\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 $$\label{e:2.4} (b(t)(a(t)x'(t))')'+\sum_{i=1}^{m}q_i(t)f(x(\sigma_i(t)))=h(t),$$ 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 $$\label{e:2.5} (b(t)(a(t)x'(t))')'+\int_{c}^{d}q(t,\xi)f(x(\sigma(t,\xi)))d\xi=0,$$ 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 $$\label{e:2.31} \int^{\infty}\frac{dt}{b(t)}=\infty,\quad \int^{\infty}\frac{dt}{a(t)}=\infty.$$ \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 $$\label{e:2.32} \int^{\infty}\Big[q(t)p(t)- \frac{b(t)(p'(t))^2}{\sum_{i=1}^{m} \frac{(\sigma_i(t)-T)}{a(\sigma_i(t))}\sigma_i'(t)4p(t)}\Big]dt=\infty,$$ where $q(t)=min \{q_1(t),q_2(t),\dots ,q_m(t)\}$, for every $T\ge 0$, and that $$\label{e:2.33} \int_{\sigma(t)}^{t}\Big[\int_{\sigma(t) }^{r}\frac{1}{a(u)}du\left(\int_{u}^{r}\frac{1}{b(v)}dv\right)\Big]\sum_{i=1}^{m}q_i(r)dr> 1,$$ 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 $$\label{e:2.34} (b(t)(a(t)x'(t))')'= -\sum_{i=1}^{m}q_i(t)x(\sigma_i(t)).$$ 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 $t\ge t_2$ $$\label{e:2.35} (a(t)x'(t))'> 0.$$ 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 $$\label{e:2.36} b(t)(a(t)x'(t))'\le b(t_3)(a(t_3)x'(t_3))'<0.$$ Dividing (\ref{e:2.36}) by $b(t)$ and then integrating between $t_3$ and $t$, we obtain $$\label{e:2.37} a(t)x'(t)-a(t_3)x'(t_3)< b(t_3)(a(t_3) x(t_3)')'\int_{t_3}^{t}\frac {1}{b(s)}ds.$$ 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 $t\ge t_4$ $$\label{e:2.38} a(t)x'(t)\le a(t_4)x'(t_4).$$ 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 $$\label{e:2.39} z'(t)\le -q(t)p(t)+\frac {p'(t)}{p(t)}z(t)-\frac{\sum_{i=1}^{m}x'(\sigma_i(t))\sigma_i'(t)}{ \sum_{i=1}^{m}x(\sigma_i(t))}z(t),$$ 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 $$\label{e:2.40} (a(t)x'(t))''\le 0.$$ Using (\ref{e:2.40}) and the equality $$\label{e:2.41} a(t)x'(t)=a(T)x'(T)+\int_{T}^{t}(a(s)x'(s))'ds,$$ we have $$\label{e:2.42} a(t)x'(t)\ge (t-T)(a(t)x'(t))'$$ for $T\ge t_2$. Since $(a(t)x'(t))'$ is non-increasing, we obtain $$\label{e:2.43} a(\sigma_i(t))x'(\sigma_i(t))\ge (\sigma_i(t)-T)(a(t)x'(t))'\quad for\quad i=1,2,\dots ,m.$$ 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 $$\label{e:2.44} \sum_{i=1}^{m} \sigma_i'(t)x'(\sigma_i (t))\ge \sum_{i=1}^{m}\frac {(\sigma_i (t)-T)}{a(\sigma_i(t))} \sigma_i'(t)(a(t)x'(t))'.$$ 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 $$\label{e:2.45} z'(t)\le -q(t)p(t)+\frac{b(t)(p')^{2}(t)}{\sum_{i=1}^{m}\frac{(\sigma_i(t)-T)}{a(\sigma_i(t))}\sigma_i'(t)4p(t)}.$$ 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 $$\label{e:2.46} -b(t)(a(t)x'(t))' +\int_{t}^{\infty}\sum_{i=1}^{m}q_i(r)x(\sigma_i(r))dr\le 0.$$ 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 $$\label{e:2.47} a(t)x'(t) +\int_{t}^{\infty}\Big(\int_{t}^{r}\frac{1}{b(u)}du\Big) \sum_{i=1}^{m}q_i(r)x(\sigma_i(r))dr\le 0.$$ Dividing (\ref{e:2.47}) by $a(t)$ and integrating again from $t$ to $\infty$ gives $$\label{e:2.48} \int_{t}^{\infty}\Big[\int_{t}^{r}\frac{1}{a(u)}du\Big(\int_{u}^{r} \frac{1}{b(v)}dv\Big)\Big]\sum_{i=1}^{m}q_i(r)x(\sigma_i(r))dr\le x(t).