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and
u
(t)
f(t),u
+
A
u(t)
−
V
∗
,
V
=
0
.
If
K
⊂
V
is a subspace, (
B.9b
) becomes
u
(t)
f(t),v
+
A
u(t)
−
V
∗
,
V
=
0
∀
v
∈
K
.
Remark B.1.3
In the weak formulation (
B.9a
)-(
B.9e
), we assumed (
B.4b
) with
α>
0,
λ
0. If (
B.4b
) holds only with
λ>
0, the function
Θ(t)
in the weak formu-
lation (
B.9a
)-(
B.9e
) has to be replaced by
=
1
2
e
−
λt
(u(t)
−
v(t))
2
H
Θ
λ
(t)
:=
t
e
−
λτ
(v
(τ )
+
+
A
u(τ )
+
λu(τ )
−
λv(τ )
−
f (τ )),
0
e
−
λτ
(u(τ )
−
v(τ))
V
∗
,
V
d
τ
;
the spaces
L
2
(J
;
V
)
and
L
2
(J
;
V
∗
)
must be replaced by spaces with weight
exp
(
−
λτ )
or, if
T
,by
L
loc
(
0
,
∞;
V
)
, etc. Then, (
B.9a
)-(
B.9e
) and all what
follows will apply also to the case when (
B.4b
) holds only with
λ>
0.
=∞
B.2 Existence
To show the existence of solutions to the PVI (
B.9a
)-(
B.9e
), we semidiscretize (
B.6
)
in time as follows: given
k>
0 with
k
=
∞
T/M
if
T<
, we define
J
km
:= [
mk, (m
+
1
)k
]
,
(B.10)
1
k
u
k,
0
=
P
u
0
,
km
:=
f(τ)
d
τ
(B.11)
J
km
and replace (
B.6
), (
B.9a
)-(
B.9e
) by the sequence of elliptic variational inequalities:
for
m
=
0
,
1
,
2
,...
, find
u
k,m
+
1
∈
K
(B.12a)
such that
u
k,m
+
1
−
u
k,m
+
k
A
u
k,m
+
1
−
f
k,m
,u
k,m
+
1
−
v
V
∗
,
V
≤
0
(B.12b)
for all
v
.
By (
B.4a
), (
B.4b
) with
λ
∈
K
=
0 and by Theorem A.8, the EVI (
B.12b
) admits a
M
m
unique solution
u
k,m
+
1
; hence
{
u
k,m
}
(
M
=∞
if
T
=∞
) is well-defined.
=
0
u
k,m
}
m
=
0
C
0
(J,
With
{
we associate a function
U
k
(t)
∈
V
)
by
U
k
(t)
J
k,m
is linear on
J
k,m
,m
=
0
,
1
,
2
,...
(B.13a)
=
U
k
(mk)
u
k,m
+
1
.
(B.13b)
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