Home Mathematics Solutions for Fractional Schrödinger Equation Involving Critical Exponent via Local Pohozaev Identities
Article Open Access

Solutions for Fractional Schrödinger Equation Involving Critical Exponent via Local Pohozaev Identities

  • Yuxia Guo , Ting Liu and Jianjun Nie EMAIL logo
Published/Copyright: November 13, 2019

Abstract

We consider the following fractional Schrödinger equation involving critical exponent:

{ ( - Δ ) s u + V ( y ) u = u 2 s * - 1 in  N , u > 0 , y N ,

where N3 and 2s*=2NN-2s is the critical Sobolev exponent. Under some suitable assumptions of the potential function V(y), by using a finite-dimensional reduction method, combined with various local Pohazaev identities, we prove the existence of infinitely many solutions. Due to the nonlocality of the fractional Laplacian operator, we need to study the corresponding harmonic extension problem.

MSC 2010: 35R11; 47J30

1 Introduction

In this paper, we are concerned with the following problem:

(1.1) { ( - Δ ) s u + V ( y ) u = u 2 s * - 1 in  N , u > 0 , y N ,

where N3 and 2s*=2NN-2s is the critical Sobolev exponent. For any s(0,1), (-Δ)s is the fractional Laplacian in N, which is a nonlocal operator defined as

(1.2) ( - Δ ) s u ( y ) = c ( N , s ) P . V . N u ( y ) - u ( x ) | x - y | N + 2 s 𝑑 x = c ( N , s ) lim ϵ 0 + N B ϵ ( x ) u ( y ) - u ( x ) | x - y | N + 2 s 𝑑 x ,

where P.V. denotes the Cauchy principal value and c(N,s) is a constant depending on N and s. This operator is well defined in Cloc1,1s, where s={uLloc1:N|u(x)|1+|x|N+2s𝑑x<}. For more details on the fractional Laplacian, we referee to [16, 22] and the references therein.

The fractional Laplacian operator appears in divers areas, including biological modeling, physics and mathematical finances, and can be regarded as the infinitesimal generator of a stable Lévy process (see, for example, [5]). From the view point of mathematics, an important feature of the fractional Laplacian operator is its nonlocal property, which makes it more challenging than the classical Laplacian operator. Thus, problems with the fractional Laplacian have been extensively studied, both for the pure mathematical research and in view of concrete real world applications, see, for example, [1, 2, 4, 3, 11, 10, 9, 8, 7, 17, 19, 29, 23, 37, 39, 40, 44] and the references therein.

Solutions of (1.1) are related to the existence of standing wave solutions to the following fractional Schrödinger equation:

{ i t Ψ + ( - Δ ) s Ψ = F ( x , Ψ ) in  N , lim | x | | Ψ ( x , t ) | = 0 for all  t > 0 .

That is, solutions with the form Ψ(x,t)=e-ictu(x), where c is a constant.

In this paper, under a weaker symmetry condition for V(y), we will construct multi-bump solutions for (1.1) through a finite-dimensional reduction method, combined with various local Pohozaev identities. More precisely, we consider V(y)=V(|y|,y′′)=V(r,y′′), y=(y,y′′)2×N-2 and assume that:

  1. V ( y ) 0 is a bounded function that belongs to C2(N), and r2sV(r,y′′) has a critical point (r0,y0′′) satisfying r0>0, V(r0,y0′′)>0 and deg((r2sV(r,y′′)),(r0,y0′′))0.

Since the fractional operator is nonlocal, we have to overcome more difficulties than the Laplace equation. Such as, we need to study the corresponding harmonic extension problem and deal with (-Δ)s(ϕφ) in an appropriate process. Hence, some new ideas and techniques are needed. We will explain these later.

Before state the main results, let us first introduce some notations. Denote Ds(N) the completion of C0(N) under the norm (-Δ)s2uL2(N), where (-Δ)s2uL2(N) is defined by (N|ξ|2s|𝒢u(ξ)|2dξ)12, and 𝒢u is the Fourier transformation of u:

𝒢 u ( ξ ) = 1 ( 2 π ) N 2 N e - i ξ x u ( x ) 𝑑 x .

We will construct the solutions in the following the energy space:

H s ( N ) = { u D s ( N ) : N V ( y ) u 2 𝑑 y < + } ,

with the norm

u H s ( N ) = ( ( - Δ ) s 2 u L 2 ( N ) 2 + N V ( y ) u 2 𝑑 y ) 1 2 .

We define the functional I on Hs(N) by

I ( u ) = 1 2 N | ( - Δ ) s 2 u | 2 d y + 1 2 N V ( | y | , y ′′ ) u 2 d y - 1 2 s * N ( u ) + 2 s * d y ,

where (u)+=max(u,0). Then the solutions of problem (1.1) correspond to the critical points of the functional I.

It is well known that the functions

U x , λ ( y ) = C ( N , s ) ( λ 1 + λ 2 | y - x | 2 ) N - 2 s 2 , λ > 0 , x N ,

where C(N,s)=2N-2s2Γ(N+2s2)/Γ(N-2s2), are the only solutions of the problem (see [32])

( - Δ ) s u = u N + 2 s N - 2 s , u > 0  in  N .

Define

H s = { u : u H s ( N ) , u ( y 1 , y 2 , y ′′ ) = u ( y 1 , - y 2 , y ′′ ) ,
u ( r cos ( θ + 2 π j k ) , r sin ( θ + 2 π j k ) , y ′′ ) = u ( r cos θ , r sin θ , y ′′ ) } .

Let

x j = ( r ¯ cos 2 ( j - 1 ) π k , r ¯ sin 2 ( j - 1 ) π k , y ¯ ′′ ) , j = 1 , , k .

To construct the solution of (1.1), we hope to use Ux,λ(y) as an approximation solution. However, the decay of Uxj,λ is not fast enough for us when N6s. So, we need to cut off this function. Let δ>0 be a small constant such that r2sV(r,y′′)>0 if |(r,y′′)-(r0,y0′′)|10δ. Let ζ(y)=ζ(r,y′′) be a smooth function satisfying ζ=1 if |(r,y′′)-(r0,y0′′)|δ, ζ=0 if |(r,y′′)-(r0,y0′′)|2δ, |ζ|C and 0ζ1. Denote

Z x j , λ = ζ U x j , λ , Z r ¯ , y ¯ ′′ , λ * = j = 1 k U x j , λ , Z r ¯ , y ¯ ′′ , λ = j = 1 k Z x j , λ .

Let

Z i , 1 = Z x i , λ λ , Z i , 2 = Z x i , λ r ¯ , Z i , k = Z x i , λ y ¯ k ′′ , k = 3 , , N .

Then a direct computation shows that

Z i , 1 = O ( λ - 1 Z x i , λ ) , Z i , l = O ( λ Z x i , λ ) , l = 2 , , N .

We obtain the following result when N>4s.

Theorem 1.1.

Suppose that N3 and 2+N-4+N24<s<min(N4,1). If V(y) satisfies condition (V), then there is an integer k0>0 such that for any integer kk0, problem (1.1) has a solution uk of the form

u k = Z r ¯ k , y ¯ k ′′ , λ k + ϕ k ,

where ϕkHs, λk[L0kN-2sN-4s,L1kN-2sN-4s], and as k, λk-N-2s2ϕkL0, (r¯k,y¯k′′)(r0,y0′′).

Remark 1.2.

  1. In this case, we always assume that k>0 is a large integer, λ[L0kN-2sN-4s,L1kN-2sN-4s] for some constants L1>L0>0, and |(r¯,y¯′′)-(r0,y0′′)|θ, with θ>0 being a small constant.

  2. Let τ=N-4s2(N-2s). When N3, the condition 2+N-4+N24<s<min(N4,1) is equivalent to τ<s<1 and N>4s+2τ, which guarantee the existence of a small constant σ>0 in the proof. Moreover, it is easy to see 13<2+N-4+N24<12 for N3.

We are also interested in the case N=4s. Since s<1, we have N=3=4s. This case is corresponding to the Laplace equation with N=4.

Theorem 1.3.

Suppose that N=3=4s. If V(y) satisfies condition (V), then there is an integer k0>0 such that for any integer kk0, problem (1.1) has a solution uk of the form

u k = Z r ¯ k , y ¯ k ′′ , λ k + ϕ k ,

where ϕkHs, λk[eL0k2s,eL1k2s], and as k, λk-N-2s2ϕkL0, (r¯k,y¯k′′)(r0,y0′′).

Remark 1.4.

In this case, we assume that k>0 is a large integer, λk[eL0k2s,eL1k2s], and |(r¯,y¯′′)-(r0,y0′′)|θ.

When N>4s, we introduce the following norms:

u * = sup y N ( j = 1 k 1 ( 1 + λ | y - x j | ) N - 2 s 2 + τ ) - 1 λ - N - 2 s 2 | u ( y ) |

and

f * * = sup y N ( j = 1 k 1 ( 1 + λ | y - x j | ) N + 2 s 2 + τ ) - 1 λ - N + 2 s 2 | f ( y ) | ,

where τ=N-4s2(N-2s). When N=3=4s, we use the following norms:

u * = sup y 3 ( j = 1 k 1 ( 1 + λ | y - x j | ) N - 2 s 2 ) - 1 λ - N - 2 s 2 | u ( y ) |

and

f * * = sup y 3 ( j = 1 k 1 ( 1 + λ | y - x j | ) N + 2 s 2 ) - 1 λ - N + 2 s 2 | f ( y ) | ,

where N-2s2=34 and N+2s2=94.

We will prove Theorems 1.1 and 1.3 by a finite-dimensional reduction method, combined with various local Pohozaev identities. The finite-dimensional reduction method has been extensively used to construct solutions for equations with critical growth. We refer to [6, 15, 14, 13, 12, 18, 21, 24, 25, 26, 28, 30, 31, 33, 34, 36, 43, 42, 41] and the references therein. Roughly speaking, the outline to carry out the reduction argument is as follows: We first construct a good enough approximation solution and linearize the original problem around the approximation solution. Then we solve the corresponding finite-dimensional problem to obtain a true solution.

To finish the second step, we have to obtain some good enough estimates in the first step. In our case, since the fractional operator is nonlocal, we have to overcome more difficulties than the Laplace equation. One of them is that we will use Zr¯,y¯′′,λ as an approximate solution, but we have to deal with (-Δ)s(ζ(y)Uxj,λ(y)) and (-Δ)s(ζ(y)Zj,t(y)). By (1.2), we can deduce that

( - Δ ) s ( ζ ( y ) U x j , λ ( y ) ) = ζ ( y ) U x j , λ 2 s * - 1 ( y ) + c ( N , s ) lim ϵ 0 + N B ϵ ( x ) ( ζ ( y ) - ζ ( x ) ) U x j , λ ( x ) | x - y | N + 2 s 𝑑 x
= : ζ ( y ) U x j , λ 2 s * - 1 ( y ) + J .

In order to obtain a good enough estimate with **, we need to deal with some concrete difficulties, and devote ourselves to calculate the last principal value very carefully (see Lemma 2.5). More precisely, when N>4s, we need to show that

| J | C λ s + σ λ N + 2 s 2 ( 1 + λ | y - x j | ) N + 2 s 2 + τ .

Before the end of this introduction, we briefly outline the proof for the case of N>4s and point out some other difficulties (the idea of the proof for the other case is similar but with different estimates). We first use Zr¯,y¯′′,λ as an approximate solution to obtain a unique function ϕ(r¯,y¯′′,λ). Then the problem of finding critical points for I(u) can be reduced to that of finding critical points of F(r¯,y¯′′,λ)=I(Zr¯,y¯′′,λ+ϕ(r¯,y¯′′,λ)). In the second step, we solve the corresponding finite-dimensional problem to obtain a solution. However, we can only obtain ϕ*Cλs+σ (see Proposition 2.3). From Lemmas B.3 and B.5, we know that

(1.3) F λ = I ( Z r ¯ , y ¯ ′′ , λ ) λ + O ( k λ - 1 ϕ * 2 ) = k ( - 2 s B 1 λ 2 s + 1 V ( r ¯ , y ¯ ′′ ) + B 3 k N - 2 s λ N - 2 s + 1 + O ( 1 λ 2 s + 1 + σ ) ) ,
(1.4) F r ¯ = I ( Z r ¯ , y ¯ ′′ , λ ) r ¯ + O ( k λ ϕ * 2 ) = k ( B 1 λ 2 s V ( r ¯ , y ¯ ′′ ) r ¯ + j = 2 k B 2 r ¯ λ N - 2 s | x 1 - x j | N - 2 s + O ( 1 λ s + σ ) )

and

(1.5) F y ¯ j ′′ = I ( Z r ¯ , y ¯ ′′ , λ ) y ¯ j ′′ + O ( k λ ϕ * 2 ) = k ( B 1 λ 2 s V ( r ¯ , y ¯ ′′ ) y ¯ j ′′ + O ( 1 λ s + σ ) ) .

