\documentclass[reqno]{amsart}
\usepackage{hyperref}


\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2014 (2014), No. 103, pp. 1--10.\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-2014/103\hfil Oscillation criteria]
{Oscillation criteria for second-order nonlinear perturbed differential equations}

\author[P. Temtek \hfil EJDE-2014/103\hfilneg]
{Pakize Temtek}  % in alphabetical order

\address{Pakize Temtek \newline
Department of Mathematics, Faculty of Science, 
Erciyes University, Kayseri, Turkey}
\email{temtek@erciyes.edu.tr}

\thanks{Submitted January 13, 2014. Published April 11, 2014.}
\subjclass[2000]{34C10, 35C15}
\keywords{Oscillation; second order nonlinear differential equation}

\begin{abstract}
 In this article, we study the oscillation of solutions to the nonlinear
 second-order differential equation
 $$
 \Big(r(t)\psi(x(t))|x'(t)|^{\alpha-1}x'(t)\Big)'
 +P(t,x'(t))\psi(x(t))+Q(t,x(t))=0.
 $$
 We obtain sufficient conditions for the oscillation of all solutions to 
 this equation.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\allowdisplaybreaks

\section{Introduction}

This article concerns the oscillation of solutions to the nonlinear
second-order differential equation
\begin{equation} \label{e1}
\Big(r(t)\psi(x(t))|x'(t)|^{\alpha-1}x'(t)\Big)'+P(t,x'(t))\psi(x(t))+Q(t,x(t))=0,
 \quad t\geq t_0
\end{equation}
where  $r\in C^1(I,\mathbb{R^{+}})$, $P,Q\in C(I\times\mathbb{R},\mathbb{R})$,
$\psi(x)\in C(\mathbb{R,\mathbb{R^{+}}})$,
$I=[T_0,\infty)\subset\mathbb{R}$, $0<\psi(x)<\gamma$ and $\alpha$
is a positive constant. Throughout this article,
 we  assume the following conditions:
\begin{itemize}
\item[(E1)] $Q\in C(I\times\mathbb{R},\mathbb{R})$ and there 
exist $f\in C^1(\mathbb{R},\mathbb{R})$ and a continuous
function $q(t)$ such that
\[
xf(x)>0 \quad\text{and}\quad \frac{Q(t,x)}{f(x)}\geq q(t)\quad\text{for }x\neq0.
\]

\item[(E2)] $P\in C(I\times\mathbb{R},\mathbb{R})$ and there exists a
continuous function $p(t)$ such that
\[
\frac{P(t,x'(t))}{|x'(t)|^{\alpha-1}x'(t)}\geq p(t)\quad\text{for }x'\neq0.
\]
\end{itemize}
We restrict our attention  to  solutions
satisfying  $\sup\{|x(t)|:t\geq T\}>0 $ for all $T\geq T_0$.

A solution of \eqref{e1} is said to be oscillatory if it has arbitrarily
large zeros, and otherwise it is said to be non-oscillatory.
If all solutions of \eqref{e1} are oscillatory, \eqref{e1}
is called oscillatory.

The oscillatory behavior of solutions of second-order ordinary
differential equations,  including the existence of oscillatory and
non-oscillatory solutions, has been the subject of intensive investigations;
see for example \cite{l1}--\cite{y1}.
Some criteria involve the behavior of the integral of alternating 
coefficients.
In this article, we give general integral criteria for the oscillation
of \eqref{e1}, which contain some of the results in the references as 
particular cases.

\section{Main results}


 Let  $h(\cdot)$ and $K(\cdot,\cdot,\cdot):\mathbb{R}\times\mathbb{R}
\times\mathbb{R^{+}}\to\mathbb{R}$ be continunous functions such that 
for each fixed $t,s$, the function $K(t,s,.)$ is nondecreasing.
Then there exists a solution to the integral equation
\begin{equation} \label{e2}
v(t)=h(t)+\int_{t_0}^t K(t,s,v(s))\,ds,\quad t\geq t_0.
\end{equation}
Furthermore there exists a ``minimal solution''
$v$ in the sense that any solution $y$ of this equation satisfies 
$v(t)\leq y(t)$ for all $t\geq t_0$.
See \cite[p. 322]{l1}.
 
\begin{lemma}[cite{w1}] \label{lem1}
If $v$ is the minimal solution of \eqref{e2} and 
\[
u(t)\geq h(t)+\int_{t_0}^{t}K(t,s,u(s))ds, \quad t\geq t_0,
\]
then $u(t)\geq v(t)$ for all $t\geq t_0$.

