1 Introduction
Let 0<ν<n and 1<θ,m<∞. The well-known classical Hardy–Littlewood–Sobolev inequality [11] states
∫
ℝ
n
∫
ℝ
n
f
(
x
)
g
(
y
)
|
x
-
y
|
ν
𝑑
x
𝑑
y
≤
C
θ
,
m
,
ν
∥
f
∥
L
m
∥
g
∥
L
θ
for any f∈Lm(ℝn), g∈Lθ(ℝn), and for 1m+1θ+νn=2.
Hardy and Littlewood also introduced the double weighted inequality, which was later generalized by Stein and Weiss in [20]. It states
(1.1)
|
∫
ℝ
n
∫
ℝ
n
f
(
x
)
g
(
y
)
|
x
|
τ
|
x
-
y
|
ν
|
y
|
t
𝑑
y
𝑑
x
|
≤
C
t
,
τ
,
θ
,
m
,
v
∥
f
∥
L
m
∥
g
∥
L
θ
,
where τ+t≥0,1θ+1m+ν+t+τn=2 and 1-1m-νn≤τn<1-1m (see also [9] for Stein–Weiss inequalities on the Heisenberg group).
One can also write the above Stein–Weiss inequality in another form. Let
T
(
g
)
(
x
)
=
∫
ℝ
n
f
(
x
)
g
(
y
)
|
x
|
τ
|
x
-
y
|
ν
|
y
|
t
𝑑
y
,
then
∥
T
(
g
)
∥
L
s
=
sup
∥
f
∥
L
m
=
1
〈
T
(
g
)
(
x
)
,
f
(
x
)
〉
≤
C
t
,
τ
,
θ
,
m
,
ν
∥
g
∥
L
θ
,
where 1θ+ν+t+τn=1+1s, 1m+1s=1.
To obtain the best constant in the weighted inequality (1.1), one can maximize the functional
J
(
f
,
g
)
=
∫
ℝ
n
∫
ℝ
n
f
(
x
)
g
(
y
)
|
x
|
τ
|
x
-
y
|
ν
|
y
|
t
𝑑
y
𝑑
x
under the constraints ∥f∥Lm=∥g∥Lθ=1. The corresponding Euler–Lagrange equations are the integral system
(1.2)
v
1
m
f
(
x
)
m
-
1
=
∫
ℝ
n
1
|
x
|
τ
|
x
-
y
|
n
-
α
|
y
|
t
g
(
y
)
𝑑
y
,
v
2
θ
g
(
x
)
θ
-
1
=
∫
ℝ
n
1
|
x
|
t
|
x
-
y
|
n
-
α
|
y
|
τ
f
(
y
)
𝑑
y
,
where f,g≥0 and v1m=v2θ=J(f,g).
Set u=c1fm-1,v=c2gθ-1,p=1m-1,q=1θ-1, then for a proper choice of constants c1 and c2, system (1.2) becomes
(1.3)
u
(
x
)
=
∫
ℝ
n
1
|
x
|
τ
|
x
-
y
|
n
-
α
|
y
|
t
v
q
(
y
)
𝑑
y
,
v
(
x
)
=
∫
ℝ
n
1
|
x
|
t
|
x
-
y
|
n
-
α
|
y
|
τ
u
p
(
y
)
𝑑
y
.
Integral system (1.3) is closely related to the following partial differential equations:
(1.4)
(
-
Δ
)
α
2
(
|
x
|
τ
u
(
x
)
)
=
|
x
|
-
t
v
q
(
x
)
,
(
-
Δ
)
α
2
(
|
x
|
t
v
(
x
)
)
=
|
x
|
-
τ
u
p
(
x
)
.
Lei, Li and Ma [10] established the equivalence between the weighted Hardy–Sobolev type system (1.3) and the corresponding Euler–Lagrange system (1.4). Chen et al. [3] obtained the symmetry, monotonicity and regularity of solutions of (1.3). In particular, they also obtained the optimal integrability of the solutions for a class of such system. The best constant in a weighted Hardy–Littlewood–Sobolev inequality related to integral system (1.3) was studied in [5] and solutions of (1.3) for α=2,t=τ,p=q were classified. For more results related to Hardy–Sobolev type equation and system, we refer the reader to [2, 15] and the references therein.
In the special cases when τ=t=0,u=v,p=q=n+αn-α, (1.3) becomes the single equation
u
(
x
)
=
∫
ℝ
n
1
|
x
-
y
|
n
-
α
u
n
+
α
n
-
α
(
y
)
𝑑
y
.
The corresponding PDE is the well-known family of semi-linear equations
(1.5)
(
-
Δ
)
α
2
u
(
x
)
=
u
n
+
α
n
-
α
(
x
)
.
In particular, when n≥3 and α=2, equation (1.5) becomes
(1.6)
(
-
Δ
)
u
(
x
)
=
u
n
+
2
n
-
2
(
x
)
.
The classification of the solutions of (1.6) has provided an important ingredient in the study of the well-known Yamable problem and the prescribing scalar curvature. Caffarelli, Gidas and Spruck [1] classified the solutions of equation (1.6). Wei and Xu [21] generalized their results to the solutions of the more general equation (1.5) with α being any even number between 0 and n.
Let 0<τ+t<α<n. We consider the following weighted Hardy–Sobolev type system:
(1.7)
u
(
x
)
=
∫
ℝ
n
1
|
x
|
τ
|
x
-
y
|
n
-
α
|
y
|
t
f
1
(
u
(
y
)
,
v
(
y
)
)
𝑑
y
,
v
(
x
)
=
∫
ℝ
n
1
|
x
|
t
|
x
-
y
|
n
-
α
|
y
|
τ
f
2
(
u
(
y
)
,
v
(
y
)
)
𝑑
y
,
where
f
1
(
u
(
y
)
,
v
(
y
)
)
=
λ
1
u
p
1
(
y
)
+
μ
1
v
q
1
(
y
)
+
γ
1
u
α
1
(
y
)
v
β
1
(
y
)
,
f
2
(
u
(
y
)
,
v
(
y
)
)
=
λ
2
u
p
2
(
y
)
+
μ
2
v
q
2
(
y
)
+
γ
2
u
α
2
(
y
)
v
β
2
(
y
)
.
Such f1(u(y),v(y)) and f2(u(y),v(y)) are considerably more complicated when none of γi, λi and μi is zero (i=1,2). Therefore, it is more difficult to deal with these cases.
In this paper, we always assume that λi,μi,γi (i=1,2) are nonnegative constants and they are not equal to zero simultaneously. We also use the following notations:
Π
1
=
{
g
(
x
)
∣
x
∈
ℝ
n
,
g
∈
L
t
11
(
ℝ
n
)
∩
L
t
12
(
ℝ
n
)
∩
L
k
0
(
ℝ
n
)
}
,
Π
2
=
{
g
(
x
)
∣
x
∈
ℝ
n
,
g
∈
L
t
21
(
ℝ
n
)
∩
L
t
22
(
ℝ
n
)
∩
L
k
0
(
ℝ
n
)
}
,
where t1i=n(pi-1)α-t-τ,t2i=n(qi-1)α-t-τ (i=1,2) and k0=n(α1+β1-1)α-t-τ=n(α2+β2-1)α-t-τ.
We first obtain the following regularity of Hardy–Sobolev type system (1.7).
Theorem 1.1
For α1,β2>0, α2,β1>1, pi,qi>1 (i=1,2), if (u,v)∈Π1×Π2 is a pair of solutions of system (1.7), then u,v∈Ls(ℝn) for any nn-α+t+τ<s<∞.
