\documentclass[reqno]{amsart}
\AtBeginDocument{{\noindent\small
Fifth Mississippi State Conference on
Differential Equations and Computational Simulations,
{\em Electronic Journal of Differential Equations},
Conference 10, 2003, pp. 79--88.\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 Southwest Texas State University.}
\vspace{1cm}}
\begin{document}
\setcounter{page}{79}
\title[\hfilneg EJDE/Conf/10\hfil Oscillation of third order functional D. E.]
{Oscillation of third order functional differential equations with delay}
\author[T. Candan \& R. S. Dahiya\hfil EJDE/Conf/10\hfilneg]
{Tuncay Candan \& Rajbir S. Dahiya}
\address{Tuncay Candan \hfill\break\indent
Department of Mathematics,
Iowa State University\hfill\break\indent
Ames, IA 50011, USA}
\email{tuncayca@iastate.edu}
\address{Rajbir S. Dahiya \hfill\break\indent
Department of Mathematics,
Iowa State University\hfill\break\indent
Ames, IA 50011, USA}
\email{dahiya@math.iastate.edu}
\thanks{Published February 28, 2003.}
\subjclass{34K11}
\keywords{Oscillation theory}
\begin{abstract}
We consider third order functional differential equations with
discrete and continuous delay. We then develop several theorems
related to the oscillatory behavior of these differential equations.
\end{abstract}
\maketitle
\newtheorem{theorem}{Theorem}[section]
\newtheorem{example}[theorem]{Example}
\numberwithin{equation}{section}
\newcommand{\ls}{\limsup_{t\rightarrow\infty}}
\newcommand{\li}{\liminf_{t\rightarrow\infty}}
\newcommand{\dst}{\displaystyle}
\section{Introduction}
Our goal in this paper is to study functional differential
equations of the form
\begin{equation}\label{e:2.4}
(b(t)(a(t)x'(t))')'+\sum_{i=1}^{m}q_i(t)f(x(\sigma_i(t)))=h(t),
\end{equation}
where $a,b,h\in C([t_0,\infty), \mathbb{R}), a(t), b(t)>0, \quad
f:\mathbb{R}\to \mathbb{R}$ continuous, $\sigma_i(t)\to \infty$,
as\newline $ t\ \to\infty,\quad i=1,2,\dots ,m$, and
\begin{equation}\label{e:2.5}
(b(t)(a(t)x'(t))')'+\int_{c}^{d}q(t,\xi)f(x(\sigma(t,\xi)))d\xi=0,
\end{equation}
where $a,b\in C([t_0,\infty), \mathbb{R})$, $f\in
C(\mathbb{R},\mathbb{R})$. The oscillations of solutions of third
order equations were studied by Rao and Dahiya \cite{rao}, Tantawy
\cite{tan}, Waltman \cite{wal} and Zafer and Dahiya \cite{dah}.
The results in this paper for equation (\ref{e:2.4}) are more
general comparing to Zafer and Dahiya \cite{dah}. The results for
equation (\ref{e:2.5}) are essentially new.
As is customary, a solution of equations (\ref{e:2.4}) and
(\ref{e:2.5}) is called oscillatory if it has arbitrarily large
zeros, otherwise it is called nonoscillatory. The solution of
equations (\ref{e:2.4}) and (\ref{e:2.5}) is called almost
oscillatory if it is oscillatory or
$\dst\lim_{t\to\infty}x^{(i)}(t)=0$, $i=0,1,2$.
\section{Main Results}
\subsection*{Oscillatory behavior of third order differential equations with discrete delay }
Assume that $xf(x)>0$, $x\ne 0$, $q_i(t)\ge 0$ is not identically zero in
any half line of the form $(\tau,\infty)$ for some $\tau\ge0$,
$i=1,2,\dots ,m$ and $\sigma_i(t) 0$, \newline
i=1,2,\ldots,m, $b'(t)\ge 0$, and
\begin{equation}\label{e:2.31}
\int^{\infty}\frac{dt}{b(t)}=\infty,\quad \int^{\infty}\frac{dt}{a(t)}=\infty.
\end{equation}
\begin{theorem}\label{t:2.3.1}
Let $f(x)=x$ and $h(t)=0$. Suppose that there exist a
differentiable function $p\in C([t_0,\infty), \mathbb{R})$,
$p(t)>0$ such that
\begin{equation}\label{e:2.32}
\int^{\infty}\Big[q(t)p(t)- \frac{b(t)(p'(t))^2}{\sum_{i=1}^{m}
\frac{(\sigma_i(t)-T)}{a(\sigma_i(t))}\sigma_i'(t)4p(t)}\Big]dt=\infty,
\end{equation}
where $q(t)=min \{q_1(t),q_2(t),\dots ,q_m(t)\}$, for every $T\ge 0$, and that
\begin{equation}\label{e:2.33}
\int_{\sigma(t)}^{t}\Big[\int_{\sigma(t)
}^{r}\frac{1}{a(u)}du\left(\int_{u}^{r}\frac{1}{b(v)}dv\right)\Big]\sum_{i=1}^{m}q_i(r)dr> 1,
\end{equation}
where $\sigma(t)=max \{\sigma_1(t),\sigma_2(t),\dots ,\sigma_m(t)\}$. Then the equation (\ref{e:2.4}) is oscillatory.