Similarly for a maximal solution $w(t)$ of \eqref{e2}:
if $u(t)\leq h(t)+\int_{t_0}^tK(t,s,u(s))ds$,
then $u(t)\leq w(t)$ for all $t\geq t_0$.
\end{lemma}

Our main results reads as follows.

\begin{theorem} \label{thm1}
Assume {\rm (E1)}, $f'(x)\geq0$, $ p(t)\leq 0$, $q(t)>0$
and  $\int_{t_0}^{\infty}(\frac{1}{r^{1/\alpha}(t)})dt=\infty$.
Also assume that there exists a positive function $\rho(t)$ such that
\begin{gather} \label{e3}
\int_{t_0}^{\infty}q(t)\rho(t)dt=\infty, \\
\label{e4}
p(t)\rho(t)\geq r(t)\rho'(t).
\end{gather}
Then every solution of \eqref{e1} is oscillatory.
\end{theorem}

\begin{proof}
 For the shake of contradiction, suppose that \eqref{e1} has a non-oscillatory
solution $x(t)$. Without loss of generality, suppose that it is an eventually
positive solution (if it is an eventually negative solution, the proof is similar),
that is, $x(t)>0$ for all $t\geq t_0$.
We consider the following three cases.
\smallskip

\textbf{Case 1.} 
Suppose that $x'(t)$ is oscillatory. Then there exists $t_1\geq t_0$ such that
 $x'(t_1)=0$. From \eqref{e1}, we have
\begin{align*}
&\Big[r(t)\psi(x(t))|x'(t)|^{\alpha-1}x'(t)\exp
\Big( \int_{t_0}^{t}\frac{p(s)}{r(s)}ds\Big)\Big]'
\\
& =[r(t)\psi(x(t))|x'(t)|^{\alpha-1}x'(t)]'\exp
\Big( \int_{t_0}^{t}\frac{p(s)}{r(s)}ds\Big)
\\
&\quad +p(t)\psi(x(t))|x'(t)|^{\alpha-1}x'(t)\exp
\Big( \int_{t_0}^{t}\frac{p(s)}{r(s)}ds\Big)
\\
&=(-P(t,x'(t))\psi(x(t))-Q(t,x(t)))\exp
\Big( \int_{t_0}^{t}\frac{p(s)}{r(s)}ds\Big)\\
&\quad +p(t)\psi(x(t))|x'(t)|^{\alpha-1}x'(t)\exp
\Big( \int_{t_0}^{t}\frac{p(s)}{r(s)}ds\Big)
\\
&\leq (-p(t)|x'(t)|^{\alpha-1}x'(t)\psi(x(t))-q(t)f(x(t)))\exp
\Big( \int_{t_0}^{t}\frac{p(s)}{r(s)}ds\Big)
\\
&\quad +p(t)\psi(x(t))|x'(t)|^{\alpha-1}x'(t)\exp
\Big(\int_{t_0}^{t}\frac{p(s)}{r(s)}ds\Big)
\\
&=-q(t)f(x(t))\exp\Big( \int_{t_0}^{t}\frac{p(s)}{r(s)}ds\Big)<0
\end{align*} %e2.5
which implies that
\begin{align*}
&r(t)\psi(x(t))|x'(t)|^{\alpha-1}x'(t)\exp
\Big( \int_{t_0}^{t}\frac{p(s)}{r(s)}ds\Big)\\
&<r(t_1)\psi(x(t_1))|x'(t_1)|^{\alpha-1}x'(t_1)\exp
\Big( \int_{t_0}^{t_1}\frac{p(s)}{r(s)}ds\Big)=0,\quad \forall t\geq t_1\,.
\end{align*} %e2.6
it follows that $x'(t)<0$ for all $t> t_1$, which contradicts to the assumption 
that $x'(t)$ is oscillatory.
\smallskip

