\documentclass[reqno]{amsart} \usepackage{amssymb} \usepackage{graphicx} \pdfoutput=1\relax\pdfpagewidth=8.26in\pdfpageheight=11.69in\pdfcompresslevel=9 \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Monograph 06, 2004, (142 pages).\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2004 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2004/Mon. 06\hfil Palais-Smale approaches] {Palais-Smale approaches to semilinear elliptic equations in unbounded domains} \author[Hwai-chiuan Wang\hfil EJDE-2004/Mon. 06\hfilneg] {Hwai-chiuan Wang} \address{Department of Mathematics\\ National Tsing Hua University\\ Hsinchu, Taiwan} \email{hwang@mail.math.nthu.edu.tw} \date{} \thanks{Submitted September 17, 2004. Published September 30, 2004.} \subjclass[2000]{35J20, 35J25} \keywords{Palais-Smale condition; index; decomposition theorem;\hfill\break\indent achieved domain; Esteban-Lions domain; symmtric Palais-Smale condition} \begin{abstract} Let $\Omega$ be a domain in $\mathbb{R}^{N}$, $N\geq1$, and $2^{\ast}=\infty$ if $N=1,2$, $2^{\ast}=\frac{2N}{N-2}$ if $N>2$, $22$, $2|x|^{2}\};\\ \mathbf{P}^{-}=\{(x,-y): (x,y)\in\mathbf{P}^{+}\};\\ \mathbf{C}=\{(x,y)\in\mathbb{R}^{N}: |x|0$, $z\in\Omega$ exists such that $B(z;r)\subset\Omega$; \newline $(i')$ We say that $\Omega$ is a strictly large domain in $\mathbb{R}^{N}$ if $\Omega$ contains an infinite cone of $\mathbb{R}^{N}$; \newline $(ii)$ We call $\Omega$ a large domain in $\mathbf{A}^{r}$ if for any positive number $m$, $a$, $b$ exist such that $b-a=m$ and $\mathbf{A} _{a,b}^{r}\subset\Omega$; \newline $(ii')$ We call $\Omega$ a strictly large domain in $\mathbf{A}^{r}$ if $\Omega$ contains a semi-strip of $\mathbf{A}^{r}$; \newline $(iii)$ We call $\Omega$ a large domain in $\mathbf{A}^{r_{1},r_{2}}$ if for any positive number $m$, $a$, $b$ exist with $a0$, $\varphi\in H^{1}(\mathbb{R}^{N})$, and $\phi\in C_{c}^{1}(\mathbb{R}^{N})$ exist such that $\Vert\varphi-\phi\Vert_{H^{1}}<\varepsilon/2(\Vert u\Vert_{H^{1}}+1).$ Let $K=\mathop{\rm supp}\phi$, then $K$ is compact. We have \begin{align*} \langle u(z+z_{n}),\phi(z) \rangle _{H^{1}} &={\int_{\mathbb{R}^{N}}} \nabla u(z+z_{n})\nabla\phi(z)dz+ {\int_{\mathbb{R}^{N}}}u(z+z_{n})\phi(z)dz\\ &={\int_{K}}\nabla u(z+z_{n})\nabla\phi(z)dz+ {\int_{K}} u(z+z_{n})\phi(z)dz\\ &\leq\| \nabla u(z+z_{n})\| _{L^{2}(K)}\| \nabla\phi\| _{L^{2}(K)}+\| u(z+z_{n})\| _{L^{2} (K)}\| \phi\| _{L^{2}(K)}\\ &= o(1)\quad \text{as } n\to\infty. \end{align*} Thus, for some $N>0$ such that $| \langle u(z+z_{n}),\phi (z)\rangle _{H^{1}}| <\frac{\varepsilon}{2}$ for $n\geq N$. In addition, \begin{align*} \langle u(z+z_{n}),\varphi(z)\rangle _{H^{1}} & =\langle u(z+z_{n}),\varphi(z)-\phi(z)\rangle _{H^{1}}+\langle u(z+z_{n}),\phi(z)\rangle _{H^{1}}\\ & \leq\Vert u(z+z_{n})\Vert_{H^{1}(\mathbb{R}^{N}) }\Vert \varphi(z)-\phi(z)\Vert_{H^{1}(\mathbb{R}^{N}) }\\ & \quad +\langle u(z+z_{n}),\phi(z)\rangle _{H^{1}}\\ & \leq\Vert u(z)\Vert_{H^{1}(\mathbb{R}^{N}) }\Vert \varphi(z)-\phi(z)\Vert_{H^{1}(\mathbb{R}^{N}) }+\frac {\varepsilon}{2}\\ & <\varepsilon\text{\ for\ }n\geq N. \end{align*} Therefore, $u(z+z_{n})\rightharpoonup0$ weakly in $H^{1}(\mathbb{R}^{N})$. \end{proof} \begin{lemma} \label{p6} For $u\in H_{0}^{1}(\mathbf{A}^{r})$ and $\{z_{n}\}$ in $\mathbf{A}^{r}$ satisfying $| z_{n}| \to\infty$ as $n\to\infty$, then $u(z+z_{n})\rightharpoonup0$ weakly in $H_{0} ^{1}(\mathbf{A}^{r})$ as $n\to\infty$. \end{lemma} The proof of this lemma is the same as the proof of Lemma \ref{p5}. Therefore, we omit it. Bounded $L^{p}(\Omega)$ sequence admits interesting convergent properties. \begin{lemma}[Br\'{e}zis-Lieb Lemma] \label{p7} Suppose $u_{n}\to u$ a.e. in $\Omega$ and there is a $c>0$ such that $\| u_{n}\| _{L^{p}(\Omega)}\leq c$ for $n=1,2,\dots$. Then \newline $(i)\| u_{n}-u\| _{L^{p}}^{p}=\| u_{n}\| _{L^{p} }^{p}-\| u\| _{L^{p}}^{p}+ o(1)$; \newline$(ii)$ $|u_{n} -u|^{p-2}(u_{n}-u) -|u_{n}|^{p-2}u_{n}+|u|^{p-2}u= o(1)$ in $L^{\frac{p}{p-1}}(\Omega)$. \end{lemma} \begin{proof} $(i)$ Let $\varphi(t) =t^{p}$ for $t>0$, then $\varphi^{\prime }(t) =pt^{p-1}$ and $| u_{n}-u| ^{p}-| u_{n}| ^{p}=\varphi(|u_{n} -u|)-\varphi(|u_{n}|)=\varphi'(t) (|u_{n} -u|-|u_{n}|) ,$ where $t=(1-\theta)|u_{n}|+\theta|u_{n}-u|\leq|u_{n}|+|u|$ for some $\theta \in[0,1]$. Thus, by the Young inequality, for $\varepsilon>0$ $| | u_{n}-u| ^{p}-| u_{n}| ^{p}| \leq p(|u_{n}|+|u|) ^{p-1}|u|\\ \leq d(|u_{n}|^{p-1}|u|) +d|u|^{p}\\ \leq\varepsilon|u_{n}|^{p}+c_{\varepsilon}|u|^{p}.$ Thus, $| | u_{n}-u| ^{p}-| u_{n}| ^{p}+|u|^{p} | \leq\varepsilon|u_{n}|^{p}+(c_{\varepsilon}+1)|u|^{p}.$ We have ${\int_{\Omega}} | | u_{n}-u| ^{p}-| u_{n}| ^{p}+|u|^{p} | \leq\varepsilon c^{p}+(c_{\varepsilon}+1) {\int_{\Omega}} |u|^{p}.$ Since $\| u\| _{L^{p}}\leq\liminf_{n\to\infty}\| u_{n}\| _{L^{p}}\leq c$. For some $\delta>0$ $|E|<\delta$ implies ${\int_{E}} |u|^{p}<\varepsilon$. In addition, $K$ in $\mathbb{R}^{N}$ exists such that $|K|<\infty$ and ${\int_{K^{c}}}|u|^{p}<\varepsilon$. Thus, \begin{gather*} {\int_{E}} | | u_{n}-u| ^{p}-| u_{n}| ^{p}+|u|^{p} | \leq(c^{p}+c_{\varepsilon}+1) \varepsilon,\\ {\int_{K^{c}}} | | u_{n}-u| ^{p}-| u_{n}| ^{p}+|u|^{p} | \leq(c^{p}+c_{\varepsilon}+1) \varepsilon. \end{gather*} Clearly, $| | u_{n}-u| ^{p}-| u_{n}| ^{p}+|u|^{p}| = o(1)$ a.e. in $\Omega$. By Theorem \ref{p16} below, ${\int_{\Omega}} | | u_{n}-u| ^{p}-| u_{n}| ^{p}+|u|^{p} | = o(1)$, or $\| u_{n}-u\| _{L^{p}}^{p}=\| u_{n}\| _{L^{p}}^{p}-\| u\| _{L^{p}}^{p}+ o(1).$ $(ii)$ Let $\varphi(t)=|t|^{p-2}t$, then $\varphi'(t)=(p-1) |t|^{p-2}$. The proof is similar to part $(i)$ \end{proof} New (PS)-sequences can be produced as follows. \begin{lemma} \label{p8} Let $u_{n}\rightharpoonup u$ weakly in $X(\Omega)$ and $J'(u_{n})=-\Delta u_{n}+u_{n}-|u_{n}|^{p-2}u_{n}= o(1)\quad\mbox{in } X^{-1}(\Omega) .$ Then \newline $(i)$ $|u_{n}-u|^{p-2}(u_{n}-u) -|u_{n}|^{p-2} u_{n}+|u|^{p-2}u= o(1)$ in $X^{-1}(\Omega)$; \newline $(ii)$ $J'(\varphi_{n})=-\Delta\varphi_{n}+\varphi_{n}-|\varphi_{n} |^{p-2}\varphi_{n}= o(1)$ in $X^{-1}(\Omega)$ where $\varphi _{n}=u_{n}-u$; \newline $(iii)$ if $\{ u_{n}\}$ is a (PS)$_{\beta}$-sequence, then $\{ \varphi_{n}\}$ is a (PS)$_{(\beta-J(u)) }$-sequence. \end{lemma} \begin{proof} $(i)$ By Lemma \ref{p7}, $\int_{\Omega}| |u_{n}-u|^{p-2}(u_{n}-u)-|u_{n}|^{p-2}u_{n} +|u|^{p-2}u| ^{\frac{p}{p-1}}= o(1).$ Now for $\varphi\in H^{1}(\Omega)$, \begin{align*} &| \langle |u_{n}-u|^{p-2}(u_{n}-u)-|u_{n}|^{p-2}u_{n} +|u|^{p-2}u,\varphi\rangle | \\ &=|{\int_{\Omega}} \varepsilon_{n}\varphi|\leq( {\int_{\Omega}}|\varepsilon_{n}|^{\frac{p}{p-1}}) ^{\frac{p-1}{p}}( {\int_{\Omega}} |\varphi|^{p}) ^{1/p}\\ &\leq c\Vert\varepsilon_{n}\Vert_{L^{\frac{p}{p-1}}}\Vert\varphi \Vert_{H^{1}}, \end{align*} where $\varepsilon_{n}=|u_{n}-u|^{p-2}(u_{n}-u)-|u_{n}|^{p-2}u_{n}+|u|^{p-2} u$. Therefore, $\| |u_{n}-u|^{p-2}(u_{n}-u)-|u_{n}|^{p-2}u_{n}+|u|^{p-2}u\| _{X^{-1}} \leq c\Vert\varepsilon_{n}\Vert_{L^{\frac{p}{p-1}}}= o(1).$ $(ii)$ Since $$J'(u_{n}) =-\Delta u_{n}+u_{n}-|u_{n}|^{p-2} u_{n}= o(1)\quad\mbox{in }X(\Omega) \label{2-1}$$ and $u_{n}\rightharpoonup u$, then by Lemma \ref{p4}, we have $J'(u)=0$, or $$-\Delta u+u-|u|^{p-2}u=0.\label{2-2}$$ Now by part $(i)$, (\ref{2-1}), and (\ref{2-2}), \begin{align*} J'(\varphi_{n}) & =-\Delta\varphi_{n}+\varphi_{n}-|\varphi_{n} |^{p-2}\varphi_{n}\\ & =-\Delta(u_{n}-u) +(u_{n}-u) -|u_{n} -u|^{p-2}(u_{n}-u) \\ & =(-\Delta u_{n}+u_{n}-|u_{n}|^{p-2}u_{n}) -(-\Delta u+u-|u|^{p-2}u) \\ & -(|u_{n}-u|^{p-2}(u_{n}-u) -|u_{n}|^{p-2} u_{n}+|u|^{p-2}u) \\ & = o(1). \end{align*} $(iii)$ Since $u_{n}\rightharpoonup u$ weakly in $X(\Omega)$ and $\{ u_{n}\}$ is a (PS)$_{\beta}$-sequence, by Lemma \ref{p4}, \ref{p7} and Theorem \ref{p21} below, a subsequence $\{ u_{n}\}$ exists such that $a(\varphi _{n}) =a(u_{n}) -a(u) + o(1)$ and $b(\varphi_{n}) =b(u_{n}) -b(u) + o(1)$. Thus, $J(\varphi_{n}) =J(u_{n}) -J(u) + o(1) =\beta-J(u)+ o(1)$. Therefore, by part $(ii)$, $\{ \varphi_{n}\}$ is a (PS)$_{(\beta-J(u)) }$-sequence. \end{proof} Define the concentration function of $|u_{n}|^{2}$ in $\mathbb{R}^{N}$ by $Q_{n}(t)=\sup_{z\in\mathbb{R}^{N}}\int_{z+B^{N}(0;t)}|u_{n}|^{2}.$ Then we have the following concentration lemma. \begin{lemma} \label{p9} Let $\{ u_{n}\}$ be bounded in $H^{1}(\mathbb{R}^{N})$ and for some $t_{0}>0$, let $Q_{n}(t_{0})= o(1)$. Then\newline $(i)$ $u_{n}= o(1)$ strongly in $L^{q}(\mathbb{R} ^{N})$ for $21$ as $r\to q$, we may choose $r$ satisfying $21$. Recall that $\Vert\{a_{n}\}\Vert_{\ell^{s}}=(\overset{\infty}{\underset{n=1}{\sum} }|a_{n}|^{s}) ^{1/s}\leq\overset{\infty}{\underset{n=1}{\sum} }|a_{n}|=\Vert\{a_{n}\}\Vert_{\ell^{1}},\quad \ell^{1}\subset\ell^{2}\subset\dots\subset\ell^{\infty}.$ Thus, \begin{align*} \sum_{i=1}^{\infty}\Big({\int_{P_{i}}} |\nabla u_{n}|^{2}+|u_{n}|^{2}\Big) ^{rt/2} & \leq\Big(\sum_{i=1}^{\infty} {\int_{P_{i}}} (|\nabla u_{n}|^{2}+|u_{n}|^{2}) \Big) ^{s}\\ & =\Big({\int_{\mathbb{R}^{N}}} (|\nabla u_{n}|^{2}+|u_{n}|^{2}) \Big) ^{s}\\ & =\Vert u_{n}\Vert_{H^{1}(\mathbb{R}^{N})}^{2s}\leq c\quad \text{for }n=1,2,\dots. \end{align*} Therefore, ${\int_{\mathbb{R}^{N}}} |u_{n}|^{q}\leq c(Q_{n}(t_{0}) )^{(1-t)}, \quad\mbox{or}\quad \int_{\mathbb{R}^{N}}|u_{n}|^{q}= o(1)\quad\text{as }n\to\infty.$ $(ii)$ In addition, if $u_{n}$ satisfies $$-\Delta u_{n}+u_{n}-|u_{n}|^{p-2}u_{n}= o(1)\quad\mbox{in }H^{-1}(\mathbb{R} ^{N}),\label{2-3}$$ then $\{u_{n}\}$ is bounded. Multiply Equation (\ref{2-3}) by $u_{n}$ and integrate it to obtain $a(u_{n})=b(u_{n})+ o(1).$ By part $(i)$, $b(u_{n})= o(1)$. Thus, $a(u_{n})= o(1)$, or $\Vert u_{n}\Vert_{H^{1}}= o(1)\quad\text{strongly in }\,H^{1}(\mathbb{R} ^{N}) .$ \end{proof} \begin{lemma} \label{p10} Let $\{u_{n}\}$ be bounded in $H_{0}^{1}(\mathbf{A}^{r})$ and for some $t_{0}>0$, $Q_{n}^{r}(t_{0})=\sup_{y\in\mathbb{R}}\int_{(0,y) +\mathbf{A} _{-t_{0},t_{0}}^{r}}|u_{n}|^{2}= o(1).$ Then\newline $(i)$ $u_{n}= o(1)$ strongly in $L^{q}(\mathbf{A} ^{r})$ for $20$, there is a $\delta>0$ such that if $v\in H_{0}^{1}(\Omega)$ solves \eqref{E1} in $\Omega$ satisfying $\Vert v\Vert_{H^{1}}\leq c$ and $\Vert v\Vert_{L^{2}}\leq\delta$, then $v\equiv0$. \end{lemma} \begin{proof} For $00$.\newline $(i)$ Let $\gamma-2\geq0$. Note that $2(1-t_{0})>0$. By (\ref{2-4}), we have $1\leq d\delta^{2(1-t_{0})}\Vert v\Vert_{H^{1}}^{\gamma-2}\leq dc^{\gamma -2}\delta^{2(1-t_{0})}.$ Let $\delta_{1}>0$ satisfy $dc^{\gamma-2}\delta_{1}^{2(1-t_{0})}<1$. If $\delta\leq\delta_{1}$, then $1\leq dc^{\gamma-2}\delta^{2(1-t_{0})}\leq dc^{\gamma-2}\delta_{1} ^{2(1-t_{0})}<1,$ which is a contradiction.\newline $(ii)$ Let $\gamma-2<0$. By (\ref{2-4}), we have $\Vert v\Vert_{H^{1}}\leq\delta^{\frac{2(1-t_{0})}{2-\gamma}}d ^{\frac{1}{2-\gamma}},$ since $\Vert v\Vert_{H^{1}}^{2}= {\int_{\Omega}} |v|^{p}\leq c_{1}\Vert v\Vert_{H^{1}}^{p}, \quad\mbox{or}\quad 1\leq c_{1}\Vert v\Vert_{H^{1}}^{p-2}.$ Thus, we have $1\leq c_{1}\Vert v\Vert_{H^{1}}^{p-2}\leq c_{2}\delta^{\frac{2(1-t_{0} )(p-2)}{2-\gamma}},$ where $c_{2}=c_{1}d^{\frac{p-2}{2-\gamma}}>0$. Note that $\frac{2(1-t_{0} )(p-2)}{2-\gamma}>0$. Let $\delta_{2}>0$ such that $c_{2}\delta_{2}^{\frac{2(1-t_{0})(p-2)}{2-\gamma}}<1.$ If $\delta\leq\delta_{2}$, then $1\leq c_{2}\delta^{\frac{2(1-t_{0} )(p-2)}{2-\gamma}}<1$, which is a contradiction.\newline Take $\delta_{0} =\min\{ \delta_{1}\text{, }\delta_{2}\}$, if $\delta\leq \delta_{0}$, from parts $(i)$ and $(ii)$, and we obtain $\Vert v\Vert_{H^{1} }=0$ or $v=0$. \end{proof} Let $\tilde{u}(z)=\begin{cases} u(z) & \text{for } z\in\Omega;\\ 0 & \text{for } z\in\mathbb{R}^{N}\backslash \Omega. \end{cases}$ Then we have the following characterization of a function in $W_{0}^{1,p}(\Omega)$. \begin{lemma} \label{p12} Let $\Omega$ be a $C^{0,1}$ domain in $\mathbb{R}^{N}$ and $u\in L^{p}(\Omega)$ with $10$ such that $| \int_{\Omega}u\frac{\partial\varphi}{\partial x_{i}}| \leq c\| \varphi\| _{L^{p}},\quad\text{for each }\varphi\in C_{c} ^{1}(\mathbb{R}^{N})\text{, }i=1,2,\dots,N;$ $(iii)$ $\tilde{u}\in W_{0}^{1,p}(\mathbb{R}^{N})$ and $\frac{\partial\widetilde{u}}{\partial z_{i}} =\frac{\widetilde{\partial u}}{\partial z_{i}}$. \end{lemma} For the proof of this lemma, see Br\'{e}zis \cite[Proposition IX.18]{B}, Gilbarg-Trudinger \cite[Theorem 7.25]{GT}, and Grisvard \cite[p26]{G}. We recall the classical compactness theorems. The Lebesgue dominated convergence theorem is a well-known compactness theorem. \begin{theorem}[Lebesgue Dominated Convergence Theorem] \label{p13} Suppose $\Omega$ is a domain in $\mathbb{R}^{N}$, $\{ u_{n}\} _{n=1}^{\infty}$ and $u$ are measurable functions in $\Omega$ such that $u_{n}\to u$ a.e. in $\Omega$. If $\varphi\in L^{1}(\Omega)$ exists such that for each $n$ $| u_{n}| \leq\varphi\quad\text{a.e. in }\Omega,$ then $u_{n}\to u$ in $L^{1}(\Omega)$. \end{theorem} The converse of the Lebesgue dominated convergence theorem fails. \begin{example} \label{p14} \rm For $n=1,2,\dots$, let $u_{n}:\mathbb{R\to R}$ be defined by $u_{n}(z)=\begin{cases} 0 & \text{for }z\leq n;\\ 2 & \text{for }z=n+1/2n;\\ 0 & \text{for }z\geq n+1/n;\\ &\text{linear otherwise.} \end{cases}$ \begin{figure}[htb] \begin{center} \includegraphics[width=0.7\textwidth]{fig03} \end{center} % \centering \resizebox{4.5in}{!}{\includegraphics{./fig03.eps}} \caption{Counter example 1.} \label{fig:fig03} \end{figure} We have ${\int_{\mathbb{R}}} u_{n}(z)dz = \frac{1}{n}<\infty\quad \quad\text{for each }n\in\mathbb{N}.$ Hence, $u_{n}\to 0$ a.e. in $\mathbb{R}$ and strongly in $L^{1} (\mathbb{R})$. Let $\varphi:\mathbb{R\to R}$ satisfy $|u_{n}| \leq\varphi$ a.e. in $\mathbb{R}$ for each $n\in\mathbb{N}$. Then $\infty=\underset{n=1}{\overset{\infty}{\sum}}\frac{1}{n}= {\int_{\mathbb{R}}} \underset{n=1}{\overset{\infty}{\sum}}u_{n}\leq {\int_{\mathbb{R}}}\varphi$. Consequently, $\varphi\notin L^{1}(\mathbb{R})$. \end{example} However, the generalized Lebesgue dominated convergence theorem is a necessary and sufficient result for $L^{1}$ convergence. \begin{theorem}[Generalized Lebesgue Dominated Convergence Theorem:] \label{p15} Suppose $\Omega$ is a domain in $\mathbb{R}^{N}$, $\{ u_{n}\} _{n=1}^{\infty}$ and $u$ are measurable functions in $\Omega$ such that $u_{n}\to u$ a.e. in $\Omega$. Then $u_{n}\to u$ in $L^{1}(\Omega)$ if and only if $\{ \varphi_{n}\} _{n=1}^{\infty },\varphi\in L^{1}(\Omega)$ exist such that $\varphi_{n}\to\varphi$ a.e. in $\Omega$, $| u_{n}| \leq\varphi_{n}$ a.e. in $\Omega$ for each $n$, and $\varphi_{n}\to\varphi$ in $L^{1}(\Omega)$. \end{theorem} \begin{proof} ($\Longrightarrow$) Suppose that $u_{n}\to u$ in $L^{1}(\Omega)$, take $\varphi_{n}=| u_{n}|$ and $\varphi=| u|$, then $\varphi_{n}\to\varphi$ in $L^{1}(\Omega)$. \newline ($\Longleftarrow$) Suppose that a sequence of measurable functions $\{ \varphi_{n}\} _{n=1}^{\infty}$ and $\varphi$ in $\Omega$ exist such that $\varphi_{n}\in L^{1}(\Omega)$, $\varphi_{n}\to\varphi$ a.e. in $\Omega$, $| u_{n}| \leq\varphi_{n}$ a.e. in $\Omega$ for each $n$, and $\varphi _{n}\to\varphi$ in $L^{1}(\Omega)$. Applying the Fatou lemma, we have $\int_{\Omega}\liminf_{n\to\infty}(\varphi_{n}-u_{n})\leq \liminf_{n\to\infty}\int_{\Omega}(\varphi_{n}-u_{n}),$ or $\int_{\Omega}u\geq\limsup_{n\to\infty}\int_{\Omega}u_{n}.$ Applying the Fatou lemma again, we have $\int_{\Omega}\liminf_{n\to\infty}(\varphi_{n}+u_{n})\leq \liminf_{n\to\infty}\int_{\Omega}(\varphi_{n}+u_{n}),$ or $\int_{\Omega}u\leq\liminf_{n\to\infty}\int_{\Omega}u_{n}.$ Thus, $\int_{\Omega}u=\lim_{n\to\infty}\int_{\Omega}u_{n}.$ \end{proof} Another necessary and sufficient result for $L^{1}$ convergence is the Vitali convergence theorem. \begin{theorem}[Vitali Convergence Theorem for $L^{1}(\Omega)$] \label{p16} Suppose $\Omega$ is a domain in $\mathbb{R}^{N}$, $\{ u_{n}\} _{n=1}^{\infty}$ in $L^{1}(\Omega)$, and $u\in L^{1}(\Omega)$. Then $\| u_{n}-u\| _{L^{1}}\to0$ if the following three conditions hold:\newline $(i)$ $u_{n}\to u$ a.e in $\Omega$; \newline $(ii)$ (Uniformly integrable) For each $\varepsilon>0$, a measurable set $E\subset\Omega$ exists such that $| E| <\infty$ and $\int_{E^{c}}| u_{n}| d\mu<\varepsilon$ for each $n\in\mathbb{N}$, where $E^{c}=\Omega\backslash E$; \newline $(iii)$ (Uniformly continuous) For each $\varepsilon>0$, $\delta>0$ exists such that $|E|<\delta$ implies $\int_{E}| u_{n}| d\mu<\varepsilon\quad \text{for each }n\in \mathbb{N}.$ Conversely, if $\| u_{n}-u\| _{L^{1}}\to0$, then conditions $(ii)$ and $(iii)$ hold and there is a subsequence $\{u_{n}\}$ such that $(i)$ holds. Furthermore, if $| \Omega| <\infty$, then we can drop condition $(ii)$. \end{theorem} \begin{proof} Assume the three conditions hold. Choose $\varepsilon>0$ and let $\delta>0$ be the corresponding number given by condition $(iii)$. Condition $(ii)$ provides a measurable set $E\subset\Omega$ with $| E| <\infty$ such that $\int_{E^{c}}| u_{n}| d\mu<\varepsilon$ for all positive integers $n$. Since $| E| <\infty$, we can apply the Egorov theorem to obtain a measurable set $B\subset E$ with $| E \backslash B| <\delta$ such that $u_{n}$ converges uniformly to $u$ on $B$. Now write $\int_{\Omega}| u_{n}-u| d\mu =\int_{B}| u_{n}-u|d\mu +\int_{E\backslash B}| u_{n}-u| d\mu+\int_{E^{c}}| u_{n}-u| d\mu.$ Since $u_{n}\to u$ uniformly in $B$, the first integral on the right can be made arbitrarily small for large $n$. The second and third integrals will be estimated with the help of the inequality $| u_{n}-u| \leq| u_{n}| +| u| .$ From condition $(iii)$, we have $\int_{E\backslash B}| u_{n}| d\mu<\varepsilon$ for all $n\in\mathbb{N}$ and the Fatou Lemma shows that $\int_{E\backslash B}| u| d\mu\leq\varepsilon$ as well. The third integral can be handled in a similar way using condition $(ii)$. Thus, it follows that $\| u_{n}-u\| _{L^{1}}\to0$. Now suppose $\| u_{n}-u\| _{L^{1}}\to0$. Then for each $\varepsilon>0$, a positive integer $n_{0}$ exists such that $\| u_{n}-u\| _{L^{1}}<\varepsilon/2$ for $n>n_{0}$, and measurable sets $A$ and $B$ of finite measure exist such that $\int_{A^{c}}| u| d\mu<\varepsilon/2\quad\text{and }\int_{B^{c} }| u_{n}| d\mu<\varepsilon \quad\text{for }n=1,2,\dots ,n_{0}.$ Minkowski's inequality implies that $\| u_{n}\| _{L^{1}(A^{c})}\leq\| u_{n}-u\| _{L^{1}(A^{c})}+\| u\| _{L^{1}(A^{c})}<\varepsilon\quad \text{for }n>n_{0.}$ Then let $E=A\cup B$ to obtain the necessity of condition $(ii)$. Similar reasoning establishes the necessity of condition $(iii)$. Convergence in $L^{1}$ implies convergence in measure. Hence, condition $(i)$ holds for a subsequence. \end{proof} There is a bounded sequence $\{u_{n}\}$ in $L^{1}(\mathbb{R})$ that violates Theorem \ref{p16} condition $(ii)$. \begin{example} \label{p17} \rm For $n=1,2,\dots$, let $u_{n}:\mathbb{R\to R}$ be defined by $u_{n}(z)=\begin{cases} 0 & \text{for }z\leq n;\\ 2 & \text{for }z=n+1/2;\\ 0 & \text{for }z\geq n+1;\\ &\text{linear otherwise,} \end{cases}$ \begin{figure}[htb] \begin{center} \includegraphics[width=0.5\textwidth]{fig04} \end{center} % \centering \resizebox{2.5in}{!}{\includegraphics{./fig04.eps}} \caption{ counter example violating Theorem \ref{p16} condition $(ii)$.} \label{fig:fig04} \end{figure} then ${\int_{\mathbb{R}}}u_{n}(z)dz=1$ for each $n\in\mathbb{N}$. Clearly, $\{u_{n}\}$ violates Theorem \ref{p16} $(ii)$. \end{example} There is a bounded sequence $\{u_{n}\}$ in $L^{1}(\mathbb{R})$ that violates Theorem \ref{p16} condition $(iii)$. \begin{example}\label{p18} \rm For $n=1,2,\dots$, let $u_{n}:\mathbb{R\to R}$ be defined by $u_{n}(z)=\begin{cases} 0 & \text{for }z\leq n;\\ 2n & \text{for }z=n+1/2n;\\ 0 & \text{for }z\geq n+1/n;\\ &\text{linear therwise.} \end{cases}$ \begin{figure}[htb] \begin{center} \includegraphics[width=0.5\textwidth]{fig05} \end{center} % \centering \resizebox{2.5in}{!}{\includegraphics{./fig05.eps}} \caption{ counter example violating Theorem \ref{p16} condition $(iii)$.} \label{fig:fig05} \end{figure} Then ${\int_{\mathbb{R}}} u_{n}(z)dz=1\quad\text{for each }n\in\mathbb{N}.$ Clearly, $\{u_{n}\}$ violates Theorem \ref{p16} condition $(iii)$. \end{example} \begin{lemma} \label{p19} In the Vitali convergence theorem \ref{p16} condition $(ii)$, the set $E$ with $|E| <\infty$ can be replaced by the condition that $E$ is bounded. \end{lemma} \begin{proof} Let $E_{n}=E\cap B^{N}(0;n)$ for $n=1,2,\dots$. Then $E_{1}\subset E_{2}\subset\dots\nearrow E$. Thus $| E_{1}| \leq| E_{2}| \leq\dots\nearrow| E|$. For $\delta>0$ as in Theorem \ref{p16} condition $(iii)$, there is an $E_{N}$ such that $| E\backslash E_{N}| <\delta$. Now $\int_{E_{N}^{c}}| u_{n}| dz=\int_{E^{c}}| u_{n}| dz+\int_{E\backslash E_{N}}| u_{n}| dz<2\varepsilon$ for each $n\in\mathbb{N}$. \end{proof} \begin{lemma} \label{p20} Let $\Omega$ be a domain in $\mathbb{R}^{N}$, $1\leq r0$, then $W_{0} ^{m,p}(\Omega)\hookrightarrow L^{q}(\Omega)$, where $q\in[p,p^{*}]$, $\frac {1}{p^{*}}=\frac{1}{p}-\frac{m}{N}$; \newline $(ii)$ If $\frac{1}{p}-\frac{m} {N}=0$, then $W_{0}^{m,p}(\Omega)\hookrightarrow L^{q}(\Omega)$, where $q\in[p,\infty)$; \newline $(iii)$ If $\frac{1}{p}-\frac{m}{N}<0$, then $W_{0}^{m,p}(\Omega)\hookrightarrow L^{\infty}(\Omega)$.\newline Moreover, if $m-\frac{N}{p}>0$ is not an integer, let $k=\left[ m-\frac{N}{p}\right]$ and $\theta=m-\frac{N}{p}-k$ $(0<\theta<1)$, then we have for $u\in W_{0}^{m,p}(\Omega)$ \begin{gather*} \| D^{\beta}u\| _{L^{\infty}}\leq c\| u\| _{W^{m,p}} \quad \text{for }|\beta|\leq k\\ | u(x)-u(y)| \leq c\| u\| _{W^{m,p}}| x-y| ^{\theta}\quad \quad\text{a.e. for }x,y\in\Omega. \end{gather*} In particular, $W_{0}^{m,p}(\Omega)\hookrightarrow C^{k,\theta}(\overline {\Omega})$. \end{theorem} For the proof ot the theorem above, see Gilbarg-Trudinger \cite[p.164]{GT}. \begin{definition} \label{p22} \rm $\Omega$ satisfies a uniform interior cone condition if a fixed cone $K_{\Omega}$ exists such that each $x\in\partial\Omega$ is the vertex of a cone $K_{\Omega}(x)\subset\overline{\Omega}$ and congruent to $K_{\Omega}$. \end{definition} \begin{theorem}[Sobolev Embedding Theorem in $W^{m,p}(\Omega)$]\label{p23} Let $\Omega$ satisfy a uniform interior cone condition, $m\in\mathbb{N}$ and $1\leq p<\infty$. Then we have the following continuous injections.\newline $(i)$ If $\frac{1}{p}-\frac{m}{N}>0$, then $W^{m,p}(\Omega)\hookrightarrow L^{q} (\Omega)$, where $q\in[p,p^{*}]$ and $\frac{1}{p^{*}}=\frac{1}{p}-\frac{m}{N}$; \newline $(ii)$ If $\frac{1}{p}-\frac{m}{N}=0$, then $W^{m,p} (\Omega)\hookrightarrow L^{q}(\Omega)$, where $q\in[p,\infty)$; \newline $(iii)$ If $\frac{1}{p}-\frac{m}{N}<0$, then $W^{m,p}(\Omega)\hookrightarrow L^{\infty}(\Omega)$.\newline Moreover, if $m-\frac{N}{p}>0$ is not an integer, let $k=\big[ m-\frac{N}{p}\big] \quad \text{and}\quad \theta=m-\frac{N}{p}-k\quad (0<\theta<1),$ then we have for $u\in W^{m,p}(\Omega)$, \begin{gather*} \| D^{\beta}u\| _{L^{\infty}}\leq c\| u\| _{W^{m,p} }\quad\text{for }\beta\quad\text{with }|\beta|\leq k\\ | D^{\beta}u(x)-D^{\beta}u(y)| \leq c\| u\| _{W^{m,p}}| x-y| ^{\theta}\quad\text{a.e. for }x,y\in\Omega\quad\text{and }|\beta|=k. \end{gather*} In particular, $W^{m,p}(\Omega)\hookrightarrow C^{k,\theta}(\overline{\Omega })$. \end{theorem} For the proof of the theorem above, see Br\'{e}zis \cite[Cor. IX.13]{B} and Gilbarg-Trudinger \cite[Theorem 7.26]{GT}. \begin{theorem}[Rellich-Kondrakov Theorem in $W_{0}^{m,p}(\Omega)$] \label{p24} Let $\Omega$ be a bounded domain, $m\in\mathbb{N}$ and $1\leq p<\infty$. Then we have the following compact injections.\newline $(i)$ If $\frac{1}{p}-\frac{m}{N}>0$, then $W_{0}^{m,p}(\Omega)\hookrightarrow L^{q}(\Omega)$, where $q\in[1,p^{*} )$, $\frac{1}{p^{*}}=\frac{1}{p}-\frac{m}{N}$; \newline $(ii)$ If $\frac{1} {p}-\frac{m}{N}=0$, then $W_{0}^{m,p}(\Omega)\hookrightarrow L^{q}(\Omega)$, where $q\in[1,\infty)$; \newline $(iii)$ If $\frac{1}{p}-\frac{m}{N}<0$, then $W_{0}^{m,p}(\Omega)\hookrightarrow C^{k}(\overline{\Omega})$, where $m-\frac{N}{p}>0$ is not an integer and $k=\left[ m-\frac{N}{p}\right]$. \end{theorem} For the proof of the aboved theroem, see Gilbarg-Trudinger \cite[Theorem 7.22]{GT}. \begin{theorem}[Rellich-Kondrakov Theorem in $W^{m,p}(\Omega)$] \label{p25} Let $\Omega$ be a bounded $C^{0,1}$ domain in $\mathbb{R}^{N}$, $m\in\mathbb{N}$ and $1\leq p<\infty$. Then we have the following compact injections.\newline$(i)$ If $\frac{1}{p}-\frac{m}{N}>0$, then $W^{m,p}(\Omega)\hookrightarrow L^{q} (\Omega)$, where $q\in[1,p^{*})$, $\frac{1}{p^{*}}=\frac{1}{p}-\frac{m}{N}$; \newline$(ii)$ If $\frac{1}{p}-\frac{m}{N}=0$, then $W^{m,p}(\Omega )\hookrightarrow L^{q}(\Omega)$, where $q\in[1,\infty)$; \newline$(iii)$ If $\frac{1}{p}-\frac{m}{N}<0$, then $W^{m,p}(\Omega)\hookrightarrow C^{k,\beta }(\overline{\Omega})$, where $m-\frac{N}{p}>0$ is not an integer, $0<\beta<\theta$, $k=\left[ m-\frac{N}{p}\right]$, and $\theta=m-\frac {N}{p}-k$ $(0<\theta<1)$. \end{theorem} For the proof of the above theorem, see Br\'{e}zis \cite[p. 169]{B} and Gilbarg-Trudinger \cite[Theorem 7.26]{GT}. For the Sobolev space $X(\Omega)$, we can drop condition $(iii)$ of the Vitali convergence theorem \ref{p16} through the interpolation results. \begin{theorem}[Rellich-Kondrakov Theorem]\label{p26} Let $\Omega$ be a domain in $\mathbb{R}^{N}$ of finite measure. Then the embedding $X(\Omega )\hookrightarrow L^{p}(\Omega)$ is compact. \end{theorem} \begin{proof} Let $\{ u_{n}\}$ be a bounded sequence in $X(\Omega)$, then by Lemma \ref{p4}, a subsequence $\{ u_{n}\}$ and $u\in X(\Omega)$ exist such that $u_{n}\to u$ a.e. in $\Omega$. By the Egorov theorem, for $\varepsilon>0$, a closed subset $F$ in $\mathbb{R}^{N}$ exists such that $F\subset\Omega$, $| \Omega\backslash F| <\varepsilon$, and $u_{n}\to u$ uniformly in $F$. Thus, $\int_{F}| u_{n}-u| ^{p}= o(1)\quad \text{as }n\to\infty.$ For $N>2$, we have \begin{align*} {\int_{\Omega\backslash F}} | u_{n}-u| ^{p} & \leq\Big( {\int_{\Omega\backslash F}}1\Big) ^{1/r} \Big({\int_{\Omega\backslash F}}| u_{n}-u| ^{ps}\Big) ^{1/s}\\ & \leq| \Omega\backslash F| ^{1/r}\Big( {\int_{\Omega}} | u_{n}-u| ^{ps}\Big) ^{1/s}\\ & \leq c\| u_{n}-u\| _{H^{1}}^{p}| \Omega\backslash F| ^{1/r} 1$to obtain the above inequality. Hence,$u_{n}\to u$strongly in$L^{p}(\Omega) $. \end{proof} \begin{theorem}[Vitali Convergence Theorem for$X(\Omega)$] \label{p27}$(i)$Let$\Omega$be a domain in$\mathbb{R}^{N}$of finite measure. Then the embedding$X(\Omega)\hookrightarrow L^{p}( \Omega) $is compact;\newline$(ii) $Let$\Omega$be a domain in$\mathbb{R}^{N}$and let$\{ u_{n}\} _{n=1}^{\infty}$be a sequence in$X(\Omega)$. Suppose that a constant$c>0$exists such that$\| u_{n}\| _{H^{1}}\leq c$for each$n$and$u_{n}\to u$a.e. in$\Omega$. Then for each$\varepsilon>0$, a measurable set$E\subset\Omega$exists such that$| E| <\infty$and$\int _{E^{c}}| u_{n}| ^{p}dz<\varepsilon$\ for each$n\in\mathbb{N} \ $if and only if$\| u_{n}-u\| _{_{L^{p}(\Omega)}}= o(1)$. \end{theorem} \begin{proof} Part$(i)$follows from Willem \cite{Wi}.$(ii)$By the Fatou lemma,$\int_{E^{c}}| u| ^{p}dz\leq\varepsilon$. Since$| E| <\infty$and$\| u_{n}\| _{H^{1}}\leq c$, by$(i) $, there is a subsequence$\{ u_{n}\} _{n=1}^{\infty}$satisfying $\int_{E}| u_{n}-u| ^{p}dz= o(1).$ Therefore, $\int_{\Omega}| u_{n}-u| ^{p}dz=\int_{E\cap\Omega}| u_{n}-u| ^{p}dz+\int_{E^{c}\cap\Omega}| u_{n}-u| ^{p}dz= o(1).$ Now suppose$\| u_{n}-u\| _{L^{p}(\Omega)}= o(1)$. Then for each$\varepsilon>0$, a positive integer$n_{0}$exists such that$\| u_{n}-u\| _{L^{p}(\Omega)}<\frac{\varepsilon^{1/p}}{2}$for$n>n_{0}$, and measurable sets$A$and$B$of finite measure exist such that $\int_{A^{c}}| u| ^{p}dz<\frac{\varepsilon}{2^{p}}\quad\text{and}\quad \int_{B^{c}}| u_{n}| ^{p}dz<\varepsilon\quad \text{for }n=1,2,\dots ,n_{0}.$ The Minkowski inequality implies $\| u_{n}\| _{L^{p}(A^{c})}\leq\| u_{n}-u\| _{L^{p}(A^{c})}+\| u\| _{L^{p}(A^{c})}<\varepsilon^{1/p}\quad\text{for }n>n_{0}.$ Then let$E=A\cup B$to obtain the conclusion. \end{proof} Let$L_{w}^{p}(\mathbb{R}^{N})=\{ u\in L_{\rm loc}^{p}(\mathbb{R}^{N}): \int_{\mathbb{R}^{N}}| u(z) | ^{p}w(z) dz<\infty\} $be a weighted Lebesgue space, where the weight$w$is nonnegative with $\| u\| _{L_{w}^{p}(\mathbb{R}^{N})}^{p}=\int_{\mathbb{R}^{N} }| u(z) | ^{p}w(z) dz.$ We denote by$Q(x,l) $the cube of the form $Q(x,l) =\{ y\in\mathbb{R}^{N}: | y_{j}-x_{j}| 2. Suppose that w\in L_{w}^{\frac{p+\delta}{\delta}}(\mathbb{R}^{N}), with 2\leq p0, and $$\lim_{| x| \to\infty}\int_{Q(x,l) }w( z) ^{^{\frac{p+\delta}{\delta}}}dz=0\label{A1}$$ for some l>0. Then H^{1}(\mathbb{R}^{N}) is compactly embedded in L_{w}^{p}(\mathbb{R}^{N}); \newline(ii) Let N=2 and suppose that w\in L_{w}^{s}(\mathbb{R}^{N}) for some s>1 and $$\lim_{|x| \to\infty}\int_{Q(x,l) }w(z) ^{^{s}}dz=0\label{A2}$$ for some l>0. Then H^{1}(\mathbb{R}^{N}) is compactly embedded in L_{w}^{p}(\mathbb{R}^{N}) for every p\geq2; \newline(iii) Let N=1 and suppose that w\in L_{\rm loc}^{1}(\mathbb{R}^{N}) and $$\lim_{| x| \to\infty}\int_{Q(x,l) }w( z) dz=0\label{A3}$$ for some l>0. Then H^{1}(\mathbb{R}^{N}) is compact embedded in L_{w} ^{p}(\mathbb{R}^{N}) for every p\geq2. \end{theorem} \begin{proof} (i) It suffices to show that for every \varepsilon>0, a R>0 exists such that $$\| u-u\chi_{Q(0,R) }\| _{L_{w}^{p}(\mathbb{R}^{N} )}<\varepsilon\label{A4}$$ for each u\in H^{1}(\mathbb{R}^{N}) such that \| u\| _{H^{1}(\mathbb{R}^{N})}\leq1, where \chi_{Q} is the characteristic function of the cube. Indeed, let \{ u_{n}\} be a bounded sequence in H^{1}(\mathbb{R}^{N}). We assume that \| u_{n}\| _{H^{1}(\mathbb{R}^{N})}\leq1 for all n\in\mathbb{N}. Consequently, a subsequence \{ u_{n}\} and a u\in H^{1}(\mathbb{R}^{N}) exist such that u_{n}\rightharpoonup u in H^{1}(\mathbb{R}^{N}) and u_{n}\to u in L^{p}(Q(0,R) ) . On the other hand, by \eqref{A4}, we have \[ \| u_{n}-u\| _{L_{w}^{p}(\mathbb{R}^{N}\backslash Q( 0,R) )}\leq\| u_{n}\| _{L_{w}^{p}(\mathbb{R}^{N}\backslash Q(0,R) )}+\| u\| _{L_{w}^{p}(\mathbb{R} ^{N}\backslash Q(0,R) )}\leq2\varepsilon.$ Combining this with the previous observation, it is easy to conclude that$u_{n}\to u$in$L_{w}^{p}(\mathbb{R}^{N}) $. To show \eqref{A4}, we cover$\mathbb{R}^{N}$with cubes$Q(\hat{z},1) $,$\hat{z}\in\mathbb{Z}^{N}$. We may assume that$(i) $holds with$l=1$. For$\eta>0$, we use$( \text{\ref{A1}}) $to find a positive constant$n_{0}$such that$\int_{Q}w(z) ^{\frac{p+\delta}{\delta}}dz<\eta$for each$Q=Q(\hat{z},1) $outside$Q(0,n_{0}) $. By the Sobolev embedding theorem, for any$u\in H^{1}(\mathbb{R}^{N}) $, a constant$c>0$exists such that $\| u\| _{L^{p}(Q) }\leq c\| u\| _{H^{1}(Q) }\quad\text{for all }2\leq p<2^{*}.$ Thus, by the H\"{o}lder inequality, we have $\int_{Q}| u| ^{p}wdz\leq\Big(\int_{Q}w^{\frac{p+\delta} {\delta}}dz\Big) ^{\frac{\delta}{p+\delta}} \Big(\int_{Q}|u| ^{p+\delta}dz\Big) ^{\frac{p}{p+\delta}} \leq c'\eta^{1/s}\| u\| _{H^{1}(Q) }^{p}$ where$c'=c^{p/(p+\delta)}$. Now, choose$c'\eta^{1/s}<\varepsilon$and add these inequalities over all$Q(\hat{z},1) $outside$Q(0,n_{0}) $to obtain$R=n_{0}$. \newline$(ii)$and$(iii)$are similar to$(i)$. \end{proof} We define$H_{r}^{1}(\Omega) =\{ u\in H_{0}^{1}(\Omega): u \quad\text{is radially symmetric}\} $. \begin{lemma}\label{p28} For$N\geq2$, every$u\in H_{r}^{1}(\mathbb{R}^{N})$is equal to a continuous function$U$a.e. in$\mathbb{R}^{N}\backslash\{0\}$such that for$z\neq0$$|U(z)|\leq(\frac{2}{\omega_{N}}) ^{1/2}|z|^{\frac{1-N}{2}} \Big(\int_{|t|\geq|z|}|u(t)|^{2}dt\Big) ^{1/4} \Big(\int_{|y|\geq|z|}|\nabla u(t)|^{2}dt\Big) ^{1/4},$ where$\omega_{N}$is the area of the unit ball in$\mathbb{R}^{N}$. \end{lemma} \begin{proof} Let$\varphi\in C_{c}^{\infty}(\mathbb{R}^{N})$be a radially symmetric function. Then for$0\leq r<\infty, \begin{align*} r^{N-1}\varphi(r)^{2} & =\int_{0}^{r}(s^{N-1}\varphi(s)^{2}) 'ds\\ & =(N-1)\int_{0}^{r}s^{N-2}\varphi(s)^{2}ds+2\int_{0}^{r}s^{N-1} \varphi(s)\varphi'(s)ds. \end{align*} Thus, $0=(N-1)\int_{0}^{\infty}s^{N-2}\varphi(s)^{2}ds+2\int_{0}^{\infty} s^{N-1}\varphi(s)\varphi'(s)ds.$ Consequently, \begin{align*} r^{N-1}\varphi(r)^{2} & \leq(N-1){\int_{0}^{\infty}} s^{N-2}\varphi(s)^{2}ds+2{\int_{0}^{r}} s^{N-1}\varphi(s)\varphi'(s)ds\\ & =-2 {\int_{r}^{\infty}} s^{N-1}\varphi(s)\varphi'(s)ds\\ & =(\frac{-2}{\omega_{N}}){\int_{|t|\geq r}} \varphi(t)\varphi'(t)dt\\ & \leq(\frac{2}{\omega_{N}}) \Big({\int_{|t|\geq r}} |\varphi(t)|^{2}dt\Big) ^{1/2}\Big({\int_{|t|\geq r}} |\nabla\varphi(t)|^{2}dt\Big) ^{1/2}. \end{align*} Foru\in H_{r}^{1}(\mathbb{R}^{N})$, take a sequence$\{\varphi_{n}\}$radially symmetric in$C_{c}^{\infty}(\mathbb{R}^{N})$, such that $\varphi_{n}\to u\quad\mbox{in }H^{1}(\mathbb{R}^{N}),$ then there is a subsequence$\{\varphi_{n}(r)\}such that \begin{align*} r^{N-1}u(r)^{2} & =\underset{n\to\infty}{\lim}r^{N-1}\varphi _{n}(r)^{2}\\ & \leq\lim_{n\to\infty}(\frac{2}{\omega_{N}}) \Big(\int_{|t|\geq r}|\varphi_{n}(t)|^{2}dt\Big) ^{1/2} \Big(\int_{|t|\geq r}|\nabla\varphi_{n}(t)|^{2}dt\Big) ^{1/2}\\ & \leq(\frac{2}{\omega_{N}}) \Big(\int_{|t|\geq r} |u(t)|^{2}dt\Big) ^{1/2} \Big(\int_{|t|\geq r}|\nabla u(t)|^{2}dt\Big)^{1/2}. \end{align*} Sinceu\in H_{r}^{1}(\mathbb{R}^{N})$, it is a function in$H^{1} (\mathbb{R})$, and there is a continuous function$U$in$\mathbb{R}$such that$u=U$a.e. and $|U(z)|\leq(\frac{2}{\omega_{N}}) ^{1/2}|z|^{\frac{1-N}{2} }\Big(\int_{|t|\geq|z|}|u(t)|^{2}dt\Big) ^{1/4} \Big(\int_{|t|\geq |z|}|\nabla u(t)|^{2}dt\Big) ^{1/4}.$ \end{proof} Let$\Theta$be an annulus, say$\Theta=\{ z\in\mathbb{R} ^{N}: 1<| z| \} $with$N\geq3$. \begin{theorem}[Rellich-Kondrakov Theorem for$H_{r}^{1}(\Theta)$] \label{p29} The embedding \\$H_{r}^{1}(\Theta)\hookrightarrow L^{p}(\Theta)$is compact. \end{theorem} \begin{proof} Let$\{ u_{n}\} $be a bounded sequence in$H_{r}^{1}(\Theta)$. Then a subsequence$\{ u_{n}\} $exists such that$u_{n} \to u$a.e. in$\Theta$and$u_{n}\rightharpoonup u$weakly in$H_{0}^{1}(\Theta)$. By Lemma \ref{p28},$\underset{|z|\to\infty} {\lim}u_{n}(z)=0$uniformly in$n$and$\underset{s\to0}{\lim} \frac{|s|^{p}}{|s|^{2}+|s|^{2^{*}}}=0$. Thus, for$\varepsilon>0$, there is a$K>0$such that if$|z|\geq K$, for each$n$, we have $| u_{n}(z)| ^{p}\leq\varepsilon(| u_{n}(z)| ^{2}+| u_{n}(z)| ^{2^{*}}) ,$ or ${\int_{\Theta_{K}^{c}}} | u_{n}| ^{p}\leq c\varepsilon,$ where$\Theta_{K}=\{ z\in\Theta : \ z|0$, then$c>0$exists such that$\| u_{n}\| _{H^{1}}\geq c$for each$n$. \end{lemma} \begin{proof} Suppose that a subsequence$\{u_{n}\}$satisfies$\underset{n\to \infty}{\lim}\| u_{n}\| _{H^{1}}=0$. Then$J(u_{n})= o(1)$, but this contradicts$\beta>0$. Thus,$c>0$exists such that$\| u_{n}\|_{H^{1}}\geq c$for each$n$. \end{proof} Let$\Omega$be an unbounded domain and$\xi_{n}$as in (\ref{1-1}), then we have the following lemma. \begin{lemma} \label{p32} Let$\{ u_{n}\} $be a (PS)$_{\beta}$-sequence in$X(\Omega) $for$J$such that $\int_{\Omega_{n}}|u_{n}|^{p}= o(1),$ where$\Omega_{n}=\Omega\cap B^{N}(0;n)$. Then for any$r\geq1$, we have\newline$(i){\int_{\Omega}} \xi_{n}^{r}|u_{n}|^{p}= {\int_{\Omega}} |u_{n}|^{p}+ o(1)=\frac{2p}{p-2}\beta+\text{o} (1)$; \newline$(ii){\int_{\Omega}} \xi_{n}^{r}(|\nabla u_{n}|^{2}+u_{n}^{2})= {\int_{\Omega}} \xi_{n}^{r}|u_{n}|^{p}+ o(1)=\frac{2p}{p-2}\beta + o(1)$; \newline$(iii){\int_{\Omega}} (\xi_{n}^{r}-1)u_{n}\varphi= o(1)\| \varphi\| _{H^{1}}$for every$\varphi\in X(\Omega)$; \newline$(iv)|{\int_{\Omega}} (\xi_{n}^{r}-1) | u_{n}| ^{p-2}u_{n}\varphi| = o(1)\| \varphi\| _{H^{1}}$for every$\varphi\in X(\Omega)$; \newline$(v)|{\int_{\Omega}} (\xi_{n}^{r}-1) \nabla u_{n}\nabla\varphi| = o(1)\| \varphi\| _{H^{1}}$for every$\varphi\in X(\Omega)$. \end{lemma} \begin{proof}$(i)$Clearly, we have $\int_{\Omega}\xi_{n}^{r}|u_{n}|^{p}=\int_{\Omega}|u_{n} |^{p}+ o(1)=\frac{2p}{p-2}\beta+ o(1).$$(ii)$Let$w_{n}=\xi_{n}^{r}u_{n}$. Since$\{w_{n}\}$is bounded in$X(\Omega), we have \begin{align*} o(1) & =\langle J'(u_{n}),w_{n}\rangle \\ & =\int_{\Omega}(\xi_{n}^{r}|\nabla u_{n}|^{2}+r\xi_{n}^{r-1} u_{n}\nabla\xi_{n}\cdot\nabla u_{n}+\xi_{n}^{r}u_{n}^{2})-\int _{\Omega}\xi_{n}^{r}|u_{n}|^{p}. \end{align*} Note that|\nabla\xi_{n}(z)|\leq\frac{c}{n}$and$\{ u_{n}\} $is bounded in$X(\Omega)$, so $\int_{\Omega}\xi_{n}^{r-1}u_{n}\nabla\xi_{n}\cdot\nabla u_{n}= o(1).$ We conclude that $\int_{\Omega}\xi_{n}^{r}(|\nabla u_{n}|^{2}+u_{n}^{2})=\int _{\Omega}\xi_{n}^{r}|u_{n}|^{p}+ o(1)=\frac{2p} {p-2}\beta+ o(1).$ Therefore, the results follow.\newline$(iii)$By the H\"{o}lder and Sobolev inequalities, we have $|{\int_{\Omega}} (\xi_{n}^{r}-1) u_{n}\varphi| \leq\Big(\int_{\Omega_{n}}|u_{n}|^{2}\Big) ^{1/2} \Big(\int_{\Omega}|\varphi|^{2}\Big) ^{1/2} \leq o(1)\| \varphi\| _{H^{1}}.$$(iv)$By the H\"{o}lder and Sobolev inequalities, we have $\Big|{\int_{\Omega}} (\xi_{n}^{r}-1) | u_{n}| ^{p-2}u_{n}\varphi\Big| \leq(\int_{\Omega_{n}}|u_{n}|^{p}) ^{\frac{p-1}{p}}( \int_{\Omega}|\varphi|^{p}) ^{1/p}\\ \leq o(1)\|\varphi\|_{H^{1}}.$$(v)$By the hypothesis and part$(i), we have \begin{align*} o(1) & =\langle J'(u_{n}),w_{n}\rangle \\ & =\langle J'(u_{n}),w_{n}\rangle -\langle J'(u_{n}),u_{n}\rangle +\langle J'(u_{n}),u_{n}\rangle \\ & =\int_{\Omega}(\xi_{n}^{r}-1) |\nabla u_{n}|^{2} +\int_{\Omega}(\xi_{n}^{r}-1) u_{n}^{2}-\int _{\Omega}(\xi_{n}^{r}-1) |u_{n}|^{p}+ o(1)\\ & =\int_{\Omega}(\xi_{n}^{r}-1) |\nabla u_{n} |^{2}+ o(1). \end{align*} Thus, $\big| \int_{\Omega}(\xi_{n}^{r}-1) |\nabla u_{n} |^{2}\big| =\int_{\Omega}(1-\xi_{n}^{r}) |\nabla u_{n}|^{2}= o(1).$ Therefore, by the H\"{o}lder inequality, \begin{align*} |{\int_{\Omega}} (\xi_{n}^{r}-1) \nabla u_{n}\nabla\varphi| & \leq\Big({\int_{\Omega}}(\xi_{n}^{r}-1) ^{2}| \nabla u_{n}| ^{2}\Big) ^{1/2}\| \varphi\| _{H^{1}}\\ & \leq\Big({\int_{\Omega}} (1-\xi_{n}^{r}) | \nabla u_{n}| ^{2}\Big)^{1/2} \| \varphi\| _{H^{1}}\\ & \leq o(1)\| \varphi\| _{H^{1}}. \end{align*} \end{proof} \begin{lemma}\label{p33}(i)$Suppose that$\{ u_{n}\} $is a sequence in$X(\Omega)$satisfying$u_{n}\rightharpoonup0$weakly in$X(\Omega)$, then there is a subsequence$\{ u_{n}\} $in$X(\Omega)$such that$\int_{\Omega_{n}}| u_{n}| ^{p} = o(1)$as$n\to\infty$; \newline$(ii)$For any$\beta>0$, suppose that$\{ u_{n}\} $is a (PS)$_{\beta}$-sequence in$X(\Omega)$for$J$satisfying$\int_{\Omega_{n}}| u_{n}| ^{p}= o(1)$as$n\to\infty$, then$\{ \xi_{n} u_{n}\} $is also a (PS)$_{\beta}$-sequence in$X(\Omega)$for$J$. \end{lemma} \begin{proof}$(i)$Since$u_{n}\rightharpoonup0$weakly in$X(\Omega)$, there is a subsequence$\{ u_{n}\} $such that$u_{n}\to u$strongly in$L_{\rm loc}^{p}(\Omega) $, or there is a subsequence$\{ u_{n}\} $such that $\int_{\Omega_{n}}| u_{n}| ^{p}= o(1),$ where$\Omega_{n}=\Omega\cap B^{N}(0;n) $. \newline$(ii)$Let$\{ u_{n}\} $be a (PS)$_{\beta}$-sequence in$X(\Omega)$for$J$satisfying$\int_{\Omega_{n}}| u_{n}| ^{p}= o(1)$as$n\to\infty. By Lemma \ref{p32}, we have \begin{align*} J(\xi_{n}u_{n}) & =\frac{1}{2} {\int_{\Omega}} \left[ |\nabla(\xi_{n}u_{n}) |^{2}+(\xi_{n}u_{n}) ^{2}\right] -\frac{1}{p} {\int_{\Omega}}| \xi_{n}u_{n}| ^{p}\\ & =\frac{1}{2} {\int_{\Omega}} \left[ |\nabla\xi_{n}|^{2}u_{n}^{2}+\xi_{n}^{2}(|\nabla u_{n}|^{2}+u_{n} ^{2})+2\xi_{n}u_{n}\nabla\xi_{n}\nabla u_{n}\right] -\frac{1}{p} {\int_{\Omega}} \xi_{n}^{p}| u_{n}| ^{p}\\ & =\frac{1}{2}a(u_{n}) -\frac{1}{p}b(u_{n}) + o(1) =\beta+ o(1) . \end{align*} Then for\varphi\in X(\Omega) , we have \begin{align*} &| \langle J'(\xi_{n}u_{n}),\varphi\rangle| \\ & =\big|\langle J'(\xi_{n}u_{n}),\varphi\rangle-\langle J' (u_{n}),\varphi\rangle+\langle J'(u_{n}),\varphi\rangle\big| \\ & =\big| \int_{\Omega}(\xi_{n}\nabla u_{n}\nabla\varphi+u_{n} \nabla\xi_{n}\nabla\varphi+\xi_{n}u_{n}\varphi-\xi_{n}^{p-1}|u_{n}|^{p-2} u_{n}\varphi) \\ &\quad -\langle J'(u_{n}),\varphi\rangle+\langle J' (u_{n}),\varphi\rangle\big| \\ & =\big| \int_{\Omega}\left[ (\xi_{n}-1)\nabla u_{n}\nabla \varphi+(\xi_{n}-1)u_{n}\varphi-(\xi_{n}^{p-1}-1)|u_{n}|^{p-2}u_{n} \varphi\right] +\langle J'(u_{n}),\varphi\rangle\big| \\ & \leq o(1)\| \varphi\| _{H^{1}} \end{align*} Thus,J'(\xi_{n}u_{n})= o(1)$. \end{proof} Moreover, we have the following lemma. \begin{lemma} \label{p34} Let$\{ u_{n}\} $be a (PS)-sequence in$H_{0} ^{1}(\Omega) $for$J$satisfying$u_{n}\rightharpoonup0$weakly in$X(\Omega) $and let$v_{n}=\xi_{n}u_{n}$. Then$\| u_{n}-v_{n}\| _{H^{1}}= o(1)$as$n\to\infty. \end{lemma} \begin{proof} Note that \begin{align*} a(u_{n}-v_{n}) & =\langle u_{n}-v_{n},u_{n}-v_{n} \rangle _{H^{1}}\\ & =a(u_{n}) +a(v_{n}) -2\langle u_{n} ,v_{n}\rangle _{H^{1}}\\ & =2a(u_{n}) -2\langle u_{n},v_{n}\rangle _{H^{1} }+ o(1) . \end{align*} Thus, it suffices to show that $a(u_{n}) =\langle u_{n},v_{n}\rangle _{H^{1}}+ o(1) .$ We have \begin{align*} \langle u_{n},v_{n}\rangle _{H^{1}} & =\int_{\Omega}\nabla u_{n}\nabla v_{n}+u_{n}v_{n}\\ & =\int_{\Omega}\xi_{n}\left[ | \nabla u_{n}| ^{2}+( u_{n}) ^{2}\right] +\int_{\Omega}u_{n}\nabla u_{n}\nabla\xi_{n}. \end{align*} Note that| \nabla\xi_{n}| \leq\frac{c}{n}$and$\{ u_{n}\} $is a (PS)-sequence in$H_{0}^{1}(\Omega) $for$J$, so $\int_{\Omega}u_{n}\nabla u_{n}\nabla\xi_{n}= o(1) .