\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2008(2008), No. 153, pp. 1--13.\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 2008 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2008/153\hfil 
Critical nonlinear elliptic systems in exterior domains]
{Critical Neumann problem for nonlinear elliptic systems
 in exterior domains}

\author[S. Deng, J. Yang\hfil EJDE-2008/153\hfilneg]
{Shengbing Deng, Jianfu Yang}  % in alphabetical order

\address{Shengbing Deng \newline
Department of Mathematics, Jiangxi Normal University,
Nanchang, Jiangxi 330022, China}
\email{shbdeng@yahoo.com.cn}

\address{Jianfu Yang \newline
Department of Mathematics, Jiangxi Normal University,
Nanchang, Jiangxi 330022, China}
\email{jfyang\_2000@yahoo.com}

\thanks{Submitted June 25, 2008. Published November 7, 2008.}
\subjclass[2000]{35J50, 35J60} 
\keywords{Neumann problem; elliptic systems; exterior domains; 
\hfill\break\indent critical Sobolev exponent; least energy solutions}

\begin{abstract}
 In this paper, we investigate the Neumann problem for a critical
 elliptic system in exterior domains. Assuming
 that the coefficient $Q(x)$ is a positive smooth function and
 $\lambda$, $\mu\geq0$ are parameters, we examine the common effect
 of the mean curvature of the boundary $\partial \Omega $ and the
 shape of the graph of the coefficient $Q(x)$ on the existence of the
 least energy solutions.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{corollary}[theorem]{Corollary}

\section{Introduction}

 In this paper, we are concerned with the
following Neumann problem for elliptic systems
\begin{equation}\label{e1.1}
\begin{gathered}
-\Delta u+\lambda
u=\frac{2\alpha}{\alpha+\beta}Q(x)|u|^{\alpha-2}u|v|^\beta \quad
\text{in } \Omega^{c},\\
-\Delta v+\mu v=
\frac{2\beta}{\alpha+\beta}Q(x)|u|^{\alpha}|v|^{\beta-2}v \quad
\text{in}  \Omega^{c},\\
 \frac{\partial u}{\partial \nu}=\frac{\partial
v}{\partial \nu}=0 \quad \text{on }  \partial \Omega, \\
\quad u,v>0 \quad \text{in }  \Omega^{c},
\end{gathered}
\end{equation}
where $\Omega\subset\mathbb{R}^N$ is a smooth bounded domain and
$\Omega^{c}={\mathbb{R}}^N\setminus\Omega$, we assume that
$\Omega^{c}$ has no bounded components. $\lambda, \mu\geq$0 are
parameters, $\alpha, \beta>1$ and $\alpha+\beta=2^{\ast}$, where
$2^{\ast}$ denotes the critical Sobolev exponent, that is,
$2^{\ast}=\frac{2N}{N-2}$ for $N \geq 3$. $\nu$ is the unit inner
normal at the boundary $\partial \Omega$. The coefficient $Q(x)$ is
H\"{o}lder continuous on $\Omega^{c}$ and $Q(x)> 0$ for all x$\ \in
\Omega^{c}$.

In\cite{BN}, critical semilinear elliptic problems for one equation
with Dirichlet boundary conditions was solved by variational
methods. Although the $(PS)$ does not hold globally, it was found in
\cite{BN} that the condition is valid locally. Critical point theory
then can be used locally to find critical points of associated
functionals.  The critical Neumann problem was considered in
\cite{WX} using the same idea as \cite{BN}. Later on, the critical
Neumann has been extensively studied. Various existence results
concerning the graph of coefficients, topology of domains etc.,  can
be found in
\cite{APY,AY,C,CR,CMW,CW,P,PW,WX} and references therein.
Particularly, the Neumann problem in exterior domains
\begin{equation} \label{e1.1a}
\begin{gathered}
-\Delta u+\lambda u=Q(x)|u|^{2^{\ast}-2}u \quad
\text{in }\quad\Omega^{c},\\
\quad \frac{\partial u}{\partial \nu}=0  \quad\text{on }\quad \partial \Omega,\\
\quad  u>0  \quad\text{in } \quad\Omega^{c},
\end{gathered}
\end{equation}
was considered in \cite{CR}. Existence results for \eqref{e1.1a}
were obtained by showing
$$
S(\Omega^{c},Q,\lambda)=\inf\big\{\int_{\Omega^{c}}(|\nabla
u|^2+\lambda u^2)\,dx, u\in
H^1(\Omega^{c}),\int_{\Omega^{c}}Q(x)|u|^{2^{\ast}}\,dx=1\big\},
$$
is achieved. The effect of the graph of $Q$ and the geometry of the
domain was taken into account on the existence of solutions of
\eqref{e1.1a}.

For the system \eqref{e1.1}, it was considered in \cite{AFS} the
existence of solutions for subcritical nonlinearities. In the
critical case, problem \eqref{e1.1} in bounded domains was
investigated in \cite{CY}, where the effect of the shape of $Q(x)$
was considered in the existence of least energy solutions. Inspired
of \cite{CR} and \cite{CY}, in this paper, we consider the existence
of solutions of problem \eqref{e1.1} in exterior domains. The
problem is both critical and setting on unbounded domains. The loss
of compactness is caused by noncompact groups of translations and
dilations. In applying variational methods, it is necessary to
figure out energy levels so that the $(PS)$ condition holds. These
energy levels are not only affected by noncompact groups of
translations and dilations but also the shape of the coefficient
$Q$. Solutions of problem \eqref{e1.1} will be found as a
minimizer of the variational problem
\begin{equation}\label{e1.2}
S_{\lambda,\mu}(\Omega^c,Q)
=\inf_{u,v\in H^1(\Omega^c)\backslash\{0\}}
\frac{\int_{\Omega^c}(|\nabla
u|^2+|\nabla v|^2+\lambda u^2+\mu v^2)\,dx}
{\big(\int_{\Omega^c}Q(x)|u|^\alpha|v|^\beta\,dx\big)^{2/(\alpha+\beta)}},
\end{equation}
which is a weak solution of  \eqref{e1.1} up to a multiple of a
constant. It was proved in \cite{AFS} that every weak solution of
problem \eqref{e1.1} is classical. As we will see, problem
\eqref{e1.2} is closely related to the problem
\begin{equation}\label{e1.3}
S_{\alpha,\beta}=\inf_{u,v\in
D^{1,2}_0(\Omega^{c})\backslash\{0\}}
\frac{\int_{\Omega^{c}}(|\nabla u|^2+|\nabla v|^2)\,dx}
{\big(\int_{\Omega^{c}}|u|^\alpha|v|^\beta\,dx\big)^{2/(\alpha+\beta)}}.
\end{equation}
We may verify as \cite{AFS} that
$$
S_{\alpha,\beta}=\big[\big(\frac{\alpha}{\beta}\big)^{\beta/(\alpha+\beta)}
+\big(\frac{\beta}{\alpha}\big)^
{\alpha/(\alpha+\beta)}\big]S:=A_{\alpha,\beta}S,
$$
where $S$ is the best Sobolev constant defined by
$$
\ S:=\inf_{u\in D_0^{1,2}(\Omega)\backslash\{0\}}
\frac{\int_{\Omega}|\nabla u|^2\,dx}{\big(\int_{\Omega}|u|^{2^*}\,dx\big)^{2/2^*}},
$$
which is achieved if and only if $\Omega=\mathbb{R}^N$ by the
function
$$
U(x)=\big[\frac{N(N-2)}{N(N-2)+|x|^2}\big]^{(N-2)/2}.
$$
The function $U$ satisfies
$$
-\Delta U=U^{2^{\ast}-1},\quad\text{in } \mathbb{R}^{N}
$$
and
$$
\int_{\mathbb{R}^N}|\nabla
U|^2\,dx=\int_{\mathbb{R}^N}|U|^{2^{\ast}}\,dx=S^{\frac{N}{2}}.
$$
Denote
$$
Q_m=\max_{\partial{\Omega}}Q(x), \quad
Q_M=\max_{\Omega^{c}}Q(x), \quad
Q_{\infty}=\lim_{|x|\to  \infty}Q(x).
$$
Suppose $Q_m, Q_M$ and $Q_\infty$ are positive, we set
$$
S_{\infty}=\min \Big\{
\frac{S_{\alpha,\beta}}{2^{2/N}Q_m^{\frac{N-2}{N}}},\frac{S_{\alpha,\beta}}{Q_M^{\frac{N-2}{N}}},
\frac{S_{\alpha,\beta}}{Q_{\infty}^{\frac{N-2}{N}}}\Big\}.
$$
Our main result is as follows.