\textbf{Case 2.} Assume that $x'(t)<0$. From \eqref{e1}, we obtain
\begin{align*}
-[r(t)\psi(x(t))|x'(t)|^{\alpha-1}x'(t)]'
&=[r(t)\psi(x(t))(-x'(t))^{\alpha}]'\\
&=P(t,x'(t))\psi(x(t))+Q(t,x(t)) \\
&\geq p(t)|x'(t)|^{\alpha-1}x'(t)\psi(x(t))+q(t)f(x(t))\\
&=-p(t)(-x'(t))^{\alpha}\psi(x(t))+q(t)f(x(t))\geq 0
\end{align*} %2.7
then there exists an $M>0$ and a $t_1\geq t_0$, such that
\begin{equation} \label{e8}
r(t)\psi(x(t))(-x'(t))^{\alpha} \geq M,\quad\forall t\geq t_1.
\end{equation}
It follows that
\begin{gather*}
\gamma(-x'(t))^{\alpha} \geq \frac{M}{r(t)},\\
x(t)\leq -\int_{t_1}^{\infty}\big(\frac{M}{\gamma}\big)^{1/\alpha}
\frac{1}{r^{1/\alpha}(t)}dt,\quad\forall t\geq t_1
\end{gather*}
which implies  $ \lim_{t\to\infty}x(t)=-\infty$;
this contradicts the assumption that $x(t)>0$.
\smallskip

\textbf{Case 3.} Suppose that $x'(t)>0$. Define  
$w(t)=\rho(t)r(t)\psi(x(t))(x'(t))^{\alpha} $.
 Differentiating $w(t)$ and using \eqref{e1},
\begin{equation} \label{e10}
w'(t)=[r(t)\psi(x(t))(x'(t))^{\alpha}]'\rho(t)+r(t)\psi(x(t))
(x'(t))^{\alpha}\rho'(t),\quad\forall t\geq t_0.
\end{equation}
Then we obtain
\begin{align*}
\frac{w'(t)}{f(x(t))}
&=-\frac{P(t,x'(t))\psi(x(t))\rho(t)}{f(x(t))}
-\frac{Q(t,x(t))\rho(t)}{f(x(t))} \\
&\quad +\frac{\rho'(t)r(t)\psi(x(t))(x'(t))^{\alpha}}{f(x(t))}, \quad
\forall t\geq t_0\,.
\end{align*}
Noticing that
\begin{align*}
\Big(\frac{w(t)}{f(x(t))}\Big)'
&=\frac{w'(t)f(x(t))-w(t)f'(x(t))x'(t)}{f^{2}(x(t))} \\
&=-\frac{P(t,x'(t))\psi(x(t))\rho(t)}{f(x(t))}
 -\frac{Q(t,x(t))\rho(t)}{f(x(t))} \\
&\quad +\frac{\rho'(t)r(t)\psi(x(t))(x'(t))^{\alpha}}{f(x(t))}
 -\frac{w(t)f'(x(t))x'(t)}{f^{2}(x(t))},\quad\forall t\geq t_0\,.
\end{align*}
Integrating the above from $t_0$ to $t$, we obtain
\begin{align*}
\frac{w(t)}{f(x(t))}
&=\frac{w(t_0)}{f(x(t_0))}
 -\int_{t_0}^{t}[\frac{P(s,x'(s))\psi(x(s))\rho(s)}{f(x(s))}
 +\frac{Q(s,x(s))\rho(s)}{f(x(s))}\\
&\quad-\frac{\rho'(s)r(s)\psi(x(s))(x'(s))^{\alpha}}{f(x(s))}
+\frac{w(s)f'(x(s))x'(s)}{f^{2}(x(s))} ]ds,
\end{align*}
\begin{align*}
\frac{w(t)}{f(x(t))}
&\leq \frac{w(t_0)}{f(x(t_0))}
-\int_{t_0}^{t}[q(s)\rho(s)
 +\frac{(\rho(s)p(s)-\rho'(s)r(s))(x'(s))^{\alpha}\psi(x(s))}{f(x(s))}\\
&\quad +\frac{w(s)f'(x(s))x'(s)}{f^{2}(x(s))} ]ds.
\end{align*}
Using \eqref{e3}, \eqref{e4} and $x'(t)>0$, we have
\begin{eqnarray*}
0\leq \lim_{t\to \infty}\frac{w(t)}{f(x(t))}=-\infty,
\end{eqnarray*}
this is a contradiction. The proof is complete.
\end{proof}

