\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2008(2008), No. 112, pp. 1--16.\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 2008 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2008/112\hfil On stability and oscillation]
{On stability and oscillation of equations
with a distributed delay which can be reduced to difference equations}

\author[E. Braverman, S. Zhukovskiy\hfil EJDE-2008/112\hfilneg]
{Elena Braverman, Sergey Zhukovskiy}

\address{Elena Braverman\newline
Department of Mathematics and Statistics,
University of Calgary,
2500 University Drive N.W., Calgary, AB, Canada T2N 1N4}
\email{maelena@math.ucalgary.ca\; Fax (403)-282-5150\; Phone (403)-220-3956}

\address{Sergey Zhukovskiy\newline
Department of Mathematics and Statistics,
University of Calgary,
2500 University Drive N.W., Calgary, AB, Canada T2N 1N4}
\email{sergey@math.ucalgary.ca; s-e-zhuk@yandex.ru}

\thanks{Submitted April 26, 2008. Published August 15, 2008.}
\thanks{Supported by an NSERC Research Grant}
\subjclass[2000]{34K20, 34K11, 34K06, 39A11}
\keywords{Piecewise constant arguments; distributed delay;
   difference equations; \hfill\break\indent oscillation; stability;
   exponential stability; integro-differential equations}

\begin{abstract}
 For the equation with a distributed delay
 $$
 x'(t) + ax(t)+ \int_0^1 x(s+[t-1])d R(s)=0
 $$
 we obtain necessary and sufficient conditions of stability,
 exponential stability and oscillation.
 These results are applied to some particular cases, such as
 integro-differential equations and equations with a piecewise
 constant argument. Well known results for equations with a
 piecewise constant argument are obtained as special cases.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}{Corollary}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{remark}[theorem]{Remark}

\section{introduction}

The study of equations with a piecewise constant delay was initiated in
1984 by Cooke and Wiener \cite{Cooke1984} and was later
continued in many other publications
\cite{AWAA,AWXProcAMS1987,Gopal1989,Huang1989,Wiener1989},
some of these results are summarized in \cite{GL}.
During the last two decades this topic has been extensively studied, see
\cite{AkhmetNA, AkhmetJMAA, Alonso2000, Cabada2003, Muroya2003,
Papa1997, Papaschinopoulos1994,  Papaschinopoulos1992,
 Rodrig, Seifert2002,
Seifert,  Shen2000, Uesugi2004, Wang2004,
Wang2007, Wang1997, Wang-Yan, Xia, Yuan1999, Yuan}
and references therein. One of the reasons of such interest is a hybrid
character of such equations \cite{AkhmetJMAA}: they incorporate properties
of continuous and discrete models. Moreover, a solution of an equation
with a piecewise constant argument at certain points also satisfies some
difference equation. Using this technique, the known results for delay
equations were applied to delay (high order) difference equations, see,
for example,
\cite{ErbeXiaYu,GP,Malygina2007}.
Another reason for attention to equations with piecewise constant
arguments is the following: such equations are semidiscretizations
of delay equations and thus are useful in numerical applications
\cite{CookeGyori, Gopalsamy1999, Gyori1991, Gyori1995, Hartung1998}.

We consider the equation with a distributed delay
\begin{equation}
\label{1}
x'(t) + ax(t)+ \int_0^1 x(s+[t-1]) d R(s)=0,
\end{equation}
where $R(s): [0,1] \to \mathbb{R}$ is a left-continuous function
of bounded variation, $a \in \mathbb{R}$. For example, if $R(s)$
is differentiable, $R'(s)= b(s)$, then \eqref{1} is the
integro-differential equation
\begin{equation}
\label{2}
x'(t) + ax(t)+ \int_0^1 b(s)x(s+[t-1]) ds =0.
\end{equation}
Let $R(s) = b \chi_{(\alpha,1]} (s)$, where
$\chi_{(a,b]}$ is the characteristic function of the interval $(a,b]$,
i.e., $\chi_{(a,b]}(x)=1$, if $x\in (a,b]$ and $\chi_{(a,b]}(x)=0$,
otherwise.
Then \eqref{1} has the form
\begin{equation}
\label{3}
x'(t) + ax(t)+ b x(\alpha+[t-1])=0,
\end{equation}
which involves equations
\begin{equation}
\label{4}
x'(t) + ax(t)+ b x([t-1])=0
\end{equation}
and
\begin{equation}
\label{5}
x'(t) + ax(t)+ b x([t])=0
\end{equation}
as particular cases
when $\alpha=0$ and $\alpha \to 1$, as well as equations where the
piecewise constant argument refers to the fractional points.

In spite of its ``continuous" form, equation \eqref{1} incorporates
properties of both continuous and discrete systems, in the next section we
will reduce its solution to the solution of a specially constructed
difference equation.

The paper is organized as follows. In Section 2 we present relevant
definitions and auxiliary results. In particular, we reduce \eqref{1} to a 
second order difference
equation. Section 3 presents necessary and sufficient oscillation and stability
conditions for \eqref{1}. The general results are applied to some special cases
of integro-differential equations and equations with piecewise constant
arguments, which allows to deduce some known results. Finally, Section 4
involves discussion and outlines some open problems and possible generalizations
of equation \eqref{1}. Some long but straightforward proofs are presented in
the Appendix.


\section{Preliminaries and Solution Representation}

We consider \eqref{1} with the initial condition
\begin{equation}
\label{initial}
x(t)=\varphi(t),\quad t \in [-1,0],
\end{equation}
%Everywhere in the following we consider \eqref{1}, (\ref{initial})
under the following assumptions:
\begin{itemize}
\item[(A1)] $R(s): [0,1] \to \mathbb{R}$ is a left-continuous function of
bounded variation which has a nonzero variation in $[0,1]$;

\item[(A2)] $\varphi:[0,1] \to \mathbb{R}$ is a Borel measurable bounded
function such that the Lebesgue Stiltjes integral ${
\int_0^1 \varphi(s-1) d R(s)}$ exists (and is finite).
\end{itemize}

\begin{definition} \label{def1} \rm
Function $x(t)$ is \emph{a solution} of
\eqref{1}, (\ref{initial}) if it satisfies \eqref{1} almost everywhere for
$t \geq 0$ and (\ref{initial}) for $t \in [-1,0]$.
\end{definition}


