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as in Sect. 3.3 with uniform mesh width
h
and uniform time steps
k
, we obtain the
matrix problem:
Find
u
m
+
1
N
∈ R
such that for
m
=
0
,...,M
−
1
kθ
A
BS
)
u
m
+
1
θ)
A
BS
)
u
m
,
(
M
+
=
(
M
−
k(
1
−
(4.15)
u
0
=
u
0
,
and
A
BS
ij
a
BS
(b
j
,b
i
)
.Let
M
be given as in (2.25).
=
=
where
M
ij
(b
j
,b
i
)
L
2
(G)
Using (3.22), we can compute
A
BS
σ
2
/
2
S
(σ
2
/
2
=
+
−
r)
B
+
r
M
explicitly with
⎛
⎞
⎛
⎞
2
−
1
01
⎝
⎠
⎝
⎠
.
.
.
.
.
.
.
.
.
.
.
.
1
h
−
1
1
2
−
1
S
=
,
B
=
.
(4.16)
.
.
.
.
.
.
.
.
.
.
.
.
−
1
1
−
12
−
10
4.4.3 Non-smooth Initial Data
As already mentioned, the advantage of finite elements is that we have low smooth-
ness assumptions on the initial data
u
0
, and therefore on the payoff function
g
.
In particular, as shown in Theorem 3.2.2, we have a unique solution for every
u
0
∈
L
2
(G)
. However, according to Theorem 3.6.5, we need
u
0
∈
H
2
(G)
to ob-
(h
2
k
r
)
where
r
−
u
N
L
2
(J
;
L
2
(G))
=
O
+
=
tain the optimal convergence rate
u
1
for
θ
∈[
1
/
2. This is due to the time discretiza-
tion since uniform time steps are used. To recover the optimal convergence rate for
u
0
∈
0
,
1
]\{
1
/
2
}
and
r
=
2for
θ
=
H
s
(G)
,0
<s<
2 we need to use
graded meshes
in time or space. We assume
for simplicity
T
=
1.
Let
λ
:[
0
,
1
]→[
0
,
1
]
be a grading function which is strictly increasing and sat-
isfies
C
0
(
C
1
((
0
,
1
)),
λ
∈
[
0
,
1
]
)
∩
λ(
0
)
=
0
,
1
)
=
1
.
We define for
M
∈ N
the algebraically graded mesh by the time points,
t
m
=
λ
m
M
,m
=
0
,
1
,...,M.
It can be shown [146, Remark 3.11] that we obtain again the optimal convergence
rate if
λ(t)
=
O
(t
β
)
where
β
depends on
r
and
s
,
β
=
β(r,s)
(algebraic grading).
Example 4.4.1
Consider the payoff functions
s
1 f
s>K,
0 .
−
K
if
s>K,
g
c
(s)
=
g
d
(s)
=
0
else,
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