# Concentrated solutions to fractional Schrödinger equations with prescribed
$L^{2}$-norm

Qing Guo, Peng Luo, Chunhua Wang and Jing Yang  College of Science, Minzu
University of China, Beijing 100081, China<EMAIL_ADDRESS>School of
Mathematics and Statistics, Central China Normal University, Wuhan 430079,
China<EMAIL_ADDRESS>School of Mathematics and Statistics and Hubei
Key Laboratory of Mathematical Sciences, Central China Normal University,
Wuhan 430079, China<EMAIL_ADDRESS>School of Science, Jiangsu
University of Science and Technology, Zhenjiang 212003, China
<EMAIL_ADDRESS>
###### Abstract.

In this work, we investigate the existence and local uniqueness of normalized
$k$-peak solutions for the fractional Schrödinger equations with attractive
interactions with a class of degenerated trapping potential with non-isolated
critical points.

Precisely, applying the finite dimensional reduction method, we first obtain
the existence of $k$-peak concentrated solutions and especially describe the
relationship between the chemical potential $\mu$ and the attractive
interaction $a$. Second, after precise analysis of the concentrated points and
the Lagrange multiplier, we prove the local uniqueness of the $k$-peak
solutions with prescribed $L^{2}$-norm, by use of the local Pohozaev
identities, the blow-up analysis and the maximum principle associated to the
nonlocal operator $(-\Delta)^{s}$.

To our best knowledge, there is few results on the excited normalized
solutions of the fractional Schrödinger equations before this present work.
The main difficulty lies in the non-local property of the operator
$(-\Delta)^{s}$. First, it makes the standard comparison argument in the ODE
theory invalid to use in our analysis. Second, because of the algebraic decay
involving the approximate solutions, the estimates, on the Lagrange multiplier
for example, would become more subtle. Moreover, when studying the
corresponding harmonic extension problem, several local Pohozaev identities
are constructed and we have to estimate several kinds of integrals that never
appear in the classic local Schrödinger problems. In addition, throughout our
discussion, we need to distinct the different cases of $p-1<\frac{4s}{N}$,
$p-1=\frac{4s}{N}$, and $p-1>\frac{4s}{N}$, which are called respectively that
the mass-subcritical, the mass-critical, and the mass-supercritical case, due
to the mass-constraint condition. Another difficulty comes from the influence
of the different degenerate rates along different directions at the critical
points of the potential $V(x).$

###### Key words and phrases:

Normalized solutions, Multi-peak solutions, Fractional Schrödinger equations,
Degenerated trapping potential

##  1\. Introduction and main results

In this paper, we consider the following fractional Schrödinger equation

$\begin{cases}(-\Delta)^{s}u+V(x)u=au^{p}+\mu u,&{x\in\mathbb{R}^{N}},\\\ u\in
H^{s}(\mathbb{R}^{N}),\end{cases}$ (1.1)

under the mass constraint

$\int_{\mathbb{R}^{N}}u^{2}(x)dx=1,$ (1.2)

where $s\in(0,1)$, $p\in(1,2_{s}^{*}-1)$ with $2_{s}^{*}=\frac{2N}{N-2s}$ is
the fractional critical Sobolev exponent.

The fractional Laplacian operator, appearing in many areas including
biological modeling, physics and mathematical finances, can be regarded as the
infinitesimal generator of a stable Levy process [1]. For any $s\in(0,1)$,
$(-\Delta)^{s}$ is the nonlocal operator defined as

$\displaystyle(-\Delta)^{s}u=c(N,s)P.V.\int_{\mathbb{R}^{N}}\frac{u(x)-u(y)}{|x-y|^{N+2s}}dy,$

where $P.V.$ is the principal value and
$c(N,s)=\pi^{2s+\frac{N}{2}}\Gamma(s+\frac{N}{2})/\Gamma(-s)$. One could refer
to [12, 22] for more details on the fractional Laplacian operator.
Particularly, this nonlocal operator $(-\Delta)^{s}$ in $\mathbb{R}^{N}$ can
be expressed as a generalized Dirichlet-to-Neumann map for a certain elliptic
boundary value problem with local differential operators defined on the upper
half-space $\mathbb{R}^{N+1}_{+}=\\{(x,t):x\in\mathbb{R}^{N},t>0\\}$.
Precisely, by [7], for any $u\in\dot{H}^{s}(\mathbb{R}^{N})$, set

$\displaystyle\tilde{u}(x,t)=\mathcal{P}_{s}[u]=\int_{\mathbb{R}^{N}}\mathcal{P}_{s}(x-z,t)u(z)dz,\
\ (x,t)\in\mathbb{R}^{N+1}_{+},$

where

$\mathcal{P}_{s}(x,t)=\beta(N,s)\frac{t^{2s}}{(|x|^{2}+t^{2})^{\frac{N+2s}{2}}}$

with a constant $\beta(N,s)$ such that
$\displaystyle\int_{\mathbb{R}^{N}}\mathcal{P}_{s}(x,1)dx=1$. Then
$\tilde{u}\in L^{2}(t^{1-2s},K)$ for any compact set $K$ in
$\overline{\mathbb{R}^{N+1}_{+}}$, $\nabla\tilde{u}\in
L^{2}(t^{1-2s},\mathbb{R}^{N+1}_{+})$ and $\tilde{u}\in
C^{\infty}(\mathbb{R}^{N+1}_{+})$. Moreover, $\tilde{u}$ satisfies

$\begin{cases}div(t^{1-2s}\nabla\tilde{u})=0,&{x\in\mathbb{R}^{N+1}_{+}},\\\\[2.84526pt]
-\displaystyle\lim_{t\rightarrow
0}t^{1-2s}\partial_{t}\tilde{u}(x,t)=\omega_{s}(-\Delta
u)^{s}u(x),&{x\in\mathbb{R}^{N}},\end{cases}$

in the distribution sense, where $\omega_{s}=2^{1-2s}\Gamma(1-s)/\Gamma(s)$.
Moreover, it holds that

$\displaystyle\|\tilde{u}\|_{L^{2}(t^{1-2s},\mathbb{R}^{N+1}_{+})}=\omega_{s}\|u\|_{\dot{H}^{s}}.$

Without loss of generality, we may assume $\omega_{s}=1$.

Problems with fractional Laplacian have been extensively studied recently, see
for example [2, 3, 4, 5, 6, 7, 17, 20, 18, 19, 24, 23, 21, 31, 34, 35, 32, 33,
36, 38] and the references therein. In particular, the existence of multiple
spike solution involving the stable critical points of the potential $V(x)$
was considered in [15], where the critical points of $V(x)$ seem to be
isolated.

We first recall the well-known results about the ground state of the following
equation

$(-\Delta)^{s}u+u=u^{p},\ \ u>0,\ \ x\in\mathbb{R}^{N},\ \
u(0)=\max_{x\in\mathbb{R}^{N}}u(x).$ (1.3)

Let $N\geq 1,s\in(0,1)$ and $1<p<2^{*}_{s}-1$. Then the following hold
(c.f.[19, 18]).
(i)(Uniqueness) The ground state solution $U\in H^{s}(\mathbb{R}^{N})$ of
(1.3) is unique.
(ii)(Symmetry, regularity, and decay) $U(x)$ is radial, positive, and strictly
decreasing in $|x|$. Moreover, $U\in H^{2s+1}(\mathbb{R}^{N})\cap
C^{\infty}(\mathbb{R}^{N})$ and satisfies

$\frac{C_{1}}{1+|x|^{N+2s}}\leq U(x)\leq\frac{C_{2}}{1+|x|^{N+2s}},\ \
for\,\,x\in\mathbb{R}^{N},$

with some constants $C_{2}\geq C_{1}>0$.
(iii)(Non-degeneracy) The linearized operator $L_{0}=(-\Delta)^{s}+1-pU^{p-1}$
is non-degenerate, i.e. its kernel is given by

$kerL_{0}=span\big{\\{}\partial_{x_{1}}U,\partial_{x_{2}}U,\ldots,\partial_{x_{N}}U\big{\\}}.$

Moreover, By Lemma C.2 of [19], for $j=1,\ldots,N$, $\partial_{x_{j}}U$ has
the decay estimate

$|\partial_{x_{j}}U|\leq\frac{C}{1+|x|^{N+2s}}.$

Throughout this paper, we assume

$N\geq\max\\{2s,2-2s\\}$

and set

$\displaystyle a_{*}=\int_{\mathbb{R}^{N}}U^{2},\ \ \
a_{0}\equiv\begin{cases}+\infty,\,\,&\text{if}\,\,p-1<\frac{4s}{N},\vspace{0.12cm}\\\
(ka_{*})^{\frac{p-1}{2}},\,\,&\text{if}\,\,p-1=\frac{4s}{N},\vspace{0.12cm}\\\
0,\,\,&\text{if}\,p-1>\frac{4s}{N}.\end{cases}$ (1.4)

Our first result is as follows.

######  Theorem 1.1.

Suppose that $u_{a}$ satisfies (1.1)-(1.2) concentrated at some points as
$a\rightarrow a_{0}$. Then it holds that $\mu_{a}\rightarrow-\infty$ as
$a\rightarrow a_{0}$ .

Moreover, if $p-1\leq\frac{4s}{N}$, there holds that

$\displaystyle
u_{a}=\Big{(}\frac{-\mu_{a}}{a}\Big{)}^{\frac{1}{p-1}}\Big{(}\sum_{i=1}^{k}U\big{(}(-\mu_{a})^{\frac{1}{2s}}(x-x_{a,i})\big{)}+\omega_{a}\Big{)},$
(1.5)

with
$\displaystyle\int_{\mathbb{R}^{N}}\Big{(}-\frac{1}{\mu_{a}}|(-\Delta)^{\frac{s}{2}}\omega_{a}|^{2}+\omega_{a}^{2}\Big{)}=o\Big{(}(-\mu_{a})^{-\frac{N}{2s}}\Big{)}$.

We call $u_{a}$ a $k$-peak solution of (1.1)–(1.2) if $u_{a}$ satisfies (1.5).
The problem has been extensively investigated if the critical points of $V(x)$
are isolated, while few is known otherwise. In this paper, we assume that
$V(x)$ obtains its local minimum or local maximum at a closed $N-1$
dimensional hyper-surface $\Gamma_{i}(i=1,\ldots,k)$, satisfying
$\Gamma_{i}\cap\Gamma_{j}=\emptyset$ if $i\neq j$. Precisely, we assume that

(V). There exist some $\delta>0$ and some $C^{2}$ compact hyper surfaces
$\Gamma_{i}(i=1,\ldots,k)$ without boundary, satisfying

$V(x)=V_{i},\ \frac{\partial V(x)}{\partial\nu_{i}}=0,\
\frac{\partial^{2}V(x)}{\partial\nu_{i}^{2}}\neq 0,\ for\ any\ x\in\Gamma_{i}\
and\ i=1,\ldots,k,$

where $V_{i}\in\mathbb{R}$, $\nu_{i}$ is the unit outward normal of
$\Gamma_{i}$ at $x\in\Gamma_{i}$. Moreover, $V(x)\in
C^{4}(\cup_{i=1}^{k}W_{\delta,i})$ with
$W_{\delta,i}:=\\{x\in\mathbb{R}^{N}:dist(x,\Gamma_{i})<\delta\\}.$

######  Remark 1.2.

We point out that the assumption (V) was first introduced in [30], where they
studied the Bose-Einstein condensates. The assumption (V) implies that the
potential $V(x)$ obtains its local minimum or local maximum on the
hypersurface $\Gamma_{i}$ for $i=1,\ldots,k$. If $\delta>0$ is small, we set
$\Gamma_{t,i}=\\{x:V(x)=t\\}\cap W_{\delta,i}$, which consists of two compact
hypersurfaces in $\mathbb{R}^{N}$ without boundary for
$t\in[V_{0},V_{0}+\sigma)$ or $t\in(V_{0}-\sigma,V_{0}]$ provided $\sigma>0$
is small enough. Moreover, the outward unit normal vector $\nu_{t,i}(x)$ and
the $j-$th principal tangential unit vector $\tau_{t,i,j}(x),j=1,\ldots,N-1$
of $\Gamma_{t,i}$ at $x$ are Lip-continuous in $W_{\delta,i}$.

We can apply the Pohozaev identities to show that a $k$-peak solution of
(1.1)-(1.2) must concentrate at some critical points of $V(x)$. The following
result explains where the concentrated points locate on
$\Gamma=\cup_{i=1}^{k}\Gamma_{i}$.

######  Theorem 1.3.

Under the condition (V), if $u_{a}$ is a $k$-peak solution of (1.1)-(1.2),
concentrating at $\\{b_{1},b_{2},\ldots,b_{k}\\}$ with $b_{i}\in\Gamma$,
$b_{i}\neq b_{j}$ if $i\neq j$, and
$|x_{a,i}-\Gamma_{i}|=O\big{(}(-\mu_{a})^{-\frac{1}{2s}}\big{)}$ as
$a\rightarrow a_{0}$, then

$\displaystyle(D_{\tau_{i,j}}\Delta V)(b_{i})=0,\ \ with\ i=1,\ldots,k\ and\
j=1,\ldots,N-1,$ (1.6)

where $\tau_{i,j}$ is the $j-$th principal tangential unit vector of $\Gamma$
at $b_{i}$.

To study the converse of Theorem 1.3, we need another non-degenerate condition
on the critical point of $V(x)$. In fact, we define that $x_{0}\in\Gamma_{i}$
is non-degenerate on $\Gamma_{i}$ if there holds that

$\frac{\partial^{2}V(x_{0})}{\partial\nu_{i}^{2}}\neq 0,\ and\
det\left(\Big{(}\frac{\partial^{2}\Delta
V(x_{0})}{\partial\tau_{i,l}\partial\tau_{i,j}}\Big{)}_{1\leq l,j\leq
N-1}\right)\neq 0.$

######  Theorem 1.4.

Assume that the condition (V) holds. If $b_{i}\in\Gamma$ are non-degenerate
critical points of $V(x)$ on $\Gamma$ for $i=1,\ldots,k$ satisfying (1.6) and
$b_{i}\neq b_{j}$ for $i\neq j,$ then (1.1)–(1.2) has a $k$-peak solution
$u_{a}$ concentrating at $b_{1},\ldots,b_{k}$ as $a\rightarrow a_{0}$.

On the other hand, if we assume the function $\Delta
V(x)|_{x\in\Gamma_{i_{0}}}$ has an isolated maximum point $b\in\Gamma_{i_{0}}$
with some $i_{0}\in\\{1,\ldots,k\\}$, that is $\Delta V(x)<\Delta V(b)$ for
all $x\in\Gamma_{i_{0}}\cap\big{(}B_{\delta}(b)\setminus b\big{)}$, then we
can obtain a $k$-peak solution concentrating at one point.

######  Theorem 1.5.

Assume ($V$) and $\frac{\partial^{2}V(x)}{\partial\nu_{i_{0}}^{2}}\neq 0$ for
any $x\in\Gamma_{i_{0}}$ with some $i_{0}\in\\{1,\ldots,k\\}$. If
$b\in\Gamma_{i_{0}}$ is an isolated maximum point of $\Delta
V(x)|_{x\in\Gamma_{i_{0}}}$ on $\Gamma_{i_{0}}$, then for any integer $k>0$,
problem (1.1)–(1.2) has a $k$-peak solution $u_{a}$ concentrating at $b$.

Another main result of this paper is the following local uniqueness result.

######  Theorem 1.6.

Suppose ($V$), and if further $N\geq 2s+4$. Let $u_{a}^{(1)}(x)$ and
$u_{a}^{(2)}(x)$ be two $k$-peak solutions of (1.1)–(1.2) concentrating at
$b_{1},\cdots,b_{k}$ with $b_{i}\in\Gamma$, and $b_{i}\neq b_{j}$ if $i\neq
j$. If $b_{i}$ is non-degenerate, $i=1,\cdots,k$,
$\displaystyle\sum^{k}_{i=1}\Delta V(b_{i})\neq 0$ when $p-1=\frac{4s}{N}$,
and

$\Big{(}\frac{\partial^{2}\Delta
V(b_{i})}{\partial\tau_{i,l}\partial\tau_{i,j}}\Big{)}_{1\leq l,j\leq
N-1}+\frac{\partial\Delta
V(b_{i})}{\partial\nu_{i}}diag\big{(}\kappa_{i,1},\cdots,\kappa_{i,N-1}\big{)},~{}\mbox{for}~{}i=1,\cdots,k$

is non-singular, where $\kappa_{i,j}$ is the $j$-th principal curvature of
$\Gamma$ at $b_{i}$ for $j=1,\cdots,N-1$, then there exists a small positive
number $\sigma$, such that

$u_{a}^{(1)}(x)\equiv u_{a}^{(2)}(x)$

for all $a$ with $0<|a-(ka_{*})^{\frac{p-1}{2}}|\leq\sigma$ if
$p-1=\frac{4s}{N}$, or $0<a\leq\sigma$ if $p-1>\frac{4s}{N}$, or
$a\geq\frac{1}{\sigma}$ if $p-1<\frac{4s}{N}$.

######  Remark 1.7.

In fact, the result of Theorem 1.6 can hold for more general $p$ and $N.$
Here, in order to avoiding writing too dispersively, we assume that $N\geq
2s+4.$

As far as we know, there are very few results on the local uniqueness for the
fractional Schrödinger equations, especially with the subcritical
nonlinearities. In particular, since the Lagrange multiplier $\mu_{a}$ in
(1.1) depends on the solution $u_{a}$, the corresponding linearized operator
has changed, which brings more nontrivial analysis, as mentioned in [30]. For
the local uniqueness results of peak (or bubbling) solutions, the classical
moving plane method is not available. If $x_{0}$ is a non-degenerate critical
point of $V(x)$, that is, $(D^{2}V)$ is non-singular at $x_{0}$, one can prove
the local uniqueness of the peak solution concentrating at $x_{0}$ either by
counting the local degree of the corresponding reduced finite dimensional
problem as in [8, 10, 27], or by using Pohozaev type identities as in [9, 16,
28, 26, 29]. One of the advantage in applying the Pohazaev identities can be
found in dealing with the degenerate case(see [9, 16, 26]). In these results,
the rate of degeneracy along each direction is the same, although the critical
point $x_{0}$ is degenerate. Following [30], under the condition ($V$), the
function $V(x)$ is non-degenerate along the normal direction $\nu_{i}$ of
$\Gamma_{i}$. But along each tangential direction of $\Gamma_{i}$, $V(x)$ is
degenerate. Such non-uniform degeneracy makes the estimates more
sophisticated.

At the end of this section, we outline the main idea of the proof and discuss
the main difficulties. For the existence result, we first consider the
following problem without constraint,

$\begin{cases}(-\Delta)^{s}w+\big{(}\lambda+V(x)\big{)}w=w^{p},&{x\in\mathbb{R}^{N}},\\\\[2.84526pt]
w\in H^{s}(\mathbb{R}^{N}),\end{cases}$ (1.7)

where $\lambda>0$ is a large parameter. For large $\lambda>0$, by the standard
reduction argument, we could construct various positive solutions
concentrating at some stable points of $V(x)$. Particularly, we can construct
positive $k-$peak solutions for (1.7) of the form

$w_{\lambda}(x)=\lambda^{\frac{1}{p-1}}\Big{(}\sum_{j=1}^{k}U\big{(}\lambda^{\frac{1}{2s}}(x-x_{\lambda,i})\big{)}+\omega_{\lambda}\Big{)},$

with
$\displaystyle\int_{\mathbb{R}^{N}}\Big{(}\frac{1}{\lambda}|(-\Delta)^{\frac{s}{2}}\omega_{\lambda}|^{2}+\omega_{\lambda}^{2}\Big{)}=o\Big{(}\lambda^{-\frac{N}{2s}}\Big{)}$.
Let
$u_{\lambda}=\frac{w_{\lambda}}{\Big{(}\displaystyle\int_{\mathbb{R}^{N}}w_{\lambda}^{2}\Big{)}^{\frac{1}{2}}}$.
Then $\displaystyle\int_{\mathbb{R}^{N}}u_{\lambda}^{2}=1$, and

$\begin{cases}(-\Delta)^{s}u_{\lambda}+\big{(}\lambda+V(x)\big{)}u_{\lambda}=a_{\lambda}u_{\lambda}^{p},&{x\in\mathbb{R}^{N}},\\\\[2.84526pt]
u_{\lambda}\in H^{s}(\mathbb{R}^{N}),\end{cases}$

with

$a_{\lambda}=\Big{(}\int_{\mathbb{R}^{N}}w_{\lambda}^{2}\Big{)}^{\frac{p-1}{2}}=\Big{(}k\lambda^{\frac{2}{p-1}-\frac{N}{2s}}\big{(}a_{*}+o(1)\big{)}\Big{)}^{\frac{p-1}{2}}.$

We note that $a_{\lambda}>0$. Moreover, as $\lambda\rightarrow+\infty$,
$a_{\lambda}\rightarrow a_{0}$, which is defined by (1.4). Therefore, we
obtain a concentrated solution with $k$-peaks for (1.7) with normalized
$L^{2}$-norm, where $\mu=-\lambda$ and some suitable $a_{\lambda}$. Hence, we
are sufficed to answer a converse question that for any $a>0$ close to
$a_{0}$, whether one can choose a suitable large $\lambda=\lambda_{a}>0$, such
that (1.1)–(1.2) hold with $\mu=-\lambda_{a},\ \
u_{a}=\frac{w_{\lambda_{a}}}{\Big{(}\displaystyle\int_{\mathbb{R}^{N}}w_{\lambda_{a}}^{2}\Big{)}^{\frac{1}{2}}},$
finally concluding Theorem 1.4. In fact, we solve the problem by discussing
the relationship between $a$ and $\mu$.

