The Existence of Solution of a Critical Fractional Equation

Hui Chen

College of Science, University of Shanghai for Science and Technology, Shanghai, China.

DOI: 10.4236/jamp.2019.71020                

Abstract

In this paper, we study the existence of solution of a critical fractional equation; we will use a variational approach to find the solution. Firstly, we will find a suitable functional to our problem; next, by using the classical concept and properties of the genus, we construct a mini-max class of critical points.

Keywords

Variational Approach, Fractional Laplacian, Minimax Principle, Genus

1. Introduction

In this paper, we focus our attention on the following problem:

$\{\begin{array}{l}{\left(-\Delta \right)}^{s}u=\lambda V\left(x\right){\left|u\right|}^{p-1}+\beta K\left(x\right){\left|u\right|}^{2_s^{*}-1}\text{in}\Omega \\ u=0\text{in}{R}^n\backslash \Omega \end{array}$ (1.1)

where $\Omega$ is a bounded domain in $R^n$ , $\lambda >0$ , $0<s<1$ and $n>2s$ , $1<p<2_s^{*}$ , $K\left(x\right)\in C\left(R^n\right)\cap L^{\infty }\left(R^n\right)$ , $V\left(x\right)\ge 0$ and $V\left(x\right)\in C\left(R^n\right)\cap L^q\left(R^n\right)$ with $q=\frac{2_s^{*}}{2_s^{*}-p}$ here ${\left(-\Delta \right)}^s$ denotes the fractional Laplace operator defined, up to a normalization factor, by

${\left(-\Delta \right)}^{s}u\left(x\right)={\displaystyle \underset{R^n}{\int }\frac{u\left(x\right)-u\left(y\right)}{{\left|x-y\right|}^{n+2s}}\text{d}y},x\in R^n$. (1.2)

The aim of this paper is to study the existence of solutions, we will see that if $1<p<2$ , then by concentration-compactness principle, together with mini-max arguments, we can prove the existence of solutions for (1.1). We now summarize the main result of the paper.

Theorem 1.1. Let $1<p<2$ , $K\left(x\right)\in C\left(R^n\right)\cap L^{\infty }\left(R^n\right)$ and $0\le V\left(x\right)\in C\left(R^n\right)\cap L^q\left(R^n\right)$ with $q=\frac{2_s^{*}}{2_s^{*}-p}$. Moreover, $V\left(x\right)>0$ is bounded on $\Omega$. Then

1) For any $\lambda >0$ , there exists $\tilde{\beta }>0$ , then for any $0<\beta <\tilde{\beta }$ , (1.1) has a consequence of weak solutions $\left\{u_n\right\}$.

2) For any $\beta >0$ , there exist $\tilde{\lambda }>0$ , then for any $0<\lambda <\tilde{\lambda }$ , (1.1) has a consequence of weak solutions $\left\{u_n\right\}$.

We denote by $H^s\left(R^n\right)$ the usual fractional Sobolev space endowed with the so-called Gagliardo norm

${\Vert u\Vert }_{H^s\left(R^n\right)}={\Vert u\Vert }_{L^2\left(R^n\right)}+{\left({\displaystyle \underset{R^n\times R^n}{\int }\frac{{\left|u\left(x\right)-u\left(y\right)\right|}^2}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}\right)}^{\frac{1}{2}},$ (1.3)

Then we defined

$X_0^s\left(\Omega \right)=\left\{u\in H^s\left(R^n\right):u=0a.e.\text{in}{R}^n\backslash \Omega \right\}$ (1.4)

endowed with the norm

${\Vert u\Vert }_{X_0^s\left(\Omega \right)}={\left({\displaystyle \underset{R^n\times R^n}{\int }\frac{{\left|u\left(x\right)-u\left(y\right)\right|}^2}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}\right)}^{\frac{1}{2}},$ (1.5)

we refer to [1] for a general definition of $X_0^s\left(\Omega \right)$ and its properties.

Observe that by [ [2] , Proposition 3.6] we have the following identity

${\Vert u\Vert }_{X_0^s\left(\Omega \right)}={\Vert {\left(-\Delta \right)}^{\frac{s}{2}}u\Vert }_{L^2\left(R^n\right)}.$ (1.6)

In this work, the Sobolev constant is given by (can be seen in [ [3] , theorem 7.58])

$S\left(n,s\right):=\underset{u\in H^s\left(R^n\right)\backslash \left\{0\right\}}{\inf }{Q}_{n,s}\left(u\right)>0,$ (1.7)

where

$Q_{n,s}\left(u\right):=\frac{{\displaystyle \underset{R^n\times R^n}{\int }\frac{\left|u\left(x\right)-u{\left(y\right)}^2\right|}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}}{{\left({\displaystyle \underset{R^n\times R^n}{\int }{\left|u\left(x\right)\right|}^{2_s^{*}}\text{d}x}\right)}^{\frac{2}{2_s^{*}}}}\text{,}u\in H^s\left(R^n\right)$ (1.8)

2. Statements of the Result

We will use a variational approach to find a solution of (1.1). Firstly, we will associate a suitable functional to our problem, the Euler-Lagrange functional related to problem (1) is given by $J:X_0^s\left(\Omega \right)\to R$ defined as follow

$J\left(u_n\right)=\frac{1}{2}{\Vert u_n\Vert }_{X_0^s\left(\Omega \right)}^2-\frac{\lambda }{p}{\displaystyle \underset{\Omega }{\int }V\left(x\right){\left|u_n\right|}^p\text{d}x}-\frac{\beta }{2_s^{*}}{\displaystyle \underset{\Omega }{\int }K\left(x\right){\left|u_n\right|}^{2_s^{*}}\text{d}x}.$ (2.1)

To proof that J satisfy the Palais Smale condition at level c, we need the following lemma.

