# Symmetry and Nonexistence of Positive Solutions for Weighted HLS System of Integral Equations on a Half Space

Authors: Linfen Cao; Zhaohui Dai
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We consider system of integral equations related to the weighted Hardy-Littlewood-Sobolev (HLS) inequality in a half space. By the Pohozaev type identity in integral form, we present a Liouville type theorem when the system is in both supercritical and subcritical cases under some integrability conditions. Ruling out these nonexistence results, we also discuss the positive solutions of the integral system in critical case. By the method of moving planes, we show that a pair of positive solutions to such system is rotationally symmetric aboutx
n-axis, which is much more general than the main result of Zhuo and Li, 2011.


---

## Body

## 1. Introduction

In [1], Jin and Li studied the weighted HLS system of nonlinear equations in R
n:
(1)
u
(
x
)
=
1
|
x
|
α
∫
R
n
‍
1
|
x
-
y
|
λ
v
q
(
y
)
|
y
|
β
d
y
,
v
(
x
)
=
1
|
x
|
β
∫
R
n
‍
1
|
x
-
y
|
λ
u
p
(
y
)
|
y
|
α
d
y
,
where 0
<
λ
<
n and 1
/
(
p
+
1
)
+
1
/
(
q
+
1
)
=
(
λ
+
α
+
β
)
/
n.By the method of moving planes in integral forms they derived symmetry and monotonicity of positive solutions of (1) under some integrability conditions.Theorem 1 (see [1]).
Let the pair(
u
,
v
) be a positive solution of system (1) with u
∈
L
p
+
1
(
R
n
), v
∈
L
q
+
1
(
R
n
) and p
,
q
≥
1, p
q
≠
1, and α
,
β
≥
0. Then u and v are radially symmetric and decreasing about some point x
0.Jin and Li [2] and Chen et al. [3] also discussed the regularity of solutions to (1).LetR
+
n be the upper half Euclidean space
(2)
R
+
n
=
{
x
=
(
x
1
,
…
,
x
n
)
∈
R
n
∣
x
n
>
0
}
.
In this paper, we want to consider the similar integral system in the half space R
+
n as (1). More precisely, we discuss the following weighted HLS type system of nonlinear equations in R
+
n:
(3)
u
(
x
)
=
1
|
x
|
α
∫
R
+
n
‍
G
(
x
,
y
,
γ
)
v
q
(
y
)
|
y
|
β
d
y
,
v
(
x
)
=
1
|
x
|
β
∫
R
+
n
‍
G
(
x
,
y
,
γ
)
u
p
(
y
)
|
y
|
α
d
y
,
where u
,
v
≥
0, 0
<
p, q
<
∞, 0
<
γ
<
n, α
+
β
≥
0, α
/
n
<
1
/
(
p
+
1
)
<
(
n
-
γ
+
α
)
/
n, and
(4)
G
(
x
,
y
,
γ
)
=
1
|
x
-
y
|
n
-
γ
-
1
|
x
*
-
y
|
n
-
γ
;
here x
* is the reflection point of x about the plane ∂
R
+
n.Similar to some integral systems or PDEs systems, the integral system (3) is usually divided into three cases according to the value of exponents (
p
,
q
). We say that system (3) is in critical case when the pair (
p
,
q
) satisfies the relation
(5)
1
p
+
1
+
1
q
+
1
=
n
-
γ
+
α
+
β
n
.
It is in supercritical case when “<” holds; and in subcritical case when “>” holds; that is
(6)
1
p
+
1
+
1
q
+
1
>
n
-
γ
+
α
+
β
n
.
In the special case, where α
=
0 and β
=
0, system (3) reduces to
(7)
u
(
x
)
=
∫
R
+
n
‍
G
(
x
,
y
,
γ
)
v
q
(
y
)
d
y
,
v
(
x
)
=
∫
R
+
n
‍
G
(
x
,
y
,
γ
)
u
p
(
y
)
d
y
,
and system (7) is closely related to the following system of PDEs with Navier boundary conditions:
(8)
(
-
Δ
)
γ
/
2
u
=
v
q
,
in
R
+
n
;
(
-
Δ
)
γ
/
2
v
=
u
p
,
in
R
+
n
;
(
-
Δ
)
k
u
=
0
on
⁡
∂
R
+
n
;
(
-
Δ
)
k
v
=
0
,
on
⁡
∂
R
+
n
;
m
n
k
=
0
,
1
,
…
,
γ
2
-
1
.
