Existence of infinitely many solutions for generalized Schrödinger-Poisson system

Abstract

We study the nonlinear generalized Schrödinger-Poisson system: $-\mathrm{\Delta }u+V(x)u+K(x)\varphi g(u)=f(x,u)$, in $R^3$, $-\mathrm{\Delta }\varphi =2K(x)G(u)$, in $R^3$, where $V(x)$ and $K(x)$ are non-negative functions. The function $f(x,u)$ is superlinear. Under appropriate assumptions on $V(x)$, $K(x)$, and $g(u)$, we prove the existence and multiplicity of nontrivial solutions by the variant fountain theorem established by Zou. Some recent results due to different authors are extended.

MSC: 35J20, 35J65, 35J60.

1 Introduction

Consider the existence and multiplicity of nontrivial solutions for the nonlinear generalized Schrödinger-Poisson system:

$$\{\begin{array}{c}-\mathrm{\Delta }u+V(x)u+K(x)\varphi g(u)=f(x,u),\text{in }{R}^3,\hfill \\ -\mathrm{\Delta }\varphi =2K(x)G(u),\text{in }{R}^3,\hfill \end{array}$$

(1.1)

where $G(u):={\int }_0^{u}g(t)dt$. Setting $g(u)=u$, (1.1) represents the well-known Schrödinger-Poisson system:

$$\{\begin{array}{c}-\mathrm{\Delta }u+V(x)u+K(x)\varphi u=f(x,u),\text{in }{R}^3,\hfill \\ -\mathrm{\Delta }\varphi =K(x)u^2,\text{in }{R}^3.\hfill \end{array}$$

(1.2)

Such a system arises in an interesting physical context. If we look for solitary solutions of the Schrödinger equation for a particle in an electrostatic field, we just need to solve (1.2). We refer the interested readers to [1], [2] for more details of the physical aspects.

With the aid of variational methods, under various hypotheses on $V(x)$, $K(x)$, and $f(x,u)$, system (1.2) has been extensively investigated over the past several decades. See for example, Benci and Fortunato [1], D’Aprile and Mugnai [2], Ambrosetti and Ruiz [3], Ambrosetti [4], and the references therein.

In recent years, there has been a lot of research on the existence of solutions for system (1.2) with $f(x,u)={|u|}^{p-1}u$ and the potential $V(x)$ being radially symmetric or nonradial. In [3], [4], the authors proved the existence of infinitely many pairs of high energy radial solutions when $2<p<5$, and also obtained some existence results for $1<p\le 2$. Sun [5] studied the existence of infinitely many solutions when $p\in (0,1)$. The authors of [6] proved the existence of positive solutions without compactness conditions if $3<p<5$. Azzollini and Pomponio [7] proved the existence of a ground state solution to (1.2) for $3<p<5$.

Furthermore, Sun et al.[8] studied system (1.2) for $f(x,u)$ being asymptotically linear and obtained ground state solutions. In [9], Huang et al. obtained the existence of at least a pair of fixed sign solutions and a pair of sign-changing solutions to system (1.2) involving a critical nonlinearity. Ding et al.[10] studied (1.2) with a nonhomogeneous term, where $f(x,u)=f(u)$ is either asymptotically linear or asymptotically 3-linear with respect to u at infinity. Very recently, Liu and Guo [11] studied (1.2) with critical growth and obtained the existence of ground state solutions via variational methods.

The problem of finding infinitely many large energy solutions is a very classical problem. There is an extensive literature concerning the existence of infinitely many large energy solutions of (1.2). Chen and Tang [12] obtained infinitely many large energy solutions by the following variant ‘Ambrosetti-Rabinowitz’ type condition (AR for short)

$$\mathrm{\exists }\mu >4\text{ such that }0\le \mu F(x,u)\le uf(x,u),\text{for every }x\in R^3,|u|\ge 1,$$

where $F(x,u):={\int }_0^{u}f(x,s)ds$. After that, in [13], the authors studied (1.2) without the (AR) condition by the variant fountain theorem established by Zou [14], Theorem 2.1]. Later, Li and Chen in [15] also obtained infinitely many large energy solutions of (1.2) with $K(x)=1$ without the (AR) condition.

By using the method of a cut-off function and the variational arguments, the authors [16] studied the following Schrödinger-Poisson system:

$$\{\begin{array}{l}-\mathrm{\Delta }u+\epsilon q\varphi f(u)=\eta {|u|}^{p-1}u,\text{in }\mathrm{\Omega },\\ -\mathrm{\Delta }\varphi =2qF(u),\text{in }\mathrm{\Omega },\\ u=\varphi =0,\text{on }\partial \mathrm{\Omega },\end{array}$$

where $\mathrm{\Omega }\subset R^3$ is a bounded domain with smooth boundary ∂ Ω, $1<p<5$, $\epsilon ,\eta =\pm 1$, $f:R\to R$ is a continuous function and $F(t):={\int }_0^{t}f(s)ds$. They proved the existence and multiplicity results assuming on f a subcritical growth condition and also they considered the existence and nonexistence results in the critical case. Lately, Li and Zhang [17] discussed (1.1) with $V(x)$ and $K(x)$ being constants and obtained the existence of a positive radially symmetric solution without compactness conditions.

