\documentclass[reqno]{amsart}
\usepackage{hyperref}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 223, pp. 1--14.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2016/223\hfil Existence and behavior of positive solutions]
{Existence and behavior of positive solutions to
elliptic system with Hardy potential}

\author[L. Wei, X. Cheng, Z. Feng \hfil EJDE-2016/223\hfilneg]
{Lei Wei, Xiyou Cheng, Zhaosheng Feng}

\address{Lei Wei \newline
School of Mathematics and Statistics,
Jiangsu Normal University, Xuzhou 221116, China}
\email{wlxznu@163.com}

\address{Xiyou Cheng \newline
School of Mathematics and Statistics,
Lanzhou University, Lanzhou 730000, China}
\email{chengxy@lzu.edu.cn}

\address{Zhaosheng Feng \newline
Department of Mathematics,
University of Texas-Rio Grande Valley,
Edinburg, TX 78539, USA}
\email{zhaosheng.feng@utrgv.edu}


\thanks{Submitted January 12, 2016. Published August 17, 2016.}
\subjclass[2010]{35B40, 35J25}
\keywords{Elliptic system; Hardy potential; existence; positive solution;
\hfill\break\indent  blow up}

\begin{abstract}
 In this article, we study a class of elliptic systems with Hardy potentials.
 We analyze the possible behavior of radial solutions to the system when
 $p,t>1$, $q,s>0$ and $\lambda,\mu>(N-2)^2/4$,
 and obtain the existence  of positive solutions to  the system with the
 Dirichlet boundary condition under certain conditions.
 When $\lambda,\mu\leq 0$, $p,t>1$ and $q,s>0$, we show that any radial
 positive solution is decreasing in $r$.
\end{abstract}

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

\section{Introduction}

We consider the elliptic system with singular potentials
\begin{equation}
\begin{gathered}
 -\Delta u=\lambda\frac{u}{|x|^2}-u^pv^q,\quad x\in B_1(0)\backslash\{0\}, \\
  -\Delta v=\mu\frac{v}{|x|^2}-u^sv^t,\quad x\in B_1(0) \backslash\{0\},
 \end{gathered}
 \label{mp}
\end{equation}
where $p>1$, $t>1$, and $B_1(0)\subset\mathbb{R}^N$ $(N\geq 3)$ is a unit
ball centered at origin.
The right hand side in \eqref{mp}
contains singular terms, which are usually called the singular inverse
square potentials or Hardy potential in the literature.
In this study, we  investigate system \eqref{mp} with the Dirichlet
boundary condition
\begin{equation} \label{eD}
 u(x)=v(x)=0,\quad x\in\partial B_1(0).
\end{equation}

Elliptic problems with Hardy potential have been an interesting topic in the
field of singular partial differential equation
for a long time. Let us briefly review some results with respect to
the single equation with Hardy potential.
Guerch and V\'eron \cite{GV} studied the equation
\begin{equation} \label{e2}
-\Delta u=V(x)u-h(u),
\end{equation}
where $V(x)=\lambda|x|^{-2}$ and $h(u)$ has the property like $u^p$.
They considered the classification of
positive solutions as $\lambda\leq\frac{(N-2)^2}{4}$, and showed the behavior
 of positive solutions as $\lambda>\frac{(N-2)^2}{4}$ under some conditions.
In a recent work \cite{C}, Cirstea studied
\begin{equation} 
-\Delta u=\lambda\frac{u}{|x|^2}-b(x)f(u),\quad \forall x\in\Omega\backslash\{0\},
\label{1.2}
\end{equation}
where $\Omega$ is a domain containing the origin, and the function $b(x)$
is a continuous positive function (may vanish at the origin) or a singular
function with a singular origin.
It is well-known that $(N-2)^2/4$ is the Hardy constant corresponding to the
 Hardy potential.
The function $f(u)$ has similar behavior like $u^p$ $(p>1)$.
When $\lambda\leq\big(\frac{N-2}{2}\big)^2$,
the qualitative properties of positive solutions at the origin were
presented and a complete classification was provided.

The supercritical case has recently been studied.
In \cite{WF} the authors considered the  singular logistic equation
\begin{equation}\label{logistic}
\begin{gathered}
-\Delta u=\lambda\frac{u}{|x|^2}-b(x)u^p,\quad x\in\Omega\backslash\{0\}, \\
 u=0,\quad  x\in \partial\Omega,
\end{gathered}
\end{equation}
where $\lambda>\frac{(N-2)^2}{4}$, $p>1$ and $b(x)$ is a nonnegative
continuous function over $\overline\Omega$.
It shows that \eqref{logistic} has a minimal positive solution
and a maximal positive solution
if $b(x)$ is a positive function.
Suppose that $b(x)$ has a vanishing
set denoted as  $\Omega_0=\{x\in\Omega:b(x)=0\}$.
If $\Omega_0\Subset\Omega$, the boundary of $\Omega$
belongs to $C^\mu$, $\Omega_0$ is a connected set and $b(0)\neq0$,
then  problem \eqref{logistic} has a minimal positive solution and a
maximal positive solution.
In \cite{WF}, the authors considered the problem
\begin{equation}
-\Delta u=\lambda\frac{u}{|x|^2}-|x|^\theta u^p,\quad  x\in\Omega\backslash\{0\},
 \label{1.3}
\end{equation}
where $\theta>-2$ and $\lambda>\frac{(N-2)^2}{4}$.
 The asymptotic estimate of positive solutions to \eqref{1.3} was
$$
\lim_{|x|\to 0}\frac{u(x)}{|x|^{-\frac{2+\theta}{p-1}}}
=\Big[\lambda+\frac{2+\theta}{p-1}\Big(\frac{2+\theta}{p-1}+2-N\Big)\Big]^{1/(p-1)}.
$$
The uniqueness of positive solutions to \eqref{1.3} with $u=0$ for all
$x\in\partial\Omega$ was shown.
In fact, a direct calculation yields
\[
\lambda+\frac{2+\theta}{p-1}(\frac{2+\theta}{p-1}+2-N)>0\quad \text{when }
 \lambda>(N-2)^2/4.
\]
 This conclusion shows that any positive solution of  \eqref{1.3} blows up
at the origin when  $\theta>-2$ and $\lambda>\frac{(N-2)^2}{4}$.
When $\theta=-2$, positive solutions are uniformly bounded near
 the origin. When $\theta<-2$, any positive solution vanishes at the origin.
So positive solutions of  \eqref{1.3} have no singularity when $\theta\leq-2$.
In addition, for  problem \eqref{logistic},
 we can prove the uniqueness of positive solutions and give the exact
behavior of the positive solution  in a similar manner.

