Bound state solutions for non-autonomous fractional Schr\"{o}dinger-Poisson equations with critical exponent

TL;DR

This paper establishes the existence of bound state solutions for a non-autonomous fractional Schrödinger-Poisson equation with critical exponent, under smallness conditions on potential functions, extending understanding of fractional quantum systems.

Contribution

It proves the existence of bound state solutions for the fractional Schrödinger-Poisson system with critical exponent, considering non-autonomous potentials and smallness conditions.

Findings

Existence of at least one bound state solution under small potential norms.

Extension of fractional Schrödinger-Poisson analysis to critical exponent case.

Conditions on potential functions for solution existence.

Abstract

In this paper, we study the fractional Schr\"{o}dinger-Poisson equation \begin{equation*} \ \left\{\begin{aligned} &(-\Delta)^{s}u+V(x)u+K(x)\phi u=|u|^{2^{\ast}_{s}-2}u, &\mbox{in} \ \mathbb{R}^{3},\\ &(-\Delta)^{s}\phi=K(x)u^{2},&\mbox{in} \ \mathbb{R}^{3}, \end{aligned}\right. \end{equation*} where $s\in (\frac{3}{4},1]$, $2^{\ast}_{s}=\frac{6}{3-2s}$ is the fractional critical exponent, $K\in L^{\frac{6}{6s-3}}(\mathbb{R}^{3})$ and $V\in L^{\frac{3}{2s}}(\mathbb{R}^{3})$ are nonnegative functions. If $\|V\|_{\frac{3}{2s}}+\|K\|_{\frac{6}{6s-3}}$ is sufficiently small, we prove that the equation has at least one bound state solution.

Equations

$$\{\begin{array}{rlr}& (-\Delta {)}^{s}u+V(x)u+K(x)\varphi u=|u{|}^{2_s^{\ast }-2}u,& \text{mbox}in{\mathbb{R}}^3,\\ & (-\Delta {)}^s\varphi =K(x)u^2,& \text{mbox}in{\mathbb{R}}^3,\end{array}$$

$$\{\begin{array}{rlr}& (-\Delta {)}^{s}u+V(x)u+K(x)\varphi u=|u{|}^{2_s^{\ast -2}}u,& \text{mbox}in{\mathbb{R}}^3,\\ & (-\Delta {)}^s\varphi =K(x)u^2,& \text{mbox}in{\mathbb{R}}^3,\end{array}$$

$$(-△{)}^{s}u(x)=c_{s}P.V.{\int }_{{\mathbb{R}}^3}\frac{u(x)-u(y)}{|x-y{|}^{3+2s}}dy,$$

$$c_s=2^{2s}{\pi }^{-\frac{3}{2}}\frac{\Gamma (\frac{3+2s}{2})}{|\Gamma (-s)|}.$$

$$\{\begin{array}{rlr}& -\Delta u+V(x)u+K(x)\varphi u=f(x,u),& \text{mbox}in{\mathbb{R}}^3,\\ & -\Delta \varphi =K(x)u^2,& \text{mbox}in{\mathbb{R}}^3,\end{array}$$

$$\{\begin{array}{rl}& -\Delta u+u+\lambda \varphi u=u^p,\\ & -\Delta \varphi =u^2,\underset{|x|\to +\infty }{\lim }\varphi (x)=0,\end{array}$$

$$\{\begin{array}{rl}& -\Delta u+V(x)u+\lambda \varphi u=\mu |u{|}^{q-1}u+u^5,\\ & -\Delta \varphi =u^2,\end{array}$$

$$\{\begin{array}{rlr}& (-\Delta {)}^{s}u+\lambda \varphi u=g(u),& \text{mbox}in{\mathbb{R}}^3,\\ & (-\Delta {)}^t\varphi =\lambda u^2,& \text{mbox}in{\mathbb{R}}^3,\end{array}$$

$$\{\begin{array}{rlr}& (-\Delta {)}^{s}u+V(x)u+\varphi u=\mu |u{|}^{p-1}+|u{|}^{2_s^{\ast }-2}u,& \text{mbox}in{\mathbb{R}}^3,\\ & (-\Delta {)}^t\varphi =u^2,& \text{mbox}in{\mathbb{R}}^3,\end{array}$$

$$\{\begin{array}{rlr}& (-\Delta {)}^{s}u+V(x)u+\varphi u=\mu |u{|}^{p-1},& \text{mbox}in{\mathbb{R}}^3,\\ & (-\Delta {)}^t\varphi =u^2,& \text{mbox}in{\mathbb{R}}^3,\end{array}$$

$$\{\begin{array}{rlr}& {\epsilon }^{2s}(-\Delta {)}^{s}u+V(x)u+\varphi u=K(x)|u{|}^{p-2}u,& \text{mbox}in{\mathbb{R}}^3,\\ & {\epsilon }^{2s}(-\Delta {)}^s\varphi =u^2,& \text{mbox}in{\mathbb{R}}^3,\end{array}$$

$$0<S_s^{-1}\Vert V{\Vert }_{\frac{3}{2s}}+S_s^{\frac{3}{2s}-3}\Vert K{\Vert }_{\frac{6}{6s-3}}^2\le 2^{\frac{2_s^{\ast }-4}{2_s\ast }}-1,$$

$$S_s=\underset{u\in D^{s,2}({\mathbb{R}}^3)\backslash \{0\}}{\inf }\frac{{\int }_{{\mathbb{R}}^3}|(-\Delta {)}^{\frac{s}{2}}u{|}^{2}dx}{({\int }_{{\mathbb{R}}^3}|u{|}^{2_s^{\ast }}{)}^{\frac{2_s^{\ast }}{2}}}$$

$$H^s({\mathbb{R}}^3)=\left\{u\in L^2({\mathbb{R}}^3):{\int }_{{\mathbb{R}}^3}(1+|\xi {|}^{2s})|\mathcal{F}u(\xi ){|}^{2}d\xi <\infty \right\}$$

$$\Vert u{\Vert }_{H^s}={\left({\int }_{{\mathbb{R}}^3}(|\mathcal{F}u(\xi ){|}^2+|\xi {|}^{2s}|\mathcal{F}u(\xi ){|}^2)d\xi \right)}^{\frac{1}{2}},$$

$$(-\Delta {)}^{s}f={\mathcal{F}}^{-1}(|\xi {|}^{2s}(\mathcal{F}f)),\forall \xi \in {\mathbb{R}}^n.$$

$$\Vert u{\Vert }_{H^s}={\left({\int }_{{\mathbb{R}}^3}(|(-\Delta {)}^{\frac{s}{2}}u{|}^2+|u{|}^2)dx\right)}^{\frac{1}{2}}.$$

$$D^{s,2}({\mathbb{R}}^3)=\left\{u\in L^{2_s^{\ast }}({\mathbb{R}}^3):|\xi {|}^s\mathcal{F}u(\xi )\in L^2({\mathbb{R}}^3)\right\},$$

$$\Vert u{\Vert }_{D^{s,2}}={\left({\int }_{{\mathbb{R}}^3}|(-\Delta {)}^{\frac{s}{2}}u{|}^{2}dx\right)}^{\frac{1}{2}}={\left({\int }_{{\mathbb{R}}^3}|\xi {|}^{2s}|\mathcal{F}u(\xi ){|}^{2}d\xi \right)}^{\frac{1}{2}}.$$