Recently, for the classical BECs problem ($s=1,p=3,N=2,3$), a complete
description of the solutions $u_{a}$ concentrating at $a_{*}$ related to
problem was given in [30]. Specifically, in this case if $u_{a}$ is a solution
of problem (1.1)–(1.2) with $s=1$, $p=3$ and $N=2,3$, which is concentrated at
some points as $a\to a_{0}$ with $a_{0}\in\mathbb{R}$, then it holds

$a_{0}=ka_{*}>0\;\;\text{and}\;\;\mu=\mu_{a}\to-\infty~{}\mbox{as}~{}a\to
a_{0},$

where $k$ is the number of peaks of the concentrated solution $u_{a}$.
Moreover, the existence and local uniqueness of this kind concentrated
solutions have been proved in [30]. Unlike the argument used in [30], a non-
existence result, which is of great importance to prove the main result, could
not be obtained by a standard comparison argument in ODE theory. We would
apply a totally different method–the sliding method–corresponding to the non-
local operator to show a counterpart result (Theorem 1.1).

On the other hand, we point out that, due to the non-local property of the
operator $(-\Delta)^{s}$, we cannot construct the local Pohozaev identities
directly, which, however, is inevitable when we carry out the blow-up analysis
especially in the proof of the local uniqueness. To this end, we are supposed
to study the corresponding harmonic extension problem (see section 4 and
section 5), and have to estimate several kind of integrals which never appear
in the classic local Schrödinger problems. Similar arguments can be found in
[25] etc..

Last but not least, throughout our discussion, we need to distinct the
different cases of $p-1<\frac{4s}{N}$, $p-1=\frac{4s}{N}$, and
$p-1>\frac{4s}{N}$, which are called respectively that the mass-subcritical,
the mass-critical, and the mass-supercritical case. Especially compared with
the problem without any constraint, in the proof of the local uniqueness
result, we have to adopt one more different local Pohozaev identity,
associated with the mass-constraint condition.

This paper is organized as follows. In section 2, we prove Theorem 1.1 using
the sliding method. Then in section 3, we estimate the Lagrange multiplier
$\mu_{a}$ in terms of $a$. The results for the location of the peaks and for
the existence of peak solutions are proved in section 4, and the local
uniqueness of peak solutions are investigated in section 5.

##  2\.  Proof of Theorem 1.1 by a non-existence result

######  Lemma 2.1.

Assume that $P(x)$ satisfies $P(x)>1$ in $B_{R}(0)\setminus B_{r}(0)$ for some
fixed $r>0$ and large $R>0$. Then the problem

$(-\Delta)^{s}u=P(x)u,\ \ u>0,\ \ in\ \mathbb{R}^{N}$

has no solution.

###### Proof.

We take $\lambda_{1}$ as the first eigenvalue of $(-\Delta)^{s}$ and $\phi$ as
the associated eigenfunction satisfying

$\displaystyle\begin{cases}(-\Delta)^{s}\phi=\lambda_{1}\phi,&x\in\
B,\\\\[2.84526pt] \phi=0,&x\in\ B^{c},\end{cases}$

where $B\subset B_{R}(0)\setminus B_{r}(0)$ is a large ball. Since $R>0$ is
large enough, it may hold that

$\lambda_{1}\leq 1<P(x)~{}~{}\mbox{in}~{}~{}B.$

Now, we set $w(x):=\displaystyle\max_{x\in B}\frac{\phi(x)}{u(x)}\cdot u(x)$.
Obviously, $w(x)\geq\phi(x)$ and $w(x_{0})=\phi(x_{0})$ with $x_{0}\in B$ the
maximal point of $\frac{\phi(x)}{u(x)}$. To get a contradiction, we are
sufficed to prove that $w(x)>\phi(x)$ by use of the maximum principle.

In fact, if we denote $\gamma=\displaystyle\max_{x\in B}\frac{\phi(x)}{u(x)}$,
then there holds that

$\displaystyle(-\Delta)^{s}w=\gamma(-\Delta)^{s}u=\gamma
P(x)u\geq\gamma\lambda_{1}u=\lambda_{1}\max_{x\in
B}\frac{\phi(x)}{u(x)}u\geq\lambda_{1}\phi=(-\Delta)^{s}\phi,$

which gives that

$(-\Delta)^{s}w\geq(-\Delta)^{s}\phi~{}\,~{}\mbox{in}~{}~{}B.$

On the other hand, $w\geq\phi$ in $B^{c}$ obviously. Moreover, since here
$w\not\equiv\phi$, the strong maximum principle implies that $w>\phi$
concluding the Lemma by contradiction. ∎

######  Remark 2.2.

For the classical BECs problem when $s=1,p=3$ and $N=2,3$ in [30], the
standard comparison method from the ODE theory is sufficed to show such a non-
existence result, which, however, is invalid in the fractional case. We take
the sliding method to overcome this obstacle, which, developed by Chen, Li and
Zhu [11], would be typically available in the study of some other similar
problems when dealing with the fractional operator.

###### Proof of Theorem 1.1.

We divide the proofs into following three steps.

Step 1. We first show $\mu_{a}\rightarrow-\infty$. By contradiction, we first
suppose $|\mu_{a}|\leq M$. Since
$\displaystyle\int_{\mathbb{R}^{N}}u_{a}^{2}=1$, we could use the argument in
[21] to get $u_{a}$ is uniformly bounded, which means that $u_{a}$ does not
blow up. Then if $\mu_{a}\rightarrow+\infty$, we set
$P(x)=\mu_{a}-V(x)+au_{a}^{p-1}$. Since $u_{a}$ concentrates at some points,
we may assume

$au_{a}^{p-1}>-1~{}~{}\mbox{in}~{}~{}\mathbb{R}^{N}\setminus
B_{r}(0)~{}~{}\mbox{for some}~{}~{}r>0.$

Hence, for any $R>0$, we have

$P(x)>1~{}~{}\mbox{for}~{}~{}x\in B_{R}(0)\setminus B_{r}(0),$

which gives a contradiction by Lemma 2.1.

Step 2. It holds that $a>0$. If not, we could obtain then

$(-\Delta)^{s}u_{a}\leq 0.$

By the maximum principle corresponding to the fractional operator [21], we
find $u_{a}\leq 0$, which is a contradiction.

Step 3. In the case of $p-1\leq\frac{4s}{N}$. Denote
$v_{a}(x)=(-\mu_{a})^{-\frac{1}{p-1}}u_{a}\big{(}(-\mu_{a})^{-\frac{1}{2s}}x\big{)}$.
Then

$\displaystyle(-\Delta)^{s}v_{a}+\Big{(}1+\frac{1}{-\mu_{a}}V\big{(}(-\mu_{a})^{-\frac{1}{2s}}x\big{)}\Big{)}v_{a}=av_{a}^{p}$

and
$\displaystyle\int_{\mathbb{R}^{N}}v_{a}^{2}=(-\mu_{a})^{\frac{N}{2s}-\frac{2}{p-1}}$.
Therefore, we apply the Moser iteration to get $v_{a}$ is bounded uniformly in
$a$. Let $y_{a}$ be a maximum point of $v_{a}$. Then we could show that $a$ is
bounded from below. Applying the blow-up argument established by [21], we
obtain that there exists some integer $k>0$ such that

$v_{a}=\sum_{i=1}^{k}U_{a_{0}}(x-y_{a,i})+\bar{\omega}_{a},$

for some $y_{a,i}\in\mathbb{R}^{N}$ with $|y_{a,i}-y_{a,j}|\rightarrow+\infty$
for $i\neq j$, where $U_{a_{0}}$ is the unique positive solution of

$\begin{cases}(-\Delta)^{s}U_{a_{0}}+U_{a_{0}}=a_{0}U_{a_{0}}^{p},\\\\[5.69054pt]
U_{a_{0}}(0)=\displaystyle\max_{x\in\mathbb{R}^{N}}U_{a_{0}}(x),\end{cases}$

and

$\displaystyle\int_{\mathbb{R}^{N}}\big{(}|(-\Delta)^{\frac{s}{2}}\bar{\omega}_{a}|^{2}+\bar{\omega}_{a}^{2}\big{)}=o(1)$

Since $\displaystyle\int_{\mathbb{R}^{N}}u_{a}^{2}=1$, we have

$\int_{\mathbb{R}^{N}}v_{a}^{2}=(-\mu_{a})^{\frac{N}{2s}-\frac{2}{p-1}}.$

In view of $U_{a}=a_{0}^{-\frac{1}{p-1}}U$, there holds that

$(-\mu_{a})^{\frac{N}{2s}-\frac{2}{p-1}}=ka_{0}^{-\frac{2}{p-1}}a_{*}+o(1),$

implying the conclusion.

∎

##  3\. Some estimates by Pohozaev identities

We use Pohozaev identities to estimate $\mu_{a}$ with respect to $a$. Let
$\varepsilon=(-\mu_{a})^{-\frac{1}{2s}}$ and
$v_{a}=(-\frac{\mu_{a}}{a})^{-\frac{1}{p-1}}u_{a}$. Then (1.1) turns to

$\displaystyle\varepsilon^{2s}(-\Delta)^{s}v_{a}+\big{(}1+\varepsilon^{2s}V(x)\big{)}v_{a}=v_{a}^{p},\,\,\,\,v_{a}\in
H^{s}(\mathbb{R}^{N}).$ (3.1)

For any $a>0$, we define the norm
$\|v\|_{a}=\Big{(}\displaystyle\int_{\mathbb{R}^{N}}\varepsilon^{2s}|(-\Delta)^{\frac{s}{2}}v|^{2}+v^{2}\Big{)}^{\frac{1}{2}}$.

Hence, a $k$-peak solution of (3.1) has the form of

$v_{a}(x)=\sum_{i=1}^{k}U_{\varepsilon,x_{a,i}}(x)+\varphi_{a}(x),$

where

$U_{\varepsilon,x_{a,i}}(x)=(1+\varepsilon^{2s}V_{i})^{\frac{1}{p-1}}U\Big{(}\frac{(1+\varepsilon^{2s}V_{i})^{\frac{1}{2s}}(x-x_{a,i})}{\varepsilon}\Big{)},\
\ \|\varphi_{a}\|_{a}=o(\varepsilon^{\frac{N}{2}}).$

We write then

$\displaystyle\varepsilon^{2s}(-\Delta)^{s}\varphi_{a}+\Big{(}\big{(}1+\varepsilon^{2s}V(x)\big{)}-p\sum_{i=1}^{k}U_{\varepsilon,x_{a,i}}^{p-1}\Big{)}\varphi_{a}=N_{a}(\varphi_{a})+l_{a}(x),$

where

$\displaystyle
l_{a}(x)=-\varepsilon^{2s}\sum_{i=1}^{k}\big{(}V(x)-V_{i}\big{)}U_{\varepsilon,x_{a,i}}+\Big{(}\sum_{i=1}^{k}U_{\varepsilon,x_{a,i}}\Big{)}^{p}-\sum_{i=1}^{k}U_{\varepsilon,x_{a,i}}^{p},$

and

$\displaystyle
N_{a}(\varphi_{a})=\Big{(}\sum_{i=1}^{k}U_{\varepsilon,x_{a,i}}+\varphi_{a}\Big{)}^{p}-\Big{(}\sum_{i=1}^{k}U_{\varepsilon,x_{a,i}}\Big{)}^{p}-p\sum_{i=1}^{k}U_{\varepsilon,x_{a,i}}^{p-1}\varphi_{a}.$
(3.2)

Set

$\displaystyle E_{a,x_{a,i}}:=\Big{\\{}u\in H^{s}(\mathbb{R}^{N}):\langle
u,\frac{\partial U_{\varepsilon,x_{a,i}}}{\partial
x_{j}}\rangle_{a}=0,j=1,\ldots,N\Big{\\}}.$

One may move $x_{a,i}$ such that the perturbing term
$\varphi_{a}\in\displaystyle\bigcap_{i=1}^{k}E_{a,x_{a,i}}$.

Let $L_{a}$ be the bounded linear operator from $H^{s}(\mathbb{R}^{N})$ to
itself, with the form of

$\displaystyle\langle
L_{a}u,v\rangle_{a}=\int_{\mathbb{R}^{N}}\varepsilon^{2s}(-\Delta)^{\frac{s}{2}}u(-\Delta)^{\frac{s}{2}}v+\big{(}1+\varepsilon^{2s}V(x)\big{)}uv-p\sum_{i=1}^{k}U_{\varepsilon,x_{a,i}}^{p-1}uv.$
(3.3)

It is standard to prove the following result.

######  Lemma 3.1.

There exists some constant $c>0$, such that for all $a$ around $a_{0}$, it
holds then

$\|L_{a}u\|_{a}\geq c\|u\|_{a},\ \ for\ all\
u\in\bigcap_{i=1}^{k}E_{a,x_{a,i}}.$

######  Lemma 3.2.

A $k$-peak solution $v_{a}$ of (3.1) concentrating at $b_{1},\ldots,b_{k}$ has
the following form

$\displaystyle
v_{a}(x)=\sum_{i=1}^{k}U_{\varepsilon,x_{a,i}}(x)+\varphi_{a}(x),$

with $\varphi_{a}\in\bigcap_{i=1}^{k}E_{a,x_{a,i}}$ and

$\displaystyle\|\varphi_{a}\|_{a}=$ $\displaystyle
O\Big{(}|\sum_{i=1}^{k}(V(x_{a,i})-V_{i})|\varepsilon^{\frac{N}{2}+2s}+|\sum_{i=1}^{k}\nabla
V(x_{a,i})|\varepsilon^{\frac{N}{2}+2s+1}+\varepsilon^{\frac{N}{2}+2s+2}\Big{)}$
$\displaystyle+\begin{cases}O\Big{(}\varepsilon^{\frac{N}{2}+\frac{p}{2}(N+2s)}\Big{)},\,&\text{if}\,\,1<p\leq
2,\\\\[5.69054pt]
O\Big{(}\varepsilon^{\frac{N}{2}+(N+2s)}\Big{)},\,&\text{if}\,\,p>2.\end{cases}$
(3.4)

###### Proof.

By calculation, we have for any $v\in H^{s}(\mathbb{R}^{N})$,

$\displaystyle\langle l_{a}(x),v\rangle_{a}=$
$\displaystyle-\varepsilon^{2s}\sum_{i=1}^{k}\int_{\mathbb{R}^{N}}\big{(}V(x)-V_{i}\big{)}U_{\varepsilon,x_{a,i}}v+\int_{\mathbb{R}^{N}}\big{(}(\sum_{i=1}^{k}U_{\varepsilon,x_{a,i}})^{p}-\sum_{i=1}^{k}U_{\varepsilon,x_{a,i}}^{p}\big{)}v$
$\displaystyle=$
$\displaystyle-\varepsilon^{2s}\sum_{i=1}^{k}\int_{\mathbb{R}^{N}}\big{(}V(x)-V(x_{a,i})\big{)}U_{\varepsilon,x_{a,i}}v-\varepsilon^{2s}\sum_{i=1}^{k}\int_{\mathbb{R}^{N}}\big{(}V(x_{a,i})-V_{i}\big{)}U_{\varepsilon,x_{a,i}}v$
$\displaystyle+\int_{\mathbb{R}^{N}}\big{(}(\sum_{i=1}^{k}U_{\varepsilon,x_{a,i}})^{p}-\sum_{i=1}^{k}U_{\varepsilon,x_{a,i}}^{p}\big{)}v:=-A_{1}-A_{2}+A_{3}.$

We estimate $A_{1}$ as follows

$\displaystyle A_{1}=$
$\displaystyle\varepsilon^{2s}\sum_{i=1}^{k}\int_{\mathbb{R}^{N}}\big{(}V(x)-V(x_{a,i})\big{)}U_{\varepsilon,x_{a,i}}v$
$\displaystyle\leq$
$\displaystyle\varepsilon^{2s}\sum_{i=1}^{k}\int_{B_{\delta}(x_{a,i})}\big{(}\nabla
V(x_{a,i})\cdot(x-x_{a,i})+O(|x-x_{a,i}|^{2})\big{)}U_{\varepsilon,x_{a,i}}v$
$\displaystyle+C\varepsilon^{2s}\Big{(}\sum_{i=1}^{k}\int_{B^{c}_{\delta}(x_{a,i})}U^{2}_{\varepsilon,x_{a,i}}\Big{)}^{\frac{1}{2}}\|v\|_{L^{2}}$
$\displaystyle\leq$ $\displaystyle
C\varepsilon^{2s+\frac{N}{2}+1}|\sum_{i=1}^{k}\nabla
V(x_{a,i})|\cdot\|v\|_{a}+\varepsilon^{2s+\frac{N}{2}+2}\|v\|_{a}.$

Similarly,

$\displaystyle A_{2}=$
$\displaystyle\varepsilon^{2s}\sum_{i=1}^{k}\int_{\mathbb{R}^{N}}\big{(}V(x_{a,i})-V_{i}\big{)}U_{\varepsilon,x_{a,i}}v\leq
C\varepsilon^{2s+\frac{N}{2}}\sum_{i=1}^{k}|V(x_{a,i})-V_{i}|\cdot\|v\|_{a}.$

Finally, since

$\displaystyle|A_{3}|=$
$\displaystyle\Big{|}\int_{\mathbb{R}^{N}}\big{(}(\sum_{i=1}^{k}U_{\varepsilon,x_{a,i}})^{p}-\sum_{i=1}^{k}U_{\varepsilon,x_{a,i}}^{p}\big{)}v\Big{|}=\begin{cases}O\Big{(}\displaystyle\int_{\mathbb{R}^{N}}\sum_{i\neq
j}U^{p-1}_{\varepsilon,x_{a,i}}U_{\varepsilon,x_{a,j}}v\Big{)},&\text{if}\,\,p>2,\vspace{0.12cm}\\\
O\Big{(}\displaystyle\int_{\mathbb{R}^{N}}\sum_{i\neq
j}U^{\frac{p}{2}}_{\varepsilon,x_{a,i}}U^{\frac{p}{2}}_{\varepsilon,x_{a,j}}v\Big{)},&\text{if}\,\,1<p\leq
2,\end{cases}$

then by Lemma A.1, we compute directly that if $p>2$

$\displaystyle\Big{|}\int_{\mathbb{R}^{N}}\sum_{i\neq
j}U^{p-1}_{\varepsilon,x_{a,i}}U_{\varepsilon,x_{a,j}}v\Big{|}\leq
C\Big{(}\int_{\mathbb{R}^{N}}\sum_{i\neq
j}U^{2(p-1)}_{\varepsilon,x_{a,i}}U^{2}_{\varepsilon,x_{a,j}}\Big{)}^{\frac{1}{2}}\|v\|_{a}$
$\displaystyle\leq C\varepsilon^{\frac{N}{2}}\sum_{i\neq
j}\Big{(}\int_{\mathbb{R}^{N}}\frac{1}{(1+|y-\frac{x_{a,j}}{\varepsilon}|)^{2(N+2s)(p-1)}}\frac{1}{(1+|y-\frac{x_{a,i}}{\varepsilon}|)^{2(N+2s)}}\Big{)}^{\frac{1}{2}}\|v\|_{a}$
$\displaystyle\leq C\varepsilon^{\frac{N}{2}}\sum_{i\neq
j}\frac{1}{|\frac{x_{a,j}-x_{a,i}}{\varepsilon}|^{N+2s}}\Big{(}\int_{\mathbb{R}^{N}}\frac{1}{(1+|y-\frac{x_{a,j}}{\varepsilon}|)^{2(N+2s)(p-1)}}+\frac{1}{(1+|y-\frac{x_{a,i}}{\varepsilon}|)^{2(N+2s)(p-1)}}\Big{)}^{\frac{1}{2}}\|v\|_{a}$
$\displaystyle\leq C\varepsilon^{\frac{N}{2}+N+2s}\|v\|_{a};$

while for $1<p\leq 2$,

$\displaystyle\Big{|}\int_{\mathbb{R}^{N}}\sum_{i\neq
j}U^{\frac{p}{2}}_{\varepsilon,x_{a,i}}U^{\frac{p}{2}}_{\varepsilon,x_{a,j}}v\Big{|}\leq
C\Big{(}\int_{\mathbb{R}^{N}}\sum_{i\neq
j}U^{p}_{\varepsilon,x_{a,i}}U^{p}_{\varepsilon,x_{a,j}}\Big{)}^{\frac{1}{2}}\|v\|_{a}$
$\displaystyle\leq C\varepsilon^{\frac{N}{2}}\sum_{i\neq
j}\Big{(}\int_{\mathbb{R}^{N}}\frac{1}{(1+|y-\frac{x_{a,j}}{\varepsilon}|)^{p(N+2s)}}\frac{1}{(1+|y-\frac{x_{a,i}}{\varepsilon}|)^{p(N+2s)}}\Big{)}^{\frac{1}{2}}\|v\|_{a}$
$\displaystyle\leq C\varepsilon^{\frac{N}{2}}\sum_{i\neq
j}\frac{1}{|\frac{x_{a,j}-x_{a,i}}{\varepsilon}|^{\frac{p}{2}(N+2s)}}\|v\|_{a}\leq
C\varepsilon^{\frac{N}{2}+\frac{p}{2}(N+2s)}\|v\|_{a}.$

To sum up, the result has been proved.