Garc\'{i}a-Meli\'{a}n and Rossi \cite{GR} considered the  competitive type system
\begin{equation}
\begin{gathered}
\Delta u=u^pv^q,\quad x\in\Omega, \\
 \Delta v=u^sv^t,\quad x\in\Omega,\\
 u=v=\infty,\quad x\in\partial\Omega,
\end{gathered}\label{1.4}
\end{equation}
where $\Omega$ is a bounded smooth domain, $p,t>1$ and $q,s>0$.
The existence, uniqueness, blow-up rate and
nonexistence of positive solutions were established under certain conditions.
Li and Wang \cite{LW} considered an elliptic system in a smooth bounded domain,
\begin{equation} 
\begin{gathered}
-\Delta u=u(a_1-b_1u^m-c_1v^n)\quad  x\in\Omega,\\
-\Delta v=v(a_2-b_2u^p-c_2v^q), \quad x\in\Omega,\\
  u=v=+\infty, \quad  x\in\partial\Omega,
 \end{gathered}
\end{equation}
 where $a_i\geq 0$, $b_i,c_i$ $(i=1,2)$ are positive constants,
$m,q>0$ and $n,p\geq 0$. Based on the construction of certain 
sub-solution and upper-solution, some
conditions on the parameters and the exponents to ensure the existence of
positive solutions were explored.
Garc\'{i}a-Meli\'{a}n \cite{G} extended the results in \cite{GR} to study
the existence and uniqueness of positive solutions of
\begin{equation}
\begin{gathered}
\Delta u=a(x)u^pv^q,\quad x\in\Omega, \\
 \Delta v=b(x)u^rv^s,\quad  x\in\Omega,\\
 u=v=\infty,\quad  x\in \partial\Omega,
\end{gathered}\label{1.4b}
\end{equation}
where $a(x)$ and $b(x)$ satisfy
\begin{gather*}
C_2d(x,\partial\Omega)^{\gamma_1}\leq a(x)
\leq C_1d(x,\partial\Omega)^{\gamma_1}\ \text{for}\ x\in\Omega, \\
C_2d(x,\partial\Omega)^{\gamma_2}\leq b(x)
\leq C_1d(x,\partial\Omega)^{\gamma_2} \ \text{for} \ x\in\Omega,
\end{gather*}
for $\gamma_1>-2$, $\gamma_2>-2$.

This article is mainly devoted to the behavior analysis of positive solutions 
to  \eqref{mp} with the boundary condition \eqref{eD}.
In our discussions, we  apply some arguments and techniques with respect 
to boundary blow-up problems,
which can be found in \cite{CR,D,DH,DM,L1,L2,MV}. 
We  also use the methods of subsolution and supersolution \cite{P}.


Let us summarize our main results into the following three theorems.
For any radial positive solution $(u(x),v(x))$, the first theorem 
reveals the behavior of $u(x)+v(x)$ at the origin.

\begin{theorem}\label{th1}
 Suppose that $\lambda,\mu>\frac{(N-2)^2}{4}$, $p,t\geq 1$, $q,s>0$ and 
$(u,v)$ is an arbitrary radial positive solution
 of \eqref{mp}. Then 
 \[
 \lim_{|x|\to0}\{u(x)+v(x)\}=\infty.
 \]
 \end{theorem}

When the positive solution $(u,v)$ satisfies $u\to\infty$ and 
$v\to\infty$ as $|x|\to0$,  $(u,v)$ is said to be a blow-up solution.
The following theorem is regarding the existence of blow-up solution 
and an estimate of blow-up solution near the origin.

\begin{theorem}\label{th2}
 Suppose that $\lambda,\mu>\frac{(N-2)^2}{4}$, $p-1>s>0$ and $t-1>q>0$. 
Then the following two statements are true.
\begin{itemize}
\item[(i)] Problem \eqref{mp} with condition \eqref{eD} has at least
one positive blow-up solution $(U,V)$.

\item[(ii)] Assume that $\lambda>\max\{4\lambda_1[B_1(0)],(N-2)^2/4\}$ and 
$(\hat u,\hat v)$ is a blow-up solution of the problem \eqref{mp}
 with condition \eqref{eD}. Then for any $\tau>0$ there exists 
$\delta:=\delta(\tau)>0$ and $C_1,C_2>0$, such that
 \begin{gather} \label{hat-u-ineq}
 C_1|x|^{-\frac{2(t-1-q)}{(p-1)(t-1)-qs}+\tau}\leq \hat u(x)
\leq C_2|x|^{-\frac{2(t-1-q)}{(p-1)(t-1)-qs}-\tau},
\quad \forall x\in B_\delta0)\backslash\{0\},\\
\label{hat-v-ineq}
 C_1|x|^{-\frac{2(p-1-s)}{(p-1)(t-1)-qs}+\tau}
\leq \hat v(x)\leq C_2|x|^{-\frac{2(p-1-s)}{(p-1)(t-1)-qs}-\tau},\quad
 \forall x\in B_\delta(0)\backslash\{0\}.
 \end{gather}
\end{itemize}
 \end{theorem}

For the parameters $\lambda,\mu\leq 0$, we may obtain the behavior of positive 
solutions.

 \begin{theorem}\label{th3}
 Suppose that $\lambda,\mu\leq 0$, $p,t>1$, $q,s>0$ and $(u,v)$ is an arbitrary
 positive radial solution of  \eqref{mp} with condition \eqref{eD}.
 Then we have
\begin{itemize}
\item[(i)]  $u'(r),v'(r)<0$ for $r\in(0,1]$. Moreover, $u(r)$ and $v(r)$ 
are convex functions; and

\item[(ii)] $u(r)\to\infty$ and $v(r)\to\infty$ when $r\to0$.
\end{itemize}
 \end{theorem}

The rest of this article is organized as follows. 
In Section 2, we introduce some preliminary results which will be
used in the proofs of our main results. 
In Section 3, some rough estimates of positive solution of the problem \eqref{mp}
are established.
 Section 4 is dedicated to the existence of positive solution to problem \eqref{mp} 
with  boundary condition \eqref{eD} and the estimate of blow-up rate 
of positive solution at the origin, i.e., the proof of Theorem \ref{th2}.
Finally, in Section 5 we show the proof of Theorem \ref{th3}.