$$S_s=\underset{u\in D^{s,2}({\mathbb{R}}^3)\backslash \{0\}}{\inf }\frac{{\int }_{{\mathbb{R}}^3}|(-\Delta {)}^{\frac{s}{2}}u{|}^{2}dx}{{\left({\int }_{{\mathbb{R}}^3}|u{|}^{2_s^{\ast }}dx\right)}^{\frac{2}{2_s^{\ast }}}}.$$

$$T_u(v)={\int }_{{\mathbb{R}}^3}K(x)u^{2}vdx.$$

$$|T_u(v)|\le \Vert K{\Vert }_{\frac{6}{6s-3}}\Vert u{\Vert }_{\frac{6}{3-2s}}^2\Vert v{\Vert }_{2_s^{\ast }}\le \Vert K{\Vert }_{\frac{6}{6s-3}}{S}_s^{-\frac{3}{2}}\Vert u{\Vert }_{D^{s,2}}^2\Vert v{\Vert }_{D^{s,2}}.$$

$$\eta (a,b)={\int }_{{\mathbb{R}}^3}(-\Delta {)}^{\frac{s}{2}}a\cdot (-\Delta {)}^{\frac{s}{2}}bdx,a,b\in D^{s,2}({\mathbb{R}}^3).$$

$${\int }_{{\mathbb{R}}^3}(-\Delta {)}^{\frac{s}{2}}{\varphi }_u^s(-\Delta {)}^{\frac{s}{2}}vdx={\int }_{{\mathbb{R}}^3}K(x)u^{2}vdx.$$

$${\varphi }_u^s(x)=c_s{\int }_{{\mathbb{R}}^3}\frac{K(y)u^2(y)}{|x-y{|}^{3-2s}}dy,$$

$$c_s=\frac{\Gamma (\frac{3-2s}{2})}{{\pi }^{3/2}{2}^{2s}\Gamma (s)}.$$

$$\Vert {\varphi }_u^s{\Vert }_{D^{s,2}}=\Vert T_u{\Vert }_{\mathcal{L}(D^{s,2},\mathbb{R})}.$$

$$\Vert {\varphi }_u^s{\Vert }_{D^{s,2}}^2$$

$$\le S_s^{-\frac{1}{2}}\Vert K{\Vert }_{\frac{6}{6s-3}}\Vert u{\Vert }_{\frac{6}{3-2s}}^2\Vert {\varphi }_u^s{\Vert }_{D^{s,2}}.$$

$$\Vert {\varphi }_u^s{\Vert }_{D^{s,2}}\le S_s^{-\frac{1}{2}}\Vert K{\Vert }_{\frac{6}{6s-3}}\Vert u{\Vert }_{\frac{6}{3-2s}}^2\le C{S}_s^{-\frac{1}{2}}\Vert K{\Vert }_{\frac{6}{6s-3}}\Vert u{\Vert }_{H^s}.$$

$$(-\Delta {)}^{s}u+V(x)u+K(x){\varphi }_u^{s}u=|u{|}^{2_s^{\ast }-2}u,x\in {\mathbb{R}}^3,$$

Full text

[TABLE]

where $s\in(\frac{3}{4},1]$, $2^{\ast}_{s}=\frac{6}{3-2s}$ is the fractional critical exponent, $K\in L^{\frac{6}{6s-3}}(\mathbb{R}^{3})$ and $V\in L^{\frac{3}{2s}}(\mathbb{R}^{3})$ are nonnegative functions. If $\|V\|_{\frac{3}{2s}}+\|K\|_{\frac{6}{6s-3}}$ is sufficiently small, we prove that the equation has at least one bound state solution.

2010 Mathematics Subjects Classification: 35Q40, 58E30

1. Introduction

In this paper, we are concerned with the existence of bound state solutions for the following fractional Schrödinger-Poisson equation

[TABLE]

where $s\in(\frac{3}{4},1]$, $2^{\ast}_{s}=\frac{6}{3-2s}$ is the fractional critical exponent, $K\in L^{\frac{6}{6s-3}}(\mathbb{R}^{3})$ and $V\in L^{\frac{3}{2s}}(\mathbb{R}^{3})$ are nonnegative functions, $(-\Delta)^{s}$ is the fractional Laplacian defined by

[TABLE]

where

[TABLE]

When $s=1$, (1.1) is the classical Schrödinger-Poisson equation or the more general one

[TABLE]

The Schrödinger-Poisson system has been introduced in [5] as a model describing a quantum particle interacting with a electromagnetic field. In recent years, (1.3) has attracted much attention, we refer to [3, 1, 17, 27, 7, 34, 39, 14] and the references therein.

When $V(x)=1$, $K(x)=\lambda$ and $f(x,u)=u^{p}$, Ruiz [22] studied the problem

[TABLE]

where $u,\phi:\mathbb{R}^{3}\rightarrow\mathbb{R}$ are positive radial functions, $\lambda>0$ and $1<p<5$. The existence and nonexistence are established with different parameters $p$ and $\lambda$. In the case that $p\in(2,3)$, Ruiz developed a new approach consists of minimizing the associated energy functional on a certain manifold which is a combination of Nehari manifold and Pohozaev equality. In the case that $p\in(1,2)$ and $\lambda$ is small enough, the existence results are obtained. Liu and Guo [21] considered the following Schrödinger-Poisson equation with critical growth

[TABLE]

where $q\in(2,5)$, $\lambda>0$. Under some assumptions on the potential $V$, they used the variational methods to prove the existence of positive ground state solutions. In [38], the authors considered (1.3) with $f(u)=u^{5}$ and applied a linking theorem to prove the existence of bound state solutions when $V(x)$, $K(x)$ satisfy some conditions. If $V(x)=1$, $f(x,u)=a(x)|u|^{p-2}+u^{5}$, $p\in(4,6)$, when $K(x)$ and $a(x)$ satisfy some assumptions, Zhang [37] studied the existence of ground state solutions and nodal solutions for (1.3).

If $\phi=0$, (1.1) reduces to a fractional Schrödinger equation, which is a fundmental equation in fractional quantum mechanics [18, 19]. In fact, if one extends the Feynman path integral from the Brownian-like to Lévy-like quantum mechanical paths, the classical Schrödinger equation will change into the fractional Schrödinger equation. In the last decade, the existence of solutions for fractional Schrödinger equations has been investigated by many authors, we refer to [16, 20, 2, 6, 15, 4, 23].

To the best of our knowledge, there are few papers which considered the equation (1.1). In [36], the authors considered a fractional Schrödinger-Poisson system

[TABLE]

where $g(u)$ satisfies the Berestucki-Lions conditions, the existence of positive solutions is proved. Teng [28] studied the existence of ground state solutions for the nonlinear fractional Schrödinger-Poisson equation with critical Sobolev exponent

[TABLE]

where $\mu>0$ is a parameter, $1<p<2_{s}^{\ast}-1=\frac{3+2s}{3-2s}$, $s,t\in(0,1)$ and $2s+2t>3$, the author prove the existence of a nontrivial ground state solutions by using the method of Nehari-Pohozaev manifold and the arguments of Brezis-Nirenberg, the monotonic trick and global compactness Lemma. Shen and Yao [24] applied Nehari-Pohozaev type manifold to prove the existence of a nontrivial least energy solution for the nonlinear fractional Schrödinger-Poisson system

[TABLE]

where $s,t\in(0,1)$, $s<t$ and $2s+2t>3$, $2<p<\frac{3+2s}{3-2s}$. In [35], the authors studied the fractional Schrödinger-Poisson system

[TABLE]

where $\varepsilon>0$, $\frac{3}{4}<s<1$, $4<p<\frac{6}{3-2s}$, $V(x)\in C(\mathbb{R}^{3})\cap L^{\infty}(\mathbb{R}^{3})$ is has positive global minimim, $K(x)\in C(\mathbb{R}^{3})\cap L^{\infty}(\mathbb{R}^{3})$ is positive and has global maximum. They proved the existence of positive ground solution and determined a concrete set related to the potentials $V$ and $K$ as the concentration position of these ground state solutions as $\varepsilon\rightarrow 0$.