Motivated by the above facts, in the present paper our aim is to study the existence of infinitely many solutions for system (1.1). To the best of our knowledge, the existence and multiplicity of nontrivial solutions to system (1.1) has never been studied by variational methods, where g is a more general function, $f(x,u)\ne {|u|}^{p-1}u$ is also general, and $V(x)$ and $K(x)$ may be non-radial symmetrical and non-periodic. Before we state our main result, we list some conditions as follows, which have a role to play.

• (V) $V(x)\in C(R^3,R)$ satisfies ${inf}_{x\in R^3}V(x)\ge a>0$, where $a>0$ is a constant. Moreover, for any $M>0$, $meas\{x\in R^3:V(x)\le M\}<\mathrm{\infty }$, where $meas(\cdot )$ denotes the Lebesgue measure in $R^3$.

• (K) $K(x)\in L^{\mathrm{\infty }}(R^3,R)$, and $K(x)\ge 0$ for any $x\in R^3$.

(g1)$g\in C(R,R)$ and there exist constants $c>0$, $\frac{5}{3}<\alpha <2$ such that

$$|g(u)|\le c(|u|+{|u|}^{\alpha -1})\text{for all }u\in R.$$

(g2)$g(-u)=-g(u)$, $\mathrm{\forall }u\in R$.

(g3)$G(u)\ge 0$ for all $u\in R$, and there exists a constant ${\vartheta }_1\ge 1$ such that

$$\tilde{G}(su)\le {\vartheta }_1\tilde{G}(u),\mathrm{\forall }s\in [0,1],$$

where $\tilde{G}(u):=G(u)-\frac{1}{4}g(u)u$.

(f1)$f\in C(R^3\times R,R)$ and there exist constants $c_1>0$, $4<\nu <2^{\ast }$ such that

$$|f(x,u)|\le c_1(1+{|u|}^{\nu -1})\text{for all }(x,u)\in R^3\times R,$$

where $2^{\ast }=6$ is the critical exponent for the Sobolev embedding in dimension 3.

(f2)${lim}_{|u|\to 0}\frac{f(x,u)}{u}=0$ uniformly for $x\in R^3$.

(f3)$F(x,0)\equiv 0$, $F(x,u)\ge 0$ and ${lim}_{|u|\to \mathrm{\infty }}\frac{F(x,u)}{u^4}=+\mathrm{\infty }$ uniformly for $x\in R^3$.

(f4) For a.e. $x\in R^3$, there exists a constant ${\vartheta }_2\ge 1$ such that

$$\tilde{F}(x,su)\le {\vartheta }_2\tilde{F}(x,u),\mathrm{\forall }(x,u)\in R^3\times R,\text{ and }s\in [0,1],$$

where $\tilde{F}(x,u)=\frac{1}{4}f(x,u)u-F(x,u)$.

(f5)$f(x,-u)=-f(x,u)$, $\mathrm{\forall }(x,u)\in R^3\times R$.

Our main result reads as follows.

Theorem 1.1

Assume that (V), (K), (g1)-(g3), and (f1)-(f5) hold; the problem (1.1) possesses infinitely many nontrivial solutions$\{(u_k,{\varphi }_k)\}$satisfying

$$\begin{array}{r}\frac{1}{2}{\int }_{R^3}({|\mathrm{\nabla }{u}_k|}^2+V(x)u_k^2)dx-\frac{1}{4}{\int }_{R^3}{|\mathrm{\nabla }{\varphi }_k|}^{2}dx\\ +{\int }_{R^3}K(x){\varphi }_{k}G(u_k)dx-{\int }_{R^3}F(x,u_k)dx\to +\mathrm{\infty },\mathit{\text{as }}k\to \mathrm{\infty }.\end{array}$$

Remark 1.1

It is well known that the (AR) condition is used to guarantee the boundedness of (P.S.) sequences of the corresponding functional. However, there are functions satisfying the assumptions (f1)-(f5), but not satisfying the (AR) condition, for instance, $f(x,t)=a(x)t^{3}ln(1+|t|)$, where $0<{inf}_{R^3}a(x)\le {sup}_{R^3}a(x)<\mathrm{\infty }$.

Remark 1.2

Li et al.[13] used

(f′4):$\tilde{F}(x,s)\le \tilde{F}(x,t)$, $\mathrm{\forall }(s,t)\in R^{+}\times R^{+}$, $s\le t$.

to solve the problem (1.2). We claim that our condition (f4) is more general than (f′4). In fact, setting ${\vartheta }_2=1$, we find that $\tilde{F}(x,t)$ is increasing in $R^{+}$ with respect to t. Moreover, the function $f(x,t)=a(x)[4t^{3}ln(1+t^4)+4sint]$ satisfies (f4) but not satisfies (f′4), where $0<{inf}_{R^3}a(x)\le {sup}_{R^3}a(x)<\mathrm{\infty }$.

