Non-radial solutions for the fractional Hénon equation with critical exponent

Abstract

We consider the following fractional Hénon type equation with critical growth:

$$\begin{aligned} {\left\{ \begin{array}{ll} (-\Delta )^{s}u=K(|y|)u^{\frac{N+2s}{N-2s}}, \hbox { }u>0, &{}\hbox { }y\in B_1(0), \\ \displaystyle u=0,&{}\hbox { } y\in B_{1}^{c}(0), \end{array}\right. } \end{aligned}$$

(0.1)

where K(|y|) is a bounded function defined in [0, 1], $B_1(0)$ is the unit ball in ${\mathbb {R}}^{N}$, $N\ge 3$ for $\frac{3}{4} \le s<1$ and $3\le N < 2s -1 +\frac{2}{3-4s}$ for $\frac{11-\sqrt{41}}{8}< s<\frac{3}{4}$. We show that if $K(1)>0$ and $K'(1)>0$, then equation (0.1) has infinitely many non-radial positive solutions, whose energy can be made arbitrarily large. The most ingredients of the paper are using the Green representation and estimating the Green function and its regular part very carefully. For this purposes, some more extra ideas and techniques are needed. We believe that our method and techniques can be applied to other related problems.

1 Introduction

In this paper, we are concerned with the following fractional Hénon type equation with critical growth:

$$\begin{aligned} {\left\{ \begin{array}{ll} (-\Delta )^{s}u=K(|y|)u^{\frac{N+2s}{N-2s}}, \hbox { }u>0, &{}\hbox { }y\in B_1(0), \\ \displaystyle u=0, &{}\hbox { }y\in B_{1}^{c}(0), \end{array}\right. } \end{aligned}$$

(1.1)

where K(|y|) is a bounded function defined in [0, 1], $B_1(0)$ is the unit ball in ${\mathbb {R}}^{N}$, $N\ge 3$ for $\frac{3}{4} \le s<1$ and $3\le N < 2s -1 +\frac{2}{3-4s}$ for $\frac{11-\sqrt{41}}{8}< s<\frac{3}{4}$. $(-\Delta )^{s}$ is the so called fractional operator defined as:

$$\begin{aligned} (-\Delta )^{s}u=C(N,s)P.V.\int _{{\mathbb {R}}^{N}}\frac{u(x)-u(y)}{|x-y|^{N-2s}}dy= C(N,s)\lim _{\epsilon \rightarrow 0^+}\int _{{\mathbb {R}}^{N} \setminus B_{\epsilon }(x)}\frac{u(x)-u(y)}{|x-y|^{N-2s}}dy, \end{aligned}$$

where P.V. is the principal value and $C(N,s) =\pi ^{-(2s+\frac{N}{2})}\frac{\Gamma (\frac{N}{2}+s)}{\Gamma (-s)}$. The fractional Laplacian operator appears in many areas including astrophysics, mathematical finances and so on, and it can be regarded as the infinitesimal generator of a stable Levy process (see [1]). For more results related to fractional operators, we refer readers to [3,4,5,6, 11, 13, 19, 20, 23] and the references therein.

The classical Hénon equation states that

$$\begin{aligned} {\left\{ \begin{array}{ll} (-\Delta )u=|y|^{\alpha }u^{q}, \hbox { }u>0, &{}\hbox { }y\in B_1(0), \\ \displaystyle u=0,&{}\hbox { } y \in \partial B_{1}(0), \end{array}\right. } \end{aligned}$$

(1.2)

where $\alpha > 0$, $q>0$ are positive constants. This equation was first introduced by Hénon in the study of astrophysics (see [10]). It has attracted lots of interests in recent years. In the subcritical case, that is $q<\frac{N+2}{N-2}$, the existence of the solution for problem (1.2) can be proved easily by variational methods. For the critical case, that is $q=\frac{N+2}{N-2}$, the loss of compactness of embedding from $H^{1}_{0}(B_{1}(0))$ to $L^{\frac{2N }{N-2}}(B_{1}(0))$ makes the problem (1.2) very difficult to study. Ni [14] observed the influence of the non-autonomous term $|y|^\alpha$ and proved that it possesses a positive radial solution when $q \in (1, \frac{N+2+2\alpha }{N-2})$.

It is natural to ask whether (1.2) has a non-radial solution. When $N=2$, Smets-Su-Willem [18] showed that the mountain pass solution is non-radial when $\alpha$ is large. When $N \ge 3$ and $q = \frac{N+2}{N-2}-\epsilon$ with $\epsilon$ is small (near critical), Cao-Peng [7] proved that the mountain pass solution is non-radial and it blows up as $\epsilon \rightarrow 0$. Using a variational method, Serra [17] proved that (1.2) has a non-radial solution when $N\ge 4$, $q=\frac{N+2}{N-2}$ and $\alpha$ is large. While Wei-Yan [21] proved there exists infinitely many non-radial solutions for $N\ge 4$, $q=\frac{N+2}{N-2}$ and any $\alpha > 0$. And this result was later extended to the polyharmonic case by Guo and Li [8].

Before the statement of the main result, we introduce some notations.

Let $H_{0}^s(\Omega )=\{ u\in H^s({\mathbb {R}}^{N}): u =0 \hbox { in } \Omega ^c \}$ with the norm:

$$\begin{aligned} ||u||_{H_{0}^s(\Omega )} = \big (\displaystyle \int _{\Omega }|(-\Delta )^{\frac{s}{2}}u|^{2}dx\big )^{\frac{1}{2}}. \end{aligned}$$

We fix a positive integer $k\ge k_0$, where $k_0$ is large enough, which will be determined later. Set $\mu =k^{\frac{N-2s+1}{N-2s}}$ be the scaling parameter, where $N\ge 3$ when $\frac{3}{4}\le s <1$ and $3\le N < 2s -1 +\frac{2}{3-4s}$ when $\frac{11-\sqrt{41}}{8}< s<\frac{3}{4}$.

Let $2^{*}_s = \frac{2N}{N-2s}$. By the transformation $u(y)\rightarrow \mu ^{-\frac{N-2s}{2}}u(\frac{y}{\mu })$, (1.1) becomes to be

$$\begin{aligned} {\left\{ \begin{array}{ll} (-\Delta )^{s}u=K\left( \frac{|y|}{\mu }\right) u^{\frac{N+2s}{N-2s}}, \hbox { }u>0, &{}\hbox { }y\in B_{\mu }(0), \\ \displaystyle u=0, &{}\hbox { }y\in B_{\mu }^{c}(0). \end{array}\right. } \end{aligned}$$

(1.3)

In the following, we just need to consider (1.3) instead of (1.1).

It is well known that the functions

$$\begin{aligned} U_{x,\Lambda }=( 4^{s}\gamma )^{\frac{N-2s}{4s}}\big ( \frac{\Lambda }{(1+\Lambda ^{2}|y-x|^{2})} \big )^{\frac{N-2s}{2}}, \hbox { }\Lambda >0, \hbox { }x\in {\mathbb {R}}^{N}, \end{aligned}$$

where $\gamma =\frac{\Gamma (\frac{N+2s}{2})}{\Gamma (\frac{N-2s}{2})}$, are the only solutions (usually called bubbles) to the following problem (see [12]):

$$\begin{aligned} (-\Delta )^{s}u=u^{\frac{N+2s}{N-2s}}, \hbox { }u>0 \hbox { in } {\mathbb {R}}^{N}. \end{aligned}$$

Noting that when $y\in B_{1}^{c}(0)$, $U_{x,\Lambda }(y)$ is not zero, we define $PU_{x,\Lambda }$ as the projection of $U_{x,\Lambda }$, that is the solution of the following problem:

$$\begin{aligned} {\left\{ \begin{array}{ll} (-\Delta )^{s}PU_{x,\Lambda }(y) = (-\Delta )^{s}U_{x,\Lambda }(y), &{}\hbox { }y \in B_{\mu }(0), \\ \displaystyle PU_{x,\Lambda }(y) =0, &{}\hbox { }y\in B_{\mu }^{c}(0). \end{array}\right. } \end{aligned}$$

(1.4)

Let $y=(y',y'')$, $y'\in {\mathbb {R}}^{2}$, $y''\in {\mathbb {R}}^{N-2}$. Define

$$\begin{aligned} H_r =&\{ u: u \text { is even in } y_h, h=2,...,N, \hbox { and} \\ \displaystyle&u(r\cos \theta ,r\sin \theta ,y'') =u\left( r\cos \left( \theta +\frac{2j\pi }{k}\right) ,r\sin \left( \theta +\frac{2j\pi }{k}\right) ,y''\right) , \\ \displaystyle&j=1,...,k \}. \end{aligned}$$

Let $x_j=( r\cos \frac{2(j-1)\pi }{k},r\sin \frac{2(j-1)\pi }{k} ,0 ),\hbox { }j=1,...,k,$ where 0 is the zero vector in ${\mathbb {R}}^{N-2}$, $r\in [\mu \big (1-\frac{r_0}{k}\big ), \mu \big (1-\frac{r_1}{k}\big ) ],\hbox { for some constants } r_1>r_0>0,$ and $L_0\le \Lambda \le L_1,$ for some constants $L_1>L_0>0$. Set

$$\begin{aligned} W_{r,\Lambda } = \displaystyle \sum _{j = 1}^k PU_{x_j,\Lambda }. \end{aligned}$$

The main result of the present paper is the following:

Theorem 1.1

Suppose that $N\ge 3$ for $\frac{3}{4} \le s<1$ and $3\le N < 2s -1 +\frac{2}{3-4s}$ for $\frac{11-\sqrt{41}}{8}< s<\frac{3}{4}.$ If K(r) is a bounded function defined in [0, 1] and statisfies $K(1)>0$ and $K'(1)>0$, then there is an integer $k_0>0$, such that for any integer $k\ge k_0$, (1.3) has a solution $u_k$ of the form

$$\begin{aligned} u_k=W_{r_{k},\Lambda _k}(y) + \omega _k, \end{aligned}$$

where $\omega _k\in H_r\cap H^s_0\big (B_{\mu }(0)\big )$, and as $k\rightarrow +\infty$, $||\omega _{k}||_{L^{\infty } }\rightarrow 0$,$L_0\le \Lambda _k \le L_1$, and $r_k\in \big [ \mu (1-\frac{r_0}{k}),\mu (1-\frac{r_1}{k}) \big ]$. As a consequence, (1.1) has infinitely many non-radial solutions.

Without loss of generality, in the following, we assume that $K(1) = 1$.

The proof of Theorem 1.1 is mainly dependent on the finite-dimensional reduction method. The main idea of the reduction argument can be found in [2, 16, 21, 22]. Roughly speaking, to carry out this reduction argument, the first step is to construct a reasonably approximation solution, so that the problem can be reduced to a finite dimensional problem. The second step is to solve the corresponding finite dimensional problem to obtain a true solution. To fulfill the second step, it is essential to obtain a good estimate for the error term in the first step. However, when one consider more complicated problems, both steps must be modified. For example, in [22], to deal with the large number of bubbles in the solution, the reduction procedure is carried out in a weighted space instead of the standard Sobolev space. Our main idea is to place a large number of bubble inside $B_{\mu }(0)$. Then the scaling parameter will be determined by the number of bubbles. We put many bubbles along a $k-$polygon inside the domain $B_{\mu }(0)$ but near the boundary. Different from [21], the non-local properties of the equation make the problem getting more complicated. In order to decide the location of the bubble points, the most ingredients of the paper are using the Green representation and estimating the Green function and its regular part very carefully. Some more extra ideas and techniques are needed. We believe that our method and techniques can be applied to other related problems.

The paper is organized as follows. In Sect. 2, we perform a finite-dimensional reduction. The proof of Theorem 1.1 will be given in Sect. 3. Energy expansion and some essential estimates are attached in Appendices.

2 Finite-dimensional reduction

In this section, we perform a finite–dimensional reduction. Let

$$\begin{aligned} \begin{aligned} ||u||_* = \sup _{y\in B_{\mu }(0)}\left( \sum _{j=1}^{k}\frac{1}{(1+|y-x_j|)^{\frac{N-2s}{2}+\tau }}\right) ^{-1}|u(y)|, \end{aligned} \end{aligned}$$

(2.1)

and

$$\begin{aligned} \begin{aligned} ||u||_{**} = \sup _{y\in B_{\mu }(0)}\left( \sum _{j=1}^{k}\frac{1}{(1+|y-x_j|)^{\frac{N+2s}{2}+\tau }}\right) ^{-1}|u(y)|, \end{aligned} \end{aligned}$$

(2.2)

where $\tau = \frac{N-2s}{N-2s+1}$, $N\ge 3$ for $\frac{3}{4} \le s<1$ and $3\le N < 2s -1 +\frac{2}{3-4s}$ for $\frac{11-\sqrt{41}}{8}< s<\frac{3}{4}$. For this choice of $\tau$, we find that

$$\begin{aligned} \sum _{j=2}^k\frac{1}{|x_j-x_1|^\tau } \le \frac{k^\tau }{\mu ^\tau }\sum _{j=2}^k\frac{1}{j^\tau }\le \frac{Ck}{\mu ^\tau }\le C'. \end{aligned}$$

Let

$$\begin{aligned} Z_{i,1} = \frac{\partial PU_{x_{i},\Lambda }}{ \partial r}, \hbox { } Z_{i,2} = \frac{\partial PU_{x_{i},\Lambda }}{ \partial \Lambda }, \hbox { }i=1,...,k. \end{aligned}$$

We consider the following problem:

$$\begin{aligned} {\left\{ \begin{array}{ll} (-\Delta )^s \phi _k - (2^*_s -1) K\big (\frac{|y|}{\mu }\big )W^{2^*_s -2}_{r,\Lambda }\phi _k = h +\displaystyle \sum _{j=1}^{2}c_j\displaystyle \sum _{i=1}^{k} U^{2^*_s -2}_{x_i,\Lambda }Z_{i,j}, &{}\hbox {in } B_{\mu }(0), \\ \displaystyle \phi _k\in H_{r}\cap H^s_0\big (B_{\mu }(0)\big ), \\ \displaystyle \langle U^{2^*_s -2}_{x_i,\Lambda }Z_{i,j},\phi _k \rangle =0,\hbox { } i=1,...,k,\hbox { } j=1,2, \end{array}\right. } \end{aligned}$$

(2.3)

for some numbers $c_i$ where $\langle u,v \rangle = \displaystyle \int _{B_{\mu }(0)}u v$.

