\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2014 (2014), No. 67, pp. 1--7.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2014 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2014/67\hfil Oscillation of solutions] {Oscillation of solutions to second-order neutral differential equations} \author[Tongxing Li, Ethiraju Thandapani \hfil EJDE-2014/67\hfilneg] {Tongxing Li, Ethiraju Thandapani} % in alphabetical order \address{Tongxing Li \newline Qingdao Technological University, Feixian, Shandong 273400, China} \email{litongx2007@163.com} \address{Ethiraju Thandapani \newline Ramanujan Institute for Advanced Study in Mathematics, University of Madras, Chennai 600 005, India} \email{ethandapani@unom.ac.in} \thanks{Submitted February 5, 2014. Published March 7, 2014.} \subjclass[2000]{34C10, 34K11} \keywords{Oscillation; neutral delay differential equation; second-order} \begin{abstract} We study the oscillatory behavior of solutions to second-order neutral differential equations. We show that under certain conditions, all solutions are oscillatory. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{example}[theorem]{Example} \newtheorem{remark}[theorem]{Remark} \allowdisplaybreaks \section{Introduction} In this article, we study the oscillation of solutions to the second-order nonlinear neutral delay differential equation \begin{equation}\label{0.1} (r(t)\psi(x(t))|Z'(t)|^{\alpha-1}Z'(t))'+q(t)f(x(\sigma(t)))=0, \end{equation} where $t\in \mathbb{I}:=[t_0, \infty)$, $Z(t):=x(t)+p(t)x(\tau(t))$, and $\alpha>0$. Throughout, we assume that the following conditions are satisfied: \begin{itemize} \item[(A1)] $r, p, q\in C(\mathbb{I}, \mathbb{R})$, $r(t)>0$, $0\leq p(t)\leq1$, $q(t)\geq0$, and $q$ is not identically zero for large $t$; \item[(A2)] $\psi\in C^1(\mathbb{R}, \mathbb{R})$, $f\in C(\mathbb{R}, \mathbb{R})$, $\psi(x)>0$, $xf(x)>0$ for all $x\neq0,$ and there exist two positive constants $k$ and $L$ such that $$ \frac{f(x)}{|x|^{\alpha-1}x}\geq k \quad\text{and}\quad \psi(x)\leq L^{-1} \quad \text{for all} \quad x\neq 0; $$ \item[(A3)] $\tau\in C(\mathbb{I}, \mathbb{R})$, $\tau(t)\leq t$, and $\lim_{t\to\infty}\tau(t)=\infty$; \item[(A4)] $\sigma\in C^1(\mathbb{I}, \mathbb{R})$, $\sigma'(t)>0$, $\sigma(t)\leq t$, and $\lim_{t\to\infty}\sigma(t)=\infty$. \end{itemize} By a solution of equation \eqref{0.1}, we mean a continuous function $x$ defined on an interval $[t_x,\infty)$ such that $r\psi(x)|Z'|^{\alpha-1}Z'$ is continuously differentiable and $x$ satisfies \eqref{0.1} for $t\in[t_x,\infty)$. We consider only solutions satisfying condition $\sup\{|x(t)|:t\geq T\geq t_x\}>0$ and tacitly assume that equation \eqref{0.1} possesses such solutions. As usual, a solution of \eqref{0.1} is called oscillatory if it is neither