\end{theorem}
\noindent
\textbf{Proof.}
Let $x(t)$ be a non-oscillatory solution of (\ref{e:2.4}). Assume $x(t)$ is
eventually positive. Since $\sigma_{i}(t)\to\infty$ as
$t\to\infty$ for $i=1,2,\dots ,m$, there exist a
$ t_1\ge t_0$ such that $x(t)>0$ and $x(\sigma_{i}(t))>0$ for $ t\ge t_1$.
From (\ref{e:2.4}), we have
\begin{equation}\label{e:2.34}
(b(t)(a(t)x'(t))')'= -\sum_{i=1}^{m}q_i(t)x(\sigma_i(t)).
\end{equation}
Since $q_{i}(t)$ is not negative and $x(\sigma_{i}(t))>0$ is positive
for $t\ge t_1$, the right-hand side becomes non-positive. Therefore, we have
\[
(b(t)(a(t)x'(t))')'\le 0
\]
for $t\ge t_1$. Thus, $x(t)$, $x'(t)$, $(a(t)x'(t))'$ are monotone
and eventually one-signed. Now we want to show that there is a
$t_2\ge t_1$ such that for $t\ge t_2$
\begin{equation}\label{e:2.35}
(a(t)x'(t))'> 0.
\end{equation}
Suppose this is not true, then $(a(t)x'(t))'\le 0$. Since
$q_i(t)$, $i=1,2,\dots ,m$ are not identically zero and $b(t)>0$,
it is clear that there is $t_3\ge t_2$ such that
$b(t_3)(a(t_3)x'(t_3))'< 0$. Then, for $t>t_3$ we have
\begin{equation}\label{e:2.36}
b(t)(a(t)x'(t))'\le b(t_3)(a(t_3)x'(t_3))'<0.
\end{equation}
Dividing (\ref{e:2.36}) by $b(t)$ and then integrating between $t_3$ and
$t$, we obtain
\begin{equation}\label{e:2.37}
a(t)x'(t)-a(t_3)x'(t_3)< b(t_3)(a(t_3)
x(t_3)')'\int_{t_3}^{t}\frac {1}{b(s)}ds.
\end{equation}
Letting $t\to\infty$ in (\ref{e:2.37}), and because of (\ref{e:2.31}) we see that
$a(t)x'(t)\to -\infty$ as $t\to\infty$. Thus there is a
$t_4\ge t_3$ such that $a(t_4)x'(t_4)<0$. Using $(a(t)x'(t))'\le 0$, we
have for $t\ge t_4$
\begin{equation}\label{e:2.38}
a(t)x'(t)\le a(t_4)x'(t_4).
\end{equation}
If we divide (\ref{e:2.38}) by $a(t)$ and integrate from $t_4$ to $t$
with $t\to\infty$, the right-hand side becomes negative. Thus, we have
$x(t)\to -\infty$. But this is a contradiction $x(t)$ being
eventually positive and therefore it proves that (\ref{e:2.35}) holds.
Now we consider two cases.
Suppose $x'(t)$ is eventually positive, say $x'(t)>0$ for
$t\ge t_2$. Define the function $z(t)$ by
\[
z(t)=\frac{ b(t)(a(t)x'(t))'}{\sum_{i=1}^{m}x(\sigma_i(t))}p(t).
\]
It is obvious that $z(t)>0$ for $t\ge t_2$ and $z'(t)$ is
\begin{eqnarray}
z'(t)=-\frac{\sum_{i=1}^{m}q_i(t)x(\sigma_i(t))}{\sum_{i=1}^{m}x(\sigma_i(t))}p(t)+\frac
{p'(t)}{p(t)}z(t)-\frac{\sum_{i=1}^{m}x'(\sigma_i(t))\sigma_i'(t)}{
\sum_{i=1}^{m}x(\sigma_i(t))}z(t). \nonumber
\end{eqnarray}
Then
\begin{equation}\label{e:2.39}
z'(t)\le -q(t)p(t)+\frac
{p'(t)}{p(t)}z(t)-\frac{\sum_{i=1}^{m}x'(\sigma_i(t))\sigma_i'(t)}{
\sum_{i=1}^{m}x(\sigma_i(t))}z(t),
\end{equation}
where $q(t)=min \{q_1(t),q_2(t),\dots ,q_m(t)\}$.
On the other hand, using \newline $(b(t)(a(t)x^{'}(t))^{'})^{'}\le 0$, $b'(t)\ge
0$ and (\ref{e:2.35}), we can find that
\begin{equation}\label{e:2.40}
(a(t)x'(t))''\le 0.
