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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 116, pp. 1--14.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu
or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2010 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2010/116\hfil Asymptotic behavior]
{Asymptotic behavior of ground state solution for
H\'enon type systems}

\author[Y. Wang, J. Yang \hfil EJDE-2010/116\hfilneg]
{Ying Wang, Jianfu Yang}  % in alphabetical order

\address{Ying Wang \newline
Department of Mathematics, Jiangxi Normal
University, Nanchang, Jiangxi 330022,  China}
\email{yingwang00@126.com}

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

\thanks{Submitted July 14, 2010. Published August 20, 2010.}
\subjclass[2000]{35J50, 35J57, 35J47}
\keywords{Asymptotic behavior; H\'enon systems; ground state solution}

\begin{abstract}
 In this article, we investigate the asymptotic behavior of
 positive ground state solutions, as
 $\alpha\to\infty$, for the following H\'enon type system
 \[
 -\Delta u=\frac{2p}{p+q}|x|^\alpha u^{p-1}v^q,\quad
 -\Delta v=\frac{2q}{p+q}|x|^\alpha u^pv^{q-1},\quad \text{in } B_1(0)
 \]
 with zero boundary condition, where $B_1(0)\subset\mathbb{R}^N$
 ($N\geq3$) is the unit ball centered at the origin,
 $p,q>1$, $p+q<2^*=2N/(N-2)$.
 We show that both components of the ground solution pair $(u, v)$
 concentrate on the same point on the boundary $\partial B_1(0)$ as
 $\alpha\to\infty$.

\end{abstract}

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

\section{Introduction}

 In this article, we investigate the asymptotic behavior of
positive ground state solution pairs of the following H\'enon type
system
\begin{equation}\label{eq:1.1}
-\Delta u=\frac{2p}{p+q}|x|^\alpha u^{p-1}v^q,\quad
-\Delta v=\frac{2q}{p+q}|x|^\alpha u^pv^{q-1},\quad \text{in } B_1(0)
\end{equation}
with zero boundary condition, where $B_1(0)\subset\mathbb{R}^N$
($N\geq3$) is the unit ball centered at the origin, $\alpha>0$,
$p,q>1$, $p+q<2^*=2N/(N-2)$.

H\'enon \cite{H} considered the so called H\'enon equation
\begin{equation}\label{eq:1.2}
\begin{gathered}
-\Delta u = |x|^\alpha u ^ {p-1}, \quad x \in \Omega, \\
u = 0, \quad x \in \partial \Omega,
\end{gathered}
\end{equation}
which stems from a research of rotating stellar structures. Such a
problem enjoys special features. As usual, for arbitrary bounded
$\Omega$, critical exponent for problem \eqref{eq:1.2} is $2^*$,
while if $\Omega$ is a ball, it was shown in \cite{Ni} that problem
\eqref{eq:1.2} has a radially symmetric solution for $p\in (2,\frac
{2(N+\alpha)}{N-2})$, the critical exponent $\frac
{2(N+\alpha)}{N-2}$ is larger than the critical Sobolev exponent
$2^*$. Moreover, even in a ball, problem \eqref{eq:1.2} possesses
non-radial solutions under some conditions, see \cite{SSW} and
references therein. This can also be seen as in \cite{CP}, where it
was shown that the ground state solution of problem \eqref{eq:1.2}
has a unique maximum point approaching to a point on $\partial
B_1(0)$ provided that $\alpha>0$ fixed,  $ p\in(2,2^*)$ and $p\to
2^*$. Similar results for $p\in(2,2^*)$ fixed and $\alpha\to\infty$
can be found in \cite{BW2,BW1,CPY}.


For system \eqref{eq:1.1}, we proved in \cite{WY} that there exists
$\alpha^*>0$ such that the ground state solution of problem
\eqref{eq:1.1} is non-radial if $\alpha>\alpha^*$, $p, q>1$ and $p+q
<2^*$; the maximum points of both components $u$ and $v$ of the
ground state solution pair $(u,v)$ concentrate at the same point on
the boundary $\partial\Omega$ as $p+q\to 2^*$.

In this paper, we investigate the asymptotic behavior of the ground
state solution pair of problem \eqref{eq:1.1} as $\alpha\to\infty$.
Our main result is as follows.

\begin{theorem}\label{th1.1}
Let $(u_\alpha,v_\alpha)$ be a positive ground state solution of
\eqref{eq:1.1} and denote   $x_0=(0,\dots,0,1)$.
Suppose $x_\alpha, y_\alpha\in B_1(0)$ is a maximum point of
   $u_\alpha$, $v_\alpha$ respectively. Then
\begin{gather*}
x_\alpha,y_\alpha\to x\in \partial {B_1(0)},\\
\lim_{\alpha\to+\infty}\alpha(1-|x_\alpha|),\;
\lim_{\alpha\to+\infty}\alpha(1-|y_\alpha|)\; \in(0,+\infty),\\
\begin{aligned}
&\alpha^{-\frac{(2-N)(p+q)+2N}{p+q-2}}\int_{B_1(0)}
  (|\nabla (u_\alpha-\alpha^{\frac{2}{p+q-2}}u(\alpha(x-x_0))|^2\\
&+  |\nabla (v_\alpha-\alpha^{\frac{2}{p+q-2}}v(\alpha(x-x_0))|^2)dx
  \to 0
\end{aligned}
\end{gather*}
as $\alpha\to+\infty$, where $(u,v)$ is a ground state solution
of the system
\begin{equation}\label{eq:1.3}
-\Delta u=\frac{2p}{p+q}e^{x_N} u^{p-1}v^q,\quad
-\Delta v=\frac{2q}{p+q}e^{x_N} u^pv^{q-1},\quad \text{in }
 \mathbb{R}^N_-
\end{equation}
with $u = v = 0$ on $\partial\mathbb{R}^N_-$.
\end{theorem}

The proof of Theorem \ref{th1.1} is inspired by that in \cite{CPY}.
In section 2, we prove that \eqref{eq:1.3} has a ground state
solution pair. We establish in section 3 an asymptotic estimate for
$S_{\alpha,p,q}$ which is defined in the section 3. Then using the
blow up argument, we show in section 4 that the maximum points of
both components of the ground state solution of \eqref{eq:1.1}
concentrate on the same point of the boundary of the domain. The
proof of Theorem \ref{th1.1} is also given in section 4.