eventually positive nor eventually negative; otherwise, we call it non-oscillatory. Equation \eqref{0.1} is termed oscillatory if all its continuable solutions oscillate. It is known that various classes of neutral differential equations are often encountered in applied problems in natural sciences and engineering; see, e.g., Hale \cite{hale}. Recently, a great deal of interest in oscillatory properties of neutral functional differential equations has been shown, we refer the reader to \cite{ag1,ag2,ba,ca,dix1,dix2,dix3,hl,ka,li,li1,li2,li3,sa,ye} and the references cited therein. Below, we briefly review the following related results that motivated our study. Ye and Xu \cite{ye} obtained several oscillation criteria for equation \eqref{0.1}, one of which we present below. For the convenience of the reader, in what follows, we use the notation $$ \varepsilon:=\left(\alpha/(\alpha+1)\right)^{\alpha+1}, \quad Q(t):=q(t)(1-p(\sigma(t)))^\alpha, \quad \pi(t):=\int_t^\infty\frac{{\rm d}s}{r^{{1}/{\alpha}}(s)}. $$ \begin{theorem}[{\cite[Theorem 2.3]{ye}}] \label{lithm1} Assume that conditions {\rm (A1)--(A4)} are satisfied and let \begin{equation}\label{nocan} \pi(t_0)<\infty. \end{equation} If $$ \int^\infty\big[Q(t)\pi^\alpha(\sigma(t))-\frac{\varepsilon}{Lk} \frac{\sigma'(t)}{\pi(\sigma(t)) r^{{1}/{\alpha}}(\sigma(t))}\big]{\rm d}t=\infty $$ and $$ \int^\infty\big[Q(t)\pi^\alpha(t)-\frac{\varepsilon}{Lk} \frac{r(\sigma(t))}{\pi(t)(\sigma'(t))^\alpha r^{{(\alpha+1)}/{\alpha}}(t)}\big]{\rm d}t=\infty, $$ then equation \eqref{0.1} is oscillatory. \end{theorem} Note that Theorem \ref{lithm1} is not valid for the differential equation \begin{equation}\label{0.21} \Big({\rm e}^{2t}\Big(x(t)+\frac{1}{2}x(t-2)\Big)'\Big)'+\big(1+\frac{{\rm e}^{2}}{2}\big){\rm e}^{2t}x(t)=0, \end{equation} where $t\geq 1$. Let $r(t)={\rm e}^{2t}$, $\psi(x(t))=1$, $p(t)=1/2$, $q(t)=(2+{\rm e}^{2}){\rm e}^{2t}/2$, $\tau(t)=t-2$, $\sigma(t)=t$, $\alpha=1$, $L=1$, and $k=1$. Then $\pi(t)={\rm e}^{-2t}/2$, $\pi(t_0)<\infty$, and $Q(t)=q(t)/2=(2+{\rm e}^{2}){\rm e}^{2t}/4$. Then, we conclude that \begin{align*} &\int^\infty\big[Q(t)\pi^\alpha(\sigma(t))-\frac{\varepsilon}{L k} \frac{\sigma'(t)}{\pi(\sigma(t)) r^{{1}/{\alpha}}(\sigma(t))}\big]{\rm d}t \\ &=\int^\infty\big[Q(t)\pi^\alpha(t)-\frac{\varepsilon}{L k} \frac{r(\sigma(t))}{ \pi(t)(\sigma'(t))^\alpha r^{{(\alpha+1)}/{\alpha}}(t)}\big]{\rm d}t \\ &= \int^\infty\frac{{\rm e}^2-2}{8}{\rm d}t=\infty. \end{align*} Hence, by Theorem \ref{lithm1}, equation \eqref{0.21} should be oscillatory. However, it is not difficult to verify that $x(t)={\rm e}^{-t}$ is a non-oscillatory solution of equation \eqref{0.21}. To amend Theorem \ref{lithm1}, Han et al. \cite{hl} established some oscillation results for \eqref{0.1} under