\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2006(2006), No. 95, pp. 1--8.\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 2006 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2006/95\hfil Existence and global attractivity]
{Existence and global attractivity  positive periodic
solutions for a discrete model}

\author[Z. Zhou,  Z. Zhang \hfil EJDE-2006/95\hfilneg]
{Zheng Zhou, Zhengqiu Zhang }

\address{Zheng Zhou\newline
Department of Applied Mathematics, Hunan University,
 Changsha 410082, China}
\email{zzzzhhhoou@yahoo.com.cn}

\address{Zhengqiu Zhang  \newline
Department of Applied Mathematics, Hunan University,
 Changsha 410082, China}
\email{z\_q\_zhang@sina.com}

\date{}
\thanks{Submitted June 14, 2006. Published August 18, 2006.}
\thanks{Supported by grant 10271044 from the NNSF of China}
\subjclass[2000]{39-99, 47H10}
\keywords{Positive periodic solution; attractivity;
          fixed point; discrete model}

\begin{abstract}
  Using a fixed point theorem in cones, we obtain  conditions
  that guarantee the existence  and attractivity of the
  unique positive periodic solution for a discrete Lasota-Wazewska
  model.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}

\section{Introduction}

Wazewska-Czyzewska and Lasota \cite{w2} investigated the delay differential
equation
$$
x'(t)=-\alpha x(t)+\beta e^{-\gamma x(t-\tau)},\quad t\geq0.
$$
as a model for the survival of red blood cells in an animal. The
oscillation and global attractivity of this equation have been
studied by Kulenovic and Ladas \cite{k1}. A few similar generalized model
were investigated by many authors, see Xu and Li \cite{x1}, Graef et
al. \cite{g2}, Jiang and Wei \cite{j2}, Gopalsamy and Trofimchuk \cite{g1}.
Recently, Liu \cite{l1} studied the existence and global attractivity of unique
positive periodic solution for the Lasota-Wazewska model
$$
x'(t)=-a(t)x(t)+\sum_{i=1}^{m}p_{i}(t)e^{-q_{i}(t)x(t-\tau_{i}(t))},
$$
by using a fixed point theorem, and got some brief conditions to
guarantee the conclusions. In \cite{j1}, the existence of one positive
periodic solution was proved by Mawhin's continuation theorem.
In \cite{z1}, the existence of multiple positive periodic solutions was
studied by employing Krasnoselskii fixed point theorem in cones.

Though most models are described with differential equations, the
discrete-time models are more appropriate than the continuous ones
when the size of the population is rarely small or the population
has non-overlapping generations \cite{a1}. To our knowledge, studies on
discrete models by using fixed point theorem are scarce, see \cite{z1}.
 In this paper, we consider the Lasota-Wazewska difference
equation
\begin{equation} \label{e1}
\Delta x(k)=-a(k)x(k)+\sum_{i=1}^{m}p_{i}(k)e^{-q_{i}(k)x(k-\tau_{i}(k))}.
\end{equation}
We will use the following hypotheses:
\begin{itemize}
\item[(H1)] $a:Z\to (0,1)$ is continuous and
$\omega$-periodic function. i.e., $a(k)=a(k+\omega)$, such that
$a(k)\not\equiv 0$, where $\omega$is a positive constant denoting
the common period of the system.

\item[(H2)] $p_{i}$ and $q_{i}$ are positive continuous
$\omega$-periodic functions, $\tau_{i}$ are continuous
$\omega$-periodic functions $(i=1,2,\dots)$.