Denote
\begin{gather}
\label{denote1}
x_n=x(n), \quad  x_{-1} = \int_0^1 \varphi(s-1)  d R(s),\\
\label{denote2}
K_n=\int_0^1 x(s+n)  d R(s), \quad  K_{-1}= \int_0^1 \varphi(s-1)  d R(s),\\
\label{denote3}
P(a)=\int _{0}^{1}e^{-as} d R(s), \quad
Q(a)=\int _{0}^{1}\frac{1-e^{-as}}{a}  d R(s).
\end{gather}

First, let us reduce the solution of \eqref{1} at integer points to a
solution of a second order difference equation. Let us notice that in the
following we will understand expressions at $a=0$ as a limit; for example,
$$
 \frac{e^{ak}-1}{a} \Big|_{a=0}=\lim_{a \to 0}
\frac{e^{ka}-1}{a} = k= \lim_{a \to 0}\frac{1-e^{-ka}}{a}.
$$


\begin{lemma}\label{lemmarepresentation}
(1) The solution of \eqref{1}, (\ref{initial}) between integer points is
\begin{equation}
\label{sol1}
x(t)= x_n e^{a(n-t)} + \frac{e^{a(n-t)}-1}{a} K_{n-1}, \quad
t\in[n,n+1),
\end{equation}
with $K_n,\,x_n$ defined by (\ref{denote1}), (\ref{denote2}),
$n=0,1,2, \dots$ .

(2) The solution of \eqref{1}, (\ref{initial}) at integer points satisfies
the second order difference equation
\begin{equation}
\label{difference}
x_{n+2}-(e^{-a}-Q(a))x_{n+1}
+\Big(\frac{1-e^{-a}}{a}P(a)-e^{-a}Q(a)\Big)x_n=0, \quad n \geq -1.
\end{equation}
\end{lemma}

\begin{proof}
 The first part is checked by a straightforward computation
and leads to
$$
x_{n+1}=e^{-a}x_n-\Big(\frac{1-e^{-a}}{a}\Big)K_{n-1},
$$
\begin{align*}
K_n&=\int _{0}^{1} x(s+n) dR(s)=
\int _{0}^{1}\Big( x_n e^{a(n-s)} + \frac{e^{a(n-s)}-1}{a} K_{n-1}
\Big) dR(s)\\
&=x_n\int _{0}^{1}e^{-as} dR(s)
-K_{n-1}\int _{0}^{1}\frac{1-e^{-as}}{a} dR(s)
= P(a)x_n - Q(a) K_{n-1}.
\end{align*}
Hence, if we denote $Y_n=(x_{n},K_{n-1})^T$, then $Y_{n+1}=A Y_n$, where
$$
A= \begin{bmatrix} e^{-a} & { -\frac{1-e^{-a}}{a}} \\
 P(a) & - Q(a)
\end{bmatrix}.
$$
Thus, $x_n$ satisfies the second order difference equation
$$
x_{n+2}- \text{tr}(A)x_{n+1} + \det(A)x_n=0;
$$
since the trace of $A$ is
$e^{-a}-Q(a)$ and the determinant is
${\frac{1-e^{-a}}{a}  P(a) - e^{-a}Q(a)}$, we immediately obtain
\eqref{difference}.
\end{proof}

\begin{remark} \label{equivalence} \rm
By Lemma \ref{lemmarepresentation} the values of \eqref{1}, (\ref{initial})
at integer points satisfy the difference equation \eqref{difference}, with
$x_0$, $x_{-1}$ defined in (\ref{denote1}).

Let us also note that for any $x_0$, $x_{-1}$ there exists
$\varphi$ satisfying (A2) which leads to these $x_0$, $x_{-1}$ in
(\ref{denote1}). Really, since $R(s)$ has a nonzero variation,
then there exists a continuous function $g:[-1,0] \to \mathbb{R}$
such that ${ \int_0^1 g(s-1) dR(s) =c \neq 0}$.
Besides, $R(s)$ is left continuous, so the relevant measure has no
atom at $x=1$, thus ${ \int_0^1 g_1(s-1) dR(s)
=\int_0^1 g(s-1) dR(s)}$, where $g_1(s)$ coincides with $g(s)$
everywhere in $[-1,0]$ but probably at $s=0$ (and the left
integral always exists). Then
$$
\varphi(s) = \begin{cases}
x_{-1} g(s)/c  , & s \in [-1,0),\\
x_0, & s=0, \end{cases}
$$
leads to any prescribed $x_{-1}$, $x_0$.
\end{remark}

\begin{definition} \label{def2} \rm
A solution of \eqref{1} \emph{oscillates} if it is neither eventually
positive nor eventually negative. Equation \eqref{1} is \emph{oscillatory}
if all its solutions oscillate.

A solution of \eqref{difference} \emph{oscillates} if the sequence
$\{ x_n \}$ is neither eventually positive nor eventually negative.
Equation \eqref{difference} is \emph{oscillatory}
if all its solutions oscillate.
\end{definition}

\begin{corollary} \label{oscil}
Equation \eqref{1} is oscillatory if and only if \eqref{difference} is
oscillatory.
\end{corollary}

\begin{proof}
Obviously if a solution of \eqref{difference} oscillates then the relevant
solution of \eqref{1} (with an appropriate initial function, see
Remark \ref{equivalence}) cannot be eventually positive or negative.
Let us notice that by (\ref{sol1}) a solution of \eqref{1} increases in
$[n,n+1)$ if ${ ax_n+K_{n-1}<0}$
and decreases if ${ ax_n+K_{n-1}>0}$.
Thus, if $x(n)$, $x(n+1)$ have the same sign, so are all the points
between $n$ and $n+1$, hence oscillation of \eqref{1} implies that
\eqref{difference} is also oscillating.
\end{proof}

According to (A2), the initial function is bounded, so we can define
the sup-norm:
$$
\| \varphi \| = \sup_{t \in [-1,0]} | \varphi (t) |.
$$

\begin{definition} \label{def3} \rm
Equation \eqref{1}  is \emph{stable} if for any $\varepsilon>0$
there exists $\delta>0$ such that for any $\varphi$ satisfying
(A2) inequality $\| \varphi \|<\delta$ implies $|x(t)|< \varepsilon$
for $t \geq 0$. Equation \eqref{1}  is \emph{asymptotically stable} if it
is stable and
${ \lim_{t  \to \infty} x(t)=0}$ for any initial conditions.
Equation \eqref{1}  is \emph{exponentially stable} if there exist
positive numbers $N, \gamma$ such that any solution satisfies
$$
|x(t)| \leq Ne^{-\gamma t} \| \varphi \|.
$$

