\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
\begin{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}
\end{equation}
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 $0<C\leq\Psi(x)\leq C_{1}$.
Let $h$, $H\in C(D,R),$ where  $D=\{(t,s): t\geq s \geq  t_0\}$,
be  such that
\begin{itemize}
\item[(i)] $ H(t,t)=0 $ for $t\geq t_0$, $ H(t,s)>0 $ 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
\begin{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
\end{equation}
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}\\
 0<C\leq \psi(x)\leq C_1 \label{e3}\,.
\end{gather}
 Suppose further there exists a continuous function $u(t)$ such
that $u(a)=u(b)=0$, $u(t)$ is differentiable on the open set $(a,b)$,
$a, b \geq t_{\star}$, and
\begin{equation} \label{e4}
 \int_{a}^{b}
\big[\big(Kq(t)-\frac{p^2(t)}{2Cr(t)}\big)
u^2(t)-2C_1r(t)(u')^2(t)\big]dt\geq 0\,.
\end{equation}
Then every solution of  \eqref{e1} has a zero in $[a,b]$.
\end{theorem}


\begin{proof} Let $x(t)$ be a solution of \eqref{e1} that has
zero on $[a,b]$.
We may assume that $x(t)>0$ for all $ t \in [a,b]$ since the case
when $x(t)<0$  can be treated analogously. Let
\begin{equation}
v(t)=-\frac{x'(t)}{x(t)}, \quad t \in [a,b]. \label{e5}
\end{equation}
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
\begin{equation} \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)
\end{equation}
Multiplying both sides of this inequality by $u^2(t)$ and integrating on
$[a,b]$.  Using integration by parts on the left
side, the condition $u(a)=u(b)=0$ and  \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 then $x(t)$ 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
\begin{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
\end{equation}
 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
\begin{equation} \label{e8}
\frac{f(x)}{x|x|}\geq K ,
\end{equation}
 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 $x(t) < 0$ for 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*}
For $t\in [c,b]$ 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
\begin{equation} \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)|}\,.
\end{equation}
 Using the H\"{o}lder inequality in \eqref{e11} we obtain
\begin{equation} \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))}\,.
\end{equation}
Multiplying both sides of this inequality  by $u^2(t)$ 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}.
Let the conditions of Theorem \ref{thm2} hold.
Then  \eqref{e7} is oscillatory.
\end{corollary}

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  \end{document}