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Proposition 9.5.4
Let Assumption
9.5.1
hold
,
and assume the finite element space
V
N
is given by
(8.19).
Then
,
the matrix
A
SV
N
×
N
∈ R
is given by
s
S
q
ii
(x
i
)
M
q
jj
(x
k
)
d
1
2
A
SV
:=
⊗
i
=
k
=
i
1
s
B
q
ij
(x
i
)
M
q
ij
(x
k
)
d
⊗
B
q
ij
(x
j
)
d
x
j
q
ij
(x
j
)
⊗
1
2
d
−
+
M
i,j
=
1
k/
∈{
i,j
}
j
=
i
s
B
M
q
ii
(x
k
)
s
B
q
ii
(x
j
)
M
q
ij
(x
k
)
d
d
1
2
1
2
d
x
i
q
ii
(x
i
)
d
+
⊗
+
⊗
k
=
i
k
=
j
i
=
1
i,j
=
1
j
=
i
s
B
μ
i
(x
i
)
M
μ
i
(x
k
)
d
M
c
k
(x
k
)
,
−
⊗
−
i
=
1
k
=
i
k
with weights
q
ij
(x
k
)
if k
=
i,
q
ij
(x
k
)
:=
d
x
k
q
ij
(x
k
)
d
if k
=
i.
Proof
The proof follows the lines which led to Proposition 8.4.2. Since the co-
efficients
q
ij
(x)
=
k
q
ij
(x
k
)
are not constant, we obtain additional terms due
to integration by parts, i.e.
b
k
a
k
=−
b
k
q
ij
(x
k
)b
i
k
b
i
k
d
x
k
a
k
(q
ik
(x
k
)b
i
k
)
b
i
k
d
x
k
=
−
b
k
a
k
q
ik
(x
k
)b
i
k
b
i
k
d
x
k
−
b
k
d
x
k
q
ik
(x
k
)
i
k
,i
k
d
B
q
ik
(x
k
)
a
k
(q
ik
)
(x
k
)b
i
k
b
i
k
d
x
k
=−
−
M
.
i
k
,i
k
Note that the matrices appearing in the representation of
A
SV
have to be imple-
mented as discussed in Sect. 3.4, since the coefficients are not constant.
κ
Example 9.5.5
Consider the transformed Heston model with operator
−
A
+
r
in
1
1
(
9.21
) and coefficients
q
11
(x
2
)
4
x
2
,
q
12
(x
2
)
q
21
(x
2
)
2
βρx
2
,
q
22
(x
2
)
β
2
,
=
=
=
=
1
1
β
2
)(
2
x
−
1
2
μ
1
(x
2
)
4
βκρ)x
2
+
r
,
μ
2
(x
2
)
2
(
2
β
2
κ
=
8
(
−
1
+
=
−
α)x
2
+
(
4
αm
−
)
and
1
2
κ(β
2
κ
α)x
2
+
c
2
(x
2
)
=
−
2
ακm
−
r
(the coefficients depending on
x
1
are equal
to 1). By Proposition
9.5.4
, we find
⊗
B
2
βρx
2
M
2
βρ
−
1
8
S
1
1
2
M
1
1
2
B
1
1
2
B
1
M
x
2
S
β
2
B
2
βρx
2
A
κ
=
⊗
+
⊗
−
+
⊗
4
αm
−
β
2
2
x
2
1
2
B
1
1
2
(
2
β
2
κ
−
α)x
2
+
M
2
βρ
M
8
(
−
1
+
4
βκρ)x
2
+
r
B
1
M
1
+
⊗
−
⊗
−
⊗
B
M
2
κ(β
2
κ
−
α)x
2
+
2
ακm
−
r
.
The above expression can be simplified to
M
1
−
⊗
1
8
S
1
M
x
2
A
κ
B
1
M
1
=
⊗
+
⊗
Y
1
+
⊗
Y
2
,
(9.32)
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