\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2003(2003), No. 45, pp. 1--8.\newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.swt.edu  (login: ftp)}
\thanks{\copyright 2003 Southwest Texas State University.}
\vspace{9mm}}

\begin{document}

\title[\hfilneg EJDE--2003/45\hfil
 Large-time dynamics of discrete-time neural networks]
{Large-time dynamics of discrete-time neural networks with
McCulloch-Pitts nonlinearity}

\author[Binxiang Dai,  Lihong Huang, \&  Xiangzhen Qian
\hfil EJDE--2003/45\hfilneg]
{Binxiang Dai, Lihong Huang, \& Xiangzhen Qian} % In alphabetical order.

\address{College of Mathematics and econometrics, Hunan University,
Changsha, Hunan 410082, China}
\email[Binxiang Dai]{bxdai@hnu.net.cn}
\email[Lihong Huang]{lhhuang@hnu.net.cn}
\email[Xiangzhen Qian]{xzqian@hnu.net.cn}

\date{}
\thanks{Submitted December 17, 2002. Published April 22, 2003.}
\thanks{Partially supported by grant 02JJY2012 from the
Hunan Province Natural Science Foundation, \newline\indent
and by grant 10071016 from the National Natural Science Foundation of China.}
\subjclass[2000]{39A10, 39A12}
\keywords{Discrete neural networks, dynamics}

\begin{abstract}
  We consider a discrete-time network system of two neurons with
  McCulloch-Pitts nonlinearity. We show that if a parameter
  is sufficiently small, then network system has a stable periodic
  solution with minimal period 4k, and if the parameter is large enough,
  then the solutions of system converge to single equilibrium.
\end{abstract}

\maketitle

\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}

\section{Introduction}

We consider the following discrete-time
neural network system
\[
\begin{gathered}
x(n)= \lambda x(n-1)+(1- \lambda)f(y(n-k)), \\
y(n)= \lambda y(n-1)-(1- \lambda)f(x(n-k),
\end{gathered}
\tag{1.1}
\]
where the signal function f is given by the following
McCulloch-Pitts nonlinearity
\[
f(\zeta)=\begin{cases} -1,& \zeta>\sigma,\\
1,& \zeta\leq\sigma.
\end{cases}\tag{1.2}
\]
in which $\lambda \in (0,1)$ represents the internal decay rate, the positive integer $k$ is
the synaptic transmission delay,and $\sigma$ is the threshold.
System (1.1) can be regarded as the discrete analog of the following
artificial neural network of two neurons with delayed feedback and
McCulloch-Pitts nonlinearity signal function
\[
\begin{gathered}
\frac{dx}{dt}=-x(t)+f(y(t-\tau)), \\
\frac{dy}{dt}=-y(t)-f(x(t-\tau)).
\end{gathered}\tag{1.3}
\]
where $\frac{dx}{dt}$ and $\frac{dy}{dt}$ are
replaced by the backward difference $x(n)-x(n-1)$ and $y(n)-y(n-1)$
respectively.

Model (1.3) has interesting applications in, for example, image
processing of moving objects, and has been extensively studied in
the literature (see [1-3] and reference herein). But, to the best
of our knowledge, the dynamics of the discrete model (1.1) are
less studied (see [4,5] ).For other discrete neural networks, we
refer to [6,7].

For the sake of convenience, let $Z$ denote the set of all
integers. For any $a,b\in Z$, $a\leq b$ define
$N(a)=\{a,a+1,\cdots\}$, $N(a,b)=\{a,a+1,\cdots ,b\}$,and
$N=N(0)$. Also, let $X=\{\phi|\phi=(\varphi,\psi):N(-k,-1)
\rightarrow R^2\}$. For the given $\sigma \in R$, let
\begin{gather*}
R^+_\sigma=\{\varphi\mid \varphi:N(-k,-1)\rightarrow R \mbox{ and } \varphi(i)-\sigma
>0, \mbox{ for } i \in N(-k,-1)\},\\
R^-_\sigma=\{\varphi\mid \varphi:N(-k,-1)\rightarrow R \mbox{ and } \varphi(i)-\sigma
\leq 0,\mbox{ for } i \in N(-k,-1)\},\\
X^{\pm,\pm}_\sigma=\{\phi\in X \mid \phi=(\varphi,\psi),\varphi\in R^\pm
_\sigma \mbox{ and } \psi \in R^\pm_\sigma\},\\
X_\sigma=X^{+,+}_\sigma \cup X^{+,-}_\sigma \cup X^{-,+}_\sigma
\cup X^{-,-}_\sigma.
\end{gather*}

