Existence and Concentration Behavior of Solutions of the Critical Schrödinger–Poisson Equation in ${\mathbb{R}}^3$

Abstract

In this paper, we study the singularly perturbed problem for the Schrödinger–Poisson equation with critical growth. When the perturbed coefficient is small, we establish the relationship between the number of solutions and the profiles of the coefficients. Furthermore, without any restriction on the perturbed coefficient, we obtain a different concentration phenomenon. Besides, we obtain an existence result.

1. Introduction

In this paper, we consider the Schrödinger–Poisson equation with critical growth:

$$\left\{\begin{array}{c}-{\epsilon }^2\Delta u+V\left(x\right)u+\varphi u=h\left(x\right)f\left(u\right)+g\left(x\right)u^5,\mathrm{in}{\mathbb{R}}^3,\hfill \\ -{\epsilon }^2\Delta \varphi =u^2,\mathrm{in}{\mathbb{R}}^3,\hfill \end{array}\right.$$

(1)

The Schrödinger–Poisson equation arises while looking for standing wave solutions of a Schrödinger equation interacting with an electrostatic field. In recent years, many researchers are interested in semiclassical states of

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

(2)

which can be used to describe the transition from quantum to classical mechanics.

When $K=0$, problem (2) reduces to the singularly perturbed problem

$$-{\epsilon }^2\Delta u+V\left(x\right)u=g\left(u\right)\mathrm{in}{\mathbb{R}}^N.$$

(3)

In the past decade, there is a lot of results on problem (3). By using variational methods, Rabinowitz [1] first obtained the existence of solutions of (3) under the assumption

$$\underset{x\to \infty }{lim\; inf}V\left(x\right)>\underset{x\in {\mathbb{R}}^N}{inf}V\left(x\right)>0.$$

(4)

In [2], Wang proved the concentration behavior of solutions of (3) as $\epsilon \to 0$. In [3], del Pino and Felmer introduced a penalization approach and obtained a localized version of the results in [1,2]. In [4], Jeanjean and Tanaka extended the results of [3] to a more general case. For other related results, see [5,6,7,8] and the reference therein.

When $K\ne 0$, a lot of research focus on the case $K\equiv 1$, $f=u^p$ $(1<p<5)$. By using the Lyapunov–Schmidt reduction method, the authors in [9,10] obtained positive bound state solutions and multi-bump solutions concentrating around a local minimum of the potential V. In [11,12], the authors proved the existence of radically symmetric solutions concentrating on the spheres. It should be pointed it out that, the Lyapunov–Schmidt reduction method is based on the uniqueness or non-degeneracy of solutions of the corresponding limiting equation. Recently, by using variational methods, he [13] considered the subcritical problem

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

(5)

Under the assumption (4), he related the number of solutions with the topology of the set where V attains its minimum and obtained the multiplicity of positive solutions. Subsequently, the authors in [14] studied the problem

$$\left\{\begin{array}{c}-{\epsilon }^2\Delta u+V\left(x\right)u+\lambda \varphi u=b\left(x\right)f\left(u\right),\mathrm{in}{\mathbb{R}}^3,\hfill \\ -{\epsilon }^2\Delta \varphi =u^2,\mathrm{in}{\mathbb{R}}^3.\hfill \end{array}\right.$$

(6)

Under suitable assumptions on $\lambda$, V, b and f, they proved the existence and concentration behavior of positive ground state solutions. For the critical case, He and Zou [15] studied the Schrödinger–Poisson equation

$$\left\{\begin{array}{c}-{\epsilon }^2\Delta u+V\left(x\right)u+\varphi u=f\left(u\right)+u^5,\mathrm{in}{\mathbb{R}}^3,\hfill \\ -{\epsilon }^2\Delta \varphi =u^2,\mathrm{in}{\mathbb{R}}^3.\hfill \end{array}\right.$$

(7)

By using (4), they obtained the ground state solution concentrating around the global minimum of the potential V. Furthermore, in [16,17], the authors considered the existence, multiplicity and concentration behavior of the critical Schrödinger–Poisson equation

$$\left\{\begin{array}{c}-{\epsilon }^2\Delta u+V\left(x\right)u+\varphi u=f\left(u\right)+u^5,\mathrm{in}{\mathbb{R}}^3,\hfill \\ -\Delta \varphi =u^2,\mathrm{in}{\mathbb{R}}^3.\hfill \end{array}\right.$$

(8)

Motivated by the above results, in this paper, we study the multiplicity and concentration behavior of positive solutions of (1). Before stating the results, we introduce the following conditions:

$\left(h_1\right)$

$h\left(x\right)\in C({\mathbb{R}}^3,\mathbb{R})$, $h\left(x\right)\ge 0$ and ${lim}_{\left|x\right|\to \infty }h\left(x\right)=h_{\infty }>0$. Moreover, there exist k points $x^1$, $x^2$, … $x^k$ in ${\mathbb{R}}^3$ such that each point is the strict global maximum of h.

$\left(V_1\right)$

$V\left(x\right)\in C({\mathbb{R}}^3,\mathbb{R})$ and ${inf}_{x\in {\mathbb{R}}^3}V\left(x\right):=V_0>0$.

$\left(V_2\right)$

$V\left(x^i\right)=V_0$, $i=1$, 2, … k and ${lim}_{\left|x\right|\to \infty }V\left(x\right)=V_{\infty }>0$.

$\left(g_1\right)$

$g\left(x\right)\in C({\mathbb{R}}^3,\mathbb{R})$ and $0\le g\left(x\right)\le 1$. Moreover, $g\left(x^i\right)=1$, $i=1$, 2, … k and ${lim}_{\left|x\right|\to \infty }g\left(x\right)=g_{\infty }>0$.

$\left(f_1\right)$

$f\in C^1({\mathbb{R}}^{+},\mathbb{R})$ and ${lim}_{u\to 0+}\frac{f\left(u\right)}{u^3}={lim}_{u\to +\infty }\frac{f\left(u\right)}{u^5}=0$. Moreover, $\frac{f\left(u\right)}{u^3}$ is increasing for $u>0$ and ${lim}_{u\to +\infty }\frac{f\left(u\right)}{u^3}=+\infty$.

Theorem 1.

Assume that $\left(h_1\right)$, $\left(V_1\right)$–$\left(V_2\right)$, $\left(g_1\right)$ and $\left(f_1\right)$ hold. Then there exists ${\epsilon }^{*}>0$ such that problem (1) has at least k different positive solutions $w_{\epsilon }^i$, $i=1$, 2, … k for $\epsilon \in (0,{\epsilon }^{*})$. Moreover, $w_{\epsilon }^i$ possesses a maximum point $y_{\epsilon }^i\in {\mathbb{R}}^3$ satisfying $h\left(y_{\epsilon }^i\right)\to {sup}_{x\in {\mathbb{R}}^3}h\left(x\right)$ as $\epsilon \to 0$. Besides, there exist $C^i$, $c^i>0$ such that for $\epsilon \in (0,{\epsilon }^{*})$,

$$\begin{array}{c}\hfill w_{\epsilon }^i\left(x\right)\le C^{i}exp\left(-c^i\frac{|x-y_{\epsilon }^i|}{\epsilon }\right),x\in {\mathbb{R}}^3.\end{array}$$

Remark 1.

In Theorem 1, we obtain the existence of spikes (multiple solutions concentrating at a single point) on the strict global maximum of h. The behavior of solutions of (1) describes the transition between quantum mechanics and classical mechanics in some sense.

Remark 2.

Some ideas to prove Theorem 1 come from [18], where the authors studied the subcritical problem

$$\begin{array}{c}\hfill -\Delta u+\mu u={Q\left(x\right)\left|u\right|}^{p-2}u\mathrm{in}{\mathbb{R}}^N.\end{array}$$

In [18], the authors imposed the condition

$$\begin{array}{c}\hfill \underset{x\in {\mathbb{R}}^N}{sup}Q\left(x\right)>\underset{\left|x\right|\to \infty }{lim\; sup}Q\left(x\right)>0,\end{array}$$

(9)

which plays an important role in proving the compactness of the Palais–Smale sequences. This type condition is first introduced by Rabinowitz in [1]. We pointed out that, when we seek multiplicity of solutions, it is crucial to prove the compactness of the Palais–Smale sequence. Many authors solved the problem by imposing the Rabinowitz type assumption, which is restrictive. Similar results can be found in [13,15,16] and the reference therein. In this paper, by estimating the Palais–Smale sequences delicately, we remove this technical condition. In fact, we use a different argument. Compared with the existing results, in this paper, we also need to study the influence of the variable coefficient of the critical term on the problem.

Inspired by Theorem 1, a natural question is whether (1) has multiple solutions without any restriction on $\epsilon$. In this paper, by using the Lusternik–Schnirelman category, we obtain a new result. We assume the following conditions:

$\left(h_2\right)$

$h\left(x\right)\in C({\mathbb{R}}^3,\mathbb{R})$ is positive and ${sup}_{x\in {\mathbb{R}}^3}h\left(x\right):=h_M<+\infty$.

$\left(Vh\right)$

${lim}_{R\to +\infty }{sup}_{\left|x\right|\ge R}\frac{h\left(x\right)}{V\left(x\right)}=0$.

$\left(g_2\right)$

$g\left(x\right)\in C({\mathbb{R}}^3,\mathbb{R})$ and $0\le g\left(x\right)\le 1$. Moreover, there exists ${\rho }_0>0$ such that $g\left(x\right)=1$ for ${\rho }_0<\left|x\right|<2{\rho }_0$.

Theorem 2.

Let $\epsilon >0$. Assume that $\left(h_2\right)$, $\left(V_1\right)$, $\left(Vh\right)$, $\left(g_2\right)$ and $\left(f_1\right)$ hold with $g\left(0\right)<1$. Then there exists $h_0>0$ such that for ${\parallel h\parallel }_{\infty }<h_0$, problem (1) has two positive solutions $u_{i,h}$, $i=1$, 2.

When $\epsilon =1$, there are an enormous amount of papers studying problem (1) or the more general form

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

(10)

Many papers, see for example, [19,20,21,22,23,24,25,26,27], focus on the case V and K being positive constant or radially symmetric functions, $f={\left|u\right|}^{p-2}u$ or $f=f\left(u\right)$. If V is non-radial, $K\equiv 1$, $f={\left|u\right|}^{p-2}u$, the authors in [20,28] obtained the existence of ground state solutions of (10) for $p\in (3,6)$. If $V\equiv 1$, $f={a\left(x\right)\left|u\right|}^{p-2}u$ is non-radial, by requiring suitable assumptions on K and a, the authors in [29] obtained ground state and bound state solutions. For other related results, see [30,31,32,33] and the reference therein. Usually, in order to ensure the boundedness of the Palais–Smale sequences, the Ambrosetti-Rabinowitz condition or some monotonicity condition on f is needed. It is natural to ask whether we can prove the boundedness of the Palais–Smale sequences without the above restrict conditions. When we study (1), we solve the problem under mild conditions and get an interesting result. Instead of $\left(g_2\right)$ and $\left(f_1\right)$, we assume the following conditions.

$\left(g_2^{\prime }\right)$

$g\left(x\right)\in C({\mathbb{R}}^3,\mathbb{R})$ and ${inf}_{x\in {\mathbb{R}}^3}g\left(x\right)>0$. Moreover, $g\left(x\right)\le 1$ and there exists ${\rho }_0>0$ such that $g\left(x\right)=1$ for ${\rho }_0<\left|x\right|<2{\rho }_0$.

$\left(f_1^{\prime }\right)$

$f\in C({\mathbb{R}}^{+},\mathbb{R})$ and $f\left(u\right)\ge 0$ for $u\ge 0$. Moreover, ${lim}_{u\to 0+}\frac{f\left(u\right)}{u^3}={lim}_{u\to +\infty }\frac{f\left(u\right)}{u^5}=0$ and ${lim}_{u\to +\infty }\frac{f\left(u\right)}{u^3}=+\infty$.

Theorem 3.

Let $\epsilon >0$. Assume that $\left(h_2\right)$, $\left(V_1\right)$, $\left(Vh\right)$, $\left(g_2^{\prime }\right)$ and $\left(f_1^{\prime }\right)$ hold. Then problem (1) has a positive solution.

The outline of this paper is as follows: in Section 2, we give some important lemmas; in Section 3, we prove Theorems 1; in Section 4, we prove Theorem 2 and 3; in Section 5, we make the conclusions.

Notations:

• ${\parallel u\parallel }_s={\left({\int }_{{\mathbb{R}}^3}{\left|u\right|}^s\mathrm{d}x\right)}^{\frac{1}{s}}$, $1\le s\le +\infty$.
• $H^1=H^1\left({\mathbb{R}}^3\right)$ denotes the Hilbert space equipped with the inner product $\langle u,v\rangle ={\int }_{{\mathbb{R}}^3}(\nabla u\nabla vs.+uv)\mathrm{d}x$ and the norm ${\parallel u\parallel }_{H^1}^2={\int }_{{\mathbb{R}}^3}{\left(\right|\nabla u|}^2+{\left|u\right|}^2)\mathrm{d}x$, $D^{1,2}=D^{1,2}\left({\mathbb{R}}^3\right)=\left\{u\in L^6\left({\mathbb{R}}^3\right):\nabla u\in L^2\left({\mathbb{R}}^3\right)\right\}$ denotes the Sobolev space equipped with the norm ${\parallel u\parallel }_{D^{1,2}}^2={\int }_{{\mathbb{R}}^3}{|\nabla u|}^2\mathrm{d}x$.
• $S={inf}_{u\in D^{1,2}\backslash \left\{0\right\}}\frac{{\int }_{{\mathbb{R}}^3}{|\nabla u|}^2\mathrm{d}x}{{\left({\int }_{{\mathbb{R}}^3}{\left|u\right|}^6\mathrm{d}x\right)}^{\frac{1}{3}}}$ denotes the best Sobolev constant.
• C denotes a positive constant (possibly different).

2. Preliminary Lemmas

Since we look for positive solutions, we assume $f\left(u\right)=0$ for $u\le 0$. Make the change of variable $x\to \epsilon x$, problem (1) becomes

$$-\Delta u+V\left(\epsilon x\right)u+{\varphi }_{u}u=h\left(\epsilon x\right)f\left(u\right)+g\left(\epsilon x\right)u^5\mathrm{in}{\mathbb{R}}^3.$$

(11)

For any $\epsilon >0$, define $H_{\epsilon }=\left\{u\in H^1:{\int }_{{\mathbb{R}}^3}V\left(\epsilon x\right){\left|u\right|}^2\mathrm{d}x<+\infty \right\}$ the Hilbert space with the inner product ${\langle u,v\rangle }_{\epsilon }={\int }_{{\mathbb{R}}^3}\left(\nabla u\nabla vs.+V\left(\epsilon x\right)uv\right)\mathrm{d}x$ and the norm ${\parallel u\parallel }_{\epsilon }=({\int }_{{\mathbb{R}}^3}{|\nabla u|}^2+{V\left(\epsilon x\right)\left|u\right|}^2{\mathrm{d}x)}^{\frac{1}{2}}$. By the Lax–Milgram theorem, for any $u\in H_{\epsilon }$, there exists a unique ${\varphi }_u\in D^{1,2}$ satisfying $-\Delta {\varphi }_u=u^2$. Moreover, by [20,26,28], we have the following results.

Lemma 1.

(i) 

${\varphi }_u\ge 0$ and ${\varphi }_{tu}=t^2{\varphi }_u$ for any $t\in \mathbb{R}$.

(ii) 

If $y\in {\mathbb{R}}^3$ and $\tilde{u}\left(x\right)=u(x+y)$, then ${\varphi }_{\tilde{u}}\left(x\right)={\varphi }_u(x+y)$ and ${\int }_{{\mathbb{R}}^3}{\varphi }_{\tilde{u}}{\tilde{u}}^2\mathrm{d}x={\int }_{{\mathbb{R}}^3}{\varphi }_u{u}^2\mathrm{d}x$.

(iii) 

If $u_n\rightharpoonup u$ weakly in $H^1$, then ${\varphi }_{u_n}\rightharpoonup {\varphi }_u$ weakly in $D^{1,2}$. Moreover, let $v_n=u_n-u$, then

$$\begin{array}{c}{\int }_{{\mathbb{R}}^3}{\varphi }_{u_n}{u}_n^2\mathrm{d}x-{\int }_{{\mathbb{R}}^3}{\varphi }_u{u}^2\mathrm{d}x={\int }_{{\mathbb{R}}^3}{\varphi }_{v_n}{v}_n^2\mathrm{d}x+o_n\left(1\right),\hfill \\ \left|{\int }_{{\mathbb{R}}^3}\left({\varphi }_{u_n}{u}_n-{\varphi }_{u}u\right)\phi \mathrm{d}x-{\int }_{{\mathbb{R}}^3}{\varphi }_{v_n}{v}_n\phi \mathrm{d}x\right|=o_n\left(1\right){\parallel \phi \parallel }_{H^1},\forall \phi \in H^1.\hfill \end{array}$$

(iv) 

$\parallel {\varphi }_u{\parallel }_{D^{1,2}}^2={\int }_{{\mathbb{R}}^3}{\varphi }_u{u}^2\mathrm{d}x\le \frac{{\parallel u\parallel }_{\frac{12}{5}}^4}{S}$.

The functional associated with (11) is

$$\begin{array}{cc}\hfill I_{\epsilon }\left(u\right)=& \frac{1}{2}{\parallel u\parallel }_{\epsilon }^2+\frac{1}{4}{\int }_{{\mathbb{R}}^3}{\varphi }_u{u}^2\mathrm{d}x-{\int }_{{\mathbb{R}}^3}h\left(\epsilon x\right)F\left(u\right)\mathrm{d}x-\frac{1}{6}{\int }_{{\mathbb{R}}^3}g\left(\epsilon x\right){\left|u\right|}^6\mathrm{d}x,\hfill \end{array}$$

(12)

where $F\left(u\right)={\int }_0^{u}f\left(s\right)\mathrm{d}s$. Obviously, $I_{\epsilon }:H_{\epsilon }\mapsto \mathbb{R}$ is of class $C^1$ and critical points of $I_{\epsilon }$ are weak solutions of (11). Let $m_{\epsilon }=inf\{I_{\epsilon }\left(u\right):u\in M_{\epsilon }\}$, where $M_{\epsilon }=\{u\in H_{\epsilon }\backslash \left\{0\right\}:(I_{\epsilon }^{\prime }\left(u\right),u)=0\}$.

From [34], we know S is attained by $\frac{{\delta }^{\frac{1}{4}}}{{\left(\delta +{\left|x\right|}^2\right)}^{\frac{1}{2}}}$, where $\delta >0$. Let $u_{\delta ,z}\left(x\right)=\frac{\psi \left(x\right){\delta }^{\frac{1}{4}}}{{\left({\delta +|x-z|}^2\right)}^{\frac{1}{2}}}$, where $\psi \in C_0^{\infty }\left(B_{2r}\left(z\right)\right)$ such that $\psi \left(x\right)=1$ on $B_r\left(z\right)$, $0\le \psi \left(x\right)\le 1$ and $|\nabla \psi |\le 2$. By the direct calculation, we have the following results.

Lemma 2.

