Springer Series in
Computational Mathematics
Elliptic Differential Equations
W. Hackbusch
Theory and Numerical Treatment
4
Springer
Wolfgang Hackbusch
Elliptic
Differential Equations
Theory and Numerical Treatment
Translated from the German
by Regine Fadiman and Patrick D.F. Ion
With 40 Figures
4
Springer
Wolfgang Hackbusch
MPI für Mathematik in den Naturwissenschaften lnselstr. 22-26 04103 Leipzig, Germany e-mail:
[email protected] Cataloging-in-Publication Data applied for A catalog record for this book is available from the Library of Congress.
Bibliographic information published by Die Deutsche Bibliothek Die Deutsche Bibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data is available in the Internet at httpd/dnb.ddb.de
Second Printing 2003
Mathematics Subject Classification (2000): 35J20, 35J25, 35J30, 35J35, 35J50, 35J55, 65N06, 65N12, 65N15, 65N25, 65N30
ISSN 0179-3632 ISBN 3-540-54822-X Springer-Verlag Berlin Heidelberg New York o B.G. Teubner, Stuttgart 1987: W. Hackbusch, Throne und Numenik elliptischer Differentialgleichungen. Mit Genehmigung des Verlagea B.G. Teubner, Stuttgart, veranstaltete, allein autorisierte engliache Obersetzung der deutschen Originalausgabe.
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Foreword
This book has developed from lectures that the author gave for mathematics students at the Ruhr-Universität Bochum and the Christian-Albrechts-UniKiel. This edition is the result of the translation and correction of the Gennan edition entitled Theorie und NtLmerik Differentialgieichungen.
The present work is restricted to the theory of partial differential equations of elliptic type, which otherwise tends to be given a treatment which is either too superficial or too extensive. The following sketch shows what the problems are for elliptic differential equations.
B: Discretisatlon: Difference Methods, finite elements, etc.
The theory of elliptic differential equations (A) is concerned with questions of existence, uniqueness, and properties of so&utions. The first problem of
iv
Foreword
numerical treatment is the description of the discretisation procedures (B), which give finite-dimensional equations for approximations to the solutions. The subsequent second part of the numerical treatment is numerical
analysis (C) of the procedure in question. In particular it is necessary to find out if, and how fast, the approximation converges to the exact solution. The solution of the finite-dimensional equations (D, E) is in general no simpie problem, since from to 106 unknowns can oceur. The discussion of this third area of numerical problems is skipped (one may find it, e.g., in Hackbusch [5] and [9)).
The descriptions of discretisation procedures and their analyses are closely connected with corresponding chapters of the theory of elliptic equations. In addition, it is not possible to undertake a well-founded numerical analysis without a basic knowledge of elliptic differential equations. Since the latter cannot, in general, be assumed of a reader, it seems to me necessary to present the numerical study along with the theory of elliptic equations. The book is conceived in the first place as an introduction to the treatment of elliptic boundary-value problems. It should, however, serve to lead the reader to further literature on special topics and and applications. It is intentional that certain topics, which are often handled rather summarily, (e.g., elgenvalue problems) are treated here in greater detail.
The exposition is strictly limited to linear elliptic equations. Thus a discussion of the Navier-Stokes equations, which are important for fluid mechanics, i8 excluded; however, one can approach these matters via the Stokes equation, which is thoroughly treated as an example of an elliptic system. In order not to exceed the limits of this book, we have not considered further discretisation methods (collocation methods, volume-element methods, spectral methods) and Integral-equation methods (boundary-element methods). The Exercises that are presented, which may be considered as remarks without proofs, are an integral part of the exposition. If this book is used as the text for a course they can be used as student problems. But the reader too should test his understanding of the subject on the exercises. The author wishes to thank his collaborators G. Hofmann, G. Wittwn and J. Burmeister for the help in reading and correcting the manuscript of this book. He thanks Thubner Verlag for their cordial collaboration in producing the first German edition. Kiel, December 1985 W. Hackbusch
This translation contains, in addition to the full text of the original edition, a short Section on the integral-equation method. The bibliography has also been expanded.
The author wishes to thank the translators, R. Fadiman and P. D. F. Ion, for their pleasant collaboration, and Springer-Verlag for their friendly cooperation. Kiel, March 1992
W. Hackbusch
Table of Contents
Foreword Thble of Contents
v
Notation 1
Partial Differential Equations and Their Classification Into Types 1.1
1.2 1.3 1.4
2
2.2 2.3 2.4
.
I .
4
.
.
.
6
.
.
.
7
Posing the Problem Singularity Function The Mean Value Property and Maximum Principle Continuous Dependence on the Boundary Data . .
12 12 14 17 .
.
The Poisson Equation 3.1
3.2 3.3 3.4 3.5 4
1
Examples Classification of Second-Order Equations into Types Type Classification for Systems of First Order Characteristic Properties of the Different Types
The Potential Equation 2.1
3
xi
Posing the Problem Representation of the Solution by the Green Function The Green Function for the Ball The Neumann Boundary Value Problem The integral Equation Method
23
27 27 .
28
34
35
36
Difference Methods for the Poisson Equation
38
Introduction: The One-Dimensional Case The Five-Point Formula M-matrices, Matrix Norms, Positive Definite Matrices Properties of the Matrix Lh Convergence
38
4.1
4.2 4.3
4.4 4.5
40 44 53 59
Vi
Table of Contents
4.6 4.7
62
Discretisations of Higher Order The Discretisation of the Neumann Boundary Value Problem 4.7.1 One-sided Difference for 9u/ôn 4.7.2 Symmetric Difference for Ou/On 4.7.3 Symmetric Difference for Ou/ôn
65 65 70 71
onanOffsetGrid Proof of the Stability Theorem 7 Discretisation in an Arbitrary Domain 4.8.1 Shortley-Weller Approximation 4.8.2 Interpolation at Points near the Boundary
72
4.7.4 4.8
5
78
78 .
General Boundary Value Problems 5.1
5.2
5.3
Dirichlet Boundary Value Problems for Linear 85 Differential Equations 85 5.1.1 Posmg the Problem 86 5.1.2 Maximum Principle 5.1.3 Uniqueness of the Solution and Continuous 87 Dependence 5.1.4 Difference Methods for the General Differential 90 Equation of Second Order 95 5.1.5 Green's Function 95 General Boundary Conditions 95 5.2.1 Fbrmulating the Boundary Value Problem . . . Difference Methods for General Boundary 5.2.2 98 Conditions 103 Boundary Problems of Higher Order 103 5.3.1 The Biharmonic Differential Equation 5.3.2 General Linear Differential Equations of Order 2m 104 5.3.3 Discretisation of the Bihaimonic Differential . 105 Equation
Ibols from Functional Analysis 6.1
6.2
83
85
.
6
.
.
.
110
Banach Spaces and Hubert Spaces 6.1.1 Normed Spaces 6.1.2 Operators 6.1.3 Banach Spaces 6.1.4 Hilbert Spaces Sobolev Spaces
110
6.2.1 L2(f1) 6.2.2 Hc(O) and
115
6.2.3 Fburier
110 111
112 114
115 116
and Hc(IRI))
119
vii
Table of Contents 6.2.4
123
Dual Spaces 6.3.1 Dual Space of a Normed Space
130
6.3.2 Adjoint Operators 6.3.3 Scales of Hilbert Spaces
132
6.4
Compact Operators
135
6.5
Bilinear Forms
137
Variational Ermulation
144
6.3
7.1
7.2 7.3 7.4 8
122
and Extension Theorems
6.2.5
7
H8(fl) for Real a 0
130 133
144
HIstorical Remarks Equations with Homogeneous Dirichiet Boundary Conditions Inhomogeneous Dirichiet Boundary Conditions Natural Boundary Conditions
145 150 152
The Method of Finite Elements 8.1
8.2 8.3
8.4
161
The Ritz-Galerkin Method Error Estimates FinIte Elements 8.3.1 Introduction: Linear Elements for 1? = (a,b) . 8.3.2 Linear Elements for a c JR2 8.3.3 Bilinear Elements for a C JR2 8.3.4 Quadratic Elements for (1 C It2 8.3.5 Elements for a C Ut3 8.3.6 Handling of Side Conditions Error Estimates for Finite Element Methods 8.4.1 H1-Estimates for Linear Elements 8.4.2 L2 and H Estimates for Linear Elements . Generalisations 8.5.1 Error Estimates for Other Elements 8.5.2 Finite Elements for Equations of Higher Order 8.5.2.1 Introduction: The One-Dimensional Btharmornc Equation 8.5.2.2 The Two-Dimensional Case 8,5.2.3 Estimating Errors Finite Elements for Non-Polygonal Regions Additional Remarks 8.7.1 Non-Conformal Elements .
8.5
8.6 8.7
161
167 171 .
171
174
178 180 182 182 185
185 .
190
193 193 .
194
194 195
196 196
199 199
viii
Thble of Contents 200
8.7.2 The Trefftz Method 8.7.3 Finite-Element Methods for Singular Solutions 8.7.4 Adaptive Thangulation 8.7.5 Hierarchical Bases
201
.
201
202 202
8.7.6 Superconvergence 8.8 9
Properties of the
203
Matrix
208
RegularIty 9.1
Solutions of the Boundary Value Problem
inH'(37),a>m 9.1.1 The Regularity Problem 9.1.2 Regularity Theorems for 9.1.3 Regularity Theorems for = 9.1.4 Regularity Theorems for General 1? C 9.1.5 Regularity for Convex Domains and Domains with Corners 9.1.6 Regularity in the Interior . Regularity Properties of Difference Equations 9.2.1 Discrete H'-Regularity 9.2.2 Consistency 9.2.3 Optimal Error Estimates 9.2.4 Hz-Regularity .
9.2
10
210 215 .
10.2
.
Differential Equations with Discontinuous Coefficients 10.1.1 Fbnnulation 10.1.2 Discretisation A Singular Perturbation Problem 10.2.1 The Convection-Diffusion Equation 10.2.2 Stable Difference Schemes 10.2.3 Finite Elements
11.2
11.3
Formulation of Eigenvalue Problems . Finite Element Discretisation 11.2.1 Discretisation 11.2.2 Qualitative Convergence Results 11.2.3 Quantitative Convergence Results 11.2.4 Complementary Problems Discretisation by Difference Methods .
219
.
226
226
232 238
240 244
Elgenvalue Problems 11.1
.
223 226
SpecIal Differential Equations 10.1
11
.
208 208
244 244
246
247 247 249 251
253 253
.
254 254 256
.
.
.
260
264 .
.
267
Table of Contents 12
ix
Stokes Equations 12.1
12.2
275
275 SyBtems of Elliptic Differential Equations 278 Variational Formulation . 278 . 12.2.1 Weak Formulation of the Stokes Equations 279 12.2.2 Saddllepo4nt Problems 12.2.3 Existence and Uniqueness of the Solution of a 282 Saddlepoint Problem 285 12.2.4 Solvability and Regularity of the Stokes Problem 12.2.5 A Vo-elliptic Variational Formulation of the Stokes 289 Problem Mixed Finite-Element Method for the Stokes Problem . 290 12.3.1 Finite-Element Discretisation of a Saddlepoint 290 Problem 291 12.3.2 Stability Conditions 12.3.3 Stable Finite-Element Spaces for the Stokes Problem 293 12.3.3.1 Stability Criterion 293 12.3.3.2 Finite-Element Discretisations with the Bubble Function 294 12.3.3.3 Stable Discretisations with Linear Elements in Vh 296 12.3.3.4 Error Estimates 297 .
12.3
Bibliography
300
Index
307
Notation
Formula Numbers. Equations in Section X.Y are numbered (X.Y.1), (X.Y.2), etc. The equation (3.2.1) is referred to within Section 3.2 8Irnply as (1). In other Sections of Chapter 3 it is called (2.1).
Theorem Numbering. All Theorems, Definitions, Lemmata etc. are numbered together. In Section 3.2 Lemma 3.2.1 is referred to as Lemma 1.
Special Symbols. The following quantities have fixed meanings:
A, B,... B, B, C
C'(D), Ck(D), C°°(D)
Gr(a) d(u, VN)
dl', diag{d1,d2,.. .}
f h
H"(i?), H'(S?), KR(y)
I L
I,
matrices boundary differential operators (cf. (5.2.la,b), (5.3.6)) the complex numbers Holder- and Lipschitz-continuousiy differentiable functions (cf. Definition 3.2.8) k-fold and infinitely continuously differentiable functions infinitely differentiable functions with compact support.a (cf. (6.2.3)) distance of the function u from the subapace VN (ci. Theorem 8.2.1) surface differentials in surface integrals diagonal matrix with the diagonal eLements d1,d2,... a function; often the right-hand side of a differential equation Green's function (cf. Section 3.2) step size (cf. Sections 4.1 and 4.2) Sobolev spaces (cf. Sections 6.2.2 and 6.2.4) open ball about y with radius R (ci. (2.2.7), Section 6.1.1) identity or inclusion (cf. Sections 6.1.2, 6.1.3) a differential operator (cf. (1.2.6)) or the operator associated with a bilinear form (ci. (7.2.9')) the stiffnees matrix (ef. Section 8.1) matrix of a discrete system of equations (cf. (4.I.9a))
Notation
linear space of bounded operators from X to Y
L(X,Y) L°°(S?) L2(J2) IN
n
n(x)
0 O(.)
P qh
Rh,Rh
supp(f) u = u(x) = u(xj, Uh
VN,Vh
x,(x,y),(x,y, z)
x=(xl,...,xn)
z I,
p(A) 1?
V
8; 8/On
)xXx' (., .) (., .)O, (, (•, (•, )a
II112 •
j
I1•1100
I
I
I
- 1k, I
lo,
I
(cf. Section 6.1.2) space of essentially bounded functions (cf. 6.1.3) space of square-integrable functions (cf. Section 6.2.1) the natural numbers {1,2,3, . . normals (ci. (2.2.3a)) the zero matrix constlg(x)I Landau symbol: 1(x) = O(g(x)) if cf. (8.1.6) a grid function, nght-hand side of the discrete equation (4.1 .9a) the real numbers, the positive real numbers restrictions (cf. (4.5.2) and (4.5.5b)) the singularity (unction (cf. Section 2.2) the support of the function f (cf. Lemma 6.2.2) a function, e.g., a solution of a differential equation a grid function (discrete solution; cf Section 4) finite-element spaces (ci. (8.1.3) and Section 8.4.1) independent variables a vector of independent variables the integers the boundary of 1? the boundary points of a grid (cf. (4.2.lb), (4.8.4)) fundamental solution function spectral radius of the matrix A or a domain (cf. DefinitIon 2.1.1) an open set In a grid (cf. (4.1.6a), (4.2.la), (4.8.2)) the Laplace operator (cf. (2.1.la)) the five-point difference operator (ci. (4.2.3)) gradient (cf. (2.2.3a)) differences (cf. (4.1.2a-c), (4.2.3)) differences in the normal direction (cf. (4.7.4)) normal derivative (cf. (2.2.3a)) acalar product (cf. (2.2.3c), (4.3.14a)) duality form (ci. Section 6.1.3) acalar product (cf. Section 6.1.4) scalar product and norms on L2(Q) scalar products and norms on the Eudidean norm (cf. (2.2.2)) the Eucidean norm and the spectral norm (cf. Section 4.3) norms equivalent to J , (cf. (6.2.15), (6.2.16b)) maximum norm (ci. (4.3.3)), row sum norm (4.3.11), or supremum norm (2.4.1), ci. also Section 6.1.1 the vector (1, 1,...) (ci. Section 4.3) J
1 Partial Differential Equations and Their Classification Into Types
1.1 Examples An ordinary differential equation describes a function which depends on only one variable. Unfortunately, for many problems it is not possible to restrict attention to a single variable. Almost all physical quantities depend on the spatial variables z, y , and z and on time t. The time dependence might be omitted for stationary processes, and one might perhaps save one spatial dimension by special geometric assumptions, but even then there would still remain at least two independent variables. Equations that contain the first partial derivatives
where (1
i
or even higher partial derivatives
etc., are called
partial differential equations. Unlike ordinary differential equations, partial differential equations cannot be analysed all together. Rather, one distinguishes between three types of equations which have different properties and also require different numerical methods. Before the characteristics for the types are defined, let us introduce some examples of partial differential equations. All of the following examples will contain only two independent variables
x,y. The first two examples are partial differential equations of first order, since only first partial derivatives occur.
Example 1.1.1. Find a solution u(x, y) of (1.1.1)
It is obvious that u(x, y) must be independent of y, i.e., the solution has the form u(x, y) = Thus u(x, y) = w(x) for some arbitrary is a solution of (1).
Equation (1) is a special case of
Example 1.1.2. Find a solution u(x, y) of (cconstant).
(1.1.2)
2
1
Partial Differential Equations and Their Types
Let u be a solution. Introduce new coordinates = x + cy, = y and define I?)) with the aid of =— = Since v(e, := ii), (chain rule) and X,3 —c, y,7 = 1, it follows from (2) that V,3 = + = 0. This equation is analogous to (1), and Example 1 shows that v(e, = co(e). If one now replaces ,g by y one obtains
u(x,y) =
(1.1.3)
Conversely, through (3) one obviously obtains a solution of Equation (2) as long as is continuously differentiable.
In order to determine uniquely the solution of an ordinary differential f(u) = 0 one needs an initial value u(x0) u0. The partial differential equation (2) can be augmented by the initial-value function equation t&' —
for x€Ut
(1.1.4)
with aconstant. The comparison of Equations (3) and on the line (4) shows that = uo(x). Thus is determined by = The unique solution of the initial value problem (2) and (4) reads
u(x, y) = tio(x
—
c(yo
—
y)).
(1.1.5)
The following three examples involve differential equations of second order.
Example 1.1.3. (Potential equation or Laplace equation) Let 0 be an open subset of
Find a solution of
infi.
(1.1.6)
If one identifies (z, y) E It2 with the complex number z = x + iy C, the solutions can be given immediately. The real and imaginary parts of any function f(z) holomorphic in flare solutions of Equation (6). Three examples are Rez0 = 1, Rez2 = x2—y2 and Relog(z—zo) if
$1. To determine the solution uniquely one needs the boundary values
for all (x,y) on theboundaryf=Ofloffl. Example 1.1.4. (Wave equation) All solutions of (1.1.7)
are given by
u(x,y) = w(x + y) +
—
y)
(1.1.8)
where
and are arbitrary twice continuously differentiable functions. Suitable initial values are, for example,
u(x,O) = uO(x), where
=
p+
u1(z),
(x E It)
(1.1.9)
and u1 are given functions. If one inserts (8) into (9), one finds tq = — , where is the derivative of and infers that
1.1 Examples
3
(t4 —tii)/2.
w'= (u1 +t4,)/2,
From this one can determine and up to constants of integration. One constant can be chosen arbitrarily, for example, by s°(O) = 0, and the other is
=
determined by ti(O, 0) =
ço(0)
+
Exercise 1.1.5. Prove that every solution of the wave equation (7) has the
Example 1.1.6. (Heat equation) Find the solution of (1.1.10)
(physical interpretation: u = temperature, y = time). The separation of variables u(z, y) = v(x)w(y) gives that for every c E lit (1.1 .lla)
sin(cx) . exp(—c2y).
u(x, y)
Another solution of (10) for y > 0 is u(x
y) =
L
(l.l.Ilb)
exp((z —
is an arbitrary continuous and bounded function. The initial condition matching Equation (10), in contrast to (9). contains only one function:
where
u(x,0) = uO(x).
(1.1.12)
The solution (lib), which initially is defined only for y > 0, can be extended continuously to y =0 and there satisfies the initial value requirement (12).
ExercIse 1.1.7. Let uo be bounded in lit and continuous at x. Then prove that the right side of Equation (lib) converges to uo(z) for y —'0. Hint: First show that tz(x, y) = 110(z) + — 110(z)) exp( — and then decompose the integral into subintegrals over [z — 6,z +5) and (—oo,z—S)U(z+6,oo). As
equations, equations of higher order can be by systems of first-order equations. In the following we give some
with ordinary differential
described
examples.
Example 1.1.8. Let the pair (u, v) be the solution of the system (1.1.13)
If u and v tions
are
+
twice
=0,
differentiable,
+
the
differentiation of
(13) yields
0, which together imply
Thus u is a solution of the wave equation
(7). The same
that
the equa-
-
=0.
be shown for v.
4
1
Partial Differential Equations and Their Types
Example 1.1.9. (Cauchy-Riemann differential equations) If u and v satisfy the system
v2—u5=O in $?cIt2,
(1.1.14)
then the same consideration as in Example 8 yIelds that both u and v satisfy the potential equation (6).
Example 1.1.10. If u and v satisfy the system
v1+u=0,
(1.1.15)
then v solves the heat equation (10). A second-order system of interest in fluid mechanics can be found in
Example 1.1.11. (Stokes equations) In the system (1.1.16a)
+
—
= 0,
(1.1.16b)
=0
(1.1.16c)
u and v denote the flow velocities in x- and y-directions, while w denotes the pressure.
1.2 Classification of Second-Order Equations into Types The general linear differential equation of second order in two variables reads
+d(x,y)u +e(x,y)uy 1-f(x,y)u-f-g(z,y) =
( 1.
Q
2. i )
DefinitIon 1.2.1. (a) Equation (I) is said to be elliptic at (x,y) if a(x,y)c(z,y) — b(x,y)2 >
0.
(1.2.2a)
(b) Equation (1) is said to be hyperbolic at (x,y), if
a(x,y)c(z, y) — b(x,y)2 2
(n—2)w,,
(2.2.la)
lit", where
=
/F(n/2),
=
2w,
= 4w,
(2.2.Ib)
with I' the Gamma function, is the surface of the n-dimensional unit sphere.
The Euclidean norm of x in IR" is denoted by 1/2
(xl
by
=
(2.2.2)
2.2.1. RrJlxed y E Ut" the potential equation in 1R"\{y} is 8olved s(x,y).
2.2 Singularity Function
15
The proof can be carried out directly. It is simplest to introduce polar coordinates with y as origin and to use (1.5), since s(x,y) depends only on
r=lx—yI. For the next theorem we need to introduce the normal derivative 8/On.
denote the outer F, i.e. n is a unit vector perpendicular to the tangential hyperplane at x and points outwards. The normal derivative of u at x E F is defined as be a domain with smooth boundary 1'. Let n(x) E
Let
normal direction at x
= (n(x), Vu(x)),
(2.2.3a)
where (2.2.3c)
is the gradient of u and (x,y) =
(2.2.3c)
is the scalar product in Ut". In the case of the sphere KR(y) (cf. (7)) the normal direction is radial, and Ou/On becomes Ou/Or with respect to r = — y11 if one uses polar coordinates with the origin at y. It follows from Os(x, y)/Or = —Ix — that Os(x,y)/On =
for x E
(2.2.4)
The first Green formula reads (cf. Green [1J)
L u
dx = —
V
E
j (Vu, Vv) + j if the domain 1? satisfies suitable condi-
tions. Here frn... dl' denotes the surface integral. Domains for which Equation (5a) holds are called normal domains. To see sufficient conditions for this refer to Kellogg (1, Chapter LVI and Hellwig 11, 1-1.21.
Functionsu,v E C2(Thinanormaldomain.Qsatisfythe second Green
formula
/ u4v =
/ viju + /
—
dl'.
(2.2.5b)
Theorem 2.2.2. Let (2 be a normal domain, and Let u E there. Then
u(y) = for all y
(2.
Here 0/On, and dl' refer to the variable x.
be harmonic
dl',
(2.2.6)
____ 16
2 The Potential Equation
PROOF. By Kr(y) := (x E
x—
2
(2.2.lla)
for E Q, =y+ — — y) and show: (a) -y is a fundamental solution in fl, (b) y(x, = x), (c) on the surface 1' = OKR(y) the following holds:
with
=
=
—
Ix — y12)Ix
E
I').
(2.2.llb)
2.3
The Mean Value Property and Maximum Principle
DefinitIon 2.3.1. A function u has the mean value property in S? if C°(fl) and if for all x E Q and all R >0 with KR(X) C a the following equation holds UE
u(x)
1
J
=
(2.3.1)
8KR(x)
dl.' = the right side in (1) is the mean value of u taken over the surface of the sphere. An equivalent diaracterisation results if Since
one averagea over the sphere KR(x).
Exercise 2.3.2. u C0(IJ) has the second mean value property in 1? if u(x) = for all x a, R> 0 with KR(x) C a. Show that this mean value property is equivalent to the mean value property (1). Hint: IKR(x)
= fR
fSK(x)
dr.
Functions with the mean value property satisfy a maximum principle, as is known from the function theory for holomorphic functions:
18
2 The Potential Equation
Theorem 2.3.3. (Maximum-minimum principle) Let (1
be a domain and let u E C°(Th be a nonconstant function which has the mean value property. Then u takes on neither a maximum nor a minimum in a.
PROOF. (a) It suffices to investigate the case of a maximum since a minimum of u is a maximum of —u, and —u also has the mean value property. (b) For an indirect proof we assume that there exists a maximum in y E a:
u(y)=Mu(x)
fora.llxEa.
In (c) we will show u(y') = M for arbitrary y' E I?, i.e. u M in contrast to the assumption u const. (c) Proof of u(y') = M. Let y' E (1. Since S? is connected, there exists a path connecting y and y' running through a, i.e. there exists a continuous (2 with y, 10,11 = y'. We set
I := {s
M for all 0 t s}.
E 10,11:
I contains at least 0, and is closed since u and ço are continuous. Thus there In (4) exists s' = max{s I}, and the definition of I shows that I = [0, it is proved that s' =lso that y= E land hence u(y') =M follows. (d) Proof of? = 1. The opposite assumption 8' 0 with C £1. Evidently, it follows that u = M in KR(x)' if it is shown that u = M on ôKr(X)' for all 0 < r < R. (e) Proof of u = Mon OKr(x'). Equation (1) in x' reads
M = u(x') =
I
JSKr(X')
In general we have u(C) M . If one had u(C') <M fore' OKr(X') and thus also u < M in a neighbourhood of one would have on the right side a mean
value smaller than M. Therefore, u = M on ÔK4x') has been proved.
•
Simple deductions from Theorem 3 are contained in
Corollary 2.3.4. Let (2 be bounded. (a) A function with the mean value property takes its maximum and its minimum on 8(2. (b) If two fsinctioni with the mean value property coincide on the boundary
8, they are identicaL PROOF. (a) The extrema are assumed on the compact set = QU Ofl. According to Theorem 3, the extremum cannot be in flif u is not constant on a connected component of fl. But in this case the assertion is also obvious. (b) If u and v with u = v satisfy the mean value property on 8$? then the latter is also satisfied for w := u v. Since w = 0 on 0(2, part (a) indicates
-
19
2.3 The Mean Value Property and Maximum Prmciple
Lemma 2.3.5. Harmonic functions have the mean value property.
