Multiplicity of Solutions for a Fractional Kirchhoff–Schrödinger Problem with Logarithmic Nonlinearity

Abstract

In this paper, we investigate the multiplicity and concentration of normalized solutions to a fractional Kirchhoff–Schrödinger problem with logarithmic nonlinearity. By combining the Pohozaev identity, the penalization technique, and the concentration–compactness principle, we overcome the twofold difficulties caused by the Kirchhoff term and the logarithmic nonlinearity and establish the validity of the (PS) condition. On this basis, we employ the Ljusternik–Schnirelmann category theory to prove the multiplicity of solutions, linking the number of solutions to the topological category of the set M in which the potential function $V\left(x\right)$ attains its minimum. Finally, we analyze the concentration behavior and algebraic decay properties of these normalized solutions as $\epsilon \to 0$.

1. Introduction

Let $s\in (0,1)$ and $2s<N<4s$. This paper is devoted to the study of the multiplicity and concentration of normalized solutions for the following class of fractional logarithmic Kirchhoff–Schrödinger problems:

$$\left\{\begin{array}{c}{\epsilon }^{2s}\left({\displaystyle a+b{\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{s/2}u\right|}^2\mathrm{d}x}\right){(-\Delta )}^{s}u+V\left(x\right)u=\lambda u+u\log u^2\mathrm{in}{\mathbb{R}}^N,\hfill \\ {\displaystyle {\int }_{{\mathbb{R}}^N}{\left|u\right|}^2\mathrm{d}x=h{\epsilon }^N,}\hfill \end{array}\right.$$

(1)

where $a,b,\epsilon ,h>0$, and $\lambda \in \mathbb{R}$ is an unknown parameter that appears as a Lagrange multiplier. Here, the fractional Laplacian operator ${(-\Delta )}^s$ is defined in the usual sense,

$$\begin{array}{c}\hfill {(-\Delta )}^{s}u\left(x\right)=c(N,s)\mathrm{P}.\mathrm{V}.{\int }_{{\mathbb{R}}^N}{\displaystyle \frac{u\left(x\right)-u\left(y\right)}{{|x-y|}^{N+2s}}}\mathrm{d}y,\end{array}$$

see [1], where $c(N,s)$ is a positive normalization constant and P.V. denotes the Cauchy principal value. In this paper, we always impose the assumption that the potential $V\in C\left({\mathbb{R}}^N\right)\cap L^{\infty }\left({\mathbb{R}}^N\right)$ fulfills the following condition:

(V) there exists a bounded set $\Lambda \subset {\mathbb{R}}^N$ such that

$$-1<V_0:=\underset{x\in \Lambda }{\inf }V\left(x\right)<\underset{x\in \partial \Lambda }{\min }V\left(x\right):=V_{\partial \Lambda }\mathrm{and}M:=\{x\in \Lambda :V\left(x\right)=V_0\}\ne \varnothing .$$

Without loss of generality, we assume that $V\left(0\right)=V_0$. In recent years, the academic community has devoted considerable attention to the study of generalized nonlinear Schrödinger equations, especially those involving fractional operators or nonlocal terms (such as Kirchhoff-type terms).

A typical example is the time-dependent fractional Kirchhoff equation of the following form:

$$\begin{array}{c}\hfill i\epsilon {\displaystyle \frac{\partial \Phi }{\partial t}}={\epsilon }^{2s}\left(a+b{\int }_{{\mathbb{R}}^N}{|{(-\Delta )}^{s/2}\Phi |}^2\mathrm{d}x\right){(-\Delta )}^s\Phi +V\left(x\right)\Phi -f(\Phi ),\mathrm{in}{\mathbb{R}}^N\times {\mathbb{R}}^{+},\end{array}$$

(2)

where $s\in (0,1)$. Such equations have significant physical applications in fields such as nonlinear optics and Bose–Einstein condensation. Usually, a valuable exploration of such equations is to seek their standing wave solutions, that is, solutions of the form $\Phi (x,t)=e^{-i\lambda t/\epsilon }u\left(x\right)$. By substituting this form into the above time-dependent equation, we obtain an elliptic equation for the function $u\left(x\right)$ that no longer depends on time. For our problem, where $f\left(u\right)=u\log u^2$, after simplification, we obtain precisely the fractional logarithmic Kirchhoff equation studied in this paper:

$${\epsilon }^{2s}\left(a+b{\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{s/2}u\right|}^2\mathrm{d}x\right){(-\Delta )}^{s}u+V\left(x\right)u=\lambda u+u\log u^2.$$

Regarding Kirchhoff–Schrödinger equations, numerous results on the existence of solutions have been obtained in recent years. For example, Lin and Zheng [2,3] related the number of solutions to the topological properties of the set of minima of the potential function on the basis of Ljusternik–Schnirelmann theory. Multiplicity and concentration properties have been established for fractional Kirchhoff–Schrödinger–Poisson systems with magnetic fields in ${\mathbb{R}}^3$, and for magnetic Kirchhoff–Schrödinger equations with critical exponents in ${\mathbb{R}}^2$. Furthermore, Bai, Costea, and Zeng [4] proved the existence of solutions for variational–hemivariational inequality systems under nonlinear coupling, while Lin and Zheng in [5] studied the problem of multiple solutions for fractional Schrödinger equations with logarithmic nonlinearity. Motivated by these works, in this paper, we investigate the multiplicity of solutions for fractional Kirchhoff–Schrödinger equations with logarithmic nonlinearity.

The Schrödinger problem involving logarithmic nonlinearity has important mathematical significance. Based on [6], let $\eta \in C^{\infty }\left({\mathbb{R}}^N\right)$ be a radial cut-off function such that $0\le \eta \le 1$, $\eta \left(x\right)=0$ for $\left|x\right|\le 2$, and $\eta \left(x\right)=1$ for $\left|x\right|\ge 3$. Define

$$u\left(x\right)=\eta \left(x\right){\displaystyle \frac{1}{{\left|x\right|}^{\frac{N}{2}}\log \left|x\right|}}(x\in {\mathbb{R}}^N),$$

and it is straightforward to get that ${\int }_{{\mathbb{R}}^N}{u}^2\log u^2\mathrm{d}x=-\infty$ is ambiguously defined in $H^1\left({\mathbb{R}}^N\right)$. To overcome this difficulty, numerous researchers have introduced various techniques (see references [7,8,9,10,11,12,13]). For example, d’Avenia, Montefusco, and Squassina in [10], by using the energy functional defined on $H_{rad}^1\left({\mathbb{R}}^N\right)$ combined with non-smooth critical point theory [14], proved the existence of multiple solutions for this type of equation:

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}-\Delta u+u=u\log u^2,\hfill & \mathrm{in}{\mathbb{R}}^N,\hfill \\ u\in H^1\left({\mathbb{R}}^N\right).\hfill \end{array}\right.\end{array}$$

Subsequently, based on the minimax theorem for lower semicontinuous functions as cited in [15], Squassina and Szulkin [16] obtained the fundamental and multiple geometrically distinct solutions for the problem below:

$$-\Delta u+V\left(x\right)u=Q\left(x\right)u\log u^2,\mathrm{in}{\mathbb{R}}^N.$$

Let $V\left(x\right)$ and $Q\left(x\right)$ be 1-periodic continuous functions such that

$$\underset{x\in {\mathbb{R}}^N}{\min }Q\left(x\right)>0\mathrm{and}\underset{x\in {\mathbb{R}}^N}{\min }\left(V+Q\right)\left(x\right)>0.$$

There are relatively few results on the existence of solutions for the following fractional logarithmic Kirchhoff–Schrödinger problem (where $s\in (0,1)$ and $N>2s$):

$$\left\{\begin{array}{cc}\left({\displaystyle a+b{\mathrm{\int }}_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{s/2}u\right|}^2\mathrm{d}x}\right){(-\Delta )}^{s}u+u=u\log u^2,\hfill & \mathrm{in}{\mathbb{R}}^N,\hfill \\ u\in H^s\left({\mathbb{R}}^N\right).\hfill \end{array}\right.$$

(3)

However, concerning the fractional logarithmic Schrödinger equation, d’Avenia, Squassina, and Zenari [8] adopted a method similar to that in [10] to prove the existence of infinitely many solutions. Through a more refined analysis, Ardila [7] demonstrated the existence and persistence of stationary solutions. Furthermore, Lv and Zheng in [11] studied a fractional p-Laplacian Schrödinger–Kirchhoff equation with both logarithmic and critical nonlinearities and obtained results on the existence of ground state solutions, where $s\in (\frac{3}{4},1),p\ge 2_s^{*}.$

On the other hand, from the perspective of physics, the existence of normalized solutions seems particularly important as it is related to the conservation of mass in physical systems. At the same time, it also accurately reflects the dynamic properties of the stationary wave solutions of Equation (2), such as stability or instability. In fact, there is already a large body of literature concerning normalized solutions for the following equation:

$$\left\{\begin{array}{cc}-\left(a+b{\mathit{\int }}_{{\mathbb{R}}^N}{|\nabla u|}^2\mathrm{d}x\right)\Delta u+V\left(x\right)u=\lambda u+{f\left(\right|u|}^2)u,\hfill & \mathrm{in}{\mathbb{R}}^N,\hfill \\ {\displaystyle {\int }_{{\mathbb{R}}^N}{\left|u\right|}^2\mathrm{d}x=h^2.}\hfill \end{array}\right.$$

(4)

These studies impose various growth assumptions concerning the nonlinear term; see [17,18,19,20,21,22]. Yang, Qi, and Zou [23] studied the existence and multiplicity of normalized solutions for Schrödinger equations with potentials. When the potential function $V\left(x\right)\not\equiv 0$, Ding and Zhong [24] studied the following mass supercritical case using the minimax structure and the properties of the Pohozaev manifold:

$$\left\{\begin{array}{cc}-\Delta u+V\left(x\right)u+\lambda u=f(x,u),\hfill & \mathrm{in}{\mathbb{R}}^N,\hfill \\ {\displaystyle {\mathit{\int }}_{{\mathbb{R}}^N}{\left|u\right|}^2\mathrm{d}x=h^2,}\hfill \end{array}\right.$$

where $V_{\infty }:={\lim }_{\left|x\right|\to \infty }V\left(x\right)\in (-\infty ,+\infty ],\mathrm{and}V\left(x\right)\le V_{\infty }\mathrm{for}\mathrm{all}x\in {\mathbb{R}}^N$. Moreover, the nonlinear term f fulfills the Berestycki–Lions-type assumptions and exhibits subcritical growth. Let $V\in C\left({\mathbb{R}}^N\right)\cap L^{\infty }\left({\mathbb{R}}^N\right)$ be a potential function satisfying

$$V\left(x\right)\ge 0=V\left(0\right)\mathrm{for}\mathrm{all}x\in {\mathbb{R}}^N\mathrm{and}\underset{\left|x\right|\to +\infty }{\lim \inf } V\left(x\right)>0.$$

Furthermore, f is a $C^1$ function that exhibits $L^2$-subcritical growth. Based on Ljusternik–Schnirelmann category theory, the existence of multiple normalized solutions was proved in [25]. The existence of solutions for the following problem was also proved by Alves and Ji [26]:

$$\left\{\begin{array}{cc}-{\epsilon }^2\Delta u+V\left(x\right)u+\lambda u={\left|u\right|}^{q-2}u\hfill & \mathrm{in}{\mathbb{R}}^N,\hfill \\ {\displaystyle {\int }_{{\mathbb{R}}^N}{\left|u\right|}^2\mathrm{d}x=h^2{\epsilon }^N,}\hfill \end{array}\right.$$

where $q\in \left(2,2+\frac{4}{N}\right)$.

Based on the above considerations, this paper is devoted to the study of multiplicity and concentration phenomena for Equation (1) under condition (V). For this purpose, given a $\delta >0$, we define $M_{\delta }$ as follows:

$$\{x\in {\mathbb{R}}^N:\mathrm{dist}(x,M)\le \delta \},$$

i.e., the set of all points whose distance to the set M is no more than $\delta$. Let Y be a closed subset of a topological space $\mathbb{X}$. Then the Ljusternik–Schnirelmann category of Y in $\mathbb{X}$, denoted by ${\mathrm{cat}}_{\mathbb{X}}\left(Y\right)$, is defined as the minimum number of closed and contractible sets in $\mathbb{X}$ required to cover Y. For more details, see [27]. In what follows, we state the main results of this paper.

Theorem 1.

Assume that the potential function $V\left(x\right)$ satisfies condition (V). Then, for any sufficiently small $\delta >0$, there exist constants $h^{*}>0$ and ${\epsilon }_0>0$ such that, for $h>h^{*}$ and $\epsilon \in (0,{\epsilon }_0)$, (1) has at least ${\mathrm{cat}}_{M_{\delta }}\left(M\right)$ weak solutions $(u_j,{\lambda }_j)$, which satisfy ${\int }_{{\mathbb{R}}^N}{\left|u_j\right|}^2\mathrm{d}x=h{\epsilon }^N$ and ${\lambda }_j<0$. Furthermore, if ${\widehat{u}}_{\epsilon }$ denotes one of the solutions and ${\eta }_{\epsilon }$ is a global maximum point of ${\widehat{u}}_{\epsilon }$, then

$$\underset{\epsilon \to 0}{\lim }V\left({\eta }_{\epsilon }\right)=V_0,$$

and there exists a positive constant C such that

$${\widehat{u}}_{\epsilon }\left(x\right)\le {\displaystyle \frac{C{\epsilon }^{N+2s}}{{\epsilon }^{N+2s}+{|x-{\eta }_{\epsilon }|}^{N+2s}}},x\in {\mathbb{R}}^N.$$

In order to transform problem (1) into an equivalent form that is easier to analyze, we perform the change of variables $x\mapsto \epsilon y$, and let $u_{\epsilon }$ be a solution of (1), defining a new function $u\left(y\right)=u_{\epsilon }\left(\epsilon y\right)$. After calculation, we obtain that u satisfies the following equivalent problem:

$$\left\{\begin{array}{cc}{\displaystyle \left(a+b{\epsilon }^{N-2s}{\int }_{{\mathbb{R}}^N}|{(-\Delta )}^{s/2}u{|}^2\mathrm{d}x\right){(-\Delta )}^{s}u+V\left(\epsilon x\right)u=\lambda u+u\log u^2,}\hfill & x\in {\mathbb{R}}^N,\hfill \\ {\displaystyle {\int }_{{\mathbb{R}}^N}{\left|u\right|}^2\mathrm{d}x=h.}\hfill \end{array}\right.$$

(5)

This means that, if u is any solution of problem (5), then $\widehat{u}\left(x\right):=u\left(\frac{x}{\epsilon }\right)$ is a solution of the original problem (1). To prove the existence of weak solutions to (5), we instead look for critical points of the energy functional $J_{\epsilon }:H^s\left({\mathbb{R}}^N\right)\to \mathbb{R}\cup \{+\infty \}$ associated with the Kirchhoff problem under the following constraint:

$$\begin{array}{cc}\hfill J_{\epsilon }\left(u\right)& :={\displaystyle \frac{a}{2}}{\int }_{{\mathbb{R}}^N}|{(-\Delta )}^{{\displaystyle \frac{s}{2}}}u{|}^2\mathrm{d}x+{\displaystyle \frac{b}{4}}{\epsilon }^{N-2s}{\left({\int }_{{\mathbb{R}}^N}|{(-\Delta )}^{\frac{s}{2}}u{|}^2\mathrm{d}x\right)}^2\hfill \\ \hfill & +\frac{1}{2}{\int }_{{\mathbb{R}}^N}V\left(\epsilon x\right){|u|}^2\mathrm{d}x-{\int }_{{\mathbb{R}}^N}F\left(u\right)\mathrm{d}x.\hfill \end{array}$$

The critical points of this functional are sought under the constraint of the following spherical surface:

$$S\left(h\right):=\left\{u\in H^s\left({\mathbb{R}}^N\right):{\int }_{{\mathbb{R}}^N}{\left|u\right|}^2\mathrm{d}x=h>0\right\}.$$

The primitive function $F\left(t\right)$ of the nonlinear term is defined as

$$F\left(t\right):={\int }_0^{t}r\log r^{2}dr=\frac{1}{2}{t}^2\log t^2-\frac{t^2}{2}.$$

As mentioned above, the logarithmic-type nonlinearity causes the energy functional $J_{\epsilon }$ to be discontinuous in the standard Sobolev space. Hence, it is essential to formulate a new function space—namely the Orlicz space—such that the energy functional associated with problem (5) is of class $C^1$ on the Orlicz space $\mathbb{X}$. Furthermore, we note that $V_0$ in (V) is not the global minimum of V. To solve this problem, one usually employs a decomposition method, as in [28]. However, in our case, this method is not directly applicable because the nonlocal operator ${(-\Delta )}^s$ and the Kirchhoff term themselves also introduce additional difficulties, so we require new techniques and a more delicate analysis. The equation studied in [25] is also generalized by our work.

