Topics in Finite Element Solution of Elliptic Problems [1st ed.] 978-3-662-39197-6;978-3-662-40197-2

179 71 8MB

English Pages [200] Year 1979

Report DMCA / Copyright

DOWNLOAD FILE

Polecaj historie

Topics in Finite Element Solution of Elliptic Problems [1st ed.]
 978-3-662-39197-6;978-3-662-40197-2

Table of contents :
Front Matter ....Pages i-v
Sobolev Spaces (Bertrand Mercier)....Pages 1-10
Abstract Variational Problems and Examples (Bertrand Mercier)....Pages 11-30
Conforming Finite Element Methods (Bertrand Mercier)....Pages 31-51
Computation of the Solution of the Approximate Problem (Bertrand Mercier)....Pages 52-68
Review of the Error Estimates for the Finite Element Method (Bertrand Mercier)....Pages 69-75
Problems with an Incompressibility Constraint (Bertrand Mercier)....Pages 76-91
Mixed Finite Element Methods (Bertrand Mercier)....Pages 92-144
Spectral Approximation for Conforming Finite Element Method (Bertrand Mercier)....Pages 145-158
Nonlinear Problems (Bertrand Mercier)....Pages 159-185
Back Matter ....Pages 186-195

Citation preview

Lectures on TOPICS IN FINITE ELEMENT SOLUTION OF ELLIPTIC PROBLEMS

By BERTRAND MEROER

Notes by G. VIJAYASUNDARAM

Published for the

TATA INSTITUTE OF FUNDAMENTAL RESEARCH. BOMBAY

i:~

I~ · .

~"

Springer-Verlag Berlin Heidelberg GmbH 1979

Author

BERTRAND MERCIER Ecole Polytechnique Centre de Mathematiques Appliquees 91128 Palaiseau (FRANCE)

©

SPRINGER-VERLAG BERLIN HEIDELBERG GMBH 1979

Originally published by Tata Institute of Fundamental Research in 1979.

ISBN 978-3-662-39197-6

ISBN 978-3-662-40197-2 (eBook)

DOI 10.1007/978-3-662-40197-2 No part of this book may be reproduced in any form by print, microfilm or any other means without written permission from the Tata Institute of Fundamental Research, Bombay 400 005

CONTENTS 1.

Sobolev Spaces

2.

Abstract Variational Problems and Examples

11

3.

Conforming Finite Element Methods

31

4.

Computation of the Solution of the Approximate Problem

53

5.

Review of Error Estimates

69

6.

Problems with an Incompressibility Constraint

76

7.

Mixed Finite Element Methods

92

8.

Eigenvalue Approximation

145

9.

Nonlinear Problems

159

BIBLIOGRAPHY

1

186

PREFACE THESE NOTF.S SUMMARISE a course on the finite element solution of Elliptic problems, which took place in August 1978, in Bangalore. I would like to thank Professor Ramanathan without whom this course would not have been possible, and Dr. K. Balagangadharan who welcomed me in Bangalore.

Mr. Vijayasundaram wrote these notes and gave them a much better form that what I would have been able to. Finally, I am grateful to all the people I met in Bangalore since they helped me to discover the smile of India and the depth of Indian civilization. Bertrand Mercier Paris, June 7, 1979.

1. SOBOLEV SPACES

We state the Sobolev imbedding theorem, Rellich

introduced.

proof of the theorems the reader is n em

Let

1.1. NOTATIONS.

n

r~ferred

= 1,?

(n

r denote the boundary of Q,

without proof.

Hl (n),

theorem, and Trace theorem for

smooth.

is

Hl(Q)

IN THIS CHAPTER the notion of Soholev space

For the

to ADAMS [1].

or 3) he an open set.

Let

it is 1ssumed to be bounded and

Let

= {f:

L2 (n)

!lfl 2 dx


1

to ~ (n)

41

n

in

in ~ (n) supp

is a

< 1

= IL

The topology chosen for ~ (Q) of elements

n. ~(n)

f:J

(Q)

is such that a sequence

converges to an element

if there exists a compact set $n'

supp

41 c I
for the duality bracket between 1!J' (0)

EXAMPLE 1. (a) If f

£

L2 (0)

~'(0).

and

~

A square integrahle function defines a distrihution: then

< f, ~ >

=!

0

f~

dx

for all

~

£

f> (0)

We identify L2 (0)

can be seen to be a distribution.

as a space

of distribution, i.e. L2 (0) C:. f)' (0) •

(b)

The dirac mass

~,

concentrated at the origin,

defined by < ~, ~ >

= ~(0)

for all

~ £~(0)

defines a distrihution.

DEFINITION.

DERIVATION OF A DISTRIBtiTION.

If f

(0) .

is a smooth function and

integration by parts we

af

obta~ a~

! .,...- ~ dx • - ! f lY dx. n xi o oxi

~

£ "

(0)

then using

3

This gives a motivatio n for defining the derivativ e of a distribut ion, If T £ ,

1

and

(o)

is a multi index then oaT£ f)' (0)

a

is defined by < DaT'

If Tn' T

£

II

'

> = ( -1) a

i) 1 (O)

< T' Da' > V ' £

then we say Tn -+

n

if

'£~(0).

is continuous since

Da:~' -+ ~~

The derivativ e mapping Tn -+ T in i) ' then < DaT ,cfl >

in I) ' (0)

forall

-+

n

if

T

i) (0) .

= (-l)lal -+ (-1) lal

= < DaT,

is defined by

The Sobolev space H1 (o)

1.3. SO~OLEV SPACE.

Hl(o) = {v

£

t2(o):

:~.

cfl £ j)(O).

'> for all

£

~

L2 (o), 1

i

~ n}

1

where the derivativ es are taken in the sense of distribut ion. • 1y _af need not 1mp axi

EXAMPLE

2.

Let

0

t2(n). u

= [-1,1] f(x)

Then f £ t2[-l,l];

£

but

=

{

if

-1

df/dx •

0

X < 0

if X > 0, ~

is not given by a locally

integrabl e function and hence not by an

L2 function.

'

4

We define an inner product (u, v) 1 • (u, v) Let

r (~u

n +

i•l oXi

I

~v

axi

(•,•) 1 in H1 (0)

) for all

u, v

£

as follows:

H1 (0)

0

II • 11 1 be the norm associated with this inner product

LE~ 1,

H1 (0)

Proof.

Let uj

?Jith

0

Then

II • 11 1 is a HilbePt space.

be a Cauchy sequence in H1 (0).

This imply

au.

{u.} 1 { ~} i = 1 1 2 1

xi

J

n

... ,

are Cluchy in L2 • Hence there exists

V1

v.

1

£

L2 (0)

1
+


Hence

Thus uj

+

u in

auJ axi

+

au axi

>+ -
1

"'
> = 0 for all

Hence - 6u + u = 0

in \J) ' (n) •

+ £ f) (Q).

6

Let

n = {X

:Rn:

£

IX I

1},


1 there are infinitely many r 1 s (r u

£ (~(n))~.

£

Sn-l)

Moreover these functions for different

linearly independent.

(~(n))l,

such that r 1s

are

(n)

in

Therefore

dimension

~ (n)Jl. ~ 2 if n

dimension

(~ (n))l.

= oo

=1

if n > 1.

This proves the claim (b). We shall define HI (n)

as the closure of

0

Hl(n).

