Multiplicity of Normalized Solutions to a Fractional Logarithmic Schrödinger Equation

Abstract

We study the existence and multiplicity of normalized solutions to the fractional logarithmic Schrödinger equation ${(-\Delta )}^{s}u+V\left(ϵx\right)u=\lambda u+ulog{u}^2\mathrm{in}{\mathbb{R}}^N,$ under the mass constraint ${\int }_{{\mathbb{R}}^N}{\left|u\right|}^{2}dx=a.$ Here, $N\ge 2$, $a,ϵ>0$, $\lambda \in \mathbb{R}$ is an unknown parameter, ${(-\Delta )}^s$ is the fractional Laplacian and $s\in (0,1)$. We introduce a function space where the energy functional associated with the problem is of class ${\mathcal{C}}^1$. Then, under some assumptions on the potential V and using the Lusternik–Schnirelmann category, we show that the number of normalized solutions depends on the topology of the set for which the potential V reaches its minimum.

1. Introduction

We focus on the following fractional Schrödinger equation [1]:

$$i\frac{\partial \Phi }{\partial t}={(-\Delta )}^s\Phi +V\left(x\right)\Phi -\Phi log{|\Phi |}^2,\mathrm{in}{\mathbb{R}}^N,$$

(1)

where $N\ge 2$, $\mathrm{i}=\sqrt{-1}$, $s\in (0,1)$, $\Phi =\Phi (t,x):\mathbb{R}\times {\mathbb{R}}^N\to \mathbb{C}$ is the complex-valued wave function and ${(-\Delta )}^s$ stands for the fractional Laplacian operator. A solution of problem (1) is called a standing wave solution if it has the form $\Phi (t,x)=e^{-i\lambda t}u\left(x\right)$. Indeed, u is a time-independent and real-valued function that satisfies the following fractional Schrödinger equation:

$${(-\Delta )}^{s}u+V\left(x\right)u=\lambda u+ulog{u}^2,\mathrm{in}{\mathbb{R}}^N,$$

(2)

where $s\in (0,1)$ and ${(-\Delta )}^s$ is defined by

$${(-\Delta )}^{s}u\left(x\right)=C(N,s)\mathrm{P}.\mathrm{V}.{\int }_{{\mathbb{R}}^N}\frac{u\left(x\right)-u\left(y\right)}{{|x-y|}^{N+2s}}dy,$$

where $C(N,s)$ is a positive real constant and $\mathrm{P}.\mathrm{V}.$ stands for the principle value (refer to [2] and the references therein). ${(-\Delta )}^s$ originates from the characterization of diverse phenomena within the realm of applied science, for example, barrier problems, phase transition phenomena, fractional quantum mechanics and Markov processes (see [3,4,5]). In recent years, an increasing number of researchers have directed their attention toward the fractional Schrödinger equation; see [6,7,8,9] and the references therein.

The solution of problem (2) can be studied from two aspects. On the one hand, one can choose the fixed frequency $\lambda \in \mathbb{R}$ and investigate the existence of nontrivial solutions of problem (2), which are obtained as critical points of the energy functional $J_{\lambda }:H^s\left({\mathbb{R}}^N\right)\to \mathbb{R}$

$$J_{\lambda }\left(u\right)=\frac{1}{2}{\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{\frac{s}{2}}u\right|}^2+(1+V\left(x\right)-\lambda )u^2-u^{2}log{u}^{2}dx.$$

Notice that $J_{\lambda }$ is not ${\mathcal{C}}^1$-smooth. Indeed, given the fractional logarithmic Sobolev inequality (see [10])

$${\int }_{{\mathbb{R}}^N}{u}^{2}log{u}^{2}dx\le \frac{c^2}{{\pi }^s}{\int }_{{\mathbb{R}}^N}{|(-\Delta )}^{\frac{s}{2}}u|dx+\left({log\parallel u\parallel }_2^2-N(1+\frac{1}{s}logc)+log\frac{s\Gamma \left(\frac{N}{2}\right)}{\Gamma \left(\frac{N}{2s}\right)}\right){\parallel u\parallel }_2^2,$$

for $c>0$ and $u\in H^s\left({\mathbb{R}}^N\right)$, it is obvious that ${\int }_{{\mathbb{R}}^N}{u}^{2}log{u}^{2}dx<+\infty$ for any $u\in H^s\left({\mathbb{R}}^N\right)$. However, there exists $u\in H^s\left({\mathbb{R}}^N\right)$ such that ${\int }_{{\mathbb{R}}^N}{u}^{2}log{u}^{2}dx=-\infty$. Thus, $J_{\lambda }$ loses ${\mathcal{C}}^1$-smoothness on $H^s\left({\mathbb{R}}^N\right)$.

To overcome this difficulty, we note that in [11], the author considered problem (2), with $s=1$ and $V\equiv 0$, in the following Banach space:

$$\widehat{W}:=\left\{u\in H^1\left({\mathbb{R}}^N\right):{\int }_{{\mathbb{R}}^N}{u}^2|log{u}^2|dx<\infty \right\},$$

and equips the norm

$${\parallel u\parallel }_{\widehat{W}}={\parallel u\parallel }_{H^1\left({\mathbb{R}}^N\right)}+inf\left\{k>0:{\int }_{{\mathbb{R}}^N}A(k^{-1}\left|u\right|)dx\le 1\right\},$$

where

$$A\left(t\right):=\left\{\begin{array}{cc}-t^{2}log{t}^2,\hfill & \mathrm{if}0\le t\le e^{-3},\hfill \\ 3t^2+4e^{-3}t-e^{-6},\hfill & \mathrm{if}{e}^{-3}\le t.\hfill \end{array}\right.$$

(3)

By Proposition 2.7 in [11], $J_{\lambda }:\widehat{W}\to \mathbb{R}$ is well defined and ${\mathcal{C}}^1$-smooth. For problem (2) with $s=1$, many authors adopt different methods to overcome the difficulty of $J_{\lambda }$ losses being ${\mathcal{C}}^1$-smooth. In addition, for the potential V satisfying different assumptions, many authors have studied the existence, concentration and multiplicity of solutions (see [6,7,12,13,14,15,16,17,18] and the references therein).

On the other hand, when the frequency $\lambda \in \mathbb{R}$ is not prescribed, we consider the prescribed $L^2$-norm solution and $\lambda$ appears as Lagrangian multipliers. For these equations, finding solutions with a prescribed $L^2$-norm is particularly relevant since this quantity is preserved along the time evolution. This approach seems to be particularly meaningful from the physical point of view, and often offers a good insight into the dynamical properties of the stationary solutions for these equations. Therefore, in this paper we consider the existence of normalized solutions for these equations.

We note that Jeanjean [19] considered problem (2) with $s=1$, $V\equiv 0$ and replaced the term $ulog{u}^2$ with g, that is,

$$-\Delta u=\lambda u+g\left(u\right),\mathrm{in}{\mathbb{R}}^N,$$

under the constraint

$$\overline{S}\left(a\right):=\left\{u\in H^1\left({\mathbb{R}}^N\right):{\int }_{{\mathbb{R}}^N}{\left|u\right|}^{2}dx=a\right\}.$$

The author uses mountain pass geometry to treat the $L^2$-supercritical condition, and the existence of a normalized solution is given. Subsequently, many scholars have paid attention to this kind of problem and studied the existence of normalized solutions of Schrödinger equations or systems (refer to [20,21,22,23,24,25]). In particular, refs. [20,21] studied the following Schrödinger problem:

$$-\Delta u+V\left(ϵx\right)u=\lambda u+g\left(u\right)\mathrm{in}{\mathbb{R}}^N,$$

under the constraint $\overline{S}\left(a\right)$, where $V\in C({\mathbb{R}}^N,\mathbb{R})$, $ϵ>0$ and $g:=ulog{u}^2$ or g satisfies $L^2$-subcritical growth. By using a series of techniques, they obtained the multiplicity of normalized solutions.

Inspired by the above paper, we seek to extend the existing results to the fractional Schrödinger equation and further explore the existence of normalized solutions of fractional Schrödinger equations with a logarithmic nonlinear term $ulog{u}^2$ and an external potential V. Precisely, we consider the following fractional Schrödinger equation:

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}{(-\Delta )}^{s}u+V\left(ϵx\right)u=\lambda u+ulog{u}^2\hfill & \mathrm{in}{\mathbb{R}}^N,\hfill \\ {\displaystyle {\int }_{{\mathbb{R}}^N}{\left|u\right|}^{2}dx=a,}\hfill \end{array}\right.\end{array}$$

(4)

where $s\in (0,1),ϵ,a>0$ and $\lambda \in \mathbb{R}$ appear as Lagrangian multipliers. Let V satisfy

$\left(\tilde{V}\right):$

$V\in C\left({\mathbb{R}}^N,\mathbb{R}\right)\cap L^{\infty }\left({\mathbb{R}}^N\right)$ and $-1\le V_0={inf}_{x\in {\mathbb{R}}^N}V\left(x\right)<{lim\; inf}_{\left|x\right|\to +\infty }V\left(x\right)=V_{\infty }$, where we assume $V\left(0\right)=V_0$.

