\documentclass[reqno]{amsart}
\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2003(2003), No. 102, pp. 1--7.\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 2003 Texas State University-San Marcos.}
\vspace{9mm}}
\begin{document}
\title[\hfilneg EJDE--2003/102\hfil Periodic solutions \dots]
{Periodic solutions for neutral nonlinear differential
equations with functional delay}
\author[Youssef N. Raffoul\hfil EJDE--2003/102\hfilneg]
{Youssef N. Raffoul}
\address{Department of Mathematics\\
University of Dayton, Dayton, OH 45469-2316, USA}
\email{youssef.raffoul@notes.udayton.edu}
\date{}
\thanks{Submitted April 22, 2003. Published October 6, 2003.}
\subjclass[2000]{34K20, 45J05, 45D05}
\keywords{Krasnoselskii, neutral, nonlinear, integral equation,
periodic solution, \hfill\break\indent unique solution}
\begin{abstract}
We use Krasnoselskii's fixed point theorem to show that
the nonlinear neutral differential equation with functional delay
$$
x'(t) = -a(t)x(t)+ c(t)x'(t-g(t))+ q\big(t, x(t), x(t-g(t)\big)
$$
has a periodic solution. Also, by transforming the problem to an
integral equation we are able, using the contraction mapping
principle, to show that the periodic solution is unique.
\end{abstract}
\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{example}[theorem]{Example}
\section{Introduction}
Motivated by the papers \cite{b1,g1,g2,p1,r1} and the references
therein, we consider the nonlinear neutral differential equation
\begin{equation}\label{e1.1}
x' (t) = -a(t)x(t) + c(t)x'(t-g(t))+ q\big(t, x(t),x(t-g(t)\big)
\end{equation}
which arises in a (food-limited) population model \cite{c1,c2,k1,g3}.
In recent years many researchers have considered a particular form
of equation \eqref{e1.1} and studied the existence of periodic solutions
and their boundedness. The authors in \cite{c1} investigated the
boundedness of solution of an equation of the form \eqref{e1.1} by using
comparison techniques. Also, the author in \cite{g2} considered the
logistic form of \eqref{e1.1} with several constants
delays where the functional $q$ is linear and showed the existence
of positive periodic solutions.
The purpose of this paper is to transform \eqref{e1.1} to an integral
equation and then use Krasnoselskii's fixed point theorem to show
the existence of a periodic solution. The obtained integral
equation is the sum of two mappings; one is a contraction and
the other is compact.
Transforming equation \eqref{e1.1} to an integral equation enables us to
show the uniqueness of the periodic solution by appealing to the
contraction mapping principle. For more on the existence and
uniqueness of periodic solutions of equations that are similar to
\eqref{e1.1} but with constant delay, we refer the reader to \cite{c1,c2}.
\section{Existence Of Periodic Solutions}
For $T>0$ define $P_T = \lbrace\phi: C(R,R), \phi (t+T)=\phi (t)\rbrace$
where $C(R,R)$ is the space of all real valued continuous functions.
Then $P_T$ is a Banach space when it is endowed with the supremum
norm
$$
\| x(t) \|=\max_{t \in [0,T]}|x(t|.
$$
In this paper we assume that
\begin{equation}\label{e2.1}
a(t+T)=a(t),\quad c(t+T)=c(t),\quad g(t+T)=g(t),\quad g(t)\geq g^{*} > 0
\end{equation}
with $c(t)$ continuously differentiable, $g(t)$ twice continuously
differentiable and $g^*$ is constant. In \cite{g2}, the author made the
assumption that $a(t)$ is positive, while here we only ask that
\begin{equation}\label{e2.2}
\int_0^T a(s)ds > 0.
\end{equation}
It is interesting to note that equation \eqref{e1.1} becomes of advanced type
when $g(t)< 0$. Since we are searching for periodic solutions, it
is natural to ask that $q(t,x,y)$ is continuous and periodic in
$t$ and Lipschitz continuous in $x$ and $y$. That is
\begin{equation}\label{e2.3}
q(t+T,x,y) = q (t,x,y)
\end{equation}
and some positive constants $L$ and $E$,
\begin{equation}\label{e2.4} |q(t,x,y) -
q(t,z,w)| \le L \|x-z\| + E\|y-w\|\,.
\end{equation}
Also, we assume that for all $t$, $0 \leq t \leq T$,
\begin{equation}\label{e2.5}
g'(t) \neq 1.