Lemma 2.1

Assume that $\phi _k$ solves (2.3) for $h =h_k,$ if $||h||_{**}$ goes to zero as k goes to infinity, so does $||\phi _k||_*$.

Proof

We argue by contradiction. Suppose that there are $k \rightarrow +\infty$, $h=h_k$, $\Lambda _k \in [L_1,L_2 ]$, $r_k \in [\mu (1-\frac{r_0}{k}), \mu (1-\frac{r_1}{k})]$, and $\phi _k$ solve (2.3) for $h= h_k$, $\Lambda = \Lambda _k$ and $r = r_k$ with $||h_k||_{**} \rightarrow 0$, and $||\phi _k||_{*}\ge c'>0$. We may assume that $||\phi _k||_{*}=1$. For simplify, we drop the subscript k. By Green representation, we rewrite (2.3) as

$$\begin{aligned} \begin{aligned} \phi (y)&= (2^*_s -1)\int _{B_{\mu }(0)} G(z,y) K\big (\frac{|y|}{\mu }\big ) W^{2^*_s -2}_{r,\Lambda }\phi (z) dz \\&\quad +\int _{B_{\mu }(0)} G(z,y)\big (h(z) +\sum _{j=1}^{2}c_j\sum _{i=1}^{k} U^{2^*_s -2}_{x_i,\Lambda }(z)Z_{i,j}(z)\big ) dz. \end{aligned} \end{aligned}$$

(2.4)

Using Lemma C.3, we have

$$\begin{aligned} \begin{aligned}&\left| (2^*_s -1)\int _{B_{\mu }(0)} G(z,y) K\left( \frac{|y|}{\mu }\right) W^{2^*_s -2}_{r,\Lambda }\phi (z) dz \right| \\&\quad \le \left| (2^*_s -1)\int _{B_{\mu }(0)} \frac{1}{|z-y|^{N-2s}} W^{2^*_s -2}_{r,\Lambda }\phi (z) dz\right| \\&\quad \le C||\phi ||_{*}\int _{{\mathbb {R}}^{N}}\frac{1}{|y-z|^{N-2s}}W^{2^*_s-2}_{r,\Lambda }(z)\sum _{j=1}^k \frac{1}{(1+|z-x_j|)^{\frac{N-2s}{2}+\tau }}dz \\&\quad \le C\sum _{j=1}^k\frac{1}{(1+|y-x_j|)^{\frac{N+2s}{2}+\tau +\theta }}. \end{aligned} \end{aligned}$$

(2.5)

It follows from Lemma C.2 that

$$\begin{aligned} \begin{aligned} \left| \int _{B_{\mu }(0)} G(z,y) h(z) dz\right|&\le C\left| \int _{B_{\mu }(0)} \frac{1}{|z-y|^{N-2s}} h(z) dz\right| \\&\le C||h||_{**}\sum _{j=1}^{k}\frac{1}{(1+|y-x_j|)^{\frac{N-2s}{2}+\tau }}, \end{aligned} \end{aligned}$$

(2.6)

and

$$\begin{aligned} \begin{aligned}&\left| \int _{B_{\mu }(0)} G(z,y) \sum _{i=1}^{k} U^{2^*_s -2}_{x_i,\Lambda }(z)Z_{i,j}(z) dz \right| \\&\quad \le C \left| \int _{B_{\mu }(0)} \frac{1}{|z-y|^{N-2s}} \sum _{i=1}^{k} U^{2^*_s -2}_{x_i,\Lambda }(z)Z_{i,j}(z) dz \right| \\&\quad \le C\sum _{j=1}^{k}\frac{1}{(1+|y-x_j|)^{\frac{N-2s}{2}+\tau }}. \end{aligned} \end{aligned}$$

(2.7)

Next, we estimate $c_l$, $l=1,2$. Multiplying (2.3) by $Z_{1,l}$ and integrating, we see that $c_{t}$ satisfies

$$\begin{aligned} \sum _{t=1}^2\sum _{i=1}^k\left\langle U^{2^*_s -2}_{x_i,\Lambda }Z_{i,t},Z_{1,l}\right\rangle c_t =\langle (-\Delta )^s \phi -(2^*_s -1)K\left( \frac{|y|}{\mu }\right) W^{ 2^*_s -2 }_{r,\Lambda }\phi ,Z_{1,l} \rangle -\langle h,Z_{1,l} \rangle .\nonumber \\ \end{aligned}$$

(2.8)

It follows from Lemma C.1 that

$$\begin{aligned} |\langle h,Z_{1,l} \rangle |\le C||h||_{**}\int _{{\mathbb {R}}^{N}}\frac{1}{(1+|z-x_1|)^{N-2s}}\sum _{j=1}^k\frac{1}{(1+|z-x_j|)^{\frac{N+2s}{2}+\tau }}dz \le C||h||_{**}. \end{aligned}$$

On the other hand, using Lemma A.1 and Lemma C.3, we can prove

$$\begin{aligned} \begin{aligned}&\left\langle (-\Delta )^{s}\phi - (2^{*}_{s}-1)K\left( \frac{|z|}{\mu }\right) W_{r,\Lambda }^{2^{*}_{s}-2} \phi ,Z_{1,l} \right\rangle = o(||\phi ||_{*}). \end{aligned} \end{aligned}$$

(2.9)

A direct computation leads to

$$\begin{aligned} \sum _{i=1}^{k}\langle U^{2^*_s -2}_{x_i,\Lambda }Z_{i,t},Z_{1,l} \rangle = {\bar{c}}\delta _{tl}+o(1), \end{aligned}$$

where ${\bar{c}} >0$ is a constant. Thus we obtain from (2.8) that

$$\begin{aligned} c_l= o(||\phi ||_*) +O(||h||_{**}). \end{aligned}$$

(2.10)

So

$$\begin{aligned} ||\phi ||_{*}\le \big ( o(1) + ||h||_{**} + \frac{\sum _{j=1}^k(1+|y-x_j|)^{-\frac{N-2s}{2}-\tau -\theta }}{\sum _{j=1}^k(1+|y-x_j|)^{-\frac{N-2s}{2}-\tau }} \big ). \end{aligned}$$

(2.11)

Since $||\phi ||_*=1$, we obtain from (2.11) that there exists $R>0$ such that

$$\begin{aligned} ||\phi ||_{L^{\infty }\big (B_{R}(x_i)\big )}\ge a >0, \end{aligned}$$

(2.12)

for some i. However, ${\bar{\phi }}=\phi (y-x_i)$ converges uniformly in any compact set to a solution u of

$$\begin{aligned} (-\Delta )^s u - (2^*_s-1)U^{2^*_s-2}_{0,\beta }u=0, \hbox { in } {\mathbb {R}}^{N}, \end{aligned}$$

(2.13)

for some $\beta \in [L_1,L_2]$, and u is perpendicular to the kernel of (2.13). Hence, $u=0$. This is a contradiction to (2.12). $\square$

From Lemma 2.1, using the same arguments as in the proof of Proposition 4.1 in [15], we can prove the following result.

Proposition 2.2

There exists $k_0 >0$ and a constant $C>0$, independent of k, such that for all $k\ge k_0$ and all $h\in L^{\infty }({\mathbb {R}}^{N})$, problem (2.3) has a unique solution $\phi \equiv L_k(h)$. Moreover

$$\begin{aligned} ||L_k(h)||_*\le C||h||_{**}, \hbox { }|c_l|\le C||h||_{**}. \end{aligned}$$

(2.14)

Now, we consider the following problem:

$$\begin{aligned} {\left\{ \begin{array}{ll} (-\Delta )^s(W_{r,\Lambda }+\phi ) = K(\frac{|y|}{\mu })(W_{r,\Lambda }+\phi )^{2^*_s-1}+\displaystyle \sum _{t=1}^2c_t\displaystyle \sum _{i=1}^k U^{2^*_s-2}_{x_i,\Lambda }Z_{i,t}, \hbox { in } B_{\mu }(0), \\ \displaystyle \phi \in H_{r}\cap H^s_0\big (B_{\mu }(0)\big ), \\ \displaystyle \langle U^{2^*_s-2}_{x_i,\Lambda }Z_{i,t},\phi \rangle =0,\hbox { } i=1,..,k,\hbox { }l=1,2. \end{array}\right. } \end{aligned}$$

(2.15)

In the rest of this section, we are devoted to the proof of the following proposition by the contraction mapping theorem.

Proposition 2.3

If $N\ge 3$ for $\frac{3}{4} \le s<1$ and $3\le N < 2s -1 +\frac{2}{3-4s}$ for $\frac{11-\sqrt{41}}{8}< s<\frac{3}{4}$, then there is an integer $k_0>0$, such that for each $k\ge k_0$, $L_0\le \Lambda \le L_1$ and $r\in [\mu (1-\frac{r_0}{k}),\mu (1-\frac{r_1}{k})]$, (2.15) has a unique solution $\phi =\phi (r,\Lambda )$, satisfying

$$\begin{aligned} ||\phi ||_{*}\le C\mu ^{-\frac{1}{2}-\sigma }, \hbox { }|c_t|\le C\mu ^{-\frac{1}{2}-\sigma },\hbox { } t=1,2, \end{aligned}$$

where $\sigma >0$ is a small constant and $\mu =k^{\frac{N-2s+1}{N-2s}}$.

We rewrite (2.15) as

$$\begin{aligned} {\left\{ \begin{array}{ll} (-\Delta )^s \phi -(2^*_s-1)K\big (\frac{|y|}{\mu }\big )W_{r,\Lambda }^{2^*_s-2}\phi = N(\phi )+l_k+\displaystyle \sum _{t=1}^2c_t\displaystyle \sum _{i=1}^k U^{2^*_s-2}_{x_i,\Lambda }Z_{i,t}, \hbox { in } B_{\mu }(0), \\ \displaystyle \phi \in H_{r}\cap H^s_0\big (B_{\mu }(0)\big ), \\ \displaystyle \langle U^{2^*_s-2}_{x_i,\Lambda }Z_{i,t},\phi \rangle =0, \hbox { }i=1,..,k,\hbox { }l=1,2, \end{array}\right. } \end{aligned}$$

(2.16)

where

$$\begin{aligned} N(\phi )=K\big (\frac{|y|}{\mu }\big )\big ( (W_{r,\Lambda }+\phi ) ^{2^*_s-1}-W_{r,\Lambda }^{2^*_s-1} -(2^*_s-1)W_{r,\Lambda }^{2^*_s-2}\phi \big ) \end{aligned}$$

and

$$\begin{aligned} l_k=K\left( \frac{|y|}{\mu }\right) W_{r,\Lambda }^{2^*_s-1}-\sum _{j=1}^k U^{2^*_s-1}_{x_j,\Lambda }. \end{aligned}$$

In order to use the contraction mapping theorem to prove that (2.16) is uniquely solvable in the set where $||\phi ||_*$ is small, we need to estimate $N(\phi )$ and $l_k$.

Lemma 2.4

If $N\ge 3$ and $\frac{11-\sqrt{41}}{8}<s<1$, then

$$\begin{aligned} || N(\phi ) ||_* \le C||\phi ||_*^{\min (2^*_s-1,2)}. \end{aligned}$$

Proof

If $2^*_s-1\le 2$, we have

$$\begin{aligned} |N(\phi )| \le C|\phi |^{2^*_s-1} . \end{aligned}$$

So

$$\begin{aligned} \begin{aligned} |N(\phi )|&\le C||\phi ||_*^{2^*_s-1}\left( \sum _{j=1}^k \frac{1}{(1+|y-x_j|)^{\frac{N-2s}{2}+\tau }}\right) ^{2^*_s-1} \\ {}&\le C||\phi ||_*^{2^*_s-1}\sum _{j=1}^k \frac{1}{(1+|y-x_j|)^{\frac{N+2s}{2}+\tau }}\left( \sum _{j=1}^k \frac{1}{(1+|y-x_j|)^{\tau }}\right) ^{\frac{4s}{N-2s}} \\&\le C||\phi ||_*^{2^*_s-1}\sum _{j=1}^k \frac{1}{(1+|y-x_j|)^{\frac{N+2s}{2}+\tau }}. \end{aligned} \end{aligned}$$

(2.17)

If $2^*_s-1 \ge 2$, we have

$$\begin{aligned} |N(\phi )| \le C W_{r,\Lambda }^{2^*_s-3} \phi ^{2}+|\phi |^{2^*_s-1}. \end{aligned}$$

Thus,

$$\begin{aligned} |N(\phi )|&\le C||\phi ||_{*}^{2}\big (\sum _{i=1}^{k}\frac{1}{(1+|y-x_i|)^{N-2s}} \big )^{2^*_s-3}\big (\sum _{i=1}^{k}\frac{1}{(1+|y-x_i|)^{\frac{N-2s}{2}+\tau }} \big )^{2} \\&\quad +C||\phi ||_{*}^{2^*_s-1}\sum _{j=1}^{k}\frac{1}{(1+|y-x_j|)^{\frac{N+2s}{2}+\tau }} \\&\le C||\phi ||_*^{2}\sum _{j=1}^k \frac{1}{(1+|y-x_j|)^{\frac{N+2s}{2}+\tau }}. \end{aligned}$$

$\square$

Next, we estimate $l_k$.

