence converging weakly, the limit of which is p, which is in the space H2'*(Q,). • The variational formulation of the Dirichlet problem is: Given / G (ff^*(n)V, 5i G H3/2(dn),

and g2 G Hl/2{dQ),

find u G F 2 -*(Q), such

that u\aa = Si, f^lan = 92, and a(u, v) =< / , v >,

Vv G

ffo'*(n).

(L53)

where (iJ 0 '*(Q) 1 is the dual space of H0'*(Q,). Theorem 23. The problem (1.53) admits a unique solution. Proof. By the inverse trace theorem there exists w0 G H2'*{Q) satisfying the boundary condition, so u — UQ satisfies homogeneous boundary conditions. Applying the Green's formula we get fat*.

A , * - f j

{( g )

2

+ 2( £ ^ j

+ ( g )

2

} * _ CMi

in iJ 0 '*(fi). Then applying the Lax-Milgram theorem the existence and uniqueness follows. • The variational formulation of the Neumann problem is: Given / G z2 (H '*(Q))', and g2 G tf-1/2^), find u G # 2 >*(Q), such 5 I G H- ' {dSl), that 2

a(u,v) =gn

+ < g2, -^ >aa + < f,v>n,

Vw G ff2'*(fi). (1.54)

34

Numerical Methods for Exterior

Problems

T h e o r e m 24. We assume that v < 1 then the sufficient and necessary condition for (1.54) admitting a solution is: dv an + <92,-^ >dn + < f,v>n=0, Vv GPi(ft). Moreover, if u\ and u 2 are two solutions to (1.54), then u\ — « 2 £ Pi(^)Proof. It is easy to see that any p G Pi(O) is a solution to the corresponding homogeneous problem. We define a space V = {u G H2>*(Q); fQ~B(^OR^pudx = 0,Vp G Pi(fi)}. | • | 2 is a norm on V, because if \11\2 = 0, then u G Pi(fi), then by the definition of V we get u = 0 by taking p = u. We replace the space P"2'*(fi) in (1.54) by V. By the definition (1.47) a(u,u) > (1 — i^Ml- Owing to the Lax-Milgram theorem there is a unique solution. If v G H2'*(fl), we divide it into two parts v = v\ + V2, such that v\ G V, and v2 G Pi(fi). It is easy to see that (1.54) holds either for v = v\ or for u = i>2. Therefore u is also a solution to the problem (1.54). • To define the variational formulation of the simply supported boundary value problem we set V = {u £ H2'*(Q,);u\dn = 0}, then the problem is; Given / G V, gi G Hz'2{dQ), and g2 G H'1/2^), find u G H2'*(Q), such that U\OQ = gi„ and a(u, v) = < 52, -~- >dn + < f, v

>Q,

VveV.

(1.55)

T h e o r e m 25. We assume that v < 1. The problem (1.55) admits a unique solution. Proof. It has no harm in assuming g\ = 0, because by the inverse trace theorem we can find UQ G H2'*(Q) such that uo|an = 5i, then consider the problem for U — UQ. We only need to verify that | • | 2 is a norm in V. If not, then there would be a sequence {«„} in V such that |u n |2 —* 0 and Vw„ \x\ In |x/i? 0 |

+

|rc|2 In

1.

(1.56)

\X/RQ\

Then by the compact imbedding, there is a subsequence, still denoted by {un}, converging strongly in i I 1 ( f 2 p ) B ( 0 , P i ) ) and converging weakly in H2'*(D,). Since |u n | 2 —> 0, it also converges in H2(Qf]B(0,Ri)), then u, the limit of {«„}, is a polynomial in Pi. By the boundary condition, u = 0 on D. f] B(0, P i ) . P i is arbitrary, so we can extract a diagonal subsequence such that un —> 0, in P 2 (fi |~) B ( 0 , P i ) ) for any P i . Then by the definition of the space H2,*(Q,), u = 0, then ||w||2,* = 0, which contradicts (1.56). •

Exterior Problems of Partial Differential

1.7 1.7.1

Equations

35

Steady Navier-Stokes equations -linearized problems Navier-Stokes

equations

The motion of viscous incompressible fluids is governed by the system of Navier-Stokes equations, dii

— +u-Vu

+ Vp = i)Au + f, V - u = 0,

(1.57) (1.58)

where u is the velocity, p is the kinematic pressure, / is the external body force, and the constant v represents the kinematic viscosity. For the steady case §£• = 0. Introducing the dimensionless Reynolds number VL —, v where L and V are reference length and velocity respectively, the equation (1.57) becomes Re=

Re • u • Vw + Vp = Aw + / ,

(1.59)

where u,p and / are new nondimensional quantities. For low Reynolds number flow the first term on the left hand side is small comparing with the first term on the right hand side, so it can be neglected. The equation (1.59) becomes - A u + Vp = / ,

(1.60)

which is the Stokes equation. Let us consider a typical problem: Let Qc be a ball B(0,1), which is a rigid body moving in a viscous incompressible fluid with a constant speed. We fix the coordinate system on this body, then the equation (1.60) is combined with the boundary conditions u = 0,

1 6 95(0,1), lim u = Woo,

(1.61) (1-62)

|x|—»oo

where — u<x, is the speed of the body. solution to the problem (1.60)-(1.62):

Stokes has derived the analytic

»w=«~-^4 x|2vx (w)H v * Vx (w 3

„ / 1

We are going to investigate the boundary value problems for general exterior domains.

36

1.7.2

Numerical Methods for Exterior

Stokes

Problems

equations

We consider the equation (1.60) with homogeneous boundary condition u = 0,

x e dfl

(1.63)

first. The dimension of the domain Q is n, n = 2,3. For the time being we don't study the behavior of the solution at the infinity, and leave the problem to later on. We introduce a Hilbert space, V={u€

(ff o 1 '*(fi)) n ;V-u = 0},

n = 2,3

and a bilinear form

a(u,v) — /

Jn

Vu:Vvdx.

Assume that / belongs to V, the dual space of V, then the weak formulation of (1.60),(1.63) is: Find uGV, such that a(u, v)=,

Vv e V.

(1.64)

Theorem 26. The problem (1.64) admits a unique solution. Proof. By Lemma 5 and Lemma 6 the bilinear form a(-,-) is coercive in V, then owing to the Lax-Milgram theorem existence and uniqueness follows. • To investigate the problems with inhomogeneous boundary conditions and the existence of p in (1.60), we need some lemmas. Lemma 19. Let g be a function in (Hl/2(dQ)) 0. Then there exists a function u £ (i71,*(f2))

that satisfies $ang-vds such that

=

u\dQ = g,

V - M = 0,

and

Mi,.
V> = 0 .

-£- = -g.T ov

Exterior Problems of Partial Differential

37

Equations

on dfl, where T is the unit clockwise tangential vector, and Let u — curl tp, then dip 0 = g • v, or and dip

< C||fl||

U- T

so u is the desired one. When g • v does not vanish on dd, let p be the solution to the Neumann problem

*-•>• £

df2

By Theorem 15 the solution p exists and belongs to the space H1'*(fl) and |p|i < C||g||i. Moreover, since g G (iJ 1 / 2 (9fi)) , by the theory of elliptic equations, p G H2 locally. Then by the expression of p in terms of single layer potentials, we see that \p\2 < C\\g\\ i. We set u\ = Vp, then V-«i = 0 , and «i • v = g • v on dfl. u — u\ can be denned as the previous case. • L e m m a 20. For any f G L2(Q,), there exists u G

(H0'*(Q))

, such that

V • u = f, and ||u||i,* < C||/||o with a constant C depending on Q.. Proof. Let {/ m } C C£°(fi) be a sequence approximating / in L 2 (fi). We define functions ipm to solve Aipm = fm on E™. ipm can be expressed in terms of the Newton potential. Owing to the theory of singular integrals (see [Calderon, A.P. and Zygmund, A. (1957)]), \1>m\2 < C\\fm\\0.

(1.65)

Let us consider the case of dim (0) = 3 first. Since fm is compactly supported, applying the expression of the Newton potential we deduce that V^m = 0 ( M ~ 2 ) as \x\ —> oo. We consider a cut-off function Ci S C°°(R), such that Ci(C) = 1 for £ < 1, C(£) = 0 for ^ > 2, and 0 < Ci(£) < 1 for all f. Let £(x) = ti{\x\/R), then d

(W„

<

^™

+

VV-r,

< da:,-•c

dxi

(VV» n

+ ^ |VW,

Therefore |CVV>m|i is bounded and Vipm can be approximated by functions in (C£°(M3))3 with respect to | • |i. Consequently Vipm € ^ ' ' ( E 3 ) ] . Applying Lemma 5 and (1.65) we obtain ||VV>m||i,* < C||/m||o- Then by the trace theorem of Sobolev spaces, ||VV>m||i an ^ C||/m||o-

38

Numerical Methods for Exterior Problems

For the case of dim (Q) = 2, we take R large enough such that B(0, R) D fic, then take constant vectors c m , such that JBI IB(0,R)

(VV'm + Cm) dx = 0.

Owing to the Poincare inequality |VV>m + Cm||0,B(O,fi) < C\Vlpm + Cm\i,B(0,R) < C\lpm\2By the definition of H1'* we see that W m + cm e ( t f 1 - " ^ ) ) . We also have |VV>m + c m ||i,* < C||/ m || 0 , and ||VV>m + c m | | i ) S n < C||/ m || 0 . ipm is the solution to the Laplace equation in Clc, so Jgn -§£*- ds = 0. Let g = — VV'm, or g = —Vipm — cm on <9fi. We apply Lemma 19 to get a function in (iJ 1, *(fi)) n , denoted by wm. Set Um = VV'm + Wm, Or Um - V ^ m + Cm + V)m.

We get um € [Hl'*{Sl)\ , V • um = fm, and ||u m ||i,* < C||/ m || 0 . m —> oo, then the conclusion follows.

Let •

Lemma 20 leads to the inequality: There exists a constant /3 > 0 such that inf

sup

. y ^ i

p6La(n),p^oug^i,.(n)j»iU940||u||i,,||p||0

>P,

(1.66)

which is the inf-sup condition, or Babuska-Brezzi condition, for this problem. We will make use of this condition later on for different problems. For the inhomogeneous boundary condition u = g,

x € dfl,

where g is a prescribed velocity field on the boundary, with g € {Hll2(d£l)) but in general fmg • yds may not be zero, by the inverse trace theorem of Sobolev spaces, there is a function u\ G (.^(fi)) with compact support on fi and satisfying ui|an = g. Then by Lemma 20 there exists «2 € (H0'*(Q,) 1 , such that V • «2 = —V • u\. Let u — u\ — 112 and p be the new variables, then they satisfy the Stokes problem with a homogeneous boundary condition, and the the external force term becomes / + A(ui + W2). It is easy to see that it belongs to the space V. By this way, the problem is reduced to (1.64). We turn now to the existence of p. Let us define b(u,p) — (p,V • u), then we deduce from the equation (1.60) that the equation for p in weak form is b(v,p)=a(u,v)-,

V v e ^ n ) ) " .

(1.67)

Exterior Problems of Partial Differential

39

Equations

Theorem 27. The equation (1-67) admits a unique solution p G L 2 (fl), and it holds that IIPIIO < C7(]|/||v + M i ) .

Proof.

(1-68)

The functional F(v) =a(u,v)-

is bounded in (H0'*(Q,)) v G (H0'*(Cl))

<

f,v>

and vanishes in V. Consider the operator A :

—> V • v G L 2 (fi). ^4 is linear and bounded. By Lemma

20 its range is L 2 (fi). Then kei(A)1- = R(A*), where ker is the kernel, * is the adjoint, and _L is annihilator. Now ker(j4) = V, therefore R{A*) = VL. Thus all functionals F are in the range of A*. On the other hand, by definition, every element in R(A*) is of the form C(Av), where £ is a functional on L2(Cl). By the Riesz representation theorem there is a unique p e L2(Q,), such that F(v) = C(Av) = / pAv dx = / pV • v dx. Jo. Jo, Thus the proof is complete. 1.7.3

Behavior

of solutions

at the

•

infinity

We need some lemmas. First of all let us prove the Wirtinger inequality. Lemma 21. Let S3 be the unit sphere inR3, S3 = 0B(0,1). and f is the average of f, then ||/-/||o
/// G

H1^) (1.69)

where V* is the projection of V on S3. Proof. We extend / by constant along all lines perpendicular to S3 to the domain fi0, the spherical shell of radii 1/2 and 1. Then |V*/| 2 = r- 2 |V/| 2 , r = |a;|. By the Poincare inequality we have

||/-/||o,no
ll/-/llU< C ll/-/lo,n 0 Thus (1.69) follows.

•

40

Numerical Methods for Exterior

Problems

Lemma 22. Let ( I c R 3 and u € Hl>*(Q), then f \u(R,s)\2ds<

\Vu\2dx,

4 / R

JS3

J\x\>R

where s € S3 and u = u(p, s), p = \x\. Proof.

It would be sufficient to prove the inequality for smooth u. Let Dr(R)=

r JR

\Vu(p,s)\2p2dsdp,

[ JS33 IS

then by the Wirtinger inequality we obtain 2

Dr(R)= f f Js3

~Js3

JR

JR

p

°P

dp

p2dpds+ f [ \V*u\2p2dsdp JR

Js3

( [ p~2 dp) \JR J

P

Therefore

ds + C~l [ JR

1

[

\u- u\2p2 dsdp.

JS3

f I '"-

u\2p2dsdp. (1.70) Js3 JR JS3 Disregarding, for the moment, the second term on the right hand side and letting R, r —> 00, we deduce that u converges strongly in L2(Ss) to some function u*. Set «o = U*, and w = u — UQ. Then D.'r(R)>Rf

U{R)\2

\u{r) -

ds + C~

lim / w(x) ds = 0. (1.71) -H°° J s3 Rewriting (1.70) with w instead of u, we recover the existence of a sequence {rm}, with rm —> 00, such that / \w(rm) - w(rm)\2 ds = 0, ° J s3 which, because of (1.71) implies lim

1_>0

lim / |w(r m )| ds = 0. -^°° J s3 Inserting it into (1.70) written with w in place of u and letting r —> 00 yields m

f \w(R,s)\2ds<^ [ \Vu\2dx. R Js3 J\x\>R Finally we show that UQ = 0. We have

(1.72)

Exterior Problems of Partial Differential

Equations

41

By Lemma 5 the first term on the right hand side is bounded, and by the estimate (1.72) the second term is bounded by O(logr) as r —> oo, so uQ = 0. • The above result shows that the space H1'*(Q) in the formulation (1.64) indeed implies a zero boundary condition at the infinity for three dimensional problems. 1.7.4

Stokes

paradox

For three dimensional problems, the result can be extended to nonhomogeneous boundary conditions at the infinity: lim

u = Utx,,

where UQO is a given constant vector. Because if the boundary condition on dfl is u = g, x € <9fi, we first solve the problem with the boundary condition lim u = 0,

U\QQ =

g-Uoo,

|rcj—>oo

to get a solution u\. Then ui + Uoo is the desired solution. The two dimensional case is different, because the unique solution to the boundary condition U\QQ, = —Moo and / = 0, is u = - « « , , which is not zero at the infinity if u^ ^ 0. As a result, the problem with boundary condition lim u = Woo,

u\en = 0,

\x\—>oo

and / = 0 is not solvable. This phenomenon is called "Stokes paradox", first discovered by Stokes. 1.7.5

Oseen

flow

Another approach to linearize the Navier-Stokes equation is the Oseen flow, u0-Vu

+ Vp = Au + f,

(1.73)

V-w = 0,

(1.74)

where UQ is a constant vector. For simplicity we assume that / S (L 2 (fi)) with a compact support. Taking a positive constant e we consider the following approximate problem: u0 • Vw + Vp + e u = Au + f,

(1.75)

42

Numerical

Methods for Exterior

Problems

V - u = 0, lim u = 0,

(1.76)

u\dn = 0.

(1.77)

|x|—>oo

We define a space Ve = {u £

(HQ(Q,))U

; V • u = 0} and a bilinear form,

a£(u, v) = (Vu, Vv) + (u 0 • Vu, u) + e(u, u). Then we have Lemma 23. The problem: Find u £ V£, such that ae(u,v) = (f,v),

Mv&Ve,

admits a unique solution. Proof.

Since (UQ • Vu, u) = 0, we have a e (u,u) = | u | f + e||u||g>min(l,e)||u||f.

(1.78)

Owing to the Lax-Milgram theorem uniqueness and existence follows.

•

To investigate the limit as e —-> 0, we define two spaces V = {u G ( i ^ ' * ( n ) ) n ; V • u = 0} and V = {u e (C£°(ft))"; V • u = 0}. Then we obtain Lemma 24. The problem: Find u £ V, such that (Vu, Vv) + (u0-Vu,v)

= (f,v),

Vv£V,

admits at least a solution. Proof.

Let a series of e tend to zero. By (1.78)

M i ^ I N o < ll/IIHMk < ci|/IM«li, so the solutions are uniformly bounded in V. Therefore we can subtract a subseries converging weakly in V. Applying Lemma 5 and Lemma 6 we get the strong convergence in Lfoc(Q). Passing to the limit the conclusion follows. • We remark that V can not be replaced by V in the weak formulation, otherwise (UQ • Vu, v) makes no sense. Moreover we remark that the space HQ'*(Q) in the formulation implies a zero boundary condition at the infinity for three dimensional problems. Using the same approach as for the Stokes equation, we can easily extend the results to more general boundary conditions, lim u = Uoo,

u|an = g.

(I- 7 9 )

Exterior Problems of Partial Differential

43

Equations

The two dimensional case is different. However we still have the following: Lemma 25. We assume that dim (f2) = 2 and UQ ^ 0. Then the solution satisfies ||w||o,r < C with r > 3. Proof. We \x\ < R, ((x) enough. Let (1.75)-(1.77).

define a cut-off function C G C 0 0 ^ 2 ) such that ((x) = 1 for = 0 for \x\ > 2R, and 0 < ((x) < 1 for all x with R large w = (1 — f)w, <j> = (1 — £)p, and (u,p) be the solution to Then (w,) solves «o • Vw + V + ew = Aw + / i ,

where A = -(u„-VC)u-(VC)p+(AC)« + 2V<-Vu + ( l - C ) / , /a = -VC-«. Noting that |VC| < 5 for R < \x\ < 2R, we have ||/i||o,, < C, ||/ 2 ||o ) 9 < C, 1 < q < oo, uniformly with respect to i?, and / i , / 2 compactly supported. Let the Fourier transform of w(x), (£), then they satisfy mo • £w + i£<j> + ew = —\£\2w + / i , i£-u> = hsome calculation we get W

l£|2/i-(£-/i)£ |£|2(™o-£+|£|2+£)

^ ; |^| 2 / 2 2

r

ing a proposition stated below we notice that |£j31 < \uo\ 1\uo • 2 £| 4- |£| and find (3 — | for the first term, and the second term is typical corresponding to the Laplace operator, where /a(0) = JQ fi dx = 0. Thus we have ||w||o,r < C with r > 3, where the constant C is independent of e. Letting R —> co and passing to the limit, we obtain the final result. • The following proposition can be found in [ Lizorkin, P.I. (1963)]. Proposition 1. Let u(£),w(£,) be the Fourier transform ofu(x),w(x), and u>(£) = $(£)«(£), and let $ be continuous together with the derivative

and all preceding derivatives for |£j| > 0, i = l , - - - n . j3 € [0,1) and M > 0 K

+ p . . . \t l icirl+/3 --^n

K

n+/3

^r1---^-

T7ien if for some

< M,

44

Numerical Methods for Exterior

Problems

where Kj is zero or one and K = Y17-1 ^j = 0,1, • • • ,n, the transform u —> w defines a bounded linear operator from Lq(Rn) into Lr(Rn), 1 < q < oo, 1/r = 1/q — (3, and we have ||w||o,r < C||Ml|o,g, with a constant C depending only on M, r, and q. Therefore the solution u also tends to zero at the infinity in the sense of this lemma. The results can be extended to more general boundary conditions (1.79). This is the essential difference between the Oseen flow and the Stokes equation, for which we already know that the boundary condition at the infinity is impossible for two dimensional problems. Uniqueness for the boundary value problems of the Oseen flow holds. For details, see [Galdi, G.P. (1994)].

1.8

Steady Navier-Stokes equations

We investigate three dimensional Navier-Stokes equations u- Wu + Vp = Au + f,

(1.80)

V - u = 0,

(1.81)

with the homogeneous boundary codition lim u = 0,

u\an = 0

(1.82)

|x|—+oo

first. This is a nonlinear problem, and to prove existence we need the Brouwer fixed point theorem: Theorem 28. Let B denote a non-void, convex and compact subset of a finite dimensional space, and let F be a continuous mapping from B to B. Then F has at least one fixed point. Corollary 1. Let H be a finite dimensional Hilbert space whose scalar product is denoted by (•,•) and corresponding norm by \ • \. Let F be a continuous mapping from H into H with the following property: There exists R > 0 such that (F(u),u)

> 0,

VueH

with |u| = R.

(1.83)

Then there exists an element u in H such that F(u) = 0 and \u\ < R.

Exterior Problems of Partial Differential

Equations

45

Proof. The proof proceeds by contradiction. Suppose that F(u) ^ 0 in the Ball B = {u G H; \u\ < R}. Then the mapping u —> —i?,F;"(, is continuous from JB into B. As the dimension of H is finite, and since B is non-void, convex and compact, we can apply the Brouwer theorem asserting that there exists u in B such that u = — R\pul\- Thus |u| = R and (F(u),u) = -R\F(u)\ < 0, since ueB. This contradicts (1.83). • We define two spaces V ={u£

(#o'*( f i ))

; V • u = 0} and V = {u G

3

(C£°(ft)) ; V • it = 0}. Assume that / G V, the dual space of V. Then the weak formulation of (1.80)- (1.82) is: Find u £V, such that (Vu,S7v) + (u-Vu,v)=,

VveV.

(1.84)

Let us introduce some lemmas first. Lemma 26. There exists a denumerable set of functions {tfk} whose linear hull is dense in V and the set has the following properties: (1)
(3) Given
0 there exists a linear combination of (p'ks, such that <£. fc=i

1,*

Proof. Since V is separable, V, as a subset of V, is also separable. Let {fk} be a basis in V of functions from V. We orthonormalise {<^fc} in V by the Schmidt procedure, to obtain another denumerable set {
(v-Vu,u) = 0,

Vu,wG(C£°(ft)) 3 .

(u'V«,to) = - ( » - V » , « ) ,

(1.85)

(1-86)

Proof. Property (1.86) is a direct consequence of (1.85) when we replace the u in it with u + w. To prove (1.85) we have / {v-Vu)-udx=

in

-

I v • V | u | 2 dx = 0.

2 JQ

Theorem 29. T/ie problem (1.84) admits at least one solution.

•

46

Numerical Methods for Exterior

Problems

Proof. We use the Galerkin method to prove existence. Let {
where Cfc satisfy (Vum,V
=< f,,

Vfc = l , - - - , m .

(1-87)

To prove the existence of Cfc, we make some estimates first. Multiplying (1.87) by Cfc and summing up with respect to A;, we obtain

) =< f,um > .

(Vu m , V u m ) + (u m • Vu

By (1.85) the second term on the left hand side vanishes. The right hand side is bounded by ||/||i/'|l u m||v- Let M be a constant such that M > H/llv, then we get (Vu m , V u m ) + (u m • Vu m , um) >< f,um>,

V|u m |i = M.

Owing to the Corollary 1 of the Brouwer theorem, the solution to (1.87) exists, and \um\i < \\f\\vLetting m —> oo, we can extract a subsequence of um, converging weakly in V, and by Lemma 5 the subsequence converges strongly in Lfoc(fl). Let u be the limit. For the second term on the left hand side of (1.87) we have (u • Vu, tpk) - {um • Vu m , ipk) - (u • V(u - um), tpk) + ((u - um) • Vum, ipk). Noting that (fk is compactly supported, we find the limit of the above expression is zero. Then we obtain (Vu, V(fk) + (u • Vu, tpk) = < f,,

V/c = 1, • • • , oo.

Since the linear hull of {(fik} is dense in V, the conclusion follows.

•

We turn now to the inhomogeneous boundary conditions, lim u = Uoo,

u\9a = g,

(1.88)

| a: | —*oo

and assume that g € (ii'1/2(<9fi)) . To prove the existence of (1.80),(1.81), (1.88) we need the Sobolev inequality. Let us prove it for a special case, the inequality for three dimensional domains. Lemma 28. For all u € C~(ft)

Ho,| < AHV«||o,i-

(1-89)

Exterior Problems of Partial Differential

Proof.

47

Equations

We have

1 fa

du dxi

dxx =

-F1(x2,x3)

and similar estimates for x2 and X3. With the obvious meaning of the symbols we then deduce \2u{x)\i <

{F1(x2,X3)F2{x1,xa)F3(x1,X2))i.

Integrating over x\, and using the Schwarz inequality, oo

\2u(x)\% dxi /

-OO

< {Fi(x2,x3))i

I /

F2(x1,x3)dx1)

I /

F3(xi,x2)dx1J

.

Integrating it successively over x2 and x3 and applying the same procedure, we find Ho,

n/..

du dxi

dx

<

aS^

du dxi

dx,

which, in turn, after employing the inequality (ai + a 2 + a3)2 < 3(af + a\ + a\) gives (1.89).

D

L e m m a 29. For all u G Cg°(n) 2 ||w||o,6 < 7/|l u li-

(1.90)

Proof. Replacing u with uA in (1.89) and then estimate the right hand side by the Holder inequality. We have

l<6<^=ll«" 30,6 |V«||0. Then (1.90) follows.

•

By the inverse trace theorem of Sobolev spaces there exists u\ € (H1^)) with a compact support on Q, such that ||ui||i < \\g — Woo||i> and u\ = g — u^ on dQ. Let R be large enough such that B(0,R) D supp Mi, and let fix = Q n B(0, R). We apply Lemma 20 to get a function u2 G ( H ^ f t i ) ) 3 such that V • u2 = - V • ui and ||w2||i < C||V • iti||o- The Lemma 20 is about an exterior domain, however it can be verified that it is

48

Numerical Methods for Exterior

Problems

also applied to bounded domains. Let us extend the function ui to zero to the exterior of B(0, R), then u-i € (HQ (&)) • We define UQ = u\ + ui + u^,, then uo satisfies the boundary condition, and V • UQ = 0, and we notice that the derivatives of UQ are compactly supported. Let w = u — wo, then the weak formulation for (1.80),(1.81),(1.88) is: Find u £ V + u0, such that (1.84) holds. Theorem 30. There is a positive constants > 0, such that if\\g — Woo||i < S, then the weak solution to (1.80), (1.81), (1.88) exists. Proof. Being analogous to Theorem 29, we set the approximation to w = u — UQ to be m

wm = y^c fc y fc , then Cfc satisfy (Vio m , Vv*;) + (wm • Vw m , (fk) + (wm • Vu 0 ,
+ (u0-'Vuo,
= -(Vu 0 ,V(^fc)+,

(1-91)

V/c = 1, ••• ,m.

Some terms in (1.91) are the same as those in (1.87). Let us estimate the other ones. By the Holder inequality, < ||Wm||o,6l|VUo||0,§-

\(wm-Vu0,Wm)\

Since Vuo is compactly supported we get \(wm • Vu0,wm)\

< C|«; m |f||Vuo||o,fii,

where we have applied Lemma 29. Being the same |(u 0 - Vu0,wm)\

< C||u 0 ||i,ni||Vuo||o|^m|i,

where we have applied the Sobolev embedding theorem to get ||«o||o,6,ni < C||wo||i,ni- By (1.85) we have (u0 •Vwm,wm)

=0.

Finally we have |(Vuo,Vu> m )| < |wo|i,f)i|w m |i. By the definition of UQ, there is a constant C\, such that C||Vuo||o,ni < C\\\g - Woo||I. We take 5 and M such that 5 < 1/Ci and (1 - 8CX)M > ll/llv' + C||wo||i,fii ||Vuo||o + |«o|i,ni- By the assumption of the theorem, llff —«oo 111 < S, then if |u>m|i = M the combination of the above inequalities

Exterior Problems of Partial Differential

49

Equations

can be summarized as (Vw m , Vw m ) + (wm • Wwm,wm)

+ (wm • Vu0, wm)

+(u0 • Vwm, wm)(u0 • V«o, wm) > - ( V u o , V w m ) + < / , wm > . The remaining part of the proof is the same as that in Theorem 29

•

The condition for g can be reduced to | fg^g • vds\ < 5, see [Galdi, G.P. (1994)]. Then for a particular case, j9ng • vds = 0, there is no restriction on the size of data. Generally speaking the solutions are not unique. An example is given in [Temam, R. (1984)]. Some results for the two dimensional case can be found in [Galdi, G.P. (1994)]. 1.9

Heat equation

To investigate evolution problems, some results of the theory of linear operators are needed. Let us introduce some definitions and some results first. Let H be a Hilbert space and A : D(A) —> H be a linear operator with domain D(A) C H. Let D(A) = H. Then the adjoint operator A* is defined by D(A*) = {ueH;3he

H, (u, Av) = (h,v),\/v

€ D(A)},

A*u = h.

A is symmetric if and only if A C A*. A is self-adjoint if and only if A = A*. Let A be a closed symmetric operator, then the Cayley transform of A is defined by UA = (A-iI)(A

+

iI)-\

where / is the identity. A bounded linear operator A : H —> H is isometric, if and only if A keeps the inner product invariant: (Au,Av) = (u,v),

\/u,v£H.

In particular if the range of A, R(A) = H, then A is called a unitary operator. A family of orthogonal projection operators E(X), —oo < A < oo, in a Hilbert space H is called a resolution of the identity if E(\)E(v)=E(mm(\,iJ,)), E{-oo) = 0,E(+oo) = I,

50

Numerical

Methods

for Exterior

Problems

where E(-oo)u

— s-lim E(X)u,

E(+oo)u = s-lim E(X)u,

A—» — oo

£(A + 0) = £(A),

A—>+oo

where E(X + 0)u = s-lim

E{y)u,

H—»A+0

and s-lim is the strong limit. Proposition 2. For ant/ u,v € H, (E(X)u,v) variation with respect to X.

is a function of bounded

Proposition 3. / / /(A) is a complex continuous function defined on (—oo, oo), and u £ H, then for — oo < a < (3 < oo we can define f{X)dE(X)u,

/i a

J a Riemann sum which is the strong limit of the

^2f(Xj)E(Xj,Xj+i]u,

a = Xi < X2 < • • • < Xn = fl,Xj e (Aj,A j+ i]

3

as max|Aj + i — Aj| —> 0, where E(Xj,Xj+i]

= E(Xj+\)

— E(Xj).

Theorem 3 1 . For a given u € H the following three conditions are equivalent:

(i) iZo fw dEWu exists> Wf^lfWfdlEMufKoo, (3) F(v) = f™ f(X)d(E(X)u,v)

defines a bounded linear functional.

Theorem 32. If f(X) is a real continuous function, then by oo

f(X)d(E(X)u,v)

/

-oo

determines a self-adjoint operator A with D(A) = D, where

D = UJ°° |/(A)|2d|£(AH2
/

Ad{E(X)u, v),

Vu G D(A), v e H,

-oo

which is denoted by oo

/

XdE(X). -oo

It is called the spectral resolution of the self-adjoint operator A.

Exterior Problems of Partial Differential

Equations

51

Theorem 33. A self-adjoint operator in a Hilbert space possesses a uniquely determined spectral resolution. Theorem 34. For a symmetric operator A there exists a closed symmetric extension A**. A closed symmetric operator A possesses a uniquely determined spectral resolution if and only if A is self-adjoint, and A is self-adjoint if and only if its Cayley transform is unitary. Let A : D{A) C H —> H be a linear operator. The resolvent set of A is defined by p(A) = {A G C; R{A - XI) — H,(A- AJ)one - to - one, (A - XI)_1 continuous}. The spectrum of A is denned by o~{A) = C \ p(A), which is further divided into three subsets. The point spectrum is o-p(A) = { A s o-(A); A — XI is not one — to — one}. The continuous spectrum is ac(A) ={A G cr(A); R(A - XI) = H, (A - A7)one - to - one, (A — A / ) - 1 is not continuous}. The residual spectrum is ar(A) = {A G a{A); R(A - XI) ^ H,{A- XI)one - to - one}. Theorem 35. Let A be a self-adjoint operator. Then X G o~p{A) if and only if E(X) — E(X — 0) 7^ 0, crr(A) is a null set, and the necessary and sufficient condition for X G cr(A) is that Ve > 0, E(I*) ^ 0, where J» = (A — £, A + e). Let us now investigate the initial boundary value problem of the heat equation. We remark that although the above definitions and results are all for complex numbers, we will assume the functions are real in the remaining part of this section, and it is easy to see that the following result is valid for both real and complex cases. The problem reads — = Au + f, u = 0,

xedCl,

(1.92) (1.93)

u(x,0) = u0(x). (1.94) We assume that dim(Q) = 3, but the results are true for two and one dimensional problems as well. We define H = L2(Q), and operator A = — A with domain D(A) = { « £ # o ( ^ ) ; A " G H}. Lemma 30. A : D(A) —> H is a self-adjoint operator .

52

Numerical Methods for Exterior

Problems

Proof. Clearly A is closed and symmetric. We consider the Cayley transform UA- Let v = (A + il)~lu, then by Theorem 17 u —> v and u —» Av are bounded in H, therefore UA is bounded in H. Besides we have UAVA = I, so UA keeps inner product invariant. From U*A = UA we get R(UA) = D(UAX) = D(UA) = H, so UA is a unitary operator. Then A is self-adjoint. • We assume that u0 G L2(fl), and / e L 2 ((0,T); L2(fl)) for T > 0. Let us write the problem (1.92)-(1.94) in the operator form: Find u : (0,T] —> .0(^4), such that dxi — +Au = f, limu(t)=«o. (1-95) at t-t+o T h e o r e m 36. The problem (1.95) admits a unique solution u{t), and ^ £ L 2 ((0,T),L 2 (fi)). Proof.

Formally we get u(t) = e~Atu0 + [ e-MeATf{r) Jo

dr.

By Theorem 17 we have cr{A) C [0, oo). Then in the notations of spectral resolution the solution to (1.95) can be expressed as / e-xtdE{X)u0+ / / e - A ( t - r ) dE{\) / ( r ) dr. Jo Jo Jo Let us verify it is the desired solution. Differential of it gives u=

J

9

/-OO

/"OO

ft

Xe Xt

= - /

~ dE(X)u0

+ /(*)"/

Ae- A ('- T ) dE(X) f(r) dr.

/

Here we notice that Ae~A* is bounded for positive t, so the above expression makes sense. For the first term we see that /»00

A/ Jo

/-OO

e-xtdE{X)u0=

/•oo

Jo

f\

Jo

fidllE(X)E(n)u0

poo

Xe~xt dE(X)u0. Jo Jo Jo The second term can be deduced in the same way. Thus u satisfies the equation. It is obvious that the initial condition is satisfied. To prove uniqueness we consider a final boundary value problem =

e~xtdx

/'OO

e~xtdx

fidllE(n)u0=

dv -— + Av — g, dt

lim v(t) = 0. t-r-o

Exterior Problems of Partial Differential

Equations

53

Then by the above argument v exists. Let u be the solution to (1.95) associated with / = 0 and UQ = 0. We have (u(t),g{t))dt=l

(u,--^+Av)dt =j

(^+Au,v)dt-

(v(T),u(T))

+ ( U (0),v(0)) = 0.

Since g is arbitrary, we obtain u(t) = 0. Thus the proof is complete.

1.10

•

Wave equation

We investigate the following initial boundary value problem of the wave equation (1.24) in this section. The initial and boundary conditions are u = 0, u(x,0) = uo(x), Let uo,ui £ L2(Q) = H. C([0, oo); H), which satisfies

L

xedfl,

(1.96)

—(x,0)=u1(x).

(1.97)

We are looking for a weak solution u €

JQu{w-Av)dxdt+lUoix)irt{x>0)dx -

/ ui(x)v(x,0)dx

= 0,

(1.98) Vw € V,

./n where V = Co([0, oo); D(A)) n C 2 ([0, oo); H), and the operator A is the same as that in the previous section. Theorem 37. The problem (1.98) admits a unique solution. Proof.

The equation (1.24) can be expressed as d2u

A

The solution to the problem can be formally expressed as u(t) = cos(A1/2t)u0

+ A~1/2

sm(A1/2t)Ul.

54

Numerical Methods for Exterior

Problems

Then in the notations of spectral resolution the solution to (1.98) can be expressed as u(t)=

[ Jo

cos(VXt) dE{X)u0 + [ Jo

sin

^ A ^ dE(X)Ul. vA

(1.99)

Let us verify that u(t), defined by (1.99), is the desired weak solution. u(t) G H is well defined because 3in(v/At)

|cos(\/A)| < 1,

In order to prove u is continuous in t, we estimate \\u(t)-u(t0)\\2 <2 ( J

| cos(v/At) - cos(\/Ato)| 2 d\\E(X)u0\\

r°° i + / Jo

- | sin(\/At) - sin(\/At 0 )| 2 d\\E{X)Ul ^

<2(x0\t-t0\2

f ° d\\E(X)uo\\2

/>oo

/-Ao

\

+ 4 / d||£(A) Uo || 2 + |t - *o|2 / d\\E(X)Ulf J\0 Jo J 2 2 2 2 <2|t - io| (A 0 ||«o|| + ||wi|| ) + 8(|| W0 || - ||E(A0)«oH2). We first take Ao large enough, then take \t — io| small. Then \\u(t) — u(to)\\ is arbitrary small, which gives u £ C([0,oo);H). To verify the weak formulation (1.98) we take v 6 V, then use Lebesgue's dominated convergence theorem to obtain /*

-

S-i

/ uo(x)~(x,0)dx

POO

=- J f00 d =J

Q

d(E(X)u0,-£(;0))

- J

= j™ 1^

f°°

r-

dv

Cos{^t)d{E{X)uo,-Q-t{;t))dt

Uos(VXt) d(E(X)u0, ^ ( - , t))

- V\sm{y/Xt)d(E(\)uo,-£(;t)))

dt,

Exterior Problems of Partial Differential

55

Equations

and analogously /•OO

0=

Jo

= / Jo

J

— dt

/>00

1 /Asin(v

Jo

/ Jo

/

At)rf( J B(A)uo,w(-^))di

{Xcos{VXt)d(E(X)u0,v{;t))

+\/Asin(\/Ai) d(E(X)u0, -^(-, t))) dt. Thus we obtain /.oo

I

/.oo

I

r

Q2V

cos(Vxt)d{E(X)u0,-^(-,t)+Av(-,t))dt

= -

Qv

uo{x) —

(x,0)dx.

Similarly we get

/°° r

sm

!/x' t) ^ E ^ ) u i > §^ <•> *)+Av(-' *)) * = - / "i^w 1 - ° ) ^

The proof of existence is thus complete. To prove uniqueness we take an arbitrary g £ CQ°(Q. X (0,OO)) and solve the equation d2v

„

-d¥~Av = 9> to get v satisfying the homogeneous boundary condition and vanishing for sufficiently large t. Let u be a solution to the problem (1.98) with homogeneous initial data, then by (1.98)

f°° f

/ Jo

(d2v

\

f°° f

I u\ -7T7 — &v dxdt — Ja \dt2 J Jo

/ uqdxdt = 0. Ja

Since CQ° is dense in L2, we obtain u = 0, which implies uniqueness. To prove the existence of v, we write down the formal expression oo

/

A~1/2

sin(A1/2{t

- s))g(s) ds.

We define a cut-off function C G C°°(R), such that C > 0, £(t) = 1 for t > 0, and £(£) = 0 for t < T, T < 0 fixed. Then we set

v(t) = ~C(t) f

J™ S l n ( V ^

S))

d(E(X)g)(; s) ds.

(1.100)

Let us verify that v(t) expressed by (1.100) is the desired solution. Following the same lines we can see v(t) is the weak solution, so the only problem is

56

Numerical Methods for Exterior

Problems

the regularity of v. For a bounded interval I C [0, oo), AE{I) is a bounded operator and AE(I)v(t)

= -C(t) j°° I MdE{j£) ULLH/y/J,) ^ I Jt J^ei Jo

ssm(y/$t i n

( ^ VA

- s)) *» d(E(X)g)(; s) ds.

Since E(/j,)E(X) = S(min(/x, A)), we have r 0 • dE($ = 0, A < fi, dE(fi)dE(X) = I dE((i), A = /x, [ 0 • dE(fx) =0,\>fi. Therefore AE(I)v(t)

= -((t)

r

S m (

f

^

S))

d(E(v)Ag)(;

s) ds.

Letting / —> [0,oo), one gets oo

/

\t-s\\\Ag(;s)\\0ds
so v(t) £ D(A). Moreover \\Av(t) -

Av(t0)\\0 /•OO

POO

+ C(M

\\Ag(;s)\\0ds+\at)-ato)\

/ Jt

\t-s\\\Ag(;s)\\0ds

s\\\Ag(-,s)\\0ds Jt0

which implies v £ Co([0, oo);D(A)). v is also second order continuously differentiable with respect to t, because

= -g(-,t)-Aj

J

V

ll

^d(£(A)As)(-,s)
Therefore v E V.

1.11

Maxwell equations

We work with complex-valued functions in this section.