$ Hence, $\langle u_{n},v_{n}\rangle _{H^{1}}=\int_{\Omega}\xi_{n}\left[ | \nabla u_{n}| ^{2}+(u_{n}) ^{2}\right] +o( 1) .$ By Lemma \ref{p32}$(i)$,$(ii) $and Lemma \ref{p33}$(i)$, we have $\langle u_{n},v_{n}\rangle _{H^{1}} =\int_{\Omega}\xi_{n}\left[ | \nabla u_{n}| ^{2}+(u_{n}) ^{2}\right] +o(1) =a(u_{n}) + o(1) .$ \end{proof} \noindent\textbf{Bibliographical notes:} The (PS)-sequences were originally introduced by Palais-Smale \cite{PS}. Lemma \ref{p3} is from Br\'{e}zis \cite[p. 35]{B}. Lemma \ref{p4} is from Zeidler \cite[II/A, p. 303]{Z}. Lemma \ref{p9} is from Bahri-Lions \cite{BaLi}. Lemma \ref{p12} is from Grisvard \cite[p. 24]{G}. \section{Palais-Smale Decomposition Theorems} In this section, we present the Palais-Smale decomposition theorem in$H_{0}^{1}(\Omega)$for$J$. This is the concentration-compactness method of P. L. Lions. \begin{theorem}[Palais-Smale Decomposition Theorem in$\mathbb{R}^{N})$] \label{d1} Let$\Omega$be strictly large domain (see Definition \ref{f4}) in$\mathbb{R}^{N}$and let$\{ u_{n}\} $be a (PS)$_{\beta}$-sequence in$H_{0}^{1}(\Omega)$for$J$. Then there are a subsequence$\{ u_{n}\} $, a positive integer$m$, sequences$\{z_{n}^{i}\}_{n=1}^{\infty}$in$\mathbb{R}^{N}$, a function$\bar{u}\in H_{0}^{1}(\Omega)$, and$0\neq w^{i}\in H^{1}(\mathbb{R}^{N}) $for$1\leq i\leq m$such that \begin{gather*} |z_{n}^{i}|\to\infty,\quad \text{for }i=1,2,\dots,m,\\ -\Delta\bar{u}+\bar{u}=|\bar{u}|^{p-2}\bar{u}\quad\mbox{in }\Omega,\\ -\Delta w^{i}+w^{i}=\mid w^{i}|^{p-2}w^{i}\quad\mbox{in }\mathbb{R}^{N}, \end{gather*} and \begin{gather*} u_{n}=\bar{u}+\sum_{i=1}^m w^{i}(\cdot-z_{n} ^{i}) + o(1)\;\text{strongly}\quad\text{}\text{in}\quad\text{}H^{1}( \mathbb{R}^{N}) ,\\ a(u_{n})=a(\bar{u})+\sum_{i=1}^m a(w^{i})+ o(1),\\ b(u_{n})=b(\bar{u})+\sum_{i=1}^m b(w^{i})+ o(1),\\ J(u_{n})=J(\bar{u})+\sum_{i=1}^m J(w^{i})+ o(1). \end{gather*} In addition, if$u_{n}\geq0$, then$\bar{u}\geq0$and$w^{i}\geq0$for each$1\leq i\leq m$. \end{theorem} \begin{proof} \textbf{Step 0}. Since$\{ u_{n}\} $is a (PS)$_{\beta}$-sequence in$H_{0}^{1}(\Omega) $for$J$, by Lemma \ref{p30} there is a$c>0$such that$\| u_{n}\| _{H^{1}}\leq c$. In the following proof of this theorem, we fix such a$c$. There is a subsequence$\{ u_{n}\} $and a$\bar{u}$in$H_{0}^{1}(\Omega)$such that$u_{n}\rightharpoonup\bar{u}$weakly in$H_{0}^{1}(\Omega)$and$\bar{u}$solves $-\Delta\bar{u}+\bar{u}=|\bar{u}|^{p-2}\bar{u}\quad\mbox{in }\Omega.$ Suppose that$u_{n}\to\bar{u}$strongly in$H_{0}^{1}(\Omega)$, then we have$u_{n}=\overline{u}+ o(1) $strongly in$H_{0}^{1}( \Omega) $,$a(u_{n}) =a(\overline{u}) + o(1) $,$b(u_{n}) =b(\overline{u}) + o(1) $,$J(u_{n}) =J(\overline{u}) + o(1) $.\newline \textbf{Step 1}. Suppose that$u_{n} \nrightarrow\bar{u}$strongly in$H_{0}^{1}(\Omega)$. Let $u_{n}^{1}=u_{n}-\bar{u}\quad \text{for }n=1,2,\dots.$ By Lemma \ref{p8},$\{u_{n}^{1}\}$is a (PS)$_{(\beta-J(\bar{u}))}$-sequence in$H_{0}^{1}(\Omega)$for$J$. \begin{itemize} \item[(1-0)]$ {\int_{B^{N}(0;1) }} |w_{n}^{1}(z)|^{2}dz\geq\frac{d_{1}}{2}$for some constant$d_{1}>0$and$n=1,2,\dots$, where$w_{n}^{1}(z)=u_{n}^{1}(z+y_{n}^{1}) $for some$\{ y_{n}^{1}\} \subset\mathbb{R}^{N}$: since$\{ u_{n}^{1}\} $is bounded,$J'(u_{n}^{1})= o(1)$, and$u_{n}^{1}\nrightarrow0$strongly in$H_{0}^{1}(\Omega)$. By Lemma \ref{p9} there is a subsequence$\{ u_{n}^{1}\} $, a constant$d_{1}>0$such that $Q_{n}^{1}=\sup_{z\in\mathbb{R}^{N}}\int_{z+B^{N}(0;1)}|u_{n}^{1}|^{2}\geq d_{1}\quad\text{for } n=1,2,\dots.$ Take$\{ y_{n}^{1}\} $in$\mathbb{R}^{N}$such that $\int_{y_{n}^{1}+B^{N}(0;1)}|u_{n}^{1}(z)|^{2}dz\geq\frac{d_{1}}{2}.$ Let$w_{n}^{1}(z)=u_{n}^{1}(z+y_{n}^{1}) $, then $\int_{B^{N}(0;1)}|w_{n}^{1}(z)|^{2}dz\geq\frac{d_{1}}{2}\quad\text{for } n=1,2,\dots.$ \item[(1-1)]$u_{n}(z)=\bar{u}(z)+w_{n}^{1}(z-y_{n}^{1}) $in$H^{1}(\mathbb{R}^{N})$. \item[(1-2)]$\Vert w_{n}^{1}\Vert_{H^{1}(\mathbb{R}^{N})}\leq c$for$n=1,2,\dots$and$\| w^{1}\| _{H^{1}}\leq c$, where$w_{n}^{1}\rightharpoonup w^{1}$weakly in$H^{1}(\mathbb{R}^{N})$: by Lemma \ref{p4}$(iii) $, $\| w_{n}^{1}\| _{H^{1}}^{2}=\| u_{n}^{1}\| _{H^{1}}^{2}=\| u_{n}\| _{H^{1}} ^{2}-\| \bar{u}\| _{H^{1}}^{2}+ o(1)\leq c^{2}+ o(1),$ we have$\Vert w_{n}^{1}\Vert_{H^{1}(\mathbb{R}^{N})}\leq c$for$n=1,2,\dots$. Then there is a subsequence$\{w_{n}^{1}\}$and a$w^{1}$in$H^{1}(\mathbb{R}^{N})$such that$w_{n}^{1}\rightharpoonup w^{1}$weakly in$H^{1}(\mathbb{R}^{N})$. By Lemma \ref{p4}$(i)$, we have $\| w^{1}\| _{H^{1}}\leq\liminf_{n\to\infty}\| w_{n} ^{1}\| _{H^{1}}\leq c.$ \item[(1-3)]$\{w_{n}^{1}\}$is a (PS)$_{(\beta-J(\bar{u})) }$-sequence in$H^{1}(\mathbb{R}^{N})$for$J$: note that$J' (u_{n}^{1})= o(1) $in$H^{-1}(\Omega) $. Because$\Omega$is a strictly large domain,$(1$-7) below and Theorem \ref{p271}, we have for every$\varphi\in H_{0}^{1}(\mathbb{R}^{N})$, $\langle J'(w_{n}^{1}) ,\varphi\rangle =\int_{\mathbb{R}^{N}}\nabla w_{n}^{1}\nabla\varphi+w_{n}^{1}\varphi -\int_{\mathbb{R}^{N}}| w_{n}^{1}| ^{p-2}w_{n}^{1}\varphi = o(1)$ Therefore,$J'(w_{n}^{1}) = o(1)$strongly in$H^{-1}(\mathbb{R}^{N}) $. Moreover, we have $J(w_{n}^{1}) =J(u_{n}^{1}(z+y_{n}^{1}) )=J( u_{n}^{1}) =(\beta-J(\bar{u})) + o(1).$ \item[(1-4)]$-\Delta w^{1}+w^{1}-|w^{1}|^{p-2}w^{1}=0$in$\mathbb{R}^{N}$: by Theorem \ref{c2}$(i)$below. \item[(1-5)]$w^{1}\not \equiv 0$: by the Rellich-Kondrakov theorem \ref{p24} and$(1-0)$, we have $\int_{B^{N}(0;1) }|w^{1}|^{2}=\lim_{n\to\infty} \int_{B^{N}(0;1) }|w_{n}^{1}|^{2}\geq\frac{d_{1}}{2},$ thus$w^{1}\not \equiv 0$. \item[(1-6)] By$(1-2)$,$(1-4)$,$(1-5)$, and Lemma \ref{p11},$\delta>0 $exists such that $\Vert w^{1}\Vert_{H^{1}(\mathbb{R}^{N})}\geq\Vert w^{1}\Vert_{L^{2} (\mathbb{R}^{N})}>\delta.$ Therefore, $J(w^{1})=(\frac{1}{2}-\frac{1}{p}) a(w^{1})>(\frac{1} {2}-\frac{1}{p}) \delta^{2}=\delta'.$ \item[(1-7)]$|y_{n}^{1}|\to\infty$: otherwise,$R>0$exists such that$y_{n}^{1}+B^{N}(0;1) \subset B^{N}(0;R) \; $for$n=1,2,\dots$. Then by$(1-0)$, we have $0=\lim_{n\to\infty}\int_{B^{N}(0;R) }|u_{n}^{1}|^{2} \geq\overline{\lim_{n\to\infty}}\int_{y_{n}^{1}+B^{N}( 0;1) }|u_{n}^{1}|^{2}\geq\frac{d_{1}}{2},$ which is a contradiction. \item[(1-8)]$a(u_{n})=a(\bar{u})+a(w_{n}^{1})+ o(1):$since$u_{n} \rightharpoonup\bar{u}$weakly in$H^{1}(\mathbb{R}^{N})$, by Lemma \ref{p4}$(iii)$, we have $a(u_{n})-a(\bar{u}) =a(u_{n}-\bar{u})+ o(1) =a(u_{n}^{1}) + o(1) =a(w_{n}^{1})+ o(1) .$ Thus,$a(u_{n})=a(\bar{u})+a(w_{n}^{1})+ o(1)$. \item[(1-9)]$b(u_{n})=b(\bar{u})+b(w_{n}^{1})+ o(1)$: since$u_{n} \to\bar{u}$a.e. in$\Omega$and$\{ u_{n}\} $is bounded in$L^{p}(\Omega) $, by Lemma \ref{p7}$(i) $, we have $b(u_{n})-b(\bar{u}) =b(u_{n}-\bar{u})+ o(1) =b(u_{n}^{1}) + o(1) =b(w_{n}^{1})+ o(1) .$ Thus,$b(u_{n})=b(\bar{u})+b(w_{n}^{1})+ o(1)$. \item[(1-10)]$J(u_{n})=J(\bar{u})+J(w_{n}^{1})+ o(1)$: by$(1-8)$and$(1-9)$, we have $J(u_{n})=J(\bar{u})+J(w_{n}^{1})+ o(1).$ \end{itemize} \textbf{Step 2}. Suppose that$w_{n}^{1}(z)\nrightarrow w^{1}(z)$strongly in$H^{1}(\mathbb{R}^{N})$. Let $u_{n}^{2}(z)=w_{n}^{1}(z)-w^{1}(z).$ We have$u_{n}^{2}\rightharpoonup0$weakly in$H^{1}(\mathbb{R}^{N})$but$u_{n}^{2}\nrightarrow0$strongly in$H^{1}(\mathbb{R}^{N})$. \begin{itemize} \item[(2-0)]$ {\int_{B^{N}(0;1) }} |w_{n}^{2}(z)|^{2}dz\geq\frac{d_{2}}{2}$for some constant$d_{2}>0$and$n=1,2,\dots$, where$w_{n}^{2}(z)=u_{n}^{2}(z+y_{n}^{2}) $for some$\{ y_{n}^{2}\} \subset\mathbb{R}^{N}$: since$\{ u_{n}^{2}\} $is bounded,$J'(u_{n}^{2})= o(1)$, and$u_{n}^{2}\nrightarrow0$strongly in$H^{1}(\mathbb{R}^{N})$, by Lemma \ref{p9} there are a subsequence$\{ u_{n}^{2}\} $, and a constant$d_{2}>0$such that $Q_{n}^{2}=\sup_{z\in\mathbb{R}^{N}}\int_{z+B^{N}(0;1) } |u_{n}^{2}(z) |^{2}dz\geq d_{2}\quad\mbox{for } n=1,2,\dots.$ \end{itemize} For$n=1,2,\dots$, take$\{ y_{n}^{2}\} $in$\mathbb{R}^{N}$such that $\int_{y_{n}^{2}+B^{N}(0;1) }|u_{n}^{2}(z)|^{2}dz\geq\frac{d_{2} }{2}\quad\text{for } n=1,2,\dots.$ Let$w_{n}^{2}(z)=u_{n}^{2}(z+y_{n}^{2})$, then $\int_{B^{N}(0;1) }|w_{n}^{2}(z)|^{2}dz\geq\frac{d_{2}} {2}\quad \text{for } n=1,2,\dots.$ As in Step 1, we have the following results. \begin{itemize} \item[(2-1)]$u_{n}(z)=\bar{u}(z)+w^{1}(z-y_{n}^{1}) +w_{n}^{2}(z-y_{n}^{1}-y_{n}^{2}) $in$H^{1}(\mathbb{R}^{N})$; \item[(2-2)]$\| w_{n}^{2}\| _{H^{1}}\leq c$for$n=1,2,\dots$and$\| w^{2}\| _{H^{1}}\leq c$, where$w_{n}^{2}\rightharpoonup w^{2}$weakly in$H^{1}(\mathbb{R}^{N})$; \item[(2-3)]$\{w_{n}^{2}\}$is a (PS)-sequence in$H^{1}(\mathbb{R}^{N})$for$J$; \item[(2-4)]$-\Delta w^{2}+w^{2}-|w^{2}|^{p-2}w^{2}=0$in$\mathbb{R}^{N}$; \item[(2-5)]$w^{2}\not \equiv 0$; \item[(2-6)]$\Vert w^{2}\Vert_{L^{2}(\mathbb{R}^{N})}>\delta$and$J(w^{2})>\delta'$; \item[(2-7)]$|y_{n}^{2}|\to\infty$; \item[(2-8)]$a(u_{n})=a(\bar{u})+a(w^{1})+a(w_{n}^{2})+ o(1)$: since $u_{n}^{2}(z) =w_{n}^{1}(z) -w^{1}(z) \rightharpoonup 0,$ we have $a(w_{n}^{2}) =a(u_{n}^{2}) =a(w_{n} ^{1}) -a(w^{1}) + o(1).$ Further by$(1-8) $, we have $a(u_{n})-a(\bar{u})=a(w_{n}^{1})+ o(1) =a(w^{1}) +a(w_{n}^{2}) + o(1) .$ \item[(2-9)]$b(u_{n})=b(\bar{u})+b(w^{1})+b(w_{n}^{2})+ o(1)$; \item[(2-10)]$J(u_{n})=J(\bar{u})+J(w^{1})+J(w_{n}^{2})+ o(1)$. \end{itemize} Continuing this process, we arrive at the$m$-th step. \begin{itemize} \item[$(m$-0)]${\int_{B^{N}(0;1) }} |w_{n}^{m}(z)|^{2}dz\geq\frac{d_{m}}{2}$for some constant$d_{m}>0$and$\;n=1,2,\dots$, where$w_{n}^{m}(z)=u_{n}^{m}(z+y_{n}^{m}) $for some$\{ y_{n}^{m}\} \subset\mathbb{R}^{N}$; \item[$(m$-1)]$u_{n}(z)=\bar{u}(z)+\underset{i=1} {\overset{m-1}{\sum}}w^{i}(z-z_{n}^{i}) +w_{n}^{m}( z-z_{n}^{m}) $in$H^{1}(\mathbb{R}^{N})$, where$z_{n}^{i}=y_{n} ^{1}+\dots+y_{n}^{i}$for$i=1,2,\dots,m:since $w_{n}^{m}(z)=u_{n}^{m}(z+y_{n}^{m})=w_{n}^{m-1}(z+y_{n}^{m})-w^{m-1} (z+y_{n}^{m}),$ thus $w_{n}^{m}(z)+w^{m-1}(z+y_{n}^{m})=w_{n}^{m-1}(z+y_{n}^{m}).$ Continuing this way, we obtain \begin{align*} &w_{n}^{m}(z)+w^{m-1}(z+y_{n}^{m})+\dots+w^{1}(z+y_{n}^{2}+\dots+y_{n}^{m})\\ &=w_{n}^{1}(z+y_{n}^{2}+\dots+y_{n}^{m})\\ &=u_{n}^{1}(z+y_{n}^{1}+y_{n}^{2}+\dots+y_{n}^{m}) \end{align*} \item[(m$-2)]$\| w_{n}^{m}\| _{H^{1}}\leq c$for$n=1,2,\dots$and$\| w^{m}\| _{H^{1}}\leq c$, where$w_{n}^{m}\rightharpoonup w^{m}$weakly in$H^{1}(\mathbb{R}^{N})$; \item[$(m$-3)]$\{w_{n}^{m}\}$is a (PS)-sequence in$H^{1}(\mathbb{R}^{N})$for$J$; \item[$(m$-4)]$-\Delta w^{m}+w^{m}-|w^{m}|^{p-2}w^{m}=0$in$\mathbb{R}^{N}$; \item[$(m$-5)]$w^{m}\not \equiv 0$; \item[$(m$-6)]$\| w^{m}\| _{L^{2}(\mathbb{R}^{N}) }>\delta$and$J(w^{m})>\delta'$; \item[$(m$-7)]$|y_{n}^{i}|=|z_{n}^{i}-z_{n}^{i-1}|\to\infty$and$|z_{n}^{i}|\to\infty$, for each$i=1,2,\dots,m:$we show it by induction on$i$. For$i=1$,$| z_{n}^{1}| =| y_{n} ^{1}| \to\infty$. Assume that$|z_{n}^{i}|\to\infty$, for$i=1,2,\dots,k$, for some$k\delta$,$R>0exists such that $z_{n}^{k+1}+B^{N}(0;R) \subset B^{N}(0;2R)$ and ${\int_{B^{N}(0;R) }}|w^{k+1}(z) |^{2}\geq(\frac{\delta}{2}) ^{2}.$ We have \begin{align*} (\frac{\delta}{2}) ^{2}&\leq \int_{B^{N}(0;R) }{\int_{B^{N}(0;R) }}|w^{k+1}(z) |^{2}\\ &=\lim_{n\to\infty}{\int_{B^{N}(0;R) }} |u_{n}^{1}(z+z_{n}^{k+1})|^{2}dz\\ &\leq\lim_{n\to\infty} {\int_{B^{N}(0;2R) }} |u_{n}^{1}(z)|^{2}=0, \end{align*} which is a contradiction. By the induction hypothesis, we have $|z_{n}^{i}|\to\infty\quad \text{for }i=1,2,\dots,m.$ \item[(m$-8)]$a(u_{n})=a(\bar{u})+\sum_{i=1}^{m-1} a(w^{i})+a(w_{n}^{m})+ o(1)$; \item[$(m$-9)]$b(u_{n})=b(\bar{u})+\sum_{i=1}^{m-1} b(w^{i})+b(w_{n}^{m})+ o(1)$; \item[$(m$-10)]$J(u_{n})=J(\bar{u})+\sum_{i=1}^{m-1} J(w^{i})+J(w_{n}^{m})+ o(1)$. \end{itemize} By the Archimedean principle,$l\in\mathbb{N}$exists such that$l\delta ^{2}>\beta$. Then after step$(l+1)$, we obtain $a(u_{n})=a(\bar{u})+a(w^{1})+a(w^{2})+\dots+a(w^{l})+a(w_{n}^{l+1})+ o(1).$ Since$a(w_{n}^{l+1})\geq0$,$a(\bar{u})>0$, and$a(w^{i})>\delta^{2}$for$i=1,2,\dots,l$, we have$\beta+ o(1)\geq l\delta^{2}>\beta$, which is a contradiction. Therefore, there is an$m\in\mathbb{N}$, such that$w_{n} ^{m}(z)=w^{m}(z)+ o(1)$strongly in$H^{1}(\mathbb{R}^{N})$,$w_{n} ^{i}(z)=w^{i}(z)+ o(1)$weakly, and$w_{n}^{i}(z)\neq w^{i}(z)+ o(1)$strongly in$H^{1}(\mathbb{R}^{N})$for$i=1,2,\dots m-1$. Then we have \begin{itemize} \item[$(sm$-0)]${\int_{B^{N}(0;1) }} |w_{n}^{m}(z)|^{2}dz\geq\frac{d_{m}}{2}$for some constant$d_{m}>0$and$n=1,2,\dots$, where$w_{n}^{m}(z)=u_{n}^{m}(z+y_{n}^{m})$for some$\{ y_{n}^{m}\} \subset\mathbb{R}^{N}$; \item[$(sm$-1)]$u_{n}(z)=\bar{u}(z)+\underset{i=1}{\overset {m}{\sum}}w^{i}(z-z_{n}^{i}) + o(1) $strongly in$H^{1}(\mathbb{R}^{N})$, where$z_{n}^{i}=y_{n}^{1}+\dots+y_{n}^{i}$for$i=1,2,\dots,m$; \item[$(sm$-2)]$\| w_{n}^{m}\| _{H^{1}}\leq c\text{for\;} n=1,2,\dots$and$\| w^{m}\| _{H^{1}}\leq c$, where$w_{n}^{m}\rightharpoonup w^{m}$weakly in$H^{1}(\mathbb{R}^{N})$; \item[$(sm$-3)]$\{w_{n}^{m}\}$is a (PS)-sequence in$H^{1}(\mathbb{R} ^{N})$for$J$; \item[$(sm$-4)]$-\Delta w^{m}+w^{m}-|w^{m}|^{p-2}w^{m}=0$in$\mathbb{R}^{N}$; \item[$(sm$-5)]$w^{m}\not \equiv 0$; \item[$(sm$-6)]$\| w^{m}\| _{L^{2}(\mathbb{R}^{N})}>\delta$and$J(w^{m})>\delta'$; \item[$(sm$-7)]$|y_{n}^{i}|=|z_{n}^{i}-z_{n}^{i-1}|\to\infty$and$|z_{n}^{i}|\to\infty$, for each$i=1,2,\dots,m$; \item[$(sm$-8)]$a(u_{n})=a(\bar{u})+\sum_{i=1}^m a(w^{i})+ o(1)$; \item[$(sm$-9)]$b(u_{n})=b(\bar{u})+\sum_{i=1}^m b(w^{i})+ o(1)$; \item[$(sm$-10)]$J(u_{n})=J(\bar{u})+\underset{i=1} {\overset{m}{\sum}}J(w^{i})+ o(1)$. \end{itemize} Finally, suppose$u_{n}\geq0$for$n=1,2,\dots$. Then\newline$(i)$Since$u_{n}\rightharpoonup\bar{u}$weakly in$H_{0}^{1}(\Omega)$. By Lemma \ref{p4}$(ii)$, there is a subsequence$\{u_{n}\}$such that$\,u_{n}\to\bar{u}$a.e. in$\Omega$. Thus,$\bar{u}\geq0$. \newline$(ii)$Since$w_{n}^{1}(z)=u_{n}(z+y_{n}^{1})-\bar{u}(z+y_{n} ^{1})\rightharpoonup w^{1}(z)$weakly in$H^{1}(\mathbb{R}^{N})$and$\bar {u}(z+y_{n}^{1})\rightharpoonup0$weakly in$H^{1}(\mathbb{R}^{N})$. Thus,$u_{n}(z+y_{n}^{1})\to w^{1}(z)$a.e. in$\Omega$, or$w^{1}\geq 0$. \newline$(iii)$Continuing this process, we obtain$w^{i}\geq0$for each$i=1,2,\dots,m$. \end{proof} We have the following useful corollary. \begin{corollary} \label{d10} Let$\Omega$be a strictly large domain in$\mathbb{R}^{N}$. If$\{ u_{n}\} $is a positive ($PS$)$_{\beta}$-sequence in$H_{0}^{1}(\Omega) $for$J$.\newline$(i)$If$\beta\neq j\alpha(\mathbb{R}^{N}) $for each$j\in\mathbb{N}$, then there is a positive solution$\overline{u}$of \eqref{E1} in$\Omega $; \newline$(ii)$If$\alpha(\mathbb{R}^{N}) <\beta <2\alpha(\mathbb{R}^{N}) $, then$\{ u_{n}\} $contains a strongly convergent subsequence. \end{corollary} \begin{proof} By Theorem \ref{d1}, we have $J(u_{n}) =J(\overline{u}) +\sum{i=1} ^{m}J(w^{i})+ o(1).$ By Corollary \ref{d4} below, the positive solutions of \eqref{E1} in$\mathbb{R}^{N}$are unique, and we obtain$J(w^{i})=\alpha( \mathbb{R}^{N}) $for each$i$. Thus, we have $\beta=J(\overline{u}) +mJ(w^{i})+ o(1).$$(i)$If$\beta\neq j\alpha(\mathbb{R}^{N}) $for each$j\in\mathbb{N}$, then$J(\overline{u})\neq0$, or$\overline{u}\neq0$. By Theorem \ref{c2}$(i)$below, there is a positive solution$\overline{u}$of \eqref{E1} in$\Omega$.\newline$(ii)$Recall that we always have$\beta\geq\alpha(\Omega) \geq\alpha(\mathbb{R} ^{N}) $. Suppose that$m\geq1$and$\alpha(\mathbb{R} ^{N}) <\beta<2\alpha(\mathbb{R}^{N}) $, then$J(\overline{u})\neq0$or$J(\overline{u})\geq\alpha(\Omega) $. Thus, $2\alpha(\mathbb{R}^{N}) >\beta+ o(1)=J(\overline {u}) +m\alpha(\mathbb{R}^{N}) \geq(m+1)\alpha( \mathbb{R}^{N}) .$ This is a contradiction. Hence,$m=0$. By the proof of Theorem \ref{d1}, we have $u_{n}=\overline{u}+ o(1)\quad\text{strongly in }H_{0}^{1}(\Omega).$ \end{proof} \begin{remark}\label{d11} \rm Note that if we replace a strictly large domain by a domain in Theorem \ref{d1}, then the theorem may fail. Let$\mathbf{A}_{0}^{r}$be an upper semi-strip with sufficiently large$r$, then$\alpha(\mathbb{R} ^{N})<\alpha(\mathbf{A}_{0}^{r})<2\alpha(\mathbb{R}^{N})$. By the Esteban-Lions theorem \ref{n3}, \eqref{E1} in$\mathbf{A}_{0}^{r}$admits only trivial solution, but if Theorem \ref{d1} holds, by Corollary \ref{d10}, \eqref{E1} in$\mathbf{A}_{0}^{r}$admits a positive solution, a contradiction. \end{remark} \begin{definition} \label{d2} \rm A domain$\Theta$in$\mathbb{R}^{N}$is a periodic domain if a partition$\{Q_{m}\}_{m=0}^{\infty}$of$\Theta$and points$\{z_{m} \}_{m=1}^{\infty}$in$\mathbb{R}^{N}$exist, satisfying the following conditions:\newline$(i)\{z_{m}\}_{m=1}^{\infty}$forms a subgroup of$\mathbb{R}^{N}$; \newline$(ii)Q_{0}$is bounded;\newline$(iii)Q_{m}=z_{m}+Q_{0}$for each$m$. \end{definition} Typical examples of periodic domains are the infinite strip$\mathbf{A}^{r}$, the infinite hollow strip$\mathbf{A}^{r_{1},r_{2}}$, and the whole space$\mathbb{R}^{N}$. Similarly, we have the Palais-Smale decomposition theorem in$H_{0}^{1}(\Omega)$for$J$in a periodic domain in$\Theta\subset \mathbb{R}^{N}$. \begin{theorem}[Palais-Smale Decomposition Theorem in a Periodic Domain] \label{d3} Let$\Omega$be a strictly large domain in$\Theta$and let$\{u_{n}\} $be a positive (PS)$_{\beta}$-sequence in$H_{0}^{1}(\Omega) $for$J$. Then there are a subsequence$\{ u_{n}\} $, a positive integer$m$, a subsequence$\{ z_{n}^{i}\} _{n=1}^{\infty}$of$\{ z_{m}\} _{m=1}^{\infty}$in$\Theta$, and a function$\bar {u}\in H_{0}^{1}(\Omega)$, and$0\neq w^{i}\in H^{1}(\Theta) $, for$1\leq i\leq m$such that \newline$(i)|z_{n}^{i}|\to,\infty\ \text{for}i=1,2,\dots,m$; \newline$(ii)-\Delta\bar{u}+\bar {u}=\mid\bar{u}\mid^{p-2}\bar{u}$in$\Omega$; \newline$(iii)-\Delta w^{i}+w^{i}=| w^{i}|^{p-2}w^{i}$in$\Theta$; \newline$(iv)u_{n}=\bar {u}+\sum_{i=1}^m w^{i}(\cdot-z_{n}^{i}) + o(1)\;\text{strongly}$in$H^{1}(\Theta)$; \newline$(v)a(u_{n})=a(\bar{u})+\sum_{i=1}^m a(w^{i})+ o(1)$; \newline$(vi)b(u_{n})=b(\bar{u})+\sum_{i=1}^{m} b(w^{i})+ o(1)$; \newline$(vii)J(u_{n})=J(\bar{u})+\sum_{i=1}^{m} J(w^{i})+ o(1)$.\newline In addition, if$u_{n}\geq0$, then$\bar{u}\geq0$and$w^{i}\geq0$for each$1\leq i\leq m$. \end{theorem} \begin{proof} The proof is similar to those of Theorem \ref{d1}: see Lien-Tzeng-Wang \cite{LTW}. Note that instead of $Q_{n}=\sup_{z\in\mathbb{R}^{N}}\int_{z+B^{N}(0;1)}|u_{n}(z)|^{2}dz$ we use $Q_{n}^{r}=\sup_{y\in\mathbb{R}}\int_{(0,y)+\mathbf{A}_{-1,1}^{r}} |u_{n}(z)|^{2}dz,$ where$\mathbf{A}_{-1,1}^{r}=\{(x,y)\in\mathbf{A}^{r}\ |\ -10$. Moreover,$\alpha_{\gamma}(\Omega)$is a (PS)-value in$X(\Omega)$for$J$. \begin{theorem} \label{i1}$\alpha_{\gamma}(\Omega)$is a (PS)-value in$X(\Omega)$for$J$. \end{theorem} \begin{proof} Let$\{u_{n}\}$in$X(\Omega)$be a maximizing sequence of$\gamma(\Omega)$. Then$a(u_{n})=1$for$n=1,2,\dots$, and ${\int_{\Omega}}|u_{n}|^{p}=\gamma(\Omega)^{p}+ o(1)\quad \text{as }n\to\infty.$ Let$v_{n}=\gamma(\Omega)^{\frac{p}{2-p}}u_{n}$for each$n=1,2,\dots. Then we have \begin{gather*} a(v_{n})={\int_{\Omega}} (|\nabla v_{n}|^{2}+v_{n}^{2})=\gamma(\Omega)^{\frac{2p}{2-p}}\quad \text{for each }n=1,2,\dots,\\ b(v_{n})= {\int_{\Omega}} |v_{n}|^{p}=\gamma(\Omega)^{\frac{2p}{2-p}}+ o(1)\quad \text{as } n\to \infty, \end{gather*} and \begin{align*} J(v_{n}) & =\frac{1}{2}a(v_{n})-\frac{1}{p}b(v_{n})\\ & =(\frac{1}{2}-\frac{1}{p})\gamma(\Omega)^{\frac{2p}{2-p}}+ o(1)\quad\text{as } n\to\infty\\ & =\alpha_{\gamma}(\Omega)+ o(1)\quad \text{as }n\to\infty. \end{align*} For eachn=1,2,\dots$and$\varphi\in X(\Omega)$, denote $l_{n}(\varphi)=\int_{\Omega}|v_{n}|^{p-2}v_{n}\varphi.$ Let$\phi\in X(\Omega)$satisfy$\Vert\phi\Vert_{H^{1}}=1$. Then$\gamma(\Omega)\geq\Vert\phi\Vert_{L^{p}}and \begin{align*} |l_{n}(\phi)| & =\big| {\int_{\Omega}}| v_{n}|^{p-2}v_{n}\phi\big|\leq\Big( {\int_{\Omega}}|v_{n}|^{p}\Big) ^{(p-1)/p} \Big({\int_{\Omega}}|\phi|^{p}\Big)^{1/p}\\ & \leq\gamma(\Omega)^{\frac{2p-2}{2-p}}\gamma(\Omega)+ o(1)=\gamma (\Omega)^{\frac{p}{2-p}}+ o(1)\quad\text{as }n\to\infty. \end{align*} Thus, $\Vert l_{n}\Vert_{X^{-1}}\leq\gamma(\Omega)^{\frac{p}{2-p}}+ o(1)\quad \text{as }n\to\infty.$ Furthermore, $l_{n}\big(\frac{v_{n}}{\Vert v_{n}\Vert_{H^{1}}}\big) =\frac{\int_{\Omega}|v_{n}|^{p}}{\Vert v_{n}\Vert_{H^{1}}}=\frac{\gamma (\Omega)^{2p/(2-p)}}{\gamma(\Omega)^{p/(2-p)}}+ o(1)=\gamma(\Omega)^{\frac {p}{2-p}}+ o(1)$ asn\to\infty$. We conclude that $\Vert l_{n}\Vert_{X^{-1}}=\gamma(\Omega)^{\frac{p}{2-p}} + o(1)\quad\mbox{as }n\to\infty.$ Since$l_{n}$is a continuous linear functional in$X(\Omega)$, by the Riesz representation theorem, for each$n$,$w_{n}\in X(\Omega)$exists such that $l_{n}(\varphi)=\langle w_{n},\varphi\rangle _{H^{1}} =\int_{\Omega}(\nabla w_{n}\cdot\nabla\varphi+w_{n}\varphi)\quad\text{for each } \varphi\in X(\Omega),$ and$\Vert w_{n}\Vert_{H^{1}}=\Vert l_{n}\Vert_{X^{-1}}. Since $\langle w_{n},v_{n}\rangle_{H^{1}}=l_{n}(v_{n}) =\int_{\Omega }| v_{n}| ^{p}=\gamma(\Omega)^{\frac{2p}{2-p}}+ o(1)\quad \text{as } n\to\infty,$ we obtain \begin{align*} \Vert v_{n}-w_{n}\Vert_{H^{1}}^{2} & =\langle v_{n},v_{n}\rangle_{H^{1}}-2\langle v_{n},w_{n}\rangle_{H^{1}} +\langle w_{n},w_{n}\rangle_{H^{1}}\\ & =\Vert v_{n}\Vert_{H^{1}}^{2}-2\langle v_{n},w_{n}\rangle_{H^{1}}+\Vert w_{n}\Vert_{H^{1}}^{2}\\ & =\gamma(\Omega)^{\frac{2p}{2-p}}-2\gamma(\Omega)^{\frac{2p}{2-p}} +\gamma(\Omega)^{\frac{2p}{2-p}}+ o(1)\\ & = o(1)\quad \text{as } n\to\infty. \end{align*} For\varphi\in X(\Omega)$satisfying$\Vert\varphi\Vert_{H^{1}}=1, we have \begin{align*} \langle J'(v_{n}),\varphi\rangle & = {\int_{\Omega}} (\nabla v_{n}\cdot\nabla\varphi+v_{n}\varphi)- {\int_{\Omega}} |v_{n}|^{p-2}v_{n}\varphi\\ & =\langle v_{n},\varphi\rangle_{H^{1}}-\langle w_{n},\varphi\rangle_{H^{1} }=\langle v_{n}-w_{n},\varphi\rangle_{H^{1}}, \end{align*} so $|\langle J'(v_{n}),\varphi\rangle|\leq\Vert v_{n}-w_{n}\Vert_{H^{1}}.$ We conclude that $J'(v_{n})= o(1)\quad\text{strongly in }X^{-1}(\Omega)\quad\text{as }n\to\infty.$ \end{proof} \noindent (B) Consider the Nehari minimizing problem $\alpha_{\mathbf{M}}(\Omega)=\inf_{v\in\mathbf{M}(\Omega)}J(v),$ where\mathbf{M}(\Omega)=\{ u\in X(\Omega)\backslash \{0\}: a(u)=b(u)\} $. Note that$\mathbf{M}(\Omega)$contains every nonzero solution of \eqref{E1}. Consider the unit sphere$\mathbf{U}(\Omega)$and the zero energy manifold$\mathbf{Z}(\Omega)$, where \begin{gather*} \mathbf{U}(\Omega)=\{u\in X(\Omega): \| u\| _{H^{1}}=1\},\\ \mathbf{Z}(\Omega)=\{u\in X(\Omega)\backslash\{0\}: \frac{1} {2}a(u)=\frac{1}{p}b(u)\}. \end{gather*}$\alpha_{\mathbf{M}}(\Omega)>0$is a consequence of part$(i)$of the following lemma. Part$(ii)$of the following lemma will be used later in Lemma \ref{i7} and Theorem \ref{i13}. \begin{lemma} \label{i2}$(i)$There is a bijective$C^{1,1}$map$m$from$\mathbf{U}(\Omega)$to$\mathbf{M}(\Omega)$. Moreover,$\mathbf{M}(\Omega)$is path-connected and a constant$c>0$exists such that for$u\in\mathbf{M} (\Omega)$,$\| u\| _{H^{1}}\geq c$and\$J(u)\geq c$;\newline$(ii)$There is a bijective$C^{1,1}$map$z$from$\mathbf{U} (\Omega)$to$\mathbf{Z}(\Omega)$. Moreover,$\mathbf{Z}(\Omega)$is path-connected and a constant$c'>0$exists such that for$u\in\mathbf{Z}(\Omega)$,$\| u\| _{H^{1}}\geq c'$. \end{lemma} \begin{proof}$(i)$For$t\geq0$,$u\in\mathbf{U}(\Omega)$, let $h_{u}(t)=J(tu) =\frac{1}{2}t^{2}-\frac{1}{p}t^{p}b(u).$ Then$h_{u}'(t)=t-t^{p-1}b(u)$. We take uniquely$s_{u}\in \mathbb{R}^{+}$such that$s_{u}>0$,$s_{u}u\in\mathbf{M}(\Omega)$, and$0=h_{u}'(s_{u})$. For$v\in\mathbf{U}(\Omega)$, a$s_{v} \in\mathbb{R}^{+}$exists such that$s_{v}v\in\mathbf{M}(\Omega)$: that is $\langle J'(s_{v}v) ,s_{v}v\rangle =s_{v}^{2}-s_{v}^{p}b(v)=0.$ Consider the function$g(t,u):\mathbb{R}^{+}\times \mathbf{U}(\Omega)\to\mathbb{R}$defined by $g(t,u) =\langle J'(tu),tu\rangle =t^{2}a(u)-t^{p}b(u).$ Note that$g(s_{v},v) =\langle J'(s_{v}v) ,s_{v}v\rangle =0$. Thus, $\frac{\partial g}{\partial t}(t,u) \big| _{(s_{v},v) } =2s_{v}-ps_{v}^{p-1}b(v)=s_{v}(2-p)<0.$ By the implicit function theorem, a neighborhood$\mathbf{W}$of$v$in$\mathbf{U}(\Omega)$and a unique function$t\in C^{1,1}$exist such that \begin{gather*} t:\mathbf{W}\to\mathbb{R}^{+},\ t(v)=s_{v},\\ g(t(u) ,u) =0\;\text{for all }u\in\mathbf{W}. \end{gather*} Therefore, for each$v\in\mathbf{U}(\Omega)$,$t:\mathbf{U}(\Omega)\to\mathbb{R}^{+}$and$m:\mathbf{U}(\Omega)\to \mathbf{M}(\Omega)$,$t$,$m\in C^{1,1}$exist such that$t(v)=s_{v}$,$m(v)=s_{v}v$. Clearly,$t$and$m$are injective. For each$u\in \mathbf{M}(\Omega)$, write$u=s_{v}v$, where$s_{v}=\| u\|_{H^{1}}$and$v=\frac{u}{\| u\| _{H^{1}}}\in\mathbf{U}(\Omega)$. Since$m(v)=u$,$m$is surjective. Since$\mathbf{U}(\Omega)$is path-connected,$\mathbf{M}(\Omega)$is path-connected. Note that$u\in\mathbf{M}(\Omega)$, so$J'(u) =0$, or$s_{v} ^{2}=\int_{\Omega}s_{v}^{p}| v| ^{p}$. By the Sobolev embedding theorem, we have$s_{v}^{2}=\int_{\Omega}s_{v}^{p}| v| ^{p}\leq ds_{v}^{p}$, or$c\leq s_{v}$, where$d$and$c$are two positive constants. Therefore,$\|u\| _{H^{1}}=\| s_{v}v\| _{H^{1} }=s_{v}\geq c$for$u\in\mathbf{M}(\Omega)$.\newline$(ii)$The proof is similar to part$(i)$. \end{proof} \begin{theorem} \label{i3} Let$\beta>0$and let$\{ u_{n}\} $in$X(\Omega )\backslash\{0\}$be a sequence for$J$such that$J(u_{n})=\beta + o(1)$and$a(u_{n})=b(u_{n})+ o(1)$. Then there is a sequence$\{s_{n}\}$in$\mathbb{R}^{+}$such that$s_{n}=1+ o(1)$,$\{s_{n}u_{n}\}$is in$\mathbf{M}(\Omega) $and$J(s_{n} u_{n})=\beta+ o(1)$. In particular, if$\{ u_{n}\} $is a (PS)$_{\beta}$-sequence for$J$, then there is a sequence$\{s_{n}\}$in$\mathbb{R}^{+}$such that$\{s_{n}u_{n}\}$is in$\mathbf{M}(\Omega) $and there is also a (PS)$_{\beta}$-sequence in$X(\Omega)$for$J$. \end{theorem} \begin{proof} By Lemma \ref{i2}, there is a sequence$\{s_{n}\}$in$\mathbb{R}^{+}$such that$\{s_{n}u_{n}\}$is in$\mathbf{M}(\Omega) :s_{n} ^{2}a(u_{n})=s_{n}^{p}b(u_{n})$for each$n$, because$a(u_{n})=b(u_{n} )+ o(1)$and$J(u_{n})=\beta+ o(1)$imply$s_{n} =1+ o(1)$. Therefore,$J(s_{n}u_{n})=\beta+ o(1)$. The last part follows from Lemma \ref{p30}. \end{proof} A minimizing sequence$\{u_{n}\}$in$\mathbf{M}(\Omega)$of$\alpha_{\mathbf{M}}(\Omega)$is a (PS)$_{\alpha_{\mathbf{M}}(\Omega)} $-sequence in$X(\Omega)$for$J$. \begin{theorem}\label{i4} Let$\{u_{n}\}$be in$X(\Omega)$. Then$\{u_{n}\}$is a (PS)$_{\alpha_{\mathbf{M}}(\Omega)}$-sequence for$J$if and only if$J( u_{n}) =\alpha_{\mathbf{M}}(\Omega)+ o(1) $and$a(u_{n}) =b(u_{n}) + o(1) $. In particular, every minimizing sequence$\{u_{n}\}$in$\mathbf{M}(\Omega)$of$\alpha_{\mathbf{M}}(\Omega)$is a (PS)$_{\alpha_{\mathbf{M}}( \Omega) } $-sequence in$X(\Omega)$for$J$. In particular,$\alpha_{\mathbf{M}}(\Omega)$is a (PS)$_{\alpha_{\mathbf{M}}(\Omega)}-$value in$X(\Omega)$for$J$. \end{theorem} \begin{proof} Suppose$\{ u_{n}\} $is a (PS)$_{\alpha_{\mathbf{M}}(\Omega)} $-sequence in$X(\Omega)$for$J$. By Lemma \ref{p30}, we have$a(u_{n}) =b(u_{n}) + o(1) $. Conversely, let$\{ u_{n}\} $satisfy$J(u_{n}) =\alpha_{\mathbf{M}}(\Omega)+ o(1) $and$a(u_{n}) =b(u_{n}) + o(1) $. Then we have $$a(u_{n}) =\frac{2p}{p-2}\alpha_{\mathbf{M}}(\Omega)+o(1) \quad \text{as }n\to\infty.\label{4-1}$$ For$n=1,2,\dots$, denote $$l_{n}(\varphi)=\int_{\Omega}|u_{n}|^{p-2}u_{n}\varphi\quad \text{for }\varphi\in X(\Omega).\label{4-2}$$ Let$\phi\in\mathbf{U}(\Omega)$. By Lemma \ref{i2},$t>0$exists such that$t\phi\in\mathbf{M}(\Omega):\| t\phi\| _{H^{1}}^{2}=\| t\phi\| _{L^{p}}^{p}$; we conclude that$t=\| \phi\| _{L^{p}}^{\frac{-p}{p-2}}$and $\alpha_{\mathbf{M}(\Omega)}\leq(\frac{1}{2}-\frac{1}{p}) \| t\phi\| _{H^{1}}^{2}=\frac{p-2}{2p}t^{2}=\frac{p-2} {2p}\| \phi\| _{L^{p}}^{\frac{-2p}{p-2}}.$ Therefore,$\| \phi\| _{L^{p}}\leq(\frac{2p}{p-2} \alpha_{\mathbf{M}}(\Omega)) ^{\frac{2-p}{2p}}$. For each$n, \begin{align*} |l_{n}(\phi)| & =\big|\int_{\Omega}|u_{n}|^{p-2}u_{n}\phi\big|\\ &\leq\Big(\int_{\Omega}|u_{n}|^{p}\Big) ^{\frac{p-1}{p}} \Big(\int_{\Omega}|\phi|^{p}\Big) ^{1/p}\\ & \leq(\frac{2p}{p-2}\alpha_{\mathbf{M}}(\Omega)) ^{\frac {p-1}{p}}(\frac{2p}{p-2}\alpha_{\mathbf{M}}(\Omega)) ^{\frac{2-p}{2p}}+ o(1)\\ & =\big(\frac{2p}{p-2}\alpha_{\mathbf{M}}(\Omega)\big) ^{1/2}+ o(1) \quad\text{as }n\to\infty, \end{align*} we have $$\Vert l_{n}\Vert_{X^{-1}}\leq(\frac{2p}{p-2}\alpha_{\mathbf{M}} (\Omega)) ^{1/2}+ o(1)\quad\text{as }n\to\infty.\label{4-3}$$ Furthermore, by \eqref{4-2}, we have \label{4-4} \begin{aligned} l_{n}(\frac{u_{n}}{\Vert u_{n}\Vert_{H^{1}}}) & =\frac {\int_{\Omega}|u_{n}|^{p}}{\Vert u_{n}\Vert_{H^{1}}}\\ &=(b(u_{n}))^{1/2}+ o(1)\\ & =(\frac{2p}{p-2}\alpha_{\mathbf{M}}(\Omega)) ^{1/2}+ o(1) \quad \text{as }n\to\infty \end{aligned} By (\ref{4-3}) and (\ref{4-4}), we conclude that $\Vert l_{n}\Vert_{X^{-1}}=(\frac{2p}{p-2}\alpha_{\mathbf{M}} (\Omega)) ^{1/2}+ o(1)\quad\text{as }n\to\infty.$ By the Riesz representation theorem, for eachn$,$w_{n}\in X(\Omega)$exists such that, for each$\varphi\in X(\Omega), $l_{n}(\varphi)=\langle w_{n},\varphi\rangle_{H^{1}}=\int_{\Omega}(\nabla w_{n}\cdot\nabla\varphi+w_{n}\varphi),$ and $$\Vert w_{n}\Vert_{H^{1}}=\Vert l_{n}\Vert_{X^{-1}}=(\frac{2p} {p-2}\alpha_{\mathbf{M}}(\Omega)) ^{1/2}+ o(1).\label{4-5}$$ Consequently, $$\langle w_{n},u_{n}\rangle_{H^{1}}=l_{n}(u_{n}) =\int_{\Omega }|u_{n}|^{p}=\frac{2p}{p-2}\alpha_{\mathbf{M}}(\Omega)+ o(1).\label{4-6}$$ By \eqref{4-1}, \eqref{4-5}, and \eqref{4-6}, we obtain \begin{align*} \Vert u_{n}-w_{n}\Vert_{H^{1}}^{2} & =\langle u_{n},u_{n}\rangle_{H^{1} }-2\langle u_{n},w_{n}\rangle_{H^{1}}+\langle w_{n},w_{n}\rangle_{H^{1}}\\ & =\Vert u_{n}\Vert_{H^{1}}^{2}-2\langle u_{n},w_{n}\rangle_{H^{1}}+\Vert w_{n}\Vert_{H^{1}}^{2}\\ & =\frac{2p}{p-2}\alpha_{\mathbf{M}}(\Omega)-2\frac{2p}{p-2}\alpha _{\mathbf{M}}(\Omega)+\frac{2p}{p-2}\alpha_{\mathbf{M}}(\Omega)+ o(1)\\ & = o(1)\quad\text{as }n\to\infty. \end{align*} For\varphi\in\mathbf{U}(\Omega), we have \begin{align*} \langle J'(u_{n}),\varphi\rangle & = {\int_{\Omega}} (\nabla u_{n}\cdot\nabla\varphi+u_{n}\varphi)- {\int_{\Omega}} |u_{n}|^{p-2}u_{n}\varphi\\ & =\langle u_{n},\varphi\rangle_{H^{1}}-\langle w_{n},\varphi\rangle_{H^{1} }=\langle u_{n}-w_{n},\varphi\rangle_{H^{1}}, \end{align*} so $\| J'(u_{n})\| _{X^{-1}}\leq\Vert u_{n}-w_{n}\Vert _{H^{1}}= o(1).$ We conclude thatJ'(u_{n})= o(1)$strongly in$X^{-1}(\Omega)$as$n\to\infty$. \end{proof} If$u$achieves$\alpha_{\mathbf{M}}(\Omega)$, then$u$is a nonzero solution of \eqref{E1}. \begin{theorem} \label{i6} Let$u\in\mathbf{M}(\Omega)$be such that$J(u)=\min_{v\in \mathbf{M}(\Omega)}J(v)$. Then$u$is a nonzero solution of \eqref{E1}. \end{theorem} \begin{proof} Set$g(v)=a(v)-b(v)$for$v\in X(\Omega)$. Note that$\langle g^{\prime }(u),u\rangle =(2-p)a(u)\neq0$. Since the minimum of$J$is achieved at$u$and is constrained in$\mathbf{M}(\Omega)$, by the Lagrange multiplier theorem,$\lambda\in\mathbb{R}$exists such that$J'(u)=\lambda g'(u)$in$X(\Omega)$. Thus, $0=\langle J'(u),u\rangle =\lambda\langle g^{\prime }(u),u\rangle ,$ or$\lambda=0$. Thus,$J'(u)=0$. Hence,$u$is a weak solution of \eqref{E1} such that$J(u)=\alpha_{\mathbf{M}}(\Omega)$. \end{proof} \noindent(C) Consider the mountain pass minimax problem $\alpha_{\Gamma}(\Omega)=\inf_{g\in\Gamma(\Omega)}\max_{t\in[0,1]}J(g(t)),$ where$e\neq0$,$J(e)=0$, and $\Gamma(\Omega)=\{g\in C([0,1],X(\Omega)): g(0)=0,g(1)=e\}.$ Then$\alpha_{\Gamma}(\Omega)>0$is a consequence of the following lemma. \begin{lemma} \label{i7} A ball$B(0;r)$in$X(\Omega)$,$c>0$, and$e\notin\overline {B(0;r)}$exist such that$J(e)=0$and$\min_{v\in\partial B(0;r)} J(v)\geq c$. \end{lemma} \begin{proof} By Lemma \ref{i2}$(ii)$, for each$u\in\mathbf{U}(\Omega)$, there is a$t>0$such that$J(tu)=0$. Let$e=tu$, then$J(e)=0$. Since for each$v\in X(\Omega)\backslash\{0\}$$J(v)=\frac{1}{2}a(v)-\frac{1}{p}b(v),$ by the Sobolev inequality, there is a constant$c_{1}>0$such that$b(v)\leq c_{1}a(v)^{p/2}$, and we have $J(v)\geq a(v)\{ \frac{1}{2}-\frac{c_{1}}{p}a(v)^{\frac{p-2}{2}}\}.$ Take$r>0$such that$e\notin\overline{B(0;r)}$and$\frac{1}{2}-\frac{c_{1} }{p}r^{p-2}\geq\frac{1}{4}$, then for$\| v\| _{H^{1}}=r$, we have $J(v)\geq c,$ where$c=\frac{1}{4}r^{2}$. \end{proof} We require the following lemma. \begin{theorem}[Ekeland variational principle]\label{i8} Let$M$be a complete metric space with metric$d$and let$F:M\to\mathbb{R}\cup\{+\infty\}$be lower semi-continuous, bounded from below, and$\not \equiv \infty$. Then for any$\varepsilon>0$and$\lambda>0$, and any$u\in M$with $F(u)\leq\inf_{M}F+\varepsilon,$ there is an element$v\in M$such that \begin{gather*} F(v)\leq F(u),\\ d(u,v)\leq\frac{1}{\lambda},\\ F(w)+\varepsilon\lambda d(v,w)>F(v)\quad \text{for }w\neq v. \end{gather*} \end{theorem} \begin{proof} It is sufficient to prove our assertion for$\lambda=1$. The general case is obtained by replacing$d$by an equivalent metric$\lambda d$. We define the relation on$M$: $w\leq v\Longleftrightarrow F(w) +\varepsilon d(v,w) \leq F(v) .$ It is easy to see that this relation define a partial ordering on$M$. We now construct inductively a sequence$\{ u_{m}\} $as follows:$u_{0}=u$; also assuming that$u_{m}$has been defined, we set $S_{n}=\{ w\in M\;|\text{\ }w\leq u_{n}\}$ and choose$u_{n+1}\in S_{n}$so that $F(u_{n+1}) \leq\underset{S_{n}}{\inf}F+\frac{1}{n+1}.$ Since$u_{n+1}\leq u_{n}$,$S_{n+1}\subset S_{n}$, and by the lower semicontinuity of$F$,$S_{n}$is closed. We now show that \textrm{diam}$S_{n}\to0$. Indeed, if$w\in S_{n+1}$, then$w\leq u_{n+1}\leq u_{n} $and consequently, $\varepsilon d(w,u_{n+1}) \leq F(u_{n+1}) -F( w) \leq \inf_{S_{n}} F+\frac{1}{n+1}-\underset{S_{n}}{\inf }F=\frac{1}{n+1}.$ This estimate implies $\mathop{\rm diam}S_{n+1}\leq\frac{2}{\varepsilon(n+1) }$ and our claim follows. The fact that$M$is complete implies that $\cap_{n\geq 0}S_{n}=\{ v\}$ for some$v\in M$. In particular,$v\in S_{0}$, that is,$v\leq u_{0}=u$. Hence, $F(v) \leq F(u) -\varepsilon d(u,v) \leq F(u) .$ Moreover, $d(u,v) \leq\varepsilon^{-1}(F(u) -F( v) ) \leq\varepsilon^{-1}\big(\inf_M F+\varepsilon-\inf_M F\big) =1.$ To complete the proof we must show$w\leq v$implies$w=v$. If$w\leq v$, then$w\leq u_{n}$for each integer$n\geq0$, that is$w\in \cap_{n\geq0} S_{n}=\{ v\} $. \end{proof} \begin{lemma} \label{i9} Let$\Gamma(\Omega)$be the complete metric space with the usual$L^{\infty}$distance$d$and$J\in C^{1}(X(\Omega),\mathbb{R}) $. Then for each$\varepsilon>0$and each$f\in\Gamma(\Omega)$such that $$\max_{s\in[0,1]} J(f(s)) \leq \alpha_{\Gamma}(\Omega)+\varepsilon,\label{4-7}$$$v\in X(\Omega)$exists such that \begin{gather*} \alpha_{\Gamma}(\Omega)-\varepsilon\leq J(v)\leq\max_{s\in[0,1]} J(f(s) ) ,\\ \mathop{\rm dist}(v,f([0,1]) ) \leq\varepsilon^{1/2},\\ | J'(v) | \leq\varepsilon^{1/2}. \end{gather*} \end{lemma} \begin{proof} Without loss of generality, we can assume that $$0<\varepsilon<\alpha_{\Gamma}(\Omega).\label{4-8}$$ Let$f\in\Gamma(\Omega)$satisfy the condition (\ref{4-7}). We define the function$\Phi:\Gamma(\Omega)\to\mathbb{R}$by $\Phi(g) =\max_{s\in[0,1]}J(g(s)) .$ Then$(i)\Phi$is bounded below:$\Phi(g)\geq\alpha_{\Gamma}(\Omega)>0$. \newline$(ii)\Phi$is continuous at each$g\in\Gamma(\Omega):$since$J$is continuous on the compact set$K=g([0,1]) $, for each$\varepsilon>0$,$u\in K$, there is a$\delta_{u}>0$such that if$w\in B(u;\delta_{u})$is an open ball in$X(\Omega)$, then$|J(w)-J(u)|<\frac{1} {2}\varepsilon$. Since$K$is compact, finite values$B(u_{i};\delta_{u_{i}} )$,$i=1,\dots,n$, exist such that $K\subset B(u_{1};\frac{\delta_{u_{1}}}{2})\cup\dots\cup B(u_{n};\frac {\delta_{u_{n}}}{2}).$ Take$\delta=\min\{ \frac{\delta_{u_{1}}}{2},\dots,\frac{\delta_{u_{n} }}{2}\} $. Let$k\in\Gamma(\Omega)$satisfy$\| k-g\| _{L^{\infty}}<\delta$. For each$s\in[0,1]$, we have $| k(s)-g(s)| <\delta,$ or$g(s)\in B(u_{i};\frac{\delta_{u_{i}}}{2})$,$k(s)\in B(u_{i};\delta _{u_{i}})$. Thus $|J(k(s))-J(g(s))|<\varepsilon, \quad\mbox{or}\quad | \Phi(k) -\Phi(g) | \leq \varepsilon.$ The Ekeland variational principle (Theorem \ref{i8}) implies the existence of$h\in\Gamma(\Omega)$such that \begin{gather*} \Phi(h) \leq\Phi(f) \leq\alpha_{\Gamma}(\Omega)+\varepsilon,\\ \max_{s\in[0,1]}| h(s) -f(s)| \leq\varepsilon^{1/2}, \end{gather*} and $$\Phi(g) >\Phi(h) -\varepsilon^{\frac{1}{2} }d(h,g) \quad\text{whenever }g\in\Gamma(\Omega)\quad\text{and }g\neq h.\label{4-9}$$ Let$A=\{ s\in[0,1]:\alpha_{\Gamma}(\Omega)-\varepsilon\leq J( h(s) ) \} $, then$A$is nonempty since $\alpha_{\Gamma}(\Omega)-\varepsilon<\alpha_{\Gamma}(\Omega)=\underset {g\in\Gamma(\Omega)}{\inf}\underset{s\in[0,1]}{\max}J(g(s))\leq\underset {s\in[0,1]}{\max}J(h(s)).$ Note that for$s\in A$, $| J'(h(s) ) | \leq \varepsilon^{1/2},$ if and only if $| \langle J'(h(s) ) ,v\rangle | \leq\varepsilon^{1/2}\quad \text{for } v\in\mathbf{U}(\Omega),$ if and only if $\langle J'(h(s) ) ,v\rangle \geq-\varepsilon^{1/2}\quad \text{for }v\in\mathbf{U}(\Omega).$ We claim that there is some$s\in A$satisfying$|J'(h(s) ) | \leq\varepsilon^{1/2}$. If this is not the case, then for each$s\in A$,$v_{s}\in\mathbf{U}(\Omega)$exists such that$\langle J'(h(s) ) ,v_{s}\rangle <-\varepsilon^{1/2}$. By the continuity of$J'$,$\delta_{s}>0$and an open ball$B_{s}$in$[0,1]$containing$s$exist such that for$t\in B_{s}$and$u\in X(\Omega)$with$| u| \leq\delta_{s}$, we have $$\langle J'(h(t) +u) ,v_{s} \rangle <-\varepsilon^{1/2}.\label{4-10}$$ Since$A$is compact, a finite subcovering$B_{s_{1}}$,$B_{s_{2}}$\dots$B_{s_{k}}$of$A$exists. We define the Lipschitz continuous functions, for each$j=1,2,\dots,k$,$\psi_{j}:[0,1]\to[0,1]$by $\psi_{j}(t) =\begin{cases} \mathop{\rm dist}(t,B_{s_{j}}^{c}) / \sum_{i=1}^k \mathop{\rm dist}(t,B_{s_{i}}^{c}) &\text{for }t\in A;\\ 0&\text{for }t\notin\cup_{i=1}^{k}B_{s_{i}}. \end{cases}$ Then \begin{gather*} \sum_{j=1}^k \psi_{j}(t) =1\text{for }t\in A;\\ \|\sum_{j=1}^k \psi_{j}(t) v_{s_{j}}\| _{H^{1}}\leq1\quad \text{for }t\in A. \end{gather*} Let$\delta=\min\{\delta_{s_{1}},\dots \delta_{s_{k}}\} $and let$\psi:[0,1]\to[0,1] $be a continuous function such that $\psi(t) =\begin{cases} 1 & \text{if }J(h(t) ) \geq\alpha_{\Gamma}(\Omega);\\ 0 & \text{if }J(h(t) ) \leq\alpha_{\Gamma}(\Omega)-\varepsilon, \end{cases}$ and let$g\in C([0,1],X(\Omega)) $be defined by $g(t) =h(t) +\delta\psi(t) \overset{k}{\underset{j=1}{\sum}}\psi_{j}(t) v_{s_{j}}.$ It follows from (\ref{4-8}) that, for$t\in\{0,1\}$, we have$J( h(t) ) =0<\alpha_{\Gamma}(\Omega)-\varepsilon$, or$\psi(t) =0$. Consequently,$g(0) =h( 0) =0$and$g(1) =h(1) =e$, that is,$g\in\Gamma(\Omega)$. The mean value theorem and (\ref{4-10}) imply that, for each$t\in A$, there is some$0<\tau<1for which \begin{aligned} &J(g(t) ) -J(h(t) ) \\ &=\langle J'(h(t) +\tau\delta\psi\big( t) \overset{k}{\underset{j=1}{\sum}}\psi_{j}(t) v_{s_{j} }\big) ,\delta\psi(t) \overset{k}{\underset{j=1}{\sum}} \psi_{j}(t) v_{s_{j}}\rangle \\ & =\delta\psi(t) \overset{k}{\underset{j=1}{\sum}}\psi _{j}(t) \langle J'\big(h(t) +\tau\delta\psi(t) \overset{k}{\underset{j=1}{\sum}}\psi _{j}(t) v_{s_{j}}\big) ,v_{s_{j}}\rangle \\ & \leq-\varepsilon^{1/2}\delta\psi(t) . \end{aligned}\label{4-11} Thus $J(g(t) ) \leq J(h(t) ) -\varepsilon^{1/2}\delta\psi(t) \leq J(h( t) ) .$ Ift\notin A$, then$\psi(t) =0$and hence$J(g( t) )=J(h(t) ) $. Let$\bar{t}\in[0,1]$satisfy$J(g(\overline{t}) )=\Phi(g) $, then we obtain $J(h(\overline{t}) )\geq J(g(\overline{t}) )\geq\alpha_{\Gamma}(\Omega),$ so that$\overline{t}\in A$and$\psi(\overline{t}) =1$. By (\ref{4-11}), we obtain $J(g(\overline{t}) )-J(h(\overline{t}) )\leq-\varepsilon^{1/2}\delta$ and in particular $\Phi(g) +\varepsilon^{1/2}\delta\leq J(h( \overline{t}) )\leq\Phi(h) ,$ so that$g\neq h$. However, by the definition of$g$, we have$d(g,h) \leq\delta$and $\Phi(g) +\varepsilon^{1/2}d(g,h) \leq \Phi(h)$ which contradicts (\ref{4-9}). The proof is complete . \end{proof}$\alpha_{\Gamma}(\Omega)$is a (PS)-value in$X(\Omega)$for$J$. \begin{theorem} \label{i10} Under the conditions of Lemma \ref{i9}, for each minimizing sequence$\{f_{k}\}\subset\Gamma(\Omega)$such that $\Phi(f_{k}) =\underset{s\in[0,1]}{\max}J(f_{k}( s) ) =\alpha_{\Gamma}(\Omega)+ o(1),$ there is a (PS)-sequence$\{v_{k}\}$in$X(\Omega)$for$J$satisfying \begin{gather*} J(v_{k})=\alpha_{\Gamma}(\Omega)+ o(1),\\ \mathop{\rm dist}(v_{k},\;f_{k}([0,1])) = o(1),\\ J'(v_{k})= o(1)\quad \text{strongly in }X^{-1}(\Omega) \end{gather*} as$k\to\infty$. In particular,$\alpha_{\Gamma}(\Omega)$is a (PS)-value in$X(\Omega)$for$J$. \end{theorem} \begin{proof} We define$\varepsilon_{k}=\underset{s\in[0,1]}{\max}J(f_{k}( s) ) -\alpha_{\Gamma}(\Omega)$if$\underset{s\in[0,1]}{\max }J(f_{k}(s) ) -\alpha_{\Gamma}(\Omega)>0$and$\varepsilon_{k}=\frac{1}{k}$in the other case. Then we apply Lemma \ref{i9} to$\varepsilon_{k}$and$f_{k}$: \begin{gather*} \alpha_{\Gamma}(\Omega)-\varepsilon_{k}\leq J(v_{k}) \leq\underset{s\in [0,1]}{\max}J(f_{k}(s) ) \leq\alpha_{\Gamma }(\Omega)+\varepsilon_{k},\\ \mathop{\rm dist}(v_{k},f_{k}([0,1]) ) \leq \varepsilon_{k}^{1/2},\\ | J'(v_{k}) | \leq\varepsilon_{k}^{\frac {1}{2}}\quad \quad\text{for each }k>0. \end{gather*} This completes the proof. \end{proof} \noindent (D) Consider the infimum of positive (PS)-values in$X(\Omega)$for$J:$$\alpha_{\mathbf{P}}(\Omega)=\inf_{\beta\in\mathbf{P}(\Omega) }\beta,$ where$\mathbf{P}(\Omega) $is the set of all positive (PS)-values in$X(\Omega)$for$J$. That$\alpha_{\mathbf{P}}(\Omega)>0$is a consequence of the following theorem. \begin{theorem} \label{i11} There is a$\beta_{0}>0$such that$\beta\geq\beta_{0}$for every positive (PS)-value$\beta$in$X(\Omega)$for$J$. \end{theorem} \begin{proof} Let$\{ u_{n}\} $be a (PS)$_{\beta}$-sequence in$X(\Omega)$for$J$with$\beta>0$. By Lemma \ref{p30}, a positive sequence$\{ c_{n}(\beta)\} $exists such that$c_{n}(\beta)= o(1)$as$n\to \infty$,$\beta\to0$, and $$a(u_{n})\leq c_{n}(\beta)^{2}.