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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 127, 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}
\thanks{\copyright 2010 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2010/127\hfil Existence of periodic solutions]
{Existence of periodic solutions for neutral nonlinear
differential equations with variable delay}

\author[D. Hafsia, D. Ahc\'ene\hfil EJDE-2010/127\hfilneg]
{Deham Hafsia, Djoudi Ahc\'ene} 

\address{Deham Hafsia \newline
Department of Mathematics, Faculty of Sciences \\
University of Annaba, P.O. Box 12 Annaba, Algeria}
\email{deh\_71@yahoo.fr}

\address{Djoudi Ahc\'ene \newline
Department of Mathematics, Faculty of Sciences \\
University  of Annaba, P.O. Box 12 Annaba, Algeria}
\email{adjoudi@yahoo.com}

\thanks{Submitted April 15, 2010. Published September 7, 2010.}
\subjclass[2000]{34K20, 45J05, 45D05}
\keywords{Periodic solution; nonlinear neutral differential equation;
\hfill\break\indent large contraction; integral equation}

\begin{abstract}
 We use a variation of Krasnoselskii fixed point theorem
 introduced by Burton to show that the nonlinear neutral differential
 equation
 \[
 x'(t)=-a(t)x^3(t)+c(t)x'(t-g(t))+G(t,x^3(t-g(t))
 \]
 has a periodic solution. Since this equation is nonlinear,
 the variation  of parameters can not be applied directly;
 we add and subtract a linear term  to transform the
 differential into an  equivalent integral equation suitable
 for applying a fixed point theorem. Our result is illustrated
 with an example.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{example}[theorem]{Example}
\allowdisplaybreaks

\section{Introduction}

We are interested in proving that the retarded scalar neutral
non linear differential equation
\begin{equation}
x'(t)=-a(t)x^3(t)+c(t)x'(t-g(t))+G(t,x^3(t-g(t)))
\label{e1.1}
\end{equation}
possesses a periodic solution. The motivation for studying
this problems comes from the problems considered in
\cite{b1,b3,b4,k1,r1}.
 Here $a(t)$ is real valued function, $c(t)$ is
continuously differentiable, $g(t)$ is twice continuously
differentiable, and $G:\mathbb{R}\times\mathbb{R}\to \mathbb{R}$
is continuous with respect to its arguments.
Clearly, the present problem is nonlinear so that the
variation of parameters can not be applied directly.
Then, we resort to the idea of adding and subtracting a
linear term. As noted by Burton in \cite{b1}, the added term
destroys a contraction already present in part of the equation but it
replaces it with the so called a large contraction mapping which
is suitable for fixed point theory. During the process we have to
transform  \eqref{e1.1} into an integral equation written as
a sum of two mappings; one is a large contraction and the other is
compact. After that, we use a variant of Krasnoselskii fixed
point theorem, due to Burton \cite{b3}, to show the existence of
a periodic solution. Our result is illustrated with an example
at the end of this article.

\section{Inversion of \eqref{e1.1}}

Let $T>0$ and define $C_{T}=\{ \varphi :\mathbb{R}\to \mathbb{R}
:\varphi \in C\text{ and }\varphi (t+T)=\varphi (t)\} $ where $C$
is the space of continuous real valued functions.
$C_{T}$ is a Banach space endowed with the
norm
\[
\| \varphi \| =\max_{0\leq t\leq T} | \varphi (t)| \,.
\]
We assume that $a,c,G$, $g$ are continuous
functions with $g(t)\geq 0$ and $c$ continuously differentiable
such that
\begin{equation}
a(t+T)=a(t),\quad c(t+T)=c(t),\quad G(t+T,x)=G(t,x), \quad g(t+T)=g(t).
 \label{e2.1}
\end{equation}
We assume further that $G(t,x)$ is globally Lipschitz continuous in $x$.
That is, there is some positive constant $k$ such that
\begin{equation}
| G(t,x)-G(t,y)| \leq k| x-y| . \label{e2.2}
\end{equation}
Also, we assume that
\begin{equation}
\int_{0}^{T}a(s)ds>0,  \label{e2.3}
\end{equation}
and for all $t$, $0\leq t\leq T$,and that  $g$ is twice
continuously differentiable and
\begin{equation}
g'(t)\neq 1.  \label{e2.4}
\end{equation}