For $\delta >0$ small,

$$\begin{array}{c}{\int }_{{\mathbb{R}}^3}|\nabla u_{\delta ,z}{|}^2\mathrm{d}x=K_1+O\left({\delta }^{\frac{1}{2}}\right),{\int }_{{\mathbb{R}}^3}{\left|u_{\delta ,z}\right|}^6\mathrm{d}x=K_2+O\left({\delta }^{\frac{3}{2}}\right),\hfill \\ {\int }_{{\mathbb{R}}^3}{\left|u_{\delta ,z}\right|}^t\mathrm{d}x=O\left({\delta }^{\frac{t}{4}}\right),t\in (1,3),\hfill \end{array}$$

where$S=\frac{K_1}{K_2^{\frac{1}{3}}}$.

3. Proof of Theorem 1

By $\left(f_1\right)$, we derive that

$$\begin{array}{c}\hfill \frac{1}{4}f\left(u\right)u\ge F\left(u\right)\ge 0,f^{\prime }\left(u\right)u^2-3f\left(u\right)u\ge 0,\forall u\in \mathbb{R}.\end{array}$$

(13)

By $\left(h_1\right)$, we have $h\left(x^i\right)=h_M$, $i=1$, 2, … k. We consider the equation:

$$-\Delta u+V_{0}u+{\varphi }_{u}u=h_{M}f\left(u\right)+u^5\mathrm{in}{\mathbb{R}}^3.$$

(14)

Let ${\parallel u\parallel }_{V_0}={\left({\int }_{{\mathbb{R}}^3}{|\nabla u|}^2+V_0{\left|u\right|}^2\mathrm{d}x\right)}^{\frac{1}{2}}$. Then the functional of (14) is

$$\begin{array}{c}\hfill \widehat{I}\left(u\right)=\frac{1}{2}{\parallel u\parallel }_{V_0}^2+\frac{1}{4}{\int }_{{\mathbb{R}}^3}{\varphi }_u{u}^2\mathrm{d}x-{\int }_{{\mathbb{R}}^3}{h}_{M}F\left(u\right)\mathrm{d}x-\frac{1}{6}{\int }_{{\mathbb{R}}^3}{\left|u\right|}^6\mathrm{d}x.\end{array}$$

Let $\widehat{m}=inf\{\widehat{I}\left(u\right):u\in \widehat{N}\}$, where $\widehat{N}=\left\{u\in H^1\backslash \left\{0\right\}:\left({\widehat{I}}^{\prime }\left(u\right),u\right)=0\right\}$. It is well known that $\widehat{m}$ is attained by w. Moreover, $\widehat{m}\in \left(0,\frac{1}{3}{S}^{\frac{3}{2}}\right)$ and ${\widehat{I}}^{\prime }\left(w\right)=0$. Let ${\parallel u\parallel }_{V_{\infty }}={\left({\int }_{{\mathbb{R}}^3}{|\nabla u|}^2+V_{\infty }{\left|u\right|}^2\mathrm{d}x\right)}^{\frac{1}{2}}$. Define the functional on $H^1$ by

$$\begin{array}{c}\hfill \breve{I}\left(u\right)=\frac{1}{2}{\parallel u\parallel }_{V_{\infty }}^2+\frac{1}{4}{\int }_{{\mathbb{R}}^3}{\varphi }_u{u}^2\mathrm{d}x-{\int }_{{\mathbb{R}}^3}{h}_{\infty }F\left(u\right)\mathrm{d}x-\frac{1}{6}{\int }_{{\mathbb{R}}^3}{g}_{\infty }{\left|u\right|}^6\mathrm{d}x.\end{array}$$

Let $m_{\infty }=inf\{\breve{I}\left(u\right):u\in N_{\infty }\}$, where $N_{\infty }=\left\{u\in H^1\backslash \left\{0\right\}:\left({\breve{I}}^{\prime }\left(u\right),u\right)=0\right\}$. For any $v\in N_{\infty }$, by $\left(f_1\right)$, we have $\breve{I}\left(v\right)={sup}_{t\ge 0}\breve{I}\left(tv\right)$. Moreover, there exists a unique $t_v>0$ satisfying $t_{v}vs.\in \widehat{N}$. Then by $V_{\infty }\ge V_0$, $h_{\infty }\le h_M$, $g_{\infty }\le 1$, we get

$$\begin{array}{c}\hfill \breve{I}\left(v\right)=\underset{t\ge 0}{sup}\breve{I}\left(tv\right)\ge \breve{I}\left(t_{v}v\right)\ge \widehat{I}\left(t_{v}v\right)\ge \widehat{m}.\end{array}$$

So $m_{\infty }\ge \widehat{m}$. Similarly, we have $m_{\epsilon }\ge \widehat{m}$.

For $\eta >0$, denote $C_{\eta }\left(x^i\right)$ the hypercube ${\mathsf{\Pi }}_{j=1}^3(x_j^i-\eta ,x_j^i+\eta )$ centered at $x^i=(x_1^i,x_2^i,x_3^i)$, $i=1$, 2, … k. Denote $\overline{C_{\eta }\left(x^i\right)}$ and $\partial C_{\eta }\left(x^i\right)$ the closure and the boundary of $C_{\eta }\left(x^i\right)$, respectively. By $\left(h_1\right)$, we have $h\left(x^i\right)=h_M$. Moreover, there exist $\eta$, $L_0>0$ such that $\overline{C_{\eta }\left(x^i\right)}$, $i=1$, 2, … k are disjoint, $h\left(x\right)<h\left(x^i\right)$ for $x\in \overline{C_{\eta }\left(x^i\right)}\backslash x^i$ and $\overline{C_{\eta }\left(x^i\right)}\subset {\mathsf{\Pi }}_{j=1}^3(-L_0,L_0)$.

Let $\epsilon \in (0,1)$. Define ${\phi }_{\epsilon }\in C_0^{\infty }\left({\mathbb{R}}^3\right)$ such that ${\phi }_{\epsilon }\left(x\right)=1$ for $\left|x\right|<\frac{1}{\sqrt{\epsilon }}-1$, ${\phi }_{\epsilon }\left(x\right)=0$ for $\left|x\right|>\frac{1}{\sqrt{\epsilon }}$, $0\le {\phi }_{\epsilon }\le 1$ and $|\nabla {\phi }_{\epsilon }|\le 2$. Let ${\zeta }_{\epsilon }^i\left(x\right)=w\left(x-\frac{x^i}{\epsilon }\right){\phi }_{\epsilon }\left(x-\frac{x^i}{\epsilon }\right)$, $i=1$, 2, … k. Let $\rho >max\left\{\right|x^1|+\eta ,|x^2|+\eta ,\dots ,|x^k|+\eta \}$. Define $\chi :{\mathbb{R}}^3\to {\mathbb{R}}^3$ such that $\chi \left(x\right)=x$ for $\left|x\right|<\rho$, $\chi \left(x\right)=\frac{\rho x}{\left|x\right|}$ for $\left|x\right|\ge \rho$. For $u\in H^1\backslash \left\{0\right\}$, define ${\beta }_{\epsilon }\left(u\right)=\frac{{\int }_{{\mathbb{R}}^3}\chi \left(\epsilon x\right){\left|u\right|}^6\mathrm{d}x}{{\int }_{{\mathbb{R}}^3}{\left|u\right|}^6\mathrm{d}x}$. Then ${\beta }_{\epsilon }$ is continuous in $H^1\backslash \left\{0\right\}$.

For simplicity, denote $C_{\eta }\left(x^i\right)=C_{\eta }^i$. Set $M_{\epsilon }^i=\left\{u\in M_{\epsilon }:u\ge 0,{\beta }_{ϵ}\left(u\right)\in C_{\eta }^i\right\}$, $\partial M_{\epsilon }^i=\left\{u\in M_{\epsilon }:u\ge 0,{\beta }_{\epsilon }\left(u\right)\in \partial C_{\eta }^i\right\}$, $i=1$, 2, … k. Let ${\gamma }_{\epsilon }^i={inf}_{u\in M_{\epsilon }^i}{I}_{\epsilon }\left(u\right)$, ${\tilde{\gamma }}_{\epsilon }^i={inf}_{u\in \partial M_{\epsilon }^i}{I}_{\epsilon }\left(u\right)$, $i=1$, 2, … k.

Lemma 3.

Let $i=1$, 2, … k. Then for any $\nu \in (0,\widehat{m})$, there exists ${\epsilon }_{\nu }>0$ such that $m_{\epsilon }\le {\gamma }_{\epsilon }^i<\widehat{m}+\nu$ for $\epsilon \in (0,{\epsilon }_{\nu })$.

Proof of Lemma 3.

It is clear that ${\gamma }_{\epsilon }^i\ge m_{\epsilon }$. Let $l_{\epsilon }\left(t\right)=I_{\epsilon }\left(t{\zeta }_{\epsilon }^i\right)$, where $t>0$. By $\left(f_1\right)$, we derive that $l_{\epsilon }\left(t\right)$ admits a unique critical point $t_{\epsilon }^i>0$ corresponding to its maximum. Then $l_{\epsilon }\left(t_{\epsilon }^i\right)={sup}_{t\ge 0}{l}_{\epsilon }\left(t{\zeta }_{\epsilon }^i\right)$, $l_{\epsilon }^{\prime }\left(t_{\epsilon }^i\right)=0$. So $t_{\epsilon }^i{\zeta }_{\epsilon }^i\in M_{\epsilon }$. Note that ${lim}_{\epsilon \to 0}{\parallel w{\phi }_{\epsilon }-w\parallel }_{V_0}=0$. By the Lebesgue dominated convergence theorem,

$$\begin{array}{c}\hfill \underset{\epsilon \to 0}{lim}{\beta }_{\epsilon }\left(t_{\epsilon }^i{\zeta }_{\epsilon }^i\right)=\underset{\epsilon \to 0}{lim}{\beta }_{\epsilon }\left({\zeta }_{\epsilon }^i\right)=x^i.\end{array}$$

Then ${\beta }_{\epsilon }\left(t_{\epsilon }^i{\zeta }_{\epsilon }^i\right)\in C_{\eta }^i$ for $\epsilon >0$ small. Since $t_{\epsilon }^i{\zeta }_{\epsilon }^i\in M_{\epsilon }$, by the definition of ${\gamma }_{\epsilon }^i$, we get ${\gamma }_{\epsilon }^i\le I_{\epsilon }\left(t_{\epsilon }^i{\zeta }_{\epsilon }^i\right)={sup}_{t\ge 0}{I}_{\epsilon }\left(t{\zeta }_{\epsilon }^i\right)$. So we just prove $I_{\epsilon }\left(t_{\epsilon }^i{\zeta }_{\epsilon }^i\right)<\widehat{m}+\nu$ for $\epsilon >0$ small. By ${lim}_{\epsilon \to 0}{\parallel w{\phi }_{\epsilon }-w\parallel }_{V_0}=0$ and Lemma 1 $\left(ii\right)$, we derive that

$$\begin{array}{c}\underset{\epsilon \to 0}{lim}{\int }_{{\mathbb{R}}^3}|\nabla {\zeta }_{\epsilon }^i{|}^2\mathrm{d}x=\underset{\epsilon \to 0}{lim}{\int }_{{\mathbb{R}}^3}|\nabla \left(w{\phi }_{\epsilon }\right){|}^2\mathrm{d}x={\int }_{{\mathbb{R}}^3}{|\nabla w|}^2\mathrm{d}x,\hfill \\ \underset{\epsilon \to 0}{lim}{\int }_{{\mathbb{R}}^3}{\varphi }_{{\zeta }_{\epsilon }^i}{\left({\zeta }_{\epsilon }^i\right)}^2\mathrm{d}x=\underset{\epsilon \to 0}{lim}{\int }_{{\mathbb{R}}^3}{\varphi }_{w{\phi }_{\epsilon }}{\left(w{\phi }_{\epsilon }\right)}^2\mathrm{d}x={\int }_{{\mathbb{R}}^3}{\varphi }_w{w}^2\mathrm{d}x.\hfill \end{array}$$

(15)

By $V\left(x^i\right)=V_0$, $h\left(x^i\right)=h_M$, $g\left(x^i\right)=1$ and the Lebesgue dominated convergence theorem, we get

$$\begin{array}{c}\underset{\epsilon \to 0}{lim}{\int }_{{\mathbb{R}}^3}V\left(\epsilon x\right)|{\zeta }_{\epsilon }^i{|}^2\mathrm{d}x=\underset{\epsilon \to 0}{lim}{\int }_{{\mathbb{R}}^3}V(\epsilon x+x^i)|w{\phi }_{\epsilon }{|}^2\mathrm{d}x={\int }_{{\mathbb{R}}^3}{V}_0{\left|w\right|}^2\mathrm{d}x,\hfill \\ \underset{\epsilon \to 0}{lim}{\int }_{{\mathbb{R}}^3}g\left(\epsilon x\right)|{\zeta }_{\epsilon }^i{|}^6\mathrm{d}x={\int }_{{\mathbb{R}}^3}{\left|w\right|}^6\mathrm{d}x,\underset{\epsilon \to 0}{lim}{\int }_{{\mathbb{R}}^3}h\left(\epsilon x\right)F\left({\zeta }_{\epsilon }^i\right)\mathrm{d}x={\int }_{{\mathbb{R}}^3}{h}_{M}F\left(w\right)\mathrm{d}x,\hfill \\ \underset{\epsilon \to 0}{lim}{\int }_{{\mathbb{R}}^3}h\left(\epsilon x\right)f\left({\zeta }_{\epsilon }^i\right){\zeta }_{\epsilon }^i\mathrm{d}x={\int }_{{\mathbb{R}}^3}{h}_{M}f\left(w\right)w\mathrm{d}x.\hfill \end{array}$$

(16)

By $t_{\epsilon }^i{\zeta }_{\epsilon }^i\in M_{\epsilon }^i$, we have

$$\begin{array}{c}{\left(t_{\epsilon }^i\right)}^2{\parallel {\zeta }_{\epsilon }^i\parallel }_{\epsilon }^2+{\left(t_{\epsilon }^i\right)}^4{\int }_{{\mathbb{R}}^3}{\varphi }_{{\zeta }_{\epsilon }^i}{\left({\zeta }_{\epsilon }^i\right)}^2\mathrm{d}x\hfill \\ ={\int }_{{\mathbb{R}}^3}h\left(\epsilon x\right)f\left(t_{\epsilon }^i{\zeta }_{\epsilon }^i\right)\left(t_{\epsilon }^i{\zeta }_{\epsilon }^i\right)\mathrm{d}x+{\left(t_{\epsilon }^i\right)}^6{\int }_{{\mathbb{R}}^3}g\left(\epsilon x\right)|{\zeta }_{\epsilon }^i{|}^6\mathrm{d}x\ge {\left(t_{\epsilon }^i\right)}^6{\int }_{{\mathbb{R}}^3}g\left(\epsilon x\right){\left|{\zeta }_{\epsilon }^i\right|}^6\mathrm{d}x.\hfill \end{array}$$

(17)

From (15)–(17), we obtain that $t_{\epsilon }^i$ is bounded. Furthermore, by (17), $\left(f_1\right)$, $\left(h_1\right)$, $\left(g_1\right)$, for $\eta =\frac{V_0}{2}$, there exists $C_{\eta }=C_{\frac{V_0}{2}}>0$ satisfying

$$\begin{array}{c}\hfill {\left(t_{\epsilon }^i\right)}^2\parallel {\zeta }_{\epsilon }^i{\parallel }_{V_0}^2\le \frac{V_0{\left(t_{\epsilon }^i\right)}^2}{2}{\int }_{{\mathbb{R}}^3}|{\zeta }_{\epsilon }^i{|}^2\mathrm{d}x+C_{\frac{V_0}{2}}{\left(t_{\epsilon }^i\right)}^6{\int }_{{\mathbb{R}}^3}{\left|{\zeta }_{\epsilon }^i\right|}^6\mathrm{d}x.\end{array}$$

Then $t_{\epsilon }^i\to t^i>0$ in view of (15) and (16). By (15)–(17),

$$\begin{array}{c}\hfill \parallel t^i{w\parallel }_{V_0}^2={\int }_{{\mathbb{R}}^3}\left(h_{M}f\left(t^{i}w\right)t^{i}w+{\left|t^{i}w\right|}^6\right)\mathrm{d}x,\end{array}$$

that is, $t^{i}w\in \widehat{N}$. By $\left(f_1\right)$, there exists a unique $t>0$ satisfying $tw\in \widehat{N}$. Since $w\in \widehat{N}$, we have $t^i=1$. Then by (15) and (16),

$$\begin{array}{c}\hfill I_{\epsilon }\left(t_{\epsilon }^i{\zeta }_{\epsilon }^i\right)=I_{\epsilon }\left({\zeta }_{\epsilon }^i\right)+O\left(\epsilon \right)=\widehat{I}\left(w\right)+O\left(\epsilon \right)=\widehat{m}+O\left(\epsilon \right).\end{array}$$

So there exists ${\epsilon }_{\nu }>0$ such that ${\gamma }_{\epsilon }^i<\widehat{m}+\nu$ for $\epsilon \in (0,{\epsilon }_{\nu })$.  □

Lemma 4.

Let $i=1$, 2, … k. Then there exist δ, ${\epsilon }_{\delta }>0$ such that ${\tilde{\gamma }}_{\epsilon }^i>\widehat{m}+\delta$ for $\epsilon \in (0,{\epsilon }_{\delta })$.

Proof of Lemma 4.