the assumptions \begin{equation}\label{contion} p'(t)\geq0, \quad \sigma(t)\leq \tau(t):=t-\tau_0, \end{equation} where $\tau_0$ is a non-negative constant. The main goal of this article is to derive new oscillation criteria for \eqref{0.1} without requiring the restrictive conditions \eqref{contion}. \section{Main results} In what follows, all functional inequalities are tacitly assumed to hold for all $t$ large enough. \begin{theorem}\label{than3.1} Assume that {\rm (A1)--(A4)} and \eqref{nocan} are satisfied and assume that $\psi(x)\geq K>0$. Suppose further that there exist two functions $\rho, m \in C^1(\mathbb{I},(0,\infty))$ such that \begin{gather}\label{hls3.0} \frac{m(t)}{(LK)^{{1}/{\alpha}}r^{{1}/{\alpha}}(t)\pi(t)}+m'(t)\leq0, \quad 1-p(t)\frac{m(\tau(t))}{m(t)}>0, \\ \label{hls3.1} \int^\infty\big[\rho(s)Q(s)- \frac{1}{Lk(\alpha+1)^{\alpha+1}}\frac{(\rho'_+(s))^{\alpha+1}r(\sigma(s))} {(\rho(s)\sigma'(s))^\alpha}\big]{\rm d}s=\infty,\\ \label{hls3.2} \int^\infty\big[k q(s)\pi^\alpha(s)\big(1-p(\sigma(s)) \frac{m(\tau(\sigma(s)))}{m(\sigma(s))}\big)^\alpha- \big(\frac{\alpha}{\alpha+1}\big)^{\alpha+1} \frac{1}{L\pi(s)r^{{1}/{\alpha}}(s)}\big]{\rm d}s=\infty, \end{gather} where $\rho'_+(t):=\max\{0,\rho'(t)\}$. Then equation \eqref{0.1} is oscillatory. \end{theorem} \begin{proof} Let $x$ be a non-oscillatory solution of \eqref{0.1}. The proofs for eventually positive and for eventually negative solutions are similar. If $y$ is a negative solution, then $x=-y$ may not be a solution of \eqref{0.1}, but $x$ satisfies key estimates such as \eqref{e2.5b} with $\psi(-x)$ instead of $\psi(x)$. Then we can use that $\psi(x)$ and $\psi(-x)$ have same bounds, $K\leq \psi(\cdot)\leq 1/L$. We assume that there exists a $t_1\geq t_0$ such that $x(t)>0$, $x(\tau(t))>0$, and $x(\sigma(t))>0$ for all $t\geq t_1$. Then $|x(t)|^{\alpha-1}x(t)= x^\alpha(t)$ and $Z(t)>0$. From \eqref{0.1} it follows that for all $t\geq t_1$, \begin{equation}\label{hls3.3} (r(t)\psi(x(t))|Z'(t)|^{\alpha-1}Z'(t))'\leq-kq(t)x^\alpha(\sigma(t))\leq0. \end{equation} Hence, there exists a $t_2\geq t_1$ such that either $Z'(t)>0$ or $Z'(t)<0$ for all $t\geq t_2$. We consider each of two cases separately. \smallskip Case 1: $Z'(t)>0$ for all $t\geq t_2$. As in the proof of \cite[Theorem 2.1]{ye}, we obtain a contradiction to \eqref{hls3.1}. \smallskip Case 2: $Z'(t)<0$ for all $t\geq t_2$. For $t\geq t_2$, we define a Riccati substitution \begin{equation}\label{hls3.4} \omega(t):=\frac{r(t)\psi(x(t))(-Z'(t))^{\alpha-1}Z'(t)}{Z^\alpha(t)}. \end{equation} Then $\omega(t)<0$ for all $t\geq t_2$. Since $(r(t)\psi(x(t))|Z'(t)|^{\alpha-1}Z'(t))'\leq0$, the function $r\psi(x)|Z'|^{\alpha-1}Z'$ is non-increasing. Thus, for all $s\geq t\geq t_2$, $$ (r(s)\psi(x(s)))^{{1}/{\alpha}}Z'(s)\leq (r(t)\psi(x(t)))^{{1}/{\alpha}}Z'(t). $$ Dividing the latter inequality by $(r(s)\psi(x(s)))^{{1}/{\alpha}}$ and integrating the resulting inequality from $t$ to $l$, for all $l\geq t\geq t_2$, we have $$ Z(l)\leq Z(t)+(r(t)\psi(x(t)))^{{1}/{\alpha}}Z'(t)\int_t^l\frac{{\rm d}s}{(r(s)\psi(x(s)))^{{1}/{\alpha}}}. $$ Since $Z'(t)<0$ and $\psi \leq 1/L$, we conclude that, for all $l\geq t\geq t_2$, $$ Z(l)\leq Z(t)+(Lr(t)\psi(x(t)))^{{1}/{\alpha}}Z'(t)\int_t^l\frac{{\rm d}s}{r^{{1}/{\alpha}}(s)}. $$ Letting $l\to\infty$ in this inequality, and using that $Z>0$, we have that for all $t\geq t_2$, $$ 0\leq Z(t)+(Lr(t)\psi(x(t)))^{{1}/{\alpha}}Z'(t)\pi(t); $$ that is, for all $t\geq t_2$, \begin{equation} \label{e2.5b} (r(t)\psi(x(t)))^{{1}/{\alpha}}\pi(t)\frac{Z'(t)}{Z(t)} \geq-\frac{1}{L^{{1}/{\alpha}}}. \end{equation} Hence, by \eqref{hls3.4}, we conclude that, for all $t\geq t_2$, \begin{equation}\label{hls3.6} -L^{-1}\leq \omega(t)\pi^\alpha(t)\leq0. \end{equation} From \eqref{e2.5b} and $K\leq \psi$, we obtain $$ \frac{Z'(t)}{Z(t)}\geq-\frac{1}{L^{{1}/{\alpha}} (r(t)\psi(x(t)))^{{1}/{\alpha}}\pi(t)}\geq -\frac{1}{(LK)^{{1}/{\alpha}}r^{{1}/{\alpha}}(t)\pi(t)}. $$ Thus, we have $$ \Big(\frac{Z(t)}{m(t)}\Big)' =\frac{Z'(t)m(t)-Z(t)m'(t)}{m^2(t)} \geq-\frac{Z(t)}{m^2(t)} \big[\frac{m(t)}{(LK)^{{1}/{\alpha}}r^{{1}/{\alpha}}(t)\pi(t)}+m'(t)\big]\geq0. $$ Hence, the function ${Z}/{m}$ is non-decreasing, and so \begin{align*} x(t)&=Z(t)-p(t)x(\tau(t))\geq Z(t)-p(t)Z(\tau(t))\\ &\geq Z(t)-p(t)\frac{m(\tau(t))}{m(t)}Z(t) =\big(1-p(t)\frac{m(\tau(t))}{m(t)}\big)Z(t). \end{align*} Differentiation of \eqref{hls3.4} yields \begin{align*} \omega'(t) &=\Big((r(t)\psi(x(t))(-Z'(t))^{\alpha-1}Z'(t))'Z^\alpha(t)\\ &\quad -\alpha r(t)\psi(x(t))(-Z'(t))^{\alpha-1}Z'(t)Z^{\alpha-1}(t)Z'(t)\Big) /Z^{2\alpha}(t). \end{align*} It follows from the latter equality and \eqref{hls3.3} that \begin{equation}\label{hls3.7} \begin{aligned} \omega'(t) &\leq -k q(t)\Big(1-p(\sigma(t))\frac{m(\tau(\sigma(t)))}{m(\sigma(t))}\Big) ^\alpha\frac{Z^\alpha(\sigma(t))}{Z^\alpha(t)} \\ &\quad -\frac{\alpha r(t)\psi(x(t))(-Z'(t))^{\alpha-1}Z'(t)Z^{\alpha-1} (t)Z'(t)}{Z^{2\alpha}(t)}. \end{aligned} \end{equation} Thus, by \eqref{hls3.4} and \eqref{hls3.7}, we have \begin{equation}\label{hls3.9} \omega'(t)+kq(t)\Big(1-p(\sigma(t))\frac{m(\tau(\sigma(t)))}{m(\sigma(t))} \Big)^\alpha+\frac{\alpha L^{{1}/{\alpha}}}{r^{{1}/{\alpha}}(t)}(-\omega(t))^{{(\alpha+1)}/{\alpha}}\leq0. \end{equation} Multiplying \eqref{hls3.9} by $\pi^\alpha(t)$ and integrating the resulting inequality from $t_3$ ($t_3> t_2$) to $t$, we deduce that \begin{equation} \label{hls4.0} \begin{aligned} &\pi^\alpha(t)\omega(t)-\pi^\alpha(t_3)\omega(t_3) +\alpha\int_{t_3}^tr^{-{1}/{\alpha}}(s) \pi^{\alpha-1}(s)\omega(s){\rm