5?J

We have the following inclugions

C

~ (n)

dense

1.5. TRACE THEOREM.

c

H1 Cn)C:H 1 (n)

dense

0

n be a bounded open subset of JRn with

Let

r:

a Lipschitz continuous boundary

i.e. there exists finite

number of local charts

aj, 1 ~ j ~ J

into JRn

B > 0 such that

and a numher J

r

=V {(y .. J= 1

I

'Y ) : Y n

n

= a J. (y

{(yl,yn):a/Y 1 ) < yn < aj(y 1 ) + {(y 1 ,y ):a.(y 1 )-s < y n

J

n

from

I ) '

e, IY I 1

{y 1

IY I I

£

:Rn-l: IY 1 I < a}

< a},

< a}CSl,l ~ j ~J,

< aj(y 1 J.IY 1 1 < alc::cn,l < j _< J. -

7 OD

It can be proved that f

CIO-

t

is dense in nl(o).

c (n)

namely yf,

we define the trace of f,

C (0)



1f OD-

£

by

(0) coo_

is continuo us and linear with norm

y : C (0} + L2 (r)

II yu II L2 (r)

f

If

~ CII u II 1 . Hence this can be extended as continuo us

linear map from H1 (n) H01 (n)

to

L2(r).

is characte rised by

THEOREM 2.

1.6, DUAL SPACF.S OF

H1 (0)

I: H1 (0)

I(v)

+

= (v,

The mapping

AND H1 (0). 0

-

(L 2(0))n+l

defined by

av .. ,, av axn) lii'

is easily seen to be an isometri c isomorphism of H1 (0) subspace of with

F(Iu)

into

(L 2(n))n+l u

= f(u)

is a continuo us linear function al on

Hence by Hahn Banach theorem F can be extended to

I(H 1 (0))

(L2(n))n+l

Therefo re, there exists f(u) = F(Iu)

= (v,u)

n

+ ~

. 1

(v., au/ax.). 1

1=

1

This represe ntation is not unique since F cannot he extended uniquely to (L2(n))n +l

For all

cf>

£

l) (n) we have

8 f(cj>)

v,u > -

Thus -

Conversely if T

where v,v 1



av. _1

ax.1 '

r

ax.1

_1

is given by

av.1

r

ax.1 '

i=l

L2 (n), 1 ~ i ~ n

then T can be extended as a

continuous linear functional on H1 (n) T(u) = (v,u)

n +

> ,

av.

1= 1 •





Hence -Au+ cu • f

in ~·(n)

(2.19)

To find the boundary condition we use Green's formula:

I

n

I

vu.vv dx •

which holds for all u

n £

-Au.vdx +

H2(n)

I

r

~u v dr,

on

and for all v

Assuming that our solution u

£

£

Hl(n).

H2(n),

from (2.19)

we have

I

n

(-Au+ cu)v •

I

n

fv

for all v

£

Hl(n).

Using Green's for.ula we obtain I·cvuvv + cuv)dx •

n

I

r

!uon v ctx + Jn fv dx.

This, together with (2.18), implies

J r

(g-

:u n

)v

dr. 0 for all v

£

nl(n).

Hence we get the desired boundary condition au an • g on r • If u

£

H2(n),

these are still valid in "some sense" which is

given in LIONS-MAGENES [29].

REMARK 2.

Even when g • 0 we cannot take the space

19 v 1 = {v

E

av H1(n): an= o on since

to be the hasic space V,

v1

r}

is not closed.

In the

Neumann problem 2.3, we obtain the boundary condition from Green's In the case of Dirichlet problem 2.1, we impose the

formula.

boundary condition in the space itself.

n

r

If

REGULARITY THEOREM (FOR DIRICHLET PROBLEM) 2.

is

c2

or

u

of

f E L2(n), then the solution

is a convex polygon and

the Dirichlet problem (2.l), (2.2) belongs to H2(n).

REGULARITY THEOREM (FOR THE NEUMANN PROBLEM) 3. n

is a convex polygon,

finer than

L 2 (n)

f

E

L 2 (n)

(for example g

If

r

is

c2

or

and g belongs to a space E H1 (r)),

then the solution

u

of the Neumann problem (2.l2J, (2.l3) belongs to H2(n). For a proof of these theorems the reader is referred to NECAS [33].

2.4. MIXED PROBLEM.

In Sections 2.1 and 2.3 we found the variational

formulation from the partial differential equation.

In the general

case it is difficult to formulate the variational problem from the p.d.e. problem.

In fact a general p.d.e. need not give rise to a variational So in this section, we will take a general variational

problem and find out the p.d.e. satisfied hy its solution.

20 Let

n

r.

be a bounded open set with boundary

Let

r = r0 U r 1 where r 0 and r 1 are disjoint. Let (2.20) It is easy to see that V is closed and hence a Hilbert space with _ _ _ __.A

II • II 1

norm.

r.

8

Fig. 2·1 We will use a(u,v) • L{v) • where a 0 > 0, aij a0

i.e.

s~~1on conven~ion hereaf~erwards.

au av

I

{aij{x)

~~+a

I

fv dx +

I

Q

n

uXi uXj

r

0

Le~

uv) dx,

{2.21) {2.22)

gvdr ,

are s.ooth and there exists two constants

and a 1 such that

~he

quadratic form aij{x) ti'J

is uniforaly continuous and

uniformly positive definite. Inequality (2.23) a{•, •)

i~lies

that the bilinear fora

is continuous and V -coercive.

Foraally we have

21

a(u,v)•/ [-"x3 (aij ~u )v +a uv]dx + Ja 1 . au n.vdr 0.

o

j

oXi

r

0

J 3Xf

(2.24)

J

Let 3u 3u .. -"-n., "oVA za lJ oXi J and 3u 3 .. -"- ) • Au = - -"- (a lJ oXi oXj If v

then the equation

£ ~(0)

=

a(u,v)

L(v)

(2.25)

becomes Au, v >




O, i • 1,2, and f

£

If u is the solution of the

t2(n).

problem a(u,v) • L(v)

for all v

£

H~(O),

and

then show that

al

au 1 -an

• a2

au 2 -an

on S •

r Fig. 2.2

Exercise 2.

Fourier Condition.

Let

V • Hl(g),

a(u,v) • L(v) •

1 vu. vv dx + 1 uvdr, r n 1 fv dx + 1 pdr, n

r

What is the boundary value problem associated with this? the proble11.

Interpret

2.5. ELASTICITY PROBLEM. 0 c lR 3

(a) 3-DIMENSIONAL CASE.

be a bounded, connected open set.

of 0 and let

r

Let

be split into two parts

r0

Let

r be the boundary and

r1 .

n

Let

be occupied by an elastic medium, which we assume to be continuous. Let the elastic material be fixed along body force acting in 0 and along

r 1 • Let

(u.(x)) 1

(g.) 1

r0 • Let (f1) be the

be the pressure load acting

denote the displacement at x.

n Fig. 2·3

In linear elasticity the stress-strain relation is (2,29)

where aij

and

£ij

denote the components of the stress and

strain tensors respectively, The problem is to find on r 1 and

(ui)

=0

a 1j

and

given

in

on r 0 •

The equations of equilibrium are (2. 30a)

25

(2.30b) ui • 0 on r0 . .

(2.30c)

We have used the summation convention in the above equations. We choose (2.31) a(u,v)

=I0

L(v) • Using (2.29),

a(u,v)

a(u,v) •

I

0

Ir

ai.(u) e .. (v)dx, J

giv.dr 1

1

+

I

o

fiv.dx.

(2.33)

1

can be written as

(A div u.div v + 2pe .. (u)e.j(v))dx, 1J

from which it is clear that a(•,•) is V -elliptic

(2.32)

1J

. is a nontrivial

refer to CIARLET [9].

1

is symmetric.

That a(•,•)

statement and the reader can

Formal application of Green's formula will

show that the boundary value problem corresponding to the variational problem (2.31) - (2.33) is (2.30). a(•,•) L(•)

can be interpreted as the interaal work and

as the work of the external loads.

a(u,v) • L(v)

Thus, the equation

for all v e V

is a reformulation of the theorem of virtual work.

26

(b) PLATE PROBLEM.

Let

2n be the thickness of the plate.

allowing

n

problem.

It will be a two dimensional problem.

