\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2005(2005), No. 65, pp. 1--14.\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}} \begin{document} \title[\hfilneg EJDE-2005/65\hfil Asymptotic behaviour of solutions] {Asymptotic behaviour of solutions to $n$-order functional differential equations} \author[S. Padhi\hfil EJDE-2005/65\hfilneg] {Seshadev Padhi} \address{Department of Applied Mathematics\\ Birla Institute of Technology\\ Mesra, Ranchi -835 215, India} \curraddr{Department of Mathematics and Statistics, Mississippi State University, MS-39762, Mississippi State, USA} \email{ses\_2312@yahoo.co.in} \date{} \thanks{Submitted September 4, 2004. Revised April 30, 2005. Published June 23, 2005.} \subjclass[2000]{34C10, 34K15} \keywords{Oscillatory solution; nonoscillatory solution; property A} \begin{abstract} We establish conditions for the linear differential equation $$ y^{(n)}(t)+p(t)y(g(t))=0 $$ to have property A. Explicit sufficient conditions for the oscillation of the the equation is obtained while dealing with the property A of the equations. A comparison theorem is obtained for the oscillation of the equation with the oscillation of a third order ordinary differential equation. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{problem}[theorem]{Problem} \newtheorem{example}[theorem]{Example} \section{Introduction} This paper concerns property A of the $n$-th order $(n\geq 2)$ delay differential equation \begin{equation}\label{first} y^{(n)}(t)+p(t)y(g(t))=0, \end{equation} under certain conditions on the coefficient function $p \in C([\sigma,\infty),[0,\infty)),\sigma\in R$, and $g \in C([\sigma,\infty),R)$ such that $g(t)\leq t$ and $g(t)\to\infty$ as $t\to\infty$. It is interesting to note that we have obtained sufficient conditions for oscillation of all solutions of \eqref{first} while dealing with property A of the equation. These sufficient conditions are easily verifiable and different from earlier ones (See \cite{pdas,erbe,gopa,gyor,kopl,ladd}). Moreover, these sufficient conditions are consistent with the situation when $p(t)$ is a constant. A continuous function $y:[g(\sigma),\infty)\to R$ is said to be a proper solution of \eqref{first} if it is absolutely continuous on $(t_{0},\infty), t_{0}\geq \sigma$ along with its derivatives up to the $(n-1)$th order and satisfies \eqref{first} almost everywhere on $(t_{0},\infty)$ and $\sup\{|y(s)|:s \geq t\}>0$ for $t\geq t_{0}$. A proper solution of \eqref{first} is called oscillatory if it has a sequence of zeros tending to infinity. Otherwise, it is called non-oscillatory. Equation \eqref{first} with $g(t)=t$ is said to be disconjugate on $[\sigma,\infty)$ if no nontrivial solution of the equation has more than $(n-1)$ zeros, counting muntiplicities. A vast body of literature exist on the oscillation of \eqref{first}. One may see the monographs due to Lakshmikantham et al \cite{ladd}, Gyori and Ladas \cite{gyor} and the references cited therein. Higher order differential equations with property A were studied by Parhi and Padhi \cite{pars} and Koplatadze \cite{kopl}. We shall see that our results are different form their results. We observe that our results do not hold for the case $g(t)=t$ (See Theorems \ref{thm2.1}-\ref{thm2.4} and \ref{thm2.25} and Corollaries \ref{coro2.5} and \ref{coro2.26}). Let $y(t)$ be a positive solution of \eqref{first} for $t\geq t_{0}\sigma$. Then there exists a $t_{1}>t_{0}$ such that $y(g(t))>0$ for $t\geq t_{1}$. Then $y^{(n)}(t)\leq 0$ for $t\geq t_{1}$,and so by a lemma due to Kiguradze \cite{kigu}, there exists an integer $\emph{l}$, $0\leq \emph{l}\leq n-1$ such that $n+\emph{l}$ odd and \begin{equation}\label{second} \begin{gathered} y^{(i)}(t) >0 , \quad i=0,1,2,\dots ,l,\\ (-1)^{i+l}y^{(i)}(t) >0 , \quad i=l+1,\dots ,n. \end{gathered} \end{equation} for large $t$. Again, for $l \in \{1,2,3,\dots ,n-1\}, n+l$ odd, the following inequality holds for large $t$, say for $t\geq t_{2}$. \begin{equation}\label{third} |y(t)| \geq\frac{(t-t_{2})^{(n-1)}}{(n-1)(n-2)\dots (n-l)}|y^{(n-1)}(2^{n-l-1}t)|, \quad t \geq t_{2}. \end{equation} Let $N$ denote the set of all nonoscillatory solutions of \eqref{first} and $N_{l}$ denote the set of all nonoscillatory solutions of \eqref{first} satisfying \eqref{second}. Then \[ N = \begin{cases} N_{0}\cup N_{2}\cup\dots \cup N_{n-1}& \mbox{if $n$ is odd},\\ N_{1}\cup N_{3}\cup\dots \cup N_{n-1}& \mbox{if $n$ is even.