Non-degeneracy of bubble solutions for higher order prescribed curvature problem

Yuxia Guo; Yichen Hu

doi:10.1515/ans-2022-0003

Article Open Access

Non-degeneracy of bubble solutions for higher order prescribed curvature problem

Yuxia Guo and Yichen Hu

Published/Copyright: March 3, 2022

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From the journal Advanced Nonlinear Studies Volume 22 Issue 1

Abstract

In this article, we are concerned with the following prescribed curvature problem involving polyharmonic operator on S N :

D m u = K ( ∣ y ∣ ) u m ∗ − 1 , u > 0 in S N , u ∈ H m ( S N ) ,

where K ( ∣ y ∣ ) is a positive function, m ∗ = 2 N N − 2 m is the Sobolev embedding critical exponent, N > 2 m + 2 . D m is the 2 m order differential operator given by

D m = ∏ l = 1 m − Δ g + 1 4 ( N − 2 l ) ( N + 2 l − 2 ) ,

where Δ g is the Laplace-Beltrami operator on S N , S N is the unit sphere with Riemann metric g . We first establish two kinds of local Pohozaev identities for polyharmonic operator, then we prove that the positive bubbling solution constructed in the study of Guo and Li is non-degenerate.

Keywords: polyharmonic operator; non-degeneracy; Pohozaev identities

MSC 2010: 35B33; 35J91

1 Introduction

We consider the following prescribed curvature problem involving polyharmonic operator on S N

(1.1) D m u = K ( ∣ y ∣ ) u m ∗ − 1 , u > 0 in S N , u ∈ H m ( S N ) ,

where K ( ∣ y ∣ ) is a positive function, m ∗ = 2 N N − 2 m is the critical exponent of Sobolev embedding, N > 2 m + 2 , D m is a 2 m order differential operator given by

D m = ∏ l = 1 m − Δ g + 1 4 ( N − 2 l ) ( N + 2 l − 2 ) ,

where Δ g is the Laplace-Beltrami operator on S N , S N is the unit sphere with Riemann metric g .

In the case of m = 1 , problem (1.1) is reduced to the following prescribed curvature problem:

(1.2) − Δ S n u + N ( N − 2 ) 2 u − K ( y ) u N + 2 N − 2 = 0 , u > 0 , on S N .

The so-called prescribed curvature problem is to find a conformal invariant metric g such that the curvature is K ( y ) .

By using the stereo-graphic projection, problem (1.2) is reduced to the following elliptic problem in R N :

(1.3) − Δ u = K ( y ) u N + 2 N − 2 , u > 0 , in R N , u ∈ D 1 , 2 ( R N ) .

Because of its geometry background, problem (1.3) has been extensively studied in the last few decades. It is known that (see [1]) (1.3) does not always admit a solution. Hence, we are more interested in the sufficient condition on the curvature function K ( y ) , under which problem (1.3) admits a solution. Indeed, there have been a lot of existence results obtained in the literature, see, for example, [2,3,4, 5,6,7] and references therein. In particular, we know that any solution of (1.3) is radially symmetric if there is an r 0 > 0 such that K ( ∣ y ∣ ) is non-increasing in ( 0 , r 0 ] and non-decreasing in [ r 0 , + ∞ ) (see [8]). It is natural to ask whether or not there are non-radial symmetric solutions to (1.3). This question was answered in the paper of [9].

In general case of m ≥ 1 , problem (1.1) also has attracted wide attention due to its geometry roots and various applications in physics during the last few decades. For instance, when m = 2 , problem (1.1) is related to the Paneitz operator, which was introduced by Paneitz [10] for smooth four-dimensional Riemannian manifolds and was generalized by Branson [11] to smooth N-dimensional Riemannian manifolds. For various existence results to problems involving polyharmonic operator and other related problems, we refer the reader to the papers [12,13,14, 15,16,17, 18,19,20, 21,22,23] and references therein. It is evident to see from these papers that compared to the problems with Laplace operator (i.e., m = 1 ), the problems involving polyharmonic operator present new and challenging features which make the problem getting more complicated and difficult.

This article is concerning on the non-degeneracy of bubbling solutions for problem (1.1). It is also known that by using the stereographic projection, the prescribed scalar curvature problem for polyharmonic operator on S N can be reduced to the following problem in R N :

(1.4) ( − Δ ) m u = K ( ∣ y ∣ ) u m ∗ − 1 , u > 0 , in R N , u ∈ D m , 2 ( R N ) ,

where m ∗ = 2 N N − 2 m is the Sobolev critical exponent of the embedding from D m , 2 ( R N ) onto L p ( R N ) , and D m , 2 ( R N ) is the completion of C 0 ∞ ( R N ) with respect to the norm induced by the scalar product:

( u , v ) = ∫ R N Δ m 2 u ⋅ Δ m 2 v , if m is even , ∫ R N ∇ Δ m − 1 2 u ⋅ ∇ Δ m − 1 2 v , if m is odd .

In particular, we are interested in the existence of non-radial solutions for (1.4). This was done in [24], where infinitely many non-radial bubble solutions were constructed under the following assumption on K ( ∣ y ∣ ) :

( K ) There are r 0 > 0 and c 0 > 0 , such that

(1.5) K ( r ) = K ( r 0 ) − c 0 ( r − r 0 ) 2 + O ( ∣ r − r 0 ∣ 3 ) , r ∈ ( r 0 − δ , r 0 + δ ) .

Without loss of generality, we may assume that K ( r 0 ) = 1 .

The main purpose of this article is to discuss the non-degeneracy of positive bubble solution constructed in [24]. We would like to mention that the non-degeneracy of the solution is very important for the further construction of new solutions for problem (1.1). For example, we may ask whether we can obtain the similar result as in [24] without the symmetry assumptions on the curvature function k ( y ) . To answer this question, among any other things, one main task is to get a better understanding of the corresponding linearized problem around the approximation solution. That is, the non-degeneracy of the linearized operator. Another application in our mind is to construct new-type bubble solutions for (1.1). We will consider these problems in our later work.

Before the statement of the main results, let us briefly introduce the bubble solutions constructed in [25].

It is known (see [9,26]) that a family of positive solutions to the following limit problem:

(1.6) ( − Δ ) m u = u m ∗ − 1 , u > 0 in R N , u ∈ D m , 2 ( R N )

are given by

(1.7) U x , μ ( y ) = C N , m μ N − 2 m 2 ( 1 + μ 2 ∣ y − x ∣ 2 ) N − 2 m 2 , x ∈ R N , μ > 0 ,

where C N , m is a constant depending on N , m .

Let k be an integer number and we consider the vertices of a regular polygon with k edges in the ( y 1 , y 2 ) -plane given by

x j = r cos 2 ( j − 1 ) π k , r sin 2 ( j − 1 ) π k , 0 , j = 1 , … , k ,

where 0 denotes the zero vector in R N − 2 and r ∈ ( r 0 − δ , r 0 + δ ) .

For any point y ∈ R N , we set y = ( y ′ , y ″ ) , y ′ ∈ R 2 , y ″ ∈ R N − 2 . Define

H s = u : u is even in y h , h = 2 , … , N , u ( r cos θ , r sin , y ″ ) = u r cos θ + 2 π j k , r sin θ + 2 π j k , y ″ ,

and set

W r , μ ( y ) = ∑ j = 1 k U x j , μ ( y ) .

Furthermore, for a function u ∈ H s ⋂ D m , 2 ( R N ) , we introduce the norm ‖ u ‖ ∗ as follows:

‖ u ‖ ∗ = sup y ∈ R N ∣ u ( y ) ∣ ∑ j = 1 k μ N − 2 m 2 ( 1 + μ ∣ y − x j ∣ ) N − 2 m 2 + τ − 1 ,

where τ is any fixed number in N − 2 m − 2 N − 2 m , 1 + θ , θ > 0 is a small constant. The result obtained in [24] states the following.

Theorem A

Suppose that K ( r ) satisfies (K) and N > 2 m + 2 . Then there is an integer k 0 > 0 , such that for any integer k ≥ k 0 , (1.4) has a solution u k of the form

u k = W r k , μ k ( y ) + ω k ,

where ω k ∈ H s ⋂ D m , 2 ( R N ) , and as k → + ∞ , ∣ r k − r 0 ∣ = O 1 μ k 1 + σ , μ k ∼ k N − 2 m N − 2 m − 2 ,

‖ ω k ‖ ∗ = O 1 μ k 1 + σ

for some σ > 0 .

Let L k be the linear operator defined by

L k ξ = ( − Δ ) m ξ − ( m ∗ − 1 ) K ( y ) u k m ∗ − 2 ξ .

The main result of the present article is the following.

Theorem 1.1

Suppose that K ( y ) satisfies ( K ), K ( y ) ∈ C 3 ( B ϑ ( r 0 ) ) for some ϑ > 0 , and K ′ ( r ) ∈ L ∞ ( R N ) . Then If N > 2 m + 2 , there exists K 0 > 0 , such that for any integer k > K 0 , the positive bubble solution u k obtained in Theorem A is non-degenerate in the sense that if ξ ∈ H s ⋂ D m , 2 ( R N ) is a solution of L k ξ = 0 , then ξ = 0 .

As a direct consequence of Theorem 1.1, we have

Theorem 1.2

Under the assumption in Theorem 1.1. There exists K 0 > 0 , such that for any integer k > K 0 , the bubble solution of the prescribed curvature problem (1.1) is non-degenerate in the sense that if ξ ∈ H s ⋂ D m , 2 ( R N ) is a solution of L k ξ = 0 , then ξ = 0 .

For the proof of Theorem 1.1, we shall proceed contrary arguments by using the local Pohozaev identities for polyharmonic equations. However, different from the case of m = 1 , the local Pohozaev identities will get more complex and there are even infinite many items (as m is large) to be dealt with. As a better understanding of different Pohozaev identities for polyharmonic equations is very important for us, some extra techniques and more detailed calculation are needed. We believe that these Pohozaev identities for polyharmonic operator can be applied for more elliptic problems involving polyharmonic operator.

The article is organized as follows. In Section 2, by delicate computations, we establish two types of local Pohozaev identities for polyharmonic. Section 3 is devoted to the proof of Theorem 1.1. For this purpose, we will first establish a finite estimate for the bubble solution u k by using the local Pohozaev identities, then we proceed a contrary argument to prove its non-degeneracy. Some essential estimates are attached in Appendices.

2 Local Pohozaev identities

In this section, we first establish the local Pohozaev identities for polyharmonic operator. Let

(2.1) ( − Δ ) m u = K ( ∣ y ∣ ) u m ∗ − 1

and

(2.2) ( − Δ ) m ξ = ( m ∗ − 1 ) K ( ∣ y ∣ ) u m ∗ − 2 ξ .

Assume that Ω is a bounded domain in R N with smooth boundary ∂ Ω . We have the following identities.

Lemma 2.1

If m = 2 l , l ∈ N + , then

(2.3) ∫ Ω u m ∗ − 1 ξ ∂ K ( ∣ y ∣ ) ∂ y s = ∫ ∂ Ω K ( ∣ y ∣ ) u m ∗ − 1 ξ ν s − ∫ ∂ Ω Δ l u Δ l ξ ν s + ∑ i = 1 l ∫ ∂ Ω Δ m − i u ∂ 2 Δ i − 1 ξ ∂ y s ∂ ν − ∑ i = 1 l ∫ ∂ Ω ∂ Δ m − i u ∂ ν ∂ Δ i − 1 ξ ∂ y s + ∑ i = 1 l ∫ ∂ Ω Δ m − i ξ ∂ 2 Δ i − 1 u ∂ y s ∂ ν − ∑ i = 1 l ∫ ∂ Ω ∂ Δ m − i ξ ∂ ν ∂ Δ i − 1 u ∂ y s

and

(2.4) ∫ Ω u m ∗ − 1 ξ ⟨ ∇ K ( ∣ y ∣ ) , y − x 0 ⟩ = ∫ ∂ Ω K ( ∣ y ∣ ) u m ∗ − 1 ξ ⟨ ν , y − x 0 ⟩ − ∫ ∂ Ω Δ l u Δ l ξ ⟨ ν , y − x 0 ⟩ − ∑ i = 1 l ∫ ∂ Ω ∂ Δ m − i u ∂ ν Δ i − 1 ⟨ ∇ ξ , y − x 0 ⟩ + ∑ i = 1 l ∫ ∂ Ω Δ m − i u ∂ Δ i − 1 ⟨ ∇ ξ , y − x 0 ⟩ ∂ ν − ∑ i = 1 l ∫ ∂ Ω ∂ Δ m − i ξ ∂ ν Δ i − 1 ⟨ ∇ u , y − x 0 ⟩ + ∑ i = 1 l ∫ ∂ Ω Δ m − i ξ ∂ Δ i − 1 ⟨ ∇ u , y − x 0 ⟩ ∂ ν + N − 2 m 2 ∑ i = 1 l ∫ ∂ Ω Δ m − i u ∂ Δ i − 1 ξ ∂ ν + ∫ ∂ Ω Δ m − i ξ ∂ Δ i − 1 u ∂ ν − ∫ ∂ Ω ∂ Δ m − i u ∂ ν Δ i − 1 ξ − ∫ ∂ Ω ∂ Δ m − i ξ ∂ ν Δ i − 1 u .