In this paper, we are devoted to establishing the existence of bounded state solutions for the fractional Schrödinger-Poisson equation (1.1). We consider the nonlinear term $f(x,u)=|u|^{2_{s}^{\ast}-2}$ without the subcritical part and we deal with the problem (1.1) in $D^{s,2}(\mathbb{R}^{3})$ instead of in $H^{s}(\mathbb{R}^{3})$. Under some assumptions on the potentials $V$ and $K$, we prove that (1.1) can not be solved by constrained minimization on the Nehari manifold. After we prove some concentration-compactness results, the existence of a bounded state solution is obtained by linking theorem.

Our main result is as folows:

Theorem 1**.**

Assume that $V\geq 0$, $V\in L^{\frac{3}{2s}}(\mathbb{R}^{3})$, $K\geq 0$, $K\in L^{\frac{6}{6s-3}}(\mathbb{R}^{3})$ and

[TABLE]

where

[TABLE]

is the best Sobolev constant for the embedding $D^{s,2}(\mathbb{R}^{3})\rightarrow L^{2_{s}^{\ast}}(\mathbb{R}^{3})$, then the problem (1.1) has at least one bound state solution.

This paper is organized as follows. In Section 2, we first present the variational setting of the problem, then we prove some lemmas. In Section 3, we study the Palais-Smale sequence and get a compactness theorem. In Section 4, we prove the existence of bounded state solutions by linking theorem.

2. Variational setting and preliminaries

For $p\in[1,\infty)$, we denote by $L^{p}(\mathbb{R}^{3})$ the usual Lebesgue space with the norm $\|u\|_{p}=\left(\int_{\mathbb{R}^{3}}|u(x)|^{p}dx\right)^{\frac{1}{p}}$. For any $r>0$ and for any $x\in\mathbb{R}^{3}$, $B_{r}(x)$ denotes the ball of radius $r$ centered at $x$. For any $s\in(0,1)$, we recall some definitions of fractional Sobolev spaces $H^{s}(\mathbb{R}^{3})$ and the fractional Laplacian $(-\Delta)^{s}$, for more details, we refer to [12]. $H^{s}(\mathbb{R}^{3})$ is defined as follows

[TABLE]

with the norm

[TABLE]

where $\mathcal{F}u$ denotes the Fourier transform of $u$. As usual $H^{-s}(\mathbb{R}^{3})$ denotes the dual of $H^{s}(\mathbb{R}^{3})$. By $\mathcal{S}(\mathbb{R}^{n})$, we denote the Schwartz space of rapidly decaying $C^{\infty}$ functions in $\mathbb{R}^{n}$. For $u\in\mathcal{S}(\mathbb{R}^{n})$ and $s\in(0,1)$, $(-\Delta)^{s}$ is defined by

[TABLE]

In fact, (2.2) is equivalent to (1.2), see [30]. By Plancherel’s theorem, we have $\|\mathcal{F}u\|_{2}=\|u\|_{2}$, $\||\xi|^{s}\mathcal{F}u\|_{2}=\|(-\Delta)^{\frac{s}{2}}u\|$. Then by (2.1), we get the equivalent norm

[TABLE]

From Theorem 6.5 and Corollary 7.2 in [12], it is known that the space $H^{s}(\mathbb{R}^{3})$ is continuously embedded in $L^{q}(\mathbb{R}^{3})$ for any $q\in[2,2^{\ast}_{s}]$ and the embedding $H^{s}(\mathbb{R}^{3})\hookrightarrow L^{q}(\mathbb{R}^{3})$ is locally compact for $q\in[1,2^{\ast}_{s})$, where $2^{\ast}_{s}=\frac{6}{3-2s}$ is the fractional critical Sobolev exponent.

For $s\in(0,1)$, the fractional Sobolev space $D^{s,2}(\mathbb{R}^{3})$ is defined as follows

[TABLE]

which is the completion of $C^{\infty}_{0}(\mathbb{R}^{3})$ with respect to the norm

[TABLE]

The best Sobolev constant is defined by

[TABLE]

By $C$, we denote the generic constants, which may change from line to line. We consider the variational setting of (1.1). For $u\in D^{s,2}(\mathbb{R}^{3})$, define the linear operator $T_{u}:D^{s,2}(\mathbb{R}^{3})\rightarrow\mathbb{R}$ as

[TABLE]

For $K\in L^{\frac{6}{6s-3}}(\mathbb{R}^{3})$, by the Hölder inequality, we have

[TABLE]

Set

[TABLE]

It is clear that $\eta(a,b)$ is bilinear, bounded and coercive. The Lax-Milgram theorem implies that for every $u\in D^{s,2}(\mathbb{R}^{3})$, there exists a unique $\phi^{s}_{u}\in D^{s,2}(\mathbb{R}^{3})$ such that $T_{u}(v)=\eta(\phi^{s}_{u},v)$ for any $v\in D^{s,2}(\mathbb{R}^{3})$, that is

[TABLE]

Therefore, $(-\Delta)^{s}\phi^{s}_{u}=K(x)u^{2}$ in a weak sense. For $x\in\mathbb{R}^{3}$, we have

[TABLE]

which is the Riesz potential [25], where

[TABLE]

Moreover,

[TABLE]

Since $K\in L^{\frac{6}{6s-3}}(\mathbb{R}^{3})$, by (2.5), (2.4),

[TABLE]

Then

[TABLE]

Substituting $\phi^{s}_{u}$ in (1.1), we have the fractional Schr$\ddot{\mbox{o}}$dinger equation

[TABLE]

The energy functional $I:D^{s,2}(\mathbb{R}^{3})\rightarrow\mathbb{R}$ corresponding to problem (2.10) is defined by

[TABLE]

It is easy to see that $I$ is well defined in $D^{s,2}(\mathbb{R}^{3})$ and $I\in C^{1}(D^{s,2}(\mathbb{R}^{3}),\mathbb{R})$, and

[TABLE]

for any $v\in D^{s,2}(\mathbb{R}^{3})$.

Definition 1**.**

*(1) We call $(u,\phi)\in D^{s,2}(\mathbb{R}^{3})\times D^{s,2}(\mathbb{R}^{3})$ is a weak solution of (1.1) if $u$ is a weak solution of (2.10).

(2)We call $u$ is a weak solution of (2.10) if*

[TABLE]

for any $v\in D^{s,2}(\mathbb{R}^{3})$.

Let $\Phi:D^{s,2}(\mathbb{R}^{3})\rightarrow D^{s,2}(\mathbb{R}^{3})$ be the operator

[TABLE]

and set

[TABLE]

Lemma 1**.**

Let $u_{n}\in D^{s,2}(\mathbb{R}^{3})$ be such that $u_{n}\rightharpoonup 0$ in $D^{s,2}(\mathbb{R}^{3})$. Then, up to subsequences, $u_{n}\rightarrow 0$ in $L_{loc}^{p}(\mathbb{R}^{3})$ for any $p\in[2,2_{s}^{\ast})$.