PROOF. Let u be harmonic in Qand y KR(y) C (2. We apply the representation (2.6) for (2 = KR(y). The value s(x,y) is coostant on 8KR(y): it be denoted by a(R). Because of (2.4), Equation (2.6) becomes
f
Ott
u(y) = cr(R) j
dl' +
1
1
udf'.
J8KR(y)
The equation agrees with (1) if the first integral vanishes. The latter follows from
Lemma 2.3.6. Let u E
C2(Th)
be harmonic in a normal domain 0. Then
we have (2.3.2)
PROOF. In Green's formula (2.5a) substitute 1 and u for uand v respectively.
S Lemma 5, Theorem 3, and Corollary 4 together imply Theorems 7 and 8:
Theorem 2.3.7. (The maximum-minimum principle for harmonic functions) Let u be harmonic in the domain $1 and nonconstant. There exists no maximum and no minimum in 11.
Theorem 2.3.8. (Uniqueness). Let $2 be bounded. A function harmonic in I? assumes its maximum and its minimum on 8$? and is uniquely determined by its values on 811. The representation (1) of u(y) by the values on OKR(y) is a special case of the following formula which will be proved at the end of this section and which provides Equation (1) for x = y.
Theorem 2.3.9. (Poi8son's integral formula) Assume we have cc E C°(OKR(y)) and n 2. The solution of the boundary value problem
4u = 0 in KR(y),
u=
cc on
OKR(y)
(2.3.3)
is given by the function u(x) =
R2_Ix_. Yl 2
f
8KR(y) IX —
which belongs to C°°(KR(y)) Ii C°(KR(y)).
for x
KR(y),
(2.3.4)
2 The Potential Equation
20
The mean value property only assumes u E C°(Th, while harmonic functions belong to C2(.O) (1C°(fl). This makes the following assertion surprising:
Theorem 2.3.10. A function is harmonic infl if and only if it has the mean value property there.
PROOF. Because of Lemma 5 it remains to be shown that a function v with the mean value property is harmonic. Let x E KR(X) C (1 be given arbitrarily. According to Theorem 9 there exists a function u harmonic in KR(X) with
inKn(x), u=v on.ÔKR(x). According to Lemma 5, u has the mean value property, as does v, and Corol-
lary 4b proves that u =
i.e. v is harmonic in KR(X). Since
v in
U
KR(X) C 11 is arbitrary, v is harmonic in (1.
An important application of Theorem 10 is
Theorem 2.3.11. (Harnack) Let monic in 1? and converging uniformly in
be a sequence of functions harUk is harmonic =
Then u
mi?. The limit process, PROOF. The limit function is continuous: u yields Equation (1) for u; i.e. applied to Uk(X) = u has the mean value property. According to Theorem 10, u is also harmonic N intl. Theorems 3 and 7 on the maximum-minimum principle pertain to global extrems. The proof of Theorem 3 does not yet exclude local extrema in the interior. It merely shows that u is then always constant in a circle KR(y) C 1?. As is known from function theory one can derive from this u(x) = M in if u is analytic, 1. e., there is a convergent power series expansion in a neighbourhood of any X E I?. Indeed, the following theorem holds whose proof can be found, for example, in Hellwig (1,11111.5]:
Theorem 2.3.12. A function harmonic in tl is analytic there. The proof of the Poisson formula still needs to be carried out. (a) First we = must show that u in (4) is a function harmonic in KR(y), i. e., it satisfies 0. Since the integrand is twice continuously differentiable and the domain of is compact, the Laplace operator commutes with the integration 1' integral sign:
=
(&i)1
j
—
Ix
—
.
Ix
—
for x (2.3.5)
2.3 The Mean Value Property and Maximum PrInciple
21
According to Exercise 2.2.4 there exists a fundamental solution i'(x,
such
that 8
8
E
Ix —
f',x E KR(y).
(2.3.6) From =0 and (5), one infers that = =0. (b) The expression (4) defines u(x) at first only for x KR(y). It still needs to be shown_that u has a continuous extension on KR(y) = KR(y) U I' (i.e., u E C°(KR(y))) and that the continuously extended values agree with the boundary values w:
forzEf'.
tim
By Equation (6), putting u R2—(x—y$2
Let z e
(2.3.7)
1 in Corollary 2.2.3 gives the identity
f Jr
foraJlxEKR(y).
(2.3.8)
I' be arbitrary. Due to Equation (8) one can then write: u(x) —
= R2—Ix—y12J
(2.3.9a) —
Figure 2.3.1. ConstructIon of We define 1'O Vfl expression (9a) into ti(x) —
(see Figure ço(z)
R2—Ix—Y12 —
= lo +
I
where
i=40
2 The Potential Equation
22
Since I
—
fro
fro
I —
x—
f,.
it follows from Equation (8) that Jo 5 max Because
of the continuity of
one
(2.3.9b)
—
F.)
can choose p >0 auth that for given e >0 0, there exist numbers N(c) and 6(c) > 0 such that the following implication holds: ip, E
n
N(c),
x€ F, y€ I',,,
The sequence un
E
6(e)
e. (2.4.4a)
C°(IZ,) converges uniformly tou E C°(.Q) if
2.4 Continuous Dependence on the Boundary Data
Remark 2.4.5.
(a)
Let K be a set which is compact (i.e. complete and E C°(K) converge C K for all n. Let
bounded) with r c K, and uniformly on K to If
on F then (4a) is satisfied. be the following (not continuous) Then (4b) is u on ii on (1 in the usual sense.
and
on
C 1? hold for all n and let (b) Let = onto Th continuation of 3quivalent to uniform convergence
and let 1 which are harmonic in 5Z, be solutions of
Theorem 2.4.6. functions
(2.4.4b)
fl I?) = 0.
— u(x)I: x
lim
25
Let $1,., C (1, with Si
-,
P. Let the
(2.4.5a)
on F,.
Then
conve!ye uniformly in the sense of(4a) to — the following as8ertwfls hold: Let
(a)
E
ifthereezistsasolutionuEC2(J?)flC°(J?) of
on!'
u=cp
(2.4.5b)
—, ti holds in the sense of (4b). then C°(a) is satisfied in the sense of (4b), then ti is —+ ti (b) if conversely the solution of (5b).
PROOF. (a) Let the continuation II,, be defined as in Remark 5b. Since u is >0 for all e >0 such that there exists uniformly continuous on
if lx
lu(x) —
Set 6(e) := exists
(2.4.6a)
—
with 6 from (4a). Because
max{Np(e), N(e/2)} (N from (4a)) we want to show that For x E S?\S1,, the estimate is trivial because for x all x c a, however, there holds —u(x)I =
— u(x)l
—ul
I'
there := — u(x)l = u(x). For —+
For n
for n
so that
(2.4.6b)
It remains to estimate withn N(e)thereexistsy €1' with
(cf. Theorem 2.3.8), because tin - u is harmonic in
forxE —
Forx€
6(e/2). Thus we obtain — u(x)j =
— u(x)l S
—
+
— u(x)l
from (4a) and (6a). Since x i',, is arbitrary it follows that ltin —til S on and (6b) proves the uniform convergence ii,, —' ti on Hence, from Remark 5b it follows that (4b) is satisfied.
26
2 The Potential Equation
(b) Let K C 1? be a compact set. Since 17,.. -+ 1' there exists a N(K) such N(K)} converges for n N(K). Thus, the sequence {u1,: n that K C uniformly in the usual sense on K to u so that one can apply Theorem 2.3.11: consequently u is harmonic in K. Since K C £1 may be chosen arbitrarily, it follows that u E C2(S?). By assumption, we already have U E C°(Th). That the
•
In Theorem 4 one was able to derive the existence of a solution t& of (5b) a as This inference is not possible for the case of just from ço, -. the following example shows. C a := K1(O)\{O} C 1R2. The boundaries are
:= Let = 0K1(O) U
and V = 0K1(O) U {O}, and satisfy
The
—p
boundary values
=
= 0 on 8K1(O),
satisfy the condition —+ (5a) can be given explicitly:
=
1
on
0) =
1
(cf. (4a) and Remark 5a). The solutions
of
u,..(x) = tog(IxI)/log(1/n). u(x) := 0 holds pointwise, but u = 0 satisfies neither (4b) Obviously, u,(x) nor the boundary value problem (5b). Conversely, one infers from Theorem 6a the following result:
solution u
a
= K1(O)\{0} c 11t2 the potential equation has In C2(S?) fl C°(J7) which assumes the boundary values u(x) =0 on
Remark 2.4.7.
8K1(O)andu(x)=linx=0.
no
3 The Poisson Equation
3.1 Posing the Problem The Poisson equation reads
au—f in!?
(3.1.la)
C°(Q). In the physical interpretation f is the source term ifor example, the charge density in the case of an electrical potential uJ. To determine the solution uniquely one needs a boundary value specification, for
with given f
example, the Dirichiet condition (3.l.lb)
DefinitIon 3.1.1. The function u is called the classical solution of the boundary value problem (la,b) if u E C2(O) fl C°(Th) satisfies the equations (la,b) pointwise. Until we introduce weak solutions in Section 7, "solution" will always mean "classical solution".
The solution of the boundary value problem (la,b) will in general no longer satisfy the mean value property and the maximum principle. But these properties still hold for the differences of two solutions u, and u2 of the equation, since =0. Thus the uniqueness of the solution of — u2) = — problem (la,b) immediately follows and Theorem 2.4.3 can be brought over:
f f
Theorem 3.1.2. Let 0 be bounded. (a) The solution of (la,b) is uniquely deterinineii (b) ffu' and u11 are solutions of the Poisson equationfor boundary values w' and then we have IIti'
—
u"II,.,
0. Show that lxi' E C'(Kft(O)), if a
otherwise lxi'
E
Hint: 1 —t' (1—i)' for O< t 1, a 0. The function u from Equation (2) can be decomposed into u1 +u2 where and ua = — f,, çoôg/On dl'; u is the solution of the boundary UI = — f0 91 value problem (1.la,b) if we are able to show that u1 and u2 are solutions of
iiuj=finf1, Theorem 3.2.10. LI the Ciw, function exists and satisfies suitable conditions then
u(x) = —
is
j
a classical solution of 4u =0 inS?, withu=
x)
(3.2.6)
on!'.
The proof goes in principle just as for Theorem 2.3.9 (cL Leis [1, p.69J).
32 Representation of the Solution by the Green Function
Theorem 3.2.11. Suppose the Green Junctiong(',x) exists, and let it be f E C'(Th). Then u(x) = —
f
31
C2(Th\{x})forx ES?
J with u = I) on I'.
in
0 for x E I' follows easily from 0 and (3). The property u C'(Th) and the representation result from the
PROOF. The boundary condition u(x) =
x be bounded and A := Exercise 3.2.12. Let (2 C E C°(A) x}. For the derivatives of / with respect to x aseume k. Prove that then with 3 0, it follows that =
. .
Coo Ca cc
> 0.
71,,7k—jEI
to Lemma 9, p(C) 0 for all
(4.3.15)
Exercise 4.3.22. Prove that (a) a symmetric matrix is positive definite if and only if all eigenvalues are positive. (b) All principal submatrices of a positive definite matrix are positive definite (cf. Exercise 19). (c) The diagonal elements a positive definite matrix are positive.
(d) A is called positive semi-definite if the inequality (15) holds with "" instead of">". A positive semidefinite matrix A has a unique positive semidefinite square root B = A"2, which has the property B2 = A. If A is positive definite, then so is A1t3. A corollary to Exercise 22a is
Lemma 4.3.23. A positive definite matrix A u nonsingular
and has a pos-
itive definite inverse. The property "A1 is positive definite" is neither necessary nor sufficient
to ensure the property "A-' 0" of an M-matrix. In both cases, however, (irreducible) diagonal dominance is a sufficient criterion (cf. Criterion 10).
Criterion 4.3.24. If a symmetric matrix with positive diagonal entries is diagonally dominant or inite.
diagonally dominant then it is positive def-
PROOF. Since
resp. the Gershgorin circles which occur in Criterion 4 do not intersect the semi-axis (—oo, 0), so that all the elgenvalues
must be positive. By Exercise 22a then A is positive definite,
Lemma 4.3.25.
be )tmin and the smallest and largest eigenvalues of a positive definite matrix A. Then there holds
UA(12 = Amex,
tIA'1j2 =
1/Arnie.
(4.3.16)
4.4 Properties of the Matrix Lh
4.4
=
and
From (5) then result p(A) =
= p(A').
p(A) and 11A'hh2
PROOF. Exercise 20a shows that hAil2
since
53
.Xmin
> 0.
U
Properties of the Matrix Lh
Theorem 4.4.1. The matr%x Lh (five-point formuLa) defined in (2.5) has the properties:
(4.4.la) (4.4.lb) (4.4.lc) (4.4.ld)
Lh is an M-matrix,
Lh is positive definite,
8h2,
1/8,
8h2 cos2(irh/2) O. M on f'h implies the
(2) Let M be the right side of (4a). —M wh inequallties —M
55
S
M on .17h.
The discrete analogue of the Grcen function
is
Let
be the scaled unit vector
h2
X
(x,
{
—
The column of the matrix
with index
E [lh).
E
(4.4.5a)
is given by
(eEQh).
(4.4.5b)
ah fixed, gh(•,E) is a grid function defined on definition is extended to x Thh:
The domain of
For
The Lh implies
are
entries of h2L':
Remark 4.4.6.
=
= for all
(4.4.5()
E
l'p,
The symmetry of
E
(ci (3.2.3)).
The representation (3.2.7) is recalled by:
Remark 4.4.7. The solution uh of the system of equations (2.4a) with boundary values 'p =0 reads uh(X)=h2
(4.4.6)
Equation (6) is the component-wise representation of the equation Uh = The factor h2 compensates for h-2 in (Sa). It was introduced so that the summation h2 E in (6) approximates the integral f0. Since in Section 3.2 we consider the Poisson equation = f, but here the equation = f, the right-hand sides in (3.2.7) and (6) differ in their signs. The discrete Green is positive also (cf. (3.2.9)):
Remark 4.4.8. 0
+
for the solution of the boundary value problem (2.2a,b) has not been mentioned until now, but will be proved in a more general context in Section 5.1.3. The maximum norm fl. in (9) can be replaced by the Euclidean norms
Jh2 V
Here, h2
and fe,, h
/h x€QA
lco(x)P.
V
and 5,, correspond to each other.
Theorem 4.4.12. Under the assumptions of Theorem 11 there holds
tIL'
—'
+
1
1
S
+
1
(4.4.11)
PROOF. (1) It suffices to consider the case of the potential equation (i.e., f 0). Let the restriction of on f'h result in the grid function
Equation
4 Difference Methods for the
58
= Ltcoh be given by the rectangular Let the mapping matrix A: uh = Açoh. According to Equation (8) the entries of A read for x E
= Since
— hn,x) =
=
A 0 (cf. Remark 8), one obtains the row sum norm IIAIk0 for the choice of ph(x) = 1 in all x fl',. as = IIAwhIko = 111U00
The column sums of A are s(e) :=
The grid function Vh boundary has the values — hn),
8(e)
The
= 1 (why?), so it follows that
then reads
solution vh =
(2) for
E fl.
XE
— hn),
h2L'l
1.
= at the points
— —
hn)
hn near the
n normal direction,
E
be a point of the left or right as is implied by Remark 7. Let e = E = 0 or 1). As mentioned in the proof of (ic), Lhwh boundary (i.e., 1 and x(l Since 0, then so is wh — z)/2. holds for wh(z,y) := hence vh < h2wh. In particular we have the following estimate
=
—
hn)
—
hn)
= h2h(1
—
h)/2
at the point near the boundary
e
or from the lower boundary one obtains the same estimate if are the row sums one uses wh(x, y) := y(l — y)/2. Since the column sums of AT, we have proved
=
E FL)
h—t/2.
(3) We have IIATAH2 = p(ATA) IIAT(IOOIIA,I Exercise 4.3.20a, (3.lOd), (3.lOa)) so that the solution Uh = satisfies the following estimate
=
u2(x) =
h2 XEOk
2 N 8x0x +
8
i— +
82
liii
j
"
+>2
I
8
+
8 j
+0.
(5.1.2)
But since, for example, the operator = + (2) can be described in the form (ib), provided the coefficients are sufficiently often differentiable. According to Definition 1.2.3, Equation (1) is elliptic in S? if all elgenvalues of A(x) have the same sign. One can assume without lose of generality that all elgenvalues are positive so that A(x) is positive definite (cf. Exercise 4.3.22a). is elliptic m (1 if
forallxEO, $,3=1
(5.1.3a)
5 Ceneral Boundary Value
86
: = 1) and it must be For any x E 1? there exists c(x) := positive (c(x) is the smallest eigenvalue of A(x)!). Hence one can also write in the form (3a'):
E
c(x) > 0 for all x E 1?,
E
(5.1.3a')
1
Definition 5.1.1. The equation (is), or the operator L, is defined to be uniformly elliptic in 1? if Q} > 0 (c(x) from (3s')).
Inf{c(x): x
(5.i.3b)
On P =81? we impose the following Dirichiet boundary value condition:
onl'.
(5.1.4)
5.1.2 Maxhnum Principle In general, the maximum principle does not hold for the equation Lu =1, nor is the solution of the boundary value problem (is), (4) uniquely determined.
Example 5.1.2. Let 1? = (O,w) x (O,w), = 0, f = 0, Lu = + 2u. = sin(x)sin(y) are solutions of the boundary Then both u = 0 and value problem. The second solution assumes its maximum at the interior point
(ir/2,ir/2) EQ. In the above example the coefficient 0(x) = sign. As soon asaO, we have
2
(ci. (ib)) has the wrong
Theorem 5.1.3. (Maximum-minimum principle) Let a(x)
0 in the
domain 1?. Assume the coefficients of the elliptic opemtor (ib) are continuous
mO. LetuEC2(Q) .atisfyLu=fandbenonconstant. Then we have: in 4?, no negative minimum of u exists in 1?; (a) if f (b) if f 0 in 1?, no positive maximum exists in 0.
PROOF. The proof is based on Hopf's lemma, which can be studied, for example, in Heliwig 11,111-1.11. Here we give a shorter proof, which however, the stronger condition 0(X) 0 in contradiction to the assumption I 0. Part (b) is proved analogously.
ExercIse 5.1.4. The trace is defined by tr(A) := (a) 0 and tr(A) 0 if A is positive semideflnite; (b) tr(AB) = tr(BA) = (c) tr(A.B) 0, if A and B are positive semidefinite.
Prove that:
Hint for (c): B112AB112 is positive semidefinite; Exercise 4.3.22d.
As in the case of the potential equation, from Theorem 3 follows the
Corollary 5.1.5. Let I? be bounded (not necessarily a domain); furthennore, as*
that the conditions of Theorem3hold JffOin (JJ and
sf there exists a negative minimum [positive maxzmurnj of u in on the boundary 8fl.
it must lie
Remark 5.1.6. The continuity of the coefficients and a of L in Theorem 3 and COrOllary 5 can be replaced by the assumption a 1) (cf. Bramble-Hubbard 11)). Layton—Morley 111 point in general out that with weaker conditions than (12) one may still obtain a matrix Lh, which, though not an M-matrix, does have a positive inverse. To obtain a method of consistency order 2, one must discretise E as in (10). The following corollary shows that —Lh is also an M-matrix when is sufficiently small.
Corollary 5.1.16. In addition following hold:
the assumptions of Theorem 14 let the
to
h
(s=1,2).
(5.1.15)
fvvm (10) together with (13') leads to a Then the discretisation of seven-point difference method of second order of consistency such that -L,, is an M-matriz.
PROOF. -Lh satisfies (4.3.la) and is irredudbly diagonally-dominant. ExercIse 5.1.17. The condition su 1a121 + 8ufflcient. Construct a counterexample with a11 = variable, and lad =6 so that L,, is singular.
•
instead of (15) is not 1, 0, h = 1/3,
=
Considerably weaker conditions for the nonsangulanty of L,, than in Theorem 14 and Corollary 16 are needed in Section 9.2 (cf. Exercise 9.2.6, Corollary 11.3.5).
is not a symmetric matrix. Symmetry of L,, is to be exin general, pected only if L is also symmetric: L = Ii. Here the formally adjoint differential operator L' which is associated to L in (2), is defined by
L=
1111
[aj
82
+
0
8
+
82
1
1
_E [(li
8
+
8
+a. (5.1.16)
It is easy to see that a 8ymmetric operator can always be written in the form
8 vxj
(5.1.17)
A difference method for this is given for the case n=2 and a12 =0 by the
five-point star
5.1 Dirichlet Boundary Value Problems for Linear Differential Equations
93
0
o —
a22(x,y
—an(x,y+
—
0
o
o
a(x,y)
0 0
o
0
0
+0
o
(5.1.18)
.
Theorem 5.1.18.
The difference method (18) is Let a22 E The associated matrix L,, is symmetric. If a.,, > 0 consistent of order (ellipticity) and a 0, = a21 0, a11 + a12 > 0, and also + h/2) + a12(x — h/2, y + h) + a12(x -f h/2, y) > 0, and a 0. Show that the difference scheme which is described by (18) and (19) has consistency order 2 and that the ansociated matrix —L,, is a symmetric, irreducibly diagonally-dominant and positive definite M-matrix. (b) What is the suitable discretisatton for the case a12 0?
a11
>
0,
Exercise 5.1.20. L reads
The difference formula from Theorem 14 for the operator
+
a when
5 General Boundary Value Problem8
94
0, 0. Let the associated matrix be Lh. Then prove that the 0, transposed matrix LZ describes a difference method for the adjoint operator V and also possesses consistency order 1.
012
In general it is possible to show for regular difference methods that a discretisation of L'. The role of regularity is demonstrated in
is
Example 5.1.21. Let Lu := u" + au' in 11 = (—1, 1) with a(x) 0 for 0 for x > 0. According to (14) au' is discretised for 0 and a(z) z and for x > 0 by a(x)8u(x). Let the associated matrix z 0 by + Lh,1, where Lh,2 and Lh,1 correspond to the terms u" and be Lh = Lh,2 should be a discretisation of au' respectively. According to the above, —(av)'. But the differences at x = 0 and x = h are h'ta(—h)vh(—h) — a(O)vh(O)
h'[a(O)vh(O) + a(h)Vh(h)
—
.-(av)' — h10(O)Vh(0) + 0(h),
—
o(2h)vh(2h)1 = —(av)'
+ h10(0)Vh(0) + 0(h);
is a possible discretisation of thus they are not consistent. Nevertheless, has order of L'v = v" — (av)', for it can be shown that the error — magnitude 0(h). To prove stability one has to show const. Obviously it is sufficient to prove this inequality for sufficiently small h. In the proof of Theorem 9 we used the fact that
—Lwl mO,
on!'
for w(x) := exp(2Ra) — — + R)). Let Dhuh(x) be the difference equations from which Lhuh results after elimination of the boundary values. We set we,, := 2Rhw; i. e., wh(x) 2w(x) for x The followrng holds:
- RhL)W - 2RhLW 2 -
-Dhwh =
Each consistent difference method satisfies fl(DhRh small ho we thus have
- RhL)w. -
-' 0.
For
sufficiently
—Dawh(x)l (XE 0h, h—D,,w,,1 holds for all grid points. Theorem 4.3.16
5.2 General Boundary Conditions
95
Theorem 5.1.22. The ducretssations from Theorem 14, Coroflary 16, Theorem 18, and Exercise 19 are stable under the conditions posed there, i.e., const for alt h H = {1/n:n E IN}. According to Theorem 4.5.3 the methods converge. The order of convergence order of consistency.
agrees
with the corresponding
5.1.5 Green's Function The idea of representing the solution by the Green function can be repeated for the general differential equation (Ia,b). The Green function (of the first x) is singular at = x and satisfies kind)
=
=0 foe
0,
x
forxEIoreEf'. Here, L' is the adjoint differential operator (16). If L L', then g is no longer x). Under suitable conditions the solution of (la-c), symmetric: g(x, (4) can be represented as
u(x) = —
x)
L
where B = Bt =
is a boundary differential operator (n, are the components of the normal vector n = F). Only when the e principal part of L agrees with is B the normal derivative. In the discrete case the inverse L' again corresponds to the Green function g(., .).
5.2
General Boundary Conditions
62.1 FormulatIng the Boundary Value Problem Let the differential equation be given by (1.la,b). The Dirichlet boundary condition (1.4) can be written in the form Bu
on I'
(5.2.la)
where B is the identity (to be precise: the trace on F). In more general settings B can be an operator — a so-called boundary differential operator — of order I: B
Ebs(x)o/ax. +bo(x),
If one introduces the vector b(x) = (bj(x), .. the form
XE F.
(5.2.lb)
By can be written in
5 General Boundary Value Problems
96
or B = bTV + b0.
Bti = (b(x), Vu(x)) +
(5.2.1W)
0 there results what is known as on F, also known as the boundary
Example 5.2.1. (a) From b = 0, b0(x)
the Dirichiet condition u = condition of the first kind.
(b) The choice b = n, ho = 0 characterises the Neumann condition, also
called the boundary condition of the second kind. (c) Equation (Ia) with (b,n) 0, ho 0, is known as the mixed, or the boundary condition of the third kind. Occasionally one just means by a mixed boundary condition: B = unTV ÷ ho = o'O/On + ho with or(x) 0. Remark 5.2.2. The case (b,n) = 0 is excluded in general. For (b,n) = 0, bTV is a tangential derivative. The boundary condition Bu = w is then very similar to a Dirichiet condition. The condition "u = on F" implies "Bu = := Bço on F" (Why is Bço defined?).
—x)/2
(a)
FIgure 5.2.1. (a) Boundary value problem with changing boundary condition type (b) Dirichiet problem in a disk with a cut
The normal derivative B = 8/8n (i.e. b = n) is important in connection with L = —4. For the general operator L in (l.lb) the so-called conormal derivative B with
b = An (A = (°ij)jJ1,...,n, aq as in (I.lb)) is of greater importance, as we will see in Section 7.4. Statements about existence and uniqueness of the solution always depend
on L and B. We have seen already that for L = —4, B = I (Dirichlet condition) uniqueness is guaranteed (cf. Theorem 3.1.2), while the problem associated to L = —4, B nTV = 8/t9n is, in general, not solvable (cf. Theorem 3.4.1).
5.2 GeneraL Boundary Conditions
97
The coefficients of B depend on position. Of course, B(x) 0, i.e., b(x) = o and bo(x) = 0, must not occur for any x E I'. But it is poesible that b(x) = 0 (and bo 0) in C I' and b(x) 0 in t'\'y. Then there is a on the piece -y and a boundary condition of fIrst Dirichiet condition u
order on the remaining boundary piece f'\-y. At the points of contact between
and I'\'y the solution generally is not smooth (it has singularities in the derivatives).
