\documentclass[reqno]{amsart}
\usepackage{hyperref}
\usepackage{graphicx}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2006(2006), No. 03, pp. 1--18.\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-2006/03\hfil Existence of positive periodic solutions]
{Existence of positive periodic solutions for a class of
nonautonomous difference equations}
\author[Z. Zeng\hfil EJDE-2006/03\hfilneg]
{Zhijun Zeng}

\address{Zhijun Zeng \hfill\break
Academy of Mathematics and Systems Sciences \\
Chinese Academy of Sciences \\
Beijing 100080, China}
\email{zthzzj@yeah.net}

\date{}
\thanks{Submitted October 25, 2005. Published January 6, 2006.}
\subjclass[2000]{34K45, 34K13, 92D25}
\keywords{Nonautonomous difference equations; periodic solution;
\hfill\break\indent
 Krasnoselskii fixed point theorem; numerical simulations}

\begin{abstract}
 In this article, we investigate the existence of positive
 periodic solutions for a class of non-autonomous difference
 equations. Using the Krasnoselskii fixed point theorem,
 we establish  sufficient criteria that are easily verifiable
 and that generalize and improve related studies in the literature.
 Numerical simulations are presented which support our theoretical
 results for some concrete models.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{example}[theorem]{Example}
\newtheorem{remark}[theorem]{Remark}

\section{Introduction}
 Let
$\mathbb{R}$ denote the set of real numbers and $\mathbb{R}_+$ the set of
nonnegative numbers. Let $\mathbb{R}_+^n =\{(x_1,\dots,x_n)^T:x_i\geq 0,\;1\leq i\leq n\}$.
Let $\mathbb{Z}$ denote the  set of integers and
$\mathbb{Z}_+$ the set of nonnegative integers.

In this paper, we apply a cone fixed point theorem due
to Krasnoselskii to investigate the existence of positive periodic
solutions for the non-autonomous difference equations
\begin{equation}
\Delta x(k)=a(k)x(k)-f(k,u(k)),\label{e1.1}
\end{equation}
and
\begin{equation}
\Delta x(k)=-a(k)x(k)+f(k,u(k)),\label{e1.2}
\end{equation}
where $\Delta x(k)=x(k+1)-x(k)$, and for $k,s\in \mathbb{Z}$
\begin{equation}
u(k)=\Big(x(g_1(k)),x(g_2(k)),\dots,x(g_{n-1}(k)),
\sum_{s=-\infty}^{k}h(k-s)x(s)\Big)\,.
\label{e1.3}
\end{equation}
It is well-known that \eqref{e1.1} includes many mathematical
ecological difference models. For example,  \eqref{e1.1} includes
the generalized discrete single species model
\begin{equation}
\Delta x(k)=x(k)[a(k)-\sum_{i=1}^{n}b_i(k)x(k-\tau_i(k))-c(k)\sum_{s=-\infty}^{k}h(k-s)
x(k)]\,.\label{e1.4}
\end{equation}
Equation \eqref{e1.1} includes also
the single species discrete periodic population
models \cite{g2,j1,j2,j4,j5,k2,l1,l2,p1,z1}
\begin{equation}
\Delta
x(k)=a(k)x(k)\big[ 1-\frac{x(k-\tau(k))}{H(k)}\big] \label{e1.5}
\end{equation}
and
\begin{equation}
\Delta x(k)=x(k)[a(k)-\sum_{i=1}^{n}b_i(k)x(k-\tau_i(k))]\,.\label{e1.6}
\end{equation}
Equation \eqref{e1.1} includes also the multiplicative delay periodic
Logistic difference equation \cite{g2,g3,j4,j5,l2,m3,p1}
\begin{equation}
\Delta x(k)=a(k)x(k)\Big[
1-\prod_{i=1}^n \frac{x(k-\tau_i(k))}{H(k)}\Big]\label{e1.7}
\end{equation}
In addition, \eqref{e1.1} includes the periodic Michaelis-Menton
discrete model \cite{j4,k4,l2}
\begin{equation}
 \Delta x(k)=a(k)x(k)\Big[
1-\sum_{i=1}^{n}\frac{a_i(k)x(k-\tau_i(k))}{1+c_i(k)x(k-\tau_i(k))}\Big]
\label{e1.8}
\end{equation}
Similarly, model \eqref{e1.2} includes many ecological equations.
 See for example the discrete Hematopoiesis model \cite{j2,j3,j5,w1,w3}
\begin{equation}
\Delta x(k)=-a(k)x(k)+b(k)\exp\{-\beta(k)x(k-\tau(k))\}\label{e1.9}
\end{equation}
and the more general discrete models of blood cell production
\cite{g3,j2,j3,j5,l2,w1}
\begin{gather}
\Delta x(k)=-a(k)x(k)+\frac{b(k)}{1+x(k-\tau(k))^n},\quad
n\in Z_+\,, \label{e1.10} \\
\Delta x(k)=-a(k)x(k)+\frac{b(k)x(k-\tau(k))}{1+x(k-\tau(k))^n},
\quad n\in Z_+\label{e1.11}
\end{gather}
Model \eqref{e1.2} includes also the  discrete Nicholson's blowflies
model \cite{g5,j3,j6,w1,w3}
\begin{equation}
\Delta
x(k)=-a(k)x(k)+b(k)x(k-\tau(k))\exp\{-\beta(k)x(k-\tau(k))\}\label{e1.12}
\end{equation}

Studying the population dynamics, especial the existence
of positive periodic solutions, has attracted much attention from
both mathematicians and mathematical biologists recently.
Many authors have investigated the existence of positive periodic
solutions for several population models; see for example
\cite{a1,f1,j1,j2,j3,j5,k1,l2,m1,r1,w1,w2,w3,w4,y1,z1,z2}
and the references therein. In \cite{f1,j4,l2}, the
existence of  one positive periodic solution was
proved by using Mawhin's continuation theorem.
In \cite{g1,j2,j3,j5,w1,w2,y1}, the existence of  multiple positive periodic
solutions was studied by employing Krasnoselskii fixed point
theorem in cones.
The author in  \cite{y1} obtained  sufficient criteria for the existence of
multiple positive periodic solutions to \eqref{e1.1} and \eqref{e1.2},
in the continuous case by applying Krasnoselskii fixed point theorem.

To the best of the author's knowledge, there are very few works on
the existence of positive periodic solutions for \eqref{e1.1} and
\eqref{e1.2}. In \cite{z1} periodic solutions of a single species discrete
population model with periodic harvest/stock was discussed. The
authors in \cite{j1,m1,w2} studied the existence of positive periodic
solutions of some discrete equations, however, they are special
cases of \eqref{e1.1} and \eqref{e1.2}.

Motivated by the work above, in  the present paper, we aim to
study systematically the existence of positive periodic solution
of \eqref{e1.1} and \eqref{e1.2} under general conditions by employing the
Krasnoselskii fixed point theorem. The conditions in our main
theorem can easily be checked in practice.

For the sake of convenience and simplicity, we will apply the
below notations throughout this paper. Let
\begin{gather*}
f^M=\max_{k\in
I_\omega}f(k),\quad f^m=\min_{k\in I_\omega}f(k),\\
|u|=\max_{1\leq j\leq n}\{u_i\}, \quad
u\in \mathbb{R}_+^n, \quad
I_\omega=\{0,1,\dots,\omega-1\},
\end{gather*}
where $f$ is an
$\omega$-periodic function from $\mathbb{Z}$ to $\mathbb{R}$.