Lemma 2.1 [4] Letting $\varphi$ be a regular function that satisfies that for some $\tilde{c}>0$

$\left|\varphi \left(x\right)\right|\le \frac{\tilde{c}}{1+{\left|x\right|}^{n+s}}\text{,}x\in R^n$ (2.2)

and

$\left|\nabla \varphi \left(x\right)\right|\le \frac{\tilde{c}}{1+{\left|x\right|}^{n+s}}\text{,}x\in R^n$ (2.3)

Let $B:X_0^{\frac{s}{2}}\left(\Omega \right)\times X_0^{\frac{s}{2}}\left(\Omega \right)\to R$ be a bilinear form defined by

$B\left(f,g\right)\left(x\right):=2{\displaystyle \underset{R}{\int }\frac{\left(f\left(x\right)-f\left(y\right)\right)\left(g\left(x\right)-g\left(y\right)\right)}{{\left|x-y\right|}^{n+s}}\text{d}y}.$ (2.4)

then, for every $s\in \left(0,1\right)$ , there exist positive constant $c_1$ and $c_2$ , such that for $x\in R^n$ , one has

$\left|{\left(-\Delta \right)}^{\frac{s}{2}}\varphi \left(x\right)\right|\le \frac{c}{1+{\left|x\right|}^{n+s}}$ and $\left|B\left(\varphi ,\varphi \right)\left(x\right)\right|\le \frac{c}{1+{\left|x\right|}^{n+s}}$. (2.5)

To establish the next auxiliary result we consider a radial, nonincreasing cut-off function

$\varphi \in C_0^{\infty }\left(R^n\right)$ and ${\varphi }_{\epsilon }\left(x\right):=\varphi \left(\frac{x}{\epsilon }\right)$ (2.6)

Lemma 2.2. [4] Letting $\left\{u_m\right\}$ be a uniformly bounded in $X_0^s\left(\Omega \right)$ and ${\varphi }_{\epsilon }\in C_0^{\infty }\left(R^n\right)$ the function defined in (2.6). Then,

$\underset{\epsilon \to 0}{\lim }\underset{m\to 0}{\lim }\left|{\displaystyle \underset{R^n}{\int }{u}_m\left(x\right){\left(-\Delta \right)}^{\frac{s}{2}}{\varphi }_{\epsilon }\left(x\right){\left(-\Delta \right)}^{\frac{s}{2}}{u}_m\left(x\right)\text{d}x}\right|=0.$ (2.7)

Lemma 2.3. [4] With the same assumptions of Lemma 2.8 we have that

$\underset{\epsilon \to 0}{\lim }\underset{m\to 0}{\lim }\left|{\displaystyle \underset{R^n}{\int }{\left(-\Delta \right)}^{\frac{s}{2}}{u}_m\left(x\right)\text{d}xB\left(u_m,{\varphi }_{\epsilon }\right)\left(x\right)}\right|=0.$ (2.8)

where B is defined in (2.4).

Lemma 2.4. [5] (Minimax principle) Assume that $E\in C\left(X,ℝ\right)$ , and $\mathcal{A}$ is a family of nonempty subset of X, denote

$c=\underset{A\in \mathcal{A}}{\inf }\underset{x\in A}{\sup }E\left(x\right)$ (2.9)

If the following conditions holds:

1) c is a finite real number;

2) there exists an $\bar{\epsilon }>0$ , such that $\mathcal{A}$ is invariant with respect to the family of mappings;

$\mathcal{T}=\left\{T\in \left(X,X\right)|T\left(x\right)=x,\text{if}E\left(x\right)<c-\bar{\epsilon }\right\}$ , (2.10)

that is, for any $T\in \mathcal{T}$ , there holds

$A\in \mathcal{A}\Rightarrow \mathcal{T}\left(A\right)\in \mathcal{A}$

Then, E possesses a ${\left(PS\right)}_c$ sequence at level c define as (6.1.1); Furthermore, if E satisfies the ${\left(PS\right)}_c$ condition (or the ${\left(PS\right)}_c$ condition at level c), then c is a critical value of E.

3. Proof of Theorem 1.1

Firstly, recalling that J is said to satisfy the Palais Smale condition at level c if any sequence $\left\{u_n\right\}\in X_0^s\left(\Omega \right)$ such that $J\left(u_n\right)\to c$ and $J^{\prime }\left(u\right)\to 0$ has a convergent subsequence.

Lemma 3.1. The ${\left(PS\right)}_c$ sequence $\left\{u_n\right\}$ for J is bounded.

Proof. Note that $\left\{u_n\right\}\subset X_0^s\left(\Omega \right)$ satisfies

$\begin{array}{l}J\left(u_n\right)=\frac{1}{2}{\Vert u_n\Vert }_{X_0^s\left(\Omega \right)}^2-\frac{\lambda }{p}{\displaystyle \underset{\Omega }{\int }V\left(x\right){\left|u_n\right|}^p\text{d}x}-\frac{\beta }{2_s^{*}}{\displaystyle \underset{\Omega }{\int }K\left(x\right){\left|u_n\right|}^{2_s^{*}}\text{d}x}\\ =c+o_n\left(1\right)\end{array}$ (3.1)

and

$\begin{array}{l}\langle J^{\prime }\left(u_n\right),\varphi \rangle ={\displaystyle \underset{\Omega }{\int }{\left(-\Delta \right)}^s{u}_n\text{d}x}-\lambda {\displaystyle \underset{\Omega }{\int }V\left(x\right){\left|u_n\right|}^{p-2}u\varphi \text{d}x}-\beta {\displaystyle \underset{\Omega }{\int }K\left(x\right){\left|u_n\right|}^{2_s^{*}-2}u\varphi \text{d}x}\\ =o_n\left(1\right){\Vert \varphi \Vert }_{X_0^s\left(\Omega \right)},\forall \varphi \in X_0^s\left(\Omega \right)\end{array}$ (3.2)

where $o_n\left(1\right)\to 0$ as $n\to \infty$. Choose $\varphi =u_n\in X_0^s\left(\Omega \right)$ as test function in (3.2), we get that