Lemma 2.5

Assume that $r\in [\mu (1-\frac{r_0}{k}),\mu (1-\frac{r_1}{k})]$. If $N\ge 3$ for $\frac{3}{4} \le s<1$ and $3\le N < 2s -1 +\frac{2}{3-4s}$ for $\frac{11-\sqrt{41}}{8}< s<\frac{3}{4}$, then

$$\begin{aligned} ||l_{k}||_{**}\le C\mu ^{-\frac{1}{2}-\sigma }. \end{aligned}$$

Proof

Define

$$\begin{aligned} \Omega _j=\left\{ y:\hbox { }y=(y',y'')\in B_{\mu }(0), \hbox { }\left\langle \frac{y'}{|y'|},\hbox { }\frac{x_j}{|x_j|} \right\rangle \ge \cos \frac{\pi }{k} \right\} . \end{aligned}$$

We have

$$\begin{aligned} l_k&=K\big (\frac{|y|}{\mu }\big )\big (W_{r,\Lambda }^{2^*_s-1} - \sum _{j=1}^{k}(PU_{x_j,\Lambda })^{2^*_s-1} \big ) \\&\quad +K\big (\frac{|y|}{\mu }\big )\big ( \sum _{j=1}^{k}(PU_{x_j,\Lambda })^{2^*_s-1} - \sum _{j=1}^{k}U_{x_j,\Lambda }^{2^*_s-1} \big )+ \sum _{j=1}^{k}U_{x_j,\Lambda }^{2^*_s-1}\big (K\big (\frac{|y|}{\mu }\big )-1\big ) \\&=:J_0+J_1+J_2. \end{aligned}$$

By the symmetry, we assume that $y\in \Omega _1$. Then for $\forall y\in \Omega _1$, $j = 2, ...,k$, we have

$$\begin{aligned} |y-x_1|\le |y-x_j|,\quad \frac{1}{1+|y-x_j|} \le \frac{C}{|x_1-x_j|}. \end{aligned}$$

For the estimate of $J_0$, we have

$$\begin{aligned} |J_0|&\le C\sum _{j=2}^k U_{x_j,\Lambda }^{2^*_s-1} + C\left| \sum _{j=2}^k PU_{x_j,\Lambda } \right| ^{2^*_s-1} +CU_{x_1,\Lambda }^{2^*_s-2}\left| \sum _{j=2}^k PU_{x_j,\Lambda } \right| \\&=: J_{00}+J_{01}+J_{02}. \end{aligned}$$

On the one hand, since $2s-\tau >\frac{1}{2}$, we have

$$\begin{aligned} | J_{00}|&\le C\sum _{j=2}^k \frac{1}{(1+|y-x_j|)^{N+2s}} \\ {}&\le \frac{ C}{(1+|y-x_1|)^{\frac{N+2s}{2}+\tau }} \sum _{j=2}^k\frac{1}{|x_j-x_1|^{\frac{N+2s}{2}-\tau }} \\ {}&\le \frac{ C}{(1+|y-x_1|)^{\frac{N+2s}{2}+\tau }} \left( \frac{k}{\mu }\right) ^{\frac{N+2s}{2}-\tau } \\ {}&\le \frac{C}{(1+|y-x_1|)^{\frac{N+2s}{2}+\tau }}\mu ^{-\frac{1}{2}-\sigma }. \end{aligned}$$

On the other hand, we have

$$\begin{aligned} |J_{01}|&\le C\left( \sum _{j=2}^{k} \frac{1}{(1+|y-x_j|)^{N-2s}}\right) ^{2^*_s-1} \\&\le \frac{C}{(1+|y-x_1|)^{\frac{N+2s}{2}+\tau }}\left( \sum _{j=2}^k \frac{1}{(1+|y-x_j|)^{\frac{N-2s}{2}-\tau \frac{N-2s}{N+2s}}}\right) ^{\frac{N+2s}{N-2s}} \\&\le \frac{C}{(1+|y-x_1|)^{\frac{N+2s}{2}+\tau }}\left( \frac{k}{\mu }\right) ^{\frac{N+2s}{2}-\tau } \\&\le \frac{C}{(1+|y-x_1|)^{\frac{N+2s}{2}+\tau }}\mu ^{-\frac{1}{2}-\sigma }, \end{aligned}$$

and

$$\begin{aligned} |J_{02}|&\le C\frac{1}{(1+|y-x_1|)^{4s}}\sum _{j=2}^{k}\frac{1}{ (1+|y-x_j|)^{N-2s} } \\&\le C\frac{1}{(1+|y-x_1|)^{\frac{N+2s}{2}+\tau }}\sum _{j=2}^{k}\frac{1}{ (1+|y-x_j|)^{\frac{N+2s}{2}-\tau } } \\&\le C\frac{1}{(1+|y-x_1|)^{\frac{N+2s}{2}+\tau }}\big (\frac{k}{\mu }\big )^{\frac{N+2s}{2}-\tau } \\&\le \frac{C}{(1+|y-x_j|)^{\frac{N+2s}{2}+\tau }}\mu ^{-\frac{1}{2}-\sigma }. \end{aligned}$$

Thus, we have

$$\begin{aligned} |J_0| \le \frac{C}{(1+|y-x_1|)^{\frac{N+2s}{2}+\tau }}\mu ^{-\frac{1}{2}-\sigma }. \end{aligned}$$

(2.18)

Now we estimate $J_1$. We have

$$\begin{aligned} |J_1|&\le C\big |(PU_{x_1,\Lambda })^{2^*_s-1} -(U_{x_1,\Lambda })^{2^*_s-1}\big | + C\sum _{j=2}^{k} U_{x_j,\Lambda } ^{2^*_s-1} \\&=:J_{10}+J_{11}. \end{aligned}$$

Similar to the estimate of $J_{00}$, we obtain

$$\begin{aligned} |J_{11}| \le \frac{C}{(1+|y-x_1|)^{\frac{N+2s}{2}+\tau }}\mu ^{-\frac{1}{2}-\sigma }. \end{aligned}$$

(2.19)

For the estimate of $J_{10}$, we have

$$\begin{aligned} |J_{10}| \le C\big ( U_{x_1,\Lambda }^{2^*_s-2}|U_{x_1,\Lambda } - PU_{x_1,\Lambda } | + |U_{x_1,\Lambda } - PU_{x_1,\Lambda } |^{2^*_s-1} \big ). \end{aligned}$$

Let $r_3=\min (r_0,1),$ if $|y-x_1|\ge \frac{r_3\mu }{8k},$ we get

$$\begin{aligned} \begin{aligned} U_{x_1,\Lambda }^{2^*_s-2}|U_{x_1,\Lambda } - PU_{x_1,\Lambda } |&\le \frac{C}{(1+|y-x_1|)^{N+2s}} \\&\le \frac{C}{(1+|y-x_1|)^{\frac{N+2s}{2}+\tau }}\mu ^{-\frac{1}{2}-\sigma }. \end{aligned} \end{aligned}$$

(2.20)

And if $|y-x_1|\le \frac{r_3\mu }{8k}$, using Lemma A.1, we have

$$\begin{aligned} U_{x_1,\Lambda }^{2^*_s-2}|U_{x_1,\Lambda } - PU_{x_1,\Lambda } |\le & {} C\frac{1}{(1+|y-x_1|)^{4s} }\big (\frac{k}{\mu }\big )^{N-2s} \nonumber \\\le & {} C\frac{1}{(1+|y-x_1|)^{\frac{N+2s}{2}+\tau }}\big (\frac{k}{\mu }\big )^{N-2s}(1+|y-x_1|)^{\frac{N+2s}{2}+\tau -4s} \nonumber \\\le & {} \frac{C}{(1+|y-x_1|)^{\frac{N+2s}{2}+\tau }}\mu ^{-\frac{1}{2}-\sigma }. \end{aligned}$$

(2.21)

So,

$$\begin{aligned} \begin{aligned} U_{x_1,\Lambda }^{2^*_s-2}|U_{x_1,\Lambda } - PU_{x_1,\Lambda } | \le \frac{C}{(1+|y-x_1|)^{\frac{N+2s}{2}+\tau }}\mu ^{-\frac{1}{2}-\sigma }. \end{aligned} \end{aligned}$$

(2.22)

Similarly, we can prove

$$\begin{aligned} \begin{aligned} |U_{x_1,\Lambda } - PU_{x_1,\Lambda } |^{2^*_s-1}\le \frac{C}{(1+|y-x_1|)^{\frac{N+2s}{2}+\tau }}\mu ^{-\frac{1}{2}-\sigma }. \end{aligned} \end{aligned}$$

(2.23)

Thus,

$$\begin{aligned} |J_{10}| \le \frac{C}{(1+|y-x_1|)^{\frac{N+2s}{2}+\tau }}\mu ^{-\frac{1}{2}-\sigma }. \end{aligned}$$

(2.24)

So

$$\begin{aligned} |J_1|\le \frac{C}{(1+|y-x_1|)^{\frac{N+2s}{2}+\tau }}\mu ^{-\frac{1}{2}-\sigma }. \end{aligned}$$

(2.25)

Now we estimate $J_2$. It is easy to check that

$$\begin{aligned} \big | \sum _{j=2}^{k}U_{x_j,\Lambda }^{2^*_s-1}\big (K\big (\frac{|y|}{\mu }\big )-1\big )\big | \le \frac{C}{(1+|y-x_1|)^{\frac{N+2s}{2}+\tau }}\mu ^{-\frac{1}{2}-\sigma }. \end{aligned}$$

(2.26)

If $|y-x_1|\ge \frac{r_0\mu }{2k}$, we have

$$\begin{aligned} \big |U_{x_1,\Lambda }^{2^*_s-1}\big (K\big (\frac{|y|}{\mu }\big )-1\big )\big |\le \frac{C}{(1+|y-x_1|)^{\frac{N+2s}{2}+\tau }}\mu ^{-\frac{1}{2}-\sigma }. \end{aligned}$$

(2.27)

And if $|y-x_1|\le \frac{r_0\mu }{2k}$, we have

$$\begin{aligned} K\big (\frac{|y|}{\mu }\big )-1 = O\big (\frac{1}{k}\big ). \end{aligned}$$

So,

$$\begin{aligned} \begin{aligned} |U_{x_1,\Lambda }^{2^*_s-1}\big (K\big (\frac{|y|}{\mu }\big )-1\big )|&\le \frac{C}{(1+|y-x_1|)^{\frac{N+2s}{2}+\tau }}\frac{1}{k} \\&\le \frac{C}{(1+|y-x_1|)^{\frac{N+2s}{2}+\tau }}\mu ^{-\frac{1}{2}-\sigma }. \end{aligned} \end{aligned}$$

(2.28)

Combining (2.26), (2.27) and (2.28), we obtain

$$\begin{aligned} |J_2|\le \frac{C}{(1+|y-x_1|)^{\frac{N+2s}{2}+\tau }}\mu ^{-\frac{1}{2}-\sigma }. \end{aligned}$$

(2.29)

Finally combining (2.18), (2.25) and (2.29), it holds

$$\begin{aligned} ||l_k||_{**} \le C\mu ^{-\frac{1}{2}-\sigma }. \end{aligned}$$

$\square$

Now, we are ready to prove proposition 2.3.

.

Proof of Proposition 2.3

Recall that $\mu = k^{\frac{N-2s+1}{N-2s}}$, $N\ge 3$ for $\frac{3}{4} \le s<1$ and $3\le N < 2s -1 +\frac{2}{3-4s}$ for $\frac{11-\sqrt{41}}{8}< s<\frac{3}{4}$.

Let

$$\begin{aligned} E&=\big \{ u: \hbox { }u \in C\big (B_{\mu }(0)\big ) \cap H_r,\hbox { } ||u||_{*}\le \big (\frac{1}{\mu }\big )^{\frac{1}{2}}, \\&\quad \int _{B_{\mu }(0)}U_{x_i,\Lambda }^{2^{*}_s-2}Z_{i,l}\phi =0,\hbox { } i=1,...,k,\hbox { } l=1,2 \big \}. \end{aligned}$$

Then (2.16) is equivalent to

$$\begin{aligned} \phi = A(\phi )=:L_{k}(N(\phi )) +L_{k}(l_k), \end{aligned}$$

where $L_k$ is defined in Proposition 2.2. We will prove that A is a contraction map from E to E. First, we have

$$\begin{aligned} \begin{aligned} ||A(\phi )||_{*}&\le C||N(\phi )||_{**}+C||l_k||_{**} \\&\le C||\phi ||_{**}^{\min (2^{*}_s-1, 2)} +C||l_k||_{**} \\&\le C\frac{1}{\mu ^{\frac{1}{2}+\sigma }}\le \big (\frac{1}{\mu }\big )^{\frac{1}{2}}. \end{aligned} \end{aligned}$$

(2.30)

On the other hand,

$$\begin{aligned} ||A(\phi _1)-A(\phi _2)||_{*}=||L_{k}(N(\phi _1))-L_{k}(N(\phi _2))||_{*}\le C||N(\phi _1)-N(\phi _2)||_{**}. \end{aligned}$$

If $2^{*}_s-2 \le 1$, then

$$\begin{aligned} |N'(t)|\le C|t|^{2^{*}_s-2}. \end{aligned}$$

As a result, we obtain

$$\begin{aligned}&|N(\phi _1)-N(\phi _2) |\\&\quad \le C( |\phi _1|^{2^{*}_s-2}+|\phi _2|^{2^{*}_s-2} )|\phi _1 -\phi _2| \\&\quad \le C( ||\phi _1||_{*}^{2^{*}_s-2}+||\phi _2||_{*}^{2^{*}_s-2} ) ||\phi _1-\phi _2||_{*}\big (\sum _{j=1}^{k} \frac{1}{(1+|y-x_j|)^{\frac{N-2s}{2} +\tau }}\big )^{2^{*}_s-1}. \end{aligned}$$

Similar to the arguments for $||N(\phi )||_*$ , we have

$$\begin{aligned} \big (\sum _{j=1}^{k} \frac{1}{(1+|y-x_j|)^{\frac{N-2s}{2}+\tau }}\big )^{2^{*}_s-1}\le C\sum _{j=1}^{k} \frac{1}{(1+|y-x_j|)^{\frac{N+2s}{2}+\tau }}. \end{aligned}$$

Thus, we obtain

$$\begin{aligned} ||A(\phi _1)-A(\phi _2)||_{*}&\le C||N(\phi _1)-N(\phi _2)||_{**} \\&\le C( ||\phi _1||_{*}^{2^{*}_s-2}+||\phi _2||_{*}^{2^{*}_s-2} )||\phi _1-\phi _2||_{*} \\&\le \frac{1}{2}||\phi _1-\phi _2||_{*}. \end{aligned}$$

As a result, A is a contraction map.

If $2^{*}_s-2 \ge 1$, then $|N'(t)|\le C|t|^{2^{*}_s-2} + CW_{r,\Lambda }^{2^{*}_s-3}|t|.$ Hence,

$$\begin{aligned}&|N(\phi _1)-N(\phi _2) | \\&\quad \le C(|\phi _1|^{2^{*}_s-2} +|\phi _1|^{2^{*}_s-2})|\phi _1-\phi _2|+C(|\phi _1|+|\phi _2|)W_{r,\Lambda }^{2^{*}_s-3}| \phi _1-\phi _2 | \\&\quad \le C(||\phi _1||_{*}^ {2^{*}_s-2}+||\phi _2||_{*}^ {2^{*}_s-2} )||\phi _1-\phi _2||_{*}\big ( \sum _{j=1}^{k}\frac{1}{(1+|y-x_j|)^{\frac{N-2s}{2}+\tau }}\big )^{2^{*}_s-1} \\&\qquad +C( ||\phi _1||_{*}+||\phi _2||_{*} )||\phi _1-\phi _2||_{*}W_{r,\Lambda }^{2^{*}_s-3}\big ( \sum _{j=1}^{k}\frac{1}{(1+|y-x_j|)^{\frac{N-2s}{2}+\tau }}\big )^{2} \\&\quad \le \frac{1}{2}||\phi _1-\phi _2||_{*}. \end{aligned}$$

Thus A is a contraction map.