Assume the following limits exist and let
\begin{gather*}
\max f_0=\lim_{|u|\downarrow 0}\max_{k\in I_\omega}\frac{f(k,u)}{|u|},
\quad
\max f_\infty=\lim_{|u|\uparrow +\infty}\max_{k\in I_\omega}
\frac{f(k,u)}{|u|},
\\
\min f_0=\lim_{|u|\downarrow 0,\;
 u_j\geq \sigma|u|,\; 1\leq j\leq n}
 \min_{k\in I_\omega}\frac{f(k,u)}{|u|},
\\
 \min f_\infty=\lim_{|u|\uparrow +\infty,\;
 u_j\geq \sigma|u|,\; 1\leq j\leq n}
 \min_{k\in I_\omega}\frac{f(k,u)}{|u|}.
\end{gather*}
The general assumptions are stated as follows:
\begin{center}
\begin{tabular}{lclc}
(P1) &$\min f_0=\infty$ & (P2) &$\min f_\infty=\infty$ \\
(P3) &$\max f_\infty=0$ & (P4) &$\max f_0=0$ \\
(P5) &$\max f_0=\alpha_1\in[0,\frac{1}{B\omega})$
  &(P6) &$\min f_\infty=\beta_1\in (\frac{1}{A\sigma\omega},\infty)$\\
(P7) &$\min f_0=\alpha_2\in (\frac{1}{A\sigma\omega},\infty)$
  &(P8) &$\max f_\infty=\beta_2\in [0,\frac{1}{B\omega})$,
\end{tabular}
\end{center} %\label{e1.13}
with $A,B,\sigma$ to be defined below.
In addition, the parameters in this paper are assumed to be not
identically  equal to zero.

 To conclude this section, we state a few concepts and
 results that will be needed in  this paper.

\subsection*{Definition}
Let $X$ be Banach space and $E$ be a closed, nonempty
 subset.  $E$ is said to be a cone if
\begin{itemize}
\item[(i)] $\alpha u+\beta v\in E$ for all $u,v\in E$ and
 all $\alpha,\beta>0$
\item[(ii)] $u,-u\in E$ imply $u=0$
\end{itemize}

\begin{lemma}[Krasnoselskii fixed point theorem] \label{lem1.2}
Let $X$ be a Banach space,and let $E$ be a
cone in $X$. Suppose $\Omega_1$ and $\Omega_2$ are open subsets of
$X$ such that $0\in\Omega_1\subset\bar{\Omega}_1\subset\Omega_2$.
Suppose that
$$
T:E\cap(\bar{\Omega}_2\setminus\Omega_1)\to E
$$
is a completely continuous operator and satisfies either
\begin{itemize}
\item[(i)] $\|Tx\|\geq\|x\|$ for any $x\in E\cap\partial\Omega_1$
and $\|Tx\|\leq\|x\|$ for any $x\in E\cap\partial\Omega_2$; or
\item[(ii)] $\|Tx\|\leq\|x\|$
for any $x\in E\cap\partial\Omega_1$ and $\|Tx\|\geq\|x\|$ for any
$x\in E\cap\partial\Omega_2$.
\end{itemize}
Then $T$ has a fixed point in
$E\cap(\bar{\Omega}_2\setminus\Omega_1)$.
\end{lemma}

\section{Positive periodic solutions of \eqref{e1.1}}

 In this section, we establish sufficient criteria for the existence
  of positive periodic solutions to
  \eqref{e1.1}. We assume the following hypotheses:
\begin{itemize}
\item[(H1)] $a:{\mathbb{Z}}\to(0,+\infty)$ is continuous and
$\omega$-periodic , 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)] $f:\mathbb{Z}\times \mathbb{R}_+^n\to \mathbb{R}_+$
is continuous and
$\omega$-periodic with respect to the first
variable, i.e.,$f(k+\omega,u_1,\dots,u_n)=f(k,u_1,\dots,u_n)$
such that $f(k,u)\not\equiv 0$;

\item[(H3)] $h:\mathbb{Z}_+\to \mathbb{R}_+$ is continuous and satisfies
$\sum_{r=0}^{\infty}h(r)=1,g_i: \mathbb{Z}\to \mathbb{Z}$ is
continuous $\omega$-periodic function and satisfies $g_i(k)<k$.
\end{itemize}

Let
$X=\{x(k):x(k+\omega)=x(k)\}$,
$\| x \| = \max\{ |x(k)| : x\in X\}$, and
\[
 \sigma=[\prod_{r=0}^{\omega-1}(1+a(r))]^{-1}\,.
 \]
Then $X$ is a Banach space endowed with the norm $\|\cdot\|$.

To prove the existence of positive solutions to \eqref{e1.1}, we first
give the following lemmas.

\begin{lemma} \label{lem2.1}
If $x(k)$ is a positive $\omega$-periodic solution of \eqref{e1.1},
then $x(k)\geq\sigma\|x\|$.
\end{lemma}

 \begin{proof} It is clear that \eqref{e1.1} is equivalent to
 $$ x(k+1)=x(k)(a(k)+1)-f(k,u(k)), $$
and that it can be written as
$$
\Delta\Big( x(k)\prod_{s=0}^{k-1}\frac{1}{1+a(s)}\Big)
=-\prod_{s=0}^{k}\frac{1}{1+a(s)}f(k,u(k)),
$$
By summing the above equation from $k=n$ to $k=n+\omega-1$,
since $x(k)=x(k+\omega)$, we obtain
\begin{equation}
x(k)=\sum_{s=0}^{\omega-1}G(k,s)f(s,u(s)), \quad
k,s\in Z\,\label{e2.1}
\end{equation}
where
$$
G(k,s)=\frac{\prod_{r=s+1}^{k+\omega-1}(1+a(r))}
 {\prod_{r=0}^{\omega-1}(1+a(r))-1}, \quad
k\leq s\leq  k+\omega-1
$$
Then, $x(k)$ is an $\omega$-periodic solution of \eqref{e1.1}
if and only if $x(k)$ is an $\omega$-periodic solution of difference
equation \eqref{e2.1}.
A direct calculation shows that
\begin{equation}
A:=\frac{1}{\prod_{s=0}^{\omega-1}(1+a(s))-1}\leq G(k,s)\leq
 \frac{\prod_{s=0}^{\omega-1}(1+a(s))}{\prod_{s=0}^{\omega-1}(1+a(s))-1}
=:B\label{e2.2}
\end{equation}
Clearly
\begin{gather*}
A=\frac{\sigma}{1-\sigma},\quad
B=\frac{1}{1-\sigma},\quad \sigma=\frac{A}{B}<1,\\
\|x\|\leq B\sum_{k=0}^{\omega-1}f(s,u(s)), \quad
x(t)\geq A\sum_{k=0}^{\omega-1}f(s,u(s)).
\end{gather*}
Therefore,
$$
x(k)\geq  A\sum_{k=0}^{\omega-1}f(s,u(s))\geq\frac{A}{B}\|x\|
=\sigma\|x\|.
$$
\end{proof}

Define a mapping $T:X\to X$ by
\begin{equation}
 (Tx)(k)=\sum_{s=0}^{\omega-1}G(k,s)f(s,u(s)),\label{e2.3}
\end{equation}
for  $x\in X,k\in Z$. Clearly, $T$ is a
 continuous and completely continuous operator on $X$. Notice that
 finding a periodic solution of \eqref{e1.1} is equivalent to
 establishing a fixed point of operator $T$.