Assume to the contrary that there exists ${\epsilon }_n\downarrow 0$ such that ${\tilde{\gamma }}_{{\epsilon }_n}^i\to c\le \widehat{m}$. Then there exists $\left\{u_n\right\}\subset \partial M_{{\epsilon }_n}^i$ satisfying $I_{{\epsilon }_n}\left(u_n\right)\to c\le \widehat{m}$. By $\left(f_1\right)$, there exists a unique $t_n>0$ such that $t_n{u}_n\in \widehat{N}$. Then

$$\begin{array}{cc}\hfill \widehat{m}\ge I_{{\epsilon }_n}\left(u_n\right)+o_n\left(1\right)=& \underset{t\ge 0}{sup}{I}_{{\epsilon }_n}\left(t{u}_n\right)+o_n\left(1\right)\hfill \\ \hfill \ge & I_{{\epsilon }_n}\left(t_n{u}_n\right)+o_n\left(1\right)\ge \widehat{I}\left(t_n{u}_n\right)+o_n\left(1\right)\ge \widehat{m}+o_n\left(1\right),\hfill \end{array}$$

from which we get $\widehat{I}\left(t_n{u}_n\right)=\widehat{m}+o_n\left(1\right)$. Moreover,

$${\int }_{{\mathbb{R}}^3}(h\left({\epsilon }_{n}x\right)-h_M)F\left(t_n{u}_n\right)\mathrm{d}x=o_n\left(1\right).$$

(18)

Since $t_n{u}_n\in \widehat{N}$, by the Ekeland’s variational principle, there exist $\left\{v_n\right\}\subset \widehat{N}$, ${\mu }_n\in \mathbb{R}$ satisfying $\parallel v_n-t_n{u}_n{\parallel }_{V_0}=o_n\left(1\right)$, $\widehat{I}\left(v_n\right)=\widehat{m}+o_n\left(1\right)$, ${\widehat{I}}^{\prime }\left(v_n\right)-{\mu }_n{\widehat{G}}^{\prime }\left(v_n\right)=o_n\left(1\right)$, where $\widehat{G}\left(v_n\right)=({\widehat{I}}^{\prime }\left(v_n\right),v_n)$. By the standard argument, we derive that ${\mu }_n\to 0$ as $n\to \infty$. Then

$$\begin{array}{c}\hfill \widehat{I}\left(v_n\right)=\widehat{m}+o_n\left(1\right),{\widehat{I}}^{\prime }\left(v_n\right)=o_n\left(1\right).\end{array}$$

(19)

By (13), we get $\parallel v_n{\parallel }_{V_0}$ is bounded. By the Lions Lemma, ${\int }_{{\mathbb{R}}^3}{\left|v_n\right|}^t\mathrm{d}x\to 0$ for any $t\in (2,6)$, or there exists $y_n\in {\mathbb{R}}^3$ such that $w_n=v_n(.+y_n)\rightharpoonup w\ne 0$ weakly in $H^1\left({\mathbb{R}}^3\right)$. If ${\int }_{{\mathbb{R}}^3}{\left|v_n\right|}^t\mathrm{d}x\to 0$ for any $t\in (2,6)$, then ${\int }_{{\mathbb{R}}^3}F\left(v_n\right)\mathrm{d}x={\int }_{{\mathbb{R}}^3}f\left(v_n\right)v_n\mathrm{d}x=o_n\left(1\right)$. Let ${\int }_{{\mathbb{R}}^3}{\left|v_n\right|}^6\mathrm{d}x\to l$. Then ${lim}_{n\to \infty }{\parallel v_n\parallel }_{V_0}^2\le l$. Since $\widehat{m}>0$, we have $l>0$. By $S\le \frac{\parallel v_n{\parallel }_{V_0}^2}{{\left({\int }_{{\mathbb{R}}^3}{\left|v_n\right|}^6\mathrm{d}x\right)}^{\frac{1}{3}}}$, we get $l\ge S^{\frac{3}{2}}$. Thus, $\widehat{m}=\widehat{I}\left(v_n\right)-\frac{1}{4}({\widehat{I}}^{\prime }\left(v_n\right),v_n)+o_n\left(1\right)\ge \frac{1}{3}{S}^{\frac{3}{2}}+o_n\left(1\right)$, a contradiction with $\widehat{m}<\frac{1}{3}{S}^{\frac{3}{2}}$. So $w_n\rightharpoonup w\ne 0$ weakly in $H^1$. By (19), we have $\widehat{I}\left(w_n\right)=\widehat{m}+o_n\left(1\right)$, ${\widehat{I}}^{\prime }\left(w_n\right)=o_n\left(1\right)$. Then ${\widehat{I}}^{\prime }\left(w\right)=0$. Moreover,

$$\begin{array}{cc}\hfill \widehat{m}+o_n\left(1\right)=& \widehat{I}\left(w_n\right)-\frac{1}{4}({\widehat{I}}^{\prime }\left(w_n\right),w_n)\hfill \\ \hfill =& \frac{1}{4}\parallel w_n{\parallel }_{V_0}^2+\frac{1}{12}{\int }_{{\mathbb{R}}^3}{\left|w_n\right|}^6\mathrm{d}x+{\int }_{{\mathbb{R}}^3}{h}_M\left(\frac{1}{4}f\left(w_n\right)w_n-F\left(w_n\right)\right)\mathrm{d}x.\hfill \end{array}$$

By Fatou’s Lemma,

$$\begin{array}{cc}\hfill \widehat{m}\ge & \frac{1}{4}{\parallel w\parallel }_{V_0}^2+\frac{1}{12}{\int }_{{\mathbb{R}}^3}{\left|w\right|}^6\mathrm{d}x+{\int }_{{\mathbb{R}}^3}{h}_M\left(\frac{1}{4}f\left(w\right)w-F\left(w\right)\right)\mathrm{d}x\hfill \\ \hfill =& \widehat{I}\left(w\right)-\frac{1}{4}({\widehat{I}}^{\prime }\left(w\right),w)=\widehat{I}\left(w\right)\ge \widehat{m}.\hfill \end{array}$$

So $w_n\to w$ in $H^1$. Note that ${\beta }_{{\epsilon }_n}\left(t_n{u}_n\right)={\beta }_{{\epsilon }_n}\left(u_n\right)\in \partial C_{\eta }^i$. By $\parallel v_n-t_n{u}_n{\parallel }_{V_0}\to 0$, we get ${\beta }_{{\epsilon }_n}\left(v_n\right)\to z_0\in \partial C_{\eta }^i$. Thus,

$$\begin{array}{c}\hfill z_0=\underset{n\to \infty }{lim}\frac{{\int }_{{\mathbb{R}}^3}\chi \left({\epsilon }_{n}x\right){\left|v_n\right|}^6\mathrm{d}x}{{\int }_{{\mathbb{R}}^3}{\left|v_n\right|}^6\mathrm{d}x}=\underset{n\to \infty }{lim}\frac{{\int }_{{\mathbb{R}}^3}\chi ({\epsilon }_{n}x+{\epsilon }_n{y}_n){\left|w_n\right|}^6\mathrm{d}x}{{\int }_{{\mathbb{R}}^3}{\left|w_n\right|}^6\mathrm{d}x}.\end{array}$$

(20)

If $|{\epsilon }_n{y}_n|\to \infty$, by $w_n\to w$ in $H^1$, we get $|z_0|=\rho$, a contradiction with $z_0\in \partial C_{\eta }^i$ and $\rho >|x^i|+\eta$. So $|{\epsilon }_n{y}_n|$ is bounded. Assume that ${\epsilon }_n{y}_n\to y_0$ as $n\to \infty$. If $|y_0|\ge \rho$, by (20), we get $|z_0|=\rho$, a contradiction. If $|y_0|<\rho$, we have $y_0=z_0\in \partial C_{\eta }^i$. Then $h\left(y_0\right)<h_M$. By $\parallel v_n-t_n{u}_n{\parallel }_{V_0}\to 0$, $w_n=v_n(.+y_n)\to w$ in $H^1$,

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}{\int }_{{\mathbb{R}}^3}{h}_{M}F\left(t_n{u}_n\right)\mathrm{d}x=\underset{n\to \infty }{lim}{\int }_{{\mathbb{R}}^3}{h}_{M}F\left(v_n\right)\mathrm{d}x={\int }_{{\mathbb{R}}^3}{h}_{M}F\left(w\right)\mathrm{d}x.\end{array}$$

Furthermore,

$$\begin{array}{cc}\hfill \underset{n\to \infty }{lim}{\int }_{{\mathbb{R}}^3}h\left({\epsilon }_{n}x\right)F\left(t_n{u}_n\right)\mathrm{d}x=& \underset{n\to \infty }{lim}{\int }_{{\mathbb{R}}^3}h({\epsilon }_{n}x+{\epsilon }_n{y}_n)F\left(v_n(.+y_n)\right)\mathrm{d}x\hfill \\ \hfill =& {\int }_{{\mathbb{R}}^3}h\left(y_0\right)F\left(w\right)\mathrm{d}x.\hfill \end{array}$$

Then by (18), we get ${\int }_{{\mathbb{R}}^3}{h}_{M}F\left(w\right)\mathrm{d}x={\int }_{{\mathbb{R}}^3}h\left(y_0\right)F\left(w\right)\mathrm{d}x$, a contradiction with $h\left(y_0\right)<h_M$.  □

By Lemmas 3 and 4, there exists ${\epsilon }_1>0$ such that $\widehat{m}\le {\gamma }_{\epsilon }^i<{\tilde{\gamma }}_{\epsilon }^i$ for $\epsilon \in (0,{\epsilon }_1)$. Furthermore, ${\gamma }_{\epsilon }^i\to \widehat{m}$ as $\epsilon \to 0$. By $\left(f_1\right)$, for $\xi =\frac{V_{\infty }}{2h_{\infty }}$, there exists $C_{\xi }=C_{\frac{V_{\infty }}{2h_{\infty }}}>0$ such that

$$\begin{array}{c}\hfill max\left\{\right|F\left(u\right)|,|f\left(u\right)u\left|\right\}\le \frac{V_{\infty }}{2h_{\infty }}{\left|u\right|}^2+C_{\frac{V_{\infty }}{2h_{\infty }}}{\left|u\right|}^6.\end{array}$$

(21)

Note that $\widehat{m}<\frac{1}{3}{S}^{\frac{3}{2}}$. We choose ${\epsilon }_0\in (0,{\epsilon }_1)$ small such that for $\epsilon \in (0,{\epsilon }_0)$,

$$\begin{array}{c}\hfill {\gamma }_{\epsilon }^i<min\left\{\frac{1}{3}{S}^{\frac{3}{2}},\frac{S^{\frac{3}{2}}}{4\sqrt{2\left(h_{\infty }{C}_{\frac{V_{\infty }}{2h_{\infty }}}+1\right)}}+\widehat{m}\right\}.\end{array}$$

(22)

Following the ideas of [18,35], we can use the implicit function theorem to get the following result. Since the proof is standard, we omit it here.

Lemma 5.

Let $\epsilon \in (0,{\epsilon }_0)$, $i=1$, 2, … k. Then for any $u\in M_{\epsilon }^i$, there exist $\rho >0$ and a differential function $s\left(w\right)>0$, where $w\in H_{\epsilon }$ and ${\parallel w\parallel }_{\epsilon }<\rho$, satisfying $s\left(0\right)=1$, $s\left(w\right)(u+w)\in M_{\epsilon }^i$. Moreover, for any $\phi \in H_{\epsilon }$,

$$\begin{array}{c}(s^{\prime }\left(0\right),\phi )\hfill \\ =\frac{-2{\langle u,\phi \rangle }_{\epsilon }-4{\int }_{{\mathbb{R}}^3}{\varphi }_{u}u\phi \mathrm{d}x+{\int }_{{\mathbb{R}}^3}(h\left(\epsilon x\right)f\left(u\right)+h\left(\epsilon x\right)f^{\prime }\left(u\right)u+6g\left(\epsilon x\right)u^5)\phi \mathrm{d}x}{{2\parallel u\parallel }_{\epsilon }^2+4{\int }_{{\mathbb{R}}^3}{\varphi }_u{u}^2\mathrm{d}x-{\int }_{{\mathbb{R}}^3}h\left(\epsilon x\right)(f\left(u\right)u+f^{\prime }\left(u\right)u^2)+6g\left(\epsilon x\right){\left|u\right|}^6\mathrm{d}x},\hfill \end{array}$$

that is, $(s^{\prime }\left(0\right),\phi )=\frac{-(G_{\epsilon }^{\prime }\left(u\right),\phi )}{(G_{\epsilon }^{\prime }\left(u\right),u)}$, where $G_{\epsilon }\left(u\right)=\left(I_{\epsilon }^{\prime }\left(u\right),u\right)$.

Lemma 6.

Let $\epsilon \in (0,{\epsilon }_0)$, $i=1$, 2, … k. Then there exists $\left\{u_n\right\}\subset M_{\epsilon }^i$ satisfying $I_{\epsilon }\left(u_n\right)\to {\gamma }_{\epsilon }^i$, $I_{\epsilon }^{\prime }\left(u_n\right)\to 0$. Moreover, $u_n$ converges strongly in $H_{\epsilon }$ up to a subsequence.

Proof of Lemma 6.

By the definition of ${\gamma }_{\epsilon }^i$, there is $u_n\in M_{\epsilon }^i$ satisfying $I_{\epsilon }\left(u_n\right)\to {\gamma }_{\epsilon }^i$. Then $\parallel u_n{\parallel }_{\epsilon }$ is bounded in view of (13). By the Ekeland’s variational principle, $I_{\epsilon }\left(u_n\right)\le {\gamma }_{\epsilon }^i+\frac{1}{n}$, $I_{\epsilon }\left(v\right)\ge I_{\epsilon }\left(u_n\right)-\frac{1}{n}{\parallel u_n-v\parallel }_{\epsilon }$ for any $v\in M_{\epsilon }^i$. By Lemma 5, there exist ${\rho }_n\downarrow 0$ and $s_n\left(w\right)>0$ satisfying $s_n\left(w\right)(u_n+w)\in M_{\epsilon }^i$ for any $w\in H_{\epsilon }$ with ${\parallel w\parallel }_{\epsilon }<{\rho }_n$. Let $w=t\varphi$, where $\varphi \in H_{\epsilon }$, $t>0$. For $t>0$ small,

$$\begin{array}{c}\frac{1}{n}\left[t{s}_n{\left(t\varphi \right)\parallel \varphi \parallel }_{\epsilon }+\left|s_n\left(t\varphi \right)-1\right|{\parallel u_n\parallel }_{\epsilon }\right]\hfill \\ \ge \frac{1}{n}{∥u_n-s_n\left(t\varphi \right)(u_n+t\varphi )∥}_{\epsilon }\hfill \\ \ge I_{\epsilon }\left(u_n\right)-I_{\epsilon }\left(s_n\left(t\varphi \right)(u_n+t\varphi )\right)\hfill \\ =\left[I_{\epsilon }\left(u_n\right)-I_{\epsilon }(u_n+t\varphi )\right]\hfill \\ +(1-s_n\left(t\varphi \right))\left(I_{\epsilon }^{\prime }({\theta }_n(u_n+t\varphi )+(1-{\theta }_n)\left(s_n\left(t\varphi \right)(u_n+t\varphi )\right)),u_n+t\varphi \right),\hfill \end{array}$$

where ${\theta }_n\in (0,1)$. Dividing by t and let $t\to 0$, $\frac{1}{n}\left[\left|(s_n^{\prime }\left(0\right),\varphi )\right|\parallel u_n{\parallel }_{\epsilon }+{\parallel \varphi \parallel }_{\epsilon }\right]\ge -(I_{\epsilon }^{\prime }\left(u_n\right),\varphi )$. By Lemma 5, $(s_n^{\prime }\left(0\right),\varphi )=\frac{-(G_{\epsilon }^{\prime }\left(u_n\right),\varphi )}{(G_{\epsilon }^{\prime }\left(u_n\right),u_n)}$. By (13),

$$\begin{array}{cc}\hfill \left(G_{\epsilon }^{\prime }\left(u_n\right),u_n\right)& =\left(G_{\epsilon }^{\prime }\left(u_n\right),u_n\right)-4\left(I_{\epsilon }^{\prime }\left(u_n\right),u_n\right)\hfill \\ & \le -2\parallel u_n{\parallel }_{\epsilon }^2-2{\int }_{{\mathbb{R}}^3}g\left(\epsilon x\right){\left|u_n\right|}^6\mathrm{d}x<0.\hfill \end{array}$$

So ${lim}_{n\to \infty }\left(G_{\epsilon }^{\prime }\left(u_n\right),u_n\right)\le 0$. If ${lim}_{n\to \infty }\left(G_{\epsilon }^{\prime }\left(u_n\right),u_n\right)=0$, we get $\parallel u_n{\parallel }_{\epsilon }\to 0$, a contradiction with $I\left(u_n\right)\to {\gamma }_{\epsilon }^i>0$. Then ${lim}_{n\to \infty }(G_{\epsilon }^{\prime }\left(u_n\right),u_n)<0$. Since $\parallel u_n{\parallel }_{\epsilon }$ is bounded, we know $\left|\right(s_n^{\prime }\left(0\right),\varphi \left)\right|$ is bounded. So $I_{\epsilon }^{\prime }\left(u_n\right)\to 0$. Assume $u_n\rightharpoonup u$ weakly in $H_{\epsilon }$.

Case 1. $u_n\rightharpoonup 0$ weakly in $H_{\epsilon }$.

Define the functional on $H^1$ by

$$\begin{array}{c}\hfill J_{\epsilon }\left(u\right)=\frac{1}{2}{\parallel u\parallel }_{V_{\infty }}^2+\frac{1}{4}{\int }_{{\mathbb{R}}^3}{\varphi }_u{u}^2\mathrm{d}x-{\int }_{{\mathbb{R}}^3}{h}_{\infty }F\left(u\right)\mathrm{d}x-\frac{1}{6}{\int }_{{\mathbb{R}}^3}g\left(\epsilon x\right){\left|u\right|}^6\mathrm{d}x.\end{array}$$

By ${lim}_{\left|x\right|\to \infty }V\left(x\right)=V_{\infty }$, ${lim}_{\left|x\right|\to \infty }h\left(x\right)=h_{\infty }$, we get ${\gamma }_{\epsilon }^i=J_{\epsilon }\left(u_n\right)+o_n\left(1\right)$, $J_{\epsilon }^{\prime }\left(u_n\right)=o_n\left(1\right)$. By the Lions Lemma, ${\int }_{{\mathbb{R}}^3}{\left|u_n\right|}^t\mathrm{d}x\to 0$ for any $t\in (2,6)$, or there exists $y_n\in {\mathbb{R}}^3$ with $|y_n|\to \infty$ satisfying $v_n=u_n(.+y_n)\rightharpoonup vs.\ne 0$ weakly in $H^1$. If ${\int }_{{\mathbb{R}}^3}{\left|u_n\right|}^t\mathrm{d}x\to 0$ for any $t\in (2,6)$, by Lemma 1 $\left(iv\right)$, we get ${\int }_{{\mathbb{R}}^3}{\varphi }_{u_n}{u}_n^2\mathrm{d}x=o_n\left(1\right)$. Moreover, ${\int }_{{\mathbb{R}}^3}F\left(u_n\right)\mathrm{d}x={\int }_{{\mathbb{R}}^3}f\left(u_n\right)u_n\mathrm{d}x=o_n\left(1\right)$. Then

$$\begin{array}{c}{\gamma }_{\epsilon }^i+o_n\left(1\right)=\frac{1}{2}\parallel u_n{\parallel }_{V_{\infty }}^2-\frac{1}{6}{\int }_{{\mathbb{R}}^3}g\left(\epsilon x\right){\left|u_n\right|}^6\mathrm{d}x,\hfill \\ \parallel u_n{\parallel }_{V_{\infty }}^2-{\int }_{{\mathbb{R}}^3}g\left(\epsilon x\right){\left|u_n\right|}^6\mathrm{d}x=o_n\left(1\right).\hfill \end{array}$$

Since ${\gamma }_{\epsilon }^i>0$, we assume ${lim}_{n\to \infty }{\parallel u_n\parallel }_{V_{\infty }}^2>0$. By $S\le \frac{\parallel u_n{\parallel }_{V_{\infty }}^2}{{\left({\int }_{{\mathbb{R}}^3}g\left(\epsilon x\right){\left|u_n\right|}^6\mathrm{d}x\right)}^{\frac{1}{3}}}$, we get ${lim}_{n\to \infty }{\parallel u_n\parallel }_{V_{\infty }}^2\ge S^{\frac{3}{2}}$. So ${\gamma }_{\epsilon }^i\ge \frac{1}{3}{S}^{\frac{3}{2}}$, a contradiction. Thus, $v_n=u_n(.+y_n)\rightharpoonup vs.\ne 0$ weakly in $H^1$ with $|y_n|\to \infty$. Define

$$\begin{array}{cc}\hfill L_{\epsilon }\left(v_n\right)=& \frac{1}{2}{\parallel v_n\parallel }_{V_{\infty }}^2+\frac{1}{4}{\int }_{{\mathbb{R}}^3}{\varphi }_{v_n}{v}_n^2\mathrm{d}x-{\int }_{{\mathbb{R}}^3}{h}_{\infty }F\left(v_n\right)\mathrm{d}x\hfill \\ & -\frac{1}{6}{\int }_{{\mathbb{R}}^3}g(\epsilon x+\epsilon y_n){\left|v_n\right|}^6\mathrm{d}x.\hfill \end{array}$$