$$ 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 $$\label{e:2.49} \int_{\sigma(t)}^{t}\Big[\int_{\sigma(t) }^{r}\frac{1}{a(u)}du \Big(\int_{u}^{r}\frac{1}{b(v)}dv\Big)\Big] \sum_{i=1}^{m}q_i(r)x(\sigma_i(r))dr\le x(\sigma(t)).$$ 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 $$\label{e:2.50} \int^{\infty}\Big[q(t)p(t)- \frac{b(t)(p'(t))^2}{\sum_{i=1}^{m} \frac{(\sigma_i(t)-T)}{a(\sigma_i(t))}\sigma_i'(t)4\lambda p(t)}\Big]dt =\infty,$$ where $q(t)=min \{q_1(t),q_2(t),\dots ,q_m(t)\}$, for every $T\ge 0$, and that $$\label{e:2.51} \limsup_{t\to\infty}\int_{\sigma(t)}^{t}\Big[\int_{\sigma(t) }^{r}\frac{1}{a(u)}du\Big(\int_{u}^{r}\frac{1}{b(v)}dv\Big)\Big] \sum_{i=1}^{m}q_i(r)dr=\infty,$$ 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 $$\label{e:2.52} z'(t)\le -q(t)p(t)+\frac {p'(t)}{p(t)}z(t)-\frac{\sum_{i=1}^{m}f'(x(\sigma_i(t))x'(\sigma_i(t))\sigma_i'(t)}{\sum_{i=1}^{m}f(x(\sigma_i(t)))}z(t),$$ 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 $$\label{e:2.53} (a(t)x'(t))''\le 0.$$ Using (\ref{e:2.53}) and the equality $$\label{e:2.54} a(t)x'(t)=a(T)x'(T)+\int_{T}^{t}(a(s)x'(s))'ds$$ will lead to $$\label{e:2.55} a(t)x'(t)\ge (t-T)(a(t)x'(t))'.$$ Now using non-increasing nature of (a(t)x'(t))', we obtain $$\label{e:2.56} a(\sigma_i(t))x'(\sigma_i(t))\ge (\sigma_i(t)-T)(a(t)x'(t))'\quad for\quad i=1,2,\dots,m.$$ 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 $$\label{e:2.57} \sum_{i=1}^{m} \sigma_i'(t)x'(\sigma_i (t))\ge \sum_{i=1}^{m}\frac {(\sigma_i (t)-T)}{a(\sigma_i(t))} \sigma_i'(t)(a(t)x'(t))^{'}.$$ 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 $$\label{e:2.58} z'(t)\le -q(t)p(t)+\frac{b(t)(p'(t))^{2}}{\lambda\sum_{i=1}^{m}\frac{(\sigma_i(t)-T)}{a(\sigma_i(t))}\sigma_i'(t)4p(t)}.$$ 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 $$\label{e:2.59} \int_{\sigma(t)}^{t}\Big[\int_{\sigma(t) }^{r}\frac{1}{a(u)}du \Big(\int_{u}^{r}\frac{1}{b(v)}dv\Big)\Big] \sum_{i=1}^{m}q_i(r)f(x(\sigma_i(r)))dr\le x(\sigma(t)).$$ 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 $$\label{e:2.60} \limsup_{t\to\infty}\int_{\sigma(t)}^{t}\Big[\int_{\sigma(t) }^{r}\frac{1}{a(u)}du\Big(\int_{u}^{r}\frac{1}{b(v)}dv\Big)\Big]\sum_{i=1}^{m}q_i(r)dr=\infty.\quad$$ 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 $$\label{e:2.61} \int^{\infty}\Big[q(t)p(t)- \frac{b(t)(p'(t))^{2}}{\sum_{i=1}^{m}\frac{(\sigma_i(t)-T)}{a(\sigma_i(t))} \sigma_i'(t)4\lambda d p(t)}\Big]dt=\infty$$ for some $d\in(0,1)$ and for every $T\ge 0$, and $$\label{e:2.62} (b(t)(a(t)\psi'(t))')'=h(t),\quad \lim_{t\to\infty}\psi^{(i)}(t)=0, \quad i=0,1,2.$$ 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 $$\label{e:2.63} y(t)=x(t)-\psi(t).$$ 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, $$\label{e:2.64} (b(t)(a(t)y'(t))')'\le 0.$$ 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 $$\label{e:2.65} z'(t)\le -q(t)p(t)+\frac {p'(t)}{p(t)}z(t)-d\lambda\frac{\sum_{i=1}^{m}y'(\sigma_i(t))\sigma_i'(t)}{ \sum_{i=1}^{m}f(dy(\sigma_i(t)))}z(t),$$ where q(t)=min \{q_1(t),q_2(t),\dots ,q_m(t)\}. We can now show that $$\label{e:2.66} \sum_{i=1}^{m} \sigma_i'(t)y'(\sigma_i (t))\ge \sum_{i=1}^{m}\frac {(\sigma_i (t)-T)}{a(\sigma_i(t))} \sigma_i'(t)(a(t)y'(t))'$$ 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 $$\label{e:2.67} z'(t)\le -q(t)p(t)+\frac{b(t)(p')^{2}(t)}{\sum_{i=1}^{m}\frac{(\sigma_i(t)-T)}{a(\sigma_i(t))}\sigma_i'(t)4\lambda d p(t)}.