\section{Preliminary results} \label{s2} 

To make our discussions in a straightforward manner, we need the 
following three technical lemmas. The first lemma is regarding the 
comparison principle which can be found in \cite{D,DM}.

\begin{lemma}\label{lem1}
 Suppose that $\Omega$ is a bounded domain in $\mathbb{R}^N$,
$\alpha(x)$ and $\beta(x)$ are continuous functions in $\Omega$ with
$\|\alpha\|_\infty<\infty$, and $\beta(x)$ is nonnegative and not
identically zero. Let $u_1, u_2\in C^1(\Omega)$ be positive in
$\Omega$ and in the weak sense satisfy 
\begin{gather*}
\Delta u_1+\alpha(x)u_1-\beta(x)g(u_1)\leq 0\leq \Delta
u_2+\alpha(x)u_2-\beta(x)g(u_2),\quad x\in \Omega, \\
\limsup_{x\to\partial \Omega}(u_2-u_1)\leq 0,
\end{gather*}
where $g(u)$ is continuous and such that $\frac{g(u)}{u}$ is
strictly increasing with respect to $u$ in the range of
$\min\{u_1,u_2\}<u<\max\{u_1,u_2\}$. Then $u_2\leq u_1$ holds.
\end{lemma}

For the Hardy potential, it is well-known that the Hardy constant
$H=\frac{(N-2)^2}{4}$ and the Hardy inequality
$$
\frac{(N-2)^2}{4}\int_\Omega\frac{\phi^2}{|x|^2}dx
\leq\int_\Omega|\nabla\phi|^2dx,\ \ \forall \phi\in W_0^{1,2}(\Omega).
$$
In addition, $\frac{(N-2)^2}{4}$ can be expressed as
\begin{equation}
\frac{(N-2)^2}{4}
=\inf_{\phi\in W_0^{1,2}(\Omega)/\{0\}}
\frac{\int_\Omega|\nabla\phi|^2dx}{\int_\Omega\frac{\phi^2}{|x|^2}dx},\label{2.1}
\end{equation}
but it is not attained.
Let $\lambda_1[a(x),\Omega]$ denote the first eigenvalue of
\begin{gather*}
 -\Delta u=\lambda a(x)u,\quad  x\in \Omega, \\
 u=0, \quad x\in\partial\Omega.
 \end{gather*}
A close relation between the Hardy constant and the first eigenvalue
of the Dirichlet eigenvalue problem is given in \cite{WF}.

\begin{lemma}\label{lem2}
 Suppose that $\Omega$ is a bounded smooth domain and $0\in\Omega$. Then we have
\begin{gather*}
\lim_{\delta\to 0}\lambda_1 \big[\frac{1}{|x|^2},\Omega_\delta\big]
=\frac{(N-2)^2}{4};\\
 \lim_{\epsilon\to0}\lambda_1\big[\frac{1}{|x|^2+\epsilon},\Omega\big]
=\frac{(N-2)^2}{4},
\end{gather*}
where $\Omega_\delta=\{x\in\Omega:|x|>\delta\}$.
\end{lemma}

By  the uniqueness and blow-up rate of positive solutions
for a single equation with the Hardy potential \cite{WF}, 
we can obtain the following conclusion.

\begin{lemma}\label{comparison-1}
 Suppose that $p>1$, $\sigma>-2$, $\lambda>(N-2)^2/4$, $\Omega$ is a 
bounded smooth domain and $0\in\Omega$.
\begin{itemize}
\item[(i)] Assume that $u$ is any positive solution of
 \[
-\Delta u=\lambda\frac{u}{|x|^2}-|x|^\sigma u^p,\ x\in \Omega\backslash\{0\}.
\]
Then $u$ satisfies
\[
\lim_{|x|\to0}|x|^{\frac{2+\sigma}{p-1}}u(x)=
\Big[\lambda+\frac{2+\sigma}{p-1}\Big(\frac{2+\sigma}{p-1}+2-N\Big)\Big]^{1/(p-1)}.
\]

\item[(ii)]  Assume that $c$ is a nonnegative constant ($c=+\infty$ is also allowed). 
Then the problem
\begin{equation}
\begin{gathered}
 -\Delta u=\lambda\frac{u}{|x|^2}-|x|^\sigma u^p,\quad x\in \Omega, \\
 u=c,\quad  x\in\partial\Omega
\end{gathered} \label{U-p-sigma}
\end{equation}
 has a unique positive solution $U_{p,\sigma}$.
\end{itemize}
\end{lemma}

\begin{proof}
Firstly, Part (i) is the direct result of \cite{WF}. Now, we simply sketch 
the proof of Part (ii).
Suppose that $c=0$. Since $\lambda>\frac{(N-2)^2}{4}$, for any sufficiently small 
$\delta>0$, by Lemma \ref{lem2} and the standard arguments of logistic equations, 
it follows that
\begin{gather*}
 -\Delta u=\lambda\frac{u}{|x|^2}-|x|^\sigma u^p,\quad x\in \Omega_\delta, \\
 u=0,\quad x\in\partial\Omega_\delta
\end{gather*}
has a unique positive solution $u_\delta$. By the comparison principle and 
regularity arguments of elliptic equations, $w=\lim_{\delta\to0}u_\delta$ 
is a positive solution of the problem \eqref{U-p-sigma}.
Suppose that $c>0$ ($=\infty$), then a similar argument shows that 
 \eqref{U-p-sigma} has a positive solution $w$.
Suppose that $\omega$ is any positive solution of \eqref{U-p-sigma}. 
By Part (i), we have
\[
\lim_{|x|\to0}|x|^{\frac{2+\sigma}{p-1}}\omega(x)=
\Big[\lambda+\frac{2+\sigma}{p-1}\Big(\frac{2+\sigma}{p-1}+2-N\Big)\Big]^{1/(p-1)}.
\]
If $c=\infty$, by the standard arguments of boundary blow-up problems, 
we can find the exact behavior near boundary for
positive solutions. By the standard method, one can see the uniqueness of 
positive solutions.
\end{proof}

\section{Blow-up behavior of positive solutions}
\label{s3} 

In this section, we present some behavior analysis of positive solutions 
of \eqref{mp} and prove Theorem \ref{th1}.

\begin{lemma} \label{sup-1}
 Suppose that $p,t\geq 1$, $q,s>0$, $\lambda,\mu>\frac{(N-2)^2}{4}$
and $(u(x),v(x))$ is an arbitrary solution of \eqref{mp}. Then 
\[
\limsup_{|x|\to0} \{u(x)+v(x)\}=\infty.
\]
That is,
\[
\limsup_{|x|\to0} u(x)=\infty\quad\text{or}\quad 
\limsup_{|x|\to0} v(x)=\infty.
\]
\end{lemma}

\begin{proof} 
By the way of contradiction, we suppose that $\limsup_{|x|\to0^+} \{u(x)+v(x)\}=M$, 
where $M\in[0,\infty)$.  Then there is a $\delta>0$ such that $0<v(x)\leq M+1$ 
for all $x\in B_\delta(0)\backslash\{0\}$.
\smallskip

\noindent\textbf{Case 1: $p>1$.} From the first equation of \eqref{mp}, we have
\begin{equation}\label{eq-1}
-\Delta u\geq\lambda\frac{u}{|x|^2}-(M+1)^qu^p\ \text{in}\ B_\delta(0)\backslash\{0\}.
\end{equation}
Let $w(x)=(M+1)^{\frac{q}{p-1}}u$, then we obtain
\[
-\Delta w\geq\lambda\frac{w}{|x|^2}-w^p\ \text{in}\ B_\delta(0)\backslash\{0\}.
\]
Since $\lambda>\frac{(N-2)^2}{4}$, by Lemma \ref{lem2}, for any sufficiently 
small $\epsilon>0$, the problem
\begin{gather*}
 -\Delta u=\lambda\frac{u}{|x|^2}-u^p,\quad
  x\in B_\delta\backslash\overline B_\epsilon(0), \\
 u=0,\quad |x|=\delta  \text {or } |x|=\epsilon
\end{gather*}
has a unique positive solution $u_\epsilon$. By the comparison principle,
the function $u_*(x):=\lim_{\epsilon\to0}u_\epsilon(x)$ is well defined in 
$B_\delta(0)\backslash\{0\}$, and hence by the regularity arguments, $u_*$ satisfies
\[
-\Delta u_*=\lambda\frac{u_*}{|x|^2}-u_*^p\ \text{in}\ B_\delta(0)\backslash\{0\}.
\]