Remark 1.3

It is easy to present many functions satisfying (g1)-(g3), for example, $g(u)=u$, $g(u)=u^{\frac{1}{3}}$, $g(u)=u^{\frac{1}{5}}$, and so on. Moreover, setting $K(x)=1$, $g(u)=u$ in (1.1), we can obtain similar results to the problem (1.1) in [15]. But our proof is different from [15]. For this reason, we use a small step.

Remark 1.4

Since we have the lack of the (AR) condition, in order to obtain the boundedness of (P.S.) sequences (see the proof of Theorem 1.1), we assume the range of α in (g1) is $\frac{5}{3}<\alpha <2$, i.e. α is subquadratic, but that of reference [17] is superquadratic and there exist some functions which satisfy the condition (f) in [17] that do not satisfy the conditions (g1) and (g3), for example, $g(u)=u^3$.

The outline of the paper is as follows: in Section 2, we present some preliminary results, which are necessary for Section 3. In Section 3, we give the proof of Theorem 1.1. Throughout the paper we shall denote by $C_i>0$ various positive constants.

2 Preliminaries

In this section we outline the variational framework for the problem (1.1) and give some preliminary lemmas. Define the function space

$$E=\{u\in H^1\left(R^3\right):{\int }_{R^3}({|\mathrm{\nabla }u|}^2+V(x)u^2)dx<+\mathrm{\infty }\}.$$

Then E is a Hilbert space equipped with the inner product and norm

$$〈u,v〉={\int }_{R^3}(\mathrm{\nabla }u\cdot \mathrm{\nabla }v+V(x)uv)dx,\parallel u\parallel ={〈u,u〉}^{1/2}.$$

Since $V(x)$ is bounded from below, the E is continuously embedded into $L^q(R^3)$ for all $q\in [2,2^{\ast }]$. Therefore, there exists a positive constant ${\eta }_q>0$ such that

$${\parallel u\parallel }_{L^q}\le {\eta }_q\parallel u\parallel ,\mathrm{\forall }u\in E,$$

(2.1)

where

$${\parallel u\parallel }_{L^q}:={\left({\int }_{R^3}{|u|}^{q}dx\right)}^{\frac{1}{q}},\text{for any }q\in [1,\mathrm{\infty })$$

is the norm of the usual Lebesgue space $L^q(R^3)$. Moreover, by (V), the embedding $E\hookrightarrow L^q(R^3)$ is also compact for any $q\in [2,2^{\ast })$[18], Lemma 3.4]. Let $D^{1,2}(R^3)$ be the completion of $C_0^{\mathrm{\infty }}(R^3)$ with respect to the norm

$${\parallel u\parallel }_{D^{1,2}}={\left({\int }_{R^3}{|\mathrm{\nabla }u|}^{2}dx\right)}^{\frac{1}{2}}.$$

It is well known that the embedding $D^{1,2}(R^3)\hookrightarrow L^6(R^3)$ is continuous (see [19]).

It is clear that system (1.1) is the Euler-Lagrange equations of the functional $J:E\times D^{1,2}(R^3)\to R$ defined by

$$J(u,\varphi )=\frac{1}{2}{\parallel u\parallel }^2-\frac{1}{4}{\int }_{R^3}{|\mathrm{\nabla }\varphi |}^{2}dx+{\int }_{R^3}K(x)\varphi G(u)dx-{\int }_{R^3}F(x,u)dx.$$

Obviously, the action functional J belongs to $C^1(E\times D^{1,2}(R^3),R)$ and its critical points are the solutions of (1.1); see for instance [1]. For any $u\in E$, by the Lax-Milgram theorem, we can obtain the result that the second equation in (1.1) has a unique solution ${\varphi }_u$. Substituting ${\varphi }_u$ to the first equation of the problem (1.1), then the problem can be transformed to a one variable equation. In fact, we firstly get the following lemma.

Lemma 2.1

For any$u\in E$, we have

• (1) ${\varphi }_u\ge 0$;

• (2) ${\parallel {\varphi }_u\parallel }_{D^{1,2}}\le C_4({\parallel u\parallel }^2+{\parallel u\parallel }^{\alpha })$;

• (3) ${\int }_{R^3}2K(x)G(u){\varphi }_{u}dx\le C_5({\parallel u\parallel }^4+{\parallel u\parallel }^{2\alpha })$.

Proof

By the condition (g1), we find that there exists $C_1>0$ such that

$$G(u)\le C_1({|u|}^2+{|u|}^{\alpha }),\mathrm{\forall }u\in E.$$

(2.2)

Then, by the Minkowski inequality and (2.1), we have

$$\begin{array}{rl}{\parallel G(u)\parallel }_{L^{\frac{6}{5}}}& \le {\left({\int }_{R^3}{\left[C_1({|u|}^2+{|u|}^{\alpha })\right]}^{\frac{6}{5}}dx\right)}^{\frac{5}{6}}\\ \le C_1({\parallel u\parallel }_{L^{\frac{12}{5}}}^2+{\parallel u\parallel }_{L^{\frac{6\alpha }{5}}}^{\alpha })\\ \le C_2({\parallel u\parallel }^2+{\parallel u\parallel }^{\alpha }).\end{array}$$