\section{A variational problem}


  We consider for $\gamma>0$ the variational problem
\begin{equation}\label{eq:2.1}
m_{\gamma,p,q}=\inf_{0\neq u,v\in
D^{1,2}_0(\mathbb{R}^N_-)}\frac{\int_{\mathbb{R}^N_-}(|\nabla
u|^2+|\nabla v|^2)dx} {(\int_{\mathbb{R}^N_-}e^{\gamma
x_N}|u|^p|v|^qdx)^{2/(p+q)}},
\end{equation}
where $\ p+q\in (2,2^*)$.  We will show that $m_{\gamma,p,q}$ is
achieved. First, we prove that the problem is well defined. For any
$u\in C^\infty_0(\mathbb{R}^N_-)$, by H\"older's inequality,
\[
|u(x',x_N)| \leq
|{x_N}|^{1/2}\Big(\int^0_{-\infty}\big|\frac{\partial
u(x',t)}{\partial t}\big|^2\,dt\Big)^{1/2}.
\]
If $p_1+q_1=2$ and $u, v\in C^\infty_0(\mathbb{R}^N_-)$, we have
\begin{align*}
&\int_{\mathbb{R}^N_-}e^{\gamma x_N}|u|^{p_1}|v|^{q_1}\,dx
\\&\leq\int_{x_N\leq0}|{x_N}|e^{\gamma
x_N}\,dx_N\int_{\mathbb{R}^{N-1}}\Big(\int^0_{-\infty}\big|\frac{\partial
u(x',t)}{\partial t}\big|^2dt\Big)^{\frac{p_1}{2}}
\Big(\int^0_{-\infty}\big|\frac{\partial v(x',t)}{\partial
t}\big|^2dt\Big)^{\frac{q_1}{2}}dx'
\\& \leq C\Big(\int_{\mathbb{R}^N_-}\big|\frac{\partial u(x',x_N)}{\partial x_N}\big|^2\,dx\Big)^{\frac{p_1}{2}}
\Big(\int_{\mathbb{R}^N_-}\big|\frac{\partial v(x',x_N)}{\partial
x_N}\big|^2\,dx\Big)^{\frac{q_1}{2}}
\\&\leq C\int_{\mathbb{R}^N_-}(|\nabla u|^2+|\nabla v|^2)\,dx.
\end{align*}
If $p_2+q_2=2^*$, again by H\"older's inequality,
\[
\int_{\mathbb{R}^N_-}e^{\gamma x_N}|u|^{p_2}|v|^{q_2}\,dx \leq
C\Big(\int_{\mathbb{R}^N_-}(|\nabla u|^2+|\nabla v|^2)\,dx\Big)^{2^*/2}.
\]
Using interpolation inequality for $p+q\in(2,2^*)$, we have
\[
\int_{\mathbb{R}^N_-}e^{\gamma x_N}|u|^{p}|v|^{q}\,dx \leq
C\Big(\int_{\mathbb{R}^N_-}(|\nabla u|^2+|\nabla v|^2)\,dx\Big)
^{(p+q)/2}.
\]
This implies $m_{\gamma,p,q}>0$. Next, for every $ R>0$,
\[
\int_{x_N\leq -R}\int_{\mathbb{R}^{N-1}}e^{\gamma x_N}|u|^{p}|v|^{q}\,dx
\leq Ce^{-\gamma R/2}\Big(\int_{\mathbb{R}^N_-}(|\nabla
u|^2+|\nabla v|^2)\,dx\Big)^{(p+q)/2}.
\]
Hence,   $\int_{x_N\leq -R}e^{\gamma x_N}|u|^{p}|v|^{q}\,dx$ is
uniformly decay in the $x_N$-direction. The variational problem
$m_{\gamma,p,q}$ is compact in the $x_N$-direction and it is
translation invariant in $x_1,\dots,x_{N-1}$. So we may prove as
the proof of \cite[Theorem 1.4]{W} the following result.


 \begin{proposition}\label{prop2.1}
Suppose $p+q\in(2,2^*)$, $\gamma>0$, $N\geq3$. Then, $m_{\gamma,p,q}$
is achieved by $(u,v)$ with positive functions
$u,v\in D^{1,2}_0(\mathbb{R}^N_-)$.
 \end{proposition}

\section{Estimate for $S_{\alpha,p,q}$}

It is known that the problem
\[
S_{\alpha,p,q}=\inf_{u,v\in
H^1_0(B_1(0))\setminus\{0\}}J_\alpha(u,v) =\inf_{u,v\in
H^1_0(B_1(0))\setminus\{0\}}\frac{\int_{B_1(0)}(|\nabla u|^2+|\nabla
v|^2)\,dx}{(\int_{B_1(0)} |x|^\alpha|u|^p|v|^q\,dx)^{2/(p+q)}}
\]
is achieved and the minimizer is a solution of problem
\eqref{eq:1.1} up to a constant. Furthermore, we have the
following result.

\begin{proposition}\label{prop3.1}
  Let $p+q\geq2$. There is $C>0$ such that
  $$
C\leq\frac{S_{\alpha,p,q}}{\alpha^{2- N+\frac{2N}{p+q}}}
\leq m_{1,p,q}+o(1),
$$
  where $o(1)\to 0$ as $\alpha\to \infty$.
 \end{proposition}


\begin{proof}
We use the idea in \cite{CPY}. We establish the upper bound first.
For any $\varepsilon>0$, there exist
$w_\varepsilon, h_\varepsilon\in C^\infty_0(\mathbb{R}^N_-)$,
$w_\varepsilon, h_\varepsilon \neq 0$,
such that
$$
J_{\varepsilon,\mathbb{R}^N_-}(w_\varepsilon,h_\varepsilon)
= \frac{\int_{\mathbb{R}^N_-}(|\nabla w_\varepsilon|^2+|\nabla
h_\varepsilon|^2)dx} {(\int_{\mathbb{R}^N_-}e^{
x_N}|w_\varepsilon|^p|h_\varepsilon|^qdx)^{2/(p+q)}}
<m_{1,p,q}+\varepsilon.
$$
Let
$$
u_\alpha(x)=w_\varepsilon(\alpha
x',\alpha(x_N+(1-|x'|^2)^{1/2})),\quad
v_\alpha(x)=h_\varepsilon(\alpha x',\alpha(x_N+(1-|x'|^2)^{1/2})),
$$
where
$x'=(x_1,x_2,\dots,x_{N-1})$. Then $u_\alpha,v_\alpha\in
H^1_0(B_1(0))$ if $\alpha>0$ is large enough. Denote
$\tilde{B}_\alpha=\{y:\alpha^{-1}y+x_0\in B_1(0)\}$, where
$x_0=(0,0,\dots,1)$. Then
\begin{equation} \label{eq:3.1}
\begin{aligned}
\int_{B_1(0)}|\nabla u_\alpha|^2dx
&=\alpha^{2-N}\int_{\tilde{B}_\alpha}(\sum^{N-1}_{i=1}\Big|D_iw_\varepsilon(x',x_N
+\alpha(1+(1-\frac{1}{\alpha^2}|x'|^2)^{1/2}))
\\
&\quad +\frac{\alpha^{-1}x_i}{(1-\alpha^{-2}|x'|^2)^{1/2}}
 D_Nw_\varepsilon(x',x_N
+\alpha(1+(1-\frac{1}{\alpha^2}|x'|^2)^{1/2}))\Big|^2
\\
&\quad +|D_Nw_\varepsilon(x',x_N
 +\alpha(1+(1-\frac{1}{\alpha^2}|x'|^2)^{1/2}))|^2)\,dx.
\end{aligned}
\end{equation}
Let $$y_i=x_i,\ i=1,2,\dots,N-1;\
y_N=x_N+\alpha(1+(1-\frac{1}{\alpha^2}|x'|^2)^{1/2}),$$ then
$|det(\frac{\partial y}{\partial x})|=1$. By \eqref{eq:3.1},
\begin{equation} \label{eq:3.2}
\begin{aligned}
\int_{B_1(0)}|\nabla u_\alpha|^2dx
&=\alpha^{2-N}\int_{\mathbb{R}^N_-}(\sum^{N-1}_{i=1}|D_iw_\varepsilon
+O(\alpha^{-1})D_Nw_\varepsilon|^2
+|D_Nw_\varepsilon|^2)\ dy
\\&=\alpha^{2-N}(\int_{\mathbb{R}^N_-}|\nabla w_\varepsilon|^2\ dy+O(\alpha^{-1})).
\end{aligned}
\end{equation}
Similarly,
\begin{equation} \label{eq:3.3}
\begin{aligned}
\int_{B_1(0)}|\nabla v_\alpha|^2dx
=\alpha^{2-N}(\int_{\mathbb{R}^N_-}|\nabla h_\varepsilon|^2\ dy+O(\alpha^{-1})).
\end{aligned}
\end{equation}
For any $x\in sptw_\varepsilon \cap spth_\varepsilon$, we have
\[
|\frac{x}{\alpha}+x_0|^\alpha=(1+\frac{2x_N}{\alpha}+O(\alpha^{-2})
)^{\alpha/2}=e^{x_N+O(\alpha^{-1})}.
\]
Therefore,
\begin{equation} \label{eq:3.4}
\begin{aligned}
&\int_{B_1(0)}|x|^\alpha|u_\alpha|^p|v_\alpha|^q\,dx\\
&=\alpha^{-N}\int_{\widetilde{B}_\alpha}|\frac{x}{\alpha}+x_0|^\alpha|w_\varepsilon(x',x_N+
\alpha(1+(1-\frac{1}{\alpha^2|x'|^2})^{1/2}))|^p\\
& \quad \times
|h_\varepsilon(x',x_N+\alpha(1+(1-\frac{1}{\alpha^2|x'|^2})^{1/2}))|^q\,dx \\
&=\alpha^{-N}\int_{\mathbb{R}^N_-}e^{x_N-\alpha(1-(1-\frac{1}{\alpha^2|x'|^2})^{1/2})
+O(\alpha^{-1})}|w_\varepsilon|^p|h_\varepsilon|^q\,dx\\
&=\alpha^{-N}(\int_{\mathbb{R}^N_-}e^{x_N}|w_\varepsilon|^p|h_\varepsilon|^q\ dy+O(\alpha^{-1})).
\end{aligned}
\end{equation}
It follows from  \eqref{eq:3.2}, \eqref{eq:3.3} and \eqref{eq:3.4}
that
\begin{align*}
J_\alpha(u_\alpha,v_\alpha)
&=\alpha^{2-N+\frac{2N}{p+q}}(J_{\varepsilon,{\mathbb{R}^N_-}}(w_\varepsilon,h_\varepsilon)+O(\alpha^{-1}))
\\&<\alpha^{2-N+\frac{2N}{p+q}}(m_{1,p,q}+\varepsilon+O(\alpha^{-1}))
\end{align*}
and then,
\[
\frac{S_{\alpha,p,q}}{\alpha^{2- N+\frac{2N}{p+q}}}\leq m_{1,p,q}+o(1).
\]