\begin{lemma} \label{lem2.1}
Suppose \eqref{e2.1}, \eqref{e2.3} and \eqref{e2.4} hold.
If $x(t)\in C_{T}$,
then $x(t)$ is a solution of \eqref{e1.1} if and only if
\begin{equation} \label{e2.5}
\begin{aligned}
x(t)&=(1-e^{-\int_{t-T}^ta(s)ds})^{-1}
\Big[\int_{t-T}^tG(u,x^3(u-g(u)))e^{-\int_{u}^ta(s)ds}du   \\
&\quad+\int_{t-T}^ta(u)( x(u)-x^3(u))
 e^{-\int_{u}^ta(s)ds}du\Big]+ \frac{c(t)x(t-g(t))}{(1-g'(t))}   \\
&\quad -(1-e^{-\int_{t-T}^ta(s)ds})^{-1}\int
_{t-T}^th(u)(x(u-g(u)))e^{-\int_{u}^ta(s)ds}du,
\end{aligned}
\end{equation}
where
\begin{equation*}
h(u)=\frac{(c'(u)+a(u)c(u))(1-g'(u))+c(u)g''(u)}{(1-g'(u))^2}.
\end{equation*}
\end{lemma}

\begin{proof}
Let $x(t)$ be a solution of \eqref{e1.1}. Rewrite  \eqref{e1.1} as
\begin{equation*}
x'(t)+a(t)x(t)=a(t)x(t)-a(t)x^3(t)+c(t)x'(t-g(t))+G(t,x^3(t-g(t))).
\end{equation*}
Multiply both sides of the above equation by $e^{\int_{0}^ta(s)ds}$
and then integrate from $t-T$ to $t$ to obtain
\begin{align*}
&\int_{t-T}^t[ x(u)e^{\int_{0}^{u}a(s)ds}] 'du \\
&=\int_{t-T}^ta(u)[ x(u)-x^3(u)] e^{\int_{0}^{u}a(s)ds}du+
\int_{t-T}^tG(u,x^3(u-g(u)))e^{\int_{0}^{u}a(s)ds}du \\
&\quad +\int_{t-T}^tc(u)x'(u-g(u))e^{\int_{0}^{u}a(s)ds}du.
\end{align*}
Rewrite the last term as
\[
\int_{t-T}^tc(u)x'(u-g(u))e^{\int_{0}^{u}a(s)ds}du
=\int_{t-T}^t\frac{c(u)x'(u-g(u))(1-g'(u))}{(1-g'(u))}
e^{\int_{0}^{u}a(s)ds}du.
\]
Using integration by parts, and that $c,g,x$ are periodic we obtain
\begin{equation} \label{e2.6}
\begin{aligned}
&\int_{t-T}^te^{\int_{0}^{u}a(s)ds}c(u)x'(u-g(u))du\\
&= \frac{c(t)}{(1-g'(t))}x(t-g(t))e^{\int_{0}^ta(s)ds}(1-e^{-
\int_{t-T}^ta(s)ds})\\
&\quad -\int_{t-T}^th(u)(x(u-g(u)))e^{\int_{0}^{u}a(s)ds}du.
\end{aligned}
\end{equation}
We arrive at
\begin{align*}
&x(t)e^{\int_{0}^ta(s)ds}-x(t-T)e^{\int_{0}^{t-T}a(s)ds} \\
&=\int_{t-T}^ta(u)[ x(u)-x^3(u)] e^{\int_{0}^{u}a(s)ds}du+
\int_{t-T}^tG(u,x^3(u-g(u)))e^{\int_{0}^{u}a(s)ds}du \\
&\quad +\frac{c(t)}{(1-g'(t))}x(t-g(t))e^{\int_{0}^ta(s)ds}
 (1-e^{-\int_{t-T}^ta(s)ds})\\
&\quad -\int_{t-T}^th(u)(x(u-g(u)))e^{\int_{0}^{u}a(s)ds}du.
\end{align*}
Now, the lemma follows by dividing both sides of the above equation by
$e^{\int_{0}^ta(s)ds}$ and using the fact that $x(t)=x(t-T)$.
\end{proof}