 Define
 $$
 E=\{x\in X:x(k)\geq 0, x(k)\geq \sigma\|x\|\}\,.
 $$
 One may readily verify that $E$ is a cone.

\begin{lemma} \label{lem2.2}
With the definitions above,  $TE\subset E$.
\end{lemma}

\begin{proof} In view of the arguments in the proof of
 Lemma \ref{lem2.1}, for each $x\in E$, we have
 $$
\|Tx\|\leq B\sum_{t=0}^{\omega-1}f(s,u(s)),
$$
By \eqref{e2.3}, one can  obtain
 $$
(Tx)(k)\geq A\sum_{k=0}^{\omega-1}f(s,u(s))\geq\frac{A}{B}\|Tx\|=\sigma\|Tx\|.
$$
 Therefore, $Tx\in E$. This completes the proof.
\end{proof}

 Now, we are in  the position to state the main
results in this section.

\begin{theorem} \label{thm2.3}
If {\rm (P1)} and {\rm (P3)} are satisfied,then \eqref{e1.1} has at
 least one positive
 $\omega$-periodic solution.
\end{theorem}

\begin{proof} By (P1), for any
$M_1>1/(A\sigma\omega)$, one can find a $r_0>0$ such that
\begin{equation}
f(k,u)\geq M_1|u|, \quad \text{for }u_j\geq\sigma|u|,1\leq j\leq n,|u|\leq r_0.
\label{e2.4}
\end{equation}
Let $\Omega_{r_0}=\{x\in X:\|x\|<r_0\}$. Note, if $x\in
E\cap\partial\Omega_{r_0}$ with $\|x\|=r_0$,
then $x(k)\geq \sigma \|x\|=\sigma r_0$.
So, from Lemma \ref{lem2.1} and $u(k)$ defined by \eqref{e1.3}, we obtain
\begin{gather*}
u_j(k)=x(g_j(k)) \geq \sigma\|x\|\geq \sigma|u|,\quad j=1,\dots,n-1,
\\
u_n(k)=\sum_{s=-\infty}^{k}h(k-s)x(s)\geq \sigma\|x\|
\sum_{s=-\infty}^{k}h(k-s)=\sigma\|x\|\geq \sigma|u|\,.
\end{gather*}
Then
$$
|u|=\max_{1\leq j\leq
n-1}\{x(g_j(k)),\sum_{s=-\infty}^{k}h(k-s)x(s)\}\geq\sigma\|x\|\geq
\sigma|u|.
$$
Therefore, by \eqref{e2.3} and \eqref{e2.4}, we have
$$
(Tx)(k)\geq A\sum_{s=0}^{\infty}f(s,u(s))
\geq AM_1\omega|u|\geq AM_1\sigma\omega r_0  \geq r_0=\|x\|.
$$
This implies that $\|Tx\|\geq\|x\|$ for any
$x\in  E\cap\partial\Omega_{r_0}$.
Again, by (P3), for any $0<\varepsilon\leq  1/(2B\omega)$,
there exists an $N_1>r_0$ such that
\begin{equation}
 f(k,u)\leq\varepsilon|u|,\quad\mbox{for }|u|\geq N_1. \label{e2.5}
\end{equation}
 Choose
 $$
r_1>N_1+1+2B\omega\max\{f(k,u): k\in I_\omega,\; |u|\leq N_1,\;
u\in \mathbb{R}_+^n \} .
$$
Let $\Omega_{r_1}=\{x\in X:\|x\|<r_1\}$.
If $x\in E\cap\partial\Omega_{r_1}$, then
\begin{align*}
(Tx)(k)
&\leq B\sum_{s=0}^{\omega-1}f(s,u(s))\\
&\leq B\sum_{s=0,\;  |u(s)|\leq N_1}^{\omega-1}f(s,u(s))
 +B\sum_{s=0,\; |u(s)|> N_1}^{\omega-1}f(s,u(s))\\
&\leq \frac{r_1}{2}+B\omega\varepsilon\|x\|\\
&\leq \frac{r_1}{2}+\frac{\|x\|}{2}=\|x\|.
\end{align*}
 This implies that $\|Tx\|\leq\|x\|$ for any $x\in  \partial\Omega_{r_1}$.

In conclusion, under the assumptions (P1) and (P3), $T$
satisfies all the requirements in Lemma \ref{lem1.2}.
Then $T$ has a fixed point $E\cap(\bar{\Omega}_{r_1}\setminus  \Omega_{r_0})$.
 Clearly, we have $r_0\leq\|x\|\leq r_1$ and
$x(k)\geq\sigma\|x\|\geq\sigma r_0>0$, which shows that $x(k)$ is a
positive $\omega$-periodic  solution of \eqref{e2.1}. By Lemma \ref{lem2.1},
$x(k)$ is a positive $\omega$-periodic  solution of \eqref{e1.1}.
This completes the proof.
\end{proof}

\begin{theorem} \label{thm2.4}
If {\rm (P2)} and {\rm (P4)}  are satisfied, then \eqref{e1.1}
has at least one positive $\omega$-periodic solution.
\end{theorem}

\begin{proof} By (P4), for any $0<\varepsilon\leq1/(B\omega)$,
there exists $r_2>0$, such that
\begin{equation}
f(k,u)\leq\varepsilon |u|,\quad\text{for }|u|\leq r_2.
\label{e2.6}
\end{equation}
Define $\Omega_{r_2}=\{x\in X: \|x\|<r_2\}$. If
$x\in E\cap\partial\Omega_{r_2}$, then by \eqref{e2.2}, \eqref{e2.3}
and \eqref{e2.6}, we have
$$
(Tx)(k)\leq B\sum_{s=0}^{\omega-1}f(s,u(s))\leq
B\varepsilon|u|\omega\leq B\varepsilon\|x\|\omega \leq\|x\|.
$$
In particular, $\|Tx\|\leq\|x\|$, for all
$x\in  E\cap\partial\Omega_{r_2}$.
Next, by (P2), for any $M_2\geq 1/(A\sigma\omega)$, there
exists a $r_3>\frac{r_2}{\sigma}$ such that
\begin{equation}
f(k,u)\geq M_{2}|u|\quad\text{for }u_j\geq\sigma|u|,\quad
 1\leq j\leq n, |u|\geq\sigma r_3.
\label{e2.7}
\end{equation}
Define $\Omega_{r_3}=\{x\in X: \|x\|<r_3\}$.
 If $x\in E\cap\partial\Omega_{r_3}$, then
\begin{gather*}
 u_j(k)=x(g_j(k))\geq
\sigma\|x\|=\sigma r_3\geq\sigma|u| \quad j=1,\dots ,n-1,\\
 u_n(k)=\sum_{s=-\infty}^{k}h(k-s)x(s)\geq
\sigma\|x\|=\sigma r_3\geq\sigma|u|,\\
|u(k)|=\max_{1\leq j\leq
n-1}\{x(g_j(k)),\sum_{s=-\infty}^{k}h(k-s)x(s)d\}\geq\sigma r_3,
\end{gather*}
Therefore, by \eqref{e2.2}, \eqref{e2.3}, and \eqref{e2.7}, we get
$$
(Tx)(k)\geq A\sum_{s=0}^{\omega-1}f(s,u(s)) \geq
AM_2\sigma\|x\|\omega \geq \|x\|.
$$
In particular, $\|Tx\|\geq\|x\|$ for all
$x\in E\cap\partial{\Omega_{r_3}}$. By Lemma \ref{lem1.2}, there exists a
fixed point  $x\in E\bigcap(\bar{\Omega}_{r_3}\setminus\Omega_{r_2})$
satisfying $r_2\leq\|x\|\leq r_3 $. That is, $x(k)$ is a positive
$\omega$-periodic solution of \eqref{e1.1}.
\end{proof}