We note that t1i,t2i and k0 cannot be compared easily, thus one cannot use interpolation to conclude what the spaces Π1 or Π2 are. Therefore, it is highly nontrivial to derive the Ls integrability of the solutions u and v in Theorem 1.1. One can also see from its proof that the above lower bound for s is the best one can get from the Hardy–Littlewood–Sobolev inequality.
Next, we will establish the C∞ of the solutions away from the origin for the solutions u and v. This improves significantly known results in the literature where only the Lipschitz continuity of the solutions of the Hardy–Sobolev system without weight has been established, except in the special case of the single equation and when τ=0 and the critical exponent p=n+α-2tn-α (see [15]).
Theorem 1.2
For α1,β2>0, α2,β1>1, nn-α+t+τ<pi,qi,αi+βi<∞ (i=1,2), if (u,v)∈Π1×Π2 is a pair of solutions of system (1.7), then |x|τu(x),|x|tv(x)∈L∞(ℝn) and u,v∈C∞(ℝn∖{0}).
It is well known that the moving plane method was invented by Alexandrov in the 1950s. Then the method was further developed by Serrin [19], Gidas, Ni and Nirenberg [8], Caffarelli, Gidas and Spruck [1] and many other authors. In [6], Chen, Li and Ou developed the method of moving plane in integral forms which is quite different from the traditional moving plane method of differential equations which relies on maximum principles. The moving plane in integral forms can be easily applied to higher order equations without using maximum principles. For more results related to the moving plane in integral forms and related symmetry results and Liouville-type theorems, see, e.g., [4, 7, 12, 13, 14, 16, 17] and the references therein.
To establish the symmetry of the solutions, we use the moving plane in integral forms and the Hardy–Littlewood–Sobolev inequality to prove the following theorem.
Theorem 1.3
For pi,qi,αi,βi>1 (i=1,2), assume that (u,v)∈Π1×Π2 is a pair of nonnegative solutions of system (1.7). Then u and v are radially symmetric and strictly decreasing about the origin.
This paper is organized as follows. In Section 2 and Section 3, we prove Theorem 1.1 and Theorem 1.2, respectively. In Section 4, we prove that each pair of solutions of (1.7) is radially symmetric and strictly decreasing about the origin, thus give the proof of Theorem 1.3.
2 Proof of Theorem 1.1
In this section, we need the following regularity lifting lemma (see [18]) and the weighted Hardy–Littlewood–Sobolev inequality to prove Theorem 1.1.
Let V be a topological vector space. Suppose there are two extended norms (i.e., the norm of an element in V might be infinity) defined on V,
∥
⋅
∥
X
,
∥
⋅
∥
Y
:
V
→
[
0
,
∞
]
.
Let
X
:=
{
f
∈
V
:
∥
f
∥
X
<
∞
}
and
Y
:=
{
f
∈
V
:
∥
f
∥
Y
<
∞
}
.
Lemma 2.1
Let T be a contraction map from X into itself and from Y into itself. Assume that for any f∈X there exists a function g∈Z:=X∩Y such that f=Tf+g∈X. Then f∈Z.
Now, we start our proof of Theorem 1.1. Denote
u
a
(
x
)
=
{
u
(
x
)
if
|
u
(
x
)
|
>
a
or
|
x
|
>
a
,
0
otherwise
,
and ub(x)=u(x)-ua(x).
Assume that (ϕ,φ)∈Ls(ℝn)×Ls(ℝn) for nn-α+t+τ<s<∞. Define
T
1
(
ϕ
,
φ
)
=
∫
ℝ
n
λ
1
u
a
p
1
-
1
(
y
)
ϕ
(
y
)
|
x
|
τ
|
x
-
y
|
n
-
α
|
y
|
t
𝑑
y
+
∫
ℝ
n
1
|
x
|
τ
|
x
-
y
|
n
-
α
|
y
|
t
f
a
(
y
)
φ
(
y
)
𝑑
y
,
T
2
(
ϕ
,
φ
)
=
∫
ℝ
n
μ
2
v
a
q
2
-
1
(
y
)
φ
(
y
)
|
x
|
t
|
x
-
y
|
n
-
α
|
y
|
τ
𝑑
y
+
∫
ℝ
n
1
|
x
|
t
|
x
-
y
|
n
-
α
|
y
|
τ
g
a
(
y
)
ϕ
(
y
)
𝑑
y
.
where
f
a
(
y
)
=
μ
1
v
a
q
1
-
1
(
y
)
+
γ
1
u
a
α
1
(
y
)
v
a
β
1
-
1
(
y
)
+
γ
1
u
a
α
1
(
y
)
v
b
β
1
-
1
(
y
)
+
γ
1
u
b
α
1
(
y
)
v
a
β
1
-
1
(
y
)
,
g
a
(
x
)
=
λ
2
u
a
p
2
-
1
(
y
)
+
γ
2
v
a
β
2
(
y
)
u
a
α
2
-
1
(
y
)
+
γ
2
v
a
β
2
(
y
)
u
b
α
2
-
1
(
y
)
+
γ
2
v
b
β
2
(
y
)
u
a
α
2
-
1
(
y
)
.
Let
F
(
x
)
=
∫
ℝ
n
λ
1
u
b
p
1
-
1
(
y
)
u
(
y
)
|
x
|
τ
|
x
-
y
|
n
-
α
|
y
|
t
𝑑
y
+
∫
ℝ
n
μ
1
v
b
q
1
-
1
(
y
)
+
γ
1
u
b
α
1
(
y
)
v
b
β
1
-
1
(
y
)
|
x
|
τ
|
x
-
y
|
n
-
α
|
y
|
t
v
(
y
)
𝑑
y
,
G
(
x
)
=
∫
ℝ
n
μ
2
v
b
q
2
-
1
(
y
)
v
(
y
)
|
x
|
t
|
x
-
y
|
n
-
α
|
y
|
τ
𝑑
y
+
∫
ℝ
n
λ
2
u
b
p
2
-
1
(
y
)
+
γ
2
v
b
β
2
(
y
)
u
b
α
2
-
1
(
y
)
|
x
|
t
|
x
-
y
|
n
-
α
|
y
|
τ
u
(
y
)
𝑑
y
.
Denote the norm in the cross product space Ls(ℝn)×Ls(ℝn) by
∥
(
ϕ
,
φ
)
∥
L
s
×
L
s
=
∥
ϕ
∥
L
s
+
∥
φ
∥
L
s
,
and define the mapping T: Ls(ℝn)×Ls(ℝn)→Ls(ℝn)×Ls(ℝn) by
T
(
ϕ
,
φ
)
=
(
T
1
(
ϕ
,
φ
)
,
T
2
(
ϕ
,
φ
)
)
.
Consider the equation
(2.1)
(
ϕ
,
φ
)
=
T
(
ϕ
,
φ
)
+
(
F
,
G
)
.
It is easy to check that (u,v) is a pair of solutions of (2.1).
In order to show (u,v)∈Ls(ℝn)×Ls(ℝn) for all s>nn-α+t+τ, we carry out the proof by two steps:
T is a contracting map from Ls(ℝn)×Ls(ℝn) to itself for sufficiently large a.
F and G belong to Ls(ℝn).
Then from Lemma 2.1, we conclude the proof.