Proof.

By the fractional Hardy inequality (see (1.1) in [32]),

[TABLE]

Let $r>0$ and let $A\subset\mathbb{R}^{3}$ be such that $A\subset B_{r}(0)$. We have

[TABLE]

this together with $u_{n}\in D^{s,2}(\mathbb{R}^{3})$ yield that $u_{n}$ is bounded in $H^{s}(A)$. Since $u_{n}\rightharpoonup 0$ in $D^{s,2}(\mathbb{R}^{3})$, then up to a subsequence, $u_{n}\rightarrow 0$ in $L^{2}(A)$. By $L^{p}$ interpolation inequality, for all $p\in[2,2_{s}^{\ast})$,

[TABLE]

where $\frac{\alpha}{2}+\frac{1-\alpha}{2_{s}^{\ast}}=\frac{1}{p}$. Then the conclusion follows. ∎

Lemma 2**.**

*Let $K\in L^{\frac{6}{6s-3}}(\mathbb{R}^{3})$ with $s>\frac{1}{2}$, then $\Phi$ and $N$ have following properties:

(1) $\Phi:D^{s,2}(\mathbb{R}^{3})\mapsto D^{s,2}(\mathbb{R}^{3})$ is continuous and maps bounded sets into bounded sets;

(2) If $u_{n}\rightharpoonup u$ in $D^{s,2}(\mathbb{R}^{3})$, then up to a subsequence $\Phi(u_{n})\rightarrow\Phi(u)$ in $D^{s,2}(\mathbb{R}^{3})$ and $N(u_{n})\rightarrow N(u)$.*

Proof.

(1) By (2.5), we see that

[TABLE]

Assume that $u_{n}\in D^{s,2}(\mathbb{R}^{3})$, $u_{n}\rightarrow u$ in $D^{s,2}(\mathbb{R}^{3})$. Since $D^{s,2}(\mathbb{R}^{3})\hookrightarrow L^{\frac{6}{3-2s}}(\mathbb{R}^{3})$, we have $||u_{n}-u||\rightarrow 0$ in $L^{\frac{6}{3-2s}}(\mathbb{R}^{3})$, then $||u_{n}-u||^{\frac{3}{2s}}\rightarrow 0$ in $L^{\frac{4s}{3-2s}}(\mathbb{R}^{3})$. Since $K\in L^{\frac{6}{6s-3}}(\mathbb{R}^{3})$, we have $\int_{\mathbb{R}^{3}}|K(x)|^{\frac{3}{2s}}|u_{n}-u|^{\frac{3}{2s}}dx\rightarrow 0$. For any $v\in D^{s,2}(\mathbb{R}^{3})$, by Hölder inequality,

[TABLE]

Since $v$ is arbitrary, we get $\lim_{n\rightarrow\infty}\|T_{u_{n}}-T_{u}\|_{\mathcal{L}(D^{s,2},\mathbb{R})}=0$. By (2.9), $\Phi$ maps bounded sets into bounded sets.

(2) By (2.12), we only need to show that

[TABLE]

Let $\varepsilon$ be any given positive number. There exists $r_{\varepsilon}>0$ large enough such that $\|K\|_{L^{\frac{6}{6s-3}}(\mathbb{R}^{3}\backslash B(0,r_{\varepsilon}))}<\varepsilon$. For any $v\in D^{s,2}(\mathbb{R}^{3})$, by Hölder inequality,

[TABLE]

Set $B_{M}=\{x\in B(0,r_{\varepsilon}):K(x)>M\}$. Since $K\in L^{\frac{6}{6s-3}}(\mathbb{R}^{3})$, then the volume $|B_{M}|\rightarrow 0$ as $M\rightarrow\infty$. By Lemma 1, up to a subsequence, $u_{n}\rightarrow u$ in $L_{loc}^{\frac{12}{3+2s}}(\mathbb{R}^{3})$. Then for large $M$, we have

[TABLE]

Therefore,

[TABLE]

Since $D^{s,2}(\mathbb{R}^{3})\hookrightarrow L^{2_{s}^{\ast}}(\mathbb{R}^{3})$ and $\phi_{u_{n}}^{s}\rightarrow\phi_{u}^{s}$ in $D^{s,2}(\mathbb{R}^{3})$, then by Hölder inequality, we have

[TABLE]

By the same argument as in (2) and (2), replacing $v$ by $\phi_{u_{n}}^{s}$, we get

[TABLE]

From (2.15), (2.16) and

[TABLE]

it follows that $N(u_{n})\rightarrow N(u)$ as $n\rightarrow\infty$. ∎

We consider the limiting problem

[TABLE]

It is known that every positive solution of (2.17) assumes the form (see [8])

[TABLE]

By [10], the best constant $S_{s}$ is only attained at $\Psi_{\delta,x_{0},\kappa}(x)$. It is known that [9] if we choose

[TABLE]

then

[TABLE]

satisfies (2.17), and

[TABLE]

The energy functional related to (2.17) is defined on $D^{s,2}(\mathbb{R}^{3})$ by

[TABLE]

Then

[TABLE]

We define the Nehari manifold corresponding to (1.1) and (2.17) by $\mathcal{N}$ and $\mathcal{N}_{\infty}$:

[TABLE]

Set

[TABLE]

Lemma 3**.**

Assume that $K\in L^{\frac{6}{6s-3}}(\mathbb{R}^{3})$ and $V\in L^{\frac{3}{2s}}(\mathbb{R}^{3})$ are nonnegative functions. Then $m=m_{\infty}$ and $m$ can’t be attained when $\|V\|_{\frac{3}{2s}}+\|K\|_{\frac{6}{6s-3}}>0$.

Proof.

Let $u\in D^{s,2}(\mathbb{R}^{3})$ and define the function

[TABLE]

where $a=\int_{\mathbb{R}^{3}}(|(-\Delta)^{\frac{s}{2}}u|^{2}+V(x)u^{2})dx$, $b=\int_{\mathbb{R}^{3}}K(x)\phi_{u}^{s}u^{2}dx$, $c=\int_{\mathbb{R}^{3}}|u|^{2_{s}^{\ast}}dx$. Since $s>\frac{3}{4}$, by a similar argument as the proof of Lemma 3.3 in [22], we know that there exist unique $t(u)>0$, $s(u)>0$ such that $t(u)u\in\mathcal{N},s(u)u\in\mathcal{N}_{\infty}$, and $I(t(u)u)=\max_{t>0}I(tu)$, $I_{\infty}(s(u)u)=\max_{t>0}I_{\infty}(tu)$. For any $u\in\mathcal{N}$,

[TABLE]

Since $u\in\mathcal{N}$, it is easy to see that $I(tu)$ attains its maximum at $t=1$, then for any $t>0$,

[TABLE]

By (2), (2.25), we have

[TABLE]

Assume that $w$ is a positive solution of (2.17) centered at zero, $(z_{n})_{n}\subset\mathbb{R}^{3}$ satisfying $|z_{n}|\rightarrow\infty$ as $n\rightarrow\infty$, $w_{n}(\cdot)=w(\cdot-z_{n})$ and $t_{n}=t(w_{n})$. It is clear that $\|w_{n}\|_{D^{s,2}}=\|w\|_{D^{s,2}}$, $\|w_{n}\|_{2_{s}^{\ast}}=\|w\|_{2_{s}^{\ast}}$ and $w_{n}\rightharpoonup 0$ in $D^{s,2}(\mathbb{R}^{3})$ as $n\rightarrow\infty$. It follows from Lemma 2 that