Next, we show the lower bound. Let $r\in (0,1], \ \omega\in S^{N-1}$.
For any  $u,v\in H^1_0(B_1(0)\setminus\{0\})$, we define $
 \varphi(r,\omega)=u(r^\beta,\omega),\psi(r,\omega)=v(r^\beta,\omega)$,
where $\beta=\frac{N}{N+\alpha}$.
 Then
\begin{equation} \label{eq:3.5}
\int_{B_1(0)}|x|^\alpha|u|^p|v|^qdx =\beta\int^1_0\int_{\omega\in
S^{N-1}}|\varphi(r,\omega)|^p|\psi(r,\omega)|^qr^{N-1}\,dr\,d\omega,
\end{equation}
and
\begin{equation} \label{eq:3.6}
\begin{aligned}
&\int_{B_1(0)}|\nabla u|^2dx\\
&=\beta\int^1_0\int_{\omega\in
S^{N-1}}(\frac{1}{\beta^2r^{2(\beta-1)}}|\varphi_r(r,\omega)|^2
+\frac{1}{r^{2\beta}}|\nabla_\omega\varphi(r,\omega)|^2)
r^{\beta(N-1)+\beta-1}\, dr\,d\omega\\
&=\frac{1}{\beta}\int^1_0\int_{\omega\in S^{N-1}}(|\varphi_r(r,\omega)|^2+\frac{\beta^2}{r^2}
|\nabla_\omega\varphi(r,\omega)|^2))r^{(2-N)(1-\beta)+N-1}\,dr\,d\omega.
\end{aligned}
\end{equation}
Similarly,
\begin{equation} \label{eq:3.8}
\begin{aligned}
&\int_{B_1(0)}|\nabla v|^2dx\\
& =\frac{1}{\beta}\int^1_0\int_{\omega\in
S^{N-1}}(|\psi_r(r,\omega)|^2+\frac{\beta^2}{r^2}
|\nabla_\omega\psi(r,\omega)|^2)r^{(2-N)(1-\beta)+N-1}\,dr\,d\omega.\\
\end{aligned}
\end{equation}
Note that
\begin{equation} \label{eq:3.9}
|\nabla_\omega\varphi|^2=\sum^{N-1}_{i=1}(\frac{\partial
\varphi}{\partial
x_i}-\frac{x_i}{(1-|x'|^2)^{1/2}}\frac{\partial
\varphi}{\partial x_N})^2
\end{equation}
and $ d\omega=(1+|x'|^2)^{-1/2}dx'$.
 Let $\bar \varphi(r,x')=\varphi(r,\beta x')$,
$\bar {\psi}(r,x')=\psi(r,\beta x')$,
$S_\beta=\{x\in S^{N-1}:|x'|\leq \eta \beta\}$, where $\eta>0$
is small. Then, we may deduce as in \cite{CPY} that for $\beta>0$
small,
\begin{equation} \label{eq:3.12}
\begin{aligned}
&\int^1_0\int_{S_\beta}(|\varphi_r(r,\omega)|^2+
\frac{\beta^2}{r^2}|\nabla_\omega\varphi(r,\omega)|^2))r^{(2-N)(1-\beta)+N-1}\,dr\,d\omega
\\
&\geq C\beta^{N-1}\int_{B_\eta}|\nabla\bar{\varphi}(r,x')|^2r^{(2-N)(1-\beta)}\,dr\,dx'
\end{aligned}
\end{equation}
and
\begin{equation} \label{eq:3.13}
\begin{aligned}
&\int^1_0\int_{S_\beta}(|\psi_r(r,\omega)|^2+\frac{\beta^2}{r^2}
|\nabla_\omega\psi(r,\omega)|^2))r^{(2-N)(1-\beta)+N-1}\,dr\,d\omega
\\
&\geq C\beta^{N-1}\int_{B_\eta}|\nabla\bar{\psi}(r,x')|^2r^{(2-N)(1-\beta)}\,dr\,dx',
\end{aligned}
\end{equation}
where $B_\eta=\{x\in B_1(0):|x'|\leq \eta\}$. Similarly,
\begin{equation} \label{eq:3.14}
\begin{aligned}
&\int^1_0\int_{S_\beta}|\varphi(r,\omega)|^p|\psi(r,\omega)|^qr^{N-1}\,dr\,d\omega \\
&\leq
C\beta^{N-1}\int_{B_\eta}|\bar{\varphi}(r,x')|^p|\bar{\psi}(r,x')|^qr^{N-1}\,dr\,dx'.
\end{aligned}
\end{equation}
Since $\bar{\varphi},\bar{\psi}=0$ on $S^{N-1}$, there exists a
constant $C>0$ such that
\begin{equation} \label{eq:3.15}
(\int_{B_\eta}|\bar{\varphi}(r,x')|^{p+q}r^{N-1}\,dr\,dx')
^{2/(p+q)}
 \leq C \int_{B_\eta}|\nabla\bar{\varphi}(r,x')|^2r^{(2-N)(1-\beta)}
\,dr\,dx'
\end{equation}
and
\begin{equation} \label{eq:3.16}
\Big(\int_{B_\eta}|\bar{\psi}(r,x')|^{p+q}r^{N-1}\,dr\,dx'\Big)^{2/(p+q)}
\leq C \int_{B_\eta}|\nabla\bar{\psi}(r,x')|^2r^{(2-N)(1-\beta)}\,dr\,dx'.
\end{equation}
Therefore, by \eqref{eq:3.15} and \eqref{eq:3.16},
\begin{equation} \label{eq:3.17}
\begin{aligned}
&\Big(\int_{B_\eta}|\bar{\varphi}(r,x')|^p|
\bar{\psi}(r,x')|^{q}r^{N-1}\,dr\,dx'\Big)^{2/(p+q)}
\\
& \leq C\int_{B_\eta}(|\bar{\varphi}(r,x')|^{p+q}r^{N-1}\,dr\,dx')^{2/(p+q)}
+C\int_{B_\eta}(|\bar{\psi}(r,x')|^{p+q}r^{N-1}\,dr\,dx')^{2/(p+q)}
\\
& \leq  C\int_{B_\eta}(|\nabla\bar{\varphi}(r,x')|^2+
|\nabla\bar{\psi}(r,x')|^2)r^{(2-N)(1-\beta)}\,dr\,dx'.
\end{aligned}
\end{equation}
We derive from \eqref{eq:3.12}-\eqref{eq:3.17} that
\begin{equation} \label{eq:3.18}
\begin{aligned}
&\Big(\int^1_0\int_{S_\beta}|\varphi(r,\omega)|^p|\psi(r,\omega)|^qr^{N-1}\,dr\,d\omega\Big)^{2/(p+q)}
\\
&\leq C\beta^{\frac{2(N-1)}{p+q}}(\int_{B_\eta}|\bar{\varphi}(r,x')|^p
|\bar{\psi}(r,x')|^qr^{N-1}\,dr\,dx')^\frac{2}{p+q}
\\
&\leq C\beta^{\frac{2(N-1)}{p+q}}\int_{B_\eta}(|\nabla\bar{\varphi}(r,x')|^2+
|\nabla\bar{\psi}(r,x')|^2)r^{(2-N)(1-\beta)}\,dr\,dx'
\\&\leq C\beta^{1-N+\frac{2(N-1)}{p+q}}\int^1_0\int_{S_\beta}(|\varphi_r(r,\omega)|^2
+\frac{\beta^2}{r^2}|\nabla_\omega\varphi(r,\omega)|^2
\\
&\quad +|\psi_r(r,\omega)|^2+\frac{\beta^2}{r^2}|\nabla_\omega
\psi(r,\omega)|^2)r^{(2-N)(1-\beta)+N-1}\,dr\,d\omega
\end{aligned}
\end{equation}
Since \eqref{eq:1.1} is rotation invariant, we may choose $\beta>0$
so that $S^{N-1}$ can be covered by finite number  $S_\beta$ up to a
rotation, that is  $S^{N-1}\subset\cup S_\beta$. Then
\begin{equation} \label{eq:3.19}
\begin{aligned}
&\Big(\int^1_0\int_{\omega\in S^{N-1}}|
 \varphi(r,\omega)|^p|\psi(r,\omega)|^qr^{N-1}\,dr\,d\omega\Big)^{2/(p+q)}
\\
&\leq  \sum\Big(\int^1_0\int_{S_\beta}|
\varphi(r,\omega)|^p|\psi(r,\omega)|^qr^{N-1}\,dr\,d\omega\Big)^{2/(p+q)}
\\
&\leq C\beta^{1-N+\frac{2(N-1)}{p+q}}
\int^1_0\int_{S_\beta}(|\varphi_r(r,\omega)|^2
+\frac{\beta^2}{r^2}|\nabla_\omega\varphi(r,\omega)|^2
\\
&\quad +|\psi_r(r,\omega)|^2+\frac{\beta^2}{r^2}|
 \nabla_\omega\psi(r,\omega)|^2)r^{(2-N)(1-\beta)+N-1}\,dr\,d\omega
\\
&\leq C\beta^{1-N+\frac{2(N-1)}{p+q}}\int^1_0\int_{S^{N-1}}
(|\varphi_r(r,\omega)|^2
+\frac{\beta^2}{r^2}|\nabla_\omega\varphi(r,\omega)|^2
\\
&\quad +|\psi_r(r,\omega)|^2+\frac{\beta^2}{r^2}
|\nabla_\omega\psi(r,\omega)|^2)r^{(2-N)(1-\beta)+N-1}\,dr\,d\omega.
\end{aligned}
\end{equation}
Hence, we deduce from \eqref{eq:3.6}-\eqref{eq:3.8} and
\eqref{eq:3.19} that
\begin{equation} \label{eq:3.20}
\frac{\int_{B_1(0)}(|\nabla u|^2+|\nabla v|^2)\,dx}{(\int_{B_1(0)}
|x|^\alpha|u|^p|v|^q\,dx)^{2/(p+q)}} \geq C
\beta^{N-2-\frac{2N}{p+q}} =C \alpha^{2-N+\frac{2N}{p+q}}.
\end{equation}
It yields
\begin{equation} \label{eq:3.21}
\alpha^{N-2-\frac{2N}{p+q}}\frac{\int_{B_1(0)}(|\nabla u|^2+|\nabla
v|^2)\,dx}{(\int_{B_1(0)} |x|^\alpha|u|^p|v|^q\,dx)^{2/(p+q)}}\geq C.
\end{equation}
The proof is complete since $u$ and $v$ are arbitrary.
\end{proof}

