\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
\begin{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}
\end{equation}
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
\begin{equation}\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}
\end{equation}
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
\begin{equation}\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}
\end{equation}
This equation was proposed by H\'{e}non in
\cite{h} when he studied rotating stellar structures.

 For $a>0$, $2<p<2^{*}=2N/(N-2)$, we
generalize  \eqref{1.2} to the case of q-Laplacian. In fact,
the weight function $\Phi_{a}(x)$ reproduces on $\Omega$ a similar
qualitative behavior of $|\cdot|^{a}$ on the unit ball $B$ in
$\mathbb{R}^{N}$.

A standard compactness argument shows that the infimum
\begin{equation}\label{1.3}
\inf_{0\neq u\in H^{1}_0(B)}\frac{\int_{B}|\nabla
u|^{2}dx}{\big(\int_{B}|x|^{a}|u|^{p}dx\big)^{2/p}}
\end{equation}
is attained for $2<p<2^{*}$ and any $a>0$.
Ni \cite{n} proved that the infimum
\begin{equation}\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}}
\end{equation}
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
\begin{equation}\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}
\end{equation}
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
\begin{equation}\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\},
\end{equation}
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 $q<\overline{p}<q^{*}$ such
that for all  $a \geq\bar{a }$ and $\bar{p}\leq p< q^{*}$,
 it results that the mountain-pass level
$c$(defined by \eqref{c})is a critical value for $R_{a ,p}$ and
it is attained by a  nonradial function.
\end{theorem}

To the best of our knowledge, the results we obtain in this article
are new. To prove Theorem \ref{thm1}, the key is to give
the energy estimates of both ground solutions and radial symmetry
solutions when $p$ is fixed. Applying these estimates, we can prove
that ground solution is not radially symmetric when $a\to
\infty $. Using the same arguments in \cite{cst}, we can prove
Theorem \ref{thm2} and Theorem \ref{thm3}. We would like to point
out that our problem is more complicated than the problem in
\cite{cst}.


 This paper is organized as follows. In Section 2 we prove our
main results. In Section 3 we describe the behavior of ground-state
solution of \eqref{1.1} when $p<q^{*}$ is fixed and
$a\to+\infty$. Although the conclusion is not as precise
as in the case $p\to q^{*}$, we can nevertheless show that a
sort of concentration near the boundary $\partial\Omega$ still
appears.

\section{Proofs of  main results}

\subsection{Proof of Theorem 1.1}
Denote
\begin{equation}\label{2.1}
 S_{a,p}=\inf_{u\in W^{1,q}_0(\Omega)\backslash\{0\}}R_{a,p}(u).
\end{equation}

It is easy to prove that up to a scaling for $p$ subcritical
$S_{a,p}$ is attained by a function $u_{a,p}$ that satisfies
\eqref{1.1}.
To prove that for $a$ is large any solution $u_{a,p}$ is
not radial  first we need an estimate from above of $S_{a,p}$.


\begin{lemma}\label{lem2.1}
Assume that $p\in(q,q^{*})$. There exists $\bar{a}$ such that for
$a\geq\bar{a}$,
\begin{equation}\label{2.2}
S_{a,p}\leq Ca^{q-N+\frac{qN}{p}}.
\end{equation}
\end{lemma}

