1 Introduction

This paper deals with the existence, multiplicity and concentration behavior of the positive solutions to the nonlinear problem of Kirchhoff type:

$$\begin{aligned} \left\{ \begin{array}{l@{\quad }l@{\quad }l} -\left(\varepsilon ^2a+\varepsilon b\int _{{\mathbb{R }^{3}}}|\nabla u|^2\right)\Delta u+V(x)u=u^{2^*-1}+\lambda f(u)&\text{ in}&{\mathbb{R }^{3}},\\ u\in H^1({\mathbb{R }^{3}}), ~u(x)>0&\text{ in}&{\mathbb{R }^{3}}, \end{array}\right. \quad \quad \quad \quad \quad \quad (\mathcal{S }\mathcal{K })_{\varepsilon } \end{aligned}$$

where \(\varepsilon \) and \(\lambda \) are positive parameters, and \(a, b>0\) are constants, \(2^*(=6)\) is the critical Sobolev exponent in \({\mathbb{R }^{3}}, f\) is a subcritical nonlinear term, and the potential \(V: {\mathbb{R }^{3}}\rightarrow \mathbb{R }\) is a continuous function satisfying

  • \((V)\) \(V_{\infty }:=\lim \inf _{|x|\rightarrow \infty }V(x)>V_0:=\inf _{x\in {\mathbb{R }^{3}}}V(x)>0\), where \(V_{\infty }\le \infty \).

This kind of hypothesis was first introduced by Rabinowitz in [40]. The nonlinearity \(f{:}\, \mathbb{R }\rightarrow \mathbb{R }\) is a function of \(C^1\) class. Since we are looking for positive solutions, we may assume that \(f(s)=0\) for \(s<0\). Furthermore, we make the following conditions:

  • \((A_1)\)   \(f(s)=o(s^3)\) as \(s\rightarrow 0\);

  • \((A_2)\)   there exists \(3<q<2^*-1=5\) such that \(\lim \limits _{|s|\rightarrow \infty }\frac{f(s)}{s^q}=0;\)

  • \((A_3)\)   there exists some \(\mu >4\) such that

    $$\begin{aligned} 0<\mu F(s)=\mu \int \limits ^s_0f(\tau )d\tau \le f(s)s \quad \text{ for} \text{ all} \ s>0; \end{aligned}$$
  • \((A_4)\)   the function \(f(s)s^{-3}\) is increasing for \(s>0\).

We note that problem \((\mathcal{SK })_{\varepsilon }\) with \(a=1,b=0\) and \({\mathbb{R }^{3}}\) replaced by \(\mathbb{R }^N\) reduces to the well-known Schrödinger equation

$$\begin{aligned} -\varepsilon ^2\Delta u+V(x)u=f(u)\quad \text{ in}~~\mathbb{R }^N. \end{aligned}$$
(1.1)

Equation (1.1) arises in different models; for example, they are involved with the existence of standing waves of the nonlinear Schrödinger equations

$$\begin{aligned} i\varepsilon \partial _tz=-\varepsilon ^2\Delta z+(V(x)+E)z-f(z)\quad \text{ for} \text{ all}\quad x\in \mathbb{R }^N, \end{aligned}$$
(1.2)

when \(f(s)=|s|^{p-2}s, 2<p<2^*:=2N/(N-2)\). A standing wave of (1.2) is a solution of the form \(z(x,t)=\exp (-iEt/\varepsilon )u(x), \) where \(u\) is a solution of (1.1).

The existence and concentration behavior of the positive solutions of (1.1) have been extensively studied in recent years, see for example, Ambrosetti et al. [4], Ambrosetti and Machiodi [6], Ambrosetti et al. [5], Bartsch and Wang [9, 10], Byeon and Jeanjean [16], Byeon and Wang [17], Chabrowski and Szulkin [19], Cingolani and Lazzo [20, 21], Del Pino and Felmer [22, 23], Ding and Lin [24], Ding and Wei [25], Floer and Weinstein [27], Oh [36, 37], Rabinowitz [40], Wang [43] and their references therein.

In \((\mathcal{SK })_{\varepsilon }\), if we set \(\varepsilon =1,V(x)=0\) and replace \({\mathbb{R }^{3}}\) and \( f(u)\) by a bounded domain \(\Omega \subset \mathbb{R }^N\) and \(f(x,u)\), respectively, it reduces to the following Dirichlet problem of Kirchhoff type:

$$\begin{aligned} \left\{ \begin{array}{l@{\quad }l@{\quad }l} -\left(a+b\int _{\Omega }|\nabla u|^2\right)\Delta u=f(x,u)&\text{ in}&\Omega , \\ u=0&\text{ on}&\partial \Omega . \end{array}\right. \end{aligned}$$
(1.3)

Problem (1.3) is related to the stationary analogue of the equation

$$\begin{aligned} u_{tt}-\left(a+b\int \limits _{\Omega }|\nabla u|^2\right)\Delta u=f(x,u). \end{aligned}$$
(1.4)

Such a hyperbolic equation is a general version of the Kirchhoff equation

$$\begin{aligned} \rho \frac{\partial ^2u}{\partial t^2}-\left(\frac{P_0}{h}+\frac{E}{2L}\int \limits ^L_0\left|\frac{\partial u}{\partial x}\right|^2\mathrm{d}x\right)\frac{\partial ^2 u}{\partial x^2}=0 \end{aligned}$$
(1.5)

proposed by Kirchhoff in [30] as an existence of the classical D’Alembert’s wave equations for free vibration of elastic strings. Kirchhoff’s model takes into account the changes in length of the string produced by transverse vibrations. The parameters in Eq. (1.5) have the following meaning: \(L\) is the length of the string, \(h\) is the area of cross-section, \(E\) denotes the Young modulus of the material, \(\rho \) is the mass density, and \(P_0\) denotes the initial tension.

Some early classical studies of Kirchhoff equations can be seen in Bernstein [13] and Pohožaev [39]. Equation (1.4) received much attention only after Lions [33] introduced an abstract framework to the problem. Some interesting results can be found, for example, in [7, 8, 18]. In [8], Arosio and Panizzi studied the Cauchy-Dirichlet type problem related to (1.4) in the Hadamard sense as a special case of an abstract second-order Cauchy problem in a Hilbert space. In [18], Cavalcanti et al. considered the question of the existence and uniqueness of regular global solutions for the Kirchhoff-Carrier equation subject to nonlinear boundary dissipation without restriction on the initial data and obtained uniform decay rates by assuming a nonlinear feedback acting on the boundary. In [7], D’Ancona and Spagnolo proved the existence of a global classical periodic solution for the degenerate Kirchhoff equation with real analytic data. In particular, in the paper [7], Kirchhoff’s equation is an example of a quasi-linear hyperbolic Cauchy problem that describes the transverse oscillations of a stretched string. We also note that several existence results have been obtained for (1.3) (on bounded domain) in recent years. For example, Ma and Rivera [34] obtained positive solutions of such problems by using variational methods. Perera and Zhang [38] obtained a nontrivial solution of (1.3) via Yang index and critical group. Recently, Alves et al. [3] considered a critical Kirchhoff-type problem of type (1.3) on bounded domain and obtained a mountain pass solution. He and Zou [28, 29] obtained infinitely many solutions by using the local minimum methods and the fountain theorems.

As far as we know, little has been done for the existence and concentration behavior of the positive solutions for Kirchhoff problems \((\mathcal{SK })_{\varepsilon }\) on \({\mathbb{R }^{3}}\) where the nonlinearity has a critical growth. As we shall see in the present paper, equation \((\mathcal{SK })_{\varepsilon }\) can be viewed as a Schrödinger equation coupled with a nonlocal term. The competing effect of the nonlocal term with the nonlinearity \(f(u)\) and the lack of compactness of the embedding of \(H^1({\mathbb{R }^{3}})\) into the space \(L^p({\mathbb{R }^{3}}), p\in (2,6),\) prevents us from using the variational methods in a standard way. Some new estimates for such a Kirchhoff equation involving Palais–Smale sequences, which are key points to apply this kinds of theory, are needed to be reestablished. The Moser iterative method (cf. [35]) has to be applied trickly. No information on the ground state solution for the critical Kirchhoff equations can be found in the existing references. To describe our main result, we first set

$$\begin{aligned} \Sigma : =\{x\in {\mathbb{R }^{3}}:~V(x)=V_0\}. \end{aligned}$$

In view of \((V)\), the set \(\Sigma \) is compact. For any \(\delta >0\), we denote by \(\Sigma _{\delta }:=\{x\in {\mathbb{R }^{3}}:~\text{ dist}(x,\Sigma )\le \delta \}\). We recall that, if \(Y\) is a closed subset of a topological space \(X\), the Ljusternik–Schnirelmann category \(cat_{X}(Y)\) is the least number of closed and contractible sets in \(X\) which cover \(Y\). The next theorem is the main result of the present paper.

Theorem 1.1

Assume that conditions \((V)\) and \((A_1)\text{--}(A_4)\) are satisfied. Then, for any \(\delta >0\), there exists \(\varepsilon _{\delta }>0\) and \(\lambda ^*>0\) such that, for any \(\varepsilon \in (0,\varepsilon _{\delta })\) and \(\lambda >\lambda ^*, (\mathcal{SK })_{\varepsilon }\) has at least \(cat_{\Sigma _{\delta }}(\Sigma )\) positive solutions. Moreover, if \(\varepsilon _n\rightarrow 0^+,~ u_{\varepsilon _n}\) is one of these solutions and \(\eta _{\varepsilon _n}\in {\mathbb{R }^{3}}\) is a point of global maximum of \(u_{\varepsilon _n}\), we have that

$$\begin{aligned} \lim _{\varepsilon _n\rightarrow 0^+}V(\eta _{\varepsilon _n})=V_0. \end{aligned}$$

The paper is organized as follows. In Section 2, we present the abstract framework of the problem as well as some preliminary results and some compactness properties of the functional associated with \((\mathcal SK )_{\varepsilon }\). In Section 3, we show that \((\mathcal SK )_{\varepsilon }\) has a ground state solution. Section 4 is devoted to the proof of Theorem 1.1.

Hereafter, we employ the following notations:

  • \(H^1({\mathbb{R }^{3}})\) is the usual Sobolev space endowed with the standard scalar product and norm

    $$\begin{aligned} (u,v)=\int \limits _{\mathbb{R }^{3}}(\nabla u\nabla v+uv)\mathrm{d}x;~~~~||u||^2=\int \limits _{\mathbb{R }^{3}}(|\nabla u|^2+u^2)\mathrm{d}x. \end{aligned}$$
  • \(\mathcal D ^{1,2}({\mathbb{R }^{3}})\) is the completion of \(C^{\infty }_0({\mathbb{R }^{3}})\) with respect to the norm

    $$\begin{aligned} ||u||^2_\mathcal{D ^{1,2}}=\int \limits _{\mathbb{R }^{3}}|\nabla u|^2\mathrm{d}x. \end{aligned}$$
  • \(H^{-1}\) denotes the dual space of \(H^1({\mathbb{R }^{3}})\).

  • \(L^q(\Omega ), 1\le q\le +\infty , \Omega \subset {\mathbb{R }^{3}},\) denotes a Lebesgue space, the norm in \(L^q(\Omega )\) is denoted by \(||u||_{q,\Omega }\), where \(\Omega \) is a proper subset of \({\mathbb{R }^{3}}\), by \(||\cdot ||_q\) when \(\Omega ={\mathbb{R }^{3}}\).

  • \(S\) is the best Sobolev constant for the Sobolev embedding \(\mathcal D ^{1,2}( {\mathbb{R }^{3}})\hookrightarrow L^{2^*}({\mathbb{R }^{3}}),\) that is,

    $$\begin{aligned} S=\inf \left\{ \int \limits _{\mathbb{R }^{3}}|\nabla u|^2\mathrm{d}x:~\Vert u\Vert _{2^*}=1\right\} . \end{aligned}$$
  • For any \(\rho >0\) and for any \(z\in {\mathbb{R }^{3}}, B_{\rho }(z)\) denotes the ball of radius \(\rho \) centered at \(z\) and \(|B_{\rho }(z)|\) denotes its Lebesgue measure.

  • We omit the symbol \(\mathrm{d}x\) in the integrals over \({\mathbb{R }^{3}}\) when no confusion arise. \( C, C_i\) denote various positive constants.

2 The variational framework and preliminary results

2.1 The mountain pass geometry and notations

Throughout this section, we suppose that the functions \(V(x)\) and \(f(u)\) satisfy conditions \((V), (A_1)\text{--}(A_4)\), respectively. Making the change of variable \(\varepsilon z=x\), we can rewrite \((\mathcal SK )_{\varepsilon }\) as the following equivalent equation

$$\begin{aligned} -\left(a+b\int \limits _{\mathbb{R }^{3}}|\nabla u|^2\right)\Delta u+V(\varepsilon x)u=u^{2^*-1}+\lambda f(u)\quad \text{ in}~~{\mathbb{R }^{3}}. \quad \quad \quad \quad \quad \quad \quad \quad \quad \,\, (\mathcal{K })_{\varepsilon } \end{aligned}$$

For any \(\varepsilon >0\), let \(W_{\varepsilon }:=\{u\in H^1({\mathbb{R }^{3}}):~\int _{{\mathbb{R }^{3}}}V(\varepsilon x)u^2<\infty \}\) be the Sobolev space endowed with the norm \( ||u||_{\varepsilon }:=\left(\int _{{\mathbb{R }^{3}}}(|\nabla u|^2+V(\varepsilon x)u^2)\right)^{1/2}\). At this step, we see that \((\mathcal K )_{\varepsilon }\) is variational and its solutions are the critical points of the functional given by

$$\begin{aligned} \mathcal I _{\varepsilon }(u):=\frac{1}{2}\int \limits _{\mathbb{R }^{3}}\left(a|\nabla u|^2+V(\varepsilon x)u^2\right) +\frac{b}{4}\left(\int \limits _{\mathbb{R }^{3}} |\nabla u|^2\right)^2 -\frac{1}{2^*}\int \limits _{\mathbb{R }^{3}}|u|^{2^*}-\lambda \int \limits _{\mathbb{R }^{3}}F(u).\nonumber \\ \end{aligned}$$
(2.1)

Moreover, \(\mathcal I _{\varepsilon }\) belongs to \(C^1(W_{\varepsilon },\mathbb{R })\). Next, we define the Nehari manifold (cf. [44]) associated with \(\mathcal I _{\varepsilon }\) as

$$\begin{aligned} \mathcal{N }_{\varepsilon }:=\{u\in W_{\varepsilon }\backslash \{0\}:~G(u)=0\}, \end{aligned}$$

where \(G:=\langle \mathcal I _{\varepsilon }^{\prime }(u),u\rangle =\int _{{\mathbb{R }^{3}}}(a|\nabla u|^2+V(\varepsilon x)u^2)+b\left(\int _{{\mathbb{R }^{3}}}|\nabla u|^2\right)^2-\int _{{\mathbb{R }^{3}}}|u|^{2^*}-\lambda \int _{{\mathbb{R }^{3}}}f(u)u\). The functional \(\mathcal I _{\varepsilon }\) satisfies the mountain pass geometry.

Lemma 2.1

For any \(\varepsilon >0, \lambda >0\) fixed, the functional \(\mathcal I _{\varepsilon }\) satisfies the following conditions.

  1. (i)

    There exist \(\alpha ,\rho >0\) such that \(\mathcal I _{\varepsilon }(u)\ge \alpha \) for \(||u||_{\varepsilon }=\rho \).