\end{equation}
Since $g(t)$ is periodic, condition (2.5) implies that
$g'(t)< 1$.
\begin{lemma} \label{lm2.1}
Suppose (2.1)-(2.2) and (2.5) hold. If $x(t)\in P_T,$ then $x(t)$ is a
solution of equation \eqref{e1.1} if and only if \vskip .1cm
\begin{equation}\label{e2.6}
\begin{aligned}
x(t)&= \frac{c(t)}{1-g'(t)}x(t-g(t))
+\big(1-e^{-\int^t_{t-T}a(s)ds}\big)^{-1} \\
&\times \int^t_{t-T}\big[-r(u)x(u-g(u))
+q(u,x(u),x(u-g(u)))\big] e^{-\int^t_{u}a(s)ds}\,du\,,
\end{aligned}
\end{equation}
where
\begin{equation}\label{e2.7}
r(u)=\frac{\Big(c'(u)-c(u)a(u)\Big)\Big(1-g'(u)\Big)
+g''(u)c(u)}{(1-g'(u))^{2}}.
\end{equation}
\end{lemma}
\begin{proof}
Let $x(t) \in P_{T}$ be a solution of
\eqref{e1.1}. Multiply both sides of \eqref{e1.1} by $\exp\int^t_{0}a(s)ds$
and then integrate from $t-T$ to $t$ to obtain
\begin{align*}
&\int^t_{t-T}\big[x(u)e^{\int^u_{0}a(s)ds}\big]' du\\
&=\int^t_{t-T}\big[c(u)x'(u-g(u))
+q(u,x(u),x(u-g(u)))\big] e^{\int^u_{0}a(s)ds}du.
\end{align*}
As a consequence, we arrive at
\begin{align*}
&x(t)e^{\int^t_{0}a(s)ds} - x(t-T)e^{\int^{t-T}_{0}a(s)ds}\\
&=\int^t_{t-T}\big[c(u)x'(u-g(u))
+q(u,x(u),x(u-g(u)))\big] e^{\int^u_{0}a(s)ds}du.
\end{align*}
By dividing both sides of the above equation by
$\exp(\int^t_{0}a(s)ds)$ and the fact that $x(t)=x(t-T)$, we obtain
\begin{equation}\label{e2.8}
\begin{aligned}
x(t)&=\big(1-e^{-\int^t_{t-T}a(s)ds}\big)^{-1}\\
&\quad \times \int^t_{t-T}\big[c(u))x'(u-g(u))
+ q(u,x(u),x(u-g(u)))\big] e^{-\int^t_{u}a(s)ds}du\,.
\end{aligned}
\end{equation}
Rewrite
\begin{align*}
&\int^{t}_{t-T}c(u)x'(u-g(u))e^{-\int^t_{u}a(s)ds}du\\
&=\int^{t}_{t-T}\frac{c(u)x'(u-g(u))(1-g'(u))}
{(1-g'(u))}e^{-\int^t_{u}a(s)ds}du.
\end{align*}
Integration by parts on the above integral with
$$
U=\frac{c(u)}{1-g'(u)}e^{-\int^t_{u}a(s)ds},
\quad\mbox{and}\quad dV=x'{(u-g(u))}(1-g'(u))du
$$
we obtain
\begin{equation}\label{e2.9}
\begin{aligned}
&\int^{t}_{t-T}c(u)x'(u-g(u))e^{-\int^{t}_{u}a(s)ds}du \\
&=\frac{c(t)}{1-g'(t)}x(t-g(t))\big(1-e^{-\int^{t}_{t-T}a(s)ds}\big)
-\int^{t}_{t-T}r(u)e^{-\int^{t}_{u}a(s)ds}x(u-g(u))du\,,
\end{aligned}
\end{equation}
where $r(u)$ is given by (2.7). Then substituting (2.9) into
(2.8) completes the proof.\end{proof}
Define the mapping $H : P_{T}\to P_{T}$ by
\begin{equation}\label{e2.10}
\begin{aligned}
(H\varphi)(t)=& \frac{c(t)}{1-g'(t)}\varphi(t-g(t))
+\big(1-e^{-\int^t_{t-T}a(s)ds}\big)^{-1}\\
&\times \int^t_{t-T}\big[-r(u)\varphi(u-g(u))
+q(u,\varphi(u),\varphi(u-g(u)))\big]
e^{-\int^t_{u}a(s)ds}du\,.
