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\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2003(2003), No. 92, pp. 1--6.\newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.swt.edu  (login: ftp)}
\thanks{\copyright 2003 Texas State University-San Marcos.}
\vspace{9mm}}

\begin{document}

\title[\hfilneg EJDE--2003/92\hfil Linear delay differential equation]
{Linear delay differential equation with a positive and a negative term}

\author[Faiz Ahmad\hfil EJDE--2003/92\hfilneg]
{Faiz Ahmad}

\address{Faiz Ahmad \hfill\break
Department of Mathematics, Faculty of Science\\
King Abdulaziz University, P. O. Box 80203\\
Jeddah 21589, Saudi Arabia}
% Telephone number: 009662-6332195
\email{faizmath@hotmail.com}

\date{}
\thanks{Submitted June 21, 2003. Published September 8, 2003.}
\subjclass{34K15, 34C10}
\keywords{Delay equation, oscillation, necessary condition, sufficient conditions}


\begin{abstract}
 We study a linear delay differential equation with a single
 positive and a single negative term. We find a necessary condition
 for the oscillation of all solutions. We also find  sufficient conditions
 for oscillation, which improve the known conditions.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}

\section{Introduction}

 Ladas and co-workers \cite{a3,g1,l1} studied the delay differential equation
 \begin{equation}
 \dot u(t)+pu(t-\tau)-qu(t-\sigma)=0, \label{e1}
 \end{equation}
where $p,q,\tau,\sigma \in R^{+}$. They obtained the following set of sufficient
conditions for the oscillation of all solutions to this equation:
\begin{equation} \label{e2}
\begin{gathered}
 p>q\,,\\
 \tau \geq \sigma\,,\\
 q(\tau-\sigma)\leq 1\,,\\
(p-q)\tau >{\frac{\rm1}{\rm e}}[1-q(\tau-\sigma)]\,.
\end{gathered}
\end{equation}
Chuanxi and Ladas \cite{c1} generalized the above conditions to a non-autonomous
equation in which $p$ and $q$ are replaced by continuous functions $P(t)$ and $Q(t)$.
The delay equation with an arbitrary number of  positive and negative terms has
been studied, for the autonomous case, by Agwo \cite{a2} and Ahmad and Alherbi \cite{a1}.
Recently Elabbasy et al. \cite{e1} have applied a technique in Li \cite{l2}
 to generalize
conditions \eqref{e2} for a non-autonomous equation having
arbitrary number of positive and negative terms.

A common feature among results pertaining to non-autonomous generalizations
of \eqref{e1} is that, in the limit of constant coefficients, they reduce to
a {\it stronger} version of conditions \eqref{e2} in which the fourth condition
is replaced by $(p-q)\tau >1$.
 Although this set of conditions underwent a gradual improvement in \cite{a3,g1,l1},
after 1991 no attempt has  been successful in improving these conditions
(as far as the author is aware).

The first and second conditions of \eqref{e2} are necessary for
oscillation \cite[p41]{g1}. However it is obvious that the third
condition is unnecessarily restrictive since, no matter how large
the coefficient of the negative term and the difference between
the delays might be,  all solutions of the equation must be
oscillatory provided the coefficient of the positive term is {\it
sufficiently large}. Ideally the oscillation conditions should
involve what Hunt and Yorke \cite{h1} have termed as the {\it torque}
associated with the delay differential equation. For \eqref{e1}
this parameter is $p\tau-q\sigma$. In this paper we shall seek
sufficient conditions which  depend either  on the difference or
the ratio of $p\tau$ and $q\sigma$. We also show that the condition
$$
p\tau-q\sigma > {\frac{\rm1}{\rm e}}
$$
is necessary for the oscillation of all solutions of \eqref{e1}.
We expect that this work will stimulate further research in the quest
of {\it necessary  and sufficient} conditions for oscillation.