0 with y radius 1. Let the differential equation and boundary conditions be given as in Figure la. The boundary condition changes its order at x = y = 0. The (cf. (2.1.3)). Check that solution in polar coordinates reads: u = r 1/2 = O(r112) and =
Example 5.2.3. Let .0 be the upper semicircle around x =
The same singularity as in Example 3 occurs in the problem described This Dirichlet problem and in Figure ib; the solution is also r112 Example 3 are closely connected with each other.
Figure 5.2.2. A domain symmetric with respect to
Example 5.2.4. Let 5? =
U Q2 U be as in Figure 2: let the reflection of S?j in -y result in S?2. If one seeks a solution of Lu = in (1, Bu on
f
OS?, and if together with u the function II reflected in 'y is also a solution, one expects u = This solution then satisfies Lu = in Ru = on
Ou/On=Oon'y.
f
While the boundary condition Bti = on OS? may be of physical origin, Example 4 shows that a Neumann condition may also have a geometric basis. Another geometrically justified boundary condition is the following. Let 5? be given as in Figure 3: and 'Y2 are parts of F =85? with
98
5 General Boundary Value Problems
= {(x,,y):yi y Y2},
Then, in addition to Bu =
on
boundary condition u(x1,y)
t4z2,y),
f'\(i'i
i =
U 72), we
uz(Zj,y) = Uz(X2,y)
1,2.
can require the periodic for
Yl y Y2,
(5.2.2)
The solution is periodically continuable in the x-direction (with and period x2 — x1). The origin of periodic boundary conditions is discussed in Example 5, on
Example 5.2.5.
(a) Let if be an annutus which is described by the p0-
lar coordinates r E (ri, r2), w E [0, 2ir). fransformation of the differential equation to polar coordinates gives as the image domain the rectangle 11 = (i-i, r2) x (0, 27r). The original boundary conditions on if become bound-
ary conditions at the upper and lower boundaries while Equation (2) describes the periodicity of the angular variable x E (0, 2ir).
(b) Instead of on (1' C Ui?, one can also define a boundary value problem on a part of the 2-dimensional surface of a 3-dimensional body. If if lies, for example, on the surface of the cylinder E e IR}, the + 0 and second inequality in (4a) agrees with (4.3.4b). The corresponding inequalities for (3c') read: Here
ct?O, C20,
(5.2.4b)
and the boundary equations (3c') 0, (4a,b) implies be combined into Ahuh =
Let the difference equations in x E given for
E
Therefore, Ah is an M-matrix if A,, is irreducible and (4.3.4a) holds for at least one x E siderations explain why the boundary discretisation should The above satisfy the conditions (4b). (4b) holds for the case b0 =0 if and only if b2/b1 E [—1,0J, (4b) can be satisfied if one interpolates between and (0, 11. If b2/b1 i.e., if the tangential and + h). If, however, — h (instead of > and component is greater, one could interpolate between k. Fbr the case b2/b1 1, however, it is more practical kh), where 1b2/bd + h) between to interpolate at the point + h) and + h,V + h). Generally one should choose as interpolation point the point of intersection of the line with the dotted straight line in Figure 4. In the preceding discussion we started with the case fl = (0, 1) x (0, 1). Iii? is a general region, two discretisation techniques offer themselves (cf. Sections 4.8.1 and 4.8.2).
5.2 General Boundary Conditions
101
F
.--
Line
—
S — h)
•: Points in U: Points in Th
FIgure 5.2.5. FIrst boundary discretisation
First option for discretisation: and lj, be chosen as in Section 4.8.1. At near-boundary points the Let differential equation is approximated by a difference scheme which in the case
of L =
corresponds to the Shortley-Weller method. For this one needs As in Figure 5 let the values of Uh at the boundary points E a boundary point. Again we can set up an equation — sjh, be E 1h has the same position as + h, in analogous to (3a—c). In Figure 5 Figure 4. In general one has to use the point of intersection of the line which is also presented by + tb (t IR), and the dotted straight line in Figure —LI
5.
Second option for discretisation: Let consist of all points of be the grid aa used above. Now let that are far from the boundary. For all (x, y) Qh difference equations (with equidistant step size) are declared which involve For each point a boundary discretisation must be found (cf. 1h := Figure 6). Equation (3a) can be set up with and instead of + and The arguments of the coefficients b1 and b2 are + h, This point results implicitly from
—1) =
—
€1'.
(5.2.5)
Remark 6.2.8. (a) At the grid points adjacent to the boundary, the first discretisation requires a difference scheme for Lu = f with nonequidistant
stepsizes. The second discretisation requires an approximation of the nonlinear problem (5). From a programming viewpoint both procedures are undesirable because of the case distinctions. (b) If the vector b from B approaches the tangential direction, both methods fail since the straight line no longer intersects the dotted straight line from Figures 5 and 6. Here, the second discretisation fails earlier than the first one.
102
5
General Boundary Value Problems TI
Line
(1+ h,
• Points in
I Points in FIgure 5.2.6.
Second
boundary dlecretisation
Another option for avoiding the difficulties described above consists in using variational difference equations, at least near the boundary (cf. Remark 8.6.1).
Occasionally it is also poesible to simplify the boundary conditions by using coordinate transformations. Let Bu = be prescribed on C P. If one finds a transformation (x, y) 0 ,) audi that the equations = and are satisfied on one obtains for the transformed problem Bu oOu/On + bou = on a vertical boundary piece (8/On = The discretisation may be earned out as described in Section 4.7. The same reasoning as in the preceding sections proves stability and
convergence: Remark 5.2.9. Let the difference method
= ía satisfy the conditions := K1 + of Theorem 5.1.22 so that 1 holds for (K> o is constant; wh is as in the proof of Theorem 5.1.22). Let the boundary discretisation (3c') satisfy
For sufficiently large K, one then obtains (Bhti,h)(x) 1 for x E Hence it follows that 1, where is the M-matrix defined immediately after Equation (4b). Since conat (cf. Theorem 4.3.16), stability has been proved. The order of convergence is the minimum of 1 (order of consistency of ; cf. Lemma 7) and the order of consistency of = Ia. In general it is not poesible to approximate bTV by symmetric differences.
To construct a discretisation of Bu = w with consistency order 2 despite this fact, set: :
—
=
E
5.3 Boundary Problems of Higher Order
103
If, in order to remove the first two are three points in = in terms in the Taylor expansion, one uses the differential equation and the tangential derivative of Bu = one obtains a discretisation of Bramble-Hubbard [2) order 2. For the special case L = 4, B = 0/On + may be chosen such that the inequalities proved that the three points (xe, hold and guarantee stability. However, in general, the + C2 ÷ 4 0, c0 where
f
f'h. Nonetheless, the do not lie in the direct neighbourhood of construction of the discretisation appears too complicated to be recommended for practical purposes. (xe,
5.3 Boundary Problems of Higher Order 5.3.1 The Biharnionic Differential Equation In elastomechanics the free vibration of rods Leads to (ordinary) differential equations of second order if longitudinal vibrations (= compression waves) or torsional vibrations are involved. By contrast, transversal vibrations ( bending waves) result in an equation of fourth order. Correspondingly, the bending vibration of a plate leads to a partial differential equation of fourth
order. This is the biharmonic equation (plate equation)
42u=f infl
(5.3.1)
where 42 = 04/8x4 + 284/8z20y2 + 04/8y4. Here u describes the deflection of the plate perpendicular to the surface. If the plate is firmly clamped at the edge, one obtains the boundary conditions
u=
and Ou/On =
on F
(5.3.2)
with = 0. A btharmonic probLem (1), (2) also results from the transformation of the Stokes equations in 11 C in.2 (cf. Remark 12.1.5). The differential equation (1) may be combined with other boundary values than (2). An example is: is
=
and
4u =
on F
(5.3.3)
(simply supported plate). Other examples can be found in (7.4.12b,e).
Exercise 5.3.1. Show that if one solves the Poisson equations 4v = f in
£7,
solution of the boundary value problem (1), (3). Why can Problem (1), (2) not be handled likewise?
Remark 5.3.2.
The solutions of L12u =
minimum pnnciple (counterexamnple: u
0
do not satisfy a maximum-
x2 + y2 in £7 = KR(0)).
5 General Boundary Value Problems
104
5.3.2 General Linear Differential Equations of Order 2m multi-index) of order IaI = The partial derivative D° (a + is defined in (3.2.5). A differential operator of order 2m has the form
(x
L=
Q)
+ (5.3.4a)
and defines the differential equation of order 2m:
Ltt
f
in (2.
(5.3.4b)
Ellipticity has been explained thus far oely for equations of second order (cf. Definition 1.2.3).
Definition 5.3.3. The differential operator L (with real-valued coefficients
aa)iasaidtobe elliptic (oforder2rn) atxE Qif (5.3.5a)
for
Here,
We set o for all
of degree is an abbreviation for the polynomial Evidently (5a) is equivalent to P(x,e) = 1. For reasons of continuity either E IR' with
P(x, 0; otherwise we scale with the factor —1 (changing from Lu(x) = 1) is compact it follows f(x) to —Lu(x) —1(x)). Since the set E min{P(x, = 1} > 0 and this justifies the formulation of that c(x) (Sa) as
c(x) >0.
(5.3.5b)
frzl.2m
Definition 5.3.4. The differential operator L is said to be uniformly elliptic in I? if inf{c(x):x E (1} >0 for c(x) from (5b). ExercIse 5.3.5. (a) L from (1.lb) into the notation of (4a). What axe the coefficients Ga for L 112? (b) Prove that the btharmonic operator 112 is uniformly elliptic. (c) Let aa be real-valued. Why are there no elliptic operators L of odd order? (d) If the coefficients Ga are sufficiently smooth one can write L from (4a) in the form L = (cf. (1.lb) and (1.2)).
For in = 1 (equation of order 2) we have used one boundary condition; for the biharmonic equation (in = 2) two boundary conditions occur. In general one needs m boundary conditions
of Higher Order
5.3 Boundary
b,0DaIL=
B,u
(j = 1,2,. ..,rn) on P
105
(5.3.6)
lcxl<m,
with boundary differential operators B1 of order 0 m3 0 arbitrary) so that the order of convergence is
(13) by i.e., almost 2.
Remark 5.3.10. Inequality (9a) shows conda(Lh) = O(h4). In general a difference method for a differential equation of order 2m leads to the condition
cond(La) =
(5.3.16)
5.3 Boundary Problems of Higher Order
This indicates greater sensitivity to
109
errors at higher orders 2m
(cf. §4.4 in Stoerili and Bulirsch—Stoer[1J). Difference methods for boundary value problems of fourth order with vailable coefficients and general domains fl are discus9ed in an paper by Zlámal
[1). There too convergence of oider 0(h3/2) is shown. By contrast, 0(h2)convergence was proved by Bramble (1J for the 13-point star (7) in a general domain if the boundary conditions are suitably discretised.
6 Tools from Functional Analysis
To enable readers without previous knowledge of functional analysis to follow
the next chapters we summarise here all definitions and results that will be needed later on.
6.1 Banach Spaces and Hubert Spaces 6.1.1 Normed Spaces Let X be a linear space (alternative term: vector space) over K where K = IR or K = C. In the following the normal case K = IR is always intended. K = C occurs only in connection with Fourier transforms. The notion of a norm . fl: X —. oo) is explained in Definition 4.3.12. The linear space X equipped with a norm is called a nor med space and is denoted by the pair (X, fl. Whenever it is clear which norm belongs to X, this norm is called Ii lix and one writes X instead of (X, fl. lix).
Example 6.1.1. (a) The Eudlidean norm H from (4.3.13) and the maximum norm (4.3.3) are norms on lit". (b) The continuous functions form the (infinite-dimensional) space If 1? is bounded, all ii E C°(fl) are bounded so that the supremum norm (2.4.1) is defined and satisfies the norm axioms. If ii is unbounded, the bounded, continuous functions form a proper subset 11C°(& of C°(fl). Instead of jJ . we use the traditional notation fl. (c) The Holder-continuous functions introduced in Definition 3.2.8 form the normed space (C'(J7), ii The norm defines a topology on X: A C X is open if for all x E A there exists an >Osothatthe "ball" in A. We write
x or x = urn
Example 6.1.2. Let I,,, I to
if
—
0.
f
The limit procees f, —' (with respect denotes the unifonn convergence known from analysis.
6.1 Banach Spaces and Hilbert Spaces
Exercise 6.1.3. The norm fl. reversed triangle inequality holds —
IlvIlI
X —* [0, oo)
fix — till
111
is continuous; in particular the
for
x,y E X.
(6.1.1)
As Example la shows, several norms can be defined on X. Two norms on X fi and ifi o < C 0,0 k < m,u E
PROOF. (a) Partial integration for
= Since also u by with
—
= I shows that
= it follows that (n + 1, one obtains in the same way
IuI? 0,
Remark 6.2.14. The set
E
k m, u
(6.2.lOe)
uE
(6.2.lOf)
C°°(47):supp(u) compact, IuIk 0. According to Lemma 7 there exists a function with a(x) = 1 E C°°(Q) with Iu — u4k <e/2. There exists a E 2. For sufficiently large R, one also has for lxi 1, a(x) = 0 for lxi <e/2. Thus there exists v(x) = a(x/R)ue(x) E
S
withlu—vlke. Since "supp(u) compact" already implies "supp(u) cc
Corollary 6.2.15.
=
we obtain
for all k 0.
The Leibniz rule for derivatives of products proves
Theorem 6.2.16. uEHk(fl).
for all a E
IlatLHHk(a)
Theorem 16 together with the substitution rule for volume integrals shows
Theorem 6.2.1T. (Transformation theorem). Let T: (1—. if be a oneto-one mappsng onto if with T and ldetdT/dxl 6>0 in f?. We write v = u o T for v(x) = n(1'(x)). Then u Hk(if) [u Hk(1l) [€ Hok(S2)j and
also impises u o T flu o
6.2.3 Fourier For tt E
(6.2.11)
TIIHb(a)
and one defines the Fourier-transformed function €i by dx.
(6.2.12)
Note that ü is described by a proper integral since the support of u is bounded.
120
6 Tools from Functional Analysis
ForR—'oo
Lemma 6.2.18. IR(u; y) :=
J
dx]
[JR'
uniformly to u(y) on supp(u).
PROOF. It suffices to discuss the case n = tion with respect to results in IR(u; y) = (1/7r)
J(x
(by Fubim's theorem). Integra-
sin(R(x — y))u(x) dx.
—
=
Then IR(1;y) =
1
1
for all R >0. Since u
C°°(LR"),
then also w(x, y) := [u(x)—n(y))/(x—y) E The estimates w(x, y) = and Wz(x,y) = O(1/x2) hold uniformly for y E supp(u). Partial integration yields —
u(y);y) = IRU
—
y)w(.,y);y) =
sin(R(x — y))w(x,y)dx
= .-(1/ir) J cos(R(x — y))Wx(X,y) dx/R
O(1/R).
The statement follows froni
IR(u;y) = u(v)IR(1;y) + IR(u() — u(y);y) = u(y) +O(1/R). Lemma 6.2.19. (s E L2(JR') and
PROOF. Lemma
18
•
= lulo for all u
shows
J [J
[J =
J
IR(u;y)u(v)dy —.
J
U
Iu(y)I2dy.
Lemma 6.2.20. The inverse Fourier transformation
=
u is defined
by(13):
J lim
J
(6.2.13)
6.2 Sobolev Spaces
PROOF. We have from Lemma
18
=
J
121
dy
J
J
IL"
= u(x) + O(1/R).
Theorem 6.2.21.
with E = = 1, i.e., 7 is an isometric mapping of L2(IR') onto itself The scalar prodtict satisfies (u,v)o = (ü,ii)o for all u,v E
PROOF. Since Cr(1R') is dense in L2(JR") (cf. Lemma 2), 7 can be contin(cf. Theorem 6.1.11). The norm estimate follows —+ ued to 7: are interchangeable (cf. (12) and from Lemma 19. The roles of 7 and (13)); thus E L(L2(JRYI), L2(IR")) also holds. The second statement results from
=
—
—
=
Exercise 6.2.22. Prove that: (a) With
+
=
—
—
.. .
=
U
there holds
for u E
(6.2.14)
(b) There exists C = C(k) such that E IR".
Lemma 6.2.23. (a) Juk = I%JEcrI 0, 1 j n, then is E Hk+1(lRn) holds. Conversely, I8hjUIk Nk+1 holds for all u E H'+1 (lit"), h > 0.
6 Tools from Functional Analysis
122
follows +Se,))(e) = PROOF. (a) From for h Hence, since 4h2s1n2((,h/2) = 1/h. Summation over j = 1,.. for I
= follows ,
n and
then gives
integration over
(1u1+i)2 =
J(1 + J
...
+
+
.
(1
+
+
+
J
J
..
1/2. The second is := IIU(e', (Fubin.i's theorem). Together because IIUIIL2RØ_I = we have:
(1 +
for
Integration over
J(i
+
E
results in =
I
=
i-
E
6.2 Sobolev Spaces
125
(cf. (16b)). In the = is proved with and the integration over is = u(0), represents already U
Thus jwI,_j/2 case n = 1, not required.
= 0. Evidently, Theorem 28 describes the restriction 14(., 0) = -yu to E 1R, with the for any other similarly we have is continuous tresp. Holderu(., same constant C8. The mapping continuous] in the following sense.
Theorem 6.2.29. For s> 1/2 the following statements hold: IR,u E
= 0 for all
—
(6.2.21a)
3/2).
PROOF. (a) Let
u E H8(IR") and The function ço,, is continuous in JR and set := since converges uniformly to — u(,x)I,_112 — t4,, for all x JR. Thus (21a) follows. = + e) — has the Fourier transform (b)
so that = = — := in the proof for Theorem 28 set first integral in the eethnate of now reads
f(i +
+
e/2)
=
(1
+
As uui
W(.,0). The
l/2—* J(i + t2)' sin2(i7t)
The decomposition of the last integral into subinte+ with ,, = grals over 1/u shows and fri The remainder of the argument follows the same lines as in the proof of Theorem 28. U Up to this point we have obtained H8(Wt") by completion of in
The next theorem shows that for suffithentiy large $ one can also complete in fl L2(IR") so that contains only claseical functions (i.e. continuous, HOlder-continuous, [HOlder-J continuously differentiable functions).
Theorem 6.2.30. (Sobolev's lemma) IN
Li {0}, s > k + n/2 and H'(IR") C
PROOF. (a) Let a
c for 0 < t
holds fork E IN, a
t + n/2.
t + n/2, 0 IaI + 1/2 there exists a restriction -1D°u E of the derivative of u E H8(Q). (b) For each uE with
s0.
U
and U' can be identified. By this one
V c U C V' (V c U continuously and densely embedded).
(6.3.7)
Corollary 6.3.10. In a Gelfand triple (7) V and U are also continuously arid densely embedded in V'.
PROOF. For U C V' see Lemma 9, for V C V' see Exercise 6.2.25.
134
6 Tools from Functional Analysis
Attention. Likewise one could identify V with V' and one would obtain U' c V' = V c U. But it is not possible to identify U with W and V with V' simultaneously. In the first case one interprets x(y) = (y, x)uxrj' for x, y E U as x)u (in particular for x, y E V U), in the second case as (y, z)v.
Exercise 6.3.11. Let (7) be true. Set W := {Jç'u:u
U} and define (x, y)w := (Jvx, Jvy)u as the scalar product on W. Show that (a) W is a Hubert space; (b) W C V is a continuous and dense embedding; (c) (v, w)v = (v, Jvvi)u for all v E V, w W; (d) l(z,y)vI HxIIuIlyIIw for all x,y E W. Because U = U' the scalar product (z, y)u can also be written in the form
y(x) = (x,y)uxu'. If z E V, then y(x) = (x,y)Vxv' also holds. That means that (x,y)u = (X,y)vxv' for all x V,71 E U C V'. Likewise one obtains (x, Y)u = (X,y)v'xv for all x U and y E V. The dense and continuous embedding U C V' proves
Remark 6.3.12. Let V C U C V' be a Gelfand triple. The continuous extension of the scalar product (., .)u to V x V' (V' x VJ results in the dual XVI. Therefore the following notation is practical form (., •)v v'
for x€V,y€V', (x,y)v'xv = (z,y)u for x E V',y E V. In connection with Sobolev spaces one always chooses U := L2((1) so that the embeddings read as follows: H0($2) c L2(Q)
c
H8($2) c L2(a) c (118(fl))'
(a 0),
(6.3.8a)
(a 0).
(6.3.8b)
ExercIse 6.3.13. Show that (8a) and (8b) are Gelfand triples. The dual space of
is also denoted by
or
:= H3(J?) :=
(a 0).
The norm of H(.Q) according to (1) reads: := sup{I(u,v)L2(fl)I/1v18:0
where (u,v)L2(a) is the dual form on
Remark 6.3.14. (a) Let (1 = :=
for a 0
v
x H'(Q) (cf. Remark 12).
The norm dual to v
€
6.4 Compact Operators
135
equivalent to and has the representation (2.16b) with —s instead of s. (b) The Fourier transform shows for all 8 IR. E (c) au E H8(Q), if u H8(fl), a C'(i7), where = Isi E IN U {O} or t > (s(. i8
6.4 Compact Operators Definition 6.4.1. A subset K of a Banach space is said to be precompact [compact] if each sequence E K (i E IN) contains a convergent subsequence
fand
E
Kl.
Another definition of compactness reads: Each open covering of K already contains a finite covering of K. Both definitions are equivalent in metric spaces (cf. Dieudooné (1, (3.16.1)]). For the terms "relatively compact" and "precompact" see also Dieudonné (1, (3.17.5)].
Remark 6.4.2. (a) K C
is precompact [compactJ if and only if K is bounded (and complete]. (b) Let X be a Banach space. The unit sphere {x X: flx(f 1} is compact
if and only if dim(X) t) are compact. (s, t E C°'1. The embeddings Hk(S?) C
(b) Fhrther, leti?
(k,l E NU TO),
k> 1) are compact.
(c) Let 0
0
exist such that
inf{sup{ia(x,y)l:yE E
= 1}:xE V,hlxiiv = 1}=e >0,
(6.5.4a)
= l}:y
(6.5.4b)
liyilv
1}
=
e'
>0;
(iii) the inequalities (4a) and (4c) hold:
sup{la(x,y)i:xe V,lJxhlv = If one of the statements (i)—(iii) holds, then
1}
>0.
(6.5.4c)
139
6.5 Bilinear Forms
(6.5.4d)
(e,e' from (4a,b)).
1/11A'IIv.._vi
C
Prom (4a) follows
l',IIxIIv =
inf{sup{Ia(x,y)I:y E V,flyIlv = 1}:x
1}
e >0.
(6.5.4e)
Conversely, (4a) follows from (4e) wsth a possibly larger e > 0. (4e) and (4c) conditions. (4e) is equivalent to are also called the
for all x
V, lIyItv = 1) cHxIIv
sup{Ia(x, y)I: y
V,
(6.5.4e')
because (4e) is equal to (4e') for all x V, IIxIIv = 1. The scaling condition tIxIIv = 1 can emdently be dropped. The left-hand side in (4e') agrees with the definition of the dual norm of Ax so that (4e) and (4e') ore also equivalent to (4e"): (6.5.4e") for all x V. hAzily'
PROOF. (a) "(i) mf{
}
Ia(x,y)I
= inf aEV
Uxllviiyiiv
.
— mf sup —
L(V',V) exist. Then (4a) follows from
(ii)": Let
i(Ax,y)l
_.
iixilvhiyhlv
I(AA'x',y)i
p€V IIA'ziivliviIv
=
— —
c
II A1 lu—i sup i(x',y)I iv' i,€v
= lIE sup 11A'x'ilv/hlx'flv'l
CV
=
C.
Because A'' =
In the same way one shows (4b) with e' =
(A')', (3.3), and V" (b) "(ii)
hIyHv
V, it follows that e = (iii)": (4c) is a weakening of (4b).
(i)": e > 0 in (4a) proves that A is injective. We wish to show (c) "(iii) with that the image W := {Ax: x V} C V' is dosed. For a sequence the From (4a) one infers W there exists zI, E V with = w1,. that via (4e) and (4e") (with x := — flwg, — WpiIv'/f. — There is Cauchy convergent, this property carries over to Since —. in V. The continuity of A E L(V, V') proves exists an x' E V with = Ax,, —+ Ax' so that w' = Ax' E W. According to Lemma 6.1.17 one Wt. If A were not surjective (thus W V'), there can decompose V' into would exist a w W1 with w 0. Then y Jv'w = E V would satisfy 0 (cf. Theorem 6.3.6, Corollary 6.3.7). Since a(z,y) = (Ax,y)v') J(x) for all z
—
2f(y)
.1(x) + CEIIZ
U
x.
The term "V-elliptic" seems to indicate that to elliptic boundary value problems correspond V-elliptic bilinear forms. In general this is not the case. Rather, V-coercive forms will be assigned to the elliptic boundary value problems. Their definition necessitates the introduction of a Gelfand triple (cf.
(6.3.7)):
V C U C V'
(U = U', V C U continuous and densely embedded).
Definition 6.5.13. Let V C U C V' be a Gelfand triple. A bilinear form is said to be V-coercive if it is continuous and if there exists oK and CE > 0 such that
a(x,x)
for
all xE V with Ck >0.
lR
(6.5.10)
142
6 Tools from Functional Analysis
a(x,y) +CK(x,y)u with CK from (10). Let
Exercise 6.5.14. Set a(x,y)
V' be the inclusion. Show that (a) the coercivity condition (10) is I: V equivalent to the V-ellipticity of a.
A + CkI to a(.,.).
L(V, V') is associated to a(.,.), then so is A Why does A L(V, V') hold?
(b) If A
The results of Riesz-Schauder theory (Theorem 6.4.12) transfer to A as soon
as the embedding V C U is not only continuous but also compact.
Theorem 6.5.15. Let V C U C V' be a Gel/and tr*ple with compact embedding V C U. Let the bilinear form a(.,.) be V-coercive with correspond:ng V' be the inclusion. operator A. Let I: V (a) For each A E C one of the foil owing alternatives holds: (i)
(A—Xf)1 €L(V',V)
(ii)
A is an eigenvo.lue.
and (A'—Xiy' EL(V',V),
f
f
in case (i) Ax — Ax = and A'x' — = are uniquely solvable for all I E V' (i.e., a(x, y) — A(x, Y)u = 1(y) and a(x',y) v)u = f(y) for all E(A) := y V). in case (ii) the,e exist finite-dimensional eigenspaces {0} such that E'(A) kernel (A' — kernel (A — Al) and {0} Ax = Ax
for x
E(A),
a(z,y) = A(x,y)u
i.e.,
for all y
V,
(6.5. lie.)
for x*EEI(A),
a*(x*,y)=X(x*,y)u forall y€V.
i.e.,
(6.5.1 ib) (b) The spectrnm o'(A) of A consists of at most countably many eigenvalues
a(A) if and only if A
which cannot accumulate in C. A
0 exists C( such that
for all x E V.
jb(x, x)f
(6.5.12a)
(b) Let the embeddings V c X and V C Y be continuous, with at Least one of them compact. Let the following hold: for all x E V.