\begin{theorem}\label{thm1.1}
If $S_{\lambda,\mu}(\Omega^{c},Q)<S_{\infty}$  for
$\lambda, \mu\geq 0$, then $S_{\lambda,\mu}(\Omega^{c},Q)$ is achieved.
\end{theorem}

In section2, we show a variant of second concentration lemma, and
then prove Theorem \ref{thm1.1}. In the rest of the paper, we will
verify the condition
\begin{equation}\label{e1.3a}
S_{\lambda,\mu}(\Omega^{c},Q)<S_{\infty}.
\end{equation}
In the case $Q_M \leq 2^{\frac{2}{N-2}}Q_m$, we assume that
\begin{itemize}
\item[(Q1)] There exists a point y $\in\partial\Omega$ such that
$Q_m=Q(y)$ and $H(y)<0$ and for $x$ near $y$,
\begin{equation}\label{e3.1}
 |Q(x)-Q(y)|=o(|x-y|)
\end{equation}
where $H(y)$ denotes the mean curvature of $\partial \Omega$ at
$y \in \partial\Omega$ with respect to the inner normal to
$\partial \Omega$ at $y$.
\end{itemize}
In the case $Q_M> 2^{\frac{2}{N-2}}Q_m$,  we assume
\begin{itemize}
\item[(Q2)] $Q_ M=Q(y)$ for some $y\in \Omega^{c}$ and for $x$ near $y$,
there holds
\begin{equation}\label{e4.1}
|Q(y)-Q(x)|=o(|x-y|^{N-2}).
\end{equation}
\end{itemize}
If there is $x\in\Omega^c$ such that $Q(x)\geq Q_\infty$, then
$$
S_{\infty}=\min \Big\{
\frac{S_{\alpha,\beta}}{2^{2/N}Q_m^{\frac{N-2}{N}}},
\frac{S_{\alpha,\beta}}{Q_M^{\frac{N-2}{N}}}\Big\}.
$$
If $Q(x)<Q_\infty$ for all $x\in \Omega^c$, we suppose
\begin{itemize}
\item[(Q3)] There exists some cone $K\subset \mathbb{R}^N$ on which the
convergence $Q(x)\to  Q_{\infty}$ holds and there exist $\bar
z\in\partial B_1(0), \delta > 0$, $C >0$ such that for
$R\to \infty$,
\begin{equation}\label{e5.1}
0<Q_{\infty}-Q(R\bar y)\leq \frac{C}{R^{p}},\quad p>\frac{N^2}{2}
\end{equation}
for every $\bar y \in \partial B_1(0)\cap B_\delta(\bar z)$.
\end{itemize}
Under conditions (Q1)-(Q3), $S_\infty$ is well defined.

\begin{theorem}\label{thm1.2}
The condition $S_{\lambda,\mu}(\Omega^{c},Q)<S_{\infty}$ holds if
one of the following conditions holds.
\begin{itemize}
\item[(i)]  $Q_M \leq 2^{\frac{2}{N-2}}Q_m$ and {\rm (Q1)} holds;

\item[(ii)] $Q_M> 2^{\frac{2}{N-2}}Q_m$ and {\rm (Q2)} holds;

\item[(iii)] $Q_{\infty}>2^{2/(N-2)}Q_m$,  $Q(x)<Q_\infty$ for all $x\in
\Omega^c$ and {\rm (Q3)} holds.
\end{itemize}
\end{theorem}

The above theorem  will be proved in sections 3, 4 and 5.


\section{Proof of Theorem \ref{thm1.1}}

Let
$$
J_{\lambda,\mu}(u,v)=\int_{\Omega^{c}}(|\nabla u|^2+|\nabla
v|^2+\lambda u^2+\mu v^2)\,dx
$$
be a functional defined on $E:=H^1(\Omega^{c})\times
H^1(\Omega^{c})$. Then
\begin{equation}
S_{\lambda,\mu}(\Omega^{c},Q)= \inf
\big\{J_{\lambda,\mu}(u,v):(u,v)\in E,
\int_{\Omega^{c}}Q(x)|u|^\alpha|v|^\beta\,dx=1\big\}.
\end{equation}

The following Br\'{e}zis-Lieb type lemma is proved in \cite{CY}.

\begin{lemma}\label{lem2.1}
Let $u_n \rightharpoonup u$ and $v_n \rightharpoonup v$ in
$H^{1}(\Omega^{c})$. Then
\begin{equation}\label{e2.1}
\lim_{n\to \infty}\int_{\Omega^{c}}|u_n|^\alpha|v_n|^\beta\,dx
=\lim_{n\to \infty} \int_{\Omega^{c}}|u_n-u|^\alpha|v_n-v|^\beta\,dx
+\int_{\Omega^{c}}|u|^\alpha|v|^\beta\,dx.
\end{equation}
\end{lemma}

Denote by $B_1(0)$ the unit ball in $\mathbb{R}^N$. We have the following
results, see \cite{CY}.