By a solution of (1.1), we mean a sequence $\{(x(n),y(n))\}$ of points in
$R^2$ that is defined for all $n\in N(-k)$ and satisfies (1.1) for $n\in N$
.Clearly, for any $\phi=(\varphi,\psi)\in X_\sigma$ ,system (1.1) has an unique
solution $(x^\phi(n),y^\phi(n))$ satisfying the initial conditions
$$
x^\phi(i)=\varphi(i),\quad y^\phi(i)=\psi(i),\quad \mbox{for } i \in N(-k,-1).
$$
Our goal is to determine the large time behaviors of
$(x^{\phi}(n),y^{\phi} (n))$ for every $\phi\in X_\sigma$.Our
analysis shows that for all $\phi= (\varphi,\psi)\in X_\sigma$,
the behaviors of $(x^\phi(n),y^\phi(n))$ as $n \rightarrow \infty$
are completely determined by the value $(\varphi(-1), \psi(-1))$
and the size of $\sigma$. \\The main results of this paper as
follows.

\begin{theorem} \label{thm1}
Let $|\sigma|\leq \frac{
1+\lambda^{2k+1}-2\lambda} {1-\lambda^{2k+1}}$,
$\phi=(\varphi,\psi)\in X_\sigma$ satisfy:
\begin{enumerate}
\item $\varphi(-1)\leq \frac{\sigma+1}{\lambda}-1$, $\psi(-1)\leq
\frac{\sigma+1-2\lambda}{\lambda^{k+1}}+1$  for $\phi \in
X^{+,+}_\sigma$;

\item $\varphi(-1) > \frac{\sigma-1}{\lambda^{k+1}}+1$,
$\psi(-1)\leq \frac{\sigma+1}{\lambda}-1$  for $\phi \in X^{-,+}_\sigma$;

\item $\varphi(-1) > \frac{\sigma-1}{\lambda}+1$, $\psi(-1) >
\frac{\sigma-1+2\lambda}{\lambda^{k+1}}-1$  for $\phi \in
X^{-,-}_\sigma$;

\item $\varphi(-1)\leq \frac{\sigma+1}{\lambda^{k+1}}-1$, $
\psi(-1)> \frac{\sigma-1}{\lambda}+1$ for  $\phi \in
X^{+,-}_\sigma$.
\end{enumerate}
Then there exists $\phi_0=(\varphi_0,\psi_0)\in X_\sigma$ such
that the solution $\{x^{\phi_{0}}(n),y^{\phi_{0}}(n)\}$ of (1.1)
with initial value $\phi_0=(\varphi_0,\psi_0)$ is $4k$ periodic.
Moreover, for any solutions $\{(x^\phi(n),y^\phi(n))\}$ of (1.1)
with initial value $\phi \in X_\sigma$, we have
\[
\lim_{n\rightarrow\infty}[x^\phi(n)-x^\phi_{0}(n)]=0 \,\quad
\lim_{n\rightarrow\infty}[y^\phi(n)-y^\phi_{0}(n)]=0.
\]
\end{theorem}

\begin{theorem} \label{thm2}
Let $|\sigma|>1$ and $\phi=(\varphi,\psi)\in X_\sigma$.
Then $\lim_{n\rightarrow\infty}(x^\phi(n),y^\phi(n))=(1,-1)$, if
$\sigma>1$;  and
$\lim_{n\rightarrow\infty}(x^\phi(n),y^\phi(n))=(-1,1)$,
if $\sigma < -1 $.
\end{theorem}

\begin{theorem} \label{thm3}
Let $\sigma=1$, Then $\lim_{n\rightarrow\infty}(x^\phi(n),y^\phi(n))=(1,-1)$,
 if $\phi\in X^{+,+}_{\sigma} \cup X^{-,+}_{\sigma} \cup
X^{-,-}_\sigma $; and $\lim_{n\rightarrow\infty}(x^
\phi(n),y^ \phi(n))=(1,1)$, if $\phi\in X^{+,-}_\sigma$.
\end{theorem}