(2.3)

For any $u\in E$, the linear functional $T_u:D^{1,2}(R^3)\to R$ is defined as

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

By the Sobolev embedding theorem, $K(x)\in L^{\mathrm{\infty }}(R^3)$, $K(x)\ge 0$, and (2.3), we have

$$\begin{array}{rl}{\int }_{R^3}2K(x)G(u)vdx& \le {\left({\int }_{R^3}{(2K(x)G(u))}^{\frac{6}{5}}dx\right)}^{\frac{5}{6}}{\left({\int }_{R^3}{|v|}^{6}dx\right)}^{\frac{1}{6}}\\ \le C_3{\parallel G(u)\parallel }_{L^{6/5}}{\parallel v\parallel }_{L^6}\\ \le C_4({\parallel u\parallel }^2+{\parallel u\parallel }^{\alpha }){\parallel v\parallel }_{D^{1,2}}.\end{array}$$

(2.4)

So, $T_u$ is continuous on $D^{1,2}(R^3)$. Hence, the Lax-Milgram theorem implies that, for every $u\in E$, there exists a unique ${\varphi }_u\in D^{1,2}(R^3)$ such that

$${\int }_{R^3}2K(x)G(u)vdx={\int }_{R^3}\mathrm{\nabla }{\varphi }_u\cdot \mathrm{\nabla }vdx,\text{for any }v\in D^{1,2}\left(R^3\right).$$

Using integration by parts, we get

$${\int }_{R^3}\mathrm{\nabla }{\varphi }_u\cdot \mathrm{\nabla }vdx=-{\int }_{R^3}v\mathrm{△}{\varphi }_{u}dx,\text{for any }v\in D^{1,2}\left(R^3\right),$$

therefore,

$$-\mathrm{△}{\varphi }_u=2K(x)G(u)$$

(2.5)

in a weak sense. We can write an integral expression for ${\varphi }_u$ in the form

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

for any $u\in C_0^{\mathrm{\infty }}(R^3)$ (see [20], Theorem 1 or Lemma 2.1 of [21]). It follows from $K(x)\ge 0$, (g2), and (g3) that ${\varphi }_u\ge 0$, and ${\varphi }_{-u}={\varphi }_u$ for any $u\in E$. By (2.4), (2.5), and $K(x)\in L^{\mathrm{\infty }}(R^3)$, for any $u\in E$ we get

$${\parallel {\varphi }_u\parallel }_{D^{1,2}}\le C_4({\parallel u\parallel }^2+{\parallel u\parallel }^{\alpha })\text{and}{\int }_{R^3}2K(x)G(u){\varphi }_{u}dx\le C_5({\parallel u\parallel }^4+{\parallel u\parallel }^{2\alpha }).$$

The proof is complete. □

We consider the functional $I:E\to R$ defined by $I(u)=J(u,{\varphi }_u)$. By (2.5), the reduced functional takes the form

$$I(u)=\frac{1}{2}{\parallel u\parallel }^2+\frac{1}{2}{\int }_{R^3}K(x){\varphi }_{u}G(u)dx-{\int }_{R^3}F(x,u)dx.$$

(2.6)

By (f1) and (f2), for any $\epsilon >0$, there exists $c(\epsilon )>0$ such that, for all $x\in R^3$, $u\in R$,

$$|f(x,u)|\le 2\epsilon |u|+\nu c(\epsilon ){|u|}^{\nu -1}$$

and

$$|F(x,u)|\le \epsilon {|u|}^2+c(\epsilon ){|u|}^{\nu }.$$

(2.7)

Therefore, by (f3), we obtain

$$0\le {\int }_{R^3}F(x,u)dx\le \epsilon {\parallel u\parallel }_{L^2}^2+c(\epsilon ){\parallel u\parallel }_{L^{\nu }}^{\nu }\le C_6{\parallel u\parallel }^2+C_7{\parallel u\parallel }^{\nu }.$$

(2.8)

Then, by Lemma 2.1, I is well defined and is a $C^1$ functional with derivative given by

$$〈I^{\prime }(u),v〉={\int }_{R^3}(\mathrm{\nabla }u\cdot \mathrm{\nabla }v+V(x)uv+K(x){\varphi }_{u}g(u)v-f(x,u)v)dx.$$

(2.9)

Now, we can apply Theorem 2.3 of [22] to our functional I and obtain the following.

Lemma 2.2

The following statements are equivalent:

• (1) $(u,\varphi )\in E\times D^{1,2}(R^3)$ is a solution of (1.1);

• (2) u is a critical point of I and $\varphi ={\varphi }_u$.

Since we do not assume the (AR) condition, the verification of the (P.S.) condition becomes complicated, so we use the following variant fountain theorem introduced in [14] without the (P.S.) condition to handle the problem (1.1).