Remark 1.

In [5], a multiplicity and concentration theory for normalized fractional logarithmic Schrödinger equations was established via penalization techniques, barycenter maps, and Ljusternik–Schnirelmann category arguments. Building on this work, the present paper extends this scheme to the fractional Kirchhoff–logarithmic setting under an $L^2$ constraint. Due to the additional nonlocal Kirchhoff term that depends on the global kinetic energy, the problem exhibits a “double nonlocality,” leading to difficulties such as the non-additivity of the energy, the appearance of cross terms in energy decompositions, and the necessity to accurately track the Kirchhoff contribution in the semiclassical limit. To overcome these challenges, we develop refined energy estimates and a compactness analysis tailored to the Kirchhoff structure while retaining the Orlicz-type framework and the penalization–barycenter construction. As a result, we obtain multiple normalized solutions and characterize their concentration behavior near the minimum set of the potential, together with uniform decay estimates.

The rest of this paper is structured as follows: Section 2 presents the necessary background on Orlicz spaces, including notations and fundamental facts. In Section 3, using the penalization method, we introduce an auxiliary problem, and, by combining the Pohozaev identity and the concentration–compactness principle, we prove the existence of a solution to this problem. In Section 4, we apply Ljusternik–Schnirelmann category theory to establish multiplicity results for Equation (8). In the final section, we analyze the concentration behavior and decay properties of these solutions, thereby completing the proof of Theorem 1.

The following notations are used in this paper:

• ${\parallel ·\parallel }_q$ is the usual norm on the space $L^q({\mathbb{R}}^N,\mathbb{R})$;
• For any $R>0$, $x\in {\mathbb{R}}^N$, $B_R\left(x\right)$ denotes the open ball centered at x with radius $R>0$, and $B_R^c\left(x\right)$ denotes the complement of $B_R\left(x\right)$ in ${\mathbb{R}}^N$, $B_R:=B_R\left(0\right)$;
• $o_n\left(1\right)$ denotes a sequence with $o_n\left(1\right)\to 0$ as $n\to \infty$;
• A generic positive constant, whose value may differ from line to line, is denoted by C or $C_i$($i=1,2,\dots$).

2. Preliminary Results

We now introduce some essential properties of fractional Sobolev spaces and Orlicz spaces (see [1,29,30]). Let

$$H^s\left({\mathbb{R}}^N\right)=\left\{u\in L^2\left({\mathbb{R}}^N\right):{\displaystyle \frac{u\left(x\right)-u\left(y\right)}{{|x-y|}^{\frac{N}{2}+s}}}\in L^2({\mathbb{R}}^N\times {\mathbb{R}}^N)\right\}.$$

This space is endowed with the norm

$${\parallel u\parallel }_{H^s\left({\mathbb{R}}^N\right)}={\left({\int }_{{\mathbb{R}}^N}\left({|(-\Delta )}^{s/2}{u|}^2+u^2\right)\mathrm{d}x\right)}^{\frac{1}{2}},$$

where

$${\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{s/2}u\right|}^2\mathrm{d}x=C_{N,s}\underset{{\mathbb{R}}^{2N}}{\int \int }{\displaystyle \frac{{|u\left(x\right)-u\left(y\right)|}^2}{{|x-y|}^{N+2s}}}\mathrm{d}x\mathrm{d}y.$$

with $C_{N,s}>0$ denoting a suitable normalization constant in the definition of the fractional Laplacian.

Associated with the potential function $V\left(\epsilon x\right)$, we define the following Banach space

$$H_{\epsilon }\left({\mathbb{R}}^N\right):=\left\{u\in H^s\left({\mathbb{R}}^N\right):{\int }_{{\mathbb{R}}^N}V\left(\epsilon x\right){\left|u\right|}^2\mathrm{d}x<\infty \right\},$$

with the norm defined by

$${\parallel u\parallel }_{\epsilon }={\left({\int }_{{\mathbb{R}}^N}\left({|(-\Delta )}^{s/2}{u|}^2+(V\left(\epsilon x\right)+1)u^2\right)\mathrm{d}x\right)}^{{\displaystyle \frac{1}{2}}}.$$

The relationship between $H^s\left({\mathbb{R}}^N\right)$ and $L^q\left({\mathbb{R}}^N\right)$ is described by the fractional Gagliardo–Nirenberg inequality, which is presented in the following fundamental proposition.

Proposition 1.

According to the fractional Gagliardo–Nirenberg inequality, the following inequality holds for any $u\in H^s\left({\mathbb{R}}^N\right)$:

$${\parallel u\parallel }_q{\le C\parallel (-\Delta )}^{s/2}{u\parallel }_2^{\theta }{\parallel u\parallel }_2^{1-\theta },\forall q\in [2,2_s^{*}],$$

where $2_s^{*}$ is the fractional Sobolev critical exponent defined by $2_s^{*}:={\displaystyle \frac{2N}{N-2s}}$, and ${\displaystyle \frac{\theta }{2_s^{*}}}+{\displaystyle \frac{1-\theta }{2}}={\displaystyle \frac{1}{q}}$.

Using an argument similar to that in the proof of [12] (Lemma 3.1), we can directly obtain the following lemma.

Lemma 1.

Let $\left\{u_n\right\}$ be a bounded sequence in $H^s\left({\mathbb{R}}^N\right)$ such that $u_n\to u\mathrm{a}.\mathrm{e}.\mathrm{in}{\mathbb{R}}^N$, and $\left\{u_n^2\log u_n^2\right\}$ is a bounded sequence in $L^1\left({\mathbb{R}}^N\right)$. Then,

$$\underset{n\to \infty }{\lim }{\int }_{{\mathbb{R}}^N}\left(u_n^2\log u_n^2-|u_n{-u|}^2\log {|u_n-u|}^2\right)\mathrm{d}x={\int }_{{\mathbb{R}}^N}{u}^2\log u^2\mathrm{d}x.$$

First, the definitions of ${\Delta }_2$ condition and N function are presented (see [29,30]). An N function is a continuous function $\Phi :\mathbb{R}\to [0,+\infty )$ that satisfies the conditions listed below:

(i)

$\Phi \left(t\right)$ is defined as an even and convex function;

(ii)

$\Phi \left(0\right)=0$, and $\Phi \left(t\right)>0$ for $t\ne 0$;

(iii)

${\lim }_{t\to 0}{\displaystyle \frac{t}{\Phi \left(t\right)}}=+\infty$ and ${\lim }_{t\to +\infty }{\displaystyle \frac{t}{\Phi \left(t\right)}}=0$.

For a function $\Phi$, to say that it satisfies the ${\Delta }_2$ condition means that there exists a constant $K>0$ such that

$$\Phi \left(2t\right)\le K\Phi \left(t\right),\mathrm{for}\mathrm{all}t\ge 0.$$

For $t\ge 0$, the conjugate function ${\Phi }^{*}$ of $\Phi$ is defined as

$${\Phi }^{*}\left(r\right)=\underset{t\ge 0}{\sup }\{rt-\Phi \left(t\right)\},r\ge 0.$$

It is easy to see that ${\Phi }^{*}\left(r\right)$ is also an N function.

Lemma 2.

Assume that the N function $\Phi \in C^1\left(\mathbb{R}\right)$ satisfies

$$1<m\le {\displaystyle \frac{{\Phi }^{\prime }\left(t\right)t}{\Phi \left(t\right)}}\le l<+\infty ,\forall t\ne 0,$$

(6)

and then both $\Phi \left(t\right)$ and ${\Phi }^{*}\left(t\right)$ satisfy the ${\Delta }_2$ condition.

Let $\mathrm{\Omega }\subset {\mathbb{R}}^N$ be an open set, and define the Orlicz space generated by the function $\Phi$ as

$$L^{\Phi }\left(\mathrm{\Omega }\right)=\left\{u\in L_{loc}^1\left(\mathrm{\Omega }\right):{\parallel u\parallel }_{\Phi }<+\infty \right\},$$

where the Luxemburg norm ${\parallel u\parallel }_{\Phi }$ is defined as

$${\parallel u\parallel }_{\Phi }=\inf \left\{\lambda >0:{\int }_{\mathrm{\Omega }}\Phi \left(\right|u|/\lambda )\mathrm{d}x\le 1\right\}.$$

It is worth noting that $\Phi \left(t\right)$ and ${\Phi }^{*}\left(r\right)$ satisfy the following two inequalities, which correspond to Young’s inequality and the Hölder-type inequality for Orlicz spaces generated by N functions, respectively. More precisely, we have

$$rt\le \Phi \left(t\right)+{\Phi }^{*}\left(r\right),\forall r,t\ge 0,$$

$$\left|{\int }_{\mathrm{\Omega }}uv\mathrm{d}x\right|\le {2\parallel u\parallel }_{\Phi }{\parallel v\parallel }_{\Phi *}\forall u\in L^{\Phi }\left(\mathrm{\Omega }\right),v\in L^{\Phi *}\left(\mathrm{\Omega }\right).$$

If $\Phi$ and ${\Phi }^{*}$ both satisfy the ${\Delta }_2$ condition, then the space $L^{\Phi }\left(\mathrm{\Omega }\right)$ is reflexive and separable. Moreover, this condition implies the following conclusions:

$$L^{\Phi }\left(\mathrm{\Omega }\right)=\left\{u\in L_{\mathrm{loc}}^1\left(\mathrm{\Omega }\right):{\int }_{\mathrm{\Omega }}\Phi \left(\right|u\left|\right)\mathrm{d}x<+\infty \right\},$$

and

$$u_n\to u\mathrm{in}{L}^{\Phi }\left(\mathrm{\Omega }\right)\iff {\int }_{\mathrm{\Omega }}\Phi \left(\right|u_n-u\left|\right)\mathrm{d}x\to 0.$$

For $u\in L^{\Phi }\left(\mathrm{\Omega }\right)$, under condition (6), we also have the following modular inequality:

$${\min \{\parallel u\parallel }_{\Phi }^m{,\parallel u\parallel }_{\Phi }^l\}\le {\int }_{\mathrm{\Omega }}\Phi \left(u\right)\mathrm{d}x\le {\max \{\parallel u\parallel }_{\Phi }^m{,\parallel u\parallel }_{\Phi }^l\}.$$

(7)

Next, we introduce two fundamental auxiliary functions $F_1$ and $F_2$. For sufficiently small $\delta >0$, we define

$$F_1\left(r\right):=\left\{\begin{array}{cc}0,\hfill & r=0,\hfill \\ -{\displaystyle \frac{1}{2}}{r}^2\log r^2,\hfill & 0<\left|r\right|<\delta ,\hfill \\ -{\displaystyle \frac{1}{2}}{r}^2(\log {\delta }^2+3)+2\delta \left|r\right|-{\displaystyle \frac{1}{2}}{\delta }^2,\hfill & \left|r\right|\ge \delta ,\hfill \end{array}\right.$$

and

$$F_2\left(r\right):=\left\{\begin{array}{cc}0,\hfill & \left|r\right|\le \delta ,\hfill \\ \frac{1}{2}{r}^2\log \left({\displaystyle \frac{r^2}{{\delta }^2}}\right)+2\delta \left|r\right|-\frac{3}{2}{r}^2-\frac{1}{2}{\delta }^2,\hfill & \left|r\right|\ge \delta .\hfill \end{array}\right.$$

Thus, $\frac{1}{2}{r}^2\log r^2$ can be rewritten as

$$\frac{1}{2}{r}^2\log r^2=F_2\left(r\right)-F_1\left(r\right),\forall r\in \mathbb{R}.$$

The following two results concerning $F_1$ and $F_2$ are straightforward and will be very useful in the analysis that follows.

Proposition 2.

As defined above, $F_1\ge 0$, and it satisfies the following properties:

(i)

$F_1$ is an N function and $F_1\in C^1\left(\mathbb{R}\right)$;

(ii)

The ${\Delta }_2$ condition is satisfied by both $F_1\mathrm{and}{F}_1^{*}$.

Proof. 

(i) can be easily verified by direct calculation. For (ii), according to the proof of [5] (Proposition 2.4), there exists a constant $l_2$ such that

$$1<l_2\le {\displaystyle \frac{F_1^{\prime }\left(r\right)r}{F_1\left(r\right)}}\le 2\forall r\ge \delta .$$

$F_1$ is an even function, which implies that (6) is satisfied for all $r\ne 0$. Hence, the proof is finished. □

We can also deduce the following properties of $F_2$ through simple calculation.

Proposition 3.

Let $\delta >0$ be a fixed constant. The function $F_2$ defined above has the following properties:

(P1)

$F_2\in C^1(\mathbb{R},\mathbb{R})\cap C^2((\delta ,+\infty ),\mathbb{R})$, and, for $p\in (2,2_s^{*})$, there exists a constant $C_p$ such that

$$|F_2^{\prime }\left(r\right)|\le C_p{\left|r\right|}^{p-1},\forall r\in \mathbb{R};$$

(P2)

${\displaystyle \frac{F_2^{\prime }\left(r\right)}{r}}$ is non-decreasing on $(0,\infty )$ and strictly increasing on $(\delta ,\infty )$;

(P3)

${\lim }_{r\to +\infty }{\displaystyle \frac{F_2^{\prime }\left(r\right)}{r}}=+\infty$.

Next, to avoid the situation where ${\int }_{{\mathbb{R}}^N}{F}_1\left(u\right)\mathrm{d}x=+\infty$ for $u\in H_{\epsilon }\left({\mathbb{R}}^N\right)$, we restrict $J_{\epsilon }$ to the space $\mathbb{X}:=H_{\epsilon }\left({\mathbb{R}}^N\right)\cap L^{F_1}\left({\mathbb{R}}^N\right)$ and denote this restriction by $J_{F_1,\epsilon }:=J_{\epsilon }{|}_{\mathbb{X}}$. This space is endowed with the norm

$$\parallel ·\parallel :={\parallel ·\parallel }_{\epsilon }+{\parallel ·\parallel }_{F_1}.$$

From Proposition 2.4 (ii), it follows that $(\mathbb{X},\parallel ·\parallel )$ is a separable and reflexive Banach space. Using standard methods, one checks that $J_{F_1,\epsilon }:\mathbb{X}\to \mathbb{R}$ is of class $C^1(\mathbb{X},\mathbb{R})$. By the proof in [31] (Lemma 2.3), the associated energy functional is continuously differentiable on the Sobolev space. The formula for its derivative $J_{F_1,\epsilon }^{\prime }$, evaluated in the direction $v\in \mathbb{X}$, is similarly given by

$$\begin{array}{cc}\hfill \langle J_{F_1,\epsilon }^{\prime }\left(u\right),v\rangle & =a{C}_{N,s}{\int }_{{\mathbb{R}}^{2N}}{\displaystyle \frac{\left(u\right(x)-u(y\left)\right)\left(v\right(x)-v(y\left)\right)}{{|x-y|}^{N+2s}}}\mathrm{d}x\mathrm{d}y\hfill \\ \hfill & +b{\epsilon }^{N-2s}{C}_{N,s}\left({\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{s/2}u\right|}^2\mathrm{d}x\right){\int }_{{\mathbb{R}}^{2N}}{\displaystyle \frac{\left(u\right(x)-u(y\left)\right)\left(v\right(x)-v(y\left)\right)}{{|x-y|}^{N+2s}}}\mathrm{d}x\mathrm{d}y\hfill \\ \hfill & +{\int }_{{\mathbb{R}}^N}(V\left(\epsilon x\right)+1)uv\mathrm{d}x+{\int }_{{\mathbb{R}}^N}{F}_1^{\prime }\left(u\right)v\mathrm{d}x-{\int }_{{\mathbb{R}}^N}{F}_2^{\prime }\left(u\right)v\mathrm{d}x,\forall u,v\in \mathbb{X}.\hfill \end{array}$$

The continuity of the embeddings $\mathbb{X}\hookrightarrow L^{F_1}\left({\mathbb{R}}^N\right)$ and $\mathbb{X}\hookrightarrow H_{\epsilon }\left({\mathbb{R}}^N\right)$ is clear.