Step 1. For any (ϕ,φ)∈Ls(ℝn)×Ls(ℝn) with s>nn-α+t+τ, by the weighted Hardy–Littlewood–Sobolev inequality and the Minkowski inequality, we have
∥
T
1
(
ϕ
,
φ
)
∥
L
s
≤
C
(
∥
u
a
p
1
-
1
ϕ
∥
L
θ
+
∥
v
a
q
1
-
1
φ
∥
L
θ
+
∥
u
a
α
1
v
a
β
1
-
1
φ
∥
L
θ
+
∥
u
a
α
1
v
b
β
1
-
1
φ
∥
L
θ
+
∥
u
b
α
1
v
a
β
1
-
1
φ
∥
L
θ
)
≤
C
(
∥
u
a
∥
L
t
11
p
1
-
1
∥
ϕ
∥
L
s
+
∥
v
a
∥
L
t
21
q
1
-
1
∥
φ
∥
L
s
+
∥
u
a
∥
L
k
0
α
1
∥
v
a
∥
L
k
0
β
1
-
1
∥
φ
∥
L
s
(2.2)
+
∥
u
a
∥
L
k
0
α
1
∥
v
b
∥
L
k
0
β
1
-
1
∥
φ
∥
L
s
+
∥
u
b
∥
L
k
0
α
1
∥
v
a
∥
L
k
0
β
1
-
1
∥
φ
∥
L
s
)
and
∥
T
2
(
ϕ
,
φ
)
∥
L
s
≤
C
(
∥
v
a
q
2
-
1
φ
∥
L
θ
+
∥
u
a
p
2
-
1
ϕ
∥
L
θ
+
∥
v
a
β
2
u
a
α
2
-
1
ϕ
∥
L
θ
+
∥
v
a
β
2
u
b
α
2
-
1
ϕ
∥
L
θ
+
∥
v
b
β
2
u
a
α
2
-
1
ϕ
∥
L
θ
)
≤
C
(
∥
v
a
∥
L
t
22
q
2
-
1
∥
φ
∥
L
s
+
∥
u
a
∥
L
t
12
p
2
-
1
∥
ϕ
∥
L
s
+
∥
v
a
∥
L
k
0
β
2
∥
u
a
∥
L
k
0
α
2
-
1
∥
ϕ
∥
L
s
(2.3)
+
∥
v
a
∥
L
k
0
β
2
∥
u
b
∥
L
k
0
α
2
-
1
∥
ϕ
∥
L
s
+
∥
v
b
∥
L
k
0
β
2
∥
u
a
∥
L
k
0
α
2
-
1
∥
ϕ
∥
L
s
)
,
where θ=nsn+(α-t-τ)s, t1i=n(pi-1)α-t-τ,t2i=n(qi-1)α-t-τ and k0=n(αi+βi-1)(α-t-τ), i=1,2.
Since u∈Lt1i(ℝn)∩Lk0(ℝn),v∈Lt2i(ℝn)∩Lk0(ℝn), we may choose sufficiently large a such that
(2.4)
{
C
∥
u
a
∥
L
t
11
p
1
-
1
≤
1
4
,
C
∥
v
a
∥
L
t
22
q
2
-
1
≤
1
4
,
C
(
∥
v
a
∥
L
t
21
q
1
-
1
+
∥
u
a
∥
L
k
0
α
1
∥
v
a
∥
L
k
0
β
1
-
1
+
∥
u
a
∥
L
k
0
α
1
∥
v
b
∥
L
k
0
β
1
-
1
+
∥
u
b
∥
L
k
0
α
1
∥
v
a
∥
L
k
0
β
1
-
1
)
≤
1
4
,
C
(
∥
u
a
∥
L
t
12
p
2
-
1
+
∥
v
a
∥
L
k
0
β
2
∥
u
a
∥
L
k
0
α
2
-
1
+
∥
v
a
∥
L
k
0
β
2
∥
u
b
∥
L
k
0
α
2
-
1
+
∥
v
b
∥
L
k
0
β
2
∥
u
a
∥
L
k
0
α
2
-
1
)
≤
1
4
.
Hence, combining (2), (2.3) and (2.4), we obtain
(2.5)
∥
T
1
(
ϕ
,
φ
)
∥
L
s
≤
1
4
(
∥
ϕ
∥
L
s
+
∥
φ
∥
L
s
)
,
∥
T
2
(
ϕ
,
φ
)
∥
L
s
≤
1
4
(
∥
ϕ
∥
L
s
+
∥
φ
∥
L
s
)
.
By (2.5), we have
∥
T
(
ϕ
,
φ
)
∥
L
s
×
L
s
≤
1
2
(
∥
ϕ
∥
L
s
+
∥
φ
∥
L
s
)
,
which implies that T is a contraction map from Ls(ℝn)×Ls(ℝn) to itself.
Step 2. Since the estimates of F and G are similar, we only consider F:
F
(
x
)
=
∫
ℝ
n
λ
1
u
b
p
1
-
1
(
y
)
u
(
y
)
|
x
|
τ
|
x
-
y
|
n
-
α
|
y
|
t
d
y
+
∫
ℝ
n
μ
1
v
b
q
1
-
1
(
y
)
+
γ
1
u
b
α
1
(
y
)
v
b
β
1
-
1
(
y
)
|
x
|
τ
|
x
-
y
|
n
-
α
|
y
|
t
v
(
y
)
d
y
=
:
J
1
+
J
2
.
For any s>nn-α+t+τ, we apply the weighted Hardy–Littlewood–Sobolev inequality and the Minkowski inequality to obtain
∥
J
1
∥
L
s
≤
C
∥
u
b
p
1
-
1
u
∥
L
θ
≤
C
∥
u
b
∥
L
k
1
p
1
-
1
∥
u
∥
L
k
2
,
∥
J
2
∥
L
s
≤
C
∥
v
b
q
1
-
1
v
∥
L
θ
+
C
∥
u
b
α
1
v
b
β
1
-
1
v
∥
L
θ
≤
C
∥
v
b
∥
L
k
3
q
1
-
1
∥
v
∥
L
k
4
+
C
∥
u
b
∥
L
k
5
α
1
∥
v
b
∥
L
k
6
β
1
-
1
∥
v
∥
L
k
7
,
where
p
1
-
1
k
1
+
1
k
2
=
q
1
-
1
k
3
+
1
k
4
=
α
1
k
5
+
β
1
-
1
k
6
+
1
k
7
=
n
+
(
α
-
t
-
τ
)
s
n
s
=
1
s
+
α
-
t
-
τ
n
.
Since ub∈Lp(ℝn) and vb∈Lp(ℝn) for 0<p≤∞, the constants k1,k3,k5,k6 can be chosen arbitrarily. In view of u,v∈Lk0(ℝn), we may choose k2=k4=k7=k0 such that
1
s
=
p
1
-
1
k
1
+
n
-
(
α
-
t
-
τ
)
k
0
n
k
0
=
q
1
-
1
k
3
+
n
-
(
α
-
t
-
τ
)
k
0
n
k
0
(2.6)
=
α
1
k
5
+
β
1
-
1
k
6
+
n
-
(
α
-
t
-
τ
)
k
0
n
k
0
.
We carry out the process by two cases.
Case 1. If k0≥nα-t-τ, for any s>nn-α+t+τ, there exist k1,k3,k5,k6>0 such that (2.6) is valid. Then we obtain
∥
F
∥
L
s
=
∥
J
1
∥
L
s
+
∥
J
2
∥
L
s
<
∞
for any
s
>
n
n
-
α
+
t
+
τ
.
Similarly, we have
∥
G
∥
L
s
<
∞
for any
s
>
n
n
-
α
+
t
+
τ
.
By Lemma 2.1, we conclude that (u,v)∈Ls(ℝn)×Ls(ℝn) for any s>nn-α+t+τ.
Case 2. If k0<nα-t-τ, for any nn-α+t+τ<s≤nk0n-(α-t-τ)k0, we have
∥
F
∥
L
s
<
∞
,
∥
G
∥
L
s
<
∞
.
By Lemma 2.1 again, we derive that (u,v)∈Ls(ℝn)×Ls(ℝn) for any nn-α+t+τ<s≤nk0n-(α-t-τ)k0. Repeating the above process with k0 replaced by k01=nk0n-(α-t-τ)k0, we can obtain that (u,v)∈Ls(ℝn)×Ls(ℝn) for any nn-α+t+τ<s≤nk01n-(α-t-τ)k01. After a few steps, there exists j0 such that k0j0≥nα-t-τ. This returns to Case 1.