In particular, when γ is an even number, the authors ([4]) proved the equivalence between the two systems (7) and (8) under some mild growth condition.Symmetry of solutions to integral system (8) was established by Zhuo and Li [5]. They proved that in critical case 1
/
(
p
+
1
)
+
1
/
(
q
+
1
)
=
(
n
-
α
)
/
n, any pair of positive solutions of (7) with u
∈
L
p
+
1
(
R
+
n
) and v
∈
L
q
+
1
(
R
+
n
) is rotationally symmetric about some line parallel to x
n-axis. Under the same integrability conditions, in [6], we obtained the nonexistence of positive solutions of (7).The general case is that, forα
≠
0 and β
≠
0 in (3), there are few results concerning symmetry and nonexistence for this doubled weighted system. In this paper, by the Pohozaev type identity in integral form, we present a Liouville type theorem when the system (3) is in both supercritical and subcritical cases under some integrability conditions. Based on these nonexistence results, we discuss the positive solutions of (3) in critical case. By the method of moving planes, we show that a pair of positive solutions to such system is rotationally symmetric about x
n-axis. To carry on the moving of planes, we explore global features of the integral equations and estimate certain integral norms. This is the essence of the method of moving planes in integral forms. The readers who are interested in the integral system and the applications of this method may consult [7–10] and the references therein.The paper is organized as follows.In Section2, by the Pohozaev type identity in integral forms, we prove the following nonexistence results.Theorem 2.
Suppose that(
u
(
x
)
,
v
(
x
)
)
∈
C
1
(
R
n
) are nonnegative solutions of (3) with u
∈
L
p
+
1
(
R
+
n
), v
∈
L
q
+
1
(
R
+
n
).(i)
Ifp and q are both supercritical, that is,
(9)
1
p
+
1
<
n
-
γ
2
n
+
α
n
,
1
q
+
1
<
n
-
γ
2
n
+
β
n
,
or
(ii)
ifp and q are both subcritical, that is,
(10)
1
p
+
1
∈
(
n
-
γ
2
n
+
α
n
,
n
-
γ
+
α
n
)
,
1
q
+
1
∈
(
n
-
γ
2
n
+
β
n
,
n
-
γ
+
β
n
)
,
then u
≡
0 and v
≡
0.Based on these results and ruling out cases where there are no solutions, we are only interested in critical case (5). In Section 3, by means of method of moving planes in integral form, we establish rotational symmetry of solutions of (3) in critical case (5) as follows.Theorem 3.
Assume thatu
∈
L
p
+
1
(
R
+
n
), v
∈
L
q
+
1
(
R
+
n
) and p
,
q satisfy (5). If (
u
,
v
) is a pair of positive solutions of (3), then (
u
,
v
) is rotationally symmetric about x
n
-axis.Remark 4.
Whenα
=
β
=
0, Theorem 3 is coincident with the result in [5].


## 2. Proof of Theorem2

In this section we will prove the nonexistence of positive solutions to the weighted HLS type system (3). These nonexistence results, known as Liouville type theorems, are useful in deriving existence, a priori estimate, regularity, and asymptotic analysis of solutions.A celebrated result of S. I. Pohozaev is known as the Pohozaev identity. This classical result has many consequences, the most immediate one being the nonexistence of nontrivial bounded solutions to PDE. Here we apply the Pohozaev type identity in integral forms to the integral system (3) (see in [9, 11]).For anyρ
≠
0, there holds
(11)
u
(
ρ
x
)
=
1
|
ρ
x
|
α
∫
R
+
n
‍
(
1
|
ρ
x
-
y
|
n
-
γ
-
1
|
ρ
x
*
-
y
|
n
-
γ
)
v
q
(
y
)
|
y
|
β
d
y
.