Theorem 2.1

Let E be a Banach space with$\parallel \cdot \parallel$and$E=\overline{{⨁}_{j\in N}{X}_j}$with$dim{X}_j<\mathrm{\infty }$for any$j\in N$. Set$Y_k={⨁}_{j=0}^k{X}_j$, $Z_k=\overline{{⨁}_{j=k+1}^{\mathrm{\infty }}{X}_j}$and$B_k=\{u\in Y_k:\parallel u\parallel \le r_k\}$.

Consider the following $C^1$ functional ${\psi }_{\lambda }:E\to R$ defined by

$${\psi }_{\lambda }(u)=A(u)-\lambda B(u),\lambda \in [1,2],$$

where $A,B:E\to R$ are two functionals. Suppose that

(F1) ${\psi }_{\lambda }$ maps bounded sets to bounded sets uniformly for $\lambda \in [1,2]$. Furthermore,

$${\psi }_{\lambda }(-u)={\psi }_{\lambda }(u)\text{for all }(\lambda ,u)\in [1,2]\times E.$$

(F2) $B(u)\ge 0$ for all $u\in E$, and $A(u)\to \mathrm{\infty }$ or $B(u)\to \mathrm{\infty }$ as $\parallel u\parallel \to \mathrm{\infty }$.

(F3) There exist $r_k>{\rho }_k>0$ such that

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

Then

$${\alpha }_k(\lambda )\le {\varsigma }_k(\lambda ):=\underset{\gamma \in {\mathrm{\Gamma }}_k}{inf}\underset{u\in B_k}{max}{\psi }_{\lambda }(\gamma (u)),\mathrm{\forall }\lambda \in [1,2],$$

where ${\mathrm{\Gamma }}_k:=\{\gamma \in C(B_k,E):\gamma \text{ is odd }\gamma {|}_{\partial B_k}=id\}$. Moreover, for a.e. $\lambda \in [1,2]$, there exists a sequence ${\{u_m^k(\lambda )\}}_{m=1}^{\mathrm{\infty }}$ such that

$$\underset{m}{sup}\parallel u_m^k(\lambda )\parallel <\mathrm{\infty },{\psi }_{\lambda }^{\prime }(u_m^k(\lambda ))\to 0\text{and}{\psi }_{\lambda }(u_m^k(\lambda ))\to {\varsigma }_k(\lambda )\text{as }m\to \mathrm{\infty }.$$

3 Proof of Theorem 1.1

In order to apply Theorem 2.1 to prove our main result, we define the functional ${\psi }_{\lambda }$ on our working space E by

$${\psi }_{\lambda }(u)=\frac{1}{2}{\parallel u\parallel }^2+\frac{1}{2}{\int }_{R^3}K(x){\varphi }_{u}G(u)dx-\lambda {\int }_{R^3}F(x,u)dx:=A(u)-\lambda B(u),$$

(3.1)

for all $u\in E$ and $\lambda \in [1,2]$. Then $B(u)\ge 0$ for all $u\in E$, $A(u)\to \mathrm{\infty }$ as $\parallel u\parallel \to \mathrm{\infty }$.

We choose a completely orthonormal basis $\{e_j:j\in N\}$ of E and let $X_j=span\{e_j\}$ for all $j\in N$. Then $Z_k$, $Y_k$ can be defined as those in Section 2. Note that ${\psi }_1=I$, where I is the functional defined in (2.6). We further need the following lemmas.

Lemma 3.1

Let (V), (K), (g3), and (f1) be satisfied, then there exist a positive integer$k_1$and a sequence${\rho }_k\to +\mathrm{\infty }$as$k\to \mathrm{\infty }$such that

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

(3.2)

where$Z_k=\overline{{⨁}_{j=k+1}^{\mathrm{\infty }}{X}_j}=\overline{span\{e_{k+1},e_{k+2},\dots \}}$for all$k\in N$.

Proof

Set

$$l_2(k)=\underset{u\in Z_k,\parallel u\parallel =1}{sup}{\parallel u\parallel }_{L^2}\text{and}{l}_{\nu }(k)=\underset{u\in Z_k,\parallel u\parallel =1}{sup}{\parallel u\parallel }_{L^{\nu }},\mathrm{\forall }k\in N.$$

(3.3)

Since E is compactly embedded into $L^2(R^3)$ and $L^{\nu }(R^3)$, we have (see [19], Lemma 3.8])

$$l_2(k)\to 0\text{and}{l}_{\nu }(k)\to 0\text{as }k\to \mathrm{\infty }.$$

(3.4)

By (V), (K), (f1), (3.1), (3.3), and the fact that ${\int }_{R^3}2K(x){\varphi }_{u}G(u)dx\ge 0$, we have

$$\begin{array}{rl}{\psi }_{\lambda }(u)& \ge \frac{1}{2}{\parallel u\parallel }^2-\lambda {\int }_{R^3}F(x,u)dx\\ \ge \frac{1}{2}{\parallel u\parallel }^2-2(\epsilon {\parallel u\parallel }_{L^2}^2+c(\epsilon ){\parallel u\parallel }_{L^{\nu }}^{\nu })\\ \ge \frac{1}{2}{\parallel u\parallel }^2-2\epsilon l_2^2(k){\parallel u\parallel }^2-2c(\epsilon )l_{\nu }^{\nu }(k){\parallel u\parallel }^{\nu }.\end{array}$$