Here, the potential V was considered in [20,21]. In particular, when $s=1$, the equation we studied returns to the problem studied in [20,21]. In a sense, our results can be seen as a generalization of [20,21]. The fractional Sobolev space $H^s\left({\mathbb{R}}^N\right)$ is defined as

$$H^s\left({\mathbb{R}}^N\right)=\{u\in L^2\left({\mathbb{R}}^N\right):{(-\Delta )}^{\frac{s}{2}}u\in L^2\left({\mathbb{R}}^N\right)\}$$

for any $s\in (0,1)$ and the norm is given by

$${\parallel u\parallel }_{H^s\left({\mathbb{R}}^N\right)}^2={\int }_{{\mathbb{R}}^N}{|(-\Delta )}^{\frac{s}{2}}{u|}^{2}dx+{\int }_{{\mathbb{R}}^N}{\left|u\right|}^{2}dx$$

with

$${\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{\frac{s}{2}}u\right|}^{2}dx={\displaystyle {\int \int }_{{\mathbb{R}}^{2N}}}\frac{{|u\left(x\right)-u\left(y\right)|}^2}{{|x-y|}^{N+2s}}dxdy.$$

Motivated by [11], the aim of this paper is to work on the following Banach space:

$$W=\left\{u\in H^s\left({\mathbb{R}}^N\right):{\int }_{{\mathbb{R}}^N}{u}^2|log{u}^2|dx<\infty \right\},$$

and equip the norm

$${\parallel u\parallel }_W={\parallel u\parallel }_{H^s\left({\mathbb{R}}^N\right)}+inf\left\{k>0:{\int }_{{\mathbb{R}}^N}A(k^{-1}\left|u\right|)dx\le 1\right\},$$

where A is defined as in (3). By variational methods, solutions to problem (4) can be obtained by finding critical points of the energy functional $I_{ϵ}\in {\mathcal{C}}^1(W,\mathbb{R})$, where

$$I_{ϵ}\left(u\right)=\frac{1}{2}{\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{\frac{s}{2}}u\right|}^{2}dx+\frac{1}{2}{\int }_{{\mathbb{R}}^N}(V\left(ϵx\right)+1)u^{2}dx-\frac{1}{2}{\int }_{{\mathbb{R}}^N}{u}^{2}log{u}^{2}dx,$$

(5)

under the constraint

$$S\left(a\right):=\left\{u\in W:{\int }_{{\mathbb{R}}^N}{\left|u\right|}^{2}dx=a\right\}.$$

(6)

Now, we give the Lusternik–Schnirelmann category for G and $G_{\delta }$, where

$$G=\left\{x:V_0=V\left(x\right)\mathrm{for}x\in {\mathbb{R}}^N\right\},G_{\delta }=\left\{x:|y-x|\le \delta ,\forall y\in G\right\}.$$

As in [26], for a subset Y of a topological space Z, the Lusternik–Schnirelmann category ${\mathrm{cat}}_Z\left(Y\right)$ is the lowest integer n such that there exists a covering of Y by n closed sets contractible in Z. In particular, if $Y=Z$, we use the notation $\mathrm{cat}\left(Z\right)$.

Theorem 1.

Suppose $\left(\tilde{V}\right)$ holds. Then there exists $a^{*}>0$ and ${ϵ}_1>0$ such that problem (4) admits at least ${\mathrm{cat}}_{G_{\delta }}\left(G\right)$ couples $\left(u_k,{\lambda }_k\right)\in W\times \mathbb{R}$ of weak solutions when $a>a^{*}$ and $0<ϵ<{ϵ}_1$, where ${\int }_{{\mathbb{R}}^N}{\left|u_k\right|}^{2}dx=a$, ${\lambda }_k\in \mathbb{R}$ and $I_{ϵ}\left(u_k\right)<0$ for $1\le k\le {\mathrm{cat}}_{G_{\delta }}\left(G\right)$. In addition, let $u_{ϵ}$ be the solution of problem (4) and ${\xi }_{ϵ}$ be the global maximum of $\left|u_{ϵ}\right|$. Then,

$$\underset{ϵ\to 0}{lim}V\left(ϵ{\xi }_{ϵ}\right)=V_0.$$

To prove Theorem 1, we also need to consider the following problem:

$$\left\{\begin{array}{cc}{(-\Delta )}^{s}u+\theta u=\lambda u+ulog{u}^2\hfill & \mathrm{in}{\mathbb{R}}^N,\hfill \\ {\displaystyle {\int }_{{\mathbb{R}}^N}{\left|u\right|}^{2}dx=a,}\hfill \end{array}\right.$$

(7)

where $s\in (0,1)$, $\theta \ge -1$, $a>0$ and $\lambda \in \mathbb{R}$. Naturally, the solutions of problem (7) correspond to critical points of the following functional:

$$I_{\theta }\left(u\right)=\frac{1}{2}{\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{\frac{s}{2}}u\right|}^{2}dx+\frac{1}{2}{\int }_{{\mathbb{R}}^N}(\theta +1)u^{2}dx-\frac{1}{2}{\int }_{{\mathbb{R}}^N}{u}^{2}log{u}^{2}dx,$$

under the constraint $S\left(a\right)$. We consider the global minimum

$$E_{\theta ,a}:=\underset{u\in S\left(a\right)}{inf}{I}_{\theta }\left(u\right).$$

We arrive at an additional conclusion as follows, which plays an important role in the proof of Theorem 1.

Theorem 2.

For any ${\theta }^{*}>-1$, there is $a^{*}\left({\theta }^{*}\right)>0$ such that $E_{\theta ,a}$ is achieved when $\theta \in [-1,{\theta }^{*}]$ and $a\ge a^{*}$. That is, problem (7) has a couple of weak solutions $(u,\lambda )$ if $\theta \in [-1,{\theta }^{*}]$ and $a\ge a^{*}$. Moreover, $u>0$ is radial, $\lambda <0$ and $a\mapsto E_{\theta ,a}$ is continuous. 

This paper is organized as follows: In Section 2, we present some definitions and subsequent requirements results. In Section 3, the proof of Theorem 2 is given. In Section 4, we consider the non-autonomous problem and give the proof of Theorem 1.

In the following, we give some notation. $L^p\left({\mathbb{R}}^N\right)$ is the usual Lebesgue space endowed with the norm ${\parallel u\parallel }_p={\left({\int }_{{\mathbb{R}}^N}{\left|u\right|}^{p}dx\right)}^{1/p}\mathrm{for}\mathrm{any}1\le p<+{\infty \mathrm{and}\parallel u\parallel }_{\infty }=\underset{x\in {\mathbb{R}}^N}{\mathrm{ess}\sup }\left|u\left(x\right)\right|.$ For $R>0$ and $x\in {\mathbb{R}}^N$, we use $B_r\left(y\right)$ to represent the opening ball of the center y and radius R in ${\mathbb{R}}^N$. We denote the Sobolev critical exponent as $2^{*}$ ( $2^{*}=+\infty$ if $N=1,2$ and $2^{*}=\frac{2N}{N-2s}$ if $N\ge 3$). For convenience, $C_1,C_2,\cdots$ denote various positive constants and represent different positive constants on different lines.

2. Preliminaries

Similar to that in [11], the function A is defined as in (3) and

$$B\left(t\right):=t^{2}log{t}^2+A\left(t\right).$$

Moreover, from Lemma 1.2 in [11], $A\in {\mathcal{C}}^1(\mathbb{R},\mathbb{R})$ and $A\ge 0$ is a convex function. For each $q\in \left(2,2+\frac{4s}{N}\right)$, there exists $C_q>0$ such that

$$\left|B\left(t\right)\right|\le C_q{t}^q\mathrm{for}\mathrm{all}t>0.$$

(8)

Set

$$X:=\left\{u\in L_{loc}^1\left({\mathbb{R}}^N\right):A\left(\right|u\left|\right)\in L^1\left({\mathbb{R}}^N\right)\right\},$$

where X is equipped with the norm ${\parallel ·\parallel }_X$ defined by

$${\parallel u\parallel }_X:=inf\left\{k>0:{\int }_{{\mathbb{R}}^N}A(k^{-1}\left|u\right|)dx\le 1\right\}.$$

Moreover, X is the Orlicz space associated with A and is a reflexive Banach space. We have the following result from Lemma 2.1 in [11].

Lemma 1.

For any $u\in X$, then $inf\left\{{\parallel u\parallel }_X,{\parallel u\parallel }_X^2\right\}\le {\int }_{{\mathbb{R}}^N}A\left(\right|u\left|\right)dx\le sup\left\{{\parallel u\parallel }_X,{\parallel u\parallel }_X^2\right\}.$ When ${\int }_{{\mathbb{R}}^N}A\left(\right|u_n\left|\right)dx\to {\int }_{{\mathbb{R}}^N}A\left(\right|u\left|\right)dx<\infty$ and $u_n\to u$ a.e. in ${\mathbb{R}}^N$, then $\parallel u_n{-u\parallel }_X\to 0$ as $n\to \infty$.

From Lemma 3.1 in [18], we have the following Brézis–Lieb-type lemma.

Lemma 2.

If $\left\{u_n\right\}$ is a bounded sequence in W, such that $\{u_n^{2}log{u}_n^2\}$ is a bounded sequence in $L^1\left({\mathbb{R}}^N\right)$ and $u_n\to u$ a.e. in ${\mathbb{R}}^N$, then

$${\int }_{{\mathbb{R}}^N}{u}_n^{2}log{u}_n^{2}dx={\int }_{{\mathbb{R}}^N}{\left|u_n-u\right|}^{2}log{\left|u_n-u\right|}^{2}dx+{\int }_{{\mathbb{R}}^N}{u}^{2}log{u}^{2}dx+o_n\left(1\right).$$

Next, we introduce the fractional Gagliardo–Nirenberg inequality [27]. If $p\in (2,\frac{2N}{N-2s})$,

$${\int }_{{\mathbb{R}}^N}{\left|u\right|}^{p}dx\le C(s,N,p){\left({\int }_{{\mathbb{R}}^N}{\left|u\right|}^{2}dx\right)}^{\frac{p}{2}-\frac{N(p-2)}{4s}}{\left({\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{\frac{s}{2}}u\right|}^{2}dx\right)}^{\frac{N(p-2)}{4s}},$$

(9)

where $C(s,N,p)>0$ is the best constant in the fractional Gagliardo–Nirenberg inequality.

3. Proof of Theorem 2

In this section, our aim is to prove Theorem 2.

Lemma 3.

Energy functional $I_{\theta }$ is coercive and bounded from below on $S\left(a\right)$.

Proof. 

