\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2005(2005), No. 134, pp. 1--18.\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-2005/134\hfil Oscillation criteria] {Oscillation criteria for first-order nonlinear neutral delay differential equations} \author[E. M. Elabbasy, T. S. Hassan, S. H. Saker\hfil EJDE-2005/134\hfilneg] {Elmetwally M. Elabbasy, Taher S. Hassan, Samir H. Saker} % in alphabetical order \address{Elmetwally M. Elabbasy\hfill\break Department of Mathematics\\ Faculty of Science\\ Mansoura University\\ Mansoura, 35516, Egypt} \email{emelabbasy@mans.edu.eg} \address{Taher S. Hassan \hfill\break Department of Mathematics\\ Faculty of Science\\ Mansoura University\\ Mansoura, 35516, Egypt} \email{tshassan@mans.edu.eg} \address{Samir H. Saker \hfill\break Department of Mathematics\\ Faculty of Science\\ Mansoura University\\ Mansoura, 35516, Egypt} \email{shsaker@mans.edu.eg} \date{} \thanks{Submitted August 2, 2005. Published November 30, 2005.} \subjclass[2000]{34K15, 34C10} \keywords{Oscillation; non-oscillation; neutral delay of differential equations} \begin{abstract} Oscillation criteria are obtained for all solutions of first-order nonlinear neutral delay differential equations. Our results extend and improve some results well known in the literature. Some examples are considered to illustrate our main results. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{example}[theorem]{Example} \allowdisplaybreaks \section{Introduction} In recent years, the literature on the oscillation of neutral delay differential equations has grown very rapidly. It is a relatively new field with interesting applications in real world life problems. In fact, neutral delay differential equations appear in modelling of the networks containing lossless transmission lines (as in high-speed computers where the lossless transmission lines are used to interconnect switching circuits), in the study of vibrating masses attached to an elastic bar, as the Euler equation in some variational problems, in the theory of automatic control and in neuro-mechanical systems in which inertia plays an important role. See Hale \cite{13ag}, Driver \cite{7ag}, Brayton and Willoughby \cite{6ag}, Popove \cite{24ag}, and Boe and Chang \cite{5ag}, and the references cited therein. Also this is evident by the number of references in the recent books by Ladde et al. \cite{b1} and by Ladas \cite{b2}. We consider a general first-order nonlinear neutral delay differential equation \begin{equation} ( x( t) -q( t) x( t-r) )'+f( t,x( \tau ( t) ) ) =0, \label{3.7} \end{equation} where for $t\geq t_0$ \begin{gather} q,\tau \in C([ t_0,\infty ) ,\mathbb{R}^+),\; q(t) \neq 1,\; r\in ( 0,\infty ) ,\; \tau (t) 0$ such that \begin{itemize} \item[(i)] $g$ is nondecreasing on $\mathbb{R}^+$ \item[(ii)] $g( -u) =g( u)$, $\lim_{u\to 0}g( u)=0$, \item[(iii)] $\int_0^\infty g( e^{-u}) du<\infty$, \item[(iv)] $\frac{1}{|u|}| f(t,u)-p( t) u| \leq p( t) g( u)$ for $t\geq t_0$ and $0<|u| <\varepsilon _0$, \item[(v)] For each $\varphi \in C([ t_0,\infty) ,R) $ with $\lim_{t\to \infty }\varphi (t) >0$, \begin{equation*} \int_{t_0}^\infty \int f( t,\varphi ( \tau ( t) ) ) dt=\infty ,\quad \int_{t_0}^\infty f( t,-\varphi ( \tau ( t) )) dt=-\infty . \end{equation*} \end{itemize} \end{itemize} As usual a solution $x( t) $ of equation \eqref{3.7} is said to be oscillatory if it has arbitrarily large zeros in $[t_0,\infty )$. Otherwise it is nonoscillatory and the equation \eqref{3.7} is called oscillatory if every solution of this equation is oscillatory. When $q( t) =0$, \eqref{3.7} reduces to \begin{equation*} x'( t) +f( t,x( \tau ( t) ) ) =0, \end{equation*} which was studied by Tang and Shen \cite{t1}. They obtained some infinite integral sufficient conditions for oscillations. The oscillatory behavior of other neutral delay differential equations have been investigated by many authors, see \cite{1ag,2ag,3ag,4ag,8ag,s9,10ag,11ag,b1,non, b2,16ag,s1,s16,17ag,18ag,19ag,20ag,21ag, B7,22ag,23ag, A21,25ag,27ag,28ag,29ag,30ag} and references therein. In recent papers Elabbasy and Saker \cite{s9}, Kubiaczyk and Saker \cite{s16} obtained an infinite integral conditions for oscillation of the linear neutral delay differential equation \begin{equation*} ( x( t) -q( t) x( t-r) )'+p( t) x( t-\tau ) =0\,. \end{equation*} Let $\delta ( t) =\max \{ \tau ( t) :t_0\leq s\leq t\} $ and $\delta ^{-1}( t) =\min \{ s\geq t_0:\delta ( s) =t\}$. Clearly, $\delta $ and $\delta ^{-1}$ are non-decreasing and satisfy \begin{itemize} \item[(A)] $\delta ( t) t$ \item[(B)] $\delta ( \delta ^{-1}( t) )=t$ and $\delta ^{-1}( \delta ( t) ) \leq t$. \end{itemize} Let $\delta ^{-k}( t) $ be defined on $[t_0,\infty) $ by \begin{equation} \delta ^{-( k+1) }( t) =\delta ^{-1}( \delta ^{-k}( t) ) ,\quad k=1,2,\dots \label{3.12} \end{equation} Throughout this paper, we use the sequence $\{ p_{k}\} $, of functions defined by \begin{gather*} p_{1}( t) =\int_t^{\delta ^{-1}( t) } p( s) ds,\quad