  2. (ii)

    There exists an \(e\in W_\varepsilon \) with \(\Vert e\Vert _{\varepsilon }>\rho \) such that \(\mathcal I _{\varepsilon }(e)<0\).

Proof

  1. (i)

    For any \(u\in W_{\varepsilon }\backslash \{0\}\) and \(\tau >0\) small, it follows from \((A_1)\text{-}(A_2)\) that there exists a \(C_{\tau }>0\) such that

    $$\begin{aligned} |f(t)|\le \tau |t|+C_{\tau }|t|^{q}, \quad |F(t)|\le \frac{\tau }{2}t^2+\frac{C_{\tau }}{q+1}|t|^{q+1},~~\forall \; t\in \mathbb{R }. \end{aligned}$$
    (2.2)

Note that the Sobolev embedding \(W_{\varepsilon }\hookrightarrow L^p({\mathbb{R }^{3}})\) for \(2\le p\le 6\) is continuous, we have

$$\begin{aligned} \mathcal I _{\varepsilon }(u) \ge \frac{\min \{1,a\}}{4}{||u||_{\varepsilon }^2}-\lambda CC_{\tau }||u||^{q+1}_{\varepsilon }-\lambda C\Vert u\Vert _{\varepsilon }^{2^*}. \end{aligned}$$

Thus, we can choose some \(\alpha >0, \rho >0\) such that \(\mathcal I _{\lambda ,\varepsilon }(u)\ge \alpha ~~\text{ for} ~||u||_{\varepsilon }=\rho \).

  1. (ii)

    By \((A_3)\), we see that \(F(t)\ge C|t|^{\mu }-C\) for all \(t\in \mathbb{R }\). Take a bounded domain \(\Omega \subset {\mathbb{R }^{3}}\) and some \(u_*\in W_{\varepsilon }\backslash \{0\}\) such that supp\(u_*\subset \Omega \). Then, we have

    $$\begin{aligned} \mathcal I _{\varepsilon }(tu_*)&\le \frac{t^2}{2}\int \limits _{\Omega }(a|\nabla u_*|^2+ V(\varepsilon x)u_*^2)+\frac{t^4b}{4 }\left(\int \limits _{\Omega }|\nabla u_*|^2\right)^2\\&-Ct^{\mu }\int \limits _{\Omega }|u_*|^{\mu }-C|\Omega | < 0 \end{aligned}$$

for \(t>0\) large enough. Hence, we can take an \(e:=t_*u_*\) for some \(t_*>0\) and (ii) follows.

\(\square \)

Hereafter, we recall that a sequence \(\{u_n\}\subset W_\varepsilon \) is called a Palais–Smale sequence (\((PS)\) sequence, for short) for \(\mathcal I _{\varepsilon }\) if \(\{ \mathcal{I }_{\varepsilon }(u_n)\}\) is bounded and \(\mathcal I _{\varepsilon }^{\prime }(u_n)\rightarrow 0\). If \(\mathcal I _{\varepsilon }(u_n)\rightarrow c\in \mathbb{R }\) and \(\mathcal I _{\varepsilon }^{\prime }(u_n)\rightarrow 0,\) then \(\{u_n\}\) is a \((PS)_c\) sequence. The functional \(\mathcal I _{\varepsilon }\) is said to satisfy the Palais–Smale condition (\((PS)\) condition, for short) (or \((PS)_c\) condition) if each Palais–Smale sequence (or \((PS)_c\) sequence) has a convergent subsequence.

By Lemma 2.1 and the mountain pass theorem without (PS) condition (cf. [44]), there exists a (PS) sequence \(\{u_n\}\subset W_{\varepsilon }\) such that \(\mathcal I _{\varepsilon }(u_n)\rightarrow c_{\varepsilon }\) and \(\mathcal I ^{\prime }_{\varepsilon }(u_n)\rightarrow 0\) in \(W^{-1}_{\varepsilon }\) at the minimax level

$$\begin{aligned} c_{\varepsilon }:=\inf _{g\in \Gamma }\sup _{t\in [0,1]}\mathcal I _{\varepsilon }(g(t))>0, \end{aligned}$$

where \(\Gamma :=\{g\in C^1([0,1],W_{\varepsilon }):~g(0)=0, \mathcal I _{\varepsilon }(g(1))<0\}\). Furthermore, we have the following conclusion.

Lemma 2.2

For every \(u\in W_{\varepsilon }\backslash \{0\}\), there exists a unique \(t(u)>0\) such that \(t(u)u\in \mathcal N _{\varepsilon }\). Moreover, \(\mathcal I _{\varepsilon }(t(u)u)=\max \limits _{t\ge 0}\mathcal I _{\varepsilon }(tu)\).

Proof

Let \(u\in W_{\varepsilon }\backslash \{0\}\) be fixed and define the function \(\pi (t)\doteq \mathcal I _{\varepsilon }(tu)\) for \(t\ge 0\). We observe that \(\pi ^{\prime }(t)=\langle \mathcal I ^{\prime }_{\varepsilon }(tu),u\rangle =0\) if and only if \(tu\in \mathcal N _{\varepsilon }\), since \(\pi ^{\prime }(t)=0\) is equivalent to

$$\begin{aligned} b\left(\int \limits _{\mathbb{R }^{3}}|\nabla u|^2\right)^2=-\frac{\int _{{\mathbb{R }^{3}}}(a|\nabla u|^2+V(\varepsilon x)u^2)}{t^2}+\lambda \int \limits _{\mathbb{R }^{3}} \frac{f(tu)u^4}{t^3u^3}+t^2\int \limits _{\mathbb{R }^{3}}|u|^{2^*}. \end{aligned}$$
(2.3)

The right side of (2.3) is an increasing function of \(t\) recalling condition \((A_4)\).

On the other hand, using a similar argument as in the proof of Lemma 2.1, it is easy to see that \(\pi (0)=0, \pi (t)>0\) for \(t>0\) small and \(\pi (t)<0\) for \(t\) large. Hence, there exists a unique \(t(u)>0\) such that \(\pi ^{\prime }(t(u))=0\), that is, \(t(u)u\in \mathcal N _{\varepsilon }\). Moreover, \(\mathcal I _{\varepsilon }(t(u)u)=\max \limits _{t\ge 0} \mathcal I _{\varepsilon }(tu)\).

\(\square \)

Next, we define the numbers

$$\begin{aligned} c^*_{\varepsilon }:=\inf _{u\in \mathcal{N }_{\varepsilon }}\mathcal I _{\varepsilon }(u),~~~c^{**}_{\varepsilon }:=\inf _{u\in W_{\varepsilon }\backslash \{0\}}\max \limits _{t\ge 0}\mathcal I _{\varepsilon }(tu). \end{aligned}$$

Lemma 2.3

\(c_{\varepsilon }=c_{\varepsilon }^*=c_{\varepsilon }^{**},\) for any fixed \(\varepsilon ,\lambda >0\)

Proof

This kind of results for Schrödinger equations has been proved in [40]. We sketch the proof as follows. It follows from Lemma 2.2 that \(c_{\varepsilon }^*=c^{**}_{\varepsilon }\). Notice that for any \(u\in W_{\varepsilon }\backslash \{0\}\), there exists some \(t_0>0\) large, such that \(\mathcal I _{\varepsilon }(t_0u)<0\). Define a path \(\gamma :~[0,1]\rightarrow W_{\varepsilon }\) by \(\gamma (t)=tt_0u\). Clearly, \(\gamma \in \Gamma \) and consequently, \(c_{\varepsilon }\le c_{\varepsilon }^{**}\). We next prove that \(c_{\varepsilon }^*\le c_{\varepsilon }\). The manifold \(\mathcal N _{\varepsilon }\) separates \(W_{\varepsilon }\) into two components. By \((A_1), (A_2)\) the component contains the origin also contains a small ball around the origin. Moreover, \(\mathcal I _{\varepsilon }(u)\ge 0\) for all \(u\) in this component, because for all \(0\le t\le t(u),\)

$$\begin{aligned} \pi ^{\prime }(t)&= \langle \mathcal I ^{\prime }_{\varepsilon }(tu),u\rangle \\&= t^3\left[b\left(\int \limits _{\mathbb{R }^{3}}|\nabla u|^2\right)^2\right.\\&\left. -\left( -\frac{\int _{{\mathbb{R }^{3}}}(a|\nabla u|^2+V(\varepsilon x)u^2)}{t^2}+\lambda \int \limits _{\mathbb{R }^{3}}\frac{f(tu)u^4}{t^3u^3} +t^2\int \limits _{\mathbb{R }^{3}}|u|^{2^*}\right)\right]\\&\ge \frac{t^3}{t(u)^4}\langle \mathcal I ^{\prime }_{\varepsilon }(t(u)u),t(u)u\rangle \\&= 0. \end{aligned}$$

Thus, every path \(\gamma \in \Gamma \) has to cross \(\mathcal N _{\varepsilon }\) and \(c^*_{\varepsilon }\le c_{\varepsilon }\). This completes the proof. \(\square \)

Lemma 2.4

Any \((PS)_c\) sequence of \(\mathcal I _{\varepsilon }\) is bounded in \(W_{\varepsilon }\), independent of \(\varepsilon \) and \(\lambda \).

Proof

Let \(\{u_n\}\subset W_{\varepsilon }\) be a \((PS)_c\) sequence for \(\mathcal I _{\varepsilon }\), that is,

$$\begin{aligned} o(1)+c&= \frac{1}{2}\int \limits _{\mathbb{R }^{3}}(a|\nabla u_n|^2+V(\varepsilon x)u^2_n)+\frac{b}{4} \left(\int \limits _{\mathbb{R }^{3}}|\nabla u_n|^2\right)^2-\frac{1}{2^*}\int \limits _{\mathbb{R }^{3}}|u_n|^{2^*}-\lambda \int \limits _{\mathbb{R }^{3}}F(u_n),\\ o(1)&= \int \limits _{\mathbb{R }^{3}}(a|\nabla u_n|^2+V(\varepsilon x)u^2_n)+b\left(\int \limits _{\mathbb{R }^{3}}|\nabla u_n|^2\right)^2 -\int \limits _{\mathbb{R }^{3}}|u_n|^{2^*}-\lambda \int \limits _{\mathbb{R }^{3}}f(u_n)u_n. \end{aligned}$$

The above two equalities together with \((A_3)\) imply that

$$\begin{aligned} o(1)+c&= \frac{1}{4}\int \limits _{\mathbb{R }^{3}}(a|\nabla u_n|^2+V(\varepsilon x)u^2_n) +\frac{\lambda }{4}\int \limits _{\mathbb{R }^{3}}(f(u_n)u_n-4F(u_n))+\frac{1}{12}\int \limits _{\mathbb{R }^{3}}|u_n|^{2^*}\\&\ge \frac{1}{4}\int \limits _{\mathbb{R }^{3}}(a|\nabla u_n|^2+V(\varepsilon x)u^2_n). \end{aligned}$$

Therefore, \(\{u_n\}\) is bounded in \(W_{\varepsilon }\) and the conclusion follows. \(\square \)

Remark 2.5

Note that for each fixed \(\lambda >0\) and any \(u\in \mathcal N _{\varepsilon }\), by (2.2) we have for \(\tau >0\) small,

$$\begin{aligned} 0&= \int \limits _{\mathbb{R }^{3}}(a|\nabla u|^2+V(\varepsilon x)u^2)+b\left(\int \limits _{\mathbb{R }^{3}}|\nabla u|^2\right)^2-\int \limits _{\mathbb{R }^{3}}|u|^{2^*}-\lambda \int \limits _{\mathbb{R }^{3}}f(u)u\\&\ge \int \limits _{\mathbb{R }^{3}}(a|\nabla u|^2+V(\varepsilon x)u^2)-\lambda \tau \int \limits _{\mathbb{R }^{3}}u^2-\lambda C_{\tau }\int \limits _{\mathbb{R }^{3}}|u|^{q+1}-\int \limits _{\mathbb{R }^{3}}|u|^{2^*}\\&\ge \frac{\min \{a,1\}}{2}||u||_{\varepsilon }^2-\lambda CC_{\tau }||u||_{\varepsilon }^{q+1}-C\Vert u\Vert _{\varepsilon }^{2^*}, \end{aligned}$$

which implies that

$$\begin{aligned} ||u||_{\varepsilon } \,\ge \, r^{*} \,>\, 0 \end{aligned}$$
(2.4)

for some \(r^*>0\).

As we shall see, it is important to compare \(c_{\varepsilon }\) with the minimax level of the autonomous problem

$$\begin{aligned} \left\{ \begin{array}{l@{\quad }l@{\quad }l} -(a+b\int _{{\mathbb{R }^{3}}}|\nabla u|^2)\Delta u+\nu u=u^{2^*-1}+\lambda f(u)&\text{ in}&{\mathbb{R }^{3}},\\ u\in H^1({\mathbb{R }^{3}}),~u(x)>0,&\text{ in}&{\mathbb{R }^{3}}, \end{array}\right. \qquad \qquad \qquad \qquad \qquad \,\,\, (\mathcal P )_{\nu } \end{aligned}$$

where \(\nu \in \mathbb{R }^+\). The solutions of \((\mathcal P )_{\nu }\) are precisely the critical points of the functional defined by

$$\begin{aligned} \mathcal E _{\nu }(u):=\frac{1}{2}\int \limits _{\mathbb{R }^{3}}(a|\nabla u|^2+\nu u^2)+\frac{b}{4 }\left(\int \limits _{\mathbb{R }^{3}}|\nabla u|^2\right)^2 -\frac{1}{2^*}\int \limits _{\mathbb{R }^{3}}|u|^{2^*}-\lambda \int \limits _{\mathbb{R }^{3}}F(u). \end{aligned}$$

Define the Nehari manifold associated with \(\mathcal E _{\nu }\) as

$$\begin{aligned} \mathcal{N }_{\nu }:=\left\{ u\in W_{\nu }\backslash \{0\}: ~~\langle \mathcal E _{\nu }^{\prime }(u),u\rangle =0\right\} , \end{aligned}$$

where \(W_{\nu }:=H^1({\mathbb{R }^{3}})\) is endowed with the norm \(||u||_{\nu }^2=\int _{{\mathbb{R }^{3}}}(|\nabla u|^2+\nu u^2)\). We define \(m_{\nu }\) by setting

$$\begin{aligned} m_{\nu }:=\inf _{u\in \mathcal{N }_{\nu }}\mathcal E _{\nu }(u). \end{aligned}$$

The number \(m_{\nu }\) and the manifold \(\mathcal{N }_{\nu }\) have properties similar to those of \(c_{\varepsilon }\) and \(\mathcal{N }_{\varepsilon }\).

2.2 Estimates for the minimax level

Arguing as in [3, 15], we are able to compare the minimax level \(m_{\nu }\) with a suitable number which involves the best constant \(S\).

Lemma 2.6

For any given \(\nu >0,\) there exist some \(u_0\in W_{\nu }\backslash \{0\}\) and constant \(\lambda ^*>0\) such that

$$\begin{aligned} \max _{t\ge 0}\mathcal E _{\nu }(tu_0)<\frac{1}{12}(aS)^{\frac{3}{2}}, \end{aligned}$$

for all \(\lambda >\lambda ^*\). In particular, \(m_{\nu }<\frac{1}{12}(aS)^{\frac{3}{2}}\).