\begin{theorem} \label{thm2} 
Assume that $f'(x)\geq 0$ and $\psi(x(t))\equiv1$. Also assume that
\begin{gather} \label{e13}
\rho_0(t)=\exp\Big(\int_{t_0}^{t}\frac{p(s)}{r(s)}ds \Big),\\
\label{e14}
\int_{t_0}^{\infty}\frac{dt}{(\rho_0(t)r(t))^{1/\alpha}}=\infty,
\end{gather}
and $\rho_0(t)$  satisfies  \eqref{e3}. 
Then every solution of \eqref{e1} is oscillatory.
\end{theorem}

\begin{proof} 
Let $x(t)$ be a non-oscillatory solution of \eqref{e1}.
 Without loss of generality, we  assume  that $x(t)$ is eventually  
positive.  Let 
  $w(t)=\rho_0(t)r(t)|x'(t)|^{\alpha-1}x'(t)$. Then  
\[
w(t)x'(t)=\rho_0(t)r(t)|x'(t)|^{\alpha-1}(x'(t))^{2}\geq 0\quad \text{for }
t\geq t_0
\]
 and
\begin{equation} \label{e15}
w'(t)=(r(t)|x'(t)|^{\alpha-1}x'(t))'\rho_0(t)+r(t)|x'(t)|^{\alpha-1}x'(t)\rho_0'(t)
\quad \forall t\geq t_0\,.
\end{equation}
In view of \eqref{e1} and \eqref{e13}, we obtain
\begin{equation} \label{e16}
\begin{gathered}
w'(t)=(-P(t,x'(t))-Q(t,x(t)))\rho_0(t)+|x'(t)|^{\alpha-1}x'(t)p(t)\rho_0(t),\\
w'(t)\leq(-p(t)|x'(t)|^{\alpha-1}x'(t)-q(t)f(x(t)))\rho_0(t)+|x'(t)|^{\alpha-1}
x'(t)p(t)\rho_0(t), \\
\frac{w'(t)}{f(x(t))}\leq -q(t)\rho_0(t)\quad\forall t\geq t_0\,.
\end{gathered}
\end{equation}
Since
\begin{align*}
\Big(\frac{w(t)}{f(x(t))}\Big)'
&=\frac{w'(t)f(x(t))-w(t)f'(x(t))x'(t)}{f^{2}(x(t))} \\
&\leq -q(t)\rho_0(t)-\frac{w(t)f'(x(t))x'(t)}{f^{2}(x(t))} \quad\forall t\geq t_0,
\end{align*}
integrating  from $t_0$ to $t$, we have
\begin{align*}
-\frac{w(t)}{f(x(t))}
\geq -\frac{w(t_0)}{f(x(t_0))}+\int_{t_0}^{t}q(s)\rho_0(s)ds
+\int_{t_0}^{t}\frac{w(s)f'(x(s))x'(s)}{f^{2}(x(s))}ds,\quad\forall t\geq t_0.
\end{align*}
By using \eqref{e3}, there exists a constant $ m>0$ and $t_1\geq t_0$ such that
\begin{equation} \label{e19}
-\frac{w(t_0)}{f(x(t_0))}+\int_{t_0}^{t}q(s)\rho_0(s)ds+
\int_{t_0}^{t_1}\frac{w(s)f'(x(s))x'(s)}{f^{2}(x(s))}ds\geq m \quad\forall t\geq t_0
\end{equation}
which means that
\begin{equation} \label{e20}
-\frac{w(t)}{f(x(t))}\geq m+\int_{t_1}^{t}\frac{w(s)f'(x(s))x'(s)}{f^{2}(x(s))}ds.
\end{equation}
Because that $x(t)$ is positive,  \eqref{e20} implies $-w(t)>0$, 
or equivalently $x'(t)<0$.