By the comparison principle, 
\[
w(x)\geq u_\epsilon(x)\quad \text{for all } x\in B_\delta(0)
\backslash\overline B_\epsilon(0).
\]
Letting $\epsilon\to0$, we find
\[
w(x)\geq u_*(x)\quad \text{for all } x\in B_\delta(0)\backslash\{0\}.
\]
By Lemma \ref{comparison-1},  $\lim_{|x|\to0^+}u(x)=\infty$,
which  contradicts  $\limsup_{|x|\to0^+} \{u(x)+v(x)\}=M$.
\smallskip

\noindent\textbf{Case 2: $p=1$.}
Since $\lambda>\frac{(N-2)^2}{4}$ and using \eqref{eq-1}, there is 
$\tau\in(0,\delta)$ such that
\[
-\Delta u\geq \frac{1}{2}(\lambda+\frac{(N-2)^2}{4})\frac{u}{|x|^2}\quad
 \text{for } x\in B_\tau(0)\backslash\{0\}.
\]
For each $\eta\in(0,\tau)$, $u$ is a positive supersolution of
\begin{gather*}
 -\Delta\phi=\frac{1}{2}(\lambda+\frac{(N-2)^2}{4})\frac{\phi}{|x|^2},\quad
  x\in B_\tau(0)\backslash\overline B_\eta(0), \\
 \phi(x)=0,\quad x\in\partial B_\tau(0)\cup\partial B_\eta(0).
\end{gather*}
So, we derive that
\[
\lambda_1\big[\frac{1}{|x|^2},\, B_\tau(0)\backslash\overline B_\eta(0)\big]
>\frac{1}{2}(\lambda+\frac{(N-2)^2}{4}),
\]
which  contradicts Lemma \ref{lem2}, namely
\[
\lim_{\eta\to 0}\lambda_1 \big[\frac{1}{|x|^2},\, B_\tau(0)\backslash
 \overline B_\eta(0) \big]=\frac{(N-2)^2}{4}.
\]
\end{proof}

\begin{remark} \label{rmk3.2} \rm
Suppose that $p,t\geq 1$, $q,s>0$ and $\lambda,\mu>(N-2)^2/4$. 
From the proof of Lemma \ref{sup-1},
it follows that $\lim_{|x|\to0}v(x)=\infty$ if $u$ is bounded near the origin,
and $\lim_{|x|\to0}u(x)=\infty$ if $v$ is bounded near the origin.
\end{remark}

\begin{proof}[Proof of Theorem \ref{th1}]
Suppose that $(u(x),v(x))$ is an arbitrary positive radial solution of  \eqref{mp}.
By Lemma \ref{sup-1}, we have
$$
\limsup_{|x|\to0}\{u(x)+v(x)\}=\infty.
$$
It suffices to show that
\[
\liminf_{|x|\to0}\{u(x)+v(x)\}=\infty.
\]
By the way of contradiction, we suppose 
$\liminf_{|x|\to0}\{u(x)+v(x)\}<\infty$. For convenience of our statement, we
denote $u(r)=u(x)$  and $v(r)=v(x)$ when $|x|=r$. In view of
 \[
 \liminf_{|x|\to0}\{u(x)+v(x)\}<\limsup_{|x|\to0}\{u(x)+v(x)\},
 \]
there is $\{r_n\}$ with $r_n\to0$ such that
\begin{gather*}
u''(r_n)+v''(r_n)\geq 0,\quad u'(r_n)+v'(r_n)=0,  \\
\lim_{n\to\infty}\{u(r_n)+v(r_n)\}=\liminf_{|x|\to0}\{u(x)+v(x)\}.
\end{gather*}
Without loss of generality, we assume that $\lambda\leq\mu$.
It follows from \eqref{mp} that
\begin{equation}\label{inequality-1}
\frac{\lambda}{r_n^2}\leq\frac{u(r_n)^pv(r_n)^q+u(r_n)^sv(r_n)^t}{u(r_n)+v(r_n)},
\end{equation}
which implies
\[
\frac{\lambda}{r_n^2}<u(r_n)^{p-1}v(r_n)^q+u(r_n)^sv(r_n)^{t-1}.
\]
So, we have
\[
u(r_n)^{p-1}v(r_n)^q+u(r_n)^sv(r_n)^{t-1}\to\infty\ \ (n\to\infty).
\]
This indicates that
$\{u(r_n)+v(r_n)\}$ is an unbounded set.
This contradicts the assumption of $\lim_{n\to\infty}(u(r_n)+v(r_n))<\infty$.
\end{proof}

We now show some analysis on the behavior of 
 positive solutions to \eqref{mp} near the origin.

\begin{proposition} \label{prop3.3}
Suppose that $\lambda,\mu>(N-2)^2/4$, $t-1>q>0$, $p-1>s>0$, and $(u(x),\,v(x))$ 
is an arbitrary positive solution of the system \eqref{mp} such that both 
$\lim_{|x|\to0}u(x)$ and $\lim_{|x|\to0}v(x)$ exist.
Then at least one of following statements holds: 
(i) both $u(x)$ and $v(x)$ blow up at the origin; 
(ii) $u(x)$ blows up and $v(x)$ vanishes at the origin; and 
(iii) $u(x)$ vanishes and $v(x)$ blows up at the origin.
\end{proposition}

\begin{proof}
 We claim that either $\lim_{|x|\to0}u(x)=\infty$ or $\lim_{|x|\to0}u(x)=0$ holds. 
Otherwise, we suppose $\lim_{|x|\to0}u(x)=m\in(0,\infty)$.
So there exists a $\delta>0$ such that
\[
m/2\leq u(x)\leq 2m\quad  \text{for } x\in B_\delta(0)\backslash\{0\}.
\]
Then, we see that
\[
-\Delta v\leq\mu\frac{v}{|x|^2}-(m/2)^sv^t\quad \text{for }
 x\in B_\delta(0)\backslash\{0\}.
\]
Let $w_1=(m/2)^\frac{s}{t-1}v$. Then 
\[
-\Delta w_1\leq\mu\frac{w_1}{|x|^2}-w_1^t\quad \text{in }B_\delta(0)\backslash\{0\}.
\]
By Lemma \ref{comparison-1} and an analogous argument described in \cite{WF}, 
it follows that
\[
v(x)\leq C|x|^{-\frac{2}{t-1}}\ \text{for } x\in B_\delta(0)\backslash\{0\}.
\]
From the first equation of \eqref{mp}, we have
\[
-\Delta u\geq \lambda\frac{u}{|x|^2}-C^q|x|^{-\frac{2q}{t-1}}u^p\quad
 \text{in } B_\delta(0)\backslash\{0\}.
\]
Let $w_2=C^{\frac{q}{p-1}}u$. Then 
\[
-\Delta w_2\geq \lambda\frac{w_2}{|x|^2}-|x|^{-\frac{2q}{t-1}}w_2^p\quad
 \text{in } B_\delta(0)\backslash\{0\}.
\]
Hence, by $t-1>q$ and Lemma \ref{comparison-1}, we deduce that
\[
u(x)\geq C_1|x|^{-\frac{2-\frac{2q}{t-1}}{p-1}}
=C_1|x|^{-\frac{2(t-1-q)}{(p-1)(t-1)}}\quad \text{on }  B_\delta(0)\backslash\{0\},
\]
which contradicts the assumption $\lim_{|x|\to0}u(x)=m\in(0,\infty)$.