$\begin{array}{c}{o}_n\left(1\right){\Vert u_n\Vert }_{X_0^s\left(\Omega \right)}=\langle J^{\prime }\left(u_n\right),u_n\rangle \\ ={\Vert u_n\Vert }_{X_0^s\left(\Omega \right)}^2-\lambda {\displaystyle \underset{\Omega }{\int }V\left(x\right){\left|u_n\right|}^p\text{d}x}-\beta {\displaystyle \underset{\Omega }{\int }K\left(x\right){\left|u_n\right|}^{2_s^{*}}\text{d}x}\\ =c+o_n\left(1\right).\end{array}$ (3.3)

therefore, by (3.1) and (3.2), we have

$\begin{array}{l}c+o_n\left(1\right)-\frac{1}{2_s^{*}}{o}_n\left(1\right){\Vert u_n\Vert }_{X_0^s\left(\Omega \right)}\\ =\frac{1}{2}{\Vert u_n\Vert }_{X_0^s\left(\Omega \right)}^2-\frac{\lambda }{p}{\displaystyle \underset{\Omega }{\int }V(x){\left|u_n\right|}^p\text{d}x}-\frac{\beta }{2_s^{*}}{\displaystyle \underset{\Omega }{\int }K\left(x\right){\left|u_n\right|}^{2_s^{*}}\text{d}x}\\ -\frac{1}{2_s^{*}}{\Vert u_n\Vert }_{X_0^s\left(\Omega \right)}^2-\frac{\lambda }{2_s^{*}}{\displaystyle \underset{\Omega }{\int }V(x){\left|u_n\right|}^p\text{d}x}-\frac{\beta }{2_s^{*}}{\displaystyle \underset{\Omega }{\int }K\left(x\right){\left|u_n\right|}^{2_s^{*}}\text{d}x}\\ \ge \frac{s}{n}{\Vert u_n\Vert }_{X_0^s\left(\Omega \right)}^2-\left(\frac{\lambda }{p}-\frac{1}{2_s^{*}}\right){\Vert V\left(x\right)\Vert }_{L^q\left(\Omega \right)}{\Vert u_n\Vert }_{L^{2_s^{*}}}^p\\ \ge \frac{s}{n}{\Vert u_n\Vert }_{X_0^s\left(\Omega \right)}^2-\left(\frac{\lambda }{p}-\frac{1}{2_s^{*}}\right)S{\left(n,s\right)}^{-\frac{p}{2}}{\Vert V\left(x\right)\Vert }_{L^q\left(\Omega \right)}{\Vert u_n\Vert }_{X_0^s\left(\Omega \right)}^p.\end{array}$ (3.4)

which yields the boundeness of $\left\{u_n\right\}$ in $X_0^s\left(\Omega \right)$ ,since $1<p<2$.

If $K\left(x\right)\in L^{\infty }\left({ℝ}^n\right)$ , then for $2<p<2_s^{*}$ , similar to the proof of $1<p<2$ , we get

$c+o_n\left(1\right)+o_n\left(1\right){\Vert u_n\Vert }_{X_0^s\left(\Omega \right)}\ge \left(\frac{p-2}{2p}\right){\Vert u_n\Vert }_{X_0^s\left(\Omega \right)}^2-\frac{\left(p-2_s^{*}\right)\beta }{2_s^{*}}{S}^{-\frac{2_s^{*}}{2}}{\Vert u_n\Vert }_{X_0^s\left(\Omega \right)}^{2_s^{*}}$

Which also yields the boundedness of ${\left(PS\right)}_c$ sequence $\left\{u_n\right\}$.

Lemma 3.2. Assume that $c<0$. Then

1) For any $\lambda >0$ , there exists ${\beta }_0>0$ , such that for any $0<\beta <{\beta }_0$ , then J satisfies ${\left(PS\right)}_c$.

2) For any $\beta >0$ there exists ${\lambda }_0>0$ such that for any $0<\lambda <{\lambda }_0$ , then J satisfies ${\left(PS\right)}_c$.

Proof. By Lemma3.1 $\left\{u_n\right\}$ is bounded in $X_0^s\left(\Omega \right)$ , up to a subsequence, we get that

$u_n\to u$ $x\in X_0^s\left(\Omega \right)$.

$u_n\to u$ $x\in L^r\left(\Omega \right)$ , $1\le r<2_s^{*}$. (3.5)

$u_n\to u$ a.e. $x\in \Omega$.