Proof

Take \(u_0=u_*\), where \(u_*\) is given in Lemma 2.1. Similar to Lemma 2.2, there exists \(t_{\lambda }>0\) such that \(\mathcal E _{\nu }(t_{\lambda }u_0)=\max \limits _{t\ge 0}\mathcal E _{\nu }(tu_0)\). Moreover,

$$\begin{aligned} t^2_{\lambda }\int \limits _{\mathbb{R }^{3}}(a|u_0|^2+\nu u_0^2)+bt^4_{\lambda }\left(\int \limits _{\mathbb{R }^{3}}|\nabla u_0|^2\right)^2&= \lambda \int \limits _{\mathbb{R }^{3}}f(t_{\lambda }u_0)t_{\lambda }u_0+t^{2^*}_{\lambda }\int \limits _{\mathbb{R }^{3}}|u_0|^{2^*}\nonumber \\&\ge t^{2^*}_{\lambda }\int \limits _{\mathbb{R }^{3}}|u_0|^{2^*}, \end{aligned}$$
(2.5)

which implies that \(\{t_{\lambda }\}\) is bounded. Thus, there exists a sequence \(\lambda _{n}\rightarrow \infty \) verifying that \(t_{\lambda _n}\rightarrow t_0\ge 0\) as \(n\rightarrow \infty \). Consequently, there exists some \(C>0\) such that

$$\begin{aligned} t_{\lambda _n}^2\int \limits _{\mathbb{R }^{3}}(a|u_0|^2+\nu u_0^2)+bt_{\lambda _n}^4\left(\int \limits _{\mathbb{R }^{3}}|\nabla u_0|^2\right)^2\le C. \end{aligned}$$

If we assume that \(t_0>0\), then by the first equality of (2.5), we have

$$\begin{aligned} \lim _{n\rightarrow \infty }t_{\lambda _n}^2\int \limits _{\mathbb{R }^{3}}(a|u_0|^2+\nu u_0^2)+bt_{\lambda _n}^4\left(\int \limits _{\mathbb{R }^{3}}|\nabla u_0|^2\right)^2=\infty , \end{aligned}$$

which leads to a contradiction. Thus, we must have \(t_0=0\), and there exists some constant \(\lambda ^*>0\) such that for \(\lambda >\lambda ^*,\) one has the following estimate

$$\begin{aligned} 0<c_{\nu }&\le \max _{t\ge 0}\mathcal E _{\nu }(tu_0)=\mathcal E _{\nu }(t_{\lambda }u_0)\\&\le \frac{t^2_{\lambda }}{2}\int \limits _{\mathbb{R }^{3}}(a|\nabla u_0|^2+\nu u_0^2)+\frac{t^4_{\lambda }b}{4}\left(\int \limits _{\mathbb{R }^{3}}|\nabla u_0|^2\right)^2\\&< \frac{1}{12}(aS)^{\frac{3}{2}}, \end{aligned}$$

where \(c_{\nu }\) is the mountain pass level of the functional \(\mathcal E _{\nu }\). This completes the proof. \(\square \)

Remark 2.7

Note that from lemma above, in case \(V_{\infty }<\infty ,\) we have \(m_{V_{\infty }}<\frac{1}{12}(aS)^{\frac{3}{2}}\) for all \(\lambda >\lambda ^*\).

Lemma 2.8

Let \(\{u_n\}\subset W_{\varepsilon }\) be a \((PS)_c\) sequence for \(\mathcal I _{\varepsilon }\) with \(c<\frac{1}{12}(aS)^{\frac{3}{2}}\) and such that \(u_n\rightharpoonup 0\) in \(W_{\varepsilon }\). Then, one of the following conclusions holds.

  1. (i)

    \(u_n\rightarrow 0\) in \(W_{\varepsilon }\);

  2. (ii)

    there exists a sequence \(\{y_n\}\subset {\mathbb{R }^{3}}\) and constants \(R,\beta >0\) such that

    $$\begin{aligned} \liminf _{n\rightarrow \infty }\int \limits _{B_R(y_n)}u_n^2\ge \beta >0. \end{aligned}$$

Moreover, we can assume that \(u_n\ge 0\) for \(n\in \mathbb{N }\).

Proof

Suppose that (ii) does not occur. By Lemma 1.1 of [32], we have \( u_n\rightarrow 0~~\text{ in}~L^t({\mathbb{R }^{3}})~~~\text{ for}~t\in (2,6)\). From (2.2), we get \( 0\le \int _{{\mathbb{R }^{3}}}f(u_n)u_n\le \tau \int _{{\mathbb{R }^{3}}}u_n^2+C_{\tau }\int _{{\mathbb{R }^{3}}}|u_n|^{q+1}, \) for \(\tau >0\) small. Since \(\tau \) can be small arbitrarily, one has \(\lambda \int _{{\mathbb{R }^{3}}}f(u_n)u_n\rightarrow 0\). Recalling that \(\langle \mathcal{I }^{\prime }_{\varepsilon }(u_n),u_n\rangle \rightarrow 0,\) we have

$$\begin{aligned} \int \limits _{\mathbb{R }^{3}}(a|\nabla u_n|^2+V(\varepsilon x) u^2)+b\left(\int \limits _{\mathbb{R }^{3}}|\nabla u_n|^2\right)^2 =\int \limits _{\mathbb{R }^{3}}|u_n|^{2^*}+o(1). \end{aligned}$$

Since \(\{u_n\}\subset W_{\varepsilon }\) is bounded, up to a subsequence, we obtain

$$\begin{aligned} \int \limits _{\mathbb{R }^{3}}(a|\nabla u_n|^2+V(\varepsilon x) u^2)+b\left(\int \limits _{\mathbb{R }^{3}}|\nabla u_n|^2\right)^2\rightarrow l\ge 0~~\text{ and}~~||u_n||^{2^*}_{2^*}\rightarrow l\ge 0. \end{aligned}$$

Assume by contradiction that \(l>0\). From

$$\begin{aligned} c+o(1)&= \mathcal I _{\varepsilon }(u_n)\\&\ge \frac{1}{4}\int \limits _{\mathbb{R }^{3}}(a|\nabla u_n|^2+V(\varepsilon x) u_n^2)+\frac{b}{4}\left(\int \limits _{\mathbb{R }^{3}}|\nabla u_n|^2\right)^2-\frac{1}{2^*}\int \limits _{\mathbb{R }^{3}}|u_n|^{2^*}\\&\rightarrow \frac{l}{4}-\frac{l}{2^*}=\frac{l}{12}, \end{aligned}$$

it follows that \(l\le 12 c\).

On the other hand, we see that

$$\begin{aligned} \int \limits _{\mathbb{R }^{3}}(a|\nabla u_n|^2+V(\varepsilon x) u^2)+b\left(\int \limits _{\mathbb{R }^{3}}|\nabla u_n|^2\right)^2&\ge a \int \limits _{\mathbb{R }^{3}}|\nabla u_n|^2\\&\ge aS||u_n||^2_{2^*}=aS(||u_n||^{2^*}_{2^*})^{\frac{2}{2^*}}. \end{aligned}$$

Taking the limit as \(n\rightarrow \infty \) at the last inequality, we obtain

$$\begin{aligned} l\ge (aS)^{\frac{3}{2}}~~\text{ or}~~c\ge \frac{1}{12}(aS)^{\frac{3}{2}}, \end{aligned}$$

which contradicts to our assumptions. Therefore, \(l=0\) and the conclusion follows.

From definition of functional \(\mathcal I _{\varepsilon }\), we have

$$\begin{aligned} \langle \mathcal I ^{\prime }_{\varepsilon }(u_n),u^{-}_n\rangle =\int \limits _{\mathbb{R }^{3}}(a|\nabla u^{-}|^2+V(\varepsilon x)|u_n^{-}|^2)+b\left(\int \limits _{\mathbb{R }^{3}}|\nabla u_n^{-}|^2\right)^2+\int \limits _{\mathbb{R }^{3}}|u_n^{-}|^{2^*}+o(1). \end{aligned}$$

Since \(\{u^{-}_n\}\) is a bounded sequence, the last equality implies

$$\begin{aligned} \int \limits _{\mathbb{R }^{3}}(a|\nabla u^{-}|^2+V(\varepsilon x)|u_n^{-}|^2)+b\left(\int \limits _{\mathbb{R }^{3}}|\nabla u_n^{-}|^2\right)^2+\int \limits _{\mathbb{R }^{3}}|u_n^{-}|^{2^*}= o(1). \end{aligned}$$

Then, we can easily compute \(\mathcal I _{\varepsilon }(u_n) = \mathcal I _{\varepsilon }(u^{+}_n)+o(1)\) and \(\mathcal I _{\varepsilon }^{\prime }(u_n) = \mathcal I _{\varepsilon }^{\prime }(u^{+}_n)+o(1)\) from where it follows that \(\{u^+_n\}\) is a \((PS)_c\) sequence. This way, we can assume without loss of generality that \(\{u_n\}\) is nonnegative. \(\square \)

Lemma 2.9

Assume that \(V_{\infty }<\infty \) and \(\{u_n\}\) is a \((PS)_d\) sequence for the functional \(\mathcal I _{\varepsilon }\) with \(d<\frac{1}{12}(aS)^{\frac{3}{2}}\) and \(u_n\rightharpoonup 0\) in \(W_{\varepsilon }\). If \(u_n\not \rightarrow 0\) in \(W_{\varepsilon },\) then \(d\ge m_{V_{\infty }},\) where \(m_{V_{\infty }}\) is the minimax level of \(\mathcal E _{V_{\infty }}\).

Proof

Let \(\{t_n\}\subset (0,\infty )\) be a sequence such that \(\{t_nu_n\}\subset \mathcal N _{V_{\infty }}\). Then, we claim that the sequence \(\{t_n\}\) satisfies \(\lim \sup _{n\rightarrow \infty } t_n\le 1\). Assume by contradiction, there exist \(\delta >0\) and a subsequence still denoted by \(\{t_n\}\) such that \( t_n\ge 1+\delta \, \text{ for} \text{ all}\, n\in \mathbb{N }\). By Lemma 2.4, \(\{u_n\}\) is bounded, and from \(\langle \mathcal I ^{\prime }_{\varepsilon }(u_n),u_n\rangle =o(1),\) we have

$$\begin{aligned} \int \limits _{\mathbb{R }^{3}}\left(a|\nabla u_n|^2+V(\varepsilon x)u^2_n\right)+b\left(\int \limits _{\mathbb{R }^{3}}|\nabla u_n|^2\right)^2=\int \limits _{\mathbb{R }^{3}}[u^6_n+\lambda f(u_n)u_n]+o(1). \end{aligned}$$
(2.6)

In view of \(t_nu_n\in \mathcal N _{V_{\infty }},\) we have

$$\begin{aligned} t^2_n\int \limits _{\mathbb{R }^{3}}(a|\nabla u_n|^2+V_{\infty }u^2_n)+t^4_nb\left(\int \limits _{\mathbb{R }^{3}}|\nabla u_n|^2\right)^2=\int \limits _{\mathbb{R }^{3}}[(t_nu_n)^6+\lambda f(t_nu_n)t_nu_n]. \end{aligned}$$
(2.7)

Combining (2.6)–(2.7), we get

$$\begin{aligned}&o(1)+\left(\frac{1}{t^2_n}-1\right)\int \limits _{\mathbb{R }^{3}}a|\nabla u_n|^2+\int \limits _{\mathbb{R }^{3}}\left(\frac{V_{\infty }}{t^2_n}-V(\varepsilon x)\right)u^2_n\\&\quad =\lambda \int \limits _{\mathbb{R }^{3}}\left(\frac{f(t_nu_n)}{t_n^3u^3_n}-\frac{f(u_n)}{u^3_n}\right)u^4_n+(t^2_n-1)\int \limits _{\mathbb{R }^{3}}u^6_n\\&\quad \ge \lambda \int \limits _{\mathbb{R }^{3}}\left(\frac{f(t_nu_n)}{t_n^3u^3_n}-\frac{f(u_n)}{u^3_n}\right)u^4_n. \end{aligned}$$

By condition \((V)\) and \(t_n>1,\) for any \(\epsilon >0\), there exists \(R=R(\epsilon )>0\) such that

$$\begin{aligned} V(\varepsilon x)\ge V_{\infty }-\epsilon >\frac{V_{\infty }}{t^2_n}-\epsilon \quad \text{ for} \text{ any}~~ |x|\ge R. \end{aligned}$$
(2.8)

Since \(||u_n||_{\varepsilon }\le C\) and \(u_n\rightarrow 0\) in \(L^2(B_R(0))\), we deduce that

$$\begin{aligned} \int \limits _{\mathbb{R }^{3}}\left(\frac{f(t_nu_n)}{(t_nu_n)^3}-\frac{f(u_n)}{u_n^3}\right)u^4_n\le \epsilon C+o(1). \end{aligned}$$
(2.9)

If \(u_n\not \rightarrow 0\) in \(W_{\varepsilon }\), we have from Lemma 2.8 that there exist \(\{y_n\}\subset {\mathbb{R }^{3}}\) and \(R^*,\beta >0\) such that

$$\begin{aligned} \int \limits _{B_{R^*}(y_n)}u_n^2\ge \beta . \end{aligned}$$
(2.10)

If we set \(\widetilde{u_n}(x)=u_n(x+y_n),\) then there exists a nonnegative function \(\widetilde{u}\) such that, up to a subsequence, \(\widetilde{u_n}\rightharpoonup \widetilde{u}\) in \(W_{\varepsilon }\). Moreover, by (2.10), there exists a subset \(\Lambda \subset B_{R^*(0)}\) with positive measure such that \(\widetilde{u}>0\) a.e. in \(\Lambda \). It follows from \((A_4)\), (2.9) and \(t_n\ge 1+\delta \) that

$$\begin{aligned} 0<\int \limits _{\Lambda }\left(\frac{f((1+\delta )\widetilde{u_n})}{((1+\delta )\widetilde{u_n})^3} -\frac{f(\widetilde{u_n})}{\widetilde{u_n}^3}\right)\widetilde{u_n}^4\le \epsilon C+o(1), \end{aligned}$$

for any \(\epsilon >0\). Taking limit in the above inequality and applying Fatou’s lemma, we get a contradiction. We next distinguish the following two cases:

Case 1 \(\lim \sup _{n\rightarrow \infty } t_n=1\). In this case, there exists a subsequence, still denoted by \(\{t_n\}\) such that \(t_n\rightarrow 1\) as \(n\rightarrow \infty \). Thus, we have

$$\begin{aligned} d+o(1)=\mathcal I _{\varepsilon }(u_n)\ge \mathcal I _{\varepsilon }(u_n)+m_{V_\infty }-\mathcal E _{V_{\infty }}(t_nu_n). \end{aligned}$$
(2.11)

Notice that

$$\begin{aligned}&\mathcal I _{\varepsilon }(u_n)-\mathcal E _{V_{\infty }}(t_nu_n)\\&\quad =\frac{a}{2}\int \limits _{\mathbb{R }^{3}}(1-t^2_n) |\nabla u_n|^2+\frac{1}{2}\int \limits _{\mathbb{R }^{3}}[V(\varepsilon x)-t^2_nV_{\infty }]u_n^2\\&\qquad \,{+}\,\frac{b(1-t^4_n)}{4}\left(\int \limits _{\mathbb{R }^{3}}|\nabla u_n|^2\right)^2+\lambda \int \limits _{\mathbb{R }^{3}}(F(t_nu_n)-F(u_n))+\frac{t_n^{6}-1}{6}\int \limits _{\mathbb{R }^{3}}u_n^{6}\\&\quad \ge \frac{1}{2}\int \limits _{\mathbb{R }^{3}}[V(\varepsilon x)-V_{\infty }]u_n^2+\lambda \int \limits _{\mathbb{R }^{3}}(F(t_nu_n)-F(u_n)) +o(1)\\&\quad =o(1)+\frac{1}{2}\left[\int \limits _{{\mathbb{R }^{3}}\backslash B_R(0)}[V(\varepsilon x)-V_{\infty }]u_n^2+\int \limits _{B_R(0)}[V(\varepsilon x)-V_{\infty }]u_n^2\right]\\&\qquad \,{+}\,\lambda \int \limits _{\mathbb{R }^{3}}(F(t_nu_n)-F(u_n))\\&\quad \ge o(1)-\frac{\epsilon }{2}\int \limits _{{\mathbb{R }^{3}}\backslash B_R(0)}u^2_n +\lambda \int \limits _{\mathbb{R }^{3}}(F(t_nu_n)-F(u_n))\\&\quad \ge o(1)-\frac{\epsilon }{2V_0}\int \limits _{\mathbb{R }^{3}}(|\nabla u_n|^2+V(\varepsilon x)u^2_n) +\lambda \int \limits _{\mathbb{R }^{3}}(F(t_nu_n)-F(u_n))\\&\quad \ge o(1)-\epsilon C+\lambda \int \limits _{\mathbb{R }^{3}}(F(t_nu_n)-F(u_n)). \end{aligned}$$

Here, we have used condition \((V),\) (2.8),\( \lim \sup _{n\rightarrow \infty }t_n=1, u_n\rightarrow 0\) in \(L^2(B_R(0))\) and \(\Vert u_n\Vert _\varepsilon \le C\). Moreover, from the mean value theorem,

$$\begin{aligned} \lambda \int \limits _{\mathbb{R }^{3}}(F(t_nu_n)-F(u_n))=o(1). \end{aligned}$$

Therefore, \(d+o(1)\ge m_{V_\infty }+o(1)-\epsilon C\), and letting \(n\rightarrow \infty , \epsilon \rightarrow 0\), we obtain \(d\ge m_{V_{\infty }}\).