\section{Basic results and preliminary lemmas}

The following properties of the exponential function are easily established.
Let $x\geq0$. Then
\begin{gather}
 e^{ax}\geq a e^x +1-a, \quad \mbox{if } a\geq1, \label{e3}\\
 e^{ax}\leq a e^x +1-a, \quad \mbox{if } a<1, \label{e4}\\
 e^{ax}\geq bx + {\frac {\rm b}{\rm a}}[1-ln\,{\frac {\rm b}{\rm a}}]\quad \mbox{if } a>0 \mbox{ and }b>0.
 \label{e5}
\end{gather}
If $q=0$ or $\tau=\sigma$, \eqref{e1} reduces to a delay equation with a single
positive term. If $\sigma=0$, then a necessary and sufficient condition for
oscillation is  \cite[p40]{g1}
$$ p\tau e^{-q\tau}\,>\,{\frac {\rm 1}{\rm e}}\,.
$$
Let $0<q<p$ and $0<\sigma<\tau$. For $x\geq0$, define
\begin{equation}\label{e6}
\begin{gathered}
g(x)= qe^{-\sigma x} -pe^{-\tau x},\\
x_1 ={\frac {\rm1}{\rm {\tau-\sigma}}}\ln({\frac{\rm p}{\rm q}}),\\
x_2={\frac{\rm1}{\rm {\tau-\sigma}}}\ln({{\frac{\rm \tau p}{\rm \sigma q}}}),\\
k=(1-{{\frac {\rm \sigma}{\rm \tau}}})q ({\frac{\rm p\tau}{\rm {q\sigma}}})^{\frac{\rm -\sigma}{\rm {\tau-\sigma}}}.
\end{gathered}
\end{equation}
A simple calculation shows that $g$ is increasing and concave on the
interval $[0,x_2]$. It vanishes at $x_1$ and attains its maximum value, $k$, at
$x_2$.

The characteristic equation corresponding to \eqref{e1} is
\begin{equation}
x+p e^{-\tau x}-q e^{-\sigma x} = 0,\label{e7}
\end{equation}
or $ x = g(x)$.

It is well-known that a delay differential equation possesses a non-oscillatory
solution if and only if its characteristic equation has a real root.
Since $p > q$, it follows that $x = 0$ does  not satisfy \eqref{e7}.
We shall investigate the existence of positive or negative roots in the following
lemmas.

\begin{lemma} \label{lm1}
 The characteristic equation does not have a positive root if $x_{1}\geq k$ i.e.
 $$
 \ln({\frac {\rm p}{\rm q}}) \geq{{\frac{\rm(\tau-\sigma)^{2}}{\rm \tau}q}}
 ({{\frac{\rm p\tau}{\rm q\sigma}}})^{{\frac{\rm -\sigma}{\rm{\tau-\sigma}}}}.
 $$
\end{lemma}

\begin{proof}
Let $f(x)= g(x)-x$. Then $f$ is negative on $[0,x_1]$ and for $c>0$,
we have
$$ f(x_1 +c)= g(x_1 +c)-x_1 -c \leq k-x_1-c< 0\,.
$$
Thus $f$ remains negative on $[0,\infty)$ implying that the characteristic equation
does not have a positive root.
\end{proof}

\begin{lemma} \label{lm2}
The characteristic equation has no positive root if
$$
\ln({{\frac{\rm p}{\rm q}}}) >{{\frac{\rm(\tau-\sigma)^{2}}{\rm \tau}}} q
({{\frac{\rm p}{\rm q}}})^{\frac{\rm-\sigma}{\rm{\tau-\sigma}}}-{{\frac{\rm(\tau-\sigma)}{\rm \tau}}}.
$$
\end{lemma}

\begin{proof}
Since $g$ is concave on $[x_1,x_2]$, there is at most one point, call it $c$,
in $[x_1,x_2]$ where the tangent to the curve $y= g(x)$ is parallel to the line
$y = x$. The intercept of the tangent with the $x-$axis is $c-g(c)$.
If this number is positive, the line $y=x$ does not intersect the curve $y=g(x)$.
This implies no positive root of the characteristic equation if
\begin{equation}
 g(c) < c,\quad\mbox{or}\quad  q e^{-\sigma c} - p e^{-\tau c} < c.
\label{e8}
\end{equation}
Since the slope at the point $(c,g(c))$ is unity, we have
\begin{equation}
-\sigma q e^{-\sigma c }+ p\tau e^{-\tau c} = 1.\label{e9}
\end{equation}
 From \eqref{e8}--\eqref{e9},  we get
\begin{equation}
-\tau c + q(\tau -\sigma )e^{-\sigma c} < 1\,. \label{e10}
\end{equation}
The expression on the left side of \eqref{e10}, as a function of $c$, is decreasing.
Hence it will hold on the entire interval $[x_1,x_2]$ provided it does so for $c=x_1$.
This gives
\begin{equation}
{{\frac{\rm-\tau}{\rm {\tau-\sigma}}}}\ln({\frac{\rm p}{\rm q}}) +q(\tau-\sigma)
e^{{{\frac{\rm -\sigma}{\rm {\tau -\sigma}}}}\ln({{\frac{\rm p}{\rm q}}})}< 1, \label{e11}
\end{equation}
or
$$
\ln({{\frac{\rm p}{\rm q}}}) >{{\frac{\rm(\tau-\sigma)^{2}}{\rm \tau}}} q
({\frac{\rm p}{\rm q}})^{{\frac{\rm -\sigma}{\rm {\tau-\sigma}}}}-{{\frac{\rm(\tau-\sigma)}{\rm \tau}}}.
$$
\end{proof}