By an elementary calculation,
(12)
d
(
|
ρ
x
|
-
α
)
d
ρ
=
-
α
2
|
ρ
x
|
-
α
-
2
·
(
2
ρ
x
·
x
)
=
(
-
α
ρ
)
|
ρ
x
|
-
α
-
2
|
x
|
2
.
d
(
|
ρ
x
-
y
|
γ
-
n
)
d
ρ
=
γ
-
n
2
|
ρ
x
-
y
|
γ
-
n
-
2
×
d
d
ρ
[
(
ρ
x
1
-
y
1
)
2
+
⋯
+
(
ρ
x
n
-
y
n
)
2
]
=
(
γ
-
n
)
|
ρ
x
-
y
|
γ
-
n
-
2
x
·
(
ρ
x
-
y
)
,
d
(
|
ρ
x
*
-
y
|
γ
-
n
)
d
ρ
=
γ
-
n
2
|
ρ
x
*
-
y
|
γ
-
n
-
2
×
d
d
ρ
[
(
ρ
x
1
-
y
1
)
2
+
⋯
+
(
ρ
x
n
-
1
-
y
n
-
1
)
2
m
m
m
m
m
m
+
(
-
ρ
x
n
-
y
n
)
2
]
=
(
γ
-
n
)
|
ρ
x
-
y
|
γ
-
n
-
2
x
*
·
(
ρ
x
*
-
y
)
.Notingu
∈
C
1
(
R
n
), differentiating both sides of (11) with respect to ρ and letting ρ
=
1, we have
(13)
x
·
∇
u
(
x
)
=
(
-
α
)
u
(
x
)
+
(
γ
-
n
)
1
|
x
|
α
∫
R
+
n
‍
[
x
·
(
x
-
y
)
|
x
-
y
|
n
-
γ
+
2
m
m
m
m
m
m
m
m
m
m
-
x
*
·
(
x
*
-
y
)
|
x
*
-
y
|
n
-
γ
+
2
]
v
q
(
y
)
|
y
|
β
d
y
.
Let B
r
+
(
0
)
=
B
r
(
0
)
∩
R
+
n be the upper half ball in the half space in R
+
n. Multiplying left side of (13) by u
p
(
x
) and integrating on B
r
+ yields
(14)
∫
B
r
+
‍
u
p
(
x
)
(
x
·
∇
u
(
x
)
)
d
x
=
1
p
+
1
∫
B
r
+
‍
x
·
∇
(
u
p
+
1
(
x
)
)
d
x
=
1
p
+
1
∫
∂
B
r
+
‍
r
u
p
+
1
(
x
)
d
σ
-
n
p
+
1
∫
B
r
+
‍
u
p
+
1
(
x
)
d
x
.
Similarly, we also have
(15)
∫
B
r
+
‍
v
q
(
x
)
(
x
·
∇
v
(
x
)
)
d
x
=
1
q
+
1
∫
∂
B
r
+
‍
r
v
q
+
1
(
x
)
d
σ
-
n
q
+
1
∫
B
r
+
‍
v
q
+
1
(
x
)
d
x
.Since(16)
∫
R
+
n
‍
u
p
+
1
(
x
)
d
x
<
∞
,
∫
R
+
n
‍
v
q
+
1
(
x
)
d
x
<
∞
.
Thus, there exists a sequence {
r
m
} such that
(17)
r
m
∫
∂
B
r
m
+
‍
u
p
+
1
(
x
)
d
σ
⟶
0
,
∫
∂
B
r
m
+
‍
r
m
v
q
+
1
(
x
)
d
σ
⟶
0
,
m
m
m
m
m
m
m
m
r
m
⟶
∞
.
Let r
m
→
∞; by (14), (15), and (17), we have
(18)
∫
R
+
n
‍
u
p
(
x
)
(
x
·
∇
u
(
x
)
)
d
x
+
∫
R
+
n
‍
v
q
(
x
)
(
x
·
∇
v
(
x
)
)
d
x
=
-
n
p
+
1
∫
R
+
n
‍
u
p
+
1
(
x
)
d
x
-
n
q
+
1
∫
R
+
n
‍
v
q
+
1
(
x
)
d
x
<
∞
.