Note that the estimate of ϕ is only good enough for the expansion (1.3), but it destroys the main terms in the expansions of (1.4) and (1.5). To overcome this difficulty, following the idea in [35], instead of studying (1.4) and (1.5), we turn to study the following local Pohozaev identities:

(1.6) - ′′ ρ + t 1 - 2 s u ~ k ν u ~ k y i + 1 2 ′′ ρ + t 1 - 2 s | u ~ k | 2 ν i = B ρ ( - V ( r , y ′′ ) u k + ( u k ) + 2 s * - 1 ) u k y i , i = 3 , , N ,

and

- ′′ ρ + t 1 - 2 s u ~ k , Y u ~ k ν + 1 2 ′′ ρ + t 1 - 2 s | u ~ k | 2 Y , ν + 2 s - N 2 ρ + t 1 - 2 s u ~ k ν u ~ k
(1.7) = B ρ ( - V ( r , y ′′ ) u k + ( u k ) + 2 s * - 1 ) y , u k ,

where uk=Zr¯,y¯′′,λ+ϕ, u~k is the extension of uk (see below (1.8)),

ρ + = { Y = ( y , t ) : | Y - ( r 0 , y 0 ′′ , 0 ) | ρ  and  t > 0 } + N + 1 ,
ρ + = { Y = ( y , t ) : | y - ( r 0 , y 0 ′′ ) | ρ , t = 0 } N ,
′′ ρ + = { Y = ( y , t ) : | Y - ( r 0 , y 0 ′′ , 0 ) | = ρ , t > 0 } + N + 1 ,
ρ + = ρ + ′′ ρ + ,
B ρ = { y : | y - ( r 0 , y 0 ′′ ) | ρ } N .

For any uDs(N), u~ is defined by

(1.8) u ~ ( y , t ) = 𝒫 s [ u ] := N P s ( y - ξ , t ) u ( ξ ) 𝑑 ξ , ( y , t ) + N + 1 := N × ( 0 , + ) ,

where

P s ( x , t ) = β ( N , s ) t 2 s ( | x | 2 + t 2 ) N + 2 s 2 ,

with constant β(N,s) such that NPs(x,1)𝑑x=1. Moreover, u~ satisfies (see [11])

(1.9) div ( t 1 - 2 s u ~ ) = 0 in  + N + 1

and

(1.10) - lim t 0 t 1 - 2 s t u ~ ( y , t ) = ω s ( - Δ ) s u ( y ) on  N ,

where ωs=21-2sΓ(1-s)/Γ(s).

Due to the nonlocality of the fractional Laplacian operator, we can not built a local Pohozaev identity for problem (1.1). So, we need to study the corresponding harmonic extension problem (1.9) and (1.10). The relationship between u and u~ is (1.8). Hence, we have to give some estimates for this kind of integrals. The local Pohozaev identities (1.6) and (1.7) are much more complicated. We have to integrate one more time than the Laplacian operator case. This is very difficult when we derive some sharp estimates for each term in (1.6) and (1.7). We need a lot of preliminary lemmas. For example, some suitable estimates on Z~xi,λ and ϕ~ are established in Lemmas A.5 and A.6.

Our paper is organized as follows. In Section 2, we perform a finite-dimensional reduction. We prove Theorem 1.1 in Section 3. Theorem 1.3 is proved in Section 4. In Appendix A, we give some essential estimates. We put the energy expansions for I(Zr¯,y¯′′,λ+ϕ(r¯,y¯′′,λ)),Zr¯,y¯′′,λλ, I(Zr¯,y¯′′,λ+ϕ(r¯,y¯′′,λ)),Zr¯,y¯′′,λr¯ and I(Zr¯,y¯′′,λ+ϕ(r¯,y¯′′,λ)),Zr¯,y¯′′,λy¯′′ in Appendix B.

2 Finite-Dimensional Reduction

In this section, we perform a finite-dimensional reduction by using Zr¯,y¯′′,λ as an approximation solution. We consider the following linearized problem:

(2.1) { ( - Δ ) s ϕ + V ( r , y ′′ ) ϕ - ( 2 s * - 1 ) Z r ¯ , y ¯ ′′ , λ 2 s * - 2 ϕ = h + l = 1 N c l i = 1 k Z x i , λ 2 s * - 2 Z i , l , u H s , i = 1 k N Z x i , λ 2 s * - 2 Z i , l ϕ = 0 , l = 1 , 2 , , N ,

for some numbers cl.

Lemma 2.1.

Suppose that 2+N-4+N24<s<min(N4,1) or N=3=4s and ϕk solves problem (2.1). If hk**0 as k, then ϕk*0 as k.

Proof.

We prove this lemma by contradiction. We first consider the case N>4s. We assume that there exist hk with hk0 as k, ϕkc>0 with λ=λk, λk[L0kN-2sN-4s,L1kN-2sN-4s] and (r¯k,y¯k′′)(r0,y0′′). Without loss of generality, we can assume that ϕk1. For simplicity, we drop the subscript k.

Firstly, we have

| ϕ ( y ) | C N 1 | y - z | N - 2 s Z r ¯ , y ¯ ′′ , λ 2 s * - 2 | ϕ | d z + C N 1 | y - z | N - 2 s [ | h | + | l = 1 N c l i = 1 k Z x i , λ 2 s * - 2 Z i , l | ] d z = : A 1 + A 2 .

For the first term A1, by Lemmas A.1 and A.2, we can deduce that

| A 1 | C ϕ N 1 | y - z | N - 2 s Z r ¯ , y ¯ ′′ , λ 2 s * - 2 i = 1 k λ N - 2 s 2 ( 1 + λ | z - x i | ) N - 2 s 2 + τ d z C ϕ λ N - 2 s 2 i = 1 k 1 ( 1 + λ | y - x i | ) N - 2 s 2 + τ + θ ,

where θ is a small constant. For the second term A2, we make use of Lemma A.2, so that

| A 2 | C h N i = 1 k λ N + 2 s 2 | y - z | N - 2 s ( 1 + λ | z - x i | ) N + 2 s 2 + τ d z + C l = 1 N | c l | N i = 1 k λ N + 2 s 2 + n l | y - z | N - 2 s ( 1 + λ | z - x i | ) N + 2 s d z
C h λ N - 2 s 2 i = 1 k 1 ( 1 + λ | y - x i | ) N - 2 s 2 + τ + C l = 1 N | c l | λ N - 2 s 2 + n l i = 1 k 1 ( 1 + λ | y - x i | ) N - 2 s 2 + τ ,

where n1=-1, nl=1 for l=2,,N. Then we have

(2.2) ( i = 1 k 1 ( 1 + λ | y - x i | ) N - 2 s 2 + τ ) - 1 λ - N - 2 s 2 | ϕ | C ϕ i = 1 k 1 ( 1 + λ | y - x i | ) N - 2 s 2 + τ + θ i = 1 k 1 ( 1 + λ | y - x i | ) N - 2 s 2 + τ + C h + C l = 1 N | c l | λ n l .

Multiplying both sides of (2.1) by Z1,t, we have

(2.3) l = 1 N c l i = 1 k N Z x i , λ 2 s * - 2 Z i , l Z 1 , t = ( - Δ ) s ϕ - V ( r , y ′′ ) ϕ - ( 2 s * - 1 ) Z r ¯ , y ¯ ′′ , λ 2 s * - 2 ϕ , Z 1 , t - h , Z 1 , t .

First of all, there exists a constant c¯>0 such that

(2.4) i = 1 k N Z x i , λ 2 s * - 2 Z i , l Z 1 , t { = ( c ¯ + o ( 1 ) ) λ 2 n t , l = t , c ¯ λ n t λ n l λ N , l t .

Since τ<s and N-2s2-τ>s, we have

| V ( r , y ′′ ) ϕ , Z 1 , t | C ϕ * N ζ λ N - 2 s + n t ( 1 + λ | y - x 1 | ) N - 2 s i = 1 k 1 ( 1 + λ | y - x i | ) N - 2 s 2 + τ
C ϕ * λ N - 2 s + n t [ N ζ ( 1 + λ | y - x 1 | ) 3 N - 6 s 2 + τ
+ i = 2 k 1 ( λ | x 1 - x i | ) τ N ζ ( 1 ( 1 + λ | y - x 1 | ) 3 N - 6 s 2 + 1 ( 1 + λ | y - x i | ) 3 N - 6 s 2 ) ]
(2.5) C ϕ * N ζ λ N - 2 s + τ + n t ( 1 + λ | y - x 1 | ) 3 N - 6 s 2 C λ n t ϕ * log λ λ min ( 2 s - τ , N - 2 s 2 - τ ) C λ n t ϕ * λ s + σ

and

(2.6) | h , Z 1 , t | C h N λ N + n t ( 1 + λ | y - x 1 | ) N - 2 s i = 1 k 1 ( 1 + λ | y - x i | ) N + 2 s 2 + τ C λ n t h .

Moreover, one has

(2.7) | ( - Δ ) s ϕ - ( 2 s * - 1 ) Z r ¯ , y ¯ ′′ , λ 2 s * - 2 ϕ , Z 1 , t | C λ n t ϕ λ s + σ .

Combining (2.3), (2.4), (2.5), (2.6) and (2.7), we have

| c t | C λ n t ( ϕ λ σ + h ) + C λ n t l t λ n l | c l | λ N .

This implies that

l = 1 N | c l | λ n l C ( ϕ λ σ + h ) .

Thus, by (2.2) and ϕ=1, there exists R>0 such that

(2.8) λ - N - 2 s 2 ϕ ( y ) L ( B R / λ ( x i ) ) a > 0

for some i. As a result, we have that ϕ~=λ-N-2s2ϕ(yλ+xi) converges uniformly, in any compact set, to a solution u of the following equation:

( - Δ ) s u - ( 2 s * - 1 ) U 0 , Λ 2 s * - 2 u = 0 in  N ,

for some 0<Λ1ΛΛ2. Since u is perpendicular to the kernel of this equation, u=0. This is a contradiction to (2.8).

When N=3=4s, we take λk[eL0k2s,eL1k2s] and τ=0 in the above proofs. We also need to alter (2.5) as follows:

| V ( r , y ′′ ) ϕ , Z 1 , t | C k ϕ * λ 3 - 2 s + n t 3 ζ ( 1 + λ | y - x 1 | ) 9 4 C k λ n t ϕ * λ 3 4 = C k λ n t ϕ * λ s .

The proof is complete. ∎

Using the same argument as in the proof of [20, Proposition 4.1], we can obtain the following proposition.

Proposition 2.2.

There exist k0>0 and a constant C>0, independent of k, such that for all kk0 and all hL(RN), problem (2.1) has a unique solution ϕ=Lk(h). Besides,

L k ( h ) * C h * * , | c l | C λ n l h * * .

Now we consider the following problem:

(2.9) { ( - Δ ) s ( Z r ¯ , y ¯ ′′ , λ + ϕ ) + V ( r , y ′′ ) ( Z r ¯ , y ¯ ′′ , λ + ϕ ) = ( Z r ¯ , y ¯ ′′ , λ + ϕ ) 2 s * - 1 + l = 1 N c l i = 1 k Z x i , λ 2 s * - 2 Z i , l in  N , ϕ H s , i = 1 k N Z x i , λ 2 s * - 2 Z i , l ϕ = 0 , l = 1 , , N .

In the rest of this section, we devote ourselves to the proof of the following proposition by using the contraction mapping theorem.

Proposition 2.3.

There exist k0>0 and a constant C>0, independent of k, such that the following hold:

  1. When 2 + N - 4 + N 2 4 < s < min ( N 4 , 1 ) for all k k 0 , L0kN-2sN-4sλL1kN-2sN-4s, |(r¯,y¯′′)-(r0,y0′′)|θ, problem (2.9) has a unique solution ϕ=ϕ(r¯,y¯′′,λ) satisfying

    ϕ * C λ s + σ , | c l | C λ s + σ ,

    where σ > 0 is a small constant.

  2. When N = 3 = 4 s for all k k 0 , eL0k2sλeL1k2s, |(r¯,y¯′′)-(r0,y0′′)|θ, problem (2.9) has a unique solution ϕ=ϕ(r¯,y¯′′,λ) satisfying

    ϕ * C λ s , | c l | C λ s .