Inequalities (8) and (9) give that

$$\begin{array}{cc}\hfill I_{\theta }\left(u\right)& =\frac{1}{2}{\int }_{{\mathbb{R}}^N}{\left(\right|(-\Delta )}^{\frac{s}{2}}u{|}^2+(\theta +1)u^2)dx-\frac{1}{2}{\int }_{{\mathbb{R}}^N}{u}^{2}log{u}^{2}dx\hfill \\ \hfill & =\frac{1}{2}{\int }_{{\mathbb{R}}^N}{\left(\right|(-\Delta )}^{\frac{s}{2}}{u|}^2+(\theta +1)u^2)dx+\frac{1}{2}{\int }_{{\mathbb{R}}^N}A\left(\right|u\left|\right)-B\left(\right|u\left|\right)dx\hfill \\ \hfill & \ge \frac{1}{2}{\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{\frac{s}{2}}u\right|}^{2}dx-C(N,s,a,q){\left({\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{\frac{s}{2}}u\right|}^{2}dx\right)}^{\frac{N(q-2)}{4s}}+\frac{1}{2}{\int }_{{\mathbb{R}}^N}A\left(\right|u\left|\right)dx.\hfill \end{array}$$

Since $q\in (2,2+\frac{4s}{N})$, we infer that $0<\frac{N(q-2)}{4s}<1$. We then conclude that $I_{\theta }$ is coercive and bounded from below on $S\left(a\right)$. □

From Lemma 3, we know that

$$E_{\theta ,a}=\underset{u\in S\left(a\right)}{inf}{I}_{\theta }\left(u\right)$$

is well defined.

Lemma 4.

For a fixed ${\theta }^{*}>-1$, then there exists $a^{*}:=a\left({\theta }^{*}\right)>0$ such that $E_{\theta ,a}<0$ for any $\theta \in [-1,{\theta }^{*}]$ and $a\ge a^{*}$.

Proof. 

Taking $\phi \in C_0^{\infty }\left({\mathbb{R}}^N\right)\setminus \left\{0\right\}$ and $r>0$, one has

$$\begin{array}{cc}\hfill I_{\theta }\left(r\phi \right)=& \frac{1}{2}{\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{\frac{s}{2}}\left(r\phi \right)\right|}^{2}dx+\frac{1}{2}{\int }_{{\mathbb{R}}^N}(\theta +1){\left(r\phi \right)}^{2}dx-{\int }_{{\mathbb{R}}^N}{\left(r\phi \right)}^{2}log{\left(r\phi \right)}^{2}dx\hfill \\ \hfill \le & \frac{r^2}{2}{\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{\frac{s}{2}}\phi \right|}^{2}dx+\frac{r^2}{2}{\int }_{{\mathbb{R}}^N}({\theta }^{*}+1){\phi }^{2}dx\hfill \\ \hfill & -\frac{r^2}{2}{\int }_{{\mathbb{R}}^N}{\phi }^{2}log{\phi }^{2}dx-r^{2}logr{\int }_{{\mathbb{R}}^N}{\phi }^{2}dx\to -\infty ,\hfill \end{array}$$

as $r\to +\infty$. Then, there exists $r_{*}=r\left({\theta }^{*}\right)>0$ satisfying

$$I_{\theta }\left(r\phi \right)<0,\mathrm{for}\mathrm{all}r\ge r_{*}.$$

Set $a^{*}=r_{*}^2{\parallel \phi \parallel }_2^2$. Hence, we conclude that

$$E_{\theta ,a}\le I_{\theta }\left(r_a\phi \right)<0,$$

for any $\theta \in [-1,{\theta }^{*}]$ and $a\ge a^{*}$. □

Lemma 5.

Let $0<a<b$. Then one has $\frac{a}{b}{E}_{\theta ,b}<E_{\theta ,a}$. In particular, for each $\theta \in [-1,{\theta }^{*}]$, if $0<a^{*}\le a<b$ we have $\frac{a}{b}{E}_{\theta ,b}<E_{\theta ,a}<0$.

Proof. 

Let $\left\{v_n\right\}\subset S\left(a\right)$ be a minimizing sequence with respect to $E_{\theta ,a}$, that is,

$$I_{\theta }\left(v_n\right)\to E_{\theta ,a},\mathrm{as}n\to +\infty .$$

Since $I_{\theta }\left(\right|u\left|\right)=I_{\theta }\left(u\right)$ for any $u\in W$, we can assume that $v_n\ge 0$ for all $n\in \mathbb{N}$. Setting $\xi =\sqrt{\frac{b}{a}}>1$, obviously, $u_n=\xi v_n\in S\left(b\right)$. Then,

$$E_{\theta ,b}\le I_{\theta }\left(u_n\right)={\xi }^2{I}_{\theta }\left(v_n\right)-\frac{1}{2}{\xi }^{2}log{\xi }^2{\int }_{{\mathbb{R}}^N}{\left|v_n\right|}^{2}dx={\xi }^2{I}_{\theta }\left(v_n\right)-\frac{1}{2}a{\xi }^{2}log{\xi }^2.$$

Letting $n\to +\infty$, it follows from $\xi >1$ that

$$E_{\theta ,b}\le {\xi }^2{E}_{\theta ,a}-\frac{1}{2}a{\xi }^{2}log{\xi }^2<{\xi }^2{E}_{\theta ,a},$$

that is,

$$\frac{a}{b}{E}_{\theta ,b}<E_{\theta ,a}.$$

Furthermore, for any fixed $\theta \in [-1,{\theta }^{*}]$ and $0<a^{*}\le a<b$, by Lemma 4, we can prove the conclusion. □

Corollary 1.

For a fixed ${\theta }^{*}>-1$, let $a>a^{*}>0$ and $-1\le {\theta }_1<{\theta }_2\le {\theta }^{*}$. If $E_{{\theta }_2,a}<0$ is achieved, then $E_{{\theta }_1,a}<E_{{\theta }_2,a}<0$.

Proof. 

Let $u_{{\theta }_2}\in S\left(a\right)$ satisfying $I_{{\theta }_2}\left(u_{{\theta }_2}\right)=E_{{\theta }_2,a}$. By Lemma 4,

$$\begin{array}{cc}\hfill E_{{\theta }_1,a}\le I_{{\theta }_1}\left(u_{{\theta }_2}\right)=& \frac{1}{2}{\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{\frac{s}{2}}{u}_{{\theta }_2}\right|}^{2}dx+\frac{({\theta }_1+1)}{2}{\int }_{{\mathbb{R}}^N}{u}_{{\theta }_2}^{2}dx-\frac{1}{2}{\int }_{{\mathbb{R}}^N}{u}_{{\theta }_2}^{2}log{u}_{{\theta }_2}^{2}dx\hfill \\ \hfill <& I_{{\theta }_2}\left(u_{{\theta }_2}\right)=E_{{\theta }_2,a}<0,\hfill \end{array}$$

and we complete the proof. □

Lemma 6.

For any $a>0$, $a\mapsto E_{\theta ,a}$ is continuous. Moreover, $a\mapsto E_{\theta ,a}$ is nonincreasing for each $\theta \in [-1,{\theta }^{*}]$ and $a\ge a^{*}$.

Proof. 

This is the discussion that is divided into two steps to prove $a\mapsto E_{\theta ,a}$ is continuous.

Step 1: $\underset{\delta \to 0^{+}}{lim}{E}_{\theta ,a-\delta }=E_{\theta ,a}$.

For each $\epsilon >0$, there exists $u\in S\left(a\right)$ satisfying

$$E_{\theta ,a}\le I_{\theta }\left(u\right)\le E_{\theta ,a}+\epsilon .$$

(10)

Set

$$r:=r\left(\delta \right)={\left(\frac{a-\delta }{a}\right)}^{\frac{1}{N}},$$

and $u^r\left(x\right)=u(x/r)$. Then,

$$\underset{\delta \to 0^{+}}{lim}r\left(\delta \right)=1and{\parallel u^r\parallel }_2^2=r^{N}a=a-\delta .$$

(11)

Moreover,

$$\begin{array}{cc}\hfill I_{\theta }\left(u^r\right)=& \frac{r^{N-2s}}{2}{\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{\frac{s}{2}}u\right|}^{2}dx+\frac{r^N(\theta +1)}{2}{\int }_{{\mathbb{R}}^N}{u}^{2}dx-\frac{r^N}{2}{\int }_{{\mathbb{R}}^N}{u}^{2}log{u}^{2}dx\hfill \\ \hfill =& \frac{r^{N-2s}}{2}{\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{\frac{s}{2}}u\right|}^{2}dx+r^N\left(I_{\theta }\left(u\right)-\frac{1}{2}{\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{\frac{s}{2}}u\right|}^{2}dx\right)\hfill \\ \hfill =& r^N{I}_{\theta }\left(u\right)+\frac{r^{N-2s}(1-r^{2s})}{2}{\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{\frac{s}{2}}u\right|}^{2}dx.\hfill \end{array}$$

By (10), (11) and Lemma 5, we have

$$\underset{\delta \to 0^{+}}{lim}{E}_{\theta ,a-\delta }\ge \underset{\delta \to 0^{+}}{lim}\frac{a-\delta }{a}{E}_{\theta ,a}=E_{\theta ,a},$$

and

$$\underset{\delta \to 0^{+}}{lim}{E}_{\theta ,a-\delta }\le \underset{\delta \to 0^{+}}{lim}{I}_{\theta }\left(u^r\right)=\underset{\delta \to 0^{+}}{lim}{r}^N{I}_{\theta }\left(u\right)+\frac{r^{N-2s}(1-r^{2s})}{2}{\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{\frac{s}{2}}u\right|}^{2}dx\le E_{\theta ,a}+\epsilon .$$

Then, Step 1 holds.

Step 2: $\underset{\delta \to 0^{+}}{lim}{E}_{\theta ,a+\delta }=E_{\theta ,a}$.