\end{itemize}
For convenience, we shall use the notation:
$$
\bar{h}=\max_{0\leq k\leq \omega}\{h(k)\},\quad
\underline{h}= \min_{0\leq k\leq \omega}\{h(k)\}.
$$
where $h$ is a continuous
$\omega$-periodic function. Also, we use
\begin{gather*}
q=\max_{1\leq i \leq m}\{\bar{q_{i}}\},\quad
\tau=\max_{1\leq i\leq m}\{\bar{\tau_{i}}\},\quad
p=\omega \sum_{i=1}^{m}p_{i}(s),\quad(k\leq s\leq k+\omega-1),
\\
A=\frac{\prod_{s=0}^{\omega-1}(1-a(s))}{1-\prod_{s=0}^{\omega-1}(1-a(s))},\\
B=\frac{1}{1-\prod_{s=0}^{\omega-1}(1-a(s))},\quad
\sigma=\prod_{s=0}^{\omega-1}(1-a(s))=\frac{A}{B}<1.
\end{gather*}
Considering the actual applications, we assume the solutions of \eqref{e1}
 with initial condition
$$
x(k)=\phi(k)>0\quad\text{for }-\tau\leq k\leq0.
$$
To prove our result, we state the following concepts and lemmas.

\noindent  \textbf{Definition.}
 Let $X$ be Banach space and $P$ be a     closed, nonempty subset,
 $P$ is said to be a cone if
\begin{itemize}
\item[(i)] $\lambda x\in P$ for all $x\in P$ and $\lambda\geq0$
\item[(ii)] $x\in P, -x\in P$ implies $x=\theta$.
\end{itemize}
The semi-order induced by the cone $P$ is denoted by $"\leq"$. That
is, $x\leq y$ if and  $y-x\in P$.

\noindent\textbf{Definition.}
A cone $P$ of $X$ is said to be normal
if there exists a positive constant $\delta$, such that
$\|x+y\|\geq\delta$ for any $x,y\in P$. $\|x\|=\|y\|=1$.

\noindent\textbf{Definition.}
Let $P$ be a cone of $X$ and $T:P\to P$ an operator. $T$ is called decreasing,
if $\theta\leq x\leq y$ implies $Tx\geq Ty$.


\begin{lemma}[Guo \cite{g3,g4}] \label{lem1.1}
 Suppose that
\begin{itemize}
\item[(i)]  $P$ is normal cone of a real Banach space $X$ and
$T:P\to P$ is decreasing and completely   continuous;

\item[(ii)] $T\theta>\theta$, $T^{2}\geq \varepsilon_{0}T\theta$, where
$\varepsilon_{0}>0$;

\item[(iii)] For any $\theta<x\leq T\theta$ and $0<\lambda<1$, there exists
$\eta=\eta(x,\lambda)>0$ such that
\begin{equation} \label{e2}
T(\lambda x)\leq[\lambda(1+\eta)]^{-1}Tx.
\end{equation}
\end{itemize}
Then $T$ has exactly one positive fixed point $\tilde{x}>\theta$.
Moreover, constructing the sequence
$x_{n}=Tx_{n-1}$ $(n=1,2,3,\dots)$ for any initial $x_{0}\in P$, it
follows that
$\|x_{n}-\tilde{x}\|\to0$ as $n\to\infty$.
\end{lemma}

 \section{Positive periodic solutions}

To apply Lemma \ref{lem1.1}, let
$X=\{x(k):x(k)=x(k+\omega)\}$,
 $\|x\|=\max\{\mid x(k)\mid:x\in X\}$. Then $X$ is a Banach space
endowed with the norm
$\|\cdot\|$.\par Define the cone
$$
P=\{x\in X:x(k)\geq0,\; x(k)\geq\sigma\|x\|\}.
$$

 \begin{lemma} \label{lem2.1}
If $x(k)$ is a positive $\omega$-periodic
    solution of \eqref{e1},then $x(k)\geq\sigma\|x\|$.
\end{lemma}