Ib(x,x)I
(6.5.12b)
(c) Let the embeddings V C X, V C Y be continuous. Let (12b) hold. For such that or Uv assume that for every e >0 there exists a or
÷
iS
for x
eIlxIIv +
V.
(6.5. 12c)
PROOF. (a) Select e = CE/2 with CE from (10). Then a(•, .)+b(.,.) satisfies the V-coercivity condition with CE/2 > 0 and CK + instead of CE and CK. (b) Lemma 6.4.13 proves (12c).
(c) Let the first inequality from (12c) hold, for example. Since the embedding V C Y is continuous, Cy exists with CyflxlIv. Choose C' = in (12c):
+
5
K
Since
I
7 Variational Formulation
'7.1 Historical Remarks In the preceding chapters it was not possible to establish even for the Dirichiet problem of the potential equation (2.1.la,b) whether, or under what conditions, a classical solution u C2(Q) n C°(fl) exists. Green took the view that his Green's function, described in 1828, always exists and that It provides the solution explicitly. This is not the case. Lebesgue proved in 1913 that for certain domains the Green function does not exist. Thomson (1847), Kelvin (1847), and Dirichiet offered a different line of
reasoning. The Dirichiet integral
J
(7.1.1)
=J
describes the energy in physics. With boundary values u = on I' given, one seeks to minImise 1(u). This variational problem is equivalent to
I(u,v) :=
=0 for all v with v =
0
on
(7.1.2)
The proof of the equivalence results from 1(u+v) = I(u)+21(u,v)+I(v) and 1(v) 0 for all v (cf. Theorem 6.5.12). Green's formula (2.2.5*) provides I(u, v)
= f0 v/lu dx =0 for all v with v =0 on F such that /lu =0 follows.
Thus, like (2), the variational problem 1(u) = miii is equivalent to the Dirichiet =0 in a, u = on I'. problem The so-called Dirichiet principle states that 1(u), since it is bounded from below by 1(u) 0, must take a minimum for sonie u. According to the above considerations this would ensure the existence of a solution of the Dirithlet problem. In 1870, WeierstraB argued against this line of reasoning, stating that while there may exist an inflinum of 1(u) over {u E fl = on r} it need not necessarily be in this set. Fur example, the Integral J(u) := fu2(x)dx in {u C°U0,11):u(0) = 0,u(1) = 1} never takes the
value inf J(u) = 0. Further, the following example due to Hadainard shows that no finite infirnum of the Dirichiet integral need exist. Let r and be the polar coordinates In the circle Q K1(0). The function = in .0 but the integral 1(u) does not exist.
7.2 Equations with Homogeneous Dirichiet Boundary Conditions
145
The above difficulties disappear if one seeks the solutions in the more suitable Sobolev spaces instead of in C2(Il) fl C°(17).
7.2 Equations with Homogeneous Dirichiet Boundary Conditions In the following we investigate the elliptic equation (7.2.la)
in Si,
Lt.i = g
L=
(7.2.lb) PIm
of order 2m (ef. Section 5.3; Exercise 5.3.5d). The principal part of L is
=
(7.2.2)
According to Definition 5.3.4, L is uniformly elliptic in Si if there exists
0
such that for all x E Si,
Attention. In the case that only
E IR".
(7.2.3)
is assumed, one needs to
replace "for all XE Si" by "for almost all x
Si".
We assume the homogeneous Dirichiet boundary conditions (7.2.4)
which are only meaningful if F =
is
sufficiently smooth. Note that in the
standard case m = 1 (an equation of second order) condition (4) becomes U =0. Since with u = 0 on F the tangential derivatives also vanish, not only the kth normal derivatives (k m — 1) but also all the derivatives of order $ m —1 are equal to zero:
inxEf forlal<m—1.
(7.2.4')
Condition (4') no longer requires the existence of a normal direction. According to Corollary 6.2.43, (4') can also be fonnulated as (7.2.4")
Let it E be a classical solution of(la) and (4). To derive the variational formulation we take an arbitrary v and consider
146
7 VariatIonal fbrmulation
(Lu, v)0 =
the integrand vanishes in the proximity of I' so that one can integrate by parts: E C8°(.O),
Dau)dx
without boundary terms occurring. Thus we have found the variational formulation
IaI,I$Im
f
dx
= Ja
g(x)v(x) dx
(7.2.5)
for all v
since Lu = g. If conversely a with boundary conditions (4) satisfies condition (5), then the partial integration can be reversed and f0(g — Lu)vdx = 0 for all v E Cr(a) proves Lu = g. This means that a classical solution of the variational problem (5) with boundary condition (4) is also a solution of the original boundary value problem. Hence the differential equation (la,b) and the variational formulation (5) are eqwvalent with respect to classical solutions. We introduce the bilinear form a(u, v) :=
J
(7.2.6)
J ag(x)v(x)dx.
(7.2.7)
and the functional
/(v)
As remarked above, the boundary condition (4) for classical solutions u means that a E Thus the "variational formulation" or "weak formulation" of the boundary value problem (1), (4) reads as follows: find
a
with a(u,v)
1(v)
for all v E
(7.2.8)
A solution of problem (8) which, according to the definition, lies in but not necessarily in is called a weak solution.
ExercIse 7.2.1. (a) Let 0 be bounded. Show that any claasical solution aE fl is also a weak solution. (b) With the aid of Example 2.4.2 show that this statement becomes false for unbounded domains.
Theorem 7.2.2.
Let
bOunded on
x
E
The bilinear form defined by (6) se
7.2 Equations with Homogeneous Dirichiet Boundary Conditions
PROOF. Let u,v E
147
The inequality (6.2.Sc) yields
const Since
to Hr(.O) x
is dense in (ci. Theorem 6.2.10), a(.,.) has an extension and is bounded by the same constant (cf. Lemma 6.5.lb). U
The function f(v) is also defined and bounded for v E if, for example, 9 E L2 ((1). According to Exercise 6.5.11, the variational formulation (8) is equivalent to the following one:
with a(u,v) = f(v)
find u
One can regain the form Lu =
for all v E
(7.2.9)
/
by applying Lemma 6.5.1. Let L E H—m(a)) and f E Hm(12) = be defined by a(u,v) = and 1(v) (f,v)ff-.,.(a)XH.,,(a) for all v E Equation (9) states that Lu—f. (7.2.9') While (la) represents an equation Lu g in C°(I2) (i.e., for a classical solution), (9') is an equation in Hm(fl). Theorem 6.5.9 guarantees unique solvability of Equation (9) if a(.,) is We first investigate the standard case m = 1 (equations of order 2tn = 2).
Theorem 7.2.3. Let 12 be bounded, m (uniform
+
)i31
a(u, u) PROOF.
1, aafi
L°°(12). Let L
(3)
elhpticity) and be equal to the principal part L0, i.e., aap = 1. Then the form a(.,.)
Since
=
=
1
e' > 0.
0 for
(7.2.10)
= one can identify a and /3 according to {1, . .• ,n}. Fbr fixed x 12 use (3)
= 8/Ox, with indices i,j with
=
=
E
Integration over $2 yields a(u, u)
= 1Vu12 dx.
Since
(cf. Lemma 6.2.11), (10) followscC0. with e' =
Corollary 7.2.4.
dx
U
The condi*ion "12 bounded" may be dropped s/for
Example 7.2.5. The Helmholtz equation the bilinear form
+u=
f in
a=
1? leads
to
148
7 Variational Fbrmulatlon
a(u, v)
liE
(x) + u(x)v(x) j dx =
ff(Vu, Vv) + uvj dx.
a(ts,v) is the scalar product tn Ha(S2) (and H'(S?)). The fact that a(u, u) = proves the (S?)-ellipticity. Exercise 7.2.6. Let the assumptions of Theorem 3 or Corollary 4 be satisfied, except for the fact that the coefficients aao and ao$ = 1) of the = first derivatives are arbitrary constants. Show that inequality (10) holds unchanged.
Theorem 3 cannot easily be extended to the case m> 1. Theorem 7.2.7. Let the coefficients of the principal part be constant8: 0afl = m. assume that constforlaj= =Oforo < 2m —1, aoo 0 for a = f3 =0. Let L be uniformly elliptic (cf. (3)). Further let either £2 be bounded or aoo >0. Then a(.,.) is
PROOF. We continue u through u =0 onto Exercise 6.2.22, and inequality (3) show that
a(u,u)-Jaoou2dx= a
Theorem 6.2.21,
>2 JaafiD0'uD6udx (aI,(PI=m
=
=
=
>2
E
>2
11
>2
ef Let aoo > 0. There exists an e' > 0, so that for all E From this follows
e' E1cz1<m
fri2 and
—
(cf. Lemma 6.2.23). IfS? is bounded, use Lemma 6.2.11.
Having shown the of the form a(., •), to apply the general proofs after Theorem 6.5.9.
we
I
are now able
Theorem 7.2.8. (Existence and uniqueness of weak solution8)If a(.,.) is Hr(fl)-eflipgic then there exists a solution u E (9) which satisfies
iUim
(CR
(6.5.5)).
of Problem
(7.2.11)
7.2 EquatIons with Homogeneous Dirichiet Boundary Conditions
Hm(.0) and u =
(11) holds for all I is equivalent to
Since
149
(cf (9')), inequaLity (11)
0 on V x V with V := {u E H'(0):u constant on f'} be defined. Show that (a) a(.,.) is V-elliptic. (b) The weak formulation: u V1 a(u,v) = for all v V corresponds to the problem —4ui- cu = g in (1,
u constant on 1',
f
fcodr.
(7.4.14)
which is also called an Adler problem. Finally we want to point out the difficulty of classically interpreting a weak solution. In the variational formulation (2a,b) the right-hand sides g and ço of the differential equation and the boundary condition are combined in the functional 1. In the variational formulation the components g and ço are
7
160
VarIational Fbrniulatlon
indistinguishable! u E H'(a) has fist derivatives in L2((J), whose restrictions to V do not have to make sense. That is why Bu cannot be defined in general;
cannot be viewed as an equality in the space H'/2(f) although
Ru =
H'/2(F)
(cf. Corollary 4b). But even if there Is a classical solution, the following paradox arises. Let u to (Ia) define be a classical solution of Lu =0 in 1?, Bu = on F.
One may also view u as a solution E (H1(fl))' by f,(v) := in 0, Bu = 0 on I. These equations may even be interpreted of Lu = classically in the following way: there exist E C°°(0) with 4 in = 0. Then (H1(i7))'. Let be the claseical solution of = converges in H'(O) to the above-mentioned classical solution u. Incorporating the boundary values Ru = in the differential equation corresponds to a modification of the discretised problem as used Lu = in Section 4. The difference equations Dhuh = fh in and the boundary on F,, resulted in the system of equations L,,u, = conditions uh (cf. (4.2.6b)). If one defines by in 0h, uh = 0 on = fh + then satisfies the equations D,,Ti,, = in iZ,, = 0 on Just as the functional f cannot be uniquely separated into g and and cannot be reconstructed from In contrast to the discrete case, the separation off into g and is possible, however, provided stronger conditions than g (H'(.Q))' are imposed on g (for example, g L2(fl)).
4
8 The Method of Finite Elements
In Chapter 7 the variational formulation was introduced only for the purpose of proving the existence of a (weak) solution. It will now turn out that the variational formulation is the foundation of a new method of discretisation.
8.1 The Ritz-Galerkin Method Suppose we have a boundary value problem in its variational formulation:
Find u€V, so that a(u,v)=f(v) foiallv€V,
(8.1.1)
and V = H'($?) (cf. where we are thiniciug, in particular, of V = Section 7.2, Section 7.4). Of course, it is assumed that a(.,.) is a bounded bilinear form defined on V x V, and that / V': Io(u,v)I
CsfItLUvIIVftv
for
'€
E
(8.1.2)
Difference methods arise through discretising the differential operators. Now we wish to leave the differential operator hidden in a(.,.) unchanged. The RitzGalerkin discretisation consists in replacing the infinite-dimensional space V with a finite-dimensional space VN:
VNCV,
dhnVN=N 0. Then the mat'*r L in (9) is nonsmgt4ar and the Ritz-Goier*in solution
E VN satisfies
(8.1.lla)
IIU?ullv
PROOF. Lisnonsingularsince
and
thus
(Lu, u) = a(Pu, Pu) CEIIPUU?, >0 and so, in particular, Lu 0. By &ercise 6.5.6a a(.,.) is also VN-eIIiptiC with the same constant CE. From Theorem 6.5.8 there holds (ha) then results from 1/CE, i.e., HU"flv
ExercIse 8.1.9. Show
II! fly' for any f
V'.
Exercise 8.1.10. Show: (a) If a(.,.) is symmetric then so is L.
(b) If a(.,.) is symmetric and V-elliptic then L is positive definite. Under the same assumptions the Ritz-Galerkin solution ut1 solves the following variational problem (cf. Theorem 6.5.12): J(uM) J(u) :=a(u,u)—2f(u)
for all u
VN.
Example 8.1.11. (Dirichiet Problem) The boundary value problem is
8.1 The Rltz-Galerkin Method
infl=(0,1)x(O,1),
.—zlu(x,y)=1
165
u=O
The weak formulation is given by (1) with V =
a(tt,v) :=
J (Vu,Vv)dxdy =
f
JvdxdY.
1(v)
The functions bj (x, y) = sin(irx) sln(wy), bs(x, y) = sin(wx) sm(3iry),
b2(x, y) = b4(x, y) = sin(3wx) sin(3,ry),
fulifi the boundary conditions and so belong to V = Hd(i7). They form a basis of V4 := span {b1,. ..,b4}. The matrix elements = can be worked
out to be L11 =
=
= 5w2/2
=
In addition the chosen basis is o(., .)-orthogonal:
=
L is diagonal. Furthermore one may calculate dxdy, getting
=
fi =4/ir2, Hence
u=
12
L'f has the components
Ui
=
u2 =
=
u4 = 8/(81ir4),
and the Ritz-Galerkin solution is then
=-.[sinwzsiniry + +
sin
+sinirzsin3iry)
sin 3iry].
The Ritz-Galerkin solution and the exact solution for x = v = 1/2 are
=
= 0.07219140...,
=
/[(i + 2v)(1 + 2p)((1 + 2p)2 + (1 +
= 0.0736713.
Example 8.1.12. (Natural boundary conditions) Let the boundary value problem be
8 The Method of Finite Elementa
166
Ou/ôn=Oon F.
= (0,1) x (0,1),
= ir2cosirx in
The solution is given by u = cos wx + const. The weak fonnulation is in terms of (1) with V = H'(s?),
a(u,v) :=
+
:=ir2Jv(x,y)cosirxdxdv.
1(v)
The boundary value problem has a unique solution in
W:={v€ V:Jvdxdv=0}. The basis functions
bi(x,y) = z are
—
b2(x,y) = (x —
1/2,
in W. The stiffness matrix L and the vector f are then L—
Ii
1/41
—2
1
9/80]'
1
so that
—L''f—1 — [—20+240/7r2 The solution is t/'(x, y) =
(3
—
60/w2)(x
—
1/2)
(20 — 240/ir2)(z
—
—
The Ritz-Galerkin solution satisfies the boundary condition öu/8n = the differential equation only approximately:
y)/On =
For x =
1/4
12
—
120/ir2
0
and
—0.16.
the approximation is v."(1/4,y) = —7/16 + 45/(4ir2) =
0.70236..., whereas u(1/4, y) = oosir/4 = 0.7071
...
is
the exact value.
In the following we shall consider the case in which o(.,.) is no longer V-elliptic, though it is V-coercive. That a(.,.) is V-coercive guarantees that either problem (1) is solvable or A = 0 is an elgenvalue. Even if one assumes V-coercivity and the solvability of the problem (1) one can not deduce the solvability of the discrete problem (4).
Example 8.1.13. a(u,v) := f(u'v' — lOuv)dx is 1)-coercive and a(u,v) = 1(v) := f01 gvdx (v E J.1o1(0, 1)) has a unique solution. Let VN spanned by bj(x) = x(1 — z) E V = Hd(0, 1) (i.e., N = problem (4) is not solvable since L = 0.
1).
Then the discrete
If one replaces the space V in Lemma 6.5.3 with VN, then there follows from Exercise 6.5.4
8.2 Error Estimates
Theorem 8.1.14.
167
The problem (4) is solvable for all f E V' and has a
unique solution, Ut", which satisfies the estimate S
(8.1.llb) EN
if and only if VN,UVIIV = 1}:u
VN,(IUIIV = 1} = CN >0. (8.1.12)
Since (4) is equivalent to the system of equations (9), one has the
Corollary 8.1.15. The matriz L is nonsingtdar if and only if (12) holds. Exercise 8.1.16. The requirement (12) is equivalent to
flullv
VN,HVUV
EN
1}
forallu€ VN.
(8.1.12')
and
= 1/EN
(LN as in (lOa)).
(8.1.12")
Warning: The condition (12) for VN does not follow from the analogous condition (6.4.5a) for V. See, however, Theorem 8.2.8.
The requirement (12") guarantees the existence of L', but it does not which is the deciding say anything about its condition, cond(L) = factor for the sensitivity of the system of equations Lu = f. For example, if (i = 1g.. .,N) then one one chooses for a(u,v) := u'v'dx the basis = iii (i+j —1). The conditioning obtains the very badly conditioned matrix of L is optimal if one chooses the basis to be a(•, )-orthogonal: a(bj,b,) = as is the case in Example 11, up to a scaling factor.
8.2 Error Estimates For difference methods the solution u and the grid function uh are defined on is, on the other hand, directly different sets. The Ritz-Galerkin solution,
comparable with is. One can measure the error due to discretisation With IIu — u?JUv or with
—
uNItu.
Let is be the solution of (1.1): a(u,v)
f(v) for v E V. Suppose — by
chance or because of a clever choice of V,1 — that u also belongs to VN; then
is also satisfies (1.4). That means: The discretisation error is zero if We shall now show: The "closer" that is is to VN the smaller is the discretisation error. is E
8 The Method of Finite Elements
168
Theorem 8.2.1. (Ceo) Assume (1.2), (1.3), and (1.12) hold. Letu E V be a solution of the problem (1.1), and let uN E VN be the Ritz-Oatev*in solution of (1.4). Then the following estimate holds: III'
— u"IIv
(1+ Cs/CN) iflf
(8.2.1)
— wIly
WEVN
with Cs from (1.2) and from (1.12). Note iflfwEyN IIu—wlIv is the distance of the function u from VN; it will be abbreviated in the following to
d(u,VN) := ml flu—wIly.
(8.2.2)
WEVN
PROOF. If u satisfies a(u,v) = /(v) for all v
V then it does in particular
for all V E VN. Since we also have a(uN, v) = f(v) for v E Vpj, it follow8 that
forall VEVN. For arbitrary v,w E VN with lIvIlv = —
= a((t/'
w, v)
1
(8.2.3)
we can therefore conclude
— tij + [u — w], v)
= a(u — w, v)
and
w,v)l 5 From (1.12') we then obtain —
— wIIvIfvIlv = CsIlu
—wily
—
why.
E VN,hIvhIv = 1} (Cs/CN)flu—wllv.
The triangle inequality then gives flu
flu
—
wily +
—
uNhiv (1+ Cs/EN)hIu —
SincewEVNisarbitrarywededucetbeaeeertion(I).
U
In Theorem I the unique solvability of the problem (1.1) was not assumed,
but only the existence of at least one solution. If the discretisation error is supposed to converge to zero one makes use of a sequence of subspaces V that converge to V In the following sense:
Theorem 8.2.2.
:=VN4 c V(iEt4) be a sequence of subspaces with lim
= 0 for all u
V.
Asumethat(1.12)holdswitheN4
(8.2.4a)
in addition assume (1.2) (continuity of a(.,.)). Then there exists a unique solutionu of the problem (1.1) and the Ritz-Galer*in solution ud := to
iIu—u'hIv--.O
for i—.oo.
169
8.2 Error Estimates
Sufficient to ensure (4a) is
UvidensemnV.
(8.24b)
PROOF. (a) Assume first the existence of a solution u. (aa) The estimate (1) implies the convergence: d(u,
ilu — u'IIv (1 +
—.0.
(ab) We now wish to show that (4a) follows from (4b). The inclusion ¾-i C Now d(u, will be a null sequence if for each implies d(u, V1) d(u,
e > 0 there exists an i such that d(u, is, for each u E V and e > 0, a w E U,
e. From the assumption (4b) there with e. Therefore we — wily have w€ for an i N. That d(u,V3) e then proves (4a). — why The convergence u hnplies the uniqueness of the solution u. (b) The next thing to show is that the image W := (Lv: v V} C V' of the operator L:V —. V1 associated to a(-, -) is closed. Foreachf E Wthereisau E V with Lu = f, so that part (a) of this proof suffices to show the convergence --. u. Since Itu'IIv it follows that hiutiv = urn 11f110'/€. Let in V' and f,, = Lug. The Cauchy E W be a sequence with —. convergence Ii!" — shows —0 so that ti/v the limit E V exists. The continuity of L L(V, V') shows that = limf,. = lirnLu,, = Lu, and thus that f E W. Hence W is closed. (c) In order to demonstrate the existence of a solution to the problem (1.1) we have to show the surjectivity of L: V —. V'. If L were not surjective (Le., W V'), there would be an I E W-1- with hIlly' = 1. Let Jv: V —. V' be the Rieaz isornorphism (ci. Corollary 6.3.7). Set v := E V. It follows that
f
f=
f(v)=
—(f,f)v' =
for all u E V. Let E satisfy v) = 0, i.e.,
0
1,
be the Ritz-Galerkin solutions. They must also a(u',v) — 1(v) =
We split up v into may
+
guarantee that
I
=
1.
where v' Vi. By (4a), —.0. This shows
with
v in
place
of u, one
/(vi)+o(ui,w)_f(wi) w1) — f(wi)
and I
=
v) — f(v)l [Cshtu'ljv + hhfik"j llw'(Iv.
Since Ilu'IIv IIfhIv'R is uniformly bounded and Hw'hIv —. (I this amounts to a contradiction. Therefore £ must be surjective, so that for each f E V'
there exists a solution u to the equation Lu =
I, i.e., to the problem (1.1). •
170
8 The Method of Finite Elements
>o from
Corollary 8.2.3. The requirement (1.12) with CN,. CR if a(.,.) is V-elliptic: a(u,u) is satisfied with
Theorem 2
is dense in V.
Exercise 8.2.4. Show that (4a) implies that
Let QN be the orthogonal projection onto VN (cf. Exercise 8.1.6c). The V V', QN:V' —' factors L:V V
(8.2.5a)
C V.
Exercise 8.2.5. Show there is also the representation SN =
PL'P'L.
(8.2.5&')
Lemma 8.2.6. SN is the projection onto VN and is c*zIled the Ritz projection. It sends the solution u of the problem (1.1) to the Ritz-Galerkin solution uN VN: uN = SNU. Assuming (1.2) and (1.12) we have (8.2.5b)
IISNIIv+-v CS/CN.
A definition of 5N equivalent to that in (5a) is SNU E VN and O(SNU,V) = a(u,v)
for all v
VN, u
(8.2.5c)
V.
PROOF. (a) Since
a(u,v) =
=
= (Lu,QNV)U = (QNLU,v)u
for all v VN, it follows that SNtI in (5c) is the Ritz-Galerkin solution for the QNLu. Conversely one may right-hand side f := QNLU E V', i.e., SNU = argue similarly, and so show the equivalence of the definitions (5a) and (5c).
projection. Inequality (5b) follows from IISNuIIv proof of (1.11)) and IILuII'v CsfIuIfv = from (1.2) (cf. Exercise 8.1.9).
(cf. the with C8
Remark 8.2.7. Let a(.,.) be V-elliptic and symmetric. IIIvIlIv := a(v,v)'/2 is a norm equivalent to f$. liv. The Ritz projection SN is, with respect to lily, an orthogonal projection onto Vp,. Thus, in particular, we have IlISp,iiiv..-v
1.
(8.2.5d)
PROOF. The scalar product associated to is a(., .), so that it is to be shown that a(SNv, w) = a(v, SNW). (5c) Implies a(SNv, Sqw) = a(v, SNW), since SNW E VN. The symmetry of a(.,.) and exchanging v and w give
8.3 Finite Elements
171
a(SNv,w) = a(w, SNV) = a(SNW, Spjv) = a(SNV, SNW), so that a(SNv, w) = a(v, SNW). (5d) results from Remark 6.3.8.
Remark 7 shows once again that the Ritz-Oalerkin solution = SNU is the best approximation to u in VN in the sense of the norm fly. This
is equivalent to the variational formulation J(uN) J(v) for all
v
VN (cf. Exercise 8,1.lOb).
The condition (1.12), with e,, e > 0, is difficult to prove, except for V-elliptic bilmear forms. However, the following theorem shows that this condition does hold for subepaces approximating well enough.
Theorem 8.2.8. Let the bilinear for,n a(.,.) be V-coercive, where V C U C V' is a continuous, dense, and compact embedding. Let Problem (1.1) be solvable for all f V'. Assume that (4a) holds for the subspaces C V. For large enough i the stability condition (1.12) is then satisfied with > 0. The proof of this will be postponed to a supplement to Lemma 11.2.7.
8.3 Finite Elements 8.3.1 Introduction: Linear Elements for 11 = (o,b) As soon as the dimension N = dim V?, becomes larger the eseential disadvantage of the general Ritz-Calerkin method becomes apparent. The matrix L is in general full, i.e., L11 0 for all i,j = 1, . ., N. Therefore one needs N2 integrations to obtain the values of = o(b,, = fe,. .., whether exactly or approximately. The final solution of the system of equations Lu = f requires 0(N3) operations. As soon as N is no longer small the general Ritz-Galerkin method therefore turns out to be unusable. A glance at the difference method shows that the matrices Lh which occur there are sparse. Thus it is natural to wonder if it is poesible to choose the basis {b1, . . ,bp,r} so that the stiffnees matrix = is also sparse. The best situation would be that the were orthogonal with respect to a(.,.): a(b,,bi) for i j. However, such a basis can be found only for special model problems such as the one in Example 8.1.11. Instead we shall base our further considerations on .
.
