1 Introduction
In this paper, we are concerned with the following Hénon-type problem with critical growth:
(1.1)
{
-
Δ
u
=
K
(
|
y
′
|
,
y
′′
)
u
2
*
-
1
,
u
>
0
in
B
1
,
u
=
0
on
∂
B
1
,
where 2*=2NN-2, N≥5, B1 is the unit sphere in ℝN, y=(y′,y′′)∈ℝ2×ℝN-2, r=|y′| and K(y)=K(r,y′′)∈C2(B1) is a bounded non-negative function.
Problem (1.1) is related to the following famous Hénon equation:
(1.2)
{
-
Δ
u
=
|
y
|
α
u
q
,
u
>
0
in
B
1
,
u
=
0
on
∂
B
1
,
where α>0 is a positive constant. Problem (1.2) was first introduced by Hénon in the study of astrophysics and it has attracted lots of interest in recent years. Ni [23] observed that the non-autonomous term |y|α changes the global homogeneity of the equation and proved that it possesses a positive radial solution when q∈(1,N+2+2αN-2). It is natural to ask whether (1.2) has a non-radial solution. The existence of a non-radial solution for 1<q<N+2N-2 was obtained by Smets, Willem and Su [30] provided α is large enough. When q=N+2N-2-ε, Cao and Peng [4] showed that the mountain-pass solution for (1.2) is non-radial and blows up near the boundary of B1(0) as ε→0. For the purely critical case q=N+2N-2, Serra [29] proved that (1.2) has at least one non-radial solution provided that N≥4 and α>0 is sufficiently large. Later then, Wei and Yan [34] showed that there are infinitely many nonradial solutions for (1.2) with N≥4 and any α>0. This result was later extended to the polyharmonic case by Guo, Li and Li [15]. For other related results, we refer the readers to [2, 3, 5, 7, 10, 11, 12, 13, 19, 18, 20, 22, 27] and the references therein.
The aim of the present paper is to show that under the weaker symmetrical assumption on K(y), still we can construct infinitely many non-radial solutions for problem (1.1). More precisely, we assume that K(y)∈C2(B1) is non-negative bounded and satisfies the following conditions:
There exists y0=(r0,y0′′)∈∂B1 satisfying 0<r0<1, 0<|y0i′′|<1, i=3,…,N and
K
(
r
0
,
y
0
′′
)
=
1
,
∇
K
(
r
0
,
y
0
′′
)
=
0
,
and
deg
(
∇
(
K
(
r
,
y
′′
)
)
,
Ω
(
r
0
,
y
0
′′
)
)
≠
0
.
K
(
r
,
y
′′
)
∈
C
3
(
B
θ
(
r
0
,
y
0
′′
)
)
and
Δ
K
(
r
0
,
y
0
′′
)
:=
∂
2
K
(
r
0
,
y
0
′′
)
∂
r
2
+
∑
i
=
3
N
∂
2
K
(
r
0
,
y
0
′′
)
∂
y
i
2
<
0
,
where θ>0 is a small constant.
It is worth to point out that the zero Dirichlet boundary condition makes it possible to construct infinitely many solutions of (1.1), although the critical point of K(y) is on the boundary of the unit ball B1.
Before the statement of the main result, let us first introduce some notations. Recall that the family of the functions
{
U
x
,
λ
(
y
)
=
(
N
(
N
-
2
)
)
N
-
2
4
(
λ
1
+
λ
2
|
y
-
x
|
2
)
N
-
2
2
:
x
∈
ℝ
N
,
λ
>
0
}
are the only radial solutions (usually called bubbles) of the following problem:
-
Δ
u
=
u
N
+
2
N
-
2
,
u
>
0
in
ℝ
N
.
Since there is no parameter in (1.1), in order to construct the non-radial solutions for (1.1), we use the scaling parameter λ as the blow-up parameter, which is originally due to [34]. The main idea is to place a large number of bubbles near the boundary of the domain, and the scaling parameter will be determined by the number of bubbles.
We define PUx,λ as the projection of Ux,λ on B1, that is,
-
Δ
P
U
x
,
λ
=
U
x
,
λ
2
*
-
1
in
B
1
,
P
U
x
,
λ
=
0
on
∂
B
1
.
Set y=(y′,y′′)∈ℝN, y′∈ℝ2, y′′∈ℝN-2, r=|y′|. Define
H
s
=
{
u
∈
ℋ
0
1
(
B
1
)
:
u
is even in
y
h
,
h
=
2
,
3
,
…
,
N
,
u
(
r
cos
θ
,
r
sin
θ
,
y
′′
)
=
u
(
r
cos
(
θ
+
2
π
j
k
)
,
r
sin
(
θ
+
2
π
j
k
)
,
y
′′
)
}
,
and
x
j
=
(
r
¯
cos
2
(
j
-
1
)
π
k
,
r
¯
sin
2
(
j
-
1
)
π
k
,
y
¯
′′
)
,
j
=
1
,
2
,
…
,
k
,
where y¯′′ is a vector in ℝN-2.
We will use PUxj,λ as an approximate solution to construct solutions concentrating at the point (r0,y0′′). Noting that PUxj,λ decays slowly when N is not large enough. For this reason, we first cut off this function. Let δ>0 be a small constant such that K(r,y′′)>0 if (r,y′′)∈B10δ(r0,y0′′)∩B¯1. Let ξ(y)=ξ(|y′|,y′′) be a smooth function defined on B1 satisfying ξ=1 if (r,y′′)∈Bδ(r0,y0′′)∩B1, ξ=0 if (r,y′′)∈B2δc(r0,y0′′)∩B1 and 0≤ξ≤1. Denote
Z
x
j
,
λ
(
y
)
=
ξ
P
U
x
j
,
λ
,
Z
r
¯
,
y
¯
′′
,
λ
*
=
∑
j
=
1
k
P
U
x
j
,
λ
,
Z
r
¯
,
y
¯
′′
,
λ
(
y
)
=
∑
j
=
1
k
Z
x
j
,
λ
(
y
)
.
In the following of the paper, we always assume that k>0 is a large integer, λ∈[L0kN-2N-4,L1kN-2N-4] for some constants L1>L0>0 and xj∈Ω(r0,y0′′), j=1,…,k, where
Ω
(
r
0
,
y
0
′′
)
:=
{
(
r
,
y
′′
)
∈
B
1
:
|
(
r
,
y
′′
)
-
(
r
0
,
y
0
′′
)
|
≤
1
λ
2
3
+
ϑ
1
,
r
0
λ
1
-
ϑ
2
≤
d
:=
1
-
r
2
+
y
′′
2
≤
r
1
λ
1
-
ϑ
2
}
,
ϑ
1
>
0
is a small constant, 1N+22-τ<ϑ2<13 is a constant and r0≪1, r1≫1 are some positive constants.
Our main result is:
Theorem 1.1.
Suppose that K(y)∈C2(B1) is non-negative bounded and satisfies (K1) and (K2) . If N≥5, then there exists an integer k0>0 such that for any integer k>k0, (1.1) has a solution uk of the form
(1.3)
u
k
=
Z
r
¯
k
,
y
¯
k
′′
,
λ
k
+
ϕ
k
,
where ϕk∈Hs, λk∈[L0kN-2N-4,L1kN-2N-4]. Moreover, (r¯k,y¯k′′)→(r0,y0′′),λk-N-22∥ϕk∥L∞→0, as k→+∞. As a consequence, problem (1.1) has infinitely many solutions, whose energy can be made arbitrarily large.
The proof of the theorem is mainly dependent on a finite reduction argument and local Pohozaev-type identities which is introduced firstly in [26]. The main idea of reduction argument can be found in [1, 28, 34, 34]. Roughly speaking, to carry out this reduction argument, the first step is to construct a reasonably approximate solution, so that the problem can be solved modula to some given functions. The second step is to solve the corresponding finite-dimensional problem to obtain a true solution. To fulfill the second step, it is essential to obtain a good estimate for the error term in the first step. However, when one consider more complicated problems, both steps must be modified. For example, Wei and Yan [34], to deal with the large number of bubbles in the solution, the reduction procedure is carried out in a weighted space instead of the standard Sobolev space. Moreover, to deal with the case where no parameter appears in the problem, the authors used k, the number of the bubbles of the solutions, as the parameter to construct infinitely many positive bubbling solutions. This idea has been used in many problems involving infinitely many solutions, see for examples [6, 9, 14, 15, 16, 17, 21, 24, 8, 34, 31, 34, 33] and the references therein. We can see from these works that in order to obtain the solutions, a key step is to find the algebraic equations which determine the location of the bubbles. This step is deliberately avoided in the previous papers, this is because the solutions, constructed in these papers, all concentrate at a local minimum point or maximum point of some functions. Otherwise, for instance, the concentrate point is a saddle point of the function, one need to calculate the main terms carefully in the corresponding algebraic equations. However, this cannot be achieved by using the methods as in [28]. In this paper, we consider the case where (r0,y0′′) may be a saddle point of K(r,y′′), inspired by the work Peng, Wang and Yan [26], we shall use the local Pohozaev identities to find algebraic equations which determine the location of the bubbles. So that we can construct solutions concentrating at some saddle points of some functions. Different from that in [25, 26], here we want to construct the bubbling solution which concentrates near the boundary of B1, but not on the boundary, the computations are more complicated.
For the reader’s convenience, let us outline the idea of the proof more precisely. We know that the functional corresponding to (1.1) is
I
(
u
)
=
1
2
∫
B
1
|
∇
u
|
2
-
1
2
*
∫
B
1
K
(
r
,
y
′′
)
(
u
)
+
2
*
.