Then

$$\begin{array}{c}\hfill |y_n|\to \infty ,{\gamma }_{\epsilon }^i=L_{\epsilon }\left(v_n\right)+o_n\left(1\right),L_{\epsilon }^{\prime }\left(v_n\right)=o_n\left(1\right).\end{array}$$

(23)

By $v_n\rightharpoonup vs.\ne 0$ weakly in $H^1$ and ${lim}_{\left|x\right|\to \infty }g\left(x\right)=g_{\infty }$, we get

$$\begin{array}{c}\hfill {\int }_{{\mathbb{R}}^3}g(\epsilon x+\epsilon y_n)v_n^5\phi \mathrm{d}x\to {\int }_{{\mathbb{R}}^3}{g}_{\infty }{v}^5\phi \mathrm{d}x,\forall \phi \in H^1.\end{array}$$

Then ${\breve{I}}^{\prime }\left(v\right)=0$. Let $w_n=v_n-v$. By Lemma $1.3$ in [36], we have

$$\begin{array}{c}\hfill {\int }_{{\mathbb{R}}^3}F\left(v_n\right)\mathrm{d}x-{\int }_{{\mathbb{R}}^3}F\left(v\right)\mathrm{d}x={\int }_{{\mathbb{R}}^3}F\left(w_n\right)\mathrm{d}x+o_n\left(1\right).\end{array}$$

(24)

By the Brezis–Lieb Lemma in [34], ${\int }_{{\mathbb{R}}^3}g(\epsilon x+\epsilon y_n)\left||v_n{|}^6-{\left|v\right|}^6-{\left|w_n\right|}^6\right|\mathrm{d}x=o_n\left(1\right)$. By the Lebesgue dominated convergence theorem, ${\int }_{{\mathbb{R}}^3}g(\epsilon x+\epsilon y_n){\left|v\right|}^6\mathrm{d}x\to {\int }_{{\mathbb{R}}^3}{g}_{\infty }{\left|v\right|}^6\mathrm{d}x$. Then

$$\begin{array}{c}\hfill {\int }_{{\mathbb{R}}^3}g(\epsilon x+\epsilon y_n)|v_n{|}^6\mathrm{d}x={\int }_{{\mathbb{R}}^3}g(\epsilon x+\epsilon y_n)|w_n{|}^6\mathrm{d}x+{\int }_{{\mathbb{R}}^3}{g}_{\infty }{\left|v\right|}^6\mathrm{d}x+o_n\left(1\right).\end{array}$$

(25)

Combining (24) and (25) and Lemma 1 $\left(iii\right)$, we get

$$\begin{array}{c}\hfill {\gamma }_{\epsilon }^i=L_{\epsilon }\left(v_n\right)+o_n\left(1\right)=L_{\epsilon }\left(w_n\right)+\breve{I}\left(v\right)+o_n\left(1\right).\end{array}$$

(26)

If $w_n\to 0$ in $H^1$, that is, $v_n\to v$ in $H^1$, by $|y_n|\to \infty$,

$$\begin{array}{c}\hfill |{\beta }_{\epsilon }\left(u_n\right)|=\left|\frac{{\int }_{{\mathbb{R}}^3}\chi \left(\epsilon x\right){\left|u_n\right|}^6\mathrm{d}x}{{\int }_{{\mathbb{R}}^3}{\left|u_n\right|}^6\mathrm{d}x}\right|=\left|\frac{{\int }_{{\mathbb{R}}^3}\chi (\epsilon x+\epsilon y_n){\left|v_n\right|}^6\mathrm{d}x}{{\int }_{{\mathbb{R}}^3}{\left|v_n\right|}^6\mathrm{d}x}\right|\to \rho ,\end{array}$$

a contradiction with $u_n\in C_{\eta }\left(x^i\right)$. So $w_n$ converges weakly and not strongly to 0 in $H^1$. By Lemma 8.9 in [34], $\left|{\int }_{{\mathbb{R}}^3}g(\epsilon x+\epsilon y_n)\left(v_n^5-v^5-w_n^5\right)\phi \mathrm{d}x\right|=o_n\left(1\right){\parallel \phi \parallel }_{V_{\infty }}$ for any $\phi \in H^1$. By the Lebesgue dominated convergence theorem, $\left|{\int }_{{\mathbb{R}}^3}(g(\epsilon x+\epsilon y_n)-g_{\infty })v^5\phi \mathrm{d}x\right|=o_n\left(1\right){\parallel \phi \parallel }_{V_{\infty }}$ for any $\phi \in H^1$. Then

$$\begin{array}{c}\hfill \left|{\int }_{{\mathbb{R}}^3}\left(g(\epsilon x+\epsilon y_n)v_n^5-g_{\infty }{v}^5-g(\epsilon x+\epsilon y_n)w_n^5\right)\phi \mathrm{d}x\right|=o_n\left(1\right){\parallel \phi \parallel }_{V_{\infty }},\forall \phi \in H^1.\end{array}$$

Similar to Lemma 8.1 in [34], we derive that

$$\begin{array}{c}\hfill \left|{\int }_{{\mathbb{R}}^3}\left(f\left(v_n\right)-f\left(v\right)-f\left(w_n\right)\right)\phi \mathrm{d}x\right|=o_n\left(1\right){\parallel \phi \parallel }_{V_{\infty }},\forall \phi \in H^1.\end{array}$$

Together with Lemma 1 $\left(iii\right)$, $L_{\epsilon }^{\prime }\left(v_n\right)=o_n\left(1\right)$, ${\breve{I}}^{\prime }\left(v\right)=0$, we get $L_{\epsilon }^{\prime }\left(w_n\right)=o_n\left(1\right)$. Thus,

$$\begin{array}{cc}\hfill o_n\left(1\right)=& \left(L_{\epsilon }^{\prime }\left(w_n\right),w_n\right)\hfill \\ \hfill =& \parallel w_n{\parallel }_{V_{\infty }}^2+{\int }_{{\mathbb{R}}^3}{\varphi }_{w_n}{w}_n^2\mathrm{d}x-{\int }_{{\mathbb{R}}^3}{h}_{\infty }f\left(w_n\right)w_n\mathrm{d}x-{\int }_{{\mathbb{R}}^3}g(\epsilon x+\epsilon y_n){\left|w_n\right|}^6\mathrm{d}x.\hfill \end{array}$$

By (21), (27) and $g(\epsilon x+\epsilon y_n)\le 1$, we derive that

$$\begin{array}{c}\hfill \parallel w_n{\parallel }_{V_{\infty }}^2\le \frac{V_{\infty }}{2}{\int }_{{\mathbb{R}}^3}|w_n{|}^2\mathrm{d}x+\left(h_{\infty }{C}_{\frac{V_{\infty }}{2h_{\infty }}}+1\right){\int }_{{\mathbb{R}}^3}{\left|w_n\right|}^6\mathrm{d}x+o_n\left(1\right).\end{array}$$

So $\frac{1}{2}\parallel w_n{\parallel }_{V_{\infty }}^2\le \frac{h_{\infty }{C}_{\frac{1}{2h_{\infty }}}+1}{S^3}{\parallel w_n\parallel }_{V_{\infty }}^6+o_n\left(1\right)$. Since $w_n$ converges weakly and not strongly to 0 in $H^1$, we get ${lim}_{n\to \infty }{\parallel w_n\parallel }_{V_{\infty }}^2\ge \frac{S^{\frac{3}{2}}}{\sqrt{2\left(h_{\infty }{C}_{\frac{V_{\infty }}{2h_{\infty }}}+1\right)}}$. Since ${\breve{I}}^{\prime }\left(v\right)=0$, we have $\breve{I}\left(v\right)\ge m_{\infty }\ge \widehat{m}$. Then by (26) and (27),

$$\begin{array}{cc}\hfill {\gamma }_{\epsilon }^i=& L_{\epsilon }\left(w_n\right)-\frac{1}{4}\left(L_{\epsilon }^{\prime }\left(w_n\right),w_n\right)+\breve{I}\left(v\right)+o_n\left(1\right)\hfill \\ \hfill \ge & \frac{1}{4}{\parallel w_n\parallel }_{V_{\infty }}^2+\widehat{m}+o_n\left(1\right)\ge \frac{S^{\frac{3}{2}}}{4\sqrt{2\left(h_{\infty }{C}_{\frac{V_{\infty }}{2h_{\infty }}}+1\right)}}+\widehat{m}+o_n\left(1\right),\hfill \end{array}$$

a contradiction with (22).

Case 2. $u_n\rightharpoonup u\ne 0$ weakly in $H_{\epsilon }$.

By $u_n\rightharpoonup u\ne 0$ weakly in $H_{\epsilon }$, we have $I_{\epsilon }^{\prime }\left(u\right)=0$. By Lemma 1.3 in [36],

$$\begin{array}{c}\hfill {\int }_{{\mathbb{R}}^3}h\left(\epsilon x\right)F\left(u_n\right)\mathrm{d}x-{\int }_{{\mathbb{R}}^3}h\left(\epsilon x\right)F\left(u\right)\mathrm{d}x={\int }_{{\mathbb{R}}^3}h\left(\epsilon x\right)F\left({\widehat{u}}_n\right)\mathrm{d}x+o_n\left(1\right),\end{array}$$

where ${\widehat{u}}_n=u_n-u$. Together with the Brezis–Lieb Lemma in [34] and Lemma 1 $\left(iii\right)$, we get ${\gamma }_{\epsilon }^i=I_{\epsilon }\left(u_n\right)+o_n\left(1\right)=I_{\epsilon }\left({\widehat{u}}_n\right)+I_{\epsilon }\left(u\right)+o_n\left(1\right)$. By Lemma 8.9 in [34], for any $\phi \in H_{\epsilon }$, $\left|{\int }_{{\mathbb{R}}^3}g\left(\epsilon x\right)\left(u_n^5-u^5-{\widehat{u}}_n^5\right)\phi \mathrm{d}x\right|=o_n\left(1\right){\parallel \phi \parallel }_{\epsilon }$. Similar to Lemma 8.1 in [34], for any $\phi \in H_{\epsilon }$, $\left|{\int }_{{\mathbb{R}}^3}h\left(\epsilon x\right)\left[f\left(u_n\right)-f\left(u\right)-f\left({\widehat{u}}_n\right)\right]\phi \mathrm{d}x\right|=o_n\left(1\right){\parallel \phi \parallel }_{\epsilon }$. Together with Lemma 1 $\left(iii\right)$, we get $I_{\epsilon }^{\prime }\left({\widehat{u}}_n\right)=o_n\left(1\right)$. Thus,

$$\begin{array}{c}\hfill {\gamma }_{\epsilon }^i=I_{\epsilon }\left({\widehat{u}}_n\right)+I_{\epsilon }\left(u\right)+o_n\left(1\right),I_{\epsilon }^{\prime }\left({\widehat{u}}_n\right)=o_n\left(1\right).\end{array}$$

(27)

We claim ${\widehat{u}}_n\to 0$ in $H_{\epsilon }$. Otherwise, ${\widehat{u}}_n$ converges weakly and not strongly to 0 in $H_{\epsilon }$. By ${lim}_{\left|x\right|\to \infty }V\left(x\right)=V_{\infty }$, ${lim}_{\left|x\right|\to \infty }h\left(x\right)=h_{\infty }$, we have

$$\begin{array}{c}\hfill {\gamma }_{\epsilon }^i=J_{\epsilon }\left({\widehat{u}}_n\right)+I_{\epsilon }\left(u\right)+o_n\left(1\right),J_{\epsilon }^{\prime }\left({\widehat{u}}_n\right)=o_n\left(1\right).\end{array}$$

(28)

By (21), $g\left(\epsilon x\right)\le 1$ and $\left(J_{\epsilon }^{\prime }\left({\widehat{u}}_n\right),{\widehat{u}}_n\right)=o_n\left(1\right)$,

$$\begin{array}{c}\hfill \parallel {\widehat{u}}_n{\parallel }_{V_{\infty }}^2\le \frac{V_{\infty }}{2}{\int }_{{\mathbb{R}}^3}|{\widehat{u}}_n{|}^2\mathrm{d}x+\left(h_{\infty }{C}_{\frac{V_{\infty }}{2h_{\infty }}}+1\right){\int }_{{\mathbb{R}}^3}{\left|{\widehat{u}}_n\right|}^6\mathrm{d}x+o_n\left(1\right).\end{array}$$

By the Sobolev embedding theorem, we get ${lim}_{n\to \infty }{\parallel {\widehat{u}}_n\parallel }_{V_{\infty }}^2\ge \frac{S^{\frac{3}{2}}}{\sqrt{2\left(h_{\infty }{C}_{\frac{V_{\infty }}{2h_{\infty }}}+1\right)}}$. Since $I_{\epsilon }^{\prime }\left(u\right)=0$, we have $I_{\epsilon }\left(u\right)\ge m_{\epsilon }\ge \widehat{m}$. Then by (28),

$$\begin{array}{cc}\hfill {\gamma }_{\epsilon }^i=& J_{\epsilon }\left({\widehat{u}}_n\right)-\frac{1}{4}\left(J_{\epsilon }^{\prime }\left({\widehat{u}}_n\right),{\widehat{u}}_n\right)+I_{\epsilon }\left(u\right)+o_n\left(1\right)\hfill \\ \hfill \ge & \frac{1}{4}{\parallel {\widehat{u}}_n\parallel }_{V_{\infty }}^2+\widehat{m}+o_n\left(1\right)\ge \frac{S^{\frac{3}{2}}}{4\sqrt{2\left(h_{\infty }{C}_{\frac{V_{\infty }}{2h_{\infty }}}+1\right)}}+\widehat{m}+o_n\left(1\right),\hfill \end{array}$$

a contradiction with (22). So ${\widehat{u}}_n\to 0$ in $H_{\epsilon }$, that is, $u_n\to u$ in $H_{\epsilon }$.  □

Lemma 7.

Let $\epsilon \in (0,{\epsilon }_0)$, Then problem (11) has at least k different positive solutions $u_{\epsilon }^i$, $i=1$, 2, … k.

Proof of Lemma 7.

Let $i=1$, 2, … k. By Lemma 6, we have $u_n^i\in M_{\epsilon }^i$, $I_{\epsilon }\left(u_n^i\right)\to {\gamma }_{\epsilon }^i$, $I_{\epsilon }^{\prime }\left(u_n^i\right)\to 0$. Moreover, $u_n^i\to u_{\epsilon }^i$ in $H_{\epsilon }$. Then $u_{\epsilon }^i\in \overline{M_{\epsilon }^i}$, $I_{\epsilon }\left(u_{\epsilon }^i\right)={\gamma }_{\epsilon }^i$, $I_{\epsilon }^{\prime }\left(u_{\epsilon }^i\right)=0$. By ${\gamma }_{\epsilon }^i<{\tilde{\gamma }}_{\epsilon }^i$, we have $u_{\epsilon }^i\in M_{\epsilon }^i$. Since $\beta \left(u_{\epsilon }^i\right)\in C_{\eta }^i$, where $C_{\eta }^i$, $i=1$, 2, … k are disjoint, we derive that $u_{\epsilon }^i$, $i=1$, 2, … k are different. Obviously, $u_{\epsilon }^i$ is non-negative. By the maximum principle, $u_{\epsilon }^i$ is positive.  □

Now we study the behavior of $u_{\epsilon }^i$ as $\epsilon \to 0$.

Lemma 8.

Let $i=1$, 2, … k. Then there exist ${\epsilon }^{*}\in (0,{\epsilon }_0)$, $\left\{x_{\epsilon }^i\right\}\subset {\mathbb{R}}^3$, $R_0$, ${\varrho }_0>0$ satisfying ${\int }_{B_{R_0}\left(x_{\epsilon }^i\right)}{\left|u_{\epsilon }^i\right|}^2\mathrm{d}x\ge {\varrho }_0$ for $\epsilon \in (0,{\epsilon }^{*})$.

Proof of Lemma 8.

Otherwise, there exists ${\epsilon }_n\downarrow 0$ such that for any $R>0$,

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}\underset{x\in {\mathbb{R}}^3}{sup}{\int }_{B_R\left(x\right)}{\left|u_{{\epsilon }_n}^i\right|}^2\mathrm{d}x=0.\end{array}$$

By the Lions Lemma, ${\int }_{{\mathbb{R}}^3}{\left|u_{{\epsilon }_n}^i\right|}^t\mathrm{d}x\to 0$ for any $t\in (2,6)$. Since $I_{{\epsilon }_n}\left(u_{{\epsilon }_n}^i\right)={\gamma }_{{\epsilon }_n}^i\to \widehat{m}$, $I_{{\epsilon }_n}^{\prime }\left(u_{{\epsilon }_n}^i\right)=0$, similar to the argument of (19), we get $\widehat{m}\ge \frac{1}{3}{S}^{\frac{3}{2}}$, a contradiction.  □

Lemma 9.

$\epsilon x_{\epsilon }^i\to x^i$ as $\epsilon \to 0$.

Proof of Lemma 9.