Remark 8.3.1. Let the bilinear form a(.,.) be given by (7.2.6). Let B be the interior of the support ri B, =
of the basis function i.e., := A sufficient condition that ensures = a(bj, b1) = 0 is
PROOF. The integration
•
= f0... can be restricted to B1 fl B,. In order to be able to apply Remark 1 the basis functions should have
as small supports as poesible. In constructing them in general one goes about
172
8 The Method of Finite Elements
it from the desired goal: one defines partitions of Q into small pieces, the socalled finite elements, from which the supports of b4 are pieced together. As an introduction let us investigate the one-dimensional boundary value problem
u(a)=u(b)=O.
—u"(x)=g(x) fora<xk (yb,7 dx are to be taken over the following Tk:
ifi=j; (i)aZZTa withx' ifi1&j. (ii) all Tk withx4 andx1 (iii) We have Lgj=O ifx' andx1 are not directly connected by the side ofa triangle.
FIgure 8.3.3. Bask
for the node (z07y0)
We get an especially regular triangulation when we first divide into squares with sides of length h, and then divide these into two triangles (0). The first and second triangulations in Figure 2 are of this sort. We call them "square grid triangulations". The corresponding basis function is depicted in Figure 3. One therefore expects that the matrix L corresponds to a 7-point formula. For the Laplaes operator, however, one finds the well-known 5-point formula (4.2.11) from Section 4.2:
8.3 FinIte Elements
177
Exercise 8.3.13. Let r be a square grid triangulation. Further let a(u, v) = Vv) dx. The basis functions are described by Figure 3. Show that one has for the entries of the stiffness matrix L
=
= —1
4,
if x' x3 = (0, ±h)
or
(±h,0),
from (4.2.8).
L agrees with
h2Lh holds in the case of the Poisson equation
Although L
=
the finite-element discretisation and the difference method still do not coincide,
since h2fh has h2g(x') as components and so differs from f
dx.
V
'7
(0,1) x3
(1,0)
(0,0)
FIgure 8.3.4. R.eferenoe triangle T
The integration ft... dx over the triangle T1 E r seems at first difficult. However, for each i one can express fT. ... dx as an integral over the reference triangle T in Figure 4. The details are in
ExercIse 8.3.14. Let x' =(x',y') (i = 1,2,3) be the vertices oITE r, and let T be the unit triangle in Figure 4. Show: (a) (b)
,j)
+
x1
(x2 —
— x') + x1)(? _yL)
—
—
x1) maps T onto T. xl) for all
(y2 —y1)(x3 —
IR.
(c) The substitution rule gives
Jv(x,Y)dxdY= (8.3.10)
In general one evaluates the integral dq over the unit triangle numerically. Examples of integration formulae can be found in Schwarz (1, §2.4.3) and Ciarlet [1, §4.1). The necessary ordering of the quadrature formulae is discussed in the same place by Ciarlet (14) and by Wztsth [1).
178
8 The Method of Finite Elements
In contrast to the difference methods finite-element discretisation offers one the possibility of changing the size of the triangles locally. The third triangulation in Figure 2 contains triangles that become smaller as they are nearer to the intruding corner. This flexibility of the finite-element method is an essential advantage. On the other hand, one does obtain systems of equations Lu = f with more complex structures, since (a) u can no longer be stored in a two-dimensional array, (b) L cannot be characterised by a star as in (4.2.12).
Remark 8.3.15. If one replaces the Dirichiet condition u =
0 on 817 by natural boundary conditions, then the following changes take place: (a) N = dim VN s the number of all nodes (inner and boundary nodes). (b) VN c H'(Q) is given by (8) without the restriction "u = 0 on 811". (c) In Remarks 9—li it should read "nodes" instead of "inner nodes". calculated in Exercise 13 are valid only for inner (d) The matrix elements nodes For a Neumann boundary value problem in the square 17= (0,1) x (0,1), L coincides with h2DhLh from Exercise 4.7.8b. (e) In calculating = one has to take account of possible boundary integrals over 811 fl if 811 fl 0.
8.3.3 Bilinear Elements for 11 c
1R2
The difference procedures In a square grid are near to partitioning 1? into squares of side h (cf. Figure 5a). If, more generally, one replaces the squares by parallelograms one obtains partitions like thoee in Figure 5a,b. An admissible
partition by parallelograms is described by the conditions (7a—d), if in (7a) the expression "triangle" is replaced by "parallelogram".
(a)
(b)
FIgure 8.3.5. PartitIon of 0 Into parallelograms If one were to define the subspace VN by the condition that u must be a linear function in each parallelogram, then there would be only three of the four vertex values that could be arbitrarily assigned. In the case of the partition in Figure 5b one can see that the only piecewise linear function u with u = 0 on 81118 the null function. Thus in each parallelogram u must
8.3 Finite Elements
179
be a function that involves four free parameters. We next consider the case of
a rectangle parallel to the axes P = (z1,x2) x (yj,y4 arid define a bilinear function on P by (8.3.lla) + a2z)(aS + a4y). u(x,y) = Then u is linear in each direction parallel to the axes — thus, in particular, along the sides of the rectangle. For an arbitrary parallelogram P such as in Figure 6a, the restriction of the function (1 la) to the side of a parallelogram is in general a quadratic function. Therefore one generalises the definition as follows. Let i-+
x1 +
— x1) +
—
(8.3.llb)
x1)
be the mapping taking the unit square (0,1) x (0,1) onto the parallelogram P (cf. Figure 6a,b). A bilinear function is defined on P by u(z, y) :=
y)),
(a +
v(e,
+ 5i).
(8.3.1 ic)
it is not
to calculate v(4r'(x1 y)) explicitly, since all the integrations can be carried out over (0, 1) x (0, 1) as the reference parallelogram (cf. Exercise 14).
y
'1
£1
FIgure
(1,1)
(0,0)
(i,0)
x
8.3.6. (a) Parallelogram
partition into parallelograms one by
VN := {u E C°(Th): on all P a bilinear ftinction} VN :=
0.
(b) Unit square as reference parallelogram
If ir = is an defines VN C H'(a) [rasp. VN C
reap.
(0,1)
u coincides with
C°(Th: ii =0 on 0S?; on all P E with a bilinear function}
Here N = dixnVN is the number of nodes [rasp. inner nodasj.
(8.3.12a) u coincides
(8.3,12b)
8 The Method of FInite Elements
180
AbilinearfünctiononPEwislinearalongeac*lsldeofP. Continuityin the node points thus already implies continuity in $1.
Remark 8.3.16. The Remarks Sa, 9, 10, and 11 hold with appropriate thanges.
Exercise 8.3.17.
Let the bilinear form aesociated to be a(u,v) = f0(Vu, Vv) dx. Aesume (1 = (0,1) x (0,1) is divided into squares of side h as in Figure 5a. What are the basis functions characterised by (9a)? Show that the matrix L coincides with the difference star
—1
—1
—1
-1
8
-1
—1
—1
—1
FIgure 8.3.7. A coinbinatlon of triangles and parallelograms
Remark 8.3.18. Thangle and parallelogram divisions can also be combined. A polygonal domain a can be divided up into both triangles and parallelograins (ci. Figure 7). In this case VN is defined as {n E
u linear on the triangles, bilinear on the parallelograma}.
8.3.4 Quadratic Elesnents for fl c Let 'r be an adinindble triangulation of a polygonal domain I?. We wish to
increase the dimension of the finite-element sulispace by allowing, instead of linear functions, quadratics
u(x, y) = afl +
+
+ ajtx2 +
+
on
Er
(8.3.13)
so that
VN={UEC°(fl):u=0000a; with a quadratic function).
u coincides (8.3.14)
8.3 FInite Elements
(a)
Figure 8.3.8.
181
(b)
Nodes for
(a) a quadratic ansatz on a triangle, (b) a quadratic a.nsatz of the serendipity clam on a parallelogram
(a) Let x1,x2,x3 be the vertices of a triangle T E r,
Lemma 8.3.19.
while x4, x5, x6 are the midpoints of the sides (cf. Figure 8a). Each function
quadratic ont is detennined by the values {u(x3):j=1,...,6}. (b) The restriction of the function (13) to a side of T E r gives a onedimensional qt&adratic function, which is uniquely deterinine4 by three of the nodes lying on this side (e.g., u(x'),u(x4),u(x2) in Figure 8a). (c) flu is quadratic on each E r and continuous at all nodes (i.e., vertices of triangles and mid-points of sides), then u is continuous on 11. PROOF. (a) The quadratic function can be obtained as the uniquely defined interpolating polynomial of the form (13). Part (b) is also elementary.
(c) Let T and T be neighbouring triangles in Figure 8a. Since UIT and coincide on x1, x4, x2, by part (b) they represent the same quadratic function
onthecocninonsideofTandT. We call all [inner] vertices of triangles and mid-poiiits of sides tinner] nodes. By Lemma 18a we can find (or each inner node a basis functIon By Lemma 18c
belongs to VN. This proves
Remark 8.3.20. The number of inner nodes is the dimension of Vjy Lfl 14. Each u E VN admits the expression u = function belonging to the node is tharacterised by (9a).
where the basis
In the sequel we wish to assume that, as in Figure 7, both triangles and parallelograms are used in the partition. On the triangles the function u E VN are quadratic. The functions on the parallelograms P must satisfy the following conditions:
(a) u(z,y)Ip must be uniquely determined by the values at the 4 vertices and 4 mid-points of the sides (cf. Figure 8b). (/3) The restriction to a side of P gIves a (one-dimensional) quadratic function (of the arc length).
By condition (a) the ansatz must contain exactly 8 coefficients. The quadratic (13) has only 6 while the biquadratic
8 The Method of Finite Elements
182
has one parameter too many. If one omits the term ansatz one obtains for the unit square the function
=
+
+ as?? +
+
in the biquadratic
+ aaq2 + are2,i +
(8.3.15)
the so-called quadratic ansatz of the serendipity class. The restric= 0, 1] give a quadratic function in tions to the sides = 0, 1 [resp. in (1 Ib) (cf. Figure 6a,b) yields the function defined on the parallelogram P, which still satisfies the conditions (a) and (f3). If one were to use the full biquadratic ansatz one would one additional node which one could choose as the barycentre of the parallelogram. Cubic ansatzes can be carried out in the same way in triangles and parallelograms n Iresp.
The mapping
tt(x,y) =
(cf. Schwarz f 1]).
8.3.5 Elements for Si C It3 In the three-dimensional case assume isapolyhedron.
(8.3.16)
The triangulation from Section 8.3.2 corresponds now to a division of .0 into tetrahedra. it is called admissible, if (7a—d) hold in the appropriate sense: (7a)
becomes "Z(1 5 iS t) areopen tetrahedra"; in(7d)itnowshouldread "For
j,
Tg ii T, is either empty, or a common vertex, side, or face of and T,". Each linear function al + + asy + a4z is uniquely determined by its values at the 4 vertices of the tetrahedron. As a basis for the space I
VN={tIEC°(Th): u=O on OS?; u islinear on each tetrahedron Tj (1
I 5 t)}
one chooses with the property (9a): = 1, = 0 (j 1). The support of consists in all tetrahedra that share x' as a vertex. The dimension N = dim VN is again the number of inner nodes (i.e., vertices of tetrahedra).
As in the two-dimensional case the linear ansatz may be replaced by a quadratic one. Instead of a tetrahedron one can use a parallelipiped or a triangular prism with corresponding ansatses for the functions (cf. Schwarz 111).
8.3.6 Handling of Side Conditions The space V which lies at the foundation of the whole matter may be a subspace of a simply discretisable space W V. A given function w W belongs to V if certain side conditions are satisfied. Before we describe this situation in full two examples will be provided as
8.3 Finite Elements
183
Example 8.3.21. If one wishes to make the Neumann boundary-value probon 1' uniquely solvable by the addition of = gin (1 and 8u/8n lem u(x) dx =0, one can choose the space V to be the side condition
V = {t&
H'(a): J u(x) dx = 0).
(8.3.17a)
For a bounded domain (1 the form a(u, v) := f0(Vtt, Vii) dx is V-elliptic. The weak formulation (1.1) with 1(v) := if9 and satisfy the integrability condition equation = g, Oufôn = /(1) = 0 (cf. (3.4.2)). But even when f(1) 0 there exists a weak solution of dx. the corrected equation —4u = := g(x) — [fag dx +
Example 8.3.22. The Adler problem of Exercise 7.4.15 uses
V = {u
H1(a): u constant on r}.
(8.3.17b)
In both cases W = H1(.Q) is a proper superset of V. Let r be a trianboundarynodes.Let WhcWbe the space of linear triangular elements (cf. Remark 15): Nh := dim Wn =
+ 4d.
(8.3.18a)
The basis functions (bi: 1 i will be described as usual by C = öjj (cf. (9a)). Astheflnita-elementsubspaceofV we define Vh := Whfl V, which isof smaller dimension: Vh:_—W4flV;
Mh:—Nh—dirnVh.
In the Examples 21 and 22 we have Mh = 1, reap. MA = difficulty in the numerical solution of the discrete problem (19),
Find i1
E V,
the
with a(,i',v) = 1(v) for all v
Vh,
for example, in Example 21 no
(8.3.18b)
- 1. The (8.3.19)
belongs to
V,andthuaalsononetoVh,amcefobidx>o. In principle, it is poesible in the case of (17a) to construct as new basis functions linear combinations := V, which have again localised (but a bit larger) supports. In the case of (17b) it is still relatively emiple to find a practical basis for Va: The basis functions b4 which belong to Inner nodes are also in since &j H01(Q) C V. As further elements one
uses bo(x) := E'bj(x),
where E' denotes the sum over all boundary nodes. In spite of this, even in this case, it would simply be easier if one could work with the standard basis of WA. In order to treat the problem (19) with the aid of E we reintroduce the notation = Pw (w a coefficient vector, of. (1.6)).
8 The Method of Finite Elements
184
Eachv€Vh,C Vh} = P'Vh. Thus problem (19) is equivalent to: Find u E V with a(Pu, Pv) = f(Pv) for all v E V.
-
From (1 8a,b) we have dim V = dim Vh = Nh Mh = dim V are described by Mh linear conditions spaces
(8.3.19')
- Mh.
The
N,,
=
0
(1 I Mh);
(8.3.20)
j=l
V = Ker C = {w E where C =
(CU)
= O},
is an Mh x Nh matrix. In the case of Example 21 we have
Mh=1,
ci,=Jb,(x)dx
For Example 22 let x0,. ..,xMh, with Mh = Then C can be defined as follows:
= -1,
1, be the boundary nodes.
CU =0 otherwise.
The variation of v over V in (19') can be IRNh
(8.3.21)
by the variation of w over
if one couples the conditions (20) by using Lagrange multipliers
(1
which can be put together to form a vector A =
(A1,... ,AM,,). The resulting formulation of the probleni is:
with Cu =0 and find u E and A a(Pu, Pw) + (A, Cw) = f(Pw) for all w where (A,
=E
(8.3.22a) (8.3.22b)
is the acalar product in
Theorem 8.3.23. The problems (19) and (22a,b)
are
equivalent in the fol-
lowing sense. IfuA is a solution pairfor(22a,b) then u is a solution of(19') and Pu is a solution of (19). Conversely, if uh = Pu isa solution of (19) u and A solve
then there exists precss ely one A E
Before we prove this theorem, we give a matrix formulation which is equivalent to (22a,b)
Remark 8.3.24. Let L and f be defined as in (1.8a,b). Further, that B := CT with C from (21). Then (22a,b) is equivalent to the system of equations
FL
Biful
IBT oJ [Aj
Ff1
[oJ
(8.3.22')
8.4 Error Estimates for Finite Element Methods
PROOF. Lu + BA = f is equivalent to (BA,w) = (f,w) = f(Pw) for all w E JR.
185
Pw) + (A, Cw) = (Lu, w) + = 0 is the same as
Note BTU
Cu=O. PROOF (of Theorem p3). (a) Assume a solution of (22a,b) is given in terns of u, A. Then Cu = 0 implies
:= Pu E Vh. Equation (22b) holds in particular for all w V, U E V and so that (19') follows since Cw 0. (b) (19') implies I — Lu E V1. Since V1 = (KerC)1 = (KerBT)'- = Im(B), there exists a A E IRMh with f — Lu = BA, so that (fl'), and thus (22a,b) are we have KerB = {0}, so satisfied. For the Nh x Mh matrix 13 with rank U that the solution is unique. Note
that in the case of Example 21 the system of equations (22') is
essentially identical to (4.7.lOa,b).
8.4 Error Estimates for Finite Element Methods 8.4.1
for Linear Elements
In this section we shall restrict ourselves to the consideration of the linear elements from Section 8.3.2 and therefore assume:
(8.4.la) (8.4.lb) (8.4.lc)
r is an admissible triangulation of 0 C VN is defined by (3.8), if V = VN is as in Remark 8.3.15b, if V = H'(i7).
By Theorem 8.2.1 one has to determine d(u, VN) = WE VN}. To do this one begins by looking at the reference triangle from Figure 8.3.4.
Lemma 8.4.1. Let T =
0, + , 2, since Vp, Hm(Q). According to Example 6.2.5,
in order to have V,, C H2(.0) it is necessary that not only the function u change continuously between elements. The ansatz but also its derivatives functions must therefore be piecewise smooth and globally in C'(.0). introduce the one-dimensional biharmonic equaAs a model problem tion
u""(x)=g(x) forO <x m 9.1.1 The Regularity Problem The weak formulation of a boundary vaLue problem
Lu—g lull, Bu=p on!'
(9.1.1)
uEV, a(u,v)=f(v) forallvEV
(9.1.2)
as
was, in Section 7, the basis upon which we were able to answer the questions of existence and uniqueness of the solution. Here, by existence of a solution we understand the existence of a weak solution u V. The error estimates in Section 8.4 made it dear that the statement u E V is not enough. Under this assumption we can only show 0. The — uhllv more interesting quantitative estimate Iu — uhIi = 0(h) as in, for example, Theorem 8.4.6 for V = HJ(a) or V = H1(i7), requires the assumption u
112(Q) n V. The assertion u E 112(Q) or, more generally u E H'(s?), is a statement of regularity, i.e., a statement about the smoothness of the solution, which will be examined in greater detail in this section. The regularity proofs in the following sections are very technical. To make the proof ideas clearer, let us sketch the proof of inequality (4) below for the
Helmholtz equation = fin Sic u =OonI'. Step 1: 5? = Since the bilinear form for this situation a(u,v) =
Vv) dx+f0 uvdx is H'(1R2)-elliptic, (4) holds for s = m = 1. Weshall prove (4) by induction for 8 = To this end we take the derivative of the differential equation with respect to x, + = If I then H3(fl) and the equation by the induction as+v= sumption, has a unique solution v with IvL_i If one sets up this inequality for v = and likewise for v = the result is IuIa IuIe_i + IuxIe_i + Iui,Ia_i 3C,—11f(8_2. Thus, (4) has been shown for S. Step 2: 5? = = lRx(O, oo). As above, we can obtain the estimate since alsosatisfies = and the boundary condition
9.1 Solutions of the Boundary Value Problem in H'(Ii), s > m
=
0
implies
on 1. This is not the case for u1. But
209 E
E
This property, however, results from the differential u— — E equation, Step 3: Let 11 be arbitrary, but sufficiently smooth. As in Section 6.2.1, Ills decomposed into (overlapping) pieces il, which can be mapped into 1R2 or
show
f
is a partition Correspondingly one splits the solution u into Exsu of unity). Then the arguments from steps I and 2 prove inequalities for which together result in (4). Note that only a sketch of the proof was given. Some of the steps of the + proof are incomplete. For example, might not the equation = have a solution in 1R2, E L2(1R2), which does not belong to H' (1R2) and +v= hence does not coincide with the solution v H'(1R2) of In the following, always let s m. The boundary value problem (2) with of problem is sald to be H3-regular if each solution u V= and satisfies the estimate (2) with f H3_2m(Sl) belongs to
(9.1.3)
+
IuI.
If L is the operator associated with a(.,
then it is also said that L is
H'-regular. Remark 9.1.1. (a) H's-regularity always holds. with (b) Let the variational problem (2) have a unique solution u I-rn for all / E H-m(Il). If the boundary value problem is H3< C01f IUIm satisfies the inequality regular, then the weak solution of (2) with f luIs
(9.1.4)
Cjf13_2m.
(c) Let L be the operator associated with a(.,.). (4) is equivalent to H8(Q)) and (4'): IlL
—1
s + n12 n/2. Then the weak jE solution u of (7) belongs to C(11t"). Hence for s 2m, the weak solution is (1180 a classsa11 solution.
PROOF. The statement results from Sobolev's lemma (Theorem 6.2.30). U
9.1 Solutiom of the Boundary Value Problem in IP(fl), 8> m
213
Let the conditions (6) and Ut Then the weak solution of problem be satssfied for all k E sotssfled in particular if f belongs The conditions (7) belongs to constant. and the coefficients to
Corollary 9.1.7. Let a(.,.) be
/
Let
Theorem 9.1.8. Let a(.,.) in (5) be
00, 1 j n
(cf. Exercise 9d). To obtain this estimate for jUIm+, we write
(i +
+
f
(i +
The second integral converges to zero for h -s 0. The first one can be estimated
by
plus the sum
c'J
(1
j=1
cuf
3
(1
+ 3
= C"
Of
+
+ (1 +
215
9.1 SolnUons of the Boundary Value Problem in H(Q), a> m
and (13) follow. As in Lemma 6.2.24, u E (2) As in the proof of Theorem 3, one carries out induction over k and inves= k. v), tigates d(u, v) := a(tL, RD1'v) —
I
Exercise 9.1.10. Generalise Corollary 6 with the aid of Theorem 8.
9.1.3 Regularity Theorems for $1 in (6.2.18) is characterised by x,, >0. As in Section 7, we The halfspace limit ourselves to the following two cases: either a Dirichiet problem is given for arbitrary m 1, or the natural boundary condition is posed for m =1.
Theorem 9.1.11. (Homogeneous Dirichiet problem). An analogue to Theorem 3 holds for the Dirichlet problem:
a(u,v) = (f,v)0 for all V E
UE
(9.1.14)
Theorem 8 can also be carried over f one eXdUdC8 the values 8=1/2,3/2,..., m — 1/2.
For the proof we need the following highly technical lemma.
The nonnl•I.
Lemma9.1.12. is equivalent to
:=
(9.1.15)
+ loI=m
PROOF. The relatively elementary case s m is left to the reader. For 0 (d) (8/ôx,2)k(4,*u)(xI,0) = 0 for k = (e) for E 6.2.43.
for a =
(f) 4,E
for x',,>
_VXn)]
0,1,..., L —2 and U E k = 0,1, L — 1. Hint: cf. Corollar . . .
1
,
—L,2—L,...,0,l,...,L.
Step 3, (provab. L results from D° E via continuation arguments from Remark 6.3.14b) so that remair to be shown. Step 4. luL I4,uI. 1114'uIIL is true according to Step 1 of the proof. Tb
inequality llJ4,uJjL Cfr., which would finish the proof, reduces to for IcwI = m,
UE
= m. For
Let
f
L
S
i.
one verifies
[(—v) "u(x
=
,
+ (—i)
u(x', —xe/v)]
As in Exercise 131 one shows that
otherwise, E
over to real a [1 + m — L, L — m] except for the cases — $ E IN (i.e a = —1/2, —3/2,...) (cf. Liona-Magenes [1, p. 54 ffJ). Let L 2m + 1 an V
one infers from
E ffm—'(JRTh) Since
(Dau,
the estimate I(Da4,u, v)oI =
for all V E
C :=
It?Im_s,
I
and therefore (16) with
=
9.1 Solutions of the Boundary Value Problem in F1'(ø), 8> m
217
U
PROOF. (of Theorem 11) (a) First let k s = 1. The proof of Theorem ,n — 1) and implies 3 can be repeated for the differences U = the existence of the derivatives Ou/Ox, E j n. Thus one has = m except for a = (0,... ,0,m). for all (b)Weset . m, is the coefficient from the bilinear form (5). The remainder of the proof runs as follows. In part (c) we will show that
where
$Fa(v)I S CIvIm_i
m, V
for
(9.1.17)
= (w,D°v)o = (_l)rn(Daw,v)o, (17) means that E and IDaWIl_m C for (at = in. According to Lemma 12 it follows that w E The coercivity of a(.,.) implies uniform ellipticity i.e., (cf. Theorem of L = Hence it follows from 7.2.13). For = (0,... ,O, 1) one obtains aoa(x) Accordand wE E J'yJ = 1, that Dàu E ((a( m,a a) belong to ing to part (a) all other derivatives anyway so that u E has been proved. = 0, (c) Proof of (17). For each a a there exists a with ('y( = 1, o < y a (component-wise inequalities). Integration by parts yields Since
Fa(V) =
dx
+ a&&(D
J
—
and thus
for v E
0, 8
one proves correspondingly 1F0(t')I
1/2,..., rn— 1/2,
CIVIm_a and hence u
The generalisation of Theorem 11 to inhomogeneous boundary values
reads as follows.
Theorem 9.1.14. Let the biliriearfortn a(.,.) from (5) be For an 8 > 0, a {1/2,. . . , m — 1/2} either let (6) hold if a = k E IN, or be the weak solution of the inhomogeneous (12), ifs N. Let u E Dirichiet problem
a(u,v) = (f,v)o 8'u/8n1=çoj
for all v E
(9.1.18a)
forl=0,1,...,m—1,
(9.1.18b)
where
f
(0 I m
(9.1.19)
and satisfies the inequality
Then it belongs to
IUIm+. C.
— 1).
[if i...m+a +
÷ IUtm].
(9.1.20)
PROOF. For m =
1 Theorem 6.2.32 guarantees the existence of ito which satisfies the boundary conditions (18b) (for m> 1 ci.
it — ito is the solution of the homogeneous Wioka [1, Theorem 8.8J). w problem a(w,v) = F(v) (f,v)o — a(uO,v) (cf. Remark 7.3.2). Theorem 11 may also be carried over to the right-hand side F(v) under discusaion here U and yields w E Hm+a(lRfl) (instead of (1'
By similar means one proves
Theorem 9.1.15.
(Natural boundary conditions) Let the bilinear
form a(.,•) from (5) be N, or (12) ifs 8 = k
N. Let it
Fbr 8 > 0 either let (6) hold if be the weak solution of the
problem
a(u,v)= 1(v) =1 9(x)v(x)dx+jp(x)v(x)df' for ally €
9.1 SolutIons of the Boundary Value Problem in H'(S?), s> m
219
where H'—1
for
( I-I
>
H' 1/2(f)
—
fors m — 1,
f f(x)v(x) dx
for all v
Hm+'(a) belongs to
and satisfies the estimate
C.
