\documentclass[reqno]{amsart} \usepackage{hyperref} \usepackage{amssymb} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2012 (2012), No. 16, pp. 1--15.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2012 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2012/16\hfil Multiple solutions] {Multiple solutions for a q-Laplacian equation on an annulus} \author[S. Tai, J. Wang \hfil EJDE-2012/16\hfilneg] {Shijian Tai, Jiangtao Wang} % in alphabetical order \address{Shijian Tai \newline Shenzhen Experimental Education Group, Shenzhen, 5182028, China} \email{363846618@qq.com} \address{Jiangtao Wang \newline School of Statistics and Mathematics, Zhongnan University of Economics and Law, Wuhan, 430073, China} \email{wjtao1983@yahoo.com.cn} \thanks{Submitted November 7, 2011. Published January 24, 2012.} \subjclass[2000]{35J40} \keywords{Ground state; minimizer; nonradial function; q-Laplacian; \hfill\break\indent Rayleigh quotient} \begin{abstract} In this article, we study the q-Laplacian equation $$-\Delta_{q}u=\big||x|-2\big|^{a}u^{p-1},\quad 1<|x|<3 ,$$ where $\Delta_{q}u=\operatorname{div}(|\nabla u|^{q-2} \nabla u)$ and $q>1$. We prove that the problem has two solutions when $a$ is large, and has two additional solutions when $p$ is close to the critical Sobolev exponent $q^{*}=\frac{Nq}{N-q}$. A symmetry-breaking phenomenon appears which shows that the least-energy solution cannot be radial function. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{remark}[theorem]{Remark} \newtheorem{definition}[theorem]{Definition} \allowdisplaybreaks \section{Introduction} This article concerns the q-Laplacian equation $$\label{1.1} \begin{gathered} -\Delta_{q}u=\Phi_{a}u^{p-1} \quad \text{in } \Omega,\\ u>0 \quad \text{in } \Omega,\\ u=0 \quad \text{on } \partial \Omega, \end{gathered}$$ where $\Delta_{q}u=div(|\nabla u|^{q-2}\nabla u)$, $\Omega=\{x\in \mathbb{R}^N |1<|x|<3\}$ is an annulus in $\mathbb{R}^{N}$, $N\geq 3$, $a>0$, $p>q>1$ and $\Phi_{a}$ is the radial function $$\Phi_{a}(x)=\big||x|-2\big|^{a}.$$ Equation \eqref{1.1} is an extension of the problem $$\label{1} \begin{gathered} -\Delta_{q}u=|x|^{a}u^{p-1} \quad \text{in } |x|<1,\\ u=0 \quad \text{on } |x|=1. \end{gathered}$$ Equation \eqref{1} can be seen as a natural extension to the annular domain $\Omega$ of the celebrated H\'{e}non equation with Dirichlet boundary conditions $$\label{1.2} \begin{gathered} -\Delta u=|x|^{a}u^{p-1} \quad \text{in } |x|<1,\\ u=0 \quad \text{on } |x|=1. \end{gathered}$$ This equation was proposed by H\'{e}non in \cite{h} when he studied rotating stellar structures. For $a>0$, $20$. Ni \cite{n} proved that the infimum $$\label{1.4} \inf_{0\neq u\in H^{1}_{0,\rm rad}(B)}\frac{\int_{B}|\nabla u|^{2}dx}{\big(\int_{B}|x|^{a}|u|^{p}dx\big)^{2/p}}$$ is attained for any $2 < p < 2^{*}+2a/(N-2)$ by a function in $H^{1}_{0,\rm rad}(B)$, the space of radial $H^{1}_0(B)$ functions. Therefore, radial solutions of \eqref{1.2} also exist for supercritical exponents $p$. Indeed, $H^{1}_{0,\rm rad}(B)$ shows a power-like decay away from the origin (as a result of the Strauss Lemma, see \cite{am,s2}) that combines with the weight $|x|^{a}$ and provides the compactness of the embedding $H^{1}_{0,\rm rad}(B)\subset L^{p}(B)$ for any $2 < p < 2^{*}+2a/(N-2)$. When $a>0$, Smets, Su and Willem obtained some symmetry-breaking results for \eqref{1.2} in \cite{ssw}. They proved that minimizers of \eqref{1.3} could not be radial, at least for $a$ sufficiently