\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2005(2005), No. 22, pp. 1--9.\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 2005 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2005/22\hfil Decay and symmetry of positive solutions]
{Decay and symmetry of positive solutions of elliptic
 systems in unbounded cylinders}
\author[K.-J. Chen\hfil EJDE-2005/22\hfilneg]
{Kuan-Ju Chen}

\address{Department of Applied Science, Chinese Naval Academy,
P. O. BOX 90175 Zuoying, Taiwan}
\email{kuanju@mail.cna.edu.tw}

\date{}
\thanks{Submitted October 26, 2004. Published February 14, 2005.}
\subjclass[2000]{35J50, 35J55}
\keywords{Symmetry of positive solutions; unbounded cylinders}

\begin{abstract}
 In this paper, we study the asymptotic behavior of positive
 solutions and apply the ``improved moving plane'' method to prove
 the symmetry of positive solutions of semilinear elliptic systems
 in unbounded cylinders.
\end{abstract}

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

\section{Introduction} \label{secIntroduction}

In studying differential equations it is often of interest to know if the
solutions have symmetry, or perhaps monotonicity, in some
direction. Questions of this kind have been investigated by
Gidas-Ni-Nirenberg \cite{g1,g2} by Nirenberg \cite{b1,b2}.
These articles are basically working on semilinear
elliptic equations with Dirichlet or Neumann boundary values.

In our previous paper \cite{c1}, we  established the existence of positive
solutions for a class of semilinear elliptic systems on unbounded domains.
In this paper, we study the  asymptotic behavior of positive
solutions and symmetry of positive solutions of the elliptic
systems of the form
\begin{equation} \label{0}
\begin{gathered}
-\Delta u+u=g(v), \quad u>0\mbox{ in }\mathbf{A}, \\
-\Delta v+v=f(u), \quad v>0\mbox{ in }\mathbf{A}, \\
u=0,\quad v=0\quad \mbox{on }\partial \mathbf{A}, \\
\lim_{|t|\to \infty }u(x,t)=0,\quad \lim_{|t|\to \infty}v(x,t)=0
\quad\mbox{uniformly in }x\in \Omega ,
\end{gathered}
\end{equation}
where $N=m+n\geq 2$, $n\geq 1$, $(x,t)$ is the generic point of
$\mathbb{R}^{N}$ with $x\in \mathbb{R}^{m}$ and $t\in \mathbb{R}^{n}$,
$\Omega  \subset \mathbb{R}^{m}$ is a smooth bounded $C^{1,1}$
domain,  and $\mathbf{A}=\Omega \times\mathbb{R}^{n}$
 is an unbounded cylinder in $\mathbb{R}^{N}$.
Our results form a further development of the work by Figueiredo and Yang
\cite{f1}.

The method of ``moving plane'' was used in \cite{f1}, as originally
introduced by Alexandroff (see Hopf \cite[Chap. 7]{h1}), later used by
Serrin \cite{s1}, and extensively used in recent times, after the
work of \cite{g1} by Gidas-Ni-Nirenberg. Generally speaking, in applying
the ``moving plane'' device it is important to first obtain the
asymptotic behavior of solutions near $\infty $ in order to get
the device started near $\infty $. In this paper, we apply the
``improved moving plane'' method given Li \cite{l1} and by
Li and Ni \cite{l2}, who make no assumption on the asymptotic behavior of
positive solutions to prove that $u(x,t)$, $v(x,t)$ are radially
symmetric in $x$ and axially symmetric in $t$.
In section 2, we  establish the asymptotic behavior of
positive solutions of the systems of semilinear elliptic equations
\eqref{0}.

\section{Asymptotic behavior}

Let $\lambda _{1}$ be the first
eigenvalue and $\phi _{1}$ the corresponding first positive
eigenfunction of the Dirichlet problem $-\Delta \phi
_{1}=\lambda _{1}\phi _{1}$ in $\mathbf{\Omega  }$, $\phi _{1}=0$ on $%
\partial \mathbf{\Omega  }$.

The basic assumptions on the functions $f$ and $g$ are as follows:
\begin{itemize}
\item[(H1)] $f, g\in C^{1}(\mathbb{R},\mathbb{R})$, with $f(t)=g(t)=0$ for
$t\leq 0$, $f(t)>0$ and $g(t)>0$ for $t>0$.

\item[(H2)] $f(t)=O(t^{p})$, $g(t)=O(t^{q})$ as $t\to 0$ for some $1<p$,
$q<\frac{N+2}{N-2}$ ($1<p,q<\infty$ if $N=2$).

\item[(H3)] $f'(t)$, $g'(t)$ are nondecreasing for $t\geq 0$.
\end{itemize}

\begin{proposition}
Assume that $f$, $g$ satisfy (H1)--(H3). Let $u(x,t)$, $v(x,t)$ be $C^{2}$
positive solutions of the elliptic systems $\eqref{0}$. Then for each
$\varepsilon >0$, there exist constants $C_{\varepsilon }$,
$\overline{C_{\varepsilon }}>0$ such that for $(x,t)\in \overline{\mathbf{A}}$,
\[
\overline{C_{\varepsilon }}\phi _{1}(x)e^{-\sqrt{1+\lambda _{1}}\,|t|}|t|^{-%
\frac{n-1}{2}-\varepsilon }\leq u(x,t),v(x,t)\leq C_{\varepsilon
}\phi _{1}(x)e^{-\sqrt{1+\lambda
_{1}}\,|t|}|t|^{-\frac{n-1}{2}+\varepsilon }.
\]
\end{proposition}