\begin{proof}
 It is clear that \eqref{e1} is equivalent to
$$
x(k+1)=(1-a(k))x(k)+\sum_{i=1}^{m}p_{i}(k)e^{-q_{i}(k)x(k-\tau_{i}(k))}.
$$
Multiplying the two sides by  $\prod_{s=0}^{k}(1-a(s))^{-1}$, we
have
$$
\Delta\Big( x(k)\prod_{s=0}^{k-1}\frac{1}{1-a(s)}\Big)
=\prod_{s=0}^{k}\frac{1}{1-a(s)}
\sum_{i=1}^{m}p_{i}(k)e^{-q_{i}(k)x(k-\tau_{i}(k))}.
$$
Summing the two sides from $k$ to $k+\omega-1$,
\begin{equation} \label{e3}
x(k)=\sum_{s=k}^{k+\omega-1}G(k,s)\sum_{i=1}^{m}p_{i}(s)
e^{-q_{i}(s)x(k-\tau_{i}(k))}.
\end{equation}
where
$$
G(k,s)=\frac{\prod_{r=s+1}^{k+\omega-1}(1-a(r))}{1-\prod_{r=0}^{\omega-1}
(1-a(r))},\quad
k\leq s\leq k+\omega-1.
$$
Then, $x(k)$ is an $\omega$-periodic solution of
\eqref{e1} if and only if $x(k)$ is $\omega$-periodic solution
of difference equation  \eqref{e3}. It is easy to calculate that
\begin{gather*}
A:=\frac{\prod_{s=0}^{\omega-1}(1-a(s))}{1-\prod_{s=0}^{\omega-1}(1-a(s))}\leq G(k,s)
\leq\frac{1}{1-\prod_{s=0}^{\omega-1}(1-a(s))}=:B,
\\
A=\frac{\sigma}{1-\sigma},\quad B=\frac{1}{1-\sigma},\quad
\sigma=\frac{A}{B}<1,
\\
\|x\|\leq B\sum_{S=k}^{k+\omega-1}\sum_{i=1}^{m}p_{i}(s)e^{-q_{i}(s)
x(s-\tau_{i}(s))},
\\
x(t)\geq A\sum_{S=k}^{k+\omega-1}\sum_{i=1}^{m}p_{i}(s)
e^{-q_{i}(s)x(s-\tau_{i}(s))}).
\end{gather*}
Therefore,
$x(k)\geq \frac{A}{B}\|x\|=\sigma\|x\|$.
\end{proof}

 Define the mapping $T:X\to X$ by
\begin{equation} \label{e4}
(Tx)(k)=\sum_{s=k}^{k+\omega-1}G(k,s)
\sum_{i=1}^{m}p_{i}(s)e^{-q_{i}(s)x(k-\tau_{i}(k))},
\end{equation}
for $x\in X$, $k\in Z$. It is not difficult to see that $T$ is a
completely continuous operator on $X$, and a periodic solution of
 \eqref{e1} is the fixed point of operator $T$.

\begin{lemma} \label{lem2.2}
Under the conditions above, $TP\subset P$.
\end{lemma}

\begin{proof}
For each $x\in P$, we have
$$
\|Tx\|\leq B\sum_{s=k}^{k+\omega-1}G(k,s)\sum_{i=1}^{m}p_{i}(s)
e^{-q_{i}(s)x(k-\tau_{i}(k))}
$$
 From  \eqref{e4}, we obtain
$$
Tx\geq A\sum_{s=k}^{k+\omega-1}G(k,s)\sum_{i=1}^{m}p_{i}(s)
e^{-q_{i}(s)x(k-\tau_{i}(k))}
\geq\frac{A}{B}\|Tx\|=\sigma\|Tx\|.
$$
Therefore, $Tx\in P$, thus $TP\subset P$.
\end{proof}

\begin{lemma} \label{lem2.3}
$x(k)$ is positive and bounded on $[0,\infty)$.
\end{lemma}

\begin{proof}
 Obviously, $x(k)$ is defined on $[-\tau,+\infty)$ and
positive on $[0,+\infty)$. Now, we prove that every solution of  \eqref{e1}
is bounded, otherwise, there exists an unbounded solution $x(k)$.
Thus, for arbitrary $M>B m\omega \overline{p}/e^{\underline{q}M}$,
there exists $N=N(M)$, when $k>N,x(k)>M$. From \eqref{e3},
we have
$$
x(k)\leq B\sum_{s=k}^{k+\omega-1}\sum_{i=1}^{m}\overline{p}e^{-\underline{q}M}
=Bm\omega \overline{p}/e^{\underline{q}M}<M.
$$
where
$$
\underline{q}=\min_{1\leq i \leq m}\{\underline{q_{i}}\},\quad
\overline{p}=\max_{1\leq i \leq m}\{\bar{p_{i}}\},
$$
which is a contradiction. Consequently, $x(k)$ is bounded.
\end{proof}

