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{
}
k>
0
by
Proposition B.3.10
Define the step-functions
u
k
(t)
u
k
(t)
J
k,m
:=
u
k,m
,m
=
0
,...,M
−
1
,
(B.55)
and analogously f
k
(t)
.
Assume
0
0
,α>
0.
Then
,
u
k
(t
+
k)
I(
0
,T )
≤
C(α)
w
0
H
+
f
k
(t)
S(
0
,T
+
k)
,
∈
K
and
(
B.4b
)
with λ
=
(B.56)
√
k
C(α)
S(
0
,T
+
k)
.
(B.57)
u
k
(t
+
k)
I(
0
,T )
≤
u
k,
1
H
+
u
k,
1
V
+
f
k
(t)
Proof
We choose 0
=
v
∈
K
in (
B.12b
). Since, for
m>
0,
u
k,m
∈
K
and we have
(
B.29
), (
B.30
) with
w
m
=
u
k,m
,
f
k,m
=
p
m
+
q
m
,(
B.56
) and (
B.57
) follow then
from (
B.53
), (
B.54
).
We also have a bound for
U
k
(t)
in (
B.13a
), (
B.13b
).
Lemma B.3.11
Assume
(
B.4b
)
with λ
.
The semi-discretization
(
B.11
), (
B.12a
), (
B.12b
)
of PVI
(
B.9a
)-(
B.9e
)
is stable in the sense that for T
=
0
and
0
∈
K
=
Mk
C
S(
0
,T
+
2
k)
.
U
k
(t)
I(
0
,T )
≤
u
0
H
+
f(t)
(B.58)
Proof
For
T
=
Mk
,wehave
f
k
(t)
S(
0
,T )
≤
f(t)
S(
0
,T
+
k)
;
U
k
(t)
I(
0
,T )
≤
u
k
(t)
I(
0
,T
+
k)
.
Inserting this into (
B.56
), we get (
B.58
).
Remark B.3.12
Note that we could also obtain an a priori bound for
U
k
(t)
from
(
B.57
): here (
B.25
) with
n
=
1 would imply
C
P
S(
0
,T
+
2
k)
.
U
k
(t)
I(
0
,T )
≤
u
0
H
+
f(t)
(B.59)
We are now able to show our first main result. To state it, we define, for
f
k
(t)
as
in (
B.55
),
f
k
(t)
−
f(t)
S(
0
,T
+
2
k)
+
f(t
+
k)
−
f(t)
S(
0
,T
+
2
k)
√
k
E(k,T )
:=
+
u
k,
2
−
u
k,
1
H
+
u
k,
2
−
u
k,
1
V
+
u
k,
1
−
P
u
k,
0
H
.
(B.60)
Lemma B.3.13
We h ave f o r T>
0,
as k
0,
kU
k
(t)
I(
0
,T )
≤
CE(k,T).
=
T/M
→
(B.61)
Proof
We note that due to its definition (
B.13a
), (
B.13b
),
kU
k
(t)
|
J
k,m
=
u
k,m
+
2
−
=
=
+
u
k,m
+
1
. Selecting
v
u
k,m
+
2
in (
B.12b
) and also
v
u
k,m
+
1
in (
B.12b
) with
m
1
in place of
m
gives the two inequalities:
u
k,m
+
1
−
u
k,m
+
k
A
u
k,m
+
1
−
f
k,m
,u
k,m
+
1
−
u
k,m
+
2
V
∗
,
V
≤
0
,
−
u
k,m
+
2
−
u
k,m
+
1
+
k
A
u
k,m
+
2
−
f
k,m
+
1
,u
k,m
+
1
−
u
k,m
+
2
V
∗
,
V
≤
0
.
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