Civil Engineering Reference
In-Depth Information
Substituting
λ
div
u
=
p,
(
3
.
30
)
and using the weak version of (3.30), we are led to the following problem:
Find
(u, p)
∈
H
1
()
×
L
2
() such that
for all v
∈
H
1
(),
2
µ(ε(u), ε(v))
0
+
(
div
v, p)
0
=
, v
(
3
.
31
)
1
λ
(p, q)
0
(
div
u, q)
0
−
=
0
for all q
∈
L
2
().
Since the bilinear form
(ε(u), ε(v))
0
is elliptic on
H
1
, (3.31) is very similar to the
Stokes problem; see Ch. III, §6. As we observed there, in the case where
0
=
∂
(more precisely if the two-dimensional measure of
1
vanishes),
pdx
=
0, and
L
2
()
should be replaced by
L
2
()/
R
.
We know from the theory of mixed problems with penalty terms that the
stability of (3.31) is the same as for the problem
2
µ(ε(u), ε(v))
0
+
(
div
v, p)
0
=
, v
,
(
div
u, q)
0
=
0
,
. The situation here is simple, since the quadratic form
(ε(v), ε(v))
0
is
coercive on the entire space
and not just for divergence-free functions. Moreover,
the penalty term is a regular perturbation. Therefore, we can solve (3.31) using
the same elements as for the Stokes problem. Since the inverse of the associated
operator
as
λ
→∞
L
:
H
1
×
L
2
→
(H
1
×
L
2
)
is bounded independently of the parameter
λ
, the finite element solution converges
uniformly in λ
.
To be more specific, consider the discretization
2
µ(ε(u
h
), ε(v))
0
+
(
div
v, p
h
)
0
=
, v
for all
v
∈
X
h
,
(
3
.
31
)
h
(
div
u
h
,q)
0
−
λ
−
1
(p, q
h
)
0
=
0
for all
q
∈
M
h
,
with
X
h
⊂
H
1
(), M
h
⊂
L
2
()
. The commonly used Stokes elements have the
following approximation property. Given
v
∈
H
1
()
and
q
∈
L
2
()
, there exist
P
h
v
∈
X
h
and
Q
h
q
∈
M
h
such that
v
−
P
h
v
1
≤
ch
v
2
,
q
−
Q
h
q
0
≤
ch
q
1
.
(
3
.
32
)
For convenience, we restrict ourselves to pure displacement boundary conditions.
Good approximation is guaranteed by the following regularity result.
If is a
convex polygonal domain, or if has a smooth boundary, then
u
2
+
λ
div
u
1
≤
c
|
f
0
;
(
3
.
33
)
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