Infinitely many weak solutions for a fractional Schrödinger equation

Abstract

In this paper we are concerned with the fractional Schrödinger equation ${(-\mathrm{\Delta })}^{\alpha }u+V(x)u=f(x,u)$, $x\in {\mathbb{R}}^N$, where $0<\alpha <1$, $N>2\alpha$, ${(-\mathrm{\Delta })}^{\alpha }$ stands for the fractional Laplacian of order α, V is a positive continuous potential, and f is a continuous subcritical nonlinearity. We obtain the existence of infinitely many weak solutions for the above problem by the fountain theorem in critical point theory.

Introduction

In this paper we consider the following fractional Schrödinger equation:

$${(-\mathrm{\Delta })}^{\alpha }u+V(x)u=f(x,u),x\in {\mathbb{R}}^N,$$

(1.1)

where $0<\alpha <1$, $N>2\alpha$, ${(-\mathrm{\Delta })}^{\alpha }$ stands for the fractional Laplacian of order α, and the potential $V:{\mathbb{R}}^N\to \mathbb{R}$ is a continuous function satisfying

• (V) $0<{inf}_{x\in {\mathbb{R}}^N}V(x)=V_0<{lim\hspace{0.17em}inf}_{|x|\to \mathrm{\infty }}V(x)=V_{\mathrm{\infty }}<\mathrm{\infty }$.

The nonlinearity $f:{\mathbb{R}}^N\times \mathbb{R}\to \mathbb{R}$ is a continuous function, satisfying the subcritical condition.

(H1) There exist $d_1>0$, $d_2>0$ and $p\in (2,2_{\alpha }^{\ast })$ such that

$$|f(x,s)|\le d_1|s|+d_2{|s|}^{p-1},\mathrm{\forall }(x,s)\in {\mathbb{R}}^N\times \mathbb{R},$$

(1.2)

where $2_{\alpha }^{\ast }=\frac{2N}{N-2\alpha }$ is the fractional critical exponent.

Recently, there have appeared plenty of works on the fractional Schrödinger equations; for example, see [[1]–[11]] and the references therein. In [[1]], Shang and Zhang considered the critical fractional Schrödinger equation

$${\epsilon }^{2\alpha }{(-\mathrm{\Delta })}^{\alpha }u+V(x)u={|u|}^{2_{\alpha }^{\ast }-2}u+\lambda f(u),x\in {\mathbb{R}}^N,$$

(1.3)

where ε and λ are positive parameters, V and f satisfy (V) and (H1), respectively. They obtained the result that (1.2) has a nonnegative ground state solution and investigated the relation between the number of solutions and the topology of the set where V attains its minimum for all sufficiently large λ and small ε. In [[2]], Shang et al. considered the existence of nontrivial solutions for (1.2) with $f(u)={|u|}^{q-2}u$, where $2<q<2_{\alpha }^{\ast }$.

In [[8]], Hua and Yu studied the critical fractional Laplacian equation

$$\{\begin{array}{c}{(-\mathrm{\Delta })}^{\frac{\alpha }{2}}u={|u|}^{2_{\alpha }^{\ast }-2}u+\mu u\text{in }\mathrm{\Omega },\hfill \\ u=0\text{on }\partial \mathrm{\Omega },\hfill \end{array}$$

(1.3)

where $0<\alpha <2$, $\mathrm{\Omega }\subset {\mathbb{R}}^N$, $N>(1+\sqrt{2})\alpha$ is a bounded domain. They obtained the result that the problem (1.3) possesses a nontrivial ground state solution for any $\mu >0$.

In [[7]], Secchi investigated the existence of radially symmetric solutions for (1.1) replacing $f(t,u)$ by $g(u)$, where g satisfies the following conditions.

(g1) $g:\mathbb{R}\to \mathbb{R}$ is of class $C^{1,\gamma }$ for some $\gamma >max\{0,1-2\alpha \}$, and odd,

(g2) $-\mathrm{\infty }<{lim\hspace{0.17em}inf}_{t\to 0^{+}}\frac{g(t)}{t}\le {lim\hspace{0.17em}sup}_{t\to 0^{+}}\frac{g(t)}{t}=-m<0$,

(g3) $-\mathrm{\infty }<{lim\hspace{0.17em}sup}_{t\to +\mathrm{\infty }}\frac{g(t)}{t^{2_{\alpha }^{\ast }-1}}\le 0$,

(g4) for some $\xi >0$ such that $G(\xi )={\int }_0^{\xi }g(t)\mathrm{d}t>0$.

Inspired by the mentioned papers, we first establish a compact embedding lemma via a fractional Gagliardo-Nirenberg inequality. Then by virtue of the fountain theorem in critical point theory, we get two existence results of infinitely many weak solutions for (1.1).

Preliminary results

In this section we offer some preliminary results which enable us to obtain the main existence theorems. First, we collect some useful facts of the fractional order Sobolev spaces.

For any $0<\alpha <1$, the fractional Sobolev space $H^{\alpha }({\mathbb{R}}^N)$ is defined by

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

endowed with the norm

$${\parallel u\parallel }_{H^{\alpha }({\mathbb{R}}^N)}={({\int }_{{\mathbb{R}}^N}{|u|}^2\mathrm{d}x+{\int }_{{\mathbb{R}}^{2N}}\frac{{|u(x)-u(y)|}^2}{{|x-y|}^{N+2\alpha }}\mathrm{d}x\mathrm{d}y)}^{\frac{1}{2}},$$

where ${[u]}_{H^{\alpha }({\mathbb{R}}^N)}={({\int }_{{\mathbb{R}}^{2N}}\frac{{|u(x)-u(y)|}^2}{{|x-y|}^{N+2\alpha }}\mathrm{d}x\mathrm{d}y)}^{1/2}$ is the so-called Gagliardo semi-norm of u. Let $\mathcal{S}$ be the Schwartz space of rapidly decaying $C^{\mathrm{\infty }}$ functions in ${\mathbb{R}}^N$, for any $u\in \mathcal{S}$ and $\alpha \in (0,1)$, and let ${(-\mathrm{\Delta })}^{\alpha }$ be defined as

$${(-\mathrm{\Delta })}^{\alpha }u(x)=k_{N,\alpha }\text{P.V.}{\int }_{{\mathbb{R}}^N}\frac{u(x)-u(y)}{{|x-y|}^{N+2\alpha }}\mathrm{d}y=k_{N,\alpha }\underset{\epsilon \to 0}{lim}{\int }_{\mathcal{C}{B}_{\epsilon }(x)}\frac{u(x)-u(y)}{{|x-y|}^{N+2\alpha }}\mathrm{d}y.$$

(2.1)

The symbol P.V. stands for the Cauchy principal value, and $k_{N,\alpha }$ is a dimensional constant that depends on N and α, precisely given by $k_{N,\alpha }={({\int }_{{\mathbb{R}}^N}\frac{1-cos{\zeta }_1}{{|\zeta |}^{N+2\alpha }}\mathrm{d}\zeta )}^{-1}$.

Indeed, the fractional Laplacian ${(-\mathrm{\Delta })}^{\alpha }$ can be viewed as a pseudo-differential operator of symbol ${|\xi |}^{2\alpha }$, as stated in the following.

