\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 $$\label{e:1.1} \left[a(t)[b(t)[x(t)+c(t)x(t-\tau)]']'\right]' +\int_{a}^{b}p(t,\xi)x(\sigma(t,\xi))d\xi=0,$$ where \begin{enumerate} \item[(a)] $a(t), a'(t), b(t), c(t)\in C([t_0,\infty),(0,\infty))$, $0 s\ge t_0\}$, $D=\{(t,s)|t\ge s\ge t_0\}$. \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 \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} and $$\label{e:2.3} \int_{t_3}^{\infty}\Big[\int_{t_3 }^{r}\frac{du}{b(u)}\Big(\int_{u}^{r}\frac{dv}{a(v)}\Big)\Big] \int_{a}^{b}p(r,\xi)d\xi dr=\infty.$$ 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 $$\label{e:2.4} y(t)=x(t)+c(t)x(t-\tau)$$ then, from (\ref{e:1.1}), we obtain $$\label{e:2.5} \left[a(t)[b(t)y'(t)]'\right]'=-\int_{a}^{b}p(t,\xi)x(\sigma(t,\xi))d\xi.$$ 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 $$\label{e:2.6} (b(t)y'(t))'> 0,\quad t\ge t_2.$$ 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 $$\label{e:2.7} a(t)(b(t)y'(t))'\le a(t_3)(b(t_3)y'(t_3))'<0,\quad t\ge t_3.$$ Dividing (\ref{e:2.7}) by $a(t)$ and integrating from $t_3$ to $t$, we obtain $$\label{e:2.8} b(t)y'(t)-b(t_3)y'(t_3)\le a(t_3)(b(t_3) y'(t_3))'\int_{t_3}^{t}\frac {ds}{a(s)}.$$ 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 $$\label{e:2.9} b(t)y'(t)\le b(t_4)y'(t_4)<0,\quad t\ge t_4 .$$ 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 \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} On the other hand, since $(a(t)(b(t)y'(t))')'\le 0$ and $a'(t)\ge 0$, we find that $$\label{e:2.11} (b(t)y'(t))''\le 0.$$ 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 $$\label{e:2.12} y'(\sigma(t,a))\ge\frac{(\sigma(t,a)-T)(b(t)y'(t))'}{b(\sigma(t,a))}.$$ 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 $$\label{e:2.13} \rho(t)\int_{a}^{b}p(t,\xi)[1-c(\sigma(t,\xi))]d\xi \le -z'(t)+\frac{\rho'(t)}{\rho(t)}z(t)-\frac{[\sigma(t,a)-T]\sigma'(t,a)z^2(t)}{a(t) \rho(t)b(\sigma(t,a))}.$$ 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 $$\label{e:2.14} -a(t)(b(t)y'(t))' +\int_{t}^{\infty}\int_{a}^{b}p(r,\xi)x(\sigma(r,\xi))d\xi dr\le 0.$$ 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 $$\label{e:2.15} b(t)y'(t) +\int_{t}^{\infty}\left(\int_{t}^{r}\frac{du}{a(u)}\right)\int_{a}^{b}p(r,\xi)x(\sigma(r,\xi))d\xi dr\le 0.$$ Dividing (\ref{e:2.15}) by $b(t)$ and integrating again from $t$ to $\infty$ gives $$\label{e:2.16} \int_{t}^{\infty}\Big[\int_{t}^{r}\frac{du}{b(u)}\Big(\int_{u}^{r}\frac{dv}{a(v)} \Big)\Big]\int_{a}^{b}p(r,\xi)x(\sigma(r,\xi))d\xi dr\le y(t)$$ for $t\ge t_3\ge t_2$. Replacing $t$ by $t_3$ in (\ref{e:2.16}), we get $$\label{e:2.17} \int_{t_3}^{\infty }\Big[\int_{t_3 }^{r}\frac{du}{b(u)}\Big(\int_{u}^{r}\frac{dv}{a(v)}\Big)\Big] \int_{a}^{b}p(r,\xi)x(\sigma(r,\xi))d\xi\, dr\le y(t_3).$$ 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 \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} and $$\label{e:2.21} \begin{gathered} \limsup_{t\to\infty}\frac{1}{H(t,t_0)}\int_{t_0}^{t}\frac{[\sigma(u,a)-T]\sigma'(u,a)\phi_+^2(u)du}{ a(u)\rho(u)b(\sigma'(u,a))}=\infty, \\ \phi_+(u)=\max_{u\ge t_0}\{\phi(u),0\}. \end{gathered}$$ 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 \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} where $M$ is a constant. On the other hand, we have \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} Let $$\label{e:2.24} v_1(t)=\frac{1}{H(t,t_1)}\int_{t_1}^{t}\frac{H(t,s)[\sigma(s,a)-T]\sigma'(s,a)z^2(s)}{a(s)b(\sigma(s,a))\rho(s)}ds,$$ and $$\label{e:2.25} v_2(t)=\frac{1}{H(t,t_1)}\int_{t_1}^{t}\sqrt{H(t,s)}h(t,s)z(s)ds.