\label{4-12}$$ By the Sobolev embedding theorem, there is a constant$d>0$such that $$b(u_{n})\leq da(u_{n})^{p/2}.\label{4-13}$$ By Lemma \ref{p30}, (\ref{4-12}), and (\ref{4-13}), we have $o(1) =a(u_{n})-b(u_{n}) \geq a(u_{n})\left[ 1-dc_{n}(\beta)^{p-2}\right] .$ Take$\beta_{0}>0$and$n_{0}>0$such that if$\beta<\beta_{0}$and$n\geq n_{0}$, then$1-dc_{n}(\beta)^{p-2}>\frac{1}{2}$. Consequently,$a(u_{n} )=b(u_{n})= o(1)$, or$J(u_{n})= o(1)$. Thus,$\beta\geq\beta_{0}$. \end{proof}$\alpha_{\mathbf{P}}(\Omega) $is a (PS)-value in$X(\Omega) $for$J$. \begin{theorem} \label{i12}$\alpha_{\mathbf{P}}(\Omega)\in\mathbf{P}(\Omega)$. \end{theorem} \begin{proof} For each$n\in\mathbb{N}$, take$u_{n}\in X(\Omega)$and$c_{n}\in \mathbf{P}(\Omega)$such that \begin{gather*} |c_{n}-\alpha_{\mathbf{P}}(\Omega)|<\frac{1}{n},\\ |J(u_{n})-c_{n}|<\frac{1}{n},\\ \| J'(u_{n})\| _{X^{-1}}<\frac{1}{n}. \end{gather*} Then$J(u_{n})=\alpha_{\mathbf{P}}(\Omega)+ o(1)$and$J'(u_{n})= o(1)$. Thus,$\alpha_{\mathbf{P}}(\Omega)\in\mathbf{P}(\Omega)$. \end{proof} The following theorem is very useful. \begin{theorem} \label{i13} Let$\beta$be a positive (PS)-value in$X(\Omega)$for$J$. Then \newline$(i)\;\beta\geq\alpha_{\gamma}(\Omega);\;\;(ii)\;\beta\geq \alpha_{\mathbf{M}}(\Omega);\;\;(iii)\;\beta\geq\alpha_{\Gamma}(\Omega)$;$(iv) \beta\geq\alpha_{\mathbf{P}}(\Omega)$. \end{theorem} \begin{proof} Let$\{ u_{n}\} $be a nonzero (PS)$_{\beta}$-sequence in$X(\Omega)$for$J$with$\beta>0$. By Lemma \ref{p30}, we have \begin{gather*} J(u_{n})=\beta+ o(1),\\ a(u_{n})-b(u_{n})= o(1). \end{gather*}$(i)$Let$w_{n}=u_{n}(a(u_{n})) ^{-\frac{1}{2}}$, then$a(w_{n})=1$and$b(w_{n})=a(u_{n})^{-p/2}b(u_{n})\leq\gamma(\Omega)^{p}$. Thus,$a(u_{n})\geq\gamma(\Omega)^{2p/(2-p)}+ o(1)$, or$\beta \geq(\frac{1}{2}-\frac{1}{p})\gamma(\Omega)^{2p/(2-p)}=\alpha_{\gamma} (\Omega)$.\newline$(ii)$By Theorem \ref{i3}, there is a sequence$\{s_{n}\}$in$\mathbb{R}^{+}$such that$\{s_{n}u_{n}\}\subset\mathbf{M}(\Omega)$and$J(s_{n}u_{n})=\beta+ o(1)$. Therefore,$\beta\geq\alpha_{\mathbf{M} }(\Omega)$.\newline$(iii)$By Theorem \ref{i3} and Lemma \ref{i2}$(ii) $, there are sequences$\{s_{n}\}$and$\{t_{n}\}$in$\mathbb{R}^{+}$such that$\{s_{n}u_{n}\}\subset\mathbf{M}(\Omega)$,$\{t_{n}u_{n}\}\subset\mathbf{Z}(\Omega)$, and$J(s_{n}u_{n})=\beta + o(1)$. Since the manifold$\mathbf{Z}(\Omega)$is path-connected, there is a path$\zeta_{n}$in$\mathbf{Z}(\Omega)$that connects$t_{n}u_{n}$to$e$. Let$\gamma_{n}'$be the line segment connecting$0$and$t_{n}u_{n}$and the path$\gamma_{n}=\gamma_{n}'\cup\zeta_{n}$, then $\alpha_{\Gamma}(\Omega)\leq\underset{0\leq t\leq1}{\max}J(\gamma _{n}(t))=J(s_{n}u_{n}) =\beta+ o(1).$ Thus,$\beta\geq\alpha_{\Gamma}(\Omega)$.\newline$(iv)$Clearly,$\beta \geq\alpha_{\mathbf{P}}(\Omega)$. \end{proof} By Theorems \ref{i1}, \ref{i4}, \ref{i10}, \ref{i12}, and \ref{i13}, we have the following theorem. \begin{theorem} \label{i14}$\alpha_{\gamma}(\Omega)=\alpha_{\mathbf{M}}(\Omega)=\alpha _{\Gamma}(\Omega)=\alpha_{\mathbf{P}}(\Omega)$. \end{theorem} \begin{definition}\label{i15} \rm By Theorem \ref{i14}, we conclude that the positive (PS)-values$\alpha_{\gamma}(\Omega)$,$\alpha_{\Gamma}(\Omega)$,$\alpha_{\mathbf{M} }(\Omega)$, and$\;\alpha_{\mathbf{P}}(\Omega)$in$X(\Omega)$for$J$are the same. Any one of them is called the index of$J$in$X(\Omega) $and denoted by$\alpha_{X}(\Omega)$. By the definition of$\alpha_{\mathbf{M} }(\Omega)$, if$u$is a nonzero solution of Equation$($\ref{E1}$)$, then$u\in\mathbf{M}(\Omega)$. Thus,$J(u)\geq\alpha_{\mathbf{M}}(\Omega )=\alpha_{X}(\Omega)$. We say that a nonzero solution$u $of Equation \eqref{E1} in$X(\Omega) $is a ground state solution if$J(u)=\alpha_{X}(\Omega)$, and is a higher energy solution if$J(u)>\alpha _{X}(\Omega)$. \end{definition} \begin{remark} \rm We denote$\alpha_{X}(\Omega)$by$\alpha(\Omega)$for$X( \Omega) =H_{0}^{1}(\Omega) $and by$\alpha_{s}(\Omega)$for$X(\Omega) =H_{s}^{1}(\Omega) $(see Definition \ref{w1}). \end{remark} \begin{remark}\label{b201} \rm By Theorem \ref{b2}, a ground state solution in$X( \Omega) $is of constant sign. Note that if$u$is a solution of \eqref{E1}, then$-u$is also a solution of \eqref{E1}. By the maximum principle, if$u$is a nonzero and nonnegative solution of \eqref{E1}, then$u$is positive. From now on, by a ground state solution in$X(\Omega) $, we mean a positive solution of \eqref{E1}. \end{remark} \begin{definition} \rm We say that a domain$\Omega$in$\mathbb{R}^{N}$is an achieved domain if there is a solution$u$in$H_{0}^{1}(\Omega)$of \eqref{E1} such that$J(u)=\alpha(\Omega)$, by Remark \ref{b201}, we may assume that$u$be positive. Otherwise, we say that$\Omega$is a nonachieved domain. \end{definition} \begin{theorem}\label{f8}$(i)$If$\Omega$is a large domain in$\mathbb{R}^{N}$, then$\alpha(\Omega)=\alpha(\mathbb{R}^{N})$; \newline$(ii)$If$\Omega$is a large domain in$\mathbf{A}^{r}$, then$\alpha(\Omega)=\alpha(\mathbf{A}^{r} )$; \newline$(iii)$If$\Omega$is a large domain in$\mathbf{A}^{r_{1},r_{2}} $, then$\alpha(\Omega)=\alpha(\mathbf{A}^{r_{1},r_{2}})$. \end{theorem} \begin{proof} It suffices to prove part$(i)$. Let$w\in H^{1}(\mathbb{R}^{N})$be a ground state solution of Equation \eqref{E1} satisfying $a(w)=\int_{\mathbb{R}^{N}}(|\nabla w|^{2}+w^{2})=b(w)=\int_{\mathbb{R}^{N} }|w|^{p}=(\frac{2p}{p-2}) \alpha(\mathbb{R}^{N}).$ For$r_{n}\to\infty$, take$\{z_{n}\}\subset\Omega$such that$B^{N}(z_{n};r_{n})\subset\Omega$. Consider the cut-off function$\eta\in C_{c}^{\infty}([ 0,\infty) ) $as in \eqref{1-2}, and for each$n$, let $w_{n}(z)=\eta(\frac{2|z-z_{n}|}{r_{n}})w(z-z_{n}).$ Then$w_{n}\in H_{0}^{1}(\Omega)$and \begin{gather*} a(w_{n})={\int_{\Omega}} (|\nabla w_{n}|^{2}+w_{n}^{2})=(\frac{2p}{p-2}) \alpha (\mathbb{R}^{N})+ o(1),\\ b(w_{n})={\int_{\Omega}} |w_{n}|^{p}=(\frac{2p}{p-2}) \alpha(\mathbb{R}^{N})+ o(1)\quad\text{ as }n\to\infty. \end{gather*} Thus, \begin{gather*} J(w_{n}) =\alpha(\mathbb{R}^{N})+ o(1) ,\\ a(w_{n})=b(w_{n})+ o(1) \quad\text{as }n\to\infty. \end{gather*} By Theorem \ref{i4},$\{ w_{n}\} $is a (PS)$_{\alpha (\mathbb{R}^{N})}$-sequence in$H_{0}^{1}(\Omega)$for$J$. Therefore,$\alpha(\Omega)\leq\alpha(\mathbb{R}^{N})$. Clearly,$\alpha(\mathbb{R} ^{N})\leq\alpha(\Omega)$, thus we have$\alpha(\Omega)=\alpha(\mathbb{R}^{N})$. \end{proof} \begin{theorem} \label{f9} Let$\Omega$be a large domain in$\mathbb{R}^{N}$. If$\beta$is a positive (PS)-value in$H_{0}^{1}(\Omega)$for$J$, then$m\beta$is also a positive (PS)-value in$H_{0}^{1}(\Omega)$for$J$, where$m=2,3,\dots$. \end{theorem} \begin{proof} It suffices to prove the case$m=2$. First embed$H_{0}^{1}(\Omega)$into$H^{1}(\mathbb{R}^{N})$. Let$\{ u_{n}\} $be a (PS)$_{\beta} $-sequence in$H_{0}^{1}(\Omega)$. Then by Lemma \ref{p30}, there is a constant$c>0$such that, for each$n$,$a(u_{n}) \leq c$and$b(u_{n}) \leq c$. For$r_{n}\to\infty$, since$\Omega\backslash B^{N}(0;5r_{n})$is also a large domain in$\mathbb{R}^{N}$,$z_{n}\in\Omega$exists such that$B^{N}(z_{n};2r_{n})\subset\Omega$and ${\int_{B^{N}(0;r_{n})^{c}}} | \nabla u_{n}| ^{2}+u_{n}^{2}<\frac{1}{n}\quad\text{and}\quad {\int_{B^{N}(0;r_{n})^{c}}}| u_{n}| ^{p}<\frac{1}{n}.$ Note that$|z_{n}|\geq5r_{n}$. Let$\eta_{n}(z)=\eta(\frac{|z|}{r_{n}}) $, where$\eta$is as in (\ref{1-2}),$v_{n}(z)=\eta_{n}(z)u_{n}(z)$and$w_{n}(z)=v_{n}(z-z_{n})$. Then we have$|\nabla\eta_{n}|\leq\frac{2}{r_{n}}$and$\mathop{\rm supp} w_{n}\subset B^{N}(z_{n};2r_{n}) $.\newline$(i)J(v_{n})=\beta+ o(1)$: note that $|\nabla v_{n}|^{2} =|\eta_{n}|^{2}|\nabla u_{n}|^{2}+|\nabla\eta_{n} |^{2}|u_{n}|^{2}+2\eta_{n}u_{n}\nabla\eta_{n}\nabla u_{n}.$ Thus, for$z\in B^{N}(0;r_{n})$, we have$|\nabla v_{n}|=|\nabla u_{n}|and \begin{align*} {\int_{\Omega}} |\nabla v_{n}|^{2} & = {\int_{B^{N}(0;r_{n})}} |\nabla v_{n}|^{2}+ {\int_{B^{N}(0;2r_{n})\backslash B^{N}(0;r_{n})}} |\nabla v_{n}|^{2}\\ & = {\int_{B^{N}(0;r_{n})}} |\nabla u_{n}|^{2}+ o(1)\\ & = {\int_{\Omega}} |\nabla u_{n}|^{2}+ o(1). \end{align*} Similarly, we have ${\int_{\Omega}}|v_{n}|^{2}= {\int_{\Omega}}|u_{n}|^{2}+ o(1),\quad {\int_{\Omega}} |v_{n}|^{p}= {\int_{\Omega}} |u_{n}|^{p}+ o(1).$ Thus,J(v_{n})=J(u_{n})+ o(1)=\beta+ o(1)$. Clearly, for each$n$,$J(w_{n})=J(v_{n})$, and hence$J(w_{n})=\beta+ o(1)$.\newline$(ii)J(v_{n}+w_{n})=2\beta+ o(1):$since the supports of$v_{n}$and$w_{n}are disjoint, we have \begin{align*} a(v_{n}+w_{n}) & ={\int_{\Omega}} | \nabla(v_{n}+w_{n}) | ^{2}+(v_{n} +w_{n}) ^{2}\\ & ={\int_{\Omega}} | \nabla v_{n}| ^{2}+v_{n}^{2}+ {\int_{\Omega}} | \nabla w_{n}| ^{2}+w_{n}^{2} +2{\int_{\Omega}} \nabla v_{n}\nabla w_{n}+2 {\int_{\Omega}} v_{n}w_{n}\\ & =a(v_{n})+a(w_{n}). \end{align*} Now, \begin{align*} &{\int_{\Omega}} | v_{n}+w_{n}| ^{p}-| v_{n}| ^{p}-|w_{n}| ^{p}\\ & ={\int_{B^{N}(0;2r_{n})}} | v_{n}+w_{n}| ^{p}-| v_{n}| ^{p}-|w_{n}| ^{p} + {\int_{B^{N}(0;2r_{n})^{c}\cap\Omega}} | v_{n}+w_{n}| ^{p}-| v_{n}| ^{p}-|w_{n}| ^{p}\\ & =0. \end{align*} Thus, $b(v_{n}+w_{n}) ={\int_{\Omega}}| v_{n}+w_{n}| ^{p} ={\int_{\Omega}}| v_{n}| ^{p}+ {\int_{\Omega}}| w_{n}| ^{p}\\ =b(v_{n})+b(w_{n}).$ Hence, $J(v_{n}+w_{n}) =\frac{1}{2}a(v_{n}+w_{n})-\frac{1}{p}b(v_{n}+w_{n}) =J(v_{n})+J(w_{n}) =2\beta+ o(1).$(iii)\| J'(v_{n}+w_{n})\| = o(1):$for$\varphi\in C_{c}^{\infty}(\Omega), we have \begin{align*} \langle J'(v_{n}) ,\varphi\rangle & =\int_{B^{N}(0;r_{n})}u_{n}(\nabla\eta_{n})\cdot\nabla\varphi+\int _{B^{N}(0;r_{n})}\eta_{n}(\nabla u_{n})\cdot\nabla\varphi+\eta_{n}u_{n} \varphi\\ & -\int_{B^{N}(0;r_{n})}| \eta_{n}u_{n}| ^{p-2}\eta_{n} u_{n}\varphi+ o(1)\\ & =\int_{B^{N}(0;r_{n})}\nabla u_{n}(z)\cdot\nabla\varphi(z)+u_{n} (z)\varphi(z)\\ &\quad -\int_{B^{N}(0;r_{n})}| u_{n}| ^{p-2}u_{n}\varphi(z)+ o(1)\\ & =\langle J'(u_{n}) ,\varphi\rangle + o(1). \end{align*} Thus,\| J'(v_{n}) \| _{H^{-1}}= o(1)$. Similarly,$\| J'(w_{n}) \| _{H^{-1} }= o(1).\newline We have \begin{align*} &{\int_{\Omega}} | v_{n}+w_{n}| ^{p-2}(v_{n}+w_{n}) \varphi-| v_{n}| ^{p-2}v_{n}\varphi-| w_{n}| ^{p-2}w_{n}\varphi\\ & ={\int_{B^{N}(0;2r_{n})}} | v_{n}+w_{n}| ^{p-2}(v_{n}+w_{n}) \varphi-| v_{n}| ^{p-2}v_{n}\varphi-| w_{n}| ^{p-2}w_{n}\varphi\\ & \quad+ {\int_{B^{N}(0;2r_{n})^{c}\cap\Omega}} | v_{n}+w_{n}| ^{p-2}(v_{n}+w_{n}) \varphi-| v_{n}| ^{p-2}v_{n}\varphi-| w_{n}| ^{p-2}w_{n}\varphi\\ & =0. \end{align*} Now for\varphi\in C_{c}^{\infty}(\Omega) , we have \begin{align*} \langle J'(v_{n}+w_{n}) ,\varphi\rangle & = {\int_{\Omega}} \nabla(v_{n}+w_{n}) \nabla\varphi+(v_{n}+w_{n}) \varphi\\ &\quad -{\int_{\Omega}} | v_{n}+w_{n}| ^{p-2}(v_{n}+w_{n})\varphi\\ & ={\int_{\Omega}} \nabla v_{n}\nabla\varphi+v_{n}\varphi+ {\int_{\Omega}} \nabla w_{n}\nabla\varphi+w_{n}\varphi\\ &\quad -{\int_{\Omega}} | v_{n}| ^{p-2}v_{n}\varphi- {\int_{\Omega}} | w_{n}| ^{p-2}w_{n}\varphi\\ & =\langle J'(v_{n}),\varphi\rangle +\langle J'(w_{n}),\varphi\rangle . \end{align*} Therefore,\| J'(v_{n}+w_{n}) \| _{H^{-1} }= o(1)$. We conclude that \begin{gather*} J(v_{n}+w_{n})=2\beta+ o(1),\\ J'(v_{n}+w_{n})= o(1)\quad \quad\text{strongly in }H^{-1}(\Omega). \end{gather*} \end{proof} The following theorem has a proof similar to that of Theorem \ref{f9}. \begin{theorem} \label{f91} Let$\Omega$be a large domain in$\mathbf{A}^{r}$. If$\beta$is a positive (PS)-value in$H_{0}^{1}(\Omega)$for$J$, then$m\beta$is also a positive (PS)-value in$H_{0}^{1}(\Omega)$for$J$, where$m=2,3,\dots$. \end{theorem} \begin{lemma} \label{f10} The set$\mathbf{P}(\Omega) $is closed. \end{lemma} The proof of this lemma is similar to the proof of Theorem \ref{i12}; so we omit it. By Lemma \ref{i2},$J(\mathbf{M}(\Omega)) $is bounded below away from zero. Actually for any domain$\Omega$in$\mathbb{R}^{N}$,$J(\mathbf{M}(\Omega)) $is unbounded above. \begin{theorem} \label{f14} If$\Omega$is a domain in$\mathbb{R}^{N}$, then$J(\mathbf{M}(\Omega)) =(\alpha(\Omega),\infty) $for a nonachieved domain$\Omega$and$J(\mathbf{M}(\Omega)) =[ \alpha(\Omega),\infty) $for an achieved domain$\Omega$. \end{theorem} \begin{proof}$(i)$Suppose that$\Omega$is bounded. By Struwe \cite[p.116 Theorem 6.6]{S}, an unbounded sequence$\{ u_{n}\} $exists in$\mathbf{M}(\Omega)$for$J$. Since$J(u_{n}) =(\frac{1}{2}-\frac{1}{p}) a(u_{n})$and$\mathbf{M}(\Omega)$is path connected, then we have$J(\mathbf{M}(\Omega)) =[\alpha(\Omega),\infty)$. \newline$(ii)$Let$\Omega$be an unbounded domain and$\Omega_{1}$be a bounded domain in$\Omega$. Then$\mathbf{M}( \Omega_{1}\mathbf{)\subset M(}\Omega)$and$\alpha( \Omega\mathbf{)\leq}\alpha(\Omega_{1})$. By part$(i)$, we have $[ \alpha(\Omega_{1}),\infty) =J(\mathbf{M}(\Omega_{1})) \subset J(\mathbf{M}(\Omega)) .$ Since$\mathbf{M}(\Omega)$is path connected, the result follows. \end{proof} \begin{theorem} \label{f15} Let$\Omega$be an Esteban-Lions domain as well as a large domain in$\mathbb{R}^{N}$. Then we have$\mathbf{P}(\Omega) =\{ \alpha(\Omega) ,\;2\alpha(\Omega) ,\;3\alpha(\Omega) ,\dots\} $. \end{theorem} \begin{proof} By Theorem \ref{f9},$\mathbf{P}(\Omega) \supset\{ \alpha(\Omega) ,\;2\alpha(\Omega) ,\;3\alpha (\Omega) ,\dots\} $. Suppose that a (PS)$_{\beta} $-sequence$\{ u_{n}\} $exists for$J$, where$k\alpha( \Omega) <\beta<(k+1) \alpha(\Omega) $for some$k\in\mathbb{N}$. By the Palais-Smale decomposition theorem \ref{d1}, Equation \eqref{E1} has a nonzero solution. This contradicts Theorem \ref{n3}. \end{proof} By Lemma \ref{p30}, if$\{ u_{n}\} $is a (PS)$_{\beta}- $sequence in$H_{0}^{1}(\Omega)$for$J$, then$a(u_{n})=b(u_{n} )+ o(1)=\frac{2p}{p-2}\beta+ o(1)$. By Theorems \ref{f14} and \ref{f15}, we have: \begin{lemma} \label{f16} Let$\Omega$be an Esteban-Lions domain as well as a large domain in$\mathbb{R}^{N}$. For each$\beta$and$m=0,1,\dots$, satisfying$m\alpha(\Omega) <\beta<(m+1)\alpha(\Omega) $, then there is a sequence$\{ u_{n}\} $in$H_{0}^{1}(\Omega)$for$J\ $satisfying $a(u_{n})=b(u_{n})+ o(1)=\frac{2p}{p-2}\beta+ o(1)$ but $J'(u_{n})\nrightarrow0\quad\text{strongly in }H^{-1}(\Omega).$ \end{lemma} Let$\Omega$be an unbounded domain in$\mathbb{R}^{N}$and$\Omega_{n} =\Omega\cap\mathbf{B}^{N}(0;r_{n}) $, then we have the following theorem. \begin{theorem} \label{f177}$\alpha_{X}(\Omega_{n}) =\alpha_{X}( \Omega) + o(1)$as$n\to\infty$. \end{theorem} \begin{proof} Suppose that$\{ u_{n}\} $in$X(\Omega) $is a minimizing sequence in$\mathbf{M}(\Omega) $of$\alpha _{X}(\Omega) $, then by Lemma \ref{p30},$\{ u_{n}\} $is bounded in$X(\Omega) $. Let$\{r_{n}\} $be a sequence of strictly increasing positive integers such that$r_{n}\geq n$, $$\int_{\Omega\cap\{ | z| \geq\frac{r_{n}}{2}\} }| \nabla u_{n}| ^{2}+| u_{n}| ^{2}<\frac{1} {n}\label{5-1}$$ and $$\int_{\Omega\cap\{ | z| \geq\frac{r_{n}}{2}\} }| u_{n}| ^{p}<\frac{1}{n}.\label{5-2}$$ Define$\eta_{n}(z) =\eta(\frac{2| z| }{r_{n}}) $, where$\eta$is as in$( \ref{1-2}) $. Then$\eta_{n}u_{n}\in X( \Omega_{n}) \subset X(\Omega) $. From the inequalities$(\ref{5-1})$and$(\ref{5-2})$, we obtain $a(\eta_{n}u_{n})=a(u_{n})+ o(1)\quad\text{and }b(\eta_{n}u_{n})=b(u_{n} )+ o(1).$ Therefore, we have $J(\eta_{n}u_{n}) =J(u_{n}) + o(1) =\alpha_{X}(\Omega) + o(1).$ and $a(\eta_{n}u_{n})=b(\eta_{n}u_{n})+ o(1).$ By Theorem \ref{i3}, there is a sequence$\{ s_{n}\} $in$\mathbb{R}^{+}$such that$s_{n}=1+ o(1)$,$\{ s_{n}\eta_{n} u_{n}\} $is in$\mathbf{M}(\Omega)$and$J(s_{n}\eta_{n}u_{n} )=\alpha_{X}(\Omega) + o(1)$. Note that$J(s_{n}\eta _{n}u_{n}) \geq\alpha_{X}(\Omega_{n}) >\alpha_{X} (\Omega)$. Hence,$\alpha_{X}(\Omega_{n}) =\alpha_{X}( \Omega) + o(1)$. \end{proof} Let$\Omega$be a domain containing zero in$\mathbb{R}^{N}$. For$\delta>0$, we define $\delta\Omega=\{ \delta z\mid z\in\Omega\} .$ Then we have the following theorem. \begin{theorem} \label{f178}$(i)\lim_{\delta\to\infty}\alpha(\delta \Omega) =\alpha(\mathbb{R}^{N}) $; \newline$(ii)$Let$\delta\Omega$be achieved for each$\delta>0$, then lim$_{\delta \to0^{+}}\alpha(\delta\Omega) =\infty$. \end{theorem} \begin{proof}$(i)$By Theorem \ref{f3} below, there is a ground state solution$u$in$H^{1}(\mathbb{R}^{N})$such that$a(u)=b(u)$and$J(u)=\alpha( \mathbb{R}^{N}) $. A sequence$\{ \delta_{n}\} $exists such that$\delta_{n}\to\infty$and$B^{N}(0;n) \subset\delta_{n}\Omega$. Consider the cut-off function$\eta$and$\eta _{n}(z)=\eta(\frac{2|z|}{n})$for$n=1,2,\dots$. Let$u_{n}(z)=\eta _{n}(z)u(z)$, then$u_{n}(z)\in H_{0}^{1}(B^{N}(0;n))$,$J(u_{n} )=\alpha(\mathbb{R}^{N}) + o(1)$, and$a(u_{n})=b(u_{n})+ o(1)$as$n\to\infty$. By Theorem \ref{i3}, there is a sequence$\{s_{n}\}$in$\mathbb{R}^{+}$such that$s_{n}=1+ o(1)$,$\{s_{n}u_{n}\}$is in$\mathbf{M}(B^{N}(0;n)) $and$J(s_{n}u_{n})=\alpha( \mathbb{R}^{N}) + o(1)$. Then we have $\lim_{n\to\infty}\alpha(\delta_{n}\Omega) \leq \lim_{n\to\infty}\alpha(B^{N}(0;n)) \leq\lim _{n\to\infty}J(s_{n}u_{n})=\alpha(\mathbb{R}^{N}) + o(1).$ However,$\alpha(\mathbb{R}^{N}) \leq\alpha(\delta \Omega) $for each$\delta>0$. Thus,$\lim_{\delta\to\infty }\alpha(\delta\Omega) =\alpha(\mathbb{R}^{N}) $.\newline$(ii)$Let$u$be a ground state solution of Equation$\eqref{E1}$in$H_{0}^{1}(\delta\Omega)$, then$a(u)=b(u)$and$J(u)=\alpha( \delta\Omega) $. Set$v(z)=u(\delta z)$, then$v\in H_{0}^{1}(\Omega)$. Note that $a(u)=\int_{\delta\Omega}(|\nabla u(z)|^{2}+u(z)^{2})dz=\delta^{N-2} \int_{\Omega}|\nabla v|^{2}+\delta^{N}\int_{\Omega}v^{2},$ and $(b(u)) ^{2/p}=(\int_{\delta\Omega}|u(z)|^{p}) ^{2/p}=\delta^{\frac{2N}{p}}(\int_{\Omega}|v|^{p}) ^{\frac{2} {p}}.$ Therefore, by the Sobolev continuous embedding theorem, $a(u)\geq\delta^{N-2}\int_{\Omega}|\nabla v|^{2}\geq c\delta^{N-2}( \int_{\Omega}|v|^{p}) ^{\frac{2}{p}}\geq c\delta^{N-2-\frac{2N}{p} }(b(u)) ^{2/p}.$ That is,$(a(u)) ^{\frac{p-2}{p}}\geq c\delta^{N-2-\frac{2N}{p} }$. Thus, $J(u)=(\frac{1}{2}-\frac{1}{p}) a(u)\geq c(\frac{1} {2}-\frac{1}{p}) (\delta^{N-2-\frac{2N}{p}}) ^{p/(p-2)}.$ Therefore,$\alpha(\delta\Omega) \geq c(\frac{1}{2} -\frac{1}{p}) (\delta^{N-2-\frac{2N}{p}}) ^{p/(p-2)}$. Since$p<\frac{2N}{N-2}$, we have$N-2-\frac{2N}{p}<0$. We conclude that$\lim_{\delta\to0+}\alpha(\delta\Omega) =\infty$. \end{proof} As a corollary of Theorem \ref{f178}, we have \begin{theorem} \label{f17}$(i)\lim_{r\to\infty}\alpha(B^{N}(0;r))=\alpha (\mathbb{R}^{N})$;$(ii)\lim_{r\to0+}\alpha(B^{N} (0;r))=\infty$. \end{theorem} Using the same argument as for the proof of Theorem \ref{f178}, we obtain the following theorem. \begin{theorem} \label{f19}$(i) \lim_{t\to\infty}\alpha( \mathbf{A}_{-t,t}^{r}) =\alpha(\mathbf{A}^{r}) $;$(ii) \lim_{t\to0^{+}}\alpha( \mathbf{A}_{-t,t}^{r}) =\infty$. \end{theorem} Let$\Omega=O\times\mathbb{R}^{l}$, where$O$is a bounded domain in$\mathbb{R}^{m}$,$m\geq1$,$n\geq1$. With the same argument of the proof in Theorem \ref{f178}, we get \begin{theorem} \label{f18}$(i)\lim_{\delta\to\infty}\alpha(\delta\Omega )=\alpha(\mathbb{R}^{N}) $;$(ii)\lim_{\delta \to0+}\alpha(\delta\Omega)=\infty$. \end{theorem} We have the following continuity property. \begin{theorem} \label{f20} We have $\underset{\delta\to1}{\lim}\alpha(\delta\mathbf{A}^{r}) =\alpha(\mathbf{A}^{r}) .$ \end{theorem} \begin{proof}$(i)$.$\underset{\delta\to1^{+}}{\lim}\alpha(\delta \mathbf{A}^{r}) =\alpha(\mathbf{A}^{r}) :$let$1<\delta<2$. By Theorem \ref{c3}$(ii) $below,$\alpha( 2\mathbf{A}^{r}) <\alpha(\delta\mathbf{A}^{r}) <\alpha(\mathbf{A}^{r}) $. In addition,$\alpha( \delta\mathbf{A}^{r}) $is increasing as$\delta$is decreasing. Let$c\equiv\lim_{\delta\to1^{+}}\alpha(\delta\mathbf{A} ^{r}) $, then$c\leq\alpha(\mathbf{A}^{r}) $. We claim that$c\geq\alpha(\mathbf{A}^{r}) $. By Theorem \ref{f3} below, a ground state solution$u_{n}$in$\mathbf{M}((1+\frac{1} {n}) \mathbf{A}^{r}) $exists such that $a(u_{n}) =b(u_{n}) \quad \text{and}\quad J(u_{n}) =\alpha\big((1+\frac{1}{n}\big) \mathbf{A} ^{r}) \quad \text{for each }n\in\mathbb{N}.$ Moreover, we have$a(u_{n}) =b(u_{n}) =( \frac{2p}{p-2}) c+ o(1) $and$J(u_{n}) =\alpha((1+\frac{1}{n}) \mathbf{A}^{r}) =c+ o(1) $. Define$v_{n}(z) =u_{n}(( 1+\frac{1}{n}) z) \in H_{0}^{1}(\mathbf{A}^{r}) $. Since $a(u_{n}) =(1+\frac{1}{n}) ^{N-2}\int _{\mathbf{A}^{r}}| \nabla v_{n}| ^{2}dz+(1+\frac{1} {n}) ^{N}\int_{\mathbf{A}^{r}}v_{n}^{2}dz$ and $b(u_{n}) =(1+\frac{1}{n}) ^{N}\int_{\mathbf{A} ^{r}}| v_{n}| ^{p}dz,$ then$\{ v_{n}\} $is bounded in$H_{0}^{1}(\mathbf{A} ^{r}) $. Thus,$a(v_{n}) =b(v_{n}) + o(1) $and$J(v_{n}) =c+ o(1) $. By Theorem \ref{i3},$s_{n}>0$exists such that$s_{n}v_{n}\in\mathbf{M}( \mathbf{A}^{r}) $,$s_{n}=1+ o(1) ,\ $and$J( s_{n}v_{n}) =c+ o(1) $. Hence,$c\geq\alpha( \mathbf{A}^{r}) $. We conclude the proof.\newline$(ii)\underset {\delta\to1^{-}}{\lim}\alpha(\delta\mathbf{A}^{r}) =\alpha(\mathbf{A}^{r}) :$by Theorem \ref{f3},$u\in \mathbf{M}(\mathbf{A}^{r}) $satisfies$a(u) =b(u) $and$J(u)=\alpha(\mathbf{A}^{r}) $. Let$v_{n}(x,y) =u((1+\frac{1}{n}) x,y) $for$n\in\mathbb{N}$. Then$v_{n}(x,y) \in H_{0}^{1}( \frac{n}{n+1}\mathbf{A}^{r}) $,$a(v_{n}) =a( u) + o(1) $, and$b(v_{n}) =b( u) + o(1) $. Thus$a(v_{n}) =b( v_{n}) + o(1) $as$n\to\infty$and $J(v_{n}) =\frac{1}{2}a(v_{n}) -\frac{1}{p}b( v_{n}) =J(u)+ o(1) .$ By Theorem \ref{i3}, for each$n$, there is an$s_{n}>0$such that$s_{n} v_{n}\in\mathbf{M}(\frac{n}{n+1}\mathbf{A}^{r}) $,$s_{n}=1+ o(1) $and$J(s_{n}v_{n}) =J( u) + o(1)$as$n\to\infty$. Moreover,$J(s_{n} v_{n}) \geq\alpha(\frac{n}{n+1}\mathbf{A}^{r}) >\alpha(\mathbf{A}^{r}) $for each$n\in\mathbb{N}$. By the squeeze theorem, $\lim_{\delta\to1^{-}}\alpha(\delta\mathbf{A}^{r}) =\alpha(\mathbf{A}^{r}) .$ \end{proof} \noindent\textbf{Bibliographical notes:} The constrained maximization problem$\alpha_{\gamma}(\Omega)$is a classical problem. Theorem \ref{i1} is from Lien-Tzeng-Wang \cite{LTW}. The Nehari minimizing problem$\alpha_{\mathbf{M} }(\Omega)$was first studied by Nehari \cite{N}. Theorem \ref{i3} is from Chen-Wang \cite[p.158]{CW}. For Theorem \ref{i4}, Stuart \cite{Stu} proved that there is a (PS)$_{\alpha_{\mathbf{M}}(\Omega)}$-sequence. However, Chen-Wang \cite[Lemma 2.1]{CW} proved that every minimizing sequence in$\mathbf{M}(\Omega) $for$\alpha_{\mathbf{M}}(\Omega)$is a (PS)$_{\alpha_{\mathbf{M}}(\Omega)}$-sequence. The mountain pass minimax problem$\alpha_{\Gamma}(\Omega)$is originally from the mountain pass lemmas \ref{i9} and \ref{i10} by Ambrosetti-Rabinowitz \cite{AR} and the new version is from Br\'{e}zis-Nirenberg \cite{BN}. Theorem \ref{i14} is due to Willem \cite{Wi} and Wang \cite[p.4241]{W}. Theorem \ref{f12} is from Lien-Tzeng-Wang \cite[Lemma 2.5]{LTW}. \section{Palais-Smale Conditions} The Palais-Smale conditions are conditions for compactness. They are useful in ascertaining the existence of solutions of \eqref{E1}. In this section, we assert that eight related (PS)-conditions in$X(\Omega)$for$J$are actually equivalent. \begin{theorem} \label{f1} The (PS)$_{\alpha_{X}(\Omega)}$-condition for$J$holds in a bounded domain$\Omega$. In particular, there is a ground state solution of \eqref{E1} in a bounded domain$\Omega$. \end{theorem} \begin{proof} Let$\{u_{n}\}$be a (PS)$_{\alpha_{X}(\Omega)}$-sequence in$X(\Omega)$for$J$, by Lemma \ref{p30},$\{u_{n}\}$is bounded and $J(u_{n})=\alpha_{X}(\Omega)+ o(1)\text{, }a(u_{n})=b(u_{n})+ o(1).$ Take a subsequence$\{u_{n}\}$and$u\in X(\Omega)$such that$u_{n} \rightharpoonup u$weakly in$X(\Omega)$. By the compactness theorem,$u_{n}\to u$strongly in$L^{p}(\Omega)$. Suppose$u=0$, then$b(u_{n})= o(1)$. Thus,$a(u_{n})= o(1)$and$J(u_{n})= o(1)$, contradicting that$\alpha(\Omega)>0$. By Theorem \ref{c2},$u$is a ground state solution in$X(\Omega) $for$J$and$u_{n}\to u$strongly in$X(\Omega) $. \end{proof} The (PS)$_{\alpha_{X}(\Omega)}$-condition holds in unbounded domains and quasibounded domains. \begin{definition}\rm A domain$\Omega$is quasibounded if $\lim_{z\in\Omega,|z|\to\infty}\text{\textrm{dist}}(z,\partial \Omega)=0.$ \end{definition} \begin{example} \rm$(i)$Let$f,g:\mathbb{R}^{N-1}\to\mathbb{R}$be two functions of$C^{1}$,$f\leq g$, $\lim_{|x|\to\infty}(g(x)-f(x))=0,$ and $\Omega=\{z=(x,y)\in\mathbb{R}^{N-1}\times\mathbb{R}: f(x)0 in \Omega^{2}, which contradicts the fact that u\in X(\Omega^{1}).\newline (ii) By part (i) and Theorem \ref{c2} (ii) , (iii) .\newline (iii) Let \{u_{n}\} in X(\Omega^{1}) satisfy J(u_{n})\to\alpha_{X}^{1} and J'(u_{n})\to0 strongly in X^{-1}(\Omega^{1}). By Theorem \ref{i3}, \{ s_{n}\} in \mathbb{R}^{+} exists such that s_{n}=1+ o(1), w_{n}=s_{n}u_{n}\in\mathbf{M}{(\Omega^{1})} and J(w_{n})=\alpha_{X}^{1}+ o(1) and J'(w_{n}) = o(1) strongly in X^{-1}(\Omega^{1}). Since \mathbf{M}( \Omega^{1}) \subset\mathbf{M}(\Omega^{2}) , \{w_{n}\}\subset\mathbf{M}(\Omega^{2}) and J( w_{n}) =\alpha_{X}^{2}+ o(1). By Theorem \ref{i4}, we have \begin{gather*} J(w_{n})=\alpha_{X}^{2}+ o(1),\\ J'(w_{n})= o(1)\quad\mbox{strongly in } X^{-1}(\Omega^{2}). \end{gather*} Suppose that J satisfies the (PS)_{\alpha_{X}^{2}}-condition. Then there is a subsequence \{w_{n}\} and a w\in X(\Omega^{2}) satisfying w_{n}\to w \ strongly in X(\Omega^{2}) and J(w)=\alpha_{X}^{2}. Hence, w\neq0. By Lemma \ref{b2} below and the maximum principle, w is a ground state solution of \eqref{E1} in \Omega^{2}. Since \{w_{n}\}\subset\mathbf{M}(\Omega^{1}) and w_{n}\to w \ strongly in X(\Omega^{2}), we have w=0 in (\Omega^{1}) ^{c}. This contradicts the fact that w is a positive solution of Equation \eqref{E1} in \Omega^{2}. Thus, J does not satisfy the (PS)_{\alpha _{X}^{2}}-condition. \end{proof} Compare Theorem \ref{c3} for X(\Omega) =H_{0} ^{1}(\Omega) and the following results. \begin{corollary} \label{c4} Let E be either \mathbb{R}^{N}, or \mathbf{A}^{r}, or \mathbf{A}^{r_{1},r_{2}} and \Omega a proper large domain of E. Then there is no any ground state solution of \eqref{E1} in \Omega. \end{corollary} The proof of this corollary follows by Theorem \ref{c3} (ii) and Theorem \ref{f8}. \begin{theorem} \label{c5} Let X(\Omega) =H_{0}^{1}(\Omega) . We have: (i) \alpha(\mathbf{A}^{r}) >\alpha( \Theta) for each domain \Theta\supsetneqq\mathbf{A}^{r}; (ii) J does not satisfy (PS)_{\alpha(\mathbf{A} ^{r}) }-condition. \end{theorem} \begin{proof} (i) Since the infinite strip \mathbf{A}^{r} is a periodic domain, by Theorem \ref{f3} below, there is a ground state solution u_{0}\in \mathbf{M}(\mathbf{A}^{r}) such that \[ J(u_{0}) =\alpha(\mathbf{A}^{r}) .$ The result follows from Theorem \ref{c3}.\newline$(ii)$Let$u_{n} =u_{0}(x,y+n) $for each$n$. Since$\mathbf{A}^{r}$is a periodic domain, we have$u_{n}\in H_{0}^{1}(\mathbf{A}^{r}) $for each$n\in\mathbb{N}$, $J(u_{n}) =\alpha(\mathbf{A}^{r}) \quad\text{\ and \ }a(u_{n}) =b(u_{n}) .$ By Theorem \ref{i4},$\{ u_{n}\} $is a (PS)$_{\alpha( \mathbf{A}^{r}) }$-sequence for$J$. Moreover, for$\varphi\in C_{c}^{\infty}(\mathbf{A}^{r}) $and$K=$\textrm{supp}$\varphi, we have \begin{align*} \langle u_{n},\varphi(z) \rangle _{H^{1}} & =\langle u_{0}(x,y+n) ,\varphi(z) \rangle _{H^{1}}\\ & = {\int_{\mathbf{A}^{r}}} \nabla u_{0}(x,y+n)\nabla\varphi(z)dz+ {\int_{\mathbf{A}^{r}}} u_{0}(x,y+n)\varphi(z)dz\\ & = {\int_{K}} \nabla u_{0}(x,y+n)\nabla\varphi(z)dz+ {\int_{K}} u_{0}(x,y+n)\varphi(z)dz\\ & = o(1)\quad\text{as}\quad n\to\infty. \end{align*} For\varepsilon>0$,$\phi\in H_{0}^{1}(\mathbf{A}^{r}) $, there is$\varphi\in C_{c}^{\infty}(\mathbf{A}^{r}) such that $\Vert\phi-\varphi\Vert_{H^{1}}<\varepsilon/(\Vert u\Vert_{H^{1}}+1)$ and \begin{align*} \langle u_{n},\phi(z)\rangle _{H^{1}} & =\langle u_{n} ,\phi(z)-\varphi(z)\rangle _{H^{1}}+\langle u_{n},\varphi (z)\rangle _{H^{1}}\\ & \leq\Vert u_{n}\Vert_{H^{1}}\Vert\phi(z)-\varphi(z)\Vert_{H^{1} }+\langle u_{n},\varphi(z)\rangle _{H^{1}}\\ & <\varepsilon\quad \text{as }n\to\infty. \end{align*} Thus,u_{n}\rightharpoonup0$weakly in$H_{0}^{1}(\mathbf{A} ^{r}) $. Suppose that$J$satisfies the (PS)$_{\alpha( \mathbf{A}^{r})}$-condition, then there is a subsequence$\{ u_{n}\} $such that$u_{n}\to0$strongly in$H_{0}^{1}( \mathbf{A}^{r}) $. This contradicts$\alpha(\mathbf{A} ^{r}) >0$. Therefore,$J$does not satisfy the (PS)$_{\alpha( \mathbf{A}^{r}) }$-condition. \end{proof} From now on$\alpha_{X}(\Omega) $is simply denoted by$\alpha_{X}$. Let$\{ u_{n}\} $in$X(\Omega)$be a (PS)$_{\alpha_{X}}$-sequence for$J$in$\Omega$. Clearly,$\{ u_{n}\} $is bounded in$X(\Omega)$. Then a subsequence$\{ u_{n}\} $and$u\in X(\Omega)$exist such that$u_{n}\rightharpoonup u $weakly in$X(\Omega)$a.e. in$\Omega$, and strongly in$L_{\rm loc}^{p}( \Omega) $. Define \begin{gather*} a_{\infty}=\lim_{R\to\infty} \limsup_{n\to\infty} \int_{\Omega\cap\{ R<| z| \} }( | \nabla u_{n}| ^{2}+u_{n}^{2}) ,\\ \\ b_{\infty}=\lim_{R\to\infty} \limsup_{n\to\infty} \int_{\Omega\cap\{ R<| z| \} }| u_{n}| ^{p}. \end{gather*} We have the following results. \begin{lemma} \label{c6} Let$\{ u_{n}\} $be a (PS)$_{\alpha_{X}}$-sequence in$X(\Omega)$for$J$, then a subsequence$\{ u_{n}\} $exists such that\newline$(i)\limsup_{n\to\infty} {\int_{\Omega}} | \nabla u_{n}| ^{2}+u_{n}^{2}=({\int_{\Omega}} | \nabla u| ^{2}+u^{2}) +a_{\infty}$; \newline$(ii)\limsup_{n\to\infty}{\int_{\Omega}}| u_{n}| ^{p}= {\int_{\Omega}}| u| ^{p}+b_{\infty}$; \newline$(iii)a_{\infty}=b_{\infty}$and$\alpha_{X}=J_{\infty}+J(u)$, where$J_{\infty}=(\frac{p-2}{2p}) b_{\infty}$. \end{lemma} \begin{proof} Since$\{ u_{n}\} $is a (PS)$_{\alpha_{X}}$-sequence in$X(\Omega)$for$J$, by Lemma \ref{p30},$\{ u_{n}\} $is bounded in$X(\Omega)$. A subsequence$\{ u_{n}\} $and$u\in X(\Omega)$exist such that$u_{n}\rightharpoonup u$weakly in$X(\Omega)$, a.e. in$\Omega$, and strongly in$L_{\rm loc}^{p}(\Omega) $. \newline$(i)$Let$\eta_{R}(z)=\eta(\frac{2| z| }{R}), then \label{6-2} \begin{aligned} &{\int_{\Omega}}| \nabla u_{n}| ^{2}+u_{n}^{2} ={\int_{\Omega\cap\{ |z|0 $o(1) =\langle J'(u_{n}) ,\eta_{R}u_{n} \rangle =\int_{\Omega}\eta_{R}(| \nabla u_{n}| ^{2}+u_{n} ^{2}) +u_{n}\nabla\eta_{R}\nabla u_{n}-\int_{\Omega}\eta_{R}| u_{n}| ^{p}.$ Thus, $\int_{\Omega}\eta_{R}\big(| \nabla u_{n}| ^{2}+u_{n} ^{2}\big) +u_{n}\nabla\eta_{R}\nabla u_{n}-\int_{\Omega}\eta_{R}| u_{n}| ^{p}= o(1).$ Then, for each $R>0$, \label{6-3} \begin{aligned} &-|{\int_{\Omega}}u_{n}\nabla\eta_{R}\nabla u_{n}| + {\int_{\Omega\cap\{ | z| 0$\limsup_{n\to\infty}\int_{\Omega}\eta_{R}(| \nabla u_{n}| ^{2}+u_{n}^{2}) \leq\frac{c}{R}+\int_{\Omega \cap\{ | z| 0$ $\limsup_{n\to\infty}\int_{\Omega}| u_{n}| ^{p}=\int_{\Omega\cap\{ | z| 0.\label{7-3} Let v_{n}=\xi_{n}u_{n}. Then v_{n}\in X(\Omega) . By (\ref{7-3}), we have $$b(v_{n})=\int_{\Omega}\xi_{n}^{p}|u_{n}|^{p}=\frac{2p}{p-2}\alpha _{X}(\Omega) + o(1).\label{7-4}$$ Since \{\xi_{n}^{2}u_{n}\} is bounded in X(\Omega), we have $$\label{7-5} o(1) =\langle J'(u_{n}),\xi_{n}^{2}u_{n}\rangle =\int_{\Omega}(\xi_{n}^{2}|\nabla u_{n}|^{2} +2\xi_{n}u_{n}\nabla \xi_{n}\cdot\nabla u_{n} +\xi_{n}^{2}u_{n}^{2})-\int_{\Omega}\xi_{n}^{2}|u_{n}|^{p}.$$ By |\nabla\xi_{n}(z)|\leq\frac{c}{n} and (\ref{7-2}), we have \[ \int_{\Omega}\xi_{n}u_{n}\nabla\xi_{n}\cdot\nabla u_{n} = o(1).$ By (\ref{7-4}) and (\ref{7-5}) we have $a(v_{n})=\int_{\Omega}\xi_{n}^{2}(|\nabla u_{n}|^{2}+u_{n} ^{2})=b(v_{n})+ o(1)=\frac{2p}{p-2}\alpha_{X}(\Omega) + o(1).$ Thus \begin{align*} J(v_{n}) & =\frac{1}{2}a(v_{n})-\frac{1}{p}b(v_{n})\\ & =\frac{1}{2}\frac{2p}{p-2}\alpha_{X}(\Omega) -\frac{1}{p} \frac{2p}{p-2}\alpha_{X}(\Omega) + o(1)\\ & =\alpha_{X}(\Omega) + o(1). \end{align*} Since $J(v_{n})=\alpha_{X}(\Omega) + o(1)$ and $a( v_{n}) =b(v_{n})+ o(1)$, by Theorem \ref{i4}, $\{v_{n}\}$ is a (PS)$_{\alpha_{X}(\Omega)}$-sequence in $X(\Omega)$ for $J$.\newline $(iii) \Longrightarrow(i)$ Let $\{ u_{n}\}$ be a (PS)$_{\alpha_{X}(\Omega) }$-sequence in $X(\Omega)$ for $J$ such that $\{ \xi _{n}u_{n}\}$ is also a (PS)$_{\alpha_{X}(\Omega) }$-sequence in $X(\Omega)$ for $J$. Let $v_{n}=\xi_{n}u_{n}$. Claim: $v_{n}\rightharpoonup0$ as $n\to\infty$. For $\phi\in C_{c} ^{1}(\Omega)$ and $K=$ \textrm{supp}$\;\phi$, $K\subset\Omega$ is compact and there is an $n_{0}$ such that $v_{n}(z) =0$ in $K$ for all $n\geq n_{0}$. We have $\langle v_{n}(z) ,\phi(z) \rangle_{H^{1}} ={\int_{\Omega}}\nabla v_{n}(z) \nabla\phi(z)dz+ {\int_{\Omega}}v_{n}(z) \phi(z)dz =0\quad \quad\text{for all }n\geq n_{0}.$ By Lemma \ref{p4}, there is a $C>0$ such that $\Vert v_{n}(z) \Vert_{H^{1}}\leq C$. For $\varepsilon>0$, $\varphi\in H_{0}^{1}(\Omega)$, $\phi\in C_{c}^{1}(\Omega)$ exists such that $\Vert\varphi-\phi\Vert_{H^{1}}<\varepsilon/2(C+1)$ Moreover, \begin{align*} \langle v_{n}(z) ,\varphi(z)\rangle _{H^{1}} &=\langle v_{n}(z) ,\varphi(z)-\phi(z)\rangle _{H^{1} }+\langle v_{n}(z) ,\phi(z)\rangle _{H^{1}}\\ & \leq\Vert v_{n}(z) \Vert_{H^{1}}\Vert\varphi(z)-\phi (z)\Vert_{H^{1}}+\langle v_{n}(z) ,\phi(z)\rangle _{H^{1}}\\ & \leq C\Vert\varphi(z)-\phi(z)\Vert_{H^{1}}\\ & <\varepsilon\text{\ for\ }n\geq n_{0}. \end{align*} This implies $v_{n}\rightharpoonup0$ weakly in $H_{0}^{1}( \Omega)$. Therefore, by Lemma \ref{c2} $J$ does not satisfy the (PS)$_{\alpha_{X}(\Omega) }$-condition in $X(\Omega)$. \end{proof} For $k\geq1$, $i=1,2,\dots,k$, let $\Omega$ be an unbounded domain and let $\Omega_{i}$ be a proper domain in $\Omega$ such that $\Omega =\cup_{i=1}^{k}\Omega_{i}$, $\Omega_{i}\cap\Omega_{j}$ is bounded, and at least one of $\Omega_{i}$ is unbounded. Let $\alpha_{X}=\alpha_{X}(\Omega)$ and $\alpha_{X}^{i}=\alpha_{X}(\Omega_{i})$, then \begin{gather*} \mathbf{M}=\{ u\in X(\Omega)\backslash\{0\}\ |\ a(u)=b(u)\} ,\\ \mathbf{M}_{i}=\{ u\in H_{0}^{1}(\Omega_{i})\backslash \{0\}\ |\ a(u)=b(u)\} \quad \text{for }i=1,2,\dots,k. \end{gather*} Since $X(\Omega_{i})\subset X(\Omega)$ and $\mathbf{M}_{i}\subset\mathbf{M}$, for $i=1,2,\dots,k$, $\alpha_{X}\leq\min\{\alpha_{X}^{1},\alpha _{X}^{2},\dots,\alpha_{X}^{k}\}$. Let $\Omega$ be an unbounded domain in $\mathbb{R}^{N}$ and $\Omega ^{0}\subsetneqq\Omega$ with the indexes $\alpha_{X}=\alpha_{X}(\Omega)$ and $\alpha_{X}^{0}=\alpha_{X}(\Omega^{0})$. Then we have $\alpha_{X}\leq \alpha_{X}^{0}$. Let \begin{gather*} \widetilde{\Omega}_{n}=\Omega\;\backslash\;\overline{B^{N}(0;n)};\\ \widetilde{\mathbf{M}}_{n}=\{ u\in H_{0}^{1}(\widetilde{\Omega} _{n})\backslash\{ 0\} \mid a(u) =b(u) \} ;\\ \widetilde{\alpha}_{X}^{n}=\alpha(\widetilde{\Omega}_{n})=\underset {u\in\widetilde{M}_{n}}{\inf}J(u) . \end{gather*} \begin{theorem} \label{c7} The following properties are equivalent:\newline $(i)$ $J$ satisfies the (PS)$_{\alpha_{X}}$-condition;\newline $(ii)$ For every (PS)$_{\alpha_{X} }$-sequence $\{u_{n}\}$ in $X(\Omega)$ for $J$, there are a subsequence $\{u_{n}\}$ and $u\neq0$ in $X(\Omega)$ such that $u_{n}\to u$ strongly in $X(\Omega)$; \newline $(iii)$ For every (PS)$_{\alpha_{X}}$-sequence $\{u_{n}\}$ in $X(\Omega)$ for $J$, there are $c>0$, a subsequence $\{u_{n} \}$, and positive integers $K$ and $n_{0}$ such that for each $n\geq n_{0}$, we have $\int_{\Omega\cap\{ |z|0, there is a measurable set E such that |E|<\infty and \int_{E^{c}} |u_{n}|^{p}dz<\varepsilon for each n\in\mathbb{N};\newline (v) \alpha _{X}<\tilde{\alpha}_{X}^{n} for each n\in\mathbb{N};\newline (vi) \alpha_{X}<\min\{\alpha_{X}^{1},\alpha_{X}^{2},\dots,\alpha_{X}^{k} \}; \newline (vii) J_{\infty}<\alpha_{X}; \newline (viii) \alpha_{X}<\alpha_{X}^{0} for each proper subdomain \Omega^{0}  of \Omega. \end{theorem} \begin{proof} (i)\Longrightarrow(ii) Suppose that J satisfies the (PS)_{\alpha_{X}}- condition. Let \{u_{n}\} be a (PS)_{\alpha_{X}}-sequence in X(\Omega)  for J. Then there are a subsequence \{u_{n}\} and a u in X(\Omega) such that u_{n}\to u strongly in X(\Omega). We conclude that J(u)=\alpha_{X}>0. Thus, u\neq0.\newline (ii)\Longrightarrow(iii) Suppose that \{u_{n}\} is a (PS)_{\alpha_{X}}-sequence in X(\Omega) for J that has a subsequence \{u_{n}\} and u\neq0 in X(\Omega) such that u_{n}\rightharpoonup u weakly in X(\Omega). By Theorem \ref{c2}, \lim_{n\to\infty}{\int_{\Omega}}|u_{n}|^{p}={\int_{\Omega}}|u|^{p}. Take K>0 and c>0 with {\int_{\Omega\cap\{ |z|0 exists such that \[ {\int_{\Omega\cap\{ |z|0, a subsequence \{u_{n}\}, positive integers K and n_{0} such that for each n\geq n_{0}, we have \[ \int_{\Omega\cap\{ |z|0, there is a set E such that |E|<\infty and \int_{E^{c}}|u_{n}|^{p}dz<\varepsilon for each n\in\mathbb{N}.\newline (iv)\Longrightarrow (v) For every (PS)_{\alpha_{X}}-sequence \{u_{n}\} in X(\Omega) for J, there is a subsequence \{u_{n}\} such that for \varepsilon>0, there is a set E such that |E|<\infty and \int_{E^{c}}|u_{n}|^{p} dz<\varepsilon for each n\in\mathbb{N}. Then \{u_{n}\} is bounded and there is a subsequence \{u_{n}\} and a u in X(\Omega) such that u_{n}\to u a.e. in \Omega. By Theorem \ref{p27}, u_{n}\to u in L^{p}(\Omega) . Note that \[ \alpha_{X}+ o(1)=J(u_{n})=(\frac{1}{2}-\frac{1}{p}) b(u_{n})+ o(1)=(\frac{1}{2}-\frac{1}{p}) b(u)+\text{o} (1).$ Thus, $u\neq0$. By Theorem \ref{c2}, $J$ satisfies the (PS)$_{\alpha_{X}}$-condition in $\Omega$. Suppose that $\widetilde{\alpha}_{X}^{n_{0}} =\alpha_{X}$ for some $n_{0}\in\mathbb{N}$, by Theorem \ref{c3} $J$ does not satisfy the (PS)$_{\alpha_{X}}$-condition in $\Omega$, which is a contradiction. Hence, we have $\alpha_{X}<\tilde{\alpha}_{X}^{n}$ for each $n$.\newline $(v)\Longrightarrow(vi)$ On the contrary, suppose that $\alpha _{X}=\min\{\alpha_{X}^{1},\alpha_{X}^{2},\dots,\alpha_{X}^{k}\}$, say $\alpha_{X}=\alpha_{X}^{1}$. Since $\Omega_{1}\subsetneqq\Omega$, by Theorem \ref{c3}, $J$ does not satisfy the (PS)$_{\alpha_{X}}$-condition in $\Omega$. By Theorem \ref{c2}, there is a (PS)$_{\alpha_{X}}$-sequence $\{u_{n}{\}}$ such that $u_{n}\rightharpoonup0$ weakly in $X(\Omega)$. There is a subsequence $\{u_{n}\}$ and a sequence $\{\Omega_{n}\}$ such that $\int_{\Omega_{n}}| u_{n}| ^{p}= o(1).$ Let $\xi_{n}$ be as in (\ref{1-1}) and $v_{n}=\xi_{n}u_{n}$. By Lemma \ref{p33}, we have \begin{gather*} J(v_{n})=\alpha_{X}+ o(1).\\ J'(v_{n})= o(1)\quad \text{strongly in }X^{-1}(\Omega). \end{gather*} Then by Theorem \ref{i3}, there is a sequence $\{s_{n}\}$ in $\mathbb{R}^{+}$ such that $w_{n}=s_{n}v_{n}$, $\{w_{n}\}\in\widetilde{\mathbf{M}}_{n}$, and $J(w_{n})=J(v_{n})+ o(1)=\alpha_{X}+ o(1)$. Note that $\widetilde{{\alpha}}_{X}^{n}\leq J(w_{n})$ for each $n\in\mathbb{N}$. Hence, $\lim_{n\to\infty}\widetilde{{\alpha}}_{X}^{n}\leq\alpha_{X}$. Since $\Omega\supset\widetilde{\Omega}_{n}\supset\widetilde{\Omega}_{n+1}$, we have $\alpha_{X}\leq\widetilde{{\alpha}}_{X}^{n}\leq\widetilde{{\alpha}} _{X}^{n+1}$ for each $n\in\mathbb{N}$. Then we can conclude that $\alpha _{X}=\widetilde{{\alpha}}_{X}^{n}$ for each $n\in\mathbb{N}$, which is a contradiction.\newline $(vi)\Longrightarrow(vii)$ Let $\{ u_{n}\}$ be a (PS)$_{\alpha_{X}}$-sequence in $X(\Omega)$. Then a subsequence $\{u_{n}\}$ and a $u$ in $X(\Omega)$ exist such that $u_{n}\rightharpoonup u$ weakly in $X(\Omega)$. By Lemma \ref{c6}, $\alpha_{X}=J_{\infty}+J(u)$. On the contrary, suppose that $J_{\infty}=\alpha_{X}$, and we have $u=0$. Thus, $u_{n}\rightharpoonup0$ weakly in $X(\Omega)$. There are a subsequence $\{u_{n}\}$ and a sequence $\{\Omega_{n}\}$ such that $\int_{\Omega_{n}}| u_{n}| ^{p}= o(1).$ Let $v_{n}=\xi_{n}u_{n}$. By Lemma \ref{p33}, $\{v_{n}\}$ is a (PS)$_{\alpha _{X}}$-sequence in $X(\Omega)$. Since $\Omega_{i}\cap\Omega_{j}$ is bounded for $i\neq j$, $n_{0}>0$, $v_{n}=0$ in $B^{N}(0;n_{0})$ exists for $n\geq n_{0}$, where $B^{N}(0;n_{0})\supset\Omega_{i}\cap\Omega_{j}$ for $i\neq j$. Set $v_{n}=v_{n}^{1}+v_{n}^{2}+\dots+v_{n}^{k}$, where $v_{n}^{i}\in H_{0}^{1}(\Omega_{i})$, and for $i=1,2,\dots,k$, $v_{n}^{i}(z)=\begin{cases} v_{n}(z) & \text{for }z\in\Omega_{i};\\ 0 & \text{otherwise.} \end{cases}$ As in the proof of Lemma \ref{p33}, we obtain $J'(v_{n}^{i})= o(1)\;\text{strongly in }X^{-1}(\Omega)\quad \text{for }i=1,2,\dots,k.$ Assume $J(v_{n}^{i})=c_{i}+ o(1)\quad \mbox{for }i=1,2,\dots,k.$ Since $J(u_{n})=\alpha_{X}+ o(1)$, we have $c_{1}+c_{2}+\dots +c_{k}=\alpha_{X}$. Since $c_{i}$ are (PS)-values in $X(\Omega)$ for $J$, by Lemma \ref{p30}, they are nonnegative. There is at least one of the $c_{i}$ that is positive, say $c_{1}>0$. By Theorem \ref{i11}, $c_{1}\geq\alpha _{X}^{1}$, thus $\alpha_{X}\geq c_{1}\geq\alpha_{X}^{1}$. This proves $\alpha_{X}\geq\min\{\alpha_{X}^{1},\alpha_{X}^{2},\dots,\alpha_{X}^{k}\}$. We conclude that $\alpha_{X}=\min\{\alpha_{X}^{1},\alpha_{X}^{2},\dots ,\alpha_{X}^{k}\}$.