\begin{lemma} \label{lm3}
The characteristic equation does not have a negative root if
 $p\tau - q\sigma > {1/b}$ or $e^{q(\tau -\sigma)-1}$,  where
$b$ is the larger root of the equation
$$
 x(1-\ln x)= {{\frac{\rm q(\tau-\sigma)}{\rm {p\tau-q\sigma}}}},
$$
and $1< b \leq e$.
\end{lemma}

\begin{proof} A negative root of the characteristic equation \eqref{e7}
 is equivalent to a positive root of the equation
\begin{equation}
  -x +p e^{\tau x} -qe^{\sigma x} = 0\,.\label{e12}
\end{equation}
In the above equation, the change of variable $y=\tau x$ yields
$$
-y + p\tau e^y - q \tau e^{{{\frac{\rm\sigma}{\rm \tau}}}y}=0\,.
$$
Suppose $y>0$ is a root of this equation. On making use of \eqref{e4}, we get
\begin{equation}
0\geq -y + p\tau e^y -q\tau({{\frac{\rm\sigma}{\rm \tau}}}\,e^y +1-{{\frac{\rm\sigma}{\rm \tau}}})
= -y +(p\tau - q \sigma) e^y -q(\tau - \sigma). \label{e13}
\end{equation}
Now we use \eqref{e5} to obtain, for arbitrary $b>0$,
\begin{equation}
0\geq -y-q(\tau-\sigma)+(p\tau-q\sigma)[by + b(1-\ln b)].\label{e14}
\end{equation}
Choose $b$ such that $1<b\leq e$ and
$$
b(1-\ln b) = {{\frac {\rm q(\tau-\sigma)}{ {p\tau-q\sigma}}}}.
$$
Inequality \eqref{e14} becomes
$0\geq y[-1+b(p\tau -q\sigma)]$.
Since $y>0$, there will be a contradiction if $p\tau -q\sigma > 1/b$.

Now define $z= y+q(\tau-\sigma)$. Inequality \eqref{e13}  becomes
\begin{align*}
0&\geq -z + (p\tau-q\sigma) e^{-q(\tau-\sigma)} e^z\\
 &\geq -z + (p\tau-q\sigma) e^{-q(\tau-\sigma)}e z\\
&=z[-1+(p\tau-q\sigma) e^{1-q(\tau-\sigma)}],
\end{align*}
which will lead to a contradiction if
$p\tau-q\sigma > e^{q(\tau-\sigma)-1}$.
\end{proof}

\section{Main Results}

It is well-known that the first two conditions \eqref{e2} are necessary for the
oscillation of all solutions of \eqref{e1}. In this section we first prove a
necessary condition which should be useful in estimating how far from the best
possible position a sufficient condition actually is.

\begin{theorem}[A necessary condition for oscillation] \label{thm1}
Let  $ p,q,\tau,\sigma \in R^+$, $p>q$ and $\tau \geq \sigma$.
If all solutions of \eqref{e1} are oscillatory then
$ p\tau-q\sigma > 1/e$.
\end{theorem}

\begin{proof}
We shall prove that otherwise a real root of the characteristic equation must
exist indicating a positive solution of the  delay equation.
Denote the left side of the characteristic equation, i.e. \eqref{e7}, by $f(x)$.
First assume $ p\tau-q\sigma = 1/e$. Since $f(0)=p-q >0$ and
$$
f(-1/\tau)=-1/\tau +pe -q e^{{\frac{\rm\sigma}{\rm \tau}}} =q[{{\frac{\rm e\sigma}{\rm \tau}}} -e^{{{\frac{\rm\sigma}{\rm\tau}}}}] \leq 0,
$$
hence $f$ has a zero in $[-1/\tau , 0)$.

