\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 99, pp. 1--5.\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-2010/99\hfil
Stability of delay differential equations]
{Stability of delay differential equations with oscillating coefficients}

\author[M. I. Gil'\hfil EJDE-2010/99\hfilneg]
{Michael I. Gil'} 

\address{Michael I. Gil' \newline
Department of Mathematics \\
Ben Gurion University of the Negev \\
P.0. Box 653, Beer-Sheva 84105, Israel}
\email{gilmi@cs.bgu.ac.il}

\thanks{Submitted April 13, 2010. Published July 22, 2010.}
\subjclass[2000]{34K20}
\keywords{Linear delay differential equation;
exponential stability}

\begin{abstract}
 We study the solutions to the delay differential equation  equation
 $$
 \dot x(t)=-a(t)x(t-h),
 $$
 where the coefficient $a(t)$ is not necessarily positive.
 It is proved that this equation is exponentially stable provided
 that $a(t)=b+c(t)$ for some positive constant $b$ less than
 $\pi/(2h)$, and the integral  $\int_0^t c(s)ds$ is sufficiently
 small for all $t>0$. In this case the $3/2$-stability theorem
 is improved.
\end{abstract}

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


\section{Introduction and preliminaries}

This article concerns the equation
\begin{equation}
\dot x(t)=-a(t)x(t-h), \label{e1.1}
\end{equation}
where $\dot x=dx/dt$, the delay $h$ is a positive constant,
and $a(t)$ a piece-wise continuous function bounded on $[0,\infty)$.
We do not require that $a(t)$ be positive, and
therefore,  the ``characteristic function'' $z+a(t)e^{-zh}$ can be
unstable for some $t\ge 0$.

The sharp stability condition
(the so called $3/2$-stability theorem) for  first-order
functional-differential equations with one variable
delay was established by Myshkis \cite{my} (see also \cite{ko}).
A similar result was established by  Lillo \cite{li}.
The  $3/2$-stability theorem asserts that \eqref{e1.1} is uniformly
stable, provided that $0< h a(t)\le 3/2$  for all $t\ge 0$.
The upper bound 3/2 is the best possible. In fact, if
$h\sup_t a(t)> 3/2$, then there are equations having unbounded
solutions. The $3/2$-theorem was generalized to nonlinear equations
and equations with unbounded delays in the very interesting
papers \cite{yo1,yo2,yo3}.
In this article, under some additional conditions we improve
the $3/2$-theorem.

We consider \eqref{e1.1} as a perturbation
of the equation
\begin{equation}
\dot y(t)=-by(t-h) \label{e1.2}
\end{equation}
with a positive constant $b<\pi /(2 h)$
satisfying a condition stated below.
The fundamental solution to \eqref{e1.2} is 
$$
F_b(t)=\frac1{2\pi i}\int_{-i\infty} ^{i\infty}
\frac{e^{zt}dz}{z+be^{-zh}}\,.
$$
For a  function $f$ defined and bounded on $[0,\infty)$
(not necessarily continuous), we introduce the norm
$\|f\|_\infty=\sup_{t\ge 0}|f(t)|$.
So $\|a\|_\infty=\sup_{t\ge 0}|a(t)|$.
In addition, put
$$
\|f\|_{L^1}=\int_0^\infty |f(t)|dt,
$$
if the integral exists.
Now we are in a position to formulate our main result.

\begin{theorem} \label{thm1.1}
Let there be a constant $b\in (0, \pi/(2h))$,
such  that
$$
w_b:=\sup_{t\ge 0}|\int_0^t (a(t)-b)dt|
$$
is finite and satisfies the inequality
\begin{equation}
w_b < \frac{1}{1+ (b+\|a\|_\infty) \|F_b\|_{L^1}}.
\label{e1.3}
\end{equation}
 Then  \eqref{e1.1} is exponentially stable.
\end{theorem}

This theorem is proved in the next section.
Its assumptions are sharp: 
if $a(t)\equiv b$, then
$w_b=0$ and condition \eqref{e1.3} is automatically fulfilled.

