\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small
2004 Conference on Diff. Eqns. and Appl. in Math. Biology, Nanaimo, BC, Canada.\newline
{\em Electronic Journal of Differential Equations},
Conference 12, 2005, pp. 47--56.\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}}
\setcounter{page}{47}

\begin{document}

\title[\hfilneg EJDE/Conf/12 \hfil Functional differential equations]
{Functional differential equations of third order}

\author[T. Candan, R. S. Dahiya \hfil EJDE/Conf/12 \hfilneg]
{Tuncay Candan,  Rajbir S. Dahiya}  % in alphabetical order

\address{Tuncay Candan \hfill\break
Department of Mathematics, Faculty of Art and
Science, Ni\u{g}de University\\
Ni\u{g}de, 51100, Turkey}
\email{tcandan@nigde.edu.tr}

\address{Rajbir S. Dahiya \hfill\break
Department of Mathematics, Iowa State University\\
Ames, IA  50011, USA}
\email{rdahiya@iastate.edu}

\date{}
\thanks{Published Arpil 20, 2005.}
\subjclass[2000]{34K11, 34K15}
\keywords{Oscillation theory; neutral equations}

\begin{abstract}
 In this paper, we consider the third-order neutral functional
 differential equation with distributed deviating arguments. We
 give sufficient conditions for the  oscillatory behavior of this
 functional differential equation.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{example}[theorem]{Example}
\allowdisplaybreaks

\section{Introduction}

 The aim of this paper is to develop some oscillation  theorems for
 a third-order equations of the form
\begin{equation}\label{e:1.1}
\left[a(t)[b(t)[x(t)+c(t)x(t-\tau)]']'\right]'
+\int_{a}^{b}p(t,\xi)x(\sigma(t,\xi))d\xi=0,
\end{equation}
where
\begin{enumerate}
\item[(a)] $a(t), a'(t), b(t), c(t)\in C([t_0,\infty),(0,\infty))$,
 $0<c(t)\le 1$, $a'(t)\ge 0$;

\item[(b)] $\int^{\infty}\frac{dt}{b(t)}=\infty$ and
$\int^{\infty}\frac{dt}{a(t)}=\infty$

\item[(c)] $p(t,\xi)\in C([t_0,\infty)\times[a,b],[0,\infty))$, and
$p(t,\xi)$ is not eventually zero on any half line $[t_m,\infty)\times[a,b]$,
$t_m\ge t_0$

\item[(d)] $\sigma(t,\xi)\in C([t_0,\infty)\times[a,b],R)$,
 $\sigma(t,\xi)+\tau\le t$, $\sigma(t,\xi)$ is nondecreasing with
 respect to $t$ and $\xi$, respectively, and
 $\liminf_{t\to\infty\,\xi \in[a,b]}\sigma(t,\xi)=\infty$.
\end{enumerate}
As is customary, a solution  of equation  (\ref{e:1.1}) is called
oscillatory if it has arbitrarily large zeros. Otherwise, it is
called nonoscillatory.

Oscillatory solutions of third-order differential equations
\[
(r_2(t)(r_1(t)x'(t))')'+q(t)f(x(g(t)))=h(t),
\]
and
\[
(b(t)(a(t)y'(t))')'+(q_1(t) y)'+ q_2(t) y'=f(t)
\]
were considered in \cite{dah} and \cite{Dah},
respectively. We refer to \cite{Agar}-\cite{Can3}
and \cite{rao}-\cite{wal} for more studies.
However,  our results are more general with
different proofs than  those works.

\section{Main Results}
Let $D_0=\{(t,s)|t> s\ge t_0\}$, $D=\{(t,s)|t\ge s\ge t_0\}$.

\begin{theorem}\label{t:2.1}
Suppose that there exist
$\frac{d}{dt}\sigma(t,a)$ and  $H(t,s)\in
C'(D;R)$, $h(t,s)\in C(D_0;R)$ and $\rho(t)\in
C'([t_0,\infty),(0,\infty))$ such that
\begin{enumerate}
\item[(i)]$H(t,t)=0$, $H(t,s)>0$
\item[(ii)]$H'_s(t,s)\le0$, and
$-H'_s(t,s)-H(t,s)\frac{\rho'(s)}{\rho(s)}=h(t,s)\sqrt {H(t,s)}$.
\end{enumerate}
If
\begin{equation}\label{e:2.2}
\begin{aligned}
 &\limsup_{t\to\infty}\frac{1}{H(t,t_0)}\int_{t_0}^{t}\Big[H(t,s)
\rho(s)\int_{a}^{b}p(s,\xi)[1-c(\sigma(s,\xi))]d\xi\\
& -\frac{a(s)\rho(s)b(\sigma(s,a))h^2(t,s)}{4[\sigma(s,a)-T]
\sigma'(s,a)}\Big]ds=\infty
\end{aligned}
\end{equation}
and
\begin{equation}\label{e:2.3}
\int_{t_3}^{\infty}\Big[\int_{t_3
}^{r}\frac{du}{b(u)}\Big(\int_{u}^{r}\frac{dv}{a(v)}\Big)\Big]
\int_{a}^{b}p(r,\xi)d\xi dr=\infty.
\end{equation}
Then every solution of (\ref{e:1.1}) is oscillatory or tends to
zero as $t\to\infty$.
\end{theorem}