On the other hand,
(19)
∫
R
+
n
‍
u
p
(
x
)
(
x
·
∇
u
(
x
)
)
d
x
=
(
γ
-
n
)
∫
‍
∫
R
+
n
‍
[
x
·
(
x
-
y
)
|
x
-
y
|
n
-
γ
+
2
-
x
*
·
(
x
*
-
y
)
|
x
*
-
y
|
n
-
γ
+
2
]
m
m
m
m
m
m
m
m
m
m
×
u
p
(
x
)
v
q
(
y
)
|
x
|
α
|
y
|
β
d
x
d
y
+
(
-
α
)
∫
R
+
n
‍
u
p
+
1
(
x
)
d
x
=
γ
-
n
2
∫
‍
∫
R
+
n
‍
[
x
·
(
x
-
y
)
|
x
-
y
|
n
-
γ
+
2
-
x
*
·
(
x
*
-
y
)
|
x
*
-
y
|
n
-
γ
+
2
]
m
m
m
m
m
m
m
m
m
×
u
p
(
x
)
v
q
(
y
)
|
x
|
α
|
y
|
β
d
x
d
y
+
γ
-
n
2
∫
‍
∫
R
+
n
‍
[
y
·
(
y
-
x
)
|
x
-
y
|
n
-
γ
+
2
-
y
*
·
(
y
*
-
x
)
|
y
*
-
x
|
n
-
γ
+
2
]
m
m
m
m
m
m
m
m
m
m
×
u
p
(
y
)
v
q
(
x
)
|
y
|
α
|
x
|
β
d
x
d
y
+
(
-
α
)
∫
R
+
n
‍
u
p
+
1
(
x
)
d
x
.
There also holds
(20)
∫
R
+
n
‍
v
q
(
x
)
(
x
·
∇
v
(
x
)
)
d
x
=
(
γ
-
n
)
∫
‍
∫
R
+
n
‍
[
x
·
(
x
-
y
)
|
x
-
y
|
n
-
γ
+
2
-
x
*
·
(
x
*
-
y
)
|
x
*
-
y
|
n
-
γ
+
2
]
m
m
m
m
m
m
m
m
m
m
×
u
p
(
y
)
v
q
(
x
)
|
y
|
α
|
x
|
β
d
x
d
y
+
(
-
β
)
∫
R
+
n
‍
v
q
+
1
(
x
)
d
x
=
γ
-
n
2
∫
‍
∫
R
+
n
‍
[
x
·
(
x
-
y
)
|
x
-
y
|
n
-
γ
+
2
-
x
*
·
(
x
*
-
y
)
|
x
*
-
y
|
n
-
γ
+
2
]
m
m
m
m
m
m
m
m
m
×
u
p
(
y
)
v
q
(
x
)
|
y
|
α
|
x
|
β
+
γ
-
n
2
∫
‍
∫
R
+
n
‍
[
y
·
(
y
-
x
)
|
x
-
y
|
n
-
γ
+
2
-
y
*
·
(
y
*
-
x
)
|
y
*
-
x
|
n
-
γ
+
2
]
m
m
m
m
m
m
m
m
m
m
×
u
p
(
x
)
v
q
(
y
)
|
x
|
α
|
y
|
β
d
x
d
y
+
(
-
β
)
∫
R
+
n
‍
v
q
+
1
(
x
)
d
x
.
Using
(21)
x
·
(
x
-
y
)
+
y
·
(
y
-
x
)
=
|
x
-
y
|
2
,
x
*
·
(
x
*
-
y
)
+
y
*
·
(
y
*
-
x
)
=
|
x
*
-
y
|
2
.
Combining the fact |
x
*
-
y
|
=
|
y
*
-
x
|, (19), and (20), we have
(22)
∫
R
+
n
‍
u
p
(
x
)
(
x
·
∇
u
(
x
)
)
d
x
+
∫
R
+
n
‍
v
q
(
x
)
(
x
·
∇
v
(
x
)
)
d
x
=
(
γ
-
n
2
-
α
)
∫
R
+
n
‍
u
p
+
1
(
x
)
d
x
+
(
γ
-
n
2
-
β
)
∫
R
+
n
‍
v
q
+
1
(
x
)
d
x
.
By (18) and (22), we have
(23)
(
γ
-
n
2
-
α
+
n
p
+
1
)
∫
R
+
n
‍
u
p
+
1
(
x
)
d
x
+
(
γ
-
n
2
-
β
+
n
q
+
1
)
∫
R
+
n
‍
v
q
+
1
(
x
)
d
x
=
0
.