large. Consequently, \eqref{1.2} had at least two solutions when $a$ was large (see also \cite{sw}). Serra \cite{s1} proved that \eqref{1.2} had at least one nonradial solution when $p=2^{*}$, and in \cite{bs} Badiale and Serra obtained the existence of more than one solutions to \eqref{1.2} also for some supercritical values of $p$. These solutions are nonradial and they are obtained by minimization under suitable symmetry constraints. Cao and Peng \cite{cp} proved that, for $p$ sufficiently close to $2^{*}$, the ground-state solutions of \eqref{1.2} possessed a unique maximum point whose distance from $\partial B$ tended to zero as $p\to 2^{*}$. And they also proved the same results in the case of q-laplacian of the H\'{e}non equation (see \cite{cpy}). This result was improved in \cite{p}, where multi-bump solutions for the H\'{e}non equation with almost critical Sobolev exponent $p$ were found, by applying a finite-dimensional reduction. These solutions are not radial, though they are invariant under the action of suitable subgroups of $O(N)$, and they concentrate at boundary points of the unit ball $B$ in $\mathbb{R}^{N}$ as $p\to 2^{*}$. However, the role of $a$ is a static one (for more results for $p\approx 2^{*}$, see also \cite{ps}). When the weight disappeared; i.e. $a=0$, Brezis and Nirenberg proved in \cite{bn} that the ground state solution of $-\Delta u=u^{p}$ in $H^{1}_0(\Omega)$ was not a radial function. Actually, they proved that both a radial and a nonradial (positive) solution arise as $p\approx 2^{*}$. When the weight in \eqref{1.1} disappeares; i.e. $a=0$, Li and Zhou \cite{lz} proved that existence of multiple solutions to the p-Laplacian type elliptic problem $$\label{1.5} \begin{gathered} -\Delta_{p} u(x)=f(x,u) \quad \text{in } \Omega,\\ u=0 \quad \text{on } \partial\Omega, \end{gathered}$$ where $\Omega$ was a bounded domain in $\mathbb{R}^{N}$ ($N > 1$) with smooth boundary $\partial\Omega$, and $f(x,u)$ went asymptotically in $u$ to $|u|^{p-2}u$ at infinity. When $q=2$ in \eqref{1.1}, Calanchi, Secchi and Terraneo \cite{cst} obtained multiple solutions for a H\'{e}non-like equation on $1<|x|<3$. For more results about asymptotic estimates for solutions of the H\'{e}non equation with $a$ large, one can see \cite{bw1,bw2}. This paper is mainly motivated by \cite{cst}. We want to extend the results in \cite{cst} to a general $q$-Laplacian problem. We consider the critical points of $$\label{1.6} R_{a,p}=\frac{\int_{\Omega}|\nabla u|^qdx}{\big(\int_{\Omega}\Phi_{a}| u|^{p}dx\big)^{q/p}},\quad u\in W^{1,q}_0(\Omega)\backslash\{0\},$$ which is the Rayleigh quotient associated with \eqref{1.1}. Our main results are as follows. \begin{theorem}\label{thm1} Assume that $p\in(q,q^{*})$. For $a$ large enough, any ground state $u_{a ,p}$ is a nonradial function. \end{theorem} \begin{theorem}\label{thm2} Assume that $a >0$. For $p$ close to $q^{*}$ the quotient $R_{a ,p}$ has at least two nonradial local minima. \end{theorem} \begin{theorem}\label{thm3} There exist $\overline{a }>0$ and $q<\overline{p}q$. As $a \to \infty$, there exist two constants $C_1$, $C_2$ depending on $p$ such that \label{2.4} 0q$and$a>1$,$C(a,p)\leq C_2$. So we obtain$S^{\rm rad}_{a,p}\leq C_2a^{q-1+\frac{q}{p}}$. To find the lower bound$C_1$, we will do some scaling. Let us define the functions$\phi_1:[1,2]\to[1,2]$and$\phi_2:[2,3]\to[2,3]$as follows: $$\phi_1(r)=2-(2-r)^{b}, \quad \phi_2(r)=2+(r-2)^{b},$$ where$b\in(0,1)$will be chosen later. It is obvious