Case 2  \(\lim \sup _{n\rightarrow \infty }t_n=t_0<1\). In this case, we may suppose, without loss of generality, that \(t_n <1\) for all \(n\in \mathbb{N }\). From (2.8), \(u_n\rightarrow 0\) in \(L^2(B_R(0))\) and \(||u||_{\varepsilon }\le C,\) we have \( \int _{{\mathbb{R }^{3}}}(V_{\infty }-V(\varepsilon x))u^2_n\le \epsilon C+o(1)\) for any given \(\epsilon >0\). Since \(\frac{1}{4}f(s)s-F(s)\) is increasing for \(s>0\), we deduce that

$$\begin{aligned} m_{V_{\infty }}&\le \mathcal E _{V_{\infty }}(t_nu_n)-\frac{1}{4}\langle \mathcal E ^{\prime }_{V_{\infty }}(t_nu_n),t_nu_n\rangle \\&= \frac{t^2_n}{4}\int \limits _{\mathbb{R }^{3}}(a|\nabla (u_n)|^2 +V_{\infty } u_n^2)+\lambda \int \limits _{\mathbb{R }^{3}}\left(\frac{1}{4}f(t_nu_n)t_nu_n-F(t_nu_n)\right)+\frac{t_n^6}{12}\int \limits _{\mathbb{R }^{3}}u_n^6\\&\le \frac{1}{4}\int \limits _{\mathbb{R }^{3}}(a|\nabla u_n|^2+V_{\infty } u_n^2) +\lambda \int \limits _{\mathbb{R }^{3}}\left(\frac{1}{4}f(u_n)u_n-F(u_n)\right)+\frac{1}{12}\int \limits _{\mathbb{R }^{3}}u_n^6\\&\le \frac{1}{4} \int \limits _{\mathbb{R }^{3}}(a|\nabla u_n|^2+V(\varepsilon x)u^2_n)+\lambda \int \limits _{\mathbb{R }^{3}}\left(\frac{1}{4}f(u_n)u_n-F(u_n)\right)\\&+\frac{1}{12}\int \limits _{\mathbb{R }^{3}}u_n^6 +\epsilon C+o(1)\\&= \mathcal I _{\varepsilon }(u_n)-\frac{1}{4}\langle \mathcal I ^{\prime }_{\varepsilon }(u_n),u_n\rangle +\epsilon C+o(1)\\&\le d+\epsilon C+o(1). \end{aligned}$$

Let \(\epsilon \rightarrow 0\) and \(n\rightarrow \infty ;\) we get \(d\ge m_{V_{\infty }}\). \(\square \)

2.3 The Palais–Smale condition

In this subsection, we establish some compactness results for the functional \(\mathcal I _{\varepsilon }\). In order to apply the Ljusternik–Schnirelmann category theory, we need \(\mathcal I _{\varepsilon }\) to satisfy Palais–Smale condition on \(\mathcal N _{\varepsilon }\). Since the Sobolev embedding \(W^{1,2}({\mathbb{R }^{3}})\hookrightarrow L^s({\mathbb{R }^{3}}), 2\le s<2^*=6\) is continuous but is not compact, in general, such a condition is not fulfilled. But we still can prove that Palais–Smale condition holds in a suitable sublevel, related to the ground energy at infinity.

Proposition 2.10

\(\mathcal I _{\varepsilon }\) satisfies the \((PS)_c\) conditions at any level \(c<m_{V_{\infty }}\) if \(V_{\infty }<\infty ,\) and at any level \(c<\frac{1}{12}(aS)^{\frac{3}{2}}\) if \(V_{\infty }=\infty \).

Proof

Let \(\{u_n\}\subset W_{\varepsilon }\) be such that \(\mathcal I _{\varepsilon }(u_n)\rightarrow c\) and \(\mathcal I ^{\prime }_{\varepsilon }(u_n)\rightarrow 0\). Since \(\{u_n\}\) is bounded in \(W_{\varepsilon }\), we may suppose that there exists \(u\in W_{\varepsilon }\) such that \(u_n\rightharpoonup u\) in \(W_{\varepsilon }\) and \(u_n\rightarrow u\) a.e. in \({\mathbb{R }^{3}}\). Moreover, \(u\) is a critical point of \(\mathcal I ^{\prime }_{\varepsilon }\). Setting \(w_n:=u_n-u,\) by a result due to Brezis–Lieb (see [14]), we have

$$\begin{aligned} \int \limits _{\mathbb{R }^{3}}|\nabla w_n|^2&= \int \limits _{\mathbb{R }^{3}}|\nabla u_n|^2-\int \limits _{\mathbb{R }^{3}}|\nabla u|^2+o(1),\\ \int \limits _{\mathbb{R }^{3}}w_n^6&= \int \limits _{\mathbb{R }^{3}}u_n^6-\int \limits _{\mathbb{R }^{3}} u^6+o(1), \end{aligned}$$

and hence,

$$\begin{aligned} \left(\int \limits _{\mathbb{R }^{3}}|\nabla w_n|^2\right)^2&= \left(\int \limits _{\mathbb{R }^{3}}|\nabla u_n|^2-\int \limits _{\mathbb{R }^{3}}|\nabla u|^2+o(1)\right)^2\\&= \left(\int \limits _{\mathbb{R }^{3}}|\nabla u_n|^2\right)^2-\left(\int \limits _{\mathbb{R }^{3}}|\nabla u|^2\right)^2\\&-2\int \limits _{\mathbb{R }^{3}}|\nabla u|^2\left(\int \limits _{\mathbb{R }^{3}}|\nabla u_n|^2-\int \limits _{\mathbb{R }^{3}}|\nabla u|^2\right)+o(1)\\&= \left(\int \limits _{\mathbb{R }^{3}}|\nabla u_n|^2\right)^2-\left(\int \limits _{\mathbb{R }^{3}}|\nabla u|^2\right)^2\\&-2\int \limits _{\mathbb{R }^{3}}|\nabla u|^2 \int \limits _{\mathbb{R }^{3}}|\nabla w_n|^2+o(1)\\&\le \left(\int \limits _{\mathbb{R }^{3}}|\nabla u_n|^2\right)^2-\left(\int \limits _{\mathbb{R }^{3}}|\nabla u|^2\right)^2+o(1). \end{aligned}$$

Arguing as in the proof of Lemma 3.1 [2], we obtain \( \int _{{\mathbb{R }^{3}}}F(w_n)=\int _{{\mathbb{R }^{3}}}(F(u_n)-F(u))+o(1)\). Therefore, we have that \( \mathcal I _{\varepsilon }(w_n)\le \mathcal I _{\varepsilon }(u_n)-\mathcal I _{\varepsilon }(u)+o(1)=c-\mathcal I _{\varepsilon }(u)+o(1):= d+o(1) \) and \(I^{\prime }_{\varepsilon }(w_n)\rightarrow 0\). By \((A_3)\), we get

$$\begin{aligned} \mathcal I _{\varepsilon }(u)=\frac{1}{4}\int \limits _{\mathbb{R }^{3}}\left(a|\nabla u|^2+V(\varepsilon x)u^2\right)+\lambda \int \limits _{\mathbb{R }^{3}}\left(\frac{1}{4}f(u)u-F(u)\right)+\frac{1}{12}\int \limits _{\mathbb{R }^{3}}u^6\ge 0. \end{aligned}$$

Therefore, if \(V_{\infty }<\infty ,\) we have \(d\le c<m_{V_{\infty }}\). It follows from Lemma 2.9 that \(w_n\rightarrow 0\) in \(W_{\varepsilon }\). Consequently, \(u_n\rightarrow u\) in \(W_{\varepsilon }\).

If \(V_{\infty }=\infty \) holds, then \(V\) is coercive and the continuous embedding \(W_{\varepsilon }\hookrightarrow L^p({\mathbb{R }^{3}})\) is compact for \(2\le p<6\). Hence, up to a subsequence, \(w_n\rightarrow 0\) in \(L^p({\mathbb{R }^{3}})\). By \((A_1)\text{-}(A_2)\), we have

$$\begin{aligned}&\int \limits _{\mathbb{R }^{3}}\left(a|\nabla w_{\varepsilon }|^2+V(\varepsilon x)w_n^2\right)+b\left(\int \limits _{\mathbb{R }^{3}}|\nabla w_n|^2\right)^2\\&\quad =\int \limits _{\mathbb{R }^{3}}(w_n^6+\lambda f(w_n)w_n)+o(1)=\int \limits _{\mathbb{R }^{3}}w_n^6+o(1). \end{aligned}$$

Assume that

$$\begin{aligned} \int \limits _{\mathbb{R }^{3}}\left(a|\nabla w_{\varepsilon }|^2+V(\varepsilon x)w_n^2\right)+b\left(\int \limits _{\mathbb{R }^{3}}|\nabla w_n|^2\right)^2\rightarrow \ell \quad \text{ and}\quad \int \limits _{\mathbb{R }^{3}}w_n^6\rightarrow \ell . \end{aligned}$$

Then, a similar argument as we have done in the proof of Lemma 2.8 yields that \(\ell =0\). Therefore, \(u_n\rightarrow u\) in \(W_{\varepsilon }\). \(\square \)

Remark 2.11

From the definition of \(m_\nu \), it is easy to see that the function \(\nu \rightarrow m_{\nu }\) is increasing for \(\nu >0\). Also, we have \(\mathcal I _{\varepsilon }(u)\ge \mathcal E _{V_0}(u)\) for all \(u\in W_{\varepsilon },\) and the characterization of \(c_{\varepsilon }\) implies that \(c_{\varepsilon }\ge m_{V_0}\) for any \(\varepsilon >0\). These properties will be used in the next section. To end this section, we state and prove below the main result of this section.

Proposition 2.12

Let \(\{u_n\}\) be a \((PS)_c\) sequence restricted in \(\mathcal{N }_{\varepsilon }\) and assume that \(c<m_{V_{\infty }}\) if \(V_{\infty }<\infty ,\) or \(c<\frac{1}{12}(aS)^{\frac{3}{2}}\) if \(V_{\infty }=\infty \). Then, \(\{u_n\}\) has a convergent subsequence in \(W_{\varepsilon }\).

Proof

Let \(\{u_n\}\subset \mathcal{N }_{\varepsilon }\) be such that \(\mathcal I _{\varepsilon }(u_n)\rightarrow c\) and \(||\mathcal I ^{\prime }_{\varepsilon }(u_n)||_{W^{-1}_{\varepsilon }}=o(1)\). Then, there exists \(\{\gamma _n\}\subset \mathbb{R }\) such that \( \mathcal I ^{\prime }_{\varepsilon }(u_n)=\gamma _n J^{\prime }_{\varepsilon }(u_n)+o(1), \) where

$$\begin{aligned} J_{\varepsilon }(u):=\int \limits _{\mathbb{R }^{3}}(a|\nabla u|^2+V(\varepsilon x)u^2)+b\left(\int \limits _{\mathbb{R }^{3}}|\nabla u|^2\right)^2-\int \limits _{{\mathbb{R }^{3}}}u^6-\lambda \int \limits _{\mathbb{R }^{3}}f(u)u. \end{aligned}$$

Using \((A_4)\), we have

$$\begin{aligned} \langle J^{\prime }_{\varepsilon }(u_n),u_n\rangle&= 2\int \limits _{\mathbb{R }^{3}}(a|\nabla u_n|^2+V(\varepsilon x)u^2_n) +4b\left(\int \limits _{\mathbb{R }^{3}}|\nabla u_n|^2\right)^2\\&- 6\int \limits _{\mathbb{R }^{3}}u^6_n-\lambda \int \limits _{\mathbb{R }^{3}}f(u_n)u_n-\lambda \int \limits _{\mathbb{R }^{3}}f^{\prime }(u_n)u^2_n\\&\le 3\lambda \int \limits _{\mathbb{R }^{3}}f(u_n)u_n-\lambda \int \limits _{\mathbb{R }^{3}}f^{\prime }(u_n)u^2_n-2\int \limits _{\mathbb{R }^{3}}u^6_n\\&\le -2\int \limits _{\mathbb{R }^{3}}u_n^{6}. \end{aligned}$$

We may suppose that \(\langle J^{\prime }_{\varepsilon }(u_n),u_n\rangle \rightarrow l\le 0\). If \(l=0,\) it follows from

$$\begin{aligned} |\langle J^{\prime }_{\varepsilon }(u_n),u_n\rangle |\ge 2\int \limits _{\mathbb{R }^{3}}u_n^{6} \end{aligned}$$

that \(u_n\rightarrow 0\) in \(L^{6}({\mathbb{R }^{3}})\). Consequently, it is easy to verify that \(u_n\rightarrow 0\) in \(W_{\varepsilon }\) by interpolations, which contradicts to (2.4). Thus, \(l\ne 0\) and \(\gamma _n=o(1),\) and we have \(\mathcal I ^{\prime }_{\varepsilon }(u_n)=o(1)\). Hence, \(\{u_n\}\) is a \((PS)_c\) sequence for \(\mathcal I _{\varepsilon }\) in \(W_{\varepsilon }\), and the conclusion follows from Proposition 2.10. \(\square \)

Corollary 2.13

The critical points of the functional \(\mathcal I _{\varepsilon }\) on \(\mathcal N _{\varepsilon }\) are critical points of \(\mathcal I _{\varepsilon }\) in \(W_{\varepsilon }\).