We rewrite (2.9) as

{ ( - Δ ) s ϕ + V ( r , y ′′ ) ϕ - ( 2 s * - 1 ) ( Z r ¯ , y ¯ ′′ , λ ) 2 s * - 2 ϕ = ( ϕ ) + l k ( y ) + l = 1 N c l i = 1 k Z x i , λ 2 s * - 2 Z i , l in  N , ϕ H s , i = 1 k N Z x i , λ 2 s * - 2 Z i , l ϕ = 0 , l = 1 , , N ,

where

( ϕ ) = ( Z r ¯ , y ¯ ′′ , λ + ϕ ) + 2 s * - 1 - Z r ¯ , y ¯ ′′ , λ 2 s * - 1 - ( 2 s * - 1 ) Z r ¯ , y ¯ ′′ , λ 2 s * - 2 ϕ

and

l k ( y ) = ( Z r ¯ , y ¯ ′′ , λ 2 s * - 1 ( y ) - ζ ( y ) j = 1 k U x j , λ 2 s * - 1 ( y ) ) - V ( r , y ′′ ) Z r ¯ , y ¯ ′′ , λ ( y ) - j = 1 k c ( N , s ) lim ϵ 0 + N B ϵ ( x ) ( ζ ( y ) - ζ ( x ) ) U x j , λ ( x ) | x - y | N + 2 s 𝑑 x
= : J 1 + J 2 + J 3 .

In order to use the contraction mapping theorem to prove Proposition 2.3, we need to estimate (ϕ) and lk(y). In the following, we assume that ϕ is small.

Lemma 2.4.

There exists a constant C>0, independent of k, such that:

  1. when 2 + N - 4 + N 2 4 < s < min ( N 4 , 1 ) , we have ( ϕ ) * * C λ 2 s ( N - 4 s ) ( N - 2 s ) 2 ϕ * min ( 2 , 2 s * - 1 ) ,

  2. when N = 3 = 4 s , we have ( ϕ ) * * C ( ln λ ) 1 s ϕ * 2 .

Proof.

We first prove (a). If 2s*3, then using the Hölder inequality, we obtain

| ( ϕ ) | C ϕ * 2 s * - 1 ( j = 1 k λ N - 2 s 2 ( 1 + λ | y - x j | ) N - 2 s 2 + τ ) 2 s * - 1
C ϕ * 2 s * - 1 λ N + 2 s 2 j = 1 k 1 ( 1 + λ | y - x j | ) N + 2 s 2 + τ ( j = 1 k 1 ( 1 + λ | y - x j | ) τ ) 4 s N - 2 s
C λ 2 s ( N - 4 s ) ( N - 2 s ) 2 ϕ * 2 s * - 1 λ N + 2 s 2 j = 1 k 1 ( 1 + λ | y - x j | ) N + 2 s 2 + τ .

When 2s*>3, we have

| ( ϕ ) | C ϕ * 2 ( j = 1 k λ N - 2 s 2 ( 1 + λ | y - x j | ) N - 2 s 2 + τ ) 2 ( j = 1 k λ N - 2 s 2 ( 1 + λ | y - x j | ) N - 2 s ) 2 s * - 3
+ C ϕ * 2 s * - 1 ( j = 1 k λ N - 2 s 2 ( 1 + λ | y - x j | ) N - 2 s 2 + τ ) 2 s * - 1
C ( ϕ * 2 + ϕ * 2 s * - 1 ) λ N + 2 s 2 ( j = 1 k 1 ( 1 + λ | y - x j | ) N - 2 s 2 + τ ) 2 s * - 1
C λ 2 s ( N - 4 s ) ( N - 2 s ) 2 ϕ * 2 λ N + 2 s 2 j = 1 k 1 ( 1 + λ | y - x j | ) N + 2 s 2 + τ .

Hence, we obtain (ϕ)**Cλ2s(N-4s)(N-2s)2ϕ*min(2,2s*-1).

Now, we prove (b):

| ( ϕ ) | C ϕ * 2 ( j = 1 k λ N - 2 s 2 ( 1 + λ | y - x j | ) N - 2 s 2 ) 2 ( j = 1 k λ N - 2 s 2 ( 1 + λ | y - x j | ) N - 2 s ) 2 s * - 3 + C ϕ * 2 s * - 1 ( j = 1 k λ N - 2 s 2 ( 1 + λ | y - x j | ) N - 2 s 2 ) 2 s * - 1
C ( ϕ * 2 + ϕ * 3 ) λ N + 2 s 2 ( j = 1 k 1 ( 1 + λ | y - x j | ) 3 4 ) 3 C ( ln λ ) 1 s ϕ * 2 λ N + 2 s 2 j = 1 k 1 ( 1 + λ | y - x j | ) 9 4 .

So, we have (ϕ)**C(lnλ)1sϕ*2. ∎

Next, we estimate lk(y).

Lemma 2.5.

There exists a constant C>0, independent of k, such that the following hold:

  1. If 2 + N - 4 + N 2 4 < s < min ( N 4 , 1 ) , then there exists a small σ > 0 such that l k * * C λ s + σ .

  2. If N = 3 = 4 s , then l k * * C λ s .

Proof.

We first prove (a). By symmetry, we can assume that yΩ1. Then |y-xj||y-x1|. We first estimate the term J1. We have

| J 1 | C [ ( j = 2 k U x j , λ ) 2 s * - 1 + U x 1 , λ 2 s * - 2 j = 2 k U x j , λ + j = 2 k U x j , λ 2 s * - 1 ]
C λ N + 2 s 2 ( j = 2 k 1 ( 1 + λ | y - x j | ) N - 2 s ) 2 s * - 1 + C λ N + 2 s 2 ( 1 + λ | y - x 1 | ) 4 s j = 2 k 1 ( 1 + λ | y - x j | ) N - 2 s .

If N-2sN+2s2-τ, then we have

1 ( 1 + λ | y - x 1 | ) 4 s j = 2 k 1 ( 1 + λ | y - x j | ) N - 2 s 1 ( 1 + λ | y - x 1 | ) N + 2 s 2 + τ j = 2 k 1 ( 1 + λ | y - x j | ) N + 2 s 2 - τ
1 ( 1 + λ | y - x 1 | ) N + 2 s 2 + τ ( k λ ) N + 2 s 2 - τ
1 λ s + σ 1 ( 1 + λ | y - x 1 | ) N + 2 s 2 + τ .

If N-2s<N+2s2-τ, then 4s>N+2s2+τ, and we obtain that

1 ( 1 + λ | y - x 1 | ) 4 s j = 2 k 1 ( 1 + λ | y - x j | ) N - 2 s 1 ( 1 + λ | y - x 1 | ) N + 2 s 2 + τ j = 2 k 1 ( λ | x 1 - x j | ) N - 2 s
1 ( 1 + λ | y - x 1 | ) N + 2 s 2 + τ ( k λ ) N - 2 s .

Using the Hölder inequality, we have

( j = 2 k 1 ( 1 + λ | y - x j | ) N - 2 s ) 2 s * - 1 j = 2 k 1 ( 1 + λ | y - x j | ) N + 2 s 2 + τ ( j = 2 k 1 ( 1 + λ | y - x j | ) N + 2 s 4 s ( N - 2 s 2 - N - 2 s N + 2 s τ ) ) 4 s N - 2 s
C j = 2 k 1 ( 1 + λ | y - x j | ) N + 2 s 2 + τ ( k λ ) N + 2 s N - 2 s ( N - 2 s 2 - N - 2 s N + 2 s τ )
C j = 2 k 1 ( 1 + λ | y - x j | ) N + 2 s 2 + τ ( 1 λ ) s + σ .

Thus,

J 1 * * C ( 1 λ ) s + σ .

Now, we estimate J2. Note that ζ=0 when |(r,y′′)-(r0,y0′′)|2δ and 1λC1+λ|y-xj| when |(r,y′′)-(r0,y0′′)|<2δ. We have

| J 2 | C λ 2 s λ N + 2 s 2 j = 1 k ζ ( 1 + λ | y - x j | ) N - 2 s C λ min ( 2 s , N - N + 2 s 2 - τ ) λ N + 2 s 2 j = 1 k 1 ( 1 + λ | y - x j | ) N + 2 s 2 + τ .

If N>4s+2τ, then J2**Cλs+σ.

We have

J 3 = j = 1 k c ( N , s ) ( lim ϵ 0 + B δ / 4 ( y ) B ϵ ( y ) ( ζ ( y ) - ζ ( x ) ) U x j , λ ( x ) | x - y | N + 2 s 𝑑 x + N B δ / 4 ( y ) ( ζ ( y ) - ζ ( x ) ) U x j , λ ( x ) | x - y | N + 2 s 𝑑 x )
= : j = 1 k c ( N , s ) ( J 31 + J 32 ) .

We first estimate J31. From the definition of function ζ, we have ζ(y)-ζ(x)=0 for x,yBδ(xj) or x,yNB2δ(xj)¯. So, J310 only if Bδ/4(y)B5/2δ(xj)B1/2δ(xj). We have

3 4 δ | y - x j | | x - y | + | x - x j | δ 4 + | x - x j | 3 2 | x - x j | 15 4 δ for  B δ / 4 ( y ) B 5 / 2 δ ( x j ) B 1 / 2 δ ( x j ) .

Furthermore, we divide J31 as follows:

J 31 = lim ϵ 0 + B δ / 4 ( y ) B ϵ ( y ) ζ ( y ) ( y - x ) U x j , λ ( x ) | x - y | N + 2 s d x + O ( lim ϵ 0 + B δ / 4 ( y ) B ϵ ( y ) U x j , λ ( x ) | x - y | N + 2 s - 2 d x ) = : J 311 + J 312 .

Note that Bδ/4(y)Bϵ(y) is a symmetric set. Then, by the mean value theorem, we get that

| J 311 | = | lim ϵ 0 + B δ / 4 ( y ) B ϵ ( y ) ζ ( y ) ( y - x ) U x j , λ ( x ) | x - y | N + 2 s |
= | C ( N , s ) lim ϵ 0 + B δ / 4 ( 0 ) B ϵ ( 0 ) ζ ( y ) z | z | N + 2 s λ N - 2 s 2 ( 1 + λ 2 | z + y - x j | 2 ) N - 2 s 2 |
= | C ( N , s ) λ N - 2 s 2 2 lim ϵ 0 + B δ / 4 ( 0 ) B ϵ ( 0 ) ζ ( y ) z | z | N + 2 s ( 1 ( 1 + λ 2 | z + y - x j | 2 ) N - 2 s 2 - 1 ( 1 + λ 2 | - z + y - x j | 2 ) N - 2 s 2 ) |
C λ N - 2 s 2 + 1 B δ / 4 ( 0 ) | ζ ( y ) | | z | N + 2 s - 2 1 ( 1 + λ | ( 2 ϑ - 1 ) z + y - x j | ) N - 2 s + 1
C λ s + σ λ N + 2 s 2 1 ( 1 + λ | y - x j | ) N + 2 s 2 + τ ,

for 0<ϑ<1 and since |(2ϑ-1)z+y-xj||y-xj|-|(2ϑ-1)z|23|y-xj| for zBδ/4(0). Similarly, we can obtain

| J 312 | C λ s + σ λ N + 2 s 2 1 ( 1 + λ | y - x j | ) N + 2 s 2 + τ .

For the term J32, we divide three cases: Case 1: If yBδ(xj), then

| J 32 | N ( B δ / 4 ( y ) B δ ( x j ) ) 1 | x - y | N + 2 s λ N - 2 s 2 ( 1 + λ | x - x j | ) N - 2 s
C λ 2 s λ N + 2 s 2 1 ( 1 + λ | y - x j | ) N - 2 s N B δ / 4 ( y ) 1 | x - y | N + 2 s
C λ s + σ λ N + 2 s 2 1 ( 1 + λ | y - x j | ) N + 2 s 2 + τ .

Case 2: If δ|y-xj|3δ, then, by Lemma A.3,

| J 32 | N B δ / 4 ( y ) 1 | x - y | N + 2 s λ N - 2 s 2 ( 1 + λ | x - x j | ) N - 2 s
C λ N + 2 s 2 N B δ λ / 4 ( λ y ) 1 | z - λ y | N + 2 s 1 ( 1 + | z - λ x j | ) N - 2 s
C λ N + 2 s 2 ( 1 ( λ | y - x j | ) N + 1 λ 2 s 1 ( λ | y - x j | ) N - 2 s )
C λ s + σ λ N + 2 s 2 1 ( 1 + λ | y - x j | ) N + 2 s 2 + τ .

Case 3: Suppose that |y-xj|>3δ. Note that |x-y||y-xj|-|x-xj|13|y-xj| when |y-xj|3δ and |x-xj|2δ. Then we have

| J 32 | B 2 δ ( x j ) 1 | x - y | N + 2 s λ N - 2 s 2 ( 1 + λ | x - x j | ) N - 2 s
C λ N - 2 s 2 B 2 δ ( x j ) 1 | x - y | N + 2 s 1 | x - x j | N - 2 s
C λ N + 2 s 2 λ N 1 | y - x j | N + 2 s 2 + τ B 2 δ ( x j ) 1 | x - x j | N - 2 s
C λ s + σ λ N + 2 s 2 1 ( 1 + λ | y - x j | ) N + 2 s 2 + τ .

Then we obtain J3**Cλs+σ.

As a result, we have proved that lk**Cλs+σ.