Proof

Proof of (2.3). Multiplying the two equations (2.1) and (2.2) by ∂ ξ ∂ y s and ∂ u ∂ y s , respectively, and subtracting the obtained two equations, we have

(2.5) ∫ Ω Δ m u ∂ ξ ∂ y s + Δ m ξ ∂ u ∂ y s = ∫ Ω K ( ∣ y ∣ ) u m ∗ − 1 ∂ ξ ∂ y s + ( m ∗ − 1 ) u m ∗ − 2 ξ ∂ u ∂ y s .

Similarly, we have

(2.6) ∫ Ω K ( ∣ y ∣ ) u m ∗ − 1 ∂ ξ ∂ y s + ( m ∗ − 1 ) u m ∗ − 2 ξ ∂ u ∂ y s = ∫ Ω K ( ∣ y ∣ ) ∂ ( u m ∗ − 1 ξ ) ∂ y s = − ∫ Ω u m ∗ − 1 ξ ∂ K ( ∣ y ∣ ) ∂ y s + ∫ ∂ Ω K ( ∣ y ∣ ) u m ∗ − 1 ξ ν s .

A direct computation shows that

(2.7) ∫ Ω Δ m u ∂ ξ ∂ y s + Δ m ξ ∂ u ∂ y s = ∫ Ω Δ m − 1 u ∂ Δ ξ ∂ y s + Δ m − 1 ξ ∂ Δ u ∂ y s − ∫ ∂ Ω Δ m − 1 u ∂ 2 ξ ∂ ν ∂ y s + ∫ ∂ Ω ∂ Δ m − 1 u ∂ ν ∂ ξ ∂ y s − ∫ ∂ Ω Δ m − 1 ξ ∂ 2 u ∂ ν ∂ y s + ∫ ∂ Ω ∂ Δ m − 1 ξ ∂ ν ∂ u ∂ y s = ∫ Ω Δ l u ∂ Δ l ξ ∂ y s + Δ l ξ ∂ Δ l u ∂ y s − ∑ i = 1 l ∫ ∂ Ω Δ m − i u ∂ 2 Δ i − 1 ξ ∂ y s ∂ ν + ∑ i = 1 l ∫ ∂ Ω ∂ Δ m − i u ∂ ν ∂ Δ i − 1 ξ ∂ y s − ∑ i = 1 l ∫ ∂ Ω Δ m − i ξ ∂ 2 Δ i − 1 u ∂ y s ∂ ν + ∑ i = 1 l ∫ ∂ Ω ∂ Δ m − i ξ ∂ ν ∂ Δ i − 1 u ∂ y s = ∫ ∂ Ω Δ l u Δ l ξ ν s − ∑ i = 1 l ∫ ∂ Ω Δ m − i u ∂ 2 Δ i − 1 ξ ∂ y s ∂ ν + ∑ i = 1 l ∫ ∂ Ω ∂ Δ m − i u ∂ ν ∂ Δ i − 1 ξ ∂ y s − ∑ i = 1 l ∫ ∂ Ω Δ m − i ξ ∂ 2 Δ i − 1 u ∂ y s ∂ ν + ∑ i = 1 l ∫ ∂ Ω ∂ Δ m − i ξ ∂ ν ∂ Δ i − 1 u ∂ y s .

This completes the proof of (2.3).

Proof of (2.4). For this formula, we multiply the two equations (2.1) and (2.2) by ⟨ ∇ ξ , y − x 0 ⟩ and ⟨ ∇ u , y − x 0 ⟩ , respectively, and again subtracting the two resulted equations, we have

(2.8) ∫ Ω ( Δ m u ⟨ ∇ ξ , y − x 0 ⟩ + Δ m ξ ⟨ ∇ u , y − x 0 ⟩ ) = ∫ Ω K ( ∣ y ∣ ) ( u m ∗ − 1 ⟨ ∇ ξ , y − x 0 ⟩ + ( m ∗ − 1 ) u m ∗ − 2 ξ ⟨ ∇ ξ , y − x 0 ⟩ ) .

On the other hand, we have

(2.9) ∫ Ω K ( ∣ y ∣ ) ( u m ∗ − 1 ⟨ ∇ ξ , y − x 0 ⟩ + ( m ∗ − 1 ) u m ∗ − 2 ξ ⟨ ∇ ξ , y − x 0 ⟩ ) = ∫ Ω K ( ∣ y ∣ ) ⟨ ∇ ( u m ∗ − 1 ξ ) , y − x 0 ⟩ = ∫ ∂ Ω K ( ∣ y ∣ ) u m ∗ − 1 ξ ⟨ ν , y − x 0 ⟩ − ∫ Ω u m ∗ − 1 ξ ⟨ ∇ K ( ∣ y ∣ ) , y − x 0 ⟩ − N ∫ Ω K ( ∣ y ∣ ) u m ∗ − 1 ξ ,

where ν is the outward unit normal of ∂ Ω at y ∈ ∂ Ω . Moreover,

(2.10) ∫ Ω ( Δ m u ⟨ ∇ ξ , y − x 0 ⟩ + Δ m ξ ⟨ ∇ u , y − x 0 ⟩ ) = ∫ Ω ( Δ l u Δ l ⟨ ∇ ξ , y − x 0 ⟩ + Δ l ξ Δ l ⟨ ∇ u , y − x 0 ⟩ ) − ∑ i = 1 l ∫ ∂ Ω ∂ Δ m − i u ∂ ν Δ i − 1 ⟨ ∇ ξ , y − x 0 ⟩ + ∑ i = 1 l ∫ ∂ Ω Δ m − i u ∂ Δ i − 1 ⟨ ∇ ξ , y − x 0 ⟩ ∂ ν − ∑ i = 1 l ∫ ∂ Ω ∂ Δ m − i ξ ∂ ν Δ i − 1 ⟨ ∇ u , y − x 0 ⟩ + ∑ i = 1 l ∫ ∂ Ω Δ m − i ξ ∂ Δ i − 1 ⟨ ∇ u , y − x 0 ⟩ ∂ ν

and

(2.11) ∫ Ω Δ l ξ Δ l ⟨ ∇ u , y − x 0 ⟩ + ∫ Ω Δ l u Δ l ⟨ ∇ ξ , y − x 0 ⟩ = ∫ Ω Δ l ξ ( Δ l − 1 ⟨ ∇ Δ u , y − x 0 ⟩ + 2 Δ u ) + ∫ Ω Δ l u ( Δ l − 1 ⟨ ∇ Δ ξ , y − x 0 ⟩ + 2 Δ ξ ) = ∫ Ω Δ l ξ Δ l − 1 ⟨ ∇ Δ u , y − x 0 ⟩ + ∫ Ω Δ l u Δ l − 1 ⟨ ∇ Δ ξ , y − x 0 ⟩ + 2 ∫ Ω Δ l ξ Δ l u = ∫ Ω Δ l ξ ⟨ ∇ Δ l u , y − x 0 ⟩ + ∫ Ω Δ l u ⟨ ∇ Δ l ξ , y − x 0 ⟩ + 4 l ∫ Ω Δ l ξ Δ l u = ∫ ∂ Ω Δ l ξ Δ l u ⟨ ν , y − x 0 ⟩ + ( 4 l − N ) ∫ Ω Δ l ξ Δ l u .

We also have

(2.12) m ∗ ∫ Ω K ( ∣ y ∣ ) u m ∗ − 1 ξ = ∫ Ω ( Δ m u ξ + Δ m ξ u ) = 2 ∫ Ω Δ l ξ Δ l u + ∑ i = 1 l ∫ ∂ Ω Δ m − i u ∂ Δ i − 1 ξ ∂ ν + ∫ ∂ Ω Δ m − i ξ ∂ Δ i − 1 u ∂ ν − ∫ ∂ Ω ∂ Δ m − i u ∂ ν Δ i − 1 ξ − ∫ ∂ Ω ∂ Δ m − i ξ ∂ ν Δ i − 1 u ,

which gives

(2.13) ∫ Ω Δ l u Δ l ξ = m ∗ 2 ∫ Ω K ( ∣ y ∣ ) u m ∗ − 1 ξ − 1 2 ∑ i = 1 l ∫ ∂ Ω Δ m − i u ∂ Δ i − 1 ξ ∂ ν + ∫ ∂ Ω Δ m − i ξ ∂ Δ i − 1 u ∂ ν − ∫ ∂ Ω ∂ Δ m − i u ∂ ν Δ i − 1 ξ − ∫ ∂ Ω ∂ Δ m − i ξ ∂ ν Δ i − 1 u .

Thus, the result follows.□

Lemma 2.2

If m = 2 l + 1 , l ∈ N + , then

(2.14) ∫ Ω u m ∗ − 1 ξ ∂ K ( ∣ y ∣ ) ∂ y s = ∫ ∂ Ω K ( ∣ y ∣ ) u m ∗ − 1 ξ ν s − ∫ ∂ Ω ⟨ ∇ Δ l u , ∇ Δ l ξ ⟩ ν s − ∑ i = 1 l ∫ ∂ Ω Δ m − i u ∂ 2 Δ i − 1 ξ ∂ y s ∂ ν + ∑ i = 1 l + 1 ∫ ∂ Ω ∂ Δ m − i u ∂ ν ∂ Δ i − 1 ξ ∂ y s − ∑ i = 1 l ∫ ∂ Ω Δ m − i ξ ∂ 2 Δ i − 1 u ∂ y s ∂ ν + ∑ i = 1 l + 1 ∫ ∂ Ω ∂ Δ m − i ξ ∂ ν ∂ Δ i − 1 u ∂ y s

and

(2.15) ∫ Ω u m ∗ − 1 ξ ⟨ ∇ K ( ∣ y ∣ ) , y − x 0 ⟩ = ∫ ∂ Ω K ( ∣ y ∣ ) u m ∗ − 1 ξ ⟨ ν , y − x 0 ⟩ − ∫ ∂ Ω ⟨ ∇ Δ l u , ∇ Δ l ξ ⟩ ⟨ ν , y − x 0 ⟩ + ∑ i = 1 l + 1 ∫ ∂ Ω ∂ Δ m − i u ∂ ν Δ i − 1 ⟨ ∇ ξ , y − x 0 ⟩ − ∑ i = 1 l ∫ ∂ Ω Δ m − i u ∂ Δ i − 1 ⟨ ∇ ξ , y − x 0 ⟩ ∂ ν + ∑ i = 1 l + 1 ∫ ∂ Ω ∂ Δ m − i ξ ∂ ν Δ i − 1 ⟨ ∇ u , y − x 0 ⟩ − ∑ i = 1 l ∫ ∂ Ω Δ m − i ξ ∂ Δ i − 1 ⟨ ∇ u , y − x 0 ⟩ ∂ ν + N − 2 m 2 ∑ i = 1 l − ∫ ∂ Ω Δ m − i u ∂ Δ i − 1 ξ ∂ ν − ∫ ∂ Ω Δ m − i ξ ∂ Δ i − 1 u ∂ ν + N − 2 m 2 ∑ i = 1 l + 1 ∫ ∂ Ω ∂ Δ m − i u ∂ ν Δ i − 1 ξ + ∫ ∂ Ω ∂ Δ m − i ξ ∂ ν Δ i − 1 u .