∎

######  Lemma 3.3.

It holds that

$\displaystyle\int_{\mathbb{R}^{N}}U^{2}=\Big{(}\frac{N}{s(p+1)}-\frac{N}{2s}+1\Big{)}\int_{\mathbb{R}^{N}}U^{p+1}.$

###### Proof.

It is directly obtained by the following Pohozaev identities:

$\displaystyle\frac{N-2s}{2}\int_{\mathbb{R}_{+}^{N+1}}t^{1-2s}|\nabla\tilde{U}|^{2}=-\frac{N}{2}\int_{\mathbb{R}^{N}}U^{2}+\frac{N}{p+1}\int_{\mathbb{R}^{N}}U^{p+1}$

and

$\displaystyle\int_{\mathbb{R}_{+}^{N+1}}t^{1-2s}|\nabla\tilde{U}|^{2}=-\int_{\mathbb{R}^{N}}U^{2}+\int_{\mathbb{R}^{N}}U^{p+1}.$

∎

######  Proposition 3.4.

There holds that

$\displaystyle
a^{\frac{2}{p-1}}=\sum_{i=1}^{k}(V_{i}-\mu)^{\frac{2}{p-1}-\frac{N}{2s}}\int_{\mathbb{R}^{N}}U^{2}+O\Big{(}|\sum_{i=1}^{k}(V(x_{a,i})-V_{i})|(-\mu)^{\frac{2}{p-1}-1-\frac{N}{2s}}$
$\displaystyle\quad\quad\quad+|\sum_{i=1}^{k}\nabla
V(x_{a,i})|(-\mu)^{\frac{2}{p-1}-1-\frac{N}{2s}-\frac{1}{2s}}+(-\mu)^{\frac{2}{p-1}-1-\frac{N}{2s}-\frac{1}{s}}\Big{)}$
$\displaystyle\quad\quad\quad+O\Big{(}\sum_{i\neq
j}(-\mu)^{\frac{2}{p-1}-\frac{N}{s}-1}|x_{a,j}-x_{a,i}|^{-(N+2s)}\Big{)}+\begin{cases}O\Big{(}(-\mu)^{\frac{2}{p-1}-\frac{N}{2s}-\frac{pN}{4s}-\frac{p}{2}}\Big{)},&\text{if}\,\,1<p\leq
2,\\\\[5.69054pt]
O\Big{(}(-\mu)^{\frac{2}{p-1}-\frac{N}{s}-1}\Big{)},&\text{if}\,\,p>2.\end{cases}$

###### Proof.

Let $u_{a}$ be a solution of (1.1)-(1.2). It holds that

$\displaystyle 1$
$\displaystyle=\int_{\mathbb{R}^{N}}|u_{a}(x)|^{2}=\int_{\mathbb{R}^{N}}\left(\sum_{i=1}^{k}(\frac{V_{i}-\mu}{a})^{\frac{1}{p-1}}U((V_{i}-\mu)^{\frac{1}{2s}}(x-x_{a,i}))+(\frac{-\mu}{a})^{\frac{1}{p-1}}\varphi_{a}\right)^{2}$
$\displaystyle=\sum_{i=1}^{k}\frac{(V_{i}-\mu)^{\frac{2}{p-1}-\frac{N}{2s}}}{a^{\frac{2}{p-1}}}\int_{\mathbb{R}^{N}}U^{2}+\sum_{i=1}^{k}\frac{(-\mu)^{\frac{1}{p-1}}}{a^{\frac{2}{p-1}}}(V_{i}-\mu)^{\frac{1}{p-1}-\frac{N}{2s}}\int_{\mathbb{R}^{N}}U(x)\varphi_{a}((V_{i}-\mu)^{-\frac{1}{2s}}x+x_{a,i})$
$\displaystyle\quad+\sum_{i\neq
j}\frac{(V_{i}-\mu)^{\frac{1}{p-1}-\frac{N}{2s}}}{a^{\frac{2}{p-1}}}(V_{j}-\mu)^{\frac{1}{p-1}}\int_{\mathbb{R}^{N}}U(x)U\Big{(}(\frac{V_{j}-\mu}{V_{i}-\mu})^{\frac{1}{2s}}x-(V_{j}-\mu)^{\frac{1}{2s}}(x_{a,j}-x_{a,i})\Big{)}$
$\displaystyle\quad+\frac{(-\mu)^{\frac{2}{p-1}}}{a^{\frac{2}{p-1}}}\int_{\mathbb{R}^{N}}\varphi_{a}^{2}$
$\displaystyle=\sum_{i=1}^{k}\frac{(V_{i}-\mu)^{\frac{2}{p-1}-\frac{N}{2s}}}{a^{\frac{2}{p-1}}}\int_{\mathbb{R}^{N}}U^{2}+\sum_{i=1}^{k}\frac{(-\mu)^{\frac{1}{p-1}}}{a^{\frac{2}{p-1}}}(V_{i}-\mu)^{\frac{1}{p-1}-\frac{N}{4s}}O(\|\varphi_{a}\|_{L^{2}})$
$\displaystyle\quad+O\Big{(}\sum_{i\neq
j}\frac{(V_{i}-\mu)^{\frac{1}{p-1}-\frac{N}{2s}}(V_{j}-\mu)^{\frac{1}{p-1}-\frac{N}{2s}-1}}{a^{\frac{2}{p-1}}}|x_{a,j}-x_{a,i}|^{-(N+2s)}\Big{)}+\frac{(-\mu)^{\frac{2}{p-1}}}{a^{\frac{2}{p-1}}}\int_{\mathbb{R}^{N}}\varphi_{a}^{2}$
$\displaystyle=\sum_{i=1}^{k}\frac{(V_{i}-\mu)^{\frac{2}{p-1}-\frac{N}{2s}}}{a^{\frac{2}{p-1}}}\int_{\mathbb{R}^{N}}U^{2}+\frac{(-\mu)^{\frac{2}{p-1}-\frac{N}{4s}}}{a^{\frac{2}{p-1}}}O\Big{(}|\sum_{i=1}^{k}(V(x_{a,i})-V_{i}))|\varepsilon^{\frac{N}{2}+2s}$
$\displaystyle\quad+|\sum_{i=1}^{k}\nabla
V(x_{a,i})|\varepsilon^{\frac{N}{2}+2s+1}+\varepsilon^{\frac{N}{2}+2s+2}\Big{)}+\frac{(-\mu)^{\frac{2}{p-1}-\frac{N}{4s}}}{a^{\frac{2}{p-1}}}\begin{cases}O\Big{(}\varepsilon^{\frac{N}{2}+\frac{p}{2}(N+2s)}\Big{)},\,\,&\text{if}\,\,1<p\leq
2,\\\\[8.53581pt]
O\Big{(}\varepsilon^{\frac{N}{2}+(N+2s)}\Big{)},\,\,&\text{if}\,\,p>2,\end{cases}$
$\displaystyle\quad+O\Big{(}\sum_{i\neq
j}\frac{(-\mu)^{\frac{2}{p-1}-\frac{N}{s}-1}}{a^{\frac{2}{p-1}}}|x_{a,j}-x_{a,i}|^{-(N+2s)}\Big{)},$

which gives the result. ∎

##  4\. Existence of the Peak Solutions

###  4.1. Locating of the peaks

Let $v_{a}$ be a peak solution of (3.1). Following the argument in section 5
of [15], we obtain that the concentrating solution $v_{a}$ to (3.1) must
satisfy that

$\displaystyle\sup_{x\in\mathbb{R}^{N}}\Big{(}\sum_{i=1}^{k}\frac{1}{\big{(}1+|\frac{x-x_{a,i}}{\varepsilon}|^{2}\big{)}^{\frac{N+2s-\theta}{2}}}\Big{)}^{-1}|v_{a}(x)|<+\infty,$
(4.1)

where $\theta$ is any constant such that $\theta\in(0,\frac{N}{2}+2s)$. Now we
quote the extension of $v_{a}$ and its equation:

$\begin{cases}div(t^{1-2s}\nabla\tilde{v}_{a})=0,&{x\in\mathbb{R}^{N+1}_{+}},\vspace{0.2cm}\\\
-\varepsilon^{2s}\displaystyle\lim_{t\rightarrow
0}t^{1-2s}\partial_{t}\tilde{v}_{a}(y,t)=v_{a}^{p}-\big{(}1+\varepsilon^{2s}V(x)\big{)}v_{a},&{x\in\mathbb{R}^{N}}.\end{cases}$
(4.2)

Denote

$\displaystyle
B_{\rho}(x_{0})=\big{\\{}y\in\mathbb{R}^{N}:|y-x_{0}|\leq\rho\big{\\}}\subseteq\mathbb{R}^{N},$
$\displaystyle\mathbf{B}_{\rho}^{+}(x_{0})=\big{\\{}Y=(y,t):|Y-(x_{0},0)|\leq\rho,t>0\big{\\}}\subseteq\mathbb{R}_{+}^{N+1},$
$\displaystyle\partial^{\prime}\mathbf{B}_{\rho}^{+}(x_{0})=\big{\\{}Y=(y,t):|Y-(x_{0},0)|\leq\rho,t=0\big{\\}}\subseteq\mathbb{R}^{N},$
$\displaystyle\partial^{\prime\prime}\mathbf{B}_{\rho}^{+}(x_{0})=\big{\\{}Y=(y,t):|Y-(x_{0},0)|=\rho,t>0\big{\\}}\subseteq\mathbb{R}_{+}^{N+1},$
$\displaystyle\partial\mathbf{B}_{\rho}^{+}(x_{0})=\partial^{\prime}\mathbf{B}_{\rho}^{+}(x_{0})\cup\partial^{\prime\prime}\mathbf{B}_{\rho}^{+}(x_{0}).$

Multiplying (4.2) by $\frac{\partial\tilde{v}_{a}}{\partial x_{j}}$ and
integrating by parts on $B_{\rho}(x_{a,i})$(or
$\mathbf{B}_{\rho}^{+}(x_{a,i})$) with some $\rho>0$, we have the following
Pohozaev identities:

$\begin{split}\varepsilon^{2s}\int_{B_{\rho}(x_{a,i})}\frac{\partial
V(x)}{\partial
x_{j}}v_{a}^{2}&=-2\varepsilon^{2s}\int_{\partial^{\prime\prime}\mathbf{B}_{\rho}^{+}(x_{a,i})}t^{1-2s}\frac{\partial\tilde{v}_{a}}{\partial\nu}\frac{\partial\tilde{v}_{a}}{\partial
y^{j}}+\varepsilon^{2s}\int_{\partial^{\prime\prime}\mathbf{B}_{\rho}^{+}(x_{a,i})}t^{1-2s}|\nabla\tilde{v}_{a}|^{2}\nu_{j}\\\
&\quad+\int_{\partial
B_{\rho}(x_{a,i})}\big{(}1+\varepsilon^{2s}V(x)\big{)}v_{a}^{2}\nu_{j}-\frac{2}{p+1}\int_{\partial
B_{\rho}(x_{a,i})}v_{a}^{p+1}\nu_{j}.\end{split}$ (4.3)

Using Lemma A.2 and Remark A.3, we estimate that for
$(y-x)^{2}+t^{2}=\rho^{2},t>0$, $\theta\in(0,\frac{N}{2}+2s)$,

$\displaystyle|\tilde{v}_{a}(y,t)|$ $\displaystyle\leq
C\sum_{i=1}^{k}\int_{\mathbb{R}^{N}}\frac{t^{2s}}{(|y-\xi|+t)^{N+2s}}\frac{1}{(1+|\frac{\xi-
x_{a,i}}{\varepsilon}|)^{N+2s-\theta}}d\xi$ $\displaystyle\leq
C\begin{cases}\varepsilon^{N+2s-\theta}\displaystyle\sum_{i=1}^{k}\frac{1}{(1+|y-x_{a,i}|)^{N+2s-\theta}},\
&\theta>2s,\\\
\varepsilon^{N}\displaystyle\sum_{i=1}^{k}\frac{1}{(1+|y-x_{a,i}|)^{N+2s-\theta}},\
&\theta<2s.\end{cases}$

Similarly, we could also prove that

$\displaystyle\Big{|}\frac{\partial}{\partial
y^{j}}\tilde{v}_{a}(y,t)\Big{|},\,\,\,\Big{|}\frac{\partial}{\partial\nu}\tilde{v}_{a}(y,t)\Big{|}$
$\displaystyle\leq
C\begin{cases}\varepsilon^{N+2s-\theta}\displaystyle\sum_{i=1}^{k}\frac{1}{(1+|y-x_{a,i}|)^{N+2s-\theta}},\
&\theta>2s,\\\
\varepsilon^{N}\displaystyle\sum_{i=1}^{k}\frac{1}{(1+|y-x_{a,i}|)^{N+2s-\theta}},\
&\theta<2s.\end{cases}$

Therefore, we estimate the boundary terms as follows: if $\theta>2s$

$\displaystyle-2\varepsilon^{2s}\int_{\partial^{\prime\prime}\mathbf{B}_{\rho}^{+}(x_{a,i})}t^{1-2s}\frac{\partial\tilde{v}_{a}}{\partial\nu}\frac{\partial\tilde{v}_{a}}{\partial
y^{j}}$ $\displaystyle\leq
C\varepsilon^{2s}\varepsilon^{2(N+2s-\theta)}\int_{\partial^{\prime\prime}\mathbf{B}_{\rho}^{+}(x_{a,i})}t^{1-2s}\Big{(}\sum_{i=1}^{k}\frac{1}{(1+|y-x_{a,i}|)^{N+2s-\theta}}\Big{)}^{2}\leq
C\varepsilon^{2(N+3s-\theta)};$

and similarly,

$\displaystyle\varepsilon^{2s}\int_{\partial^{\prime\prime}\mathbf{B}_{\rho}^{+}(x_{a,i})}t^{1-2s}|\nabla\tilde{v}_{a}|^{2}\nu_{j}\leq
C\varepsilon^{2(N+3s-\theta)};$ $\displaystyle\int_{\partial
B_{\rho}(x_{a,i})}\big{(}1+\varepsilon^{2s}V(x)\big{)}v_{a}^{2}\nu_{j}\leq
C\varepsilon^{2(N+2s-\theta)};$

$\displaystyle\frac{2}{p+1}\int_{\partial
B_{\rho}(x_{a,i})}v_{a}^{p+1}\nu_{j}\leq C\varepsilon^{(p+1)(N+2s-\theta)}.$

While if $\theta<2s$,

$\displaystyle-2\varepsilon^{2s}\int_{\partial^{\prime\prime}\mathbf{B}_{\rho}^{+}(x_{a,i})}t^{1-2s}\frac{\partial\tilde{v}_{a}}{\partial\nu}\frac{\partial\tilde{v}_{a}}{\partial
y^{j}}$ $\displaystyle\leq
C\varepsilon^{2s}\varepsilon^{2N}\int_{\partial^{\prime\prime}\mathbf{B}_{\rho}^{+}(x_{a,i})}t^{1-2s}\Big{(}\sum_{i=1}^{k}\frac{1}{(1+|y-x_{a,i}|)^{N+2s-\theta}}\Big{)}^{2}$
$\displaystyle\leq C\varepsilon^{2(N+s)};$
$\displaystyle\varepsilon^{2s}\int_{\partial^{\prime\prime}\mathbf{B}_{\rho}^{+}(x_{a,i})}t^{1-2s}|\nabla\tilde{v}_{a}|^{2}\nu_{j}\leq
C\varepsilon^{2(N+s)};$ $\displaystyle\int_{\partial
B_{\rho}(x_{a,i})}\big{(}1+\varepsilon^{2s}V(x)\big{)}v_{a}^{2}\nu_{j}\leq
C\varepsilon^{2(N+2s-\theta)};$

and

$\displaystyle\int_{\partial B_{\rho}(x_{a,i})}v_{a}^{p+1}\nu_{j}\leq
C\varepsilon^{(p+1)(N+2s-\theta)}.$

Hence (4.3) gives then for $\theta\in(0,\frac{N}{2}+2s)$,

$\displaystyle\int_{B_{\rho}(x_{a,i})}\frac{\partial V(x)}{\partial
x_{j}}v_{a}^{2}=\begin{cases}O\big{(}\varepsilon^{2(N+s-\theta)}\big{)},\
&\theta>2s,\\\\[5.69054pt]
O\big{(}\varepsilon^{\min\\{2N,2(N+s-\theta)\\}}\big{)},\
&\theta<2s,\end{cases}$ (4.4)

which implies the first necessary condition for the concentrated points
$b_{i}$:

$\nabla V(b_{i})=0,\ \ for\ i=1,\ldots,k.$

######  Remark 4.1.

In our discussion, since we are sufficed to consider certain small
$\theta\in(0,\frac{N}{2}+2s)$ fixed, then for simplicity, we may as well
suppose

$\theta<s.$

Hence (4.4) turns to be

$\displaystyle\int_{B_{\rho}(x_{a,i})}\frac{\partial V(x)}{\partial
x_{j}}v_{a}^{2}=O(\varepsilon^{2N}).$ (4.5)

###### Proof of Theorem 1.3.

In view of $x_{a,i}\rightarrow b_{i}\in\Gamma_{i}$, if $\Gamma_{i}$ is a local
minimum set of $V(x)$, there exists some $t_{a}\in[V_{i},V_{i}+\theta]$; while
if $\Gamma_{i}$ is a local maximum set of $V(x)$, there exists some
$t_{a}\in[V_{i}-\theta,V_{i}]$, such that $x_{a,i}\in\Gamma_{t_{a},i}$.

Let $H(x)=\langle\nabla V(x),\tau_{a,i}\rangle$ and $\tau_{a,i}$ be the unit
tangential vector of $\Gamma_{t_{a},i}$ at $x_{a,i}$. Then

$H(x_{a,i})=0.$

On the one hand, we have the expansion for $x\in B_{\rho}(x_{a,i})$,

$\displaystyle H(x)=\langle\nabla
H(x_{a,i}),x-x_{a,i}\rangle+\frac{1}{2}\big{\langle}\langle\nabla^{2}H(x_{a,i}),x-x_{a,i}\rangle,x-x_{a,i}\big{\rangle}+o\big{(}|x-x_{a,i}|^{2}\big{)},$

and then,

$\begin{split}&\int_{B_{\rho}(x_{a,i})}H(x)U_{\varepsilon,x_{a,i}}^{2}(x)\\\
&=-2\int_{B_{\rho}(x_{a,i})}H(x)U_{\varepsilon,x_{a,i}}(x)\varphi_{a}-\int_{B_{\rho}(x_{a,i})}H(x)\varphi_{a}^{2}+O(\varepsilon^{2(N+s-\theta)})\\\
&=O\left((\varepsilon^{\frac{N}{2}+1}|\nabla
H(x_{a,i})|+\varepsilon^{\frac{N}{2}+2})\|\varphi_{a}\|_{a}+\varepsilon|\nabla
H(x_{a,i})|\cdot\|\varphi_{a}\|_{a}^{2}\right)+O\Big{(}\varepsilon^{2(N+s-\theta)}\Big{)}\\\
&=O\Big{(}\varepsilon^{N+2s+2}\Big{)},\end{split}$ (4.6)

where we used the fact that, by assumption,

$|V(x_{a,i})-V_{i}|=O(\varepsilon)~{}~{}\mbox{and}~{}~{}\|\varphi_{a}\|_{a}=O\big{(}\varepsilon^{\frac{N}{2}+2s+1}\big{)}.$

On the other hand, since $H(x_{a,i})=0$,

$\displaystyle\int_{B_{\rho}(x_{a,i})}H(x)U_{\varepsilon,x_{a,i}}^{2}(x)=\frac{1}{2N}\varepsilon^{N+2}\Delta
H(x_{a,i})\int_{\mathbb{R}^{N}}|x|^{2}U^{2}+O(\varepsilon^{N+4})$

which, combined with (4.6), gives that

$\Delta H(x_{a,i})=O(\varepsilon^{2s}).$

Hence from the condition (V), we get (1.6) concluding Theorem 1.3. ∎

###  4.2. Existence of the normalized peak solutions

For the existence result of problem (1.7) with large $\lambda>0$, we set
$\eta=\lambda^{-\frac{1}{2s}}$ and $w(x)\mapsto\lambda^{\frac{1}{p-1}}w(x)$,
and (1.7) turns to

$\begin{cases}\eta^{2s}(-\Delta)^{s}w+\big{(}1+\eta^{2s}V(x)\big{)}w=w^{p},&{x\in\mathbb{R}^{N}},\\\\[5.69054pt]
w\in H^{s}(\mathbb{R}^{N}),\end{cases}$ (4.7)

which is equivalent to

$\begin{cases}div(t^{1-2s}\nabla\tilde{w})=0,&{x\in\mathbb{R}^{N+1}_{+}},\\\\[2.84526pt]
-\eta^{2s}\displaystyle\lim_{t\rightarrow
0}t^{1-2s}\partial_{t}\tilde{w}(y,t)=w^{p}-\big{(}1+\eta^{2s}V(x)\big{)}w,&{x\in\mathbb{R}^{N}}.\end{cases}$

We denote $\langle
u,v\rangle_{\eta}=\int_{\mathbb{R}^{N}}\big{(}\eta^{2s}(-\Delta)^{\frac{s}{2}}u(-\Delta)^{\frac{s}{2}}v+uv\big{)}$,
and $\|u\|_{\eta}=\langle u,u\rangle_{\eta}^{\frac{1}{2}}$. In the following,
for $\eta>0$ small, we construct a $k$-peak solution $u_{\eta}$ to (4.7)
concentrating at $b_{1},\ldots,b_{k}$.