3. The Auxiliary Problem

The main purpose of this section is to construct an auxiliary problem using the penalization method [32] and to prove the existence of its solution. We note that, due to the specific nature of the nonlinear term $u\log u^2$, the common methods used in [32] are not applicable here. Specifically, by Proposition 3, we can choose sufficiently small $\tau >0$ and $h_0>\delta$ such that $V_0+1>2\tau$ and ${\displaystyle \frac{F_2^{\prime }\left(h_0\right)}{h_0}}=\tau$ (where $V_0$ is given by condition (V)). Accordingly, we define the modified function ${\widehat{F}}_2^{\prime }\left(t\right)$ as follows:

$${\widehat{F}}_2^{\prime }\left(t\right)=F_2^{\prime }\left(t\right),\mathrm{for}\left|t\right|\le h_0;{\widehat{F}}_2^{\prime }\left(t\right)=\tau t,\mathrm{for}\left|t\right|>h_0.$$

Following the approach of [33], we also fix a function $\vartheta \in C_c^{\infty }(\mathbb{R},\mathbb{R})$ and establish constants $\delta ,t_{h_0},h_0,T_{h_0}$ satisfying the inequalities $0<\delta <t_{h_0}<h_0<T_{h_0}$ such that

(${\theta }_1$)

for all $t\in [t_{h_0},T_{h_0}]$, one has $\vartheta \left(t\right)\le {\widehat{F}}_2^{\prime }\left(t\right)$;

(${\theta }_2$)

$\vartheta \left(t_{h_0}\right)={\widehat{F}}_2^{\prime }\left(t_{h_0}\right)$, $\vartheta \left(T_{h_0}\right)={\widehat{F}}_2^{\prime }\left(T_{h_0}\right)$, ${\vartheta }^{\prime }\left(t_{h_0}\right)={\widehat{F}}_2^{″}\left(t_{h_0}\right)$, and ${\vartheta }^{\prime }\left(T_{h_0}\right)={\widehat{F}}_2^{″}\left(T_{h_0}\right)$;

(${\theta }_3$)

the mapping $t\mapsto {\displaystyle \frac{\vartheta \left(t\right)}{t}}$ is increasing on the interval $[t_{h_0},T_{h_0}]$.

Using the above notation, and noting that $F_2\left(t\right)\in C^2(\mathbb{R},\mathbb{R})$, we define the following function:

$${\tilde{F}}_2^{\prime }\left(t\right):=\left\{\begin{array}{cc}{\widehat{F}}_2^{\prime }\left(t\right),\hfill & t\notin [t_{h_0},T_{h_0}],\hfill \\ \vartheta \left(t\right),\hfill & t\in [t_{h_0},T_{h_0}].\hfill \end{array}\right.$$

Based on this, we introduce the penalized nonlinear term $G_2^{\prime }:{\mathbb{R}}^N\times \mathbb{R}\to \mathbb{R}$.

$$G_2^{\prime }(x,t):={\chi }_{\Lambda }{F}_2^{\prime }\left(t\right)+(1-{\chi }_{\Lambda }){\tilde{F}}_2^{\prime }\left(t\right),$$

where ${\chi }_{\Lambda }$ is the characteristic function of the set $\Lambda$. Subsequently, our primary objective is to prove that solutions exist for the auxiliary problem presented below. This problem is obtained by combining the modified Kirchhoff–Schrödinger equation with the penalization function:

$$\left\{\begin{array}{cc}M{(-\Delta )}^{s}u+(V\left(\epsilon x\right)+1)u=\lambda u+G_2^{\prime }(\epsilon x,u)-F_1^{\prime }\left(u\right),\hfill & x\in {\mathbb{R}}^N,\hfill \\ {\int }_{{\mathbb{R}}^N}{\left|u\right|}^2\mathrm{d}x=h.\hfill \end{array}\right.$$

(8)

Here, $M=a+b{\epsilon }^{N-2s}X\left(u\right)$ and $X\left(u\right)={\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{s/2}u\right|}^2\mathrm{d}x$. Notice that the energy functional corresponding to the auxiliary problem belongs to $C^1(\mathbb{X},\mathbb{R})$ and is given by

$$\begin{array}{cc}\hfill I_{\epsilon }\left(u\right)& :={\displaystyle \frac{a}{2}}{\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{s/2}u\right|}^2\mathrm{d}x+{\displaystyle \frac{b{\epsilon }^{N-2s}}{4}}{\left({\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{s/2}u\right|}^2\mathrm{d}x\right)}^2\hfill \\ \hfill & +\frac{1}{2}{\int }_{{\mathbb{R}}^N}(V\left(\epsilon x\right)+1){\left|u\right|}^2\mathrm{d}x-{\int }_{{\mathbb{R}}^N}{G}_2(\epsilon x,u)\mathrm{d}x+{\int }_{{\mathbb{R}}^N}{F}_1\left(u\right)\mathrm{d}x.\hfill \end{array}$$

The energy functional corresponding to the auxiliary problem is constrained on the sphere $S\left(h\right)$, where $G_2(x,t):={\int }_0^t{G}_2^{\prime }(x,r)\mathrm{d}r$. Let ${\Lambda }_{\epsilon }:=\{x\in {\mathbb{R}}^N:\epsilon x\in \Lambda \}$, and we observe that, if $u\left(x\right)$ is a solution of the auxiliary problem (8) and satisfies $\left|u\left(x\right)\right|\le t_{h_0}$ on ${\Lambda }_{\epsilon }^c$, then $u\left(x\right)$ is a solution of the original problem (5). From now on, the problem is reduced to study the auxiliary problem (8). To this end, we introduce the following autonomous problem:

$$\left\{\begin{array}{cc}{\displaystyle \left(a+b{\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{\frac{s}{2}}u\right|}^2\mathrm{d}x\right){(-\Delta )}^{s}u+\mu u=\lambda u+u\log u^2,}\hfill & x\in {\mathbb{R}}^N,\hfill \\ {\displaystyle {\int }_{{\mathbb{R}}^N}{\left|u\right|}^2\mathrm{d}x=h.}\hfill \end{array}\right.$$

(9)

Here, $\mu >0$. We define the following Banach space to study this problem:

$$H_{\mu }^s\left({\mathbb{R}}^N\right):=\left\{u\in H^s\left({\mathbb{R}}^N\right):{\int }_{{\mathbb{R}}^N}\mu {\left|u\right|}^2\mathrm{d}x<\infty \right\}.$$

The space is endowed with the norm

$${\parallel u\parallel }_{H_{a,\mu }^s}={\left({\int }_{{\mathbb{R}}^N}{\left(a\right|(-\Delta )}^{s/2}u{|}^2+(\mu +1)u^2)\mathrm{d}x\right)}^{\frac{1}{2}}.$$

We restrict the self-consistent functional $J_{\mu }$ to the space ${\mathbb{X}}_{\mu }:=H_{\mu }^s\left({\mathbb{R}}^N\right)\cap L^{F_1}\left({\mathbb{R}}^N\right)$, for which the norm is defined as

$${\parallel ·\parallel }_{\mu }:={\parallel ·\parallel }_{H_{a,\mu }^s}+{\parallel ·\parallel }_{F_1}.$$

The corresponding energy functional $J_{\mu }\left(u\right):{\mathbb{X}}_{\mu }\to \mathbb{R}$ is

$$J_{\mu }\left(u\right):={\displaystyle \frac{a}{2}}X\left(u\right)+{\displaystyle \frac{b}{4}}{X}^2\left(u\right)+\frac{1}{2}{\int }_{{\mathbb{R}}^N}(\mu +1)u^2\mathrm{d}x+{\int }_{{\mathbb{R}}^N}{F}_1\left(u\right)\mathrm{d}x-{\int }_{{\mathbb{R}}^N}{F}_2\left(u\right)\mathrm{d}x.$$

The functional is constrained on the sphere $S\left(h\right)$.

The following lemma is introduced to establish the well-definedness of $Y_{\epsilon ,h}:={\inf }_{u\in S\left(h\right)}{I}_{\epsilon }\left(u\right)$ and $Y_{\mu ,h}:={\inf }_{u\in S\left(h\right)}{J}_{\mu }\left(u\right)$.

Lemma 3.

$I_{\epsilon }\left(u\right)$ and $J_{\mu }\left(u\right)$ are functionals that have a lower bound and satisfy a coercivity condition on $S\left(h\right)$.

Proof. 

It is sufficient to establish that $I_{\epsilon }\left(u\right)$ is both bounded and coercive from below on $S\left(h\right)$. The case of $J_{\mu }\left(u\right)$ can be proved in a similar manner. First, its auxiliary energy functional is such that

$$\begin{array}{cc}\hfill I_{\epsilon }\left(u\right)& ={\displaystyle \frac{a}{2}}{\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{s/2}u\right|}^2\mathrm{d}x+{\displaystyle \frac{b}{4}}{\epsilon }^{N-2s}{\left({\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{s/2}u\right|}^2\mathrm{d}x\right)}^2\hfill \\ \hfill & +\frac{1}{2}{\int }_{{\mathbb{R}}^N}(V\left(\epsilon x\right)+1){\left|u\right|}^2\mathrm{d}x-{\int }_{{\mathbb{R}}^N}{G}_2(\epsilon x,u)\mathrm{d}x+{\int }_{{\mathbb{R}}^N}{F}_1\left(u\right)\mathrm{d}x.\hfill \end{array}$$

The definition of $G_2$ and property $\left(P_1\right)$ (Propostion 3) imply that, for a given $p\in \left(2,2+\frac{4s}{N}\right)$, a positive constant $C_p$ exists satisfying $G_2(\epsilon x,s)\le F_2\left(s\right)$ and $|F_2^{\prime }\left(s\right)|\le C_p{\left|s\right|}^{p-1}$, which implies that $G_2(\epsilon x,s)\le C{\left|s\right|}^p$. Since $F_1\left(u\right)\ge 0$ and $V\left(\epsilon x\right)$ is bounded, we can obtain a lower bound for $I_{\epsilon }\left(u\right)$:

$$I_{\epsilon }\left(u\right)\ge {\displaystyle \frac{a}{2}}{\int |(-\Delta )}^{s/2}{u|}^2\mathrm{d}x+{\displaystyle \frac{b}{4}}{\epsilon }^{N-2s}{\left({\int |(-\Delta )}^{s/2}{u|}^2\mathrm{d}x\right)}^2-C_1\int {|u|}^p\mathrm{d}x-C_2.$$

By using the fractional Gagliardo–Nirenberg inequality, we have

$${\parallel u\parallel }_p\le C_3{\parallel (-\Delta )}^{s/2}{u\parallel }_2^{{\theta }_p}{\parallel u\parallel }_2^{1-{\theta }_p}.$$

Since $u\in S\left(h\right)$, we have ${\parallel u\parallel }_2^2=h$, where h is a constant. Therefore,

$${\int |u|}^p\mathrm{d}x={\parallel u\parallel }_p^p\le C_4{\left({\left({\int |(-\Delta )}^{s/2}{u|}^2\mathrm{d}x\right)}^{{\displaystyle \frac{{\theta }_p}{2}}}\right)}^p=C_4{\left({\int |(-\Delta )}^{s/2}{u|}^2\mathrm{d}x\right)}^{{\displaystyle \frac{p{\theta }_p}{2}}}.$$

Substituting this estimate into the lower bound expression of $I_{\epsilon }\left(u\right)$, we obtain

$$I_{\epsilon }\left(u\right)\ge {\displaystyle \frac{a}{2}}X\left(u\right)+{\displaystyle \frac{b}{4}}{\epsilon }^{N-2s}{X}^2\left(u\right)-C_{5}X{\left(u\right)}^{{\displaystyle \frac{p{\theta }_p}{2}}}-C_2.$$

According to the analysis in the original paper, when $p\in (2,2+\frac{4s}{N})$, the exponent satisfies $p{\theta }_p<2$, which implies that $\frac{p{\theta }_p}{2}<1$.

Therefore, as ${\parallel u\parallel }_{H_{a,\mu }^s}\to \infty$, it follows that $X\left(u\right)\to \infty$. In the above inequality, the positive quartic term ${\displaystyle \frac{b}{4}}{\epsilon }^{N-2s}{X}^2\left(u\right)$ grows much faster than the negative sublinear term $-C_5{X}^{{\displaystyle \frac{p{\theta }_p}{2}}}$. This ensures that $I_{\epsilon }\left(u\right)\to +\infty$ as $X\left(u\right)\to \infty$, and hence $I_{\epsilon }$ is coercive and bounded from below on $S\left(h\right)$. □

Proposition 4.

We can find a constant $h^{*}>0$ such that $Y_{\mu ,h}<0$ for all $h\ge h^{*}$. Furthermore, if $h_1,h_2$ satisfy $h^{*}\le h_1<h_2$, then $h_1{Y}_{\mu ,h_2}<h_2{Y}_{\mu ,h_1}<0$.

Proof. 

We prove this in two steps.

Step 1. Prove that $Y_{\mu ,h}<0$ when $h\ge h^{*}$.