Proof

Proof of (2.14). Using the similar arguments as in the proof of (2.3), we have

(2.16) ∫ Ω − Δ m u ∂ ξ ∂ y s − Δ m ξ ∂ u ∂ y s = ∫ Ω K ( ∣ y ∣ ) u m ∗ − 1 ∂ ξ ∂ y s + ( m ∗ − 1 ) u m ∗ − 2 ξ ∂ u ∂ y s .

And

(2.17) ∫ Ω K ( ∣ y ∣ ) u m ∗ − 1 ∂ ξ ∂ y s + ( m ∗ − 1 ) u m ∗ − 2 ξ ∂ u ∂ y s = ∫ Ω K ( ∣ y ∣ ) ∂ ( u m ∗ − 1 ξ ) ∂ y s = − ∫ Ω u m ∗ − 1 ξ ∂ K ( ∣ y ∣ ) ∂ y s + ∫ ∂ Ω K ( ∣ y ∣ ) u m ∗ − 1 ξ ν s .

Further computation leads to

(2.18) ∫ Ω − Δ m u ∂ ξ ∂ y s − Δ m ξ ∂ u ∂ y s = ∫ Ω − Δ l + 1 u ∂ Δ l ξ ∂ y s − Δ l + 1 ξ ∂ Δ l u ∂ y s + ∑ i = 1 l ∫ ∂ Ω Δ m − i u ∂ 2 Δ i − 1 ξ ∂ y s ∂ ν − ∑ i = 1 l ∫ ∂ Ω ∂ Δ m − i u ∂ ν ∂ Δ i − 1 ξ ∂ y s + ∑ i = 1 l ∫ ∂ Ω Δ m − i ξ ∂ 2 Δ i − 1 u ∂ y s ∂ ν − ∑ i = 1 l ∫ ∂ Ω ∂ Δ m − i ξ ∂ ν ∂ Δ i − 1 u ∂ y s = ∫ Ω ∇ Δ l u , ∇ Δ l ∂ ξ ∂ y s + ∫ Ω ∇ Δ l ξ , ∇ Δ l ∂ u ∂ y s + ∑ i = 1 l ∫ ∂ Ω Δ m − i u ∂ 2 Δ i − 1 ξ ∂ y s ∂ ν − ∑ i = 1 l + 1 ∫ ∂ Ω ∂ Δ m − i u ∂ ν ∂ Δ i − 1 ξ ∂ y s + ∑ i = 1 l ∫ ∂ Ω Δ m − i ξ ∂ 2 Δ i − 1 u ∂ y s ∂ ν − ∑ i = 1 l + 1 ∫ ∂ Ω ∂ Δ m − i ξ ∂ ν ∂ Δ i − 1 u ∂ y s = ∫ ∂ Ω ⟨ ∇ Δ l u , ∇ Δ l ξ ⟩ ν s − ∑ i = 1 l ∫ ∂ Ω Δ m − i u ∂ 2 Δ i − 1 ξ ∂ y s ∂ ν + ∑ i = 1 l ∫ ∂ Ω ∂ Δ m − i u ∂ ν ∂ Δ i − 1 ξ ∂ y s − ∑ i = 1 l ∫ ∂ Ω Δ m − i ξ ∂ 2 Δ i − 1 u ∂ y s ∂ ν + ∑ i = 1 l ∫ ∂ Ω ∂ Δ m − i ξ ∂ ν ∂ Δ i − 1 u ∂ y s .

And (2.14) is proved.

Proof of (2.15). We have

(2.19) ∫ Ω ( − Δ m u ⟨ ∇ ξ , y − x 0 ⟩ − Δ m ξ ⟨ ∇ u , y − x 0 ⟩ ) = ∫ Ω K ( ∣ y ∣ ) ( u m ∗ − 1 ⟨ ∇ ξ , y − x 0 ⟩ + ( m ∗ − 1 ) u m ∗ − 2 ξ ⟨ ∇ ξ , y − x 0 ⟩ ) .

It is also easy to check that

(2.20) ∫ Ω K ( ∣ y ∣ ) ( u m ∗ − 1 ⟨ ∇ ξ , y − x 0 ⟩ + ( m ∗ − 1 ) u m ∗ − 2 ξ ⟨ ∇ ξ , y − x 0 ⟩ ) = ∫ Ω K ( ∣ y ∣ ) ⟨ ∇ ( u m ∗ − 1 ξ ) , y − x 0 ⟩ = ∫ ∂ Ω K ( ∣ y ∣ ) u m ∗ − 1 ⟨ ν , y − x 0 ⟩ − ∫ Ω u m ∗ − 1 ξ ⟨ ∇ K ( ∣ y ∣ ) , y − x 0 ⟩ − N ∫ Ω K ( ∣ y ∣ ) u m ∗ − 1 ξ ,

where ν is the outward unit normal of ∂ Ω at y ∈ ∂ Ω . Moreover,

(2.21) ∫ Ω ( − Δ m u ⟨ ∇ ξ , y − x 0 ⟩ − Δ m ξ ⟨ ∇ u , y − x 0 ⟩ ) = − ∫ Ω ( Δ l + 1 u Δ l ⟨ ∇ ξ , y − x 0 ⟩ + Δ l + 1 ξ Δ l ⟨ ∇ u , y − x 0 ⟩ ) + ∑ i = 1 l ∫ ∂ Ω ∂ Δ m − i u ∂ ν Δ i − 1 ⟨ ∇ ξ , y − x 0 ⟩ − ∑ i = 1 l ∫ ∂ Ω Δ m − i u ∂ Δ i − 1 ⟨ ∇ ξ , y − x 0 ⟩ ∂ ν + ∑ i = 1 l ∫ ∂ Ω ∂ Δ m − i ξ ∂ ν Δ i − 1 ⟨ ∇ u , y − x 0 ⟩ − ∑ i = 1 l ∫ ∂ Ω Δ m − i ξ ∂ Δ i − 1 ⟨ ∇ u , y − x 0 ⟩ ∂ ν = ∫ Ω ⟨ ∇ Δ l u , Δ l ⟨ ∇ ξ , y − x 0 ⟩ ⟩ + ∫ Ω ⟨ ∇ Δ l ξ , Δ l ⟨ ∇ u , y − x 0 ⟩ ⟩ + ∑ i = 1 l ∫ ∂ Ω ∂ Δ m − i u ∂ ν Δ i − 1 ⟨ ∇ ξ , y − x 0 ⟩ − ∑ i = 1 l + 1 ∫ ∂ Ω Δ m − i u ∂ Δ i − 1 ⟨ ∇ ξ , y − x 0 ⟩ ∂ ν + ∑ i = 1 l ∫ ∂ Ω ∂ Δ m − i ξ ∂ ν Δ i − 1 ⟨ ∇ u , y − x 0 ⟩ − ∑ i = 1 l + 1 ∫ ∂ Ω Δ m − i ξ ∂ Δ i − 1 ⟨ ∇ u , y − x 0 ⟩ ∂ ν

and

(2.22) ∫ Ω ⟨ ∇ Δ l u , Δ l ⟨ ∇ ξ , y − x 0 ⟩ ⟩ + ∫ Ω ⟨ ∇ Δ l ξ , Δ l ⟨ ∇ u , y − x 0 ⟩ ⟩ = ∫ Ω ⟨ ∇ Δ l u , ⟨ ∇ Δ l ξ , y − x 0 ⟩ ⟩ + ∫ Ω ⟨ ∇ Δ l ξ , ⟨ ∇ Δ l u , y − x 0 ⟩ ⟩ + 4 l ∫ Ω ⟨ ∇ Δ l u , ∇ Δ l ξ ⟩ = ∫ Ω ⟨ ∇ Δ l u , ∇ Δ l ξ ⟩ ⟨ ν , y − x 0 ⟩ + ( 4 l + 2 − N ) ∫ Ω ⟨ ∇ Δ l u , ∇ Δ l ξ ⟩ .

On the other hand, we have

(2.23) m ∗ ∫ Ω K ( ∣ y ∣ ) u m ∗ − 1 ξ = − ∫ Ω ( Δ m u ξ + Δ m ξ u ) = 2 ∫ Ω ⟨ ∇ Δ l u , ∇ Δ l ξ ⟩ − ∑ i = 1 l ∫ ∂ Ω Δ m − i u ∂ Δ i − 1 ξ ∂ ν + ∫ ∂ Ω Δ m − i ξ ∂ Δ i − 1 u ∂ ν + ∑ i = 1 l + 1 ∫ ∂ Ω ∂ Δ m − i u ∂ ν Δ i − 1 ξ + ∫ ∂ Ω ∂ Δ m − i ξ ∂ ν Δ i − 1 u ,

which gives

(2.24) ∫ Ω ⟨ ∇ Δ l u , ∇ Δ l ξ ⟩ = m ∗ 2 ∫ Ω K ( ∣ y ∣ ) u m ∗ − 1 ξ − 1 2 − ∑ i = 1 l ∫ ∂ Ω Δ m − i u ∂ Δ i − 1 ξ ∂ ν + ∫ ∂ Ω Δ m − j ξ ∂ Δ i − 1 u ∂ ν + ∑ i = 1 l + 1 ∫ ∂ Ω ∂ Δ m − i u ∂ ν Δ i − 1 ξ + ∫ ∂ Ω ∂ Δ m − i ξ ∂ ν Δ i − 1 u .

Thus, the result follows.□

3 Non-degeneracy of the bubble solutions

In this section, we first use Lemmas 2.1 and 2.2 to establish a fine estimate on the k -bubble solution u k obtained in Theorem A. Then we prove a non-degeneracy result by using a contradiction argument. We introduce following norms by:

‖ u ‖ ∗ = sup y ∈ R N ∣ u ( y ) ∣ ∑ j = 1 k μ k N − 2 m 2 ( 1 + μ k ∣ y − x k , j ∣ ) N − 2 m 2 + τ − 1

and

‖ u ‖ ∗ ∗ = sup y ∈ R N ∣ u ( y ) ∣ ∑ j = 1 k μ k N + 2 m 2 ( 1 + μ k ∣ y − x k , j ∣ ) N + 2 m 2 + τ − 1 ,

where x k , j = r k cos 2 ( i − 1 ) π k , r k sin 2 ( i − 1 ) π k , 0 , and τ is any fixed number in N − 2 m − 2 N − 2 m , 1 + θ , θ > 0 is a small constant. Noting that μ k ∼ k N − 2 m N − 2 m − 2 and the choice of x k , j and τ , we find

∑ j = 2 k 1 ( μ k ∣ x k , 1 − x k , j ∣ ) τ ≤ C k τ μ k τ ∑ j = 1 k 1 j τ ≤ C 1 k μ k τ ≤ C ′ .

It is known that (see [15]), λ 1 = 1 , λ 2 = λ 3 = … = λ N + 2 are the eigenvalues of the following eigenvalue problem:

( − Δ ) m v = λ U x i , μ m ∗ − 2 v , for v ∈ D m , 2 ( R N ) ,

and the corresponding linear independent eigenvectors v 1 , v 2 , … , v N + 2 are given by

v 1 = U x i , μ , v j + 1 = ∂ U x i , μ ∂ y j , j = 1 , 2 , … , N , v N + 2 = x ⋅ ∇ U x i , μ + N − 2 m 2 U x i , μ .

Let

Ω j = y = ( y ′ , y ″ ) ∈ R 2 × R N : ( y ′ , 0 ) ∣ y ′ ∣ , x k , j ∣ x k , j ∣ ≥ cos π k .

We define the linear operator L k by

(3.1) L k ξ = ( − Δ ) m ξ − ( m ∗ − 1 ) K ( ∣ y ∣ ) u k m ∗ − 2 ξ .