Krasnoselskii \cite{b2,s1} combined the contraction mapping
theorem and Shauder's theorem and formulated the following
hybrid result.

\begin{theorem} \label{thm2.1}
Let $M$ be a closed convex non-empty subset
of a Banach space $(S,\| \cdot\| )$. Suppose that $A$ and $B$ map $M$
into $S$ such that the following conditions hold
\begin{itemize}
\item[(i)] $Ax+By\in M$, for all $x,y\in M$;

\item[(ii)] $A$ is continuous and $AM$ is contained in a compact set;

\item[(iii)] $B$ is a contraction with $\alpha <1$.
\end{itemize}
Then there is a $z\in M$, with $z=Az+Bz$.
\end{theorem}

This is a captivating result and has a number of interesting
applications. In recent year much attention has been paid
to this theorem.  Burton \cite{b2}
observed  that Krasnoselskii result can be more interesting in
applications with certain changes and formulated in
Theorem \ref{thm2.3}
below (see \cite{b3} for the proof).

Let $(M,d)$ be a metric space and $B:M\to M$. $B$ is said to be a
large contraction if $\varphi ,\psi \in M$, with
$\varphi \neq \psi $ then $d(B\varphi ,B\psi )<d(\varphi ,\psi )$
and if for all $\varepsilon >0$ there exists $\delta <1$ such that
\begin{equation*}
[ \varphi ,\psi \in M,\;d(\varphi ,\psi )\geq \varepsilon ]
\Rightarrow d(B\varphi ,B\psi )\leq \delta d(\varphi ,\psi ).
\end{equation*}

\begin{theorem} \label{thm2.2}
Let $(M,d)$ be a complete metric space and $B$ be a
large contraction. Suppose there is an $x\in M$ and an $L>0$,
such that $d(x,B^{n}x)\leq L$ for all $n\geq 1$. Then $B$ has
a unique fixed point in $M$.
\end{theorem}

\begin{theorem} \label{thm2.3}
Let $M$ be a bounded convex non-empty subset
of a Banach space $(S,\|\cdot\| )$. Suppose that $A$, $B$ map $M$
into $M$ and that
\begin{itemize}
\item[(i)] for all $x,y\in M\Rightarrow Ax+By\in M$,

\item[(ii)] $A$ is continuous and $AM$ is contained in a compact
 subset of $M$,

\item[(iii)] $B$ is a large contraction.
\end{itemize}
Then there is a $z\in M$ with $z=Az+Bz$.
\end{theorem}

We will use this theorem to prove the existence of periodic
solutions for \eqref{e1.1}.

\section{Existence of periodic solutions}

To apply Theorem \ref{thm2.3}, we need to define a Banach space $S$, a
bounded convex subset $M$ of $S$ and construct two mappings, one is a
large contraction and the other is completely continuous. So, we
let $(S,\| \cdot\| )=(C_{T},\| \cdot\| )$ and
$M=\{\varphi \in S:\| \varphi \| \leq L,\varphi '\text{ is bounded}\}$,
where $L=\sqrt{3}/3$. We express  \eqref{e1.1} as
\begin{equation*}
\varphi (t)=(B\varphi )(t)+(A\varphi )(t):=(H\varphi )(t),
\end{equation*}
where $A,B:S\to S$ are defined by
\begin{equation}
(B\varphi )(t):=(1-e^{-\int_{t-T}^ta(s)ds})^{-1}\int_{t-T}^ta(u)(
\varphi (u)-\varphi ^3(u)) e^{-\int_{u}^ta(s)ds}du,  \label{e3.1}
\end{equation}
and
\begin{equation} \label{e3.2}
\begin{aligned}
(A\varphi )(t)
&:=(1-e^{-\int_{t-T}^ta(s)ds})^{-1}\int_{t-T}^tG(u,\varphi
^3(u-g(u)))e^{-\int_{u}^ta(s)ds}du\\
&\quad +  \frac{c(t)\varphi (t-g(t))}{(1-g'(t))}\\
 &\quad -(1-e^{-
\int_{t-T}^ta(s)ds})^{-1}\int_{t-T}^th(u)(\varphi(u-g(u)))
e^{-\int_{u}^ta(s)ds}du.
\end{aligned}
\end{equation}
We need the following assumptions
\begin{gather}
( kL^3+| G(t,0)| ) \leq \beta La(t), \label{e3.3} \\
| h(t)| \leq \delta a(t),  \label{e3.4} \\
\max_{t\in [ 0,T] }| \frac{c(t)}{(1-g'(t))} | =\alpha ,
\label{e3.5} \\
J(\beta +\alpha +\delta )\leq 1,  \label{e3.6}
\end{gather}
where $\alpha ,\beta ,\delta $ and $J$ are constants with $J\geq 3$.