We first prove $|\epsilon x_{\epsilon }^i|$ is bounded. Assume to the contrary that there exists ${\epsilon }_n\downarrow 0$ satisfying $|{\epsilon }_n{x}_{{\epsilon }_n}^i|\to \infty$. By $I_{{\epsilon }_n}\left(u_{{\epsilon }_n}^i\right)={\gamma }_{{\epsilon }_n}^i\to \widehat{m}$, $I_{{\epsilon }_n}^{\prime }\left(u_{{\epsilon }_n}^i\right)=0$, we derive that $\parallel u_{{\epsilon }_n}^i{\parallel }_{{\epsilon }_n}$ is bounded. Let $v_{{\epsilon }_n}^i=u_{{\epsilon }_n}^i(.+x_{{\epsilon }_n}^i)$. By Lemma 8, we get $v_{{\epsilon }_n}^i\rightharpoonup v^i\ne 0$ weakly in $H^1$. Let

$$\begin{array}{cc}\hfill T_{\epsilon }\left(u\right)=& \frac{1}{2}{\int }_{{\mathbb{R}}^3}\left({|\nabla u|}^2+V(\epsilon x+\epsilon x_{\epsilon }^i){\left|u\right|}^2\right)\mathrm{d}x+\frac{1}{4}{\int }_{{\mathbb{R}}^3}{\varphi }_u{u}^2\mathrm{d}x\hfill \\ \hfill & -{\int }_{{\mathbb{R}}^3}h(\epsilon x+\epsilon x_{\epsilon }^i)F\left(u\right)\mathrm{d}x-\frac{1}{6}{\int }_{{\mathbb{R}}^3}g(\epsilon x+\epsilon x_{\epsilon }^i){\left|u\right|}^6\mathrm{d}x,u\in H_{\epsilon }.\hfill \end{array}$$

Then $T_{{\epsilon }_n}\left(v_{{\epsilon }_n}^i\right)\to \widehat{m}$, $T_{{\epsilon }_n}^{\prime }\left(v_{{\epsilon }_n}^i\right)=0$. Furthermore,

$$\begin{array}{cc}\hfill & {\int }_{{\mathbb{R}}^3}\left(\nabla v_{{\epsilon }_n}^i\nabla v^i+V({\epsilon }_{n}x+{\epsilon }_n{x}_{{\epsilon }_n}^i)v_{{\epsilon }_n}^i{v}^i\right)\mathrm{d}x+{\int }_{{\mathbb{R}}^3}{\varphi }_{v_{{\epsilon }_n}^i}{v}_{{\epsilon }_n}^i{v}^i\mathrm{d}x\hfill \\ \hfill & ={\int }_{{\mathbb{R}}^3}h({\epsilon }_{n}x+{\epsilon }_n{x}_{{\epsilon }_n}^i)f\left(v_{{\epsilon }_n}^i\right)v^i\mathrm{d}x+{\int }_{{\mathbb{R}}^3}g({\epsilon }_{n}x+{\epsilon }_n{x}_{{\epsilon }_n}^i){\left(v_{{\epsilon }_n}^i\right)}^5{v}^i\mathrm{d}x.\hfill \end{array}$$

By $|{\epsilon }_n{x}_{{\epsilon }_n}^i|\to \infty$, we derive that

$$\begin{array}{c}\hfill \parallel v^i{\parallel }_{V_{\infty }}^2+{\int }_{{\mathbb{R}}^3}{\varphi }_{v^i}|v^i{|}^2\mathrm{d}x={\int }_{{\mathbb{R}}^3}{h}_{\infty }f\left(v^i\right)v^i\mathrm{d}x+{\int }_{{\mathbb{R}}^3}{g}_{\infty }{\left|v^i\right|}^6\mathrm{d}x,\end{array}$$

that is, $v^i\in \widehat{N}$. By $T_{{\epsilon }_n}\left(v_{{\epsilon }_n}^i\right)\to \widehat{m}$, $T_{{\epsilon }_n}^{\prime }\left(v_{{\epsilon }_n}^i\right)=0$,

$$\begin{array}{cc}\hfill \widehat{m}& =T_{{\epsilon }_n}\left(v_{{\epsilon }_n}^i\right)-\frac{1}{4}\left(T_{{\epsilon }_n}^{\prime }\left(v_{{\epsilon }_n}^i\right),v_{{\epsilon }_n}^i\right)+o_n\left(1\right)\hfill \\ \hfill & =\frac{1}{4}{\int }_{{\mathbb{R}}^3}\left(|\nabla v_{{\epsilon }_n}^i{|}^2+V({\epsilon }_{n}x+{\epsilon }_n{x}_{{\epsilon }_n}^i){\left|v_{{\epsilon }_n}^i\right|}^2\right)\mathrm{d}x\hfill \\ \hfill & +{\int }_{{\mathbb{R}}^3}h({\epsilon }_{n}x+{\epsilon }_n{x}_{{\epsilon }_n}^i)\left(\frac{1}{4}f\left(v_{{\epsilon }_n}^i\right)v_{{\epsilon }_n}^i-F\left(v_{{\epsilon }_n}^i\right)\right)\mathrm{d}x\hfill \\ \hfill & +\frac{1}{12}{\int }_{{\mathbb{R}}^3}g({\epsilon }_{n}x+{\epsilon }_n{x}_{{\epsilon }_n}^i){\left|v_{{\epsilon }_n}^i\right|}^6\mathrm{d}x+o_n\left(1\right).\hfill \end{array}$$

(29)

Then by Fatou’s Lemma and $|{\epsilon }_n{x}_{{\epsilon }_n}^i|\to \infty$,

$$\begin{array}{cc}\hfill \widehat{m}\ge & \frac{1}{4}\parallel v^i{\parallel }_{V_{\infty }}^2+{\int }_{{\mathbb{R}}^3}{h}_{\infty }\left(\frac{1}{4}f\left(v^i\right)v^i-F\left(v^i\right)\right)\mathrm{d}x+\frac{1}{12}{\int }_{{\mathbb{R}}^3}{g}_{\infty }{|v^i|}^6\mathrm{d}x\hfill \\ \hfill =& \breve{I}\left(v^i\right)-\frac{1}{4}\left({\breve{I}}^{\prime }\left(v^i\right),v^i\right)=\breve{I}\left(v^i\right).\hfill \end{array}$$

(30)

Since $v^i\in \widehat{N}$, we have $\breve{I}\left(v^i\right)\ge m_{\infty }\ge \widehat{m}$. Combining (29) and (30), we get $v_{{\epsilon }_n}^i\to v^i$ in $H^1$. Note that

$$\begin{array}{c}\hfill \beta \left(u_{{\epsilon }_n}^i\right)=\frac{{\int }_{{\mathbb{R}}^3}\chi \left({\epsilon }_{n}x\right){\left|u_{{\epsilon }_n}^i\right|}^6\mathrm{d}x}{{\int }_{{\mathbb{R}}^3}{\left|u_{{\epsilon }_n}^i\right|}^6\mathrm{d}x}=\frac{{\int }_{{\mathbb{R}}^3}\chi ({\epsilon }_{n}x+{\epsilon }_n{x}_{{\epsilon }_n}^i){\left|v_{{\epsilon }_n}^i\right|}^6\mathrm{d}x}{{\int }_{{\mathbb{R}}^3}{\left|v_{{\epsilon }_n}^i\right|}^6\mathrm{d}x}.\end{array}$$

Then by $|{\epsilon }_n{x}_{{\epsilon }_n}^i|\to \infty$, we get $\rho ={lim}_{n\to \infty }\left|\beta \left(u_{{\epsilon }_n}^i\right)\right|$, a contradiction with $\beta \left(u_{{\epsilon }_n}^i\right)\in C_{\eta }^i$.

Now we prove $\epsilon x_{\epsilon }^i\to x^i$ as $\epsilon \to 0$. Since $|\epsilon x_{\epsilon }^i|$ is bounded, we assume $\epsilon x_{\epsilon }^i\to x_0^i$ as $\epsilon \to 0$. Let $v_{\epsilon }^i=u_{\epsilon }^i(.+x_{\epsilon }^i)$. By Lemma 8, we get $v_{\epsilon }^i\rightharpoonup v^i\ne 0$ weakly in $H^1$. Then by $I_{\epsilon }^{\prime }\left(u_{\epsilon }^i\right)=0$, we derive that

$$\begin{array}{cc}\hfill & {\int }_{{\mathbb{R}}^3}\left(|\nabla v^i{|}^2+V\left(x_0^i\right){\left|v^i\right|}^2\right)\mathrm{d}x+{\int }_{{\mathbb{R}}^3}{\varphi }_{v^i}{\left|v^i\right|}^2\mathrm{d}x\hfill \\ \hfill & ={\int }_{{\mathbb{R}}^3}h\left(x_0^i\right)f\left(v^i\right)v^i\mathrm{d}x+{\int }_{{\mathbb{R}}^3}g\left(x_0^i\right){\left|v^i\right|}^6\mathrm{d}x.\hfill \end{array}$$

Let

$$\begin{array}{cc}\hfill L_0\left(u\right)=& \frac{1}{2}{\int }_{{\mathbb{R}}^3}\left({|\nabla u|}^2+V\left(x_0^i\right){\left|u\right|}^2\right)\mathrm{d}x-{\int }_{{\mathbb{R}}^3}h\left(x_0^i\right)F\left(u\right)\mathrm{d}x\hfill \\ \hfill & -\frac{1}{6}{\int }_{{\mathbb{R}}^3}g\left(x_0^i\right){\left|u\right|}^6\mathrm{d}x,u\in H^1.\hfill \end{array}$$

Then $\left(L_0^{\prime }\left(v^i\right),v^i\right)=0$. By $I_{\epsilon }\left(u_{\epsilon }^i\right)={\gamma }_{\epsilon }^i\to \widehat{m}$, $I_{\epsilon }^{\prime }\left(u_{\epsilon }^i\right)=0$,

$$\begin{array}{cc}\hfill \widehat{m}=& I_{\epsilon }\left(u_{\epsilon }^i\right)-\frac{1}{4}\left(I_{\epsilon }^{\prime }\left(u_{\epsilon }^i\right),u_{\epsilon }^i\right)+o_{\epsilon }\left(1\right)\hfill \\ \hfill =& \frac{1}{4}{\int }_{{\mathbb{R}}^3}\left(|\nabla v_{\epsilon }^i{|}^2+V(\epsilon x+\epsilon x_{\epsilon }^i){\left|v_{\epsilon }^i\right|}^2\right)\mathrm{d}x+\frac{1}{12}{\int }_{{\mathbb{R}}^3}g(\epsilon x+\epsilon x_{\epsilon }^i){\left|v_{\epsilon }^i\right|}^6\mathrm{d}x\hfill \\ \hfill & +{\int }_{{\mathbb{R}}^3}h(\epsilon x+\epsilon x_{\epsilon }^i)\left(\frac{1}{4}f\left(v_{\epsilon }^i\right)v_{\epsilon }^i-F\left(v_{\epsilon }^i\right)\right)\mathrm{d}x+o_{\epsilon }\left(1\right).\hfill \end{array}$$

(31)

Then by Fatou’s Lemma,

$$\begin{array}{cc}\hfill \widehat{m}& \ge \frac{1}{4}{\int }_{{\mathbb{R}}^3}\left(|\nabla v^i{|}^2+V\left(x_0^i\right){\left|v^i\right|}^2\right)\mathrm{d}x+{\int }_{{\mathbb{R}}^3}h\left(x_0^i\right)\left(\frac{1}{4}f\left(v^i\right)v^i-F\left(v^i\right)\right)\mathrm{d}x\hfill \\ \hfill & +\frac{1}{12}{\int }_{{\mathbb{R}}^3}g\left(x_0^i\right){\left|v^i\right|}^6\mathrm{d}x\hfill \\ \hfill & \ge L_0\left(v^i\right)-\frac{1}{4}\left(L_0^{\prime }\left(v^i\right),v^i\right)=L_0\left(v^i\right).\hfill \end{array}$$

(32)

Since $\left(L_0^{\prime }\left(v^i\right),v^i\right)=0$, by $\left(f_1\right)$, we have $L_0\left(v^i\right)={sup}_{t\ge 0}{L}_0\left(t{v}^i\right)$. Moreover, there exists a unique ${\check{t}}^i>0$ satisfying ${\check{t}}^i{v}^i\in \widehat{N}$. Then

$$\begin{array}{c}\hfill L_0\left(v^i\right)=\underset{t\ge 0}{sup}{L}_0\left(t{v}^i\right)\ge L_0\left({\check{t}}^i{v}^i\right)\ge \widehat{I}\left({\check{t}}^i{v}^i\right)\ge \widehat{m}.\end{array}$$

(33)

Combining (31)–(33), we get $h\left(x_0^i\right)=h_M$ and $v_{\epsilon }^i\to v^i$ in $H^1$. Then $h\left(\epsilon x_{\epsilon }^i\right)\to h_M$. Note that

$$\begin{array}{c}\hfill \beta \left(u_{\epsilon }^i\right)=\frac{{\int }_{{\mathbb{R}}^3}\chi \left(\epsilon x\right){\left|u_{\epsilon }^i\right|}^6\mathrm{d}x}{{\int }_{{\mathbb{R}}^3}{\left|u_{\epsilon }^i\right|}^6\mathrm{d}x}=\frac{{\int }_{{\mathbb{R}}^3}\chi (\epsilon x+\epsilon x_{\epsilon }^i){\left|v_{\epsilon }^i\right|}^6\mathrm{d}x}{{\int }_{{\mathbb{R}}^3}{\left|v_{\epsilon }^i\right|}^6\mathrm{d}x}.\end{array}$$

If $|x_0^i|\ge \rho$, then ${lim}_{\epsilon \to 0}\left|\beta \left(u_{\epsilon }^i\right)\right|=\rho$, a contradiction with $\beta \left(u_{\epsilon }^i\right)\in C_{\eta }^i$. So $|x_0^i|<\rho$, from which we derive that ${lim}_{\epsilon \to 0}\beta \left(u_{\epsilon }^i\right)=x_0^i\in \overline{C_{\eta }^i}$. Since $\epsilon x_{\epsilon }^i\to x_0^i$, $h\left(x_0^i\right)=h_M$, we get $\epsilon x_{\epsilon }^i\to x^i$ as $\epsilon \to 0$.  □

Proof of Theorem 1.

By Lemma 7, problem (11) has at least k different positive solutions $u_{\epsilon }^i$, $i=1$, 2, … k. Let $v_{\epsilon }^i=u_{\epsilon }^i(.+x_{\epsilon }^i)$. Then

$$\begin{array}{c}\hfill -\Delta v_{\epsilon }^i+V(\epsilon x+\epsilon x_{\epsilon }^i)v_{\epsilon }^i+{\varphi }_{v_{\epsilon }^i}{v}_{\epsilon }^i=h(\epsilon x+\epsilon x_{\epsilon }^i)f\left(v_{\epsilon }^i\right)+g(\epsilon x+\epsilon x_{\epsilon }^i){\left(v_{\epsilon }^i\right)}^5.\end{array}$$

(34)

By Lemma 9, $v_{\epsilon }^i\to v^i\ne 0$ in $H^1$, $\epsilon x_{\epsilon }^i\to x^i$. By the argument of Lemmas 3.8 and 3.11 in [15], ${lim}_{\left|x\right|\to \infty }{v}_{\epsilon }^i\left(x\right)=0$ uniformly for $\epsilon \in (0,{\epsilon }^{*})$ and there exists ${\acute{C}}^i>0$ independent of $\epsilon \in (0,{\epsilon }^{*})$ satisfying $\parallel v_{\epsilon }^i{\parallel }_{\infty }\le {\acute{C}}^i$. Furthermore, there exist ${\stackrel{`}{C}}^i$, $c^i>0$ such that $v_{\epsilon }^i\left(x\right)\le {\stackrel{`}{C}}^{i}exp\left(-c^i\left|x\right|\right)$ uniformly for $\epsilon \in (0,{\epsilon }^{*})$.

We claim there exists ${\gamma }_0>0$ such that $\parallel v_{\epsilon }^i{\parallel }_{\infty }\ge {\gamma }_0$ uniformly for $\epsilon \in (0,{\epsilon }^{*})$. Otherwise, $\parallel v_{\epsilon }^i{\parallel }_{\infty }\to 0$ as $\epsilon \to 0$. By (34) and $\left(f_1\right)$, we get $\parallel v_{\epsilon }^i{\parallel }_{V_0}^2\le \frac{V_0}{2}{\int }_{{\mathbb{R}}^3}|v_{\epsilon }^i{|}^2\mathrm{d}x+C{\int }_{{\mathbb{R}}^3}{\left|v_{\epsilon }^i\right|}^6\mathrm{d}x$. Then $\parallel v_{\epsilon }^i{\parallel }_{V_0}^2\le 2C\parallel v_{\epsilon }^i{\parallel }_{\infty }^4{\int }_{{\mathbb{R}}^3}{\left|v_{\epsilon }^i\right|}^2\mathrm{d}x\to 0$, a contradiction with $v_{\epsilon }^i\to v^i\ne 0$ in $H^1$. Let $z_{\epsilon }^i$ be the maximum point of $v_{\epsilon }^i$. Then $|v_{\epsilon }^i\left(z_{\epsilon }^i\right)|\ge {\gamma }_0$. By ${lim}_{\left|x\right|\to \infty }{v}_{\epsilon }^i\left(x\right)=0$ uniformly for $\epsilon \in (0,{\epsilon }^{*})$, we derive that there exists $Z^i>0$ such that $|z_{\epsilon }^i|\le Z^i$.

Since $v_{\epsilon }^i=u_{\epsilon }^i(.+x_{\epsilon }^i)$, we know $y_{\epsilon }^i:=x_{\epsilon }^i+z_{\epsilon }^i$ is the maximum point of $u_{\epsilon }^i$. By $\epsilon x_{\epsilon }^i\to x^i$ as $\epsilon \to 0$ and $|z_{\epsilon }^i|\le Z^i$, we get $\epsilon (x_{\epsilon }^i+z_{\epsilon }^i)\to x^i$ as $\epsilon \to 0$. Moreover, for $\epsilon \in (0,{\epsilon }^{*})$,

$$\begin{array}{c}\hfill u_{\epsilon }^i\left(x\right)=v_{\epsilon }^i(.-x_{\epsilon }^i)\le {\stackrel{`}{C}}^{i}exp\left(-c^i|x-x_{\epsilon }^i|\right)\le C^{i}exp\left(-c^i|x-(x_{\epsilon }^i+z_{\epsilon }^i)|\right).\end{array}$$

Let $w_{\epsilon }^i=u_{\epsilon }^i\left(\frac{.}{\epsilon }\right)$, $y_{\epsilon }^i=\epsilon (x_{\epsilon }^i+z_{\epsilon }^i)$. Then $w_{\epsilon }^i$ is the positive solution of (1). Furthermore, $y_{\epsilon }^i$ is the maximum point of $w_{\epsilon }^i$, $h\left(y_{\epsilon }^i\right)\to {sup}_{x\in {\mathbb{R}}^3}h\left(x\right)$ and there exist $C^i$, $c^i>0$ such that $w_{\epsilon }^i\left(x\right)\le C^{i}exp\left(-c^i\frac{|x-y_{\epsilon }^i|}{\epsilon }\right)$ for $\epsilon \in (0,{\epsilon }^{*})$.  □

4. Proof of Theorems 2 and 3

For simplicity, let $\epsilon =1$. Denote $H=\left\{u\in H^1:{\int }_{{\mathbb{R}}^3}V\left(x\right){\left|u\right|}^2\mathrm{d}x<+\infty \right\}$ the Hilbert space with the norm $\parallel u\parallel ={\left({\int }_{{\mathbb{R}}^3}{|\nabla u|}^2+V\left(x\right){\left|u\right|}^2\mathrm{d}x\right)}^{\frac{1}{2}}$. Let

$$\begin{array}{c}\hfill I\left(u\right)=\frac{1}{2}{\parallel u\parallel }^2+\frac{1}{4}{\int }_{{\mathbb{R}}^3}{\varphi }_u{u}^2\mathrm{d}x-{\int }_{{\mathbb{R}}^3}h\left(x\right)F\left(u\right)\mathrm{d}x-\frac{1}{6}{\int }_{{\mathbb{R}}^3}g\left(x\right){\left|u\right|}^6\mathrm{d}x.\end{array}$$

Let $m=inf\left\{I\right(u):u\in M\}$, where $M=\{u\in H\backslash \left\{0\right\}:(I^{\prime }\left(u\right),u)=0\}$.

We first prove Theorem 3. Let X be the Banach space. Recall that $\left\{u_n\right\}\subset X$ is a ${\left(C\right)}_c$ sequence for the functional I if $I\left(u_n\right)\to c$ and $(1+\parallel u_n{\parallel }_X)\parallel I^{\prime }\left(u_n\right)\parallel \to 0$ as $n\to \infty$.