Therefore, combining Case 1 and Case 2, we conclude that (u,v)∈Ls(ℝn)×Ls(ℝn) for any s>nn-α+t+τ. This completes the proof of Theorem 1.1.
3 Proof of Theorem 1.2
In this section, we adopt ideas from Lu and Zhu [15] to prove Theorem 1.2. The proof is carried out by two parts.
Part I. We show that |x|τu(x),|x|tv(x)∈L∞(ℝn). Since the estimates of |x|τu(x) and |x|tv(x) are similar, we only discuss |x|τu(x). The function |x|τu(x) can be written as
|
x
|
τ
u
(
x
)
=
∫
ℝ
n
λ
1
u
p
1
(
y
)
|
x
-
y
|
n
-
α
|
y
|
t
𝑑
y
+
∫
ℝ
n
μ
1
v
q
1
(
y
)
|
x
-
y
|
n
-
α
|
y
|
t
𝑑
y
+
∫
ℝ
n
γ
1
u
α
1
(
y
)
v
β
1
(
y
)
|
x
-
y
|
n
-
α
|
y
|
t
𝑑
y
=
:
λ
1
A
(
x
)
+
μ
1
B
(
x
)
+
γ
1
C
(
x
)
.
We claim that ∥A(x)∥L∞<∞. Write
(3.1)
|
A
(
x
)
|
≤
∫
B
a
(
0
)
|
u
(
y
)
|
p
1
|
x
-
y
|
n
-
α
|
y
|
t
d
y
+
∫
ℝ
n
∖
B
a
(
0
)
|
u
(
y
)
|
p
1
|
x
-
y
|
n
-
α
|
y
|
t
d
y
=
:
A
1
(
x
)
+
A
2
(
x
)
.
We first estimate A1(x). If x∈ℝn∖B2a(0), by the Hölder inequality and u∈Ls for any s>nn-α+t+τ, we have
(3.2)
A
1
(
x
)
≤
∫
B
a
(
0
)
|
u
(
y
)
|
p
1
|
x
-
y
|
n
-
α
|
y
|
t
𝑑
y
≤
∫
B
a
(
0
)
|
u
(
y
)
|
p
1
|
y
|
n
-
α
+
t
𝑑
y
<
∞
.
If x∈B2a(0), we also have
(3.3)
A
1
(
x
)
≤
∫
B
a
(
0
)
|
u
(
y
)
|
p
1
|
y
|
n
-
α
+
t
𝑑
y
+
∫
B
3
a
(
x
)
|
u
(
y
)
|
p
1
|
x
-
y
|
n
-
α
+
t
𝑑
y
<
∞
.
Combining (3.2) and (3.3), for any x∈ℝn, we derive that
(3.4)
A
1
(
x
)
=
∫
B
a
(
0
)
|
u
(
y
)
|
p
1
|
x
-
y
|
n
-
α
|
y
|
t
𝑑
y
<
∞
.
Next, it suffices to estimate A2(x). For x∈ℝn, we have
(3.5)
∫
ℝ
n
∖
B
a
(
0
)
|
u
(
y
)
|
p
1
|
x
-
y
|
n
-
α
|
y
|
t
𝑑
y
≤
1
a
t
∫
B
a
(
x
)
|
u
(
y
)
|
p
1
|
x
-
y
|
n
-
α
𝑑
y
+
∫
(
ℝ
n
∖
B
a
(
0
)
)
∩
(
ℝ
n
∖
B
a
(
x
)
)
|
u
(
y
)
|
p
1
|
x
-
y
|
n
-
α
|
y
|
t
𝑑
y
.
In virtue of the Hölder inequality, it is easy to see that
(3.6)
1
a
t
∫
B
a
(
x
)
|
u
(
y
)
|
p
1
|
x
-
y
|
n
-
α
𝑑
y
<
∞
.
We also have
(3.7)
∫
(
ℝ
n
∖
B
a
(
0
)
)
∩
(
ℝ
n
∖
B
a
(
x
)
)
|
u
(
y
)
|
p
1
|
x
-
y
|
n
-
α
|
y
|
t
𝑑
y
<
∞
since
∫
(
ℝ
n
∖
B
a
(
0
)
)
∩
(
ℝ
n
∖
B
a
(
x
)
)
|
u
(
y
)
|
p
1
|
x
-
y
|
n
-
α
|
y
|
t
𝑑
y
≤
∫
ℝ
n
∖
B
a
(
0
)
|
u
(
y
)
|
p
1
|
y
|
n
-
α
+
t
𝑑
y
+
∫
ℝ
n
∖
B
a
(
x
)
|
u
(
y
)
|
p
1
|
x
-
y
|
n
-
α
+
t
𝑑
y
.
Combining (3.5), (3.6) and (3.7), we deduce that
(3.8)
A
2
(
x
)
=
∫
ℝ
n
∖
B
a
(
0
)
|
u
(
y
)
|
p
1
|
x
-
y
|
n
-
α
|
y
|
t
𝑑
y
<
∞
.
Therefore, from (3.1), (3.4) and (3.8), we can derive that A(x)∈L∞(ℝn). Similarly, we can also show that
B
(
x
)
∈
L
∞
(
ℝ
n
)
and
C
(
x
)
∈
L
∞
(
ℝ
n
)
.
Thus we accomplish the estimate of Part I.
Part II. We show that u(x),v(x)∈C∞(ℝn∖{0}). One only need to prove that |x|τu(x)∈C∞(ℝn∖{0}). For any x∈ℝn∖{0}, we choose a ball B3r(x) with radius 3r such that {0}∩B¯3r(x)=∅, then
|
x
|
τ
u
(
x
)
=
∫
ℝ
n
λ
1
u
p
1
(
y
)
|
x
-
y
|
n
-
α
|
y
|
t
𝑑
y
+
∫
ℝ
n
μ
1
v
q
1
(
y
)
|
x
-
y
|
n
-
α
|
y
|
t
𝑑
y
+
∫
ℝ
n
γ
1
u
α
1
(
y
)
v
β
1
(
y
)
|
x
-
y
|
n
-
α
|
y
|
t
𝑑
y
=
:
λ
1
A
(
x
)
+
μ
1
B
(
x
)
+
γ
1
C
(
x
)
.
We only show that A(x), C(x)∈C∞(ℝn∖{0}). We split A(x) into two parts, i.e. A(x) can be written as
(3.9)
A
(
x
)
=
∫
ℝ
n
∖
B
3
r
(
x
)
λ
1
u
p
1
(
y
)
|
x
-
y
|
n
-
α
|
y
|
t
𝑑
y
+
∫
B
3
r
(
x
)
λ
1
u
p
1
(
y
)
|
x
-
y
|
n
-
α
|
y
|
t
𝑑
y
.
We show that
P
(
x
)
=
∫
ℝ
n
∖
B
3
r
(
x
)
u
p
1
(
y
)
|
x
-
y
|
n
-
α
|
y
|
t
𝑑
y
∈
C
∞
(
ℝ
n
∖
{
0
}
)
.
Let
Q
(
x
,
y
)
=
u
p
1
(
y
)
|
x
-
y
|
n
-
α
|
y
|
t
χ
ℝ
n
∖
B
3
r
(
x
)
.