Take $\delta =\frac{1}{n}\mathrm{for}n\in N$ and $u_n\in S(a+\frac{1}{n})$ such that $I_{\theta }\left(u_n\right)\le E_{\theta ,a+\frac{1}{n}}+\frac{1}{n}$. From Lemmas 1 and 3, $\left\{u_n\right\}$ is bounded in W. Set

$$v_n\left(x\right):=\sqrt{\frac{na}{na+1}}{u}_n\left(x\right).$$

Then, we have

$$\parallel v_n{\parallel }_2^2=\frac{na}{na+1}{\parallel u_n\parallel }_2^2=\frac{na}{na+1}\left(a+\frac{1}{n}\right)=a,$$

and

$$\left|\right|v_n-u_n{\left|\right|}_W=\left(1-\sqrt{\frac{na}{na+1}}\right)\left|\right|u_n{\left|\right|}_W\to 0\mathrm{as}n\to +\infty .$$

Thus,

$$E_{\theta ,a}\le I_{\theta }\left(v_n\right)=I_{\theta }\left(u_n\right)+o_n\left(1\right)\le E_{\theta ,a+\delta }+o_n\left(1\right),$$

that is,

$$\underset{\delta \to 0^{+}}{lim}{E}_{\theta ,a+\delta }\ge E_{\theta ,a}.$$

Moreover, it follows from Lemma 5 that

$$\underset{\delta \to 0^{+}}{lim}{E}_{\theta ,a+\delta }\le \underset{\delta \to 0^{+}}{lim}\frac{a+\delta }{a}{E}_{\theta ,a}=E_{\theta ,a}.$$

Thus, Step 2 holds.

By Step 1 and Step 2, we complete the proof that $a\mapsto E_{\theta ,a}$ is continuous. Next, by Lemmas 4 and 5, it is clear that the map $a\mapsto E_{\theta ,a}$ is nonincreasing for $\theta \in [-1,{\theta }^{*}]$ and $a\ge a^{*}>0$. □

Lemma 7.

For a fixed ${\theta }^{*}>-1$, let $a>a^{*}$, $\theta \in (-1,{\theta }^{*})$ and $\left\{u_n\right\}\subset S\left(a\right)$ be a minimizing sequence of $E_{\theta ,a}$. Then, one of the following conclusions is true:

(1)

There exists $u\in S\left(a\right)$ such that $u_n\to u$ in W;

(2)

There are $\left\{y_n\right\}\subset {\mathbb{R}}^N$ and $|y_n|\to \infty$ such that ${\tilde{u}}_n\left(x\right):=u_n(x+y_n)\to \tilde{u}\in S\left(a\right)$. Moreover, $I_{\theta }\left(\tilde{u}\right)=E_{\theta ,a}$.

Proof. 

According to Lemmas 1 and 3, $\left\{u_n\right\}$ is bounded in W. Then, $u_n\rightharpoonup u$ in W up to a subsequence. Next, we will discuss it in three cases.

Case 1: ${\parallel u\parallel }_2^2=a$.

It is clear that

$${\int }_{{\mathbb{R}}^N}|u_n{|}^{2}dx\to {\int }_{{\mathbb{R}}^N}{\left|u\right|}^{2}dx.$$

Interpolation theorem and (8) lead to

$${\int }_{{\mathbb{R}}^N}B\left(u_n\right)dx\to {\int }_{{\mathbb{R}}^N}B\left(u\right)dx.$$

Since $A\ge 0$ and ${\displaystyle E_{\theta ,a}=\underset{n\to +\infty }{lim}{I}_{\theta }\left(u_n\right)}$, we have

$$\begin{array}{cc}\hfill E_{\theta ,a}=& \underset{n\to +\infty }{lim}\frac{1}{2}{\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{\frac{s}{2}}{u}_n\right|}^2+(\theta +1)u_n^{2}dx+\frac{1}{2}{\int }_{{\mathbb{R}}^N}A\left(u_n\right)dx-\frac{1}{2}{\int }_{{\mathbb{R}}^N}B\left(u_n\right)dx\hfill \\ \hfill \ge & \frac{1}{2}{\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{\frac{s}{2}}u\right|}^2+(\theta +1)u^{2}dx+\frac{1}{2}{\int }_{{\mathbb{R}}^N}A\left(u\right)dx-\frac{1}{2}{\int }_{{\mathbb{R}}^N}B\left(u\right)dx\hfill \\ \hfill =& I_{\theta }\left(u\right).\hfill \end{array}$$

In fact, $u\in S\left(a\right)$. So, $E_{\theta ,a}=I_{\theta }\left(u\right)$, and then

$$A\left(u_n\right)\to A\left(u\right)\mathrm{in}{L}^1\left({\mathbb{R}}^N\right)$$

and

$${\int }_{{\mathbb{R}}^N}{|(-\Delta )}^{\frac{s}{2}}{u}_n{|}^{2}dx\to {\int }_{{\mathbb{R}}^N}{|{(-\Delta )}^{\frac{s}{2}}u|}^{2}dx,$$

which implies that $u_n\to u$ in W.

Case 2: ${\parallel u\parallel }_2^2=b\in (0,a)$.

Let $v_n:=u_n-u$. From [7,26] one has

$$\begin{array}{cc}\hfill {\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{\frac{s}{2}}{u}_n\right|}^{2}dx=& {\int }_{{\mathbb{R}}^N}{|(-\Delta )}^{\frac{s}{2}}{u|}^{2}dx+{\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{\frac{s}{2}}{v}_n\right|}^{2}dx+o_n\left(1\right).\hfill \end{array}$$

Furthermore, by Lemma 2,

$${\int }_{{\mathbb{R}}^N}{u}_n^{2}log{u}_n^{2}dx={\int }_{{\mathbb{R}}^N}|v_n{|}^{2}log{|v_n|}^{2}dx+{\int }_{{\mathbb{R}}^N}{u}^{2}log{u}^{2}dx+o_n\left(1\right).$$

Set $d_n:={\parallel v_n\parallel }_2^2$. Then by using

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

we have $\parallel v_n{\parallel }_2^2\to d\in (0,a)$. Thus,

$$\begin{array}{cc}\hfill E_{\theta ,a}=& I_{\theta }\left(u_n\right)+o_n\left(1\right)+o_n\left(1\right)\hfill \\ \hfill =& \frac{1}{2}{\int \int }_{{\mathbb{R}}^{2N}}\frac{|v_n\left(x\right)-v_n{\left(y\right)|}^2}{{|x-y|}^{N+2s}}dxdy+\frac{\theta +1}{2}{\parallel v_n\parallel }_2^2-\frac{1}{2}{\int }_{{\mathbb{R}}^N}{v}_n^{2}log{v}_n^{2}dx\hfill \\ \hfill & +\frac{1}{2}{\int \int }_{{\mathbb{R}}^{2N}}\frac{{|u\left(x\right)-u\left(y\right)|}^2}{{|x-y|}^{N+2s}}dxdy+\frac{\theta +1}{2}{\parallel u\parallel }_2^2-\frac{1}{2}{\int }_{{\mathbb{R}}^N}{u}^{2}log{u}^{2}dx+o_n\left(1\right)\hfill \\ \hfill \ge & E_{\theta ,b}+E_{\theta ,d_n}+o_n\left(1\right).\hfill \end{array}$$

Letting $n\to +\infty$, from Lemmas 5 and 6, one has

$$\begin{array}{cc}\hfill E_{\theta ,a}\ge & \frac{d}{a}{E}_{\theta ,a}+E_{\theta ,b}.\hfill \end{array}$$

Since $a=b+d$, again by using Lemma 5, we obtain

$$E_{\theta ,a}>\frac{b}{a}{E}_{\theta ,a}+\frac{d}{a}{E}_{\theta ,a}=\left(\frac{b}{a}+\frac{d}{a}\right)E_{\theta ,a}=E_{\theta ,a},$$

which is a contradiction and implies that Case 2 is impossible.

Case 3: ${\parallel u\parallel }_2^2=0$.

This case implies that $u_n\rightharpoonup 0$ in W. Next, we show

$$\delta :=\underset{n\to +\infty }{lim\; inf}\underset{y_n\in {\mathbb{R}}^N}{sup}{\int }_{B_R\left(y_n\right)}{\left|u_n\right|}^{2}dx>0.$$

(12)

If not, one has $u_n\to 0$ in $L^l\left({\mathbb{R}}^N\right)$ for any $l\in (2,2^{*})$ (see [28]). Then, (8) leads to

$${\int }_{{\mathbb{R}}^N}B\left(u_n\right)dx\to 0.$$

Moreover,

$$\begin{array}{cc}\hfill 0>E_{\theta ,a}+o_n\left(1\right)=I_{\theta }\left(u_n\right)& =\frac{1}{2}{\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{\frac{s}{2}}{u}_n\right|}^{2}dx+\frac{\theta +1}{2}{\int }_{{\mathbb{R}}^N}{u}_n^{2}dx-\frac{1}{2}{\int }_{{\mathbb{R}}^N}{u}_n^{2}log{u}_n^{2}dx\hfill \\ \hfill & \ge -\frac{1}{2}{\int }_{{\mathbb{R}}^N}B\left(u\right)dx\to 0.\hfill \end{array}$$

We arrive at a contradiction. Hence, (12) holds. It implies that there are $\left\{y_n\right\}$ such that

$${\int }_{B_R\left(y_n\right)}{\left|u_n\right|}^{2}dx\ge \frac{\delta }{2}.$$

Since ${\parallel u\parallel }_2^2=0$, the Sobolev embedding and (12) imply that $|y_n|\to +\infty$. Set ${\tilde{u}}_n\left(x\right):=u(x+y_n)$, which is also a bounded minimizing sequence of $E_{\theta ,a}$. Then, there exists $\tilde{u}\in W\setminus \left\{0\right\}$ such that ${\tilde{u}}_n\left(x\right)\rightharpoonup \tilde{u}$ in W. Based on Case 1 and Case 2, we have ${\tilde{u}}_n\left(x\right)\to \tilde{u}$ in W and $I_{\theta }\left(\tilde{u}\right)=E_{\theta ,a}$. □

Proof of Theorem 2.