Proof

Since the proof is standard, we just sketch it for completeness. Let \(\tilde{\mathcal{I }}_{\varepsilon }:= \mathcal I _{\varepsilon }|_\mathcal{N _{\varepsilon }}\). From the proof of Proposition 2.12, we see that if \(u\in \mathcal N _{\varepsilon }\), then \(\langle J_{\varepsilon }^{\prime }(u),u\rangle <0\), and hence, \(J_{\varepsilon }^{\prime }(u)\not = 0\) on \(\mathcal N _{\varepsilon }\). Then,

$$\begin{aligned} \tilde{\mathcal{I }}_{\varepsilon }^{\prime }(u)= \mathcal I _{\varepsilon }^{\prime }(u)-\frac{\langle \mathcal I _{\varepsilon }^{\prime }(u), u\rangle }{\Vert J_{\varepsilon }^{\prime }(u)\Vert ^2}J_{\varepsilon }^{\prime }(u) \end{aligned}$$

and \( \langle \tilde{\mathcal{I }}_{\varepsilon }^{\prime }(u),u\rangle =\langle \mathcal I _{\varepsilon }^{\prime }(u),u\rangle \left(1-\frac{\langle J_{\varepsilon }^{\prime }(u), u\rangle }{\Vert J_{\varepsilon }^{\prime }(u)\Vert ^2}\right)\). So, \( \tilde{\mathcal{I }}_{\varepsilon }^{\prime }(u)=0 \) iff \( \mathcal I _{\varepsilon }^{\prime }(u)=0\). \(\square \)

3 Existence of a ground state solution to \((\mathcal K )_{\varepsilon }\)

In this section, we prove the existence of a ground state solution to \((\mathcal{K })_{\varepsilon }\), that is, a critical point \(u_{\varepsilon }\) of \(\mathcal I _{\varepsilon }\) satisfying \(\mathcal I _{\varepsilon }(u_{\varepsilon })=c_{\varepsilon }\).

Theorem 3.1

Assume that conditions \((V)\) and \((A_1)\text{-}(A_4)\) hold. Then, there exists \(\varepsilon ^{*}\!>\!0, \lambda ^*>0\) such that for any \(\varepsilon \in (0,\varepsilon ^*), \lambda >\lambda ^*\) problem \((\mathcal{K })_{\varepsilon }\) has a ground state solution

Proof

By Lemma 2.1, the functional satisfies the mountain pass geometry. Then using a version of the mountain pass theorem without \((PS)\) condition (see [44]), there exists \(\{u_n\}\subset W_{\varepsilon }\) satisfying

$$\begin{aligned} \mathcal I _{\varepsilon }(u_n)\rightarrow c_{\varepsilon }~~\text{ and} ~~\mathcal I ^{\prime }_{\varepsilon }(u_n)\rightarrow 0. \end{aligned}$$

If \(V_{\infty }<\infty ,\) we may assume without loss of generality that \(V_0=V(0)=\inf _{x\in {\mathbb{R }^{3}}}V(x)\). For a fix \(\nu \in \mathbb{R }^+\) such that \(V_0<\nu <V_{\infty },\) we have \(m_{V_0}<m_{\nu }<m_{V_{\infty }},\) and there exists a nonnegative function \(w\in W_{\varepsilon }\) with compact support such that \(\mathcal E _{\nu }(w)=\max _{t\ge 0}\mathcal E _{\nu }(tw)\) and \(\mathcal E _{\nu }(w)<m_{V_{\infty }}\). The condition \((V)\) implies that for some \(\varepsilon ^*>0,\)

$$\begin{aligned} V(\varepsilon x)\le \nu \quad \text{ for} \text{ any}~\varepsilon \in (0,\varepsilon ^*)\quad \text{ and}\quad x\in \text{ supp}~w. \end{aligned}$$

Thus,

$$\begin{aligned} \mathcal I _{\varepsilon }(tw)\le \mathcal E _{\nu }(tw)\quad \text{ for} \text{ any}~\varepsilon \in (0,\varepsilon ^*)~~\text{ and}\quad t\ge 0 \end{aligned}$$

and

$$\begin{aligned} \max _{t\ge 0}\mathcal I _{\varepsilon }(tw)\le \max _{t\ge 0}\mathcal E _{\nu }(tw)= \mathcal{E }_{\nu }(w)<m_{V_{\infty }}\quad \text{ for} \text{ any}\quad \varepsilon \in (0,\varepsilon ^*). \end{aligned}$$

It follows from Lemma 2.3 that \(c_{\varepsilon }<m_{V_{\infty }}\) for \(\varepsilon \in (0,\varepsilon ^*),\) and the theorem follows from Proposition 2.11, immediately.

If \(V_{\infty }=\infty ,\) for any given \(\nu >V_0,\) arguing as in the proof of Lemma 2.6, we can show that there exists some \(v^*\in W_{\nu }\) and \(\lambda ^*>0\) such that \(\max _{t\ge 0}\mathcal I _{\varepsilon }(t w)\le \mathcal E _{\nu }(w)\le \mathcal E _{\nu }(v^*)<\frac{1}{12}(aS)^{\frac{3}{2}}\) for all \(\lambda >\lambda ^*\). Consequently, \(c_{\varepsilon }<\frac{1}{12}(aS)^{\frac{3}{2}}\). So, from Proposition 2.10, there exists some \(u\in W_{\varepsilon }\) (the weak limit of \(\{u_n\}\) in \(W_{\varepsilon }\)) such that

$$\begin{aligned} \mathcal I _{\varepsilon }(u)= c_{\varepsilon }~~\text{ and}~~\mathcal I ^{\prime }_{\varepsilon }(u)= 0. \end{aligned}$$

Using the standard arguments as in [12, 41], we have that \(u\in L^{\infty }({\mathbb{R }^{3}}),\) and \(u\in C^{1,\alpha }_{\mathrm{loc}}({\mathbb{R }^{3}})\) with \(0<\alpha <1\). Moreover, by Harnack’s inequality [42], \(u(x)>0\) for all \(x\in {\mathbb{R }^{3}}\). This completes the proof. \(\square \)

4 Multiplicity of solutions to \((\mathcal{K })_{\varepsilon }\)

In this section, our goal is to show the multiplicity of solutions and study the behavior of its maximum points concentrating on the set \(\Sigma \) of global minima of \(V\) given in Section 1. The main result of this section is equivalent to Theorem 1.1 and it can be stated as follows.

Theorem 4.1

Suppose that conditions \((V)\) and \((A_1)\text{-}(A_4)\) are satisfied. Then, for any \(\delta >0\) given, there exists \(\varepsilon _{\delta }>0, \lambda ^*>0\) such that, for any \(\varepsilon \in (0,\varepsilon _{\delta })\) and \(\lambda >\lambda ^*\), problem \((\mathcal{K })_{\varepsilon }\) has at least \(cat_{\Sigma _{\delta }}(\Sigma )\) positive solutions. Moreover, if \(u_{\varepsilon }\) denotes one of these positive solutions and \(z_{\varepsilon }\in {\mathbb{R }^{3}}\) its global maximum, then

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0}V(\varepsilon z_{\varepsilon })=V_0. \end{aligned}$$

In the following subsections, we give some preliminary lemmas that are useful to prove Theorem 4.1

4.1 Preliminary lemmas

Let \(w\) be a ground state solution of problem \((\mathcal P )_{V_0}\) and \(\eta \) be a smooth nonincreasing function defined in \([0,\infty )\) such that \(\eta (s)=1\) if \(0\le s\le 1/2\) and \(\eta (s)=0\) if \(s\ge 1\). For any \(y\in \Sigma ,\) we define

$$\begin{aligned} \Psi _{\varepsilon ,y}(x)=\eta (|\varepsilon x-y|)w\left(\frac{\varepsilon x-y}{\varepsilon }\right) \end{aligned}$$

and \(t_{\varepsilon }>0\) satisfying \( \max _{t\ge 0}\mathcal I _{\varepsilon }(t\Psi _{\varepsilon ,y})=\mathcal I _{\varepsilon }(t_{\varepsilon }\Psi _{\varepsilon ,y})\) and define \(\Phi _{\varepsilon }:~\Sigma \rightarrow \mathcal{N }_{\varepsilon }\) by \( \Phi _{\varepsilon }(y):=t_{\varepsilon }\Psi _{\varepsilon ,y}\). By the construction, \(\Phi _{\varepsilon }(y) \) has a compact support for any \(y\in \Sigma \).

Lemma 4.2

The function \(\Phi _{\varepsilon }\) has the following property: \( \lim \nolimits _{\varepsilon \rightarrow 0} \mathcal I _{\varepsilon }(\Phi _{\varepsilon }(y))\!=\!m_{V_0}~~\text{ uniformly} \text{ in}~y\in \Sigma \).

Proof

Suppose by contradiction that there exist some \(\delta _0>0, \{y_n\}\subset \Sigma \) and \(\varepsilon _n\rightarrow 0\) such that

$$\begin{aligned} |\mathcal I _{\varepsilon _n}(\Phi _{\varepsilon _n}(y_n))-m_{V_0}|\ge \delta _0. \end{aligned}$$
(4.1)

Now, we claim that \(\lim _{n\rightarrow \infty } t_{\varepsilon _n}=1\). In fact, by the definition of \(t_{\varepsilon _n}\) and (2.4), we have

$$\begin{aligned} \min \{1,a\}r^*&\le \int \limits _{\mathbb{R }^{3}}\left(a|\nabla (t_{\varepsilon _n}\Psi _{\varepsilon _n,y_n})|^2+V(\varepsilon _n x)(t_{\varepsilon _n}\Psi _{\varepsilon _n,y_n})^2\right)\nonumber \\&+\,t^4_{\varepsilon _n}b\left(\int \limits _{\mathbb{R }^{3}}|\nabla \Psi _{\varepsilon _n,y_n}|^2\right)^2\nonumber \\&= \int \limits _{\mathbb{R }^{3}}\left[\lambda f(t_{\varepsilon _n} \Psi _{\varepsilon _n,y_n})t_{\varepsilon _n}\Psi _{\varepsilon _n,y_n}+(t_{\varepsilon _n} \Psi _{\varepsilon _n,y_n})^6\right]. \end{aligned}$$
(4.2)

By (2.2), we have \(|f(s)s|\le \tau s^2+C_{\tau }|s|^{q+1}\) for any \(\tau >0\). We see that \(t_{\varepsilon _n}\) cannot go zero, that is to say, \(t_{\varepsilon _n}\ge t_0>0\) for some \(t_0>0\). By \((A_3),\) we have \(sf(s)>4F(s), \forall s>0\). If \(t_{\varepsilon _n}\rightarrow \infty ,\) we get by the boundedness of \(\Psi _{\varepsilon _n,y_n},\)

$$\begin{aligned}&\frac{1}{t_{\varepsilon _n}^2}\int \limits _{\mathbb{R }^{3}}(a|\nabla \Psi _{\varepsilon _n,y_n}|^2 +V(\varepsilon _n x)\Psi ^2_{\varepsilon _n,y_n})+b\left(\int \limits _{\mathbb{R }^{3}}|\nabla \Psi _{\varepsilon _n,y_n}|^2\right)^2 \\&\quad =\lambda \int \limits _{\mathbb{R }^{3}}\frac{f(t_{\varepsilon _n}\Psi _{\varepsilon _n,y_n})}{(t_{\varepsilon _n} \Psi _{\varepsilon _n,y_n})^3}\Psi ^4_{\varepsilon _n,y_n}+t_{\varepsilon _n}^2\int \limits _{\mathbb{R }^{3}}\Psi ^6_{\varepsilon _n,y_n}\\&\quad >\lambda \int \limits _{B_{\frac{1}{2}}(0)}\frac{f(t_{\varepsilon _n}\eta (|\varepsilon _nz|)w(z))}{(t_{\varepsilon _n}\eta (|\varepsilon _nz|)w(z))^3}(\eta (|\varepsilon _nz|)w(z))^4\\&\quad =\lambda \int \limits _{B_{\frac{1}{2}}(0)}\frac{f(t_{\varepsilon _n}w)}{(t_{\varepsilon _n}w)^3}w^4\\&\quad \ge \lambda \int \limits _{B_{\frac{1}{2}}(0)}\frac{f(t_{\varepsilon _n}\varpi )}{(t_{\varepsilon _n}\varpi )^3}\mu ^4 \rightarrow \infty \end{aligned}$$

in view of \(f(t_{\varepsilon _n}\varpi )/(t_{\varepsilon _n}\varpi )^3\rightarrow \infty \) as \(n\rightarrow \infty \). Here, \(\varpi :=\inf _{x\in B_{\frac{1}{2}}(0)}w(x)\). But the left hand side of the above inequality tends to \(b(\int _{{\mathbb{R }^{3}}}|\nabla w|^2)^2\) since \(t_{\varepsilon _n}\rightarrow \infty \) as \( n\rightarrow \infty \). This yields a contradiction. Hence, \(0<t_0<t_{\varepsilon _n}\le C\). Assume that \(t_{\varepsilon _n}\rightarrow T\). Now, we claim that \(T=1\). By using the Lebesgue’s theorem, we can verify that

$$\begin{aligned} \lim _{n\rightarrow \infty }||\Psi _{\varepsilon _n,y_n}||^2_{\varepsilon _n}=||w||^2_{V_0},\quad \lim _{n\rightarrow \infty }\int \limits _{\mathbb{R }^{3}}F(\Psi _{\varepsilon _n,y_n})=\int \limits _{\mathbb{R }^{3}}F(w), \end{aligned}$$

and

$$\begin{aligned} \lim _{n\rightarrow \infty } \int \limits _{\mathbb{R }^{3}}f(\Psi _{\varepsilon _n,y_n})\Psi _{\varepsilon _n,y_n}=\int \limits _{\mathbb{R }^{3}}f(w)w,\quad \lim _{n\rightarrow \infty }\int \limits _{\mathbb{R }^{3}}|\Psi _{\varepsilon _n,y_n}|^6=\int \limits _{\mathbb{R }^{3}}w^6. \end{aligned}$$

Therefore,

$$\begin{aligned} \frac{1}{T^2}\int \limits _{\mathbb{R }^{3}}(a|\nabla w|^2+V_0w^2)+b\left(\int \limits _{\mathbb{R }^{3}}|\nabla w|^2\right)^2=\lambda \int \limits _{\mathbb{R }^{3}}\frac{f(Tw)}{T^3w^3}w^4+T^2\int \limits _{\mathbb{R }^{3}}w^6. \end{aligned}$$
(4.3)

Since \(w\) is a ground state solution of \((\mathcal P )_{V_0}\), then

$$\begin{aligned} \int \limits _{\mathbb{R }^{3}}(a|\nabla w|^2+V_0w^2)+b\left(\int \limits _{\mathbb{R }^{3}}|\nabla w|^2\right)^2=\lambda \int \limits _{\mathbb{R }^{3}}f(w)w+\int \limits _{\mathbb{R }^{3}}w^6. \end{aligned}$$
(4.4)

From (4.3)–(4.4), we have

$$\begin{aligned} \left(\frac{1}{T^2}-1\right)\int \limits _{\mathbb{R }^{3}}(a|\nabla w|^2+V_0w^2)=\lambda \int \limits _{\mathbb{R }^{3}}\left(\frac{f(Tw)}{(Tw)^3}-\frac{f(w)}{w^3}\right)w^4+(T^2-1)\int \limits _{\mathbb{R }^{3}}w^6.\qquad \end{aligned}$$
(4.5)

