Civil Engineering Reference
In-Depth Information
Since
c
comes from Friedrichs' inequality and depends only on
, the saddle point
problem (5.2) is stable.
It is easy to construct suitable finite elements for a triangulation
T
h
. Choose
k
≥
1, and set
k
−
1
)
d
={
σ
h
∈
L
2
()
d
X
h
:
=
(
M
;
σ
h
|
T
∈
P
k
−
1
for
T
∈
T
h
}
,
k
0
,
0
={
v
h
∈
H
0
()
;
v
h
|
T
∈
P
k
M
h
:
=
M
for
T
∈
T
h
}
.
Note that only the functions in
M
h
are continuous. Since
∇
M
h
⊂
X
h
,wecan
verify the inf-sup condition in the same way as for the continuous problem.
The saddle point problem with a different pairing is more important for prac-
tical computations. It refers to the space encountered in Problem II.5.14:
={
τ
∈
L
2
()
d
H(
div
,)
:
;
div
τ
∈
L
2
()
}
with the graph norm of the divergence operator,
2
0
2
0
)
1
/
2
.
τ
H(
div
,)
:
=
(
τ
+
div
τ
(
5
.
4
)
We seek
(σ, u)
∈
H(
div
,)
×
L
2
()
such that
(σ, τ )
0
,
+
(
div
τ,u)
0
,
=
0
for all
τ
∈
H(
div
, ),
(
5
.
5
)
(
div
σ, v)
0
,
=−
(f, v)
0
,
for all
v
∈
L
2
().
To apply the general theory, we set
=
H(
div
, ),
=
L
2
(),
X
:
M
:
a(σ, τ)
:
=
(σ, τ )
0
,
,
b(τ, v)
:
=
(
div
τ,v)
0
,
.
Clearly, the linear forms are continuous. Then since div
τ
=
0 for
τ
in the kernel
V
,wehave
2
0
2
0
2
0
2
a(τ, τ)
=
τ
=
τ
+
div
τ
=
τ
H(
div
,)
.
This establishes the ellipticity of
a
on the kernel. Moreover, for given
v
∈
L
2
there
exists
w
∈
C
0
()
with
1
2
v
−
w
0
,
≤
v
0
,
. Set
ξ
:
=
inf
{
x
1
;
x
∈
}
and
x
1
τ
1
(x)
=
w(t, x
2
,...,x
n
)dt,
ξ
τ
i
(x)
=
0
for
i>
1
.
Then obviously div
τ
=
∂τ
1
/∂x
1
=
w
, and the same argument as in the proof of
Friedrichs' inequality gives
τ
≤
c
w
0
. Hence,
0
b(τ, v)
τ
H(
div
,)
≥
(w, v)
0
,
1
+
c)
w
0
,
≥
+
c)
v
,
(
1
2
(
1
and so the inf-sup condition is satisfied.
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