•

Exterior Problems of Partial Differential

Equations

57

The Maxwell equations in vacuum are: div£=^,

(1.101)

an

curl£ = - — ,

(1.102)

div£ = 0,

(1.103)

cuilB = p (e^

+j \,

(1.104)

where E = (E1,E2,E3)T is the electric field, B = (B1,B2,B3)T is the magnetic flux density, e is the dielectric constant, (i is the permeability, p is the electric charge density, and j is the current density. The relation between p and j is | ? + d i v j = 0.

(1.105)

Let H = (H\,H2,H3)T be the magnetic field, then B = p,H. For given p and j there are six unknowns, E and B, in the system (1.101)-(1.104). It seems the system is over determined with eight equations. However (1.101) and (1.103) are not independent. If (1.103) is satisfied at time t = 0, then applying the operator div to the both sides of (1.102), we find that (1.103) is satisfied at all time t. Being the same, applying div to the both sides of (1.104) and noting (1.105) we can get d_

di

(divE

-?\=0.

Therefore if (1.101) is satisfied at the initial time, then it is satisfied at all times. Let us consider the total reflection boundary condition, E x j/|an = 0, -gjB • V\QQ = 0, and the initial conditions E\t=o = Eo, H\t=o = Ho. To investigate the initial boundary value problem, we introduce some Hilbert spaces first. Define ff (curl; fi) = {v G (L2{n)f; curl v £ (L 2 (fi)) 3 }, equipped with norm \\v\\H(curW) =

{\\v\\20+\\cm\v\\l}1'2.

58

Numerical Methods for Exterior

Problems

Lemma 31. The mapping v —> v x v\en, defined on (C°°(Sl))3 can be extended by continuity to a linear continuous mapping from i7(curl; SI) onto ( J ff- 1 /2(sn))3. Then we define #o(curl; Si) = {v € ff (curl; SI); v x i/\da = 0}. We choose H=

{L2{Sl))3x(L2(Sl))3

with the weight matrix

\0 pi J i

such that

(U,V)H

= {u,Mv) and ||u||# = (u,u)fj.

«-(!)• —

We define

'(;T).

with domain D(A) = H0(cuil; fi) x /f(curl; fi). The initial boundary value problem can be written in the form of - ^ + iAu = / ,

u(0) = uo-

(1.106)

Lemma 32. A : -D(^4) —* H is a self-adjoint operator. Proof,

clearly A is closed and symmetric. By definition D(A*) = {v£ H;3g € H,(Au,v)H

= (u,fl)j/,Vu e D(A)}.

We denote v = (vi,V2) T and g = (
V£ £

ff0(curl;fi).

Therefore g\ = ificurl vi and the domain for v2 is H(cui\;Sl). letting E = 0, we have V - X ( V x H,Vl)

= (H,g2),

Similarly

Vtf £ tf (curl; ft).

Therefore g% — —iecurl v\ and the domain for v\ is i^o(curl; SI). Hence D{A*) = ff0(curl; SI) x ff (curl; SI) = D(A).

Exterior Problems of Partial Differential

59

Equations

We assume that / £ C([0, oo); H). The formulation of a weak solution to (1.106) reads: Find u £ C([0, oo); H) such that

/

/ \U \ ~df~

+ iA

^)

dxdt

~ ^M

~ I Mx)v>(x,0)dx

= 0, (1.107)

for all ip £ C o ([0, oo); D(A)) n C 1 ^ , oo); ff). The weak formulation (1.107) is equivalent to the classical formulation (1.101)-(1.105) with the total reflection boundary condition and the initial condition. Just like the previous argument claiming that (1.101) and (1.103) are not independent, we take = C(*)(VVi> V ( P2) T J where (pi £ Co°(fi), ip2 e C°°(0), and C € Co°([0,oo)). Then by the definition of A we find iuATp = 0, and consequently

/ 1U V T T ) ~ ^ f

/

dxdt

~

/ u0(x)<j>(x,0) dx = 0.

Since
\\u(t)\\H < \\U0\\H + / Wf(T)\\HdT Jo

(1.108)

holds. Proof.

Formally we get u(t) = e-iAtu0

e-iA^-T^f(T)dT.

+ [ Jo

In the notations of spectral resolution the solution can be expressed as pt

OO

/

e-iXtdE{X)u0+ -oo

/

/>00

/

Jo J —oo

e-

iA(

*- T > dE{\) / ( r ) dr.

£

Numerical Methods for Exterior

60

Problems

Let us verify that it is the desired solution. We have oo

d{E(X)uo,
-oo -oo />oo

= -J

e-iXtd{E{X)uQ,V{t))Hdt

jtJ

/»oo

/*oo

= /

iXe-iXtd(E(X)u0,f(t))Hdt

/

-^oof

r°° /"

/•OO

JO

fj-iXtd(E(X)u0,^-)Hdt

J- OO /•OO OO

ie-iXtd(E(X)u0,A
pC pOO

,-i\tMi?f\\..

Jo

<*¥>(*) d{E{\)uQ,^j^) Hdt.

J-t

Similarly we have /•OO

(/(r), ^(r))„ = / JT

/*00

/

ie-W-^

d(E(X)f(r),

A
J — OO

' FII

e iX{t T) WA)/(T)

~

~

' ^r)H c

Then (1.107) follows by noting f£° dr J^° dt = f£° dtfQ dr. It is easy to see that u is continuous with respect to t. Then by (1.107) ^ exists in the sense of distributions and (1.106) holds, therefore u is continuously differentiable. Taking the inner product of (1.106) with u we get

l^L+i(Au,u)H

= (f,u)H.

For the real part we have ld\\ufH 2 dt

= Re(/,«) ff

<||/||HHH.

Then (1.108) follows, which implies uniqueness.

•

Up to now we have investigated three kinds of evolution problems. Applying some similar argument, the results for some other problems, for example the evolution elasticity problem, can also be obtained.

Exterior Problems of Partial Differential

1.12

61

Equations

Darwin model

The Darwin model is a simplified model [Hewett, D.W. and Nielson, C. (1978)] [Li, T.-t. and Qin, T. (1997)] for the Maxwell equations (1.101)(1.105). It was shown in [Degond, P. and Raviart, P.A. (1992)] that the Darwin model approximates the Maxwell equations up to the second order for B and to the third order for E, provided 77 = v/c is small, where v is a characteristic velocity, and c is the light velocity. Let us introduce the Darwin model for bounded domains first. For a domain Qo let I \ , 0 < i < m, be the connected components of the boundary dCto, and TQ be the outer boundary. Note that the electric field E can be written as the sum of a transverse component ET and a longitudinal component EL, such that E = ET + EL, and V • ET = 0,

V x EL = 0.

The equation (1.104) becomes / 3EL

dET

\

In order to get the Darwin model, we neglect the term for ET and get cuxW = n ( e ^

+j \.

(1.109)

Then the system (1.101)-(1.103),(1.109) with the total reflection boundary condition and the initial condition will be reduced to (a) EL = — V, and <j> satisfies

(/>|r; = oii,

0 < i < m.

a — {ai} satisfies „

d a

1 f • „

a(0) = a 0 ,

,

62

Numerical Methods for Exterior

Problems

where OJO depends on Eo, C = {cy} is the capacitance matrix. Here Cyi r ~~dv ^ s ' anc ^ X = {Xi} i s t n e solution of Axi = 0, Xi|r\,

=Sij.

(b) 5 satisfies - A B = fiW x j ,

(1.110)

V - 5 = 0,

(1-111)

B-v\dn0=B0-v\dn0,

(1.112)

(V x B) x ^|an 0 = /A7 x Han 0 ,

(1.113)

AET = T T V x 5 ,

(1-114)

V - £ r = 0,

(1.115)

£ r x Han 0 = 0,

(1.116)

where B 0 = HHQ. (c) £ r satisfies

I ET-vds

= 0,

\
(1-117)

Now we turn to the exterior domain $7 and assume that dim(fi) = 2. We investigate problem (c) as an example, the method to problem (b) is similar. In order to figure out the suitable functional setting for the problem, we introducing some spaces. The space #(curl,div; Q 0 ) = {u£ (L 2 (fi 0 )) 2 ; V • u, V x u e L 2 (fto)}, provided with the norm ||u||o,curl,div = ( H o + IIV • U||g + || V X «||g)* ,

Exterior Problems of Partial Differential

63

Equations

is a Hilbert space. We take a cut-off function £ £ C°°(f2), which satisfies: C = 1 near the boundary dfl, £ = 0 near the infinity and 0 < C < 1. Let V be the space of distributions. Define tfoc(ft) = { « £ V; ||Cu||o,curi,div < oo, (1 - 0 « e (#o'*( f i )) 2 >«

x

Han = 0},

and ||u||, = {||V x u||§ + ||V • u||g+ I n } ^ However || • ||* does not provide a norm in Hoc(Q), and actually we have the following result: Lemma 33. Let V0 = {u £ Hoc(ty', ||u||* = 0}, then Vo ={u = ( U I , U 2 ) ; U I = -~—,u 2 = - - — ,

ox 2

oxi

4> = a(x2 + f(x)) + b{Xl + g(x)), (a, b) £ M2}, where f,g £ H1'*^)

satisfy - A V' = 0, dlj) ~dv~

dQ

iell,

dx dv

x

where x{^\-,^'i) — %i, \ separately. Proof.

We take any u £ Vo, then it will satisfy V - u = 0,

(1.118)

V x u = 0,

(1.119)

u x i/| fin = 0,

(1.120)

/ u-vds = 0. (1.121) Jan From (1.119)-(1.121), J r —v,2dx\ + U\ dxi = 0 is true for any closed curve r in Q.. Thus we can define

4>(x) = f (-«2 dxi

+ u\ dx2),

Jxn

where Xo can be any but fixed point in fi. Notice now ui = J^jf-, «2 = — J^~-

64

Numerical Methods for Exterior

Problems

From (1.119),(1.120), we can get

A

^ * ' ^ - f l = v * « = °-

and d d<{>

x Han = u x Han = 0. 0x2' dx\ da Thus starting from every u G Vo, we will end up with a problem: find a function <j>, satisfying ~dv

A
xeQ (1.122)

=0.

du

an Besides, u is a bounded harmonic function, which can be developed in a neighborhood of the infinity as follows:

E

ZOk

fc=0

2

w h e r e zofc = Cfc + ibk, (dk, bk) G M , z = x + iy = re%e. T h a t is / ^ > a,k cos k6 + bk sin fc# ^ ^

(«1,«2)= n

^

afc

sin k6 — bk cos k6

,2^

\fc=0

^

fc=0

/

From (1.118),(1.121) we know ai = 0, so m —iu2=

[a0 + \

bi sin k9 r '

b\ cos A;# \ r J

^ ^

z

ok zk

Hence the asymptotic expansion of (p(x) near the infinity will be cj){x) -» a0{x2-X2o)+bo(x1-xw)

+ b1(lnr~lnr0)

+ 0(-)

^

h(x)+c+0(-),

where h(x) = 0,0X2 + boxi + bi lnr. Introduce a new function ip, such that (/> = I/J + h. Then ip will satisfy A v> =0, chp_ dv

iefl dh

an Notice that it suffices to consider the solution in H1'*(Q.)/M. now.

(1.123)

Exterior Problems of Partial Differential

65

Equations

Knowing that the well-posedness of the problem (1.123) is equivalent to

/. so let flr = flfl B(0,r), /

then by the Green's formula we get

— ds = \

Jandv

9h , — ds = 0,

Ahdx — /

Jnr

—— ds = — 2itb\.

JdD(r) °v

Therefore b\ = 0 is sufficient and necessary in order that (1.123) admits a unique solution in # 1 -*(ft)/R. Once bi = 0, there exist f,g£ i7 1 '*(n)/M, such that i> = a0f(x)

- b0g{x),

where / , g are the functions to (1.123) with h(x) = x-i,x\ Consequently

respectively.

4>(x) = a0(x2 + f(x)) + b0(xi + g(x)). On the other hand it is easy to see that the function u given in the Lemma for all a, b are in Vo. • Let V = {v e Ho c (ft); < v • u, 1 > a n = 0}, and let the closure of the quotient space V/VQ with respect to the norm || • ||* be W. Besides, let Q = {p G L2(Q); supp p C C il}, equipped with norm IIPIIB = (||PIIO + I ^ P ^ I

:

Then we take closure to obtain a Hilbert space Q. lip £ Q and limpn = p, pn S Q, then we define f^pdx = lim fnpndx. Notice this can be seen as a generalized integral. The subspace of Q, {p £ Q; j„p dx — 0} is denoted byQoWe define bilinear forms on W x W and W x Qo, d(u, v) = / (V x u) • (V x v) dx+ / (V • u)(V • v) dx

Jn

Jn

+ a n < v-v,l

b(u,p) — \ (V • u)pdx,

>an,

u € W,p e Q0.

u,v

€W,

66

Numerical Methods for Exterior

Problems

Now the variational formulation for the problem will be: Find u G W, p 6 Q o ) such that d( u,v)+b(v,p)=

I S - ( V xv)dx, Jn b(u,q)=Q,

VveW,

VgeQo,

(1.124) (1.125)

In order to prove that there is a unique solution to the problem (1.124),(1.125), we introduce one result on the inf-sup condition, which appears in Section 1.7, and will be used later on in some different problems. [Girault, V. and Raviart, P.A. (1988)] Let W, Q be two Hilbert spaces, and b(u,p) be a bounded linear form defined on W x Q. Define Z={w&

W;b(w,q) = 0,Wq € Q},

z° = {/ eW-,< f,w >=oyw e z}, Z± = {we W; {w,v)w

=

0,Vv€Z},

where (•, -)w is the inner product in W. Define two operators B : W —• Q' and B' : Q —» W, satisfying < Bw,q >= b(w,q),

\/w

£W,q£Q,

< B'q,w>=b(w,q),

Vw

£W,q£Q,

then ker B = Z. Theorem 39. The three following properties are equivalent: (a) there exists a constant f3 > 0, such that inf

sup

^ V > / 3 ;

(1-126)

(b) the operator B' is an isomorphism from Q onto Z° and \\B'q\\w> > P\\q\\Q,

V96Q;

(c) the operator B is an isomorphism from Z \\Bw\\Q. > P\\w\\w,

(1.127)

onto Q' and

Vw e W.

(1.128)

Exterior Problems of Partial Differential

Proof, to

Equations

67

(i) Let us show that (a)<=>(b). The statement (a) is equivalent

sup

< B'q,w > ^ ... .. ——i: > p\\q\\Q,

TIT

P

_ q£Q,

Iff

that is, (1.126) is equivalent to (1.127). It remains to prove that B' is an isomorphism. (1.127) implies that B' is a one-to-one operator from Q onto its range R(B'). Moreover, it also implies that the inverse of B' is continuous. Hence B' is an isomorphism from Q onto R(B'). Therefore R(B') is a closed subspace of W, since B' is an isomorphism. We can apply the closed range theorem of Banach spaces (see [Yosida, K. (1974)]) which says that R(B') = (ker(B)) 0 = Z°.

(ii) (b)«=Kc). We observe that Z° can be identified with (Z-1)'. Indeed, for w € W let w1 denote the orthogonal projection of w on Z1-. Then with each / G {Z1-)1 we associate the element / of W defined by < f,w >=< f,w1

>,

Vw e W.

Obviously / e Z° and it is easy to check that the correspondence / —> / maps isometrically (Z1)' onto Z°. This permits to identity (Z^)' and Z°. As a consequence, statements (b) and (c) are equivalent. • Besides, we need an auxiliary lemma. Lemma 34. For a given q £ Qo there is a (yq + Vo) £ W such that V • vg = q,

V x vq = 0,

vq x

V\9Q

= 0,

consequently

\\vq\\l = h\\l = h\\l Proof. For a given e > 0 we have qe £ Q so that \\q — qe\\$ < e. Then consider the following problem: Find G H0'*(Q), such that

By Lax-Milgram theorem there is a unique solution, and € flrli*(fi) fl

HIMTake a cut-off function ( G C^°(M2) such that C = l f o r | x | < l , C = 0 for |x| > 2, and 0 < £ < 1. Define Ca(^) = C(x/a) a n d ^a = <^Ca- Let qea =

68

Numerical Methods for Exterior

Problems

Acf>a, then q£a £ Q. When a is large enough, we have C a A0 = Acj> = qe therefore for sufficiently large a it holds that \\Qea - qe\\t =\\&a ~ AJWf = \\ACa + 2VCa • W + CaA
2

|| ACa4> + 2VCa • W | | g + | / ( A C a 0 + 2VCa • V
Jn (1.129) Let us estimate the right hand side of (1.129). Since \x\2 In2 \x\ < |a;|3 < 8a 3 in the domain D = {x; a < \x\ < 2a}, we have

||AC*||§ <4l|AC||^ )00 / \4>(x)\2dx
JD

JD \X\ \D.

\x\

QS~t

< —l|AC||3, O ollW||3, n ->0, a

(a-»oo).

Analogously we have l|2VCa-V^||g<^||VC||§ i O O ||V0||g i n -»O. For the second term of (1.129) we notice that / A^dx Jn

= - / VCaV^cb, Jn

therefore / ( A C o ^ + 2VCa-V0)da:| Jn in <J

|VCa • V<£| dx < |||VC||o,n||V^|| 0 ,D < C\\V\\0,D - 0.

To conclude there is a ao such that ||g ea — <7e||j < £ for a > ao. We fix such an a£. Let vq£ = V = fQ qeae dx. Consequently i

|V • vq£ - a||o + \
:

>9n - / qdx\ Jn

- q\\i < \We - q\% + ||fca. - 9 e | | | <

2e

-

Exterior Problems of Partial Differential

Equations

69

Therefore vqE converges as £ —> 0. Let vq be the limit, then vq + Vo is the element in W we are looking for. • Now we are ready to prove the well-posedness. Theorem 40. The problem (1.124), (1.125) admits a unique solution and p = 0. Proof.

(u,p),

Because

d(u, u) = / |V x u\2 dx+ / |V • u\2 dx = \\u\\l, Jo. Jo. the bilinear form d(-, •) is coercive. By Lemma 34 SUP v€W

> "^

INI*

[j— -

-TT—rp- -

IKII*

g

•

(1.13U)

Ikllit

Therefore the inf-sup condition with /3 = 1 holds. We define W0 = {u G W;6(p, M) = 0,Vp G QO}, then being analogous to Theorem 26, the problem: find u G WQ, such that d(u, w) = f B • (V x v) dx,

VuGWo,

(1.131)

admits a unique solution. To prove the existence and uniqueness of p, we note that the equation (1.124) defines a linear operator F(v) = -d(u, v)+

/ B • (V x v) dx Ja.

on W, and by the equation (1.131) it is in the space Z°. Then by Theorem 39, p is determined by the inverse of B' uniquely. Finally let us show p = 0. According to Lemma 34 there is vp corresponding to the solution p. We take v = vp in (1.124), then we have d(u,vp) — / (V • u)(V • vp)dx — / (V -u)pdx = b(u,p) = 0, Ju Jo. I B • (V x vp) dx = 0, Jn and b(vP,p) = (P,p)-

70

Numerical Methods for Exterior

Problems

Equation (1.124) implies (p,p) = 0, which gives p = 0.

D

Remark 1. The solutions are not unique. They may differ from a function in Vo- It seems the most natural way to make the solution unique is to impose a boundary condition at the infinity, u||,,.|=00 = 0, because a function in Vo satisfying this condition is zero. Unfortunately we still can not prove the existence if this boundary condition is imposed.

Chapter 2

Boundary Element Method

The boundary element method is based on a reduction of a boundary value problem on a domain to an equivalent problem defined on the boundary. The dimension of the problem is thus deducted by one, and it does not care very much if the underlying domain is interior or exterior, so it is suitable to deal with exterior problems. The potential theory is classical. We will derive the boundary equations by using the potential theory. On the other hand we will introduce the approach by Feng and his students, using the theory of singular integrations. The advantage of this approach is that the energy norm of it is exactly the same as that of the original boundary value problem. For inhomogeneous equations a Newton potential is usually applied to reduce it to homogeneous ones, and a numerical scheme of integration can be applied to the Newton potential. Therefore we consider homogeneous equations only in this chapter.

2.1

Some typical domains

For some typical domains the solutions to exterior problems can be expressed explicitly. We work with complex-valued functions in this section. 2.1.1

Harmonic

equation

Let us begin with the exterior domain fi of the unit disk B(0,1) Dirichlet boundary value problem of the harmonic equation. In polar coordinates the problem is

iiL (

du

l d2u

\

-o

r2 d82

r dr \ dr J 71

and the

72

Numerical

Methods for Exterior

Problems

« | r = l = 9(0)-

The solution can be expressed in terms of Fourier series, oo

Pn(r)ein9-

u = Y, n = —oo

Plug it into the equation and obtain

y

n2pn\

fid fdPn\

^

1 r dr \ dr J v

n= — oo

N

ine_Q

2

r J

'

'

Therefore each term vanishes: \d_( d£n\ _ n2pn _ r dr \ dr J r2 The general solutions to these ordinary differential equations are: p n = anr~n + bnrn, Vn ^ 0.

Po — a0 + &o lnr,

Since u £ H1'*(Q), we get the solution oo

u=

anr-^ein9.

Y, n= — oo

By the boundary condition we can determine an:

^j\-in*g{
an = Therefore the solution is

u

r \ S r-Me*0-Ag(
=t J

®

l,n= — oo

(2.1

J

The power series in (2.1) are equal to: oo

1

E

X

n

in(e-
Finally we get n=0

- 1 V r-|nLi"(0-¥>) = _

~~ l-.r-lei(6-V>)'

Z ^ n = —oo

(-27T

U =

1 /""r 2^y0 r2

+

r^ - 1 l-2rcos(0-
r-lp-i(9-ip)

!

l_r-le-t(fl-v)'

(2.2)

Boundary

Element

73

Method

This is the Poisson formula for harmonic equations. If we replace the boundary condition by u\r=R = g(0), and solve the problem exterior to B(0,R), let v(^,0) = u(r,0). Replacing r by -^ in the Poisson formula we get for the general case, 27r

2Wo

2 2 ~ r - R " 2

r2 +

g(p.

R -2rRcos{6-
(2.3)

Under a conformal mapping a harmonic function is transformed to a new harmonic function. Applying this property one can get some more solutions for other exterior domains. For example, if we define w = az H—, a > 6 > 0, z where z = x\ + ix^ = re10, w = £1 + i& = pezlf', then the exterior of the unite circle r = 1 is transformed to fii, the exterior of an ellipse, dQ.\ = {(£1162); £1 = (a + b)cos#,£ 2 = (a — b)sm6}. The function v(p,ip) =u{r,6) can be expressed in terms of the Poisson formula, 1 r2ir r2 - 1 (P,
Av = o,

b pezv = are16 + — ^ , re

(6,6)efii,

vlecit =g(0)There is another approach to derive the Poisson formula (2.2), which is the use of the potential theory. Let us investigate the integral equation (1.11). Since 917 is the unit circle, we have d

1

duy

_ c o s ( ( x - y),vy)

\x-y\

_

\x-y\

so the equation is

1 If a x o ( ) ~ TZ \ 2

47T Jdn

a

(v) dsy = g(x).

1 2'

74

Numerical Methods for Exterior

Problems

We replace the boundary value g(x) by g(x) = g(x) — •— j d n g(y) dsy, then the integral equation 1

1 /*

admits a solution cr(x) = 2g(x). The double layer potential V{x) =

^La(y)

\x-y\

dS

«

satisfies (1.1),(1.4), and V\QQ = g. Therefore If 27r

cos({x-y),vy)

1 A 27r

J do.

F ~ 2/1 ./an 2\x - y\ cos((a: - y), vy) + \x- y\2 |2 27r7aan F-J/l' ;

9{y)asy

Noting that \x — y\2 = r2 + 1 — 2r cos(# — y>) and 2|a; — y\ cos((a; — y), vy) -f \x — y\2 = r2 — 1, we derive the Poisson formula (2.2) again. The above approach can be applied to derive the three dimensional Poisson formula of the Dirichlet problem. Because the boundary dQ, is the unite sphere S3 = dB(0,1), the integral equation (1.11) is

r{x)~iL3a{y)\^v\dSy=9{x)We replace the boundary value g(x) by If ~g{x) = g{x) - — J^

1 g(y)j——-^dsy,

then <J{X) = 2g(x) is the solution. We notice that the difference between g and g is nothing but a single layer potential, so the solution u is I f , . cos((x — yy),vv) u(x)=—
=J- / 4TT JS3

,

1 f , . +~ g(y)4TT Jan

zi» - yl«»((» - y). ^») +1» - yl2

( }

1 -dsy \x-y\

d,

\x - y\A

Hence we get the Poisson formula l [ r2 - 1 U X ( ) = JZ 77--; 1 7l9{y)dsy, 47r ^Sa (r 2 + 1 - 2r cos 95) 2

Boundary

Element

75

Method

where r = \x\ and ip is the angle between the vectors x and y & S3. If we replace the boundary condition by u\r=R = g(x), and solve the problem exterior to B(0, R), let v(^, 9) = u(r,6). Replacing r by ^ in the Poisson formula we get for the general case, 1 /• r2 - R2 «(*) = 7 z / ! 2 ^ m 7i9(y)Rdsv. 0 p 47T Js3 (r 2 + R2 - 2rR cos 9?) 2 2.1.2

Bi-harmonic

(2.4)

equation

Let us solve the bi-harmonic equation next. The problem is: 1 d ( 8_\ r dr \ dr)

u\r=i = gi(0),

1 d2 \ r2 d82 J

—

2

= 92(0). r=l

The solution is expressed in terms of Fourier series, 00

u=

Y,

Pn(r)ein9.

Then we get 1 d ( d\ r dr \ dr I

n2s2 r2

Therefore we have I d f d , --r [r-r- ) po = ao + bo In r, r dr \ dr '

Solving these equations again and noting that u € H2'*(Q), we get Po = ao + M l + 21nr),

pn = anr~M

+ bnr-^+2,

Vn ^ 0.

76

Numerical Methods for Exterior

Problems

Taking the boundary conditions into account we solve the coefficients as follows: $1 (0) = J > „ + bn)ein9 + (ao + b0), njtO

- (\n\ ~ 2)bn)einB + 260.

92(6) = ^2(-\n\an Then i

r* e-m<egi{
an + bn = —l

-\n\an

- (\n\ - 2)bn = i - f

e'^g2{V)

dtp.

Then the solution is <.27r

-J_ /i u=—

in

^-f)

e

• {(51 - \92 + \92T2)r-^ 1

+

2n

f2n

- M S 1 ( 1 - r*)r-W} dtp

(9i+92^r)d(p.

After some calculation we finally get r2T

r* + 1 -2r Je(0 -
=i l 1 +

/-2,r 2r2 - r(r 2 + 1) cos(0 -
2^y0 1

+

Ji

27r/

/"27r

a

(r2 + l - 2 , C o s ( / - y ) ) 2 ( 1 - r ) g l ( y ) ^ 1

^lnr+ 2 ~

1

2r^92^d
^

Boundary Element

77

Method

If we replace the boundary condition by 011

= 92(0). r=R

and solve the problem exterior to B(0, R), let v(^,6) the boundary condition, v\r=i = gi{6),

dv dr

= u(r, 6). v satisfies

= RtoW). r=l

Then for this problem the solution is ,2TT

_j_ r ~2Wo +

r

2 _

^2

r» + R* - 2rRcoS(6 I ^ j <*<*> " f *<*> + h .

1 f2n 2R2r2 - Rr(r2 + R2) cos(0 - cp) M 2^J0 (r 2 + R? - 2rRcos(9 - &)* ( 1 " 2n 1 f r 1 1 r2

r2 ,

9

,

* ^ *P

., *"

&

)9l(ip)

(2.5) 2.1.3

Stokes

equation

For the exterior problem of the Stokes equation, - A u + Vp = 0,

(2.6) (2.7)

u\r=R = g{6),

(2.8)

we introduce the stream function 1p(x) = <j> {-U2
78

Numerical Methods for Exterior

Problems

Applying the operator x T

d

d

dx2'

dx\

to the equation (2.6) we get A2ip = 0, so the stream function ip is the solution to the bi-harmonic equation. We assume

I

2TT

g • v d6 = 0

first, then in the polar coordinate ip(R,2n) = tp(R,0). By this assumption ip is single valued. Let (ur, ug) be the components of u in polar coordinates, that is ur\ ug J

( cos6 s i n 8 \ (u\ \ — sin 9 cos 9 J \ u^

Under the polar coordinates it holds that /(r,e) ip(r,9) = (b (-uedr + urrd9), •Wo) The boundary condition (2.8) becomes Ur\r=R = 9r{9),

1 dip ur = - — , r 06

dip ue = - — . Or

U6\r=R = ge(0).

The boundary condition for ip is lp\r=R =R r=R =

9r(v)dr),

=

-fr

-9e(9).

r=R

Applying the expression for the solutions to the bi-harmonic equation we obtain r2 - R2 ^=W 0 ri + R>-2rRcos(9-v){RJ0 R 1 \ d
+

f2"

(

[* *<*>*>

J_ r2" 2R2r2 - Rr{r2 + R2) cos(9 - tp) r2 2 2 2 ( 2n J0 (r + R - 2rRcos(9 - tp)) R?' 2 2 rv> 1 f* r 1 1 r

(RJ

gr(r))dr})dtp - — J

(\n- +

-~-^)Rg9{ip)d
Boundary

Element

79

Method

Then after some calculation we obtain the solution u: 1

f2"

_ J_ f2n

+ 3i? 2 )sin(6>-y)

[*

(r 2 - R2)2 sin(0 - ip) 2Tg9(tp)dip (r * {I- + -f R n~2-2rRcos(6 — zrrtcos^c- - y>)) sin(6> - p) (r2 + R2-2rRcos(6-
2Wo 27r

+

R2(-r2

2^7o

(2.9)

r2 fv (1-^2)/ gr(v)dvd
^r2w 2(r2-iZ2)(r-J?cos(fl-y)),n ^

,

2^J0 (ri + R2-2rRcoS(0-v))2iRJo R 1 + -^geiv) - ^9e(
1 f(-27T

+ 2^J0

ri

2r R -2rRcoS(6-lp){RJ0 2

+

^

fv ^

^

) d n

R 1 + -^9e{9) - ^99(
_2

1 r 2TT 7

_

^2

. - 2ri? cos((9 -

r r2 + # 2 ip) H^ _ J _ f ^ ( 2 i ? V - Rr(r2 + R2) cos(6 - ip))(r - Rcos(6 - ip)) 2n J0 (r 2 + R? - 2rRcos(6 - ip)f 2 r [* • ( 1 _ ^ 2 " ) - R / 9r(v)dr]dip 0 2

+

1 /-^ <±R3r - 4Rr3 + (5r 4 - R4) cos(6> - ip) fv 2^/o (r-2+^-2ri?cos(0-„))2 /0 ^

1 f2n R

,

x

, , < ^

r

-2^1 ( 7 - ^ M ^ . (2.10) If /•27T

1 /"^ — y gr(6)d0 = gjLO,

80

Numerical Methods for Exterior

Problems

we define a particular solution

to the equations (2.6),(2.7). Then /•27T

/ Jo

(u- u0)\r=R • vdQ = 0.

u — UQ can be expressed in term of the above expressions. 2.1.4

Plane

elasticity

For the plane elasticity problem, —fi A u — (A + /i)grad divu = 0, UllN = l =flli«2||a:| = l =

iefi.

92,

we introduce the complex expression for the displacement u = (u\,U2)T first. Let f(z) be an analytic function, / = p + iq, z — x + iy. Let L — —jU A — (A + ^i)grad div. By the Cauchy-Riemann equation, we have

""(A)-0-

div

C)-0'

and Ap = Aq = 0, therefore

For the function zf(z) = (xp + yq) + i{yp - xq), we have

div^

+ y
\yp-xqj

g r addivf^

+

^)=2f|V

\yp-xq)

and

-(rs)-(i-i)-

y-^J

Boundary

Element

Method

81

Therefore

£

+

£-S)~°* *"(i)

Let af(z) - zf'(z) = U + iV, then U L( )

= -2a(\ + n) (^]+2(X

We set a = ^ f ,

then L (^

+ 2^ ( j f M .

) = 0.

Let us return to the Lame equation(1.35). If Ui+iu2 = aip(z) — z(p'(z) — tp{z), where
00

i>{z) = J2

n=0

bnr-ne-in9.

n=0

Thus we get the expression, 00

00

«! + iu2 = a J2 anr-ne-inf> n=0

+^

00

na^r-neiin+2'>9

n=0

- J2

Kr~nein9.

n—0

Applying the boundary conditions we get 00

00

9 = 9x + i92 = a Y^ "ne-™9 + £ n=0

00

na^e^+V9

- £

n=0

KeinB'.

n=0

The coefficients can be evaluated with 60 = 0, 1

1 /"27r e in ~2^J ~ ^(4>)d
—

bn =

1

bn =

~2^J

C2w

/- 2 7 r

77 — 2

e

~

in

*

9(<W#+

~2^~/

n = l,2,

/" 27r

e-^-WgWty,

n = 3, • • • .

82

Numerical Methods for Exterior

Problems

Finally we get /"27r f °°

1

•'°

°°

ln=0

n=0

+ J2r-nein{e-*)g() n=l 27T 7 0

—n

- J2 n=3

e^e^-W-Vgit)

a

I
+

\l_r-le-i(0-¥>)

a (1 _ r - i e ^ - v ) ) 2 27r

2TT JO 2W0

- —

' r2 - 1 I r 2 + 1 - 2r cos(0 - ip) ff(¥>)

r ( r _ e i ( e -, ) ) 2 3(
<**•

If we replace the boundary condition by Ul\\x\=R

= 9l,U2\\x\

= R = 32,

then 1 Ul+lU2=

2^J0

/2w ( {ri +

r2-R2 Ri-2rRCos(e-V)gM

e2i9 ei(9-V)R(r2

a 2.1.5

Helmholtz

_ #2)

ReW-v))2

r(r -

(2-11)

-g(4>) } dc/>.

equation

In order to solve the exterior problem of the Helmholtz equation let us introduce the definition and some basic properties of the Bessel functions. The Bessel equation is dx2

x dx

x2 J

\

One solution of it is the Bessel function of the first kind of order s, which can be developed into a series, ^

,

1

/a:\ 2 f c + s

'•M-B-U'rjrnE+yk) fc=0

'

Boundary

Element

83

Method

where T is the Gamma function. The asymptotic behavior of it as x —» oo is S

Js(x) =

]ffxCos(x-

^-^+0(x^).

Another linear independent solution of the Bessel equation is usually denoted by Ys(x), the Bessel function of the second kind of order s, which possesses a singular point x = 0. We omit its expression, but give the asymptotic expression of it here:

y.W.yX*,(x-f-J) + 0(x-n Moreover the Bessel functions of the third kind of order s are called the Hankel functions of the first kind of order s and the second kind of order s. The definitions of them are: Hi1\x) = Js(x)+iYs(x), H?\x) = Js(x)-iYs{x). Therefore the asymptotic behavior of them is the following: HM(x) = A /A e <*-¥-i) + 0(x-V2), V ixx

(2.12)

H™(x) = . / l e - ' ( - ¥ - ? ) + 0(x-*/2).

(2.13)

and

V TTX

Multiplying them by a factor e~luJt, we find Hs (x)e~"lu't represents an outgoing plane wave and Hs (x)e~ZUJt represents an incoming plane wave. For the Dirichlet boundary value problem of the homogeneous Helmholtz equation (A + w2)u = 0, u = g,

zefi,

xedfl,

(2.14) (2.15)

84

Numerical Methods for Exterior

Problems

we consider the case of two dimension first and let <9f2 = {x; r = R}. The solution can be expressed in terms of Fourier series, oo

u(r,0)=

J2

anH^(ujr)ein9.

(2.16)

n= — oo

Taking into account of the boundary condition we get oo

g(6)=

anHil\uR)ein9,

£

(2.17)

n = —oo

therefore

2Wo

g(6)e-in9de/Hi1\u;R).

Lemma 35. If g £ #2(dft), the series (2.16) converges and solves the problem (2.14),(2.15). Proof. By (2.17) the boundary condition (2.15) is satisfied, provided the series converges. Making use of separation of variables it is easy to see that each term of (2.16) is a solution to the equation (2.14). It represents an outgoing sphere wave, so the Sommerfield radiation condition is satisfied. It remains to prove convergence. We recall the estimate H I v < Cll/llo,,,

(2.18)

where / is the right hand side of the equation (1.25) and the support of / is bounded. For the problem (2.14),(2.15) we apply the inverse trace theorem of Sobolev spaces to take a function UQ in H2(Q) with a compact support, so that t*o|an = g, and ||ito||2 < C\\g\\ 3. Let u\ = u — u0, then it is the solution to (A + w 2 ) Wl = - ( A + w 2 )u 0 . By the estimate (2.18) we have \Hv<

||«o||v + | | « i | | v < C | | s | | | .

Therefore the convergence of the series (2.17) in H^(dil) vergence of (2.16) in V.

implies the con•

Boundary Element

85

Method

The treatment for three dimensional case is analogous. The solution can be represented as a series of spherical functions: oo

n

u(r,9) = ^2

E

^nm^ll2(u>r)Ynm(e,
(2.19)

n=0 m=—n

where Cgi(z) = y ^ < + i W .

P m)

n (x)

2.2

PlT\cos6)eim*,

Ynm{9,tp) =

(1 — x2)^ dn+m = dx^+m^2 2"n

~

1) -

"

General domains

Let us consider the harmonic equation first. The integral equations (1.11) and (1.13) can be regarded as a boundary reduction, so the corresponding boundary element method is based on the approximation to these integral equations. However for Dirichlet boundary value problems we have shown that the integral equations are not in fact equivalent to the original problems, which causes some inconvenience. Another choice is representing the solutions of Dirichlet problems in terms of single layer potentials. We start from the two dimensional case and consider the problem (1.1),(1.2),(1.4). Let the solution u be expressed as

<x) = rrif

"(y)ln T-^—\ dsy + c -

( 2 - 2 °)

to Jan F - V\ then the boundary condition (1.2) leads to a Predholm integral equation of the first type: g{x) = -!- / w(y) In — ! — dsy + C, x€ dQ, to Jan F - V\ where the density w and the constant C are to be determined.

(2.21)

Theorem 41. If g e C2(dQ) then the equation (2.21) admits a unique solution (w,C) satisfying f u(y)dsy Jan

= 0.

(2.22)

86

Numerical Methods for Exterior Problems

Proof. Owing to the theory of elliptic equations the solution u to the problem (1.1),(1.2) satisfies u £ C 1 (fi) under the hypothesis of this theorem. Therefore f^ e C(dfi). By Theorem 7 u can be expressed in terms of a double layer potential plus a constant. It can be derived from (1.7) that Vu = 0(|:r|~ 2 ) for large |a;|, then we notice J n Audx = 0 and make use of the Green's formula to obtain fm | ^ ds = 0. Let us regard | ^ as a boundary value, then by Theorem 8 u can be expressed in terms of a single layer potential plus one constant and the density satisfies f9n uj(y) dsy = 0, which proves the existence. To prove uniqueness, we assume that g = 0 in (2.21), then define u by (2.20). By (2.21) u is a constant on the boundary dQ. By uniqueness of the Dirichlet problem u is identity to a constant on the entire plane. Then by Lemma 3w = 0. Then the equation (2.21) implies C = 0. • To consider more general g's, we need to study weak solutions and the variational formulation. We will see that the variational formulation is convenient to implementation by using the finite element method. Letting X 6 C(d£l) and fm x{x) dsx = 0, we multiply the equation (2.21) by x a n d integrate the equation on dQ to obtain b(u>,x) = — 2?r

/

u;{y)x{x)\n-

JOQ J9Q

\X

-dsydsx= - y\

g{x)x{x)dsx. Jdn

(2.23) To study the equation (2.23), we take v 6 Co°(R 2 ), and consider the bilinear form a(u,v) = / Wu-Wvdx= Jv

h»

V f — / ui(y)\n-. \27r Jan \x-y\

rdsv

+ C ) • Vvdx. J

By exchanging the order of integrations we get a(u,v) = / ijj(y)dsy \ V x — h i : 27r Jan hi \x-y\

:-Vxvdx.

Let / = — Av, then by the property of Newton potential (see Section 1.2) f 1 1 v(y) —In: -f(x)dx= Ky = / > JR22ir \x-y\JK '

f 1 1 / V ^ — In:S7v(x)dx. Jw x2n \x - y\

Therefore a(u,v)=

f u(y)v(y)dsy. Jan

(2.24)

Boundary Element

87

Method

Now we set v G F 1 '*(E 2 ) and take a series {vn} c Co°(IR2) tending to v. Passing to the limit in (2.24) we find that (2.24) holds for all v G Hl'*(R2). Letting it be an equation for u, we will prove it is well-posed. By the trace theorem of Sobolev spaces, v G H1^2 on the boundary dft, so we may assume that w G H~1^2(dfl), the dual space of H1/2(dQ,). The equation (2.24) is thus written in the form of distributions: Find u G H1'*($?), such that a(u, v) =< u, v >an,

Vv G

ff1-*(R2).

(2.25)

We define W = {w G H~1/2{dVl)\ < w, 1 >dn= 0}, then we have Lemma 36. The problem (2.25) admits a solution for all w G W, and the difference of any two solutions is a constant. Proof. We define V = {u e Hl>*(M.2); JB(0 Ro)udx and consider the problem: Find u G V, such that a(u,v) =< UJ,V >an,

= 0} with R0 > 0,

V « e V.