We will show that any nontrivial solution has negative energy, which directly implies that the infimum of the energy $Y_{\mu ,h}$ is negative. For this purpose, let u be a solution to problem (9) with corresponding Lagrange multiplier $\lambda$. Multiplying the equation by u and integrating over the domain, we get

$$aX\left(u\right)+b{X}^2\left(u\right)+{\int }_{{\mathbb{R}}^N}\mu u^2\mathrm{d}x=\lambda h+{\int }_{{\mathbb{R}}^N}{u}^2\log u^2\mathrm{d}x.$$

(10)

(10) is a Nehari-type identity obtained by testing (9) with u. Using the above identity, we can express the energy functional $J_{\mu }\left(u\right)$ as follows:

$$J_{\mu }\left(u\right):={\displaystyle \frac{a}{2}}X\left(u\right)+{\displaystyle \frac{b}{4}}X{\left(u\right)}^2+\frac{1}{2}{\int }_{{\mathbb{R}}^N}(\mu +1)u^2\mathrm{d}x+{\int }_{{\mathbb{R}}^N}{F}_1\left(u\right)\mathrm{d}x-{\int }_{{\mathbb{R}}^N}{F}_2\left(u\right)\mathrm{d}x.$$

(11)

Consider the Pohozaev manifold

$${\mathcal{P}}_h:=\left\{u\in S\left(h\right):aX\left(u\right)+b{X}^2\left(u\right)=\frac{N}{2s}h\right\}.$$

In particular, any solution u of (9) belongs to ${\mathcal{P}}_h$ thanks to (14). Next, we use the Pohozaev identity to prove that the energy of u is negative. The corresponding Pohozaev identity for this equation is as follows (see [34] (Lemma 3.7)):

$${\displaystyle \frac{N-2s}{2}}\left(aX\left(u\right)+b{X}^2\left(u\right)\right)+{\displaystyle \frac{N}{2}}{\int }_{{\mathbb{R}}^N}\mu u^2\mathrm{d}x={\displaystyle \frac{N}{2}}\lambda h+{\displaystyle \frac{N}{2}}{\int }_{{\mathbb{R}}^N}\left(u^2\log u^2-u^2\right)\mathrm{d}x.$$

(12)

By combining (10) and (11), we obtain

$$J_{\mu }\left(u\right)=-{\displaystyle \frac{b}{4}}{X}^2\left(u\right)+{\displaystyle \frac{h}{2}}(\lambda +1).$$

(13)

By combining (10) and (12), we obtain

$$aX\left(u\right)+b{X}^2\left(u\right)=\frac{N}{2s}h.$$

(14)

And, by combining (13) and (14), we obtain

$$Y_{\mu ,h}-h{\displaystyle \frac{{\lambda }_h}{2}}={\displaystyle \frac{sa}{N}}X\left(u\right)+{\displaystyle \frac{4s-N}{4N}}bX{\left(u\right)}^2>0.$$

(15)

Using a scaled test function $u_{h,t}\left(x\right)=\sqrt{h}{t}^{N/2}\phi \left(tx\right)\in S\left(h\right)$ and the corresponding scaling estimates, we obtain

$${\displaystyle \frac{Y_{\mu ,h}}{h}}\le {\displaystyle \frac{J_{\mu }\left(u_{h,t}\right)}{h}}\le -c\log h+C\to -\infty (h\to \infty ),$$

where $c=\frac{1}{2}\left(1-\frac{N}{4s}\right)>0$ since $2s<N<4s$. On the other hand, by (3.8), we have ${\lambda }_h<2Y_{\mu ,h}/h$. Hence, ${\lambda }_h\to -\infty$ as $h\to \infty$, and there exists $h^{*}>0$ such that ${\lambda }_h+1\le 0$ for all $h\ge h^{*}$. Therefore, for $h\ge h^{*}$, by (3.6), we have

$$Y_{\mu ,h}=J_{\mu }\left(u\right)=-{\displaystyle \frac{b}{4}}{X}^2\left(u\right)+{\displaystyle \frac{h}{2}}({\lambda }_h+1)\le -{\displaystyle \frac{b}{4}}{X}^2\left(u\right)<0.$$

The proof of Step 1 is now complete.

Step 2. When $h^{*}\le h_1<h_2$, it holds that $h_1{Y}_{\mu ,h_2}<h_2{Y}_{\mu ,h_1}$.

To prove this inequality, it suffices to show that the function $g\left(h\right)=Y_{\mu ,h}/h$ is strictly decreasing on $[h^{*},\infty )$. According to [35], $Y_{\mu ,h}$ is differentiable with respect to h, and its derivative is $Y_{\mu ,h}^{\prime }={\displaystyle \frac{{\lambda }_h}{2}}$ (where ${\lambda }_h$ is the corresponding Lagrange multiplier). Therefore, the derivative of $g\left(h\right)$ satisfies

$$g^{\prime }\left(h\right)={\displaystyle \frac{Y_{\mu ,h}^{\prime }·h-Y_{\mu ,h}}{h^2}}=\frac{h{\lambda }_h/2-Y_{\mu ,h}}{h^2}.$$

To prove that $g^{\prime }\left(h\right)<0$, we only need to show that its numerator $h{\lambda }_h/2-Y_{\mu ,h}$ is negative, that is, to prove $Y_{\mu ,h}>h{\lambda }_h/2$. We once again make use of the Pohozaev identity. For (15), $\frac{sa}{N}X\left(u\right)$ is strictly positive since $s\in (0,1)$, $a>0$, $N>0$, and $X\left(u\right)>0$ for any nontrivial u. Moreover, under the standing assumption $2s<N<4s$, we have $4s-N>0$, and hence $\frac{4s-N}{4N}b{X}^2\left(u\right)>0$. Therefore,

$$\frac{sa}{N}X\left(u\right)+\frac{4s-N}{4N}b{X}^2\left(u\right)>0.$$

Therefore, we have $Y_{\mu ,h}>h{\lambda }_h/2$, which indicates that the numerator of $g^{\prime }\left(h\right)$, namely $h{\lambda }_h/2-Y_{\mu ,h}$, is strictly negative. Since $h^2>0$, it follows that $g^{\prime }\left(h\right)<0$. Since $g\left(h\right)={\displaystyle \frac{Y_{\mu ,h}}{h}}$ is strictly decreasing, for $h_1<h_2$, we have $g\left(h_2\right)<g\left(h_1\right)$; that is,

$${\displaystyle \frac{Y_{\mu ,h_2}}{h_2}}<{\displaystyle \frac{Y_{\mu ,h_1}}{h_1}}.$$

Since we have already proved that $Y_{\mu ,h_1}<0$, multiplying both sides of the above inequality by $h_1{h}_2>0$, we have $h_1{Y}_{\mu ,h_2}<h_2{Y}_{\mu ,h_1}$. The proposition is thus proved. □

Lemma 4.

Define $h_1^{*}:=\inf \{h>0:Y_{\mu ,h}<0\}$. When $h\le h_1^{*}$, we obtain $Y_{\mu ,h}\ge 0$.

Proof. 

We argue by contradiction. Suppose that there exists $h_0\in (0,h_1^{*})$ such that $Y_{\mu ,h_0}<0$. Then, $h_0\in \{h>0:Y_{\mu ,h}<0\}$, and, hence, by the definition of the infimum, we can obtain

$$h_1^{*}=\inf \{h>0:Y_{\mu ,h}<0\}\le h_0,$$

which contradicts $h_0<h_1^{*}$. Therefore, $Y_{\mu ,h}\ge 0$ for all $0<h<h_1^{*}$. □

Proposition 5.

Let $h\ge h^{*}$, and let $\left\{u_n\right\}\subset S\left(h\right)$ be a minimizing sequence for the functional $J_{\mu }$. In this case, one of the following two situations must occur:

(i)

$\left\{u_n\right\}$ converges strongly in ${\mathbb{X}}_{\mu }$;

(ii)

There exists a sequence $\left\{y_n\right\}\subset {\mathbb{R}}^N$ with $|y_n|\to +\infty$ such that the translated sequence ${\tilde{u}}_n\left(x\right):=u_n(x+y_n)$ converges strongly in ${\mathbb{X}}_{\mu }$ to $\tilde{u}\left(x\right)\in S\left(h\right)$, and $J_{\mu }\left(\tilde{u}\right)=Y_{\mu ,h}$.

Proof. 

According to Lemma 3, the coercivity of $J_{\mu }$ on $S\left(h\right)$ guarantees that the minimizing sequence $\left\{u_n\right\}$ is bounded in ${\mathbb{X}}_{\mu }$. Therefore, we can extract a subsequence (still denoted by $\left\{u_n\right\}$) such that it converges weakly in ${\mathbb{X}}_{\mu }$ to an element $u\in {\mathbb{X}}_{\mu }$. The proof proceeds by considering two separate cases.

Case 1. When $u\not\equiv 0$, we define $v_n:=u_n-u$. According to the Brezis–Lieb lemma and Lemma 1, we have the following splitting property:

1.

$\parallel u_n{\parallel }_2^2={\parallel u\parallel }_2^2+{\parallel v_n\parallel }_2^2+o_n\left(1\right)$;

2.

${\int }_{{\mathbb{R}}^N}{|(-\Delta )}^{s/2}{u}_n{|}^2\mathrm{d}x={\int }_{{\mathbb{R}}^N}{|(-\Delta )}^{s/2}{u|}^2\mathrm{d}x+{\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{s/2}{v}_n\right|}^2\mathrm{d}x+o_n\left(1\right)$.

Let $X\left(v_n\right)={\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{s/2}{v}_n\right|}^2\mathrm{d}x$. Using the above splitting property, we can obtain a key inequality as follows:

$$J_{\mu }\left(u_n\right)\ge J_{\mu }\left(u\right)+J_{\mu }\left(v_n\right)+o_n\left(1\right).$$

Since $\left\{u_n\right\}$ is a minimizing sequence, we have $\underset{n\to \infty }{\lim }{J}_{\mu }\left(u_n\right)=Y_{\mu ,h}$. Let $g={\parallel u\parallel }_2^2\in (0,h]$ and $d_n={\parallel v_n\parallel }_2^2\to d=h-g$. By the definition of $Y_{\mu ,h}$, it is clear that $J_{\mu }\left(u\right)\ge Y_{\mu ,g}$ and $J_{\mu }\left(v_n\right)\ge Y_{\mu ,d_n}$. Taking the limit as $n\to \infty$, we obtain $Y_{\mu ,h}\ge Y_{\mu ,g}+Y_{\mu ,d}$.

Suppose that dichotomy occurs; i.e., $g\in (0,h)$ and $d=h-g\in (0,h)$. When $g>h^{*}$ and $d<h_1^{*}$, according to the proof in Proposition 4 and Lemma 4, we have

$$Y_{\mu ,g}>{\displaystyle \frac{g}{h}}{Y}_{\mu ,h}and{Y}_{\mu ,d}>{\displaystyle \frac{d}{h}}{Y}_{\mu ,h}.$$

Adding these two inequalities yields

$$Y_{\mu ,g}+Y_{\mu ,d}>{\displaystyle \frac{g+d}{h}}{Y}_{\mu ,h}={\displaystyle \frac{h}{h}}{Y}_{\mu ,h}=Y_{\mu ,h},$$

which contradicts $Y_{\mu ,h}\ge Y_{\mu ,g}+Y_{\mu ,d}$. Therefore, dichotomy cannot occur.

Since $u\not\equiv 0$, we have $g>0$, which forces $d=0$. This implies that $|v_n{|}_2^2\to 0$. In other words, $u_n\to u$ strongly in $L^2\left({\mathbb{R}}^N\right)$. Furthermore, using the boundedness of $\left\{u_n\right\}$ in $H^s\left({\mathbb{R}}^N\right)$, it follows from interpolation inequalities that $u_n\to u$ strongly in $L^p\left({\mathbb{R}}^N\right)$ for all $p\in [2,2_s^{*})$. This ensures the convergence of all nonlinear integral terms in the energy functional. The strong $L^2$ convergence of $u_n$ shows that ${\left|u\right|}_2^2=h$, and thus $u\in S\left(h\right)$. The weak lower semicontinuity of the functional $J_{\mu }$ guarantees that

$$J_{\mu }\left(u\right)\le \underset{n\to \infty }{\lim \inf } J_{\mu }\left(u_n\right)=Y_{\mu ,h}.$$

On the other hand, since $u\in S\left(h\right)$ and $Y_{\mu ,h}$ is the infimum on $S\left(h\right)$, we have $J_{\mu }\left(u\right)\ge Y_{\mu ,h}$. Combining these two, we conclude that $J_{\mu }\left(u\right)=Y_{\mu ,h}$.

Finally, we need to prove that $u_n\to u$ strongly in ${\mathbb{X}}_{\mu }$. We already know that $u_n\rightharpoonup u$ weakly in ${\mathbb{X}}_{\mu }$, and $u_n\to u$ strongly in $L^p\left({\mathbb{R}}^N\right)$ for $p\in [2,2_s^{*})$. Starting from the energy convergence ${\lim }_{n\to \infty }{J}_{\mu }\left(u_n\right)=J_{\mu }\left(u\right)$ and substituting the definition of $J_{\mu }$, we have

$$\begin{array}{cc}\hfill \underset{n\to \infty }{\lim }& \left[{\displaystyle \frac{a}{2}}X\left(u_n\right)+{\displaystyle \frac{b}{4}}{X}^2\left(u_n\right)+{\displaystyle \frac{1}{2}}{\int }_{{\mathbb{R}}^N}(\mu +1)u_n^2\mathrm{d}x+{\int }_{{\mathbb{R}}^N}{F}_1\left(u_n\right)\mathrm{d}x-{\int }_{{\mathbb{R}}^N}{F}_2\left(u_n\right)\mathrm{d}x\right]\hfill \\ \hfill & ={\displaystyle \frac{a}{2}}X\left(u\right)+{\displaystyle \frac{b}{4}}{X}^2\left(u\right)+{\displaystyle \frac{1}{2}}{\int }_{{\mathbb{R}}^N}(\mu +1)u^2\mathrm{d}x+{\int }_{{\mathbb{R}}^N}{F}_1\left(u\right)\mathrm{d}x-{\int }_{{\mathbb{R}}^N}{F}_2\left(u\right)\mathrm{d}x,\hfill \end{array}$$

(16)

where $X\left(u_n\right)=\int {\left|{(-\Delta )}^{s/2}{u}_n\right|}^2\mathrm{d}x$. Since $u_n\to u$ strongly in $L^p$ ($p\in [2,2_s^{*})$) and $F_1$ satisfies the ${\Delta }_2$ condition, we have

$$\begin{array}{cc}\hfill & \underset{n\to \infty }{\lim }{\int }_{{\mathbb{R}}^N}(\mu +1)u_n^2\mathrm{d}x={\int }_{{\mathbb{R}}^N}(\mu +1)u^2\mathrm{d}x,\hfill \\ \hfill & \underset{n\to \infty }{\lim }{\int }_{{\mathbb{R}}^N}{F}_1\left(u_n\right)\mathrm{d}x={\int }_{{\mathbb{R}}^N}{F}_1\left(u\right)\mathrm{d}x,\underset{n\to \infty }{\lim }{\int }_{{\mathbb{R}}^N}{F}_2\left(u_n\right)\mathrm{d}x={\int }_{{\mathbb{R}}^N}{F}_2\left(u\right)\mathrm{d}x.\hfill \end{array}$$

(17)

Substituting (17) into (16), we obtain a limit that involves only the kinetic energy term

$$\underset{n\to \infty }{\lim }\left({\displaystyle \frac{a}{2}}X\left(u_n\right)+{\displaystyle \frac{b}{4}}{X}^2\left(u_n\right)\right)={\displaystyle \frac{a}{2}}X\left(u\right)+{\displaystyle \frac{b}{4}}{X}^2\left(u\right).$$

From the weak lower semicontinuity of the norm, we know that ${\displaystyle X\left(u\right)\le \underset{n\to \infty }{\lim \inf } X\left(u_n\right)}$. We define $f\left(x\right)={\displaystyle \frac{a}{2}}x+{\displaystyle \frac{b}{4}}{x}^2$. The conditions $a>0$ and $b>0$ ensure that $f\left(x\right)$ is strictly increasing for $x\ge 0$. Using the fact that ${\displaystyle \underset{n\to \infty }{\lim }f\left(X\left(u_n\right)\right)=f\left(X\left(u\right)\right)}$, combined with ${\displaystyle X\left(u\right)\le \underset{n\to \infty }{\lim \inf } X\left(u_n\right)}$ and the strict monotonicity of f, a standard argument in real analysis (e.g., by considering subsequences) allows us to deduce that the entire sequence $X\left(u_n\right)$ converges to $X\left(u\right)$; that is,

$$\underset{n\to \infty }{\lim }{\int }_{{\mathbb{R}}^N}{|(-\Delta )}^{s/2}{u}_n{|}^2\mathrm{d}x={\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{s/2}u\right|}^2\mathrm{d}x.$$

Since $u_n\rightharpoonup u$ weakly in $H^s\left({\mathbb{R}}^N\right)$ and their $H^s$ seminorms (i.e., $X{\left(u_n\right)}^{1/2}$) converge to $X{\left(u\right)}^{1/2}$, together with the strong convergence $u_n\to u$ in $L^2\left({\mathbb{R}}^N\right)$, it follows that $u_n\to u$ strongly in $H^s\left({\mathbb{R}}^N\right)$. Therefore, $\parallel u_n{-u\parallel }_{H_{a,\mu }^s}\to 0$. Finally, noting that $F_1$ satisfies the ${\Delta }_2$ condition, ${\int }_{{\mathbb{R}}^N}{F}_1\left(u_n\right)\mathrm{d}x\to {\int }_{{\mathbb{R}}^N}{F}_1\left(u\right)\mathrm{d}x$ implies $\parallel u_n{-u\parallel }_{F_1}\to 0$. Combining this with the strong convergence in the $H_{a,\mu }^s$ norm, we conclude that $u_n\to u$ strongly in the ${\mathbb{X}}_{\mu }$ norm.