\end{equation}
Using (\ref{e:2.40}) and the equality
\begin{equation}\label{e:2.41}
a(t)x'(t)=a(T)x'(T)+\int_{T}^{t}(a(s)x'(s))'ds,
\end{equation}
we have
\begin{equation}\label{e:2.42}
a(t)x'(t)\ge (t-T)(a(t)x'(t))'
\end{equation}
for $T\ge t_2$. Since $(a(t)x'(t))'$ is non-increasing, we obtain
\begin{equation}\label{e:2.43}
a(\sigma_i(t))x'(\sigma_i(t))\ge (\sigma_i(t)-T)(a(t)x'(t))'\quad for\quad i=1,2,\dots ,m.
\end{equation}
Multiplying both sides of (\ref{e:2.43}) by
\[
\frac{\sigma_i'(t)}{a(\sigma_i(t))}
\]
and taking the summation from $1$ to $m$, we have
\begin{equation}\label{e:2.44}
\sum_{i=1}^{m} \sigma_i'(t)x'(\sigma_i (t))\ge \sum_{i=1}^{m}\frac
{(\sigma_i (t)-T)}{a(\sigma_i(t))} \sigma_i'(t)(a(t)x'(t))'.
\end{equation}
Then, using (\ref{e:2.44}) in (\ref{e:2.39}), it follows that
\[
z'(t)\le -q(t)p(t)+\frac
{p'(t)}{p(t)}z(t)-\frac{\sum_{i=1}^{m}\frac{(\sigma_i(t)-T)}{a(\sigma_i(t))}\sigma_i'(t)}{b(t)p(t)}z^{2}(t),
\]
and completing the square will leads to
\begin{equation}\label{e:2.45}
z'(t)\le
-q(t)p(t)+\frac{b(t)(p')^{2}(t)}{\sum_{i=1}^{m}\frac{(\sigma_i(t)-T)}{a(\sigma_i(t))}\sigma_i'(t)4p(t)}.
\end{equation}
Integrating (\ref{e:2.45}) between $T$ and t and letting
$t\to\infty$, we see that $\dst\lim_{t\to\infty} z(t)=-\infty$.
This contradicts $z(t)$ being eventually positive.
If $x'(t)$ is eventually negative. We integrate
(\ref{e:2.4}) from $t$ to $\infty$ and since
\[
b(t)(a(t)x'(t))'>0,
\]
we have
\begin{equation}\label{e:2.46}
-b(t)(a(t)x'(t))'
+\int_{t}^{\infty}\sum_{i=1}^{m}q_i(r)x(\sigma_i(r))dr\le
0.
\end{equation}
Now integrating (\ref{e:2.46}) from $t$ to $\infty$ after dividing by $b(t)$
and using $a(t)x'(t)<0$, will lead to
\begin{equation}\label{e:2.47}
a(t)x'(t)
+\int_{t}^{\infty}\Big(\int_{t}^{r}\frac{1}{b(u)}du\Big)
\sum_{i=1}^{m}q_i(r)x(\sigma_i(r))dr\le
0.
\end{equation}
Dividing (\ref{e:2.47}) by $a(t)$ and integrating again from $t$ to $\infty$ gives
\begin{equation}\label{e:2.48}
\int_{t}^{\infty}\Big[\int_{t}^{r}\frac{1}{a(u)}du\Big(\int_{u}^{r}
\frac{1}{b(v)}dv\Big)\Big]\sum_{i=1}^{m}q_i(r)x(\sigma_i(r))dr\le
x(t).
\end{equation}
Replacing $t$ by $\sigma(t)$ in (\ref{e:2.48}),
where $\sigma(t)=\max\{\sigma_1(t),\sigma_2(t),\dots ,\sigma_m(t)\}$, will give
\begin{equation}\label{e:2.49}
\int_{\sigma(t)}^{t}\Big[\int_{\sigma(t) }^{r}\frac{1}{a(u)}du
\Big(\int_{u}^{r}\frac{1}{b(v)}dv\Big)\Big]
\sum_{i=1}^{m}q_i(r)x(\sigma_i(r))dr\le x(\sigma(t)).
\end{equation}
Using the fact that $\sigma_i(t)0$, and $h(t)=0$.
Suppose that there exist a differentiable function $p\in
C([t_0,\infty), \mathbb{R})$, $p(t)>0$ such that
\begin{equation}\label{e:2.50}
\int^{\infty}\Big[q(t)p(t)- \frac{b(t)(p'(t))^2}{\sum_{i=1}^{m}
\frac{(\sigma_i(t)-T)}{a(\sigma_i(t))}\sigma_i'(t)4\lambda
p(t)}\Big]dt =\infty,
\end{equation}
where $q(t)=min \{q_1(t),q_2(t),\dots ,q_m(t)\}$,
for every $T\ge 0$, and that
\begin{equation}\label{e:2.51}
\limsup_{t\to\infty}\int_{\sigma(t)}^{t}\Big[\int_{\sigma(t)
}^{r}\frac{1}{a(u)}du\Big(\int_{u}^{r}\frac{1}{b(v)}dv\Big)\Big]
\sum_{i=1}^{m}q_i(r)dr=\infty,
\end{equation}
where $\sigma(t)=max\{\sigma_1(t),\sigma_2(t),\ldots,\sigma_m(t)\}$. Then the equation (\ref{e:2.4}) is oscillatory.