We begin with the following proposition (see \cite{b1,b2}).

\begin{proposition} \label{prop3.1}
Let $\|\cdot\| $ be the supremum norm,
\begin{equation*}
M=\{ \varphi :\mathbb{R\to R}:\varphi \in C,\;
\| \varphi\| \leq \sqrt{3}/3,\| \varphi '\| \leq L'\} ,
\end{equation*}
and define ($B\varphi )(t):=\varphi (t)-\varphi ^3(t)$.
Then $B$ is a large contraction of the set $M$.
\end{proposition}

\begin{proof}
For each $t\in \mathbb{R}$ we have, for $\varphi ,\psi $ real functions,
\begin{align*}
| (B\varphi )(t)-(B\psi )(t)|
&=| \varphi (t)-\varphi ^3(t)-\psi (t)+\psi ^3(t)| \\
&=| \varphi (t)-\psi (t)| | 1-( \varphi ^2(t)+\varphi
(t)\psi (t)+\psi ^2(t)) | .
\end{align*}
Then for
\[
| \varphi (t)-\psi (t)| ^2=\varphi ^2(t)-2\varphi (t)\psi
(t)+\psi ^2(t)\leq 2(\varphi ^2(t)+\psi ^2(t))
\]
and for $\varphi ^2(t)+\psi ^2(t)<1$, we have
\begin{align*}
| (B\varphi )(t)-(B\psi )(t)|
&=| \varphi (t)-\psi (t)| [1-( \varphi ^2(t)+\psi ^2(t))
+| \varphi (t)\psi (t)| ] \\
&\leq | \varphi (t)-\psi (t)| [ 1-( \varphi
^2(t)+\psi ^2(t)) +\frac{\varphi ^2(t)+\psi ^2(t)}{2}] \\
&\leq | \varphi (t)-\psi (t)| [ 1-\frac{\varphi ^2(t)+\psi
^2(t)}{2}] .
\end{align*}
Thus, $B$ is pointwise a large contraction. But application $B$ is
still a large contraction for the supremum norm. For, let
$\varepsilon \in ( 0,1) $ be given and let
$\varphi,\psi \in M$ with $\| \varphi -\psi \| \geq \varepsilon $.

(a) Suppose that for some $t$ we have
$\varepsilon /2\leq | \varphi (t)-\psi (t)|$.
Then
\[
( \varepsilon /2) ^2\leq | \varphi (t)-\psi (t)|
^2\leq 2(\varphi ^2(t)+\psi ^2(t));
\]
that is,
$\varphi ^2(t)+\psi ^2(t)\geq \varepsilon ^2/8$.
For such $t$ we have
\[
| (B\varphi )(t)-(B\psi )(t)| \leq | \varphi (t)-\psi
(t)| [ 1-\frac{\varepsilon ^2}{8}]
\leq \| \varphi -\psi \| [ 1-\frac{\varepsilon ^2}{8}
].
\]

(b) Suppose that for some $t$,
$| \varphi (t)-\psi (t)| \leq \varepsilon /2$.
Then
\[
| (B\varphi )(t)-(B\psi )(t)| \leq | \varphi (t)-\psi (t)| \leq
(1/2)\| \varphi -\psi \| .
\]
Consequently,
\[
\| B\varphi -B\psi \| \leq \min [ 1/2,1-\frac{ \varepsilon
^2}{8}] \| \varphi -\psi \| .
\]
\end{proof}