\begin{theorem} \label{thm4}
Let $\sigma=-1$, Then $ \lim_{n
\rightarrow\infty}(x^\phi(n),y^\phi(n))=(-1,1), $ if $\phi \in
X^{+,+}_\sigma \cup X^{+,-}_\sigma \cup X^{-,-}_\sigma$; and
$\lim_{n \rightarrow \infty
}(x^\phi(n),y^\phi(n))=(-1,-1)$.  if $\phi \in X^{-,+}_\sigma $.
\end{theorem}

For the sake of simplicity, in the remaining part of this paper, for a given
$n \in N $ and a sequence $z(n)$ defined on $N(-k)$, we define $z_{n}:N(-k,-1)
\rightarrow R$ by $z_{n}(m)=z(n+m)$ for all $m \in N(-k,-1)$.

\section{Preliminary Lemmas}

 In this section, we establish several
technical lemmas, important in the proofs of our main results.
Assume $n_0 \in N$, we first note the difference equation
\[
x(n)=\lambda x(n-1)-1+\lambda, \quad n \in N(n_0) \tag{2.1}
\]
with initial condition $x(n_{0}-1)=a$ is given by
\[
x(n)=(a+1)\lambda^{n-n_{0}+1}-1, \quad n \in N(n_0). \tag{2.2}
\]
And that the solution of the difference equation
\[
x(n)=\lambda x(n-1)+1-\lambda,\quad n \in N(n_0) \tag{2.3}
\]
with initial condition $x(n_{0}-1)=a$ is given by
\[
x(n)=(a-1)\lambda^{n-n_{0}+1}+1,\quad n \in N(n_0). \tag{2.4}
\]
Let $(x(n),y(n))$ be a solution of (1.1) with a given initial value
$\phi=(\varphi,\psi)\in X_\sigma$.Then we have the following:

\begin{lemma} \label{lm1}
Let $-1<\sigma\leq 1$. If there exists $n_0 \in N$
such that $( x_{n_{0}},y_{n_{0}})\in X^{+,+}_{\sigma}$, then there
exists $n_1 \in N(n_0)$ such that $(x_{n_{1}+k},y_{n_{1}+k})\in
X^{-,+} _{\sigma}$. Moreover, if $x(n_{0}-1)\leq
\frac{\sigma+1}{\lambda}-1$,then $(x_{n_{0}+k},y_{n_{0}+k})\in
X^{-,+}_{\sigma}$.
\end{lemma}