Roughly speaking, the problem of finding a critical point for I(u) with the form (1.3) can be reduced to that of finding a critical point of the following perturbed function:
F
(
r
¯
,
y
¯
′′
,
λ
)
:=
I
(
Z
r
¯
,
y
¯
′′
,
λ
+
ϕ
r
¯
,
y
¯
′′
,
λ
)
,
where λ∈[L0kN-2N-4,L1kN-2N-4] for some constants L1>L0>0 and (r¯,y¯′′)∈Ω(r0,y0′′). Follow the idea of [26], the novelty of the present paper is, instead of estimating the derivatives of the function F(r¯,y¯′′,λ) with respect to r¯ and y¯k, k=3,…,N directly, we turn to prove that if (r¯,y¯′′) satisfies, in a suitable closed region Dρ of (r0,y0′′), that
∫
D
ρ
(
-
Δ
u
k
-
K
(
r
,
y
′′
)
(
u
k
)
+
2
*
-
1
)
〈
y
,
∇
u
k
〉
=
0
and
∫
D
ρ
(
-
Δ
u
k
-
K
(
r
,
y
′′
)
(
u
k
)
+
2
*
-
1
)
∂
u
k
∂
y
i
=
0
,
i
=
3
,
…
,
N
,
then ∂F∂r¯=0 and ∂F∂y¯i′′=0, where uk=Zrk¯,y¯k′′,λk+ϕrk¯,y¯k′′,λk is the function obtained by the reduction argument.
Our paper is organized as follows. We will carry out the finite reduction procedure in Section 2. In Section 3, we will give the asymptotic expansions for the derivatives of the energy functional of problem (1.1). We will study the finite-dimensional problem via local Pohozaev-type identities and give the proof of Theorem 1.1 in Section 4. We put some basic estimates in the Appendix.
2 Finite-Dimensional Reduction
In this section, we will use Zr¯,y¯′′,λ as the approximation solution and consider the linearization of problem (1.1) around Zr¯,y¯′′,λ. Let
∥
u
∥
*
=
sup
y
∈
B
1
(
∑
j
=
1
k
λ
N
-
2
2
(
1
+
λ
|
y
-
x
j
|
)
N
-
2
2
+
τ
)
-
1
|
u
(
y
)
|
and
∥
f
∥
*
*
=
sup
y
∈
B
1
(
∑
j
=
1
k
λ
N
+
2
2
(
1
+
λ
|
y
-
x
j
|
)
N
+
2
2
+
τ
)
-
1
|
f
(
y
)
|
,
where τ=N-4N-2. Denote
Z
j
,
1
=
∂
Z
x
j
,
λ
∂
λ
,
Z
j
,
2
=
∂
Z
x
j
,
λ
∂
r
¯
,
Z
j
,
l
=
∂
Z
x
j
,
λ
∂
y
¯
l
′′
,
l
=
3
,
…
,
N
.
Consider the linearized problem of (1.1) around Zr¯,y¯′′,λ:
(2.1)
{
-
Δ
ϕ
-
(
2
*
-
1
)
K
(
r
,
y
′′
)
Z
r
¯
,
y
¯
′′
,
λ
2
*
-
2
ϕ
=
h
+
∑
l
=
1
N
c
l
∑
j
=
1
k
Z
x
j
,
λ
2
*
-
2
Z
j
,
l
in
B
1
,
ϕ
∈
H
s
,
∑
j
=
1
k
∫
B
1
Z
x
j
,
λ
2
*
-
2
Z
j
,
l
ϕ
=
0
,
l
=
1
,
…
,
N
,
for some real numbers cl. Then, we have:
Lemma 2.1.
Assume ϕk solves (2.1) for hk. Then ∥hk∥** goes to zero as k→+∞ implies that ∥ϕk∥* goes to zero as k→+∞.
Proof.
We proceed the proof by contradiction arguments. Assume that there exist k→+∞, r¯k→r0, y¯k′′→y0′′, λk∈[L0kN-2N-4,L1kN-2N-4] and ϕk solves problem (2.1) for h=hk,λ=λk,r¯=r¯k,y¯′′=y¯k′′ with ∥hk∥**→0 and ∥ϕk∥*≥c>0. Without loss of generality, we may assume that ∥ϕk∥*=1. For simplicity, in the following, we drop the subscript k.
We have
|
ϕ
(
y
)
|
≤
C
∫
B
1
K
(
|
z
′
|
,
z
′′
)
|
y
-
z
|
N
-
2
Z
r
¯
,
y
¯
′′
,
λ
2
*
-
2
(
z
)
|
ϕ
(
z
)
|
𝑑
z
+
C
∫
B
1
1
|
y
-
z
|
N
-
2
(
|
h
|
+
|
∑
l
=
1
N
c
l
∑
j
=
1
k
Z
x
j
,
λ
2
*
-
2
Z
j
,
l
|
)
𝑑
z
.
By Lemma A.3, it holds
∫
B
1
K
(
|
z
′
|
,
z
′′
)
|
y
-
z
|
N
-
2
Z
r
¯
,
y
¯
′′
,
λ
2
*
-
2
(
z
)
|
ϕ
(
z
)
|
𝑑
z
≤
∥
ϕ
∥
*
λ
N
-
2
2
∫
B
1
K
(
|
z
′
|
,
z
′′
)
|
y
-
z
|
N
-
2
Z
r
¯
,
y
¯
′′
,
λ
2
*
-
2
(
z
)
(
∑
j
=
1
k
1
(
1
+
λ
|
z
-
x
j
|
)
N
-
2
2
+
τ
)
𝑑
z
≤
C
∥
ϕ
∥
*
λ
N
-
2
2
∑
j
=
1
k
1
(
1
+
λ
|
y
-
x
j
|
)
N
-
2
2
+
τ
+
θ
,
where θ>0 is a small constant.
By Lemma A.2, we obtain
∫
B
1
1
|
y
-
z
|
N
-
2
|
h
(
z
)
|
𝑑
z
≤
∥
h
∥
*
*
λ
N
+
2
2
∫
B
1
1
|
y
-
z
|
N
-
2
(
∑
j
=
1
k
1
(
1
+
λ
|
z
-
x
j
|
)
N
+
2
2
+
τ
)
𝑑
z
=
∥
h
∥
*
*
λ
N
+
2
2
∑
j
=
1
k
∫
B
1
1
|
y
-
z
|
N
-
2
1
(
1
+
λ
|
z
-
x
j
|
)
N
+
2
2
+
τ
𝑑
z
≤
C
∥
h
∥
*
*
λ
N
-
2
2
∑
j
=
1
k
1
(
1
+
λ
|
y
-
x
j
|
)
N
-
2
2
+
τ
,
and
∫
B
1
1
|
y
-
z
|
N
-
2
|
∑
j
=
1
k
Z
x
j
,
λ
2
*
-
2
Z
j
,
l
|
𝑑
z
≤
C
λ
n
l
∑
j
=
1
k
∫
B
1
1
|
y
-
z
|
N
-
2
U
x
j
,
λ
2
*
-
1
𝑑
z
≤
C
λ
n
l
∑
j
=
1
k
∫
B
1
1
|
y
-
z
|
N
-
2
λ
N
+
2
2
(
1
+
λ
|
z
-
x
j
|
)
N
+
2
𝑑
z
≤
C
λ
N
-
2
2
+
n
l
∑
j
=
1
k
1
(
1
+
λ
|
y
-
x
j
|
)
N
≤
C
λ
N
-
2
2
+
n
l
∑
j
=
1
k
1
(
1
+
λ
|
y
-
x
j
|
)
N
-
2
2
+
τ
,
where n1=-1 and nj=1, j=2,…,N.
Next we will estimate cl, l=1,…,N. Multiplying (2.1) by Z1,l(l=1,…,N) and integrating in B1, we have
(2.2)
∑
h
=
1
N
c
h
∑
j
=
1
k
∫
B
1
Z
x
j
,
λ
2
*
-
2
Z
j
,
h
Z
1
,
l
=
〈
-
Δ
ϕ
-
(
2
*
-
1
)
K
(
r
,
y
′′
)
Z
r
¯
,
y
¯
′′
,
λ
2
*
-
2
ϕ
,
Z
1
,
l
〉
-
〈
h
,
Z
1
,
l
〉
.
It follows from Lemma A.1,
(2.3)
|
〈
h
,
Z
1
,
l
〉
|
≤
C
λ
n
l
∥
h
∥
*
*
∫
B
1
λ
N
(
1
+
λ
|
y
-
x
1
|
)
N
-
2
∑
j
=
1
k
1
(
1
+
λ
|
y
-
x
j
|
)
N
+
2
2
+
τ
≤
C
λ
n
l
∥
h
∥
*
*
(
C
+
C
∑
j
=
2
k
1
(
λ
|
x
j
-
x
i
|
)
τ
)
≤
C
λ
n
l
∥
h
∥
*
*
.
On the other hand, a direct computation shows that
(2.4)
|
〈
-
Δ
ϕ
-
(
2
*
-
1
)
K
(
r
,
y
′′
)
Z
r
¯
,
y
¯
′′
,
λ
2
*
-
2
ϕ
,
Z
1
,
l
〉
|
=
O
(
λ
n
l
∥
ϕ
∥
*
λ
1
+
ϵ
)
.
Combining (2.3) and (2.4), we have
(2.5)
〈
-
Δ
ϕ
-
(
2
*
-
1
)
K
(
r
,
y
′′
)
Z
r
¯
,
y
¯
′′
,
λ
2
*
-
2
ϕ
,
Z
1
,
l
〉
-
〈
h
,
Z
1
,
l
〉
=
O
(
λ
n
l
(
∥
ϕ
∥
*
λ
1
+
ϵ
+
∥
h
∥
*
*
)
)
.