Lemma 2.1

(see [[12]])

Let$\alpha \in (0,1)$and${(-\mathrm{\Delta })}^{\alpha }:\mathcal{S}\to L^2({\mathbb{R}}^N)$be the fractional Laplacian operator defined by (2.1). Then for any$u\in \mathcal{S}$,

$${(-\mathrm{\Delta })}^{\alpha }u(x)={\mathcal{F}}^{-1}({|\xi |}^{2\alpha }(\mathcal{F}u))(x),\mathrm{\forall }\xi \in {\mathbb{R}}^N,$$

where ℱ is the Fourier transform, i.e.,

$$\mathcal{F}(\varphi )(\xi )=\frac{1}{{(2\pi )}^{N/2}}{\int }_{{\mathbb{R}}^N}exp\{-2\pi i\xi \cdot x\}\varphi (x)\mathrm{d}x.$$

Now we can see that an alternative definition of the fractional Sobolev space $H^{\alpha }({\mathbb{R}}^N)$ via the Fourier transform is as follows:

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

It can be proved that

$$2k_{N,\alpha }^{-1}{\int }_{{\mathbb{R}}^N}{|\xi |}^{2\alpha }{|\mathcal{F}u|}^2\mathrm{d}\xi =2k_{N,\alpha }^{-1}{\parallel {(-\mathrm{\Delta })}^{\frac{\alpha }{2}}u\parallel }_{L^2({\mathbb{R}}^N)}^2={[u]}_{H^{\alpha }({\mathbb{R}}^N)}^2.$$

(2.2)

As a result, the norms on $H^{\alpha }({\mathbb{R}}^N)$,

$$\begin{array}{r}u\mapsto {\parallel u\parallel }_{H^{\alpha }({\mathbb{R}}^N)},\\ u\mapsto {({\parallel u\parallel }_{L^2({\mathbb{R}}^N)}^2+{\parallel {(-\mathrm{\Delta })}^{\frac{\alpha }{2}}u\parallel }_{L^2({\mathbb{R}}^N)}^2)}^{\frac{1}{2}},\\ u\mapsto {({\parallel u\parallel }_{L^2({\mathbb{R}}^N)}^2+{\int }_{{\mathbb{R}}^N}{|\xi |}^{2\alpha }{|\mathcal{F}u|}^2\mathrm{d}\xi )}^{\frac{1}{2}},\end{array}$$

(2.3)

are all equivalent.

In this paper, in view of the presence of potential $V(x)$, we consider its subspace

$$E=\{u\in H^{\alpha }\left({\mathbb{R}}^N\right):{\int }_{{\mathbb{R}}^N}V(x)u^2\mathrm{d}x<\mathrm{\infty }\}.$$

We define the norm in E by

$${\parallel u\parallel }_E={({\int }_{{\mathbb{R}}^N}({|\xi |}^{2\alpha }{\hat{u}}^2+{\hat{u}}^2)\mathrm{d}\xi +{\int }_{{\mathbb{R}}^N}V(x)u^2\mathrm{d}x)}^{\frac{1}{2}},$$

where $\hat{u}=\mathcal{F}(u)$. Moreover, by [[6]], E is a Hilbert space with the inner product

$${〈u,v〉}_E={\int }_{{\mathbb{R}}^N}({|\xi |}^{2\alpha }\hat{u}(\xi )\hat{v}(\xi )+\hat{u}(\xi )\hat{v}(\xi ))\mathrm{d}\xi +{\int }_{{\mathbb{R}}^N}V(x)u(x)v(x)\mathrm{d}x,\mathrm{\forall }u,v\in E.$$

Note that by (2.2) and (2.3), together with the condition (V), we know that ${\parallel \cdot \parallel }_E$ is equivalent to the norm

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

(2.4)

The corresponding inner product is

$$〈u,v〉={\int }_{{\mathbb{R}}^N}({(-\mathrm{\Delta })}^{\frac{\alpha }{2}}u(x){(-\mathrm{\Delta })}^{\frac{\alpha }{2}}v(x)+V(x)u(x)v(x))\mathrm{d}x.$$

Throughout out this paper, we will use the norm $\parallel \cdot \parallel$ in E.

Definition 2.2

We say that $u\in E$ is a weak solution of (1.1), if

$${\int }_{{\mathbb{R}}^N}({(-\mathrm{\Delta })}^{\frac{\alpha }{2}}u{(-\mathrm{\Delta })}^{\frac{\alpha }{2}}\varphi +V(x)u\varphi )\mathrm{d}x={\int }_{{\mathbb{R}}^N}f(x,u)\varphi \mathrm{d}x,\mathrm{\forall }\varphi \in E.$$

Lemma 2.3

(see [[7]] and [[12]])

E is continuously embedded into$L^p({\mathbb{R}}^N)$for$p\in [2,2_{\alpha }^{\ast }]$and compactly embedded into$L_{\mathrm{loc}}^p({\mathbb{R}}^N)$for$p\in [2,2_{\alpha }^{\ast })$.

Lemma 2.4

E is compactly embedded into$L^p({\mathbb{R}}^N)$for$p\in [2,2_{\alpha }^{\ast })$with$2_{\alpha }^{\ast }=\frac{2N}{N-2\alpha }$.

Proof

By [[4]], we know E is compactly embedded into $L^2({\mathbb{R}}^N)$, i.e., if there exists a sequence $\{u_n\}\subset E$ and $u_0\in E$ such that $u_n\rightharpoonup u_0$ weakly in E, passing to a subsequence if necessary, we have $u_n\to u_0$ strongly in $L^2({\mathbb{R}}^N)$. Therefore, we only consider $p\in (2,2_{\alpha }^{\ast })$. In order to do this, we need the following fractional Gagliardo-Nirenberg inequality, see [[13], Corollary 2.3]. Let $1\le p,p_2<\mathrm{\infty }$, $0<s<p<\mathrm{\infty }$, $0<\alpha <N$ and $1<p_1<N/\alpha$. Then

$${\parallel u(x)\parallel }_{L^p({\mathbb{R}}^N)}\le {\eta }^{\frac{s}{p}}{\parallel {(-\mathrm{\Delta })}^{\alpha /2}u(x)\parallel }_{L^{p_1}({\mathbb{R}}^N)}^{\frac{s}{p}}{\parallel u(x)\parallel }_{L^{p_2}({\mathbb{R}}^N)}^{1-\frac{s}{p}}$$

(2.5)

with

$$s(\frac{1}{p_1}-\frac{\alpha }{N})+\frac{p-s}{p_2}=1\text{and}\eta =2^{-\alpha }{\pi }^{-\alpha /2}\frac{\mathrm{\Gamma }((N-\alpha )/2)}{\mathrm{\Gamma }((N+\alpha )/2)}{\left(\frac{\mathrm{\Gamma }(N)}{\mathrm{\Gamma }(N/2)}\right)}^{\alpha /N}.$$

Note that the dimension $N>2\alpha$, we can take $p_1=p_2=2$, and then $s(\frac{1}{2}-\frac{\alpha }{N})+\frac{p-s}{2}=1$ whence $s=\frac{(p-2)N}{2\alpha }\in (0,p)$ as $p\in (2,2_s^{\ast })$. Consequently, from (2.5) we have

$${\parallel u\parallel }_{L^p({\mathbb{R}}^N)}\le {\eta }^{\frac{s}{p}}{\parallel {(-\mathrm{\Delta })}^{\alpha /2}u\parallel }_{L^2({\mathbb{R}}^N)}^{\frac{s}{p}}{\parallel u\parallel }_{L^2({\mathbb{R}}^N)}^{1-\frac{s}{p}}.$$