By \((A_4)\), we see that \(T=1\) and claim is proved. On the other hand,

$$\begin{aligned} \mathcal I _{\varepsilon _n}(\Phi _{\varepsilon _n}(y_n))&= \frac{t^2_{\varepsilon _n}}{2} \int \limits _{\mathbb{R }^{3}}\left(a|\nabla (\eta (|\varepsilon _n z|)w)|^2+V(\varepsilon _nz+y_n)|\eta (|\varepsilon _nz|)w|^2\right)\\&\,{+}\,\frac{t^4_{\varepsilon _n}b}{4}\left(\int \limits _{\mathbb{R }^{3}}|\nabla (\eta (|\varepsilon _nz|)w)|^2\right)^2- \lambda \int \limits _{\mathbb{R }^{3}}F(t_{\varepsilon _n}\eta (|\varepsilon _nz|)w)\\&\,{-}\,\frac{t^6_{\varepsilon _n}}{6}\int \limits _{\mathbb{R }^{3}}|\eta (\varepsilon _nz)w|^6. \end{aligned}$$

Let \(n\rightarrow \infty ,\) we get \(\lim _{n\rightarrow \infty }\mathcal I _{\varepsilon _n}(\Phi _{\varepsilon _n}(y_n))=E_{V_0}(w)=m_{V_0},\) which contradicts to (4.1). This completes the proof. \(\square \)

For any \(\delta >0,\) let \(\rho =\rho (\delta )>0\) be such that \(\Sigma _{\delta }\subset B_{\rho }(0)\). Define the map \(\chi :{\mathbb{R }^{3}}\rightarrow {\mathbb{R }^{3}}\) as \(\chi (x)=x\) for \(|x|\le \rho \) and \(\chi (x)=\rho x/|x|\) for \(|x|\ge \rho \). Finally, we define the map \(\beta _{\varepsilon }:~\mathcal{N }_{\varepsilon }\rightarrow {\mathbb{R }^{3}}\) by

$$\begin{aligned} \beta _{\varepsilon }(u):=\frac{\int \nolimits _{{\mathbb{R }^{3}}}\chi (\varepsilon x)u^2}{ \int \nolimits _{{\mathbb{R }^{3}}}u^2}. \end{aligned}$$

Since \(\Sigma \subset B_{\rho }(0),\) by the definition of \(\chi \) and Lebesgue’s theorem, we conclude that \( \lim _{\varepsilon \rightarrow 0}\beta _{\varepsilon }(\Phi _{\varepsilon }(y))=y~~\text{ uniformly} \text{ in}~y\in \Sigma \). To go on, we need the following compactness lemma.

Lemma 4.3

Let \(\{u_n\}\subset \mathcal{N }_{\nu }\) be a sequence satisfying \(\mathcal E _{\nu }(u_n)\rightarrow m_{\nu }\). Then,

  1. (a)

    \(\{u_n\}\) has a subsequence strongly convergent in \(H^1({\mathbb{R }^{3}})\); or

  2. (b)

    there exists \(\{\widetilde{y_n}\}\subset {\mathbb{R }^{3}}\) such that the sequence \(v_n(x)=u_n(x+\widetilde{y_n})\) converges strongly in \(H^1({\mathbb{R }^{3}})\).

In particular, there exists a minimizer \(u\) for \(\mathcal E _{\nu }\) with \(\mathcal E _{\nu }(u)=m_{\nu }\).

Proof

Using well known arguments, we have that the sequence \(\{u_n\}\) is bounded in \(H^1({\mathbb{R }^{3}})\). Hence, there exists an \(u\in H^1({\mathbb{R }^{3}})\) such that \(u_n\rightharpoonup u\) in \(H^1({\mathbb{R }^{3}}),\) up to some subsequence. We may assume that \(\{u_n\}\) satisfies the following limits

$$\begin{aligned} \mathcal E _{\nu }(u_n)\rightarrow m_{\nu }\quad \text{ and}\quad \mathcal E ^{\prime }_{\nu }(u_n)\rightarrow 0. \end{aligned}$$
(4.6)

In fact, using the Ekeland’s variational principle in [26], there exists a sequence \(\{w_n\}\subset \mathcal{N }_{\nu }\) satisfying \( w_n=u_n+o(1),~\mathcal E _{\nu }(w_n)\rightarrow m_{\nu }~~\text{ and}~~\mathcal E ^{\prime }_{\nu }(w_n)-\gamma _nJ^{\prime }_{\nu }(w_n)=o(1),\) where \(\gamma _n\) is a real number and \(J_{\nu }(w)=\langle \mathcal E ^{\prime }_{\nu }(w), w\rangle , \forall \, w\in H^1({\mathbb{R }^{3}})\). We claim that there exists a \(\delta >0\) such that \( |\langle J^{\prime }_{\nu }(w_n),w_n\rangle |\ge \delta ,~\forall \, n\in \mathbb{N }\). Indeed, using similar arguments as we have done in the proof of Proposition 2.12, we can prove that \(\gamma _n=o(1),\) which yields that \( \mathcal E _{\nu }(w_n)\rightarrow m_{\nu }~~\text{ and}~\mathcal E ^{\prime }_{\nu }(w_n)\rightarrow 0\). Consequently, without loss of generality, we may assume that \( \mathcal E _{\nu }(u_n)\rightarrow m_{\nu }~~\text{ and}~\mathcal E ^{\prime }_{\nu }(u_n)\rightarrow 0\). We next continue our arguments by distinguishing two cases: \(u\ne 0\) and \(u= 0\).

Case 1 \(u\ne 0\). In this case, since \(f\) has subcritical growth, it is easy to see that for some subsequence, still denoted by \(\{u_n\}, \nabla u_n(x)\rightarrow \nabla u(x)~~\text{ a.e.} \text{ in}~{\mathbb{R }^{3}}\). Using the fact that \(\langle \mathcal E ^{\prime }_{\nu }(u_n),u\rangle =o(1)\), we conclude that \(\langle \mathcal E ^{\prime }_{\nu }(u),u\rangle =0\). By semi-continuity of norm, we have

$$\begin{aligned} \lim _{n\rightarrow \infty }\int \limits _{\mathbb{R }^{3}}(a|\nabla u_n|^2+\nu u^2_n)\ge \int \limits _{\mathbb{R }^{3}}(a|\nabla u|^2+\nu u^2). \end{aligned}$$
(4.7)

Observe that we must have the equality in (4.7). Otherwise, by Fatou’s lemma,

$$\begin{aligned} m_{\nu }&\le \mathcal E _{\nu }(u)=\mathcal E _{\nu }(u)-\frac{1}{4}\langle \mathcal E ^{\prime }_{\nu }(u),u\rangle \\&= \frac{1}{4}\int \limits _{\mathbb{R }^{3}}(a|\nabla u|^2+\nu u^2)+\lambda \int \limits _{\mathbb{R }^{3}}\left(\frac{1}{4}f(u)u-F(u)\right)+\frac{1}{12}\int \limits _{\mathbb{R }^{3}}u^6\\&< \liminf _{n\rightarrow \infty }\left\{ \frac{1}{4}\int \limits _{\mathbb{R }^{3}}(a|\nabla u_n|^2+\nu u^2_n) +\lambda \int \limits _{\mathbb{R }^{3}}\left(\frac{1}{4}f(u_n)u_n-F(u_n)\right)+\frac{1}{12}\int _{{\mathbb{R }^{3}}}u^6_n\right\} \\&= \liminf _{n\rightarrow \infty }\left(\mathcal E _{\nu }(u_n)-\frac{1}{4} \langle \mathcal E ^{\prime }_{\nu }(u_n),u_n\rangle \right)\\&\le m_{\nu }, \end{aligned}$$

which leads to a contradiction. Thus, up to subsequences, we conclude that \( \int _{{\mathbb{R }^{3}}}(a|\nabla u_n|^{2}\!+\!\nu u^2_n)\rightarrow \int _{{\mathbb{R }^{3}}}(a|\nabla u|^{2}+\nu u^2)\). Hence, \(u_n\rightarrow u\) in \(H^1({\mathbb{R }^{3}})\).

Case 2 \(u=0\).  In this case, since \(\{u_n\}\subset \mathcal N _{\nu }\), applying a similar argument as we have done in (2.4), we see that \(\Vert u_n\Vert _{\nu }\not \rightarrow 0\). So, from Lemma 2.8, there exist \(R,\eta >0\) and \(y_n\in {\mathbb{R }^{3}}\) such that \( \limsup _{n\rightarrow \infty }\int _{B_R(y_n)}u^2_n\ge \eta \). Let \(v_n(x)=u_n(x+\widetilde{y}_n),\) then \( \mathcal E _{\nu }(v_n)\rightarrow m_{\nu }~~\text{ and}~\mathcal E ^{\prime }_{\nu }(v_n)\rightarrow 0\). It is clear that \(\{v_n\}\) is bounded in \(H^1({\mathbb{R }^{3}}) \) and there exists \(v\in H^1({\mathbb{R }^{3}})\) with \(v\ne 0\) such that \( v_n\rightharpoonup v~~\text{ in}~H^1({\mathbb{R }^{3}})\). Then, the proof follows from the arguments used in Case 1. \(\square \)

Proposition 4.4

Let \(\varepsilon _n\rightarrow 0\) and \(\{u_n\}\subset \mathcal{N }_{\varepsilon _n}\) be such that \(\mathcal I _{\varepsilon _n}(u_n)\rightarrow m_{V_0}\). Then, there exists a sequence \( \{\widetilde{y_n}\}\subset {\mathbb{R }^{3}}\) such that \(v_n(x)=u_n(x+\widetilde{y_n})\) has a convergent subsequence in \(H^1({\mathbb{R }^{3}})\). Moreover, up to a subsequence, \(y_n:=\varepsilon _n\widetilde{y_n}\rightarrow y\in \Sigma \).

Proof

Note that \(m_{V_0}>0\), and since \(||u_n||_{\varepsilon _n}\rightarrow 0\) would imply \(\mathcal I _{\varepsilon _n}(u_n)\rightarrow 0\), arguing as in the proof of Lemma 2.8, we obtain a sequence \(\{\widetilde{y_n}\}\subset {\mathbb{R }^{3}}\) and constants \(R, \beta >0\) such that \( \liminf _{n\rightarrow \infty }\int _{B_R(\widetilde{y_n})}u^2_n\ge \beta >0\). Define \(v_n(x)=u_n(x+\widetilde{y_n}),\) then along a subsequence, we have \(v_n\rightharpoonup v\ne 0~~\text{ in}~H^1({\mathbb{R }^{3}})\).

Let \(t_n>0\) be such that \(\widetilde{v_n}:=t_nv_n\in \mathcal N _{V_0}\) and put \(y_n=\varepsilon _n \widetilde{{y_n}}\). Since \(u_n\in \mathcal{N }_{\varepsilon _n},\) we have

$$\begin{aligned} \mathcal E _{V_0}(\widetilde{v_n})&\le \frac{1}{2}\int \limits _{\mathbb{R }^{3}}\left(a|\nabla \widetilde{v_n}|^2)+ V(\varepsilon _n(x+\widetilde{y_n}))\widetilde{v_n}^2\right) +\frac{b}{4}\left(\int \limits _{\mathbb{R }^{3}}|\nabla \widetilde{v_n}|^2\right)^2\\&-\frac{1}{6}\int \limits _{\mathbb{R }^{3}}v^6_n -\lambda \int \limits _{\mathbb{R }^{3}}F(\widetilde{v_n})\\&= \mathcal I _{\varepsilon _n}(t_nu_n)\\&\le \mathcal I _{\varepsilon _n}(u_n)=m_{V_0}+o(1). \end{aligned}$$

Note that \(m_{V_0}\le \mathcal E _{V_0}(\widetilde{v_n})\); thus, \(\lim _{n\rightarrow \infty } \mathcal E _{V_0}(\widetilde{v_n})=m_{V_0}\).

We claim that, up to subsequence, \(t_n\rightarrow t^{\star }>0\). Indeed, since \(v_n(x)\not \rightarrow 0\) in \(H^1({\mathbb{R }^{3}})\), there exists \(\delta >0\) such that \(0<\delta \le ||v_n||\). Hence, \(0<\delta ^\star \le ||v_n||_{V_0}\) with \(\delta ^\star =\delta \min \{a,V_0\}^{1/2}\). Therefore, we have

$$\begin{aligned} 0\le t_n \delta ^\star \le ||t_nv_n||_{V_0}=||\widetilde{v_n}||_{V_0}\le C, \end{aligned}$$

from which \(\{t_n\}\) is bounded, and we can assume that \(t_n\rightarrow t^\star \ge 0\). If \(t^\star =0,\) we have \(\widetilde{v_n}=t_nv_n\rightarrow 0\) in \(H^1({\mathbb{R }^{3}})\) and \(\mathcal E _{V_0}(\widetilde{v_n})\rightarrow 0,\) which contradicts to \(m_{V_0}>0\). Thus, \(t^\star >0\) and the weak limit of \(\{\widetilde{v_n}\}\) is different from zero. Let \(\widetilde{v}\) be the weak limit of \(\{\widetilde{v_n}\}\) in \(H^1({\mathbb{R }^{3}})\). Since \(t_n\rightarrow t^\star >0\) and \(v_n\rightharpoonup v\not \equiv 0\), we have from the uniqueness of the weak limit that \(\widetilde{v}=t^\star v\not \equiv 0\). From Lemma 4.3, \(\widetilde{v_n}\rightarrow \widetilde{v}\) in \(H^1({\mathbb{R }^{3}}),\) and so, \(v_n\rightarrow v\) in \(H^1({\mathbb{R }^{3}})\). This proves the first part of the lemma.

We next show that \(\{y_n\}\) has a bounded subsequence. In fact, suppose by contradiction that \(|y_n|\rightarrow \infty \). We first consider the case \(V_{\infty }=\infty \). Note that the following inequality

$$\begin{aligned} \int \limits _{\mathbb{R }^{3}}V(\varepsilon _nx+y_n)v_n^2&\le \int \limits _{\mathbb{R }^{3}}(a|\nabla v_n|^2 +V(\varepsilon _nx+y_n)v_n^2)+ \frac{b}{4}\left(\int \limits _{\mathbb{R }^{3}}|\nabla v_n|^2\right)^2\\&= \int \limits _{\mathbb{R }^{3}}(\lambda f(v_n)v_n+v^6_n), \end{aligned}$$

together with Fatou’s lemma implies \( \infty =\liminf _{n\rightarrow \infty }\int _{{\mathbb{R }^{3}}}(\lambda f(v_n)v_n+v^6_n),\) which makes nonsense since the sequence \(\{\lambda f(v_n)v_n+v^6_n\}\) is bounded in \(L^1({\mathbb{R }^{3}})\). For the case \(V_{\infty }<\infty ,\) we have from \(\widetilde{v_n}\rightarrow \widetilde{v}\) in \(H^1({\mathbb{R }^{3}})\) and \(V_0<V_{\infty }\) that

$$\begin{aligned} m_{V_0}&= \mathcal E _{V_0}(\widetilde{v})<\mathcal E _{V_{\infty }}(\widetilde{v})\\&\le \liminf _{n\rightarrow \infty }\left\{ \frac{1}{2}\int \limits _{\mathbb{R }^{3}}(a|\nabla \widetilde{v_n}|^2+V(\varepsilon _nx+y_n)\widetilde{v_n}^2)\right.\\&\left.+\frac{b}{4}\left(\int \limits _{\mathbb{R }^{3}}|\nabla \widetilde{v_n}|^2\right)^2-\frac{1}{6}\int \limits _{\mathbb{R }^{3}}\widetilde{v_n}^6-\lambda \int \limits _{\mathbb{R }^{3}}F(\widetilde{v_n})\right\} \\&= \liminf _{n\rightarrow \infty }\left\{ \frac{t^2_n}{2}\int \limits _{\mathbb{R }^{3}}\left(a|\nabla u_n|^2+V(\varepsilon _nz)u^2_n\right)\right.\\&\left.+\frac{t^4_nb}{4}\left(\int \limits _{\mathbb{R }^{3}}|\nabla u_n|^2\right)^2-\frac{t_n^6}{6}\int \limits _{\mathbb{R }^{3}}{u_n}^6-\lambda \int \limits _{\mathbb{R }^{3}}F(t_nu_n)\right\} \\&= \liminf _{n\rightarrow \infty } \mathcal I _{\varepsilon _n}(t_nu_n) \le \liminf _{n\rightarrow \infty } \mathcal I _{\varepsilon _n}(u_n)=m_{V_0}, \end{aligned}$$

which is impossible. Therefore, \(\{y_n\}\) is bounded and up to a subsequence, \(y_n\rightarrow y\) in \({\mathbb{R }^{3}}\). If \(y\not \in \Sigma ,\) then \(V(y)>V_0\), and we obtain a contradiction by the same arguments made above. So, \(y\in \Sigma \) and the conclusion follows.