Eq. \eqref{difference} is \emph{stable} if for any $\varepsilon>0$
there exists $\delta>0$ such that $\max \{ |x_0|, |x_{-1}| \}<\delta$
implies $|x_n|<\varepsilon$ for any $n \geq 0$. Equation
\eqref{difference} is \emph{asymptotically stable} if it is stable and
${ \lim_{n \to \infty} x_n=0}$ for any initial conditions.
Equation \eqref{difference} is \emph{exponentially stable} if there exist
positive numbers $N, \gamma$ such that any solution satisfies
$$
|x_n| \leq Ne^{-\gamma n} \max \{ |x_0|, |x_{-1}| \}.
$$
\end{definition}

\begin{corollary} \label{stab}
Equation \eqref{1} is stable (asymptotically stable, exponentially stable)
if and only if \eqref{difference} is stable (asymptotically
stable, exponentially stable).
\end{corollary}

\begin{proof}
As in the previous corollary, for any solution of
\eqref{1}, $\max_{t \in [n,n+1]} |x(t)|$
is attained at the ends and equals either $|x(n)|=|x_n|$ or
$|x(n+1)|=|x_{n+1}|$. Thus any type of stability
of \eqref{1} is equivalent to the appropriate stability kind for
\eqref{difference}.
\end{proof}


\section{Stability and Oscillation Tests}

In this section we will obtain necessary and sufficient conditions for
oscillation, stability and exponential stability of
equation \eqref{1} with a distributed delay.
In the following we will also apply the well known result for second
order difference equations with constant coefficients (see,
for example, \cite[p. 53--65]{gan}).

\begin{lemma} \label{secondorder}
A difference equation with constant coefficients
\begin{equation}
\label{second}
x_{n+1}-p_1 x_{n+1}+p_2 x_n=0, \quad n=-1,0,1, \dots  ,
\end{equation}
is stable if both roots of the characteristic equation
$\lambda^2-p_1 \lambda +p_2=0$ are in the unit circle and
is exponentially stable if the roots are inside the unit circle.
The latter condition is satisfied if $|p_1|<p_2+1<2$ and is also
equivalent to the asymptotic stability of \eqref{second}.
The former condition is satisfied if $|p_1|\leq p_2+1 \leq 2$.

Equation \eqref{second} is oscillatory if and only if its characteristic
equation has no positive roots, which is valid if either the
discriminant is negative ($p_1^2<4p_2$) or all coefficients are
nonnegative ($p_1 \leq 0$, $p_2 \geq 0$).
\end{lemma}

Lemma \ref{secondorder} together with the form of \eqref{difference} and
Corollaries \ref{oscil}, \ref{stab} imply the following oscillation and
stability tests for equation \eqref{1}.

\begin{theorem} \label{oscillationtest}
Suppose (A1)-(A2) are satisfied. Equation \eqref{1} is oscillatory
if and only if at least one of the two following conditions holds:
\begin{gather} \label{oscil1}
\frac{1}{4} \left( e^{-a}+Q(a)\right)^2 < \frac{1-e^{-a}}{a}P(a), \\
\label{oscil2}
e^{-a} \leq Q(a)  \leq \frac{e^{a}-1}{a}P(a).
\end{gather}
\end{theorem}

\begin{proof}
By Lemma \ref{secondorder} equation \eqref{difference} is oscillatory
if and only if  either
$$
\frac{1}{4} \left( e^{-a}-Q(a)\right)^2 <\frac{1-e^{-a}}{a}P(a)-e^{-a}Q(a)
$$
or
$$ e^{-a} \leq Q(a) \leq \frac{1-e^{-a}}{a}e^a P(a),
$$
where the former inequality
is equivalent to \eqref{oscil1} and the latter to \eqref{oscil2}.
\end{proof}

\begin{theorem} \label{stabilitytest}
Suppose (A1)-(A2) are satisfied. Equation \eqref{1}
is stable if and only if
\begin{equation}
\label{stability1}
\left| Q(a)-e^{-a} \right| \leq \frac{1-e^{-a}}{a}P(a)-e^{-a}Q(a)+1 \leq 2
\end{equation}
and is exponentially stable if and only if
\begin{equation}
\label{stability2}
\left| Q(a)-e^{-a} \right| < \frac{1-e^{-a}}{a}P(a)-e^{-a}Q(a)+1 < 2.
\end{equation}
\end{theorem}

To illustrate Theorems \ref{oscillationtest},
\ref{stabilitytest}, let us
consider two particular cases of \eqref{1}.
First, let $a,b \in \mathbb{R}$. Consider the integro-differential
equation
\begin{equation}
\label{eq2}
x'(t)+ax(t)+b \int _{[t-1]}^{[t]} x(s) ds = 0,
\end{equation}
which is a special case of \eqref{1}, with $R(s)=bs$.
Then $P(a),Q(a)$ in (\ref{denote3}) have the form
\begin{equation}
\label{integralcoef}
P(a)=\int_0^1 e^{-as} b ds= b  \frac{1-e^{-a}}{a}, \quad
Q(a)=\int_0^1  \frac{1-e^{-as}}{a} b ds=  b  \frac{a-1+e^{-a}}{a^2}.
\end{equation}

The following results are corollaries of Theorems
\ref{oscillationtest}, \ref{stabilitytest}, where $P(a)$ and $Q(a)$ are
substituted from (\ref{integralcoef}).
However, the straightforward computation is long and thus is presented in
the Appendix.

\begin{theorem}\label{constantoscil}
The following two statements are equivalent.
\begin{itemize}
 \item[(1)] Equation
(\ref{eq2}) is oscillatory.

\item[(2)] If $a\neq 0$ then
         \begin{equation}\label{osc2}
           b>a^2 \left(\frac{|1-e^{-a}| -
\sqrt{1-e^{-a}-ae^{-a}}}{a-1+e^{-a}}\right)^2  ;
         \end{equation}
       if $a=0$ then
         \begin{equation}\label{osc2zer}
            b>6-4\sqrt{2}.
         \end{equation}
\end{itemize}
\end{theorem}

The domain of parameters $a,b$ where (\ref{eq2}) oscillates is illustrated
in Fig \ref{fig-osc}.