\newline $(vii)\Longrightarrow(viii)$ Let $\{ u_{n}\}$ be a (PS)$_{\alpha}$-sequence in $\Omega$. Then a subsequence $\{u_{n}\}$ and a $u$ in $X(\Omega)$ exist such that $u_{n}\rightharpoonup u$ weakly in $X(\Omega)$. By Lemma \ref{c6}, $J(u)=\alpha_{X}-J_{\infty}$. Suppose that $J_{\infty}<\alpha_{X}$, then $u\neq0$. By Theorem \ref{c2}, $J$ satisfies (PS)$_{\alpha_{X}}$-condition in $X(\Omega)$. By Theorem \ref{c3}, $\alpha_{X}<\alpha_{X}^{0}$ for each proper subdomain $\Omega^{0}$ of $\Omega$.\newline $(viii) \Longrightarrow(i)$ Suppose that $J$ does not satisfy the (PS)$_{\alpha_{X}}$-condition in $\Omega$. By Lemma \ref{p33} and Theorem\ref{c3}, $\{ u_{n}\}$ exists in $X(\Omega)$ and is a (PS)$_{\alpha_{X}}$-sequence for $J$ such that $\{ \xi_{n}u_{n}\}$ also is a (PS)$_{\alpha_{X}}$-sequence in $X(\Omega)$for $J$. Let $v_{n}=\xi_{n}u_{n}$, then \begin{gather*} J(v_{n}) = \alpha+ o(1),\\ J'(v_{n}) = o(1)\quad \text{strongly in }X^{-1}(\Omega). \end{gather*} By Theorem \ref{i3}, there is a sequence $\{ s_{n}\}$ in $\mathbb{R}^{+}$ such that $w_{n}=s_{n}v_{n}$, $\{ w_{n}\} \in\mathbf{M}(\Omega\backslash \overline{B^{N}(0;\frac{n} {2}) })$ and $J(w_{n})=J(v_{n})+ o(1)=\alpha _{X}+ o(1)$. $n_{0}>0$ exists such that $\Omega\ \backslash \ \overline{B^{N}(0;n_{0}) }\subsetneqq\Omega$. Let $\Omega ^{0}=\Omega\backslash \overline{B^{N}(0;n_{0}) }$. Then $w_{n}\in\mathbf{M}(\Omega^{0})$ for $n\geq n_{0}$. Since $\mathbf{M}(\Omega^{0}) \subset\mathbf{M}(\Omega)$ and $J(w_{n})=\alpha_{X}+ o(1)$, thus, $\alpha_{X}^{0}=\alpha_{X}$, which is a contradiction. \end{proof} \noindent\textbf{Bibliographical notes:} Theorem \ref{c2} is from Chen-Lee-Wang \cite[Lemma 19]{CLW}. Theorem \ref{c7} is from Chen-Lin-Wang \cite[Theorem 23]{CLWW}. \section{Symmetric Palais-Smale Conditions} In this section, we focus on the symmetric Palais-Smale conditions which will be used in Section 13. \begin{definition}\rm \label{w1} $(i)$ Suppose that $(x,y)\in\Omega$ if and only if $(x,-y)\in \Omega$, then we call $\Omega$ a $y$-symmetric domain$\,$;\newline$( ii)$ Let $\Omega$ be a $y$-symmetric domain and $\Theta$ be a $y$-symmetric bounded domain in $\mathbb{R}^{N}$. If two disjoint subdomains $\Omega_{1}$ and $\Omega_{2}$ of $\Omega$ exist such that \begin{gather*} (x,y)\in\Omega_{2}\quad \quad\text{if and only if }\quad (x,-y)\in\Omega_{1},\\ \Omega\backslash \bar{\Theta}=\Omega_{1}\cup\Omega_{2}, \end{gather*} then we say that $\Omega$ is separated by $\Theta$;\newline$(iii)$ Let $\Omega$ be a $y$-symmetric domain in $\mathbb{R}^{N}$. If a function $u:$ $\Omega\to\mathbb{R}$ satisfies $u(x,y)=u(x,-y)$ for $(x,y)\in\Omega$, then we call $u$ a $y$-symmetric (axially symmetric) function;\newline$(iv)$ Let $\Omega$ be a $y$-symmetric domain in $\mathbb{R}^{N}$ and denote the space $H_{s}^{1}(\Omega)$ by the $H^{1}$-closure of the space $\{u\in C_{0}^{\infty}(\Omega): \mbox{$u$is$y$-symmetric}\}$. \end{definition} \begin{remark} \rm \label{w2} $(i)$ Note that $H_{s}^{1}(\Omega)$ is a closed linear subspace of $H_{0}^{1}(\Omega)$. Let $H_{s}^{-1}(\Omega)$ be the dual space of $H_{s}^{1}(\Omega)$;\newline$(ii)$ Let $\Omega$ be a $y$-symmetric domain in $\mathbf{A}^{r}$ and let $B^{N}(0;r+1)$ be a $N-$ball. Then clearly $\Omega$ is separated by $B^{N}(0;r+1)$. \end{remark} \begin{example} \rm \label{w3} $(i)$ For each $\rho>0$, let $\Omega=(\mathbb{R} ^{N}\backslash\overline{K_{\rho}}) \cup\mathbf{A}^{r}$. Then $\Omega$ is a $y$-symmetric large domain in $\mathbb{R}^{N}$ separated by a bounded domain $\mathbf{A}_{\rho}^{r}$;\newline$(ii)$ Let $\Omega=[ \mathbf{P}^{+}\mathbf{+}(0,\frac{R}{2})] \cup B^{N}(0;R)\cup[ \mathbf{P}^{-}\mathbf{-}(0,\frac{R}{2}) ]$, then $\Omega$ is a $y$-symmetric large domain in $\mathbb{R}^{N}$ separated by the bounded domain $B^{N}(0;R)$. \end{example} \begin{theorem} \label{w4} $(i)$ $\alpha_{s}(B^{N}(0;R))=\alpha(B^{N}(0;R))$; \newline $(ii)$ $\alpha_{s}(\mathbb{R}^{N})=\alpha(\mathbb{R}^{N})$; \newline $(iii)$ $\alpha_{s}(\mathbf{A}_{-t,t}^{r})=\alpha(\mathbf{A}_{-t,t}^{r})$; \newline $(iv)$ $\alpha_{s}(\mathbf{A}^{r})=\alpha(\mathbf{A}^{r})$. \end{theorem} \begin{proof} By Lien-Tzeng-Wang \cite{LTW} and Theorem \ref{b3} below, there is a ground state solution of \eqref{E1} in $B^{N}(0;R)$, $\mathbb{R}^{N}$, $\mathbf{A}_{-t,t}^{r}$, and $\mathbf{A}^{r}$. By Gidas-Ni-Nirenberg \cite{GNN1} and \cite{GNN2} and Chen-Chen-Wang \cite{CCW}, every positive solution of \eqref{E1} in $B^{N}(0;R)$, $\mathbb{R}^{N}$, $\mathbf{A}_{-t,t}^{r}$, and $\mathbf{A}^{r}$ is $y$-symmetric. \end{proof} The following symmetric results are required to assert our main result. \begin{theorem} \label{w5} Suppose that $\Omega$ is a $y$-symmetric large domain in $\mathbb{R}^{N}$ separated by a $y$-symmetric bounded domain, then $\alpha _{s}(\Omega) \leq2\alpha(\Omega)$. \end{theorem} \begin{proof} First, by Lien-Tzeng-Wang \cite{LTW} and Gidas-Ni-Nirenberg \cite{GNN2}, there is a positive solution $u_{0}$ of Equation \eqref{E1} with radial symmetry such that $J(u_{0}) =\alpha( \mathbb{R}^{N})$. Since $\Omega$ is a $y$-symmetric proper large domain in $\mathbb{R}^{N}$, for $n=1,2,\dots$, sequences $\{ z_{n}\}$ and $\{ r_{n}\}$ exist such that $B^{N}( z_{n};r_{n}) \subset\Omega$ and $r_{n}\to\infty$ as $n\to\infty$. Let $\eta_{n}(z) =\eta( \frac{2| z-z_{n}| }{r_{n}})$ as in (\ref{1-2}), and $u_{n}(z) =\eta_{n}(z) u_{0}(z-z_{n})$. Then $u_{n}(z) \in H_{0}^{1}(\Omega)$, and \begin{gather*} J(u_{n}) =J(u_{0}) + o(1) =\alpha(\mathbb{R}^{N}) + o(1) \\ a(u_{n}) =b(u_{n}) + o(1) . \end{gather*} By Theorem \ref{i4} and Theorem \ref{f8}, $\{ u_{n}\}$ is a (PS)$_{\alpha(\Omega) }$-sequence in $H_{0}^{1}(\Omega)$ for $J$. Moreover, if we let $w_{n}=u_{n}(x,-y)$, then $w_{n}$ is also a (PS)$_{\alpha(\Omega) }$-sequence in $H_{0}^{1}(\Omega)$ for $J$ such that $\text{supp}w_{n}\cap\text{supp}u_{n}=\emptyset$ and $\{ u_{n}+w_{n}\} \subset H_{s}^{1}(\Omega)$. We have \begin{align*} a(u_{n}+w_{n}) & = {\int_{\Omega}} | \nabla(u_{n}+w_{n}) | ^{2}+(u_{n} +w_{n}) ^{2}\\ & ={\int_{\Omega}}| \nabla u_{n}| ^{2}+u_{n}^{2}+ {\int_{\Omega}}| \nabla w_{n}| ^{2}+w_{n}^{2}\\ &\quad +2{\int_{\Omega}}\nabla u_{n}\nabla w_{n}+2 {\int_{\Omega}}u_{n}w_{n}\\ & =a(u_{n})+a(w_{n}), \end{align*} and $b(u_{n}+w_{n})= {\int_{\Omega}} | u_{n}+w_{n}| ^{p}= {\int_{\Omega}} | u_{n}| ^{p}+ {\int_{\Omega}} | w_{n}| ^{p}=b(u_{n})+b(w_{n})$. Hence, \begin{align*} J(u_{n}+w_{n}) & =\frac{1}{2}a(u_{n}+w_{n})-\frac{1}{p}b(u_{n}+w_{n})\\ & =J(u_{n})+J(w_{n})\\ & =2\alpha(\Omega) + o(1). \end{align*} Moreover, for $\varphi\in C_{c}^{\infty}(\Omega)$ with $y$-symmetry, we have \begin{align*} &| \langle J'(u_{n}+w_{n}) ,\varphi\rangle | \\ & =big|{\int_{\Omega}}\nabla(u_{n}+w_{n}) \nabla\varphi+(u_{n}+w_{n}) \varphi -{\int_{\Omega}}| u_{n}+w_{n}| ^{p-2}(u_{n}+w_{n})\varphi\big|\\ & =\big|{\int_{\Omega}}\nabla u_{n}\nabla\varphi+u_{n}\varphi+ {\int_{\Omega}}\nabla w_{n}\nabla\varphi+w_{n}\varphi -{\int_{\Omega}}| u_{n}| ^{p-2}u_{n}\varphi-{\int_{\Omega}} | w_{n}| ^{p-2}w_{n}\varphi\big| \\ & =| \langle J'(u_{n}),\varphi\rangle | +| \langle J'(w_{n}),\varphi\rangle | \\ & \leq\| J'(u_{n})\| _{H^{-1}}+\| J'(w_{n})\| _{H^{-1}}. \end{align*} Therefore, $\| J'(u_{n}+w_{n}) \|_{H_{s}^{-1}}= o(1)$. We conclude that $\{ u_{n}+w_{n}\}$ is a (PS)$_{2\alpha(\Omega) }$-sequence in $H_{s}^{1}(\Omega)$ for $J$. By Theorems \ref{i13} and \ref{i14}, $\alpha_{s}(\Omega) \leq2\alpha(\Omega)$. \end{proof} We have the following symmetric Palais-Smale condition. \begin{theorem} \label{w6} Suppose that $\Omega$ is a $y$-symmetric large domain in $\mathbb{R}^{N}$ separated by a $y$-symmetric bounded domain. Then $\alpha _{s}(\Omega) <2\alpha(\Omega)$ if and only if $J$ satisfies the (PS)$_{\alpha_{s}(\Omega)}$-condition in $H_{s}^{1}(\Omega)$. \end{theorem} \begin{proof} Let $\alpha_{s}(\Omega) <2\alpha(\Omega)$. Suppose $J$ does not satisfy the (PS)$_{\alpha_{s}(\Omega)}$-condition. By Theorem \ref{c6-1}, a (PS)$_{\alpha_{s}(\Omega) }$-sequence $\{ u_{n}\}$ in $H_{s}^{1}(\Omega)$ for $J$ exists such that $\{ \xi_{n}u_{n}\}$ is also a (PS)$_{\alpha _{s}(\Omega) }$-sequence in $H_{s}^{1}(\Omega)$ for $J$, where $\xi_{n}$ is as at (\ref{1-1}). Let $w_{n}=\xi_{n}u_{n}$, then by Lemma \ref{p2}, we obtain $$\begin{gathered} J(w_{n})=\alpha_{s}(\Omega)+ o(1),\\ J'(w_{n})= o(1)\quad\mbox{in }H^{-1}(\Omega) . \end{gathered}\label{7}$$ Since $\Omega$ is a $y$-symmetric domain in $\mathbb{R}^{N}$ separated by a bounded domain $Q$, $n_{0}>0$, exists such that $w_{n}=0$ in $\overline{Q}$ for $n\geq n_{0}$ and two disjoint subdomains $\Omega_{1}$ and $\Omega_{2}$ of $\Omega$ exist such that \begin{gather*} (x,y)\in\Omega_{2}\quad\text{if and only if }(x,-y)\in\Omega_{1},\\ \Omega\backslash \overline{Q}=\Omega_{1}\cup\Omega_{2}. \end{gather*} Note that, for $n\geq n_{0}$, $w_{n}=w_{n}^{1}+w_{n}^{2}$ and $w_{n}^{1}(x,y) =w_{n}^{2}(x,-y)$, where for $i=1,2$, $w_{n}^{i}(x)=\begin{cases} w_{n}(x) & \text{for }x\in\Omega_{i},\\ 0 & \text{for }x\notin\Omega_{i}. \end{cases}$ Then $w_{n}^{i}\in H_{0}^{1}(\Omega_{i})$. We obtain $J(w_{n}^{1}) =J(w_{n}^{2})$ and $\alpha_{s}(\Omega) + o(1) =J(w_{n}) =J(w_{n}^{1}) +J(w_{n}^{2}) =2J(w_{n} ^{i}) \quad\text{for }i=1,2,$ or $J(w_{n}^{i}) =\frac{1}{2}\alpha_{s}(\Omega) + o(1) \quad\text{for }i=1,2.$ By (\ref{7}), we have $J'(w_{n}^{i})= o(1)\quad\mbox{in }H_{0}^{1}(\Omega_{i})\quad\text{for }i=1,2.$ Therefore $\frac{1}{2}\alpha_{s}(\Omega)$ is a (PS)-value in $H_{0}^{1}(\Omega)$ for $J$. By Theorems \ref{i13} and \ref{i14}, $\frac{1}{2}\alpha_{s}(\Omega) \geq\alpha(\Omega _{i}) .$ Since $\Omega$ and $\Omega_{i}$ are large domains of $\mathbb{R}^{N}$, by Theorem \ref{f8}, we have $\alpha(\Omega_{i}) =\alpha(\mathbb{R}^{N}) =\alpha(\Omega) .$ Thus $\alpha_{s}(\Omega) \geq2\alpha(\Omega)$, which is a contradiction.\newline Conversely, suppose that $J$ satisfies the (PS)$_{\alpha_{s}(\Omega)}$-condition in $H_{s}^{1}(\Omega)$. By Theorem \ref{w5}, we have $\alpha_{s}(\Omega) \leq2\alpha( \Omega)$. Suppose that $\alpha_{s}(\Omega) =2\alpha(\Omega)$. By the definition of the large domain in $\mathbb{R}^{N}$, we may take a domain $\tilde{\Omega}=\Omega\;\backslash \;\overline{B^{N}(0;\tilde{r}) }$ for some $\tilde{r}>0$ such that $\tilde{\Omega}\subsetneqq\Omega$, and $\tilde{\Omega}$ is a proper $y$-symmetric large domain in $\mathbb{R}^{N}$ separated by a $y$-symmetric bounded domain. By Theorem \ref{c3}, we have $2\alpha( \mathbb{R}^{N}) =2\alpha(\Omega) =\alpha_{s}( \Omega) <\alpha_{s}(\tilde{\Omega})$. By Theorem \ref{w5}, $\alpha_{s}(\tilde{\Omega}) \leq2\alpha( \tilde{\Omega}) =2\alpha(\mathbb{R}^{N})$. Thus, $2\alpha(\mathbb{R}^{N}) <2\alpha(\mathbb{R}^{N})$, which is a contradiction. \end{proof} As a consequence of Theorem \ref{w6}, we have the following result. \begin{theorem} \label{w7} If $\Omega$ is a $y$-symmetric large domain in $\mathbb{R}^{N}$ separated by a $y$-symmetric bounded domain, then $\alpha(\Omega) <\alpha_{s}(\Omega)$. \end{theorem} \begin{proof} Since $\Omega\subsetneqq\mathbb{R}^{N}$, we have $\alpha_{s}( \mathbb{R}^{N}) \leq\alpha_{s}(\Omega)$. Assume that $\alpha_{s}(\mathbb{R}^{N}) =\alpha_{s}(\Omega)$. Then by Theorem \ref{c3}, $J$ does not satisfy the (PS)$_{\alpha_{s}(\Omega )}$-condition in $H_{s}^{1}(\Omega)$ for $J$. By Theorem \ref{f8}, $\alpha(\mathbb{R}^{N}) =\alpha(\Omega)$, by Theorem \ref{w4}, $\alpha_{s}(\mathbb{R}^{N}) =\alpha(\mathbb{R}^{N})$, and by Theorem \ref{w6}, $2\alpha(\Omega) \leq\alpha_{s}(\Omega)$. We conclude that $2\alpha(\mathbb{R}^{N}) =2\alpha(\Omega) \leq\alpha_{s}(\Omega) =\alpha_{s}(\mathbb{R}^{N}) =\alpha(\mathbb{R}^{N}) ,$ which is a contradiction. \end{proof} Consider the $y$-symmetric large domain $\Omega_{R}$ in $\mathbb{R}^{N}$ separated by a $y$-symmetric bounded domain, where $\Omega_{R}=\left[ \mathbf{P}^{+}\mathbf{+}(0,\frac{R}{2}) \right] \cup B^{N}(0;R)\cup\left[ \mathbf{P}^{-}\mathbf{-}(0,\frac {R}{2}) \right]$. Then we have the following existence result. \begin{theorem} \label{w8} An $R_{0}>0$ exists such that for $R\geq R_{0}$, there is a positive $y$-symmetric solution of \eqref{E1} in $\Omega_{R}$. \end{theorem} \begin{proof} By Lien-Tzeng-Wang \cite{LTW}, $\alpha(B^{N}(0,R) )$ is strictly decreasing as $R$ is strictly increasing and $\alpha(B^{N}(0,R) ) \searrow\alpha( \mathbb{R}^{N}) \quad\text{as }R\to\infty.$ By Theorem \ref{w4}, $\alpha(B^{N}(0,R) ) =\alpha_{s}(B^{N}(0,R) )$ for each $R$. Thus, there is a $R_{0}>0$ such that $\alpha_{s}(\Omega_{R}) \leq\alpha_{s}(B^{N}(0,R) ) <2\alpha( \mathbb{R}^{N}) =2\alpha(\Omega_{R})$ for each $R\geq R_{0}$. By Theorem \ref{w5} and Theorem \ref{w6}, there is a $y$-symmetric positive solution of Equation \eqref{E1} in $\Omega_{R}$ for each $R\geq R_{0}$. \end{proof} \noindent\textbf{Bibliographical notes:} The results of this section are from Wang-Wu \cite{WW1}. \section{Symmetric Palais-Smale Decomposition Theorems} In this section, we present the symmetric Palais-Smale decomposition theorem in $\mathbf{A}^{r}$. \begin{lemma} \label{u1} Let $\Theta_{1}\subset\Theta_{2}\subset\Theta_{3}\subset\dots$, where $\overset{\infty}{\underset{n=1}{\cup}}\Theta_{n}=\mathbf{A}^{r}$. If $f_{n}(z) =\begin{cases} g_{n}(z) -h_{n}(z) , & \text{for }z\in\Theta_{n},\\ 0, & \text{otherwise,} \end{cases}$ $f_{n}\to f$ a.e., $g_{n}\geq0$, and $h_{n}\to0$ a.e., then $f\geq0$. \end{lemma} \begin{proof} For $z\in\mathbf{A}^{r}$, we have $z\in\Theta_{m}$ for some $m\in\mathbb{N}$, then $z\in\Theta_{m+i}$ for $i=0,1,2,\dots$. Since $g_{m+i}(z)=f_{m+i}(z) +h_{m+i}(z)$, $f_{m+i}\to f$ a.e., $h_{m+i}\to 0$ a.e. for $i\to\infty$, hence $g_{m+i}\to f$ a.e., and since $g_{m+i}\geq0$, we have $f\geq0$. \end{proof} \begin{theorem}[Symmetric Palais-Smale Decomposition Theorem in $\mathbf{A}^{r}$] \label{u2} Let $\{ u_{n}\}$ be a (PS)$_{\beta}$-sequence in $H_{s}^{1}(\mathbf{A}^{r})$ for $J$. Then there are a subsequence $\{ u_{n}\}$, a positive integer $m$, sequences $\{ \tilde{z}_{n}^{i,j}\} _{n=1}^{\infty}$ in $\mathbf{A}^{r}$,\ a function $\bar{u}\in H_{s}^{1}(\mathbf{A}^{r})$, and $0\neq w^{i,j}\in H_{0}^{1}(\mathbf{A}^{r})$ for $1\leq i\leq m$, $j=1,2$ such that \begin{gather*} w^{i,1}(x,y) =w^{i,2}(x,-y) ,\\ |\tilde{z}_{n}^{i,j}|\to\infty\quad \text{for }i=1,2,\dots,m,\\ -\Delta\bar{u}+\bar{u}=\mid\bar{u}\mid^{p-2}\bar{u}\quad\mbox{in }\mathbf{A} ^{r},\\ -\Delta w^{i,j}+w^{i,j}=\mid w^{i,j}|^{p-2}w^{i,j}\quad\mbox{in }\mathbf{A}^{r}, \end{gather*} and \begin{gather*} u_{n}=\bar{u}+\underset{j=1}{\overset{2}{\sum}}\underset{i=1}{\overset{m} {\sum}}w^{i,j}(\cdot-\tilde{z}_{n}^{i,j}) + o(1)\;\text{strongly}\quad\text{}\text{in}\quad\text{}H_{0}^{1}( \mathbf{A}^{r}) ,\\ a(u_{n})=a(\bar{u})+2\sum_{i=1}^m a(w^{i,j})+ o(1)\quad \text{for some }j=1,2,\\ b(u_{n})=b(\bar{u})+2\sum_{i=1}^m b(w^{i,j})+ o(1)\quad\text{for some }j=1,2,\\ J(u_{n})=J(\bar{u})+2\sum_{i=1}^m J(w^{i,j})+ o(1)\quad \text{for some }j=1,2. \end{gather*} In addition, if $u_{n}\geq0$, then $\bar{u}\geq0$ and $w^{i,j}\geq0$ for each $1\leq i\leq m$, $j=1,2$. \end{theorem} \begin{proof} \textbf{Step 0}. Since $\{ u_{n}\}$ is a (PS)$_{\beta}$-sequence in $H_{s}^{1}(\mathbf{A}^{r})$ for $J$, by Lemma \ref{p30} there is a $c>0$ such that $\| u_{n}\| _{H^{1}}\leq c$. In the following proof of this theorem, we fix the value of $c$. There is a subsequence $\{ u_{n}\}$ and a $\bar{u}$ in $H_{s} ^{1}(\mathbf{A}^{r})$ such that $u_{n}\rightharpoonup\bar{u}$ weakly in $H_{s}^{1}(\mathbf{A}^{r})$ and $\bar{u}$ solves $-\Delta\bar{u}+\bar{u}=|\bar{u}|^{p-2}\bar{u}\quad\mbox{in }\mathbf{A}^{r}.$ \textbf{Step 1}. Suppose that $u_{n}\nrightarrow\bar{u}$ strongly in $H_{s}^{1}(\mathbf{A}^{r})$. Let $u_{n}^{1}=u_{n}-\bar{u}\quad \text{for }n=1,2,\dots.$ By Lemma \ref{p8}, $\{ u_{n}^{1}\}$ is a (PS)$_{( \beta-J(\bar{u}) ) }$-sequence in $H_{s}^{1} (\mathbf{A}^{r})$ for $J$. Let $v_{n}^{1}=\xi_{n}u_{n}^{1}$, where $\xi_{n}$ are as in \ref{1-1}. Note that $u_{n}^{1}\rightharpoonup0$ weakly in $H_{s}^{1}(\mathbf{A}^{r})$ and $u_{n}^{1}\nrightarrow0$ strongly in $H_{s}^{1}(\mathbf{A}^{r})$. By Lemma \ref{p33}, $\{ v_{n}^{1}\}$ is also a (PS)$_{(\beta-J(\bar{u}) ) }$-sequence in $H_{s}^{1}(\mathbf{A}^{r})$ for $J$. Moreover, $J(u_{n}^{1}) =J(v_{n}^{1}) + o(1),\;v_{n}^{1}\rightharpoonup0$ weakly in $H_{s}^{1}(\mathbf{A}^{r})$ and $v_{n}^{1}\nrightarrow0$ strongly in $H_{s}^{1}(\mathbf{A}^{r})$. Let $K=2(\left[ r\right] +1)$ and $Q_{K}=\mathbf{A}^{r}\cap B^{N}(0;K)$. Then $v_{n}^{1}=0$ in $\overline{Q_{K}}$ for $n\geq2K$. Two disjoint strictly large domains $\Omega^{1}=\mathbf{A}_{K}^{r}$ and $\Omega^{2}=\widetilde{\mathbf{A}} _{-K}^{r}$ in $\mathbf{A}^{r}$ exist such that \begin{gather*} (x,y)\in\Omega^{2}\quad \text{if and only if}\quad (x,-y)\in\Omega^{1},\\ \Omega\backslash \overline{Q_{K}}=\Omega^{1}\cup\Omega^{2}. \end{gather*} For $j=1,2$, let $v_{n}^{1,j}(z)=\begin{cases} v_{n}^{1}(z) & \text{for }z\in\Omega^{j},\\ 0 & \text{for }z\notin\Omega^{j}. \end{cases}$ Then $v_{n}^{1,j}\in H_{0}^{1}(\Omega^{j})$, $v_{n}^{1,1}(x,y) =v_{n}^{1,2}(x,-y)$, $v_{n}^{1}=v_{n}^{1,1}+v_{n}^{1,2}$, and $J(v_{n}^{1,1}) =J(v_{n}^{1,2})$. We claim that $\{v_{n}^{1,j}\}$ is a (PS)$_{\frac{1}{2}(\beta-J(\bar {u}) ) }$-sequence in $H_{0}^{1}(\Omega^{j})$ for $J$. In fact, $J(v_{n}^{1,j})=\frac{1}{2}J(v_{n}^{1}) =\frac{1}{2}J( u_{n}^{1}) + o(1) =\frac{1}{2}\left[ \beta-J(\bar {u})\right] + o(1),$ and for $\varphi\in C_{c}^{\infty}(\Omega^{j})$, $\| \varphi\| _{H^{1}}=1$, we have $| \langle J'(v_{n}^{1,j}),\varphi\rangle | =| \langle J'(v_{n}^{1}),\varphi\rangle | \leq\| J'(v_{n}^{1})\| _{H^{-1}}\| \varphi\| _{H^{1}}.$ Thus, $\| J'(v_{n}^{1,j})\| _{H^{-1}}\leq\| J^{\prime }(v_{n}^{1})\| _{H^{-1}}= o(1).$ Note that $v_{n}^{1}\rightharpoonup0$ weakly in $H_{0}^{1}(\Omega)$ and $v_{n}^{1}\nrightarrow0$ strongly in $H_{0}^{1}(\Omega)$, so we have $v_{n}^{1,j}\rightharpoonup0$ weakly in $H_{0}^{1}(\Omega^{j})$ and $v_{n}^{1,j}\nrightarrow0$ strongly in $H_{0}^{1}(\Omega^{j})$. \begin{itemize} \item[(1-0)] ${\int_{A_{-1,1}^{r}}} |w_{n}^{1,j}(z)|^{2}dz\geq\frac{d_{1}}{2}$ for some constant $d_{1}>0$, $n=1,2,\dots$, and $j=1,2$, where $w_{n}^{1,j}(z)=v_{n} ^{1,j}(z+z_{n}^{1,j})$ for some $\{ z_{n}^{1,j}\} \subset\mathbf{A}^{r}$ : for $j=1,2$, since $\{ v_{n}^{1,j}\}$ is bounded, $J'(v_{n}^{1,j})= o(1)$, and $v_{n}^{1,j}\nrightarrow0$ strongly in $H_{0}^{1}(\Omega^{j})$. By Lemma \ref{p10}, there is a subsequence $\{ v_{n}^{1,j}\}$, and a constant $d_{1}>0$ such that $Q_{n}^{r,1}=\sup_{y\in\mathbb{R}}\int_{(0,y)+A_{-1,1}^{r}}|v_{n}^{1,j}( z) |^{2}dz\geq d_{1}\quad\mbox{for } n=1,2,\dots.$ For$\;n=1,2,\dots,$\ take $z_{n}^{1,1}=(0,y_{n}^{1})$ and $z_{n}^{1,2}=(0,-y_{n}^{1})$ in $\mathbf{A}^{r}$ such that $\int_{z_{n}^{1,j}+A_{-1,1}^{r}}|v_{n}^{1,j}(z)|^{2}dz\geq\frac{d_{1}} {2}\;\text{for}\quad n=1,2,\dots.$ Let $w_{n}^{1,j}(z)=v_{n}^{1,j}(z+z_{n}^{1,j})$ then $\int_{A_{-1,1}^{r}}|w_{n}^{1,j}(z)|^{2}dz\geq\frac{d_{1}}{2}\;\text{for} \quad n=1,2,\dots.$ \item[(1-1)] $u_{n}(z)=\bar{u}(z)+\sum_{j=1}^{2}w_{n}^{1,j}( z-z_{n}^{1,j}) + o(1)$ strongly in $H_{0}^{1}(\mathbf{A}^{r})$. By Lemma \ref{p34}, we have the following equalities in the strong sense in $H_{0} ^{1}(\mathbf{A}^{r})$ $\sum_{j=1}^{2}w_{n}^{1,j}(z-z_{n}^{1,j}) =\sum_{j=1}^{2} v_{n}^{1,j}(z) =v_{n}^{1}(z) =u_{n}^{1}( z) + o(1)=u_{n}(z) -\overline{u}(z) + o(1),$ or $u_{n}(z) =\overline{u}(z) +\sum_{j=1}^{2} w_{n}^{1,j}(z-z_{n}^{1,j}) + o(1)\quad \text{strongly in }H_{0} ^{1}(\mathbf{A}^{r}).$ \item[(1-2)] $\| w_{n}^{1,j}\| _{H^{1}}\leq c$ for $n=1,2,\dots$, and $\| w^{1,j}\| _{H^{1}}\leq c$, where $w_{n}^{1,j}\rightharpoonup w^{1,j}$ weakly in $H_{0}^{1}(\mathbf{A}^{r})$ for $j=1,2$. By Lemma \ref{p4}$( iii)$, \begin{align*} \| w_{n}^{1,j}\| _{H^{1}}^{2} & =\| v_{n}^{1,j}\| _{H^{1}}^{2}=\frac{1}{2} \| v_{n}^{1}\| _{H^{1}}^{2}=\frac{1}{2}\| u_{n}^{1}\| _{H^{1}}^{2}+ o(1)\\ & =\frac{1}{2}(\| u_{n}\| _{H^{1}}^{2}-\| \bar{u}\| _{H^{1}}^{2}) + o(1)\\ & \leq\frac{1}{2}c^{2}+ o(1), \end{align*} we have $\Vert w_{n}^{1,j}\Vert_{H_{0}^{1}(\mathbf{A}^{r})}\leq c$ for $n=1,2,\dots$. Then there is a subsequence $\{w_{n}^{1,j}\}$ and $w^{1,j}$ in $H_{0}^{1}(\mathbf{A}^{r})$ such that $w_{n}^{1,j}\rightharpoonup w^{1,j}$ weakly in $H_{0}^{1}(\mathbf{A}^{r})$. In addition, $w^{1,1}( x,y) =w^{1,2}(x,-y)$. By Lemma \ref{p4} $(i)$, we have $\| w^{1,j}\| _{H^{1}}\leq\liminf_{n\to\infty}\| w_{n} ^{1,j}\| _{H^{1}}\leq c\;\text{for\ }j=1,2.$ \item[(1-3)] $\{w_{n}^{1,j}\}$ is a (PS)$_{\frac{1}{2}(\beta -J(\bar{u})) }$-sequence in $H_{0}^{1}(\mathbf{A}^{r})$ for $J$: note that $J'(v_{n}^{1,j})= o(1)$ in $H^{-1}(\Omega^{j})$. Because $\Omega^{j}$ is a half infinite strip, $(1-7)$ below and Theorem \ref{p271}, we have for every $\varphi\in H_{0}^{1}(\mathbf{A}^{r})$, $\langle J'(w_{n}^{1,j}) ,\varphi\rangle =\int_{\mathbf{A}^{r}}\nabla w_{n}^{1,j}\nabla\varphi+w_{n}^{1,j}\varphi -\int_{\mathbf{A}^{r}}| w_{n}^{1,j}| ^{p-2}w_{n}^{1,j}\varphi = o(1) .$ Therefore, $J'(w_{n}^{1,j}) = o(1)$ in $H^{-1}( \mathbf{A}^{r})$. Moreover, we have $J(w_{n}^{1,j}) =J(v_{n}^{1,j}(z+z_{n}^{1,j}) )=\frac{1}{2}J(v_{n}^{1}) =\frac{1}{2}(\beta-J(\bar {u})) + o(1).$ \item[(1-4)] $-\Delta w^{1,j}+w^{1,j}-|w^{1,j}|^{p-2}w^{1,j}=0$ in $\mathbf{A}^{r}:$ by Theorem \ref{c2} $(i)$. \item[(1-5)] $w^{1,j}\not \equiv 0:$ by the Rellich-Kondrakov theorem \ref{p24} and $(1-0)$, we have $\int_{A_{-1,1}^{r}}|w^{1,j}|^{2}=\lim_{n\to\infty}\int_{A_{-1,1}^{r} }|w_{n}^{1,j}|^{2}\geq\frac{d_{1}}{2},$ thus $w^{1,j}\not \equiv 0$. \item[(1-6)] By (1-2), (1-4), (1-5), and Lemma \ref{p11}, there is a $\delta>0$ such that $\Vert w^{1,j}\Vert_{H_{0}^{1}(\mathbf{A}^{r})}\geq\Vert w^{1,j}\Vert _{L^{2}(\mathbf{A}^{r})}>\delta.$ Therefore, $J(w^{1,j})=(\frac{1}{2}-\frac{1}{p}) a(w^{1,j})>(\frac {1}{2}-\frac{1}{p}) \delta^{2}=\delta'.$ \item[(1-7)] $|z_{n}^{1,j}|\to\infty$ : otherwise, there is a $R>0$ such that $z_{n}^{1,j}+A_{-1,1}^{r}\subset A_{-R,R}^{r}$ for $n=1,2,\dots$. Then by (1-0), we have $0=\lim_{n\to\infty}\int_{A_{-R,R}^{r}}|v_{n}^{1,j}|^{2}\geq \overline{\lim_{n\to\infty}}\int_{z_{n}^{1,j}+A_{-1,1}^{r}} |v_{n}^{1,j}|^{2}\geq\frac{d_{1}}{2},$ which is a contradiction. \item[(1-8)] $a(u_{n})=a(\bar{u})+2a(w_{n}^{1,j})+ o(1)$ for $j=1,2:$ since $u_{n}\rightharpoonup\bar{u}$ weakly in $H_{0}^{1}(\mathbf{A}^{r})$, by Lemma \ref{p4}$(iii)$, \begin{align*} a(u_{n})-a(\bar{u}) & =a(u_{n}-\bar{u})+ o(1)\\ & =a(u_{n}^{1}) + o(1) \\ & =a(v_{n}^{1}) + o(1) \\ & =a(v_{n}^{1,1}) +a(v_{n}^{1,2}) +o( 1) \\ & =a(w_{n}^{1,1})+a(w_{n}^{1,2})+ o(1) \\ & =2a(w_{n}^{1,j})+ o(1) \;\quad\text{for }j=1,2, \end{align*} thus, $a(u_{n})=a(\bar{u})+2a(w_{n}^{1,j})+ o(1)$ for $j=1,2$. \item[(1-9)] $b(u_{n})=b(\bar{u})+2b(w_{n}^{1,j})+ o(1)$ for $j=1,2:$ since $u_{n}\to\bar{u}$ a.e. in $\Omega$ and $\{ u_{n}\}$ is bounded in $L^{p}(\Omega)$, by Lemma \ref{p7}$(i)$, we have \begin{align*} b(u_{n})-b(\bar{u}) & =b(u_{n}-\bar{u})+ o(1)\\ & =b(u_{n}^{1}) + o(1) \\ & =b(v_{n}^{1}) + o(1) \\ & =b(v_{n}^{1,1}) +b(v_{n}^{1,2}) +o( 1) \\ & =b(w_{n}^{1,1})+b(w_{n}^{1,2})+ o(1) \\ & =2b(w_{n}^{1,j})+ o(1) \;\quad\text{for }j=1,2 \end{align*} thus $b(u_{n})=b(\bar{u})+2b(w_{n}^{1,j})+ o(1)$ for $j=1,2$. \item[(1-10)] $J(u_{n})=J(\bar{u})+2J(w_{n}^{1,j})+ o(1)$ for $j=1,2:$ by $(1-8)$, $(1-9)$ and $J(w^{1,1})=J(w^{1,2})$, we have \begin{align*} J(u_{n}) & =J(\bar{u})+J(w_{n}^{1,1})+J(w_{n}^{1,2}) + o(1)\\ & =J(\bar{u})+2J(w_{n}^{1,j})+ o(1)\;\text{for }j=1,2. \end{align*} \end{itemize} \noindent\textbf{Step 2}. Suppose that $w_{n}^{1,j}(z)\nrightarrow w^{1,j}(z)$ strongly in $H_{0}^{1}(\mathbf{A}^{r})$. Let $v_{n}^{2,j}(z)=w_{n}^{1,j}(z)-w^{1,j}(z).$ We have $v_{n}^{2,j}\rightharpoonup0$ weakly in $H_{0}^{1}(\mathbf{A}^{r})$ but $v_{n}^{2,j}\nrightarrow0$ strongly in $H_{0}^{1}(\mathbf{A}^{r})$. \begin{itemize} \item[(2-0)] ${\int_{A_{-1,1}^{r}}} |w_{n}^{2,j}(z)|^{2}dz\geq\frac{d_{2}}{2}$ for some constant $d_{2}>0$, $n=1,2,\dots$, and $j=1,2$, where $w_{n}^{2,j}(z)=v_{n}^{2,j}( z+z_{n}^{2,j})$ for some $\{ z_{n}^{2,j}\} \subset \mathbf{A}^{r}$: for $j=1,2$, since $\{ v_{n}^{2,j}\}$ is bounded, $J'(v_{n}^{2,j})= o(1)$, and $v_{n}^{2,j}\nrightarrow0$ strongly in $H_{0}^{1}(\mathbf{A}^{r})$, by Lemma \ref{p10}, there is a subsequence $\{ v_{n}^{2,j}\}$, a constant $d_{2}>0$ such that $Q_{n}^{r,2}=\sup_{y\in\mathbb{R}}\int_{(0,y)+A_{-1,1}^{r}}|v_{n}^{2,j}( z) |^{2}dz\geq d_{2}\quad\mbox{for } n=1,2,\dots.$ For $n=1,2,\dots$, take $z_{n}^{2,1}=(0,y_{n}^{2})$ and $z_{n}^{2,2}=(0,-y_{n}^{2})$ in $\mathbf{A}^{r}$ such that $\int_{z_{n}^{2,j}+A_{-1,1}^{r}}|v_{n}^{2,j}(z)|^{2}dz\geq\frac{d_{2}}{2} \quad\mbox{for } n=1,2,\dots.$ Let $w_{n}^{2,j}(z)=v_{n}^{2,j}(z+z_{n}^{2,j})$, then $\int_{A_{-1,1}^{r}}|w_{n}^{2,j}(z)|^{2}dz\geq\frac{d_{2}}{2}\quad \text{for}\quad n=1,2,\dots.$ As in Step 1, we have the following results. \item[(2-1)] $u_{n}(z)=\bar{u}(z)+\sum_{j=1}^{2}w^{1,j}(z-z_{n} ^{1,j}) +\sum_{j=1}^{2}w_{n}^{2,j}(z-z_{n}^{1,j}-z_{n} ^{2,j}) + o(1)$ strongly in $H_{0}^{1}(\mathbf{A}^{r})$ ; \item[(2-2)] $\| w_{n}^{2,j}\| _{H^{1}}\leq c$ for $n=1,2,\dots$ and $\| w^{2,j}\| _{H^{1}}\leq c$, where $w_{n}^{2,j}\rightharpoonup w^{2,j}$ weakly in $H_{0}^{1}(\mathbf{A}^{r})$ for $j=1,2$; \item[(2-3)] $\{w_{n}^{2,j}\}$ is a (PS)-sequence in $H_{0}^{1} (\mathbf{A}^{r})$ for $J$; \item[(2-4)] $-\Delta w^{2,j}+w^{2,j}-|w^{2,j}|^{p-2}w^{2,j}=0$ in $\mathbf{A}^{r}$; \item[(2-5)] $w^{2,j}\not \equiv 0$; \item[(2-6)] $\Vert w^{2,j}\Vert_{L^{2}(\mathbf{A}^{r})}>\delta$ and $J(w^{2,j})>\delta'$; \item[(2-7)] $|z_{n}^{2,j}|\to\infty$; \item[(2-8)] $a(u_{n})=a(\bar{u})+2a(w^{1,j})+2a(w_{n}^{2,j})+ o(1)$ : since $v_{n}^{2,j}(z) =w_{n}^{1,j}(z) -w^{1,j}( z) \rightharpoonup0,$ we have $a(w_{n}^{2,j}) =a(v_{n}^{2,j}) =a( w_{n}^{1,j}) -a(w^{1,j}) + o(1).$ Further, by $(1-8)$, we have \begin{align*} a(u_{n})-a(\bar{u}) & =a(w_{n}^{1,1})+a(w_{n}^{1,2})+ o(1) \\ & =2a(w^{1,j}) +2a(w_{n}^{2,j}) +o( 1) . \end{align*} \item[(2-9)] $b(u_{n})=b(\bar{u})+2b(w^{1,j})+2b(w_{n}^{2,j})+ o(1)$; \item[(2-10)] $J(u_{n})=J(\bar{u})+2J(w^{1,j})+2J(w_{n}^{j})+ o(1)$. \end{itemize} Continuing this process, we arrive at the $m$-th step \begin{itemize} \item[$(m$-0)] ${\int_{A_{-1,1}^{r}}} |w_{n}^{m,j}(z)|^{2}dz\geq\frac{d_{m}}{2}$ for some constant $d_{m}>0$, $n=1,2,\dots$and $j=1,2$, where $w_{n}^{m,j}(z)=v_{n}^{m,j}( z+z_{n}^{m,j})$ for some $\{ z_{n}^{m,j}\}\subset\mathbf{A}^{r}$; \item[$(m$-1)] $u_{n}(z)=\bar{u}(z)+\sum_{j=1}^{2}\underset{i=1}{\overset {m-1}{\sum}}w^{i,j}(z-\tilde{z}_{n}^{i,j}) +\sum_{j=1}^{2} w_{n}^{m,j}(z-\tilde{z}_{n}^{m,j}) + o(1)$ strongly in $H_{0} ^{1}(\mathbf{A}^{r})$, where $\tilde{z}_{n}^{i,j}=z_{n}^{1,j}+\dots +z_{n}^{i,j}$ for $i=1,2,\dots,m$ and $j=1,2$; \item[$(m$-2)] $\| w_{n}^{m,j}\| _{H^{1}}\leq c$ for $n=1,2,\dots$ and $\| w^{m,j}\| _{H^{1}}\leq c$, where $w_{n}^{m,j}\rightharpoonup w^{m,j}$ weakly in $H_{0}^{1}(\mathbf{A}^{r})$; \item[$(m$-3)] $\{w_{n}^{m,j}\}$ is a (PS)-sequence in $H_{0}^{1} (\mathbf{A}^{r})$ for $J$; \item[$(m$-4)] $-\Delta w^{m,j}+w^{m,j}-|w^{m,j}|^{p-2}w^{m,j}=0$ in $\mathbf{A}^{r}$; \item[$(m$-5)] $w^{m,j}\not \equiv 0$; \item[$(m$-6)] $\| w^{m,j}\| _{L^{2}(\mathbf{A} ^{r}) }>\delta$ and $J(w^{m,j})>\delta'$; \item[$(m$-7)] $|z_{n}^{i,j}|=$ $|\tilde{z}_{n}^{i,j}-\tilde{z}_{n} ^{i-1,j}|\to\infty$ and $|\tilde{z}_{n}^{i,j}|\to\infty$, for each $i=1,2,\dots,m$: we show this by induction on $i$. For $i=1$, $|\tilde{z}_{n}^{i,j}| =| z_{n}^{1,j}| \to\infty$. Assume that $|\tilde{z}_{n}^{i,j}|\to\infty$, for $i=1,2,\dots,k$, for some $k\delta$, $R>0$ exists such that $\tilde{z}_{n}^{k+1,j}+A_{-R,R}^{r}\subset A_{-2R,2R}^{r}$ and ${\int_{A_{-R,R}^{r}}} |w^{k+1,j}|^{2}\geq(\frac{\delta}{2}) ^{2}.$ We have \begin{align*} (\frac{\delta}{2}) ^{2} &\leq{\int_{A_{-R,R}^{r}}}|w^{k+1,j}|^{2}\\ &=\lim_{n\to\infty}{\int_{A_{-R,R}^{r}}} |v_{n}^{1,j}(z+\tilde{z}_{n}^{k+1,j})|^{2}\\ &\leq\lim_{n\to\infty}{\int_{A_{-2R,2R}^{r}}}|v_{n}^{1,j}(z)|^{2}=0, \end{align*} which is a contradiction. By the induction hypothesis, we have $|\tilde{z}_{n}^{i,j}|\to\infty$ for $i=1,2,\dots,m$. \item[$(m$-8)] $a(u_{n})=a(\bar{u})+2\sum_{i=1}^{m-1} a(w^{i,j})+2a(w_{n}^{i,j})+ o(1)$; \item[$(m$-9)] $b(u_{n})=b(\bar{u})+2\sum_{i=1}^{m-1} b(w^{i,j})+2b(w_{n}^{i,j})+ o(1)$; \item[$(m$-10)] $J(u_{n})=J(\bar{u})+2\sum_{i=1}^{m-1} J(w^{i,j})+2J(w_{n}^{i,j})+ o(1)$. \end{itemize} By the Archimedean principle, $k\in\mathbb{N}$ exists such that $k\delta^{2}>\beta$. Take $l=[ \frac{k}{2}] +1$, then after step $(l+1)$, we obtain $a(u_{n})=a(\bar{u})+2a(w^{1,j})+2a(w^{2,j})+\dots+2a(w^{l,j})+2a(w_{n} ^{l+1,j})+ o(1).$ Since $a(w_{n}^{l+1,j})\geq0$, $a(\bar{u})>0$, and $a(w^{i,j})>\delta^{2}$ for $i=1,2,\dots,l$, we have $\beta+ o(1)\geq2l\delta^{2}>k\delta^{2} >\beta$, which is a contradiction. Therefore, there is an $m\in\mathbb{N},$ such that $w_{n}^{m,j}(z)=w^{m,j}(z)+ o(1)$ strongly in $H_{0}^{1} (\mathbf{A}^{r})$, $w_{n}^{i,j}(z)=w^{i,j}(z)+ o(1)$ weakly in $H_{0} ^{1}(\mathbf{A}^{r})$, and $w_{n}^{i,j}(z)\neq w^{i,j}(z)+ o(1)$ strongly in $H_{0}^{1}(\mathbf{A}^{r})$ for $i=1,2,\dots,m-1$. Then we have \begin{itemize} \item[$(sm$-0)] ${\int_{A_{-1,1}^{r}}} |w_{n}^{m,j}(z)|^{2}dz\geq\frac{d_{m}}{2}$ for some constant $d_{m}>0$, $n=1,2,\dots$, and $j=1,2$, where $w_{n}^{m,j}(z)=v_{n}^{m,j}( z+z_{n}^{m,j})$ for some $\{ z_{n}^{m,j}\} \subset \mathbf{A}^{r}$; \item[$(sm$-1)] $u_{n}(z)=\bar{u}(z)+\sum_{j=1}^{2}\underset{i=1}{\overset {m}{\sum}}w^{i,j}(z-\tilde{z}_{n}^{i,j}) + o(1)$ strongly in $H_{0}^{1}(\mathbf{A}^{r})$, where $\tilde{z}_{n}^{i,j}=z_{n}^{1,j} +\dots+z_{n}^{i,j}$, for $i=1,2,\dots,m$ and $j=1,2$; \item[$(sm$-2)] $\| w_{n}^{m,j}\| _{H^{1}}\leq c$ for $n=1,2,\dots$ and $\| w^{m,j}\| _{H^{1}}\leq c$, where $w_{n}^{m,j}\rightharpoonup w^{m,j}$ weakly in $H_{0}^{1}(\mathbf{A}^{r})$; \item[$(sm$-3)] $\{w_{n}^{m,j}\}$ is a (PS)-sequence in $H_{0} ^{1}(\mathbf{A}^{r})$ for $J$; \item[$(sm$-4)] $-\Delta w^{m,j}+w^{m,j}-|w^{m,j}|^{p-2}w^{m,j}=0$ in $\mathbf{A}^{r}$; \item[$(sm$-5)] $w^{m,j}\not \equiv 0$; \item[$(sm$-6)] $\| w^{m,j}\| _{L^{2}(\mathbf{A} ^{r}) }>\delta$ and $J(w^{m,j})>\delta'$; \item[$(sm$-7)] $|z_{n}^{i,j}|=$ $|\tilde{z}_{n}^{i,j}-\tilde{z}_{n} ^{i-1,j}|\to\infty$ and $|\tilde{z}_{n}^{i,j}|\to\infty$, for each $i=1,2,\dots,m$; \item[$(sm$-8)] $a(u_{n})=a(\bar{u})+2\underset{i=1}{\overset{m}{\sum} }a(w^{i,j})+ o(1)$; \item[$(sm$-9)] $b(u_{n})=b(\bar{u})+2\underset{i=1}{\overset{m}{\sum} }b(w^{i,j})+ o(1)$; \item[$(sm$-10)] $J(u_{n})=J(\bar{u})+2\underset{i=1} {\overset{m}{\sum}}J(w^{i,j})+ o(1)$. \end{itemize} Finally, suppose $u_{n}\geq0$ for $n=1,2,\dots$. Then\newline$(i)$ Since $u_{n}\rightharpoonup\bar{u}$ weakly in $H_{0}^{1}(\Omega)$. By Lemma \ref{p4}$(ii)$, there is a subsequence $\{u_{n}\}$ such that $\,u_{n}\to\bar{u}$ a.e. in $\Omega$. Thus, $\bar{u}\geq0$. \newline$(ii)$ For $j=1,2$, let $z\in\Omega^{j}-z_{n}^{1,j}$, then $w_{n}^{1,j}(z)=v_{n}^{1,j}(z+z_{n}^{1,j}) =v_{n}^{1}( z+z_{n}^{1,j}) =\xi_{n}(z+z_{n}^{1,j}) u_{n}^{1}( z+z_{n}^{1,j}) .$ Thus, $w_{n}^{1,j}(z)=\begin{cases} \xi_{n}(z+z_{n}^{1,j}) u_{n}^{1}(z+z_{n}^{1,j}) & \text{if }z\in\Omega^{j}-z_{n}^{1,j}\\ 0 & \text{otherwise.} \end{cases}$ Let $h_{n}(z) =\xi_{n}(z+z_{n}^{1,j}) \bar {u}(z+z_{n}^{1,j})$ and $g_{n}(z) =\xi _{n}(z+z_{n}^{1,j}) u_{n}(z+z_{n}^{1,j}) \geq0$. Since $w_{n}^{1,j}(z)\rightharpoonup w^{1,j}(z)$ weakly in $H_{0} ^{1}(\mathbf{A}^{r})$ and $h_{n}(z) \rightharpoonup0$ weakly in $H_{0}^{1}(\mathbf{A}^{r})$, we have $w_{n}^{1,j}(z)\to w^{1,j}(z)$ a.e. in $\mathbf{A}^{r}$ and $h_{n}(z) \to0$ a.e. in $\mathbf{A}^{r}$. By Lemma \ref{u1}, we have $w^{1,j}\geq0$.\newline $(iii)$ In fact, we have $w_{n}^{2,j}(z) =v_{n}^{2,j}(z+z_{n} ^{2,j}) =w_{n}^{1,j}(z+z_{n}^{2,j}) -w^{1,j}( z+z_{n}^{2,j})$. By Lemma \ref{p5}, $w^{1,j}(z+z_{n} ^{2,j}) \rightharpoonup0$ weakly in $H_{0}^{1}(\mathbf{A}^{r})$. Moreover, $w_{n}^{2,j}(z)\rightharpoonup w^{2,j}(z)$ weakly in $H_{0} ^{1}(\mathbf{A}^{r})$, hence $w_{n}^{1,j}(z+z_{n}^{2,j})\rightharpoonup w^{2,j}(z)$ weakly in $H_{0}^{1}(\mathbf{A}^{r})$, since $w_{n}^{1,j}(z+z_{n}^{2,j})=\begin{cases} \xi_{n}(z+\tilde{z}_{n}^{2,j}) u_{n}^{1}(z+\tilde{z} _{n}^{2,j}) & \text{if }z\in\Omega^{j}-z_{n}^{1,j},\\ 0 & \text{otherwise.} \end{cases}$ Similar to $(ii)$, $w^{2,j}\geq0.\newline(iv)$ Continuing this process, we obtain $w^{i,j}\geq0$ for each $i=1,2,\dots,m$. \end{proof} Similarly, we have \begin{theorem} [Symmetric Palais-Smale Decomposition Theorem in $\mathbb{R}^{N}$] \label{u3} Let $\Omega$ be a $y$-symmetric large domain in $\mathbb{R}^{N}$ separated by a $y$-symmetric bounded domain, and let $\{ u_{n}\} \subset H_{s}^{1}(\Omega)$ be a (PS)$_{\beta}$-sequence in $H_{0}^{1}(\Omega)$ for $J$. Then there are a subsequence $\{ u_{n}\}$, a positive integer $m$, sequences $\{ \widetilde {z}_{n}^{i,j}\} _{n=1}^{\infty}$ in $\mathbb{R}^{N}$, a function $\bar{u}\in H_{s}^{1}(\Omega)$, and $0\neq w^{i,j}\in H^{1}( \mathbb{R}^{N})$ for $1\leq i\leq m$, $j=1,2$ such that \begin{gather*} w^{i,1}(x,y) =w^{i,2}(x,-y) ,\\ |\widetilde{z}_{n}^{i,j}|\to\infty\quad \quad\text{for }i=1,2,\dots,m,\\ -\Delta\bar{u}+\bar{u}=\mid\bar{u}\mid^{p-2}\bar{u}\quad\mbox{in }\Omega,\\ -\Delta w^{i,j}+w^{i,j}=\mid w^{i,j}|^{p-2}w^{i,j}\quad\mbox{in }\mathbb{R}^{N}, \end{gather*} and \begin{gather*} u_{n}=\bar{u}+\sum_{i=1}^m w^{i,1}(\cdot -\widetilde{z}_{n}^{i,1}) +\sum_{i=1}^m w^{i,2}(\cdot-\widetilde{z}_{n}^{i,2}) + o(1)\quad\text{strongly in } H^{1}(\mathbb{R}^{N}) ,\\ a(u_{n})=a(\bar{u})+2\sum_{i=1}^m a(w^{i,j})+ o(1)\quad \text{for some }j=1,2,\\ b(u_{n})=b(\bar{u})+2\sum_{i=1}^m b(w^{i,j})+ o(1)\quad \text{for some }j=1,2,\\ J(u_{n})=J(\bar{u})+2\sum_{i=1}^m J(w^{i,j})+ o(1)\quad \text{for some }j=1,2. \end{gather*} In addition, if $u_{n}\geq0$, then $\bar{u}\geq0$ and $w^{i,j}\geq0$ for each $1\leq i\leq m$, $j=1,2$. \end{theorem} \begin{corollary} \label{w14} If $\Omega=\mathbf{A}^{r}\backslash\omega$, where $\omega$ is an axially symmetric bounded set in $\mathbf{A}^{r}$ and $\{ u_{n}\}$ is a (PS)$_{\beta}$-sequence in $H_{s}^{1}(\Omega)$ for $J$ and $0<\beta<2\alpha(\mathbf{A}^{r})$, then the sequence $\{ u_{n}\}$ contains a strongly convergent subsequence and there is a positive solution $\overline{u}$ of \eqref{E1} in $\Omega$. \end{corollary} \begin{proof} By Theorem \ref{u2}, we have $J(u_{n})=J(\bar{u})+2\sum_{i=1}^m J(w^{i,j})+ o(1)\quad\text{ for some }j=1,2.$ Note that $J(w^{i,j})\geq\alpha(\mathbf{A}^{r})$ and $J(\bar {u})\geq0$. If $m\geq1$, then we have $2\alpha(\mathbf{A}^{r}) >\beta+ o(1)=J(u_{n})\geq J(\bar {u})+2m\alpha(\mathbf{A}^{r}) ,$ which is a contradiction. Thus, $m=0$. By Theorem \ref{u2}, we have $u_{n}=\bar{u}+ o(1)\;\text{strongly in }H_{0}^{1}(\mathbf{A}^{r}).$ Since $J(\bar{u})=\beta$ and $\beta>0$, we have $\bar{u}\neq0$. \end{proof} \begin{corollary}\label{w15} If $\Omega$ is a $y$-symmetric strictly large domain in $\mathbb{R}^{N}$ separated by a $y$-symmetric bounded domain, and $\{ u_{n}\}$ is a positive (PS)$_{\beta}$-sequence in $H_{s}^{1}( \Omega)$ for $J$, with $0<\beta<3\alpha(\mathbb{R}^{N})$ but $\beta\neq2\alpha(\mathbb{R}^{N})$, then $\{ u_{n}\}$ contains a strongly convergent subsequence, and there is a positive solution $\bar{u}$ of Equation \eqref{E1} in $\Omega$. In particular, if $\{ u_{n}\}$ is a (PS)$_{\beta}$-sequence in $H_{s}^{1}(\Omega)$ for $J$ with $0<\beta <2\alpha(\mathbb{R}^{N})$, then $\{ u_{n}\}$ contains a strongly convergent subsequence. \end{corollary} \begin{proof} By Theorem \ref{u3}, we have $J(u_{n})=J(\bar{u})+2\sum_{i=1}^m J(w^{i,j})+ o(1),\quad \text{for some }j=1,2.$ By the uniqueness of the positive solution for Equation $( \ref{E1})$ in $\mathbb{R}^{N}$, we have $J(w^{i,j})=\alpha( \mathbb{R}^{N})$ and $J(\bar{u})\geq0$. If $m\geq1$, then $3\alpha(\mathbb{R}^{N}) >\beta+ o(1)=J(u_{n})=J(\bar{u} )+2m\alpha(\mathbb{R}^{N}) ,$ thus, $m=0,1$. Suppose that $m=1$, then $2\alpha(\mathbb{R}^{N}) \neq\beta+ o(1)=J(u_{n})=J(\bar {u})+2\alpha(\mathbb{R}^{N}) ,$ which implies that $J(\bar{u})\neq0$ or $J(\bar{u})>0$, or $J(\bar{u} )\geq\alpha(\mathbb{R}^{N})$. Therefore, $3\alpha(\mathbb{R}^{N}) >\beta+ o(1)=J(u_{n})\geq J(\bar {u})+2m\alpha(\mathbb{R}^{N}) \geq3m\alpha(\mathbb{R} ^{N}) ,$ which is a contradiction. Thus, $m=0$. By Theorem \ref{u3} again, we have $u_{n}=\bar{u}+ o(1)\quad\text{strongly in }H_{0}^{1}(\mathbb{R}^{N}) .$ Since $J(\bar{u})=\beta$ and $\beta>0$, we have $\bar{u}\neq0$. \end{proof} \noindent\textbf{Bibliographical notes:} The results of this section are from Wang-Wu \cite{WW4}. \section{Fundamental Properties, Regularity, and Asymptotic Behavior of Solutions} In this section we study the fundamental properties, regularity, and asymptotic behavior of solutions of \eqref{E1}. \subsection{Fundamental Properties of Solutions} \begin{theorem} \label{b1} Let $u\in H_{0}^{1}(\Omega)$ be a positive symmetric and radially decreasing solution of \eqref{E1}. Then $u(0)\geq1$. \end{theorem} \begin{proof} Since $u$ is symmetric and radially decreasing, we have $-\Delta u(0)\geq0$. Thus, $u(0)\leq-\Delta u(0)+u(0)=u^{p-1}(0),$ or $u(0)\geq1$. \end{proof} A ground state solution in $X(\Omega)$ is of constant sign. \begin{lemma} \label{b2} Let $u$ in $X(\Omega)$ be a solution of \eqref{E1} that changes sign , and let $\alpha_{X}(\Omega)$ be the index of $J$ in $\Omega$. Then $J(u)>2\alpha_{X}(\Omega)$. \end{lemma} \begin{proof} Let $u^{-}=\max\{ -u,0\}$. Then $u^{-}$ is nonzero. Multiply \eqref{E1} by $u^{-}$ and integrate to obtain ${\int_{\Omega}} \nabla u\nabla u^{-}+ {\int_{\Omega}} uu^{-}= {\int_{\Omega}} |u|^{p-2}uu^{-}.$ Consequently, ${\int_{\Omega}} | \nabla u^{-}| ^{2}+ {\int_{\Omega}} | u^{-}| ^{2}= {\int_{\Omega}} |u^{-}|^{p}.$ Thus, $u^{-}\in\mathbf{M}(\Omega)$ and hence $J(u^{-})\geq\alpha_{X}( \Omega)$. Suppose that $J(u^{-})=\alpha_{X}(X)$. By Theorem \ref{i6}, $u^{-}$ is a nonzero solution of \eqref{E1}. By the maximum principle, $u=u^{-}$, which contradicts the sign assumption on $u$. Thus $J(u^{-})>\alpha_{X}(\Omega)$. Similarly, $J(u^{+} )>\alpha_{X}(\Omega)$, where $u^{+}=\max\{ u,0\}$. Thus, $J(u)=J(u^{+})+J(u^{-})>2\alpha_{X}(\Omega) .$ \end{proof} The positive solution of Equation $\eqref{E1}$ in $\mathbb{R}^{N}$ is unique. \begin{theorem} \label{b4}$(i)$ There is a ground state solution of Equation $\eqref{E1}$ in $\mathbb{R}^{N}$;\newline$(ii)$ The only positive solutions of Equation $\eqref{E1}$ in $\mathbb{R}^{N}$ are ground state solutions;\newline$(iii)$ Every positive ground state solution $\bar{u}\in H^{1}(\mathbb{R}^{N})$ of Equation $\eqref{E1}$ is spherically symmetric about some point $x_{0}$ in $\mathbb{R}^{N}$, $\bar{u}'(r)<0$ for $r=| x-x_{0}|$, and \begin{align*} \underset{r\to\infty}{\lim}r^{\frac{N-1}{2}}e^{r}\bar{u}( r) & =\gamma>0,\\ & \\ \underset{r\to\infty}{\lim}r^{\frac{N-1}{2}}e^{r}\bar{u}^{\prime }(r) & =-\gamma; \end{align*} $(iv)$ The positive solution of Equation $\eqref{E1}$ in $\mathbb{R}^{N}$ is unique. \end{theorem} \begin{proof} $(i)$ By Lien-Tzeng-Wang \cite{LTW}. For $(ii)$ and $(iii)$ see Gidas-Ni-Nirenberg \cite{GNN2}. For $(iv)$ See Kwong \cite{K}. \end{proof} \subsection{Regularity of Solutions} In addition to the study of \eqref{E1}, we also study Equation (\ref{E2}), a perturbation of \eqref{E1}: associated with $( \ref{E2})$. We consider the energy functionals $J_{h}$ for $u\in H_{0}^{1}(\Omega):$ $J_{h}(u)=\frac{1}{2}a(u)-\frac{1}{p}b(u) - {\int_{\Omega}} hu.$ We let $J_{0}=J$. We first recall some fundamental estimates for elliptic equations. Let us first consider the classical $C^{\beta}-$setting: Schauder estimates. \begin{theorem} \label{s1} Let $\Omega$ be a bounded $C^{2,\beta}-$domain, $h\in C^{\beta }(\overline{\Omega})$. Then the Dirichlet problem (\ref{E2}) has a unique classical solution $u\in C^{2,\beta}(\overline{\Omega})$. \end{theorem} For the proof of the above theorem see Gilbarg-Trudinger \cite[Theorem 6.14]{GT}. We have the following Kato regularity. \begin{theorem} \label{s2} Let $\Omega$ be a domain in $\mathbb{R}^{N}$ and let $f:\Omega \times\mathbb{R}\to\mathbb{R}$ be a Caratheodory function such that for almost every $z\in\Omega$ $|f(z,u)|\leq a(z)(1+|u|)$ with a nonnegative function $a\in L_{\rm loc}^{N/2}(\Omega)$. In addition, let $u\in H_{\rm loc}^{1}(\Omega)$ be a weak solution of Equation (\ref{E5}). Then $u\in L_{\rm loc}^{q}(\Omega)$ for any $1\leq q<\infty$. If $a\in L^{N/2} (\Omega)\cap$ $L^{2}(\Omega)$ and $u\in H_{0}^{1}(\Omega)$, then $u\in L^{q}(\Omega)$ for any $2\leq q<\infty$. \end{theorem} \begin{proof} Recall that if $u\in H_{\rm loc}^{1}(\Omega)$ is a weak solution of Equation $(\ref{E5})$ in $\Omega$, then $u$ satisfies ${\int_{\Omega}}\nabla u\nabla\varphi={\int_{\Omega}}f(z,u)\varphi$ for each $\varphi\in C_{c}^{\infty}(\Omega)$. Choose $\eta\in C_{c}^{\infty }(\Omega)$ and for $s\geq0$, $L\geq1$, let $\varphi=\varphi_{s,L}=u\min\{|u|^{2s},L^{2}\}\eta^{2}\in H_{0}^{1}(\Omega),$ with \textrm{supp\ }$\eta=F\subset\subset\Omega$. Note that $\nabla\varphi=\nabla u\min\{|u|^{2s},L^{2}\}\eta^{2}+2s\chi_{\{|u|^{s}\leq L\}}|u|^{2s-2}u^{2}\nabla u\eta^{2}+2u\min\{|u|^{2s},L^{2}\}\eta\nabla\eta.