By the orthogonality, we get
(2.6)
∑
j
=
1
k
〈
Z
x
j
,
λ
2
*
-
2
Z
j
,
h
,
Z
1
,
l
〉
=
(
c
0
+
o
(
1
)
)
δ
h
l
λ
2
n
l
,
for some constant c0>0.
Inserting (2.5) and (2.6) into (2.2), we conclude that
c
l
=
1
λ
n
l
(
o
(
∥
ϕ
∥
*
)
+
O
(
∥
h
∥
*
*
)
)
.
Thus,
(2.7)
∥
ϕ
∥
*
≤
(
o
(
1
)
+
∥
h
k
∥
*
*
+
∑
j
=
1
k
1
(
1
+
λ
|
y
-
x
j
|
)
N
-
2
2
+
τ
+
θ
∑
j
=
1
k
1
(
1
+
λ
|
y
-
x
j
|
)
N
-
2
2
+
τ
)
.
From (2.7) and ∥ϕ∥*=1, we obtain that there is R>0 such that
(2.8)
∥
λ
-
N
-
2
2
ϕ
∥
L
∞
(
B
R
/
λ
(
x
j
)
)
≥
a
>
0
for some j. Furthermore, for this particular j, the dilations ϕ~(y)=λ-N-22ϕ(λ-1y+xj) converges uniformly, on any compact set, to a solution u of the equation
(2.9)
-
Δ
u
-
(
2
*
-
1
)
U
0
,
1
2
*
-
2
u
=
0
in
ℝ
N
,
and u is perpendicular to the kernel of (2.9). Hence u=0, which contradicts to the inequality (2.8). ∎
With the help of Lemma 2.1, the following result is a direct consequence of [8, Proposition 4.1].
Lemma 2.2.
There exist an integer k0>0 and a constant C>0, independent of k, such that for all k≥k0 and all h∈L∞(B1), problem (2.1) has a unique solution ϕ≡Lk(h). Moreover,
(2.10)
∥
L
k
(
h
)
∥
*
≤
C
∥
h
∥
*
*
,
|
c
l
|
≤
C
λ
n
l
∥
h
∥
*
*
.
Now we consider the following perturbation problem:
(2.11)
{
-
Δ
(
Z
r
¯
,
y
¯
′′
,
λ
+
ϕ
)
=
K
(
r
,
y
′′
)
(
Z
r
¯
,
y
¯
′′
,
λ
+
ϕ
)
+
2
*
-
1
+
∑
l
=
1
N
c
l
∑
j
=
1
k
Z
x
j
,
λ
2
*
-
2
Z
j
,
l
in
B
1
,
ϕ
∈
H
s
,
∑
j
=
1
k
∫
B
1
Z
x
j
,
λ
2
*
-
2
Z
j
,
l
ϕ
=
0
,
l
=
1
,
…
,
N
.
We have:
Proposition 2.3.
There exists an integer k0>0 such that for each k≥k0, λ∈[L0kN-2N-4,L1kN-2N-4], r¯∈[r0-θ,r0+θ], y¯′′∈Bθ(y0′′), where θ>0 small, (2.11) has a unique solution ϕ=ϕr¯,y¯′′,λ∈Hs satisfying
(2.12)
∥
ϕ
∥
*
≤
C
(
1
λ
)
1
+
ϵ
,
|
c
l
|
≤
C
(
1
λ
)
1
+
n
l
+
ϵ
,
where ϵ>0 is a small constant.
Note that problem (2.11) can be written as
(2.13)
{
-
Δ
ϕ
-
(
2
*
-
1
)
K
(
r
,
y
′′
)
Z
r
¯
,
y
¯
′′
,
λ
2
*
-
2
ϕ
=
ℱ
(
ϕ
)
+
l
k
+
∑
l
=
1
N
c
l
∑
j
=
1
k
Z
x
j
,
λ
2
*
-
2
Z
j
,
l
in
B
1
,
ϕ
∈
H
s
,
∑
j
=
1
k
∫
B
1
Z
x
j
,
λ
2
*
-
2
Z
j
,
l
ϕ
=
0
,
l
=
1
,
…
,
N
,
where
ℱ
(
ϕ
)
=
K
(
r
,
y
′′
)
(
(
Z
r
¯
,
y
¯
′′
,
λ
+
ϕ
)
+
2
*
-
1
-
Z
r
¯
,
y
¯
′′
,
λ
2
*
-
1
-
(
2
*
-
1
)
Z
r
¯
,
y
¯
′′
,
λ
2
*
-
2
ϕ
)
and
l
k
=
(
K
(
r
,
y
′′
)
Z
r
¯
,
y
¯
′′
,
λ
2
*
-
1
-
∑
j
=
1
k
ξ
U
x
j
,
λ
2
*
-
1
)
+
Z
r
¯
,
y
¯
′′
,
λ
*
Δ
ξ
+
2
∇
ξ
⋅
∇
Z
r
¯
,
y
¯
′′
,
λ
*
=
:
J
0
+
J
1
+
J
2
.
In the following, we will make use of the Contraction Mapping Theorem to prove that equation (2.13) is uniquely solvable under the condition that ∥ϕ∥* is small enough. Therefore, we have to estimate ℱ(ϕ) and lk.
Lemma 2.4.
If N≥5, then
∥
ℱ
(
ϕ
)
∥
*
*
≤
C
∥
ϕ
∥
*
m
i
n
{
2
*
-
1
,
2
}
.
Proof.
Note that K(r,y′′) is bounded, it is easy to see that
|
ℱ
(
ϕ
)
|
≤
{
C
|
ϕ
|
2
*
-
1
if
N
≥
6
,
C
|
Z
r
¯
,
y
¯
′′
,
λ
|
1
3
ϕ
2
+
C
|
ϕ
|
7
3
if
N
=
5
.
In the case of N≥6, by the Hölder inequality, we have
|
ℱ
(
ϕ
)
|
≤
C
∥
ϕ
∥
*
2
*
-
1
(
∑
j
=
1
k
λ
N
-
2
2
(
1
+
λ
|
y
-
x
j
|
)
N
-
2
2
+
τ
)
2
*
-
1
=
C
∥
ϕ
∥
*
2
*
-
1
λ
N
+
2
2
(
∑
j
=
1
k
1
(
1
+
λ
|
y
-
x
j
|
)
N
-
2
2
+
τ
)
2
*
-
1
≤
C
∥
ϕ
∥
*
2
*
-
1
λ
N
+
2
2
(
∑
j
=
1
k
1
(
1
+
λ
|
y
-
x
j
|
)
N
+
2
2
+
τ
)
(
∑
j
=
1
k
1
(
1
+
λ
|
y
-
x
j
|
)
τ
)
4
N
-
2
≤
C
∥
ϕ
∥
*
2
*
-
1
λ
N
+
2
2
(
∑
j
=
1
k
1
(
1
+
λ
|
y
-
x
j
|
)
N
+
2
2
+
τ
)
.
Thus, ∥ℱ(ϕ)∥**≤C∥ϕ∥*2*-1 for N≥6. Similarly, for the case of N=5, we have
ℱ
(
ϕ
)
≤
C
|
Z
r
¯
,
y
¯
′′
,
λ
|
1
3
ϕ
2
+
C
|
ϕ
|
7
3
≤
C
(
∥
ϕ
∥
*
2
+
∥
ϕ
∥
*
7
3
)
λ
7
2
(
∑
j
=
1
k
1
(
1
+
λ
|
y
-
x
j
|
)
3
2
+
τ
)
7
3
≤
C
∥
ϕ
∥
*
2
∑
j
=
1
k
λ
7
2
(
1
+
λ
|
y
-
x
j
|
)
7
2
+
τ
.
Hence ∥ℱ(ϕ)∥**≤C∥ϕ∥*2 for N=5. ∎
Lemma 2.5.
If N≥5, then there exists a constant ϵ>0 such that
∥
l
k
∥
*
*
≤
C
λ
1
+
ϵ
.
Proof.
Define
Ω
j
=
{
y
:
y
=
(
y
′
,
y
′′
)
∈
B
1
,
〈
y
′
|
y
′
|
,
x
j
′
|
x
j
′
|
〉
≥
cos
π
k
}
.
By symmetry, we can assume that y∈Ω1; then |y-xj|≥|y-x1| for all y∈Ω1.
We will estimate those terms in lk one by one. For J0, we have
J
0
=
K
(
r
,
y
′′
)
Z
r
¯
,
y
¯
′′
,
λ
2
*
-
1
-
∑
j
=
1
k
ξ
U
x
j
,
λ
2
*
-
1
=
K
(
r
,
y
′′
)
(
Z
r
¯
,
y
¯
′′
,
λ
2
*
-
1
-
∑
j
=
1
k
ξ
U
x
j
,
λ
2
*
-
1
)
+
(
K
(
r
,
y
′′
)
-
1
)
∑
j
=
1
k
ξ
U
x
j
,
λ
2
*
-
1
=
:
J
01
+
J
02
.
Noting that
J
01
=
K
(
r
,
y
′′
)
(
Z
r
¯
,
y
¯
′′
,
λ
2
*
-
1
-
(
∑
j
=
1
k
ξ
U
x
j
,
λ
)
2
*
-
1
)
+
K
(
r
,
y
′′
)
(
(
∑
j
=
1
k
ξ
U
x
j
,
λ
)
2
*
-
1
-
∑
j
=
1
k
ξ
U
x
j
,
λ
2
*
-
1
)
,
by Lemma A.4, we have
|
J
01
|
≤
C
|
(
∑
j
=
1
k
ξ
P
U
x
j
,
λ
)
2
*
-
1
-
(
∑
j
=
1
k
ξ
U
x
j
,
λ
)
2
*
-
1
|
+
C
U
x
1
,
λ
4
N
-
2
∑
j
=
2
k
U
x
j
,
λ
+
C
(
∑
j
=
2
k
U
x
j
,
λ
)
2
*
-
1
≤
C
U
x
1
,
λ
4
N
-
2
ψ
x
1
,
λ
+
C
U
x
1
,
λ
4
N
-
2
∑
j
=
2
k
U
x
j
,
λ
+
C
(
∑
j
=
2
k
U
x
j
,
λ
)
2
*
-
1
≤
C
U
x
1
,
λ
4
N
-
2
H
(
y
,
x
1
)
λ
N
-
2
2
+
C
U
x
1
,
λ
4
N
-
2
∑
j
=
2
k
U
x
j
,
λ
+
C
(
∑
j
=
2
k
U
x
j
,
λ
)
2
*
-
1
.
