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\markboth{\hfil Positive periodic solutions \hfil EJDE--2002/55}
{EJDE--2002/55\hfil Youssef N. Raffoul \hfil}
\begin{document}

\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations}, Vol. {\bf
2002}(2002), No. 55, 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)}
 \vspace{\bigskipamount} \\
 %
  Positive periodic solutions of nonlinear functional difference equations
 %
\thanks{ {\em Mathematics Subject Classifications:} 39A10, 39A12.
\hfil\break\indent 
{\em Key words:} Cone theory, positive, periodic, functional 
difference equations. 
\hfil\break\indent
\copyright 2002 Southwest Texas State University.
\hfil\break\indent Submitted December 22, 2001. Published June 13, 2002.} }
\date{}
%
\author{Youssef N. Raffoul}
\maketitle

\begin{abstract}
 In this paper, we apply a cone theoretic fixed point theorem
 to obtain sufficient conditions for the existence
 of multiple positive periodic solutions to the nonlinear
 functional difference equations
 $$
 x(n+1)=a(n)x(n)\pm \lambda h(n) f(x(n-\tau(n))).
 $$
\end{abstract}

\newtheorem{theorem}{Theorem}[section]
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{lemma}[theorem]{Lemma}
\numberwithin{equation}{section}

\section{Introduction}

Let $\mathbb{R}$ denote the real numbers, $\mathbb{Z}$ the
integers and $\mathbb{R}^{+}$ the positive real numbers. Given
$a<b$ in $\mathbb{Z}$,  let $[a, b]=\{ a, a+1,\dots ,b\}$. In this
paper, we investigate the existence of multiple positive periodic
solutions for the nonlinear delay functional difference equation
\begin{equation}\label{e1.1}
 x(n+1)=a(n)x(n)+\lambda h(n) f(x(n-\tau(n)))
\end{equation}
where $a(n), h(n)$ and $ \tau(n)$ are $T$-periodic for $T$ is an
integer with $T \geq 1$. We assume that $\lambda$, $a(n)$, $f(x)$
and $h(n)$ are nonnegative with $0< a(n) < 1 $ for all $n \in
[0,T-1]$.

The existence of multiple  positive periodic solutions of
nonlinear functional differential equations have been studied
extensively in recent years. We cite some appropriate references
here \cite{sc} and \cite{y}. We are particularly motivated by the
work of Cheng and Zhang \cite{sc} on functional differential
equations and the work of Eloe, Raffoul and others \cite{er} on a
boundary value problem involving functional difference equation.
It is customary when working with boundary value problems, whether
in differential or difference equations, to display the desired
solution in terms of a suitable Green function and then apply cone
theory \cite{dh,hp,aw2,er,hl,hh,m}. Since our equation
\eqref{e1.1} is not of the type of boundary value we obtain a
variation of parameters formula and then try to find a lower and
upper estimates for the kernel inside the summation. Once those
estimates are found we use Krasnoselskii's fixed point theorem to
show the existence of multiple positive periodic solutions. In
\cite{r}, the author studied the existence of periodic solutions
of an equation similar to equation \eqref{e1.1} using Schauder's
Second fixed point theorem. Throughout this paper, we denote the
product of $y(n)$ from $n=a$ to $n=b$ by $\prod^{b}_{n=a}y(n)$
with the understanding that $\prod^{b}_{n=a}y(n) = 1$ for all
$a>b$,

\section{Positive periodic solutions}
 We now state Krasnosel'skii fixed point theorem
 \cite{k}.

\begin{theorem}[Krasnosel'skii] \label{thm2.1}
 Let $\mathcal{B}$ be a Banach
space, and let $\mathcal{P}$ be a cone in $\mathcal{B}$. Suppose
$\Omega_{1}$ and $\Omega_2$ are open subsets of $\mathcal{B}$ such
that $0 \in \Omega_1 \subset \overline{\Omega}_1 \subset \Omega_2$
and suppose that
$$ T:\mathcal{P} \cap ( \overline{\Omega}_2  \backslash \Omega_1)
\to  \mathcal{P}
$$
is a completely continuous operator such that
\begin{itemize}
\item [(i)] $\|Tu \|\leq \|u \|$, $u \in  \mathcal{P} \cap \partial \Omega_1$,
 and
$\|Tu \|\geq \|u \|$, $u\in {\mathcal{P}} \cap \partial\Omega_2$;
or