\begin{proof}
We divide the proof into the following steps:\\
(1) We claim that for any $0<\delta <1+\lambda _{1}$, there exists
$\alpha _{1} >0$ such that
\[
u(x,t)+v(x,t)\leq \alpha _{1} \phi _{1}(x)e^{-\sqrt{1+\lambda _{1}-\delta }\,|t|},
\quad\mbox{for }(x,t)\in \overline{\mathbf{A}}.
\]
Without loss of generality, we assume $\delta <1$. Since
$\lim_{|t|\to \infty }u(x,t)=0$ and $\lim_{|t|\to \infty }v(x,t)=0$
uniformly in $x\in \mathbf{\Omega  }$, we may choose $R_{0}>0$ large enough
such that
\begin{equation}
\frac{f(u(x,t))}{u(x,t)}\leq \delta,\quad
\frac{g(v(x,t))}{v(x,t)}\leq \delta , \quad \mbox{for }(x,t)\in
\overline{\mathbf{A}},|t|\geq R_{0}.   \label{7}
\end{equation}
Let $(z_{x},z_{t})\in \partial \mathbf{A}$, and $B$ be a small
ball in $\mathbf{A}$ such that
$(z_{x},z_{t})\in \partial B$. Since $\phi _{1}(x)>0$ for
$x\in \mathbf{\Omega  }$, $\phi _{1}(z_{x})=0$, $u(x,t)>0$, $v(x,t)>0$
for $(x,t)\in B$, $u(z_{x},z_{t})=0$, and $v(z_{x},z_{t})=0$, by the
 Hopf boundary point lemma (see Gilbarg and Trudinger \cite{g3}),
 $\frac{\partial \phi _{1}}{\partial x}(z_{x})<0$,
$\frac{\partial u}{\partial \nu }(z_{x},z_{t})<0$, and
$\frac{\partial v}{\partial \nu }(z_{x},z_{t})<0$, where $\nu $ is the outward unit
normal vector at $(z_{x},z_{t})$. Thus
\[
\lim_{(x,t)\to (z_{x}, z_{t})}
\frac{u(x,t)+v(x,t)}{\phi _{1}(x)}=\frac{\frac{\partial u}{\partial \nu }
(z_{x},z_{t})+\frac{\partial v}{\partial \nu }(z_{x},z_{t})}{\frac{\partial
\phi _{1}}{\partial x}(z_{x})}>0\,,
\]
where $(x,t)\in \mathbf{A}$ and $(x,t)\to (z_{x}, z_{t})$ normally.
Note that $(u(x,t)+v(x,t))\phi _{1}^{-1}(x)>0$ for $(x,t)\in
\mathbf{A}$, thus
\[
(u(x,t)+v(x,t))\phi _{1}^{-1}(x)>0\quad \mbox{for }(x,t)\in
\overline{\mathbf{A}}.
\]
Since $\phi _{1}(x)e^{-\sqrt{1+\lambda _{1}-\delta }\,|t|}$,
$u(x,t)$, and $v(x,t)$ are $C^{1}(\overline{\mathbf{A}})$, if we
set
\[
\alpha _{1}=\sup_{(x,t)\in \overline{\mathbf{A}},\, |t|\leq R_{0} }
\big\{ (u(x,t)+v(x,t))\phi _{1}^{-1}(x)e^{\sqrt{1+\lambda _{1}-\delta }%
\,|t|}\big\},
\]
then $\alpha _{1}>0$ and
\begin{equation}
\alpha _{1}\phi _{1}(x)e^{-\sqrt{1+\lambda _{1}-\delta }\,|t|}\geq
u(x,t)+v(x,t), \quad \mbox{for }(x,t)\in
\overline{\mathbf{A}}, |t|\leq R_{0}.    \label{20}
\end{equation}
Let $\Phi _{1}(x,t)=\alpha _{1}\phi _{1}(x)e^{-\sqrt{1+\lambda _{1}-\delta }%
\,|t|}$ for $(x,t)\in \overline{\mathbf{A}}$. Then for $(x,t)\in
\overline{\mathbf{A}}$, $|t|\geq R_{0}$, from (\ref{7}), we have
\begin{align*}
& \Delta (\Phi _{1}-u-v)(x,t)-(\Phi _{1}-u-v)(x,t) \\
&=  [-\delta -\frac{\sqrt{1+\lambda _{1}-\delta }(n-1)}{|t|}]\Phi
_{1}(x,t)+g(v)+f(u) \\
&\leq  -\delta \Phi _{1}+\delta (u+v) \\
&=  -\delta (\Phi _{1}-u-v).
\end{align*}
Hence $\Delta (\Phi _{1}-u-v)(x,t)-(1-\delta )(\Phi
_{1}-u-v)(x,t)\leq 0$ for $(x,t)\in \overline{\mathbf{A}}$,
$|t|\geq R_{0}$. Then by the strong maximum principle, we obtain
\begin{equation}
u(x,t)+v(x,t)\leq \Phi _{1}(x,t), \quad \mbox{for } (x,t)\in
\overline{\mathbf{A}},\; |t|\geq R_{0}.  \label{21}
\end{equation}
Hence from (\ref{20}) and (\ref{21}), we get the claim. \smallskip

