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Rem ark B.2.1 In ( B.13b ), u k,m + 1 is needed, as by ( B.11 ) u k, 0 =
V
for general
u 0 K H ; the choice U k (mk) = u k,m in ( B.13b ) would then imply U k (t) V
u 0 /
for
0 <t<k .
For every k> 0, U k is well-defined.
Theorem B.2.2
(i) The mapping T k
:{
u 0 ,f
}→
U k (t) is bounded and Lipschitz-continuous
L 2 (J
; V )
L 2 (J
from
H ×
; V
) , uniformly in k . Fo r a n y
{
u 0 ,f
}∈ H ×
L 2 ( 0 ,T
; V ) , the family
U k } k> 0 is Cauchy in L 2 ( 0 ,T
) and its limit
u L 2 (J, V ) is the unique weak solution of the PVI ( B.9a )-( B.9e ), which
satisfies ( B.8 ) and , moreover ,
{
; V
u C 0 (J,
H
).
(B.14)
(ii) If , moreover , f
S( 0 ,T) ( cf .( B.2 )), then T k
is bounded and Lipschitz-
continuous from ( cf .( B.3 ))
T k : H ×
S( 0 ,T)
I( 0 ,T)
U k } k> 0 is Cauchy in I( 0 ,T) .
(iii) Finally , assuming f
and
{
=
g
+
h with
H 1 (J
; V ),
g
BV (J
; H
),
h
(B.15a)
P u 0 K ,A P u 0 h( 0 ) H ,
(B.15b)
U k } k> 0 is uniformly bounded in I( 0 ,T) , and u
{
=
also
lim k 0 U k satisfies
u I( 0 ,T) and the second line in ( B.6 ).
B.3 Proof of the Existence Result
We prove the existence result by establishing a priori estimates for time-semidiscrete
approximate solutions in ( B.12a ), ( B.12b ). We start by analyzing ( B.12a ), ( B.12b )
for m
=
0, k> 0 and write u 1 for u k, 1 and f 0 in place of f k, 0 in ( B.11 ). We assume
that f
S( 0 ,T) ,i.e.
f
L 1 (J
L 2 (J
; V ),
=
g
+
h, g
; H
), h
and set
k
k
1
k
1
k
g 0 :=
g(τ) d τ, h 0 :=
h(τ ) d τ.
(B.16)
0
0
Then, ( B.12a ), ( B.12b ) reads: find u 1 K
such that for all v K
:
k
u 1 +
k
A
u 1 ,u 1
v
V , V
(G 0 ,u 1
v)
+
H 0 ,u 1
v
V , V
(B.17)
where
kh 0 V .
G 0 :=
u 0 +
kg 0 H
,H 0 :=
(B.18)
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