\end{theorem}
\noindent
\textbf{Proof.}
The beginning part of the proof is similar to the proof of
Theorem~\ref{t:2.3.1} until we reach at two possible cases.
Suppose $x'(t)$ is eventually positive. Then, we can define
\[
z(t)=\frac{ b(t)(a(t)x'(t))'}{\sum_{i=1}^{m}f(x(\sigma_i(t)))}p(t)>0.
\]
It is obvious that $z(t)>0$ for $t\ge t_2$ and $z'(t)$ is
\begin{eqnarray}
z'(t)&=&-\frac{\sum_{i=1}^{m}q_i(t)f(x(\sigma_i(t)))}{\sum_{i=1}^{m}
f(x(\sigma_i(t)))}p(t) \nonumber\\[2pt]
&&+\frac
{p'(t)}{p(t)}z(t)-\frac{\sum_{i=1}^{m}f'(x(\sigma_i(t)))x'(\sigma_i(t))\sigma_i'(t)}{
\sum_{i=1}^{m}f(x(\sigma_i(t)))}z(t). \nonumber
\end{eqnarray}
Then
\begin{equation}\label{e:2.52}
z'(t)\le -q(t)p(t)+\frac
{p'(t)}{p(t)}z(t)-\frac{\sum_{i=1}^{m}f'(x(\sigma_i(t))x'(\sigma_i(t))\sigma_i'(t)}{\sum_{i=1}^{m}f(x(\sigma_i(t)))}z(t),
\end{equation}
where $q(t)=\min \{q_1(t),q_2(t),\dots ,q_m(t)\}$. On the other hand, since\newline $(b(t)(a(t)x^{'}(t))^{'})^{'}\le 0$, (\ref{e:2.35})
holds and $b'(t)\ge 0$, we can obtain
\begin{equation}\label{e:2.53}
(a(t)x'(t))''\le 0.
\end{equation}
Using (\ref{e:2.53}) and the equality
\begin{equation}\label{e:2.54}
a(t)x'(t)=a(T)x'(T)+\int_{T}^{t}(a(s)x'(s))'ds
\end{equation}
will lead to
\begin{equation}\label{e:2.55}
a(t)x'(t)\ge (t-T)(a(t)x'(t))'.
\end{equation}
Now using non-increasing nature of $(a(t)x'(t))'$, we obtain
\begin{equation}\label{e:2.56}
a(\sigma_i(t))x'(\sigma_i(t))\ge (\sigma_i(t)-T)(a(t)x'(t))'\quad for\quad i=1,2,\dots,m.
\end{equation}
Multiplying both sides of (\ref{e:2.56}) by
\[
\frac{\sigma_i'(t)}{a(\sigma_i(t))}
\]
and taking the summation from $1$ to $m$, we have
\begin{equation}\label{e:2.57}
\sum_{i=1}^{m} \sigma_i'(t)x'(\sigma_i (t))\ge \sum_{i=1}^{m}\frac
{(\sigma_i (t)-T)}{a(\sigma_i(t))} \sigma_i'(t)(a(t)x'(t))^{'}.
\end{equation}
Then, using (\ref{e:2.57}) in (\ref{e:2.52}), it follows that
\[
z'(t)\le -q(t)p(t)+\frac
{p'(t)}{p(t)}z(t)-\lambda\frac{\sum_{i=1}^{m}\frac{(\sigma_i(t)-T)}{a(\sigma_i(t))}\sigma_i'(t)}{
b(t)p(t)}z^{2}(t),
\]
and then completing the square leads to
\begin{equation}\label{e:2.58}
z'(t)\le
-q(t)p(t)+\frac{b(t)(p'(t))^{2}}{\lambda\sum_{i=1}^{m}\frac{(\sigma_i(t)-T)}{a(\sigma_i(t))}\sigma_i'(t)4p(t)}.
\end{equation}
Integrating (\ref{e:2.58}) between $T$ to t and letting
$t\to\infty$, we see that $\dst\lim_{t\to\infty} z(t)=-\infty$.
This contradicts $z(t)$ being eventually positive.
If $x'(t)$ is eventually negative and proceeding as in the proof of Theorem~\ref{t:2.3.1} we will end up with
\[
\int_{\sigma(t)}^{\infty}\Big[\int_{\sigma(t)
}^{r}\frac{1}{a(u)}du \Big(\int_{u}^{r}\frac{1}{b(v)}dv\Big)\Big]
\sum_{i=1}^{m}q_i(r)f(x(\sigma_i(r)))dr\le x(\sigma(t)),
\]
where $\sigma(t)=\max\{\sigma_1(t),\sigma_1(t),\dots ,\sigma_n(t)\}$.