Furthermore, note that (2.4); we see that

$${\parallel u\parallel }_{L^p({\mathbb{R}}^N)}\le {\eta }^{\frac{s}{p}}{\parallel u\parallel }^{\frac{s}{p}}{\parallel u\parallel }_{L^2({\mathbb{R}}^N)}^{1-\frac{s}{p}}.$$

(2.6)

Then by (2.6) and $E\hookrightarrow \hookrightarrow L^2({\mathbb{R}}^N)$, we find

$$\begin{array}{rl}{\parallel u_n-u_0\parallel }_{L^p({\mathbb{R}}^N)}& \le {\eta }^{\frac{s}{p}}{\parallel u_n-u_0\parallel }^{\frac{s}{p}}{\parallel u_n-u_0\parallel }_{L^2({\mathbb{R}}^N)}^{1-\frac{s}{p}}\\ \le {\eta }^{\frac{s}{p}}({\parallel u_n\parallel }^{\frac{s}{p}}+{\parallel u_0\parallel }^{\frac{s}{p}}){\parallel u_n-u_0\parallel }_{L^2({\mathbb{R}}^N)}^{1-\frac{s}{p}}\to 0.\end{array}$$

Therefore, E is compactly embedded into $L^p({\mathbb{R}}^N)$ for $p\in [2,2_{\alpha }^{\ast })$ with $2_{\alpha }^{\ast }=\frac{2N}{N-2\alpha }$, as required. This completes the proof. □

The functional associated with (1.1) is defined by

$$J(u)=\frac{1}{2}{\int }_{{\mathbb{R}}^N}({\left|{(-\mathrm{\Delta })}^{\frac{\alpha }{2}}u\right|}^2+V(x)u^2)\mathrm{d}x-{\int }_{{\mathbb{R}}^N}F(x,u)\mathrm{d}x,\mathrm{\forall }u\in E,$$

(2.7)

where $F(x,u)={\int }_0^{u}f(x,s)\mathrm{d}s$.

Now, we list our assumptions on f and F.

(H2) ${lim}_{|s|\to \mathrm{\infty }}\frac{F(x,s)}{{|s|}^2}=+\mathrm{\infty }$ uniformly for $x\in {\mathbb{R}}^N$.

(H3) There exist $d_3>0$ and ${\phi }_1>0$ such that ${\int }_{{\mathbb{R}}^N}{\phi }_1(x)\mathrm{d}x<+\mathrm{\infty }$ and

$$tf(x,t)-2F(x,t)\le sf(x,s)-2F(x,s)+d_3{\phi }_1,\mathrm{\forall }0<t<s\text{ or }s<t<0,x\in {\mathbb{R}}^N.$$

(H4) $F(x,s)\ge 0$ for $(x,s)\in {\mathbb{R}}^N\times \mathbb{R}$.

(H5) There exist $\beta >2$, $r_0>0$ such that

$$-\beta F(x,s)+sf(x,s)\ge 0,|s|\ge r_0,\text{uniformly for }x\in {\mathbb{R}}^N.$$

(H6) $F(x,s)=F(x,-s)$ for all $(x,s)\in {\mathbb{R}}^N\times \mathbb{R}$.

Remark 2.5

• (1) Let $F(x,s)=s^{2}ln(|s|+1)$, for all $x\in {\mathbb{R}}^N$ and $s\in \mathbb{R}$. Then (H1), (H2), (H4), and (H6) hold. Moreover, we easily have $f(x,s)=2sln(|s|+1)+s^3{[|s|(|s|+1)]}^{-1}$ and $sf(x,s)-2F(x,s)={|s|}^3{(|s|+1)}^{-1}$, so (H3) is satisfied.

However, we can see that $F(x,s)$ does not satisfy the Ambrosetti-Rabinowitz condition (see [[6], ($f_4$)]):

(AR) there is a constant $\mu >2$ such that

$$0<\mu F(x,s)\le sf(x,s)\text{for all }x\in {\mathbb{R}}^N\text{ and }s\in \mathbb{R}\mathrm{\setminus }\{0\}.$$

Indeed, $sf(x,s)-\mu F(x,s)=s^2[(2-\mu )ln(|s|+1)+|s|{(|s|+1)}^{-1}]\ge 0$ is impossible for all $x\in {\mathbb{R}}^N$ and $s\in \mathbb{R}\mathrm{\setminus }\{0\}$.

• (2) Let $\beta >2$ and $F(x,s)={|s|}^{\beta }ln(|s|+1)$, for all $x\in {\mathbb{R}}^N$ and $s\in \mathbb{R}$. Then (H1), (H2), (H4), and (H6) hold. Moreover, from $-\beta F(x,s)+sf(x,s)={|s|}^{\beta +1}{(|s|+1)}^{-1}$, and (H5) holds.

Note that from Theorem 4 in [[14]] we have (H4) and (H5) imply (H2).

Lemma 2.6

(see [[15], Lemma 1])

Let (V) and (H1) hold. Then$J\in C^1(E,\mathbb{R})$and its derivative

$$(J^{\prime }(u),\varphi )={\int }_{{\mathbb{R}}^N}({(-\mathrm{\Delta })}^{\frac{\alpha }{2}}u{(-\mathrm{\Delta })}^{\frac{\alpha }{2}}\varphi +V(x)u\varphi )\mathrm{d}x-{\int }_{{\mathbb{R}}^N}f(x,u)\varphi \mathrm{d}x,\mathrm{\forall }u,\varphi \in E.$$

Moreover, the critical points of J are weak solutions of (1.1).

To complete the proofs of our theorems, we need the following critical point theorems in [[16]–[19]].

Definition 2.7

Let $(X,\parallel \cdot \parallel )$ be a real Banach space, $J\in C^1(X,\mathbb{R})$. We say that J satisfies the (${\mathrm{C}}_{\mathrm{c}}$) condition if any sequence $\{u_n\}\subset X$ such that $J(u_n)\to c$ and $\parallel J^{\prime }(u_n)\parallel (1+\parallel u_n\parallel )\to 0$ as $n\to \mathrm{\infty }$ has a convergent subsequence.

Let X be a Banach space equipped with the norm $\parallel \cdot \parallel$ and $X=\overline{{⨁}_{j\in \mathbb{N}}{X}_j}$, where $dim{X}_j<\mathrm{\infty }$ for any $j\in \mathbb{N}$. Set $Y_k={⨁}_{j=1}^k{X}_j$ and $Z_k=\overline{{⨁}_{j=k}^{\mathrm{\infty }}{X}_j}$.