Since |• 11 is a norm of V, it admits a unique solution u. For any v G H1'*(M.2) we can divide it into two parts: v = v\ + C, where v\ G V and C is a constant. Noting that < w, 1 >aa= 0, we find u is also a solution to the problem (2.25). If u is a solution to the homogeneous problem, then because a(u, u) = 0, u is a constant. • We have shown that a classical solution u can be expressed in terms of a single layer potential (2.20), and satisfies (2.25). Now we set w G W, then solve (2.25) to get u G H1'*(M2). The mapping from w to u is continuous. Therefore we can take the closure in (2.20) and define a generalized single layer potential u(x) = < 57I11 urzq,w >an, which defines a continuous operator from W to .ff 1 '*(R 2 ). Let us return to the problem (2.23). b(uj,x) is n o w ready to be defined OTIW xW. The problem is: For a given g G H1/2(dCl), find UJ €W, such that b{u, x) =< 9, X >dn,

VX G W.

(2.26)

Theorem 42. The problem (2.26) admits a unique solution. Proof. We consider a single layer potential v(x) =< ^ In T^ITT, X >9n, then b(uj,x) =< v,v >sn- By (2.25) we get b(w,x) = a(u,v). For a given v G H1'2(dD.), we consider the solutions to the interior and exterior

88

Numerical Methods for Exterior

Problems

problems with Dirichlet boundary value v, still denoted by v, then v £ / ^ ' " ( R 2 ) , and ||V||I,*,R2 < M||w|| 1 / 2 ,an with a constant M. Hence for all UJ £ W it holds that ||w||_l/2=

SUp

-r—r

vem/2 <M

<M

SUp

\\v\\i/2,esi <w,v>«o dU

sup

—rr-r

ugjfV! ,, =M

IMIi,*,n a(u,v)

sup

| |

, ,, , < M|M|I.

We get b(u>,u>) > -j^j\W\\^-i/2- Owing to the Lax-Milgram theorem uniqueness and existence follows. • Using the boundary element method to solve (1.1),(1.2),(1.4), we first solve (2.26) for a given g, then evaluate u(x) —< <^ln urrq,w >aa +C, where the constant C is determined by g(x) = < ^ In T^T,W >an + C , x £ dil. It is easy to verify that u is the desired solution. For three dimensional problems the situation is similar. The solution to (1.1),(1.2),(1.4) can be expressed in terms of a single layer potential,

<^ = -k I ^y)u^7\dsv 4?r Jan

( 2 - 27 )

\x - y\

The kernel u> satisfies the problem: Find UJ £ H~1^2(dQ), b(oj, x) =< 9, X >an,

VX £ H-V2(dn),

such that (2.28)

where &(w,x) = T " / / UJ(y)x(x)] -dsydsx. 47r Jen Jan \x ~ V\ Because u,x € H~1^2(dD,), the bilinear form &(•,•) should be understood in the sense of distributions. Being the same, (2.27) is also in the sense of distributions. For a given g £ H1^2(dCl) the problem (2.28) admits a unique solution u. It is easy to verify that u defined by (2.27) is the desired solution. For the Neumann problem (1.1),(1.3),(1.4) if we express the solution in terms of a double layer potential, then u x

()

= TT" / 2TT J9n

<J

(2 / ) ~a~

ln

1

1

dvv \x - y\ for two dimensional case. Then the kernel a satisfies Lai case. Then the kernel a satisfies V

'

^Jan

Ky

'dvxdvy

\x-y\

yi

ds

y

Boundary

Element

89

Method

The kernel is hyper-singular. We will discuss this kind of integral equations later on. Next let us consider the Dirichlet boundary value problem of the biharmonic equation, (1.53). The fundamental solution K(r) = -^r2lnr, r = | a; |, satisfies A 2 u = 5{x), and K depends on r only. The solution u to the problem (1.53) is expressed in terms of a potential form, u(x) =

w(y)K(x-y)dsy+ Jan

Jan

a(y)—K(x-y)dsy+p(x), °vv

p 6 ?i(n).

(2.29) Being analogous to (2.22) we require that the densities w, a satisfy the condition,

J ("(y)v(y) + °(y)^r)

ds

v = °.

Vv e p

i•

(2-3°)

We verify that under the condition (2.30) the function u determined by (2.29) is in the space H2'*{Q). Then we derive the variational formulations for u and (w,«r). Finally we verify that u is the solution to (1.53). L e m m a 37. If u> and a are continuous functions and satisfy (2.30), then u G H2'*{Sl). Proof.

Let XQ £ Q c , then by the condition (2.30) we have

/ (w(y)(K(x - x0)+VK(x Jan +VK(x

- xQ) • (x0 - y)) + a(y) — (K(x - x0) CVy - XQ) • (XQ - y))) dsy = 0.

Then we get u(x) = / u(y){K(x Jan + / Jan

av

y

-y)-

K(x - a;0) - VK(x - x0) • {x0 - y)} dsy o-{y)-—{K(x-y)-K(x-xo)-VK(x-x0)-(x0-y)}dsy+p(x).

Applying Taylor's expansion we find K(x -y)-

K(x - x0) - VK{x - x0) • (x0 - y) = 0(\D2K\

• \x0 - y\2),

90

Numerical Methods for Exterior

Problems

as i —> oo, with xo being a fixed point and y bounded. expression of K we easily find u2

Then by the

f

/ -A dx < oo. Jn |x| 4 ln 4 |a:/i?o| The other two terms in (1.52) can be verified in the same way. • 2 We substitute (2.29) into (1.47) with v G C^°(R ). Let us evaluate the first term as the following: f vA\ f w(y)K{x-y)dsy JVL2

KJdQ

+ /

er(y)~—K(x - y) dsv +p(x) > • Av(x) dx

JdCl

"Vy

J

= / u(y) dsy / 2 uAxK(x Jan Jw + /

er(2/)-— /

J do.

Wy

— y)Av(x)

i/AxK(x-y)

dx

•

Av(x)dxdsy

JW-

= / ijj{y) dsy I vK{x — y)A v(x) dx Jan JM2 + / cr(y)^— / 2 i>K(x— y) • Jan vvy Jm.

A2v(x)dxdsy.

By the property of the Newton potential JR2 K(x — y)A2v(x) the first term of (1.47) is

dx = v(y), so

i> oj{yHy)+aiy)

L{

^')dsy

The second term can be evaluated in the same way, and we get

a(U,V) =

Ian (^{y)v^

+a

^^~)

ds

v

( 2 - 31 )

Lemma 38. Let w £ F " 3 / 2 ( 5 f i ) , u G H^^idCl), and (2.30) be satisfied, 2 2 then the solution u G # -*(R ) to the equation (2.31) for all v G H2'*(R2) exists, and the difference of any two solutions belongs to P\. Proof. We define V = {u G H2'*{R2); JR2 updx = 0,Vp G P i } , then follow the same line as the proof of Lemma 36 to obtain the result. • We remark that some integrals are understood in the sense of distributions. Since there is no misunderstanding we will not distinguish them with normal integrals.

Boundary Element Method

91

From the boundary conditions of (1.53) and the expression (2.29) we get gi(x) = Jan

uj{y)K{x-y)dsy+

a{y)—K{x-y)dsy+p{x),

ov

Jan

y

x£dQ, and 92{x)=-—\l

u{y)K{x-y)dsy

+ /

cr

(y)-z—K(x-y)dsv+p(x)\>

Jan

xedn.

)

°vy

We define v(y) = / x(x)K(x-y)dsxJan

I Jan

T(X)-—K{x-y)dsx,

OXJ

x

with dq{x) I ( X(x)q{x) + T{X)dvx Jan \

dsx = 0,

VgePi,

then we consider the dual products <9i,X> + / Jan +

+ < 92,T>=

\ Jan vJan

w{y)K{x-y)dsy

a

°vy

I 4 r \ ( Jan avx i J an

(.y)-^—K(x-y)dsy+p{x)\x(x)dsx )

(2.32)

»(v)K{x-y)dsv

+ / a(y)^— K{x-y)dsy+p{x) Jan My

J

}T{x)dsx.

By the expression of v we have < 5 i , X > + <92,T

>=

+

.

(2.33)

92

Numerical Methods for Exterior

Problems

With regard to (2.32) we define a bilinear form on (H~3/2(dft) H-^2(dn)) x (H~3/2{dfl) x H-l'2{dQ)) as follows:

x

b((co,a),(X,r)) =

\ u(y)K(x-y)dsy+ Jen Udn +

Jan

-S—\ u(y)K(x-y)dsy+ Jan °vx Udn

a(y)-—K{x °vy Jan

- y) dsy\ x{x) dsx )

a(y)-—K(x 0v y

- y)dsy\

)

r(x)dsx

We define a space

w = {(u,a) e(H-3/2(dn) x H-^2(dn))<w,v>dn

dv + dn= 0,Vv G P i } .

Theorem 43. TTie problem: find (w,cr) G W, suc/i i/iat b((w,a),(x,T))=

+ <92,T>,

V(X,T)eW,

(2.34)

admits a unique solution. Proof. By (2.31) we see that b((u),a), (x> r )) = a ( u i y )> then following the same lines as the proof of Theorem 42 we get the result. D With the above argument the problem (1.53) is reduced to a boundary equation (2.34). Then we use (W,P £ -Pi} on dfl, which is three dimensional. Then by the definition of W the dual products of the solutions to (2.34), (uj,a) € W, applied to S is zero. We set (X,T) G 5 in (2.32), then we get three equations to determine the three coefficients of p. To see (2.29) is the solution to (1.53) we only need to verify the boundary conditions, that is, (2.32) should be satisfied for all x a n d T. Let (X,T) £ (L2(dQ.))2, then we can divide it into two parts, one is in W and the other one is in S, then it is straight forward to see that (2.32) holds for all (Xi T ) £ (L2{dtt))2, so u is the solution to (1.53).

Boundary

2.3

Element

93

Method

Subdivision of the domain

If the domain Q is in a complicated shape, the simplest way to deal with exterior problems is to truncate the domain. Let us consider the problem Au = 0,

u = g,

(2.35)

(2.36)

x G d£l,

in three dimension first. Let f2o be a domain with a simple shape, and Jlo 3 fic- We define Qi = Qo H fi. The domain ft is replaced by fii, and some artificial boundary condition is imposed on the artificial boundary d£l0.

•X

X

The most self evident attempt is to impose u = 0,

x G 3f2o.

(2.37)

Let Uh be the solution to (2.35),(2.36),(2.37) on £l\. It is indeed an approximation, because we have the following estimate.

94

Numerical Methods for Exterior

Problems

Lemma 39. We suppose that dim (fi) = 3. Let R = min{|a;|;a; e then \u-uh\

dflo},

=O

Proof. By the expression in Section 1.1 we get u = 0(1/R) on dttoThen applying the maximum principle for u — Uh the conclusion follows.• For two dimensional problems the situation is worse, because the solution u does not tend to zero at the infinity. However we still have the following: Lemma 40. We suppose that dim(Q) = 2. Let g £ Hl/2(dfl). Then the approximate solution Uh converges to u on any bounded domains weakly in H1 and strongly in L2, as R —> oo. Proof. Owing to the inverse trace theorem of Sobolev spaces, we can take a function UQ e iJ 1 (fi) with a compact support, such that uo|sn = g and ||«o||i < C||g||i. Let R be large enough, so that uo = 0 on dfto. Let Mi = Uh — UQ, then by the weak formulation (1.20) it holds that a(ui,v) = -a(uo,v),

Vv £ ffo(fli).

u

Let v = Mi, then |wi|f < |uo|il i|ii which gives |ui|i < |wo|i < C||fl||ii where the constant C is independent of the domain Q\. We extend Uh by zero on D.2 = K2 \ ^ , then Uh G Hl'*{Q). Uh satisfies

KHi,.
Vveflo1^!),

a(u,v) = 0,

Wei^fti).

to obtain

Since R —> oo, u is the solution to the problem (2.35),(2.36). Because u is unique, Uh converges to u as R —» oo. • Although the above strategy generates convergent approximate solutions, the rate is very slow. The artificial boundary condition (2.37) should be improved. Using the theory in Section 1.1 and Section 2.2 the solution

Boundary

Element

Method

95

u on the domain SI2 can be expressed in terms of single or double layer potentials. For definiteness, let us consider two dimensional domains. By (2.20)

"(y)ln 7-^—r dsv + c '

«(*) = TT I 2

/

u

ds

(y)

(2-38)

v = °.

x

^ Jen2 \ ~y\ Jdn2 so we can impose a boundary condition on dflo instead of (2.37), u x

()

=

7T / 27r

w

(2/) m 1

7afi 2

F dsv + C, x € 8Q.Q,

F - 2/1

Here w and C are unknown, so (1.9) 1 du(x0) = - - w ( x/ 0 )x + dv+ 2

/

uj(y) dsy = 0.

JdQ0

(2.39) we need another boundary condition. By

1 /f w(j/)^— , x d ,In, 1 rdsy, — 2TT Jano dvx \XQ - y\ for XQ S 9J7O- Therefore we get another boundary condition, du dv+

W -nu(x) + 7T (y)-S— l n l \dsv x€dn0. (2. 40) w 2 2TT 7 a n o 5z/x |x-y| Theorem 44. Let g £ Hll2{dSl). The problem (2.35), (2.S6), (2.39), (2.40) with unknowns U,OJ,C on the domain fii is equivalent to the problem (2.35), (2.36) on the domain fi.

Proof. Let u be the solution to (2.35),(2.36), then u is a harmonic function, so it is analytic. By Theorem 42 u can be expressed by (2.38) on SI2, so u is the solution to (2.35),(2.36),(2.39),(2.40). Conversely, if u is a solution to (2.35),(2.36),(2.39),(2.40), then we can define u on ^2 by (2.38). Then u £ Hl,*{ST), u is a harmonic function on Q,\ and Q.2 respectively, and u satisfies the boundary condition (2.36). We only need to show that u is a harmonic function on the entire domain SI. The weak formulations are: /

Vu- Vvdx=<

--u+

— /

u>(y)—\nVveH1(n1),v\dil

;dsy,v>dQo, = 0.

and / Vu- Vvdx=< 7n2

-w- — / w(y)— In, r dsj,,v >an 0 , 2 2TT J a n o Si/ I - —2/|

96

Numerical Methods for Exterior

Problems

Now we take v G H0'*(Q) and add them together, then we get a(u, v) = I VM • Vv dx + Vu • Vv dx = 0. Jsix Jn2 Therefore u is a harmonic function on Q,.

•

The artificial boundary conditions (2.39) (2.40) are exact, so they yield better results.

2.4

Dirichlet to Neumann operator

The above mapping U\QQ0 —• (w,C) -+ | ^ | e n defines an operator if : Hl'2{dQ,o) -> l/'- 1 / 2 (ano), which is called a "Dirichlet to Neumann operator" . Using the operator an exact artificial boundary condition is designed to get a problem equivalent to (2.35)(2.36): Aw = 0, u = g, du — = Ku, av

x G fli, x £ dfl, x G d£l0.

The operator K depends only on the the equation and the artificial boundary dtto- To obtain K, one needs to solve an integral equation for ui. However the choosing of flo is quite flexible. If on some occasions fio = B(0, R), then using the results in Section 2.1 K can be expressed explicitly. This idea is due to Feng and named after "natural boundary reduction". See [ Feng, K. (1982)] [ Feng, K. (1983)] [Feng, K. (1984)] [Feng, K. (1994)] [Feng, K. and Yu, D. (1983)], also see [Han, H. and Ying, L. (1980)]. Let us consider the exterior problem of harmonic functions first. Differential of (2.3) gives du _ 1 f ~dr ~ 2TT J0

27r

ArR2 - 2R(r2 + R2) cos(6 -
Let r —> R, then we get the following formula formally, du dr

= T^ r 4nR J0

—Te^9^)dV. 2

Sm

^

(2.41)

Boundary

Element

Method

97

Let v denote the outward unit normal vector along the boundary, then we get an explicit expression for K:

For the Helmholtz equation differential of (2.16) gives oo

ur(r,0)=

Y,

anuHV'(ujr)eine,

(2.42)

a

(2.43)

n— — oo

then formally we get oo

ur(R,6)=

n"Hiiy'{wR)einf>'.

J2 n = —oo

Let v be directed to the exterior of the domain Clc, we define tr{u){6) = u(R,6) = g(0), tr{uv)(6) = uv{R,6), then (2.16) and (2.43) together defines an operator K, such that tr{uv) =

K(tr(u)),

or explicitly (•27T

-ur(R,6)

= K(6) *u(R,6)

= / Jo

K{6 - 6')u(R,0')d8',

(2.44)

where the function K is

The integral in (2.44) is hyper-singular. K is in fact a pseudo-differential operator of order 1 on the boundary manifold Oil, and defines a linear continuous map

K

:Hs{dn)-^Hs-1{dn).

Parallel to the association of the operator A + w2 in the domain D, with the bilinear functional a(u,v) = / ( y u • y v + to2uv) dx, Jn

98

Numerical

Methods

for Exterior

Problems

there is also an association of the operator K on the boundary dd with the bilinear functional /•27T

b(
Jo

/»27T

tp(9) dB / Jo

K{9 - 0')v(6')

dO',

which is inherently related t o a(-, •) by t h e equality, a(u,v) =

b{tr(u),tr(v)).

Being t h e same, for three dimensional case differential of (2.19) gives oo

n

ur{r,0) = J2

Yl

uanm^i±{ur)Ynm{6,
(2.45)

n = 0 m= — n

t h e n (2.19) a n d (2.45) together define a n operator K. We will provide a rigorous mathematical setting of t h e operator K in t h e next section.

2.5

Finite part of divergent integrals

Sometimes some integrals in t h e expressions of some D t N operators are divergent in t h e sense of Riemann integral. In fact they are singular integrals in t h e sense of Cauchy or H a d a m a r d . To provide a rigorous meaning of these integrals, let us introduce some basic definitions. [ Feng, K. (1982)] Let xm, x>0, 0, x < 0, ' where m is a complex number. Letting

/

/-OO

xr£
/ x^ip(x)dx, (2.46) •oo Jo which converges in t h e sense of Riemann if R e ( m ) > —1. Therefore x™ is a distribution, denoted by [x™]. It is easy t o see t h a t )

dx^\T{m

r rn-\-p-i I J

+ p+l)[X+

J

^

r ml

T(m+l)[X+1,

for positive p . Let us define distributions for R e ( m ) < — 1. T h e dual p r o d u c t ([x™], tp) defines a n analytic function for Re(m) > — 1. T h e analytic continuation can

Boundary Element

99

Method

be defined as the following: For m ^ negative integers, and a non-negative integer p, satisfying Re(m) +p > — 1, define a distribution, PfTm J

1

dP

=

+

[

Wm+P]

{m + p)---(m+l)dxP

+

J

T(m + 1) d? T(m + p+l)dxP[

=

+

J

+

'

The above definition is independent of the choosing of the parameter p, and Pfx™ = [x^] provided Re(m) > - 1 . By the definition of distributions one gets a finite part of (2.46), (Pfxy,
(2.47)

where ip^ = ^ £ . It is easy to check that (Pfx™, ip) is an analytic function with first order poles m = negative integers. One explicit expression of (2.47) is derived as follows. For e > 0, we consider J°° xmtp(x) dx. Let a > e be a fixed number, and

(fc)

^ =£^E ^(oLrfe^ fc=0

then /»oo

/ Je

pa

xm
pa

/ xm(ip(x)-tpp(x))dx+ Je

/•oo

xmcpp(x)dx+ Je

If rn is not a negative integer, then

, .. , _

k=o

xmV{x)dx. Ja

k\

m+ k+1

fc=o

m P(x) dx = If /m xis a
0
^ ^

v(fc)(0)

,,

fc!

a™+fc+i

;—7 1+

m + fc + 1

^ ( 0 ) £ m + f c +! k\ m + k + 1

w(-" , - 1 )(0) ( - m - 1 ) ! In a (/ p(-

m 1

- )(0) (-m-1)!

\i k ^ —m — 1, the limit of em+k+x is either zero or infinity as e —+ 0. Naturally the finite part of the divergent integral is defined by: If m is not

100

Numerical Methods for Exterior

Problems

a negative integer, then xmcp{x)dx=

Pf

xm(ip(x)-
J0

xm
I

J0

Ja (2 48)

A y W ( Q ) a"+fc+i ^ k\ m + k + 1'

|

-

fc=0

If m is a negative integer, then /»oo

Pf / Jo

/>a

/»oo

xm(ip(x) -
x"V(:r) cte = / Jo

y> +

^

xm
Ja y(fc)(Q) am+k+l

fc!

m +fc+ 1

y(-m-l)(Q)

+

(-m-1)!

n

°'

(2.49)

Lemma 41. 7/m zs not a negative integer, then /•OO

xmcp(x)dx = (Pfx™,lf).

Pf Proof.

(2.50)

It is easy to see that

I

xmtp(x)dx = ([x+l], — 1. However both sides are analytic functions with respect to m, and both sides of (2.50) are analytic continuations, therefore by the uniqueness they are equal. D Jo

It remains to prove that the definition of (2.48)(2.49) is independent of the choosing of parameters p and a. If m is not a negative integer, the assertion is already contained in the previous lemma. Here we give another proof which is valid for all complex numbers m. Lemma 42. The definition of (2.48), (2.49) is independent of p and a. Proof. Letfc*be the smallest non-negative integer p satisfying Re(m) + p + 1 > 0. If k > k* and e —> 0, then

m+k + 1 ~~*

Boundary

Element

101

Method

By the definition, if m is not a negative integer, Pf

7 /

V

/.oo

xmip(x) dx = lim

/

£-•0

Je

= lim

/ A

Jo

e-tO

<^fc)(0) em+k+1 k\ m + k + 1 fc=0

xm
£^(fc)(0)

xmip(x) dx + y^k=0

k\

m+k+1 £

m + k+1

The right hand side is independent of p and a. Noting that if m is a negative integer, k* = —m — 1, then we get /*oo

Pf /

z " M x)dx

Jo

— lim £->0

/»oo j /s

fc* y <^fc)(0) e^+fc+i x cp{x) dx+^2 A;! m + k + l fc=o m

+

kJ

which is also independent of p and a.

lne

a

We have derived (2.41) formally in the previous section. Now we are ready to prove it rigorously. Let us assume that g € Cco(dQ) first. By (2.1) we get

dr ~

2ir J0

I ^

""

|n

r

-|n|-lem(9-^)l5^-)^

)

^ n = —oo

where for simplicity we take R= I. We define a function 1

°°

(2.51)

then for a fixed r € (1, oo) the right hand side can be expressed in terms of a convolution product, or a dual product on the space Cco(dQ): du or

—

=

- K

r

* g ••

.

For this boundary condition the solution u to the exterior problem belongs to C°°(fi), Therefore | ^ converges as r —> 1. The functions Kr is thus convergent weakly with respect to r in the dual space C°°(fi)'. Let the limit be K £ C°°(d)', which is a distribution by definition. We obtain the derivative at r = 1,

fr=-.

(2.52)

102

Numerical

Methods

for Exterior

Problems

Now let us derive the expression of K. Let us define 1

°°

r-l

n

l-1ei™e

(2.53)

n = —oo

and

**<«> = -2ns n =<
(2.54)

ra^O then fl#ir 06

Kr

OK-2r "

=

06

Kir,

for r £ ( l , o o ) . Consequently < tfr(0 - •), = < A"lr(« " O.fl' > = < ^ 2 r ( 0 " O.fl" > Letting r —> 1, by (2.51) .ft^r converges in L 2 (9fi) t o 1 YimK2r{6) r-+l

=

°°

1

==

- Y—

27T

In 2 s i n -

2

7T

n—~ oo

\-K t-^ Ira which is a function in the classical sense. Therefore

<9u Or

p27T

i r Wo

-r=\

In 2 sin- - V

g"{
(2.55)

This is the expression of the distribution —K. Taking classical derivatives on 6 £ (0, 27r) we get

A/I In d0 \ ^

2sin2

27rCOt2'

d f 1 6»1 —I —cot- } =

i6\2TT d0

2J

1 47rsir 2

e

Noting (2.53),(2.54) we can define some distributions as the limit of some divergent series, K =

Pf-

47r sin 2

e

'Alt 2rc

\n\e

Boundary

Element

103

Method

and

K1

=

-.

f 1 9\ Pf \ 7 - cot - } \2TT 2J

^^

^ 27U

I

I

^ n n = -co

>.

The derivative of u is the finite part of a hyper-singular integral, du dr

1

2TT

Pf- 4TT JO

sin2 6-^

;9{
(2.56)

For general R, the formulas are, ne

in&

and du dr

2.6

p, r=R

iv)dr

{-ssr^'

.

Numerical approximation

We investigate the implementation of the boundary element method in this section. Some meshes on the boundaries are required in the numerical computation of the above integral equations, and if the exterior domains are truncated, some meshes on some bounded domains are required as well. First of all let us consider the integral equation (1.11) for two dimensional problems as an example. Let us fix a point on dSl and let s = 0 at this point, s £ [0, L] is denned by the arc length along anticlockwise direction. The curve dfl is divided into some arc segments by some nodes s\ = 0, s%, • • • , sjv, and the elements, Ci, i = 1,2, •, N, are the arc segments s{s2, S2S3, • • • , srfsi. Let the approximate solution be
Si

—crh(Si+i) H

Si+i

r

-
104

Numerical Methods for Exterior

Problems

on SiS'i+i, where hi is the arc length of this element. Usually there are two methods to solve a^. 1. Collocation method Let us write the function K by K(s,t), s G [0,L],t e [0,L]. We assume that the equation (1.11) is satisfied at nodes, 2crh(si)+

ah(t)K{si,t)dt

= g(si),

i = l,2,---,N,

(2.57)

which is an algebraic system with N equations and N unknowns ah(si). We can define a finite dimensional space S(dQ) = {ah G C([o,L]);ah(0)

= ah(L),ah\ei

G Pi(eO,t = l , 2 , - - - ,N},

where Pi is the set of linear functions. Then we can define a set of basis functions in Sh'. UeSidU),

Li(si) = l,Li(sj)=0,j^i,

i=

l,2,---,N.

The function ah can be developed in terms of the basis functions: N

Vh(s) = ^2Msi)Li(s).

(2.58)

Then the equation (2.57) can be written as N

1

a

( fL

\

2 h{si) + X ! I /

Lj(t)K(si,t)dt

3=

J ah(sj) = g(si) t),

i=

l,2,---,N.

2. Galerkin method Multiply the equation by r G C(dQ.) and integrate the equation over dfi,, then we have - /

a(x)r(x)dx+

* Jdo.

/

a(y)K(x,y)T(x)dsydsx=

Jan J an

g(x)r(x)dsx.

Jan

The Galerkin scheme is: Find ah G S(dfl), such that - / ah(x)Th(x) * Jon = f g(x)Th{x)dsx Jan

dx + / / Jen i s n

ah(y)K(x,y)Th(x)dsydsx C2

KQ\

\/Th£S(dn).

Since S(dQ) is a iV-dimensional linear space, the equation (2.59) is a N x N algebraic system. Let the unknown function ah be developed in terms of

Boundary

Element

105

Method

the basis functions as (2.58), and N T

h(s) = y"Vh(Si)Li(s), i=l

then the equation (2.59) can be written as N

Y^kijah{sj)

= gi,

i = 1,2, ••• ,N,

3=1

where -i

/»£

pL

kij = - / Lj(s)Li(s)ds+ 2 Jo

pL

/ Vo Vo

Lj(t)K(s,t)Li(s)dtds,

and /

./o

9 ( 5 ) ^ ( 5 )Ids. ,

The strategy for solving the integral equation (1.11) for three dimensional problems is similar. However, it is difficult to construct a continuous and piecewise polynomial function on a closed smooth surface dQ, so usually the boundary dQ, is approximated by an surface dflh, which causes an additional error. Next let us consider the problem (2.26). Let us define a subspace of S(dfi): S0(d£l) = I u € S(dtl); I

u{s) ds = 0 L ,

then the approximation to (2.26) is: Find u>h & So(dil), such that & K , Xh) =< g, Xh >aa,

VX G So{dSl).

The bilinear form b(-, •) defined in (2.23), b

/ w(y)x{x)\n-dsydsx, 2TT Jan Jan F - V\ is symmetric, so the equation (2.60) is equivalent to (u,x)

= 7T

2—

1 2b(UJh,UJh">~

< 9 Wh

'

>m

I

=

°'

(2.60)

106

Numerical Methods for Exterior

Problems

With respect to the constraint fQ Uh(s) ds = 0, we introduce a Lagrangian multiplier A and define a function -b(uh,wh)2

< g,tOh >dn +A /

u!h(s)ds.

The problem (2.60) is equivalent to: Find (uh, -M G S(d£l) x R, such that H"h,Xh) + *

Jo

Xh(s)ds=

Jo

g(s)Xh(s)ds,

VX € S(dfi),

and fL / u>h(s) ds =0. Jo This is a (N + 1) x (N + 1) algebraic system. Finally let us consider the exterior problem of the Poisson equation, -Au

= f,

u = 0,

rrefi, x€dQ

(2.61) (2.62)

in two dimensional space, where / G Co(fi). For simplicity of the domain partition, let us assume that Clc is a polygon. Let f2o be a disk B(0, R) with R large enough, so that supp / C B{0,R). Let fii = flo f~l fi, then by the argument in Section 2.4 the problem is equivalent to -Aw = /, u = 0,

i£f!i, z G dCl,

— - Ku, x£ dfto, a;/ where K is the Dirichlet to Neumann operator. Let a(u,v) = /

(2.63) (2.64) (2.65)

Vu-Vvdx,

JQI

then the variational formulation of the problem is: Find u G Hl{£l\), that U\QQ = 0 and a(u,v) - (Ku,v)da0

= (f,v)ni,

W G tf^fii^Hafi = °-

such

Boundary Element

107

Method

Applying the finite element method to solve this variational problem, we set a mesh. For example, the domain fii is divided into triangular elements {ej}, and linear interpolation is designed on each element with respect to the vertices. Since dflo is a circle, each element neighboring the circle possesses an arc edge. The finite element space is denned by, S(fii) = {u € ^ ( O i ) ; ^ , G Pi(eO,Vidian = 0}. The finite element formulation is: Find Uh € S(D.i), such that a(uh, v) - (Kuh, v)dno

= (/, v)Ql,

Vu £ S^).

(2.66)

The evaluation of a{uh, v) and (/, v) is routine. However the evaluation of (Kuh, V)SQ0 requires some special consideration. There are two kinds of approaches: 1. Truncation r-2-K

p

(Kuh,v)dno=j

•,

(

[-2TT

f \ - ^ l

°°

"I

\n\ein^-^uh(R,lp)diP\v(R,6)de.

E

Let N be a finite number, then approximately 2TT J"

/

I

p2TT

\ - ^ j

N

1

\n\ein{9-'fi)uh(R,
S

2. Integration by parts If Uh and v are infinitely differentiable, we denote g(6)=uh(R,9),1(e)=v(R,6), and

W ) = -^ln

«2 s •i n -e 2

then by (2.55) it holds that Kuh{6) =

-

Numerical Methods for Exterior

108

Problems

Consequently we have PZTT

(Kuh,v)ano=-

/ Jo

< K2{6 - •),

i(0)RM

/•2ir

:/ Jo

1{6)Rde r2-K

I

i{6)Rd6,

Jo Jo

that is r2n />Z7T

-i

(Kuh,v)dQo

= - / "• 7o

2ir r/*Z7T

/ Jo

Q

In 2 sin •

_

^

g'(
(2.68)

The definition of S(fii) implies that U\QQ0 S iJ^dfio), so applying the fact that C°°(d£lo) is dense in H1(dQo) we can show that (2.68) holds for Uh, v € S'(fii). The integrals in (2.68) are convergent, and can be evaluated using a numerical scheme of integration. 2.7

Error estimates

Let us prove convergence and estimate errors to some of the above schemes. The first example is (2.60). We assume that the exact solution w £ H2(dd), and let II denote the interpolation operator. Let h be the maximum length of all elements. Then owing to the error estimates of interpolation: [Ciarlet, P.G. (1978)] ||IL; — w||_i/ 2 < C||na> — w||o < Ch2\u>\2Let eh =< HLJ, 1 > / / af2 ds, then | < IIu; - <J, 1 > |

|ino;-^l|_1/2l[l||1/2 ~ < Oft \u\2? Ja-n ds Jan ds Let u>i = IIw — eh, then u>j £ So(dtt), and \eh\ =

< C

\\ui -u}\\-i/2

<

Ch2\u>\2.

The following error estimate is routine: a

w w - Uh\\_i/

2

-ujh) — b(u> -u>h,oJ — wj)

<||w - w^||-i/ 2 ||w - w/H-i/2-

Boundary Element

109

Method

Consequently we have the following: Theorem 45. It holds that < Ch2\u\2.

\\u-Uh\\-i/2

The second example is the scheme (2.66). We assume that the partition is regular, that is the interior angles of all elements possess a common lower bound 9Q > 0. First of all let us show the existence and uniqueness of the solution to the problem (2.66). We extend a function Uh € S(fli) to D,2, the exterior of dfi,o, by solving a boundary value problem of the harmonic equation on $72 with the boundary value u/i|an 0 i then Uh G -ff1'*(fi), and the application of Green's formula leads to a(uh, v) - (Kuh, v)an0 = / Vuh • Vv dx, Vv e H1'*^). Jn Owing to the estimate for exterior problems of the harmonic equation and the trace theorem of Sobolev spaces, we obtain |"h|i,n 2 ^ c\\uh\\i/2,dn0

< C'lluhUi,^

Consequently it holds that \a(uh,v) - (Kuh,v)9n0\

< Clluhlli^illulli.fi!,

Vuh,v G 5(fii),

and \a(uh,uh)

- (Kuh,uh)da0\

> a\\uh\\jni,

Vuh G S(Cli),

with a > 0. Owing to the Lax-Milgram theorem there is a unique solution. Let u be the exact solution to (2.61),(2.62), then a standard argument leads to the following: (see [Ciarlet, P.G. (1978)]) Theorem 46. It holds that ||w-"/i||i,ni ^ Ch\u\2,niFinally let us consider the error of the truncation (2.67). Let u be the solution to (2.63)-(2.65), and UN be the solution to —AUJV = / , UJV

= 0,

x £

fli,

x G dQ,

110

Numerical Methods for Exterior

KNUN,

~&7

Problems

x e d£l0,

where JV

,2TT

KNuN

= -—

27r

\n\ein(e-^uN{R,V)dip.

V

Jo

n=-JV Jr'K,

The variational formulation is: Find UJV G - ^ ( ^ l ) ) such that wjv|an = 0, and bN{uN,v)

= (f,v),

W G ^ ( n ^ . u l a n = 0,

(2.69)

where /•27T

i>N(uN,v) = a(uN,v) — / Jo

KNU^(0)v(R,0)Rd6

and a(uN,v)

= = /

Vujv • Vv da:,

where we notice that although the problem and functions are real, the Fourier series are expressed in terms of complex functions. Define b(u,v) = a(u,v) -

(Ku,v)dno,

then the variational formulation of the problem (2.63)-(2.65) is: Find u G i7 1 (Oi) such that U\QQ, = 0, and W G H1^),

b(u, v) = (/, v),

(2.70)

v\dn = 0.

Let us define a norm: \\WN\\\

=bN(uN,uN)* r2n

= \ l"Ar|i,ni +

N

2^

2

inv

h= i^ E \/W\e- uN(R,ip)dV 27r

Jo

„=_JV

then \bN(uN,v)\

< \\\uN\\\ • \\\v\\

Owing to the Lax-Milgram theorem the problem (2.69) admits a unique solution.

Boundary Element

111

Method

We turn now to prove the convergence of (2.69). We have bjv(u — UN, u — UN) =6JV(U, u — UJV) — b(u,u

— UJV) + b(u,u

— UN) . — 6JV(UJV; w — UN)

=6JV(W, u — ujv) — b{u, u — UN)-

By the definition of b and 6JV, 6JV(W,W —

«jv) - b(u,u — UN) = (Ku -

KNU,U

- UN)an0-

We apply the Schwarz inequality to get \bN(u,u - uN) - b(u,u - uN)\ < \\Ku - KNu\\0taa0\\u

- UN\\o,ano- ( 2 - 71 )

Let us estimate \\u — Ujv||o,9n0- Let RQ be the maximum positive constant, such that B(0, RQ) C fic, Ri be the minimum constant, such that supp (/) C B(0,Ri), and let the constant R satisfy R> R\. Extend the function u — UN by zero on Clc, then we get R

(u-uN)(R,6)=

d(u — dr

f JRo

UN)

dr.

By the Schwarz inequality we obtain ((u-uN)(R,0))2

p2ir

T)

((u-uN)(R,e))2d6<—

d(u — dr

UN)

dr.

JRo

Hence we have / Jo

f

-fto

pR

flir

rdr JR0

JO

d(u — dr

R

UN)

d6 <

-£-\u-uN\ln

that is u

N\\lidn0

R2 < -5~\u - uN\ln

R2 <

-p-\\\u-uN\

(2.72)

Next let us estimate \\Ku—K Nu\\otan0- The solution u can be developed into a series u=

J^

anr-{nleine,

with DI«I

1 2TT

r2ir

/ Jo

e-^uiR^^dtp,

112

Numerical

Methods

for Exterior

Problems

for r > R\. Therefore du dr

Kv

Y^ |n|ani?-l"l-1e™e, r=R

and ^ InlanR-W-1^inO

Ku-KNu=-

\n\>N

The norm is equal to \\Ku-KNu\\ldUo=

\Ku-KNu\2ds=

f JdQ.0

\Ku-KNu\2R dO

[ JO

=2nR J2 | n | 2 K | 2 i T 2 H - 2 |n|>AT

(2.73)

Integrating by parts gives \n\

r2n

R[;

2TT ni

|„|2|_

Jo

|2 _ p2|"l 2TT

\n\

(2.74)

dip,

dip

\an\

J0

dip

— rtj

2

Plugging it into (2.73) gives

\-^MR^)d^ dip

\n\>N -2(N+1)

2

*

<±R-> (2•2TT \RI

n\>N 2(JV+1)

x

<-R~ -2?r

_inv,du(RUip) dip d
oo

r2n

y

(# \Ri

f\-^
f r ^ Jo

-2(N+1)

9p

r2TT

a
R_y*"-° Ri

f

I

(du{RUip)

(•

ds

ds.

Boundary

Element

113

Method

Therefore \\Ku-KNu\\0,da0

< ( ^-J

Mi.aBCo.flO-

(2-75)

The combination of (2.71),(2.72),(2.75) gives the following: Theorem 47. It holds that |||U-UJV||| < - r f ^ - j

\u\i,aFHo,R1y

R§ Moreover, the solution u is infinitely differentiable on Q. If we replace (2.74) by integrating by parts for k times, then the result is generalized to Theorem 48. It holds that

|||«-^|||<^-(]^T) 2.8

(§)

+2

\u\k,aBio,Rl),

Vfc = l , 2 , . . . .

Domain decomposition

The stiffness matrices of the finite element method are usually sparse, while the matrices of the natural boundary element method are usually circulant. It is known that effective numerical methods to solve these two kinds of matrices are different. For the coupling of these two methods one strategy to separate these two matrices is using an iterative procedure. For example the problem (2.66) can be solved by « K , v) - (K\n, v)dno

= (/, u)m,

to

e S(fii),

A"+1 = enunh + (i - en)xn, where n = 1,2, •••, and 6n £ (0,1) is the relaxation factor. This is the conventional Dirichlet-Neumann alternating scheme in the non-overlapping domain decomposition method. Another approach is the overlapping domain decomposition method. Again for the problem (2.66) let R > Ri, B(0,R±) D supp (/), and define fii = Qf]B{0,R), n2 = {x;r > Rr}. The scheme reads: u? h G S(fii), n —'1 uih\r=R = IIu22hh \r=R, and a « h ) v) = (/, v)ni,

Vv e 5(fii), v\r=R = 0,

Numerical Methods for Exterior

114

Problems

and U2h is given by the Poisson formula on Q.2'^

=

1 t o j

f2v

r2 - R2 Ri-2rR11coS(0-v)U^R^)dif>

r> +

0

where n = 1, 2, • • •, and II is the interpolation operator to the finite element space. Convergence proof is essentially standard. For details, see [Yu, D. (2002)]. 2.9

Boundary perturbation

For two dimensional case if the boundary is a small perturbation of a circle: d£l = {(r,6);r = Ro + £7(0)}, then the solution to the exterior problem can be developed into a power series in terms of the parameter e. Following the argument by [Bruno, O.P. and Reitich, F. (1996)] let us introduce this approach for the exterior boundary value problem of the Helmholtz equation: (A + w 2 )u = 0,

ieO,

(2.76)

with the boundary conditions, u = g,

x£dQ,

(2.77)

£-iu;« =0(l). Let B(0,R) D JF, and U\ = {(r,9);R0 developed as

(2.78)

< r < R}. The solution u can be

00

u=Y/Mr,0)en.

(2.79)

n=0

The functions un satisfy {A + LU2)un = Q,

un = gn, dun —— = Kun, av

XGQ.I,

r=

RQ,

r — R,

Boundary

Element

Method

115

where K is the Dirichlet to Neumann operator given by (2.44), and gn are equal to gn{9) - 6n0g(9) - ^ 1=0

1—— v

^ - j

.

'

It is observed that the above algorithm suffers from ill conditioning due to significant cancelations [Nicholis, D.P. and Reitich, F. (2004a)][Nicholis, D.P. and Reitich, F. (2004b)]. There is another approach. First of all a mapping is denned to transfer the domain $7 Q B(0, R) to fii: ,

(R - R0)r - Re^e) (R - Ro) - e 7 (0)

& = 6.