} \end{cases} \] \subsection*{Definition} We say that \eqref{first} has property A if any of its solution is oscillatory when $n$ is even and either is oscillatory or satisfies $N_{0}$ when $n$ is odd. The following conjecture is given in \cite[pp.29, Problem 1.14]{kigu}, which we state as a problem. \begin{problem} \label{prob1.1} Let $M_{n^{*}}=\max(\lambda(\lambda-1)(\lambda-2)\dots (\lambda-n+1))$. If \[ \int^{\infty}t^{n-1}\big[p(t)-\frac{M_{n^{*}}}{t^{n}}\big]\,dt=\infty, \] then \eqref{first} with $g(t)=t$ has property A. \end{problem} Our Theorem \ref{thm2.20} gives a partial answer to the above problem for the case $n=2$ and $g(t)=t$ in \eqref{first}. The following lemma, due to Kiguradze \cite{kigu}, is needed for our use in the sequel. \begin{lemma} \label{lem1.2} Let for a certain $l \in \{1,2,3,\dots ,n-1\}$, the inequality \eqref{second} hold. Then \begin{equation}\label{fourth} \int^{\infty}_{t_{1}}s^{n-l-1}|y^{(n)}(s)|\,ds <\infty, \end{equation} \begin{equation}\label{fifth} y^{(i)}(t)\geq y^{(i)}(t_{1})+\frac{1}{(l-i-1)!}\int^{t}_{t_{1}}(t-s)^{l-i-1}y^{(i)}(s)\,ds \end{equation} for $t\geq t_{1},i=0,1,2,\dots ,l-1$ and \begin{equation}\label{sixth} y^{(l)}(t)\geq \frac{1}{(l-i-1)!}\int^{\infty}_{t}(s-t)^{n-l-1}|y^{(n)}(s)|\,ds \end{equation} for $t\geq t_{1}$. If in addition \begin{equation}\label{seventh} \int^{\infty}_{t_{1}}s^{n-l}|y^{(n)}(s)|\,ds = \infty, \end{equation} then there exists $t_{2}\geq t_{1}$ such that \begin{equation}\label{eighth} y^{(l-1)}(t) \geq \frac{t}{(n-l)!}\int^{\infty}_{t}s^{n-l-1}|y^{(n)}(s)|\,ds \end{equation} for $t\geq t_{2}$ and \begin{equation}\label{nineth} iy^{(l-1)}\geq ty^{(l-i+1)}(t)\geq (i-1)y^{(l-i)}(t) \end{equation} for $t\geq t_{2}$, $i\in \{1,2,\dots ,l\}$. \end{lemma} \section{Main Results} \begin{theorem} \label{thm2.1} Let $g(t) (n-l-1)!.l1 \end{equation} hold. Then \eqref{first} has property A. \end{theorem} \begin{proof} Let $y(t)$ be a nonoscillatory solution of \eqref{first}. Without loss of generality, we may assume that $y(t)>0$ for $t\geq t_{0}>\sigma$. Thus there exists a $T_{1}\geq t_{0}$ such that $y(g(t))>0$ for $t\geq T_{1}$. Consequently, from \eqref{first}, it follows that $y^{(n)}(t)\leq 0$ for $t\geq T_{1}$. Then, there exists a $l\in\{0,1,2,\dots ,n-1\}$ and $n+l$ odd such that \eqref{second} holds for some $t\geq t_{1}>T_{1}$. We claim that $l=0$. If not, then $l\in\{1,2,\dots ,n-1\}$. Putting $i=0$ in \eqref{fifth}, we get \begin{equation}\label{one} y(t)\geq \frac{1}{(l-1)!}\int^{t}_{t_{1}}(t-s)^{l-1}y^{(l)}(s)\,ds, \quad t\geq t_{1}. \end{equation} We can find a $t_{2}\geq t_{1}$ such that $g(t)>t_{1}$ for $t\geq t_{2}$. Hence, for $t\geq t_{2}$ \begin{equation*} y(t)\geq \frac{y^{(l)}(t)}{(l-1)!}\int^{t}_{g(t)}(t-s)^{l-1}\,ds \geq \frac{y^{(l)}(t)}{(l-1)!}.\frac{(t-g(t))^{l}}{l}; \end{equation*} that is, \begin{equation}\label{eleventh} y(t)\geq \frac{(t-g(t))^{l}}{l!}y^{(l)}(t). \end{equation} Using \eqref{sixth} in \eqref{eleventh}, we obtain \begin{equation*} \begin{split} y(t) & \geq \frac{(t-g(t))^{l}}{l!}.\frac{1}{(n-l-1)!}\int^{\infty}_{t}(s-t)^{n-l-1}|y^{(n)}(s)|\,ds\\ & \geq \frac{(t-g(t))^{l}}{l!}.\frac{1}{(n-l-1)!}\int^{\infty}_{g^{-1}(t)}(s-t)^{n-l-1}|y^{(n)}(s)|\,ds\\ & \geq \frac{(t-g(t))^{l}}{l!}.\frac{1}{(n-l-1)!}\int^{\infty}_{g^{-1}(t)}(s-t)^{n-l-1}p(s)y(g(s))\,ds\\ & \geq \frac{(t-g(t))^{l}}{l!}.\frac{1}{(n-l-1)!