######  Proposition 4.2.

There exists an $\eta_{0}>0$, such that for any $\eta\in(0,\eta_{0}]$, and
$d(z_{i},\Gamma_{i})=O(\eta)$, there exist $\omega_{\eta,z}$ with
$z=(z_{1},\ldots,z_{k})$, such that

$\displaystyle\int_{\mathbb{R}^{N}}\big{(}\eta^{2s}(-\Delta)^{\frac{s}{2}}w_{\eta}(-\Delta)^{\frac{s}{2}}\psi+(1+\eta^{2s}V(x))w_{\eta}\psi\big{)}=\int_{\mathbb{R}^{N}}w_{\eta}^{p}\psi,\
\ for\ all\ \psi\in F_{\eta,z},$

where $w_{\eta}(x)=\displaystyle\sum_{i=1}^{k}U_{\eta,z}+\omega_{\eta,z}(x)$,
and

$F_{\eta,z}=\Big{\\{}\psi(x)\in H^{s}(\mathbb{R}^{N}):\langle v,\frac{\partial
U_{\eta,z}}{\partial
x_{j}}\rangle_{\eta}=0,j=1,\ldots,N,i=1,\ldots,k\Big{\\}}.$

Moreover, for $N\geq\max\\{2-2s,2s\\}$

$\displaystyle\begin{split}\|\omega_{\eta,z}\|_{\eta}=&O\Big{(}\sum_{i=1}^{k}|V(z_{i})-V_{i}|\eta^{\frac{N}{2}+2s}+\sum_{i=1}^{k}|\nabla
V(z_{i})|\eta^{\frac{N}{2}+2s+1}+\eta^{\frac{N}{2}+2s+2}\Big{)}\\\
&+\begin{cases}O\big{(}\eta^{\frac{N}{2}+\frac{p}{2}(N+2s)}\big{)},\,&\text{if}\
1<p\leq 2\\\\[5.69054pt]
O\big{(}\eta^{\frac{N}{2}+(N+2s)}\big{)},\,&\text{if}\
p>2\end{cases}=O\Big{(}\eta^{\frac{N}{2}+2s+\gamma}\Big{)}\end{split}$ (4.8)

with $\gamma=\max\\{0,1-2s\\}$.

######  Remark 4.3.

The proof of the existence result is standard, and we omit the details. The
last inequality in (4.8) holds because that $N\geq\max\\{2-2s,2s\\}$ implies
$\frac{p}{2}(N+2s)>2s+\gamma$.

It is standard to know that, to obtain a true solution of (4.7), we need to
choose $z$ such that

$\displaystyle-\eta^{2s}\int_{\partial^{\prime\prime}\mathbf{B}_{\rho}^{+}(x_{a,i})}t^{1-2s}\frac{\partial\tilde{\omega}_{\eta,z}}{\partial\nu}\frac{\partial\tilde{\omega}_{\eta,z}}{\partial
y^{j}}+\frac{1}{2}\eta^{2s}\int_{\partial^{\prime\prime}\mathbf{B}_{\rho}^{+}(x_{a,i})}t^{1-2s}|\nabla\tilde{\omega}_{\eta,z}|^{2}\nu_{j}$
$\displaystyle=-\int_{\partial
B_{\rho}(x_{a,i})}\big{(}1+\eta^{2s}V(x)\big{)}\omega_{\eta,z}\frac{\partial\omega_{\eta,z}}{\partial
y^{j}}+\int_{\partial
B_{\rho}(x_{a,i})}\omega_{\eta,z}^{p}\frac{\partial\omega_{\eta,z}}{\partial
y^{j}},$

which, similar to the proof of (4.5), is equivalent to

$\displaystyle\int_{B_{\rho}(x_{a,i})}\frac{\partial V(x)}{\partial
x_{j}}w_{\eta,z}^{2}=O\big{(}\eta^{2N}\big{)}.$

We consider $z=(z_{1},\ldots,z_{k})$, $z_{i}$ close to $b_{i}$,
$z_{i}\in\Gamma_{t,i}$, for some $t$ close to $V_{i}$. Denote $\nu_{i}$ as the
unit normal vector of $\Gamma_{t,i}$ at $z_{i}$, while $\tau_{i,j}$ as the
principal directions of $\Gamma_{t,i}$, with $i=1,\ldots,k,\ j=1,\ldots,N-1$.
Then, at $z_{i}$, there holds that

$D_{\tau_{i,j}}V(z_{i})=0,~{}~{}\mbox{for}~{}~{}j=1,\ldots,N-1,~{}~{}\mbox{and}~{}~{}|\nabla
V(z_{i})|=|D_{\nu_{i}}V(z_{i})|.$

Moreover, we have the following results.

######  Lemma 4.4.

Under the assumption (V), then

$\displaystyle\int_{B_{\rho}(x_{a,i})}D_{\nu_{i}}V(x)w_{\eta,z}^{2}=O(\eta^{2N})$

is equivalent to

$\displaystyle D_{\nu_{i}}V(z_{i})=O(\eta^{2s+\gamma}).$

###### Proof.

On the one hand,

$\displaystyle\int_{B_{\rho}(x_{a,i})}D_{\nu_{i}}V(x)U_{\eta,z_{i}}^{2}(x)$
$\displaystyle=-2\int_{B_{\rho}(x_{a,i})}D_{\nu_{i}}V(x)U_{\eta,z_{i}}(x)\omega_{\eta,z}-\int_{B_{\rho}(x_{a,i})}D_{\nu_{i}}V(x)\omega_{\eta,z}^{2}+O(\eta^{2N})$
$\displaystyle=O(\eta^{N+2s+\gamma}).$ (4.9)

On the other hand,

$\displaystyle\int_{B_{\rho}(x_{a,i})}D_{\nu_{i}}V(x)U_{\eta,z_{i}}^{2}(x)=a_{*}\eta^{N}D_{\nu_{i}}V(z_{i})+O(\eta^{N+2})$

which, combined with (4.2), gives that

$D_{\nu_{i}}V(z_{i})=O(\eta^{2s+\gamma}).$

∎

######  Lemma 4.5.

Under the assumption (V), then

$\displaystyle\int_{B_{\rho}(x_{a,i})}D_{\tau_{i}}V(x)w_{\eta,z}^{2}=O(\eta^{2N})$

is equivalent to

$\displaystyle(D_{\tau_{i}}\Delta)V(z_{i})=O\Big{(}\sum_{i=1}^{k}|V(z_{i})-V_{i}|\eta^{2s-1}+\eta^{2s}\Big{)}.$

###### Proof.

Let $H(x)=\langle\nabla V(x),\tau_{i}\rangle$. On the one hand,

$\displaystyle\int_{B_{\rho}(x_{a,i})}H(x)U_{\eta,z_{i}}^{2}(x)$
$\displaystyle=-2\int_{B_{\rho}(x_{a,i})}H(x)U_{\eta,z_{i}}(x)\omega_{\eta,z}-\int_{B_{\rho}(x_{a,i})}H(x)\omega_{\eta,z}^{2}+O(\eta^{2N})$
$\displaystyle=O\left((\eta^{\frac{N}{2}+1}|\nabla
H(z_{i})|+\eta^{\frac{N}{2}+2})\|\omega_{\eta,z}\|_{\eta}+|\nabla
H(z_{i})|\|\omega_{\eta,z}\|_{\eta}^{2}\right)+O(\eta^{2N}).$ (4.10)

On the other hand, since $H(x_{a,i})=0$, we set

$\displaystyle B=\frac{1}{N}\int_{\mathbb{R}^{N}}|x|^{2}U^{2},$ (4.11)

and then

$\displaystyle\int_{B_{\rho}(x_{a,i})}H(x)U_{\eta,z_{i}}^{2}(x)=\frac{B}{2}\eta^{N+2}\Delta
H(z_{i})+O(\eta^{N+4})$

which, combined with (4.2), gives the result.

∎

######  Theorem 4.6.

For $\lambda>0$ large, the problem (1.7) has a solution $w_{\lambda}$ of the
form

$\displaystyle
w_{\lambda}(x)=\sum_{i=1}^{k}\lambda^{\frac{1}{p-1}}U\Big{(}\lambda^{\frac{1}{2s}}(x-x_{\lambda,i})\Big{)}+\omega_{\lambda},$

where $x_{\lambda,i}\rightarrow b_{i}$ and
$\displaystyle\int_{\mathbb{R}^{N}}\big{(}|(-\Delta)^{\frac{s}{2}}\omega_{\lambda}|^{2}+\omega_{\lambda}^{2}\big{)}\rightarrow
0$ as $\lambda\rightarrow+\infty$.

###### Proof.

We are sufficed to solve (4.7). From Lemma 4.4 and Lemma 4.5, we know that
solving (4.7) is equivalent to solving $d(z_{i},\Gamma_{i})=O(\eta)$,

$D_{\nu_{i}}V(z_{i})=O\big{(}\eta^{2s+\gamma}\big{)},\ \ \
(D_{\tau_{i}}\Delta)V(z_{i})=O\Big{(}\sum_{i=1}^{k}|V(z_{i})-V_{i}|\eta^{2s-1}+\eta^{2s}\Big{)}.$

Now we take $\bar{z}_{i}\in\Gamma_{i}$ to be the point such that
$z_{i}-\bar{z}_{i}=\alpha_{i}\nu_{i}$ for some $\alpha_{i}\in\mathbb{R}$.
Then, we have $D_{\nu_{i}}V(\bar{z}_{i})=0$. Then,

$D_{\nu_{i}}V(z_{i})=D_{\nu_{i}}V(z_{i})-D_{\nu_{i}}V(\bar{z}_{i})=D^{2}_{\nu_{i}\nu_{i}}V(\bar{z}_{i})\big{\langle}z_{i}-\bar{z}_{i},\nu_{i}\big{\rangle}+O\big{(}|z_{i}-\bar{z}_{i}|^{2}\big{)}.$

By the non-degenerate assumption, it holds that

$D_{\nu_{i}}V(z_{i})=O\big{(}\eta^{2s+\gamma}\big{)}\Leftrightarrow\langle
z_{i}-\bar{z}_{i},\nu_{i}\rangle=O(\eta^{2s+\gamma}+|z_{i}-\bar{z}_{i}|^{2}),$

which means that

$\displaystyle|z_{i}-\bar{z}_{i}|=O(\eta^{2s+\gamma})\ \Rightarrow\
d(z_{i},\Gamma_{i})=|z_{i}-\bar{z}_{i}|=O(\eta).$ (4.12)

Let $\bar{\tau}_{i,j}$ be the $j$-th tangential unit vector of $\Gamma_{i}$ at
$\bar{z}_{i}$. By the assumption (V), we have

$(D_{\tau_{i,j}}\Delta)V(z_{i})=(D_{\bar{\tau}_{i,j}}\Delta)V(\bar{z}_{i})+O(|z_{i}-\bar{z}_{i}|)=(D_{\bar{\tau}_{i,j}}\Delta)V(\bar{z}_{i})+O(\eta^{2s+\gamma}),$

and

$\begin{split}(D_{\bar{\tau}_{i,j}}\Delta)V(\bar{z}_{i})=&(D_{\bar{\tau}_{i,j}}\Delta)V(\bar{z}_{i})-(D_{\tau_{i,j,0}}\Delta)V(b_{i})\\\
=&\big{\langle}(\nabla_{T}D_{\tau_{i,j,0}}\Delta
V)(b_{i}),\bar{z}_{i}-b_{i}\big{\rangle}+O\big{(}|\bar{z}_{i}-b_{i}|^{2}\big{)},\end{split}$

where $\nabla_{T_{i}}$ is the tangential gradient on $\Gamma_{i}$ at
$b_{i}\in\Gamma_{i}$, and $\tau_{i,j,0}$ is the $j$-th tangential unit vector
of $\Gamma_{i}$ at $b_{i}$. Therefore, we have

$(D_{\tau_{i,j}}\Delta)V(z_{i})=O\Big{(}\sum_{i=1}^{k}|V(z_{i})-V_{i}|\eta^{2s-1}+\eta^{2s}\Big{)},$

which implies that

$\displaystyle\Big{\langle}(\nabla_{T}D_{\tau_{i,j,0}}\Delta
V)(b_{i}),\bar{z}_{i}-b_{i}\Big{\rangle}=O\Big{(}\sum_{i=1}^{k}|V(z_{i})-V_{i}|\eta^{2s-1}+\eta^{2s}\Big{)}=O\Big{(}\eta^{2s}\Big{)}.$
(4.13)

Hence, we can solve (4.12) and (4.13) to obtain $z_{i}=x_{\eta,i}$ with
$x_{\eta,i}\rightarrow b_{i}$. ∎

Now we are in a position to prove Theorem 1.4.

###### Proof of Theorem 1.4.

Let $w_{\lambda}$ be a peak solution obtained in Theorem 4.6. Define
$u_{\lambda}=w_{\lambda}/\Big{(}\displaystyle\int_{\mathbb{R}^{N}}w_{\lambda}^{2}\Big{)}^{\frac{1}{2}}$.
Then $\displaystyle\int_{\mathbb{R}^{N}}u_{\lambda}^{2}=1$ and

$\displaystyle(-\Delta)^{s}u_{\lambda}+V(x)u_{\lambda}=a_{\lambda}u_{\lambda}^{p}-\lambda
u_{\lambda},\ \ a_{\lambda}=\int_{\mathbb{R}^{N}}w_{\lambda}^{2}.$ (4.14)

Similar to the proof of Proposition 3.4, there holds that

$\displaystyle\lambda^{\frac{N}{2s}-\frac{2}{p-1}}\Big{(}\int_{\mathbb{R}^{N}}w_{\lambda}^{2}\Big{)}^{\frac{2}{p-1}}=ka_{*}+o(1),\
as\ a\rightarrow a_{0}$

with $a_{0}=0$ if $\frac{N}{2s}-\frac{2}{p-1}>0$,
$a_{0}=(ka_{*})^{\frac{p-1}{2}}$ if $\frac{N}{2s}-\frac{2}{p-1}=0$, while
$a_{0}=+\infty$ if $\frac{N}{2s}-\frac{2}{p-1}<0$.

Take $\lambda_{0}>0$ large and
$b_{0}=\displaystyle\int_{\mathbb{R}^{N}}w_{\lambda_{0}}^{2}$. Then, if
$\frac{N}{2s}-\frac{2}{p-1}>0$, for any $a\in(0,b_{0})$, by the mean value
theorem, there exists some large $\lambda_{a}$, such that the solution $u_{a}$
to (1.1) with $\lambda=\lambda_{a}$ satisfies
$\displaystyle\int_{\mathbb{R}^{N}}w_{\lambda}^{2}=a$, and for such $a$, we
obtain a $k$-peak solution for (1.7), with $\mu_{a}=-\lambda_{a}$.

Similarly, if $\frac{N}{2s}-\frac{2}{p-1}=0$, for any $a$ between $b_{0}$ and
$(ka_{*})^{\frac{p-1}{2}}$, while if $\frac{N}{2s}-\frac{2}{p-1}<0$, for any
$a>b_{0}$ large enough, there exists $\lambda_{a}$, such that the solution
$u_{a}$ to (1.1) with $\lambda=\lambda_{a}$ satisfies
$\displaystyle\int_{\mathbb{R}^{N}}w_{\lambda}^{2}=a$, and for such $a$, we
obtain a $k$-peak solution for (1.7), with $\mu_{a}=-\lambda_{a}$.

∎

###  4.3. Clustering peak solutions

In the end of this section, we are concerned with the clustering multi-peak
solutions to problem (1.1)–(1.2), and sketch the proof of Theorem 1.5.

The function $\Delta V(x)|_{x\in\Gamma_{i}}$ has a minimum point and a maximum
point. Let us assume that the function $\Delta V(x)|_{x\in\Gamma_{i_{0}}}$ has
isolated maximum point $b\in\Gamma_{i_{0}}$, that is $\Delta V(x)<\Delta V(b)$
for all $x\in\Gamma_{i_{0}}\cap(B_{\delta}(b)\setminus b)$. Let
$x_{\eta,j}\rightarrow b,j=1,\ldots,k$,
$\frac{|x_{\eta,i}-x_{\eta,j}|}{\eta}\rightarrow+\infty,i\neq j$ as
$\eta\rightarrow 0$. We take $\displaystyle\sum_{j=1}^{k}U_{\eta,x_{\eta,j}}$
as the approximate solution of (4.7) and sufficed to show the following
result.

######  Proposition 4.7.

Assume ($V$) and $\frac{\partial^{2}V(x)}{\partial\nu_{i_{0}}^{2}}\neq 0$ for
any $x\in\Gamma_{i_{0}}$ with some $i_{0}\in\\{1,\ldots,k\\}$. If
$b\in\Gamma_{i_{0}}$ is an isolated maximum point of $\Delta
V(x)|_{x\in\Gamma_{i_{0}}}$ on $\Gamma_{i_{0}}$, then for any integer $k>0$,
there exists an $\eta_{0}>0$, such that for any $\eta\in(0,\eta_{0}]$, problem
(4.7) has a solution $w_{\eta}$ of the form

$\displaystyle w_{\eta}(x)=\sum_{j=1}^{k}U_{\eta,x_{\eta,j}}+\omega_{\eta},$

where $x_{\eta,j}\to b$, $j=1,\cdots,k$,
$\frac{|x_{\eta,i}-x_{\eta,j}|}{\eta}\to+\infty$, $i\neq j$, and
$\displaystyle\int_{\mathbb{R}^{N}}\bigl{(}\eta^{2s}|(-\Delta)^{\frac{s}{2}}\omega_{\eta}|^{2}+\omega_{\eta}^{2}\bigr{)}=o(\eta^{\frac{N}{2}})$
as $\eta\to 0$.

###### Proof.

Define the energy functional

$I(u)=\frac{1}{2}\int_{\mathbb{R}^{N}}\Bigl{(}\eta^{2s}|(-\Delta)^{\frac{s}{2}}u|^{2}+\big{(}1+\eta^{2s}V(x)\big{)}u^{2}\Bigr{)}-\frac{1}{p+1}\int_{\mathbb{R}^{N}}u^{p+1}.$

We obtain the energy expansion:

$\begin{split}&I\Bigl{(}\sum_{j=1}^{k}U_{\eta,x_{\eta,j}}\Bigr{)}\\\
=&(1+\eta^{2s}V_{i_{0}})^{-\frac{N}{2s}}\Big{(}kE_{1}\eta^{N}+E_{2}\eta^{N+2s}\sum_{j=1}^{k}\frac{\partial^{2}V(\bar{x}_{\eta,j})}{\partial\nu_{i,j}^{2}}r_{\eta,j}^{2}+E_{3}\eta^{N+2s+2}\sum_{j=1}^{k}\Delta
V(\bar{x}_{\eta,j})\\\
&-\sum_{j>m}(c_{0}+o(1))\eta^{N}\Bigl{(}\frac{\eta}{|x_{\eta,m}-x_{\eta,j}|}\Bigr{)}^{N+2s}\\\
&+O\Big{(}\eta^{N+2s+3}+\eta^{N}\sum_{j>m}\bigl{(}\frac{\eta}{|x_{\eta,m}-x_{\eta,j}|}\bigr{)}^{N+2s+\tau}+\eta^{N+2s}r_{\eta,j}^{3})\Big{)},\end{split}$
(4.15)

where $E_{l},c_{0}>0,l=2,3$,

$E_{1}=(\frac{1}{2}-\frac{1}{p+1})\displaystyle\int_{\mathbb{R}^{N}}U^{p+1}>0~{}~{}\mbox{and}~{}~{}r_{\eta,j}=|x_{\eta,j}-\bar{x}_{\eta,j}|,$

with $\bar{x}_{\eta,j}\in\Gamma_{i_{0}}$ the point such that
$|x_{\eta,j}-\bar{x}_{\eta,j}|=d(x_{\eta,j},\Gamma_{i_{0}})$.

In fact, it is easy to get

$\begin{split}&I\Bigl{(}\sum_{j=1}^{k}U_{\eta,x_{\eta,j}}\Bigr{)}\\\
=&(1+\eta^{2s}V_{i_{0}})^{-\frac{N}{2s}}\Big{(}k(\frac{1}{2}-\frac{1}{p+1})\displaystyle\int_{\mathbb{R}^{N}}U^{p+1}+E_{2}^{\prime}\eta^{N+2s}\sum_{j=1}^{k}\big{(}V(x_{\eta,j})-V(\bar{x}_{\eta,j})\big{)}\\\
&+E_{3}^{\prime}\eta^{N+2s+2}\sum_{j=1}^{k}\Delta
V(x_{\eta,j})-\sum_{j>m}\big{(}c_{0}+o(1)\big{)}\eta^{N}\Bigl{(}\frac{\eta}{|x_{\eta,m}-x_{\eta,j}|}\Bigr{)}^{N+2s}\\\
&+O(\eta^{N+2s+3}+\eta^{N}\sum_{j>m}\Bigl{(}\frac{\eta}{|x_{\eta,m}-x_{\eta,j}|}\Bigr{)}^{N+2s+\tau})\Big{)},\end{split}$
(4.16)

and

$V(x_{\eta,j})=1+\frac{\partial^{2}V(\bar{x}_{\eta,j})}{\partial\nu_{i,j}^{2}}r_{\eta,j}^{2}+O(r_{\eta,j}^{3})~{}\mbox{and}~{}\Delta
V(x_{\eta,j})=\Delta V(\bar{x}_{\eta,j})+O(r_{\eta,j}).$ (4.17)

So, (4.15) follows from (4.16) and (4.17).