Let H(y,x) be the regular part of the Green function for -Δ in B1 with zero boundary condition. Let x* be the reflection point of x with respect to ∂B1. Then it holds in [34] that
H
(
y
,
x
1
)
λ
N
-
2
=
C
(
λ
|
y
-
x
1
*
|
)
N
-
2
≤
C
(
1
+
λ
|
y
-
x
1
|
)
N
-
2
and
H
(
y
,
x
1
)
=
C
|
y
-
x
1
*
|
N
-
2
≤
C
d
N
-
2
.
Therefore, we have
U
x
1
,
λ
4
N
-
2
H
(
y
,
x
1
)
λ
N
-
2
2
≤
C
λ
N
+
2
2
(
1
+
λ
|
y
-
x
1
|
)
4
+
(
1
-
t
)
(
N
-
2
)
(
H
(
y
,
x
1
)
λ
N
-
2
)
t
≤
C
λ
N
+
2
2
(
1
+
λ
|
y
-
x
1
|
)
4
+
(
1
-
t
)
(
N
-
2
)
1
(
λ
d
)
(
N
-
2
)
t
≤
C
λ
ϑ
2
(
N
-
2
)
t
λ
N
+
2
2
(
1
+
λ
|
y
-
x
1
|
)
4
+
(
1
-
t
)
(
N
-
2
)
≤
C
λ
1
+
ε
λ
N
+
2
2
(
1
+
λ
|
y
-
x
1
|
)
N
+
2
2
+
τ
,
where we choose t satisfying 1ϑ2(N-2)<t≤1N-2(N+22-τ). On the other hand, by using Lemma A.1 and following the estimates in [25], taking now N-22<β≤min{N+22-τ,N-2}, we obtain that for any y∈Ω1,
U
x
1
,
λ
4
N
-
2
∑
j
=
2
k
U
x
j
,
λ
≤
C
λ
N
+
2
2
(
1
+
λ
|
y
-
x
1
|
)
4
∑
j
=
2
k
1
(
1
+
λ
|
y
-
x
j
|
)
N
-
2
≤
C
λ
N
+
2
2
(
1
+
λ
|
y
-
x
1
|
)
N
+
2
-
β
∑
j
=
2
k
1
|
λ
(
x
j
-
x
1
)
|
β
≤
C
λ
N
+
2
2
(
1
+
λ
|
y
-
x
1
|
)
N
+
2
-
β
(
k
λ
)
β
≤
C
λ
1
+
ε
λ
N
+
2
2
(
1
+
λ
|
y
-
x
1
|
)
N
+
2
2
+
τ
,
and by the Hölder inequality, we have
(
∑
j
=
2
k
U
x
j
,
λ
)
2
*
-
1
≤
C
(
∑
j
=
2
k
λ
N
+
2
2
(
1
+
λ
|
y
-
x
j
|
)
N
+
2
2
+
τ
)
(
∑
j
=
2
k
1
(
1
+
λ
|
y
-
x
j
|
)
N
+
2
4
(
N
-
2
2
-
τ
N
-
2
N
+
2
)
)
4
N
-
2
≤
C
(
k
λ
)
N
+
2
4
(
N
-
2
2
-
τ
N
-
2
N
+
2
)
4
N
-
2
∑
j
=
2
k
λ
N
+
2
2
(
1
+
λ
|
y
-
x
j
|
)
N
+
2
2
+
τ
≤
C
λ
1
+
ε
∑
j
=
2
k
λ
N
+
2
2
(
1
+
λ
|
y
-
x
j
|
)
N
+
2
2
+
τ
,
since
N
+
2
N
-
2
(
1
-
2
τ
N
+
2
)
>
1
.
Thus, we have proved
∥
J
01
∥
*
*
≤
C
λ
1
+
ε
.
For J02, in the region y∈B1 and |y-y0|≤δλ12+ε, by Taylor expansion, we have
|
J
02
|
≤
C
|
y
-
y
0
|
2
∑
j
=
1
k
ξ
λ
N
+
2
2
(
1
+
λ
|
y
-
x
j
|
)
N
+
2
≤
C
λ
1
+
ε
∑
j
=
1
k
λ
N
+
2
2
(
1
+
λ
|
y
-
x
j
|
)
N
+
2
2
+
τ
.
On the other hand, in the region y∈B1 and δλ12+ε≤|y-y0|≤2δ, we have
1
1
+
λ
|
y
-
x
j
|
≤
C
λ
1
2
-
ε
,
which implies
|
J
02
|
≤
C
∑
j
=
1
k
ξ
λ
N
+
2
2
(
1
+
λ
|
y
-
x
j
|
)
N
+
2
≤
C
λ
1
+
ε
∑
j
=
1
k
λ
N
+
2
2
(
1
+
λ
|
y
-
x
j
|
)
N
+
2
2
+
τ
ξ
λ
1
+
ε
(
1
+
λ
|
y
-
x
j
|
)
N
+
2
2
-
τ
≤
C
λ
1
+
ε
∑
j
=
1
k
λ
N
+
2
2
(
1
+
λ
|
y
-
x
j
|
)
N
+
2
2
+
τ
.
Thus, we obtain that
∥
J
02
∥
*
*
≤
C
λ
1
+
ε
.
For J1, noting that the cut-off function ξ, we just need to consider the region y∈B1 and |y-y0|≤2δ, which induces
1
λ
≤
C
1
+
λ
|
y
-
x
j
|
.
Therefore, we obtain that
|
J
1
|
≤
C
∑
j
=
1
k
λ
N
-
2
2
|
Δ
ξ
|
(
1
+
λ
|
y
-
x
j
|
)
N
-
2
=
C
λ
1
+
ε
∑
j
=
1
k
λ
N
+
2
2
|
Δ
ξ
|
λ
1
-
ε
(
1
+
λ
|
y
-
x
j
|
)
N
-
2
≤
C
λ
1
+
ε
∑
j
=
1
k
λ
N
+
2
2
(
1
+
λ
|
y
-
x
j
|
)
N
+
2
2
+
τ
,
since N-1-ε≥N+22+τ for ε>0 small. Thus, we obtain
∥
J
1
∥
*
*
≤
C
λ
1
+
ε
.
Similarly, for J2, we have that
|
J
2
|
≤
C
∑
j
=
1
k
λ
N
2
|
∇
ξ
|
(
1
+
λ
|
y
-
x
j
|
)
N
-
1
=
C
λ
1
+
ε
∑
j
=
1
k
λ
ε
λ
N
+
2
2
|
∇
ξ
|
(
1
+
λ
|
y
-
x
j
|
)
N
-
1
≤
C
λ
1
+
ε
∑
j
=
1
k
λ
N
+
2
2
(
1
+
λ
|
y
-
x
j
|
)
N
+
2
2
+
τ
,
since we used the fact that in the region y∈B1 and δ≤|y-y0|≤2δ, the assumption xj∈Ω(r0,y0′′) induces
1
1
+
λ
|
y
-
x
j
|
≤
C
λ
.
Thus, we obtain
∥
J
2
∥
*
*
≤
C
λ
1
+
ε
.
As a result of the above estimates, we finish the proof. ∎
Now we are ready to prove Proposition 2.3.
Proof of Proposition 2.3.
Recall that λ∈[L0kN-2N-4,L1kN-2N-4]. Set
𝒩
=
{
w
:
w
∈
C
(
B
1
)
∩
H
s
,
∥
w
∥
*
≤
1
λ
,
∑
j
=
1
k
∫
B
1
Z
x
j
,
λ
2
*
-
2
Z
j
,
l
w
=
0
,
l
=
1
,
…
,
N
}
.
By Lemma 2.2, we can regard hk as the sum ℱ(ϕ)+lk. Then (2.13) is equivalent to the following fixed point problem:
(2.14)
ϕ
=
A
(
ϕ
)
:=
L
k
(
ℱ
(
ϕ
)
)
+
L
k
(
l
k
)
.
Hence, it is sufficient to prove that A is a contraction map from 𝒩 to 𝒩. We only prove the case of N≥6 and the case of N=5 is similar. In fact, for all ϕ∈𝒩,
∥
A
(
ϕ
)
∥
*
≤
∥
L
k
(
ℱ
(
ϕ
)
)
∥
*
+
∥
L
k
(
l
k
)
∥
*
≤
C
[
∥
ℱ
(
ϕ
)
∥
*
*
+
∥
l
k
∥
*
*
]
≤
C
[
∥
ϕ
∥
*
min
{
2
*
-
1
,
2
}
+
1
λ
1
+
ϵ
]
≤
1
λ
for
k
large enough
,
which shows that A maps 𝒩 to itself and 𝒩 is invariant under the operator A. On the other hand, for all ϕ1,ϕ2∈𝒩, we have
∥
A
(
ϕ
1
)
-
A
(
ϕ
2
)
∥
*
=
∥
L
k
(
ℱ
(
ϕ
1
)
)
-
L
k
(
ℱ
(
ϕ
2
)
)
∥
*
≤
C
∥
ℱ
(
ϕ
1
)
-
ℱ
(
ϕ
2
)
∥
*
*
.