Now, we prove (b). Since 4s>94, we have

1 ( 1 + λ | y - x 1 | ) 4 s j = 2 k 1 ( 1 + λ | y - x j | ) N - 2 s 1 ( 1 + λ | y - x 1 | ) 9 4 j = 2 k 1 ( λ | x 1 - x j | ) N - 2 s
1 ( 1 + λ | y - x 1 | ) 9 4 ( k λ ) 3 2 .

Using the Hölder inequality, we have

( j = 2 k 1 ( 1 + λ | y - x j | ) N - 2 s ) 2 s * - 1 j = 2 k 1 ( 1 + λ | y - x j | ) 9 4 ( j = 2 k 1 ( 1 + λ | y - x j | ) N + 2 s 4 s 3 4 ) 4 s N - 2 s
C j = 2 k 1 ( 1 + λ | y - x j | ) 9 4 ( k λ ) 9 4 .

Thus,

J 1 * * C ( 1 λ ) s .

For the term J2, we have

| J 2 | C λ s λ 9 4 j = 1 k 1 ( 1 + λ | y - x j | ) N - s = C λ s λ 9 4 j = 1 k 1 ( 1 + λ | y - x j | ) 9 4 .

Now, we estimate the term J3:

| J 31 | C λ 2 s - 1 λ 9 4 B δ / 4 ( 0 ) | ζ ( y ) | | z | N + 2 s - 2 1 ( 1 + λ | ( 2 ϑ - 1 ) z + y - x j | ) 2 s + 1
+ C λ 2 s λ 9 4 B δ / 4 ( 0 ) 1 | z | N + 2 s - 2 1 ( 1 + λ | z + y - x j | ) N - 2 s
C λ s λ 9 4 1 ( 1 + λ | y - x j | ) 9 4 .

Similar to (a), we can deduce that

| J 32 | C λ s λ 9 4 1 ( 1 + λ | y - x j | ) 9 4 .

So, we obtain J3**Cλs.

As a result, we have proved that lk**Cλs. ∎

Proof of Proposition 2.3.

Let y=(y,y′′), y2, y′′N-2. If N>4s, we set

E = { u : u C ( N ) H s , u 1 λ s , i = 1 k N Z x i , λ 2 s * - 2 Z i , l u = 0 , l = 1 , , N } .

By Proposition 2.2, the solution ϕ of (2.9) is equivalent to the following fixed point problem:

ϕ = A ( ϕ ) = : L k ( ( ϕ ) ) + L k ( l k ) .

Hence, it is sufficient to prove that the operator A is a contraction map from the complete space E to itself. In fact, if ϕL(N), then, by [38, Proposition 2.9], we can obtain ϕC(N). For any ϕE, by Proposition 2.2, Lemma 2.4 and Lemma 2.5, we have

A ( ϕ ) C L k ( ( ϕ ) ) + C L k ( l k ) C [ ( ϕ ) + l k ]
C [ λ 2 s ( N - 4 s ) ( N - 2 s ) 2 λ s × min ( 2 , 2 s * - 1 ) + 1 λ s + σ ] C λ s ,

since 2s(N-4s)(N-2s)2<s×min(1,2s*-2). This shows that A maps E to E itself and E is invariant under A operator.

If 2s*3, then for all ϕ1,ϕ2E, we have

A ( ϕ 1 ) - A ( ϕ 2 ) * = L k ( ( ϕ 1 ) - ( ϕ 2 ) ) C ( ϕ 1 ) - ( ϕ 2 )
C ( | ϕ 1 | + | ϕ 2 | ) 2 s * - 2 | ϕ 1 - ϕ 2 | 1 2 ϕ 1 - ϕ 2 .

The case 2s*>3 can be discussed in a similar way.

Hence, A is a contraction map. The Banach fixed point theorem tells us that there exists a unique solution ϕE for problem (2.9).

Finally, by Proposition 2.2, we have

ϕ C ( 1 λ ) s + σ and | c l | C ( ϕ ) + l k C ( 1 λ ) s + σ .

When N=3=4s, we set

E = { u : u C ( N ) H s , u C 0 λ s , i = 1 k N Z x i , λ 2 s * - 2 Z i , l u = 0 , l = 1 , , N } ,

where C0> is a large constant such that

A ( ϕ ) C L k ( ( ϕ ) ) + C L k ( l k ) C [ C ( ln λ ) 1 s ϕ * 2 + C λ s ] C 0 λ s .

By the process of the case N>4s, we can obtain the result. ∎

3 Proof of the Main Theorem: The Case N>4s

Let ϕ be the function obtained in Proposition 2.3 and uk=Zr¯,y¯′′,λ+ϕ. In order to use local Pohozaev identities, we quote the extension of uk, that is, u~k=Z~r¯,y¯′′,λ+ϕ~. Z~r¯,y¯′′ and ϕ~ are extensions of Zr¯,y¯′′ and ϕ, respectively. Then we have

(3.1) { div ( t 1 - 2 s u ~ k ) = 0 in  + N + 1 , - lim t 0 + t 1 - 2 s t u ~ k = ω s ( - V ( r , y ′′ ) u k + ( u k ) + 2 s * - 1 + l = 1 N c l j = 1 k Z x j , λ 2 s * - 2 Z j , l ) on  N .

Without loss of generality, we may assume ωs=1. Multiplying (3.1) by u~kyi (i=3,,N) and u~k,Y, respectively, and then integrating by parts, we have the following two Pohozaev identities:

- ′′ ρ + t 1 - 2 s u ~ k ν u ~ k y i + 1 2 ′′ ρ + t 1 - 2 s | u ~ k | 2 ν i
(3.2) = B ρ ( - V ( r , y ′′ ) u k + ( u k ) + 2 s * - 1 + l = 1 N c l j = 1 k Z x j , λ 2 s * - 2 Z j , l ) u k y i , i = 3 , , N ,

and

- ′′ ρ + t 1 - 2 s u ~ k , Y u ~ k ν + 1 2 ′′ ρ + t 1 - 2 s | u ~ k | 2 Y , ν + 2 s - N 2 ρ + t 1 - 2 s u ~ k ν u ~ k
(3.3) = B ρ ( - V ( r , y ′′ ) u k + ( u k ) + 2 s * - 1 + l = 1 N c l j = 1 k Z x j , λ 2 s * - 2 Z j , l ) y , u k .

In the following, assume ρ(2δ,5δ). We have the following lemma.

Lemma 3.1.

Suppose that (r¯,y¯′′,λ) satisfies

(3.4) - ′′ ρ + t 1 - 2 s u ~ k ν u ~ k y i + 1 2 ′′ ρ + t 1 - 2 s | u ~ k | 2 ν i = B ρ ( - V ( r , y ′′ ) u k + ( u k ) + 2 s * - 1 ) u k y i , i = 3 , , N ,
- ′′ ρ + t 1 - 2 s u ~ k , Y u ~ k ν + 1 2 ′′ ρ + t 1 - 2 s | u ~ k | 2 Y , ν + 2 s - N 2 ρ + t 1 - 2 s u ~ k ν u ~ k
(3.5) = B ρ ( - V ( r , y ′′ ) u k + ( u k ) + 2 s * - 1 ) y , u k

and

(3.6) N ( ( - Δ ) s u k + V ( r , y ′′ ) u k - ( u k ) + 2 s * - 1 ) Z r ¯ , y ¯ ′′ , λ λ = 0 .

Then we have cl=0, l=1,,N.

Proof.

By (3.2), (3.3), (3.4) and (3.5), we have

(3.7) l = 1 N c l j = 1 k B ρ Z x j , λ 2 s * - 2 Z j , l u k y i = 0 , i = 3 , , N , l = 1 N c l j = 1 k B ρ Z x j , λ 2 s * - 2 Z j , l y , u k = 0 .

Note that ζ=0 in NBρ. By (3.6) and (3.7), we have

(3.8) l = 1 N c l j = 1 k N Z x j , λ 2 s * - 2 Z j , l v = l = 1 N c l j = 1 k B ρ Z x j , λ 2 s * - 2 Z j , l v = 0

for v=ukyi, v=uk,y and v=Zr¯,y¯′′,λλ.

By direct calculations, we have

j = 1 k B ρ Z x j , λ 2 s * - 2 Z j , 2 y , y Z r ¯ , y ¯ ′′ , λ = k λ 2 ( a 1 + o ( 1 ) ) ,
j = 1 k B ρ Z x j , λ 2 s * - 2 Z j , i Z r ¯ , y ¯ ′′ , λ y i = k λ 2 ( a 2 + o ( 1 ) ) , i = 3 , , N ,

and

(3.9) j = 1 k B ρ Z x j , λ 2 s * - 2 Z j , 1 Z r ¯ , y ¯ ′′ , λ λ = k λ 2 ( a 3 + o ( 1 ) ) ,

where a1>0, a2>0 and a3>0.

Furthermore, we have

l = 1 N c l j = 1 k B ρ Z x j , λ 2 s * - 2 Z j , l y , Z r ¯ , y ¯ ′′ , λ = j = 1 k B ρ Z x j , λ 2 s * - 2 Z j , 2 y , y Z r ¯ , y ¯ ′′ , λ c 2 + O ( k λ N - 2 | c 2 | ) + o ( k λ 2 l = 3 N | c l | ) + o ( k | c 1 | )
(3.10) = k λ 2 ( a 1 + o ( 1 ) ) c 2 + o ( k λ 2 l = 3 N | c l | ) + o ( k | c 1 | )

and

l = 1 N c l j = 1 k B ρ Z x j , λ 2 s * - 2 Z j , l Z r ¯ , y ¯ ′′ , λ y i = j = 1 k B ρ Z x j , λ 2 s * - 2 Z j , i Z r ¯ , y ¯ ′′ , λ y i c i + o ( k λ 2 l 1 , i | c l | ) + o ( k | c 1 | )
(3.11) = k λ 2 ( a 2 + o ( 1 ) ) c i + o ( k λ 2 l 1 , i | c l | ) + o ( k | c 1 | ) , i = 3 , , N .

Since ϕ is a solution to (2.9), by fractional elliptical equation estimates (see, for example, [38, Proposition 2.9] and [16, Theorem 12.2.1]), we can obtain ϕC1(Bρ). By integrating by parts and using ϕ*Cλs+σ, we have

l = 1 N c l j = 1 k B ρ Z x j , λ 2 s * - 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¯′′,λyi.

It follows from (3.8) that

(3.12) l = 1 N c l j = 1 k B ρ Z x j , λ 2 s * - 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¯′′,λyi.

By (3.10), (3.11) and (3.12), we have

(3.13) c l = o ( 1 λ 2 | c 1 | ) , l = 2 , , N .

From (3.8), (3.9) and (3.13), we deduce that

0 = l = 1 N c l j = 1 k B ρ Z x j , λ 2 s * - 2 Z j , 1 Z r ¯ , y ¯ ′′ , λ λ = j = 1 k B ρ Z x j , λ 2 s * - 2 Z j , 1 Z r ¯ , y ¯ ′′ , λ λ c 1 + o ( k λ 2 ) c 1 = k ( a 3 + o ( 1 ) ) c 1 + o ( k λ 2 ) c 1 ,

which implies that c1=0. We also have cl=0, l=2,,N.

Note that

2 s - N 2 ρ + t 1 - 2 s u ~ k ν u ~ k = 2 s - N 2 ′′ ρ + t 1 - 2 s u ~ k ν u ~ k + 2 s - N 2 B ρ ( - V ( r , y ′′ ) u k + ( u k ) + 2 s * - 1 + l = 1 N c l j = 1 k Z x j , λ 2 s * - 2 Z j , l ) u k ,
B ρ ( - V ( r , y ′′ ) u k + ( u k ) + 2 s * - 1 ) y , u k = B ρ ( - 1 2 V ( r , y ′′ ) y , u k 2 + 1 2 s * y , ( u k ) + 2 s * )
= - 1 2 B ρ V ( r , y ′′ ) u k 2 y , ν + 1 2 B ρ ( N V ( r , y ′′ ) + V ( r , y ′′ ) , y ) u k 2
+ 1 2 s * B ρ ( u k ) + 2 s * y , ν + 2 s - N 2 B ρ ( u k ) + 2 s *

and l=1NclBρj=1kZxj,λ2s*-2Zj,lϕ=0. We find that (3.5) is equivalent to

B ρ ( s V ( r , y ′′ ) + 1 2 V ( r , y ′′ ) , y ) u k 2
= - ′′ ρ + t 1 - 2 s u ~ k , Y u ~ k ν + 1 2 ′′ ρ + t 1 - 2 s | u ~ k | 2 Y , ν + 2 s - N 2 ′′ ρ + t 1 - 2 s u ~ k ν u ~ k
(3.14)     + 1 2 B ρ V ( r , y ′′ ) u k 2 y , ν - 1 2 s * B ρ ( u k ) + 2 s * y , ν + 2 s - N 2 l = 1 N c l B ρ j = 1 k Z x j , λ 2 s * - 2 Z j , l Z r ¯ , y ¯ ′′ , λ .

