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

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2005(2005), No. 142, pp. 1--11.\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 2005 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2005/142\hfil Periodicity and stability]
{Periodicity and stability in neutral nonlinear
differential equations with functional delay}
\author[Y. M. Dib, M. R.  Maroun, Y. N. Raffoul\hfil EJDE-2005/142\hfilneg]
{Youssef M. Dib, Mariette R.  Maroun, Youssef N. Raffoul}  % in alphabetical order

\address{Youssef M. Dib \hfill\break
Department of Mathematics \\
University of Louisiana at Lafayette \\
Lafayette, LA 70504-1010, USA}
\email{youssef@louisiana.edu}

\address{Mariette R.  Maroun \hfill\break
Department of Mathematics\\
Baylor University\\
Waco, TX 76798-7328, USA}
\email{Mariette\_Maroun@baylor.edu}

\address{Youssef N. Raffoul \hfill\break
Department of Mathematics\\
University of Dayton \\
Dayton, OH 45469-2316, USA}
\email{youssef.raffoul@notes.udayton.edu}

\date{}
\thanks{Submitted November 30, 2004. Published December 6, 2005.}
\subjclass[2000]{34K20, 45J05, 45D05}
\keywords{Krasnoselskii; contraction; neutral differential equation;
\hfill\break\indent
 integral equation; periodic solution; asymptotic stability}

\begin{abstract}
 We study the existence and uniqueness of periodic
 solutions and the stability of the zero solution of the nonlinear
 neutral differential equation
 $$
 \frac{d}{dt}x(t) = -a(t)x(t)+ \frac{d}{dt}Q(t, x(t-g(t)))
 +G(t,x(t), x(t-g(t))).
 $$
 In the process we use  integrating factors and
 convert the given neutral differential equation into an equivalent
 integral equation. Then we construct appropriate mappings and
 employ Krasnoselskii's fixed point theorem to show the existence
 of a periodic solution of this neutral differential equation. We
 also use the contraction mapping principle to show the existence
 of a unique periodic solution and the asymptotic stability of the
 zero solution provided that $Q(0,0)= G(t, 0,0) = 0$.
\end{abstract}

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

\section{Introduction}

Theory of functional differential equations with
delay has undergone a rapid development in the previous fifty years.
We refer the readers to \cite{a1,b1,h1,h2,g4,l1}
 and the references therein
for a wealth of reference materials on the subject. More recently
researchers have given special attentions to the study of
equations in which the delay argument occurs in the derivative of
the state variable as well as in the independent variable,
so-called neutral differential equations. In particular,
qualitative analysis such as periodicity and stability of
solutions of neutral differential equations has been studied
extensively by many authors. We refer to
\cite{b2,g1,g2,g3,k1,m1,r1,r2,r3,r4} for some
recent work on the subject of periodicity and stability of neutral
equations. This reference list is not complete by any
means.

Neutral differential equations have many applications. For
example, these equations arise in the study of two or more simple
oscillatory systems with some interconnections between them
\cite{c1,s1}, and in modelling physical problems such as vibration of
masses attached to an elastic bar \cite{s1}.

In the current paper, we study the existence of periodic solutions
of the nonlinear system of differential equations
\begin{equation}\label{e1.1}
\frac{d}{dt}x(t) = -a(t)x(t)+ \frac{d}{dt}Q(t, x(t-g(t)))+G(t,
x(t), x(t-g(t)))
\end{equation}
where $a(t)$ is a continuous real-valued function. The functions
$Q: \mathbb{R}\times \mathbb{R} \to \mathbb{R}$ and $G:
\mathbb{R}\times \mathbb{R} \times \mathbb{R} \to
\mathbb{R}$ are continuous in their respective arguments. In the
analysis we use the idea of integrating factor and convert
equation \eqref{e1.1} into an integral equation.  Then we employ
Krasnoselskii's fixed point theorem and show the existence of a
periodic solution of \eqref{e1.1} in Theorem \ref{thm2.5}.  We also obtain
the existence of a unique periodic solution in Theorem \ref{thm2.6}
employing the contraction mapping principle as the basic
mathematical tool.

While most of the existing research on the qualitative analysis of
neutral differential equations deal with constant delay, equation
\eqref{e1.1} contains a non-constant function $g(t)$ as the delay term.
Related to our present work are articles \cite{m1,r1}, where in
\cite{r1}
the author has studied the periodic solutions of a scalar neutral
differential equation, and in \cite{m1} the authors considered the
discrete analogue of \cite{r1}. We remark that equation \eqref{e1.1} has a
couple of features that are distinct from the features of the
equation in \cite{r1}. The neutral term $\frac{d}{dt}Q(t, x(t-g(t)))$
of \eqref{e1.1} allows nonlinearity in the derivative term
$x'(t-g(t))$ (see  Example \ref{ex2.8}).  On the other hand, the
neutral term $x'(t-g(t))$ in \cite{r1} enters linearly.  As a
result of these differences, the mathematical analysis used in
this research to construct the mappings to employ fixed point
theorems are different than that of \cite{r1}. In addition to the study
of periodicity, in this research we obtain sufficient conditions
for the asymptotic stability of the zero solution.