Thus we have
\begin{equation}\label{e:2.59}
\int_{\sigma(t)}^{t}\Big[\int_{\sigma(t) }^{r}\frac{1}{a(u)}du
\Big(\int_{u}^{r}\frac{1}{b(v)}dv\Big)\Big]
\sum_{i=1}^{m}q_i(r)f(x(\sigma_i(r)))dr\le x(\sigma(t)).
\end{equation}
Using the fact that $\sigma_i(t)0$. Suppose
$\dst\lim_{t\to\infty}x(t)=0$, then
\[
\lim_{t\to\infty}\frac{x(\sigma(t))}{f(x(\sigma(t)))}=
\lim_{t\to\infty}\frac{1}{f'(x(\sigma(t)))}=\frac{1}{f'(0)}\le\frac{1}{\lambda}.
\]
This is a contradiction to (\ref{e:2.51}) . Therefore, the proof is complete.
%%%%%%%%%%%%%%%
%%%%%%%%%%%%%
%%%%555%%%%%%%%%%
\begin{theorem}\label{t:2.3.3}
Suppose that $f'(x)\ge \lambda$ for some $\lambda>0$ and
\begin{equation}\label{e:2.60}
\limsup_{t\to\infty}\int_{\sigma(t)}^{t}\Big[\int_{\sigma(t)
}^{r}\frac{1}{a(u)}du\Big(\int_{u}^{r}\frac{1}{b(v)}dv\Big)\Big]\sum_{i=1}^{m}q_i(r)dr=\infty.\quad
\end{equation}
In addition, suppose that there exist a continuously
differentiable function $p\in C([t_0,\infty), \mathbb{R})$,
$p(t)>0$ and an oscillatory function $\psi(t)$ such that
\begin{equation}\label{e:2.61}
\int^{\infty}\Big[q(t)p(t)-
\frac{b(t)(p'(t))^{2}}{\sum_{i=1}^{m}\frac{(\sigma_i(t)-T)}{a(\sigma_i(t))}
\sigma_i'(t)4\lambda d p(t)}\Big]dt=\infty
\end{equation}
for some $d\in(0,1)$ and for every $T\ge 0$, and
\begin{equation}\label{e:2.62}
(b(t)(a(t)\psi'(t))')'=h(t),\quad \lim_{t\to\infty}\psi^{(i)}(t)=0,
\quad i=0,1,2.
\end{equation}
Then the equation (\ref{e:2.4}) is almost oscillatory.
\end{theorem}
\noindent \textbf{Proof.} Let $x(t)$ be a non-oscillatory solution
of (\ref{e:2.4}). Without loss of generality we may assume that
$x(t)$ is eventually positive. Consider
\begin{equation}\label{e:2.63}
y(t)=x(t)-\psi(t).
\end{equation}
Obviously $y(t)$ is eventually positive, otherwise, $x(t)<\psi(t)$ and
it is a contradiction with oscillatory behavior of $\psi(t)$. We know that,
\begin{equation}\label{e:2.64}
(b(t)(a(t)y'(t))')'\le 0.
\end{equation}
Proceeding as in the proof of Theorem~\ref{t:2.3.1}, there is a $t_1\ge 0$ such that for
$t\ge t_1$
\[
(a(t)y'(t))'> 0\quad and\quad (a(t)y'(t))''\le 0.
\]
Consider again two cases. Suppose that $y'(t)$ is eventually
positive, then $y(t)$ is increasing and eventually positive. On
the other hand, since $\psi(t)\to 0\quad as \quad t\to\infty$ and
$y(t)=x(t)-\psi(t)$, there exists a $t_2\ge t_1$ such that
\[
x(\sigma_i(t))\ge d y(\sigma_i(t))\quad for \quad t\ge t_2\quad
and\quad d\in (0,1), \quad i=1,2,\dots,m.
\]
Since $f$ is an increasing function, we obtain
\[
f(x(\sigma_i(t)))\ge f(d y(\sigma_i(t)))\quad for \quad t\ge t_2,
\quad i=1,2,\dots,m.