\begin{proof}
Since $(x_{n_{0}},y_{n_{0}})\in X^{+,+}_{\sigma}$, for
$n \in N(n_0,n_0+k-1)$ we have
\[
\begin{gathered}
x(n)=\lambda x(n-1)-1+\lambda ,\\
y(n)=\lambda y(n-1)+1-\lambda,
\end{gathered}\tag{2.5}
\]
By (2.2) and (2.4), for $n \in N(n_0,n_0+k-1)$, we get
\[ \begin{gathered}
x(n)=[x(n_0-1)+1]\lambda^{n-n_{0}+1}-1,\\
y(n)=[y(n_0-1)-1]\lambda^{n-n_{0}+1}+1.
\end{gathered} \tag{2.6}
\]
We claim that there exists a $n_1 \in N(n_0)$ such that $x(n)>\sigma$
for $n \in N(n_0-k,n_1-1)$ and $x(n_1) \leq \sigma$. Assume, for the sake
of contradiction, that $x(n)>\sigma$ for all $n \in N(n_0-k)$. From (1.1)
and (1.2), we have
$$
y(n)=\lambda y(n-1)+1-\lambda,\quad n \in N(n_0),
$$
which yield that
$$ y(n)=[y(n_0-1)-1]\lambda^{n-n_{0}+1}+1
>(\sigma-1)\lambda^{n-n_{0}+1}+1>\sigma,\quad n \in N(n_0).
$$
Therefore, for all $n \in N(n_0-k) $, we have $y(n)>\sigma$.
By(1.1), then
$$ x(n)=\lambda x(n-1)-1+\lambda ,\quad n \in N(N_0),
$$
which implies that
$$ x(n)=[x(n_0-1)+1]\lambda^{n-n_{0}+1}-1,\quad n \in N(N_0).
$$
Therefore, $\lim_{n \rightarrow \infty}x(n)=-1$, which
contradicts  $\lim_{n \rightarrow \infty}x(n) \geq
\sigma>-1$. This proofs our claim.
>From (1.1) and (1.2), we have
$$
y(n)=\lambda y(n-1)+1-\lambda,\quad n\in N(n_0,n_1+k-1),
$$
which implies that
$$
y(n)=[y(n_{0}-1)-1]\lambda^{n-n_0+1}+1,\quad n\in N(n_0,n_1+k-1).
$$
Note that $y_{n_{0}} \in R^+_\sigma$ and $\sigma<1 $ implies
\[
y(n)>\sigma,\quad n\in N(n_{0}-k,n_{1}+k-1), \tag{2.7}
\]
that is $y_{n_1+k}\in R^+_\sigma$.  This, together with (2.1) and (2.2),
implies that $x(n) \leq \sigma $ for
$n \in N(n_1,n_1+2k-1)$, that is $x_{n_1+k}\in R^-_\sigma$. So $(x_{n_1+k},
y_{n_1+k})\in X^{-,+}_\sigma$.
In addition, if $x(n_0-1) \leq \frac{\sigma +1}{\lambda}-1$,
then from (2.6) we get $y_{{n_0}+k}\in R^+_\sigma$ and
$x(n_0)=(x(n_0-1)+1)\lambda -1 \leq \sigma$, Note that
$x(n_0-1)+1>\sigma+1>0,$(2.6) implies that
$$
x(n_0+k-1)\leq x(n_0+k-2)\leq \cdots \leq x(n_0) \leq \sigma,
$$
that is $x_{n_{0}+k} \in R^-_\sigma$. So $(x_{n_{0}+k},y_{n_{0}+k}) \in
 X^{-,+}_\sigma$.
This completes the proof. \end{proof}

\begin{lemma} \label{lm2}
Let $\sigma>-1$. If there exists $n_0 \in N$ such
that $(x_{n_{0}},y_{n_{0}}) \in X^{-,+}_\sigma$,then there exists
$n_1 \in N(n_0)$,such that $(x_{n_{1}+k},y_{n_{1}+k})\in
X^{-,-}_\sigma$.  Moreover, if
$y(n_0-1)\leq \frac{\sigma+1}{\lambda}-1$, then
$(x_{n_{0}+k},y_{n_{0}+k})\in X^{-,-}_\sigma$.
\end{lemma}

\begin{proof}
Since $(x_{n_{0}},y_{n_{0}})\in X^{-,+}_\sigma$, from (1.1)
and (1.2), it follows that  for $n \in N(n_0,n_0+k-1)$,
\[ \begin{gathered}
x(n)=\lambda x(n-1)-1+\lambda,\\
y(n)=\lambda y(n-1)-1+\lambda.
\end{gathered}\tag{2.8}
\]
So
\[\begin{gathered}
x(n)=[x(n_0-1)+1]\lambda^{n-n_{0}+1}-1,\\
y(n)=[y(n_0-1)+1]\lambda^{n-n_{0}+1}-1.
\end{gathered}\tag{2.9}
\]
Note that $(x_{n_0},y_{n_0})\in X^{-,+}_\sigma$ implies $x(n_0-1) \leq \sigma$,
$y(n_0-1)> \sigma$. Similar to the proof of Lemma \ref{lm1}, we know that there exists
$n_1 \in N(n_0)$ such that $y(n)>\sigma$ for $n\in N(n_0-k,n_1-1)$ and
$y(n_1) \leq \sigma$. Then (2.8) and (2.9) hold for $n \in N(n_0,n_1+k-1)$.
So $(x_{n_{1}+k},y_{n_{1}+k})\in X^{-,-}_\sigma$.