Now, we  introduce two extra
assumptions to be used in the next theorems.
\begin{itemize}
\item[(P9)] There exists  $d_1>0$ such that
$f(k,u)>{d_1}/{(A\omega)}$, for $|u|\in[\sigma d_1,d_1]$.

\item[(P10)] There exists $d_2>0$ such that
$f(k,u)<{d_2}/{(B\omega)}$, for $|u|\leq d_2$.
\end{itemize}

\begin{theorem} \label{thm2.5}
If {\rm (P3), (P4), (P9)} are satisfied, then \eqref{e1.1} has at
least two positive $\omega$-periodic solutions  $x_1$ and $x_2$
 satisfying $0<\|x_1\|<d_1<\|x_2\|$.
\end{theorem}

\begin{proof} By assumption (P4), for any
$0<\varepsilon\leq 1/(B\omega)$, there exists a $r_4<d_1$ such
that
\begin{equation}
f(k,u)\leq\varepsilon |u|,\quad\text{for }|u|\leq r_4
\label{e2.8}
\end{equation}
Define $\Omega_{r_4}=\{x\in X:\|x\|<r_4\}$. Then for any
$x\in E\cap\partial\Omega_{r_4}$, we have $\|x\|=r_4$. From
\eqref{e2.2}, \eqref{e2.3} and \eqref{e2.8}, we obtain
$$
(Tx)(k) \leq B\sum_{s=0}^{\omega-1}f(s,u(s)) \leq
B\varepsilon r_4\omega \leq r_4 =\|x\|,
$$
which implies $\|Tx\|\leq\|x\|$ for all
$x\in E\cap\partial\Omega_{r_4}$.
Likewise, from (P3), for any $0<\varepsilon\leq 1/(2B\omega)$,
there exists an $N_2>d_1$ such that
\begin{equation}
f(k,u)\leq\varepsilon |u|,\quad\text{for }|u|\geq N_2.
\label{e2.9}
\end{equation}
Choose
\begin{equation}
r_5>N_2+1+2B\omega \max\{f(k,u) :k\in I_\omega,\; |u|\leq N_2,\;
u\in \mathbb{R}_+^n\}  \label{e2.10}
\end{equation}
Let $\Omega_{r_5}=\{x\in X:\|x\|<r_5\}$. If $x\in
E\cap\partial\Omega_{r_5}$, then by \eqref{e2.3}, \eqref{e2.4},
 and \eqref{e2.10}, we have
\begin{align*}
(Tx)(k)&\leq B\sum_{s=0}^{\omega-1}f(s,(u(s))\\
&= B\sum_{s=0,\; |u(s)|\leq N_2}^{\omega-1} f(s,u(s))
+ B\sum_{s=0,\;  |u(s)|> N_2}^{\omega-1} f(s,u(s))\\
&\leq \frac{r_5}{2}+\frac{\|x\|}{2}=\|x\|.
\end{align*}
Which shows that $\|Tx\|\leq \|x\|$ for all
$x\in E\cap\partial\Omega_{r_5}$.

Set $\Omega_{d_1}=\{x\in X:\|x\|<d_1\}$. Then, for any $x\in
E\cap\partial\Omega_{d_1}$, we have $x(k)\geq\sigma\|x\|=\sigma
d_1$.
Consequently,
\begin{gather*}
u_j(k)=x(g_j(k))\geq \sigma\|x\|=\sigma d_1 \quad j=1,\dots, n-1,
\\
 u_n(k)=\sum_{s=-\infty}^{k}h(k-s)x(s)\geq
\sigma\|x\|=\sigma d_1,
\end{gather*}
That is
$$
|u(k)|=\max_{1\leq j\leq
n-1}\{x(g_j(k)),\sum_{s=-\infty}^{k}h(k-s)x(s)\}\geq\sigma
d_1.
$$
Thus, by \eqref{e1.3}, \eqref{e2.3}, (P9), we have
$$
(Tx)(k)\geq A\sum_{s=o}^{\omega-1}f(s,u(s)) >A\frac{d_1}
{A\omega}\omega =d_1=\|x\|.
$$
This yields $\|Tx\|>\|x\|$ for all $x\in
E\cap\partial\Omega_{d_1}$. By Lemma \ref{lem1.2}, there exist two
positive $\omega$-periodic solutions  $x_1$ and $x_2$ satisfying
$0<\|x_1\|<d_1<\|x_2\|$. This completes the proof.
\end{proof}

 From the arguments in the above proof, we
have the following result immediately.

\begin{corollary} \label{coro2.6}
If {\rm (P1), (P2), (P10)} are satisfied, then \eqref{e1.1} has at
least two $\omega$-periodic solutions $ x_1$ and $x_2$ satisfying
$0<\|x_1\|<d_2<\|x_2\|$.
\end{corollary}

To obtain better results in this section, we give a
more general criterion in the following, which plays an important
role later.

\begin{theorem} \label{thm2.7}
Suppose that {\rm (P9)} and {\rm (P10)}  are satisfied,
then \eqref{e1.1} has at least one positive
$\omega$-periodic solution $x$ with $\|x\|$ lying between  $d_2$ and
$d_1$, where $d_1$ and $d_2$ are defined in {\rm(P9)} and
{\rm (P10)}, respectively.
\end{theorem}

\begin{proof} Without loss of generality, we
 assume that $d_2<d_1$. Set $\Omega_{d_2}=\{x\in X:\|x\|<d_2\}$.
 If $x\in E\cap\partial\Omega_{d_2}$, then from
\eqref{e2.2}, \eqref{e2.3} and (P10), we  get
$$
(Tx)(k)\leq B\sum_{s=0}^{\omega-1}f(s,u(s))
<B\frac{d_2}{B\omega}\omega =d_2 =\|x\|,
$$
In particular, $\|Tx\|<\|x\|$ for all $x\in E\cap\partial\Omega_{d_2}$.