}y(t)\int^{\infty}_{g^{-1}(t)}(s-t)^{n-l-1}p(s)\,ds \end{split} \end{equation*} for $t\geq t_{2}$, which is a contradiction to the hypothesis of the theorem. Hence \eqref{first} has property A. This completes the proof of the theorem. \end{proof} \begin{theorem} \label{thm2.2} Suppose that for every $l\in \{1,2,3,\dots ,n-1\}$ , $n+l$ is odd,, \begin{equation}\label{twelfth} \limsup_{t\to\infty}t^{n-1}\int^{\infty}_{g^{-1}(t)}p(s)\,ds > (n-1)\dots (n-l)2^{(n-1)(n-l)}, \end{equation} holds. Then \eqref{first} has property A. \end{theorem} \begin{proof} Let $y(t)$ be a non-oscillatory solution of \eqref{first}. Without any loss of generality, we may assume that $y(t)>0$ for $t\geq t_{0}>\sigma$. Then there exists a $t_{1}\geq t_{0}$ such that $y(g(t))>0$ for $t\geq t_{1}$. Consequently, it follows from \eqref{first} that $y^{(n)}(t)\leq 0$ for $t\geq t_{1}$ and \eqref{second} holds. If possible, suppose that \eqref{first} has not property A. Then $l\in\{1,2,3,\dots ,n-1\}$. Clearly \eqref{third} holds for some $t\geq t_{2}\geq t_{1}$. Since $y'(t)>0$, then for $t>t.2^{l+1-n} \geq t_{2}$, we have \begin{equation}\label{thirteen} y(t)\geq y(2^{l+1-n}t) \geq \frac{1}{(n-1)\dots (n-l).2^{(n-1)(n-l)}}t^{n-1}y^{(n-1)}(t). \end{equation} On the other hand, integrating \eqref{first} from $t(\geq t_{2})$ to $\infty$,we have \begin{equation*} y^{(n-1)}(t) > \int^{\infty}_{t}p(s)y(g(s))\,ds > \int^{\infty}_{g^{-1}(t)}p(s)y(g(s))\,ds >y(t)\int^{\infty}_{g^{-1}(t)}p(s)\,ds. \end{equation*} Then \eqref{thirteen} gives \begin{equation*} 1\geq \frac{1}{(n-1)\dots (n-l).2^{(n-1)(n-l)}}t^{n-1}\int^{\infty}_{g^{-1}(t)}p(s)\,ds \end{equation*} for $t\geq t_{2}$, which contradicts \eqref{twelfth}. Hence \eqref{first} has property A. The Theorem is proved. \end{proof} \begin{theorem} \label{thm2.3} Suppose that $g(t)(l-1)!.(n-l-1)! \end{equation} holds. Then \eqref{first} has property A. \end{theorem} \begin{proof} Let $y(t)$ be a nonoscillatory solution of \eqref{first}. Without loss of generality, we may assume that $y(t)>0$ and $y(g(t))>0$ for $t\geq t_{0}>\sigma$. Thus \eqref{second} holds for some $t\geq t_{1}>t_{0}$. Suppose that $l\in\{1,2,\dots ,n-1\}$. Putting $i=0$ in \eqref{fifth}, we get \begin{equation}\label{fifteen} y(t)\geq \frac{1}{(l-1)!}\int^{t}_{t_{1}}(t-s)^{l-1}y^{(l)}(s)\,ds. \end{equation} From \eqref{fifth}, we obtain \begin{equation}\label{sixteen} y^{(l)}(t)\geq \frac{1}{(n-l-1)!}\int^{\infty}_{t}(s-t)^{n-l-1}p(s)y(g(s))\,ds. \end{equation} Then from \eqref{fifteen} and \eqref{sixteen}, we obtain \begin{equation}\label{sixtin} y(t)\geq \frac{1}{(n-l-1)!.(l-1)!}\int^{t}_{t_{1}}(t-s)^{l-1} \int^{\infty}_{s}(u-s)^{n-l-1}p(u)y(g(u))\,du\,ds. \end{equation} We can find a $t_{2}\geq t_{1}$ such that $g(t)>t_{1}$ for $t\geq t_{2}$. Thus, for $t\geq t_{2}$ \begin{equation*} y(t)\geq \frac{1}{(n-l-1)!.(l-1)!}\int^{t}_{g(t)}(t-s)^{l-1} \int^{\infty}_{g^{-1}(g^{-1}(s))}(u-s)^{n-l-1}p(u)y(g(u))\,du\,ds \end{equation*} which in turn, yields \begin{equation*} 1\geq \frac{1}{(n-l-1)!.(l-1)!}\int^{t}_{g(t)}(t-s)^{l-1} \int^{\infty}_{g^{-1}(g^{-1}(s))}(u-s)^{n-l-1}p(u)\,du\,ds. \end{equation*} Taking limit sup., we obtain a contradiction. Consequently, \eqref{first} has property A. Hence the theorem is proved. \end{proof} \begin{theorem} \label{thm2.4} Let $g(t) (n-1)!. \end{equation} Then \eqref{first} has no solution satisfying the property $(-1)^{i}y^{(i)}(t)>0$ for large $t$. \end{theorem} \begin{proof} If possible, suppose that \eqref{first} has a nonoscillatory solution $y(t)$ satisfying the property $(-1)^{i}y^{(i)}(t)>0$ for large $t$. Then $l=0$ in \eqref{second}. Suppose that $y(g(t))>0$ and $y(t)>0$ for some $t\geq t_{1}>\sigma$. From Lemma 1.2 due to Kiguradze and Chanturia \cite{kigu}, it follows for $i=0$, that \begin{equation*} \begin{split} y(t) &\geq \frac{1}{(n-1)!}\int^{\infty}_{t}(s-t)^{n-1}p(s)y(g(s))\,ds\\ &\geq \frac{1}{(n-1)!}\int^{g^{-1}(t)}_{t}(s-t)^{n-1}p(s)y(g(s))\,ds\\ &\geq \frac{y(t)}{(n-1)!