By the contraction mapping theorem, there exists a unique $\phi \in E$, such that $\phi =A(\phi ).$ Moreover, it follows from Proposition 2.2 that

$$\begin{aligned} ||\phi ||_{*}\le C||l_{k}||_{**}+C||N(\phi )||_{**}\le C||l_k||_{**}+C||\phi ||_{*}^{ \min (2^{*}_s-1,2) }, \end{aligned}$$

which leads to

$$\begin{aligned} ||\phi ||_{*}\le C\big ( \frac{1}{\mu } \big )^{\frac{1}{2}+\sigma }. \end{aligned}$$

The estimate of $c_t$ follows from (2.14). $\square$

3 Proof of theorem 1.1

Let

$$\begin{aligned} F(d,\Lambda ) = I(W_{r,\Lambda } +\phi ) \end{aligned}$$

where $r=|x_1|$, $d=1-\frac{r}{\mu }$, $\phi$ is the function obtained in Proposition 2.3 and

$$\begin{aligned} I(u) = \frac{1}{2}\int _{B_{\mu }(0)}|(-\Delta )^{\frac{s}{2}}u|^{2}dy -\frac{1}{2^{*}_{s}}\int _{B_{\mu }(0)}K \left( \frac{|y|}{\mu }\right) |u|^{2^{*}_{s}}. \end{aligned}$$

Let ${\bar{x}}_j=\frac{1}{\mu }x_j$, G(x, y) be the Green function of $(-\Delta )^{s}$ in $B_{1}(0)$ with homogenous Dirichlet boundary condition and H(x, y) be the part of the Green function.

Proposition 3.1

If $N\ge 3$ for $\frac{3}{4} \le s<1$ and $3\le N < 2s -1 +\frac{2}{3-4s}$ for $\frac{11-\sqrt{41}}{8}< s<\frac{3}{4}$, then

$$\begin{aligned} F(d,\Lambda )&= I(W_{r,\Lambda }) +kO\big (\frac{1}{\mu ^{1+\sigma }}\big ) \nonumber \\&=k\big (A+\frac{B_{1}H(\bar{x_1},\bar{x_1})}{\Lambda ^{N-2s}\mu ^{N-2s}} +B_2K'(1)d -\sum _{j=2}^{k}\frac{B_{1}G(\bar{x_j},\bar{x_1})}{\Lambda ^{N-2s}\mu ^{N-2s}} + O\big (\frac{1}{\mu ^{1+\sigma }}\big ) \big ) \nonumber \\&=k\big ( A + \frac{A_1}{\Lambda ^{N-2s}\mu ^{N-2s}d^{N-2s}} +A_2 d -\frac{A_3k^{N-2s}}{\Lambda ^{N-2s}\mu ^{N-2s}} \nonumber \\&\quad \times \sum _{i=2}^{+\infty }\frac{1}{((i-1)\pi )^{N-2s}}\int _{1}^{\big ( \frac{(dk)^{2}}{((i-1)\pi )^{2}} +1 \big )^{\frac{1}{2}}} (v^2-1)^{s-1}v^{1-N} dv + O\big (\frac{1}{\mu ^{1+\sigma }}\big )\big ),\nonumber \\ \end{aligned}$$

(3.1)

where A, $A_1$, $A_2$, $A_3$, $B_1$ and $B_2$ are positive constants, and $\sigma >0$ is a small constant.

Proof

Since

$$\begin{aligned} \langle I'(W_{r,\Lambda }+\phi ),\phi \rangle = 0,\hbox { } \forall \phi \in E, \end{aligned}$$

there is $t\in (0,1)$ such that

$$\begin{aligned} F(d,\Lambda )&= I(W_{r,\Lambda }) - \frac{1}{2}D^{2}I(W_{r,\Lambda }+t\phi )(\phi ,\phi ) \\&=I(W_{r,\Lambda }) - \frac{1}{2}\int _{B_{\mu }(0)} \big ( |(-\Delta )^{\frac{s}{2}}\phi |^{2} -(2^{*}_s-1)K \big (\frac{|y|}{\mu }\big )( W_{r,\Lambda }+t\phi )^{ 2^{*}_s-2 } \phi ^{2}\big ) \\&=I(W_{r,\Lambda })+\frac{2^{*}_s-1}{2}\int _{B_{\mu }(0)}K\big (\frac{|y|}{\mu }\big ) \big (( W_{r,\Lambda }+t\phi )^{ 2^{*}_s-2 } -W_{r,\Lambda } ^{ 2^{*}_s-2 }\big ) \phi ^{2} \\&\quad -\frac{1}{2}\int _{B_{\mu }(0)}(N(\phi ) +l_k)\phi \\&=I(W_{r,\Lambda })+ O\big (\int _{B_{\mu }(0)}(|\phi |^{2^{*}_s} +|N(\phi )||\phi | +|l_k||\phi |)\big ). \end{aligned}$$

However,

$$\begin{aligned}&\int _{B_{\mu }(0)}|N(\phi )||\phi | +|l_k||\phi | \\&\quad \le C(||N(\phi )||_{**}+||l_k||_{**})||\phi ||_{**} \int _{B_{\mu }(0)}\sum _{j=1}^{k}\frac{1}{(1+|y-x_{j}|)^{\frac{N+2s}{2}+\tau }}\\&\qquad \times \sum _{j=1}^{k}\frac{1}{(1+|y-x_{j}|)^{\frac{N-2s}{2}+\tau }}. \end{aligned}$$

By Lemma C.1, for $N\ge 3$,

$$\begin{aligned}&\sum _{j=1}^{k}\frac{1}{(1+|y-x_{j}|)^{\frac{N+2s}{2}+\tau }}\sum _{j=1}^{k}\frac{1}{(1+|y-x_{j}|)^{\frac{N-2s}{2}+\tau }} \\&\quad =\sum _{j=1}^{k}\frac{1}{(1+|y-x_j|)^{N+2\tau }}+\sum _{j=1}^{k}\sum _{i\ne j}\frac{1}{(1+|y-x_j|)^{\frac{N+2s}{2}+\tau }}\frac{1}{(1+|y-x_i|)^{\frac{N-2s}{2}+\tau }} \\&\quad \le C\sum _{j=1}^{k}\frac{1}{(1+|y-x_j|)^{N+2\tau }}+C\sum _{j=1}^{k}\frac{1}{(1+|y-x_j|)^{N+\tau }}\sum _{j=2}^k\frac{1}{|x_j-x_1|^{\tau }} \\&\quad \le C\sum _{j=1}^{k}\frac{1}{(1+|y-x_j|)^{N+\tau }}. \end{aligned}$$

Thus, we obtain

$$\begin{aligned} \int _{B_{\mu }(0)}|N(\phi )||\phi | +|l_k||\phi | \le Ck( ||N(\phi )||_{**}+||l_k||_{**})||\phi ||_{**} )\le Ck\left( \frac{1}{\mu ^{1+\sigma }}\right) . \end{aligned}$$

On the other hand,

$$\begin{aligned} \int _{B_{\mu }(0)}|\phi |^{2^{*}_{s}}\le C||\phi ||_{**}\int _{B_{\mu }(0)} \left( \sum _{j=1}^{k}\frac{1}{(1+|y-x_j|)^{\frac{N-2s}{2}+\tau }}\right) ^{2^{*}_{s}}. \end{aligned}$$

Noting that for $y\in \Omega _1$ and $N\ge 3$

$$\begin{aligned}&\sum _{j=2}^{k}\frac{1}{(1+|y-x_j|)^{\frac{N-2s}{2}+\tau }} \\ {}&\quad \le C\frac{1}{(1+|y-x_1|)^{\frac{N-2s}{2}}}\sum _{j=2}^{k}\frac{1}{|x_1-x_j|^{\tau }}\le C\frac{1}{(1+|y-x_1|)^{\frac{N-2s}{2}}}. \end{aligned}$$

Thus, we have

$$\begin{aligned} \left( \sum _{j=1}^{k}\frac{1}{(1+|y-x_j|)^{\frac{N-2s}{2}+\tau }}\right) ^{2^{*}_{s}}\le \frac{C}{(1+|y-x_1|)^{N}}, \,\, y\in \Omega _1, \end{aligned}$$

which leads to

$$\begin{aligned} \int _{B_{\mu }(0)}\left( \sum _{j=1}^{k}\frac{1}{(1+|y-x_j|)^{\frac{N-2s}{2}+\tau }}\right) ^{2^{*}_{s}}\le Ck\ln k. \end{aligned}$$

So, we have proved

$$\begin{aligned} \int _{B_{\mu }(0)}|\phi |^{2^{*}_{s}}\le Ck\ln k||\phi ||_{*}^{2^{*}_{s}}\le Ck\ln k\left( \frac{1}{\mu ^{2^{*}_{s}\left( \frac{1}{2}+\sigma \right) }}\right) , \hbox { for } N\ge 3. \end{aligned}$$

Thus,

$$\begin{aligned} F(d,\Lambda )&= I(W_{r,\Lambda }) +kO\left( \frac{1}{\mu ^{1+\sigma }}\right) . \end{aligned}$$

And (3.1) follows from Proposition B.1. $\square$

Proposition 3.2

If $N\ge 3$ for $\frac{3}{4} \le s<1$ and $3\le N < 2s -1 +\frac{2}{3-4s}$ for $\frac{11-\sqrt{41}}{8}< s<\frac{3}{4}$, then we have

$$\begin{aligned} \begin{aligned} \frac{\partial F(d,\Lambda )}{\partial \Lambda }&= kB_1(N-2s)\big ( -\frac{ H(\bar{x_1},\bar{x_1})}{\Lambda ^{N-2s+1}\mu ^{N-2s}} +\sum _{j=2}^{k}\frac{ G(\bar{x_j},\bar{x_1})}{\Lambda ^{N-2s+1}\mu ^{N-2s}} + O\big (\frac{1}{\mu ^{1+\sigma }}\big )\big ) \\&= k \big ( -\frac{A_1(N-2s)}{\Lambda ^{N-2s+1}\mu ^{N-2s}d^{N-2s}} + \frac{A_3k^{N-2s}(N-2s)}{\Lambda ^{N-2s+1}\mu ^{N-2s}} \\&\quad \times \sum _{i=2}^{+\infty }\frac{1}{((i-1)\pi )^{N-2s}}\int _{1}^{\big ( \frac{(dk)^{2}}{((i-1)\pi )^{2}} +1 \big )^{\frac{1}{2}}} (v^2-1)^{s-1}v^{1-N} dv + O\big (\frac{1}{\mu ^{1+\sigma }}\big )\big ) , \end{aligned}\nonumber \\ \end{aligned}$$

(3.2)

and

$$\begin{aligned} \begin{aligned} \frac{\partial F(d,\Lambda )}{\partial d}&=k\big ( \frac{B_{1}\frac{\partial H(\bar{x_1},\bar{x_1})}{ \partial d}}{\Lambda ^{N-2s}\mu ^{N-2s}} +B_2K'(1)-\sum _{j=2}^{k}\frac{B_1\frac{\partial G(\bar{x_j},\bar{x_1})}{ \partial d}}{{\Lambda ^{N-2s}\mu ^{N-2s}} }+ O\big (\frac{1}{\mu ^{\sigma }}\big )\big ) \\&= k \big ( -\frac{A_1(N-2s)}{\Lambda ^{N-2s}\mu ^{N-2s}d^{N-2s+1}} +A_2 - \frac{A_3k^{N-2s}(N-2s)}{\Lambda ^{N-2s}\mu ^{N-2s}} \\&\quad \times \sum _{i=2}^{+\infty }\frac{1}{((i-1)\pi )^{N}} \big ( \frac{(dk)^{2}}{((i-1)\pi )^{2}} +1 \big )^{\frac{-N}{2}}dk^{2}(dk)^{2(s-1)} + O\big (\frac{1}{\mu ^{\sigma }} \big )\big ), \end{aligned} \end{aligned}$$

(3.3)

where $A_1$, $A_2$, $A_3$, $B_1$ and $B_2$ are the same constants as in Proposition 3.1, and $\sigma > 0$ is a small constant.