\]
Define $z(t)$ by
\[
z(t)=\frac{ b(t)(a(t)y'(t))'}{\sum_{i=1}^{m}f(dy(\sigma_i(t)))}p(t),
\]
then obviously $z(t)>0$ for $t\ge t_2$ and $z'(t)$ is
\begin{eqnarray}
z'(t)&=&-\frac{\sum_{i=1}^{m}q_i(t)f(x(\sigma_i(t)))}{\sum_{i=1}^{m}f(dy(\sigma_i(t)))}p(t)+\frac
{p'(t)}{p(t)}z(t)
\nonumber\\[2pt]
&&-d\frac{\sum_{i=1}^{m}f'(dy(\sigma_i(t)))y'(\sigma_i(t))\sigma_i'(t)}{
\sum_{i=1}^{m}f(dy(\sigma_i(t)))}z(t). \nonumber
\end{eqnarray}
Then, using $f'(x)\ge \lambda>0$, we obtain
\begin{equation}\label{e:2.65}
z'(t)\le -q(t)p(t)+\frac
{p'(t)}{p(t)}z(t)-d\lambda\frac{\sum_{i=1}^{m}y'(\sigma_i(t))\sigma_i'(t)}{
\sum_{i=1}^{m}f(dy(\sigma_i(t)))}z(t),
\end{equation}
where $q(t)=min \{q_1(t),q_2(t),\dots ,q_m(t)\}$. We can now show that
\begin{equation}\label{e:2.66}
\sum_{i=1}^{m} \sigma_i'(t)y'(\sigma_i (t))\ge \sum_{i=1}^{m}\frac
{(\sigma_i (t)-T)}{a(\sigma_i(t))} \sigma_i'(t)(a(t)y'(t))'
\end{equation}
as in proof of Theorem~\ref{t:2.3.1}.
Using (\ref{e:2.65}) and (\ref{e:2.66}), we have
\[
z'(t)\le -q(t)p(t)+\frac
{p'(t)}{p(t)}z(t)-d\lambda\frac{\sum_{i=1}^{m}\frac{(\sigma_i(t)-T)}{a(\sigma_i(t))}\sigma_i'(t)}{
b(t)p(t)}z^{2}(t).
\]
Completing the square in the above equation leads to
\begin{equation}\label{e:2.67}
z'(t)\le
-q(t)p(t)+\frac{b(t)(p')^{2}(t)}{\sum_{i=1}^{m}\frac{(\sigma_i(t)-T)}{a(\sigma_i(t))}\sigma_i'(t)4\lambda
d p(t)}.
\end{equation}
Integrating (\ref{e:2.67}) from $T$ to $t$ and letting
$t\to\infty$, we see that $\dst\lim_{t\to\infty} z(t)=-\infty$.
This contradicts $z(t)$ being eventually positive.
Now suppose $y'(t)$ is eventually negative. Since $y$ is
eventually positive and decreasing, $\dst\lim_{t\to\infty}y(t)=c$,
where $c$ is a nonnegative number. Therefore,
$\dst\lim_{t\to\infty}x(t)=c$. Integrating (\ref{e:2.4}) three
times as we did in the proof of Theorem~\ref{t:2.3.1}, we will end
up with
\[
\int_{\sigma(t)}^{\infty}\Big[\int_{\sigma(t)
}^{r}\frac{1}{a(u)}du \Big(\int_{u}^{r}\frac{1}{b(v)}dv\Big)\Big]
\sum_{i=1}^{m}q_i(r)f(x(\sigma_i(r)))dr\le y(t),
\]
where $\sigma(t)=\max\{\sigma_1(t),\sigma_1(t),\dots ,\sigma_n(t)\}$.
Thus we have
\begin{equation}\label{e:2.68}
\int_{\sigma(t)}^{t}\Big[\int_{\sigma(t) }^{r}\frac{1}{a(u)}du
\Big(\int_{u}^{r}\frac{1}{b(v)}dv\Big)\Big]
\sum_{i=1}^{m}q_i(r)f(x(\sigma_i(r)))dr\le y(t).
\end{equation}
Hence, we conclude that $\dst\liminf_{t\to\infty} x(t)=0$. But
$x(t)$ is monotone, so we have\newline
$\dst\lim_{t\to\infty}x(t)=0$. Thus $c=0$ and by (\ref{e:2.62})
and (\ref{e:2.63}) $\dst\lim_{t\to\infty}x^{(i)}(t)=0$, $i=0,1,2$,
which means that $x(t)$ is almost oscillatory. This completes the
proof.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{Oscillatory behavior of third order differential equations with continuous
deviating arguments}
Suppose that the following conditions hold unless stated otherwise
\begin{enumerate}
\item[(a)] $a(t)>0$, $b(t)>0$, $b'(t)\ge 0$,
$\int^{\infty}\frac{dt}{a(t)}=\infty$,
$\int^{\infty}\frac{dt}{b(t)}=\infty$, \item[(b)] $q(t,\xi)\in
C([t_0,\infty)\times[c,d],\mathbb{R})$, $q(t,\xi)> 0$,
\item[(c)]$\frac{f(x)}{x}\ge\epsilon>0$, for $x\ne0$, $\epsilon\quad is\quad
a\quad constant$, \item [(d)] $\sigma(t,\xi)\in
C([t_0,\infty)\times[c,d],\mathbb{R})$, $\sigma(t,\xi)< t$,
$\xi\in [c,d]$, $\sigma(t,\xi)$ is nondecreasing with respect to
$t$ and $\xi$ and
\[\lim_{t\to\infty}\min_{\xi\in [c,d]}\sigma(t,\xi)=\infty.