$ Testing Equation $(\ref{E5})$ with $\varphi$, we obtain \begin{align*} &{\int_{\Omega}} |\nabla u|^{2}\min\{|u|^{2s},L^{2}\}\eta^{2}+\frac{s}{2} {\int_{\{|u(x)|^{s}\leq L\}}} |\nabla(|u|^{2})|^{2}|u|^{2s-2}\eta^{2}\\ &{\int_{\Omega}} 2u\nabla u\min\{|u|^{2s},L^{2}\}\eta\nabla\eta\\ &={\int_{\Omega}}f(\cdot,u)u\min\{|u|^{2s},L^{2}\}\eta^{2}. \end{align*} Note that $2xy=2\sqrt{1/2}x\sqrt{2}y\leq\frac{1}{2}x^{2}+2y^{2},$ if $|u|\leq1$ we have $(1+|u|)|u|\min\{|u|^{2s},L^{2}\}\leq2,$ and if $|u|\geq1$, we have $(1+|u|)|u|\min\{|u|^{2s},L^{2}\}\leq2|u|^{2}\min\{|u|^{2s},L^{2}\}.$ Thus, $|a|(1+|u|)|u|\min\{|u|^{2s},L^{2}\}\eta^{2}\leq2|a\| u|^{2}\min\{|u|^{2s} ,L^{2}\}\eta^{2}+2|a|\eta^{2}.$ Hence, \begin{aligned} &{\int_{\Omega}}|\nabla u|^{2}\min\{|u|^{2s},L^{2}\}\eta^{2}+\frac{s}{2} {\int_{\{|u(x)|^{s}\leq L\}}} |\nabla(|u|^{2})|^{2}|u|^{2s-2}\eta^{2}\\ &\leq-2{\int_{\Omega}}u\nabla u\min\{|u|^{2s},L^{2}\}\eta\nabla\eta+ {\int_{\Omega}}|a|(1+|u|)|u|\min\{|u|^{2s},L^{2}\}\eta^{2}\\ &\leq\frac{1}{2}{\int_{\Omega}} |\nabla u|^{2}\min\{|u|^{2s},L^{2}\}\eta^{2}+2 {\int_{\Omega}}|u|^{2}\min\{|u|^{2s},L^{2}\}|\nabla\eta|^{2}\\ &\quad+2{\int_{\Omega}} a|u|^{2}\min\{|u|^{2s},L^{2}\}\eta^{2}+2 {\int_{\Omega}}a\eta^{2}. \end{aligned}\label{10-1} Suppose that $u\in L_{\rm loc}^{2s+2}(\Omega)$ for some $s\geq0$. Let \begin{gather*} c_{s}= {\int_{F}} |u|^{2s+2}<\infty;\quad d=\max\{1,\max|\nabla\eta|^{2}\};\\ \epsilon(M)=\Big({\int_{\{a\geq M\}}}|a|^{N/2}\Big)^{2/N}= o(1). \end{gather*} We have ${\int_{\Omega}}| u| ^{2}\min\{ | u| ^{2s},L^{2}\} | \nabla\eta| ^{2} \leq d {\int_{F}}| u| ^{2s+2}=dc_{s},$ and \begin{align*} &{\int_{\Omega}}a| u| ^{2}\min\{ | u| ^{2s},L^{2}\}\eta^{2}\\ &={\int_{\{ a\geq M\} \cap F}}a| u| ^{2}\min\{ | u| ^{2s},L^{2}\} \eta^{2}+{\int_{\{ a0($independent of$u)$such that $\| u\| _{W^{2,q}(\Omega)}\leq c\| -\Delta u+u\| _{L^{q}(\Omega)}$ for each$u\in W^{2,q}(\Omega)\cap W_{0}^{1,q}(\Omega)$,$1N$and$\theta=2-\frac{N}{s}-[2-\frac{N}{s}]$, then$0<\theta<1$,$u\in C^{1,\theta}(\overline{\Omega})\cap W^{2,s}(\Omega)$, and $\| u\| _{L^{\infty}(\Omega)}\leq\| u\| _{C^{1,\theta}(\overline{\Omega})}\leq c\| u\| _{W^{2,s}(\Omega )};$ \newline$(iii)$Let$h\in C^{\theta}(\overline{\Omega})\cap L^{N/2} (\Omega)\cap L^{s}(\Omega)\cap L^{2}(\Omega)$for$s$as defined in$(ii)$, then$u\in C^{2}(\overline{\Omega})$. \end{theorem} \begin{proof}$(i)N=1,2$follows from the Sobolev inequality. Suppose that$N\geq3$. For$d\geq0$,$l\geq1$, let$\varphi=\varphi_{d,l}=u\min\{|u|^{2d},l^{2}\}$, then $\nabla\varphi=\min\{|u|^{2d},l^{2}\}\nabla u+\chi_{\{ |u|^{d}\leq l\} }2d|u|^{2d}\nabla u.$ Since$u\in H_{0}^{1}(\Omega), we have \begin{align*} {\int_{\Omega}}|\nabla\varphi|^{2} & ={\int_{\Omega}} | \min\{|u|^{2d},l^{2}\}\nabla u+\chi_{\{ |u|^{d}\leq l\} }2d|u|^{2d}\nabla u| ^{2}\\ & \leq2\Big({\int_{\Omega}} | \min\{|u|^{2d},l^{2}\}\nabla u| ^{2}+| \chi_{\{ |u|^{d}\leq l\} }2d|u|^{2d}\nabla u| ^{2}\Big) \\ & \leq2l^{4}{\int_{\Omega}}|\nabla u|^{2}+8d^{2}l^{4} {\int_{\{|u|^{d}\leq l\}}}|\nabla u|^{2}\\ & \leq c{\int_{\Omega}}|\nabla u|^{2}<\infty. \end{align*} Clearly,\varphi\in L^{2}(\Omega) $. Thus,$\varphi\in H_{0}^{1}(\Omega)$. Note that we have$h\varphi\leq| h| | u| $for$| u| \leq1$and$h\varphi\leq| h| | u| ^{2}\min\{ | u| ^{2d},l^{2}\} $for$| u| >1$. Suppose$u\in L^{2d+p}$. Multiplying and integrating \eqref{E2} with$\varphi, we have \begin{align*} &{\int_{\Omega}}\nabla u\nabla\varphi\\ &=-{\int_{\Omega}}u\varphi+{\int_{\Omega}} | u| ^{p-2}u\varphi+{\int_{\Omega}}h\varphi\\ &\leq-{\int_{\Omega}}| u| ^{2}\min\{ | u| ^{2d},l^{2}\} + {\int_{\Omega}}| u| ^{2d+p}+{\int_{\Omega}} |h| | u| +{\int_{\Omega}}| h| | u| ^{2}\min\{ | u|^{2d},l^{2}\} \\ &\leq\| u\| _{L^{2d+p}}^{2d+p}+\| h\| _{L^{2}}^{{}}\| u\| _{L^{2}}^{{}}+ {\int_{\{ h0 be such that $(1+\frac{d}{2})S\varepsilon(M) <\frac {1}{2}$. Then we have \begin{align*} &\| \nabla(u\min\{ | u| ^{d},l\}) \| _{L^{2}}^{2}\\ &\leq(2+d)(\| u\| _{L^{2d+p}}^{2d+p}+\| h\| _{L^{2}}\| u\| _{L^{2}}+M\| u\| _{L^{2} }^{\alpha(2d+2)}\| u\| _{L^{2d+p}}^{(1-\alpha)(2d+2) }) . \end{align*} By (\ref{10-7}), we have \begin{align*} &{\int_{\{|u|^{d}\leq l\}}}| \nabla(|u|^{d+1})| ^{2}\\ &\leq {\int_{\Omega}}| \nabla(u\min\{| u| ^{d},l\})| ^{2}\\ &\leq(2+d)(\| u\| _{L^{2d+p}}^{2d+p}+\| h\| _{L^{2}}\| u\| _{L^{2}}+M\| u\| _{L^{2} }^{\alpha(2d+2)}\| u\| _{L^{2d+p}}^{(1-\alpha) (2d+2) }) . \end{align*} Letting $l\to\infty$, we obtain \begin{align*} &\| \nabla(| u| ^{d+1}) \| _{L^{2}}^{2}\\ &\leq(2+d)\Big(\| u\| _{L^{2d+p}}^{2d+p}+\| h\| _{L^{2}}\| u\| _{L^{2}}+M\| u\| _{L^{2} }^{\alpha(2d+2)}\| u\| _{L^{2d+p}}^{(1-\alpha) (2d+2) }\Big) . \end{align*} Since $\| u\| _{L^{(d+1)2^{*}}}^{2(d+1)}=\| | u| ^{d+1}\| _{L^{2^{*}}}^{2}\leq S\| \nabla(| u| ^{d+1}) \| _{L^{2}}^{2},$ we have \begin{align*} &\| u\| _{L^{(d+1)2^{*}}}^{2(d+1)}\\ &\leq S(2+d)\Big[ \| u\| _{L^{2d+p}}^{2d+p}+\| h\| _{L^{2} }\| u\| _{L^{2}}+M\| u\| _{L^{2}}^{\alpha (2d+2)}\| u\| _{L^{2d+p}}^{(1-\alpha) (2d+2) }\Big] . \end{align*} Let $d_{0}=0$ and $2d_{i}+p=(d_{i-1}+1) 2^{*}$ for $i\geq1$. Since $p<2^{*}$, then $d_{1}>0$ and $d_{i}\geq(\frac{2^{*}}{2}) ^{i-1}d_{1}$. Hence, $\lim_{n\to\infty}d_{i}=\infty$ and $\| u\| _{L^{2d_{i}+p}}=\| u\| _{L^{(d_{i-1}+1)2^{*} }}\leq p_{i-1}(\| u\| _{L^{2d_{i-1}+p}}) .$ By iterating, we conclude that $\| u\| _{L^{2d_{i}+p}}\leq p_{i}(\| u\| _{H^{1}}) .$ By the interpolation property, for every $q\in[ 2,\infty)$, we have $u\in L^{q}$ that satisfies $\| u\| _{L^{q}}\leq p_{q}(\| u\| _{H^{1}}) .$ $(ii)$ Let $g(z,u)=|u|^{p-2}u+h(z)$, since $(p-1)s>s>N$, then by $(i)$, $|u|^{p-2}u\in L^{s}(\Omega)\cap L^{2}(\Omega)$. Thus, $g(z,u)\in L^{s} (\Omega)\cap L^{2}(\Omega)$. By Theorem \ref{s3}, the Dirichlet problem \begin{gather*} -\Delta v+v=g(z,u)\quad\mbox{in }\Omega,\\ v\in W_{0}^{1,s}(\Omega)\cap H_{0}^{1}(\Omega), \end{gather*} has a unique strong solution $v\in W^{2,s}(\Omega)\cap W_{0}^{1,s}(\Omega)\cap H_{0}^{1}(\Omega)$ and $\| v\| _{W^{2,s}(\Omega)}\leq c\| g(z,u)\| _{L^{s}(\Omega)}.$ Since $v\in H_{0}^{1}(\Omega)$ satisfies Equation $(\ref{E3})$, we have for each $\varphi\in C_{c}^{\infty}(\Omega)$, $\int_{\Omega}(\nabla v\nabla\varphi+v\varphi) =\int_{\Omega }g(z,u)\varphi.$ Thus $u$ and $v$ satisfy weakly \begin{gather*} -\Delta v+v=g(z,u)\quad\mbox{in }\Omega,\\ -\Delta u+u=g(z,u)\quad\mbox{in }\Omega. \end{gather*} Let $w=v-u$. Then $-\Delta w+w=0$ in $\Omega$. By Theorem \ref{s7}, $w=0$, or $u=v\in W^{2,s}(\Omega)\cap W_{0}^{1,s}(\Omega)\cap H_{0}^{1}(\Omega)$ and $\| u\| _{W^{2,s}(\Omega)}\leq c\| g(z,u)\| _{L^{s}(\Omega)}.$ Now $s>N$ and $\theta=2-\frac{N}{s}-[ 2-\frac{N}{s}]$. Then by the Sobolev embedding theorem \ref{p23}, $u\in C^{1,\theta}(\overline{\Omega })$ and $\| u\| _{L^{\infty}(\Omega)}\leq\| u\| _{C^{1,\theta}(\overline{\Omega})}\leq c\| u\| _{W^{2,s}(\Omega)}.$ $(iii)$ By $(ii)$, we know that $u\in C^{1,\theta}(\overline{\Omega })$. Since $u$ is bounded, then $| u| ^{p-2}u\in C^{\theta}(\overline{\Omega})$. By Theorem \ref{s6}, we have $u\in C^{2,\theta}(\overline{\Omega})$. \end{proof} \subsection{Asymptotic Behavior of Solutions} By Theorem \ref{s8}, we obtain the following three results about asymptotic behavior of solutions. We define the generalized infinite strip $\mathbf{S}^{r}=B^{m}(0;r)\times\mathbb{R}^{n}$, where $N\geq4$, $m\geq2$, $n\geq1$ and $m+n=N$. Let $\lambda_{1}$ be the first eigenvalue of $-\Delta$ in $B^{m}(0;r)$ with the Dirichlet problem, and $\phi_{1}$ the corresponding positive eigenfunction to $\lambda_{1}$. \begin{theorem} \label{s9} $(i)$ Let $h\in L^{N/2}(\mathbf{S}^{r})\cap L^{s}(\mathbf{S} ^{r})\cap L^{2}(\mathbf{S}^{r})$ for $s>N$. If $u$ in $H_{0}^{1} (\mathbf{S}^{r})$ is a weak solution of \eqref{E2} in $\mathbf{S} ^{r}$, then $u\in C^{1}(\overline{\mathbf{S}^{r}})$ and $\lim_{|y|\to\infty}u(x,y)=0\;\quad\text{uniformly in}x\in B^{m}(0;r);$ $(ii)$ Let $h\in L^{N/2}(\mathbf{F}_{s}^{r})\cap L^{s}(\mathbf{F}_{s}^{r})\cap L^{2}(\mathbf{F}_{s}^{r})$ for $s>N$. If $u$ in $H_{0}^{1}(\mathbf{F}_{s} ^{r})$ is a weak solution of \eqref{E2} in $\mathbf{F}_{s}^{r}$, then $u\in C^{1}(\overline{\mathbf{F}_{s}^{r}})$ and $\lim_{y\to\infty}u(x,y)=0\quad \text{uniformly in} x\in B^{m}(0;r);$ $(iii)$ Let $\Omega$ be an unbounded domain in $\mathbb{R}^{N}$, $h\in L^{N/2}(\Omega)\cap L^{s}(\Omega)\cap L^{2}(\Omega)$ for $s>N$. If $u$ in $H_{0}^{1}(\Omega)$ is a weak solution of \eqref{E2} in $\Omega$, then $u\in C^{1}(\overline{\Omega})$ and $\lim_{|z|\to\infty}u(z)=0$. \end{theorem} \begin{proof} $(i)$ By Theorem \ref{s8}, $u\in C^{1}(\overline{\mathbf{S}^{r}})\cap W^{2,s}(\mathbf{S}^{r})$. For each $t>0$, apply Theorem \ref{s8} $(ii)$ to obtain $\| u\|_{L^{\infty}(\mathbf{S}_{t}^{r})}\leq c\| u\|_{W^{2,N}(\mathbf{S}_{t}^{r})},$ where $\mathbf{S}_{t}^{r}=\{z=(x,y)\in\mathbf{S}^{r}: |y|>t\}$ . Since $\| u\|_{W^{2,N}(\mathbf{S}_{t}^{r})}= o(1)$ as $t\to\infty$, we obtain $\lim_{|y|\to\infty}u(x,y)=0\quad \text{uniformly in }x\in B^{m}(0;r).$ The proofs of $(ii)$ and $(iii)$ are similar to $(i)$. \end{proof} By Lien-Tzeng-Wang \cite{LTW}, there is a positive solution of \eqref{E1} in $\mathbf{S}^{r}$. Such a solution admits exponential decay in $y$. \begin{theorem} \label{s10} Let $u$ be a positive solution of Equation $\eqref{E1}$ in $\mathbf{S}^{r}$. Then for every $0<\delta<1+\lambda_{1}$ $\gamma>0$ and $\beta>0$ exist such that $\gamma\phi_{1}(x)e^{-\sqrt{1+\lambda_{1}+\delta}\,|y|}\leq u(z)\leq\beta \phi_{1}(x)e^{-\sqrt{1+\lambda_{1}-\delta}\,|y|}\quad\text{for }\,z=(x,y)\in \mathbf{S}^{r}.$ \end{theorem} \begin{proof} By Theorem \ref{s8} $(iii)$, $u\in C^{2}(\overline {\Omega})$.\newline $(i)$ $\gamma\phi_{1}(x)e^{-\sqrt{1+\lambda _{1}+\delta}\,|y|}\leq u(z)$ for$\,\,\,z=(x,y)\in\mathbf{S}^{r}:$ define $w_{\delta}(z)=\phi_{1}(x)e^{-\sqrt{1+\lambda_{1}+\delta}\,|y|}\quad \text{for }z=(x,y) \in\overline{B^{m}(0;r) } \times\mathbb{R}^{n}=\overline{\mathbf{S}^{r}}.$ For $0<\delta<1+\lambda_{1}$, take $R>0$ such that $\delta-\frac {\sqrt{1+\lambda_{1}+\delta}(n-1) }{| y| }\geq0$ for $| y| \geq R$ (for $n=1$, take $R=1$). Set $\gamma =\inf_{z\in\mathbf{S}^{r},\,|y|\leq R } \frac{u(x,y)}{w_{\delta}(x,y)}.$ Note that $w_{\delta}(x,y)$ and $u(x,y)$ are radially symmetric in $x$ and $y$, and decreasing in $|y|$ for a fixed $x$. Thus ${ \frac{u(x,y)}{w_{\delta}(x,y)}=}\frac{u(x,y)}{\phi_{1}( x) }e^{\sqrt{1+\lambda_{1}+\delta}| y| }\geq\frac{u( x,Re_{1}) }{\phi_{1}(x) },\quad \quad\text{for }| y| \leq R, \; x\in B^{m}(0;r) ,$ where $e_{1}=(1,0,\dots,0) \in\mathbb{R}^{n}$. Therefore $\gamma =\inf_{z\in\mathbf{S}^{r},\, |y|\leq R } \frac{u(z)}{w_{\delta}(z)}\geq\inf_{x\in B^{m}(0;r) } \frac{u(x,Re_{1}) }{\phi_{1}(x) }=\inf_{x\in L}\frac{u(x,Re_{1}) }{\phi_{1}(x) },$ where $L$ is a fixed diameter of $B^{m}(0;r)$. Note that ${ \frac{u(x,Re_{1}) }{\phi_{1}(x)}>0}\quad \text{for } x\in L.$ Furthermore, for each $x_{0}\in\partial L\subset\partial B^{m}( 0;r)$, take a small ball $B^{1}$ in $B^{m}(0;r)$ such that $x_{0}\in\partial B^{1}$. Note that $\phi_{1}(x) >0$ for $x\in B^{1}$, $\phi_{1}(x_{0}) =0$, and $\phi_{1}( x) \in C^{2}(\overline{B^{m}(0;r) })$. Then by Lemma \ref{a1} below, $\frac{\partial\phi_{1}}{\partial\nu}(x_{0})<0$, where $\nu$ is the outward unit normal vector of $B^{1}$ at $x_{0}$. Let $u_{1}( x) =u(x,Re_{1})$, and for each $z_{1}=( x_{0},Re_{1}) \in\partial L\times\mathbb{R}^{n}\subset\partial \mathbf{S}^{r}$, take a small ball $B^{2}$ in $\mathbf{S}^{r}$ such that $z_{1}\in\partial B^{2}$. Note that $u(z) >0$ for $z\in B^{2}$, and $u(z_{1}) =0$. By Theorem \ref{s9}, $u(x) \in C^{2}(\overline{\mathbf{S}^{r}})$, then by Lemma \ref{a1} below, $\frac{\partial u}{\partial\bar{\nu}}(z_{1})<0$, where $\bar{\nu}=( \nu,0)$ is the outward unit normal vector of $B^{2}$ at $z_{1}$. Thus, $\frac{\partial u_{1}}{\partial\nu}(x_{0}) =\nabla u_{1}( x_{0}) \cdot\nu=\nabla u(z_{1}) \cdot\bar{\nu} =\frac{\partial u}{\partial\bar{\nu}}(z_{1})<0$. By L'H\^{o}pital's rule, we have $\lim_{\substack{x\in B^{m}(0;r) \\x\to x_{0} \quad\text{normally} }}\frac{u(x,Re_{1}) }{\phi_{1}( x) }=\lim_{h\to0^{-}}\frac{u_{1}(x_{0}+h\nu)}{\phi_{1} (x_{0}+h\nu)}=\frac{\frac{\partial u_{1}}{\partial\nu}(x_{0})}{\frac {\partial\phi_{1}}{\partial\nu}(x_{0})}>0.$ Define ${ \frac{u(x,Re_{1}) }{\phi_{1}(x_{0})}=} \lim_{\substack{x\in B^{m}(0;r) \\x\to x_{0} \quad\text{normally} }} \frac{u(x,Re_{1}) }{\phi_{1}( x) }.$ Thus $\frac{u(x,Re_{1}) }{\phi_{1}(x)}>0$ for $x\in\overline{L}$. Since $\frac{u(x,Re_{1}) }{\phi_{1}(x)}:\overline {L}\to\mathbb{R}$ is continuous, we have $\gamma{ \geq\inf_{x\in\bar{L}}\frac{u(x,Re_{1}) }{\phi_{1}(x) }>0.}$ Let $v(z)=\gamma w_{\delta}(z)$ for $z\in\overline{\mathbf{S}^{r}}$, then we have $\ v(z)\leq u(z)\;\text{for}\text{\;}z\in\overline{\mathbf{S}^{r}},\ |y|\leq R.$ For $z\in\mathbf{S}^{r},\ |y|>R$, we have \begin{align*} -\Delta(u-v)(z)+(u-v)(z) & =(-\Delta u(z)+u(z))+(\Delta v(z)-v(z))\\ & =u^{p-1}+v(z) (\delta-\frac{\sqrt{1+\lambda_{1}+\delta }(n-1) }{| y| }) \geq0. \end{align*} (For $n=1$, we only consider the domain $\{ z\in\mathbf{S} ^{r}|\ y>R\} .$) Since $u(z)-v(z)\geq0$ on $\partial A\cup \partial\mathbf{S}^{r}$ where $A=\{ z\in\mathbf{S}^{r} : |y|>R\}$, by Lemma \ref{a3} below, we have $v(z)\leq u(z)$ for $z=(x,y)\in\mathbf{S}^{r}$, $|y|>R$. Thus, we conclude that $v(z)\leq u(z)$ for$\text{\;}z\in\mathbf{S}^{r}$, or $\gamma\phi _{1}(x)e^{-\sqrt{1+\lambda_{1}+\delta}\,|y|}\leq u(z)$.\newline $(ii)$ $u(z)\leq\beta\phi_{1}(x)e^{-\sqrt{1+\lambda_{1}-\delta}\,|y|}$, for $z=(x,y)\in\mathbf{S}^{r}:$ for $0<\delta<1+\lambda_{1}$. By Theorem \ref{s9}, $\underset{|y|\to\infty}{\lim}u(x,y)=0$ uniformly in $x$. Take $R'>0$ such that$\$ $u^{p-2}\leq\frac{\delta}{2+[\lambda_{1} ]}\text{\;}$for$\ |y|\geq R'$, $x\in B^{m}(0;r)$. Define \begin{gather*} w^{\delta}(z)=\phi_{1}(x)e^{-\sqrt{1+\lambda_{1}-\delta}\,|y|}\quad\text{for } z=(x,y) \in\overline{B^{m}(0;r) } \times\mathbb{R}^{n}\mathbb{=}\overline{\mathbf{S}^{r}};\\ \beta =\sup_{z\in\mathbf{S}^{r},\, |y|\leq R' } \frac{u(z)}{w^{\delta}(z)}>0{ ;} \\ \mu(z)=\beta w^{\delta}(z)\quad \text{for }z\in\overline{\mathbf{S}^{r}}. \end{gather*} Fix a diameter $L$ of $B^{m}(0;r)$. We have $\frac{u(z)}{w^{\delta}(z)}={ \frac{u(x,y)}{\phi_{1}(x)} e^{\sqrt{1+\lambda_{1}-\delta}\,|y|}\leq\frac{u(x,0)}{\phi_{1}(x)} e^{\sqrt{1+\lambda_{1}-\delta}\,R'},}$ for $| y|\leq R'$, $x\in B^{m}(0;r)$, and $\beta =\sup_{z\in\mathbf{S}^{r},\,|y|\leq R' } \frac{u(z)}{w^{\delta}(z)}{ \leq \underset{_{x\in B^{m}(0;r) }}{\sup}\frac{u(x,0)}{\phi_{1} (x)}e^{\sqrt{1+\lambda_{1}-\delta}\,R'}=\underset{_{x\in L}}{\sup }\frac{u(x,0)}{\phi_{1}(x)}e^{\sqrt{1+\lambda_{1}-\delta}\,R'}.}$ Similarly to part $(i)$, for $x_{0}\in\partial L\subset\partial B^{m}(0;r)$, we have $\lim_{\substack{x\in B^{m}(0;r) \\x\to x_{0} \quad\text{normally} }}\frac{u(x,0)}{\phi_{1}(x) }=\lim _{h\to0^{-}}\frac{u_{0}(x_{0}+h\nu)}{\phi_{1}(x_{0}+h\nu)}=\frac {\frac{\partial u_{0}}{\partial\nu}(x_{0})}{\frac{\partial\phi_{1}} {\partial\nu}(x_{0})}<\infty,$ where $\nu$ is the outward unit normal vector of $B^{m}(0;r)$ at $x_{0}$ and $u_{0}(x) =u(x,0) .$\ Define ${ \frac{u(x_{0},0)}{\phi_{1}(x_{0}) }=} \lim_{\substack{x\in B^{m}(0;r) \\x\to x_{0} \quad\text{normally} }} \frac{u(x,0)}{\phi_{1}(x) }<\infty.$ Thus, $\beta{ \leq\underset{_{x\in L}}{\sup}\frac{u(x,0)}{\phi_{1} (x)}e^{\sqrt{1+\lambda_{1}-\delta}\,R'}\leq\underset{_{x\in\bar{L}} }{\sup}\frac{u(x,0)}{\phi_{1}(x)}e^{\sqrt{1+\lambda_{1}-\delta}\,R' }<\infty.}$ Therefore, $\mu(z)\geq u(z)$ for $z\in\mathbf{S}^{r}\mathbf{,\;}|y|\leq R'$. For $z\in\mathbf{S}^{r}\mathbf{,\;}|y|>R'$ we have \begin{align*} -\Delta(u-\mu)(z)+(u-\mu)(z) & =(-\Delta u(z)+u(z))+(\Delta\mu(z)-\mu(z))\\ & =u^{p-1}(z)+\big(-\delta-\frac{\sqrt{1+\lambda_{1}-\delta}( n-1) }{| y| }\big) \mu(z)\\ & \leq\frac{\delta}{2+[\lambda_{1}]}(u-\mu)(z), \end{align*} therefore, $-\Delta(u-\mu)(z)+(1-\frac{\delta}{2+[\lambda_{1}]})(u-\mu)(z)\leq0.$ Since $1-\frac{\delta}{2+[\lambda_{1}]}>0$, $u(z)-\mu(z)\leq0$ on $\partial B$, where $B=\{ z\in\mathbf{S}^{r}: |y|>R'\}$, by Lemma \ref{a3} below, $u(z)-\mu(z)\leq0$ in $B$. Thus, we conclude that $u(z)\leq\mu(z)$ for $z\in\mathbf{S}^{r}$. \end{proof} We similarly present the asymptotic behavior of each solution of \eqref{E1} in the interior flask domains $\mathbf{F}_{s}^{r}$, where $s>r$. \begin{theorem} \label{s11} Let $u$ be a positive solution of \eqref{E1} in $\mathbf{F}_{s}^{r}$. Then for any $0<\delta<1+\lambda_{1}$, $\gamma>0$, $\beta>0$, and $R>s$ exist such that for$\,\,\,z=(x,y)\in\mathbf{A}_{R}^{r}$, $\gamma\phi_{1}(x)e^{-\sqrt{1+\lambda_{1}}\,y}\leq u(z)\leq\beta\phi _{1}(x)e^{-\sqrt{1+\lambda_{1}-\delta}\,y}.$ \end{theorem} \begin{proof} $(i)$ $\gamma\phi_{1}(x)e^{-\sqrt{1+\lambda_{1}}\,y}\leq u(z)$ for$\,\,\,z=(x,y)\in\mathbf{A}_{R}^{r}:$ by Theorem \ref{s9} $(ii)$, $\underset{y\to\infty}{\lim}u(x,y)=0$ uniformly in $x$, where $( x,y) \in\mathbf{F}_{s}^{r}$. For $0<\delta<1+\lambda_{1}$, take $R>s$ such that $u^{p-2}(x,y)\leq\frac{\delta}{2+[ \lambda_{1}]}$ for $y\geq R$. In the remaining proofs, we fix such $R$. Define $w(z)=\phi_{1}(x)e^{-\sqrt{1+\lambda_{1}}\,y}\quad\text{for } z=(x,y) \in\overline{\mathbf{A}_{s}^{r}}.$ Set $\gamma =\inf_{z\in\mathbf{A}_{s}^{r},\, y=R } \frac{u(x,y)}{w(x,y)}.$ Similarly to Theorem \ref{s10}, $\gamma>0$. Let $v(z)=\gamma w(z)$ for $z\in\overline{\mathbf{A}_{s}^{r}}$, then we have $\ v(z)\leq u(z)$ for$\text{\;}z\in\overline{\mathbf{A}_{s}^{r}},\ y=R$. For $z\in \mathbf{A}_{R}^{r}$, we have $-\Delta(u-v)(z)+(u-v)(z)=(-\Delta u(z)+u(z))+(\Delta v(z)-v(z))=u^{p-1}\geq0.$ Since $u-v\geq0$ on $\partial\mathbf{A}_{R}^{r}$, by the strong maximum principle, we have $v(z)\leq u(z)$ for $z=(x,y)$, $z\in\mathbf{A}_{R}^{r}$.\newline$(ii)$ $u(z)\leq\beta\phi_{1}(x)e^{-\sqrt{1+\lambda_{1}-\delta} \,y}$, for$\,\,\,z=(x,y)\in\mathbf{A}_{R}^{r}:$ define \begin{gather*} w_{\delta}(z)=\phi_{1}(x)e^{-\sqrt{1+\lambda_{1}-\delta}\,y}\quad \text{for }z=(x,y) \in\overline{\mathbf{A}_{s}^{r}};\\ \beta =\sup_{z\in\mathbf{A}_{s}^{r},\, y=R } \frac{u(z)}{w_{\delta}(z)}>0{ ;}\\ \mu(z)=\beta w_{\delta}(z)\quad \text{for }z\in\overline{\mathbf{A}_{s}^{r}}. \end{gather*} Similarly to Theorem \ref{s10}, $\beta<\infty$. Therefore, $u(z)\leq\mu (z)$ for$\;z\in\overline{\mathbf{A}_{s}^{r}}\mathbf{,\;}y=R$. For $z\in\mathbf{A}_{s}^{r}\mathbf{,\;}y\geq R$, we have \begin{align*} -\Delta(u-\mu)(z)+(u-\mu)(z) & =(-\Delta u(z)+u(z))+(\Delta\mu(z)-\mu(z)) \\ & =u^{p-1}(z)-\delta\mu(z)\ \\ & \leq\frac{\delta}{2+[ \lambda_{1}] }(u-\mu)(z); \end{align*} therefore, $-\Delta(u-\mu)(z)+(1-\frac{\delta}{2+\left[ \lambda_{1}\right] } )(u-\mu)(z)\leq0.$ Since $1-\frac{\delta}{2+\left[ \lambda_{1}\right] }>0$ and $u-\mu\leq0$ on $\partial\mathbf{A}_{R}^{r}$, by the strong maximum principle, we obtain $u(z)\leq\mu(z)$ for $z\in\mathbf{A}_{R}^{r}$. \end{proof} \noindent\textbf{Bibliographical notes:} Theorem \ref{b2} is from Benci-Cerami \cite{BeCe}. Theorem \ref{b4} is from Lien-Tzeng-Wang \cite{LTW} and Gidas-Ni-Nirenberg \cite{GNN2}.The asymptotic behavior results are from Wang \cite{W} and Chen-Chen-Wang \cite{CCW}. \section{Symmetry of Solutions} We use the asymptotic behavior of solutions developed in Section 8 and apply the moving plane'' method to prove the symmetry of solutions to \eqref{E1} in the infinite strip $\mathbf{A}^{r}$. Our approach is similar to those in Gidas-Ni-Nirenberg \cite[Theorem 1]{GNN1} and \cite[Theorem 2]{GNN2} but is more complicated. Before proving our main results, we first establish a version of the Hopf boundary point lemma and the strong maximum principle that will be used in our case. \begin{lemma}[Hopf Boundary Point Lemma]\label{a1} Let $\Omega$ be a domain (possibly unbounded) in $\mathbb{R}^{n}$. Let $L$ be a differential operator given by $Lu=\sum_{i,j=1}^{n}a^{ij}(x)D_{ij}u+\sum_{i=1}^{n}b^{i}(x)D_{i} u+c(x)u,\;\;a^{ij}(x)=a^{ji}(x),$ which is uniformly elliptic, $\frac{|b_{i}(x)|}{\lambda(x)}$ and $\frac {c(x)}{\lambda(x)}$ are uniformly bounded, where $\lambda(x)$ is the minimum eigenvalue of $[a^{ij}(x)]$. Assume $Lu\leq0$. Let $x_{0}\in\partial\Omega$ satisfy:\newline$(i)$ $u$ is continuous at $x_{0}$;\newline$(ii)$ $u(x_{0})u(x)$ for all $x\in\Omega$;\newline$(iii)$ A ball $B\subset\Omega$ exists with $x_{0}\in\partial B$. \newline Suppose that one of the following conditions holds:\newline$(i)$ $u(x_{0})>0$ and $c(x)\leq0$;\newline$(ii)$ $u(x_{0})=0$;\newline$(iii)$ $c(x)\equiv0$.\newline If the outer normal derivative $\frac{\partial u}{\partial\nu}(x_{0})$ of $u$ at $x_{0}$ exists, then $\frac{\partial u}{\partial\nu}(x_{0})>0$. \end{lemma} For the proof of the above lemma, see Gilbarg-Trudinger \cite[Lemma 3.4]{GT}. \begin{lemma}[Strong Maximum Principle]\label{a3} Let $L$ be uniformly elliptic, $c=0$ and $Lu\geq0$ $(\leq0)$ in a domain $\Omega$ (not necessarily bounded). Then if $u$ achieves its maximum (minimum) in the interior of $\Omega$, $u$ is constant. If $c\leq0$ and $c/\lambda$ is bounded, then $u$ cannot achieve a nonnegative maximum (nonpositive minimum) in the interior of $\Omega$ unless it is constant. \end{lemma} For the proof of the above lemma, see Gilbarg-Trudinger \cite[Theorem 3.5]{GT}. We define the generalized infinite strip by $\mathbf{S}^{r} =B^{m}(0;r) \times\mathbb{R}^{n}$, where $m\geq2$, $n\geq1$, and $m+n=N$, and suppose that\newline$(g1)$ $g(u)>0$ as $u>0$; \newline$(g2)$ $g(u)=O(u^{p})$ as $u\to0$ for some $p>1$.\newline Now we consider the equation $$\begin{gathered} -\Delta u+u=g(u)+h(z) \quad \text{in }\mathbf{S}^{r},\\ u>0 \quad \text{in }\mathbf{S}^{r},\\ u=0 \quad \text{on }\partial\mathbf{S}^{r},\\ \lim{|y|\to\infty}u(x,y)=0 \quad \text{uniformly in }x\in B^{m}(0;r) . \end{gathered} \label{E6}$$ We apply the moving plane'' method to prove the symmetry of solutions of (\ref{E6}). \begin{theorem} \label{a5} Assume that $g\in C^{1}$ satisfies $(g1)$ and $h$ is radially symmetric in $x$ and $y$ and strictly decreasing in $| x|$ and $| y|$ . Let $u(x,y)$ be a $C^{2}$ solution of Equation $($\ref{E6}$)$. Then $u$ is radially symmetric in $x$ and in $y$; that is to say, $u(x,y)=u(|x|,|y|)$. \end{theorem} \noindent\textbf{Part I:} $u$ is radially symmetric in $y$.\\ \textbf{Notation:} \begin{gather*} S_{\theta}=\{(x,y_{1},y_{2},\dots,y_{n})\in\mathbf{S}^{r}\mid x\in B^{m}(0;r) ,\;y_{1}=\theta\};\\ \Gamma_{\theta}=\{(x,y_{1},y_{2},\dots,y_{n})\in\mathbf{S}^{r}\mid x\in B^{m}(0;r),\; y_{1}<\theta\}. \end{gather*} For any $(x,y)\in\mathbf{S}^{r}$, set $(x,y^{\theta})=(x,2\theta-y_{1} ,y_{2},\dots,y_{n})$; that is, $(x,y^{\theta})$ is the reflection of $(x,y)$ with respect to $S_{\theta}$; Let $\Theta$ be the collection of all $\theta\in\mathbb{R}$ such that the following statements hold: \begin{gather*} u(x,y)0 \quad \text{on }S_{\theta}. \end{gather*} \begin{lemma} \label{a6} There exists $\theta_{0}>0$ such that either $(-\infty,-\theta_{0}]\subset\Theta$ or $u(x,y)\equiv u(x,y^{-\theta_{0}})$ in $\Gamma_{-\theta_{0}}$. \end{lemma} \begin{proof} Given $\theta<0\mathbf{,}$ set $w^{\theta}(x,y)=u(x,y)-u(x,y^{\theta})$ for $(x,y)\in\Gamma_{\theta}$, and $w^{\theta}(x,y)$ satisfies $$\Delta w^{\theta}(x,y)+c_{\theta}(x,y)w^{\theta}(x,y)=h(x,y^{\theta }) -h(x,y) \geq0,\label{11-1}$$ where $c_{\theta}(x,y)=\frac{g(u(x,y))-g(u(x,y^{\theta}))} {u(x,y)-u(x,y^{\theta})}-1=g'(\xi_{\theta})-1$ and $\xi_{\theta}$ is in between $u(x,y)$ and $u(x,y^{\theta})$. We claim that $\theta_{0}>0$ exists such that if $\theta\leq-\theta_{0}$, then $w^{\theta}(x,y)\leq0$ in $\Gamma_{\theta}$. Otherwise, suppose $w^{\theta}(x,y)>0$ for some $(x,y)\in\Gamma_{\theta}$. Since $\lim{|y| \to\infty}w^{\theta}(x,y)=0$ uniformly in $x$, $w^{\theta }(x,y)$ achieves its maximum at $(x_{\theta},y_{\theta})\in\Gamma_{\theta}$. Then $\nabla w^{\theta}(x_{\theta},y_{\theta})=0,\\quad \{w_{ij}^{\theta}(x_{\theta },y_{\theta})\}\leq0.$ Note that by $(g2)$, $\lim{t\to0^{+}}g'(t)=0$. Take $t_{0}>0$ such that if $00$ such that if $y_{1}\leq-\theta_{0}$, then $u(x,y)\leq t_{0}$, therefore, $g'(u(x,y))<1$. For $\theta\leq-\theta_{0}$, $(x_{\theta},y_{\theta })\in\Gamma_{\theta}$, then $\Delta w^{\theta}(x_{\theta},y_{\theta})\leq0,\;\;\;c_{\theta}(x,y)w^{\theta }(x_{\theta},y_{\theta})=(g'(\xi_{\theta})-1)w^{\theta}(x_{\theta },y_{\theta})<0,$ contradicting $($\ref{11-1}$)$. As a consequence of the maximum principle and the Hopf boundary point lemma, either $w^{-\theta_{0}}(x,y)\equiv0$ in $\Gamma_{-\theta_{0}}$ or $w^{\theta}(x,y)<0$ in $\Gamma_{\theta}$ and $w_{y_{1}}^{\theta}(x,y)>0$ for $(x,y)\in S_{\theta}$ for $\theta\leq -\theta_{0}$, or $u_{y_{1}}(x,y)>0$ for $(x,y)\in S_{\theta}$. \end{proof} \begin{lemma} \label{a7} If $(-\infty,\theta]\subset\Theta$, then there exists $\varepsilon>0$ such that $[\theta,\theta+\varepsilon)\subset\Theta$. \end{lemma} \begin{proof} Suppose instead that a decreasing sequence $\theta_{k}\to\theta$ and a sequence $\{(x^{k},y^{k})\}$ of points in $\Gamma_{\theta_{k}}$ exist such that $w^{\theta_{k}}(x^{k},y^{k})=u(x^{k},y^{k})-u(x^{k},y^{\theta_{k}})>0$, where $(x^{k},y^{\theta_{k}})$ is the reflection of $(x^{k},y^{k})$ with respect to $S_{\theta_{k}}$. There is a subsequence $\{(x^{k},y^{k})\}$ such that $x^{k}\to\overline{x}$ as $k\to \infty$. Two possibilities may arise, as shown in Case 1 and 2. \noindent Case 1. $| y^{k}| \to\infty$. As shown in Lemma \ref{a6}, we assume $w^{\theta_{k}}(\widetilde{x^{k}},\widetilde{y^{k}})=\max{(x,y)\in \overline{\Gamma}_{\theta_{k}}}w^{\theta_{k}}(x,y),$ $$\nabla w^{\theta_{k}}(\widetilde{x^{k}},\widetilde{y^{k}})=0,\;\{w_{ij} ^{\theta_{k}}(\widetilde{x^{k}},\widetilde{y^{k}})\}\leq0.\label{11-2}$$ From $\lim{| y| \to\infty}u(x,y)=0$, as in Lemma \ref{a6}, we obtain a contradiction. \noindent Case 2. $y^{k}\to\overline{y}$. We have $(x^{k} ,y^{k})\to(\overline{x},\overline{y})\in\overline{\Gamma_{\theta}}$ , thus $w^{\theta}(\overline{x},\overline{y})\geq0$. Clearly $(\overline {x},\overline{y})\notin\Gamma_{\theta}$. If $(\overline{x},\overline{y})\in S_{\theta}$ , then $u_{y_{1}}(\overline{x},\overline{y})<0$, which contradicts $\theta\in\Theta$. Moreover, $(\overline{x},\overline{y})\notin\partial \mathbf{S}^{r}\cap\overline{\Gamma_{\theta}}$. Note that $w^{\theta}(x,y)$ satisfies Equation $($\ref{11-1}$)$, and by the Hopf boundary point lemma, we obtain $\frac{\partial}{\partial\nu}w^{\theta}(\overline{x},\overline{y})>0$. On the other hand, by taking limits in $($\ref{11-2}$)$, we obtain $\nabla w^{\theta}(\overline{x},\overline{y})=0$, which is a contradiction. We conclude that either Case 1 or Case 2 is impossible. \end{proof} \begin{proof}[Proof of Part I] Let $\sigma=\sup\{\theta \in\mathbb{R}:(-\infty,\theta)\subset\Theta\}$. Then $\sigma\notin\Theta$. If not, by Lemma \ref{a7} we would have $[\sigma,\sigma+\epsilon)\subset \Theta$, which contradicts the definition of $\sigma$. We claim that $\sigma=0$. Suppose instead that $\sigma\in(-\infty,0)$. By continuity, $u(x,y)\leq u(x,y^{\sigma})$ for all $(x,y)\in\Gamma_{\sigma}$, then by the maximum principle, we have $u(x,y)\equiv u(x,y^{\sigma})$ for all $(x,y)\in\Gamma_{\sigma}$. This implies that $h(x,y) =h( x,y^{\sigma})$ for all $(x,y)\in\Gamma_{\sigma}$, which is a contradiction. This proves $u(x,y)$ is symmetric with respect to the hyperplane $y_{1}=0$ for all $(x,y)\in\mathbf{S}^{r}$. By reversing the $y_{1}$ axis, we conclude that $u(x,y)$ is symmetric with respect to the hyperplane $S_{0}$. Since the $y_{1}$ direction can be chosen arbitrarily, we conclude that $u(x,y)$ is radially symmetric in $\mathbb{R}^{n}$. \end{proof} \noindent\textbf{Part II:} $u$ is radially symmetric in $B^{m}(0;r)$.\\ \textbf{Notation:} \begin{gather*} T_{\lambda}=\{(x,y)=(x_{1},x_{2},\dots,x_{N-1},y)\in\mathbf{S}^{r}: x_{1}=\lambda\};\\ \Sigma_{\lambda}=\{(x,y)=(x_{1},x_{2},\dots,x_{N-1},y)\in\mathbf{S}^{r}: x_{1}<\lambda\}. \end{gather*} For any $(x,y)=(x_{1},x_{2},\dots,x_{N-1},y)\in\mathbf{S}^{r}$, set $(x^{\lambda},y)=(2\lambda-x_{1},x_{2},\dots,x_{N-1},y)$, that is, $(x^{\lambda},y)$ is the reflection of $(x,y)$ with respect to $T_{\lambda}$. Let $\Lambda$ be the collection of all $\lambda\in(-r,0)$ such that the following statements hold: \begin{gather*} u(x,y)0 \quad \text{on }T_{\lambda}. \end{gather*} \begin{lemma} \label{a8} For some $\delta$ such that $0<\delta0$ for some $(x,y)\in\Sigma _{\lambda}$. Since $\lim{| y| \to\infty}v^{\lambda}(x,y)=0$ uniformly in $x$, $v^{\lambda}(x,y)$ achieves its maximum at $(x_{\lambda},y_{\lambda})\in\Sigma_{\lambda}$. Then $\nabla v^{\lambda}(x_{\lambda},y_{\lambda})=0,\;\;\;\{v_{ij}^{\lambda }(x_{\lambda},y_{\lambda})\}\leq0.$ Note that by $(g2)$, $\lim{t\to0^{+}}g'(t)=0$. Take $t_{0}>0$ such that if $00$ for $(x,y)\in T_{\lambda}$. Hence, $u_{x_{1}}(x,y)>0$ for $(x,y)\in T_{\lambda}$. Then $(-r,-r+\delta)\subset\Lambda$. \end{proof} \begin{lemma} \label{a9} If $(-r,\lambda]\subset\Lambda$, then there is a $\tau>0$ such that $[\lambda,\lambda+\tau)\subset\Lambda$. \end{lemma} \begin{proof} Suppose that a decreasing sequence $\lambda_{k}\to\lambda$ and a sequence $\{(x^{k},y^{k})\}$ of points in $\Sigma_{\lambda_{k}}$ exist such that $v^{\lambda_{k}}(x^{k},y^{k})=u(x^{k},y^{k})-u(x^{\lambda_{k}},y^{k})>0$, where $(x^{\lambda_{k}},y^{k})$ is the reflection of $(x^{k},y^{k})$ with respect to $T_{\lambda}$. There is a subsequence $\{(x^{k},y^{k})\}$ such that $x^{k}\to\overline{x}\in\overline {B^{m}(0;r) }$. Two possibilities may arise as shown in Case 1 and 2: \noindent Case 1. $|y^{k}|\to\infty$. As shown in Lemma \ref{a8}, we assume \begin{gather*} v^{\lambda_{k}}(\widetilde{x^{k}},\widetilde{y^{k}})=\max{(x,y)\in \overline{\Sigma_{\lambda_{k}}}}v^{\lambda_{k}}(x,y),\\ \nabla v^{\lambda_{k}}(\widetilde{x^{k}},\widetilde{y^{k}})=0,\;\;\;\{v_{ij} ^{\lambda_{k}}(\widetilde{x^{k}},\widetilde{y^{k}})\}\leq0. \end{gather*} From $\lim{|y|\to\infty}$ $u(x,y)=0$, as in Lemma \ref{a8}, we obtain a contradiction. \noindent Case 2. $y^{k}\to\overline{y}$. We have $(x^{k},y^{k})\to (\overline{x},\overline{y})\in\overline{\Sigma_{\lambda}}$. Thus, $v^{\lambda }(\overline{x},\overline{y})\geq0$. Clearly $(\overline{x},\overline{y} )\notin\Sigma_{\lambda}$. If $(\overline{x},\overline{y})\in T_{\lambda}$ then $u_{x_{1}}(\overline{x},\overline{y})<0$, which contradicts $\lambda\in \Lambda$. Moreover, $(\overline{x},\overline{y})\notin\partial\mathbf{S} ^{r}\cap\overline{\Sigma_{\lambda}}$, since if $(\overline{x},\overline{y} )\in\partial\mathbf{S}^{r}\cap\overline{\Sigma_{\lambda}}$ then $0=u(\overline {x},\overline{y})\geq u(\overline{x}^{\lambda},\overline{y})>0$, a contraction. We conclude that either Case 1 or Case 2 is impossible. \end{proof} \begin{proof}[Proof of Part II] Let $\mu=\sup\{\lambda \in(-r,0):(-r,\lambda)\subset\Lambda\}$. Then $\mu\notin\Lambda$. If not, by Lemma \ref{a9}, we would have $[\mu,\mu+\epsilon)\subset\Lambda$, which contradicts the definition of $\mu$. We claim that $\mu=0$. Suppose instead that $\mu\in(-r,0)$. By continuity we have $u(x,y)\leq u(x^{\mu},y)$ for all $(x,y)\in\Sigma_{\mu}$. Then by the maximum principle we have $u(x,y)\equiv u(x^{\mu},y)$ for all $(x,y)\in\Sigma_{\mu}$, which is impossible. Thus $\mu=0$. By reversing the $x_{1}$ axis, we conclude that $u(x,y)$ is symmetric with respect to the hyperplane $T_{0}$ and $u_{x_{1}}(x,y)<0$ for $x_{1}>0$. Since the $x_{1}$ direction can be chosen arbitrarily, we conclude that $u(x,y)$ is radially symmetric in $B^{m}(0;r)$. \end{proof} \begin{corollary} \label{a10} Assume that $g\in C^{1}$ satisfies property $( g1)$ and $h(x) \equiv0$. Let $u(x,y)$ be a $C^{2}$ solution of Equation $(\ref{E6})$. Then $u$ is radially symmetric in $x$ and in $y$; that is to say, $u(x,y)=u(|x|,|y|)$. \end{corollary} \begin{proof}[Proof of Part I] $u$ is radially symmetric in $y$. Similarly to Theorem $\ref{a5}$, let $\sigma_{1}=\sup\{\theta\in\mathbb{R}\mid(-\infty ,\theta)\subset\Theta\}$. Note that $\sigma_{1}$ is not necessarily zero. Similarly, $u(x,y)$ is symmetric with respect to the hyperplane $y_{1} =\sigma_{1}$ for all $(x,y)\in\mathbf{S}^{r}$. The same conclusion holds for the other coordinate direction, and we conclude that $u$ is symmetric about each of $n$ planes $y_{j}=\sigma_{j}$ and $\nabla u=0$ only at their intersection. We may now take their intersection as the origin. The same argument may be applied to any unit direction $\gamma$ and we infer that $u$ is symmetric about some plane $B^{m}(0;r) \times\{ y\in\mathbb{R}: y\cdot\gamma=c(\gamma) =\text{const}.\}$ At the point on this plane where $u$ achieves its maximum we have $\nabla u=0$ (since the derivative normal to the plane is zero at every point of the plane). It follows that $c(\gamma) =0$. Thus $u$ is symmetric about every plane through the origin, that is, $u$ is radially symmetric in $y$. In addition, we also conclude that $u_{\rho}<0$ where $\rho=| y|$. \noindent\textbf{Proof of Part II:} Since $u$ is radially symmetric in $B^{m}(0;r)$: the proof is similarly to the proof of Theorem \ref{a5}. \end{proof} \subsection{Open Question:} Are positive solutions of \eqref{E1} in the generalized infinite strip by $\mathbf{S}^{r}$ unique up to a translation? \noindent\textbf{Bibliographical notes:} The results of this section are from Chen-Chen-Wang \cite{CCW}. \section{Nonachieved Domains and Esteban-Lions Domains} In this section we also characterize Esteban-Lions domains. We prove that proper large domains, Esteban-Lions domains, and some interior flask domains are nonachieved. \begin{theorem} \label{n14} Let $\Omega_{2}$ be one of $\mathbf{A}^{r}$, $\mathbf{A} ^{r_{1},r_{2}}$, and $\mathbb{R}^{N}$, and $\Omega_{1}$ a proper large domain of $\Omega_{2}$. Then $\alpha_{1}=\alpha_{2}$, $J$ does not satisfy the (PS)$_{\alpha_{1}}$-condition, and Equation $\eqref{E1}$ does not admit any ground state solution in $\Omega_{1}$. In particular, a proper large domain $\Omega_{1}$ of $\Omega_{2}$ is nonachieved. \end{theorem} \begin{proof} Since $\Omega_{1}$ is a proper large domain of $\Omega_{2}$, by Theorem \ref{f8}, $\alpha_{1}=\alpha_{2}$. Then by Theorem \ref{c3} $(i)$ and $(ii)$, $J$ does not satisfy the (PS)$_{\alpha_{1}}$-condition, and Equation $\eqref{E1}$ does not admit any ground state solution in $\Omega_{1}$. \end{proof} The only solution in an Esteban-Lions domain is trivial. The following lemmas are required. The first lemma is from Protter \cite{Pr}. \begin{lemma} \label{n101} Let $u$ be a $C^{2}$ real-valued function in $\Omega$. Suppose that $\delta_{0}$, $\rho_{0}>0$ and $z_{0}\in\Omega$ exist such that $0<\delta_{0}<\rho_{0}<1$ and $u(z)=0\quad\mbox{in }\{ z\in\mathbb{R}^{N}: \delta_{0} \leq| z-z_{0}| \leq\rho_{0}\} ^{c}.$ Then $m_{0}>0$ and $c>0$ exist such that if $m\geq m_{0}$, then $\int_{\Omega}\rho^{-2m-2}e^{2\rho^{-m}}u^{2}dz\leq\frac{c}{m^{4}}\int_{\Omega }\rho^{m+2}e^{2\rho^{-m}}(\Delta u)^{2}dz,$ where $\rho=| z-z_{0}|$. \end{lemma} \begin{proof} Without loss of generality, we assume that $z_{0}=0$, so that $\rho=| z|$. Let $v=e^{\rho^{-m}}u$. By the inequality $(A+B+C) ^{2}\geq2B(A+C)$, we have \begin{align*} (\Delta u)^{2} & =\big[ e^{-\rho^{-m}}\Delta v+2\nabla v\nabla( e^{-\rho^{-m}}) +v\Delta(e^{-\rho^{-m}}) \big] ^{2}\\ & =[e^{-\rho^{-m}}\Delta v+2me^{-\rho^{-m}}\rho^{-m-2}\sum_{i=1}^{N}z_{i} \frac{\partial v}{\partial z_{i}}\\ &\quad +mv\rho^{-m-2}e^{-\rho^{-m}}(m\rho^{-m}-m-2+N) ]^{2}\\ & \geq4me^{-2\rho^{-m}}\rho^{-m-2}\sum_{i=1}^{N}z_{i}\frac{\partial v}{\partial z_{i}}\big[ \Delta v+mv\rho^{-m-2}(m\rho^{-m} -m-2+N) \big] \end{align*} and \begin{align*} & \rho^{m+2}e^{2\rho^{-m}}(\Delta u)^{2}\\ & \geq4m\sum_{i=1}^{N}z_{i}\frac{\partial v}{\partial z_{i}}\left[ \Delta v+m^{2}v\rho^{-2m-2}-(m+2-N) mv\rho^{-m-2}\right] . \end{align*} Thus, ${\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} \rho^{m+2}e^{2\rho^{-m}}(\Delta u)^{2}dz\geq(I) +( II) -(III) ,$ where \begin{gather*} (I) =4m\sum{i=1}^{N} {\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} z_{i}\frac{\partial v}{\partial z_{i}}\Delta v\,dz\,, \\ (II) =4m^{3} {\sum{i=1}^{N}} {\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} \rho^{-2m-2}vz_{i}\frac{\partial v}{\partial z_{i}}\,dz\,,\\ (III) =4m^{2}{\sum{i=1}^{N}} {\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} \rho^{-m-2}(m+2-N) vz_{i}\frac{\partial v}{\partial z_{i}}\,dz\,. \end{gather*} We claim that $(i)$ $(I) =2m(N-2) \int_{\{ \delta_{0}\leq |z|\leq\rho_{0}\} }| \nabla v| ^{2}dz>0;$ $(ii)$ There exists $m_{1}>0$ such that $(II) \geq m^{4} {\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} \rho^{-2m-2}v^{2}dz\quad\text{for }m\geq m_{1};$ and $(iii)$ for $0<\varepsilon<1$, there exists $m_{2}>0$ such that $(III) \leq\varepsilon m^{4}\big( {\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} \rho^{-2m-2}v^{2}dz\big) \quad \text{for }m\geq m_{2}.$ $(i)$ Since $\frac{\partial v}{\partial z_{i}}=0$ on $\partial\{ \delta_{0}\leq|z|\leq\rho_{0}\}$ for all $i=1,2,\dots,N$ and $\sum_{i=1}^{N} {\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} z_{i}\frac{\partial v}{\partial z_{i}}\triangle vdz=\sum_{i=1}^{N}\sum _{j=1}^{N} {\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} z_{i}\frac{\partial v}{\partial z_{i}}\frac{\partial^{2}v}{\partial z_{j}^{2} }dz.$ For $i\neq j$, by integration by parts \begin{align*} {\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} z_{i}\frac{\partial v}{\partial z_{i}}\frac{\partial^{2}v}{\partial z_{j}^{2} }dz & =- {\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} \frac{\partial v}{\partial z_{j}}\frac{\partial}{\partial z_{j}}( z_{i}\frac{\partial v}{\partial z_{i}}) dz\\ & =- {\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} z_{i}\frac{\partial v}{\partial z_{j}}\frac{\partial^{2}v}{\partial z_{i}\partial z_{j}}dz\\ & = {\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} \big[ (\frac{\partial v}{\partial z_{j}}) ^{2}+z_{i} \frac{\partial v}{\partial z_{j}}\frac{\partial^{2}v}{\partial z_{i}\partial z_{j}}\big] dz, \end{align*} we have ${\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} z_{i}\frac{\partial v}{\partial z_{j}}\frac{\partial^{2}v}{\partial z_{i}\partial z_{j}}dz=-\frac{1}{2} {\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} (\frac{\partial v}{\partial z_{j}}) ^{2}dz.$ Thus, \begin{align*} \sum_{\substack{j=1 \\j\neq i }}^{N} {\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} z_{i}\frac{\partial v}{\partial z_{i}}\frac{\partial^{2}v}{\partial z_{j}^{2}}dz & =\frac{1}{2}\sum_{\substack{j=1 \\j\neq i }}^{N} {\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} (\frac{\partial v}{\partial z_{j}}) ^{2}dz\\ & =\frac{1}{2} {\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} | \nabla v| ^{2}-(\frac{\partial v}{\partial z_{i} }) ^{2}dz. \end{align*} For $i=j$, ${\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} z_{i}\frac{\partial v}{\partial z_{i}}\frac{\partial^{2}v}{\partial z_{j}^{2} }dz=- {\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} \big[ z_{i}\frac{\partial v}{\partial z_{i}}\frac{\partial^{2}v}{\partial z_{i}^{2}}+(\frac{\partial v}{\partial z_{i}}) ^{2}\big]\,dz,$ then we have ${\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} z_{i}\frac{\partial v}{\partial z_{i}}\frac{\partial^{2}v}{\partial z_{j}^{2}}dz=-\frac{1}{2} {\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} (\frac{\partial v}{\partial z_{i}}) ^{2}dz.$ Hence, \begin{align*} &\sum_{i=1}^{N} {\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} z_{i}\frac{\partial v}{\partial z_{i}}\triangle v\,dz \\ & =\sum_{i=1}^{N}\Big( \frac{1}{2} {\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} | \nabla v| ^{2}dz- {\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} (\frac{\partial v}{\partial z_{i}}) ^{2}dz\Big) \\ & =(\frac{N}{2}-1) \int| \nabla v| ^{2}>0. \end{align*} $(ii)$ $m_{1}>0$ exists such that $4m^{4}+4m^{3}-2m^{3}N\geq m^{4}$ for $m\geq m_{1}$. Since $v=0$ on $\partial\{ \delta_{0} \leq|z|\leq\rho_{0}\}$, we have \begin{align*} &{\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} \rho^{-2m-2}vz_{i}\frac{\partial v}{\partial z_{i}}dz \\ & =-{\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} \frac{\partial}{\partial z_{i}}(\rho^{-2m-2}vz_{i}) vdz\\ & =(2m+2){\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} \rho^{-2m-4}v^{2}z_{i}^{2}dz\\ &\quad - {\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} \rho^{-2m-2}vz_{i}\frac{\partial v}{\partial z_{i}}dz- {\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} \rho^{-2m-2}v^{2}dz \end{align*} and \begin{align*} &{\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} \rho^{-2m-2}vz_{i}\frac{\partial v}{\partial z_{i}}dz\\ & =(m+1) {\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} \rho^{-2m-4}v^{2}z_{i}^{2}dz- {\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} \rho^{-2m-2}v^{2}dz. \end{align*} Thus, for $m\geq m_{1}$ we have \begin{align*} (II) & =4m^{3} {\sum_{i=1}^{N}} {\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} \rho^{-2m-2}vz_{i}\frac{\partial v}{\partial z_{i}}dz\\ & =4m^{3}(m+1){\sum_{i=1}^{N}} {\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }}\rho^{-2m-4}v^{2}z_{i}^{2}dz -2m^{3} {\sum_{i=1}^{N}}{\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} \!