Similarly, we can prove that either $\lim_{|x|\to0}v(x)=\infty$ 
or $\lim_{|x|\to0}v(x)=0$ holds.
\end{proof}


\begin{proposition} \label{prop3.4}
Suppose that $p,t\geq1$, $q,s>0$ and $\lambda,\mu>(N-2)^2/4$ and 
$(u(x),v(x))$ is an arbitrary positive solution of \eqref{mp}.
Then both $u^{p-1}v^q$ and $u^sv^{t-1}$ are unbounded near the origin. 
In particular, when $p=1$ and $t=1$, both $u$ and $v$ are unbounded near the origin.
\end{proposition}

\begin{proof}
Suppose that $u^{p-1}v^q$ is bounded near the origin. Using the first equation
 in \eqref{mp}, we have
\[
-\Delta u+(u^{p-1}v^q )u=\lambda\frac{u}{|x|^2}\quad \text{in }
 B_1(0)\backslash\{0\}.
\]
Since $u^{p-1}v^q$ is bounded near the origin, there is a sufficiently small 
$\delta_0>0$ such that for any $\delta\in(0,\delta_0]$,
\[
-\Delta\geq \frac{1}{2} \big[\lambda+\frac{(N-2)^2}{4} \big]\frac{u}{|x|^2}\quad
 \text{in } B_{\delta_0}(0).
\]
This implies
\[
\lambda_1\big[\frac{1}{|x|^2},\,B_{\delta_0}(0)\backslash B_\delta(0)\big]
\geq \frac{1}{2}\big[\lambda+\frac{(N-2)^2}{4} \big]\quad \text{for }
 \delta\in(0,\delta_0),
\]
which is a contradiction to 
$$
\lim_{\delta\to0}\lambda_1
\big[\frac{1}{|x|^2},B_{\delta_0}(0)\backslash B_\delta(0)\big]
=\frac{(N-2)^2}{4}.
$$
\end{proof}

\section{Existence and estimates of blow-up solutions}
\label{s4} 

In this section, we deal with the existence of positive blow-up solutions 
of the problem \eqref{mp} with the boundary condition \eqref{eD}
and establish estimates of positive blow-up solutions near the origin.

Suppose that $\lambda>4\lambda_1[B_1(0)]$ and $\alpha\in(0,1)$ satisfies
 $\lambda\frac{\alpha^2}{(1+\alpha)^2}>\lambda_1[B_1(0)]$. 
For the case $\sigma\geq 0$, let $u_{0,\alpha}$ denote the unique positive
 solution of
\begin{gather*}
 -\Delta u=\lambda\frac{\alpha^2}{(1+\alpha)^2}u-\alpha^2(1+\alpha)^\sigma u^p, 
\quad x\in B_1(0) \\
  u(x)=0,\quad  x\in\partial B_1(0).
\end{gather*}
For convenience, for the case of $-2<\sigma<0$, by $u_{0,\alpha}$ we denote 
the unique positive solution of
\begin{gather*}
 -\Delta u=\lambda\frac{\alpha^2}{(1+\alpha)^2}u-\alpha^2(1-\alpha)^\sigma u^p, 
\quad x\in B_1(0) \\
  u(x)=0,\quad  x\in\partial B_1(0).
\end{gather*}
When $\sigma\geq 0$, let $u_{\infty,\alpha}$ denote the unique positive solution of
\begin{gather*}
 -\Delta u=\lambda\frac{\alpha^2}{(1-\alpha)^2}u-\alpha^2(1-\alpha)^\sigma u^p,
\quad  x\in B_1(0) \\
  u(x)=\infty,\quad  x\in\partial B_1(0).
\end{gather*}
For the case of $-2<\sigma<0$, let $u_{\infty,\alpha}$ denote the unique 
positive solution of
\begin{gather*}
 -\Delta u=\lambda \frac{\alpha^2}{(1-\alpha)^2}u-\alpha^2(1+\alpha)^\sigma u^p, 
\quad x\in B_1(0) \\
  u(x)=\infty,\quad  x\in\partial B_1(0).
\end{gather*}
In fact, there is a relation between $u_{0,\alpha},u_{\infty,\alpha}$ and $\alpha$. 
For our convenience, we  denote
$u_{0,\alpha}$ and $u_{\infty,\alpha}$ by $u_0$ and $u_\infty$, respectively. 
In some places, we will write
$u_\infty(x;\lambda)$ and $u_0(x;\lambda)$  instead of $u_\infty$ and $u_0$,
 to clearly indicate the relation depending on
the parameter $\lambda$.

\begin{lemma}\label{lem-in-3}
 Let $p>1$, $\sigma>-2$, $\lambda>4\lambda_1[B_1(0)]$ and $U$
 be a positive solution of
\[
-\Delta U=\lambda\frac{U}{|x|^2}-|x|^\sigma U^p\quad \text{on }
  B_R(0)\backslash\{0\}.
\]
Then we have
\begin{equation} \label{inequality-2}
u_0(0)|x|^{-\frac{2+\sigma}{p-1}}\leq U(x)
\leq u_{\infty}(0)|x|^{-\frac{2+\sigma}{p-1}}\quad \text{for }
 x\in B_{R/2}(0)\backslash\{0\}.
\end{equation}
\end{lemma}