Similarly, (3.4) is equivalent to

1 2 B ρ V ( r , y ′′ ) y i ′′ u k 2 = ′′ ρ + t 1 - 2 s u ~ k ν u ~ k y i - 1 2 ′′ ρ + t 1 - 2 s | u ~ k | 2 ν i
(3.15) + 1 2 B ρ V ( r , y ′′ ) u k 2 ν i + 1 2 s * B ρ u k 2 s * ν i , i = 3 , , N .

Lemma 3.2.

Relations (3.14) and (3.15) are, respectively, equivalent to

(3.16) B ρ ( s V ( r , y ′′ ) + 1 2 V ( r , y ′′ ) , y ) u k 2 = O ( k λ 2 s + σ )

and

(3.17) B ρ V ( r , y ′′ ) y i u k 2 = O ( k λ 2 s + σ ) , i = 3 , , N .

Proof.

We only give the proof for (3.16). The proof of (3.17) is similar.

Note that u~k=Z~r¯,y¯′′,λ+ϕ~. We have

′′ ρ + t 1 - 2 s u ~ k , Y u ~ k ν = ′′ ρ + t 1 - 2 s Z ~ r ¯ , y ¯ ′′ , λ , Y Z ~ r ¯ , y ¯ ′′ , λ ν + ′′ ρ + t 1 - 2 s ϕ ~ , Y ϕ ~ ν
+ ′′ ρ + t 1 - 2 s Z ~ r ¯ , y ¯ ′′ , λ , Y ϕ ~ ν + ′′ ρ + t 1 - 2 s ϕ ~ , Y Z ~ r ¯ , y ¯ ′′ , λ ν .

Using Lemma A.5, we obtain

| ′′ ρ + t 1 - 2 s Z ~ r ¯ , y ¯ ′′ , λ , Y Z ~ r ¯ , y ¯ ′′ , λ ν | C λ N - 2 s ′′ ρ + t 1 - 2 s ( i = 1 k 1 ( 1 + | y - x i | ) N - 2 s + 1 ) 2
(3.18) C k 2 λ N - 2 s ′′ ρ + t 1 - 2 s ( 1 + | y - x 1 | ) 2 N - 4 s + 2 C k 2 λ N - 2 s .

By (A.2) in Lemma A.6,

(3.19) | ′′ ρ + t 1 - 2 s ϕ ~ , Y ϕ ~ ν | C ′′ ρ + t 1 - 2 s | ϕ ~ | 2 C k ϕ * 2 λ τ .

By the process of the proof of (3.18) and (3.19), we also have

| ′′ ρ + t 1 - 2 s Z ~ r ¯ , y ¯ ′′ , λ , Y ϕ ~ ν + ′′ ρ + t 1 - 2 s ϕ ~ , Y Z ~ r ¯ , y ¯ ′′ , λ ν | C k ϕ * λ N - 2 s 2 .

Note that N>4s. So we have proved that

| ′′ ρ + t 1 - 2 s u ~ k , Y u ~ k ν | C k λ 2 s + σ .

Similarly, we can prove

| ′′ ρ + t 1 - 2 s | u ~ k | 2 Y , ν | C k λ 2 s + σ .

Next, we estimate the term ′′ρ+t1-2su~kνu~k:

′′ ρ + t 1 - 2 s u ~ k ν u ~ k = ′′ ρ + t 1 - 2 s Z ~ r ′′ , y ¯ ′′ , λ ν Z ~ r ′′ , y ¯ ′′ , λ + ′′ ρ + t 1 - 2 s ϕ ~ ν ϕ ~
+ ′′ ρ + t 1 - 2 s Z ~ r ′′ , y ¯ ′′ , λ ν ϕ ~ + ′′ ρ + t 1 - 2 s ϕ ~ ν Z ~ r ′′ , y ¯ ′′ , λ .

By Lemma A.5,

| ′′ ρ + t 1 - 2 s Z ~ r ′′ , y ¯ ′′ , λ ν Z ~ r ′′ , y ¯ ′′ , λ | C λ N - 2 s ′′ ρ + t 1 - 2 s i = 1 k 1 ( 1 + | y - x i | ) N - 2 s + 1 × j = 1 k 1 ( 1 + | y - x j | ) N - 2 s
C k 2 λ N - 2 s ′′ ρ + t 1 - 2 s ( 1 + | y - x 1 | ) 2 N - 4 s + 1 C k 2 λ N - 2 s .

It follows from (A.4) that

(3.20) ′′ ρ + t 1 - 2 s | ϕ ~ | 2 C k ϕ * 2 λ τ .

By (3.20) and (A.2) in Lemma A.6, one has

| ′′ ρ + t 1 - 2 s ϕ ~ ν ϕ ~ 𝑑 S | ( ′′ ρ + t 1 - 2 s | ϕ ~ | 2 ) 1 2 ( ′′ ρ + t 1 - 2 s ϕ ~ 2 ) 1 2 C k ϕ * 2 λ τ .

Similarly,

| ′′ ρ + t 1 - 2 s Z ~ r ′′ , y ¯ ′′ , λ ν ϕ ~ + ′′ ρ + t 1 - 2 s ϕ ~ ν Z ~ r ′′ , y ¯ ′′ , λ | C k ϕ * 2 λ τ .

We have proved that

| ′′ ρ + t 1 - 2 s u ~ k ν u ~ k | C k λ 2 s + σ .

Since ζ=0 on Bρ, uk=ϕ on Bρ, we deduce that

| B ρ V ( r , y ′′ ) u k 2 y , ν | C ϕ * 2 B ρ ( j = 1 k λ N - 2 s 2 ( 1 + λ | y - x j | ) N - 2 s 2 + τ ) 2 C k 2 ϕ * 2 λ 2 τ C k λ 2 s + τ

and

| B ρ ( u k ) + 2 s * y , ν | C k 2 s * ϕ * 2 s * λ 2 s * τ C k λ 2 s + τ .

From Proposition 2.3, we know the following estimate for cl:

| c l | C ( 1 λ ) s + σ .

On the other hand,

j = 1 k B ρ Z x j , λ 2 s * - 2 Z j , l Z r ¯ , y ¯ ′′ , λ = j = 1 k B ρ Z x j , λ 2 s * - 1 Z j , l + j = 1 k B ρ i j Z x j , λ 2 s * - 2 Z j , l Z x i , λ = O ( 1 λ N ) + O ( k λ 2 s ) .

These imply that

| l = 1 c l j = 1 k B ρ Z x j , λ 2 s * - 2 Z j , l Z r ¯ , y ¯ ′′ , λ | C k λ 2 s + σ .

Combining the above estimates, we find that (3.14) is equivalent to

B ρ ( s V ( r , y ′′ ) + 1 2 V ( r , y ′′ ) , y ) u k 2 = O ( k λ 2 s + σ ) .

The proof is complete. ∎

Lemma 3.3.

For any function g(r,y′′)C1(RN), we have

B ρ g ( r , y ′′ ) u k 2 = k ( 1 λ 2 s g ( r ¯ , y ¯ ′′ ) N U 0 , 1 2 + o ( 1 λ 2 s ) ) .

Proof.

We have

B ρ g ( r , y ′′ ) u k 2 = D ρ g ( r , y ′′ ) Z r ¯ , y ¯ ′′ , λ 2 + 2 D ρ g ( r , y ′′ ) Z r ¯ , y ¯ ′′ , λ ϕ + D ρ g ( r , y ′′ ) ϕ 2 .

Note that

| 2 B ρ g ( r , y ′′ ) Z r ¯ , y ¯ ′′ , λ ϕ + B ρ g ( r , y ′′ ) ϕ 2 | C ( ϕ * B ρ i = 1 k ζ λ N - 2 s ( 1 + λ | y - x i | ) N - 2 s j = 1 k 1 ( 1 + λ | y - x j | ) N - 2 s 2 + τ
+ ϕ * 2 B ρ ( i = 1 k λ N - 2 s 2 ( 1 + λ | y - x i | ) N - 2 s 2 + τ ) 2 )
C k ϕ * λ s + C k ϕ * 2 λ 2 τ C k λ 2 s + σ

and

B ρ g ( r , y ′′ ) Z r ¯ , y ¯ ′′ , λ 2 = j = 1 k ( B ρ g ( r , y ′′ ) Z x j , λ 2 + i j B ρ g ( r , y ′′ ) Z x i , λ Z x j , λ ) = k ( 1 λ 2 s g ( r ¯ , y ¯ ′′ ) N U 0 , 1 2 + o ( 1 λ 2 s ) ) ,

and we get the result. ∎

Proof of Theorem 1.1.

By (3.16) and (3.17), we deduce that

B ρ ( s V ( r , y ′′ ) + 1 2 r V ( r , y ′′ ) r ) u k 2 = O ( k λ 2 s + σ ) .

That is

(3.21) B ρ 1 r 2 s - 1 ( r 2 s V ( r , y ′′ ) ) r u k 2 = O ( k λ 2 s + σ ) .

Applying Lemma 3.3 to (3.17) and (3.21), we obtain

k ( 1 λ 2 s V ( r ¯ , y ¯ ′′ ) y ¯ i N U 0 , 1 2 + o ( 1 λ 2 s ) ) = o ( k λ 2 s )

and

k ( 1 λ 2 s 1 r ¯ 2 s - 1 ( r ¯ 2 s V ( r ¯ , y ¯ ′′ ) ) r ¯ N U 0 , 1 2 + o ( 1 λ 2 s ) ) = o ( k λ 2 s ) .

Therefore, the equations to determine (r¯,y¯′′) are

(3.22) ( r ¯ 2 s V ( r ¯ , y ¯ ′′ ) ) y ¯ i = o ( 1 ) , i = 3 , , N ,

and

( r ¯ 2 s V ( r ¯ , y ¯ ′′ ) ) r ¯ = o ( 1 ) .

From (3.6) and (B.1), the equation to determine λ is

(3.23) - B 1 λ 2 s + 1 V ( r ¯ , y ¯ ′′ ) + B 3 k N - 2 s λ N - 2 s + 1 = O ( 1 λ 2 s + 1 + σ ) .

Let λ=tkN-2sN-4s. Then t[L0,L1]. It follows from (3.23) that

(3.24) - B 1 t 2 s + 1 V ( r ¯ , y ¯ ′′ ) + B 3 t N - 2 s + 1 = o ( 1 ) , t [ L 0 , L 1 ] .

Define

H ( t , r ¯ , y ¯ ′′ ) = ( r ¯ , y ¯ ′′ ( r ¯ 2 s V ( r ¯ , y ¯ ′′ ) ) , - B 1 t 2 s + 1 V ( r ¯ , y ¯ ′′ ) + B 3 t N - 2 s + 1 ) .

Then

deg ( H ( t , r ¯ , y ¯ ′′ ) , [ L 0 , L 1 ] × B θ ( ( r 0 , y 0 ′′ ) ) ) = - deg ( r ¯ , y ¯ ′′ ( r ¯ 2 s V ( r ¯ , y ¯ ′′ ) ) , B θ ( ( r 0 , y 0 ′′ ) ) 0 .

Hence, (3.22), (3.22) and (3.24) have a solution tk[L0,L1] and (r¯k,y¯k′′)Bθ((r0,y0′′)). ∎

4 Proof of the Main Theorem: The Case N=3=4s

Lemma 4.1.

When N=3=4s, relations (3.14) and (3.15) are, respectively, equivalent to

(4.1) B ρ ( s V ( r , y ′′ ) + 1 2 V ( r , y ′′ ) , y ) u k 2 = O ( k 2 λ 2 s )

and

(4.2) B ρ V ( r , y ′′ ) y i u k 2 = O ( k 2 λ 2 s ) , i = 3 .

Proof.

By using (A.3) in Lemma A.6, we can prove this result as Lemma 3.2. ∎

Lemma 4.2.

For any function g(r,y′′)C1(R3), we have

B ρ g ( r , y ′′ ) u k 2 = k ( 4 π C 2 ( N , s ) g ( r ¯ , y ¯ ′′ ) ln λ λ 2 s + o ( ln λ λ 2 s ) ) .

Proof.

We have

| 2 B ρ g ( r , y ′′ ) Z r ¯ , y ¯ ′′ , λ ϕ + B ρ g ( r , y ′′ ) ϕ 2 | C ( ϕ * B ρ i = 1 k ζ λ N - 2 s ( 1 + λ | y - x i | ) N - 2 s j = 1 k 1 ( 1 + λ | y - x j | ) N - 2 s 2
+ ϕ * 2 B ρ ( i = 1 k λ N - 2 s 2 ( 1 + λ | y - x i | ) N - 2 s 2 ) 2 )
C k 2 ϕ * λ s + C k 2 ϕ * 2 C k 2 λ 2 s

and

B ρ g ( r , y ′′ ) Z r ¯ , y ¯ ′′ , λ 2 = j = 1 k ( B ρ g ( r , y ′′ ) Z x j , λ 2 + i j B ρ g ( r , y ′′ ) Z x i , λ Z x j , λ )
= k ( g ( r ¯ , y ¯ ′′ ) B ρ ( x j ) U x j , λ 2 + O ( k λ 2 s ) )
= k ( g ( r ¯ , y ¯ ′′ ) 4 π C 2 ( N , s ) λ 2 s 0 λ ρ r 2 ( 1 + r 2 ) 3 2 𝑑 r + O ( k λ 2 s ) )
= k ( 4 π C 2 ( N , s ) g ( r ¯ , y ¯ ′′ ) ln λ λ 2 s + o ( ln λ λ 2 s ) ) .