Following [6] it is easy to prove that $X_0^s\left(\Omega \right)$ could also be the $X_0^s\left(\Omega \right)$ -norm. Applying [ [7] , Theorem1.5], we have that the exist an index. Set $I\subseteq N$ a sequence of point ${\left\{x_k\right\}}_{x\in I\subset \Omega }$ and two sequences of nonnegative real numbers ${\left\{{\mu }_k\right\}}_{k\in I},{\left\{v_k\right\}}_{k\in I}$ , such that

${\left|{\left(-\Delta \right)}^{\frac{s}{2}}{u}_n\right|}^2\to \mu {\left|{\left(-\Delta \right)}^{\frac{s}{2}}u\right|}^2+{\displaystyle \underset{k\in I}{\sum }{\mu }_k{\delta }_{x_k}}$. (3.6)

moreover

${\left|u_n\right|}^{2_s^{*}}\to \mu {\left|u\right|}^{2_s^{*}}+{\displaystyle \underset{k\in I}{\sum }{v}_k{\delta }_{x_k}}$. (3.7)

in the sense of measures, with

$v_k\le S{\left(s,n\right)}^{-\frac{2_s^{*}}{2}}{\mu }_k^{\frac{2_s^{*}}{2}}$ for every $k\in I$ (3.8)

here ${\delta }_{x_k}$ denotes the Dirac Delta at $x_k$ , while $S\left(n,s\right)$ is the constant given in (1.7), we consider $\varphi \in C_0^{\infty }\left(R^n\right)$ a nonincreasing cut-off function satisfying

$\varphi =1\text{in}{B}_1\left(x_{k_0}\right)$ and $\varphi =0\text{in}{B}_2{\left(x_{k_0}\right)}^c$ (3.9)

Set ${\varphi }_{\epsilon }\left(x\right)=\varphi \left(\frac{x}{\epsilon }\right),x\in R^n$ taking the derivative of (1.6), for any $u,\varphi \in X_0^s\left(\Omega \right)$. We obtain that

${\displaystyle \underset{R^n\times R^n}{\int }\frac{\left(u\left(x\right)-u\left(y\right)\right)\left(\varphi \left(x\right)-\varphi \left(y\right)\right)}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}={\displaystyle \underset{R^n}{\int }\varphi \left(x\right){\left(-\Delta \right)}^{s}u\left(x\right)\text{d}x}$ (3.10)

Then, taking ${\varphi }_{\epsilon }{u}_n$ as a test function in $J^{\prime }\left(u_n\right)\to 0$

$\underset{n\to 0}{\lim }{\displaystyle \underset{{ℝ}^n}{\int }{\varphi }_{\epsilon }{u}_n\left(-\Delta \right)u_n\text{d}x}-\left(\lambda {\displaystyle \underset{B_{2\epsilon }\left(x_{k_0}\right)}{\int }V\left(x\right)u_n^p{\varphi }_{\epsilon }\text{d}x}+\beta {\displaystyle \underset{B_{2\epsilon }\left(x_{k_0}\right)}{\int }K\left(x\right)u_n^{2_s^{*}}{\varphi }_{\epsilon }\text{d}x}\right)=0$ (3.11)

by (3.10), we have

$\begin{array}{l}\underset{n\to \infty }{\lim }{\displaystyle \underset{{ℝ}^n}{\int }{u}_n\left(x\right){\left(-\Delta \right)}^{\frac{s}{2}}{u}_n\left(x\right){\left(-\Delta \right)}^{\frac{s}{2}}{\varphi }_{\epsilon }\left(x\right)\text{d}x}\\ -2{\displaystyle \underset{{ℝ}^n}{\int }{\left(-\Delta \right)}^{\frac{s}{2}}{u}_n\left(x\right)}{\displaystyle \underset{{ℝ}^n}{\int }\frac{\left({\varphi }_{\epsilon }\left(x\right)-{\varphi }_{\epsilon }\left(y\right)\right)\left(u_n\left(x\right)-u_n\left(y\right)\right)}{{\left|x-y\right|}^{n+s}}\text{d}x\text{d}y}\\ =\underset{n\to \infty }{\lim }\lambda {\displaystyle \underset{B_{2\epsilon }\left(x_{k_0}\right)}{\int }V\left(x\right){\left|u_n\right|}^p\left(x\right){\varphi }_{\epsilon }\left(x\right)\text{d}x}+\beta {{\displaystyle \underset{B_{2\epsilon }\left(x_{k_0}\right)}{\int }K\left(x\right)\left|u_n\right|}}^{2_s^{*}}\left(x\right){\varphi }_{\epsilon }\left(x\right)\text{d}x\\ -{\displaystyle \underset{B_{2\epsilon }\left(x_{k_0}\right)}{\int }{\left({\left(-\Delta \right)}^{\frac{s}{2}}{u}_n\right)}^2{\varphi }_{\epsilon }\left(x\right)\text{d}x}.\end{array}$ (3.12)

therefore, by (3.5) (3.6) and (3.7) we get

$\begin{array}{l}\underset{\epsilon \to 0}{\lim }\underset{n\to \infty }{\lim }{\displaystyle \underset{{ℝ}^n}{\int }{u}_n\left(x\right){\left(-\Delta \right)}^{\frac{s}{2}}{u}_n\left(x\right){\left(-\Delta \right)}^{\frac{s}{2}}{\varphi }_{\epsilon }\left(x\right)\text{d}x}\\ -2{\displaystyle \underset{{ℝ}^n}{\int }{\left(-\Delta \right)}^{\frac{s}{2}}{u}_n\left(x\right)}{\displaystyle \underset{{ℝ}^n}{\int }\frac{\left({\varphi }_{\epsilon }\left(x\right)-{\varphi }_{\epsilon }\left(y\right)\right)\left(u_n\left(x\right)-u_n\left(y\right)\right)}{{\left|x-y\right|}^{n+s}}\text{d}x\text{d}y}\\ =\underset{\epsilon \to 0}{\lim }\lambda {\displaystyle \underset{B_{2\epsilon }\left(x_{k_0}\right)}{\int }V\left(x\right){\left|u_n\right|}^p\left(x\right){\varphi }_{\epsilon }\left(x\right)\text{d}x}+{\displaystyle \underset{B_{2\epsilon }\left(x_{k_0}\right)}{\int }{\varphi }_{\epsilon }\left(x\right)\text{d}v}\\ -\beta {\displaystyle \underset{B_{2\epsilon }\left(x_{k_0}\right)}{\int }K\left(x\right){\varphi }_{\epsilon }\left(x\right)\text{d}\mu }.\end{array}$ (3.13)