Lemma 2.8

Let$(X,\parallel \cdot \parallel )$be a real reflexive Banach space, $J\in C^1(X,\mathbb{R})$satisfies the (${\mathrm{C}}_{\mathrm{c}}$) condition for any$c>0$and J is even. If for each sufficiently large$k\in \mathbb{N}$, there exist${\rho }_k>r_k>0$such that the following conditions hold:

• (i) $a_k:={inf}_{\{u\in Z_k,\parallel u\parallel =r_k\}}J(u)\to +\mathrm{\infty }$ as $k\to \mathrm{\infty }$,

• (ii) $b_k:={max}_{\{u\in Y_k,\parallel u\parallel ={\rho }_k\}}J(u)\le 0$,

then the functional J has an unbounded sequence of critical values, i.e., there exists a sequence$\{u_k\}\subset X$such that$J^{\prime }(u_k)=0$and$J(u_k)\to +\mathrm{\infty }$as$k\to \mathrm{\infty }$.

In the following, we will introduce a variant fountain theorem by Zou [[16]]. Let X and the subspace $Y_k$ and $Z_k$ be defined above. Consider the following $C^1$-functional $J_{\lambda }:X\to \mathbb{R}$ defined by

$$J_{\lambda }(u):=A(u)-\lambda B(u),\lambda \in [1,2].$$

(2.8)

Lemma 2.9

If the functional $J_{\lambda }$ satisfies

(T1) $J_{\lambda }$maps bounded sets to bounded sets uniformly for$\lambda \in [1,2]$, and, moreover, $J_{\lambda }(-u)=J_{\lambda }(u)$for all$(\lambda ,u)\in [1,2]\times X$,

(T2) $B(u)\ge 0$for all$u\in X$; moreover, $A(u)\to \mathrm{\infty }$or$B(u)\to \mathrm{\infty }$as$\parallel u\parallel \to \mathrm{\infty }$,

(T3) there exist$r_k>{\rho }_k>0$such that

$$a_k(\lambda ):=\underset{u\in Z_k,\parallel u\parallel ={\rho }_k}{inf}{J}_{\lambda }(u)>b_k(\lambda ):=\underset{u\in Y_k,\parallel u\parallel =r_k}{max}{J}_{\lambda }(u),\mathrm{\forall }\lambda \in [1,2],$$

then

$$a_k(\lambda )\le {\zeta }_k(\lambda )=\underset{\gamma \in {\mathrm{\Gamma }}_k}{inf}\underset{u\in B_k}{max}{J}_{\lambda }(\gamma (u)),\mathrm{\forall }\lambda \in [1,2],$$

where$B_k=\{u\in Y_k:\parallel u\parallel \le r_k\}$and${\mathrm{\Gamma }}_k=\{\gamma \in C(B_k,X):\gamma \mathit{\text{ is odd}},\gamma {|}_{\partial B_k}=\mathit{\text{id}}\}$. Moreover, for a.e. $\lambda \in [1,2]$, there exists a sequence${\{u_n^k(\lambda )\}}_{n=1}^{\mathrm{\infty }}$such that

$$\underset{n}{sup}\parallel u_n^k(\lambda )\parallel <\mathrm{\infty },J_{\lambda }^{\prime }(u_n^k(\lambda ))\to 0\mathit{\text{and}}{J}_{\lambda }(u_n^k(\lambda ))\to {\zeta }_k(\lambda )\mathit{\text{as }}n\to \mathrm{\infty }.$$

Remark 2.10

As mentioned in [[6]], E is a Hilbert space. Let $\{e_j\}$ be an orthonormal basis of E and define $X_j:=span\{e_j\}$, $Y_k:={⨁}_{j=1}^k{X}_j$, and $Z_k:=\overline{{⨁}_{j=k+1}^{\mathrm{\infty }}{X}_j}$, $k\in \mathbb{N}$. Clearly, $E=\overline{{⨁}_{j\in \mathbb{N}}{X}_j}$ with $dim{X}_j<\mathrm{\infty }$ for all $j\in \mathbb{N}$.

Existence of weak solutions for (1.1)

Theorem 3.1

Assume that (V), (H1)-(H4), and (H6) hold. Then (1.1) has infinitely many weak solutions$\{u_k\}$satisfying

$$\frac{1}{2}{\int }_{{\mathbb{R}}^N}({\left|{(-\mathrm{\Delta })}^{\frac{\alpha }{2}}{u}_k\right|}^2+V(x)u_k^2)\mathrm{d}x-{\int }_{{\mathbb{R}}^N}F(x,u_k)\mathrm{d}x\to +\mathrm{\infty }\mathit{\text{as }}k\to \mathrm{\infty }.$$

Proof

We first prove that J satisfies the (${\mathrm{C}}_{\mathrm{c}}$) condition for any $c>0$. Let $\{u_n\}\subset E$ be a (${\mathrm{C}}_{\mathrm{c}}$) sequence, i.e.,

$$J(u_n)\to c>0,\parallel J^{\prime }(u_n)\parallel (1+\parallel u_n\parallel )\to 0\text{when }n\to \mathrm{\infty },$$

(3.1)

which implies that

$$c=J(u_n)+o(1),J^{\prime }(u_n)u_n=o(1)\text{as }n\to \mathrm{\infty }.$$

(3.2)

In what follows, we shall show that $\{u_n\}$ is bounded. Otherwise, up to a subsequence, $\{u_n\}$ is unbounded in E, and we may assume that $\parallel u_n\parallel \to \mathrm{\infty }$ as $n\to \mathrm{\infty }$. We define the sequence $\{w_n\}$ by $w_n=\frac{u_n}{\parallel u_n\parallel }$, $n=1,2,\dots$ . Clearly, $\{w_n\}\subset E$ and $\parallel w_n\parallel =1$ for any n. Going over, if necessary, to a subsequence, we may assume that

$$\begin{array}{r}{w}_n\rightharpoonup w\text{ weakly in }E,\\ w_n\to w\text{ strongly in }{L}^p\left({\mathbb{R}}^N\right)\text{ for }p\in [2,2_s^{\ast }),\\ w_n(x)\to w(x)\text{ a.e. }x\in {\mathbb{R}}^N.\end{array}$$

(3.3)

Suppose that $w\ne 0$ in E. Dividing by ${\parallel u_n\parallel }^2$ in both sides of (2.7), noting that $J(u_n)\to c$, we obtain

$${\int }_{{\mathbb{R}}^N}\frac{F(x,u_n)}{{\parallel u_n\parallel }^2}\mathrm{d}x=\frac{1}{2}+o\left({\parallel u_n\parallel }^{-2}\right)<+\mathrm{\infty }.$$

(3.4)

On the other hand, denote ${\mathrm{\Omega }}_{\ne }:=\{x\in {\mathbb{R}}^N:w(x)\ne 0\}$, by (H2), for all $x\in {\mathrm{\Omega }}_{\ne }$, and we find

$$\frac{F(x,u_n)}{{\parallel u_n\parallel }^2}=\frac{F(x,u_n)}{{|u_n|}^2}\frac{{|u_n|}^2}{{\parallel u_n\parallel }^2}=\frac{F(x,u_n)}{{|u_n|}^2}{|w_n|}^2\to +\mathrm{\infty }\text{when }n\to \mathrm{\infty }.$$

If $|{\mathrm{\Omega }}_{\ne }|>0$, using Fatou’s lemma, we obtain

$${\int }_{{\mathbb{R}}^N}\frac{F(x,u_n)}{{\parallel u_n\parallel }^2}\mathrm{d}x\to +\mathrm{\infty }\text{as }n\to \mathrm{\infty }.$$