\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.6\textwidth]{fig1} %NEW_osc.eps
\end{center}
\caption{Illustration for the Theorem \ref{constantoscil}. The shaded
area designates the set of parameters $(a,b)$ where any solution
of the initial value problem (\ref{eq2}), (\ref{initial}) oscillates.}
\label{fig-osc}
\end{figure}


\begin{theorem} \label{constantstab}
The following two statements are equivalent.
\begin{itemize}
\item[(1)]
Equation (\ref{eq2}) is exponentially stable.
\item[(2)] If $a<0$ then
\begin{equation} \label{integrneg}
-a < b<\frac{a^2}{1-e^{-a}-ae^{-a}};
\end{equation}
if $a=0$ then
\begin{equation}\label{integrzer}
0<b<2;
\end{equation}
       if $a>0$ then
         \begin{equation}\label{integrpos}
            -a< b < \min \left\{
        -\frac{a^2(1+e^{-a})}{2-2e^{-a}-ae^{-a}-a},
        \frac{a^2}{1-e^{-a}-ae^{-a}} \right\}.
        \end{equation}
\end{itemize}
\end{theorem}

Figure \ref{fig-stab} illustrates Theorem \ref{constantstab}. For any
value of the parameters $a,b$ such that point $(a,b)$ is in the grey
area, equation (\ref{eq2}) is exponentially stable. For any value of
the parameters $a,\,b$ such that point $(a,b)$ is in the white area
the equation is unstable and it is stable (but not exponentially) on the 
boundary.

\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.75\textwidth]{fig2} % NEW_stab.eps
\end{center}
\caption{The shaded area designates the set of parameters $(a,b)$ where
(\ref{eq2}) is exponentially stable, the equation is stable (but not
asymptotically) for values $(a,b)$ on the boundary.}
\label{fig-stab}
\end{figure}

Next, let $R(s)$ be a step function $R(s)= b \chi_{(r,1]}(t)$,
$0 \leq r<1$. Then \eqref{1} has the form
\begin{equation}
\label{piececonst}
x' (t)+ax(t)+bx(r+[t-1])=0,
\end{equation}
equations for $r=0$ or $r=1$ were considered in
\cite{AWAA,AWXProcAMS1987,Cooke1984}.
Then
$$
P(a)=be^{-ar}, \quad  Q(a)=b \frac{1-e^{-ar}}{a}.
$$
Thus, Theorems \ref{oscillationtest} and \ref{stabilitytest} immediately imply
the following oscillation and stability criteria for \eqref{piececonst}.

Equation \eqref{piececonst} is oscillatory if at least one of the following two inequalities
holds
\begin{equation}
\label{osc1ex1}
\frac{1}{4} \Big( e^{-a}+ b \frac{1-e^{-ar}}{a}\Big)^2 <
b \frac{1-e^{-a}}{a}  e^{-ar},
\end{equation}
\begin{equation}
\label{osc2ex1}
e^{-a} \leq b \frac{1-e^{-ar}}{a}  \leq b \frac{e^a-1}{a}  e^{-ar}.
\end{equation}


Equation \eqref{piececonst} is stable if and only if
\begin{equation}
\label{32}
\big| b \frac{1-e^{-ar}}{a}-e^{-a} \big| \leq
b \frac{e^{-ar}-e^{-a}}{a}+1
\leq 2
\end{equation}
and is exponentially stable if and only if
\begin{equation}
\label{33}
\big| b \frac{1-e^{-ar}}{a}-e^{-a} \big| <
b \frac{e^{-ar}-e^{-a}}{a}+1 <2.
\end{equation}


\begin{theorem} \label{oscilpiececonst}
Let $0<r<1$. Equation \eqref{piececonst} is oscillatory if and
only if
\begin{equation}
\label{osc3ex1}
b> \Big( \frac{a}{1-e^{-ar}} \Big)^2 \Big[ \sqrt{ \frac{e^{-ar} -
e^{-a(r+1)}}{a} } - \sqrt{ \frac{e^{-ar}-e^{-a}}{a} } \Big]^2.
\end{equation}
\end{theorem}

\begin{proof}
Using (\ref{osc1ex1}), (\ref{osc2ex1}), we will obtain explicit
conditions for $b$, if $a$ is given. Since ${ \frac{1-e^{-ar}}{a}
\geq 0}$, then the left inequality in (\ref{osc2ex1}) becomes
${ b \geq \frac{ae^{-a}}{1-e^{-ar}} }$, while the right inequality
${ b  \frac{1-e^{a(1-r)}}{a} \leq 0 }$ is $b \geq 0$. Thus,
(\ref{osc2ex1}) has the form
$$
{ b \geq \max \big\{ 0,
\frac{ae^{-a}}{1-e^{-ar}} \big\} =\frac{ae^{-a}}{1-e^{-ar}} }.
$$
Inequality (\ref{osc1ex1}) can be rewritten as a quadratic inequality in
$b$:
\begin{equation}
\label{auxil1}
\Big( \frac{1-e^{-ar}}{a} \Big)^2 b^2 + 2  \frac{e^{-a} -
2e^{-ar}+e^{-a(r+1)}}{a}  b + e^{-2a} <0.
\end{equation}
The discriminant of the above quadratic inequality in $b$ is
\begin{align*}
D&= \frac{(e^{-a} -2e^{-ar}+e^{-a(r+1)})^2-(e^{-a}-e^{-a(r+1)})^2}{a^2}\\
&= 4 \frac{e^{-ar}-e^{-a(r+1)}}{a} \frac{e^{-ar}-e^{-a}}{a}
\end{align*}
which is positive as a product of two positive factors.
A solution of inequality (\ref{auxil1}) is between the two roots $b_1<b_2$ of
the relevant quadratic equation, the largest of them is
\begin{align*}
b_2 & =  \Big( \frac{a}{1-e^{-ar}} \Big)^2
\Big[ \frac{2e^{-ar}-e^{-a}
-e^{-a(r+1)} }{a} + 2 \sqrt{\frac{e^{-ar}-e^{-a(r+1)}}{a} }
\sqrt{\frac{e^{-ar}-e^{-a}}{a} } \Big]
\\
& =  \Big( \frac{a}{1-e^{-ar}} \Big)^2 \Big[
\sqrt{\frac{e^{-ar}-e^{-a(r+1)}}{a} } + \sqrt{\frac{e^{-ar}-e^{-a}}{a}}
\Big]^2,
\end{align*}
similarly,
\begin{equation}
\label{auxil2}
b_1 = \Big( \frac{a}{1-e^{-ar}} \Big)^2 \Big[
\sqrt{\frac{e^{-ar}-e^{-a(r+1)}}{a} } - \sqrt{\frac{e^{-ar}-e^{-a}}{a}}
\Big]^2
\end{equation}
is obviously nonnegative.