\begin{proof} Assume, for the sake of contradiction,
that equation (\ref{e:1.1}) has an eventually positive solution
$x(t)$. That is, there exists a $t_0\ge 0$ such that $x(t)>0$ for
$t\ge t_0$. If we put
\begin{equation}\label{e:2.4}
y(t)=x(t)+c(t)x(t-\tau)
\end{equation}
then, from (\ref{e:1.1}), we obtain
\begin{equation}\label{e:2.5}
\left[a(t)[b(t)y'(t)]'\right]'=-\int_{a}^{b}p(t,\xi)x(\sigma(t,\xi))d\xi.
\end{equation}
 Since $x(t)$ is an eventually positive solution of (\ref{e:1.1}) and
 $\sigma(t,\xi)\to\infty$ as $t\to\infty$, $\xi\in[a,b]$, there exist a
$t_1\ge t_0$ such that
 $x(t-\tau)>0$ and $x(\sigma(t,\xi))>0$ for $ t\ge t_1$,
$\xi\in[a,b]$. Thus in view of (\ref{e:2.4}) and  (\ref{e:2.5}),
we have $y(t)>0$, $t\ge t_1$  and
$[a(t)[b(t)y'(t)]']'\le 0$, $t\ge t_1$. Thus, $y(t)$,
$y'(t)$, $(b(t)y'(t))'$ are monotone and eventually one-signed. We
want to show that there is a $t_2\ge t_1$ such that
\begin{equation}\label{e:2.6}
(b(t)y'(t))'> 0,\quad t\ge t_2.
\end{equation}
Suppose on the contrary,  $(b(t)y'(t))'\le 0$. Since the right
hand side of (\ref{e:2.5}) is not identically zero and $a(t)>0$,
it is clear that there exists  a $t_3\ge t_2$ such that
$a(t_3)(b(t_3)y'(t_3))'< 0$. Then we have
\begin{equation}\label{e:2.7}
a(t)(b(t)y'(t))'\le a(t_3)(b(t_3)y'(t_3))'<0,\quad t\ge t_3.
\end{equation}
Dividing (\ref{e:2.7}) by $a(t)$ and  integrating from $t_3$ to
$t$, we obtain
\begin{equation}\label{e:2.8}
   b(t)y'(t)-b(t_3)y'(t_3)\le a(t_3)(b(t_3)
y'(t_3))'\int_{t_3}^{t}\frac {ds}{a(s)}.
\end{equation}
Letting $t\to\infty$ in (\ref{e:2.8}), and because of
$(b)$ we see that $b(t)y'(t)\to -\infty$ as
$t\to\infty$. Thus,  there is a $t_4\ge t_3$ such that
$b(t_4)y'(t_4)<0$. By making use of $(b(t)y'(t))'\le 0$, we obtain
\begin{equation}\label{e:2.9}
   b(t)y'(t)\le b(t_4)y'(t_4)<0,\quad t\ge t_4 .
\end{equation}
If we divide (\ref{e:2.9}) by $b(t)$ and integrate from $t_4$ to
$t$ with $t\to\infty$, the right-hand side becomes
negative. Thus, we have $y(t)\to -\infty$. But this is a
contradiction, since $y(t)$ is eventually positive, which
therefore  proves that (\ref{e:2.6}) holds. Now we  have two
possibilities: $(I)$ $y'(t)>0$ for $t\ge t_2$, $(II)$ $y'(t)<0$
for $t\ge t_2$.


