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1
8
(y∂
x
ϕ,y∂
x
φ)
+
1
2
β
2
(∂
y
ϕ,∂
y
φ)
+
1
2
ρβ(y∂
x
ϕ,∂
y
φ)
a
κ
(ϕ, φ)
=
1
2
[
ρβ
−
1
8
[
+
2
r
]
(∂
x
ϕ,φ)
+
1
−
4
βκρ
]
(y∂
x
ϕ,yφ)
1
2
[
1
2
[
2
β
2
κ
β
2
(y
−
1
∂
y
ϕ,φ)
−
−
α
]
(y∂
y
ϕ,φ)
−
4
αm
−
]
1
2
κ
β
2
κ
−
[
−
α
]
(yϕ, yφ)
−[
2
καm
−
r
]
(ϕ, φ)
9
=:
b
k
(ϕ, φ).
(9.22)
k
=
1
Define the weighted Sobolev space
(G)
·
V
,
C
0
V
:=
(9.23)
where the closure is taken with respect to the norm
1
2
2
2
2
y
2
v
v
V
:=
y∂
x
v
L
2
(G)
+
∂
y
v
L
2
(G)
+
+
L
2
(G)
.
(9.24)
Theorem 9.3.1
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 one has
a
κ
(ϕ, φ)
|
|≤
C
1
ϕ
V
φ
V
,
a
κ
(ϕ, ϕ)
2
2
≥
C
2
ϕ
V
−
C
3
ϕ
L
2
(G)
.
C
0
(G)
. We first show continuity of the bilinear form
a
κ
(
Proof
Let
ϕ,φ
)
.
The only terms in (
9.22
) which cannot be estimated directly by an application of
Cauchy-Schwarz are
(∂
x
ϕ,φ)
and
(y
−
1
∂
y
ϕ,φ)
. For both terms, the continuity fol-
lows from the Hardy inequality (4.26)
y
−
1
φ
L
2
(G)
≤
∈
·
,
·
2
∂
y
φ
L
2
(G)
.
(9.25)
Indeed,
2
[
]
y
−
1
∂
y
ϕ,φ
1
β
2
4
αm
−
(
9.25
)
≤
∂
y
ϕ
L
2
(G)
∂
y
φ
L
2
(G)
,
≤
C
∂
y
ϕ
L
2
(G)
y
−
1
φ
L
2
(G)
and similarly
1
2
[
]
∂
x
ϕ,φ
≤
y
−
1
φ
ρβ
−
2
r
C
y∂
x
ϕ
L
2
(G)
L
2
(G)
≤
C
y∂
x
ϕ
L
2
(G)
∂
y
φ
L
2
(G)
.
a
κ
(ϕ, φ)
To this end, it follows from the triangle inequality
V
.To
prove the Gårding inequality, we consider each term in (
9.22
) separately. We have,
since
(y∂
y
ϕ,ϕ)
=
|
|≤
C
1
ϕ
V
φ
1
2
(∂
y
(ϕ
2
), y)
=−
1
2
2
ϕ
L
2
(G)
,
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