\section{Asymptotic behavior of ground state solution}

 Let $(U_\alpha,V_\alpha)$ be a minimizer of
$S_{\alpha,p,q}$. Choosing
$\lambda_\alpha=(\frac{2}{S_{\alpha,p,q}})^{\frac{1}{2-(p+q)}}$ and
defining $u_\alpha=\lambda_\alpha U_\alpha,v_\alpha=\lambda_\alpha
V_\alpha$, we see that $(u_\alpha, v_\alpha)$ is a solution pair of
\eqref{eq:1.1}, which is also a minimizer of $S_{\alpha,p,q}$. That
is,
\begin{equation} \label{eq:4.1}
\int_{B_1(0)}(|\nabla u_\alpha|^2+|\nabla v_\alpha|^2)\,dx
=2\int_{B_1(0)}|x|^\alpha|u_\alpha|^{p}|v_\alpha|^{q}\,dx
\end{equation}
 and
\begin{equation} \label{eq:4.2}
S_{\alpha,p,q}=\frac{\int_{B_1(0)}(|\nabla u_\alpha|^2+|\nabla
v_\alpha|^2)\,dx}
{(\int_{B_1(0)}|x|^\alpha|u_\alpha|^{p}|v_\alpha|^{q}\,dx)^{2/(p+q)}}.
\end{equation}
It yields
\begin{equation} \label{eq:4.3}
\int_{B_1(0)}(|\nabla u_\alpha|^2+|\nabla v_\alpha|^2)\,dx
=2\int_{B_1(0)}|x|^\alpha|u_\alpha|^{p}|v_\alpha|^{q}\,dx=
2^{-\frac{2}{p+q-2}}S_{\alpha,p,q}^{\frac{p+q}{p+q-2}}.
\end{equation}
By Proposition \ref{prop3.1}
\begin{equation} \label{eq:4.4}
C\alpha^{\frac{(2-N)(p+q)+2N}{p+q-2}} \leq \int_{B_1(0)}(|\nabla
u_\alpha|^2+|\nabla v_\alpha|^2)\,dx \leq
C'\alpha^{\frac{(2-N)(p+q)+2N}{p+q-2}}.
\end{equation}
Let
$$\bar{u}_\alpha(x)=\alpha^{-\frac{2}{p+q-2}}u_\alpha(\frac{x}{\alpha}),\
\
\bar{v}_\alpha(x)=\alpha^{-\frac{2}{p+q-2}}v_\alpha(\frac{x}{\alpha}),\
\ x\in B_\alpha(0).$$ Then
\begin{equation} \label{eq:4.5}
C\leq\int_{B_\alpha(0)}(|\nabla \bar{u}_\alpha|^2+|\nabla
\bar{v}_\alpha|^2)\,dx\leq C'.
\end{equation}
Choose $p_1, q_1 >0$ such that $p>p_1,\, q>q_1$ and $p_1+q_1=2$.