If $b_1 <b<b_2$, then (\ref{osc1ex1}) is
satisfied. Let us demonstrate that ${ b_2 \geq
\frac{ae^{-a}}{1-e^{-ar}} }$, then for $b \geq b_2$ inequality (\ref{osc2ex1})
is satisfied, thus for $b>b_1$ all solutions are oscillatory. Hence
$b>b_1$, where $b_1$ is defined in
(\ref{auxil2}), immediately implies oscillation condition (\ref{osc3ex1}).
Consider
\begin{align*}
&b_2 - \frac{ae^{-a}}{1-e^{-ar}}\\
& = \Big( \frac{a}{1-e^{-ar}} \Big)^2
\Big[\Big( \sqrt{\frac{e^{-ar}-e^{-a(r+1)}}{a} }
+ \sqrt{\frac{e^{-ar}-e^{-a}}{a}} \Big)^2 -
 \frac{e^{-a}-e^{-a(r+1)}}{a} \Big] \\
&=\Big( \frac{a}{1-e^{-ar}} \Big)^2
\Big( \sqrt{\frac{e^{-ar}-e^{-a(r+1)}}{a}} +
\sqrt{\frac{e^{-ar}-e^{-a}}{a}} + \sqrt{\frac{e^{-a}-e^{-a(r+1)}}{a}} \Big)
\\
&\times
\Big( \sqrt{\frac{e^{-ar}-e^{-a(r+1)}}{a}}  +
\sqrt{\frac{e^{-ar}-e^{-a}}{a}} - \sqrt{\frac{e^{-a}-e^{-a(r+1)}}{a}} \Big)
\geq 0
\end{align*}
as a product of two nonnegative terms, the latter term is nonnegative since
$\sqrt{x+y} \leq \sqrt{x}+\sqrt{y}$ for any nonnegative $x$, $y$.
Consequently, (\ref{osc3ex1}) is necessary and sufficient for oscillation, which
completes the proof.
\end{proof}

\noindent
\textbf{Remark.} However, Theorem \ref{oscilpiececonst} does not consider the
cases $r=0$, $r \to 1$, which correspond to equations (\ref{4}) and (\ref{5}),
respectively. First, let $r=0$. Then (\ref{osc2ex1}) is never valid
(it involves $e^{-a} \leq 0b= 0$), we stay with (\ref{osc1ex1}) which
has
the form ${ \frac{1}{4} e^{-2a} < b  \frac{1-e^{-a}}{a} }$.
Hence
$$
b> \frac{ae^{-2a}}{4(1-e^{-a})} = \frac{ae^{-a}}{4(e^a-1)},
$$
which was obtained in \cite{AWXProcAMS1987} as a necessary and sufficient
oscillation condition for (\ref{4}).

If $r \to 1$, then (\ref{osc2ex1}) tends to
${ e^{-a} \leq b  \frac{1-e^{-a}}{a}}$, while (\ref{osc1ex1})
has the form
$$
\Big(  e^{-a}+b  \frac{1-e^{-a}}{a} \Big)^2 < 4b
\Big(\frac{1-e^{-a}}{a}\Big) e^{-a}, \quad \text{or}\quad
\Big(  e^{-a}-b   \frac{1-e^{-a}}{a} \Big)^2 < 0,
$$
which is impossible. The inequality ${ b>\frac{a}{e^a-1} }$
is sufficient for oscillation of (\ref{5}), see \cite{AWAA}.
\smallskip

Now let us proceed to stability of \eqref{piececonst}.

\begin{theorem}\label{stabpiececonst}
Let $0<r<1$. Equation \eqref{piececonst} is stable if and
only if
\begin{equation}
\label{stab1ex1}
-a \leq b \leq C,
\end{equation}
where
\begin{equation}
\label{stab1ex1a}
 C=  \begin{cases}
 \min \big\{ \frac{a(1+e^{-a})}{1+e^{-a}-2e^{-ar}},
\frac{a}{e^{-ar}-e^{-a}} \big\}, & \text{if }
{ \frac{1+e^{-a}-2e^{-ar}}{a} >0,} \\[3pt]
{ \frac{a}{e^{-ar}-e^{-a}}}, & \text{if }
{ \frac{1+e^{-a}-2e^{-ar}}{a} \leq 0, }
\end{cases}
\end{equation}
and is exponentially stable if and only if
\begin{equation}
\label{stab2ex1}
-a  < b <  C.
\end{equation}
\end{theorem}

\begin{proof}
 By (\ref{33}) exponential stability is equivalent to
the following inequalities
\begin{gather}
\label{stab1a}
-b  \frac{e^{-ar}-e^{-a}}{a}-1 < b   \frac{1-e^{-ar}}{a} - e^{-a}
< b   \frac{e^{-ar}-e^{-a}}{a}+1, \\
\label{stab1b}
b   \frac{e^{-ar}-e^{-a}}{a}<1.
\end{gather}
The latter inequality can be rewritten as
${ b <\frac{a}{e^{-ar}-e^{-a}} }$, while the left inequality of
(\ref{stab1a}) is
$$
b   \frac{1-e^{-a}}{a}>e^{-a}-1, \quad\text{or}\quad
b > \frac{e^{-a}-1}{1-e^{-a}} a=-a.$$
Further, consider the right inequality in
(\ref{stab1a}) which is equivalent to
\begin{equation}
\label{stab1abc}
\frac{1+e^{-a}-2e^{-ar}}{a}b < 1+e^{-a}.
\end{equation}
The right hand side is positive, so if the left hand side is nonpositive
then (\ref{stab1abc}) holds. Thus, to prove that (\ref{stab2ex1}) is sufficient
for exponential stability, it is enough to consider the case
${ \frac{1+e^{-a}-2e^{-ar}}{a}<0}$, $b<0$. Since $b>-a$, then we
deduce $a<0$. We have $|b|/|a|<1$ and
$$
\big| \frac{1+e^{-a}-2e^{-ar}}{a}b \big|
=\big| \frac{b}{a}\big| \big|1+e^{-a}-2e^{-ar} \big|
< \big|1+e^{-a}-2e^{-ar} \big| \leq 1+e^{-a},
$$
since $2e^{-ar}\leq 2$ for $a>0$, $r \geq 0$, which completes the proof
for the exponential stability. Stability is considered similarly.
\end{proof}

\noindent
\textbf{Remark.} In the case $r=0$ we have ${C=\frac{a}{1-e^{-a}} }$;
the exponential stability condition
${ -a<b<\frac{a}{1-e^{-a}}}$ is well known for (\ref{4}),
see \cite{Cooke1984}.
If $r \to 1$, then ${ C=\frac{a(1+e^{-a})}{1-e^{-a}}}$ and the
exponential stability condition ${-a<b<\frac{a(1+e^{-a})}{1-e^{-a}}}$
for (\ref{5}) is also known
\cite{Cooke1984}.