Moreover, if $y(n_0-1)\leq \frac{\sigma+1}{\lambda}-1$, then $x(n)\leq \sigma$
for $n \in N (n_0,n_0+k-1)$, that is $x_{n_{0}+k} \in R^-_\sigma$, and
$$
y(n_0)=(y(n_0-1)+1)\lambda -1 \leq \sigma.
$$
By (2.9) we get
$$
y(n_0+k-1)\leq y(n_0+k-2) \leq \cdots \leq y(n_0) \leq \sigma,
$$
which implies $y_{n_{0}+k}\in R^-_\sigma$. So
$(x_{n_{0}+k},y_{n_{0}+k})\in X^{-,-}_\sigma$.
\end{proof}

By a similar argument as that in the proofs of Lemmas \ref{lm1} and \ref{lm2},
 we obtain the following result.

\begin{lemma} \label{lm3}
Let $-1 \leq \sigma <1$, if there exists $n_0 \in N$ such that
$(x_{n_{0}}, y_{n_{0}})\in X^{-,-}_\sigma$, then
there exists $n_1 \in N(n_0)$, such that
$(x_{n_{1}+k},y_{n_{1}+k}) \in X^{+,-}_\sigma$. Moreover, if
$x(n_0-1)> \frac{\sigma-1}{\lambda}+1$, then
$(x_{n_{0}+k},y_{n_{0}+k}) \in X^{+,-}_\sigma$.
\end{lemma}

\begin{lemma} \label{lm4}
Let $\sigma<1$,if there exists $n_0 \in N$ such
that $(x_{n_{0}},y_{n_{0}}) \in X^{+,-}_\sigma$, then there exists
$n_1 \in N(n_0)$, such that $(x_{n_{1}+k},y_{n_{1}+k}) \in
X^{+,+}_\sigma$.  Moreover, if
$y(n_0-1)>\frac{\sigma-1}{\lambda}+1$, then
$(x_{n_{0}+k},y_{n_{0}+k}) \in X^{+,+}_\sigma$.
\end{lemma}

\section{Proofs of Main Results}

\begin{proof}[Proof of Theorem \ref{thm1}]
 In view of Lemmas 1-4, it suffices to consider the solution $\{(x(n),y(n))\}$ of
(1.1) with initial value $\phi =(\varphi,\psi)\in X^{+,+}_\sigma$.
>From Lemma1, we obtain $(x_k,y_k)\in X^{-,+}_\sigma$,
which implies that for $n \in N(0,k-1)$,
\[ \begin{gathered}
x(n)=[\varphi(-1)+1]\lambda^{n+1}-1,\\
y(n)=[\psi(-1)-1]\lambda^{n+1}+1\,.
\end{gathered} \tag{3.1}
\]
It follows that
\begin{gather*}
x(k-1)=[\varphi(-1)+1]\lambda^k-1,\\
y(k-1)=[\psi(-1)-1]\lambda^k+1.
\end{gather*}
Using $\psi(-1) \leq \frac{\sigma+1-2\lambda}{\lambda^{k+1}}$, then
$y(k-1) \leq \frac{\sigma+1}{\lambda}-1$.