\noindent (2) We claim that for any $\varepsilon >0$, there exists
$C_{\varepsilon }>0$ such that
\[
u(x,t)+v(x,t)\leq C_{\varepsilon }\phi _{1}(x)e^{-\sqrt{1+\lambda _{1}}
\,|t|}|t|^{-\frac{n-1}{2}+\varepsilon }, \quad \mbox{for}(x,t)\in
\overline{\mathbf{A}}.
\]
Without loss of generality, we assume that $0<\varepsilon <\frac{n-1}{2}$.
Now, given $\varepsilon >0$ and fixed $\delta >0$ as in part
(1). By part (1), there exists $\alpha _{1}>0$ such that
\[
u(x,t)+v(x,t)\leq \alpha _{1}\phi _{1}(x)e^{-\sqrt{1+\lambda _{1}-\delta }
\,|t|}, \quad \mbox{for }(x,t)\in \overline{\mathbf{A}}.
\]
 From (\ref{7}),
\begin{equation}
f(u)+g(v)\leq \delta \alpha _{1}\phi _{1}(x)e^{-\sqrt{1+\lambda _{1}-\delta }
\,|t|}, \quad \mbox{for }(x,t)\in \overline{\mathbf{A}},\;
|t|\geq R_{0}.   \label{22}
\end{equation}
Let $m_{\varepsilon }=\frac{n-1}{2} -\varepsilon $ and
\[
h(t)=-2\varepsilon \sqrt{1+\lambda _{1}}|t|^{-m_{\varepsilon
}-1}+m_{\varepsilon }(m_{\varepsilon }-n+2)|t|^{-m_{\varepsilon
}-2}.
\]
We can choose $R_{1}>R_{0}$ such that for $|t|\geq R_{1}$,
\begin{equation}
e^{-\sqrt{1+\lambda _{1}}\,|t|}h(t)
+\delta \alpha _{1}e^{-\sqrt{1+\lambda _{1}-\delta }\,|t|}\leq 0.   \label{28}
\end{equation}
As in part (1), if we set
\[
\alpha _{2}=\sup_{(x,t)\in \overline{\mathbf{A}},\, |t|\leq R_{1}}
\big\{ (u(x,t)+v(x,t))\phi _{1}^{-1}(x)e^{\sqrt{1+\lambda _{1}}
\,|t|}|t|^{m_{\varepsilon }} \big\} +1,
\]
then $\alpha _{2}>1$. Let $\Phi _{2}(x,t)=\alpha _{2}\phi _{1}(x)e^{-\sqrt{1+\lambda _{1}}%
\,|t|}|t|^{-m_{\varepsilon }}$ for $(x,t)\in
\overline{\mathbf{A}}$, then
\begin{equation}
u(x,t)+v(x,t)\leq \Phi _{2}(x,t), \quad\mbox{for } (x,t)\in
\overline{\mathbf{A}},\; |t|\leq R_{1}.  \label{25}
\end{equation}
For $(x,t)\in \overline{\mathbf{A}}$, $|t|\geq R_{1}$, from
(\ref{22}) and (\ref{28}), we have
\begin{align*}
& \Delta (\Phi _{2}-u-v)(x,t)-(\Phi _{2}-u-v)(x,t) \\
&= h(t)|t|^{m_{\varepsilon }}\Phi _{2}(x,t)+g(v)+f(u) \\
&\leq  \alpha _{2}\phi _{1}(x)e^{-\sqrt{1+\lambda _{1}}\,|t|}h(t)+\delta
\alpha _{1}\phi _{1}(x)e^{-\sqrt{1+\lambda _{1}-\delta }\,|t|} \\
&\leq \phi _{1}(x)(e^{-\sqrt{1+\lambda _{1}}\,|t|}h(t)+\delta
\alpha _{1}e^{-\sqrt{1+\lambda _{1}-\delta }\,|t|})
\leq  0\,.
\end{align*}
Then by the strong maximum principle,
\begin{equation}
u(x,t)+v(x,t)\leq \Phi _{2}(x,t), \quad \mbox{for } (x,t)\in
\overline{\mathbf{A}},\; |t|\geq R_{1}.  \label{26}
\end{equation}
Hence from (\ref{25}) and (\ref{26}), and since $u(x,t)$, $v(x,t)$
are positive solutions, it is straightforward that for any $%
\varepsilon >0$, there exists $C_{\varepsilon }>0$ such that
\[
u(x,t),v(x,t)\leq C_{\varepsilon }\phi _{1}(x)e^{-\sqrt{1+\lambda _{1}}%
\,|t|}|t|^{-\frac{n-1}{2}+\varepsilon }, \;\;\mbox{for
}(x,t)\in \overline{\mathbf{A}}.
\] \smallskip

\noindent(3) For a given $\varepsilon >0$, let
$\overline{m_{\varepsilon }}=\frac{n-1}{2}+\varepsilon $ and
\[
k(t)=2\varepsilon \sqrt{1+\lambda _{1}}|t|^{-1}+\overline{m_{\varepsilon }}
(\overline{m_{\varepsilon }}-n+2)|t|^{-2}.
\]
We can choose $R_{2}>0$ such that $k(t)\geq 0$ for $|t|\geq R_{2}$. As in
part (1), if we set
\[
\beta =\inf_{(x,t)\in \overline{\mathbf{A}},\, |t|\leq R_{2}}
\big\{u(x,t)\phi _{1}^{-1}(x)e^{\sqrt{1+\lambda _{1}}\,|t|}
|t|^{\overline{m_{\varepsilon }}}\big\} ,
\]
then $\beta >0$ and
\begin{equation}
\beta \phi _{1}(x)e^{-\sqrt{1+\lambda _{1}}\,|t|}
|t|^{-\overline{m_{\varepsilon }}}\leq u(x,t), \quad\mbox{for }(x,t)\in
\overline{\mathbf{A}}, \;|t|\leq R_{2}.  \label{23}
\end{equation}
Let $\Psi (x,t)=\beta \phi _{1}(x)e^{-\sqrt{1+\lambda _{1}}\,|t|}|t|^{-%
\overline{m_{\varepsilon }}}$ for $(x,t)\in
\overline{\mathbf{A}}$. Then for $(x,t)\in \overline{\mathbf{A}}$,
$|t|\geq R_{2}$, we have
\[
 \Delta (\Psi -u)(x,t)-(\Psi -u)(x,t)
=  k(t)\Psi (x,t)+g(v) \geq  0.
\]
Then by the strong maximum principle,
\begin{equation}
u(x,t)\geq \Psi (x,t), \quad \mbox{for } (x,t)\in
\overline{\mathbf{A}},\; |t|\geq R_{2}.   \label{24}
\end{equation}
Hence from (\ref{23}) and (\ref{24}), for each $\varepsilon >0$, there exists
$\overline{C_{\varepsilon }}>0$ such that
\[
u(x,t)\geq \overline{C_{\varepsilon }}\phi
_{1}(x)e^{-\sqrt{1+\lambda
_{1}}\,|t|}|t|^{-\frac{n-1}{2}-\varepsilon }, \quad \mbox{for }
(x,t)\in \overline{\mathbf{A}}.
\]
For the positive solution $v(x,t)$ in the elliptic systems
\eqref{0}, we can get the same results as for $u(x,t)$.
\end{proof}