\section{Discussion and Open Problems}

We have obtained sharp oscillation and stability conditions
for equation \eqref{1} and some of its particular cases.
After some straightforward computations, we have the following relation
between the properties of equations (\ref{eq2}), (\ref{4}) and (\ref{5}).

\begin{enumerate}
\item
Exponential stability of (\ref{4}) implies exponential stability
of (\ref{5}) and (\ref{eq2}). However, we cannot compare conditions of
exponential stability of (\ref{5}) and (\ref{eq2}). For example, if
$a=-1$, $b=2$, then
equation (\ref{5}) is exponentially stable and (\ref{eq2}) is not stable,
while for $a=2$, $b=6$ equation (\ref{eq2})  is exponentially  stable,
unlike (\ref{5}).
\item
Oscillation domains of (\ref{4}), (\ref{5}) and (\ref{eq2}) in
$(a,b)$-plane also cannot be compared: for each pair of equations there
are two examples when one oscillates while the other does not for the same
values of parameters $a,b$.
\end{enumerate}

Let us discuss some possible applications and generalizations of
our results, as well as relevant open problems.

\begin{enumerate}
\item
Apply the results of the present paper to obtain sharp stability and
oscillation conditions for the equation
$$
x'(t) + ax(t)+ \sum_{j=1}^k b_j x(\alpha_j+[t-1])=0,\quad
0 \leq \alpha_j <1, \quad j=1, \dots , k,
$$
which is a partial case of \eqref{1} with ${
R(s)= \sum_{j=1}^k b_j \chi_{(\alpha_j,1]} (s) }$.
\item
The present paper contains a comprehensive analysis of \eqref{1} which
can be reduced to an autonomous second order difference equation. Using
the same method, reduce
\begin{equation}
\label{1nonautonom}
x'(t) + ax(t)+ \int_0^1 x(s+[t-1])d_s R(t,s)=0,
\end{equation}
to the second order nonautonomous difference equation: in
(\ref{denote3}) we will have $P_n(a)$, $Q_n(a)$ rather than $P(a)$ and
$Q(a)$.
Deduce sufficient stability, oscillation and nonoscillation conditions
for (\ref{1nonautonom}).
\item
Consider the equation, where the derivative depends on the solution in
some previous intervals
\begin{equation}
\label{1deep}
x'(t) + ax(t)+ \sum_{j=1}^k \int_0^1 x(s+[t-j])d_s R_j(s)=0
\end{equation}
and its nonautonomous version. Reduce (\ref{1deep}) to a high order
difference equation, establish oscillation and stability conditions.
\item
Equations with piecewise constant delays are sometimes considered as
a semidiscretization of delay equations \cite{Gopalsamy1999}. If
\eqref{1} is a semidiscretization of the integro-differential equation
$$
x'(t) + ax(t)+ \int_0^1 x(s+t-1)d_s R(s)=0,
$$
study the relation between oscillation and stability conditions of two
equations. Consider a more accurate semidiscretization of type
(\ref{1deep})
\begin{equation}
\label{semidiscr}
x'(t) + ax(t)+ \sum_{j=1}^k \int_0^r x\Big( s+ r\big[
\frac{t-j}{r} \big] \Big)  d_s R_j(s)=0
\end{equation}
and study the convergence of solutions. If $a=0$ and $R_j(s)$ are step
functions then we obtain the well known convergence problem for a finite
difference approximation.
\end{enumerate}



\section{Appendix}

In the proof of Theorems \ref{constantoscil} and \ref{constantstab}
we will apply the following obvious result.

\begin{lemma}
\label{pr-ineq}
For any real number $x$ the following inequalities hold:\\
    1. If $x\neq 0$ then $(1-e^{-x})/x>0$;\\
    2. If $x\neq 0$ then $xe^{-x}+e^{-x}-1<0$;\\
    3. If $x<0$ then $2-xe^{-x}-2e^{-x}-x>0,$ if $x>0$ then
$2-xe^{-x}-2e^{-x}-x<0$;\\
    4. $xe^{-x}-1<0$;\\
    5. If $x<0$ then $1-xe^{-x}-e^{-x}-x>0,$ if $x>0$ then
    $1-xe^{-x}-e^{-x}-x<0$;\\
    6. If $x\neq 0$ then $e^{-x}+x-1>0$.
\end{lemma}