It is easy to know that $a\mapsto E_{\theta ,a}$ is continuous by Lemma 6.

By Lemmas 1, 3 and 4, there exists a bounded minimizing sequence $\left\{u_n\right\}\subset S\left(a\right)$ such that $I_{\theta }\left(u_n\right)\to E_{\theta ,a}<0$. From Lemma 7, there exists $u\in S\left(a\right)$ such that $I_{\theta }\left(u\right)=E_{\theta ,a}$. In addition, according to the Lagrange multiplier,

$$I_{\theta }^{\prime }\left(u\right)=\lambda {\Psi }^{\prime }\left(u\right)\mathrm{in}{W}^{\prime }\mathrm{for}\lambda \in \mathbb{R},$$

(13)

where $\Psi :W\to \mathbb{R}$ is defined as

$$\frac{1}{2}\Psi \left(u\right)={\parallel u\parallel }_2^2,u\in W.$$

From (13), we have

$${(-\Delta )}^{s}u+\theta u=\lambda u+ulog{u}^2,\mathrm{in}{\mathbb{R}}^N.$$

Since $I_{\theta }\left(u\right)=E_{\theta ,a}<0$, we obtain

$$\begin{array}{cc}\hfill \lambda {\int }_{{\mathbb{R}}^N}{u}^{2}dx=& {\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{\frac{s}{2}}u\right|}^2+\theta u^{2}dx-{\int }_{{\mathbb{R}}^N}{u}^{2}log{u}^{2}dx\hfill \\ \hfill \le & {\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{\frac{s}{2}}u\right|}^2+(\theta +1)u^2-u^{2}log{u}^{2}dx\hfill \\ \hfill =& 2I_{\theta }\left(u\right)=2E_{\theta ,a}<0,\hfill \end{array}$$

which shows that $\lambda <0$. Since

$$E_{\theta ,a}\le I_{\theta }\left(\right|u\left|\right)=I_{\theta }\left(u\right)=E_{\theta ,a},$$

we can assume $u\ge 0$. Moreover, let $u^{*}$ represent the Schwarz rearrangement of u. Then, from [29] one has

$$\begin{array}{c}\hfill {\int }_{{\mathbb{R}}^N}{|(-\Delta )}^{\frac{s}{2}}{u}^{*}{|}^{2}dx\le {\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{\frac{s}{2}}u\right|}^{2}dx\end{array}$$

and

$${\int }_{{\mathbb{R}}^N}{\left|u\right|}^{2}dx={\int }_{{\mathbb{R}}^N}{\left|u^{*}\right|}^{2}dx.$$

From Chapter 3.3 in [30], we have

$${\int }_{{\mathbb{R}}^N}A\left(u^{*}\right)dx={\int }_{{\mathbb{R}}^N}A\left(u\right)dx\mathrm{and}{\int }_{{\mathbb{R}}^N}B\left(u^{*}\right)dx={\int }_{{\mathbb{R}}^N}B\left(u\right)dx.$$

Therefore, $u^{*}\in S\left(a\right)$ and

$$E_{\theta ,a}=I_{\theta }\left(u\right)\ge I_{\theta }\left(u^{*}\right)\ge E_{\theta ,a},$$

Then, we replace u with $u^{*}$. In addition, similar to [7], we have $u\left(x\right)>0$ for all $x\in {\mathbb{R}}^N$. □

4. Proof of Theorem 1

Define the following functionals:

$$\begin{array}{c}\hfill I_{V_0}\left(u\right)=\frac{1}{2}{\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{\frac{s}{2}}u\right|}^{2}dx+\frac{1}{2}{\int }_{{\mathbb{R}}^N}(V_0+1)u^{2}dx-\frac{1}{2}{\int }_{{\mathbb{R}}^N}{u}^{2}log{u}^{2}dx\end{array}$$

and

$$I_{\infty }\left(u\right)=\frac{1}{2}{\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{\frac{s}{2}}u\right|}^{2}dx+\frac{1}{2}{\int }_{{\mathbb{R}}^N}(V_{\infty }+1)u^{2}dx-\frac{1}{2}{\int }_{{\mathbb{R}}^N}{u}^{2}log{u}^{2}dx.$$

Moreover, define $E_{V_0,a}$, $E_{\infty ,a}$ and $E_{ϵ,a}$ as following:

$$E_{V_0,a}=\underset{u\in S\left(a\right)}{inf}{I}_{V_0}\left(u\right),E_{\infty ,a}=\underset{u\in S\left(a\right)}{inf}{I}_{\infty }\left(u\right)\mathrm{and}{E}_{ϵ,a}=\underset{u\in S\left(a\right)}{inf}{I}_{ϵ}\left(u\right),$$

where $I_{ϵ}\left(u\right)$ is given by (5). Since $-1\le V_0<V_{\infty }<+\infty$, we choose ${\displaystyle {\theta }^{*}=\underset{x\in {\mathbb{R}}^N}{sup}V\left(x\right)>-1}$ in Lemma 4, and then Corollary 1 gives that

$$E_{V_0,a}<E_{\infty ,a}<0,\mathrm{if}a>a^{*}>0.$$

(14)

Lemma 8.

Let $\left(\tilde{V}\right)$ hold and $a>a^{*}>0$. Then $\underset{ϵ\to 0^{+}}{lim\; sup}{E}_{ϵ,a}\le E_{V_0,a}<E_{\infty ,a}<0$.

Proof. 

For any $a>a^{*}>0$ and by Theorem 2, we can choose $u_0\in S\left(a\right)$ satisfying $I_{V_0}\left(u_0\right)=E_{V_0,a}$. Then,

$$\begin{array}{c}\hfill E_{ϵ,a}\le I_{ϵ}\left(u_0\right)=\frac{1}{2}{\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{\frac{s}{2}}{u}_0\right|}^{2}dx+\frac{1}{2}{\int }_{{\mathbb{R}}^N}(V\left(\epsilon x\right)+1)u_0^{2}dx-\frac{1}{2}{\int }_{{\mathbb{R}}^N}{u}_0^{2}log{u}_0^{2}dx.\end{array}$$

Taking $ϵ\to 0^{+}$, we derive

$$\underset{ϵ\to 0^{+}}{lim\; sup}{E}_{ϵ,a}\le \underset{ϵ\to 0^{+}}{lim}{I}_{ϵ}\left(u_0\right)=I_{V_0}\left(u_0\right)=E_{V_0,a}.$$

(15)

The inequality $\underset{ϵ\to 0^{+}}{lim\; sup}{E}_{ϵ,a}\le E_{V_0,a}<E_{\infty ,a}$ follows from (14) and (15). □

From Lemma 8, there exists ${ϵ}_1>0$ such that $E_{ϵ,a}<E_{\infty ,a}$ for any $ϵ\in (0,{ϵ}_1)$. Now, we set $0<{\rho }_1:=\frac{1}{2}\left(E_{\infty ,a}-E_{V_0,a}\right)$.

Lemma 9.

Let $\left(\tilde{V}\right)$ hold, $a>a^{*}>0$, $ϵ\in \left(0,{ϵ}_1\right)$ and $c<E_{V_0,a}+{\rho }_1<0$. If $\left\{u_n\right\}\subset S\left(a\right)$ such that $I_{ϵ}\left(u_n\right)\to c$ as $n\to +\infty$. Then

$$\underset{n\to \infty }{lim\; inf}\underset{y\in {\mathbb{R}}^N}{sup}{\int }_{B_1\left(y\right)}{\left|u_n\left(x\right)\right|}^{2}dx\ge \delta >0.$$

(16)

Moreover, $u_n\rightharpoonup u$ in W implies $u\not\equiv 0$.

Proof. 

If (16) is not true, one has $u_n\to 0$ in $L^p\left({\mathbb{R}}^N\right)$ for $p\in (2,2^{*})$. Thus, (8) leads to

$${\int }_{{\mathbb{R}}^N}B\left(u_n\right)dx\to 0.$$

By $\left(\tilde{V}\right)$, we have

$$\begin{array}{cc}\hfill E_{V_0,a}+{\rho }_1+o_n\left(1\right)>I_{\epsilon }\left(u_n\right)=& \frac{1}{2}{\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{\frac{s}{2}}{u}_n\right|}^2+(V\left(\epsilon x\right)+1)u_n^{2}dx-\frac{1}{2}{\int }_{{\mathbb{R}}^N}{u}_n^{2}log{u}_n^{2}dx\hfill \\ \hfill \ge & -{\int }_{{\mathbb{R}}^N}B\left(u_n\right)dx\to 0,\hfill \end{array}$$

which is a contradiction.

Assume that $u\equiv 0$. Then, there exists $\left\{y_n\right\}$ and $|y_n|\to \infty$. Set ${\tilde{u}}_n=u_n(·+y_n)$. One has

$$\begin{array}{cc}\hfill E_{V_0,a}+{\rho }_1+o_n\left(1\right)>& I_{ϵ}\left(u_n\right)\hfill \\ \hfill =& \frac{1}{2}{\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{\frac{s}{2}}{u}_n\right|}^2+(V\left(ϵx\right)+1)u_n^{2}dx-\frac{1}{2}{\int }_{{\mathbb{R}}^N}{u}_n^{2}log{u}_n^{2}dx\hfill \\ \hfill =& \frac{1}{2}{\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{\frac{s}{2}}{\tilde{u}}_n\right|}^2+(V(ϵx+ϵy)+1){\tilde{u}}_n^{2}dx-\frac{1}{2}{\int }_{{\mathbb{R}}^N}{\tilde{u}}_n^{2}log{\tilde{u}}_n^{2}dx\hfill \\ \hfill =& I_{\infty }\left({\tilde{u}}_n\right)+\frac{1}{2}{\int }_{{\mathbb{R}}^N}(V(ϵx+ϵy_n)-V_{\infty }){\tilde{u}}_n^{2}dx+o_n\left(1\right)\hfill \\ \hfill =& I_{\infty }\left({\tilde{u}}_n\right)+o_n\left(1\right)\ge E_{\infty ,a}+o_n\left(1\right).\hfill \end{array}$$

This contradicts $E_{V_0,a}+{\rho }_1<E_{\infty ,a}<0$. □

Lemma 10.

