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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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