\begin{proof}
Suppose that $x_0\in B_{R/2}(0)\backslash\{0\}$ is an arbitrary point.
Since $\lambda>4\lambda_1[B_1(0)]$, we can choose $\alpha\in(0,1)$ close to $1$
such that $\lambda\frac{\alpha^2}{(1+\alpha)^2}>\lambda_1[B_1(0)]$.
Let
\[
\tilde U(x)=|x_0|^{\frac{2+\sigma}{p-1}}U(x_0+{\alpha|x_0|}x),\quad x\in B_1(0).
\]
So, for all $x\in B_1(0)$, we obtain
\[
(1-\alpha)|x_0|\leq|(x_0+\alpha|x_0|x)|\leq(1+\alpha)|x_0|.
\]
When $\sigma\geq 0$, a straightforward calculation leads to
\begin{gather*}
-\Delta\tilde U\leq\lambda\frac{\alpha^2}{(1-\alpha)^2}
\tilde U-\alpha^2(1-\alpha)^\sigma\tilde U^p\quad \text{in } x\in B_1(0), \\
-\Delta\tilde U\geq\lambda\frac{\alpha^2}{(1+\alpha)^2}
 \tilde U-\alpha^2(1+\alpha)^\sigma\tilde U^p\quad \text{in } x\in B_1(0).
\end{gather*}
We find that if $0>\sigma>-2$, then 
\begin{gather*}
-\Delta\tilde U\leq\lambda\frac{\alpha^2}{(1-\alpha)^2}\tilde U
-\alpha^2(1+\alpha)^\sigma\tilde U^p\quad x\in B_1(0), \\
-\Delta\tilde U\geq\lambda\frac{\alpha^2}{(1+\alpha)^2}\tilde U
-\alpha^2(1-\alpha)^\sigma\tilde U^p\quad x\in B_1(0).
\end{gather*}
By the comparison principle, we obtain
\[
u_0(x)\leq \tilde U(x)\leq u_\infty(x)\quad \text{for all } x\in B_1(0).
\]
Particularly, choosing $x=0$ and by the arbitrariness of $x_0$, 
we arrive at \eqref{inequality-2}.
\end{proof}

\begin{lemma}\label{lem-in-4} 
Suppose that $C>0$, $R>0$ and $\lambda>\max\{4\lambda_1[B_1(0)],\;(N-2)^2/4\}$.
\begin{itemize}
\item[(i)]  If $\mathcal{U}>0$ satisfies
\[
 -\Delta \mathcal{U}\geq \lambda\frac{\mathcal{U}}{|x|^2}
-C|x|^\sigma \mathcal{U}^p,\quad x\in B_R(0)\backslash\{0\},
\]
then 
\begin{equation} \label{in-4}
\mathcal{U}(x)\geq C^{-\frac{1}{p-1}}u_0(0)|x|^{-\frac{2+\sigma}{p-1}}\quad
 \text{for } x\in B_{R/2}(0)\backslash\{0\}.
\end{equation}

\item[(ii)]
If $\mathcal{U}>0$ satisfies
\begin{equation} \label{sub-1}
 -\Delta \mathcal{U}\leq \lambda\frac{\mathcal{U}}{|x|^2}-C|x|^\sigma \mathcal{U}^p,
\quad x\in B_R(0)\backslash\{0\},
\end{equation}
then
\begin{equation} \label{in-5}
\mathcal{U}(x)\leq C^{-\frac{1}{p-1}}u_\infty(0)|x|^{-\frac{2+\sigma}{p-1}}\quad
 \text{for } x\in B_{R/2}(0)\backslash\{0\}.
\end{equation}
\end{itemize}
\end{lemma}

\begin{proof} 
From the condition in  case (i), it follows that $C^\frac{1}{p-1}\mathcal{U}$ 
is a supersolution of
\begin{equation}
\begin{gathered}
-\Delta u=\lambda\frac{u}{|x|^2}-|x|^\sigma u^p,\quad
 x\in B_R(0)\backslash\{0\},\\
 u=0, \quad x\in\partial B_R(0).
\end{gathered} \label{equation-1}
\end{equation}
Let $U$ be the unique positive solution of \eqref{equation-1}, then
$$
U(x)=\lim_{\delta\to0}U_\delta(x),
$$
 where $U_\delta$ is the unique positive solution of
\begin{equation}
\begin{gathered}
-\Delta u=\lambda\frac{u}{|x|^2}-|x|^\sigma u^p,\quad
 x\in B_R(0)\backslash\overline B_\delta(0), \\
 u=0, \quad x\in\partial B_R(0)\cup \partial B_\delta(0).
\end{gathered} \label{equation-7}
\end{equation}
So, the comparison principle yields
\[
U_\delta(x)\leq C^\frac{1}{p-1}\mathcal{U}(x)\quad
 \text{for all } x\in B_R(0)\backslash B_\delta(0).
\]
Further, letting $\delta\to0$, we have
\[
U(x)\leq C^\frac{1}{p-1}\mathcal{U}(x)\quad \text{for all }  x\in B_R(0).
\]
In view of Lemma \ref{lem-in-3}, we arrive at \eqref{in-4}.

Now, we prove conclusion (ii). Suppose that $\mathcal{U}>0$ satisfies \eqref{sub-1}. 
Let $V_\delta$ be the unique positive solution of 
\begin{equation}
\begin{gathered}
-\Delta u=\lambda\frac{u}{|x|^2}-|x|^\sigma u^p,\quad
 x\in B_R(0)\backslash\overline B_\delta(0), \\
 u=\infty,\quad x\in\partial B_R(0)\cup \partial B_\delta(0).
\end{gathered}
\label{equation-2}
\end{equation}
From \cite{WF}, we know that
\[
\hat U(x):=\lim_{\delta\to0}V_\delta(x)\quad
\text{for all $x\in B_R(0)\backslash\{0\}$ is well defined},
\]
and that $\hat U$ satisfies
\begin{gather*}
-\Delta\hat U=\lambda\frac{\hat U}{|x|^2}-|x|^\sigma\hat U^p,\quad
 x\in B_R(0)\backslash\{0\}, \\
 \hat U=\infty,\quad x\in\partial B_R(0).
\end{gather*}
So by the comparison principle, it is easy to see
\[
C^{\frac{1}{p-1}}\mathcal{U}(x)\leq\hat U(x)\quad \text{for }
 x\in B_R(0)\backslash\{0\}.
\]
Using Lemma \ref{lem-in-3} again, we arrive at \eqref{in-5}.
\end{proof}

Next, we give the proof of Theorem \ref{th2}, i.e., we explore the existence 
of blow-up solution of the problem \eqref{mp}
with the boundary condition \eqref{eD}, and give the estimate of positive 
solutions near the origin.