Proof

We estimate $\frac{\partial F(d,\Lambda )}{\partial \Lambda }$ first. We have

$$\begin{aligned} \frac{\partial F(d,\Lambda )}{\partial \Lambda }&=\left\langle I'(W_{r,\Lambda } +\phi ), \frac{\partial W_{r,\Lambda }}{\partial \Lambda } + \frac{\partial \phi }{\partial \Lambda } \right\rangle \\ {}&=\left\langle I'(W_{r,\Lambda } +\phi ), \frac{\partial W_{r,\Lambda }}{\partial \Lambda } \right\rangle +\sum _{l=1}^{2}\sum _{i=1}^{k}c_l\left\langle U_{x_{i},\Lambda }^{2^*_s-2} Z_{i,l}, \frac{\partial \phi }{\partial \Lambda } \right\rangle . \end{aligned}$$

Noting that

$$\begin{aligned} \left\langle U_{x_{i},\Lambda }^{2^*_s-2} Z_{i,l},\frac{\partial \phi }{\partial \Lambda } \right\rangle = -\left\langle \frac{\partial (U_{x_{i},\Lambda }^{2^*_s-2} Z_{i,l})}{\partial \Lambda }, \phi \right\rangle , \end{aligned}$$

we have

$$\begin{aligned}&\sum _{i=1}^{k}c_l\left\langle U_{x_{i},\Lambda }^{2^*_s-2} Z_{i,l},\frac{\partial \phi }{\partial \Lambda } \right\rangle \\&\quad \le C|c_l|||\phi ||_{*}\int _{{\mathbb {R}}^{N}}\sum _{i=1}^{k}\frac{1}{(1+|y-x_i|)^{N+2s}}\sum _{i=1}^{k}\frac{1}{(1+|y-x_i|)^{\frac{N-2s}{2}+\tau }} \\ {}&\quad \le Ck\frac{1}{\mu ^{1+\sigma }}. \end{aligned}$$

On the other hand,

$$\begin{aligned}&\int _{B_{\mu }(0)}(-\Delta )^{\frac{s}{2}}( W_{r,\Lambda } +\phi )(-\Delta )^{\frac{s}{2}}\left( \frac{\partial W_{r,\Lambda }}{\partial \Lambda }\right) \\&\quad = \int _{B_{\mu }(0)}(-\Delta )^{\frac{s}{2}}( W_{r,\Lambda } )(-\Delta )^{\frac{s}{2}}\left( \frac{\partial W_{r,\Lambda }}{\partial \Lambda }\right) +O\left( \frac{k}{\mu ^{1+\sigma }}\right) . \end{aligned}$$

If $2^{*}_s-1 \le 2$, we have

$$\begin{aligned}&\int _{B_{\mu }(0)}K\left( \frac{|y|}{\mu }\right) ( W_{r,\Lambda } +\phi )^{ 2^{*}_s-1}\frac{\partial W_{r,\Lambda }}{\partial \Lambda } \\&\quad =\int _{B_{\mu }(0)}K\left( \frac{|y|}{\mu }\right) W_{r,\Lambda }^{ 2^{*}_s-1}\frac{\partial W_{r,\Lambda }}{\partial \Lambda }+(2^{*}_s-1 )\int _{B_{\mu }(0)}K\left( \frac{|y|}{\mu }\right) W_{r,\Lambda }^{ 2^{*}_s-2}\frac{\partial W_{r,\Lambda }}{\partial \Lambda }\phi \\&\qquad +O\left( \int _{B_{\mu }(0)}|\phi |^{2^{*}_s-1}\left| \frac{\partial W_{r,\Lambda }}{\partial \Lambda }\right| \right) \\&\quad =\int _{B_{\mu }(0)}K\left( \frac{|y|}{\mu }\right) W_{r,\Lambda }^{ 2^{*}_s-1}\frac{\partial W_{r,\Lambda }}{\partial \Lambda }+(2^{*}_s-1 )\\&\qquad \times \int _{B_{\mu }(0)}K\left( \frac{|y|}{\mu }\right) W_{r,\Lambda }^{ 2^{*}_s-2}\frac{\partial W_{r,\Lambda }}{\partial \Lambda }\phi +O\left( \frac{k}{\mu ^{1+\sigma }}\right) . \end{aligned}$$

If $2^{*}_s-1 \ge 2$, we have

$$\begin{aligned}&\int _{B_{\mu }(0)}K\big (\frac{|y|}{\mu }\big )( W_{r,\Lambda } +\phi )^{ 2^{*}_s-1}\frac{\partial W_{r,\Lambda }}{\partial \Lambda } \\ {}&\quad =\int _{B_{\mu }(0)}K\big (\frac{|y|}{\mu }\big ) W_{r,\Lambda }^{ 2^{*}_s-1}\frac{\partial W_{r,\Lambda }}{\partial \Lambda }+(2^{*}_s-1 )\int _{B_{\mu }(0)}K\big (\frac{|y|}{\mu }\big )W_{r,\Lambda }^{ 2^{*}_s-2}\frac{\partial W_{r,\Lambda }}{\partial \Lambda }\phi \\&\qquad +O\big (\int _{B_{\mu }(0)}(|\phi |^{2^{*}_s-1}\big |\frac{\partial W_{r,\Lambda }}{\partial \Lambda }\big | + |W_{r,\Lambda }|^{2^{*}_s-3}|\phi |^{2}\big |\frac{\partial W_{r,\Lambda }}{\partial \Lambda }\big | )\big ) \\&\quad =\int _{B_{\mu }(0)}K\big (\frac{|y|}{\mu }\big )W_{r,\Lambda }^{ 2^{*}_s-1}\frac{\partial W_{r,\Lambda }}{\partial \Lambda }+(2^{*}_s-1 )\int _{B_{\mu }(0)}K\big (\frac{|y|}{\mu }\big )W_{r,\Lambda }^{ 2^{*}_s-2}\frac{\partial W_{r,\Lambda }}{\partial \Lambda }\phi +O\big (\frac{k}{\mu ^{1+\sigma }}\big ). \end{aligned}$$

Thus, we have

$$\begin{aligned}&\int _{B_{\mu }(0)}K\left( \frac{|y|}{\mu }\right) ( W_{r,\Lambda } +\phi )^{ 2^{*}_s-1}\frac{\partial W_{r,\Lambda }}{\partial \Lambda } \\&\quad =\int _{B_{\mu }(0)}K\left( \frac{|y|}{\mu }\right) W_{r,\Lambda }^{ 2^{*}_s-1} \frac{\partial W_{r,\Lambda }}{\partial \Lambda }+(2^{*}_s-1 )\\&\quad \times \int _{B_{\mu }(0)}K\left( \frac{|y|}{\mu }\right) W_{r,\Lambda }^{ 2^{*}_s-2}\frac{\partial W_{r,\Lambda }}{\partial \Lambda } \phi +O\left( \frac{k}{\mu ^{1+\sigma }}\right) . \end{aligned}$$

Noting that $\phi \in E$, so we have

$$\begin{aligned}&\int _{B_{\mu }(0)}K\left( \frac{|y|}{\mu }\right) W_{r,\Lambda }^{ 2^{*}_s-2}\frac{\partial W_{r,\Lambda }}{\partial \Lambda }\phi \\ {}&\quad =\int _{B_{\mu }(0)}K\left( \frac{|y|}{\mu }\right) ( W_{r,\Lambda }^{ 2^{*}_s-2} \frac{\partial W_{r,\Lambda }}{\partial \Lambda } - \sum _{j=1}^{k}U_{x_j,\Lambda }^{2^{*}_s-2} \frac{\partial U_{x_j,\Lambda } }{\partial \Lambda }\phi ) \\ {}&\qquad +\int _{B_{\mu }(0)}\sum _{j=1}^{k} \left( K\left( \frac{|y|}{\mu }\right) -1 \right) U_{x_j,\Lambda } ^{2^{*}_s-2}\frac{\partial U_{x_j,\Lambda } }{\partial \Lambda }\phi \\ {}&\quad =k\int _{\Omega _1}K \left( \frac{|y|}{\mu }\right) \left( W_{r,\Lambda }^{ 2^{*}_s-2} \frac{\partial W_{r,\Lambda }}{ \partial \Lambda } - \sum _{j=1}^{k}U_{x_j,\Lambda }^{2^{*}_s-2} \frac{\partial U_{x_j,\Lambda } }{\partial \Lambda }\right) \phi \\ {}&\qquad +k\int _{B_{\mu }(0)}\left( K\left( \frac{|y|}{\mu }\right) -1 \right) U_{x_1,\Lambda } ^{2^{*}_s-2}\frac{\partial U_{x_1,\Lambda } }{\partial \Lambda }\phi , \end{aligned}$$

and using Lemma A.1, if $2^{*}_s-2\le 1$, we have

$$\begin{aligned}&\int _{\Omega _1} K\left( \frac{|y|}{\mu }\right) (W_{r,\Lambda }^{ 2^{*}_s-2} \frac{\partial W_{r,\Lambda }}{ \partial \Lambda } - \sum _{j=1}^{k}U_{x_j,\Lambda }^{2^{*}_s-2}\frac{\partial U_{x_j,\Lambda } }{\partial \Lambda })\phi \\ {}&\quad \le C\int _{\Omega _1}(|W_{r,\Lambda }|^{ 2^{*}_s-2}| \frac{\partial U_{x_1,\Lambda } }{\partial \Lambda } -\frac{\partial PU_{x_1,\Lambda } }{\partial \Lambda } ) | +U_{x_1,\Lambda }^{2^{*}_s-2 }|\frac{\partial U_{x_1,\Lambda } }{\partial \Lambda } -\frac{\partial PU_{x_1,\Lambda } }{\partial \Lambda } | \\ {}&\qquad +U_{x_1,\Lambda }^{2^{*}_s-2 }\sum _{j=2}^{k}U_{x_j,\Lambda } +\sum _{j=2}^{k}U_{x_j,\Lambda }^{2^{*}_s-1 })|\phi |\le C\frac{1}{\mu ^{1+\sigma }}. \end{aligned}$$

If $2^{*}_s-2\ge 1 ,$ we have

$$\begin{aligned}&\int _{\Omega _1} K\left( \frac{|y|}{\mu }\right) \left( W_{r,\Lambda }^{ 2^{*}_s-2} \frac{\partial W_{r,\Lambda }}{ \partial \Lambda } - \sum _{j=1}^{k}U_{x_j,\Lambda }^{2^{*}_s-2}\frac{\partial U_{x_j,\Lambda } }{\partial \Lambda }\right) \phi \\&\quad \le C\int _{\Omega _1}\left( |W_{r,\Lambda }|^{ 2^{*}_s-2}| \frac{\partial U_{x_1,\Lambda } }{\partial \Lambda } -\frac{\partial PU_{x_1,\Lambda } }{\partial \Lambda } \right) |+U_{x_1,\Lambda }^{2^{*}_s-2 }\left| \frac{\partial U_{x_1,\Lambda } }{\partial \Lambda } -\frac{\partial PU_{x_1,\Lambda } }{\partial \Lambda } \right| \\ {}&\qquad +U_{x_1,\Lambda }^{2^{*}_s-2 }\sum _{j=2}^{k}U_{x_j,\Lambda } +\sum _{j=2}^{k}U_{x_j,\Lambda }^{2^{*}_s-1 } + \left( \sum _{j=2}^{k}U_{x_j,\Lambda }\right) ^{ 2^{*}_s-2 }U_{x_1,\Lambda })|\phi | \\ {}&\quad \le C\frac{1}{\mu ^{1+\sigma }}. \end{aligned}$$

On the other hand,

$$\begin{aligned}&\int _{B_{\mu }(0)} \left( K\left( \frac{|y|}{\mu }\right) -1 \right) U_{x_1,\Lambda } ^{2^{*}_s-2}\frac{\partial U_{x_1,\Lambda } }{\partial \Lambda }\phi \\ {}&\quad \le \left| \int _{B_{\frac{r_3\mu }{2k}}(x_{1})} \left( K\left( \frac{|y|}{\mu }\right) -1 \right) U_{x_1,\Lambda } ^{2^{*}_s-2} \frac{\partial U_{x_1,\Lambda } }{\partial \Lambda }\phi \right| \\ {}&\qquad +\int _{{\mathbb {R}}^{N}\setminus B_{\frac{r_3\mu }{2k}}(x_{1})} \left( K\left( \frac{|y|}{\mu }\right) -1 \right) U_{x_1,\Lambda } ^{2^{*}_s-2} \frac{\partial U_{x_1,\Lambda } }{\partial \Lambda }\phi | \\ {}&\quad \le C\frac{1}{\mu ^{1+\sigma }}, \end{aligned}$$

where $r_3=\min (r_0,1)$. Thus, we have proved

$$\begin{aligned} \frac{\partial F(d,\Lambda )}{\partial \Lambda } = \frac{\partial I(W_{r,\Lambda })}{\partial \Lambda } +k O\left( \frac{1}{\mu ^{1+\sigma }}\right) , \end{aligned}$$

and (3.2) follows from Proposition B.2.

Finally using Proposition B.2, we can estimate $\frac{\partial F(d,\Lambda )}{\partial d}$ in a similar way. $\square$

Now we are ready to prove Theorem 1.1.

Proof of Theorem 1.1

Note that $d =1 -\frac{r}{\mu }$ and $\mu = k^{\frac{N-2s+1}{N-2s}}$. Define $D =dk$. Then from (3.2) and (3.3), $\frac{\partial F(d,\Lambda )}{\partial \Lambda } =0$ and $\frac{\partial F(d,\Lambda )}{\partial d} =0$ are equivalent to

$$\begin{aligned}&-\frac{A_1(N-2s)}{\Lambda ^{N-2s+1}D^{N-2s}} + \frac{A_3(N-2s)}{\Lambda ^{N-2s+1}} \nonumber \\&\quad \times \sum _{i=2}^{+\infty }\frac{1}{((i-1)\pi )^{N-2s}}\int _{1}^{\big ( \frac{(D)^{2}}{((i-1)\pi )^{2}} +1 \big )^{\frac{1}{2}}} (v^2-1)^{s-1}v^{1-N} dv + O\big (\frac{1}{\mu ^{\sigma }} \big ) =0, \end{aligned}$$

(3.4)

and

$$\begin{aligned} \begin{aligned}&-\frac{A_1(N-2s)}{\Lambda ^{N-2s}D^{N-2s+1}} +A_2 - \frac{A_3(N-2s)}{\Lambda ^{N-2s}} \\ {}&\quad \times \sum _{i=2}^{+\infty }\frac{1}{((i-1)\pi )^{N}}\big ( \frac{D^{2}}{((i-1)\pi )^{2}} +1 \big )^{\frac{-N}{2}}D^{2s-1} + O\big (\frac{1}{\mu ^{\sigma }} \big ) =0 , \end{aligned} \end{aligned}$$

(3.5)

respectively.

Let

$$\begin{aligned} \begin{aligned} f_1(D,\Lambda ) =&-\frac{A_1(N-2s)}{\Lambda ^{N-2s+1}D^{N-2s}} + \frac{A_3(N-2s)}{\Lambda ^{N-2s+1}} \\&\times \sum _{i=2}^{+\infty }\frac{1}{((i-1)\pi )^{N-2s}}\int _{1}^{\big ( \frac{(D)^{2}}{((i-1)\pi )^{2}} +1 \big )^{\frac{1}{2}}} (v^2-1)^{s-1}v^{1-N} dv \\ :=&\frac{1}{\Lambda ^{N-2s+1}}f_3(D), \end{aligned} \end{aligned}$$

(3.6)

and

$$\begin{aligned} \begin{aligned} f_2( D,\Lambda ) =&-\frac{A_1(N-2s)}{\Lambda ^{N-2s}D^{N-2s+1}} +A_2 - \frac{A_3(N-2s)}{\Lambda ^{N-2s}} \\ {}&\times \sum _{i=2}^{+\infty }\frac{1}{((i-1)\pi )^{N}}\big ( \frac{D^{2}}{((i-1)\pi )^{2}} +1 \big )^{\frac{-N}{2}}D^{2s-1}. \end{aligned} \end{aligned}$$

(3.7)

Since $\displaystyle \lim _{D\rightarrow 0}f_3(D) =-\infty$ and $\displaystyle \lim _{D\rightarrow +\infty }f_3(D) > 0$, there is a positive constant $D_0$, such that $f_3(D_0) =0$. So for any $\Lambda >0$, $f_1(D_0,\Lambda ) =0$.