that we can obtain a piecewise$C^{1}$homeomorphism$\phi:[1,3]\to[1,3]$by gluing$\phi_1$and$\phi_2$. Now, for any radial function$u\in W^{1,q}_0(\Omega)$, setting$\upsilon(\rho)=u(\phi(\rho))$and choosing$b=1/(a+1), we obtain \begin{align*} &\int_{\Omega}\Phi_{a}(x)\big|u(|x|)\big|^{p}dx\\ &= \omega_{N-1}\int_1^3\Phi_{a}(r)|u(r)|^{p}r^{N-1}dr\\ &\leq 3^{N-1}\omega_{N-1}\int_1^3\Phi_{a}(r)|u(r)|^{p}dr \\ &= 3^{N-1}\omega_{N-1} \Big(\int_1^{2} \Phi_{a}(\phi_1(\rho))|\upsilon(\rho)|^{p}\phi'_1(\rho)d\rho + \int_2^3\Phi_{a}(\phi_2(\rho))|\upsilon(\rho)|^{p} \phi'_2(\rho)d\rho\Big)\\ &= 3^{N-1}\omega_{N-1}b\int_1^3|\upsilon(\rho)|^{p}d\rho \end{align*} and \begin{align*} \int_{\Omega}|\nabla u|^qdx &= \omega_{N-1}\int_1^3|u'(r)|^qr^{N-1}dr\\ &\geq\omega_{N-1}\int_1^3|u'(r)|^qdr\\ &= \omega_{N-1} \Big(\int_1^{2}|\upsilon' (\rho)|^q\frac{1}{|\phi'_1(\rho)|^{q-1}}d\rho +\int_2^3|\upsilon'(\rho)|^q\frac{1}{|\phi'_2(\rho)|^{q-1}} d\rho\Big)\\ &= \omega_{N-1}\frac{1}{b^{q-1}}\int_1^3|\upsilon'(\rho)|^q|\rho -2|^{(1-b)(q-1)}d\rho \\ &\geq \omega_{N-1}\frac{1}{b^{q-1}}\int_1^3|\upsilon'(\rho)|^q| \rho-2|^{(q-1)}d\rho. \end{align*} Therefore, $$\label{2.5} R_{a,p}(u)\geq Ca^{q-1+\frac{q}{p}}\inf_{\upsilon\in W^{1,q}_0(\Omega)\backslash\{0\}} \frac{\int_1^3|\upsilon'(\rho)|^q |\rho-2|^{(q-1)}d\rho}{\big(\int_1^3|\upsilon(\rho)|^{p}d\rho \big)^{q/p}},$$ whereC$depends only on$N$. To end the proof, we will show that the right-hand side of \eqref{2.5} is greater than zero. This follows from some general Hardy-type inequality (see \cite[Theorem 11.4]{ok}), but we present here an elementary proof for the sake of completeness. Indeed, given$\upsilon\in W^{1,p}_{0,\rm rad}(\Omega)$, for$\rho\in[1,2], we can write \begin{align*} |\upsilon(\rho)| &= |\upsilon(\rho)-\upsilon(1)|\\ &\leq \int_1^{\rho}|\upsilon'(t)||2-t|^{(q-1)/q} \frac{1}{|2-t|^{(q-1)/q}}dt\\ &\leq \Big(\int_1^{\rho}|\upsilon'(t)|^q|2-t|^{q-1}dt\Big)^{1/q} \Big(\int_1^{\rho}\frac{1}{|2-t|}dt\Big)^{(q-1)/q}\\ &\leq \Big(\int_1^3|\upsilon'(t)|^q|2-t|^{q-1}dt\Big)^{1/q} \Big(-\ln|2-\rho|\Big)^{(q-1)/q}. \end{align*} Hence, \begin{align*} \int_1^{2}|\upsilon(\rho)|^{p}d\rho &\leq \Big(\int_1^3|\upsilon'(t)|^q|2-t|^{q-1}dt\Big)^{p/q} \int_1^{2}\Big(-\ln|2-\rho|\Big)^{\frac{p(q-1)}{q}}d\rho\\ &= \Big(\int_1^3|\upsilon'(t)|^q|2-t|^{q-1}dt\Big)^{p/q} \int_0^{\infty}t^{\frac{p(q-1)}{q}}e^{-t}dt\\ &\leq \Gamma\Big(\frac{p(q-1)+q}{q}\Big) \Big(\int_1^3|\upsilon'(t)|^q|2-t|^{q-1}dt\Big)^{p/q} \end{align*} and in a similar way, $$\int_2^3|\upsilon(\rho)|^{p}d\rho\leq\Gamma\Big(\frac{p(q-1)+q}{q}\Big) \Big(\int_1^3|\upsilon'(t)|^q|2-t|^{q-1}dt\Big)^{p/q}.$$ Therefore, $$\int_1^3|\upsilon'(t)|^q|2-t|^{q-1}dt \geq \Big(\int_1^3|\upsilon(\rho)|^{p}d\rho\Big)^{q/p} \frac{1}{2^{q/p}\Gamma\big(\frac{p(q-1)+q}{q}\big)^{q/p}}.$$ This implies that the infimum in \eqref{2.5} is strictly positive. There exists a constantC_1 >0$such that$S^{\rm rad}_{a,p}\geq C_1a^{q-1+\frac{q}{p}}$. \end{proof} We are now in position to prove Theorem \ref{thm1}. \begin{proof}[Proof of Theorem \ref{thm1}] Note that$q-N+\frac{q}{p}N0$. Any minimum$u_{a,p}$of$R_{a,p}(u)$in$W^{1,q}_0\backslash\{0\}$satisfies for some$x_0\in \partial \Omega$, \begin{itemize} \item[(1)]$|\nabla u_{a,p}|^q\to \mu \delta_{x_0}$weakly in sense of measure as$p \to q^{*}$; \item[(2)]$|u_{a,p}|^{q^{*}}\to \nu \delta_{x_0}$weakly in sense of measure as$p \to q^{*}$, \end{itemize} where$\mu > 0$and$\nu >0$are such that$\mu\geq S_{0,q^{*}}\nu^{q/q^{*}}$and$\delta_{x}$is