Now, we are in position to state the main results in this
section.

\begin{theorem} \label{thm2.1}
Assume that (H1)-(H2) hold and  $Bpq\leq1$. Then \eqref{e1}
 has a unique $\omega$-periodic     positive solution $\tilde{x}(t)$.
Moreover,
$$
\|x(k)-\tilde{x}\|   \to0(k\to\infty)m
$$
where $x(k)=Tx(k-1)(k=1,2,\dots)$ for any initial $x_{0}\in P$.
\end{theorem}

\begin{proof}
 Firstly, it is clear that the cone $P$ is  normal. By an easy calculation,
we know that $T$ is decreasing,  in fact
\begin{align*}
&(Tx)(k)-(Ty)(k)\\
&=\sum_{s=k}^{k+\omega-1}G(k,s)\sum_{i=1}^{m}p_{i}(s)
(e^{-q_{i}(s)x(s-\tau_{i}(s))}-e^{-q_{i}(s)y(s-\tau_{i}(s))})\\
&=\sum_{s=k}^{k+\omega-1}G(k,s)\sum_{i=1}^{m}p_{i}(s)e^{-q_{i}(s)x(s-\tau_{i}(s))}
[1-e^{-q_{i}(s)(y(s-\tau_{i}(s))-x(s-\tau_{i}(s)))}]
\geq 0
\end{align*}
when $\theta\leq x\leq y$,  i.e.,
$y(s-\tau_{i}(s))-x(s-\tau_{i}(s))\geq0$.

Secondly, we will show that the condition (ii) of Lemma \ref{lem1.1} is
satisfied. From  \eqref{e4}, we have
$$
Bp\geq(T\theta)(k)=\sum_{s=k}^{k+\omega-1}G(k,s)\sum_{i=1}^{m}p_{i}(s)
\geq Ap>0.
$$
Thus, $T\theta>\theta$, and
\begin{align*}
(T^{2}\theta)(k)
&= \sum_{s=k}^{k+\omega-1}G(k,s)\sum_{i=1}^{m}p_{i}(s)
e^{-q_{i}(s)(T\theta)(s-\tau_{i}(s))}\\
&\geq  e^{-Bpq}\sum_{s=k}^{k+\omega-1}G(k,s)\sum_{i=1}^{m}p_{i}(s)\\
&= e^{-Bpq}(T\theta)(k).
\end{align*}
So that $T^{2}\theta\geq\varepsilon_{0}T\theta$, where
$\varepsilon_{0}=e^{-Bpq}>0$.

Finally, we prove that the condition (iii) of Lemma \ref{lem1.1} is also
satisfied. For any $\theta<x<T\theta$ and $0<\lambda<1$, we have
$\|x\|\leq\|T\theta\|\leq Bp$  and
\begin{equation} \label{e5}
\begin{aligned}
T(\lambda x)(k)
&=\sum_{s=k}^{k+\omega-1}G(k,s)\sum_{i=1}^{m}p_{i}(s)
e^{-\lambda q_{i}(s)x(s-\tau_{i}(s))}\\
&=\sum_{s=k}^{k+\omega-1}G(k,s)\sum_{i=1}^{m}p_{i}(s) e^{-
q_{i}(s)x(s-\tau_{i}(s))}e^{(1-\lambda)q_{i}(s)x(s-\tau_{i}(s))}\\
&\leq e^{(1-\lambda)Bpq}(Tx)(k)\\
&=\lambda^{-1}\lambda e^{(1-\lambda)Bpq}(Tx)(k).
\end{aligned}
\end{equation}