Hence, if
(24)
γ
-
n
2
-
α
+
n
p
+
1
>
0
,
γ
-
n
2
-
β
+
n
q
+
1
>
0
or
(25)
γ
-
n
2
-
α
+
n
p
+
1
<
0
,
γ
-
n
2
-
β
+
n
q
+
1
<
0
,
hold, it follows that u
≡
0 and v
≡
0.This completes the proof of Theorem2.Remark 5.
In [11], the authors consider another weighted HLS type integral system
(26)
u
(
x
)
=
∫
R
+
n
‍
G
(
x
,
y
,
γ
)
|
y
|
-
s
v
q
(
y
)
d
y
,
v
(
x
)
=
∫
R
+
n
‍
G
(
x
,
y
,
γ
)
|
y
|
-
t
u
p
(
y
)
d
y
,
mmmmmmmmmmmmm
∀
x
∈
R
+
n
and showed the Liouville type theorem as follows.Theorem 6 (see [11]).
Suppose thatu
(
x
)
,
v
(
x
)
∈
C
1
(
R
n
) are positive solutions of (26) when p and q are both subcritical; that is 1
/
(
p
+
1
)
>
(
n
-
γ
)
/
2
(
n
-
t
) and 1
/
(
q
+
1
)
>
(
n
-
γ
)
/
2
(
n
-
s
). If ∫
R
+
n
‍
(
u
p
+
1
/
|
x
|
t
)
d
x
<
∞, ∫
R
+
n
‍
(
v
q
+
1
/
|
x
|
s
)
d
x
<
∞ and γ
-
s
>
1, γ
-
t
>
1, then u
≡
0 and v
≡
0.Whens
=
t
=
0 in system (26) or α
=
β
=
0 in system (3), the two systems reduce to the simple integral system (7). In this special case, we can find that Theorem 6 is coincident with case (ii) in Theorem 2.


## 3. Proof of Theorem3

In this section, we will consider rotational symmetry of weighted HLS type system (3) in critical case (5).Firstly, we need the following weighted HLS inequality.Lemma 7 (see [12]).
Let1
<
l
,
m
<
∞, 0
<
γ
<
n, τ
+
β
≥
0, 1
/
l
+
1
/
m
+
(
γ
+
τ
+
β
)
/
n
=
2, and 1
-
1
/
m
-
γ
/
n
<
τ
/
n
<
1
-
1
/
m. Then
(27)
|
∫
‍
∫
R
n
‍
f
(
x
)
g
(
y
)
|
x
|
τ
|
x
-
y
|
γ
|
y
|
β
d
x
d
y
|
≤
C
∥
f
∥
m
∥
g
∥
l
.
One can also write the weighted HLS inequality in another form. Let
(28)
T
g
(
x
)
=
∫
R
n
‍
g
(
y
)
|
x
|
τ
|
x
-
y
|
γ
|
y
|
β
d
y
.
Then
(29)
∥
T
g
(
x
)
∥
L
μ
=
Sup
⁡
∥
f
∥
m
=
1
⁡
<
T
g
(
x
)
,
f
(
x
)
>
≤
∥
g
∥
L
l
,
where 1
/
l
+
(
γ
+
τ
+
β
)
/
n
=
1
+
1
/
μ, 1
/
μ
+
1
/
m
=
1.For a given real numberλ, define
(30)
Σ
λ
=
{
x
=
(
x
1
,
x
2
,
…
,
x
n
)
∈
R
+
n
∣
x
1
<
λ
}
,
T
λ
=
{
x
∈
R
+
n
∣
x
1
=
λ
}
.
Let x
λ
=
(
2
λ
-
x
1
,
x
2
,
…
,
x
n
-
1
,
x
n
) be the reflection of the point x
=
(
x
1
,
x
2
,
…
,
x
n
) about the plane T
λ. Set
(31)
u
λ
(
x
)
=
u
(
x
λ
)
,
v
λ
(
x
)
=
v
(
x
λ
)
.Lemma 8 (see [8, 13]).
Forx
,
y
∈
Σ
λ, x
≠
y, one has
(32)
G
(
x
,
y
,
γ
)
≥
G
(
x
λ
,
y
,
γ
)
.Lemma 9.