d }s\\ &+k\int_{t_3}^tq(s)\left(1-p(\sigma(s)) \frac{m(\tau(\sigma(s)))}{m(\sigma(s))}\right)^\alpha\pi^\alpha(s){\rm d}s \\ &+\alpha L^{{1}/{\alpha}}\int_{t_3}^t\frac{\pi^\alpha(s)} {r^{{1}/{\alpha}}(s)}(-\omega(s))^{{(\alpha+1)}/{\alpha}}{\rm d}s\leq0. \end{aligned} \end{equation} Let $p:=(\alpha+1)/\alpha$, $q:=\alpha+1$, \begin{gather*} a:=L^{{1}/{(\alpha+1)}}(\alpha+1)^{{\alpha} /{(\alpha+1)}}\pi^{{\alpha^2}/{(\alpha+1)}}(t)\omega(t), \\ b:=L^{-{1}/{(\alpha+1)}}\frac{\alpha}{(\alpha+1)^{{\alpha} /{(\alpha+1)}}}\pi^{-{1}/{(\alpha+1)}}(t). \end{gather*} Using Young's inequality, $$ |ab|\leq\frac{1}{p}|a|^p+\frac{1}{q}|b|^q, \quad \text{where} \quad a,\ b\in \mathbb{R},\ p>1,\ q>1,\ \frac{1}{p}+\frac{1}{q}=1, $$ we have $$ -\alpha\pi^{\alpha-1}(t) \omega(t)\leq \alpha L^{{1}/{\alpha}} \pi^{\alpha}(t)(-{\omega(t)})^{{(\alpha+1)}/{\alpha}} + \big(\frac{\alpha}{\alpha+1}\big)^{\alpha+1}\frac{1}{L\pi(t)}, $$ and hence $$ -\alpha\frac{\pi^{\alpha-1}(t)\omega(t)}{r^{{1}/{\alpha}}(t)}\leq \alpha L^{{1}/{\alpha}}\frac{\pi^\alpha(t)(-{\omega(t)})^{{(\alpha+1)} /{\alpha}}}{r^{{1}/{\alpha}}(t)}+ \big(\frac{\alpha}{\alpha+1}\big)^{\alpha+1}\frac{1}{L\pi(t)r^{{1}/{\alpha}}(t)}. $$ Therefore, it follows from \eqref{hls3.6} and \eqref{hls4.0} that \begin{align*} &\int_{t_3}^t\Big[k q(s)\pi^\alpha(s)\Big(1-p(\sigma(s))\frac{m(\tau(\sigma(s)))}{m(\sigma(s))} \Big)^\alpha- \big(\frac{\alpha}{\alpha+1}\big)^{\alpha+1} \frac{1}{L\pi(s)r^{{1}/{\alpha}}(s)}\Big]{\rm d}s \\ &\leq\pi^\alpha(t_3)\omega(t_3)-\pi^\alpha(t)\omega(t)\\ &\leq L^{-1}+\pi^\alpha(t_3)\omega(t_3), \end{align*} which contradicts \eqref{hls3.2}. This completes the proof. \end{proof} \begin{remark}\label{rem31} \rm A function $m$ in Theorem \ref{than3.1} can be obtained by setting $m(t):=\pi(t)$ in the case $LK\geq1$. \end{remark} It may happen that the restriction $\psi(x)\geq K>0$ is not satisfied and Theorem \ref{than3.1} cannot be applied. For example when $$ \psi(x)=\frac{1}{x^2+1}, $$ in which case the following result proves to be useful. \begin{theorem}\label{than3.2} Assume that conditions {\rm (A1)--(A4)} and \eqref{nocan} hold. Let $\psi$ be non-increasing for all $x>0$, and non-decreasing for all $x<0$. Suppose further that there exist two functions $\rho,m\in C^1(\mathbb{I},(0,\infty))$ such that, for any fixed constant $l>0$, \begin{equation}\label{litx1} \frac{m(t)}{(L\psi(l))^{{1}/{\alpha}}r^{{1}/{\alpha}}(t)\pi(t)}+m'(t)\leq0, \quad 1-p(t)\frac{m(\tau(t))}{m(t)}>0, \end{equation} and such that conditions \eqref{hls3.1} and \eqref{hls3.2} are satisfied. Then equation \eqref{0.1} is oscillatory. \end{theorem} \begin{proof} As in the proof of Theorem \ref{than3.1}, we only need to prove the case where $Z'(t)<0$. In this case, there exists a constant $l>0$ such that $0