Let
\begin{equation} \label{e21}
u(t)=-w(t)=- \rho_0(t)r(t)|x'(t)|^{\alpha-1}x'(t)=\rho_0(t)r(t)(-x'(t))^{\alpha},
\end{equation}
thus \eqref{e20} can be written as
\begin{equation} \label{e22}
u(t)\geq m f(x(t))+\int_{t_1}^{t}\frac{f(x(t))f'(x(s))(-x'(s))}{f^{2}(x(s))}u(s)ds.
\end{equation}
Define
\begin{equation} \label{e23}
K(t,s,u)=\frac{f(x(t))f'(x(s))(-x'(s))}{f^{2}(x(s))}u.
\end{equation}
Then, for any fixed $t$ and $s$, $K(t,s,u)$ is nondecreasing in $u$. 
Let $v(t)$ be the minimal solution of the equation
\begin{equation} \label{e24}
v(t)= m f(x(t))+\int_{t_1}^{t}\frac{f(x(t))f'(x(s))(-x'(s))}{f^{2}(x(s))}v(s)ds.
\end{equation}
Applying Lemma \ref{lem1}, we obtain
\begin{equation} \label{e25}
u(t)\geq v(t)\quad\forall t\geq t_0.
\end{equation}
Dividing both sides of \eqref{e24} by $f(x(t))$ and deriving both sides of
 \eqref{e24},
\begin{equation}
\begin{aligned}
\Big(\frac{v(t)}{f(x(t))}\Big)'
=\Big(m+\int_{t_1}^{t}\frac{f'(x(s))(-x'(s))}{f^{2}(x(s))}v(s)ds\Big)'
=\frac{f'(x(t))(-x'(t))}{f^{2}(x(t))}v(t).
\end{aligned}  \label{e26}
\end{equation}
On the other hand
\begin{equation} \label{e27}
\Big(\frac{v(t)}{f(x(t))}\Big)'
=\frac{v'(t)}{f(x(t))}+\frac{f'(x(t))(-x'(t))}{f^{2}(x(t))}v(t).
\end{equation}
Combining \eqref{e26} and \eqref{e27}, it follows that
\begin{equation} \label{e28}
v'(t)\equiv 0.
\end{equation}
So $v(t)=v(t_1)=mf(x(t_1))$, $ t\geq t_0$. From \eqref{e25}, we obtain
\begin{equation} \label{e29}
-x'(t)\geq (mf(x(t_1)))^{1/\alpha}\frac{1}{(\rho_0(t)r(t))^{1/\alpha}},\quad
\forall t\geq t_1.
\end{equation}
Integrating both sides of this inequality above from $t_1$ to $t$, we have
\[ % 2.30
-x(t)+x(t_1)\geq (mf(x(t_1)))^{1/\alpha}\int_{t_1}^{t}
\frac{ds}{(\rho_0(s)r(s))^{1/\alpha}}.
\]
Letting $t\to\infty$, and using \eqref{e14}, it follows 
that $\lim_ {t\to\infty} x(t)\leq -\infty$, which contradicts to that 
$x(t)$ is eventually positive. The proof is complete.
\end{proof}

In what follows, we always assume that 
$H(t)\in C^{2}(\mathbb{R};\mathbb{R})$ and it satisfies the following 
two conditions:
\begin{itemize}
\item[(H1)]  $H(t)>0$ for all $t\geq t_0$, $H(t)$ is a bounded;