For fixed x∈ℝn, if h is small enough, we have
|
Q
(
x
+
h
e
i
,
y
)
-
Q
(
x
,
y
)
h
|
=
|
u
p
1
(
y
)
|
y
|
t
(
χ
ℝ
n
∖
B
3
r
(
x
+
h
e
i
)
|
x
+
h
e
i
-
y
|
n
-
α
-
χ
ℝ
n
∖
B
3
r
(
x
)
|
x
-
y
|
n
-
α
)
h
|
≤
C
|
u
(
y
)
|
p
1
χ
ℝ
n
∖
B
3
r
(
x
+
θ
¯
h
e
i
)
|
y
|
t
|
x
+
θ
¯
h
e
i
-
y
|
n
-
α
+
1
(3.10)
≤
C
|
u
(
y
)
|
p
1
|
y
|
t
r
n
-
α
+
1
χ
B
ϵ
(
0
)
+
C
|
u
(
y
)
|
p
1
|
y
|
t
|
x
-
y
|
n
-
α
+
1
χ
ℝ
n
∖
(
B
r
(
x
)
∪
B
ϵ
(
0
)
)
,
where ei={0,…,1,…,0} is the i-th unit vector, 0<θ¯<1 and ϵ is sufficiently small such that B3r(x)∩Bϵ(0)=∅. For such fixed r and ϵ, it is easy to verify
(3.11)
∫
B
ϵ
(
0
)
|
u
(
y
)
|
p
1
|
y
|
t
r
n
-
α
+
1
<
∞
since u∈Ls(ℝn) for any s>nn-α+t+τ. Consider that
∫
ℝ
n
∖
(
B
r
(
x
)
∪
B
ϵ
(
0
)
)
|
u
(
y
)
|
p
1
|
y
|
t
|
x
-
y
|
n
-
α
+
1
𝑑
y
≤
∫
ℝ
n
∖
B
r
(
x
)
|
u
(
y
)
|
p
1
|
x
-
y
|
n
-
α
+
1
+
t
𝑑
y
+
∫
ℝ
n
∖
B
ϵ
(
0
)
|
u
(
y
)
|
p
1
|
y
|
n
-
α
+
1
+
t
𝑑
y
,
and u∈Ls(ℝn) for any s>nn-α+t+τ. By the Hölder inequality, we have
∫
ℝ
n
∖
B
r
(
x
)
|
u
(
y
)
|
p
1
|
x
-
y
|
n
-
α
+
1
+
t
𝑑
y
<
∞
and
∫
ℝ
n
∖
B
ϵ
(
0
)
|
u
(
y
)
|
p
1
|
y
|
n
-
α
+
1
+
t
𝑑
y
<
∞
.
Thus, we conclude that
(3.12)
∫
ℝ
n
∖
(
B
r
(
x
)
∪
B
ϵ
(
0
)
)
|
u
(
y
)
|
p
1
|
y
|
t
|
x
-
y
|
n
-
α
+
1
𝑑
y
<
∞
.
Combining (3.10), (3.11), (3.12) and Lebesgue’s dominated convergence theorem, we obtain that P(x)∈C1(ℝn∖{0}). Continuing this process, we can derive that
(3.13)
P
(
x
)
∈
C
∞
(
ℝ
n
∖
{
0
}
)
.
By the standard singular integral estimate (see [11]), for any σ<α, we have
(3.14)
∫
B
2
r
(
x
)
∪
B
ϵ
(
0
)
|
u
(
y
)
|
p
1
|
y
|
t
|
x
-
y
|
n
-
α
𝑑
y
∈
C
σ
(
ℝ
n
∖
{
0
}
)
.
Combining (3.9), (3.13) and (3.14), we obtain that A(x)∈Cσ(ℝn∖{0}). By the bootstrap technique, we can prove that A(x)∈C∞(ℝn∖{0}).
For the estimate of C(x), similarly, we also can split C(x) into two parts, i.e. C(x) can be written as
(3.15)
C
(
x
)
=
∫
ℝ
n
∖
B
3
r
(
x
)
μ
1
u
α
1
(
y
)
v
β
1
(
y
)
|
x
-
y
|
n
-
α
|
y
|
t
𝑑
y
+
∫
B
3
r
(
x
)
μ
1
u
α
1
(
y
)
v
β
1
(
y
)
|
x
-
y
|
n
-
α
|
y
|
t
𝑑
y
.
We show that
P
*
(
x
)
=
∫
ℝ
n
∖
B
3
r
(
x
)
u
α
1
(
y
)
v
β
1
(
y
)
|
x
-
y
|
n
-
α
|
y
|
t
𝑑
y
∈
C
∞
(
ℝ
n
∖
{
0
}
)
.
Let
Q
*
(
x
,
y
)
=
u
α
1
(
y
)
v
β
1
(
y
)
|
x
-
y
|
n
-
α
|
y
|
t
χ
ℝ
n
∖
B
3
r
(
x
)
.
For fixed x∈ℝn, if h is small enough, we use the same estimate as (3.10) to obtain
|
Q
*
(
x
+
h
e
i
,
y
)
-
Q
*
(
x
,
y
)
h
|
<
C
{
|
u
(
y
)
|
α
1
|
v
(
y
)
|
β
1
|
y
|
t
r
n
-
α
+
1
χ
B
ϵ
(
0
)
+
|
u
(
y
)
|
α
1
|
v
(
y
)
|
β
1
|
y
|
t
|
x
-
y
|
n
-
α
+
1
χ
ℝ
n
∖
(
B
r
(
x
)
∪
B
ϵ
(
0
)
)
}
.
For such fixed r and ϵ, it is easy to verify that
∫
B
ϵ
(
0
)
|
u
(
y
)
|
α
1
|
v
(
y
)
|
β
1
|
y
|
t
r
n
-
α
+
1
<
∞
.
Note that
(3.16)
∫
ℝ
n
∖
(
B
r
(
x
)
∪
B
ϵ
(
0
)
)
|
u
(
y
)
|
α
1
|
v
(
y
)
|
β
1
|
y
|
t
|
x
-
y
|
n
-
α
+
1
𝑑
y
≤
∫
ℝ
n
∖
B
r
(
x
)
|
u
(
y
)
|
α
1
|
v
(
y
)
|
β
1
|
x
-
y
|
n
-
α
+
1
+
t
𝑑
y
+
∫
ℝ
n
∖
B
ϵ
(
0
)
|
u
(
y
)
|
α
1
|
v
(
y
)
|
β
1
|
y
|
n
-
α
+
1
+
t
𝑑
y
and u,v∈Ls(ℝn) for any s>nn-α+t+τ. We have
∫
ℝ
n
∖
B
r
(
x
)
|
u
(
y
)
|
α
1
|
v
(
y
)
|
β
1
|
x
-
y
|
n
-
α
+
t
+
1
𝑑
y
≤
(
∫
ℝ
n
∖
B
r
(
x
)
(
|
u
(
y
)
|
α
1
|
v
(
y
)
|
β
1
)
m
1
𝑑
y
)
1
/
m
1
(
∫
ℝ
n
∖
B
r
(
x
)
1
|
x
-
y
|
m
2
(
n
-
α
+
t
+
1
)
𝑑
y
)
1
/
m
2
(3.17)
≤
∥
u
∥
L
m
1
(
α
1
+
β
1
)
α
1
∥
v
∥
L
m
1
(
α
1
+
β
1
)
β
1
(
∫
ℝ
n
∖
B
r
(
x
)
1
|
x
-
y
|
m
2
(
n
-
α
+
t
+
1
)
𝑑
y
)
1
/
m
2
<
∞
,
where 1m1+1m2=1.
Similarly, we can also obtain
(3.18)
∫
ℝ
n
∖
B
ϵ
(
0
)
|
u
(
y
)
|
α
1
|
v
(
y
)
|
β
1
|
y
|
n
-
α
+
1
+
t
𝑑
y
<
∞
.
Combining (3.16), (3.17) and (3.18), we can derive that
∫
ℝ
n
∖
(
B
r
(
x
)
∪
B
ϵ
(
0
)
)
|
u
(
y
)
|
α
1
|
v
(
y
)
|
β
1
|
y
|
t
|
x
-
y
|
n
-
α
+
1
𝑑
y
<
∞
.
