\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2008(2008), No. 03, pp. 1--6.\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 2008 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2008/03\hfil Oscillation of solutions] {Sufficient conditions for the oscillation of solutions to nonlinear second-order \\ differential equations} \author[A. Berkane\hfil EJDE-2008/03\hfilneg] {Ahmed Berkane} \address{Ahmed Berkane \newline Facult\'e des Sciences, D\'epartement de Math\'ematiques, Universit\'e Badji Mokhtar, 23000 Annaba, Algeria} \email{aberkane20052005@yahoo.fr} \thanks{Submitted September 7, 2007. Published January 2, 2008.} \subjclass[2000]{34C15} \keywords{Nonlinear differential equations; oscillation} \begin{abstract} We present sufficient conditions for all solutions to a second-order ordinary differential equations to be oscillatory. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{corollary}[theorem]{Corollary} \section{Introduction} Kirane and Rogovchenko \cite{k3} studied the oscillatory solutions of the equation $$\big[r(t)\psi(x(t))x'(t)\big]' +p(t) x'(t)+q(t)f(x(t))=g(t), \quad t \geq t_0 , \label{eE}$$ where $t_0\geq 0$, $r(t) \in C^{1}([t_0, \infty); (0,\infty))$, $p(t)\in C([t_0, \infty); \mathbb{R})$, $q(t)\in C([t_0,\infty);\break (0, \infty))$, $q(t)$ is not identical zero on $[t_\ast,\infty)$ for some $t_\ast \geq t_0, f(x), \psi(x) \in C(\mathbb{R},\mathbb{R})$ and $\psi(x) > 0$ for $x\neq 0$. Their results read as follows \begin{theorem} \label{thmA} Case $g(t)\equiv0$: Assume that for some constants $K,C,C_{1}$ and for all $x\neq0$, $f(x)/x\geq K >0$ and $00$ in $D_0=\{(t,s): t\geq s \geq t_0\}$ \item[(ii)] $H$ has a continuous and non-positive partial derivative in $D_0$ with respect to the second variable, and $-\frac{\partial H}{\partial s }=h(t,s)\sqrt{H(t,s)}$ for all $(t,s)\in D_0$. \end{itemize} If there exists a function $\rho \in C^{1}([t_0, \infty); (0, \infty))$ such that $$\limsup_{t \rightarrow +\infty} \frac{1}{H(t,t_0)}\int_{t_0}^{t}[H(t,s) \Theta(s) -\frac{C_1}{4}\rho(s)r(s)Q^{2}(t,s)]ds=\infty \,,$$ where \begin{gather*} \Theta(t)=\rho(t)\Big( Kq(t)-\big(\frac{1}{C}-\frac{1}{C_1}\big) \frac{p^2(t)}{4r(t)}\Big),\\ Q(t,s)=h(t,s)+\big[\frac{p(s)}{C_1r(s)}-\frac{\rho'(s)}{\rho(s)}\big] \sqrt{H(t,s)}, \end{gather*} then \eqref{eE} is oscillatory. \end{theorem} \begin{theorem} \label{thmB} Case $g(t)\neq 0$: Let the assumptions of theorem 1 be satisfied and suppose that the function $g(t) \in C([t_0, \infty); \mathbb{R})$ satisfies $$\int^{\infty}\rho(s)|g(s)|ds=N<\infty\,.$$ Then any proper solution $x(t)$ of \eqref{eE}; i.e, a non-constant solution which exists for all $t\geq t_0$ and satisfies $\sup_{t\geq t_0}|x(t)|>0$, satisfies $$\liminf_{t \to \infty}|x(t)|=0.