\begin{proof}[Proof of Theorem \ref{constantoscil}]
We recall that we have to prove that a solution of
(\ref{eq2}), (\ref{initial}) oscillates for any initial function satisfying
(A2) if and only if
\begin{gather*}
b>a^2 \left(\frac{|1-e^{-a}|
- \sqrt{1-e^{-a}-ae^{-a}}}{a-1+e^{-a}} \right)^2, \quad\text{if }  a \neq 0,\\
\text{and}\quad  b>6-4\sqrt{2},  \quad \text{if } a=0.
\end{gather*}
First, consider $a\neq 0$. By Theorem \ref{oscillationtest}, the solution
of the initial value problem
(\ref{eq2}), (\ref{initial}) oscillates for any $\varphi$ if and
only if either \eqref{oscil1} or \eqref{oscil2} holds. Substitute
$P(a)$, $Q(a)$
from (\ref{integralcoef}) into \eqref{oscil1}, \eqref{oscil2} and obtain
that at least one of the following two inequalities should hold:
\begin{gather*}
      { b \frac{1-e^{-a}}{a}>\frac{a}{4}\Big(e^{-a}+b
  \frac{a-1+e^{-a}}{a^2}\Big)^2/(1-e^{-a})},\\
      { e^{-a} \leq b   \frac{a-1+e^{-a}}{a^2} \leq b
 \frac{1-e^{-a}}{a} \frac{e^{a}-1}{a}}.
\end{gather*}
The first inequality above
is equivalent to
  $$
    4 b \Big(\frac{1-e^{-a}}{a}\Big)^2 > \Big(e^{-a}+b
  \frac{a-1+e^{-a}}{a^2}\Big)^2
$$
which is a quadratic inequality in $b$
  $$
(a-1+e^{-a})^2  b^2 + 2 a^2 (e^{-a}(a-1+e^{-a})-2(1-e^{-a})^2)
 b + (a^2 e^{-a})^2 < 0.
  $$
The corresponding quadratic equation has the discriminant
\begin{align*}
&4 a^4 (e^{-a}(a-1+e^{-a})-2(1-e^{-a})^2)^2 - 4 (a-1+e^{-a})^2 (a^2
    e^{-a})^2 \\
&= 4a^4\Big[e^{-2a}(a-1+e^{-a})^2-4(1-e^{-a})^2(a-1+e^{-a})e^{-a}
  +4(1-e^{-a})^4\\
&\quad -e^{-2a}(a-1+e^{-a})^2\Big]\\
&= 16 a ^4 (1-e^{-a})^2 ((1-e^{-a})^2-(a-1+e^{-a})e^{-a})\\
&= 16a^4(1-e^{-a})^2\,(1-e^{-a}-ae^{-a}).
\end{align*}
Note that by Lemma \ref{pr-ineq}, Part 2, we have $1-e^{-a}-ae^{-a}>0$ for
any $a\neq 0$. Therefore, the quadratic equation has two real
solutions, $b_1<b_2$, given by the quadratic formula
\begin{equation}
\begin{aligned}
b & =  \frac{2a^2 \,(2(1-e^{-a})^2-e^{-a}(a-1+e^{-a}))
    \pm 4a^2 |1-e^{-a}| \sqrt{1-e^{-a}-ae^{-a}}}{2(a-1+e^{-a})^2}\\
  &  =  a^2 \frac{(1-e^{-a})^2 + (1-e^{-a}-ae^{-a})
    \pm 2 |1-e^{-a}|\sqrt{1-e^{-a}-ae^{-a}}}{(a-1+e^{-a})^2} \\
  &  =   a^2 \Big(\frac{|1-e^{-a}| \pm
    \sqrt{1-e^{-a}-ae^{-a}}}{a-1+e^{-a}}\Big)^2.
\end{aligned} \label{add4}
\end{equation}
Then the solution $b$ of the quadratic inequality satisfies
$b_1<b<b_2$.

Consider the second inequality of the system:
$$
    e^{-a} \leq b   \frac{a-1+e^{-a}}{a^2} \leq b  \frac{1-e^{-a}}{a}
\frac{e^{a}-1}{a}
$$
if and only if
$$
 a^2 e^{-a}\leq b  (a-1+e^{-a}) \leq b  (e^{a}+e^{-a}-2).
$$
Since $a-1+e^{-a}>0$ by Lemma \ref{pr-ineq}, Part 6,
then the latter inequality is equivalent to
${b \geq \frac{a^2 e^{-a}}{a-1+e^{-a}} }$.
Moreover,
  ${ b (a-1+e^{-a}) \leq b \, (e^{a}+e^{-a}-2)}$ can be rewritten as
$b (e^{a}-a -1) \geq 0$, which is equivalent to
$b\geq 0$,
since $e^{a}-a-1>0$ by Lemma \ref{pr-ineq}, Part 6.

Lemma \ref{pr-ineq}, Part 6, implies $(a^2 e^{-a})/(a-1+e^{-a})>
0$ for $a \neq 0,$ so
  $$
    e^{-a} \leq b   \frac{a-1+e^{-a}}{a^2} \leq b   \frac{1-e^{-a}}{a}
\frac{e^{a}-1}{a}
    \quad \Leftrightarrow \quad
    b \geq \frac{a^2 e^{-a}}{a-1+e^{-a}}  .
  $$
Thus, the solution of the initial value problem
(\ref{eq2}), (\ref{initial}) oscillates for any $\varphi$ if and
only if
  $$
\text{either}\quad  b_1<b<b_2 \quad \text{or}\quad  b \geq  (a^2
e^{-a})/(a-1+e^{-a}),
  $$
where $b_1, b_2$ are defined in (\ref{add4}). To
simplify this system let us prove that $b_2 > a^2 e^{-a} /
(a-1+e^{-a})$. In fact,
\begin{align*}
 b_2-\frac{a^2e^{-a}}{a-1+e^{-a}}
&= a^2 \Big(\frac{|1-e^{-a}| +
    \sqrt{1-e^{-a}-ae^{-a}}}{a-1+e^{-a}}\Big)^2 -
    \frac{a^2e^{-a}}{a-1+e^{-a}} \\
&= a^2 \Big(|1-e^{-a}|^2 + 2|1-e^{-a}|\sqrt{1-e^{-a}-ae^{-a}} +1-e^{-a} \\
&\quad -ae^{-a} - e^{-a}(1-e^{-a}-ae^{-a})\Big)
 \big/\big(1-e^{-a}-ae^{-a}\big)^2 \\
&\leq a^2 \frac{|1-e^{-a}|^2 +  1-e^{-a}-ae^{-a} -
    e^{-a}(1-e^{-a}-ae^{-a})}{(1-e^{-a}-ae^{-a})^2} \\
&=    2a^2 \frac{1-e^{-a}-ae^{-a}}{(a-1+e^{-a})^2} .
\end{align*}
By Lemma \ref{pr-ineq}, Part 6, we have
$e^{a}-a-1>0$ , so
${2a^2 \frac{1-e^{-a}-ae^{-a}}{(a-1+e^{-a})^2}>0}$
and thus ${ b_2 > \frac{a^2e^{-a}}{a-1+e^{-a}} }$. Therefore,
the oscillation condition becomes
  $$
    b>a^2 \Big(\frac{|1-e^{-a}| -
\sqrt{1-e^{-a}-ae^{-a}}}{a-1+e^{-a}}\Big)^2.
  $$

Next let $a=0$. By Theorem \ref{oscillationtest}, the solution
of the initial value problem
(\ref{eq2}), (\ref{initial}) oscillates for any $\varphi$ if and
only if either \eqref{oscil1} or \eqref{oscil2} holds. Substituting
$P(0)=b$, $Q(0)=b/2$ from (\ref{integralcoef}) into
\eqref{oscil1}, \eqref{oscil2},
we obtain
\begin{align*}
 & b> \frac{1}{4} \big(\frac{b}{2}+1\big)^2 \text{ or }
      1 \leq \frac{b}{2} \leq b\\
&\Leftrightarrow \quad  b^2-12b+4<0 \text{ or }    b \geq 2\\
&\Leftrightarrow \quad
  6-4\sqrt{2}< b<6+4\sqrt{2}
 \text{ or } {b \geq 2}\\
&\Leftrightarrow \quad    b>6-4\sqrt{2},
\end{align*}
which completes the proof.
\end{proof}

\begin{proof}[Proof of Theorem \ref{constantstab}]
We remark that we have to prove that the exponential stability of
(\ref{eq2}) is equivalent to the following systems (in each of the three
cases $a<0$, $a=0$, $a>0$):
         \begin{equation}\label{stab2neg}
   {b>-a}\text{ and } { b<\frac{a^2}{1-e^{-a}-ae^{-a}}}
    \quad  \text{if } a<0,
\end{equation}
\begin{equation}\label{stab2zer}
            0<b<2, \quad \text{if } a=0,
\end{equation}
\begin{equation}\label{stab2pos}
            {b>-a}, \quad
            { b<-\frac{a^2(1+e^{-a})}{2-2e^{-a}-ae^{-a}-a}},\quad
            { b<\frac{a^2}{1-e^{-a}-ae^{-a}}}
         \quad \text{if } a>0.
\end{equation}