\begin{proof}
We use the same techniques as in \cite{ssw}.  Let $\phi$ be
a smooth function with support in the unit ball $B$. Let us consider
$\phi_{a}(x)=\phi(a(x-x_{a}))$, where
$x_{a}=(3-\frac{1}{a},0,\dots,0)$. Since $\phi_{\alpha}$ has
support in the ball $B(x_{a},\frac{1}{a})$, by the change of
variable $y=a(x-x_{a})$, we obtain
\begin{equation}\label{2.1.1}
\int_{\Omega}\Phi_{a}\phi_{a}^{p}(x)dx
=\int_{B(x_{a},\frac{1}{a})}\big||x|-2\big|^{a}\phi_{a}^{p}(x)dx\geq
\big(1-\frac{2}{a}\big)^{a}a^{-N}\int_{B}\phi^{p}(y)dy.
\end{equation}
Furthermore,
\begin{equation}\label{2.1.2}
\int_{\Omega}|\nabla\phi_{a}|^qdx=a^q\int_{\Omega}|\nabla\phi(a(x-x_{a}))|^qdx=
a^{q-N}\int_{B}|\nabla\phi|^qdx.
\end{equation}
It follows from \eqref{2.1.1} and \eqref{2.1.2} that
$$
R_{a,p}(\phi_{a})=\frac{\int_{\Omega}|\nabla\phi_{a}|^qdx}
{\big(\int_{\Omega}\Phi_{a}(x)\phi_{a}^{p}dx\big)^{q/p}}
\leq Ca^{q-N+\frac{qN}{p}}.
$$
Hence we obtain
$$
S_{a,p}\leq R_{a,p}(\phi_{a})\leq Ca^{q-N+\frac{qN}{p}}
$$
for all $a$ large enough.
\end{proof}

Let $W^{1,q}_{0,\rm rad}(\Omega)$ be the space of radially symmetric
functions of $W^{1,q}_0(\Omega) $. In the sequel, we denote
$u(x)=u(|x|)$ for $u\in W^{1,q}_{0,\rm rad}(\Omega)$.

Consider the minimization problem
\begin{equation}\label{2.3}
S^{\rm rad}_{a,p}=\inf_{u\in
W^{1,q}_{0,\rm rad}(\Omega)\backslash\{0\}}R_{a,p}(u).
\end{equation}
 It is well known that any
minimizers of \eqref{2.3} can be scaled so as to be solutions of
\eqref{1.1}. Thus, we will use freely this fact in the sequel.

In the following lemma, we obtain an estimate of the energy
$S^{\rm rad}_{a,p}$ as $a\to\infty$.

\begin{lemma}\label{lem2.2}
Let $p>q$. As $a \to \infty$, there exist two constants
$C_1$, $C_2$ depending on $p$ such that
\begin{equation}\label{2.4}
0<C_1 \leq \frac{S^{\rm rad}_{a,p}}{a^{q-1+\frac{q}{p}}}
\leq C_2 < +\infty .
\end{equation}
\end{lemma}

\begin{proof}
Let $\phi\in C_0^{\infty}(\Omega)$ be a positive, radial function,
and set
$$\phi_{a}(x)=\phi_{a}(|x|)=\phi(a(|x|-3+3/a)).$$
Then
 $$
\int_{\Omega}\Phi_{a}\phi_{a}^{p}dx\geq
\big(1-\frac{2}{a}\big)^{a}a^{-1}\int_{\Omega}\phi^{p}dx
 $$
and
\begin{align*}
\int_{\Omega}|\nabla\phi_{a}|^qdx
&=  \omega_{N-1}\int_{3-\frac{2}{a}}^3(\phi'_{a}(r))^qr^{N-1}dr\\
&=  \omega_{N-1}\int_1^3a^q(\phi'(s))^q
 \Big(\frac{s}{a}+3-\frac{3}{a}\Big)^{N-1}a^{-1}ds\\
&=  a^{q-1}\omega_{N-1}\int_1^3(\phi'(s))^q
 \Big(\frac{s+3a-3}{sa}\Big)^{N-1}s^{N-1}ds \\
&\leq   3^{N-1}a^{q-1}\int_{\Omega}|\nabla\phi|^qdx,
\end{align*}
 since  $1\leq\frac{s+3a-3}{sa}\leq 3$.
Therefore,
$$
R^{\rm rad}_{a,p}(\phi_{a})=
\frac{\int_{\Omega}|\nabla\phi_{a}|^qdx}{\big(\int_{\Omega}
\Phi_{a}(x)\phi_{a}^{p}dx\big)^{q/p}} \leq
C(a,p)a^{q-1+\frac{q}{p}},
$$
where
$$
C(a,p)=
\frac{3^{N-1}\int_{\Omega}|\nabla\phi|^qdx}{(1-\frac{2}{a})^{qa/p}
\big(\int_{\Omega}\phi^{p}dx\big)^{q/p}},
$$
for any $p>q$ 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,
\begin{equation}\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}},
\end{equation}
where $C$ 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 constant $C_1 >0$ 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 $q-N+\frac{q}{p}N<q-1+\frac{q}{p}$. Then it follows from
\eqref{2.2} and \eqref{2.4} that $S_{a,p}<S^{\rm rad}_{a,p}$ when
$a$ is large. So any ground state $u_{a,p}$ is a nonradial function.
\end{proof}