Theorem 4

([37]). Let X be the Banach space and $I\in C^1(X,\mathbb{R})$ satisfying

$$max\{I\left(0\right),I\left(u_1\right)\}\le {\alpha }_2<{\alpha }_1\le \underset{{\parallel u\parallel }_X=\rho }{inf}I\left(u\right)$$

for some $\rho >0$ and $u_1\in X$ with $\parallel u_1{\parallel }_X>\rho$. Let $c={inf}_{\gamma \in \Gamma }{max}_{0\le t\le 1}I\left(\gamma \left(t\right)\right)$, where $\Gamma =\{\gamma \in C([0,1],X):\gamma \left(0\right)=0,\gamma \left(1\right)=u_1\}$. Then there exists a ${\left(C\right)}_c$ sequence $\left\{u_n\right\}$ for the functional I satisfying $c\ge {\alpha }_1$.

For any $y\in {\mathbb{R}}^3$ with $\left|y\right|=1$, by $\left(h_2\right)$, there exist $\sigma$, $\xi >0$ such that $h\left(x\right)\ge \xi$ for $|x-\frac{3}{2}{\rho }_{0}y|\le \sigma$. Let $r<min\{\frac{1}{4}{\rho }_0,\frac{1}{2}\sigma \}$. Define $u_{\delta ,\tilde{y}}:=u_{\delta ,\frac{3}{2}{\rho }_{0}y}\left(x\right)=\frac{\psi \left(x\right){\delta }^{\frac{1}{4}}}{{\left(\delta +|x-\frac{3}{2}{\rho }_0{y|}^2\right)}^{\frac{1}{2}}}$, where $\psi \in C_0^{\infty }\left(B_{2r}\left(\frac{3}{2}{\rho }_{0}y\right)\right)$ such that $\psi \left(x\right)=1$ for $|x-\frac{3}{2}{\rho }_{0}y|<r$, $0\le \psi \left(x\right)\le 1$, $|\nabla \psi |\le 2$.

Lemma 10.

Assume that $\left(h_2\right)$, $\left(V_1\right)$, $\left(g_2^{\prime }\right)$ and $\left(f_1^{\prime }\right)$ hold. Then there exists ${\delta }_0>0$ independent of $y\in {\mathbb{R}}^3$ with $\left|y\right|=1$ such that for any $\delta \in (0,{\delta }_0)$,

$$\begin{array}{c}\hfill \underset{t\ge 0}{sup}I\left(t{u}_{\delta ,\tilde{y}}\right)\le \frac{1}{3}{S}^{\frac{3}{2}}-{\delta }^{\frac{1}{2}}.\end{array}$$

Proof of Lemma 10.

From Lemma 1 $\left(iv\right)$, there exists $C_0>0$ such that

$$\begin{array}{c}\hfill I\left(t{u}_{\delta ,\tilde{y}}\right)\le \frac{t^2}{2}\parallel u_{\delta ,\tilde{y}}{\parallel }^2+\frac{C_0{t}^4}{4}\parallel u_{\delta ,\tilde{y}}{\parallel }^4-\frac{t^6}{6}{\int }_{{\mathbb{R}}^3}g\left(x\right){\left|u_{\delta ,\tilde{y}}\right|}^6\mathrm{d}x.\end{array}$$

By Lemma 2 and $\left(g_2^{\prime }\right)$, there exists ${\delta }_1>0$ independent of y such that for $\delta \in (0,{\delta }_1)$,

$$\begin{array}{c}\hfill \parallel u_{\delta ,\tilde{y}}{\parallel }^2\le \frac{3K_1}{2},{\int }_{{\mathbb{R}}^3}g\left(x\right)|u_{\delta ,\tilde{y}}{|}^6\mathrm{d}x={\int }_{{\mathbb{R}}^3}{\left|u_{\delta ,\tilde{y}}\right|}^6\mathrm{d}x\ge \frac{K_2}{2}.\end{array}$$

Then there exist a small $t_1>0$ and a large $t_2>0$ independent of $\delta \in (0,{\delta }_1)$ satisfying

$$\begin{array}{c}\hfill \underset{t\in [0,t_1]\cup [t_2,+\infty )}{sup}I\left(t{u}_{\delta ,\tilde{y}}\right)<\frac{1}{6}{S}^{\frac{3}{2}}.\end{array}$$

(35)

By Lemmas 1 and 2, we get

$$\begin{array}{cc}\hfill \underset{t\in [t_1,t_2]}{sup}I\left(t{u}_{\delta ,\tilde{y}}\right)\le & \underset{t\ge 0}{sup}\left(\frac{t^2}{2}\parallel u_{\delta ,\tilde{y}}{\parallel }^2-\frac{t^6}{6}{\parallel u_{\delta ,\tilde{y}}\parallel }_6^6\right)+\frac{t_2^4}{4S}{\parallel u_{\delta ,\tilde{y}}\parallel }_{\frac{12}{5}}^4\hfill \\ \hfill & -\xi \underset{t\in [t_1,t_2]}{inf}{\int }_{{\mathbb{R}}^3}F\left(t{u}_{\delta ,\tilde{y}}\right)\mathrm{d}x\hfill \\ \hfill =& \frac{1}{3}{\left(\frac{\parallel u_{\delta ,\tilde{y}}{\parallel }^2}{\parallel u_{\delta ,\tilde{y}}{\parallel }_6^2}\right)}^{\frac{3}{2}}+O\left(\delta \right)-\xi \underset{t\in [t_1,t_2]}{inf}{\int }_{{\mathbb{R}}^3}F\left(t{u}_{\delta ,\tilde{y}}\right)\mathrm{d}x\hfill \\ \hfill \le & \frac{1}{3}{S}^{\frac{3}{2}}+{\delta }^{\frac{1}{2}}-\xi \underset{t\in [t_1,t_2]}{inf}{\int }_{{\mathbb{R}}^3}F\left(t{u}_{\delta ,\tilde{y}}\right)\mathrm{d}x.\hfill \end{array}$$

By $\left(f_1^{\prime }\right)$, we have ${lim}_{u\to +\infty }\frac{F\left(u\right)}{u^4}=+\infty$. Then for $L>\frac{2}{\xi t_1^4{\int }_{\left|x\right|\le 1}\mathrm{d}x}$, there exists $R_L>0$ satisfying $F\left(u\right)\ge {L\left|u\right|}^4$ for $\left|u\right|\ge R_L$. Let $r<min\{\frac{1}{4}{\rho }_0,\frac{1}{2}\sigma \}$. Note that $u_{\delta ,\tilde{y}}\left(x\right)\ge {\delta }^{-\frac{1}{4}}$ for $|x-\frac{3}{2}{\rho }_{0}y|\le {\delta }^{\frac{1}{2}}\le r$. Let ${\delta }_0=min\{{\delta }_1,r^2\}$ and $\delta \in (0,{\delta }_0)$. Then for $t\in [t_1,t_2]$ and $|x-\frac{3}{2}{\rho }_{0}y|\le {\delta }^{\frac{1}{2}}$, we have $F\left(t{u}_{\delta ,\tilde{y}}\right)\ge L{t}_1^4{u}_{\delta ,\tilde{y}}^4\ge L{t}_1^4{\delta }^{-1}$. Since $F\left(t{u}_{\delta ,\tilde{y}}\right)\ge 0$, we derive that

$$\begin{array}{cc}\hfill \underset{t\in [t_1,t_2]}{inf}{\int }_{{\mathbb{R}}^3}F\left(t{u}_{\delta ,\tilde{y}}\right)\mathrm{d}x\ge & \underset{t\in [t_1,t_2]}{inf}{\int }_{|x-\frac{3}{2}{\rho }_{0}y|\le {\delta }^{\frac{1}{2}}}F\left(t{u}_{\delta ,\tilde{y}}\right)\mathrm{d}x\ge L{t}_1^4{\delta }^{\frac{1}{2}}{\int }_{\left|x\right|\le 1}\mathrm{d}x.\hfill \end{array}$$

Thus, for $\delta \in (0,{\delta }_0)$,

$$\begin{array}{c}\hfill \underset{t\in [t_1,t_2]}{sup}I\left(t{u}_{\delta ,\tilde{y}}\right)\le \frac{1}{3}{S}^{\frac{3}{2}}+{\delta }^{\frac{1}{2}}-\xi L{t}_1^4{\delta }^{\frac{1}{2}}{\int }_{\left|x\right|\le 1}\mathrm{d}x\le \frac{1}{3}{S}^{\frac{3}{2}}-{\delta }^{\frac{1}{2}}.\end{array}$$

(36)

Combining (35) and (36), we get Lemma 10.  □

Define the functional on H by

$$\begin{array}{c}\hfill J\left(u\right)=\frac{1}{2}{\parallel u\parallel }^2+\frac{1}{4}{\int }_{{\mathbb{R}}^3}{\varphi }_u{u}^2\mathrm{d}x-\frac{1}{6}{\int }_{{\mathbb{R}}^3}g\left(x\right){\left|u\right|}^6\mathrm{d}x.\end{array}$$

Lemma 11.

Assume that $\left(h_2\right)$, $\left(V_1\right)$, $\left(Vh\right)$, $\left(g_2^{\prime }\right)$ and $\left(f_1^{\prime }\right)$ hold. If $\left\{u_n\right\}\subset H$ such that $u_n\rightharpoonup u$ weakly in H, $I\left(u_n\right)\to c\in \left(0,\frac{1}{3}{S}^{\frac{3}{2}}\right)$ and $I^{\prime }\left(u_n\right)\to 0$, then $I^{\prime }\left(u\right)=0$ and $u_n\rightharpoonup u\ne 0$ weakly in H. Moreover, if $I\left(u\right)\ge 0$, then $u_n\to u$ in H.

Proof of Lemma 11.

By $u_n\rightharpoonup u$ weakly in H, we have $I^{\prime }\left(u\right)=0$. Let $v_n=u_n-u$. By Lemma 1.3 in [36], we obtain that ${\int }_{{\mathbb{R}}^3}h\left(x\right)F\left(v_n\right)\mathrm{d}x={\int }_{{\mathbb{R}}^3}h\left(x\right)F\left(u_n\right)\mathrm{d}x-{\int }_{{\mathbb{R}}^3}h\left(x\right)F\left(u\right)\mathrm{d}x+o_n\left(1\right)$. By $\left(h_2\right)$ and $\left(f_1^{\prime }\right)$, for any $\epsilon >0$, there exists $C_{\epsilon }>0$ such that $|h\left(x\right)F\left(v_n\right)|\le \epsilon |v_n{|}^6+C_{\epsilon }h\left(x\right){\left|v_n\right|}^2$. Then for any $R>0$,

$$\begin{array}{c}\hfill {\int }_{\left|x\right|\ge R}|h\left(x\right)F\left(v_n\right)|\mathrm{d}x\le \epsilon {\int }_{{\mathbb{R}}^3}|v_n{|}^6\mathrm{d}x+C_{\epsilon }\underset{\left|x\right|\ge R}{sup}\frac{h\left(x\right)}{V\left(x\right)}\times {\int }_{{\mathbb{R}}^3}V\left(x\right){\left|v_n\right|}^2\mathrm{d}x.\end{array}$$

Since $\parallel v_n\parallel$ is bounded, by $\left(Vh\right)$, for any $\epsilon >0$, there exists $R_{\epsilon }>0$ such that

$$\begin{array}{c}\hfill \underset{\left|x\right|\ge R_{\epsilon }}{sup}\frac{h\left(x\right)}{V\left(x\right)}\times {\int }_{{\mathbb{R}}^3}V\left(x\right){\left|v_n\right|}^2\mathrm{d}x\le \epsilon .\end{array}$$

By $v_n\rightharpoonup 0$ weakly in H, we have ${\int }_{\left|x\right|\le R_{\epsilon }}h\left(x\right)F\left(v_n\right)\mathrm{d}x=o_n\left(1\right)$. Thus,

$$\begin{array}{c}\hfill {\int }_{{\mathbb{R}}^3}h\left(x\right)F\left(u_n\right)\mathrm{d}x-{\int }_{{\mathbb{R}}^3}h\left(x\right)F\left(u\right)\mathrm{d}x={\int }_{{\mathbb{R}}^3}h\left(x\right)F\left(v_n\right)\mathrm{d}x+o_n\left(1\right)=o_n\left(1\right).\end{array}$$

(37)

By the Brezis–Lieb Lemma in [34], we get $\parallel v_n{\parallel }^2=\parallel u_n{\parallel }^2-{\parallel u\parallel }^2+o_n\left(1\right)$ and ${\int }_{{\mathbb{R}}^3}g\left(x\right)|v_n{|}^6\mathrm{d}x={\int }_{{\mathbb{R}}^3}g\left(x\right)|u_n{|}^6\mathrm{d}x-{\int }_{{\mathbb{R}}^3}g\left(x\right){\left|u\right|}^6\mathrm{d}x+o_n\left(1\right)$. Together with Lemma 1 $\left(iii\right)$, we have

$$\begin{array}{c}\hfill c-I\left(u\right)=I\left(u_n\right)-I\left(u\right)+o_n\left(1\right)=J\left(v_n\right)+o_n\left(1\right).\end{array}$$

(38)

By Lemma 8.9 in [34], we have $\left|{\int }_{{\mathbb{R}}^3}g\left(x\right)\left(u_n^5-u^5-v_n^5\right)\phi \mathrm{d}x\right|=o_n\left(1\right)\parallel \phi \parallel$ for any $\phi \in H$. Similar to Lemma 8.1 in [34], we derive for any $\phi \in H$, there holds

$\left|{\int }_{{\mathbb{R}}^3}h\left(x\right)\left[f\left(u_n\right)-f\left(u\right)-f\left(v_n\right)\right]\phi \mathrm{d}x\right|=o_n\left(1\right)\parallel \phi \parallel$. Together with Lemma 1 $\left(iii\right)$, we get $I^{\prime }\left(v_n\right)=o_n\left(1\right)$. So $\left(I^{\prime }\left(v_n\right),v_n\right)=o_n\left(1\right)$. Similar to (37), we get ${\int }_{{\mathbb{R}}^3}h\left(x\right)f\left(v_n\right)v_n\mathrm{d}x=o_n\left(1\right)$. Thus, we have $\left(J^{\prime }\left(v_n\right),v_n\right)=o_n\left(1\right)$.

Assume to the contrary that $u_n\rightharpoonup 0$ weakly in H. Then $v_n=u_n$. Assume ${\int }_{{\mathbb{R}}^3}g\left(x\right){\left|v_n\right|}^6$

$\mathrm{d}x\to l$. Then ${lim}_{n\to \infty }{\parallel v_n\parallel }^2\le l$. If $l>0$, by $S\le \frac{\parallel v_n{\parallel }^2}{{\left({\int }_{{\mathbb{R}}^3}g\left(x\right){\left|v_n\right|}^6\mathrm{d}x\right)}^{\frac{1}{3}}}$, we get $l\ge S^{\frac{3}{2}}$. So $c-I\left(u\right)=J\left(v_n\right)-\frac{1}{4}(J^{\prime }\left(v_n\right),v_n)+o_n\left(1\right)\ge \frac{1}{3}{S}^{\frac{3}{2}}+o_n\left(1\right)$. Since $I\left(u\right)=0$, we get $c\ge \frac{1}{3}{S}^{\frac{3}{2}}$, a contradiction. So $l=0$. By $\left(J^{\prime }\left(v_n\right),v_n\right)=o_n\left(1\right)$, we get $u_n=v_n\to 0$ in H, a contradiction with $I\left(u_n\right)\to c>0$. So $u_n\rightharpoonup u\ne 0$ weakly in H.

If $I\left(u\right)\ge 0$, similar to the above argument, we can derive that $u_n\to u$ in H. We omit the proof here.  □

Proof of Theorem 3.

From $\left(h_2\right)$, $\left(f_1^{\prime }\right)$, for $\epsilon =\frac{V_0}{4}$, there exists $C_{\epsilon }=C_{\frac{V_0}{4}}>0$ such that $\left|h\left(x\right)F\left(u\right)\right|\le \frac{V_0}{4}{\left|u\right|}^2+C_{\frac{V_0}{4}}{\left|u\right|}^6$. By the Sobolev embedding theorem,

$$\begin{array}{c}\hfill I\left(u\right)\ge \frac{1}{4}{\parallel u\parallel }^2-\left(C_{\frac{V_0}{4}}+\frac{1}{6}\right){\int }_{{\mathbb{R}}^3}{\left|u\right|}^6\mathrm{d}x\ge \frac{1}{4}{\parallel u\parallel }^2-\frac{\left(C_{\frac{V_0}{4}}+\frac{1}{6}\right){\parallel u\parallel }^6}{S^3}.\end{array}$$

Then there exist ${\rho }_0$, ${\gamma }_0>0$ such that $I\left(u\right)\ge {\gamma }_0$ for $\parallel u\parallel ={\rho }_0$. Furthermore, $I\left(0\right)=0$. By the argument of Lemma 4.1, ${lim}_{t\to +\infty }I\left(t{u}_{\delta ,\tilde{y}}\right)=-\infty$. So by Lemma 10 and Theorem 4, there is $\left\{u_n\right\}\subset H$ such that

$$\begin{array}{c}\hfill I\left(u_n\right)\to c\in \left(0,\frac{1}{3}{S}^{\frac{3}{2}}\right),(1+\parallel u_n\parallel )\parallel I^{\prime }\left(u_n\right)\parallel \to 0.\end{array}$$

If $\parallel u_n\parallel$ is bounded, by Lemma 11, we have $u_n\rightharpoonup u\ne 0$ weakly in H and $I^{\prime }\left(u\right)=0$, that is, (1) has a positive solution. So we just need to prove that $\parallel u_n\parallel$ is bounded.

Otherwise, we have $\parallel u_n\parallel \to \infty$. Let $v_n=\frac{u_n}{\parallel u_n\parallel }$. Then $v_n\rightharpoonup v$ weakly in H.

Case 1. $v\left(x\right)=0$ a.e. $x\in {\mathbb{R}}^3$. Let $\theta \in (4,6)$ and ${inf}_{x\in {\mathbb{R}}^3}g\left(x\right):=g_0$. By $\left(h_2\right)$, $\left(f_1^{\prime }\right)$ and $\left(g_2^{\prime }\right)$, for $\epsilon \in \left(0,g_0\times \left(\frac{1}{\theta }-\frac{1}{6}\right)\right)$, there exists $C_{\epsilon }>0$ such that

$$\begin{array}{c}\hfill \left|\frac{1}{\theta }h\left(x\right)f\left(u_n\right)u_n-h\left(x\right)F\left(u_n\right)\right|\le \epsilon |u_n{|}^6+C_{\epsilon }h\left(x\right){\left|u_n\right|}^2.\end{array}$$

Then

$$\begin{array}{c}\hfill \left(I\left(u_n\right)-\frac{1}{\theta }\left(I^{\prime }\left(u_n\right),u_n\right)\right)\ge \left(\frac{1}{2}-\frac{1}{\theta }\right)\parallel u_n{\parallel }^2-C_{\epsilon }{\int }_{{\mathbb{R}}^3}h\left(x\right){\left|u_n\right|}^2\mathrm{d}x.\end{array}$$

(39)

From (39), we derive that

$$\begin{array}{c}\hfill \frac{1}{\parallel u_n{\parallel }^2}\left(I\left(u_n\right)-\frac{1}{\theta }\left(I^{\prime }\left(u_n\right),u_n\right)\right)\ge \left(\frac{1}{2}-\frac{1}{\theta }\right)-C_{\epsilon }{\int }_{{\mathbb{R}}^3}h\left(x\right){\left|v_n\right|}^2\mathrm{d}x.\end{array}$$

Similar to (37), we have ${\int }_{{\mathbb{R}}^3}h\left(x\right){\left|v_n\right|}^2\mathrm{d}x\to 0$. Then by $I\left(u_n\right)\to c$, $(I^{\prime }\left(u_n\right),u_n)\to 0$ and $\parallel u_n\parallel \to \infty$, we get $0\ge \left(\frac{1}{2}-\frac{1}{\theta }\right)$, a contradiction.