\section{Existence of Periodic Solutions}

 For $T>0$ let  $P_T$ be the set of all continuous scalar functions
$x(t)$, periodic in $t$ of period $T$. Then $(P_T, \|\cdot\|)$ is
a Banach space with the supremum norm
$$
\| x \|= \sup_{t \in \mathbb{R}}|x(t)|=\sup_{t \in [0,T]}|x(t)|.
$$
Since we are
searching for the existence of periodic solutions for system
\eqref{e1.1}, it is natural to assume that
\begin{equation}\label{e2.1}
 a(t+T)=a(t),\quad  g(t+T)=g(t)
\end{equation}
with $g(t)$ being scalar, continuous, and $g(t)\geq g^{*}> 0$. Also,
we assume
\begin{equation} \label{e2.2}
 \int_0^T a(s)ds > 0.
 \end{equation}
Functions $Q(t,x)$ and $G(t,x,y)$ are  periodic in $t$ of period
$T$.  They are also globally Lipschitz continuous in $x$ and in
$x$ and $y$, respectively. That is
\begin{equation}\label{e2.3}
Q(t+T,x) = Q (t,x), \; G(t + T, x, y) = G(t, x, y),
\end{equation}
and there are positive constants $E_1, E_2,  E_3$ such that
 \begin{equation}\label{e2.4}
|Q(t,x) -Q(t,y)| \le E_1 \|x-y\|
 \end{equation}
 and
 \begin{equation}\label{e2.5}
|G(t,x, y)- G(t,z, w))| \le E_2 \|x-z\| + E_3 \|y-w\|.
\end{equation}
The next lemma is crucial to our results.

\begin{lemma} \label{lem2.1}
Suppose \eqref{e2.1} and \eqref{e2.3}
hold. If $x(t)\in P_T$, then $x(t)$ is a solution of equation
\eqref{e1.1} if and only if
\begin{equation}\label{e2.6}
\begin{aligned}
  x(t) &= Q(t, x(t-g(t)))\\
&\quad + (1-e^{-\int^t_{t-T}a(s)ds})^{-1}\int^t_{t-T}\Big[-a(u)Q(u,x(u-g(u)))
\\
&\quad +G(u, x(u),x(u-g(u)))\Big] e^{-\int^t_{u}a(s)ds}du.
\end{aligned}
\end{equation}
\end{lemma}

\begin{proof} Let $x(t) \in P_{T}$ be a solution of
\eqref{e1.1}.  First we write this equation as
\begin{align*}
 \frac{d}{dt}\{x(t)-Q(t, x(t-g(t)))\}
&=-a(t)\{x(t)-Q(t, x(t-g(t)))\}\\
&\quad -a(t)Q(t, x(t-g(t)))+G(t,x(t),x(t-g(t))).
\end{align*}
 Multiply both sides of \eqref{e1.1} with $e^{\int^t_{0}a(s)ds}$
and then integrate from $t-T$ to $t$ to obtain
\begin{align*}
&\int^t_{t-T}\Big[\big(x(u)-Q(u,
x(u-g(u)))\big)e^{\int^u_{0}a(s)ds}\Big]' du\\
&=\int^t_{t-T}\Big[-a(u)Q(u, x(u-g(u)))
+G(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*}
&\Big(x(t)-Q(t, x(t-g(t)))\Big)e^{\int^t_{0}a(s)ds}\\
&-\Big(x(t-T)-Q(t-T, x(t-T-g(t-T)))\Big)e^{\int^{t-T}_{0}a(s)ds}\\
&=\int^t_{t-T}\Big[-a(u)Q(u,
x(u-g(u)))+G(u, x(u),x(u-g(u)))\Big] e^{\int^u_{0}a(s)ds}du.
\end{align*}
Dividing both sides of the above equation by
 $\exp(\int^t_{0}a(s)ds)$ and the fact that $x(t)=x(t-T)$ and \eqref{e2.1},
 we obtain
\begin{align*}
&x(t) - Q(t, x(t-g(t)) \\
&=(1-e^{-\int^t_{t-T}a(s)ds})^{-1}\int^t_{t-T}\Big[-a(u)Q(u,
x(u-g(u)))\\
&+G(u, x(u),x(u-g(u)))\Big] e^{-\int^t_{u}a(s)ds}du.
\end{align*}
\end{proof}

Define a mapping $H $ by
\begin{equation}\label{e2.7}
\begin{aligned}
 (H\varphi)(t)&=Q(t,\varphi(t-g(t))) \\
&\quad + (1-e^{-\int^t_{t-T}a(s)ds})^{-1}\int^t_{t-T}
\Big[-a(u)Q(u,\varphi(u-g(u)))\\
&\quad +G(u,\varphi(u),\varphi(u-g(u)))\Big] e^{-\int^t_{u}a(s)ds}du.
\end{aligned}
\end{equation}
It is clear form \eqref{e2.7}  that $H : P_{T}\to P_{T}$
by the way it was constructed in Lemma \ref{lem2.1}.