This contradicts (3.4). Hence, ${\mathrm{\Omega }}_{\ne }$ has zero measure, i.e., $w=0$ a.e. in ${\mathbb{R}}^N$. Let $t_n\in [0,1]$ such that

$$J(t_n{u}_n)=\underset{t\in [0,1]}{max}J(t{u}_n).$$

Then we claim $J(t_n{u}_n)$ is bounded. If $t_n=0$, $J(0)=0$; if $t_n=1$, $J(t_n{u}_n)=J(u_n)\to c$. Therefore, $J(t_n{u}_n)$ is bounded when $t_n=0,1$. If $0<t_n<1$ for n large enough

$$\begin{array}{r}{\int }_{{\mathbb{R}}^N}({(-\mathrm{\Delta })}^{\frac{\alpha }{2}}{t}_n{u}_n{(-\mathrm{\Delta })}^{\frac{\alpha }{2}}{t}_n{u}_n+V(x)t_n{u}_n\cdot t_n{u}_n)\mathrm{d}x-{\int }_{{\mathbb{R}}^N}f(x,t_n{u}_n)t_n{u}_n\mathrm{d}x\\ =(J^{\prime }(t_n{u}_n),t_n{u}_n)=t_n\frac{\mathrm{d}}{\mathrm{d}t}{|}_{t=t_n}J(t{u}_n)=0.\end{array}$$

Consequently, by (H3), noting that (3.1) and (3.2) hold, we have

$$\begin{array}{rl}J(t{u}_n)& \le J(t_n{u}_n)-\frac{1}{2}(J^{\prime }(t_n{u}_n),t_n{u}_n)\\ =\frac{1}{2}{\int }_{{\mathbb{R}}^N}f(x,t_n{u}_n)t_n{u}_n\mathrm{d}x-{\int }_{{\mathbb{R}}^N}F(x,t_n{u}_n)\mathrm{d}x\\ \le {\int }_{{\mathbb{R}}^N}[(\frac{1}{2}{u}_{n}f(x,u_n)-F(x,u_n))+\frac{d_3}{2}{\phi }_1(x)]\mathrm{d}x\\ =J(u_n)-\frac{1}{2}(J^{\prime }(u_n),u_n)+{\int }_{{\mathbb{R}}^N}\frac{d_3}{2}{\phi }_1(x)\mathrm{d}x\\ \le d_4,\mathrm{\forall }t\in [0,1],\end{array}$$

(3.5)

where $d_4$ is a positive constant. But fixing any $m>d_4$, we let ${\overline{w}}_n=\sqrt{2m}\frac{u_n}{\parallel u_n\parallel }=\sqrt{2m}{w}_n$. Note that from (H1) we see that there exist $d_5>0$, $d_6>0$ such that

$$F(x,u)\le d_5{|u|}^2+d_6{|u|}^p,\mathrm{\forall }(x,u)\in {\mathbb{R}}^N\times \mathbb{R}.$$

(3.6)

Then by (3.3) we have

$$\underset{n\to \mathrm{\infty }}{lim}{\int }_{{\mathbb{R}}^N}F(x,{\overline{w}}_n)\mathrm{d}x\le \underset{n\to \mathrm{\infty }}{lim}{\int }_{{\mathbb{R}}^N}(d_5{|{\overline{w}}_n|}^2+d_6{|{\overline{w}}_n|}^p)\mathrm{d}x=0.$$

Then for n large enough,

$$J(t_n{u}_n)\ge J\left(\frac{\sqrt{2m}}{\parallel u_n\parallel }{u}_n\right)=J({\overline{w}}_n)=m-{\int }_{{\mathbb{R}}^N}F(x,{\overline{w}}_n)\mathrm{d}x\ge m.$$

This also contradicts (3.5).

Now the sequence $\{u_n\}$ is bounded, as required. Next, we verify that $\{u_n\}$ has a convergent subsequence. Without loss of generality, we assume that

$$\begin{array}{r}{u}_n\rightharpoonup u\text{ weakly in }E,\\ u_n\to u\text{ strongly in }{L}^p\left({\mathbb{R}}^N\right)\text{ for }p\in [2,2_s^{\ast }).\end{array}$$

(3.7)

Combining this with (H1) and the Hölder inequality, we see

$$\begin{array}{r}|{\int }_{{\mathbb{R}}^N}[f(x,u_n)-f(x,u)](u_n-u)\mathrm{d}x|\\ \le {\int }_{{\mathbb{R}}^N}[d_1(|u_n|+|u|)+d_2({|u_n|}^{p-1}+{|u|}^{p-1})]|u_n-u|\mathrm{d}x\\ \le d_1({\parallel u_n\parallel }_2+{\parallel u\parallel }_2){\parallel u_n-u\parallel }_2+d_2({\parallel u_n\parallel }_p^{p-1}+{\parallel u\parallel }_p^{p-1}){\parallel u_n-u\parallel }_p\to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

Consequently,

$${\parallel u_n-u\parallel }^2=(J^{\prime }(u_n)-J^{\prime }(u),u_n-u)+{\int }_{{\mathbb{R}}^N}[f(x,u_n)-f(x,u)](u_n-u)\mathrm{d}x\to 0$$

with the fact that $(J^{\prime }(u_n)-J^{\prime }(u),u_n-u)\to 0$ when $n\to \mathrm{\infty }$. Therefore, we prove that J satisfies the (${\mathrm{C}}_{\mathrm{c}}$) condition for any $c>0$.

Clearly, $J(u)=J(-u)$ by (H6). It remains to prove that the conditions (i) and (ii) of Lemma 2.8 hold. Let ${\beta }_r(k):={sup}_{\{u\in Z_k,\parallel u\parallel =1\}}{\parallel u\parallel }_r$ with $r\in [2,2_{\alpha }^{\ast })$, where $Z_k$ is defined in Remark 2.10. Then by Lemma 3.8 of [[19]], ${\beta }_r(k)\to 0$ as $k\to \mathrm{\infty }$ for the fact that $E\hookrightarrow \hookrightarrow L^r({\mathbb{R}}^N)$.

Now for $u\in Z_k$ with $\parallel u\parallel =r_k={({\beta }_2(k)+{\beta }_p(k))}^{-1}$, we obtain

$$\begin{array}{rl}J(u)& \ge \frac{1}{2}{\parallel u\parallel }^2-{\int }_{{\mathbb{R}}^N}(d_5{|u|}^2+d_6{|u|}^p)\mathrm{d}x=\frac{1}{2}{\parallel u\parallel }^2-d_5{\parallel u\parallel }_2^2-d_6{\parallel u\parallel }_p^p\\ \ge \frac{1}{2}{\parallel u\parallel }^2-d_5{\beta }_2^2(k){\parallel u\parallel }^2-d_6{\beta }_p^p(k){\parallel u\parallel }^p\ge \frac{1}{2}{r}_k^2-d_5-d_6\to \mathrm{\infty }\text{as }k\to \mathrm{\infty }.\end{array}$$

Hence,

$$a_k:=\underset{u\in Z_k,\parallel u\parallel =r_k}{inf}J(u)\to +\mathrm{\infty }\text{as }k\to \mathrm{\infty }.$$

(3.8)