We shall prove that the mapping $H$ has a fixed point which
solves \eqref{e1.1}, whenever its derivative exists.

\begin{lemma} \label{lem3.1}
For $A$ defined in \eqref{e3.2}, suppose that
\eqref{e2.1}-\eqref{e2.3} and \eqref{e3.3}-\eqref{e3.6} hold.
Then $A:M\to M$ is continuous in
the supremum norm and maps $M$ into a compact subset of $M$.
\end{lemma}

\begin{proof}
Clearly, if $\varphi $ is continuous then $A\varphi $ is. A change of
variable in \eqref{e3.2} shows that $(A\varphi )(t+T)=\varphi (t)$.
Observe that
\[
| G(t,x)| \leq | G(t,x)-G(t,0)| +| G(t,0)|
\leq k| x| +| G(t,0)| .
\]
So, for any $\varphi \in M$, we have
\begin{align*}
| (A\varphi )(t)|
&\leq (1-e^{-\int_{t-T}^ta(s)ds})^{-1}\int_{t-T}^t| G(u,\varphi
^3(u-g(u)))| e^{-\int_{u}^ta(s)ds}du \\
&\quad +| \frac{c(t)\varphi (t-g(t))}{(1-g'(t))}|\\
&\quad +(1-e^{-\int_{t-T}^ta(s)ds})^{-1}\int_{t-T}^t|
 h(u)(\varphi (u-g(u)))|
 e^{-\int_{u}^ta(s)ds}du \\
&\leq (1-e^{-\int_{t-T}^ta(s)ds})^{-1}\int_{t-T}^t(kL^3+|
G(u,0)| )e^{-\int_{u}^ta(s)ds}du+\alpha L \\
&\quad +(1-e^{-\int_{t-T}^ta(s)ds})^{-1}\int_{t-T}^t\delta
 a(u)Le^{-\int_{u}^ta(s)ds}du \\
&\leq \beta
L(1-e^{-\int_{t-T}^ta(s)ds})^{-1}\int_{t-T}^ta(u)e^{-
\int_{u}^ta(s)ds}du+\alpha L \\
&\quad +\delta
L(1-e^{-\int_{t-T}^ta(s)ds})^{-1}\int_{t-T}^ta(u)e^{-
\int_{u}^ta(s)ds}du \\
&\leq (\beta +\alpha +\delta )L\leq \frac{L}{J}<L.
\end{align*}
That is $A\varphi \in M$.