}\int^{g^{-1}(t)}_{t}(s-t)^{n-1}p(s)\,ds, \end{split} \end{equation*} that is, \begin{equation*} (n-1)! \geq \int^{g^{-1}(t)}_{t}(s-t)^{n-1}p(s)\,ds, \end{equation*} for some $t\geq t_{2} \geq t_{1}$. Then there exists a $t_{3}\geq t_{2}$ such that $g(t)>t_{2}$ for $t\geq t_{3}$. Hence for $t\geq t_{3}$, we have \begin{equation*} (n-1)! \geq \int^{t}_{g(t)}(s-g(t))^{n-1}p(s)\,ds. \end{equation*} Taking limit sup., we obtain a contradiction. Hence $l\neq 0$. The theorem is proved. \end{proof} \begin{corollary} \label{coro2.5} Suppose that $g(t)0$ be a constant and $g(t)=t-\tau$, $\tau>0$ be a constant. Then \eqref{first} becomes \begin{equation}\label{twentyfirst} y^{(n)}(t) +py(t-\tau)=0. \end{equation} Clearly, the conditions of \eqref{tenth},\eqref{twelfth} and \eqref{fourteen} are consistent with $p(t)=p$ and $g(t)=t-\tau$. Hence form Corollary \ref{coro2.5}, it follows that, if \begin{equation}\label{twentysecond} p\tau^{n} >n!, \end{equation} then \eqref{twentyfirst} is oscillatory. \smallskip The characteristic equation associated with \eqref{twentyfirst} is given by \begin{equation}\label{twentythird} \lambda^{n}+pe^{-\tau\lambda}=0. \end{equation} Setting $F(\lambda)= \lambda^{n}+pe^{-\tau\lambda}$, we see that $F(\lambda)>0$ for $\lambda \geq 0$. Suppose that $\lambda <0$. We claim that $F(\lambda)>0$ for $\lambda < 0$. If possible suppose that $F(\lambda)\leq 0$ for $\lambda<0$. Then $\lambda^{n}\leq -pe^{-\tau\lambda}$. Then $\lambda^{n}\tau^{n}\leq -n!.e^{-\tau\lambda}$. If $n$ is even, then $\lambda^{n}\tau^{n}\leq 0$, a contradiction. Hence $n$ must be odd. Let $\lambda=-\gamma,\gamma>0$. Then $\gamma^{n}\tau^{n}\geq n!.e^{\tau\gamma}$. Setting $\tau\gamma = \beta$, we see that $\beta^{n}\geq n!.e^{\beta}$, a contradiction. Hence our claim holds, that is, $F(\lambda)>0$ for $\lambda<0$. Thus \eqref{twentysecond} implies that all solutions of \eqref{twentyfirst} are oscillatory. \textbf{Remark:} Although the conditions in Theorems \ref{thm2.1} and \ref{thm2.1} are legitimate, these are not efficient. When $g(t)$ is close to $t$, the conditions \eqref{tenth} and \eqref{fourteen} fails to hold. This is evident from the following examples : If we replace $g(t)=\frac{t}{3}$ in \eqref{nineteen} by $g(t)=\frac{3t}{4}$, then the equation becomes \begin{equation}\label{twentithird} y'''(t)+\frac{82}{t^{3}}y(\frac{3t}{4})=0 , t\geq 1. \end{equation} Condition \eqref{tenth} fails to hold and hence Theorem \ref{thm2.1} cannot be applied to \eqref{twentithird}. Similarly,consider the equation \begin{equation}\label{twenthird} y'''(t)+\frac{46}{t^{3}}y(\frac{t}{2})=0,t\geq1. \end{equation} Theorem \ref{thm2.3} can be applied to this example. On the other hand, if $g(t)=\frac{t}{2}$ in \eqref{twenthird} is replaced by $g(t)=\frac{10t}{11}$, then \eqref{twenthird} becomes \begin{equation}\label{tenthird} y'''(t)+\frac{46}{t^{3}}y(\frac{10t}{11})=0,t\geq1, \end{equation} then\eqref{fourteen} fails and hence Theorem \ref{thm2.3} cannot be applied. The following theorems provides sufficient conditions for \eqref{first} to have property A when $g(t)$ is close to $t$. \begin{theorem} \label{thm2.9} Assume that $g(t)(n-l-1)!.l! \end{equation} holds, then \eqref{first} has property A. \end{theorem} \begin{proof} We can find a $t_{2}>t_{1}$ such that $t-g(t)>t_{1}$ for $t\geq t_{2}$. Hence for $t\geq t_{2}$, \eqref{one} gives \begin{equation*} y(t)\geq \frac{y^{(l)}(t)}{(l-1)!}\int^{t}_{t-g(t)}(t-s)^{l-1}\,ds \geq \frac{g^{l}(t)}{l!}y^{(l)}(t). \end{equation*} using \eqref{sixth}in the above inequality, we obtain a contradiction. The proof is complete. \end{proof} \begin{corollary} \label{coro2.10} Suppose that the conditions of Theorems \ref{thm2.4} and \ref{thm2.9} are satisfied. then all solutions of \eqref{first} oscillates. \end{corollary} \begin{example} \label{exa2.11} \rm By Theorem \ref{thm2.9}, \eqref{twentithird} has property A. \end{example} \begin{theorem} \label{thm2.12} Let $g(t)(l-1)!.(n-l-1)! \end{equation} holds, then \eqref{first} has property A. \end{theorem} \begin{proof} Proceeding as in the proof of Theorem \ref{thm2.3}, we arrive at \eqref{sixtin} for $t\geq t_{1}$. Then we can find a $t_{2}\geq t_{1}$ such that $t-g(t)>t_{1}$ for $t\geq t_{2}$. Hence from \eqref{sixtin}, we obtain \begin{equation*} y(t)\geq \frac{1}{(n-l-1)!.(l-1)!}\int^{t}_{t-g(t)}(t-s)^{l-1}\int^{\infty}_{g^{-1}(g^{-1}(s))}(u-s)^{n-l-1}p(u)y(g(u))\,du\,ds \end{equation*} which further yields \begin{equation*} 1\geq \frac{1}{(n-l-1)!.(l-1)!