Choose $\Omega_{d_1}=\{x\in X:\|x\|<d_1\}$. For any
$x\in E\cap\partial\Omega_{d_1}$, we have
$x(k)\geq\sigma\|x\|=\sigma d_1 $. Thus,
\begin{gather*}
\sigma d_1=\sigma\|x\|\leq |u|=\max_{1\leq j\leq
n-1}\{x(g_j(k)),\sum_{s=-\infty}^{k}h(k-s)x(s)\}\leq\|x\|=d_1, \\
u_j\geq\sigma\|x\|\geq\sigma|u| \quad (j=1, 2,\dots, n).
\end{gather*}
 From \eqref{e2.3} and (P9), one has
 $$
(Tx)(k)\geq A\sum_{s=0}^{\omega-1}f(s,u(s))
> A\frac{d_1}{A\omega}\omega =d_1 =\|x\|.
$$
This implies $\|Tx\|>\|x\|$ for all
$x\in E\cap\partial\Omega_{d_1}$. Therefore, by Lemma \ref{lem1.2}, we can
obtain the conclusion and this completes the proof.
\end{proof}

\begin{theorem} \label{thm2.8}
If {\rm (P5)} and
{\rm (P6)} are satisfied, then \eqref{e1.1} has at least one positive
$\omega$-periodic solution.
\end{theorem}

\begin{proof} By  assumption (P5), for
any $\varepsilon=\frac{1}{B\omega}-\alpha_1>0$, there exists a
sufficiently small $d_2>0$ such that
$$
\max_{k\in I_\omega}\frac{f(k,u)}{|u|}<\alpha_1+\varepsilon=\frac{1}{B\omega},
\quad\text{for }|u|\leq d_2;
$$
that is,
$$
f(k,u)<\frac{1}{B\omega}|u|\leq\frac{d_2}{B\omega}\quad\text{for }
|u|\leq d_2,\; k\in I_\omega
$$
  So, (P10) is satisfied.
By the assumption (P6), for
$\varepsilon=\beta_1-\frac{1}{A\sigma\omega}>0$, there exists a
sufficiently large $d_1>0$  such that
$$
\min_{k\in I_\omega}\frac{f(k,u)}{|u|}>\beta_1-\varepsilon
=\frac{1}{A\sigma\omega}, \quad\text{for }
|u|\geq\sigma d_1,\; u_j\geq\sigma |u|,
$$
Which leads to
$$
f(k,u)>\frac{1}{A\sigma\omega}\sigma d_1=\frac{d_1}{A\omega},
\quad\text{for }|u|\in [\sigma d_1, d_1], u_j\geq\sigma|u|
$$
That is, (P9) holds. By Theorem \ref{thm2.7} we complete the proof.
\end{proof}

\begin{theorem} \label{thm2.9}
If {\rm (P{7})} and {\rm (P8)} are satisfied,
then \eqref{e1.1} has  at least one positive $\omega$-periodic
solution.
\end{theorem}

 \begin{proof} By  assumption  (P7), for
any $\varepsilon= \alpha_2-\frac{1}{A\sigma\omega}>0$,
there exists a sufficiently small $d_1>0$ such that
$$
\min_{k\in I_\omega}\frac{f(k,u)}{|u|}>\alpha_2-\varepsilon
=\frac{1}{A\sigma\omega}\quad\text{for }0\leq |u|\leq
d_1,\; u_j\geq\sigma |u|
$$
Therefore,
$$
f(k,u)>\frac{1}{A\sigma\omega}\sigma
d_1=\frac{d_1}{A\omega}\quad\text{for }|u|\in[\sigma d_1,d_1],
u_j\geq\sigma|u|
$$
for $j=1, 2,\dots, n$ and $k\in I_\omega$;
that is, (P9) holds.
By  assumption (P8), for
$\varepsilon=\frac{1}{B\omega} -\beta_2>0$, there
exists a sufficiently large $d$  such that
\begin{equation}
\max_{k\in I_\omega}\frac{f(k,u)}{|u|}<\beta_2+\varepsilon
=\frac{1}{B\omega}\quad\text{for }|u|>d.
 \label{e2.13}
\end{equation}
In the following, we consider two cases to prove (P10):
$\max_{k\in I_\omega}f(k,u)$ bounded and
unbounded. The bounded case is clear.
If $\max_{k\in I_\omega}f(k,u)$ is
unbounded, then there exists $u^{*}\in\mathbb{R}^n_+, |u^{*}|=d_2>d$
and $k_0\in I_\omega$ such that
\begin{equation}
f(k,u)\leq
f(k_0,u^{*})\quad\text{for }0<|u|\leq |u^{*}|=d_2.\label{e2.14}
\end{equation}
Since
$|u^{*}|=d_2>d$, by \eqref{e2.13} and \eqref{e2.14}, we have
$$
f(k,u)\leq f(k_0,u^{*})<\frac{1}{B\omega}|u^{*}|
=\frac{d_2}{B\omega}\quad\text{for }0<|u|\leq d_2, k\in I_\omega
$$
Which implies  condition (P10) holds. Therefore, using
Theorem \ref{thm2.7} we complete the proof.
\end{proof}


\begin{theorem} \label{thm2.10}
 Suppose that {\rm (P6), (P{7}), (P10)} are satisfied,
then \eqref{e1.1} has at least two positive  $\omega$-periodic
solutions $x_1$ and $x_2$ satisfying $0<\|x_1\|<d_2<\|x_2\|$,
 where $d_2$ is defined in {\rm (P10)}.
 \end{theorem}

\begin{proof} From (P6) and the proof of
Theorem \ref{thm2.8}, we know that there exists a sufficiently large
$d_1>d_2$, such that $f(k,u)>{d_1}/{(A\omega)}$ for
$|u|\in[\sigma d_1,d_1]$, $u_j\geq\sigma|u|$ ($j=1, 2,\dots, n$).
 From (P7) and  the proof of Theorem \ref{thm2.9}, we can find a
sufficiently small $d_1^{*}\in(0,d_2)$ such that
$f(k,u)>{d_1^{*}}/{(A\omega)}$ for $|u|\in[\sigma d_1^{*},d_1^{*}]$,
$u_j\geq\sigma|u|$ ($j=1, 2,\dots, n$). Therefore, from the proof of
Theorem \ref{thm2.7}, there exists two positive solutions $x_1$ and $x_2$
satisfying $d_1^{*}<\|x_1\|<d_2<\|x_2\|<d_1$.
 \end{proof}

 From the arguments in the above proof, we have the following
statement.

\begin{corollary} \label{coro2.11}
Suppose that {\rm (P5)}, {\rm (P8)} and  {\rm (P9)} are satisfied,
then \eqref{e1.1} has at least two positive  $\omega$-periodic
solutions $x_1$ and $x_2$ satisfying $0<\|x_1\|<d_2<\|x_2\|$,
where $d_1$ is defined in
{\rm (P9)}.
\end{corollary}


\begin{theorem} \label{thm2.12}
If {\rm (P1)} and {\rm (P8)} are satisfied, then \eqref{e1.1}
has at least one positive $\omega$-periodic solution.
\end{theorem}


\begin{proof} Let $\Omega_{r_0}=\{x\in X:\|x\|<r_0\}$.
 From  assumption (P1) and the proof of
Theorem \ref{thm2.5}, we know that $\|Tx\|\geq\|x\|$ for all
$x\in E\cap\partial\Omega_{r_0}$.
Choose $\Omega_{r_1}=\{x\in X:\|x\|<r_1\}$. From (P8) and
Theorem \ref{thm2.9}, as $|u|\leq r_1$,
we know that $ f(k,u)<\frac{r_1}{B\omega} $
and
$$
(Tx)(k)\leq B\sum_{s=0}^{\omega-1}f(s,u(s))
<B\frac{r_1}{B\omega}\omega =r_1=\|x\|.
$$
Which implies $\|Tx\|<\|x\|$ for all $x\in E\cap\partial\Omega_{r_1}$.
 This completes the proof.
 \end{proof}