$$ \end{theorem} Note that localization of the zeros is not given in the work by Kirane and Rogovchenko \cite{k3}. Here we intend to give conditions that allow us to localize the zeros of solutions to $\eqref{eE}$. Observe that in contrast to \cite{k3} where a Ricatti type transform, $$v(t)=\rho\frac{r(t)\psi(x(t))x'(t)}{x(t)},$$ is used, here we simply use a usual Ricatti transform. \section{Main Results} \subsection*{Differential equation without a forcing term} Consider the second-order differential equation $$\label{e1} \big[r(t)\psi(x(t))x'(t)\big]'+p(t)x'(t)+q(t)f(x(t))=0, \quad t\geq t_0$$ where $t_0 \geq 0$, $r(t) \in C^{1}([t_0, \infty); (0,\infty))$, $p(t) \in C([t_0, \infty)); \mathbb{R})$, $q(t)\in C([t_0,\infty)); R)$, $p(t)$ and $q(t)$ are not identical zero on $[t_{\star}, \infty)$ for some $t_{\star}\geq t_0, f(x),\psi(x)\in C(\mathbb{R}, \mathbb{R})$ and $\psi(x) >0$ for $x\neq 0$. The next theorem follows the ideas in Nasr \cite{n1}. Assume that there exists an interval $[a,b]$, where $a,b \geq t_{\star}$, such that $e(t)\geq0$. \begin{theorem} \label{thm1} Assume that for some constants $K, C, C_1$ and for all $x\neq 0$, \begin{gather} \frac{f(x)}{x}\geq K \geq 0, \label{e2}\\ 00$for all$ t \in [a,b]$since the case when$x(t)<0$can be treated analogously. Let $$v(t)=-\frac{x'(t)}{x(t)}, \quad t \in [a,b]. \label{e5}$$ Multiplying this equality by$r(t)\psi(x(t))and differentiate the result. Using \eqref{e1} we obtain \begin{align*} (r(t)\psi(x(t))v(t))' &=-\frac{(r(t)\psi(x(t))x'(t))'}{x(t)} +r(t)\psi(x(t))v^2(t) \\ &=-p(t)v(t)+q(t)\frac{f(x(t))}{x(t)}+r(t)\psi(x(t))v^2(t)\\ &=\frac{(r(t)\psi(x(t))}{2}v^2(t)+\frac{r(t)\psi(x(t))}{2} \big(v^2(t)-2\frac{p(t)}{r(t)\psi(x(t))v(t)}\big) \\ &\quad +q(t) \frac{f(x(t))}{x(t)}\\ & =\frac{(r(t)\psi(x(t))}{2}v^2(t)+ \frac{r(t)\psi(x(t))}{2}\big(v(t)-\frac{p(t)}{(r(t)\psi(x(t))}\big)^2\\ &\quad -\frac{p^2(t)}{2r(t)\psi(x(t))} + q(t)\frac{f(x(t))}{x(t)}\,. \end{align*} Using \eqref{e2}-\eqref{e3} and the fact that $$\frac{(r(t)\psi(x(t))}{2}\big(v(t)-\frac{p(t)}{r(t)\psi(x(t))}\big)^2\geq 0,$$ we have $$\label{e6} (r(t)\psi(x(t))v(t))'\geq \frac{(r(t)\psi(x(t))}{2}v^2(t)-\frac{p^2(t)}{2 Cr(t)}+Kq(t)$$ Multiplying both sides of this inequality byu^2(t)$and integrating on$[a,b]$. Using integration by parts on the left side, the condition$u(a)=u(b)=0and \eqref{e3}, we obtain \begin{align*} 0&\geq\int_{a}^{b}\frac{r(t)\psi(x(t))}{2}v^2(t)u^2(t)dt +2 \int_{a}^{b}r(t)\psi(x(t))v(t)u(t)u'(t)dt \\ &\quad +\int_{a}^{b}Kq(t)u^2(t)dt-\int_{a}^{b}\frac{p^2(t)}{2Cr(t)}u^2(t)dt \\ &\geq \int_{a}^{b}\frac{r(t)\psi(x(t))}{2}(v^2(t)u^2(t)+4v(t)u(t)u'(t))dt\\ &\quad + \int_{a}^{b}Kq(t)u^2(t)dt-\int_{a}^{b}\frac{p^2(t)u^2(t)}{2Cr(t)}dt\\ &\geq \int_{a}^{b}\frac{r(t)\psi(x(t))}{2}[v(t)u(t)+2u'(t)]^2dt -2\int_{a}^{b}r(t)\psi(x(t))u^{\prime 2}(t)dt \\ &\quad +\int_{a}^{b}Kq(t)u^2(t)dt-\int_{a}^{b}\frac{p^2(t)}{2 