\item [(ii)] $\|Tu \|\geq \|u \|$, $u\in {\mathcal{P}} \cap \partial
\Omega_1$, and $\|Tu \|\leq \|u \|$, $u\in {\mathcal{P}} \cap
\partial \Omega_2$.
\end{itemize}
Then $T$ has a fixed point in ${\mathcal{P}} \cap
(\overline{\Omega}_2 \backslash \Omega_1)$.
\end{theorem}

Let $\mathcal{X}$ be the set of all real $T$-periodic sequences.
This set endowed with the maximum norm $\|x\|=
\max_{n\in[0,T-1]}|x(n)|$, $\mathcal{X}$ is a Banach space. The
next Lemma is essential in obtaining our results.

\begin{lemma} \label{lm2.2}
$x(n)\in \mathcal{X}$ is a solution of equation \eqref{e1.1} if
and only if \vskip .1cm
\begin{equation}\label{e2.1}
 x(n)=\lambda
\sum^{n+T-1}_{u=n}G(n,u)h(u)f(x(u-\tau(u)))
\end{equation}
where
\begin{equation}
\label{e2.2} G(n,u)=\frac{\prod^{n+T-1}_{s=u+1}a(s)}
{1-\prod^{n+T-1}_{s=n}a(s)}, \quad  u \in [n, n+T-1].
\end{equation}
\end{lemma}

Note that the denominator in $G(n,u)$ is not zero since $0 <
a(n)<1$ for $n \in [0,T-1]$. The proof of Lemma 2.1 is easily
obtained by noting that \eqref{e1.1} is equivalent to
$$\triangle \Big(\prod^{n-1}_{s=-\infty}a^{-1}(s)x(n)\Big)
=\lambda h(n)f(x(n-\tau(n))\prod^{n}_{s=-\infty}a^{-1}(s).
$$
By summing the above equation from $u=n$ to $u=n+T-1$ we obtain
\eqref{e2.1}. Note that since $0< a(n)< 1$ for all $n\in [0,T-1]$,
we have
$$ N \equiv G(n,n)\leq G(n,u)\leq G(n,n+T-1)=G(0,T-1)\equiv M
$$
for $n\leq u \leq n+T-1$ and
$$ 1\geq \frac{G(n,u)}{G(n,n+T-1)}\geq
\frac{G(n,n)}{G(n,n+T-1)}=\frac{N}{M}>0.
$$
For each $x\in X$, define a cone by
$$
\mathcal{P} = \big\{ y \in \mathcal{X}: y(n)\geq 0, n\in
\mathbb{Z} \quad\mbox{and}\quad y(n) \geq  \eta \|y \|\big\},
$$
where $\eta = N/M$. Clearly, $\eta \in(0,1)$. Define a mapping
$T:\mathcal{X}\to \mathcal{X}$ by
$$
(Tx)(n)=\lambda \sum^{n+T-1}_{u=n}G(n,u)h(u)f(x(u-\tau(u))
$$
 where $G(n,u)$ is
given by \eqref{e2.2}. By the nonnegativity of $\lambda$, $f$,
$a$, $h$,
 and $G$, $Tx(n) \geq 0$  on $[0,T-1]$. It is clear
that $(Tx)(n+T)=(Tx)(n)$ and $T$ is completely continuous on
bounded subset of $\mathcal{P}$. Also, for any $x \in \mathcal{P}$
we have
\begin{eqnarray*}
(Tx)(n)&=& \lambda
\sum^{n+T-1}_{u=n}G(n,u)h(u)f(x(u-\tau(u)))\\
&\leq &\lambda \sum^{T-1}_{u=0}G(0,T-1)h(u)f(x(u-\tau(u)).
\end{eqnarray*}
Thus,
$$ \|Tx \|= \max_{n \in [0, T-1]} |Tx(n)|
\leq \lambda \sum_{u=0}^{T-1} G(0,T-1) h(u)f(x(u-\tau(u))).
$$
Therefore,
\begin{eqnarray*}
 Tx(n)&=&  \lambda \sum_{u=n}^{n+T-1}
   G(n,u) h(u) f(x(u-\tau(u))) \\
                     &\geq& \lambda N \sum_{u=0}^{T-1}
    h(u)f(x(u-\tau(u)))  \\
                     &=& \lambda N  \sum_{u=0}^{T-1}
    \frac{G(0,T-1)}{M}h(u)f(x(u-\tau(u))) \\&\geq& \eta \|Tx \|.
\end{eqnarray*}
That is, $T\mathcal{P}$ is contained in $\mathcal{P}$.
\hfill$\diamondsuit$
\smallskip