the Dirac mass at$x$. \end{proposition} Since the result can be proved by using the same arguments in \cite{cp}, with some minor modifications, we omit its proof here. \begin{remark}\label{rem2.7} \rm Proposition \ref{prop2.3} implies that any ground state solution concentrates in a single point at the boundary as$p\to q^{*}$and consequently this solution is not radial. This symmetry breaking can be also proved by using a continuation argument as in \cite{bn}. Indeed,$\lim_{p\to q^{*}}S_{a,p}=S_{0,q^{*}}$, and since$S_{0,q^{*}} < S^{\rm rad}_{0,q^{*}}$we conclude as in \cite{bn} that ground state of$S_{a,p}$cannot be radially symmetric as$p\to q^{*}$. \end{remark} Let \begin{gather*} \Omega^{-}=\big\{x\in \mathbb{R}^{N}: 1<|x|<2\big\} , \quad \Omega^{+}=\big\{x\in \mathbb{R}^{N}: 2<|x|<3\big\}, \\ \Sigma=\big\{u\in W^{1,p}_0(\Omega)\backslash\{0\}: \int_{\Omega^{-}}|\nabla u|^qdx=\int_{\Omega^{+}}|\nabla u|^qdx\big\},\\ \wedge=\big\{u\in W^{1,p}_0(\Omega): \int_{\Omega^{-}}|\nabla u|^qdx>\int_{\Omega^{+}}|\nabla u|^qdx\big\}. \end{gather*} We already know that any global minimizer of$R_{a,p}$yields a first solution$u_{a,p}$. From Proposition \ref{prop2.3} we know that this solution concentrates at precisely one point of the boundary$\partial \Omega$. Noting that this boundary has two connected components, we will minimize$R_{a,p}$over the set$\Lambda$of$W^{1,q}_0$functions which concentrate at the opposite component of boundary. We need a careful estimate to show that minimizers fall inside the interior of$\Lambda$. To obtain a second local minimizer, we assume without loss of generality that any$u_{a,p}$concentrates at some point on the sphere$|x|=3$. After a rotation, we can even assume that any$u_{a,p}$concentrates at the point$(3,0,\dots,0)$. \begin{lemma}\label{lem2.4} If$a> 0$, then there exists$ \delta > 0 $such that $$\label{2.6} \liminf_{p\to q^{*}}T_{a,p} > S_{0,q^{*}}+\delta,$$ where$T_{a,p}=\inf_{u\in \Sigma}R_{a,p}(u). $\end{lemma} \begin{proof} First we prove that$T_{a,p}$is attained by a function$\upsilon_{a,p} \in \Sigma$. Consider a minimizing sequence$\{u_{n}\}$for$T_{a,p}$. We can apply the homogeneity of$R_{a,p}$and assume that$\int_{\Omega}|\nabla u_{n}|^qdx=1$. Passing to a subsequence,$u_{n}$converges to$\upsilon =\upsilon_{a,p}$weakly in$W^{1,q}_0(\Omega)$and strongly in$L^{s}(\Omega)$, for all$s\in (q,q^{*})$. What we have to check is that$\upsilon \in \Sigma$(proving that the convergence of$u_{n}$to$\upsilon$is strong). From the strong convergence in$L^{s}(\Omega)$we have that $$\label{2.7} R_{a,p}(\upsilon)\leq \frac{1}{\big(\int_{\Omega}\Phi_{a}(x)|\upsilon|^{p}dx\big)^{q/p}} =T_{a,p}$$ and particularly$\upsilon \neq 0$. We may assume that$\upsilon \geq 0$in$\Omega$. For the sake of contradiction, assume that $$\int_{\Omega^{+}}|\nabla \upsilon|^qdx < \frac{1}{2}.$$ Fix a nonnegative smooth function$\psi_1 \in C^{\infty}_0(\Omega^{+})$,$\psi_1\neq 0$and$\delta\geq 0$. Setting$u=\upsilon+\delta\psi_1$from the positivity of$\upsilon$and$\psi_1$, we have, for$\delta>0$, $$\label{2.8} \int_{\Omega}\Phi_{a}(x)|\upsilon|^{p}dx < \int_{\Omega}\Phi_{a}(x)|u|^{p}dx.