+

0 in (a) we obtain the equations for the plate

We have to find the bending moments Mij displacement

By

(u 1).

and

These two satisfy the equations; (2.34)

(2.35) u •

o on r,

(2.36)

and au an

=0

if the plate is clamped,

(2.37)

M1 jninj • 0 if the plate is simply suppo,ted

(2.37a)

We take

v. {

H2(n) • {v o H2(n)()

£

H2: v • __ av an

H~(n),

=o

on r}, if the plate is clamped;

(2.38)

if the plate is simply supported

Formally, using Green's formula we obtain

• nI M.1 j

a2v a a dx xi xj

-Ir Mijnj

av aMij --a dr+ fa vn1dr xi r xj

for all v

£

V.

(2.39)

27 But

3M ..

Ir ~ vn. oXj 1

dr

=0

for all v

£

V,

since v • 0 on r,

and

I f

where

M•• n.

= If

av dr -;--

lJ J aXi

av Mijn. (ni -;J

an

+

av )dr, si -;aS

3v/3n denotes the normal derivative of v and

denotes the tangential derivative. av J M•• n. -;dr • r lJ J aXi

av/3s

By (2.37) and (2.37a) we have

o.

Hence a2M..

lJ f f VdX • OI aXiaXj O ~

~

VdX •

I

~2 v

u

r1, •

~

a ~

O lJ aXiaXj

dX f or a 11

V £

V,

We therefore choose (2 .40)

and L(v) = a(•,•)

f

fv dx.

0

can be proved to be V-coercive if and

REGULARITY TIIEOREM 4.

a >

the solution u of the problem

u

=o

in 0, on

8

~

0

o.

When 0 is smooth cmd f

-6u • f

(2.41)

r,

£

L2 (0) ,

then

28

belongs to H2(n).

have

Mozeeoveze~ '-'6

whezee C is a constant. This proves the coerciveness of a(u,v) S

=0

and a

>

0.

The motion of an incompressible, viscous fluid

2.6. STOKES PROBLEM. in a region

n

is governed by the equations

-Au

n, n,

(2.42)

o in r;

(2.44)

vp • f

+

in

div u • 0 in u

where u

above for

= (u 1).t= 1, ••• , n

p denotes the pressure.



(2.43)

denotes the velocity of the fluid and We have to solve for u and p, given f.

We impose the condition (2.43) in the space V itself. That is, we define (2.45) Taking the scalar product on both sides of equation (2.42) with v

£

V and integrating, we obtain

J f. v Sl since

dx

au. av.

~ =J Sl oXj

~ oXj

,

29

and Ivp.v

0

as v

V.

£

= QI

~

~

aXi

avi P ---

I 0

v. • 1

3Xi

+

Ir

pvini • 0

Therefore we define (2.46)

L(v) •

I

n

f.v d11.

(2.47)

We now have the technical lemma.

LE~

5.

The space ~ • {v

£

(~(n))n: div v •

0}

is dense in V.

If

The proof of this Lemma can be found in LADYZHENSKAYA (27].

The equation a(u,v)

= L(v)

for all v e V with

a( , ), L( ), v defined by (2,45) - (2,47) is then equivalent to Au




• 0

for all

+

e,,

(2.47)

denotes the duality bracket between

(f) (n)) n.

+

(!} 1 (0))n and

Notice that (2. 47) is not valid for all

(i) (n)) n

is not contained in 1r.

+e

(~ (n)) n

To prove conversely

that the solution of (2.47) satisfies (2.42), we need

THEOREM 6,

v.L. = {v

The annihilator

,..1. of "' in

: there e:rists a

p

£

(i) 1 (n))n is given by

~I (n) such that

v

= Vp}

II

30

Theorem 2.6 and Equation (2.47) imply that there exists a p

£

~ 1 (n)

such that 6u

+ f

a:

6p.

Since

Therefore Vp

£

(H

-1

(O))

n

.

We now state

THEOREM 7.

If

p

£

~ 1 (n) and Vp

£

(H-l(n))n,

then p

£

L2 (0)

and

fJhere

C

is a constant.

From this Theorem we obtain that p f

£ t2(n)

and

n

£

L2 (n).

Thus, if

is smooth, we have proved that the problem

(2.45) - (2.47) has a solution u

£

V and p

£

L2 (n).

3. CONFORMING FINITE ELEMENT METHODS IN CHAPTER 2 WE dealt with the abstract variational problems and some examples.

In all our examples the function-

space V is infinite dimensional.

Our aim is to approximate V

by means of finite dimensional subspaces Vh and study the problem in Vh.

Solving the variational problem in Vh will

correspond to solving some system of linear equations.

In this

Chapter we will study an error estimate, the construction of Vh and examples of finite elements.

3.1. APPROXIMATE PROBLEM. find

u

The abstract variational problem is: E

V such that

a(u,v) • L(v). for all v

V,

E

(3.1)

where a(•,•), L(•), V are as in Chapter 2. Let Vh be a finite dimensional subspace of V.

Then

the approximate problem corresponding to (3.1) is: find

~ E

a(~,v)

Vh such that

• L(v)

for all v

£

Vh.

(3.2)

By the Lax-Milgram Lema& (Chapter 2, Theorem 2.1), (3.2) has a unique solution. Let dimension (Vh)= N(h) be a basis of Vh.

Let

and let

(wi)i=l, •.• ,N(h)

32

~ where ui,vj e R, 1

N(h)

• I

i•l

~

i,j

~

N(h) uiwi,vh • N(h),

I

j•l

vjwj,

Substituting these in (3.2),

we obtain (3 .3)

Let T

A • (a(wi,wj))i,j'U • (ui)i' V • (vi)i,b •(L(wi))i Then (3.3) can be written as V1 AU • V1 b.

This is true for all V e RN(h) • Hence AU • b.

(3.4)

If the linear system (3.4) is solved, then we know the solution

~

of (3.2).

This approximation method is called the

Rayleigh-Galerkin method. A is positive definite since

A is symmetric if the bilinear form a(•,•)

is symmetric.

From the computational point of view it is desirable to have A as a sparse matrix, i.e.

A has many zero elements.

33

Usually

will be given by an integral and the matrix A

a(·~·)

wUl be sparse if the support of the basis functions is "small". For exuple, if a(u,v)

~

J vu.vv

n

dx,

then a(wi,wj} • 0 if supp win supp wj • •·

Now we will prove a theorem regarding the error committed when the approximate solution

~

is taken instead of the exact

solution u,

ntEOREM 1.

(3,2}

Proof,

If u and

~espectioet.y~

~

dtmote the so Z.utions of ( 3, lJ and

then we have

We have a(u,v} • L(v}

for all v

t

V,

so a(u -

~,v)

• 0 for all v

By the V ~coerciness of a(•,•)

II u

- t;.

112

~ 1/m a(u - t;.•u ..