In this paper we shall make the following assumptions.
\begin{itemize}
\item[(A1)] the function $f: \mathbb{R^{+}}\to \mathbb{R^{+}}$ is continuous

\item [(A2)] $h(n)> 0$ for $n\in \mathbb{Z}$

\item [(L1)] $\lim_{x\to 0}\frac{f(x)}{x}=\infty $

\item [(L2)] $\lim_{x\to \infty}\frac{f(x)}{x}=\infty $

\item [(L3)] $\lim_{x\to 0}\frac{f(x)}{x}=0 $

\item [(L4)] $\lim_{x\to \infty}\frac{f(x)}{x}=0 $

\item [(L5)] $\lim_{x\to 0}\frac{f(x)}{x}=l $ with $0<l<\infty$

\item [(L6)] $lim_{x\to \infty}\frac{f(x)}{x}=L $ with $0<L<\infty$.
\end{itemize}
For the next theorem we let
\begin{equation}\label{e2.3}
A=\max_{0\leq n \leq T-1}\sum^{T-1}_{u=0}G(n,u)h(u)
\end{equation}
and
\begin{equation}\label{e2.4}
B=\min_{0\leq n \leq T-1}\sum^{T-1}_{u=0}G(n,u)h(u).
\end{equation}

\begin{theorem} \label{thm2.4}
Assume that (A1), (A2), (L5), and (L6) hold. Then, for each
$\lambda$ satisfying
\begin{equation}\label{e2.5}
\frac{1}{\eta BL}<\lambda <\frac{1}{Al}
\end{equation}
or
\begin{equation}\label{e2.6}
\frac{1}{\eta Bl}<\lambda <\frac{1}{AL}
\end{equation}
equation \eqref{e1.1} has at least one positive periodic solution.
\end{theorem}

\paragraph{Proof} Suppose \eqref{e2.5} hold. We construct
the sets $\Omega_1$ and $\Omega_2$ in order to apply Theorem
\ref{thm2.1}. Let $\epsilon > 0$ be such that
$$ \frac{1}{\eta
B(L-\epsilon)}\leq\lambda \leq\frac{1}{A(l+\epsilon)}.
$$
By condition (L5), there exists $H_1>0$ such that $f(y)\leq
(l+\epsilon)y$ for $0<y\leq H_1$. Define
$$
\Omega_{1}=\{x\in \mathcal{P}:\|x\|<H_1\}
$$
Then, if $x\in\mathcal{P}\cap \partial\Omega_1$,
\begin{eqnarray*}
(Tx)(n)&\leq& \lambda(l+\epsilon)
\sum^{n+T-1}_{u=n}G(n,u)h(u)x(u-\tau(u)))\\
&\leq&\lambda(l+\epsilon)\|x\|\sum^{T-1}_{u=o}G(n,u)h(u)\\
&\leq&\lambda A(l+\epsilon)\|x\|\leq \|x\|.
 \end{eqnarray*}
 In particular, $\|Tx\|\leq \|x\|$, for all $x\in \mathcal{P}\cap \partial
\Omega_1$.