Case 2. When $u\equiv 0$, in this case, $u_n\rightharpoonup 0$. We prove by contradiction that the sequence is non-vanishing. Assume, on the contrary, that the sequence vanishes. Then, by Lions’ concentration–compactness lemma [36], we have $u_n\to 0$ strongly in $L^t\left({\mathbb{R}}^N\right)$ for all $t\in (2,2_s^{*})$. This leads to

$${\int }_{{\mathbb{R}}^N}{F}_1\left(u_n\right)\mathrm{d}x\to 0\mathrm{and}{\int }_{{\mathbb{R}}^N}{F}_2\left(u_n\right)\mathrm{d}x\to 0.$$

Considering that the kinetic and Kirchhoff terms in $J_{\mu }$ are non-negative, we get

$$Y_{\mu ,h}=\underset{n\to \infty }{\lim }{J}_{\mu }\left(u_n\right)\ge 0.$$

However, this contradicts Propostion 4 (where $Y_{\mu ,h}<0$). Therefore, the sequence $\left\{u_n\right\}$ must be non-vanishing.

By the concentration–compactness principle [37], there must exist a sequence $\left\{y_n\right\}\subset {\mathbb{R}}^N$ and constants $R,\beta >0$ such that ${\int }_{B_R\left(y_n\right)}{\left|u_n\right|}^2\mathrm{d}x\ge \beta$. We then define the translated sequence ${\tilde{u}}_n\left(x\right):=u_n(x+y_n)$. It is clear that this new sequence is also a minimizing sequence in $S\left(h\right)$. Since the sequence is bounded in ${\mathbb{X}}_{\mu }$, we may assume that it converges weakly to some limit $\tilde{u}$. By the above integral inequality, it can be shown that $\tilde{u}\not\equiv 0$. Now, the sequence $\left\{{\tilde{u}}_n\right\}$ falls exactly into the same situation as in Case 1. Therefore, we conclude that $\left\{{\tilde{u}}_n\right\}$ converges strongly in ${\mathbb{X}}_{\mu }$ to a ground state solution $\tilde{u}\in S\left(h\right)$, and $J_{\mu }\left(\tilde{u}\right)=Y_{\mu ,h}$.

Finally, we prove by contradiction that $|y_n|\to +\infty$. Assume $\left\{y_n\right\}$ is bounded. Then, up to a subsequence, $y_n\to y_0$ for some $y_0$. This would imply that $\left\{u_n\right\}$ has a strongly convergent subsequence (with limit $\tilde{u}(x-y_0)\not\equiv 0$), which contradicts our premise that $u_n\rightharpoonup 0$ in this case. Therefore, $\left\{y_n\right\}$ must be unbounded; i.e., $|y_n|\to +\infty$. This completes the proof of the proposition. □

Theorem 2.

There exists a constant $h^{*}>0$ such that, for every $h\ge h^{*}$, the problem (9) admits a solution pair $(u,\lambda )$, where $u>0$ and $\lambda <0$.

Proof. 

A minimizing sequence $\left\{u_n\right\}\subset S\left(h\right)$ exists, converging strongly to some $u\in S\left(h\right)$ such that $J_{\mu }\left(u\right)=Y_{\mu ,h}$ according to Lemma 3 and Proposition 5. According to the principle of Lagrange multipliers, u must be a critical point of $J_{\mu }$ on the constraint manifold $S\left(h\right)$; that is, there exists a ${\lambda }_h\in \mathbb{R}$ such that

$$J_{\mu }^{\prime }\left(u\right)={\lambda }_h{\Psi }^{\prime }\left(u\right)\mathrm{in}{\mathbb{X}}_{\mu }^{\prime }.$$

(18)

Here, $\Psi :{\mathbb{X}}_{\mu }\to \mathbb{R}$ is defined by $\Psi \left(u\right)=\frac{1}{2}{\int }_{{\mathbb{R}}^N}{\left|u\right|}^2\mathrm{d}x$. Substituting the expressions of $J_{\mu }^{\prime }\left(u\right)$ and ${\Psi }^{\prime }\left(u\right)$ into Equation (18), we obtain the weak formulation satisfied by u, that is, the problem (9). According to Proposition 4, for all $h\ge h^{*}$, we have $J_{\mu }\left(u\right)=Y_{\mu ,h}<0$.

From the proof of Proposition 4, we have $Y_{\mu ,h}-\frac{h{\lambda }_h}{2}>0.$ Combining $Y_{\mu ,h}<0$, we can obtain ${\lambda }_h<0$.

To prove the positivity of the solution, we note that the structure of the functional $J_{\mu }$, together with the fractional “diamagnetic inequality” ${\parallel (-\Delta )}^{s/2}{\left|u\right| \parallel }_2\le {\parallel {(-\Delta )}^{s/2}u\parallel }_2$ from [38], implies $J_{\mu }\left(\right|u\left|\right)\le J_{\mu }\left(u\right)$. Since u is a minimizer for $Y_{\mu ,h}$ (i.e., $J_{\mu }\left(u\right)=Y_{\mu ,h}$), and $J_{\mu }\left(\right|u\left|\right)$ must be greater than or equal to the infimum $Y_{\mu ,h}$, it follows that

$$J_{\mu }\left(\right|u\left|\right)=J_{\mu }\left(u\right)=Y_{\mu ,h}.$$

This means $\left|u\right|$ is also a minimizer and therefore a solution. Thus, we can assume without loss of generality that $u\ge 0$. Finally, standard regularity theory and the strong maximum principle in [39] (Proposition 3.1) guarantee that u is strictly positive on ${\mathbb{R}}^N$. □

Proposition 6.

For every $h\ge h^{*}>0$, we define

$$Y_{0,h}:=\underset{u\in S\left(h\right)}{\inf }{J}_{V_0}\left(u\right)and{Y}_{\partial \Lambda ,h}:=\underset{u\in S\left(h\right)}{\inf }{J}_{V_{\partial \Lambda }}\left(u\right).$$

Here, $J_{\mu }$ denotes the Kirchhoff energy functional. Then the following conclusions hold:

(i)

$Y_{0,h}<Y_{\partial \Lambda ,h}<0$;

(ii)

There exists ${\epsilon }_0>0$ such that, for all $\epsilon \in (0,{\epsilon }_0)$, we have $Y_{0,h}<Y_{\partial \Lambda ,h}$. Here, $Y_{\epsilon ,h}:={\inf }_{u\in S\left(h\right)}{I}_{\epsilon }\left(u\right)$, where $I_{\epsilon }$ is the energy functional corresponding to the equivalent problem (5).

Proof. 

(i) The strict monotonicity of $J_{\mu }\left(u\right)$ with respect to $\mu$ (since ${\int }_{{\mathbb{R}}^N}{u}^2\mathrm{d}x=h>0$) ensures $J_{V_0}\left(u\right)<J_{V_{\partial \Lambda }}\left(u\right)$ (for all $u\in S\left(h\right)$), which implies $Y_{0,h}\le Y_{\partial \Lambda ,h}$. This inequality can be strengthened to a strict one by considering the ground state solutions. Furthermore, since Propostion 4 states that $Y_{\mu ,h}<0$, it follows that $Y_{0,h}<Y_{\partial \Lambda ,h}<0$.

(ii) According to Theorem 2, we select a ground state solution $u_0\in S\left(h\right)$ for $J_{V_0}$ (such that $J_{V_0}\left(u_0\right)=Y_{0,h}$) and pick $x_0\in \Lambda$ such that $V\left(x_0\right)=V_0$. We construct the test function $v_{\epsilon }\left(y\right):=u_0(y-x_0/\epsilon )$. It is easy to see that $v_{\epsilon }\in S\left(h\right)$. For the functional $I_{\epsilon }$, its infimum $Y_{\epsilon ,h}$ necessarily satisfies $Y_{\epsilon ,h}\le I_{\epsilon }\left(v_{\epsilon }\right)$. We will now compute the value of $I_{\epsilon }\left(v_{\epsilon }\right)$ as $\epsilon \to 0^{+}$. Recall the definition of $I_{\epsilon }$:

$$I_{\epsilon }\left(u\right)={\displaystyle \frac{a}{2}}X\left(u\right)+{\displaystyle \frac{b{\epsilon }^{N-2s}}{4}}{X}^2\left(u\right)+{\displaystyle \frac{1}{2}}{\int }_{{\mathbb{R}}^N}(V\left(\epsilon y\right)+1)u^2\mathrm{d}y+{\int }_{{\mathbb{R}}^N}{F}_1\left(u\right)\mathrm{d}y-{\int }_{{\mathbb{R}}^N}{G}_2(\epsilon y,u)\mathrm{d}y.$$

Therefore, we can obtain

$$\underset{\epsilon \to 0^{+}}{\lim \sup } Y_{\epsilon ,h}\le \underset{\epsilon \to 0^{+}}{\lim }{I}_{\epsilon }\left(v_{\epsilon }\right)=Y_{0,h}-{\displaystyle \frac{b}{4}}{X}^2\left(u_0\right).$$

Since $u_0$ is a nontrivial solution, we have $X\left(u_0\right)>0$ and $b>0$; hence, $Y_{0,h}-\frac{b}{4}{X}^2\left(u_0\right)<Y_{0,h}$. Combining this with the conclusion in (i), $Y_{0,h}<Y_{\partial \Lambda ,h}$, we obtain

$$\underset{\epsilon \to 0^{+}}{\lim \sup } Y_{\epsilon ,h}<Y_{0,h}<Y_{\partial \Lambda ,h}.$$

This shows that, for sufficiently small $\epsilon >0$ (for example, $\epsilon \in (0,{\epsilon }_0)$), we necessarily have $Y_{\epsilon ,h}<Y_{\partial \Lambda ,h}.$ The proposition is thus proved. □

Lemma 5.

Let $\left\{u_n\right\}\subset S\left(h\right)$ satisfy $I_{\epsilon }\left(u_n\right)\to c$, where $c<Y_{\partial \Lambda ,h}<0$. If $u_n\rightharpoonup u$ weakly in $H^s\left({\mathbb{R}}^N\right)$, then $u\not\equiv 0$.

Proof. 

According to the definition of the penalization function $G_2$, we derive the lower bound of the energy functional $I_{\epsilon }\left(u_n\right)$ as follows:

$$\begin{array}{cc}\hfill I_{\epsilon }\left(u_n\right)& ={\displaystyle \frac{a}{2}}{\int }_{{\mathbb{R}}^N}|{(-\Delta )}^{s/2}{u}_n{|}^2\mathrm{d}x+{\displaystyle \frac{b{\epsilon }^{N-2s}}{4}}{\left({\int }_{{\mathbb{R}}^N}|{(-\Delta )}^{s/2}{u}_n{|}^2\mathrm{d}x\right)}^2\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}{\int }_{{\mathbb{R}}^N}\left(V\left(\epsilon x\right)+1\right){\left|u_n\right|}^2\mathrm{d}x-{\int }_{{\mathbb{R}}^N}{G}_2(\epsilon x,u_n)\mathrm{d}x+{\int }_{{\mathbb{R}}^N}{F}_1\left(u_n\right)\mathrm{d}x\hfill \\ \hfill & \ge {\displaystyle \frac{a}{2}}{\int }_{{\mathbb{R}}^N}{|(-\Delta )}^{s/2}{u}_n{|}^2\mathrm{d}x+{\displaystyle \frac{b{\epsilon }^{N-2s}}{4}}{\left({\int }_{{\mathbb{R}}^N}|{(-\Delta )}^{s/2}{u}_n{|}^2\mathrm{d}x\right)}^2\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}{\int }_{{\mathbb{R}}^N}(V_0+1)|u_n{|}^2\mathrm{d}x-\tau {\int }_{{\Lambda }_{\epsilon }^c}|u_n{|}^2\mathrm{d}x-C{\int }_{{\Lambda }_{\epsilon }}{|u_n|}^p\mathrm{d}x\hfill \\ \hfill & \ge {\displaystyle \frac{1}{2}}{\int }_{{\mathbb{R}}^N}(V_0+1)|u_n{|}^2\mathrm{d}x-\tau h-C{\int }_{{\Lambda }_{\epsilon }}{|u_n|}^p\mathrm{d}x.\hfill \end{array}$$

We proceed by contradiction and assume that $u\equiv 0$. Since $\left\{u_n\right\}$ is bounded in $\mathbb{X}$, by [5] (Theorem 2.6), the embedding $\mathbb{X}\hookrightarrow L^p\left({\Lambda }_{\epsilon }\right)$ is compact (with $p<2_s^{*}$ and ${\Lambda }_{\epsilon }$ bounded). Hence, up to a subsequence, $u_n\to 0$ strongly in $L^p\left({\Lambda }_{\epsilon }\right)$; i.e.,

$${\int }_{{\Lambda }_{\epsilon }}{\left|u_n\right|}^p\mathrm{d}x\to 0.$$

Next, taking the limit inferior on both sides of the inequality, we have

$$c=\underset{n\to \infty }{\lim }{I}_{\epsilon }\left(u_n\right)\ge \underset{n\to \infty }{\lim \inf } \left[{\displaystyle \frac{1}{2}}\int (V_0+1){\left|u_n\right|}^2\mathrm{d}x\right]-\tau h={\displaystyle \frac{1}{2}}(V_0+1)h-\tau h.$$

Let $K_L={\displaystyle \frac{1}{2}}(V_0+1)h$. Since $V_0+1>0,h>0$, it is clear that $K_L>0$. Then we have

$$c\ge K_L-\tau h.$$

Since $K_L>0$ is a fixed value, we can choose sufficiently small $\tau$ such that $K_L-\tau h\ge 0$.

Both scenarios lead to a contradiction: the first one is a direct contradiction, and the second one ($c\ge 0$) contradicts the lemma’s assumption that $c<Y_{\partial \Lambda ,h}<0$. Therefore, the assumption $u\equiv 0$ is false. □

Lemma 6.

The functional $I_{\epsilon }{|}_{S\left(h\right)}$ satisfies the ${\left(PS\right)}_c$ condition at any level $c<Y_{\partial \Lambda ,h}$ as long as the parameter $\epsilon >0$ is small enough.

Proof. 