The system is modified, upon dropping primes, to du \ r-— dr Vr dr )

d2u — 2 +r'iLj2u = F(r,6,u), d6

R0 < r < R,

Ro, 3n — =Ku + J{6),

,r = R,

where F and J are some known functions. Then following the same steps u is developed into a series (2.79), and un are solved by recursion. A spectralGalerkin method is given to solve this problem. For details, see [Nicholis, D.P. and Shen, J.].

Chapter 3

Infinite Element Method

Using an infinite number of degree of freedom, one can make a direct discretization of the original exterior problem. The problem is the implementation of the approximate problem which contains infinite unknowns. The introducing of a structured mesh can solve the problem. It is possible to analyze the discrete problem and eliminate infinite number of unknowns in advance. Then only a very small amount of unknowns are needed to be taken into account in the real computation to result in a solution with infinite unknowns exactly. We will present this approach and study the related mathematical problems in this chapter.

3.1 3.1.1

Harmonic equation-two dimensional problems Infinite

element

formulation

We start with the problem (1.1),(1.2) and (1.4). We assume that flc is a polygon in a plane, and it is in star-shape with respect to the point O, that is, the line segment linking each point on dfl with the point O is included entirely in the interior of dQ. Applying the approach in Section 1.3, we have the variational formulation: For a given g € H1/2(dil) find u G iJ 1 '*(fi) such that u\on = 9, and Vveffo'*( fi )>

a{u,v)=0, where a(u, v) = I

Jn 117

S/u-Vvdx.

C3-1)

118

Numerical Methods for Exterior

Problems

To solve (3.1) fl is divided into an infinite number of triangular elements. Let us denote dd by r 0 . Taking a constant £ > 1, we draw the similar curves of To with center O and the constants of proportionality £>£2>' •' i£ fc >''' i which are denoted by T i ^ , • • • ,Tfc, • • • respectively. The domain between two polygons is named a "layer". Afterwards each layer is further divided into elements. It may be proceeded in such a way: some points on To are selected as nodes, where the vertices of To must be nodes and some other nodes on the line segments can be properly selected according to our requirement, then rays are drawn from the origin to the nodes, consequently every layer is divided into some quadrilaterals which are similar to each other among the layers, finally each quadrilateral is divided into two triangles such that the manner of partition for each layer is the same. The triangular elements are denoted by e,, i = 1,2, • • •. We denote £kQ. the exterior of Tk and Q.k = (£ fc-1 fi) \ (£fcft).

• * i

Fig. 4

Sometimes the domain Qc may not be in star-shape, and the curve To may be a polygon in very complicated shape. Then we may decompose

Infinite Element

Method

119

the exterior domain fi to an unbounded domain with regular shape and a bounded domain with complicated shape like Section 2.3. We still denote by n the unbounded domain, which can be divided in the above way, while the bounded domain can be divided into finite elements in conventional way. We need the two styles of partition to be conform to each other, that is, the nodes and line segments of the elements are coincide to each other along the interior boundary. We still first analyze the domain fl for this situation. We will state an algorithm for calculating the combined stiffness matrix of 0,, which is a matrix with finite order, and which is applied either to solve a boundary value problem directly, or to solve a boundary value problem for a domain with complicated shape by assembling it with the stiffness matrices of other elements. For the later case 0, is treated as one element. we consider the infinite element subspaces of i71'*(f2). Let the nodes be the vertices of all triangular elements, and linear interpolation is applied in all elements. Then let the interpolating functions belong to the space iJ 1 '*(Q). The infinite element spaces is defined as the following: S(Q) = {ue

H^ffl;

u\et G P^eO, t = 1,2, • • • } ,

So(n) = {ueS(n);u\dQ

S(rk)

=

= 0},

{u\rk;ueS(Q)}.

Let gj G S(To) be an approximation of g. The formulation of the infinite element scheme is: find Uh G S(Q) such that U\QQ = gj, and a{uh,v)=0,

VD e 5 0 (0).

(3.2)

L e m m a 43. The problem (3.2) admits a unique solution. Proof. We define a function «o £ S(Q), such that uo\an = 9i and uo = 0 for all x G £Q, which can be obtained by interpolation with respect to all nodes. Let Wh = Uh — «oi then u>h is the solution to: find Wh G SQ(Q), such that a(wh, v) = -a(u0, v),

Vv G So(fi).

Owing to the Lax-Milgram theorem, existence and uniqueness follows.

•

120

3.1.2

Numerical Methods for Exterior

Tranfer

Problems

matrix

We now turn to the implementation of the above scheme. In order to carry out the real computation, we deduce an equivalent and finite form of it. We consider the polygon Tk, and starting from one node, we arrange the nodes of Tk in an order according to the anticlockwise direction. Let the number of nodes be n. The nodal values of the solution of (3.2) are also arranged in an order, which are denoted by yk ,yk , ••• ,yk , and they form a n-dimensional column vecter yk = (yk ,yk ,••• ,yk ) T .Let us denote yk = BkUh, where Bk is the trace operator. We will identify yk & K n with u\„ rather than distinguish them. ' >- k

We consider the fc-th layer Clk. The element stiffness matrix of each triangular element can be evaluated by conventional way (for instance, see [Ciarlet, P.G. (1978)]). Upon assembling the element stiffness matrices by the nodes in the A;-th layer, we obtain a stiffness matrix of one layer, which is denoted by

that is,

/„^-vw"("K-r^)(r> where zk-i = Bk-\v and zk = Bkv, and KQ, K0, and A are n x n matrices. It is known by the fundamental theory of the finite element method that (3.3) is a symmetric matrix, therefore AT and A are the transpose of each other, and KQ,KQ are symmetric. By the equation (1.1) and the similarity of the partition it is easy to see that the stiffness matrices for all layers are the same. Upon assembling the layer stiffness matrices by the nodes, we obtain an infinite by infinite total stiffness matrix. Noting the equation and the boundary condition, we get an algebraic system of equations of infinite order. Let K = KQ + KQ, then it is -Ay0 + Kyi - ATy2 = 0, -Ayk-i

+ Kyk - ATyk+x

= 0,

Infinite Element

121

Method

which takes the form of a block tri-diagonal Toeplitz matrix. If the boundary condition is the Neumann boundary condition (1.3), there is another equation for yo, K0yo - ATVl

= /o,

(3.5)

where /o is the "load vecter", which can be evaluated from the boundary value g by conventional finite element method. Now our aim is giving the algorithm of the combined stiffness matrix on ft, thus we ignore the difference between these two boundary conditions at this occasion. In order to reduce the algebraic system of equations of infinite order, we need a careful study of the properties of the solutions. Lemma 44. Ifu £ S(fl), yk = BkU, then there is a constant C independent of u, such that .

oo

.

2

oo

2

/ \Vu\ dx
-J2\yk-yk-i\ < fc=l

^

n

(3.6)

fc=0

Proof. Let ej C fl be an arbitrary triangular element. Let the length of the side pointing to the origin O be L, and the values of u at the both ends of it be yk and Uk_v Let measej denote the area of the triangle ej, and p the least height of e-j. Then

ly^-Vk-il L

^ , „ i ^ \yk\ + \yk-i\

< |Vu| <

Therefore it holds that iJ

L2

-\v{?-VkU2
|V«|2dx< = £ ^ 1 + ^ 1 ) 2 . Jet

P

By similarity me^ei a n d me^S63' depend only on the position of ej in fifc and are independent of k. Summing up the above inequalities with respect to all triangles ej we obtain (3.6). • Lemma 45. KQ and K'0 are positive definite matrices. Proof. We take an arbitrary yo ^ 0 and set y\ = 0. The interpolating function u of them is not a constant, hence a(u,u) = /

in,

|VM| 2 dx > 0.

122

Numerical Methods for Exterior

Problems

On the other hand a(u,u)

-AT K'n

K0 -A

( « )

T

Vo T

Thus VoKoyo > 0,

Vy0 ^ 0,

so Ko is a positive definite matrix. The argument for K'Q is similar.

•

By Lemma 43 the problem (3.2) admits a unique solution Uh for any yo G K™ such that B$Uh = 2/o- Let yk = BkUh, then 2/1 is determined by yo uniquely. It is easy to verify that the mapping 2/0 —• 2/1 is linear. Therefore there is a real matrix X such that 2/1 = Xy 0 Let tu(:r) = Ufe(£fc:r), then w G S,(fi), Bow = 2/fc and w satisfies the problem (3.2), hence 2/fc+i =

Xyk.

We obtain by induction that yk = Xky0,

k = l,2,---.

(3.7)

The matrix X is the transfer matrix for the problem. We will make use of the complex function solutions to the Laplace equation in the remaining part of this section. The above argument can be applied for this case without any modification except we regard that 2/i, 2/2, • • •, yk, • • • are vectors over the field of complex numbers. We observe that the transfer matrix X is real as we have proved. Substituting (3.7) into (3.4) we obtain {-A + KX-

ATX2)y0

= 0.

2/o is an arbitrary vector, hence X satisfies the equation ATX2 -KX

+ A = 0.

(3.8)

The solutions to (3.8) are certainly not unique. We need the following lemmas to identify a unique solution. L e m m a 46. 7/A is an eigenvalue of the transfer matrix X and |A| > 1, then A = 1.

Infinite Element Method

123

Proof. Let g be t h e eigenvector associated with A. Taking yo = g as t h e b o u n d a r y value of problem (3.2), we get j / i , 2/2> •• •, 2/fc, •• •, a n d by (3.7) yk = Xkg = Xkg. Hence oo

£ \yk - yk^

oo

oo

= £ \Xk - \*-*\*\g\* = £ |A|2fc-2|A - 1|2|5|2.

fc=i fe=i fc=i By L e m m a 44, t h e series on the right h a n d side converges. We have \g\ ^ 0, hence A = 1. • L e m m a 4 7 . The elementary divisor corresponding of the transfer matrix X is linear.

to an eigenvalue

X—1

Proof. Suppose it were not t h e case, then there would be nonvanishing vectors g a n d go such t h a t (X-

7)3o=0,

(X-

I)g = g0,

where 7 is a unit matrix, i.e. Xg = go + gBy induction we would get Xkg

= kg0 + g.

Taking yo = 9 as the b o u n d a r y value of the problem (3.2), we obtain j / i , 2/2,--•, 2/fc,..-, then by Vk = kg0 + g we would get oo

oo

2

X ] 12/fc - 2/fc-i I = X ] l^ol2 = +°°> fe=l fc=l which contradicts u € 771'*(J7) according t o L e m m a 44. L e m m a 4 8 . There is an eigenvalue X = 1 of the transfer the associated eigenvector is g\ = ( 1 , 1 , . . . , 1)T.

• matrix X , and

124

Numerical Methods for Exterior

Problems

Proof. We take yo =
Xgi=gi.

n

Lemma 49. As k —+ oo, Xkyo —> j/oo for any complex vector yo, where 2/oo = agi, and a is a constant. Proof. On the basis of Lemma 46 and Lemma 47, we conclude that the k limit of X exists as k —> oo. Denote by J/QQ the limit of Xkyo, which is (j/oo , J/oo , • • •, I/oo1 ) T in component form. It will suffice to prove yio' — yio = • • • = y£ . If it were not true, we might assume that y^ / J/oo , which correspond to the i-th. node and the (i + l)-th node respectively. There must be one element ej in fifc such that the connecting line segment of the above two nodes is one side of ej. Let the length of this side be L, then we have .„

/

,

mease^. a\

fi+ii.o

| V ul9| 2 c f c > — ^ y W - y J T 1 ^ 2 .

J en

By similarity me^ei [s independent of k. Summing up the above inequality with respect to k, we obtain / fc=i

The general term of the series on the right hand side would take \y,O(0 O ylo | 2 7^ 0 as its limit as k —» oo, hence 2

/ |Vu| dx = oo, in JQ

which leads to a contradiction.

•

Lemma 50. There is a unique eigenvalue X = 1 of the transfer matrix X, and the absolute values of all other eigenvalues are less than 1. Proof. If it were not true, then by Lemma 46, Lemma 47 and Lemma 48, there would be another eigenvector g' associated with the eigenvalue A = 1. According to Lemma 49 we have Xkg' —> ag\ (k —> oo), but Xkg' — g', hence g' = ag\, which contradicts the fact that g\ and g' are linearly independent. •

Infinite Element

125

Method

Let

(3 9)

* = l'/"o")- * - ( ? ? ' ) •

'

where I is the unit matrix. The equation (3.4) can be written as Rl(

y* )

= R

KVk-\J

Jy^).

(3.10)

\ Vk

Let A, g be a couple of eigenvalue and eigenvector, then Xg = Xg.

(3.11)

Taking k = 1 and setting y0 = g in (3.10), and noting (3.7),(3.11) we get

Therefore A and I 1 are the generalized eigenvalue and generalized eigenV 9 J vector of the matrices bundle R\ — XR2. To find matrix X, we first solve the above generalized eigenvalue problem to get all 2n eigenvalues and eigenvectors, then pick out the n eigenvalues and eigenvectors of X. We will always denote by det the determinant of a matrix. Now we evaluate det(i?i - \R2) = det .

'K-\AT

-A

This matrix is split into blocks. We add the second column by the product of the first column and A, then obtain , ,n ^r,•, , (K-XAT det(Ri - XR2) = det (

-A +

= ( - l ) " d e t ( - A + XK-

XK-\2AT\ j X2AT).

K is symmetric, so this expression is a symmetric polynomial with the independent variable A. Therefore if A ^ 0 is an eigenvalue, then so is 1/A. Therefore we only need to pick out the eigenvalues, which satisfy |A| < 1, of the matrices bundle Ri — XR2, then supply a A = 1. Let the eigenvalues be Ai, A2, • • • , An, and the corresponding eigenvectors be g\, g2, • • • ,gn, and we set T = (gi,g2, • • • , gn), A = diag(Ai, A2, • • • , A n ), where diag means a diagonal matrix or a block diagonal matrix. By (3.11) we have

XT = TA,

126

Numerical Methods for Exterior

Problems

that is X = TAT'1.

(3.13)

(3.13) is the formula for evaluating the transfer matrix X. Once X is obtained, then (3.7) gives the solution on the entire domain ft. If A is invertible, then the above generalized eigenvalue problem can be reduced to a conventional eigenvalue problem. Multiplying R^1 on the left of (3.12), we get

( A A

rr- T x ;)-e; which is a conventional eigenvalue problem. Eigenvalue A may be a complex number, thus the above calculation may encounter complex numbers. To prevent from this inconvenience, we can vary (3.13) slightly. Eigenvalues and eigenvectors must appear in conjugate pairs. Let A = a ± i(3, g = p ± iq be a pair of eigenvalues and eigenvectors, then by (3.11) we have X(p ± iq) = (a ± i(3)(p ± iq). Separating the real and imaginary parts we obtain Xp = ap - (3q, Xq = /3p + aq, that is

*<»>«)-(")(",!! We substitute the corresponding two columns in matrices T and A, and let Ti

=

(•••,?,

*.-""(••••(-"„£).•••)• It is easy to see X = TiAiTf 1 . After substituting with respect to all complex eigenvalues, the calculation of the transfer matrix X becomes real.

Infinite Element Method

3.1.3

Further discussion

127

for the transfer

matrix

We need to prove that the above conditions for X are not only necessary but sufficient. The argument is the following. Let
+ C2(f2 H

h Ciipi

oftp'iS, then So(iY) C So(£l). Each function in So(fi) has a bounded support. Theorem 49. So(Q) is dense in SQ(£1). Proof. We take an arbitrary u £ S0 (fi). Since S0(il) C Hi\'*(n), there exists MI G C Q ° ( Q ) , such that ||M — MI||I,* < £ for any £ > 0. We consider the following problem: Find u2 G So(fi), s u c n that a(u2,v) = a(ui,v),

Vv G S0(il).

(3.14)

We can prove that the problem (3.14) admits a unique solution as we did for the conclusion of Lemma 43. By the positive definiteness of the following quadratic functional we obtain -a(u2,u2)-a(ui,U2)=

min

\-a(v,v)

-

a(ui,v)\,

hence -a(u2,u2)

-a{ui,u2)

< -a(u,u)

-a(ui,u).

By adding |a(ui,Mi) to the both sides of the above inequality we obtain a(u2 — Mi,t*2 — Mi) < a(u — MI,M — Mi), thus a(u2

— Mi, u2 — Mi) < £.

Applying the inequality (1.19) we get ||"2 - W l | | l , * < 5£.

128

Numerical Methods for Exterior

Problems

But «i € Co°(f2), so there exists a natural number fco such that u\ = 0 on the domain £fc°fi. The equation (3.14) imply that VvG5o(^on),

a(u2,u)=0,

where we accept that every function in So(£k°Q) is extended to zero on ft, \ £fc°fi. According to the conclusion of Lemma 49, u 2 tends to a constant a as |a;| —> oo, therefore u 2 is a bounded function. We construct truncated functions «2,fc 6 So(Q), fc = 1,2,... such that _ J u 2 , for x e fi \ £fcfi, ' \ 0 , for re G £ fc+1 n.

U2 fc_

Let yfc = BkU2, then by Lemma 44 we have [ \Vu2M\2dx<

Jn

f

Jn

\Vu2\2dx+hyk\2.

°

Therefore {u2,k} is bounded in So(fi). Taking any v e So(^) and making use of the Schwarz inequality we obtain / Vv • V(u 2 Jn

f < ( / \Jnk+1

u2,k)dx

Vt; V(u 2 - u2,k)dx + \Wv\2dx) J

f

Vv Vw

2

cfa;

([ |V(u2-w2,fc)|2^] \JQk+1 J

+

/ -V+ifi

Wv-Vu2dx

Because both u2 and u2]fe are bounded functions, JQ |V(u 2 ~ w2,fc)|2 dx are uniformly bounded with respect to fc. And on account of v, u2 6 So(Q,), /

\Vv\2dx->0,

/

Vv-Vu 2 ote

as fc —> oo, therefore /

Wv • V(u 2 - u2,k) dx —y 0

as fc —* oo, which means u2,fc weakly converges to u2. By the Mazur Theorem [Yosida, K. (1974)], there is a linear combination, U 3 = Ci 1*2,1 + C 2 U 2 ,2 "I

1"
Infinite Element

129

Method

such that ||u 3 - it 2 ||i,* < £• Thus ||u - U3II1,* < 7e. But u 3 £ S0(Q), which leads to the conclusion. • Now we continue to discuss the Laplace equation. We give another formulation for the infinite element method: Find u G S(Cl), such that u h\an = 9i &nd a(uh,
i = l,2, •••.

(3.15)

Theorem 50. The formulation (3.15) is equivalent to (3.2). Proof. It will suffice to prove that (3.15) implies (3.2). Let Uh be the solution to (3.15), then Uh satisfies a(uh,v) = 0,

Vv € 5o(fi),

which yields (3.2 by noting Theorem 49.

•

Expressing (3.15) in terms of matrices, we get the system of equations (3.4), and the equation (3.8) for the transfer matrix X. We should bear in mind that the solutions to (3.8) are not unique. Theorem 51. The necessary and sufficient conditions to determine the transfer matrix X are (a) X satisfies (3.8); (b) X admits an eigenvalue A = 1 which corresponds to a linear elementary divisor and the eigenvector g\; (c) The absolute values of all other eigenvalues of X are less than 1. Proof. The lemmas from Lemma 46 to Lemma 50 and the equation (3.8) show that those conditions are necessary. It remains to prove they are sufficient. If a matrix X satisfies (a),(b),(c), let us prove it is the transfer matrix. We take an arbitrary g £ W1, and set yk = Xkg, k = 0 , 1 , . . . , then yk satisfy (3.4). Let Uh be the interpolating function. If we can prove Uh 6 S(Q), then owing to Theorem 50, Uh is the solution to the problem (3.2), which means X is the desired one. By (b) and (c), the limit of Xkyo exists as k —> 00. Let it be y^. We have Xyoo = X lim Xky0 k—>oo

= lim Xk+1y0

= yx,

k—>oo

therefore y ^ = agi. Let y'k = yk — y^, then the interpolating function of y'k is just u' = Uh — a. We only need to show u' £ S(fl).

130

Numerical Methods for Exterior

Problems

We express X in the Jordan canonical form, X = TJT~\

(3.16)

where 1 where J\ consists of those Jordan blocks, the absolute values of the eigenvalues corresponding to which are less than 1. By the definition of yk, Xy'k_x = X(yk-i

- t/oo) = Vk - ?/oo = y'k-

(3-17)

Let zk = T~ly'k, then (3.16) and (3.17) yield zk = Jzk-i,

k = 1,2,-•• .

By induction we obtain Zk = J

Z0,

that is

Zfc=

0./x fc ) ZO '

(3 18)

'

Because y'k —> 0 as k —> oo, we have Zk —> 0. (3.18) implies that the first component of ZQ is zero, otherwise the limit of zk would not be zero. Thus

Let || • || be the spectral norm of a matrix. Denote by p\ = p(J\) the spectral radius of the matrix J\, and p the highest order among the Jordan blocks of J\ whose spectral radius are equal to p\, then [ Varga, R.S. (1962)]

Pi fc U
(3.19)

where k > p — 1, c(p) is a constant depending only on p, and Ck~ combinational number. We have oo

oo

2

oo

fc=p—1

2

2

p+1

2

E I^I ^ E II^II M < E c (p){crvr } M 2 . k=p— 1

2

is the

fc=p—1

Infinite Element

131

Method

It is known that p\ < 1, hence the series on the right hand side converges. Therefore oo

oo

oo

fc=0 fc=0 fc=0

converges too. On account of Lemma 44 /

Vu'| 2 dx < +oo. n

1, Applying Lemma 7 we have u' G i?n,* *(fi).

D

We remark that using the Jordan canonical form (3.16) and (3.7) we can get the limit of the solution at the infinity, yoo=Xooy0-T('10)r-1.

3.1.4

Combined

stiffness

matrix

The combined stiffness matrix is the Dirichlet to Neumann operator in discrete form. Substituting (3.7) into (3.5) we obtain (K0 - ATX)y0 T

We define Kz = KQ — A X. role of this matrix.

= /o.

The following theorem indicates the important

Theorem 52. Let Uh be the solution to the problem (3.2), then for an arbitrary v £ S(Q) it holds that a(uh,v)

= ZoKzy0,

yo = B0uh,z0

= B0v.

(3.20)

Proof. We decompose v = v\ + v2 in S(D,), where BoVi = BQV = z 0 , BkV\ = 0, k = 1,2, • • •, and B0V2 = 0, BkV2 = B^v, k = 1,2, • • •. Then V2 S So(n), and by (3.2) we get a(u^,v 2 ) = 0. By (3.3) we have a(uh,v) =a(uh,vi)

= /

Vuh • Vvi dx = (j/
Since the stiffness matrix is symmetric, we have

*»•»>- w")(-VO(i)which gives (3.20).

D

132

Numerical Methods for Exterior

Problems

Therefore Kz is defined as a combined stiffness matrix. We should note that the combined stiffness matrix is different from the total stiffness matrix. At this circumstance, the total stiffness matrix is an infinite by infinite matrix, by which we can get the strain energy arising from any nodal values, but the combined stiffness matrix is only a n x n matrix, by which we can only get the strain energy when the nodal values yo, y\, • • • satisfy the equations (3.4). The procedure from the the total stiffness matrix to the combined stiffness matrix can be viewed as a procedure of elimination. By this procedure, y\, 1/2, • • • are all eliminated. Lemma 51. The eigenvectors of the transfer matrix X associated with the eigenvalue A = 1 are necessarily the null eigenvectors of the combined stiffness matrix Kz. Proof. Let g be such an eigenvector. Since it corresponds to a constant solution Uh, we have

Here I fore I

,

I is a symmetric and semi-positive definite matrix, there-

I is its null eigenvector, that is

-A

K'Q )

\g)

We obtain from the first row that (K0 - AT)g = 0, then by substituting g = Xg into it we get Kzg = (K0 - ATX)g = 0. Theorem 53. Kz is a symmetric and semi-positive definite matrix. Proof. a(-,-) is symmetric, hence by (3.20) we have ZQKzy0 = y%Kzz0 provided v is a solution to the problem (3.2). yo and zo are arbitrary, so Kz is symmetric. Because a(u,u) > 0 for any u € S(Q), Kz is semi-positive definite. D

Infinite Element

133

Method

If the domain ftc is not in star-shape, or the curve To is in complicated shape, we can decompose the exterior domain ft to an unbounded domain with regular shape and a bounded domain with complicated shape, as we have stated previously. Then we can find the combined stiffness matrix for the unbounded domain, which can be regarded as a single element, owing to the conclusion of Theorem 52. Upon assembling the combined stiffness matrix with the stiffness matrices of the other elements, we get an algebraic system of equations. Then the problem can be solved.

3.2

General elements

The pattern of partition in Section 3.1 can be generalized. On the one hand the elements are not restricted to be linear triangular elements, on the other hand each layer may not be divided by rays. We should notice that we have not made any other assumption to the stiffness matrices in the preceding argument except that they are the same for all layers. Therefore the previous argument is applicable for more general partitions. The triangular elements can be of higher order, and quadrilateral elements can also be applied. There are only two restrictions, namely, one to one correspondence of the nodes between Tk-i and Tfc and the geometry of the layer mesh is similar to each other. It should be noticed that there may be some nodes which do not belong to Tfc or Tfc_i. It is needed to eliminate those nodal values from the systems and obtain the combined stiffness matrix for one layer. Let yfe_ j. be a column vector consisting of the interior nodal values, then we get the matrix expression for the general cases: oo (Kn Ki2Ki3\ a(u,v) = £ (Vk-i Vk-i Vk) K2i K22 K23 fc=1

2

\K31K32K33J

f

z

k-i\ zk_,_ = 0,

(3.21)

\ zk )

in terms of the total stiffness matrix, v is arbitrary in So (ft). Let it vanish on all sub-domains except one ftfc for a fixed k. Then

(yl-ivLxvl)

I K22 \zk_h = 0.

134

Numerical Methods for Exterior

Besides, zk_i

is arbitrary. Therefore J/fc-1^12 + yl-1^22

yk_i

Problems

+ VkKz2 = 0.

can be solved explicitly: 2/fc-i = -K^iK^yk-i

+

Kj2yk).

Upon substituting it into (3.21), and rearranging the terms, we can get

«->=£(*.*) (_*r£)(t')This expression is the same as that of the linear triangular elements, except the stiffness matrices may be different. Then all above algorithm can be applied here without any change. .

3.3

Harmonic equation-three dimensional problems

The discussion for the Laplace equation in three space dimension is similar to that for the two dimensional case. Let To be a closed convex surface in a three dimensional space, it is stitched up with space quadrilaterals, and the point O is in the interior of I V Taking £ > 1, we draw the similar surfaces to To with center O and the constants of proportionality £, £ 2 , • • • , £fc, • • •, which are denoted by Ti, I V • • • , Tk, • • • respectively. The domain between Tfc_i and Tk may be divided into many sorts of elements, for instance eight nodes hexahedron isoparametric elements. Let

(K0-AT\ \-A K0 j be the stiffness matrix of the layer between To and T\, where the degree of freedom with respect to yi has already been eliminated, as we have done in Section 3.2. Then the stiffness matrices of the other layers are

« " ( - ! " « ) •

* - ' • » • • • • •

Let K = £,*Ko + £~?Kb, then the equation corresponding to (3.4) is -Ayk-i

+ ^Kyk

- £ATyk+i

= 0.

Infinite Element Method

135

Hence the transfer matrix X satisfies the equation

£ATX2-^KX

+ A = 0.

If we set £ 5 X = Y, then Y satisfies the equation ATY2 - KY + A = 0, the form of which is the same as (3.8). The solutions are not unique, so it is necessary to study the properties of X. Lemma 52. IfuG S(Cl), yk = BkU, then there is a constant C independent of u, such that 1

„

oo

^n

fc=i

oo

fc=o

The proof of it is analogous to that of Lemma 44, thus omitted. Being analogous to two dimensional cases, KQ and K'0 are symmetric and positive definite matrices. Moreover we have Lemma 53. Each eigenvalue A of the transfer matrix X satisfies |A| < £~ i . Proof. Let g be the eigenvector associated with A. We take yo = 9 as the boundary value of the problem (3.2) (three dimensional case), and get 2/1,2/2, • • •, Vk, • • •, then yk = Xkg. Hence oo

oo

fc=l fc=l By Lemma 52 the above series converges. If A ^ 1, then £fc|A|2fc < 1, t h a t is, |A| < £~ 2. If A = 1, then y0 = y1 = •••= yk = •••=: g. We have

Jnk

^4d >ce, \x X

thus lull,* = oo, which is unsuited to present needs. Therefore A = 1 is not an eigenvalue. • By this result all eigenvalues A of the matrix Y satisfy | A| < 1. We define matrices R\, i?2 by (3.9), then we can evaluate the general eigenvalues and eigenvectors of the matrices bundle R\ — \Ri- Finally we get the matrix Y by the approach in Section 3.1.

Numerical Methods for Exterior

136

Problems

Theorem 54. The necessary and sufficient conditions to determine the transfer matrix X are (a) X satisfies the equation ZATX2-^KX

+ A = 0;

(b) The absolute values of all eigenvalues of X are less than £ - 5 .

3.4

Inhomogeneous equations

We study the infinite element method for the Poisson equation - A u = /,

x £fl,

(3.22)

in two space dimension in this section. We assume that / is compactly supported, otherwise the computing procedure can not be terminated in finite steps. The variational formulation for the Dirichlet problem is: For given / G L2{Q) and g G ^^(dfl) find u G F x '*(fi) such that u\an = g, and a(u, v) = (/, v),

Vw G Ho1*(CI).

(3.23)

For simplicity we still use the linear triangular elements and the same partition as that in Section 3.1. The infinite element formulation is: find Uh G S(il) such that Uh\dQ =
= (f,v),

(3-24)

VUGSO(Q).

Being analogous to (3.3) there are vectors fk,gk such that /

Vuh • Vw dx — / /

/ • vdx TX

/

/

N

(3-25)

Let {y£} be a solution of the infinite element scheme to the equation, but it does not necessary satisfy the boundary conditions. Let jjk = Vk — J/)t, then yk+i = Xjjk- We have 2/o =yo + Vo, 2/1=^1 + 2/1= Xy0 + 2/i = X(y0 - y*) + y* = Xy0 + qlt

Infinite Element

Method

Vk = Xyk-i

+qk,

137

Let W

«= E

\\ I \^h\2dx-

2 /iff =fc+ii 1 ^

f f-uhdx\

• / n«

J

= \ylKzyk + ylhk. (3.26) 2

On the other hand

(3.27)

+ (»* ^M-i) ( ^ J

+ Ivk+iK.yk+i + yl+ihk+ij

We take the partial derivatives with respect to yk+i, then let it be zero to obtain (K'0 + Kz)yk+i

- Ayk + gk+i + hk+i = 0.

We solve yk+i from this equation then substitute it into (3.27) and compare (3.27) with (3.26). It is deduced that hk = AT(K'0 + Kzr\gk+i

+ hk+i) + fk+i,

(3.28)

and we have qk = -(K^ + Kzy1(gk

+ hk).

(3.29)

Thus the procedure to solve inhomogeneous equations is: (1) evaluate hk from the recurrence formula (3.28); (2) evaluate qk from the formula (3.29); (3) then get yk. The recurrence is backward in the first step. Since / is compactly supported, hk+x = 0 for k large enough, then it is the initial point.

Numerical Methods for Exterior

138

3.5

Problems

Plane elasticity

We consider plane stress or plane strain here. The Lame equation for the displacement u = (wi,W2)T is -fiAu-(X

+ /i)grad divu = / ,

x G

fi.

(3.30)

The only difference for these two cases lies in the constants. We consider the Dirichlet boundary condition, u = 0,

xedfl,

(3.31)

as an example, and the approach for the Neumann problem is similar. Inhomogeneous problems can be dealt with applying the approach in Section 3.4, so we consider homogeneous problems only in this section. We will work in the spaces V = (.ff1,*(fi)) , and Vo = {u G V;U\QQ = 0}, and the corresponding bilinear form is

f

2

a(u,v) = i 2_, crij(u)€ij{u)

dx 2

• /

^ A div u div v + 2/u Y J eij(u)eij(v) > dx,

/n Ja

i,j = l

where the strains are

and the stresses are <Tij(u) = o-ji{u) = A Y^e^(u)

J Sij + 2fj,eij(u).

The variational formulation of the problem (3.30),(3.31) is: For a given g G Hll2(d£l), find u G V, such that U\QQ = g, and a(u,v)=0,

VVGV0.

(3.32)

For simplicity we consider linear triangular elements here. The infinite element formulation is: Find u G (S(Q))2, such that U\QQ = gi, and a(u, v) = 0,

Vv G (S0(fl))2.

(3.33)

Infinite Element

Method

139

Let the number of nodes on Tfc be n. According to the anticlockwise direction the displacements on the nodes are u\k', u2k,u\k\u2k>''' >uifc >u2fc successively, and they form a (2n)- dimensional column vector y^. Now we can evaluate the stiffness matrix and get

a(u,v) = g (l/ZLi Vl) (K°A ~£, ) ( ^ the form of which is the same as that for the Laplace equation. In parallel with the propositions in Section 3.1, we establish the following lemmas and theorems. The proof of some propositions is the same as the previous one, thus omitted. Lemma 54. KQ and K0 are positive definite matrices. Proof. We take j / o ^ 0 and yk = 0, k = 1, 2, • • •, then the corresponding interpolating function is not a rigid body motion, hence we get a(u, u) > 0. It follows that KQ is a positive definite matrix. The proof for K0 is the same. • Lemma 55. / / A is an eigenvalue of the transfer matrix X and |A| > 1, then A = 1. Lemma 56. The elementary divisor associated with an eigenvalue A = 1 of the transfer matrix X is linear. Lemma 57. As k —> oo, Xkyo —> y ^ for any complex vector yo, where 2/oo = agi + /3g2, where g\ = (1,0,1,0, •• • , 1,0) T , g2 = (0,1,0,1, • • • , 0 , 1 ) T , and a, (3 are constants. Lemma 58. If an eigenvalue A of the transfer matrix X is equal to 1, then the associated eigenvector of it is a linear combination of g\ and g2 • Theorem 55. The necessary and sufficient conditions to determine the transfer matrix X are (a) X satisfies the equation, ATX2 -KX

+ A = 0;

(b) There are two eigenvalues A = 1 of X, the associated elementary divisors are linear, and the corresponding eigenvectors are g\, g2; (c) The absolute values of all other eigenvalues of X are less than 1.

140

Numerical Methods for Exterior

Problems

For three dimensional elastic problems the approach in Section 3.3 can be applied. The algorithm is similar.

3.6

Bi-harmonic equations

According to Section 1.6 the bi-harmonic equations are A2u = /,

(3.34)

and the corresponding bilinear form is a(u,v) = / DAu-

Jn

Avdx

fn _ - l f — — JQ \dx\dx\

d2^ d2v dx\dx2dxidx2

d2ud2v\ dx\dx\) (3.35)

We consider homogeneous equation here. For inhomogeneous the algorithm in Section 3.4 can be applied here without change. Then variational formulation of the Dirichlet problem G H3/2(dty, and g2 € H^2(dn), find u G # 2 ' * ( 0 ) , gi u\,dn = gi,$\aa =92, and a{u,v) = 0,

equations significant is: Given such that

VvGffo'*(°)-

( 3 - 36 )

For definiteness we use the Morley triangular elements here [Ciarlet, P.G. (1978)]. The values of u are given at the vertices and the normal derivatives of u are given at the middle points. Quadratic functions are applied as the interpolation functions and six coefficients are determined uniquely by the six nodal values. Here it should be noticed that at each middle point we must fix one normal direction and all layers must be similar, including the direction. Let the six nodes of an element e* be -X.- , j = 1,2,3, and xff,

1 < j s be S i , and let the

set of all Xjl s be £2- Then the infinite element space is denned as S(Q) = { « £ P2(ei)>Vi; u continuous on S i , Vu continuous on S2, 2 , Iwl2,ei < °°}-

Infinite Element

So(fi) = {uG S{tt);u(x)

141

Method

= 0,Va; £ To n S i , ^ ^ - = 0,Va; e T 0 D E 2 } .

The corresponding bilinear form is a,h(u,v) =2_\ /

7n

vAu-Avdx

\<9a;2 5a;2

dx\dx2dx\dx2

L e m m a 59. Let ||u||s(0) = (J2i Mi.eJ 2 > then ^

dx\dx\)

J5 a norm

over

dx.

SoCty-

Proof. If ||u||s(n) = 0, then u £ Pi(ej) for all i. By the continuity of u and Vu at nodes, u e Pi(fi). Then by the zero boundary condition on To, u = 0. D Let g\j and g2J be the approximation of g\ and <72- The infinite element formulation is: Find UH € S(Q), such that u(a;) = 31/(2;), Vx £ To n S i , ^i=52(x),Va;eronS2,and ah(u,v)=0,

VVGS'O(O).

(3.37)

Theorem 56. The problem (3.37) admits a unique solution. Proof. It is routine to transfer the problem into a problem with homogeneous boundary condition. Applying the Lax-Milgram theorem we need to verify that a/»(-, •) is coercive. Clearly a,h(u,u) is equivalent to ||u||s(n)) thus the conclusion follows. • We turn now to the implementation. We consider the fc-th layer between Tk-i and Tfc. If the number of triangular elements is N, then there are N nodes on Tk-i (or Tk) and inside the A;-th layer there are other iV nodes. Let yk-i (or yk) be a vector with components of A or | ^ at the nodes on Tfc-i (or Tfc), and J/J._I be a vector with components of | ^ at the interior nodes. Notice that we use the value of nr then by (3.35) the total stiffness matrices of all layers are just the same. Upon introducing some matrices, we have the following, which is in the same form as (3.21), 00

ah(u,v)

= ^ fc

=i

(Kn

( j / L i 2/Li Vl)

Ki2

K13\

K21 K22 K2z

\K31 K32K33)

(

z

*>-i\

z fc _i

= 0. (3.38)

\ zk J

Then following the same lines as those of Section 3.2, y ^ - i nated. Then there is a transfer matrix X.

can

be elimi-

142

Numerical Methods for Exterior

Problems

Lemma 60. All eigenvalues of X which are not equal to 1 have a magnitude less than 1. When A = 1, A is a double eigenvalue and its eigenvectors correspond to u = x and u = y. Proof. Let A be an eigenvalue of X with an associated eigenvector g. suppose that |A| > 1, then yk = Xkyo- 2/fc-i a r e then derived, which depend linearly on yk and yk-i- If . /KuK12K13\

Wl = (yl y\ y[)

(yo\

K21 K22 K2Z U i U 0, V ^ i K$2 K33J \ 2/1 /

then ah(uh,uh)

= ^Afc

1

Wi=oo,

fc=i

which is impossible. If W\ — 0, then ||M/I||S(Q) = 0, which implies that Uh = x, or Uh = y, 01 Uh = 1, or their linear combination. It can be seen that Uh = 1 corresponds to A = j < 1 by noting that the ratio of |x| between Tk and Tk-i is £ and A is the variable defined on nodes. If Uh is not the linear combination of x, y, then the magnitude of A has to be less than 1. This completes the proof. D

3.7

Stokes equation

We study two dimensional problems first. As before we consider homogeneous equations only. The system of equations is: - A u + Vp = 0,

(3.39)

V • u = 0,

(3.40)

with the boundary condition, u = g,

xedfl.

The corresponding bilinear forms are: a(u,v) = / Vw : Vvdx,

b(u,p) = (p, VM).

(3.41)

Infinite Element

143

Method

The corresponding variational problem is: For a given g G (tf 1/2 (ft)) ,find u G (H1,*{$!)') and p G L2(fl), such that U\SQ = g, and a(u,v)-b(v,p)=0,

6(u,g) = 0,

toe

(FQ1'*^))2,

VgGL 2 (fi).

(3.42)

(3.43)

The problem is essentially equivalent to (1.64) and (1.67) with a slightly difference, where the equation is inhomogeneous and the boundary condition is homogeneous. In Section 1.7 it has been explained how a problem with inhomogeneous boundary condition can be reduced to a problem with homogeneous boundary condition. Therefore the problem (3.42),(3.43) admits a unique solution. We consider the infinite element method for the problem (3.42),(3.43). We use P2 — Po elements as an example, where u is approximated by piecewise quadratic polynomials and p is approximated by piecewise constants. The stability and convergence proof of this kind of elements for interior problems can be found in [Girault, V. and Raviart, P.A. (1988)]. The infinite element spaces are S(Q) = {ue

(ff x '*(fi)) 2 ; u\ei G (P 2 (e0) 2 ,Vt} ,

M(Q) = {pe L2(n);p\ei

G Po(ei), V*} ,

5„(fl) = { « e S ( f i ) ; U | f f i = 0},

S(rk) =

{u\rk;ues(n)}.

Let gi G S(To) be the approximation of g. The infinite element formulation is: Find Uh G S(fl) and ph G M(Q), such that u/j|en = gi, and a(uh, v) - b(v,ph) = 0, b(uh,q) = 0,

Vv G So(fi),

VgGM(fi).

(3.44) (3.45)

To prove the well-posedness of (3.44),(3.45) we need to verify the inf-sup condition for this problem.