$$ Thus, from (\ref{e:2.22}) and (\ref{e:2.23}), we see that $$\label{e:2.26} \liminf_{t\to\infty}[v_1(t)+v_2(t)]<\infty.$$ Now we want to show that $$\label{e:2.27} \int_{t_1}^{\infty}\frac{[\sigma(s,a)-T]\sigma'(s,a)z^2(s)}{a(s)\rho(s) b(\sigma(s,a))}ds<\infty,\quad t> t_1.$$ Assume that $$\label{e:2.28} \int_{t_1}^{\infty}\frac{[\sigma(s,a)-T]\sigma'(s,a)z^2(s)}{a(s)\rho(s) b(\sigma(s,a))}ds=\infty,\quad t> t_1.$$ Because of (\ref{e:2.18}), there exists a constant $L>0$ such that $$\label{e:2.29} \inf_{s\ge t_0}\Big[\liminf_{t\to\infty}\frac{H(t,s)}{H(t,t_0)}\Big]> L>0.$$ From (\ref{e:2.28}) one can see that for any positive number $\lambda>0$, there exists a $T>t_1$ such that $$\label{e:2.30} \int_{t_1}^{t}\frac{[\sigma(s,a)-T]\sigma'(s,a)z^2(s)}{a(s)\rho(s)b(\sigma(s,a))}ds\ge \frac{\lambda}{L},\quad t> T.$$ On the other hand, \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} 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 $$\label{e:2.32} \frac{H(t,T)}{H(t,t_0)}\ge L,\quad t\ge t_2.$$ 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 $$\label{e:2.33} \lim_{t\to\infty}v_1(t)=\infty.$$ 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 $$\label{e:2.34} \lim_{n\to\infty}[v_1(t_n)+v_2(t_n)]=\liminf_{t\to\infty}[v_1(t)+v_2(t)]<\infty.$$ As a result of (\ref{e:2.34}) there exists a positive integer $n_1$ and constant $k$ such that $v_1(t_n)+v_2(t_n)n_1$ and from (\ref{e:2.33}), we have $$\label{e:2.35} \lim_{n\to\infty}v_1(t_n)=\infty.$$ Thus (\ref{e:2.34}) and (\ref{e:2.35}) will give $$\label{e:2.36} \lim_{n\to\infty}v_2(t_n)=-\infty.$$ 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 $$\label{e:2.37} \frac{v_2(t_n)}{v_1(t_n)}<\epsilon-1<0,\quad n>n_2.$$ Thus (\ref{e:2.36}) and (\ref{e:2.37}) give $$\label{e:2.38} \lim_{n\to\infty}\frac{v_2(t_n)}{v_1(t_n)}v_2(t_n)=\infty.$$ 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 $$\label{e:2.39} \frac{v_2^2(t_n)}{v_1(t_n)}\le\Big[\frac{1}{H(t_n,t_1)}\int_{t_1}^{t_n}\frac{a(s) \rho(s)b(\sigma(s,a))h^2(t_n,s)}{[\sigma(s,a)-T]\sigma'(s,a)}ds\Big].$$ Using (\ref{e:2.32}), we see that $$\label{e:2.40} \frac{1}{H(t_n,t_1)}\le\frac{1}{H(t_n,T)}=\frac{H(t_n,t_0)}{H(t_n,T)}\frac{1}{H(t_n,t_0)} \le\frac{1}{L H(t_n,t_0)},\quad T>t_1.$$ Therefore, we see from (\ref{e:2.39}) and (\ref{e:2.40}) that $$\label{e:2.41} \frac{v_2^2(t_n)}{v_1(t_n)}\le\frac{1}{L H(t_n,t_0)}\int_{t_1}^{t_n}\frac{a(s)\rho(s)b(\sigma(s,a))h^2(t_n,s)}{[\sigma(s,a)-T]\sigma'(s,a)}ds.$$ 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 $$\label{e:2.42} \liminf_{t\to\infty}\frac{1}{H(t,t_0)}\int_{t_0}^{t}H(t,s) \rho(s)\int_{a}^{b}p(s,\xi)[1-c(\sigma(s,\xi))]d\xi ds<\infty.$$ If there exists a function $\phi(t)\in C([t_0,\infty),R)$ satisfying \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} 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 \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} 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} \begin{thebibliography}{99} \bibitem{Agar} R. Agarwal and S. 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