(I) Assume  $y'(t)>0$ for $t\ge t_2$. From (\ref{e:2.4}),
$y(t)\ge x(t)$ for $t\ge t_2$ and
\[
y(\sigma(t,\xi))\ge y(\sigma(t,\xi)-\tau)\ge
x(\sigma(t,\xi)-\tau),\quad t\ge t_3\ge t_2.
\]
Thus from (\ref{e:2.4}) and (\ref{e:2.5}),
\begin{align*}
\left[a(t)[b(t)y'(t)]'\right]'
&=-\int_{a}^{b}p(t,\xi)[y(\sigma(t,\xi))-c(\sigma(t,\xi))x(\sigma(t,\xi)-\tau)]d\xi
\\
&\le -\int_{a}^{b}p(t,\xi)[y(\sigma(t,\xi))-c(\sigma(t,\xi))y(\sigma(t,\xi))]d\xi
\\
&=-\int_{a}^{b}p(t,\xi)[1-c(\sigma(t,\xi))]y(\sigma(t,\xi))d\xi \\
&\le -y(\sigma(t,a))\int_{a}^{b}p(t,\xi)[1-c(\sigma(t,\xi))]d\xi.
\end{align*}
Then, we have
\[
\frac{\left[a(t)[b(t)y'(t)]'\right]'}{y(\sigma(t,a))}
\le-\int_{a}^{b}p(t,\xi)[1-c(\sigma(t,\xi))]d\xi.
\]
Now set
\[
z(t)=\frac{ a(t)(b(t)y'(t))'\rho(t)}{y(\sigma(t,a))}.
\]
It is obvious that $z(t)>0$ for $t\ge t_3$ and  the derivative of
$z(t)$ is
\begin{equation}\label{e:2.10}
\begin{aligned}
z'(t)&=\frac{ [a(t)(b(t)y'(t))']'\rho(t)}{y(\sigma(t,a))}+
\frac{\rho'(t)}{\rho(t)}z(t)-\frac{[a(t)(b(t)y'(t))']\rho(t)y'(\sigma(t,a))\sigma'(t,a)}{y^2(\sigma(t,a))}
\\
&\le -\rho(t)\int_{a}^{b}p(t,\xi)[1-c(\sigma(t,\xi))]d\xi
+\frac{\rho'(t)}{\rho(t)}z(t)-\frac{y'(\sigma(t,a))\sigma'(t,a)z(t)}{y(\sigma(t,a))}.
\end{aligned}
\end{equation}
On the other hand, since  $(a(t)(b(t)y'(t))')'\le
0$ and $a'(t)\ge 0$, we find that
\begin{equation}\label{e:2.11}
(b(t)y'(t))''\le 0.
\end{equation}
Using the above inequality and
\[
b(t)y'(t)=b(T)y'(T)+\int_{T}^{t}(b(s)y'(s))'ds,
\]
we obtain
\[
b(t)y'(t)\ge (t-T)(b(t)y'(t))',\quad t\ge T\ge t_3.
\]
 Since $(by')'$ is non-increasing, we have
\[
b(\sigma(t,a))y'(\sigma(t,a))\ge (\sigma(t,a)-T)(b(t)y'(t))',\quad
t\ge t_4\ge t_3.
\]
Thus, we have
\begin{equation}\label{e:2.12}
y'(\sigma(t,a))\ge\frac{(\sigma(t,a)-T)(b(t)y'(t))'}{b(\sigma(t,a))}.
\end{equation}
Then, substituting  (\ref{e:2.12}) in  (\ref{e:2.10}), it
follows that
\[
z'(t)\le-\rho(t)\int_{a}^{b}p(t,\xi)[1-c(\sigma(t,\xi))]d\xi+
\frac{\rho'(t)}{\rho(t)}z(t)-\frac{[\sigma(t,a)-T]\sigma'(t,a)z^2(t)}{a(t)\rho(t)
b(\sigma(t,a))},
\]
and
\begin{equation}\label{e:2.13}
  \rho(t)\int_{a}^{b}p(t,\xi)[1-c(\sigma(t,\xi))]d\xi  \le
  -z'(t)+\frac{\rho'(t)}{\rho(t)}z(t)-\frac{[\sigma(t,a)-T]\sigma'(t,a)z^2(t)}{a(t)
\rho(t)b(\sigma(t,a))}.
\end{equation}
Multiplying both sides of equation (\ref{e:2.13}) by $H(t,s)$, and
integrating by parts from $T^*$ to $t$, and using the properties
$(i)$ and $(ii)$,  we obtain
\begin{align*}
&\int_{T^*}^{t}H(t,s)
\rho(s)\int_{a}^{b}p(s,\xi)[1-c(\sigma(s,\xi))]d\xi ds
\\
&\le -\int_{T^*}^{t}H(t,s)z'(s)ds+
\int_{T^*}^{t}\frac{H(t,s)\rho'(s)z(s)}{\rho(s)}ds \\
&\quad -\int_{T^*}^{t}\frac{H(t,s)[\sigma(s,a)-T]\sigma'(s,a)z^2(s)}{a(s)\rho(s)
b(\sigma(s,a))}ds
\\
&= H(t,T^*)z(T^*)+\int_{T^*}^{t}\Big[\frac{dH(t,s)}{ds}+H(t,s)\frac{\rho'(s)}{\rho(s)}\Big]
z(s)ds\\
&\quad -\int_{T^*}^{t}\frac{H(t,s)[\sigma(s,a)-T]\sigma'(s,a)z^2(s)}{a(s)\rho(s)b(\sigma(s,a))}ds
\\
&= H(t,T^*)z(T^*)-\int_{T^*}^{t}h(t,s)\sqrt{H(t,s)}z(s)ds\\
&\quad -\int_{T^*}^{t}\frac{H(t,s)[\sigma(s,a)-T]\sigma'(s,a)z^2(s)}{a(s)\rho(s)b(\sigma(s,a))}ds
\\
&= H(t,T^*)z(T^*)-\int_{T^*}^{t}\Big[\sqrt{\frac{H(t,s)[\sigma(s,a)-T]
\sigma'(s,a)}{a(s)b(\sigma(s,a))\rho(s)}}z(s)\\
&\quad +\frac{\sqrt{a(s)
b(\sigma(s,a))\rho(s)}h(t,s)}{2\sqrt{[\sigma(s,a)-T]\sigma'(s,a)}}\Big]^2ds
+\int_{T^*}^{t}\frac{a(s)\rho(s)b(\sigma(s,a))h^2(t,s)}{4[\sigma(s,a)-T]
\sigma'(s,a)}ds,
\end{align*}
$t>T^*\ge t_4$. As a result of this, we get
\begin{align*}
&\int_{T^*}^{t}\Big[H(t,s)
\rho(s)\int_{a}^{b}p(s,\xi)[1-c(\sigma(s,\xi))]d\xi-
\frac{a(s)\rho(s)b(\sigma(s,a))h^2(t,s)}{4[\sigma(s,a)-T]\sigma'(s,a)}\Big]ds
\\
&= H(t,T^*)z(T^*)-\int_{T^*}^{t}\Big[\sqrt{\frac{H(t,s)[\sigma(s,a)-T]\sigma'(s,a)}
{a(s)b(\sigma(s,a))\rho(s)}}z(s)\\
&\quad +\frac{\sqrt{a(s)b(\sigma(s,a))\rho(s)}h(t,s)}{2\sqrt{[\sigma(s,a)-T]
\sigma'(s,a)}}\Big]^2 ds \,.
\end{align*}
 From $(ii)$ $H'_s(t,s)\le 0$, we have $H(t,t_4)\le H(t,t_0)$,
$t_4\ge t_0$ and therefore
\begin{align*}
&\int_{t_4}^{t}\Big[H(t,s)
\rho(s)\int_{a}^{b}p(s,\xi)[1-c(\sigma(s,\xi))]d\xi-
\frac{a(s)\rho(s)b(\sigma(s,a))h^2(t,s)}{4[\sigma(s,a)-T]\sigma'(s,a)}\Big]ds
\\
&\le H(t,t_4)z(t_4)\le H(t,t_0)z(t_4)
\end{align*}
which implies that
\begin{align*}
&\frac{1}{H(t,t_0)}\int_{t_0}^{t}\Big[H(t,s)
\rho(s)\int_{a}^{b}p(s,\xi)[1-c(\sigma(s,\xi))]d\xi
-\frac{a(s)\rho(s)b(\sigma(s,a))h^2(t,s)}{4[\sigma(s,a)-T]\sigma'(s,a)}\Big]ds
\\
&= \frac{1}{H(t,t_0)}\Big[\int_{t_0}^{t_4}+\int_{t_4}^{t}\Big]
\Big[H(t,s) \rho(s)\int_{a}^{b}p(s,\xi)[1-c(\sigma(s,\xi))]d\xi\\
&\quad -\frac{a(s)\rho(s)b(\sigma(s,a))h^2(t,s)}{4[\sigma(s,a)-T]\sigma'(s,a)}\Big]ds
\\
&\le z(t_4)+\int_{t_0}^{t_4}\frac{H(t,s)}{H(t,t_0)}\rho(s)
\int_{a}^{b}p(s,\xi)[1-c(\sigma(s,\xi))]d\xi ds
\\
&\le z(t_4)+\int_{t_0}^{t_4}\rho(s)\int_{a}^{b}p(s,\xi)[1-c(\sigma(s,\xi))]d\xi
 ds.
\end{align*}
Now taking upper limits  as ${t\to\infty}$, we obtain
\begin{align*}
&\limsup_{t\to\infty}\frac{1}{H(t,t_0)}\int_{t_0}^{t}\Big[H(t,s)
\rho(s)\int_{a}^{b}p(s,\xi)[1-c(\sigma(s,\xi))]d\xi\\
&-\frac{a(s)\rho(s)b(\sigma(s,a))h^2(t,s)}{4[\sigma(s,a)-T]\sigma'(s,a)}\Big]ds
\\
&\le z(t_4)+\int_{t_0}^{t_4}\rho(s)\int_{a}^{b}p(s,\xi)[1-c(\sigma(s,\xi))]d\xi
ds =M<\infty,
\end{align*}
where $M$ is a constant. Hence, this result leads to a
contradiction to  $(2)$. \smallskip