144

Numerical Methods for Exterior

Problems

Lemma 61. There exists a constant (3 > 0 independent ofh such that inf sup -^PL->p. peM(n) lP ^o„ eS . o(n):U ^o IMIi,*l|Pl|o

(3.46)

Proof. By Lemma 20, for any p G M(Cl), there exists u G (H0'*(Cl)) , such that V • u = p, and ||w||i,* < C||p||o with a constant C depending on fi. We apply the Clement interpolation operator Ph [Clement, P. (1975)] on u. P^u satisfies the following: Let e^ be an arbitrary element, and Ei be a "macro element" consisting of e, and all elements neighboring e^, then \U - Phu\ltei

+

ff_1||u

- Phu\\o>et

<

C\U\hEi,

where H is the longest length of diameters of elements in Ei. Let Xj, j = 1, 2,3 be the vertices of e, and Xji, 1 < j < I < 3, be the middle points of the edges Sji. One interpolation u —» w is defined (see [Girault, V. and Raviart, P.A. (1988)]) such that w £ S{Q) and w(Xj) =Phu(Xj),j = 1,2,3,

/ (u-w)ds = 0,1 < j < I < 3. Jsjl

Clearly / V • wdx = I i/ei

w • uds = /

J dei

u • v ds = / V • u ete.

Jdei

J ei

It is proved in [Girault, V. and Raviart, P.A. (1988)] that \PhU~w\1<ei

< C ( / l _ 1 | | u - P f t u | | o , e i + Iw-PfeWll.eJ,

where h is the diameter of e*. Then |« - w|i, ei < C(l + — )|w|l,£ i . The constant C depends on the shape of elements. However the number of elements in one layer is finite, and the elements are similar to each other on different layers, so there is a uniform constant C for all elements. Certainly C depends on the mesh. Being the same j - is bounded by a uniform constant. Hence |u;|i < C|w|i, and b{w,p)

IHIi,*lbllo which gives (3.46).

_

b(u,p)

IIHIi,*INIo

>

1_

c" H

Infinite Element

Method

145

By Theorem 39, we obtain that Theorem 57. The problem (3.44), (3-45) admits a unique solution. We consider the k-th layer flk- If the number of triangular elements is N, then there are N nodes on Tfc_i (or Tfc). Inside the A;-th layer there are other N nodes. The total number of nodes on one layer is 3iV. There are 2N degrees of freedom of the velocity components on Fk-i (or Tk), which are arranged as a 2N dimensional vector yk-i (or yk). There is an analogous 2N dimensional vector t/j._i associated with the interior nodes. The values of pressure in the k-th layer form a N dimensional vector, which is denoted by q. We impose an additional constraint for the implementation of (3.44),(3.45): [ gj-i>ds = 0. Jan

(3.47)

The reason is interpreted as follows. Let fio be a sub-domain consisting of a finite number of elements. We take q = 1 in f2o and q = 0 outside fioi then by (3.45) Jn V • Uhdx = 0, and then by the Green's formula Ian uh • v ds = 0. That is, the total flux on dflo vanishes and the nodal values are constrained by a linear relationship. Therefore the basis functions in V do not have local supports. In order to get a block tri-diagonal Toeplitz matrix like (3.4) the property of local supports of the basis functions is essential. That is the difficulty. On the other hand, if (3.47) is satisfied, then one can see that there is a vector 7 such that 7Tj/o = 0. We take a particular function q e M(fl) at the equation (3.45) as follows: q = 0 on £fcf2, and q = 1 on fi\£fcfi, then we get lTVk = 0.

(3.48)

We normalize 7 to a unit vector, then construct an orthogonal matrix Tr = (7G), and set yk = GTyk, then there is an one to one correspondence between yk and yk- There is no restriction between yk-i and yk, so it is easy to construct stiffness matrices on Qk and get a block tri-diagonal Toeplitz matrix with respect to ykGiven j/o and j/i, satisfying (3.48), we solve the finite element approximation of the Stokes equation on Q\ with boundary data yo,Vi and the given mesh. Let the approximate solution be u/,, then there are matrices

146

KO,KQ,

Numerical Methods for Exterior

Problems

and A, such that

where Zk = GTZk, and Zfc is associated with v S V. The above expression is valid for all layers Qfc, hence the infinite element scheme is deduced to the a system of infinite equations, which takes the same form as (3.4). Then the algorithm is the same as that for the Laplace equation. The transfer matrix X and the combined stiffness matrix Kz can be evaluated. If Jdn g • v ds = 0 then it is easy to get an approximation gi to satisfy Ion 9i • vds = 0. If Jdn g • v ds ^ 0, we can set Mi = u+ — V In |x| / 2TT Jen

g-vds,

then u\ satisfies the same equation and the condition (3.47). The link between the combined elements and other conventional elements is as follows: If the conventional elements occupy a domain Q*, then we can define a(u, v) = I

Vu : Vvdx +

y^Kzzo,

and b(u,p) = /

in*

pV • udx,

where yo = GTyo and ZQ = GTZQ, and To is the boundary of the domain Q. Define spaces S{W) = ( u e {H\n*))2;u\ei M ( S T ) =
€ (P 2 (ei)) 2 ,Vi},

G Po(e<),Vt, f

pdx = 0

Let the union of Q and fi* be Q<x,. The numerical computation for the problem, — A u + Vp = 0, V - u = 0,

x e Hoo,

i6fl(

Infinite Element

u = g,

147

Method

x £ dCloo,

is reduced to a conventional finite element scheme on fi*. The above algorithm can be extended to three dimensional problems and the axial symmetric flow. For the later the system of equations is 1 „,

„

.

1

dp

OC-t

0X~i

V(ziVui) + - j u i + ~

X\

-—V(iiVu2)

= 0,

+ ^ - = 0, OX2

Xi

d . . d , . - — (XiUl) + -z—{XiU2) = 0. 8x\ OX2

The corresponding bilinear forms are a(u, v) =

xi ( Vui • Vvi + V«2 • Vt/2 H

6(w,p) =

P[ g—(xiui)

j - 1
+ -g—{x1u2) J dx.

The corresponding Hilbert spaces are equipped with norms lbllo= I

xlP2(x)dx,

JA

IMI? = jf*i(|V« a (*)| a + ^ ) c f a : 1

For details, see [Ying, L. (1986)]. 3.8

Darwin model

Let us recall some definitions in Section 1.12. ff(curl,div;fi0)

= { « € (L 2 (ft 0 )) 2 ; V • «, V x « e L 2 (fi 0 )},

||«||0,curl,div = ( N l o + IIV • «||§ + || V X U|| 2 ) * ,

148

Numerical Methods for Exterior

Problems

H0c(n) = {UG V; ||C«||o,curi,div < oo, (1 - ()u e (fi"o'* ( n ) ) 2 . u

x

"Ian = 0},

ll«ll* = {||Vx u||§ + ||V • u\\20+ I f i } i V = {v £ # 0 c (ft); < « • i/, 1 > a n = 0} Let the closure of the quotient space V/VQ with respect to the norm || • ||„ be W. Let Q = {p £ L 2 (fi);suppp CC fi}, equipped with norm

blllt=(lbll§+I^P^|2)a. Then take closure to obtain a Hilbert space Q. If p £ Q and limpn = P,Pn £ Q, then define JQpdx = limjnpndx. The subspace of Q, {p £ Q; Jnpdx = 0} is denoted by Qo. The bilinear forms are d(u, v) = / (V x u) • (V x v) dx + / (V • w)(V • v) dx Jo, Jo. + aa< v-i/,1 >an,

u,v

eW,

and 6(u,p) = / (V • u)pdx,

u £ W,p £ Q 0 .

The problem: Find w £ W, p £ Qo, such that d(u, v) + b{v,P) = f 5 • (V xv)dx, &(«,
Vv £ W,

(3.50)

V?€Q0,

(3.51)

admits a unique solution. We consider the infinite element method for (3.50),(3.51). Being the same as the Stokes problem, Pi — PQ elements are applied for u and p. The mesh restricted on a bounded domain 0 \ £kQ is also a mesh with finite number of elements. In general let fio be a union of element, finite or infinite, we define E(n0) = {uG (C(Th))\u\et

£ (P 2 ( e i )) 2 ,V e i C fio}.

Let D{n\£kfL)

= lu£E(n\£kn);ux

v\ro = 0 , J u-vds

=0

Infinite Element

149

Method

and let u

ll.,n\«*n = ( / \Jn\£kn

\V-u\2dx+

J Jn\?>=n

Lemma 62. || • ||*,n\^fcn *s a norm on D(D. Proof.

\Vxu\2dx >

\£kfl).

If ||u||^tQ\^kQ = 0, then Au = 0, so u is a harmonic function.

u G P2 on individual elements, hence u G P 2 on Q \ £fcfi. We extend u to the interior of To analytically. Then we define the stream function tp, such that u = (Q£-, — gjf-) , then Aip = 0. By the boundary condition of u, -g^lr0 = 0 , so ip is a constant in the interior of To. Therefore u = 0.

•

We return now to the exterior domain Q. Lemma 63. The space Wh(Q) = {u G E(Q); v • u, V x u G L 2 (fi),u x ^|an = 0, < M • v, 1 >afi= 0} is a Hilbert space under the norm \\ • ||*. Proof, because VoC\E(n) = {0}, || • ||* is a norm on Wh(Q). Let us prove it is complete. Let {«„} be a Cauchy sequence with limit u. We are going to prove u G E(Q). {un} is also a Cauchy sequence on D(fl \ £kfl). By Lemma 62, u G D(fl \ £fcfi). Since A; is arbitrary, u G J5(fi). • Lemma 64. Wh{n) c ^ ' " ( f i ) . Proo/. Let u G Wh(fl). We define a function < G C°°(ft), such that £ = 0 near To, C = 1 n e a r the infinity, and 0 < C < 1- We take an arbitrary
Jn

Jn

(CM))

dx + [ (V x

K«|i =

sup

Jn

, ,

v w

< ||c«||„

therefore £u G # o ' * ( n ) > t h a t is, u S H 1 ' * ^ ^ ) for large k. Then by u G F 1 on any bounded domain we have u G i7 1, *(fi). • Lemma 65. The norm || • ||* is equivalent to || • ||i t , on Wh(Q). Proof. It is the direct consequence of the closed graph theorem and Lemma 64. •

150

Numerical Methods for Exterior

Problems

We assume an inhomogeneous boundary condition, u x V\QQ = g, and let gi be an approximation of g, then the formulation of the infinite element method is: Find Uh G E(Q.)nH1'*(Cl), ph G M(fi), such that uh x v\r-0 = gi, Jr Uh • uds = 0, and d(uh,v) + b(v,Ph)=

I B-(Vxv)dx,

V« e W h (n),

(3.52)

Jn b(uh, q) = 0,

Vq G M(fi).

(3.53)

Theorem 58. T7ie problem (3.52), (3.53) admits a unique solution. Proof. d(-, •) is coercive. To verify the inf-sup condition we apply the inf-sup condition of the infinite element method for the Stokes equation, sup ^ ^ v e So(fi) , v | 1 Now Wh(fi) D E(Q) n fl'o'*^). sup

»e^)

* ^ |w|1

> P\\p\\0,

Vp G M(Q).

thus

> /3||P||o,

V^ G M(Q).

Moreover we have ||t>||ij| < 2|u|i, thus sup v€Wh(il)

b IHI

-^Pl>±\\p\\0, * ^

VpGM(fi).

Then the conclusion follows, and the proof is the same as the that in Theorem 40. • The relationship between the solutions of the infinite element method to the Stokes equation and to the Darwin model is the following: Theorem 59. The solution to (3.52),(3.53) is a solution to (3.44),(3.45) with appropriate boundary value and inhomogeneous term. Proof. We take v in (3.52) such that v\r0 = 0, then v G H0'*(n). Let a series {vn} C Co°(fi) tend to v in HQ'*^), then by the Green's formula, d(uh,vn) = a(uh,vn). Letting n - » o o w e verify that Uk,Ph satisfy (3.44),(3.45). •

Infinite Element

151

Method

Now let us go to the algorithm for the Darwin model. For the formulation (3.52),(3.53), we take a boundary value 7/j for the Stokes equation such that 7/j x v = gi, and / r ^h-vds — 0. Then solve the following problem: Find Wh £ S(Q), rh € M(fl), such that Wh\r0 = 7h, a n d a(wh,v) + b(v,rh)=

[ B-{Vxv)dx,

6(«; h ,9) = 0,

V«eS0(fi),

VgeM(fi).

(3. 54) (3.55)

The solutions to (3.52),(3.53) also satisfy a(uh, v) + b(v, rh)=

/ 5 • (V x v) dx,

b{uh,q) = 0,

W e S'o(fi),

V?eM(fi).

(3.56)

(3.57)

Let Uh — Wh = Uh and Ph — fh — Ph, then we have a(Uh,v) + b(v,Ph) = 0,

b(Uh,q) = 0,

VveS0{n),

V9eM(Q).

(3.58)

(3.59)

Let the vectors yk = BkUh and z/t = GTyk, then it holds that Zfc+i = X^fc, which yields yk+1 = G I G r | / , .

(3.60)

Let the function form of (3.60) be Uh\vx = YUh\r0i where F is an operator. So far the problem (3.52),(3.53) is deduced to a finite element approximation on a bounded domain fti for the following system of equations: - A « = VxB, V • u = 0, u x

xefii, i e fii,

Hr 0 = 3 h ,

/ u • v ds = 0, •/r0 (u - w/i)|rx = Y(u - Wh)|r0,

152

Numerical Methods for Exterior

Problems

where Wh is given. We solve the finite element problem on fii with the same mesh, and get the solution Uh to (3.52),(3.53) on Sl\. Then we use the formula (3.60) and the solution Wh to (3.54),(3.55) to get the solution Uh on the entire domain Q. Finally we need to show that (3.52),(3.53) with homogeneous boundary condition is an approximation to (3.50),(3.51). If the solution to (3.50),(3.51) satisfies u G i7 1, *(fi), then we can modify the formulation to: Find u G # 1 , *(fi) nW,pG M(fi), such that d{u,v) + b(v,p)=

[ B-(V

Jo,

xv)dx,

b(u,q) = 0,

W€#1,*(fi)nW)

Vg G M(fi).

(3.61)

(3.62)

This is because the solution u satisfies V • u = 0, b(u,p) = 0 for all p G L 2 (fi)., Because H1'*^) n W is dense in W, uniqueness holds for the problem (3.61),(3.62). The spaces in (3.52),(3.53) are subspaces of those in (3.61),(3.62), so the solutions to (3.52),(3.53) are indeed approximate solutions.

3.9 3.9.1

Elliptic equations with variable coefficients A homogeneous

equation

Let us consider an equation, V • (fc(z)Vu) = 0

(3.63)

on the domain f2. Let A > 1 be a parameter. We make a similarity transformation Ax —> x. Under the transformation the equation (3.63) becomes V • (Jfc(Ax)Vu) = 0,

(3.64)

which is still considered on Q. The bilinear form associated with (3.64) is thus defined by a$u,v)

= / k(\x)VuVvdx.

(3.65)

We assume that (3.63) is elliptic, that is, there is a constant fco > 0 such that k(x) > k0 for all x G fi. In addition we assume that k(x) is continuous

Infinite Element

153

Method

and approaches to a constant as x —> oo. It has no harm in assuming Hindoo k(x) = 1. We still use linear triangular elements for the infinite element mesh. The total stiffness matrix on f2i is defined by

(K0-AT\ \-A K'Q ) ' that is, k(Xx)Vw • Vv dx = ( y% yj) (K0

/

\-A

JO.!

-A'\

K'Q

z0\

\zj>

where u,v £ S(Q), and the space S(Q) is defined in Section 3.1. The infinite formulation for the equation (3.64) is: for a given gi G S(To), find UH G S(Q), such that ax(uh, v) = 0,

Vv G So(fi).

(3.66)

The problem (3.66) admits a unique solution. Then a combined stiffness matrix KZ(X) is defined by y^Kz(X)zo = a\(v,h,v). Kz{$ is real and symmetric. The transfer matrix X(X) is defined by y\ = X(X)yo. Let the domain Q be regarded as a union of Q\ and £fi. The bilinear form a\ can also be written as ax(uh,v)

= (yl $X{XF)

(f° ^

)

(*)

+!#X(A)3X(AO*1

Since z\ is arbitrary, we have yl{-A

+ X{X)TK'0 + X(X)KZ(X£))

= 0.

yo is arbitrary too, so X(X) = (K^ +

Kz(X0)-1A.

On the other hand, being analogous to Theorem 52 it holds that KZ{X) =K0-

ATX{X).

Therefore KZ(X) = K0-

AT{K'Q + tf^AO)-1 A

(3.67)

154

Numerical Methods for Exterior

Problems

To solve an equation with variable coefficients like (3.64), it is convenient to consider the critical case £ = 1 + . We set y\ = j/o + (£ — 1)771, then

K0 -AT\ fy0 -A K'Q ) U K0 - A - AT + K'Q - ( £ - 1)(AT - K'Q) $y0 -{£-${A-K'0) (d-l)2K>0 ){m

L0 -#r\ fy0 -B L'0 ) W By (3.67) we get

KZ(X) = Kz(XO + L0 '

BT

- Kz(XO) (—^-,

+ Kz(XO)

(-^-

- Kz(XO

Dividing it by £ — 1, letting £ —> 1 and defining LQ

= lim - — - , £-i£-l'

B = lim - — - , C-U-l'

L'0 = lim ° ° C-U-l'

to obtain A

^A^

=

"L° +

(i?T

" K*(X))(L'o)~l(B ~ K*W)'

(3-68)

which is the ordinary differential equation for KZ(X). Letting A —> 00, the equation (3.64) approaches to the Laplace equation, Aw = 0 asymptotically. We denote by KZQ the combined stiffness matrix associated with the Laplace equation, which is the initial data of (3.68) at A = 00. If we know more properties of the function k{x) near |a;| = 00 we can get more accurate asymptotic expression for KZ(X). Let us assume that k{x) = l + ^ + o ( ± ) , \x\ \\x\J

0= tan-^, X!

then t(A

,)

= 1 +

^ )

A

- .+

0

(^

Infinite Element

Method

155

accordingly we have L0 = Loo + LoiX'1 + o(A~1), L'0 = L'00 + B^Bo

L'01X~l+o(X~l), + oiX-1).

+ BxX^

Formally we set KZ(X) = Kzo + KziX~l we obtain a linear equation for Kz±: Kzl - K^L'^iBo

and substitute it into (3.68), then

=Loi - (5 0 T - K^iL'wr'L'M^iBo - B^{L'mr\Bo

Kzo){L'o0rlKzl

- Kz0) - (B% -

- Kz0) T

(3.69)

1

- Kzo) - (5 0 - ^ 0 ) ( ^ o ) - B i .

We can take a constant f slightly larger than 1, then get an approximation of Kzo- Using (3.69) we get Kz\. KZO + KZ\XQ1 is taken as an approximation to Kz{Xo) for a large Ao. Then we solve (3.68) backward to get KZ{X) for all A G [1, Ao]. KZ{1) is our desired combined stiffness matrix for the equation (3.63). 3.9.2

An inhomogeneous

equation

Next let us consider an equation, - V • (Jfc(a;)V«) = f{x).

(3.70) 2

To guarantee well-posedness it is assumed that /(x)|a;| log \x\ is bounded for large \x\. Under the transformation Arc —• x it becomes - V • {k(\x)Vu)

= X2f(Xx).

Fx(v) = X2 [ Ja

f(Xx)vdx,

(3.71)

We define

and J\(u,v)

= a\(u,v)

-F\(v).

The infinite element formulation of the Dirichlet boundary value problem to (3.71) is: Find Uh G S(fi) for given gi G S(To) such that uh\ro = 91 and Jx(uh,v)=0,

Vv€S0(fl).

(3.72)

156

Numerical Methods for Exterior

Problems

If vi,v2 € S(fl) and u i | r o = v 2 |r 0 = 2 o, then v = vi — v2 € S0(fl). (3.72) implies that J\(u,vi) = J\(u,V2). Therefore as a functional on the space S(fi), J\{u,-) depends on zQ only. We write J\ as the sum of a bilinear part and a linear part: Jx(u,v)

= y%Kz(X)z0 - H{X)z0,

(3.73)

where H(X) is a row vector. Because (3.73) is valid for all homogeneous equations, KZ(X) is just the combined stiffness matrix associated with (3.64). Let us derive the equation for H(X). We divide the domain fi into two parts: fii and £fi for given £ > 1, then we have y%Kz(X)z0 - H(X)z0 = y?X(A£)zi

+ (vi vi) ( 5 l ~KI) ( Z ) ~ H(x^Zl' A2 £ f{Xx)vdxThe last term depends on (

) linearly. We define

($o(A) $i(A)) (Z°)

= A2 J

f(Xx)vdx.

Noting (3.67) we get y%(K0 - AT(K'Q + Kz(XO)-1A)z0

+

- H(X)z0 =

y^K^X^

^^)(-rO(*)- F(A0 * i - ( *° (A) * i(A)) (*)(3.74)

Since Uh is the solution, (3.74) holds for all z\, we have yjKz(XO

- yTQA + y?K'0 - H(\Q - $j(A) = 0,

which gives yi

= {K + K^XOr'iAyo

+ HT(X£) + *f(A)).

(3.75)

Infinite Element

157

Method

We substitute (3.75) into (3.74) to obtain -H(X)z0

= - H(\£)(K'0

+

K^XOT'Azo

- *i(A)(tf£ + K^XOT'Azo

- $o(A)z 0 .

Since ZQ is arbitrary, it holds that -H(X)

= -(H(XO

+ *i(\))(K'0

+ K^XOr'A

- $ 0 (A).

We set z\ = ZQ + (£ — l)Ci then (*o(A) *i(A)) ( j )

= ($ 0 (A) + ^ ( A ) ( ^ - 1 ) $ ! ( A ) ) (*?

We define *o(A) - *„(A) + $!(A),

*i(A) = (£ - l)*i(A),

then it holds that

ff (A) = H(\£) +

*i(A)\

Z-l

^+177^+^(^)1 f-^T-^(AoU+*o v ^_1)2 *v ^ y ^ _ ! "yJ (

*'

C-l (3.76)

Let V>o (A) = lim ^ ,

Vi (A) = lim 7JZTJ2 •

We divide (3.76) by £ — 1 and let £ —• 1 to obtain A

^ T

=

-HWiL'^iB

- KZ(X)) - Vo(A),

which is the ordinary differential equation for H(X). If f(x) = / o (0)|z|- 3 +o(|:r|- 3 ), then Vo(A) = ^oX*1+o{X~1). we set H(X) = H0 + HiX'1, then by (3.77) we have

(3.77)

Formally

-H0(L'oo)-1(B0-KzO)=0, -Hi = - H0(L'00)~

L'0l(L'00)~ 1

(B0 -

Kz0)

- F 1 ( ^ 0 ) - ( B o - KzQ) - H0{L'M)-l{Bi

- KzX) - V>oo,

158

Numerical Methods for Exterior

Problems

which gives

I \-lfD. ri(B V-K M iPoo{I-(L' oo 0 z0)}

H1 = 3.9.3

General

multiply

connected

l

domains

Let Dj c Q, j = 1, • • • , N, where Dj are simply-connected domains with boundary jj. We assume that Dj are disjoint to each other and set do = n \ U j = i A/> thenfio is a multiply connected domain. Let us consider the equation (3.70) on fioFirst of all we consider the Neumann boundary condition: du

fa

= 0,

j = l,;N.

(3.78)

1i

The boundary condition on To is flexible as usual. If the boundaries jj are suitable regular, say, satisfying the Lipschitz condition, then functions in Hloc(Q0) can be extended into Dj , still in H1. Therefore if we assume w

"\o,

xen\n0,

to-{™'*

efi0,

and consider the equation - V • (fc(x)Vu) = f(x) on Q, then it is equivalent to (3.70) on fio- We define a seminorm Hull!,. = ([ k(x) (\Vu(x)\2 \Jn V

+

f{x) ) dx) |z| log \x\J J 2

on tf1 •*(£}) and let T = {uG/r 1 '*(n);||u||i,» = 0} )

(3.79)

Infinite Element

Method

159

then for a given function UQ on To we consider the variational problem: find 1, u e i7 *(fi)/T, such that u|r 0 = gi, and i>*/ VveHo'*{Q).

Jx(u,v)=0,

The problem is thus defined on fi, and the approaches in the previous section can be applied here. We provide more details about the mathematical formulations in the next subsection. Secondly we consider the Dirichlet boundary conditions u|7i=0,

j = l,---,N.

(3.80)

We use penalty method and consider the problem: find u £ such that u|r 0 = gi and

H1'*(Q)/T

N

f k{x)Vu -VvdxJn

f f(x)vdx Jn

+ ~ ^ / uvds = 0, e j=i I'J

Vv £ #o'*(fi),

where e > 0 is a small constant. We denote by A - 1 7j the image of 7,- under the transformation Xx —> x, then define N

J\(u,v)

2

= / fc(Aa;)Vw • Vvdx — A / f(Xx)vdx Jn Jn

-\—YJ / £ ~[J(\-i-iJ)r\n

uvds,

(3.81) where we notice that (A - 1 7,) fl ft may not be a closed curve. Then we use the approach in the previous section to solve this problem. Concerning the last term of (3.81) the scheme to evaluate KZ{X) should be modified. Let us consider one domain Dj as an example. Let T\ be the similar figures of To with the center O and the constant of proportionality A > 1. We assume that 7j consists of some parts, some of which lie in Tx1 and r^ 2 , and some are transversal to T\ (Fig. 5). The matrix KZ(X) is discontinuous at A = Aj and A2: y g X ( A 2 - 0)z0 = j / 2 X ( A 2 + 0)z 0 + — f uvds, £ ./(AjSiJnro

j/gX(Ai - 0)^o = s/2X(Ai + 0)z0 + ^

£

/

uvds.

1 (A-'T^nro Jixr

The other parts of jj have some contribution to the matrix LQ. Let us consider "element stiffness matrices".

160

Numerical Methods for Exterior

Fig. 5

rAl

r\2

Problems

°

Fig. 6

Suppose 1 — 2 — 3 — 4 is a combination of two elements, the intersection of A _1 7j with it is approximated by a line segment 7' with an oblique angle #.(Fig 6) 7'- divides the line segment 1 — 2 into two parts with lengths li, I2 respectively. Let i : / i = / 4 = l,/2 = / 3 = 0 , V>2 : h = h = l , / i = /4 = 0 , V 3 : / 4 = e - l , / l = /2 = / 3 = 0 , V 3 : / 3 = ^ - l , / l = /2 = /4 = 0.

(^3 and y>4 have no contribution to the stiffness matrices, and the length of 7J is r(£ — l ) / s i n # , therefore

lim

1 / vf „, _ hra _i_ f fc y ^ >

€-»i € - 1 /y;.

€ - i £ - l V'l+^y

=

sine

' f_A_y. sinflVji+W

The other entries can be evaluated in the same way. We get

esinfl I hh \ (h+h)2

( h \2 I ' \h+h] )

We assemble all matrices K of one layer then add the sum to Lo in the equation (3.68).

Infinite Element

3.9.4

Transfer

161

Method

matrices

To solve a boundary value problem after getting Kz{\) matrices X(Ao,A) and Xi(Ao,A) by yx=X(\o,X)yx0+X1(X0,\),

we define transfer

A > A0,

(3.82)

where y\ = w|rx- If Ux0 is given, then we regard it as a boundary data. By the well-posedness of the problem y\ is determined uniquely, hence the matrix X and X\ are well defined. By some calculation we get from (3.75) that

* = < 7 + ((^52 + *«(*>) (^k - K>w1 > y° L

'°

+KZ(XO)

1

(HT(XO + *I{X)

Therefore lim ^ | ^

= ( L i ) - 1 ^ - ^ ( A ) ) y 0 + (Lj,)" 1 ff r (A).

We define an infinitesimal transfer matrix Y(X) = (L'0)~1(B — KZ(X)) and an infinitesimal transfer vector Y\(X) = (L'Q)"1 HT(X). We then take Ao > 1, AAo > 0, and make a similarity transformation Xox —» x, then (Ao + AA0):r -> (1 + ^)x. Let £ = 1 + ^ , then approximately 2AO+AAO

= yx0 + ^-l){Y(X0)yXo+Y1{X0))

= yXo +

—l(Y(X0)yx0+Y1(X0)),

and yx=X(X0,X)yx0+X1(X0,X) =X(X0 + AA0, X)yx0+AX0 + *i(Ao + AA0, A) =X(X0 + AAo, A) (yXo + ^(Y(X0)yXo

+

^(AQ)))

+ Xi(A 0 + AA0, A).

Since yx0 is arbitrary, we get approximately X(X0, A) = X(A 0 + AAo, A) (i + \

*i(Ao, A ) = ^X(X0

^Y(X0) A0

+ AAo, A)yi(Ao) + XX(A0 + AA0, A).

Dividing them by AAo and letting AAo —> 0 we get -£-X(X0, CAo

X) = -A 0 - 1 X(A 0 , A)Y(Ao),

(3.83)

162

Numerical Methods for Exterior

Problems

— X i ( A 0 , A ) = -Ao 1 X(A 0 ,A)y 1 (A 0 ).

(3.84)

We have initial value X(X,X) = I,

X1(X,X) = 0,

so we can solve (3.83),(3.84) to get X and Xi, then get the solution from a given boundary condition. 3.10

Convergence

All schemes in this chapter can be proved convergent. We investigate one of them for example. Let us consider the Poisson equation, -Aw = /,

ie(l,

(3.85)

with homogeneous boundary condition, u\dn = 0,

(3.86)

where dim (Cl) = 2. We assume that / G Hm~1(Cl) for a positive integer m, and / is compactly supported. Moreover we assume that u G H^l(Cl). The partition of the domain CI is described in Section 3.1. Let h be a reference length of the mesh, for example we can set h to be the maximum length of the diameters of all elements in fii. Some assumptions are made on the mesh: (A) The partition is regular, that is the interior angles of all elements possess a common lower bound 8o > 0. (B) O $ Si, Vi. And let d(0,ei) be the distance from e, to the point O, then there is a constant x, such that meas e, < \h2(d(0,

e,)) 2 ,

Vi.

For instance the similar partition given in Section 3.1 satisfies the condition (B). Let m-th order elements are employed in the scheme. Then the infinite element subspaces are S(Cl) = { « £ H1'*^);

M * G P m (ei), Vt},

and S0(Cl) = {u G S(Cl);u\an

= 0}.

Infinite Element

Method

163

The infinite element formulation of the problem is: Find Uh £ that a(uh,v)

= (f,v),

such

SQ(Q,),

VveS0(Q).

(3.87)

By the expressions of the solutions of the Laplace equation in Section 1.1, we find the weighted semi-norm |u|m+l(m) "

/n J is bounded, where D u denotes the (m + l)-th order derivatives of u. Let II be the interpolating operator with respect to S{Q), then we have m+1

Theorem 60. If the conditions (A),(B) hold, u G #,7 c +1 (f2), \u\m+i{m) < oo, then we have the interpolating estimate, |u-nu|i
U

f j^(u(x)-Uu(x))2dx\2

and (3.89)

< Chm+l\u\m+1{m).

(3.90)

Proof. By the embedding theorem of Sobolev spaces, u is continuous, so II u makes sense. According to the interpolating estimate on triangular elements [Ciarlet, P.G. (1978)], |u-n«|.,e. |m+1,ei,

s = 0,l,

(3.91)

where hi is the diameter of e,. By the condition (A) we have meas e, > h\ sin 3 9Q COS 9Q , then by the condition (A),(B) we have 2

> 2 (d(Q,e,)) 2 sin t'ocosfc'o

Substituting it into (3.91) we get \U - Uu\s,ei < Chm+1-S(d(0, We have d(0,ei)

ei))

m+1 S

- |«U+1,e,.

< \x\ on the triangle e,, therefore

(d(0,ei))m+1-s\u\m+1,ez

< (^O.ei))1-^!^!^),^.

We obtain

(d(o, ei )) s -V-nu| S;ei <

chm+l-s\u\m+l{mUi.

Upon summing them up with respect to i we get (3.89) and (3.90).

D

164

Numerical Methods for Exterior

Problems

We turn to estimate the error u — it/,. Theorem 61. / / the conditions (A),(B) hold, and the solution u to the problem (3.85),(3.86) satisfies |u| m +i( m ) < oo, then |«-Ufc|i
(3.92)

The proof is routine. We take an arbitrary v € S(£l), then a(u — Uh,v) = 0.

Thus a(u — Uh,u — Uh) = a(u — ith,u — Uu), which gives \u - Uh\f <\u-

uh\\\u~

Ylu\i.

(3.92) follows by applying Theorem 60.

•

Thus the error estimates are optimal. Using Nitsche's trick, we can prove L 2 -norm estimate. However, because the exterior domain Q can not be convex, and the boundary dfi, is not smooth, the result is not optimal. There are some approaches to deal with reentrant corners, but it is another story. Let R > 0 be large enough, such that B(0,R) D flc. We have the theorem: Theorem 62. Under the assumptions of Theorem 61 it holds that II" - Uh\\o,ar\B(o,R) < Chs\u - uh\i,

(3.93)

where 0 < s < 1, depending on Q, and the constant C depends on Q, and R. Proof.

We set ( u - Uh, \x\ < R, ;

[0,

ja;j>i2,

and consider an auxiliary problem: find ip S H0'*(Q), such that a(v,
veH^*(Q).

Let us estimate ip first. By Theorem 14

IMIi,.
(3.94)

Infinite Element

165

Method

If xo is a corner point on dfl, we consider two small domains Q^ = Q fl B(x0,S), fl2s = O n B(x0,2S) with S > 0. Then by the separation of variables, the solution (p in Q-28 can be expressed explicitly, and it can be estimated that IMIi+ s ,n 4 < C(||/|| 0 ,n, + IM|o,n2J> where 0 < s < 1, depending on the angle. Let f2i = fl n B(0, R + 1), and fl2 = {z;]-Xl > # + ! } > then using the interior estimate of elliptic equations [ Gilbarg, D. and Trudinger, N.S. (1977)], we obtain

IMIi+.,n 1
Jn The second inequality is due to the embedding theorem of Sobolev spaces. Thus we obtain

!MI1+..0! +

(J

\x\2\D^{x)\2 dx\ ' < C||/||o.

Letting v = u — Uh in (3.94) and noting that a(u - uh,Uf)

= 0,

we get \u- uh\i\
a(u-uh,ip-Ilip)

=

(f,u-Uh).

Using the normal estimate for interpolating operators on fii and Theorem 60 on Q2> we have

lv-n v |i

we

obtain finally

H/llo^Cfc'lu-Ufclx, and the proof is complete. Other kinds of error estimates can be found in [Ying, L. (1995)].

HI

Chapter 4

Artificial Boundary Conditions

Some artificial boundary conditions, exact or approximate, have been given in the previous chapters, for example, the application of "Dirichlet to Neumann" operator or the combined stiffness matrix. In this chapter we will investigate more artificial boundary conditions. A pseudo-differential operator is introduced for the wave equation to reduce the boundary condition at the infinity to an interface. Since this condition is global in space, some works have been done to localize the boundary condition on the analytic level. The problem of stability for some approximations is still open, because there are some higher order derivatives in the conditions. In this chapter we will also show some works on the artificial boundary conditions of the Navier-Stokes equations.

4.1

Absorbing boundary conditions

Absorbing boundary conditions are applied for wave problems. Let Q be an exterior domain, and let CIQ be a bounded domain such that do ^> ^ c The truncated domain is fix = fl D fio and dilo is the artificial boundary. Suppose there is no incoming waves from the exterior of CIQ. The artificial boundary condition on dflo is denned to absorb all outgoing waves from Qi, so that the waves do not reflect from dflo. It is named as an "absorbing boundary condition". We start from one dimensional case and consider the wave equation,

^

= ^ .

at2

dx2 167

(41) {

'

168

Numerical Methods for Exterior

Problems

Suppose the domain is (0, oo), and the initial and boundary conditions are du ^7

u\t=o = wo,

ui,

x > 0,

(4.2)

ot w|x=o = 0,

t > 0,

(4.3)

where M0(0) = 0 , m(0) = 0 . The equation (4.1) can be factorized to

, d !:l«-o\dt - ! : dxJ ) ( \dtl + dx

<«)

It yields two equations

5-S-°.

<"'

du

,, „. (4 6)

and du

+

m o-x=°-

-

Obviously the solutions to (4.5),(4.6) are solutions to (4.1). Conversely, if we make a transform of variables £ = x — t and r\ = x +1, then the equation (4.1) is transformed to d2u _ Q d£dr] and the general solution is u = /(£) + g(rj) = f(x — t) + g(x + t), where / , g are arbitrary functions. Therefore the general solution to (4.1) is the linear combination of the solutions to (4.5) and (4.6). Using the expression of general solutions one can get the solution to the initial-boundary value problem (4.1),(4.2),(4.3): \{ua(x -t) + u0(x + *)) + ! / * + / u i ( 0 d£ | ( - u 0 ( * -x)+

u0(x + *)) + § / t ! i * « i ( 0 C i < *.

If wo and u\ vanish for x > a, then on the domain {(x,t);x solution is ,,,„. ts _ / 5"o(z - 0 + | C t MO « , 1

x>t,

> a} the

a: > *,

' j " \ -§«o(t - a:) + I /;_, «i(0 d6, x < t.

Thus only outgoing waves appear in this domain. The solution u satisfies (4.5) on the line x = a. (4.5) is the exact absorbing boundary condition

Artificial Boundary

169

Conditions

on x = a. With this boundary condition the exterior problem is reduced to an initial boundary value problem on the interval (0,a), and the DtN (Dirichlet to Neumann) operator is thus K = J^. We turn now to investigate two dimensional problems and consider the equation = A«-

W

(4-7)

Let the plane x\ — a be the artificial boundary, and there is no incoming waves from xi > a. Like (4.4), formally we can make the factorization:

/

V

d dxi

e2 d

Qx\

0.

di\ i - aF

(4.8)

Then the absorbing boundary condition at x = a is o

I

~&*~\ (4.9)

and the DtN operator is K

a2

d_ dt\

82

aF In order to interpret (4.9) precisely, we introduce some basic definitions of pseudo-differential operators briefly.[Taylor, M. (1974)] The Fourier transform is defined on the spaces: S(Rn) = {
lim xaD^if(x)

= 0,Va,/3},

|x|—>oo

where a, (3 are multiple indices, and C£°(R n ) C <S(R") C C°°(K n ). The dual space of C£° (E"), «S(Rn), and C°°(E n ) are denote by X>'(En), <S'(En), and £'(E") respectively. They are spaces of distributions, and the relation of them is D'(M") D <S'(En) D £'(E"). Letting / € 5 ( E " ) , the Fourier transform / of it is denned by

/(0= f f(x)e-i^dx,

170

Numerical Methods for Exterior

Problems

where £ € K™. The inverse Fourier transform is

Using duality the Fourier transform F : <S(Rn) —> <S(Mn) can be extended to: F : <S'(R") -> S'(Rn). Differentiating by parts one can find that if / = F(f), then i£jf = F (•%*-'), that is

df(x)

kpL*fw*

dxj

(2

In general, if P(£) is a polynomial of the independent variables £ with constant coefficients, then P(D)f(x)

= -±- f [Z1T)

^ e r e D = \ (•&,

p(iomyx
JRn

&,--,•&;).

If p(x,£) is a function defined on ft x K™, satisfying

\D2Dfp(x,S)\
P(x,D)f(x) = -^— [ p(x,0f(&<**<%, which is a linear continuous mapping from £'(Mn) to V(M.n). p(x,£) is called the "symbol" of the pseudo-differential operator P(x,D). Let us now return to the equation (4.7). Let u be the Fourier transform of u, then u(Xl,x2,t)

= - ± -

[ u(Z1,b,u>)ei^+t>x'+ut)d£1dZ2
(4.10)

(2TT)™ JR3

The formula (4.10) can be explained as follows. For a fixed constant vector (fii&iw), the function w(x,t) == u(£i,6,w)e i ( C l ! B 1 + € a : r a + ''" ) is a particular solution to the equation (4.7). w is a constant on the plane Ci^i + £2^2 + u>t = c, where c is a constant, so it represents a plane wave traveling along a direction (£1, £2) on the x plane with a speed of s = w/|£|. The formula (4.10) shows that u is the linear combination of all plane waves.

Artificial Boundary

171

Conditions

Applying the Fourier transform operator to the equation (4.7), we get

(u2-ei-e2)u=o.

(4.H)

Therefore u vanishes if w2 — £2 — £| ¥" 0- The support of u lies in a cone: £i ± wWl — (77)

= 0 . Since ^

= 1 on this cone, the wave speed is 1.

Restricted on this cone the integral (4.10) can be expressed by u xi,x2,t)

=-±-

f

MMe^-ft^+to^dkdu,

(2TT)" J\(2\<\U,\

+ T^vT / f2(Z2,u)eil-uJ1-&?Xl+t>x>+ut)dZ2(L;. 27r ( ) J\(2\<M In the first term £i/w > 0, so it represents the combination of those waves traveling towards to the positive x\ direction, the outgoing waves. On the other hand the second term represents the combination of incoming waves. The equation (4.11) can be factorized to

6 - ^ / i - i-S i Ui+«Ji-(|Vifi=o. It yields two equations 6 - ^ / 1 - 1 - )

| « = 0,

(4.12)

£x+w\\l-[Q\

| « = 0.

(4.13)

and

They are the equations for outgoing waves and incoming waves respectively. Returning to the physical variables, they are ^ - K u ox\

= 0,

^ + Ku = 0, ox\

(4.14)

respectively, where

(4.15)

172

Numerical Methods for Exterior

Problems

The first equation in (4.14) is the absorbing boundary condition for the artificial boundary on the right, while the second one is that for the artificial boundary on the left.