\end{aligned}
\end{equation}
Next we state Krasnoselskii's fixed point theorem which enables us
to prove the existence of a periodic solution. For its proof
we refer the reader to \cite{k2}.
\begin{theorem}[Krasnoselskii] \label{thm2.2}
Let $\mathbb{M}$ be a closed convex nonempty subset of a Banach space
$(\mathbb{B}, \| \cdot\|)$. Suppose that $A$ and $B$ maps $\mathbb{M}$ into
$\mathbb{B}$
such that
\begin{itemize}
\item[(i)] $x,y \in \mathbb{M}$, implies $Ax + By \in \mathbb{M}$,
\item[(ii)] $A$ is compact and continuous,
\item[(iii)] $B$ is a contraction mapping.
\end{itemize}
Then there exists $z \in \mathbb{M}$ with $z=Az+Bz$.
\end{theorem}
Note that to apply the above theorem we need to construct two
mappings; one is a contraction and the other is
compact. Therefore, we express equation (2.10) as
$$
(H\varphi)(t)=(B\varphi)(t)+(A\varphi)(t)
$$
where $A,B:P_{T}\to P_{T}$ are given by
\begin{equation}\label{e2.11}
(B\varphi)(t)= \frac{c(t)}{1-g'(t)}\varphi(t-g(t))
\end{equation}
and
\begin{equation}\label{e2.12}
\begin{aligned}
(A\varphi)(t)\
&= (1-e^{-\int^t_{t-T}a(s)ds})^{-1}\\
&\quad\times \int^t_{t-T}\big[-r(u)\varphi(u-g(u))
+q(u,\varphi(u),\varphi(u-g(u)))\big] e^{-\int^t_{u}a(s)ds}du.
\end{aligned}
\end{equation}
\begin{lemma} \label{lm2.3}
Suppose (2.1)-(2.5) hold. Then $A:P_{T}\to P_{T}$,
as defined by (2.12), is compact provided that
\begin{equation}\label{e2.13}
\max_{t \in [0,T]}\Big|\frac{c'(t)}{1-g'(t)}+
\frac{g'' (t)}{(1-g'(t))^{2}}\Big|\leq Q
\end{equation}
for some positive constant $Q$.
\end{lemma}
\begin{proof}
A change of variable in (2.12) shows that
$(A\varphi)(t+T)=(A\varphi)(t)$.
To see that $A$ is continuous, we let $\varphi,\psi\in P_{T}$
with $\|\varphi\|\leq C$ and $\|\psi\|\leq C$. Let
\begin{equation}\label{e2.14}
\eta = \max_{t \in [0,T]}
|(1-e^{-\int^t_{t-T}a(s)ds})^{-1}|,\quad
\beta = \max_{t \in [0,T]}|r(t)|,\quad
\gamma =\max_{u \in [t-T,t]} e^{-\int^t_{u}a(s)ds}.
\end{equation}
Given $\epsilon > 0$, take $\delta =\epsilon/M$ such that
$\|\varphi-\psi\|<\delta$. By making use of (2.4) into
(2.12) we get
\begin{align*}
\big\|\big(A\varphi(t)\big)-\big(A\psi(t)\big)\big\|
&\leq \gamma\eta\int^t_{t-T}\big[L\|\varphi-\psi\|+E\|\varphi-\psi\|
+ \beta\|\varphi-\psi\|\big]du\\
&\leq M\|\varphi-\psi\|<\epsilon
\end{align*}
where $L$, $E$ are given by (2.4) and $M = T\gamma \eta[\beta + L
+E]$. This proves $A$ is continuous. To show $A$ is compact, we
let $\varphi_{n}\in P_{T}$ with $\|\varphi_{n}\|\leq R,$ where $n$
is a positive integer and $R>0$. Observe that in view of (2.4) we arrive at
\begin{align*}
|q(t,x,y)| &= |q(t,x,y)- q(t,0,0)+ q(t,0,0)|\\
&\leq |q(t,x,y)- q(t,0,0)|+ |q(t,0,0)|\\
&\leq L\|x\| + E\|y\| + \alpha
\end{align*}
where $\alpha = |q(t,0,0)|$. Hence, if $A$ is given by (2.12) we
obtain that
$$
\|A(\varphi_{n}(t))\| \leq D
$$
for some positive constant $D$. Now, it can be easily checked that
\begin{align*}
(A\varphi_{n})'(t)
&=-a(t)\varphi_{n}(t)-\big[\frac{c{\prime}(t)}{1-g'(t)}
+\frac{g''(t)}{(1-g'(t))^{2}}\big]
\varphi_{n}(t-g(t))\\
&\quad +q(t,\varphi_{n}(t),\varphi_{n}(t-g(t))).