 Similar to Theorem \ref{thm2.12}, one immediately has the
following statements.

\begin{theorem} \label{thm2.13}
If {\rm (P2)} and {\rm (P5)} are satisfied, then \eqref{e1.1}
 has at least one positive $\omega$-periodic solution.
\end{theorem}

\begin{theorem} \label{thm2.14}
If {\rm (P3)} and {\rm (P7)} are satisfied, then
\eqref{e1.1} has at least one positive $\omega$-periodic solution.
\end{theorem}

\begin{theorem} \label{thm2.15}
If {\rm (P4)} and {\rm (P6)} are satisfied, then
\eqref{e1.1} has at least one positive $\omega$-periodic solution.
\end{theorem}

\begin{theorem} \label{thm2.16}
If {\rm (P1)}, {\rm (P6)}
and {\rm (P10)} are satisfied, then \eqref{e1.1} has at least two
positive $\omega$-periodic solutions $x_1$ and $x_2$ satisfying
$0<\|x_1\|< d_2<\|x_2\|$, where $d_2$ is defined in
{\rm (P10)}.
\end{theorem}

\begin{proof} Let $\Omega_{r_*}=\{x\in X:\|x\|< r_*\}$, where $r_*<d_2$.
By  assumption (P1) and
the proof of Theorem \ref{thm2.3}, we know $\|Tx\|\geq\|x\|$ for all
$x\in E\cap\partial\Omega_{r_*}$.
Let $\Omega_{d_1}=\{x\in X:\|x\|<d_1\}$. By the assumption (P6)
and the proof of Theorem \ref{thm2.7}, we see that
$f(k,u)>\frac{d_1}{A\omega}$ for $|u|\in[\sigma d_1, d_1]$.
 From (P10) and the proof of Theorem \ref{thm2.7}, we know that
there exist two positive $\omega$-periodic solutions $x_1$ and $x_2$
satisfying $0<\|x_1\|< d_2<\|x_2\|$.
\end{proof}

\begin{theorem} \label{thm2.17}
If {\rm (P2), (P7), (P10)} are satisfied, then \eqref{e1.1} has
at least two positive $\omega$-periodic solutions $x_1$ and $x_2$
satisfying $0<\|x_1\|< d_2<\|x_2\|$, where $d_2$ is defined in
{\rm (P10)}.
\end{theorem}

\begin{theorem} \label{thm2.18}
If {\rm (P3), (P5), (P9)} are satisfied, then \eqref{e1.1}
has at least two positive $\omega$-periodic solutions $x_1$ and
$x_2$ satisfying
$0<\|x_1\|< d_1<\|x_2\|$, where $d_1$ is defined in
{\rm (P9)}.
\end{theorem}


\begin{theorem} \label{thm2.19}
If {\rm (P4), (P8), (P9)} are satisfied, then \eqref{e1.1} has at
least two positive $\omega$-periodic solutions $x_1$ and $x_2$
satisfying $0<\|x_1\|<d_1<\|x_2\|$, where $d_1$ is defined in
{\rm (P9)}.
\end{theorem}

 \section{Existence of periodic solution to \eqref{e1.2}}

In this section, we prove  the existence of positive
 periodic solution to \eqref{e1.2}. By carrying out
similar arguments as in Section 2, one can easily obtain sufficient
criteria for the existence of positive periodic solutions of
\eqref{e1.2}.
We assume that  (H2) and (H3) hold. Moreover we will use the assumption
\begin{itemize}
\item[(H1')] $a:\mathbb{Z}\to (0,1)$ 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.
\end{itemize}
Define
\begin{gather}
\sigma=\prod_{s=0}^{\omega-1}(1-a(s)), \label{e3.1}
\\
H(k,s)=\frac{\prod_{r=s+1}^{k+\omega-1}(1-a(r))}{
1-\prod_{r=0}^{\omega-1}(1-a(r))},\; k,s\in Z,
\end{gather}
From the definition of $H(k,s)$, for any $ s\in[k,k+\omega-1]$, we have
\begin{equation}
A:=\frac{\prod_{s=0}^{\omega-1}(1-a(s))}{
1-\prod_{s=0}^{\omega-1}(1-a(s))}\leq H(k,s)\leq
\frac{1}{1-\prod_{s=0}^{\omega-1}(1-a(s))}:=B \label{e3.2}
\end{equation}
Clearly,
$$
 A=\frac{\sigma}{1-\sigma},\quad
 B=\frac{1}{1-\sigma},\quad
 \sigma=\frac{A}{B}<1.
$$

\begin{lemma} \label{lem3.1}
$x(k)$ is an $\omega$-periodic solution of \eqref{e1.2} if and only if it
 is an $\omega$-periodic solution of the  difference equation
\begin{equation}
x(k)=\sum_{s=k}^{k+\omega-1}H(k,s)f(s,u(s)).
\label{e3.3}
\end{equation}
\end{lemma}

Similarly, we can establish sufficient criteria
for the existence of periodic solutions of \eqref{e1.2}. Now we list the
corresponding criteria without proof.

\begin{theorem} \label{thm3.2}
Suppose that one of the following pairs of conditions holds:
{\rm (P1)} and {\rm (P3)},
or {\rm (P1)} and {\rm (P8)},
or {\rm (P2)} and {\rm (P4)},
or {\rm (P2)} and {\rm (P5)},
or {\rm (P3)} and {\rm (P7)},
or {\rm (P4)} and {\rm (P6)},
or {\rm (P5)} and {\rm (P6)},
or {\rm (P7)} and {\rm (P8)},
or {\rm (P9)} and {\rm (P10)}.
Then \eqref{e1.2} has at least one
positive $\omega$-periodic solution.
\end{theorem}

\begin{theorem} \label{thm3.3}
Suppose that {\rm (P9)}holds and one of the following
pairs of conditions is satisfied:
{\rm (P3)} and {\rm (P4)},
or {\rm (P3)} and {\rm (P5)},
or {\rm (P4)} and {\rm (P8)},
or {\rm (P5)} and {\rm (P8)}.
Then \eqref{e1.2} has at least two $\omega$-periodic solutions
$x_1$ and $x_2$ satisfying $0<\|x_1\|<d_1<\|x_2\|$, where $d_1$ is
defined in {\rm (P9)}.
\end{theorem}


\begin{theorem} \label{thm3.4}
Suppose that {\rm (P10)} and one of the following pairs of conditions
is satisfied:
{\rm (P1)} and {\rm (P2)},
or {\rm (P1)} and {\rm (P6)},
or {\rm (P2)} and {\rm (P7)},
or {\rm (P6)} and {\rm (P7)}.
Then \eqref{e1.2} has at least two
$\omega$-periodic solutions  $x_1$ and $x_2$ satisfying
$0<\|x_1\|<d_2<\|x_2\|$, where $d_2$ is defined in {\rm (P10)}
\end{theorem}

\begin{remark} \label{rmk3.5} \rm
In this section, the values of $A,B,\sigma$ in (P1)--(P8) should be
replaced by corresponding values  defined in \eqref{e3.1} and \eqref{e3.2}.
\end{remark}

\section{Examples and numerical simulations}

 In  this section, we apply our main results to investigate some
classical biological models and test our
 theoretical results.