Next assume $p\tau-q\sigma < 1/e. $  If $q\sigma=0$, it is easy to see that a
zero of $f$ will exist in $[-1/\tau,0)$. Let $q\sigma >0$ and define $a=q\sigma e$.
There exists an $\epsilon >0$ such that
$$
p\tau-q\sigma = {{\frac{\rm 1-2\epsilon}{\rm e}}} < {{\frac{\rm 1-\epsilon}{\rm e}}}.
$$
Without loss of generality we can choose $\epsilon < a$. Thus
\begin{equation}
 p\tau < q\sigma +{{\frac{\rm1-\epsilon}{\rm e}}}
= {{\frac{\rm a+1-\epsilon}{\rm e}}}\,.\label{e15}
\end{equation}
We write $ f(x)=f_1(x) + f_2(x)$, where
\begin{gather*}
f_1(x)= (a+1-\epsilon)x +pe^{-\tau x},
f_2(x)= -(a-\epsilon)x -q e^{-\sigma x}.
\end{gather*}
Since an equation of the form $x+ c e^{-dx}=0$, has a real root if and only if
$cd \leq 1/e$, it follows from \eqref{e15} that $f_1(x)=0$ has a real root, say $x_0$,
while
$$
q\sigma =a/e >{{\frac{\rm a-\epsilon}{\rm e}}},
$$
shows that $f_2(x)=0$ does not have any real root. Since $f_2(0)<0$
it follows that $f_2(x_0)<0$. Now
$$
f(x_0)=f_1(x_0)+f_2(x_0)<0\,.
$$
Also $f(0)=p-q >0, $  hence the characteristic equation possesses a root in
$(0,x_0)$. This proves  the necessity of the condition $p\tau-q\sigma > 1/e$
for the oscillation of all solutions of \eqref{e1}.
The  proof of Theorem 1 is complete.
\end{proof}

Putting the results of Lemmas 2.1-2.3 together, we obtain the following Theorem.

\begin{theorem}[Sufficient conditions for oscillation] \label{thm2}
All solutions of \eqref{e1} will be oscillatory if
\begin{gather*}
p > q > 0\,, \\
\tau > \sigma \geq 0\,,\\
\ln({{\frac{\rm p}{\rm q}}}) > \min \lbrace {{\frac{\rm(\tau-\sigma)^{2}}{\rm \tau}}}
q ({{\frac{\rm p\tau}{\rm q\sigma}}})^{{\frac{\rm -\sigma}{\rm{\tau-\sigma}}}}, \;{{\frac{\rm(\tau-\sigma)^{2}}{\rm \tau}}}
q ({{\frac{\rm p}{\rm q}}})^{{\frac{\rm -\sigma }{\rm{\tau-\sigma}}}}-{{\frac{\rm(\tau-\sigma)}{\rm \tau}}} \rbrace, \\
p\tau - q\sigma > \min \lbrace 1/b \, ,e^{q(\tau -\sigma)-1} \rbrace ,
\end{gather*}
where $b$ is the larger root of the equation
$$
x(1-\ln x)= {{\frac{\rm q(\tau-\sigma)}{\rm {p\tau-q\sigma}}}},
$$
and $1< b \leq e$. If $q=0$ and/or $\tau = \sigma$ then $b=e$.
\end{theorem}

A glance at \eqref{e11} indicates that it can be satisfied for arbitrary
$q(\tau-\sigma)$ if $p$ is large enough. For example if $q=3, \tau=4, \sigma=2$,
the inequality of Lemma \ref{lm2} will hold for $p \geq 6.82$.
Also $p\tau- q\sigma = (p-q)\tau +q(\tau-\sigma)$, hence the condition
$$
p\tau -q\sigma > e^{q(\tau-\sigma)-1},
$$
is equivalent to
\begin{equation}
(p-q)\tau > e^{q(\tau -\sigma)-1} - q(\tau-\sigma).\label{e16}
\end{equation}
Since on $[0,1]$, $e^x -ex \leq 1-x$, it is clear that when conditions
\eqref{e2} hold, the number on the right side of the fourth condition is
larger than its counterpart on the right side of \eqref{e16}.

\begin{thebibliography}{0}

\bibitem{a1} Ahmad, F. and  Alherbi, R. A.,
\emph{Oscillation of solutions of a delay differential equation with positive and
negative coefficients}, submitted for publication.

\bibitem{a2} Agwo H, A.,
\emph{On the oscillation of delay differential equations with real coefficients},
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\bibitem{a3} Arino, O.,  Ladas, G., and Sficas, Y. G.;
\emph{On oscillation of some retarded differential equations}
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\bibitem{c1} Chuanxi, Q. and Ladas, G.
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\bibitem{e1} Elabbasy, El M.,  Hegazi, A. S., and Saker, S. H.;
\emph{Oscillation of solutions to delay differential equations with positive and
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{\textbf {2000}} (2000), No.13 , 1-13.

\bibitem{g1} Gyori, I. and Ladas, G.
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\bibitem{h1} Hunt, B. R. and Yorke, J. A.,
\emph{When all solutions of $ x^{'}= -\sum _{j=1}^{n}q_j (t) x(t-T_j(t))$ oscillate},
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\bibitem{l1} Ladas, G., and  Sficas, Y. G.,
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\bibitem{l2} Li, B. \emph{Oscillation of first order delay differential equations},
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\end{thebibliography}


 \end{document}