Case 2. $v\left(x\right)\ne 0$. Let $\mathsf{\Omega }=\{x\in {\mathbb{R}}^3:v\left(x\right)\ne 0\}$. Then the measure of $\mathsf{\Omega }$ is positive. For $x\in \mathsf{\Omega }$, by $v_n\left(x\right)=\frac{u_n\left(x\right)}{\parallel u_n\parallel }\to v\left(x\right)$, we get ${lim}_{n\to \infty }\left|u_n\left(x\right)\right|=+\infty$. Let $q\in (4,6)$. By $\left(h_2\right)$, $\left(f_1^{\prime }\right)$ and Fatou’s Lemma,

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}{\int }_{\mathsf{\Omega }}\frac{h\left(x\right)F\left(u_n\right)+\frac{1}{6}g\left(x\right){\left|u_n\right|}^6}{\parallel u_n{\parallel }^q}\mathrm{d}x\ge \frac{g_0}{6}\underset{n\to \infty }{lim}{\int }_{\mathsf{\Omega }}|v_n{|}^q{\left|u_n\right|}^{6-q}\mathrm{d}x=+\infty .\end{array}$$

(40)

Furthermore, for $\epsilon \in \left(0,\frac{g_0}{6}\right)$, there exists $C_{\epsilon }>0$ such that $|h\left(x\right)F\left(u_n\right)|\le \epsilon |u_n{|}^6+C_{\epsilon }{\left|u_n\right|}^2$. Then

$$\begin{array}{cc}\hfill & {\int }_{{\mathbb{R}}^3\backslash \mathsf{\Omega }}h\left(x\right)F\left(u_n\right)\mathrm{d}x+\frac{1}{6}{\int }_{{\mathbb{R}}^3\backslash \mathsf{\Omega }}g\left(x\right){\left|u_n\right|}^6\mathrm{d}x\hfill \\ \hfill & \ge -C_{\epsilon }{\int }_{{\mathbb{R}}^3\backslash \mathsf{\Omega }}|u_n{|}^2\mathrm{d}x+\left(\frac{g_0}{6}-\epsilon \right){\int }_{{\mathbb{R}}^3\backslash \mathsf{\Omega }}|u_n{|}^6\mathrm{d}x\ge -C_{\epsilon }{\int }_{{\mathbb{R}}^3\backslash \mathsf{\Omega }}{\left|u_n\right|}^2\mathrm{d}x,\hfill \end{array}$$

from which we get

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}{\int }_{{\mathbb{R}}^3\backslash \mathsf{\Omega }}\frac{h\left(x\right)F\left(u_n\right)+\frac{1}{6}g\left(x\right){\left|u_n\right|}^6}{\parallel u_n{\parallel }^q}\mathrm{d}x\ge -C_{\epsilon }\underset{n\to \infty }{lim}\frac{1}{\parallel u_n{\parallel }^{q-2}}{\int }_{{\mathbb{R}}^3\backslash \mathsf{\Omega }}{\left|v_n\right|}^2\mathrm{d}x\ge 0.\end{array}$$

(41)

Combining (40) and (41), we derive that

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}{\int }_{{\mathbb{R}}^3}\frac{h\left(x\right)F\left(u_n\right)+\frac{1}{6}g\left(x\right){\left|u_n\right|}^6}{\parallel u_n{\parallel }^q}\mathrm{d}x=+\infty .\end{array}$$

(42)

On the other hand, by Lemma 1, $\left(iv\right)$,

$$\begin{array}{c}\hfill I\left(u_n\right)+{\int }_{{\mathbb{R}}^3}\left(h\left(x\right)F\left(u_n\right)+\frac{1}{6}g\left(x\right){\left|u_n\right|}^6\right)\mathrm{d}x\le \frac{1}{2}\parallel u_n{\parallel }^2+\frac{C_0}{4}{\parallel u_n\parallel }^4.\end{array}$$

So $\underset{n\to \infty }{lim}{\int }_{{\mathbb{R}}^3}\frac{h\left(x\right)F\left(u_n\right)+\frac{1}{6}g\left(x\right){\left|u_n\right|}^6}{\parallel u_n{\parallel }^q}\mathrm{d}x=0$, a contradiction with (42).  □

Now we prove Theorem 2. By the Lagrange multipliers Theorem, we can derive the following result. Since the proof is standard, we omit it here.

Lemma 12.

Assume that $\left(h_2\right)$, $\left(V_1\right)$, $\left(g_2\right)$ and $\left(f_1\right)$ hold. Let $\left\{u_n\right\}\subset M$ such that $I\left(u_n\right)\to c\in \left(0,\frac{1}{3}{S}^{\frac{3}{2}}\right)$ and ${I|}_M^{\prime }\left(u_n\right)\to 0$. Then $I^{\prime }\left(u_n\right)\to 0$. Moreover, $\parallel u_n\parallel$ is bounded.

Lemma 13.

Assume that $\left(V_1\right)$, $\left(g_2\right)$ and $\left(f_1\right)$ hold. Then there exists ${\eta }_0>0$ such that ${\int }_{{\mathbb{R}}^3}\frac{x}{\left|x\right|}{|\nabla u|}^2$ $\mathrm{d}x\ne 0$ for $u\in M_0$ with $J\left(u\right)\le \frac{1}{3}{S}^{\frac{3}{2}}+{\eta }_0$, where $M_0=\{u\in H\backslash \left\{0\right\}:(J^{\prime }\left(u\right),u)=0\}$.

Proof of Lemma 13.

Assume to the contrary that there exists $u_n\in M_0$ such that $J\left(u_n\right)\to \frac{1}{3}{S}^{\frac{3}{2}}$ and ${\int }_{{\mathbb{R}}^3}\frac{x}{\left|x\right|}{|\nabla u_n|}^2\mathrm{d}x=0$. Then ${\int }_{{\mathbb{R}}^3}|\nabla u_n{|}^2\mathrm{d}x\le {\int }_{{\mathbb{R}}^3}g\left(x\right){\left|u_n\right|}^6\mathrm{d}x$. By $S\le \frac{{\int }_{{\mathbb{R}}^3}{|\nabla u_n|}^2\mathrm{d}x}{{\left({\int }_{{\mathbb{R}}^3}g\left(x\right){\left|u_n\right|}^6\mathrm{d}x\right)}^{\frac{1}{3}}}$, we get ${\int }_{{\mathbb{R}}^3}g\left(x\right){\left|u_n\right|}^6\mathrm{d}x\ge S^{\frac{3}{2}}$. So

$$\begin{array}{cc}\hfill \frac{1}{3}{S}^{\frac{3}{2}}+o_n\left(1\right)& =J\left(u_n\right)-\frac{1}{6}(J^{\prime }\left(u_n\right),u_n)\hfill \\ \hfill & \ge \frac{1}{3}{\int }_{{\mathbb{R}}^3}\left(|\nabla u_n{|}^2+V\left(x\right){\left|u_n\right|}^2\right)\mathrm{d}x\hfill \\ \hfill & \ge \frac{S}{3}{\left({\int }_{{\mathbb{R}}^3}{\left|u_n\right|}^6\mathrm{d}x\right)}^{\frac{1}{3}}\ge \frac{S}{3}{\left({\int }_{{\mathbb{R}}^3}g\left(x\right){\left|u_n\right|}^6\mathrm{d}x\right)}^{\frac{1}{3}}\ge \frac{1}{3}{S}^{\frac{3}{2}},\hfill \end{array}$$

from which we derive that ${lim}_{n\to \infty }{\int }_{{\mathbb{R}}^3}V\left(x\right){|u_n|}^2\mathrm{d}x=0$,

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}{\int }_{{\mathbb{R}}^3}|\nabla u_n{|}^2\mathrm{d}x=S^{\frac{3}{2}},\underset{n\to \infty }{lim}{\int }_{{\mathbb{R}}^3}|u_n{|}^6\mathrm{d}x=\underset{n\to \infty }{lim}{\int }_{{\mathbb{R}}^3}g\left(x\right){\left|u_n\right|}^6\mathrm{d}x=S^{\frac{3}{2}}.\end{array}$$

(43)

Let $v_n=\frac{u_n}{{\left({\int }_{{\mathbb{R}}^3}{\left|u_n\right|}^6\mathrm{d}x\right)}^{\frac{1}{6}}}$. We get ${\int }_{{\mathbb{R}}^3}V\left(x\right){\left|v_n\right|}^2\mathrm{d}x\to 0$, ${\int }_{{\mathbb{R}}^3}{|\nabla v_n|}^2\mathrm{d}x\to S$, ${\int }_{{\mathbb{R}}^3}\frac{x}{\left|x\right|}{|\nabla v_n|}^2$

$\mathrm{d}x=0$. By Theorem 1.41 in [34], there exist $y_n\in {\mathbb{R}}^3$ and ${\mu }_n\in (0,+\infty )$ such that

$$\begin{array}{c}\hfill {∥v_n-\frac{1}{{\mu }_n^{\frac{1}{2}}}{v}_0\left(\frac{x-y_n}{{\mu }_n}\right)∥}_{D^{1,2}}\to 0.\end{array}$$

(44)

So ${\int }_{{\mathbb{R}}^3}{|\nabla v_0|}^2\mathrm{d}x=S$, ${\int }_{{\mathbb{R}}^3}{\left|v_0\right|}^6\mathrm{d}x=1$, that is, S is attained by $v_0$. By [38], $v_0=\frac{c_0}{(1+|d_0(x-x_0){{|}^2)}^{\frac{1}{2}}}$, where $c_0\ne 0$, $d_0>0$, $x_0\in {\mathbb{R}}^3$. Thus,

$$\begin{array}{c}\hfill {∥v_n-\frac{c_0{\mu }_n^{\frac{1}{2}}}{{\left({\mu }_n^2+d_0^2{|·-y_n-x_0{\mu }_n|}^2\right)}^{\frac{1}{2}}}∥}_{D^{1,2}}\to 0.\end{array}$$

(45)

By $g\left(0\right)<1$, there exist ${\varrho }_0>0$ and $g_m\in (0,1)$ such that $g\left(x\right)\le g_m$ for $\left|x\right|\le {\varrho }_0$.

Case 1. ${\mu }_n\to +\infty$ as $n\to \infty$.

Let $x={\mu }_{n}z+y_n$. By $V\left(x\right)\ge V_0$, we get ${\int }_{{\mathbb{R}}^3}|v_n{\left(x\right)|}^2\mathrm{d}x={\mu }_n^2{\int }_{{\mathbb{R}}^3}{|{\mu }_n^{\frac{1}{2}}{v}_n({\mu }_{n}z+y_n)|}^2\mathrm{d}z\to 0$. Then ${\int }_{{\mathbb{R}}^3}{|{\mu }_n^{\frac{1}{2}}{v}_n({\mu }_{n}z+y_n)|}^2\mathrm{d}z\to 0$. By (44), we have ${\mu }_n^{\frac{1}{2}}{v}_n({\mu }_{n}x+y_n)\to v_0$ a.e. By Fatou’s Lemma, we get ${\int }_{{\mathbb{R}}^3}{\left|v_0\right|}^2\mathrm{d}z=0$, a contradiction.

Case 2. ${\mu }_n\to \tilde{\mu }\ne 0$ as $n\to \infty$.

Similar to Case 1, we get a contradiction.

Case 3. ${\mu }_n\to 0$ as $n\to \infty$ and $|y_n|\le {\varrho }_0$ for large n.

Assume $y_n\to y_0$. Then $|y_0|\le {\varrho }_0$. By (43), we have ${\int }_{{\mathbb{R}}^3}(1-g\left(x\right)){\left|v_n\right|}^6\mathrm{d}x=o_n\left(1\right)$. Then by (44),

$$\begin{array}{c}\hfill o_n\left(1\right)={\int }_{{\mathbb{R}}^3}(1-g\left(x\right)){\left|\frac{1}{{\mu }_n^{\frac{1}{2}}}{v}_0\left(\frac{x-y_n}{{\mu }_n}\right)\right|}^6\mathrm{d}x={\int }_{{\mathbb{R}}^3}(1-g({\mu }_{n}x+y_n)){\left|v_0\right|}^6\mathrm{d}x.\end{array}$$

By the Lebesgue dominated convergence theorem, we derive that $0={\int }_{{\mathbb{R}}^3}(1-g\left(y_0\right)){\left|v_0\right|}^6\mathrm{d}x>0$, a contradiction.

Case 4. ${\mu }_n\to 0$ as $n\to \infty$ and there exists a subsequence of ${\left\{y_n\right\}}_{n=1}^{\infty }$ (still denoted by ${\left\{y_n\right\}}_{n=1}^{\infty }$) satisfying $|y_n|>{\varrho }_0$. By ${\int }_{{\mathbb{R}}^3}\frac{x}{\left|x\right|}{|\nabla v_n|}^2\mathrm{d}x=0$ and (45),

$$\begin{array}{cc}\hfill o_n\left(1\right)=& {\int }_{{\mathbb{R}}^3}\left(\frac{x}{\left|x\right|}-\frac{y_n+x_0{\mu }_n}{|y_n+x_0{\mu }_n|}\right)\frac{{\mu }_n{|x-y_n-x_0{\mu }_n|}^2}{{\left({\mu }_n^2+d_0^2{|x-y_n-x_0{\mu }_n|}^2\right)}^3}\mathrm{d}x\hfill \\ \hfill & +\frac{y_n+x_0{\mu }_n}{|y_n+x_0{\mu }_n|}{\int }_{{\mathbb{R}}^3}\frac{{\mu }_n{|x-y_n-x_0{\mu }_n|}^2}{{\left({\mu }_n^2+d_0^2{|x-y_n-x_0{\mu }_n|}^2\right)}^3}\mathrm{d}x.\hfill \end{array}$$

(46)

By ${\mu }_n\to 0$ and $|y_n|>{\varrho }_0$, we have $|y_n+x_0{\mu }_n|\ge \frac{|y_n|}{2}\ge \frac{{\varrho }_0}{2}$ for large n. For any x, $z\in {\mathbb{R}}^3\backslash \left\{0\right\}$,

$$\begin{array}{c}\hfill \left|\frac{x}{\left|x\right|}-\frac{z}{\left|z\right|}\right|=\frac{\left|x\right(\left|z\right|-\left|x\right|)+|x\left|\right(x-z\left)\right|}{\left|x\right|\left|z\right|}\le \frac{2|x-z|}{\left|z\right|}.\end{array}$$

Thus,

$$\begin{array}{cc}\hfill & {\int }_{|x-y_n-x_0{\mu }_n|\le {\mu }_n}\left|\frac{x}{\left|x\right|}-\frac{y_n+x_0{\mu }_n}{|y_n+x_0{\mu }_n|}\right|\frac{{\mu }_n{|x-y_n-x_0{\mu }_n|}^2}{{\left({\mu }_n^2+d_0^2{|x-y_n-x_0{\mu }_n|}^2\right)}^3}\mathrm{d}x\hfill \\ \hfill & \le {\int }_{|x-y_n-x_0{\mu }_n|\le {\mu }_n}\frac{2|x-y_n-x_0{\mu }_n|}{|y_n+x_0{\mu }_n|}\frac{{\mu }_n{|x-y_n-x_0{\mu }_n|}^2}{{\left({\mu }_n^2+d_0^2{|x-y_n-x_0{\mu }_n|}^2\right)}^3}\mathrm{d}x\hfill \\ \hfill & \le \frac{4{\mu }_n}{{\varrho }_0}{\int }_{{\mathbb{R}}^3}\frac{{\left|x\right|}^2}{{\left(1+d_0^2{\left|x\right|}^2\right)}^3}\mathrm{d}x\le C_2{\mu }_n.\hfill \end{array}$$

(47)

Furthermore,

$$\begin{array}{cc}\hfill & {\int }_{|x-y_n-x_0{\mu }_n|\ge {\mu }_n}\left|\frac{x}{\left|x\right|}-\frac{y_n+x_0{\mu }_n}{|y_n+x_0{\mu }_n|}\right|\frac{{\mu }_n{|x-y_n-x_0{\mu }_n|}^2}{{\left({\mu }_n^2+d_0^2{|x-y_n-x_0{\mu }_n|}^2\right)}^3}\mathrm{d}x\hfill \\ \hfill & \le \frac{2}{|y_n+x_0{\mu }_n|}{\int }_{|x-y_n-x_0{\mu }_n|\ge {\mu }_n}\frac{{\mu }_n{|x-y_n-x_0{\mu }_n|}^3}{{\left({\mu }_n^2+d_0^2{|x-y_n-x_0{\mu }_n|}^2\right)}^3}\mathrm{d}x\hfill \\ \hfill & \le \frac{4}{{\varrho }_0}{\int }_{\left|x\right|\ge 1}\frac{{\mu }_n{\left|x\right|}^3}{{\left(1+d_0^2{\left|x\right|}^2\right)}^3}\mathrm{d}x\le C_3{\mu }_n.\hfill \end{array}$$

(48)

From (46)–(48), we get

$$\begin{array}{c}\hfill 0=\underset{n\to \infty }{lim}\left|{\int }_{{\mathbb{R}}^3}\frac{x}{\left|x\right|}\frac{{\mu }_n{|x-y_n-x_0{\mu }_n|}^2}{{\left({\mu }_n^2+d_0^2{|x-y_n-x_0{\mu }_n|}^2\right)}^3}\mathrm{d}x\right|={\int }_{{\mathbb{R}}^3}\frac{{\left|x\right|}^2}{{\left(1+d_0^2{\left|x\right|}^2\right)}^3}\mathrm{d}x,\end{array}$$

a contradiction.  □

Lemma 14.

Assume that $\left(h_2\right)$, $\left(V_1\right)$, $\left(g_2\right)$ and $\left(f_1\right)$ hold. Then there exists $h_0>0$ such that for ${\parallel h\parallel }_{\infty }<h_0$ and $u\in M$ with $I\left(u\right)<\frac{1}{3}{S}^{\frac{3}{2}}$, there holds ${\int }_{{\mathbb{R}}^3}\frac{x}{\left|x\right|}{|\nabla u|}^2\mathrm{d}x\ne 0$.

Proof of Lemma 14.