Lemma 3.1

There exists a constant C > 0 such that for all y ∈ R N ,

∣ u k ( y ) ∣ ≤ C ∑ j = 1 k μ k N − 2 m 2 ( 1 + μ k ∣ y − x k , j ∣ ) N − 2 m .

Proof

For convenience, we set u ^ k = μ k − N − 2 m 2 u k ( μ k − 1 y ) , then

( − Δ ) m u ^ k = K ( μ k − 1 y ) u ^ k m ∗ − 1 .

Using Green’s formula, we know that

u ^ k ( y ) = ∫ R N C N , m ∣ z − y ∣ N − 2 m K ( μ k − 1 z ) u ^ k m ∗ − 1 ( z ) d z .

Recall that the bubble solution u k = ∑ j = 1 k U x j , μ k + ω k with ‖ ω k ‖ ∗ = O 1 μ k 1 + σ for some σ > 0 , we find

∣ u ^ k ( y ) ∣ ≤ C ∫ R N 1 ∣ z − y ∣ N − 2 m ∣ u ^ k m ∗ − 1 ( z ) ∣ d z ≤ C ∫ R N 1 ∣ z − y ∣ N − 2 m ∑ j = 1 k 1 ( 1 + ∣ z − x ^ k , j ∣ ) N − 2 m 2 + τ m ∗ − 1 d z ≤ C ∫ R N 1 ∣ z − y ∣ N − 2 m ∑ j = 1 k 1 ( 1 + ∣ z − x ^ k , j ∣ ) N + 2 m 2 + τ + 4 m ( τ − τ 1 ) N − 2 m ∑ j = 1 k 1 ( 1 + ∣ z − x ^ k , j ∣ ) τ 1 4 m N − 2 m d z ≤ C ∫ R N 1 ∣ z − y ∣ N − 2 m ∑ j = 1 k 1 ( 1 + ∣ z − x ^ k , j ∣ ) N + 2 m 2 + τ + 4 m ( τ − τ 1 ) N − 2 m d z ≤ C ∑ j = 1 k 1 ( 1 + ∣ y − x ^ k , j ∣ ) N − 2 m 2 + τ + 4 m ( τ − τ 1 ) N − 2 m ,

where x ^ k , j = μ k x k , j and τ 1 ∈ N − 2 m − 2 N − 2 m , τ . Noting that

N − 2 m 2 + τ + 4 m ( τ − τ 1 ) N − 2 m > N − 2 m 2 + τ ,

we can choose τ 1 such that N − 2 m 2 − τ 4 m ( τ − τ 1 ) N − 2 m is not an integer.

Let η = 4 m ( τ − τ 1 ) N − 2 m and l = N − 2 m 2 − τ 4 m ( τ − τ 1 ) N − 2 m , then we have

∣ u ^ k ( y ) ∣ ≤ C ∑ j = 1 k 1 ( 1 + ∣ y − x ^ k , j ∣ ) N − 2 m 2 + τ + η .

Continuing this process, we have

∣ u ^ k ( y ) ∣ ≤ C ∑ j = 1 k 1 ( 1 + ∣ y − x ^ k , j ∣ ) N − 2 m 2 + τ + η + 4 m ( τ + η − τ 2 ) N − 2 m ,

where τ 2 ∈ N − 2 m − 2 N − 2 m , τ + η . We can choose τ 2 such that 4 m ( τ + η − τ 2 ) N − 2 m = η , then we have

∣ u ^ k ( y ) ∣ ≤ C ∑ j = 1 k 1 ( 1 + ∣ y − x ^ k , j ∣ ) N − 2 m 2 + τ + 2 η .

Repeating this process again, we have

∣ u ^ k ( y ) ∣ ≤ C ∑ j = 1 k 1 ( 1 + ∣ y − x ^ k , j ∣ ) N − 2 m 2 + τ + l η .

∣ u ^ k ( y ) ∣ ≤ C ∫ R N 1 ∣ z − y ∣ N − 2 m ∑ j = 1 k 1 ( 1 + ∣ z − x ^ k , j ∣ ) N + 2 m 2 + τ + l η + η d z ≤ C ∑ j = 1 k 1 ( 1 + ∣ y − x ^ k , j ∣ ) N − 2 m .

And the desired result follows immediately.□

Lemma 3.2

There exists a constant C > 0 such that for all y ∈ R N and r = 1 , … , 2 m − 1 ,

( ∂ ) r u k ( ∂ ) y 1 i 1 … ( ∂ ) y N i N ( y ) ≤ C ∑ j = 1 k μ k N − 2 m 2 + j ( 1 + μ k ∣ y − x k , j ∣ ) N − 2 m + j ,

where ∑ s = 1 N i s = r .

Proof

Let u ^ k = μ − N − 2 m 2 u k ( μ k − 1 y ) , then

( − Δ ) m u ^ k = K ( μ k − 1 y ) u ^ k m ∗ − 1 .

We have

u ^ k ( y ) = ∫ R N 1 ∣ z − y ∣ N − 2 m K ( μ k − 1 z ) u ^ k m ∗ − 1 ( z ) d z .

Hence,

( ∂ ) r u ^ k ( ∂ ) y 1 i 1 … ( ∂ ) y N i N ( y ) = C ∫ R N ( ∂ ) r ( ∂ ) y 1 i 1 … ( ∂ ) y N i N 1 ∣ z − y ∣ N − 2 m K ( μ k − 1 z ) u ^ k m ∗ − 1 ( z ) d z ≤ C ∫ R N 1 ∣ z − y ∣ N − 2 m + j ∣ K ( μ k − 1 z ) u ^ k m ∗ − 1 ( z ) ∣ d z .

Similar to the proof of Lemma 3.1, we can prove

( ∂ ) r u ^ k ( ∂ ) y 1 i 1 … ( ∂ ) y N i N ( y ) ≤ C ∑ j = 1 k 1 ( 1 + ∣ y − x ^ k , j ∣ ) N − 2 m + j .

Therefore,

( ∂ ) j u k ( ∂ ) y 1 i 1 … ( ∂ ) y N i N ( y ) ≤ C ∑ j = 1 k μ k N − 2 m 2 + j ( 1 + μ k ∣ y − x k , j ∣ ) N − 2 m + j . □

In the following, we shall prove Theorem 1.1 by using contradiction arguments. Suppose that there are k n → + ∞ , satisfying

(3.2) L k n ξ n = 0 ,

but ξ n ≠ 0 . Without loss of generality, we may assume ‖ ξ n ‖ ∗ = 1 and obtain the contradictions through the following steps. Set

(3.3) ξ ^ n ( y ) = μ k n − N − 2 m 2 ξ n ( μ k n − 1 y + x k n , 1 ) .

Lemma 3.3

It holds

(3.4) ξ ^ n → b 0 ψ 0 + b 1 ψ 1 ,

uniformly in C m ( B R ( 0 ) ) for any R > 0 , where b 0 and b 1 are some constants,

ψ 0 = ∂ U 0 , μ ∂ μ μ = 1 , ψ i = ∂ U 0 , 1 ∂ y i , i = 1 , … , N .

Proof

In view of ∣ ξ ^ n ∣ ≤ C , we may assume that ξ ^ n → ξ in C loc m ( R N ) . Then ξ satisfies

(3.5) ( − Δ ) m ξ = ( m ∗ − 1 ) U 0 , 1 m ∗ − 1 ξ , in R N ,

which gives

(3.6) ξ = ∑ i = 0 N b i ψ i .

Since ξ n is even in y i , i = 2 , … , N , it holds b i = 0 , i = 2 , … , N .□

We decompose

ξ n ( y ) = b 0 , n μ k n ∑ j = 1 k n ∂ U x k n , j , μ k n ∂ μ k n + b 1 , n μ k n − 1 ∑ j = 1 k n ∂ U x k n , j , μ k n ∂ r + ξ n ∗ ,

where ξ n ∗ satisfies

∫ R N U x k n , j , μ k n m ∗ − 2 ∂ U x k n , j , μ k n ∂ μ k n ξ n ∗ = ∫ R N U x k n , j , μ k n m ∗ − 2 ∂ U x k n , j , μ k n ∂ r ξ n ∗ = 0 .

It follows from Lemma 2.3 that b 0 , n and b 1 , n are bounded.

Lemma 3.4

It holds

(3.7) ‖ ξ n ∗ ‖ ≤ C μ k n − 1 − σ ,

where σ > 0 is a small constant.

Proof

Since L k n ξ n = 0 . We can deduce that

L k n ξ n ∗ ≔ ( − Δ ) m ξ n ∗ − ( m ∗ − 1 ) K ( ∣ y ∣ ) u k n m ∗ − 2 ξ n ∗ = − ( m ∗ − 1 ) ( K ( ∣ y ∣ ) − 1 ) u k n m ∗ − 2 ∑ j = 1 k n b 0 , n μ k n ∂ U x k n , j , μ k n ∂ μ k n + b 1 , n μ k n − 1 ∂ U x k n , j , μ k n ∂ r − ( m ∗ − 1 ) ∑ j = 1 k n ( u k n m ∗ − 2 − U x k n , j , μ k n m ∗ − 2 ) b 0 , n μ k n ∂ U x k n , j , μ k n ∂ μ k n + b 1 , n μ k n − 1 ∂ U x k n , j , μ k n ∂ r .

Then we can prove

(3.8) ( K ( ∣ y ∣ ) − 1 ) u k n m ∗ − 2 ∑ j = 1 k n b 0 , n μ k n ∂ U x k n , j , μ k n ∂ μ k n + b 1 , n μ k n − 1 ∂ U x k n , j , μ k n ∂ r ∗ ∗ ≤ ( K ( ∣ y ∣ ) − 1 ) ∑ j = 1 k n U x k n , j , μ k n m ∗ − 1 ∗ ∗ ≤ C μ k n − 1 − σ .

In fact, without loss of generality, we may assume y ∈ Ω 1 , then we compute

( K ( ∣ y ∣ ) − 1 ) ∑ j = 1 k n U x k n , j , μ k n m ∗ − 1 ≤ C μ k n N + 2 m 2 ( 1 + μ k n ∣ y − x k n , 1 ∣ ) N + 2 m + C ∑ j = 2 k n μ k n N − 2 m 2 ( 1 + μ k n ∣ y − x k n , j ∣ ) N − 2 m N + 2 m N − 2 m ≕ I 1 + I 2 .

Since N − 2 m 2 − τ N − 2 m N + 2 m > 1 , we have

I 2 ≤ C μ k n N + 2 m 2 ( 1 + μ k n ∣ y − x k n , 1 ∣ ) N + 2 m 2 + τ k n μ k n N + 2 m 2 − τ ≤ C μ k n N + 2 m 2 ( 1 + μ k n ∣ y − x k n , 1 ∣ ) N + 2 m 2 + τ 1 μ k n 2 N − 2 m N + 2 m 2 − τ ≤ C μ k n N + 2 m 2 ( 1 + μ k n ∣ y − x k n , 1 ∣ ) N + 2 m 2 + τ μ k n − 1 − σ .

Now, we turn to consider the first term I 1 . We split the slice Ω 1 into two parts, namely,

J 1 ≔ { y ∈ Ω 1 ∣ ∣ ∣ y ∣ − r 0 ∣ ≥ δ } , J 2 ≔ { y ∈ Ω 1 ∣ ∣ ∣ y ∣ − r 0 ∣ ≤ δ } ,

where δ > 0 is a fixed constant.

For the region J 1 , we have

∣ y − x k n , 1 ∣ ≥ ∣ ∣ y ∣ − ∣ x k n , 1 ∣ ∣ ≥ ∣ ∣ y ∣ − r 0 ∣ − ∣ r 0 − ∣ x k n , 1 ‖ ≥ 1 2 δ .

Hence,

I 1 ≤ C μ k n N + 2 m 2 ( 1 + μ k n ∣ y − x k n , 1 ∣ ) N + 2 m 2 + τ 1 μ k n N + 2 m − τ ≤ C μ k n N + 2 m 2 ( 1 + μ k n ∣ y − x k n , 1 ∣ ) N + 2 m 2 + τ μ k n − 1 − σ .