Since $\varphi$ is regular function with compact support, it is easy to see that it satisfies the hypothesis of Lemma 2.1, by Lemma 2.2 and Lemma 2.3 applied to the sequence $\left\{u_n\right\}$ , it follows that the left hand side of (3.13) goes to zero. We obtain that

$\begin{array}{l}\underset{\epsilon \to 0}{\lim }\left(\lambda {\displaystyle \underset{B_{2\epsilon }\left(x_{k_0}\right)}{\int }V\left(x\right){\left|u_n\right|}^p\left(x\right){\varphi }_{\epsilon }\left(x\right)\text{d}x}+{\displaystyle \underset{B_{2\epsilon }\left(x_{k_0}\right)}{\int }{\varphi }_{\epsilon }\left(x\right)\text{d}v}-\beta {\displaystyle \underset{B_{2\epsilon }\left(x_{k_0}\right)}{\int }K\left(x\right){\varphi }_{\epsilon }\left(x\right)}\right)\text{d}\mu \\ =\beta K\left(x_{k_0}\right)v_{k_0}-{\mu }_{k_0}=0.\end{array}$ (3.14)

Clearly, if $K\left(x\right)\le 0$ , we get ${\mu }_{k_0}=v_{k_0}=0$ ; if $K\left(x_{k_0}\right)>0$ , by (3.8), we get $v_{k_0}=0$ or $v_{k_0}\ge {\left[\frac{S\left(n,s\right)}{\beta K\left(x_{k_0}\right)}\right]}^{\frac{n}{2s}}$.

suppose that $v_{k_0}\ne 0$ , we know that

$0>c=\underset{n\to \infty }{\lim }\left[J\left(u_n\right)-\frac{1}{2_s^{*}}\langle J^{\prime }\left(u_n\right),u_n\rangle \right]$ (3.15)

according to the embedded theorem, we have

$\begin{array}{l}0>c\ge \left(\frac{1}{2}-\frac{1}{2_s^{*}}\right){\Vert u_n\Vert }_{X_0^s\left(\Omega \right)}^2-\left(\frac{\lambda }{p}-\frac{\lambda }{2_s^{*}}\right){\displaystyle \underset{\Omega }{\int }V\left(x\right){\left|u_n\right|}^p}\text{d}x\\ =\frac{s}{n}{\Vert u_n\Vert }_{X_0^s\left(\Omega \right)}^2-\left(\frac{\lambda }{p}-\frac{\lambda }{2_s^{*}}\right){\displaystyle \underset{\Omega }{\int }V\left(x\right){\left|u_n\right|}^p}\text{d}x\\ \ge \frac{s}{n}{S}^{-1}\left(n,s\right){\Vert u_n\Vert }_{L^{2_s^{*}}\left(\Omega \right)}^2-\left(\frac{\lambda }{p}-\frac{\lambda }{2_s^{*}}\right)S^{-\frac{p}{2}}\left(n,s\right){\Vert V\left(x\right)\Vert }_{L^q\left(\Omega \right)}{\Vert u_n\Vert }_{L^{2_s^{*}}}^p.\end{array}$ (3.16)

This yields that

${\Vert u\Vert }_{L^{2_s^{*}}\left(\Omega \right)}\le C{\lambda }^{\frac{1}{2-p}}$. (3.17)

Thus, if $v_{k_0}\ge {\left[\frac{S\left(n,s\right)}{\beta K\left(x_{k_0}\right)}\right]}^{\frac{n}{2s}}$ , we get that

$\begin{array}{l}0>c=\underset{n\to \infty }{\lim }\left[J\left(u_n\right)-\frac{1}{2_s^{*}}\langle J^{\prime }\left(u_n\right),u_n\rangle \right]\\ \ge \left(\frac{1}{2}-\frac{1}{2_s^{*}}\right){\Vert u\Vert }_{X_0^s\left(\Omega \right)}^2+\frac{s}{n}{\mu }_{k_0}-\left(\frac{\lambda }{p}-\frac{\lambda }{2_s^{*}}\right){\displaystyle \underset{\Omega }{\int }V\left(x\right){\left|u\right|}^p}\text{d}x\\ \ge \frac{s}{n}{S}^{-1}\left(n,s\right){\Vert u\Vert }_{L^{2_s^{*}}\left(\Omega \right)}^2+\frac{s}{n}{\mu }_{k_0}-\left(\frac{\lambda }{p}-\frac{\lambda }{2_s^{*}}\right)S^{-\frac{p}{2}}\left(n,s\right){\Vert V\left(x\right)\Vert }_{L^q\left(\Omega \right)}{\Vert u\Vert }_{L^{2_s^{*}}}^p\\ \ge \frac{s}{n}{S}^{-1}\left(n,s\right){\Vert u\Vert }_{L^{2_s^{*}}\left(\Omega \right)}^2+\frac{s}{n}{\mu }_{k_0}-\left(\frac{\lambda }{p}-\frac{\lambda }{2_s^{*}}\right)S^{-\frac{p}{2}}\left(n,s\right){\Vert V\left(x\right)\Vert }_{L^q\left(\Omega \right)}{\Vert u\Vert }_{L^{2_s^{*}}}^p\\ \ge \frac{s}{n}S\left(n,s\right)v_{k_0}^{-\frac{2_s^{*}}{2}}-\left(\frac{\lambda }{p}-\frac{\lambda }{2_s^{*}}\right)S^{-\frac{p}{2}}\left(n,s\right){\Vert V\left(x\right)\Vert }_{L^p\left(\Omega \right)}{\Vert u\Vert }_{L^{2_s^{*}}}^p\\ \ge \frac{s}{n}{S}^{\frac{n}{2s}}\left(n,s\right){\left[\beta K\left(x_{k_0}\right)\right]}^{\frac{2s-n}{2s}}-C{\lambda }^{\frac{2}{2-p}}.\end{array}$ (3.18)