By Lebesgue’s dominated convergence theorem, we conclude that
(3.19)
P
*
(
x
)
∈
C
1
(
ℝ
n
∖
{
0
}
)
.
Combining (3.15), (3.19) and the standard singular integral estimate (see [11]), for any σ<α, we have
∫
B
2
r
(
x
)
∪
B
ϵ
(
0
)
|
u
(
y
)
|
α
1
|
v
(
y
)
|
β
1
|
y
|
t
|
x
-
y
|
n
-
α
𝑑
y
∈
C
σ
(
ℝ
n
∖
{
0
}
)
.
Thus we derive that C(x)∈Cσ(ℝn∖{0}). By the bootstrap technique, we can prove that C(x)∈C∞(ℝn∖{0}). Then the proof of Part II is accomplished.
4 Proof of Theorem 1.3
In this section, we will use the method of moving plane in integral forms introduced in [6] to prove that each pair (u,v) of solutions of (1.7) is radially symmetric and strictly decreasing about the origin. In order to prove our theorem, we first introduce some notations. For any real number λ, let the moving plane be Tλ={x∣x1=λ} and denote Σλ={x=(x1,x2,…,xn)∣x1≥λ}. Let xλ=(2λ-x1,x2,…,xn) be the reflection of the point x about the moving plane Tλ, and define uλ(x)=u(xλ) and vλ(x)=v(xλ).
Lemma 4.1
If (u,v) is a pair of nonnegative solutions of (1.7), for any x∈Σλ and x≠y, we have
u
(
x
)
-
u
(
x
λ
)
=
∫
Σ
λ
(
1
|
x
λ
|
τ
|
x
λ
-
y
|
n
-
α
-
1
|
x
|
τ
|
x
-
y
|
n
-
α
)
(
1
|
y
λ
|
t
-
1
|
y
|
t
)
f
1
(
u
(
y
)
,
v
(
y
)
)
𝑑
y
+
∫
Σ
λ
(
1
|
x
|
τ
-
1
|
x
λ
|
τ
)
(
1
|
x
-
y
|
n
-
α
|
y
λ
|
t
+
1
|
x
λ
-
y
|
n
-
α
|
y
λ
|
t
)
f
1
(
u
λ
(
y
)
,
v
λ
(
y
)
)
𝑑
y
+
∫
Σ
λ
(
1
|
x
λ
|
τ
|
x
λ
-
y
|
n
-
α
|
y
λ
|
t
-
1
|
x
|
τ
|
x
-
y
|
n
-
α
|
y
λ
|
t
)
⋅
(
f
1
(
u
λ
(
y
)
,
v
λ
(
y
)
)
-
f
1
(
u
(
y
)
,
v
(
y
)
)
d
y
and
v
(
x
)
-
v
(
x
λ
)
=
∫
Σ
λ
(
1
|
x
λ
|
t
|
x
λ
-
y
|
n
-
α
-
1
|
x
|
t
|
x
-
y
|
n
-
α
)
(
1
|
y
λ
|
τ
-
1
|
y
|
τ
)
f
2
(
u
(
y
)
,
v
(
y
)
)
𝑑
y
+
∫
Σ
λ
(
1
|
x
|
t
-
1
|
x
λ
|
t
)
(
1
|
x
-
y
|
n
-
α
|
y
λ
|
τ
+
1
|
x
λ
-
y
|
n
-
α
|
y
λ
|
τ
)
f
2
(
u
λ
(
y
)
,
v
λ
(
y
)
)
𝑑
y
+
∫
Σ
λ
(
1
|
x
λ
|
t
|
x
λ
-
y
|
n
-
α
|
y
λ
|
τ
-
1
|
x
|
t
|
x
-
y
|
n
-
α
|
y
λ
|
τ
)
⋅
(
f
2
(
u
λ
(
y
)
,
v
λ
(
y
)
)
-
f
2
(
u
(
y
)
,
v
(
y
)
)
d
y
.
Proof.
By (1.7), we have
u
(
x
)
=
∫
Σ
λ
1
|
x
|
τ
|
x
-
y
|
n
-
α
|
y
|
t
f
1
(
u
(
y
)
,
v
(
y
)
)
𝑑
y
+
∫
Σ
λ
1
|
x
|
τ
|
x
-
y
λ
|
n
-
α
|
y
λ
|
t
f
1
(
u
λ
(
y
)
,
v
λ
(
y
)
)
𝑑
y
and
u
(
x
λ
)
=
∫
Σ
λ
1
|
x
λ
|
τ
|
x
λ
-
y
|
n
-
α
|
y
|
t
f
1
(
u
(
y
)
,
v
(
y
)
)
𝑑
y
+
∫
Σ
λ
1
|
x
λ
|
τ
|
x
λ
-
y
λ
|
n
-
α
|
y
λ
|
t
f
1
(
u
λ
(
y
)
,
v
λ
(
y
)
)
𝑑
y
.
Since |xλ-y|=|x-yλ| and |xλ-yλ|=|x-y|, we have
u
(
x
)
-
u
(
x
λ
)
=
∫
Σ
λ
(
1
|
x
|
τ
|
x
-
y
|
n
-
α
|
y
|
t
-
1
|
x
λ
|
τ
|
x
λ
-
y
|
n
-
α
|
y
|
t
)
f
1
(
u
(
y
)
,
v
(
y
)
)
𝑑
y
+
∫
Σ
λ
(
1
|
x
|
τ
|
x
-
y
λ
|
n
-
α
|
y
λ
|
t
-
1
|
x
λ
|
τ
|
x
λ
-
y
λ
|
n
-
α
|
y
λ
|
t
)
f
1
(
u
λ
(
y
)
,
v
λ
(
y
)
)
𝑑
y
=
∫
Σ
λ
(
1
|
x
λ
|
τ
|
x
λ
-
y
|
n
-
α
-
1
|
x
|
τ
|
x
-
y
|
n
-
α
)
(
1
|
y
λ
|
t
-
1
|
y
|
t
)
f
1
(
u
(
y
)
,
v
(
y
)
)
𝑑
y
+
∫
Σ
λ
(
1
|
x
|
τ
-
1
|
x
λ
|
τ
)
(
1
|
x
-
y
|
n
-
α
|
y
λ
|
t
+
1
|
x
λ
-
y
|
n
-
α
|
y
λ
|
t
)
f
1
(
u
λ
(
y
)
,
v
λ
(
y
)
)
𝑑
y
+
∫
Σ
λ
(
1
|
x
λ
|
τ
|
x
λ
-
y
|
n
-
α
|
y
λ
|
t
-
1
|
x
|
τ
|
x
-
y
|
n
-
α
|
y
λ
|
t
)
⋅
(
f
1
(
u
λ
(
y
)
,
v
λ
(
y
)
)
-
f
1
(
u
(
y
)
,
v
(
y
)
)
d
y
.
Similarly, we can also obtain the second formula. This completes the proof of Lemma 4.1. ∎
Now we start to prove Theorem 1.3, the proof will be carried out by two steps.
Step 1. We compare the values of uλ(x) and u(x), vλ(x) and v(x). For λ sufficiently negative, we are going to show that
(4.1)
u
λ
(
x
)
-
u
(
x
)
≤
0
,
v
λ
(
x
)
-
v
(
x
)
≤
0
for all
x
∈
Σ
λ
.
Define
Σ
λ
u
=
{
x
∈
Σ
λ
∣
u
λ
(
x
)
-
u
(
x
)
>
0
}
,
Σ
λ
v
=
{
x
∈
Σ
λ
∣
v
λ
(
x
)
-
v
(
x
)
>
0
}
.