We show that $A$ is continuous in the supremum norm.
Let $\varphi ,\psi \in M $, and let
\begin{equation}
\begin{gathered}
\alpha '=\max_{t\in [ 0,T]} (1-e^{-\int_{t-T}^ta(s)ds})^{-1},\quad
\beta '=\max_{u\in[ t-T,t]} e^{-\int_{u}^ta(s)ds}, \\
\sigma =\max_{t\in [ 0,T]} \{ a(t)\} ,\quad
\rho = \max_{t\in [ 0,T]} | G(t,0)| ,\\
\mu =\max_{t\in [ 0,T]} | \frac{c'(t)}{(1-g'(t))}| , \quad
\upsilon =\max_{t\in [ 0,T]} | \frac{g''(t)c( t) }{( 1-g'(t)) ^2} | .
\end{gathered}   \label{e3.7}
\end{equation}
Then
\begin{align*}
&| (A\varphi )(t)-(A\psi )(t)|  \\
&\leq (1-e^{-\int_{t-T}^ta(s)ds})^{-1}\int_{t-T}^t| G(u,\varphi
 ^3(u-g(u)))-G(u,\psi ^3(u-g(u)))| \\
 &\quad\times e^{-\int_{u}^ta(s)ds}du
+| \frac{c(t)\varphi (t-g(t))}{(1-g'(t))}-\frac{c(t)\psi
(t-g(t))}{(1-g'(t))}| \\
&\quad + (1-e^{-\int_{t-T}^ta(s)ds})^{-1}\int_{0}^t| h(u)| |
(\varphi (u-g(u))-\psi (u-g(u))|\\
&\quad\times e^{-\int_{u}^ta(s)ds}du \\
&\leq k(1-e^{-\int_{t-T}^ta(s)ds})^{-1}\| \varphi ^3-\psi
 ^3\| \int_{t-T}^te^{-\int_{u}^ta(s)ds}
 +\alpha \| \varphi -\psi \| +\delta \| \varphi -\psi \|  \\
&\leq (3kT\alpha '\beta 'L^2+\alpha +\delta )\| \varphi -\psi \|.
\end{align*}
Let $\varepsilon >0$ be arbitrary. Define
$\eta =\frac{\varepsilon }{K}$, with $K=3kT\alpha '\beta 'L^2+\alpha
+\delta $, where $k$ is given by \eqref{e2.2}. Then, for
$\| \varphi -\psi \| <\eta $, we obtain
\[
\| A\varphi -A\psi \| \leq K\| \varphi -\psi \| <\varepsilon .
\]
It is left to show that $A$ is compact.
 Let $\varphi _{n}\in M$, where $n$ is a
positive integer. Then, as above, we see that
\begin{equation}
\| A\varphi _{n}\| \leq L.  \label{e3.8}
\end{equation}
Moreover, a direct calculation shows that
\begin{align*}
&(A\varphi _{n})'(t)\\
&=( G(t,\varphi _{n}^3(t-g(t)))-h(t) \varphi _{n}(t-g(t))
-a(t)(1-e^{-\int_{t-T}^ta(s)ds})^{-1}\\
&\quad\times \int_{t-T}^t[ G(u,\varphi
_{n}^3(u-g(u)))-h(u)\varphi _{n}(u-g(u))] e^{-\int_{u}^ta(s)ds}du
\\
&\quad +\frac{c'( t) \varphi _{n}(t-g(t))+c( t)
\varphi _{n}{}'(t-g(t))}{1-g'(t)}
+\frac{g''(t)c( t) \varphi _{n}(t-g(t))}{( 1-g'(t))^2}.
\end{align*}
By invoking the conditions \eqref{e2.2}, \eqref{e3.3}-\eqref{e3.5},
\eqref{e3.7} and \eqref{e3.8} we obtain
\begin{align*}
| (A\varphi _{n})'(t)|
&\leq kL^3+\rho +\delta a(t)L+a(t)L+\alpha L+\mu L+\alpha L'
+\upsilon L \\
&\leq kL^3+\rho +(\delta +1)\sigma L+(\alpha +\mu +\upsilon )L
+\alpha L'
\leq D,
\end{align*}
for some positive constant $D$. Hence the sequence $(A\varphi _{n})$
is uniformly bounded and equicontinuous. The Ascoli-Arzela theorem
implies that the subsequence $(A\varphi _{n_{k}})$ of $(A\varphi _{n})$
converges uniformly to a continuous $T-$periodic function.
Thus, $A$ is continuous and $AM$ is a compact set.
\end{proof}

\begin{lemma} \label{lem3.2}
Suppose \eqref{e2.1}-\eqref{e2.3} and \eqref{e3.3} hold. For $A,B$
defined in \eqref{e3.2} and \eqref{e3.1}, if $\varphi ,\psi \in M$
are arbitrary, then
\begin{equation*}
A\varphi +B\psi :M\to M.
\end{equation*}
Moreover, $B$ is a large contraction on $M$ with a unique fixed
point in $M$.
\end{lemma}