[TABLE]

Since $D^{s,2}(\mathbb{R}^{3})\hookrightarrow L^{2_{s}^{\ast}}(\mathbb{R}^{3})$, then $w_{n}^{2}\rightharpoonup 0$ in $L^{\frac{2_{s}^{\ast}}{2}}(\mathbb{R}^{3})$, this together with $V\in L^{\frac{3}{2s}}(\mathbb{R}^{3})$ yield that

[TABLE]

By (2.27), (2.28), we have

[TABLE]

Since $w_{n}\in\mathcal{N}_{\infty}$ and $t_{n}w_{n}\in\mathcal{N}$, then

[TABLE]

and

[TABLE]

From (2.30) and (2.31), it follows that

[TABLE]

By (2.27), (2.28) and (2.31), for sufficiently large $n$, we have

[TABLE]

Then $(t_{n})_{n}$ is bounded. This together with $\|w_{n}\|_{D^{s,2}}=\|w\|_{D^{s,2}}$ and (2.32) yield that $\lim_{n\rightarrow\infty}t_{n}=1$. By (2.29), we see that $\lim_{n\rightarrow\infty}I(u_{n})=m_{\infty}$. Therefore by (2.22) and (2.26), $m=m_{\infty}$.

Now we will prove that $m$ can’t be attained. Arguing by contradiction, we let $\bar{u}\in\mathcal{N}$ be a function such that $I(\bar{u})=m=m_{\infty}$. Then, since $s(\bar{u})\bar{u}\in\mathcal{N}_{\infty}$, we have

[TABLE]

We deduce that

[TABLE]

then $\bar{u}\in\mathcal{N}_{\infty}$, $I_{\infty}(\bar{u})=m_{\infty}$. Therefore $\bar{u}$ admits the form (2.18), contradicting $\int_{\mathbb{R}^{3}}V(x)\bar{u}^{2}dx=0$ and $N(\bar{u})=0$. ∎

3. A compactness theorem

Lemma 4**.**

(see Lemma 2.5 in [33]) Let $\{u_{n}\}\subset H_{loc}^{s}(\mathbb{R}^{N})$ be a bounded sequence of functions such that $u_{n}\rightharpoonup 0$ in $H^{s}(\mathbb{R}^{N})$. Suppose that there exist a bounded open set $Q\subset\mathbb{R}^{N}$ and a positive constant $\gamma>0$ such that

[TABLE]

Moreover suppose that

[TABLE]

where $\chi_{n}\in H^{-s}(\mathbb{R}^{N})$ and

[TABLE]

where $U$ is an open neighborhood of $Q$ and $\varepsilon_{n}$ is a sequence of positive numbers converging to 0. Then there exist a sequence of points $\{y_{n}\}\subset\mathbb{R}^{N}$ and a sequence of positive numbers $\{\sigma_{n}\}$ such that

[TABLE]

converges weakly in $D^{s,2}(\mathbb{R}^{N})$ to a nontrivial solution of

[TABLE]

morover,

[TABLE]

Lemma 5**.**

Let $\{u_{n}\}$ be a Palais-Smale sequence for $I$, such that $u_{n}\in C_{0}^{\infty}(\mathbb{R}^{3})$ and

[TABLE]

Then there exist a sequence of points $\{y_{n}\}\subset\mathbb{R}^{3}$ and a sequence of positive numbers $\{\sigma_{n}\}$ such that

[TABLE]

converges weakly in $D^{s,2}(\mathbb{R}^{3})$ to a nontrival solution of (2.17) and

[TABLE]

[TABLE]

Proof.

Since $u_{n}\rightharpoonup 0\ in\ D^{s,2}(\mathbb{R}^{3})$, by (2.28) and (2) of Lemma 2, we get

[TABLE]

For any $h\in D^{s,2}(\mathbb{R}^{3})$, by the Hölder inequality,

[TABLE]

From $u_{n}\rightharpoonup 0$ in $L^{2_{s}^{\ast}}(\mathbb{R}^{3})$, we have $u_{n}^{\frac{6}{3+2s}}\rightharpoonup 0$ in $L^{\frac{3+2s}{3-2s}}(\mathbb{R}^{3})$. This together with $V^{\frac{6}{3+2s}}\in L^{\frac{3+2s}{4s}}(\mathbb{R}^{3})$ yield that

[TABLE]

Since $D^{s,2}(\mathbb{R}^{3})\hookrightarrow L^{2_{s}^{\ast}}(\mathbb{R}^{3})$, then $\|h\|_{2_{s}^{\ast}}\leq C\|h\|_{D^{s,2}}$. So

[TABLE]

By the similar argument as that in the proof of (2) in Lemma 2, we have

[TABLE]

Therefore

[TABLE]

[TABLE]

Put $\sigma_{n}=\|u_{n}\|_{2}^{1/s}$ and define

[TABLE]

From Proposition 3.6 in [12], up to a positive constant, it follows that

[TABLE]

Then we can get

[TABLE]

By the invariance of these two norms with respect to rescaling $u\rightarrow r^{\frac{3-2s}{2}}u(r\cdot)$, note that (3.7), we have

[TABLE]

By the scale change,

[TABLE]

If $\tilde{u}_{n}\rightharpoonup\tilde{u}\neq 0$ in $D^{s,2}(\mathbb{R}^{3})$, we are done: the wanted functions $v_{n}$ and $v$ are precisely $\tilde{u}_{n}$ and $\tilde{u}$, and $y_{n}=0$. In fact, since $\tilde{u}_{n}\rightharpoonup\tilde{u}$ in $D^{s,2}(\mathbb{R}^{3})$, by the Br$\acute{\mbox{e}}$zis-Lieb Lemma [40],

[TABLE]

From (3.6), (3.7), (3.10) and (3.12), it follows that

[TABLE]

By (3.9),

[TABLE]

If $\tilde{u}_{n}\rightharpoonup 0$ in $D^{s,2}(\mathbb{R}^{3})$. We cover $\mathbb{R}^{3}$ with balls of radius 1 in such a way that each point of $\mathbb{R}^{3}$ is contained in at most 4 balls. Set

[TABLE]

We claim that there exists $\gamma>0$ such that

[TABLE]

Otherwise, by Lemma 2.2 in [11], we have

[TABLE]

By (3.7), $\langle I^{\prime}_{\infty}(\tilde{u}_{n}),\tilde{u}_{n}\rangle=o(1)$. This together with (3.15) yield that

[TABLE]

Then by (3.9), we have $\|u_{n}\|_{D^{s,2}}=o(1)$. This contradicts $u_{n}\nrightarrow 0$ strongly in $D^{s,2}(\mathbb{R}^{3})$.