Next we state
Krasnoselskii's fixed point theorem which enables us to prove the
existence of a periodic solution. For the proof of Krasnoselskii's
fixed point theorem we refer the reader to \cite{d1}.

\begin{theorem}[Krasnoselskii] \label{thm2.2}
Let $\mathbb{M}$ be a closed convex nonempty subset of a Banach
space $(\mathbb{B}, \| \cdot\|)$. Suppose that $C$ and $B$ map
$\mathbb{M}$ into $\mathbb{B}$
such that
\begin{itemize}
\item[(i)] $x,y \in \mathbb{M}$, implies $Cx + By \in \mathbb{M}$,
\item[(ii)] $C$ is continuous and $C\mathbb{M}$ is contained in a compact set,
\item[(iii)] $B$ is a contraction mapping.
\end{itemize}
Then there exists $z \in \mathbb{M}$ with $z=Cz+Bz$.
\end{theorem}

We note that to apply the above theorem we need to construct two
mappings; one is contraction and the other is
compact.  Therefore, we express equation \eqref{e2.7} as
$$
(H\varphi)(t)=(B\varphi)(t)+(C\varphi)(t),
$$
where $C,B:P_{T}\to P_{T}$ are given by
\begin{equation}\label{e2.8}
(B\varphi)(t)= Q(t, \varphi(t-g(t)))
\end{equation}
and
\begin{equation}\label{e2.9}
\begin{aligned}
(C\varphi)(t)
&=(1-e^{-\int^t_{t-T}a(s)ds})^{-1}\int^t_{t-T}
\Big[-a(u)Q(u,\varphi(u-g(u))) \\
&\quad +G(u, \varphi(u),\varphi(u-g(u)))\Big]
e^{-\int^t_{u}a(s)ds}du.
\end{aligned}
\end{equation}
To simplify notations, we introduce the following constants.
\begin{equation}\label{e2.10}
  \eta = \max_{t \in [0,T]}  |(1-e^{-\int^t_{t-T}a(s)ds})^{-1}|,\quad
    \rho = \max_{t \in [0,T]}|a(t)|,\quad
\gamma =\max_{u \in [t-T,t]} e^{-\int^t_{u}a(s)ds}.
\end{equation}

\begin{lemma} \label{lem2.3}
 If $C$ is defined by \eqref{e2.9}, then $C$ is
continuous and the image of $C$ is contained in a compact set.
\end{lemma}

 \begin{proof} Let $C$ be defined by \eqref{e2.9}
and let $\varphi,\psi\in P_{T}$.
 Given $\epsilon > 0$, take $\delta =\epsilon/N$ with
$N= \eta\gamma T(\rho E_1 + E_2 + E_3)$, where $E_1$, $E_2$ and
$E_3$ are given by \eqref{e2.4} and \eqref{e2.5}.  Now, for
 $\|\varphi-\psi\|<\delta$. we obtain
  \[
\big\|C\varphi-C\psi\big\|
\leq \eta\gamma\int^T_{0}\big[\rho E_1\|\varphi-\psi\|+(E_2 +
E_3)\|\varphi-\psi\|  \big]du
\leq  N\|\varphi-\psi\|<\epsilon.
\]
This proves that $C$ is continuous. To show that the image of $C$
is contained in a compact set, we consider $D = \{\varphi \in
P_T:\|\varphi\| \leq R\}$, where $R$ is a fixed positive constant.
Let $\varphi_{n}\in D$ where $n$
is a positive integer. Observe that in view of (2.4) and (2.5) we have
\begin{align*}
|Q(t,x)| &= |Q(t,x)- Q(t,0)+Q(t,0)|\\
&\leq |Q(t,x)- Q(t,0)|+ |Q(t,0)|\\
&\leq E_1\|x\| +  \alpha.
\end{align*}
Similarly,
\begin{align*}
|G(t,x,y)| &= |G(t,x,y)- G(t,0,0)+ G(t,0,0)|\\
&\leq |G(t,x,y)- G(t,0,0)|+ |G(t,0,0)|\\
&\leq E_2\|x\| + E_3\|y\| + \beta
\end{align*}
where $\alpha =  \sup_{t\in[0,T]}|Q(t,0)|$ and
$\beta=\sup_{t\in[0,T]}|G(t,0,0)|$. Hence, if $C$ is given by \eqref{e2.9}
we obtain that
$$ \|C\varphi_{n}\| \leq L
$$
for some positive constant $L$. Next we calculate
$(C\varphi_{n})'(t)$ and show
that it is uniformly bounded. By making use of \eqref{e2.2} and
\eqref{e2.3} we obtain by taking the derivative in \eqref{e2.9} that
\[
(C\varphi_{n})'(t)= -a(t)C(\varphi_{n}(t)) -a(t)Q(t,
\varphi_n(t-g(t)))+ G(t,\varphi_n(t), \varphi_n(t-g(t))).
\]
Thus, the above expression yields
 $\|(C\varphi_{n})'\| \leq F$, for some positive constant $F$.
Thus the sequence
${C\varphi_{n}}$ is uniformly bounded and equi-continuous. Hence
by Ascoli-Arzela's theorem $C(D)$ is compact.
\end{proof}