\begin{proof}
Let $\varphi ,\psi \in M$ be arbitrary. Note first that
$| \psi (t)| \leq \sqrt{3}/3$ implies
\begin{equation*}
| \psi (t)-\psi ^3(t)| \leq (2\sqrt{3})/9.
\end{equation*}
Using the definition of $B$, and the result of Lemma \ref{lem3.1},
we obtain
\begin{align*}
&| (A\varphi )(t)+(B\psi )(t)| \\
&\leq |(1-e^{-\int_{t-T}^ta(s)ds})^{-1}\int_{t-T}^tG(u,\varphi
^3(u-r(u)))e^{-\int_{u}^ta(s)ds}du \\
&\quad +\frac{c(t)\varphi (t-g(t))}{(1-g'(t))}-(1-e^{-
\int_{t-T}^ta(s)ds})^{-1}\int_{t-T}^th(u)(\varphi
(u-g(u)))e^{-\int_{u}^ta(s)ds}du| \\
&\quad +|  (1-e^{-\int_{t-T}^ta(s)ds})^{-1}\int_{t-T}^ta(u)|
\psi (u)-\psi
^3(u)| e^{-\int_{u}^ta(s)ds}du | \\
&\leq \frac{\sqrt{3}}{3J}+\frac{2\sqrt{3}}{9}
\leq L.
\end{align*}
Thus, $A\varphi +B\psi \in M$. Left to show that $B$ is a large
contraction with a unique fixed point in $M$.
Proposition \ref{prop3.1} shows that $\psi -\psi^3 $ is a large contraction
in the supremum norm. For any $\varepsilon $,
from the proof of that proposition, we have found a $\delta <1$,
such that
\begin{align*}
| (B\psi )(t)-(B\varphi )(t)|
&\leq (1-e^{-\int_{t-T}^ta(s)ds})^{-1}\int_{t-T}^ta(u)\delta \|
\psi -\varphi \| e^{-\int_{u}^ta(s)ds}du \\
&\leq \delta \| \psi -\varphi \| .
\end{align*}
Further, since $0\in M$ the above inequality shows, when $\varphi
=0$, we see that $B:M\to M$. This completes the proof.
\end{proof}

\begin{theorem} \label{thm3.1}
Let $(S,\|\cdot\| )$ be the Banach space of
continuous $T$-periodic real functions and
$M=\{ \varphi \in S:\| \varphi \| \leq L,\varphi'\text{ is bounded}\}$,
where $L=\sqrt{3}/3$. Suppose \eqref{e2.1}- \eqref{e2.3}
and \eqref{e3.3}-\eqref{e3.6} hold.
Then equation \eqref{e1.1} possesses a periodic solution $\varphi $ in
the subset $M$.
\end{theorem}

\begin{proof}
By Lemma \ref{lem2.1}, $\varphi $ is a solution of \eqref{e1.1} if
\begin{equation*}
\varphi =A\varphi +B\varphi ,
\end{equation*}
where $A$ and $B$ are given by \eqref{e3.2}, \eqref{e3.1} respectively.
By Lemma \ref{lem3.1}, $A:M\to M$ is continuous and $AM$ is
contained in compact subset of $M $. By Lemma \ref{lem3.2},
$A\varphi +B\psi \in M$ whenever
$\varphi ,\psi \in M$. Moreover, $B:M\to M$ is a large
contraction. Clearly, all the hypotheses of Theorem \ref{thm2.3} of
Krasnoselskii are satisfied. Thus, there exists a fixed point
$\varphi \in M$ such that $\varphi =A\varphi +B\varphi $. Hence
\eqref{e1.1} has a T-periodic solution.
\end{proof}

\begin{example} \label{exa3.1}\rm
Let $S$ and $M$ as in Theorem \ref{thm3.1} with $T=2\pi $ and
consider the neutral nonlinear equation
\[
x'(t)=-4.10^{-1}x^3(t)+10^{-3}\sin t.x'(t-1)+10^{-3}( \cos t+x^3(t-1)) .
\]
Then
\[
T=2\pi ,\quad a(t)=4.10^{-1},\quad c(t)=10^{-3}\sin t, \quad
G(t,x)=10^{-3}(\cos t+x),\quad  g(t)=1.
\]
Doing straightforward computations, we obtain
\[
k=\rho =\alpha =\mu =10^{-3},\quad
\upsilon =0,\quad \alpha '=(1-e^{-0.8\pi }) ^{-1}, \quad \beta '=1.
\]
By replacing, in \eqref{e3.3} and \eqref{e3.4},
$\beta =\delta =10^{-2}$, then any $J\in [3,47] $ satisfies
\eqref{e3.6}. All hypotheses of Theorem \ref{thm3.1} are
fulfilled and so the equation have at least a
$2\pi$-periodic solution belonging to $M$.
\end{example}

\subsection*{Acknowledgments}
The authors would like to thank the anonymous referee for
his/her valuable remarks.

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\end{document}