 \begin{lemma}\label{lemma4.1}
  As $\alpha\to+\infty$, we have
\[
0<C \leq C\max_{x\in
B_\alpha(0)}|\bar{u}_\alpha|^{p-p_1}\max_{x\in
B_\alpha(0)}|\bar{v}_\alpha|^{q-q_1}.
\]
\end{lemma}

\begin{proof}
 By Proposition \ref{prop3.1},
\[
C\alpha^2\leq S_{\alpha,p_1,q_1}=\frac{\int_{B_1(0)}(|\nabla
u_\alpha|^2+|\nabla v_\alpha|^2)\,dx}
{\int_{B_1(0)}|x|^\alpha|u_\alpha|^{p_1}|v_\alpha|^{q_1}\,dx}.
\]
Equation \eqref{eq:4.5} implies
\begin{equation} \label{eq:4.6}
\int_{B_\alpha(0)}|\frac{x}{\alpha}|^\alpha|\bar{u}_\alpha|^{p_1}
|\bar{v}_\alpha|^{q_1}\,dx
\leq C\int_{B_\alpha(0)}(|\nabla \bar{u}_\alpha|^2 +|\nabla
\bar{v}_\alpha|^2)\,dx\leq C.
\end{equation}
Hence, by \eqref{eq:4.3} and \eqref{eq:4.4},
\begin{align*}
0<C &\leq
\int_{B_\alpha(0)}|\frac{x}{\alpha}|^\alpha|\bar{u}_\alpha|^{p}|\bar{v}_\alpha|^{q}\,dx
\\&\leq\max_{x\in B_\alpha(0)}(|\bar{u}_\alpha|^{p-p_1}|\bar{v}_\alpha|^{q-q_1})
\int_{B_\alpha(0)}|\frac{x}{\alpha}|^\alpha|\bar{u}_\alpha|^{p_1}|\bar{v}_\alpha|^{q_1}\,dx
\\&\leq C\max_{x\in B_\alpha(0)}|\bar{u}_\alpha|^{p-p_1}\max_{x\in
B_\alpha(0)}|\bar{v}_\alpha|^{q-q_1}.
\end{align*}
The assertion follows.
\end{proof}

\begin{lemma}\label{lemma4.2}
 There is $C>0$ such that
 $|\bar{u}_\alpha(x)|\leq C$,
$|\bar{v}_\alpha(x)|\leq C$ for $x\in B_\alpha(0)$.
  \end{lemma}

\begin{proof}
Since $(u_\alpha,v_\alpha)$ is a solution pair of \eqref{eq:1.1},
then for $x\in B_\alpha(0)$ we have
\begin{equation} \label{eq:4.7}
-\Delta \bar{u}_\alpha(x)
=\frac{2p}{p+q}|\frac{x}{\alpha}|^\alpha
\bar{u}^{p-1}_\alpha(x)\bar{v}^{q}_\alpha(x)\leq
\frac{2p}{p+q}
\bar{u}^{p-1}_\alpha(x)\bar{v}^{q}_\alpha(x)
\end{equation}
and
\[
-\Delta \bar{v}_\alpha(x)
=\frac{2q}{p+q}|\frac{x}{\alpha}|^\alpha
\bar{u}^{p}_\alpha(x)\bar{v}^{q-1}_\alpha(x)\leq
\frac{2q}{p+q} \bar{u}^{p}_\alpha(x)\bar{v}^{q-1}_\alpha(x).
\]
Now we use the Moser iteration to prove the result. Without
confusion, we use $(u,v)$ to denote $(\bar u_\alpha, \bar
v_\alpha)$. Let $s\geq 1$. Multiplying \eqref{eq:4.7} by $u^{2s}$
and integrating by parts, we obtain
\[
s^{-2}(2s -1)\int_{B_\alpha(0)}|\nabla u^s|^2\,dx \leq \frac{2p}{p+q}\int_{B_\alpha(0)}u^{p-1+2s} v^q\,dx.
\]
Since $s^{-2}(2s -1)\geq s^{-1}$ if $s\geq 1$,
\[
\int_{B_\alpha(0)}|\nabla u^s|^2\,dx \leq \frac{2sp}{p+q}\int_{B_\alpha(0)}u^{p-1+2s} v^q\,dx.
\]
By Sobolev  inequality and H\"older's inequality, we deduce
\begin{equation}\label{eq:4.8}
\begin{aligned}
&\Big(\int_{B_\alpha(0)}u^{2^*s}\,dx\Big)^{2/2^*}\\
&\leq \frac{2sp}{p+q}\int_{B_\alpha(0)}u^{p-1+2s} v^q\,dx \\
&\leq\frac{2sp}{p+q}\Big(\int_{B_\alpha(0)}u^{p+q-1+2s}\,dx\Big)^{\frac{p-1+2s}{p+q-1+2s}}\Big(\int_{B_\alpha(0)}v^{p+q-1+2s}\,dx\Big)^{\frac{q}{p+q-1+2s}}\\
&\leq\frac{sp}{p+q}\int_{B_\alpha(0)}(u^{p+q-1+2s} + v^{p+q-1+2s})\,dx.
\end{aligned}
\end{equation}
Similarly, we have
\begin{equation}\label{eq:4.9}
\Big(\int_{B_\alpha(0)}v^{2^*s}\,dx\Big)^{2/2^*}
\leq\frac{sq}{p+q}\int_{B_\alpha(0)}(u^{p+q-1+2s} + v^{p+q-1+2s})\,dx.
\end{equation}
Therefore,
\begin{equation}\label{eq:4.10}
\begin{aligned}
\Big(\int_{B_\alpha(0)}(u^{2^*s} + v^{2^*s})\,dx\Big)^{2/2^*}
&\leq \Big(\int_{B_\alpha(0)}u^{2^*s}\,dx\Big)^{2/2^*}
 + \Big(\int_{B_\alpha(0)}v^{2^*s}\,dx\Big)^{2/2^*}\\
&\leq s\int_{B_\alpha(0)}(u^{p+q-1+2s} + v^{p+q-1+2s})\,dx.
\end{aligned}
\end{equation}
Now we define $\{s_j\}$ by induction. Let $p+q-1+2s_0 =2^*$
and $p+q-1+2s_{j+1} =2^*s_j$, $j= 0,1, 2,\dots$. We also
 define $M_0 = 1$, $M_{j+1} = (s_jM_j)^{\frac {2^*}2}$,
$j= 0,1, 2,\dots $.
We claim that for all $j\geq 0$,
\begin{equation}\label{eq:4.11}
\int_{B_\alpha(0)}(u^{p+q-1+2s_{j}} + v^{p+q-1+2s_{j}})\,dx\leq C M_j
\end{equation}
and
\begin{equation}\label{eq:4.12}
M_j\leq e^{ms_{j-1}},
\end{equation}
where $C, m>0$. \eqref{eq:4.11} and \eqref{eq:4.12} imply
\[
\Big(\int_{B_\alpha(0)}(u^{2^*s_{j}} + v^{2^*s_{j}})\,dx\Big)
^{\frac 1{2^*s_j}}\leq C M_j^{\frac 1{2^*s_j}}
\leq e^{\frac {ms_{j-1}}{2^*s_j}}\leq C
\]
for all $j$. The assertion then follows. Now, we show \eqref{eq:4.11}.
Obviously, if $j = 0$, \eqref{eq:4.11} holds. Suppose it holds
for $j$, we deduce it holds for $j+1$. Indeed,
\begin{align*}
&\int_{B_\alpha(0)}(u^{p+q-1+2s_{j+1}} + v^{p+q-1+2s_{j+1}})\,dx\\
&= \int_{B_\alpha(0)}(u^{2^*s_{j}} + v^{2^*s_{j}})\,dx\\
&\leq s_j^{2^*/2}\Big(\int_{B_\alpha(0)}(u^{p+q-1+2s_{j}}
 + v^{p+q-1+2s_{j}})\,dx\Big)^{2^*/2}\\
&\leq \Big(s_jM_j\Big)^{2^*/2} = M_{j+1}.
\end{align*}
Inequality \eqref{eq:4.12} can be proved, as in \cite{LNT}.
\end{proof}