\begin{proof}[Proof of Theorem \ref{th2}]
Since $s<p-1$ and $q<t-1$, we may choose $\gamma,\sigma\in(-2,0)$ such that
\begin{equation}\label{t-s}
-\frac{\gamma}{q}=\frac{2+\sigma}{t-1} \quad \text{and}\quad
 -\frac{\sigma}{s}=\frac{2+\gamma}{p-1}.
\end{equation}
Denote by $U_{p,\gamma}$ the unique positive solution of \eqref{equation-1}
 with $\sigma$ being replaced by $\gamma$ and $R=1$, and denote by $U_{t,\sigma}$ 
the unique positive solution of \eqref{equation-1}
with $p$ being replaced by $t$ and $R=1$.
Choose $\eta\in(s/(t-1),(p-1)/q)$.
A simple calculation shows that for sufficiently small $\epsilon>0$ 
and sufficiently large $M>0$, we have
\begin{gather*}
-\Delta(\epsilon U_{p,\gamma})
\leq\lambda\frac{\epsilon U_{p,\gamma}}{|x|^2}
-(\epsilon U_{p,\gamma})^p(\epsilon^{-\eta}U_{t,\sigma})^q\quad
 \text{in } B_1(0)\backslash\{0\}, \\
-\Delta(\epsilon^{-\eta} U_{t,\sigma})
\geq\lambda\frac{\epsilon^{-\eta} U_{t,\sigma}}{|x|^2}
-(\epsilon U_{p,\gamma})^s(\epsilon^{-\eta}U_{t,\sigma})^t\quad
 \text{in } B_1(0)\backslash\{0\}, \\
-\Delta(MU_{p,\gamma})\geq\lambda\frac{MU_{p,\gamma}}{|x|^2}-(M U_{p,\gamma})^p
(M^{-\eta}U_{t,\sigma})^q\ \text{in}\ B_1(0)\backslash\{0\}, \\
-\Delta(M^{-\eta} U_{t,\sigma})\leq\lambda\frac{M^{-\eta} U_{t,\sigma}}{|x|^2}
-(MU_{p,\gamma})^s(M^{-\eta}U_{t,\sigma})^t\ \text{in}\ B_1(0)\backslash\{0\}.
\end{gather*}
So, $(\epsilon U_{p,\gamma},M^{-\eta}U_{t,\sigma})$ and 
$(MU_{p,\gamma},\epsilon^{-\eta}U_{t,\sigma})$ are a pair of subsolution 
and supersolution of
\begin{equation} \begin{gathered}
-\Delta u=\lambda\frac{ u}{|x|^2}-u^pv^q,\quad
 x\in B_1(0)\backslash\overline B_{1/n}(0), \\
 -\Delta v=\mu\frac{v}{|x|^2}- u^s v^t,\quad
x\in B_1(0)\backslash\overline B_{1/n}(0),\\
 u=v=0, \quad x\in\partial B_1(0),\\
 u=\epsilon U_{p,\gamma},\quad v=M^{-\eta}U_{t,\sigma},\quad
 x\in\partial B_{1/n}(0).
 \end{gathered} \label{system-1}
 \end{equation}
By the supersolution and subsolution method, we know that \eqref{system-1}
 has at least one solution $(U_n(x),V_n(x))$ satisfying
\begin{gather*}
\epsilon U_{p,\gamma}\leq U_n(x)\leq MU_{p,\gamma}(x)\quad
 \text{for } x\in B_1(0)\backslash\overline B_{1/n}(0), \\
 M^{-\eta}U_{t,\sigma}(x)\leq V_n(x)\leq \epsilon^{-\eta}U_{t,\sigma}(x)\quad
 \text{for } x\in B_1(0)\backslash\overline B_{1/n}(0).
\end{gather*}

In view of the regularity arguments of elliptic equations, 
$(U_n,V_n)$ has a convergent subsequence in 
$C^2_{\rm loc}(\overline B_1(0)\backslash\{0\})\times C^2_{\rm loc}(\overline B_1(0)
\backslash\{0\})$.
We by $(u,v)$ denote the limit functions, and hence $(u,v)$ is a solution of 
the problem \eqref{mp}.
Clearly, since both $U_{p,\gamma}$ and $U_{t,\sigma}$ blow up at the origin, 
we can see that both $u$ and $v$ blow up at the origin.
The proof of (i) is complete.

Now, we prove the conclusion (ii).
By hypothesis, both $\hat{u}(x)$ and $\hat{v}(x)$ have positive bound 
from below in $B_{1/2}(0)$.
Denote $m_0=\inf\{\hat{v}(x):x\in B_{1/2}(0)\}$, and hence $\hat u$ satisfies
\[
-\Delta \hat u\leq\lambda\frac{\hat u}{|x|^2}-m_0^q\hat u^p,\quad
 x\in B_{1/2}(0)\backslash\{0\}.
\]
In view of Lemma \ref{lem-in-4}, we have
\[
\hat u(x)\leq m_0^{-\frac{q}{p-1}}u_\infty(0;\lambda)|x|^{-\frac{2}{p-1}}\quad
 \text{for } x\in B_{1/4}(0)\backslash\{0\}.
\]
Let $a_1=m_0^{-\frac{q}{p-1}}u_\infty(0;\lambda)$ and $\alpha_0=\frac{2}{p-1}$. 
Then we find that
\[
\hat u(x)\leq a_1|x|^{-\alpha_0}\quad \text{for } x\in B_{1/4}(0)\backslash\{0\}.
\]
By the second equation in \eqref{mp}, it follows that
\[
-\Delta\hat v\geq\mu\frac{\hat v}{|x|^2}-a_1^s|x|^{-\frac{2s}{p-1}}\hat v^t,\quad
x\in B_{1/4}(0)\backslash\{0\}.
\]
Since $s<p-1$, we have $-\frac{2s}{p-1}>-2$. Take $a_2=a_1^{-\frac{s}{t-1}}$. 
In view of Lemma \ref{lem-in-4} again, we obtain
\[
\hat v(x)\geq a_2|x|^{-\frac{2-s\alpha_0}{t-1}}\quad \text{for }
 x\in B_{2^{-3}}(0)\backslash\{0\}.
\]
Choose $\beta_1=\frac{2-s\alpha_0}{t-1}$, and hence we see that
\[
\hat v(x)\geq a_2|x|^{-\beta_1}\quad \text{for }  x\in B_{2^{-3}}(0)\backslash\{0\}.
\]
Let $a_3=a_2^{-\frac{q}{p-1}}$ and $\alpha_2=\frac{2-q\beta_1}{p-1}$.
Using the first equation in \eqref{mp} and Lemma \ref{lem-in-4} again, we have
\[
\hat u\leq a_3|x|^{-\frac{2-q\beta_1}{p-1}}\quad \text{in }
  B_{2^{-4}}(0)\backslash\{0\}.
\]
Proceeding in this way inductively, we obtain
\begin{gather*}
\hat u(x)\leq a_{2n-1}|x|^{-\frac{2-q\beta_{n-1}}{p-1}},\quad
 x\in B_{2^{-2n}}(0)\backslash\{0\}, \\
\hat v(x)\geq a_{2n}|x|^{-\frac{2-q\alpha_{n-1}}{t-1}},\ \ x\in B_{2^{-(2n+1)}}(0)\backslash\{0\},
\end{gather*}
where $n\geq 1$, and
\begin{gather*}
\alpha_0=\frac{2}{p-1},\quad \beta_0=0, \quad \alpha_n=\frac{2-q\beta_{n}}{p-1},
\quad \beta_n=\frac{2-s\alpha_{n-1}}{t-1}, \\
a_1=m_0^{-\frac{q}{p-1}}u_\infty(0;\lambda),\quad a_{2n}=a_{2n-1}^{-\frac{s}{t-1}},
\quad a_{2n+1}=a_{2n}^{-\frac{q}{p-1}}.
\end{gather*}
From the relation between $\alpha_n$ and $\beta_n$, we can deduce that 
$\{\alpha_n\}$ is a decreasing sequence
and $\{\beta_n\}$ is an increasing sequence. By letting $n\to\infty$, 
it follows that
\[
\lim_{n\to\infty}\alpha_n=\frac{2(t-1-q)}{(p-1)(t-1)-qs} \quad \text{and}\quad
\lim_{n\to\infty}\beta_n=\frac{2(p-1-s)}{(p-1)(t-1)-qs}.
\]
From the relation between $a_n$ and $a_{n+1}$, we can deduce that
\[
a_{2n+1}=a_1^{\frac{q^ns^n}{(t-1)^n(p-1)^n}}.
\]
In view of $(p-1)(t-1)>qs$ and $\min\{a_1,1\}\leq a_{2n+1}\leq \max\{a_1,1\}$ 
for all $n$, we have
\[
\min\{a_1,a_1^{-\frac{s}{t-1}}\}\leq a_n\leq \max\{a_1,a_1^{-\frac{s}{t-1}}\}\quad
 \text{for all } n.
\]
Thus, for any $\tau>0$ there are $\delta:=\delta(\tau)$ and $C_1,\,C_2>0$ such that
\begin{gather*}
C_1|x|^{-\frac{2(p-1-s)}{(p-1)(t-1)-qs}+\tau}\leq \hat v(x)\quad \text{for }
 x\in B_\delta(0)\backslash\{0\}, \\
\hat u(x)\leq C_2|x|^{-\frac{2(t-1-q)}{(p-1)(t-1)-qs}-\tau}\quad \text{for }
 x\in B_\delta(0)\backslash\{0\}.
\end{gather*}
Consequently, the second inequality of \eqref{hat-u-ineq} and the first 
inequality of \eqref{hat-v-ineq} hold.
By a similar argument, we can prove the remaining cases of \eqref{hat-u-ineq} 
and \eqref{hat-v-ineq}.
\end{proof}