Again by Lemma \ref{lm2}, we get $(x_{2k},y_{2k})\in X^{-,-}_\sigma$, which implies
that for $n\in N(k,2k-1)$,
\[\begin{gathered}
x(n)=[x(k-1)+1]\lambda^{n-k+1}-1,\\
y(n)=[y(k-1)+1]\lambda^{n-k+1}-1\,.
\end{gathered} \tag{3.2}
\]
It follows that
\begin{gather*}
x(2k-1)=[x(k-1)+1]\lambda^k-1,\\
y(2k-1)=[y(k-1)+1]\lambda^k-1.
\end{gather*}
Note that $x(k-1)>(\sigma +1)\lambda^k-1$ and
$\sigma \leq \frac{1+\lambda^{2k+1}-2\lambda}{1-\lambda^{2k+1}}$
yield
$$
x(2k-1)>(\sigma+1)\lambda^{2k}-1 \geq \frac{\sigma-1}{\lambda}+1.
$$
By Lemma \ref{lm3}, we obtain $(x_{3k},y_{3k})\in X^{+,-}_\sigma$, which implies
that for $n \in N(2k,3k-1)$,
\[\begin{gathered}
x(n)=[x(2k-1)-1]\lambda^{n-2k+1}+1,\\
y(n)=[y(2k-1)+1]\lambda^{n-2k+1}-1\,.
\end{gathered} \tag{3.3}
\]
It follows that
\begin{gather*}
x(3k-1)=[x(2k-1)-1]\lambda^k+1,\\
y(3k-1)=[y(2k-1)+1]\lambda^k-1.
\end{gather*}
Note that $y(2k-1)>(\sigma+1)\lambda^k-1$ and $\sigma \leq
\frac{1+\lambda^{2k+1}-2\lambda}{1-\lambda^{2k+1}}$, we have
$$
y(3k-1)>(\sigma+1)\lambda^{2k}-1 \geq \frac{\sigma-1}{\lambda}+1.
$$
By Lemma \ref{lm4}, we obtain $(x_{4k},y_{4k})\in X^{+,+}_\sigma$, which implies
that for $n\in N(3k,4k-1)$,
\[\begin{gathered}
x(n)=[x(3k-1)-1]\lambda^{n-3k+1}+1,\\
y(n)=[y(3k-1)-1]\lambda^{n-3k+1}+1\,.
\end{gathered}\tag{3.4}
\]
It follows that
\begin{gather*}
x(4k-1)=[x(3k-1)-1]\lambda^{k}+1,\\
y(4k-1)=[y(3k-1)-1]\lambda^{k}+1.
\end{gather*}
Note that $x(3k-1)\leq (\sigma -1)\lambda^k+1$ and $\sigma \geq
-\frac{1+\lambda^{2k+1}-2\lambda}{1-\lambda^{2k+1}}$, we have
$$
x(4k-1)\leq (\sigma -1)\lambda^{2k}+1 \leq \frac{\sigma+1}{\lambda}-1.
$$
Again by Lemma1, we obtain $(x_{5k},y_{5k})\in X^{-,+}_\sigma$, which implies
that for $n\in N(4k,5k-1)$,
\[\begin{gathered}
x(n)=[x(4k-1)+1]\lambda^{n-4k+1}-1,\\
y(n)=[y(4k-1)-1]\lambda^{n-4k+1}+1\,.
\end{gathered}\tag{3.5}
\]
It follows that
\begin{gather*}
x(5k-1)=[x(4k-1)+1]\lambda^k-1,\\
y(5k-1)=[y(4k-1)-1]\lambda^k+1.
\end{gather*}
In general, for $i \in N(1)$, we can get:
\begin{gather*}
x(n)=[\varphi (-1)+1]\lambda^{n+1}+
2\lambda^{n+1}\frac{\lambda^{-4(i-1)k}-1}{\lambda^{2k}+1}-1,\\
y(n)=[\psi (-1)-1]\lambda^{n+1}+
2\lambda^{n+k+1}\frac{\lambda^{-(4i-2)k}+1}{\lambda^{2k}+1}-1
\end{gather*}
for $n\in N((4i-3)k,(4i-2)k-1)$;
\begin{gather*}
x(n)=[\varphi (-1)+1]\lambda^{n+1}-
2\lambda^{n+1}\frac{\lambda^{-(4i-2)k}+1}{\lambda^{2k}+1}+1,\\
y(n)=[\psi (-1)-1]\lambda^{n+1}+
2\lambda^{n+k+1}\frac{\lambda^{-(4i-2)k}+1}{\lambda^{2k}+1}-1
\end{gather*}
for $n\in N((4i-2)k,(4i-1)k-1)$;
\begin{gather*}
x(n)=[\varphi (-1)+1]\lambda^{n+1}-
2\lambda^{n+1}\frac{\lambda^{-(4i-2)k}+1}{\lambda^{2k}+1}+1,\\
y(n)=[\psi (-1)-1]\lambda^{n+1}-
2\lambda^{n+k+1}\frac{\lambda^{-4ik}-1}{\lambda^{2k}+1}+1,
\end{gather*}
for $n \in N((4i-1)k,4ik-1)$;
\begin{gather*}
x(n)=[\varphi (-1)+1]\lambda^{n+1}+
2\lambda^{n+1}\frac{\lambda^{-4ik}-1}{\lambda^{2k}+1}-1,\\
y(n)=[\psi(-1)-1]\lambda^{n+1}-
2\lambda^{n+k+1}\frac{\lambda^{-4ik}-1}{\lambda^{2k}+1}+1,
\end{gather*}
for $n\in N(4ik,(4i+1)k-1)$.\\