\end{align*}
Hence by invoking (2.13) we obtain $\|(A\varphi_{n})'(t)\|\leq F$, for some
positive constant $F$. Thus the sequence
${(A\varphi_{n})}$ is uniformly bounded and equi-continuous. The
Arzela-Ascoli theorem implies that ${(A\varphi_{{n}_{k}})}$
uniformly converges to a continuous $T$-periodic function $\varphi^*$.
Thus A is compact.
\end{proof}
\begin{lemma} \label{lm2.4}
Let $B$ be defined by (2.11) and
\begin{equation}\label{e2.15}
\Big\|\frac{c(t)}{1-g'(t)}\Big\| \leq \zeta <1.
\end{equation}
Then $B$ is a contraction.
\end{lemma}
\begin{proof} For $\varphi, \psi \in P_{T}$, we have
\begin{align*}
\|B(\varphi)-B(\psi)\|&=\max_{t\in[0,T]}|B(\varphi)-B(\psi)|\\
&=\max_{t\in[0,T]}\big|\frac{c(t)}{1-g'(t)}\big|
\big|\varphi(t-g(t))-\psi(t-g(t)\big| \\
&\leq \zeta\|\varphi-\psi\|.
\end{align*}
Hence $B$ defines a contraction mapping with contraction constant
$\zeta$.
\end{proof}
\begin{theorem} \label{thm2.5}
Let $\alpha =\|q(t,0,0)\|$. Let $\eta, \beta$ and $\gamma$ be given by
\eqref{e2.14}. Suppose \eqref{e2.1}--\eqref{e2.5}, \eqref{e2.13} and
\eqref{e2.15} hold. Suppose there is a
positive constant $G$ such that all solutions $x(t)$ of \eqref{e1.1},
$x(t)\in P_{T}$ satisfy $|x(t)| \leq G,$ the inequality
\begin{equation}\label{e2.16}
\big\{\zeta + \eta \gamma T (\beta + L + E )\big\}G
+\eta \gamma T \alpha \leq G
\end{equation}
holds. Then equation\eqref{e1.1} has a $T-$periodic solution.
\end{theorem}
\begin{proof} Define $\mathbb{M} = \{ \varphi \in P_{T}: \|\varphi \| \leq G\}$.
Then Lemma 2.3 implies $A:P_{T}\to P_{T}$ and $A$ is compact and
continuous.
Also, from Lemma 2.4, the mapping $B$ is a contraction and it is
clear that $B:P_{T}\to P_{T}$ . Next, we show that if
$\varphi, \psi \in \mathbb{M} $, we have $\|A\phi +B\psi\|\leq G$.
Let $\varphi, \psi \in \mathbb{M} $ with $\|\varphi\|, \|\psi\|\leq G$.
Then from (2.11)-(2.12) and the fact that $|q(t,x,y)| \leq
L\|x\| + E\|y\| + \alpha$, we have
\begin{align*}
\big\|\big(A\varphi(t)\Big)+\Big(B\psi(t)\big)\big\|
&\leq \gamma\eta\int^t_{t-T}\big[L\|\varphi\|+E\|\varphi\|
+ \beta\|\varphi\|+ \alpha \big]du +\zeta \|\psi\|\\
&\leq \big\{\zeta + \eta \gamma T (\beta + L + E )\big\}G + \eta \gamma T \alpha
\leq G.
\end{align*}
We see that all the conditions of Krasnoselskii's theorem are
satisfied on the set $\mathbb{M}$. Thus there exists a fixed point
$z$ in $\mathbb{M}$ such that $z=Az+Bz$. By Lemma 2.1, this fixed
point is a solution of \eqref{e1.1}. Hence \eqref{e1.1} has a
$T-$periodic solution.
\end{proof}
\begin{theorem} \label{thm2.6}
Suppose \eqref{e2.1}--eqref{e2.5}, \eqref{e2.13} and
\eqref{e2.15} hold. Let $\eta, \beta$ and $\gamma$ be given by \eqref{e2.14}. If
$$
\zeta + T \gamma \eta \big(\beta + L + E\big) < 1,
$$
then equation \eqref{e1.1} has a unique $T$-periodic solution.
\end{theorem}
\begin{proof} Let the mapping $H$ be given by (2.10). For
$\varphi,\psi\in P_{T}$, in view of (2.10), we have
\begin{align*}
&\big\|\big(H\varphi(t)\big)-\big(H\psi(t)\big)\big\| \\
&\leq \zeta\|\varphi-\psi\|
+\gamma\eta\int^t_{t-T}\big[L\|\varphi-\psi\|+E\|\varphi-\psi\|
+ \beta\|\varphi-\psi\|\big]du\\
&\leq \big[\zeta + T \gamma \eta \big(\beta + L + E\big)\big]\|\varphi-\psi\|.