\]
\end{enumerate}
\begin{theorem}\label{t:2.4.1}
If
\begin{equation}\label{e:2.69}
\int_{t_1}^{\infty}\int_{c}^{d}q(s,\xi)d\xi ds=\infty
\end{equation}
and
\begin{equation}\label{e:2.70}
\epsilon\int_{g(t)}^{t}\Big[\int_{g(t)
}^{r}\frac{1}{a(u)}du\Big(\int_{u}^{r}\frac{1}{b(v)}dv\Big)\Big]
\int_{c}^{d}q(r,\xi)d\xi dr>1,
\end{equation}
where $g(t)=\sigma(t,d)$.
Then the equation (\ref{e:2.5}) is oscillatory.
\end{theorem}
\noindent
\textbf{Proof.}
Suppose that $x(t)$ is non-oscillatory solution of
(\ref{e:2.5}). Without loss of generality we may assume that $x(t)$ is
eventually positive. (If $x(t)$ is eventually negative solution, it can
be proved by the same arguments). From (\ref{e:2.5}), we have
\begin{equation}\label{e:2.71}
(b(t)(a(t)x'(t))')'=-\int_{c}^{d}q(t,\xi)f(x(\sigma(t,\xi)))d\xi.
\end{equation}
Proceeding as in the proof of Theorem~ \ref{t:2.3.1}, we have
\[
(b(t)(a(t)x'(t))')'\le 0,
\]
\[
(a(t)x'(t))'> 0\quad\mbox{and}\quad (a(t)x'(t))''\le 0
\]
for large enough $t$. Thus, $x(t)$, $x'(t)$ and $(a(t)x'(t))'$ are monotone and eventually one-signed.
From condition $(c)$,
\[
f(x(\sigma(t,\xi)))\ge\epsilon x(\sigma(t,\xi))>0.
\]
Therefore,
\begin{equation}\label{e:2.72}
0\ge
(b(t)(a(t)x'(t))')'+\epsilon\int_{c}^{d}q(t,\xi)x(\sigma(t,\xi))d\xi.
\end{equation}
Now consider again two cases.
Suppose that
$x'(t)$ is eventually positive, say $x'(t)>0$ for $t>t_2$. Now we can choose a constant $k>0$ such that $x(k)>0$. By $(d)$,
there exist a sufficiently large $T$ such that $\sigma(t,\xi)>k$ for $t>T$,
$\xi\in [c,d]$. Therefore,
\[
x(\sigma(t,\xi))\ge x(k).
\]
Thus,
\begin{equation}\label{e:2.73}
%\label{e:2.1.3}
(b(t)(a(t)x'(t))')'+\epsilon x(k)\int_{c}^{d}q(t,\xi)d\xi\le 0.
\end{equation}
Integrating this last equation from $t_1$ to $t$, we get
\begin{equation}\label{e:2.74}
b(t)(a(t)x'(t))'\le b(t_1)(a(t_1)x'(t_1))' -\epsilon
x(k)\int_{t_1}^{t}\int_{c}^{d}q(s,\xi)d\xi ds.
\end{equation}
Taking the limit of both sides as $t\to\infty$ and using (\ref{e:2.69}),
the last inequality above leads to a contradiction to $(a(t)x'(t))'>0$.
Now suppose $x'(t)$ is eventually negative. Proceeding as in the proof of
Theorem~\ref{t:2.3.1} and integrating equation (\ref{e:2.5}) three
times, we get
\begin{equation}\label{e:2.75}
\int_{t}^{\infty}\Big[\int_{t}^{r}\frac{1}{a(u)}du
\Big(\int_{u}^{r}\frac{1}{b(v)}dv\Big)\Big]
\int_{c}^{d}q(r,\xi)f(x(\sigma(r,\xi)))d\xi dr\le x(t)
\end{equation}
Using ({c}) in (\ref{e:2.75}), we obtain
\begin{equation}\label{e:2.76}
\int_{t}^{\infty}\Big[\int_{t}^{r}\frac{1}{a(u)}du
\Big(\int_{u}^{r}\frac{1}{b(v)}dv\Big)\Big]\epsilon\int_{c}^{d}q(r,\xi)
x(\sigma(r,\xi))d\xi dr\le x(t).
\end{equation}
Replace $t$ by $g(t)$ in (\ref{e:2.76}), where $g(t)=\sigma(t,d)$, then we have
\begin{equation}\label{e:2.77}
\epsilon\int_{g(t)}^{t}\Big[\int_{g(t)}^{r}\frac{1}{a(u)}du
\Big(\int_{u}^{r}\frac{1}{b(v)}dv\Big)\Big]
\int_{c}^{d}q(r,\xi)x(\sigma(r,\xi))d\xi dr\le x(g(t)).
\end{equation}
Since $x(t)$ is decreasing and positive,
\begin{equation}\label{e:2.78}
\epsilon\int_{g(t)}^{t}\Big[\int_{g(t)}^{r}\frac{1}{a(u)}du
\Big(\int_{u}^{r}\frac{1}{b(v)}dv\Big)\Big]\int_{c}^{d}q(r,\xi)d\xi
dr\le 1.