Let $M_\alpha=\max_{x\in {B_1(0)}}u_\alpha$, $N_\alpha=\max_{x\in
{B_1(0)}}v_\alpha$.

 \begin{lemma}\label{lemma4.3}
There holds
$$
C_2\alpha^{\frac{2}{p+q-2}}\leq M_\alpha,N_\alpha\leq
C_3\alpha^{\frac{2}{p+q-2}}.
$$
  \end{lemma}

\begin{proof}
By Lemmas \ref{lemma4.1} and \ref{lemma4.2}, we have
\[
0<C_1 \leq C\max_{x\in
B_\alpha(0)}|\bar{u}_\alpha|^{p-p_1}\quad \text{and}\quad
0<C_1 \leq C\max_{x\in B_\alpha(0)}|\bar{v}_\alpha|^{q-q_1}.
\]
This yields the result.
\end{proof}

Let $x_\alpha\in B_1(0)$ be a maximum point of $u_\alpha$ and
$y_\alpha\in B_1(0)$ be a maximum point of $v_\alpha$.

\begin{lemma}\label{lemma4.4}
 The following hold
$\lim_{\alpha\to+\infty}\alpha(1-|x_\alpha|)$ and
$\lim_{\alpha\to+\infty}\alpha(1-|y_\alpha|)$
 are in $\in(0,+\infty)$.
\end{lemma}

 \begin{proof}
We only prove $\lim_{\alpha\to+\infty}\alpha(1-|x_\alpha|)= L
$ and $L\in(0,+\infty)$. The other case can be done in the same way.
Let $B_\alpha(-x_\alpha)=\{x:\frac{x}{\alpha}+x_\alpha\in B_1(0)\}$
and define
$$
\tilde{u}_\alpha(x)=\alpha^{-\frac{2}{p+q-2}}
 u_\alpha(\frac{x}{\alpha}+x_\alpha), \quad
\tilde{v}_\alpha(x)=\alpha^{-\frac{2}{p+q-2}}
 v_\alpha(\frac{x}{\alpha}+x_\alpha).
$$
Then, for $x\in B_\alpha(-x_\alpha)$,
\[
-\Delta \tilde{u}_\alpha(x)
=\frac{2p}{p+q}\Big|\frac{x}{\alpha}+x_\alpha\Big|^\alpha\tilde{u}^{p-1}_\alpha(x)
\tilde{v}^q_\alpha(x).
\]
By Lemma \ref{lemma4.3},
$$
\tilde{u}_\alpha,\tilde{v}_\alpha\leq C,\quad
\tilde{u}_\alpha(0)=\max_{x\in
B_\alpha(-x_\alpha)}\tilde{u}_\alpha(x) \geq C_1>0
$$
and
\[
\int_{B_\alpha(-x_\alpha)}(|\nabla \tilde{u}_\alpha|^2+|\nabla
\tilde{v}_\alpha|^2)\,dx\leq C.
\]
Suppose that $\alpha(1-|x_\alpha|)\to +\infty$, we assume
that there are $\tilde{u},\tilde{v}\in D^{1,2}(\mathbb{R}^N)$ such
that
\begin{gather*}
\tilde{u}_\alpha\rightharpoonup \tilde{u},\quad
\tilde{v}_\alpha\rightharpoonup \tilde{v},\quad \text{in }
D^{1,2}(\mathbb{R}^N)\\
\tilde{u}_\alpha\to \tilde{u},\quad
\tilde{v}_\alpha\to \tilde{v},\quad \text{in }
C^1_{\rm loc}(\mathbb{R}^N).
\end{gather*}
Now, we distinguish two cases:
(i) $|x_\alpha|\leq l<1$;
(ii) $|x_\alpha|\to 1$ as $\alpha\to+\infty$.
For any $x$ with $|x|\leq C$, in case (i), we have
\[
|\frac{x}{\alpha}+x_\alpha|^\alpha\leq
(\frac{C}{\alpha}+|x_\alpha|)^\alpha\leq
(\frac{C}{\alpha}+l)^\alpha\leq(l+\varepsilon)^\alpha\to 0,
\quad\text{as } \alpha\to+\infty.
\]
In case (ii), since $\alpha(1-|x_\alpha|)\to +\infty$,
\[
|\frac{x}{\alpha}+x_\alpha|^\alpha \leq e^{\alpha \ln
(\frac{|x|}{\alpha}+|x_\alpha|-1+1)} =O(e^{\alpha
(\frac{|x|}{\alpha}+|x_\alpha|-1)})
=O(e^{|x|+\alpha(|x_\alpha|-1)})\to 0.
\]
So $\tilde{u}$ satisfies
\[
-\Delta \tilde{u}=0, \quad \tilde{u}\in  D^{1,2}(\mathbb{R}^N).
\]
This implies $\tilde{u}=0$, a contradiction to
$\tilde{u}(0)=\lim_{\alpha\to+\infty}\tilde{u}_\alpha(0)\geq C>0$.
Therefore, $\alpha(1-|x_\alpha|)\to L<+\infty$.