To obtain a solution $w_{\eta}$ of the form
$\displaystyle\sum_{j=1}^{k}U_{\eta,x_{\eta,j}}+\omega_{\eta}$, we can first
carry out the reduction argument as in Proposition 4.2 to obtain
$\omega_{\eta}$, satisfying

$\|\omega_{\eta}\|_{\eta}=O\Big{(}\eta^{\frac{N}{2}+2s+1}+\eta^{\frac{N}{2}}\sum_{i\neq
j}\big{(}\frac{\eta}{|x_{\eta,i}-x_{\eta,j}|}\big{)}^{\min\\{\frac{p}{2},1\\}(N+2s)}\Big{)}.$
(4.18)

Define

$K(x_{\eta,1},\cdots,x_{\eta,k})=I\bigl{(}U_{\eta,x_{\eta,j}}+\omega_{\eta}\bigr{)}.$

From (4.18), we get the same expansion (4.15) for
$K(x_{\eta,1},\cdots,x_{\eta,k})$. Now we set

$\mathcal{B}=\bigl{\\{}\bigl{(}r,\bar{x}):r\in(-\delta\eta^{\frac{1}{2}},\delta\eta^{\frac{1}{2}}),\;\bar{x}\in
B_{\delta}(b_{i})\cap\Gamma_{i}\bigr{\\}}.$

If $\Gamma_{i_{0}}$ is a local maximum set of $V(x)$ and
$\frac{\partial^{2}V(\bar{x})}{\partial\nu_{i_{0}}^{2}}<0$ for any
$\bar{x}\in\Gamma_{i_{0}}$, then it is easy to prove that
$K(x_{\eta,1},\cdots,x_{\eta,k})$ has a critical point, which is a maximum
point of $K$ in

$S_{\eta,k}:=\Bigl{\\{}(x_{\eta,1},\cdots,x_{\eta,k}):x_{\eta,j}\in\mathcal{B},~{}~{}\frac{|x_{\eta,j}-x_{\eta,m}|}{\eta}>\eta^{\sigma-1},\,m\neq
j\Bigr{\\}},$

where $\sigma\in(\frac{1}{2},\frac{N-1}{N+2s})$ is some constant such that
both $\sigma>\frac{1}{2}$ and $(1-\sigma)(N+2s)>2s+1$ hold.

If $\Gamma_{i_{0}}$ is a local minimum set of $V(x)$ and
$\frac{\partial^{2}V(\bar{x})}{\partial\nu_{i_{0}}^{2}}>0$ for any
$\bar{x}\in\Gamma_{i_{0}}$, then

$E_{2}\frac{\partial^{2}V(\bar{x})}{\partial\nu_{i_{0}}^{2}}r^{2}+E_{3}\eta^{2}\Delta
V(\bar{x})$

has a saddle point $(0,b)$ in $\mathcal{B}$. We can use a topological argument
as in [13, 14] to prove that $K(x_{\eta,1},\cdots,x_{\eta,k})$ has a critical
point in $S_{\eta,k}$.

∎

##  5\. Local Uniqueness

###  5.1. Precise estimates by Pohozaev identities

From Lemma 3.2, we know that a $k$-peak solution $v_{a}$ to equation (3.1) can
be written as

$\displaystyle
v_{a}(x)=\sum_{i=1}^{k}U_{\varepsilon,x_{a,i}}(x)+\varphi_{a}(x),$

where
$|x_{a,i}-b_{i}|=o(1),d(x_{a,i},\Gamma_{i})=O(\varepsilon),\varepsilon=(-\mu_{a})^{-\frac{1}{2s}}$,

$U_{\varepsilon,x_{a,i}}(x)=\big{(}1+\varepsilon^{2s}V_{i}\big{)}^{\frac{1}{p-1}}U\Big{(}\frac{(1+\varepsilon^{2s}V_{i})^{\frac{1}{2s}}(x-x_{a,i})}{\varepsilon}\Big{)},$

$\varphi_{a}\in\displaystyle\bigcap_{i=1}^{k}E_{a,x_{a,i}}$. For $N\geq 2s+2$,
it holds that $\frac{p}{2}(N+2s)>2s+1$, which implies then

$\displaystyle\|\varphi_{a}\|_{a}=\|\varphi_{a}\|_{\varepsilon}=$
$\displaystyle
O\Big{(}|\sum_{i=1}^{k}\big{(}V(x_{a,i})-V_{i})\big{)}|\varepsilon^{\frac{N}{2}+2s}+|\sum_{i=1}^{k}\nabla
V(x_{a,i})|\varepsilon^{\frac{N}{2}+2s+1}+\varepsilon^{\frac{N}{2}+2s+1}\Big{)}.$
(5.1)

Moreover, $x_{a,i}\in\Gamma_{t_{a},i}$ with some $t_{a}\rightarrow V_{i}$.
Denote $\nu_{a,i}$ and $\tau_{a,i,j}$ as the unit normal vector and the
principal direction of $\Gamma_{t_{a},i}$ at $x_{a,i}$ respectively. Then at
$x_{a,i}$, there hold that

$D_{\tau_{a,i,j}}V(x_{a,i})=0,\ \ |\nabla
V(x_{a,i})|=|D_{\nu_{a,i}}V(x_{a,i})|.$

######  Lemma 5.1.

Under the assumption (V), it holds that

$\displaystyle
D_{\nu_{a,i}}V(x_{a,i})=O\Big{(}\varepsilon^{\min\\{2,2s+1\\}}\Big{)}.$

###### Proof.

Using (4.5), we obtain that

$\displaystyle\int_{B_{\rho}(x_{a,i})}D_{\nu_{a,i}}V(x)v_{a}^{2}=O(\varepsilon^{2N}).$

On the one hand,

$\begin{split}&\int_{B_{\rho}(x_{a,i})}D_{\nu_{a,i}}V(x)U_{\varepsilon,x_{a,i}}^{2}(x)\\\
&=-2\int_{B_{\rho}(x_{a,i})}D_{\nu_{a,i}}V(x)U_{\varepsilon,x_{a,i}}(x)\varphi_{a}-\int_{B_{\rho}(x_{a,i})}D_{\nu_{a,i}}V(x)\varphi_{a}^{2}+O(\varepsilon^{2N})\\\
&=O\big{(}\varepsilon^{\frac{N}{2}}|D_{\nu_{a,i}}V(x_{a,i})|\|\varphi_{a}\|_{a}+\varepsilon^{\frac{N}{2}+1}\|\varphi_{a}\|_{a}+\|\varphi_{a}\|_{a}^{2}\big{)}+O(\varepsilon^{2N})\\\
&=O\Big{(}\sum_{i=1}^{k}|V(x_{a,i})-V_{i}|\varepsilon^{N+2s}+|\sum_{i=1}^{k}\nabla
V(x_{a,i})|\varepsilon^{N+2s+1}+\varepsilon^{N+2s+2}\Big{)}=O(\varepsilon^{N+2s+1}).\end{split}$
(5.2)

On the other hand, using the definition (4.11)

$\begin{split}&\int_{B_{\rho}(x_{a,i})}D_{\nu_{a,i}}V(x)U_{\varepsilon,x_{a,i}}^{2}(x)\\\
&=a_{*}\varepsilon^{N}(1+\varepsilon^{2s}V_{i})^{\frac{2}{p-1}}D_{\nu_{a,i}}V(x_{a,i})+\frac{B}{2}\varepsilon^{N+2}\Delta
D_{\nu_{a,i}}V(x_{a,i})+O(\varepsilon^{N+4}),\end{split}$ (5.3)

which, combined with (5.2), gives that

$D_{\nu_{a,i}}V(x_{a,i})=O\big{(}\varepsilon^{\min\\{2,2s+1\\}}\big{)}.$

∎

Now we take $\bar{x}_{a,i}\in\Gamma_{i}$ be the point such that
$x_{a,i}-\bar{x}_{a,i}=\alpha_{a,i}\nu_{a,i}$ for some
$\alpha_{a,i}\in\mathbb{R}$.

######  Lemma 5.2.

Under the condition (V), we have that

$\begin{cases}\bar{x}_{a,i}-b_{i}=L_{i}\varepsilon^{2}+O(\varepsilon^{2s+2}),\\\\[5.69054pt]
x_{a,i}-\bar{x}_{a,i}=-\frac{B}{2a_{*}}\Delta
D_{\nu_{a,i}}V(b_{i})\Big{(}\frac{\partial^{2}V(b_{i})}{\partial\nu_{i}^{2}}\Big{)}^{-1}\varepsilon^{2}+O(\varepsilon^{2s+2}),\end{cases}$
(5.4)

where B is as in (4.11), $L_{i}$ is a vector depending on $b_{i}$ and
$i=1,...,k.$

###### Proof.

From (5.3), and

$\displaystyle\int_{B_{\rho}(x_{a,i})}D_{\nu_{a,i}}V(x)U_{\varepsilon,x_{a,i}}^{2}(x)$
$\displaystyle=-2\int_{B_{\rho}(x_{a,i})}D_{\nu_{a,i}}V(x)U_{\varepsilon,x_{a,i}}(x)\varphi_{a}-\int_{B_{\rho}(x_{a,i})}D_{\nu_{a,i}}V(x)\varphi_{a}^{2}+O(\varepsilon^{2N})$
$\displaystyle=O\Big{(}\varepsilon^{\frac{N}{2}}|D_{\nu_{a,i}}V(x_{a,i})|\cdot\|\varphi_{a}\|_{a}+\varepsilon^{\frac{N}{2}+1}\|\varphi_{a}\|_{a}+\|\varphi_{a}\|_{a}^{2}\Big{)}+O(\varepsilon^{2N})$
$\displaystyle=O\big{(}\varepsilon^{N+2s+2}\big{)},$

we obtain that

$\displaystyle\big{(}a_{*}+$ $\displaystyle
O(\varepsilon^{2s})\big{)}D_{\nu_{a,i}}V(x_{a,i})+\frac{B}{2}\varepsilon^{2}\Delta
D_{\nu_{a,i}}V(x_{a,i})$ $\displaystyle=$ $\displaystyle
O\Big{(}\varepsilon^{2s+2}+\varepsilon^{2s+1}\sum_{i=1}^{k}|V(x_{a,i})-V_{i}|\Big{)}$
$\displaystyle=$ $\displaystyle
O\Big{(}\varepsilon^{2s+2}+\varepsilon^{2s+1}\sum_{i=1}^{k}|x_{a,i}-\bar{x}_{a,i}|^{2}\Big{)}.$

Since $\frac{\partial^{2}V(b_{i})}{\partial\nu_{i}^{2}}\neq 0$, the outward
vector $\nu_{a,i}(x)$ and the tangential unit vector $\tau_{a,i}(x)$ of
$\Gamma_{a,i}$ at $x_{a,i}$ are Lip-continuous in $W_{\delta,i}$, we obtain
from the above that

$\displaystyle x_{a,i}-\bar{x}_{a,i}=-\frac{B}{2a_{*}}\Delta
D_{\nu_{a,i}}V(b_{i})\Big{(}\frac{\partial^{2}V(b_{i})}{\partial\nu_{i}^{2}}\Big{)}^{-1}\varepsilon^{2}+O\Big{(}\varepsilon^{2s+2}+\varepsilon^{2s+1}\sum_{i=1}^{k}|x_{a,i}-\bar{x}_{a,i}|^{2}\Big{)}.$
(5.5)

From (5.1) and (5.5), for $\frac{4s+4}{N+2s}<p<\frac{N+2s}{N-2s}$ (which can
be satisfied by $N\geq 2s+4$),

$\displaystyle\|\varphi_{a}\|_{a}$
$\displaystyle=O\Big{(}\sum_{i=1}^{k}|x_{a,i}-\bar{x}_{a,i}|\varepsilon^{\frac{N}{2}+2s}+\varepsilon^{\frac{N}{2}+4s+2}+\varepsilon^{\frac{N}{2}+2s+2}\Big{)}=O\big{(}\varepsilon^{\frac{N}{2}+2s+2}\big{)}.$
(5.6)

Let $H(x)=\langle\nabla V(x),\tau_{a,i}\rangle$. Then for $N\geq 2s+4$,

$\displaystyle\int_{B_{\rho}(x_{a,i})}H(x)U_{\varepsilon,x_{a,i}}^{2}(x)$
$\displaystyle=-2\int_{B_{\rho}(x_{a,i})}H(x)U_{\varepsilon,x_{a,i}}(x)\varphi_{a}-\int_{B_{\rho}(x_{a,i})}H(x)\varphi_{a}^{2}$
$\displaystyle=-2\int_{B_{\rho}(x_{a,i})}H(x)U_{\varepsilon,x_{a,i}}(x)\varphi_{a}+O(\|\varphi_{a}\|_{a}^{2})+O(\varepsilon^{2N})$
$\displaystyle=-2\int_{B_{\rho}(x_{a,i})}\langle\nabla
H(x_{a,i}),x-x_{a,i}\rangle
U_{\varepsilon,x_{a,i}}(x)\varphi_{a}+O(\varepsilon^{N+2s+4})$ (5.7)

On the one hand, since $\nabla V=0$, for $x\in\Gamma_{i}$,

$\displaystyle\nabla
H(x_{a,i})=\big{\langle}\nabla^{2}V(x_{a,i}),\tau_{a,i}\big{\rangle}=\big{\langle}\nabla^{2}V(\bar{x}_{a,i}),\bar{\tau}_{a,i}\big{\rangle}+O\big{(}|x_{a,i}-\bar{x}_{a,i}|\big{)}=O\big{(}|x_{a,i}-\bar{x}_{a,i}|\big{)},$
(5.8)

where $\bar{x}_{a,i}\in\Gamma_{i}$ is the point such that
$x_{a,i}-\bar{x}_{a,i}=\beta_{a,i}\nu_{a,i}$ for some
$\beta_{a,i}\in\mathbb{R}$ and $\bar{\tau}_{a,i}$ is the tangential vector of
$\Gamma_{i}$ at $\bar{x}_{a,i}\in\Gamma_{i}$. Hence,

$\displaystyle\int_{B_{\rho}(x_{a,i})}\langle\nabla
H(x_{a,i}),x-x_{a,i}\rangle U_{\varepsilon,x_{a,i}}(x)\varphi_{a}$
$\displaystyle=O\big{(}\varepsilon^{\frac{N}{2}+1}|\nabla
H(x_{a,i})|\cdot\|\varphi_{a}\|_{a}\big{)}=O\big{(}\varepsilon^{N+2s+3}|x_{a,i}-\bar{x}_{a,i}|\big{)}=O\big{(}\varepsilon^{N+2s+5}\big{)}.$

Then by (5.1) and (5.8), it follows

$\displaystyle\int_{B_{\rho}(x_{a,i})}H(x)U_{\varepsilon,x_{a,i}}^{2}(x)=O\big{(}\varepsilon^{N+2s+4}\big{)}.$
(5.9)

On the other hand, applying the Taylor’s expansion, we obtain that

$\begin{split}\int_{B_{\rho}(x_{a,i})}H(x)U_{\varepsilon,x_{a,i}}^{2}(x)=&\frac{B\varepsilon^{N+2}}{2}\big{(}1+V_{i}\varepsilon^{2s}\big{)}(D_{\tau_{a,i}}\Delta
V)(x_{a,i})\\\
&+\frac{H_{\tau_{i}}\varepsilon^{N+4}}{24}+O\big{(}\varepsilon^{N+2s+4}\big{)},\end{split}$
(5.10)

where

$H_{\tau_{i}}=\sum_{l=1}^{2}\sum_{m=1}^{2}\frac{\partial^{4}H(b_{i})}{\partial
x_{l}^{2}\partial x_{m}^{2}}\int_{\mathbb{R}^{N}}x_{l}^{2}x_{m}^{2}U^{2}.$

Combining (5.9) and (5.10), we obtain that

$\displaystyle(D_{\tau_{a,i}}\Delta
V)(x_{a,i})=-\frac{H_{\tau_{i}}\varepsilon^{2}}{12B}+O(\varepsilon^{2s+2}).$
(5.11)

By (5.5), we get

$\begin{split}\big{(}D_{\tau_{a,i}}\Delta\big{)}V(x_{a,i})&=\big{(}D_{\bar{\tau}_{a,i}}\Delta\big{)}V(\bar{x}_{a,i})+\langle
A_{\tau_{i}},x_{a,i}-\bar{x}_{a,i}\rangle+O\big{(}|x_{a,i}-\bar{x}_{a,i}|^{2}\big{)}\\\
&=(D_{\bar{\tau}_{a,i}}\Delta)V(\bar{x}_{a,i})+B_{\tau_{i}}\varepsilon^{2}+O\big{(}\varepsilon^{2s+2}\big{)},\end{split}$
(5.12)

where $A_{\tau_{i}}$ is a vector depending on $b_{i}$ and $B_{i}$ is a
constant depending on $b_{i}$. Moreover,

$\displaystyle\big{(}D_{\bar{\tau}_{a,i}}\Delta\big{)}V(\bar{x}_{a,i})=\big{(}(D^{2}_{\tau_{a,i}}\Delta)V(b_{i})\big{)}(\bar{x}_{a,i}-b_{i})+O\big{(}|\bar{x}_{a,i}-b_{i}|^{2}\big{)}.$
(5.13)

Therefore, by (5.11), (5.12) and (5.13), there hold that

$\displaystyle\big{(}(D^{2}_{\tau_{a,i}}\Delta)V(b_{i})(x_{a,i})\big{)}(\bar{x}_{a,i}-b_{i})=-\big{(}\frac{H_{\tau_{i}}}{12B}+B_{\tau_{a,i}}\big{)}\varepsilon^{2}+O\big{(}\varepsilon^{2s+2}\big{)}+O\big{(}|\bar{x}_{a,i}-b_{i}|^{2}\big{)}.$

Since $\big{(}D^{2}_{\tau_{a,i}}\Delta\big{)}V(b_{i})$ is non-singular, we
conclude the proof. ∎

###  5.2. Local uniqueness

We assume $\frac{4s+4}{N+2s}<p<\frac{N+2s}{N-2s}$, which can be satisfied by
$N\geq 2s+4$. Set

$\displaystyle\delta_{a}=\begin{cases}\Big{(}\frac{a^{\frac{2}{p-1}}}{ka_{*}}\Big{)}^{\frac{1}{N-\frac{4s}{p-1}}},\,\,\,&p-1\neq\frac{4s}{N},\\\\[5.69054pt]
\Big{(}\frac{2s}{(s+2)B_{1}}|ka_{*}-a^{\frac{2}{p-1}}|\Big{)}^{\frac{1}{2(s+1)}},\,\,\,&p-1=\frac{4s}{N},\end{cases}$

where $B_{1}=\frac{1}{N}\displaystyle\sum_{i=1}^{k}\Delta
V(b_{i})\displaystyle\int_{\mathbb{R}^{N}}|x|^{2}U^{2}(x).$

######  Proposition 5.3.

Under the assumption (V), there hold that

$\displaystyle-\mu_{a}\delta_{a}^{2s}=1+\gamma_{1}\delta_{a}^{2s}+O(\delta_{a}^{2s+2}),$
(5.14)

and

$\displaystyle x_{a,i}-b_{i}=\bar{L}_{i}\delta_{a}^{2}+O(\delta_{a}^{2s+2}),\
\ i=1,\ldots,k,$ (5.15)

where $\gamma_{1}$ and the vector $\bar{L}_{i}$ are constants.

###### Proof.

(a) Firstly, if $p-1\neq\frac{4s}{N}$, we follow the proof of Proposition 3.4
to find

$\displaystyle\frac{a^{\frac{2}{p-1}}}{(-\mu_{a})^{\frac{2}{p-1}-\frac{N}{2s}}}$
$\displaystyle=\sum_{i=1}^{k}(1+\frac{V_{i}}{-\mu_{a}})^{\frac{2}{p-1}-\frac{N}{2s}}\int_{\mathbb{R}^{N}}U^{2}+O(|\sum_{i=1}^{k}(V(x_{a,i})-V_{i})|(-\mu_{a})^{-1}$
$\displaystyle\quad+|\sum_{i=1}^{k}\nabla
V(x_{a,i})|(-\mu_{a})^{-1-\frac{1}{2s}}+(-\mu_{a})^{-1-\frac{1}{s}})+\begin{cases}O((-\mu_{a})^{-\frac{pN}{4s}-\frac{p}{2}}),\,&\text{
if}\,\,1<p\leq 2,\\\ O((-\mu_{a})^{-\frac{N}{2s}-1}),\,&\text{
if}\,\,p>2,\end{cases}$
$\displaystyle\quad+O\Big{(}(-\mu_{a})^{-{\frac{N}{2s}-1}}|x_{a,j}-x_{a,i}|^{-(N+2s)}\Big{)},$

which implies that

$\displaystyle\frac{a^{\frac{2}{p-1}}}{ka_{*}(-\mu_{a})^{\frac{2}{p-1}-\frac{N}{2s}}}$
$\displaystyle=1+\sum_{i=1}^{k}(\frac{2}{p-1}-\frac{N}{2s})\frac{V_{i}}{-\mu_{a}}+O\Big{(}|\sum_{i=1}^{k}(V(x_{a,i})-V_{i})|(-\mu_{a})^{-1}$
$\displaystyle\quad+|\sum_{i=1}^{k}\nabla
V(x_{a,i})|(-\mu_{a})^{-1-\frac{1}{2s}}+(-\mu_{a})^{-1-\frac{1}{s}}\Big{)}+\begin{cases}O((-\mu_{a})^{-\frac{pN}{4s}-\frac{p}{2}}),&\text{
if}\,1<p\leq 2,\\\ O((-\mu_{a})^{-\frac{N}{2s}-1}),&\text{
if}\,p>2.\end{cases}$
$\displaystyle\quad+O\Big{(}(-\mu_{a})^{-{\frac{N}{2s}-1}}|x_{a,j}-x_{a,i}|^{-(N+2s)}\Big{)}.$

Set
$\delta_{a}=(\frac{a^{\frac{2}{p-1}}}{ka_{*}})^{\frac{1}{N-\frac{4s}{p-1}}}.$
Then we obtain (5.14). Finally, (5.15) can be obtained by (5.14) and Lemma
5.2.