\end{align*}
This completes the proof.
\end{proof}
We finish this paper with an example in which we give criteria
that enable us to determine the a priori bound $G$ in Theorem \ref{thm2.5}.
\begin{example} \label{ex2.7} \rm
Consider \eqref{e1.1} along with
conditions (2.1)--(2.5) and (2.13)--(2.15). Suppose that $a(t)\neq
0$ for all $t\in[0,T]$. Set
\begin{equation*}
\rho = \min_{t \in [0,T]}|a(t)|,\quad \delta =\max_{t \in [0,T]}k(t),
\end{equation*}
where
$$
k(t)=\frac{c'(t)\big(1-g'(t)\big) +g''(t)c(t)}{(1-g'(t))^{2}}.
$$
Suppose $1-\|c\|>0$. If
$$
\rho(1-\|c\|)>(1-\|c\|)(\delta+L+E)+T \rho (\|a\|+L+E)
$$
holds and $G$ which is defined by
$$
G=\frac{\alpha(1-\|c\|+T\rho)}{\rho
(1-\|c\|)-(1-\|c\|)(\delta+L+E)-T \rho (\|a\|+L+E)}
$$
satisfies inequality (2.16), then \eqref{e1.1} has a T-periodic solution.
\end{example}
\begin{proof} Let the mappings $A$ and $B$ be defined by
(2.12) and (2.11), respectively. Let $x(t)\in P_{T}$. An
integration of equation \eqref{e1.1} from $0$ to $T$ yields,
\begin{equation}\label{e2.17}
x(T)-x(0)=\int^T_{0}[-a(t)x(t) +
c(t)x' (t-g(t))+q(t,x(t), x(t-g(t)))]dt.
\end{equation}
Rewrite and then integrate by parts to obtain,
\begin{align*}
\int^{T}_{0}c(t)x'(t-g(t))dt
&=\int^{T}_{0}\frac{c(t)x'(t-g(t))(1-g'(t))}{(1-g'(t))}dt\\
&=-\int^{T}_{0}k(t)x(t-g(t))dt.
\end{align*}
As a consequence, (2.17) becomes
\[
\int^T_{0}a(t)x(t)dt = \int^T_0 [k(t)x(t-g(t))-q(t,x(t), x(t-g(t)))]dt,
\]
from which, it implies that there exists a $t^{*}\in (0,T)$ such
that
$$
Ta(t^{*})x(t^{*}) = \int^T_0 [k(t)x(t-g(t))-q(t,x(t), x(t-g(t)))]dt.
$$
Then
$$
T|a(t^{*})\|x(t^{*})| \leq
\int^T_0 [|k(t)\|x(t-g(t))|+|q(t,x(t), x(t-g(t)))|]dt.
$$
By taking the maximum over $t\in [0,T]$, from the above inequality, we obtain
\begin{equation}\label{e2.18}
\|x(t^*)\|\leq \frac{1}{\rho}\Big(\delta +L +E \Big)\|x\|
+\frac{\alpha}{\rho}.
\end{equation}
Since for all $t \in [0,T]$,
$$
x(t) = x(t^*) + \int^{t}_{t^*}x'(s)ds,
$$
we have
$$
|x(t)|\leq |x(t^*)| + \int^{t}_{0}|x'(s)|ds.
$$
Taking the maximum over $t\in [0,T]$ and using (2.18), yield
\begin{equation}\label{e2.19}
\begin{aligned}
\|x(t)\|&\leq \|x(t^*)\| + \int^{t}_{0}\|x'(s)\|ds \\
&\leq \frac{1}{\rho}\big(\delta +L +E\big)\|x\|+\frac{\alpha}{\rho}+ T\|x'\|.
\end{aligned}
\end{equation}
Taking the norm in \eqref{e1.1} yields
\begin{equation}\label{e2.20}
\|x'\|\leq \frac{(\|a\|+L+E)\|x\|+\alpha}{1-\|c\|}.
\end{equation}
Substitution of (2.20) into (2.19), yields that for all $x(t)\in
P_T$,
$\|x(t)\|\leq G$.
Define $\mathbb{M} = \{ \varphi \in P_{T}: \|\varphi \| \leq G\}$.
Then by Theorem \ref{thm2.5}, \eqref{e1.1} has a T-periodic solution.
\end{proof}
\subsection*{Remark} The author of this paper has
studied the asymptotic stability of the zero solution of \eqref{e1.1}
using fixed point theory; see \cite{r1}. However, the question of uniform
asymptotic stability of the zero solution of \eqref{e1.1} remains open.
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\end{document}