\begin{lemma} \label{lem2.4}
If $B$ is given by \eqref{e2.8}
with $E_1 <1$, then $B$ is a contraction.
\end{lemma}

 \begin{proof}
Let $B$ be defined by \eqref{e2.8}. Then for $\varphi, \psi \in  P_{T}$
 we have
 \begin{align*}
\|B(\varphi)-B(\psi)\|&=\sup_{t\in[0,T]}|B(\varphi)-B(\psi)|\\
&\leq  E_1 \sup_{t\in[0,T]}|\varphi(t-g(t))-\psi(t-g(t)|\\
&\leq E_1\|\varphi-\psi\|.
\end{align*}
Hence $B$ defines a contraction.
\end{proof}

\begin{theorem} \label{thm2.5}
 Suppose the hypothesis of Lemma \ref{lem2.4}. Let $\alpha = \sup_{t\in[0,T]}|Q(t,0)|$ and
$\beta =\sup_{t\in[0,T]}|G(t,0,0)|$. Suppose \eqref{e2.1}-\eqref{e2.5}
hold. Let $J$ be a positive constant satisfying the inequality
\begin{equation}\label{e2.11}
\alpha+ E_1 J + \eta T \gamma\big[\rho (E_1 J +\alpha) +(E_2 +
E_3)J+\beta\big] \leq J.
\end{equation}
Let $\mathbb{M} = \{ \varphi \in P_{T}: \|\varphi \| \leq J\}$.
Then equation \eqref{e1.1} has a solution in $M$.
\end{theorem}

\begin{proof} Define $\mathbb{M} = \{ \varphi \in P_{T}:
\|\varphi \| \leq  J\}$. By Lemma \ref{lem2.3}, $C$ is continuous and $C
\mathbb{M}$ is contained in a compact set. Also, from Lemma \ref{lem2.4},
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 $\|C\phi +B\psi\|\leq J$. Let $\varphi,
\psi \in \mathbb{M} $ with $\|\varphi\|, \|\psi\|\leq J$. Then
\begin{align*}
\big\|C\varphi+B\psi\big\|
 &\leq E_1 \|\psi\| + \alpha +  \eta\gamma\int^T_0[|a(u)|(\alpha
+E_1\|\varphi\|)+E_2
 \|\varphi\| + E_3 \|\varphi\| +\beta]du\\
 &\leq \alpha+ E_1 J + \eta T \gamma\Big[\rho(E_1 J +\alpha) +(E_2 +
E_3)J+\beta\Big]  \leq J .
 \end{align*}
We now see that all the conditions of Krasnoselskii's theorem are
satisfied. Thus there exists a fixed point $z$ in $\mathbb{M}$ such
that $z=Az+Bz$. By Lemma \ref{lem2.1}, 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} hold. If
$$
E_1 + \eta \gamma T(\rho E_1 + E_2 + E_3)< 1,
$$
then equation \eqref{e1.1} has a unique $T$-periodic solution.
\end{theorem}

\begin{proof} Let the mapping $H$ be given by
\eqref{e2.7}. For $\varphi,\psi\in P_{T}$, in view of
\eqref{e2.7}, we have
\[
\big\|H\varphi-H\psi\big\| \leq \big(E_1 +
\eta \gamma T(\rho E_1 + E_2 + E_3)\big)\|\varphi-\psi\|.
\]
 This completes the proof by invoking the contraction mapping
principle.
\end{proof}

It is worth noting that Theorems \ref{thm2.5} and \ref{thm2.6}
 are not applicable to functions such as
\begin{align*}
G(t,\varphi(t),\varphi(t-g(t)))= f_1(t)\varphi^2(t)+
f_2(t)\varphi^2(t-g(t))),
 \end{align*}
 where $f_1(t), f_2(t)$ and $g(t) > 0$ are continuous and
 periodic. To accommodate such functions, we state the following corollary,
in which  the functions $G$ and $Q$ are required to satisfy
local Lipschitz conditions.