(b) Secondly, we turn to deal with the case of $p-1=\frac{4s}{N}$. From (5.4),
it holds that

$x_{a,i}-b_{i}=\bar{L}_{i}(-\mu_{a})^{-\frac{1}{s}}+O((-\mu_{a})^{-\frac{1}{s}-1}),$

with some vector $\bar{L}_{i}$. Moreover, in view of (5.6),
$\|\varphi_{a}\|_{a}=O(\varepsilon^{\frac{N}{2}+2s+2})$.

We know that $u_{a}$ is the $k$-peak solutions of (1.1)-(1.2), and we have the
following Pohozaev identity:

$\displaystyle\begin{split}&-\int_{\partial^{\prime\prime}{\bf
B}_{\rho}(x_{a,i})}t^{1-2s}\frac{\partial\tilde{u}_{a}}{\partial\nu}\langle
Y-X_{a,i},\nabla\tilde{u}_{a}\rangle+\frac{1}{2}\int_{\partial^{\prime\prime}{\bf
B}_{\rho}(x_{a,i})}t^{1-2s}|\nabla\tilde{u}_{a}|^{2}\langle
Y-X_{a,i},\nu\rangle\\\ &-\frac{N-2s}{2}\int_{\partial^{\prime\prime}{\bf
B}_{\rho}(x_{a,i})}t^{1-2s}\frac{\partial\tilde{u}_{a}}{\partial\nu}\tilde{u}_{a}\\\
&-\int_{\partial
B_{\rho}(x_{a,i})}\Big{(}\frac{(u_{a})^{p+1}}{p+1}-\frac{-\mu_{a}+V(x)}{2}(u_{a})^{2}\Big{)}\langle
y-x_{a,i},\nu\rangle\\\
=&\int_{B_{\rho}(x_{a,i})}\Big{(}\frac{N-2s}{2}-\frac{N}{p+1}\Big{)}|u_{a}|^{p+1}+s(-\mu_{a}+V(x))(u_{a})^{2}\\\
&+\frac{1}{2}\langle y-x_{a,i},\nabla V(x)\rangle(u_{a})^{2},\end{split}$

which can be rewritten as

$\displaystyle\begin{split}&\int_{B_{\rho}(x_{a,i})}sV(x)(u_{a})^{2}+\frac{1}{2}\langle
y-x_{a,i},\nabla V(x)\rangle(u_{a})^{2}\\\
&=\int_{B_{\rho}(x_{a,i})}s\mu_{a}(u_{a})^{2}+a\Big{(}\frac{N}{p+1}-\frac{N-2s}{2}\Big{)}|u_{a}|^{p+1}+\int_{\partial^{\prime\prime}\bf{B}_{\rho}(x_{a,i})}W_{1}+\int_{\partial
B_{\rho}(x_{a,i})}W_{1},\end{split}$ (5.16)

where

$\displaystyle W_{1}=-t^{1-2s}\frac{\partial\tilde{u}_{a}}{\partial\nu}\langle
Y-X_{a,i},\nabla\tilde{u}_{a}\rangle+\frac{1}{2}t^{1-2s}|\nabla\tilde{u}_{a}|^{2}\langle
Y-X_{a,i},\nu\rangle-\frac{N-2s}{2}t^{1-2s}\frac{\partial\tilde{u}_{a}}{\partial\nu}\tilde{u}_{a},$
$\displaystyle
W_{2}=\Big{(}-\frac{a(u_{a})^{p+1}}{p+1}+\frac{(-\mu_{a}+V(x))}{2}(u_{a})^{2}\Big{)}\langle
y-x_{a,i},\nu\rangle.$

Recall that

$\displaystyle
u_{a}(x)=\sum_{i=1}^{k}(\frac{-\mu_{a}+V_{i}}{a})^{\frac{1}{p-1}}\left(U\big{(}(-\mu_{a}+V_{i})^{\frac{1}{2s}}(x-x_{a,i})\big{)}+\varphi_{a}\right).$

Then,

$\displaystyle\begin{split}&\int_{B_{\rho}(x_{a,i})}sV(x)(u_{a})^{2}+\frac{1}{2}\langle
y-x_{a,i},\nabla V(x)\rangle(u_{a})^{2}\\\
&=(\frac{-\mu_{a}+V_{i}}{a})^{\frac{2}{p-1}}\int_{B_{\rho}(x_{a,i})}\left(s(V(x)-V_{i})+\frac{1}{2}\langle
y-x_{a,i},\nabla
V(x)\rangle\right)U^{2}((-\mu_{a}+V_{i})^{\frac{1}{2s}}(x-x_{a,i}))\\\
&\quad+sV_{i}\int_{B_{\rho}(x_{a,i})}u_{a}^{2}+O\big{(}(-\mu_{a})^{-2-\frac{2}{s}}\big{)}\\\
&=\frac{\frac{s}{2}+1}{(-\mu_{a}+V_{i})^{\frac{1}{s}}}\Delta
V(x_{a,i})\int_{\mathbb{R}^{N}}|x|^{2}U^{2}(x)+sV_{i}\frac{a_{*}}{a^{\frac{2}{p-1}}}+o\big{(}(-\mu_{a}+V_{i})^{-\frac{1}{s}}\big{)}.\end{split}$

Moreover, we calculate more precisely that for some constants $c_{l}$,
$l=1,2,3$,

$\Delta V(x_{a,i})=\Delta
V(b_{i})+c_{1}(-\mu_{a})^{-\frac{1}{s}}+O\big{(}(-\mu_{a})^{-\frac{1}{s}-1}\big{)}.$

Hence,

$\displaystyle\begin{split}&\int_{B_{\rho}(x_{a,i})}sV(x)(u_{a})^{2}+\frac{1}{2}\langle
y-x_{a,i},\nabla V(x)\rangle(u_{a})^{2}\\\
&=\frac{\frac{s}{2}+1}{(-\mu_{a}+V_{i})^{\frac{1}{s}}}\Delta
V(b_{i})\int_{\mathbb{R}^{N}}|x|^{2}U^{2}(x)+sV_{i}\frac{a_{*}}{a^{\frac{2}{p-1}}}+c_{2}(-\mu_{a})^{-\frac{1}{s}-1}+O\big{(}(-\mu_{a})^{-\frac{2}{s}-1}\big{)}.\end{split}$

Now summing (5.16) from $i=1$ to $i=k$, since
$\frac{N}{p+1}-\frac{N-2s}{2}=\frac{2s^{2}}{N+2s}>0$, we have

$\displaystyle\begin{split}&s\mu_{a}+\frac{2as^{2}}{N+2s}\int_{\mathbb{R}^{N}}|u_{a}|^{p+1}\\\
&=\sum_{i=1}^{k}\frac{\frac{s}{2}+1}{(-\mu_{a})^{\frac{1}{s}}}\Delta
V(b_{i})\int_{\mathbb{R}^{N}}|x|^{2}U^{2}(x)+\sum_{i=1}^{k}sV_{i}\frac{a_{*}}{a^{\frac{2}{p-1}}}+c_{2}(-\mu_{a})^{-\frac{1}{s}-1}+O\big{(}(-\mu_{a})^{-\frac{2}{s}-1}\big{)}.\end{split}$
(5.17)

On the other hand, by the orthogonality
$\displaystyle\int_{\mathbb{R}^{N}}U^{p}\big{(}(-\mu_{a}+V_{i})^{\frac{1}{2s}}(x-x_{a,i})\big{)}\varphi_{a}=0$
for $i=1,\ldots,k$, there holds that

$\displaystyle\begin{split}\int_{\mathbb{R}^{N}}u_{a}^{p+1}&=\sum_{i=1}^{k}(\frac{-\mu_{a}+V_{i}}{a})^{\frac{p+1}{p-1}}\Big{(}(-\mu_{a}+V_{i})^{-\frac{N}{2s}}\int_{\mathbb{R}^{N}}U^{p+1}\\\
&\quad\quad+(p+1)\int_{\mathbb{R}^{N}}(U^{p}((-\mu_{a}+V_{i})^{\frac{1}{2s}}(x-x_{a,i}))\varphi_{a}+O(|\varphi_{a}|^{p+1}))\Big{)}\\\
&=\frac{N+2s}{2s}\frac{a_{*}}{a^{\frac{p+1}{p-1}}}\sum_{i=1}^{k}(-\mu_{a}+V_{i})^{\frac{p+1}{p-1}-\frac{N}{2s}}+O(\mu_{a}^{\frac{p+1}{p-1}}\|\varphi_{a}\|_{a}^{2})\\\
&=\frac{N+2s}{2s}\frac{a_{*}}{a^{\frac{p+1}{p-1}}}\sum_{i=1}^{k}(-\mu_{a}+V_{i})^{\frac{p+1}{p-1}-\frac{N}{2s}}+O((-\mu_{a})^{-1-\frac{2}{s}})\\\
&=\frac{N+2s}{2s}\frac{a_{*}}{a^{\frac{p+1}{p-1}}}\sum_{i=1}^{k}(-\mu_{a}+V_{i})+O\big{(}(-\mu_{a})^{-1-\frac{2}{s}}\big{)},\end{split}$

which, combined with (5.17), gives then

$\displaystyle\begin{split}&s(-\mu_{a})^{1+\frac{1}{s}}(ka_{*}-a^{\frac{2}{p-1}})=(\frac{s}{2}+1)\sum_{i=1}^{k}\Delta
V(b_{i})\int_{\mathbb{R}^{N}}|x|^{2}U^{2}(x)+c_{3}(-\mu_{a})^{-1}+O\big{(}(-\mu_{a})^{-\frac{1}{s}-1}\big{)}.\end{split}$

If we set
$\delta=\left(\frac{2s}{(s+2)B_{1}}|ka_{*}-a^{\frac{2}{p-1}}|\right)^{\frac{1}{2(s+1)}},$
then we obtain (5.14), and (5.15) can be found by (5.14) and Lemma 5.2.

∎

By a change of variable, the problem (1.1)-(1.2) can be rewritten as

$\displaystyle\delta_{a}^{2s}(-\Delta)^{s}u+\big{(}-\mu_{a}\delta_{a}^{2s}+\delta_{a}^{2s}V(x)\big{)}u=u^{p},\
\ u\in H^{s}(\mathbb{R}^{N}),$ (5.18)
$\displaystyle\int_{\mathbb{R}^{N}}u^{2}=(a\delta_{a}^{2s})^{\frac{2}{p-1}}.$
(5.19)

Similar to Lemma 3.2, the $k$-peak solution of (5.18)–(5.19) concentrating at
$b_{1},\ldots,b_{k}$ can be written as

$u=\sum_{i=1}^{k}\bar{U}_{\delta_{a},x_{a,i}}+\bar{\varphi}_{a}$

with $|x_{a,i}-b_{i}|=o(1)$,
$\|\bar{\varphi}_{a}\|_{\delta_{a}}=o(\delta_{a}^{\frac{N}{2}})$. Moreover,

$\bar{\varphi}_{a}\in\bigcap_{i=1}^{k}\bar{E}_{a,x_{a,i}}:=\Big{\\{}\phi\in
H^{s}(\mathbb{R}^{N}):\langle\phi,\frac{\partial\bar{U}_{\delta_{a},x_{a,i}}}{\partial
x_{j}}\rangle_{\delta_{a}}=0,j=1,\ldots,N\Big{\\}},$

where

$\bar{U}_{\delta_{a},x_{a,i}}(x):=(1+(\gamma_{1}+V_{i})\delta_{a}^{2s})^{\frac{1}{p-1}}U\Big{(}\frac{(1+(\gamma_{1}+V_{i})\delta_{a}^{2s})^{\frac{1}{2s}}(x-x_{a,i})}{\delta_{a}}\Big{)},$
$\|\phi\|_{\delta_{a}}^{2}=\int_{\mathbb{R}^{N}}(\delta_{a}^{2}|(-\Delta)^{\frac{s}{2}}\phi|^{2}+\phi^{2}).$

Now, we write the equation (5.18) as

$\bar{L}_{a}(\bar{\varphi}_{a})=N_{a}(\bar{\varphi}_{a})+\bar{l}_{a}(x),$

where

$\bar{L}_{a}(\bar{\varphi}_{a})=\delta_{a}^{2s}(-\Delta)^{s}\bar{\varphi}_{a}+\big{[}1+(-\mu_{a}\delta^{2s}_{a}-1)+\delta_{a}^{2s}V\big{]}\bar{\varphi}_{a}-p\sum_{i=1}^{k}\bar{U}^{p-1}_{\delta_{a},x_{a,i}}\bar{\varphi}_{a},$

$N_{a},L_{a}$ are defined by (3.2),(3.3), and

$\displaystyle\bar{l}_{a}(x)=\big{(}\mu_{a}\delta_{a}^{2s}+1+\gamma_{1}\delta^{2s}_{a}-\delta_{a}^{2s}V(x)+V_{i}\delta^{2s}_{a}\big{)}\sum_{i=1}^{k}\bar{U}_{\delta_{a},x_{a,i}}+\Big{(}\sum_{i=1}^{k}\bar{U}_{\delta_{a},x_{a,i}}\Big{)}^{p}-\sum_{i=1}^{k}\bar{U}_{\delta_{a},x_{a,i}}^{p}.$

######  Lemma 5.4.

It holds that

$\displaystyle\|\bar{\varphi}_{a}\|_{\delta_{a}}=O\big{(}\delta_{a}^{\frac{N}{2}+2s+2}\big{)}.$
(5.20)

###### Proof.

Similar to Lemma 3.2, we obtain (5.20) only from proving

$\displaystyle\|\bar{l}_{a}\|_{\delta_{a}}=O\Big{(}\sum_{i=1}^{k}|V(x_{a,i})-V_{i}|\delta_{a}^{\frac{N}{2}+2s}+|\sum_{i=1}^{k}\nabla
V(x_{a,i})|\delta_{a}^{\frac{N}{2}+2s+1}+\delta_{a}^{\frac{N}{2}+2s+2}\Big{)}=O\big{(}\delta_{a}^{\frac{N}{2}+2s+2}\big{)}.$

∎

Let $u_{a}^{(1)}$ and $u_{a}^{(2)}$ be two $k$-peak solutions to (5.18) and
(5.19) concentrating at k points $b_{1},\ldots,b_{k}$, of the form

$\displaystyle
u_{a}^{(l)}=\sum_{i=1}^{k}\bar{U}_{\delta_{a},x_{a,i}^{(l)}}+\bar{\varphi}_{a}^{(l)},\
\ \bar{\varphi}_{a}^{(l)}\in\bigcap_{i=1}^{k}\bar{E}_{a,x_{a,i}^{(l)}}\ \
l=1,2.$ (5.21)

Set
$\xi_{a}(x)=\frac{u_{a}^{(1)}(x)-u_{a}^{(2)}(x)}{\|u_{a}^{(1)}-u_{a}^{(2)}\|_{L^{\infty}(\mathbb{R}^{N})}}$.
Then $\|\xi_{a}\|_{L^{\infty}(\mathbb{R}^{N})}=1$. We get from (5.18) that

$\displaystyle\delta^{2s}(-\Delta)^{s}\xi_{a}+C_{a}(x)\xi_{a}=g_{a}(x),$

where

$\displaystyle
C_{a}(x)=\delta^{2s}V(x)-\delta^{2s}\mu_{a}^{(1)}-p\int_{0}^{1}(tu_{a}^{(1)}+(1-t)u_{a}^{(2)})^{p-1}dt,$
$\displaystyle
g_{a}(x)=\frac{\delta_{a}^{2s}(\mu_{a}^{(1)}-\mu_{a}^{(2)})}{\|u_{a}^{(1)}-u_{a}^{(2)}\|_{L^{\infty}(\mathbb{R}^{N})}}u_{a}^{(2)}.$

Following [15] and (4.1), we can prove that

$\displaystyle|\xi_{a}(x)|+|\nabla\xi_{a}(x)|\leq
C\sum_{i=1}^{k}\frac{1}{\left(1+|\frac{x-x_{a,i}}{\delta_{a}}|^{2}\right)^{\frac{N+2s-\theta}{2}}}.$
(5.22)

Applying blowing-up technique, we set

$\bar{\xi}_{a,i}(x)=\xi_{a,i}\Big{(}\frac{\delta_{a}}{P_{i}^{\frac{1}{2s}}}x+x_{a,i}^{(1)}\Big{)},\
\ P_{i}=1+(\gamma_{1}+V_{i})\delta^{2s}.$

The $\bar{\xi}_{a,i}$ satisfies that

$\displaystyle(-\Delta)^{s}\bar{\xi}_{a,i}+\frac{1}{P_{i}}C_{a}(\frac{\delta_{a}}{P_{i}^{\frac{1}{2s}}}x+x_{a,i}^{(1)})\bar{\xi}_{a,i}=\frac{1}{P_{i}}g_{a}(\frac{\delta_{a}}{P_{i}^{\frac{1}{2s}}}x+x_{a,i}^{(1)}).$
(5.23)

######  Lemma 5.5.

For $x\in B_{\delta_{a}^{-1}P_{i}^{\frac{1}{2s}}\rho}(0)$, it holds that

$\displaystyle\frac{1}{P_{i}}C_{a}(\frac{\delta_{a}}{P_{i}^{\frac{1}{2s}}}x+x_{a,i}^{(1)})=1-pU^{p-1}(x)+O\Big{(}\delta_{a}^{2s}+\sum_{l=1}^{2}\varphi_{a}^{(l)}(\frac{\delta_{a}}{P_{i}^{\frac{1}{2s}}}x+x_{a,i}^{(1)})\Big{)},$
(5.24)

and

$\displaystyle\frac{1}{P_{i}}g_{a}(\frac{\delta_{a}}{P_{i}^{\frac{1}{2s}}}x+x_{a,i}^{(1)})=-\frac{p-1}{ka_{*}}U(x)\sum_{j=1}^{k}\int_{\mathbb{R}^{N}}U^{p}(x)\bar{\xi}_{a,j}(x)+O\Big{(}\delta_{a}^{2s+2}+\sum_{l=1}^{2}\varphi_{a}^{(l)}(\frac{\delta_{a}}{P_{i}^{\frac{1}{2s}}}x+x_{a,i}^{(1)})\Big{)}.$
(5.25)

###### Proof.

Since

$\displaystyle C_{a}(\frac{\delta_{a}}{P_{i}^{\frac{1}{2s}}}x+x_{a,i}^{(1)})=$
$\displaystyle\delta^{2s}V(\frac{\delta_{a}}{P_{i}^{\frac{1}{2s}}}x+x_{a,i}^{(1)})-\delta^{2s}\mu_{a}^{(1)}$
$\displaystyle-p\int_{0}^{1}(tu_{a}^{(1)}+(1-t)u_{a}^{(2)})^{p-1}(\frac{\delta_{a}}{P_{i}^{\frac{1}{2s}}}x+x_{a,i}^{(1)})dt,$

we can obtain (5.24) directly.