\begin{lemma}\label{lem2.2}
Let\ $\tilde{B}=B_1(0)\cap \{x_N>h(x')\}$ and $h(x')$ be a $C^1$
function defined on $\{x'\in\mathbb{R}^{N-1}:|x'|<1\}$ with $h$,
$\nabla h$ vanishing at 0. For every $u$, $v \in H^1(B_1(0))$ with
$\mathop{\rm supp} u, \mathop{\rm supp} v \subset \tilde B$, we
have
\begin{itemize}
\item[(A)] If $h\equiv0$, then
\[
\int_{\tilde{B}}(|\nabla u|^2+|\nabla v|^2)\,dx\geq 2^{-2/N}
S_{\alpha,\beta}\Big(\int_{\tilde{B}}
|u|^{\alpha}|v|^{\beta}\Big)^{2/2^{\ast}}.
\]
\item[(B)] For every $\varepsilon>0$ there exists a $\delta>0$ depending
only on $\varepsilon$ such that if $|\nabla h|\leq \delta$, then
\begin{equation*}
\int_{\tilde{B}}(|\nabla u|^2+|\nabla v|^2)\,dx \geq
\Big(\frac{S_{\alpha,\beta}}{2^{2/N}}-\varepsilon\Big)
\Big(\int_{\tilde{B}}|u|^{\alpha}|v|^{\beta}\Big)^{2/2^{\ast}}.
\end{equation*}
\end{itemize}
\end{lemma}

To show the compactness of a $(PS)$ sequence, we need a
concentration - compactness lemma. In \cite{PL}, it gave a
remarkably characterization of non-compactness of the injection of
$W^{1,q}(\Omega)$ into $L^{q^*}(\Omega)$ for $1\leq q<n$ and
$q^*=\frac{qn}{n-q}$. The proof of the following results are
essentially in spirit of \cite{PL}, see also \cite{E}.

\begin{lemma}\label{lem2.3}
Let $u_n\rightharpoonup u$ and $v_n\rightharpoonup v$ in
$H^1(\Omega^{c})$. Suppose that $(|\nabla u_n|^2+|\nabla v_n|^2)
\rightharpoonup\mu$,  $|u_n|^{\alpha}|v_n|^{\beta}
\rightharpoonup\nu$  in the sense of measure, and denote
\begin{gather*}
\lim_{R\to \infty}
\limsup_{n\to \infty} \int_{\Omega^{c}\cap\{|x|\geq
R\} }(|\nabla u_n|^2+|\nabla v_n|^2)\,dx=\mu_{\infty},
\\
\lim_{R\to \infty}
\limsup_{n\to \infty}\int_{\Omega^{c}\cap\{|x|\geq
R\} }|u_n|^{\alpha}|v_n|^{\beta} \,dx=\nu_{\infty}.
\end{gather*}
Then there exist an at most countable index set $J$ and sequences
$\{x_{j}\}\subset\Omega^{c}\cup\partial\Omega, \{\mu_{j}\},
\{\nu_{j}\}\subset(0,\infty), j\in J$,such that
\begin{gather}\label{e2.2}
\nu=|u|^{\alpha}|v|^{\beta}+\sum_{j \in
J}\nu_{j}\delta_{x_{j}}+\nu_{\infty}\delta_{\infty}, \\
\label{e2.3}
\mu\geq |\nabla u|^2+|\nabla v|^2+\sum_{j \in
J}\mu_{j}\delta_{x_{j}}+\mu_{\infty}\delta_{\infty},
\end{gather}
and
\begin{gather}\label{e2.4}
S_{\alpha,\beta}\nu_{\infty}^{2/2^{\ast}} \leq \mu_{\infty}, \\
\label{e2.5}
S_{\alpha,\beta}\nu_{j}^{2/2^{\ast}} \leq \mu_{j},\quad\text{if }x_{j}\in
\Omega^{c}, \\
\label{e2.6}
\frac{S_{\alpha,\beta}}{2^{2/N}}\nu_{j}^{2/2^{\ast}} \leq \mu_{j},\
\quad\text{if }x_{j}\in \partial \Omega^{c},
\end{gather}
where $\delta_x$ denotes the Dirac-mass of mass 1 concentrated at $x$.
\end{lemma}

\begin{proof}
We consider first the case $u=v=0$. Since $\mu$ is a finite measure,
the set $F:=\{x\in \overline{\Omega^{c}}| \mu(\{x\})>0\}$ is at most
countable. We can therefore write $F=\{x_j\}_{j\in J}$,
$\mu_j:=\mu(x_j), \,j\in J $ so that
\[
\mu\geq\sum_{j\in J}\mu_j\delta_{x_j} + \mu_\infty\delta_\infty.
\]
If $x_j\in \Omega^c$, for any $\xi\in C_0^\infty(\Omega^{c})\cap
L^\infty(\Omega^{c})$, we have
\begin{equation}\label{e2.6a}
\begin{aligned}
\int_{\Omega^{c}}|\xi|^{2^{\ast}}\,d \nu
&= \lim_{n\to \infty}\int_{\Omega^{c}}|\xi|^{2^{\ast}}|u_n|^{\alpha}
 |v_n|^{\beta}\,dx \\
&\leq \lim_{n\to \infty}S_{\alpha,\beta}^{-2^{\ast}/2}
\Big(\int_{\Omega^{c}}|\nabla(\xi
u_n)|^{2}+|\nabla(\xi v_n)|^{2}\,dx\Big)^{2^{\ast}/2}.
\end{aligned}
\end{equation}
Since $u_n\to 0$, $v_n\to 0$ in $L^2_{\rm loc}(\Omega^c)$, we deduce
\begin{equation}\label{e2.6b}
\int_{\Omega^{c}}|\xi|^{2^{\ast}}\,d \nu \leq
S_{\alpha,\beta}^{-2^{\ast}/2}\Big(\int_{\Omega^{c}}|\xi|^2\,d\mu\Big)^{2^{\ast}/2}.
\end{equation}
By approximation, for any Borel set $E\in \Omega^{c}$ we have
\begin{equation}\label{e2.6c}
\nu(E) \leq S_{\alpha,\beta}^{-2^{\ast}/2} \mu(E)^{2^{\ast}/2}
\end{equation}
as well as particularly, (\ref{e2.4}) and (\ref{e2.5}) hold.
Because (\ref{e2.6c}) implies $\nu\ll \mu$, we have for $E\in
\Omega^{c}$ being Borel set that
\begin{equation}\label{e2.6d}
\nu(E)= \int_E D_\mu\nu d\mu,
\end{equation}
where
\begin{equation}\label{e2.6e}
D_\mu\nu(x) = \lim_{r\to 0}\frac{\nu(B_r(x))}{\mu(B_r(x))},
\end{equation}
this limit exists for $\mu$-a.e. $x\in \mathbb{R}^N$. From
(\ref{e2.6c}), we have
\begin{equation}\label{e2.6f}
D_\mu\nu = 0,\quad \mu-\text{a.e. }x\in \Omega^c\setminus F.
\end{equation}
Define $\nu_j = D_\mu\nu(x_j)\mu_j$, we see from
(\ref{e2.6c})-(\ref{e2.6f}) that (\ref{e2.2}) holds for the
case $u=v=0$.