\section{Symmetry of Positive Solutions}
Let
$$
\mathbf{S}=\{(x,t)\in
 B^{N-1}(0;R)\times \mathbb{R}:x=(x_{1},\dots ,x_{N-1})\in
 B^{N-1}(0;R), t\in \mathbb{R}\},
$$
where $B^{N-1}(0;R)$ is a ball  with center at the origin and of radius $R$
in $\mathbb{R}^{N-1}$.
Now we consider the systems of semilinear elliptic equations
\begin{equation} \label{1}
\begin{gathered}
-\Delta u+u=g(v), \quad u>0 \mbox{ in }\mathbf{S}, \\
-\Delta v+v=f(u), \quad v>0 \mbox{ in }\mathbf{S}, \\
u=0,\quad v=0\quad \mbox{on }\partial \mathbf{S}, \\
\lim_{|t|\to \infty }u(x,t)=0,\quad
\lim_{|t|\to \infty }v(x,t)=0\quad \mbox{uniformly in } x\in B^{N-1}(0;R).
\end{gathered}
\end{equation}
The purpose of this section is to apply the ``improved moving
plane'' method to prove the symmetry of positive solutions of the
elliptic systems \eqref{1} which makes no assumption on the
asymptotic behavior of positive solution.

\begin{theorem}\label{T1.9}
Assume that $f$, $g$ satisfy (H1)--(H3). Let $u(x,t)$, $v(x,t)$ be $C^{2}$
positive solutions of the elliptic
systems \eqref{1}. Then $u(x,t)$, $v(x,t)$ are radially
symmetric in $x$ and axially symmetric in $t$; that is to say,
$u(x,t-\sigma
)=u(|x|,|t-\sigma |)$, $v(x,t-\sigma )=v(|x|,|t-\sigma |)$ for some $\sigma $.
\end{theorem}

\subsection*{Part I}
  $u(x,t)$, $v(x,t)$ are radially symmetric in $x\in B^{N-1}(0;R)$.\\
\textbf{Notation:}
\begin{itemize}
\item $T_{\lambda }=\{(x,t)=(x_{1},x_{2},\dots ,x_{N-1},t)\in
\mathbf{S}:x_{1}=\lambda \}$

\item $\Sigma _{\lambda }=\{(x,t)\in \mathbf{S}:x_{1}<\lambda \}$

\item For  $(x,t)=(x_{1},x_{2},\dots ,x_{N-1},t)\in \mathbf{S}$, set
$(x^{\lambda },t)=(2\lambda -x_{1},x_{2},\dots ,x_{N-1},t)$; that
is to say, $(x^{\lambda },t)$ is the reflection of $(x,t)$ with respect to
$T_{\lambda }$

\item Let $\Lambda $ be the collection of all $\lambda \in (-R,0)$ such that
\begin{gather*}
u(x,t)<u(x^{\lambda },t),\quad v(x,t)<v(x^{\lambda },t)\quad
\mbox{for all $(x,t)\in \Sigma _{\lambda }$}, \\
u_{x_{1}}(x,t)>0,\quad v_{x_{1}}(x,t)>0\quad \mbox{on }\mathbf{S}\cap T_{\lambda }.
\end{gather*}
\end{itemize}
In the sequel, we take $\lambda \leq 0$. On the set $\overline{\Sigma_{\lambda }}$
we define the functions
\[
U^{\lambda }(x,t)=u(x,t)-u(x^{\lambda
},t)\quad \mbox{and}\quad V^{\lambda }(x,t)=v(x,t)-v(x^{\lambda },t).
\]
Then
\begin{gather*}
-\Delta U^{\lambda }(x,t)+U^{\lambda }(x,t)=g(v(x,t))-g(v(x^{\lambda },t)), \\
-\Delta V^{\lambda }(x,t)+V^{\lambda }(x,t)=f(u(x,t))-f(u(x^{\lambda },t)).
\end{gather*}
By the mean value theorem,
\begin{gather*}
g(v(x,t))-g(v(x^{\lambda },t))=g'(\psi _{\lambda }(x,t))V^{\lambda
}(x,t), \\
f(u(x,t))-f(u(x^{\lambda },t))=f'(\varphi _{\lambda
}(x,t))U^{\lambda }(x,t),
\end{gather*}
where $\psi _{\lambda }(x,t)$ is a real number between $v(x,t)$ and $%
v(x^{\lambda },t)$, $\varphi _{\lambda }(x,t)$ is some value
between $u(x,t)$ and $u(x^{\lambda },t)$, respectively. Let us
denote $g'(\psi _{\lambda }(x,t))=c_{\lambda }(x,t)$ and
$f'(\varphi _{\lambda }(x,t))=d_{\lambda }(x,t)$. So
\begin{equation} \label{2}
\begin{gathered}
-\Delta U^{\lambda }(x,t)+U^{\lambda }(x,t)=c_{\lambda }(x,t)V^{\lambda
}(x,t)\quad \mbox{in }\Sigma _{\lambda }, \\
-\Delta V^{\lambda }(x,t)+V^{\lambda }(x,t)=d_{\lambda
}(x,t)U^{\lambda }(x,t)\quad \mbox{in }\Sigma _{\lambda }.
\end{gathered}
\end{equation}
To prove Part I, we need the following lemmas.

\begin{lemma}\label{L1.11}
For some $0<\delta <R$, $(-R,-R+\delta )\subset \Lambda$.
\end{lemma}

\begin{proof}
Note that by (H2), $\lim_{t\to 0^{+}}g'(t)=0$ and $\lim_{t\to 0^{+}}f'(t)=0$.
Take $t_{0}>0$ such that if $0<t\leq t_{0}$, then $g'(t)<1$ and $f'(t)<1$.
Since $\lim_{|x|\to R}u(x,t)=0$ and $\lim_{|x|\to R}v(x,t)=0$
uniformly in $t$, we can choose $\delta $, $R>\delta >0$ such that if
$R-\delta <|x|<R$, $u(x,t)\leq t_{0}$ and $v(x,t)\leq t_{0}$ uniformly in $t$.