So, we get the result. ∎

Proof of Theorem 1.3.

By (4.1) and (4.2), we can deduce that

(4.3) B ρ 1 r 2 s - 1 ( r 2 s V ( r , y ′′ ) ) r u k 2 = O ( k 2 λ 2 s ) .

Applying Lemma 4.2 to (4.2) and (4.3), we obtain

k ( ln λ λ 2 s 4 π C 2 ( N , s ) V ( r ¯ , y ¯ ′′ ) y ¯ i + o ( ln λ λ 2 s ) ) = O ( k 2 λ 2 s )

and

k ( ln λ λ 2 s 4 π C 2 ( N , s ) 1 r ¯ 2 s - 1 ( r ¯ 2 s V ( r ¯ , y ¯ ′′ ) ) r ¯ + o ( ln λ λ 2 s ) ) = O ( k 2 λ 2 s ) .

This is

( r ¯ 2 s V ( r ¯ , y ¯ ′′ ) ) y ¯ i = o ( 1 ) , i = 3 ,

and

( r ¯ 2 s V ( r ¯ , y ¯ ′′ ) ) r ¯ = o ( 1 ) .

By (B.2), we have

- D 1 ln λ λ 2 s + 1 V ( r ¯ , y ¯ ′′ ) + D 3 k 2 s λ 2 s + 1 = o ( ln λ λ 2 s + 1 ) .

Similar to the proof of Theorem 1.1, we can prove Theorem 1.3. ∎


Communicated by Laurent Veron


Award Identifier / Grant number: 11771235

Award Identifier / Grant number: 11801545

Funding statement: This work is supported by NSFC (11771235, 11801545).

A Some Estimates

In this section, we give some essential estimates. For xi,xj,yN, define gij(y)=1(1+|y-xi|)α(1+|y-xj|)β, where xixj, α>0 and β>0 are two constants.

Lemma A.1.

For any constant γ(0,min(α,β)], we have

g i j ( y ) C ( 1 + | x i - x j | ) γ ( 1 ( 1 + | y - x i | ) α + β - γ + 1 ( 1 + | y - x j | ) α + β - γ ) .

Proof.

See the proof of [42, Lemma A.1]. ∎

Lemma A.2.

For any constant 0<ϑ<N-2s, there is a constant C>0 such that

N 1 | y - z | N - 2 s 1 ( 1 + | z | ) 2 s + ϑ 𝑑 z C ( 1 + | y | ) ϑ .

Proof.

See the proof of [26, Lemma 2.1]. ∎

Lemma A.3.

Let μ>0. For any constants 0<β<N, there exists a constant C>0, independent of μ, such that

N B μ ( y ) 1 | y - z | N + 2 s 1 ( 1 + | z | ) β 𝑑 z C ( 1 ( 1 + | y | ) β + 2 s + 1 μ 2 s 1 ( 1 + | y | ) β ) .

Proof.

Without loss of generality, we set |y|2, and let d=|y|2. Then we have

N B μ ( y ) 1 | y - z | N + 2 s 1 ( 1 + | z | ) β 𝑑 z B d ( 0 ) + B d ( y ) B μ ( y ) + N ( B d ( 0 ) B d ( y ) ) 1 | y - z | N + 2 s 1 ( 1 + | z | ) β 𝑑 z .

By direct computation, we have

B d ( 0 ) d z | y - z | N + 2 s ( 1 + | z | ) β C d N + 2 s 0 d r N - 1 d r ( 1 + r ) β C d β + 2 s

and

B d ( y ) B μ ( y ) d z | y - z | N + 2 s ( 1 + | z | ) β C d β B d ( y ) B μ ( y ) d z | y - z | N + 2 s C μ 2 s d β .

For zN(Bd(0)Bd(y)), we have |y-z||y|2 and |z||y|2. If |z|2|y|, then |y-z||z|-|y||z|2, and if |z|<2|y|, then |y-z||y|2>|z|4. Thus, we have

N ( B d ( 0 ) B d ( y ) ) d z | y - z | N + 2 s ( 1 + | z | ) β C N B d ( 0 ) d z ( 1 + | z | ) β | z | N + 2 s C d β + 2 s .

The proof is complete. ∎

Lemma A.4.

Let ρ>0. Suppose that (y-x)2+t2=ρ2, t>0 and α>N. Then, for 0<β<N, we have

(A.1) N 1 ( t + | z | ) α 1 | y - z - x | β 𝑑 z C ( 1 ( 1 + | y - x | ) β 1 t α - N + 1 ( 1 + | y - x | ) α + β - N ) .

Proof.

The proof is similar to that of [27, Lemma A.3]. ∎

Lemma A.5.

Suppose that (y-x)2+t2=ρ2. Then there exists a constant C>0 such that

| Z ~ x i , λ | C λ N - 2 s 2 1 ( 1 + | y - x i | ) N - 2 s 𝑎𝑛𝑑 | Z ~ x i , λ | C λ N - 2 s 2 1 ( 1 + | y - x i | ) N - 2 s + 1 .

Proof.

By Lemma A.4, we have

| Z ~ x i , λ ( y , t ) | = | β ( N , s ) N t 2 s ( | y - ξ | 2 + t 2 ) N + 2 s 2 ζ ( ξ ) U x i , λ ( ξ ) 𝑑 ξ |
= | β ( N , s ) C ( N , s ) N t 2 s ( | y - ξ | 2 + t 2 ) N + 2 s 2 ζ ( ξ ) ( λ 1 + λ 2 | ξ - x i | 2 ) N - 2 s 2 𝑑 ξ |
C λ N - 2 s 2 N 1 ( 1 + | z | ) N + 2 s 1 ( λ - 1 + | y - t z - x i | ) N - 2 s 𝑑 z
C λ N - 2 s 2 N t 2 s ( t + | z | ) N + 2 s 1 ( λ - 1 + | y - z - x i | ) N - 2 s 𝑑 z
C λ N - 2 s 2 1 ( 1 + | y - x i | ) N - 2 s .

Note that, for l=1,,N,

y l N t 2 s ( | y - ξ | 2 + t 2 ) N + 2 s 2 ζ ( ξ ) ( λ 1 + λ 2 | ξ - x i | 2 ) N - 2 s 2 𝑑 ξ
= 1 λ N - 2 s 2 y l N 1 ( 1 + | z | 2 ) N + 2 s 2 ζ ( y - t z ) ( 1 λ - 2 + | y - t z - x i | 2 ) N - 2 s 2 𝑑 z
= 2 s - N λ N - 2 s 2 N 1 ( 1 + | z | 2 ) N + 2 s 2 ζ ( y - t z ) ( y - t z - x i ) l ( λ - 2 + | y - t z - x i | 2 ) N - 2 s 2 + 1 𝑑 z
    + 1 λ N - 2 s 2 N 1 ( 1 + | z | 2 ) N + 2 s 2 ζ ( y - t z ) y l 1 ( λ - 2 + | y - t z - x i | 2 ) N - 2 s 2 𝑑 z ,

and

t N t 2 s ( | y - ξ | 2 + t 2 ) N + 2 σ 2 ( λ 1 + λ 2 | ξ - x i | 2 ) N - 2 s 2 𝑑 ξ
= N - 2 s λ N - 2 s 2 N 1 ( 1 + | z | 2 ) N + 2 s 2 ζ ( y - t z ) l = 1 N ( y - t z - x k , L ) l z l ( λ - 2 + | y - t z - x i | 2 ) N - 2 s 2 + 1 𝑑 z
    + 1 λ N - 2 s 2 N 1 ( 1 + | z | 2 ) N + 2 s 2 ζ ( y - t z ) z ( λ - 2 + | y - t z - x i | 2 ) N - 2 s 2 𝑑 z .

Then, by the definition of ζ and (A.1), we have

| Z ~ x i , λ | C λ N - 2 s 2 N 1 ( 1 + | z | ) N + 2 s - 1 1 ( 1 + | y - t z - x i | ) N - 2 s + 1 𝑑 z
C λ N - 2 s 2 N t 2 s - 1 ( t + | z | ) N + 2 s - 1 1 ( 1 + | y - z - x i | ) N - 2 s + 1 𝑑 z
C λ N - 2 s 2 1 ( 1 + | y - x i | ) N - 2 s + 1 .

The proof is complete. ∎

For any δ>0, we define the following two sets

D 1 = { Y = ( y , t ) : δ < | Y - ( r 0 , y 0 ′′ , 0 ) | < 6 δ , t > 0 }

and

D 2 = { Y = ( y , t ) : 2 δ < | Y - ( r 0 , y 0 ′′ , 0 ) | < 5 δ , t > 0 } .

Lemma A.6.

For any δ>0, there exists ρ=ρ(δ)(2δ,5δ) such that when N>4s,

(A.2) ′′ ρ + t 1 - 2 s | ϕ ~ | 2 𝑑 S C k ϕ * 2 λ τ ,

and when N=3=4s,

(A.3) ′′ ρ + t 1 - 2 s | ϕ ~ | 2 𝑑 S C k 2 ϕ * 2 ,

where C is a constant, dependent on δ.

Proof.

We first consider the case N>4s. By (A.1), for (y,t)D1, we have

| ϕ ~ ( y , t ) | = | N β ( N , s ) t 2 s ( | y - ξ | 2 + t 2 ) N + 2 s 2 ϕ ( ξ ) 𝑑 ξ |
C ϕ t 2 s λ τ i = 1 k N 1 ( | z | + t ) N + 2 s 1 | y - z - x i | N - 2 s 2 + τ 𝑑 z
C ϕ t 2 s λ τ i = 1 k ( 1 ( 1 + | y - x i | ) N - 2 s 2 + τ 1 t 2 s + 1 ( 1 + | y - x i | ) N + 2 s 2 + τ )
(A.4) C ϕ λ τ i = 1 k 1 ( 1 + | y - x i | ) N - 2 s 2 + τ .

Let φC0(N+1) be a function with φ(y,t)=1 in D2, φ(y,t)=0 in N+1D1 and |φ|C. Note that ϕ~ satisfies

- div ( t 1 - 2 s ϕ ~ ) = 0 in  + N + 1 ,
- lim t 0 t 1 - 2 s t ϕ ~ ( y , t ) = - V ( r , y ′′ ) ϕ + ( 2 s * - 1 ) ( Z r ¯ , y ¯ ′′ , λ ) 2 s * - 2 ϕ + ( ϕ ) + l k + l = 1 N c l i = 1 k Z x i , λ 2 s * - 2 Z i , l , in  N .

Multiplying φ2ϕ~ on the both sides of the equation and integrating by parts over D1, we have

0 = D 1 - div ( t 1 - 2 s ϕ ~ ) φ 2 ϕ ~ d y d t = D 1 t 1 - 2 s ϕ ~ ( φ 2 ϕ ~ ) d y d t
= D 1 t 1 - 2 s ϕ ~ ( φ 2 ϕ ~ + 2 φ φ ϕ ~ ) 𝑑 y 𝑑 t .

For any ϵ>0, we have

D 1 t 1 - 2 s ϕ ~ φ φ ϕ ~ d y d t ϵ D 1 t 1 - 2 s | ϕ ~ | 2 φ 2 𝑑 y 𝑑 t + C ( ϵ ) D 1 t 1 - 2 s ϕ ~ 2 | φ | 2 𝑑 y 𝑑 t .

Taking ϵ=14 and using (A.4), we obtain that

D 2 t 1 - 2 s | ϕ ~ | 2 𝑑 y 𝑑 t C D 1 t 1 - 2 s ϕ ~ 2 | φ | 2 C ϕ * 2 λ 2 τ D 1 t 1 - 2 s ( i = 1 k 1 ( 1 + | y - x i | ) N - 2 s 2 + τ ) 2
C ϕ * 2 λ 2 τ D 1 t 1 - 2 s k 2 ( 1 + | y - x 1 | ) N - 2 s + 2 τ C k ϕ * 2 λ τ .

By using the mean value theorem of integrals, there exists ρ=ρ(δ)(2δ,5δ) such that

′′ ρ + t 1 - 2 s | ϕ ~ | 2 𝑑 S C k ϕ * 2 λ τ .