Next we shall prove that, for any finite dimensional subspace $\mathcal{X}\subset E$, we have

$$J(u)\to -\mathrm{\infty }\text{as }\parallel u\parallel \to \mathrm{\infty },u\in \mathcal{X}.$$

(3.9)

Suppose the contrary. For some sequence $\{u_n\}\subset \mathcal{X}$ with $\parallel u_n\parallel \to \mathrm{\infty }$, there is a $M>0$ such that $J(u_n)\ge -M$ for all $n\in \mathbb{N}$. Put $v_n=\frac{u_n}{\parallel u_n\parallel }$ and then $\parallel v_n\parallel =1$. Up to a subsequence, assume that $v_n\rightharpoonup v$ weakly in E. Since $dim\mathcal{X}<\mathrm{\infty }$, $v_n\to v\in \mathcal{X}$ in E, $v_n\to v$ a.e. on ${\mathbb{R}}^N$, and $\parallel v\parallel =1$. Denote $\mathrm{\Omega }:=\{x\in {\mathbb{R}}^N:v(x)\ne 0\}$, then $meas(\mathrm{\Omega })>0$ and for a.e. $x\in \mathrm{\Omega }$, ${lim}_{n\to \mathrm{\infty }}|u_n(x)|\to \mathrm{\infty }$. It follows from (2.7) that

$$\underset{n\to \mathrm{\infty }}{lim}\frac{{\int }_{{\mathbb{R}}^N}F(x,u_n)\mathrm{d}x}{{\parallel u_n\parallel }^2}=\underset{n\to \mathrm{\infty }}{lim}\frac{\frac{1}{2}{\parallel u_n\parallel }^2-J(u_n)}{{\parallel u_n\parallel }^2}\le d_7\text{with a constant }{d}_7>0.$$

(3.10)

But, for large n, on account of F being nonnegative, (H2) and Fatou’s Lemma enable us to obtain

$$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{lim}\frac{{\int }_{{\mathbb{R}}^N}F(x,u_n)\mathrm{d}x}{{\parallel u_n\parallel }^2}& \ge \underset{n\to \mathrm{\infty }}{lim}{\int }_{\mathrm{\Omega }}\frac{F(x,u_n)v_n^2}{u_n^2}\mathrm{d}x\ge \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{\int }_{\mathrm{\Omega }}\frac{F(x,u_n)v_n^2}{u_n^2}\mathrm{d}x\\ \ge {\int }_{\mathrm{\Omega }}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\frac{F(x,u_n)v_n^2}{u_n^2}\mathrm{d}x\\ ={\int }_{\mathrm{\Omega }}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\frac{F(x,u_n)}{u_n^2}[{\chi }_{\mathrm{\Omega }}(x)]v_n^2\mathrm{d}x\to \mathrm{\infty }\end{array}$$

as $n\to \mathrm{\infty }$. This contradicts (3.10). Consequently, (3.9) holds, as required. Note that $dim{Y}_k<\mathrm{\infty }$ in Remark 2.10, and there exist positive constants $d_k$ such that

$$J(u)\le 0,\text{for each }u\in Y_k\text{ and }\parallel u\parallel \ge d_k.$$

(3.11)

Combining this and (3.8), we can take ${\rho }_k:=max\{d_k,r_k+1\}$, and thus $b_k:={max}_{\{u\in Y_k,\parallel u\parallel ={\rho }_k\}}J(u)\le 0$. Until now, we have proved the functional J satisfies all the conditions of Lemma 2.8. Hence, J has an unbounded sequence of critical values, i.e., there exists a sequence $\{u_k\}\subset E$ such that $J^{\prime }(u_k)=0$ and $J(u_k)\to +\mathrm{\infty }$ as $k\to \mathrm{\infty }$. This completes the proof. □

We prove that there exists $a_0=a_0(r_0)>0$ ($r_0$ is determined in (H5)) such that

$$|-\beta F(x,s)+sf(x,s)|\le a_0{|s|}^2,|s|\le r_0,\text{ for all }x\in {\mathbb{R}}^N.$$

(3.12)

Indeed, by (3.6) we see

$$|F(x,s)|\le d_5{|s|}^2+d_6{|s|}^p\le (d_5+d_6{r}_0^{p-2}){|s|}^2,|s|\le r_0,\mathrm{\forall }x\in {\mathbb{R}}^N.$$

This, together with (H1), implies that

$$\begin{array}{rl}|-\beta F(x,s)+sf(x,s)|& \le \beta |F(x,s)|+|sf(x,s)|\le \beta (d_5+d_6{r}_0^{p-2}){|s|}^2+\left|s(d_1|s|+d_2{|s|}^{p-1})\right|\\ \le (d_1+\beta d_5+(d_2+\beta d_6)r_0^{p-2}){|s|}^2,\text{if }|s|\le r_0.\end{array}$$

Clearly, (3.12) holds true with $a_0=d_1+\beta d_5+(d_2+\beta d_6)r_0^{p-2}$. In the following theorem, we make the following assumption instead of (V):

(V′): $V\in C({\mathbb{R}}^N,\mathbb{R})$, ${inf}_{x\in {\mathbb{R}}^N}V(x)=V_0\ge {[\frac{1}{2a_0+1}(\frac{\beta }{2}-1)]}^{-1}>0$, where $a_0$ in (3.12), β in (H5).Especially, by (V′), we obtain

$${\parallel u\parallel }_2^2\le \frac{1}{2a_0+1}(\frac{\beta }{2}-1){\parallel u\parallel }^2,u\in E.$$

(3.13)

Now, we define a class of functionals on E by

$$J_{\lambda }(u)=\frac{1}{2}{\parallel u\parallel }^2-\lambda {\int }_{{\mathbb{R}}^N}F(x,u)\mathrm{d}x=A(u)-\lambda B(u),\lambda \in [1,2].$$

It is easy to know that $J_{\lambda }\in C^1(E,\mathbb{R})$ for all $\lambda \in [1,2]$ and the critical points of $J_1$ correspond to the weak solutions of problem (1.1). Note that $J_1=J$, where J is the functional defined in (2.7).

Theorem 3.2

Assume that (V′), (H1), and (H4)-(H6) hold. Then (1.1) possesses infinitely many weak solutions.

Proof

We first prove that there exist a positive integer $k_1$ and two sequences $r_k>{\rho }_k\to \mathrm{\infty }$ as $k\to \mathrm{\infty }$ such that

$$a_k(\lambda )=\underset{u\in Z_k,\parallel u\parallel ={\rho }_k}{inf}{J}_{\lambda }(u)>0,\mathrm{\forall }k\ge k_1,$$

(3.14)

$$b_k(\lambda )=\underset{u\in Y_k,\parallel u\parallel =r_k}{max}{J}_{\lambda }(u)<0,\mathrm{\forall }k\in \mathbb{N},$$

(3.15)

where $Y_k$ and $Z_k$ are defined in Remark 2.10.

Step 1. We claim that (3.14) is true.