}\int^{t}_{t-g(t)}(t-s)^{l-1}\int^{\infty}_{g^{-1}(g^{-1}(s))}(u-s)^{n-l-1}p(u)\,du\,ds. \end{equation*} Taking limit sup. both sides in the above inequality, we obtain a contradiction. This completes the proof of the theorem. \end{proof} \begin{corollary} \label{coro2.13} Suppose that the conditions of Theorem \ref{thm2.4} and \ref{thm2.12} are satisfied. Then all solutions of \eqref{first} are oscillatory. \end{corollary} \begin{example} \label{exa2.14} \rm By Theorem \ref{thm2.12}, \eqref{tenthird} has property A. \end{example} Let $y(t)$ be a nonoscillatory solution of \eqref{first} such that \eqref{one} holds for $t\geq t_{1}$. Then for $t>t_{2}\geq 2t_{1}$, \eqref{one} gives \begin{equation*} y(t)\geq \frac{1}{(l-1)!}\int^{t}_{t/2}(t-s)^{l-1}y^{(l)}(s)\,ds, \quad t\geq t_{1}. \end{equation*} Using \eqref{sixth} and the above inequality, we obtain the following theorem. \begin{theorem} \label{thm2.15} Let $g(t)\leq t$. If for every $l\in \{1,2,\dots ,n-1\}$ such that $n+l$ is odd, \begin{equation*} \limsup_{t\to\infty}t^{l}\int^{\infty}_{g^{-1}(t)}(s-t)^{n-l-1}p(s)\,ds >(n-l-1)!.l!.2^{l} \end{equation*} holds, then \eqref{first} has property A. \end{theorem} \begin{theorem} \label{thm2.16} Let $g(t)\leq t$ and for every $l\in\{1,2,\dots ,n-1\}$ such that $n+l$ is odd, \begin{equation} \limsup_{t\to\infty}\int^{t}_{t/2}(t-s)^{l-1}\int^{\infty}_{g^{-1}(g^{-1}(s))}(u-s)^{n-l-1}p(u)\,du\,ds >(l-1)!.(n-l-1)! \end{equation} holds, then \eqref{first} has property A. \end{theorem} \begin{proof} Proceeding as in the proof of Theorem \ref{thm2.3}, we obtain \eqref{sixtin}. Then for $t\geq t_{2}>2t_{1}$, \eqref{sixtin} yields a contradiction. This completes the proof of the theorem. \end{proof} We note that when $g(t)=t/2$, then Theorem \ref{thm2.3}, \ref{thm2.12} and \ref{thm2.16} give same sufficient conditions to have property A of \eqref{first}. \begin{corollary} \label{coro17} Suppose that the conditions of Theorem \ref{thm2.4} are satisfied. If either of the conditions of Theorem \ref{thm2.15} or \ref{thm2.16} hold, then all solutions of \eqref{first} are oscillatory. \end{corollary} \begin{example} \label{exa2.18} \rm Consider \begin{equation*} y'''(t)+\frac{44}{t^{3}}y(\frac{3t}{5})=0, t\geq1. \end{equation*} Theorem \ref{thm2.1} and Theorem \ref{thm2.9} can be applied to this example, whereas Theorem \ref{thm2.15} cannot be applied to this example. \end{example} \begin{example} \label{exa2.19} \rm Consider \begin{equation*} y'''(t)+\frac{160}{t^{3}}y(\frac{t}{3})=0, t\geq1. \end{equation*} By Theorem \ref{thm2.15} this equation has property A, whereas Theorem \ref{thm2.9} fails. \end{example} \begin{theorem} \label{thm2.20} Let $g'(t)>0$. If for every $l\in\{1,2,3,\dots ,n-1\}$ such that $n+l$ is odd, \begin{equation}\label{twentyfourth} \int^{\infty}H_{l}(t)\,dt = \infty, \end{equation} then then for $n$ even every solution of \eqref{first} oscillates and for $n$ odd every solution of \eqref{first} is either oscillates or tend to zero as $t\rightarrow\infty$,in particular, \eqref{first} has property A, where \begin{equation}\label{twentyfifth} H_{n-1}(t)=t^{n-1}p(t)-\frac{(n-1)!.(n-1)2^{n-4}t^{n-3}}{g'(t)g^{n-2}(t)} \end{equation} and \begin{equation}\label{twentysixth} H_{l}(t) = \frac{t^{l}}{(n-l-2)!}\int^{\infty}_{t}(s-t)^{n-l-2}p(s)\,ds-\frac{l!.l.2^{l-3}t^{l-2}}{g'(t)g^{l-1}(t)}, \end{equation} for $l = 1,2,3,\dots ,n-2$. \end{theorem} \noindent \textbf{Remark:} Let $g(t)=t$ and $n=2$. From Theorem \ref{thm2.20}, it follows that, if \begin{equation}\label{twentyseventh} \int^{\infty}[tp(t)-\frac{1}{4t}]\,dt=\infty, \end{equation} then \begin{equation}\label{twentyeighth} y''+p(t)y=0 \end{equation} is oscillatory. This gives a partial answer to Problem \ref{prob1.1}. Further, our result improves the results due to Kneser \cite[pp.45]{swan} and Hille and Kneser \cite[Theorem 2.41]{swan}. We note that Theorem \ref{thm2.20} holds for \eqref{first} with $g(t)=t$ for $n=2$ and $n=3$. however, the theorem cannot be applied to higher order ordinary differential equations, viz., \eqref{first} with $g(t)=t$ and $n\geq 4$, because of the conditions \eqref{twentyfourth} and \eqref{twentysixth}. Now, suppose that $n=3$ and $g(t)=t$. then Theorem \ref{thm2.20} yields that, if \begin{equation*} \int^{\infty}[t^{2}p(t)-\frac{2}{t}]\,dt=\infty, \end{equation*} then \begin{equation}\label{twentynineth} y'''+p(t)y=0 \end{equation} has property A. On the other hand, from Hanan \cite[Theorem 5.7]{hana}, and Kiguradze and Chanturia \cite[Theorem 1.1]{kigu}, it follows that \eqref{twentynineth} has property A if \begin{equation}\label{thirty} \int^{\infty}[t^{2}p(t)-\frac{2}{3\sqrt{3}t}]\,dt=\infty. \end{equation} hence Theorem \ref{thm2.20} is yet to be improved. \begin{proof}[Proof of Theorem \ref{thm2.20}] If possible, suppose that \eqref{first} dose not have property A. Then \eqref{first} admits a nonoscillatory solution $y(t)$ such that $y\in N_{l}$ where $l\in \{1,2,3,\dots ,n-1\}$. We may assume, without any loss of generality, that $y(t)>0$ and $y(g(t))>0$ for $t\geq t_{1}>\sigma$. Clearly, \eqref{second} holds, where $l\in\{0,1,2,3,\dots ,n-1\}$ and $n+l$ odd. Let $l=n-1$. Set $z(t)=\frac{t^{n-1}y^{(n-1)}(t)}{y(g(t))}$. Then \begin{equation}\label{thirtyfirst} z'(t)=-t^{n-1}p(t)+\frac{n-1}{t}z(t)-g'(t) \frac{y'(g(t))}{y(g(t))}z(t). \end{equation} Putting $i=1$, $l=n-1$ in \eqref{fifth}, we obtain, for $t\geq t_{1}$ \begin{equation*} y'(t)\geq \frac{1}{(n-2)!}(t-t_{1})^{n-2}y^{(n-1)}(t). \end{equation*} Hence for $t\geq 2t_{1}$, we get \begin{equation*} y'(t)\geq \frac{t^{n-2}}{(n-2)!.2^{n-2}}y^{(n-1)}(t). \end{equation*} Thus, for $t\geq t_{2}>2t_{1}$, \begin{equation*} y'(g(t))\geq\frac{(g(t))^{n-2}}{(n-2)!.2^{n-2}}y^{(n-1)}(t) \end{equation*} Using the above inequality, \eqref{thirtyfirst} yields \begin{equation}\label{thirtysecond} z'(t)\leq -F_{n-1}(t), \end{equation} where \begin{equation*} F_{n-1}(t)= t^{n-1}p(t)-\frac{n-1}{t}z(t)+\frac{g'(t)(g(t))^{n-2}}{(n-2)!2^{n-2}t^{n-1}}z^{2}(t), \end{equation*} which as a function of $z$, attains the minimum $H_{n-1}(t)$ given in \eqref{twentyfifth}. Now, the integration of \eqref{thirtysecond} from $t_{2}$ to $t$ yields $z(t)<0$ for large $t$, a contradiction. Next, suppose that $l\in\{1,2,3,\dots ,n-2\}$. Setting $z_{1}(t)=\frac{t^{l}y^{(l)}(t)}{y(g(t))} , t\geq t_{1}$, we see that $z_{1}(t)>0$ for $t\geq t_{1}$ and \begin{equation}\label{thirtythird} z'_{1}(t)=\frac{t^{l}y^{(l+1)}(t)}{y(g(t))}+\frac{l}{t}z_{1}(t)- g'(t)\frac{y'(g(t))}{y(g(t))}z_{1}(t). \end{equation} Putting $i=1$ in \eqref{fifth}, we get \begin{equation*} y'(t)\geq \frac{1}{(l-1)!}(t-t_{1})^{l-1}y^{(l)}(t). \end{equation*} Thus, for $t\geq t_{2}\geq 2t_{1}$, \begin{equation*} y'(t)\geq \frac{1}{(l-1)!.2^{l-1}}t^{l-1}.y^{(l)}(t). \end{equation*} We can find a $t_{3}>t_{2}$ such that $g(t)>t_{2}$ for $t\geq t_{3}$. Hence \begin{equation}\label{thirtyfourth} y'(g(t))\geq \frac{1}{(l-1)!.2^{l-1}}(g(t))^{l-1}y^{(l)}(g(t))> \frac{1}{(l-1)!.2^{l-1}}(g(t))^{l-1}y^{(l)}(t) \end{equation} for $t\geq t_{3}$. Putting $i=l+1, k=n$ and $s >t\geq t_{3}$ in the inequality \begin{equation}\label{thirtyfifth} y^{(i)}(t)=\sum^{k-1}_{j=i}\frac{(t-s)^{j-i}}{(j-i)!}y^{(j)}(s)+\frac{1} {(k-i-1)!}\int^{t}_{s}(t-u)^{k-i-1}y^{(k)}(u)\,du, \end{equation} and letting $s\rightarrow\infty$, we obtain \begin{equation}\label{thirtysixth} y^{(l+1)}(t)\leq -\frac{y(g(t))}{(n-l-2)!}\int^{\infty}_{t}(s-t)^{n-l-2}p(s)\,ds. \end{equation} Making the use of \eqref{thirtyfourth} and \eqref{thirtysixth} in \eqref{thirtythird}, we have \begin{equation}\label{thirtyseventh} z'_{1}(t)\leq -F_{l}(t), \end{equation} where \begin{equation*} F_{l}(t)=\frac{g'(t).g^{l-1}(t)}{(l-1)!.2^{l-1}.t^{l}}z^{2}_{1}(t) -\frac{l}{t}z_{1}(t)+\frac{t^{l}}{(n-l-2)!}\int^{\infty}_{t}(s-t)^{n-l-2}p(s)\,ds, \end{equation*} which as a function of $z_{1}$, attains the minimum $H_{l}(t)$ given in \eqref{twentysixth}. In view of the conditions \eqref{twentyfourth} and \eqref{twentysixth}, integration of \eqref{thirtyseventh} yields a contradiction. Hence \eqref{first} has property A, that is $l=0$ for $t\geq t_{2}\geq t_{1}$.. Thus the theorem is proved when $n$ is even. Now $l=0$ implies that $n$ is odd. Our theorem will be proved if we can show that $y(t)\rightarrow 0 $ as $t\rightarrow\infty$. Since $l=0$ then $\lim y(t)=\lambda, 0\leq\lambda <\infty$ exists. We claim that $\lambda=0$. If