\begin{corollary} \label{coro2.7}
 Suppose \eqref{e2.1}-\eqref{e2.3} hold and let
$\alpha$ and $ \beta $ be the constants defined in Theorem \ref{thm2.5}. Let
$J$ be a positive constant and define $\mathbb{M} = \{ \varphi \in
P_{T}: \|\varphi \| \leq J\}$. Suppose there are positive constants
$E_1^{*}, E_2^{*}$ and $E_3^{*}$ so that for $x, y, z$ and $w  \in
\mathbb{M}$ we have
 \begin{gather*}
|Q(t,x) - Q(t,y)| \le E_1^{*} \|x-y\|, \\
|G(t,x, y)- G(t,z, w)| \le E_2^{*} \|x-z\| + E_3^{*} \|y-w\|.
\end{gather*}
If
$E_1^{*} < 1 $ and $\|H\varphi\| \leq J$, for
$\varphi \in \mathbb{M}$, then  \eqref{e1.1} has a $T$-periodic
solution in $\mathbb{M}$.
Moreover, if
\begin{equation}
\label{e2.12} E_1^{*} + \eta\gamma T(\rho E_1^{*} + E_2^{*} + E_3^{*})< 1,
\end{equation}
then \eqref{e1.1} has a unique solution in $\mathbb{M}$.
\end{corollary}

\begin{proof} Let
$\mathbb{M} = \{ \varphi \in P_{T}:\|\varphi \| \leq  J\}$.
Let the mapping $H$ be given by \eqref{e2.7}. Then the results
follow immediately from Theorem \ref{thm2.5} and Theorem \ref{thm2.6}.
\end{proof}

We remark that the constants $E_j^{*}$, $j =1,2,3$ may depend on $J$.
Now we display an example.

\begin{example} \label{ex2.8}
For small positive $\varepsilon_1$ and $\varepsilon_2$,
 we consider the perturbed Van Der Pol   equation
  \begin{equation}\label{e2.13}
  x'= -(2 + \sin(\omega t))x(t) +
  \varepsilon_1\frac{d}{dt}\big(\sin(\omega t)x^2(t-g(t))\big)
+\varepsilon_2 \big(\cos(\omega t)+ x^2(t)\big)
  \end{equation}
where $g(t)$ is nonnegative, continuous and
$\frac{2\pi}{\omega}$-periodic for $\omega$ is positive. So we have
$$
a(t) =2 + \sin(\omega t), \quad  Q(t, x(t-g(t)))
= \varepsilon_1\sin(\omega t)x^2(t-g(t))
$$
and
$$
G(t,x(t), x(t-g(t)))=\varepsilon_2 \big(\cos(\omega t)+x^2(t)\big).
$$
Define $\mathbb{M} = \{ \phi \in
P_{\frac{2\pi}{\omega}}: \|\phi \| \leq J\}$, where $J$ is a
positive constant. For  $\varphi \in \mathbb{M}$, we have
\begin{align*}
\|H\varphi\|&=\|Q(t,\varphi(t-g(t)))\\
&\quad +(1-e^{-\int^t_{t-T}a(s)ds})^{-1}\int^t_{t-T}
\Big[-a(u)Q(u,\varphi(u-g(u)))\\
&\quad +G(u,\varphi(u),\varphi(u-g(u)))\Big] e^{-\int^t_{u}a(s)ds}du\|\\
&\leq \varepsilon_1J^2 +\frac{2 \pi}{\omega}\Big(1 - e^{-\frac{4
\pi}{\omega}}\Big)^{-1} \Big[3\varepsilon_1 J^2+\varepsilon_2J^2
+\varepsilon_2 \Big].
\end{align*}
Thus, the inequality
\begin{equation}\label{e2.14}
\varepsilon_1J^2+\frac{2 \pi}{\omega}
\big(1 - e^{-\frac{4 \pi}{\omega}}\big)^{-1}
\big[3\varepsilon_1 J^2+\varepsilon_2J^2 +\varepsilon_2 \big] \leq
J,
\end{equation}
which is satisfied for small
$\omega,\,\varepsilon_1$ and $\varepsilon_2$, implies
$\|H\varphi\|\leq J$. Hence,  \eqref{e2.13} has a
$\frac{2\pi}{\omega}$-periodic solution, by Corollary \ref{coro2.7}.

For the uniqueness of the solution we let
$ \varphi, \psi \in \mathbb{M}$. From  (2.13) we see that
$$
\eta = \big(1 -e^{-\frac{4 \pi}{\omega}}\big)^{-1}, \quad
\rho \leq 3 ,\quad \gamma \leq 1.
$$
 Also
$\alpha = 0$, $\beta =\varepsilon_2$,
$E_1^{*} = 2\varepsilon_1  J$,
$E_3^{*}= 0$, $E_2^{*} =2\varepsilon_2J$,
where $J$ is given by \eqref{e2.14}. If
$$
2\varepsilon_1 J +\frac{2\pi}{\omega}
\big(1 - e^{-\frac{4 \pi}{\omega}}\big)^{-1}
\big[6\varepsilon_1 J+2\varepsilon_2J \big] <1
$$
is satisfied for small $\varepsilon_1$ and $\varepsilon_2$,
then  \eqref{e2.13} has a unique $\frac{2\pi}{\omega}$-periodic solution.
\end{example}

\section{Stability}

Lyapunov functions and functionals have been successfully used to
obtain boundedness, stability and the existence of periodic
solutions of differential equations, differential equations with
functional delays and functional differential equations. In the
study of differential equations with functional delays by using
Lyapunov functionals, many difficulties arise if the delay is
unbounded or if the differential equation in question has
unbounded terms. In \cite{r3}, the third author using fixed point
theory, studied the asymptotic stability of the zero solution  of
the scalar neutral differential equation
\begin{equation}\label{e3.1}
 x'(t) = -a(t)x(t)+ c(t)x'(t-g(t))+
q\big(t,x(t), x(t-g(t)\big),
\end{equation}
where $a(t), b(t), g(t)$ and $q$ are continuous in their
respective arguments. It is clear that \eqref{e1.1} is more
general than \eqref{e3.1}.