$$ Now, $$\label{2.9} \int_{\Omega^{+}}|\nabla u|^qdx=\int_{\Omega^{+}}|\nabla \upsilon+\delta\nabla\psi_1|^qdx,$$ if we define$f_1:[0,+\infty]\to \mathbb{R}$by $$f_1(\delta)=\int_{\Omega^{+}}|\nabla \upsilon+\delta\nabla\psi_1|^qdx,$$ we know that$f_1$is continuous and$f_1(0)< \frac{1}{2}$,$\lim_{\delta \to \infty}f_1(\delta)=+\infty$. Hence there exists$\delta_1> 0$with$f_1(\delta_1)=1/2$. We can reason in an analogous way if$\int_{\Omega^{-}}|\nabla \upsilon|^qdx < \frac{1}{2}$in order to find$\delta_2\geq 0$and$\psi_2\geq0$such that$\int_{\Omega^{-}}|\nabla (\upsilon+\delta_2\psi_2)|^qdx =1/2$. From \eqref{2.8}, this implies that there exists$\omega=\upsilon+\delta_1\psi_1+\delta_2\psi_2 \in \Sigma$such that$R_{a,p}(\omega)< T_{a,p}$, which yields a contradiction. Finally we must have that$\upsilon_{a,p}\in \Sigma$is a minimum point. Moreover for any$a > 0 $and$q0,\; y\in \mathbb{R}^{N}. $$We set$$ U^{i}_{\varepsilon}(x)=\varepsilon^{-\frac{N-q}{q}}U \Big(\frac{x-x_{i}}{{\varepsilon^{(q-1)/q}}}\Big) =\frac{1}{(\varepsilon+|x-x_{i}|^{\frac{q}{q-1}})^{\frac{N-q}{q}}} $$and denote by \psi_{i}(i=0,1) two cut-off functions such that 0\leq\psi_{i}\leq 1, |\nabla\psi_{i} |\leq C|\ln\varepsilon| for some constant C>0, and$$ \psi_{i}=\begin{cases} 1 & \text{if } |x-x_{i}|< \frac{1}{2|\ln\varepsilon|},\\ 0 & \text{if } |x-x_{i}|\geq \frac{1}{|\ln\varepsilon|}. \end{cases} $$The following lemma shows that the truncated functions $$\label{2.10} u^{i}_{\varepsilon}=\psi_{i}(x)U^{i}_{\varepsilon}(x), \quad i=0,1,$$ are almost minimizers for S_{0,q^{*}}. Since it is an easy modification of the arguments of \cite{cp}, we omit the proof of this fact. \begin{lemma}\label{lem2.5} If a > 0, then $$\label{2.11} \lim_{p\to q^{*}}R_{a,p}(u^{i}_{\varepsilon})=S_{o,q^{*}}+K(\varepsilon),$$ with \lim_{\varepsilon\to 0}K(\varepsilon)=0. \end{lemma} As a direct consequence of Lemma \ref{lem2.5}, we obtain the following result. \begin{corollary}\label{lem2.6} S_{0,q^{*}}=S_{\alpha,q^{*}}. \end{corollary} \begin{proof} On one hand, S_{0,q^{*}}\leq S_{a,q^{*}} since \Phi_{a}(|x|)\leq 1. On the other hand by Lemma \ref{lem2.5}, we have$$ R_{a,q^{*}}(u^{i}_{\varepsilon}) =\lim_{p\to q^{*}}R_{a,p}(u^{i}_{\varepsilon}) =S_{0,q^{*}}+K(\varepsilon), $$which implies that S_{0,q^{*}}+K(\varepsilon)\geq S_{a,q^{*}} for every \varepsilon>0. Letting \varepsilon\to 0, we infer S_{0,q^{*}}\geq S_{a,q^{*}}. Therefore S_{0,q^{*}}= S_{a,q^{*}}. \end{proof} Now we are ready to prove Theorem \ref{thm2}. \begin{proof}[Proof of Theorem \ref{thm2}] Let u_{a,p} be a ground state solution. Let us suppose that it concentrates on the outer boundary. The infimun of R_{a,p} on \bar{\Lambda} is attained. However it cannot be attained on the boundary \partial \Lambda = \Sigma. In fact, from Lemma \ref{lem2.4}, we obtain$$ \inf_{\Sigma}R_{a,p}(u)> S_{0,q^{*}}+\delta, \quad\text{as } p\to q^{*} $$and$$ \inf_{\Lambda}R_{a,p}(u)\leq R_{a,p}(u^{1}_{\varepsilon})\to S_{0,q^{*}}+K_1(\varepsilon), \quad \text{as } p\to q^{*}, $$since u^{1}_{\varepsilon} \in \Lambda for \varepsilon sufficiently small. Then the infimum is attained in an interior point of \Lambda and is therefore a critical point of R_{a,p}. \end{proof} \subsection{Proof of Theorem \ref{thm3}} Now we prove the existence of a third nonradial solution, in the previous section we proved the existence of two solutions of \eqref{1.1} which were local minima of Rayleigh quotient for p near q^{*}. We would expect another critical point of R_{a,p} located in some sense between these minimum points. For \varepsilon  sufficiently small let u^{i}_{\varepsilon}=\psi_{i}(x)U^{i}_{\varepsilon}(x), i\in\{0,1\}, be defined as in \eqref{2.10}. We will verify that R_{a,p} has the mountain-pass geometry. Let us introduce the mountain-pass level $$\label{c} c=c(a,p)=\inf_{\gamma\in\Gamma}\max_{t\in[0,1]}R_{a,p}(\gamma(t)),$$ where$$ \Gamma=\big\{\gamma\in C([0,1],W^{1,q}_0(\Omega)): \gamma(0) =u^{0}_{\varepsilon},\gamma(1)=u^{1}_{\varepsilon}\big\} $$is the set of continuous paths joining u^{0}_{\varepsilon} with u^{1}_{\varepsilon}. We claim that c is a critical value for R_{a,p}. We start to prove that c is larger, uniformly with respect to \varepsilon, than the values of the functional R_{a,p} at the points u^{0}_{\varepsilon} and u^{1}_{\varepsilon}. \begin{lemma}\label{lem2.8} Set M_{\varepsilon}=\max\{R_{a,p}(u^{0}_{\varepsilon}), R_{a,p}(u^{1}_{\varepsilon})\}. There exists \sigma > 0 such that c \geq M_{\varepsilon}+ \sigma uniformly with respect to \varepsilon. \end{lemma} \begin{proof} We prove that there exists \sigma such that for all \gamma \in \Gamma,$$ \max R_{a,p}(\gamma(t))\geq M_{\varepsilon}+ \sigma. $$A simple continuity argument shows that for every \gamma \in \Gamma there exists t_{\gamma} such that \gamma(t_{\gamma}) \in \Sigma, where$$ \Sigma=\big\{u\in W^{1,p}_0(\Omega)\backslash\{0\}: \int_{\Omega^{-}}|\nabla u|^qdx=\int_{\Omega^{+}}|\nabla u|^qdx\big\}. $$In fact the map$$ t \in [0,1] \mapsto\int_{\Omega^{+}}|\nabla \gamma(t)|^qdx-\int_{\Omega^{-}}|\nabla \gamma(t)|^qdx $$is continuous and it takes a negative value at t=0 and a positive value at t=1. It follows from Lemma \ref{lem2.4} that for p near q^{*} there exists \delta > 0 such that$$ \max_{t\in [0,1]}R_{a,p}(\gamma(t))\geq R_{a,p}(\gamma(t_{\gamma})) \geq \inf_{u\in\Sigma}R_{a,p}(u) \geq S_{0,q^{*}}+\delta. $$On the other hand, for \varepsilon small enough, we have$$ M_{\varepsilon} < S_{0,q^{*}}+\frac{\delta}{2}. $$This completes the proof. \end{proof} By the previous estimates, we can show that c is a critical level for R_{a,p}. As a result a further nonradial solution to \eqref{1.1} arises. \begin{proof}[Proof of Theorem \ref{thm3}] By the previous results, we can apply a deformation argument (see \cite{am,s3}) to prove that c is a critical level and it is achieved ( since the PS condition is satisfied ) by a function \upsilon. By the asymptotic estimate \eqref{2.4} for the radial level S_{a,p}^{\rm rad}, we know that there exists a constant C independent of p such that$$ S_{a,p}^{\rm rad} \geq Ca^{q-1+\frac{q}{p}}. $$Particularly, we obtain  S_{a,p}^{\rm rad} \to +\infty as a \to +\infty.  Therefore we can choose a_0 such that$$ S_{a,p}^{\rm rad}\geq 3S_{0,q^{*}} \quad\forall a\geq a_0. $$Define \zeta\in\Gamma by \zeta(t)=tu^{1}_{\varepsilon}+(1-t)u^{0}_{\varepsilon} for all t \in [0,1], and let \tau \in [0,1] be such that$$ R_{a,p}(\zeta(\tau))=\max_{t\in[0,1]}R_{a,p}(\zeta(t)). Noting that u^{0}_{\varepsilon} and u^{1}_{\varepsilon} have disjoint support one has, for \varepsilon small enough, we have \begin{align*} R_{a,p}(\upsilon) &=c\leq R_{a,p}(\zeta(\tau))\\ &= \frac{\int_{\Omega}|\nabla(\tau u^{1}_{\varepsilon} +(1-\tau)u^{0}_{\varepsilon})|^qdx} {\big(\int_{\Omega}\Phi_{a}|\tau u^{1}_{\varepsilon} +(1-\tau)u^{0}_{\varepsilon}|^{p}dx\big)^{q/p}}\\ &= \frac{\int_{\Omega}\tau^q|\nabla u^{1}_{\varepsilon}|^qdx+ \int_{\Omega}(1-\tau)^q |\nabla u^{0}_{\varepsilon}|^qdx} {\big(\tau^{p} \int_{\Omega}\Phi_{a}|u^{1}_{\varepsilon}|^{p}dx +(1-\tau)^{p} \int_{\Omega}\Phi_{a}|u^{0}_{\varepsilon}|^{p}dx \big)^{q/p}}\\ &\leq \frac{\tau^q\int_{\Omega}|\nabla u^{1}_{\varepsilon}|^qdx} {\big(\tau^{p}\int_{\Omega}\Phi_{a}|u^{1}_{\varepsilon}|^{p}dx\big)^{q/p}}+ \frac{(1-\tau)^q\int_{\Omega} |\nabla u^{0}_{\varepsilon}|^qdx} {\big((1-\tau)^{p} \int_{\Omega}\Phi_{a}|u^{0}_{\varepsilon}|^{p} dx\big)^{q/p}}\\ &= R_{a,p}(u^{1}_{\varepsilon})+R_{a,p}(u^{0}_{\varepsilon})\\ &\leq 2M_{\varepsilon}< 3S_{0,q^{*}}\\ &\leq S^{\rm rad}_{a,p}. \end{align*} \end{proof} \section{Behavior of the ground-state solution for a large} In this section, we mainly analyze a ground state solution as a\to +\infty. Even in this case this solution tends to concentrate" at the boundary \partial\Omega. However, this concentration is much weaker than concentration as p\to q^{*}. We use the notation C(\rho_1,\rho_2)=\{x\in \mathbb{R}^{N}| \rho_1 <|x| < \rho_2 \}. Let \delta be sufficiently small (say \delta< \frac{1}{2} ) and \varphi be a smooth cut-off function such that 0 \leq \varphi \leq 1 with $$\label{3} \varphi(x)=\begin{cases} 1 , & x\in C(1,1+\delta)\cup C(3-\delta,3),\\ 0 , & x\in C(2-\delta,2+\delta). \end{cases}$$ From now on, since p\in(q,q^{*}) is fixed we denote a ground state solution of problem \eqref{1.1} u_{a,p} with u_{a}. \begin{lemma}\label{lem3.1} Let u_{a} be such that R_{a,p}(u_{a})= S_{a,p}. If \varphi is defined in \eqref{3}, then $$\label{3.1} R_{a,p}(\varphi u_{a})= S_{a,p}+o(S_{a,p}) as a \to +\infty.$$ \end{lemma} \begin{proof} By the homogeneity of R_{a,p}, we may assume \int_{\Omega}|\nabla u_{a}|^qdx=1. We will prove it by two steps. Step 1. We claim that $$\label{3.2} \int_{\Omega}\Phi_{a}(\varphi u_{a})^{p}dx=\int_{\Omega}\Phi_{a}u_{a}^{p}dx+o\Big(\int_{\Omega}\Phi_{a}u_{a}^{p}dx\Big).$$ Actually, if we assume \limsup_{\alpha\to \infty}\frac{\int_{\Omega}\Phi_{a}u_{a}^{p}(1-\varphi^{p})dx} {\int_{\Omega}\Phi_{a}u_{a}^{p}dx}=b>0, $$which implies that, up to some subsequence,$$ \frac{\int_{\Omega}\Phi_{a}u_{a}^{p}(1-\varphi^{p})dx} {\int_{\Omega}\Phi_{a}u_{a}^{p}dx}>\frac{b}{2}>0. Since 1-\varphi^{p}\equiv0 on C(1,1+\delta)\cup C(3-\delta,3), we have \begin{align*} \int_{\Omega}\Phi_{a}u_{a}^{p}(1-\varphi^{p})dx &= \int_{C(1+\delta,3-\delta)}\Phi_{a}u_{a}^{p}(1-\varphi^{p})dx\\ &\leq (1-\delta)^{a}\int_{\Omega}u_{a}^{p}(1-\varphi^{p})dx\\ &\leq (1-\delta)^{a}\int_{\Omega}u_{a}^{p}dx. \end{align*} Hence, \int_{\Omega}u_{a}^{p}dx\geq(1-\delta)^{-a} \int_{\Omega}\Phi_{a}u_{a}^{p}(1-\varphi^{p})dx. $$Thus, $$\label{4} \frac{\int_{\Omega}u_{a}^{p}dx}{\int_{\Omega}\Phi_{a}u_{a}^{p}dx} \geq(1-\delta)^{-a}\frac{\int_{\Omega}\Phi_{a}u_{a}^{p}(1-\varphi^{p})dx}{\int_{\Omega}\Phi_{a}u_{a}^{p}dx} \geq(1-\delta)^{-a}\frac{b}{2}.$$ Since S_{a,p}^{p/q}=(\int_{\Omega}\Phi_{a}u_{a}^{p}dx)^{-1}, \eqref{4} can be written as$$ S_{a,p}^{p/q} \geq \frac{b}{2}\frac{(1-\delta)^{-a}}{\int_{\Omega}u_{a}^{p}dx} \geq \frac{b}{2} (1-\delta)^{-a}S_{0,p}^{p/q}, $$where$$ S_{0,p}=\inf_{u\neq0}\frac{\int_{\Omega}|\nabla u|^qdx}{\big(\int_{\Omega}u^{p}dx\big)^{q/p}}. $$On the other hand, from \eqref{2.2}, it follows that$$ S_{a,p}^{p/q}\leq Ca^{p-\frac{pN}{q}+N}, $$which gives a contradiction for a large. Hence \eqref{3.2} is true. Step 2. Now we prove that $$\label{3.3} \int_{\Omega}|\nabla(\varphi u_{a})|^qdx=\int_{\Omega}|\nabla u_{a}|^qdx+o(1)=1+o(1).