\begin{remark} \label{rmk4.3} \rm
From the proof of  (i) in Theorem \ref{th2}, it follows that \eqref{mp} 
has a positive solution $(u,v)$ satisfying
\begin{gather*}
C_1|x|^{-\frac{2+\gamma}{p-1}}\leq u(x)
 \leq C_2|x|^{-\frac{2+\gamma}{p-1}}\quad \text{for } x\in\Omega\backslash\{0\},\\
C_1|x|^{-\frac{2+\sigma}{t-1}}\leq v(x)
\leq C_2|x|^{-\frac{2+\sigma}{t-1}}\quad \text{for } x\in\Omega\backslash\{0\},
\end{gather*}
where $\gamma$ and $\sigma$ satisfies \eqref{t-s}. In fact, a simple calculation 
shows that
\[
\frac{2+\gamma}{p-1}=\frac{2(t-1-q)}{(p-1)(t-1)-qs}\quad \text{and}\quad
\frac{2+\sigma}{t-1}=\frac{2(p-1-s)}{(p-1)(t-1)-qs}.
\]
\end{remark}

\section{Proof of Theorem \ref{th3}} \label{s5} 

\begin{proof}
 Suppose that $(u(x),v(x))$ is a positive radial solution of  \eqref{mp} with 
 condition \eqref{eD}.
For convenience, we denote $u(r)=u(x)$ and $v(r)=v(x)$ as $|x|=r$.
Define
\[
t_0=\inf\{r_0\in(0,1):u'(r)<0\text{ and } v'(r)<0\text{ for } r\in(r_0,1)\}.
\]
By Hopf's lemma, when $r<1$ is close to $1$, we have $u'(r)<0$ and $v'(r)<0$. 
So, this leads to $t_0<1$. We claim $t_0=0$. Otherwise, we will see  that
\begin{itemize}
\item[Case (1)] $ u'(t_0)=0$, $u'(r)<0$ and $v'(r)<0$ for $r\in(t_0,1)$, and

\item[Case (2)] $v'(t_0)=0$, $v'(r)<0$ and $u'(r)<0$ for $r\in(t_0,1)$.
\end{itemize}
Without loss of generality, we assume that Case (1) occurs.
Denote
\[
\tilde u=u(t_0)-u(x)\quad \text{for } t_0<|x|<1.
\]
Then, $\tilde u(x)>0$ when $t_0<|x|<1$ and $\tilde u'(t_0)=\tilde u(t_0)=0$. 
A simple calculation gives
\[
-\Delta\tilde u-\lambda\frac{\tilde u}{|x|^2}
=-\lambda\frac{u(t_0)}{|x|^2}+u^pv^q>0 \quad \text{when }  t_0<|x|<1.
\]
By Hopf's lemma, there holds $\tilde u'(t_0)>0$, which  contradicts 
$\tilde u'(t_0)=0$. The conclusion $t_0=0$ implies that
\[
u'(r)<0\ \text{and}\ v'(r)<0\ \text{for}\ r\in(0,1).
\]

Since $u'(r)<0$ in $(0,1)$ and $\lambda\leq0$, we have
 $u''\geq u^pv^q$ in $(0,1)$, which implies that
$u(r)$ is a convex function in $(0,1)$. Analogously, we may show that $v$ 
is a convex function. So,
the conclusion (i) is proved.

Now we prove the conclusion (ii). From  \eqref{mp}, we see that
\[
-(r^{N-1}u')'=\lambda r^{N-3}u-r^{N-1}u^pv^q\ \text{for}\ r\in(0,r).
\]
Integrating  on $[r,1]$ gives
\[
r^{N-1}u'(r)-u'(1)=\lambda\int_r^1\tau^{N-3}u(\tau)d\tau
-\int_r^1\tau^{N-1}u(\tau)^pv(\tau)^qd\tau\quad \text{ for } r\in(0,1).
\]
That is,
\begin{align*}
u'(r)
&= r^{1-N}u'(1)+\lambda r^{1-N}\int_r^1\tau^{N-3}u(\tau)d\tau\\
&\quad -r^{1-N}\int_r^1\tau^{N-1}u(\tau)^pv(\tau)^qd\tau\quad \text{for } r\in(0,1).
\end{align*}
Integrating  on $[r,1]$ yields
\begin{align*}
u(r)&=-\int_r^1s^{1-N}u'(1)ds-\lambda\int_r^1s^{1-N}
 \int_s^1\tau^{N-3}u(\tau)d\tau ds\\
&\quad +\int_r^1s^{1-N}\int_s^1\tau^{N-1}u(\tau)^pv(\tau)^qd\tau ds\\
&\geq -\int_r^1s^{1-N}u'(1)ds.
\end{align*}
This inequality implies that $u(r)\to\infty$ as $r\to 0$. 
In a similar manner, one can see that $v(r)\to\infty$ as $r\to0$.
\end{proof}

\subsection*{Acknowledgements}
This work is supported by NSF of China, No. 11471148.



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