Let(
u
,
v
) be any pair of positive solutions of (3) in critical case (5); for any x
∈
Σ
λ and |
x
|
>
|
x
λ
|, one has
(33)
u
(
x
)
-
u
λ
(
x
)
≤
1
|
x
|
α
∫
Σ
λ
‍
[
G
(
x
,
y
,
γ
)
-
G
(
x
λ
,
y
,
γ
)
]
v
q
(
y
)
-
v
λ
q
(
y
)
|
y
|
β
d
y
,
v
(
x
)
-
v
λ
(
x
)
≤
1
|
x
|
β
∫
Σ
λ
‍
[
G
(
x
,
y
,
γ
)
-
G
(
x
λ
,
y
,
γ
)
]
u
p
(
y
)
-
u
λ
p
(
y
)
|
y
|
α
d
y
.Proof.
Through the calculation, we have(34)
u
(
x
)
=
1
|
x
|
α
∫
Σ
λ
‍
G
(
x
,
y
,
γ
)
v
q
(
y
)
|
y
|
β
d
y
mmmm
+
1
|
x
|
α
∫
Σ
λ
‍
G
(
x
λ
,
y
,
γ
)
v
λ
q
(
y
)
|
y
λ
|
β
d
y
,
u
λ
(
x
)
=
1
|
x
λ
|
α
∫
Σ
λ
‍
G
(
x
λ
,
y
,
γ
)
v
q
(
y
)
|
y
|
β
d
y
mmmm
+
1
|
x
λ
|
α
∫
Σ
λ
‍
G
(
x
,
y
,
γ
)
u
λ
p
(
y
)
|
y
λ
|
β
d
y
.
By the assumption |
x
|
>
|
x
λ
|, we have
(35)
u
(
x
)
-
u
λ
(
x
)
≤
1
|
x
|
α
∫
Σ
λ
‍
[
G
(
x
,
y
,
γ
)
-
G
(
x
λ
,
y
,
γ
)
]
m
m
m
m
m
m
×
(
v
q
(
y
)
|
y
|
β
-
v
λ
q
(
y
)
|
y
λ
|
β
)
d
y
≤
1
|
x
|
α
∫
Σ
λ
‍
[
G
(
x
,
y
,
γ
)
-
G
(
x
λ
,
y
,
γ
)
]
v
q
(
y
)
-
v
λ
q
(
y
)
|
y
|
β
d
y
.
Similarly, we have(36)
v
(
x
)
-
v
λ
(
x
)
≤
1
|
x
|
β
∫
Σ
λ
‍
[
G
(
x
,
y
,
γ
)
-
G
(
x
λ
,
y
,
γ
)
]
mmmmmm
×
u
p
(
y
)
-
u
λ
p
(
y
)
|
y
|
α
d
y
.Proof of Theorem3.
Step 1. We will show that for sufficiently negative λ,
(37)
u
λ
(
x
)
≥
u
(
x
)
,
v
λ
(
x
)
≥
v
(
x
)
,
a
.
e
.
∀
x
∈
Σ
λ
.
Define
(38)
Σ
λ
u
=
{
x
∈
Σ
λ
,
u
(
x
)
>
u
λ
(
x
)
}
,
Σ
λ
v
=
{
x
∈
Σ
λ
,
v
(
x
)
>
v
λ
(
x
)
}
.
We prove that, for sufficiently negative λ, both Σ
λ
u and Σ
λ
v must be empty and thus (37) holds.