For the region J 2 , we have

∣ K ( ∣ y ∣ ) − 1 ∣ ≤ C ∣ ∣ y ∣ − r 0 ∣ 2 ≤ C ( ∣ ∣ y ∣ − ∣ x k n , 1 ∣ ∣ + ∣ ∣ x k n , 1 ∣ − r 0 ∣ ) 2 ≤ C ( ∣ ∣ y ∣ − ∣ x k n , 1 ∣ ∣ 2 + μ k n − 2 − σ ) .

Thus, for σ > 0 small such that N + 2 m 2 − τ − 1 − σ ≥ 2 m + 1 2 − τ − σ > 0 , we have

I 1 ≤ C μ k n N + 2 m 2 ( 1 + μ k n ∣ y − x k n , 1 ∣ ) N + 2 m 2 + τ ∣ ∣ y ∣ − ∣ x k n , 1 ∣ ∣ 2 + μ − 2 − σ ( 1 + μ k n ∣ y − x k n , 1 ∣ ) N + 2 m 2 − τ ≤ C μ k n N + 2 m 2 ( 1 + μ k n ∣ y − x k n , 1 ∣ ) N + 2 m 2 + τ ( μ k n N + 2 m 2 − τ + μ k n − 2 − σ ) ≤ C μ k n N + 2 m 2 ( 1 + μ k n ∣ y − x k n , 1 ∣ ) N + 2 m 2 + τ μ k n − 1 − σ .

In the following, for simplicity, without loss of generality, we assume y ∈ Ω 1 .

Case 1. N ≥ 6 m : We have

∑ j = 1 k n U x k n , j , μ k n m ∗ − 2 − U x k n , 1 , μ k n m ∗ − 2 b 0 , n μ k n ∂ U x k n , j , μ k n ∂ μ k n + b 1 , n μ k n − 1 ∂ U x k n , j , μ k n ∂ r ≤ C U x k n , 1 , μ k n m ∗ − 1 ∑ j = 2 k n U x k n , j , μ k n ≤ C μ k n N + 2 m 2 ( 1 + μ k n ∣ y − x k n , 1 ∣ ) N + 2 m 2 + τ ∑ j = 2 k n 1 ( μ k n ∣ y − x k n , j ∣ ) N + 2 m 2 − τ ≤ C μ k n N + 2 m 2 ( 1 + μ k n ∣ y − x k n , 1 ∣ ) N + 2 m 2 + τ μ k n − 1 − σ .

Case 2. N < 6 m : We have

∑ j = 1 k n U x k n , j , μ k n m ∗ − 2 − U x k n , 1 , μ k n m ∗ − 2 b 0 , n μ k n ∂ U x k n , j , μ k n ∂ μ k n + b 1 , n μ k n − 1 ∂ U x k n , j , μ k n ∂ r ≤ C U x k n , 1 , μ k n m ∗ − 2 ∑ j = 2 k n U x k n , j , μ k n + U x k n , 1 , μ k n ∑ j = 2 k n U x k n , j , μ k n m ∗ − 2 ≤ C μ k n N + 2 m 2 ( 1 + μ k n ∣ y − x k n , 1 ∣ ) N + 2 m 2 + τ ∑ j = 2 k n 1 ( μ k n ∣ y − x k n , j ∣ ) N + 2 m 2 − τ + μ k n N + 2 m 2 ( 1 + μ k n ∣ y − x k n , 1 ∣ ) N + 2 m 2 + τ ∑ j = 2 k n 1 ( μ k n ∣ y − x k n , j ∣ ) ( N − 2 m ) m + N 2 − τ 4 m 4 m N − 2 m ≤ C μ k n N + 2 m 2 ( 1 + μ k n ∣ y − x k n , 1 ∣ ) N + 2 m 2 + τ μ k n − 1 − σ .

So we have proved

(3.9) ∑ j = 1 k n U x k n , j , μ k n m ∗ − 2 − U x k n , 1 , μ k n m ∗ − 2 b 0 , n μ k n ∂ U x k n , j , μ k n ∂ μ k n + b 1 , n μ k n − 1 ∂ U x k n , j , μ k n ∂ r ≤ C μ k n N + 2 m 2 ( 1 + μ k n ∣ y − x k n , 1 ∣ ) N + 2 m 2 + τ μ k n − 1 − σ .

On the other hand, we have

(3.10) ∑ j = 2 k n ∑ i = 1 k n U x k n , i , μ k n m ∗ − 2 − U x k n k n , j , μ k n m ∗ − 2 b 0 , n μ k n ∂ U x k n , j , μ k n ∂ μ k n + b 1 , n μ k n − 1 ∂ U x k n , j , μ k n ∂ r ≤ C ∑ j = 2 k n U x k n , 1 , μ k n m ∗ − 2 + ∑ i = 2 k n U x k n , i , μ k n m ∗ − 2 U x k n , j , μ k n ≤ C ∑ j = 2 k n U x k n , 1 , μ k n m ∗ − 2 + C μ k n 2 m ( 1 + μ k n ∣ y − x k n , 1 ∣ ) 4 m − 4 m τ N − 2 m U x k n , j , μ k n ≤ C μ k n N + 2 m 2 ( 1 + μ k n ∣ y − x k n , 1 ∣ ) N + 2 m 2 + τ μ k n − 1 − σ .

Combining (3.9) and (3.10), we obtained

∑ j = 1 k n ∑ i = 1 k n U x k n , i , μ k n m ∗ − 2 − U x k n , j , μ k n m ∗ − 2 b 0 , n μ k n ∂ U x k n , j , μ k n ∂ μ k n + b 1 , n μ k n − 1 ∂ U x k n , j , μ k n ∂ r ∗ ∗ ≤ μ k n − 1 − σ .

Moreover, for N ≥ 6 m , we have

∑ j = 1 k n ∑ i = 1 k n U x k n , i , μ k n m ∗ − 2 − u k n m ∗ − 2 b 0 , n μ k n ∂ U x k n , j , μ k n ∂ μ k n + b 1 , n μ k n − 1 ∂ U x k n , j , μ k n ∂ r ∗ ∗

≤ C ∑ j = 1 k n ∑ i = 1 k n U x k n , i , μ k n m ∗ − 2 − u k n m ∗ − 2 U x k n , j , μ k n ∗ ∗ ≤ C ∑ j = 1 k n ∑ i = 1 k n U x k n , i , μ k n m ∗ − 3 ω k n U x k n , j , μ k n ∗ ∗ ≤ C ∑ i = 1 k n U x k n , i , μ k n m ∗ − 2 ω k n ∗ ∗ ≤ C ‖ ω k n ‖ ∗ ≤ C μ k n − 1 − σ .

The penultimate term follows the fact that

∑ j = 1 k n μ k n N − 2 m 2 ( 1 + μ k n ∣ y − x k n , j ∣ ) N − 2 m 2 4 m N − 2 m ∑ j = 1 k n μ k n N − 2 m 2 ( 1 + μ k n ∣ y − x k n , j ∣ ) N − 2 m 2 + τ ≤ C μ k n N + 2 m 2 ( 1 + μ k n ∣ y − x k n , 1 ∣ ) ( N − 2 m − η ) 4 m N − 2 m + N − 2 m 2 + τ − η = C μ k n N + 2 m 2 ( 1 + μ k n ∣ y − x k n , 1 ∣ ) N + 2 m 2 + τ + 2 m − η N + 2 m N − 2 m ≤ C ∑ j = 1 k n μ k n N − 2 m 2 ( 1 + μ k n ∣ y − x k n , j ∣ ) N + 2 m 2 + τ ,

where η = N − 2 m − 2 N − 2 m and the last inequality follows the fact that 2 m − η N + 2 m N − 2 m > 0 .

For N < 6 m , we can also prove

(3.11) ∑ j = 1 k n ∑ i = 1 k n U x k n , i , μ k n m ∗ − 2 − u k n m ∗ − 2 b 0 , n μ k n ∂ U x k n , j , μ k n ∂ μ k n + b 1 , n μ k n − 1 ∂ U x k n , j , μ k n ∂ r ∗ ∗ ≤ C ∑ i = 1 k n U x k n , i , μ k n m ∗ − 2 ω k n ∗ ∗ + ∑ i = 1 k n U x k n , i , μ k n ( ω k n ) m ∗ − 2 ∗ ∗ ≤ C ( ‖ ω k n ‖ ∗ + ‖ ω k n ‖ ∗ m ∗ − 2 ) ≤ C μ k n − 1 − σ .

In fact, noting that

N − 2 m 2 − N − 2 m − 2 N − 2 m N + 2 m N − 2 m + τ 4 m N − 2 m − τ ≥ N − 2 m 2 − N − 2 m − 2 N − 2 m N + 2 m N − 2 m + N − 2 m − 2 N − 2 m 4 m N − 2 m − N − 2 m − 2 N − 2 m = N − 2 m 2 − 2 N − 2 m − 2 N − 2 m > 0 ,

the inequality (3.11) follows from

∑ j = 1 k n μ k n N − 2 m 2 ( 1 + μ k n ∣ y − x k n , j ∣ ) N − 2 m 2 4 m N − 2 m ∑ j = 1 k n μ k n N − 2 m 2 ( 1 + μ k n ∣ y − x k n , j ∣ ) N − 2 m 2 + τ ≤ C ∑ j = 1 k n μ k n N − 2 m 2 ( 1 + μ k n ∣ y − x k n , j ∣ ) N + 2 m 2 + τ ,

and

∑ j = 1 k n μ k n N − 2 m 2 ( 1 + μ k n ∣ y − x k n , j ∣ ) N − 2 m 2 ∑ j = 1 k n μ k n N − 2 m 2 ( 1 + μ k n ∣ y − x k n , j ∣ ) N − 2 m 2 + τ 4 m N − 2 m ≤ C μ k n N + 2 m 2 ( 1 + μ k n ∣ y − x k n , 1 ∣ ) N − N − 2 m − 2 N − 2 m N + 2 m N − 2 m + τ 4 m N − 2 m ≤ C ∑ j = 1 k n μ k n N + 2 m 2 ( 1 + μ k n ∣ y − x k n , j ∣ ) N + 2 m 2 + τ .

As a consequence, we have proved

∑ j = 1 k n ∑ i = 1 k n U x k n , i , μ k n m ∗ − 2 − u k n m ∗ − 2 b 0 , n μ k n ∂ U x k n , j , μ k n ∂ μ k n + b 1 , n μ k n − 1 ∂ U x k n , j , μ k n ∂ r ∗ ∗ ≤ C μ k n − 1 − σ .

On the other hand, from

∫ R N U x k n , j , μ k n m ∗ − 2 ∂ U x k n , j , μ k n ∂ μ k n ξ n ∗ = ∫ R N U x k n , j , μ k n m ∗ − 2 ∂ U x k n , j , μ k n ∂ r ξ n ∗ = 0

and Lemma 3.1, we can see that there exists ρ > 0 , such that

‖ L k n ξ n ∗ ‖ ∗ ∗ ≥ ρ ‖ ξ n ∗ ‖ ∗ .

Thus, the result follows.□

Lemma 3.5

There exists a constant C > 0 such that for all y ∈ R N and r = 1 , … , 2 m − 1 ,

( ∂ ) r ξ n ∗ ( ∂ ) y 1 i 1 … ( ∂ ) y N i N ( y ) ≤ C ∑ j = 1 k n μ k n N − 2 m 2 + j ( 1 + μ k n ∣ y − x k , j ∣ ) N − 2 m 2 + τ + j μ k n − 1 − σ ,

where ∑ s = 1 N i s = r .

Proof

Since

L k n ξ n ∗ ≔ ( − Δ ) m ξ n ∗ − ( m ∗ − 1 ) K ( ∣ y ∣ ) u k n m ∗ − 2 ξ n ∗ ,

we have

( ∂ ) r ξ n ∗ ( ∂ ) y 1 i 1 … ( ∂ ) y N i N ( z ) = C ∫ R N ( ∂ ) r ( ∂ ) y 1 i 1 … ( ∂ ) y N i N 1 ∣ z − y ∣ N − 2 m ( ( m ∗ − 1 ) K ( ∣ z ∣ ) u k n m ∗ − 2 ξ n ∗ + L k n ξ n ∗ ) d z ≤ C ∫ R N 1 ∣ z − y ∣ N − 2 m + j ∣ ( m ∗ − 1 ) K ( ∣ z ∣ ) u k n m ∗ − 2 ξ n ∗ + L k n ξ n ∗ ∣ d z .