First, consider $a\neq 0$. By Theorem \ref{stabilitytest},  equation
(\ref{eq2})
is exponentially stable if and only if inequalities (\ref{stability2})
hold. After substituting $P(a),\, Q(a)$ from (\ref{integralcoef}) into
(\ref{stability2}) we obtain
\begin{gather*}
{b \frac{1-e^{-a}}{a}>-a\Big(b \frac{a-1+e^{-a}}{a^2}+1\Big)},\\
{ b \frac{1-e^{-a}}{a}>a
\Big(b \frac{a-1+e^{-a}}{a^2}-1\Big) \frac{1+e^{-a}}{1-e^{-a}}},\\
  { b \frac{1-e^{-a}}{a}< \frac{a}{1-e^{-a}}
\Big(1+e^{-a}b \frac{a-1+e^{-a}}{a^2}\Big)},
\end{gather*}
which can be rewritten as
\begin{gather*}
{ b\Big(\frac{1-e^{-a}}{a}+\frac{a-1+e^{-a}}{a}\Big)>-a},\\
{ b\Big(\frac{1-e^{-a}}{a}-\frac{a-1+e^{-a}}{a} \cdot
\frac{1+e^{-a}}{1-e^{-a}}\Big)>-a\frac{1+e^{-a}}{1-e^{-a}}},\\
{b\Big(\frac{1-e^{-a}}{a}-\frac{e^{-a}(a-1+e^{-a})}{a(1-e^{-a})}\Big)}<
{ \frac{a}{1-e^{-a}} }.
\end{gather*}
These inequalities can be simplified to the form
  \begin{equation} \label{add2}
\begin{gathered}
      { b>-a},\\
      { b  \frac{2-2e^{-a}-a-ae^{-a}}{a(1-e^{-a})} > -a
\frac{1+e^{-a}}{1-e^{-a}} },\\
      { b  \frac{1-e^{-a}-ae^{-a}}{a(1-e^{-a})} < \frac
{a}{1-e^{-a}}}.
\end{gathered}
\end{equation}

Consider $a<0$. For negative $a$ we have $1-e^{-a}<0$ and
$a(1-e^{-a})>0$. Moreover,
$2-2e^{-a}-a-ae^{-a}>0$ and $1-e^{-a}-ae^{-a}>0$
by Lemma \ref{pr-ineq}, Parts 3 and 2. 
Inequalities  (\ref{add2}) are equivalent to
\begin{gather*}
      {b>-a},\\
      {  b>-\frac{a^2(1+e^{-a})}{2-2e^{-a}-ae^{-a}-a}},\\
      {  b<\frac{a^2}{1-e^{-a}-ae^{-a}}}.
\end{gather*}
To simplify this system let us compare $-a$ and
$-(a^2(1+e^{-a}))/(2-2e^{-a}-ae^{-a}-a)$.
Since
$$
-\frac{a^2(1+e^{-a})}{2-2e^{-a}-ae^{-a}-a} - (-a)=
    2a \frac{1-e^{-a}-a-ae^{-a}}{2-2e^{-a}-ae^{-a}-a} ,
$$
then by Lemma \ref{pr-ineq}, Parts 3 and 5, $a<0$ implies
$1-e^{-a}-a-ae^{-a}>0$ and $2-2e^{-a}-ae^{-a}-a>0$.
Hence
${a   \frac{1-e^{-a}-a-ae^{-a}}{2-2e^{-a}-ae^{-a}-a}<0  }$
and
  ${-\frac{a^2(1+e^{-a})}{2-2e^{-a}-ae^{-a}-a} < -a}$;
therefore,
\begin{gather*}
      {b>-a},\\
      { b>-\frac{a^2(1+e^{-a})}{2-2e^{-a}-ae^{-a}-a}},\\
      { b<\frac{a^2}{1-e^{-a}-ae^{-a}}}
 \end{gather*}
is equivalent to
 \begin{gather*}
      {b>-a},\\
      {  b<\frac{a^2}{1-e^{-a}-ae^{-a}}}.
 \end{gather*}
Thus, if $a<0$ then (\ref{eq2}) is exponentially stable if and
only if system (\ref{stab2neg}) holds.

Consider $a>0$. If $a>0$ then $1-e^{-a}>0$ and $a(1-e^{-a})>0$.
Moreover, by Lemma \ref{pr-ineq}, Parts 2 and 3, we have
$2-ae^{-a}-2e^{-a}-a<0$ and $1-e^{-a}-ae^{-a}>0$.
Applying these inequalities, we obtain that (\ref{add2}) is equivalent to
\begin{gather*}
      {b>-a},\\
      { b<-\frac{a^2(1+e^{-a})}{2-2e^{-a}-ae^{-a}-a}},\\
      { b<\frac{a^2}{1-e^{-a}-ae^{-a}}}.
    \end{gather*}
So, if $a>0$ then (\ref{eq2}) is exponentially stable if and
only if inequalities (\ref{stab2pos}) hold.

Finally, let $a=0$. By Theorem \ref{stabilitytest} equation (\ref{eq2})
is exponentially stable if and only if inequalities (\ref{stability2})
hold. Substitute $P(0)=b$, $Q(0)=b/2$ from (\ref{integralcoef}) into
(\ref{stability2}):
\[
      {b > 0},\quad
      {b > b - 2},\quad
      { b < \frac {b}{2} +1}.
\]
which can be rewritten as
\[
      {b > 0},\quad
      {b < 2} .
\] 
So, if $a=0$ then (\ref{eq2}) is exponentially stable if and
only if inequalities (\ref{stab2zer}) hold.
\end{proof}


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