\noindent (II) Assume  $y'(t)<0$ for $t\ge t_2$. We integrate
(\ref{e:1.1}) from $t$ to $\infty$ and since
\[
a(t)(b(t)y'(t))'>0,\quad t\ge t_2,
\]
we have
\begin{equation}\label{e:2.14}
-a(t)(b(t)y'(t))'
+\int_{t}^{\infty}\int_{a}^{b}p(r,\xi)x(\sigma(r,\xi))d\xi dr\le 0.
\end{equation}
Now integrating (\ref{e:2.14}) from $t$ to $\infty$ after dividing
by $a(t)$ and using $b(t)y'(t)<0$, will lead to
\begin{equation}\label{e:2.15}
b(t)y'(t)
+\int_{t}^{\infty}\left(\int_{t}^{r}\frac{du}{a(u)}\right)\int_{a}^{b}p(r,\xi)x(\sigma(r,\xi))d\xi
dr\le 0.
\end{equation}
Dividing (\ref{e:2.15}) by $b(t)$ and integrating again from $t$
to $\infty$ gives
\begin{equation}\label{e:2.16}
\int_{t}^{\infty}\Big[\int_{t}^{r}\frac{du}{b(u)}\Big(\int_{u}^{r}\frac{dv}{a(v)}
\Big)\Big]\int_{a}^{b}p(r,\xi)x(\sigma(r,\xi))d\xi dr\le y(t)
\end{equation}
for $t\ge t_3\ge t_2$. Replacing $t$ by $t_3$ in
(\ref{e:2.16}), we get
\begin{equation}\label{e:2.17}
\int_{t_3}^{\infty }\Big[\int_{t_3
}^{r}\frac{du}{b(u)}\Big(\int_{u}^{r}\frac{dv}{a(v)}\Big)\Big]
\int_{a}^{b}p(r,\xi)x(\sigma(r,\xi))d\xi\, dr\le
y(t_3).
\end{equation}
 On the other hand, since $y$ is monotonically decreasing function
 in the interval $[t_3, \infty]$, we get $\lim_{t\to\infty} y(t)
 =\lim_{t\to\infty}[x(t)+c(t)x(t-\tau)]=K\ge 0$.
 Suppose that $K>0$, then $[x(t)+c(t)x(t-\tau)]\ge \frac{K}{2}>0$ for
 $t\ge t_4\ge t_3$. From this we can observe that there exists $K_1$
 such that $x(\sigma(r,\xi))\ge K_1>0$. Thus from (\ref{e:2.17})
 \[
\int_{t_3}^{\infty }\Big[\int_{t_3
}^{r}\frac{du}{b(u)}\Big(\int_{u}^{r}\frac{dv}{a(v)}\Big)\Big]
\int_{a}^{b}p(r,\xi)K_1 d\xi dr\le y(t_3).
\]
 From the previous  equation, we have
\[
\int_{t_3}^{\infty}\Big[\int_{t_3}^{r}\frac{du}{b(u)}
\Big(\int_{u}^{r}\frac{dv}{a(v)}\Big)\Big]\int_{a}^{b}p(r,\xi)
d\xi dr< \infty.
\]
This is a contradiction to (\ref{e:2.3}). Therefore,
$\lim_{t\to\infty}[x(t)+c(t)x(t-\tau)]=0$. Then
$\lim_{t\to\infty}x(t)=0$, so the proof is complete.
\end{proof}

