\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2017 (2017), No. 19, pp. 1--15.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu} \thanks{\copyright 2017 Texas State University.} \vspace{8mm}} \begin{document} \title[\hfilneg EJDE-2017/19\hfil Perturbed biharmonic equations] {Semiclassical solutions of perturbed biharmonic equations with critical nonlinearity} \author[Y. He, X. Tang, W. Zhang \hfil EJDE-2017/19\hfilneg] {Yubo He, Xianhua Tang, Wen Zhang} \address{Yubo He \newline School of Mathematics and Statistics, Central South University, Changsha, 410083 Hunan, China. \newline Department of Mathematics, Huaihua College, Huaihua, 418008 Hunan, China} \email{hybmath@163.com} \address{Xianhua Tang \newline School of Mathematics and Statistics, Central South University, Changsha, 410083 Hunan, China} \email{tangxh@mail.csu.edu.cn} \address{Wen Zhang (corresponding author)\newline School of Mathematics and Statistics, Hunan University of Commerce, Changsha, 410205 Hunan, China} \email{zwmath2011@163.com} \dedicatory{Communicated by Claudianor O. Alves} \thanks{Submitted November 15, 2016. Published January 16, 2017.} \subjclass[2010]{35J35, 35J60, 58E05, 58E50} \keywords{Perturbed biharmonic equation; semiclassical solution; \hfill\break\indent critical nonlinearity} \begin{abstract} We consider the perturbed biharmonic equations \[ \varepsilon^4\Delta^2u+V(x)u=f(x,u),\quad x\in\mathbb{R}^N \] and \[ \varepsilon^4\Delta^2u+V(x)u=Q(x)|u|^{2^{\ast\ast}-2}u+f(x,u), \quad x\in\mathbb{R}^N \] where $\Delta^2$ is the biharmonic operator, $N\geq 5$, $2^{\ast\ast}=\frac{2N}{N-4}$ is the Sobolev critical exponent, $Q(x)$ is a bounded positive function. Under some mild conditions on $V$ and $f$, we show that the above equations have at least one nontrivial solution provided that $\varepsilon \leq \varepsilon_0$, where the bound $\varepsilon_0$ is formulated in terms of $N, V, Q$ and $f$. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \allowdisplaybreaks \section{Introduction} We study the perturbed biharmonic equations with subcritical nonlinearity \begin{equation}\label{1.1} \begin{gathered} \varepsilon^4\Delta^2u+V(x)u=f(x,u), \quad x\in \mathbb{R}^N,\\ u\in H^2(\mathbb{R}^N),\quad u(x)\to 0, \quad \text{as }|x|\to \infty, \end{gathered} \end{equation} and with critical nonlinearity \begin{equation}\label{1.2} \begin{gathered} \varepsilon^4\Delta^2u+V(x)u=Q(x)|u|^{2^{\ast\ast}-2}u+f(x,u), \quad x\in \mathbb{R}^N,\\ u\in H^2(\mathbb{R}^N),\quad u(x)\to 0,\quad \text{as }|x|\to \infty, \end{gathered} \end{equation} where $\varepsilon>0$ is small, $\Delta^2$ is the biharmonic operator, $N\geq5$, $2^{\ast\ast}=2N/(N-4)$ denotes the Sobolev critical exponent; $V, Q: \mathbb{R}^N\to \mathbb{R}\in C(\mathbb{R}^N, \mathbb{R})$. In this paper, we are interested in the existence of semiclassical solutions for the above equations. When $\Omega$ is a bounded domain of $\mathbb{R}^N$, the problem \begin{gather*} \Delta^2 u +c\Delta u =f(x,u)\quad \text{in } \Omega,\\ u=\Delta u=0 \quad\text{on } \partial\Omega, \end{gather*} has been extensively investigated in recent years. This problem arises in the study of traveling waves in suspension bridges (see \cite{CM,LM,MW}) and the study of the static deflection of an elastic plate in a fluid. For results on multiple nontrivial and sign changing solutions of problem \eqref{1.2} we refer the readers to \cite{AN, AN1, AE,FW,PWT,Tang, WZZ,W,YZ, Z,ZL,ZTCZ, ZW1, ZW2} and the references therein. Problems in the whole space $\mathbb{R}^N$ have been considered in several works; see for example \cite{CM1,CBT,LCW,NSY,PS,W2,WS,YS,YT1,YT2,YW,ZTZ,ZTZ1,ZTZ2,ZTZ3}. To our knowledge, there are only two papers \cite{PS,W2} on the singularly perturbation problem. In \cite{PS}, the authors dealt with the autonomous problem \begin{gather*} \varepsilon^4\Delta^2 u +V(x) u =f(u)\quad \text{in }\mathbb{R}^N,\\ u\in H^2(\mathbb{R}^N), \end{gather*} where $\varepsilon>0$, $N\geq5$, and $V:\mathbb{R}^N\to \mathbb{R}$ is such that there exists a bounded domain $\Omega\subset\mathbb{R}^N$ and $x_0\in\Omega$ with $00$ such that $\mathcal{V}_{b}:=\{x\in \mathbb{R}^N: V(x)0$ and $c>0$ such that \[ |f(x, t)| \leq c(1+|t|^{p_0-1}), \quad \forall (x, t)\in \mathbb{R}^N\times \mathbb{R}; \] \item[(A4)] $f(x, t)=o(|t|)$, as $|t|\to 0$ uniformly in $x$. \item[(A5)] There exist $c_0>0$ and $p>2$ such that $F(x, t) \geq c_0|t|^{p}$ for all $(x,t)$. \item[(A6)] There exists $2<\mu <2^{\ast\ast}$ such that \[ \mu F(x, t) \leq f(x, t)t\quad \text{for all $(x, t)$, where } F(x, t)=\int_0^{t}f(x, s)ds \] \end{itemize} It is worth pointing out that a crucial technique from \cite{W2} is used in the process of proof: for any $(PS)_{c}$ sequence $\{u_{n}\}$ for $I_{\lambda}$ with $u_{n}\rightharpoonup u$, where $\lambda=\varepsilon^{-2}$ and \begin{equation*} I_{\lambda}(u)=\frac{1}{2}\int_{\mathbb{R}^N}(|\Delta u|^2+\lambda|u|^2)dx -\frac{\lambda}{2^{\ast\ast}}\int_{\mathbb{R}^N}Q(x)|u|^{2^{\ast\ast}}dx -\lambda\int_{\mathbb{R}^N}F(x, u)dx, \end{equation*} the author constructed a new sequence $\{v_{n}\}$ such that $I_{\lambda}$, $I'_{\lambda}$ satisfy BL-splits, i.e., \begin{equation*} I_{\lambda}(v_{n})\to c-I_{\lambda}(u),\quad I'_{\lambda}(v_{n})\to0. \end{equation*} With the aid of this property, the author showed that $I_{\lambda}$ satisfies the $(PS)$-condition at the levels less than $\alpha_0\lambda^{1-\frac{N}{4}}$ with some $\alpha_0>0$ independent of $\lambda$. Based on such arguments, there are many works devote to semilinear Schr\"{o}inger equations, to quasilinear Schr\"{o}inger equations and elliptic system, we refer readers to \cite{CST,DS,DW,YD,YSD,YW1,ZZX} and the references therein. Inspired by \cite{W2,LT}, we consider problems \eqref{1.1} and \eqref{1.2}. The main ingredients of our work are two aspects. On the one hand, our aim is to weaken the above conditions to generalize and improve the result in \cite{W2}; on the other hand, we will develop a more direct and simpler approach. The novel approach not only makes such an extension possible but also lead to some better results. For example, it enable us to give an explicit upper bound for the parameter $\varepsilon$. To state our results, we make the following assumptions which are considerably weaker than the ones in the previous work: \begin{itemize} \item[(A7)] $F(x,t)\geq0$ and $\lim_{t\to \infty}|F(x,t)|/|t|^2=\infty$ uniformly in $x$, and there exist $a_0>0, T_0>0$ and $q\in (2, 2^{\ast\ast})$ such that \begin{gather*} F(x, t) \geq a_0|t|^{q}, \quad \forall (x, t)\in \mathbb{R}^N\times [-T_0, T_0], \\ t^{-2}h^{6-N}\int_{|x|\leq h}F(\lambda ^{-1/4}x, t/h)dx \geq \frac{(4N^2+2)S_{N}}{N(1-2^{-N})^2}, \quad \forall h\geq 1,\; \lambda\geq 1,\; t\geq hT_0, \end{gather*} where and in the sequel, $S_{N}=\operatorname{meas}(B_{1}(0)) =\frac{2\pi N/2}{N\Gamma (N/2)}$; \item[(A8)] $\mathcal{F}(x, t):=\frac{1}{2}tf(x, t)-F(x, t)\geq 0$ for all $(x, t)\in \mathbb{R}^N\times \mathbb{R}$, and there exist $R_0>0, a_{1}>0$ and $\kappa>\max\{1, \frac{N}{4}\}$ such that \begin{gather*} tf(x, t)\le \frac{b}{3}|t|^2, \quad \forall (x, t)\in \mathbb{R}^N\times \mathbb{R}, \; |t|\le R_0, \\ |f(x, t)|^{\kappa}\le a_1|t|^{\kappa}\mathcal{F}(x, t), \quad \forall (x, t)\in \mathbb{R}^N\times \mathbb{R}, \; |t|\ge R_0; \end{gather*} \item[(A9)] $tf(x, t)\ge 2F(x, t)$ for all $(x, t)\in \mathbb{R}^N\times \mathbb{R}$ and there exist $a_*>0$, $T_{1}>0$ and $q\in (2, 2^{\ast\ast})$ such that \[ \frac{1}{2^{\ast\ast}}Q(x)|t|^{2^{\ast\ast}}+F(x, t)\ge a_*|t|^q, \quad \text{for } (x, t)\in \mathbb{R}^N\times [-T_1, T_1]. \] \end{itemize} In light of (A3)--(A4), there exist $R_*>0$ and $a_2>0$ such that \begin{gather}\label{1.3} Q(x)|t|^{2^{\ast\ast}}+tf(x, t)\le \frac{b}{3}|t|^2, \quad \forall (x, t)\in \mathbb{R}^N\times \mathbb{R}, \; |t|\le R_*, \\ \label{1.4} tf(x, t)\le a_2Q(x)|t|^{2^{\ast\ast}}, \quad \forall (x, t)\in \mathbb{R}^N\times \mathbb{R}, \; |t|\ge R_*. \end{gather} \begin{remark} \label{rmk1.1}\rm It is easy to check that the conditions (A7), (A8) and (A9) are weaker than (A5) and (A6). It is well known that many nonlinearities such as \begin{equation}\label{1.01} f(x,t)=t\ln(1+|t|), \end{equation} do not satisfy (A6). A crucial role that (A6) plays is to ensure the boundedness of Palais-Smale sequences. \end{remark} Now