Set $f(\lambda)=\lambda e^{Bpq(1-\lambda)}$; therefore,
$f'(\lambda)=(1-Bpq\lambda)e^{Bpq(1-\lambda)}>0$  for
$\lambda\in(0,1)$. Thus $0<f(\lambda)<f(1)=1$. so set
$f(\lambda)=(1+\eta)^{-1}$, where $\eta=\eta(\lambda)>0$. From \eqref{e5},
we have
$$
T(\lambda x)\leq\lambda^{-1}f(\lambda)Tx=\lambda^{-1}(1+\eta)^{-1}Tx
=[\lambda(1+\eta)]^{-1}Tx.
$$
By Lemma \ref{lem1.1}, we see that $T$ has
exactly one positive fixed point $\tilde{x}>\theta$. Moreover,
$\|x(k)-\tilde{x}\|\to0(n\to\infty)$, where
$x(k)=Tx(k-1)(k=1,2,\dots)$ for any initial $x_{0}\in P$ for $k\in
N$.
\end{proof}

\begin{remark} \label{rmk2.1} \rm
 Theorem \ref{thm2.1} not only gives the sufficient
    conditions for the existence of unique positive periodic
    solution of \eqref{e1}, but also contains the conclusion of
    convergence of $x(k)$ to $\tilde{x}$.
\end{remark}

\begin{remark} \label{rmk2.2} \rm
From the statements above, we have
\begin{gather}
\tilde{x}(k)=(T\tilde{x})(k)=\sum_{s=k}^{k+\omega-1}G(k,s)
\sum_{i=1}^{m}p_{i}(s)
e^{-q_{i}(s)\tilde{x}(s-\tau_{i}(s))}
\geq Ape^{-q\|\tilde{x}\|}>0,
 \label{e6}\\
Ape^{-Bpq}\leq\tilde{x}(k)\leq Bp \geq0. \label{e7}
\end{gather}
which will be used in the following section.
\end{remark}

\section{Global attractivity of the solution to \eqref{e1}}
\begin{theorem} \label{thm3.1}
 Assume that (H1)-(H2) hold  and $Bpq\leq1$. Then the unique
$\omega$-periodic solution $\tilde{x}(k)$ of \eqref{e1}
is a global attractor of all other positive solutions of
\eqref{e1}.
\end{theorem}

\begin{proof}
 Let $y(k)=x(k)-\tilde{x}(k)$, where $x(k)$ is arbitrary solution
of  \eqref{e1}, Then it is easy to obtain
\begin{equation} \label{e8}
\begin{aligned}
\triangle y(k)&=\triangle(x(k)-\tilde{x}(k))\\
&=\triangle x(k)-\triangle\tilde{x}(k)\\
&= -a(k)y(k)+\sum_{i=1}^{m}p_{i}(s)
e^{-q_{i}(s)\tilde{x}(s-\tau_{i}(s))}(e^{-q_{i}(s)y(s-\tau_{i}(s))}-1).
\end{aligned}
\end{equation}
Now, we shall prove $\lim_{k\to\infty}y(k)=0$ in the
following three cases:

\noindent\emph{Case 1.}
Suppose that $y(t)$ is eventually positive solution of
\eqref{e8}. It is easy to see that $\triangle y(k)<0$ for all
sufficiently large $k$, so $\lim_{k\to\infty}y(k)=l\geq0$.
we claim that $l=0$. If $l>0$, then there exists $N>0$ such that
$\triangle y(k)<-la(k),k\geq N$. Summing the two sides of the
inequality from $N$ to $\infty$, we have
$$
l-y(N)=\sum_{k=N}^{\infty}\triangle y(k)<-l\sum_{k=N}^{\infty}a(k)=-\infty.
$$
which is a contradiction, so $l=0$.

\noindent\emph{Case 2.} Suppose that $y(k)$ is eventually negative.
By similar proof as above we obtain that $l=0$.