By (3.6) and (H4) we have

$$\begin{array}{rl}{J}_{\lambda }(u)& =\frac{1}{2}{\parallel u\parallel }^2-\lambda {\int }_{{\mathbb{R}}^N}F(x,u)\mathrm{d}x\ge \frac{1}{2}{\parallel u\parallel }^2-2{\int }_{{\mathbb{R}}^N}F(x,u)\mathrm{d}x\\ \ge \frac{1}{2}{\parallel u\parallel }^2-2{\int }_{{\mathbb{R}}^N}(d_5{|u|}^2+d_6{|u|}^p)\mathrm{d}x\ge \frac{1}{2}{\parallel u\parallel }^2-2d_5{\parallel u\parallel }_2^2-2d_6{\parallel u\parallel }_p^p.\end{array}$$

(3.16)

Since $E\hookrightarrow \hookrightarrow L^r({\mathbb{R}}^N)$ with $r\in [2,2_{\alpha }^{\ast })$, and from Theorem 3.1 we have

$$J_{\lambda }(u)\ge \frac{1}{2}{\parallel u\parallel }^2-2d_5{\beta }_2^2(k){\parallel u\parallel }^2-2d_6{\beta }_p^p(k){\parallel u\parallel }^p.$$

Let ${\rho }_k=\frac{1}{{\beta }_2(k)+{\beta }_p(k)}\to \mathrm{\infty }$ as $k\to \mathrm{\infty }$. Then there exists $k_1$ such that $\frac{1}{2}{\rho }_k^2-2d_5-2d_6>0$, $\mathrm{\forall }k\ge k_1$. Therefore,

$$a_k(\lambda )=\underset{u\in Z_k,\parallel u\parallel ={\rho }_k}{inf}{J}_{\lambda }(u)\ge \frac{1}{2}{\rho }_k^2-2d_5-2d_6>0,\mathrm{\forall }k\ge k_1.$$

Step 2. We show that (3.15) is true.

We apply the method in Lemma 2.6 of [[20]] to verify the claim. First, we prove that there exists $\epsilon >0$ such that

$$meas(x\in {\mathbb{R}}^N:|u(x)|\ge \epsilon \parallel u\parallel )\ge \epsilon ,\mathrm{\forall }u\in \mathcal{X}\mathrm{\setminus }\{0\},\mathrm{\forall }\mathcal{X}\subset E\text{ and }dim\mathcal{X}<\mathrm{\infty }.$$

(3.17)

There would otherwise exist a sequence ${\{u_n\}}_{n\in \mathbb{N}}\subset \mathcal{X}\mathrm{\setminus }\{0\}$ such that

$$meas(x\in {\mathbb{R}}^N:|u_n(x)|\ge \frac{\parallel u_n\parallel }{n})<\frac{1}{n},\mathrm{\forall }n\in \mathbb{N}.$$

(3.18)

For each $n\in \mathbb{N}$, let $v_n:=\frac{u_n}{\parallel u_n\parallel }\in \mathcal{X}$. Then $\parallel v_n\parallel =1$, $\mathrm{\forall }n\in \mathbb{N}$ and

$$meas(x\in {\mathbb{R}}^N:|v_n(x)|\ge \frac{1}{n})<\frac{1}{n},\mathrm{\forall }n\in \mathbb{N}.$$

(3.19)

Passing to a subsequence if necessary, we may assume $v_n\to v_0$ in E for some $v_0\in \mathcal{X}$ since $\mathcal{X}$ is of finite dimension. We easily find $\parallel v_0\parallel =1$. Consequently, there exists a constant ${\sigma }_0>0$ such that

$$meas(x\in {\mathbb{R}}^N:|v_0(x)|\ge {\sigma }_0)\ge {\sigma }_0.$$

(3.20)

Indeed, if not, then we have

$$meas(x\in {\mathbb{R}}^N:|v_0(x)|\ge \frac{1}{n})=0,\mathrm{\forall }n\in \mathbb{N},$$

(3.21)

which implies

$$0\le {\int }_{{\mathbb{R}}^N}{|v_0(x)|}^3\mathrm{d}x\le \frac{{\parallel v_0\parallel }_2^2}{n}\to 0\text{as }n\to \mathrm{\infty }.$$

This leads to $v_0=0$, contradicting $\parallel v_0\parallel =1$. In view of $E\hookrightarrow \hookrightarrow L^p({\mathbb{R}}^N)$ and the equivalence of any two norms on $\mathcal{X}$, we have

$${\int }_{{\mathbb{R}}^N}{|v_n-v_0|}^2\mathrm{d}x\to 0\text{as }n\to \mathrm{\infty }.$$

(3.22)

For every $n\in \mathbb{N}$, denote

$$\mathcal{N}:=\{x\in {\mathbb{R}}^N:|v_n(x)|<\frac{1}{n}\}\text{and}{\mathcal{N}}^c:=\{x\in {\mathbb{R}}^N:|v_n(x)|\ge \frac{1}{n}\},$$

and ${\mathcal{N}}_0:=\{x\in {\mathbb{R}}^N:|v_0(x)|\ge {\sigma }_0\}$, where ${\sigma }_0$ is defined by (3.20). Then for n large enough, by (3.20), we see

$$meas(\mathcal{N}\cap {\mathcal{N}}_0)\ge meas({\mathcal{N}}_0)-meas\left({\mathcal{N}}^c\right)\ge {\sigma }_0-\frac{1}{n}\ge \frac{3{\sigma }_0}{4}.$$

Consequently, for n large enough, we find

$$\begin{array}{rl}{\int }_{{\mathbb{R}}^N}{|v_n-v_0|}^2\mathrm{d}x& \ge {\int }_{\mathcal{N}\cap {\mathcal{N}}_0}{|v_n-v_0|}^2\mathrm{d}x\ge \frac{1}{4}{\int }_{\mathcal{N}\cap {\mathcal{N}}_0}{|v_0|}^2\mathrm{d}x-{\int }_{\mathcal{N}\cap {\mathcal{N}}_0}{|v_n|}^2\mathrm{d}x\\ \ge (\frac{{\sigma }_0^2}{4}-\frac{1}{n^2})meas(\mathcal{N}\cap {\mathcal{N}}_0)\ge \frac{9{\sigma }_0^3}{64}>0.\end{array}$$

This contradicts (3.22). Therefore, (3.17) holds. For the ε given in (3.17), we let

$${\mathcal{N}}_u:=\{x\in {\mathbb{R}}^N:|u(x)|\ge \epsilon \parallel u\parallel \},\mathrm{\forall }u\in \mathcal{X}\mathrm{\setminus }\{0\}.$$

Then by (3.17), we find

$$meas({\mathcal{N}}_u)\ge \epsilon ,\mathrm{\forall }u\in \mathcal{X}\mathrm{\setminus }\{0\}.$$

(3.23)

As is well known, (H5) implies (H2), and hence for any $k\in \mathbb{N}$, there is a constant $S_k>0$ such that

$$F(x,u)\ge \frac{{|u|}^2}{{\epsilon }^3},\mathrm{\forall }|u|\ge S_k,$$

where ε is determined in (3.17). Therefore,

$$\begin{array}{rl}{J}_{\lambda }(u)& =\frac{1}{2}{\parallel u\parallel }^2-\lambda {\int }_{{\mathbb{R}}^N}F(x,u)\mathrm{d}x\le \frac{1}{2}{\parallel u\parallel }^2-{\int }_{{\mathcal{N}}_u}\frac{{|u|}^2}{{\epsilon }^3}\mathrm{d}x\le (\frac{1}{2}-1){\parallel u\parallel }^2.\end{array}$$