not, them for $0<\epsilon<\lambda$, there exists a $t_{3}\geq t_{2}$ such that $y(g(t)) > \lambda-\epsilon$ for $t\geq t_{3}$. Now putting $i=0,k=n$ and $s>t=t_{3}$ and letting $s\rightarrow\infty$ in \eqref{thirtyfifth}, we obtain \begin{equation*} y(t_{3})>(\lambda-\epsilon)\int^{\infty}_{t_{3}}(u-t_{3})^{n-1}p(u)\,du \end{equation*} which further gives \begin{equation}\label{thirtiseveth} \int^{\infty}_{t_{3}}(u-t_{3})^{n-1}p(u)\,du <\infty. \end{equation} On the other hand, the condition \eqref{twentyfourth} with $l=n-1$ yields that $\int^{\infty}_{t_{3}}t^{n-1}p(t)\,dt=\infty$ which contradicts to \eqref{thirtiseveth}. Hence $\lambda=0$. This completes the proof of the theorem. \end{proof} \begin{example} \label{exa2.21} \rm Consider \begin{equation}\label{thirtyeight} y'''(t)+\frac{24(t-1)^{2}}{t^{5}}y(t-1)=0 , \quad t\geq 2. \end{equation} All the conditions of Theorem \ref{thm2.20} are satisfied. Hence \eqref{thirtyeight} has property A. In particular, $y(t)=1/t^{2}$ is a nonoscillatory solution of \eqref{thirtyeight}. \end{example} \begin{corollary} \label{coro2.22} Suppose that the conditions of Theorems \ref{thm2.4} and \ref{thm2.20} are satisfied. Then all solutions of \eqref{first} are oscillatory. \end{corollary} Now, we consider the following ordinary differential equations associated with the delay differential equations \eqref{eighteen}, \eqref{nineteen}, \eqref{twentith}, and \eqref{thirtyeight}. \begin{gather}\label{thirtinineth} y'''+\frac{30}{t^{3}}y=0, \quad t\geq 2. \\ \label{fourty} y'''+\frac{82}{t^{3}}y=0, \quad t\geq 1. \\ \label{fortyfirst} y'''+\frac{63}{t^{3}}y=0, \quad t\geq 1. \\ \label{fortysecond} y'''+\frac{24(t-1)^{2}}{t^{3}}y=0, \quad t\geq 2. \end{gather} From Hanan \cite[Theorem 5.7]{hana}, it follows that \eqref{thirtinineth}-\eqref{fortysecond} are oscillatory. We note that a third order ordinary differential equation is said to be oscillatory if it has an oscillatory solution ; otherwise,it is called nonoscillatory. However, all solutions of \eqref{thirtinineth}-\eqref{fortysecond}are not oscillatory. This is because, \eqref{thirtinineth}-\eqref{fortysecond} are of Class I or $C_{I}$ and hence admits a nonoscillatory solution (see Lemma 2.2 and Theorem 3.1 in \cite{pari}). We may note that Eq.\eqref{twentynineth} is said to be of Class I or $C_{I}$ if any of its solution $y(t)$ for which $y(t_{0})=y'(t_{0})=0$ and $y''(t_{0})>0, (\sigma0$ for $t\in[\sigma,t_{0})$. It seems that the presence of delay in \eqref{nineteen} and \eqref{twentith} is responsible for the change in the qualitative behaviour of solutions of the equations. It is easy to construct an example of a third order delay differential equation all solutions of which are oscillatory but it is not difficult to construct such an example of a third order ordinary differential equation. It is evident from the following examples due to Dolan \cite{dola} and Parhi and the author \cite{parh} respectively. \begin{example}Dolan \cite{dola}] \label{2.23}\rm All solutions of \begin{equation*} \{[z'-\frac{r'(t)}{r(t)}z]+r(t)z\}'=0 \end{equation*} are oscillatory, where $r(t)=[1+\sqrt{2}\epsilon\sin(t+\frac{\pi}{4})]^{-1}>0$, $t\geq 0, 0<\epsilon<\frac{1}{\sqrt{2}}$. \end{example} To the best of the authors knowledge, the following is the only explicit example of which all solutions are oscillatory. \begin{example}[Parhi and Padhi \cite{parh}] \label{exa2.24} \rm All solutions of \begin{equation*} y'''-y''+\big(\frac{1}{1.0000004}+\frac{1}{t}\big)y'-\frac{k}{t^{2}}y=0, \quad t\geq 2 \end{equation*} are oscillatory, where $k$ is a constant. \end{example} \begin{theorem} \label{thm2.25} Let $n\geq3$. Suppose that for any $\mu\in(0,1/2)$, each of the the third order ordinary differential equation \begin{equation}\label{fortythird} u'''+G_{l}(t)u=0 , \quad i\in \{1,2,\dots ,n-1\}, \; n+l \mbox{ odd} \end{equation} admits an oscillatory solution, where \begin{equation}\label{fortyfourth} G_{n-1}(t)=\frac{\mu}{(n-3)!}(g(t)-g(g(t)))^{n-3}\big(\frac{g(t)}{t}\big)^{2}p(t) \end{equation} and \begin{equation}\label{fortyfifth} \begin{aligned} G_{l}(t)&=\frac{\mu}{(n-l-2)!.