On the other hand, when $\Lambda _0\! =\! \big (\frac{\frac{A_1(N-2s)}{D_0^{N-2s+1}} +A_3(N-2s)\sum _{i=2}^{+\infty }\frac{1}{((i-1)\pi )^{N}}\big ( \frac{D_0^{2}}{((i-1)\pi )^{2}} +1 \big )^{\frac{-N}{2}}D_0^{2s-1} }{A_2}\big )^{\frac{1}{N-2s}}$, $f_2(D_0,\Lambda _0) = 0$. Moreover, it is easy to check that

$$\begin{aligned} \frac{\partial f_1(D_0,\Lambda _0)}{ \partial \Lambda } =0, \quad \frac{\partial f_1(D_0,\Lambda _0)}{ \partial d} >0, \end{aligned}$$

and

$$\begin{aligned} \frac{\partial f_2(D_0,\Lambda _0)}{ \partial \Lambda }>0. \end{aligned}$$

Thus the linear operator of $f_1=0$ and $f_2=0$ at $(D_0,\Lambda _0)$ is invertible. As a result, (3.4) and (3.5) have a solution near $( D_0,\Lambda _0)$. $\square$

Appendices

Appendix A. The estimate of the projection

In both appendices, we always assume that

$$\begin{aligned} x_j=\left( r\cos \frac{2(j-1)\pi }{k},r\sin \frac{2(j-1)\pi }{k},0\right) ,\hbox { }j=1,...,k, \end{aligned}$$

where 0 is the zero vector in ${\mathbb {R}}^{N-2}$, and $r\in [ \mu (1-\frac{r_0}{k}), \mu (1-\frac{r_1}{k}) ]$. Recall that ${\bar{x}}_j=\frac{1}{\mu }x_j.$ G(x, y) be the Green function of $(-\Delta )^{s}$ in $B_{1}(0)$ with homogenous Dirichlet boundary condition. H(x, y) be the part of the Green function. $PU_{x,\Lambda }$ is the solution of (1.4) and $r_3 =\min ( r_0,1)$.

For $l=1,..,N$, we denote $\partial _l PU_{x,\Lambda }(y) = \frac{\partial PU_{x,\Lambda } (y) }{\partial x_l}$, $\partial _l U_{x,\Lambda }(y) = \frac{\partial U_{x,\Lambda } (y) }{\partial x_l}$, and $\partial _l H(x,y) = \frac{\partial H }{\partial x_l}(x,y)$. For $l=N+1$, we set $\partial _l PU_{x,\Lambda }(y) = \frac{\partial PU_{x,\Lambda } (y) }{\partial \Lambda }$, and $\partial _l U_{x,\Lambda }(y) = \frac{\partial U_{x,\Lambda } (y) }{\partial \Lambda }$.

Lemma A.1

Assume $N\ge 3$. For any $i=1,...,k$, if $y\in B_{{\frac{\mu r_3}{8k}}}(x_j)$ and $j\ne i$, then

$$\begin{aligned}&U_{x_i,\Lambda }(y) - PU_{x_i,\Lambda }(y) \nonumber \\&\quad = \frac{A_{N,s}H(\bar{x_i},\mu ^{-1}y)}{ \Lambda ^{\frac{N-2s}{2}}\mu ^{N-2s}} + O\left( \frac{1}{\mu ^{N}|\bar{x_i} -\mu ^{-1} y |^{N} } +\frac{k^{2s}}{\mu ^{N}|\bar{x_i} -\mu ^{-1}y |^{N-2s}}\right) , \end{aligned}$$

(A.1)

$$\begin{aligned}&\partial _l U_{x_i,\Lambda }(y) - \partial _l PU_{x_i,\Lambda }(y) \nonumber \\ {}&\quad = \frac{A_{N,s}\partial _l H(\bar{x_i},\mu ^{-1}y)}{\Lambda ^{\frac{N-2s}{2}}\mu ^{N-2s+1}} + O\left( \frac{1}{\mu ^{N+1}|\bar{x_i} -\mu ^{-1} y |^{N+1} } +\frac{k^{2s+1}}{\mu ^{N+1}|\bar{x_i} -\mu ^{-1}y |^{N-2s}}\right) , \nonumber \\ {}&\quad l=1,..,N, \end{aligned}$$

(A.2)

and

$$\begin{aligned} \begin{aligned}&\partial _{N+1} U_{x_i,\Lambda }(y) - \partial _{N+1} PU_{x_i,\Lambda }(y) \\ {}&\quad = \frac{-(N-2s)A_{N,s} H(\bar{x_i},\mu ^{-1}y)}{2\Lambda ^{\frac{N-2s+2}{2}}\mu ^{N-2s}} + O\left( \frac{1}{\mu ^{N}|\bar{x_i} -\mu ^{-1} y |^{N} } +\frac{k^{2s}}{\mu ^{N}|\bar{x_i} -\mu ^{-1}y |^{N-2s}}\right) , \end{aligned}\nonumber \\ \end{aligned}$$

(A.3)

where $A_{N,s}$ is a constant only depend on N and s. If $y\in B_{{\frac{\mu r_3}{8k}}}(x_i)$, then

$$\begin{aligned} U_{x_i,\Lambda }(y) - PU_{x_i,\Lambda }(y)= & {} \frac{A_{N,s}H(\bar{x_i},\mu ^{-1}y)}{\Lambda ^{\frac{N-2s}{2}}\mu ^{N-2s}} + O\left( \frac{k^{N}}{\mu ^{N}}\right) , \end{aligned}$$

(A.4)

$$\begin{aligned} \partial _l U_{x_i,\Lambda }(y) - \partial _l PU_{x_i,\Lambda }(y)= & {} \frac{A_{N,s}\partial _l H(\bar{x_i},\mu ^{-1}y)}{\Lambda ^{\frac{N-2s}{2}}\mu ^{N-2s+1}} + O(\frac{k^{N+1}}{\mu ^{N+1}}), \quad l=1,..,N,\qquad \end{aligned}$$

(A.5)

and

$$\begin{aligned} \partial _{N+1} U_{x_i,\Lambda }(y) - \partial _{N+1} PU_{x_i,\Lambda }(y) = \frac{-(N-2s)A_{N,s} H(\bar{x_i},\mu ^{-1}y)}{2\Lambda ^{\frac{N-2s+2}{2}} \mu ^{N-2s}} + O\left( \frac{k^{N}}{\mu ^{N}}\right) .\qquad \end{aligned}$$

(A.6)

Proof

Let ${\bar{PU}}_{x_i,\Lambda }(x) = \mu ^{\frac{N-2s}{2}}PU_{x_i,\Lambda }(\mu x)$, then

$$\begin{aligned} {\left\{ \begin{array}{ll} (-\Delta )^{s}{\bar{PU}}_{x_i,\Lambda }(x) = \big (\mu ^{\frac{N-2s}{2}}U_{x_i,\Lambda }(\mu x)\big )^{2^{*}_{s}-1}, &{}\hbox { in } B_{1}(0), \\ \displaystyle {\bar{PU}}_{x_i,\Lambda }(x) = 0,&{} x \in B^{c}_{1}(0). \end{array}\right. } \end{aligned}$$

(A.7)

So,

$$\begin{aligned} \begin{aligned}&\mu ^{\frac{N-2s}{2}}U_{x_i,\Lambda }(\mu y) - {\bar{PU}}_{x_i,\Lambda }(y) \\&\quad = \int _{{\mathbb {R}}^{N}}\frac{C_{1,N,s}}{|x-y|^{N-2s}}\frac{C_{2,N,s}(\Lambda \mu )^{\frac{N+2s}{2}}}{ (1 + (\Lambda \mu |x-\bar{x_i}|)^{2})^{ \frac{N+2s}{2}} }\\&\qquad - \int _{B_{1}(0)}G(x,y) \frac{C_{2,N,s}(\Lambda \mu )^{\frac{N+2s}{2}}}{ (1 + (\Lambda \mu |x-\bar{x_i}|)^{2})^{ \frac{N+2s}{2}} } \\&\quad = \int _{{\mathbb {R}}^{N}\setminus B_{1}(0)}\frac{C_{1,N,s}}{|x-y|^{N-2s}}\frac{C_{2,N,s}(\Lambda \mu )^{\frac{N+2s}{2}}}{ (1 + (\Lambda \mu |x-\bar{x_i}|)^{2})^{ \frac{N+2s}{2}} } \\ {}&\qquad + \int _{B_{1}(0)}H(x,y) \frac{C_{2,N,s}(\Lambda \mu )^{\frac{N+2s}{2}}}{ (1 + (\Lambda \mu |x-\bar{x_i}|)^{2})^{ \frac{N+2s}{2}} } \\ {}&\quad =I_1 +I_2. \end{aligned} \end{aligned}$$

(A.8)

Case 1: $y\in B_{\frac{r_3}{8k}}(\bar{x_i})$, it is easy to check

$$\begin{aligned} |I_1| = O\left( \frac{k^{N}}{\mu ^{\frac{N+2s}{2} } }\right) , \end{aligned}$$

(A.9)

and

$$\begin{aligned} I_2 =&\int _{B_{1}(0)\setminus B_{\frac{r_0}{2k}}(\bar{x_i}) } H(x,y) \frac{C_{2,N,s}(\Lambda \mu )^{\frac{N+2s}{2}}}{ (1 + (\Lambda \mu |x-\bar{x_i}|)^{2})^{ \frac{N+2s}{2}} } \nonumber \\ {}&+ \int _{ B_{\frac{r_0}{2k}(\bar{x_i})} }H(x,y) \frac{C_{2,N,s}(\Lambda \mu )^{\frac{N+2s}{2}}}{ (1 + (\Lambda \mu |x-\bar{x_i}|)^{2})^{ \frac{N+2s}{2}} } \nonumber \\ =&\int _{ B_{\frac{r_0}{2k}(\bar{x_i})} }H(\bar{x_i},y) \frac{C_{2,N,s}(\Lambda \mu )^{\frac{N+2s}{2}}}{ (1 + (\Lambda \mu |x-\bar{x_i}|)^{2})^{ \frac{N+2s}{2}} } \nonumber \\&+ O\left( \int _{ B_{\frac{r_0}{2k}(\bar{x_i})} }|\bar{x_{i}} -x|^2|\nabla ^2 H(\bar{x_i} +t(\bar{x_i}-x),y) |\frac{C_{2,N,s}(\Lambda \mu )^{\frac{N+2s}{2}}}{ (1 + (\Lambda \mu |x-\bar{x_i}|)^{2})^{ \frac{N+2s}{2}} }\right) \nonumber \\ {}&+O\left( \frac{k^{N}}{\mu ^{\frac{N+2s}{2} }}\right) \nonumber \\=&\frac{A_{N,s}H(\bar{x_i},y)}{\Lambda ^{\frac{N-2s}{2}}\mu ^{\frac{N-2s}{2}}} + O\left( \frac{k^{N}}{\mu ^{\frac{N+2s}{2} }}\right) . \end{aligned}$$

(A.10)

So,

$$\begin{aligned} \mu ^{\frac{N-2s}{2}}U_{x_i,\Lambda }(\mu y) - {\bar{PU}}_{x_i,\Lambda }(y) = \frac{A_{N,s}H(\bar{x_i},y)}{\Lambda ^{\frac{N-2s}{2}}\mu ^{\frac{N-2s}{2}}} + O(\frac{k^{N}}{\mu ^{\frac{N+2s}{2} }}). \end{aligned}$$

(A.11)

That is, if $y\in B_{{\frac{\mu r_3}{8k}}}({\bar{x}}_i)$, then

$$\begin{aligned} U_{x_i,\Lambda }(y) - PU_{x_i,\Lambda }(y) = \frac{A_{N,s}H(\bar{x_i},\mu ^{-1}y)}{\Lambda ^{\frac{N-2s}{2}}\mu ^{N-2s}} + O\left( \frac{k^{N}}{\mu ^{N}}\right) . \end{aligned}$$

Case 2: $y \in B_{{\frac{r_3}{8k}}}(\bar{x_j})$, where $j\ne i$. In this case, it is easy to check

$$\begin{aligned} I_1 =O\left( \frac{1}{\mu ^{\frac{N+2s}{2}}|\bar{x_i} -y |^{N} } +\frac{k^{2s}}{\mu ^{\frac{N+2s}{2}}|\bar{x_i} -y |^{N-2s}}\right) , \end{aligned}$$

and

$$\begin{aligned} I_2&=\int _{B_{1}(0)\setminus B_{\frac{|y-\bar{x_i}|}{2}}(\bar{x_i}) } H(x,y) \frac{C_{2,N,s}(\Lambda \mu )^{\frac{N+2s}{2}}}{ (1 + (\Lambda \mu |x-\bar{x_i}|)^{2})^{ \frac{N+2s}{2}} } \nonumber \\&\quad + \int _{ B_{\frac{|y-\bar{x_i}|}{2}(\bar{x_i}) } \cap B_{1}(0)}H(x,y) \frac{C_{2,N,s}(\Lambda \mu )^{\frac{N+2s}{2}}}{ (1 + (\Lambda \mu |x-\bar{x_i}|)^{2})^{ \frac{N+2s}{2}} } \nonumber \\&=\int _{ B_{\frac{r_3}{8k}(\bar{x_i})} }H(\bar{x_i},y) \frac{C_{2,N,s}(\Lambda \mu )^{\frac{N+2s}{2}}}{ (1 + (\Lambda \mu |x-\bar{x_i}|)^{2})^{ \frac{N+2s}{2}} } \nonumber \\&\quad + O( \int _{B_{\frac{|y-\bar{x_i}|}{2}(\bar{x_i})} }|\bar{x_{i}} -x|^2|\nabla ^2 H(\bar{x_i} +t(\bar{x_i}-x),y) |\frac{C_{2,N,s}(\Lambda \mu )^{\frac{N+2s}{2}}}{ (1 + (\Lambda \mu |x-\bar{x_i}|)^{2})^{ \frac{N+2s}{2}} }) \nonumber \\&\quad +O\left( \int _{ B_{\frac{|y-\bar{x_i}|}{2}(\bar{x_i}) } \setminus B_{\frac{r_3}{8k}(\bar{x_i})} } |H(x,y)| \frac{C_{2,N,s}(\Lambda \mu )^{\frac{N+2s}{2}}}{ (1 + (\Lambda \mu |x-\bar{x_i}|)^{2})^{ \frac{N+2s}{2}} }\right) \nonumber \\&\quad +O\left( \frac{1}{\mu ^{\frac{N+2s}{2}}|\bar{x_i} -y |^{N} }\right) \nonumber \\&=\frac{A_{N,s}H(\bar{x_i},y)}{\Lambda ^{\frac{N-2s}{2} }\mu ^{\frac{N-2s}{2}}} + O(\frac{1}{\mu ^{\frac{N+2s}{2}}|\bar{x_i} -y |^{N} }+\frac{k^{2s}}{\mu ^{\frac{N+2s}{2}}|\bar{x_i} -y |^{N-2s}}). \end{aligned}$$