Furthermore,  let
\begin{equation}
ehb<1. \label{e1.4}
\end{equation}
Then $F_b(t)\ge 0$ and  \eqref{e1.2} is exponentially stable,
cf. \cite{gi} and references therein.
Now, integrating \eqref{e1.2}, we have
$$
1=F_b(0)=b\int_0^\infty F_b(t-h)dt=b\int_h^\infty  F_b(t-h)dt=b\|F_b\|_{L^1}.
$$
Thus, Theorem \ref{thm1.1} implies the following result.

\begin{corollary} \label{coro1.2}
Let \eqref{e1.4} and

\begin{equation}
w_b< \frac{b}{2b+\|a\|_\infty} \label{e1.5}
\end{equation}
hold. Then \eqref{e1.1} is exponentially stable.
\end{corollary}

Now for a positive constant $\omega$, let
\begin{equation}
a(t)=b+u(\omega t), \label{e1.6}
\end{equation}
where $u(t)$ is a piece-wise continuous function such that
$$
\nu_u:=\sup_t \big|\int_0^t u(s)ds\big|<\infty.
$$
Then
$$
w_b=\sup_t |\int_0^t u(\omega s)ds|= \nu_u/\omega.
$$
For example, when $u(t)=\sin\;(t)$, then $\nu_u=2$.
Now Theorem \ref{thm1.1} and \eqref{e1.5} imply our next result.

\begin{corollary} \label{coro1.3}
Let  \eqref{e1.4}, \eqref{e1.6} and
\begin{equation}
\omega > \frac{\nu_u(3b+\|u\|_\infty)}{b} \label{e1.7}
\end{equation}
hold. Then \eqref{e1.1} is exponentially stable.
\end{corollary}

\begin{example} \label{exa1.4}
Consider the equation
\begin{equation}
\dot x=-bx(t-1) + c_2\sin (\omega t)x(t-1), \label{e1.8}
\end{equation}
where $b,c_2$ are positive constant with $b<e^{-1}$.
Then $\nu_u=2c_2$ and \eqref{e1.7} has the form
\begin{equation}
\omega> \frac{2c_2(3b+c_2)}{b}. \label{e1.9}
\end{equation}
\end{example}
In summary, for each $c_2$ there exists an $\omega$, such that
 \eqref{e1.8} is exponentially stable.
Meanwhile, the $3/2$-stability theorem requires the
additional condition $c_2+b<3/2$.
Therefore, Theorem \ref{thm1.1} supplements the interesting
results obtained in \cite{be2}.

\section{Proof of Theorem \ref{thm1.1}}

For simplicity, we put $F_b(t)=F(t)$.
Due to the Variation of Constants Formula the equation
$$
\dot x(t) =-bx(t-h) + f(t)\quad (t\geq  0),
$$
with a  given function $f$ and the zero initial condition
$x(t)=0$ ($t \le 0$)
is equivalent to the equation
\begin{equation}
x(t)=\int_0^t F(t-s)f(s)ds. \label{e2.1}
\end{equation}
Recall that a  function $G(t,s)$, ($t\ge s\ge 0$) differentiable
in $t$, is the fundamental solution to \eqref{e1.1} if it satisfies
that equation in $t$ and the initial conditions
$$
G(s, s)=1,\quad G(t,s)=0\quad ( t < s,\;s\geq 0).
$$
Put $G(t,0)=G(t)$.
Subtracting \eqref{e1.2} from \eqref{e1.1} we have
$$
\frac{d}{dt}(G(t)-F(t))= -b(G(t-h)-F(t-h))+c(t)G(t-h)
$$
where $c(t)=-(a(t)-b)$.
Now \eqref{e2.1} implies
\begin{equation}
G(t)=F(t) +\int_0^t   F(t-s)c(s)G(s-h)ds .
\label{e2.2}
\end{equation}
We need the following simple lemma.