However, if $\beta >0$ is given, we can choose ${\lambda }_0>0$ so small for every $0<\lambda <{\lambda }_0$ that last term on the right-hand side above is greater than 0 which is contradiction when $2<p<2_s^{*}$

$\begin{array}{c}0>c=\underset{n\to \infty }{\lim }\left[J\left(u_n\right)-\frac{1}{p}\langle J^{\prime }\left(u_n\right),u_n\rangle \right]\\ =\left(\frac{1}{2}-\frac{1}{p}\right){\Vert u\Vert }_{X_0^s\left(\Omega \right)}^2-\left(\frac{\beta }{2_s^{*}}-\frac{\beta }{p}\right){\displaystyle \underset{\Omega }{\int }K\left(x\right){\left|u\right|}^{2_s^{*}}\text{d}x}\\ \ge \left(\frac{1}{2}-\frac{1}{p}\right){\Vert u\Vert }_{X_0^s\left(\Omega \right)}^2-\left(\frac{\beta }{2_s^{*}}-\frac{\beta }{p}\right){\displaystyle \underset{\Omega \cap \left\{K\left(x\right)<0\right\}}{\int }K\left(x\right){\left|u\right|}^{2_s^{*}}\text{d}x}\\ \ge \left(\frac{1}{2}-\frac{1}{p}\right){\Vert u\Vert }_{X_0^s\left(\Omega \right)}^2-\left(\frac{\beta }{2_s^{*}}-\frac{\beta }{p}\right){\Vert K\left(x\right)\Vert }_{L^{\infty }}{\Vert u\Vert }_{L^{2_s^{*}}}^{2_s^{*}}\end{array}$

$\beta$ is the same as $\lambda$ greater than 0. We see that $v_{k_0}\ge {\left[\frac{S\left(n,s\right)}{\beta K\left(x_{k_0}\right)}\right]}^{\frac{n}{2s}}$ cannot occur if ${\lambda }_0$ or ${\beta }_0$ are choose properly. Thus ${\mu }_k=v_k=0$. As consequence, we obtain that ${\left(u_n\right)}_{+}-u\to 0$ in $L^{2_s^{*}}\left(\Omega \right)$ , that is $\underset{n\to \infty }{\lim }{\displaystyle \underset{R^n}{\int }{\left|{\left(u_n\right)}_{+}\right|}^{2_s^{*}}}\text{d}x={\displaystyle \underset{R^n}{\int }{\left|u\right|}^{2_s^{*}}}\text{d}x$. This implies convergence of $\lambda V\left(x\right){\left|u_n\right|}^{p-1}+\beta K\left(x\right){\left|u_n\right|}^{2_s^{*}-1}$ in $L^{2_s^{*}}\left(\Omega \right)$. Finally using the continuity of the inverse operator ${\left(-\Delta \right)}^s$. We obtain strong convergence of $u_n$ in $X_0^s\left(\Omega \right)$. #

Next, by using the classical concept and properties of the genus, we construct a min-max class of the critical point.

For a Banach space X, We define the set

$\mathcal{A}=\left\{A\subset X\backslash \left\{0\right\}:A\text{is closed in}X\text{and symmetric with respect to the orign}\right\}$

For $\mathcal{A}\in A$ , define

$\gamma \left(A\right):=\inf \left\{m\in N,\exists \varphi \in C\left(A,R^m\backslash \left\{0\right\}\right),\varphi \left(x\right)=-\varphi \left(-x\right)\right\}$ (3.19)

If there is no mapping $\varphi$ as above for any $m\in N$ , there $\gamma \left(A\right)=+\infty$. we refer to [8] for the properties of the genus.

Proposition 3.3. [8] Let $A,B\subset {\rm A}$ ,

1) If there exists an odd map $f\in C\left(A,B\right)$ , then $\gamma \left(A\right)\le \gamma \left(B\right)$ ;

2) If $A\subset B$ , then $\gamma \left(A\right)\le \gamma \left(B\right)$ ;

3) $\gamma \left(A\cup B\right)\le \gamma \left(A\right)+\gamma \left(B\right)$ ;

4) If S is a sphere centered at the origin in $R^m$ , then $\gamma \left(s\right)=m$ ;

5) If A is compact, there exists a symmetric Neighborhood N of A, such that $\gamma \left(\bar{N}\right)=\gamma \left(A\right)$.