In the case of N≥6, by the Hölder inequality, we obtain
|
ℱ
(
ϕ
1
)
-
ℱ
(
ϕ
2
)
|
≤
|
ℱ
′
(
ϕ
1
+
θ
(
ϕ
2
-
ϕ
1
)
)
|
|
ϕ
1
-
ϕ
2
|
≤
C
(
|
ϕ
1
|
2
*
-
2
+
|
ϕ
2
|
2
*
-
2
)
|
ϕ
1
-
ϕ
2
|
≤
C
(
∥
ϕ
1
∥
*
2
*
-
2
+
∥
ϕ
2
∥
*
2
*
-
2
)
∥
ϕ
1
-
ϕ
2
∥
*
(
∑
j
=
1
k
λ
N
-
2
2
(
1
+
λ
|
y
-
x
j
|
)
N
-
2
2
+
τ
)
2
*
-
1
≤
C
(
∥
ϕ
1
∥
*
2
*
-
2
+
∥
ϕ
2
∥
*
2
*
-
2
)
∥
ϕ
1
-
ϕ
2
∥
*
∑
j
=
1
k
λ
N
+
2
2
(
1
+
λ
|
y
-
x
j
|
)
N
+
2
2
+
τ
,
that is,
∥
ℱ
(
ϕ
1
)
-
ℱ
(
ϕ
2
)
∥
*
*
≤
C
(
∥
ϕ
1
∥
*
2
*
-
2
+
∥
ϕ
2
∥
*
2
*
-
2
)
∥
ϕ
1
-
ϕ
2
∥
*
.
Thus,
∥
A
(
ϕ
1
)
-
A
(
ϕ
2
)
∥
*
≤
C
(
∥
ϕ
1
∥
*
2
*
-
2
+
∥
ϕ
2
∥
*
2
*
-
2
)
∥
ϕ
1
-
ϕ
2
∥
*
≤
1
2
∥
ϕ
1
-
ϕ
2
∥
*
for
k
large enough
.
Hence A is a contraction map. By the contraction mapping theorem, there exists a unique solution ϕ∈𝒩 for problem (2.14). Moreover, by Lemmas 2.2, 2.4 and 2.5, we have
∥
ϕ
∥
*
≤
∥
L
k
(
ℱ
(
ϕ
)
)
∥
*
+
∥
L
k
(
l
k
)
∥
*
≤
C
[
∥
ℱ
(
ϕ
)
∥
*
*
+
∥
l
k
∥
*
*
]
≤
C
λ
1
+
ϵ
.
Similarly, we can get estimate (2.12) of cl from (2.10). ∎
3 The Energy Expansion
Recall that, the functional corresponding to (1.1) is
I
(
u
)
=
1
2
∫
B
1
|
∇
u
|
2
-
1
2
*
∫
B
1
K
(
r
,
y
′′
)
(
u
)
+
2
*
.
Lemma 3.1.
If N≥5, then
∂
I
(
Z
r
¯
,
y
¯
′′
,
λ
)
∂
λ
=
k
(
-
B
0
λ
3
+
∑
j
=
2
k
B
2
λ
N
-
1
|
x
j
-
x
1
|
N
-
2
+
O
(
1
λ
3
+
ε
)
)
,
where B0,B2 are some positive constants.
Proof.
Direct computation shows that
∂
I
(
Z
r
¯
,
y
¯
′′
,
λ
)
∂
λ
=
∂
I
(
Z
r
¯
,
y
¯
′′
,
λ
*
)
∂
λ
+
O
(
k
λ
3
+
ε
)
=
∫
B
1
(
1
-
K
(
r
,
y
′′
)
)
(
Z
r
¯
,
y
¯
′′
,
λ
*
)
2
*
-
1
∂
Z
r
¯
,
y
¯
′′
,
λ
*
∂
λ
-
∫
B
1
(
(
Z
r
¯
,
y
¯
′′
,
λ
*
)
2
*
-
1
-
∑
j
=
1
k
U
x
j
,
λ
2
*
-
1
)
∂
Z
r
¯
,
y
¯
′′
,
λ
*
∂
λ
+
O
(
k
λ
3
+
ε
)
=
:
I
1
-
I
2
+
O
(
k
λ
3
+
ε
)
.
By symmetry and Taylor’s expansion, we have
(3.1)
I
1
=
k
∫
Ω
1
(
1
-
K
(
r
,
y
′′
)
)
(
Z
r
¯
,
y
¯
′′
,
λ
*
)
2
*
-
1
∂
Z
r
¯
,
y
¯
′′
,
λ
*
∂
λ
=
k
(
∫
Ω
1
(
1
-
K
(
r
,
y
′′
)
)
P
U
x
1
,
λ
2
*
-
1
∂
P
U
x
1
,
λ
∂
λ
+
O
(
1
λ
∫
Ω
1
U
x
1
,
λ
2
*
-
1
∑
j
=
2
k
U
x
j
,
λ
)
)
=
k
(
∫
Ω
1
(
1
-
K
(
r
,
y
′′
)
)
P
U
x
1
,
λ
2
*
-
1
∂
P
U
x
1
,
λ
∂
λ
+
O
(
1
λ
3
+
ε
)
)
.
Noting that
∫
Ω
1
(
1
-
K
(
r
,
y
′′
)
)
P
U
x
1
,
λ
2
*
-
1
∂
P
U
x
1
,
λ
∂
λ
=
∫
Ω
1
(
1
-
K
(
r
,
y
′′
)
)
U
x
1
,
λ
2
*
-
1
∂
U
x
1
,
λ
∂
λ
-
(
2
*
-
1
)
∫
Ω
1
(
1
-
K
(
r
,
y
′′
)
)
U
x
1
,
λ
2
*
-
2
ψ
x
1
,
λ
∂
U
x
1
,
λ
∂
λ
-
∫
Ω
1
(
1
-
K
(
r
,
y
′′
)
)
U
x
1
,
λ
2
*
-
1
∂
ψ
x
1
,
λ
∂
λ
+
(
2
*
-
1
)
∫
Ω
1
(
1
-
K
(
r
,
y
′′
)
)
U
x
1
,
λ
2
*
-
2
ψ
x
1
,
λ
∂
ψ
x
1
,
λ
∂
λ
+
O
(
1
λ
∫
Ω
1
ψ
x
1
,
λ
2
*
-
1
U
x
1
,
λ
)
,
by Lemma A.4, choosing 0<t<1 fixed, by Taylor expansion, we have
|
(
2
*
-
1
)
∫
Ω
1
(
1
-
K
(
r
,
y
′′
)
)
U
x
1
,
λ
2
*
-
2
ψ
x
1
,
λ
∂
U
x
1
,
λ
∂
λ
|
≤
C
λ
∫
Ω
1
∩
B
δ
(
x
1
)
λ
N
|
y
-
y
0
|
2
(
1
+
λ
|
y
-
x
1
|
)
N
+
2
(
H
(
y
,
x
1
)
λ
N
-
2
)
t
(
H
(
y
,
x
1
)
λ
N
-
2
)
1
-
t
+
C
λ
∫
Ω
1
∩
B
δ
c
(
x
1
)
λ
N
(
1
+
λ
|
y
-
x
1
|
)
N
+
2
H
(
y
,
x
1
)
λ
N
-
2
≤
C
λ
1
(
λ
d
)
(
N
-
2
)
t
∫
Ω
1
λ
N
-
2
(
1
+
λ
|
y
-
x
1
|
)
N
+
(
N
-
2
)
(
1
-
t
)
+
C
λ
3
(
λ
d
)
N
-
2
≤
C
λ
3
+
ε
,
where δ is a fixed small constant. The other terms can be estimated similarly. Thus we have
(3.2)
∫
Ω
1
(
1
-
K
(
r
,
y
′′
)
)
P
U
x
1
,
λ
2
*
-
1
∂
P
U
x
1
,
λ
∂
λ
=
∫
Ω
1
(
1
-
K
(
r
,
y
′′
)
)
U
x
1
,
λ
2
*
-
1
∂
U
x
1
,
λ
∂
λ
+
O
(
1
λ
3
+
ε
)
.