t\geq t_0, \\ p_{k+1}( t) =\int_t^ {\delta ^{-1}( t) } p( s) p_{k}( s) ds,\quad t\geq t_0,\;k=1,2,\dots \end{gather*} Our main results are the following. \begin{theorem} \label{thm1} Assume that \eqref{3.8}, \eqref{1.22**}, \eqref{3.9}, and \textrm{(H)} hold, and there exist a bounded positive function $\sigma(t)$ such that \begin{equation} \int_{\tau ( t)}^t B( s) ds> \frac{1}{e}, \label{3.13*} \end{equation} and \begin{equation} \int_{t_0}^\infty p( t) \sigma (t)\Big[ \exp \Big( \int_{\tau ( t) }^t p( s) ds-\frac{\sigma ( t) }{e}\Big) -1\Big] dt=\infty , \label{3.14*} \end{equation} where $B( t) = p( t)/ \sigma ( t)$. Then every solution of \eqref{3.7} oscillates. \end{theorem} \begin{theorem} \label{thm2} Assume that \eqref{3.8}, \eqref{1.22**}, \eqref{3.9} and \textrm{(H)} hold, and that \begin{equation} \liminf_{t\to \infty } \int_{\tau ( t) } ^t p( s) ds\geq 0. \label{3.15*} \end{equation} and suppose that there exists a positive integer $n$ such that \begin{equation} \int_{t_0}^\infty p( t) \ln (e^{n-1}p_{n}( t) +1) dt=\infty . \label{3.16} \end{equation} Then every solution of \eqref{3.7} oscillates. \end{theorem} \begin{corollary} \label{coro1} Assume that \eqref{3.8} \eqref{3.9}, \eqref{1.22**}, \eqref{3.15*} and \textrm{(H)} hold, and that \begin{equation} \int_{t_0}^\infty p( t)\Big[ \exp \Big( \int_{\tau ( t) }^t p( s) dsBig) -1\Big] dt=\infty . \label{3.14} \end{equation} Then every solution of \eqref{3.7} oscillates. \end{corollary} \begin{corollary} \label{coro2} Assume that \eqref{3.8}, \eqref{3.9}, \eqref{1.22**}, \eqref{3.15*} and \textrm{(H)} hold, and that \begin{equation*} \int_{t_0}^\infty p( t) \ln \Big( \int_t ^{\delta ^{-1}( t) } p( s) ds+1\Big) dt=\infty . \end{equation*} Then every solution of \eqref{3.7} oscillates. \end{corollary} \begin{corollary} \label{coro3} Assume that \eqref{3.8}, \eqref{3.9}, \eqref{1.22**}, \eqref{3.15*} and \textrm{(H)} hold, and suppose that there exists a positive integer $n$ such that \begin{equation*} \int_{t_0}^\infty p( t) \ln (e^{n}p_{n}( t) ) dt=\infty . \end{equation*} Then every solution of \eqref{3.7} oscillates. \end{corollary} %\textbf{Remark.\ } Note that if \begin{equation*} \limsup_{t\to \infty } \int_{\tau ( t) }^t p( s) ds>2, \end{equation*} then by Lemma \ref{lem4} every solution of \eqref{3.7} oscillates. Thus, we will consider the case \begin{equation*} \limsup_{t\to \infty } \int_{\tau ( t) }^t p( s) ds\leq 2. \end{equation*} This implies that for some $\epsilon >0$ and large $t$, \begin{equation*} \int_{\tau ( t) }^t p( s) ds\leq 2+\epsilon \,. \end{equation*} Thus we have \begin{equation*} \liminf_{t\to \infty } p_{k}( t) \leq ( 2+\epsilon ) ^{k-1}\liminf_{t\to \infty } \int_{t} ^{\delta ^{-1}( t) } p( s) ds \leq ( 2+\epsilon ) ^{k-1}\liminf_{t\to \infty } \int_{\tau ( t)}^t p( s) ds. \end{equation*} As a result, by Theorem \ref{thm2} we have \begin{corollary} \label{coro4} Assume that \eqref{3.8}, \eqref{3.9}, \eqref{1.22**} and \textrm{(H)} hold, and that there exists a positive integer $n$ such that \begin{equation*} \lim_{t\to \infty }\inf p_{n}( t) >0. \end{equation*} Then every solution of \eqref{3.7} oscillates. \end{corollary} The proofs of the above Theorems and also some Lemmas to be used in these proofs will be given in the next two sections. Some examples which illustrate and the advantage of our results will be given in section 4. \section{Preliminary Lemmas} \begin{lemma} \label{lem1} Assume that \eqref{3.8}, \eqref{1.22**} and \eqref{3.9} hold. Let $x(t) $ be an eventually positive solution of \ref{3.7} and set \begin{equation} z( t) =x( t) -q( t) x( t-r) . \label{1.22*} \end{equation} Then $z( t) $ is eventually nonincreasing and positive function. \end{lemma} \begin{proof} From \eqref{3.7}), \eqref{3.9}, we have $z'( t) =-f( t,x( \tau (t) ) ) \leq 0$ eventually. We prove that $z(t) $ is a positive function. If not, then there exist $T\geq t_0$ and $\alpha <0$ such that $z( t) <\alpha $ for $t\geq T$. Then from \eqref{1.22*}, we have $x( t) <\alpha +q( t) x( t-r)$ which implies \begin{equation*} x( t+r) <\alpha +q( t+r) x( t) . \end{equation*} Now we choose $k$ such that $t_{k}=t^{*}+kr>T$. Then $x(t_{k+1}) <\alpha +q( t_{k+1}) x( t_{k}) $. Applying this inequality by induction, it gives \begin{equation*} x( t_{n}) <\alpha\Big[ 1+\sum_{i=k+2}^{n}\prod_{j=0}^{n-i}q% ( t_{n-j}) \Big] +\prod_{i=k+1}^{n}q( t_{i}) x(t_{k}) . \end{equation*} Now define $q_{n}$ and $d_{n}$ by \begin{equation*} q_{n}=1+\sum_{i=k+2}^{n}\prod_{j=0}^{n-i}q( t_{n-j}), \quad d_{n}=\prod_{i=k+1}^{n}q( t_{i}) , \end{equation*} and let \begin{equation*} s_{n}=\sum_{i=1}^{n}\prod_{j=1}^{i}\frac{1}{q( t_{j}) }. \end{equation*} Then \begin{equation*} s_{n}^{*}=\frac{q_{n}}{d_{n}}=\Big( s_{n}-\sum_{i=1}^{k+1}\prod_{j=1}^{i}% \frac{1}{q( t_{j}) }\Big) q( t_{k+1}) \dots q(t_{1}) \to \infty \quad \text{as }n\to \infty , \end{equation*} by condition \eqref{1.22**}. Using the above inequality, \begin{equation*} x( t_{n}) <\big[ s_{n}^{*}+\frac{x( t_{k}) }{\alpha }% \big] \alpha d_{n}\to -\infty \quad \text{as }n\to \infty , \end{equation*} and this contradicts the assumption that $x( t) >0$. Then $z( t) $ must be positive function. The proof is complete. \end{proof} Note that the proof of Lemma \ref{lem1} is similar to that in \cite[Lemma 1]{N8}; we state it here for the sake of completeness. \begin{lemma} \label{lem2} Assume that \eqref{3.8}, \eqref{3.9}, \eqref{1.22**} and \textrm{(H)} hold. Then every non-oscillatory solution of \eqref{3.7} converges to zero monotonically for large $t$ as $t\to \infty $. \end{lemma} \begin{proof} Suppose that $x( t)$ is a non-oscillatory solution of equation \eqref{3.7} which we shall assume to be eventually positive [If $x( t) $ is eventually negative the proof is similar]. From Lemma \ref{lem1}, we have $z( t) $ is eventually non-increasing and positive function. Choose a $t_{1}\geq t_0$ such that $x( t) >0$, $z( t)>0$ for $t\geq t_{1}$. It follows from equations \eqref{3.7}-\eqref{3.9} and \textrm{(H)} that there exists $t_{2}>t_{1}$ such that $\tau ( t) \geq t_{1}$ and $z'( t) \leq 0$ for $t>t_{2}$. Hence the following limits exist and \begin{equation*} \lim_{t\to \infty }x( t) \geq \lim_{t\to \infty }z( t) =\alpha \geq 0\,. \end{equation*} If $\alpha >0$, then from \eqref{3.7} we have \begin{equation*} z( t) -z(t_0)=-\int_{t_0} ^t f(t,x(\tau ( s) )) ds. \end{equation*} It follows from assumption \textrm{(H)(v)} that $\lim_{t\to \infty }z( t) =-\infty $, which contradicts that $z(t)$ being positive function, then $\alpha =0$, from \eqref{3.8}, we have $\lim_{t\to \infty }x(t) =0$. The proof of Lemma \ref{lem2} is complete. \end{proof} \begin{lemma} \label{lem3} Assume that \eqref{3.8}, \ref{3.9}, \eqref{1.22**} and \textrm{(H)} hold. If $x( t) $ is a non-oscillatory solution of \eqref{3.7}, then there exist $A>0$, $\varepsilon >0$ and $T\in ( 0,\infty ) $ such that for $t\geq T$, \begin{equation} | x( t) | \leq A\exp \Big( -\frac{1}{2} \int_T^t p( s) ds\Big) +\varepsilon , \label{4.1} \end{equation} \end{lemma} \begin{proof} We shall assume $x( t) $ to be eventually positive [If $x(t)$ is eventually negative the proof is similar]. By Lemma \ref{lem2}, there exists $t_{1}>0$ such that \begin{equation*} 0t_{1}$ such that for $t\geq T$, \begin{equation*} f( t,x(\tau ( t) )) \geq \frac{1}{2}p( t) x( \tau ( t) ) \geq \frac{1}{2}p( t) x( t) , \end{equation*} and it follows from \eqref{3.7} that for $t\geq T$, \[ ( x( t) -q( t) x( t-r) )'+\frac{1}{2}p( t) x( t) \leq 0, \\ z'( t) +\frac{1}{2}p( t) z( t) \leq 0, \] where $z( t) =x( t) -q( t) x(t-r) $. This yields, for $t\geq T$, \begin{gather*} z( t) \leq A\exp\Big[ -\frac{1}{2}\int_T^t p( s) ds\Big] , \\ | x( t) | \leq A\exp \Big( -\frac{1}{2}\int_T^t p( s) ds\Big) +\varepsilon , \end{gather*} where $A=x( T) -q( T) x( T-r) $. \end{proof} \begin{lemma} \label{lem4} Assume that \eqref{3.8}, \ref{3.9}, \eqref{1.22**} and \textrm{(H)} hold. If equation \eqref{3.7}) has a nonoscillatory solution, then \begin{equation} \int_{\tau ( t) }^t p( s) ds\leq 2\quad \text{and}\quad p_{k}( t) \leq 2^{k},\quad k=1,2,\dots \label{4.2} \end{equation} eventually. \end{lemma} \begin{proof} Suppose that $x(t)$ is a nonoscillatory solution of equation \eqref{3.7} which we shall assume to be eventually positive [if $x(t)$ is eventually negative the proof is similar]. By Lemma \ref{lem2}, there exists $T\geq 0$ such that \begin{gather} x( \tau ( t) ) \geq x( t) >0\quad \text{for }t\geq T, \nonumber\\ ( x( t) -q( t) x( t-r) )'+\frac{1}{2}p( t) x( \tau ( t)) \leq 0, \nonumber\\ z'( t) +\frac{1}{2}p( t) z( \tau ( t) ) \leq 0\quad \text{for }t\geq T. \label{@} \end{gather} Integrating both sides from $\tau ( t) $ to $t$ yields \begin{equation*} z( t) -z( \tau ( t) ) +\frac{1}{2}\int_{\tau ( t) }^t p( s) z( \tau ( s) ) ds\leq 0\quad \text{for }t\geq T. \end{equation*} By the decreasing nature of $z( t) $ for large $t$ and the increasing nature of $\tau ( t)$, there exists $T_{1}\geq T$ such that \begin{equation*} z( t) -z( \tau ( t) ) +\frac{1}{2}z( \tau ( t) ) \int_{\tau(t)}^t p( s) ds\leq 0\quad \text{for }t\geq T_{1}. \end{equation*} Then, $ \int_{\tau ( t) }^t p( s) ds\leq 2$. Also, integrating both sides of equation \eqref{@} from $t$ to $\delta ^{-1}( t) $ yields \begin{equation*} z( \delta ^{-1}( t) ) -z( t) +\frac{1}{2} \int_t {\delta ^{-1}( t) } p( s) z( \tau ( s) ) ds\leq 0\quad \text{for }t\geq T. \end{equation*} By the decreasing nature of $z( t) $ for large $t$ and the increasing nature of $\tau (t) $, there exists $T_{1}\geq T$ such that \begin{equation*} z( \delta ^{-1}( t) ) -z( t) +\frac{1}{2} \Big( \int_t^{\delta ^{-1}( t)} p(s) ds\Big) z( \tau ( \delta ^{-1}( t) )) \leq 0\quad \text{for } t\geq T_{1}. \end{equation*} or \begin{equation*} z( \delta ^{-1}( t) ) -z( t) +\frac{1}{2} \Big( \int_t^{\delta ^{-1}( t) } p(s) ds\Big) z( t) \leq 0\quad \text{for } t\geq T_{1}. \end{equation*} Then, we have \begin{equation*} p_{1}( t) =\int_t^{\delta ^{-1}( t) } p( s) ds\leq 2. \end{equation*} By