In fact, by Lemma9 and the mean value theorem, we have, for x
∈
Σ
λ
u,
(39)
0
<
u
(
x
)
-
u
λ
(
x
)
≤
1
|
x
|
α
∫
Σ
λ
‍
[
G
(
x
,
y
,
γ
)
-
G
(
x
λ
,
y
,
γ
)
]
v
q
(
y
)
-
v
λ
q
(
y
)
|
y
|
β
d
y
≤
∫
Σ
λ
v
‍
1
|
x
|
α
[
G
(
x
,
y
,
γ
)
-
G
(
x
λ
,
y
,
γ
)
]
v
q
(
y
)
-
v
λ
q
(
y
)
|
y
|
β
d
y
≤
1
|
x
|
α
∫
Σ
λ
v
‍
G
(
x
,
y
,
γ
)
v
q
(
y
)
-
v
λ
q
(
y
)
|
y
|
β
d
y
≤
q
|
x
|
α
∫
Σ
λ
v
‍
1
|
x
-
y
|
n
-
γ
ψ
λ
q
-
1
(
y
)
v
(
y
)
-
v
λ
(
y
)
|
y
|
β
d
y
≤
q
∫
Σ
λ
v
‍
1
|
x
|
α
|
x
-
y
|
n
-
γ
|
y
|
β
v
q
-
1
(
y
)
[
v
(
y
)
-
v
λ
(
y
)
]
d
y
,
where ψ
λ
(
y
) is valued between v
(
y
) and v
λ
(
y
); therefore, on Σ
λ
v, we have
(40)
0
≤
v
λ
(
y
)
≤
ψ
λ
(
y
)
≤
v
(
y
)
.
By Lemma 7 and the Hölder inequality, we have
(41)
∥
u
λ
-
u
∥
L
p
+
1
(
Σ
λ
u
)
≤
C
∥
v
q
-
1
(
v
λ
-
v
)
∥
L
T
1
(
Σ
λ
v
)
≤
C
∥
v
∥
L
q
+
1
(
Σ
λ
v
)
q
-
1
∥
v
λ
-
v
∥
L
q
+
1
(
Σ
λ
v
)
,
(42)
∥
v
λ
-
v
∥
L
q
+
1
(
Σ
λ
v
)
≤
C
∥
u
p
-
1
(
u
λ
-
u
)
∥
L
T
2
(
Σ
λ
u
)
≤
C
∥
u
∥
L
p
+
1
(
Σ
λ
u
)
p
-
1
∥
u
λ
-
u
∥
L
p
+
1
(
Σ
λ
u
)
,
where T
1
=
n
(
p
+
1
)
/
(
n
+
(
γ
-
α
-
β
)
(
p
+
1
)
) and T
2
=
n
(
q
+
1
)
/
(
n
+
(
γ
-
α
-
β
)
(
q
+
1
)
). It easy to show that T
1
,
T
2
>
1. Combining (41) and (42), we arrive
(43)
∥
u
λ
-
u
∥
L
p
+
1
(
Σ
λ
u
)
≤
C
∥
v
∥
L
q
+
1
(
Σ
λ
v
)
q
-
1
∥
u
∥
L
p
+
1
(
Σ
λ
u
)
p
-
1
∥
u
λ
-
u
∥
L
p
+
1
(
Σ
λ
u
)
.
The conditions u
∈
L
p
+
1
(
R
+
n
) and v
∈
L
q
+
1
(
R
+
n
) make us able to choose sufficiently negative λ, so that
(44)
C
∥
v
∥
L
q
+
1
(
Σ
λ
v
)
q
-
1
∥
u
∥
L
p
+
1
(
Σ
λ
u
)
p
-
1
≤
1
2
.
Now inequality (43) implies
(45)
∥
u
λ
-
u
∥
L
p
+
1
(
Σ
λ
u
)
=
0
,
and therefore Σ
λ
u must be measure zero. Similarly, one can show that Σ
λ
v is measure zero. Therefore (37) holds.
Step 2. Inequality (37) provides a starting point to move the plane T
λ
=
{
x
∈
R
+
n
∣
x
1
=
λ
}. Now we start from the neighborhood of x
1
=
-
∞ and move the plane to the right as long as (37) holds to the limiting position. More precisely, define
(46)
λ
0
=
sup
⁡
{
Σ
μ
λ
∣
u
(
x
)
≤
u
μ
(
x
)
,
m
m
v
(
x
)
≤
v
μ
(
x
)
,
μ
≤
λ
,
∀
x
∈
Σ
μ
}
.
We will prove that λ
0
=
0. On the contrary, we suppose λ
0
<
0. We show that u
(
x
) and v
(
x
) are symmetric about the plane T
λ
0; that is
(47)
u
λ
0
(
x
)
≡
u
(
x
)
,
v
λ
0
(
x
)
≡
v
(
x
)
,
a
.
e
.
∀
x
∈
Σ
λ
0
.