Assume that $\left(\tilde{V}\right)$ holds, $a>a^{*}>0$, $ϵ\in \left(0,{ϵ}_1\right)$ and $\left\{u_n\right\}\subset S\left(a\right)$ is a ${\left(PS\right)}_c$ sequence of $I_{ϵ}$ and $c<E_{V_0,a}+{\rho }_1<0$. If $u_n\rightharpoonup u_{ϵ}$ and $u_n\nrightarrow u_{ϵ}$ in W, then there is $\tau >0$ such that

$$\underset{n\to +\infty }{lim\; inf}{\int }_{{\mathbb{R}}^N}{|u_n-u_{ϵ}|}^{2}dx\ge \tau .$$

Proof. 

Let $\Phi :W\to \mathbb{R}$ define as

$$\Phi \left(u\right)=\frac{1}{2}{\int }_{{\mathbb{R}}^N}{\left|u\right|}^{2}dx,$$

Then, $S\left(a\right)={\Phi }^{-1}\left(\{a/2\}\right)$. From [26], there are $\left\{{\lambda }_n\right\}\subset \mathbb{R}$

$$\left|\right|I_{ϵ}^{\prime }\left(u_n\right)-{\lambda }_n{\Phi }^{\prime }\left(u_n\right){\left|\right|}_{W^{\prime }}\to 0\mathrm{as}n\to +\infty .$$

(17)

Because $\left\{u_n\right\}$ is bounded in W, (17) implies that $\left\{{\lambda }_n\right\}$ is bounded in $\mathbb{R}$. There exists ${\lambda }_{ϵ}$ satisfying ${\lambda }_n\to {\lambda }_{ϵ}$ in $\mathbb{R}$. Combined with (17), one has

$$I_{ϵ}^{\prime }\left(u_{ϵ}\right)-{\lambda }_{ϵ}{\Phi }^{\prime }\left(u_{ϵ}\right)=0\mathrm{in}{W}^{\prime }.$$

Set $v_n:=u_n-u_{ϵ}$, and it follows from Lemma 7 that

$$\left|\right|I_{ϵ}^{\prime }\left(v_n\right)-{\lambda }_n{\Phi }^{\prime }\left(v_n\right){\left|\right|}_{W^{\prime }}\to 0\mathrm{as}n\to +\infty .$$

(18)

By calculation, we obtain

$$\begin{array}{cc}\hfill E_{V_0,a}+{\rho }_1>& I_{ϵ}\left(u_n\right)+o_n\left(1\right)\hfill \\ \hfill =& I_{ϵ}\left(u_n\right)-\frac{1}{2}{I}_{ϵ}^{\prime }\left(u_n\right)u_n+\frac{1}{2}{\lambda }_{n}a+o_n\left(1\right)\hfill \\ \hfill =& \frac{1}{2}{\int }_{{\mathbb{R}}^N}{u}_n^{2}dx+\frac{1}{2}{\lambda }_{n}a+o_n\left(1\right),\hfill \end{array}$$

which implies that

$${\lambda }_{ϵ}\le \frac{2(E_{V_0,a}+{\rho }_1)}{a}<0,\mathrm{for}\mathrm{all}ϵ\in (0,{ϵ}_1).$$

(19)

From (18), we have

$$\begin{array}{c}\hfill {\int }_{{\mathbb{R}}^N}{|(-\Delta )}^{\frac{s}{2}}{v}_n{|}^2+(V\left(ϵx\right)+1-{\lambda }_{ϵ})|v_n{|}^{2}dx={\int }_{{\mathbb{R}}^N}B\left(\right|v_n\left|\right)-A\left(\right|v_n\left|\right)dx+o_n\left(1\right).\end{array}$$

(20)

Combined with (19), (20) and $A\left(s\right)\ge 0$ for $s>0$, one has

$${\int }_{{\mathbb{R}}^N}{|(-\Delta )}^{\frac{s}{2}}{v}_n{|}^2+(V\left(ϵx\right)+1)|v_n{|}^{2}dx-\frac{2(E_{V_0,a}+{\rho }_1)}{a}{\parallel v_n\parallel }_2^2\le {\int }_{{\mathbb{R}}^N}B\left(v_n\right)dx+o_n\left(1\right),$$

Then, (8) implies

$${\int }_{{\mathbb{R}}^N}{|(-\Delta )}^{\frac{s}{2}}{v}_n{|}^{2}dx-\frac{2(E_{V_0,a}+{\rho }_1)}{a}\parallel v_n{\parallel }_2^2\le C_1{\parallel v_n\parallel }_q^q+o_n\left(1\right),$$

where $C_1>0$. Since the Sobolev embedding $H^s\left({\mathbb{R}}^N\right)\hookrightarrow L^q\left({\mathbb{R}}^N\right)$ is continuous, we obtain

$${∥v_n∥}_{H^s\left({\mathbb{R}}^N\right)}^2\le C_2{\parallel v_n\parallel }_q^q+o_n\left(1\right)\le C_3{∥v_n∥}_{H^s\left({\mathbb{R}}^N\right)}^q+o_n\left(1\right)$$

(21)

where $C_2,C_3>0$ are independent of $ϵ$. We claim that

$${\displaystyle \underset{n\to +\infty }{lim\; inf}{∥v_n∥}_{H^s\left({\mathbb{R}}^N\right)}>0.}$$

If not, one has $B\left(v_n\right)\to 0$ in $L^1\left({\mathbb{R}}^N\right)$. Combined with (20) and Lemma 1, we deduce that $u_n\to u$ in W, which is a contradiction. Then, (21) implies that

$$\underset{n\to +\infty }{lim\; inf}{∥v_n∥}_{H^s\left({\mathbb{R}}^N\right)}\ge {\left(\frac{1}{C_3}\right)}^{\frac{1}{q-2}}.$$

By (21), there exists $C>0$ independent of $ϵ$ such that

$$\underset{n\to +\infty }{lim\; inf}{\parallel v_n\parallel }_q^q\ge C.$$

(22)

By (9), we have

$$\begin{array}{cc}\hfill \underset{n\to +\infty }{lim\; inf}{\int }_{{\mathbb{R}}^N}{\left|v_n\right|}^{q}dx\le & C(s,N,q){\left({\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{\frac{s}{2}}{v}_n\right|}^{2}dx\right)}^{\frac{N(q-2)}{4s}}{\left(\underset{n\to +\infty }{lim\; inf}{\int }_{{\mathbb{R}}^N}{\left|v_n\right|}^{2}dx\right)}^{\frac{q}{2}-\frac{N(q-2)}{4s}}\hfill \\ \hfill \le & C(s,N,q){\mathcal{K}}^{\frac{N(q-2)}{4s}}{\left(\underset{n\to +\infty }{lim\; inf}{\int }_{{\mathbb{R}}^N}{\left|v_n\right|}^{2}dx\right)}^{\frac{q}{2}-\frac{N(q-2)}{4s}},\hfill \end{array}$$

(23)

where $\mathcal{K}$ is a positive real number independent of $ϵ$ satisfying ${\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{\frac{s}{2}}{v}_n\right|}^{2}dx\le \mathcal{K}$. From (22) and (23), we can obtain the desired result. □

Lemma 11.

Let $\left(\tilde{V}\right)$ hold, $a>a^{*}>0$, $ϵ\in \left(0,{ϵ}_1\right)$ and $0<\rho <min\left\{\frac{1}{2},\frac{\tau }{a}\right\}\left(E_{\infty ,a}-E_{V_0,a}\right)\le {\rho }_1$. Then, $I_{ϵ}$ satisfies the ${\left(PS\right)}_c$ condition restricted to $S\left(a\right)$ with $c<E_{V_0,a}+\rho$.

Proof. 

Suppose that $\left\{u_n\right\}$ is a ${\left(PS\right)}_c$ sequence for $c<E_{V_0,a}+\rho$. Similar to Lemmas 1 and 3, we know that $\left\{u_n\right\}$ is bounded in W. Lemma 9 implies $u_n\rightharpoonup u_{ϵ}\not\equiv 0$ in W. If $u_n\to u_{ϵ}$ in W, this concludes the proof. Otherwise, by Lemma 10

$$\underset{n\to +\infty }{lim\; inf}{\parallel v_n\parallel }_2^2\ge \tau >0,$$

where $v_n:=u_n-u_{ϵ}$. Let $d_n={\parallel v_n\parallel }_2^2$, then $\parallel v_n{\parallel }_2^2\to d\ge \tau$. Using the Brezis–Lieb Lemma (see [26]) and Lemma 7, we obtain $a=b+d$ and

$$c=I_{ϵ}\left(u_n\right)+o_n\left(1\right)=I_{ϵ}\left(u_{ϵ}\right)+I_{ϵ}\left(v_n\right)+o_n\left(1\right).$$

Since $v_n\rightharpoonup 0$ in W, from Lemmas 5 and 9 we have

$$\begin{array}{c}\hfill E_{V_0,a}+\rho +o_n\left(1\right)\ge I_{ϵ}\left(u_n\right)\ge E_{\infty ,d_n}+E_{V_0,b}\ge \frac{d_n}{a}{E}_{\infty ,a}+\frac{b}{a}{E}_{V_0,a}.\end{array}$$