we only show that $f(x,t)$ satisfies (A7) and (A8). Indeed, by a straightforward computation, \begin{gather*} F(x,t)=\frac{t^2-1}{2}\ln(1+|t|)+\frac{1}{4}|t|(2-|t|), \\ \mathcal{F}(x,t)=\frac{1}{2}tf(x,t)-F(x,t) =\frac{1}{2}\ln(1+|t|)+\frac{1}{4}|t|(|t|-2). \end{gather*} Observe that, letting $h\geq1$, $t\geq hT_0$ for some $T_0\geq2$, we have \begin{align*} &t^{-2}h^{6-N}\int_{|x|\leq h} F(\lambda^{-1/4}x,t/h)dx\\ &=t^{-2}h^{6-N}\int_{|x|\leq h}\big[\frac{(\frac{t}{h})^2-1}{2} \ln\big(1+\frac{t}{h}\big)+\frac{\frac{t}{h}(2-\frac{t}{h})}{4}\big]dx\\ &=\frac{1}{N}S_{N}h^Nt^{-2}h^{6-N} \big[\frac{t^2-h^2}{2h^2}\ln\big(1+\frac{t}{h}\big)+\frac{t(2h-t)}{4h^2}\big]\\ &=\frac{1}{2N}S_{N}h^4\big[\big(1-(\frac{h}{t})^2\big) \ln\big(1+\frac{t}{h}\big)+\frac{1}{2}\big(\frac{2h}{t}-1\big)\big]\\ &\geq \frac{1}{2N}S_{N}\big[(1-T_0^{-2})\ln(1+T_0)+\frac{1}{T_0}-\frac{1}{2} \big]\\ &\geq \frac{1}{2N}S_{N}\big[\frac{3}{4}\ln(1+T_0)-\frac{1}{2}\big]. \end{align*} This implies that \[ t^{-2}h^{6-N}\int_{|x|\leq h} F(\lambda^{-1/4}x,t/h)dx \geq\frac{(4N^2+2)S_{N}}{N(1-2^{-N})^2},\quad \forall h\geq1, t\geq hT_0 \] for suitable large $T_0$. When $t\in[-T_0,T_0]$, it is easy to see that there exist $\theta>0$ such that \[ \theta|t|\leq \ln(1+|t|)\leq e^{-1}|t|, \] then \begin{align*} F(x,t)&=\frac{t^2-1}{2}\ln(1+|t|)+\frac{1}{4}|t|(2-|t|)\\ &\geq \frac{\theta}{2}|t|^{3}-\frac{1}{2}|t|^2 +\big(\frac{1}{2}-\frac{e^{-1}}{2}\big)|t|. \end{align*} Thus, there exist $a_0>0$ and $q\in (2,2^{**})$ such that \[ F(x,t)\geq a_0|t|^{q},\quad t\in[-T_0,T_0]. \] From the above fact, we deduce that \eqref{1.01} satisfies (A7). On the other hand, we note that \begin{align*} \mathcal{F}(x,t)&=\frac{1}{2}tf(x,t)-F(x,t)=\frac{1}{2}\ln(1+|t|) +\frac{1}{4}|t|(|t|-2)\\ &\geq \frac{1}{2}|t|-\frac{1}{4}|t|^2+\frac{1}{4}|t|^2-\frac{1}{2}|t|=0. \end{align*} By a straightforward computation, there exist $R_0>0$, $a_{1}>0$ and $\kappa>\max\{1,\frac{N}{4}\}$ such that \[ tf(x, t)=t^2\ln(1+|t|)\le \frac{b}{3}|t|^2, \quad |t|\le R_0, \] and \[ |\frac{f(x, t)}{t}|^{\kappa}=\big(\ln(1+|t|)\big)^{\kappa} \le a_1\big(\frac{1}{2}\ln(1+|t|)+\frac{1}{4}|t|(|t|-2)\big) = a_1\mathcal{F}(x, t), \quad |t|\ge R_0. \] This shows that \eqref{1.01} satisfies (A8). The main results of this article are the following theorems. \begin{theorem} \label{thm1.2} Assume that {\rm (A1), (A3), (A4), (A7), (A8)} are satisfied. Then there exists $\varepsilon_0>0$, such that for $0<\varepsilon\le \varepsilon_0$, equation \eqref{1.1} has a solution $u_{\varepsilon}$ satisfying \begin{gather*} 0<\Phi_{\varepsilon^{-4}}(u_{\varepsilon}) \le \frac{b^{(4\kappa-4)/4}}{3^{\kappa}a_1 \left(\gamma_{2^{\ast\ast}}\gamma_0\right)^{N/2}}\varepsilon^{N-4}, \\ \int_{\mathbb{R}^N}\mathcal{F}(x, u_{\varepsilon})\mathrm{d}x \le \frac{b^{(4\kappa-4)/4}}{3^{\kappa}a_1 \big(\gamma_{2^{\ast\ast}}\gamma_0\big)^{N/2}}\varepsilon^N. \end{gather*} \end{theorem} \begin{theorem} \label{thm1.3} Assume that {\rm (A1)--(A4), (A9)} are satisfied. Then there exists $\varepsilon_{\ast}>0$, such that for $0<\varepsilon\le \varepsilon_{\ast}$, equation \eqref{1.2} has a solution $u_{\varepsilon}$ satisfying \begin{gather*} 0<\Phi_{\varepsilon^{-4}}(u_{\varepsilon}) \le \frac{Q_2}{[3(1+a_2)Q_2]^{N/4}N(\gamma_{2^{\ast\ast}} \gamma_0)^{N/2}}\varepsilon^{N-4}, \\ \int_{\mathbb{R}^N}\mathcal{F}(x, u_{\varepsilon})\mathrm{d}x +\frac{2}{N}\int_{\mathbb{R}^N}Q(x)|u_{\varepsilon}|^{2^{\ast\ast}}\mathrm{d}x\le \frac{Q_2}{[3(1+a_2)Q_2]^{N/4}N(\gamma_{2^{\ast\ast}}\gamma_0)^{N/2}} \varepsilon^N. \end{gather*} \end{theorem} Next, instead of handling \eqref{1.1} and \eqref{1.2} directly, but handle the equivalent problems. Let $\lambda=\varepsilon^{-4}$, then equations \eqref{1.1} and \eqref{1.2} are equivalent to the following equations respectively \begin{equation}\label{1.5} \begin{gathered} \Delta^2u+\lambda V(x)u=\lambda f(x,u), \quad x\in \mathbb{R}^N,\\ u\in H^2(\mathbb{R}^N),\quad u(x)\to 0, \quad \text{as } |x|\to \infty, \end{gathered} \end{equation} and \begin{equation}\label{1.6} \begin{gathered} \Delta^2u+\lambda V(x)u=\lambda Q(x)|u|^{2^{\ast\ast}-2}u+\lambda f(x,u), \quad x\in \mathbb{R}^N,\\ u\in H^2(\mathbb{R}^N),\quad u(x)\to 0,\quad \text{as } |x|\to \infty, \end{gathered} \end{equation} Therefore, Theorems \ref{thm1.2} and \ref{thm1.3} are equivalent to the following theorems. \begin{theorem} \label{thm1.4} Assume that {\rm (A1), (A3), (A4), (A7), (A8)} are satisfied. Then there exists $\lambda_0>1$, such that for $\lambda\ge \lambda_0$, equation \eqref{1.5} has a solution $u_{\lambda}$ satisfying \begin{gather*} 0<\Phi_{\lambda}(u_{\lambda}) \le \frac{b^{(4\kappa-4)/4}}{3^{\kappa}a_1 \left(\gamma_{2^{\ast\ast}}\gamma_0\right)^{N/2}}\lambda^{1-N/4}, \\ \int_{\mathbb{R}^N}\mathcal{F}(x, u_{\lambda})\mathrm{d}x \le \frac{b^{(4\kappa-4)/4}}{3^{\kappa}a_1 \left(\gamma_{2^{\ast\ast}}\gamma_0\right)^{N/2}}\lambda^{-N/4}. \end{gather*} \end{theorem} \begin{theorem} \label{thm1.5} Assume that {\rm (A1)--(A4), (A9)} are satisfied. Then there exists $\lambda_*>1$, such that for $\lambda\ge \lambda_*$, equation \eqref{1.6} has a solution $u_{\lambda}$ satisfying \begin{gather*} 0<\Phi_{\lambda}(u_{\lambda}) \le \frac{Q_2}{[3(1+a_2)Q_2]^{N/4} N(\gamma_{2^{\ast\ast}}\gamma_0)^{N/2}}\lambda^{1-N/4}, \\ \int_{\mathbb{R}^N}\mathcal{F}(x, u_{\lambda})\mathrm{d}x +\frac{2}{N}\int_{\mathbb{R}^N}Q(x)|u_{\lambda}|^{2^{\ast\ast}}\mathrm{d}x \le \frac{Q_2}{[3(1+a_2)Q_2]^{N/4}N(\gamma_{2^{\ast\ast}} \gamma_0)^{N/2}}\lambda^{-N/4}. \end{gather*} \end{theorem} In the next section, we provide some preliminaries and then prove these theorems. \section{Proof of the main results} To prove our results, first, we introduce the working space $$ E=\big\{u\in H^2(\mathbb{R}^N) : \int_{\mathbb{R}^N} V(x)|u|^2\mathrm{d}x< +\infty \big\} $$ and the associated norm $$ \|u\|=\Big(\int_{\mathbb{R}^N} [|\Delta{u}|^2+\lambda V(x)|u|^2]\mathrm{d}x\Big)^{1/2}, \quad u\in E. $$ By using (A1) and the Sobolev inequality, we can demonstrate that there exists a constant $\gamma_0>0$ independent of $\lambda$ such that \begin{equation}\label{2.1} \|u\|_{H^2(\mathbb{R}^N)}\le \gamma_0\|u\|, \quad \forall u\in E, \; \lambda\ge 1. \end{equation} This shows that $(E, \|\cdot\|)$ is a Banach space for $\lambda \ge 1$. Furthermore, by the Sobolev embedding theorem, we have \begin{equation}\label{2.2} \|u\|_{s}\le \gamma_s\|u\|_{H^2(\mathbb{R}^N)}\le \gamma_s\gamma_0\|u\|, \quad \forall u\in E, \; \lambda\ge 1, \; 2\le s\le 2^{\ast\ast}, \end{equation} where and in the sequel, by $\|\cdot\|_{s}$ we denote the usual norm in space $L^{s}(\mathbb{R}^N)$. Let \begin{equation}\label{2.3} \Phi_{\lambda}(u)=\frac{1}{2}\int_{\mathbb{R}^N}\left(|\Delta u|^2+\lambda V(x)|u|^2\right) \mathrm{d}x-\lambda\int_{\mathbb{R}^N}F(x, u)\mathrm{d}x \end{equation} and \begin{equation}\label{2.4} \begin{aligned} \Psi_{\lambda}(u) &=\frac{1}{2}\int_{\mathbb{R}^N}\left(|\Delta u|^2+\lambda V(x)|u|^2\right)\mathrm{d}x -\frac{\lambda}{2^{\ast\ast}}\int_{\mathbb{R}^N}Q(x)|u|^{2^{\ast\ast}} \mathrm{d}x \\ &\quad -\lambda\int_{\mathbb{R}^N}F(x, u)\mathrm{d}x. \end{aligned} \end{equation} It is well known that $\Phi_{\lambda}$ and $\Psi_{\lambda}$ are of $C^1(E, \mathbb{R})$, and \begin{equation}\label{2.5} \langle\Phi'_{\lambda}(u), v\rangle=\int_{\mathbb{R}^N}\left(\Delta u\Delta v +\lambda V(x)uv\right)\mathrm{d}x-\lambda\int_{\mathbb{R}^N}f(x, u)v\mathrm{d}x, \quad \forall u, v\in E \end{equation} and \begin{equation}\label{2.6} \begin{aligned} \langle\Psi'_{\lambda}(u), v\rangle & = \int_{\mathbb{R}^N}\left(\Delta u\Delta v +\lambda V(x)uv\right)\mathrm{d}x \\ & -\lambda \int_{\mathbb{R}^N} \left[Q(x)|u|^{2^{\ast\ast}-2}u+f(x, u)\right]v\mathrm{d}x, \quad \forall u, v\in E. \end{aligned} \end{equation} Observe that, since $(q-2)N-4q<0$, we can let $h_0\ge 1$ and $h_*\ge 1$ be such that \begin{equation}\label{2.7} \begin{aligned} & \frac{(q-2)S_N}{2Nq(qa_0)^{2/(q-2)}}\Big\{\frac{4N^{3}+2} {(N+4)(1-2^{-N})^2}\Big\}^{q/(q-2)}h_0^{[(q-2)N-4q]/(q-2)} \\ & = \frac{b^{(4\kappa-N)/4}}{3^{\kappa}a_1\left(\gamma_{2^{\ast\ast}} \gamma_0\right)^{N/2}} \end{aligned} \end{equation} and \begin{equation}\label{2.8} \begin{aligned} & \frac{(q-2)S_N}{2qN(qa_*)^{2/(q-2)}}\Big\{\frac{4N^{3}+2(N+4)} {(N+4)(1-2^{-N})^2}\Big\}^{q/(q-2)}h_*^{[(q-2)N-4q]/(q-2)} \\ & = \frac{Q_2}{[3(1+a_2)Q_2]^{N/4}N(\gamma_{2^{\ast\ast}}\gamma_0)^{N/2}}. \end{aligned} \end{equation} Let $x_0\in \mathbb{R}^N$ be such that $V(x_0)=0$. From now on, we assume without loss of generality that $x_0 = 0$, that is $V(0)=0$, then we can choose $\lambda_0>1$ and $\lambda_*>1$ such that \begin{gather}\label{2.9} \sup_{\lambda^{1/4}|x|\le 2h_0}|V(x)|\le h_0^{-4}, \quad \forall \lambda\ge \lambda_0, \\ \label{2.10} \sup_{\lambda^{1/4}|x|\le 2h_*}|V(x)|\le h_*^{-4}, \quad \forall \lambda\ge \lambda_*. \end{gather} Next, we give the proofs of Theorems \ref{thm1.2}--\ref{thm1.5}. Subsection 2.1 considers the subcritical cases Theorems \ref{thm1.2} and \ref{thm1.4}, while Subsection 2.2 considers the critical cases Theorems \ref{thm1.3} and \ref{thm1.5}. \subsection{Subcritical case} In view of the definition of the norm $\|\cdot\|$, we can re-write $\Phi_{\lambda}$ in the form \begin{equation}\label{2.11} \Phi_{\lambda}(u)=\frac{1}{2}\|u\|^2-\lambda\int_{\mathbb{R}^N}F(x, u)\mathrm{d}x, \quad \forall u\in E. \end{equation} Let \begin{equation}\label{2.12} \vartheta(x): = \begin{cases} \frac{1}{h_0}, & |x|\le h_0,\\[4pt] \frac{h_0^{N-1}}{1-2^{-N}}[|x|^{-N}-(2h_0)^{-N}], & h_0<|x|\le 2h_0,\\[4pt] 0, & |x|>2h_0. \end{cases} \end{equation} Then $\vartheta\in H^2(\mathbb{R}^N)$, moreover, \begin{gather}\label{2.13} \|\Delta\vartheta\|_2^2 = \int_{\mathbb{R}^N}|\Delta \vartheta(x)|^2\mathrm{d}x \le \frac{4N^2S_Nh_0^{N-6}}{(N+4)(1-2^{-N})^2}, \\ \label{2.14} \|\vartheta\|_2^2 = \int_{\mathbb{R}^N}|\vartheta(x)|^2\mathrm{d}x \le \frac{2S_Nh_0^{N-2}}{(1-2^{-N})^2N}. \end{gather} Let $e_{\lambda}(x)=\vartheta(\lambda^{1/4}x)$. Then we can prove the following lemma which is very important. \begin{lemma} \label{lem2.1.1} Suppose that {\rm (A1), (A3), (A4), (A7)} are satisfied. Then \begin{equation}\label{2.15} \sup\{\Phi_{\lambda}(se_{\lambda}) : s\ge 0\} \le \frac{b^{(4\kappa-N)/4}}{3^{\kappa}a_1 (\gamma_{2^{\ast\ast}}\gamma_0)^{N/2}} \lambda^{1-N/4}, \quad \forall \lambda\ge \lambda_0. \end{equation} \end{lemma} \begin{proof} From (A7), \eqref{2.3}, \eqref{2.7}, \eqref{2.9}, \eqref{2.12}, \eqref{2.13} and \eqref{2.14}, we obtain \begin{equation}\label{2.16} \begin{aligned} &\Phi_{\lambda}(se_{\lambda}) \\ & = \frac{s^2}{2}\int_{\mathbb{R}^N}\left(|\Delta e_{\lambda}|^2 +\lambda V(x)|e_{\lambda}|^2\right)\mathrm{d}x -\lambda\int_{\mathbb{R}^N}F(x, se_{\lambda})\mathrm{d}x \\ & = \lambda^{1-N/4}\Big[\frac{s^2}{2}\int_{\mathbb{R}^N} \Big(|\Delta \vartheta|^2+V(\lambda^{-1/4}x)|\vartheta|^2\Big)\mathrm{d}x -\int_{\mathbb{R}^N}F(\lambda^{-1/4}x, s\vartheta)\mathrm{d}x\Big] \\ & \le \lambda^{1-N/4}\Big[\frac{s^2}{2}\Big(\|\Delta \vartheta\|_2^2 +\|\vartheta\|_2^2\sup_{|x|\le 2h_0}|V(\lambda^{-1/4}x)|\Big) \\ &\quad -\int_{|x|\le h_0}F(\lambda^{-1/4}x, s/h_0)\mathrm{d}x\Big] \\ & \le \lambda^{1-N/4}\Big[\frac{s^2}{2} \left(\|\Delta \vartheta\|_2^2+h_0^{-4}\|\vartheta\|_2^2\right) -\int_{|x|\le h_0}F(\lambda^{-1/4}x, s/h_0)\mathrm{d}x\Big], \end{aligned} \end{equation} for all $s\ge 0$ and $\lambda\ge \lambda_0$, \begin{equation}\label{2.17} \begin{aligned} & \frac{s^2}{2}\left(\|\Delta \vartheta\|_2^2+h_0^{-4}\|\vartheta\|_2^2\right) -\int_{|x|\le h_0}F(\lambda^{-1/4}x, s/h_0)\mathrm{d}x \\ & \le \frac{s^2}{2}\left[\|\Delta \vartheta\|_2^2+h_0^{-4}\|\vartheta\|_2^2 -\frac{(4N^2+2)S_N}{N\left(1-2^{-N}\right)^2}h_0^{N-6}\right]\le 0, \end{aligned} \end{equation} for all $s\ge h_0T_0$ and $\lambda\ge \lambda_0$, and \begin{equation}\label{2.18} \begin{aligned} & \frac{s^2}{2}\left(\|\Delta \vartheta\|_2^2+h_0^{-4}\|\vartheta\|_2^2\right) -\int_{|x|\le h_0}F(\lambda^{-1/4}x, s/h_0)\mathrm{d}x \\ & \le \frac{s^2}{2}\left(\|\Delta \vartheta\|_2^2 +h_0^{-4}\|\vartheta\|_2^2\right)-\frac{a_0S_N}{N}s^qh_0^{N-q} \\ & \le \frac{(q-2)\left(\|\Delta \vartheta\|_2^2+h_0^{-4}\|\vartheta\|_2^2\right)^{q/(q-2)}} {2q\big(\frac{qa_0S_N}{N}h_0^{N-q}\big)^{2/(q-2)}} \\ & \le \frac{(q-2)S_N}{2Nq(qa_0)^{2/(q-2)}}\Big\{\frac{4N^{3}+2} {(N+4)(1-2^{-N})^2}\Big\}^{q/(q-2)}h_0^{[(q-2)N-4q]/(q-2)} \\ & = \frac{b^{(4\kappa-N)/4}}{3^{\kappa}a_1 \left(\gamma_{2^{\ast\ast}}\gamma_0\right)^{N/2}}, \quad \forall 0\le s\le h_0T_0, \; \lambda\ge \lambda_0. \end{aligned} \end{equation} Now the conclusion of Lemma \ref{lem2.1.1} follows from \eqref{2.16}, \eqref{2.17} and \eqref{2.18}. \end{proof} Applying the mountain-pass lemma without the (PS) condition, by standard arguments, we can prove the following lemma. \begin{lemma} \label{lem2.1.2} Suppose that {\rm (A1), (A3), (A4), (A7)} are satisfied. Then there exist a constant $c_{\lambda}\in (0, \sup_{s\ge 0} \Phi_{\lambda}(se_{\lambda})]$ and a sequence $\{u_n\}\subset E$ satisfying \begin{equation}\label{2.19} \Phi_{\lambda}(u_n)\to c_{\lambda}, \quad \|\Phi_{\lambda}'(u_n)\|_{E^{\ast}}(1+\|u_n\|)\to 0. \end{equation} \end{lemma} \begin{lemma} \label{lem2.1.3} Suppose that {\rm (A1), (A3), (A4), (A7), (A8)} are satisfied. Then any sequence $\{u_n\}\subset E$ satisfying \eqref{2.19} is bounded in $E$. \end{lemma} \begin{proof} To prove the boundedness of $\{u_n\}$, arguing by contradiction, suppose that $\|u_n\| \to \infty$. Let $v_n=u_n/\|u_n\|$. Then $\|v_n\|=1$. If $$ \delta:=\limsup_{n\to\infty}\sup_{y\in \mathbb{R}^N}\int_{B(y,1)}|v_n|^2\mathrm{d}x=0, $$ then by Lions' concentration compactness principle \cite{Lio} or \cite[Lemma 1.21]{W1}, $v_n\to 0$ in $L^{s}(\mathbb{R}^N)$ for $20$. Going to a subsequence, if necessary, we may assume the existence of $k_n\in \mathbb{Z}^N$ such that $\int_{B_{1+\sqrt{N}}(k_n)}|v_n|^2dx > \frac{\delta}{2}$. Let $w_n(x)=v_n(x+k_n)$. Then \begin{equation}\label{2.24} \int_{B_{1+\sqrt{N}}(0)}|w_n|^2dx> \frac{\delta}{2}. \end{equation} Now we define $\tilde{u}_n(x)=u_n(x+k_n)$, then $\tilde{u}_n/\|u_n\|=w_n$ and $\|w_n\|_{2}^2 =\|v_n\|_{2}^2$. Passing to a subsequence, we have $w_n\rightharpoonup w$ in $H^2(\mathbb{R}^N)$, $w_n\to w$ in $L^{s}_{\mathrm{loc}}(\mathbb{R}^N)$, $2\le s<2^{\ast\ast}$ and $w_n\to w$ a.e. on $\mathbb{R}^N$. Obviously, \eqref{2.24} implies that $w\ne 0$. For a.e. $x\in \{z\in \mathbb{R}^N : w(z)\ne 0\}$, we have $\lim_{n\to\infty}|\tilde{u}_n(x)|=\infty$. Hence, it follows from \eqref{2.11}, \eqref{2.19}, (A7) and Fatou's lemma that \begin{align*} 0 & = \lim_{n\to\infty}\frac{c_{\lambda}+o(1)}{\|u_n\|^2} = \lim_{n\to\infty}\frac{\Phi_{\lambda}(u_n)}{\|u_n\|^2}\\ & = \lim_{n\to\infty}\Big[\frac{1}{2}\|v_n\|^2 -\lambda\int_{\mathbb{R}^N}\frac{F(x+k_n, \tilde{u}_n)} {|\tilde{u}_n|^2}|w_n|^2\mathrm{d}x\Big]\\ & \le \frac{1}{2}-\lambda\int_{\mathbb{R}^N}\liminf_{n\to\infty} \frac{F(x+k_n, \tilde{u}_n)}{|\tilde{u}_n|^2}|w_n|^2\mathrm{d}x = -\infty. \end{align*} This contradiction shows that $\{\|u_n\|\}$ is bounded. \end{proof} \begin{proof}[Proof of Theorem \ref{thm1.4}] Applying Lemmas \ref{lem2.1.1}, \ref{lem2.1.2} and \ref{lem2.1.3}, we deduce that there exists a bounded sequence $\{u_n\}\subset E$ satisfying \eqref{2.20} with \begin{equation}\label{2.25} c_{\lambda}\le \frac{b^{(4\kappa-N)/4}}{3^{\kappa}a_1 \left(\gamma_{2^{\ast\ast}}\gamma_0\right)^{N/2}}\lambda^{1-N/4}, \quad \forall \lambda\ge \lambda_0. \end{equation} Going if necessary to a subsequence, we can assume that $u_n\rightharpoonup u_{\lambda}$ in $(E, \|\cdot\|)$ and $\Phi_{\lambda}'(u_n)\to 0$. Next, we prove that $u_{\lambda}\ne 0$. Arguing by contradiction, suppose that $u_{\lambda}=0$, i.e. $u_n\rightharpoonup 0$ in $E$, and so $u_n\to 0$ in $L^{s}_{\mathrm{loc}}(\mathbb{R}^N)$, $2\le s<2^{\ast\ast}$ and $u_n\to 0$ a.e. on $\mathbb{R}^N$. Since $\mathcal{V}_{b}$ is a set of finite measure and $u_n\rightharpoonup 0$ in $E$, it holds \begin{equation}\label{2.26} \|u_n\|_2^2=\int_{\mathbb{R}^N\setminus \mathcal{V}_{b}}|u_n|^2\mathrm{d}x +\int_{\mathcal{V}_{b}}|u_n|^2\mathrm{d}x \le \frac{1}{\lambda b}\|u_n\|^2+o(1). \end{equation} For $s\in (2, 2^{\ast\ast})$, from \eqref{2.2}, \eqref{2.26} and the H\"older inequality it follows that \begin{equation}\label{2.27} \begin{aligned} \|u_n\|_s^s & \le \|u_n\|_2^{2(2^{\ast\ast}-s)/(2^{\ast\ast}-2)} \|u_n\|_{2^{\ast\ast}}^{2^{\ast\ast}(s-2)/(2^{\ast\ast}-2)} \\ & \le (\gamma_{2^{\ast\ast}}\gamma_0)^{2^{\ast\ast}(s-2)/(2^{\ast\ast}-2)} (\lambda b)^{-(2^{\ast\ast}-s)/(2^{\ast\ast}-2)}\|u_n\|^s+o(1). \end{aligned} \end{equation} According to (F4) and \eqref{2.26}, one can obtain \begin{equation}\label{2.28} \lambda\int_{|u_n|\le