£

vh

(3 .5)

we obtain

tJbl

• 1/m a(u-

~,u-v+v-~),for

• 1/m a(u-

~,u-v),

~

M/m

II u

-

~~~

by

II u -vII

all v

(3.5)

£

Vh

34

This proves the theorem with C •

.

M/~

3. 2. INTERNAL APPROXIMATION OF HI (0).

nc=. ~2

Let

Let Th be a triangulation of

polygonal domain.

n:

be a that is Th

a finite collection of triangles such that

n

u

= K

Let that

P(K)C

E

K and Kn KI

for K, KI

£

Th, K ~ KI



P(K)

be a function space defined on K such

nl (K). Usually we take P(K) to be the space of We have

If

vh bJhel'e



Th

polynomials of some degree.

THEOREM 2.



= {vh

t:

P(K)C.Hl(K),

c•(n): vhiK

£

P(K),K

t:

Th}

then VhC:H 1 (n).

Proof. Let u t: Vh and vi be a function defined on 0 such that vi IK = a!. (u IK) • This makes sense since u IK £ H1 (K) • Moreover 1

L2 (n),

vi

E

vi

= ax. au

since v 1 1K • a!i (uiKl

in ~

I (

E

0) •

1

For any

+ £ f) (0),

we have

L2 (K).

We will show that

35

• where n.IC is the 1

1.th

component of t he outward drawn normal to

arc. so (3.6) The second term on the right hand side of (3.6) is zero since u is continuous in 0 and if IC

adjacent triangles then ni 1

ni

a+

u.-r--dx•
..>. ..

i ,j=l

EXAMPLE 3. (Finite Element of Degree 3).

Thus dim

at

can be written in the form

p(ai)>.i(2>.i- 1)

PK = P3 (K) =Span

1/4

>. 1>. 2

a 12 and zero at the other nodes.



p

rK

>.1( 2AI - 1)

a 1, a 2 , a 13 , a 23 , a 3 and takes the value

vanishes at a 12 .

=0

The

at other nodes.

The equations of the lines >. 2

a 13 a 12 are

a 3 a 2 and

denote the vertices of K and

43

Fig. 3.5

It is easy to see that p.1

= 1/2

- 1)

~.(3~. 1

1

(3~i-

2),

= 9/2 ).. ~. (3).. - 1), 1

1 ~ i,j

~

Moreover, is

3,

J

1

is a basis of P3 (K).

P.1 is

1

at the node a.1

and zero at the other nodes;

at the node a ... and vanishes at the other nodes; llJ P123 is zero at all no4es except a 123 where its value is 1, REMARK 3.

1

In the above three examples

rK contains only Dirac

masses and not derivatives of Dirac masses.

All the above three

finite elements are called LagPange finite etements. Let n be a polygonal domain and let Th be a triangulation of

n,

Polynomials of degree for each

K e:: Th.

i.e. ~

t.

n= Let

l) Y. K e:: Th

Let PK

(K,PK,rK)

= P1 (K)

consists of

be a finite element

44

FroM the definition of finite eleaent it follows that a function in V is uniquely determined by the distributions in h

REMARK 4. Eh

In the example

= {~ a. : 1

a.1 is a vertex of a triangle in the triangulation}.

we proved in Sec. 3.2 that VhC: Hl(n).

In this case we say that

the finite element is confoming.

REMARK nl(n).

s.

The Vh so constructed above need not be contained in

If vh~Hl(n)

we say that the finite element aethod is

non-oonfoming. Let

K be a triangle, E

K

Then

(K,P 1 ,EIC)

={~

aij

PK

= P1 (K)

and

1 ~ i < j ~ 3} •

will be a finite element, but the space

vh¢ H1 (n).

Fig.3·6

,

45

(K,PK,EK)

When

is as in Example 2, we will prove

c• (0). This together with theorem

that vhc:

3.2 implies

Thus the finite element in Example 2 is conforming.

vhc: H1 (0).

To prove that Vhc:

c•

(0), let K1 and

K2 be two

adjacent triangJes in the triangulation. Oz

Fig. 3·7

A polynomial of degree 2 in x and

y

when restricted

to a line in the plane is a polynomial of degree 2 in a single variable and hence can be determined on the line if the value of the polynomial at three distinct points on the line are known. Let vh vhiK

1

£

Vh.

Let

and vhiK

2

v1 to

and Kl

v2 and

be the continuous extension of

K2 respectively;

vl

v2

and

are polynomials of degree 2 in one variable along the common side; vl

and

v2

agree at the two common vertices and at the midpoint

of the common side. shows vh

Exercise 1. vhc:H 1 (n).

Hence

is continuous.

Taking

v1 = v2

on the common side.

Hence Vh c:.

(K,PK,EK)

This

c• (n).

as in Example 3, show that

46 3.4. INTERNAL APPROXIMATION OF

H2(n).

In this section we give

an example of a finite element which is such that the associated space Vh

is contained in H2(n).

This finite element can be

used to solve some fourth order problems.

We need

THEOREM 3.

is contained in cl(n)

then vh is contained in H2(n).

The proof of this theorem is similar to that of theorem 2 of this section.

EXAMPLE 4.

Let K be a triangle PIC • P3 (JC) • Span {l,x,y,x2,xy,y2,x3,x2y,xy2,y3}, tiC •

{6ai' !x g 6ai' ~ gy 6ai , 6a123 , 1 -< i -< 3}

where ai are vertices of IC, a 123 is the centroid of IC and dim PIC • Card

tiC •

10.

Fig. 3·8

The arrows in the figure denote that the values of the derivatives at the vertices are given.

Using the formula

47 3

p • i!l (-2Al+ 3A~- 7A 1A2A3)p(ai)+27A 1A2A3p(a 123) +

where

We obtain that p : 0 if

Hence

(K,PK,IK)

is a finite element.

The corresponding Vh is in C1 {ii).

This shows that Vh ¢ H2(n).

c• (0)

but not in

Hence this is .not a

conforming finite element for fourth order problems. We now give an example of a finite element with

EXAMPLE S.

nfE ARGYRIS TRIANGLE.

degrees of freedom.

The Argyris triangle has 21

Here the values of the polynomial, its first

and second derivatives are specified at the vertices; the normal derivative is given at the mid points. In the figure we denote the derivatives by circles and normal derivative by a straight lines.

48

F"eg. 3 .g We take Pr • P5 • Space of polyno.ials of degree less than or equal to

s.

dia Pr • 21;

{aaf ,

t. • A

a -r-

a2

•X

aai ,

a

~

•r

a2

1'i""1Y aai' 1?' aai'l

a2 aai , -ax£ aai

~ i ~ 3,

ana aaij

where ai denote the vertices of K, aij line joining a1 and aj, Let p prove that p

J

£

Pr

0 in

and a/an,

be such that

r.

1 ~ i < j ~ 3},

the midpoint of the

the noraal derivative.

L(p) • o, L c t 1

p is a polynomial of

variable along the side a 1 a 2,

derr~e

By assumption p,

We will

s

in one

and its

first and second derivatives vanish at a 1 and a 2, Hence p • o along a 1 a 2• Consider ap/an along a 1 a 2• By assu.ption ap/an vanishes at a 1 a 2 and a 12 • Since the second derivatives of p vanish at a 1, a 2 we have the first derivatives of ap/an vanish at a 1 and a 2;

ap/an is

polyn•ial of degree 4 in one variable along a 1 a 2, Hence ap/an • 0 along a 1 a 2• Since p • 0 aleng a 1 a 2, ap/&t • 0 along at a2, where a/at denote the tangential derivative, i.e. dertyattve along a 1 a 2• Tberefore we have p and its

49

first derivatives zero along a 1 a 2.

Fig. 3.10

The equation of a 1 a 2 is

A3 • 0.

a line perpendicular to a 1 a 2 as the

Hence we can choose

A3 axis.

Let T denote