By $\left(f_1\right)$, for any $u\in M$, there exists a unique $t_u>0$ such that $t_{u}u\in M_0$. Then

$$\begin{array}{cc}\hfill & {\parallel u\parallel }^2+{\int }_{{\mathbb{R}}^3}{\varphi }_u{u}^2\mathrm{d}x={\int }_{{\mathbb{R}}^3}h\left(x\right)f\left(u\right)u\mathrm{d}x+{\int }_{{\mathbb{R}}^3}g\left(x\right){\left|u\right|}^6\mathrm{d}x,\hfill \\ \hfill & t_u^2{\parallel u\parallel }^2+t_u^4{\int }_{{\mathbb{R}}^3}{\varphi }_u{u}^2\mathrm{d}x=t_u^6{\int }_{{\mathbb{R}}^3}g\left(x\right){\left|u\right|}^6\mathrm{d}x.\hfill \end{array}$$

If $t_u<1$, then

$$\begin{array}{cc}\hfill t_u^4\left({\parallel u\parallel }^2+{\int }_{{\mathbb{R}}^3}{\varphi }_u{u}^2\mathrm{d}x\right)\le & t_u^6\left({\int }_{{\mathbb{R}}^3}h\left(x\right)f\left(u\right)u\mathrm{d}x+{\int }_{{\mathbb{R}}^3}g\left(x\right){\left|u\right|}^6\mathrm{d}x\right)\hfill \\ \hfill =& t_u^6\left({\parallel u\parallel }^2+{\int }_{{\mathbb{R}}^3}{\varphi }_u{u}^2\mathrm{d}x\right),\hfill \end{array}$$

that is, $t_u\ge 1$, a contradiction. So $t_u\ge 1$. By (49),

$$\begin{array}{c}\hfill t_u^2\le \frac{{\parallel u\parallel }^2+{\int }_{{\mathbb{R}}^3}{\varphi }_u{u}^2\mathrm{d}x}{{\int }_{{\mathbb{R}}^3}g\left(x\right){\left|u\right|}^6\mathrm{d}x}.\end{array}$$

(49)

Since $u\in M$, by $\left(h_2\right)$, $\left(f_1\right)$, for $\epsilon =V_0$, there exists $C_{\epsilon }=C_{V_0}>0$ such that ${\parallel u\parallel }^2\le {\parallel h\parallel }_{\infty }{\int }_{{\mathbb{R}}^3}{V}_0{\left|u\right|}^2\mathrm{d}x+C_{V_0}{\parallel h\parallel }_{\infty }{\int }_{{\mathbb{R}}^3}{\left|u\right|}^6\mathrm{d}x+{\int }_{{\mathbb{R}}^3}g\left(x\right){\left|u\right|}^6\mathrm{d}x$. Then for ${\parallel h\parallel }_{\infty }<1$,

$$\begin{array}{c}\hfill S{\left({\int }_{{\mathbb{R}}^3}{\left|u\right|}^6\mathrm{d}x\right)}^{\frac{1}{3}}\le C_{V_0}{\parallel h\parallel }_{\infty }{\int }_{{\mathbb{R}}^3}{\left|u\right|}^6\mathrm{d}x+{\int }_{{\mathbb{R}}^3}g\left(x\right){\left|u\right|}^6\mathrm{d}x.\end{array}$$

(50)

Since $g\left(x\right)\le 1$, we obtain that there exists ${\eta }_0>0$ such that ${\int }_{{\mathbb{R}}^3}{\left|u\right|}^6\mathrm{d}x\ge {\eta }_0$. By $I\left(u\right)<\frac{1}{3}{S}^{\frac{3}{2}}$ with $u\in M$, we get ${\parallel u\parallel }^2\le \frac{4}{3}{S}^{\frac{3}{2}}$. Then ${\int }_{{\mathbb{R}}^3}{\left|u\right|}^6\mathrm{d}x\le \frac{{\parallel u\parallel }^6}{S^3}\le \frac{64}{27}{S}^{\frac{3}{2}}$. So ${\eta }_0\le {\int }_{{\mathbb{R}}^3}{\left|u\right|}^6\mathrm{d}x\le \frac{64}{27}{S}^{\frac{3}{2}}$. By (50), there exists $h_0^{\prime }\in (0,1)$ such that for ${\parallel h\parallel }_{\infty }<h_0^{\prime }$,

$$\begin{array}{c}\hfill \frac{S}{2}{\left({\int }_{{\mathbb{R}}^3}g\left(x\right){\left|u\right|}^6\mathrm{d}x\right)}^{\frac{1}{3}}\le \frac{S}{2}{\left({\int }_{{\mathbb{R}}^3}{\left|u\right|}^6\mathrm{d}x\right)}^{\frac{1}{3}}\le {\int }_{{\mathbb{R}}^3}g\left(x\right){\left|u\right|}^6\mathrm{d}x.\end{array}$$

Then ${\int }_{{\mathbb{R}}^3}g\left(x\right){\left|u\right|}^6\mathrm{d}x\ge {\left(\frac{S}{2}\right)}^{\frac{3}{2}}$. Together with (49) and ${\parallel u\parallel }^2\le \frac{4}{3}{S}^{\frac{3}{2}}$, we derive that there exists $T_0>0$ such that $t_u\le T_0$. By $\left(f_1\right)$, we obtain that for $u\in M$ with $I\left(u\right)<\frac{1}{3}{S}^{\frac{3}{2}}$,

$$\begin{array}{c}\hfill \frac{1}{3}{S}^{\frac{3}{2}}>I\left(u\right)=\underset{t\ge 0}{sup}I\left(tu\right)\ge I\left(t_{u}u\right)\ge J\left(t_{u}u\right)-{\parallel h\parallel }_{\infty }{\int }_{{\mathbb{R}}^3}F\left(t_{u}u\right)\mathrm{d}x.\end{array}$$

By $t_u\le T_0$ and ${\parallel u\parallel }^2\le \frac{4}{3}{S}^{\frac{3}{2}}$, there exists $h_0\in (0,h_0^{\prime })$ such that $J\left(t_{u}u\right)\le \frac{1}{3}{S}^{\frac{3}{2}}+{\eta }_0$ for ${\parallel h\parallel }_{\infty }<h_0$. Since $t_{u}u\in M_0$, by Lemma 13, we get ${\int }_{{\mathbb{R}}^3}\frac{x}{\left|x\right|}{|\nabla \left(t_{u}u\right)|}^2\mathrm{d}x\ne 0$, that is, ${\int }_{{\mathbb{R}}^3}\frac{x}{\left|x\right|}{|\nabla u|}^2\mathrm{d}x\ne 0$.  □

We introduce the Lusternik–Schnirelman category.

Definition 1.

For a topological space X, a nonempty, closed subset $A\subset X$ is contractible to a point y in X if and only if there exists a continuous mapping $\eta :[0,1]\times A\to X$ such that $\eta (0,x)=x$ for $x\in A$ and $\eta (1,x)=y$ for $x\in A$.

Definition 2.

Define

$$\begin{array}{cc}\hfill \mathrm{cat}\left(X\right)=min\{k\in \mathbb{N}:& \mathrm{there}\mathrm{exist}\mathrm{closed}\mathrm{subsets}{A}_1,\dots ,A_k\subset X\mathrm{such}\mathrm{that}\hfill \\ \hfill & A_i\mathrm{is}\mathrm{contractible}\mathrm{to}\mathrm{a}\mathrm{point}\mathrm{in}X\mathrm{for}\mathrm{all}i\mathrm{and}\hfill \\ \hfill & {\cup }_{i=1}^k{A}_i=X\}.\hfill \end{array}$$

In particular, if there does not exist finitely many closed subsets $A_1,\dots ,A_k\subset X$ such that ${\cup }_{i=1}^k{A}_i=X$ and $A_i$ is contractible to a point in X for all i, denote $\mathrm{cat}\left(X\right)=\infty$.

The following two lemmas are introduced to prove Theorem 2.

Lemma 15

(Lemma 2.5 in [39]). Let X be a topological space. Assume there exist two continuous mapping

$$P:{\mathbb{S}}^2=\{y\in {\mathbb{R}}^3:\left|y\right|=1\}\to X,Q:X\to {\mathbb{S}}^2$$

such that $Q\circ P$ is homotopic to identity, that is, there is a continuous mapping $\sigma :[0,1]\times {\mathbb{S}}^2\to {\mathbb{S}}^2$ such that $\sigma (0,x)=(Q\circ P)\left(x\right)$ for $x\in {\mathbb{S}}^2$ and $\sigma (1,x)=x$ for $x\in {\mathbb{S}}^2$. Then $\mathrm{cat}\left(X\right)\ge 2$.

Lemma 16

(Proposition 2.4 in [39]). Let M be a Hilbert manifold and $I\in C^1(M,\mathbb{R})$. If there exist $c_0\in \mathbb{R}$ and $k\in \mathbb{N}$ such that $I\left(u\right)$ satisfies the Palais–Smale condition for $c\le c_0$ and $\mathrm{cat}\left(\{u\in M:I\left(u\right)\le c_0\}\right)\ge k$, then $I\left(u\right)$ admits at least k critical points in $\{u\in M:I\left(u\right)\le c_0\}$.

Proof of Theorem 2.

We note that $I\left(0\right)=0$. By the proof of Theorem 3, there exist ${\rho }_0$, ${\gamma }_0>0$ such that

$$\begin{array}{c}\hfill I\left(u\right)\ge {\gamma }_0,\forall \parallel u\parallel ={\rho }_0.\end{array}$$

By the argument of Lemma 10, ${lim}_{t\to +\infty }I\left(t{u}_{\delta ,\tilde{y}}\right)=-\infty$. Then $I\left(t{u}_{\delta ,\tilde{y}}\right)$ attained its maximum at a $t_y>0$. So $\frac{d}{dt}I\left(t{u}_{\delta ,\tilde{y}}\right){|}_{t=t_y}=0$. We note that

$$\begin{array}{c}\hfill \frac{d}{dt}I\left(tu\right)=t^3\left[\frac{1}{t^2}{\parallel u\parallel }^2+{\int }_{{\mathbb{R}}^3}{\varphi }_u{u}^2\mathrm{d}x-\frac{1}{t^3}{\int }_{{\mathbb{R}}^3}h\left(x\right)f\left(tu\right)u\mathrm{d}x-t^2{\int }_{{\mathbb{R}}^3}g\left(x\right){\left|u\right|}^6\mathrm{d}x\right].\end{array}$$

By $\left(f_1\right)$, we get $t_y$ is unique. Moreover, $t_y{u}_{\delta ,\tilde{y}}\in M$. By Lemma 10, for $\delta \in (0,{\delta }_0)$,

$$\begin{array}{c}\hfill \underset{t\ge 0}{sup}I\left(t{u}_{\delta ,\tilde{y}}\right)=I\left(t_y{u}_{\delta ,\tilde{y}}\right)\le \frac{1}{3}{S}^{\frac{3}{2}}-{\delta }^{\frac{1}{2}}.\end{array}$$

(51)

Define $P:{\mathbb{S}}^2\to M$ by $P\left(y\right)=t_y{u}_{\delta ,\tilde{y}}$. Then P is continuous and

$P\left({\mathbb{S}}^2\right)\subset \left\{u\in M:I\left(u\right)\le \frac{1}{3}{S}^{\frac{3}{2}}-{\delta }^{\frac{1}{2}}\right\}$. Define

$$\begin{array}{c}\hfill Q:\left\{u\in M:I\left(u\right)\le \frac{1}{3}{S}^{\frac{3}{2}}-{\delta }^{\frac{1}{2}}\right\}\to {\mathbb{S}}^2\end{array}$$

by $Q\left(u\right)=\frac{{\int }_{{\mathbb{R}}^3}\frac{x}{\left|x\right|}{|\nabla u|}^2\mathrm{d}x}{\left|{\int }_{{\mathbb{R}}^3}\frac{x}{\left|x\right|}{|\nabla u|}^2\mathrm{d}x\right|}$. By Lemma 14, we know Q is continuous. Define $\sigma (\theta ,y):[0,1]\times {\mathbb{S}}^2\to {\mathbb{S}}^2$ such that $\sigma (\theta ,y)=Q\left((1-2\theta )P\left(y\right)+2\theta u_{\delta ,\tilde{y}}\right)$ for $\theta \in [0,\frac{1}{2})$, $\sigma (\theta ,y)=Q\left(u_{2(1-\theta )\delta ,\tilde{y}}\right)$ for $\theta \in [\frac{1}{2},1)$ and $\sigma (\theta ,y)=y$ for $\theta =1$. By the argument of Case 4 in Lemma 13, we have

$$\begin{array}{c}\hfill \left|\frac{x}{\left|x\right|}-\frac{\frac{3}{2}{\rho }_{0}y}{|\frac{3}{2}{\rho }_{0}y|}\right|\le \frac{2|x-\frac{3}{2}{\rho }_{0}y|}{|\frac{3}{2}{\rho }_{0}y|}.\end{array}$$

By a direct calculation, for $|x-\frac{3}{2}{\rho }_{0}y|\le r$,

$$\begin{array}{c}\hfill |\nabla u_{2(1-\theta )\delta ,\tilde{y}}{|}^2=\frac{{\left(2(1-\theta )\delta \right)}^{\frac{1}{2}}{|x-\frac{3}{2}{\rho }_{0}y|}^2}{{\left(2(1-\theta )\delta +|x-\frac{3}{2}{\rho }_0{y|}^2\right)}^3}.\end{array}$$

Then

$$\begin{array}{cc}\hfill & {\int }_{|x-\frac{3}{2}{\rho }_{0}y|\le r}\left|\frac{x}{\left|x\right|}-\frac{\frac{3}{2}{\rho }_{0}y}{|\frac{3}{2}{\rho }_{0}y|}\right|{|\nabla u_{2(1-\theta )\delta ,\tilde{y}}|}^2\mathrm{d}x\hfill \\ \hfill & \le {\int }_{{\mathbb{R}}^3}\frac{4|x-\frac{3}{2}{\rho }_{0}y|}{3{\rho }_0}\frac{{\left(2(1-\theta )\delta \right)}^{\frac{1}{2}}{|x-\frac{3}{2}{\rho }_{0}y|}^2}{{\left(2(1-\theta )\delta +|x-\frac{3}{2}{\rho }_0{y|}^2\right)}^3}\mathrm{d}x\hfill \\ \hfill & =\frac{4{\left(2(1-\theta )\delta \right)}^{\frac{1}{2}}}{3{\rho }_0}{\int }_{{\mathbb{R}}^3}\frac{{\left|x\right|}^3}{{\left(1+{\left|x\right|}^2\right)}^3}\mathrm{d}x.\hfill \end{array}$$

By a direct calculation, for $r\le |x-\frac{3}{2}{\rho }_{0}y|\le 2r$,

$$\begin{array}{cc}\hfill |\nabla u_{2(1-\theta )\delta ,\tilde{y}}{|}^2& \le \frac{{2|\nabla \psi |}^2{\left(2(1-\theta )\delta \right)}^{\frac{1}{2}}}{2(1-\theta )\delta +|x-\frac{3}{2}{\rho }_0{y|}^2}+\frac{2{\psi }^2{\left(2(1-\theta )\delta \right)}^{\frac{1}{2}}{|x-\frac{3}{2}{\rho }_{0}y|}^2}{{\left(2(1-\theta )\delta +|x-\frac{3}{2}{\rho }_0{y|}^2\right)}^3}\hfill \\ \hfill & \le \frac{10{\left(2(1-\theta )\delta \right)}^{\frac{1}{2}}}{2(1-\theta )\delta +|x-\frac{3}{2}{\rho }_0{y|}^2}.\hfill \end{array}$$

Then

$$\begin{array}{cc}\hfill & {\int }_{r\le |x-\frac{3}{2}{\rho }_{0}y|\le 2r}\left|\frac{x}{\left|x\right|}-\frac{\frac{3}{2}{\rho }_{0}y}{|\frac{3}{2}{\rho }_{0}y|}\right|{|\nabla u_{2(1-\theta )\delta ,\tilde{y}}|}^2\mathrm{d}x\hfill \\ \hfill & \le {\int }_{r\le |x-\frac{3}{2}{\rho }_{0}y|\le 2r}\frac{40|x-\frac{3}{2}{\rho }_{0}y|}{3{\rho }_0}\frac{{\left(2(1-\theta )\delta \right)}^{\frac{1}{2}}}{2(1-\theta )\delta +|x-\frac{3}{2}{\rho }_0{y|}^2}\mathrm{d}x\hfill \\ \hfill & \le \frac{40{\left(2(1-\theta )\delta \right)}^{\frac{1}{2}}}{3{\rho }_{0}r}{\int }_{r\le \left|x\right|\le 2r}\mathrm{d}x.\hfill \end{array}$$

Thus, we have ${lim}_{\theta \to 1-}{\int }_{{\mathbb{R}}^3}\left(\frac{x}{\left|x\right|}-\frac{y}{\left|y\right|}\right){|\nabla u_{2(1-\theta )\delta ,\tilde{y}}|}^2\mathrm{d}x=0$. By Lemma 2,

$$\begin{array}{c}\hfill \underset{\theta \to 1-}{lim}{\int }_{{\mathbb{R}}^3}\frac{x}{\left|x\right|}|\nabla u_{2(1-\theta )\delta ,\tilde{y}}{|}^2\mathrm{d}x=\underset{\theta \to 1-}{lim}\frac{y}{\left|y\right|}{\int }_{{\mathbb{R}}^3}{|\nabla u_{2(1-\theta )\delta ,\tilde{y}}|}^2\mathrm{d}x=K_{1}y.\end{array}$$

Then by the continuity of Q, we obtain that $\sigma (\theta ,y)\in C([0,1]\times {\mathbb{S}}^2,{\mathbb{S}}^2)$, $\sigma (0,y)=Q\circ P\left(y\right)$ for $y\in {\mathbb{S}}^2$ and $\sigma (1,y)=y$ for $y\in {\mathbb{S}}^2$. By Lemma 15, we have

$$\begin{array}{c}\hfill \mathrm{cat}\left(\left\{u\in M:I\left(u\right)\le \frac{1}{3}{S}^{\frac{3}{2}}-{\delta }^{\frac{1}{2}}\right\}\right)\ge 2.\end{array}$$

By Lemmas 11 and 12, we know I satisfies the $\left(PS\right)$ condition for $c\in \left(0,\frac{1}{3}{S}^{\frac{3}{2}}\right)$. Then by Lemma 16, we obtain that I has two nonnegative critical points $u_{i,h}$, $i=1$, 2. By the maximum principle, $u_{i,h}$ is positive.  □

5. Conclusions

We first study multiplicity of solutions of the singularly perturbed Schrödinger–Poisson equation with critical growth. When the perturbed coefficient is small, we establish the relationship between the number of solutions and the profiles of the coefficients, which is different from the existing results. We pointed out that, when we seek multiplicity of solutions, it is crucial to prove the compactness of the Palais–Smale sequence. Many authors solved the problem by imposing the Rabinowitz type assumption, which is restrictive. In this paper, we remove the technical assumption. Furthermore, we study multiplicity of solutions without any restriction on the perturbed coefficient. By using the Lusternik–Schnirelman category and developing some techniques, we obtain a multiplicity result. Besides, we study the existence of solutions of non-autonomous Schrödinger–Poisson equations without the classical (AR) condition or the monotony condition. We introduce a new argument to solve the problem.