Letting $n\to \infty$, we obtain

$$\begin{array}{c}\hfill E_{V_0,a}+\rho \ge \frac{b}{a}{E}_{V_0,a}+\frac{d}{a}{E}_{\infty ,a}\ge E_{V_0,a}+\frac{\tau }{a}(E_{\infty ,a}-E_{V_0,a}).\end{array}$$

Since $\rho <\frac{\tau }{a}(E_{\infty ,a}-E_{V_0,a})$, we can obtain $v_n\to 0$ in W, that is, $\parallel u_{ϵ}{\parallel }_2^2=a$ and

$$-\Delta u_{ϵ}+V\left(ϵx\right)u_{ϵ}={\lambda }_{ϵ}{u}_{ϵ}+u_{ϵ}log{u}_{ϵ}^2,\mathrm{in}{\mathbb{R}}^N.$$

The lemma proof is completed. □

From Theorem 2, let w be a positive solution of problem (7) with $\theta =V_0$. Then, $I_{V_0}\left(w\right)=E_{V_0,a}$. Set $\delta >0$ as fixed and $\chi$ as a cut-off function satisfying $\eta \left(t\right)=0$ when $t\ge \delta$ and $\eta \left(t\right)=1$ when $0\le t\le \frac{\delta }{2}$. For all $y\in G$, we define

$${\Psi }_{ϵ,z}\left(y\right)=\eta \left(\right|ϵy-z\left|\right)w\left(y-\frac{z}{ϵ}\right),$$

and

$${\Phi }_{ϵ}\left(z\right)=\sqrt{a}{\displaystyle \frac{{\Psi }_{ϵ,z}\left(y\right)}{\parallel {\Psi }_{ϵ,z}{\parallel }_2}}.$$

Obviously, ${\Phi }_{ϵ}:G\mapsto S\left(a\right)$.

Lemma 12.

The function ${\Phi }_{ϵ}$ satisfying

$$\underset{ϵ\to 0}{lim}{E}_{V_0,a}-I_{ϵ}\left({\Phi }_{ϵ}\left(y\right)\right)=0,forally\in G.$$

Proof. 

If not, there exists ${ϵ}_n\to 0$, $\delta >0$ and $\left\{y_n\right\}\subset G$ with $y_n\to y$ such that

$$\left|E_{V_0,a}-I_{{ϵ}_n}\left({\Phi }_{{ϵ}_n}\left(y_n\right)\right)\right|\ge \delta ,\mathrm{for}\mathrm{all}n\in \mathbb{N}.$$

Then, we have

$$\begin{array}{c}\hfill {\int }_{{\mathbb{R}}^N}|{\Phi }_{{ϵ}_n}\left(y_n\right){|}^{2}dx={\int }_{{\mathbb{R}}^N}{\left|{\Psi }_{{ϵ}_n,y_n}\left(x\right)\right|}^{2}dx={\int }_{{\mathbb{R}}^N}{\left|\eta \left({ϵ}_{n}z\right)w\left(z\right)\right|}^{2}dz={\parallel w\parallel }_2^2+o_n\left(1\right);\end{array}$$

$${\int }_{{\mathbb{R}}^N}A\left({\Phi }_{{ϵ}_n}\left(y_n\right)\right)dx={\int }_{{\mathbb{R}}^N}A\left(\sqrt{a}\frac{{\Psi }_{ϵ,y}\left(x\right)}{\parallel {\Psi }_{{ϵ}_n,y_n}{\parallel }_2}\right)dx={\int }_{{\mathbb{R}}^N}A\left(w\left(z\right)\right)dz+o_n\left(1\right);$$

$${\int }_{{\mathbb{R}}^N}B\left({\Phi }_{{ϵ}_n}\left(y_n\right)\right)dx={\int }_{{\mathbb{R}}^N}B\left(\sqrt{a}\frac{{\Psi }_{ϵ,y}\left(x\right)}{\parallel {\Psi }_{{ϵ}_n,y_n}{\parallel }_2}\right)dx={\int }_{{\mathbb{R}}^N}B\left(w\left(z\right)\right)dz+o_n\left(1\right);$$

$$\begin{array}{cc}\hfill {\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{\frac{s}{2}}{\Phi }_{{ϵ}_n}\left(y_n\right)\right|}^{2}dx& =\frac{a}{\parallel {\Psi }_{{ϵ}_n,y}{\parallel }_2^2}{\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{\frac{s}{2}}\left(\eta \left({ϵ}_{n}z\right)w\left(z\right)\right)\right|}^{2}dz\hfill \\ & ={\int }_{{\mathbb{R}}^N}{\left|{(-\Delta )}^{\frac{s}{2}}w\left(z\right)\right|}^{2}dz+o_n\left(1\right);\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill {\int }_{{\mathbb{R}}^N}V\left({ϵ}_{n}x\right){\left|{\Phi }_{{ϵ}_n}\left(y_n\right)\right|}^{2}dx& ={\int }_{{\mathbb{R}}^N}\frac{V\left({ϵ}_{n}z+y_n\right)a}{\parallel {\Psi }_{{ϵ}_n,y}{\parallel }_2^2}{\left|\eta \left({ϵ}_{n}z\right)w\left(z\right)\right|}^{2}dz\hfill \\ \hfill & =V_0{\int }_{{\mathbb{R}}^N}{\left|w\left(z\right)\right|}^{2}dz+o_n\left(1\right).\hfill \end{array}$$

Consequently,

$$I_{{ϵ}_n}\left({\Phi }_{{ϵ}_n}\left(y_n\right)\right)-E_{V_0,a}=I_{V_0}\left(w\right)-E_{V_0,a}+o_n\left(1\right)=o_n\left(1\right),$$

which is a contradiction, and it follows that the conclusion of the lemma is correct. □

Define $\chi :{\mathbb{R}}^N\to {\mathbb{R}}^N$ satisfying $\chi \left(x\right)=x$ for $\left|x\right|\le R$ and $\chi \left(x\right)=\frac{Rx}{\left|x\right|}$ for $\left|x\right|\ge R$, where $R=R\left(\delta \right)>0$ satisfying $G_{\delta }\subset B_R\left(0\right)$ for any $\delta >0$. Set ${\beta }_{ϵ}:S\left(a\right)\to {\mathbb{R}}^N$ as follows:

$${\beta }_{ϵ}\left(u\right)=\frac{{\int }_{{\mathbb{R}}^N}\chi \left(ϵx\right){\left|u\right|}^{2}dx}{a}.$$

The proof of the following lemma is standard, refer to Lemma 4.2 in [21].

Lemma 13.

The function ${\Phi }_{ϵ}$ satisfying

$$\underset{ϵ\to 0}{lim}{\beta }_{ϵ}\left({\Phi }_{ϵ}\left(y\right)\right)=y,\mathit{uniformly}\mathit{in}y\in G.$$

Lemma 14.

If $\left(\tilde{V}\right)$ holds and $a>a^{*}>0$, set ${ϵ}_n\to 0$ and $\left\{u_n\right\}\subset S\left(a\right)$ with $I_{{ϵ}_n}\left(u_n\right)\to E_{V_0,a}$. Then, there exists $\left\{{\tilde{y}}_n\right\}\subset {\mathbb{R}}^N$ such that $v_n(·)=u_n\left(·+{\tilde{y}}_n\right)\to v\in W\setminus \left\{0\right\}$ in W. In addition, $y_n={ϵ}_n{\tilde{y}}_n\to y\in G$.

Proof. 

As in Lemma 9, there exist $R_0,\delta >0$ and $\left\{y_n\right\}$ satisfying

$${\int }_{B_{R_0\left({\overline{y}}_n\right)}}{\left|u_n\right|}^{2}dx\ge \delta ,\forall n\in \mathbb{N}.$$

Set $v_n\left(x\right):=u_n\left(x+{\tilde{y}}_n\right)$, there exists $v\in W\setminus \left\{0\right\}$ such that $v_n\rightharpoonup v$ in W up to a subsequence. Obviously, $\left\{v_n\right\}\subset S\left(a\right)$ and

$$E_{V_0,a}+o_n\left(1\right)=I_{{ϵ}_n}\left(u_n\right)\ge I_{V_0}\left(u_n\right)=I_{V_0}\left(v_n\right)\ge E_{V_0,a},$$

It follows that $I_{V_0}\left(v_n\right)\to E_{V_0,a}$. By Lemma 7, we have $v\in S\left(a\right)$ and $v_n\to v$ in W.

Claim: $\left\{y_n\right\}$ is bounded in ${\mathbb{R}}^N$.

If not, one has

$$\begin{array}{cc}\hfill E_{V_0,a}& =\underset{n\to +\infty }{lim}\left(\frac{1}{2}{\int }_{{\mathbb{R}}^N}{|(-\Delta )}^{\frac{s}{2}}{u}_n{|}^{2}dx+{\int }_{{\mathbb{R}}^N}(V\left({ϵ}_{n}x\right)+1){\left|u_n\right|}^{2}dx-\frac{1}{2}{\int }_{{\mathbb{R}}^N}{u}_n^{2}log{u}_n^{2}dx\right)\hfill \\ \hfill & =\underset{n\to +\infty }{lim}\left(\frac{1}{2}{\int }_{{\mathbb{R}}^N}{|(-\Delta )}^{\frac{s}{2}}{v}_n{|}^{2}dx+{\int }_{{\mathbb{R}}^N}(V({ϵ}_{n}x+y_n)+1){\left|v_n\right|}^{2}dx-\frac{1}{2}{\int }_{{\mathbb{R}}^N}{v}_n^{2}log{v}_n^{2}dx\right)\hfill \\ \hfill & \ge \frac{1}{2}{\int }_{{\mathbb{R}}^N}{|(-\Delta )}^{\frac{s}{2}}{v|}^{2}dx+\frac{1}{2}(V_{\infty }+1){\int }_{{\mathbb{R}}^N}{\left|v\right|}^{2}dx-\frac{1}{2}{\int }_{{\mathbb{R}}^N}{v}^{2}log{v}^{2}dx\ge E_{\infty ,a},\hfill \end{array}$$

which contradicts (14). So, the claim holds.