Otherwise, on Σ
λ
0,
(48)
u
(
x
)
≤
u
λ
0
(
x
)
,
v
(
x
)
≤
v
λ
0
(
x
)
,
but
u
(
x
)
≢
u
λ
0
(
x
)
or
v
(
x
)
≢
v
λ
0
(
x
)
.
We show that the plane can be moved further to the right. More precisely, there exists an ϵ
>
0 such that, for ∀
λ
∈
[
λ
o
,
λ
o
+
ϵ
),
(49)
u
(
x
)
≤
u
λ
(
x
)
,
v
(
x
)
≤
v
λ
(
x
)
,
a
.
e
.
∀
x
∈
Σ
λ
.
Without loss of generality, we assume(50)
v
(
x
)
≢
v
λ
0
(
x
)
,
on
⁡
Σ
λ
0
.
by Lemma 9, we have in fact u
(
x
)
<
u
λ
0
(
x
) in the interior of Σ
λ
0. Let
(51)
Σ
λ
0
u
¯
=
{
x
∈
Σ
λ
0
∣
u
(
x
)
≥
u
λ
0
(
x
)
}
,
Σ
λ
0
v
¯
=
{
x
∈
Σ
λ
0
∣
v
(
x
)
≥
v
λ
0
(
x
)
}
.
Then obviously Σ
λ
0
u
¯ has measure zero and lim
⁡
λ
→
λ
0
Σ
λ
u
⊂
Σ
λ
0
u
¯. The same argument above is also true for the other solution v of (3). From (41) and (42), we deduce
(52)
∥
u
λ
-
u
∥
L
p
+
1
(
Σ
λ
u
)
≤
C
∥
v
∥
L
q
+
1
(
Σ
λ
v
)
q
-
1
∥
u
∥
L
p
+
1
(
Σ
λ
u
)
p
-
1
∥
u
λ
-
u
∥
L
p
+
1
(
Σ
λ
u
)
.
Again the conditions that u
∈
L
p
+
1
(
R
+
n
) and v
∈
L
q
+
1
(
R
+
n
) ensure that one can choose ϵ sufficiently small, so that, for all λ in [
λ
o
,
λ
o
+
ϵ
),
(53)
C
∥
v
∥
L
q
+
1
(
Σ
λ
v
)
q
-
1
∥
u
∥
L
p
+
1
(
Σ
λ
u
)
p
-
1
≤
1
2
.
The method to verify this inequality is standard and the proofs of the rest are similar to the proof in paper [6, 11, 14].
Now by (52) and (53), we have ∥
u
λ
-
u
∥
L
p
(
Σ
λ
u
)
=
0, and therefore Σ
λ
u must be measure zero. Similarly, Σ
λ
v must also be measure zero. Hence, for these values of λ
>
λ
o, we have
(54)
u
λ
(
x
)
≥
u
(
x
)
,
v
λ
(
x
)
≥
v
(
x
)
,
a
.
e
.
∀
x
∈
Σ
λ
0
.
This (47) must hold and therefore both u
(
x
) and v
(
x
) are symmetric about the plane T
λ
0.
Now we show that the plane cannot stop before hitting the origin. Otherwise, assume that the plane stops atx
1
=
λ
0
<
0. By the fact that |
y
|
>
|
y
λ
0
|, we have
(55)
u
(
x
)
-
u
λ
0
(
x
)
≤
1
|
x
|
α
∫
Σ
λ
0
‍
[
G
(
x
,
y
,
γ
)
-
G
(
x
λ
0
,
y
,
γ
)
]
m
m
m
m
m
×
[
v
q
(
y
)
|
y
|
β
-
v
λ
0
q
(
y
)
|
y
λ
0
|
β
]
d
y
<
1
|
x
|
α
∫
Σ
λ
0
‍
[
G
(
x
,
y
,
γ
)
-
G
(
x
λ
0
,
y
,
γ
)
]
m
m
m
m
m
m
×
[
v
q
(
y
)
-
v
λ
0
q
(
y
)
|
y
|
β
]
d
y
=
0
.
This contradicts with (47).
As the direction ofx
1 can be chosen arbitrarily, we derive that (
u
(
x
)
,
v
(
x
)
) is rotationally symmetric about x
n
-axis. This completes the proof of Theorem 3.



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*Source: 593210-2014-04-30.xml*