the variable along a 1 a 2 • Changing the coordinates from to

(x,y)

we can write the polynomial p as

(y3 ,T)

P

where q.(T) 1

i

5

= .L

1•0

A3 qi(T)

is a polynomial in T of degree


II vII

we obtain, using Brezzi's condition, b(v, A ) S II A£ II ~ Sup II v 1i ~ v £ v

II f II *

+

Ia Ill u£11



This implies (7.7) Fro• (7.6) and (7.7) we obtain

94

Hence there exists a subsequence {t'},

where C is a constant. u

£

V,

~ £

M,

such that u

£

Obviously {u,>.}

t

~

u and >. £ , ~ >.

is a solution of (7.1).

a(u 1 - u2,v)

b(v,>. 1 - >. 2) • 0

+

b(u 1 - u 2,p) • 0 Taking v

= u1 -

all ul

u2 and

-ll •

v l1

>. 1 - >.2'

£

v v £ v, (7. 8)

M.

we obtain

- u2112 ~ a(ul - u2,ul - u2) = 0.

Therefore

Since u 1 • u 2,

using Brezzi's condition , we obtain

from (7.8) that

Hence the solution of (7.1) is unique.

REMARK 1.

We now give an error estiaate for the solution of (7.1)

and the regulariz ed problem (7.4). a(u - u £ ,v)

+

ell>-- >-£11

~

We have

b(v,>. - >. £ )

=0

Yv

£

V.

Hence

I• I· II u- u£11.

From (7.lb) and (7.4b) we obtain

(7.9)

95

=u

Choosing v

- u£, ~

=A-

A£,

we get

Thus, using (7.7) llu-u£i!. • 0 in n ,

ll. • 0 on r , an

"

• 0

(7 .14)

on r .

In the variational form of the biharmonic equation, V = Hl(n)

notice that

~

(7.11) has one solution. H3(n)r\

H~(n)

L2(n) =II.

w~

It is easy to see when (7.10),

If a solution

>. of (7.14) is in

then u defined by (7.13) is in Hl(n).

Moreover,

one can check that this {u,>.}

is a solution of (7.10), (7.11).

7. 2. ntE APPROXIMATE PROBLEM.

Let Vh C V and

~C

M be two

families of finite-dimensional spaces approximating V and M. We shall study the approximate problem: Find

such that (7.1Sa)

99

(7.15b)

Exercise 1. Show that the problem (7.15) leads to solving a linear system with matrix

where A is a m x m positive definite matrix and matrix where m = dim Vh Since

II • IIH · on S(h)

Vh

Vh

and

B is

n

x

m

Mb·

n • dim

is finite-dimensional, the norms

II • II v

and

are equivalent, that is there exists a function

such that (7 .16)

We introduce the affine spaces Zh(')

Z(')

Exercise 2.

Let

= {vh

= {v

£

£

V:

'= 0.

Vh: b(vh,~h) b(v,~)

= (''~h)

= (~,~)

V ~ £

V ~h

£

Mb}

M}

Show that (7.1) is equivalent to the

problem: Find

= Z(O) such a(u,v) = (f,v) V v

u

£

Z

that £

}

Z

In the same way show that (7.15) is equivalent to: Find

(7 .17)

100

'it {

£

Zh

= Zh(O)

such that (7 .18)

a(uh,vh)= (f,vh) The present framework allows us to deal with

Zh~z

and hence we can consider non-conforming approximations of (7.17). We now give an error bound in H -norm.

THEORE~1 2.

Assuming that the aontinuous problem has at Zeast one

solution

{u,~}

Proof.

Let wh

one has the

= vh

-

~

ePPOP

bound

and we have

From (7.la) and (7.15a), we obtain

and hence

Using the H -coercivene ss of a(.,.) of a(.,.)

and

b(.,.)

we obtain

and the continuity

101

Hence

We get the desired result by noticing that

#

REMARK 4.

If

Zh(O)C: Z(O)

then the error estimate (7.19) reduces

to (7.20)

When

~

= 0,

inview of exercise 2.

the error estimate (7.20) is obvious, Then

(7.20)

is the error bound obtained

in the conforming case.

7.3. APPLICATION TO THE STOKES PROBLEM.

In what follows

Th

will

denote a regular family of triangulations of the polygonal domain

n, '\Th

will denote the set of vertices of Th'

side points and

~h

the set of edges.

mh

the set of mid

102

We consider the Stokes problem (example 1) where u is the velocity and A is the pressure. We shall choose for Vh a conforming for

P 2 space and

a piecewise constant space, namely

~

and

We notice that dim Vh

=n

(I

internal vertices

+ I

internal edges)

for the set of degrees of freedom in each component

and choose rh where

M (p) Y

=~ J p IYI y

ds

denoting the average on the edge.

With this choice of rh,

the interpolation operator

defined by

will have some nice properties. on a subset of V since u(N) (H~(n))n;

wh

Note that

wh is defined only

is not defined for all u in

is defined on V n (H2(n))n since the functions

103 in

(H2(n))

n

are continuo us,

K be a triangle .

Let

Exercise 3,

are edges of K.

where a 1 are the vertices of X and yi

M

Y·1

is defined by M (p) = Yf

Show that

rK

~

f

IYtl Y·

s.

p d

1

PK unisolv ent.

is

We have

LEMMA 3.

One has

J div

K

Proof.

(whv) dx

a

J div

K

v dx V v

£

(H2(o))n

Indeed, by Green's formula,

J div

K

(whv) dx

=J

3K

•J

aK

(whv.n) ds v .n ds

since n is constan t on each side of K. Applying again Green's formula we get the desired result. II

104 II u-

ERROR ESTIMATES FOR

~

11 1 •

when o is convex) then, since u

Mb

If u £

Z(O)

£

H2(o)

(which is true

(i.e. div u

= 0)

and

contains functions which are piecewise constant, we have, by ,..hu

Lenna 3,

£

Zh(O).

Hence we obtain

If the solution u

£

H3(o)

(which is unlikely since o

is a polygon) the error bound becomes ch211 u 11 3 • interpolation operator ,..h P 2 (K)

on each element

Chapter

s.

Indeed, the

leaves invariant the polynomial space

K and the above error bound follows from

However, we have

provided that the pressure

A£ Hl(o).

Finally, Theorem 2 gives

due to the low degree of approximation for the

which is only O(h) multiplier.

othe~

Choices Let

fo~

vh

and

Mb·

105

where A1 A2 A3 is called the bubble dimension of Q(K)

is

function.

Note that the

7.

The choice

leads to the error estimates

(See CROUZEIX- RAVIART [14]). The choice

= {vh

£

- 2 :vh IK £ (C •(O))

~ • {JJh

£

c•(n): l1hiK

Vh

in which

.Mb

£

(

P 2 (K) ,K

£

Th' vh

=0

on

30}

.P 1 (K), K £ Th} ,

contains continuous piecewise linear functions, leads

to the same error estimates.

(See BERCOVIER-PIRONNEAU [3]).

This last method, due to TAYLOR-HOOD [42], is widely used by engineers.

7.4. DUAL ERROR ESTIMATES FOR u -

We denote by



Let

the norms in V.1 (i

= 0,2)

and

II • 11 1 the

106 nor. ln M1• We assu.e that Y0

v0

:

v0•

(In practical applications

will be a L2 space) and that H • V (i.e.

a(,,.)

is

V -coercive). Let g

£

Y0 and {w,•}

£

Y x M satisfy

a(v,w) + b(v,•> • (g,v) V v b(w,p) • 0 Y p

£

£

V,

(7.21a) (7.21b)

M.

We assuae (regularity result) that (7 .22)

llwll2+ ll•lll~cll&llo and vh

J.lh

inf £

II w \(+)

vh lly ~ e(h) II w.11 2,

(7.24)

inf II J.l- ph liM!. e(h) II+ 11 1• £

(7.23)

~

A v0- error estimate is given by 111EOREM 4.

Proof.

T.Jndn the abmJs

assunrptiorur~

we

hatle

We know that

II u-~

(g,u-~)

llo •

sup g £ vo

II II

g 0

(7 .25)