iteration we deduce, from this, that $p_{k}( t) \leq 2^{k}$ which shows that \eqref{4.2} holds for $t\geq T_{1}$. The proof of Lemma \ref{lem4} is complete. \end{proof} \begin{lemma} \label{lem5} Assume that \eqref{3.8}, \eqref{3.9}, \eqref{3.15*}, \eqref{1.22**} and \textrm{(H)} hold. If $x( t) $ is a nonoscillatory solution of equation \eqref{3.7}), then $\frac{z( \tau ( t) ) }{z( t) }$ is well defined for large $t$ and is bounded. \end{lemma} \begin{proof} Suppose that $x(t)$ is a nonoscillatory solution of equation \eqref{3.7} which we shall assume to be eventually positive [if $x(t)$ is eventually negative the proof is similar]. By the same argument as in the proof of Lemma \ref{lem3}, there exists $T>0$, such that \begin{gather*} x( \tau ( t) ) \geq x( t) >0\quad \text{for } t\geq T, \\ ( x( t) -q( t) x( t-\sigma ) )'+\frac{1}{2}p( t) x( \tau ( t)) \leq 0, \\ z'( t) +\frac{1}{2}p( t) z( \tau ( t) ) \leq 0\quad \text{for }t\geq T. \end{gather*} The rest of the proof is similar to in \cite[Lemma 5]{B7}, and thus it is omitted. \end{proof} \section{Proofs of Theorems} \subsection*{Proof of Theorem \ref{thm1}} Assume that \eqref{3.7} has a nonoscillatory solution $x( t) $ which will be assumed to be eventually positive (if $x( t) $ is eventually negative the proof is similar). By Lemma \ref{lem2}, there exists $t_{1}\geq t_0$ such that \begin{equation} 0t_{1}$ be such that $\tau ( t) \geq t_{1}$ for $t\geq t_{2}$. Integrating both sides of \eqref{7.3} from $\tau( t) $ to $t$, we obtain \begin{equation} \omega ( t) \geq \sigma ( t) \exp \Big( \int_{\tau ( t) }^t B( s) \omega (s)[ 1-g( x( \tau ( s) ) ) ] ds\Big) ,\quad t\geq t_{2}. \label{7.4} \end{equation} By \eqref{3.13*}, for $t\geq t_{2}$, we have \begin{equation} \int_{\delta ( t) }^t p( s) ds= \int_{\tau ( t^{\ast }) }^t p( s)ds\geq \int_{\tau ( t^{\ast })}^{t^{\ast }} p( s) ds\geq e^{-1}, \label{7.5} \end{equation} where $t^{\ast }\in[ t_0,t] $ with $\tau ( t^{\ast}) =\delta ( t) $. From \eqref{3.13*} and \eqref{7.4}), we find that for $t\geq t_{2}$, \begin{align*} \omega ( t) &\geq \sigma ( t) \exp \Big( \int_{\tau ( t) }^t B( s) ( \omega ( s) -\sigma ( t) ) ds+\frac{\sigma (t) }{e}\Big) \\ &\quad \times \exp \Big( \int_{\tau ( t) }^t p( s) ds-\frac{\sigma ( t) }{e}\Big) \exp \exp\Big( -\int_{\tau ( t) }^t B( s) \omega( s) g( x( \tau ( s) ) ) ds\Big) \\ &\geq \sigma ( t) \Big( e\int_{\delta ( t) }^t B( s) ( \omega ( s) -\sigma ( t) ) ds+\sigma ( t) \Big) \exp \Big(\int_{\tau ( t) }^t p( s) ds- \frac{\sigma ( t) }{e}\Big) \\ &\quad \times \exp \Big( -\int_{\tau(t)}^t B( s) \omega ( s) g( x( \tau ( s) ) ) ds\Big) . \end{align*} Let $\upsilon ( t) =\omega ( t) -\sigma (t) $ for $t\geq t_{1}$. Then $\upsilon ( t) \geq 0$ for $t\geq t_{1}$, and so for $t\geq t_{2}$, \begin{align*} &\upsilon ( t) -e\int_{\delta(t)}^t B( s) \upsilon ( s) ds\\ &\geq\Big( e\int_{\delta ( t)}^t B( s) \upsilon (s) ds+\sigma ( t)\Big) \Big[\sigma ( t) \exp\Big(\int _{\tau(t)}^t p( s) ds -\frac{\sigma (t) }{e}\Big) \\ &\quad\times \exp \Big( -\int_{\tau(t)}^tB( s) \omega ( s) g( x( \tau (s) ) ) ds\Big) -1\Big] , \end{align*}% that is, for $t\geq t_{2}$, \begin{equation} \begin{aligned} &B( t) \upsilon ( t) -B( t) e\int_{\delta ( t) }^t B( s) \upsilon (s) ds\\ &\geq B( t) \Big( e\int_{\delta(t)}^t B( s) \upsilon ( s) ds+\sigma ( t) \Big) \Big[ \sigma ( t) \exp \Big( \int_{\tau(t)}^t p( s) ds-\frac{\sigma ( t) }{e}\Big)\\ &\quad \times \exp \Big( -\int_{\tau(t)}^t B( s) \omega ( s) g( x( \tau ( s) ) ) ds\Big) -1\Big] . \end{aligned} \label{7.6} \end{equation} By Lemmas \ref{lem1}--\ref{lem5}, there exist $T>t_{2}$, $A>0$, $\varepsilon >0$ and $M>0$ such that for $t\geq T$, \begin{gather} x( \tau ( t) ) \leq A\exp \Big( -\frac{1}{2}\int_{T}^{\tau ( t) } p( s) ds\Big) +\varepsilon , \label{7.7} \\ \int_{\tau(t)}^t p( s) ds\leq 2, \label{7.8} \\ \omega ( t) \leq \sigma ( t) M,\quad \sigma (t) \leq \eta . \label{7.9} \end{gather} Let \begin{equation*} \alpha ( t) =\frac{1}{2}\int_{T}^t p(s) ds,\quad t\geq T. \end{equation*} Clearly, \eqref{3.13*} implies that $\alpha ( t)\to \infty $ as $t\to \infty $. For $t\geq t_{2}$, set \begin{equation} \begin{aligned} D( t) &=p( t) \Big( e\int_{\delta(t)}^t B( s) \upsilon ( s) ds+\sigma ( t) \Big) \exp \Big( \int_{\tau(t)}^t p( s) ds-\frac{\sigma ( t) }{e}|Big) \\ &\quad \times\Big[ 1-\exp \Big( -\int_{\tau(t)}^t B( s) \omega ( s) g( x( \tau ( s) ) ) ds\Big) \Big] . \end{aligned}\label{7.10} \end{equation} One can easily see that \begin{equation} 0\leq 1-e^{-c}\leq c\quad \text{for }c\geq 0. \label{7.11} \end{equation} It follows from \eqref{7.10} that for $t\geq t_{2}$, \begin{equation} \begin{aligned} D( t) &\leq p( t) \Big( e\int_{\delta(t)}^t B( s) \upsilon ( s) ds+\sigma ( t) \Big) \exp \Big( \int_{\tau(t)}^t p( s) ds-\frac{\sigma ( t) }{e}% \Big) \\ &\times \int_{\tau(t)}^t B(s) \omega ( s) g( x( \tau ( s) )) ds. \end{aligned} \label{7.12} \end{equation} Therefore, \begin{align*} &D( t) \\ &\leq p( t) \Big( e\int_{\delta(t)}^t B( s) \upsilon ( s) ds+\sigma ( t) \Big) \exp \Big(\int_{\tau(t)}^t p( s) ds\Big) \int_{\tau(t)}^t B(s) \omega ( s) g( x( \tau ( s) )) ds. \end{align*} Let $T^{\ast }>T$ be such that $\tau ( \tau ( t) )\geq T$ for $t\geq T^{\ast }$ and $\alpha ( T^{\ast }) >2+\ln A$. Set $M_{1}=e^{2}\eta M[ 2e( M-1) +\eta ] $ and $A_{1}=eA$. Noting that \begin{equation*} e\int_{\delta(t)}^t B( s) \upsilon ( s) ds+\sigma ( t) \leq 2e( M-1) +\eta \quad \text{for }t\geq T. \end{equation*}% from \eqref{7.7}--\eqref{7.9}, \eqref{7.12}, and assumption $\textrm{(H)}$, we obtain $N\geq T^{\ast }$, \begin{align*} &\int_{T^ast}^N D( t) dt \\ &\leq M_{1} \int_{T^ast}^N p( t) \int_{\tau(t)}^t p( s) g\Big( A \exp \Big( \frac{1}{2}\int_T^{\tau ( t) } p(s) ds\Big) +\varepsilon \Big) \,ds\,dt \\ &=M_{1}\int_{T^ast}^N p( t) \int_{\tau ( t)}^t p( s) g\Big( A\exp \Big( -\frac{1}{2}\int_T^s p( \mu ) d\mu + \frac{1}{2}\int_{\tau ( s) }^s p( \mu) d\mu \Big) +\varepsilon \Big) \,ds\,dt \\ &\leq M_{1}\int_{T^ast}^N p( t) \int_{\tau(t)}^t p( s) g\Big(A_{1}e^{-\alpha ( s) }+\varepsilon \Big) \,ds\,dt \\ &=2M_{1}\int_{T^ast}^N p( t) \int_{\alpha ( \tau ( t) ) }^{\alpha ( t)} g( A_{1}e^{-u}+\varepsilon ) du\,dt \\ &=2M_{1}\int_{T^ast}^N p( t) \int_{\alpha ( t) -\beta ( t) }^{\alpha (t) } g( A_{1}e^{-u}+\varepsilon ) \,du\,dt,\quad \beta( t) =\frac{1}{2}\int_{\tau(t)}^t p( s) ds \\ &\leq 4M_{1}\int_{\alpha ( T^{\ast }) }^{\alpha( N) } p( t) \int_{v-1}^v g( A_{1}e^{-u}+\varepsilon ) du\,dv \\ &\leq 4M_{1}\int_{\alpha ( T^{\ast }) -1}^{\alpha( N) } g( A_{1}e^{-u}+\varepsilon ) du \\ &= 4M_{1}\int_{\ln ( A_{1}e^{1-\alpha ( T^{\ast }) }+\varepsilon ) ^{-1}} ^{\ln ( A_{1}e^{-\alpha ( N) }+\varepsilon ) ^{-1}} g( e^{-u}) \frac{e^{-u}}{e^{-u}-\varepsilon }du \\ &\leq 4M_{1}\int_0 ^{\alpha ( N) } g(e^{-u}) \frac{e^{-u}}{e^{-u}-\varepsilon }du\\ &\leq 4M_{1}\int_0^\infty g( e^{-u}) du<\infty . \end{align*} and \begin{equation} \int_T^\infty D( t) dt<\infty . \label{7.13} \end{equation} Substituting \eqref{7.10} into \eqref{7.6}, for $t\geq t_{2}$, we obtain \begin{align*} &B( t) \upsilon ( t) -eB( t) \int_{\delta ( t) }^t B( s) \upsilon (s) ds \\ & \geq p( t) \Big( e\int_{\delta(t)}^t B( s) \upsilon ( s) ds +\sigma ( t)\Big)\Big[ \exp \Big( \int_{\tau(t)}^t p( s) ds -\frac{\sigma ( t) }{e}\Big) -1\Big] -D( t) , \end{align*} \begin{align*} &B( t) \upsilon ( t) -eB( t) \int_{\delta ( t) }^t B( s) \upsilon (s) ds \\ &\geq p( t) \sigma ( t)\Big[ \exp \Big( \int_{\tau ( t) }^t p( s) ds-\frac{\sigma ( t) }{e}\Big) -1\Big] -D( t) . \end{align*} % \label{7.14} Integrating both sides from $T^{\ast }$ to $N>\tau ^{-1}( T^{\ast }) $, we have \begin{equation} \begin{aligned} &\int_{T^ast}^N B( t) \upsilon (t) dt -e\int_{T^ast}^N B( t) \int_{\delta(t)}^t B( s) \upsilon ( s) \,ds\,dt \\ &\geq \int_{T^ast}^N p( t) \sigma ( t)\Big[ \exp \Big(\int _{\tau(t)}^{t} p( s) ds -\frac{\sigma ( t) }{e}\Big) -1\Big] dt -\int_{T^ast}^N D( t) dt. \end{aligned}\label{7.15} \end{equation} By interchanging the order of integrations and by \eqref{7.5}, we have \begin{equation} \begin{aligned} e\int_{T^ast}^N B( t) \int_{\delta(t)}^t B( s) \upsilon ( s) \,ds\,dt &\geq e\int_{T^{\ast }}^{\delta ( N) } B( t) \upsilon ( t) \int_{t}^{\delta ^{-1}( t) } B( s) \,ds\,dt \\ &\geq \int_{T^{\ast }}^{\delta ( N) } B(t) \upsilon ( t) dt. \end{aligned} \label{7.16} \end{equation} From this and \eqref{7.15}, it follows that \begin{equation} \int_{\delta ( N) }^N B( t) \upsilon ( t) dt\geq \int_{T^ast}^N p( t) \sigma ( t) \Big[ \exp \Big( \int_{\tau(t)}^t p( s) ds-\frac{\sigma ( t) }{e}\Big) -1\Big] dt-\int_{T^ast}^N D( t) dt. \label{7.17} \end{equation} By \eqref{7.8} and \eqref{7.9}, \begin{equation*} \int_{\delta ( N) }^N B( t) \upsilon ( t) dt\leq ( M-1) \int_{\delta ( N) }^N p( t) dt\leq ( M-1) \int_{\tau ( N) }^N p( t) dt\leq 2( M-1) , \end{equation*} and so by \eqref{7.17}, \begin{equation*} 2( M-1) \geq \int_{T^ast}^N p(t) \sigma ( t) \Big[ \exp \Big( \int_{\tau(t)}^t p( s) ds -\frac{\sigma (t) }{e}\Big) -1\Big] dt -\int_{T^ast}^N D( t) dt. \end{equation*} This implies that \begin{equation*} 2( M-1) \geq \int_{T^{\ast }}^\infty p( t) \sigma ( t) \Big[ \exp \Big( \int_{\tau(t)}^t p( s) ds -\frac{\sigma (t) }{e}\Big) -1\Big] dt -\int_{T^{\ast }}^\infty D( t) dt, \end{equation*} which together with \eqref{7.13} yields \begin{equation*} \int_{T^\ast }^\infty p( t) \sigma (t)\Big[ \exp \Big( \int_{\tau(t)}^t p( s) ds-\frac{\sigma ( t) }{e}\Big) -1\Big] dt<\infty . \end{equation*} This contradicts \eqref{3.14*} and so the proof is complete. \subsection*{Proof of Theorem \ref{thm2}} Assume that \eqref{3.7} has a nonoscillatory solution $x( t) $ which will be assumed to be eventually positive (if $x( t) $ is eventually negative the proof is similar). By Lemma \ref{lem1} and assumption $\textrm{(H)} $, there exists $t_0^{\ast }\geq t_0$ such that \begin{equation} 0t_{2}$, $A>0$, $\varepsilon >0$ and $M>0$ such that for $t\geq T$, \begin{gather} x( \tau ( t) ) \leq A\exp \Big( -\frac{1}{2} \int_T^{\tau ( t)} p( s) ds)+\varepsilon , \label{7.22} \\ \int_{\delta(t)}^t p( s) ds\leq \int_{\tau(t)}^t p( s) ds\leq 2,\quad p_{k}( t) \leq 2^{k},\quad k=1,2,\dots , \label{7.23} \\ \frac{z( \delta ( t) ) }{z( t) }\leq \frac{z( \tau ( t) ) }{z( t) }\leq M. \label{7.24} \end{gather} Let $t_{k}=\delta ^{-k}( T)$, $k=1,2,\dots $ Clearly $t_{k}\to \infty $ as $k\to \infty $. Set $\lambda ( t) =-z'( t)/z(t)$, for $t\geq T$. Then \begin{equation*} \dfrac{z( \delta ( t) ) }{z( t) }=\exp\int_{\delta(t)}^t \lambda (s) ds,\quad t\geq t_{1}, \end{equation*} and from \eqref{7.21}, for $t\geq t_{1}$, we have \begin{equation} \lambda ( t) \geq p( t) \exp \int_{\delta(t)}^t \lambda ( s) ds-p( t) g( x( \tau ( t) ) ) \frac{z( \delta ( t) ) }{z( t) }. \label{7.25} \end{equation} It follows from \eqref{7.22}--\eqref{7.25} that for $t\geq t_{1}$, \begin{equation} \begin{aligned} &\lambda ( t)\\ & \geq p( t) \exp \int_{\delta(t)}^t \lambda ( s) ds -Mp( t)g\Big( A\exp \Big( -\frac{1}{2}\int_T^{\tau ( t) } p( s) ds\Big) +\varepsilon \Big) \\ &\geq p( t) \exp \int_{\delta(t)}^t \lambda ( s) ds -Mp( t) g\Big( A_{1}\exp \Big(-\frac{1}{2}\int {T}^t p( s) ds\Big) +\varepsilon \Big) , \end{aligned} \label{7.26} \end{equation} where $A_{1}=eA$. By the inequality $e^{c}\geq ec$ for $c\geq 0$, we have for $t\geq t_{1}$, \begin{equation} \lambda ( t) \geq e p( t) \int_{\delta(t)}^t \lambda ( s) ds -Mp( t) g\Big( A_{1}\exp \Big( -\frac{1}{2}\int_{T}^t p(s) ds\Big) +\varepsilon \Big) . \label{7.27} \end{equation} Set \begin{equation} \alpha ( t) =\frac{1}{2}\int_{T}^t p(s) ds,\quad t\geq T, \label{7.28} \end{equation} and \begin{equation} \begin{gathered} \lambda _0( t) =\lambda ( t) , \quad t\geq T, \\ \lambda _{k}( t) =p( t) \int_{\delta(t)}^t \lambda _{k-1}( s) ds, \quad t\geq t_{k},\;k=1,2,\dots ,n, \end{gathered} \label{7.29} \end{equation} and \begin{equation} \begin{gathered} G_0(t)=0, \quad t\geq T, \\ \begin{aligned} G_{k}( t) =& ep( t)\int_{\delta(t)}^t G_{k-1}( s) ds &+Mp( t) g( A_{1}\exp ( -\alpha ( t) ) +\varepsilon ) , \end{aligned} \end{gathered}\label{7.30} \end{equation} for $t\geq t_{k}$, $k=1,2,\dots ,n$. Clearly \eqref{3.15*} implies that $\alpha ( t) $ is nondecreasing on $[ T,\infty ) $ and $\alpha ( t)\to \infty $ as $t\to \infty $. By iteration we deduce from \eqref{7.27} that \begin{equation} \lambda ( t) \geq e^{k}\lambda _{k}( t) -G_{k}( t) ,\quad t\geq t_{k},\;k=1,2,\dots n-1, \label{7.31} \end{equation} and so by \eqref{7.26}, \begin{equation} \lambda ( t) \geq p( t) \exp \Big( e^{n-1} \int_{\delta ( t) }^t \lambda _{n-1}( s) ds\Big) \exp \Big( -\int_{\delta(t)}^t G_{n-1}( s) ds\Big) -G_{1}( t), \label{7.32} \end{equation} for $t\geq t_{n}$. From \eqref{7.30}, one can easily obtain \begin{equation} G_{k+1}( t) -G_{k}( t) =ep( t) \int_{\delta ( t) }^t [ G_{k}( s) -G_{k-1}( s) ] ds, \label{7.33} \end{equation} for $t\geq t_{k+1}$, $k=1,2,\dots ,n-1$. By \eqref{7.23}, \eqref{7.28} and \eqref{7.30}, for $t\geq t_{2}$, we have \begin{equation} \begin{aligned} \int_{\delta(t)}^t G_{1}( s) ds &=M\int _{\delta(t)}^t p(s) g( A_{1}\exp ( -\alpha ( s) ) +\varepsilon ) ds \\ &= 2M\int_{\alpha ( \delta ( t) ) }^{\alpha ( t) } g( A_{1}e^{-u}+\varepsilon ) du \\ &\leq 2M \int_{\alpha ( t) -1} ^{\alpha (t) } g( A_{1}e^{-u}+\varepsilon ) du. \end{aligned} \label{7.34} \end{equation} Thus, from \eqref{7.33}, we get \[[ G_{2}( t) -G_{1}( t) = ep( t) \int_{\delta ( t) }^t G_{1}( s) ds \\ \leq 2eMp( t) \int_{\alpha ( t) -1}^{\alpha ( t) } g( A_{1}e^{-u}+\varepsilon )du,\quad t\geq t_{2}, \] \begin{align*} G_{3}( t) -G_{2}( t) &= ep( t) \int_{\delta ( t) }^t [ G_{2}( s) -G_{1}( s) ] ds \\ &\leq 2e^{2}Mp( t) \int_{\delta(t)}^t p( s) \int_{\alpha ( s) -1}^{\alpha ( s) } g( A_{1}e^{-u}+\varepsilon ) du\,ds \\ &=4e^{2}Mp( t) \int_{\alpha ( \delta ( t)) }^{\alpha ( t) } \int_{v-1}^v g( A_{1}e^{-u}+\varepsilon ) du\,dv \\ &\leq 4e^{2}Mp( t) \int_{\alpha ( t) -1}^{\alpha ( t) } \int_{v-1}^v g(A_{1}e^{-u}+\varepsilon ) du\,dv \\ &\leq 4e^{2}Mp( t) \int_{\alpha ( t) -2}^{\alpha ( t) } g( A_{1}e^{-u}+\varepsilon )du,\quad t\geq t_{3}. \end{align*} By induction, one can prove in general that for $k=2,3,\dots ,n-1,$% \[ G_{k}( t) -G_{k-1}( t) \leq ( 2e) ^{k-1}( k-2) !Mp( t) \int_{\alpha ( t) -( k-1)} ^{\alpha ( t)} g( A_{1}e^{-u}+\varepsilon ) du,\quad t\geq t_{k}, \] and so \begin{equation} \begin{aligned} G_{n-1}( t) &=\sum_{k=1}^{n-1}[ G_{k}( t) -G_{k-1}( t) ]\\ &\leq G_{1}( t) +Mp( t) \sum_{k=2}^{n-1}( 2e) ^{k-1}( k-2) !