If $x_j\in \partial\Omega^c$, by Lemma \ref{lem2.2},
\begin{align*}
&\int_{\Omega^{c}\cap B_\varepsilon(x_j)\cap \{x_N>h(x')\}}(|\nabla
u|^2+|\nabla v|^2)\,dx\\
&\geq \Big(\frac{S_{\alpha,\beta}}{2^{2/N}}-\varepsilon\Big)
\Big(\int_{\Omega^{c}\cap B_\varepsilon(x_j)\cap
\{x_N>h(x')\}}|u|^{\alpha}|v|^{\beta}\Big)^{2/2^{\ast}}
\end{align*}
implying similarly that
$$
\frac{S_{\alpha,\beta}}{2^{2/N}}\nu_{j}^{2/2^{\ast}} \leq \mu_{j}.
$$

Next, in the general case, let $\hat{u}_n= u_n-u$ and $\hat{v}_n =
v_n-v$. We may apply above results to $\hat{u}_n$ and $\hat{v}_n$.
Moreover, in terms of Lemma \ref{lem2.1},
\begin{gather*}
|\nabla \hat u_n|^2 + |\nabla \hat v_n|^2\rightharpoonup \mu +
|\nabla u|^2 + |\nabla v|^2, \\
|\nabla \hat u_n|^\alpha|\nabla \hat v_n|^\beta \rightharpoonup \nu
+ |\nabla u|^\alpha|\nabla v|^\beta
\end{gather*}
in the sense of measures. The proof is complete.
\end{proof}


\begin{proof}[Proof of Theorem \ref{thm1.1}]
 Let $\{u_n,v_n\}$ be a
minimizing sequence for $ S_{\lambda,\mu}(\Omega^{c},Q)$; that is,
$$
\int_{\Omega^{c}}(|\nabla u_n|^2+|\nabla v_n|^2+\lambda u_n^2+\mu
v_n^2)\,dx \to  S_{\lambda,\mu}(\Omega^{c},Q),\quad
\int_{\Omega^{c}}Q(x)|u_n|^\alpha|v_n|^\beta\,dx=1.
$$
We may assume that $u_n\rightharpoonup u$, $v_n\rightharpoonup v$
in $H^1(\Omega^{c})$. By Lemma \ref{lem2.3},
\begin{gather*}
|\nabla u_n|^2+|\nabla v_n|^2 \rightharpoonup \mu \geq |\nabla
u|^2+|\nabla v|^2+\sum_{j \in
J}\mu_{j}\delta_{x_{j}}+\mu_{\infty}\delta_{\infty},\\
Q(x)|u_n|^\alpha|v_n|^\beta \rightharpoonup
\nu=Q(x)|u|^{\alpha}|v|^{\beta}+\sum_{j \in
J}\nu_{j}Q(x_j)\delta_{x_{j}}+\nu_{\infty}Q_\infty\delta_{\infty}
\end{gather*}
and
\begin{equation}\label{eS1}
1=\int_{\Omega^{c}}Q(x)|u|^\alpha|v|^\beta\,dx+\sum_{j\in J} \nu_j
Q(x_j)+\nu_{\infty}Q_{\infty}.
\end{equation}
Therefore, using (\ref{e2.4})-(\ref{e2.6}) we obtain
\begin{equation}\label{eS}
\begin{aligned}
&S_{\lambda,\mu}(\Omega^{c},Q)\\
& \geq \int_{\Omega^{c}}(|\nabla
u|^2+|\nabla v|^2+\lambda u^2+\mu v^2)\,dx+\sum_{j\in J} \mu_j
+\mu_{\infty}\\
&\geq S_{\lambda,\mu}(\Omega^{c},Q)
 \Big(\int_{\Omega^{c}}Q(x)|u|^\alpha|v|^\beta\,dx\Big)^{2/(\alpha+\beta)}+\sum_{x_j\in \Omega^{c}}
S_{\alpha,\beta} \nu_j^{2/(\alpha+\beta)}\\
&\quad + \sum_{x_j\in
\partial \Omega} \frac{S_{\alpha,\beta}}{2^{2/N}}
\nu_j^{2/(\alpha+\beta)}+S_{\alpha,\beta}
\nu_{\infty}^{2/(\alpha+\beta)}\\
&= S_{\lambda,\mu}(\Omega^{c},Q)
\Big(\int_{\Omega^{c}}Q(x)|u|^\alpha|v|^\beta\,dx\Big)^{{(N-2)/N}}\\
&\quad  +\sum_{x_j\in \Omega^{c}}
\frac{S_{\alpha,\beta}}{Q(x_j)^{(N-2)/N}}\left(\nu_j
Q(x_j)\right)^{(N-2)/N}\\
 &\quad + \sum_{x_j\in
\partial \Omega} \frac{S_{\alpha,\beta}}{2^{2/N}Q(x_j)^{(N-2)/N}}
(\nu_j
Q(x_j))^{(N-2)/N}+\frac{S_{\alpha,\beta}}{Q_{\infty}^{(N-2)/2}}
(\nu_{\infty}Q_{\infty})^{(N-2)/N}\\
&\geq S_{\lambda,\mu}(\Omega^{c},Q)
 \Big(\int_{\Omega^{c}}Q(x)|u|^\alpha|v|^\beta\,dx\Big)^{(N-2)/N}
 +\sum_{x_j\in \Omega^{c}} \frac{S_{\alpha,\beta}}{Q_M^{(N-2/N)}} (\nu_j
 Q(x_j))^{(N-2)/N}\\
&\quad +\sum_{x_j\in
\partial \Omega} \frac{S_{\alpha,\beta}}{2^{2/N}Q_m^{(N-2)/N}}
(\nu_j Q(x_j))^{(N-2)/N}+\frac{S_{\alpha,\beta}}{Q_{\infty}^{(N-2)/2}}
(\nu_{\infty}Q_{\infty})^{(N-2)/N}.
\end{aligned}
\end{equation}
Since $S_{\lambda,\mu}(\Omega^{c},Q)<S_{\infty}$, we deduce that
$\nu_j=0$ for all $j \in J\cup\{\infty\}$. Indeed, otherwise, we
infer from (\ref{eS1}) and (\ref{eS}) that
\begin{align*}
&S_{\lambda,\mu}(\Omega^{c},Q) \\
&> S_{\lambda,\mu}(\Omega^{c},Q)
\Big(\int_{\Omega^{c}}Q(x)|u|^\alpha|v|^\beta\,dx+\nu_j Q(x_j)+ \nu_j Q(x_j)+
\nu_{\infty}Q_{\infty}\Big)^{(N-2)/N}\\
&=S_{\lambda,\mu}(\Omega^{c},Q),
\end{align*}
which is a contradiction. Hence,
$\int_{\Omega^{c}}Q(x)|u|^\alpha|v|^\beta\,dx=1$,
and then
$$
 \int_{\Omega}(|\nabla u|^2+|\nabla v|^2+\lambda u^2+\mu
v^2)\,dx \leq S_{\lambda,\mu}(\Omega^{c},Q).
$$
The assertion follows.
\end{proof}

To verify condition (\ref{e1.3a}), we need some preliminaries.
Set
$$
S(\Omega^{c},\lambda)=\inf_{u\in
H^1(\Omega^{c})\backslash\{0\}} \frac{\int_{\Omega^{c}}(|\nabla
u|^2+\lambda u^2)\,dx}{(\int_{\Omega^{c}}|u|^{2^{\ast}}\,dx)^{2/2^{\ast}}},
$$
The following result was proved in \cite{PW}.

\begin{lemma}\label{lem2.4}
Assume $\Omega$ is a smooth bounded domain in $\mathbb{R}^{N}$ such
that $\Omega^{c}$ has no bounded components. Then we have
\begin{itemize}
\item[(i)] $S(\Omega^{c},\lambda)$ is nondecreasing in $\lambda$;