We will prove that for sufficiently negative λ, both Σλu and Σλv must be empty, and hence (4.1) holds.
According to the mean value theorem and Lemma 4.1, for any x∈Σλu, we have
u
(
x
λ
)
-
u
(
x
)
≤
∫
Σ
λ
(
1
|
x
|
τ
|
x
-
y
|
n
-
α
|
y
λ
|
t
-
1
|
x
λ
|
τ
|
x
λ
-
y
|
n
-
α
|
y
λ
|
t
)
⋅
(
f
1
(
u
λ
(
y
)
,
v
λ
(
y
)
)
-
f
1
(
u
(
y
)
,
v
(
y
)
)
)
𝑑
y
≤
λ
1
p
1
∫
Σ
λ
u
u
λ
p
1
-
1
(
y
)
(
u
λ
(
y
)
-
u
(
y
)
)
|
x
|
τ
|
x
-
y
|
n
-
α
|
y
|
t
𝑑
y
+
μ
1
q
1
∫
Σ
λ
v
v
λ
q
1
-
1
(
y
)
(
v
λ
(
y
)
-
v
(
y
)
)
|
x
|
τ
|
x
-
y
|
n
-
α
|
y
|
t
𝑑
y
+
γ
1
α
1
∫
Σ
λ
u
u
λ
α
1
-
1
(
y
)
v
λ
β
1
(
y
)
(
u
λ
(
y
)
-
u
(
y
)
)
|
x
|
τ
|
x
-
y
|
n
-
α
|
y
|
t
𝑑
y
+
γ
1
β
1
∫
Σ
λ
v
u
α
1
(
y
)
v
λ
β
1
-
1
(
y
)
(
v
λ
(
y
)
-
v
(
y
)
)
|
x
|
τ
|
x
-
y
|
n
-
α
|
y
|
t
𝑑
y
=
:
A
1
(
x
)
+
A
2
(
x
)
+
A
3
(
x
)
+
A
4
(
x
)
.
For any s>nn-α+t+τ, we apply the weighted Hardy–Littlewood–Sobolev inequality and the Hölder inequality to the above inequality and obtain
∥
A
1
∥
L
s
(
Σ
λ
u
)
≤
C
∥
u
λ
p
1
-
1
(
u
λ
-
u
)
∥
L
n
s
n
+
(
α
-
t
-
τ
)
s
(
Σ
λ
u
)
∥
A
2
∥
L
s
(
Σ
λ
v
)
≤
C
∥
v
λ
q
1
-
1
(
v
λ
-
v
)
∥
L
n
s
n
+
(
α
-
t
-
τ
)
s
(
Σ
λ
v
)
(4.2)
≤
C
∥
u
λ
∥
L
t
11
(
Σ
λ
u
)
p
1
-
1
∥
u
λ
-
u
∥
L
s
(
Σ
λ
u
)
,
≤
C
∥
u
λ
α
1
-
1
v
λ
β
1
∥
L
n
α
-
t
-
τ
(
Σ
λ
u
)
∥
u
λ
-
u
∥
L
s
(
Σ
λ
u
)
∥
A
3
∥
L
s
(
Σ
λ
u
)
≤
C
∥
u
λ
α
1
-
1
v
λ
β
1
(
u
λ
-
u
)
∥
L
n
s
n
+
(
α
-
t
-
τ
)
s
(
Σ
λ
u
)
(4.3)
≤
C
∥
v
λ
∥
L
t
21
(
Σ
λ
v
)
q
1
-
1
∥
v
λ
-
v
∥
L
s
(
Σ
λ
v
)
,
(4.4)
≤
C
∥
u
λ
∥
L
k
0
(
Σ
λ
u
)
α
1
-
1
∥
v
λ
∥
L
k
0
(
Σ
λ
u
)
β
1
∥
u
λ
-
u
∥
L
s
(
Σ
λ
u
)
and
∥
A
4
∥
L
s
(
Σ
λ
u
)
≤
C
∥
u
α
1
v
λ
β
1
-
1
(
v
λ
-
v
)
∥
L
n
s
n
+
(
α
-
t
-
τ
)
s
(
Σ
λ
v
)
≤
C
∥
u
α
1
v
λ
β
1
-
1
∥
L
n
α
-
t
-
τ
(
Σ
λ
v
)
∥
v
λ
-
v
∥
L
s
(
Σ
λ
v
)
(4.5)
≤
C
∥
u
∥
L
k
0
(
Σ
λ
v
)
α
1
∥
v
λ
∥
L
k
0
(
Σ
λ
v
)
β
1
-
1
∥
v
λ
-
v
∥
L
s
(
Σ
λ
v
)
,
where t11=nα-t-τ(p1-1), t21=nα-t-τ(q1-1) and k0=nα-t-τ(α1+β1-1).
Combining (4.2), (4.3), (4.4) and (4.5), we obtain
∥
u
λ
-
u
∥
L
s
(
Σ
λ
u
)
≤
C
∥
u
λ
∥
L
t
11
(
Σ
λ
u
)
p
1
-
1
∥
u
λ
-
u
∥
L
s
(
Σ
λ
u
)
+
C
∥
v
λ
∥
L
t
21
(
Σ
λ
v
)
q
1
-
1
∥
v
λ
-
v
∥
L
s
(
Σ
λ
v
)
+
C
∥
u
λ
∥
L
k
0
(
Σ
λ
u
)
α
1
-
1
∥
v
λ
∥
L
k
0
(
Σ
λ
u
)
β
1
∥
u
λ
-
u
∥
L
s
(
Σ
λ
u
)
+
C
∥
u
∥
L
k
0
(
Σ
λ
v
)
α
1
∥
v
λ
∥
L
k
0
(
Σ
λ
v
)
β
1
-
1
∥
v
λ
-
v
∥
L
s
(
Σ
λ
v
)
≤
C
(
∥
u
λ
∥
L
t
11
(
Σ
λ
u
)
p
1
-
1
+
∥
u
λ
∥
L
k
0
(
Σ
λ
u
)
α
1
-
1
∥
v
λ
∥
L
k
0
(
Σ
λ
u
)
β
1
)
∥
u
λ
-
u
∥
L
s
(
Σ
λ
u
)
(4.6)
+
C
(
∥
v
λ
∥
L
t
21
(
Σ
λ
v
)
q
1
-
1
+
∥
u
∥
L
k
0
(
Σ
λ
v
)
α
1
∥
v
λ
∥
L
k
0
(
Σ
λ
v
)
β
1
-
1
)
∥
v
λ
-
v
∥
L
s
(
Σ
λ
v
)
.
Similarly, we can also deduce that
∥
v
λ
-
v
∥
L
s
(
Σ
λ
v
)
≤
C
(
∥
u
λ
∥
L
t
12
(
Σ
λ
u
)
p
2
-
1
+
∥
u
λ
∥
L
k
0
(
Σ
λ
u
)
α
2
-
1
∥
v
λ
∥
L
k
0
(
Σ
λ
u
)
β
2
)
∥
u
λ
-
u
∥
L
s
(
Σ
λ
u
)
(4.7)
+
C
(
∥
v
λ
∥
L
t
22
(
Σ
λ
v
)
q
2
-
1
+
∥
u
∥
L
k
0
(
Σ
λ
v
)
α
2
∥
v
λ
∥
L
k
0
(
Σ
λ
v
)
β
2
-
1
)
∥
v
λ
-
v
∥
L
s
(
Σ
λ
v
)
,
where t12=nα-t-τ(p2-1), t22=nα-t-τ(q2-1) and k0=nα-t-τ(α2+β2-1).