From (5.18) and (5.19), for $l=1,2$, we get that

$\displaystyle\mu_{a}^{(l)}\delta_{a}^{\frac{4s}{p-1}+2s}a^{\frac{2}{p-1}}=\delta_{a}^{2s}\int_{\mathbb{R}^{N}}\Big{(}|(-\Delta)^{\frac{s}{2}}u_{a}^{(l)}|^{2}+V(x)(u_{a}^{(l)})^{2}\Big{)}-\int_{\mathbb{R}^{N}}|u_{a}^{(l)}|^{p+1},$

which implies that

$\displaystyle\frac{\delta_{a}^{\frac{4s}{p-1}+2s}a^{\frac{2}{p-1}}(\mu_{a}^{(1)}-\mu_{a}^{(2)})}{\|u_{a}^{(1)}-u_{a}^{(2)}\|_{L^{\infty}(\mathbb{R}^{N})}}$
$\displaystyle=$
$\displaystyle\delta_{a}^{2s}\int_{\mathbb{R}^{N}}((-\Delta)^{\frac{s}{2}}u_{a}^{(1)}+(-\Delta)^{\frac{s}{2}}u_{a}^{(2)})\cdot(-\Delta)^{\frac{s}{2}}\xi_{a}+V(x)(u_{a}^{(1)}+u_{a}^{(2)})\xi_{a}$
$\displaystyle-\int_{\mathbb{R}^{N}}\frac{(|u_{a}^{(1)}|^{p+1}-|u_{a}^{(2)}|^{p+1})}{\|u^{(1)}_{a}-u^{(2)}_{a}\|_{L^{\infty}(\mathbb{R}^{N})}}$
$\displaystyle=$
$\displaystyle\delta_{a}^{2s}\int_{\mathbb{R}^{N}}((-\Delta)^{\frac{s}{2}}u_{a}^{(1)}+(-\Delta)^{\frac{s}{2}}u_{a}^{(2)})\cdot(-\Delta)^{\frac{s}{2}}\xi_{a}+V(x)(u_{a}^{(1)}+u_{a}^{(2)})\xi_{a}$
$\displaystyle-\int_{\mathbb{R}^{N}}\frac{|u_{a}^{(1)}|^{p+1}-|u_{a}^{(2)}|^{p+1}}{\|u_{a}^{(1)}-u_{a}^{(2)}\|_{L^{\infty}(\mathbb{R}^{N})}}-\mu_{a}^{(2)}\delta_{a}^{2s}\int_{\mathbb{R}^{N}}(u_{a}^{(1)}+u_{a}^{(2)})\xi_{a}$
$\displaystyle=$
$\displaystyle(\mu_{a}^{(1)}-\mu_{a}^{(2)})\delta^{2s}_{a}\int_{\mathbb{R}^{N}}u_{a}^{(1)}\xi_{a}+\int_{\mathbb{R}^{N}}(|u_{a}^{(1)}|^{p}+|u_{a}^{(2)}|^{p})\xi_{a}-(p+1)\int_{\mathbb{R}^{N}}\xi_{a}\int_{0}^{1}(tu_{a}^{(1)}+(1-t)u_{a}^{(2)})^{p}dt.$
(5.26)

Hence, in view of (5.14), (5.15), (5.20), (5.2) gives that

$\displaystyle\frac{\delta_{a}^{2s}(\mu_{a}^{(1)}-\mu_{a}^{(2)})}{\|u_{a}^{(1)}-u_{a}^{(2)}\|_{L^{\infty}(\mathbb{R}^{N})}}=$
$\displaystyle-\frac{p-1}{ka_{*}\delta_{a}^{N}}\Big{(}\int_{\mathbb{R}^{N}}(|u_{a}^{(1)}|^{p}+|u_{a}^{(2)}|^{p})\xi_{a}-(p+1)\xi_{a}\int_{0}^{1}(tu_{a}^{(1)}+(1-t)u_{a}^{(2)})^{p}dt\Big{)}$
$\displaystyle+O(\delta_{a}^{2s+2}),$

which implies (5.25). ∎

Since $|\bar{\xi}_{a,i}|\leq 1$, we suppose that
$\bar{\xi}_{a,i}\rightarrow\bar{\xi}_{i}(x)$ in $C^{1}_{loc}(\mathbb{R}^{N})$.
Lemma 5.5 actually shows the following result.

######  Lemma 5.6.

There hold that

$\displaystyle(-\Delta)^{s}\bar{\xi}_{i}+(1-pU^{p-1})\bar{\xi}_{i}=-\frac{p-1}{ka_{*}}U(x)\sum_{j=1}^{k}\int_{\mathbb{R}^{N}}U^{p}(x)\bar{\xi}_{j}(x),\
\ i=1,\ldots,k.$ (5.27)

In order to show $\bar{\xi}_{i}=0$, we write

$\displaystyle\bar{\xi}_{a,i}(x)=\sum_{j=0}^{N}\gamma_{a,i,j}\Psi_{j}+\bar{\zeta}_{a,i}(x),\
\ in\ H^{s}(\mathbb{R}^{N}),$ (5.28)

where $\Psi_{j}(j=0,1,\ldots,N)$ are functions in (A.3),
$\bar{\zeta}_{a,i}\in\bar{E}$ and

$\bar{E}=\\{u\in H^{s}(\mathbb{R}^{N}),\langle u,\Psi_{j}\rangle=0,\ for\
j=0,1,\ldots,N\\}.$

It is standard to show the following problem.

######  Lemma 5.7.

For any $u\in\bar{E}$, there holds that

$\displaystyle\|\bar{L}(u)\|\geq c_{0}\|u\|,$

where $c_{0}>0$ is a constant, and the linear operator is defined by

$\displaystyle\bar{L}(u)=(-\Delta)^{s}u+(1-pU^{p-1})u+\frac{p-1}{a_{*}}U(x)\int_{\mathbb{R}^{N}}U^{p}(x)u(x)dx.$
(5.29)

######  Proposition 5.8.

The $\bar{\xi}_{a,i}(x)$ defined by (5.28) satisfies that

$\displaystyle\|\bar{\zeta}_{a,i}\|=O(\delta_{a}^{2s}).$ (5.30)

###### Proof.

Using (5.23)-(5.27), for any $\psi\in H^{s}(\mathbb{R}^{N})$ we have by Lemma
5.5 and (5.20) that

$\displaystyle\langle\bar{L}(\bar{\zeta}_{a,i}),\psi\rangle$
$\displaystyle=\int_{\mathbb{R}^{N}}\big{[}(1-pU^{p-1}-\frac{1}{P_{i}}C_{a}(\frac{\delta_{a}}{P_{i}^{\frac{1}{2s}}}x+x_{a,i}^{(1)}))\bar{\xi}_{a,i}\big{]}\psi$
$\displaystyle\quad+\int_{\mathbb{R}^{N}}\Big{[}\frac{1}{P_{i}}g_{a}(\frac{\delta_{a}}{P_{i}^{\frac{1}{2s}}}x+x_{a,i}^{(1)})+\frac{p-1}{ka_{*}}U(x)\sum_{j=1}^{k}\int_{\mathbb{R}^{N}}U^{p}(x)\bar{\xi}_{a,j}(x)\Big{]}\psi$
$\displaystyle=O(\delta_{a}^{2s})\|\psi\|+O\Big{(}\int_{\mathbb{R}^{N}}\sum_{l=1}^{2}\varphi_{a}^{(l)}(\frac{\delta_{a}}{P_{i}^{\frac{1}{2s}}}x+x_{a,i}^{(1)}))|\psi|\Big{)}$
$\displaystyle=O(\delta_{a}^{2s})\|\psi\|+O(\delta_{a}^{-\frac{N}{2}})\|\varphi_{a}^{(l)}\|_{\delta_{a}}\|\psi\|=O(\delta_{a}^{2s})\|\psi\|,$

which implies (5.30) by Lemma 5.7.

∎

In the following we prove

$\gamma_{a,i,j}=o(1)~{}~{}\mbox{for}~{}~{}j=0,1,\ldots,N.$

First, we have the estimate for $\xi_{a}$.

######  Lemma 5.9.

It holds that

$\displaystyle\begin{split}&\delta_{a}^{2s}\sum_{i=1}^{k}\int_{B_{\rho}(x_{a,i}^{(1)})}\left(sV(x)+\frac{1}{2}\langle
y-x_{a,i}^{(1)},\nabla V(x)\rangle\right)(u_{a}^{(1)}+u_{a}^{(2)})\xi_{a}\\\
=&(\mu_{a}^{(1)}-\mu_{a}^{(2)})\delta_{a}^{2s}\int_{\mathbb{R}^{N}}u_{a}^{(1)}\xi_{a}+\int_{\mathbb{R}^{N}}((u_{a}^{(1)})^{p}+(u_{a}^{(2)})^{p})\xi_{a}\\\
&-(p+1)\int_{\mathbb{R}^{N}}\int_{0}^{1}(tu_{a}^{(1)}+(1-t)u_{a}^{(2)})^{p}\xi_{a}\\\
&+(\frac{N}{p+1}-\frac{N-2s}{2})(p+1)\int_{B_{\rho}(x_{a,i}^{(1)})}\int_{0}^{1}(tu_{a}^{(1)}+(1-t)u_{a}^{(2)})^{p}\xi_{a}+O(\delta_{a}^{2N+2s}).\end{split}$
(5.31)

###### Proof.

Since $u_{a}^{(l)}$ are the $k$-peak solutions of (5.18)–(5.19), we have the
following Pohozaev identity:

$\displaystyle\begin{split}&-\delta_{a}^{2s}\int_{\partial^{\prime\prime}{\bf
B}_{\rho}(x_{a,i}^{(1)})}t^{1-2s}\frac{\partial\tilde{u}_{a}^{(l)}}{\partial\nu}\langle
Y-X_{a,i},\nabla\tilde{u}_{a}^{(l)}\rangle+\frac{\delta_{a}^{2s}}{2}\int_{\partial^{\prime\prime}{\bf
B}_{\rho}(x_{a,i}^{(1)})}t^{1-2s}|\nabla\tilde{u}_{a}^{(l)}|^{2}\langle
Y-X_{a,i},\nu\rangle\\\
&-\frac{N-2s}{2}\delta_{a}^{2s}\int_{\partial^{\prime\prime}{\bf
B}_{\rho}(x_{a,i}^{(1)})}t^{1-2s}\frac{\partial\tilde{u}_{a}^{(l)}}{\partial\nu}\tilde{u}_{a}^{(l)}\\\
&-\int_{\partial
B_{\rho}(x_{a,i}^{(1)})}\Big{(}\frac{(u_{a}^{(l)})^{p+1}}{p+1}-\frac{\delta_{a}^{2s}(-\mu_{a}^{(l)}+V(x))}{2}(u_{a}^{(l)})^{2}\Big{)}\langle
y-x_{a,i},\nu\rangle\\\
=&\int_{B_{\rho}(x_{a,i}^{(1)})}(\frac{N-2s}{2}-\frac{N}{p+1})(u_{a}^{(l)})^{p+1}+s\delta_{a}^{2s}(-\mu_{a}^{(l)}+V(x))(u_{a}^{(l)})^{2}\\\
&+\frac{\delta_{a}^{2s}}{2}\langle y-x_{a,i}^{(1)},\nabla
V(x)\rangle(u_{a}^{(l)})^{2},\end{split}$

which can be rewritten as

$\displaystyle\begin{split}&\delta_{a}^{2s}\int_{B_{\rho}(x_{a,i}^{(1)})}\big{(}sV(x)(u_{a}^{(l)})^{2}+\frac{1}{2}\langle
y-x_{a,i}^{(1)},\nabla V(x)\rangle(u_{a}^{(l)})^{2}\big{)}\\\
&=\int_{B_{\rho}(x_{a,i}^{(1)})}\delta^{2s}_{a}s\mu_{a}^{(l)}(u_{a}^{(l)})^{2}+(\frac{N}{p+1}-\frac{N-2s}{2})(u_{a}^{(l)})^{p+1}\\\
&+\delta_{a}^{2s}\int_{\partial^{\prime\prime}\bf
B_{\rho}(x_{a,i}^{(1)})}W_{1}^{(l)}+\delta_{a}^{2s}\int_{\partial
B_{\rho}(x_{a,i}^{(1)})}W_{2}^{(l)},\end{split}$

where

$\displaystyle
W_{1}^{(l)}=-t^{1-2s}\frac{\partial\tilde{u}_{a}^{(l)}}{\partial\nu}\langle
Y-X_{a,i},\nabla\tilde{u}_{a}^{(l)}\rangle+\frac{1}{2}t^{1-2s}|\nabla\tilde{u}_{a}^{(l)}|^{2}\langle
Y-X_{a,i},\nu\rangle-\frac{N-2s}{2}t^{1-2s}\frac{\partial\tilde{u}_{a}^{(l)}}{\partial\nu}\tilde{u}_{a}^{(l)},$
$\displaystyle
W_{2}^{(l)}=\Big{(}-\frac{(u_{a}^{(l)})^{p+1}}{\delta_{a}^{2s}(p+1)}+\frac{(-\mu_{a}^{(l)}+V(x))}{2}(u_{a}^{(l)})^{2}\Big{)}\langle
y-x_{a,i},\nu\rangle.$

Then, the above local Pohozaev identity implies that

$\displaystyle\begin{split}&\delta_{a}^{2s}\int_{B_{\rho}(x_{a,i}^{(1)})}\Big{(}sV(x)+\frac{1}{2}\langle
y-x_{a,i}^{(1)},\nabla V(x)\rangle\Big{)}(u_{a}^{(l)}+u_{a}^{(2)})\xi_{a}\\\
=&\frac{(\mu_{a}^{(1)}-\mu_{a}^{(2)})\delta_{a}^{2s}}{\|u_{a}^{(l)}-u_{a}^{(2)}\|_{L^{\infty}}}\int_{B_{\rho}(x_{a,i}^{(1)})}(u_{a}^{(1)})^{2}+\mu_{a}^{(2)}\delta_{a}^{2s}\int_{B_{\rho}(x_{a,i}^{(1)})}(u_{a}^{(1)}+u_{a}^{(2)})\xi_{a}\\\
&+(\frac{N}{p+1}-\frac{N-2s}{2})(p+1)\int_{B_{\rho}(x_{a,i}^{(1)})}\int_{0}^{1}(tu_{a}^{(1)}+(1-t)u_{a}^{(2)})^{p}\xi_{a}+J_{a,i,1}+J_{a,i,2},\end{split}$
(5.32)

where

$\displaystyle\begin{split}&J_{a,i,1}=\int_{\partial^{\prime\prime}\bf{B}_{\rho}(x_{a,i}^{(1)})}\frac{\delta_{a}^{2s}(W_{1}^{(1)}-W_{1}^{(2)})}{\|u_{a}^{(l)}-u_{a}^{(2)}\|_{L^{\infty}}}\\\
&=-\delta_{a}^{2s}\int_{\partial^{\prime\prime}\bf{B}_{\rho}(x_{a,i}^{(1)})}t^{1-2s}\frac{\partial\tilde{u}_{a}^{(1)}}{\partial\nu}\langle
Y-X_{a,i},\nabla\tilde{\xi}_{a}\rangle-\delta_{a}^{2s}\int_{\partial^{\prime\prime}\bf{B}_{\rho}(x_{a,i}^{(1)})}t^{1-2s}\frac{\partial\tilde{\xi}_{a}}{\partial\nu}\langle
Y-X_{a,i},\nabla\tilde{u}_{a}^{(2)}\rangle\\\
&+\frac{\delta_{a}^{2s}}{2}\int_{\partial^{\prime\prime}\bf{B}_{\rho}(x_{a,i}^{(1)})}t^{1-2s}\nabla(\tilde{u}_{a}^{(1)}+\tilde{u}_{a}^{(2)})\cdot\nabla\tilde{\xi}_{a}\langle
Y-X_{a,i},\nu\rangle\\\
&-\frac{N-2s}{2}\delta_{a}^{2s}\int_{\partial^{\prime\prime}\bf{B}_{\rho}(x_{a,i}^{(1)})}t^{1-2s}(\frac{\partial\tilde{\xi}_{a}}{\partial\nu}\tilde{u}_{a}^{(1)}-\frac{\partial\tilde{u}_{a}^{(2)}}{\partial\nu}\tilde{\xi}_{a})=O(\delta_{a}^{2s+2N}),\end{split}$

and since $p+1>2$

$\displaystyle\begin{split}&J_{a,i,2}=\int_{\partial
B_{\rho}(x_{a,i}^{(1)})}\frac{\delta_{a}^{2s}(W_{2}^{(1)}-W_{2}^{(2)})}{\|u_{a}^{(l)}-u_{a}^{(2)}\|_{L^{\infty}}}=O(\delta_{a}^{2s+2(N+2s-\theta)}+\delta_{a}^{(p+1)N+2ps})=O(\delta_{a}^{2s+2(N+2s-\theta)}).\end{split}$

Summing (5.32) from $i=1$ to $i=k$ and considering the decay (4.1) and (5.22),

$\displaystyle\begin{split}&\sum_{i=1}^{k}\delta_{a}^{2s}\int_{B_{\rho}(x_{a,i}^{(1)})}\Big{(}sV(x)+\frac{1}{2}\langle
y-x_{a,i}^{(1)},\nabla V(x)\rangle\Big{)}(u_{a}^{(l)}+u_{a}^{(2)})\xi_{a}\\\
=&\frac{(\mu_{a}^{(1)}-\mu_{a}^{(2)})\delta_{a}^{2s}}{\|u_{a}^{(l)}-u_{a}^{(2)}\|_{L^{\infty}}}\int_{\mathbb{R}^{N}}(u_{a}^{(1)})^{2}+\mu_{a}^{(2)}\delta_{a}^{2s}\int_{\mathbb{R}^{N}}(u_{a}^{(1)}+u_{a}^{(2)})\xi_{a}\\\
&+(\frac{N}{p+1}-\frac{N-2s}{2})(p+1)\int_{B_{\rho}(x_{a,i}^{(1)})}\int_{0}^{1}(tu_{a}^{(1)}+(1-t)u_{a}^{(2)})^{p}\xi_{a}+O(\delta_{a}^{2N+2s}),\end{split}$

which, combined with (5.2), gives (5.31), concluding the proof.

∎

In the following, we first show $\gamma_{a,i,0}=o(1)$ for $i=1,\ldots,k$.

######  Lemma 5.10.

It holds that

$\displaystyle\gamma_{a,i,0}=o(1),\ \ i=1,\ldots,k.$

###### Proof.

From (5.15), (5.20), (5.21) and

$\displaystyle
u_{a}^{(2)}(\frac{\delta_{a}}{P_{i}^{\frac{1}{2s}}}x+x_{a,i}^{(1)})=P_{i}U(x)+O(\frac{P_{i}^{\frac{1}{2s}}|x_{a,i}^{(1)}-x_{a,i}^{(2)}|}{\delta_{a}})+\bar{\varphi}_{a}(\frac{\delta_{a}}{P_{i}^{\frac{1}{2s}}}x+x_{a,i}^{(1)}),$

we obtain that

$\displaystyle\begin{split}&\int_{B_{\rho}(x_{a,i}^{(1)})}\Big{(}sV(x)+\frac{1}{2}\langle
y-x_{a,i}^{(1)},\nabla V(x)\rangle\Big{)}(u_{a}^{(l)}+u_{a}^{(2)})\xi_{a}\\\
=&\int_{B_{\rho}(x_{a,i}^{(1)})}\Big{(}s(V(x)-V_{i})+\frac{1}{2}\langle
y-x_{a,i}^{(1)},\nabla
V(x)\rangle\Big{)}\Big{(}\sum_{i=1}^{k}(\bar{U}_{\delta_{a},x_{a,i}^{(1)}}+\bar{U}_{\delta_{a},x_{a,i}^{(2)}})+\bar{\varphi}_{a}^{(1)}+\bar{\varphi}_{a}^{(2)})\Big{)}\xi_{a}\\\
&+\int_{B_{\rho}(x_{a,i}^{(1)})}sV_{i}(u_{a}^{(l)}+u_{a}^{(2)})\xi_{a}\\\
=&2\int_{B_{\rho}(x_{a,i}^{(1)})}\Big{(}s(V(x)-V_{i})+\frac{1}{2}\langle
y-x_{a,i}^{(1)},\nabla
V(x)\rangle\Big{)}\bar{U}_{\delta_{a},x_{a,i}^{(1)}}\xi_{a}\\\
&+\int_{B_{\rho}(x_{a,i}^{(1)})}sV_{i}(u_{a}^{(l)}+u_{a}^{(2)})\xi_{a}+O(\delta_{a}^{N+2s+2}).\end{split}$

From (5.4) (5.15), (5.28) and (5.30), we obtain that

$\displaystyle\begin{split}&\int_{B_{\rho}(x_{a,i}^{(1)})}(V(x)-V_{i})\bar{U}_{\delta_{a},x_{a,i}^{(1)}}\xi_{a}\\\
=&\delta_{a}^{N}\int_{\mathbb{R}^{N}}(V(\frac{\delta_{a}}{P_{i}^{\frac{1}{2s}}}x+x_{a,i}^{(1)})-V_{i})U(x)(\sum_{j=0}^{N}\gamma_{a,i,j}\Psi_{j})+O(\delta_{a}^{N+2s+2})\\\
=&-\frac{B}{2}\Delta
V(b_{i})\gamma_{a,i,0}\delta_{a}^{N+2}+O(\delta_{a}^{N+2s+2}+\delta_{a}^{N+3}),\end{split}$

where $B$ is the constant as defined in (4.11).