\item[(ii)] $0<S(\Omega^{c},\lambda)\leq (1/2)^{2/N}S$
for all $\lambda\geq0$;

\item[(iii)] If the mean curvature of $\partial \Omega$ is negative at
some point, then for all $\lambda\geq0$,
$S(\Omega^{c},\lambda)<(1/2)^{2/N}S$;

\item[(iv)] If $\lambda\geq0$ and $S(\Omega^{c},\lambda)<
(1/2)^{2/N}S$, then $S(\Omega^{c},\lambda)$ is
achieved.
\end{itemize}
\end{lemma}

Let
$$
U_{\varepsilon,y}(x)=\varepsilon^{-\frac{N-2}{2}}U
\big(\frac{x-y}{\varepsilon}\big),
$$
for $y\in\mathbb{R}^{N}$, $\varepsilon>0$, and denote
$U_{\varepsilon}=U_{\varepsilon,0}$.


\begin{corollary}\label{coro2.1}
Assume $\Omega$ is a smooth bounded domain in $\mathbb{R}^{N}$ such
that $\Omega^{c}$ has no bounded components. Then there hold
\begin{itemize}
\item[(i)] $S_{\lambda,\mu}(\Omega^{c},1)$ is nondecreasing in
$\lambda,\mu$;

\item[(ii)] $0<S_{\lambda,\mu}(\Omega^{c},1)\leq
(1/2)^{2/N}S_{\alpha,\beta}$ for all $\lambda,
\mu\geq0$;

\item[(iii)] If the mean curvature of $\partial \Omega$ is negative at
some point, then for all $\lambda, \mu\geq0$,
$S_{\lambda,\mu}(\Omega^{c},1)<(1/2)^{2/N}S_{\alpha,\beta}$.
\end{itemize}
\end{corollary}

\begin{proof}
(i) is obvious.
Now we show (ii) and (iii) only. Since $S_{0,0}(\Omega^{c},1) =
S_{\alpha, \beta}>0$, $S_{\lambda,\mu}(\Omega^{c},1)>0$ follows from
(i). Now we show that
$S_{\lambda,\mu}(\Omega^{c},1)\leq (1/2)^{2/N}S_{\alpha,\beta}$
for all $\lambda, \mu\geq0$. Let $y$ be a point on $\partial\Omega$.
Let $\Phi$ be the
diffeomorphism from a small ball $B_{\delta}(0)$ centered at the
origin to a neighborhood $\omega$ of $y$ so that  $\Phi:
B_{\delta}^{+}(0)\to \bar{\omega}\cap\Omega^{c}$ and
$\Phi:B_{\delta}(0)\cap\{y_N=0\}\to
\bar{\omega}\cap\partial\Omega^{c}$. We denote $\Psi:=\Phi^{-1}$.
Take a radial cut-off function $\eta$ such that $\eta(r)\equiv1$ for
$r\leq\frac{\delta}{2}$, $\eta(r)=0$ for $r\geq\delta$ and
$0\leq\eta\leq1$. Define
\begin{equation*}
\tilde U_{\varepsilon}(x)= \begin{cases}
(\eta U_{\varepsilon})(\Psi(x))
& \text{if } x\in\bar{\omega}\cap\Omega^{c};\\
0 & \text{if } x\in\Omega^{c}\backslash\bar{\omega}.
  \end{cases}
\end{equation*}
It has shown in \cite{PW} that
\begin{equation}\label{e2.14}
\frac{\int_{\Omega^{c}}[|\nabla \tilde U_{\varepsilon}(x)|^2+\lambda
\tilde U_{\varepsilon}(x)^2]\,dx}{\{\int_{\Omega^{c}}|\tilde
U_{\varepsilon}(x)|^{2^{\ast}}\,dx\}^{2/2^{\ast}}}
=\frac{S}{2^{2/N}}+A_NH(y)\beta_1(\varepsilon)
+O(\beta_2(\varepsilon)),
\end{equation}
where $A_{N}>0$ is a constant and $H(y)$ denotes the mean curvature
of $\partial\Omega$ at $y$, when viewed from inside, and
\begin{equation}\label{e2.15}
\beta_1(\tau)=  \begin{cases}
\tau\log \frac({1}{\tau})  &\text{if } N=3;\\
\tau \quad &\text{if } N\geq 4,
  \end{cases}
\quad
 \beta_2(\tau)= \begin{cases}
\tau   & \text{if } N=3;\\
\tau^2 \log (\frac{1}{\tau})  & \text{if } N=4;\\
\tau^2   & \text{if } N\geq 5.
 \end{cases}
\end{equation}
Choosing $s$ and $t$ such that
\begin{equation}\label{e2.16}
\frac{2\alpha}{\alpha+\beta}s^{-(\alpha-2)}t^{-\beta}=1 \quad
\text{and}\quad
\frac{2\beta}{\alpha+\beta}s^{-\alpha}t^{-(\beta-2)}=1;
\end{equation}
that is, $\frac{s^2}{t^2}=\frac{\beta}{\alpha}$, we have
\begin{equation}\label{e2.17}
\frac{s^2+t^2}{(s^\alpha t^\beta)^{2/2^{\ast}}}=A_{\alpha,\beta}.
\end{equation}
By (\ref{e2.14}),
\begin{equation}\label{e2.18}
\begin{aligned}
S_{\lambda,\mu}(\Omega^c,1)
&\leq \frac{J_{\lambda,\mu}(s\tilde U_{\epsilon}(x),t\tilde
U_{\epsilon}(x))}{\big(\int_{\Omega^{c}}|s\tilde
U_{\epsilon}(x)|^{\alpha}
|t\tilde U_{\epsilon}(x)|^{\beta} \,dx\big)^{2/2^{\ast}}} \\
&\leq \frac{s^2+t^2}{(s^\alpha t^\beta)^{2/2^{\ast}}}
\frac{\int_{\Omega^{c}}[|\nabla \tilde
U_{\epsilon}(x)|^2+\max\{\lambda,\mu\}\tilde U_{\epsilon}(x)^2]\,dx}
{\big(\int_{\Omega^{c}}\tilde U_{\epsilon}(x)^{2^{\ast}}\big)^{2/2^{\ast}}} \\
&\leq A_{\alpha,\beta}\frac{\int_{\Omega^{c}}[|\nabla \tilde
U_{\epsilon}(x)|^2+(\lambda+\mu)\tilde U_{\epsilon}(x)^2]\,dx}
{\big(\int_{\Omega^{c}}\tilde U_{\epsilon}(x)^{2^{\ast}}\big)^{2/2^{\ast}}}  \\
&= \frac{S_{\alpha,\beta}}{2^{2/N}}+B_{N}H(y)\beta_1(\varepsilon)
+O(\beta_2(\varepsilon)),
\end{aligned}
\end{equation}
where $B_{N}>0$ is a constant. Let $\epsilon\to 0$ in
\eqref{e2.18}, we obtain (ii). Equation \eqref{e2.18} also implies (iii)
since we may assume $H(y)<0$.
\end{proof}


