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v
0
)e
−
η
, where
g(e
x
)
, and where
η(x,
y
2
.
:=
−
v
0
=
=
=
w
(
v
v(
0
,x,
y)
y)
κ/
2
Thus, the transformed time value of the option
w
solves
B
f
κ
∂
t
w
−
A
κ
w
+
rw
=
in
J
× R × R
≥
0
,
(15.15)
w(
0
,x,
y)
=
0in
R × R
≥
0
,
e
−
κ/
2
y
2
(
A
where
f
κ
B
B
κ
H
κ
+
A
H
κ
:=
v
0
−
r
v
0
)
and the operator
A
:=
A
, where
A
is
as in (9.21).
As for the Heston model, we drop “
”, let
G
:= R × R
≥
0
and denote by
(
·
,
·
)
=
G
ϕφ
d
x
d
y
. We associate to
the
L
2
(G)
-inner product, i.e.
(ϕ, φ)
B
κ
−
A
+
r
the
bilinear form
a
κ
(
)
via
a
κ
(ϕ, φ)
:=
(
−
A
·
,
·
κ
+
r)ϕ,φ
,ϕ,φ
∈
C
0
(G).
B
We find, since
1,
a
κ
(ϕ, φ)
=
a
κ
(ϕ, φ)
+
λ
0
κ
(∂
x
ϕ,φ)
+
λ
0
(ϕ, φ)
−
λ
0
ν
0
(
d
ζ)
=
R
ϕ(x
+
ζ,y)ν
0
(
d
ζ),φ
,
(15.16)
R
(G)
·
V
, where the
C
0
where the bilinear form
a
κ
(
·
,
·
)
is given in (9.22). Let
V
:=
closure is taken with respect to norm
·
V
defined in (9.24).
Theorem 15.3.1
Assume that
0
<κ<α/β
2
and that
4
αm/β
2
|
>ρ
2
.
1
−
2
|
−
1
Then
,
there exist constants C
i
>
0,
i
=
1
,
2
,
3,
such that for all ϕ,φ
∈
V there holds
a
κ
(ϕ, φ)
|
|≤
C
1
V
V
,
ϕ
φ
a
κ
(ϕ, ϕ)
2
2
≥
C
2
ϕ
V
−
C
3
ϕ
L
2
(G)
.
Proof
The continuity of
a
κ
(
)
follows from the continuity of
a
κ
(
·
,
·
·
,
·
)
,bytheHardy
y
−
1
ϕ
inequality
|
λ
0
κ
(∂
x
ϕ,φ)
|≤|
λ
0
||κ|
y∂
x
ϕ
L
2
(G)
L
2
(G)
≤
C
y∂
x
ϕ
L
2
(G)
×
R
∂
y
ϕ
L
2
(G)
, and Young's inequality
ϕ(x
+
ζ,y)ν
0
(
d
ζ)
L
2
(G)
≤
ϕ
L
2
(G)
. Fur-
thermore,
a
κ
(ϕ, ϕ)
2
2
2
≥
c
1
y∂
x
ϕ
L
2
(G)
+
c
2
∂
y
ϕ
L
2
(G)
+
c
3
yϕ
L
2
(G)
+
c
4
ϕ
L
2
(G)
,
with
c
1
,
c
2
,
c
3
as in the proof of Theorem 9.3.1, and
c
4
=
r
−
2
λ
0
−
2
καm
.Now
deduce as in the proof of Theorem 9.3.1.
Consequently, the weak formulation to the (transformed) Bates model (
15.15
)
L
2
(J
H
1
(J
L
2
(G))
such that
∈
;
∩
;
Find
w
V)
a
κ
(w, v)
f
κ
,v
(15.17)
(∂
t
w,v)
+
=
V
∗
,V
,
∀
v
∈
V,
a.e. in
J,
w(
0
)
=
0
admits a unique solution for every
f
κ
∈
V
∗
.
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