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Now,
u
m
+
1
N
)
w
m
+
1
v
w
m
+
1
≥
(
u
m
+
1
N
)
w
m
+
1
(
v
−
=
−
,
0
=
0
hence,
(
v
−
u
m
+
1
N
=
(
v
−
u
m
+
1
N
)
(
M
(
u
m
+
1
N
−
u
N
)
+
k
A
BS
u
m
+
1
)
w
m
+
1
−
k
f
)
≥
0
N
N
≥
∈ R
for all
v
0
, which is the second line of (
5.9
).
N
≥
'
⇒
': From the second line of (
5.9
), we have
∀
v
∈ R
0
v
w
m
+
1
≥
(
u
m
+
1
N
)
w
m
+
1
.
Now suppose
(
w
m
+
1
)
k
<
0forsome
k
∈{
1. Then the
left hand side becomes arbitrarily small, which is a contradiction. Hence
w
m
+
1
1
,...,N
}
, and let
(
v
)
k
≥
0
,
which is the inequality
B
u
m
+
1
N
≥
F
m
.Now
u
m
+
1
N
N
≥
(
u
m
+
1
N
)
w
m
+
1
∈ R
0
⇒
≥
0
.
0
, .e.
(
−
u
m
+
1
N
)
w
m
+
1
Set
in
the
second
line
of
(
5.9
)
v
=
≥
0,
hence
)
B
u
m
+
1
N
(
u
m
+
1
N
0. We conclude
(
u
m
+
1
N
0, which is
(
u
m
+
1
N
)
w
m
+
1
)
w
m
+
1
≤
=
−
F
m
=
0.
u
N
}
M
m
We give a convergence result for the price approximated by FEM. Let
{
=
1
be the solution of (
5.9
) and let
u
m
u(t
m
,x)
be the solution of (
5.4
) at time level
t
m
. The proof of the next convergence result is shown in [121, Theorem 22].
:=
C
0
((
0
,T
H
2
(G))
C
1
,
1
(J
L
2
(G))
.
Then
,
there ex-
Theorem 5.3.2
Assume u
∈
];
∩
;
ists a constant C
=
C(u) >
0
such that the following error bound holds
k
1
/
2
M
u
m
u
N
L
2
(G)
+
u
m
u
N
2
H
1
(G)
max
m
−
1
−
≤
C(k
+
h).
m
=
Thus, as in the European case (compare with Theorem 3.6.5), we obtain first
order convergence in the energy norm
u
M
u
N
H
1
(G)
=
O
−
+
(k
h)
, provided that
u(t, x)
is sufficiently smooth.
5.4 Numerical Solution of Linear Complementarity Problems
In the following, two methods for solving the derived LCPs are described. Both
methods are iterative approaches.
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