R_0}f(x, u_n)u_n\mathrm{d}x \le \frac{\lambda b}{3}\int_{|u_n|\le R_0}|u_n|^2\mathrm{d}x \le \frac{1}{3}\|u_n\|^2+o(1). \end{equation} By \eqref{2.3}, \eqref{2.5} and \eqref{2.19}, we have \begin{equation}\label{2.29} \Phi_{\lambda}(u_n)-\frac{1}{2}\langle\Phi_{\lambda}'(u_n), u_n\rangle =\lambda\int_{\mathbb{R}^N}\mathcal{F}(x, u_n)\mathrm{d}x =c_{\lambda}+o(1). \end{equation} Using (A8), \eqref{2.25}, \eqref{2.27} with $s=2\kappa/(\kappa-1)$ and \eqref{2.29}, we obtain \begin{align} & \lambda\int_{|u_n|\geq R_0}f(x, u_n)u_n\mathrm{d}x \nonumber \\ & \le \lambda\Big(\int_{|u_n|\geq R_0}\Big(\frac{|f(x, u_n)|}{|u_n|} \Big)^{\kappa}\mathrm{d}x\Big)^{1/\kappa}\|u_n\|_s^2 \nonumber\\ & \le a_1^{1/\kappa}(\gamma_{2^{\ast\ast}}\gamma_0)^{2\cdot 2^{\ast\ast} (s-2)/s(2^{\ast\ast}-2)} \lambda^{1-1/\kappa}(\lambda b)^{-2(2^{\ast\ast}-s)/s(2^{\ast\ast}-2)} c_{\lambda}^{1/\kappa}\|u_n\|^2+o(1) \nonumber \\ & \le a_1^{1/\kappa}(\gamma_{2^{\ast\ast}}\gamma_0)^{N/2\kappa} \lambda^{1-1/\kappa}c_{\lambda}^{1/\kappa} (\lambda b)^{-(4\kappa-N)/4\kappa}\|u_n\|^2+o(1) \label{2.30}\\ & = \frac{a_1^{1/\kappa}(\gamma_{2^{\ast\ast}}\gamma_0)^{N/2\kappa}} {b^{(4\kappa-N)/4\kappa}}[\lambda^{(N-4)/4}c_{\lambda}]^{1/\kappa} \|u_n\|^2+o(1) \nonumber \\ & \le \frac{1}{3}\|u_n\|^2+o(1), \nonumber \end{align} which, together with \eqref{2.5}, \eqref{2.19} and \eqref{2.28}, yields \begin{equation}\label{2.31} o(1) = \langle\Phi_{\lambda}'(u_n), u_n\rangle = \|u_n\|^2-\lambda\int_{\mathbb{R}^N}f(x, u_n)u_n\mathrm{d}x \ge \frac{1}{3}\|u_n\|^2+o(1); \end{equation} this results in the fact that $\|u_n\|\to 0$. Consequently, from (A3), \eqref{2.11} and \eqref{2.19} it follows that $$ 02h_*. \end{cases} \end{equation} Then we can prove the following lemma in the same way as the proof of Lemma \ref{lem2.1.1}. \begin{lemma} \label{lem2.2.1} Suppose that {\rm (A1), (A3), (A4), (A9)} are satisfied. Then \begin{equation}\label{2.35} \sup\left\{\Psi_{\lambda}(se^*_{\lambda}) : s\ge 0\right\} \le \frac{Q_2}{[3(1+a_2)Q_2]^{N/4}N(\gamma_{2^{\ast\ast}} \gamma_0)^{\frac{N}{2}}}\lambda^{1-N/4}, \quad \forall \lambda\ge \lambda_*. \end{equation} \end{lemma} Applying the mountain-pass lemma without the (PS) condition, by standard arguments, we can also prove the following lemma. \begin{lemma} \label{lem2.2.2} Suppose that {\rm (A1), (A3), (A4), (A9)} are satisfied. Then there exist a constant $c_{\lambda}\in (0, \sup_{s\ge 0}\Psi_{\lambda}(se^*_{\lambda})]$ and a sequence $\{u_n\}\subset E$ satisfying \begin{equation}\label{2.36} \Psi_{\lambda}(u_n)\to c_{\lambda}, \quad \|\Psi_{\lambda}'(u_n)\|_{E^*}(1+\|u_n\|)\to 0. \end{equation} \end{lemma} \begin{lemma} \label{lem2.2.3} Suppose that {\rm (A1), (A3), (A4), (A9)}t are satisfied. Then any sequence $\{u_n\}\subset E$ satisfying \eqref{2.36} is bounded in $E$. \end{lemma} \begin{proof} To prove the boundedness of $\{u_n\}$, arguing by contradiction, suppose that $\|u_n\| \to \infty$. Let $v_n=u_n/\|u_n\|$. Then $\|v_n\|=1$. In view of (A2) and (A4), we can choose $R_{\lambda}\in (0, 1)$ such that \begin{equation}\label{2.37} |Q(x)|t|^{2^{\ast\ast}-2}t+f(x, t)| \le \frac{1}{3\lambda(\gamma_{2^{\ast\ast}}\gamma_0)^2}|t|, \quad \forall x\in \mathbb{R}^N, \; |t|\le R_{\lambda}. \end{equation} Hence, by \eqref{2.2}, \eqref{2.37} and the H\"older inequality, it holds \begin{equation}\label{2.38} \begin{aligned} & \frac{\lambda}{\|u_n\|}\int_{|u_n|\le R_{\lambda}}| [Q(x)|u_n|^{2^{\ast\ast}-2}+f(x, u_n)]v_n|\mathrm{d}x \\ & \le \frac{1}{3(\gamma_{2^{\ast\ast}}\gamma_0)^2\|u_n\|} \int_{|u_n|\le R_{\lambda}}|u_n||v_n|\mathrm{d}x \\ & \le \frac{1}{3(\gamma_{2^{\ast\ast}}\gamma_0)^2\|u_n\|}\|u_n\|_2\|v_n\|_{2} \le \frac{1}{3}. \end{aligned} \end{equation} From (A2), (A9), \eqref{2.6}, \eqref{2.33} and \eqref{2.36}, one has \begin{equation}\label{2.39} \begin{aligned} c_{\lambda}+o(1) &= \lambda\int_{\mathbb{R}^N} \big[\frac{2}{N}Q(x)|u_n|^{2^{\ast\ast}}+\mathcal{F}(x, u_n)\big]\mathrm{d}x\\ &\ge \frac{2\lambda Q_1}{N}\int_{|u_n|\ge R_{\lambda}}|u_n|^{2^{\ast\ast}} \mathrm{d}x. \end{aligned} \end{equation} Sing (A3), (A2), \eqref{2.39} and the H\"older inequality, we obtain \begin{equation}\label{2.40} \begin{aligned} & \frac{\lambda}{\|u_n\|}\int_{|u_n|\ge R_{\lambda}}| \big[Q(x)|u_n|^{2^{\ast\ast}-2}u_n +f(x, u_n)\big]v_n|\mathrm{d}x \\ & \le \frac{\lambda C_{\lambda}Q_2}{\|u_n\|}\int_{|u_n|\ge R_{\lambda}}|u_n|^{2^{\ast\ast}-1}|v_n|\mathrm{d}x \\ & \le \frac{\lambda C_{\lambda}Q_2}{\|u_n\|}\|v_n\|_{2^{\ast\ast}} \Big(\int_{|u_n|\ge R_{\lambda}}|u_n|^{2^{\ast\ast}}\mathrm{d}x \Big)^{(2^{\ast\ast}-1)/2^{\ast\ast}} = o(1), \end{aligned} \end{equation} where $C_{\lambda}$ is a constant depend on $\lambda$. Combining \eqref{2.38} with \eqref{2.40} and using \eqref{2.6} and \eqref{2.36}, we have \begin{align*} 1 +o(1) & = \frac{\|u_n\|^2-\langle\Psi_{\lambda}'(u_n), u_n\rangle}{\|u_n\|^2}\\ & = \frac{\lambda}{\|u_n\|}\int_{\mathbb{R}^N} \big[Q(x)|u_n|^{2^{\ast\ast}-2}u_n+f(x, u_n)\big]v_n\mathrm{d}x\\ & \le \frac{\lambda}{\|u_n\|}\int_{|u_n|< R_{\lambda}}| \big[Q(x)|u_n|^{2^{\ast\ast}-2}u_n+f(x, u_n) \big]v_n|\mathrm{d}x\\ & \quad +\frac{\lambda}{\|u_n\|}\int_{|u_n|\ge R_{\lambda}}| \big[Q(x)|u_n|^{2^{\ast\ast}-2}u_n+f(x, u_n)\big]v_n|\mathrm{d}x \\ & \le \frac{1}{3}+o(1), \end{align*} which is a contradiction. Thus the sequence $\{u_n\}$ is bounded in $E$. \end{proof} \begin{proof}[Proof of Theorem \ref{thm1.5}] Applying Lemmas \ref{lem2.2.1}, \ref{lem2.2.2} and \ref{lem2.2.3}, we deduce that there exists a bounded sequence $\{u_n\}\subset E$ satisfying \eqref{2.36} with \begin{equation}\label{2.41} c_{\lambda}\le \frac{Q_2}{[3(1+a_2)Q_2]^{N/4} N(\gamma_{2^{\ast\ast}}\gamma_0)^{N/2}}\lambda^{1-N/4}, \quad \forall \lambda\ge \lambda_*. \end{equation} Going to a subsequence,if necessary, we can assume that $u_n\rightharpoonup u_{\lambda}$ in $(E, \|\cdot\|)$ and $\Psi_{\lambda}'(u_n)\to 0$. Next, we prove that $u_{\lambda}\ne 0$. Arguing by contradiction, suppose that $u_{\lambda}=0$, i.e. $u_n\rightharpoonup 0$ in $E$, and so $u_n\to 0$ in $L^{s}_{\mathrm{loc}}(\mathbb{R}^N)$, $2\le s<2^{\ast\ast}$ and $u_n\to 0$ a.e. on $\mathbb{R}^N$. Since $\mathcal{V}_{b}$ is a set of finite measure and $u_n\rightharpoonup 0$ in $E$, \begin{equation}\label{2.42} \|u_n\|_2^2=\int_{\mathbb{R}^N\setminus \mathcal{V}_{b}}|u_n|^2\mathrm{d}x +\int_{\mathcal{V}_{b}}|u_n|^2\mathrm{d}x \le \frac{1}{\lambda b}\|u_n\|^2+o(1), \end{equation} which, together with \eqref{1.3}, yields \begin{equation}\label{2.43} \begin{aligned} &\lambda\int_{|u_n|\le R_*}\big[Q(x)|u_n|^{2^{\ast\ast}}+f(x, u_n)u_n\big] \mathrm{d}x \\ &\le \frac{\lambda b}{3}\int_{|u_n|\le R_*}|u_n|^2\mathrm{d}x \le \frac{1}{3}\|u_n\|^2+o(1). \end{aligned} \end{equation} By \eqref{2.6}, \eqref{2.33} and \eqref{2.36}, we have \begin{equation}\label{2.44} \begin{aligned} \Psi_{\lambda}(u_n)-\frac{1}{2}\langle\Psi_{\lambda}'(u_n), u_n\rangle &=\lambda\int_{\mathbb{R}^N}\left[\frac{2}{N}Q(x)|u_n|^{2^{\ast\ast}} +\mathcal{F}(x, u_n)\right]\mathrm{d}x\\ &=c_{\lambda}+o(1). \end{aligned} \end{equation} Using \eqref{2.2}, \eqref{1.4}, \eqref{2.41}, \eqref{2.44} and the H\"older inequality, we obtain \begin{align} &\lambda\int_{|u_n|>R_*}\left[Q(x)|u_n|^{2^{\ast\ast}} +f(x, u_n)u_n\right]\mathrm{d}x \nonumber \\ & \le (1+a_2)\lambda\int_{|u_n|>R_*}Q(x)|u_n|^{2^{\ast\ast}}\mathrm{d}x \nonumber\\ & \le (1+a_2)\lambda( Q_2)^{2/2^{\ast\ast}} \Big(\int_{|u_n|>R_*}Q(x)|u_n|^{2^{\ast\ast}}\mathrm{d}x\Big)^{4/N} \Big(\int_{|u_n|>R_*}|u_n|^{2^{\ast\ast}}\mathrm{d}x\Big)^{2/2^{\ast\ast}} \nonumber\\ & = (1+a_2)(\lambda Q_2)^{2/2^{\ast\ast}} \Big(\int_{|u_n|>R_*}Q(x)|u_n|^{2^{\ast\ast}}\mathrm{d}x\Big)^{4/N} \|u_n\|_{2^{\ast\ast}}^2 \label{2.45}\\ & = (1+a_2)Q_2(\gamma_{2^{\ast\ast}}\gamma_0)^2\big(\frac{N}{Q_2}\big)^{4/N} (\lambda^{\frac{N-4}{4}}c_{\lambda})^{4/N}\|u_n\|^2+o(1) \nonumber\\ & \le \frac{1}{3}\|u_n\|^2+o(1), \nonumber \end{align} which, together with \eqref{2.6} and \eqref{2.43}, yields \begin{equation}\label{2.46} \begin{aligned} o(1) & = \langle\Psi_{\lambda}'(u_n), u_n\rangle \\ & = \|u_n\|^2-\lambda\int_{\mathbb{R}^N} \big[Q(x)|u_n|^{2^{\ast\ast}}+f(x, u_n)u_n\big]\mathrm{d}x \\ & \ge \frac{1}{3}\|u_n\|^2+o(1); \end{aligned} \end{equation} this results in the fact that $\|u_n\|\to 0$. 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