Now, we consider the case N=3=4s. We have

| ϕ ~ ( y , t ) | C ϕ t 2 s i = 1 k 3 1 ( | z | + t ) N + 2 s 1 | y - z - x i | N - 2 s 2 𝑑 z
C ϕ t 2 s i = 1 k ( 1 ( 1 + | y - x i | ) N - 2 s 2 1 t 2 s + 1 ( 1 + | y - x i | ) N + 2 s 2 )
C ϕ i = 1 k 1 ( 1 + | y - x i | ) N - 2 s 2 .

This gives

D 2 t 1 - 2 s | ϕ ~ | 2 𝑑 y 𝑑 t C D 1 t 1 - 2 s ϕ ~ 2 | φ | 2 C k 2 ϕ * 2 .

So, we have

′′ ρ + t 1 - 2 s | ϕ ~ | 2 𝑑 S C k 2 ϕ * 2 .

The proof is complete. ∎

B Energy Expansion

In this section, we give some estimates of the energy expansions for I(Zr¯,y¯′′,λ+ϕ(r¯,y¯′′,λ)),Zr¯,y¯′′,λλ, I(Zr¯,y¯′′,λ+ϕ(r¯,y¯′′,λ)),Zr¯,y¯′′,λr¯ and I(Zr¯,y¯′′,λ+ϕ(r¯,y¯′′,λ)),Zr¯,y¯′′,λy¯′′.

Lemma B.1.

If N>4s, then

I ( Z r ¯ , y ¯ ′′ , λ ) λ = k ( - 2 s B 1 λ 2 s + 1 V ( r ¯ , y ¯ ′′ ) + j = 2 k B 2 λ N - 2 s + 1 | x j - x 1 | N - 2 s + O ( 1 λ 2 s + 1 + σ ) ) ,

where B1 and B2 are two positive constants.

Proof.

By a direct computation, we have

I ( Z r ¯ , y ¯ ′′ , λ ) λ = I ( Z r ¯ , y ¯ ′′ , λ * ) λ + O ( k λ 2 s + 1 + σ )
= N V ( y ) Z r ¯ , y ¯ ′′ , λ * Z r ¯ , y ¯ ′′ , λ * λ - N ( ( Z r ¯ , y ¯ ′′ , λ * ) 2 s * - 1 - j = 1 k U x j , λ 2 s * - 1 ) Z r ¯ , y ¯ ′′ , λ * λ + O ( k λ 2 s + 1 + σ )
= I 1 - I 2 + O ( k λ 2 s + 1 + σ ) .

For the term I1, by Lemma A.1, we can check that

I 1 = k ( N V ( y ) U x 1 , λ U x 1 , λ λ + O ( 1 λ N U x 1 , λ j = 2 k U x j , λ ) )
= k ( V ( r ¯ , y ¯ ′′ ) 2 λ N U x 1 , λ 2 𝑑 y + O ( 1 λ 2 s + 1 j = 2 k 1 ( λ | x 1 - x j | ) N - 4 s - σ ) + O ( 1 λ 2 s + 1 + σ ) )
= k ( - s V ( r ¯ , y ¯ ′′ ) λ 2 s + 1 N U 0 , 1 2 𝑑 y + O ( 1 λ 2 s + 1 + σ ) )
= k ( - 2 s B 1 V ( r ¯ , y ¯ ′′ ) λ 2 s + 1 + O ( 1 λ 2 s + 1 + σ ) ) ,

where B1=12NU0,12𝑑y>0 and N-4s-σ-N-4sN-2s>0.

Next, we estimate I2:

I 2 = k Ω 1 ( ( Z r ¯ , y ¯ ′′ , λ * ) 2 s * - 1 - j = 1 k U x j , λ 2 s * - 1 ) Z r ¯ , y ¯ ′′ , λ * λ
= k ( Ω 1 ( 2 s * - 1 ) U x 1 , λ 2 s * - 2 j = 2 k U x j , λ U x 1 , λ λ + O ( 1 λ 2 s + 1 + σ ) )
= k ( - j = 2 k B 2 λ N - 2 s + 1 | x j - x 1 | N - 2 s + O ( 1 λ 2 s + 1 + σ ) ) ,

for some constant B2>0.

Thus, we obtain that

I ( Z r ¯ , y ¯ ′′ , λ ) λ = k ( - 2 s B 1 λ 2 s + 1 V ( r ¯ , y ¯ ′′ ) + j = 2 k B 2 λ N - 2 s + 1 | x j - x 1 | N - 2 s + O ( 1 λ 2 s + 1 + σ ) ) .

The proof is complete. ∎

Lemma B.2.

If N=3=4s, then

I ( Z r ¯ , y ¯ ′′ , λ ) λ = k ( - 2 s D 1 ln λ λ 2 s + 1 V ( r ¯ , y ¯ ′′ ) + j = 2 k D 2 λ N - 2 s + 1 | x j - x 1 | N - 2 s + o ( ln λ λ 2 s + 1 + σ ) ) ,

where D1 and D2 are two positive constants.

Proof.

We have

I ( Z r ¯ , y ¯ ′′ , λ ) λ = 3 V ( y ) Z r ¯ , y ¯ ′′ , λ Z r ¯ , y ¯ ′′ , λ λ - 3 ( ( Z r ¯ , y ¯ ′′ , λ * ) 2 s * - 1 - j = 1 k U x j , λ 2 s * - 1 ) Z r ¯ , y ¯ ′′ , λ * λ + O ( k λ 2 s + 1 )
= I 1 - I 2 + O ( k λ 2 s + 1 ) .

For the term I1, By Lemma A.1, we can check that

I 1 = k ( B ρ ( x 1 ) V ( y ) U x 1 , λ U x 1 , λ λ + O ( 1 λ B ρ ( x 1 ) U x 1 , λ j = 2 k U x j , λ + 1 λ 2 s + 1 ) )
= k ( V ( r ¯ , y ¯ ′′ ) s C 2 ( 3 , s ) λ 2 s + 1 B ρ ( x 1 ) λ 3 ( 1 - λ 2 | y - x 1 | 2 ) ( 1 + λ 2 | y - x 1 | 2 ) 5 2 + O ( k λ 2 s + 1 ) )
= k ( V ( r ¯ , y ¯ ′′ ) 4 π s C 2 ( 3 , s ) λ 2 s + 1 0 ρ λ ( 1 - r 2 ) r 2 ( 1 + r 2 ) 5 2 + O ( k λ 2 s + 1 ) )
= k ( - 2 s D 1 ln λ λ 2 s + 1 V ( r ¯ , y ¯ ′′ ) + o ( ln λ λ 2 s + 1 ) ) ,

where

D 1 = 2 π C 2 ( 3 , s ) and lim λ 0 ρ λ ( 1 - r 2 ) r 2 ( 1 + r 2 ) 5 / 2 ln λ = - 1 .

Similar to the proof in Lemma B.1, we have

I 2 = k ( - j = 2 k D 2 λ N - 2 s + 1 | x j - x 1 | N - 2 s + O ( 1 λ 2 s + 1 ) ) ,

for some constant D2>0.

Thus, we obtain that

I ( Z r ¯ , y ¯ ′′ , λ ) λ = k ( - 2 s D 1 ln λ λ 2 s + 1 V ( r ¯ , y ¯ ′′ ) + j = 2 k D 2 λ N - 2 s + 1 | x j - x 1 | N - 2 s + o ( ln λ λ 2 s + 1 ) ) .

The proof is complete. ∎

Lemma B.3.

If N>4s, then we have

I ( Z r ¯ , y ¯ ′′ , λ + ϕ ) , Z r ¯ , y ¯ ′′ , λ λ = k ( - 2 s B 1 λ 2 s + 1 V ( r ¯ , y ¯ ′′ ) + j = 2 k B 2 λ N - 2 s + 1 | x j - x 1 | N - 2 s + O ( 1 λ 2 s + 1 + σ ) )
(B.1) = k ( - 2 s B 1 λ 2 s + 1 V ( r ¯ , y ¯ ′′ ) + B 3 k N - 2 s λ N - 2 s + 1 + O ( 1 λ 2 s + 1 + σ ) ) ,

where B1 and B2 are the same constants as in Lemma B.1, B3>0.

Proof.

By symmetry, we have

I ( Z r ¯ , y ¯ ′′ , λ + ϕ ) , Z r ¯ , y ¯ ′′ , λ λ = N ( ( - Δ ) s u k + V ( r , y ′′ ) u k - ( u k ) + 2 s * - 1 ) Z r ¯ , y ¯ ′′ , λ λ
= I ( Z r ¯ , y ¯ ′′ , λ ) , Z r ¯ , y ¯ ′′ , λ λ + k ( - Δ ) s ϕ + V ( r , y ′′ ) ϕ - ( 2 s * - 1 ) Z r ¯ , y ¯ ′′ , λ 2 s * - 2 ϕ , Z x 1 , λ λ
- N ( ( Z r ¯ , y ¯ ′′ , λ + ϕ ) + 2 s * - 1 - Z r ¯ , y ¯ ′′ , λ 2 s * - 1 - ( 2 s * - 1 ) Z r ¯ , y ¯ ′′ , λ 2 s * - 2 ϕ ) Z r ¯ , y ¯ ′′ , λ λ
= : I ( Z r ¯ , y ¯ ′′ , λ ) , Z r ¯ , y ¯ ′′ , λ λ + k J 1 - J 2 .

By (2.5) and (2.7), we have

J 1 = O ( ϕ * λ 1 + s + σ ) = O ( 1 λ 2 s + 1 + σ ) .

Note that (1+t)+γ-1-γt=O(t2) for all tN if 1<γ2, and |(1+t)+γ-1-γt|C(t2+|t|γ) for all tN if γ>2. So, if 2s*3, we have

| J 2 | = | N ( ( Z r ¯ , y ¯ ′′ , λ + ϕ ) + 2 s * - 1 - Z r ¯ , y ¯ ′′ , λ 2 s * - 1 - ( 2 s * - 1 ) Z r ¯ , y ¯ ′′ , λ 2 s * - 2 ϕ ) Z r ¯ , y ¯ ′′ , λ λ |
C N Z r ¯ , y ¯ ′′ , λ 2 s * - 3 ϕ 2 | Z r ¯ , y ¯ ′′ , λ λ |
C ϕ * 2 λ N ( j = 1 k λ N - 2 s 2 ( 1 + λ | y - x j | ) N - 2 s ) 2 s * - 2 ( i = 1 k λ N - 2 s 2 ( 1 + λ | y - x i | ) N - 2 s 2 + τ ) 2
C ϕ * 2 λ N λ N j = 1 k 1 ( 1 + λ | y - x j | ) 4 s i = 1 k 1 ( 1 + λ | y - x i | ) N - 2 s + τ
C k ϕ * 2 λ = O ( k λ 2 s + 1 + σ ) .

If 2s*>3, we have

| J 2 | C N ( Z r ¯ , y ¯ ′′ , λ 2 s * - 3 ϕ 2 | Z r ¯ , y ¯ ′′ , λ λ | + | ϕ | 2 s * - 1 | Z r ¯ , y ¯ ′′ , λ λ | ) = O ( k λ 2 s + 1 + σ ) .

Thus, we obtain

I ( Z r ¯ , y ¯ ′′ , λ + ϕ ) , Z r ¯ , y ¯ ′′ , λ λ = I ( Z r ¯ , y ¯ ′′ , λ ) , Z r ¯ , y ¯ ′′ , λ λ + O ( k λ 2 s + 1 + σ ) .

Combining this with Lemma B.1, we finish the proof. ∎

Lemma B.4.

If N=3=4s, then we have

I ( Z r ¯ , y ¯ ′′ , λ + ϕ ) , Z r ¯ , y ¯ ′′ , λ λ = I ( Z r ¯ , y ¯ ′′ , λ ) , Z r ¯ , y ¯ ′′ , λ λ + O ( k λ 2 s + 1 )
(B.2) = k ( - 2 s D 1 ln λ λ 2 s + 1 V ( r ¯ , y ¯ ′′ ) + D 3 k N - 2 s λ N - 2 s + 1 + o ( ln λ λ 2 s + 1 ) ) ,

where D1 is the same constants as in Lemma B.1 and D3>0.

Proof.

The proof is similar to that of Lemma B.3 . ∎

Note that Zi,l=O(λZxi,λ), l=2,,N. Similarly, we can prove the following lemma.

Lemma B.5.

If N>4s, then we have

I ( Z r ¯ , y ¯ ′′ , λ + ϕ ) , Z r ¯ , y ¯ ′′ , λ r ¯ = I ( Z r ¯ , y ¯ ′′ , λ ) , Z r ¯ , y ¯ ′′ , λ r ¯ + O ( k λ s + σ )
= k ( B 1 λ 2 s V ( r ¯ , y ¯ ′′ ) r ¯ + j = 2 k B 2 r ¯ λ N - 2 s | x 1 - x j | N - 2 s + O ( 1 λ s + σ ) )

and

I ( Z r ¯ , y ¯ ′′ , λ + ϕ ) , Z r ¯ , y ¯ ′′ , λ y ¯ j ′′ = I ( Z r ¯ , y ¯ ′′ , λ ) , Z r ¯ , y ¯ ′′ , λ y ¯ j ′′ + O ( k λ s + σ ) = k ( B 1 λ 2 s V ( r ¯ , y ¯ ′′ ) y ¯ j ′′ + O ( 1 λ s + σ ) ) ,

where B1 and B2 are the same constants as in Lemma B.1.