We claim that if $-R<\lambda <-R+\delta $, then $U^{\lambda }(x,t)\leq 0$ and
$V^{\lambda }(x,t)\leq 0$ in $\Sigma _{\lambda }$.
On the contrary, suppose there exists $\lambda $ such that
$-R<\lambda <-R+\delta $, $U^{\lambda }(x,t)>0$ for some
$(x,t)\in \Sigma _{\lambda }$, or $V^{\lambda }(x,t)>0$ for some
$(x,t)\in \Sigma _{\lambda }$. Since $\lim_{| t| \to \infty }U^{\lambda }(x,t)=0$
and $\lim_{| t| \to \infty }V^{\lambda }(x,t)=0$ uniformly in
$x\in B^{N-1}(0;R)$, $U^{\lambda }(x,t)$ achieves its maximum at
$(x_{0},t_{0})\in \Sigma _{\lambda }$ and $V^{\lambda }(x,t)$ achieves its
maximum at $(x_{1},t_{1})\in \Sigma _{\lambda }$. Then
\begin{gather*}
\nabla U^{\lambda }(x_{0},t_{0})=0,\quad \{U_{ij}^{\lambda}(x_{0},t_{0})\}\leq 0, \\
\nabla V^{\lambda }(x_{1},t_{1})=0,\quad \{V_{ij}^{\lambda}(x_{1},t_{1})\}\leq 0.
\end{gather*}
Since $\Delta U^{\lambda }(x_{0},t_{0})\leq 0$ and
$\Delta V^{\lambda }(x_{1},t_{1})\leq 0,$ from the elliptic systems
\eqref{2} it follows that
\begin{gather*}
c_{\lambda }(x_{0},t_{0})V^{\lambda }(x_{0},t_{0})\geq U^{\lambda
}(x_{0},t_{0})>0, \\
d_{\lambda }(x_{1},t_{1})U^{\lambda }(x_{1},t_{1})\geq V^{\lambda
}(x_{1},t_{1})>0.
\end{gather*}
 From (H1)--(H3) it follows that $c_{\lambda }(x_{0},t_{0})\geq 0$ and
$d_{\lambda }(x_{1},t_{1})\geq 0$. Then we obtain that
$V^{\lambda }(x_{0},t_{0})>0$ and $U^{\lambda }(x_{1},t_{1})>0$.
Moreover, if $-R<\lambda <-R+\delta $, since $V^{\lambda }(x_{0},t_{0})>0$,
$U^{\lambda }(x_{1},t_{1})>0$, then from (H3),
$c_{\lambda }(x_{0},t_{0})<1$, $d_{\lambda }(x_{1},t_{1})<1$, and get
\begin{equation} \label{3}
\begin{gathered}
V^{\lambda }(x_{0},t_{0})>c_{\lambda }(x_{0},t_{0})V^{\lambda
}(x_{0},t_{0})\geq U^{\lambda }(x_{0},t_{0}), \\
U^{\lambda }(x_{1},t_{1})>d_{\lambda }(x_{1},t_{1})U^{\lambda
}(x_{1},t_{1})\geq V^{\lambda }(x_{1},t_{1}).
\end{gathered}
\end{equation}
Since $V^{\lambda }(x_{1},t_{1})\geq V^{\lambda }(x_{0},t_{0})$, from
\eqref{3} it follows that $U^{\lambda }(x_{1},t_{1})>U^{\lambda}(x_{0},t_{0})$,
we come to a contradiction. So for $-R<\lambda <-R+\delta $,
$U^{\lambda }(x,t)\leq 0$ and $V^{\lambda }(x,t)\leq 0$ in $\Sigma _{\lambda }$.
Applying the maximum principle and the Hopf boundary point lemma, for
$-R<\lambda <-R+\delta $, we get $U^{\lambda }(x,t)<0$ in $\Sigma _{\lambda }$,
$U_{x_{1}}^{\lambda }(x,t)>0$ for $(x,t)\in \mathbf{S}\cap T_{\lambda }$,
$V^{\lambda }(x,t)<0$ in $\Sigma _{\lambda }$, and
$V_{x_{1}}^{\lambda}(x,t)>0$ for $(x,t)\in \mathbf{S}\cap T_{\lambda }$.
Hence $u_{x_{1}}(x,t)>0 $ and $v_{x_{1}}(x,t)>0$ for $(x,t)\in \mathbf{S}\cap
T_{\lambda }$. Then $(-R,-R+\delta )\subset \Lambda $.
\end{proof}

\begin{lemma} \label{L1.12}
If $(-R,\lambda ]\subset \Lambda $, then there exists $\tau >0$
such that $[\lambda ,\lambda +\tau )\subset \Lambda $.
\end{lemma}

\begin{proof}
Suppose not. Then there exist a decreasing sequence $\lambda_{k}\to \lambda $
and a sequence $\{(x_{k0},t_{k0})\}$ of
points in $\Sigma _{\lambda _{k}}$ such that $U^{\lambda
_{k}}(x_{k0},t_{k0})=u(x_{k0},t_{k0})-u(x_{k0}^{\lambda
_{k}},t_{k0})>0$, or a sequence $\{(x_{k1},t_{k1})\}$ of points in
$\Sigma _{\lambda _{k}}$ such that $V^{\lambda
_{k}}(x_{k1},t_{k1})=v(x_{k1},t_{k1})-v(x_{k1}^{\lambda
_{k}},t_{k1})>0$. There exists a subsequence $\{(x_{k0},t_{k0})\}$
such that $x_{k0}\to \overline{x}_{0}\in
\overline{B^{N-1}(0;R)}$ or a subsequence $\{(x_{k1},t_{k1})\}$ such that
$x_{k1}\to \overline{x}_{1}\in \overline{B^{N-1}(0;R)}$. There may arise
two possibilities: Cases 1 and 2 below.