(3.5)

It follows from (3.4) that there exists a positive integer $k_1$ such that $2\epsilon l_2^2(k)\le \frac{1}{8}$, $\mathrm{\forall }k\ge k_1$. Then we have

$$\begin{array}{rl}{\psi }_{\lambda }(u)& \ge (\frac{1}{2}-\frac{1}{8}){\parallel u\parallel }^2-2c(\epsilon )l_{\nu }^{\nu }(k){\parallel u\parallel }^{\nu }\\ =\frac{3}{8}{\parallel u\parallel }^2-2c(\epsilon )l_{\nu }^{\nu }(k){\parallel u\parallel }^{\nu }.\end{array}$$

(3.6)

For each $k\ge k_1$, choose

$${\rho }_k:={(16c(\epsilon )l_{\nu }^{\nu }(k))}^{\frac{1}{2-\nu }}.$$

(3.7)

Then

$${\rho }_k\to +\mathrm{\infty }\text{as }k\to \mathrm{\infty }$$

(3.8)

since $4<\nu <2^{\ast }$. By (3.6) and (3.7), direct computation shows

$${\alpha }_k(\lambda ):=\underset{u\in Z_k,\parallel u\parallel ={\rho }_k}{inf}{\psi }_{\lambda }(u)\ge \frac{{\rho }_k^2}{4}>0,\mathrm{\forall }k\ge k_1.$$

(3.9)

The proof is complete. □

Lemma 3.2

Under the assumptions of (V), (K) and (f1)-(f3), then for the positive integer$k_1$and the sequence$\{{\rho }_k\}$obtained in Lemma 3.1, there exists$r_k>{\rho }_k$for each$k\ge k_1$such that

$${\beta }_k(\lambda ):=\underset{u\in Y_k,\parallel u\parallel =r_k}{max}{\psi }_{\lambda }(u)<0\mathit{\text{for }}\lambda \in [1,2],\mathrm{\forall }k\ge k_1,$$

where$Y_k={⨁}_{j=1}^k{X}_j=span\{e_1,\dots ,e_k\}$for all$k\in N$.

Proof

It follows from (f3) that, for any $M>0$, there exists $\delta =\delta (M)>0$ such that, for all $x\in R^3$, $|u|>\delta$,

$$F(x,u)\ge M{|u|}^4.$$

(3.10)

From (f1) and (f2), there exists $C_8=C_8(M)>0$ such that, for all $x\in R^3$ and $0<|u|\le \delta$,

$$\frac{|f(x,u)u|}{{|u|}^2}\le C_8.$$

Then, by the mean value theorem, for all $x\in R^3$, $0<|u|\le \delta$, we obtain

$$|F(x,u)|\le \frac{C_8}{2}{|u|}^2.$$

(3.11)

Set $\tilde{C}=M\delta +\frac{C_8}{2}$, combining (3.10) with (3.11), we get

$$F(x,u)\ge M{|u|}^4-\tilde{C}{|u|}^2,\mathrm{\forall }(x,u)\in R^3\times R.$$

(3.12)

For $u\in Y_k$, by Lemma 2.1 and (3.12), we have

$$\begin{array}{rl}{\psi }_{\lambda }(u)& \le \frac{1}{2}{\parallel u\parallel }^2+\frac{C_5}{4}({\parallel u\parallel }^4+{\parallel u\parallel }^{2\alpha })-{\int }_{R^3}F(x,u)dx\\ \le \frac{1}{2}{\parallel u\parallel }^2+\frac{C_5}{4}({\parallel u\parallel }^4+{\parallel u\parallel }^{2\alpha })-M{\parallel u\parallel }_{L^4}^4+\tilde{C}{\parallel u\parallel }_{L^2}^2\\ \le \frac{1}{2}{\parallel u\parallel }^2+\frac{C_5}{4}({\parallel u\parallel }^4+{\parallel u\parallel }^{2\alpha })-M{C}_9{\parallel u\parallel }^4+C_9\tilde{C}{\parallel u\parallel }^2,\end{array}$$

(3.13)

where in the last inequality we use the equivalence of all norms on the finite dimensional subspace $Y_k$. Let us choose M large enough such that $\frac{C_5}{4}-M{C}_9<0$. Then, when M is fixed, $\tilde{C}$ is also fixed. Since $\frac{5}{3}<\alpha <2$, we can choose $\parallel u\parallel =r_k>{\rho }_k>0$ such that

$${\beta }_k(\lambda ):=\underset{u\in Y_k,\parallel u\parallel =r_k}{max}{\psi }_{\lambda }(u)<0\text{for }\lambda \in [1,2],\mathrm{\forall }k\ge k_1.$$

