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and, by the Cauchy-Schwarz inequality,
s∂
s
ϕφ
d
s
≤
ϕ
ρ
s
2
−
2
ρ
φ
2
d
s
R
R
1
2
≤
ϕ
ρ
R
1
−
ρ
φ
L
2
(G)
.
0
0
C
0
1
Thus, for
ϕ,φ
∈
(G)
,
ρ
=
2
, one has
|
a
CEV
ρ
(ϕ, φ)
|≤
C(ρ,σ,r)
ϕ
ρ
φ
ρ
.
Hence, we may extend the bilinear form
a
CEV
ρ
(
·
,
·
)
from
C
0
(G)
to
W
ρ
by continuity
1
∈[
]\{
2
}
for
ρ
0
,
1
. Furthermore, we have
2
ρσ
2
R
1
2
σ
2
1
a
CEV
ρ
s
ρ
∂
s
ϕ
2
s
2
ρ
−
1
∂
s
(ϕ
2
)
d
s
(ϕ, ϕ)
=
L
2
(G)
+
0
2
r
r
R
R
1
s∂
s
(ϕ
2
)
d
s
ϕ
2
d
s.
−
+
0
0
1
Integrating by parts, we get, for 0
≤
ρ
≤
2
,
1
)
R
R
s
2
ρ
−
1
∂
s
(ϕ
2
)
d
s
s
2
ρ
−
2
ϕ
2
d
s
=−
(
2
ρ
−
≥
0
.
0
0
2
0
2
0
1
1
1
2
s∂
s
(ϕ
2
)
d
s
=−
ϕ
2
d
s
, hence we get for 0
Analogously,
≤
ρ
≤
1
2
σ
2
3
2
r
1
2
min
a
CEV
s
ρ
∂
s
ϕ
2
σ
2
,
3
r
≥
L
2
(G)
+
L
2
(G)
≥
{
}
ρ
.
(ϕ, ϕ)
ϕ
ϕ
ρ
By Theorem 3.2.2, we deduce
L
2
(J
Corollary 4.5.2
Problem
(
4.23
)
admits a unique solution V
∈
;
W
ρ
)
∩
H
1
(J
L
2
(G)) for
0
;
≤
ρ<
1
/
2.
ρ<
2
. The case
1
2
The previous result addressed only the case 0
≤
≤
ρ<
1
1
(which includes, for
ρ
2
, the Heston model and the CIR process) requires a mod-
ified variational framework due to the failure of the Hardy inequality (
4.26
)for
ε
=
=
1. Let us develop this framework. We multiply the first equation in (
4.19
)byan
s
2
μ
w
, where
μ
is a parameter to be selected and
w
C
0
∈
(G)
is a test function, and
integrate from
s
=
0to
s
=
R
. We get from (
4.19
)
(∂
t
v,s
2
μ
w)
a
CEV
C
0
+
ρ,μ
(v, w)
=
0
,
∀
w
∈
(G),
(4.27)
where the bilinear form
a
CEV
ρ,μ
(
·
,
·
)
is defined by
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