This section is mainly concerned with the asymptotic stability of
the zero solution of \eqref{e1.1}. We assume that the functions
$Q$ and $G$ are continuous, as before. Also, we assume that $g(t)$
is continuous, $g(t)\geq g^{*}
> 0$ and $Q(t,0) = G(t,0,0) = 0$. The techniques used in this
section are adapted from the paper of \cite{b1}.

To arrive at the correct mapping, we rewrite \eqref{e1.1} as in
the proof of Lemma \ref{lem2.1}, multiply both sides by
$e^{\int^t_{0}a(s)ds}$ and then integrate from $0$ to $t$ to
obtain
\begin{equation}\label{e3.2}
\begin{aligned}
  x(t) &= Q(t, x(t-g(t)))+ \Big(x(0) - Q(0,
x(-g(0)))\Big)e^{-\int^t_{0}a(s)ds}\\
&\quad + \int^t_{0}\Big[-a(u)Q(u,x(u-g(u)))\\
&\quad + G(u, x(u),x(u-g(u)))\Big] e^{-\int^t_{u}a(s)ds}du.
\end{aligned}
\end{equation}
Thus, we see that $x(t)$ is a solution of \eqref{e1.1} if and only
if it satisfies \eqref{e3.2}.
Let $\psi (t):(-\infty ,0]\to \mathbb{R}$ be a given continuous
bounded initial function. We say $x(t):=x(t,0,\psi)$ is a solution
of \eqref{e1.1} if $x(t) = \psi(t) $ for $t \leq 0 $ and satisfies
 \eqref{e1.1} for $t \geq 0$.

We say the zero solution of \eqref{e1.1} is stable at $t_0$ if for
each $\varepsilon>0$, there is a $\delta = \delta(\varepsilon)>0$
such that $\big[\psi:(-\infty, t_{0}]\to \mathbb{R}$ with
$\|\psi\|< \delta$ on $(-\infty, t_{0}]$,
implies $|x(t,t_0,\psi)|< \varepsilon$.

Without loss of generality, we will state and prove our results by
starting at $t_0 = 0$.
Let $C$ be the space of all continuous functions from
$\mathbb{R}\to \mathbb{R}$ and define the set $S$ by
\begin{align*}
S=\Big\{&\varphi :\mathbb{R}\to \mathbb{R}:
\varphi(t)=\psi(t) \mbox{ if } t\leq 0,
  \varphi(t)\to 0, \mbox{ as } t\to \infty,\\
& \varphi \in C  \mbox{ and } \varphi  \mbox{ is bounded} \Big\}.
\end{align*}
Then, $(S,\|\cdot\|)$ is a complete metric space
 where $\|\cdot\|$ is the supremum norm.
 For the next theorem we impose the following conditions.
\begin{equation}\label{e3.3}
\int^{t}_{0}a(s)\;ds > 0\quad\mbox{and}\quad
e^{-\int^{t}_{0}a(s)ds}\to 0,\quad \mbox{as }
t\to \infty,
\end{equation}
there is an $\alpha > 0$ such that
\begin{gather}\label{e3.4}
E_1+ \int^t_{0}\Big[|a(u)|E_1 + E_2 +
E_3\Big]e^{-\int^t_{u}a(s)ds}du \leq \alpha < 1, \quad t \geq 0,\\
\label{e3.5}
t - g(t) \to \infty, \quad \mbox{as } t\to \infty, \\
\label{e3.6}
Q(t, 0) \to 0, \quad \mbox{as } t\to \infty.
\end{gather}

\begin{theorem} \label{thm3.1}
If \eqref{e2.4},\eqref{e2.5}, \eqref{e3.3} - \eqref{e3.6} hold,
then every solution $x(t,0,\psi)$ of \eqref{e1.1} with small continuous
initial function $\psi (t)$, is bounded and approaches  zero as
$t \to \infty$. Moreover, the zero solution is stable at
$t_0=0$.
\end{theorem}