According Holder inequality, we get that

$\begin{array}{l}J\left(u\right)=\frac{1}{2}{\Vert u\Vert }_{X_0^s}^2-\frac{\lambda }{p}{{\displaystyle \underset{\Omega }{\int }V\left(x\right)\left|u\right|}}^p\text{d}x-\frac{\beta }{2_s^{*}}{\displaystyle \underset{\Omega }{\int }K\left(x\right){\left|u\right|}^{2_s^{*}}}\text{d}x\\ \ge \frac{1}{2}{\Vert u\Vert }_{X_0^s}^2-C_1\lambda {\Vert u\Vert }_{X_0^s}^p-C_2\beta {\Vert u\Vert }_{X_0^s}^{2_s^{*}}\end{array}$ (3.20)

We define the function

$Q\left(t\right):=\frac{1}{2}{t}^2-C_1\lambda t^p-C_2\beta t^{2_s^{*}}$ (3.21)

Then it is easy to see that given $\beta >0$ , there exists ${\lambda }_1>0$ so small that for every $0<\lambda <{\lambda }_1$ , there exists $0<T_0<T_1$ such that $Q\left(t\right)<0$ for $0\le t\le T_0$ , $Q\left(t\right)>0$ for $T_0<t<T_1$. and $Q\left(t\right)<0$ $t>T_1$. Analogously, for given $\lambda >0$ , we can choose ${\beta }_1>0$ with the property that $T_0,T_1$ as above for each $0<\beta <{\beta }_1$. Clearly, $Q\left(T_0\right)=Q\left(T_1\right)=0$.

As in [9] , Let $\tau :{ℝ}^{+}\to \left[0,1\right]$ be a nonincreasing $C^{\infty }$ function such that $\tau \left(t\right)=1$ if $0\le \tau \le T_0$ and $\tau \left(t\right)=0$. if $\tau \ge T_0$. Set $\Psi \left(u\right)=\tau \left({\Vert u\Vert }_{X_0^s\left(\Omega \right)}\right)$ , we make the following truncation of the function J:

$\tilde{J}\left(u\right)=\frac{1}{2}{\Vert u\Vert }_{X_0^s}^2-\frac{\lambda }{p}{{\displaystyle \underset{\Omega }{\int }V\left(x\right)\left|u\right|}}^p\text{d}x-\frac{\beta }{2_s^{*}}\psi \left(u\right){\displaystyle \underset{\Omega }{\int }K\left(x\right){\left|u\right|}^{2_s^{*}}}\text{d}x$ (3.22)

then

$\tilde{J}\left(u\right)\ge \tilde{Q}{\Vert u\Vert }_{X_0^s\left(\Omega \right)}.$ (3.23)

where $\tilde{Q}\left(t\right):=\frac{1}{2}{t}^2-C_1\lambda t^p-C_2\beta t^{2_s^{*}}\psi \left(t\right)$.

It is clear that $\tilde{J}\left(u\right)\in C^1$ and is bounded from below.

Lemma 3.4. [10] 1) For any $\lambda >0$ and $0<\beta <{\beta }_1$ or any $\beta >0$ and $0<\lambda <{\lambda }_1$ , if $\tilde{J}\left(u\right)<0$ , then ${\Vert u\Vert }_{X_0^s\left(\Omega \right)}<T_0$ and $\tilde{J}\left(u\right)=J\left(u\right)$.

2) For any $\lambda >0$ , there exists such that if $0<\beta <\bar{\beta }$ and $c<0$ then $\tilde{J}$ satisfies ${\left(PS\right)}_c$.

3) For any $\beta >0$ ,there exists $\tilde{\lambda }>0\left(\tilde{\lambda }\le {\lambda }_1\right)$ such that if $0<\lambda <\tilde{\lambda }$ and $c<0$ then $\tilde{J}$ satisfies ${\left(PS\right)}_c$.

Lemma 3.5. Denote ${\tilde{J}}^{\alpha }:=\left\{u\in X_0^s\left(\Omega \right),\tilde{J}\left(u\right)\le \alpha \right\}$. Then for any $m\in N$ , there is ${\epsilon }_m<0$ such that $\gamma \left({\tilde{J}}^{{\epsilon }_m}\right)\ge m$.

Proof. Denote by $X_0^s\left(\Omega \right)$ the closure of $C_0^{\infty }\left(\Omega \right)$ with the respect to norm ${\Vert u\Vert }_{X_0^s\left(\Omega \right)}={\left({\displaystyle \underset{\Omega }{\int }\frac{{\left|u\left(x\right)-u\left(y\right)\right|}^2}{{\left|x-y\right|}^{n+2s}}\text{d}x\text{d}y}\right)}^{\frac{1}{2}}$ , $V\left(x\right)>0$ in $\Omega$. Extending functions in

$X_0^s\left(\Omega \right)$ by 0 outside $\Omega$. Let $X_m$ be a m-dimensional subspace of $X_0^s\left(\Omega \right)$. For any $u\in X_m,u\ne 0$. We write $u=r_{m}w$ with $w\in X_m$ and ${\Vert w\Vert }_{X_0^s\left(\Omega \right)}=1$. From the assumptions of $V\left(x\right)$ , it is easy to see for every $w\in X_m$ with ${\Vert w\Vert }_{X_0^s\left(\Omega \right)}=1$ that there exists $d_m>0$ such that

${\displaystyle \underset{\Omega }{\int }V\left(x\right){\left|w\right|}^p}\text{d}x\ge d_m$ (3.24)

For $0<r_m<T_0$. Since all the norms are equivalent, we get

$\begin{array}{l}\tilde{J}\left(u\right)=J\left(u\right)=\frac{1}{2}{\Vert u\Vert }_{X_0^s\left(\Omega \right)}^2-\frac{\lambda }{p}{\displaystyle \underset{\Omega }{\int }V\left(x\right){\left|u\right|}^p}\text{d}x-\frac{\beta }{2_s^{*}}{\displaystyle \underset{\Omega }{\int }K\left(x\right){\left|u\right|}^{2_s^{*}}}\text{d}x\\ \le \frac{1}{2}{\Vert u\Vert }_{X_0^s\left(\Omega \right)}^2-\frac{\lambda }{p}{\displaystyle \underset{\Omega }{\int }V\left(x\right){\left|u\right|}^p}\text{d}x+\frac{\beta }{2_s^{*}}\left|{\displaystyle \underset{\Omega }{\int }K\left(x\right){\left|u\right|}^{2_s^{*}}}\text{d}x\right|\\ \le \frac{1}{2}{r}_m^2-\lambda c{d}_m+c\beta r_m^{2_s^{*}}:={\epsilon }_m.\end{array}$