By using the fact ‖ L k n ξ n ∗ ‖ ∗ ∗ ≤ C μ k n − 1 − σ , we obtain

(3.12) ∫ R N 1 ∣ z − y ∣ N − 2 m + j ∣ L k n ξ n ∗ ∣ d z

≤ C μ k n − 1 − σ ∫ R N 1 ∣ z − y ∣ N − 2 m + j ∑ j = 1 k n μ k n N + 2 m 2 ( 1 + μ k n ∣ z − x k , j ∣ ) N + 2 m 2 + τ d z ≤ C ∑ j = 1 k n μ k n N − 2 m 2 + j ( 1 + μ k n ∣ y − x k , j ∣ ) N − 2 m 2 + τ + j μ k n − 1 − σ .

Note that ‖ ξ n ∗ ‖ ∗ ≤ C μ k n − 1 − σ , we have

(3.13) ∫ R N 1 ∣ z − y ∣ N − 2 m + j ∣ ( m ∗ − 1 ) K ( ∣ y ∣ ) u k n m ∗ − 2 ξ n ∗ ∣ d z ≤ C μ k n − 1 − σ ∫ R N 1 ∣ z − y ∣ N − 2 m + j ∑ j = 1 k n μ k n N − 2 m 2 ( 1 + μ k n ∣ z − x k , j ∣ ) N − 2 m m ∗ − 2 ∑ j = 1 k n μ k n N − 2 m 2 ( 1 + μ k n ∣ z − x k , j ∣ ) N − 2 m 2 + τ d z ≤ C μ k n − 1 − σ ∑ j = 1 N ∫ Ω j 1 ∣ z − y ∣ N − 2 m + j μ k n N + 2 m 2 ( 1 + μ k n ∣ z − x k , j ∣ ) N + 2 m 2 + 2 m + τ − η ( m ∗ − 1 ) d z ≤ C μ k n − 1 − σ ∑ j = 1 N 1 ∣ z − y ∣ N − 2 m + j μ k n N − 2 m 2 ( 1 + μ k n ∣ y − x k , j ∣ ) N + 2 m 2 + 2 m + τ − η ( m ∗ − 1 ) ≤ C μ k n − 1 − σ ∑ j = 1 k n μ k n N − 2 m 2 + j ( 1 + μ k n ∣ y − x k , j ∣ ) N − 2 m 2 + τ + j .

Combining (3.12) and (3.13), we obtain

( ∂ ) r ξ n ∗ ( ∂ ) y 1 i 1 … ( ∂ ) y N i N ( y ) ≤ C ∑ j = 1 k n μ k n N − 2 m 2 + j ( 1 + μ k n ∣ y − x k , j ∣ ) N − 2 m 2 + τ + j μ k n − 1 − σ . □

Lemma 3.6

There exists a constant C > 0 such that for all y ∈ R N and r = 1 , … , 2 m − 1 ,

( ∂ ) r ω k n ( ∂ ) y 1 i 1 … ( ∂ ) y N i N ( y ) ≤ C ∑ j = 1 k n μ k n N − 2 m 2 + j ( 1 + μ k n ∣ y − x k , j ∣ ) N − 2 m 2 + τ + j μ k n − 1 − σ ,

where ∑ s = 1 N i s = r .

Proof

From (3.2) of [24] and similar to the proof of Lemma 3.5, we can prove

( ∂ ) r ω k n ( ∂ ) y 1 i 1 … ( ∂ ) y N i N ( y ) ≤ C ∑ j = 1 k n μ k n N − 2 m 2 + j ( 1 + μ k n ∣ y − x k , j ∣ ) N − 2 m 2 + τ + j μ k n − 1 − σ . □

Lemma 3.7

ξ ^ n → 0 uniformly in C m ( B R ( 0 ) ) for any R > 0 .

Proof

The proof consists of the following steps.

Step 1. Recall that

Ω j = y = ( y ′ , y ″ ) ∈ R 2 × R N − 2 : ( y ′ , 0 ) ∣ y ′ ∣ , x k n , j ∣ x k n , j ∣ ≥ cos π k n .

To prove b 1 , n → 0 , we apply the identities in Lemmas 2.1 and 2.2 in the domain Ω 1 .

Case 1. m is even: By Lemma 2.1, we have

(3.14) ∫ R N u k n m ∗ − 1 ξ n ⟨ ∇ K ( ∣ y ∣ ) , y ⟩ = 0 .

Thus,

∫ Ω 1 u k n m ∗ − 1 ξ n ⟨ ∇ K ( ∣ y ∣ ) , y ⟩ = 0 ,

which implies that

(3.15) r k n ∫ Ω 1 u k n m ∗ − 1 ξ n ∂ K ( ∣ y ∣ ) ∂ y 1 = − ∫ Ω 1 u k n m ∗ − 1 ξ n ⟨ ∇ K ( ∣ y ∣ ) , y − x k n , 1 ⟩ .

Case 2. If m is odd: By Lemma 2.2, similar to Case 1, we also get (3.15).

Since ∇ K ( x k n , 1 ) = O ( ‖ x k n , 1 ∣ − r 0 ∣ ) = O ( μ k n − 1 − σ ) and

(3.16) ∫ Ω 1 u k n m ∗ − 1 ξ n = ∫ ( Ω 1 ) x k n , 1 , μ k n μ k n − N − 2 m 2 u k n ( μ k n − 1 y + x k n , 1 ) m ∗ − 1 ξ ^ n = ∫ ( Ω 1 ) x k n , 1 , μ k n U 0 , 1 m ∗ − 1 ( b 0 , n ψ 0 + b 1 , n ψ 1 ) + O ( μ k n − 1 − σ ) = ∫ R N U 0 , 1 m ∗ − 1 ( b 0 , n ψ 0 + b 1 , n ψ 1 ) + O ( μ k n − 1 − σ ) = O ( μ k n − 1 − σ ) ,

where ( Ω 1 ) x k n , 1 , μ k n = { y : μ k n − 1 y + x k n , 1 ∈ Ω 1 } , we find

(3.17) ∫ Ω 1 u k n m ∗ − 1 ξ n ∂ K ( ∣ y ∣ ) ∂ y 1 = ∫ Ω 1 u k n m ∗ − 1 ξ n ∂ K ( ∣ y ∣ ) ∂ y 1 − ∂ K ( ∣ x k n , 1 ∣ ) ∂ y 1 + O ( μ k n − 2 − 2 σ ) = ∫ B r k n sin π k n ( x k n , 1 ) u k n m ∗ − 1 ξ n ∂ K ( ∣ y ∣ ) ∂ y 1 − ∂ K ( ∣ x k n , 1 ∣ ) ∂ y 1 + O ( μ k n − 2 − 2 σ ) = ∫ B r k n sin π k n ( x k n , 1 ) u k n m ∗ − 1 ξ n ∇ ∂ K ( ∣ x k n , 1 ∣ ) ∂ y 1 , y − x k n , 1 + 1 2 ( ∇ ) 2 ∂ K ( ∣ x k n , 1 ∣ ) ∂ y 1 ( y − x k n , 1 ) , y − x k n , 1 + O ( ∣ y − x k n , 1 ∣ 3 ) + O ( μ k n − 2 − 2 σ ) = ∫ R N U 0 , 1 m ∗ − 1 ( b 0 , n ψ 0 + b 1 , n ψ 1 ) ∇ ∂ K ( ∣ x k n , 1 ∣ ) ∂ y 1 , y μ k n + 1 2 ( ∇ ) 2 ∂ K ( ∣ x k n , 1 ∣ ) ∂ y 1 y μ k n , y μ k n + O ( μ k n − 2 − 2 σ ) = K ″ ( ∣ x k n , 1 ∣ ) b 1 , n μ k n ∫ R N U 0 , 1 m ∗ − 1 ψ 1 y 1 + ∂ Δ K ( x k n , 1 ) ∂ y 1 b 0 , n 2 N μ k n 2 ∫ R N U 0 , 1 m ∗ − 1 ψ 0 ∣ y ∣ 2 + O ( μ k n − 2 − 2 σ ) .

On the other hand, we have

(3.18) ∫ Ω 1 u k n m ∗ − 1 ξ n ⟨ ∇ K ( ∣ y ∣ ) , y − x k n , 1 ⟩

= ∫ Ω 1 u k n m ∗ − 1 ξ n ⟨ ∇ K ( ∣ y ∣ ) − ∇ K ( ∣ x k n , 1 ∣ ) , y − x k n , 1 ⟩ + O ( μ k n − 2 − 2 σ ) = ∫ B r k n sin π k n ( x k n , 1 ) u k n m ∗ − 1 ξ n ⟨ ∇ K ( ∣ y ∣ ) − ∇ K ( ∣ x k n , 1 ∣ ) , y − x k n , 1 ⟩ + O ( μ k n − 2 − 2 σ ) = ∫ B r k n sin π k n ( x k n , 1 ) u k n m ∗ − 1 ξ n ⟨ ( ∇ ) 2 K ( ∣ x k n , 1 ∣ ) ( y − x k n , 1 ) , y − x k n , 1 ⟩ + O ( μ k n − 2 − 2 σ ) = ∫ R N U 0 , 1 m ∗ − 1 ( b 0 , n ψ 0 + b 1 , n ψ 1 ) ⟨ ( ∇ ) 2 K ( ∣ x k n , 1 ∣ ) μ k n − 1 y , μ k n − 1 y ⟩ + O ( μ k n − 2 − 2 σ ) = b 0 , n Δ K ( ∣ x k n , 1 ∣ ) N μ k n 2 ∫ R N U 0 , 1 m ∗ − 1 ψ 0 ∣ y ∣ 2 + O ( μ k n − 2 − 2 σ ) .

Therefore, (3.17) and (3.18) give

(3.19) b 1 , n = − ∫ R N U 0 , 1 m ∗ − 1 ψ 0 ∣ y ∣ 2 μ k n K ″ ( ∣ x k n , 1 ∣ ) ∫ R N U 0 , 1 m ∗ − 1 ψ 1 y 1 K ″ ( ∣ x k n , 1 ∣ ) r k n N + ∂ Δ K ( x k n , 1 ) ∂ y 1 2 N b 0 , n + O ( μ k n − 1 − 2 σ ) .

Step 2. Next we prove that b 0 , n = o ( 1 ) .

Case 1. If m is even: We apply Pohozaev identity (2.15) in Lemma 2.1 to u k n with x 0 = x k n , 1 and d ∼ μ k n − 1 + α , where max 2 N , 1 N − 2 m 2 + τ < α < 2 N − 2 m : we have

(3.20) ∫ B d ( x k n , 1 ) u k n m ∗ − 1 ξ ⟨ ∇ K ( ∣ y ∣ ) , y − x 0 ⟩ = ∫ ∂ B d ( x k n , 1 ) K ( ∣ y ∣ ) u k n m ∗ − 1 ξ n ⟨ ν , y − x 0 ⟩ − ∫ ∂ B d ( x k n , 1 ) Δ m 2 u k n Δ m 2 ξ n ⟨ ν , y − x 0 ⟩ + ∑ i = 1 m 2 ∫ ∂ B d ( x k n , 1 ) − ∂ Δ m − i u k n ∂ ν Δ i − 1 ⟨ ∇ ξ n , y − x 0 ⟩ + Δ m − i u k n ∂ Δ i − 1 ⟨ ∇ ξ n , y − x 0 ⟩ ∂ ν − ∂ Δ m − i ξ n ∂ ν Δ i − 1 ⟨ ∇ u k n , y − x 0 ⟩ + Δ m − i ξ n ∂ Δ i − 1 ⟨ ∇ u k n , y − x 0 ⟩ ∂ ν + N − 2 m 2 ∑ i = 1 m 2 ∫ ∂ B d ( x k n , 1 ) Δ m − i u k n ∂ Δ i − 1 ξ n ∂ ν + Δ m − i ξ n ∂ Δ i − 1 u k n ∂ ν − ∂ Δ m − i u k n ∂ ν Δ i − 1 ξ n − ∂ Δ m − i ξ n ∂ ν Δ i − 1 u k n .

