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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
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 2 ), y) =−
1
2
2 ϕ
L 2 (G) ,
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