 \begin{proof} Define the mapping $P:S \to S$ by
\[
\big(P\varphi\big)(t)=
\begin{cases}
\psi (t)& \mbox{if }t \leq 0,\\[4pt]
Q(t, \varphi(t-g(t)))+ \big[\psi(0) - Q(0,
\psi(-g(0)))\big]e^{-\int^t_{0}a(s)ds} \\
+\int^t_{0}\Big[-a(u)Q(u,\varphi(u-g(u))) \\
+G(u, \varphi(u),\varphi(u-g(u)))\Big]
e^{-\int^t_{u}a(s)ds}du, &\mbox{if }t \geq 0\,.
\end{cases}
\]
It is clear that for $\varphi \in S$, $P\varphi$ is continuous.
Let $\varphi \in S$ with $\|\varphi\|\leq K$, for some positive
constant $K$. Let $\psi(t)$ be a small given continuous initial
function with $\|\psi\|<\delta, \delta >0$. Then using
\eqref{e3.4} in the definition of $(P\varphi)(t)$, we have
\begin{equation}\label{e3.7}
\begin{aligned}
\|(P\varphi\big)(t) \|
&\leq  E_1 K+ \big|\big(\psi(0) - Q(0,
\psi(-g(0)))\big)\big|e^{-\int^t_{0}a(s)ds} \\
&\quad+ \int^t_{0}\big[|a(u)|E_1 + E_2 +
E_3\big]e^{-\int^t_{u}a(s)ds}du K \\
&\leq \big(1+E_1\big)\delta +E_1 K + \int^t_{0}\big[|a(u)|E_1 + E_2 +
E_3\big]e^{-\int^t_{u}a(s)ds}du K \\
&\leq \big(1 +E_1\big)\delta +  \alpha K,
\end{aligned}
\end{equation}
which implies $\|\big(P\varphi\big)(t) \| \leq K$, for the
right $\delta$. Thus, \eqref{e3.7} implies that $(P\varphi)(t)$ is
bounded. Next we show that $\big(P\varphi\big)(t) \to 0$
as $t \to \infty$. The second term on the right side of
$\big(P\varphi\big)(t)$ tends to zero, by condition \eqref{e3.3}.
Also, the first term on the right side tends to zero, because of
\eqref{e3.5}, \eqref{e3.6}  and the fact that $\varphi \in S$.
Left to show that the integral term goes to zero as $t\to \infty$.

Let $\epsilon >0$ be given and  $\varphi \in S$ with
$\|\varphi\|\leq K$, $K > 0$. Then, there exists a $t_{1} >0$ so
that for $t> t_{1}$, $|\varphi(t-g(t))| < \epsilon$. Due to
condition \eqref{e3.3}, there exists a $t_{2} > t_{1}$ such that
for $t>t_{2}$ implies that
$\exp(-\int^t_{t_{1}}a(s)ds) < \frac{\epsilon}{\alpha K}$.
Thus for $t>t_{2}$, we have

\begin{align*}
&\Big|\int^t_{0}\Big[-a(u)Q(u,\varphi(u-g(u)))+G(u,
\varphi(u),\varphi(u-g(u)))\Big]
e^{-\int^t_{u}a(s)ds}du\Big|\\
&\leq K\int^{t_{1}}_{0}\big[|a(u)|E_1 + E_2 +
E_3\big]e^{-\int^t_{u}a(s)ds}du\\
&\quad+ \epsilon \int^t_{t_{1}}\big[|a(u)|E_1 + E_2 +
E_3\big]e^{-\int^t_{u}a(s)ds}du\\
&\leq Ke^{-\int^{t}_{t_1}a(s)ds}\int^{t_{1}}_{0}\big[|a(u)|E_1 + E_2 +
E_3\big]e^{-\int^{t_{1}}_{u}a(s)ds}du + \alpha \epsilon \\
&\leq \alpha K e^{-\int^{t}_{t_1}a(s)ds} + \alpha \epsilon
\leq \epsilon +\alpha \epsilon.
\end{align*}
Hence, $\big(P\varphi\big)(t) \to 0$ as $t \to
\infty$. It remains to show that $\big(P\varphi\big)(t)$ is a
contraction under the supremum norm. Let $\zeta, \eta \in S$. Then
\begin{align*}
 \big|(P\zeta)(t)-(P\eta)(t)\big| &\leq \big\{E_1 +
\int^t_{0}\big[|a(u)|E_1 + E_2 +
E_3\big]e^{-\int^t_{u}a(s)ds}du\big\}\|\zeta - \eta\|\\
&\leq \alpha \|\zeta - \eta\|.
\end{align*}
Thus, by the contraction mapping principle, $P$ has a unique fixed
point in $S$ which solves \eqref{e1.1}, bounded and tends to zero
as t tends to infinity. The stability of the zero solution at $t_0
=0$ follows from the above work by simply replacing $K$ by
$\epsilon$. This completes the proof.
\end{proof}