(If I—m+. +
(9.1.23)
For inhomogeneous boundary conditions
=
with
the statement a and the estimate (20).
instead of u
•,m —1)
Hm(a) implies the statement a
PROOF. (a) Let {U':i = 0,1,...,N} with U' CS? be a covering of S? as with in Lemma 6.2.36. Let C U', be E the associated partition of unity from Lemma 6.2.37. There exist derivatives
which map U' (i 1) into such that a'(OU' n I) c By contrast, U° lies in the interior of 5? so that In 81)° =0. The solution u
a'
Regularity
220
can be written as part (c) xøL for i 1.
In part (b) of the proof we will treat xou, and in
(b) We set
(i=O,l,...,N)
d1(u,v) :=a(Xgu,v)—a(u,Xsv) and wish to show the estimate (u E
Ido(u, v)l CdIUImIVlm_a for
8
VE
8
J dx.
= XoDatj+ lower denvatives of u, one has
Since
=
—
161
(9.1.24)
= 1. Each suminand of do(u,v) has the form —
with
1)
with
161 2m—1. One integrates
161
= m by parts, and obteins a bound CIUImIVIm_i, whence (24) follows.
a way that the correThe coefficients Denote the resulting bilinear form sponding condition (22) Ia satisfied on v). Since Xo E C°°(.Q) has a support supp(xo) C U°, the extension of by poses no probleres. We may formally Xou through Xou(x) =0 for x since only the restriction of v to U° is of define do(u,v) for v any consequence. By (24) do(u, v) can be written, for a fixed u E in the form
do(u,v) =
With d0
Idols_rn
CdlUlm.
Xou is the weak solution of
v) = a(xou, v) = a(tt, xov) + do(u, v) = (f, xov)o + (d0, v)o for ally Hm(R"). = (thus too
Theorem 3 (resp. 8J proves xou IXoulrn÷.
CEIXOII_m+. +
l4I-m+. + iXOt&Irnl
0.
if t$I = m. Then the
must belong to of the problem
/ f(x)v(x) dx
belongs to
for dlv
E
and satisfies the estimate (23).
The condition of can be replaced by that of uniform ellipticity (7.2.3) (cf. Theorems 7.2.11, 7.2.13). The statement of Theorem 21 cannot be extended to 8 1/2 since then u Hr'(Si) would contain another boundary condition.
9.1 Solutions of the Boundary Value Problem in H'($2), 8> m
223
(cf. The proof of Theorem 21 uses an isomorphism R = R related to and also between Hr(fl) and proof of Theorem 8) between such that the form b(u,v) := a(Ru,Rv) is and It is necessary to prove that b(u, v) := a(u, R2v) is also
/
We know f H_rn+8(.11) implies R2f E a(u, v) = v)o is also a solution of a(u, = b(u, so that u E follows.
Eath solution of = (1,
(f,
9.1.5 Regularity for Convex Domains and Domains with Corners A domain .0 is convex if with x', x" E.0, x' + t(x" — x') belongs to I? for all 0 t 1 Convex domains in particular belong to C°", but permit stronger regularity statements than Theorem 21. .
Theorem 9.1.22. (Kadlec [1]) Let I? be bounded and convex. Let the bilinear form (5) be Lipschitz-continuous:
Let the coefficients of the principal part be
for all
=
=
1;
for the remaining ones let the following hol&
for all a,f3,-y withy
E
Then every weak solution u E a(u, v)
of the problem
clx for all v E = j f(x)v(x)
L2(Q) belongs to
with f
IaI+I/31 0. One calls the diffusion term and = (c,Vu) the convection term. For small e the convection term dominates.
248
Differential Equations
10
Exercise 10.2.1. Let the coefficients transformed by v(x) elliptic equation
=I
+
Equation (1) is elliptic for all
be
constants. Equation (1) can be into the symmetric Hd(fl)-
in a, v = 0 on 1.
(10.2.1')
> 0. That there is a unique solution
follows from Exercise 1 (cf. also Theorem 5.1.8). Denote the solution by
Remark 10.2.2. ug(x) and
cannot converge uniformly on Th for
—40.
PROOF. If
and 8u,/8x4 were continuous in one would be able to take the limit —+0 in Equation (1) and one would obtain the first-order differential equation
intl
(10.2.2)
:= This equation is of hyperbolic type, but not compatible with the boundary condition u =0 on I' (ci. Section 1.4), i.e., Equation (2) has, in general, no solution with =0 on 1'. U for
Equation (2) is called the "reduced equation". Equations (1) and (2) differ in the "perturbation term" -€ilu. Since the equations, (1) and (2), are of different types one speaks of a "singular" perturbation. The following example will show that *10 = limc..oue exists and satisfies Equation (2), but that the boundary condition mio =0 is only satisfied on a part Vr, of the boundary r.
Example 10.2.3.
(a) The solution of the ordinary boundary value problem
—at" + u'
=
1
in (0,1),
ti(0) = u(1) = 0
(10.2.3a)
= x_(ea/( 1). On [0,1), converges to *10(z) =x. This function satisfies the reduced equation (2), u' = 1, and the left boundary condition =0, but not uo(1) 0. (b) The solution of —at" — u'
= 0 in (0, 1),
u(0) = 0, *41) =
1
(l0.2.3b)
is Ue(X) = (1 — e_x/e)/(1 — In this case, u0(x) := = 1 satisfies Equation (2), —u' = 0, and uo(1) = 1, but not the boundary condition
at z =
0.
Which boundary condition is fulfilled depends for Equations (3a,b) on the sign of the convection term ±tt'. In the many-dimensional case the decisive factor is the direction of the vector c = (Cl,...
249
10.2 A Singular Perturbation Problem
tAO
1-.
1
1x 0-
0
I-
0
1
FIgure 10.2.1. (a) Solution of Example 3a; (b) Solution of Example 3b
In Figure la,b the solutions of Example 3 are sketched. In the interior Ue only in the neighbourhood of x = 1 (Figure la) [resp. x = 0 close to in order to satisfy the second boundary differ from (Figure lb)i does attain the condition. These neighbourhoods, in which the derivatives of is
order O(1/€), are called boundary layers. For Example 3 the thidmees of the boundary layer is of order Exercise 10.2.4. The interval [1 — ij, in which the function —1) exceeds the value 1, i.e., even the maximum principle does not hold. In addition one can see that Ue,h(X, 1/2) converges to 1 — x for the even multiples x = iih, and thus not to uo(x,y) = sin icy. For the intervening odd multiples x = vh an oscillation of amplitude O(h2/2€) develops.
The resulting difficulty is similar to that for initial value problems for stiff (ordinary) differential equations: If one keeps constant and leta it go to zero, the convergence assertions of Section 5.1.4 hold. However, if is very small, the condition it < 2€, without which one does not obtain a reasonable solution, cannot in practice be satisfied. One way out of this difficulty has already been described in Section 5.1.4. One must apprsdmate the convection term by suitable one-sided differences (5.1.14)
-t +
= h2 —t— h4 4€ + hIcjI + hIc2I
-€-h4
wherect :=max{0,4}, c
—
+ hcj
(10.2.6)
10.2 A Singular Perturbation Problem
251
Remark 10.2.6. If one discretises Equation (1) using (6) then one obtains, for all e > 0 and h > 0, an M-matnx Lh. For fixed e, the scheme has consistency order 1.
Exercise 10.2.7. The discretisation of Equation (3b) corresponding to (6) is 2e +h — hi and this gives the discrete solution uh(x) = = — (1 + — (1 + If one applies the difference operator (6) to a smooth function one obtains the Taylor series
= Lu — { The
0(h) term hIcjIu
+ Ic2luyp}h + 0(h2).
(10.2.7)
+ hIc2Iu..fl, is called the numerical viscosity (or
the numerical ellipticity) since it amplilles the principal part. A second remedy consists in replacing the parameter e by a discretisation If one chooses using Ch :=
hIcjI/2, hlc2I/2} or Ch
e
+
(10.2.8)
max
then the symmetric difference method
-1 =
—1
—1 +
4
1
02
0
cj
(10.2.9)
C2
leads
to an M-matrix. It is true that the convection term has been discretised
to a second order of consistency, but the error of the diffusion term is O(eh —c),
which is relevant in practice, amounts to
which, for the case h > 0(h). Instead of (7) one has Lu
The difference
—
—
—
c)zlu + 0(h2).
(10.2. 10)
is called the artificial viscosity.
In the one.dimensional case the methods using numerical and artificial viscosities do not differ:
Remark 10.2.8. If, in the one-dimensional case, on chooses
according to the second alternative in (8), then the difference formulae (6) and (9) coincide.
10.2.3 Finite Elements The difficulties described in the previous section are not restricted to difference methods.
Exercise 10.2.9. Show that
10 Special Differential Equations
252
(a) Linear finite elements on a square grid triangulation applied in the case of Equation (4) give a discretisation that is identical with the difference method —1
L,, =
4
—1
+h
—1
—1
0 —1/3 —1/6
—1/6 0
1/6 1/3
1/6
0
(1O.2.lla)
(cf. Exercise 8.3.13). (b) For bilinear elements (cf. Exercise 8.3.17) one obtains
=
1
—1
—1
—1
8
—1
—1
—1
—1
h
+— 12
0 0
+1
—4 —1
a
+1
1
+4
(c) One-dimensional linear elements for —eu" + u' = difference formula
=
th' [—1
2
—11+1—1
0
.
(10.2.llb)
f lead to the central (10.2.llc)
1).
The Exercises 9b,c show that finite-element methods correspond to central difference formulas, and thus can equally well lead to instability. The method of artificial viscosity corresponds to the finite-element solution of the equation (c, grad *4 = for appropriate As in Exercise 9b one may show
f
Remark 10.2.10. If one sets
:= IciIh, Ic2Ih} and uses bilinear elements then the discretisation of Equation (1) leads to an M-matrix.
On the other hand the matrix (ha) has different signs in the sub-diagonal and in the super-diagonal, so that it is not possible to have an M-matrix for any value of €h. The analogues of one-sided differences are more difficult to construct. One
approach is to combine a finite-element method for the diffusion term with a (one-sided) difference method for the convection term (cf. Thomasset [1, §2.4]).
A second possibility is the generalisation of the Galerkin method to the
Petrov-Galerkin method, in which the discrete solution of the general equation (8.1.1) is defined by problem (12): Find u E V,1, so that a(u,v) = 1(v) for ally e W,,
(cf. Fletcher [1, §7.2), Thomassetll, §2.2]). Here we have
dimVh = dimWh (but in general Vh
Wh).
(10.2.12)
11 Eigenvalue Problems
11.1 Formulation of Elgenvalue Problems The classical formula*;ion of an eigenva)ue problem reads Le
Xe
in .17,
B,e = 0 on F (j
1,•• ,m).
(11.1.1)
Here L is an elliptic differential operator of order 2rn, and B, are boundary operators. A solution e of (1) is called an eigenfunction if e 0. In this case, A is the elgenvalue associated with e. As in Section 7, one can replace the classical representation (1) by a variational formulation, with a suitable bilinear form a(., .): V xV —+ IR taking the place of {L,B,}: Find e
V with a(e,v) = A(e, v)0
for all v
V.
(11.1.2a)
dx is the L2($?)-scalar product. Strictly speaking one ought = where U is the Hubert space of the Gelfand triple to replace (., .)o by (., V c U c V1 (cf. Section 6.3.3). But here we limit ourselves to the standard case U = L2(fl). The adjoint eigenvalue problem is formulated as (ti,
Find e E V with a(v, e) =
for all V E V.
(11.1.2b)
Definition 11.1.1. Let A E e
C. By E(A) one denotes the subepace of all E(A) is called the eigenspace V which Satisfy Equation (1)[ resp.
for A. With E(A) one denotes the corresponding eigenspace of Equation (2b). A is called an eigenvalue if dim E(A) 1.
Theorems 6.5.15 and 7.2.14 already contain the following statements:
Theorem 11.1.2. Let V C L2(.17) be continuously, densely and compactly embedded (for example, let V = with bounded.1?). Let a(.,.) be Vcoercive. Then the problems (2a,b) have countably many eigenvalues A E C which may only have an accumulation point at oo. For all A C we have dimE(A) = dimE(A) 0 in C A and have and holomorphic. Caucby'a integral formula says (L —
)J)' = 2ws
—
A)'(L — CI)'dC
for all A E Ke(A).
From this one infers —
ES
C)IIvv
A')
(ci. Exercise 4(i)). Thus, w(A) cannot
i.e., w(A) min{w(():C E
For wh(A) the conclusion is the same.
aseume a proper minimum in
The converse of Theorem 8 is contained in
Theorem 11.2.10. Let (2a-c) hold. Let A0 be the eigenvalue of(1.2a). Then Ah = A0. of(la) (for all h) stich that there exist discrete eigenvalnes
PROOF. Let e > 0 be arbitrary. Aceording to Theorem 11.1.2, Ao is an isolated eigenvalue: >0 for 0 < IA — Aol e (e sufficiently small). is compact, we have that is continuous and Since positive. Because of (5) and w(Ao) = 0 one obtains is IA — Aol = for sufficiently small h Wh(A)
Cw(A)
—
Cp, —
> v1(h)/C
must have a proper minimum iii Ke(A0). Thus By Lemma 9 the minimal value is zero. Thus there exists a Ah Ke(A0) which is a discrete eigenvalue, Wh(Ah) 0. for all A E
I
The convergence of the eigenfunctions is obtained from Theorem 11.2.11. Let (2a—c) hold. Let e1'
be discrete eigenfunc= 1 and A0. Then there exists a subsequence e1" which converges in V to an eigenfunction eE(A0): Lions with
e
E(Ao),
— dlv
0
(i
oo),
lIellv
1.
PROOF. The functions
are uniformly bounded in V. Since V C L2(fl) is compactly embedded (ci. (2b)), there exists a subeequence which in L2(Q) to an e L2(Q): lie —
—. 0
(i
oo).
(11.2.7a)
11 Elgenvalue Problems
260
We define z = Z(Ao)e, = Theorem 8.2.2 there exists an h1(e) > 0
<e/2
liz —
for
1 holds if and only if there exists a solution v V for (L—AoI)v = e. According to Theorem 6.5.15c this equation has a solution If and only if (e,e)o 0.
•
Let E(Ao) = span {e}, E'(Ao) = span {e'}. Under the aesumptlon (8) e and e can be normalised so that
(e,e')o = 1.
(11.2.8')
261
11.2 FnIte Element Dscretiaation
= {v' e V':(v',e')o We define V := {v E V:(v,e')o = 0}, be the dual norm for = 11 liv. Fbr Problem (9):
For f
V', find u
(f,v)o for all V
aA(u,v)
O}. Let
(11.2.9)
E
one defines the variable corresponding to (3b)
ml
:=
sup
(11.2.10)
laA(u,v)i.
uEc'
Lt"IIv=l 11v11v1
Lemma 11.2.14. Let (2a,b), (8), and dim E(Ao) = 1 hold. Then there exists e. Problem (9) has exactly C > 0 for all IA — an e > 0 such that one solution u E V with ilulIv Ill Ilv'/2z(A), if
> 0.
C. x PROOF. Let L: 1' be the operator aesociated with a(., .): e (e sufficiently small) (L — AI)u = f has a unique solution For 0 < — Aol u E V. From f V' follows
0=
= ([L — A.1]u,e')o = (u, [L — M)'e')o =
—
X)(u, c')0,
— AI)': —+ Thus there exists as the restriction of For A = Ac Problem (9) has a unique solution cording to Theorem 6.5.15c. Then there exists (L — Al)-1 for all A
i.e., u E (L —
)J)-' to V' C V'.
According to Remark 5 (with V instead of V) , c2i(A) must be positive in Kd(Ao). The continuity of proves C > 0. In analogy to (4) one results from The bound by has llullv =
I
Exercise 11.2.15. Show that
Lemrnk 11.2.16. Let (2a-c), dim E(Ao) = 1, and condition (8) hold. Let A,, be discrete eigenvalues with urn,,....0 A,, = According to Exercise 12b there exists an e1' E,,(A,,) and e'1' with e" —' e E E'(7.o), (e,e')o = 1. This enables one to construct the space := {v" E V,,: (v", = 0} and the variable
inf
sup IaA(u,v)l.
UuIIv=1
there exists a C > 0 independent of h and A E ( such that Wh(A) Cw,,(A). For sufficiently smallc >0 and h, Wh(A) irj> Ofor c. Then
PROOF. (a) First statement: there exists h0 >0 and C such that
min{iiv+ae"llv:aEC}IIvIlv/C
262
11 Elgenvalue Problems
The proof is carried out indirectly. The negation reads: there exists a se0. = 1 and with C, hj —4 0, + quence E —* ? in L2(a). Evi—. a', Thus there exist subsequences with must have the limit w' = liznwj = v' + a'e' dently, := v, + = 0, it follows that w = 0, in L2(a). Since limIIw4IL2(n) lim(vj,e*h)o and thus v' = —a'e'. From 0 = = (v',e')o = one infers a' = 0. Thus the contradiction follows from 1 = tim 11v41v = + IIwiIIv = (b) Second statement: L21,,(A) Ccüh(A) with C from (a). Because aj(u,v)
foruEVh we have
=
inf
sup IaA(u,v)I/(IIuIIvIIvIIv)
C inf
sup max(aA(u,v+ae")I/[(ItLIIvIjv+ae"flv] aEC
=C inf
sup
IaA(u,w)I/[IIuhIvIIwUv]
O#ioEV,.
C
in!
sup
(c) Let e > 0 be chosen such that Ao is the only elgenvalue in For sufficiently small h, Ah is the only discrete eigenvalue in Ke(A0). In the proof of
From Part b follows that > 0 for A E According i := to Exercise 12a, aA(u,v) = (f,v)o (v E is solvable for each 1€ and all A E Ke(A0) such that = 0 is excluded. Lemma 9 shows that
I
Exercise 11.2.17. Let (2a,b) hold. The functions u, v V (u, v)o 0. Let d(eA, Vh) be defined as in (8.2.2). Show that if d(u, Vh) is sufficiently small then there exists a u1'
Vh with
IIu_uhllv 2d(u,Vh), Lemmk 11.2.18.
Let (2a—c) hOZCL Let Ao be an eigenvalue with (8) and dimE(Ao) =1. For sufficientty h there exists Eh(Ah) with —
C[IAo — AhI + d(e, Vh)].
PROOF. Let zh := Zh(AO)e be the solution of (6b). Since e = Z(Ao)e, one has n11e—zh
For all v
Vh we have
IIvCId(e,Vh).
(1l.2.lIa)
11.2 Finite Element Dlscretisation
v) =
= so
(.\o
—
(Ao
.&)(e, v)0
—
— (Ah
263
v)0
—
—
)th)(e,v)o + (Ah —
ehv)0 + (Ah —
—
that
=
+(Ah —.p)(e —z',v)o for ally
—
Vh.
(11.2.llb)
According to Lemma 16 we have f(e, — eh can be scaled so that = 0. (lib) corresponds to Problem (9). Lemma —
+ lie
Wh(Ah) 'Cu)10 —
Together with (1 la)
one
Theorem 11.2.19. =
1.
Let
14 proves
z"Iivi C'[
+ d(e, Vh)J.
—
obtains the statement.
Let
(2a—c) hold. Let
e
lAo —
U
Ao be
an esgenvalue with
E'(10), itchy = 1, (e,e)o =
E(A0), e there exist discrete eigenvalues Xh
dimE(Ao)
—
> 0 for sufficiently small h.
(8) and 1.
Then
with
)'hl Cd(e, Vh)d(e', Vh).
(11.2.12)
PROOF. Choose according to &ercise 17 such that 1Ie — uhhlv 2d(e,Vh), (e — u",e)o = 0. Discrete elgenvalues exist by Theo—i
rem 10. From
0=
a,10(eh,es)
=
= aAh(e",e) —(Ao _Ah)(e!l,e*)o —u") —()1.o —
_tiI)_(Ao_Ah)(eh,e*)o = follows
—
exists an
— e,
e — u")
—
— eilv[11e
Ahi
(Ao
e,
—
+
—
—
AhiJ.
e)o]
By Lemma 18 there
Eh(Ah) such that
P10 —
C'C"[(Ao
From this one obtains (12)
—
Ahl + d(e, Vh)J[I)1o — Ahi
with
C = 3C'C"
for
+ 2d(e*, Vh)J.
sufficiently small h, since
U
d(e,Vh)—40.
Theorem 11.2.20.
asswnptions of Theorem 19 there exiet for
e
elgenfunctions
E(Ao), e' E
Under the E()10) discrete
E Ea(Ah),
with lie
—
Cd(e, Vh),
lie*
— e"lIv Cd(e, Vh).
(11.2.13)
PROOF. Insert (12) with d(e, Vh) const into Lemma 18. The second estimate in (13) follows analogously. U
11 Eigenvalue Problems
264
In the following, let V C H'(fl). Theorem 11.1.5 proves C H14'(37).
C
(11.2.14a)
Also, let (14b) hold (cf. (8.4.10)): for all u
d(u, Vh) Ch'IIUIIHl+.(n)
E
U
(l1.2.14b)
be the cigenvalue Corollary 11.2.21. Let (2a), (2c), (14a,b) hold. Let with(8) and dimE(Ao) = 1. Then there exists Ah,e" Eh(Ah), esh E
such
that
IAo
— AhJ Ch2',
Ch',
lie —
lie5 — e*hIlv CM.
(11.2.15)
Occasionally eigenfunctions may have better regularity than is proven for ordinary boundary value problems. For example, let —& = Ac be in the rectangle (1 = (0,1) x (0, 1) with e= 0 on V. First, Theorem 11.1.5 implies thus e C°(17) (cf. Theorem 6.2.30). Thus e = 0 holds e E H2(.Q) ii in the corners of .0. AccordIng to Example 9.1.25 it follows that e (j7)
for s 1, one must replace C'h2 by
11.2.4
Complementary Problems
In Problem (9) we have already encountered a singular equation which neverthelees was solvable. In the following let A0 be the only eigenvalue in the disc Kr(A0). The equation
aA(u,v)=(f,v)o forallv€V
(11.2.16a)
8 singular for A = Ao. For A A0 Equation (16a) is ill-conditioned. In the following we are going to show that Equation (16a) is well-defined and well-
conditioned if the right-hand side f lies in the orthogonal complement of E(Ao):
11.2 Finite Element Discretisation
f
,i.e.,
(f,e')a =0
for all e' E
265
(11.2.16b)
In the case of A = A0
, with it, it + e (e E E(Ao)) is also the solution. The uniqueness of the solution is obtained under the conditions (8) and (16c):
E*(A0)
(11.2.16c)
Remark 11.2.23. Let (2a,b) and (8) hold. Let A0 be the only eigenvalue in Kr(A0). Then (16a,b) has exactly one solution it for all IA — A0I r which satisfies (16c). There exists a C independent of f and A such that IIuIIv CII! liv' PROOF. This follows from Lemma 14 in which the assumption E(Ao) = 1 is not necessary. U The finite element discretisation of Equation (16a) reads:
Find U"
V,,
= (f,v)o for all v E Vh.
with
(11.2.17)
In general, Equation (17) need not be well-defined, even assuming (16b). For the sake of simplicity we limit ourselves in the following to simple eigenvalues: d.imE(A0) = 1. Equation (17) is replaced by (18a):
Find
E
with aA(u",v) =
for all v
V,,
with f(s) .1.
(11.2.18a)
(11.2.18b) (11.2.18c)
Exercise 11.2.24. Show that is equivalent to: Find it" v) = (f(h), v)o for all v V,, with fl
SA(U1',
=
with as in Lemma 16.
Lemma 16 proves the
Remark 11.2.25. Let (2a-c), (8), dim E(A0) = 1 hold. Let A0 be the only eigenvalue in Kr(Ao). Then there exists an h0 >0 auth that) for all h ho and all A Kr(A0), the Problem (18a,b) has a unique solution it" u"(A) which satisfies the additional conditions (18c). Further there exists a C independent of h, A, and 1(h) such that CII! If E'(A0), f from (16b) need not satisfy condition (18b). If Eh(Ah) and E with (e",e"')o = 1 are known, one can define
f satisfies (18b) since Qh represents the projection on
(11.2.19)
(As)'.
11 Ezgenvalue Problems
266
=
=
Exercise 11.2.26. Letu 1.
1,
= 1,
= 1. Show that
d(u,
E
v" J..
+ llullvinf{11e8 —
= i hold. Let dim E(Ao) = dim Let h be sufficiently small such that Ao be the only eigenvatue in (following Remav* 25) the Problem (18a-c) is solvable. For the solutions u and of (i6a-c) and (iSa—c) the ermr estimate
Theorem 11.2.27. Let (2a-c),
llu—u'iiv
Cfd(u, Vh)+llfllvI inf(lle*
Iv': e E
—1 lIv'i
(il.2.20)
holds, with C independent off, f(h), and h. PROOF. Repeat the proof of Theorem 8.2.1 for aA(-,-) instead of a(-, .). Here one must choose w E Vh with w 1. Furthermore, (8.2.3) becomes
u,v) =
—
— f,v)o
for all v E Vh.
CN agrees with
> 0 (A E (cf. Lemma 16). flu — Why is estimated with the aid of Exercise 26, with llullv il/ liv' being added.
I
Corollary 11.2.28. If 1(h) is defined by (19), then inequality (20) becomes flu —
PROOF.
C'[d(u, Vh) + hilly' inf{hle* — e*hIlv: e' E E'(Ao)}J. (i2.2.2ia)
—lily' Cj(f,e")ol = Cl(f,e Cflf liv' hie* —
Corollary 11.2.29. If additionally the assumptions u E (14a), arid then (21a) yields the estimate d(u,Vh) S Chitihi÷1 hold forts E flu
—
Ch'JIulIffl+.(a).
(11.2.21b)
It remains to add the
PROOF. (Theorem 22) For e E(.Ao) there exists e1' Eh(Ah) with f := e — 1. E(Ao) and Ifhi = Il/ liv Ch = Ch. According to Remark 25 the problem OA0(v,w) = (v,f)o has a solution w I E(A0) for all v E V. The assumption of regularity yields w E H2(I?), 1w12 Cl/k such that E Vh exists with I. 1w — Chlwi2 C'hIflo. The value
11.3 Discretisation by Difference Methods
267
=0—aA0(e5,w")
=
(Ao
—
Aa)(e",w")o
—
=
(cf. (15)). From
be bounded by
can
= (f, f)o
—
a.x0(f,w) = a.x0(f,w — w") + (Ao
—
0). Show that this system of three equations is uniformly elliptic: IdetL"(e)I = (&,
For the treatment of Stokes equations we will use a variational formulation
in the next section. For reasons of completeneas we point out the following transformation.