 \subsection{Proof of Theorem \ref{thm2}}
Now we consider any minimum $u_{a,p}$. The following proposition
describes the profile of $u_{a,p}$ as $p\to q^{*}$.

\begin{proposition}\label{prop2.3}
 Let $p\in(q,q^{*})$ and $a>0$. 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
\begin{equation}\label{2.6}
\liminf_{p\to q^{*}}T_{a,p} > S_{0,q^{*}}+\delta,
\end{equation}
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
\begin{equation}\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}
\end{equation}
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 $\delta>0$,
\begin{equation}\label{2.8}
\int_{\Omega}\Phi_{a}(x)|\upsilon|^{p}dx <
\int_{\Omega}\Phi_{a}(x)|u|^{p}dx.
\end{equation}
Now,
\begin{equation}\label{2.9}
\int_{\Omega^{+}}|\nabla u|^qdx=\int_{\Omega^{+}}|\nabla
\upsilon+\delta\nabla\psi_1|^qdx,
\end{equation}
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 $q<p<q^{*}$, we have
$T_{a,p}\geq S_{a,p}$. We want to prove that the inequality
is strict at least for $p\to q^{*}$. In fact assume on the
contrary that
$$
\liminf_{p\to q^{*}}T_{a,p}=\liminf_{p\to
q^{*}}R_{a,p}(\upsilon_{a,p})=S_{0,q^{*}}.
$$
By the definition of $S_{0,q^{*}}$ and H\"{o}lder inequality we obtain,
for a subsequence $p=p_{k}\to q^{*}$,
\begin{align*}
S_{0,q^{*}}
&\leq \frac{\int_{\Omega}|\nabla
 \upsilon_{a,p}|^qdx}{\big(\int_{\Omega}|\upsilon_{a,p}|^{q^{*}}
 dx\big)^{\frac{q}{q^{*}}}}\\
&\leq  \big|\Omega \big|^\frac{(q^{*}-p)q}{q^{*}p}
 \frac{\int_{\Omega}|\nabla
 \upsilon_{a,p}|^qdx}{\big(\int_{\Omega}|\upsilon_{a,p}|^{p}dx
 \big)^{q/p}}\\
&\leq \big|\Omega\big|^\frac{(q^{*}-p)q}{q^{*}p}
\frac{\int_{\Omega}|\nabla \upsilon_{a,p}|^qdx}
{\big(\int_{\Omega}\Phi_{a}(x)|\upsilon_{a,p}|^{p}dx\big)^{q/p}}
= S_{0,q^{*}}+o(1),
\end{align*}
since the weight satisfies $\Phi_{a}\leq 1$. In particular, we obtain
$$
\frac{\int_{\Omega}|\nabla
\upsilon_{a,p}|^qdx}{\big(\int_{\Omega}|\upsilon_{a,p}|^{q^{*}}dx\big)^{\frac{q}{q^{*}}}}\to
S_{0,q^{*}}
$$
and $\upsilon_{a,p}$ is a minimizing sequence of $S_{0,q^{*}}$.