\rho^{-2m-2}v^{2}dz\\ & =(4m^{4}+4m^{3}-2m^{3}N) {\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} \rho^{-2m-2}v^{2}dz\\ & \geq m^{4} {\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} \rho^{-2m-2}v^{2}dz\quad\text{for }m\geq m_{1}. \end{align*} \newline$(iii)$ Since $\rho_{0}<1$, for $0<\varepsilon<1$ there is an $m_{2}>0$ such that \begin{gather*} 2m^{2}(m+2-N) ^{2}\leq2m^{4}\quad\text{for }m\geq m_{2},\\ \int\rho^{-m-2}v^{2}dz\leq\frac{\varepsilon}{2}\int\rho^{-2m-2}v^{2}dz\quad\text{ for }m\geq m_{2}. \end{gather*} Similarly to part $(ii)$, we have \begin{align*} &{\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} \rho^{-m-2}vz_{i}\frac{\partial v}{\partial z_{i}}dz \\ & =(m+2){\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} \rho^{-m-4}v^{2}z_{i}^{2}dz\\ &\quad -{\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} \rho^{-m-2}vz_{i}\frac{\partial v}{\partial z_{i}}dz -{\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} \rho^{-m-2}v^{2}dz \end{align*} and ${\sum_{i=1}^{N}} {\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} \rho^{-m-2}vz_{i}\frac{\partial v}{\partial z_{i}}dz =\frac{1}{2}(m+2-N) {\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} \rho^{-m-2}v^{2}dz.$ Thus, for $m\geq m_{2}$ we have \begin{align*} (III) & =4m^{2}(m+2-N) {\sum_{i=1}^{N}} {\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} \rho^{-m-2}vz_{i}\frac{\partial v}{\partial z_{i}}dz\\ & =2m^{2}(m+2-N) ^{2} {\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} \rho^{-m-2}v^{2}dz\\ & \leq m^{4} {\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} \rho^{-m-2}v^{2}dz\\ & \leq\varepsilon m^{4} {\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} \rho^{-2m-2}v^{2}dz. \end{align*} Take $m_{0}=\max\{ m_{1},m_{2}\}$. For $m\geq m_{0}$, since $u\in C^{2}$ and $u=0$ on the set $\{ z\in\mathbb{R}^{N}:\delta_{0}\leq|z|\leq\rho_{0}\} ^{c}$, we have \begin{align*} & {\int_{\Omega}} \rho^{m+2}e^{2\rho^{-m}}(\Delta u)^{2}dz\\ & = {\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} \rho^{m+2}e^{2\rho^{-m}}(\Delta u)^{2}dz\\ & \geq(I) +(II) -(III) \\ & \geq m^{4}( {\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} \rho^{-2m-2}v^{2}dz) -\varepsilon m^{4}( {\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} \rho^{-2m-2}v^{2}dz) \\ & =(1-\varepsilon) m^{4}( {\int_{\{ \delta_{0}\leq|z|\leq\rho_{0}\} }} \rho^{-2m-2}v^{2}dz) \\ & =(1-\varepsilon) m^{4}( {\int_{\Omega}} \rho^{-2m-2}v^{2}dz) . \end{align*} Hence, for $m\geq m_{0}$, we have $\int\rho^{-2m-2}e^{2\rho^{-m}}u^{2}dz\leq\frac{c}{m^{4}} {\int} \rho^{m+2}e^{2\rho^{-m}}(\Delta u)^{2}dz.$ \end{proof} Recall the uniqueness of the analytic function: suppose that $f$ is analytic in a domain $\Omega$ in $\mathbb{R}^{2}$. If $f(z_{n})=0$ for some sequence $\{ z_{n}\}$ of distinct points such that $z_{n}\to z_{0}\in\Omega$, then $f\equiv0$ in $\Omega$. We know that $f=u+iv$, where $u$ and $v$ are harmonic functions, but the uniqueness of harmonic functions is not elegant, taking the form below. Let $\delta( z_{0}) =dist(z_{0},\partial\Omega)$. Then we have the following uniqueness result (see Heinz \cite{He}). \begin{theorem} \label{n102} Let $u$ be a $C^{2}$ real-valued function on $\Omega$. Suppose that $u(z)=0$ is in the neighborhood of a point $z_{0}\in\Omega$ and that $M>0$ exists such that $(\Delta u) ^{2}\leq Mu^{2}\quad\text{for any }z\in\Omega.$ Then $u(z)=0$ for any $z\in\Omega$. \end{theorem} \begin{proof} Let $R=\min\{ \frac{1}{2},\frac{1}{4}\delta(z_{0}) \}$ and $\Phi(t) \in C_{c}^{2}([0,\infty) )$ satisfy $\Phi(t) =\begin{cases} 1 & \text{for }0\leq t\leq R,\quad\text{}\\ 0 & \text{for }\frac{3}{2}R\leq t<\infty. \end{cases}$ Let $\widetilde{u}(z) =u(z)\Phi(| z-z_{0}| )$. Note that $2R<$ $\delta(z_{0})$ and $B^{N}( z_{0};2R) \subset\Omega$. Thus, $\widetilde{u}(z)$ is well-defined on $B^{N}(z_{0};2R)$ and $\widetilde{u}( z) \in C_{c}^{2}\big(\overline{B^{N}(z_{0};2R)}\big)$. We also have $\widetilde{u}(z) =\begin{cases} u(z) & \mbox{in }\{ z\in\mathbb{R}^{N}:| z-z_{0}| \leq R\} ,\\ 0 & \mbox{in }\{ z\in\mathbb{R}^{N}:| z-z_{0}|\geq\frac{3}{2}R\} , \end{cases}$ with $\frac{3}{2}R<1$. By Lemma \ref{n101}, $m_{0}>0$ and $c_{0}>0$ exist such that $R^{-2m-2}-c_{0}Mm^{-4}R^{m+2}>0,$ and for all $m\geq m_{0}$ $\int\rho^{-2m-2}e^{2\rho^{-m}}\tilde{u}^{2}dz\leq\frac{c_{0}}{m^{4}}\int \rho^{m+2}e^{2\rho^{-m}}(\Delta\tilde{u})^{2}dz.$ We have \begin{align*} &R^{-2m-2}{\int_{\{ | z-z_{0}| \leq R\} }} e^{2\rho^{-m}}u^{2}dz\\ & \leq {\int_{\{ | z-z_{0}| \leq R\} }} \rho^{-2m-2}e^{2\rho^{-m}}u^{2}dz\\ & \leq {\int_{\{ | z-z_{0}| \leq2R\} }} \rho^{-2m-2}e^{2\rho^{-m}}\widetilde{u}^{2}dz\\ & \leq c_{0}m^{-4}{\int_{\{ | z-z_{0}| \leq2R\} }} \rho^{m+2}e^{2\rho^{-m}}(\Delta\widetilde{u}) ^{2}dz\\ & \leq c_{0}Mm^{-4} {\int_{\{ | z-z_{0}| \leq R\} }} \rho^{m+2}e^{2\rho^{-m}}u^{2}dz +c_{0}m^{-4}{\int_{\{ R\leq| z-z_{0}| \leq2R\} }} \!\rho^{m+2}e^{2\rho^{-m}}(\Delta\widetilde{u}) ^{2}dz\\ & \leq c_{0}Mm^{-4}R^{m+2}{\int_{\{ | z-z_{0}| \leq R\} }} e^{2\rho^{-m}}u^{2}dz\\ &\quad +c_{0}m^{-4}(2R) ^{m+2}e^{2R^{-m}} {\int_{\{ R\leq| z-z_{0}| \leq2R\} }} (\Delta\widetilde{u}) ^{2}dz, \end{align*} or \begin{align*} & (R^{-2m-2}-c_{0}Mm^{-4}R^{m+2}) {\int_{\{ | z-z_{0}| \leq R\} }}e^{2\rho^{-m}}u^{2}dz\\ & \leq c_{0}m^{-4}(2R) ^{m+2}e^{2R^{-m}} {\int_{\{ R\leq| z-z_{0}| \leq2R\} }}(\Delta\widetilde{u}) ^{2}dz. \end{align*} Since $\widetilde{u}(z) \in C_{c}^{2}(\overline {B^{N}(z_{0};2R) })$, we have \begin{align*} & (R^{-2m-2}-c_{0}Mm^{-4}R^{m+2}) e^{2R^{-m}} {\int_{\{ | z-z_{0}| \leq R\} }} u^{2}dz\\ & \leq(R^{-2m-2}-c_{0}Mm^{-4}R^{m+2}) {\int_{\{ | z-z_{0}| \leq R\} }} e^{2\rho^{-m}}u^{2}dz\\ & \leq c_{0}m^{-4}(2R) ^{m+2}e^{2R^{-m}} {\int_{\{ R\leq| z-z_{0}| \leq2R\} }} (\Delta\widetilde{u}) ^{2}dz\\ & \leq c_{0}m^{-4}(2R) ^{m+2}e^{2R^{-m}}C. \end{align*} Thus, ${\int_{\{ | z-a| \leq R\} }} u^{2}(z)dz\leq\frac{c_{0}m^{-4}(2R) ^{m+2}C}{R^{-2m-2} -c_{0}Mm^{-4}R^{m+2}}.$ Let $m\to\infty$, then $u(z)=0$ for any $z\in B^{N}( z_{0};R)$. We claim that $u(\tilde{z}) =0$ for any $\tilde{z}\in\Omega$. For $\tilde{z}\in\Omega$, let $h:\left[ 0,1\right] \to\Omega$ be a path satisfying $h(0) =z_{0}$, $h( 1) =\tilde{z}$. Since $h$ and $\delta$ are continuous functions, then $A=\min\big\{\frac{1}{2},\frac{1}{4}\underset{0\leq t\leq1}{\inf} \delta(h(t) ) \big\} >0.$ Since $\{h(t) : 0\leq t\leq1 \}$ is a compact set, $0=t_{0}0$, and $u\in W^{2,p}(\Omega\cap B^{N}(0;R) )$ be a solution of $$\begin{gathered} -\triangle u=f(u)\quad\mbox{in }\Omega;\\ u=0\quad\mbox{on }\partial\Omega, \end{gathered} \label{13-1}$$ where $f$ is locally Lipschitz continuous on $\mathbb{R}$, and $\Omega$ is a smooth unbounded domain. Let $F(t) =\int_{0}^{t}f(s)ds$, and assume that $\nabla u\in L^{2}(\Omega)$ and $F(u) \in L^{1}(\Omega)$. Then a sequence $\{ R_{k}\} \nearrow\infty$ exists such that \newline$(i)$ (Pohozaev identity) $\int_{\Omega}\{ NF(u)+(1-\frac{N}{2}) | \nabla u| ^{2}\} dz=\lim_{k\to\infty} \frac{1}{2} \int_{\partial\Omega\cap B^{N}(0;R_{k}) }(z\cdot \nu(z) ) | \nabla u| ^{2}ds$ $(ii)$ $\lim_{k\to\infty} \int_{\partial\Omega\cap B^{N}(0;R_{k})} \nu_{i}(z) | \nabla u| ^{2}ds=0\quad\text{for each }1\leq i\leq N,$ where $\nu(z) =(\nu_{1}(z) ,\nu_{2}( z) ,\dots,\nu_{N}(z) )$ is the outward unit normal vector at $z$. \end{lemma} \begin{proof} $(i)$ See Esteban-Lions \cite[Proposition I.1]{EL} .\newline$(ii)$ Let $B_{R}=B^{N}(0;R)$. Then we multiply $(\ref{13-1})$ by $\frac{\partial u}{\partial z_{i}}$ and use integration by parts over $\Omega\cap B_{R}$ to obtain \begin{align*} \int_{\Omega\cap B_{R}}(-\triangle u) \frac{\partial u}{\partial z_{i}} & =\int_{\Omega\cap B_{R}}f(u)\frac{\partial u}{\partial z_{i}}=\int_{\Omega\cap B_{R}}\frac{\partial F(u)}{\partial z_{i}}\\ & =\int_{\partial(\Omega\cap B_{R}) }F(u)\nu_{i}ds\\ & =\int_{\Omega\cap\partial B_{R}}F(u)\frac{z_{i}}{| z| }ds. \end{align*} Note that $\nabla u=\frac{\partial u}{\partial\nu}\nu$ on $\partial\Omega$ and $\nu=\frac{z}{| z| }$ on $\partial B_{R}$. We use the Green first identity and integrate by parts to obtain \begin{align*} \int_{\Omega\cap B_{R}}(-\triangle u) \frac{\partial u}{\partial z_{i}} & =\int_{\Omega\cap B_{R}}\nabla u\nabla( \frac{\partial u}{\partial z_{i}}) -\int_{\partial(\Omega\cap B_{R}) }\frac{\partial u}{\partial\nu}\frac{\partial u}{\partial z_{i} }ds\\ & =\sum{j=1}^{N}\int_{\Omega\cap B_{R}}\frac{\partial u}{\partial z_{j}}\frac{\partial^{2}u}{\partial z_{i}\partial z_{j}}-\int_{\partial( \Omega\cap B_{R}) }\frac{\partial u}{\partial\nu}\frac{\partial u}{\partial z_{i}}ds\\ & =\sum{j=1}^{N}\frac{1}{2}\int_{\Omega\cap B_{R}}\frac{\partial }{\partial z_{i}}(\frac{\partial u}{\partial z_{j}}) ^{2} -\int_{\partial(\Omega\cap B_{R}) }\frac{\partial u}{\partial \nu}\frac{\partial u}{\partial z_{i}}ds\\ & =\sum{j=1}^{N}\frac{1}{2}\int_{\partial(\Omega\cap B_{R}) }(\frac{\partial u}{\partial z_{j}}) ^{2}\nu _{i}ds-\int_{\partial(\Omega\cap B_{R}) }\frac{\partial u}{\partial\nu}\frac{\partial u}{\partial z_{i}}ds\\ & =\frac{1}{2}\int_{\partial\Omega\cap B_{R}}| \nabla u| ^{2} \nu_{i}ds+\frac{1}{2}\int_{\Omega\cap\partial B_{R}}| \nabla u| ^{2}\frac{z_{i}}{| z| }ds\\ & -\int_{\partial\Omega\cap B_{R}}(\frac{\partial u}{\partial\nu }) ^{2}\nu_{i}ds-\int_{\Omega\cap\partial B_{R}}\frac{\partial u}{\partial\nu}\frac{\partial u}{\partial z_{i}}ds\\ & =-\frac{1}{2}\int_{\partial\Omega\cap B_{R}}| \nabla u| ^{2} \nu_{i}ds+\frac{1}{2}\int_{\Omega\cap\partial B_{R}}| \nabla u| ^{2}\frac{z_{i}}{| z| }ds\\ & -\int_{\Omega\cap\partial B_{R}}\frac{\partial u}{\partial\nu}\frac{\partial u}{\partial z_{i}}ds. \end{align*} Thus, we have \begin{align*} \big| -\frac{1}{2}\int_{\partial\Omega\cap B_{R}}| \nabla u| ^{2}\nu_{i}ds\big| & =\big| \int_{\Omega\cap\partial B_{R}}\big[ F(u)\frac{z_{i}}{| z| }-\frac{1}{2}| \nabla u| ^{2}\frac{z_{i}}{| z| }+\frac{\partial u}{\partial\nu} \frac{\partial u}{\partial z_{i}}\big] ds\big| \\ & \leq\int_{\Omega\cap\partial B_{R}}\big(| F(u)| +\frac{1} {2}| \nabla u| ^{2}+| \nabla u| ^{2}\big)ds. \end{align*} Since $F(u)\in L^{1}(\Omega)$ and $\nabla u\in L^{2}( \Omega)$, \begin{align*} \infty >{\int_{\Omega}} (| F(u)| +\frac{3}{2}| \nabla u| ^{2})dz & =\lim_{R\to\infty}{\int_{\Omega\cap B_{R}}} (| F(u)| +\frac{3}{2}| \nabla u| ^{2})dz\\ & ={\int_{0}^{\infty}} r^{N-1}\big[{\int_{\Omega\cap\partial B_{r}}} (| F(u)| +\frac{3}{2}| \nabla u| ^{2})ds\big] dr\\ & ={\int_{0}^{\infty}}r^{N-1}M(r)dr, \end{align*} where $M(r)= {\int_{\Omega\cap\partial B_{r}}} (| F(u)| +\frac{3}{2}| \nabla u| ^{2})ds$. Suppose that for every sequence $R_{k}\to\infty$ we have $M(R_{k})\nrightarrow 0$ as $k\to\infty$, that is, a subsequence $\{ R_{k}\}$ and $c>0$ exist such that $M(R_{k})\to c$ as $n\to\infty$. Then for sufficiently large $k$ we obtain $M(R_{k})\geq\frac{c}{2}>0$. Thus, ${\int_{0}^{\infty}} r^{N-1}M(r)dr\geq\frac{c}{2}\int_{k}^{\infty}r^{N-1}dr=\infty.$ This is a contradiction. We conclude that $R_{k}\to\infty$ exists such that $\lim_{k\to\infty}{\int_{\Omega\cap\partial B_{R_{k}}}} (| F(u)| +\frac{3}{2}| \nabla u| ^{2}) ds=0,$ that is, $\lim_{k\to\infty}{\int_{\Omega\cap\partial B_{R_{k}}}} | \nabla u| ^{2}\nu_{i}ds=0\quad\text{for }1\leq i\leq N.$ \end{proof} \begin{lemma} \label{n105} Let $Lu=a^{ij}(z)D_{ij}u+b^{i}(z)D_{i}u+c(z)u$ and $u\in W_{\rm loc}^{2,p}(\Omega)$ be a solution of the elliptic equation $Lu=f$ in a domain $\Omega$, where the coefficients of $L$ belong to $C^{k-1,\alpha}(\Omega)$, $f\in C^{k-1,\alpha}( \Omega)$ with $10$. Thus, $\delta>0$ exists such that $\nu( z) \cdot\chi>0$ for $z\in\partial\Omega\cap B^{N}(z_{0};\delta)$. Then we have $\nabla u=0$ on $\partial\Omega\cap B^{N}(z_{0};\delta)$. Let $\widetilde{u}(z)=\begin{cases} u(z) & \text{for }z\in\overline{\Omega};\\ 0 & \text{for }z\in B^{N}(z_{0};\delta) \backslash \overline{\Omega}. \end{cases}$ \begin{figure}[htb] \begin{center} \includegraphics[width=0.7\textwidth]{fig06} \end{center} % \centering \resizebox{2.5in}{!}{\includegraphics{./fig06.eps}} \caption{ Esteban-Lions domain with a small ball.} \label{fig:fig06} \end{figure} We claim that $\widetilde{u}$ is twice weakly differentiable. Since $u\in C^{2}(\overline{\Omega})$ and $u=0$ on $\partial\Omega$, for each $\varphi\in C_{c}^{1}(\Omega\cup B^{N}(z_{0};\delta) )$, we have \begin{align*} \int_{\Omega\cup B^{N}(z_{0};\delta) }\widetilde{u} \frac{\partial\varphi}{\partial z_{i}} & =\int_{\Omega}u\frac{\partial \varphi}{\partial z_{i}}\\ & =-\int_{\Omega}\frac{\partial u}{\partial z_{i}}\varphi+\int_{\partial \Omega}u\varphi\nu_{i}ds\\ & =-\int_{\Omega\cup B^{N}(z_{0};\delta) }v_{i}\varphi, \end{align*} where $v_{i}(z)=\begin{cases} \frac{\partial u}{\partial z_{i}} & \text{for }z\in\Omega;\\ 0 & \text{for }z\in B^{N}(z_{0};\delta) \backslash\Omega. \end{cases}$ Thus, $v_{i}=\frac{\partial\widetilde{u}}{\partial z_{i}}$. Similarly, we have $w_{ij}=\frac{\partial^{2}\widetilde{u}}{\partial z_{i}\partial z_{j}}$, where $w_{ij}(z)=\begin{cases} \frac{\partial^{2}u}{\partial z_{i}\partial z_{j}} & \text{for }z\in \Omega;\\ 0 & \text{for }z\in B^{N}(z_{0};\delta) \backslash\Omega. \end{cases}$ Therefore, $\widetilde{u}\in W_{\rm loc}^{2,2}(\Omega)$. Next, since $u\in C^{2}(\overline{\Omega})$ and by the Green first identity, for each $\varphi\in C_{c}^{\infty}(\Omega\cup B^{N}(z_{0} ;\delta) )$, we have \begin{align*} \int_{\Omega\cup B^{N}(z_{0};\delta) }\nabla\widetilde{u} \nabla\varphi & =\int_{\Omega}\nabla u\nabla\varphi\\ & =-\int_{\Omega}(\triangle u) \varphi+\int_{\partial\Omega }\frac{\partial u}{\partial\nu}\varphi ds\\ & =\int_{\Omega}f(u)\varphi+\int_{\partial\Omega\cap B^{N}(z_{0} ;\delta) }\frac{\partial u}{\partial\nu}\varphi ds+\int_{\partial \Omega\backslash B^{N}(z_{0};\delta) }\frac{\partial u} {\partial\nu}\varphi ds\\ & =\int_{\Omega\cup B^{N}(z_{0};\delta) }f(\widetilde{u} )\varphi. \end{align*} Hence, $\widetilde{u}$ is a weak solution of \begin{gather*} -\triangle\widetilde{u}=f(\widetilde{u}) \quad \text{in }\Omega\cup B^{N}( z_{0};\delta) ;\\ \widetilde{u}=0 \quad \text{on }\partial(\Omega\cup B^{N}(z_{0};\delta) ) . \end{gather*} We claim that $f\circ\widetilde{u}$ is locally H\"{o}lder continuous with exponent $\alpha$ in $\Omega\cup B^{N}(z_{0};\delta)$, where $0<\alpha<1$. That is, for each compact set $K\subset\Omega\cup B^{N}( z_{0};\delta)$, we must show that a constant $C(K) >0$ exists such that $$| f\circ\widetilde{u}(z) -f\circ\widetilde{u}( \overline{z}) | \leq C(K) | z-\overline {z}| ^{\alpha}\label{13-2}$$ for all $z,\overline{z}\in K$. Since $f$ is locally Lipschitz with $f(0)=0$, and $u\in C^{2}(\overline{\Omega})$, for $z,\overline{z}\in K$, we have \newline$(i)$ (\ref{13-2}) holds for $z,\overline{z}\in\{ B^{N}(z_{0};\delta) \backslash\Omega\} \cap K$ ;\newline$(ii)$ if $z\in\Omega$ and $\overline{z}\in\{ B^{N}( z_{0};\delta) \backslash\Omega\} \cap K$, then $\widehat{z} \in\partial\Omega$ exists such that $| z-\widehat{z}| \leq| z-\overline{z}|$. Thus, \begin{align*} | f\circ\widetilde{u}(z) -f\circ\widetilde{u}( \overline{z}) | & \leq| f\circ\widetilde{u}( z) -f\circ\widetilde{u}(\widehat{z}) | +| f\circ\widetilde{u}(\widehat{z}) -f\circ\widetilde{u}( \overline{z}) | \\ & \leq C_{1}(K) | \widetilde{u}(z)-\widetilde{u}(\widehat{z}) | \\ & \leq C_{2}(K) | z-\widehat{z}| \leq C_{2}( K) | z-\overline{z}| \\ & \leq C_{2}(K) | z-\overline{z}| ^{1-\alpha}| z-\overline{z}| ^{\alpha}\\ & \leq C_{3}(K)| z-\overline{z}| ^{\alpha}; \end{align*} $(iii)$ it is clear that both $z$ and $\overline{z}$ are in $\Omega\cap K$.\newline By Lemma \ref{n105}, we have $\widetilde{u}\in C^{2}( \Omega\cup B^{N}(z_{0};\delta) )$. Finally, since $f$ is locally Lipschitz and $f(0)=0$, then $(\triangle\widetilde{u}) ^{2}=| f(\widetilde{u})| ^{2}\leq C(K) ( \widetilde{u}(z)) ^{2}$ on each compact subset $K$ of $\Omega\cup B^{N}(z_{0};\delta)$. By Theorem \ref{n102}, $\widetilde {u}\equiv0$ on $K$, and $\widetilde{u}\equiv0$ on $\Omega\cup B^{N}( z_{0};\delta)$. Otherwise, if there is a $z\in\Omega$ such that $\widetilde{u}(z)\not =0$, then a bounded domain $\Omega_{1}$ exists such that $z\in\Omega_{1}\subset\overline{\Omega}_{1}\subset\Omega\cup B^{N}( z_{0};\delta)$, and by the previous argument, $\widetilde{u}\equiv0$ on $\overline{\Omega}_{1}$, which is a contradiction. Hence, $u\equiv0$. \end{proof} Esteban-Lions \cite[Theorem I.1]{EL} proved the following result. \begin{theorem} \label{n3} Equation \eqref{E1} in an Esteban-Lions domain $\Omega$ does not admit any nontrivial solution. In particular, an Esteban-Lions domain is a nonachieved domain. \end{theorem} We have the following lemma. \begin{lemma} \label{n4} $(i)$ Esteban-Lions domains are invariant under any rigid motions;\newline$(ii)$ If $\Omega$ is an Esteban-Lions domain, then $\overline{\Omega}^{c}$ is also an Esteban-Lions domain. \end{lemma} \begin{proof} $(i)$ Clearly.\newline$(ii)$ Note that $n(x)\cdot\chi=-n(x)\cdot-\chi$. \end{proof} In fact, in a star-shaped domain $\Omega\subset\mathbb{R}^{N}$, there is a fixed point $z_{0}\in\Omega$ such that the segment $\overline{zz_{0}} \$is contained in $\Omega$ for each $z\in\Omega$. We assert that an Esteban-Lions domain $\Omega$ is an infinite star-shaped domain in the sense that it is $v-$convex for some direction $v$: if $z_{1},z_{2}\in\Omega$ with $z_{1}-$ $z_{2}=tv$ for some $t\in\mathbb{R}$, then the segment $\overline {z_{1}z_{2}}$ is contained in $\Omega$. This is a consequence of the following lemma. \begin{lemma} \label{n5} An Esteban-Lions domain $\Omega$ with $\chi$ as in Definition \ref{f11} in $\mathbb{R}^{N}$ is $\chi-$convex. \end{lemma} \begin{proof} Let $\Omega$ be an Esteban-Lions domain in $\mathbb{R}^{N}$. Without loss of generality, we may assume that $\chi=(0,\dots,0,-1)\in\mathbb{R}^{N}$ satisfying $n(z) \cdot\chi\geq0$, and $n(z) \cdot\chi\not \equiv 0$ for each $z\in\partial\Omega$. For each $z_{1} \in\Omega$, we claim that $\{ z_{1}-\lambda\chi| \lambda \geq0 \} \subset\Omega$. Otherwise, set $\lambda_{0} =\inf\{ \lambda\geq0| z_{1}-\lambda\chi\not \in \Omega \}$. Then $\lambda_{0}>0$ and the point $z_{0}=z_{1}-\lambda_{0} \chi\in\partial\Omega$. There are only two possibilities: \newline$(i)$ the curve $\partial\Omega$ is transverse with the $x_{N}-$axis at $z_{0}$, and then $n(z_{0}) \cdot\chi<0$;\newline$(ii)$ the curve $\partial\Omega$ is tangent with the $x_{N}-$axis at $z_{0}$, and then $z\in\partial\Omega$ near $z_{0}$ exists such that $n(z) \cdot\chi<0$.\newline Both $(i)$ and $(ii)$ contradict the definition of $\chi$. We conclude that an Esteban-Lions domain $\Omega$ in $\mathbb{R} ^{N}$ is $\chi-$convex. \end{proof} In $\mathbb{R}^{2}$, let an upper semi-strip $\Omega_{1}$, a first quadrant $\Omega_{2}$, a second quadrant $\Omega_{3}$, and an upper half plane $\Omega_{4}$ be defined as follows: \begin{gather*} \Omega_{1}=\{ z=(x,y)\in\mathbb{R}^{2}: a0$,$x\in\Omega$exists such that$B^{N}(x,r) \subset\Omega$which implies$\tilde{B}^{N}(x,r) \subset\tilde{\Omega}$, where$\tilde{B}^{N}(x,r) $and$\tilde{\Omega}$are the projections of$B^{N}(x,r) \ $and$\Omega$, respectively. On the other hand, suppose$\tilde{\Omega}$is a large domain in$\mathbb{R}^{N-1}$. Then for$r>0$,$x\in\Omega$exists such that$\tilde{B}^{N}(x,r) \subset\tilde{\Omega}$, which means that for any$\tilde{y}\in\tilde{B} ^{N}(x,r) $,$y\in\Omega$exists and$\tilde{y}$is the projection of$y$. By Lemma \ref{n5},$\lambda>0$exists such that$\{ \tilde{y}-\chi t: t\geq\lambda\} \subset\Omega$. Let $\bar{\lambda}=\inf{\tilde{y}\in\tilde{B}^{N}(x,r) }\{ \lambda>0: \tilde{y}-\chi t\in\Omega\quad\text{for } t\geq\lambda\} ,$ then$B^{N}(z,r) \subset\Omega$, where$z=\tilde{x}-( r+1+\bar{\lambda}) \chi$. Thus,$\Omega$is a large domain in$\mathbb{R}^{N}$. \end{proof} In$\mathbb{R}^{3}$, there is an Esteban-Lions domain that is not a large domain. \begin{example} \label{n12}\rm Let$\Omega$be a domain in$\mathbb{R}^{3}$such that$\Omega$contains the point$(0,0,1)$with the boundary $\partial\Omega=\{ ((u+2) \cos v,\quad\text{}\sin v,\quad\text{}u) \in\mathbb{R}^{3}\ | \ 0\leq v\leq2\pi,\quad\text{} u\geq0 \} .$ Then$\Omega$is an Esteban-Lions domain in$\mathbb{R}^{3}$with$\chi=(0,0,-1) $, but it is not a large domain in either$\mathbb{R}^{3}$or$\mathbf{A}^{r}$. \end{example} \begin{figure}[htb] \begin{center} \includegraphics[width=0.5\textwidth]{fig10} \end{center} % \centering \resizebox{2.5in}{!}{\includegraphics{./fig10.eps}} \caption{ Esteban-Lion domain but not a large domain.} \label{fig:fig10} \end{figure} \begin{proof} Let$p(u,v) =((u+2) \cos v,\sin v,u) $. Then $p_{v}=(-(u+2) \sin v,\cos v,0)$ and$p_{u}=(\cos v,0,1) $. For$x\in\partial\Omega$, we have the outer normal vector$n(x) =p_{v}\times p_{u}=(\cos v,(u+2) \sin v,-\cos^{2}v) $, and$\chi\cdot n( x) =\cos^{2}v\geq0$. Therefore, we have$\chi\cdot n(x) \geq0$. Clearly,$B(x;2) \nsubseteq\Omega$for any$x\in\Omega$, and thus,$\Omega$is not a large domain in$\mathbb{R}^{3}$. \end{proof} Let$\mathbf{F}_{s}^{r}=\mathbf{A}_{0}^{r}\cup B^{N}(0;s)$be an interior flask domains. Interior flask domains are achieved for large$s$, but are nonachieved for small$s$. By Theorem \ref{h1} below, we have the following result. \begin{theorem} \label{n15}$s_{0}>0$exists such that the interior flask domains$\mathbf{F}_{s}^{r}$nonachieved if$sr$and$\mathbf{B=}B^{N}((0,h);r/2)$, let$\Omega=\Omega_{h}=( \mathbf{A}_{0}^{r}\cup B^{N}(0;r)) \backslash\overline{\mathbf{B}}$be the upper half strip with a hole. By Theorem \ref{n14}, there are no ground state solutions of Equation \eqref{E1} in$\Omega$. However, in this section, we prove that a positive higher energy solution of Equation \eqref{E1} exists in$\Omega$. Let$\overline{u}$be a ground state solution of Equation$( \ref{E1}) $in$\mathbf{A}^{r}$,$\overline{h}=(0,h)\in\mathbf{A}^{r}$and$\phi:\mathbf{A}^{r}\mathbf{\to[}0,1\mathbf{]}$, a$C^{\infty}$cut-off function such that$0\leq\phi\leq1$and \begin{gather*} \phi(z)=\begin{cases} 0& text{for }z\in\mathbf{B}\cup(\mathbf{A}^{r}\backslash( \mathbf{A}_{0}^{r}\cup B^{N}(0;r)) \mathbb{)},\\ 1 &\text{for }z\in(\mathbf{A}_{0}^{r}\cup B^{N} (0;r)) \backslash(B^{N}(0;\frac{2}{3}r)\\ & \hspace{12mm} \cup\{z=(x,y)\in\mathbf{A}^{r}|y\leq r\}), \end{cases} \\ I=\{0\}\times[-\frac{r}{2},\frac{r}{2}],\quad\text{}I_{h}=\overline{h}+I,\\ v_{t}(z)=\phi(z)\bar{u}(z-t-2\overline{h})\quad\text{for}\;\;z\in \mathbf{A}^{r},\quad\text{}t\in I. \end{gather*} Then$v_{t}\in H_{0}^{1}(\Omega)$. Furthermore, we have the following lemma. \begin{lemma} \label{y1} For$t\in I$or$\overline{t}\in I_{h}$, where$\overline {t}=\overline{h}+t$, then\newline$(i)\| v_{t}(z)-\bar{u}(z-t-2\overline {h})\| _{L^{p}(\mathbf{A}^{r})}= o(1)$as$h\to\infty$; \newline$(ii)\| v_{t}(z)-\bar{u}(z-t-2\overline{h})\| _{H^{1}(\mathbf{A}^{r})}= o(1)$as$h\to\infty$; \newline$(iii)J(v_{t})=\alpha(\mathbf{A}^{r})+ o(1)$as$h\to\infty$. \end{lemma} \begin{proof}$(i)\begin{align*} \| v_{t}(z)-\bar{u}(z-t-2\overline{h})\| _{L^{p}(\mathbf{A}^{r})}^{p} &={\int_{\mathbf{A}^{r}}}| \phi(z)-1| ^{p}| \bar{u}(z-t-2\overline{h})| ^{p}\\ &\leq{\int_{(\mathbf{A}_{h+r}^{r}) ^{c}}} | \bar{u}(z-t-2\overline{h})| ^{p}\\ &= o(1)\quad\text{as }h\to\infty. \end{align*}(ii)We have \begin{align*} &\| v_{t}(z)-\bar{u}(z-t-2\overline{h})\| _{H^{1}(\mathbf{A}^{r})}^{2}\\ &=\| (\phi(z)-1)\bar{u}(z-t-2\overline{h})\| _{H^{1}(\mathbf{A}^{r})}^{2}\\ &\leq c(\frac{1}{r^{2}}+1) {\int_{(\mathbf{A}_{h+r}^{r}) ^{c}}} (| \nabla\bar{u}(z-t-2\overline{h})| ^{2}+| \bar {u}(z-t-2\overline{h})| ^{2}) \\ &= o(1)\quad \text{as } h\to\infty. \end{align*}(iii)$This follows from$(i)$,$(ii)$, Theorem \ref{f3}, and $\alpha(\mathbf{A}^{r})=J(\overline{u})=\frac{1}{2}a(\bar{u})-\frac{1}{p} b(\bar{u})=\frac{1}{2}a(v_{t})-\frac{1}{p}b(v_{t})+ o(1)=J(v_{t})+ o(1).$ \end{proof} From Lemma \ref{y1}, since$\| \overline{u}\| _{H^{1}(\mathbf{A}^{r})} ^{2}=\| \overline{u}\| _{L^{p}(\mathbf{A}^{r})}^{p}$, we have \begin{gather*} \| v_{t}\| _{H^{1}(\mathbf{A}^{r})}^{2} =\| \overline{u}\| _{H^{1}(\mathbf{A}^{r} )}^{2}+ o(1)\quad\text{as } h\to\infty,\\ \| v_{t}\| _{L^{p}(\mathbf{A}^{r})}^{p} =\| \overline{u}\| _{L^{p}(\mathbf{A}^{r} )}^{p}+ o(1)\quad \text{as } h\to\infty. \end{gather*} Therefore,$\| v_{t}\| _{H^{1}(\mathbf{A}^{r})}^{2}=\| v_{t}\| _{L^{p} (\mathbf{A}^{r})}^{p}+ o(1)$as$h\to\infty$. By Lemma \ref{i2}, there exists$\lambda_{t}>0$such that$u_{t}=\lambda_{t}v_{t}$in$\mathbf{M}$:$\| u_{t}\| _{H^{1}(\mathbf{A}^{r})}^{2}=\| u_{t}\| _{L^{p} (\mathbf{A}^{r})}^{p}$. Therefore,$\lambda_{t}\to 1$as$h\to\infty$, or$J(u_{t})=\alpha(\mathbf{A}^{r})+ o(1)$as$h\to\infty$. For$u\in H_{0}^{1}(\Omega)$, define the center of mass function by $j(u)=\| u\| _{L^{p}(\mathbf{A}^{r})}^{-p}\int_{\mathbf{A}^{r}}(\overline {h}+\frac{r}{2}\frac{z}{| z| })| u(x,y)| ^{p}dxdy.$ Let $\beta_{0}=\inf\{ J(u): u\in\mathbf{M}(\Omega) ,\, u\geq0,\, j(u)=\overline{h}\} .$ \begin{proposition} \label{y2}$\alpha(\mathbf{A}^{r})=\alpha(\Omega)<\beta_{0}$. \end{proposition} \begin{proof} By Theorem \ref{n14},$\alpha(\mathbf{A}^{r}\mathbb{)}=\alpha(\Omega)$. Clearly$\alpha(\mathbf{A}^{r})\leq\beta_{0}$. Suppose$\alpha(\mathbf{A} ^{r})=\beta_{0}$. By Theorem \ref{i4}, there is a sequence$\{u_{k}\}$in$\mathbf{M}(\Omega) $,$u_{k}\geq0$,$j(u_{k})=\overline{h}$for each$k$, such that \begin{gather*} J(u_{k})=\alpha(\mathbf{A}^{r})+ o(1) \quad \text{as }k\to\infty,\\ J'(u_{k})= o(1) \quad \text{strongly}\quad\text{in }H^{-1} (\Omega)\quad \text{as }k\to\infty. \end{gather*} By Theorem \ref{d3}, there is an unbounded sequence$\{(0,y_{k})\}$in$\mathbf{A}^{r}$such that $u_{k}(x,y)=\overline{u}(x,y-y_{k})+ o(1)\quad \text{strongly in } H_{0}^{1}(\mathbf{A}^{r}),$ where$\overline{u}$is a ground state solution of Equation$( \ref{E1}) $in$\mathbf{A}^{r}$. Assume$(\overline{h}+\frac{r}{2} \frac{(0,y_{k})}{| (0,y_{k})| })=\varsigma+ o(1)$as$k\to\infty$, where$\varsigma\in\partial I_{h}. Then by the Lebesgue Dominated Convergence Theorem, we have \begin{align*} \overline{h}=j(u_{k}) & =\| u_{k}\|_{L^{p}(\mathbf{A}^{r})}^{-p} {\int_{\mathbf{A}^{r}}} (\overline{h}+\frac{r}{2}\frac{z}{| z| })| u_{k} (x,y)| ^{p}dxdy\\ & =\|\overline{u}\|_{L^{p}(\mathbf{A}^{r})}^{-p} {\int_{\mathbf{A}^{r}}} (\overline{h}+\frac{r}{2}\frac{(x,y+y_{k})}{| (x,y+y_{k})| })| \overline{u}(x,y)| ^{p}dxdy+ o(1)\\ & =\varsigma+ o(1)\quad \text{as } k\to\infty, \end{align*} which is a contradiction. Therefore\alpha(\mathbf{A}^{r})=\alpha (\Omega)<\beta_{0}$. \end{proof} Let \begin{gather*} V=\{u\in\mathbf{M}(\Omega) : u\geq0\};\\ \Gamma=\{k:I_{h}\to V\;\text{continuous}: k(\overline{t} )=u_{t}\;\;\text{for}\;\;t\in\partial I\};\\ \beta_{1}=\inf_{k\in\Gamma}\max_{\overline{t}\in I_{h}}J(k(\overline{t})). \end{gather*} \begin{proposition} \label{y3} There is an$h_{0}>0$such that for$h\geq h_{0}$,\newline$(i)\alpha(\mathbf{A}^{r})0$, for$t\in\partial I$. \end{proposition} \begin{proof}$(i)$and$(ii)$follow from Theorem \ref{n14}, Proposition \ref{y2}, and Lemma \ref{y1}.$(iii)c_{1},c_{2}>0$exist such that$c_{1} \leq\|\phi(z)\bar{u}(z-t-2\overline{h})\|_{L^{p}(\mathbf{A} ^{r})}\leq c_{2}$. For$t\in\partial I$with$z+t+2\overline{h}\neq0, we have \begin{align*} (\frac{z+t+2\overline{h}}{| z+t+2\overline{h}| },t) &=|z+t+2\overline{h}|-\frac{1}{| z+t+2\overline{h}| }(z+t+2\overline{h},z+2\overline{h})\\ &\geq| z+t+2\overline{h}| -| z+2\overline {h}| \geq| t| -2| z+2\overline{h}| =\frac {r}{2}-2| z+2\overline{h}| . \end{align*} Then there are constantsc_{3}$,$c_{4}$,$c_{5}$,$c_{6}$,$h_{0}>0$such that for$h\geq h_{0}\begin{align*} (j(u_{t}),\overline{h}+t) & =\| u_{t}(z)\| _{L^{p}(\mathbf{A}^{r}\mathbb{)} }^{-p} {\int_{\mathbf{A}^{r}}} (\overline{h}+\frac{r}{2}\frac{z}{| z| },\overline{h}+t)| u_{t}(z)| ^{p}dz\\ & =c_{3} {\int_{\mathbf{A}^{r}}} (\overline{h}+\frac{r}{2}\frac{z}{| z| },\overline{h}+t)| \phi(z)\bar{u}(z-t-2\overline{h})| ^{p}dz\\ & =c_{3}{\int_{\mathbf{A}^{r}}} (\overline{h}+\frac{r}{2}\frac{z+t+2\overline{h}}{| z+t+2\overline {h}| },\overline{h}+t)| \phi(z+t+2\overline{h})\bar{u}(z)|^{p}dz\\ & \geq c_{3}(h^{2}c_{5}-hc_{4}c_{5}-hc_{5}-2c_{6}-4hc_{5} )>0\quad \text{since } h\geq h_{0}, \end{align*} where {\int_{\mathbf{A}^{r}}} | \phi(z+t+2\overline{h})\bar{u}(z)| ^{p}dz\geq c_{5}$and$ {\int_{\mathbf{A}^{r}}} |z\| \phi(z+t+2\overline{h})\bar{u}(z)| ^{p}dz\geq c_{6}$. By Theorem \ref{s10},$c_{4}<\infty$. \end{proof} \begin{proposition} \label{y4} For$h\geq h_{0}$, we have$\alpha(\mathbf{A}^{r})< \beta_{0}=\beta_{1}<2^{\frac{p-2}{p}} \alpha(\mathbf{A}^{r})$. \end{proposition} \begin{proof} We claim that:$(i)\beta_{0}=\beta_{1}:$for any$k\in\Gamma$, consider the homotopy$H(\lambda,\overline{t}):[0,1]\times I_{h} \to\mathbb{R}^{N}$defined by $H(\lambda,\overline{t})=(1-\lambda)j(k(\overline{t}))+\lambda i(\overline {t}),$ where$i$denotes the identity map. Note that$j(k(\overline{t}))=j(u_{t})$for$\overline{t}\in\partial I_{h}$. By Proposition \ref{y3}$(iii)$,$H(\lambda,\overline{t})\neq\overline{h}$for$\overline{t}\in\partial I_{h}$and$\lambda\in[0,1]$. Therefore $\deg(j\circ k,I_{h},\overline{h})=\deg(i,I_{h},\overline{h})=1.$$t_{0}\in I_{h}$exists such that $j(k(t_{0}))=\overline{h}.$ Hence, for each$k\in\Gamma, \begin{align*} \beta_{0} & =\inf\{ J(u): u\in\mathbf{M},\quad\text{}u\geq0,\quad\text{ }j(u)=\overline{h}\} \\ & \leq J(k(t_{0}))\\ & \leq\max_{\overline{t}\in I_{h}}J(k(\overline{t})). \end{align*} We have\beta_{0}\leq\beta_{1}$. On the other hand, by Proposition \ref{y3}$(i)$, for$t\in I$, we have$u_{t}\in V$and$J(u_{t})<\beta_{0}$. Thus,$\max_{t\in I}J(u_{t})\leq\beta_{0}$, or$\beta_{1}\leq\beta_{0}$. \newline$(ii)\beta_{1}<2^{\frac{p-2}{p}}\alpha(\mathbf{A}^{r}):$by Proposition \ref{y3}$(ii)$,$J(u_{t})<2^{\frac{p-2}{p}}\alpha(\mathbf{A} ^{r})$for$t\in I$. Thus $\max_{t\in I}J(u_{t})<2^{\frac{p-2}{p}}\alpha(\mathbf{A}^{r}).$ We have$\beta_{1}<2^{\frac{p-2}{p}}\alpha(\mathbf{A}^{r})$. By Proposition \ref{y3}$(i)$, we have $\alpha(\mathbf{A}^{r})<\beta_{0}=\beta_{1}<2^{\frac{p-2}{p}}\alpha (\mathbf{A}^{r}).$ \end{proof} Now we assert that there is a higher energy solution of Equation \eqref{E1} in$\Omega$. \begin{theorem} \label{y5} Suppose that the positive solution of Equation \eqref{E1} in the infinite strip$\mathbf{A}^{r}$is unique up to$y$-translations.$h_{0}>0$exists such that if$h\geq h_{0}$, then there is a positive higher energy solution$v$of Equation \eqref{E1} in the upper half strip with a hole$\Omega$such that$\alpha(\mathbf{A}^{r})0$in$\mathbf{A}^{r}$,$1\leq i\leq\ell, satisfying \begin{gather*} \begin{aligned} u_{k}(z)&=u^{0}(z)+[u^{1}(z-z_{k}^{1})+u^{2}(z-z_{k}^{2})+\dots\\ &\quad +u^{\ell}(z-z_{k}^{\ell})]+ o(1)\quad \text{strongly in} H_{0}^{1}(\mathbf{A}^{r}),\end{aligned} \\ -\triangle u^{0}+u^{0}=(u^{0})^{p-1}\quad\text{in }\Omega,\\ -\triangle u^{i}+u^{i}=(u^{i})^{p-1}\;\quad\text{in }\mathbf{A} ^{r},\; 1\leq i\leq\ell,\\ J(u_{k})=J(u^{0})+\sum_{i=1}^{\ell}J(u^{i})+ o(1)\quad \text{as } k\to\infty. \end{gather*} Suppose that the solution of \eqref{E1} in the infinite strip\mathbf{A}^{r}$is unique up to$y$-translations. From Theorems \ref{d3} and \ref{a5}, we find that$u^{i}$are the same and$J(u^{i})=\alpha (\mathbf{A}^{r})$for$i=1,2,\dots,l$. Therefore $\beta_{0}=J(u^{0})+l\alpha(\mathbf{A}^{r}).$ Since$\alpha(\mathbf{A}^{r})<\beta_{0}<2^{\frac{p-2}{p}}\alpha(\mathbf{A} ^{r})$. We conclude that$u^{0}$is nonzero and$l=0$. Thus, there is a positive higher energy solution$v=u^{0}$of Equation \eqref{E1} in the upper half strip with a hole$\Omega$such that$\alpha(\mathbf{A}^{r})0$,$a_{j}=0,\;j=2,3,\dots$. Thus,$|\nabla u_{k}|^{2} dz=c\delta_{z_{1}}+ o(1)$. Clearly,$z_{1}=0$. \end{proof} Similarly, let$\{w_{k}\}$be the solutions of Equation \eqref{E1} in the interior flask domains$\mathbf{F}_{s}^{r}$, where$s>s_{0}$. Then we have dynamic systems of$\{w_{k}\}$as follows: \begin{theorem} \label{y8} Let$\overline{w}$be a ground state solution of Equation$( \ref{E1}) $in$\mathbb{R}^{N}$. Then$w_{k}\to\overline{w}$strongly in$H^{1}(\mathbb{R}^{N})$as$k\to\infty$. \end{theorem} \begin{proof} Note that \begin{gather*} J(w_{k})=\alpha(\mathbb{R}^{N})+ o(1),\\ J'(w_{k})= o(1)\quad\text{as}\;\;k\to\infty. \end{gather*} By Theorem \ref{d1}, we prove that an integer$\ell\geq0$and sequences$\{z_{k}^{i}\}\subset\mathbb{R}^{N}$for$0\leq i\leq\ell$exist such that for some subsequence$\{w_{k}\}$, there are$w^{i}\in H^{1}(\mathbb{R} ^{N}),\;w^{i}>0$in$\mathbb{R}^{N}$for$0\leq i\leq\ell, satisfying \begin{gather*} \begin{aligned} w_{k}(z)&=w^{0}(z)+[w^{1}(z-z_{k}^{1})+w^{2}(z-z_{k}^{2})+\dots\\ &\quad +w^{\ell}(z-z_{k}^{\ell})]+ o(1)\quad\text{strongly in} H^{1}(\mathbb{R}^{N}), \end{aligned}\\ -\triangle w^{i}+w^{i}=(w^{i})^{p-1}\quad text{in }\mathbb{R}^{N},\;0\leq i\leq\ell,\\ J(w_{k})=\sum_{i=0}^{\ell}J(w^{i})+ o(1)\quad \text{as } k\to\infty. \end{gather*} Then, sinceJ(w_{k})=\alpha(\mathbb{R}^{N})+ o(1)$, we conclude that$w_{k}(z)=\overline{w}(z)+ o(1)$strongly in$H^{1}(\mathbb{R}^{N})$. \end{proof} \noindent\textbf{Bibliographical notes:} The results of this section are from Wang \cite{W}. \section{Achieved Domains} In this section we assert that the bounded domains, the quasibounded domains, the periodic domains, some interior flask domains, some flat interior flask domains, some canal domains, and some manger domains are achieved. We begin with the following lemma. \begin{lemma} \label{l1}$\gamma(\Omega)$is achieved if and only if$\alpha(\Omega)$is achieved. \end{lemma} \begin{proof} Recall that$\alpha(\Omega)=(\frac{1}{2}-\frac{1}{p}) \gamma(\Omega)^{\frac{2p}{2-p}}$. Suppose that there is a$u\inH_{0} ^{1}(\Omega)$such that $J(u)=\alpha(\Omega),\ \langle J'(u),u\rangle =a(u)-b(u)=0.$ Then we have$a(u)^{(\frac{1}{p}-\frac{1}{2})}=\gamma(\Omega)$. Let$v=\frac{u}{\| u\| _{H^{1}}}$. Then $\| v\| _{L^{p}}=\frac{b(u)^{1/p}}{a(u)^{1/2}}=a(u)^{\frac{1}{p}-\frac{1}{2} }=\gamma(\Omega).$ Thus,$\gamma(\Omega)$is achieved by$v$. On the other hand, let$\gamma(\Omega)$be achieved by some function$u$where$a(u)=\| u\| _{H^{1} }^{2}=1$and$b(u)=\| u\| _{L^{p}}^{p}=\gamma(\Omega)^{p}$. By the Lagrange multiplier theorem there is a$\lambda$such that$b'(u)=\lambda a'(u)$. It is easy to see that$\lambda=\frac{p}{2}\gamma(\Omega)^{p}$, so we have $b'(u)=\frac{p}{2}\gamma(\Omega)^{p}a'(u).$ This implies $\gamma(\Omega)^{-p}( {\int} | u| ^{p-2}u\varphi) =( {\int} \nabla u\nabla\varphi+u\varphi) .$ Thus,$u$is a weak solution of $-\Delta u+u=\gamma(\Omega)^{-p}| u| ^{p-2}u.$ Let$v=\gamma(\Omega)^{\frac{p}{2-p}}u$. Then $-\Delta v+v=| v| ^{p-2}v.$ We have$a(v)=b(v)=\gamma(\Omega)^{\frac{2p}{2-p}}$,$\langle J^{\prime }(v),\varphi\rangle =0$for each$\varphi\in C_{c}^{\infty}(\Omega)$, and $J(v)=(\frac{1}{2}-\frac{1}{p}) \gamma(\Omega)^{\frac{2p}{2-p} }=\alpha(\Omega).$ \end{proof} \begin{remark}\label{l2}\rm Note that if$u$is a ground state solution for$J$in$\Omega$, then$u$solves the semilinear elliptic \eqref{E1} and$J(u)=\alpha (\Omega)$. By the Kato regularity,$L^{p}-$regularity and Schauder regularity, the ground state solution$u$of \eqref{E1} is classical. \end{remark} \begin{theorem} \label{b3} A bounded domain$\Omega$is an achieved domain. \end{theorem} For the proof of this theorem follows from Theorem \ref{f1}. An unbounded domain may be achieved. \begin{theorem} \label{l103} A$C^{1}$quasibounded domain is achieved. \end{theorem} The statment of this theorem follows from Theorem \ref{f2-1}. A periodic domain in$\mathbb{R}^{N}$is achieved. In Theorem \ref{a1}, we proved that if a (PS)$_{\alpha}$-sequence for$J$admits a nonzero weak limit$u$, then$u$is a ground state solution for$J$. However, even though the weak limit is zero we can still obtain a ground state solution for$J$if the domain is periodic. \begin{theorem} \label{f3} A periodic domain in$\mathbb{R}^{N}$is achieved. In particular, there is a ground state solution of Equation$\eqref{E1}$in$\mathbf{A}^{r}$,$\mathbf{A}^{r_{1},r_{2}}$, and$\mathbb{R}^{N}$. \end{theorem} \begin{proof} It suffices to prove the case$\Omega=\mathbf{A}^{r}$. Let$\{u_{n}\}$be a (PS)$_{\alpha(\mathbf{A}^{r})}$-sequence such that $J(u_{n})=\alpha(\mathbf{A}^{r})+ o(1)\text{, }J'(u_{n})= o(1).$ By Lemma \ref{p30}, there are a subsequence$\{u_{n}\}$and a$u\in H_{0} ^{1}(\mathbf{A}^{r})$such that $u_{n}\rightharpoonup u\quad\text{weakly in }H_{0}^{1}(\mathbf{A}^{r}).$ Suppose that$u$is nonzero, then by Theorem \ref{c2}, we are done. Suppose that$u_{n}\rightharpoonup0$weakly in$H_{0}^{1}(\mathbf{A}^{r})$. Since$\alpha(\Omega)$is positive, we have$u_{n}\nrightarrow0$strongly in$H_{0}^{1}(\Omega)$. By Lemma \ref{p9}, there is a subsequence$\{ u_{n}\} $, and a constant$\alpha>0$such that for$n=1,2,\dots$, $Q_{n}=\sup_{y\in\mathbb{R}}\int_{(0,y)+\mathbf{A}_{-2,2}^{r}}|u_{n} (z)|^{2}dz>\alpha>0.$ Take$\{ z_{n}\} $in$\mathbf{A}^{r}$, where$z_{n}=(0,y_{n})$such that$\int_{z_{n}+\mathbf{A}_{-2,2}^{r}}|u_{n}(z)|^{2}dz\geq\alpha/2$, and let$w_{n}(z)=u_{n}(z+z_{n}) $. Then for$n=1,2,\dots$, \begin{gather*} {\int_{\mathbf{A}_{-2,2}^{r}}} |w_{n}(z)|^{2}dz= {\int_{z_{n}+\mathbf{A}_{-2,2}^{r}}} |u_{n}(z)|^{2}dz\geq\alpha/2,\\ \Vert w_{n}\Vert_{H^{1}(\mathbf{A}^{r})}=\Vert u_{n}\Vert_{H^{1} (\mathbf{A}^{r})}\leq c, \end{gather*} so$w\in H_{0}^{1}(\mathbf{A}^{r}) $exists such that$w_{n}\rightharpoonup w$weakly in$H_{0}^{1}(\mathbf{A}^{r}) $. Clearly,$\{w_{n}\}$is a (PS)-sequence in$H_{0}^{1}(\mathbf{A} ^{r}) $for$J$. By Theorem \ref{p24}, $\int_{\mathbf{A}_{-2,2}^{r}}|w|^{2}=\lim_{n\to\infty}\int _{\mathbf{A}_{-2,2}^{r}}|w_{n}|^{2}\geq\alpha/2,$ so$w\not \equiv 0$. By Theorem \ref{c2}, there is a ground state solution of Equation$\eqref{E1}$in$H_{0}^{1}(\Omega) $. \end{proof} Moreover,$\mathbb{R}$is an achieved domain: there is a classical solution$u$of the Equation $$u''=u-\gamma(\mathbb{R)}^{-1}| u| ^{p-2}u.\label{E4}$$ By Berestycki-Lions \cite{BL}, such a solution is unique. The solution can be constructed as follows by routine computations. \begin{theorem} \label{l5} With$\mu=2/(p-2)$, we have \begin{gather*} u(r)=\big(\frac{p\gamma(\mathbb{R)}}{2}\big) ^{\frac{\mu}{2}}\{ \cosh\ (r/\mu) \} ^{-\mu};\\ \gamma(\mathbb{R)}=\big[ \frac{(2\mu+1)\Gamma(2\mu)}{\mu\Gamma(\mu)^{2} }\big] ^{\mu^{-1}}(\frac{\mu}{4}) (\mu+1) ^{-\frac{p}{2}}, \end{gather*} to solve Equation (\ref{E4}). In particular,$\mathbb{R}$is an achieved domain. \end{theorem} Next we present achieved domains from the perturbations of nonachieved domains. By Theorem \ref{n3}, the upper half strip$\mathbf{A} _{0}^{r}$and the upper half hollow strip$\mathbf{A}_{0}^{r_{1},r_{2}}$are nonachieved. However, the perturbed domains of$\mathbf{A}_{0}^{r}$and$\mathbf{A}_{0}^{r_{1},r_{2}}$may be achieved. Let$\mathbf{F}_{s} ^{r}=\mathbf{A}_{0}^{r}\cup B^{N}(0;s)$be an interior flask domain. Interior flask domains are achieved for large$s$, but are nonachieved for small$s$. \begin{theorem} \label{h1}$s_{0}>0$exists such that Equation \eqref{E1} has a ground state solution in$\mathbf{F}_{s}^{r}$if$s>s_{0}$, but does not have any ground state solution if$ss_{0}$, while$\mathbf{F} _{s}^{r}$are nonachieved if$s\alpha(\mathbb{R}^{N})$. By Theorem \ref{f8}$(ii)$, we have$\alpha (\mathbf{A}^{r})=\alpha(\mathbf{A}_{0}^{r})$and by Theorem \ref{f17},$\lim_{s\to\infty}\alpha(B^{N}(0;s))=\alpha(\mathbb{R}^{N})$. Take$s $large enough so that $\alpha(B^{N}(0;s))<\alpha(\mathbf{A}^{r})=\alpha(\mathbf{A}_{0}^{r}).$ By Theorem \ref{f1}, there is a ground state solution of Equation$( \ref{E1}) $in$B^{N}(0;s)$. Then by Theorem \ref{c3}$(ii)$, we have $\alpha(\mathbf{F}_{s}^{r})<\alpha(B^{N}(0;s)).$ We conclude that $\alpha(\mathbf{F}_{s}^{r})<\alpha(B^{N}(0;s))<\alpha(\mathbf{A}_{0}^{r}),$ or $\alpha(\mathbf{F}_{s}^{r})<\min\{ \alpha(B^{N}(0;s)),\alpha (\mathbf{A}_{0}^{r})\} .$ By the equivalence of$(i)$and$(vi)$in Theorem \ref{c7}, Equation$( \text{\ref{E1}}) $has a ground state solution in$\mathbf{F}_{s}^{r}$for large$s$. If Equation \eqref{E1} has a ground state solution in$\mathbf{F}_{s_{1}}^{r}$and$s_{1}r: \text{Equation \eqref{E1} has a ground state solution in }\mathbf{F}_{s}^{r}\}. \] We then conclude that Equation \eqref{E1} has a ground state solution in $\mathbf{F}_{s}^{r}$ if $s>s_{0}$, and Equation $( \text{\ref{E1}})$ does not have any ground state solution in $\mathbf{F}_{s}^{r}$ if $ss_{0}$. In fact, if we replace $\mathbf{A}_{0}^{r}\cup B^{N}(0;s)$ by $\mathbf{A}_{0}^{r}\cup\Omega$, where $\Omega$ is a bounded domain containing $B^{N}(0;s)$, the theorem still holds. \end{remark} For $\delta>0$, there is a $\varepsilon(\delta)>0$ such that a flat interior flask domain $\Omega_{\varepsilon}$ is an achieved domain, where $E_{\varepsilon}=\{(x,y)\in\mathbb{R}^{N}: (x,\varepsilon y)\in \mathbf{B}(0;r+\delta)\};\quad \Omega_{\varepsilon}=\mathbf{A}_{0}^{r}\cup E_{\varepsilon}.