Let $\left\{u_n\right\}\subset S\left(h\right)$ be a ${\left(\mathrm{PS}\right)}_c$ sequence of the functional $I_{\epsilon }$ restricted to $S\left(h\right)$, where $c<Y_{\partial \Lambda ,h}<0$ and $u_n\rightharpoonup u_{\epsilon }$ weakly in $\mathbb{X}$. That is,

$$I_{\epsilon }\left(u_n\right)\to c\mathrm{and}\parallel I_{\epsilon }^{\prime }{{|}_{S\left(h\right)}\left(u_n\right)\parallel }_{{\mathbb{X}}^{\prime }}\to 0,asn\to +\infty .$$

We define the functional $\Psi \left(u\right):={\displaystyle \frac{1}{2}}{\int }_{{\mathbb{R}}^N}{\left|u\right|}^2\mathrm{d}x$. According to Willem’s monograph [27] (Proposition 5.12), there exists a sequence $\left\{{\lambda }_n\right\}\subset \mathbb{R}$ such that

$$\parallel I_{\epsilon }^{\prime }\left(u_n\right)-{\lambda }_n{\Psi }^{\prime }\left(u_n\right){\parallel }_{{\mathbb{X}}^{\prime }}\to 0,\mathrm{as}n\to \infty .$$

(19)

The boundedness of $\left\{u_n\right\}$ in $\mathbb{X}$ implies the boundedness of $\left\{{\lambda }_n\right\}$. Thus, up to a subsequence, we may assume ${\lambda }_n\to {\lambda }_{\epsilon }$ (as $n\to \infty$). Together with Equation (19), we conclude that $u_{\epsilon }$ is a solution to the following Euler–Lagrange equation:

$$I_{\epsilon }^{\prime }\left(u_{\epsilon }\right)-{\lambda }_{\epsilon }{\Psi }^{\prime }\left(u_{\epsilon }\right)=0in{\mathbb{X}}^{\prime }.$$

(20)

We need to show that ${\lambda }_{\epsilon }<0$. The Pohozaev identity (for its derivation, see the proof of Proposition 4) indicates a relationship between ${\lambda }_{\epsilon }$ and $I_{\epsilon }\left(u_{\epsilon }\right)$. Using the condition $I_{\epsilon }\left(u_{\epsilon }\right)\le c<0$, a calculation shows that, for sufficiently small $\epsilon >0$, ${\lambda }_{\epsilon }$ must be negative. Therefore, there exists a constant ${\lambda }^{*}<0$ such that

$${\lambda }_{\epsilon }\le {\lambda }^{*}<0.$$

(21)

Now, we prove the strong convergence of the sequence by contradiction. Suppose that $v_n:=u_n-u_{\epsilon }$ does not converge to 0 in $\mathbb{X}$. We consider $\langle I_{\epsilon }^{\prime }\left(u_n\right)-{\lambda }_n{\Psi }^{\prime }\left(u_n\right),v_n\rangle$. Since $I_{\epsilon }^{\prime }\left(u_n\right)-{\lambda }_n{\Psi }^{\prime }\left(u_n\right)\to 0$ strongly in ${\mathbb{X}}^{\prime }$ and $\left\{v_n\right\}$ is bounded in $\mathbb{X}$, we have

$$\underset{n\to \infty }{\lim }\langle I_{\epsilon }^{\prime }\left(u_n\right)-{\lambda }_n{\Psi }^{\prime }\left(u_n\right),v_n\rangle =0.$$

Expanding the above expression and combining the Brezis–Lieb theorem, we obtain

$$\begin{array}{cc}\hfill & a{\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{s/2}{v}_n\right|}^2\mathrm{d}x+b{\epsilon }^{N-2s}X\left(u_n\right){\int }_{{\mathbb{R}}^N}{(-\Delta )}^{s/2}{u}_n{(-\Delta )}^{s/2}{v}_n\mathrm{d}x\hfill \\ \hfill & +{\int }_{{\mathbb{R}}^N}(V\left(\epsilon x\right)+1){\left|v_n\right|}^2\mathrm{d}x+{\int }_{{\mathbb{R}}^N}{F}_1^{\prime }\left(v_n\right)v_n\mathrm{d}x\hfill \\ \hfill & ={\lambda }_{\epsilon }{\int }_{{\mathbb{R}}^N}{\left|v_n\right|}^2\mathrm{d}x+{\int }_{{\mathbb{R}}^N}{G}_2^{\prime }(\epsilon x,v_n)v_n\mathrm{d}x+o_n\left(1\right).\hfill \end{array}$$

(22)

Using $u_n=u_{\epsilon }+v_n$ and ${\langle u_{\epsilon },v_n\rangle }_s=o_n\left(1\right)$, we obtain

$$\begin{array}{cc}X\left(u_n\right){\int }_{{\mathbb{R}}^N}{(-\Delta )}^{s/2}{u}_n{(-\Delta )}^{s/2}{v}_n\mathrm{d}x\hfill & =\left(X\left(u_{\epsilon }\right)+X\left(v_n\right)+o_n\left(1\right)\right)\left(X\left(v_n\right)+o_n\left(1\right)\right)\hfill \\ \hfill & =X^2\left(v_n\right)+X\left(u_{\epsilon }\right)X\left(v_n\right)+o_n\left(1\right)\hfill \\ \hfill & \ge X^2\left(v_n\right)+o_n\left(1\right).\hfill \end{array}$$

(23)

From (22) and (23), we obtain

$$\begin{array}{cc}\hfill & aX\left(v_n\right)+b{\epsilon }^{N-2s}{X}^2\left(v_n\right)+{\int }_{{\mathbb{R}}^N}(V\left(\epsilon x\right)+1){\left|v_n\right|}^2\mathrm{d}x+{\int }_{{\mathbb{R}}^N}{F}_1^{\prime }\left(v_n\right)v_n\mathrm{d}x\hfill \\ \hfill & \le {\lambda }_{\epsilon }{\int }_{{\mathbb{R}}^N}{|v_n|}^2\mathrm{d}x+{\int }_{{\mathbb{R}}^N}{G}_2^{\prime }(\epsilon x,v_n)v_n\mathrm{d}x+o_n\left(1\right)\hfill \\ \hfill & \le {\lambda }^{*}{\int }_{{\mathbb{R}}^N}|v_n{|}^2\mathrm{d}x+\tau h+C{\int }_{{\Lambda }_{\epsilon }}{|v_n|}^p\mathrm{d}x+o_n\left(1\right).\hfill \end{array}$$

(24)

Now we rescale and rearrange the above inequality. Since $V\left(\epsilon x\right)\ge V_0>-1$, we obtain

$$a{\int }_{{\mathbb{R}}^N}{|(-\Delta )}^{s/2}{v}_n{|}^2\mathrm{d}x+(V_0+1-{\lambda }^{*}){\int }_{{\mathbb{R}}^N}|v_n{|}^2\mathrm{d}x\le \tau h+C{\int }_{{\Lambda }_{\epsilon }}{\left|v_n\right|}^p\mathrm{d}x+o_n\left(1\right).$$

(25)

Our contradictory hypothesis $v_n\nrightarrow 0$ implies that $\lim \inf |v_n{|}_2^2\ge \beta >0$ (for some $\beta >0$) and ${\lim \inf \int |(-\Delta )}^{s/2}{v}_n{|}^2\mathrm{d}x=:K>0$. At the same time, $v_n\rightharpoonup 0$ and [5] (Theorem 2.6) guarantee that ${\int }_{{\Lambda }_{\epsilon }}{\left|v_n\right|}^p\mathrm{d}x\to 0$. Taking the limit inferior as $n\to \infty$ on both sides of (25), we get

$$aK+(V_0+1-{\lambda }^{*})\beta \le \tau h.$$

Since $(V_0+1-{\lambda }^{*})\beta >0$ and $\tau$ is an arbitrarily small positive number, we can always choose $\tau$ small enough such that $\tau h<\alpha K+(V_0+1-{\lambda }^{*})\beta$.

In this case, we obtain the contradiction. It follows that $v_n\to 0$ in $\mathbb{X}$; that is, $u_n\to u_{\epsilon }$ strongly in $\mathbb{X}$. This completes the proof of the lemma. □

4. Multiple Solutions of Problem (8)

We now apply Ljusternik–Schnirelmann category theory to obtain multiple solutions for problem (8). The core idea is to link the number of solutions with the topological structure of the set M (from (V)). To this end, we fix $\delta >0$ such that $M_{\delta }\subset \Lambda$ and define a smooth non-increasing cut-off function $\eta \in C_c^{\infty }({\mathbb{R}}^{+},[0,1])$ as follows:

$$\eta \left(t\right)=\left\{\begin{array}{cc}1,\hfill & \mathrm{for}0\le t\le {\displaystyle \frac{\delta }{2}},\hfill \\ 0,\hfill & \mathrm{for}t\ge \delta .\hfill \end{array}\right.$$

Let $y\in M$ be arbitrary; we define the localized function as follows:

$${\Psi }_{\epsilon ,y}\left(x\right):=\eta \left(\right|\epsilon x-y\left|\right)\omega ((\epsilon x-y)/\epsilon ).$$

By Theorem 2, problem (3.2) admits a positive solution $\omega \in \mathbb{X}$. We define the mapping ${\Phi }_{\epsilon }:M\to S\left(h\right)$ as follows:

$${\Phi }_{\epsilon }\left(y\right):=\sqrt{h}{\Psi }_{\epsilon ,y}/{\parallel {\Psi }_{\epsilon ,y}\parallel }_2.$$

By definition, for any $y\in M$, ${\Phi }_{\epsilon }\left(y\right)$ has a compact support.

Lemma 7.

For the test function ${\Phi }_{\epsilon }\left(y\right):=\sqrt{h}{\Psi }_{\epsilon ,y}/{\parallel {\Psi }_{\epsilon ,y}\parallel }_2$, the following limit holds:

$$\underset{\epsilon \to 0}{\lim }{I}_{\epsilon }\left({\Phi }_{\epsilon }\left(y\right)\right)=Y_{0,h}-{\displaystyle \frac{b}{4}}{\left({\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{s/2}\omega \left(z\right)\right|}^2\mathrm{d}z\right)}^2.$$

Moreover, this convergence is uniform with respect to $y\in M$.

Proof. 

We proceed by contradiction. Suppose the convergence is not uniform. Then there exist a constant ${\delta }_0>0$, a sequence ${\epsilon }_n\to 0$, and a sequence $\left\{y_n\right\}\subset M$ such that

$$|I_{{\epsilon }_n}\left({\Phi }_{{\epsilon }_n}\left(y_n\right)\right)-L|\ge {\delta }_0,\forall n\in \mathbb{N}.$$

Let $L=Y_{0,h}-\frac{b}{4}{\left({\int |(-\Delta )}^{s/2}{\omega |}^2\mathrm{d}z\right)}^2$ be the target limit value. Our goal is to derive a contradiction by computing ${\displaystyle \underset{n\to \infty }{\lim }{I}_{{\epsilon }_n}\left({\Phi }_{{\epsilon }_n}\left(y_n\right)\right)}$. This calculation is based on the definition of $I_{\epsilon }$ (corresponding to (8)),

$$I_{\epsilon }\left(u\right)={\displaystyle \frac{a}{2}}X\left(u\right)+{\displaystyle \frac{b{\epsilon }^{N-2s}}{4}}{X}^2\left(u\right)+{\displaystyle \frac{1}{2}}{\int }_{{\mathbb{R}}^N}(V\left(\epsilon x\right)+1)u^2\mathrm{d}x+{\int }_{{\mathbb{R}}^N}{F}_1\left(u\right)\mathrm{d}x-{\int }_{{\mathbb{R}}^N}{G}_2(\epsilon x,u)\mathrm{d}x.$$

By the convergence $\parallel \eta \left({\epsilon }_{n}z\right)\omega \left(z\right)-\omega \left(z\right){\parallel }_{H^s}\to 0$, which can be established using an argument similar to [40] (Lemma 5), we obtain

$$\underset{n\to \infty }{\lim }X\left({\Psi }_{{\epsilon }_n,y_n}\right)=\underset{{\mathbb{R}}^{2N}}{\int \int }{\displaystyle \frac{|\omega \left(z_1\right)-\omega \left(z_2\right){|}^2}{|z_1-z_2{|}^{N+2s}}}\mathrm{d}{z}_1\mathrm{d}{z}_2=X\left(\omega \right),$$

from the proof of [41] (Lemma 12). In view of the convergence of the norm and the nonlinear term, we can similarly obtain

$$\underset{n\to \infty }{\lim }{\displaystyle \frac{1}{2}}\int V\left({\epsilon }_{n}x\right)|{\Phi }_{{\epsilon }_n}\left(y_n\right){|}^2\mathrm{d}x={\displaystyle \frac{1}{2}}{\int }_{{\mathbb{R}}^N}{V}_0{\left|\omega \left(z\right)\right|}^2\mathrm{d}z,$$

$$\underset{n\to \infty }{\lim }{\int }_{{\mathbb{R}}^N}{F}_1\left({\Phi }_{{\epsilon }_n}\left(y_n\right)\right)\mathrm{d}x={\int }_{{\mathbb{R}}^N}{F}_1\left(\omega \left(z\right)\right)\mathrm{d}z,$$

Then we get that

$$\begin{array}{cc}\underset{n\to \infty }{\lim }{I}_{{\epsilon }_n}\left({\Phi }_{{\epsilon }_n}\left(y_n\right)\right)\hfill & ={\displaystyle \frac{a}{2}}X\left(\omega \right)+0+{\displaystyle \frac{1}{2}}{\int }_{{\mathbb{R}}^N}(V_0+1){\left|\omega \right|}^2\mathrm{d}z+{\int }_{{\mathbb{R}}^N}{F}_1\left(\omega \right)\mathrm{d}z-{\int }_{{\mathbb{R}}^N}{F}_2\left(\omega \right)\mathrm{d}z\hfill \\ \hfill & =J_{V_0}\left(\omega \right)-{\displaystyle \frac{b}{4}}{X}^2\left(\omega \right)=Y_{0,h}-{\displaystyle \frac{b}{4}}{X}^2\left(\omega \right)=L.\hfill \end{array}$$

Our calculation shows that, for any sequence ${\epsilon }_n\to 0$ and any $\left\{y_n\right\}\subset M$, we have $I_{{\epsilon }_n}\left({\Phi }_{{\epsilon }_n}\left(y_n\right)\right)\to L$. This contradicts the (contradiction) assumption that there exists a subsequence such that $|I_{{\epsilon }_n}\left({\Phi }_{{\epsilon }_n}\left(y_n\right)\right)-L|\ge {\delta }_0$. Therefore, the lemma is proved.

For any $\delta >0$, there exists an $R=R\left(\delta \right)>0$ such that $M_{\delta }\subset B_R\left(0\right)$. Now we consider the barycenter map ${\beta }_{\epsilon }:S\left(h\right)\to {\mathbb{R}}^N$ and an auxiliary map $\kappa :{\mathbb{R}}^N\to {\mathbb{R}}^N$, defined as follows:

$$\kappa \left(x\right):=\left\{\begin{array}{cc}x,\hfill & \left|x\right|\le R,\hfill \\ {\displaystyle \frac{Rx}{\left|x\right|}},\hfill & \left|x\right|>R,\hfill \end{array}\right.{\beta }_{\epsilon }\left(u\right):={\displaystyle \frac{{\int }_{{\mathbb{R}}^N}\kappa \left(\epsilon x\right){\left|u\right|}^2\mathrm{d}x}{h}}.$$

The proof of the next lemma is similar to the argument used in [25] (Lemma 4.2) and can be easily verified. □

Lemma 8.

The following limit holds:

$$\underset{\epsilon \to 0}{\lim }{\beta }_{\epsilon }\left({\Phi }_{\epsilon }\left(y\right)\right)=y.$$

This convergence is uniform with respect to $y\in M$.

Now we choose a positive function $\gamma :[0,+\infty )\to [0,+\infty )$ such that $\gamma \left(\epsilon \right)\to 0$ as $\epsilon \to 0$ and define the following set:

$$\tilde{S}\left(h\right):=\{u\in S\left(h\right):I_{\epsilon }\left(u\right)\le Y_{0,h}+\gamma \left(\epsilon \right)\}.$$

Lemma 7 guarantees that ${\lim }_{\epsilon \to 0}{I}_{\epsilon }\left({\Phi }_{\epsilon }\left(y\right)\right)=L=Y_{0,h}-{\displaystyle \frac{b}{4}}X{\left(\omega \right)}^2<Y_{0,h}$. Thus, for any $\gamma \left(\epsilon \right)\to 0^{+}$ (e.g., $\gamma \left(\epsilon \right)=\sqrt{\epsilon }$), when $\epsilon >0$ is sufficiently small,

$$I_{\epsilon }\left({\Phi }_{\epsilon }\left(y\right)\right)<L+{\displaystyle \frac{Y_{0,h}-L}{2}}={\displaystyle \frac{Y_{0,h}+L}{2}}<Y_{0,h}\le Y_{0,h}+\gamma \left(\epsilon \right).$$

Hence, for any $y\in M$, we have ${\Phi }_{\epsilon }\left(y\right)\in \tilde{S}\left(h\right)$ for sufficiently small $\epsilon$, which implies $\tilde{S}\left(h\right)$ is non-empty. In order to prove Theorem 1, we also need the following crucial compactness result.

Lemma 9.