From (7.2la) we have (7.26)

107

Moreover, we have

and (7.28) Fro. (7.26), (7.2lb), (7.27) and (7.28) we obtain

Using the continuity of a(.,.) and b(., •)

we get

Taking the infimum over vh & ~(+)

and Ph,vh

&~

and using (7.23), (7.24), we obtain (g,u-~) ~

c[ll w 11 2 cll u-~ llv+

inf

II

>.-ph liM>

+

~&~

+II u-'it llv II -; 11 1]

e(h)

Finally, using the regularity result (7.22) and (7.25) we get the desired result. #

APPLICATION TO STOKES PROBLEM.

We choose

108

Prom the error estimates of section 3, we have

and

The regularity result (7.22) is nothing but the regularity result for the Stokes problem. Hence applying Theorem 4, we obtain L2 -error estimate.

7. 5. NONCONFORMING FINITE ELEMENT METIIOD FOR DIRICHLET PROBLEM

We recall that the variational formulation of the Dirichlet problem - 6u • f

u. 0

in

n }

on an

(7.29)

is:

Find u e H!(n)

such that

a(u,v) • (f,v)

Y v e H~(n),

where a(u,v) •

(f,v) •

I vu.vv • I

n

I

n

K e Th

fv ,

Ivu.vv ,

K

(7.30)

109 is a triangulation of

and Th

~.

We like to consider the nonconforming finite element approximation of (7o30), namely Find

~ £

such that

zh

(7 31) 0

where

across the rnidside points of internal vh

=o

on

an}.

Zh ¢

Notice that

edges~

H~ (O)

o

We will construct a mixed finite element which is Multiplying the first equation in (7o29)

equivalent to (7o3l)o by v

£

n Hl(K) and integrating we obtain, using integration by K

parts in each triangle

If

K K

K,

vu oVV

-

If

K aiC

(VU on) v

=f fv 0



This suggests

where

lJ

a(u,v)

=~

b(v,p)

= - I J (p.n)

(7. 32)

~ VUoVV ,

KK

v,

belongs to some suitable spaoe. We have to construct finite-dimensional subspaces

and

Mb

(7.33)

such that the problem

Vh

110

(7.34)

is equivalent to (7.31).

Here a(•,•), b(•,•)

are as in (7.32)

and (7.33). We take Vh • n P 1 (IC). IC It is easy to see that if

(ph.n)

is constant and

continuous along internal edges, then b(vh,ph) •

0 Y

vh

£

Zh.

Define

by

If ax

+ BY • 1

is the equation of an edge

constant on y for q The

£

y

then q,n is

Q(K), where n is normal to y.

set

I IC

• {(q. n) (aij) : aij

are mid points of the sides of IC}

-unisolvent . Hence

is Q(K)

• {q

£

(L2 )2

:

ql 1

£

Q(IC), K £ Th, q.n is cOntinuous

across the edges of Th} serves our purpose.

111

Exercise 4.

With the above constructed Vh and

(7.31) and (7.34) are equivalent.

Mb

show that

Further show that Zh(O) • Zh.

I

The continuous problem corresponding to (7.34) can be obtained as follows: It is natural to take V

When

~

= n Hl(K). K

is smooth we can write b(v,~) • -

rK aKJ

(~.n)v

• - rK (/ div ~.v

+

~

f

K

~.VV)

Hence we take

Thus the continuous problem is: Find

{u,A}

a(u,v)

+

b(u,~)

=0

E

V x M such that

b(v,A)

where a(u,v)

V~

=

(f,v) V v

£

M,

= r J Vu.Vv K K

.

We have the characterisation:

£

V,

(7.35)

112

LE*A S.

z• Proof.

{v £ V: b(v,p) •

ovp

£ M} • Hl(o). 0

Let v £ f) (0) • Then b(v,p) • -

rK (/K div p.V

• -(/ div p.v +

o

f

+

f

o

K

p.Vv)

p.Vv)

• - < div p,v > - < p, vv > • -
+


• o. is eontinuous on

Sinee b( ·, •) in the

H! (0)

II • II 1

V x M and i) (0)

non topology 1 we obtain

We have to prove the other inelusion. Define

Then vi£ L2(o). Let

+ £ ~ (0) •


is dense in

KK

Let v

£

Z.

113

=Since v

£

Z,

b(v,~)




1

~ £

rK I3K v • nt.

M.

Taking

o = b(v,~) = -(r

ldiv ~.v KK

=- r1

K 3K

• (+,0), we obtain

~

+ lvv.~)

K

since ~

(~.n>v

is smooth

=Therefore,
n

llvllv

(Vtjl)

I

l.l2

dx

II Vtjl II v ~ ~ II ll II o •

! II

ll

II o.

As approximate spaces, we choose

= {v

Vh

V: viK

£

t

Q(K), v.n is

continuous across the sides of Th} • (See Section 5 for the definition of Q(K)).

As

div: Vh

~

Mb,

we see that

equilibrium elements: div uh • •

for



Zh(O)C: Z(O).

(Hence the name

will satisfy equilibrium equations

~

piecewise constant).

We may then apply the error estimate derived in Section 2 and use the improvement given in Remark 4, since Zh(O)C: Z(O). We shall choose vh

where

wh

is the interpolation operator

= J v.n

ds

for each edge

= whv

defined in Section 5. Indeed, we have /Cwh v) .n ds

y

y

y

of Th.

130

Therefore ,

=l J

K aK

= J ~h

(v.n) ~h dr

n

div v dx V ~h

£



Finally we get

where we have used the estimate given in Theorem 6. To get an error estimate for

II ). -

).h

II 0 ,

we shall

make use of the results in Section 6 and construct the operator occuring in Lemma 9. Let v in V be given and

6+ = div v + = o on

in

+

satisfy

n,

an .

Let

be defined by Av =

V+ •

We have (regulari ty result) 11Avll 1 ~clldiv vll 0 , so that

~

131

satisfies

and b(~h,wh

Av) •

b(~h'

Av) •

b(~h'v)

,

where we used the definitions of wh and Av. Thus

~

satisfies the conditions required in Lemma 9.

Hence we have

by Theorem 8. For further details about equilibrium elements the reader can refer the thesis of J.M. THOMAS, 1977. REMARIC 8.

If we replace Q(K)

by

(P 1 (K)) 2 we get a finite

element with 6 degrees of freedom instead of 3 (2 values of v.n on each side).

The interpolation operator wh is defined with the

help of the degrees of freedom and has the same properties.

o2.-------~-------+------__.a3

0 223

°332

Fig. 7·1

132

In fact,

wh

is defined by

/p(whv).nds =f p(v.n)ds y

y

V p

£

P 1 (y)

However, the error estimates now become

II

u - ~

II ). REMARK 9.

).h

II 0 ~

ch 2

II 0 ~

ch

The present finite element method can be extended to

the elasticity equation where v represents the stress tensor aij.

The difficulty lies in the required symmetry of aij See C. JOHNSON-B. MERCIER [25]

but can be surmounted.

and

AMARA-THOMAS [2].

REMARK 10.

APOSTERIORI ERROR ESTIMATE.

Let us consider the

following optimization problem: lnf J(V,JJ)

where J (v, JJ)

v

£

z(+)

JJ

£

M

= 1/2 II

v - VJJII 2

zero and corresponds to v Since v e Z(,),

=u

Clearly the optimal value is



and

JJ

= ).

solution of (7 .54).

we also have

Since J(u,).)

~

J(vh,).)

V vh e Z(+),

we obtain

133

1/211 u 112

~ 1/211 vh

112.

Adding

to both sides, where IJh

£

M,

we g.et

That is,

Suppose that

IJh

is a solution of Dirichlet problem

with a conforming finite element method, then an upper bound for the error in the energy norm is given by

where

is arbitrary.

One can choose vh • uh,

a solution of the present

equilibrium finite element approximation to Dirichlet problem.

7.11. EQUILIBRIUM ELEMENTS FOR THE PLATE PROBLEM.

We recall that

the equations of the plate problem are: Find a .. lJ

o .. ,w lJ •• = >.n w