4.2

Some approximations

The conventional differential operator is local in the sense that the derivatives of a function at one point depends on the function value in a neighborhood, as small as desired. On the other hand the DtN operator K defined above is a pseudo-differential operator, and it is global. It is not so convenient to carry out any numerical computation for a pseudo-differential operator. Many approximations have been developed to localize the operator (4.15). The idea is the following: The symbol u)\ 1 — f *j J is approximated by a rational function p(s)/q(s), where s = *jj, then the equation (4.12) is reduced to

fr-w(p(§)/,(|)))«

= 0,

that is

6
U)

Multiply the both sides by ujm, where m is a suitable power, then it becomes a conventional differential equation applied on the artificial boundary. If the domain fio is rectangular, then similar conditions are applied corresponding to X2 on the other two line segments . 1. Clayton-Engquist-Majda's boundary conditions [ Engquist, B. and Majda, A. (1977)] [Engquist, B. and Majda, A. (1979)] Using Pade's approximation, the function %/l — s2 is replaced by rational functions, r

(DW

=

l,

r (fc + i)( s) =

i _ _ ^ _ ,

f c

> l .

The absorbing boundary conditions of different orders are: du dt

du dx\

Artificial Boundary

82U

Conditions

82U

173

1 d2U _

2U

~~d¥~ dtd^ ~ 204 ~ '

BN+iu

d Id2 = -JTBNU - - ^ - 2 - B J V - I W = 0.

Pade's approximation is developed near the origin, so it gives satisfactory results if s is small, that is, if the incidence angles of the outgoing waves are small. 2. Halpern-Trefethen's boundary conditions [ Halpern, L. and Trefethen, L.N. (1988)] Approximations are made for different s's corresponding to different incidence angles, which can be large. Take 2K symmetric points on the interval [—1,1]: ±sx = ± s i n # i , • • • ,±SK = ± s i n # # , then define a rational function r(s), such that r(s) — y/1 — s2 at the 2K points. The formulas are the following:

^) = (r ? -0^)...(7 ? - v /l-4), r(s) = (p(v) + P(-V))V

-p{v) + p{-v) '

where r\ = y/1 — s2. Because the numerator and denominator are even functions of 77, r(s) is a rational function. Besides, because |p(—ry)| > \p(ri)\ for rj > 0 and p'(0) ^ 0, r(s) has neither poles nor zero points on [—1,1]. Therefore r(s) is indeed an interpolation function of y/l — s2. There are several ways to define 6k'a. Pade; b. Chebyshev; c. Least squares; d. Chebyshev-Pade; e. Interpolation in Newman points. We omit the details here. 3. Lindman's boundary conditions [Lindman, E.L. (1975)] A rational function r(s) is introduced to approximate , * ±: N

r(*) = l + £

2

T

* - i -

^

174

Numerical Methods for Exterior Problems

Applying this approximation in (4.12) yields N

0

__ .2

M^MfeM*Ml

*;=i Then applying the inverse Fourier transform to get the boundary conditions: du

du

r—v

1

d2hk dt2

d2hk dx\

fc=i

cP_(du_ dx\ \dx\

= a

4. Higdon's boundary conditions [Higdon, R.L. (1986)] [ Higdon, R.L. (1987)] A general boundary condition is

n( cosa 4-£9 M=o '

fc=i and all the above boundary conditions are particular cases. Plane waves with incidence angles ak satisfy this condition. 5. Safjan's boundary conditions [Safjan, A.J. (1998)] In order to implement some high order absorbing boundary conditions, some approximations are introduced. For example, the second order Clayton-Engquist-Majda's boundary condition, d2u

d2u

1 d2u =

W ~ dtdxx ~ 2dxJ is approximated by a two steps scheme:

dt

~ dxi

~

'

dV2>

aVa> _ i d2u^

~dt2

dtdxx

~ 2

dx\

That is dt

dx\

0,

Artificial Boundary

dt

dxi

175

Conditions

~ 2 dx\

'

where V,(xi,X2,t)

= / Jo

u(Xi,X2,T)dT.

The operator on the left hand side is always first order. Likely the third order boundary condition, d3u dx\dt2

d3u dt3

+

3 d3u 4dx?,dt

1

d3u kdx\dx\

can be approximated by dt

dxi

dt du& dt

dxi 3d2u& 4 dx\

du^ dxi

2 dx\ 1

' d3u^ Adxidx^

where u(Xi,X2,t)

4.3

= / JO

u(xi,X2,T)d,T.

Bayliss-Turkel radiation boundary conditions

Consider the equation (4.7) for dim (f2) = 2,3. The radiation boundary conditions in [Bayliss, A. and Turkel, E. (1980)] is based on the expansions of solutions near the infinity. For three dimensional cases the solutions u can be expressed in terms of independent variables t — r,r,Q,4>, where (r, 9,(f>) is the spherical coordinate system. For large r, the solutions propagating in directions that are outward from the origin can be expanded in a convergent series of the form

U=

g"j(«-J.g.*), j=i

(4.16)

176

Numerical Methods for Exterior

Problems

For two dimensional cases, the expansion expression is Uj (t — r, 6)

_

E 3=0

s !yj+2 i^.

(4,7)

The leading term for the solutions are 0{r~x) and 0(rz) respectively. This fact can be seen from the fundamental solutions of the wave equations, which are 47rt

Vl

'

h

2TTV*2 - M 2

\l,|a:|
for dimension three or two, where J is the Dirac function. The fundamental solutions satisfy (4.7) and initial data, w|t=o = 0,

du -gr\t=o = S(x).

For the proof of (4.16) and (4.17), see [Friedlander, F.G. (1962)][Wilcox, C.H. (1956)] [Karp, S.N. (1961)]. Define an operator d_ dt

d_ dx'

then substitute (4.16) in Lu to get Lu = 0(r~2). To improve the error let B\ = L + i , then B\u = 0(r~3). B\u = 0 is then applied as the absorbing boundary condition o n r = i? with R large. To improve the operator B\, let Bi = (L + | ) i ? i , then B-^u = 0(r~5). In general B3 = (L + ^ - ^ ) Bi-U

BjU =

Oir-V'1).

For two dimensional problems substitute (4.17) in Lu to get Lu = O ( r - i ) . Then define Bi = L + ±, and B

i=(L+^r)B^

4.4

B3u = 0(r-V-h

A lower order absorbing boundary condition

In the previous sections some absorbing boundary conditions are of higher order. That is, the order of the boundary condition is higher than that of the

Artificial Boundary

Conditions

177

equation. Some of them may cause unstable problems. One second order in derivatives, but higher order in precision absorbing boundary condition is given in the following. [Zhang, G. (1985)] [Zhang, G. (1993)] [Zhang, G. and Wei, S.(1998)] Noting that VT^£2 = 1 - - [ \A-r2-^ dr, 7T i - l 1 ~ r2£2 the expression (4.15) can be written as Ku

=TnZ

I «&.&(«;)ia;(l-± f VT^T* 2 ^ 2c2 dr)

Let £ ( 6 , 6 , w , r ) = -= LJZ

£2 272^' — r^t,2

then the first equation of (4.14) is equivalent to du dx\

du dt d2v

d I— I dt 2

r

2d

v

=

v 1 — r2v(x,t,r)

d2u

W~ '~dx\ ~dx\'

r

dr >

e(-M).

One can get an approximate boundary condition with n-th order accuracy: du dx\

du d ^ dt = dt -^2_^a-kVk\x,t,rk), fe=i

d2vk dt2

2

2d k

vk dx\

d\ dx\'

afc =

1 . 2 sin n+1

where rk = cos

kn , n+1

kn n+1

178

4.5

Numerical Methods for Exterior

Problems

Liao extrapolation in space and time

Suppose that the solution for t < to and x\ < a is known and x\ = a is an artificial boundary. Liao introduced an extrapolation scheme to evaluate u(a, x^, X3, to + At). [Liao, Z., Wong, H., Yang, B. and Yuan, Y. (1984)] [ Liao, Z. and Wong, H. (1984)] Let uo = u(a,X2,xs,to + At), u\ = u(a — h,X2,X3,to), u
A

Ul

= J2 (~l)k+lct-i
~ kh, x2,2:3*0 - (k -

l)At),

fe=i

where C™ is the binomial coefficient defined by k

{m-k)\k\'

uo is unknown and evaluated by means of extrapolation. uo = MI + A ^ i + A 2 «i H

h AL_1ui.

Shao and Lan introduced a modified scheme of this kind. [Shao, X. and Lan, Z.][ Shao, X.] 4.6

Maxwell equations

For the initial-boundary value problems of Maxwell equations (1.106) we notice that

A2 = £ -v- 1 \f v x f 10v x

y x°v x y)

We assume that p = 0 and j = 0 in (1.101),(1.104). Then V • E = 0 and V • H = 0, and we get „ _ i _ i / A 0

42

For the homogeneous equation (1.106) we apply the operator ^ to obtain 92u w

= s

_i

p

_i .

Au,

Artificial Boundary

Conditions

179

which is the wave equation. Therefore all the above artificial boundary conditions for wave equations can be applied to Maxwell equations. For two dimensional cases we assume that E and H are independent of x3, then the system of equation (1.106) are separated to two independent system of equations: 1. The transverse magnetic mode (TM): 0HX

-XdEz

~0T 0H2 .idE3 =H Ot dx\ ' 0H2 0E3 Ot dxi

(4.18) dHi dx2

2. The transverse electric mode (TE): dEi _ _xdHz =e Ot dx2 ' _i OH, 0E2 = —s Oxi Ot OEi 0H3 -l =M Ot Oxn

(4.19) 0E2 0x\

We set u = {Hi H2 E3)T and u = (Ei E2 H3)T for the TM mode and the TE mode respectively, then both of the systems of equations can be written in the form of du + iAu = 0, ~dl where

A = -i

and

(4.20)

180

Numerical Methods for Exterior

Problems

We eliminate H\,H2 in (4.18) by apply the operator / d d \dx2 ' dxi'

jT dt,

on the right, then we get the wave equation for E3,

Analogously the system (4.19) yields & Hs

-1

-1

A

rr

Therefore the above artificial boundary conditions can be applied here as well. We turn now to investigate the initial-boundary value problems of (4.20). Let the initial conditions be u\t=o = wo,

(4.21)

satisfying V • E = 0 and V • H = 0. Being the same as that in Section 1.11 we consider the total reflection boundary condition, E x u\dci = 0, §tB • v\an = 0. Then for the TM mode (4.18) dE3 dE3 8x2 ' dxi

v

= 0, an

and E3V1 = E3U2 = 0. The combination of these two conditions leads to the Dirichlet boundary value, E3\da=0.

(4.22)

Being analogous the Neumann boundary condition,

dH3, QV

fdH3 lan

v QX2 '

dH3 gXl

x v

= 0,

(4.23)

an

is generated for the TE mode (4.19) . To study well-posedness, we introduce some spaces, f 5(curl;fi) = \E = (E^E*)

BE BE e (L 2 (fi)) 2 ; - ^ - - ^ €

S 0 (curl; fl) = { £ € S(curl; Q); E x u\da = 0}.

1 L2(n)^

Artificial Boundary

181

Conditions

For the TM mode, we work in the space VTM = 5(curl; CI) x H^(Q), and for the TE mode the space is VTE = S0 (curl; fi) x

ff^fi).

Then the initial-boundary value problems are (4.20),(4.21),(4.22), where the domain of A is VTM, and (4.20),(4.21),(4.23), where the domain of A is VTE, respectively. Following the argument in Section 1.11 let us introduce weighted L2spaces H with the weights //i0 0\ M = 0 / x 0 \00e/

or

/e00\ M = I 0£0 I \ 0 0/x/

and prove that Lemma 66. A : VTM —> ( £ 2 ( ^ ) ) 3 ana" A : V r B —> (Z,2(f2))3 are self-adjoint operators in H. Proof. We consider the TM mode first. Clearly A is closed and symmetric. By definition D(A*) = {ve (L 2 (fi)) 3 ; 3g e (L2(n))3,

(Au,v)H

We denote v = (i>i,U2)T and g = (gi,g2)Thave

= (u,g)H,Vu

e VTM}.

Letting # i = #2 = 0 in u, we

-^-'((Mp-ir) ••)-<*•*>•

V£3€ff (!!)

« '

Therefore g 3 = * ^ ~ 1 ( i ^ ~ f f i ) and the domain for v is 5(curl; O). Similarly letting Ez = 0, we have ~ i £ _ 1 ("tof " ^ " '

V 3

j

=

UHi>H*)>9),

V(F 1 ; H 2 ) e S(curl;fi).

Therefore g = i £ - 1 ( | ^ - , — f^|) and the domain for v3 is ^ ( f i ) . Hence £>(A*) = 5(curl;fi) x H%(Q) = VTMThe proof for the TE mode is analogous.

•

Numerical Methods for Exterior

182

Problems

Following the same lines as that in Section 1.11, we can prove that the problems (4.20),(4.21),(4.22) and (4.20),(4.21),(4.23) admit unique solutions. The details are thus omitted.

4.7

F i n i t e difference schemes

After truncating the domain ft and imposing some artificial boundary conditions, numerical schemes for bounded domains can be applied. For wave problems one finite difference scheme is the second order Yee's scheme. [Yee, K.S. (1966)] For example the TM mode can be approximated by the following scheme: Let {Ez)^ ; = Ez{kAx\, lAx%, jAt), and the same notations for Hi and Hi, then the approximate solutions are given by

At

^

Ax2

M

At

An

MfcUi-wi-i At i

Arri

Ax2

The finite difference schemes for three dimensional problems and for the TE mode are the same. Mur introduced a finite difference scheme on the artificial boundary which is consistent with the Yee's scheme .[Mur, G. (1981)] For example for the Clayton-Engquist-Majda's boundary condition, B2U

~

d2u dt*

d2u ld2u dtdx, ~ 2 dx2 ~

'

the derivatives are approximated by Vj+1

c2 2

du

dtdxi

.

u

fc+i,;+i

-Uj

+ 1

u

U

k+l,l+i 2AtAa;i

k,i+\

+

fc,/+i

Artificial Boundary

d2u . (ui+i,i+l 2 dt ~ \ \

d2u ^ Ox2,

4.8 4.8.1

l\

183

Conditions

-H,i+i+ui,iU2

2

At2

At

(ui+i,i+i -2uk+i,i+i+uk+i,i-i

u

fe,/+f

2

-H«+i+uM-i Ax22

Ax 2

Stationary Navier-Stokes equations Homogeneous

boundary

condition

at the

infinity

The governing system of equations with boundary conditions for stationary incompressible viscous flow is the following: u • Vu + Vp = v A u + / ,

(4.24)

V • u = 0,

(4.25)

u = g,

xedfl,

lim u = Uoc.

(4.26) (4.27)

\x\ — • o c

An artificial boundary condition for three dimensional problems with homogeneous boundary condition at the infinity, UQO = 0, is investigated by [Nazarov, S.A. and Specovius-Neugebauer, M. (2003)], where the incompressible condition (4.25) is generalized to - V - u = /i.

(4.28)

For UQO = 0, the system behaves like the Stokes equation near the infinity. A "power-law solution" for the associated homogeneous system is defined by (u,p) = (\x\~1U(s), |a;| _ 2 P(s)), where x = \x\s, s G S 3 , and S3 is the unit sphere in M.3. It is proved that there exists a "power-law solution", which is the leading term of the solution for the problem under some smallness conditions. The "power-law solution" is applied as the artificial boundary condition.

184

Numerical Methods for Exterior

Problems

Let the artificial boundary be 0B(O, R) with R large enough, so that B(0,R) D ttc. Introducing spherical coordinates, x\ = rcos<j)sm6, X2 = rsinsin#, xz = rcos6, and letting u = (u r ,w^,,ue), the homogeneous system of equations in the coordinates is dur dr

ug dur u$ dur Ug + «0 r d6 rsin.6 d<j> r dp / A 2 2 d(u9smd) v r - 2-^u 2 '•£>[ Au ' r r dr \ r r sin 6 86 1 d{r2ur) r2 dr

1 fd(uesmQ) rsin# \ dO

2 du4 * r sin 0 d<j> 2

du$\ _ dcj) J

where we write down the radial part only, which is needed later on. We plug the power-law solution (u,p) — (r~xU(s),r~2P(s)), U = (Ur,Ug,Uj,), 3 2 into them and multiply them by r and r respectively to obtain

2P +

" , ( A ^ - 2 , ^

Sia6d(l> +

^

\6

^

^ d6

X

+

- ^ ^ suit/ 8
(4 29) '

(

where A is the angular part of the Laplacian— the Laplace-Beltrami operator on the unit sphere, -

d2

nd

d92

d6

I d

2

sin 9 d
Applying (4.30) the equation (4.29) turns into dUr Uj, dUr 86 sin# d

(4-31)

Artificial

Boundary

185

Conditions

where d Kd6'

1 d sin6 d4>y

The Stokes equation gives a hint of the Neumann operator, /-i>-^ur+ps T(u,p) = I -y^u9 and for the power-law solution, d 1 - — u= -u. or r Then (4.31) is applied to replace p in the Neumann operator to define an artificial Neumann boundary condition at r = R, namely,

-V-fr=RUe>

-V-bV=RU*-

(4 33)

'

This defines an approximate DtN operator. Introduce weighted Holder spaces hxa(£l), a £ (0,1), equipped with the norm,

ii/ii A ^=^su P |^r'- Q + fc iD fc /(^)i +

sup x,y £ f2 2\x-y\<\x\

\x-y\-a\\xfDkf(x)-\yfDkf(y)\.

We denote by (u,p) the solution to (4.24)-(4.27), and by {uh,Ph) the solution to (4.24)-(4.26),(4.32),(4.33). Uniqueness and error estimates are summarized in the following theorem. Theorem 63. Let I be a positive integer, a G (0,1), /3 S (l + l + a,l + 2+a), and 7 € (I + 2 + a, I + 3 + a). Then there exist positive constants p, C\ and R\, such that if I A ^ + IISIIA!;'' Ri, then the problem (4-24)-(4-26),(4-32),(4-33) solution (uh,Ph) in the ball

admits a unique

and Uh satisfies the error estimate, | | « - « h | | A ^ < C 2 i 2 M l l / l l A ! ; » + NlAV»). Where C% is independent of R> R\ and 4.8.2

Inhomogeneous

boundary

(f,g).

conditions

at the

infinity

An artificial boundary condition for three dimensional problems (4.24)(4.27) with inhomogeneous boundary condition at the infinity, Moo ^ 0, is investigated by [Deuring, P. and Kracmar, S (2004)]. Without losing generality we choose coordinates so that M ^ is parallel to the xi-axis. Replacing u by u — Moo and introducing the Reynolds number R, the Navier-Stokes equations can be transformed into a dimensionless form ~ du

R—-

~

+ Ru • Vw + Vp = AM + / .

(4.34)

OXi

For Moo 7^ 0, the flow behaves like the Oseen flow near the infinity. A more general equation, ~ 3ii

R—

+ TU • VM + Vp = AM + / ,

(4.35)

OXi

is investigated, where r £ [0, R], and if T = 0, it is the Oseen flow. The following artificial boundary condition is imposed at |x| = R with R large enough: du x T , . (1 RR — xi\ _, ,A „„. This defines an approximate DtN operator. Letting fix = Q n B(0, R), two bilinear forms are defined for the problem:

a(u, v)=

( Vw : Wv + R—^- • v + R{u •

b(u,p) = - /

V)M

Vu-pdx.

• v j dx

Artificial Boundary

187

Conditions

The following results are proved: Theorem 64. There is a constant S > 0, such that for g £ lH^(dQ) J , / e (•£%c(^))

tui£/i | / a n 5 • vds\ < 5, there exists a pair of functions

1

x L 2 (fii) iwi/i V • u = 0, u\dQ = g, and satisfying

(u,p) 6 (ff (fii)) a(u,v)+b(v,p)

Vve(H1(n1))3,v\dn

= (f,v)Ql,

= 0.

(4.37)

We denote by (u,p) the solution to (4.24)-(4.27), and by (uh,Ph) the solution to (4.24)-(4.26),(4.36). Uniqueness and error estimates require stronger conditions. Theorem 65. We assume the hypotheses of Theorem 64 to be satisfied. Moreover, assume that there are positive constants 7 and a > 4, such that f € ( i ? ( f i ) ) 3 , and \f(x)\

< j\x\-°.

Let T = ||/|| 0 ,6/5 + 7 + II&II1/2 and

Tm = m a x ( l , R ) . Then there exist positive constants a.\,a,2,C\,Ci, such that if r m a x ( r , r 2 ) < CiT^1, then the solution to (4-37) is unique, and the following estimates are satisfied: ||V(tX - Ufc)||0 + (R~X + R)1/2\\u
4.8.3

A linear boundary

12

• R- '

- Ufc||o,flB(0,fl)

.minil^RR)-1'2).

condition

For two dimensional symmetric flow with inhomogeneous boundary conditions at the infinity a linear artificial boundary condition is proposed by [B6nisch,S., Heuveline, V. and Wittwer, P. (2005)], which is based on a theorem about the asymptotic behavior of solutions at the infinity [Wittwer, P. (2002)]. The flow is governed by the problem (4.24)-(4.27). Consider the homogeneous equation, that is / = 0, and assume that u^ ^ 0. The coordinates are chosen so that u^ is parallel to the a;i-axis as before. By a scaling argument the problem can be transformed into a dimensionless form, u • Vw + Vp = Au, V - w = 0,

(4.38) (4.39)

188

Numerical Methods for Exterior Problems

u = 0, lim

x&dQ,,

(4.40)

M = (1,0).

(4.41)

\x\—*oo

Regard the x\ variable as "time", then the equations behave like parabolic equations at long times. Consider an "initial condition" u k = i = (l,0) +
(4.42)

and the corresponding "initial value problem" on x\ > 1. Let v = (^1,^2) — u — (1,0), then the following holds: Theorem 66. Under some smallness conditions there exists a (locally unique) solution to (4-38),(4-39),(4.4%) and the boundary condition (4-41) on x\ = 0 0 . Furthermore there exist functions , .

. .

c 1 __^L d x\ I^K ^fx{ ir\x\2 c

x2

V02{x) = —=^--re 6 2 4i/7T

x

'

-A. ixi

d x2

+-T-T0, Z IT \X\

such that c

lim x\/2 ( sup $vi -v01)(xi,x2)$

i- > °°

lim xi

Vx2eK

sup \(v2 - v02)(xi,x2)\

= 0, /

= 0,

and oo

/

/»oo

vi (x) dx2 = / vol (x) dx2 = c + d. J—00 a potential flow with a source or sink The first terms in VQ2 represent at the origin. The second terms represent a wake, within which the vorticity of the fluid is concentrated. The asymptotic behavior VQ = (^01,^02) is universal in the sense that this asymptotic behavior is independent of the geometry of the domain fi and Qc, except for the amplitudes c and d. In the original coordinates the parameter c should be multiplied by a factor y/v\uoo\, and d by v. We now consider the exterior domain Q,. Suppose the body fic is symmetric with respect to the xi-axis, and the flow is also symmetric with respect to the Zi-axis, that is, ui(xi,x2) = ui(xi,—x2), u2(xi,x2) = —142(^1,-3:2), and p(xi,x2) = p(xx,-x2). Let a square, Q0 = {a:;|:ri| < -00 and VQI

Artificial Boundary

Conditions

189

a, \x21 < a}, be large enough, so that fio 3 ^ c - The system is solved on Qi = fi n fio, with the following artificial boundary condition: u = v0 + (1,0), « = - • r ^ 2 +(1,0),

xi > 0, a:1<0.

There are two parameters c and
I "H

pv)ds=

lim /

(UQUQ

+p)vds,

(4.43)

/ u-vds= lim / uo-vds, (4.44) Jan °^°° Jdn0 where UQ = vo + (1,0), and we notice that u = 0 on dfi and the viscous term tends to zero as a —» oo. Noting that the flow is symmetric, the x% component of the equation (4.43) makes no contribution, so there are exact two equations. We remark that for high Reynolds number symmetric flow is unstable, so the above algorithm can be applied to low Reynolds number flow only.

Chapter 5

Perfectly Matched Layer Method

The artificial boundary conditions in the previous chapter can be regarded as an artificial material with zero thickness. Another approach to absorb the outgoing waves is to put a layer with a positive thickness. A special medium, not necessary existing in the real world, is designed to be filled in this layer, which can absorb most of the waves. This layer should bear two important features: 1. The interface must be penetrable so that no wave is reflected to the real underlying domain. 2. The special medium must have a damping behavior. We investigate some kinds of layers in this chapter. Functions in this chapter are complex.

5.1

Wave equations

Like Section 4.1, we start from one dimensional case and consider the wave equation, ~W=d^-

(5 1}

'

Suppose the domain is (0, oo), and the initial and boundary conditions are du -Q-

u\t=o = wo,

= MI,

x > 0,

(5.2)

t=o

u\x=0 = 0,

t > 0,

(5.3)

where uo(0) = 0 and ui(0) = 0. Let | ^ = p, and ^ = q, then the equation (5.1) is equivalent to dp^ dt

=

dq^ dx' 191

dq_ = dp_ dt 3x

K

' '

192

Numerical Methods for Exterior

Problems

We see that djp + q) dt

djp + q) _ dx

djp - q) dt

djp - q) dx

Therefore p + q are constants along characteristics x + t = constant, which are left going waves. On the other hand, p—q are constants along characteristics x — t = constant, which are right going waves. The solutions to (5.4) are the superposition of the right going waves (p, q) = (/(x — t), —fix — t)) and the left going waves (p, q) = ((7(2: + t),gix + t)), where / and g are arbitrary functions. By the initial condition (5.2), p\t=o = Po = u'o a n d q\t=o = qo = " i . We have / = i ( p 0 - q0) and g = \(pQ + q0). The left going waves touches the boundary x = 0, then the boundary reflects the wave to create a right going wave (p, q) = ( *' ^'^ \(fl(a; + t) + (7(t-a:), f l (a;-|-t)-ff(t-a:)), a; < t.

f5 5^ ";

l

We assume that uo and u\ vanish for x > a and put some damping medium on the interval (a, a + d). Artificial boundary condition is imposed at x = a + d. Then the approximate problem is: dp dq d-t = dx> dp

dq

m+ap=dx->

dq dp di = dx-> dq

q\t=o = 9o,

p\t=o = q\t=o = 0,
dp

-d-t+aq=dx->

p\t=o = Po,

*G(°'a)' , x

*(a>a

,.

+ d><

x G (0, a),

xe ia,a + d),

p | x = a - 0 = P\x=a+0,

P\x=a+d = 0,

where a > 0 is the damping constant. Noting that the interface x = a is penetrable, because p and q are continuous across the interface, so are the right going waves p — q and left going waves p + q- Therefore a wave can go through the interface without any reflection.

193

Perfectly Matched Layer Method

We set p = peat and q = qeat, then the equations on (a, a + d) become dp dt

dq dx'

dq dt

dp dx'

which are the same as (5.4). The right going waves satisfy (p-q){a

+ d,t) = (p-q)(a,t-d),

t>d,

that is (p -q)(a + d, t) = e~(Td{p -q)(a,t-d),

t > d.

Consequently a right going wave passing through the medium decays by a factor e~ad. The same is true for a left going wave. Let (p, q) = (f(x — t), —f(x — t)), t > 0, be a right going wave on (0,a). The interface x = a is penetrable, so (p, q) = eat(f{a — t), —f(a — t)), t > 0, at x = a + 0. As the wave is reflected at a; = a + d, and the left going wave reaches x = a, it is (p,q) = e^'^(-f (a + 2d - t), -f(a + 2d - t)), t > 2d, then (p, q) = e-2ad(-f(a + 2d - t), -f(a + 2d - t)), t > 2d, at x = a - 0. We see that the amplitude of the wave decays by a factor e~2"d. By (5.5) the exact solution at x = 0 is p(0,t) = 2g(t) = Po{t) + qo(t). Since po(t) = qo(t) = 0 for t > a, the exact solution p(0,t) vanishes for t > a. The value of p(0, t) is not disturbed by the reflection wave from the artificial boundary x = a+d when t < a+2d, so still we have p(0, t) — 2g(t), but it is disturbed when t > a + 2d. By induction we can see p(0,t) = {-l)me-2madp(0,t-2m{a+d)),

t e

(2m(a+d),2(m+l){a+d)),

for m = 1, 2, • • •. It seems the bigger ad the better, but if d large, the truncated domain is large, and the computing cost is high, while if a is too large, the difference of equations between two sides of x = a is too large, which would cause a large computational error and instability. There is another point of view. Let the Fourier transform of p, q with respect to t be p{x, —UJ) = F(p(x,t)), and q(x, —LJ) — F(q(x,t)), then the equations are ~iL°P=

sf >

~iuq=

g|,

x e (0,a),

(5.6)

and (-iw + a)p = —,

(-iuj + a)q = —,

x€(a,a

+ d),

(5.7)

194

Numerical Methods for Exterior

Problems

respectively. They are known as the time harmonic forms. By eliminating q (or p) they become second order equations d2p -^+u>2p

= 0,

xe(0,a),

and d2p —^-(-iu)

+ a)2p = Q,

x G (a,a + d).

The general solutions for them are p = ci(w)e^ x + c a ^ e - " * ,

x e (0, a),

(5.8)

and p = c 3 (w)e-(- i a , + t r ) a : + c 4 (w)e ( -^ + < 7 ) x ,

x G (a, a + d).

(5.9)

By multiplying a factor e~lut we find that the first term is a right going wave and the second term is a left going wave. From (5.9) we see that the first term decays to zero exponentially a s u +oo, and so does the second term as x —> —oo. By (5.8) and (5.6) we get

and analogously by (5.9) and (5.7) we get q = - c 3 ( w ) e - ( - ^ + C T ) x + C4(w)e(-*"+(ciHe"°,-CiHe"a). Simply we set

C%(UJ)

= ci(u>)e(Ta, then the right going wave on x > a is (ci(^)eaae-(--iu'+'T)x,-c1(uj)eaae{-iu'+'7)x).

(P. 9) =

p and q are continuous across the interface x = a without any reflection. Another observation is useful to design the damping layer. If we define a transform of independent variables: ( x,

\a+(l

X

1

•i^)(x-a),x>a,

then the equation (5.6) and (5.7) can be written in a uniform form

(K 1 f ) \

195

Perfectly Matched Layer Method

The solutions to (5.8),(5.9) can be expressed by p = di^e^

+

c2(u)e-iui,

which satisfies the equation (5.11). p is certainly continuous at x = a, therefore a right going wave (5.8) can pass across the interface x = a without any reflection, which again leads to the conclusion that the interface is penetrable. We turn now to investigate the two dimensional wave equation d2u _ d2u

d2u

.

+

[b U)

'W-dxJ dxJ-

'

Let p — (pi,P2) = Vw and q = § | , then the equation is equivalent to

We define the Fourier transform with respect to t and x2, p(^i,^2, —w) = F(p(xi,X2,t)) and q(xi, £2,— w) = F(q(xi, X2, t)). The equations for p, q are . -uoq=

dpj,

h«6P2,

OXi

By eliminating p we get a second order equation for q:

S + (^2-^)? = 0. dx2

(5.14)

The general solution is q = cl{w^2)ei^Z^x'

+ c2(u,,6)e-iv/^if*i,

(5.15)

and Pi = y i - ^

+c2(W,6)e-iv/^Ii?-i) ,

(-Cifa&jV*^*!

P2 = - § ( d ^ K ^ ^ *

1

+c2(W,£2)e-H/^Z^Xl) .

(5.I6)

(5.17)

The first term is a right going wave and the second term is left going.

196

Numerical Methods for Exterior

Problems

In the damping layer x\ € (a,a + d) the equations are dq

dp

T-,

_

(-iu> + a)q = — - + if 2 p 2 , 02; i

^2 + ("H" +

CT

) - « ) $ = 0.

(5.18)

The general solution is q = C3(w, ^ 2 ) e V ( - + - ) 2 - « 2 2 - i

+ C4(W)

^2)e-V^+^)2-«22-i;

Pi

19)

(5.20) i

+ c 4 (o;,6)e- ^ P2 =

(5

+iff)2

Xl

-^ ) ,

7 ^ - (c 3 (w,6)eV("+-) 2 -«2*i + c 4 ( W ,6)e- i V / (-+-) 2 -«2 2 -i) .

(5.21) We see the system for p, q consists of one algebraic equation and two first order differential equations for q and pi, thus the interface condition is q\x!=a-0

= Q~\x1=a+0,

Pllx^a-0 =

Pl\xi=a+0-

For a right going wave on x\ < a, the necessary condition for a right going wave on xi > a is

-Jly (

f. , c (^2K^ (ui + lay2 3

+ig)2

c -* 2Q = -A/l-4 i(^^2)eiV^Z«f«. V w^

This is an over-determined system for the coefficient C3, which causes difficulty for two dimensional problems. The interface x\ = a is not fully penetrable in this setting, and there must be a reflection wave going to the left.

197

Perfectly Matched Layer Method

On the other hand if we introduce a new variable x\ as the one dimensional case, -

J 3-1) 1 _

\a+

(l+i£)(xi

-a),

x

i

< ai

IK

Xl

>a,

^ '

no")

and replace x\ in (5.15) by xi,

which satisfies

S

+ (W2-^22)g = 0.

(5.23)

Then the interface is fully penetrable. Taking inverse Fourier transform, we can get the equation on the physical domain. However the equation (5.23) leads to a convolution product. To prevent this inconvenience some approaches have been given. We will discuss them at Section 5.2 and Section 5.4. If £2 = 0, these two equations(5.18) and (5.23) are the same. By the discussion in Section 4.1 we know that £2 represents the incidence angle. So if the incidence angle is small, the reflection in (5.18) is weak, otherwise the reflection is strong.

5.2

Berenger's perfectly matched layers

We investigate the two dimensional wave equation again. Following Berenger's idea [Berenger, J.P. (1994)], q is divided into two parts: q = Qi + a n d the system of equations (5.13) is equivalent to

dpi dt

dgi _ dpi_ dt dxi'

dq^ _ dp2 dt 8x2'

(5.24)

<%i + 92) dx\

dp2 dt

(5.25)

d(qi + q2) 8x2

In the perfectly matched layer x\ € (a,a + d) the system of equations is dqi

-m

dpi

+aq =

'

av

dq2

dp2

=

-m dx-2'

. . (5 26)

-

198

Numerical Methods for Exterior

dpi -df

+ api=

d(qi + q2) Bxx '

dP2 ~dt

Problems

<9(gi + g2) dx2 •

=

,_ 0 7 , (5 27) "

Applying the Fourier transform to the system, we get (-iw + a)qi = -—,

-iwq2 = i&P2,

(5.28)

(-iLJ + a)px = d(qi+q2\ -iwp2 = i^(qx + q2). (5.29) dx\ If we apply the transform of independent variables (5.22), we find (5.28),(5.29) is just the Fourier transform of (5.24),(5.25) with x\ replaced by x\. Following the argument in the previous section, we conclude that the interface x\ = a is penetrable. Now let Q, be an exterior domain as usual. Suppose a square fio = {x € 2 K , |:ri| < a, \x2\ < a} is large enough, so that f2o 3 Qc. We extend it to a larger square, Qd = {x £ E 2 , |xi| < a + d, \x2\ < a + d}. Then fid \ flo is the PML domain. The system of equations in the PML domain is designed by dqi

dpi

+a

-m ^ dpi . -d7 + aiPl

=

dq2

= dx-i'

d(qi+q2) ^x^>

dp2

+a

-m ™ dp2 -dt+a^

= dx-2' =

d{qx + q2) ^d^2—

(5 30)

'

(5 31)

-

Two damping parameters o\ and a2 are designed by a,=0, forkil 0, for \xj\ > a )

=

1|2.

If flo is a disk, {x; \x\ < a}, polar coordinates can be used. The equation is

d2u _ 1 d ( du\

1 (Pu

Introducing du

du

1 du

Perfectly Matched Layer Method

199

•

Xj

Fig. 7 we get the equivalent system of equations, dq 1 f d dt = r \d-r{rPl)

dpi ~~dt

=

+

dp2 \ ^9)

dq dp2 1 dq dv' ~~dT = rd~0'

Define the Fourier transform of (q,pi,P2) by {q,pi,p~2)(—u,r,6), system is

. .

i /s

ap2\

r \dr

06

then the

200

Numerical Methods for Exterior

^

1

=

dq , _— or,

Problems

1 dq --u« ^ , =- - - .

Introducing a new independent variable, r,

r < a,

a+(l

+ i^){r-a),r>a,

'

we replace the independent variable r by f to obtain

- ^ = Hl */ 0( ,f -p _~,l ) + ^dp2\J ' iiVpl

~

=

dg dr'>

iu,p2

-

1 dq ?de-

=

~

(5 33) (5 34)

'

The variable q is divided into two parts: q = q\ + qi, and q\ satisfies -iuqi

= -—r. (5.35) or Returning to the independent variable r, we deduce from the equations (5.33)-(5.35) to get -M[1+l-T-^)q=r{d-rirPl)

+

1

d

((

^e)+

r

~^^°PA

•

f:

-(

1 +

~aJ i

£) r -

•CT\ - = dpi -lu(l /-. + l-)q —, l

for r > a. We notice that f-(l

+ i^)r - — r

IUJ

=

aa — . r

Returning to the original independent variable (t, r, 6), we get the equations in the PML domain, dq

r- a

aa

1 / d ,

.

dp2 \

201

Perfectly Matched Layer Method

dq d-r>

dpi +<JPl

dt dp2 dt

r —a r

dqi -dt+(Tqi 5.3

=

1 dq r dt) dpi -dr--

=

Stability analysis

The system (5.26),(5.27) is not symmetric. It is not trivial to see if it is stable. Using the energy estimate, we study the problem in this section. We begin with a simpler case: the system (5.24),(5.25) with initial value imposed on M2. This system is equivalent to the system (5.13) plus an additional equation: dq_ _ cHn_ dp2 dt dx\ dx2' dpi dt

dq_ dxi'

(5.36)

dp2 _ dq dt dx2'

dqi dt

dpi dxi

(5.38)

Let us define the energy norm. Let U = (q,Pi,P2)T,

E(t) = [

du dt

(5.37)

2

+

du

2

dxi

+

dU dx2

and dx,

and 5i (

lo= / lldxJUL2

Lemma 67. / / the solutions to the system (C([0,co); J ff 1 (R 2 ))) 3 n(C 1 ([0,oo);L 2 (E 2 ))) 3 , then the following are valid: E(t) = E(0),

qi

(5.36)-(5.38) are U G e C([0,oo);i 2 (E 2 )) ;

(5.39)

202

Numerical Methods for Exterior

Problems

||9i(i)llo<||5i(0)||o+ I E(r)L2dT. (5.40) Jo Proof. We may assume that the solutions are sufficiently smooth, then use a series of smooth functions tending to the solutions to achieve the conclusion. Taking first order derivatives in the equations (5.36),(5.37) gives d2q _ d2pi dt2 dxidt d2pi dt2 d2p2 dt2

d2p2 _ d2q dx2dt dx\

d2q dxidt d2g dx2dt

=

d2pi dx\ =

d2q dx\'

d2pi dx\dx2^

d2Pl dx\dx2

d2p2 dx\ '

Multiplying the equations by ^f, -J^, and -j^- respectively, then integrating on E 2 , we get

±± [ (^Ydx = -1-± f (°L)2dx-l± 2 dt JR2 \dt)

2dt JR2 yOxJ

[

J®2

dt

ox1dx2

^l^^dx

JR2 dt

R

2dt Jw V dt J

(*Xdx

2 dt Jm \dx2J

1 ±± I (^)2dx = -2dtJ -±2\dxiJ [ (^Ydx+ 2dtJ 2\dtJ U

f

2dt JR2

dxidx2

\dx2)

Applying the equations (5.37) and integrating by parts gives r dp! d2p2 7H2 dt dx\dx2

dx_

r d2px dp2 Jyp dxzdt dx\

dx

JR2 oxiOt oxi

=

r

2 dt yK2 \ox\ J

Analogously it holds that

JR2 at dxidx2

2dtJR2

d2q dp2 J^pdxidxzdxi

\dx2J

dx

203

Perfectly Matched Layer Method

Adding these three equations together yields 1 dEit) = 0, 2 dt which implies (5.39). Then (5.40) is a direct consequence of the equation (5.38). • The estimate (5.40) is classified as a "weak stability", since the L 2 -norms of qi, q2 are not only bounded by the L 2 -norms of the initial data, but also by the L 2 -norms of the derivatives of the initial data. This property may cause instabilities in numerical computation. The energy estimate for the system (5.26),(5.27) is more technical. The Gronwall inequality is needed here: Theorem 67. / / a function E(t) > 0 for t>0,

and satisfies

E(t)
(5.41)

Jo with two constants a, b, then E(t) < beat for Proof.

t>0.

We multiply the inequality (5.41) by e~at and get e-at(E(t)

-a

f E{r)dT) < be~at.

Jo

It is dt{e~at

f E(r)dT) Jo

Then we have • f E(r)dT
e-aTdT=-(l a

[ Jo

We substitute it into (5.41) and the conclusion follows.

•

The system (5.26),(5.27) is equivalent to

dl+aqi d

Pi

, „

n

-W+(7Pl

=

dq =

d^'

d^

+

(5 42)

^ dp2

-

dq

~dt=8^'

(5 43)

-

204

Numerical Methods for Exterior

Problems

% + -*-&

<"«>

Theorem 68. / / the solutions to the system (5.42)-(5.44) are U G {C({0,oo);Hl(R2)))3f](Cl([0,oo);L2(R2)))3, e C([0,oo);L 2 (R 2 )), Ql then the following are valid: E(t) < C(a)e C r t (i?(0) + \\qi(0)\\20),

llftWIlo < e- fft ||?i(0)|| 0 + / E(r)i

(5.45)

dr,

(5.46)

Jo

where C{a) depends on a. Proof.