We choose $\tilde{y}_{n}$ such that

[TABLE]

and let

[TABLE]

By (3.14), (3.16) and (3.17),

[TABLE]

If $w_{n}\rightharpoonup w\neq 0$ in $D^{s,2}(\mathbb{R}^{3})$, we are done: the wanted functions $v_{n}$ and $v$ are precisely $w_{n}$ and $w$, and $y_{n}=\sigma_{n}\tilde{y}_{n}$. Suppose that $w_{n}\rightharpoonup 0$ in $D^{s,2}(\mathbb{R}^{3})$. By Lemma 1, $w_{n}\rightarrow 0$ in $L_{loc}^{2}(\mathbb{R}^{3})$. Then by (3.18), we have

[TABLE]

By (3.10), (3.17), the invariance of the $H^{s}(\mathbb{R}^{3})$ and $L^{2_{s}^{\ast}}(\mathbb{R}^{3})$ norms by translation, we apply Lemma 4 to $w_{n}$ and to claim that there exists a sequence $\eta_{n}\rightarrow 0$ of numbers and a sequence $y_{n}$ of points: $y_{n}\rightarrow\bar{y}\in\overline{B_{1}(0)}$ such that

[TABLE]

By (3.9), it is easy to see that

[TABLE]

Since $v_{n}\rightharpoonup v$ in $D^{s,2}(\mathbb{R}^{3})$, by the Br$\acute{\mbox{e}}$zis-Lieb Lemma [40], (3.6), (3.7) and (3.19), we have

[TABLE]

So $v_{n}$ and $v$ are the wanted functions. ∎

Theorem 2**.**

Suppose that $V\in L^{\frac{3}{2s}}(\mathbb{R}^{3})$, $V\geq 0$ and $K\in L^{\frac{6}{6s-3}}(\mathbb{R}^{3})$, $K\geq 0$. Let $\{u_{n}\}$ be a Palsis-Smale sequence of $I$ at level $c$. Then up to a subsequence, $u_{n}\rightharpoonup\bar{u}$. Moreover, there exists a number $k\in\mathbb{N}$, $k$ sequences of points $\{y_{n}^{i}\}_{n}\subset\mathbb{R}^{3}$, $1\leq i\leq k$, $k$ sequences of positive numbers $\{\sigma_{n}^{i}\}_{n}$, $1\leq i\leq k$, $k$ sequences of functions $\{u_{n}^{i}\}\subset D^{s,2}(\mathbb{R}^{3})$, $1\leq i\leq k$, such that

[TABLE]

where $u^{i}$, $1\leq i\leq k$, are solutions of (2.17). Moreover as $n\rightarrow\infty$,

[TABLE]

[TABLE]

Proof.

Since $u_{n}$ is a Palais-Smale sequence for $I$, it is easy to show that $u_{n}$ is bounded in $D^{s,2}(\mathbb{R}^{3})$ and in $L^{2_{s}^{\ast}}(\mathbb{R}^{3})$. Up to a subsequence, we assume that $u_{n}\rightharpoonup\bar{u}$ in $D^{s,2}(\mathbb{R}^{3})$ and in $L^{2_{s}^{\ast}}(\mathbb{R}^{3})$ as $n\rightarrow\infty$. By the weak convergence, we have $I^{\prime}(\bar{u})=0$. Note that $V\in L^{\frac{3}{2s}}(\mathbb{R}^{3})$, $V\geq 0$ and $K\in L^{\frac{6}{6s-3}}(\mathbb{R}^{3})$, $K\geq 0$, by Proposition 5.1.1 in [13] and Lemma 2.7 in [29], we have $\bar{u}\in L^{\infty}(\mathbb{R}^{3})$. Set $z_{n}^{1}=u_{n}-\bar{u}$, then $z_{n}^{1}\rightharpoonup 0$ in $D^{s,2}(\mathbb{R}^{3})$. If $z_{n}^{1}\rightarrow 0$ in $D^{s,2}(\mathbb{R}^{3})$, we are done. Suppose that $z_{n}^{1}=u_{n}-\bar{u}\nrightarrow 0$ in $D^{s,2}(\mathbb{R}^{3})$, for any $v\in C_{0}^{\infty}(\mathbb{R}^{3})$,

[TABLE]

Since $\bar{u}\in L^{\infty}(\mathbb{R}^{3})$, by arguments similar to those of Lemma 8.9 in [40], we get

[TABLE]

Since $u_{n}\rightharpoonup\bar{u}$ in $D^{s,2}(\mathbb{R}^{3})$, it is easy to see that

[TABLE]

Then

[TABLE]

From (3), (3.21) and (3.22), it follows that $\langle I^{\prime}(z_{n}^{1}),v\rangle=o(1)v$, then $z_{n}^{1}$ is a Palais-Smale sequence for $I$.

By the definition of $D^{s,2}(\mathbb{R}^{3})$, for any $n$, there exists $\tilde{z}_{n}^{1}\in C_{0}^{\infty}(\mathbb{R}^{3})$ such that

[TABLE]

Thus $\bar{z}_{n}^{1}$ is a Palais-Smale sequence, and

[TABLE]

[TABLE]

By a similar argument to that in proof of Lemma 2, we get that as $n\rightarrow\infty$,

[TABLE]

By (3.23), (3.24), (3.25) and the Br$\acute{\mbox{e}}$zis-Lieb Lemma, we conclude that

[TABLE]

[TABLE]

[TABLE]

Then $\tilde{z}_{n}^{1}$ is a Palais-Smale sequence for $I$. Since $z_{n}^{1}\rightharpoonup 0$ in $D^{s,2}(\mathbb{R}^{3})$ and $z_{n}^{1}\nrightarrow 0$, we have that $\tilde{z}_{n}^{1}\rightharpoonup 0$ in $D^{s,2}(\mathbb{R}^{3})$ and $\tilde{z}_{n}^{1}\nrightarrow 0$. Thus $\tilde{z}_{n}^{1}$ satisfies the assumptions of Lemma 5, so there exist a sequence of points $\{x_{n}^{1}\}$ and a sequence of positive numbers $\eta_{n}^{1}$ such that

[TABLE]

converges weakly in $D^{s,2}(\mathbb{R}^{3})$ to a nontrival solution $u^{1}$ of (2.17) and by (3.1), (3.2), (3.26) and (3.27), we have

[TABLE]

[TABLE]

Iterating this procedure, we get Palais-Smale sequences of functions

[TABLE]

and

[TABLE]

such that $z_{n}^{j}=\tilde{z}_{n}^{j}+(z_{n}^{j}-\tilde{z}_{n}^{j})$ and

[TABLE]

We can obtain sequences of points $\{x_{n}\}^{j}\subset\mathbb{R}^{3}$ and sequences of numbers $\eta_{n}^{j}$ such that

[TABLE]

converges weakly in $D^{s,2}(\mathbb{R}^{3})$ to a nontrivial solution $u^{j}$ of (2.17). Furthermore, by (3.29), (3.30) and (3.31), we have

[TABLE]

and

[TABLE]

By the definition of $S_{s}$, we have $\|u^{j}\|_{D^{s,2}}^{2}\geq S_{s}\|u^{j}\|_{2_{s}^{\ast}}^{2}$. Since $u^{j}$ is a solution of (2.17), then we have

[TABLE]

This together with the boundedness of $u_{n}$ and (3) yield that the iteration must terminate at some $k>0$. The proof is complete. ∎

Corollary 1**.**

Assume that $\{u_{n}\}\subset D^{s,2}(\mathbb{R}^{3})$ satisfies the assumption of the Theorem 2 with $c\in(\frac{s}{3}S_{s}^{\frac{3}{2s}},\frac{2s}{3}S_{s}^{\frac{3}{2s}})$, then $\{u_{n}\}$ contains a subsequence strongly convergent in $D^{s,2}(\mathbb{R}^{3})$.

Proof.