\begin{lemma}\label{lem2.6}
For every $\lambda,\mu \geq 0$ we have
$$
\frac{S_{\lambda,\mu}(\Omega^{c},1)}{Q_M^{\frac{N-2}{N}}} \leq
S_{\lambda,\mu}(\Omega^{c},Q)\leq S_{\infty}.
$$
If $Q(x)<Q_{\infty}$ for every $x\in \Omega^{c}$, $Q_M$ on the left
side should be replaced by $Q_{\infty}$.
\end{lemma}

\begin{proof} The first inequality is obvious. To show the second
inequality,  first, since $Q_{\infty}=\lim_{|x|\to \infty}Q(x)$,
for any $\varepsilon>0$ there exists $R>0$ such that
$\Omega \subset B_R(0)$, and $Q(x)\geq Q_{\infty}-\varepsilon$ for
$x\in \mathbb{R}^{N}\setminus B_R(0)$. Hence,
\[
S_{\lambda,\mu}(\Omega^{c},Q)\leq\frac{\int_{\Omega^{c}}(|\nabla
u|^2+|\nabla v|^2+\lambda u^2+\mu v^2)\,dx}
{(Q_{\infty}-\varepsilon)^{\frac{N-2}{N}}
\big(\int_{\mathbb{R}^{N}\setminus
B_R(0)}|u|^{\alpha}|v|^{\beta}\,dx\big)^{2/2^{\ast}}}
\]
for all $u$,$v\in H^1(\Omega^{c})$. Taking infimum over
$u,v\in H_0^1(\mathbb{R}^N\setminus B_R(0))$ and noting that
$S_{\alpha,\beta}$ is independent of $\Omega^c$, since $\epsilon>0$
is arbitrary, we obtain from the above inequality that
$$
S_{\lambda,\mu}(\Omega^{c},Q)\leq
\frac{S_{\alpha,\beta}}{Q_{\infty}^{\frac{N-2}{N}}}.
$$
Next, if $Q(y)=Q_M$ for some $y\in \Omega^{c}$,  using
$(sU_{\varepsilon,y}, tU_{\varepsilon,y})$ as test function in the
expression of $S_{\lambda,\mu}(\Omega^{c},Q)$, where $s$ and $t$
satisfy (\ref{e2.14}), we obtain
$$
S_{\lambda,\mu}(\Omega^{c},Q)\leq
\frac{S_{\alpha,\beta}}{Q_M^{\frac{N-2}{N}}}.
$$
Finally, if $y\in\partial\Omega$ is such that $Q_m=Q(y)$, we deduce
as \eqref{e2.18} that
$$
S_{\lambda,\mu}(\Omega^{c},Q)\leq
\frac{S_{\alpha,\beta}}{2^{2/N}Q_m^{\frac{N-2}{N}}}.
$$
The proof is complete.
\end{proof}

\section{Case $Q_M \leq 2^{2/(N-2)} Q_m $}

We will prove Theorem \ref{thm1.2}.

\begin{proposition}\label{prop3.1}
Assume $Q_M \leq 2^{2/(N-2)} Q_m $ and {\rm (Q1)}. Then there holds
\begin{equation}\label{e3.2}
S_{\lambda,\mu}(\Omega^{c},Q) <
\frac{S_{\alpha,\beta}}{2^{2/N}Q_m^{(N-2)/N}}.
\end{equation}
\end{proposition}

\begin{proof}
If $N\geq 5$, let $s$, $t>0$ be chosen as (\ref{e2.16}). Then
\begin{equation}\label{e3.3}
\frac{J_{\lambda,\mu}(sU_{\varepsilon,y},tU_{\varepsilon,y})}
{\big(\int_{\Omega^{c}}Q(x)|sU_{\varepsilon,y}|^{\alpha}
|tU_{\varepsilon,y}|^{\beta} \,dx\big)^{2/2^{\ast}}} \leq
A_{\alpha,\beta}\frac{\int_{\Omega^{c}}[|\nabla
U_{\varepsilon,y}|^2+(\lambda+\mu)U_{\varepsilon,y}^2]\,dx}
{\big(\int_{\Omega^{c}}Q(x)U_{\varepsilon,y}^{2^{\ast}}\big)^{2/2^{\ast}}}.
\end{equation}
By the assumption (Q1),
\begin{equation}\label{e3.4}
\int_{\Omega^{c}}Q(x)U_{\varepsilon,y}^{2^{\ast}}\,dx
=Q_m\int_{\Omega^{c}}U_{\varepsilon,y}^{2^{\ast}}\,dx+o(\varepsilon).
\end{equation}
Corollary \ref{coro2.1}, (\ref{e3.3}) and (\ref{e3.4}) yield
\begin{align*}
S_{\lambda,\mu}(\Omega^{c},Q)
&\leq\frac{J_{\lambda,\mu}(sU_{\varepsilon,y},tU_{\varepsilon,y})}
{\big(\int_{\Omega^{c}}Q(x)|sU_{\varepsilon,y}|^{\alpha}
|tU_{\varepsilon,y}|^{\beta} \,dx\big)^{2/2^{\ast}}} \\
&<\frac{SA_{\alpha,\beta}}{2^{2/N}Q_m^{(N-2)/N}}=
\frac{S_{\alpha,\beta}}{2^{2/N}Q_m^{(N-2)/N}}.
\end{align*}
If $N=3, 4$, we replace $U_{\varepsilon,y}$ by
$U_{\varepsilon,y}\phi_R$, where $\phi_R\in C^1(\mathbb{R}^N)$,
 $\phi_R(x)=1$ for $x\in B_R(0)$, $\phi_R(x)=0$ for $x\in
\mathbb{R}^N\backslash B_{R+1}(0)$, and $0\leq \phi_R(x)\leq$ 1 on
$\mathbb{R}^N$. Then, (\ref{e3.2}) can be proved in the same way.
\end{proof}


\section{Case $Q_M > 2^{2/(N-2)} Q_m $}

In this section, we show (ii) of Theorem \ref{thm1.2}.

\begin{proposition}\label{prop4.1}
Suppose $Q_M > 2^{2/(N-2)} Q_m $ and {\rm (Q2)}, then there exists
$\Lambda>0$ such that
$$
S_{\lambda,\mu}(\Omega^{c},Q) <
\frac{S_{\alpha,\beta}}{Q_M^{\frac{N-2}{N}}},
$$
for all $0\leq \lambda,\ \mu < \Lambda$.
\end{proposition}