On the other hand, we compute
∫
Ω
1
(
1
-
K
(
r
,
y
′′
)
)
U
x
1
,
λ
2
*
-
1
∂
U
x
1
,
λ
∂
λ
=
∫
{
y
∈
Ω
1
:
|
y
-
x
1
|
≤
1
λ
σ
}
(
1
-
K
(
r
,
y
′′
)
)
U
x
1
,
λ
2
*
-
1
∂
U
x
1
,
λ
∂
λ
+
∫
{
y
∈
Ω
1
:
|
y
-
x
1
|
≥
1
λ
σ
}
(
1
-
K
(
r
,
y
′′
)
)
U
x
1
,
λ
2
*
-
1
∂
U
x
1
,
λ
∂
λ
=
-
1
2
∫
{
y
∈
Ω
1
:
|
y
-
x
1
|
≤
1
λ
σ
}
∂
2
K
(
y
0
)
∂
y
i
∂
y
j
(
y
-
y
0
)
i
(
y
-
y
0
)
j
U
x
1
,
λ
2
*
-
1
∂
U
x
1
,
λ
∂
λ
+
O
(
1
λ
∫
{
y
∈
Ω
1
:
|
y
-
x
1
|
≤
1
λ
σ
}
|
y
-
y
0
|
3
U
x
1
,
λ
2
*
)
+
O
(
1
λ
3
+
ε
)
=
-
1
2
∫
{
y
∈
Ω
1
:
|
y
-
x
1
|
≤
1
λ
σ
}
∂
2
K
(
y
0
)
∂
y
i
∂
y
j
(
y
-
y
0
)
i
(
y
-
y
0
)
j
U
x
1
,
λ
2
*
-
1
∂
U
x
1
,
λ
∂
λ
+
O
(
1
λ
3
+
ε
)
=
-
1
2
∫
ℝ
N
∂
2
K
(
y
0
)
∂
y
i
∂
y
j
(
y
-
y
0
)
i
(
y
-
y
0
)
j
U
x
1
,
λ
2
*
-
1
∂
U
x
1
,
λ
∂
λ
+
O
(
1
λ
∫
{
y
∈
Ω
1
:
|
y
-
x
1
|
≤
1
λ
σ
}
c
|
y
-
y
0
|
2
U
x
1
,
λ
2
*
)
+
O
(
1
λ
3
+
ε
)
=
-
1
2
∫
ℝ
N
∂
2
K
(
y
0
)
∂
y
i
∂
y
j
(
y
-
y
0
)
i
(
y
-
y
0
)
j
U
x
1
,
λ
2
*
-
1
∂
U
x
1
,
λ
∂
λ
+
O
(
1
λ
3
+
ε
)
(3.3)
=
1
2
1
2
*
1
λ
3
Δ
K
(
y
0
)
N
∫
ℝ
N
z
2
U
0
,
1
2
*
+
O
(
1
λ
3
+
ε
)
=
-
B
0
λ
3
+
O
(
1
λ
3
+
ε
)
,
where we choose σ satisfying 0<σ<min{23+ϑ1,N-22N,1-23N} and B0>0 is a constant. Combining (3.1), (3.2), (3.3), we obtain that
I
1
=
k
(
-
B
0
λ
3
+
O
(
1
λ
3
+
ε
)
)
.
For I2, we have
I
2
=
∫
B
1
(
(
∑
j
=
1
k
U
x
j
,
λ
)
2
*
-
1
-
∑
j
=
1
k
U
x
j
,
λ
2
*
-
1
)
∂
Z
r
¯
,
y
¯
′′
,
λ
*
∂
λ
+
∫
B
1
(
(
∑
j
=
1
k
P
U
x
j
,
λ
)
2
*
-
1
-
(
∑
j
=
1
k
U
x
j
,
λ
)
2
*
-
1
)
∂
Z
r
¯
,
y
¯
′′
,
λ
*
∂
λ
=
k
∫
Ω
1
(
(
∑
j
=
1
k
U
x
j
,
λ
)
2
*
-
1
-
∑
j
=
1
k
U
x
j
,
λ
2
*
-
1
)
∂
Z
r
¯
,
y
¯
′′
,
λ
*
∂
λ
+
O
(
k
λ
3
+
ε
)
=
k
∫
Ω
1
(
2
*
-
1
)
U
x
1
,
λ
2
*
-
2
∑
j
=
2
k
U
x
j
,
λ
∂
U
x
1
,
λ
∂
λ
+
O
(
k
λ
3
+
ε
)
=
k
(
-
∑
j
=
2
k
B
2
λ
N
-
1
|
x
j
-
x
1
|
N
-
2
+
O
(
1
λ
3
+
ε
)
)
,
for some constant B2>0. The result follows. ∎
By using the similar arguments, we can prove the following lemma.
Lemma 3.2.
If N≥5, then
∂
I
(
Z
r
¯
,
y
¯
′′
,
λ
)
∂
r
¯
=
k
(
∑
j
=
2
k
B
2
r
¯
λ
N
-
2
|
x
j
-
x
1
|
N
-
2
+
O
(
1
λ
1
+
ε
)
)
and
∂
I
(
Z
r
¯
,
y
¯
′′
,
λ
)
∂
y
¯
k
′′
=
k
O
(
1
λ
1
+
ε
)
,
k
=
3
,
…
,
N
,
where B2>0 is a constant.
4 Proof of the Main Result
In this section, we will choose suitable (r¯,y¯′′,λ) such that Zr¯,y¯′′,λ+ϕr¯,y¯′′,λ is a solution of (1.1). For this purpose, we first prove the following:
Proposition 4.1.
Suppose that (r¯,y¯′′,λ) satisfies
(4.1)
∫
D
ρ
(
-
Δ
u
k
-
K
(
r
,
y
′′
)
(
u
k
)
+
2
*
-
1
)
〈
y
,
∇
u
k
〉
=
0
,
(4.2)
∫
D
ρ
(
-
Δ
u
k
-
K
(
r
,
y
′′
)
(
u
k
)
+
2
*
-
1
)
∂
u
k
∂
y
i
=
0
,
i
=
3
,
…
,
N
,
and
(4.3)
∫
B
1
(
-
Δ
u
k
-
K
(
r
,
y
′′
)
(
u
k
)
+
2
*
-
1
)
∂
Z
r
¯
,
y
¯
′′
,
λ
∂
λ
=
0
,
where uk=Zr¯,y¯′′,λ+ϕr¯,y¯′′,λ and Dρ={(r,y′′):|(r,y′′)-(r0,y0′′)|≤ρ}∩B¯1 with ρ∈(2δ,5δ). Then cl=0 for l=1,…,N.
Proof.
Noting that Zxj,λ=0 in B1\Dρ, we conclude from (4.1), (4.2) and (4.3),
(4.4)
∑
l
=
1
N
c
l
∑
j
=
1
k
∫
B
1
Z
x
j
,
λ
2
*
-
2
Z
j
,
l
v
=
0
for
v
=
〈
y
,
∇
u
k
〉
,
v
=
∂
u
k
∂
y
i
and
v
=
∂
Z
r
¯
,
y
¯
′′
,
λ
∂
λ
.
A direct computation leads to
(4.5)
∑
j
=
1
k
∫
B
1
Z
x
j
,
λ
2
*
-
2
Z
j
,
2
〈
y
′
,
∇
y
′
Z
r
¯
,
y
¯
′′
,
λ
〉
=
k
λ
2
(
a
1
+
o
(
1
)
)
,
(4.6)
∑
j
=
1
k
∫
B
1
Z
x
j
,
λ
2
*
-
2
Z
j
,
i
∂
Z
r
¯
,
y
¯
′′
,
λ
∂
y
i
=
k
λ
2
(
a
2
+
o
(
1
)
)
,
i
=
3
,
…
,
N
,
(4.7)
∑
j
=
1
k
∫
B
1
Z
x
j
,
λ
2
*
-
2
Z
j
,
1
∂
Z
r
¯
,
y
¯
′′
,
λ
∂
λ
=
k
λ
2
(
a
3
+
o
(
1
)
)
for some constants a1≠0,a2≠0 and a3>0. Integrating by parts , we get
(4.8)
∑
l
=
1
N
c
l
∑
j
=
1
k
∫
B
1
Z
x
j
,
λ
2
*
-
2
Z
j
,
l
v
=
o
(
k
λ
2
)
∑
l
=
2
N
|
c
l
|
+
o
(
k
|
c
1
|
)
for v=〈y,∇ϕr¯,y¯′′,λ〉 and v=∂ϕr¯,y¯′′,λ∂yi. Then (4.8) together with (4.4) leads to
(4.9)
∑
l
=
1
N
c
l
∑
j
=
1
k
∫
B
1
Z
x
j
,
λ
2
*
-
2
Z
j
,
l
v
=
o
(
k
λ
2
)
∑
l
=
2
N
|
c
l
|
+
o
(
k
|
c
1
|
)
for v=〈y,∇Zr¯,y¯′′,λ〉 and v=∂Zr¯,y¯′′,λ∂yi. Noting that
〈
y
,
∇
Z
r
¯
,
y
¯
′′
,
λ
〉
=
〈
y
′
,
∇
y
′
Z
r
¯
,
y
¯
′′
,
λ
〉
+
〈
y
′′
,
∇
y
′′
Z
r
¯
,
y
¯
′′
,
λ
〉
,
by direct calculations, we obtain
(4.10)
∑
l
=
1
N
c
l
∑
j
=
1
k
∫
B
1
Z
x
j
,
λ
2
*
-
2
Z
j
,
l
∂
Z
r
¯
,
y
¯
′′
,
λ
∂
y
i
=
c
i
∑
j
=
1
k
∫
B
1
Z
x
j
,
λ
m
*
-
2
Z
j
,
i
∂
Z
r
¯
,
y
¯
′′
,
λ
∂
y
i
+
o
(
k
λ
2
)
∑
t
≠
1
,
i
|
c
t
|
+
o
(
k
|
c
1
|
)
,
i
=
3
,
…
,
N
,
and
(4.11)
∑
l
=
1
N
c
l
∑
j
=
1
k
∫
B
1
Z
x
j
,
λ
2
*
-
2
Z
j
,
l
〈
y
,
∇
Z
r
¯
,
y
¯
′′
,
λ
〉
=
c
2
∑
j
=
1
k
∫
B
1
Z
x
j
,
λ
2
*
-
2
Z
j
,
2
〈
y
′
,
∇
y
′
Z
r
¯
,
y
¯
′′
,
λ
〉
+
o
(
k
|
c
1
|
)
+
O
(
k
λ
2
)
∑
t
=
3
N
|
c
t
|
.
Combining (4.6), (4.9) and (4.10), we have
(4.12)
c
i
=
o
(
1
λ
2
)
c
1
,
i
=
3
,
…
,
N
.
Similarly, from (4.5), (4.9), (4.11) and (4.12), we obtain that
c
2
=
o
(
1
λ
2
)
c
1
.