In virtue of the condition (u,v)∈Π1×Π2, we can choose sufficiently negative λ such that
(4.8)
{
C
∥
u
λ
∥
L
t
11
(
Σ
λ
u
)
p
1
-
1
+
C
∥
u
λ
∥
L
k
0
(
Σ
λ
u
)
α
1
-
1
∥
v
λ
∥
L
k
0
(
Σ
λ
u
)
β
1
<
1
4
,
C
∥
v
λ
∥
L
t
21
(
Σ
λ
v
)
q
1
-
1
+
C
∥
u
∥
L
k
0
(
Σ
λ
v
)
α
1
∥
v
λ
∥
L
k
0
(
Σ
λ
v
)
β
1
-
1
<
1
4
,
C
∥
u
λ
∥
L
t
12
(
Σ
λ
u
)
p
2
-
1
+
C
∥
u
λ
∥
L
k
0
(
Σ
λ
u
)
α
2
-
1
∥
v
λ
∥
L
k
0
(
Σ
λ
u
)
β
2
<
1
4
,
C
∥
v
λ
∥
L
t
22
(
Σ
λ
v
)
q
2
-
1
+
C
∥
u
∥
L
k
0
(
Σ
λ
v
)
α
2
∥
v
λ
∥
L
k
0
(
Σ
λ
v
)
β
2
-
1
<
1
4
.
By (4.6), (4.7) and (4.8), we have
∥
u
λ
-
u
∥
L
s
(
Σ
λ
u
)
=
0
,
∥
v
λ
-
v
∥
L
s
(
Σ
λ
v
)
=
0
,
which implies that Σλu and Σλv must be of measure zero. Therefore, both Σλu and Σλv must be empty sets.
Step 2. Inequality (4.1) provides a starting point to move the plane Tλ={x∈ℝn∣x1=λ}. Now we start from the negative infinity of the x1-axis and move the plane to the right as long as (4.1) holds. Define
λ
0
=
sup
{
λ
∣
u
μ
(
x
)
≤
u
(
x
)
,
v
μ
(
x
)
≤
v
(
x
)
,
μ
≤
λ
for all
x
∈
Σ
μ
\
{
0
}
}
.
We will prove
λ
0
=
0
.
Suppose on the contrary that λ0<0. We will then show that u and v must be symmetric about the plane Tλ0, that is,
(4.9)
u
λ
0
(
x
)
≡
u
(
x
)
,
v
λ
0
(
x
)
≡
v
(
x
)
for all
x
∈
Σ
λ
0
.
Otherwise, we may assume on Σλ0,
u
(
x
)
≥
u
λ
0
(
x
)
,
v
(
x
)
≥
v
λ
0
(
x
)
,
but
u
λ
0
(
x
)
≢
u
(
x
)
or
v
λ
0
(
x
)
≢
v
(
x
)
.
In the case of uλ0(x)≢u(x) on Σλ0, by Lemma 4.1, we can obtain vλ0(x)≢v(x). Furthermore, in virtue of Lemma 4.1, we can derive u(x)>uλ0(x) and v(x)>vλ0(x) in the interior of Σλ0.
Next, we will show that the plane can be moved further to the right. More precisely, there exists an ε>0 such that for any λ∈[λ0,λ0+ε),
u
(
x
)
≥
u
λ
(
x
)
,
v
(
x
)
≥
v
λ
(
x
)
for all
x
∈
Σ
λ
\
{
0
}
.
Let
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
)
}
.
Obviously, both Σλ0u¯ and Σλ0v¯ have measure zero, and
lim
λ
→
λ
0
Σ
λ
u
⊂
Σ
λ
0
u
¯
,
lim
λ
→
λ
0
Σ
λ
v
⊂
Σ
λ
0
v
¯
.
This together with the integrability condition (u,v)∈Π1×Π2 ensures that one can choose ε small enough such that, for all λ∈[λ0,λ0+ε),
C
∥
u
λ
∥
L
t
11
(
Σ
λ
u
)
p
1
-
1
+
C
∥
u
λ
∥
L
k
0
(
Σ
λ
u
)
α
1
-
1
∥
v
λ
∥
L
k
0
(
Σ
λ
u
)
β
1
<
1
4
,
C
∥
v
λ
∥
L
t
21
(
Σ
λ
v
)
q
1
-
1
+
C
∥
u
∥
L
k
0
(
Σ
λ
v
)
α
1
∥
v
λ
∥
L
k
0
(
Σ
λ
v
)
β
1
-
1
<
1
4
,
C
∥
u
λ
∥
L
t
12
(
Σ
λ
u
)
p
2
-
1
+
C
∥
u
λ
∥
L
k
0
(
Σ
λ
u
)
α
2
-
1
∥
v
λ
∥
L
k
0
(
Σ
λ
u
)
β
2
<
1
4
,
C
∥
v
λ
∥
L
t
22
(
Σ
λ
v
)
q
2
-
1
+
C
∥
u
∥
L
k
0
(
Σ
λ
v
)
α
2
∥
v
λ
∥
L
k
0
(
Σ
λ
v
)
β
2
-
1
<
1
4
.
Using the similar estimates as (4.6) and (4.7), we can derive that
∥
u
-
u
λ
∥
L
s
(
Σ
λ
u
)
=
|
v
-
v
λ
∥
L
s
(
Σ
λ
v
)
=
0
,
thus Σλu and Σλv must be of measure zero, which verifies (4.9). Finally, we show that the plane cannot stop before hitting the origin. On the contrary, if the plane stops at x1=λ0<0, then u(x) and v(x) must be symmetric about the plane x1=λ0, that is
(4.10)
u
(
x
)
=
u
λ
0
(
x
)
,
v
(
x
)
=
v
λ
0
(
x
)
for all
x
∈
Σ
λ
0
.
In virtue of |xλ0-y|>|x-y|, |x-yλ0|>|x-y|, |xλ0|>|x| and |yλ0|>|y|, we have
u
(
x
)
-
u
(
x
λ
0
)
=
∫
Σ
λ
0
(
1
|
x
|
τ
|
x
-
y
|
n
-
α
|
y
|
t
-
1
|
x
λ
0
|
τ
|
x
λ
0
-
y
|
n
-
α
|
y
|
t
)
f
1
(
u
(
y
)
,
v
(
y
)
)
𝑑
y
+
∫
Σ
λ
0
(
1
|
x
|
τ
|
x
-
y
λ
0
|
n
-
α
|
y
λ
0
|
β
-
1
|
x
λ
0
|
τ
|
x
-
y
|
n
-
α
|
y
λ
0
|
t
)
f
1
(
u
λ
0
(
y
)
,
v
λ
0
(
y
)
)
𝑑
y
>
∫
Σ
λ
0
1
|
x
|
τ
(
1
|
x
-
y
|
n
-
α
|
y
|
t
-
1
|
x
λ
0
-
y
|
n
-
α
|
y
|
t
)
f
1
(
u
(
y
)
,
v
(
y
)
)
𝑑
y
+
∫
Σ
λ
0
1
|
x
|
τ
(
1
|
x
-
y
λ
0
|
n
-
α
|
y
λ
0
|
t
-
1
|
x
-
y
|
n
-
α
|
y
λ
0
|
t
)
f
1
(
u
λ
0
(
y
)
,
v
λ
0
(
y
)
)
𝑑
y
=
∫
Σ
λ
0
1
|
x
|
τ
(
1
|
x
-
y
|
n
-
α
-
1
|
x
λ
0
-
y
|
n
-
α
)
(
1
|
y
|
t
-
1
|
y
λ
0
|
t
)
f
1
(
u
(
y
)
,
v
(
y
)
)
𝑑
y
>
0
,
which contradicts (4.10). Since the x1 direction can be chosen arbitrary, we conclude that both u(x) and v(x) are radially symmetric and strictly decreasing about the origin. This completes the proof of Theorem 1.3.