$ \begin{theorem} \label{l8} Given $\delta>0$, $\varepsilon_{0}>0$ exists such that if $\varepsilon\leq\varepsilon_{0}$, then the flat interior flask domain $\Omega_{\varepsilon}$ is an achieved domain. \end{theorem} \begin{proof} By Theorem \ref{f3}, the infinite strip $\mathbf{A}^{r}$ admits a ground state solution. Since $\mathbf{A}^{r}\subsetneq\mathbf{A}^{r+\delta}$, by Theorem \ref{a5} we have $\alpha(\mathbf{A}^{r+\delta})<\alpha(\mathbf{A}^{r})$. Since $E_{\varepsilon}\subset\mathbf{A}^{r+\delta}$ and $\lim_{\varepsilon \to0}\alpha(E_{\varepsilon})=\alpha(\mathbf{A}^{r+\delta})$, there is an $\varepsilon_{0}>0$ such that if $\varepsilon\leq\varepsilon_{0}$, then $\alpha(E_{\varepsilon})<\alpha(\mathbf{A}^{r})$. If $\varepsilon$, $\varepsilon\leq\varepsilon_{0}$ is fixed, a large $N\in\mathbb{N}$ exists such that $\alpha((\widetilde{\Omega_{\varepsilon}})_{N})=\alpha(\mathbf{A}_{N} ^{r})=\alpha(\mathbf{A}^{r}).$ Thus, $\alpha(\Omega_{\varepsilon})\leq\alpha(E_{\varepsilon})<\alpha(\mathbf{A} ^{r})=\alpha((\widetilde{\Omega_{\varepsilon}})_{N}).$ By Theorem \ref{a6}, a ground state solution $u$ of \eqref{E1} exists. By Theorem \ref{l1}, $\Omega_{\varepsilon}$ is an achieved domain. \end{proof} Fix a number $1\leqq l\leqq N-1$ and write $\mathbb{R}^{N}=\mathbb{R} ^{l}\times\mathbb{R}^{N-l}$, so that a generic $z\in\mathbb{R}^{N}$ is written as $z=(x,y)$ with $x\in\mathbb{R}^{l}$ and $y\in\mathbb{R} ^{N-l}$. Let $\Omega$ be a domain in $\mathbb{R}^{N}$. For $y\in \mathbb{R}^{N-l}$, we denote by $\Omega^{y}\subset\mathbb{R}^{l}$ the $y$-section of $\Omega$, that is, $\Omega^{y}=\{ x\in\mathbb{R}^{l}| \quad\text{}(x,y) \in\Omega \} .$ We consider the following canal properties:\newline$(\Omega1)$ $\Omega$ is a smooth domain in $\mathbb{R}^{N}$ and the sections $\Omega^{y}$ are contained in a bounded set for each $y\in\mathbb{R}^{N-l}$; \newline$(\Omega2)$ there is a smooth domain $O$ in $\mathbb{R}^{l}$ such that $O\subset\Omega^{y}\quad\text{for each }y\in\mathbb{R}^{N-l};$ \newline$(\Omega3)$ for each $\delta>0$ there is an $M>0$ such that $\Omega^{y}\subset\{ x\in\mathbb{R}^{l}| \mathop{\rm dist}(x,O) <\delta \} \quad\text{for each }| y| \geqq M.$ \begin{figure}[htb] \begin{center} \includegraphics[width=0.6\textwidth]{fig11} \end{center} % \centering \resizebox{2.5in}{!}{\includegraphics{./fig11.eps}} \caption{ perturbed infinite strip domain.} \label{fig:fig11} \end{figure} \begin{theorem} \label{h2} Assume that $\Omega$ satisfies $(\Omega_{1})$, $(\Omega_{2})$ and $(\Omega_{3})$. Then Equation \eqref{E1} admits a ground state solution in the canal domain $\Omega$. \end{theorem} \begin{proof} $(i)$ $N-l=1:$ $(\Omega_{2})$ gives $\widehat{O}=O\times\mathbb{R}^{N-l}\subsetneqq\Omega.$ By Theorem \ref{f3}, there is a ground state solution of \eqref{E1} in $\widehat{O}$. By Theorem \ref{c3} $(ii)$, we have $$\alpha(\Omega) <\alpha(\widehat{O}) .\label{12-1}$$ Let \begin{gather*} \Omega_{+}=\{ z=(x,y) \in\Omega: y>-1 \} ,\\ \Omega_{-}=\{ z=(x,y) \in\Omega: y<1 \} ,\\ \widehat{O}_{+}=\{ z=(x,y) \in\widehat{O} : y>-1 \} ,\\ \widehat{O}_{-}=\{ z=(x,y) \in\widehat{O}: y<1 \} . \end{gather*} Then $\Omega=\Omega_{+}\cup\Omega$ and$_{-}\widehat{O}=\widehat{O}_{+} \cup\widehat{O}$. Moreover, both $\Omega_{+}\cap\Omega_{-}$ and $T,\widehat {O}_{+}\cap\widehat{O}$ are bounded. Since $\widehat{O}_{+}\subset\Omega_{+}$, then $\alpha(\Omega_{+}) \leq\alpha(\widehat{O} _{+}) =\alpha(\widehat{O})$. \newline$(a)$ Suppose that $\alpha(\Omega_{+}) =\alpha(\widehat{O}_{+}) =\alpha(\widehat{O})$. By (\ref{12-1}), we have $\alpha( \Omega) <\alpha(\Omega_{+})$.\newline$(b)$ Suppose that $\alpha(\Omega_{+}) <\alpha(\widehat{O}_{+}) =\alpha(\widehat{O})$. By Theorem \ref{f20}, $\lim_{\delta\to 1} \alpha(\delta\widehat{O})=\alpha(\widehat{O}) ,$ and a $\delta_{0}>1$ exists such that $\alpha(\Omega_{+}) <\alpha(\delta_{0}\widehat{O})$. From $(\Omega _{3})$, there is $n_{0}>0$ such that $\Omega_{+}\backslash \overline{B^{N}(0;n_{0})}\subset\delta_{0}\widehat{O}$. Thus, $\alpha( \delta_{0}\widehat{O}) \leq\alpha(\Omega_{+}\backslash \overline{B^{N}(0;n)})$ for $n\geq n_{0}$. Therefore, $\alpha( \Omega_{+}) <\alpha(\Omega_{+}\backslash\overline{B^{N} (0;n)})$ for $n\geq n_{0}$. From the proof of Theorem \ref{c7}, if we assume that $(v)$ holds for each $n\geq n_{0}$, then we obtain $(i)$. Thus $J$ satisfies the $(PS)_{\alpha( \Omega_{+}) }$-condition. By Theorem \ref{c3} $(i)$, $\alpha(\Omega) <\alpha(\Omega_{+})$.\newline From Case $(i)$ and Case $(ii)$, we conclude that $\alpha(\Omega) <\alpha(\Omega_{+})$. Similarly, we have $\alpha(\Omega) <\alpha( \Omega _{-})$. Finally, we have $\alpha(\Omega) <\min\{ \alpha(\Omega_{+}) ,\alpha(\Omega_{-}) \} .$ By the equivalence of $(i)$ and $(vi)$ in Theorem \ref{c7}, $J$ satisfies the $(PS)_{\alpha(\Omega) }$-condition. By Theorem \ref{c2} $(iii)$, \eqref{E1} admits a ground state solution in $\Omega$.\newline$(ii)$ $N-l\geq2:$ $(\Omega2)$ gives $\widehat{O}=O\times\mathbb{R}^{N-l}\subsetneqq\Omega,$ and by Theorem \ref{f3} and \ref{c3} $(ii)$, we have $\alpha( \Omega) <\alpha(\widehat{O})$. By Lemma \ref{f20}, $\lim_{\gamma\to1}\alpha(\gamma\widehat{O}) =\alpha(\widehat{O}) .$ Thus, take a $\gamma$ close to $1$ such that $\alpha(\Omega) <\alpha(\gamma\widehat{O})$. By $(\Omega3)$, there is an $n_{0}>0$ such that $\Omega\backslash B^{N}(0;n) \subset\gamma\widehat{O}$ for $n\geq n_{0}$. Hence, $\alpha( \gamma\widehat{O}) \leq\tilde{\alpha}_{n}$ for $n\geq n_{0}$. Thus $\alpha(\Omega) <\tilde{\alpha}_{n}$ for each $n\geq n_{0}$. The result follows from the proof of $(v)$ to $(i)$ in Theorem \ref{c7}. \end{proof} Assume $z=(x,y)\in\mathbb{R}^{N-1}\times\mathbb{R}$ for $N\geq3$. For $r_{2}>r_{1}>0$ and $t>0$, we consider a manger domain $D_{t}=\mathbf{A} _{0}^{r_{1},r_{2}}\cup\mathbf{A}_{0,t}^{r_{2}}$. We have the following result. \begin{theorem} \label{h3} $\widetilde{t}\geq0$ exists such that Equation $( \ref{E1})$ admits a ground state solution in the manger domain $D_{t}$ if $t>\widetilde{t}$, and does not admit any ground state solution in $D_{t}$ if $t<\widetilde{t}$. \end{theorem} \begin{proof} $(i)$ We claim that there is a $t_{0}>0$ such that Equation $( \ref{E1})$ admits a ground state solution in $D_{t}$ if $t\geq t_{0}$. \noindent\textbf{Method (I)}: let $\{ u_{n}\}$ be a (PS)$_{\alpha(D_{t}) }$-sequence in $H_{0}^{1}( D_{t})$ for $J$. By Lemma \ref{c6} $(iii)$, we have $a_{\infty }=b_{\infty}$ and $J_{\infty}\leq\alpha(D_{t})$, where $J_{\infty}=\frac{p-2}{2p}b_{\infty}$. We claim that there is a $t_{0}>0$ such that $J_{\infty}<\alpha(D_{t})$ for $t\geq t_{0}$. On the contrary, suppose that $J_{\infty}=\alpha(D_{t_{n}})$ for a sequence $\{t_{n}\}$ such that $t_{n}\to\infty$ as $t\to \infty$. Then $J_{\infty}=\alpha(D_{t})$ for each $t\geq t_{1}$. Let $\xi(z)$ be as in (\ref{1-1}) and $\xi_{R}(z)=\xi(\frac{2|z|} {R})$. Then there is an $R_{0}>0$ such that $\xi_{R}u_{n}\in H_{0}^{1}( \mathbf{A}_{0}^{r_{1},r_{2}})$ for $R\geq R_{0}$. Let $\lambda_{n}=(\frac{a(\xi_{R}u_{n}) }{b(\xi _{R}u_{n}) }) ^{1/(p-2)}.$ Then we have $a(\lambda_{n}\xi_{R}u_{n}) =b(\lambda_{n}\xi_{R}u_{n})$. For $R\geq R_{0}$, we have $J(\lambda_{n}\xi_{R}u_{n})\geq\alpha(\mathbf{A}_{0}^{r_{1},r_{2} }) ,$ or $(\frac{1}{2}-\frac{1}{p}) \frac{a(\xi_{R}u_{n}) ^{p/(p-2)}}{b(\xi_{R}u_{n}) ^{2/(p-2)}}\geq\alpha( \mathbf{A}_{0}^{r_{1},r_{2}}) .$ Letting $R\to\infty$ and $n\to\infty$ and using $a_{\infty }=b_{\infty}$, we obtain $\alpha(D_{t}) =J_{\infty}=\frac{p-2}{2p}b_{\infty}\geq \alpha(\mathbf{A}_{0}^{r_{1},r_{2}}) .$ Thus we have $$\alpha(D_{t}) \geq\alpha(\mathbf{A}_{0}^{r_{1},r_{2} }) \quad \text{for each }t>t_{1}.\label{12-2}$$ Since $\mathbf{A}_{0}^{r_{2}}$ and $\mathbf{A}_{0}^{r_{1},r_{2}}$ are a large domain of $\mathbf{A}^{r_{2}}$ and $\mathbf{A}^{r_{1,}r_{2}}$, respectively, then by Theorem \ref{f8}, we have $\alpha(\mathbf{A}_{0}^{r_{2} }) =\alpha(\mathbf{A}^{r_{2}})$ and $\alpha( \mathbf{A}_{0}^{r_{1,}r_{2}}) =\alpha(\mathbf{A}^{r_{1,}r_{2} })$. Moreover, by Theorem \ref{f3}, $\alpha(\mathbf{A}^{r_{2} }) <\alpha(\mathbf{A}^{r_{1,}r_{2}})$. We conclude that $$\alpha(\mathbf{A}_{0}^{r_{2}}) <\alpha(\mathbf{A} _{0}^{r_{1,}r_{2}}) .\label{12-3}$$ By Theorem \ref{f177}, we have $\alpha((\mathbf{A}_{0}^{r_{2}}) _{n}) =\alpha(\mathbf{A}_{0}^{r_{2}}) + o(1)$. Hence, there is an $n_{0}>0$ such that $\alpha(\mathbf{A}_{0}^{r_{2}}) <\alpha(( \mathbf{A}_{0}^{r_{2}}) _{n}) <\alpha(\mathbf{A} _{0}^{r_{1,}r_{2}}) \quad\text{for }n\geq n_{0}.$ Then we have $\alpha(D_{n}) \leq\alpha((\mathbf{A}_{0}^{r_{2} }) _{n}) <\alpha(\mathbf{A}_{0}^{r_{1,}r_{2}}) \quad\text{for }n\geq n_{0}.$ This contradicts (\ref{12-2}). Thus, there is a $t_{0}>0$ such that $J_{\infty}<\alpha(D_{t})$ for $t\geq t_{0}$. By the equivalence of $(i)$ and $(vii)$ in Theorem \ref{c7}, Equation $( \ref{E1})$ admits a ground state solution in $D_{t}$ for $t\geq t_{0}$. \noindent \textbf{Method (II):} Since $\mathbf{A}_{0}^{r_{2}}$ and $\mathbf{A}_{0}^{r_{1},r_{2}}$ are large domains of $\mathbf{A}^{r_{2}}$ and $\mathbf{A}^{r_{1,}r_{2}}$, respectively, then by Theorem \ref{f8}, we have $\alpha(\mathbf{A}_{0}^{r_{2}}) =\alpha(\mathbf{A} ^{r_{2}})$ and $\alpha(\mathbf{A}_{0}^{r_{1,}r_{2}}) =\alpha(\mathbf{A}^{r_{1,}r_{2}})$. Moreover, by Theorem \ref{f3}, $\alpha(\mathbf{A}^{r_{2}}) <\alpha( \mathbf{A}^{r_{1,}r_{2}})$. We conclude that $\alpha(\mathbf{A}_{0}^{r_{2}}) <\alpha(\mathbf{A} _{0}^{r_{1,}r_{2}}) .$ By Theorem \ref{f177}, we have $\alpha((\mathbf{A}_{0}^{r_{2} }) _{n}) =\alpha(\mathbf{A}_{0}^{r_{2}}) + o(1)$. Hence, there is an $n_{0}>0$ such that $\alpha(\mathbf{A}_{0}^{r_{2}}) <\alpha(( \mathbf{A}_{0}^{r_{2}}) _{n}) <\alpha(\mathbf{A} _{0}^{r_{1,}r_{2}}) \quad\text{for }n\geq n_{0}.$ Then we have $\alpha(D_{n}) <\alpha((\mathbf{A}_{0}^{r_{2} }) _{n}) <\alpha(\mathbf{A}_{0}^{r_{1,}r_{2}}) \quad\text{for }n\geq n_{0}.$ By the equivalence of $(i)$ and $(vi)$ in Theorem \ref{c7}, Equation \eqref{E1} admits a ground state solution in $D_{n}$ for $n\geq n_{0}$.\newline$(ii)$ If Equation $( \ref{E1})$ admits a ground state solution in $D_{t_{2}}$ and $t_{2}0: \text{Equation \eqref{E1} has a ground state solution in }D_{t}\}. \] Then$\widetilde{t}\geq0$such that Equation \eqref{E1} admits a ground state solution in$D_{t}$if$t>\widetilde{t}$and does not admit any ground state solution in$D_{t}$if$t<\widetilde{t}$. \end{proof} \subsection{Open Question:} in Theorem \ref{h1}, is$s_{0}=r$? \noindent\textbf{Bibliographical notes:} Theorem \ref{f3} is from Lien-Tzeng-Wang \cite{LTW}. Theorem \ref{h1} is from Chen-Lee-Wang \cite{CLW} and Chen-Wang \cite{CW}. Theorem \ref{h1} is from Lien-Tzeng-Wang \cite{LTW}, Chen- Lee-Wang \cite[Lemma 19]{CLW}, and Chen-Wang \cite[Proposition 2.10]{CW}. Theorem \ref{h2} is from del Pino-Felmer \cite{PiFe2}. Theorem \ref{h3} is from Chabrowski \cite{Ch3}. \section{Multiple Solutions} In Section 12 we prove that there is a ground state solution in an achieved domain. In this section we prove that if we perturb \eqref{E1} or perturb the achieved domain by adding or taking out a domain, then we obtain multiple solutions. \subsection{Multiple Solutions for a Perturbed Equation} \subsubsection{Introduction} Let$N\geq2$and$20$such that$\lambda u\in \mathbf{M}(\Omega) $, and then the computations is routine. \end{proof} Throughout this section, we assume that$h(z) $satisfies$0<\| h\| _{L^{2}}0\quad \text{for each } u\in\mathbf{M}_{h}.\label{15-1} Suppose that there is a $w\in\mathbf{M}_{h}$ such that $\langle \psi'(w) ,w\rangle =0$. Then we have $a( w) =(p-1) b(w)$ and ${\int_{\Omega}} hw=a(w) -b(w) =(p-2) b( w)$. Now \begin{align*} 0 0\} ;\\ \mathbf{M}_{h}^{-} =\{ u\in\mathbf{M}_{h}: a(u) -(p-1) b(u) <0\} . \end{gather*} Consider the Nehari minimization problems for Equation $(\ref{E2} )$. Let \begin{gather*} \alpha_{h}(\Omega) =\inf_{u\in\mathbf{M}_{h}}J_{h}(u) ;\\ \alpha_{h}^{+}(\Omega) =\inf_{u\in\mathbf{M}_{h}^{+}}J_{h}(u) ;\\ \alpha_{h}^{-}(\Omega) =\inf_{u\in\mathbf{M}_{h}^{-}}J_{h}(u) . \end{gather*} Let $\alpha_{h}(\Omega) =\alpha(\Omega)$, $\mathbf{M}_{h}=\mathbf{M}(\Omega)$ and $J_{h}(u) =J(u)$ for $h=0$. For each $u\in H_{0}^{1}( \Omega) \backslash\{ 0\}$, we write $t_{\rm max}=\big[ \frac{a(u) }{(p-1) b(u) }\big] ^{\frac{1}{p-2}}.$ Clearly, $t_{\rm max}>0$. Note that $$\begin{gathered} J_{h}(tu) =\frac{1}{2}t^{2}a(u) -\frac{1}{p} t^{p}b(u) -t\int_{\Omega}hu;\\ \frac{d}{dt}J_{h}(tu) =ta(u) -t^{p-1}b(u) -{\int_{\Omega}}hu;\\ \frac{d^{2}}{dt^{2}}J_{h}(tu) =a(u) -(p-1) t^{p-2}b(u) ;\\ \langle J_{h}'(tu) ,tu\rangle=t^{2}a(u)-t^{p}b(u) -t{\int_{\Omega}}hu. \end{gathered}\label{15-2}$$ The following lemma is required. \begin{lemma} \label{g4} For each $u\in H_{0}^{1}(\Omega) \backslash\{ 0\}$, \newline$(i)$ there is a unique $t^{-}=t^{-}(u) >t_{\rm max}>0$ such that $t^{-}u\in\mathbf{M}_{h}^{-}$ and $J_{h}( t^{-}u) =\max_{t\geq t_{\rm max}}J_{h}(tu)$; \newline$(ii)$ $t^{-}(u)$ is a continuous function for nonzero $u$ ;\newline$(iii)$ $\mathbf{M}_{h}^{-}=\{ u\in H_{0}^{1}( \Omega) \backslash\{ 0\} : \frac{1}{\| u\| _{H^{1}}}t^{-}(\frac{u}{\| u\| _{H^{1}}}) =1\}$; \newline$(iv)$ if $\int_{\Omega}hu>0$, then there is a unique $0t_{\rm max}$ such that $s( t^{-}) =\int_{\Omega}hu$ and $s'(t^{-}) <0$. Now, \begin{align*} a(t^{-}u) -(p-1) b(t^{-}u) & =(t^{-}) ^{2}\Big[ a(u) -(p-1) (t^{-}) ^{p-2}b(u) \Big] \\ & =(t^{-}) ^{2}s'(t^{-}) <0, \end{align*} and \begin{align*} \langle J_{h}'(t^{-}u) ,t^{-}u\rangle & =( t^{-}) ^{2}a(u) -(t^{-}) ^{p}b( u) -t^{-}\int_{\Omega}hu\\ & =t^{-}\Big[ t^{-}a(u) -(t^{-}) ^{p-1}b( u) -\int_{\Omega}hu\Big] \\ & =t^{-}\Big[ s(t^{-}) -\int_{\Omega}hu\Big] =0. \end{align*} Thus, $t^{-}u\in\mathbf{M}_{h}^{-}$, since for $t>t_{\rm max}$, we have \begin{gather*} a(tu) -(p-1) b(tu) <0;\\ \frac{d^{2}}{dt^{2}}J_{h}(tu) =\frac{1}{t^{2}}[ a( tu) -(p-1) b(tu) ] <0;\\ \frac{d}{dt}J_{h}(tu) =ta(u) -t^{p-1}b(u) - {\int_{\Omega}} hu=0\quad \quad\text{if }t=t^{-}. \end{gather*} Thus, $J_{h}(t^{-}u) =\max_{t\geq t_{\rm max}}J_{h}(tu)$. \noindent Case $(b):$ ${\int_{\Omega}} hu>0$. By (\ref{15-3}), $s(0) =0<\int_{\Omega}hu\leq\| h\| _{L^{2}}\| u\| _{H^{1}}<\| u\| _{H^{1}}d(p,\alpha) \leq s(t_{\rm max}) ,$ there are unique $t^{+}$ and $t^{-}$ such that $00>s'(t^{-}) . \end{gather*} This case is similar to Case$(a)$: We have$t^{+}u\in\mathbf{M} _{h}^{+}$,$t^{-}u\in\mathbf{M}_{h}^{-}$, and$J_{h}(t^{-}u) \geq J_{h}(tu) \geq J_{h}(t^{+}u) $for each$t\in\left[ t^{+},t^{-}\right] $, and$J_{h}(t^{+}u) \leq J_{h}(tu) $for each$t\in\left[ 0,t^{+}\right] $. Thus, \begin{gather*} J_{h}(t^{-}u) =\max_{t\geq t_{\rm max}}J_{h}(tu) ,\\ J_{h}(t^{+}u) =\min_{0\leq t\leq t^{-}}J_{h}(tu) . \end{gather*}$(ii)$By the uniqueness of$t^{-}(u) $and the extremity property of$t^{-}(u) $,$t^{-}(u) $is a continuous function for nonzero$u$.\newline$(iii)$For$u\in\mathbf{M} _{h}^{-}$, let$v=\frac{u}{\| u\| _{H^{1}}}$. By part$(i)$, there is a unique$t^{-}(v) >0$such that$t^{-}( v) v\in\mathbf{M}_{h}^{-}$or$t^{-}(\frac{u}{\| u\| _{H^{1}}}) \frac{1}{\| u\| _{H^{1}}} u\in\mathbf{M}_{h}^{-}$. Since$u\in\mathbf{M}_{h}^{-}$, we have$t^{-}( \frac{u}{\| u\| _{H^{1}}}) \frac{1}{\| u\| _{H^{1}}}=1$, implying $\mathbf{M}_{h}^{-}\subset\big\{ u\in H_{0}^{1}(\Omega) \backslash\{ 0\} : \frac{1}{\| u\| _{H^{1}}} t^{-}(\frac{u}{\| u\| _{H^{1}}}) =1\big\} .$ Conversely, let$u\in H_{0}^{1}(\Omega) \backslash\{ 0\} $such that$\frac{1}{\| u\| _{H^{1}}}t^{-}( \frac{u}{\| u\| _{H^{1}}}) =1$. Then $t^{-}(\frac{u}{\| u\| _{H^{1}}}) \frac{u}{\| u\| _{H^{1}}}\in\mathbf{M}_{h}^{-}.$ Thus, $\mathbf{M}_{h}^{-}=\big\{ u\in H_{0}^{1}(\Omega) \backslash\{ 0\} : \frac{1}{\| u\| _{H^{1}}} t^{-}(\frac{u}{\| u\| _{H^{1}}}) =1\big\} .$$(iv)$By Case$(b)$of part$(i)$. \end{proof} We have the following results. \begin{lemma} \label{g5}$(i) $For each$u\in\mathbf{M}_{h}^{+}$, we have$\int_{\Omega}hu>0$and$J_{h}(u) <0$. In particular,$\alpha_{h}(\Omega) \leq\alpha_{h}^{+}(\Omega) <0$; \newline$(ii) J_{h}$is coercive and bounded below on$\mathbf{M}_{h}; $\newline$(iii) $For each minimizing sequence$\{ u_{n}\} $in$\mathbf{M}_{h}$for$J_{h}$, we have $0<\limsup_{n\to\infty}| \langle \psi^{\prime }(u_{n}) ,u_{n}\rangle | <\infty.$ \end{lemma} \begin{proof}$(i) $For each$u\in\mathbf{M}_{h}^{+}$,$a(u) -(p-1) b(u) >0$and$a(u) =b( u) +\int_{\Omega}hu. Thus, $\int_{\Omega}hu =a(u) -b(u) >(p-2) b(u) >0,$ and \begin{align*} J_{h}(u) & =\frac{1}{2}a(u) -\frac{1}{p}b( u) -\int_{\Omega}hu\\ & =(\frac{1}{2}-\frac{1}{p}) b(u) -\frac{1}{2} \int_{\Omega}hu\\ & <\frac{p-2}{2p}b(u) -\frac{p-2}{2}b(u) \\ & =-\frac{(p-1) (p-2) }{2p}b(u) <0. \end{align*} Then\alpha_{h}(\Omega) =\underset{u\in\mathbf{M}_{h}}{\inf }J_{h}(u) \leq\underset{u\in\mathbf{M}_{h}^{+}}{\inf} J_{h}(u) =\alpha_{h}^{+}(\Omega) <0$. \newline$(ii) $For$u\in\mathbf{M}_{h}$, we have$a( u) -b(u) -\int_{\Omega}hu=0. Then \begin{align*} J_{h}(u) & =(\frac{1}{2}-\frac{1}{p}) a( u) -(1-\frac{1}{p}) \int_{\Omega}hu\\ & \geq(\frac{1}{2}-\frac{1}{p}) \| u\| _{H^{1}} ^{2}-(1-\frac{1}{p}) \| h\| _{L^{2}}\| u\| _{H^{1}}\\ & =(\frac{1}{2}-\frac{1}{p}) \big(\| u\|_{H^{1}}-\frac{p-1}{p-2}\| h\| _{L^{2}}\big) ^{2} -\frac{1}{2p(p-2) }\left[ (p-1) \| h\| _{L^{2}}\right] ^{2}\\ & \geq-\frac{1}{2p(p-2) }\left[ (p-1) \| h\| _{L^{2}}\right] ^{2}. \end{align*} Thus,J_{h}$is coercive and bounded below on$\mathbf{M}_{h}$. \newline$(iii) $Let$\{ u_{n}\} $be a minimizing sequence in$\mathbf{M}_{h}$for$J_{h}$. Since$J_{h}$is coercive on$\mathbf{M}_{h} $, we can assume$\{ u_{n}\} $is bounded in$\mathbf{M}_{h}$. By the Sobolev embedding theorem, a$c>0$exists such that$| \langle \psi'(u_{n}) ,u_{n}\rangle | =| a(u_{n}) -(p-1) b( u_{n}) | \leq c$. Thus,$\limsup_{n\to\infty}| \langle \psi'(u_{n}) ,u_{n}\rangle | <\infty$. Suppose that there is a minimizing sequence$\{ w_{n}\} $in$\mathbf{M}_{h}$for$J_{h}$such that$\langle \psi'(w_{n}) ,w_{n}\rangle = o(1) $. Since$J_{h}$is a continuous function with$J_{h}(0) =0$, by part$(i) $,$\alpha_{h}(\Omega) <0$. We claim that there is a$\delta>0 $such that$\| w_{n}\| _{H^{1}} >\delta$for each$n$. Otherwise, a subsequence$\{ w_{n}\} $exists such that$\| w_{n}\| _{H^{1}}= o(1) $. Then$J_{h}(w_{n}) = o(1) $, which is a contradiction. Since$\| w_{n}\| _{H^{1}}>\delta$for each$n$and $$o(1) =\langle \psi'(w_{n}) ,w_{n}\rangle =a(w_{n}) -(p-1) b( w_{n}) ,\label{15-4}$$ there is a$\gamma>0$such that$b(w_{n}) \geq\gamma$for each$n$, and $\big(\frac{a(w_{n}) ^{p-1}}{b(w_{n}) }\big) ^{\frac{1}{p-2}}=(p-1) ^{\frac{^{p-1}}{p-2}}b( w_{n}) + o(1) .$ Since$w_{n}\in\mathbf{M}_{h}, by (\ref{15-1}) and (\ref{15-4}), we have $I(w_{n}) \geq\| w_{n}\| _{H^{1}}(d( p,\alpha) -\| h\| _{L^{2}}) \geq\delta( d(p,\alpha) -\| h\| _{L^{2}})$ and $\int_{\Omega}hw_{n}=a(w_{n}) -b(w_{n}) =( p-2) b(w_{n}) + o(1) .$ Now we have \begin{align*} 0 & <\delta(d(p,\alpha) -\| h\| _{L^{2}}) \leq I(w_{n}) \\ & =(\frac{1}{p-1}) ^{\frac{p-1}{p-2}}(p-2) \big( \frac{a(w_{n}) ^{p-1}}{b(w_{n}) }\big) ^{\frac{1}{p-2}}-\int_{\Omega}hw_{n}\\ & =(\frac{1}{p-1}) ^{\frac{p-1}{p-2}}(p-2) ( p-1) ^{\frac{^{p-1}}{p-2}}b(w_{n}) -(p-2) b(w_{n}) + o(1) \\ & = o(1) , \end{align*} which is a contradiction. \end{proof} \begin{lemma} \label{g6} Letu$be in$\mathbf{M}_{h}$such that$J_{h}(u) =\underset{v\in\mathbf{M}_{h}}{\min}J_{h}(v) =\alpha_{h} (\Omega)$. Then\newline$(i)\int_{\Omega}hu>0$; \newline$(ii)u$is a solution of Equation \eqref{E2} in$\Omega$. \end{lemma} \begin{proof}$(i)$By Lemma \ref{g5}$(i)$, we have $0>J_{h}(u) =(\frac{1}{2}-\frac{1}{p}) a(u)-(1-\frac{1}{p}) \int_{\Omega}hu.$ Thus, $\int_{\Omega}hu>0.$$(ii)$By Lemma \ref{g3} $\langle \psi'(v),v\rangle =a(v)-(p-1)b(v)\neq0\quad \text{for each } v\in\mathbf{M}_{h}.$ Since$J_{h}(u) =\min_{v\in\mathbf{M}_{h}} J_{h}(v) $, by the Lagrange multiplier theorem, there is a$\lambda\in\mathbb{R}^{N}$such that$J'(u)=\lambda\psi'(u) $in$H^{-1}(\Omega) $. Then we have $0=\langle J_{h}'(u),u\rangle =\lambda\langle \psi'(u),u\rangle .$ Thus,$\lambda=0$and$J_{h}'(u)=0$in$H^{-1}(\Omega) $. Therefore,$u$is a solution of Equation \eqref{E2} in$\Omega$with$J_{h}(u)=\alpha_{h}(\Omega) $. \end{proof} The following Lemma is required to prove the existence of the (PS)$_{\alpha_{h}(\Omega) }$- sequence for$J_{h}$. \begin{lemma} \label{g7} Given$u\in\mathbf{M}_{h}$, then a$\delta>0$and a differentiable functional$l:B(0;\delta) \subset H_{0}^{1}( \Omega) \to\mathbb{R}^{+}$exist such that$l(0) =1$,$l(v) (u-v) \in\mathbf{M}_{h}$for$v\in B(0;\delta) $and $\langle l'(v) ,\varphi\rangle \big|_{(l,v)=(1,0)} =\frac{\langle \psi'(u) ,\varphi\rangle }{\langle \psi'(u) ,u\rangle }\quad\text{for }\varphi\in C_{c}^{\infty}(\Omega).$ \end{lemma} \begin{proof} For$u\in\mathbf{M}_{h}$, let$G:\mathbb{R}\times H_{0}^{1}( \Omega) \to\mathbb{R}$be given by $G(l,v) =\psi(l(u-v) ) .$ Note that$G(1,0) =\psi(u)=\langle J_{h}^{\prime }(u),u\rangle =0. Then by Lemma \ref{g3} \begin{align*} D_{l}G(1,0) & =\frac{\partial}{\partial l}\big[ l^{2}a(u-v)-| l| ^{p}b(u-v)-l\int_{\Omega}h(u-v) \big] \big| _{(1,0)} \\ & =\big[ 2la(u-v)-p| l| ^{p-2}lb(u-v)-\int_{\Omega }h(u-v)\big] \big| _{(1,0)} \\ & =2a(u)-pb(u)-(a(u)-b(u))\\ & =a(u) -(p-1) b(u) \neq0. \end{align*} By the implicit function theorem, there exist\delta>0$and a differentiable functional$l:B(0;\delta) \subset H_{0}^{1}(\Omega)\to\mathbb{R}$such that$l(0) =1$and $G(l(v),v)=0\;\text{for }v\in B(0;\delta) ,$ Thus,$l(v) (u-v) \in\mathbf{M}_{h}$for$v\in B(0;\delta) $. Moreover,$\varphi\in C_{c}^{\infty}(\Omega)\begin{align*} D_{\varphi}l(v)| _{(1,0)} & =\langle l'( v) ,\varphi\rangle | _{(1,0)} =-\frac{G_{v} (l,v)}{G_{l}(l,v)}| _{(1,0)} \\ & =-\frac{-2 {\int_{\Omega}} \nabla u\nabla\varphi+u\varphi+p {\int_{\Omega}} | u| ^{p-2}u\varphi+ {\int_{\Omega}} h\varphi}{a(u) -(p-1) b(u) }\\ & =\frac{\langle \psi'(u),\varphi\rangle }{\langle \psi'(u),u\rangle }. \end{align*} \end{proof} \begin{proposition} \label{g8}(i) $A (PS)$_{\alpha_{h}(\Omega) } $-sequence$\{ u_{n}\} $exists in$\mathbf{M}_{h}$for$J_{h} $; \newline$(ii) $A (PS)$_{\alpha_{h}^{+}(\Omega) }$-sequence$\{ u_{n}\} $exists in$\mathbf{M}_{h}^{+}$for$J_{h}$; \newline$(iii) $A (PS)$_{\alpha_{h}^{-}( \Omega) }$-sequence$\{ u_{n}\} $exists in$\mathbf{M} _{h}^{-}$for$J_{h}$. \end{proposition} \begin{proof}$(i) $Let$\{ v_{n}\} $be a minimizing sequence in$\mathbf{M}_{h}$for$J_{h}$. Since$J_{h}$is continuous and bounded below on$\mathbf{M}_{h}$, by the Ekeland variational principle we have a minimizing sequence$\{ u_{n}\} $in$\mathbf{M}_{h}$such that \newline$(a) \;J_{h}(u_{n}) \leq J_{h}( v_{n}) <\alpha_{h}(\Omega) +\frac{1}{n^{2}}$; \newline$(b) \;\| u_{n}-v_{n}\| _{H^{1}}=o( 1) $; \newline$(c) \;J_{h}(w) \geq J_{h}(u_{n}) -\frac{1}{n}\| u_{n}-w\| _{H^{1}}$for each$w\in\mathbf{M}_{h}$.\newline Assume that there is an$n_{0}>0$such that$\| J_{h}'(u_{n}) \| _{H^{-1}}>0$for$n\geq n_{0}$, otherwise we are done. For$n\geq n_{0}$, by the Riesz representation theorem, a unique$\phi_{n}\in H_{0}^{1}(\Omega)$exists such that$\| \phi_{n}\| _{H^{1}}=1$and $\langle \frac{J_{h}'(u_{n})}{\| J_{h}' (u_{n})\| _{H^{-1}}},\varphi\rangle =\langle \phi _{n},\varphi\rangle _{H^{1}}\quad\text{for each }\varphi\in H_{0}^{1} (\Omega).$ Let$t_{n}(\varepsilon) =l_{n}(\varepsilon\phi _{n}) $. Applying Lemma \ref{g7}, we have $w_{\varepsilon}=t_{n}(\varepsilon) \left[ u_{n}-\varepsilon \phi_{n}\right] \in\mathbf{M}_{h}.$ Now, $t_{n}'(0) =\underset{\varepsilon\to0}{\lim}\frac {l_{n}(\varepsilon\phi_{n}) -l_{n}(0) }{\varepsilon} =\langle l_{n}'(0) ,\phi_{n}\rangle =\frac{\langle \psi'(u_{n}) ,\phi_{n}\rangle }{\langle \psi'(u_{n}) ,u_{n}\rangle }.$ By Lemma \ref{g5}$(iii)$, we have$0<\limsup_{n\to\infty}| \langle \psi'(u_{n}),u_{n}\rangle | <\infty$. Thus there is a subsequence$\{u_{n}\} $and$c_{1}>0$such that $| \langle \psi'(u_{n}) ,u_{n}\rangle | \geq c_{1}.$ By the H\"{o}lder inequality and$\| \phi_{n}\| _{H^{1}}=1, we obtain \begin{align*} | \langle \psi'(u_{n}) ,\phi_{n}\rangle | & =| 2\langle u_{n},\phi_{n}\rangle _{H^{1}} -p\int_{\Omega}| u_{n}| ^{p-2}u_{n}\phi_{n}-\int_{\Omega} h\phi_{n}| \\ & \leq2\| u_{n}\| _{H^{1}}+p\| u_{n}\| _{L^{p} }^{p-1}\| \phi_{n}\| _{L^{p}}+\| h\| _{L^{2}}\\ & \leq c_{1}+\| h\| _{L^{2}}. \end{align*} Thus,| t_{n}'(0) | \leq c_{2}$for each$n\geq n_{0}$. Moreover,$\varepsilon_{0}$and$c_{3}>0$exist such that for$\varepsilon\leq\varepsilon_{0}\begin{align*} \frac{\| u_{n}-w_{\varepsilon}\| _{H^{1}}}{\varepsilon} & =\frac{1}{\varepsilon}\| (1-t_{n}(\varepsilon)) u_{n}+\varepsilon t_{n}(\varepsilon)\phi_{n}\| _{H^{1}}\\ & \leq\big[ \frac{| t_{n}(0)-t_{n}(\varepsilon)| }{\varepsilon }\| u_{n}\| _{H^{1}}+t_{n}(\varepsilon)\| \phi_{n}\| _{H^{1}}\big] \\ & =| t_{n}'(0) | \| u_{n}\| _{H^{1}}+1+ o(1)\quad \quad\text{as }\varepsilon\to0.\\ & \leq c_{3}. \end{align*} Note that\{ u_{n}\} $is bounded in$H_{0}^{1}(\Omega), so \begin{align*} &\| t_{0}u_{n}+(1-t_{0})w_{\varepsilon}-u_{n}\| _{H^{1}} \\ &=\| (1-t_{0})(t_{n}(\varepsilon)-1) u_{n}-\varepsilon (1-t_{0})t_{n}(\varepsilon)\phi_{n}\| _{H^{1}}\\ & \leq(1-t_{0})| t_{n}(\varepsilon)-1| \| u_{n}\| _{H^{1}}+\varepsilon(1-t_{0})t_{n}(\varepsilon)\\ & = o(1)\quad \text{as }\varepsilon\to0, \end{align*} SinceJ_{h}\in C^{1}, we have \begin{align*} & \frac{1}{\varepsilon}| \langle J_{h}'(t_{0} u_{n}+(1-t_{0})w_{\varepsilon}) ,(u_{n}-w_{\varepsilon}) \rangle -\langle J_{h}'(u_{n}) ,( u_{n}-w_{\varepsilon}) \rangle | \\ & \leq\| J_{h}'(t_{0}u_{n}+(1-t_{0})w_{\varepsilon }) -J_{h}'(u_{n}) \| _{H^{-1}} \frac{\| u_{n}-w_{\varepsilon}\| _{H^{1}}}{\varepsilon}\\ & = o(1)\quad\text{as }\varepsilon\to0. \end{align*} Thus, $\langle J_{h}'(t_{0}u_{n}+(1-t_{0})w_{\varepsilon}) ,(u_{n}-w_{\varepsilon}) \rangle =\langle J_{h}'(u_{n}) ,(u_{n}-w_{\varepsilon}) \rangle +o(\varepsilon),$ where\frac{o(\varepsilon)}{\varepsilon}\to0$as$\varepsilon \to0$. By condition$(c) and the mean value theorem, we have \begin{align*} \frac{1}{n}\| u_{n}-w_{\varepsilon}\| _{H^{1}} & \geq J_{h}(u_{n}) -J_{h}(w_{\varepsilon}) \\ & =\langle J_{h}'(t_{0}u_{n}+(1-t_{0})w_{\varepsilon }) ,(u_{n}-w_{\varepsilon}) \rangle, \end{align*} wheret_{0}\in(0,1)$. Then we obtain $$\frac{1}{n}\| u_{n}-w_{\varepsilon}\| _{H^{1}}\geq\langle J_{h}'(u_{n}) ,(u_{n}-w_{\varepsilon}) \rangle +o(\varepsilon).\label{15-5}$$ We divide$(\ref{15-5})$by$\varepsilon>0and obtain \begin{align*} \frac{c_{3}}{n} & \geq\frac{1}{n\varepsilon}\| u_{n}-w_{\varepsilon }\| _{H^{1}}\\ &\geq\frac{(1-t_{n}(\varepsilon)) }{\varepsilon}\langle J_{h}'(u_{n}) ,u_{n} \rangle +t_{n}(\varepsilon)\langle J_{h}'( u_{n}) ,\phi_{n}\rangle +\frac{o(\varepsilon)}{\varepsilon}\\ & =\frac{(1-t_{n}(\varepsilon)) }{\varepsilon}\langle J_{h}'(u_{n}) ,u_{n}\rangle +t_{n}(\varepsilon )\| J_{h}'(u_{n}) \| _{H^{-1}} +\frac{o(\varepsilon)}{\varepsilon}. \end{align*} Since\{u_{n}\}\subset\mathbf{M}_{h}$, let$\varepsilon\to0$to obtain $0\geq-t_{n}'(0) \langle J_{h}'( u_{n}) ,u_{n}\rangle +\| J_{h}'( u_{n}) \| _{H^{-1}}=\| J_{h}'(u_{n}) \| _{H^{-1}}\text{, }$ which is a contradiction. Therefore,$\| J_{h}'( u_{n}) \| _{H^{-1}}\to0$as$n\to\infty$.\newline$(ii) $and$(iii) $can be proved similarly. \end{proof} \subsubsection{Existence of a Local Minimum} By Proposition \ref{g8}$(i)$, there is a (PS)$_{\alpha_{h}( \Omega) }$-sequence$\{ u_{n}\} $in$\mathbf{M}_{h}$for$J_{h}$. Then we have the following (PS)$_{\alpha_{h}(\Omega) }$-condition. \begin{proposition} \label{m1} Let$\{ u_{n}\} $in$\mathbf{M}_{h}$be a (PS)$_{\alpha_{h}(\Omega) }$-sequence for$J_{h}$. Then a subsequence$\{ u_{n}\} $and$u_{0}$in$H_{0}^{1}( \Omega) $exist such that$u_{n}\to u_{0}$strongly in$H_{0}^{1}(\Omega) $. Furthermore,$u_{0}$is a solution of Equation$(\ref{E2}) $such that$J_{h}(u_{0}) =\alpha_{h}(\Omega) $. \end{proposition} \begin{proof} By Lemma \ref{g5}$(ii) $,$\{ u_{n}\} $is bounded in$H_{0}^{1}(\Omega)$. Take a subsequence$\{ u_{n}\} $and$u_{0}\in H_{0}^{1}(\Omega) $such that$u_{n}\rightharpoonup u_{0}$weakly in$H_{0}^{1}(\Omega) $. By Lemma \ref{p4},$u_{0} $is a nonzero solution of Equation$(\ref{E2}) $in$\Omega$. Since $$\begin{gathered} J_{h}(u_{n}) =\frac{1}{2}a(u_{n}) -\frac{1} {p}b(u_{n}) -\int_{\Omega}hu_{n}=\alpha_{h}(\Omega) + o(1) ,\\ \langle J_{h}'(u_{n}) ,u_{n}\rangle =a(u_{n}) -b(u_{n}) -\int_{\Omega}hu_{n}= o(1). \end{gathered}\label{15-6}$$ By (\ref{15-6}), we have $(\frac{1}{2}-\frac{1}{p}) a(u_{n}) -( 1-\frac{1}{p}) \int_{\Omega}hu_{n}=\alpha_{h}(\Omega) + o(1) .$ Since the functional$a$is weakly lower semicontinuous and$\int_{\Omega }hu_{n}\to\int_{\Omega}hu_{0}$as$n\to\infty, we have \begin{align*} \alpha_{h}(\Omega) & \leq J_{h}(u_{0}) =( \frac{1}{2}-\frac{1}{p}) a(u_{0}) -(1-\frac{1} {p}) \int_{\Omega}hu_{0}\\ & \leq(\frac{1}{2}-\frac{1}{p}) \liminf_{n\to\infty}a(u_{n}) -(1-\frac{1}{p}) \lim_{n\to\infty}\int_{\Omega}hu_{n}\\ & =\liminf_{n\to\infty}\big[ (\frac{1}{2}-\frac {1}{p}) a(u_{n}) -(1-\frac{1}{p})\int_{\Omega}hu_{n}\big] \\ & =\alpha_{h}(\Omega) \end{align*} orJ_{h}(u_{0}) =\alpha_{h}(\Omega) $. Let$p_{n}=u_{n}-u_{0}. By Lemma \ref{p4} and \ref{p7}, we have \begin{aligned} J_{h}(p_{n}) & =\frac{1}{2}a(p_{n}) -\frac{1} {p}b(p_{n}) -\int_{\Omega}hp_{n} \\ & =\frac{1}{2}a(u_{n}) -\frac{1}{2}a(u_{0}) -\frac{1}{p}b(u_{n}) +\frac{1}{p}b(u_{0}) -\int_{\Omega}hu_{n}+\int_{\Omega}hu_{0}+ o(1) \\ & =J_{h}(u_{n}) -J_{h}(u_{0}) + o(1) = o(1) \end{aligned} .\label{15-7} By Lemma \ref{p4}, \ref{p7},\int_{\Omega}hp_{n}= o(1) $and$u_{0}$is a solution of Equation$(\ref{E2}) , so \begin{aligned} \langle J_{h}'(p_{n}) ,p_{n}\rangle & =a(p_{n}) -b(p_{n}) -\int_{\Omega}hp_{n} \\ & =a(u_{n}) -a(u_{0}) -b(u_{n}) +b(u_{0}) -\int_{\Omega}hu_{n}+\int_{\Omega}hu_{0}+o(1)\\ & =\langle J_{h}'(u_{n}) ,u_{n}\rangle -\langle J_{h}'(u_{0}) ,u_{0}\rangle = o(1) \end{aligned} .\label{15-8} Thus, by(\ref{15-7}) $,$(\ref{15-8}) $and$\int_{\Omega}hp_{n}= o(1) $, we have $\frac{p-2}{2p}a(p_{n}) = o(1)$ or$u_{n}\to u_{0}\text{\ strongly in }H_{0}^{1}(\Omega)$. \end{proof} The following result is required to prove that$u_{0}$is the unique critical point of$J_{h}(u) $in$B(r_{0}) $. \begin{lemma} \label{m2} Let$r_{0}=(\frac{1}{p-1}) ^{\frac{1}{p-2}}( \frac{2p}{p-2}) ^{1/2}\alpha(\Omega) ^{\frac {1}{2}}$. Then \newline$(i) \;\mathbf{M}_{h}^{+}\subset B( r_{0}) =\{ u\in H_{0}^{1}(\Omega) : \| u\| _{H^{1}}(p-1) b(u) $and$a(u) =b( u) +\int_{\Omega}hu. Thus, $a(u) <\frac{1}{p-1}a(u) +\| h\| _{L^{2}}\| u\| _{H^{1}}.$ This implies \begin{align*} \| u\| _{H^{1}} & <(\frac{p-1}{p-2}) \|h\| _{L^{2}}\\ & <(\frac{p-1}{p-2}) (p-2) (\frac{1} {p-1}) ^{\frac{p-1}{p-2}}(\frac{2p}{p-2}) ^{\frac{1}{2} }\alpha(\Omega) ^{1/2}\\ & =(\frac{1}{p-1}) ^{\frac{1}{p-2}}(\frac{2p} {p-2}) ^{1/2}\alpha(\Omega) ^{1/2}=r_{0}. \end{align*}(ii) $Similarly to Adachi-Tanaka \cite{AT}, we have $J_{h}''(u) (v,v) =a(v) -(p-1) \int_{\Omega}| u| ^{p-2}v^{2}\quad\text{for all }v\in H_{0}^{1}(\Omega) .$ Thus, by Lemma \ref{g2} for$u\in H_{0}^{1}(\Omega)\backslash\{ 0\} $\big[ \frac{a(u)^{p/2}}{b(u)}\big] ^{\frac{1}{p-2}}\geq( \frac{2p}{p-2}) ^{1/2}\alpha(\Omega) ^{1/2},$ then \begin{align*} J_{h}''(u) (v,v) & \geq a( v) -(p-1) \| u\| _{L^{p}}^{p-2}\| v\| _{L^{p}}^{2}\\ & \geq a(v) -(p-1) \big[ a(u)^{\frac{p-2}{2} }(\frac{p-2}{2p}) ^{\frac{p-2}{2}}\alpha(\Omega) ^{-\frac{(p-2)^{2}}{2p}}\big] \\ & \quad\times \big[ a(v)(\frac{p-2}{2p}) ^{\frac{p-2}{p}}\alpha( \Omega) ^{\frac{-(p-2) }{p}}\big] \\ & \geq a(v)\big[ 1-(p-1) (\frac{2p}{p-2}\alpha( \Omega) ) ^{\frac{2-p}{2}}\| u\| _{H^{1}} ^{p-2}\big] \\ & >0\quad \text{for }u\in B(r_{0}) . \end{align*} Thus,J_{h}''(u) $is positive definite for$u\in B(r_{0}) $and$J_{h}$is strictly convex in$B( r_{0}) $. \end{proof} By Proposition \ref{m1}, a solution$u_{0}\in\mathbf{M}_{h}$of Equation$(\ref{E2}) $exists such that$J_{h}(u_{0})=\alpha _{h}(\Omega)$. Furthermore, we have the following theorem. \begin{theorem} \label{m3}$(i)u_{0}\in\mathbf{M}_{h}^{+}$and$J_{h}(u_{0})=\alpha_{h} ^{+}(\Omega)=\alpha_{h}(\Omega)$; \newline$(ii)u_{0}$is the unique critical point of$J_{h}(u) $in$B(r_{0}) $, where$r_{0}$is as in Lemma \ref{m2}; \newline$(iii)J_{h}(u_{0})$is a local minimum in$H_{0}^{1}(\Omega) $. \end{theorem} \begin{proof}$(i)$By Lemma \ref{g6}$(i)$,$\int_{\Omega}hu_{0}>0$. We claim that$u_{0}\in\mathbf{M}_{h}^{+}$. Otherwise, if$u_{0}\in\mathbf{M}_{h}^{-}$, then by Lemma \ref{g4} a unique$t^{-}(u_{0})=1>t^{+}(u_{0})>0$exists such that$t^{+}(u_{0})u_{0}\in\mathbf{M}_{h}^{+}$and $\alpha_{h}(\Omega) \leq\alpha_{h}^{+}(\Omega) \leq J_{h}(t^{+}(u_{0})u_{0})0 and a differentiable functional l(w) >0 exist such that l(0) =1 and $$l(w) (u_{0}+w) \in\mathbf{M}_{h}\quad\text{for }\| w\| _{H^{1}}<\delta_{1}.\label{15-9}$$ By Lemma \ref{g4}, 1=t^{+}(u_{0})\delta_{2}>0, such that l(w)\delta>0 such that \[ 10, then u_{0} is a nonnegative solution of Equation \eqref{E1} in \Omega. Moreover, a positive solution of equation (\text{\ref{E2} }) exists for h\gvertneqq0; \newline(ii) If \int_{\Omega}h| u_{0}| <0, then u_{0} is a nonpositive solution of Equation ( \text{\ref{E2}}) in \Omega. Moreover, a negative solution of Equation (\text{\ref{E2}}) exists for h\lvertneqq0; \newline(iii) If \int_{\Omega}h| u_{0}| =0, then u_{0} is a solution of Equation (\text{\ref{E2}}) in \Omega that changes sign. \end{theorem} \begin{proof} (i) If \int_{\Omega}h| u_{0}| >0, then by Lemma \ref{g4} t_{\rm max}(| u_{0}| )>t^{+}(| u_{0}| ) >0 exists such that t^{+}(| u_{0}| ) | u_{0}| \in\mathbf{M}_{h}^{+}. Since t_{\rm max}(| u_{0}| )=t_{\rm max}(u_{0})>1, we have \[ \alpha_{h}(\Omega) =\alpha_{h}^{+}(\Omega) \leq J_{h}(t^{+}(| u_{0}| ) | u_{0}| ) \leq J_{h}(| u_{0}| ) \leq J_{h}( u_{0}) =\alpha_{h}(\Omega) ,$ or$J_{h}(| u_{0}| ) =\alpha_{h}( \Omega) $. By Lemma \ref{g6}$(ii)$and Lemma \ref{m3}$(ii)$,$u_{0}=| u_{0}| $. Thus, we can take$u_{0}\geq0$. Moreover, if$h\gvertneqq0$, we apply the maximum principle and obtain$u_{0}>0$. \newline$(ii)$The proof is similar to$(i)$.\newline$(iii)$Let$u_{0} ^{+}=\max\{ u_{0},0\} $and$u_{0}^{-}=\max\{ -u_{0},0\} $. Since$\int_{\Omega}h| u_{0}| =0$, then $\int_{\Omega}hu_{0}^{+}+\int_{\Omega}hu_{0}^{-}=0.$ By Lemma \ref{g6}$(i) $, we have$\int_{\Omega}hu_{0}^{+}>0$and$\int_{\Omega}hu_{0}^{\_}<0$. Thus,$u_{0}^{+}\gvertneqq0$and$u_{0} ^{-}\gvertneqq0$. Hence,$u_{0}$is a solution of equation$( \ref{E2}) $in$\Omega$that changes sign. \end{proof} \subsubsection{Existence of Two Solutions} Let$u_{0}$be the local minimum for$J_{h}$in$H_{0}^{1}( \Omega) $in Theorem \ref{m3}. Then we have the following restricted (PS)-condition. \begin{proposition} \label{m5} If$\{ u_{n}\} $is a (PS)$_{\beta}$-sequence in$H_{0}^{1}(\Omega) $for$J_{h}$with$\beta<\alpha_{h}( \Omega) +\alpha(\Omega) $, then a subsequence$\{ u_{n}\} $and$u^{0}$in$H_{0}^{1}(\Omega) $exist such that$u_{n}\to u^{0}$strongly in$H_{0}^{1}(\Omega) $and$J_{h}(u^{0}) =\beta$. \end{proposition} \begin{proof} Let$\{u_{n}\}$be a (PS)$_{\beta}$-sequence in$H_{0}^{1}( \Omega) $for$J_{h}$. By Lemma \ref{g5}$(ii) $,$\{ u_{n}\} $is bounded. As in the proof of Proposition \ref{m1}, a subsequence$\{u_{n}\}$and a solution$u^{0}$of Equation \eqref{E2} exist such that$u_{n}\rightharpoonup u^{0}$weakly in$H_{0} ^{1}(\Omega)$. Suppose that$u_{n}\nrightarrow u^{0}$strongly in$H_{0} ^{1}(\Omega)$. Let$p_{n}=u_{n}-u^{0}$for$n=1,2,\dots $. By Lemma \ref{p4} and \ref{p7}, we have $$\label{15-12} \begin{gathered} a(p_{n}) =a(u_{n}) -a(u^{0})+0(1) ,\\ b(p_{n}) =b(u_{n}) -b(u^{0}) +0(1) . \end{gathered}$$ Since$a(u_{n}) -b(u_{n}) -\int_{\Omega} hu_{n}= o(1) $,$a(u^{0}) -b(u^{0}) -\int_{\Omega}hu^{0}=0$and$u_{n}\rightharpoonup u^{0}$weakly in$H_{0} ^{1}(\Omega), we have \begin{align*} a(p_{n}) & =a(u_{n}) -a(u^{0})+0(1) \\ & =b(u_{n}) +\int_{\Omega}hu_{n}-b(u^{0}) -\int_{\Omega}hu^{0}+ o(1) \\ & =b(p_{n}) + o(1) . \end{align*} Sincep_{n}\nrightarrow0$, we have $J(p_{n}) =\frac{1}{2}a(p_{n}) -\frac{1}{p}b(p_{n}) =(\frac{1}{2}-\frac{1}{p}) a(p_{n}) +o(1) >0.$ By Theorem \ref{i3}, a sequence$\{ s_{n}\} $in$\mathbb{R}^{+}$exists such that$\{ s_{n}p_{n}\} $in$\mathbf{M}( \Omega) $and$J(s_{n}p_{n}) =J(p_{n}) + o(1) $. Thus, by$(\ref{15-12}) $and$u_{n}\rightharpoonup u^{0}$weakly in$H_{0}^{1}(\Omega), we have \begin{align*} \alpha(\Omega) & \leq J(s_{n}p_{n}) =J( p_{n}) + o(1) \\ & =J_{h}(p_{n}) + o(1) \\ & =J_{h}(u_{n}) -J_{h}(u^{0}) + o(1) \\ & =\beta-J_{h}(u^{0}) + o(1) \\ & <\alpha_{h}(\Omega) +\alpha(\Omega) -J_{h}(u^{0}) + o(1) . \end{align*} Then\alpha_{h}(\Omega) >J_{h}(u^{0}) \geq \alpha_{h}(\Omega) $, which is a contradiction. Thus,$u_{n}\to u^{0}$strongly in$H_{0}^{1}(\Omega)$. \end{proof} Throughout this section, let$\Omega$be an achieved domain in$\mathbb{R}^{N}$. \begin{lemma} \label{m6} Let$\overline{u}$be a positive solution of Equation$($\ref{E1}$)$in$\Omega$such that$J(\overline{u})=\alpha( \Omega) $and$u_{0}$is the local minimum in Theorem \ref{m3}. Then\newline$(i) $If$\int_{\Omega}h| u_{0}| >0, then we have \underset{t\geq0}{\sup}J_{h}(u_{0}+t\overline{u})0, by Theorem \ref{m4} (i), u_{0} is a nonnegative solution of Equation (\ref{E2}). Let f(s)=s^{p-1} and F(u)= {\int_{0}^{u}} f(s)ds=\frac{1}{p}b(u), then \label{15-13} \begin{aligned} &J_{h}(u_{0}+t\overline{u}) \\ & =\frac{1}{2}a(u_{0}+t\overline{u})-\frac{1} {p}b(u_{0}+t\overline{u})-\int_{\Omega}h(u_{0}+t\overline{u})\\ & =\frac{1}{2}\left[ a(u_{0})+a(t\overline{u})+2\langle u_{0} ,t\overline{u}\rangle _{H^{1}}\right] -\frac{1}{p}b(u_{0}+t\overline {u})-\int_{\Omega}h(u_{0}+t\overline{u})\\ & =J_{h}(u_{0})+J(t\overline{u})+\langle u_{0},t\overline{u}\rangle _{H^{1}}+\frac{1}{p}\left[ b(u_{0})+b(t\overline{u})-b(u_{0}+t\overline {u})\right] -\int_{\Omega}ht\overline{u}\\ & =J_{h}(u_{0})+J(t\overline{u})+t(\int_{\Omega}u_{0}^{p-1}\overline {u}+h\overline{u}) -\int_{\Omega}ht\overline{u}\\ & +\frac{1}{p}\left[ b(u_{0})+b(t\overline{u})-b(u_{0}+t\overline{u})\right] \\ & =J_{h}(u_{0})+J(t\overline{u})-\int_{\Omega}\big\{ \int_{0}^{t\overline {u}}[ f(u_{0}+s)-f(s)-f(u_{0})] ds\big\} . \end{aligned} For v>0 and w>0, we have $$f(v+w) =(v+w) ^{p-1} =(v+w) ^{p-2}v+(v+w) ^{p-2}w >v^{p-1}+w^{p-1}=f(v)+f(w).\label{15-14}$$ Thus, J_{h}(u_{0}+t\overline{u})\leq J_{h}(u_{0})+J(t\overline{u}). Since J(t\overline{u})\to-\infty as t\to\infty, there is a t_{0}>0 such that J_{h}(u_{0}+t\overline{u})0 there is a t_{0}>t_{1}>0 such that g_{1}(t)t\geq0. Then \[ \underset{0\leq t\leq t_{1}}{\sup}J_{h}(u_{0}+t\overline{u}) \leq J_{h}(u_{0})+\frac{1}{2}\alpha(\Omega) 0 such that t^{-}(u) u\in\mathbf{M}_{h}^{-} and \[ J_{h}(t^{-}(u) u) =\max_{t\geq t_{\rm max}} J_{h}(tu) . By Lemma \ref{g4}(ii)$and$(iii)$, we have that$t^{-}(u) $is a continuous function for nonzero$u$and $\mathbf{M}_{h}^{-}=\big\{ u\in H_{0}^{1}(\Omega) \backslash\{ 0\} : \frac{1}{\| u\| _{H^{1}}} t^{-}(\frac{u}{\| u\| _{H^{1}}}) =1\big\} .$ Let \begin{gather*} A_{1}=\big\{ u\in H_{0}^{1}(\Omega) \backslash\{ 0\} :\frac{1}{\| u\| _{H^{1}}}t^{-}( \frac{u}{\| u\| _{H^{1}}}) >1 \big\} \cup\{ 0\} \\ A_{2}=\big\{ u\in H_{0}^{1}(\Omega) \backslash\{ 0\} : \frac{1}{\| u\| _{H^{1}}}t^{-}( \frac{u}{\| u\| _{H^{1}}}) <1 \big\} . \end{gather*} Then$H_{0}^{1}(\Omega) \backslash\mathbf{M}_{h}^{-}=A_{1}\cup A_{2}$. For each$u\in\mathbf{M}_{h}^{+}$, we have $10 exists such that u_{0}+t_{0}\bar{u}\in A_{2}. First, we find a constant c>0 such that 0c^{2}>\left[ t^{-}\big(\frac{u_{0}+t_{0}\overline{u}}{\| u_{0}+t_{0}\overline{u}\| _{H^{1}}}\big) \right] ^{2},$ that is,$u_{0}+t_{0}\overline{u}\in A_{2}$. Define a path$\gamma( s) =u_{0}+st_{0}\overline{u}$for$s\in[ 0,1] $where$t_{0}>1$, then $\gamma(0)=u_{0}\in A_{1}, \quad \gamma(1)=u_{0}+t_{0}\overline{u}\in A_{2}.$ Since$\frac{1}{\| u\| _{H^{1}}}t^{-}(\frac{u}{\| u\| _{H^{1}}}) $is a continuous function for nonzero$u$and$\gamma([0,1])$is connected, a$s_{0}\in(0,1) $exists such that$u_{0}+s_{0}t_{0}\overline{u}\in\mathbf{M}_{h}^{-}. Thus, by Lemma \ref{m4} and Theorem \ref{m6} we have $\alpha_{h}^{-}\leq J_{h}(u_{0}+s_{0}t_{0}\overline{u})\leq\underset {s\in\left[ 0,1\right] }{\max}J_{h}(\gamma(s) ) 0.$ Similarly, we also have $\alpha_{h}^{-}0. By Lemma \ref{g4}, t^{+}(u^{0})>0 exists such that t^{+}(u^{0})u^{0} \in\mathbf{M}_{h}^{+} and \[ \alpha_{h}^{+}(\Omega)\leq J_{h}(t^{+}(u^{0})u^{0})0 exists such that \[ t^{-}(| u^{0}| ) | u^{0}| \in\mathbf{M}_{h}^{-}\text{, }t^{-}(| u^{0}| )>t_{\rm max}(| u^{0}| )=t_{\rm max}(u^{0})$ and \begin{align*} \alpha_{h}^{-}(\Omega) & \leq J_{h}(t^{-}(| u^{0}| ) | u^{0}| ) \leq J_{h}(t^{-}(| u^{0}| ) u^{0}) \\ & \leq\max_{t\geq t_{\rm max}(u^{0})}J_{h}(tu^{0}) =J_{h} (u^{0})=\alpha_{h}^{-}(\Omega). \end{align*} Thus, $J_{h}(t^{-}(| u^{0}| ) | u^{0}| )=J_{h}(t^{-}(| u^{0}| ) u^{0}) =\alpha_{h}^{-}(\Omega).