\begin{example} \label{ex2} \rm
Consider the functional differential equation
\[
[x(t)+\frac{1}{2}x(t-\pi)]'''
+\int_{1/5\pi}^{2/7\pi}\frac{(2-e^{-\pi})e^{1/\xi}}{\xi^2}x(t-\frac{1}{\xi})d\xi=0
\]
so that $a(t)=b(t)=1$, $c(t)=\frac{1}{2}$, $\tau=\pi$,
$p(t,\xi)=\frac{(2-e^{-\pi})e^{1/\xi}}{\xi^2}$,
 $\sigma(t,\xi)=t-\frac{1}{\xi}$, $\sigma(t)=\sigma(t,b)=t-\frac{7\pi}{2}$,
 $\rho(s)=s$, $H(t,s)=(t-s)^2$, $h(t,s)=[2-\frac{(t-s)}{s}]$.

We can see that the conditions of Theorem~\ref{t:2.1} are
satisfied. It is easy to verify that $x(t)=e^t\sin t$ is a
solution of this problem, which is oscillatory.
\end{example}

\begin{theorem}\label{t:2.2}
Suppose that  the conditions of Theorem~\ref{t:2.1}, and condition
(\ref{e:2.3}) holds, and
\begin{gather}\label{e:2.18}
 0<\inf_{s\ge
 t_0}\Big[\liminf_{t\to\infty}\frac{H(t,s)}{H(t,t_0)}\Big]\le\infty, \\
\label{e:2.19}
 \limsup_{t\to\infty}\frac{1}{H(t,t_0)}\int_{t_0}^{t}\frac{
a(s)\rho(s)b(\sigma(s,a))h^2(t,s)}{[\sigma(s,a)-T)]\sigma'(s,a)}ds<\infty.
\end{gather}
If there exists a function $\phi(t)\in C([t_0,\infty),R)$
satisfying
\begin{equation}\label{e:2.20}
\begin{aligned}
 &\limsup_{t\to\infty}\frac{1}{H(t,u)}\int_{u}^{t}\Big[H(t,s)
\rho(s)\int_{a}^{b}p(s,\xi)[1-c(\sigma(s,\xi))]d\xi\\
&-\frac{a(s)\rho(s)b(\sigma(s,a))h^2(t,s)}{4[\sigma(s,a)-T)]\sigma'(s,a)}\Big]ds
\\
&\ge \phi(u),\quad u\ge t_0,
\end{aligned}
\end{equation}
and
\begin{equation}\label{e:2.21}
\begin{gathered}
\limsup_{t\to\infty}\frac{1}{H(t,t_0)}\int_{t_0}^{t}\frac{[\sigma(u,a)-T]\sigma'(u,a)\phi_+^2(u)du}{
a(u)\rho(u)b(\sigma'(u,a))}=\infty, \\
\phi_+(u)=\max_{u\ge t_0}\{\phi(u),0\}.
\end{gathered}
\end{equation}
Then every solution of functional differential equation
(\ref{e:1.1}) is oscillatory or tends to zero as $t\to\infty$.
\end{theorem}