The proof is complete. □

Proof of Theorem 1.1

It follows from Lemma 2.1 and (2.8) that ${\psi }_{\lambda }$ maps bounded sets to bounded sets uniformly for $\lambda \in [1,2]$. By (g2) and (f5), ${\psi }_{\lambda }(-u)={\psi }_{\lambda }(u)$ for all $(\lambda ,u)\in [1,2]\times E$. Thus, it follows from Lemmas 3.1 and 3.2 that the conditions of Theorem 2.1 are satisfied. Hence, for a.e. $\lambda \in [1,2]$, there exists a sequence ${\{u_m^k(\lambda )\}}_{m=1}^{\mathrm{\infty }}\subset E$ such that

$$\begin{array}{r}\underset{m}{sup}\parallel u_m^k(\lambda )\parallel <\mathrm{\infty },{\psi }_{\lambda }^{\prime }(u_m^k(\lambda ))\to 0\text{and}\\ {\psi }_{\lambda }(u_m^k(\lambda ))\to {\varsigma }_k(\lambda )\ge {\alpha }_k(\lambda )\ge {\tilde{\alpha }}_k\text{as }m\to \mathrm{\infty },\end{array}$$

(3.14)

where

$${\varsigma }_k(\lambda ):=\underset{\gamma \in {\mathrm{\Gamma }}_k}{inf}\underset{u\in B_k}{max}{\psi }_{\lambda }(\gamma (u)),\mathrm{\forall }\lambda \in [1,2],{\tilde{\alpha }}_k={\rho }_k^2/4\to \mathrm{\infty },\text{as }k\to \mathrm{\infty }$$

with $B_k=\{u\in Y_k:\parallel u\parallel \le r_k\}$ and ${\mathrm{\Gamma }}_k:=\{\gamma \in C(B_k,E):\gamma \text{ is odd},\gamma {|}_{\partial B_k}=id\}$.

Furthermore, it follows from the proof of Lemma 3.1 that

$${\varsigma }_k(\lambda )\in [{\tilde{\alpha }}_k,{\tilde{\varsigma }}_k]\text{for all }\lambda \in [1,2],$$

(3.15)

where ${\tilde{\varsigma }}_k:={max}_{u\in B_k}{\psi }_1(u)$.

In view of (3.14), we can choose ${\lambda }_n\to 1$ with ${\lambda }_n\in [1,2]$ and obtain the corresponding sequences $\{u_m^k({\lambda }_n)\}$ (denoted by $\{u_m\}$) satisfying

$$\underset{m}{sup}\parallel u_m\parallel <\mathrm{\infty },{\psi }_{{\lambda }_n}^{\prime }(u_m)\to 0\text{as }m\to \mathrm{\infty }.$$

(3.16)

Claim 1. The sequence $\{u_m\}$ has a strong convergent subsequence.

Set $\mathrm{\Phi }(u)={\int }_{R^3}F(x,u)dx$. It follows from (2.8) and the Sobolev embedding theorem that Φ is well defined and ${\mathrm{\Phi }}^{\prime }:E\to E^{\ast }$ is compact, where $〈{\mathrm{\Phi }}^{\prime }(u),v〉={\int }_{R^3}f(x,u)vdx$ for $u,v\in E$. By (3.16), without loss of generality, we may assume

$$u_m\rightharpoonup u_0\text{as }m\to \mathrm{\infty },$$

(3.17)

for $u_0\in E$. By virtue of the Riesz representation theorem, ${\mathrm{\Phi }}^{\prime }:E\to E^{\ast }$ can be viewed as ${\mathrm{\Phi }}^{\prime }:E\to E$. We obtain from (2.6) and (3.1) that

$$u_m={\psi }_{{\lambda }_n}^{\prime }(u_m)+{\lambda }_n{\mathrm{\Phi }}^{\prime }(u_m),\mathrm{\forall }m\in \mathrm{N}.$$

(3.18)

By (3.16), (3.17), and the compactness of ${\mathrm{\Phi }}^{\prime }:E\to E$, the right-hand side of (3.18) converges strongly in E. Hence $u_m\to u_0$ in E. We may suppose $u_m\to u^k({\lambda }_n)$ (denoted by $\{u_n\}$) as $m\to \mathrm{\infty }$; then we obtain from (3.15) and (3.16)

$${\psi }_{{\lambda }_n}^{\prime }(u_n)=0,{\psi }_{{\lambda }_n}(u_n)\in [{\tilde{\alpha }}_k,{\tilde{\varsigma }}_k].$$

(3.19)

Claim 2. The sequence $\{u_n\}$ is bounded.