By the same argument as Cao and Peng
\cite[Theorem 1.1]{cp}, we can prove that $\upsilon_{a,p}$
concentrates at precisely one point of the boundary $\partial\Omega$.
This yields a contradiction since
$\int_{\Omega^{+}}|\nabla
\upsilon_{a,p}|^qdx=\int_{\Omega^{-}}|\nabla
\upsilon_{a,p}|^qdx$.
\end{proof}

Now we consider the points
\[
x_{0,\varepsilon}=x_0=\big(3-\frac{1}{|\ln\varepsilon|},0,
\dots,0\big), \quad
x_{1,\varepsilon}=x_1=\big(1+\frac{1}{|\ln\varepsilon|},0,
\dots,0\big),
\]
and the function
\[
U(x)=\frac{1}{(1+|x|^{\frac{q}{q-1}})^{(N-q)/q}}.
\]
We recall that $S_{0,q^{*}}$ is not achieved on any proper subset of
$\mathbb{R}^{N}$, and that it is independent of $\Omega$. However, it is
known that $S_{0,q^{*}}(\mathbb{R}^{N})$ is achieved, and all the minimizers
can be written in the form
$$
\mathcal{U}_{\theta,y}=\frac{1}{(\theta^{2}+|x-y|
^{\frac{q}{q-1}})^{(N-q)/q}},\quad \theta>0,\; y\in \mathbb{R}^{N}.
$$
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
\begin{equation}\label{2.10}
u^{i}_{\varepsilon}=\psi_{i}(x)U^{i}_{\varepsilon}(x), \quad   i=0,1,
\end{equation}
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
\begin{equation}\label{2.11}
\lim_{p\to
q^{*}}R_{a,p}(u^{i}_{\varepsilon})=S_{o,q^{*}}+K(\varepsilon),
\end{equation}
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
\begin{equation}\label{c}
c=c(a,p)=\inf_{\gamma\in\Gamma}\max_{t\in[0,1]}R_{a,p}(\gamma(t)),
\end{equation}
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
\begin{equation}\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}
\end{equation}

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
\begin{equation}\label{3.1}
R_{a,p}(\varphi u_{a})= S_{a,p}+o(S_{a,p})  as a \to
+\infty.
\end{equation}
\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
\begin{equation}\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).
\end{equation}
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,
\begin{equation}\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}.
\end{equation}
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
\begin{equation}\label{3.3}
\int_{\Omega}|\nabla(\varphi u_{a})|^qdx=\int_{\Omega}|\nabla
u_{a}|^qdx+o(1)=1+o(1).
\end{equation}
It is easy to prove that $u_{a}$ satisfies the problem
\begin{equation}\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}
\end{equation}
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
\begin{equation}\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,
\end{equation}
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 $q<s<q^{*}$. Hence
$\int_{\Omega}|\nabla u|^{q-2}\nabla u\nabla
\phi dx=0$ for all $\phi\in C_0^{\infty}(\Omega)$.
Since $u\in W_0^{1,q}(\Omega)$, this implies that $u=0$.

Note that
\begin{equation}\label{3.6}
\begin{split}
&\Big|\int_{\Omega}|\nabla u_{a}|^qdx-\int_{\Omega}|\nabla(\varphi
u_{a})|^qdx\Big|\\
&=\Big|\int_{\Omega}|\nabla u_{a}|^qdx-\int_{\Omega}|\nabla
u_{a}\varphi+\nabla\varphi u_{a}|^qdx\Big|\\
& \leq\int_{\Omega}|\nabla u_{a}|^q(1-\varphi^q)dx+
  \int_{\Omega}|\nabla\varphi u_{a}|^qdx\\
&\quad +C\int_{\Omega}\Big||\nabla u_{a}\varphi|^{q-1}\nabla\varphi u_{a} +
  \nabla u_{a}\varphi|\nabla\varphi u_{a}|^{q-1}\Big|dx.
\end{split}
\end{equation}
Due to the strong convergence in $L^{s}$ for
all $s\in (q,q^{*})$, the last terms tend to zero.
To estimate the term
$\int_{\Omega}|\nabla(u_{a})|^q(1-\varphi^q)dx$,
we multiply \eqref{3.4} by $u_{a}(1-\varphi^q)=u_{a}\eta$ and
integrate. Since
$\operatorname{supp}\eta=\operatorname{supp}(1-\varphi^q)\Subset \Omega$, we have
$$
\int_{\Omega}|\nabla u_{a}|^{q-2}\nabla u_{a}\nabla(
u_{a}\eta)dx=\int_{\Omega}S_{a,p}^{p/q}\Phi_{a}u_{a}^{p}\eta dx.
$$
Therefore,
\begin{align*}
\big|\int_{\Omega}|\nabla u_{a}|^q\eta dx\big|
&\leq \big|\int_{\Omega}|\nabla u_{a}|^{q-2}\nabla u_{a}\nabla
 \eta \cdot u_{a}dx\big|
+ \big|\int_{\Omega}S_{a,p}^{p/q} \Phi_{a}u_{a}^{p}\eta dx\big|\\
&\leq  \|\nabla\eta\|_{\infty}\int_{\operatorname{supp} \eta}|\nabla
u_{a}|^{q-1}|u_{a}|dx+
\big|\int_{\operatorname{supp} \eta}S_{a,p}^{p/q}\Phi_{a}u_{a}^{p}\eta
dx\big|\to 0.
\end{align*}
Hence \eqref{3.3} holds.
\end{proof}