\begin{proof} Assume, for the sake of contradiction, that
equation \eqref{e:1.1} has positive solution, say $x(t)>0$, $t\ge t_0$.
Then, proceeding as in the proof of Theorem \ref{t:2.1}, we obtain
\begin{align*}
&\frac{1}{H(t,u)}\int_{u}^{t}\Big[H(t,s)
\rho(s)\int_{a}^{b}p(s,\xi)[1-c(\sigma(s,\xi))]d\xi-
\frac{a(s)\rho(s)b(\sigma(s,a))h^2(t,s)}{4[\sigma(s,a)-T]\sigma'(s,a)}\Big]ds
\\
&\le z(u)-\frac{1}{H(t,u)}\int_{u}^{t}\Big[\sqrt{\frac{H(t,s)[\sigma(s,a)-T]
\sigma'(s,a)}{a(s)b(\sigma(s,a))\rho(s)}}z(s)\\
&+\frac{\sqrt{a(s)b(\sigma(s,a))\rho(s)}h(t,s)}{2\sqrt{[\sigma(s,a)-T]
\sigma'(s,a)}}\Big]^2ds,
\end{align*}
$t>u\ge t_1\ge t_0$. Thus taking upper limit as $t\to\infty$ and
using (\ref{e:2.20}), we have
\begin{align*}
z(u)&\ge\phi(u)
+\liminf_{t\to\infty}\frac{1}{H(t,u)}\int_{u}^{t}\Big[\sqrt{\frac{H(t,s)[\sigma(s,a)-T]
\sigma'(s,a)}{a(s)b(\sigma(s,a))\rho(s)}}z(s)\\
&\quad +\frac{\sqrt{a(s)b(\sigma(s,a))\rho(s)}h(t,s)}{2\sqrt{[\sigma(s,a)-T]
\sigma'(s,a)}}\Big]^2ds.
\end{align*}
Then from the last inequality we see that  $z(u)\ge \phi(u)$, and
\begin{equation}\label{e:2.22}
\begin{aligned}
&\liminf_{t\to\infty}\frac{1}{H(t,u)}\int_{u}^{t}
\Big[\sqrt{\frac{H(t,s)[\sigma(s,a)-T]\sigma'(s,a)}{a(s)b(\sigma(s,a))
\rho(s)}}z(s)\\
&+\frac{\sqrt{a(s) b(\sigma(s,a))\rho(s)}h(t,s)}{2\sqrt{[\sigma(s,a)-T]
\sigma'(s,a)}}\Big]^2 ds\\
&\le z(u)-\phi(u)=M<\infty,
\end{aligned}
\end{equation}
 where $M$ is a constant. On the other hand, we have
\begin{equation}\label{e:2.23}
\begin{aligned}
&\liminf_{t\to\infty}\frac{1}{H(t,t_1)}\int_{t_1}^{t}\Big[\sqrt{\frac{H(t,s)[\sigma(s,a)-T]
\sigma'(s,a)}{a(s)b(\sigma(s,a))\rho(s)}}z(s)\\
&+\frac{\sqrt{a(s)
b(\sigma(s,a))\rho(s)}h(t,s)}{2\sqrt{[\sigma(s,a)-T]\sigma'(s,a)}}\Big]^2 ds
\\
&\ge\liminf_{t\to\infty}\Big[\frac{1}{H(t,t_1)}\int_{t_1}^{t}\frac{H(t,s)[\sigma(s,a)-T]
\sigma'(s,a)z^2(s)}{a(s)b(\sigma(s,a))\rho(s)}ds\\
&\quad +\frac{1}{H(t,t_1)}\int_{t_1}^{t}\sqrt{H(t,s)}h(t,s)z(s)ds\Big],
\quad t>t_1.\end{aligned}
\end{equation}
 Let
\begin{equation}\label{e:2.24}
v_1(t)=\frac{1}{H(t,t_1)}\int_{t_1}^{t}\frac{H(t,s)[\sigma(s,a)-T]\sigma'(s,a)z^2(s)}{a(s)b(\sigma(s,a))\rho(s)}ds,
\end{equation}
and
\begin{equation}\label{e:2.25}
v_2(t)=\frac{1}{H(t,t_1)}\int_{t_1}^{t}\sqrt{H(t,s)}h(t,s)z(s)ds.
\end{equation}
Thus, from (\ref{e:2.22}) and (\ref{e:2.23}), we see that
\begin{equation} \label{e:2.26}
\liminf_{t\to\infty}[v_1(t)+v_2(t)]<\infty.
\end{equation}
Now we want to show  that
\begin{equation}\label{e:2.27}
 \int_{t_1}^{\infty}\frac{[\sigma(s,a)-T]\sigma'(s,a)z^2(s)}{a(s)\rho(s)
 b(\sigma(s,a))}ds<\infty,\quad t> t_1.
\end{equation}
Assume that
\begin{equation}\label{e:2.28}
 \int_{t_1}^{\infty}\frac{[\sigma(s,a)-T]\sigma'(s,a)z^2(s)}{a(s)\rho(s)
 b(\sigma(s,a))}ds=\infty,\quad  t> t_1.
\end{equation}
Because of (\ref{e:2.18}), there exists a constant $L>0$ such that
\begin{equation}\label{e:2.29}
 \inf_{s\ge
 t_0}\Big[\liminf_{t\to\infty}\frac{H(t,s)}{H(t,t_0)}\Big]> L>0.
\end{equation}
 From (\ref{e:2.28}) one can see that for any
positive number $\lambda>0$, there exists a
$T>t_1$ such that
\begin{equation}\label{e:2.30}
 \int_{t_1}^{t}\frac{[\sigma(s,a)-T]\sigma'(s,a)z^2(s)}{a(s)\rho(s)b(\sigma(s,a))}ds\ge
 \frac{\lambda}{L},\quad t> T.
\end{equation}
On the other hand,
\begin{equation}\label{e:2.31}
\begin{aligned}
v_1(t)&= \frac{1}{H(t,t_1)}\int_{t_1}^{t}\frac{H(t,s)[\sigma(s,a)-T]\sigma'(s,a)z^2(s)}{a(s)b(\sigma(s,a))\rho(s)}ds
\\
&= \frac{1}{H(t,t_1)}\int_{t_1}^{t}H(t,s)d\left[\int_{t_1}^{s}\frac{[\sigma(u,a)-T]\sigma'(u,a)z^2(u)}{a(u)b(\sigma(u,a))\rho(u)}\right]ds
\\
&= \frac{1}{H(t,t_1)}\int_{t_1}^{t}\left[\int_{t_1}^{s}\frac{[\sigma(u,a)-T]\sigma'(u,a)z^2(u)}{a(u)b(\sigma(u,a))\rho(u)}du\right]\left[-\frac{\partial
H(t,s)}{\partial s}\right]ds
\\
&\ge \frac{1}{H(t,t_1)}\int_{T}^{t}\left[\int_{t_1}^{s}\frac{[\sigma(u,a)-T]\sigma'(u,a)z^2(u)}{a(u)b(\sigma(u,a))\rho(u)}du\right]\left[-\frac{\partial
H(t,s)}{\partial s}\right]ds
\\
&\ge \frac{\lambda}{L H(t,t_1)}\int_{T}^{t}\left[-\frac{\partial
H(t,s)}{\partial s}\right]ds
\\
&= \frac{\lambda}{L}\frac{H(t,T)}{H(t,t_1)}\ge
\frac{\lambda}{L}\frac{H(t,T)}{H(t,t_0)},\quad t\ge T>t_1.
\end{aligned}
\end{equation}
Moreover, it follows from (\ref{e:2.29}) that
\[
 \liminf_{t\to\infty}\frac{H(t,s)}{H(t,t_0)}>L>0,\quad s\ge  t_0.
\]
Therefore, there exists a $t_2> T$  such that
\begin{equation}\label{e:2.32}
 \frac{H(t,T)}{H(t,t_0)}\ge L,\quad t\ge  t_2.
\end{equation}
It follows from (\ref{e:2.31}) and (\ref{e:2.32}) that
$ v_1(t)\ge \lambda$, $t\ge t_2$.
Since $\lambda$ is arbitrary , we  have
\begin{equation}\label{e:2.33}
\lim_{t\to\infty}v_1(t)=\infty.
\end{equation}
Moreover from  (\ref{e:2.26}), there exists a convergence
subsequence $\{t_n\}_{1}^{\infty}$ on $[t_1,\infty)$ such that
$\lim_{n\to\infty}t_n=\infty$ and
\begin{equation}\label{e:2.34}
\lim_{n\to\infty}[v_1(t_n)+v_2(t_n)]=\liminf_{t\to\infty}[v_1(t)+v_2(t)]<\infty.
\end{equation}
As a result of (\ref{e:2.34}) there exists a positive integer
$n_1$ and constant $k$ such that
\[
v_1(t_n)+v_2(t_n)<k , \quad n>n_1
\]
and from (\ref{e:2.33}), we have
\begin{equation}\label{e:2.35}
\lim_{n\to\infty}v_1(t_n)=\infty.
\end{equation}
Thus (\ref{e:2.34}) and (\ref{e:2.35}) will give
\begin{equation}\label{e:2.36}
\lim_{n\to\infty}v_2(t_n)=-\infty.
\end{equation}
Moreover, for any $\epsilon\in(0,1)$, there exists a positive
integer $n_2$ such that
\[
\frac{v_2(t_n)}{v_1(t_n)}+1<\epsilon,\quad n>n_2;
\]
then
\begin{equation}\label{e:2.37}
\frac{v_2(t_n)}{v_1(t_n)}<\epsilon-1<0,\quad n>n_2.
\end{equation}
Thus (\ref{e:2.36}) and (\ref{e:2.37}) give
\begin{equation}\label{e:2.38}
\lim_{n\to\infty}\frac{v_2(t_n)}{v_1(t_n)}v_2(t_n)=\infty.
\end{equation}
By using the Cauchy-Schwartz inequality,  we obtain
\begin{align*}
 0&\le v_2^2(t_n)=\frac{1}{H^2(t_n,t_1)}
 \Big[\int_{t_1}^{t_n}\sqrt{H(t_n,s)}h(t_n,s)z(s)ds\Big]^2 \\
&\le
\Big[\frac{1}{H(t_n,t_1)}\int_{t_1}^{t_n}\frac{H(t_n,s)[\sigma(s,a)-T]
\sigma'(s,a)}{a(s)b(\sigma(s,a))\rho(s)}z^2(s)ds\Big]\\
&\quad \Big[\frac{1}{H(t_n,t_1)}\int_{t_1}^{t_n}\frac{
a(s)\rho(s)b(\sigma(s,a))h^2(t_n,s)}{[\sigma(s,a)-T]\sigma'(s,a)}ds\Big]
\\
&= v_1(t_n)\Big[\frac{1}{H(t_n,t_1)}\int_{t_1}^{t_n}\frac{a(s)\rho(s)
b(\sigma(s,a))h^2(t_n,s)}{[\sigma(s,a)-T]\sigma'(s,a)}ds\Big],\quad
t\ge t_1
\end{align*}
and
\begin{equation}\label{e:2.39}
\frac{v_2^2(t_n)}{v_1(t_n)}\le\Big[\frac{1}{H(t_n,t_1)}\int_{t_1}^{t_n}\frac{a(s)
\rho(s)b(\sigma(s,a))h^2(t_n,s)}{[\sigma(s,a)-T]\sigma'(s,a)}ds\Big].
\end{equation}
Using (\ref{e:2.32}), we see that
\begin{equation}\label{e:2.40}
\frac{1}{H(t_n,t_1)}\le\frac{1}{H(t_n,T)}=\frac{H(t_n,t_0)}{H(t_n,T)}\frac{1}{H(t_n,t_0)}
\le\frac{1}{L H(t_n,t_0)},\quad T>t_1.
\end{equation}
Therefore, we see  from (\ref{e:2.39}) and (\ref{e:2.40}) that
\begin{equation}\label{e:2.41}
\frac{v_2^2(t_n)}{v_1(t_n)}\le\frac{1}{L
H(t_n,t_0)}\int_{t_1}^{t_n}\frac{a(s)\rho(s)b(\sigma(s,a))h^2(t_n,s)}{[\sigma(s,a)-T]\sigma'(s,a)}ds.
\end{equation}
Then, it follows from (\ref{e:2.38}) and (\ref{e:2.41}) that
\[
\lim_{n\to\infty}\frac{1}{H(t_n,t_0)}\int_{t_0}^{t_n}
\frac{a(s)\rho(s)b(\sigma(s,a))h^2(t_n,s)}{[\sigma(s,a)-T]\sigma'(s,a)}ds=\infty,
\]
and then
\[
\limsup_{t\to\infty}\frac{1}{H(t,t_0)}\int_{t_0}^{t}\frac{a(s)\rho(s)b(\sigma(s,a))h^2(t,s)}{[\sigma(s,a)-T]\sigma'(s,a)}ds=\infty,
\]
which contradicts (\ref{e:2.19}). Thus, using $z(s)\ge \phi (s)$
with (\ref{e:2.27}), we obtain
\[
 \int_{t_1}^{\infty}\frac{[\sigma(s,a)-T]\sigma'(s,a)\phi_+^2(s)ds}{
a(s)\rho(s)b(\sigma'(s,a))}\le\int_{t_1}^{\infty}\frac{[\sigma(s,a)-T]\sigma'(s,a)z^2(s)ds}{
a(s)\rho(s)b(\sigma'(s,a))}<\infty,
\]
which leads to  a contradiction to (\ref{e:2.21}). This completes
the proof.
\end{proof}