\begin{proof}
 First we consider the case
$N \geq5$. For any $\delta > 0$, using (\ref{e4.1}) we have
$$
\int_{\Omega^{c}}Q(x)|sU_{\varepsilon,y}|^\alpha|tU_{\varepsilon,y}|^\beta\,dx
=\int_{\Omega^{c}}s^{\alpha}t^{\beta}Q_MU_{\varepsilon,y}^{2^{\ast}}\,dx
+\int_{\Omega^{c}}s^{\alpha}t^{\beta}(Q(x)-Q_M)U_{\varepsilon,y}^{2^{\ast}}\,dx
$$
 Since
$$
\int_{\Omega^{c}}s^{\alpha}t^{\beta}Q_MU_{\varepsilon,y}^{2^{\ast}}\,dx
=Q_M\int_{\mathbb{R}^{N}}s^{\alpha}t^{\beta}U_{\varepsilon,y}^{2^{\ast}}\,dx
-Q_M\int_{\Omega}s^{\alpha}t^{\beta}U_{\varepsilon,y}^{2^{\ast}}\,dx
$$
and
\begin{align*}
&\int_{\Omega^{c}}s^{\alpha}t^{\beta}(Q(x)-Q_M)
 U_{\varepsilon,y}^{2^{\ast}}\,dx \\
&=  \int_{\Omega^{c}\cap
 B_{\delta}(y)}s^{\alpha}t^{\beta}(Q(x)-Q_M)U_{\varepsilon,y}^{2^{\ast}}\,dx
 +\int_{\Omega^{c}\backslash
 B_{\delta}(y)}s^{\alpha}t^{\beta}(Q(x)-Q_M)U_{\varepsilon,y}^{2^{\ast}}\,dx\\
&=  \int_{\Omega^{c}\cap
 B_{\delta}(y)}s^{\alpha}t^{\beta}o(|x-y|^{N-2})U_{\varepsilon,y}^{2^{\ast}}\,dx+o(\varepsilon^N),\\
\end{align*}
we have
\begin{equation}
\int_{\Omega^{c}}Q(x)|sU_{\varepsilon,y}|^\alpha|tU_{\varepsilon,y}|^\beta\,dx\\
=s^\alpha t^\beta Q_MK_2+o(\varepsilon^{N-2}),
\end{equation}
where $K_2=\int_{\mathbb{R}^{N}}U^{2^{\ast}}\,dx$. Since
$y \in \Omega^{c}$, there exists a constant $C_1>0$ such that
$$
\int_{\Omega^{c}}|\nabla U_{\varepsilon,y}|^2\,dx=
\int_{\mathbb{R}^{N}}|\nabla U_{\varepsilon,y}|^2\,dx-\int_{\Omega}|\nabla U_{\varepsilon,y}|^2\,dx \leq
K_1-C_1\varepsilon^{N-2},
$$
where $K_1=\int_{\mathbb{R}^{N}}|\nabla U|^2\,dx$, and
$\frac{K_1}{(K_2)^{2/2^{\ast}}}=S$. Hence,
\begin{align*}% \label{e4.4}
&\frac{\int_{\Omega^{c}}[|\nabla(sU_{\varepsilon,y})|^2+
|\nabla(tU_{\varepsilon,y})|^2+\lambda(sU_{\varepsilon,y})^2+
\mu(tU_{\varepsilon,y})^2]\,dx}
{\big(\int_{\Omega^{c}}s^{\alpha}t^{\beta}Q(x)U_{\varepsilon,y}^{2^{\ast}}\big)^{2/2^{\ast}}}\\
&\leq \frac{s^2+t^2}{(s^\alpha t^\beta)^{2/2^{\ast}}}
\frac{\int_{\Omega^{c}}[|\nabla
U_{\varepsilon,y}|^2+\max\{\lambda,\mu\}U_{\varepsilon,y}^2]\,dx}{(\int_{\Omega^{c}}Q(x)U_{\varepsilon,y}^{2^{\ast}})^{2/2^{\ast}}}\\
&\leq \frac{s^2+t^2}{(s^\alpha
t^\beta)^{2/2^{\ast}}}\frac{K_1-C_1\varepsilon^{N-2}+K_3\max\{\lambda,\mu\}\varepsilon^2}
{(Q_MK_2+o(\varepsilon^{N-2}))^{2/2^{\ast}}}\\
&= A_{\alpha,\beta}\big(K_1-C_1\varepsilon^{N-2}+K_3\max\{\lambda,\mu\}
\varepsilon^2\big)
\big\{(Q_MK_2)^{-(N-2)/N}\\
&\quad -\frac{N-2}{N}(Q_MK_2)^{-(2N+2)/N}o(\varepsilon^{N-2})\big\}\\
&< \frac{A_{\alpha,\beta}S}{Q_M^{(N-2)/N}}
 =\frac{S_{\alpha,\beta}}{Q_M^{(N-2)/N}}
\end{align*}
for $\varepsilon>0$, $\lambda$ and $\mu\geq 0$ sufficiently small,
where $K_3$ is a constant independent of $\varepsilon$. If $N=3, 4$,
we replace $U_{\varepsilon,y}$ by $U_{\varepsilon,y}\phi_R$, where
$\phi_R$ is a $C^1$ function such that $\phi_R=1$ if $x\in B_R(y)$,
$\phi_R=0$ if $x\in\mathbb{R}^{N}\subset B_{R+1}(y)$ with $R>0$
large, and we may proceed in the same way.
\end{proof}

\section{Case $S_{\infty}=\frac{S_{\alpha,\beta}}{Q_{\infty}^{(N-2)/N}}$}

In this section,  we show (iii) of Theorem \ref{thm1.2}.

\begin{proposition}\label{prop5.1}
Suppose \eqref{e5.1} holds, then there exists $\Lambda>0$ such that for
$0\leq\lambda$, $\mu<\Lambda$ there holds
$$
S_{\lambda,\mu}(\Omega^c, Q)<\frac{S_{\alpha,\beta}}{Q_{\infty}^{(N-2)/N}}.
$$
\end{proposition}

\begin{proof}
Let $R>0$ be such that $\Omega\subset B_{R/2}(0)$ and
$K_\delta$ denote the cone $K_\delta=\{\tau \bar y; \bar
y\in\partial B_1(0)\cap B_\delta(\bar z), \tau>0\}$, then for $s$
and $t$ satisfying (\ref{e2.16}) we have