\(\square \)

Let \(h:~\mathbb{R }^+\rightarrow \mathbb{R }^+\) be any positive function satisfying \(h(\varepsilon )\rightarrow 0^+\) as \(\varepsilon \rightarrow 0^+\). Define the set

$$\begin{aligned} {\widetilde{\mathcal{N }}}_{{\widetilde{\varepsilon }}}:=\{u\in \mathcal{N }_{\varepsilon }:~\mathcal I _{\varepsilon }(u)\le m_{V_0}+h(\varepsilon )\}. \end{aligned}$$

Given \(y\in \Sigma ,\) we can use Lemma 4.1 to deduce that \(h(\varepsilon )=|\mathcal I _{\varepsilon }(\Phi _{\varepsilon }(y))-m_{V_0}| \rightarrow 0\) as \(\varepsilon \rightarrow 0^+\). Thus, \(\Phi _{\varepsilon }(y)\in \widetilde{\mathcal{N }_{\varepsilon }}\) and \(\widetilde{\mathcal{N }_{\varepsilon }}\ne \emptyset \) for any \(\varepsilon >0\).

Lemma 4.5

For any \(\delta >0,\) there holds that \(\lim _{\varepsilon \rightarrow 0}\sup \limits _{u\in \widetilde{\mathcal{N }_{\varepsilon }}} \text{ dist}(\beta _{\varepsilon }(u),\Sigma _{\delta })=0\)

Proof

Let \(\{\varepsilon _n\}\subset \mathbb{R }^+\) be such that \(\varepsilon _n\rightarrow 0\). By definition, there exists \(\{u_n\}\subset \widetilde{\mathcal{N }_{\varepsilon _n}}\) such that

$$\begin{aligned} \text{ dist}(\beta _{\varepsilon _n}(u_n),\Sigma _{\delta })= \sup _{u\in \widetilde{\mathcal{N }_{\varepsilon _n}}}\text{ dist}(\beta _{\varepsilon _n}(u),\Sigma _{\delta })+o(1). \end{aligned}$$

So, it suffices to find a sequence \(\{y_n\}\subset \Sigma _{\delta }\) satisfying \( |\beta _{\varepsilon _n}(u_n)-y_n|=o(1)\). From \(\mathcal E _{V_0}(tu_n)\le \mathcal I _{\varepsilon }(tu_n)\) for \(t\ge 0\) and \(\{u_n\}\subset \widetilde{\mathcal{N }_{\varepsilon _n}}\subset \mathcal{N }_{\varepsilon _n},\) we get \( m_{V_0}\le c_{\varepsilon _n}\le \mathcal I _{\varepsilon _n}(u_n)\le m_{V_0}+h(\varepsilon _n)\). This leads to \(\mathcal I _{\varepsilon }(u_n)\rightarrow m_{V_0}\). Thus, we can invoke Proposition 4.4 to obtain a sequence \(\{\widetilde{{y_n}}\}\subset {\mathbb{R }^{3}}\) such that \( y_n=\varepsilon _n \widetilde{{y_n}} \in \Sigma _{\delta }\) for \(n\) sufficiently large. Hence,

$$\begin{aligned} \beta _{\varepsilon _n}(u_n)=y_n+\frac{\int _{{\mathbb{R }^{3}}}(\chi (\varepsilon _nz+y_n)-y_n)u^2_n(z+ \widetilde{{y_n}})}{\int _{{\mathbb{R }^{3}}}u^2_n(z+\widetilde{{y_n}})}. \end{aligned}$$

Since \(\varepsilon _nz+y_n\rightarrow y\in \Sigma ,\) we have that \(\beta _{\varepsilon _n}(u_n)=y_n+o(1)\), and therefore, the sequence \(\{y_n\}\) is what we desire. \(\square \)

The following lemma plays a fundamental role in the study the behavior of the maximum points of the solutions. We sketch its proof by using some arguments explored in [1, 31], which are initiated in the Moser iterative method [35].

Lemma 4.6.

Assume that conditions \((V)\) and \((A_1)\text{-}(A_4)\) hold and let \(v_n\) be a solution of the following problem

$$\begin{aligned} \left\{ \begin{array}{l@{\quad }l@{\quad }l} -(a+b\int \nolimits _{{\mathbb{R }^{3}}}|\nabla v_n|^2)\Delta v_n+V_n(x)v_n=v_n^{2^*-1}+\lambda f(v_n)&\text{ in}&{\mathbb{R }^{3}},\\ v_n\in H^1({\mathbb{R }^{3}}),~ v_n>0&\text{ in}&{\mathbb{R }^{3}},\end{array} \right. \end{aligned}$$
(4.8)

where \(V_n(x):=V(\varepsilon _n(x+\widetilde{{y_n}}))\). If \(v_n\rightarrow v\) in \(H^1({\mathbb{R }^{3}})\) with \(v\ne 0,\) then \(v_n\in L^{\infty }({\mathbb{R }^{3}})\) and \(||v_n||_{L^{\infty }({\mathbb{R }^{3}})}\le C\) for all \(n\in \mathbb{N }\). Furthermore, \( \lim \limits _{|x|\rightarrow \infty }v_n(x)=0~~\text{ uniformly} \text{ in}~n\).

Proof

For any \(R>0, 0<r\le R/2,\) let \(\eta \in C^{\infty }({\mathbb{R }^{3}}), 0\le \eta \le 1\) with \(\eta (x)=1\) if \(|x|\ge R\) and \(\eta (x)=0\) if \(|x|\le R-r\) and \(|\nabla \eta |\le 2/r\). For each \(n\in \mathbb{N }\) and \(l>0\), put

$$\begin{aligned} v_{l,n}(x):=\left\{ \begin{array}{l@{\quad }l} v_n(x),&v_n(x)\le l,\\ l,&v_n(x)\ge l,\end{array}\right. \end{aligned}$$

and \( z_{l,n}:=\eta ^2v^{2(\beta -1)}_{l,n}v_n~~\text{ and}~w_{l,n}:=\eta v_nv^{\beta -1}_{l,n} \) with \(\beta >1\) to be determined later. Taking \(z_{l,n}\) as the test function, we have

$$\begin{aligned} 0=\int \limits _{\mathbb{R }^{3}}(a \nabla v_n\nabla z_{l,n}+V_nv_nz_{l,n}) +b\int \limits _{\mathbb{R }^{3}}|\nabla v_n|^2\int \limits _{\mathbb{R }^{3}}\nabla v_n\nabla z_{l,n}-\int \limits _{\mathbb{R }^{3}}(v_n^5+\lambda f(v_n))z_{l,n}, \end{aligned}$$

that is,

$$\begin{aligned}&\left(a+b\int \limits _{\mathbb{R }^{3}}|\nabla v_n|^2\right)\int \limits _{\mathbb{R }^{3}}\left(|\nabla v_n|^2\eta ^2v_{l,n}^{2(\beta -1)}+2\nabla v_n\nabla \eta \eta v_nv_{l,n}^{2(\beta -1)}\right.\\&\qquad \left. \,{+}\, 2(\beta -1)\nabla v_n\nabla v_{l,n}v_nv_{l,n}^{2\beta -3}\eta ^2\right) +\int \limits _{\mathbb{R }^{3}}V_nv_n^2\eta ^2 v_{l,n}^{2(\beta -1)}\\&\quad =\int \limits _{\mathbb{R }^{3}}v_n^6\eta ^2v_{l,n}^{2(\beta -1)}+\lambda \int \limits _{\mathbb{R }^{3}}f(v_n)\eta ^2v_nv_{l,n}^{2(\beta -1)}. \end{aligned}$$

For simplicity, we denote by \(L_n:=a+b\int _{{\mathbb{R }^{3}}}|\nabla v_n|^2\). Since \(v_n\rightarrow v\) in \(H^1({\mathbb{R }^{3}})\) with \(v\ne 0,\) we see that \(a\le L_n\le a^\star \) for some constant \(a^\star >0\). Thus, we can rewrite the above equality as

$$\begin{aligned} L_n\int \limits _{\mathbb{R }^{3}}\eta ^2v^{2(\beta -1)}_{l,n}|\nabla v_n|^2&= \lambda \int \limits _{\mathbb{R }^{3}}f(v_n)\eta ^2v_n v^{2(\beta -1)}_{l,n} -2L_n\int \limits _{\mathbb{R }^{3}}\eta v^{2(\beta -1)}_{l,n}v_n\nabla v_n\nabla \eta \\&-2L_n(\beta -1) \int \limits _{\mathbb{R }^{3}}\eta ^2v^{2\beta -3}_{l,n}v_n\nabla v_n\nabla v_{l,n}\\&-\int \limits _{\mathbb{R }^{3}}V_n(x)v_n^2\eta ^2v^{2(\beta -1)}_{l,n}+\int \limits _{\mathbb{R }^{3}}v_n^6\eta ^2v_{l,n}^{2(\beta -1)}. \end{aligned}$$

By \((A_1)\text{--}(A_2)\), we have that for any \(\tau >0,\) there exists \(D_{\tau }>0\) such that \( f(t)\le \tau t+D_{\tau }t^5 \; \text{ for} \text{ all}~ t\ge 0\). Since \(\tau \) can be small sufficiently, we have the following inequality:

$$\begin{aligned} L_n\int \limits _{\mathbb{R }^{3}}\eta ^2v^{2(\beta -1)}_{l,n}|\nabla v_n|^2\le D_{\tau }\int \limits _{\mathbb{R }^{3}}v^6_n\eta ^2v^{2(\beta -1)}_{l,n}-2L_n\int \limits _{\mathbb{R }^{3}}\eta v^{2(\beta -1)}_{l,n}v_n\nabla v_n\nabla \eta . \end{aligned}$$

For each \(\zeta >0,\) using Young’s inequality, we get

$$\begin{aligned} L_n\int \limits _{\mathbb{R }^{3}}\eta ^2v^{2(\beta -1)}_{l,n}|\nabla v_n|^2&\le D_{\tau }\int \limits _{\mathbb{R }^{3}}v^6_n\eta ^2v^{2(\beta -1)}_{l,n} +2L_n\zeta \int \limits _{\mathbb{R }^{3}}\eta ^2v^{2(\beta -1)}_{l,n}|\nabla v_n|^2\\&\,{+}\, 2L_nC_{\zeta }\int \limits _{\mathbb{R }^{3}}v^2_n|\nabla \eta |^2v^{2(\beta -1)}_{l,n}. \end{aligned}$$

Choosing \(\zeta \in (0,1/2)\), we have

$$\begin{aligned} \int \limits _{\mathbb{R }^{3}}\eta ^2v^{2(\beta -1)}_{l,n}|\nabla v_n|^2\le C\int \limits _{\mathbb{R }^{3}}v^6_n\eta ^2v^{2(\beta -1)}_{l,n}+ C\int \limits _{\mathbb{R }^{3}}|\nabla \eta |^2v^2_nv^{2(\beta -1)}_{l,n}. \end{aligned}$$
(4.9)

On the other hand, by Sobolev inequality and Hölder inequality, we have

$$\begin{aligned} \Vert w_{l,n}\Vert ^2_6&\le C \int \limits _{\mathbb{R }^{3}}|\nabla w_{l,n}|^2=C \int \limits _{\mathbb{R }^{3}}|\nabla (\eta v^{\beta -1}_{l,n}v_{n})|^2\nonumber \\&\le C \beta ^2\left(\int \limits _{\mathbb{R }^{3}}|\nabla \eta |^2v^2_nv^{2(\beta -1)}_{l,n}+\int \limits _{\mathbb{R }^{3}}\eta ^2v^{2(\beta -1)}_{l,n}|\nabla v_n|^2\right). \end{aligned}$$
(4.10)

Combining (4.9)–(4.10), we have

$$\begin{aligned} \Vert w_{l,n}\Vert ^2_6\le C \beta ^2\left(\int \limits _{\mathbb{R }^{3}}|\nabla \eta |^2v^2_nv^{2(\beta -1)}_{l,n}+\int \limits _{\mathbb{R }^{3}}v^6_n\eta ^2v^{2(\beta -1)}_{l,n} \right). \end{aligned}$$
(4.11)

Taking \(\beta =\frac{2^*}{2}=3,\) we have from (4.11) that

$$\begin{aligned} \Vert w_{l,n}\Vert ^2_6&\le C\beta ^2\left(\int \limits _{\mathbb{R }^{3}}v^2_n|\nabla \eta |^2v^4_{l,n}+\int \limits _{\mathbb{R }^{3}}v^6_n\eta ^2v^4_{l,n}\right)\\&\le C\beta ^2\int \limits _{\mathbb{R }^{3}}v^2_n|\nabla \eta |^2v^4_{l,n}+C\beta ^2 \left(\int \limits _{\mathbb{R }^{3}}(v_n\eta v^2_{l,n})^6\right)^{\frac{1}{3}}\left(\int \limits _{|x|>R/2}v^6_n\right)^{\frac{2}{3}}. \end{aligned}$$

By the definition of \(w_{l,n}\), we can rewrite the last inequality as

$$\begin{aligned} \left(\int \limits _{\mathbb{R }^{3}}(\eta v_nv^{2}_{l,n})^6\right)^{\frac{1}{3}}&\le C\beta ^2\int \limits _{\mathbb{R }^{3}}v^2_n|\nabla \eta |^2v^4_{l,n}\\&+\,C\beta ^2 \left(\int \limits _{\mathbb{R }^{3}}(v_n\eta v^2_{l,n})^6\right)^{\frac{1}{3}}\left(\int \limits _{|x|>R/2}v^6_n\right)^{\frac{2}{3}}. \end{aligned}$$

Notice that \(v_n\rightarrow v\) in \(H^1({\mathbb{R }^{3}}),\) for \(R\) large enough, we conclude that \( \int _{|x|\ge \frac{R}{2}}v^6_n\le \tau ~\text{ uniformly} \text{ in}~n\). Thus,

$$\begin{aligned} \left(\int \limits _{|x|\ge R}(v_nv^2_{l,n})^6\right)^{\frac{1}{3}}&\le \left(\int \limits _{|x|\ge R-r}(\eta v_nv^2_{l,n})^6\right)^{\frac{1}{3}}\\&\le C\beta ^2\int \limits _{\mathbb{R }^{3}}v^2_n|\nabla \eta |^2v^4_{l,n}\\&\le \frac{9C}{r^2}\int \limits _{\mathbb{R }^{3}}v^6_n\le \frac{C}{r^2}. \end{aligned}$$

Aplying Fatou’s lemma in the variable \(l\), we have

$$\begin{aligned} \int \limits _{|x|\ge R}v_n^{18}<\int \limits _{|x|\ge R-r}\eta ^6 v_n^{18}<\frac{C}{r^6}<\infty . \end{aligned}$$
(4.12)

This implies that \(v_n\in L^{18}(|x|\ge R)\) for \(R\) large enough.