\item[(H2)]  $H'(t)=h(t)$ is a bounded.
\end{itemize}

\begin{theorem} \label{thm3}
 Assume that $f'(x)\geq 0$, 
$\int_{t_0}^{\infty}\frac{dt}{(r(t))^{1/\alpha}}=\infty$, $\psi(x(t))\equiv1$,
and
\begin{equation} \label{e31}
p(t)\leq 0, \quad q(t)> 0,
\end{equation}
or
\begin{equation} \label{e32}
p(t)\leq 0, \quad q(t)\leq 0, \quad \lim_{t\to\infty}\frac{p(t)}{q(t)}=M>0.
\end{equation}
Suppose further that there exists a function $H(t)$ that satisfies
{\rm (H1), (H2)}, and such that
\begin{gather} \label{e33}
\int_{t_0}^{\infty}H(t)\varphi(t)dt=\infty, \\
\label{e34}
\limsup_{t\to\infty}v(t)r(t)<\infty,
\end{gather}
where
\begin{gather} \label{e35}
\varphi(t)=v(t)(q(t)-p(t)h(t)-(r(t)h(t))'), \\
 \label{e36}
v(t)=\exp\Big(\int_{t_0}^{t}\big(\frac{p(s)}{r(s)}-\frac{h(s)}{H(s)}\big)ds\Big).
\end{gather}
Then every solution of \eqref{e1} is oscillatory.
\end{theorem}

\begin{proof} 
Assume to the contrary that \eqref{e1} has a non-oscillatory solution $x(t)$.
 Without loss of generality, we may assume that $x(t)>0 $ for all $t\geq t_0 $.
Define
\begin{equation} \label{e37}
u(t)=v(t)r(t)\Big(\frac{|x'(t)|^{\alpha-1}x'(t)}{f(x(t))}+h(t)\Big).
\end{equation}
Differentiating, we obtain
\begin{gather*}
\begin{aligned}
u'(t)&=\Big(\frac{p(t)}{r(t)}-\frac{h(t)}{H(t)}\Big)u(t)
 +v(t)\Big[-\frac{P(t,x'(t))}{f(x(t))}\\
&\quad -\frac{Q(t,x(t))}{f(x(t))}
-\frac{r(t)|x'(t)|^{\alpha-1}(x'(t))^{2}f'(x(t))}{f^{2}(x(t))}+(r(t)h(t))'\Big],
\end{aligned} \\
\begin{aligned}
u'(t)&\leq \Big(\frac{p(t)}{r(t)}-\frac{h(t)}{H(t)}\Big)u(t)
 +v(t)\Big[-\frac{p(t)|x'(t)|^{\alpha-1}x'(t)}{f(x(t))}-q(t)\\
&\quad -\frac{r(t)|x'(t)|^{\alpha-1}(x'(t))^{2}f'(x(t))}{f^{2}(x(t))}+(r(t)h(t))'\Big]
\\
&\leq p(t)v(t)h(t)-\frac{h(t)}{H(t)}u(t)-q(t)v(t)+v(t)(r(t)h(t))'
\\
&=-\frac{h(t)}{H(t)}u(t)-v(t)[q(t)-p(t)h(t)-(r(t)h(t))']
\end{aligned} \\
u'(t)\leq -\frac{h(t)}{H(t)}u(t)-\varphi(t).
\end{gather*}
Multiplying  by $H(t)$, it follows that
\begin{equation} \label{e39}
\varphi(t)H(t)\leq -H(t)u'(t)-h(t)u(t).
\end{equation}
We consider the following three cases.
\smallskip

\textbf{Case 1.} $u(t)$ is oscillatory. 
Then there exists a sequence $\{t_n\}$, $(n=1,2,\dots)$, $t_n\to\infty$ as 
$n\to\infty$ and such that $u(t_n)=0$ $(n=1,2,\dots)$. Integrating both 
sides of \eqref{e39} from $t_0$ to $t_n$, we obtain
\begin{align*}
\int_{t_0}^{t_n}H(t)\varphi(t)dt
&\leq -\int_{t_0}^{t_n}H(t)u'(t)dt-\int_{t_0}^{t_n}h(t)u(t)dt\\
&= -H(t)u(t)\mid_{t_0}^{t_n}-\int_{t_0}^{t_n}(-H'(t)u(t)+h(t)u(t))dt\\
&= H(t_0)u(t_0)-H(t_n)u(t_n)=H(t_0)u(t_0);
\end{align*}
that is,
\[ % \label{e41}
\lim_{t_n\to\infty}\int_{t_0}^{t_n}H(t)\varphi(t)dt\leq H(t_0)u(t_0),
\]
which contradicts \eqref{e33}.
\smallskip