Let $\left\{u_n\right\}\subset S\left(h\right)$ and ${\displaystyle \underset{n\to \infty }{\lim }{I}_{{\epsilon }_n}\left(u_n\right)=L=Y_{0,h}-\frac{b}{4}{X}^2\left(\omega \right)}$ (where ω is a ground state solution of $J_{V_0}$, and $X\left(\omega \right)={\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{s/2}\omega \right|}^2\mathrm{d}x$). Then there exists a sequence $\left\{{\tilde{y}}_n\right\}\subset {\mathbb{R}}^N$ such that ${\tilde{u}}_n\left(x\right):=u_n(x+{\tilde{y}}_n)$ has a convergent subsequence in $\mathbb{X}$. By passing to a subsequence, we have $y_n={\epsilon }_n{\tilde{y}}_n\to y_0$, where $y_0\in M$.

Proof. 

We divide the proof into the following three steps.

Step 1. Prove that ${\tilde{u}}_n\left(x\right)=u_n(x+{\tilde{y}}_n)$ has a convergent subsequence. We apply the concentration–compactness principle (CCP) [42].

(a) Ruling out vanishing. We first claim that the sequence does not vanish; i.e., there exist constants $\beta >0$ and ${\tilde{y}}_n\subset {\mathbb{R}}^N$ such that

$${\int }_{B_R\left({\tilde{y}}_n\right)}{\left|u_n\right|}^2\mathrm{d}x\ge \beta .$$

If the sequence vanishes, then $u_n\to 0$ strongly in $L^p\left({\mathbb{R}}^N\right)$ (for $p\in (2,2_s^{*})$), which implies ${\int }_{{\mathbb{R}}^N}{F}_1\left(u_n\right)\mathrm{d}x\to 0$ and ${\int }_{{\mathbb{R}}^N}{F}_2\left(u_n\right)\mathrm{d}x\to 0$. Considering the non-negativity of the other terms in $I_{{\epsilon }_n}\left(u_n\right)$, we obtain ${\lim \inf }_{n\to \infty }{I}_{{\epsilon }_n}\left(u_n\right)\ge 0$. However, this contradicts the premise that $I_{{\epsilon }_n}\left(u_n\right)\to L<Y_{0,h}<0$. Therefore, the sequence must be non-vanishing. Combined with the proof of Proposition 5, we can further conclude that ${\tilde{u}}_n\left(x\right)$ is also non-vanishing.

(b) Ruling out dichotomy. From the proof of Proposition 5, we can get that dichotomy cannot occur.

According to the CCP, after ruling out the “vanishing” and “dichotomy” cases, the sequence $\left\{{\tilde{u}}_n\right\}$ must be tight. This implies that, up to a subsequence, $\left\{{\tilde{u}}_n\right\}$ converges strongly in the space $\mathbb{X}$ to its weak limit $\tilde{u}$. Therefore, $\tilde{u}$ is a solution to (9) (corresponding to $V_0$) and $J_{V_0}\left(\tilde{u}\right)=Y_{0,h}$. This proves that $\tilde{u}$ is a ground state solution $\omega$.

Step 2. Prove that ${\lim }_{n\to \infty }\mathrm{dist}({\epsilon }_n{\tilde{y}}_n,\overline{\Lambda })=0$. We proceed by contradiction. Suppose that there exists a subsequence such that ${\epsilon }_n{\tilde{y}}_n\to y_0$ with $y_0\notin \overline{\Lambda }$. Since ${\tilde{u}}_n\to \tilde{u}$ strongly in $\mathbb{X}$ and ${\epsilon }_n^{N-2s}\to 0$, we can compute the limit of the energy:

$$\begin{array}{cc}\underset{n\to \infty }{\lim }{I}_{{\epsilon }_n}\left(u_n\right)=\underset{n\to \infty }{\lim }{I}_{{\epsilon }_n}\left({\tilde{u}}_n\right)\hfill & ={\displaystyle \frac{a}{2}}X\left(\tilde{u}\right)+{\displaystyle \frac{1}{2}}{\int }_{{\mathbb{R}}^N}(V\left(y_0\right)+1){\left|\tilde{u}\right|}^2\mathrm{d}x\hfill \\ \hfill & +{\int }_{{\mathbb{R}}^N}{F}_1\left(\tilde{u}\right)\mathrm{d}x-\underset{n\to \infty }{\lim }{\int }_{{\mathbb{R}}^N}{G}_2({\epsilon }_{n}x+y_n,{\tilde{u}}_n)\mathrm{d}x.\hfill \end{array}$$

(26)

Moreover, since $y_n\to y_0\notin \overline{\Lambda }$, one has ${\chi }_{\Lambda }({\epsilon }_{n}x+y_n)\to 0$ a.e. in ${\mathbb{R}}^N$; hence, by the growth bound of $G_2$ and the strong convergence ${\tilde{u}}_n\to \tilde{u}$ in $L^p\left({\mathbb{R}}^N\right)$, we have

$${\int }_{{\mathbb{R}}^N}{G}_2({\epsilon }_{n}x+y_n,{\tilde{u}}_n)dx\to {\int }_{{\mathbb{R}}^N}{\tilde{F}}_2\left(\tilde{u}\right)dx.$$

Therefore,

$$\underset{n\to \infty }{\lim }{I}_{{\epsilon }_n}\left({\tilde{u}}_n\right)={\tilde{J}}_{V\left(y_0\right)}\left(\tilde{u}\right)-{\displaystyle \frac{b}{4}}{X}^2\left(\tilde{u}\right).$$

By the assumption, ${\displaystyle \underset{n\to \infty }{\lim }{I}_{{\epsilon }_n}\left(u_n\right)=L=Y_{0,h}-\frac{b}{4}{X}^2\left(\omega \right)}$. Since the first step showed that $\tilde{u}$ is $\omega$ (up to translation), we have $X\left(\tilde{u}\right)=X\left(\omega \right)$. Thus,

$$Y_{0,h}-{\displaystyle \frac{b}{4}}{X}^2\left(\tilde{u}\right)={\tilde{J}}_{V\left(y_0\right)}\left(\tilde{u}\right)-{\displaystyle \frac{b}{4}}{X}^2\left(\tilde{u}\right),$$

and hence $Y_{0,h}={\tilde{J}}_{V\left(y_0\right)}\left(\tilde{u}\right)$. Noting that ${\tilde{F}}_2\le F_2$, we have ${\tilde{J}}_{V\left(y_0\right)}\left(\tilde{u}\right)\ge J_{V\left(y_0\right)}\left(\tilde{u}\right)$. Since $y_0\notin \overline{\Lambda }$, one has $V\left(y_0\right)\ge V_{\partial \Lambda }$, and therefore

$$Y_{0,h}={\tilde{J}}_{V\left(y_0\right)}\left(\tilde{u}\right)\ge J_{V\left(y_0\right)}\left(\tilde{u}\right)\ge J_{V_{\partial \Lambda }}\left(\tilde{u}\right)\ge Y_{\partial \Lambda ,h}.$$

This contradicts Proposition 2 that $Y_{0,h}<Y_{\partial \Lambda ,h}$. Hence, $y_0\in \overline{\Lambda }$, and the proof is complete.

Step 3. To prove $y_0\in M$, we use $y_0\in \overline{\Lambda }$ and $Y_{0,h}=J_{V\left(y_0\right)}\left(\tilde{u}\right)$ from Step 2. From $y_0\in \overline{\Lambda }$, we know $V\left(y_0\right)\ge V_0$. Because $J_{\mu }$ is monotone with respect to $\mu$ and $\tilde{u}$ is a ground state solution of $J_{V_0}$ ($J_{V_0}\left(\tilde{u}\right)=Y_{0,h}$), it follows that $J_{V\left(y_0\right)}\left(\tilde{u}\right)\ge J_{V_0}\left(\tilde{u}\right)=Y_{0,h}$. Comparing this with $Y_{0,h}=J_{V\left(y_0\right)}\left(\tilde{u}\right)$, we get

$$Y_{0,h}=J_{V\left(y_0\right)}\left(\tilde{u}\right)\ge J_{V_0}\left(\tilde{u}\right)=Y_{0,h}.$$

This forces all the inequalities to become equalities; that is, $J_{V\left(y_0\right)}\left(\tilde{u}\right)=J_{V_0}\left(\tilde{u}\right)$, which implies

$${\int }_{{\mathbb{R}}^N}(V\left(y_0\right)-V_0){\left|\tilde{u}\right|}^2\mathrm{d}x=0.$$

Since $\tilde{u}\not\equiv 0$ and $\tilde{u}>0$ (by Theorem 2) and $V\left(y_0\right)\ge V_0$ almost everywhere on the support of $\tilde{u}$, we can obtain $V\left(y_0\right)=V_0$. According to the definition of the set M, we therefore have $y_0\in M$. The proof is complete. □

Lemma 10.

For any $\delta >0$, the following limit holds:

$$\underset{\epsilon \to 0}{\lim }\underset{u\in \tilde{S}\left(h\right)}{\sup }\underset{z\in M_{\delta }}{\inf }|{\beta }_{\epsilon }\left(u\right)-z|=0.$$

Lemma 11.

Consider the sequence $\left\{{\tilde{u}}_n\right\}$ from Lemma 9. This sequence is bounded in $L^{\infty }\left({\mathbb{R}}^N\right)$; i.e., ${\sup }_{n\in \mathbb{N}}{\left|{\tilde{u}}_n\right|}_{\infty }\le K$ (for some $K>0$). Furthermore,

$${\tilde{u}}_n\left(x\right)\to 0as\left|x\right|\to +\infty ,$$

The convergence is uniform with respect to $n\in \mathbb{N}$.

Proof. 

We first define a cut-off function.

$$\phi \left(t\right):=\left\{\begin{array}{cc}0,\hfill & t<0,\hfill \\ t^{\beta },\hfill & 0\le t\le T,\hfill \\ \beta T^{\beta -1}t-(\beta -1)T^{\beta },\hfill & t>T.\hfill \end{array}\right.$$

The constants $T>0$ and $\beta >1$ are to be determined subsequently. By a straightforward computation, we obtain its derivative as follows:

$${\phi }^{\prime }\left(t\right)=\left\{\begin{array}{cc}\beta t^{\beta -1},\hfill & t\le T,\hfill \\ \beta T^{\beta -1},\hfill & t>T.\hfill \end{array}\right.$$

When ${\tilde{u}}_n\ge 0$, we have

$$\begin{array}{cc}\hfill \phi \left({\tilde{u}}_n\right)& \ge 0,{\phi }^{\prime }\left({\tilde{u}}_n\right)\ge 0,\mathrm{and}{\tilde{u}}_n{\phi }^{\prime }\left({\tilde{u}}_n\right)\le \beta \phi \left({\tilde{u}}_n\right).\hfill \end{array}$$

(27)

We substitute $\phi \left({\tilde{u}}_n\right){\phi }^{\prime }\left({\tilde{u}}_n\right)$ as a test function into the Kirchhoff equation and obtain

$$\begin{array}{cc}\hfill & M_n{\int }_{{\mathbb{R}}^N}\phi \left({\tilde{u}}_n\right){\phi }^{\prime }\left({\tilde{u}}_n\right){(-\Delta )}^s{\tilde{u}}_n\mathrm{d}x+{\int }_{{\mathbb{R}}^N}(V({\epsilon }_{n}x+{\epsilon }_n{\tilde{y}}_n)+1){\tilde{u}}_n\phi \left({\tilde{u}}_n\right){\phi }^{\prime }\left({\tilde{u}}_n\right)\mathrm{d}x\hfill \\ \hfill & ={\int }_{{\mathbb{R}}^N}{G}_2^{\prime }({\epsilon }_{n}x+{\epsilon }_n{\tilde{y}}_n,{\tilde{u}}_n)\phi \left({\tilde{u}}_n\right){\phi }^{\prime }\left({\tilde{u}}_n\right)\mathrm{d}x-{\int }_{{\mathbb{R}}^N}{F}_1^{\prime }\left({\tilde{u}}_n\right)\phi \left({\tilde{u}}_n\right){\phi }^{\prime }\left({\tilde{u}}_n\right)\mathrm{d}x.\hfill \end{array}$$

(28)

Here, $M_n=a+b{\epsilon }_n^{N-2s}X\left({\tilde{u}}_n\right)$ is a bounded sequence of positive numbers (and $\lim M_n=a$). According to the definition of the penalized function $G_2^{\prime }$, we have the following growth estimate:

$$G_2^{\prime }(x,t)\le \tau t+C{t}^{2_s^{*}-1},\forall (x,t)\in {\mathbb{R}}^N\times {\mathbb{R}}^{+}.$$

Utilizing the condition $F_1^{\prime }\left(s\right)\ge 0$ alongside (27) and (28), we rescale Equation (28) and obtain the following inequality:

$$\begin{array}{cc}\hfill & M_n{\int }_{{\mathbb{R}}^N}\phi \left({\tilde{u}}_n\right){\phi }^{\prime }\left({\tilde{u}}_n\right){(-\Delta )}^s{\tilde{u}}_n\mathrm{d}x+{\int }_{{\mathbb{R}}^N}(V({\epsilon }_{n}x+{\epsilon }_n{\tilde{y}}_n)+1){\tilde{u}}_n\phi \left({\tilde{u}}_n\right){\phi }^{\prime }\left({\tilde{u}}_n\right)\mathrm{d}x\hfill \\ \hfill & \le C_1\beta {\int }_{{\mathbb{R}}^N}{\left|{\tilde{u}}_n\right|}^{2_s^{*}-2}{\phi }^2\left({\tilde{u}}_n\right)\mathrm{d}x+o_n\left(1\right).\hfill \end{array}$$

(29)

On the other hand, from [43], we know the following key inequality:

$${(-\Delta )}^s\phi \left({\tilde{u}}_n\right)\le {\phi }^{\prime }\left({\tilde{u}}_n\right){(-\Delta )}^s{\tilde{u}}_n.$$

Combining this inequality with the Sobolev inequality and (29), we derive the following:

$$S\parallel \phi \left({\tilde{u}}_n\right){\parallel }_{2_s^{*}}^2\le {\int }_{{\mathbb{R}}^N}\phi \left({\tilde{u}}_n\right){(-\Delta )}^s\phi \left({\tilde{u}}_n\right)\mathrm{d}x\le {\int }_{{\mathbb{R}}^N}\phi \left({\tilde{u}}_n\right){\phi }^{\prime }\left({\tilde{u}}_n\right){(-\Delta )}^s{\tilde{u}}_n\mathrm{d}x.$$

Since the sequence $M_n$ has a positive lower bound and the potential function V is bounded from below, we can isolate the following kinetic term from Equation (29):

$$\begin{array}{cc}{\int }_{{\mathbb{R}}^N}\phi \left({\tilde{u}}_n\right){\phi }^{\prime }\left({\tilde{u}}_n\right){(-\Delta )}^s{\tilde{u}}_n\mathrm{d}x\hfill & \le {\displaystyle \frac{1}{M_n}}\left[C_1\beta {\int }_{{\mathbb{R}}^N}{\left|{\tilde{u}}_n\right|}^{2_s^{*}-2}{\phi }^2\left({\tilde{u}}_n\right)\mathrm{d}x-{\int }_{{\mathbb{R}}^N}{V}_n\left(x\right)\mathrm{d}x\right]\hfill \\ \hfill & \le C_2\beta {\int }_{{\mathbb{R}}^N}{\left|{\tilde{u}}_n\right|}^{2_s^{*}-2}{\phi }^2\left({\tilde{u}}_n\right)\mathrm{d}x,\hfill \end{array}$$

where $V_n\left(x\right):=(V({\epsilon }_{n}x+{\epsilon }_n{\tilde{y}}_n)+1){\tilde{u}}_n\phi \left({\tilde{u}}_n\right){\phi }^{\prime }\left({\tilde{u}}_n\right)$. Then, we obtain the following key inequality:

$$S\parallel \phi \left({\tilde{u}}_n\right){\parallel }_{2_s^{*}}^2\le C_2\beta {\int }_{{\mathbb{R}}^N}{\left|{\tilde{u}}_n\right|}^{2_s^{*}-2}{\phi }^2\left({\tilde{u}}_n\right)\mathrm{d}x.$$

(30)

