\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 $$\label{e1} \Delta x(k)=-a(k)x(k)+\sum_{i=1}^{m}p_{i}(k)e^{-q_{i}(k)x(k-\tau_{i}(k))}.$$ 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 $\theta0$ such that $$\label{e2} T(\lambda x)\leq[\lambda(1+\eta)]^{-1}Tx.$$ \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 $k+\omega-1$, $$\label{e3} x(k)=\sum_{s=k}^{k+\omega-1}G(k,s)\sum_{i=1}^{m}p_{i}(s) e^{-q_{i}(s)x(k-\tau_{i}(k))}.$$ 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 $$\label{e4} (Tx)(k)=\sum_{s=k}^{k+\omega-1}G(k,s) \sum_{i=1}^{m}p_{i}(s)e^{-q_{i}(s)x(k-\tau_{i}(k))},$$ 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 $$x(k)\leq B\sum_{s=k}^{k+\omega-1}\sum_{i=1}^{m}\overline{p}e^{-\underline{q}M} =Bm\omega \overline{p}/e^{\underline{q}M}0.$$ 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 $\theta0$ for $\lambda\in(0,1)$. Thus $00$. 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 \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} 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 $$\label{e9} \lim_{k\to\infty}\sup y(k)=c\geq0 \quad \text{and} \quad \lim_{k\to\infty}\inf y(k)=d\leq0.$$ 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 \label{e10} d-\epsilon