$$ It is easy to prove that u_{a} satisfies the problem $$\label{3.4} \begin{gathered} -\Delta_{q}u_{a}=S_{a,p}^{p/q}\Phi_{a}u_{a}^{p-1} \quad \text{in } \Omega,\\ u_{a}>0 \quad \text{in } \Omega,\\ u_{a}=0 \quad \text{on } \partial\Omega. \end{gathered}$$ Since \int_{\Omega}|\nabla u_{a}|^qdx=1, up to subsequences, as a\to \infty, we have that:$$ \text{$u_{a}\to u$ weakly in $W_0^{1,q}(\Omega)$, and strongly in $L^{s}(\Omega)$ \a. e. in $\Omega$}. $$Now we prove that u=0. In fact, multiplying problem \eqref{3.4} by a smooth function \phi with \operatorname{supp}\phi\Subset \Omega and integrate, we obtain $$\label{3.5} \int_{\Omega}|\nabla u_{a}|^{q-2}\nabla u_{a}\nabla \phi dx=\int_{\Omega}S_{a,p}^{p/q}\Phi_{a}u_{a}^{p-1}\phi dx \to 0, \quad \text{as } a\to +\infty,$$ since, by \eqref{2.2}, S_{a,p}^{p/q}\Phi_{a} \to 0 uniformly on \operatorname{supp}\phi and u_{a} is uniformly bounded in L^{s} for q0. We get $$\label{3.9} \begin{split} &R_{a,p}(\varphi u_{a})\\ &= \frac{\int_{\Omega}|\nabla u_{a,1}|^qdx+\int_{\Omega}|\nabla u_{a,2}|^qdx} {\big(\lambda_{a}\int_{\Omega}\Phi_{a}u_{a,2}^{p}dx +\int_{\Omega}\Phi_{a}u_{a,2}^{p}dx\big)^{q/p}}\\ &= \frac{\int_{\Omega}|\nabla u_{a,1}|^qdx}{(\lambda_{a}+1)^{q/p} \big(\int_{\Omega}\Phi_{a}u_{a,2}^{p}dx\big)^{q/p}} +\frac{\int_{\Omega}|\nabla u_{a,2}|^qdx}{(\lambda_{a}+1)^{q/p} \big(\int_{\Omega}\Phi_{a}u_{a,2}^{p}dx\big)^{q/p}}\\ &= \frac{\lambda_{a}^{q/p}\int_{\Omega}|\nabla u_{a,1}|^qdx}{(\lambda_{a}+1)^{q/p} \big(\int_{\Omega}\Phi_{a}u_{a,1}^{p}dx\big)^{q/p}} +\frac{\int_{\Omega}|\nabla u_{a,2}|^qdx}{(\lambda_{a}+1)^{q/p} \big(\int_{\Omega}\Phi_{a}u_{a,2}^{p}dx\big)^{q/p}}. \end{split}$$ By the definition of S_{a,p} each quotient R_{a,p}(u_{a,1}) and R_{a,p}(u_{a,2}) in the last term is greater than or equal to S_{a,p}. Therefore by Lemma \ref{lem3.1} and equation \eqref{3.8} we obtain $$\label{3.10} S_{a,p}+o(S_{a,p})\geq \frac{1+\lambda_{a}^{q/p}}{(\lambda_{a}+1)^{q/p}}S_{a,p}.$$ We notice that the function g(x)=\frac{1+x^{q/p}}{(x+1)^{q/p}} is strictly greater than 1 for every x>0, g(0)=1 and g(x)\to 1 as x\to +\infty. Further it is increasing in [0,1] and decreasing in  [ 1, +\infty) and \max_{x>0}g(x)=g(1)=2^{1-q/p}. Let L\in \Lambda and \{a_{n}\} a sequence such that \lambda_{a_{n}}\to L as n\to +\infty. Taking to the limit in \eqref{3.10}, we obtain that 1\geq\frac{1+L^{q/p}}{(L+1)^{q/p}} and so either L=+\infty or L=0. \end{proof} \begin{corollary}\label{cor3.3} With the notation of Proposition \ref{prop3.2}, for any sequence \{a_{n}\} such that \lambda_{a_{n}}\to 0 one has $$\label{3.11} \lim_{n\to +\infty}\frac{\int_{\Omega}|\nabla u_{a_{n},1} |^qdx}{\int_{\Omega}|\nabla u_{a_{n},2} |^qdx} = 0 .$$ \end{corollary} \begin{proof} Denote$$ \frac{\int_{\Omega}|\nabla u_{a,1} |^qdx}{\int_{\Omega}|\nabla u_{a,2} |^qdx}=\kappa_{a} and suppose that \limsup_{n\to \infty}\kappa_{a_{n}}>0. Passing to subsequences, \kappa_{a_{n}}>\kappa>0 for some \kappa. Therefore we have \begin{align*} S_{a_{n},p}+o(S_{a_{n},p}) &= \frac{\int_{\Omega}|\nabla u_{a_{n},1}|^qdx+\int_{\Omega}|\nabla u_{a_{n},2}|^qdx} {\big(\int_{\Omega}\Phi_{a_{n}}u_{a_{n},1}^{p}dx +\int_{\Omega}\Phi_{a_{n}}u_{a_{n},2}^{p}dx\big) ^{q/p}}\\ &= \frac{(1+\kappa_{a_{n}})\int_{\Omega}|\nabla u_{a_{n},2}|^qdx} {(\int_{\Omega}\Phi_{a_{n}}u_{a_{n},2}^{p}dx)^{q/p} (1+\lambda_{a_{n}})^{q/p}}\\ &\geq R_{a_{n},p}(u_{a_{n},2})\frac{1+\kappa}{1+o(1)}\\ &\geq (1+\kappa)S_{a_{n},p}+o(S_{a_{n},p}), \end{align*} which is a contradiction. Therefore, \kappa_{a_{n}}= \frac{\int_{\Omega}|\nabla u_{a_{n},1} |^qdx}{\int_{\Omega}|\nabla u_{a_{n},2} |^qdx}\to 0.  \end{proof} The following result is an immediate consequence of the previous results. \begin{proposition}\label{prop3.4} For any $a_{n}$ such that $\lambda_{a_{n}}\to 0$, $$\label{3.12} \lim_{n\to +\infty}\int_{\Omega}|\nabla u_{a_{n},1}|^q dx=0 .$$ \end{proposition} \begin{thebibliography}{99} \bibitem{am} A. Ambrosetti, A. 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