Let $\phi_0=(\varphi_0,\psi_0)\in X^{+,+}_\sigma$, with
$$
\varphi_0(-1)=\frac{1-\lambda^{2k}}{1+\lambda^{2k}},
\psi_0(-1)=\frac{1+\lambda^{2k}-2\lambda^k}{1+\lambda^{2k}}.
$$
Then
\begin{gather*}
x^{\phi_{0}}(n)=\frac{2}{1+\lambda^{2k}}\lambda^{n-4(i-1)k+1}-1,\\
y^{\phi_{0}}(n)=\frac{2}{1+\lambda^{2k}}\lambda^{n-(4i-3k)+1}-1
\end{gather*}
for $n\in N((4i-3)k,(4i-2)k-1)$;
\begin{gather*}
x^{\phi_{0}}(n)=-\frac{2}{1+\lambda^{2k}}\lambda^{n-(4i-2)k+1}+1,\\
y^{\phi_{0}}(n)=\frac{2}{1+\lambda^{2k}}\lambda^{n-(4i-3k)+1} -1
\end{gather*}
for $n \in N((4i-2)k,(4i-1)k-1)$;
\begin{gather*}
x^{\phi_{0}}(n)=-\frac{2}{1+\lambda^{2k}}\lambda^{n-(4i-2)k+1}+1,\\
y^{\phi_{0}}(n)=-\frac{2}{1+\lambda^{2k}}\lambda^{n-(4i-1)k+1}+1
\end{gather*}
for $n \in N((4i-1)k,4ik-1)$;
\begin{gather*}
x^{\phi_{0}}(n)=\frac{2}{1+\lambda^{2k}}\lambda^{n-4ik+1}-1,\\
y^{\phi_{0}}(n)=-\frac{2}{1+\lambda^{2k}}\lambda^{n-(4i-1)k+1}+1,
\end{gather*}
for $n \in N(4ik,(4i+1)k-1)$.

 Clearly, $\{(x^{\phi_0}(n),y^{\phi_0}(n))\}$ is periodic with
minimal period $4k$, and as $n \rightarrow \infty$,
\begin{gather*}
x^\phi (n)-x^{\phi_0}(n)=[\varphi(-1)+1]\lambda^{n+1}-
\frac{2\lambda^{n+1}}{1+\lambda^{2k}}\rightarrow 0, \\
y^\phi(n)-y^{\phi_{0}}(n)=[\psi(-1)-1]\lambda^{n+1}+
\frac{2\lambda^{n+k+1}}{1+\lambda^{2k}} \rightarrow 0.
\end{gather*}
This completes the proof of Theorem \ref{thm1}.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm2}]
  We  prove only the case where $\sigma > 1$, the
case where $\sigma < -1$ is similar. We distinguish several cases.