\textbf{Case 2.} $u(t)$ is eventually positive. Integrating both sides 
of \eqref{e39} from $t_0$ to $\infty$, we obtain
\[
\int_{t_0}^{\infty}H(t)\varphi(t)dt\leq H(t_0)u(t_0)-\lim_{t\to\infty}H(t)u(t)\leq H(t_0)u(t_0),
\]
which also contradicts\eqref{e33}.
\smallskip

\textbf{Case 3.} $u(t)$ is eventually negative. 
If  $\limsup_{t\to\infty}u(t)>-\infty $, then there exists a sequence 
$\{\bar{t}_{n}\}$, (n=1,2,\dots), that satisfies $\{\bar{t}_{n}\}\to\infty$  
as  $n\to\infty$ and such that 
$\lim_{\bar{t}_n\to\infty}u(\bar{t}_n)=\limsup_{t\to\infty}u(t)=M_1>-\infty$. 
Because $H(t)$ is a bounded function, then there exists a  $M_2>0$ such that 
$H(\bar{t}_n)\leq M_2$, (n=1,2,\dots). According to \eqref{e39}, we obtain
\begin{equation} \label{e43}
\int_{t_0}^{\bar{t}_n}H(t)\varphi(t)dt
\leq H(t_0)u(t_0)-H(\bar{t}_n)u(\bar{t}_n)\leq H(t_0)u(t_0)-M_2u(\bar{t}_n).
\end{equation}
Using \eqref{e33} and taking limit as $\bar{t}_{n}\to\infty$,
 it is easy to show that
\begin{align*}
\infty&=\lim_{\bar{t}_n\to\infty}\int_{t_0}^{\bar{t}_n}H(t)\varphi(t)dt\\
&\leq H(t_0)u(t_0)-\lim_{\bar{t}_n\to\infty}H(\bar{t}_n)u(\bar{t}_n)\\
&\leq H(t_0)u(t_0)-M_1M_2<\infty,
\end{align*}
which is obviously a contradiction.

If $\limsup_{t\to\infty}u(t)=-\infty$, then
 $\lim_{t\to\infty}u(t)=-\infty$. From the definition of $h(t)$, 
combining \eqref{e34} and \eqref{e37}, it follows that $x'(t)<0$ and
\[
\lim_{t\to\infty}(|x'(t)|^{\alpha-1}x'(t)/f(x(t)))=-\infty, 
\]
which implies that
 $\lim_{t\to\infty}((-x'(t))^{\alpha}/f(x(t)))=\infty$. 
Owing to $p(t)\leq 0$, $q(t)\geq 0$, or $p(t)\leq 0$, $q(t)\leq 0$ 
and $\lim_{t\to\infty}(p(t)/q(t))=M>0$, using the similar method of the proof 
of Case 2 in Theorem \ref{thm1}, we will derive a contradiction. 
The proof is complete.
\end{proof}


\begin{theorem} \label{thm4} 
Assume that \eqref{e34} holds, $f'(x)\geq 0$, 
$\int_{t_0}^{\infty}\frac{dt}{(r(t))^{1/\alpha}}=\infty$, 
and \eqref{e31} or \eqref{e32} hold.
Suppose further that there exists a function $H(t)$ that satisfies 
{\rm (H1), (H2)}, and such that
\begin{equation} \label{e45}
\int_{t_0}^{\infty}H(t)\bar{\varphi}(t)dt=\infty,
\end{equation}
where
\begin{equation} \label{e46}
\bar{\varphi}(t)=v(t)(q(t)+p(t)h(t)+(r(t)h(t))'),
\end{equation}
and $v(t)$ is defined in \eqref{e36}. 
Then every solution of \eqref{e1} is oscillatory when $\psi(x(t))\equiv1$.
\end{theorem}