Similarly,

$\displaystyle\begin{split}&\int_{B_{\rho}(x_{a,i}^{(1)})}\langle
y-x_{a,i}^{(1)},\nabla
V(x)\rangle\bar{U}_{\delta_{a},x_{a,i}^{(1)}}\xi_{a}=-\frac{B}{2}\Delta
V(b_{i})\gamma_{a,i,0}\delta_{a}^{N+2}+O(\delta_{a}^{N+2s+2}+\delta_{a}^{N+3}).\end{split}$

Therefore, it holds

$\displaystyle\begin{split}LHS\ of\ \eqref{5.41}&=-\frac{2s+1}{4}B\Delta
V(b_{i})\gamma_{a,i,0}\delta_{a}^{N+2+2s}+O(\delta_{a}^{N+6s}+\delta_{a}^{N+3+2s})\\\
&\quad+s\delta_{a}^{2s}V_{i}\int_{B_{\rho}(x_{a,i}^{(1)})}(u_{a}^{(l)}+u_{a}^{(2)})\xi_{a}.\end{split}$
(5.33)

Moreover, we have

$\displaystyle\begin{split}&\int_{B_{\rho}(x_{a,i}^{(1)})}(u_{a}^{(l)}+u_{a}^{(2)})\xi_{a}\\\
&=2\gamma_{a,i,0}\delta_{a}^{N}\int_{\mathbb{R}^{N}}U(U+\frac{p-1}{2s}x\cdot\nabla
U)+O(|x_{a,i}^{(1)}-x_{a,i}^{(1)}|\delta_{a}^{N-1}+\delta_{a}^{\frac{N}{2}}\|\bar{\varphi}_{a}^{(2)}\|_{\delta_{a}})\\\
&=2\gamma_{a,i,0}\delta_{a}^{N}(1-\frac{N(p-1)}{4s})\int_{\mathbb{R}^{N}}U^{2}+O(\delta_{a}^{N+2s+2}+\delta_{a}^{N+3}),\end{split}$

which, combined with (5.33), gives that

$\displaystyle\begin{split}LHS\ of\ \eqref{5.41}&=-\frac{2s+1}{4}B\Delta
V(b_{i})\gamma_{a,i,0}\delta_{a}^{N+2+2s}+O(\delta_{a}^{N+6s}+\delta_{a}^{N+3+2s})\\\
&+sV_{i}a_{*}\gamma_{a,i,0}(1-\frac{N(p-1)}{4s})\delta_{a}^{N+2s}.\end{split}$

On the other hand, for the RHS of (5.31),

$\displaystyle\begin{split}&\int_{\mathbb{R}^{N}}((u_{a}^{(1)})^{p}+(u_{a}^{(2)})^{p})\xi_{a}-(p+1)\int_{\mathbb{R}^{N}}\int_{0}^{1}(tu_{a}^{(1)}+(1-t)u_{a}^{(2)})^{p}\xi_{a}\\\
&+(\frac{N}{p+1}-\frac{N-2s}{2})(p+1)\int_{B_{\rho}(x_{a,i}^{(1)})}\int_{0}^{1}(tu_{a}^{(1)}+(1-t)u_{a}^{(2)})^{p}\xi_{a}\\\
=&\Big{(}N-(p-1)-\frac{(N-2s)(p+1)}{2}\Big{)}\Big{(}1-\frac{N}{p+1}\Big{)}\gamma_{a,i,0}\int_{\mathbb{R}^{N}}U^{p+1}+O(\delta^{N+2+4s}_{a})\\\
=&\Big{(}N-(p-1)-\frac{(N-2s)(p+1)}{2}\Big{)}\Big{(}1-\frac{N(p-1)}{2s(p+1)}\Big{)}\frac{1}{(\frac{N}{s(p+1)}-\frac{N}{2s}+1)}a_{*}\gamma_{a,i,0}\\\
=&-(\tilde{\gamma}+o(1))a_{*}\gamma_{a,i,0}\delta_{a}^{N}\end{split}$

with some constant $\tilde{\gamma}>0$ since $N\geq 2s+2$ and
$p<\frac{N+2s}{N-2s}$. Moreover, by (5.14),(5.15) and (5.20), it holds that

$\displaystyle\begin{split}&(\mu_{a}^{(1)}-\mu_{a}^{(2)})\delta_{a}^{2s}\int_{\mathbb{R}^{N}}u_{a}^{(1)}\xi_{a}=O(\delta_{a}^{N+6s}).\end{split}$

To sum up, we finally obtain that

$\displaystyle\begin{split}&-\frac{2s+1}{4}B\Delta
V(b_{i})\gamma_{a,i,0}\delta_{a}^{N+2+2s}+O(\delta_{a}^{N+6s}+\delta_{a}^{N+3+2s}+\delta_{a}^{2N+2s})\\\
&+sV_{i}a_{*}\gamma_{a,i,0}(1-\frac{N(p-1)}{4s})\delta_{a}^{N+2s}=-(\tilde{\gamma}+o(1))a_{*}\gamma_{a,i,0}\delta_{a}^{N},\end{split}$

which implies $\gamma_{a,i,0}=o(1)$.

∎

######  Lemma 5.11.

It holds that $\gamma_{a,i,j}=o(1)$, for $i=1,\ldots,k$ and $j=1,\ldots,N$.

###### Proof.

This Lemma can be proved just following that in [30], so we only sketch the
main steps as follows.

Step 1: To prove $\gamma_{a,i,N}=O(\delta_{a})$ for $i=1,\cdots,k$.

On one hand, using (4.5), we deduce

$\displaystyle\int_{B_{\rho}(x_{a,i}^{(1)})}\frac{\partial
V(x)}{\partial\nu_{a,i}}A_{a}(x)\xi_{a}=O\big{(}\delta_{a}^{2N}\big{)},$
(5.34)

where $\nu_{a,i}$ is the outward unit vector of $\partial
B_{\rho}(x_{a,i}^{(1)})$ at $x$,
$A_{a}(x)=\displaystyle\sum^{2}_{l=1}u_{a}^{(l)}(x)$.

On the other hand, by (5.15), and the Taylor expansions

$\begin{split}\displaystyle\int_{B_{\rho}(x_{a,i}^{(1)})}&\frac{\partial
V(x)}{\partial\nu_{a,i}}A_{a}(x)\xi_{a}\\\
=&\int_{\mathbb{R}^{N}}\frac{\partial
V(x_{a,i}^{(1)})}{\partial\nu_{a,i}}A_{a}(x)\xi_{a}+\int_{\mathbb{R}^{N}}\Bigl{\langle}\nabla\frac{\partial
V(x_{a,i}^{(1)})}{\partial\nu_{a,i}},x-x_{a,i}^{(1)}\Bigr{\rangle}A_{a}(x)\xi_{a}+O\big{(}\delta_{a}^{N+2}\big{)}\\\
=&-\frac{\partial^{2}V(x_{a,i}^{(1)})}{\partial\nu^{2}_{a,i}}a_{*}\gamma_{a,i,N}\delta_{a}^{N+1}+O\big{(}\delta_{a}^{N+2}\big{)}.\end{split}$
(5.35)

Then (5.34) and (5.35) imply $\gamma_{a,i,N}=O(\delta_{a})$.

Step 2: To prove $\gamma_{a,i,j}=o(1)$ for $i=1,\cdots,k$ and
$j=1,\cdots,N-1$.

Similar to (5.34), we first have

$\int_{B_{\rho}(x_{a,i}^{(1)})}\frac{\partial
V(x)}{\partial\tau_{a,i,j}}A_{a}(x)\xi_{a}=O\big{(}\delta_{a}^{2N}\big{)},~{}\mbox{for}~{}i=1,\cdots,k,~{}~{}j=1,\cdots,N-1.$
(5.36)

Using suitable rotation, we assume that $\tau_{a,i,1}=(1,0,\cdots,0),\cdots$,
$\tau_{a,i,N-1}=(0,\cdots,0,1,0)$ and $\nu_{a,i}=(0,\cdots,0,1)$. Under the
condition ($V$), we know

$\begin{split}\frac{\partial
V(\delta_{a}x+x^{(1)}_{a,i})}{\partial\tau_{a,i,j}}=&\delta_{a}\sum^{N}_{l=1}\frac{\partial^{2}V(x^{(1)}_{a,i})}{\partial
x_{l}\partial\tau_{a,i,j}}x_{l}+\frac{\delta_{a}^{2}}{2}\sum^{N}_{m=1}\sum^{N}_{l=1}\frac{\partial^{3}V(x^{(1)}_{a,i})}{\partial
x_{m}\partial x_{l}\partial\tau_{a,i,j}}x_{m}x_{l}\\\
&+\frac{\delta^{3}_{a}}{6}\sum^{N}_{s=1}\sum^{N}_{m=1}\sum^{N}_{l=1}\frac{\partial^{4}V(x^{(1)}_{a,i})}{\partial
x_{s}\partial x_{l}\partial
x_{m}\partial\tau_{a,i,j}}x_{s}x_{l}x_{m}+o\big{(}\delta_{a}^{3}|x|^{3}\big{)},~{}\mbox{in}~{}B_{d{\delta_{a}^{-1}}}(0).\end{split}$
(5.37)

By (1.6), (5.4), (5.20), (5.28), and the symmetry of $\varphi_{j}(x)$, we
find, for $j=1,\cdots,N-1$,

$\begin{split}\sum^{N}_{m=1}\sum^{N}_{l=1}&\frac{\partial^{3}V(x^{(1)}_{a,i})}{\partial
x_{m}\partial
x_{l}\partial\tau_{a,i,j}}\int_{B_{d\delta_{a}^{-1}}(0)}A_{a}(\delta_{a}x+x^{(1)}_{a,i})\bar{\xi}_{a,i}x_{m}x_{l}\\\
=&B\gamma_{a,i,0}\frac{\partial\Delta
V(x^{(1)}_{a,i})}{\partial\tau_{a,i,j}}+O(\delta_{a}^{2s})=O\big{(}|x^{(1)}_{a,i}-b_{i}|\big{)}+O(\delta_{a}^{2s})=O(\delta_{a}^{2s}).\end{split}$

Also from Step 1, we get

$\begin{split}&\sum^{N}_{s=1}\sum^{N}_{m=1}\sum^{N}_{l=1}\frac{\partial^{4}V(x^{(1)}_{a,i})}{\partial
x_{s}\partial x_{l}\partial
x_{m}\partial\tau_{a,i,j}}\int_{B_{d\delta_{a}^{-1}}(0)}A_{a}(\delta_{a}x+x^{(1)}_{a,i})\bar{\xi}_{a,i}x_{s}x_{l}x_{m}\\\
=&-3B\Big{(}\sum_{q=1}^{N-1}\frac{\partial^{2}\Delta
V(b_{i})}{\partial\tau_{a,i,q}\partial\tau_{a,i,j}}\gamma_{a,i,q}\Big{)}+o\big{(}1\big{)}.\end{split}$

By (A.2), we estimate

$\frac{\partial^{2}V(x^{(1)}_{a,i})}{\partial
x_{l}\partial\tau_{a,i,j}}=-\frac{\partial
V(x^{(1)}_{a,i})}{\partial\nu_{a,i}}\kappa_{i,l}(x^{(1)}_{a,i})\delta_{lj},\quad
l,j=1,\cdots,N-1.$

Since $\frac{\partial V(\bar{x}^{(1)}_{a,i})}{\partial\nu_{a,i}}=0$, from
(5.4), we find

$\begin{split}\frac{\partial^{2}V(x^{(1)}_{a,i})}{\partial
x_{l}\partial\tau_{a,i,j}}=\frac{B}{2a_{*}}\frac{\partial\Delta
V(b_{i})}{\partial\nu_{i}}\delta_{a}^{2}\kappa_{i,l}(b_{i})\delta_{lj}+o(\delta_{a}^{2}).\end{split}$
(5.38)

Therefore from (5.28), (5.20) and (5.38), we get

$\begin{split}\sum^{N}_{l=1}&\frac{\partial^{2}V(x^{(1)}_{a,i})}{\partial
x_{l}\partial\tau_{a,i,j}}\int_{B_{d\delta_{a}^{-1}}(0)}A_{a}(\delta_{a}x+x^{(1)}_{a,i})\bar{\xi}_{a,i}x_{l}=-\frac{B}{2}\frac{\partial\Delta
V(b_{i})}{\partial\nu_{i}}\delta_{a}^{2}\kappa_{i,j}(b_{i})\gamma_{a,i,j}+o(\delta_{a}^{2}).\end{split}$
(5.39)

Combining (5.37)–(5.39), we obtain

$\begin{split}\int_{B_{\rho}(x_{a,i}^{(1)})}&\frac{\partial
V(x)}{\partial\tau_{a,i,j}}A_{a}(x)\xi_{a}\\\
=&-\frac{B}{2}\frac{\partial\Delta
V(b_{i})}{\partial\nu_{i}}\delta_{a}^{N+3}\kappa_{i,j}(b_{i})\gamma_{a,i,j}-\frac{B}{2}\Big{(}\sum_{l=1}^{N-1}\frac{\partial^{2}\Delta
V(b_{i})}{\partial\tau_{i,l}\partial\tau_{i,j}}\gamma_{a,i,j}\Big{)}\delta_{a}^{N+3}+o(\delta_{a}^{N+3}).\end{split}$
(5.40)

From (5.36) and (5.40), we find

$\frac{\partial\Delta
V(b_{i})}{\partial\nu_{i}}\kappa_{i,j}(b_{i})\gamma_{a,i,j}+\Big{(}\sum_{l=1}^{N-1}\frac{\partial^{2}\Delta
V(b_{i})}{\partial\tau_{i,l}\partial\tau_{i,j}}\gamma_{a,i,l}\Big{)}=o(1),$

which implies $\gamma_{a,i,j}=o(1)$ for $i=1,\cdots,k$ and $j=1,\cdots,N-1$. ∎

Now we are ready to prove Theorem 1.6.

###### Proof of Theorem 1.6:.

First, for large fixed $R$, (5.24) and (5.25) give

$C_{a}(x)\geq\frac{1}{2},\;\;|g_{a}(x)|\leq
C\sum^{k}_{i=1}\frac{\delta_{a}^{N+2s}}{|x-x^{(1)}_{a,i}|^{N+2s}},\quad
x\in\mathbb{R}^{N}\backslash\bigcup^{k}_{i=1}B_{R\delta_{a}}(x^{(1)}_{a,i}).$

Using the comparison principle associated to the nonlocal operator
$(-\Delta)^{s}$, we get

$\xi_{a}(x)=o(1),~{}\mbox{in}~{}\mathbb{R}^{N}\backslash\bigcup^{k}_{i=1}B_{R\delta_{a}}(x^{(1)}_{a,i}).$

On the other hand, it follows from (5.28), (5.30), Lemmas 5.10 and 5.11 that

$\xi_{a}(x)=o(1),~{}\mbox{in}~{}\bigcup^{k}_{i=1}B_{R\delta_{a}}(x^{(1)}_{a,i}).$

This is in contradiction with $\|\xi_{a}\|_{L^{\infty}(\mathbb{R}^{N})}=1$. So
$u^{(1)}_{a}(x)\equiv u^{(2)}_{a}(x)$ for $a\rightarrow a_{0}$. ∎

## Appendix

##  A Basic estimates

######  Lemma A.1 (c.f. [37]).

Let $\alpha,\beta\geq 1$ be two constants. For any
$0<\delta\leq\min\\{\alpha,\beta\\}$, and $x_{1}\neq x_{2}$, it holds that

$\displaystyle\frac{1}{(1+|y-x_{1}|)^{\alpha}(1+|y-x_{2}|)^{\beta}}\leq\frac{C}{|x_{1}-x_{2}|^{\delta}}\Big{(}\frac{1}{(1+|y-x_{1}|)^{\alpha+\beta-\delta}}+\frac{1}{(1+|y-x_{2}|)^{\alpha+\beta-\delta}}\Big{)}.$

######  Lemma A.2 (c.f. [25]).

Let $\rho>\theta>0$ be two constants. Suppose
$(y-x)^{2}+t^{2}\geq\rho^{2},t>0$ and $\alpha>N$. Then, when $0<\beta<N$, it
holds that

$\displaystyle\int_{\mathbb{R}^{N}}\frac{1}{(t+|z|)^{\alpha}|y-z-x|^{\beta}}\leq
C\left(\frac{1}{(1+|y-x|)^{\beta}}\frac{1}{t^{\alpha-N}}+\frac{1}{(1+|y-x|)^{\alpha+\beta-N}}\right),$
(A.1)

and

$\displaystyle\int_{\mathbb{R}^{N}\setminus
B_{\theta}(0)}\frac{1}{(t+|z|)^{\alpha}|y-z-x|^{\beta}}\leq\frac{C}{(1+|y-x|)^{\beta}}\frac{1}{\theta^{\alpha-N}};$

when $N<\beta$, we have that

$\displaystyle\int_{\mathbb{R}^{N}\setminus
B_{\theta}(y-x)}\frac{1}{(t+|z|)^{\alpha}|y-z-x|^{\beta}}\leq
C\left(\frac{1}{(1+|y-x|)^{\beta}}\frac{1}{t^{\alpha-N}}+\frac{1}{(1+|y-x|)^{\beta}}\frac{1}{\theta^{\beta-N}}\right),$

where $C>0$ is a constant independent of $\theta$.

######  Remark A.3.

For some $\epsilon\rightarrow 0$, in the case of $\beta<N$, (A.1) implies
directly that

$\displaystyle\begin{split}\int_{\mathbb{R}^{N}}\frac{1}{(t+|z|)^{\alpha}(1+\frac{|y-z-x|}{\epsilon})^{\beta}}&\leq
C\epsilon^{\beta}\int_{\mathbb{R}^{N}}\frac{1}{(t+|z|)^{\alpha}|y-z-x|^{\beta}}\\\
&\leq
C\epsilon^{\beta}\left(\frac{1}{(1+|y-x|)^{\beta}}\frac{1}{t^{\alpha-N}}+\frac{1}{(1+|y-x|)^{\alpha}}\right).\end{split}$

While in the case of $\beta>N$, following the proof in [25], we can get indeed
that,

$\displaystyle\begin{split}\int_{\mathbb{R}^{N}}\frac{1}{(t+|z|)^{\alpha}(1+\frac{|y-z-x|}{\epsilon})^{\beta}}&\leq
C\left(\frac{\epsilon^{\beta}}{(1+|y-x|)^{\beta}}\frac{1}{t^{\alpha-N}}+\frac{\epsilon^{N}}{(1+|y-x|)^{\alpha}}\right)\\\
&\leq
C\epsilon^{N}\left(\frac{1}{(1+|y-x|)^{\beta}}\frac{1}{t^{\alpha-N}}+\frac{1}{(1+|y-x|)^{\alpha}}\right).\end{split}$

Now let $\Gamma\in C^{2}$ be a closed hypersurface in $\mathbb{R}^{N}$. For
$y\in\Gamma$, let $\nu(y)$ and $T(y)$ denote respectively the outward unit
normal to $\Gamma$ at $y$ and the tangent hyperplane to $\Gamma$ at $y$. The
curvatures of $\Gamma$ at a fixed point $y_{0}\in\Gamma$ are determined as
follows. By a rotation of coordinates, we can assume that $y_{0}=0$ and
$\nu(0)$ is the $x_{N}$-direction, and $x_{j}$-direction is the $j$-th
principal direction. In some neighborhood $\mathcal{N}=\mathcal{N}(0)$ of $0$,
we have

$\Gamma=\bigl{\\{}x:x_{N}=\varphi(x^{\prime})\bigr{\\}},$

where $x^{\prime}=(x_{1},\cdots,x_{N-1})$,

$\varphi(x^{\prime})=\frac{1}{2}\sum_{j=1}^{N-1}\kappa_{j}x_{j}^{2}+O\big{(}|x^{\prime}|^{3}\big{)},$

where $\kappa_{j}$, is the $j$-th principal curvature of $\Gamma$ at $0$. The
Hessian matrix $[D^{2}\varphi(0)]$ is given by

$[D^{2}\varphi(0)]=diag[\kappa_{1},\cdots,\kappa_{N-1}].$

Suppose that $W$ is a smooth function, such that $W(x)=a$ for all
$x\in\Gamma$.

######  Lemma A.4 (c.f. [30]).

We have

$\frac{\partial W(x)}{\partial
x_{l}}\Bigr{|}_{x=0}=0,~{}\mbox{~{}}l=1,\cdots,N-1,$
$\frac{\partial^{2}W(x)}{\partial x_{m}\partial
x_{l}}\Bigr{|}_{x=0}=-\frac{\partial W\big{(}x\big{)}}{\partial
x_{N}}\Bigr{|}_{x=0}\kappa_{i}\delta_{ml},~{}\mbox{for}~{}m,l=1,\cdots,N-1,$
(A.2)

where $\kappa_{1},\cdots,\kappa_{N-1}$, are the principal curvatures of
$\Gamma$ at $0$.

##  B Linearization

######  Lemma B.1.

Let $\xi=(\xi_{1},\xi_{2},\ldots,\xi_{k})$ be the solution of the following
problem

$\displaystyle(-\Delta)^{s}\xi_{i}+(1-pU^{p-1})\xi_{i}=-\frac{p-1}{ka_{*}}U(x)\sum_{j=1}^{k}\int_{\mathbb{R}^{N}}U^{p}(x)\xi_{j}(x),\
\ i=1,\ldots,k.$ (A.1)

Then, there holds that

$\displaystyle\xi_{i}(x)=\sum_{j=0}^{N}\gamma_{i,j}\Psi_{j},$ (A.2)

where $\gamma_{i,j}$ are constants, and

$\displaystyle\Psi_{0}=U+\frac{p-1}{2s}x\cdot\nabla U,\ \
\Psi_{j}=\frac{\partial U}{\partial x_{j}},\ j=1,\ldots,N.$ (A.3)

Moreover, $\gamma_{i,0}=\gamma_{l,0}$ for $i,l=1,\ldots,k$.

###### Proof.

Setting

$\bar{L}(u)=(-\Delta)^{s}u+(1-pU^{p-1})u~{}~{}\mbox{and}~{}~{}\Psi^{j}=(\Psi_{j},\ldots,\Psi_{j}),$

then it is easy to find that $\bar{L}(\Psi^{j})=0$ and $\Psi^{j}$ solve (A.1)
for $j=1,\ldots,N$.

On the other hand, we denote

$\Psi=(U+\frac{p-1}{2s}x\cdot\nabla U,\ldots,U+\frac{p-1}{2s}x\cdot\nabla U),$

and then by checking the extension equation associated with (A.1), we could
obtain that $\Psi$ solves (A.1) too. Noting that $\Psi,\Psi^{j}$ are linearly
independent, we obtain (A.2).

Moreover, putting (A.2) into (A.1), we obtain that $\gamma_{i,0}=\gamma_{l,0}$
for any $i,l=1,\ldots,k$.

∎

Acknowledgments The authors would like to thank Professor Wenxiong Chen for
the helpful discussion with him. Guo was supported by NSFC grants
(No.11771469). Luo and Wang were supported by the Fundamental Research Funds
for the Central Universities(No.KJ02072020-0319). Luo was supported by NSFC
grants (No.11701204, No.11831009). Wang was supported by NSFC grants
(No.12071169). Yang was supported by NSFC grants (No.11601194).

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