Therefore for given $\lambda$ and $\beta$. we can choose $r_m\in \left(0,T_0\right)$ sufficiently small so that $\tilde{J}\left(u\right)\le {\epsilon }_m<0$.#

Let $S_{r_m}=\left\{u\in X_0^s\left(\Omega \right):{\Vert u\Vert }_{X_0^s\left(\Omega \right)}=r_m\right\}$. Then $S_{r_m}\cap X_m\subset {\tilde{J}}^{{\epsilon }_m}$ , Hence by proposition 3.3 (2) and (4) $r\left({\tilde{J}}^{{\epsilon }_m}\right)\ge r\left(S_{r_m}\cap X_m\right)\ge m$.

We denote ${\Gamma }_m=\left\{A\in {\rm A}:\gamma \left(A\right)\ge m\right\}$ and let

$C_m:=\underset{A\in {\Gamma }_m}{\inf }\underset{u\in A}{\sup }J\left(u\right)$ (3.25)

then

$-\infty <C_m\le {\epsilon }_m<0\text{,}m\in N$ (3.26)

because ${\tilde{J}}^{{\epsilon }_m}\in {\Gamma }_m$ and $\tilde{J}$ is bounded from below.

Proposition 3.6. Let $\lambda ,\beta$ be as in Lemma 3.5 (2) and (3). Then all $c_m$ given by (3.25) are critical values of $\tilde{J}$ and $c_m\to 0$ as $m\to 0$.

Proof. Denote $K_{\epsilon }=\left\{u\in X_0^s\left(\Omega \right):\tilde{J}\left(u\right)=c,{\tilde{J}}^{\prime }\left(u\right)=0\right\}$. Then by Lemma 3.4 (2) and (3), if $c<0$ , $K_c$ is compact. It is clear that $C_m\le C_{m+1}$. By (3.26) $C_m<0$. Hence $C_m\to \bar{C}\le 0$. Moreover, since ${\left(PS\right)}_c$ satisfied, it follows from a standard argument (see [11] ) that all $C_m$ are critical values of $\tilde{J}$. Now, we claim that $\bar{c}=0$. If $\bar{c}<0$ because $K_{\bar{c}}$ is compact and $K_{\bar{c}}\in A$ , it follows from Proposition 3.3 (5) that $\gamma \left(K_{\bar{c}}\right)=m_0<+\infty$ and there exists $\delta >0$ such that $\gamma \left(K_{\bar{c}}\right)=\gamma \left(N_{\delta }\left(K_{\bar{c}}\right)\right)=m_0$. By the deformation Lemma [9] , there exists $\epsilon >0\left(\bar{c}+\epsilon <0\right)$ and an odd homeomorphism $\varsigma (\cdot ):X_0^s\left(\Omega \right)\to X_0^s\left(\Omega \right)$ such that

$\varsigma \left({\tilde{J}}^{\bar{c}+\epsilon }\backslash N_{\delta }\left(K_{\bar{c}}\right)\right)\subset {\tilde{J}}^{\bar{c}-\epsilon }$ (3.27)

Since $c_m$ is increasing anad converges to $\bar{c}$. there exists $m\in N$ such that

$c_m>\bar{c}-\epsilon$. (3.28)

And exists a $A\in {\Gamma }_{m+m_0}$ such that

$\underset{u\in A}{\sup }\tilde{J}\left(u\right)<\bar{c}+\epsilon$ (3.29)

By Proposition 3.3 (3), we obtain

$\gamma \left(\overline{A\backslash N_{\delta }\left(K_{\stackrel{¯}{c}}\right)}\right)\ge \gamma \left(A\right)-\gamma \left(N_{\delta }\left(K_{\bar{c}}\right)\right)\ge m$ (3.30)

By Proposition 3.3 (1), we obtain

$\gamma \left(\overline{\varsigma \left(A\backslash N_{\delta }\left(K_{\stackrel{¯}{c}}\right)\right)}\right)\ge m$ (3.31)

therefore

$\varsigma \left(A\backslash N_{\delta }\left(K_{\bar{c}}\right)\right)\in {\Gamma }_m$

consequently, from (3.28), we get

$\underset{u\in \varsigma \left(A\backslash N_{\delta }\left(K_{\bar{c}}\right)\right)}{\sup }\tilde{J}\left(u\right)\ge c_m>\bar{c}-\epsilon$ (3.32)

on the other hand, by (3.27) and (3.29)

$\varsigma \left(A\backslash N_{\delta }\left(K_{\bar{c}}\right)\right)\subset \varsigma \left({\tilde{J}}^{\bar{c}+\epsilon }\backslash N_{\delta }\left(K_{\bar{c}}\right)\right)\subset {\tilde{J}}^{\bar{c}-\epsilon }$ (3.33)

which implies that

$\underset{u\in \varsigma \left(A\backslash N_{\delta }\left(K_{\bar{c}}\right)\right)}{\sup }\tilde{J}\left(u\right)\le \bar{c}-\epsilon$ (3.34)

this contradicts to (3.32).Hence $c_m\to 0$. #

By (1) of Lemma 3.4 $\tilde{J}\left(u\right)=J\left(u\right)$ if $\tilde{J}\left(u\right)<0$. This and Proposition 3.6 give Theorem1.1.

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

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