Now, we claim $L>0$. Indeed, we have
$\tilde{u}(0)=\lim_{\alpha\to+\infty}\tilde{u}_\alpha(0)>0$. Since
\eqref{eq:1.1} is invariant under the rotations. After suitably
rotating the coordinate system, we may assume that
$x_\alpha=(0,\dots,0,x^\alpha_N)$, where $x^\alpha_N\to 1$, as
$\alpha\to+\infty$. Then $(\tilde{u},\tilde{v})$ is a positive
solution pair of \eqref{eq:1.3} in
$\Omega=\mathbb{R}^N_-+(0,\dots,0,L)$ with $\tilde{u}=\tilde{v}=0$
on $\partial\Omega$. If $L=0$, we would have
$\Omega=\mathbb{R}^N_-$, and then we obtain $\tilde{u}(0)=0$, a
contradiction.  The proof is complete.
\end{proof}


By Lemma \ref{lemma4.4}, we know that
$x_\alpha\to x_0\in\partial{B_1(0)}$,
$y_\alpha\to y_0\in\partial{B_1(0)}$ if
$\alpha\to+\infty$. In the following, we show that $x_0=y_0$.

\begin{lemma}\label{lemma4.6}
Both $x_\alpha$ and  $y_\alpha$ converge to a point
$x_0\in \partial B_1(0)$ as $\alpha\to+\infty$.
\end{lemma}

\begin{proof}
 We argue by contradiction.
Suppose $x_0\neq y_0$, then there is a $\delta>0$ such that
$B_\delta(x_0)\cap B_\delta(y_0)=\emptyset$. After suitably rotating
the coordinate system, we may assume that $x_0=(0,\dots,0,1)$.
Applying the blow up argument for
$$
\tilde{u}_\alpha(x)=\alpha^{-\frac{2}{p+q-2}}u_\alpha
(\frac{x}{\alpha}+x_0),\quad
\tilde{v}_\alpha(x)=\alpha^{-\frac{2}{p+q-2}}
v_\alpha(\frac{x}{\alpha}+x_0)
$$
in $B_1(0)\cap B_\delta(x_0)$, since
$\tilde{u}_\alpha,\tilde{v}_\alpha$ are bounded in
$D^{1,2}_0(\mathbb{R}^N_-)$, we may assume that there are
$\tilde{u},\tilde{v}\in D^{1,2}_0(\mathbb{R}^N_-)$ such that
\begin{gather*}
\tilde{u}_\alpha\rightharpoonup \tilde{u},\quad
\tilde{v}_\alpha\rightharpoonup \tilde{v},\quad\text{in }
D^{1,2}_0(\mathbb{R}^N_-),\\
\tilde{u}_\alpha\to \tilde{u},\quad
\tilde{v}_\alpha\to \tilde{v},\quad \text{in }
C^1_{\rm loc}(\mathbb{R}^N_-).
\end{gather*}
Moreover, $(\tilde{u},\tilde{v})$ with
$\tilde u,\tilde{v}\in D^{1,2}_0(\mathbb{R}^N_-)$ is a positive
solution of \eqref{eq:1.3}. In the same way, we may assume
$y_0=(0,\dots,0,1)$. Define
$$
\bar{u}_\alpha(x)=\alpha^{-\frac{2}{p+q-2}}u_\alpha(\frac{x}{\alpha}
+ y_0),\quad
\bar{v}_\alpha(x)=\alpha^{-\frac{2}{p+q-2}}v_\alpha(\frac{x}{\alpha}+
y_0).
$$
Then
\begin{gather*}
\bar u_\alpha\rightharpoonup \bar u,\quad
 \bar v_\alpha\rightharpoonup \bar v,
\quad \text{in } D^{1,2}_0(\mathbb{R}^N_-),\\
\bar u_\alpha\to \bar u,\quad
\bar v_\alpha\to \bar v,\quad \text{in }
 C^1_{\rm loc}(\mathbb{R}^N_-),
\end{gather*}
and
 $(\bar{u},\bar{v})$ is a positive solution of
 \eqref{eq:1.3}. It implies
\[
\int_{\mathbb{R}^N_-}(|\nabla \tilde{u} |^2+|\nabla \tilde{v}
|^2)dx,\quad
\int_{\mathbb{R}^N_-}(|\nabla\bar{u} |^2+|\nabla \bar{v}
|^2)dx \geq 2^{-\frac{2}{p+q-2}}m^{\frac{p+q}{p+q-2}}_{1,p,q}
\]
By Proposition \ref{prop3.1},
\begin{align*}
I(u_\alpha,v_\alpha)
&=(\frac{1}{2}-\frac{1}{p+q})\int_{B_1(0)}(|\nabla
u_\alpha|^2+|\nabla v_\alpha |^2)dx\\
&=(\frac{1}{2}-\frac{1}{p+q})2^{-\frac{2}{p+q-2}}
 S^{\frac{p+q}{p+q-2}}_{\alpha,p,q}\\
& \leq (\frac{1}{2}-\frac{1}{p+q})2^{-\frac{2}{p+q-2}}
\alpha^{\frac{(2-N)(p+q)+2N}{p+q-2}}
(m^{\frac{p+q}{p+q-2}}_{1,p,q}+o(1)).
\end{align*}
On the other hand,
\begin{align*}
I(u_\alpha,v_\alpha)
&\geq (\frac{1}{2}-\frac{1}{p+q})\int_{B_1(0)\cap B_\delta(x_0)}
(|\nabla u_\alpha|^2+|\nabla v_\alpha |^2)dx
\\
&\quad +(\frac{1}{2}-\frac{1}{p+q})\int_{B_1(0)\cap B_\delta(
y_0)}(|\nabla u_\alpha|^2+|\nabla v_\alpha |^2)dx
\\
&\geq\alpha^{\frac{(2-N)(p+q)+2N}{p+q-2}}(\frac{1}{2}-\frac{1}{p+q})
\int_{B_\alpha(-x_0)\cap B_{\alpha\delta}(0)}(|\nabla
\tilde{u}_\alpha |^2 +|\nabla \tilde{v}_\alpha |^2)dx
\\
&\quad +
\alpha^{\frac{(2-N)(p+q)+2N}{p+q-2}}(\frac{1}{2}-\frac{1}{p+q})\int_{B_\alpha(-
y_0)\cap B_{\alpha\delta}(0)}(|\nabla \bar{u}_\alpha
|^2+|\nabla \bar{v}_\alpha |^2)dx.
\end{align*}
So we obtain
\begin{align*}
&\int_{B_\alpha(-x_0)\cap B_{\alpha\delta}(0)}(|\nabla
\tilde{u}_\alpha |^2+|\nabla \tilde{v}_\alpha |^2)dx +
\int_{B_\alpha(-\theta y_0)\cap B_{\alpha\delta}(0)}(|\nabla
\bar{u}_\alpha |^2+|\nabla \bar{v}_\alpha |^2)dx \\&\leq
2^{-\frac{2}{p+q-2}}(m^{\frac{p+q}{p+q-2}}_{1,p,q}+o(1)).
\end{align*}
Therefore,
\begin{align*}
2^{-\frac{2}{p+q-2}}(m^{\frac{p+q}{p+q-2}}_{1,p,q}+o(1))
&\geq
\int_{\mathbb{R}^N_-}(|\nabla \tilde{u} |^2+|\nabla \tilde{v} |^2)dx
 + \int_{\mathbb{R}^N_+}(|\nabla \bar{u} |^2+|\nabla \bar{v} |^2)dx
\\
&\geq 2^{-\frac{2}{p+q-2}}(m^{\frac{p+q}{p+q-2}}_{1,p,q}+M^{\frac{p+q}{p+q-2}}_{1,p,q}),
\end{align*}
which is impossible. The proof is complete.
 \end{proof}

Now, we may assume that $x_0=(0,\dots,0,1)$. Let
 \begin{equation} \label{eq:4.13}
 \hat{u}_\alpha(x)=\alpha^{-\frac{2}{p+q-2}}u_\alpha(\frac{x}{\alpha}+x_0),\ \
 \hat{v}_\alpha(x)=\alpha^{-\frac{2}{p+q-2}}v_\alpha(\frac{x}{\alpha}+x_0),
 \end{equation}
which, as before, satisfies
\begin{gather*}
\hat u_\alpha\rightharpoonup \hat u,\quad
\hat v_\alpha\rightharpoonup \hat v, \quad \text{in }
  D^{1,2}_0(\mathbb{R}^N_-), \\\
\hat u_\alpha\to \hat u,\quad \hat v_\alpha\to \hat v,\quad \text{in }
C^1_{\rm loc}(\mathbb{R}^N_-)
\end{gather*}
and $(\hat u,\hat v)\neq (0,0)$ is a
positive solution of \eqref{eq:1.3}.