By Lemma 3, any nontrival solution $u$ have energy $I(u)>\frac{s}{3}S_{s}^{\frac{3}{2s}}$. From (3.15) in [9], any sign-changing solution $v$ of (2.17) satisfies $I_{\infty}(v)\geq\frac{2s}{3}S_{s}^{\frac{3}{2s}}$. Since any positive solution of (2.17) has energy $\frac{s}{3}S_{s}^{\frac{3}{2s}}$, by Theorem 2, we get the conclusion. ∎

Corollary 2**.**

If $\{u_{n}\}$ is a minimizing sequence for $I$ on $N$, then there exist a sequence of points $\{y_{n}\}\subset\mathbb{R}^{3}$, a sequence of positive numbers $\{\delta_{n}\}\subset\mathbb{R}^{+}$ and a sequence $\{w_{n}\}\subset D^{s,2}(\mathbb{R}^{3})$ such that

[TABLE]

where $\psi_{\delta_{n},y_{n}}(x)$ are functions defined in (2.19) and $w_{n}\rightarrow 0$ strongly in $D^{s,2}(\mathbb{R}^{3})$.

Proof.

The result follows from Theorem 2 and Lemma 3. ∎

4. Proof of the main result

Set

[TABLE]

and define

[TABLE]

[TABLE]

where

[TABLE]

Lemma 6**.**

If $|y|\geq\frac{1}{2}$, then

[TABLE]

Proof.

From (2.19), it follows that

[TABLE]

By (3.8) and Proposition 2.2 in [12], we have

[TABLE]

By (4), for every $\varepsilon>0$, there exists a $\hat{\delta}$ such that $\forall\delta\in(0,\hat{\delta}]$,

[TABLE]

Then

[TABLE]

If $\varepsilon$ is small enough, for $|y|\geq\frac{1}{2}$ and $x\in B_{\varepsilon}(y)$,

[TABLE]

Then by (2.20),

[TABLE]

By (4.2) and (4), we have

[TABLE]

∎

Define

[TABLE]

and define

[TABLE]

Lemma 7**.**

Suppose that $V\geq 0$, $V\in L^{\frac{3}{2s}}(\mathbb{R}^{3})$, $K\geq 0$, $K\in L^{\frac{6}{6s-3}}(\mathbb{R}^{3})$ and $\|V\|_{\frac{3}{2s}}+\|K\|_{\frac{6}{6s-3}}>0$. Then

[TABLE]

Proof.

It is obvious that $c_{0}\geq\frac{s}{3}S_{s}^{\frac{3}{2s}}$. To prove $c_{0}>\frac{s}{3}S_{s}^{\frac{3}{2s}}$, we argue by contradiction. Suppose that there exists a sequence $\{u_{n}\}\subset\mathcal{N}$ such that

[TABLE]

[TABLE]

By Corollary 2, there exist a sequence of points $\{y_{n}\}\subset\mathbb{R}^{3}$, a sequence of positive numbers $\{\delta_{n}\}\subset\mathbb{R}^{+}$ and a sequence of functions $\{w_{n}\}\subset D^{s,2}(\mathbb{R}^{3})$ converging to 0 in $D^{s,2}(\mathbb{R}^{3})$ such that

[TABLE]

Since $w_{n}\rightarrow 0$ in $D^{s,2}(\mathbb{R}^{3})$, then for $n$ big enough, we have

[TABLE]

From (4.4) and (4), it follows that

[TABLE]

[TABLE]

For $\delta_{n}$, up to a sunsequence, one of these cases occurs

(a) $\delta_{n}\rightarrow\infty$ as $n\rightarrow\infty$

(b) $\delta_{n}\rightarrow\tilde{\delta}\neq 0$ as $n\rightarrow\infty$

(c) $\delta_{n}\rightarrow 0$ and $y_{n}\rightarrow\bar{y}$, $\bar{y}<\frac{1}{2}$ as $n\rightarrow\infty$

(d) $\delta_{n}\rightarrow 0$ as $n\rightarrow\infty$ and $|y_{n}|\geq\frac{1}{2}$ for $n$ large.

We will prove that none of the possibilities (a)-(d) can be true. If (a) holds,

[TABLE]

This contradicts (ii). If (b) holds, then $|y_{n}|\rightarrow+\infty$, otherwise $\psi_{\delta_{n},y_{n}}$ would converge strongly in $D^{s,2}(\mathbb{R}^{3})$, so $u_{n}$ would converge strongly in $D^{s,2}(\mathbb{R}^{3})$ against Lemma 3. Then we have

[TABLE]

This contradicts (ii). If (c) holds, then

[TABLE]

This contradicts (ii). If (d) holds, by Lemma 6, we have

[TABLE]

This contradicts (i). ∎

Define $\theta:D^{s,2}(\mathbb{R}^{3})\backslash\{0\}\rightarrow\mathcal{N}$:

[TABLE]

where $t(u)$ is the unique positive number such that $t(u)u\in\mathcal{N}$. Let $T$ be the operator

[TABLE]

defined by

[TABLE]

Lemma 8**.**

Assume that $V\geq 0$, $V\in L^{\frac{3}{2s}}(\mathbb{R}^{3})$, $K\geq 0$, $K\in L^{\frac{6}{6s-3}}(\mathbb{R}^{3})$. Then for any $\varepsilon>0$, there exist $\delta_{1}=\delta_{1}(\varepsilon)$, $\delta_{2}=\delta_{2}(\varepsilon)$ such that

[TABLE]

for any $y\in\mathbb{R}^{3}$ and $\delta\in(0,\delta_{1}]\cup[\delta_{2},+\infty)$.

Proof.

Since $V\in L^{\frac{3}{2s}}(\mathbb{R}^{3})$, then for any $\varepsilon>0$, there exists $R>0$ such that

[TABLE]

From $\lim_{\delta\rightarrow+\infty}\sup_{y\in\mathbb{R}^{3}}|\psi_{\delta,y}|=0$, it follows that $\lim_{\delta\rightarrow+\infty}\int_{B_{R}(0)}|\psi_{\delta,y}|^{2_{s}^{\ast}}dx=0$. Therefore there exists $\delta_{2}=\delta_{2}(\varepsilon)$ such that

[TABLE]

By (4.7) and (4.8), we have

[TABLE]

Since $V\in L^{\frac{3}{2s}}(\mathbb{R}^{3})$, for any $\varepsilon>0$, there exists $r>0$ small enough such that

[TABLE]

By (2.19),

[TABLE]

so there is a $\delta_{1}=\delta_{1}(\varepsilon)$ such that for any $\delta\in(0,\delta_{1}]$, there holds

[TABLE]

By (4.10) and (4.11), we get

[TABLE]

From (4) and (4), it follows that

[TABLE]

Similarly, we can get that

[TABLE]

Since $\psi_{\delta,y}\in\mathcal{N}_{\infty}$, it is easy to show that there exists $t_{\delta,y}=t(\psi_{\delta,y})$ such that $t_{\delta,y}\psi_{\delta,y}\in\mathcal{N}$. By an argument analogous to (2.27)-(2.32), we verify that for any $y\in\mathbb{R}^{3}$,

[TABLE]

By (4.13), (4.14) and (4.15), we obtain

[TABLE]

for any $y\in\mathbb{R}^{3}$ and $\delta\in(0,\delta_{1}]\cup[\delta_{2},+\infty)$. ∎

Lemma 9**.**

Assume that $V\geq 0$, $V\in L^{\frac{3}{2s}}(\mathbb{R}^{3})$, $K\geq 0$, $K\in L^{\frac{6}{6s-3}}(\mathbb{R}^{3})$. Then for any fixed $\delta>0$,

[TABLE]

Proof.

