179 71 8MB
English Pages [200] Year 1979
Table of contents :
Front Matter ....Pages iv
Sobolev Spaces (Bertrand Mercier)....Pages 110
Abstract Variational Problems and Examples (Bertrand Mercier)....Pages 1130
Conforming Finite Element Methods (Bertrand Mercier)....Pages 3151
Computation of the Solution of the Approximate Problem (Bertrand Mercier)....Pages 5268
Review of the Error Estimates for the Finite Element Method (Bertrand Mercier)....Pages 6975
Problems with an Incompressibility Constraint (Bertrand Mercier)....Pages 7691
Mixed Finite Element Methods (Bertrand Mercier)....Pages 92144
Spectral Approximation for Conforming Finite Element Method (Bertrand Mercier)....Pages 145158
Nonlinear Problems (Bertrand Mercier)....Pages 159185
Back Matter ....Pages 186195
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~ · .
~"
SpringerVerlag Berlin Heidelberg GmbH 1979
Author
BERTRAND MERCIER Ecole Polytechnique Centre de Mathematiques Appliquees 91128 Palaiseau (FRANCE)
©
SPRINGERVERLAG BERLIN HEIDELBERG GMBH 1979
Originally published by Tata Institute of Fundamental Research in 1979.
ISBN 9783662391976
ISBN 9783662401972 (eBook)
DOI 10.1007/9783662401972 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))~.
£
Snl)
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
£
:Rnl: 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 LIONSMAGENES [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) 3DIMENSIONAL 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 stressstrain 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 Vcoercive 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
£
(Hl(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 LaxMilgram 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
RayleighGalerkin 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
~,uv+v~),for
• 1/m a(u
~,uv),
~
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.rdx•
..>. ..
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.
3·
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
nonoonfoming. 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) lluu£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 finitedimensional 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 finitedimensional, 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 nonconforming 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 BERCOVIERPIRONNEAU [3]).
This last method, due to TAYLORHOOD [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 finitedimensional 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. JOHNSONB. MERCIER [25]
but can be surmounted.
and
AMARATHOMAS [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 BREZZIRAVIART [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 BREZZIRAVIART (7]). The above method is called HermannJohnson 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 HermannJohnson 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. 73
REMARK 11.
We note that the Morley element has advantage over
HerrmannJohnson 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.
LaxMilgram 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 selfadjoint 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 selfadjoint, 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 lls1,0 •
(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 nonzero
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 hemicontinuous 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 BROWDERPETRYSHN [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
= (xy)
>..(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 oneone.
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),uv)
>
alluv
11 2
V u, v
£
C
(9. 40)
and Lipshitzidn1
II A(u) then the for all
 A(v)
a~orithm
0