\noindent\emph{Case 3.} Suppose that $y(k)$ is oscillatory, from
Lemma \ref{lem2.3},
 we know $y(k)$ is bounded. We set
\begin{equation} \label{e9}
\lim_{k\to\infty}\sup y(k)=c\geq0 \quad \text{and} \quad
\lim_{k\to\infty}\inf y(k)=d\leq0.
\end{equation}
For arbitrarily small positive constant $\epsilon$, $d-\epsilon<0$ and
$c+\epsilon>0$. In view of \eqref{e9}, there exists $N_{\epsilon}>0$, such
that
\begin{equation} \label{e10}
d-\epsilon<y(k)<c+\epsilon\quad\text{for all }
k\geq N_{\epsilon}-\tau.
\end{equation}
From \eqref{e8}and \eqref{e10},we have
$$
y(k+1)-(1-a(k))y(k)=\sum_{i=1}^{m}p_{i}(s)
e^{-q_{i}(s)\tilde{x}(s-\tau_{i}(s))}(e^{-q_{i}(s)y(s-\tau_{i}(s))}-1).
$$
Multiplying the two sides by $\prod_{s=0}^{k}(1-a(s))^{-1}$, we
have
\begin{equation} \label{e11}
\begin{aligned}
&\triangle(y(k)\prod_{s=0}^{k-1}\frac{1}{1-a(s)})\\
&= \prod_{s=0}^{k}\frac{1}{1-a(s)}\sum_{i=1}^{m}p_{i}(s)
e^{-q_{i}(s)\tilde{x}(s-\tau_{i}(s))}(e^{-q_{i}(s)y(s-\tau_{i}(s))}-1)\\
&\leq (e^{-q(d-\epsilon)}-1)\prod_{s=0}^{k}\frac{1}{1-a(s)}\sum_{i=1}^{m}p_{i}(s)
e^{-q_{i}(s)\tilde{x}(s-\tau_{i}(s))}\\
&= (e^{-q(d-\epsilon)}-1)\triangle(\tilde{x}(k)\prod_{s=0}^{k-1}\frac{1}{1-a(s)}).
\end{aligned}
\end{equation}
Summing the two sides from $N_{\epsilon}$ to$\infty$,  for $k\geq N_{\epsilon}$,
we have
\begin{equation} \label{e12}
\begin{aligned}
&y(k+1)\prod_{s=0}^{k}\frac{1}{1-a(s)}-
y(N_{\epsilon})\prod_{s=0}^{N_{\epsilon}-1} \frac{1}{1-a(s)}\\
&\leq(e^{-q(d-\epsilon)}-1)[\tilde{x}(k+1)\prod_{s=0}^{k}\frac{1}{1-a(s)}-
\tilde{x}(N_{\epsilon})\prod_{s=0}^{N_{\epsilon}-1}\frac{1}{1-a(s)}].
\end{aligned}
\end{equation}
Thus
\begin{equation} \label{e13}
y(k+1)\leq
y(N_{\epsilon})\prod_{s=N_{\epsilon}}^{k}(1-a(s))+(e^{-q(d-\epsilon)}-1)
[\tilde{x}(k+1)-\tilde{x}(N_{\epsilon})\prod_{s=N_{\epsilon}}^{k}(1-a(s))].
\end{equation}
 From \eqref{e9}, \eqref{e13} and Remark \ref{rmk2.2}, we have
$$c\leq Bp(e^{-q(d-\epsilon)}-1).\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$$
As $\epsilon$ is arbitrary small, we have that
\begin{equation} \label{e14}
c\leq Bp(e^{-qd}-1).
\end{equation}
By the similar method as above, we obtain
\begin{equation} \label{e15}
d\geq Bp(e^{-qc}-1).
\end{equation}
 From results in \cite{l1,x1}, $Bpq\leq1$ implies that \eqref{e14}, \eqref{e15}
have a unique solution $c=d=0$. Therefore,
$$
\lim_{k\to\infty}y(k)=\lim_{k\to\infty}[x(k)-\tilde{x}(k)]=0.
$$
The proof is complete.
\end{proof}

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\end{document}