If not, without loss of generality, we suppose that $\parallel u_n\parallel \to \mathrm{\infty }$. Let $w_n=\frac{u_n}{\parallel u_n\parallel }$, then, up to a sequence, in view of the compact embedding of E into $L^q(R^3)$, $2\le q<2^{\ast }$, we have

$$\begin{array}{r}{w}_n\rightharpoonup w_0\text{weakly in }E,\\ w_n\to w_0\text{strongly in }{L}^q\left(R^3\right),2\le q<2^{\ast },\\ w_n(x)\to w_0(x)\text{a.e. }x\in R^3.\end{array}$$

(3.20)

Case 1. $w_0=0$ in E. As in [23], we choose $\{t_n\}\subset [0,1]$ such that

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

For any $D>0$, we set ${\tilde{w}}_n=\sqrt{4D}{w}_n$. By (2.8) and (f3), we have

$$0\le {\int }_{R^3}F(x,{\tilde{w}}_n)dx\le {\int }_{R^3}(\epsilon {|{\tilde{w}}_n|}^2+c(\epsilon ){|{\tilde{w}}_n|}^{\nu })dx\to 0\text{as }n\to \mathrm{\infty }.$$

Then, choosing n sufficiently large such that

$${\psi }_{{\lambda }_n}(t_n{u}_n)\ge {\psi }_{{\lambda }_n}({\tilde{w}}_n)=2D+\frac{1}{2}{\int }_{R^3}K(x){\varphi }_{{\tilde{w}}_n}G({\tilde{w}}_n)dx-{\lambda }_n{\int }_{R^3}F(x,{\tilde{w}}_n)dx\ge D.$$

Thus, ${lim}_{n\to \mathrm{\infty }}{\psi }_{{\lambda }_n}(t_n{u}_n)=\mathrm{\infty }$. In view of the choice of $t_n$, we know that $〈{\psi }_{{\lambda }_n}^{\prime }(t_n{u}_n),t_n{u}_n〉=0$. Then, by (g3) and (f4), we have

$$\begin{array}{rl}\mathrm{\infty }\leftarrow {\psi }_{{\lambda }_n}(t_n{u}_n)=& {\psi }_{{\lambda }_n}(t_n{u}_n)-\frac{1}{4}〈{\psi }_{{\lambda }_n}^{\prime }(t_n{u}_n),t_n{u}_n〉\\ =& \frac{1}{4}{\parallel t_n{u}_n\parallel }^2+\frac{1}{2}{\int }_{R^3}K(x){\varphi }_{t_n{u}_n}[G(t_n{u}_n)-\frac{1}{4}g(t_n{u}_n)t_n{u}_n]dx\\ +{\lambda }_n{\int }_{R^3}[\frac{1}{4}f(x,t_n{u}_n)t_n{u}_n-F(x,t_n{u}_n)]dx\\ \le & \frac{1}{4}{\parallel u_n\parallel }^2+\frac{1}{2}{\vartheta }_1{\int }_{R^3}K(x){\varphi }_{u_n}[G(u_n)-\frac{1}{4}g(u_n)u_n]dx\\ +{\lambda }_n{\vartheta }_2{\int }_{R^3}[\frac{1}{4}f(x,u_n)u_n-F(x,u_n)]dx\\ \le & \frac{1}{4}{\parallel u_n\parallel }^2+\frac{1}{2}\vartheta {\int }_{R^3}K(x){\varphi }_{u_n}[G(u_n)-\frac{1}{4}g(u_n)u_n]dx\\ +{\lambda }_n\vartheta {\int }_{R^3}[\frac{1}{4}f(x,u_n)u_n-F(x,u_n)]dx\\ =& {\psi }_{{\lambda }_n}(u_n)-\frac{1}{4}\vartheta 〈{\psi }_{{\lambda }_n}^{\prime }(u_n),u_n〉\in [{\tilde{\alpha }}_k,{\tilde{\varsigma }}_k],\end{array}$$

where $\vartheta =max\{{\vartheta }_1,{\vartheta }_2\}$. This is a contradiction according to (3.19).

Case 2. $w_0\ne 0$ in E. On the subspace $\mathrm{\Omega }:=\{x\in R^3:w_0(x)\ne 0\}$, combining Lemma 2.1, (3.1), (3.20) with (f3), by Fatou’s lemma, we have

$$\begin{array}{r}\frac{1}{2{\parallel u_n\parallel }^2}+\frac{{\int }_{R^3}K(x)G(u_n){\varphi }_{u_n}dx}{2{\parallel u_n\parallel }^4}-\frac{{\psi }_{{\lambda }_n}(u_n)}{{\parallel u_n\parallel }^4}\\ ={\lambda }_n{\int }_{R^3}\frac{F(x,u_n)}{{\parallel u_n\parallel }^4}dx\\ \ge {\lambda }_n{\int }_{\mathrm{\Omega }}{|w_n|}^4\frac{F(x,u_n)}{{|u_n|}^4}dx\to +\mathrm{\infty }\end{array}$$

as $n\to \mathrm{\infty }$, a contradiction to (3.19) again. Then $\{u_n\}$ is bounded in E.

In view of Claim 2 and (3.19), using similar arguments to the proof of Claim 1, we can also show that the sequence $\{u_n\}$ has a strong convergent subsequence with the limit $u^k$ being just a critical point of $I={\psi }_1$. Obviously, $I(u^{(k)})\in [{\tilde{\alpha }}_k,{\tilde{\varsigma }}_k]$. Since ${\tilde{\alpha }}_k\to +\mathrm{\infty }$ as $k\to \mathrm{\infty }$, we know that ${\{u^{(k)}\}}_{k=1}^{\mathrm{\infty }}$ is an unbound sequence of critical points of functional I. Thus, the proof of Theorem 1.1 is complete. □