Cr(t)}u^2(t)dt \\ &\geq\int_{a}^{b}\big[\big(Kq(t)-\frac{p^2(t)}{2 Cr(t)}\big)u^2(t)-2r(t)\psi(x(t))u^{\prime 2}(t)\big]dt\\ &\quad +\int_{a}^{b}\frac{r(t)\psi(x(t))}{2}[v(t)u(t)+2u'(t)]^2dt\,. \end{align*} Now, from \eqref{e3} we have \begin{align*} 0&\geq\int_{a}^{b}\big[\big(Kq(t)-\frac{p^2(t)}{2 Cr(t)}\big)u^2(t)-2r(t)C_1u^{\prime 2}(t)\big]dt\\ &\quad +\int_{a}^{b}\frac{r(t)\psi(x(t))}{2}[v(t)u(t)+2u'(t)]^2dt. \end{align*} If the first integral on the right-hand side of the inequality is greater than zero, then we have directly a contradiction. If the first integral is zero and the second is also zero thenx(t)$has the same zeros as$u(t)$at the points$a$and$b$;$(x(t)=ku^2(t))$, which is again a contradiction with our assumption. \end{proof} \begin{corollary} \label{coro1} Assume that there exist a sequence of disjoint intervals$[a_{n},b_{n}]$, and a sequence of functions$u_n(t)$defined and continuous an$[a_{n},b_{n}]$, differentiable on$(a_{n},b_{n})$with$u_n(a_n)=u_n(b_n)=0$, and satisfying assumption \eqref{e4}. Let the conditions of Theorem \ref{thm1}. hold. Then \eqref{e1} is oscillatory. \end{corollary} \subsection*{Differential equation with a forcing term} Consider the differential equation $$\label{e7} \big[r(t)\psi(x(t))x'(t)\big]'+p(t)x'(t)+q(t)f(x(t))=g(t), \quad t\geq t_0$$ where$t_0 \geq 0$,$g(t) \in C([t_0, \infty); \mathbb{R})r(t) \in C^{1}([t_0, \infty); (0,\infty))$,$p(t) \in C([t_0,\infty)); R)$,$q(t)\in C([t_0, \infty)); R)$,$p(t)$and$ q(t)$are not identical zero on$[t_{\star},\infty[$for some$t_{\star}\geq t_0$,$ f(x),\psi(x)\in C(\mathbb{R}, \mathbb{R}) $and$\psi(x) >0$for$x\neq 0$. Assume that there exists an interval$[a,b]$, where$a,b \geq t_{\star}$, such that$g(t)\geq 0 $and there exists$c\in (a,b)$such that$g(t)$has different signs on$[a,c]$and$[c,b]$. Without loss of generality, let$g(t)\leq 0$on$[a,c]$and$g(t)\geq0$on$[c,b]$. \begin{theorem} \label{thm2} Let \eqref{e3} hold and assume that $$\label{e8} \frac{f(x)}{x|x|}\geq K ,$$ for a positive constant$K$and for all$x\neq 0$. Furthermore assume that there exists a continuous function$u(t)$such that$u(a)=u(b)=u(c)=0$,$u(t)$differentiable on the open set$(a,c)\cup(c,b)$, and satisfies the inequalities \begin{gather} \int_{a}^{c} \big[ \big(\sqrt{Kq(t)g|(t)|}-\frac{p^2(t)}{2Cr(t)}\big)u^2-2C_1r(t)(u')^2 (t)\big]d(t)\geq 0, \label{e9} \\ \int_{c}^{b} \big[\big(\sqrt{Kq(t)g|(t)|}-\frac{p^2(t)}{2Cr(t)}\big)u^2 -2C_1r(t)(u')^2 (t)\big]d(t)\geq 0\,. \label{e10} \end{gather} Then every solution of equation \eqref{e7} has a zero in$[a,b]$. \end{theorem} \begin{proof} Assume to the contrary that$x(t)$, a solution of \eqref{e7}, has no zero in$[a,b]$. Let$x(t) < 0for example. Using the same computations as in the first part, we obtain: \begin{align*} (r(t)\psi(x(t))v(t))' &= -\frac{(r(t)\psi(x(t))x'(t))'}{x(t)}+r(t)\psi(x(t))v^2(t) -\frac{g(t)}{x(t)} \\ &=-p(t)v(t)+q(t)\frac{f(x(t))}{x(t)}+r(t)\psi(x(t))v^2(t) -\frac{g(t)}{x(t)} \\ &= \frac{(r(t)\psi(x(t))}{2}v^2(t)+\frac{r(t)\psi(x(t))}{2} \big(v^2(t)-2\frac{p(t)v(t)}{r(t)\psi(x(t))}\big) \\ &\quad +q(t) \frac{f(x(t))}{x(t)}-\frac{g(t)}{x(t)} \\ &=\frac{(r(t)\psi(x(t))}{2}v^2(t)+\frac{r(t)\psi(x(t))}{2} \big(v(t)-\frac{p(t)}{r(t)\psi(x(t))v(t)}\big)^2 \\ &\quad -\frac{p^2(t)}{2r(t)\psi(x(t))} + q(t)\frac{f(x(t))}{x(t)}-\frac{g(t)}{x(t)} \end{align*} Fort\in [c,b]we have \begin{align*} (r(t)\psi(x(t))v(t))' &= \frac{r(t)\psi(x(t))}{2}v^2(t) +\frac{r(t)\psi(x(t))}{2}\big(v(t)-\frac{p(t)}{r(t)\psi(x(t))}\big)^2 \\ &\quad -\frac{p^2(t)}{2 r(t)\psi(x(t))}+ q(t)\frac{f(x(t))}{x(t)|x(t)|}|x(t)|+\frac{|g(t)|}{|x(t)|} \end{align*} From \eqref{e8}, and using the fact that $$\frac{r(t)\psi(x(t))}{2}\big(v(t)-\frac{p(t)}{r(t)\psi(x(t))}\big)^2\geq0$$ we deduce $$\label{e11} (r(t)\psi(x(t))v(t))'\geq \frac{(r(t)\psi(x(t))}{2}v^2(t)-\frac{p^2(t)}{2 r(t)\psi(x(t))}+ Kq(t)|x(t)|+\frac{|g(t)|}{|x(t)|}\,.$$ Using the H\"{o}lder inequality in \eqref{e11} we obtain $$\label{e12} (r(t)\psi(x(t))v(t))'\geq \frac{(r(t)\psi(x(t))}{2}v^2(t) +\sqrt{Kq(t)|g(t)|}-\frac{p^2(t)}{2 r(t)\psi(x(t))}\,.$$ Multiplying both sides of this inequality byu^2(t)$and integrating on$[c,b]$, we obtain after using integration by parts on the left-hand side and the condition$u(c)=u(b)=0, \begin{align*} 0&\geq\int_{c}^{b}\frac{r(t)\psi(x(t))}{2}v^2(t)u^2(t)dt+ \int_{c}^{b}\sqrt{Kq(t)|g(t)|}u^2(t)dt \\ &\quad -\int_{c}^{b}\frac{p^2(t)u^2(t)}{2 r(t)\psi(x(t))}dt+2\int_{c}^{b}r(t)\psi(x(t))v(t)u(t)u'(t)dt \\ &\geq\int_{c}^{b}\frac{r(t)\psi(x(t))}{2} [v(t)u(t)-2u'(t)]^2dt-2 \int_{c}^{b}r(t)\psi(x(t))u^{\prime 2}(t)dt \\ &\quad +\int_{c}^{b}\sqrt{Kq(t)|g(t)|}u^2(t)dt -\int_{c}^{b}\frac{p^2(t)u^2(t)}{2 r(t)\psi(x(t))}dt. \end{align*} Assumption \eqref{e3} allows us to write \begin{align*} 0&\geq\int_{c}^{b}\frac{r(t)\psi(x(t))}{2} [v(t)u(t)+2u'(t)]^2dt-2\int_{c}^{b}C_1r(t)(u')^2(t)dt \\ &\quad +\int_{c}^{b}\sqrt{Kq(t)|g(t)|}u^2(t)dt -\int_{c}^{b}\frac{p^2(t)u^2(t)}{2 C r(t)}dt \\ &\geq\int_{c}^{b}\frac{r(t)\psi(x(t))}{2} [v(t)u(t)+2u'(t)]^2dt \\ &\quad + \int_{c}^{b} \big[\big(\sqrt{Kq(t)g|(t)|}-\frac{p^2(t)}{2Cr(t)} \big)u^2(t)-2C_1r(t)(u')^2(t)\big]dt. \end{align*} This leads to a contradiction as in Theorem \ref{thm1}; the proof is complete. \end{proof} \begin{corollary} \label{coro2} Assume that there exist a sequence of disjoint intervals[a_n,b_n]$a sequences of points$c_n\in (a_n,c_n)$, and a sequence of functions$u_n(t)$defined and continuous on$[a_n,b_n]$, differentiable on$(a_n,c_n)\cup(c_n,b_n) $with$u_n(a_n)=u_n(b_n)=u_n(c_n)=0$, and satisfying assumptions \eqref{e9}-\eqref{e10}. 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