\begin{theorem} \label{thm4.1}
Assume that $a(k),c(k),b_i(k)\in C(Z,(0,+\infty))$,
$\tau_i(k)\in C(Z,Z)$ ($i=1,\dots,n$) are all $\omega$-periodic,
then \eqref{e1.4} has at least one positive $\omega$-periodic solution.
\end{theorem}

\begin{proof} Note that
$$
f(k,u)=u_0(\sum_{i=1}^{n}b_i(k)u_i+c(k)u_{n+1}), \quad
u=(u_0,\dots,u_{n+1})\in\mathbb{R}_+^{n+2}.
$$
Clearly, conditions (H1)--(H3) are satisfied. Moreover, when
$|u|\to 0$, we have
$$
\max_{k\in I_\omega}\frac{|f(k,u)|}{|u|}
\leq(\sum_{i=1}^{n}b_i^M+c^M)|u|\to 0,
$$
Hence, $\max f_0=0$. In addition, if $u\in\mathbb{R}_+^{n+2}$ and
$u_i>\sigma |u|$, then
$$
\min_{k\in I_\omega}\frac{|f(k,u)|}{|u|}\geq
\sigma^2(\sum_{i=1}^{n}b_i^m+c^m)|u|\to +\infty,
\quad\text{as }|u|\to +\infty\,;
$$
that is, $\min f_{\infty}=\infty$. Therefore {\rm (P2)} and
{\rm (P4)} are  satisfied. The claim follows from Theorem \ref{thm2.4}.
 \end{proof}

 \begin{corollary} \label{coro4.2}
Assume that  $a(k),b(k),H(k)\in C(Z,(0,+\infty))$,
$\tau_i(k)\in C(Z,Z)$ ($i=1,\dots,n$)
are all $\omega$-periodic, then \eqref{e1.5} has at least  one
positive $\omega$-periodic solution.
\end{corollary}

 \begin{corollary} \label{coro4.3}
Assume that   $a(k),b(k),H(k)\in C(Z,(0,+\infty))$,
$\tau_i(k)\in C(Z,Z)$ ($i=1,\dots,n$)
are all  $\omega$-periodic, then \eqref{e1.6} has at least  one
positive $\omega$-periodic solution.
\end{corollary}

\begin{corollary} \label{coro4.4}
Assume that   $a(k), H(k)\in C(Z,(0,+\infty))$,
$\tau_i(k)\in C(Z,Z)$ ($i=1,\dots,n$)
 are all $\omega$-periodic, then  \eqref{e1.7} has at least  one positive
$\omega$-periodic solution.
\end{corollary}

\begin{theorem} \label{thm4.5}
Assume that $a(k),a_i(k), c_i(k)\in C(Z,(0,+\infty))$,
$\tau_i(k)\in C(Z,Z)$ ($i=1,\dots,n$) are all $\omega$-periodic.
Also assume that
$$
a^m \sum_{i=1}^{n}\frac{a_i^m}{c_i^M}>\frac{1-\sigma}{\sigma^2\omega},
$$
where $\sigma=\prod_{k=0}^{\omega-1}(1+a(k))^{-1}$.
Then \eqref{e1.8} has at least  one positive $\omega$-periodic solution.
\end{theorem}

\begin{proof} Note that
$$
f(k,u)=a(k)u_0\sum_{i=1}^{n}\frac{a_i(k)u_i}{1+c_i(k)u_i},
$$
It is clear that (H1)--(H3) are satisfied. Moreover,
$$
\max_{k\in I_\omega}\frac{|f(k,u)|}{|u|}
\leq a^M\sum_{i=1}^{n}\frac{a_i^M|u|}{1+c_i^m|u|}\to 0, \quad
\text{as }|u|\to 0\,.
$$
Thus $\max f_0=0$. In addition, when $|u|\to +\infty$, we have
$$
\min_{k\in I_\omega}\frac{|f(k,u)|}{|u|}\geq
a^m \sum_{i=1}^{n}\frac{a_i^m|u|}{1+c_i^M|u|}\to
a^m \sum_{i=1}^{n}\frac{a_i^m}{c_i^M},
$$
 By the assumption, one has $\min f_{\infty}>\frac{1}{A\sigma\omega}$.
 Therefore, by Theorem \ref{thm2.15},the proof is completed.
\end{proof}

\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.7\textwidth]{fig1}
\end{center}
\caption{Existence of periodic solutions of \eqref{e4.1}}
\end{figure}

In [\cite{z1}, the authors studied the periodic solution of a single species
discrete population model  with periodic harvest. Their model is
$$
x(k+1)=\mu
x(k)\big[1-\frac{x(k)}{T}\big]+b(k), \quad k\in \mathbb{Z},
$$
where $\mu>0,T>0,b(t)\in C(Z,R),a(k)= a(k+\omega)$. Now we
consider
$$
\Delta x(k)=x(k)\big[ a(k)-\frac{b(k)x(k)}{1+c x(k)}\big],
$$
whose  growth law obeys Michaelis-Menton type growth equation.
Moreover, we assume   that the population subjects to harvesting.
Under the catch-per-unit-effort hypothesis, the harvest population's
growth equation can be written as
\begin{equation}
\Delta x(k)=x(k)\big[ a(k)-\frac{b(k)x(k)}{1+c
x(k)}\big]-qEx(k),\label{e4.1}
\end{equation}
where $a(k),b(k)\in C(Z,(0,+\infty))$ are $\omega$-periodic,
$c$ is positive constant,
$q$ and $E$ are positive constants denoting the
catch-ability-coefficient and the harvesting effort,respectively.

\begin{theorem} \label{thm4.6}
If
$$
0<qE<\frac{1-\sigma}{\omega},\quad \frac{b^m}{c}+qE>\frac{1-\sigma}{\sigma^2\omega},
$$
Then \eqref{e4.1} has at least one positive $\omega$-periodic solution,
where
$$
\sigma=\prod_{k=0}^{\omega-1}(1+a(k))^{-1}.
$$
\end{theorem}

\begin{proof} Note that
$$
f(k,u)=\frac{b(k)u^2}{1+cu}+qEu,\quad u\geq 0.
$$
It is not difficult to show that
$$
\max f_0=qE,\quad \min f_{\infty}=\frac{b^m}{c}+qE.
$$
The conditions in Theorem \ref{thm4.6} guarantee that {\rm (P5)} and
{\rm (P6)} hold. Then by Theorem \ref{thm3.2}, the proof is
complete.
\end{proof}