\begin{theorem}\label{t:2.3}
Suppose that  the conditions of Theorem \ref{t:2.1}, and
conditions  \eqref{e:2.3},
 \eqref{e:2.18}, \eqref{e:2.21} hold, in addition to
\begin{equation}\label{e:2.42}
\liminf_{t\to\infty}\frac{1}{H(t,t_0)}\int_{t_0}^{t}H(t,s)
\rho(s)\int_{a}^{b}p(s,\xi)[1-c(\sigma(s,\xi))]d\xi ds<\infty.
\end{equation}
If there exists a function $\phi(t)\in C([t_0,\infty),R)$
satisfying
\begin{equation}\label{e:2.43}
\begin{aligned}
 &\liminf_{t\to\infty}\frac{1}{H(t,u)}\int_{u}^{t}\Big[H(t,s)
\rho(s)\int_{a}^{b}p(s,\xi)[1-c(\sigma(s,\xi))]d\xi\\
&-\frac{a(s)\rho(s)b(\sigma(s,a))h^2(t,s)}{4[\sigma(s,a)-T)]\sigma'(s,a)}\Big]ds
\\
&\ge \phi(u),\quad u\ge t_0,
\end{aligned}
\end{equation}
then every solution of (\ref{e:1.1}) is oscillatory or tends to
zero as $t\to\infty$.
\end{theorem}