(l-1)!} \Big(\int^{g^{-1}(t)}_{t}(s-t)^{n-l-2}p(s)\,ds\Big) \\ &\quad\times \big(g(t)-g(g(t))\big)^{l-2}\big(\frac{g(g(t))}{t}\big)^{2} \end{aligned} \end{equation} for $l\in\{1,2,3,\dots ,n-2\}$. Then \eqref{first} has property A. \end{theorem} \begin{proof} If \eqref{first} has not property A, then it admits a non-oscillatory solution $y(t)$ such that \eqref{second} is satisfied for $l\in\{1,2,3,\dots ,n-1\}$. We may assume, without any loss of generality, that $y(t)>0$ and $y(g(t))>0$ for some $t\geq t_{1}>t_{0}>\sigma$. Let $l=n-1$. Setting $x(t)=y^{(n-3)}(t)$, we see that $x(t)>0,x'(t)>0,x''(t)>0$ and $x'''(t)<0$ for $t\geq t_{2}\geq t_{1}$. For any $\mu\in(0,1/2)$, there exists a $T\_{\mu}\geq>t_{2}$ such that \begin{equation}\label{fortysixth} \frac{x(g(t))}{x(t)} \geq \mu \big(\frac{g(t)}{t}\big)^{2} \end{equation} for $t\geq T_{\mu}$ (See Theorem 2.2 in \cite{erbe}). Setting $z(t)=x'(t)/x(t)$ for $t\geq T_{\mu}$, we get \begin{equation}\label{fortyseventh} z'(t)=\frac{x''(t)}{x(t)}-z^{2}(t). \end{equation} Further, assuming $u(t)=\exp\big(\int^{t}_{T_{\mu}}z(s)\, ds\big)$ and using \eqref{fortysixth}, \eqref{fortyseventh} and the inequality \begin{equation*} y(t)\geq \frac{y^{(n-3)}(t)}{(n-3)!}(t-g(t))^{n-3}, \end{equation*} we obtain \begin{equation*} u'''(t)+\frac{\mu}{(n-3)!}(g(t)-g(g(t)))^{n-3}\big(\frac{g(t)}{t}\big)^{2} p(t)u(t)\leq 0 \end{equation*} for $t\geq T_{\mu}$. From Lemma 4 in \cite{greg}, it follows that \eqref{fortythird} with $l=n-1$ is disconjugate on $[T_{\mu},\infty)$, a contradiction. Next let $l\in\{1,2,3,\dots ,n-2\}$. Putting $i=l+1,k=n$ and $s=g^{-1}(t)>t_{1}$ in \eqref{thirtyfifth}, we get \begin{equation*} y^{(l+1)}(t)+\frac{1}{(n-l-2)!}\Big(\int^{g^{-1}(t)}_{t}(s-t)^{n-l-2}p(s)\,ds \Big)y(g(t))\leq 0. \end{equation*} for $t\geq T\geq t_{1}$, which further gives, for $t\geq T$ \begin{equation}\label{fortynineth} \begin{aligned} y^{(l+1)}(t)+\frac{1}{(n-l-2)!.(l-1)!} \Big(\int^{g^{-1}(t)}_{t}(s-t)^{n-l-2}p(s)\,ds\Big)&\\ \times (g(t)-g(g(t)))^{l-2}y^{(l-2)}(w(t))&\leq 0 \end{aligned} \end{equation} where $g(g(t))=w(t)$. Let $x_{1}(t)=y^{(l-2)}(t)$. Then $x_{1}(t)>0$, $x'_{1}(t)>0$, $x''_{1}(t)>0$ and $x'''_{1}(t)<0$ for $t\geq T$ and hence we can find a $t\geq T_{\mu}>T$ such that \begin{equation*} \frac{x_{1}(w(t))}{x_{1}(t)}\geq \mu \big(\frac{w(t)}{t}\big)^{2}; \end{equation*} that is, \begin{equation}\label{fifty} \frac{y^{(l-2)}(w(t))}{y^{(l-2)}(t)}\geq \mu \big(\frac{w(t)}{t}\big)^{2}, \end{equation} for $t\geq T_{\mu'}$. Then $z'_{1}(t)=\frac{x''_{1}(t)}{x_{1}(t)}-z^{2}_{1}(t)$. Further, setting $v(t)=e^{\left(\int^{t}_{T_{\mu'}}z_{1}(s)\,ds\right)}$ and using \eqref{fifty},\eqref{fortynineth} gives \begin{equation*} \begin{aligned} v'''(t)+ \frac{\mu}{(n-l-2)!.(l-1)!}\Big(\int^{g^{-1}(t)}_{t}(s-t)^{n-l-2}p(s)\,ds\Big)&\\ \times (g(t)-g(g(t)))^{l-2}\big(\frac{w(t)}{t}\big)^{2}v(t)&\leq 0 \end{aligned} \end{equation*} for $t\geq T_{\mu'}$.This in turn implies that \eqref{fortythird} is disconjugate, by in \cite[Lemma 4 ]{greg}, a contradiction to the hypothesis of the theorem for the case $l\in \{1,2,3,\dots ,n-2\}$. Hence \eqref{first} has property A. This completes the proof of the theorem. \end{proof} \begin{corollary} \label{coro2.26} Suppose that $g(t)\frac{2}{3\sqrt{3}}$, then, for every $\mu\in(0,1/2)$, the equation \begin{equation*} u'''+\mu e^{-1}\big(\frac{t-1}{t}\big)^{2}u=0 , \quad t\geq 2 \end{equation*} admits an oscillatory solution by Theorem 5.7 of \cite{hana}. from Theorem \ref{thm2.25}, it follows that \eqref{fiftyone} has property A. In particular, $y(t)=e^{-t}$ is a solution of \eqref{fiftyone} for $t\geq 2$. \end{example} \noindent\textbf{Remark:} Consider Equations \eqref{nineteen} and \eqref{twentith}. For $0<\mu <\frac{\sqrt{3}}{82}$, it follows that $\lim_{t\to\infty}t^{3}\mu \frac{82}{9t^{3}}<\frac{2}{\sqrt{3}}$ and hence $u'''+\mu\frac{82}{9t^{3}}u=0$ is nonoscillatory, by \cite[Theorem 5.7]{hana}. Similarly, for $0<\mu <\frac{4}{189\sqrt{3}}$, the equation $u'''+\mu\frac{63}{4t^{3}}u=0$is nonoscillatory. 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