Similar to (3.18), we obtain

(3.21) LHS of (3.20) = b 0 , n Δ K ( x k n , 1 ) N μ k n 2 ∫ U 0 , 1 m ∗ − 1 ψ 0 ∣ y ∣ 2 + O 1 μ k n 2 + σ = a ˜ N , m , K b 0 , n μ k n 2 + O 1 μ k n 2 + σ ,

where a ˜ N , m , K = K ″ ( r 0 ) N ∫ U 0 , 1 m ∗ − 1 ψ 0 ∣ y ∣ 2 = − ( N − 2 m ) K ″ ( r 0 ) 2 N C N , m m ∗ ∫ ∣ y ∣ 2 ( ∣ y ∣ 2 − 1 ) ( 1 + ∣ y ∣ 2 ) N + 1 .

In the following, we estimate the right-hand side (RHS) of (3.20).

A direct computation leads to

(3.22) ∫ ∂ B d ( x k n , 1 ) K ( ∣ y ∣ ) u k n m ∗ − 1 ξ n ⟨ y − x k n , 1 , ν ⟩ = O 1 ( μ k n d ) N = O 1 μ k n 2 + σ .

Integrating by parts, we have

(3.23) − ∫ ∂ B d ( x k n , 1 ) Δ m 2 u k n Δ m 2 ξ n ⟨ ν , y − x 0 ⟩ + ∑ i = 1 m 2 ∫ ∂ B d ( x k n , 1 ) − ∂ Δ m − i u k n ∂ ν Δ i − 1 ⟨ ∇ ξ n , y − x 0 ⟩ + Δ m − i u k n ∂ Δ i − 1 ⟨ ∇ ξ n , y − x 0 ⟩ ∂ ν − ∂ Δ m − i ξ n ∂ ν Δ i − 1 ⟨ ∇ u k n , y − x 0 ⟩ + Δ m − i ξ n ∂ Δ i − 1 ⟨ ∇ u k n , y − x 0 ⟩ ∂ ν + N − 2 m 2 ∑ i = 1 m 2 ∫ ∂ B d ( x k n , 1 ) Δ m − i u k n ∂ Δ i − 1 ξ n ∂ ν + Δ m − i ξ n ∂ Δ i − 1 u k n ∂ ν − ∂ Δ m − i u k n ∂ ν Δ i − 1 ξ n − ∂ Δ m − i ξ n ∂ ν Δ i − 1 u k n = − ∫ B d ( x k n , 1 ) ( − Δ ) m u k n ⟨ ∇ ξ n , y − x k n , 1 ⟩ + ( − Δ ) m ξ n ⟨ ∇ u k n , y − x k n , 1 ⟩ − N − 2 m 2 ∫ B d ( x k n , 1 ) ( − Δ ) m u k n ξ n + ( − Δ ) m ξ n u k n .

Define

I ( u , v ) ≔ − ∫ B d ( x k n , 1 ) ( − Δ ) m u ⟨ ∇ v , y − x k n , 1 ⟩ + ( − Δ ) m v ⟨ ∇ u , y − x k n , 1 ⟩ − N − 2 m 2 ∫ B d ( x k n , 1 ) ( − Δ ) m u v + ( − Δ ) m v u .

Noting that

u k n = ∑ j = 1 k n U x k n , j , μ k n + w k n

and

ξ n = b 0 , n ∑ j = 1 k n μ k n ∂ U x k n , j , μ k n ∂ μ k n + b 1 , n ∑ j = 1 k n μ k n − 1 ∂ U x k n , j , μ k n ∂ r k n + ξ n ∗ ,

we have

(3.24) I ( u k n , ξ n ) = b 0 , n I ∑ j = 1 k n U x k n , j , μ k n , ∑ j = 1 k n μ k n ∂ U x k n , j , μ k n ∂ μ k n + b 0 , n I w k n , ∑ j = 1 k n μ k n ∂ U x k n , j , μ k n ∂ μ k n + b 1 , n I ∑ j = 1 k n U x k n , j , μ k n , ∑ j = 1 k n μ k n − 1 ∂ U x k n , j , μ k n ∂ r k n + b 1 , n I w k n , ∑ j = 1 k n μ k n − 1 ∂ U x k n , j , μ k n ∂ r k n + I ∑ j = 1 k n U x k n , j , μ k n , ξ n ∗ + I ( w k n , ξ n ∗ ) .

We can compute that

(3.25) I ∑ j = 1 k n U x k n , j , μ k n , ∑ j = 1 k n μ k n ∂ U x k n , j , μ k n ∂ μ k n = ∫ B d ( x k n , 1 ) − N − 2 m 2 U x k n , 1 , μ k n m ∗ − 1 ∑ j = 2 k n μ k n ∂ U x k n , j , μ k n ∂ μ k n + ( m ∗ − 1 ) U x k n , 1 , μ k n m ∗ − 2 μ k n ∂ U x k n , 1 , μ k n ∂ μ k n ∑ j = 2 k n U x k n , j , μ k n − U x k n , 1 , μ k n m ∗ − 1 ∇ μ k n ∂ U x k n , 1 , μ k n ∂ μ k n , y − x k n , 1 − ( m ∗ − 1 ) U x k n , 1 , μ k n m ∗ − 2 μ k n ∂ U x k n , 1 , μ k n ∂ μ k n ⟨ ∇ U x k n , 1 , μ k n , y − x k n , 1 ⟩ − N U x k n , 1 , μ k n m ∗ − 1 μ k n ∂ U x k n , 1 , μ k n ∂ μ k n + O 1 μ k n 2 + σ

= − N − 2 m 2 μ k n ∂ ∂ μ k n ∫ B d ( x k n , 1 ) U x k n , 1 , μ k n m ∗ − 1 ∑ j = 2 k n U x k n , j , μ k n + μ k n ∫ ∂ B d ( x k n , 1 ) U x k n , 1 , μ k n m ∗ − 1 ∂ U x k n , 1 , μ k n ∂ μ k n ⟨ ν , y − x k n , 1 ⟩ + O 1 μ k n 2 + σ = − N − 2 m 2 μ k n ∫ B μ k n d ( 0 ) U 0 , 1 m ∗ − 1 ∂ ∂ μ k n ∑ j = 2 k n U μ k n ( x k n , j − x k n , 1 ) , 1 + O ( ( μ k n d ) − N ) + O 1 μ k n 2 + σ = ( N − 2 m ) 2 2 C N , m ∫ B μ k n d ( 0 ) U 0 , 1 m ∗ − 1 ∑ j = 2 k n 1 ( μ k n ∣ x k n , j − x k n , 1 ∣ ) N − 2 m + O 1 μ k n 2 + σ .

And from [24], we know that

∑ j = 2 k n 1 ( μ k n ∣ x k n , 1 − x k n , j ∣ ) N − 2 m = − K ″ ( r 0 ) ∫ ∣ y ∣ 2 U 0 , 1 m ∗ C N , m N 2 μ k n 2 ∫ U 0 , 1 m ∗ − 1 + O 1 μ k n 2 + σ .

Thus,

(3.26) I ∑ j = 1 k n U x k n , j , μ k n , ∑ j = 1 k n μ k n ∂ U x k n , j , μ k n ∂ μ k n = a N , m , K μ k n 2 + O 1 μ k n 2 + σ ,

where

(3.27) a N , m , K = − N − 2 m 2 N 2 C N , m m ∗ ∫ ∣ y ∣ 2 ( 1 + ∣ y ∣ 2 ) N ∫ N ∣ y ∣ 2 − 2 m ( 1 + ∣ y ∣ 2 ) N + 2 m 2 + 1 ∫ 1 ( 1 + ∣ y ∣ 2 ) N + 2 m 2 K ″ ( r 0 ) .

Similarly, noting that μ k n − 1 ∂ U x k n , 1 , μ k n ∂ r k n , ∇ μ k n − 1 ∂ U x k n , 1 , μ k n ∂ r k n , y − x k n , 1 is odd with respect to ( y − x k n , 1 ) 1 , we have that

(3.28) I ∑ j = 1 k n U x k n , j , μ k n , ∑ j = 1 k n μ k n − 1 ∂ U x k n , j , μ k n ∂ r k n = − ∫ B d ( x k n , 1 ) U x k n , 1 , μ k n m ∗ − 1 ∇ μ k n − 1 ∂ U x k n , 1 , μ k n ∂ r k n , y − x k n , 1 − ∫ B d ( x k n , 1 ) ( m ∗ − 1 ) U x k n , 1 , μ k n m ∗ − 2 μ k n − 1 ∂ U x k n , 1 , μ k n ∂ r k n × ⟨ ∇ U x k n , 1 , μ k n , y − x k n , 1 ⟩ − N ∫ B d ( x k n , 1 ) U x k n , 1 , μ k n m ∗ − 1 μ k n − 1 ∂ U x k n , 1 , μ k n ∂ r k n + O 1 μ k n 1 + σ = O 1 μ k n 1 + σ .

For the term I w k n , ∑ j = 1 k n μ k n ∂ U x k n , j , μ k n ∂ μ k n , integrating by parts, we obtain

I w k n , ∑ j = 1 k n μ k n ∂ U x k n , j , μ k n ∂ μ k n = − ∫ ∂ B d ( x k n , 1 ) Δ m 2 w k n Δ m 2 ∑ j = 1 k n μ k n ∂ U x k n , j , μ k n ∂ μ k n ⟨ ν , y − x 0 ⟩ + ∑ i = 1 m 2 ∫ ∂ B d ( x k n , 1 ) − ∂ Δ m − i w k n ∂ ν Δ i − 1 ∇ ∑ j = 1 k n μ k n ∂ U x k n , j , μ k n ∂ μ k n , y − x 0

+ Δ m − i w k n ∂ Δ i − 1 ∇ ∑ j = 1 k n μ k n ∂ U x k n , j , μ k n ∂ μ k n , y − x 0 ∂ ν − ∂ Δ m − i ∑ j = 1 k n μ k n ∂ U x k n , j , μ k n ∂ μ k n ∂ ν Δ i − 1 ⟨ ∇ w k n , y − x 0 ⟩ + Δ m − i ∑ j = 1 k n μ k n ∂ U x k n , j , μ k n ∂ μ k n ∂ Δ i − 1 ⟨ ∇ w k n , y − x 0 ⟩ ∂ ν + N − 2 m 2 ∑ i = 1 m 2 ∫ ∂ B d ( x k n , 1 ) Δ m − i w k n ∂ Δ i − 1 ∑ j = 1 k n μ k n ∂ U x k n , j , μ k n ∂ μ k n ∂ ν + Δ m − i ∑ j = 1 k n μ k n ∂ U x k n , j , μ k n ∂ μ k n ∂ Δ i − 1 w k n ∂ ν − ∂ Δ m − i w k n ∂ ν Δ i − 1 ∑ j = 1 k n μ k n ∂ U x k n , j , μ k n ∂ μ k n − ∂ Δ m − i ∑ j = 1 k n μ k n ∂ U x k n , j , μ k n ∂ μ k n ∂ ν Δ i − 1 w k n .