(A.12)

So,

$$\begin{aligned} \begin{aligned}&\mu ^{\frac{N-2s}{2}}U_{x_i,\Lambda }(\mu y) - {\bar{PU}}_{x_i,\Lambda }(y) \\ {}&\quad = \frac{A_{N,s}H(\bar{x_i},y)}{\Lambda ^{\frac{N-2s}{2} }\mu ^{\frac{N-2s}{2}}} + O \left( \frac{1}{\mu ^{\frac{N+2s}{2}}|\bar{x_i} -y |^{N} }+\frac{k^{2s}}{\mu ^{\frac{N+2s}{2}}|\bar{x_i} -y |^{N-2s}}\right) . \end{aligned} \end{aligned}$$

(A.13)

That is, if $y\in B_{{\frac{\mu \min (r_0,1)}{8k}}}(x_j)$, then

$$\begin{aligned}&U_{x_i,\Lambda }(y) - PU_{x_i,\Lambda }(y) \nonumber \\&\quad = \frac{A_{N,s}H(\bar{x_i},\mu ^{-1}y)}{ \Lambda ^{\frac{N-2s}{2}}\mu ^{N-2s}} + O \left( \frac{1}{\mu ^{N}|\bar{x_i} -\mu ^{-1} y |^{N} } +\frac{k^{2s}}{\mu ^{N}|\bar{x_i} -\mu ^{-1}y |^{N-2s}}\right) . \end{aligned}$$

(A.14)

Since for $l=1,...,N+1$,

$$\begin{aligned} {\left\{ \begin{array}{ll} (-\Delta )^{s}\partial _l PU_{x_i,\Lambda }(x) = \partial _l (U_{x_i,\Lambda }(\mu x))^{2^{*}_{s}-1}, &{}\hbox { in } B_{1}(0), \\ \displaystyle {\bar{PU}}_{x_i,\Lambda }(x) = 0, &{}x \in B^{c}_{1}(0). \end{array}\right. } \end{aligned}$$

(A.15)

Similar to (A.1) and (A.2), we can prove (A.3)-(A.6). $\square$

Appendix B. Energy expansion

In this section, we will give the expansion of the energy $I(W_{r,\Lambda })$. Recall that $\mu =k^{\frac{N-2s+1}{N-2s}}$, $d=1-\frac{r}{\mu }$,

$$\begin{aligned} I(u)&=\frac{1}{2}\int _{B_{\mu }(0)}|(-\Delta )^{\frac{s}{2}} u|^{2} -\frac{1}{2^{*}_s}\int _{B_{\mu }(0)}K\left( \frac{|y|}{\mu }\right) |u|^{2^{*}_s},\\ U_{x_j,\Lambda }(y)&= C_{N,s}\frac{\Lambda ^{\frac{N-2s}{2}}}{(1+\Lambda ^{2}|y-x_j|^{2})^{\frac{N-2s}{2}}}, \end{aligned}$$

and

$$\begin{aligned} W_{r,\Lambda }(y)=\sum _{j=1}^{k}PU_{x_j,\Lambda }(y), \end{aligned}$$

where $PU_{x_j,\Lambda }(y)$ is the solution of (1.4).

Proposition B.1

If $N\ge 3$ for $\frac{3}{4} \le s<1$ and $3\le N < 2s -1 +\frac{2}{3-4s}$ for $\frac{11-\sqrt{41}}{8}< s<\frac{3}{4}$, then we have

$$\begin{aligned} I(W_{r,\Lambda })= & {} k\big (A+\frac{B_{1}H(\bar{x_1},\bar{x_1})}{\Lambda ^{N-2s}\mu ^{N-2s}} +B_2K'(1)d -\sum _{j=2}^{k}\frac{B_{1}G(\bar{x_j},\bar{x_1})}{\Lambda ^{N-2s}\mu ^{N-2s}} + O\big (\frac{1}{\mu ^{1+\sigma }}\big ) \big )\nonumber \\= & {} k\big ( A + \frac{A_1}{\Lambda ^{N-2s}\mu ^{N-2s}d^{N-2s}} +A_2 d -\frac{A_3k^{N-2s}}{\Lambda ^{N-2s}\mu ^{N-2s}} \nonumber \\&\times \sum _{i=2}^{+\infty }\frac{1}{((i-1)\pi )^{N-2s}}\int _{1}^{\big ( \frac{(dk)^{2}}{((i-1)\pi )^{2}} \!+\!1 \big )^{\frac{1}{2}}} (v^2-1)^{s-1}v^{1-N} dv \!+\! O\big (\frac{1}{\mu ^{1+\sigma }}\big )\big ),\qquad \nonumber \\ \end{aligned}$$

(B.1)

where A, $A_1$, $A_2$, $A_3$, $B_1$ and $B_2$ are positive constants.

Proof

By the symmetry, we have

$$\begin{aligned} \begin{aligned}&\int _{B_{\mu }(0)}|(-\Delta )^{\frac{s}{2}} W_{r,\Lambda ]}|^{2}\\&\quad =\sum _{j=1}^{k}\sum _{i=1}^{k}\int _{B_{\mu }(0)}U_{x_j,\Lambda }^{2^{*}_s-1}PU_{x_i,\Lambda } \\ {}&\quad = k\big ( \int _{B_{\mu }(0)} U_{x_1,\Lambda }^{2^{*}_s} -\int _{B_{\mu }(0)} U_{x_1,\Lambda }^{2^{*}_s-1}(U_{x_1,\Lambda }-PU_{x_1,\Lambda } ) +\sum _{i=2}^{k}\int _{B_{\mu }(0)} U_{x_1,\Lambda }^{2^{*}_s-1}PU_{x_i,\Lambda } \big ) \\ {}&\quad = k\big ( \int _{{\mathbb {R}}^{N}} U_{x_1,\Lambda }^{2^{*}_s} - \frac{\bar{B_1}H(\bar{x_1}, \bar{x_1} )}{(\Lambda \mu )^{N-2s}} + \frac{\bar{B_1}\sum _{i=2}^{k}G( \bar{x_i}, \bar{x_1} )}{(\Lambda \mu )^{N-2s}} +O\left( \frac{1}{\mu ^{1+\sigma }}\right) \big ), \end{aligned}\nonumber \\ \end{aligned}$$

(B.2)

where $\bar{B_1} =\displaystyle \int _{{\mathbb {R}}^{N}} U_{x_1,\Lambda }^{2^{*}_s-1}.$

Let

$$\begin{aligned} \Omega _j =\left\{ y:\hbox { }y=(y',y'')\in B_{\mu }(0),\hbox { } \left\langle \frac{y'}{|y'|},\frac{x_j}{|x_j|} \right\rangle \ge \cos \frac{\pi }{k} \right\} . \end{aligned}$$

Then,

$$\begin{aligned} \begin{aligned}&\int _{B_{\mu }(0)}K\big (\frac{|y|}{\mu }\big ) |W_{r,\Lambda }|^{2^{*}_s}\\&\quad =k\int _{\Omega _1} K\big (\frac{|y|}{\mu }\big )|W_{r,\Lambda }|^{2^{*}_s} \\ {}&\quad =k\big ( \int _{\Omega _1}K\big (\frac{|y|}{\mu }\big ) (PU_{x_1,\Lambda })^{2^{*}_s} -2^{*}_s \int _{\Omega _1}\sum _{i=2}^{k}(PU_{x_{1,\Lambda }})^{2^{*}_s-1}PU_{x_i,\Lambda } \\ {}&\qquad +O\big (\int _{\Omega _1}\big | K\big (\frac{|y|}{\mu }\big )-1 \big | \sum _{i=2}^{k}U_{x_{1,\Lambda }}^{2^{*}_s-1}U_{x_i,\Lambda } \\ {}&\qquad + \int _{\Omega _1} (U_{x_{1,\Lambda }})^{2^{*}_s-2} \big (\sum _{i=2}^{k}U_{x_i,\Lambda }\big )^{2} + \int _{\Omega _1} \big (\sum _{i=2}^{k}U_{x_i,\Lambda }^{2^{*}_s} \big ) \big ). \end{aligned} \end{aligned}$$

(B.3)

Note that for $y \in \Omega _1$, $|y-x_i|\ge |y-x_1|$. Thus

$$\begin{aligned} \begin{aligned} (U_{x_{1,\Lambda }})^{2^{*}_s-2}\left( \sum _{i=2}^{k}U_{x_i,\Lambda }\right) ^{2} \le \frac{C}{(1+|y-x_1|)^{N+\theta }}\sum _{i=2}^{k}\frac{1}{|x_i-x_1|^{N-\theta }}, \end{aligned} \end{aligned}$$

(B.4)

and

$$\begin{aligned} \begin{aligned} \left( \sum _{i=2}^{k}U_{x_i,\Lambda }\right) ^{2^{*}_s} \le \frac{C}{(1+|y-x_1|)^{N+\theta }}\sum _{i=2}^{k}\frac{1}{|x_i-x_1|^{N-\theta }}. \end{aligned} \end{aligned}$$

(B.5)

where $\theta$ is a small positive constant. We can chose $\theta$ small enough, then

$$\begin{aligned} \int _{\Omega _1} (U_{x_{1,\Lambda }})^{2^{*}_s-2}\big (\sum _{i=2}^{k}U_{x_i,\Lambda }\big )^{2} + \int _{\Omega _1} \big (\sum _{i=2}^{k}U_{x_i,\Lambda }\big )^{2^{*}_s} = O\big (\big (\frac{k}{\mu }\big )^{N-\theta }\big ) =O\big ( \frac{1}{\mu ^{1+\sigma }}\big ).\qquad \end{aligned}$$

(B.6)

On the other hand, it is easy to show that

$$\begin{aligned} \begin{aligned} \int _{\Omega _1}\sum _{i=2}^{k}(PU_{x_{1,\Lambda }})^{2^{*}_s-1}PU_{x_i,\Lambda }&= \sum _{i=2}^{k}\frac{{\bar{B}}_1G(\bar{x_i},\bar{x_1} )}{(\Lambda \mu )^{N-2s}} +O\big (\big (\frac{k}{\mu }\big )^{N}\big ) \\ {}&=\sum _{i=2}^{k}\frac{{\bar{B}}_1G(\bar{x_i},\bar{x_1} )}{(\Lambda \mu )^{N-2s}}+O\big ( \frac{1}{\mu ^{1+\sigma }}\big ), \end{aligned} \end{aligned}$$

(B.7)

and

$$\begin{aligned} \begin{aligned} \int _{\Omega _1}| K\big (\frac{|y|}{\mu }\big )-1 | \sum _{i=2}^{k}U_{x_{1,\Lambda }}^{2^{*}_s-1}U_{x_i,\Lambda } = O\big ( \frac{1}{\mu ^{1+\sigma }}\big ). \end{aligned} \end{aligned}$$

(B.8)

Moreover

$$\begin{aligned} \begin{aligned}&\int _{\Omega _1}K\big (\frac{|y|}{\mu }\big )\big (PU_{x_1,\Lambda }\big )^{2^{*}_s} \\&\quad = \int _{B_{\frac{r_3 \mu }{8k}}(x_2)}K\big (\frac{|y|}{\mu }\big )\big (PU_{x_1,\Lambda }\big )^{2^{*}_s} + O\big ( \frac{1}{\mu ^{1+\sigma }}\big ) \\&\quad =\int _{B_{\frac{r_3 \mu }{8k}}(x_2)}(PU_{x_1,\Lambda })^{ 2^{*}_s }+ \int _{B_{\frac{r_3 \mu }{8k}}(x_2)}\big (K\big (\frac{|y|}{\mu }\big ) -1 \big )U_{x_1,\Lambda }^{ 2^{*}_s } \\ {}&\qquad +O\big ( \int _{B_{\frac{r_3 \mu }{8k}}(x_2)} \big |K\big (\frac{|y|}{\mu }\big ) -1\big |U_{x_1,\Lambda }^{ 2^{*}_s -1}\frac{H({\bar{x}}_{1},\mu ^{-1}y)}{\mu ^{N-2s}} \big ) + O\big ( \frac{1}{\mu ^{1+\sigma }}\big ) \\ {}&\quad =\int _{{\mathbb {R}}^{N}}U_{0,1}^{ 2^{*}_s } -2^{*}_s\frac{{\bar{B}}_1H({\bar{x}}_1,{\bar{x}}_1)}{ (\Lambda \mu )^{N-2s} } + \int _{B_{\frac{r_3 \mu }{8k}}(x_2)}\big (K\big (\frac{|y|}{\mu }\big ) -1 \big )U_{x_1,\Lambda }^{ 2^{*}_s } +O\big ( \frac{1}{\mu ^{1+\sigma }}\big ). \end{aligned} \end{aligned}$$

(B.9)

However,

$$\begin{aligned} \begin{aligned} \int _{B_{\frac{r_3 \mu }{8k}}(x_2)}\big (K\big (\frac{|y|}{\mu }\big ) -1 \big )U_{x_1,\Lambda }^{ 2^{*}_s }&= \big ( K(|{\bar{x}}_1|) -1 \big )\int _{{\mathbb {R}}^{N}}U_{0,1}^{ 2^{*}_s } + O\big (\frac{1}{k^{2}}\big ) \\ {}&= -K'(1)d\int _{{\mathbb {R}}^{N}}U_{0,1}^{ 2^{*}_s } + O\big ( \frac{1}{\mu ^{1+\sigma }}\big ). \end{aligned} \end{aligned}$$

(B.10)

Thus, we have proved

$$\begin{aligned} \begin{aligned} \int _{B_{\mu }(0)} K\big (\frac{|y|}{\mu }\big ) |W_{r,\Lambda }|^{2^{*}_s}&= k\big (\int _{{\mathbb {R}}^{N}}U_{0,1}^{ 2^{*}_s } -K'(1)d\int _{{\mathbb {R}}^{N}}U_{0,1}^{ 2^{*}_s } - 2^{*}_s \frac{{\bar{B}}_1H({\bar{x}}_1,{\bar{x}}_1)}{ (\Lambda \mu )^{N-2s} } \\&\quad + \sum _{i=2}^{k}\frac{{\bar{B}}_1G({\bar{x}}_i,{\bar{x}}_1)}{ (\Lambda \mu )^{N-2s} } + O\big ( \frac{1}{\mu ^{1+\sigma }}\big ) \big ). \end{aligned} \end{aligned}$$

(B.11)

Combining (B.2) and (B.11), we can get

$$\begin{aligned} \begin{aligned} I(W_{r,\Lambda })=k\big ( A+\frac{B_{1}H(\bar{x_1},\bar{x_1})}{\Lambda ^{N-2s}\mu ^{N-2s}} +B_2K'(1)d -\sum _{i=2}^{k}\frac{B_{1}G(\bar{x_i},\bar{x_1})}{\Lambda ^{N-2s}\mu ^{N-2s}} + O(\frac{1}{\mu ^{1+\sigma }})\big ), \end{aligned}\nonumber \\ \end{aligned}$$

(B.12)

where A, $B_1$ and $B_2$ are positive constants.