\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.7\textwidth]{fig2}
\end{center}
\caption{Existence of periodic solutions of \eqref{e1.9}}
\end{figure}


\begin{theorem} \label{thm4.7}
 Assume that $a(k)\in C(Z,(0,1)),b(k),\beta(k)\in C(Z,(0,\infty))$,
 $\tau\in C(Z,Z)$ are all $\omega$-periodic, then
\eqref{e1.9} has at least one positive $\omega$-periodic solution.
\end{theorem}

\begin{proof} It is clear that
(H1')--(H3) are satisfied. Note
that
$$
f(k,u)=b(k)\exp\{-\beta(k)u\},\quad u\in \mathbb{R}_+
$$
Then
\begin{gather*}
 \min_{k\in I_\omega]}\frac{|f(k,u)|}{|u|}=\min_{k\in I_\omega}
 \frac{b(k)}{u\exp\{\beta(k) u\}}\geq\frac{b^m}{u\exp\{\beta^M u\}}\to+\infty,
 \quad u\to 0,\\
 \max_{k\in I_\omega}\frac{|f(k,u)|}{|u|}\leq\frac{b^M}{u\exp\{\beta^m u\}}\to
0,\quad u\to +\infty.
\end{gather*}
Thus, $\min f_0=\infty$ and $\max f_{\infty}=0$. Thus
(P1) and (P3) are satisfied. Theorem \ref{thm2.3} proves the claim.
\end{proof}

\begin{example} \label{ex4.8} \rm
Consider the system \eqref{e4.1} with $a(k)=1+\sin(k\pi )$,
$ b(k)=2+\cos(k\pi)$, $qE=0.2$, $c=1$.
Then a sketch of the existence of periodic solutions is shown in
figure 1.
\end{example}

\begin{example} \label{ex4.9} \rm
Consider again the system \eqref{e1.9} with
$a(k)=\frac{1}{8}+\frac{1}{16}\sin \frac {k\pi}{2}$,
$b(k)=\frac{1}{16}+\frac{1}{32}\cos\frac{k\pi}{2}$,
$\beta(k)\equiv 1$, $\tau(k)\equiv 1$. Periodic
solutions of \eqref{e1.9} is shown in figure 2.
\end{example}

\begin{theorem} \label{thm4.10}  Assume that
$a(k)\in C(Z,(0,1))$, $b(k)\in C(Z,(0,+\infty))$ are all
$\omega$-periodic. Moreover, $b^m>\frac{1-\sigma}{\sigma^2\omega}$,
then \eqref{e1.11} has at least  one positive $\omega$-periodic solution.
where $\sigma=\prod_{k=0}^{\omega-1}(1+a(k))^{-1}$. 
\end{theorem}

\begin{proof} Note that
$f(t,u)=b(k)u/(1+u^n)$. Then
$$
\max_{k\in I_\omega}\frac{|f(k,u)|}{|u|}=\frac{b^M}{1+u^n},\quad
\min_{k\in I_\omega}\frac{|f(k,u)|}{|u|}=b^m, \quad
u\geq 0.
$$
In view of  $b^m>\frac{1-\sigma}{\sigma^2\omega}$, then
$\max f_\infty=0$, $\min f_0=b^m>\frac{1}{A\sigma\omega}$.
Therefore, by Theorem \ref{thm3.2}, we conclude that \eqref{e1.11} has
at least one positive $\omega$-periodic solution.
\end{proof}

\begin{corollary} \label{coro4.11}
Assume that $a(k)\in C(Z,(0,1))$, $b(k)\in C(Z,(0,+\infty))$ are all
$\omega$-periodic, then \eqref{e1.10} has at least one positive
$\omega$-periodic solution.
\end{corollary}

\begin{theorem} \label{thm4.12}
Assume that $a(k)\in C(Z,(0,1))$, $b(k),\beta(k)\in C(Z,(0,\infty))$,
$\tau(k)\in C(Z,Z)$ are all $\omega$-periodic.
If $b^m>\frac{1-\sigma}{\sigma^2\omega}$, then \eqref{e1.12} has at
least one positive $\omega$-periodic solution.
\end{theorem}

The proof is exactly the same as that of Theorem \ref{thm4.10}; so we
omit it here.

\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.8\textwidth]{fig3}
\end{center}
\caption{Periodic solutions of \eqref{e1.11}}
\end{figure}

\begin{example} \label{ex4.13} \rm
Consider the difference equation
\begin{equation}
 \Delta x(k)=-\frac{7}{8}\sin\frac{k\pi}{2}
x(k)+\big(\frac{1}{16}+\frac{1}{32}\cos\frac{k\pi}{2}
\big)(x^{\alpha}(k)+x^{\beta}(k)), \label{e4.2}
\end{equation}
where $0<\alpha<1$ and $\beta>1$ are constants.
It is cleat that
\begin{gather*}
f(k,u)=\big(\frac{1}{16}+\frac{1}{32}\cos(\frac{k\pi}{2}
)\big)(u^{\alpha}+u^{\beta}),\quad u\geq 0
\\
\min_{k\in [0,3]}\frac{|f(k,u)|}{|u|}=\frac{1}{32}(u^{\alpha-1}+u^{\beta-1}).
\end{gather*}
Then,it follows that $\min f_0=\min f_{\infty}=\infty$. That is
(P1) and (P2) are valid. Let $r_2=1$, then for any
$0\leq u\leq r_2$, we have
$$
f(k,u)\leq\frac{3}{16}<\frac{r_2}{B\omega}=\frac{49}{256},
$$
Hence,
(P10) is satisfied. By Theorem \ref{thm3.4}, there exist two positive
periodic solutions $x^*_1(t)$  and  $x^*_2(t)$ satisfying
$0<\|x^*_1\|<1<\|x^*_2\|$.
\end{example}

\begin{example} \label{ex4.14} \rm
Consider  the system \eqref{e1.11} with
$a(k)=\frac{1}{8}+\frac{1}{16}\sin \frac {k\pi}{2}$,
$b(k)=2+\frac{1}{2}\cos\frac{k\pi}{2}$, $\tau(k)=1$,
when $n=2,3$. The sketches of positive periodic
solutions are shown in figure 3.
\end{example}

\begin{example} \label{ex4.15} \rm
Consider the difference equation \eqref{e4.2} with $\alpha=1/4$, $\beta=2$.
The solution curves satisfying $x(0)=0.01$, $x(0)=0.03$ and
$x(0)=0.05$ are illustrated in extended phase space and the
periodic solutions is shown in Figure 4.
\end{example}

\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.7\textwidth]{fig4}
\end{center}
\caption{Periodic solutions of \eqref{e4.2}}
\end{figure}


\subsection*{Concluding remarks}
In this paper, we employed Krasnoselskii fixed
point theorem  to investigate systematically the existence and
multiplicity of positive periodic solutions of difference
equations \eqref{e1.1} and \eqref{e1.2}. From our arguments,
the famous theorem is effective in dealing with the  difference
equations. However, all of the considered difference equations are
those equations with no delays or  retarded types. It still
remains open to test it with forward or neutral or mixed types.

Though we  discuss the existence and multiplicity of positive
periodic solutions of difference equations \eqref{e1.1} and \eqref{e1.2} in
detail based on four  key numbers, i.e., $\max f_0$, $\min f_0$,
$\max f_{\infty}$, $\min f_{\infty}$, there are some cases not
covered; for example, the cases of $\min f_\infty=0$,
$\max f_\infty=\infty$, $\min f_0=0$, $\max f_0=\infty$.
In fact, solving these cases is beyond our ability by
using Krasonselskii fixed point theorem.
In \cite{w2}, the author established the non-existence
criteria of periodic solutions, then whether or not can we
establish corresponding non-existence criteria of periodic
solutions for the rest cases by employing the same method applied
in \cite{w2}.

Our numerical simulations strongly support the analytical
achievements. From the above figures, we find that the positive
periodic solutions are stable, although we concern about the
existence of periodic solutions only. Therefore, we leave an open
question, whether or not our concise criteria guarantee the
stability of positive periodic solutions.


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