\noindent Case 1. $|t_{k0}|\to \infty $. As shown in Lemma \ref{L1.11}, we
assume that
\begin{gather*}
U^{\lambda _{k}}(x_{k0},t_{k0})=\max_{(x,t)\in
\overline{\Sigma} _{\lambda _{k}}}U^{\lambda _{k}}(x,t),\\
\nabla U^{\lambda_{k}}(x_{k0},t_{k0})=0, \;\;\;\{U_{ij}^{\lambda
_{k}}(x_{k0},t_{k0})\}\leq 0.
\end{gather*}
 From $\lim_{|t_{k0}|\to \infty } u(x_{k0},t_{k0})=0$, as in Lemma
\ref{L1.11}, we obtain a contradiction. The same argument applies to
$V^{\lambda _{k}}(x_{k1},t_{k1})$.

\noindent Case 2. $t_{k0}\to \bar{t}_{0}$. We have $(x_{k0},t_{k0})\to
(\overline{x}_{0},\bar{t}_{0})\in \overline{\Sigma _{\lambda }}$. Thus
$U^{\lambda }(\overline{x}_{0},\bar{t}_{0})\geq 0$. Clearly
$(\overline{x}_{0},\bar{t}_{0})\notin \Sigma _{\lambda }$ since
$U^{\lambda }(x,t)<0$ in $\Sigma _{\lambda }$.
If $(\overline{x}_{0},\bar{t}_{0})\in T_{\lambda }$,
then $u_{x_{1}}(\overline{x}_{0},\bar{t}_{0})<0$, which contradicts to
$\lambda \in \Lambda $. Moreover,
$(\bar{x}_{0},\bar{t}_{0})\notin \partial
\mathbf{S}\cap \overline{\Sigma _{\lambda }}$ since if
 $(\bar{x}_{0},\bar{t}_{0})\in \partial \mathbf{S}\cap \overline{\Sigma _{\lambda }}$
then $0=u(\bar{x}_{0},\bar{t}_{0})\geq u(\bar{x}_{0}^{\lambda },\bar{t}_{0})>0$, a
contraction. We conclude that Case 2 is impossible. The same argument
applies to $V^{\lambda _{k}}(x_{k1},t_{k1})$.
\end{proof}

\begin{proof}[Proof of Part I]
Let $\mu =\sup \{\lambda \in (-R,0):(-R,\lambda )\subset \Lambda \}$.
Then $\mu \notin \Lambda $. If not, by Lemma \ref{L1.12} we would
have $[\mu ,\mu +\tau )\subset \Lambda $, which contradicts to the
definition of $\mu $.
We claim that $\mu =0$. Suppose not, $\mu \in (-R,0)$. By
continuity we have $u(x,t)\leq u(x^{\mu },t)$ and $v(x,t)\leq
v(x^{\mu },t)$ for all $(x,t)\in \Sigma _{\mu }$, then
by the maximum principle we have $u(x,t)\equiv u(x^{\mu },t)$ and
$v(x,t)\equiv v(x^{\mu },t)$ for all $(x,t)\in \Sigma _{\mu }$, which is
impossible. Thus $\mu =0$. By reversing the $x_{1}$ axis, we conclude that
$u(x,t)$ and $v(x,t)$ are symmetric with respect to the hyperplane $T_{0}$,
$u_{x_{1}}(x,t)<0$ and $v_{x_{1}}(x,t)<0$ for $x_{1}>0$. Since the $x_{1}$
direction can be chosen arbitrarily, we conclude that $u(x,t)$ and $v(x,t)$
are radially symmetric in $x\in B^{N-1}(0;R)$.
\end{proof}

\subsection*{Part II}  $u(x,t)$, $v(x,t)$ are axially symmetric with respect
to some hyperplane $t=\sigma $.
\\
\textbf{Notation:}
\begin{itemize}
\item $S_{\theta }=\{(x,t)\in \mathbf{S}:x\in B^{N-1}(0;R), t=\theta \}$
\item $\Gamma _{\theta }=\{(x,t)\in \mathbf{S}:x\in B^{N-1}(0;R),$ $t<\theta \}$
\item For any $(x,t)\in \mathbf{S}$, set
$(x,t^{\theta})=(x,2\theta -t)$
 that is to say, $(x,t^{\theta })$ is the reflection of $ (x,t)$ with respect
to $S_{\theta }$
\item  Let $\Theta $ be the collection of all $\theta \in \mathbb{R}$ such that
\begin{gather*}
u(x,t)<u(x,t^{\theta }),\quad v(x,t)<v(x,t^{\theta })
\quad \mbox{for all } (x,t)\in \Gamma _{\theta }, \\
u_{t}(x,t)>0,\quad v_{t}(x,t)>0\quad \mbox{on }\mathbf{S}\cap S_{\theta }.
\end{gather*}
\end{itemize}
On $\overline{\Gamma _{\theta }}$, we define the functions
\[
M^{\theta }(x,t)=u(x,t)-u(x,t^{\theta})\quad
\mbox{and}\quad N^{\theta }(x,t)=v(x,t)-v(x,t^{\theta }).
\]
Then
\begin{gather*}
-\Delta M^{\theta }(x,t)+M^{\theta }(x,t)=g(v(x,t))-g(v(x,t^{\theta })),\\
-\Delta N^{\theta }(x,t)+N^{\theta }(x,t)=f(u(x,t))-f(u(x,t^{\theta })).
\end{gather*}
By the mean value theorem,
\begin{gather*}
g(v(x,t))-g(v(x,t^{\theta }))=g'(\xi _{\theta }(x,t))N^{\theta }(x,t), \\
f(u(x,t))-f(u(x,t^{\theta }))=f'(\zeta _{\theta }(x,t))M^{\theta
}(x,t),
\end{gather*}
where $\xi _{\theta }(x,t)$ is a real number between $v(x,t)$ and
$v(x,t^{\theta })$, $\zeta _{\theta }(x,t)$ is some value between
$u(x,t)$ and $u(x,t^{\theta })$, respectively. Let us denote
$g'(\xi _{\theta }(x,t))=e_{\theta }(x,t)$ and
$f'(\zeta _{\theta }(x,t))=f_{\theta }(x,t)$. So
\begin{equation} \label{4}
\begin{gathered}
-\Delta M^{\theta }(x,t)+M^{\theta }(x,t)=e_{\theta }(x,t)N^{\theta
}(x,t)\quad \mbox{in }\Gamma _{\theta }, \\
-\Delta N^{\theta }(x,t)+N^{\theta }(x,t)=f_{\theta
}(x,t)M^{\theta }(x,t)\quad \mbox{in }\Gamma _{\theta }.
\end{gathered}
\end{equation}
To prove Part II, we  need the following lemmas.