\noindent{\bf Case 1} $\phi = (\varphi,\psi) \in X^{-,-}_\sigma$.
In view of (1.1), for $n \in N(0,k-1)$ we have
\[
\begin{gathered}
x(n)=\lambda x(n-1)+1-\lambda,\\
y(n)=\lambda y(n-1)-1+\lambda\,.
\end{gathered} \tag{3.6}
\]
which yields that for $n \in N(0,k-1)$,
\[ \begin{gathered}
x(n)=[\varphi(-1)-1]\lambda^{n+1}+1,\\
y(n)=[\psi(-1)+1]\lambda^{n+1}-1\,.
\end{gathered}\tag{3.7}
\]
This implies that $x_k(m) \leq \sigma, y_k(m) \leq \sigma $ for $m
\in N(-k,-1)$, therefore $(x_k,y_k)\in X^{-,-}_\sigma$. Repeating
the above argument on $N(0,k-1),N(k,2k-1),\cdots ,$consecutively,
we can obtain that $(x_n,y_n)\in X^{-,-}_\sigma$ for all $n \in
N$. Therefore, (3.7) holds for all $n \in N$, and hence
$$
\lim_{n \rightarrow \infty}(x(n),y(n))=(1,-1).
$$
{\bf Case 2} $\phi=(\varphi, \psi)\in X^{-,+}_\sigma \cup
X^{+,-}_\sigma \cup X^{+,+}_\sigma$. By (1.1), for $n\in N$, we
have
\begin{gather*}
x(n)\leq \lambda x(n-1)+1-\lambda,\\
y(n)\leq \lambda y(n-1)+1-\lambda\,.
\end{gather*}
By induction, this implies
\[ \begin{gathered}
x(n)\leq [\varphi (-1)-1]\lambda^{n+1}+1,\\
y(n)\leq [\psi (-1)-1]\lambda^{n+1}+1\,.
\end{gathered} \tag{3.8}
\]
Since
\begin{gather*}
\lim_{n \rightarrow \infty}[(\varphi(-1)-1)\lambda^{n+1}+1]=1< \sigma,\\
\lim_{n \rightarrow \infty}[(\psi(-1)-1)\lambda^{n+1}+1]=1< \sigma,
\end{gather*}
then there exists $m \in N(1)$,  such that $x(n)< \sigma ,y(n)< \sigma $ for
$n \in N(m)$. This implies that $(x_{n+k},y_{n+k})\in X^{-,-}_\sigma$ for all
$n \in N(m)$. Thus, by case 1, we have
$$
\lim_{n \rightarrow \infty}(x(n),y(n))=(1,-1).
$$
This completes the proof of Theorem \ref{thm2}.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm3}]
  We distinguish several cases.

\noindent{\bf Case 1} $\phi =(\varphi, \psi)\in X^{-,-}_\sigma$.
Using a similar argument to that in Case 1 for the proof of Theorem \ref{thm2},
we can show the conclusion is true.

\noindent{\bf Case 2} $\phi =(\varphi,\psi)\in X^{-,+}_\sigma$.
By lemma 2, there exists $n_0 \in N$ such that $(x_{n_{0}+k},y_{n_{0}+k})\in
X^{-,-}_\sigma$. Thus, it follows from Case 1 that conclusion is true.

\noindent{\bf Case 3} $\phi=(\varphi, \psi)\in X^{+,+}_\sigma $.
By Lemma \ref{lm1}, there exists $n_0 \in N$, such that
$(x_{n_{0}+k},y_{n_{0}+k})\in  X^{-,+}_\sigma $.
Thus, it follows from Case 2 that the conclusion is true.

\noindent{\bf Case 4} $\phi =(\varphi, \psi)\in X^{+,-}_\sigma $.
By (1.1) and (1.2) we have that for $n\in N(0,k-1)$,
\begin{gather*}
x(n)=\lambda x(n-1)+1-\lambda,\\
y(n)=\lambda y(n-1)+1-\lambda
\end{gather*}
which implies that for $i\in N(-k,-1)$,
\[ \begin{gathered}
x_k(i)=[\varphi(-1)-1]\lambda^{i+k+1}+1,\\
y_k(i)=[\psi(-1)-1]\lambda^{i+k+1}+1\,.
\end{gathered}\tag{3.9}
\]
Since $\varphi (-1) >\sigma =1, \psi (-1)\leq \sigma =1$, then (3.9) implies
that $x_k(i)>1$, $y_k(i)\leq 1$ for $i \in N(-k,-1)$, and so
$(x_k,y_k)\in X^{+,-}_\sigma$. Repeating the above argument on
$N(k,2k-1),N(2k,3k-1),\dots$, consecutively, we can get, for all $n\in N$,
\begin{gather*}
x(n)=[\varphi (-1)-1]\lambda^{n+1}+1,\\
y(n)=[\psi (-1)-1]\lambda^{n+1}+1\,.
\end{gather*}
Therefore, $\lim_{n \rightarrow \infty}(x(n),y(n))=(1,1)$. This completes the
proof of Theorem \ref{thm3}.
\end{proof}

The proof of Theorem \ref{thm4} is similar to that of Theorem \ref{thm3} and we
omit it.

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\end{document}