\begin{proof} Assume, for the sake of contradiction, that
equation \eqref{e:1.1} has positive solution, say $x(t)>0$, $t\ge t_0$.
It follows from (\ref{e:2.43}) that
\begin{equation}\label{e:2.44}
\begin{aligned}
&\phi(t_0)\le\liminf_{t\to\infty}\frac{1}{H(t,t_0)}\int_{t_0}^{t}\Big[H(t,s)
\rho(s)\int_{a}^{b}p(s,\xi)[1-c(\sigma(s,\xi))]d\xi\\
&-\frac{a(s)\rho(s)b(\sigma(s,a))h^2(t,s)}{4[\sigma(s,a)-T]\sigma'(s,a)}\Big]ds
\\
&\le \liminf_{t\to\infty}\frac{1}{H(t,t_0)}\int_{t_0}^{t}\Big[H(t,s)
\rho(s)\int_{a}^{b}p(s,\xi)[1-c(\sigma(s,\xi))]d\xi\Big]ds \\
&-\limsup_{t\to\infty}\frac{1}{H(t,t_0)}\int_{t_0}^{t}
\frac{a(s)\rho(s)b(\sigma(s,a))h^2(t,s)}{4[\sigma(s,a)-T]\sigma'(s,a)}ds.
\end{aligned}
\end{equation}
Then, from (\ref{e:2.42}) and (\ref{e:2.44})
\[
\limsup_{t\to\infty}\frac{1}{H(t,t_0)}
\int_{t_0}^{t}\frac{a(s)\rho(s)
b(\sigma(s,a))h^2(t,s)}{4[\sigma(s,a)-T]\sigma'(s,a)}ds<\infty.
\]
Thus (\ref{e:2.19}) holds in Theorem \ref{t:2.2}. Since  the
remaining part of the proof is similar to the proof of
Theorem \ref{t:2.2}, it is omitted.
\end{proof}

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\end{document}