Next we construct the set $\Omega_2$. Apply condition $(L6)$ and
find $H$ such that $f(y)\geq (L-\epsilon)y,$ for all $y\geq H$.
Let $H_{2}=\max\{2H_{1}, \eta H\}$. Define
$$ \Omega_{2}=\{x\in \mathcal{P}:\|x\|<H_2\}
$$
Then, if $x\in\mathcal{P}\cap \partial \Omega_2,$
\begin{eqnarray*}
(Tx)(n)&\geq& \lambda(L-\epsilon)
\sum^{n+T-1}_{u=n}G(n,u)h(u)x(u-\tau(u)))\\
&\geq&\lambda(L-\epsilon)\eta\|x\|\sum^{T-1}_{u=o}G(n,u)h(u)\\
&\geq&\lambda (L-\epsilon)\eta B\|x\|\geq \|x\|.
 \end{eqnarray*}
 In particular, $\|Tx\|\geq \|x\|$, for all $x\in \mathcal{P}\cap \partial
\Omega_2$. Apply condition $(i)$ of Theorem \ref{thm2.1}, and this
completes the proof. When condition \eqref{e2.6} holds, the proof
can be similarly obtained by invoking condition $(ii)$ of Theorem
\ref{thm2.1}.

\begin{theorem} \label{thm2.5}
Assume that (A1) and (A2) hold. Also, if either  (L1) and (L4)
hold, or, (L2) and (L3) hold, then \eqref{e1.1} has at least one
positive periodic solution for any $\lambda >0$.
\end{theorem}

\paragraph{Proof:} Apply (L1) and choose $H_{1}>0$ such that
if $0<y<H_{1}$, then
$$ f(y)\geq \frac{y}{\lambda \eta B}.
$$
Define $\Omega_{1}=\{x\in \mathcal{P}:\|x\|<H_1\}$. If
$x\in\mathcal{P}\cap \partial \Omega_1,$ then
\begin{eqnarray*}
(Tx)(n)&\geq& \lambda \frac{1}{\lambda \eta B}
\sum^{T-1}_{u=0}G(n,u)h(u)x(u-\tau(u)))\\
&\geq&\frac{1}{\lambda \eta B}\lambda\|x\|\sum^{T-1}_{u=o}G(n,u)h(u)\\
&\geq& \|x\|.
 \end{eqnarray*}
 In particular, $\|Tx\|\geq \|x\|$, for all $x\in \mathcal{P}\cap \partial
\Omega_1$. In order to construct $\Omega_2,$ we consider two
cases, $f$ bounded and $f$ unbounded. The case where $f$ is
bounded is straight forward. If $f(y)$ is bounded by $Q>0$, set
$$
H_{2}=\max\{2H_{1}, \lambda QA\}.
$$
Then if $x\in \mathcal{P}$ and $\|x\|=H_{2}$,
\begin{eqnarray*}
Tx(n)&\leq& \lambda N \sum^{T-1}_{u=0}G(n,u)h(u)\\&\leq& \lambda Q
A \leq H_{2}.
\end{eqnarray*}
Now assume $f$ is unbounded. Apply condition $(L4)$ and set
$\epsilon_1>0$ such that if $x>\epsilon_1$, then
$$ f(y)<\frac{y}{\lambda A}.
$$
Set $H_{2}=\max\{2H_1, \epsilon_1\}$ and define $\Omega_{2}=\{x\in
\mathcal{P}:\|x\|<H_2\}$. If $x\in\mathcal{P}\cap \partial
\Omega_2,$ then
\begin{eqnarray*}
(Tx)(n)&\leq& \lambda \frac{1}{\lambda A}
\sum^{T-1}_{u=0}G(n,u)h(u)x(u-\tau(u))\\
&\leq& \frac{1}{\lambda A}\lambda H_{2} \sum^{T-1}_{u=o}G(n,u)h(u)\\
&\leq& H_{2}=\|x\|.
 \end{eqnarray*}
In particular, $\|Tx\|\leq \|x\|$, for all $x\in \mathcal{P}\cap
\partial \Omega_2$. Apply condition $(ii)$ of Theorem \ref{thm2.1}, and this
completes the proof. The proof of the other part, follows
similarly by invoking condition $(ii)$ of Theorem \ref{thm2.1}.
The next two corollaries are consequence of the previous two
theorems.