= b(v.T) b(•.•)

2)

Y v

£ H~(o).

T

£

v.

is continuous over V x M where

so that (7.57) (clamped case)· is equivalent

to:

Find {a.w} a(a.Tl

+

£

V x M such that

b(T.w) • 0 V T

V v

b(a.v) • -(f.v)

£

£

v. (7.59)

M.

We take

The Brezzi condition holds only on Hl(o). 0

is smooth and

then

Tij • v6ij•

Mn (T)



v

t

since if v

137

and b (v, T)

=

f

n

IVv l 2dx ~ a II v II

f ~ c II v 11 1

II T II •

For the proof of existence of solutions of (7.59) and modified error estimates see BREZZI-RAVIART [7]. We choose Vh • {T: T

£

(

P (K)) 4 , K

o

s

t

Th, Mn (T)

is continuous} •

Since

we find that after integration by parts on each of

3K, b(T,V)

involves only the values of v at the vertices of

Th~

b(T,V)

= rIM

K 3K ns

(T) :v s

= r R(T,N) N

v(N)

Notice that only the value of

v

v

T

£

vh •

(7.59b)

at the vertices has to

be taken; therefore, we choose

SO

that, if T

Therefore,

Here

£

Vh'

then

Zh(O)C: Z(O).

{ah,wh}

£

Vh x

~

Hence

is the solution of the approximate problem.

138

a(ah,Th) {

+

r R(Th,N) wh(N) = 0

r R(ah,N)

INTERPOLATION OPERATOR.

vh(N)

= -(f,vh)

V Th

Y vh

£

£

Vh'

Mb

(7 .60)

The interpo lation operato r

is defined by

f

Y

for each edge

y

M (wh,a)ds n

=f

Y

M (a) ds , n

of the triangu lation.

We have the estimate

TIIEOREM 12.

'l'he~e

e:dsts a, constan t

c

such

indepen dent of h

that (7 .61)

and (7 .62)

The proof of this is found in C. JOHNSON [24]. PROPERTIES OF Th.

We have

'

139 Hence (7.63) Therefore

wh maps

into Zh(+).

Z(+)

Equations (7.61) and (7.63) imply that the discrete Brezzi condition is satisfied. We have the error estimate

(See BREZZI-RAVIART (7]). The above method is called Hermann-Johnson method.

MORLEY NONCONFORMING ME'niOD.

Let

vh continuous at the vertices, continuous at the mid side point, vh

=0

at the boundary vertices,

av anh

=0

at the mid point of boundary edges}

The space Wh makes use of the Morley finite element which has 6 degrees of freedom, namely, values at the three vertices and the values of the normal points.

derivat~ves

at the three mid side

140

Fig. 1.2

We consider, for simplicity, the case ~

= 1/2

~

=0

and

so that

= nf Ti.J

a(o,T) Let

L(v)

= r fN N

oiJ" dx.

v(N) ;

that is

L is a linear combination of Dirac masses (concentrated

loads).

Then

ntEORF.M 13.

'l'he prob Zem:

Find ~ e wh

such that (7.64)

is equivalent to (7.60) (f,v)

~hsn

= r fN N

v(N)

in ths stmStJ that ~(N)

and

= Wh(N)

at the

ve~tices

N,

141 Proof.

Let

~

be a solution of (7.64).

Define

We will show that ~

{oh(N), wh}

is the solution of (7.60).

Since

is a solution (7.64), we have (7.65)

Using Green's formula, we obtain

If bi

is one mid side point, then substituting vh

satisfying: vh

=0

3vh an

at the vertices { 1 at bi

=

0 at the other nodes

in the above equation we obtain that Mn(oh) bi

(by using 7.59b). Since

oh

£

r I M (oh) K 3K ns But

This proves that

Vh, 3vh

L fNvh(N) v

N

~

i

is continuous at £

Vh.

equation (7.66) gives

-"lo- ..

95

h

bj, j

vh

£

wh

142

Hence (7.67) Let vh

t

Mb·

Consider vh

t

Vh defined by

Then (7.67) gives

Therefore

This is nothing but the second equation in (7.60) with ah replaced by ah .

Now

I cah> lJ.. =r K K = Kr IK

t ..

lJ

a2~ t ••

axi axj

1J

auh

+ I M (t) -a= rK IK Mn (t) -as K ns n

a~

lf

t

£

vh

by Green's formula. The first term in the right side is zero since and

T £

vh.

The second term equals

-

r R(t,N) uh(N).

N

obtain a(ah,t)

+

r R(t,N)

N

wh(N)

=0

V t

£

Vh ,

~

e Wh

Hence we

143

Thus

{oh, wh}

= oh

we have oh

is a solution of (7.60).

and wh

= wh.

Thus we have proved that (7.64) be the solution of (7.60). ~(N) =

By uniqueness

~

(7.60).

We will show that

wh(N)

~

Let

{oh,wh}

defined by

for each vertex N

(7.68)

is the solution of (7.64). It is easy to see that (7.68) and (7.69) define a unique uh

such that uhiK

this uh

£

P 2(K)

for each

IC

£

Th.

We will prove that

wh.

£

From the first equation in (7.60), we obtain

lf

M (T)

JC aJC n

This implies

and

auh/an

=0

Let ...

vh(N) • vh(N).

a~ = 0

an

3~/3n

V

is continuous at mid side points

at the boundary mid side points. vh

£

Wh.

Hence uh

Then there exists vh e Mh

This proves (7.60)

~

(7.64).

Wh.

such that

Hence the second equation in (7.60) gives

This shows

£

144 Thus the Hermann-Johnson method and the Morley nonconforming method are equivalent, in this particular case where the load is a sum of concentrated loads.

Exercise 7.

Let

K be a triangle.

rK = { ~

ai

,

Let

-!-~ ,1< an aij

I'K = P 2 (K)

i < j

~

and

3}

where a. 's denote the vertices of K and a .. 's denote the 1 lJ

mid points of the sides of K.

Show that

tK

is

PK -unisolvent.

The above finite element is called the Morley finite element.

Fig. 7-3

REMARK 11.

We note that the Morley element has advantage over

Herrmann-Johnson method, since in Morley's method we get a positive definite matrix and we have no constraints.

8, SPECTRAL APPROXIMATION FOR CONFORMING FINITE ELEMENT METHOD 1. THE EIGEN VALUE PROBLEM. Let V and H be Hilbert spaces such that V~H. Let by

We also assume that this imbedding is compact.

II • II 1 denote the norm in V. The norm II • II or II • II 0 and the scalar product

in H is denoted in H is

( •, •) .

We identify H with its dual H'. Let a(•,•) : V ~ V +

be a continuous, symmetric

~

bilinear form which is V -coercive with a as the coercive constant. We shall consider the eigen value problem: Find u

£

V,

1J £

~

such that

V v

a(u,v) • p(u,v)

t

V

(8 .1)

In the following, for an operator T : H + H,

II T II • 2•. THE OPERATOR T If f

£

H then

f

Sup £ H,f

~

0

W•

T : H + V is defined as follows.

The operator

I

we write

Tf is defined to be the unique solution of the

variational equation a(Tf,v) • (f,v) By

v

v

£

v.

Lax-Milgram Lenuaa Tf is well defined for all

As the imbedding

V~

H is compact we obtain that

T,

f

£

H.

considered

146

as an operator fro- H iDto H, a(•,•)

is compact. The symmetry of

implies that T is s,.-etric.

It is easy to see that

(8.1) is equivalent to: Find u

£

A £ a such that

V and

(8. 2)

TU • AU

The

and

~

A in (8.1) and (8.2) have the relation A~

• 1.

From the Spectral Theorem for compact self-adjoint operators we have: Sp(T) other than zero.

is a countable set with no accumulation point Every point in Sp(T)

other than zero is an

eigenvalue of T with finite multiplicity.

3. EXAMPLE.

where

n

The model problem for .(8.1) is