As a result,

I w k n , ∑ j = 1 k n μ k n ∂ U x k n , j , μ k n ∂ μ k n ≤ C μ k n − 1 − σ ∫ ∂ B d ( x k n , 1 ) ∑ j = 1 k n μ k n N − 2 m 2 + m ( 1 + μ k n ∣ y − x k n , j ∣ ) N − 2 m 2 + τ + m ∑ j = 1 k n μ k n N − 2 m 2 + m ( 1 + μ k n ∣ y − x k n , j ∣ ) N − 2 m + m + ∑ i = 1 m 2 ∫ ∂ B d ( x k n , 1 ) ∑ j = 1 k n μ k n N − 2 m 2 + 2 ( m − i ) + 1 ( 1 + μ k n ∣ y − x k n , j ∣ ) N − 2 m 2 + τ + 2 ( m − i ) + 1 ∑ j = 1 k n μ k n N − 2 m 2 + 2 ( i − 1 ) + 1 ( 1 + μ k n ∣ y − x k n , j ∣ ) N − 2 m + 2 ( i − 1 ) + 1 + ∑ j = 1 k n μ k n N − 2 m 2 + 2 ( m − i ) ( 1 + μ k n ∣ y − x k n , j ∣ ) N − 2 m 2 + τ + 2 ( m − i ) ∑ j = 1 k n μ k n N − 2 m 2 + 2 ( i − 1 ) + 2 ( 1 + μ k n ∣ y − x k n , j ∣ ) N − 2 m + 2 ( i − 1 ) + 2 d + ∑ i = 2 m 2 ∫ ∂ B d ( x k n , 1 ) ∑ j = 1 k n μ k n N − 2 m 2 + 2 ( m − i ) + 1 ( 1 + μ k n ∣ y − x k n , j ∣ ) N − 2 m 2 + τ + 2 ( m − i ) + 1 ∑ j = 1 k n μ k n N − 2 m 2 + 2 ( i − 1 ) ( 1 + μ k n ∣ y − x k n , j ∣ ) N − 2 m + 2 ( i − 1 ) + ∑ j = 1 k n μ k n N − 2 m 2 + 2 ( m − i ) ( 1 + μ k n ∣ y − x k n , j ∣ ) N − 2 m 2 + τ + 2 ( m − i ) ∑ j = 1 k n μ k n N − 2 m 2 + 2 ( i − 1 ) + 1 ( 1 + μ k n ∣ y − x k n , j ∣ ) N − 2 m + 2 ( i − 1 ) + 1 + ∑ i = 1 m 2 ∫ ∂ B d ( x k n , 1 ) ∑ j = 1 k n μ k n N − 2 m 2 + 2 ( i − 1 ) + 1 ( 1 + μ k n ∣ y − x k n , j ∣ ) N − 2 m 2 + τ + 2 ( i − 1 ) + 1 ∑ j = 1 k n μ k n N − 2 m 2 + 2 ( m − i ) + 1 ( 1 + μ k n ∣ y − x k n , j ∣ ) N − 2 m + 2 ( m − i ) + 1 + ∑ j = 1 k n μ k n N − 2 m 2 + 2 ( i − 1 ) + 2 ( 1 + μ k n ∣ y − x k n , j ∣ ) N − 2 m 2 + τ + 2 ( i − 1 ) + 2 ∑ j = 1 k n μ k n N − 2 m 2 + 2 ( m − i ) ( 1 + μ k n ∣ y − x k n , j ∣ ) N − 2 m + 2 ( m − i ) d + ∑ i = 2 m 2 ∫ ∂ B d ( x k n , 1 ) ∑ j = 1 k n μ k n N − 2 m 2 + 2 ( i − 1 ) + 1 ( 1 + μ k n ∣ y − x k n , j ∣ ) N − 2 m 2 + τ + 2 ( i − 1 ) + 1 ∑ j = 1 k n μ k n N − 2 m 2 + 2 ( m − i ) ( 1 + μ k n ∣ y − x k n , j ∣ ) N − 2 m + 2 ( m − i ) + ∑ j = 1 k n μ k n N − 2 m 2 + 2 ( i − 1 ) + 1 ( 1 + μ k n ∣ y − x k n , j ∣ ) N − 2 m 2 + τ + 2 ( i − 1 ) + 1 ∑ j = 1 k n μ k n N − 2 m 2 + 2 ( m − i ) ( 1 + μ k n ∣ y − x k n , j ∣ ) N − 2 m + 2 ( m − i ) + ∑ i = 1 m 2 ∫ ∂ B d ( x k n , 1 ) ∑ j = 1 k n μ k n N − 2 m 2 + 2 ( m − i ) ( 1 + μ k n ∣ y − x k n , j ∣ ) N − 2 m 2 + τ + 2 ( m − i ) ∑ j = 1 k n μ k n N − 2 m 2 + 2 ( i − 1 ) + 1 ( 1 + μ k n ∣ y − x k n , j ∣ ) N − 2 m + 2 ( i − 1 ) + 1

+ ∑ j = 1 k n μ k n N − 2 m 2 + 2 ( m − i ) + 1 ( 1 + μ k n ∣ y − x k n , j ∣ ) N − 2 m 2 + τ + 2 ( m − i ) + 1 ∑ j = 1 k n μ k n N − 2 m 2 + 2 ( i − 1 ) ( 1 + μ k n ∣ y − x k n , j ∣ ) N − 2 m + 2 ( i − 1 ) + ∑ j = 1 k n μ k n N − 2 m 2 + 2 ( i − 1 ) ( 1 + μ k n ∣ y − x k n , j ∣ ) N − 2 m 2 + τ + 2 ( i − 1 ) ∑ j = 1 k n μ k n N − 2 m 2 + 2 ( m − i ) + 1 ( 1 + μ k n ∣ y − x k n , j ∣ ) N − 2 m + 2 ( m − i ) + 1 + ∑ j = 1 k n μ k n N − 2 m 2 + 2 ( i − 1 ) + 1 ( 1 + μ k n ∣ y − x k n , j ∣ ) N − 2 m 2 + τ + 2 ( i − 1 ) + 1 ∑ j = 1 k n μ k n N − 2 m 2 + 2 ( m − i ) ( 1 + μ k n ∣ y − x k n , j ∣ ) N − 2 m + 2 ( m − i ) ≤ C μ k n − 1 − σ ∫ ∂ B d ( x k n , 1 ) μ k n N d ( 1 + μ k n ∣ y − x k n , 1 ∣ ) N − 2 m 2 + τ + N + μ k n N − 1 ( 1 + μ k n ∣ y − x k n , 1 ∣ ) N − 2 m 2 + τ + N − 1 ≤ C μ k n − 1 − σ ( μ k n d ) − N − 2 m 2 + τ ≤ C μ k n − 2 − σ .

As a consequence, we obtain

(3.29) I w k n , ∑ j = 1 k n μ k n ∂ U x k n , j , μ k n ∂ μ k n = O 1 μ k n 2 + σ .

Similarly, we have

(3.30) I w k n , ∑ j = 1 k n μ k n − 1 ∂ U x k n , j , μ k n ∂ r k n = O 1 μ k n 1 + σ ,

(3.31) I ∑ j = 1 k n U x k n , j , μ k n , ξ n ∗ = O 1 μ k n 2 + σ ,

and

(3.32) ∣ I ( w k n , ξ n ∗ ) ∣ ≤ C μ k n − 2 − σ ∫ ∂ B d ( x k n , 1 ) μ k n N d ( 1 + μ k n ∣ y − x k n , 1 ∣ ) N + 2 τ + μ k n N − 1 ( 1 + μ k n ∣ y − x k n , 1 ∣ ) 2 τ + N − 1 ≤ C μ k n − 2 − σ .

Combining (3.26)–(3.32) and b 1 , n = O 1 μ k n in Step 1, (3.24) becomes

I ( u k n , ξ n ) = a N , m , K μ k n 2 b 0 , n + O 1 μ k n 2 + σ .

Combining (3.22)–(3.24), we have

(3.33) RHS of (3.20) = a N , m , K μ k n 2 b 0 , n + O 1 μ k n 2 + σ .

As a consequence, we have that

a ˜ N , s , K b 0 , n μ k n 2 + O 1 μ k n 2 + σ = a N , m , K b 0 , n μ k n 2 + O 1 μ k n 2 + σ ,

where a N , m , K is defined in (3.27).

A direct computation shows that a N , m , K a ˜ N , s , K = N − 2 m 2 , due to N > 2 m + 2 .

Thus, b 0 , n = o ( 1 ) .

Case: m is odd. Appling (2.15) and similar to Case 1, we also have b 0 , n = o ( 1 ) .

Noting that b 0 , n → b 0 , b 1 , n → b 1 , as n → + ∞ and Lemma 3.3, the result follows.□

Now, we give the proof of Theorem 1.1.

Proof

With the aid of the above prepared lemmas, it is sufficient to get a contradiction with ‖ ξ n ‖ ∗ = 1 .

In fact, we have

(3.34) ξ n ( y ) = ( m ∗ − 1 ) ∫ R N 1 ∣ z − y ∣ N − 2 m K ( z ) u k n m ∗ − 2 ( z ) ξ n ( z ) d z

and

∫ R N 1 ∣ z − y ∣ N − 2 m K ( z ) u k n m ∗ − 2 ( z ) ξ n ( z ) d z ≤ C ‖ ξ n ‖ ∗ ∫ R N 1 ∣ z − y ∣ N − 2 m K ( z ) u k n m ∗ − 2 ( z ) ∑ j = 1 k n μ k n N − 2 m 2 ( 1 + μ k n ∣ z − x k n , j ∣ ) N − 2 m 2 + τ d z ≤ C ‖ ξ n ‖ ∗ ∑ j = 1 k n μ k n N − 2 m 2 ( 1 + μ k n ∣ y − x k n , j ∣ ) N − 2 m 2 + τ + θ ,

for some θ > 0 . So we obtain

∣ ξ n ( y ) ∣ ∑ j = 1 k n μ k n N − 2 m 2 ( 1 + μ k n ∣ y − x k , j ∣ ) N − 2 m 2 + τ ≤ C ‖ ξ n ‖ ∗ ∑ j = 1 k n μ k n N − 2 m 2 ( 1 + μ k n ∣ y − x k n , j ∣ ) N − 2 m 2 + τ + θ ∑ j = 1 k n μ k n N − 2 m 2 ( 1 + μ k n ∣ y − x k n , j ∣ ) N − 2 m 2 + τ .

Since ξ n → 0 in B R μ k n − 1 ( x k m , j ) and ‖ ξ n ‖ ∗ = 1 , we know that

∣ ξ n ( y ) ∣ ∑ j = 1 k n μ k n N − 2 m 2 ( 1 + μ k n ∣ y − x k n , j ∣ ) N − 2 m 2 + τ

attains its maximum in R N ⧹ ∪ j = 1 k n B R μ k n − 1 ( x k n , j ) . Thus,

‖ ξ n ‖ ∗ ≤ o ( 1 ) ‖ ξ n ‖ ∗ .

So ‖ ξ n ‖ ∗ → 0 as m → + ∞ . This is a contradiction to ‖ ξ n ‖ ∗ = 1 .□

Funding information: This work was supported by NSFC (No. 11771235 and 12031015).
Conflict of interest: Authors state no conflict of interest.

Appendix Basic estimates

In this section, we give some essential estimates which can be found in [27,1]. Define

g i j ( y ) ≔ 1 ( 1 + ∣ y − x i ∣ ) α ( 1 + ∣ y − x j ∣ ) β ,

where x i ≠ x j , α > 0 , β > 0 are some constants.

Lemma A.1

For any constant γ satisfying 0 < γ ≤ min { α , β } , we have

g i j ( y ) ≤ C ∣ x i − x j ∣ γ 1 ( 1 + ∣ y − x i ∣ ) α + β − γ + 1 ( 1 + ∣ y − x i ∣ ) α + β − γ .

Lemma A.2

For any constant κ > 0 with κ ≠ N − 2 m , there is a constant C > 0 such that

∫ R N 1 ∣ y − z ∣ N − 2 m 1 ( 1 + ∣ z ∣ ) 2 m + κ d z ≤ C ( 1 + ∣ y ∣ ) min { κ , N − 2 m } .

For κ = N − 2 m , there exists a constant C > 0 such that

∫ R N 1 ∣ y − z ∣ N − 2 m 1 ( 1 + ∣ z ∣ ) N d z ≤ C max { 1 , log ∣ y ∣ } ( 1 + ∣ y ∣ ) min { κ , N − 2 m } .

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Received: 2021-10-27

Revised: 2022-01-19

Accepted: 2022-02-04

Published Online: 2022-03-03

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

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Keywords for this article

polyharmonic operator; non-degeneracy; Pohozaev identities

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