 \begin{example} \label{ex3.2} \rm
Let $\psi$ be a small continuous initial
  function, $\psi : (-\infty,0] \to \mathbb{R}$ with
$\|\psi\| \leq \delta$, for positive
$\delta$. For small $\varepsilon_1$ and $\varepsilon_2$, we
consider the nonlinear neutral differential equation,
\begin{equation}\label{e3.8}
    x' (t) = -2x(t)+ \varepsilon_1 \frac{d}{dt} x^2(t- (\frac{t}{2})) +
    \varepsilon_2\big(x^2(t) + x^2(t- (\frac{t}{2}))\big).
\end{equation}
Suppose
\begin{equation}\label{e3.9}
0 < 4(2\varepsilon_1 +
\varepsilon_2)\delta(1 + \varepsilon_1\;\delta) <1,
\end{equation}
and define the set $S$ by
\begin{equation*}
 S=\big\{\varphi :\mathbb{R}\to \mathbb{R}|\;
\varphi(t)=\psi(t)\; \mbox{if}\; t\leq 0, \varphi(t)\to 0,
\mbox{as}\; t\to \infty, \varphi \in C \; \mbox{and} \;
\|\varphi\| \leq K\big\},
\end{equation*}
for positive constant $K$ satisfying the inequality
\begin{equation}
\label{e3.10}\frac{1 -
\sqrt{1-4(2\varepsilon_1 + \varepsilon_2)\delta(1 +
\varepsilon_1\;\delta)}}{2(2\varepsilon_1 + \varepsilon_2)} < K <
\frac{1}{2(2\varepsilon_1 + \varepsilon_2)}.
\end{equation}
Then every solution $x(t, 0, \psi)$ of \eqref{e3.8} is bounded and
approaches  0 as $t \to \infty$.
\end{example}

 \begin{proof} It is
clear from \eqref{e3.9} that inequality \eqref{e3.10} is well
defined. Define
\[
(P\varphi)(t)= \begin{cases}
\psi (t) &\mbox{if } t \leq 0 ,\\[4pt]
=\varepsilon_1 \varphi^2(t/2)+
\big(\psi(0)- \varepsilon_1\;\psi^2(0)\big)e^{-2t}
+\int^t_{0}\big(-2\varepsilon_1 \varphi^2(u/2) \\
+\varepsilon_2(\varphi^2(u) + \varphi^2(u/2))\big)
e^{-2(t-u)}du &\mbox{if } t\geq 0]\,.
\end{cases}
\]
  Then, for $\varphi \in S $ with $\|\varphi \| \leq K$,
we have
\begin{equation}\label{e3.11}
\begin{aligned}
\|P\varphi)\| &\leq \varepsilon_1\; K^2 + (1
+\varepsilon_1\;\delta)\delta +
2(\varepsilon_1+\varepsilon_2)K^2\;\int^t_0e^{-2(t-u)}\;du\nonumber\\
&\leq (2\varepsilon_1 + \varepsilon_2)K^2 + \delta(1 +
\varepsilon_1\;\delta).
\end{aligned}
\end{equation}
In order for $P$ to map $S$ into itself, we need to ask that,
using \eqref{e3.11},
\begin{equation}\label{e3.12}
(2\varepsilon_1 + \varepsilon_2)K^2 + \delta(1 +
\varepsilon_1\;\delta) \leq K.
\end{equation}
But inequality
\eqref{e3.12} is satisfied for $K$ satisfying \eqref{e3.10}, by
noting that
$$
\frac{1}{2(2\varepsilon_1 + \varepsilon_2)}< \frac{1
+ \sqrt{1-4(2\varepsilon_1 + \varepsilon_2)\delta(1 +
\varepsilon_1\;\delta)}}{2(2\varepsilon_1 + \varepsilon_2)}.
$$
 Thus, we have shown that if $\varphi \in S$, then $\|\big(P\varphi\big)
\| \leq K$. It is obvious that conditions \eqref{e3.3},
\eqref{e3.5}  and \eqref{e3.6} are satisfied. Left to show that
that $P$ defines a contraction mapping on the metric  space $S$.
Let $\zeta, \eta \in S$. Then
\begin{align*}
 \big|(P\zeta)(t)-(P\eta)(t)\big|
&\leq 2\varepsilon_1\;K\;\|\zeta - \eta\|
  + \big(4\varepsilon_1 K + 4\varepsilon_2 K\big)\int^t_0e^{-2(t-u)}\;du \;\|\zeta - \eta\|\\
&\leq \big[2\varepsilon_1 K +
  2K(\varepsilon_1+\varepsilon_2)\big]\|\zeta - \eta\|\\
&=2(2\varepsilon_1+\varepsilon_2)K\|\zeta - \eta\|.
\end{align*}
  By condition \eqref{e3.10}, we have
\[
 \Big|(P\zeta)(t)-(P\eta)(t)\Big|\leq \alpha\|\zeta - \eta\|,\quad
 \alpha\in (0,1).
\]
Hence, by Theorem \ref{thm3.1}, every solution $x(t,0,\psi)$ of
\eqref{e3.8} with small continuous initial function
$\psi(t):(-\infty,\;0]\to \mathbb{R}$, is in $S$, bounded and
approaches zero as $t \to \infty$.
\end{proof}

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\end{document}