$ We conclude that\int_{\Omega}hu^{0}=\int_{\Omega}h| u^{0}| $. Let$u_{+}^{0}=\max\{ u^{0},0\} $and$u_{-}^{0}=\max\{ -u^{0},0\} $, then$\int_{\Omega}hu_{-}^{0}=0$. Since$h\gvertneqq0$and$u_{-}^{0}\geq0$, we have$u_{-}^{0}=0$. Hence,$u^{0}\geq0$. By the maximum principle,$u^{0}>0$. \end{proof} \begin{remark} \label{m8} \rm By Theorems \ref{m3} and \ref{m7}, there is a unique solution$u_{0}$of Equation$(\ref{E2}) $in$\Omega$such that$J_{h}(u_{0}) =\alpha_{h}^{+}(\Omega) =\alpha _{h}(\Omega) $. \end{remark} \noindent\textbf{Bibliographical notes:} The results of this section are from Lin-Wang-Wu \cite{LiWaWu3}. \subsubsection{Three Solutions} Throughout this section, we consider a$C^{1,1}$domain$\Omega$to be$\mathbf{A}^{r}$,$\mathbb{R}^{N}$,$\mathbf{A}^{r}\backslash \overline{D}$, or$ \mathbb{R}^{N}\backslash \overline{D}$, where$D$is a bounded domain in$\mathbb{R}^{N}$and assume that$h\in L^{\frac{N}{2}}( \Omega) \cap L^{s}(\Omega) \cap L^{2}( \Omega) $for some$s>N$,$0<\| h\| _{L^{2}}0$such that$D\subset B^{N}(0;R_{1}) $. For any$0<\delta<\min\{ \varepsilon,1\} $, we choose$R_{2}>R_{1}>0$such that $$(1+\delta) -\frac{\sqrt{1+\delta}(N-1) }{| z| }\geq1\quad \text{for }| z| \geq R_{2}.\label{15-28}$$ Let$\beta=\sqrt{1+\delta}$and$v_{1}(z) =\mu\exp( -\beta(| z| -R_{2}) ) $, where$\mu=\min_{| z| =R_{2}}u(z) >0$. Then$\min_{| z| =R_{2}} (u-v_{1}) (z) \geq0$. By \eqref{15-28}, for$|z| >R_{2}\begin{align*} \triangle(u-v_{1}) (z) & =u-| u| ^{p-2}u-h(z) -\big(\beta^{2}-\frac{\beta(N-1) }{| z| }\big) v_{1}\\ & \leq u-\big(\beta^{2}-\frac{\beta(N-1) }{| z|}\big) v_{1}\\ & \leq(u-v_{1}) (z) . \end{align*} By Lemma \ref{a3}, for| z| >R_{2}, $u(z) -v_{1}(z) \geq\underset{| z| =R_{2}}{\min}(u-v_{1}) (z) \geq0.$ Thus, we have \label{15-29} \begin{aligned} u(z) & \geq v_{1}(z) \\ & =\mu\exp(-\beta(| z| -R_{2}) ) \\ & =\mu\exp(R_{2}\sqrt{1+\delta}) \exp(-\beta|z| ) \\ & \geq c_{\delta}^{1}\exp(-(1+\delta) | z|) \quad \text{for }| z| \geq R_{2}. \end{aligned} We know that there exist positive numbers\varepsilon$,$c$such that $0\leq h(z) \leq c\exp(-(1+\varepsilon) | z| ) \quad \text{for any }z\in\Omega.$ For any$0<\delta<\min\{ \varepsilon,1\} $, by$( \text{\ref{15-29}}) $, there is$R_{3}>R_{2}>0$such that $$\frac{\delta}{2}u(z) \geq h(z) \quad \text{for }| z| \geq R_{3}. \label{15-30}$$ Since$\lim_{| z| \to\infty}u(z)=0$, there is$R>R_{3}>0$such that $$1-u^{p-2}\geq1-\frac{\delta}{2}\quad \text{for }| z| \geq R.\label{15-31}$$ Let$\gamma=\sqrt{1-\delta}$and$v_{2}(z) =\nu\exp(-\gamma(| z| -R) ) $, where$\nu=\max_{|z|=R}u(z) >0$. Thus$\min{|z|=R}(v_{2}-u) (z) \geq0$. By$(\text{\ref{15-30}}) $and$( \text{\ref{15-31}}) $, for$| z| >R\begin{align*} \triangle(v_{2}-u) (z) & =(\gamma^{2}-\frac{\gamma(N-1) }{| z| }) v_{2}(z) -u+| u| ^{p-2}u+h(z) \\ & \leq\gamma^{2}v_{2}-(1-\frac{\delta}{2}) u+h(z)\\ & =(1-\delta) (v_{2}(z) -u(z)) -\frac{\delta}{2}u+h(z) \\ & \leq(1-\delta) (v_{2}(z) -u(z) ) . \end{align*} By Lemma \ref{a3}, for| z| >R$v_{2}(z) -u(z) \geq\underset{| z| =R}{\min}(v_{2}-u) (z) \geq0.$ Thus, we have \begin{align*} u(z) & \leq v_{2}(z) =\nu\exp( -\gamma(| z| -R) ) \\ & =\nu\exp(R\sqrt{1-\delta}) \exp(-\gamma| z| ) \\ & \leq c_{\delta}^{2}\exp(-(1-\delta) | z| ) \quad\text{for }| z| \geq R. \end{align*} \end{proof} By Lien-Tzeng-Wang \cite{LTW}, there is a positive ground state solution\bar{u}$of the Equation \eqref{E1} in$\mathbb{R}^{N}$such that$J(\bar{u}) =\alpha( \mathbb{R}^{N}) $. By Gidas-Ni-Nirenberg \cite{GNN2}, we have that$\overline{u}$is radially symmetric about$0$in$\mathbb{R}^{N}$. Similarly to Lemma \ref{m16}, for any$\delta'>0$, positive constants$c_{\delta'}^{1}$and$c_{\delta'}^{2}$exist such that $c_{\delta'}^{1}\exp(-(1+\delta') | z| ) \leq\overline{u}(z)\leq c_{\delta'}^{2}\exp( -(1-\delta') | z| ) \quad\text{for } z\in\mathbb{R}^{N}.$ By Lemma \ref{m16}, there is a$R>0$such that$D\subset B(0;R)$. For such$R $, let$\psi_{R}:\mathbb{R}^{N}\to\left[ 0,1\right] $be a$C^{\infty}-$function on$\mathbb{R}^{N}$such that$0\leq\psi_{R}\leq1$, $\psi_{R}(z) =\begin{cases} 1 & \text{for }| z| \geq R+1;\\ 0 & \text{for }| z| \leq R. \end{cases}$ For$\overline{z}\in\mathbb{R}^{N}$, we define $v_{\overline{z}}(z) =\psi_{R}(z) \bar{u}( z-\overline{z}) .$ Clearly,$v_{\overline{z}}(z) \in H_{0}^{1}( \Omega) $. \begin{lemma} \label{m17}$(i) \;a(v_{\overline{z}}) =b( v_{\overline{z}}) + o(1) $as$| \overline {z}| \to\infty$; \newline$(ii)J(v_{\overline{z} }) =\alpha(\Omega) + o(1) $as$| \overline{z}| \to\infty$; \newline$(iii)v_{\overline{z} }\rightharpoonup0$weakly in$H_{0}^{1}(\Omega)$as$| \overline {z}| \to\infty$. \end{lemma} \begin{proof}$(i$-1)\quad$a(v_{\overline{z}}) =a(\bar {u}) + o(1) $as$| \overline{z}| \to\infty:$since$\overline{u}\in H_{0}^{1}(\mathbb{R} ^{N}) , we have \begin{align*} &\| v_{\overline{z}}(z) -\bar{u}(z-\overline {z}) \| _{H^{1}}^{2}\\ &={\int_{\mathbb{R}^{N}}}| \nabla[ (\psi_{R}(z) -1) \bar {u}(z-\overline{z}) ] | ^{2}dz+ {\int_{\mathbb{R}^{N}}}| (\psi_{R}(z) -1) \bar{u}( z-\overline{z}) | ^{2}dz\\ &\leq2 {\int_{\mathbb{R}^{N}}}| \nabla(\psi_{R}(z) -1) | ^{2}| \bar{u}(z-\overline{z}) | ^{2}dz+2 {\int_{\mathbb{R}^{N}}}| (\psi_{R}(z) -1) | ^{2}| \nabla\bar{u}(z-\overline{z}) | ^{2}dz\\ &\quad +{\int_{\{ | z| \leq R+1\} }}| \bar{u}(z-\overline{z}) | ^{2}dz\\ &\leq2 {\int_{\{ R\leq| z| \leq R+1\} }} | \bar{u}(z-\overline{z}) | ^{2}dz+2 {\int_{\{ | z| \leq R+1\} }}(| \nabla\bar{u}(z-\overline{z}) | ^{2}+| \bar{u}(z-\overline{z}) | ^{2})dz= o(1) . \end{align*} Thus,a(v_{\overline{z}}) =a(\bar{u}( z-\overline{z}) ) + o(1) =a(\bar{u}) + o(1) $as$| \overline{z}| \to\infty$. \noindent$(i$-2)\quad$b(v_{\overline{z}}) =b(\bar{u}) + o(1) $as$| \overline{z}| \to\infty$: since$\overline{u}\in H_{0}^{1}(\Omega) , we have \begin{align*} \| v_{\overline{z}}(z) -\bar{u}(z-\overline {z}) \| _{L^{p}}^{p} & =\int_{\mathbb{R}^{N}}| (\psi_{R}(z) -1) | ^{p}| \bar{u}( z-\overline{z}) | ^{p}dz\\ & \leq\int_{\{ | z| \leq R+1\} }| \bar{u}(z-\overline{z}) | ^{p}dz = o(1) . \end{align*} Thus,b(v_{\overline{z}}) =b(\bar{u}( z-\overline{z}) ) + o(1) =b(\bar{u}) + o(1) $as$| \overline{z}| \to\infty$. By$(i$-1),$(i$-2)and that$\bar{u}$is a solution of the Equation \eqref{E1}, we have $a(v_{\overline{z}}) =a(\bar{u}) +o( 1) =b(\bar{u}) + o(1) =b( v_{\overline{z}}) + o(1) \quad\text{as }| \overline {z}| \to\infty.$$(ii)$By Lemma \ref{f8},$J(v_{\overline{z}}) =J(\bar {u}) + o(1) =\alpha(\mathbb{R}^{N}) + o(1) =\alpha(\Omega) + o(1) $as$| \overline{z}| \to\infty$. \newline$(iii) $For$\phi\in C_{c}^{1}(\Omega) $with$K=\mathop{\rm supp}\phi$, then$K\subset\Omegais compact. \begin{align*} | \langle v_{\overline{z}},\phi\rangle _{H^{1}}| & =\big|{\int_{\Omega}}\nabla v_{\overline{z}}(z) \nabla\phi(z)dz+ {\int_{\Omega}}v_{\overline{z}}(z) \phi(z)dz\big| \\ & =\big|{\int_{K}} \nabla\left[ \psi_{R}(z) \bar{u}(z-\overline{z}) \right] \nabla\phi(z)dz+{\int_{K}} \psi_{R}(z) \bar{u}(z-\overline{z}) \phi (z)dz\big| \\ & \leq\| \nabla\left[ \psi_{R}(z) \bar{u}( z-\overline{z}) \right] \| _{L^{2}(K)}\| \nabla \phi\| _{L^{2}(K)}\\ &\quad +\| \psi_{R}(z) \bar{u}(z-\overline{z}) \| _{L^{2}(K)}\| \phi\| _{L^{2}(K)}\\ & = o(1)\quad \text{as }| \overline{z}| \to\infty. \end{align*} By part(i) $, there is a$c>0$such that$\Vert v_{\overline {z}}\Vert_{H^{1}}\leq c$. For$\varepsilon>0$and$\varphi\in H_{0}^{1} (\Omega)$, there exist$\phi\in C_{c}^{1}(\Omega) $and$l_{0}>0such that \begin{gather*} \Vert\varphi-\phi\Vert_{H^{1}} <\varepsilon/2c\\ | \langle v_{\overline{z}},\phi\rangle _{H^{1}}| <\varepsilon/2\quad\text{for }| \overline{z}| \geq l_{0}. \end{gather*} Thus, \begin{align*} \langle v_{\overline{z}},\varphi\rangle _{H^{1}} & =\langle v_{\overline{z}},\varphi-\phi\rangle _{H^{1}}+\langle v_{\overline{z}},\phi\rangle _{H^{1}}\\ & \leq\Vert v_{\overline{z}}\Vert_{H^{1}}\Vert\varphi-\phi\Vert_{H^{1} }+\langle v_{\overline{z}},\phi\rangle _{H^{1}}\\ & 0, \] then $(\int_{D}f(z)e^{-\sigma| z-\overline{z}| }dz) e^{\sigma| \overline{z}| }=\int_{D}f(z)e^{\sigma\frac {\langle z,\overline{z}\rangle }{| \overline{z}| } }dz+ o(1)\quad\text{as }| \overline{z}| \to\infty.$ \end{lemma} \begin{proof} Since\sigma| \overline{z}| \leq\sigma| z| +\sigma| z-\overline{z}| $, we have $| f(z)e^{-\sigma| z-\overline{z}| }e^{\sigma| \overline{z}| }| \leq| f(z)e^{\sigma| z|}| .$ Since$-\sigma| z-\overline{z}| +\sigma| \overline {z}| =\sigma\frac{\langle z,\overline{z}\rangle }{| \overline{z}| }+ o(1)$as$| \overline{z}| \to \infty$, then the lemma follows from Theorem \ref{p13}. \end{proof} Since$\Omega$here is nonachieved, we need a delicate result. \begin{lemma} \label{m19}$l_{0}>0$exists such that for$| \overline{z}| \geq l_{0}$\sup_{t\geq0} J_{h}(u_{0}+tv_{\overline{z}}) 0 such that for 0\leq t0 such that for | \overline{z}| \geq l_{0} \[ \underset{t\geq t_{0}}{\sup}J_{h}(u_{0}+tv_{\overline{z}}) 0 independent of a and b such that \[ (a+b) ^{p}\geq a^{p}+b^{p}+p(a^{p-1}b+ab^{p-1})-ca^{p/2}b^{p/2}.$ Hence, we get \begin{align*} &\int_{\Omega}(u_{0}+tv_{\overline{z}}) ^{p}dz \\ & \geq \int_{\Omega}(u_{0}^{p}+(tv_{\overline{z}}) ^{p} +ptu_{0}^{p-1}v_{\overline{z}}+pu_{0}(tv_{\overline{z}})^{p-1}) dz -c\int_{\Omega}u_{0}^{p/2}(tv_{\overline{z}})^{p/2}dz. \end{align*} By Lemma \ref{i2} and Theorem \ref{i3}, there exists at^{0}( v_{\overline{z}}) >0$such that$t^{0}(v_{\overline{z}}) v_{\overline{z}}\in\mathbf{M}(\Omega) $and $$\max {t\geq0} J(tv_{\overline{z}}) =J( t^{0}(v_{\overline{z}}) v_{\overline{z}}) =\alpha( \Omega) + o(1)\quad\text{as }| \overline{z}| \to \infty.\label{15-32}$$ By$(\ref{15-32}) $and Lemma \ref{m17}, we deduce for$t\geq t_{0}\begin{align*} J_{h}(u_{0}+tv_{\overline{z}}) & =\frac{1}{2}\int_{\Omega }\big[ | \nabla(u_{0}+tv_{\overline{z}}) | ^{2}+(u_{0}+tv_{\overline{z}}) ^{2}\big] dz\\ &\quad -\frac{1}{p}\int_{\Omega}(u_{0}+tv_{\overline{z}}) ^{p} dz-\int_{\Omega}h(u_{0}+tv_{\overline{z}}) dz\\ & \leq J_{h}(u_{0}) +\alpha(\Omega) -t^{\frac {p}{2}}\big(t_{0}^{\frac{p-2}{2}}\int_{\Omega}v_{\overline{z}}^{p-1} u_{0}dz-\frac{c}{p}\int_{\Omega}u_{0}^{p/2}v_{\overline{z}}^{\frac {p}{2}}dz\big) + o(1). \end{align*} Let0<\delta'<\min\{ \varepsilon,\frac{p-2}{p+2}\} $, then$\frac{(1-\delta') p}{2}-(1+\delta ') >0$. We choose $0<\delta<\delta'<(p-2) +(p-1) \delta.$ By Lemma \ref{m16}, \ref{m17} and$u_{0}$is a positive solution of the Equation$(\ref{E2}) , we have \begin{align*} \int_{\Omega}v_{\overline{z}}^{p-1}u_{0}dz & =\int_{\Omega}(\nabla v_{\overline{z}}\nabla u_{0}+v_{\overline{z}}u_{0}) dz+ o(1)\\ & =\int_{\Omega}(v_{\overline{z}}u_{0}^{p-1}+hv_{\overline{z}}) dz+ o(1)\\ & =\int_{\Omega}v_{\overline{z}}u_{0}^{p-1}dz+ o(1)\\ & \geq c_{1}\int_{\{ | z| \geq R+1\} }e^{( -(1+\delta') | z-\overline{z}| ) }e^{(p-1) (-(1+\delta) | z| ) }dz+ o(1), \end{align*} and $\int_{\Omega}u_{0}^{p/2}v_{\overline{z}}^{p/2}dz\leq c_{2} \int_{\{ | z| \geq R\} }e^{(-\frac{( 1-\delta') p}{2}| z-\overline{z}| ) }e^{(-\frac{(1-\delta) p}{2}| z| ) }dz,$ wherec_{1}=c_{1}(\delta,\delta') $and$c_{2} =c_{2}(\delta,\delta') $. Since $\int_{\{ | z| \geq R+1\} }e^{(p-1) (-(1+\delta) | z| ) }e^{( 1+\delta') | z| }dz<\infty$ and $\int_{\{ | z| \geq R\} }e^{(-\frac{( 1-\delta) p}{2}| z| ) }e^{\frac{( 1-\delta') p}{2}| z| }dz<\infty,$ by Lemma \ref{m18}, we deduce that as$| \overline{z}| \to\infty\begin{align*} \int_{\Omega}v_{\overline{z}}^{p-1}u_{0}dz & \geq c_{1}\int_{\{ | z| \geq R+1\} }e^{(-(1+\delta') | z-\overline{z}| ) }e^{(p-1) ( -(1+\delta) | z| ) }dz+ o(1)\\ & =c_{1}\big\{ \int_{\{ | z| \geq R+1\} }e^{(p-1) (-(1+\delta) | z| ) }e^{(1+\delta') \frac{\langle z,\overline{z}\rangle }{| \overline{z}| }}dz+ o(1)\big\} e^{-(1+\delta') | \overline{z}| } \end{align*} and \begin{align*} \int_{\Omega}u_{0}^{p/2}v_{\overline{z}}^{p/2}dz & \leq c_{2}\int_{\{ | z| \geq R\} }e^{( -\frac{(1-\delta') p}{2}| z-\overline{z}| ) }e^{(-\frac{(1-\delta) p}{2}| z| ) }dz\\ & =c_{2}\big\{ \int_{\{ | z| \geq R\} }e^{( -\frac{(1-\delta) p}{2}| z| ) } e^{\frac{(1-\delta') p}{2}\frac{\langle z,\overline{z}\rangle }{| \overline{z}| }}dz+ o(1)\big\} e^{-\frac{(1-\delta') p}{2}| \overline{z}|}. \end{align*} Thus, %by \eqref{15-33}, \begin{align*} &\frac{\int_{\Omega}v_{\overline{z}}^{p-1}u_{0}dz}{\int_{\Omega}u_{0}^{\frac {p}{2}}v_{\overline{z}}^{p/2}dz} \\ & \geq\frac{c_{1}\{\int_{\{ | z| \geq R+1\} }e^{(p-1) (-(1+\delta) | z| ) }e^{( 1+\delta') \frac{\langle z,\overline{z}\rangle }{| \overline{z}| }}dz+ o(1)\} }{c_{2}\{ \int_{\{ | z| \geq R\} }e^{(-\frac{( 1-\delta) p}{2}| z| ) }e^{\frac{( 1-\delta') p}{2}\frac{\langle z,\overline{z}\rangle }{| \overline{z}| }}dz+ o(1)\} }e^{(\frac {(1-\delta')p}{2}-(1+\delta') ) | \overline{z}| } \end{align*} which approaches\infty$as$| \overline{z}| \to\infty$. Then$l_{0}>0$exists such that for$| \overline{z}| \geq l_{0}$$t_{0}^{\frac{p-2}{2}}\int_{\Omega}v_{\overline{z}}^{p-1}u_{0}dz-\frac{c} {p}\int_{\Omega}u_{0}^{p/2}v_{\overline{z}}^{p/2}dz>0.$ Hence, we have that for$| \overline{z}| \geq l_{0}$, $\sup_{t\geq t_{0}} J_{h}(u_{0}+tv_{\overline{z}}) 0 such that for n\geq n_{0} \[ \underset{t\geq0}{\sup}J_{h}(u_{0}+tv_{n}) 0 $$\mathop{\rm cat}([ J_{h}\leq\alpha_{h}(\Omega) +\alpha(\mathbb{R}^{N}) -\sigma] ) \geq 2.\label{15-36}$$ To prove \eqref{15-36}, we need some preliminaries. Recall the definition of Lusternik-Schnirelman category. \begin{definition} \rm \label{m9} (i) For a topological space X, we say a non-empty, closed subset A\subset X is contractible to a point in X if and only if there exists a continuous mapping \[ \eta:\left[ 0,1\right] \times A\to X$ such that for some$x_{0}\in X$\begin{gather*} \eta(0,x) =x\quad\text{for all }x\in A,\\ \eta(1,x) =x_{0}\quad\text{for all }x\in A. \end{gather*}$(ii) We define \begin{align*} \mathop{\rm cat}(X) & =\min\big\{ k\in\mathbb{N}:\text{there exist closed subsetsA_{1},\dots , A_{k}\subset X$such that }\\ & \text{$A_{j}$is contractible to a point in$X$for all$jand} \cup_{j=1}^k A_{j}=X\} . \end{align*} \end{definition} When we do not have finitely many closed subsetsA_{1},\dots ,A_{k}\subset X$such that$A_{j}$is contractible to a point in$X$for all$j$and$\underset{j=1}{\overset{k}{\cup}}A_{j}=X$, we say$\mathop{\rm cat}(X)=\infty$. For fundamental properties of Lusternik-Schnirelman category, we refer to Ambrosetti \cite{A} and Schwartz \cite{S}. Here we use the following property: \begin{theorem} \label{m10} Suppose that$X$is a Hilbert manifold and$\Psi\in C^{1}(X,\mathbb{R}) $. Assume that there are$c_{0}\in\mathbb{R}$and$k\in\mathbb{N}$, \newline$(i)\Psi(x) $satisfies the (PS)$_{c}$-condition for$c\leq c_{0}$;\newline$(ii)\mathop{\rm cat}(\{ x\in X: \Psi(x) \leq c_{0}\}) \geq k$. \newline Then$\Psi(x)$has at least$k$critical points in$\{ x\in X;\Psi(x) \leq c_{0}\} $. \end{theorem} \begin{theorem} \label{m11} Let$N\geq1$,$S^{N-1}=\{ x\in\mathbb{R}^{N};| x| =1\} $, and let$X$be a topological space. Suppose that there are two continuous maps $F:S^{N-1}\to X,\quad\text{}G:X\to S^{N-1}$ such that$G\circF$is homotopic to the identity map of$S^{N-1}$, that is, a continuous map$\zeta:\left[ 0,1\right] \times S^{N-1}\to S^{N-1}exists such that \begin{align*} \zeta(0,x) & =(G\circ F) (x) \quad\text{for each }x\in S^{N-1},\\ \zeta(1,x) & =x\quad\text{for each }x\in S^{N-1}. \end{align*} Then\mathop{\rm cat}(X) \geq2$. \end{theorem} \begin{proof} We argue indirectly and suppose that$\mathop{\rm cat}(X) =1$, that is, that$X$is contractible to a point in itself. Thus, a continuous map$\eta:\left[ 0,1\right] \times X\to X$exists such that for some$x_{0}\in X$\begin{gather*} \eta(0,x) =x\quad\text{for all }x\in X,\\ \eta(1,x) =x_{0}\quad\text{for all }x\in X. \end{gather*} Consider a homotopy$\beta:[0,1] \times S^{N-1}\to S^{N-1}$defined by $\beta(s,x) =G(\eta(s,F(x) )) .$ Then \begin{gather*} \beta(0,x) =(G\circ F) (x) \quad\text{for all }x\in X,\\ \beta(1,x) =G(x_{0}) \quad\text{for all }x\in X. \end{gather*} Thus$G\circ F$is homotopic to a constant map. However,by assumption,$G\circ F$is homotopic to the identity. Thus$S^{N-1}$is contractible to a point in$S^{N-1}$, which is a contradiction. Therefore$\mathop{\rm cat}(X)\geq2$. \end{proof} Let \begin{gather*} A_{1}=\big\{ u\in H_{0}^{1}(\Omega) \backslash\{ 0\} : \frac{1}{\| u\| _{H^{1}}}t^{-}(\frac{u}{\| u\| _{H^{1}}}) >1 \big\} \cup\{ 0\} \\ A_{2}=\big\{ u\in H_{0}^{1}(\Omega) \backslash\{0\} : \frac{1}{\| u\| _{H^{1}}}t^{-}(\frac{u}{\| u\| _{H^{1}}}) <1 \} . \end{gather*} \begin{lemma} \label{m12} We have the following results:\newline$(i)H_{0}^{1}(\Omega) \backslash\mathbf{M}_{h}^{-}=A_{1}\cup A_{2}$; \newline$(ii)\mathbf{M}_{h}^{+}\subset A_{1}$; \newline$(iii)t_{0}>1$and$n_{1}\geq n_{0}$exist such that$u_{0}+t_{0}v_{n}\in A_{2}$for each$n\geq n_{1}$, where$n_{0}$is defined as in Remark \ref{m20}; \newline$(iv)$a sequence$\{ s_{n}\} \subset(0,1) $exists such that$u_{0}+s_{n}t_{0}v_{n}\in\mathbf{M}_{h}^{-}$for each$n\geq n_{1}$. \end{lemma} \begin{proof}$(i)$By Lemma \ref{g4}$(iii)$, we have $\mathbf{M}_{h}^{-}=\big\{ u\in H_{0}^{1}(\Omega) \backslash\{ 0\} : \frac{1}{\| u\| _{H^{1}}} t^{-}(\frac{u}{\| u\| _{H^{1}}}) =1\big\} .$ Then$H_{0}^{1}(\Omega) \backslash\mathbf{M}_{h}^{-}=A_{1}\cup A_{2}$.\newline$(ii)$For each$u\in\mathbf{M}_{h}^{+}, we have $10 such that 00. By Lemma \ref{m17}, we have \[ a(v_{n}) =b(v_{n}) + o(1) =\frac {2p}{p-2}\alpha(\Omega) + o(1) .$ Thus, \begin{align*} b(w_{n}) & =\frac{1}{\| \frac{u_{0}}{t_{n}} +v_{n}\| _{H^{1}}^{p}} {\int_{\Omega}} (\frac{u_{0}}{t_{n}}+v_{n}) ^{p}\\ & \geq\frac{b(v_{n}) }{2^{p-1}\big(\| \frac{u_{0} }{t_{n}}\| _{H^{1}}^{p}+\| v_{n}\| _{H^{1}}^{p}\big)}\\ & =\frac{\frac{2p}{p-2}\alpha(\Omega) }{2^{p-1}\big( \frac{\| u_{0}\| _{H^{1}}^{p}}{c_{0}^{p}}+\big(\frac{2p} {p-2}\alpha(\Omega) \big) ^{p/2}\big) }+o( 1) . \end{align*} Case(b) :t_{n}\to\infty$as$n\to\infty$. The proof is similar to Case$(a) $. \noindent Case$(c):t_{n}= o(1)$as$n\to\infty$. By Lemma \ref{m17}, we have $\| u_{0}+t_{n}v_{n}\| _{H^{1}}^{2} =\| u_{0}\|_{H^{1}}^{2}+t_{n}^{2}\| v_{n}\| _{H^{1}}^{2}+2t_{n}\langle v_{n},u_{0}\rangle _{H^{1}} =\| u_{0}\| _{H^{1}}^{2}+ o(1).$ Thus, $b(w_{n}) \geq\frac{1}{\| u_{0}+t_{n}v_{n}\| _{H^{1}}^{p}}{\int_{\Omega}}u_{0}^{p} =\frac{1}{\| u_{0}\| _{H^{1}}^{p}} {\int_{\Omega}}u_{0}^{p}+ o(1).$ From Case$(a) $,$(b) $and$(c) $,$b(w_{n}) $is bounded below away from zero. Since$t^{-}(w_{n}) w_{n}\in\mathbf{M}_{h}^{-}\subset \mathbf{M}_{h}$, we have $J_{h}(t^{-}(w_{n}) w_{n}) =\frac{1}{2}[t^{-}(w_{n})] ^{2} -\frac{1}{p}[ t^{-}(w_{n})] ^{p}b(w_{n}) -t^{-}(w_{n}) \int_{\Omega}hw_{n}$ which approaches$-\infty$as$n\to\infty$. However,$J_{h}$is bounded below on$\mathbf{M}_{h}, which is a contradiction. Let $t_{0}=(\frac{p-2}{2p\alpha(\Omega) }| c^{2}-a(u_{0}) | ) ^{1/2}+1,$ then \begin{align*} \| u_{0}+t_{0}v_{n}\| _{H^{1}}^{2} & =a(u_{0}) +t_{0}^{2}(\frac{2p}{p-2}) \alpha(\Omega) + o(1)\\ & >c^{2}+ o(1)\geq\left[ t^{-}(\frac{u_{0}+t_{0}v_{n}}{\| u_{0}+t_{0}v_{n}\| _{H^{1}}}) \right] ^{2}+ o(1). \end{align*} Thus, there is ann_{1}\geq n_{0}$, where$n_{0}$is defined as in Remark \ref{m20}, such that, or$n\geq n_{1}$, $\frac{1}{\| u_{0}+t_{0}v_{n}\| _{H^{1}}}t^{-}(\frac {u_{0}+t_{0}v_{n}}{\| u_{0}+t_{0}v_{n}\| _{H^{1}}}) <1,$ or$u_{0}+t_{0}v_{n}\in A_{2}$.\newline$(iv)$Define a path$\gamma_{n}( s) =u_{0}+st_{0}v_{n}$for$s\in\left[ 0,1\right] $and each$n\geq n_{1}$where$t_{0}>1$, then $\gamma_{n}(0)=u_{0}\in A_{1}\text{, }\gamma_{n}(1)=u_{0}+t_{0}v_{n}\in A_{2}.$ Since$\frac{1}{\| u\| _{H^{1}}}t^{-}(\frac{u}{\| u\| _{H^{1}}}) $is a continuous function for nonzero$u$and$\gamma_{n}(\left[ 0,1\right] ) $is connected, a sequence$\{ s_{n}\} \subset(0,1) $exists such that$u_{0}+s_{n}t_{0}v_{n}\in\mathbf{M}_{h}^{-}$. \end{proof} Define a map$F_{n}:S^{N-1}\to H_{0}^{1}( \Omega) $by, for$\overline{z}\in S^{N-1}$, $F_{n}(\overline{z}) (z) =u_{0}(z) +s_{n}t_{0}v_{n}(z)\quad\text{for }n\geq n_{1},$ where$v_{n}(z)=\psi_{R}(z) \overline{u}(z-t_{n} \overline{z}) $and$n_{1}$is defined as in Lemma \ref{m12}. Then we have the following proposition. \begin{proposition} \label{m13} A sequence$\{ \sigma_{n}\} \subset\mathbb{R}^{+}$exists such that $F_{n}(S^{N-1}) \subset\left[ J_{h}\leq\alpha_{h}( \Omega) +\alpha(\mathbb{R}^{N}) -\sigma_{n}\right] .$ \end{proposition} \begin{proof} By Lemma \ref{m12}$(iv) $and Remark \ref{m20}, we have that for each$n\geq n_{1}u_{0}+s_{n}t_{0}v_{n}\in\mathbf{M}_{h}^{-}$and$J_{h}(u_{0}+s_{n}t_{0}v_{n}) \leq\alpha_{h}( \Omega) +\alpha(\mathbb{R}^{N}) -\sigma_{n}$, the conclusion holds. \end{proof} For$c>0$, we define \begin{gather*} b_{c}(u) =\int_{\Omega}c| u| ^{p};\\ I_{c}(u) =\frac{1}{2}a(u) -\frac{1}{p}b_{c}(u) ;\\ \mathbf{M}_{I_{c}}=\{ u\in H_{0}^{1}(\Omega) \backslash\{ 0\} :\langle I_{c}'(u),u\rangle =0\} . \end{gather*} Recall that a unique$t^{-}=t^{-}(u) >0$and$t^{1}=t^{1}( u) >0$exist such that$t^{-}u\in\mathbf{M}_{h}^{-}$and$t^{1} u\in\mathbf{M}(\Omega) $. \begin{lemma} \label{m14} For$u\in\Sigma=\{ u\in H_{0}^{1}(\Omega) \mid\| u\| _{H^{1}}=1\} $, we have the following results:\newline$(i) $a unique$t^{c}(u) >0$exists such that$t^{c}(u) u\in\mathbf{M}_{I_{c}}$and $\underset{t\geq0}{\max}I_{c}(tu) =I_{c}(t^{c}( u) u) =(\frac{1}{2}-\frac{1}{p}) b_{c}( u) ^{-\frac{2}{p-2}};$ \newline$(ii) $for$0<\mu<1$,$d_{1}(\mu) >0$exists such that for$\| h\| _{L^{2}}0. \] Then $f'(t^{c}(u) ) =0$, $t^{c}( u) u\in\mathbf{M}_{I_{c}}$ and \begin{align*} (t^{c}(u) ) ^{2}f''(t^{c}(u) ) & =a(t^{c}(u) u)-(p-1) b_{c}(t^{c}(u) u) \\ & =(2-p) (t^{c}(u) ) ^{2}a(u) <0. \end{align*} Thus, a unique $t^{c}(u) >0$ exists such that $t^{c}( u) u\in\mathbf{M}_{I_{c}}$ and $\max_{t\geq0} I_{c}(tu) =I_{c}(t^{c}(u) u) =(\frac{1}{2}-\frac{1}{p}) b_{c}(u) ^{-\frac{2}{p-2}}.$ $(ii)$ Let $c=\frac{1}{1-\mu}$, $t^{c}=t^{\frac {1}{1-\mu}}>0$ and $t^{1}=t^{1}(u) >0$ such that $t^{c}u\in M_{I_{c}}$ and $t^{1}u\in\mathbf{M}(\Omega)$. For $\mu \in(0,1) ,$ we have $| \int_{\Omega}ht^{c}udz| \leq\| t^{c}u\| _{H^{1} }\| h\| _{L^{2}}\leq\frac{\mu}{2}\| t^{c}u\| _{H^{1}}^{2}+\frac{1}{2\mu}\| h\| _{L^{2}}^{2}.$ Then by part $(i)$, \begin{align*} \underset{t\geq0}{\sup}J_{h}(tu) & \geq J_{h}( t^{c}u) \geq\frac{1-\mu}{2}\| t^{c}u\| _{H^{1}}^{2} -\frac{1}{p}b(t^{c}u) -\frac{1}{2\mu}\| h\| _{L^{2}}^{2}\\ & =(1-\mu) \Big[ \frac{1}{2}\| t^{c}u\| _{H^{1} }^{2}-\frac{1}{(1-\mu) p}\int_{\Omega}| t^{c}u| ^{p}\Big] -\frac{1}{2\mu}\| h\| _{L^{2}}^{2}\\ & =(1-\mu) I_{\frac{1}{1-\mu}}(t^{c}u) -\frac {1}{2\mu}\| h\| _{L^{2}}^{2}\\ & =(1-\mu) ^{\frac{p}{p-2}}(\frac{1}{2}-\frac{1} {p}) b(u) ^{-\frac{2}{p-2}}-\frac{1}{2\mu}\| h\| _{L^{2}}^{2}\\ & =(1-\mu) ^{\frac{p}{p-2}}J(t^{1}u) -\frac {1}{2\mu}\| h\| _{L^{2}}^{2}\\ & \geq(1-\mu) ^{\frac{p}{p-2}}\alpha(\Omega) -\frac{1}{2\mu}\| h\| _{L^{2}}^{2}. \end{align*} For $\mu\in(0,1)$, there exists $d_{1}(\mu) >0$ such that for $\| h\| _{L^{2}}0. \] By Lemma \ref{g4}, there exists$t^{-}=t^{-}(u)>0$such that$t^{-}u\in \mathbf{M}_{h}^{-}$and $\underset{t\geq0}{\sup}J_{h}(tu) =J_{h}(t^{-}u) .$ Thus, for$\| h\| _{L^{2}}0$exists such that if$u\in\mathbf{M}(\Omega) $and$J(u) \leq\alpha(\mathbb{R}^{N}) +\delta_{0}$, then $\int_{\mathbb{R}^{N}}\frac{z}{| z| }(| \nabla u| ^{2}+u^{2}) dz\neq0.$ \end{lemma} \begin{proof} If not, a sequence$\{ u_{n}\} \subset\mathbf{M}( \Omega) $exists such that$J(u_{n}) =\alpha( \mathbb{R}^{N}) + o(1) $and $\int_{\mathbb{R}^{N}}\frac{z}{| z| }(| \nabla u_{n}| ^{2}+u_{n}^{2}) dz=0.$ By Theorem \ref{i4},$\{ u_{n}\} $is a (PS)$_{\alpha( \mathbb{R}^{N}) }$-sequence in$H_{0}^{1}(\Omega) $for$J$. By Theorem \ref{f8},$\inf_{v\in\mathbf{M}(\Omega)}J(v) =\alpha(\Omega) =\alpha(\mathbb{R}^{N})$is not achieved. Let$\overline{u}$be the unique positive solution of Equation \eqref{E1} in$\mathbb{R}^{N}$. It follows from Theorem \ref{d1} that a sequence$\{ z_{n}\} $exists in$\mathbb{R}^{N}$such that$| z_{n}| \to\infty$as$n\to\infty$and $u_{n}(z) =\overline{u}(z-z_{n})+ o(1) \quad\text{strongly in }H^{1}(\mathbb{R}^{N}) .$ Assume$\frac{z_{n}}{| z_{n}| }\to z_{0}$as$n\to\infty$, where$z_{0}$is a unit vector in$\mathbb{R}^{N}. Then by Theorem \ref{p13}, we have \begin{align*} 0 & =\int_{\mathbb{R}^{N}}\frac{z}{| z| }(| \nabla u_{n}| ^{2}+u_{n}^{2}) dz\\ & =\int_{\mathbb{R}^{N}}\frac{z+z_{n}}{| z+z_{n}| }( | \nabla\overline{u}| ^{2}+\overline{u}^{2}) dz+o( 1) \\ & =(\frac{2p}{p-2}) z_{0}\alpha(\mathbb{R}^{N}) + o(1) , \end{align*} which is a contradiction. \end{proof} \begin{lemma} \label{m15}d_{0}>0$exists such that for$\| h\| _{L^{2}} 0$such that$t^{0}u/\| u\| _{H^{1}}\in\mathbf{M.}$By Lemma \ref{m14}$(ii) $, we have for any$\mu\in(0,1) $and$\| h\| _{L^{2}} d_{0}>0 $and$\delta_{0}>0$exist such that for$\| h\| _{L^{2}}N$, then by Theorem \ref{s8} we have$u_{0}\in C^{1,\theta}(\overline{\Omega }) $. First, we claim that$\lim_{\theta\to1^{-}} \zeta_{n}(\theta,\overline{z}) =\overline{z}$and$\lim_{\theta\to\frac{1}{2}^{-}}\zeta_{n}(\theta,\overline{z}) =G(\overline{u}(z-t_{n}\overline{z}) ) $. \newline$(a)\underset{\theta\to1^{-}}{\lim }\zeta_{n}(\theta,\overline{z}) =\overline{z}:since \begin{align*} & \int_{\mathbb{R}^{N}}\frac{z}{| z| }\Big(\big| \nabla\overline{u}\big(z-\frac{t_{n}}{2(1-\theta) } \overline{z}\big) \big| ^{2}+\overline{u}\big(z-\frac{t_{n}}{2( 1-\theta) }\overline{z}\big) ^{2}) dz\\ & =\int_{\mathbb{R}^{N}}\frac{z+\frac{t_{n}}{2(1-\theta) }\overline{z}}{| z+\frac{t_{n}}{2(1-\theta) }\overline {z}| }(| \nabla\overline{u}(z) | ^{2}+\overline{u}(z) ^{2}) dz\\ & =(\frac{2p}{p-2}) \alpha(\mathbb{R}^{N}) \overline{z}+ o(1)\quad\text{as }\theta\to1^{-}, \end{align*} then\underset{\theta\to1^{-}}{\lim}\zeta_{n}(\theta ,\overline{z}) =\overline{z}$.\newline$(b) \underset{\theta\to\frac{1}{2}^{-}}{\lim}\zeta_{n}( \theta,\overline{z}) =G(\overline{u}(z-t_{n}\overline {z}) ) :$since$\overline{u}$and$u_{0}\in C^{1,\theta}( \overline{\Omega}) $, then $\| (1-2\theta) F_{n}(\overline{z}) +2\theta\overline{u}(z-t_{n}\overline{z}) \| _{H^{1} }=\| \overline{u}(z-t_{n}\overline{z}) \| + o(1)\quad\text{as }\theta\to\frac{1}{2}^{-}.$ By the continuity of$G$, we obtain$\underset{\theta\to\frac{1} {2}^{-}}{\lim}\zeta_{n}(\theta,\overline{z}) =G( \overline{u}(z-t_{n}\overline{z}) ) $. Thus,$\zeta _{n}(\theta,\overline{z}) \in C(\left[ 0,1\right] \times S^{N-1},S^{N-1}) $and \begin{gather*} \zeta_{n}(0,\overline{z}) =G(F_{n}(\overline {z}) ) \quad\text{for\ all }\overline{z}\in S^{N-1},\\ \zeta_{n}(1,\overline{z}) =\overline{z}\quad\text{for\ all }\overline{z}\in S^{N-1}, \end{gather*} provided$n\geq n_{1}$and$\| h\| _{L^{2}}N$,$0<\| h\| _{L^{2}}0$exists such that for$t\geq t_{0}$, Equation$( \ref{E1}) $on$\Omega_{t}$has three positive solutions, one of which is$y$-symmetric while the other two are nonaxially symmetric. \subsubsection{Existence of Three Solutions} \begin{example}[$y$-symmetric large domain] \label{e1} \rm \quad$(i)$For$00. \] Then $\Omega_{l}$ is a $y$-symmetric large domain in $\mathbf{A}^{r}$; \newline$(ii)$ Let $00$ exists such that $w_{n}=0$ in $\overline{Q_{K}}$, and two disjoint large domains $\Omega^{1}$ and $\Omega^{2}$ in $\mathbf{A}^{r}$ exist such that \begin{gather*} (x,y)\in\Omega^{2}\quad\text{if and only if }\quad (x,-y)\in\Omega^{1},\\ \Omega\backslash \overline{Q_{K}}=\Omega^{1}\cup\Omega^{2}\quad \text{where }Q_{K}=\Omega\cap B^{N}(0;K) . \end{gather*} Let $w_{n}^{i}(x)=\begin{cases} w_{n}(x) & \text{for }x\in\Omega^{i},\\ 0 & \text{for }x\notin\Omega^{i}, \end{cases}$ for $i=1,2$. Then $w_{n}^{i}\in H_{0}^{1}(\Omega^{i})$, $w_{n}^{1}( x,y) =w_{n}^{2}(x,-y)$, $w_{n}=w_{n}^{1}+w_{n}^{2}$, and $J(w_{n}^{1}) =J(w_{n}^{2})$. Moreover, we have $\alpha_{s}(\Omega) + o(1) =J(w_{n}) =J(w_{n}^{1}) +J(w_{n}^{2}) =2J(w_{n} ^{i}) \quad\text{for }i=1,2,$ or $J(w_{n}^{i}) =\frac{1}{2}\alpha_{s}(\Omega) + o(1) \quad\text{for }i=1,2.$ By (\ref{15-15}), we have $J'(w_{n}^{i})= o(1)$ in $H^{-1}(\Omega^{i})$ for $i=1,2$. Therefore, $\frac{1}{2}\alpha_{s}(\Omega)$ is a positive (PS)-value in $H_{0}^{1}(\Omega^{i})$ for $J$. By Lemma \ref{i13}, Theorem \ref{i14}, and Definition \ref{i15}, $\frac{1}{2}\alpha_{s}(\Omega) \geq\alpha_{\mathbf{M}}( \Omega^{i}) =\alpha(\Omega^{i}) .$ Since $\Omega^{i}$ is a large domain in $\mathbf{A}^{r}$, by Lemma \ref{f8} $(ii)$, we have $\alpha(\Omega^{i}) =\alpha(\mathbf{A}^{r}) .$ Thus, $\alpha_{s}(\Omega) \geq2\alpha(\mathbf{A}^{r})$. \noindent $(ii)$ Clearly, we have $\alpha(\mathbf{A}^{r}) \leq\alpha _{s}(\Omega)$. Assume that $\alpha(\mathbf{A} ^{r}) =\alpha_{s}(\Omega)$, then by Theorem \ref{c3}, $J$ does not satisfy the (PS)$_{\alpha_{s}(\Omega)}$-condition in $H_{s} ^{1}(\Omega)$ for $J$. By Lemma \ref{e4}, $2\alpha( \mathbf{A}^{r}) \leq\alpha_{s}(\Omega) =\alpha( \mathbf{A}^{r})$, which is a contradiction. \end{proof} \begin{theorem} \label{e5} Let $00$ $\Theta_{t}=\mathbf{A}^{r}\;\backslash\;\left[ \overline{B^{N}(( x,t+r_{1}) ;r_{1}) }\cup\overline{B^{N}(( x,-(t+r_{1})) ;r_{1}) }\right] .$ Then $t_{0}>0$ exists such that $\alpha_{s}(\Theta_{t}) <2\alpha(\mathbf{A}^{r})$ for all $t\geq t_{0}$. In particular, there is a $y$-symmetric positive ground state solution of Equation $( \ref{E1})$ in $\Omega_{t}$. \end{theorem} \begin{proof} By Lien-Tzeng-Wang \cite{LTW}, $\alpha(\mathbf{A}_{-t,t}^{r})$ is strictly decreasing as $t$ is strictly increasing and $\alpha(\mathbf{A}_{-t,t}^{r}) \searrow\alpha( \mathbf{A}^{r}) \quad\text{as }t\to\infty.$ Thus, there is a $t_{0}>0$ such that $\alpha(\mathbf{A}_{-t,t} ^{r}) <2\alpha(\mathbf{A}^{r})$ for each $t\geq t_{0}$. By Proposition \ref{e3}, $\alpha(\mathbf{A}_{-t,t}^{r}) =\alpha_{s}(\mathbf{A}_{-t,t}^{r})$ for each $t$. Thus, $\alpha_{s}(\mathbf{A}_{-t,t}^{r}) <2\alpha( \mathbf{A}^{r})$ for each $t\geq t_{0}$. Since $\Theta_{t} \supset\mathbf{A}_{-t,t}^{r}$ for each $t\geq t_{0}$. Therefore, we have $\alpha_{s}(\Theta_{t}) \leq\alpha_{s}(\mathbf{A} _{-t,t}^{r})$ for each $t\geq t_{0}$. We then conclude that $\alpha_{s}(\Theta_{t}) \leq\alpha_{s}(\mathbf{A} _{-t,t}^{r}) <2\alpha(\mathbf{A}^{r}) \quad\text{for each }t\geq t_{0}.$ Since $\Theta_{t}$ is a $y$-symmetric large domain in $\mathbf{A}^{r}$, by Theorem \ref{e4}, $J$ satisfies the (PS)$_{\alpha_{s}(\Omega)}$-condition in $H_{s}^{1}(\Omega)$, or telse here is a $y$-symmetric positive ground state solution of Equation \eqref{E1} in $\Theta_{t}$ for each $t\geq t_{0}$. \end{proof} \begin{figure}[hbt] \begin{center} \includegraphics[width=0.4\textwidth]{fig12} \quad \includegraphics[width=0.4\textwidth]{fig13} \end{center} \caption{the finite strip with a hole} \end{figure} Let $00$ exists such that for $t\geq t_{0}$, Equation $( \ref{E1})$ on $\Omega_{t}$ has three positive solutions of which one is $y$-symmetric and the other two are nonaxially symmetric. \end{theorem} \begin{proof} Let $\Omega=\mathbf{A}^{r}\;\backslash\;\overline{B^{N}(( x,0) ;r_{1}) }$. Then $\Omega$ is a $y$-symmetric large domain in $\mathbf{A}^{r}$. By Theorem \ref{e4}, we have $\alpha( \mathbf{A}^{r}) <\alpha_{s}(\Omega)$. By Lien-Tzeng-Wang \cite{LTW}, \eqref{E1} admits a ground state solution in $\mathbf{A}_{0,t}^{r}$ and in $\mathbf{A}^{r}$, and $\alpha( \mathbf{A}_{0,t}^{r})$ is strictly decreasing as $t$ is strictly increasing and $\alpha(\mathbf{A}_{0,t}^{r}) \searrow\alpha( \mathbf{A}^{r}) \quad\text{as }t\to\infty.$ Take $t_{1}>0$ such that for $t\geq t_{1}$, $$\alpha(\mathbf{A}^{r}) <\alpha(\mathbf{A}_{0,t} ^{r}) <\alpha_{s}(\Omega) .\label{15-16}$$ Note that $\mathbf{A}_{r_{1},t_{1}+r_{1}}^{r}\subsetneqq$ $\Omega _{t}\subsetneqq\mathbf{A}^{r}$ for $t\geq t_{0}=t_{1}+r_{1}$. By Theorem \ref{c3}, we conclude that $$\alpha(\mathbf{A}^{r}) <\alpha(\Omega_{t}) <\alpha(\mathbf{A}_{r_{1},t_{1}+r_{1}}^{r}).\label{15-17}$$ By Lien-Tzeng-Wang \cite{LTW}, if $\Omega$ is a domain of $\mathbb{R}^{N}$, then $\alpha(\Omega)$ is invariant by rigid motions. Thus, $$\alpha(\mathbf{A}_{r_{1},t_{1}+r_{1}}^{r})=\alpha(\mathbf{A}_{0,t_{1} }^{r}) .\label{15-18}$$ Therefore, by (\ref{15-16})-(\ref{15-18}) $$\alpha(\mathbf{A}^{r}) <\alpha(\Omega_{t}) <\alpha(\mathbf{A}_{0,t_{1}}^{r}) <\alpha_{s}( \Omega) .\label{15-19}$$ Since $\Omega_{t}\subset\Omega$, we have $$\alpha_{s}(\Omega) \leq\alpha_{s}(\Omega_{t}) .\label{15-20}$$ By (\ref{15-19}) and (\ref{15-20}), we obtain $$\alpha(\Omega_{t}) <\alpha_{s}(\Omega_{t}) .\label{15-21}$$ By Theorem \ref{b3}, there are a $y-$symmetry solution $u_{1}$ and a solution $u_{2}$ of Equation \eqref{E1} in domain $\Omega_{t}$ such that \begin{gather*} J(u_{1})=\alpha_{s}(\Omega_{t}) ,\\ J(u_{2})=\alpha(\Omega_{t}) . \end{gather*} By Theorem \ref{c3}, we may take $u_{1}$ and $u_{2}$ to be positive. Let $u_{3}(x,y)=u_{2}(x,-y).$ Then $u_{3}$ is the third solution. By (\ref{15-21}), $u_{1},u_{2}$ and $u_{3}$ are different. Moreover, $u_{1}$ is a $y$-symmetric solution while both $u_{2}$ and $u_{3}$ are nonaxially symmetric solutions of Equation \eqref{E1} in domain $\Omega_{t}$. \end{proof} \noindent\textbf{Bibliographical notes:} The results of this section are from Wang-Wu \cite{WW1}. \subsection{Multiple Solutions in Domains with Two Bumps} That the existence of solutions of \eqref{E1} is affected by the shape of the domain $\Omega$ has been the focus of a great deal of research in recent years . By the Rellich compactness theorem, it is easy to obtain a solution of \eqref{E1} in a bounded domain. For general unbounded domains $\Omega$, because of the lack of compactness, the existence of solutions of Equation \eqref{E1} is an important open question. Recently, there has been some progress in determining the existence and multiplicity of solutions as follows: Bahri-Lions \cite{BL}, Coti Zelati \cite{Co}, Chabrowski \cite{Ch3}, Chen-Lee-Wang \cite{CLW}, Chen-Wang\cite{CW} , Chen-Lin-Wang \cite{CLWW}, Lien-Tzeng-Wang \cite{LTW}, del Pino-Felmer \cite{PiFe1}, \cite{PiFe2}, and Wang \cite{W} used the (PS)$-$theory to treat the existence of solutions of \eqref{E1}. Byeon \cite{By}, Chen-Ni-Zhou \cite{CNZ}, Dancer \cite{Da}, and Wang-Wu \cite{WW1} asserted the existence of three positive solutions of semilinear elliptic equations in a dumbbell domain. Jimbo \cite{J1} and \cite{J2} asserted the existence of solutions depending on the width of the corridor of the dumbbell. In this section we assert that there is a $R_{0}>0$ such that for $R>R_{0}$ Equation \eqref{E1} on the two bumps domain $D_{R}$ has three positive solutions in which one is $y$-symmetric and other two are nonaxially symmetric. (see Theorem \ref{x3}). Since finite dumbbell is a two bumps domain, the results of Byeon \cite{By}, Chen-Ni-Zhou \cite{CNZ}, and Dancer \cite{Da} are the consequences of our Theorem \ref{x3}. \subsubsection{Existence of Three Solutions} We have the following results. \begin{theorem} \label{x101} $(i)$ The bounded domains in $\mathbb{R}^{N}$ are the achieved domains in $\mathbb{R}^{N}$; \newline$(ii)$ The $C^{1}$ quasi-bounded domains are the achieved domains in $\mathbb{R}^{N}$; \newline$(iii)$ $\mathbb{R}^{N}$ is an achieved domain in $\mathbb{R}^{N}$; \newline$(iv)$ The periodic domains in $\mathbb{R}^{N}$ are the achieved domains. In particular, the infinite strip $\mathbf{A}^{r}$ is an achieved domain in $\mathbb{R}^{N}$;\newline$(v)$ $s_{0}>0$ exists such that $\mathbf{F}_{s}^{r}$ is an achieved domain in $\mathbb{R}^{N}$ if $s>s_{0}$. \end{theorem} \begin{proof} $(i)$ By Theorem \ref{b3}.\newline$(ii)$ By Theorem \ref{f2-1}.\newline$(iii)$ and $(iv)$ follows from Lien-Tzeng-Wang \cite{LTW}.\newline $(v)$ By Theorem \ref{h1}. \end{proof} Throughout this section, let $\Theta$ be a proper achieved $y$-symmetric domain in $\mathbb{R}^{N}$ bounded in the $x-$direction, such as bounded domains and the infinite strip $\mathbf{A}^{r}$. For $R>0$, let $\Omega_{R}^{1}$ and $\Omega_{R}^{2}$ be two bounded domains in $\mathbb{R} ^{N}$ such that \begin{gather*} R=\mathop{\rm dist}\{ 0,\Omega_{R}^{1}\} ,\\ \Omega_{R}^{2}=\{ (x,y): (x,-y)\in\Omega_{R}^{1}\} . \end{gather*} and the $y$-symmetric domain $D_{R}=\Omega_{R}^{1}\cup\Theta\cup\Omega_{R}^{2}.$ We call $D_{R}$ a two-bumps domain. Here are some examples of two-bumps domains. \begin{example} \label{x1} \rm $(i)$ For $t>R>r>0$. The bounded dumbbell domain $D_{R}$ is a two-bumps domain, where $D_{R}=B^{N}((0,-t) ,r) \cup\mathbf{A}_{-t,t} ^{r}\cup B^{N}((0,t) ,r) ;$ $(ii)$ For $t>R>r>0$. The unbounded dumbbell domain $D_{R}$ is a two-bumps domain, where $D_{R}=B^{N}((0,-t) ,r) \cup\mathbf{A}^{r}\cup B^{N}((0,t) ,r) ;$ $(iii)$ For $t>R>r>0$. The curved dumbbell domain $D_{R}=\Omega _{R}^{1}\cup\Theta\cup\Omega_{R}^{2}$ is a two bumps domain, where $\Omega _{R}^{1}$ and $\Omega_{R}^{2}$ are two bounded domains in $\mathbb{R}^{N}$ such that \begin{gather*} R=\mathop{\rm dist}\{ 0,\Omega_{R}^{1}\} ,\\ \Omega_{R}^{2}=\{ (x,y): (x,-y)\in\Omega_{R}^{1}\} . \end{gather*} and $\Theta$ is a curved bounded $y$-symmetric domain in $\mathbb{R}^{N}$. \end{example} Then we have the following assertion. \begin{theorem} \label{x2} Let $D_{R}$ be a two-bumps domain, where $D_{R}=\Omega_{R}^{1} \cup\Theta\cup\Omega_{R}^{2}$, and let $\Theta$ be a proper achieved $y$-symmetric domain in $\mathbb{R}^{N}$ bounded in the $x-$direction. Then we have, for all $R>0$,\newline $(i)$ $\alpha(\Theta)\geq\alpha(D_{R}) >\alpha(\mathbb{R}^{N})$; \newline $(ii)$ $J$ satisfies the (PS)$_{\alpha_{X}(D_{R})}$-condition in $X(D_{R})$. \end{theorem} \begin{proof} By Theorem \ref{c3} and Theorem \ref{b3}, it suffices to assume that $\Theta$ is unbounded.\newline$(i)$ Since $\Theta\subset D_{R} \subsetneqq\mathbb{R}^{N}$, we have $\alpha(\Theta) \geq \alpha(D_{R}) \geq\alpha(\mathbb{R}^{N})$. Suppose that $\alpha(D_{R}) =\alpha(\mathbb{R} ^{N})$, by Theorem \ref{c3}, $J$ does not satisfy the (PS)$_{\alpha (D_{R}) }$-condition. By Theorem \ref{c6-1}, a sequence $\{ u_{n}\}$ in $H_{0}^{1}(D_{R})$ exists such that $\{ u_{n}\}$ and $\{ \xi_{n}u_{n}\}$ are the (PS)$_{\alpha(D_{R}) }$-sequences for $J$, where $\xi_{n}$ is defined as in (\ref{1-1}). Let $w_{n}=\xi_{n}u_{n}$. Then \begin{gather*} J(w_{n})=\alpha(D_{R})+ o(1),\\ J'(w_{n})= o(1)\quad\mbox{in }H^{-1}(D_{R}) . \end{gather*} Since $D_{R}=\Omega_{R}^{1}\cup\Theta\cup\Omega_{R}^{2}$ is a $y$-symmetric domain in $\mathbb{R}^{N}$ separated by a bounded domain, $n_{0}>0$ exists such that for $n\geq n_{0}$, $w_{n}\in H_{0}^{1}(\Theta)$, $J(w_{n})=\alpha(D_{R})+ o(1)$, and $a(w_{n}) =b( w_{n}) + o(1)$. By Theorem \ref{i3}, there is a sequence $\{s_{n}\}$ in $\mathbb{R}^{+}$ such that $\{s_{n}w_{n}\}$ is in $\mathbf{M}(\Omega)$ and $\{s_{n}w_{n}\}$ is a (PS)$_{\alpha (D_{R})}$-sequence in $X(\Omega)$ for $J$. Thus $\alpha( \Theta) \leq\alpha(D_{R})$. We then conclude that $\alpha(\Theta) =\alpha(D_{R}) =\alpha( \mathbb{R}^{N})$. However, since $\Theta$ is a proper achieved $y$-symmetric domain in $\mathbb{R}^{N}$, by Theorem \ref{c3}, $\alpha( \Theta) >\alpha(\mathbb{R}^{N})$. This is a contradiction. Thus, $\alpha(\Theta) \geq\alpha( D_{R}) >\alpha(\mathbb{R}^{N})$ for all $R>0$. \newline$(ii)$ It suffices to prove the case $X( D_{R}) =H_{0}^{1}(D_{R})$. Since $\Omega_{R}^{1}$, $\Theta$, and $\Omega_{R}^{2}$ are achieved, by Theorem \ref{c3}, $\alpha(D_{R}) <\min\{ \alpha(\Omega_{R}^{1}),\alpha( \Theta) ,\alpha(\Omega_{R}^{2})\} .$ By Theorem \ref{c7}, $J$ satisfies the (PS)$_{\alpha(D_{R})}$-condition in $H_{0}^{1}(D_{R})$. \end{proof} We apply Theorems \ref{w7} and \ref{x2} to prove the following result. \begin{theorem} \label{x3} There is an $R_{0}>0$ such that for $R>R_{0}$ Equation $( \ref{E1})$ on $D_{R}$ has three positive solutions, of which one is $y$-symmetric and other two are nonaxially symmetric. \end{theorem} \begin{proof} Take $\rho>0$ such that $\Omega=(\mathbb{R}^{N}\ \backslash \ \overline{\mathbb{R}_{-\rho,\rho}^{N}}) \cup\Theta$ is connected. Then $\Omega$ is a $y$-symmetric large domain in $\mathbb{R}^{N}$ separated by a bounded domain. By Theorem \ref{w7}, we have $\alpha(\mathbb{R} ^{N}) =\alpha(\Omega) <\alpha_{s}(\Omega)$. By Lien-Tzeng-Wang \cite{LTW}, $\alpha(B^{N}(0,R) )$ is strictly decreasing as $R$ is strictly increasing and $\alpha(B^{N}(0,R) ) \searrow\alpha( \mathbb{R}^{N}) \quad\text{as }R\to\infty.$ Take $R_{0}>0$, such that for $R>R_{0}$ $$\alpha(\mathbb{R}^{N}) <\alpha(B^{N}(0,R) ) <\alpha_{s}(\Omega) .\label{15-22}$$ Since $B^{N}((x_{R},y_{R}) ,R)$ $\subsetneqq$ $D_{R}\subsetneqq\mathbb{R}^{N}$, by Theorem \ref{c3} and Theorem \ref{x2}, we conclude that $$\alpha(\mathbb{R}^{N}) <\alpha(D_{R}) <\alpha(B^{N}((x_{R},y_{R}) ,R) )=\alpha( B^{N}(0,R) ) .\label{15-23}$$ Therefore, by (\ref{15-22}) and (\ref{15-23}) and $D_{R}\subset\Omega$, we have $$\alpha(D_{R}) <\alpha(B^{N}(0,R) ) <\alpha_{s}(\Omega) \leq\alpha_{s}(D_{R}).\label{15-24}$$ Thus, $$\alpha(D_{R}) <\alpha_{s}(D_{R}) .\label{15-25}$$ By Theorem \ref{b3}, there are a $y-$symmetry positive solution $u_{1}$ and a positive solution $u_{2}$ of Equation \eqref{E1} in domain $D_{R}$ for $R>R_{0}$ such that \begin{gather*} J(u_{1})=\alpha_{s}(D_{R}) ,\\ J(u_{2})=\alpha(D_{R}) . \end{gather*} Let $u_{3}(x,y)=u_{2}(x,-y)$, then $u_{3}$ is the third positive solution. 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