It is obvious that $\lim_{|y|\rightarrow+\infty}\int_{\mathbb{R}^{3}}V(x)\psi_{\delta,y}^{2}dx=0$, $\lim_{|y|\rightarrow+\infty}N(\psi_{\delta,y})=0$. By a similar argument to (2.29)-(2.32), we have

[TABLE]

where $t_{\delta,y}$ is the unique positive number such that $t_{\delta,y}\psi_{\delta,y}\in\mathcal{N}$. Then

[TABLE]

∎

By Lemma 7, there exist a positive number $\mu$ such that $\frac{s}{3}S_{s}^{\frac{3}{2s}}+\mu<c_{0}$.

Lemma 10**.**

*There is a $\delta_{1}:0<\delta<\frac{1}{2}$ such that

$(a)$ $I(\theta\circ T(y,\delta))<\frac{s}{3}S_{s}^{\frac{3}{2s}}+\mu$,   $\forall y\in\mathbb{R}^{3}$

$(b)$ $\gamma(\theta\circ T(y,\delta))<\frac{1}{2}$,   $\forall y:|y|<\frac{1}{2}$

$(c)$ \big{|}\beta(\theta\circ T(y,\delta))-\frac{y}{|y|}\big{|}<\frac{1}{4},   $\forall y:|y|\geq\frac{1}{2}$.*

Proof.

By Lemma 8, $(a)$ holds. By (3.8) and Proposition 2.2 in [12],

[TABLE]

this together with (4.15) yield that $(b)$ holds. By Lemma 6 and (4.15), we get $(c)$. ∎

Lemma 11**.**

*There exists a $\delta_{2}>\frac{1}{2}$ such that

$(a)$ $I(\theta\circ T(y,\delta))<\frac{s}{3}S_{s}^{\frac{3}{2s}}+\mu$,   $\forall y\in\mathbb{R}^{3}$

$(b)$ $\gamma(\theta\circ T(y,\delta))>\frac{1}{2}$,   $\forall y:|y|>\frac{1}{2}$.*

Proof.

By Lemma 8, $(a)$ holds. Since

[TABLE]

and

[TABLE]

we have

[TABLE]

then $(b)$ holds. ∎

Lemma 12**.**

*There exists $R>0$ such that

$(a)$ $I(\theta\circ T(y,\delta))<\frac{s}{3}S_{s}^{\frac{3}{2s}}+\mu$,   $\forall y:|y|\geq R$ and $\delta\in[\delta_{1},\delta_{2}]$

$(b)$ $(\beta(\theta\circ T(y,\delta))\cdot y)_{\mathbb{R}^{3}}>0$,   $\forall|y|\geq R$ and $\delta\in[\delta_{1},\delta_{2}]$.*

Proof.

Since

[TABLE]

by Lemma 9 and the compactness of $[\delta_{1},\delta_{2}]$, we can get $R_{1}$ big enough such that

[TABLE]

For $|y|$ large enough, by (4.16) and a similar argument as that in the proof of Lemma 6, the result follows. ∎

Let $\delta_{1}$, $\delta_{2}$ and $R$ be the constant in Lemma 10, Lemma 4.4 and Lemma 12, respectively. Define a bounded domain $D\subset\mathbb{R}^{3}\times\mathbb{R}$ by

[TABLE]

and define the map $\vartheta:D\rightarrow\mathbb{R}^{3}\times\mathbb{R}^{+}$ by

[TABLE]

Lemma 13**.**

Assume that $V\geq 0$, $V\in L^{\frac{3}{2s}}(\mathbb{R}^{3})$, $K\geq 0$, $K\in L^{\frac{6}{6s-3}}(\mathbb{R}^{3})$ and $\|V\|_{\frac{3}{2s}}+\|K\|_{\frac{6}{6s-3}}>0$. Then

[TABLE]

Proof.

We consider the homotopy

[TABLE]

By the homotopy invariance of the topological degree, and by the fact that

[TABLE]

we need to prove that

[TABLE]

We have

[TABLE]

where

[TABLE]

If $(y,\delta)\in\Gamma_{1}$, by Lemma 10 $(b)$,

[TABLE]

If $(y,\delta)\in\Gamma_{2}$, by Lemma 10 $(c)$,

[TABLE]

then $\forall s\in[0,1]$,

[TABLE]

If $(y,\delta)\in\Gamma_{3}$,

[TABLE]

If $(y,\delta)\in\Gamma_{4}$, by Lemma 12 $(b)$,

[TABLE]

Therefore,

[TABLE]

∎

Proof of Theorem 1. In order to apply the linking theorem in [26], we define

[TABLE]

We claim that $\mathcal{M}$ links $\partial Q$, that is,

$(a)$ $\partial Q\cap\mathcal{M}=\varnothing$;

$(b)$ $h(Q)\cap\mathcal{M}\neq\varnothing$ for any $h\in\Gamma=\{h\in C(Q,\mathcal{N}):h(\partial Q)=id\}$.

In fact, if $u\in\theta\circ T(\partial D)$, by Lemma 10 $(a)$, Lemma 11 $(a)$, Lemma 12 $(a)$,

[TABLE]

then $u\notin\mathcal{M}$.

To prove $(b)$, for any $h\in\Gamma$, we define $\eta:D\rightarrow\mathbb{R}^{3}\times\mathbb{R}+$ by

[TABLE]

Since $h(\partial D)=id$, then

[TABLE]

This together with Lemma 13 yield that

[TABLE]

Then there exists $(y^{\prime},\delta^{\prime})\in D$ such that $h\circ\theta\circ T(y^{\prime},\delta^{\prime})\in\mathcal{M}$.

Define

[TABLE]

By linking theorem, $d\geq c_{\mathcal{M}}>\frac{s}{3}S_{s}^{\frac{3}{2s}}$. From $Q=\theta\circ T(D)$ and (4.17), it follows that

[TABLE]

By $t_{\delta,y}\psi_{\delta,y}\in\mathcal{N}$, we have

[TABLE]

this together with $\|\psi_{\delta,y}\|_{2_{s}^{\ast}}^{2_{s}^{\ast}}=\|\psi_{\delta,y}\|_{D^{s,2}}^{2}$ yield that

[TABLE]

Since $\psi_{\delta,y}>0$, $V(x)\geq 0$, $K(x)\geq 0$ and $\|V\|_{\frac{3}{2s}}+\|K\|_{\frac{6}{6s-3}}>0$, by (4.19), we get that $t_{\delta,y}>1$. From (4.18) and the Hölder inequality, it follows that

[TABLE]

By (2.4) and (2.7),

[TABLE]

By (4.19), (4.21) and $\|\psi_{\delta,y}\|_{2_{s}^{\ast}}^{2_{s}^{\ast}}=\|\psi_{\delta,y}\|_{D^{s,2}}^{2}=S_{s}^{\frac{3}{2s}}$, we have

[TABLE]

Then, by (2.11), (4.18), (4.22) and (1.4),

[TABLE]

Therefore, $\frac{s}{3}S_{s}^{\frac{3}{2s}}<d<\frac{2s}{3}S_{s}^{\frac{3}{2s}}$. By Corollary 1, $d$ is the critical value of $I$. The proof is complete.

5. Acknowledgements

This work is partially supported by National Natural Science Foundation of China under the contract No.11571269, China Postdoctoral Science Foundation Funded Project under contracts No.2015M572539 and No.2016T90899 and Shaanxi Province Postdoctoral Science Foundation Funded Project.