Lemma B.6.

If N=3=4s, then we have

I ( Z r ¯ , y ¯ ′′ , λ + ϕ ) , Z r ¯ , y ¯ ′′ , λ r ¯ = k ( D 1 ln λ λ 2 s V ( r ¯ , y ¯ ′′ ) r ¯ + j = 2 k D 2 r ¯ λ N - 2 s | x 1 - x j | N - 2 s + O ( 1 λ s ) )

and

I ( Z r ¯ , y ¯ ′′ , λ + ϕ ) , Z r ¯ , y ¯ ′′ , λ y ¯ j ′′ = k ( D 1 λ 2 s V ( r ¯ , y ¯ ′′ ) y ¯ j ′′ + O ( 1 λ s ) ) ,

where D1 and D2 are the same constants as in Lemma B.2.

References

[1] V. Ambrosio, Ground states for a fractional scalar field problem with critical growth, Differential Integral Equations 30 (2017), no. 1–2, 115–132. 10.57262/die/1484881222Search in Google Scholar

[2] V. Ambrosio, Multiplicity of positive solutions for a class of fractional Schrödinger equations via penalization method, Ann. Mat. Pura Appl. (4) 196 (2017), no. 6, 2043–2062. 10.1007/s10231-017-0652-5Search in Google Scholar

[3] V. Ambrosio, Periodic solutions for critical fractional problems, Calc. Var. Partial Differential Equations 57 (2018), no. 2, Article ID 45. 10.1007/s00526-018-1317-ySearch in Google Scholar

[4] V. Ambrosio and T. Isernia, Concentration phenomena for a fractional Schrödinger–Kirchhoff type equation, Math. Methods Appl. Sci. 41 (2018), no. 2, 615–645. 10.1002/mma.4633Search in Google Scholar

[5] D. Applebaum, Lévy Processes and Stochastic Calculus, 2nd ed., Cambridge Stud. Adv. Math. 116, Cambridge University, Cambridge, 2009. 10.1017/CBO9780511809781Search in Google Scholar

[6] A. Bahri and J.-M. Coron, The scalar-curvature problem on the standard three-dimensional sphere, J. Funct. Anal. 95 (1991), no. 1, 106–172. 10.1016/0022-1236(91)90026-2Search in Google Scholar

[7] B. Barrios, E. Colorado, A. de Pablo and U. Sánchez, On some critical problems for the fractional Laplacian operator, J. Differential Equations 252 (2012), no. 11, 6133–6162. 10.1016/j.jde.2012.02.023Search in Google Scholar

[8] C. Brändle, E. Colorado, A. de Pablo and U. Sánchez, A concave-convex elliptic problem involving the fractional Laplacian, Proc. Roy. Soc. Edinburgh Sect. A 143 (2013), no. 1, 39–71. 10.1017/S0308210511000175Search in Google Scholar

[9] X. Cabré and Y. Sire, Nonlinear equations for fractional Laplacians, I: Regularity, maximum principles, and Hamiltonian estimates, Ann. Inst. H. Poincaré Anal. Non Linéaire 31 (2014), no. 1, 23–53. 10.1016/j.anihpc.2013.02.001Search in Google Scholar

[10] X. Cabré and J. Tan, Positive solutions of nonlinear problems involving the square root of the Laplacian, Adv. Math. 224 (2010), no. 5, 2052–2093. 10.1016/j.aim.2010.01.025Search in Google Scholar

[11] L. Caffarelli and L. Silvestre, An extension problem related to the fractional Laplacian, Comm. Partial Differential Equations 32 (2007), no. 7–9, 1245–1260. 10.1080/03605300600987306Search in Google Scholar

[12] D. Cao, E. S. Noussair and S. Yan, On the scalar curvature equation -Δu=(1+ϵK)u(N+2)/(N-2) in N, Calc. Var. Partial Differential Equations 15 (2002), no. 3, 403–419. 10.1007/s00526-002-0137-1Search in Google Scholar

[13] S.-Y. A. Chang and P. C. Yang, A perturbation result in prescribing scalar curvature on Sn, Duke Math. J. 64 (1991), no. 1, 27–69. 10.1215/S0012-7094-91-06402-1Search in Google Scholar

[14] C.-C. Chen and C.-S. Lin, Estimate of the conformal scalar curvature equation via the method of moving planes. II, J. Differential Geom. 49 (1998), no. 1, 115–178. 10.4310/jdg/1214460938Search in Google Scholar

[15] C.-C. Chen and C.-S. Lin, Prescribing scalar curvature on SN. I. A priori estimates, J. Differential Geom. 57 (2001), no. 1, 67–171. 10.4310/jdg/1090348090Search in Google Scholar

[16] W. Chen, Y. Li and P. Ma, The Fractional Laplacian, World Scientific, Singapore, 2019. 10.1142/10550Search in Google Scholar

[17] E. Cinti, J. Davila and M. Del Pino, Solutions of the fractional Allen–Cahn equation which are invariant under screw motion, J. Lond. Math. Soc. (2) 94 (2016), no. 1, 295–313. 10.1112/jlms/jdw033Search in Google Scholar

[18] J. Dávila, M. del Pino and I. Guerra, Non-uniqueness of positive ground states of non-linear Schrödinger equations, Proc. Lond. Math. Soc. (3) 106 (2013), no. 2, 318–344. 10.1112/plms/pds038Search in Google Scholar

[19] J. Dávila, M. del Pino and J. Wei, Concentrating standing waves for the fractional nonlinear Schrödinger equation, J. Differential Equations 256 (2014), no. 2, 858–892. 10.1016/j.jde.2013.10.006Search in Google Scholar

[20] M. del Pino, P. Felmer and M. Musso, Two-bubble solutions in the super-critical Bahri–Coron’s problem, Calc. Var. Partial Differential Equations 16 (2003), no. 2, 113–145. 10.1007/s005260100142Search in Google Scholar

[21] M. del Pino, J. Wei and W. Yao, Intermediate reduction method and infinitely many positive solutions of nonlinear Schrödinger equations with non-symmetric potentials, Calc. Var. Partial Differential Equations 53 (2015), no. 1–2, 473–523. 10.1007/s00526-014-0756-3Search in Google Scholar

[22] E. Di Nezza, G. Palatucci and E. Valdinoci, Hitchhiker’s guide to the fractional Sobolev spaces, Bull. Sci. Math. 136 (2012), no. 5, 521–573. 10.1016/j.bulsci.2011.12.004Search in Google Scholar

[23] P. Felmer, A. Quaas and J. Tan, Positive solutions of the nonlinear Schrödinger equation with the fractional Laplacian, Proc. Roy. Soc. Edinburgh Sect. A 142 (2012), no. 6, 1237–1262. 10.1017/S0308210511000746Search in Google Scholar

[24] Y. Guo and B. Li, Infinitely many solutions for the prescribed curvature problem of polyharmonic operator, Calc. Var. Partial Differential Equations 46 (2013), no. 3–4, 809–836. 10.1007/s00526-012-0504-5Search in Google Scholar

[25] Y. Guo, T. Liu and J. Nie, Construction of solutions for the polyharmonic equation via local Pohozaev identities, Calc. Var. Partial Differential Equations 58 (2019), no. 4, Article ID 123. 10.1007/s00526-019-1569-1Search in Google Scholar

[26] Y. Guo and J. Nie, Infinitely many non-radial solutions for the prescribed curvature problem of fractional operator, Discrete Contin. Dyn. Syst. 36 (2016), no. 12, 6873–6898. 10.3934/dcds.2016099Search in Google Scholar

[27] Y. Guo, J. Nie, M. Niu and Z. Tang, Local uniqueness and periodicity for the prescribed scalar curvature problem of fractional operator in N, Calc. Var. Partial Differential Equations 56 (2017), no. 4, Article ID 118. 10.1007/s00526-017-1194-9Search in Google Scholar

[28] Y. Guo, S. Peng and S. Yan, Local uniqueness and periodicity induced by concentration, Proc. Lond. Math. Soc. (3) 114 (2017), no. 6, 1005–1043. 10.1112/plms.12029Search in Google Scholar

[29] T. Jin, Y. Li and J. Xiong, On a fractional Nirenberg problem, part I: Blow up analysis and compactness of solutions, J. Eur. Math. Soc. (JEMS) 16 (2014), no. 6, 1111–1171. 10.4171/JEMS/456Search in Google Scholar

[30] Y. Li, Prescribing scalar curvature on Sn and related problems. II. Existence and compactness, Comm. Pure Appl. Math. 49 (1996), no. 6, 541–597. 10.1002/(SICI)1097-0312(199606)49:6<541::AID-CPA1>3.0.CO;2-ASearch in Google Scholar

[31] Y. Li and W.-M. Ni, On conformal scalar curvature equations in 𝐑n, Duke Math. J. 57 (1988), no. 3, 895–924. 10.1215/S0012-7094-88-05740-7Search in Google Scholar

[32] E. H. Lieb, Sharp constants in the Hardy–Littlewood–Sobolev and related inequalities, Ann. of Math. (2) 118 (1983), no. 2, 349–374. 10.1007/978-3-642-55925-9_43Search in Google Scholar

[33] M. Niu, Z. Tang and L. Wang, Solutions for conformally invariant fractional Laplacian equations with multi-bumps centered in lattices, J. Differential Equations 266 (2019), no. 4, 1756–1831. 10.1016/j.jde.2018.08.008Search in Google Scholar

[34] E. S. Noussair and S. Yan, The scalar curvature equation on N, Nonlinear Anal. 45 (2001), no. 4, 483–514. 10.1016/S0362-546X(99)00428-9Search in Google Scholar

[35] S. Peng, C. Wang and S. Yan, Construction of solutions via local Pohozaev identities, J. Funct. Anal. 274 (2018), no. 9, 2606–2633. 10.1016/j.jfa.2017.12.008Search in Google Scholar

[36] R. Schoen and D. Zhang, Prescribed scalar curvature on the n-sphere, Calc. Var. Partial Differential Equations 4 (1996), no. 1, 1–25. 10.1007/BF01322307Search in Google Scholar

[37] R. Servadei and E. Valdinoci, Mountain pass solutions for non-local elliptic operators, J. Math. Anal. Appl. 389 (2012), no. 2, 887–898. 10.1016/j.jmaa.2011.12.032Search in Google Scholar

[38] L. Silvestre, Regularity of the obstacle problem for a fractional power of the Laplace operator, Comm. Pure Appl. Math. 60 (2007), no. 1, 67–112. 10.1002/cpa.20153Search in Google Scholar

[39] J. Tan, The Brezis–Nirenberg type problem involving the square root of the Laplacian, Calc. Var. Partial Differential Equations 42 (2011), no. 1–2, 21–41. 10.1007/s00526-010-0378-3Search in Google Scholar

[40] J. Tan and J. Xiong, A Harnack inequality for fractional Laplace equations with lower order terms, Discrete Contin. Dyn. Syst. 31 (2011), no. 3, 975–983. 10.3934/dcds.2011.31.975Search in Google Scholar

[41] J. Vétois and S. Wang, Infinitely many solutions for cubic nonlinear Schrödinger equations in dimension four, Adv. Nonlinear Anal. 8 (2019), no. 1, 715–724. 10.1515/anona-2017-0085Search in Google Scholar

[42] J. Wei and S. Yan, Infinitely many solutions for the prescribed scalar curvature problem on 𝕊N, J. Funct. Anal. 258 (2010), no. 9, 3048–3081. 10.1016/j.jfa.2009.12.008Search in Google Scholar

[43] S. Yan, Concentration of solutions for the scalar curvature equation on 𝐑N, J. Differential Equations 163 (2000), no. 2, 239–264. 10.1006/jdeq.1999.3718Search in Google Scholar

[44] S. Yan, J. Yang and X. Yu, Equations involving fractional Laplacian operator: Compactness and application, J. Funct. Anal. 269 (2015), no. 1, 47–79. 10.1016/j.jfa.2015.04.012Search in Google Scholar

Received: 2019-08-09
Revised: 2019-10-02
Accepted: 2019-10-04
Published Online: 2019-11-13
Published in Print: 2020-02-01

© 2019 Walter de Gruyter GmbH, Berlin/Boston

This work is licensed under the Creative Commons Attribution 4.0 International License.

Downloaded on 9.1.2026 from https://www.degruyterbrill.com/document/doi/10.1515/ans-2019-2067/html
Scroll to top button