is a smooth bounded open subset of a(u,v) •JVu.VV

n

The compactness of the imbedding nl(n)~ L2(n) Problem (8.1) corresponds to: Find u e H01 (n), ~ e 1R such that

is well known.

147

{

u •

n,

u in

- AU •

(8,3)

o on an

We note that T is the inverse of

4. APPROXIMATE PROBLEM.

of V.

Let

Vh~V

be a finite element subspace

We consider the approximate eigenvalue problem: Pind

~ t

Vh, ph

t



such that (8.4)

Here again, we introduce an operator Th: H ~ H where rhf is the unique solution of

As in the continuous case, we have shows that

Th

is uniformly bounded.

Find uh

t

Vh and

~h

It is obvious that Th

II Th II ~ c/a which

Again (8.4) is equivalent to:

• 1/ph such that

is a self-adjoint, compact operator.

We assume that (8,7)

and (8. 8)

for all smooth f 0

and rf.

~

e(h)

~

Further, we assume

e(h)

and e(h)

~

0

(8.9)

EXAMPLE.

Let

where Th

is a regular family of triangulations of

n.

We have

(cf. Chapter 5)

provided that n is a convex polygon and that

II Tf lls+l

~ell f lls-1,0 •

(8.11)

From GRISVARD [22] this is atleast true for which shows that

= 1~

e(h) • ch~ and e(h) • O(hk+l).

5. CONVERGENCE AND ERROR ESTIMATE FOR THE EIGEN SPACE. (8.7)

s

(8.9) show that Th

~

Assumption

T in norm.

From KATO (26] (Chapter V. Section 4.3) we know that the spectrum of Th converges to the spectrum of T in the following sense:

For all non-zero

m and for each h such that d •

A'

e(h)

A £ Sp(T)
= 0 V v

•- b(v,l) V v £ C £

C,

which is equivalent to (9.1).

Thus we proved the equivalence of (9.1) and (9.12) (9.13). A natural approximation Ch

where

~b

approximates

M.

to C will be

In this case

ch 4 c.

164

EXAMPLE 1. NONLINEAR DIRICHLET PROBLEM.

=! I

J(v)

P

n

lv viP dx -

I

n

fv dx,

where f £ Lq, 1/p + 1/q

For

1


o 0

=~ I

n

lvvl 2 dx -

I

n

a.e.

on n} ,

fv dx.

Existence and uniqueness of the solution of the minimization problem are straightforward . Let Vh be the standard Lagrange finite element space of degree 1 and

ch

=en

it seems that one gets

vh.

One has (9.10) with

y = 2;

therefore

165

since the interpolate

wh u

Ch

t

as long as u

£

c.

However, one

has

Hence

Therefore

II u

II 1 =

- ~

O(h). II

EXAMPLE 3. ELASTO - PLASTIC TORSION. C

Vh

= {v

t

same as in Example

H~(O): 2

lvvl

and Ch

J

~ 1

is in C.

One gets

O(h 1/2 -

£

a.e. on 0},

CA Vh.

=

In this case the interpolate u

and V as in Example 2,

whu

is not in Ch whereas

). #

EXAMPLE 4.

THE FLOW OF A BINGHAM FLUID IN A CYLINDRICAL PIPE.

is a particular case of Exercise 2 with J,V as above and ~(v)

2. GENERALIZATION.

= f lvvl n

Note that J'

dx.

V + V'

satisfies

This

166

(J'(u) - J'(v),u- v)

0 V u,v e V.

~

An operator A: V + V' (Au - Av, u - v)

~

is said to be monotone if 0 V u, v e V.

(9.15)

A is bounded if A maps bounded sets of V into bounded sets of V'. A is hemi-continuous if lim

(A(u

+

A+ 0

A is

Aw),v)

coe~ive

(j{!)l()

+ •

= (A(u),v)

V u,v,w

£

V.

(9.16)

if if

II v II

+ •

(9.17)

for v e C.

We have 'IHEOREM 2.

If A is a

coe~ve ope~tor

monotone~

then the

bounded hemicontinuous and

p~blem:

Find u e C such that

(A(u), v- u) > 0 V v e C

(9 .18)

has at least one so Zution. For a proof of this Theorem see LIONS [28]. (9.18) has at most one solution if A is there exists a,y

all u

>

st~ngZy

The problem

monotone, i.e.

0 such that

- v IIY ~ (A(u) - A(v),u - v) V u,v

£

C

The error analysis can be carried out in the same way.

(9 .19)

167 Let T: C ~ C be a mapping, where C

3. CONTRACTIVE OPERATORS.

The scalar product

is a closed, convex subset of a Hilbert space H. in H is denoted by

(•,•).

We call T

aon~tive

iff

II Tx - Ty II ~ II x - y II , V x, y T is st.riatly

aont~tive

£

(9. 20)

C.

0 < e < 1

iff there exists a e with

such that II Tx - Tyll ~ e llx - Yll

v

x, y

£

c.

(9.21)

We say that T is fin'flly aontmative iff (cf BROWDER-PETRYSHN [8]) II Tx - Ty

II 2

~ (Tx - Ty, x - y) V x,y

£

(9. 22)

C

T is quasi firmly aontmative iff there exists a e, 0


0. #

176

EXAMPLE 5.

.Let

A: V-+ V'

satisfy (9.15) - (9.17) with

v ~H

~v·

dense

Then the restriction of A to D(A) = {v

£

V: Av

£

H}

is a maximal monotone operator.

Exercise 5.

Use Theorem 6 to prove that the operator defined in

Example 5 is monotone. Ne have

LEMMA 8.

If A is maximal monotone then T = (I + >.A)-l

is firmly contractive.

Proof.

Let (I + >.. A)x = (I + >..A)y.

Then

= -(x-y)

>..(A(x) - A(y)) Therefore

-II x Hence

x

= y.

y

11 2 = >..(A(x) -

This proves

(I

+

A(y),

x - y) ~ 0.

>..A) is one-one.

177

From Theorem 6, we obtain

R(I + lA)

= H.

Hence

is well defined on H.

(I + AA)-l

Let u.

1

= Tx.1 , X.1

£

H, i

= 1 , 2.

Then

u.1

+

AAu.1 = x1.•

We have to prove that

i.e.

i.e.

i.e.

which is true since A is monotone.

COROLLARY 1.

The atgozoithm (9 .34)

conve~es ~eakly

to a solution of

A(u)

=0

Note that algozoithm (9. 34) can be woitten as

(9. 35)

178 n

n+l

X

and

co~~~sponds

- X

A

to an implicit

= (I

Since T

+ AA)

(9.36)

sch~m~ fo~

= 0.

au + A(u) at Proof.

=O

+ A(xn+l)

-1

(9.37)

is firmly contractive, algorithm

(9.34) converges weakly to a fixed point of T which is a solution of (9.35).

REMARK 2.

, Algorithm (9.34) is called a

· Note t hat comput1ng

xn+l

p~imal

point

a~o~thm.

· ht be as diffi cu 1t as at each step m1g

the original problem except in some special cases.

RF.MARK 3.

If A: V ~ V'

hetter to choose H = V.

where V is a Hilbert space, then it is Let

J: V'

Then one has to replace A by JA.

~

V be the Riesz isometry.

Then algorithm (9.34) is an

implicit scheme for au at

+ JA(u)

= 0.

5. APPLICATION TO PROBLEMS WITH CONSTRAINT.

We want to solve the

problem (A(u),v - u) If u have

~ 0

V

v £C.

is a solution of (9.38) then for any

(9.38) A > 0 we

179

(u which fmplies

~A(u)

u

- u, v - u) < 0 V v

= PcSu,

£

where Su

= u-

~A(u).

Conversely if u is a fixed point of PeS, of (9.38).

C

then u is a solution

We like to solve (9.38) via the algorithm n+l x

= Pc Sxn = Pc (xn

Note that if J

-

~A(x

n

)) •

is a convex,

differentiable function and A = J'

(9. 39)

t.s.c., Gateaux

then

(9.38) is the gradient

algorithm with projection for solving v

Infc J(v). £

We will now give some conditions on A and

~

which

will ensure the convergence of the algorithm (9.39) to a solution of (9.38).

ntEOREM 9.

If

is strongly monotOM, i.e.

A

(A(u)-A(v),u-v)

>

allu-v

11 2

V u, v

£

C

(9. 40)

and Lipshitzidn1

II A(u) then the for all

- A(v)

a~orithm

0