Remark 12.1.5. Let n = 2 and thus u = exists a so-called stream function
(01,02). Because
with uj =
divu = 0 there
02 =
12 Stokes Equations
278
= Insertion in Equation (lat,2) results in the biharmonic equation = 0 on F. This 8f2/Ox1 — Ofj/0x2. The boundary condition (3) means =0 and 845/& =0 on F where ô/8t is the tangential is equivalent to = 0 implies = const on P. Since the constant may be derivative. =0 on P. chosen arbitrarily, one sets =
12.2 Variational Formulation 12.2.1 Weak Formulation of the Stokes Equations Since u = (u1,. •. , u,1) is a vector-valued function, we introduce
Ho'(a) x
x
(n-fold product).
x
A corresponding definition holds for H'(S?), H2(Q), etc. The norm associin the following. will again be denoted by ated with According to Remark 12.1.1 the pressure component p of the Stokes problem is not uniquely determined. In order to determine uniquely the constant in p = fi+const, we standardise p by the requirement p dx =0. That is the reason why in the following p will always belong to the subspace C
:= {p E L2(.O): J p(x) dx = O}. To derive the weak formulation we proceed as in Section 7.1 and assume that u and p are classical solutions of the Stokes problem (1.2a,b). Multiplication of the ith equation in (1.2a) with Vj E Cr(s?) and subsequent integration implies that
j =
f
d(x) =
+
{(Vui(x), Vvj(x)) —
dx
dx for
E
(12.2.la)
1
Summation over i now gives
J{(Vu(x), Vv(x))
—
p(x) div, v(x)} dx
=
L (1(x), v(x)) dx,
(12.2. la')
where the abbreviation
=
(Vu,Vv) i=1
*3=1
used. Equation (1 .2b) is then multiplied with some q grated, giving is
and inte-
12.2 Va atlonal Formulation
_fq(x)divti(x)dx = 0
279
(12.2.lb)
for all q E
With the bilinear forms
J(Vu(x)7Vv(x))dx for u,v E
a(u,v) b(p, v) —
J p(x)div v(x) dx
(12.2.2a) (12.2.2b)
vE
for p
we obtain the lVweak formulation of the Stokes problem as (3a-c):
Find u
such that
and p E
a(u,v) + b(p,v) b(q,u)
1(v) :=
J(f,v)dx
(12.2.3a)
for all v E
(12.2.3b)
for all q
0
In (3b) we first replace "v E
(12.2.3c)
Since both sides of and since Cr(I1) is dense in RW?),
by "v
(3b) depend continuously on v E (3b) follows for all v E
Remark 12.2.1. A classical solution u E
pE
of the Stokes problem (1.2a,b), (1.3) is also a weak solution, i.e., a solution of (3a-c). If conversely (3a-c) has a solution with u C2(Th, p C1 (1?), then it is also the classical solution of the boundary value problem (i.2a,b), (1.3).
PROOF. (a) The above considerations prove the first part. (b) Equation (3c) implies divu = 0. Let i {1,.. . ,n}. In Equation (3b) one can choose v with vj =0 for j i. Integration by parts recovers (la) and hence the ith equation in (1.2a).
I
12.2.2 Saddlepoint Problems The situation in (3a—'c) is a special case of the following problem. We replace the spaces and in (3a—'c) by two Hilbert spaces V and W. Let o(., .): V x V b(., .): W x
—' IR
V
JR
on V x
V,
(12.2.4a)
a continuous bilinear form on W x
V,
(12.2.4b)
a continuous bilinear form
fi
E
V',
12
E W'.
In generalisation of (6.5.1) we call b(., .): W xV
if there exists a Cb
JR
(12.2.4e) JR
continuous (or bounded),
such that CbIjwIIwIIvIIv
for all w E W,v E
The objective of this chapter is to solve the problem (5a-c):
V.
12 Stokes Equations
280
Find v V and w W with a(v,x) + b(w,x) = fi(x) for all x E V, b(y,v) = f2(Y) for all y W.
(12.2.5a)
(12.2.5b) (12.2.5c)
Formally, (5a—c) can be transformed to the form
Find u E X with c(u,z) = f(z) for all z E X
(12.2.6a)
if one sets:
X =V x W,
c(u,z) :=a(v,x)+b(to,x)+b(y,v) and
(x).
1(z) =fl(x)+f2(y) for u=
(12.2.6b)
Exercise 12.2.2. Show that (a) c(., .):X x X —+ IRis a continuous bilinear form.
(b) Problems (5a—c) and (6a,b) are equivalent.
That the variational problems (5) and (6) must be handled differently than in Section 7 is made clear by the following remark.
Remark 12.2.3. The bilinear form c(•,.) in (6b) cannot be X-elliptic.
(0).
PROOF. We have c(u,u)=0 for all tL= In analogy to (6.5.9) we set
J(v,w) := a(v,v) + 2b(w,v) and
—
2fi(v) — 212(w),
therefore J(v, w) = c(u, u) — 21(u) for u =
(t').i'or v E V, w E
W
we know J(v, w) is neither bounded below nor above. Therefore the solution v,w' of (5a—c) does not give a minimum of J; however, under suitable con-
ditions, v,w' may be a saddle point. Theorem 12.2.4. Let (4a-c) hold. Let a(.,.) be symmetrsc and V-elliptic. The pair
E V,
J(v,w)
E W is a solution of the problem (5a-c) if and only if
J(v,w') J(v,w)
for all v E V,w
W.
(12.2.7)
Another equivalent characterisation is
J(v*,w*) = mini(v,w) = max mmJ(v,w). VEV
WEWvEV
PROOF. (aa) Let v',w solve (5). The expreselon in brackets in
(12.2.8)
12.2 VariatIonal Formulation
w) = a(v
J(v, w) —
v,
—
—
281
v)
_v)+b(w*,v* —v) —fi(v —v)J vanishes because of (5b). Since a(v' — v,v — v) >0 for all v v E V, the second inequality in (7) follows. One also proves the converse as for Theorem
6.5.12: If J(v,w) is minimal for v = (ab) If v is a solution of (Sc), then
J(v',w) — J(v', to) =
then (5b) holds.
21b(w*
—
to,?) — f2(W' — to))
vanishes for all to, which proves the first part of (7) in the stronger form J(v,w) = J(v,w). If, however, v is not a solution of (Sc), there exists a WE W such that 6 := — J(v,w) 0. Since J(v',w') — J(v',ii,) = —5 for tis := 2w — to, the first inequality cannot be valid. For the converse = define to. The first part of (7) implies
J(ve,w) _J(v*,w±) =
0
for both signs, and so b(w,v') (Sc).
f2(W)J
= 12(w). Since to W is arbitrary, one obtains
(ba) We set j(w) := mInVEv J(v,w). According to Theorem 6.5.12, j(w) = to), where E V is the solution of (5b). If and are the solutions for to and to', it follows that
—v,,x) = F(s) := b(w—w',x) for all x Since IIFIlv'
— w'Ijy and IIvw
IIvw — vwsflv 5
(bb) By using the definition of
S C'IIFIIv' one obtains
—
— w'IIv
V.
for all w,w" E W.
(12.2.9a)
in (5b) we can write:
J(v*,w*)_J(vv,w)__a(v*,v*)+2b(w*,v*)_211(v)_212(w*) + 2b(w,
—
=
—
+
v*)
—
b(w,
—
—
—
v')
—
2f2(w)J
v'))
to,?) — f2(w — to)] _w,v*)_12(w* w)j.
.—?) + 2jb(w
—
—
(12.2.9b)
(be) Let v', wa be a solution of (5a-c). Because of (Sc) the expression in brackets in (9b) vanishes and we have
J(v*,w*) = (5b) gives
=
+
—
—?)
= j(w).
and so
J(v,w) =j(w') = maxj(w); wEW
(12.2.9c)
282
12 Stokss Equations
i.e., (8) holds. (bd) Now let v', w be a solution of (8). If in (9b) one sets w = w',
=
oneobtainsfromJ(v',w)=j(w')thatv =vw..Henoea(vw_v*,vw_vs)= depends quadratically on — — WIIV (ef. (9a)). The — variation over w := w' — Ày (A IR, y W arbitrary) givee
0=
= 21b(y,v)
—
—
1(y)],
and so (Sc) is proved. (Sb) has already been established with v =
v,L,..
12.2.3 Existence and Uniqueness of the Solution of a Saddlepoint
Problem To make the saddlepoint problem (5a-c) somewhat more transparent we introduce the operators associated with the bilinear forms: A
L(V,V') with a(v,x) =
for v,x
B*EL(V,Wl) with b(w,x) = C E L(X, X') with c(u, z) = (Cu,
V, (12.2.IOa)
BEL(W,V'),
= (w,Bz)wxw', (12.2.lOb) for u, z
Thus problem (6a,b) now has the form Cu =
X.
(12.2.1(k)
while (5a-c) can be written
as
Ày + Bw = B4v = 12.
If one assumes the existence of A1 v= and
fi
(12.2.lla) (12.2.llb)
L(V', V)1 one can solve (ha) for v:
A'(f1
—
Bw)
(12.2.12a)
substitute in (lib):
=
BA'f1
—
12.
(12.2.12b)
The invortibility of A is in no way necessary for the solvability of the saddlepoint problem. However, it does simplify the analysis, and does hold true in the case of the Stokes problem.
Remark 12.2.5. (a) Under the assumptions E L(V',V),
E L(W',W)
(12.2.13)
the saddlepoint problem (5a—c) jresp. Equations (1 la,b)] are uniquely solvable.
(b) A necessary condition for the existence of (BA'B)' is BE
V') is injective.
(12.2.14)
12.2 Variational
Formulation
283
PROOF. (a) Under the assumption of (BA'B)' E L(W',W), (12b) is uniquely solvable for w, and then (12a) yields v. (b) The mjectivity of BA 1B implies (14).
I
Attention. In general, B: W —+ V' is not bijective so that the representation is not possible. The example of the 3x 3 matrix of(BA'B)' as A
=
and B =
=
shows
that a system
of the form (1 la,b) can be solvable even with a singular matrix A. Therefore A closer look reveals the the assumption A—1 E L(V', V) is not subspace V0 := ker
B = {v
V: B'v = O} = {v
V:b(y,v) =
0
for all yE W} C V, (12.2.15)
as we noted before, in general is not trivial. The kernel of a continuous mapping is closed so that V, according to Lemma 6.1.17, can be represented as the sum of orthogonal spaces: which,
V=
V0
:= (V0)1.
with
V1
(12.2.16a)
ExercIse 12.2.6. Let (16a) hold. Show that (a) V' can be represented as where
:= {t/
(12.2.16b)
V':v'(v) = 0 for all V E Vj.}, (12.2.16c)
:= {v' E V':v'(v) = 0 for all v E Vo}.
AsanormonV01andVf oneusesll.IIv'. (b)The Riesz isomorphismJv:V—' V' maps V0onto V0' and V1 ontoVf (d) The following holds:
=
E
E Vf
(12.2.16d)
The decompositions (16a,b) of V and V' define a block decomposition of the operator A:
Aoj1 I
A
L.tilo A00
L(Vo,V01),
A01
L(Vj,Vo'),
A
I)
It11J
A10
L(Vo,VI),
A11
L(V1,Vfl.
Here, for example, A00 is defined as follows:
forvoEVo The corresponding decomposition of B into (Ba, B1) is written as (0, B), since =0 to the definition of V0. Conversely, we have range (B) C so that B = System (lla,b) thus becomes
284
12
Stokes Equations Aoovo
(12.2.17a)
+ A01v1 = ho
A10v0 + Ajjvj + Bw = B*VJ. = /2 where v =
+vJ.1
E V0, vj E Vj,
= ho +
(12.2.1Th) (12.2.17c)
fu, hoE
E
Theorem 12.2.7. Let (4a—c) hold. Let V0 be defined by (15). A necessary and sufficient condition for the unique solvability of the saddkpoint problem (5a-c) for all Ii V' is the aristence of the inverses E
E
L(Vf,W).
(12.2.18)
PROOF. (a) (17a—c) represents a staggered system of equations. (18) implies B*_l = (B—') L(W',V1) so that one can solve (17c) for vj = B''f2. v0 E V0 is obtained from (17a): vo = — Ao.jvj). Finally, w results from (17b). (b) In order to show that (18) is neceesary, we take ho V01 arbitrary, /1.1. = 0 and 12 = 0. By hypothesis here we have a solution (vo, vl, w) K Vo x V1 x W. = 0 because V0 n V1 = {0}. Thus = 0 implies Vj K Vo, so that Aoovo = ho has a unique solution vo E Vo for each 110 V01. Since A00: Vo is bijective and bounded, Theorem 6.1.13 shows that A&J' L(VO', Vo). If arbitrary and ito = 0, 12 = 0, one infers vj = 0 and one takes hi E = 0, so that Bw = has a unique solution w K W. As we did for A00,
Iii
one also infers that B1 E L(Vf, W). The formulation of conditions (18) in terms of the bilinear forms results in the conditions: inf{sup{Ia(vo,xo)I:zo E Vo, IIxoIIv = 1}:vo
Vo,flvoIIv = 1)
? a >0, (12.2.19a)
sup{Ia(xo,vo)I:xo K
Vo, IIxoIIv
= 1} >0 for all 0
V,HxIIv = 1}:w€ W,flwIIw
vO E V0,
(12.2.lOb)
1}fl>0. (12.2.19c)
ExercIse 12.2.8. Show that (19a) [resp. (19c)J are equivalent to (19a') [reap. (19c')j:
sup{ia(vo,zo)I:xo K Vo,IIxoIIv = 1} ajIvoIIv
for all v0
V0,
(12.2. 19a')
sup{Ib(w,x)I:z E V, Dxliv
LemmR 12.2.9.
1} I3Dwllw
for all w€ W. (12.2.19c')
Let (4a,b) hold. Let V0 be defined by (15). Then conditions (18) and (19a—c) ore equivalent. Here we have 1/a, 1/13.
12.2 Variational Formulation
= 0 for z E Vo one can write (19c) in
PROOF. Because of (16d) and the form inf{sup{Ib(w, x)j: x E V1. HxItv = For 0
285
l}:w
W, llwIIw = 1}
> 0. (12.2. 19d)
Vo, therefore according to (15):
x E V1 one has that x
= 1) >0 for all 0
E
xE V1.
(12.2.19e)
As in the proof of Lemma 6.5.3, we obtain the equivalence of (19a,b) with U and of (19d,e) with B' e E
Corollary 12.2.10.
(a) Condition (19b) becomes superfluous if a(.,.) is
symmetric on Vo x Vo or tf Lemma 6.5.17 applies. (b) Each of the following conditions is sufficient for (19a,b) and hence aLso
for
E L(V01, V0):
a(., .): Vo x Vo
JR
is V0-eliiptic:
for all vO E Vo,
a(vo,vo) a(., .): V x V
JR
(12.2.20a) (12.2.20b)
is V-elliptic.
PROOF. (a) As in Lemma 6.5.17. (b) (20b) implies (20a); (20a) yields U
(19a,b).
(4a—c), (18) is also
ExercIse 12.2.11. Show that under the equivalent to the existence of C1
L(X',X) (cf. (lOc)). Find a bound for
flC'flx._x' in terms of flAoo'flv0*_v0', IIAUv'4_v', and
12.2.4 Solvability and Regularity of the Stokes Problem Conditions (19a,b) (i.e., problem:
E L(V0", Vo)) are easy to 8atisfy for the Stokes
Lemma 12.2.12. Let.0 be bounded. Then the forms (2a,b) describing the Stokes problem
the conditions (4a,b) and (19a,b).
PROOF. (4a,b) is self-evident. According to Example 7.2.10, (.0)-elliptic. From this follows the H,?j(Q)-ellipticity of is
lobproves(19a,b).
Vv) dx Corollary U
It remains to prove condition (19c), which for the Stokes problem assumes the form
sup{I f w(x)divu(x) dxl: n E
luli =
l}
I3lwk
for all w E
(12.2.21)
286
12 Stokes Equations
or
(12.2.21')
for all w E
IIVWIIH-1(n)
Lemma 12.2.13. Sufficient and necessary for (21) is that for eachw E such that there exists u E w
PROOF. (a) For w The
=divu,
luli
fl'Iwk.
(12.2.21")
select u such that (21") holds and set ii
E
left-hand side in (21) is faw(x)divü(x)dx =
(b) If (21) holds one infers as in Section 12.2.3 the bijectivity of B: Vj —, W B'1w satisfies condition (21"). 1/ft. Therefore u with
Necäs [2j proves
Theorem 12.2.14. Condition (21) is satisfied If 11 E C°" is bounded. Under this assumption, the Stokes problem —4u + Vp = f,
—
div u = g in $1,
has a unique solution (u,p) luli + Info
u =0 on F (12.2.22) forf E H'(Q) andg E (12.2.23)
I/I—i + Igk].
Remark 12.2.15. Under the conditions that n = 2 and that 1? E C2 is a bounded domain1 the existence proof can be carried out as follows. PROOF. We need to prove (21"). Fbr w solve = win .0, go =0 on F. Theorem 9.1.16 shows that go E Since Vgo H1(f1) and n(x) C2(F) it follows that g := ôu/8n H'/2(f') (cf. Theorem 6.2.40a). From (3.4.2) one infers that = fawdx = 0 since w Integration of g over F yields C with OG/Ot = g, where is the tangential derivative. There exists a function %b E with = C and = 0 on F and CIGI3,2 C"Iwk. We set
Clearly, ul,u2
E
H'(.0). Let the nonnal direction at x
(nI(x),n2(x))T. The tangent direction is thus t(x) u = (u1,u2)T one obtains
(u, n) =
=
—cP5fli
(u,t) =
—goxn2
—
+ g011n1 —
F be n(x) =
(n9(x), _ni(x))T. For
= —g-i-ÔC/8t =0,
= —&p/8t
—
= 0,
since go = 0 on F implies Ogo/àt = 0. (u, a) = (u, t) = 0 yields u = 0 on 1' such that u = (ui, u2) E is proved. One verifies that
12.2 VariationaL Formulation
divlh = with
+
=
—
-
=W
+
.
k012 +
IUIi
287
The above proof uses the H2-regularity of the Poisson problem and requires corresponding assumptions on (1. Theorem 14 also assumes Q C°'1 Since the Poisson equation = I is solvable for any domwn .0 which is
contained in the disk KR(0), or at least in a strip {x E < R}, reone might conjecture that a similar sulting in the inequality CRlf result holds for the Stokes problem.
E (0, 1) let = {(x,y): —1 0 such that (21") holds for all 0, w e
Counterexample 12.2.16. For
0
function is defined by :=
—
—
i)
in T,
u
0 otherwise.
(12.3.13a)
The name derives from the fact that u is positive only in T and vanishes on (cf. Exercise 8.3.14) OT and outside. The map -. T to a general T results in the expression ü.1(x,y) :=
(12.3.13b)
for the bubble function on
Exercise 12.3.10. Let
be a quasiuniform triangulation. Show that there dzdy. exists a C >0 independent of h such that dx dy
295
12.3 Mixed Finite-Element Method for the Stokes Problem
We set
linear combinations of the linear elements E H0t (fl)
and the bubble functions for f'
Wa: linear elements E
Vh :=
For the side condition have
(12.3.14)
E
see Section 8.3.6. Since ü..p e H0t(fl), we
C
C
be the quasiunifonn tnangulatson on a bounded and Wh be given by (14). Then the stability conpolygonal domain £7. Let and that the dition (8) is satisfieaL Under the further conditions that Si E Pois8on problem be H2(Si)-regt4ar the Brezzi condition (6b,c) also holds.
Theorem 12.3.11. Let rh
PROOF. (a)
On every T E Ta, Vp is constant:
an arbitrary p We set
Vp = v := >
E
so that
ii := v/lyle
(ü.jr. bubble function (13b) on
= 1. Exercise 10 yields
b(p, v)
J (Vp, v) dx = TEm
+
= 0
f
T
Since lVplo and lpI' are equivalent norris on the subapace H'(Q) fl
it follows that lb(p,v)l
and
C'lplilVpk/jvk. In a
similar way one shows that lvlo
C"(VpIo and obtains 43 := C'/C" independent of h. The left-hand side in (8) is
with so
that
(8) follows.
(b) (9a) is satisfied for Vh (cf. Theorem 8.4.5). The same holds for the inverse estimate (9b) (cf. Theorem 8.8.5). Therefore (6b,c) follows from Theorem 8. U
A general result on the stabilisation by bubble functions can be found in Brezzi-Pitkäranta [11.
12 Stokes Equations
296
12.3.3.3 Stable Discretisations with Linear Elements in Vh If one wants to avoid bubble functions, one must increase the dimension of V,, in some other way. In this section we shall consider for Vh and Wh two different triangulations and Th. By decomposing each T rh as in Figure 1, through halving the sides, into four similar triangles, one obtains Th12. We define:
: linear elements for triangulation Th/2,
V,, c
Wh c
H'(fl) fl
(12 3 15)
: linear elements for triangulation
or
V,, c
:
quadratic elements for triangulation Th,
W,, C H'(Q) fl
linear elements for triangulation r11.
FIgure 12.3.1. Theorem 12.3.12.
Th
'12
16
and Th/2
Theorem 11 holds analogowsig for V,, x Wh in (15) or
(16).
PROOF. (a) Let Vh x Wh be given by (15). Fbr each inner triangular side of the triangulation m there exist two triangles T17,T21, E with = (cf. Figure 2). Let 8/Ot be the derivative in the direction of 'y, let 8/On be the directional derivative perpendicular to it. There exists and with
+
=
1,
8/Ox = a78/On + b78/Ot,
O/Oy = b.1O/On
—
a70/Ot.
In contrast to Op/On, Op/Otis constant on T17 UT27 U We denote its value by pu,. The mid-pomt of is a node of rh/2. We define the piecewise linear function over by its values at the nodes u.1(x7) =
w,(x') = 0 at the remaining nodes
(12.3.17a)
and set v
:= E (
E Vh,
v/lv$o.
The sum E7 extends over all interior sides of Tj,. In
(12.3 ITh)
U T2-, we have
12.3 Mixed Finite—Element Method for the Stokes Problem
297
(Vp,
so that
J (Vp, (
I
dx =
T17UT2.1
Ch2>1pt1.y12. are the sides of T
Th (r5 is quasiunifonn!), then
From this one infers that b(p, v) 11, and finishes the proof analogously.
dx
i IVpIo, as in the proof of Theorem
(b) In the case of quadratic elements given by (16) one has the same nodes as in (a) (cf. Figure 8.3.8a and Figure 1). Use (17a,b) to define the quadratic function V E Vh and carry out the proof as in (a). U
FIgure 12.3.2.
Ti.,, Ta.-,
12.3.3.4 Error Estimates In the following, the condition (8.1.12) should be replaced by the stability condition (6a.-c). In the place of the approximation property (8.4.6') we now have the inequalities inf{Iv —
inf{Ip —
Vh}
E W,j
Chfi42 for all V E V =
for all p
W=
(12.3.18a)
(12.3.18b)
Condition (18a) is equivalent to condition (9a) in Theorem 8. The following theorem applies to general saddlepoint problems, and can be reduced to Theorem 8.2.1.
Theorem 12.3.13. Let u =
X = V x W be the solution of (2.5a—c) [resp. (6a)1. Let the discrete problem (la-c) with Xh = x C X satiify
12 Stokes Equations
298
the Brezzi condition (6a.-c) and have the solution
= (v's, w"). Then there
exssts a varsable C independent of h such that —
uhjlx Cinf{IIu —
(12.3.19)
Xh}.
PROOF. The Brezzi condition (cf. Theorem 6) yields S lix' for all right1-hand sides f X' in (2.5a--c), in particular for all f E Xh = The above inequality means t' for the operator Lh:Xh —, which belongs to c(., .): Xh x Xh -. IR. According to Exercise 8.1.16 is equivalent to Condition (8.1.12) for c(., .): X x X IR [instead of a(., .): V x V IRj with 1/n. Theorem 8.2.1 yields the statement (19).
For the Stokes problem with (),
instead of u =
inequality (19) is now rewritten as follows: uhi2 + [p
—
C2inf{lu
+
—
_phlo2:uh
(),
=
Vh,p" E Wh}
or lu
+
—
—
uhfi + [p
Wh}.
(12.3.20)
Theorem 12.3.14.
Let the Stokes equation (1.2a,b) have a solution u E (ci. Theorem 12.2.19). For the suhspaces
p
and Wh C l€t the Brezzi condition (6a—c) and the approximation conditions (18a,b) be Then the discrete solution satisfies the estimate C
lu —
+
—
P"lo
C'hflul9 + Ipli].
PROOF. Combine inequalities (20) and (18a,b).
(12.3.21)
N
Using the same reasoning as in the second proof for Theorem 8.4.11 with
c(..) instead of a(.,.) one proves
Theorem 12.3.15. For each f
L2(a), g E
fl H'(fl) let the Stokes
problem ÷ Vp f, —divu = g have a solution u E p H1(Q) fl with 1u12 + C[1f10 + 1gb]. Under the asstImptlon8 of Theorem 14 we then have the estimates —
+ [p — p'i—i
C'h[Ju(i + Iplol,
(12.3.22a)
—
÷ [p
C"h2[Iu12 + [p1']
(12.3.22b)
—
p41.i
for the finite element solutions. Here lpf—t is the dual norm
Corollary 12.3.16. Combining (22b) and (21) one obtains
12.3 MIxed Finite-Element Method for the Stokes Problem —
+
—p"Io
+
299
(12.3.22c)
A new exposition of the mixed finite-element discretisatiori can be found in Brezzi-Fbrtin [1).
Bibliography
Adams, R. A. RI Sobolev spaces, Academic Press, New York (1975).
Agmon, S., A. Douglis, and L. Nirenberg Estimates near the boundary for solutions of elliptic partial differential equations satisfying general boundary conditions II, Comm. AppI. Math., 17 (1964) 35—92.
Aziz, A. K. (ad.)
(11 The mathematical found at wn of the finite element method with applications to partial differential equations, Academic Press, New York (1972).
and A. K. Aziz [1) Survey lectures on the mathematical foundations of the finite element method, In: Aziz (1) 3-363.
Bank, R. E., T. F. Dupont, and H. Yserentant [1) The hierarthical basis multigrid method, Numer. Math., 52 (1988) 427458.
Blum, H. and M. Dobrowolaki [1) On finite element methods for elliptic equations on domains with corners, Computing, 28 (1982) 53-63.
Blum, H. and R. Eannacher [1J On the boundary value problem of the biharmonic operator on domains with angular corners, Math. Meth. App!. Sci., 2(1980)556-581. [2) Extrapolation techniques for reducing the pollution effect of reentrant corners in the finite element method, Numer. Math., 52 (1988) 539564.
Bramble, J. H. [1J A second order finite difference analog of the first biharmonic boundary value problem, Numer. Math., 9 (1966) 236-249.
Bramble, J. H. and B. E. Hubbard [1) A finite difference analog of the Neumann problem for Poisson's equation, SIAM J. Numer. AnaL., 2 (1964) 1-14.