\begin{corollary} \label{coro2.6}
 Assume that $(A1)$ and
(A2) hold. Also, if either  (L1) and (L6) hold, or, (L2) and (L5)
hold, then \eqref{e1.1} has at least one positive periodic
solution if $\lambda$ satisfies either $0<\lambda<1/(AL)$, or,
$0<\lambda<1/(Al)$.
\end{corollary}

\begin{corollary} \label{coro2.7}
Assume that (A1) and (A2) hold. Also, if either  (L3) and (L6)
hold, or, (L4) and (L5) hold, then \eqref{e1.1} has at least one
positive periodic solution if $\lambda$ satisfies either $1/(\eta
BL)<\lambda <\infty$, or, $1/(\eta Bl)<\lambda <\infty$.
\end{corollary}

Next we turn our attention to the equation
\begin{equation}\label{e2.7}
x(n+1)=a(n)x(n)-\lambda h(n) f(x(n-\tau(n)))
\end{equation}
where $\lambda, a(n), f(x)$ and $h(n)$ satisfy the same
assumptions stated for \eqref{e1.1} except that  $ a(n) > 1 $ for
all $n \in [0,T-1]$. In view of \eqref{e2.7} we have that
\begin{equation}\label{e2.8}
 x(n)=\lambda \sum^{n+T-1}_{u=n}K(n,u)h(u)f(x(u-\tau(u)))
\end{equation}
where
\begin{equation} \label{e2.9}
K(n,u)=\frac{\prod^{n+T-1}_{s=u+1}a(s)}
{\prod^{n+T-1}_{s=n}a(s)-1}, \quad \quad u \in [n, n+T-1].
\end{equation}
Note that the denominator in $G(n,u)$ is not zero since $ a(n)>1$
for $n \in [0,T-1]$. Also, it is easily seen that since $a(n)> 1$
for all $n\in [0,T-1]$, we have
$$
M \equiv K(n,n)\geq K(n,u)\geq K(n,n+T-1)=K(0,T-1)\equiv N
$$
for $n\leq u \leq n+T-1$ and
$$ 1\geq \frac{K(n,u)}{K(n,n)}\geq
\frac{K(n,n+T-1)}{K(n,n)}=\frac{N}{M}>0.
$$
Finally, by defining
$$ A^{1}=\underset{0\leq n \leq T-1}{\max}\sum^{T-1}_{u=0}K(n,u)h(u)
$$
and
$$
B^{1}=\underset{0\leq n \leq T-1}{\min}\sum^{T-1}_{u=0}K(n,u)h(u)
$$
similar theorems and corollaries can be easily stated and proven
regarding equation \eqref{e2.7}.

 We conclude this paper with the following open problems.
   Assume that (A1) and (A2) hold. In view of this paper, what can be said
   about equations \eqref{e1.1} and \eqref{e2.7} when:
\begin{enumerate}
\item The conditions $(L1)$ and $(L2)$ hold?
\item The conditions $(L3)$ and $(L4)$ hold?
\item $0<a(n)<1$ in \eqref{e2.7} and $a(n) > 1$ in \eqref{e1.1}
 for all $n\in[0, T-1]$?
\end{enumerate}

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\noindent\textsc{Youssef N. Raffoul }\\
Department of Mathematics,
University of Dayton \\
Dayton, OH 45469-2316 USA \\
e-mail:youssef.raffoul@notes.udayton.edu
\end{document}