\begin{equation}\label{e5.2}
\begin{aligned}
&\int_{\Omega^{c}}Q(x)|sU_{\varepsilon,R \bar
z}|^{\alpha}|tU_{\varepsilon,R\bar z}|^{\beta}\,dx \\
&= s^{\alpha}t^{\beta}\Big\{\int_{B_{R/2}(0)\setminus\Omega}Q(x)U_{\varepsilon,R
\bar z}^{2^{\ast}}\,dx+\int_{\mathbb{R}^N\setminus
B_{R/2}(0)}(Q(x)-Q_{\infty})U_{\varepsilon,R \bar
z}^{2^{\ast}}\,dx  \\
&\quad +\int_{\mathbb{R}^N\setminus
B_{R/2}(0)}Q_{\infty}U_{\varepsilon,R \bar
z}^{2^{\ast}}\,dx \Big\}  \\
&\geq s^{\alpha}t^{\beta}\Big\{\int_{\mathbb{R}^N\setminus(B_{R/2}(0)\cup
K_{\delta})}(Q(x)-Q_{\infty})U_{\varepsilon,R\bar z}^{2^{\ast}}\,dx\\
&\quad  + \int_{K_{\delta}\setminus
B_{R/2}(0)}(Q(x)-Q_{\infty})U_{\varepsilon,R
\bar z}^{2^{\ast}}\,dx
+\int_{\mathbb{R}^N}Q_{\infty}U_{\varepsilon,R\bar z}^{2^{\ast}}\,dx\\
&\quad -\int_{B_{R/2}(0)}Q_{\infty}U_{\varepsilon,R \bar z}^{2^{\ast}}\,dx \Big\}  \\
&= s^{\alpha}t^{\beta}(I_{1}+I_{2}+Q_{\infty}K_{2}+I_{3}).
\end{aligned}
\end{equation}
Since $|x-R\bar z|>\delta R$ if
$x\in B_{2R}(0)\setminus(B_{R/2}(0))\cup K_\delta)$, and
$|x-R\bar z|\geq |x|/2$ if $x\in \mathbb{R}^N\setminus B_{2R}(0)$,
\begin{equation}\label{e5.3}
\begin{aligned}
I_1 &\geq
-Q_{\infty}\int_{\mathbb{R}^N\setminus(B_{R/2}(0)\cup
K_{\delta})}\frac{\varepsilon^NC_N^{2^{\ast}}}{(N(N-2)\varepsilon^2+|x-R\bar
z|^2)^N}\,dx  \\
&\geq -Q_{\infty}\varepsilon^NC_N^{2^{\ast}}
 \int_{R/2}^{2R}\frac{r^{N-1}}{(\delta
R)^{2N}}\,dr-Q_{\infty}\varepsilon^NC_N^{2^{\ast}}\int_{2R}^{\infty}
 \frac{r^{N-1}}{(\frac{r}{2})^{2N}}\,dr  \\
&= -C\frac{\varepsilon^N}{R^N}.
\end{aligned}
\end{equation}
Next, by  assumption \eqref{e5.1},
\begin{equation}\label{e5.4}
I_2=\int_{K_{\delta}\setminus
B_{R/2}(0)}(Q(x)-Q_{\infty})U_{\varepsilon,R\bar
z}^{2^{\ast}}\,dx \geq -\frac{C2^p}{R^p}
\int_{\mathbb{R}^N}U_{\varepsilon,R\bar z}^{2^{\ast}}\,dx=-\frac{CK_2}{R^p}.
\end{equation}
Finally, by the fact that $|x-R\bar z|\geq \frac{R}{2}$ for $x\in
B_{R/2}(0)$,
\begin{equation}\label{e5.5}
I_3=\int_{B_{R/2}(0)}Q_{\infty}U_{\varepsilon,R\bar
z}^{2^{\ast}}\,dx\geq
-\frac{Q_{\infty}C_N^{2^{\ast}}2^{2N}}{R^{2N}}
\int_0^{\frac{R}{2}}r^{N-1}\,dr=-\frac{CK_2}{R^N}
\end{equation}
Consequently,
\begin{equation}\label{e5.6}
\int_{\Omega^{c}}Q(x)|sU_{\varepsilon,R\bar
z}|^{\alpha}|tU_{\varepsilon,R\bar z}|^{\beta}\,dx \geq
s^{\alpha}t^{\beta}
\Big(K_2Q_{\infty}-C\frac{\varepsilon^N}{R^p}-\frac{CK_2}{R^p}\Big).
\end{equation}
On the other hand,
\begin{equation}\label{e5.7}
\begin{aligned}
\int_{\Omega^{c}}|\nabla U_{\varepsilon,R\bar z}|^2\,dx
&= \int_{\mathbb{R}^N}|\nabla U_{\varepsilon,R\bar z}|^2\,dx
  -\int_{\Omega}|\nabla U_{\varepsilon,R\bar z}|^2\,dx  \\
&\leq  K_1-C_N^2(N-2)\varepsilon^{N-2}\int_{\Omega}\frac{|x-R\bar
  z|^2}{(\varepsilon^2N(N-2)+|x-R\bar z|^2)^N}\,dx  \\
&\leq  K_1-\frac{C\varepsilon^{N-2}}{R^{2N-2}}.
\end{aligned}
\end{equation}
Therefore,
\begin{equation}\label{e5.8}
\frac{J_{\lambda,\mu}(sU_{\varepsilon,R\bar z},tU_{\varepsilon,R\bar
z})}{(\int_{\Omega^{c}}Q(x)|sU_{\varepsilon,R\bar z}|^{\alpha}
|tU_{\varepsilon,R\bar z}|^{\beta} \,dx)^{2/2^{\ast}}}  \\
\leq A_{\alpha,\beta}\frac
{K_1-C\frac{\varepsilon^{N-2}}{R^{2N-2}}+\max\{\lambda,\mu\}C\varepsilon^2}
{(K_2Q_{\infty}-C\frac{\varepsilon^N}{R^N}-\frac{CK_2}{R^p})^{(N-2)/N}}.
\end{equation}
Hence, there exist constants $A>0$, $B>0$, $C>0$ and $D>0$ such that
\begin{align*}
&\frac{J_{\lambda,\mu}(sU_{\varepsilon,R\bar z},tU_{\varepsilon,R\bar
z})}{(\int_{\Omega^{c}}Q(x)|sU_{\varepsilon,R\bar z}|^{\alpha}
|tU_{\varepsilon,R\bar z}|^{\beta} \,dx)^{2/2^{\ast}}}\\
&\leq \frac{SA_{\alpha,\beta}}{Q_{\infty}^{(N-2)/N}}
 -\frac{A\varepsilon^{N-2}}{R^{2N-2}}+\max\{\lambda,\mu\}B\varepsilon^2
+\frac{C\varepsilon^N}{R^N}\frac{D}{R^p}.
\end{align*}
If $\lambda=\mu=0$,  we choose $\varepsilon=\varepsilon(R)$ such
that
$$
\frac{A\varepsilon^{N-2}}{2R^{2N-2}}=\frac{C\varepsilon^N}{R^N},
\quad\text{i.e., }
\frac{1}{\varepsilon^2}=\frac{2CR^{N-2}}{A}.
$$
 Then, we choose $R>0$ so that
$$
\frac{A\varepsilon^{N-2}}{2R^{2N-2}}=\frac{A}{2R^{2N-2}}
\Big(\frac{A}{2CR^{N-2}}\Big)^{(N-2)/2}>\frac{D}{R^p};
$$
that is,
$$
\frac{A^{N/2}}{2^{N/2}R^{N^2/2}}>\frac{D}{R^p}\,,
$$
which is possible if $p>\frac{N^2}{2}$. Hence for this choice of $R$
we have
$$
\frac{J_{\lambda,\mu}(sU_{\varepsilon,R\bar z},tU_{\varepsilon,R\bar
z})}{(\int_{\Omega^{c}}Q(x)|sU_{\varepsilon,R\bar z}|^{\alpha}
|tU_{\varepsilon,R\bar z}|^{\beta} \,dx)^{2/2^{\ast}}}<\frac{S_{\alpha,\beta}}{Q_{\infty}^{(N-2)/N}}.
$$
If $\lambda,\mu > 0$. we can similarly choose $R$ such that the
above inequality holds. The proof is complete.
\end{proof}

\subsection*{Acknowledgements}
This work is supported  by grants  10571175 and 10631030
 from the by National Natural Sciences Foundations of
China.

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