301
BIbliography
12) Approximation of solutions of mixed boundary value problems for Poisson's equation by finite differences, J. Assoc. Comput. Mach., 12 (1965) 114—123.
Bramble, J. H. and A. H. Schatz 111 Higher order local accuracy by averaging in the finite element method, Math. Comp., 31 (1977) 94—111. Brezzi, F. [1) On the existence, uniqueness and approximation of saddle-point problems arising from Lagrangian multipliers, RAIRO AnaL Nurner., 8 (1974) 129—151.
Brezzi, F. and M. (1) Mixed and hybrid finite element methods, Springer, Berlin (1991).
Brezzi, F. and J. Pitkäranta [11 On the stabilization of finite element approximations of the Stokes equations, In: Hackbusch 161 11—19.
Chatelin, F. [1] Spectral approximation of lsnear operators, Academic Press, New York (1983).
Ciarlet, Ph. 0. Ill The finite element method for elliptic prvbiems, North-Holland, Amsterdam (1978) (paperback edition 1980). Collatz, L. 11] The numerical treatment of differential equations, (1966).
Springer,
Berlin
Coener, C. [1J On the definition of ellipticity for systems of partial differential equations, J. Math. Anal. Appl., 158 (1991) 80-93. Dieudonné, 3. [1) Foundations of Modern Analysis, Academic Press, New York (1960).
Erikason, K. and C. Johnson An adaptive finite element method for linear elliptic problems, Math. Comp., 50 (1988) 361-383.
Fleteher, C. A. 3. Computationo4 Cale,*in methods, Springer, New York (1984). Gantmacher, F. R.. [1) The theory of matrices. VoL 1, Chelsea Publishing, New York (1977) German edition: Matrizenredinung. Band I. Deutactier Verlag der Wissensthaften, Berlin, 1958
Gladwell, I. and R. Wait (ads.) (1J A survey of numerical methods for partial differential equations, Clarendon Press, Oxford (1979).
302
Bibliography
G. Lube, H.-G. Roos, and L. Tobiska Goering, H., A. [1] Singularly perturbed differential equations, Mathematische Forschung, Vol. 13, Akademie-Verlag, Berlin (1983). Green, C. [1) An essay on the applioation of mathematiail analysis to the theorse8 of electricity and magnetism, Nottingham (1828). Grisvard, P. [1J Elliptic problems in nonamooth domains, Pitman, Boston (1985). Hackbusch, W. 1] On the regularity of difference schemes, Ark. Mat., 19 (1981) 71—95. 12] On the regularity of difference schemes. Part H: Regularity estimates for linear and nonlinear problems, Ark. Mat., 21 (1983) 3-28.
[3] A note on the penalty correction method, Zeitachrift fur Analysis und i/ire Anwendungen, 2 (1982) 59-69. [4] Local defect correction method and domain decomposition techniques, Computing, 5 Supplement(1984) 89—113.
[5] Multi-grid methods and applications, Springer, Berlin (1985). [61 Efficient solvers for elliptic systems, Notes on Numerical Fluid Mechanics, 10, Vieweg, Braunachweig (1984). [7] Integrulgleidiungen. Theorie und Numerik, Thubn&, Stuttgart (1989). [8J On first and second order box schemes, Computing, 41(1989)277-296. [9] Iterutive Lösting grofier schwachbesetzter Gleichungasysteme, Ibubner, Stuttgart (1991) English translation: Iterative solution of large sparse systems. Springer, Berlin (to appear in 1992) [10] Parallel algorithms for PDEs, Proceedings. Kiel, January 1990. Notes on Numerical Fluid Mechanics, 31, Vieweg, Braunathweig (1991). Hackbusch, W. and R. Rannacher (eds.) [1] Numerical treatment of the Namer-Stokes equations, Proceedings. Kiel, January 1989. Notes on Numerical Fluid Mechanics, 30, Vieweg, Braunschweig (1990).
Heinnch, B. (1] Finite difference methods on irregular networks. A generalized approach to second order problems, ISNM, Vol. 82, Birkhäuser, Basel (1987). Heliwig, G. [1] Partial differential equations, 2nd ed., Thubner, Stuttgart (1977) German edition: Partielle Differentialgleichungen. Thubuer, Stuttgart, 1960 Heuser, H. [1] Ftinktionalanalysis, 2nd ed., Teubuer, Stuttgart (1986). Kadlec, J. [1] 0 reguljarnosti zadth Puassona na oblasti s granicej, lokoi'no podobno granice vypukloj oblasti (On the regularity of the solution
Bibliography
303
of the Poisson equation on a domain with boundary locally similar to the boundary of a convex domain), Czecho8lovak Math. J., 14 (1964) 386—393.
Kato, T. Lii Perturbation theory for linear openztors, Springer, Berlin (1966). Kellogg, 0. D. [11 Foundations of potential theory, Springer, Berlin (1929).
Kellogg, R. B. and J. B. Osborn 111 A regularity result for the Stokes problem in a convex polygon, J. Flrnct. Anal., 21 (1976) 397—431.
Kevorkian, J. and J. D. Cole [iJ Perturbation methods in applied mathematics,
Springer,
New York
(1981).
Kondrat'ev, V. A. LII Boundary value problems for elliptic equations in domains with conical or angular points, Thins. Moscow Math. Society, 16 (1967) 227—313.
Ladytenskaja, 0. A. Lii Fimktionalanalytische Untersuchungen der Navier-Stokessehen Olcichungen, Akademie-Verlag, Berlin (1965). Ladytenskaja. 0. A. and N. N. lJral'ceva Linear and qwisihnear elliptic equations, Academic Press, New York (1968).
Layton, W. and T. D. Morley Lii On central difference approximations to general second order elliptic equations, LAA, 97 (1987) 65—75. Leis, Ft.
Voriesungen uberpariielle Differentialgicichungen zweiter Oninung, Bibliographiaches Institut, Mannheim (1967). 121 Initial boundary value problems in mathematical physics, Thubner, Stuttgart and Wiley, Chichester (1986).
Lesaint, P. and M. Zlámal [1] Superconvergence of the gradient of finite element solutions, RAIRO Arwlyse Numer., 13 (1979) 139—166.
Lions, J. L. and E. Magenes (1] Non-homogeneous boundary value problems and applications. Springer, Berlin (1972).
Vol. I,
Louis, A. [1] Acceleration of converenge for finite element solutions of the Poisson equation, Numer. Math., 33 (1979) 43—53.
Marchuk, C. I. and V. V. Shaidurov Lii Difference methods and their extnzpolation, Springer, New York (1983).
304
Bibliography
Meis, Th. and U. Marcowitz Numerscal solution of partial differebtial equations, Springer, New York (1981) German edition: Numerische Behandlung partieller Differentialgleichungen. Springer, Berlin, 1978
Miranda, C. (11 Partial differential equations of elliptic type, Springer, Berlin (1970). Mitchell, A. R. (l} Computational methods in partial differential equations, Wiley, London (1969).
Mitchell, A. R. and R. Wait (11 The finite element method in partial differential equations, Wiley, London (1977). Mizohata, S. (11 The theory of partial differential equations, University Press, Cambridge (1973).
J.
[1] Sur Ia coercivité des formes sesqui-linéaires elliptiques, Rev. Roumaine Math. Put-es AppL, 9(1964)47-69. (2] Equations aux dérivée partielles, Presses de l'Universit4 de Montréal. (31 Lea Méthodes directes en théorie des equations etlsptiques, Masson, Paris (1967).
Nitsthe, J. (1] Bin Kriteriwn für die Quasi-Optimalitãt des
Verfahrens, Nu-
met-. Math., 11 (1968) 346-348.
Oden, J. R, and J. N. Reddy [1] An introduction to the mathematical theory of finste elements, Wiley, New York (1976).
Pereyra, V. numerical solution of quasi-linear elliptic systems in n [lj Highly dimensions, Math. Comp., 24 (1970) 771-783.
Pereyra, V., W. Proskurowski, and 0. Widlund [11 High order fast Laplace solvers for the Dirichiet problem on general regions, Math. Comp., 31 (1977) 1—16.
Rheinboldt, W. C. (1] Adaptive mesh refinement processes for finite element solutions, mt. .1. Numer. Meth. Engty., 17(1981) 649-662.
Sthatz, A. H. [1] A weak discrete maximum principle and stability in the finite element method in L°° on plane polygonal domains, Math. Comp., 34(1980) 77—91.
Bibliography
305
Schatz, A. H. and L. B. Wahibin [11 Maximum norm estimates in the finite element method on plane polygonal domains. Part 2: Refinements, Math. Comp., 33 (1979) 465-492. Schwarz, H. R. [1) Finite element methods, Academic Press, London (1988) German edi-
tion: Methode der finiten Eleinente. 3rd edition. Thubner, Stuttgart 1991.
Shortley, G. H. and R. Weller [1] Numerical solution of Laplace's equation, J. Appl. Phy8., 9 (1938) 334— 348.
Stoer, J. [1) Einfiihrung in die Nume.rische Mathematik 1., 5 Aufi., Springer, Berlin (1989).
Stoer, J. and R. Buhrsch Introduction to numerical analysis, Springer, Berlin (1980).
Strang,G. andG.J.Fix (1] An analysis of the finite element method,
Prentice-Hall,
Englewood
Cliffs, N.J. (1973).
Stummel, F. [1] Diskrete Konvergenz linearer Operatoren. I, Math. Ann., 190 (1970) 45—92.
12) The generalized patth test, SIAM J. Numer. Anal., 16 (1979) 449-471.
Thomasset, F. [1) Implementation of finite element methods for Namer-Stokes equations, Springer, New York (1981). Thomée, V.
[1) Discrete interior Schauder estimates for elliptic difference operators, SIAM J. Namer. Anal., 5(1968)626-645. 121 High order local approximations to derivatives in the finite element method1 Math. Comp., 31(1977)652-660. Varga, R. S. ji) Matrix iterative analysis, Prentice Hall, Englewood Clifs, N. J. (1962). Velte, W. [1) Direkte Methoden der Variationsrechnung, Teubner, Stuttgart (1976).
Walter, W. [1] Einfiihrung in die Potenholtheorie, Bibliographisdies Institut, Mannheim (1971).
Wathen A. J. [1] Realistic eigenvalue bounds for the Galerkin mass matrix, J. of Numer. Analysis, 7' (1987) 449-457.
306
Bibliography
Whiteman, J. R. (ed.) [1] The mathematics of finite elements and applitxdion8, VII - MAFELAP London 1990. Proceedings. Uxbndge, April 1990, Academic (1991).
Wioka, J. Cambridge [1] Partial differential equations, Cambridge University (1987) Cennan edition: Partielle Differentialgleithungen. Teubner,
Stuttgart,
1982
Yosida, K.
Rinctional analysis, Springer, Berlin (1974). Yserentant, H. Lii On the multi-level splitting of finite element spaces, Numer. Math., 49 (1986) 379—412.
Zienkiewicz, 0. C. [11 The finite element method in engineering science, McGraw-Hill, New York (1977) German translation: Methode der finiten Elemente. Carl Hanser Verlag, München, 1984 Zlámal, M.
Discretization and error estimates for elliptic boundary value problems of the fourth order, SIAM J. Numer. Anal., 4 (1967) 626-639.
Index
Note: The page numbers given in bold-face refer to occurrences of the term
printed emphatically on the page indicated. Adjoint problem, 141, 190, 196 Adler problem, 159, 183 Agmon, condition of, 158 AnalytIc, 20 condItion, 139 BabuMca-Brezzi condition, 284 Banach space, 110, 111, 112, 113, 131, 135, 137, 161 Basis, 161, 165, 166, 171, 182, 183, 204, 291 hIerarchical, 202 Basis condition, 171, 172, 173, 175, 176, 177, 180, 181, 183, 200
Beltrami operator, 14 Biharmonic equation. Ses Differential equation Bilinear ftrin, 137, 154, 155, 227, 279
adjoint, 138, 190, 191 continuous or bounded, 138, 146, 157, 161, 185, 279, 280 corresponding to a pde of order 2m, 146, 210 corresponding to the Helmholtz equation, 147, 154, 208 corresponding to the Poisson equation, 149, 154, 165, 176, 180, 183
extension of a, 138 operator associated to a. See Operator, 138, 140, 142, 163, 209
symmetrIc, 138, 141, 149, 152, 164, 170, 200, 204, 208, 280, 285
V-elliptic (H"-elliptic, etc.), 140, 141, 147, 148, 149, 152, 153, 154, 159, 164, 170, 183, 200, 204, 212, 228, 280 V-coercive etc.), 141, 142, 143, 149, 150, 153, 154, 157, 158, 166, 171, 185, 210, 211, 212, 213, 218, 219, 222, 223, 226, 227, 228, 229, 256, 267
Block, block matrix, block partitionlog, 283 Boundary condition, 158, 226, 275 Dirichlet, 27, 35, 38, 65, 85, 86, 96, 145, 150, 159, 164, 174, 178, 215, 222, 229 discretisation of the, 100
first, 35,96 mixed, 96 natural, 152, 154, 155, 159, 168, 178, 218, 221 Neumann, 36, 65, 96, 154, 174, 183
periodic, 98 second, 96 third, 96 Boundary element method (BEM),
37 Boundary layer, 249
Index
308
Boundary value, 7, 9 10, 12, 26 Boundary value problem, 13, 19,
27,
28,85
Brezzi condition, 292, 293, 295, 298 Bubble function, 294, 295, 296 Cauchy-Riemann equations, 4, 7 Chequer-board. See Ordering Coefficient vector, 161, 255 Compact (cf. embedding, nine-point formula, operator), 135, 201, 257, 269 Complementary problem, 264 Complete, 24, 112 Completion, 112, 117, 122, 125 CondItion, 108, 204 Cone condition, 136 Conservation form, 244 Convex. See Domain Consistent, consistency, 59, 64, 69, 93, 94, 200, 228, 232, 233, 235, 239, 242, 267 Convection-diffusion equation, 247 Convergence, 168 of the difference method, 59, 60, 61,
69, 83, 84, 95, 102, 108,
239
of the eigenfunction, 255, 259, 260, 271, 272 of the 255, 258, 260 super, 202, 203 Coordinate transformation. See Domain Corner, re-entrant (cf. domaIn), 34, 223, 225 Covering, 127
10, 112, 113, 114, 115, 117, 119, 123, 141, 169, 170, 224
Dense,
Derivative, 104 conormal, 96, 201 normal, 15, 65, 95, tangential,
96,
96, 145
99, 103, 145,
224,
246, 278 weak,
115,
116, 172
Diagonalisable, real, 6, 7 Diagonally dominant, 47, 48, 49, 52 Irreducibly, 47, 48, 49, 52, 53, 92, 93, 250
Difference (divided), 38, 49, 62, 79, 240
backward or left-sided, 38,
269
108, 109 of higher order, 62 Difference operator, 10, 44, 121, 210, 13-point,
213,
228, 233
Difference
star (stencil), 44, 62, 91
Differential equation(s) biharmonic, 74, 95, 103, 104, 105, 106, 108, 158, 194, 218, 278
223,
order, 1, 3, 4, 6, 277 of order 2m, 104, 108, 145, 158 of second order, 2, 5, 8, 12, 85, of
first
90,150
ordinary, 1, 2, 3, 38, 248, 254 partial, 1, 2, 7 system of, 3, 4, 6, 7, 105, 275, 276
Differential operator, 5, 59, 60, 75, 85, 89, 104, 253, 276 formally ad.joint, 92, 95 boundary, 95, 154, 155, 156 Dirichiet integral, 144 Dirichiet principle, 144 Divergence (operator), 275 12 convex, 79, 197, 223, 224, 225, 231, 235, 236, 242, 288, 293 exterior, 36 general, 78, 100, 109, 196, 234, 241, 243 L-shaped, 13, 34, 127, 201, 243 normal, 15, 16, 19, 28, 33 transformation of the, 6, 89 unbounded, 23, 146, 147 Double-layer potential. See Potential Dual form, 131, 134 Dual space, 130 Eigenfunctlon, 222, 253, 254, 259, Domain,
263, 271 5, 6, 7, 10, 47, 51, 52, 53, 137, 142, 150, 258, 254,
Elgenvalue,
255, 256, 258, 259, 260, 261, 262,
264, 265, 271,
Eigenvalue problem, 7, 56, 65,
72, 91, 250
forward or right-sIded, 38, 227, 250 second, 38, 80, 238, 240
symmetric, 38, 70, 71, 102, 233, 249, 251, 252 Difference method, 38, 59, 65, 90, 91, 93, 94, 102, 108, 173, 177, 178, 202, 204, 247, 249, 267,
91,
274 137, 150,
11,
253
adjoint, 253, 256 elliptic, 253, 254, 255, 274 Ethptic,
4,
104,
5, 6, 7, 9, 10, 11, 12, 85,
276
Index
uniformly, 86, 87, 88, 89, 104, 145, 147, 148, 149, 150, 155, 222, 229, 276, 277 V-. See Bilinear form, 248, 256, 280, 285, 289, 293 Embedding compact, 185, 136, 142, 143, 150, 153, 171, 186, 253, 256 continuous, 118, 131, 133, 134, 136, 137, 141, 143, 171, 185, 253, 256 dense, 131, 133, 134, 137, 141, 171, 253, 256
Error estimates for difference methods, 190, 238, 239
for eigenvalue problems, 260, 263, 264
for finite element methods, 185, 190, 193, 196, 246, 266, 297, 298
for Ritz-Galerkln methods, 167 Estimate. See Error estimate, inverse estimate Existence. See Solution Extension. See Bilinear form, operator of a function, 123, 129, 215, 237 Extrapolation (method), 61, 84 Finite elements, 171, 172, 174, 251, 254
constant (bi)cubic, 182, 194, 195, 196, 296 (bi)Ilnear, 172, 174, 175, 178, 179, 182, 183, 185, 190, 194, 199, 204, 246, 247, 252, 295, 296 isoparainetrlc, 198,247 mixed, 290, 299 nonconforjnal, 199, 200, 291 of the serendipity class, 181, 182, 193
quadratIc, 181, 182, 193 Finite element method, 24 Five-point formula, 40, 41, 53, 60, 62, 92, 176, 190, 231 Form. See Bilinear form, dual form, 8esquilinear form Fourier expansion, 9 Fourier transform, 110, 119, 120, 124, 125, 126, 135, 213 FunctIonal, 132, 140, 146, 152 Fundamental solution, 16, 17, 28 Galerkin method. See RJtz-Galerkln, Petrov-Galerkln G6rding, Theorem of, 150
309
Gei'fand triple, 41, 133, 134, 136, 141, 163, 207, 253 (Ierschgorin, Theorem of, 46, 48 Gradient, 15 Green's formula, 15, 29, 144, 153 Green's function of the first kind, 28, 29, 30, 31, 33, 34, 95, 105, 144
discrete, 55, 72, 95 Green's function of the second kind, 35 Grid, 39, 40, 226 Grid function, 39, 44 Grid points close to the boundary, 63, 79, 83 far from the boundary, 42,63, 79 80 neighbouring, 41, 42, Grid size (width, step sIze), 39 Harmonic, 13, 15, 16, 19, 20, 22, 23, 24, 25, 144 Harnack, Theorem of, 20 Heat equation, 3, 4, 5, 7, 8, 10 Hehnholtz equation, 147, 154, 208 hubert space, 110, 114, 115, 117, 122, 132, 134, 279, 289 Hökler continuous, 29, 30, 110, 123, 125
Hyperbolic, 4, 5, 6, 7, 8, 9, 10, 11, 248
Improperly posed. See Well-posed Inclusion, 135, 136, 142 Initial-boundary value problem, 8, 9, 207 Initial value problem, 2, 3, 7, 8, 9, 38
Instationaiy problem, 10, 11 Integral equation method, 36, 37 Interpolation, 84, 124 Hermite, 194, 196, 207 spline, 194, 196, 207 Inverse estimate, 206, 221, 240, 293 Irreducible, 45, 54 Irreducibly diagonally dominant. See Diagonally dominant Lagranglan (6.ctor), 184, 290 Lamé differential equations, 277 Laplace equation (of potential equations), 2, 12 Laplace operator, 12, 14, 131, 176 Lexicographical ordering. See Ordering
Lipschitz continuity, 23, 30 Map, mapping. See Operator Mass matrix, 205
Index
310
Maximum (-minimum) prmciple, 17, 18, 19, 20, 27, 86, 103, 250, 258
Mean-value property, 17, 18, 19, 20,
27,54 Mehrstallen method, 64, 83 M-matnx, 45, 49, 50, 51, 52, 53, 64, 68, 81, 83, 84, 91, 92, 93, 100, 106, 249, 250, 251, 252 Multi-index, 30, 104 Neighbour. See Grid points Neumann condition. See Boundary condition Nine-point formula, 90, 180 compact, 63, 63 Nodal points, 172, 175, 178, 181, 203
Nodal values, 172, 195 Norm, 29, 30, 50 dual, 131, 132, 134, 163, 194, 215, 227, 298 Euclldean, 14, 51, 57, 110, 163, 204, 226 equIvalent, 111, 117, 123, 204, 295
matrix, 50 assocIated, 50, 204, 227 maximum, 46, 110 operator, 111 row sum, 50, 59, 170, 242 Sobolev, 117, 122 Sobolev-Slobodeckil, 122, 247
spectral, 51, 59, 204 supremum, 23, 110, 112 vector, 50 Normal (dIrection), 15, 116, 145, 224
Normal system, 105 Normed space, 110
Operator, ill adjoint, 132, 163, 213, 216, 242 associated. See Bilinear form, 209, 256, 282 compact, 135, 136, 137 difference. See Difference operator differential. See Differential operator
dual, 131 selfadjoint, 132 Ordering (of the grid poInts), 42, 43, 45 lexicographical, 42, 67, 70 chequer-board, 43 Orthogonal (ef. projection), 114 Orthogonal space, 114, 163, 283
Parabolic, 4,5, 5,7,8, 9, 10, 11 Partition of unity, 128, 130, 150, 209, 219
Petrov-Galerkin method, 252 Plate equation (cf. blharmonic), 85, 103 Poisson equation, 27, 34, 35, 38, 41, 57, 103, 149, 155, 177, 201, 225, 231, 254, 276, 287, 289, 293, 295 Poisson's integral formula, 19, 20, 22
Polar coordinates, 13, 14, 15, 97, 98, 144
Positive definite, 5, 52, 53, 93, 106, 164, 204, 255
Positive semidefinite, 52, 87 Potential, double-layer (dipole), 36 sIngle-layer, 36 volume, 36 Potential equation (of. Laplace equatIon), 2, 4, 5, 7, 9, 12, 13, 14, 16, 28, 35 Precompact, 135 Principal part, 6, 12, 95, 145, 147, 148, 150, 223, 228, 251, 276 Projection, 132, 170 orthogonal, 132, 133, 163, 170, 225, 294
Ritz, 170, 189, 191, 192, 193, 225 Rayleigh-Ritz. See Ritz Reduced equation, 248, 250 Reference triangle (reference element), 10, 177, 179, 185, 198, 204, 294
Regularity, 190, 192, 196, 208, 209, 215, 219, 222, 223, 225, 232, 246, 254, 264, 267, 285, 287, 288, 293, 295 discrete, 227, 228, 229, 230, 231, 239, 240, 241, 242, 243, 271 Riesz reprssent.atlon theorem, 132 Riesz isomorphism, 132, 283 Rlesz-Schauder theory, 1ST, 142 Ritz-Galerkln method (cf. solution), 161, 171, 290 Robin problem, 159 Saddle-point (problem), 279, 280, 282, 284, 290, 297 Scaler product, 52, 114, 115, 117, 121, 132, 148, 162, 170, 227 Separation (of variables), 3, 11 Serendipity class. See Finite elements Sesquilinear form, 138
Index
Seven-point formula, 91, 92, 176 Shortley-Weiler discretisation, 78, 80, 81, 83, 229, 231, 237 Side condition, 182, 183, 290, 291 Single-layer potential. See Potential Singular perturbation, 247, 248 Singularity function, 14, 16, 56 discrete Sobolev, lemma of, 125, 212, 222 Sobolev space (cf. norm), 114, 117, 122, 128, 133, 134, 136, 145 Solution classical, 27, 30, 32, 145, 146, 147, 154, 212, 222, 246, 279
existence of a, 9,13,23,29, 32, 35, 37, 67, 137, 142, 144, 151, 155, 161, 286, 292 existence and uniqueness of a, 137, 141, 142, 147, 148, 149, 150, 151, 152, 153, 162, 164, 166, 167, 168, 171, 183, 188, 222, 248, 282, 284, 285, 286, 288, 289, 292 Ritz-Galerkin, 164, 165, 166, 168, 170
uniqueness of the, 18, 19, 26, 27, 35, 86, 87, 88, 148, 149 weak, 27, 146, 150, 159, 190, 208, 209, 210, 212, 213, 218, 219. 222, 223, 226, 246, 279 Sparse matrices, 44, 171, 173 Spectral radIus, 47 Spectrum, 137, 142 Stable, stabilIty, 59, 64, 70, 90, 94, 95, 102, 103, 227, 228, 242, 249, 252, 291, 293 Star (stencil), 178, 180 Stationary problem, 10 Steklov problem, 254 Step size. See Grid size Stiffness matrIx, 102, 163, 165, 171, 177, 191, 200, 202, 203, 255 Stokes equations, 4, 103, 275, 278, 279, 282, 285, 286, 288, 290, 291, 292, 293, 298 Support, 115, 171, 173, 176, 182, 183, 194, 195, 202 Time, 8, 9, 10, 11 Trace (of a function), 3, 123, 129 Trace (of a matrix), 8? Transformation. See Domain, Fourier Transformation theorem, 119, 129 Transition condition, 245, 246, 247 Trefftz'method, 200
311
Triangulation, 174 adaptive, 201 admissible, 174, 175, 179, 185, 190
quasi-unifurm, 188, 190, 204, 208, 294, 295
regular, 188 uniform, 188
Type (of a pde), 1,4,5,6,7,10 Type invariance, 6 Uniqueness. See Solution Variational formulation (weak formulation), 144, 146, 149, 150, 158, 161, 165, 166, 173, 183, 194, 208, 244, 253, 254, 278, 288
Variational problem, 141, 152, 159, 184, 171 dual/complementary, 200 Viscosity, artificial, 251, 252 numerIcal, 251 Wave equation, 2, 3, 5, 8, 10 Weak formulation. See Variational formulation Well-posed (problems), 23 Wilson's rectangle, 199
This book offers a simultaneous treatment of the theory and the numerical treatment of elliptic problems. The subjects dealt with include the classical theory (Green' s function, maximum principle, etc.) as well as the variational formulation. The author describes and analyses finite difference and finite element methods. Specific chapters
are devoted to the eigenvalue problem and the Stokes problem.
ISSN 0179-3632 ISBN 3-540-54822-X
111111
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