\begin{lemma}\label{L1.9a}
There exists $\theta _{0}>0$, such that either
$(-\infty,-\theta _{0}]\subset \Theta $ or $u(x,t)\equiv u(x,t^{-\theta _{0}})$
and $v(x,t)\equiv v(x,t^{-\theta _{0}})$ in $\Gamma _{-\theta _{0}}$.
\end{lemma}

\begin{proof}
Note that by (H2), $\lim_{t\to 0^{+}}g'(t)=0$ and
$\lim_{t\to 0^{+}}f'(t)=0$. Take $t_{0}>0$ such that if
$0<t\leq t_{0}$, then $g'(t)<1$ and $f'(t)<1$. Since
$\lim_{|t|\to \infty }u(x,t)=0$ and
$\lim_{|t|\to \infty}v(x,t)=0$ uniformly in $x\in B^{N-1}(0;R)$,
we can choose $\theta _{0}>0$
such that if $t\leq -\theta _{0}$, $u(x,t)\leq t_{0}$ and $v(x,t)\leq t_{0}$
uniformly in $x\in B^{N-1}(0;R)$.

We claim that if $\theta \leq -\theta _{0}$, then $M^{\theta }(x,t)\leq 0$ and
$N^{\theta }(x,t)\leq 0$ in $\Gamma _{\theta }$.

On the contrary, suppose that there exists $\theta $ such that
$\theta \leq -\theta _{0} $, $M^{\theta }(x,t)>0$ for some
$(x,t)\in \Gamma _{\theta }$, or $N^{\theta }(x,t)>0$ for some
$(x,t)\in \Gamma _{\theta }$. Since $\lim_{t\to -\infty }M^{\theta }(x,t)=0$
and $\lim_{t\to -\infty }N^{\theta }(x,t)=0$ uniformly in $x\in B^{N-1}(0;R)$,
$M^{\theta}(x,t)$ achieves its maximum at $(x_{2},t_{2})\in \Gamma _{\theta }$
and $N^{\theta }(x,t)$ achieves its maximum at
 $(x_{3},t_{3})\in \Gamma _{\theta} $. Then
\begin{gather*}
\nabla M^{\theta}(x_{2},t_{2})=0, \quad
\{M_{ij}^{\theta}(x_{2},t_{2})\}\leq 0, \\
\nabla N^{\theta}(x_{3},t_{3})=0, \quad
\{N_{ij}^{\theta}(x_{3},t_{3})\}\leq 0.
\end{gather*}
Since $\Delta M^{\theta }(x_{2},t_{2})\leq 0$ and
$\Delta N^{\theta}(x_{3},t_{3})\leq 0$, from  elliptic systems
\eqref{4} it follows that
\begin{gather*}
e_{\theta }(x_{2},t_{2})N^{\theta }(x_{2},t_{2})\geq M^{\theta
}(x_{2},t_{2})>0, \\
f_{\theta }(x_{3},t_{3})M^{\theta }(x_{3},t_{3})\geq N^{\theta
}(x_{3},t_{3})>0.
\end{gather*}
 From (H1)--(H3) it follows that $e_{\theta }(x_{2},t_{2})\geq 0$ and
$f_{\theta}(x_{3},t_{3})\geq 0$. Then we obtain that
$N^{\theta }(x_{2},t_{2})>0$ and $M^{\theta }(x_{3},t_{3})>0$.
Moreover, if $\theta \leq -\theta _{0}$, since
$N^{\theta }(x_{2},t_{2})>0$, $M^{\theta }(x_{3},t_{3})>0$, then from (H3),
$e_{\theta }(x_{2},t_{2})<1$, $f_{\theta }(x_{3},t_{3})<1$, and get
\begin{equation} \label{5}
\begin{gathered}
N^{\theta }(x_{2},t_{2})>e_{\theta }(x_{2},t_{2})N^{\theta
}(x_{2},t_{2})\geq M^{\theta }(x_{2},t_{2}), \\
M^{\theta }(x_{3},t_{3})>f_{\theta }(x_{3},t_{3})M^{\theta
}(x_{3},t_{3})\geq N^{\theta }(x_{3},t_{3}).
\end{gathered}
\end{equation}
Since $N^{\theta }(x_{3},t_{3})\geq N^{\theta }(x_{2},t_{2})$, from
\eqref{5} it follows that $M^{\theta }(x_{3},t_{3})>M^{\theta }(x_{2},t_{2})$.
We come to a contradiction. So for $\theta \leq -\theta _{0}$,
$M^{\theta }(x,t)\leq 0$ and $N^{\theta }(x,t)\leq 0$ in $\Gamma
_{\theta }$. As a consequence of the maximum principle and the
Hopf boundary point lemma,
either $M^{-\theta _{0}}(x,t)\equiv 0,$ $N^{-\theta _{0}}(x,t)\equiv 0$ in
$\Gamma _{-\theta _{0}}$ or for $\theta \leq -\theta _{0}$,
$M^{\theta }(x,t)<0$ in $\Gamma _{\theta }$, $M_{t}^{\theta}(x,t)>0$
for $(x,t)\in \mathbf{S}\cap S_{\theta }$, $N^{\theta}(x,t)<0$
in $\Gamma _{\theta }$, and $N_{t}^{\theta }(x,t)>0$ for
$(x,t)\in \mathbf{S}\cap S_{\theta }$. Hence either
$(-\infty,-\theta _{0}]\subset \Theta $ or
$u(x,t)\equiv u(x,t^{-\theta_{0}})$ and $v(x,t)\equiv v(x,t^{-\theta _{0}})$
in $\Gamma_{-\theta _{0}}$.
\end{proof}