\begin{lemma} \label{lem2.1}
 Assume that on each  finite segment of the real axis,
functions $f(t)$ and $v(t)$ are boundedly differentiable
and $w(t)$ is integrable.
Then with the notation
$$
j_w(t, \tau)=\int_\tau^t w(s)ds\quad (t>\tau>-\infty),
$$
the equality
$$
\int_\tau ^t f(s)w(s)v(s)ds=f(t)j_w(t,\tau)v(t)
-\int_\tau ^t [f'(s)j_w(s,\tau)v(s)+ f(s)j_w(s)v'(s)]ds
$$
is valid.
\end{lemma}

\begin{proof}
Clearly,
$$
\frac{d}{dt}f(t)j_w(t,\tau)v(t)=f'(t)j_w(t,\tau)v(t)
+f(t)w(t)v(t)+f(t)j_w(t,\tau)v'(t).
$$
Integrating, this equality and taking into account that
 $j_w(\tau,\tau)=0$,
we arrive at the required result.
\end{proof}


Put
$J(t):=\int_0^t c(s)ds$.
By  the previous lemma,
\begin{align*}
&\int_0^t   F(t-\tau)c(\tau)G(\tau-h)d\tau\\
&=F(0)J(t) G(t-h)-
\int_0^t \Big[\frac{dF(t-\tau)}{d\tau} J(\tau)G(\tau-h)
 + F(t-\tau)J(\tau)\frac{dG(\tau-h)}{d\tau}\Big]d\tau.
\end{align*}
However,
$$
\frac{dG(\tau-h)}{d\tau}=-a(\tau-h) G(\tau-2h)\quad\text{and}\quad
\frac{dF(t-\tau)}{d\tau}=-\frac{dF(t-\tau)}{dt}=bF(t-\tau-h).
$$
Thus,
\begin{align*}
&\int_0^t   F(t-\tau)c(\tau)G(\tau-h)d\tau\\
&=J(t) G(t-h)+ \int_0^t J(\tau)\Big[-b F(t-\tau-h) G(\tau-h)\\
&\quad + F(t-\tau)a(\tau-h)G(\tau-2h)\Big]d\tau.
\end{align*}
Now \eqref{e2.2} implies the following result.

\begin{lemma} \label{lem2.2}
It holds that
\begin{align*}
G(t)&=F(t) + J(t) G(t-h)+ \int_0^t J(\tau)\Big[-b F(t-\tau-h) G(\tau-h)\\
&\quad + F(t-\tau)a(\tau-h)G(\tau-2h)\Big]d\tau\,.
\end{align*}
\end{lemma}

 From the previous lemma,
$$
\|G\|_\infty\le \|F\|_\infty  + \|G\|_\infty
w_b [1+(b+\|a\|_\infty) \|F\|_{L^1}]\,.
$$
If condition \eqref{e1.3} holds, then
$$
\theta:= w_b [1+ (b+\|a\|_\infty) \|F\|_{L^1}]<1
$$
and therefore,
\begin{equation}
\|G\|_\infty\le \frac{\|F\|_\infty}{1-\theta}. \label{e2.3}
\end{equation}
So the stability of \eqref{e1.1} is proved.
Substituting
\begin{equation}
x_\epsilon(t)=e^{\epsilon t} x(t) \label{e2.4}
\end{equation}
with $\epsilon>0$ into \eqref{e1.1}, we have  the equation
\begin{equation}
\dot x_\epsilon(t)=\epsilon x_\epsilon(t) - a(t)e^{\epsilon h}x_\epsilon(t-h). \label{e2.5}
\end{equation}
If $\epsilon>0$ is sufficiently small, then 
considering (2.5) as a perturbation of the equation 
$\dot y(t)=\epsilon y(t) - be^{\epsilon h}y(t-h)$, and applying our above arguments,
according to \eqref{e2.3}
we obtain $\|x_\epsilon\|_\infty<\infty$ for
any solution $x_\epsilon$  of \eqref{e2.5}.
Hence \eqref{e2.4} implies $|x(t)|\le  e^{-\epsilon t}\|x_\epsilon\|_\infty$
for any solution $x$ of \eqref{e1.1}.


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\end{thebibliography}

\end{document}