Thus,
0
=
∑
l
=
1
N
c
l
∑
j
=
1
k
∫
B
1
Z
x
j
,
λ
2
*
-
2
Z
j
,
l
∂
Z
r
¯
,
y
¯
′′
,
λ
∂
λ
=
c
1
∑
j
=
1
k
∫
B
1
Z
x
j
,
λ
m
*
-
2
Z
j
,
1
∂
Z
r
¯
,
y
¯
′′
,
λ
∂
λ
+
∑
l
=
2
N
c
l
∑
j
=
1
k
∫
B
1
Z
x
j
,
λ
2
*
-
2
Z
j
,
l
∂
Z
r
¯
,
y
¯
′′
,
λ
∂
λ
=
k
λ
2
(
a
3
+
o
(
1
)
)
c
1
+
o
(
k
λ
2
)
c
1
,
which leads to c1=0. ∎
Lemma 4.2.
We have
∫
B
1
(
-
Δ
u
k
-
K
(
r
,
y
′′
)
(
u
k
)
+
2
*
-
1
)
∂
Z
r
¯
,
y
¯
′′
,
λ
∂
λ
=
k
(
-
B
0
λ
3
+
∑
j
=
2
k
B
2
λ
N
-
1
|
x
j
-
x
1
|
N
-
2
+
O
(
1
λ
3
+
ε
)
)
=
k
(
-
B
0
λ
3
+
B
3
k
N
-
2
λ
N
-
1
+
O
(
1
λ
3
+
ε
)
)
,
where B0,B2,B3 are some positive constants.
Proof.
By symmetry, we have
∫
B
1
(
-
Δ
u
k
-
K
(
r
,
y
′′
)
(
u
k
)
+
2
*
-
1
)
∂
Z
r
¯
,
y
¯
′′
,
λ
∂
λ
=
〈
I
′
(
Z
r
¯
,
y
¯
′′
,
λ
)
,
∂
Z
r
¯
,
y
¯
′′
,
λ
∂
λ
〉
+
k
〈
-
Δ
ϕ
-
(
2
*
-
1
)
K
(
r
,
y
′′
)
Z
r
¯
,
y
¯
′′
,
λ
2
*
-
2
ϕ
,
∂
Z
x
1
,
λ
∂
λ
〉
-
∫
B
1
K
(
r
,
y
′′
)
(
(
Z
r
¯
,
y
¯
′′
,
λ
+
ϕ
)
+
2
*
-
1
-
Z
r
¯
,
y
¯
′′
,
λ
2
*
-
1
-
(
2
*
-
1
)
Z
r
¯
,
y
¯
′′
,
λ
2
*
-
2
ϕ
)
∂
Z
r
¯
,
y
¯
′′
,
λ
∂
λ
=
:
〈
I
′
(
Z
r
¯
,
y
¯
′′
,
λ
)
,
∂
Z
r
¯
,
y
¯
′′
,
λ
∂
λ
〉
+
k
I
1
-
I
2
.
From (2.4), we have
I
1
=
O
(
∥
ϕ
∥
*
λ
2
+
ε
)
=
O
(
1
λ
3
+
ε
)
.
If N≥6, then
|
I
2
|
=
|
∫
B
1
K
(
r
,
y
′′
)
(
(
Z
r
¯
,
y
¯
′′
,
λ
+
ϕ
)
+
2
*
-
1
-
Z
r
¯
,
y
¯
′′
,
λ
2
*
-
1
-
(
2
*
-
1
)
Z
r
¯
,
y
¯
′′
,
λ
2
*
-
2
ϕ
)
∂
Z
r
¯
,
y
¯
′′
,
λ
∂
λ
|
≤
C
∫
B
!
Z
r
¯
,
y
¯
′′
,
λ
2
*
-
3
ϕ
2
|
∂
Z
r
¯
,
y
¯
′′
,
λ
∂
λ
|
≤
C
λ
∫
B
1
Z
r
¯
,
y
¯
′′
,
λ
2
*
-
2
ϕ
2
≤
C
∥
ϕ
∥
*
2
λ
∫
B
1
(
∑
j
=
1
k
λ
N
-
2
2
(
1
+
λ
|
y
-
x
j
|
)
N
-
2
)
2
*
-
2
(
∑
j
=
1
k
λ
N
-
2
2
(
1
+
λ
|
y
-
x
j
|
)
N
-
2
2
+
τ
)
2
=
O
(
k
λ
3
+
ε
)
.
Similarly, for N=5, we have
|
I
2
|
≤
C
∫
B
1
(
Z
r
¯
,
y
¯
′′
,
λ
1
3
ϕ
2
|
∂
Z
r
¯
,
y
¯
′′
,
λ
∂
λ
|
+
|
ϕ
|
7
3
|
∂
Z
r
¯
,
y
¯
′′
,
λ
∂
λ
|
)
=
O
(
k
λ
3
+
ε
)
.
Hence
〈
I
′
(
Z
r
¯
,
y
¯
′′
,
λ
+
ϕ
)
,
∂
Z
r
¯
,
y
¯
′′
,
λ
∂
λ
〉
=
〈
I
′
(
Z
r
¯
,
y
¯
′′
,
λ
)
,
∂
Z
r
¯
,
y
¯
′′
,
λ
∂
λ
〉
+
O
(
k
λ
3
+
ε
)
.
Combining Lemma 3.1, we finish the proof. ∎
Similarly, we can prove the following:
Lemma 4.3.
We have
(4.13)
〈
I
′
(
Z
r
¯
,
y
¯
′′
,
λ
+
ϕ
)
,
∂
Z
r
¯
,
y
¯
′′
,
λ
∂
r
¯
〉
=
k
(
∑
j
=
2
k
B
2
r
¯
λ
N
-
2
|
x
j
-
x
1
|
N
-
2
+
O
(
1
λ
1
+
ε
)
)
and
(4.14)
〈
I
′
(
Z
r
¯
,
y
¯
′′
,
λ
+
ϕ
)
,
∂
Z
r
¯
,
y
¯
′′
,
λ
∂
y
¯
i
′′
〉
=
k
O
(
1
λ
1
+
ε
)
,
where B2>0 is a constant.
It is worth pointing out that we cannot determine the parameters r¯,y¯′′ by (4.13) and (4.14). Instead, we consider the local Pohozaev-type identities (4.1) and (4.2). Next, we estimate (4.1) and (4.2) and find their equivalent identities.
By integrating by parts and using (2.12), we obtain that (4.1) and (4.2) are equivalent to
(4.15)
1
2
*
∫
D
ρ
r
∂
K
(
r
,
y
′′
)
∂
r
(
u
k
)
+
2
*
=
o
(
k
λ
2
)
+
O
(
∫
∂
D
ρ
|
∇
ϕ
|
2
+
|
ϕ
|
2
*
+
|
ϕ
|
2
)
and
(4.16)
∫
D
ρ
∂
K
(
r
,
y
′′
)
∂
y
i
(
u
k
)
+
2
*
=
O
(
∫
∂
D
ρ
|
∇
ϕ
|
2
+
|
ϕ
|
2
*
)
,
i
=
3
,
…
,
N
.
In order to estimate (4.15) and (4.16), we first prove:
Lemma 4.4.
We have
∫
B
1
|
∇
ϕ
|
2
+
|
ϕ
|
2
*
+
|
ϕ
|
2
=
O
(
k
λ
2
+
ε
)
.
Proof.
Multiplying (2.11) by ϕ and integrating by parts in B1, we have
(4.17)
∫
B
1
|
∇
ϕ
|
2
=
∫
B
1
(
K
(
r
,
y
′′
)
(
Z
r
¯
,
y
¯
′′
,
λ
+
ϕ
)
+
2
*
-
1
+
Δ
Z
r
¯
,
y
¯
′′
,
λ
)
ϕ
=
∫
B
1
K
(
r
,
y
′′
)
(
(
Z
r
¯
,
y
¯
′′
,
λ
+
ϕ
)
+
2
*
-
1
-
Z
r
¯
,
y
¯
′′
,
λ
2
*
-
1
)
ϕ
+
∫
B
1
(
K
(
r
,
y
′′
)
Z
r
¯
,
y
¯
′′
,
λ
2
*
-
1
+
Δ
Z
r
¯
,
y
¯
′′
,
λ
)
ϕ
=
:
J
¯
1
+
J
¯
2
and
(4.18)
|
J
¯
1
|
≤
C
∫
B
1
Z
r
¯
,
y
¯
′′
,
λ
2
*
-
2
|
ϕ
|
2
+
|
ϕ
|
2
*
≤
C
∥
ϕ
∥
*
2
∫
B
1
Z
r
¯
,
y
¯
′′
,
λ
2
*
-
2
(
∑
j
=
1
k
λ
N
-
2
2
(
1
+
λ
|
y
-
x
j
|
)
N
-
2
2
+
τ
)
2
+
C
∥
ϕ
∥
*
2
*
∫
B
1
(
∑
j
=
1
k
λ
N
-
2
2
(
1
+
λ
|
y
-
x
j
|
)
N
-
2
2
+
τ
)
2
*
≤
C
(
∥
ϕ
∥
*
2
+
∥
ϕ
∥
*
2
*
)
∫
B
1
(
∑
j
=
1
k
λ
N
-
2
2
(
1
+
λ
|
y
-
x
j
|
)
N
-
2
2
+
τ
)
2
*
≤
C
∥
ϕ
∥
*
2
k
∫
Ω
1
(
∑
j
=
1
k
λ
N
-
2
2
(
1
+
λ
|
y
-
x
j
|
)
N
-
2
2
+
ν
)
2
*
(
1
+
∑
j
=
1
k
1
(
λ
|
x
j
-
x
1
|
)
τ
-
ν
)
2
*
≤
C
k
λ
2
+
ε
,
where ν>0 is any fixed small constant. Using Lemma 2.5, we obtain that
(4.19)
|
J
¯
2
|
≤
C
∥
ϕ
∥
*
∥
l
k
∥
*
*
∫
B
1
∑
j
=
1
k
λ
N
+
2
2
(
1
+
λ
|
y
-
x
j
|
)
N
+
2
2
+
τ
∑
j
=
1
k
λ
N
-
2
2
(
1
+
λ
|
y
-
x
j
|
)
N
-
2
2
+
τ
≤
C
k
∥
ϕ
∥
*
∥
l
k
∥
*
*
≤
C
k
λ
2
+
ε
.