\begin{lemma} \label{L1.10}
If $(-\infty ,\theta ]\subset \Theta $, then there exists $\varepsilon >0$
such that $[\theta ,\theta +\epsilon )\subset \Theta $.
\end{lemma}

\begin{proof}
Suppose not. Then there exist a decreasing sequence
$\theta _{k}\to \theta $ and a sequence $\{(x_{k2},t_{k2})\}$ of points
in $\Gamma _{\theta_{k}}$ such that
$M^{\theta_{k}}(x_{k2},t_{k2})=u(x_{k2},t_{k2})-u(x_{k2},
t_{k2}^{\theta _{k}})>0$, or
a sequence $\{(x_{k3},t_{k3})\}$ of points in $\Gamma _{\theta _{k}}$ such
that
$N^{\theta_{k}}(x_{k3},t_{k3})=v(x_{k3},t_{k3})-v(x_{k3},t_{k3}^{\theta _{k}})>0$.
There exists a subsequence $\{(x_{k2},t_{k2})\}$ such that $%
x_{k2}\to \bar{x}_{2}\in \overline{B^{N-1}(0;R)}$ or a
subsequence $\{(x_{k3},t_{k3})\}$ such that $x_{k3}\to
\overline{x}_{3}\in \overline{B^{N-1}(0;R)}$. There may arise two
possibilities: Cases 1 and 2 below.
\\
Case 1. $t_{k2}\to -\infty $. As shown in Lemma \ref{L1.9a}, we
assume
\begin{equation}  \label{6}
\begin{gathered}
M^{\theta _{k}}(x_{k2},t_{k2})=\max_{(x,t)\in
 \overline{\Gamma}_{\theta _{k}}} M^{\theta _{k}}(x,t),\\
\nabla M^{\theta _{k}}(x_{k2},t_{k2})=0,\quad
\{M_{ij}^{\theta _{k}}(x_{k2},t_{k2})\}\leq 0.
\end{gathered}
\end{equation}
 From $\lim_{t_{k2}\to -\infty }u(x_{k2},t_{k2})=0$, as in Lemma
 \ref{L1.9a}, we obtain a contradiction. The same argument applies to
 $N^{\theta_{k}}(x_{k3},t_{k3})$.

\noindent Case 2. $t_{k2}\to \bar{t}_{2}$. We have
$(x_{k2},t_{k2})\to (\bar{x}_{2},\bar{t}_{2})\in \overline{\Gamma
_{\theta }}$. Thus $M^{\theta }(\bar{x}_{2},\bar{t}_{2})\geq 0$. Clearly
$(\bar{x}_{2},\bar{t}_{2})\notin \Gamma _{\theta }$ since
$M^{\theta }(x,t)<0$ in $\Gamma _{\theta }$.
If $(\bar{x}_{2},\bar{t}_{2})\in S_{\theta }$, then
$u_{t}(\bar{x}_{2},\bar{t}_{2})<0$, which contradicts
$\theta \in \Theta$. Moreover,
$(\bar{x}_{2},\bar{t}_{2})\notin
\partial {\bf S}\cap \overline{\Gamma _{\theta }}$.
Note that $M^{\theta }(x,t)$ satisfies the elliptic
systems \eqref{4}, and by the Hopf boundary point lemma, we obtain
$\frac{\partial }{\partial \nu }M^{\theta }(\bar{x}_{2},\bar{t}_{2})<0$. On
the other hand, taking the limit in $($\ref{6}$)$, we obtain $\nabla
M^{\theta }(\bar{x}_{2},\bar{t}_{2})=0$, a contradiction. We conclude that
Case 2 is impossible. The same argument applies to
$N^{\theta_{k}}(x_{k3},t_{k3})$.
\end{proof}

\begin{proof}[Proof of Part II]
Let $\sigma =\sup \{\theta \in \mathbb{R}:(-\infty ,\theta )\subset \Theta \}$.
Then $\sigma \notin \Theta $. If not, by Lemma \ref{L1.10} we would have
$[\sigma,\sigma +\epsilon )\subset \Theta $, which contradicts to the
definition of $\sigma $. By continuity we have
$u(x,t)\leq u(x,t^{\sigma })$ and $v(x,t)\leq v(x,t^{\sigma })$ for all
$(x,t)\in \Gamma _{\sigma }$, then by the maximum principle we
have $u(x,t)\equiv u(x,t^{\sigma })$ and
$v(x,t)\equiv v(x,t^{\sigma })$ for all $(x,t)\in \Gamma _{\sigma }$. This
proves $u(x,t)$ and $v(x,t)$ are symmetric with respect to the
hyperplane $t=\sigma $ for all $(x,t)\in \mathbf{S}$.
\end{proof}

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\end{document}