Finally, we have following result.

\begin{proposition}\label{prop4.1}
The pair $(\hat u,\hat v)$ is a minimizer of
$m_{1,p,q}$, which satisfies
$$
\int_{R^N_-}(|\nabla (\hat{u}_\alpha-\hat u )|^2+
|\nabla (\hat{v}_\alpha-\hat v)|^2)dx\to 0,\quad\text{as }
\alpha\to+\infty.
$$
\end{proposition}

\begin{proof}
 By \eqref{eq:1.3}, we have
\[
\int_{\mathbb{R}^N_-}(|\nabla \hat u|^2+|\nabla \hat
v|^2)dx=2\int_{\mathbb{R}^N_-}e^{x_N}\hat u^p\hat v^qdx,
\]
 and
  $$
m_{1,p,q}\leq\frac{\int_{\mathbb{R}^N_-}(|\nabla u|^2+|\nabla v|^2)dx}
{(\int_{\mathbb{R}^N_-}e^{ x_N}|u|^p|v|^qdx)^{2/(p+q)}}.
$$ So
we obtain
\[
\int_{\mathbb{R}^N_-}(|\nabla u|^2+|\nabla v|^2)dx
\geq 2^{-\frac{2}{p+q-2}}m^{\frac{p+q}{p+q-2}}_{1,p,q}.
\]
For $R>0$ define
$$
B_{R,\alpha}=\{x:\frac{x}{\alpha}+x_0\in B_{\frac{R}{\alpha}(x_0)}\cap B_1(0)\},\ \
\Omega_{\alpha}=\{x:\frac{x}{\alpha}+x_0\in B_1(0)\}.
$$
By Proposition \ref{prop3.1}, $(u_\alpha,v_\alpha)$ is a minimizer of
$S_{\alpha,p,q}$ and satisfies \eqref{eq:1.1}, then
\begin{equation} \label{eq:4.14}
\int_{B_1(0)}(|\nabla u_\alpha|^2+|\nabla v_\alpha|^2)dx \leq
2^{-\frac{2}{p+q-2}}\alpha^{\frac{(2-N)(p+q)+2N}{p+q-2}}
(m^{\frac{p+q}{p+q-2}}_{1,p,q}+o(1)).
\end{equation}
Moreover,
\begin{equation} \label{eq:4.15}
\begin{aligned}
&\int_{B_1(0)}(|\nabla u_\alpha|^2+|\nabla v_\alpha|^2)dx\\
&=\int_{B_{\frac{R}{\alpha}(x_0)}\cap B_1(0)}(|\nabla u_\alpha|^2+|\nabla v_\alpha|^2)dx
+\int_{ B_1(0)/{B_{\frac{R}{\alpha}(x_0)}}}
(|\nabla u_\alpha|^2+|\nabla v_\alpha|^2)dx\\
&=\alpha^{\frac{(2-N)(p+q)+2N}{p+q-2}}\Big(\int_{B_{R,\alpha}}
(|\nabla \hat{u}_\alpha|^2+|\nabla \hat{v}_\alpha|^2)dx+
\int_{\Omega_\alpha/{B_{R,\alpha}}}(|\nabla \hat{u}_\alpha|^2+
|\nabla  \hat{v}_\alpha|^2)dx\Big)\\
&\geq \alpha^{\frac{(2-N)(p+q)+2N}{p+q-2}}\Big(\int_{\mathbb{R}^N_-\cap B_R(0)}
(|\nabla \hat{u}_\alpha|^2+|\nabla \hat{v}_\alpha|^2)dx+o(1)\Big)
\\
&\geq \alpha^{\frac{(2-N)(p+q)+2N}{p+q-2}}\Big(\int_{\mathbb{R}^N_-\cap B_R(0)}
(|\nabla u|^2+|\nabla v|^2)dx+o(1)\Big)
\\
&\geq \alpha^{\frac{(2-N)(p+q)+2N}{p+q-2}}(2^{-\frac{2}{p+q-2}}
m^{\frac{p+q}{p+q-2}}_{1,p,q}+o(1))
\\
&= 2^{-\frac{2}{p+q-2}}\alpha^{\frac{(2-N)(p+q)+2N}{p+q-2}}
((m^{\frac{p+q}{p+q-2}}_{1,p,q}+o(1)).
\end{aligned}
\end{equation}
By \eqref{eq:4.9} and \eqref{eq:4.15},
\[
\int_{\Omega_\alpha/{B_{R,\alpha}}}(|\nabla \hat{u}_\alpha|^2+
|\nabla \hat{v}_\alpha|^2)dx=o(1)+o_R(1),
\]
\begin{align*}
\int_{B_{R,\alpha}}(|\nabla \hat{u}_\alpha|^2+|\nabla
\hat{v}_\alpha|^2)dx &=\int_{\mathbb{R}^N_-\cap B_R(0)}(|\nabla
u|^2+|\nabla v|^2)dx+o(1)
\\&=2^{-\frac{2}{p+q-2}}m^{\frac{p+q}{p+q-2}}_{1,p,q}+o(1)+o_R(1).
\end{align*}
Let $R\to+\infty$, the above equation yields
\begin{equation} \label{eq:4.17}
\int_{\mathbb{R}^N_-}(|\nabla u|^2+|\nabla v|^2)dx
=2^{-\frac{2}{p+q-2}}m^{\frac{p+q}{p+q-2}}_{1,p,q}.
\end{equation}
An application of the Brezis-Lieb's Lemma gives
\[
\int_{\mathbb{R}^N_-}(|\nabla (\hat{u}_\alpha-u)|^2+|\nabla
(\hat{v}_\alpha-v)|^2)dx\to 0
\]
as $\alpha\to+\infty$. On the other hand, by \eqref{eq:1.3}
and \eqref{eq:4.17},
\[
\frac{\int_{\mathbb{R}^N_-}(|\nabla u|^2+|\nabla v|^2)dx}
{(\int_{\mathbb{R}^N_-}e^{ x_N}|u|^p|v|^qdx)^{2/(p+q)}}
=2^{2/(p+q)}(\int_{\mathbb{R}^N_-}(|\nabla u|^2+|\nabla
v|^2)dx)^{\frac{p+q-2}{p+q}} =m_{1,p,q}.
\]
This implies that $(u,v)$ achieves $m_{1,p,q}$. As a consequence, we
have
\begin{align*}
&\alpha^{-\frac{(2-N)(p+q)+2N}{p+q-2}}\int_{B_1(0)}
  (|\nabla (u_\alpha-\alpha^{\frac{2}{p+q-2}}u(\alpha(x-x_0))|^2\\
&+ |\nabla (v_\alpha-\alpha^{\frac{2}{p+q-2}}
v(\alpha(x-x_0))|^2)dx\to 0,
\end{align*}
as $\alpha\to+\infty$.
\end{proof}

 Now the the proof of Theorem \ref{th1.1}
is completed by Lemmas \ref{lemma4.4}, \ref{lemma4.6} and
Proposition \ref{prop4.1}.


\subsection*{Acknowledgements}
This research was partially supported by
grants N10961016 and 10631030 from the NNSF of China, and
2009GZS0011 from  NSF of Jiangxi.

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\end{document}