Combining (4.17), (4.18) and (4.19), we have
(4.20)
∫
B
1
|
∇
ϕ
|
2
=
O
(
k
λ
2
+
ε
)
.
In (4.18), we have proved
(4.21)
∫
B
1
|
ϕ
|
2
*
=
O
(
k
λ
2
+
ε
)
.
On the other hand, by Lemma A.1
(4.22)
∫
B
1
|
ϕ
|
2
≤
C
∥
ϕ
∥
*
2
∫
B
1
(
∑
j
=
1
k
λ
N
-
2
2
(
1
+
λ
|
y
-
x
j
|
)
N
-
2
2
+
τ
)
2
=
C
k
∥
ϕ
∥
*
2
∫
B
1
λ
N
-
2
(
1
+
λ
|
y
-
x
j
|
)
N
-
2
+
2
τ
+
C
k
∥
ϕ
∥
*
2
(
∑
j
=
2
k
∫
B
1
λ
N
-
2
(
1
+
λ
|
y
-
x
j
|
)
N
-
2
2
+
τ
(
1
+
λ
|
y
-
x
j
|
)
N
-
2
2
+
τ
)
≤
C
k
∥
ϕ
∥
*
2
(
1
λ
2
τ
+
∑
j
=
2
k
1
(
λ
|
x
j
-
x
1
|
)
τ
∫
B
1
λ
N
-
2
(
1
+
λ
|
y
-
x
1
|
)
N
-
2
+
τ
)
≤
C
k
∥
ϕ
∥
*
2
λ
τ
≤
C
k
λ
2
+
ε
.
The result follows from (4.20), (4.21) and (4.22). ∎
From Lemma 4.4, we have that
∫
D
4
δ
\
D
3
δ
|
∇
ϕ
|
2
+
|
ϕ
|
2
*
+
|
ϕ
|
2
=
O
(
k
λ
2
+
ε
)
,
which implies that there exists a ρ∈(3δ,4δ), such that
(4.23)
∫
∂
D
ρ
|
∇
ϕ
|
2
+
|
ϕ
|
2
*
+
|
ϕ
|
2
=
O
(
k
λ
2
+
ε
)
=
o
(
k
λ
2
)
.
Next, we prove:
Lemma 4.5.
For any C1 bounded function g(r,y′′), it holds
∫
D
ρ
g
(
r
,
y
′′
)
(
u
k
)
+
2
*
=
k
(
g
(
r
¯
,
y
¯
′′
)
∫
ℝ
N
U
0
,
1
2
*
+
o
(
1
)
)
,
where o(1) denotes a quantity that goes to zero when λ goes to infinity.
Proof.
Since uk=Zr¯,y¯′′,λ+ϕ, we have
(4.24)
∫
D
ρ
g
(
r
,
y
′′
)
(
u
k
)
+
2
*
=
∫
D
ρ
g
(
r
,
y
′′
)
Z
r
¯
,
y
¯
′′
,
λ
2
*
+
O
(
∫
D
ρ
Z
r
¯
,
y
¯
′′
,
λ
2
*
-
1
|
ϕ
|
)
.
A direct computation leads to
(4.25)
∫
D
ρ
Z
r
¯
,
y
¯
′′
,
λ
2
*
-
1
|
ϕ
|
≤
C
∥
ϕ
∥
*
∫
D
ρ
(
∑
j
=
1
k
λ
N
-
2
2
(
1
+
λ
|
y
-
x
j
|
)
N
-
2
)
2
*
-
1
(
∑
j
=
1
k
λ
N
-
2
2
(
1
+
λ
|
y
-
x
j
|
)
N
-
2
2
+
τ
)
≤
C
∥
ϕ
∥
*
∫
D
ρ
(
∑
j
=
1
k
λ
N
-
2
2
(
1
+
λ
|
y
-
x
j
|
)
N
-
2
2
+
τ
)
2
*
-
1
(
∑
j
=
1
k
λ
N
-
2
2
(
1
+
λ
|
y
-
x
j
|
)
N
-
2
2
+
τ
)
≤
C
∥
ϕ
∥
*
∫
D
ρ
(
∑
j
=
1
k
λ
N
+
2
2
(
1
+
λ
|
y
-
x
j
|
)
N
+
2
2
+
τ
)
(
∑
j
=
1
k
λ
N
-
2
2
(
1
+
λ
|
y
-
x
j
|
)
N
-
2
2
+
τ
)
≤
C
k
∥
ϕ
∥
*
(
∫
D
ρ
λ
N
(
1
+
λ
|
y
-
x
1
|
)
N
+
2
τ
+
∑
j
=
2
k
1
(
λ
|
x
j
-
x
1
|
)
τ
∫
D
ρ
λ
N
(
1
+
λ
|
y
-
x
1
|
)
N
+
τ
)
≤
C
k
λ
1
+
ε
.
Similarly, we can prove that
(4.26)
∑
i
≠
j
∫
D
ρ
g
(
r
,
y
′′
)
Z
x
i
,
λ
Z
x
j
,
λ
2
*
-
1
≤
C
∑
i
≠
j
∫
λ
N
(
1
+
λ
|
y
-
x
i
|
)
N
-
2
(
1
+
λ
|
y
-
x
j
|
)
N
+
2
≤
C
∑
i
≠
j
1
(
λ
|
x
j
-
x
i
|
)
N
-
2
∫
λ
N
(
1
+
λ
|
y
-
x
1
|
)
N
+
2
≤
C
k
(
k
λ
)
N
-
2
=
o
(
k
λ
1
+
ε
)
.
On the other hand, noting that Zxj,λ=ξPUxj,λ=ξUxj,λ-ξψxj,λ, we have
(4.27)
∫
D
ρ
g
(
r
,
y
′′
)
Z
x
j
,
λ
2
*
=
∫
D
ρ
g
(
r
¯
,
y
¯
′′
)
Z
x
j
,
λ
2
*
+
∫
D
ρ
(
g
(
r
,
y
′′
)
-
g
(
r
¯
,
y
¯
′′
)
)
Z
x
j
,
λ
2
*
=
g
(
r
¯
,
y
¯
′′
)
∫
D
ρ
(
ξ
U
x
j
,
λ
)
2
*
+
O
(
∫
D
ρ
U
x
j
,
λ
2
*
-
1
ψ
x
j
,
λ
)
+
o
(
1
λ
1
-
ε
)
=
g
(
r
¯
,
y
¯
′′
)
∫
D
ρ
(
ξ
U
x
j
,
λ
)
2
*
+
o
(
1
λ
1
-
ε
)
=
g
(
r
¯
,
y
¯
′′
)
∫
ℝ
N
U
0
,
1
2
*
+
o
(
1
)
,
where in the last equality we used the scaling transformation z=λ(y-xj).
The result follows from (4.24), (4.25), (4.26) and (4.27). ∎
Applying Lemma 4.5 and (4.23) to (4.15) and (4.16), we obtain
k
(
1
2
*
r
¯
∂
K
(
r
¯
,
y
¯
′′
)
∂
r
¯
∫
ℝ
N
U
0
,
1
2
*
+
o
(
1
)
)
=
o
(
k
λ
2
)
,
k
(
∂
K
(
r
¯
,
y
¯
′′
)
∂
y
¯
i
∫
ℝ
N
U
0
,
1
2
*
+
o
(
1
)
)
=
o
(
k
λ
2
)
,
i
=
3
,
…
,
N
.
Therefore, the equations to determine (r¯,y¯′′) are
∂
K
(
r
¯
,
y
¯
′′
)
∂
r
¯
=
o
(
1
)
,
∂
K
(
r
¯
,
y
¯
′′
)
∂
y
¯
i
=
o
(
1
)
,
i
=
3
,
…
,
N
.
Proof of Theorem 1.1.
Through the discussion above, we have proved that (4.1), (4.2) and (4.3) are equivalent to
(4.28)
∂
K
(
r
¯
,
y
¯
′′
)
∂
r
¯
=
o
(
1
)
,
(4.29)
∂
K
(
r
¯
,
y
¯
′′
)
∂
y
¯
i
=
o
(
1
)
,
i
=
3
,
…
,
N
,
(4.30)
-
B
0
λ
3
+
B
3
k
N
-
2
λ
N
-
1
=
O
(
1
λ
3
+
ε
)
.
Let λ=tkN-2N-4, t∈[L0,L1]. Then from (4.30), we have
(4.31)
-
B
0
t
3
+
B
3
t
N
-
1
=
o
(
1
)
,
t
∈
[
L
0
,
L
1
]
.
Let
F
(
t
,
r
¯
,
y
¯
′′
)
=
(
∇
r
¯
,
y
¯
′′
K
(
r
¯
,
y
¯
′′
)
,
-
B
0
t
3
+
B
3
t
N
-
1
)
.
Then
deg
(
F
(
t
,
r
¯
,
y
¯
′′
)
,
[
L
0
,
L
1
]
×
Ω
(
r
0
,
y
0
′′
)
)
=
-
deg
(
∇
r
¯
,
y
¯
′′
K
(
r
¯
,
y
¯
′′
)
,
Ω
(
r
0
,
y
0
′′
)
)
≠
0
.
Thus, (4.28), (4.29) and (4.31) have a solution tk∈[L0,L1], (r¯k,y¯k′′)∈Ω(r0,y0′′). ∎
and
Ting Liu