Now, we estimate $H( {\bar{x}}_1,{\bar{x}}_1)$ and $G( {\bar{x}}_i,{\bar{x}}_1)$, $i\ge 2$. Let ${\bar{x}}_1^{*} = (\frac{1}{1-d},0,...,0 )$ be the reflection of ${\bar{x}}_1$ withe respect to the unit sphere. Then

$$\begin{aligned} H( {\bar{x}}_1,{\bar{x}}_1) = \frac{1}{2^{N-2s}d^{N-2s}}\big (1+O(d)\big ). \end{aligned}$$

(B.13)

We can compute that,

$$\begin{aligned}&\sum _{i=2}^{k} G( {\bar{x}}_i,{\bar{x}}_1) \nonumber \\&\quad = C\sum _{i=2}^{[\frac{k}{2}]}\frac{1}{\big (2( 1-d )\sin \frac{(i-1)\pi }{k} \big )^{N-2s}} \int _{1}^{ \big ( \frac{ d^{2}(2-d)^2 }{ \big ( 2( 1-d )\sin \frac{(i-1)\pi }{k} \big )^{2} } +1 \big )^{\frac{1}{2}} } (v^2-1)^{s-1}v^{1-N} dv\nonumber \\&\qquad +O(k^{N-2s-1})\nonumber \\&\quad = Ck^{N-2s}\sum _{i=2}^{+\infty }\frac{1}{2^{N-2s}((i-1)\pi )^{N-2s}}\int _{1}^{\big ( \frac{(dk)^{2}}{((i-1)\pi )^{2}} +1 \big )^{\frac{1}{2}}} (v^2-1)^{s-1}v^{1-N} dv \nonumber \\&\qquad + O(k^{N-2s-1}). \end{aligned}$$

(B.14)

In fact, for $i\le k^{\alpha }$, where $\alpha \in (\frac{1}{N-2s-1},1)$ is a fix constant, we have

$$\begin{aligned} \begin{aligned}&\frac{1}{\big (2( 1-d )\sin \frac{(i-1)\pi }{k} \big )^{N-2s}}\\&\quad = \frac{k^{N-2s}}{ (1-d)^{N-2s}2^{N-2s}((i-1)\pi )^{N-2s}}+ {\bar{O}}\big (\frac{k^{N-2s-2}}{(i-1)^{N-2s-2}}\big ) \\ {}&\quad =\frac{k^{N-2s}}{ 2^{N-2s}((i-1)\pi )^{N-2s}}+ {\bar{O}}\big (\frac{k^{N-2s-2}}{(i-1)^{N-2s-2}}\big ) +{\bar{O}}\big (\frac{k^{N-2s-1}}{(i-1)^{N-2s}}\big ) , \end{aligned} \end{aligned}$$

(B.15)

where ${\bar{O}}(f(i,k))$ means that, there is a constant C and $k_0$ , for any $k >k_0$ and any $2 \le i \le k^{\alpha }$,

$$\begin{aligned} | {\bar{O}}(f(i,k)) | \le C |f(i,k)|. \end{aligned}$$

On the other hand,

$$\begin{aligned}&\int _{1}^{ \big ( \frac{ d^{2}(2-d)^2 }{ \big ( 2( 1-d )\sin \frac{(i-1)\pi }{k} \big )^{2} } + \big )^{\frac{1}{2}} } (v^2-1)^{s-1}v^{1-N} dv \\ {}&\quad =\int _{1}^{ \big ( \frac{ d^{2}(2-d)^2 }{ \big ( 2( 1-d ) \big (\frac{(i-1)\pi }{k} + {\bar{O}}\big (\frac{(i-1)^3)}{k^3}\big ) \big ) \big )^{2} } +1 \big )^{\frac{1}{2}} } (v^2-1)^{s-1}v^{1-N} dv \\ {}&\quad =\int _{1}^{ \big ( \frac{ d^{2}(2-d)^2 }{ \big ( \big (2( 1-d ) \big (\frac{(i-1)\pi }{k}\big )^2 + {\bar{O}} \big (\frac{(i-1)^4)}{k^4}\big ) \big ) \big ) } +1 \big )^{\frac{1}{2}} } (v^2-1)^{s-1}v^{1-N} dv \\ {}&\quad =\int _{1}^{ \big ( \frac{ d^{2} }{ \big (\frac{(i-1)\pi }{k})^2 \big ) } +1 \big )^{\frac{1}{2}} } (v^2-1)^{s-1}v^{1-N} dv +{\bar{O}}\big (\frac{(i-1)^{2(1-s)}}{k^2}\big ) +{\bar{O}}\big (\frac{1}{(i-1)^{2s}k} \big ). \end{aligned}$$

Noting $N \ge 3$, by direct computation, we have

$$\begin{aligned} \begin{aligned}&\sum _{i=2}^{[k^{\alpha }]}\frac{1}{\big (2( 1-d )\sin \frac{(i-1)\pi }{k} \big )^{N-2s}} \int _{1}^{ \big ( \frac{ d^{2}(2-d)^2 }{ \big ( 2( 1-d )\sin \frac{(i-1)\pi }{k} \big )^{2} } +1 \big )^{\frac{1}{2}} } (v^2-1)^{s-1}v^{1-N} dv \\ {}&\quad =k^{N-2s}\sum _{i=2}^{+\infty }\frac{1}{2^{N-2s}((i-1)\pi )^{N-2s}}\int _{1}^{\big ( \frac{(dk)^{2}}{((i-1)\pi )^{2}} +1 \big )^{\frac{1}{2}}} (v^2-1)^{s-1}v^{1-N} dv + O(k^{N-2s-1}) \\ {}&\qquad +O\big (\sum _{i=2}^{[k^{\alpha }]}\frac{1}{(i-1)^{N-2s}}\big ) \\ {}&\quad =k^{N-2s}\sum _{i=2}^{+\infty }\frac{1}{2^{N-2s}((i-1)\pi )^{N-2s}}\int _{1}^{\big ( \frac{(dk)^{2}}{((i-1)\pi )^{2}} +1 \big )^{\frac{1}{2}}} (v^2-1)^{s-1}v^{1-N} dv + O(k^{N-2s-1}) \\ {}&\qquad +O(k^{(N-2s)-\alpha (N-2s-1)}) \\ {}&\quad =k^{N-2s}\sum _{i=2}^{+\infty }\frac{1}{2^{N-2s}((i-1)\pi )^{N-2s}}\int _{1}^{\big ( \frac{(dk)^{2}}{((i-1)\pi )^{2}} +1 \big )^{\frac{1}{2}}} (v^2-1)^{s-1}v^{1-N} dv + O(k^{N-2s-1}). \end{aligned} \end{aligned}$$

(B.16)

Since

$$\begin{aligned} \sin \big (\frac{i-1}{k}\big ) \ge C|\frac{i-1}{k}|,\hbox { for } i=1,...,\big [\frac{k}{2}\big ], \end{aligned}$$

and

$$\begin{aligned} \int _1^{+\infty }(v^2-1)^{s-1}v^{1-N} dv < +\infty , \end{aligned}$$

it is easy to check

$$\begin{aligned}&\sum _{[k^{\alpha }]+1}^{\big [\frac{k}{2}\big ]}\frac{1}{\big (2( 1-d )\sin \frac{(i-1)\pi }{k} \big )^{N-2s}} \int _{1}^{ \big ( \frac{ d^{2}(2-d)^2 }{ \big ( 2( 1-d )\sin \frac{(i-1)\pi }{k} \big )^{2} } +1 \big )^{\frac{1}{2}} } (v^2-1)^{s-1}v^{1-N} dv\nonumber \\&\quad = O(k^{(N-2s)-\alpha (N-2s-1)}) =O(k^{N-2s-1}). \end{aligned}$$

(B.17)

Combining (B.16) and (B.17), we have (B.14).

Finally, combining (B.12), (B.13) and (B.14), we get (B.1).$\square$

Proposition B.2

If $N\ge 3$ for $\frac{3}{4} \le s<1$ and $3\le N < 2s -1 +\frac{2}{3-4s}$ for $\frac{11-\sqrt{41}}{8}< s<\frac{3}{4}$, then we have

$$\begin{aligned} \begin{aligned} \frac{\partial I(W_{r,\Lambda })}{\partial \Lambda }&=kB_1(N-2s)\big ( -\frac{ H(\bar{x_1},\bar{x_1})}{\Lambda ^{N-2s+1}\mu ^{N-2s}} +\sum _{j=2}^{k}\frac{ G(\bar{x_j},\bar{x_1})}{\Lambda ^{N-2s+1}\mu ^{N-2s}} + O\big (\frac{1}{\mu ^{1+\sigma }}\big )\big ) \\&= k \big ( -\frac{A_1(N-2s)}{\Lambda ^{N-2s+1}\mu ^{N-2s}d^{N-2s}} + \frac{A_3k^{N-2s}(N-2s)}{\Lambda ^{N-2s+1}\mu ^{N-2s}} \\&\quad \times \sum _{i=2}^{+\infty }\frac{1}{((i-1)\pi )^{N-2s}}\int _{1}^{\big ( \frac{(dk)^{2}}{((i-1)\pi )^{2}} +1 \big )^{\frac{1}{2}}} (v^2-1)^{s-1}v^{1-N} dv + O\big (\frac{1}{\mu ^{1+\sigma }}\big )\big ) , \end{aligned} \end{aligned}$$

(B.18)

and

$$\begin{aligned} \begin{aligned} \frac{\partial I(W_{r,\Lambda })}{\partial d}&=k\big ( \frac{B_{1}\frac{\partial H(\bar{x_1},\bar{x_1})}{ \partial d}}{\Lambda ^{N-2s}\mu ^{N-2s}} +B_2K'(1)-\sum _{j=2}^{k}\frac{B_1\frac{\partial G(\bar{x_j},\bar{x_1})}{ \partial d}}{{\Lambda ^{N-2s}\mu ^{N-2s}} }+ O(\frac{1}{\mu ^{\sigma }})\big ) \\&= k \big ( -\frac{A_1(N-2s)}{\Lambda ^{N-2s}\mu ^{N-2s}d^{N-2s+1}} +A_2 - \frac{A_3k^{N-2s}(N-2s)}{\Lambda ^{N-2s}\mu ^{N-2s}} \\&\quad \times \sum _{i=2}^{+\infty }\frac{1}{((i-1)\pi )^{N}}\big ( \frac{(dk)^{2}}{((i-1)\pi )^{2}} +1 \big )^{\frac{-N}{2}}dk^{2}(dk)^{2(s-1)} + O\big (\frac{1}{\mu ^{\sigma }} \big )\big ). \end{aligned} \end{aligned}$$

(B.19)

where $A_1$, $A_2$, $A_3$, $B_1$ and $B_2$ are the same positive constants as in Proposition B.1.

Proof

We use $\partial$ to denote either $\frac{\partial }{\partial \Lambda }$ or $\frac{\partial }{\partial d}$. Using the symmetry, we have

$$\begin{aligned} \partial I(W_{r,\Lambda })&= k\big (\frac{2^{*}_s-1}{2}\int _{B_{\mu }(0)}\sum _{i=1}^{k}U_{x_1,\Lambda }^{ 2^{*}_s-2 }\partial U_{x_1,\Lambda } PU_{x_i,\Lambda } + \frac{1}{2}\int _{B_{\mu }(0)}\sum _{i=1}^{k}U_{x_1,\Lambda }^{ 2^{*}_s-1 }\partial PU_{x_i,\Lambda } \\&\quad - \int _{\Omega _1}K\big (\frac{|y|}{\mu }\big ) W_{r,\Lambda }^{ 2^{*}_s-1} \partial W_{r,\Lambda } \big ). \end{aligned}$$

Then the proof of this proposition is similar to the proof of Proposition B.1, so we omit it. $\square$

Appendix C Basic estimates

In this section, we list some lemmas, whose proof can be found in [9] and [22].

For each fixed i and j, $i \ne j$, consider the function

$$\begin{aligned} g_{i,j}(y) = \frac{1}{(1+|y-x_i|)^{\alpha }}\frac{1}{(1+|y-x_j|)^{\beta }}, \end{aligned}$$

(C.1)

where $\alpha \ge 1$ and $\beta \ge 1$ are constants.

Lemma C.1

For any constant $0< \sigma <\min (\alpha ,\beta )$, there is a constant $C>0$, such that

$$\begin{aligned} g_{i,j}(y)\le \frac{C}{|x_i-x_j|^{\sigma }}\left( \frac{1}{(1+|y-x_i|)^{\alpha +\beta -\sigma }} +\frac{1}{(1+|y-x_j|)^{\alpha +\beta -\sigma }} \right) . \end{aligned}$$

Lemma C.2

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

$$\begin{aligned} \int _{{\mathbb {R}}^{N}}\frac{1}{|y-z|^{N-2s}}\frac{1}{(1+|z|)^{2s+\sigma }}dz \le \frac{C}{(1+|y|)^{\sigma }}. \end{aligned}$$

Lemma C.3

Suppose that $N\ge 3$, then there is a small $\theta >0$, such that

$$\begin{aligned}&\int _{{\mathbb {R}}^{N}}\frac{1}{|y-z|^{N-2s}}W^{2^*_s-2}_{r,\Lambda }(z)\sum _{j=1}^k \frac{1}{(1+|z-x_j|)^{\frac{N-2s}{2}+\tau }}dz \\&\quad \le C\sum _{j=1}^k\frac{1}{(1+|y-x_j|)^{\frac{N+2s}{2}+\tau +\theta }}, \end{aligned}$$

where $W_{r,\Lambda } = \sum _{j=1}^k PU_{x_j,\Lambda }.$