In Lemma \ref{lem3.1} we proved that the infimum of the Rayleigh
quotient $R_{a,p}$ is essentially achieved by the function $\phi
u_{a}$. From the definition of $\phi$, we can decompose $\phi
u_{a}=u_{a,1}+u_{a,2}$, where $u_{a,1}$ vanishes in $C(2-\delta,3)$
and $u_{a,2}$ vanishes in $C(1,2+\delta)$. The following proposition
is the key step in order to prove that the function $\phi u_{a}$
concentrates at the boundary.

\begin{proposition}\label{prop3.2}
Let  $\phi u _{a} = u_{a,1}+ u_{a,2}$, where $\operatorname{supp}
  u_{a,1} \subset C(1,2-\delta)$ and
$\operatorname{supp}  u_{a,2} \subset C(2+\delta,3)$,  and
$\lambda_{a}=\int_{\Omega}\Phi_{a}u_{a,1}^{p}dx/
\int_{\Omega}\Phi_{a}u_{a,2}^{p}dx$.
If  $\lim_{n\to\infty} \lambda_{a_{n}}=L$ for a
sequence $a_{n}\to \infty$, then either $L = 0 $ or $
L=\infty $.
\end{proposition}

\begin{remark} \rm
For the quantity
$\lambda_{a}=\int_{\Omega}\Phi_{a}u_{a,1}^{p}dx/\int_{\Omega}\Phi_{a}u_{a,2}^{p}dx$,
we cannot exclude the case $\limsup_{a\to\infty}
\lambda_{a}=+\infty$ and $\liminf_{a\to\infty}
\lambda_{a}=0$. If a uniqueness result for the minimizer $u_{a}$
were known, then it would be easy to infer that $a\mapsto u_{a}$ is
continuous. Hence $\lambda_{a}$ would be continuous too, and we
could replace both the lower and the upper limit by a unique limit.
Generally, one does not expect such a uniqueness property for any
$p$ and any $a$. However, when $p\approx q^{*}$ we conjecture that
the uniqueness argument of \cite{ps} may be used to our setting.
\end{remark}

\begin{proof}
By the definition of $u_{a,1}$ and $u_{a,2}$ we have
\begin{equation}\label{3.8}
R_{a,p}(\varphi u_{a}) = \frac{\int_{\Omega}|\nabla
u_{a,1}|^qdx+\int_{\Omega}|\nabla u_{a,2}|^qdx}
{\big(\int_{\Omega}\Phi_{a}u_{a,1}^{p}dx
+\int_{\Omega}\Phi_{a}u_{a,2}^{p}dx\big)^{q/p}}.
\end{equation}
Since $u_{a}$ is a positive solution, we can say that
$\lambda_{a}>0$. We get
\begin{equation}\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}
\end{equation}
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
\begin{equation}\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}.
\end{equation}
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
\begin{equation}\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{equation}
\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 $\lambda_{a_{n}}\to 0$,
\begin{equation}\label{3.12}
\lim_{n\to +\infty}\int_{\Omega}|\nabla u_{a_{n},1}|^q dx=0 .
\end{equation}
\end{proposition}

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\end{document}