Next, we note that if \(\beta =\frac{3(t-1)}{t}\) with \(t=\frac{9}{2},\) then \(\beta >1, \frac{2t}{t-1}<6\). Now suppose that \(t=\frac{9}{2}\), and \(v_n\in L^{2\beta t/(t-1)}(|x|\ge R-r)\) for some \(\beta \ge 1\). Returning to (4.11) and using the Hölder inequality with exponent \(\frac{t}{t-1}\) and \(t\), we have

$$\begin{aligned} \Vert w_{l,n}\Vert ^2_6&\le C\beta ^2\left(\int \limits _{|x|\ge R-r} v^{2\beta }_n|\nabla \eta |^2+\int \limits _{|x|\ge R-r}\eta ^2v^4_nv^{2\beta }_n\right)\\&\le C\beta ^2\left\{ \left(\int \limits _{|x|\ge R-r} v_n^{\frac{2\beta t}{t-1}}\right)^{\frac{t-1}{t}} \left(\int \limits _{R-r\le |x|\le R}|\nabla \eta |^{2t}\right)^{\frac{1}{t}}\right.\\&\left.+\left(\int \limits _{|x|\ge R-r}(\eta ^2v^4_n)^t\right)^{\frac{1}{t}}\left(\int \limits _{|x|\ge R-r}v^{\frac{2\beta t}{t-1}}_n\right)^{\frac{t-1}{t}}\right\} \\&\le C\beta ^2\left\{ \frac{[R^3-(R-r)^3]^{\frac{1}{t}}}{r^2}\left(\int \limits _{|x|\ge R-r} v_n^{\frac{2\beta t}{t-1}}\right)^{\frac{t-1}{t}} \right.\\&\left.+\left(\int \limits _{|x|\ge R-r}(\eta ^{4/3}v^{4}_n)^t\right)^{\frac{1}{t}}\left(\int \limits _{|x|\ge R-r}v^{\frac{2\beta t}{t-1}}_n\right)^{\frac{t-1}{t}}\right\} . \end{aligned}$$

By (4.12), we deduce that

$$\begin{aligned} \Vert w_{l,n}\Vert ^2_6\le C\beta ^2\left(\frac{1}{r^{6/t}}+\frac{R^{3/t}}{r^2}\right) \left(\int \limits _{|x|\ge R-r}v_n^{\frac{2\beta t}{t-1}}\right)^{\frac{t-1}{t}}. \end{aligned}$$

Using Fatou’s lemma, we obtain

$$\begin{aligned} \Vert v_{n}\Vert ^{2\beta }_{6\beta ,(|x|\ge R)} \le C\beta ^2\left(\frac{1}{r^{6/t}}+\frac{R^{3/t}}{r^2}\right) \Vert v_n\Vert _{\frac{2\beta t}{t-1},(|x|>R-r)}^{2\beta }. \end{aligned}$$

If we set \(\psi =3(t-1)/t,~ s=2t/(t-1),\) then

$$\begin{aligned} \Vert v_n\Vert _{\beta \psi s,(|x|\ge R)}\le C^{1/\beta }\beta ^{1/\beta }\left(\frac{1}{r^{6/t}}+\frac{R^{3/t}}{r^2}\right)^{1/2\beta }\Vert v_n\Vert _{\beta s,(|x|\ge R-r)}. \end{aligned}$$
(4.13)

Put \(\beta =\psi ^m~(m=1,2,\ldots ),\) we obtain

$$\begin{aligned} \Vert v_n\Vert _{\psi ^{m+1} s,(|x|\ge R)}\le C^{\psi ^{-m}}\psi ^{m\psi ^{-m}}\left(\frac{1}{r^{6/t}}+\frac{R^{3/t}}{r^2}\right)^{\frac{1}{2\psi ^m}}\Vert v_n\Vert _{\psi ^m s,(|x|\ge R-r)}. \end{aligned}$$

Note that \(2>6/t>3/t=2/3\). Therefore, if we take \(r_m:=2^{-(m+1)}R,\) then it follows from the last inequality and (4.13) that

$$\begin{aligned}&\Vert v_n\Vert _{\psi ^{m+1}s,(|x|\ge R)}\\&\quad \le \Vert v_n\Vert _{\psi ^{m+1}s,(|x|\ge R-r_{m+1})}\\&\quad \le C^{\psi ^{-m}}\psi ^{m\chi ^{-m}}\left(\frac{2^{\frac{4(m+2)}{3}}}{R^{4/3}}+2^{2(m+2)} \right)^{\frac{1}{2\psi ^m}}\Vert v_n\Vert _{\psi ^m s,(|x|\ge R-r_m)}\\&\quad \le C^{\psi ^{-m}}\psi ^{m\chi ^{-m}}\left(2\times 2^{2(m+2)} \right)^{\frac{1}{2\psi ^m}}\Vert v_n\Vert _{\psi ^m s,(|x|\ge R-r_m)}\\&\quad \le C^{\psi ^{-m}+\psi ^{-(m-1)}}\psi ^{m\chi ^{-m}+(m-1)\chi ^{-(m-1)}}\times \left(2\times 2^{2(m+2)} \right)^{\frac{1}{2\psi ^m}}\left(2\times 2^{2(m+1)} \right)^{\frac{1}{2\psi ^{m-1}}}\\&\qquad \times \Vert v_n\Vert _{\psi ^{m-1} s,(|x|\ge R-r_{m-1})}\\&\quad \vdots \\&\quad \le C^{\Sigma _{i=1}^{m}\psi ^{-i}}\psi ^{\Sigma _{i=1}^{m}i\psi ^{-i}}\exp \left(\sum _{i=1}^{m}\frac{\ln (2\times 2^{2(i+2)})}{2\psi ^i}\right)\Vert v_n\Vert _{\psi s,(|x|\ge R-r_{1})}\\&\quad \le C\Vert v_n\Vert _{6,(|x|\ge R/2)}. \end{aligned}$$

Letting \(m\rightarrow \infty \) in the last inequality, we get

$$\begin{aligned} ||v_n||_{L^{\infty }(|x|\ge R)}\le C \Vert v_n\Vert _{6,(|x|\ge R/2)}. \end{aligned}$$

Again by the convergence of \(\{v_n\}\) to \(v\) in \(H^1({\mathbb{R }^{3}})\), for any \(\tau >0\) fixed, there exists an \(R>0\) such that \(||v_n||_{L^{\infty }(|x|\ge R)}\le \tau ~\text{ for} \text{ all}~n\in \mathbb{N }\). Hence, \(\lim _{|x|\rightarrow \infty } v_n(x)=0\) uniformly in \(n\) and the conclusion follows. \(\square \)

4.2 Proof of Theorem 4.1

We shall divide the proof into two parts.

Part 1: Multiplicity of positive solutions.  For any fixed \(\delta >0,\) we can invoke Lemmas 4.2 and 4.5 to obtain some \(\varepsilon _{\delta }>0\) such that, for any \(\varepsilon \in (0,\varepsilon _{\delta }),\) the diagram \( \Sigma \stackrel{\Phi _{\varepsilon }}{\longrightarrow }\widetilde{ \mathcal{N }_{\varepsilon }}\stackrel{\beta _{\varepsilon }}{\longrightarrow }\Sigma _{\delta }\) is well defined. By the arguments in the paragraph just before Lemma 4.3, we see that

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0}\beta _{\varepsilon }(\Phi _{\varepsilon }(y))=y~~\text{ uniformly} \text{ in}\quad y\in \Sigma . \end{aligned}$$

For \(\varepsilon \) small enough, we denote \( \beta _{\varepsilon }(\Phi _{\varepsilon }(y)):= y+\theta (y)\) for \(y\in \Sigma ,\) where \(\Vert \theta (y)\Vert <\delta /2\) uniformly for \(y\in \Sigma \). Define \(\mathcal H (t, y)=y+(1-t)\theta (y)\). Then, \(\mathcal H : [0, 1]\times \Sigma \rightarrow \Sigma _\delta \) is continuous. Obviously, \(\mathcal H (0, y)= \beta _{\varepsilon }(\Phi _{\varepsilon }(y)), \mathcal H (1, y)=y\) for all \(y\in \Sigma \). That is, the composite map \(\beta _{\varepsilon }\circ \Phi _{\varepsilon }\) is homotopic to the inclusion map \(Id:~ \Sigma \rightarrow \Sigma _{\delta } \) (similar conclusion can be observed in [1] and [21]). Using homotopy and by the same arguments contained in the proof of Lemmas 4.2 and 4.3 of [11], we obtain that

$$\begin{aligned} \text{ cat}_{\widetilde{ \mathcal{N }_{\varepsilon }}} (\widetilde{ \mathcal{N }_{\varepsilon }})\ge \text{ cat}_{\Sigma _{\delta }}(\Sigma ). \end{aligned}$$

On the other hand, using the definition of \(\widetilde{ \mathcal{N }_{\varepsilon }}\) and choosing \(\varepsilon _{\delta }\) small if necessary, we see that \(\mathcal I _{\varepsilon }\) satisfies the Palais–Smale condition in \(\widetilde{ \mathcal{N }_{\varepsilon }}\) for \(\lambda >\lambda ^*\) by recalling Proposition 4.4, Lemma 4.3, Proposition 2.12 and Lemma 2.6. Therefore, standard Ljusternik–Schnirelmann theory provides at least \(cat_{\widetilde{ \mathcal{N }_{\varepsilon }}}(\widetilde{\mathcal{N }_{\varepsilon }})\) critical points of \(\mathcal I _{\varepsilon }\) restricted to \( \mathcal{N }_{\varepsilon }\). Consequently, using Corollary 2.13, we conclude that \(\mathcal I _{\varepsilon }\) has at least \(\text{ cat}_{\Sigma _{\delta }}(\Sigma )\) critical points in \(W_{\varepsilon }\).

Part 2: Concentration of the maximum points. Let \(u_{\varepsilon _n}\) be a solution of \((\mathcal K )_{\varepsilon _n}\); then, \(v_n(x)=u_{\varepsilon _n}(x+ \widetilde{{y_n}})\) is a solution of the problem

$$\begin{aligned} \left\{ \begin{array}{l@{\quad }l@{\quad }l} -(a+b\int _{{\mathbb{R }^{3}}}|\nabla v_n|^2)\triangle v_n+V_n(x)v_n=v_n^{2^*-1}+\lambda f(v_n)&\text{ in}&{\mathbb{R }^{3}},\\ v _n\in H^1({\mathbb{R }^{3}}),~ v_n>0&\text{ in}&{\mathbb{R }^{3}},\end{array} \right. \end{aligned}$$

with \(V_n(x)=V(\varepsilon _nx+\varepsilon _n \widetilde{{y_n}})\) and \(\{\widetilde{ {y_n}}\}\subset {\mathbb{R }^{3}}\) given in Proposition 4.4. Furthermore, up to a subsequence, \(v_n\rightarrow v\) in \(H^1({\mathbb{R }^{3}})\) and \(y_n\rightarrow y\) in \(\Sigma \) with \(y_n=\varepsilon _n \widetilde{{y_n}}\). We claim that there exists a \(\rho >0\) such that \(||v_n||_{L^{\infty }({\mathbb{R }^{3}})}\ge \rho , \forall n\in \mathbb{N }\). In fact, suppose that \(||v_n||_{L^{\infty }({\mathbb{R }^{3}})}\rightarrow 0\). Let \(\varepsilon _0= \frac{V_0}{2},\) it follows from \((A_1)\) that there exists an \(n_0\in \mathbb{N }\) such that

$$\begin{aligned} \lambda \frac{f(||v_n||_{L^{\infty }({\mathbb{R }^{3}})})}{||v_n||_{L^{\infty }({\mathbb{R }^{3}})}^{3}}<\frac{V_0}{2}~~\text{ for}~~n\ge n_0. \end{aligned}$$

Hence, by \((A_4)\), we conclude that

$$\begin{aligned}&\int \limits _{\mathbb{R }^{3}}(a|\nabla v_n|^2+V_0v^2_n)+b\left(\int \limits _{\mathbb{R }^{3}}|v_n|^2\right)^2\\&\quad \le \lambda \int \limits _{\mathbb{R }^{3}}\frac{f(v_n)}{v^3_n}v^4_n+\int \limits _{\mathbb{R }^{3}}v^6_n\\&\quad \le \lambda \int \limits _{\mathbb{R }^{3}}\frac{f(||v_n||_{L^{\infty }({\mathbb{R }^{3}})})}{||v_n||_{L^{\infty }({\mathbb{R }^{3}})}^{3}}v^4_n+\int \limits _{\mathbb{R }^{3}}v^6_n\\&\quad \le \frac{V_0}{2}||v_n||^2_{L^{\infty }({\mathbb{R }^{3}})}\int \limits _{\mathbb{R }^{3}}v^2_n+||v_n||_{L^{\infty }}^4\int \limits _{\mathbb{R }^{3}}v_n^2. \end{aligned}$$

This implies that \(||v_n||_{V_0}=0\) for \(n\ge n_0\), which is impossible because \(v_n\rightarrow v\) in \(H^1({\mathbb{R }^{3}})\) and \(v\ne 0\). Then, the claim is true. Considering \(p_n\) the global maximum of \(v_n\), by Lemma 4.6 and the claim above, we see that \(p_n\in B_R(0)\) for some \(R>0\). Thus, the global maximum of \(u_{\varepsilon _n}\) given by \(z_{\varepsilon _n}=p_n+ \widetilde{{y_n}}\) satisfies \( \varepsilon _nz_{\varepsilon _n}=\varepsilon _n p_n+y_n\). Since \(\{p_n\}\) is bounded, it follows that \(\varepsilon _nz_{\varepsilon _n}\rightarrow y\), thus \( \lim _{n\rightarrow \infty }V(\varepsilon _nz_{\varepsilon _n})=V_0\). \(\Box \)

4.3 Proof of Theorem 1.1

If \(u_{\varepsilon }\) is a positive solution of \((\mathcal K )_{\varepsilon },\) the function \(w_{\varepsilon }(x)=u_{\varepsilon }(x/\varepsilon )\) is a positive solution of \((\mathcal SK )_{\varepsilon }\). Thus, the maximum point \(\eta _{\varepsilon }\) and \(z_{\varepsilon }\) of \(w_{\varepsilon }\) and \(u_{\varepsilon }\), respectively, satisfy the equality \(\eta _{\varepsilon }=\varepsilon z_{\varepsilon }\). Hence, \( \lim _{\varepsilon \rightarrow 0}V(\eta _{\varepsilon })=\lim _{n\rightarrow \infty }V(\varepsilon _n z_{\varepsilon _n})=V_0\). Consequently, Theorem 1.1 follows from Theorem 4.1 \(\square \)