Being analogous to the previous lemma we obtain equations, d2q

d2pi dt2

d\

d\

d2pi dx\

=

d2p2 dt2

=

dPl

,

d2p2 _ ^dqi_ _ ^dpx dx\dx2 dxi dt ' d2Pl +(Pp±_ dx\dx
^ 8x2

Multiplying the equations by |f, ^ - , and ^p- respectively, then integrating on Rl22, we gt get 1 dE(t) 2 dt

f f 2adpi9q dxi

7K2 \

/'c^iVdt )

dqi dp2 dx2 dt

dt

I a2qLdqdt a9qidxi dp2 dpi dxi dx2

dpi

dt

-(£)>*£*•

Then we have 1

-(E(t)-E(Q)) = J2[th.

1

(5.47)

k=iJo

Let us estimate the terms on the right hand side of (5.47). We have the following:

205

Perfectly Matched Layer Method

-2af( Jo CT

d£

^ d x d t < OX! dt

2

JR

2

Jo JR \oxiJ

Jo JR R \ 2

ot

dxdt, J

dq^2

/ / 1*7T dxdt < — I Jo JR2 ot 2 J 0 JK2

all Jo i s

OX! OX2

dxdt % [ I ( ^ ) , * Jo R V dt 2

JR2

fl^.)dp2 dxdt +i

p.p.dxdt
2

qjdxdt-

2 J0

f

[

(OElXdxdt

JM2\dX!j

Noting the equation (5.44) and taking integration by parts, we obtain -a

j / ———^dxdt Jo JR2 ox! ot

d ^ A \ +a f f/ d2-——pidxdt ^ A A+ = -cr f -—p!dx 2 JR ox! o Jo JR2 ox!dt

f ^ / x =o I
2

~

\2eJRR

1

f

( °^ ^ M A A+ \-cr\-^-T]p1dxdt Jo JR2 \ ox! dxf J 9 f f ( P1 {dPA2\ A A, 2 dx Jo JR I i \oxiJ I

2 y E2

[ /

(dpiX \dx!j qi dx dt,

where e > 0 is a small constant. Analogously we have -a

/ —-^-—^-dxdt Jo Jm.2 0x2 ot

f dqi , * f* f d2qi J J = - a I 2 -—p2 dx + er / / 2 - — — p 2 dx dt JR 0x2 Jo JR ox2ot 0 =a

f ^ , ' f f ( °
=v JR2

9

P ^ A \ f f ( °P* qi-x~dx +a / [aq! ox2 J0 JR2 \ dx2 0

d

Pl°PA A A, ^—-—) dxdt ox! dx2)

206

Numerical Methods for Exterior

Problems

<
+

. r r (P)'**+* r

ftp)'**.

Jo JR* \dxxj 4 Jo Jw We plug these estimates in (5.47) and obtain

\dx2J

+ a)E(0) + 2a\\qi(0)\\20+4a

(l-ea)E{t)<(l

+ 2a3

/

qfdt+

Jo J«L2

—

f E(r)dT Jo

(5.48)

q\ dx.

e

Jm.2

We apply the equation (5.44) to estimate q\. qi(x,t)

=qi(x,0)e-^+

f' e^(t-r)dPi(^r) dxi Jo

^

Therefore we obtain dpi

ll«i(t)||o < |ki(0)||oe-'*+ Te—(*—> dx\ W Jo

dr,

(5.49)

and by the Schwarz inequality we get \qi(t)\\l<2\\qi(0)\\le-^

dpi (r) dxi

+ 2 ^ e , -- " °( * - T )

•t

<2\\qi{0)\\20e-2at + 2 [ e-^-^dr Jo <2||«i

2„-2
f Jo

dpi (r) dxi

+
dpi (r) dxi

e^'^

dr

dr.

Then we have rt

f \\qi(s)\\2Qds<-hi(0)\\o Jo a

+ l I ds <* Jo

Jo

OXi

['e-*-^

OXi

dr

207

Perfectly Matched Layer Method

We plug these inequalities into (5.48) and set s = l/2a to obtain -E(t)

< (1 + a)E{0) + 16a

Li

f

E{T)dT + (2a + 10ar2)\\qi

Jo Then applying the Gronwall inequality we obtain (5.45), and (5.46) follows from (5.49). •

The fact of weak stability may cause unstable finite difference schemes. The second order Yee's scheme in Section 4.7 is stable for Maxwell equations. However following an analysis of [Abarbanel, S. and Gottlieb, D. (1997)] we will show that it is unstable as it is applied to the system (5.24),(5.25). The system of finite difference equations is + +5 K + 1 =- P? + Atf 1 (<£ '+92 ).

9i

n+i Q2 '

= 9i

AtS2(q?+*+qZ+*),

••P2 +

P2+1

n—i

92

'

+ At62p21,

n+i

where p™ = pa\t=nAt, qa 2 = qa\t=(n+$)Au a = !> 2 > a n d <*<* is the finite difference operator A/Aa; Q . The matrix form of the system is /10-Atfi 0 1 -AtS2 00 1

Voo

-AtSi\ -AtS2 0

o

i

Pn2+1

/

/ '

1 0 AtSi

0 1 0

V 0

0 0 \ ( Pi \ P? 00 n-I 10

At82Ql)

v?r*y

The Fourier symbols of d\ and S2 are, respectively 2i sin 6 Axx

Si

where -n/2

<6,^<

o

At'

n/2. The amplification matrix is

/ 1 0 -iki -ih \ 0 1 — ik2 —ik2 G = 00 1 0

\oo

2i sin < Ax2

. k\ ~ ~At' l

i J

/ 1 0 iki

0 00\ 1 00 0 10

V 0

ik20lj

(\ — k\ —k\k2 iki iki \ —kik2 1 — k•22 ik2 ik2 0 1 0 iki

V

0

;/c2

o i /

A necessary, but not sufficient, condition for the stability of the difference scheme is that all the eigenvalues of G are less than or equal to 1 in magnitude (the von Neumann condition). The eigenvalues X's of G are A = l,l,l

7

7 = M,Ai,A2,

208

Numerical Methods for Exterior

Problems

where 7 = kf + k\. Under the CFL condition, 7 < 4, then X's satisfy |A| = 1, so the scheme meets the necessary condition of stability. However it is unstable. The Jordan canonical form of G is: /Ai 0 0 V0

J

0 0 0\ A2 0 0 0 11 0 01/

That is, G = TJT 1, where T is the "Jordanizing" matrix. After n steps, we have Gn = TJnT~\ and

n

J

fX^ 0 0 0 ^ 0 A£ 0 0 = 0 0 In

V0 0 0 1 / The L 2 -norm of J " is | | J l = m a x | | J " £ | | = (n 2 + l ) i Thus ||G n || grows with n without a finite bound, and the scheme is unstable. 5.4

Uniaxial perfectly matched layers

Since Berenger's perfectly matched layer method yields a system which is weakly stable only, it is desirable to replace it to a form without splitting the field. Let

u

+ i-, xi > a, x\ < a,

then the equations (5.28),(5.29) are equivalent to . -iuq= . -

-tujpi

-

1 dpi . \-t&P2, s oxi

1 dq

--—, S OXi

. .

-iujp2

=

(5.50) ...

i&q,

(5.51)

where q — <ji + q~i- The important matter is to derive the corresponding equations in the physical domain. To this end, introduce two variables P =

Pi

Q =

209

Perfectly Matched Layer Method

and replace the equations (5.50), (5.51) by dP -iujq = —— + i£2p2, ox i -iwpi = -—,

( 5 - 52 )

-icjp2 - ibq-

(5.53)

OX i

Taking inverse Fourier transform, we get the equations on PML: dt

dxx

dpi _ dQ_ dt dxi' dq

dQ

V

dx2 dp2 _ dq dt dx2'

„

dj>!

'

(5.55)

dP

This is the "uniaxial perfectly matched layer" (UPML). We remark that the system (5.52),(5.53) is different from (5.50),(5.51). Although they are the same for x\ < a and x\ > a, s is a piecewise constant function of x\, so the interface boundary conditions at x\ = a for them are different. Let us derive the interface boundary conditions to make a comparison. We multiply (5.50) by — iuj and obtain -w

q = -*«>--

h 0J&P2,

s dx\

then by (5.51) we get the equation for q,

On the other hand by (5.52),(5.53) we can get 2-_

d

(I

d

(q\\

e~ c2

-^*-srU^wr *

<558)

-

There are two first order differential equations for pi and q, and one algebraic equation, in (5.50),(5.51), therefore the interface condition is: P l | x i = a + 0 = Pl\x1=a-0,

9 | T 1 = O + 0 =
On the other hand the interface condition for (5.52),(5.53) is: P\xi=a+0

= P\xi=a-0,

Q|xi=a+0 =

Q\xl=a-0-

Therefore (5.52),(5.53) represents a model different from (5.50),(5.51). This fact has to be taken into account when a finite difference scheme, or a finite element scheme, is designed on the entire truncated domain, including the PML domain.

210

5.5

Numerical Methods for Exterior

Problems

Maxwell equations

For two dimensional TM mode (4.18), if we replace (e/z)5:r by x and set q = f-J

E3, pi = H2, and P2 = —Hi, then it is identity to (5.13).

Analogously, for the TE mode (4.19) we replace {e^i)^x by x and set q = — ( M ) 2 H3, pi = —E2, and P2 = Ei, then it is identity to (5.13), too. Therefore the above approaches can be applied. We turn now to consider three dimensional Maxwell equations. Let Clo be a cube {x; \xk\ < a,k = 1,2,3} with the constant a large enough so that Qo 3 ^ c - Then take d > 0 and define a cube, fi 0, k = 1,2,3, for \xk\ £ (a,a + d), with the relationship of /i '

£

Let us assume ak = cr£ = 0 for \xk\ < a, then the system of Berenger's perfectly matched layer scheme is defined by s§-t+a2^Ei2

e§-t+a3)Ei3

=

^-(H3i+H32),

= -1^(H23

+ H2i),

4t+ £

l

+

^ E 2 1 =

s§-t+a^E3i

-^(H31+H32),

= ^-i(H23

+ H2i),

»—, +t + ^ 2Hl2 ) #12 = = - — ( £ 3 1 + £32),

•m V

-tk'

Perfectly Matched Layer Method

211

4+^)^13 = ^(^3 +^ ) ,

4t+°3)^

= -~(En + E13),

4t+^Hn

= ^-i(E31+E32),

+<*) ^ 2 = ^ 1 2 + E13).

4 Let

Oh

Oi

We define Fourier transform with respect to t: E(x, —u>) = F(E(x, t)), and H(x, — u>) = F(H(x, t)). Define new independent variables Xk = Lk Sk dxk, and V - (— \ dii'

— — dx2' dx3

Analogous to (5.50)(5.51) the equations can be written by -ILJEE

= V x H,

(5.59)

-icjfiH = - V x E,

(5.60)

without splitting the field. We turn now to the UPML scheme. Define Ek = Sfc-Efc and Hk = SkHkEk and Ek, Ek and Ek, are the same if Sk = 1, but they are different in the PML. Replace the equations (5.59) and (5.60) by -lue \

'S2S3SI"1

°

0

\

0

ssSiS^1

0

\E = VxH,

0 S2S3S^ ILJH

0 0 0

0 0 S3S1S2 S3SIS21

0

(5.61)

si^sJT/ 0 0 0

\H

S1S2S3

1

= - V x E.

(5.62)

212

Numerical Methods for Exterior

Problems

In order to take inverse Fourier transform we introduce some new variables. For example for the first row we set D\ = es^s^Ei, then the equations are dDi dt

_ 8H3 dxn

|

8H2 dxz '

dD1 a, 8E, -57- + —Dt = e--- + azEx. at e at 5.6

Helmholtz equations

Following the analysis by [Collino, F. and Monk, P.B. (1998)] we consider the exterior boundary value problem of the Helmholtz equations: (A + w 2 )u = / ,

isfl,

(5.63)

with the boundary conditions, u = 0,

xGdQ,

^ _ ^

= 0(1),

(5.64)

(5.65)

in this section. We assume that / £ L 2 (Q), and it is compactly supported. Let R be large enough, so that B(0, R) D supp / . The solution for \x\ > R is thus expressed by (2.16). Introduce a complex radius f = JQ a(p)dp = r(3(r), where a(r) = 1 + icr(r), and the real function a{r) satisfies w

\>0,r>it,

r-+*>JR

We replace r in (2.16) by f then we get another function, 00

u=

Y,

anH^(cjf)eine,

(5.66)

n = —oo

for r > R. By (2.12) the asymptotic behavior of the Hankel function is

V 7rwr and iujf = iuir — wlm f. So H^ '(u>f) decays to zero as r —> 00.

Perfectly Matched Layer Method

213

Obviously u satisfies the equation 1 ( d („du\ . ld2u\ r\6r\f8f)+-fW)+^

,

2_ = 0

-

^

Returning to the original coordinates (r, 6), we have the equation, 1 ( d (p

r

du\ +

a d2u\ +a U

a

2_

r{0-r{a ^) -pW) ^

n

= °-

In Cartesian coordinates the equation is V • (AVu) + a(3uj2u = / ,

i£(l,

(5.68)

where £ cos2 6 + | sin 2 6 cos<9sin0 ( § - | Y c o s 6 » s i n 0 f | - | ) £ sin2 61 + | cos2 61 The function u is the solution to the equation (5.68) and the boundary condition (5.64), and it satisfies u bounded as \x\ —+ oo,

(5.69)

by noting the above asymptotic behavior of the Hankel functions. Obviously we have Lemma 68. The solution u of (5.68), (5.64), (5.69) is equal to u of the problem (5.63),(5.64),(5.65) on B(0,R)f]fl. Moreover we have the following Lemma 69. The solution u to the problem (5.68), (5.64),(5.69) is unique. Proof. we get

If / = 0, by the Green's formula on the domain fix = QnB(0, f

(-du

du 1 ,

i, rt {"*-5:}*-

a

R),

„

( 0)

"

For r > R, u is the solution to (5.67), so it can be expressed in terms of Hankel functions, oo

*{r,8) = £ > n f t f > ( u ; f ) +

bnH£\wf))eine.

214

Numerical

Methods

for Exterior

Problems

Since we assume that a is chosen so that Im (ur) —* oo as r —> oo, the Hankel functions Hn {uif) are unbounded. Therefore bn = 0. By the integral (5.70) we get oo

\an\2{H£\ujR)Hnl)\wR)

^2

- Hk1](wR)H^'{wR))

= 0.

n= — oo

The Wronskian of Hankel functions does not vanish, so an — 0 for all n. Then u = 0 for r > R. u is an analytic function, so u = 0 for r < R, too.

• Since the solutions to (5.68),(5.64),(5.69) decay to zero as |a;| —• oo, the domain can be truncated. We assume that the thickness of the perfectly matched layer be d and denote fi<j = fl (1 B(0, R + d) and f^2 = B(0, R + d) \ B(0, R), then we get the approximate problem: V • (AWuh) + aj3uj2uh = / , uh = 0,

x G Q.d,

(5.71)

xG dQ,

uh = 0,

\x\=R

(5.72)

+ d.

(5.73)

We are going to investigate the well-posedness of the problem (5.71)- (5.73). Theorem 69. For almost all LJ2 £ (0, oo), except for a point spectrum set a' the problem (5.71)-(5.73) admits a unique solution. Proof. We consider the operator L = —V • (-AV) + A. We claim that for positive A the corresponding sesqui-linear form, a(uh,v) = (AVuh,Vv)

+

(Xuh,v),

is coercive. To see that, for any u £ H^Cld), \(AVu,S7u)\

>Re{AVu,Vu)=

[

\Vu\2 dx +

Re(Vu • AVu) dx.

By the expression of A,

J

Re(Vu • AVu) dx R+d

JR

r-2-K

JO

r | Re ( £ a

du dr

2 +

I t e |

^

du ~d~e

2>

dBdr.

215

Perfectly Matched Layer Method

It suffices to show that min(Re(a//?),Re(/?/a)) > 8 > 0, where 6 is independent of r. Let (3(r) = 1 + i 1 + m a x j j < r < f i + ( i Re((^Vu, Vu) + \(u, u)) > S\u\l + A||u||o > min(<J, A)||u||i. By the Lax-Milgram theorem there is an inverse operator of L. Let A be a fixed number, the equation (5.71) can be written as: uh = L~\[X + a(3uj2)uh - / ) = (A + a(3u2)L-luh

-

L~xj.

L"1 is a bounded operator from L2(Qi) to ii" 1 (Qi), then by the compact embedding of Sobolev spaces, it is a compact operator from L2(fli) to L 2 (f2i). Applying Theorem 4, the spectrum includes point spectrum av only. Therefore if \ + a0u2 is not included in ap, the problem (5.71)-(5.73) admits a unique solution. We define a'p = {ui2 £ (0, oo); A + a(3cj2 G crp}, then the conclusion follows. • Generally speaking 2 is the eigenvalue of the Laplace operator. For small d the spectrum is a small perturbation of the spectrum of the Laplace operator. However for large d there exists a unique solution. See [Bramble, J.H. and Pasciak, J.E.]. Error estimates of this scheme can be found in [Chen, Z. and Liu, X.].[Bramble, J.H. and Pasciak, J.E.]. The analysis of this scheme for the time-harmonic Maxwell equations can be found in [Bramble, J.H. and Pasciak, J.E.][Bao, G. and Wu, H. (2005)].

Chapter 6

Spectral Method

The spectral method aims at using an orthogonal set of smooth basis functions to construct approximate solutions. The fast Fourier transform (FFT) is a powerful tool in scientific computing. One of the advantage of the spectral method is the application of FFT. Another advantage is that the method enjoys a theoretical spectral convergence rate. Sometimes the order of error estimates can be as high as desired provided the exact solution is smooth enough. According to the chosen of basis functions the method has the following ingredients: trigonometric functions, orthogonal polynomials, or orthogonal rational functions. For a comprehensive study of the spectral method we refer to the books [Guo, B. (1998)] and [Shen, J. and Tang, T. (2006)], and for the application of the spectral method to exterior problems we refer to the survey [Guo, B. (2002)], and references therein. Following the analysis of Guo, Shen, and their co-workers, we will restrict ourselves to some examples in this chapter to show this method in the simplest form. 6.1

Introduction

For simplicity we denote d£u = -^u(x) in this chapter. To show how the spectral method works for exterior problems, let us see a model problem first. Let A = (0,oo). We consider the following ordinary differential equation on A with a boundary condition at x = 0: -dlu + u = e~x,

u(0)=0.

(6.1)

The solution is u=-±xe-*

+

C(e-x -ex),

217

(6.2)

218

Numerical Methods for Exterior

Problems

with an arbitrary constant C. Let us impose one boundary condition at the infinity by two means: either we assume that u is bounded, or we define the corresponding variational formulation: Find u G HQ(A), such that />oo

a(u,v) = / Jo

e~xv(x) dx,

Vv G

HQ(A),

where a(u,v) = J0°°(dxu(x)dxv(x) + u(x)v(x))dx. Then C — 0 and the unique solution is u = — \xe~x. We look for a polynomial p G PAT (A) to approximate u. The first glance is that it is impossible, because all polynomials, except constants, tend to the infinity as x —> oo, and the solution u decays exponentially. However it is still possible if we introduce some weights. Letting (3 > 0, multiplying the equation (6.1) by e~l3xv, and applying the Green's formula, we get /•OO

y*oo

/ e-px(dxudxv-/3dxuv + uv)dx= / e-((3+1)xvdx. (6.3) Jo Jo Thus we introduce a weight function, LJ(X) — e~®x and define a weighted L 2 -space, /•OO

L 2 (A) = {u G LL(A); / Jo

o;W2^
with norm ||u||o,w = ||w3u||oFor any non-negative integer m and weight x, we define the weighted Sobolev space #™(A) = {v;d*v G L£(A),0 < fc < m} equipped with the following inner product, semi-norm and norm (u,u)m,x,A=

5Z

(^S'^Kx.A,

Mm,x,A = | l # > l k x , A >

0
|M|m,X,A = (^)m, X ,AFor any real r > 0, the space H^(A) and its norm ||v||r,x,A are defined by space interpolation. [Adams, R.A. (1975)] We return to the expression (6.3). With regard to the boundary condition, a subspace is defined by tf«L(A) = { u e f f i ( A ) ; u ( 0 ) = 0 } .

219

Spectral Method

We define a new bilinear form, /•OO

a(u,v) = / Jo

e~/3x(dxudxv

-/3dxuv+

uv)dx,

then the new variational formulation is: Find u € HQ^{K),

(6.4) such that

/•OO

e~(l3+1)xvdx,

a(u,v)=

to

(6-5)

€ #£ W (A).

We notice that / Jo

e-^xf3dxuudx

<^ / 2 70

e-/3x(dxu)2dx+^

f 2 70

e'^u2

dx.

Therefore owing to the Lax-Milgram theorem if /? < 2, the problem (6.5) admits a unique solution. The space P/v (A) is a finite dimensional subspace of H^(A), so the following spectral scheme also admits a unique solution: Find p e PJV(A), such that p(0) = 0, and /•OO

e~^+^xvdx,

(6.6) to € PJV(A),u(0) = 0. Jo The condition /? < 2 is natural, because if /? > 2, there are infinitely many solutions (6.2) in the space HQW(A). To solve (6.6) numerically, one needs a basis of the space PN(A). This is the subject of the next section. One will see the Laguerre polynomials are suitable for this problem. Therefore applying the spectral method to solve an exterior problem, one needs a finite set of basis functions, a functional space which involves the analytic solution and the the given finite set, and a variational formulation in which the analytical problem and the approximate problem are both well posed. The solution p to the problem (6.6) can be a good approximation to u near x = 0, but it is a poor approximation on the far field. To improve the result, one can make a transform of variables. Let u — ex/2u, then it is the solution to a(p,v)=

-dlux + dxu + -u = e~x/2. 4 We define a weight w = e~x, and a new bilinear form, /•oo

a(u,v) = / Jo

2

e~x(dxudxv+-uv)dx, 4

220

Numerical Methods for Exterior

Problems

then it is easy to see that the corresponding variational problem admits a unique solution in HQUI(A) with w = e~x, and the spectral scheme is: Find p G PJV(A), such that p(0) = 0, and /•OO

a(p, v)

e~3x/2vdx,

/ Jo

to

which also admits a unique solution in variable u we have ||U - e-*'M\

G PN(A),v{0) PJV(A).

= 0,

(6.7)

Returning to the original

= R ( e - * / 2 « - e~*'2p)\\l + \\e~x'2u -

e^2p\\2

= R ( £ - P ) | | L + J||«-P||L> so it is just the approximation in the space H1(A). Another approach is to transfer the exterior domain into a finite one. Let z = - 2 1 n ( l - £ ) + 21n2, then the domain is transferred to A = (—1,1) and the equation (6.1) is transferred to 1 4

-diu+T^id*u

+

(i_g)2M = L

We can define a bilinear form, a(u, v) =

((1 - £)2d£iid£V - (1 - Qd^uv + Auv) de,

and define a weighted Sobolev space iJj(A) in accordance with the inner product, <(» • • » = / ' , ((1 - Z)2d€ud(V + uv) d£. Since

the corresponding variational problem admits a unique solution, that is: Find u G #£(A), such that u ( - l ) = 0, and a(u, v)=

f

v d£,

to

G ^ ( A ) ' v ( - 1 ) = °-

221

Spectral Method

The spectral scheme is: Find p 6 PAT (A), such that p ( - l ) = 0, and a(p,v)=

[

vd£,

VwGPjv(A),t;(-l) = 0)

which also admits a unique solution. We will see that the Jacobi polynomials are suitable for an orthogonal basis with some weight. The third approach is applying rational and irrational functions. Let us introduce a weight UJ = , K^ and define a bilinear form in analogous to (6.4), f°° 1 / -~-——^{dxudxv

a(u,v)=

2 —dxuv

The corresponding variational formulation is: Find u 6 Z"00 Jo

+ uv)dx. such that

HQU(A),

1 x

\ + -U

L e m m a 70. There exists a > 0, swc/i t/iat o(u,u)>a||u||5 i ( i ; ) Proof.

V«e^u(A).

We may assume that u € C 1 (A). Let e € (0,1), then

Jo /o

-dxuudx (^ + 1) 3 T00 1

1

f°°

1

Let c > 0, then a(u,u)>e Jo +

-—-—=(dxu)2dx {x + l)2

-——-=u2 dx {x + iy

+ Jo

1 f°° 1 2'da; i-£y0 (x+i)4u ( l - £ ) ( z + l)2

uf + u2

dx

(l-e)(a; + l)2

dx.

For the second term on the right, we take a large c, such that n-eVc+n 2 < 1, that is c > (1 — e)~ 2 — 1, i.e. for a positive constant ci > 0, c > cie. For

222

Numerical Methods for Exterior

Problems

the first term we have

I jdwu2

dx=

f idw (fdtU(t) dt)

dx

* f jdw [{dtu{t)fdt dx=IoidMt))2 r2

<

^

dt

[ idwdx

J\dtu{t)fdt.

We take a small c, such that 2{i-e) < Tc+W ^ a t *s °2 < £ ( c + i ^ ' *-e- ^ or a positive constant C2 > 0, c < C2£1^2. Let e be sufficient small, so that the chosen of c is possible. Thus a(-, •) is coercive. • Owing to the Lax-Milgram theorem, the variational problem admits a unique solution. Polynomials are no longer in this space HQ W (A). However it is possible to use rational functions. We define a finite dimensional space,

VN = (R^MX) = v ( f ^ j ) ." e fiv(-i, i)}, then Vivo = {u G Viv;u(0) = 0} is a subspace of scheme is: Find u^ £ Vivo, such that

f°°

The spectral

1 e Tucfa:,

a(Mjv,w)= / 7o

HQW(A).

Vv e V/vo,

(.^ + J-J

which also admits a unique solution. We will see the Legendre rational functions, Ri(x) = y/2Li ( f r y ) - generate an orthogonal basis in Vjv with the weight w. The definitions of Legendre functions L\ will be given in the next section. Another approach is introducing the weight w(x) = /^,lx+1-. • We will see in the next section the rational functions generated by Chebyshev polynomials form an orthogonal basis of functions. Multiplying the weight u and a test function v to the equation (6.1), and integrating by parts, we define a bilinear form a(u, v) =

dxudxvuj dx + 2

Jo ^

rOO

(vdxu — udxv)dxw Jo

/(> g\

,-00

— — / uvdxoj dx + 2 Jo Jo

dx l

uvu> dx.

b

-

8

;

223

Spectral Method

The corresponding variational formulation is: Find u £

HQ!U;(A),

such that

/•OO

e~xvujdx,

a(u,v)=

Vu € f #

(A).

(6.9)

Jo The point x = 0 is a singular point in the integrals in (6.8). To verify these integrals are well defined, we need the following lemma. Lemma 71. Ifve

then

HQ^A),

Proof. It is not obvious that the left hand side is well defined either. We will prove the convergence of this integral in the process of estimation. It has no harm in assuming v € CQ°(A) first. We notice that 3x + l OT-UJ

=

;

(jj,

2{x-\-l)x

'

then we have

M„{x)=£at(m^dt

v2{x)

-a

2VMdtv{tMt)

_ ^ ) w ( t ) + ^-Qdtw(t) ) dt

Using the Schwarz inequality we find v2{x)

fx v2(t) 5t + 3

X

<

r ^^(t) dt + r (dtv{t))2u(t) dt.

Letting x —> oo, we get the desired inequality. Since #3iU,(A), the results hold for v <= H^(A). Using the Schwarz inequality and the inequality 2

*

W

15x2 + lOx + 3 15 w_ ~ ~ 4(x + l) 2 x 2 ^ - T ^ 2

CQ°(A)

is dense in D

224

Numerical Methods for Exterior

Problems

we can estimate the terms in (6.8). For example f°° / vdxudxu

( f°° \ i ( r™ (dxu>)2 ^ 2 dx )'?»* At the same time we have proved that a(-, •) is bounded, |a(u,v)| < Af||«||i,„||vjli,„. To verify that the problem (6.9) admits a unique solution, we need to prove a(u,u)>a\\u\\liU,,

a > 0,

which needs a little more careful investigation. Let us verify 15a;2 + 10a; + 3 14 , 2 2 8(a: + l )-r— -< — 27 x -\

15 32',

15a;2 + 10a;+ 3 /(*) = 8(x + l ) 2

14 27

x > 0.

(6.10)

Let 2

15 32'

then we have

f

J[)( x )

-

5x + 1

~2(x

1f»(x) [

+ l)3

'

=-

5 x

2

-

1

X

27' + 1

(a; + l)4

f>tl)=0

I[

2>

'

-^<0 27

Hence ± is the only root of f(x) in [0,oo). Thus f(x) < f{\) which implies (6.10). By (6.8) we get /•OO

-1

/*00

< 0, Va; > 0,

/*00

a(u, u) = / ( ^ u ) 2 ^ dx — - I u2d2u> dx + / u 2 w da; Jo 2 J0 Jo 1 Z-00 13 f°° 1 > — / (dxu)2uj dx + — j u2u;dx>— WuW2^.

225

Spectral Method

Owing to the Lax-Milgram theorem the problem (6.9) admits a unique solution. The finite dimensional space VJvo defined above is also a subspace of HQ W ( A ) , and the spectral scheme is: Find WJV G ^ivoi such that /•OO

a(uN,v) = / ./o

e~xvu>dx,

Vu € Vjvo,

which also admits a unique solution. We will investigate the spectral method for more complicated exterior problems in the following sections.

6.2

Orthogonal systems of polynomials

Let A = (—1,1). The Legendre polynomial of degree I is

Ll(x)={-^fdUl~x2)1. It is the Z-th eigenfunction of the singular Sturm-Liouville problem, dx((l-x2)dxv)

+Xv = 0,

xeA,

related to the l-th eigenvalue A/ = 1(1 + 1). Clearly Lo(x) = 1, Li(x) = x, and they satisfy the recurrence relations,

(21 + l)Li(x) = dxLl+1(x)

- dxLi-!(x),

It can be checked that Li(l) = 1, Li(-l) dxLi(-l)

=

I > 1.

= (-1)1, dxLt(l)

= \l(l + 1), and

+i

. III + I 1). 1 I l\/l A f A / - n r A T « '^^1(1 Moreover

\Li(x)\<\,

dxLt(x)<

^1(1 + 1),

16A.

The set of Legendre polynomials is the L 2 -orthogonal system in A, i.e.,

L

Lt(x)Lm(x)

dx=(l

/A

v

+1 ) 2

By integrating by parts, we derive that j (dxLi(x)f JA

dx = 1(1 + 1).

Slm.

(6.11)

226

Numerical Methods for Exterior

Problems

The Legendre expansion of a function v € L2(A) is oo

v(x) =

^viLi(x), 1=0

with + -J

vt=[l

v(x)Li(x)dx.

For the rational functions Ri(x) = y/2Li ( f x j ) , defined in Section 6.1, we make a variable transform in (6.11) to obtain

i / Jo

„ . _ , . .

Ri(x)Rm(x)

/.

dx = [I + -

v-1

<5/m.

Therefore the set of i?/ is a weighted orthogonal system. Let the weight w(x) = (1 — a; 2 ) - 2. The Chebyshev polynomial of the first kind of degree I is Ti(x) = cos(Zarccosx). It is the l-th eigenfunction of the singular Sturm-Liouville problem, dx ((l - x2)^dxv\

+ A(l - x2)-iv

= 0,

i£A,

related to the Z-th eigenvalue A/ = I2. Clearly To(x) = 1, T\{x) = x, and they satisfy the recurrence relations, Tl+1(x)=2xTi(x)-Tl-1(x), 2Tt{x) = j^dxTi+^x)

I > 1,

- jL-d^i^ix),

I > 1.

It can be checked that T ; (l) = 1, T j ( - l ) = ( - 1 ) ' , dxTi(l) dxTi(-l) = {-l)l+1l2. Moreover \TL{X)\ < 1,

dxTi{x)
= I2, and

xG A.

The set of Chebyshev polynomials is the L^-orthogonal system in A, i.e., / Ti(x)Tm(x)uj(x)

dx =

-ciSim,

Spectral

227

Method

where c0 = 2 and c; = 1 for I > 1. The Chebyshev expansion of a function v e Ll(\) is oo

1=0

with v; = — / net JA

v(x)Ti(x)u)(x)dx.

We can define rational functions, Ri(x) =Ti( f^y J, then they generate a weighted L 2 -orthogonal system on the interval (0,oo), i.e., f°° I n / Ri(x)Rm(x)-j=-——— dx = -ci5im. Jo y/x[x + i) l Let the weight UJ(X) = (1 — x)a(l + x)13 for a,/? > —1, the Jacobi polynomial J ; ( a ; ) is the /-th eigenfunction of the singular Sturm-Liouville problem, (1 - x)-a(l

+ x)~pdx ((1 - x)a+1(l

+ xf+1dxv)

+\v

= 0,

ieA,

related to the Z-th eigenvalue A; = Z (Z + 1 + a + f3), and normalized by T<*,fim_

'

U j

r(M-oH-i)

~

Z!r(a+1) •

The Jacobi polynomials satisfy the recurrence relations, Jg'p(x) = 1,

J^{x)

= \{a + (3 + 2)1 + ± ( a - /?),

./£?(*) = ( a f ^ s - b^)J^(x)

- cfJjtfix),

where a,fj = V

(21 + a + /? + 1)(2Z + a + ff + 2) " 2(i + l)(Z + a + / 3 + l )

,a,/3 = Q3 2 -a 2 )(2Z + a + / ? + l ) ' 2(Z + l)(Z + a + /? + l)(2Z + a + / 3)' a , / 3 = (l + a)(l + (3)(2l + a + p + 2) " (Z + 1)(Z + a + (3 + 1)(2Z + a + PY

C/

I > 1,

228

Numerical Methods for Exterior

Problems

The set of Jacobi polynomials is the L^-orthogonal system in A, i.e.

L We also have the following derivative recurrence relation

dxJ^(x) = ±(l + a + /3 + l)J?+^+1(x). By integrating by parts, we deduce that

L

dxJ?'0(x)dxJ%P(x)(l

- x)a+1(l

+ xf+1 dx = 0,

l^m.

/A

Both Legendre and Chebyshev polynomials are special cases of Jacobi polynomials, namely, the Legendre polynomials Li(x) correspond to a = f3 = 0 with the normalization L/(l) — 1 and the Chebyshev polynomials Ti{x) correspond to a = f3 = — \ with the normalization T;(l) = 1. For a < — I or (3 < — 1 the weight function to(x) is not in L^(A), so the Jacobi polynomials are defined for a,P > — 1. However, the generalized Jacobi polynomials are defined for the cases where the indexes a and /3 can be negative integers.

Q,

4 "(*)

' (1 - x)~a(l + x^J^-^x), (l-x)-<*Jr_atf(x), (1 + x)-^llp{x),

if a,/3 < -1, ifa<-l,/J>-l if a > - 1 , (3 < - 1 ,

where I > IQ with IQ = — a—/?, —a, — (3 respectively for the above three cases. The generalized Jacobi polynomials J; £ Pi and they are orthogonal with the generalized weight, [ J?-(,(x)J%e(x)u(x)dx

= 0,

l^m.

JA

The generalized Jacobi polynomials with negative integer indexes can be expressed in terms of Legendre polynomials, i

Jf^(x)= Yl aiLAx),

l>-a-p.

j=l+a+/3

Finally, by the definition the generalized Jacobi polynomials satisfy some zero boundary conditions, OiJ^(l)=0,

i = 0,1, . . . , - a - l ,

Spectral

Method

5 i ^T«,0 "(-l)=0,

229

i = 0,l,--- , - j S - l .

Let A = (0,oo) and the weight u>(x) = e~x. The Laguerre polynomial of degree I is

£,(*) = ±e*4(o:'e-*). It is the /-th eigenfunction of the singular Sturm-Liouville problem, dx (xe~xdxv)

+ Xe~xv = 0,

ieA,

related to the l-th eigenvalue A; = I. Clearly CQ{X) = 1, C\{x) = l — x, and they satisfy the recurrence relations, {I + l)£i+i{x)

= (2Z + 1 — x)Ci{x) - lCi-i{x),

Ci(x) = dxCi(x) - dxCi+i(x),

I > 1,

I > 0.

It can be checked that £i(0) = 1, dxCi{0) = —l for l > 1, and |£j(a:)|<e*,

i£A.

The set of Laguerre polynomials is the L^-orthogonal system in A, i.e.,

L

Ci(x)Cm(x)u>(x) dx = Sim.

IA

By integrating by parts, we deduce that / dxCi(x)dxCm(x)xu>(x)dx

= l6im.

JA

The Laguerre expansion of a function v £ L^ (A) is 00

v{x) =

Y^vi£i(x), 1=0

with vi = / v(x)£i(x)u(x)

dx.

JA

The generalized Laguerre polynomial of degree I is defined by C\as\x)

= ^-ae^^(ii+ae-^),

I = 0,1, • • • .

230

Numerical Methods for Exterior

Problems

If /? = 1 we drop the superscript f3 and denote the generalized Laguerre polynomial by C\a'(x). It is straightforward to derive the following properties: Lt

(U)-£-;

W -

dxC^P\x)

r ( a

+ 1)r(j + 1)>

= -l3C£?>f,\x),

= (2l + a + l - px)C\a
(I + l)c\^\x)

=fi-l(dxc\a
^ ° >

1>1,

(x) -(1 + a)C\Zf

- dxc\lf\x)),

(x),

l> 1,

I > 1.

Let uatp(x) = xae~l3x, a > — 1, /3 > 0. In particular, we denote u)a(x) = w 0 ] i(i) = xae~x. The set of Cf (x) is the complete L^a (A)-orthogonal system, namely, (Aa,0)

6.3 6.3.1

Wa./Jh

_

Mr

a,/5 _ r(Z + a + 1)

Laguerre spectral methods Mixed

Laguerre-Fourier

spectral

method

First of all let us consider the mixed Laguerre-Fourier spectral method for the Helmholtz exterior problem (1.25)(1.26) - ( A + A)u = / , u = 0,

(6.13)

x e dfi,

(6.14)

with a real A < 0, dim (fi) = 2, and fi = {x; \x\ > 1}. Owing to Theorem 17, the problem (6.13)(6.14) admits a unique solution in HQ(Q). Under the following polar transformation: xi = (p+l)cos6,

X2 =

(p+l)sm6,

the equation (6.13) becomes

-fti9"^+l)d"u) ~ jp~hyd2eU ~Xu=zf-

(6 15)

-

231

Spectral Method

We use the following change of variables

u=(p+l)~h%u,

f=(p+l)§e$f,

to transform (6.15) to -(p+l)2d2pu

+ (p2-l)dpu-d2u+(-\(p+l)2

+ ± + ^~p2)u

= f. (6.16)

To solve the boundary value problem of (6.16) we introduce some spaces. Let / = (0,2TT) and H™(I) be the subspace of Hm(I) consisting of all functions whose derivatives of order up to m— 1 are periodic with the period 27r. Let M be any positive integer, and VM(I) =span {e%w; \l\ < M}. We denote by VM(I) the subset of VM(I) consisting of all real-valued functions. Let A = (0, oo), D = A x I and LX{D) be the weighted Sobolev space with the following inner product and norm, (u,v)0,x=

/ u(p,6)v(p,8)x(p)dpd9,

\\v\\o,x = (v,v)o,x-

JD

The weighted Sobolev spaces H^(D) and its norm \\v\\rx and semi-norm \v\r,x are defined in the usual manner. In particular we set KPJD)

= { » ^ Hi(D);v(p,0

+ 2v) = v(p,0),v(0,0)

= O,V0 e I,p £ A}.

With the weights u>(p) = e~p and r](p) = (p + l)2e~p isotropic space

we define the non-

= {^e InocW;Nil,,,* < oo},

KP^JD) where

N i l , , , - = (M?,„, u + hWljK

\v\i,v,u, = (IIVIIo,, +

Let us denote PO,JV(A) = { U € PN(A);V(0) VN,M(D) =

\\dev\\ljl.

= 0} and PO,N(^VM{I)).

Let us define the variational formulation of the boundary value problems of (6.16). For this purpose we define a bilinear form, a(u,v) = / (p+l)2e JD

p

dpudpvdpd8 + / e JD

p

dgudev dpdO

+ JD e~p (-A(p + l) 2 -\p2 + \p+\)uv

dpdB.

232

Numerical Methods for Exterior

The variational formulation is: Find u G HQ a(u, v) = (/, „)„,„,

Problems

W{D), such that

Vv e Hlp^{D).

(6.17)

To investigate the well-posedness of (6.17), we need some lemmas. Lemma 72. For any v e #o]W2 (A) f] L^,0 it Zio/ds t/iat ll^llL 2 ,A^ 8 l^lL 2 ,A + 8ll^llo;.o,A' where oja with a = 0,2 is defined in Section 6.2. Proof.

We have ./o Jo

Jo

Jo Letting p —» oo and using the Schwarz inequality we obtain IMIiU.A < 5llHlL2,A + 2 ll^llo,. 2 ,A + 2llVllo,Wl,AThus IMIiU.A < 4||0„i;|liU,A + 4 | | V | | ^ i A .