\int_{\alpha ( t) -(k-1) }^{\alpha ( t) } g(A_{1}e^{-u}+\varepsilon ) du, \end{aligned} \label{7.35} \end{equation} for $t\geq t_{n-1}$. By \eqref{7.23}, \eqref{7.24} and \eqref{7.29}, we obtain \begin{equation} \begin{gathered} \begin{aligned} \lambda _{1}( t) &=p( t) \int_{\delta(t)}^t \lambda ( s) ds=p( t) \ln\big[ \frac{z( \delta ( t) ) }{z( t) }\big] \\ &\leq p( t) \ln M,\quad t\geq t_{1}, \end{aligned} \\ \begin{aligned} \lambda _{2}( t) &=p( t) \int_{\delta(t)}^t \lambda _{1}( s) ds \leq p(t) \ln M\int _{\delta(t)}^t p(s) ds \\ &\leq 2p( t) \ln M,\quad t\geq t_{2}, \end{aligned}\\ \dots \\ \lambda _{n-1}( t) \leq 2^{n-2}p( t) \ln M,\quad t\geq t_{n-1}. \end{gathered} \label{7.36} \end{equation} For $t\geq t_{n}$, set \begin{align*} &D( t) &=p( t) \exp \Big( e^{n-1}\int_{\delta(t)}^t \lambda _{n-1}( s) ds \Big)\Big[ 1-\exp\Big( -\int_{\delta(t)}^t G_{n-1}(s) ds\Big) \Big] +G_{1}( t). \end{align*} From \eqref{7.11}, \eqref{7.23}, \eqref{7.34}, \eqref{7.35} and \eqref{7.36}, we have \begin{equation} \begin{aligned} D( t) &\leq p( t) \exp \Big( e^{n-1}\int_{\delta(t)}^t \lambda _{n-1}( s) ds\Big)\int_{\delta(t)}^t G_{n-1}( s) ds+G_{1}( t) \\ &\leq G_{1}( t) +p( t) \exp \Big( 2^{n-2}e^{n-1}\ln M \int_{\delta(t)}^t p( s) ds\Big) \\ &\quad \times \int_{\delta(t)}^t \Big[ G_{1}( s) +Mp( s) \sum_{k=2}^{n-1}( 2e) ^{k-1}( k-2) !\int_{\alpha ( s) -( k-1)}^{\alpha ( s)} g( A_{1}e^{-u}+\varepsilon ) du\Big] ds \\ &\leq G_{1}( t) +2Mp( t) \exp ( ( 2e) ^{n-1}\ln M) \int_{\alpha ( t) -1}^{\alpha ( t) } g( A_{1}e^{-u}+\varepsilon ) du \\ &\quad +Mp( t) \exp ( ( 2e) ^{n-1}\ln M) \\ &\quad \times \sum_{k=2}^{n-1}( 2e) ^{k-1}( k-2) ! \int_{\delta ( t) }^t p( s) \int_{\alpha ( s) -( k-1) } ^{\alpha ( s) } g( A_{1}e^{-u}+\varepsilon ) du\,ds \\ &\leq G_{1}( t) +M_{1}p( t) \sum_{k=1}^{n-1}( 2e) ^{k-1}( k-1) ! \int_{\alpha ( t) -k}^{\alpha ( t) } g( A_{1}e^{-u}+\varepsilon ) du,\quad t\geq t_{n}, \end{aligned} \label{7.37} \end{equation} where $M_{1}=2M\exp ( ( 2e) ^{n-1}\ln M) $. Let $T^{\ast }>t_{n}$ be such that $\alpha ( T^{\ast }) >n+\ln A_{1}$. It follows from \eqref{7.37} and (H) that \begin{align*} &\int_{T^\ast }^\infty D( t) dt\\ &\leq \int_{T^\ast }^\infty G_{1}( t) dt+M_{1}\sum_{k=1}^{n-1}( 2e) ^{k-1}( k-1) ! \int_{T^{\ast }}^\infty p( t) \int_{\alpha (t) -k}^{\alpha ( t) } g( A_{1}e^{-u}) \,du\,dt \\ &\leq 2M\int_{\alpha ( T^{\ast }) }^\infty g( A_{1}e^{-u}) du +2M_{1}\sum_{k=1}^{n-1}( 2e)^{k-1}( k-1) ! \int_{\alpha ( T^{\ast }) }^\infty \int_{v-k}^v g( A_{1}e^{-u}) du\,dv \\ &\leq 2M\int_{\alpha ( T^{\ast }) }^\infty g( A_{1}e^{-u}) du+2M_{1}\sum_{k=1}^{n-1}( 2e) ^{k-1}k! \int_{\alpha ( T^{\ast }) -( k+1) }^\infty g( A_{1}e^{-u}) du \\ &\int_{T^\ast }^\infty D( t) dt\leq 2M \int_0^\infty g( e^{-u})du+2M_{1}\sum_{k=1}^{n-1}( 2e) ^{k-1}k! \int_0^\infty g( e^{-u}) du<\infty . \end{align*} \label{7.38} Since \begin{align*} &p( t) \exp \Big( e^{n-1}\int_{\delta(t)}^t \lambda _{n-1}( s) ds\Big) \exp \Big( -\int_{\delta(t)}^t G_{n-1}( s)ds\Big) -G_{1}( t) \\ &= p( t) \exp \Big( e^{n-1}\int_{\delta(t)}^t \lambda _{n-1}( s) ds\Big) -D( t),\quad t\geq t_{n}, \end{align*} it follows from \eqref{7.32} that \begin{equation} \lambda ( t) \geq p( t) \exp \Big( e^{n-1} \int_{\delta ( t) }^t \lambda _{n-1}( s) ds\Big) -D( t) , \quad t\geq t_{n}. \label{7.39} \end{equation} One can easily show that $\gamma e^{x}\geq x+\ln ( \gamma +1) $ for $\gamma >0$, and so for $t\geq t_{n}$, \begin{align*} p_{n}( t) \lambda ( t) &\geq p( t) e^{1-n}( e^{n-1}p_{n}( t) ) \exp \Big( e^{n-1}\int_{\delta(t)}^t \lambda _{n-1} (s) ds\Big) -p_{n}( t) D( t) \\ &\geq p( t) \int_{\delta(t)}^t \lambda _{n-1}( s) ds +e^{1-n}p( t) \ln (e^{n-1}p_{n}( t) +1) -p_{n}( t) D( t), \end{align*} that is, for $t\geq t_{n}$, \begin{equation} p_{n}( t) \lambda ( t) -p( t) \int_{\delta ( t) }^t \lambda _{n-1}( s) ds\geq e^{1-n}p( t) \ln ( e^{n-1}p_{n}( t) +1) -p_{n}( t) D( t) . \label{7.40} \end{equation} For $N>\delta ^{-n}( T^{\ast }) $, we have \begin{equation} \begin{aligned} &\int_{T^ast}^N p_{n}( t) \lambda (t) dt -\int_{T^ast}^N p( t) \int_{\delta(t)}^t \lambda _{n-1}(s) \,ds\,dt \\ &\geq e^{1-n}\int_{T^ast}^N p( t) \ln ( e^{n-1}p_{n}( t) +1) dt -\int_{T^{\ast }}^N p_{n}( t) D( t) dt. \end{aligned} \label{7.41} \end{equation} Let $\delta ^{1}( t) =\delta ( t)$, $\delta^{k+1}( t) =\delta ( \delta ^{k}( t) )$, $k=1,2,\dots ,n$. Then by interchanging the order of integration, we have \begin{align*} \int_{T^ast}^N p( t) \int_{\delta(t)}^t \lambda _{n-1}( s) \,ds\,dt &\geq \int_{T^{\ast }}^{\delta ( N) } \lambda_{n-1}( t) \int_{t}^{\delta ^{-1}( t) } p( s) \,ds\,dt \\ &=\int_{T^{\ast }}^{\delta ( N) } p(t) p_{1}( t) \int_{\delta(t)}^t \lambda _{n-2}( s) \,ds\,dt \\ &\geq \int_{T^{\ast }}^{\delta ^{2}( N) } \lambda _{n-2}( t) \int_t^{\delta ^{-1}(t) } p( s) p_{1}( s) \,ds\,dt \\ &=\int_{T^{\ast }}^{\delta ^{2}( N) } p(t) p_{2}( t) \int_{\delta(t)}^t \lambda _{n-3}( s) \,ds\,dt \\ &\dots \\ &\geq \int_{T^{\ast }}^{\delta ^{n}( N) } \lambda ( t) p_{n}( t) dt. \end{align*} From this and \eqref{7.41}, we have \begin{equation} \int_{\delta ^{n}( N)}^N p_{n}(t) \lambda ( t) dt \geq e^{1-n}\int_{T^\ast}^{N} p( t) \ln ( e^{n-1}p_{n}( t) +1) dt-\int_{T^ast}^N p_{n}( t) D(t) dt, \label{7.42} \end{equation} which together with \eqref{7.23} yields \begin{equation*} 2^{n}\int_{\delta ^{n}( N) }^{N} \lambda( t) dt\geq e^{1-n}\int_{T^ast}^N p( t) \ln ( e^{n-1}p_{n}( t) +1) dt -2^{n}\int_{T^ast}^N D( t) dt, \end{equation*}% or \begin{equation} \ln \frac{x( \delta ^{n}( N) ) }{x( N) } \geq 2^{-n}e^{1-n}\int_{T^ast}^N p( t) \ln ( e^{n-1}p_{n}( t) +1) dt-\int_{T^\ast}^N D( t) dt. \label{7.43} \end{equation} In view of \eqref{3.16} and \eqref{7.38}, we have \begin{equation} \lim_{N\to \infty }\frac{x( \delta ^{n}( N) )}{x( N) }=\infty . \label{7.44} \end{equation} On the other hand, \eqref{7.24} implies that \begin{equation*} \frac{x( \delta ^{n}( N) ) }{x( N) } =\frac{x( \delta ^{1}( N) ) }{x( N) }. \frac{x( \delta ^{2}( N) ) }{x( \delta ^{1}(N) ) }\cdots \frac{x( \delta ^{n}( N) ) }{x( \delta ^{n-1}( N) ) }\leq M^{n}. \end{equation*} This contradicts \eqref{7.44} and completes the proof. \section{Examples} In this section we introduce some examples to illustrate our main results. \begin{example} \label{exam1} \rm Consider the neutral delay differential equation \begin{equation} \big( x( t) -( \frac{5}{2}+\sin t) x( t-\pi ) \big) '+f( t,x( \tau ( t) ) ) =0,\quad t\geq 3. \label{8.1} \end{equation} For $f( t,u) =p( t) f( u)$, with \begin{gather} f( u) =\begin{cases} u[ 1+( 1+\ln ^{2}| u| ) ^{-1}] , & u\neq 0, \\ 0,& u=0, \end{cases} \label{L} \\ g( u) =\begin{cases} 1,& | u| >1, \\ ( 1+\ln ^{2}| u| ) ^{-1}, & 0<|u| \leq 1, \\ 0, & u=0, \end{cases} \label{LL} \\ p( t) =\frac{1}{et\ln 2}+\frac{1}{t\ln t},\quad \tau ( t) =\frac{t}{2}, \end{gather} with $\int_3^\infty p( t) dt=\infty $. It is easily seen that condition (H) holds. We check that the conditions \eqref{3.15*} and \eqref{3.14} in Corollary \ref{coro1} hold. In fact, for $t\geq 3$, \[ \int_{\frac{t}{2}}^t p( s) ds =\int_{\frac{t}{2}}^t \big[ \frac{1}{es\ln 2}+\frac{1}{s\ln s} \big] ds =\frac{1}{e}-\ln\big[ 1-\frac{\ln 2}{\ln t}\big] \geq \frac{1}{e}, \] $\liminf_{t\to \infty } \int_{t/2}^t p( s) ds=1/e$, and \begin{align*} \int_3^\infty p( t)\big[ \exp\big[ \int_{t/2}^t p( s) ds-\frac{1}{e}\big] -1\big] dt &\geq \int_3\infty p(t)\big[ \int_{t/2}^t p( s)ds -\frac{1}{e}\big] dt \\ &\geq -\frac{1}{e\ln 2}\int_3^\infty \frac{1}{t}\ln [ 1-\frac{\ln 2}{\ln t}] dt=\infty , \end{align*} because \begin{equation*} \int_3^\infty \frac{1}{t\ln t}dt=\infty \quad \text{and}\quad \lim_{t\to \infty }( \ln t) \ln[ 1-\frac{\ln 2}{\ln t}] =-\ln 2. \end{equation*} By Theorem \ref{thm1} every solution of \eqref{8.1} oscillates. \end{example} \begin{example} \label{ex2} \rm Consider the neutral delay differential equation \begin{equation} \big( x( t) -( \frac{5}{2}+\sin t) x( t-\frac{% \pi }{2}) \big) '+f( t,x( \tau ( t) ) ) =0,\quad t\geq 3, \label{8.1*} \end{equation} For \[ p( t) =\frac{\delta }{t},\quad \tau ( t) =\frac{t}{\lambda}, \quad \delta <\frac{1}{e\ln \lambda },\quad \lambda >1, \quad f( t,u) =p( t) f( u) , \] where $f( u) $ and $g( u) $ are defined by \eqref{L} and \eqref{LL} and with $\int_3^\infty p( t) dt=\infty $, and \begin{equation*} \int_{t/\lambda }^t p( s) ds= \int_{t/\lambda }^t \frac{\delta }{s}ds=\delta \big( \ln t-\ln \frac{t}{\lambda }\big) =\delta \ln \lambda <\frac{1}{e} \end{equation*} It is easily seen that condition $\textrm{(H)} $ holds. We check that the conditions \eqref{3.13*} and \eqref{3.14} in Theorem \ref{thm2} hold. In fact, for $t\geq 3$, \[ \lim_{t\to \infty }\inf \int_{t/\lambda }^t p( s) ds = \lim_{t\to \infty }\inf \int_{t/\lambda }^t \frac{\delta }{s}ds = \delta ( \ln t-\ln \frac{t}{\lambda }) =\delta \ln \lambda >0 \] and \begin{align*} \int_3^\infty p( t)[ \exp (\int_{t/\lambda }^t p( s) ds) -1] dt &\geq \int_3^\infty p( t)( \int_{t/\lambda }^t p( s)ds) dt \\ &\geq \int_3^\infty \frac{\delta ^{2}\ln \lambda }{t}dt =\delta ^{2}\ln \lambda ( \infty ) =\infty \end{align*} By Corollary \ref{coro1} every solution of \eqref{8.1*} oscillates. \end{example} \begin{example} \label{ex3} \rm Consider the neutral delay differential equation \begin{equation} \big( x( t) -( \frac{5}{2}+\sin t) x( t-\pi ) \big) '+f( t,x( \tau ( t) ) ) =0,\quad t\geq 3, \label{8.2} \end{equation} where \begin{equation*} \tau ( t) =t-1\quad \text{and\ \ \ }f( t,u) =[ \exp 3( \sin t-1) +| u| ] ^{1/3}u. \end{equation*} Let $p( t) =\exp ( \sin t) -0.1$ and $g( u) =e^{2}| u| ^{1/3}$. It is easy to see that assumption (H) holds. Clearly \begin{gather*} \liminf_{t\to \infty } \int_{t-1}^t p(s) ds<\frac{1}{e}, \\ \int_0^\infty p( t) \ln \Big( \int_t^{t+1} p( s) ds+1\Big) dt \geq \int_0^\infty \exp ( \sin t-1) \ln \Big( \int_t^{t+1} \exp ( \sin s) ds\Big) dt \end{gather*} By Jensen's inequality, \begin{align*} \int_0^\infty p( t) \ln \Big( \int_t^{t+1}p( s) ds+1\Big) dt &\geq \int_0^\infty \exp ( \sin t-1) \int_t^{t+1}\sin s\,ds\,dt \\ &=\frac{2\sin 2^{-1}}{e}\int_0^\infty \exp (\sin t) \sin ( t+\frac{1}{2}) dt. \end{align*} On the other hand, it is easy to see that $\int_0^t \exp ( \sin s) \cos s\,ds$ is bounded and \begin{equation*} \int_0^{2\pi } \exp ( \sin t) \sin tdt>0. \end{equation*} Thus \begin{equation*} \int_0^\infty p( t) \ln \Big( \int_t^{t+1} p( s) ds+1\Big) dt=\infty . \end{equation*} By Corollary \ref{coro2}, every solution of \eqref{8.2} oscillates. \end{example} \begin{thebibliography}{99} \bibitem{1ag} R. P. Agarwal, S. R. Grace and D. 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