We assume that $y_n\to y$ in ${\mathbb{R}}^N$, and thus

$$E_{V_0,a}\ge \frac{1}{2}{\int }_{{\mathbb{R}}^N}{|(-\Delta )}^{\frac{s}{2}}{v|}^2+(V\left(y\right)+1){\left|v\right|}^{2}dx-\frac{1}{2}{\int }_{{\mathbb{R}}^N}{v}^{2}log{v}^{2}dx\ge E_{V\left(y\right),a}.$$

(24)

From Corollary 1, $E_{V\left(y\right),a}>E_{V_0,a}$ if $V\left(y\right)>V_0$. Combined with (24) and $V\left(y\right)\ge V_0$ for all $y\in {\mathbb{R}}^N$, one has $V\left(y\right)=V_0$, namely, $y\in G$. □

Let $g:[0,+\infty )\to [0,+\infty )$ be a positive function satisfying $g\left(ϵ\right)\to 0$ as $ϵ\to 0$. In addition, define

$$\tilde{S}\left(a\right):=\left\{u\in S\left(a\right):I_{ϵ}\left(u\right)\le E_{V_0,a}+g\left(ϵ\right)\right\}.$$

(25)

In particular,

$$g\left(ϵ\right)=\underset{y\in M}{sup}\left|I_{ϵ}\left({\Phi }_{ϵ}\left(y\right)\right)-E_{V_0,a}\right|.$$

By Lemma 12, ${\Phi }_{ϵ}\left(y\right)\in \tilde{S}\left(a\right)$ for any $y\in G$.

Lemma 15.

Set $\delta >0$, and then

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

Proof. 

Let $u_n\in \tilde{S}\left(a\right)$ and ${ϵ}_n\to 0$ satisfy

$$\underset{z\in G_{\delta }}{inf}\left|{\beta }_{{ϵ}_n}\left(u_n\right)-z\right|=\underset{u\in \tilde{S}\left(a\right)}{sup}\underset{y\in G_{\delta }}{inf}\left|{\beta }_{{ϵ}_n}\left(u_n\right)-z\right|+o_n\left(1\right).$$

Next, we find $\left\{y_n\right\}\subset G_{\delta }$ satisfying

$$\left|{\beta }_{ϵ}\left(u_n\right)-y_n\right|\to 0\mathrm{as}n\to \infty .$$

Since $u_n\in \tilde{S}\left(a\right)$, we have

$$E_{V_0,a}\le I_{V_0}\left(u_n\right)\le I_{{ϵ}_n}\left(u_n\right)\le E_{V_0,a}+g\left({ϵ}_n\right),\mathrm{for}\mathrm{all}n\in \mathbb{N}.$$

Then,

$$I_{{ϵ}_n}\left(u_n\right)\to E_{V_0,a}\mathrm{as}n\to \infty .$$

By Lemma 14, $\left\{{\tilde{y}}_n\right\}\subset {\mathbb{R}}^N$ satisfies $v_n(·)=u_n\left(·+{\tilde{y}}_n\right)\to v\in W$ with $v\ne 0$ and $y_n={ϵ}_n{\tilde{y}}_n\to y$ for some $y\in G$. Thus, $\left\{y_n\right\}\subset G_{\delta }$ when $n\in \mathbb{N}$ is large enough. Moreover,

$$\begin{array}{cc}\hfill {\beta }_{{ϵ}_n}\left(u_n\right)& =\frac{{\int }_{{\mathbb{R}}^N}\chi \left({ϵ}_{n}x+y_n\right){\left|v_n\right|}^{2}dx}{a}=\frac{{\int }_{{\mathbb{R}}^N}\left(\chi \left({ϵ}_{n}z+y_n\right)-y_n\right){\left|v_n\right|}^{2}dz}{a}+y_n,\hfill \end{array}$$

(26)

and

$$\frac{{\int }_{{\mathbb{R}}^N}\left(\chi \left({ϵ}_{n}z+y_n\right)-y_n\right){\left|v_n\right|}^{2}dz}{a}\to 0\mathrm{as}n\to +\infty .$$

(27)

By combining (26) and (27), we complete the proof. □

Proof of Theorem 1.

For $c\in \left(E_{V_0,a},E_{V_0,a}+g\left(ϵ\right)\right)$, we know that $I_{ϵ}$ satisfies the ${\left(PS\right)}_c$ condition by Lemma 11. Then, we derive that $I_{ϵ}$ has at least ${cat}_{\tilde{S}\left(a\right)}\left(\tilde{S}\left(a\right)\right)=cat\left(\tilde{S}\left(a\right)\right)$ critical points on $S\left(a\right)$ (see [26,31]).

Let $ϵ\in \left(0,{ϵ}_1\right)$. By Lemmas 12, 13 and 15, we can follow the argument in [32] to conclude that ${\beta }_{ϵ}\circ {\Phi }_{ϵ}:G\to G_{\delta }$ is homotopic to the inclusion map. Thus, we have

$$cat\left(\tilde{S}\left(a\right)\right)\ge {cat}_{G_{\delta }}\left(G\right).$$

Therefore, we obtain at least ${\mathrm{cat}}_{G_{\delta }}\left(G\right)$ couples $\left(u_k,{\lambda }_k\right)\in W\times \mathbb{R}$ of weak solutions to problem (4) with ${\int }_{{\mathbb{R}}^N}{\left|u_k\right|}^{2}dx=a$, ${\lambda }_k\in \mathbb{R}$ and $I_{ϵ}\left(u_k\right)<0$ for $1\le k\le {\mathrm{cat}}_{G_{\delta }}\left(G\right)$.

Let $u_{ϵ}$ be a solution of (4) with $I_{ϵ}\left(u_{ϵ}\right)\le E_{V_0,a}+g\left(ϵ\right)$, where g was given in (25). By Lemma 14, let ${ϵ}_n\to 0$, and there exists ${\tilde{y}}_n\in {\mathbb{R}}^N$ satisfying $y_n={ϵ}_n{\tilde{y}}_n\to y\in G$ and ${\tilde{u}}_n\left(x\right)=u_{{ϵ}_n}\left(x+{\tilde{y}}_n\right)\to v\in W\setminus \left\{0\right\}$. Since ${\tilde{u}}_n$ satisfying

$${(-\Delta )}^s{\tilde{u}}_n+V\left({ϵ}_{n}x+y_n\right){\tilde{u}}_n={\lambda }_n{\tilde{u}}_n+{\tilde{u}}_{n}log{\tilde{u}}_n^2,\mathrm{in}{\mathbb{R}}^N,$$

with

$$\underset{ϵ\to 0}{lim\; sup}{\lambda }_n\le \frac{{\rho }_1+E_{V_0,a}}{a}<0,$$

${\tilde{u}}_n\to v$ in W is similar to the discussion in Theorem 1.1 in [6]. Then,

$$\underset{\left|x\right|\to +\infty }{lim}{\tilde{u}}_n\left(x\right)=0,\mathrm{uniformly}\mathrm{for}n\in \mathbb{N}.$$

(28)

We show that $\parallel {\tilde{u}}_n{\parallel }_{\infty }\nrightarrow 0$. If $\parallel {\tilde{u}}_n{\parallel }_{\infty }\to 0$, ${\tilde{u}}_n\to 0$ in X, which is contradictory to $\parallel {\tilde{u}}_n{\parallel }_2^2=a$. Let $\parallel {\tilde{u}}_n{\parallel }_{\infty }\ge 2\tau$ for fixed $\tau >0$ and set $z_n\in {\mathbb{R}}^N$ such that $\left|{\tilde{u}}_n\left(z_n\right)\right|={\parallel {\tilde{u}}_n\parallel }_{\infty }$ for any $n\in \mathbb{N}$. Moreover, (28) implies that there are $R_1>0$ and $n_0\in \mathbb{N}$ satisfying

$$\left|{\tilde{u}}_n\left(x\right)\right|\le \tau \mathrm{for}\left|x\right|\ge R_1\mathrm{and}n\ge n_0.$$

Therefore $\left|z_n\right|\le R_1$ for any $n\in \mathbb{N}$. In addition, we set ${\xi }_n\in {\mathbb{R}}^N$ satisfying $\left|u_n\left({\xi }_n\right)\right|={\parallel u_n\parallel }_{\infty }$ for all $n\in \mathbb{N}$. Then, ${\xi }_n=z_n+{\tilde{y}}_n$ and

$$V\left({ϵ}_n{\xi }_n\right)=V\left({ϵ}_n{z}_n+{ϵ}_n{\tilde{y}}_n\right)\to V\left(y\right)=V_0\mathrm{as}n\to +\infty .$$

This completes the proof. □

5. Conclusions

The main purpose of this paper is to study the multiplicity of the normalized solution of the fractional logarithmic Schrödinger equation. First, we introduce the fractional logarithmic Schrödinger equation and its related results. We then consider the existence of a normalized solution of the corresponding equation in the autonomous case, which is useful in the subsequent proof of the main result. Finally, the main results of Theorem 1 are proved by means of a variational method, the Lusternik–Schnirelmann category and some analytical techniques. The result shows that the number of normalized solutions depends on the topology of the set for which the potential V reaches its minimum.

Based on the current results and those of Alves [33], for the following equation

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

(29)

where h satisfies the following assumptions:

• $h\in \mathcal{C}({\mathbb{R}}^N,{\mathbb{R}}^{+}),0<h_{\infty }={lim}_{\left|x\right|\to +\infty }h\left(x\right)<{max}_{x\in {\mathbb{R}}^N}h\left(x\right)=h\left(a_i\right)\mathrm{for}1\le i\le kwith{a}_1=0\mathrm{and}{a}_j\ne a_i\mathrm{if}i\ne j,$

we believe that we can obtain the multiplicity of the normalization solution of Equation (29) through a similar proof to that in this paper.