Then, by applying [5] (Lemma 4.5), we can obtain

$$\parallel {\tilde{u}}_n{\parallel }_{\infty }\le {\left({\displaystyle \frac{C}{S}}{k}^{2_s^{*}-3}\right)}^{{\displaystyle \frac{1}{2(\beta -1)}}}{\beta }^L{\parallel {\tilde{u}}_n\parallel }_{2_s^{*}}\le K,\forall n\in \mathbb{N}.$$

Next, we reformulate the equation fulfilled by the limiting function $\tilde{u}$. The function $\tilde{u}$ satisfies the Euler–Lagrange equation:

$$a{(-\Delta )}^s\tilde{u}+(V_0+1)\tilde{u}={\lambda }_0\tilde{u}+\tilde{u}\log {\tilde{u}}^2.$$

Let $M=a>0$ be a constant. Then the equation becomes

$$M{(-\Delta )}^s\tilde{u}+(V_0+1-{\lambda }_0)\tilde{u}=\tilde{u}\log {\tilde{u}}^2.$$

The above equation can be rewritten in the form ${(-\Delta )}^s\tilde{u}+K_0\tilde{u}=g\left(\tilde{u}\right)$. Here, $K_0=(V_0+1-{\lambda }_0)/M$, and the nonlinear term is $g\left(\tilde{u}\right)={\displaystyle \frac{1}{M}}\tilde{u}\log {\tilde{u}}^2$. To compare the growth order of the nonlinear term $g\left(\tilde{u}\right)$ and the linear term $K_0\tilde{u}$, we turn to examine the asymptotic behavior of $g\left(\tilde{u}\left(x\right)\right)$ versus $K_0\tilde{u}\left(x\right)$ as $\left|x\right|\to \infty$, specifically their ratio:

$${\displaystyle \frac{g\left(\tilde{u}\left(x\right)\right)}{K_0\tilde{u}\left(x\right)}}={\displaystyle \frac{\log \tilde{u}{\left(x\right)}^2}{V_0+1-{\lambda }_0}}.$$

(31)

Since $V_0>-1$, and, for the ground state solution, ${\lambda }_0$ is negative, $V_0+1-{\lambda }_0$ is strictly positive. As $\left|x\right|\to \infty$, we have $\tilde{u}\left(x\right)\to 0^{+}$. We then compute the limit of the above (31):

$$\underset{\left|x\right|\to \infty }{\lim }{\displaystyle \frac{g\left(\tilde{u}\left(x\right)\right)}{K_0\tilde{u}\left(x\right)}}=\underset{t\to 0^{+}}{\lim }{\displaystyle \frac{\log t^2}{V_0+1-{\lambda }_0}}=-\infty .$$

(32)

According to the equation ${(-\Delta )}^s\tilde{u}=g\left(\tilde{u}\right)-K_0\tilde{u}$, this is equivalent to proving that $g\left(\tilde{u}\right)-K_0\tilde{u}\le -{\displaystyle \frac{K_0}{2}}\tilde{u}$. Then, from Equation (32), we can see that there exists a value ${\displaystyle \frac{K_0}{2}}$ and some $R>0$ such that, for all $x\in B_R^c\left(0\right)$,

$${\displaystyle \frac{g\left(\tilde{u}\left(x\right)\right)}{\tilde{u}\left(x\right)}}<{\displaystyle \frac{K_0}{2}}.$$

Therefore, substituting $g\left(\tilde{u}\right)<{\displaystyle \frac{K_0}{2}}\tilde{u}$ into the equation ${(-\Delta )}^s\tilde{u}+K_0\tilde{u}=g\left(\tilde{u}\right)$, we have

$${(-\Delta )}^s\tilde{u}\left(x\right)+{\displaystyle \frac{K_0}{2}}\tilde{u}\left(x\right)<0,for\left|x\right|>R.$$

The solution ${\tilde{u}}_n$ can be expressed in the convolution form with the Bessel kernel $\mathcal{K}$ as ${\tilde{u}}_n=\mathcal{K}*g_n$, where $g_n$ contains all the lower-order terms. Since $\left\{g_n\right\}$ is uniformly bounded in $L^{\infty }$ and possesses certain compactness properties, the Bessel kernel $\mathcal{K}$ satisfies the following properties [44]:

(i)

$\mathcal{K}$ is positive, radially symmetric, and smooth on ${\mathbb{R}}^N\setminus \left\{0\right\}$;

(ii)

There exists a constant $C>0$ such that $\mathcal{K}\left(x\right)\le {\displaystyle \frac{C}{{\left|x\right|}^{N+2s}}}$;

(iii)

For any $r\in [1,\frac{N}{N-2s})$, we have $\mathcal{K}\in L^r\left({\mathbb{R}}^N\right)$.

Similar to the argument in [45] (Lemma 2.6) (utilizing convolution and the decay properties of the kernel), we can obtain that ${\tilde{u}}_n\left(x\right)\to 0$ (as $\left|x\right|\to \infty$), and this convergence is uniform in $n\in \mathbb{N}$. The proof is thus complete. □

5. Proof of Theorem 1

The proof is organized into three parts.

(1) To obtain multiple solutions for (1), we apply Lusternik–Schnirelmann theory [27]. Lemmas 3 and 6 establish the boundedness (from below) of $I_{\epsilon }{|}_{\tilde{S}\left(h\right)}$, as well as the ${\left(\mathrm{PS}\right)}_c$ condition (at levels $c<Y_{\infty ,h}$). And the theory then guarantees that $I_{\epsilon }$ has at least ${\mathrm{cat}}_{\tilde{S}\left(h\right)}\left(\tilde{S}\left(h\right)\right)$ critical points on $\tilde{S}\left(h\right)$.

Applying Lemmas 7–10 and following the method in [46] (Section 6) for a given $\delta >0$, we obtain ${\epsilon }_{\delta }>0$ such that, for all $\epsilon \in (0,{\epsilon }_{\delta })$, the mapping

$$M\stackrel{{\Phi }_{\epsilon }}{\to }\tilde{S}\left(h\right)\stackrel{{\beta }_{\epsilon }}{\to }{M}_{\delta }$$

is well defined, and the composite mapping ${\beta }_{\epsilon }\circ {\Phi }_{\epsilon }$ is homotopic to the inclusion map $\iota :M\to M_{\delta }$. According to [47] (Lemma 2.2), this fact implies that ${\mathrm{cat}}_{\tilde{S}\left(h\right)}\left(\tilde{S}\left(h\right)\right)\ge {\mathrm{cat}}_{M_{\delta }}\left(M\right)$. Hence, there exists at least ${\mathrm{cat}}_{M_{\delta }}\left(M\right)$ critical points for problem (8).

Let $u_{\epsilon }$ solve problem (3.1) with $I_{\epsilon }\left(u_{\epsilon }\right)\le Y_{0,h}+\gamma \left(\epsilon \right)$. We claim that

$$u_{\epsilon }\left(x\right)<t_{h_0},\forall x\in {\overline{\Lambda }}_{\epsilon }^c,\mathrm{where}{\overline{\Lambda }}_{\epsilon }:=\left\{x\in {\mathbb{R}}^N:\epsilon x\in \Lambda \right\}.$$

This means that, for all $\epsilon \in (0,{\epsilon }_{\delta })$, $u_{\epsilon }$ is a solution of problem (1). In fact, for each sequence ${\epsilon }_n\to 0$, let $u_{{\epsilon }_n}$ be a solution of problem (8) with $I_{{\epsilon }_n}\left(u_{{\epsilon }_n}\right)\le Y_{0,h}+\gamma \left({\epsilon }_n\right)$. Suppose by contradiction that

$$u_{{\epsilon }_n}\left(x\right)\ge t{h}_0,\forall x\in {\Lambda }_{{\epsilon }_n}^c.$$

(33)

Lemma 9 provides $y_n={\epsilon }_n{\tilde{y}}_n\to y_0\in M$ and the strong convergence of ${\tilde{u}}_{{\epsilon }_n}\left(x\right)=u_{{\epsilon }_n}(x+{\tilde{y}}_n)$ in $\mathbb{X}$. Lemma 11 then guarantees the uniform decay of $u_{{\epsilon }_n}$ (i.e., $\exists R_1>0$ s.t. $u_{{\epsilon }_n}\left(x\right)<t_{h_0}$ in $B_{R_1}^c\left({\tilde{y}}_n\right)$ for large n). By appropriately choosing sufficiently small r and ${\epsilon }_n$, we can have ${\Lambda }_{{\epsilon }_n}^c\subset B_{R_1}^c\left({\tilde{y}}_n\right)$, which implies $u_{{\epsilon }_n}\left(x\right)<t_{h_0}$ for all $x\in \Lambda {{\epsilon }_n}^c$. This contradicts the contradiction hypothesis. Hence, we have demonstrated the multiplicity of solutions for (1).

(2) Concentration of the maximum points. We now study the behavior of the maximum points of the solution $|{\widehat{u}}_{\epsilon }|$. Let $u_{{\epsilon }_n}$ be a sequence of solutions to problem (5) with ${\epsilon }_n\to 0$. According to Lemma 11, we can find $R>0$ and $\gamma >0$ such that

$$\parallel u_{{\epsilon }_n}{\parallel }_{L^{\infty }\left(B_R^c\left({\tilde{y}}_n\right)\right)}<\gamma ,\mathrm{for}n\mathrm{large}\mathrm{enough}.$$

(34)

We claim that $\parallel u_{{\epsilon }_n}{\parallel }_{L^{\infty }\left(B_R\left({\tilde{y}}_n\right)\right)}\ge \gamma$. By contradiction, assume that $\parallel u_{{\epsilon }_n}{\parallel }_{L^{\infty }\left({\mathbb{R}}^N\right)}<\gamma$. If $\gamma \in (0,t_{h_0})$ is sufficiently small, then $G_2^{\prime }({\epsilon }_{n}x,t)t\le \tau t^2$. The solution $u_{{\epsilon }_n}$ satisfies $\langle I_{{\epsilon }_n}^{\prime }\left(u_{{\epsilon }_n}\right),u_{{\epsilon }_n}\rangle ={\lambda }_{n}h$. For the Kirchhoff equation, this expands as follows:

$$aX\left(u_{{\epsilon }_n}\right)+b{\epsilon }_n^{N-2s}X{\left(u_{{\epsilon }_n}\right)}^2+{\int }_{{\mathbb{R}}^N}(V\left({\epsilon }_{n}x\right)+1){\left|u_{{\epsilon }_n}\right|}^2\mathrm{d}x={\lambda }_{n}h+{\int }_{{\mathbb{R}}^N}(G_2^{\prime }-F_1^{\prime })u_{{\epsilon }_n}\mathrm{d}x.$$

(35)

Since the linear term satisfies ${\int }_{{\mathbb{R}}^N}(G_2^{\prime }-F_1^{\prime })u_{{\epsilon }_n}\mathrm{d}x\le {\int }_{{\mathbb{R}}^N}\left(\tau u_{{\epsilon }_n}\right)u_{{\epsilon }_n}\mathrm{d}x=\tau h$ and the potential term satisfies $\int V\left({\epsilon }_{n}x\right){\left|u_{{\epsilon }_n}\right|}^2\mathrm{d}x\ge (V_0+1)h$, it follows that Equation (35) can be rewritten as follows:

$$aX\left(u_{{\epsilon }_n}\right)+b{\epsilon }_n^{N-2s}X{\left(u_{{\epsilon }_n}\right)}^2\le {\lambda }_{n}h+\tau h-(V_0+1)h.$$

(36)

Since $X\left(u_{{\epsilon }_n}\right)\ge 0$, $b{\epsilon }_n^{N-2s}\ge 0$, $a>0$, ${\lambda }_n\le {\lambda }^{*}<0$ and $V_0>-1$, by choosing sufficiently small $\tau$ and $\gamma$, the right-hand side of (36) can be made strictly negative while the left-hand side is non-negative. This would imply $u_{{\epsilon }_n}\equiv 0$, which contradicts $\parallel u_{{\epsilon }_n}{\parallel }_2^2=h$. Therefore, the assertion holds.

Combining (34) and the above assertion, the maximum point ${\xi }_n$ of $u_{{\epsilon }_n}$ must belong to $B_R\left({\tilde{y}}_n\right)$; that is, ${\xi }_n={\tilde{y}}_n+z_n$ with $z_n\in B_R$. The maximum point of the original solution ${\widehat{u}}_{{\epsilon }_n}\left(x\right)=u_{{\epsilon }_n}(x/{\epsilon }_n)$ is ${\eta }_{{\epsilon }_n}={\epsilon }_n{\xi }_n={\epsilon }_n{\tilde{y}}_n+{\epsilon }_n{z}_n$. Using ${\epsilon }_n{\tilde{y}}_n\to y_0\in M$, ${\epsilon }_n{z}_n\to 0$ and the continuity of V, we obtain

$$\underset{n\to \infty }{\lim }V\left({\eta }_{{\epsilon }_n}\right)=V\left(y_0\right)=V_0.$$

(3) Decay estimates. We focus on the decay estimates of $|{\widehat{u}}_{ϵ}|$. According to Lemma 11, the strong limit $\tilde{u}$ satisfies the equation $M{(-\Delta )}^s\tilde{u}+(V_0+1-{\lambda }_0)\tilde{u}=\tilde{u}\log {\tilde{u}}^2$, where $M=a+bX\left(\tilde{u}\right)>0$. Since $\tilde{u}\left(x\right)\to 0$ as $\left|x\right|\to \infty$, there exist constants $R_1>0$ and $K_0>0$, and, by referring to the proof steps in Lemma 11, we can obtain that

$${(-\Delta )}^s\tilde{u}+K_0\tilde{u}\le 0,$$

where $K_0=(V_0+1-{\lambda }_0)/M>0$. Since the sequence ${\tilde{u}}_n$ is uniformly bounded in $L^{\infty }$ and decays uniformly, by invoking [44] (Lemma 4.3), we can choose a function $\omega \left(x\right)$ and an appropriate $R_2>0$ such that

$${(-\Delta )}^s\omega \left(x\right)+K_0\omega \left(x\right)\ge 0\mathrm{and}0<\omega \left(x\right)\le {\displaystyle \frac{C}{1+{\left|x\right|}^{N+2s}}},x\in B_{R_2}^c\left(0\right).$$

(37)

Let $R_3=\max \{R_1,R_2\}$, define $c:={\inf }_{B_{R_3}\left(0\right)}\omega \left(x\right)>0$, and let ${\tilde{\omega }}_n\left(x\right):=(d+1)\omega \left(x\right)-c{\tilde{u}}_n\left(x\right)$, where $d={\sup }_n{\parallel {\tilde{u}}_n\parallel }_{L^{\infty }}<\infty$. It can be verified that

$${(-\Delta )}^s{\tilde{\omega }}_n\left(x\right)+K_0{\tilde{\omega }}_n\left(x\right)\ge 0,\forall x\in B_{R_3}^c\left(0\right).$$

(38)

Subsequently, we assume that ${\tilde{\omega }}_n\left(x\right)\ge 0$ in ${\mathbb{R}}^N$. Its proof is similar to the argument used in the final part of [5] concerning the decay estimate.

Combining the definition of ${\tilde{\omega }}_n\left(x\right)$ with (37), we obtain a constant $\tilde{C}>0$ satisfying

$$0\le {\tilde{u}}_n\left(x\right)\le {\displaystyle \frac{d+1}{c}}\omega \left(x\right)\le {\displaystyle \frac{\tilde{C}}{1+{\left|x\right|}^{N+2s}}},\forall n\in \mathbb{N},x\in {\mathbb{R}}^N.$$

As ${\widehat{u}}_n\left(x\right)$ is defined, it follows that

$$\begin{array}{cc}\hfill {\widehat{u}}_n\left(x\right)=u_n(x/{\epsilon }_n)={\tilde{u}}_n(x/{\epsilon }_n-{\tilde{y}}_n)& \le {\displaystyle \frac{\tilde{C}{\epsilon }_n^{N+2s}}{{\epsilon }_n^{N+2s}+{|x-{\eta }_{{\epsilon }_n}|}^{N+2s}}}.\hfill \end{array}$$

The proof of Theorem 1 is completed.