Now for any $k\in \mathbb{N}$, if we take $r_k>max\{{\rho }_k,\frac{S_k}{\epsilon }\}$, so $\parallel u\parallel =r_k$ is large enough, we have

$$b_k(\lambda )=\underset{u\in Y_k,\parallel u\parallel =r_k}{max}{J}_{\lambda }(u)<0,\mathrm{\forall }k\in \mathbb{N}.$$

Step 3. Clearly, $J_{\lambda }\in C^1(E,\mathbb{R})$ implies that $J_{\lambda }$ maps bounded sets to bounded sets uniformly for $\lambda \in [1,2]$. In view of (H6), $J_{\lambda }(-u)=J_{\lambda }(u)$ for all $(\lambda ,u)\in [1,2]\times E$. Thus the condition (T1) of Lemma 2.9 holds. Besides, the condition (T2) of Lemma 2.9 holds for the fact that $A(u)=\frac{1}{2}{\parallel u\parallel }^2\to \mathrm{\infty }$ as $\parallel u\parallel \to \mathrm{\infty }$ and $B(u)\ge 0$ since $F(x,u)\ge 0$. Evidently, Step 1 and Step 2 imply that the condition (T3) of Lemma 2.9 also holds for all $k\ge k_1$. Consequently, Lemma 2.9 implies that for any $k\ge k_1$ and a.e. $\lambda \in [1,2]$, there exists a sequence ${\{u_n^k(\lambda )\}}_{n=1}^{\mathrm{\infty }}$ such that

$$\underset{n}{sup}\parallel u_n^k(\lambda )\parallel <\mathrm{\infty },J_{\lambda }^{\prime }(u_n^k(\lambda ))\to 0\text{and}{J}_{\lambda }(u_n^k(\lambda ))\to {\zeta }_k(\lambda )\text{as }n\to \mathrm{\infty },$$

where

$$\begin{array}{c}{B}_k=\{u\in Y_k:\parallel u\parallel \le r_k\},{\mathrm{\Gamma }}_k=\{\gamma \in C(B_k,W):\gamma \text{ is odd},\gamma {|}_{\partial B_k}=\text{id}\},\hfill \\ {\zeta }_k(\lambda )=\underset{\gamma \in {\mathrm{\Gamma }}_k}{inf}\underset{u\in B_k}{max}{J}_{\lambda }(\gamma (u)),\mathrm{\forall }\lambda \in [1,2].\hfill \end{array}$$

Furthermore, we easily have ${\zeta }_k(\lambda )\in [{\overline{a}}_k,{\overline{\zeta }}_k]$, $\mathrm{\forall }k\ge k_1$, where ${\overline{\zeta }}_k:={max}_{u\in B_k}{J}_{\lambda }(\gamma (u))$ and ${\overline{a}}_k:=\frac{1}{2}{\rho }_k^2-2d_5-2d_6\to \mathrm{\infty }$ as $k\to \mathrm{\infty }$.

Claim 1. ${\{u_n^k(\lambda )\}}_{n=1}^{\mathrm{\infty }}\subset E$ possesses a strong convergent subsequence in E, a.e. $\lambda \in [1,2]$ and $k\ge k_1$. In fact, by the boundedness of ${\{u_n^k(\lambda )\}}_{n=1}^{\mathrm{\infty }}$, passing to a subsequence, as $n\to \mathrm{\infty }$, we may assume $u_n^k(\lambda )\rightharpoonup u^k(\lambda )$ in E. By the method of Theorem 3.1, we easily prove that $u_n^k(\lambda )\to u^k(\lambda )$ strongly in E.

Thus, for each $k\ge k_1$, we can choose ${\lambda }_l\to 1$ such that for the sequence ${\{u_n^k({\lambda }_l)\}}_{n=1}^{\mathrm{\infty }}$ we have obtained a convergent subsequence, and passing again to a subsequence, we may assume

$$\underset{n\to \mathrm{\infty }}{lim}{u}_n^k({\lambda }_l)=u_l^k\text{ in }E,\mathrm{\forall }l\in \mathbb{N}\text{ and }k\ge k_1.$$

Thus we obtain

$$J_{{\lambda }_l}^{\prime }\left(u_l^k\right)=0\text{and}{J}_{{\lambda }_l}\left(u_l^k\right)\in [{\overline{a}}_k,{\overline{\zeta }}_k],\mathrm{\forall }l\in \mathbb{N}\text{ and }k\ge k_1.$$

(3.24)

Claim 2. $\{u_l^k\}$ is bounded in E and has a convergent subsequence with the limit $u^k\in E$ for all $k\ge k_1$. For convenience, we set $u_l^k=u_l$ for all $l\in \mathbb{N}$. Consequently, (3.12) and (H5) imply that

$$\begin{array}{r}\beta J_{{\lambda }_l}(u_l)-(J_{{\lambda }_l}^{\prime }(u_l),u_l)\\ =(\frac{\beta }{2}-1){\parallel u_l\parallel }^2+{\lambda }_l{\int }_{{\mathbb{R}}^N}[-\beta F(x,u_l(x))+f(x,u_l(x))u_l(x)]\mathrm{d}x\\ =(\frac{\beta }{2}-1){\parallel u_l\parallel }^2+{\lambda }_l{\int }_{|u_l|\le r_0}[-\beta F(x,u_l(x))+f(x,u_l(x))u_l(x)]\mathrm{d}x\\ +{\lambda }_l{\int }_{|u_l|>r_0}[-\beta F(x,u_l(x))+f(x,u_l(x))u_l(x)]\mathrm{d}x\\ \ge (\frac{\beta }{2}-1){\parallel u_l\parallel }^2-{\lambda }_l{\int }_{|u_l|\le r_0}|-\beta F(x,u_l(x))+f(x,u_l(x))u_l(x)|\mathrm{d}x\\ \ge (\frac{\beta }{2}-1){\parallel u_l\parallel }^2-{\lambda }_l{a}_0{\parallel u_l\parallel }_2^2\ge (\frac{\beta }{2}-1){\parallel u_l\parallel }^2-2a_0{\parallel u_l\parallel }_2^2\\ \ge (\frac{\beta }{2}-1){\parallel u_l\parallel }^2-\frac{2a_0}{2a_0+1}(\frac{\beta }{2}-1){\parallel u_l\parallel }^2=\frac{1}{2a_0+1}(\frac{\beta }{2}-1){\parallel u_l\parallel }^2.\end{array}$$

Therefore, ${\{u_l\}}_{l=1}^{\mathrm{\infty }}$ is bounded in E. By Claim 1, we see that ${\{u_l\}}_{l=1}^{\mathrm{\infty }}$ has a convergent subsequence, which converges to an element $u^k\in W$ for all $k\ge k_1$.

Hence, passing to the limit in (3.24), we see $J_1^{\prime }(u^k)=0$ and $J_1(u^k)\in [{\overline{a}}_k,{\overline{\zeta }}_k]$, $\mathrm{\forall }l\in \mathbb{N}$ and $k\ge k_1$. Since ${\overline{a}}_k\to \mathrm{\infty }$ as $k\to \mathrm{\infty }$, we get infinitely many nontrivial critical points of $J_1=J$. Therefore (1.1) possesses infinitely many nontrivial solutions by Lemma 2.9. This completes the proof. □