$$ 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 $$\label{e:2.68} \int_{\sigma(t)}^{t}\Big[\int_{\sigma(t) }^{r}\frac{1}{a(u)}du \Big(\int_{u}^{r}\frac{1}{b(v)}dv\Big)\Big] \sum_{i=1}^{m}q_i(r)f(x(\sigma_i(r)))dr\le y(t).$$ 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 $$\label{e:2.69} \int_{t_1}^{\infty}\int_{c}^{d}q(s,\xi)d\xi ds=\infty$$ and $$\label{e:2.70} \epsilon\int_{g(t)}^{t}\Big[\int_{g(t) }^{r}\frac{1}{a(u)}du\Big(\int_{u}^{r}\frac{1}{b(v)}dv\Big)\Big] \int_{c}^{d}q(r,\xi)d\xi dr>1,$$ 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 $$\label{e:2.71} (b(t)(a(t)x'(t))')'=-\int_{c}^{d}q(t,\xi)f(x(\sigma(t,\xi)))d\xi.$$ 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, $$\label{e:2.72} 0\ge (b(t)(a(t)x'(t))')'+\epsilon\int_{c}^{d}q(t,\xi)x(\sigma(t,\xi))d\xi.$$ 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, $$\label{e:2.73} %\label{e:2.1.3} (b(t)(a(t)x'(t))')'+\epsilon x(k)\int_{c}^{d}q(t,\xi)d\xi\le 0.$$ Integrating this last equation from $t_1$ to $t$, we get $$\label{e:2.74} b(t)(a(t)x'(t))'\le b(t_1)(a(t_1)x'(t_1))' -\epsilon x(k)\int_{t_1}^{t}\int_{c}^{d}q(s,\xi)d\xi ds.$$ 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 $$\label{e:2.75} \int_{t}^{\infty}\Big[\int_{t}^{r}\frac{1}{a(u)}du \Big(\int_{u}^{r}\frac{1}{b(v)}dv\Big)\Big] \int_{c}^{d}q(r,\xi)f(x(\sigma(r,\xi)))d\xi dr\le x(t)$$ Using ({c}) in (\ref{e:2.75}), we obtain $$\label{e:2.76} \int_{t}^{\infty}\Big[\int_{t}^{r}\frac{1}{a(u)}du \Big(\int_{u}^{r}\frac{1}{b(v)}dv\Big)\Big]\epsilon\int_{c}^{d}q(r,\xi) x(\sigma(r,\xi))d\xi dr\le x(t).$$ Replace $t$ by $g(t)$ in (\ref{e:2.76}), where $g(t)=\sigma(t,d)$, then we have $$\label{e:2.77} \epsilon\int_{g(t)}^{t}\Big[\int_{g(t)}^{r}\frac{1}{a(u)}du \Big(\int_{u}^{r}\frac{1}{b(v)}dv\Big)\Big] \int_{c}^{d}q(r,\xi)x(\sigma(r,\xi))d\xi dr\le x(g(t)).$$ Since $x(t)$ is decreasing and positive, $$\label{e:2.78} \epsilon\int_{g(t)}^{t}\Big[\int_{g(t)}^{r}\frac{1}{a(u)}du \Big(\int_{u}^{r}\frac{1}{b(v)}dv\Big)\Big]\int_{c}^{d}q(r,\xi)d\xi dr\le 1. \nonumber$$ 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 $$\label{e:2.79} \int^{\infty}\Big[\Gamma(t)p(t)- \frac{a(\sigma(t,c))b(t)(p'(t))^2}{(\sigma(t,c)-T)\sigma'(t,c)4p(t)}\Big]dt =\infty,$$ 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 $(c)$, $f(x(\sigma(t,\xi)))\ge\epsilon x(\sigma(t,\xi))>0.$ $$\label{e:2.80} (b(t)(a(t)x'(t))')'+\epsilon\int_{c}^{d}q(t,\xi)x(\sigma(t,\xi))d\xi\le 0.$$ 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 $$a(t)x'(t)\ge (t-T)(a(t)x'(t))'. \nonumber$$ Since $(a(t)x(t))'$ is non-increasing, we have $$a(\sigma(t,c))x'(\sigma(t,c))\ge (\sigma(t,c)-T)(a(t)x'(t))', \nonumber$$ then $$\label{e:2.82} x'(\sigma(t,c))\ge \frac{(\sigma(t,c)-T)(a(t)x'(t))'}{a(\sigma(t,c))}.$$ 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 $$\label{e:2.83} z'(t)\le -\Gamma(t)p(t)+\frac{b(t)a(\sigma(t,c))(p'(t))^2}{(\sigma(t,c)-T)\sigma'(t,c))4p(t)}.$$ Integrating (\ref{e:2.83}) from $T$ to t and letting $t\to\infty$, we see that $\dst\lim_{t\to\infty} z(t)=-\infty$. 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