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;
H
ρ
(G))
L
2
(J
H
1
(J
L
2
(G))
such that
Find
u
R
∈
∩
;
a
J
R
(u
R
,v)
∈
H
ρ
(G),
a.e. in
J,
(∂
t
u
R
,v)
+
=
0
,
∀
v
(10.23)
u
R
(
0
)
=
u
0
|
G
.
By Theorem
10.4.3
and the inclusion
H
ρ
(G)
⊂
H
ρ
(
R
)
, we conclude that
a
R
(
·
,
·
)
satisfies a Gårding inequality and is continuous. Therefore, the problem (
10.23
)is
well-posed by Theorem 3.2.2, i.e. there exists a unique solution
u
R
of (
10.23
).
10.6 Discretization
=
Since the diffusion part has been already discussed in Sect. 4.4,weset
σ
0 and
only consider pure jump models. The main problem is the singularity of the Lévy
measure at
z
0. Therefore, we integrate the jump generator by parts twice, to
obtain for
f
∈
H
2
(G)
with
G
=
(
−
r, r)
.
∞
=
zf
(x))k(z)
d
z
(f (x
+
z)
−
f(x)
−
0
z
=∞
z
zf
(x))k
(
−
1
)
(z
)
=
+
−
−
|
(f (x
z)
f(x)
=
0
=
0
∞
(f
(x
+
z)
−
f
(x))k
(
−
1
)
(z)
d
z
−
0
∞
(f
(x
f
(x))k
(
−
2
)
(z
)
z
=∞
z
=
f
(x
z)k
(
−
2
)
(z)
d
z,
=−
+
z)
−
|
+
+
0
0
=
0
where
k
(
−
i)
(z)
is the
i
th antiderivative of
k
vanishing at
±∞
,i.e.
z
−∞
k
(
−
i
+
1
)
(x)
d
x
if
z<
0
,
k
(
−
i)
(z)
=
−
z
k
(
−
i
+
1
)
(x)
d
x
if
z>
0
.
Therefore, we obtain
J
f )(x)
=
f
(x
+
z)k
(
−
2
)
(z)
d
z.
(
A
(10.24)
R
H
0
(G)
as
Similarly, we can write the bilinear form for
ϕ,φ
∈
a
J
(ϕ, φ)
ϕ
(y)φ
(x)k
(
−
2
)
(y
:=
−
x)
d
y
d
x.
(10.25)
G
G
The second antiderivative
k
(
−
2
)
(z)
may still be singular at
z
=
0, but the singularity
is integrable, i.e.
−∞
|
k
(
−
2
)
(z)
. We again use the finite element and the
finite difference method to discretize the partial integro-differential equation.
|
d
z<
∞
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