\nonumber
\end{equation}
This is a contradiction to (\ref{e:2.70}). Therefore, the proof is complete.
%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%
\begin{example} \rm
Consider the following functional differential equation
\[
x'''+\int_{2/7\pi}^{1/2\pi}\frac{2e^{-1/\xi}}{\xi^2}x(t-\frac{1}{\xi})d\xi=0
\]
so that $a(t)=1$, $b(t)=1$, $f(x)=x$, $q(t,\xi)=\frac{2e^{-1/\xi}}{\xi^2}$,
$\sigma(t,\xi)=t-\frac{1}{\xi}$.
We can easily see that the conditions of Theorem~\ref{t:2.4.1} are
satisfied. It is easy to verify that
$x(t)=e^{-t}\sin t$ is a solution of this problem.
\end{example}
%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%
\begin{theorem}\label{t:2.4.2}
Suppose (\ref{e:2.70}) holds. In addition to that suppose there
exist $p\in C([t_0,\infty), \mathbb{R})$, $p(t)>0$ such that
\begin{equation}\label{e:2.79}
\int^{\infty}\Big[\Gamma(t)p(t)-
\frac{a(\sigma(t,c))b(t)(p'(t))^2}{(\sigma(t,c)-T)\sigma'(t,c)4p(t)}\Big]dt
=\infty,
\end{equation}
where $\Gamma(t)=\epsilon\int_{c}^{d}q(t,\xi)d\xi$.
Then the equation (\ref{e:2.5}) is oscillatory.
\end{theorem}
\noindent
\textbf{Proof.}
Suppose that $x(t)$ is non-oscillatory solution of (\ref{e:2.5}). We
can assume that $x(t)$ is eventually positive. The case of $x(t)$ is
eventually negative can be proved by the same arguments. Proceeding as
in the proof of Theorem~\ref{t:2.3.1}, we have
\[
(b(t)(a(t)x'(t))')'\le 0,
\]
\[
(a(t)x'(t))'> 0\quad and\quad (a(t)x'(t))''\le 0.
\]
Thus, $x(t)$, $x'(t)$ and $(a(t)x'(t))'$ are monotone and eventually one-signed.
From condition $(c)$,
\[
f(x(\sigma(t,\xi)))\ge\epsilon x(\sigma(t,\xi))>0.
\]
\begin{equation}\label{e:2.80}
(b(t)(a(t)x'(t))')'+\epsilon\int_{c}^{d}q(t,\xi)x(\sigma(t,\xi))d\xi\le 0.
\end{equation}
If $x'(t)$ is eventually positive, then we can define
\[
z(t)=\frac{ b(t)(a(t)x'(t))'}{x(\sigma(t,c))}p(t).
\]
It is obvious that $z(t)>0$ for $t\ge t_2$ and $z'(t)$ is
\begin{eqnarray}\label{e:2.81}
z'(t)=\frac{(b(t)(a(t)x'(t))')'}{x(\sigma(t,c))}p(t)+\frac
{p'(t)}{p(t)}z(t)-\frac{x'(\sigma(t,c))\sigma'(t,c)}{
x(\sigma(t,c))}z(t).
\end{eqnarray}
From proof of Theorem~\ref{t:2.3.1}, we have
\begin{equation}
a(t)x'(t)\ge (t-T)(a(t)x'(t))'.
\nonumber
\end{equation}
Since $(a(t)x(t))'$ is non-increasing, we have
\begin{equation}
a(\sigma(t,c))x'(\sigma(t,c))\ge (\sigma(t,c)-T)(a(t)x'(t))',
\nonumber
\end{equation}
then
\begin{equation}\label{e:2.82}
x'(\sigma(t,c))\ge \frac{(\sigma(t,c)-T)(a(t)x'(t))'}{a(\sigma(t,c))}.
\end{equation}
Plug (\ref{e:2.82}) in (\ref{e:2.81}), then we obtain
\begin{eqnarray}
z'(t)=\frac{(b(t)(a(t)x'(t))')'}{x(\sigma(t,c))}p(t)+\frac
{p'(t)}{p(t)}z(t)-\frac{(\sigma(t,c)-T)\sigma'(t,c)}{
p(t)b(t)a(\sigma(t,c)))}z^2(t). \nonumber
\end{eqnarray}
Completing the square leads to
\begin{equation}\label{e:2.83}
z'(t)\le -\Gamma(t)p(t)+\frac{b(t)a(\sigma(t,c))(p'(t))^2}{(\sigma(t,c)-T)\sigma'(t,c))4p(t)}.
\end{equation}
Integrating (\ref{e:2.83}) from $T$ to t and letting $t\to\infty$,
we see that $\dst\lim_{t\to\infty} z(t)=-\infty$. This contradicts
$z(t)$ being eventually positive.
If $x'(t)$ is eventually negative, the proof is exactly the same as
in the second part of the proof of previous Theorem.
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\smallskip
\end{document}