(6.18)

An integration by parts yields IMIiU.A = / Pe~"v(p)dpv(p) dp + \\v\\l>U0>A. J\ The Schwarz inequality yields 2 / pe-"v(p)dpv(p) dp < |v|f A + ||v||jjiUOiA. J\ Substituting the above into (6.18) we obtain the result. Lemma 73. For any u , » £

HQpn^(D),

a(v, v)> f (p + l)2e-"{dpv)2 dpdQ + f e-p(dev)2 dpdQ JD

JD

_ (A + 4I) f (p + JD

l)2e-pv2

dpdQ +\ 4

f e~pv2 dpd6, JD

•

Spectral Method

233

and \a(u,v)\ < C||u||i i77iU ,||u||i^ >w . Proof.

Obviously 1 2

+ P+

1

1

!/

{p+1)

^2

-4P 2 2^-A

3

+

4'

which leads to the first result, and the second result follows from Lemma 72. • Theorem 70. If\<-\ and ( p + l ) - 1 / e l £ ( £ ) , then the problem (6.17) admits a unique solution with

N k w ^ q K p + irVllo.u,. Proof. HQPVU1(D).

Due to A < — | and Lemma 73, a(-, •) is coercive on Hlpr)U1{D) Moreover by Lemma 72,

Thus the conclusion follows from the Lax-Milgram theorem.

x

•

Next, we consider the mixed Laguerre-Fourier approximation for (6.17): Find UN,M € V/v.Mi such that a(uN,M'v)

= i.f,v)o,u,,

Vu e VJV.M-

(6.19)

The following result is a direct consequence of the Lax-Milgram theorem. Theorem 71. If\<-\ and (p + l ) " 1 / e Ll(D), admits a unique solution with

then the problem (6.19)

||UJV,M||I,^
We next describe the implementation for (6.19). For simplicity, we denote C\ by C and set V'i(p) = ^i-i(p) ~ A(p), 1 < I < A7", a n d ^ ( / j . f l ) = ipi(p)cosm9, vflip,9)

= ipi(p) sinmi9,

l
234

Numerical Methods for Exterior

Problems

Since ^(O) = 0, v\\^ can be used as basis functions for

V)V,M-

Hence we

can expend UJV,M as N

UNMP^)

/ M

=£

M

u

\

£ <>S>>0) + £ «?,m«£(p,e) .

i=l \m=0

m=l

/

On the other hand we write oo

oo

f(p, 0) = J2 £

(/j'm^(p) cosm« + /, 2 m A(p) sinmfl).

(=0 m = 0

Let us denote ZM = {{j,n);j

= l,n = 0,1,-•• , M; j = 2,n = 1, 2, • • • , M } .

Taking v = vj? n in (6.19) for (j, n) £ ZM, we derive by using the orthogonality of the trigonometric functions that (6.19) is equivalent to the following 2M + 1 linear systems:

f V

[(p+lfe-rdpiPid^kdp + j f e-"(-A(p + l ) 2 + n 2 - ^ + P- + \ ) ^

k

dp^j u{m = £

m>

l
ai,k =

bi
JA

Q,k=

e~ptpi(p)-ipk(p)dp,

JA

/ pe~p^i(p)ipk{p)dp,

diik=

/

JA

p2e~pipi(p)ijjk(p)dp.

JA

Furthermore, let Xn

— ( w l,n> u 2,n> " ' > UN,n)'

^n

=

(Sl.ni 02,n' " ' ' 5jV,n)>

J = 1,2.

We obtain the system of equations, J 4 + (_ A + n 2 + i ) B + (-2A + i ) C - (A + ±)I>} X>n = G{, (j,n) £ ZM. Indeed, A and C are five-diagonal matrices, B is a three-diagonal matrix and D is a seven-diagonal matrix. Moreover all matrices are symmetric and positive definite.

235

Spectral Method

6.3.2

Spherical method

harmonic-generalized

Laguerre

spectral

We consider the problem (6.13),(6.14) in three space dimension and the spherical harmonic-generalized Laguerre spectral method to solve this problem. Let (f> and 9 be the longitude and the latitude respectively. Then the spherical surface is S3 = {(<j>, 0);O < < 2n, - § < 9 < § } . The Laplace operator with spherical coordinates is given by A = \d2r(r2dr) r

+ -^L-de(cos6d e) r2 cos 8

+

* 8%. rz cos J 9 v

Being Analogous to the investigation of (6.13),(6.14) we make the variable transformation r = p+l,u

= e^u, f={p +

ife^Pf.

Then the equation (6.13) is changed to -{p + l)2d2ppu - (2(p + 1) - (3(p + 1 „ , ~ ^9dB(coe9m)

lf)dpu

1 ~-^M8**

2

+ (-X(p+l) +(3(p

(6-20) 2

+ l)-^(p

+ l) )u = f.

We define some spaces. Let D = A x S3 and the weights be u>o^{p) = e-Pp and 77,3 (p) = (p+ l ) 2 e - ^ . The spaces L\{D) and H^{D) are defined in the usual way. We denote ffo,P,«-o.„(D) = {ve H^(D);

v(p, 4> + 2n, 9) = v(p, cj>, 9), v(0, , 9)=0},

and (D) = {ve Hlp>U0JD);

||«||i,,fllWOl(J,D < 00},

where \v\l,Vt>,<*o,fi,D = {\\9Pv\\o,Vf„D

\\vh,v^o,p,D

+ H^ga*VHo,<-o,fl,D + ll^llo.oio.fl.D)*-.

= (\v\i,r,0,UO:l3,D +

\\V\\Q,UO,P,D)^

•

Let Ym,9), 0 < m < 00, |2| < m, be the normalized spherical harmonic functions. They form the complete normalized L2(S,3)-orthogonal system.

236

Numerical Methods for Exterior

Problems

Let M be any positive integer, and define the finite dimensional space VM(S3)

as

VM(S3) = span{ym,((,0);m < M, \l\ < m } . Denote by VM(SS) Let VNM

=

the subset of VM(SZ) containing all real-valued functions. PO,N(A;VM{S3)).

For ease of notations, let Vsu = ( -d$v,dev \cos9 We define a bilinear form a(u,v)=

/ {p+\)2e~l3pdpudpvdsdp

+ I

JD

e _ / 3 p V s uV s vdsdp

JD

+ f (-X(p + l ) 2 + (3(p + 1) - £4 - ( p + l)2)e-P>>uv dsdp. JD

Obviously /32 \a{u,v)\ <max(-A+Y,l)||w||i,^,u;o,0,£»lk||i,^, W o,^,r>>

( 6 - 21 )

and for A < — *|?2

(6-22)

a(«,«)>(-A-^)||«HU,«o. f l ,iJThe weak formulation of the problem is: Find u € HotP>r] that a(u,u) = (f,v)0

Vv€HlPiVfliU0JD).

UQ

(D), such

(6.23)

If A < - f and / G ( ^ ^ . ^ ^ ( D ) ) ' , then by (6.21),(6.22) and the LaxMilgram theorem, (6.23) admits a unique solution. The mixed spectral scheme for (6.23) is: Seek uw,M £ V/v,M> such that a{uN,M>v) = {f,v)o,u>0if),D,

Vw G

V(V,M-

(6.24)

Under the same conditions the problem (6.24) also admits a unique solution. We next describe the implementation for (6.24). Set ipk(p) = - Q - i \P)~ C™(p) and Zml(
Z2m^<j>,9) = - L cos(lcf>)C(sm9). VZTT

237

Spectral Method

Moreover, let v£>mi,(/3,<M) = ipk(p)Z3mtl(
N

M

m

j = \ k=l m—0 / = — m 2

M

m

j=l m=0l= — m

Take v — vJk m t(p, <j>,6), j = 1,2, in (6.24). Then we derive a linear system for the coefficients u3kml. More precisely, let the matrices A = (an>k), B = (6„,fc), C = (cn>fc) and D = (Dntk) with the following entries: an,k = I p2e~0pdpi)n{p)dpi})k{p)dp, cn,k=

p2e~ppipn(p)tpk(p)dp,

6„,fc = /

JA

JA

/ pe~0ptpn{p)'4'k(p) dp,

dnM=

I

JA

e~0ptpn(p)il>k(p)dp.

JA

Furthermore, let ^m,l

=

( M l,m,Z' U 2,m,;> ' ' ' iUN,m,l)>

^m,l

=

\9\,m,V

92,mp

' ' ' '3jV,m,i)>

j = 1,2, where

9{,m,i=

f

JA

e~^fmtl(p)Mp)dp.

We obtain the system of equations, ^A-(X+^)B

+ pC + m(m+l)D^X3mil

= Gltl,

j = 1,2.

Indeed, A and C are five-diagonal matrices, B is a seven-diagonal matrix and D is a three-diagonal matrix. Moreover all matrices are symmetric and positive definite. 6.3.3

Generalized

Laguerre pseudo-spectral

method

We consider the spherically symmetric solution of the equation (6.20), which is reduced to -(p + l)2d2u - (2(p + 1) - (3{p + g2 + (-X(p + l ) 2 + B(p + l)-^(p+

l)2)dpu (6.25) lf)u

= /.

Numerical Methods for Exterior

238

Problems

Let cra,f){p) = (p + \)ae-PP, and denote H^ p(A) = {v e H^a (A);v(0) = 0}. The variational formulation of the boundary value problems of (6.25) is: Find u e H^„2 0{A) f\Ll (A), such that VveHla2JA)f)L2aiJA),

a(u,v) = (f,v)o,u,o,0,

(6.26)

where the bilinear form is defined by a{u,v) = (dpu,dpv)ota2tl3

- (A + —)(u,v)0t<72if) 4

+

P(u,v)0t„lifl.

One can verify readily that |«C«, «)l < ^ ( ( H - ^ H ^ I I l . o r a , , , -H/?» ll«l|0,cr 1 . /s )CCl-|->S)||vHi, er31 ^ -4-^9* llvllo.or^,,,),

and for X < ——-, \a(v,v)\>C(\\v\\la2ff+0\\v\\l«2JHence, if / e ( H ^ (A)p|L£ 1 / 3 (A)j , then (6.26) admits a unique solution. To define the generalized Laguerre pseudo-spectral scheme for (6.26) we need to introduce the generalized Laguerre-Gauss-Radan interpolation. Let €NJ b e t n e z e r o s °f xdx£}j+i(x)They a r e arranged in ascending order. Denote the corresponding weights, wjy*'. , 0 < j < N, such that N

r

The weights can be evaluated as (a,/3) w N,j -

(a+l)r2(q+l)r(JV+l) I /3°+lT(iV+ a +2) \ r(N+<x+l)

3=0, 1

1 < 7 < AT

The discrete inner product and norm are defined by

By the exactness of (6.27) (u,v)0,u

PtN

= (u,v)o,u,a,0,

VUV€P2N-

(6-28)

Spectral

239

Method

In particular ||w||o,w a ,„,.W = ||w||o,u;„, f l ,

Vw G

PN.

The generalized Laguerre-Gauss-Radan interpolation IN,a ,pv G PN is defined by

lN,aM&f)

= v(&f),

0<j
The generalized Laguerre pseudo-spectral scheme for (6.26) is: Seek WJV £ Po,N = { « £ ^V; M(0) = 0}, such that a(uN,v)

= (f,v)o>u,O0>N,

\/v G P0,N,

(6-29)

where a(u,v) =(dxu,dxv)0

+ 2(dxu,dxv)OtU,li0,N

- (A + ±P2)(u,

U)O,U2I/J>JV

+

(dxu,dxv)0

- (2A + - / 3 2 - /?)(«, U)O,U,1,/>,JV

- ( A + i/3 2 -/?)( M , v ) 0 , WO] „iv. According to (6.28), the formulation (6.29) is equivalent to a(uN,v) 6.3.4

Nonlinear

= (lN,o,pf,v)o,u0il3,

Vv G P0,JV-

equations

As an example, we consider the spectral method for the initial boundary value problem of the Burgers equation, dtu+-dx(u2)

= iJ,dlu + f{x,t),

i€(0,oo),

(6.30)

u\x=0 = 0,

(6.31)

u\t=o = u0(x),

(6.32)

where fi > 0 is the viscosity and the conditions on / and UQ will be specified later on. Let u = exl2u , f = ex/2f, and UQ = ex/2u0, then the equation and the initial condition are transferred to, dtu + ^e*'2dx{eTxu2)

= n(d2u - dxu + ifi) + / ,

240

Numerical Methods for Exterior

Problems

and u t=o = wox

Let the weight be w = e , and let UN : HQ^A) —> Po,w(A) be an orthogonal projection operator, defined by: Tl^u £ Po,iv(A), (dx(UNu-

u),dxv)0,uj + (UNu-u,v)0tU,

= 0,

\/v G P 0iA r(A).

We assume that u0 G #o, w (A) and / G L°°((0,T); L^(A)), then multiply the equation by wv, and take integral on A = (0, oo) to get dt(u,v)0tU+-

/ e~x/2dx(e~xu2)vdx+fi(dxu,dxv)o,u--{u,v)0^

— (/,u)o,w

(6.33) Thus the Laguerre spectral scheme is: Find UN(t) G Pjv(A) for £ > 0, such that WAf|a:=o = 0, and ^ -

_

J(MN,V)0,W

(6-34)

= {f,v)0,oj,

Vw G

PJV(A),V|T=0

= 0,

and «/v|t=o = ILvwo. (6.35) We develop the function UJV into a series in terms of Laguerre polynomials: N

u

N(t) = ^2ci(t)Cl{x), 1=0

then the system (6.34),(6.35) is an initial value problem of ordinary differential equations for ci(t). The function v is also developed in the same way, then thanks to the orthogonal property of Laguerre polynomials, the matrix associated with (UJV,*0O,U> is the identity. Therefore the system (6.34),(6.35) at least admits a local solution. Next let us prove global existence of (6.34),(6.35). Set v = uN in (6.34), then we get d

t\\uN\\ltU,+ 2 / e~x/2dx(e~xu2N)uNdx+fi\uN\liUJ--\\uN\\l^

=

We have |

f e-x/2dx(e-xu2N)uN

dx = ^fdx

(e-*x'2u%)

dx = 0,

(f,uN)oiU.

Spectral

241

Method

and (/,«Jv)o 1 u,<^||/||o lW + ^ll«Jv||o,u-

Therefore it holds that

dt\\uN\\20iU + H«*|?,w < (^ + I ) IKIliU + ^||/||g,w.

(6.36)

Integrating (6.36) over (0, t), t
ll«jv(t)||2,u+/i / K l ? , ^ t

t

udt.

( 6 - 37 )

Set £(£) = ||ujv(t)llo,u<- Thanks to the Gronwall inequality (Theorem 67) with a = f + i and b = ||«Ar(0)||§ita; + § / 0 T ||/||§ ifc , dt, it holds that + IJQT

\\f\\l„dtj e< *+*>«.

E{T) < (\\uN{0)\\l„ + \ [

Il/IIL dtj e<«+*>r.

E{t) < (\\uN{Q)\\l,„ Let t = T then

Since T is arbitrary, we replace T by £ and obtain finally that

£(*) <

(||«JV(0)||L

+ \£

||/||g,wdt) e^ + i)*.

(6.38)

-B(i) 1 / 2 plays the role of a norm on PJV(A), so the solution to (6.34),(6.35) is bounded uniformly with respect to t. The solution can be extended continuously to T, which proves the global existence. 6.4

Jacobi spectral methods

As an example we consider the spectral method for the initial-boundary value problem of the Klein-Gordon equation, d?u + u3 = d%u + f(x,t), «|x=o = 0,

xe(0,oo),

(6.39) (6.40)

242

Numerical Methods for Exterior

u\t=o = u0(x),

Problems

dtu\t=o = ui{x),

(6.41)

where the conditions on / , u\ and ui will be specified later on. We introduce a new variable £ and set x = — 2 ln(l — £) + 2 In 2, then the equation (6.39) is changed to d2tu + u3 = 1(1 - O0«((l " 0 ^ « ) + / .

C G (-1,1).

Letting A = (-1,1) andw = ( l - £ ) 2 , we define V = { « £ £ 4 (A); ||^u|| 0 , w < oo,w(—1) = 0}, then the weak formulation is: Find u(t) G V such that u\t=o = u0, dtu\t=o = ui, and (%u,v) + (u3,v) + 1((1 - 0 ^ « , ^ ( ( 1 - Of)) = (/,*)>

V« G V,

where (•,•) is the L 2 -inner product. The Jacobi spectral scheme is: Find uN(t) G Pjv(A) such that Ujv(0k=-i = °> uiv|*=o = "OJV, dtuN\t=0 = uiN, and ( 5 2 W N , t ; ) + ( ^ ) i ; ) + i ( ( l - C ) ^ U J v , 5 £ ( ( l - £ » ) - (f,v),

W G Pjv,v(-1) = 0,

where UQN,UIN G PJV(A) are some approximations of uo and « i . Being analogous to the investigation of (6.34),(6.35), to prove global existence we only need to verify a global estimate. Let v = dtu^, then we get

J \^dt{dtuNf

+ ^dt(u%)\

dt+±{{l-t)dtuN,dt((l-£)dtUN))

=

(f,dtuN).

Let us estimate the terms as follows:

\(0.-t)dtuN,dttti-oatuN)) = -. / (1 - Q2dzuNd£dtuN 4

d£ - - / (1 - £)d/:uNdtuN d£

JA

^ JA 2

2

= 1 / ( l - 0 « t ( ^ « J v ) d e - 7 f (I - Od(UNdtuN dC 4

° JA

2

JA

2

> 1 1 (i - o ^(«««iv) # - 1 J a - O2(«C«JV)2 # - 1 j[($ u ") 2 #• (/, dtuN)
243

Spectral Method

We set E(t) = JA{\(9tuN)2

+ \(u%) + \(1-

0 2 ( ^ « i v ) 2 } de,

then it follows that

dtE(t)<2E(t) + l [ f2d£. 1

JA By integrating over the interval (0, t) it yields

E(t) < 2 J

E(T) dr + E(0)

+ \ J J' f' d^

We assume that E(0) < oo and JQ JA f2 d£ < oo, then apply the Gronwall inequality to obtain the boundedness of E(t), then global existence follows. 6.5

Rational and irrational spectral methods

The rational spectral method can be applied to the initial boundary value problem of the Klein-Gordon equation (6.39)-(6.41). The weight is u> — T ^ W , and we set A = (0, oo) and V = HQIU)(A) f]Z£(A). We multiply the equation (6.39) by u>v, then integrate the equation over the interval A, thus the weak formulation of the problem is derived as: Find u(t) £ V, such that u\t=o = u0, dtu\t=o = «i, and / T -r^ < d?uv + u3v 2 7A {* + I) I

-dxuv + dxudxv \ dx a: + 1 J

= / 7—rr^fvdx> J A (X + l ) 2 We define a finite dimensional subspace of V: VN = L=

V2v ( ~ j \

VVGV.

W G PN(A),v(-l)

= ol,

where A = (—1,1). The rational spectral spectral scheme is: Find ujv(i) € Vjv, such that uN\t=0 = U0N, <9tUjv|t=o = "IJV, and / 7—T~7T? \ dtuNV + u3Nv J A {* + I) 2 I =

—dxuNv x+ 1

/ 7—rr&f v d x > J A (X + l ) 2

+ dxuNdxv

VV^VN,

\ dx J

244

Numerical Methods for Exterior

where

€

UON,UIN

VN

are

Problems

the orthogonal projection of uo and u±, namely,

(dx(u0N - u0),dxv)otUJ + (u0N - u0,v)oiU = 0,

Vt; e VN,

(dx(u1N

Vu e VN.

and - ui),dxv)0
Being analogous to the investigation of (6.34),(6.35), to prove global existence we only need to verify a global estimate. Let v = dtUN, then we get

L

(X

+ I)2 I

dfupjdtUN + u3NdtUN

—dxUNdtUN x+1

+ dxUNdxdtUN > dx

/ 7—7T^f d t U N d x > J A (x + I) 2 where

A(^ + l)3

d'fuNdtuN

1„,„ x, = ^dt(dtuN)2,

dxu^dtUN

dx <

„ „ u3NdtuN

L jdw{dxUN)2

dxuNdxdtuN

=

1.

=

-dt(u%),

dx+

L ¥hy{dtUN)2

-dt(dxuN)2

and

L {* +1;

-^fdtuNdx

<

(x + 1)2

(dtuN)2

dx •

JA^

+ 1}

We set

E(t) = J

l 2{x +1)2

{(dtuN)2 + \{u%) + (<W) 2 } dx,

then it follows that dtE(t) < AE{t) + - /

21 / A IK

By integrating over the interval (0, t) it yields

fzdx.

;fdx.

dx

'

245

Spectral Method

We assume that E(0) < oo and / 0 / A ( T ^ 1 ) 2 f2 d£ < oo, then apply the Gronwall inequality to obtain the boundedness of E(t), then global existence follows.

6.6

Error estimates

The error estimates of the spectral method rely on the approximation theory of orthogonal sets of functions. Here we study the error estimates for the Laguerre spectral methods as examples. To begin with, we prove some properties of weighted Sobolev spaces. For definiteness we denote A = (0, oo), u>(x) = e~@x and u>a,i3{x) = xae~^x in this section. Lemma 74. We assume that /3 £ (0,2), then it holds that ||e-^«||o,oo < N k u , , Proof.

W 6 HlJA).

(6.42)

Thanks to the embedding theorem , v € C(A). We have [X Jo

(er^v{x)f=

dt(e-^v{t))2dt

= f e-pt2v(t)dtv(t),dt-l3 Jo

e-f3tv2(t)dt

f Jo

< [ e-0t{dtv{t))2dt+{l-(3) Jo

f Jo

e-*3tv2(t)dt.

Thus the proof is complete. Lemma 75. Let

P0,AT

={p£

inf \\vN -vW^ vNeP0,N+l

D PN;P(0)

= 0}, then

V^GF0

1 i

^(A)fl^(A),r>0

for N > r, where the constant C depends on (3 only.

246

Numerical Methods for Exterior

Proof.

Problems

We have / Jo

e~'3xv2(x)dx

Jo

/>oc

= / Jo

e~0x

= /*oo

2v(t)dtv(t) dt / Jt /

e-ptv2(t)dt\

2v(t)dtv{t) dt Jo />oo

e-0x dx = 2/T x / Jo (

0t

e-

v(t)dtv(t)

e-P\dtv{t)fdt\

dt

.

Therefore /

(/

/-co

\ 5

e-^« 2 (t)dtj

/ /-oo

<W-lU

\ 5

e-e\dtv{t)fdt\

,

that is ||v||o,u,<2/3- 1 ||a :cV ||o, w .

(6.44)

We develop the function dxv in terms of generalized Laguerre polynomials: oo -(0,/3)/

dxv(x)=T,c
M'

1=0

where ••OO

c, = /3 / e ^ d X ^ ./o

0

' ^ (a:) c£r.

Let N ™JV(Z) = X ^ c ' ^ z=o

'(0,0)/ ' W.

then wjv(:r) - dxv(x) = -

c;£J 0,/3) (:r).

^

Owing to the orthogonal property, it holds that oo i=JV+l

247

Spectral Method

where jf'^ are given in (6.12). On the other hand oo

,r+l„||2

_

N„r

oo

V " ,, r(°'V

|,2

_ || V % , &•&

=0

II?

= V"

(=r

r ? ^

/=r

Thanks to (6.12) we obtain

tf'V _F{l-r)\

^Jfi

ll — r

Therefore

\\WN-dxv\\i^<^J\\d:+'v\\i^0 Letting follows.

VN(X)

= J*

we apply (6.44) for

WN(X)CIX,

VJV

— v, then (6.43) •

Having proved the above primary results, the error estimates for the schemes (6.6) and (6.7) are routine. Because a\\u - p | | i , w < a.(u -p,u-p)

= a(u -p,u-

q),

Vg e Po,N,

and a(u-p,u-q)

< M||M-p||i,w||u-g||iiU,,

we obtain that ||«-p||ilW<— a

inf

qePo.N

||«-9||i,fblJV
provided the right hand side is bounded. We turn now to the error estimate of the scheme (6.34),(6.35) for the problem (6.30)-(6.32). The assumptions on the functions u 0 and / will be given in the process of estimation. We first assume that uo and / are bounded functions, then by the maximum principle of parabolic equations the exact solution u is bounded, \u\ < M. Then it yields that |e~^u| < M. As N —> oo IIjv«o converges to u0 in H^ with /? = 1. Owing to Lemma 75 and Lemma 74 with (3 = 1, |e~i(UNUQ — «o)| < M ii N is large enough. We notice that there are only finite terms in the spectral expression of MJV, hence by continuity there exists T* > 0, such that |e~^(ujv — u)\ < M for t € [0, T*]. Let us derive the energy estimate on the interval t G [0,T*] first.

248

Numerical

Methods

for Exterior

Problems

Let WN — Iljvu, then being analogous to the estimates for (6.6),(6.7) we get (6.45)

•WNWI^^CN-'W^UWO,^,

provided the right hand side is bounded. It remains to estimate e^ = UJV — to AT- By (6.33) and (6.34) we take v — e^ to get / e-x/2{dx(e-xu2N)

(dt(uN-u),eN)0,„+-

+ n(dx(uN

- dx(e-xu2)}eN

dx

- J(UN - u, ejv)o,w = 0.

- u), dxeN)0,u

We note that \e~%(v,N + u)\ < 3M, so the nonlinear term can be estimated as the following: I [ e-x'2{dx{e-xu2N) 1 JA

- dx{e-xu2)}eN

- - / e~x/2{e~xu2N 2 7A

L

e

3x/2

^

x e „—x-2 u }e/v dx

f

+ -eN)(u2N

{--dxeN

/• _ <3M I e x\JA -x,

1

e~xu2}dxeNdx

-

-a„,2

+

1 -dxeN 2 a

dx

- u2) dx

1 + -eN\ • \uN - u\ dx 4 ,

1

x2 j

,

9 M 2

f

-x,

- N 2 J

<£ / e x ( - - o x e ] v + -ejv) dx + —— / e x(uN JA 2 4 4e 7A

-uydx,

where e > 0 is a constant to be determined. Noting that UN - « = e« + (u>;v — u) we get 1 o ii

n2

,

i

|2

A4 ii

n2

r>

where \R\ < e\eN\l^

+ C{\\dt(wN

- u)\\20iU + (1 +e-1)(\\eN\\lu,

+ \\wN -

u\\lJ}.

Being analogous to (6.45) we have the estimate \\dt{u-wN)\\liU

< C^-5||at^+1u||olWr,1,

(6.46)

Spectral Method

249

provided the right hand side is bounded. Let e = ^, then we obtain

^ t l M l L + %\eN\lu < C\\eN\\l,u + CN-r. Integrate over the interval (0, t) to get ^ l l ^ | | ^ + | ^ t | e i V | ? , a ; ^ < c £ | | e j v | | ^ ^ + i | | e i v ( 0 ) | | ^ + CtiV- r . The inequality (6.45) implies

IMo)||gita, < ^ - r Applying the Gronwall inequality we get an estimate on (0,T*), \\\eN\\iu

+ f fQ \eN\l„ dt < Cec\\

+ t)N~r.

Along with (6.45) we obtain |||«JV - u\\ltU + | J

\uN - u\l„ dt < Cect{l

+ t)N-r.

(6.47)

Then we take v = dte^ in (6.33) and (6.34) to get + -z / e"xl2{dx{e~xu2N) 1

{dt{uN-u),dteN)0,uj

- dx{e-xu2)}dteN

dx

JK

+ fi{dx{uN - u),dxdteN)o,ui - j(uN

- u,dteN)0tUJ

= 0. (6.48)

The linear terms can be estimated as follows: {dt{uN -u),dteN)0:U>

- \\dteN\\l!U1 - {dt{u -

>\\dteN\\20iU - {\\dt{u - WN)^

V-{dx{uN =n{dx{uN

+

wN),dteN)0tU

\\\dteN\\U

u),dxdteN)0^

-u),dxdt{uN

-u))o,u

+ n{dx{uN

-u),dxdt{u 2

>^dt\uN

- fi|?iW - | | U J V - «)|fiW - ^\dt{u - wN)\

>^dt\uN

- u\lu - ^\uN - u)\l„ -

1>u

CN-r\\dtdl+'u\\l^

-wN))QiU)

250

Numerical Methods for Exterior

•|(UJV

- u,dteN)0,u,

> -Cect{l

Problems

+ t)N~r -

-^d^Wl^

where we employ the estimates (6.46) and (6.47). The nonlinear term can be estimated as follows: \ [ e-*'2{dx{e-*u2N) 1

dx{e-xu2)}dteNdx

-

JK

= — / e~ x' {u — uN + 2UN9XUN 2

— 2udxu}dteN

<\\\9teN\\l,u,

+ ^ f e~2x{u2 -u2N + 2uNdxuN

- 2udxu}2 dx * I.

We assume that the exact solution u satisfies \e~x!2dxu\ e x{u — uN + 2u^dxUN — 2udx„~.l2 u] - 2 i

dx

JA

< M\, then

{(u + UN)(U — ujsr) + 2dxii(uN — u) + 2ui\[dx(upj — w)}2

<3e-x{(3M)2(u

- uN)2 + AMl(uN

- it)2 + A{2M)2{dx{uN

-

u))2}.

Then

I<\\\teN\ft
+ C\\uN-ufXtll>.

Substituting these estimates into (6.48) gives

?<

|2 ^-dt\uN -u\2luJ <

//-r|„...

,-.|2 C\uN -u\i„ + CN-

Integrate over the interval (0, t) to get ft r ol«iV - «ll,u, < £l(«N - u)(0)\2liU + CJQ \uN-u\\^dt + CtN- . 21 21

Applying the Gronwall inequality we get another estimate on (0, T*), \uN-u\lu
+ t)N-r.

(6.49)

Thanks to Lemma 74 it holds that \e-x/2(uN

- u)\2 < CeCT(l

+ T)N~r,

(6.50)

for t < T*. Let N be large enough so that \e~x/2[uN - u)\ < f on [0,T% then by the definition of T* this interval can be extended. Since the constant C in (6.50) is independent of T*, we can always extend this interval up to T, then the global error estimates (6.47) and (6.49) are valid for all t € [0,T].

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Index

absorbing boundary condition, 167-169, 172, 174, 176, 177 amplification matrix, 207 analytic function, 21, 22, 80, 81, 98-100, 214 artificial boundary condition, 167, 179, 180, 182, 183, 186, 187, 189, 191 asymptotic behavior, 20, 83, 187, 188, 212, 213

Cayley transform, 49, 51, 52 CFL condition, 208 Chebyshev approximation, 173 Chebyshev polynomial, 222, 226, 228 Clayton-Engquist-Majda's boundary condition, 172, 182 closed graph theorem, 16, 149 closed range theorem, 67 coercive, 18, 36, 69 collocation method, 104 combinational number, 130 combined stiffness matrix, 119, 121, 131-133, 146, 153-156, 167 convolution product, 101, 197 current density, 57 cut-off function, 15, 21-23, 28, 32, 33, 37, 43, 55, 63, 67

Berenger's perfectly matched layer, 197, 210 Babuska-Brezzi condition, 38 Banach space, 9, 67 bending moment load, 31 Bessel equation, 18, 20, 82, 83 Bessel function, 20, 82, 83 bi-harmonic equation, 29, 75, 78, 89, 140 bilinear form, 26, 29, 36, 42, 65, 69, 86, 88, 92, 105, 138, 140-142, 147, 148, 152, 153, 186, 219-222, 231, 236, 238 boundary element method, 1, 71, 85, 88, 103, 113 Brouwer fixed point theorem, 44, 46 Burgers equation, 239

Darwin model, 61, 147, 150, 151 deflection, 29 determinant, 125 dielectric constant, 57 Dirac function, 5, 176 Dirichlet boundary condition, 2, 5, 26, 31, 138, 159, 180 Dirichlet boundary value problem, 71, 83, 85, 89, 155 Dirichlet problem, 3-5, 11, 12, 33, 74, 85, 86, 136, 140 Dirichlet-Neumann alternating scheme, 113 displacement, 25, 29, 80, 138, 139

Cauchy inequality, 14, 32 Cauchy sequence, 25, 149 Cauchy-Riemann equation, 80 265

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distribution, 60, 63, 87, 88, 90, 98, 99, 101, 102, 169 domain decomposition, 113 non-overlapping, 113 overlapping, 113 double layer potential, 5, 6, 8, 10, 11, 74, 86, 88, 95 dual product, 15, 91, 92, 98, 101 dual space, 26, 33, 36, 45, 87, 101, 169 eigenfunction, 20, 225-227, 229 eigenvalue, 122-126, 129, 130, 132, 135, 136, 139, 142, 207, 215, 225-227, 229 eigenvector, 123-126, 129, 132, 135, 139, 142 electric charge density, 57 electric field, 57 element stiffness matrix, 120, 159 elementary divisor, 123 elliptic differential equation, 21, 24, 86, 152 equilibrium equation, 25 existence, 5, 10-12, 18, 22, 25, 26, 29, 33, 36, 38, 40, 42, 44, 46, 55, 59, 69, 70, 86, 88, 109, 119, 240-245 exterior problem, 1, 2, 8-10, 13, 18, 71, 77, 82, 88, 93, 96, 106, 109, 114, 117, 169, 217, 219, 225, 230 external body force, 35 finite element method, 86, 107, 113, 120, 121 Fourier series, 72, 75, 84, 110 Fourier transform, 43, 169, 170, 193, 195, 198, 199, 209, 211, 212, 217 Fredholm alternative theorem, 8-10 fundamental solution, 5, 89, 176 Galerkin method, 1, 46, 104, 115 Gamma function, 83 generalized Laguerre-Gauss-Radan interpolation, 238 Green's formula, 21, 27, 33, 65, 86, 109, 145, 149, 150, 189, 213, 218

Problems

Gronwall inequality, 203, 207, 241, 243, 245, 249, 250 Holder inequality, 23, 47, 48 Holder space, 185 Hankel function, 83, 212-214 harmonic equation, 2, 71, 73, 85, 109, 117, 134 harmonic function, 3, 4, 20, 64, 73, 95, 96, 149, 235 heat equation, 49, 51 Helmholtz equation, 17, 18, 82, 83, 97, 114, 212, 230 Hilbert space, 13-15, 19, 32, 36, 44, 49, 51, 57, 63, 65, 66, 147-149 Hooke's law, 25, 28, 29 incidence angle, 197 incoming wave, 19, 167, 169, 171 plane, 83 inf-sup condition, 38, 66, 69, 143, 150 infinite element formulation, 117, 136, 138, 141, 143, 155, 163 infinite element method, 1, 117, 129, 136, 143, 148, 150 infinite element space, 119, 140, 143 initial boundary value problem, 51, 53, 57, 58, 169, 239, 243 inner product, 13, 19, 49, 52, 60, 66, 218, 220, 231, 238, 242 integral equation, 8-11, 73, 74, 85, 89, 96, 103, 105 Fredholm, 8, 85 interior problem, 8-10, 143 inverse Fourier transform, 170, 174, 197 inverse trace theorem, 15, 33, 34, 36, 38, 47, 84, 94 isometric, 49 isomorphism, 66, 67 isotropic, 25 Jacobi polynomial, 221, 227, 228 Jordan canonical form, 130, 131, 208 kinematic pressure, 35

267

Index kinematic viscosity, 35 Klein-Gordon equation, 241, 243 Korn's inequality, 26-28 Kronecker's symbol, 25 Laguerre polynomial, 219, 229, 230, 240, 246 Lame coefficient, 25 Lame equation, 26, 81, 138 Laplace equation, 2, 12, 38, 129, 134, 139, 146, 154, 163 lateral shearing force, 31 Laurent series, 81 Lax-Milgram theorem, 13, 15, 16, 18, 27, 28, 33, 34, 36, 42, 67, 88, 109, 110, 119, 141, 215, 219, 222, 225, 233, 236 Lebesgue integral, 15 Lebesgue's dominated convergence theorem, 54 Legendre polynomial, 222, 225, 228 linear elasticity, 25 linear functional, 13, 50 linear hull, 45, 46 Lipschitz continuous, 1 load vector, 121 longitudinal component, 61 magnetic flux density, 57 maximum principle, 3, 4, 94, 247 Maxwell equation, 56, 57, 61, 178, 179, 207, 210 natural boundary reduction, 96 Navier-Stokes equation, 1, 35, 41, 44, 167, 183, 186 Neumann boundary condition, 3, 5, 27, 31, 121, 158, 180, 185 Neumann problem, 3-5, 12, 15, 29, 33, 37, 88, 138 Newton potential, 12, 22, 37, 71, 86, 90 Nitsche's trick, 164 norm, 13-17, 19, 24, 28, 32, 34, 44, 57, 62, 63, 65, 71, 87, 94, 110, 112, 130, 141, 147-149, 164, 185, 203,

208, 218, 231, 238, 241 energy, 201 weighted, 19 operator, 18, 39, 51-53, 57, 66, 78, 96-98, 151, 172, 175, 176, 178, 180, 214, 215 adjoint, 49 bounded, 56, 215 Clement interpolation, 144 compact, 9, 215 continuous, 87 differential, 172 Dirichlet to Neumann, 96, 98, 106, 115, 131, 167, 169, 172, 185, 186 finite difference, 207 Fourier transform, 171 interpolation, 108, 114, 163, 165 Laplace, 43, 215, 235 Laplace-Beltrami, 184 linear, 44, 49, 51, 69 Neumann, 185 one-to-one, 67 orthogonal projection, 49, 240 pseudo-differential, 97, 167, 169, 170, 172 self-adjoint, 50, 51, 58, 181 sphere harmonic, 20 symmetric, 49, 51 trace, 120 unitary, 49, 52 ordinary differential equation, 72, 154, 217, 240 orthogonal polynomial, 217, 225 orthogonal projection, 67, 244 orthogonal rational function, 217 Oseen flow, 41, 44, 186 outgoing wave, 19, 167, 168, 171, 173, 191 plane, 83 sphere, 84 outward normal vector, 7, 30 Pade's approximation, 172, 173 partial differential equation, 1, 13

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penalty method, 159 perfectly matched layer, 198, 200, 209-211 perfectly matched layer method, 191 permeability, 57 plane elasticity, 29, 31, 80, 138 plane strain, 29 plane stress, 28, 138 Poincare inequality, 39 Poisson equation, 12, 13, 106, 136, 162 Poisson formula, 73-75, 114 Poisson integral, 4 Poisson ratio, 25, 31 polar coordinates, 71, 78, 198 positive definite matrix, 121, 122, 139 potential theory, 1, 2, 5, 71, 73 power series, 72, 114 quotient space, 65, 148 recurrence relation, 225-229 Rellich's estimate, 19, 22 Rellich's selection theorem, 24 resolution of the identity, 49 Reynolds number, 35, 186, 189 Riemann integral, 98 Riemann sum, 50 Riesz representation theorem, 39 Schmidt procedure, 45 Schwarz inequality, 47, 111, 128, 206, 223, 232 semi-norm, 13, 33, 218, 231 weighted, 163 semi-positive definite matrix, 132 sesqui-linear form, 13, 18, 214 simply supported boundary condition, 31 single layer potential, 5-11, 37, 85-88, 95 singular integral, 37, 98 hyper-singular, 103 Sobolev embedding theorem, 48, 163, 165, 245 Sobolev inequality, 46

Sobolev space, 13, 15, 36-38, 47, 84, 87, 94, 109, 215 weighted, 218, 220, 231, 245 Sommerfield radiation condition, 19, 84 space interpolation, 218 spectral method, 1, 217, 219, 225, 239, 241, 245 generalized Laguerre pseudo, 237 Jacobi, 241 Laguerre, 230, 245 mixed Laguerre-Fourier, 230 rational, 243 rational and irrational , 243 spherical harmonic-generalized Laguerre, 235 spectral resolution, 59 spectral scheme, 219-222, 225, 236, 238-240, 242, 243 spectrum, 18, 51, 215 continuous, 51 point, 51, 214, 215 residual, 51 sphere coordinates, 20, 175 spherical function, 85 star-shape, 117, 118 Stokes equation, 35, 36, 42, 44, 77, 142, 145, 150, 151, 183, 185 Stokes paradox, 41 Stokes problem, 31, 38, 148 strain, 25, 132, 138 stress, 28, 138 Sturm-Liouville problem, 225-227, 229

Tayer's expansion, 89 Toeplitz matrix, 121, 145 total reflection boundary condition, 57, 59, 61, 180 total stiffness matrix, 132 transfer matrix, 120, 122-124, 126, 127, 129, 132, 135, 136, 139, 141, 146, 153, 161 transpose, 25, 120 transverse component, 61

269

Index transverse electric mode, 179-182, 210 transverse magnetic mode, 179-182, 210 uniaxial perfectly matched layer, 208, 209, 211 uniqueness, 3, 9-12, 18, 19, 25, 26, 29, 33, 36, 42, 44, 52, 55, 59, 60, 69, 86, 88, 100, 109, 119, 152, 185, 187 variational formulation, 1, 5, 13, 15, 16, 33, 34, 66, 86, 89, 106, 110, 117, 136, 138, 140, 218, 219, 221, 223, 231, 232 velocity, 35, 38, 61, 77, 145 viscosity, 239 von Neumann condition, 207

wave equation, 17, 19, 53, 167, 176, 179, 180, 191, 195, 197 weak formulation, 26, 36, 42, 45, 48, 54, 59, 94, 95, 236, 242, 243 weak stability, 203, 207 weakly singular, 6 weight, 58, 181, 218, 219, 221, 222, 226-229, 231, 235, 238, 240, 243 well-posedness, 19, 27, 65, 69, 143, 155, 161, 180, 232 Wirtinger inequality, 39, 40 Wronskian, 214 Yee's scheme, 182, 207 Young's modulus, 25

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