\begin{proof} 
For the sake of contradiction, let \eqref{e1} have a non-oscillatory solution. 
Without loss of generality, we may assume that \eqref{e1} has an eventually 
positive  $x(t)>0 $ for all $t\geq t_0 $.
Define
\[
u(t)=v(t)r(t)\Big(\frac{|x'(t)|^{\alpha-1}x'(t)}{f(x(t))}-h(t)\Big).
\]
The rest of proof is similar to Theorem \ref{thm3} and is omitted.
\end{proof}

\begin{theorem} \label{thm5} 
Assume \eqref{e34}, $p(t)\leq0$, $q(t)>0$, $f'(x)\geq 0$ and  
$\int_{t_0}^{\infty}\frac{dt}{(r(t))^{1/\alpha}}=\infty$.
Suppose further that there exists a function $H(t)$ that satisfies 
{\rm (H1), (H2)}, and such that
\begin{equation} \label{e48}
\int_{t_0}^{\infty}H(t){\phi}(t)dt=\infty,
\end{equation}
where
\begin{equation} \label{e49}
{\phi}(t)=v(t)(-p(t)h(t)-(r(t)h(t))'),
\end{equation}
where $v(t)$ is defined in \eqref{e36}.
 Then every solution of \eqref{e1} is oscillatory when $\psi(x(t))\equiv1$.
\end{theorem}

\begin{proof} 
To the contrary, assume that \eqref{e1} has a non-oscillatory solution $x(t)$. 
Without loss of generality, we may assume that \eqref{e1} has an eventually 
positive  $x(t)>0 $ for all $t\geq t_0 $. Define
\begin{equation} \label{e50}
u(t)=v(t)r(t)\Big(\frac{|x'(t)|^{\alpha-1}x'(t)}{x(t)}+h(t)\Big).
\end{equation}
We use (E1)  and  noting that $xf(x)\geq0$ for $x\neq0$, so 
$\frac{f(x)}{x}\geq0$ for $x\neq0$. Differentiating \eqref{e50}, we obtain
\begin{align*}
u'(t)&=\Big(\frac{p(t)}{r(t)}-\frac{h(t)}{H(t)}\Big)u(t)
 +v(t)\Big[-\frac{P(t,x'(t))}{x(t)}-\frac{Q(t,x(t))}{x(t)}\\
&\quad -\frac{r(t)|x'(t)|^{\alpha-1}(x'(t))^{2}}{x^{2}(t)}+(r(t)h(t))'\Big]
\\
&\leq\Big(\frac{p(t)}{r(t)}-\frac{h(t)}{H(t)}\Big)u(t)
 +v(t)\Big[-\frac{p(t)|x'(t)|^{\alpha-1}x'(t)}{x(t)}-\frac{q(t)f(x(t))}{x(t)}\\
&\quad -\frac{r(t)|x'(t)|^{\alpha-1}(x'(t))^{2}}{x^{2}(t)}+(r(t)h(t))'\Big]
\\
&\leq p(t)v(t)h(t)-\frac{h(t)}{H(t)}u(t)+v(t)(r(t)h(t))'
\\
&=-\frac{h(t)}{H(t)}u(t)-v(t)[-p(t)h(t)-(r(t)h(t))']
\\
&=-\frac{h(t)}{H(t)}u(t)-\phi(t).
\end{align*}
Multiplying  by $H(t)$, it follows that
\[
H(t)\phi(t)\leq -H(t)u'(t)-h(t)u(t).
\]
The rest of the proof is similar to Theorem \ref{thm3}, and it is omitted.
\end{proof}

\begin{theorem} \label{thm6} 
Assume \eqref{e34}, $p(t)\leq0$, $q(t)>0$, $f'(x)\geq 0$ and 
$\int_{t_0}^{\infty}\frac{dt}{(r(t))^{1/\alpha}}=\infty$.
Suppose further that there exists a function $H(t)$ satisfying 
{\rm (H1), (H2)}, and such that
\begin{equation} \label{e53}
\int_{t_0}^{\infty}H(t)\overline{{\phi}}(t)dt=\infty,
\end{equation}
where
\begin{equation} \label{e54}
\overline{{\phi}}(t)=v(t)(p(t)h(t)+(r(t)h(t))'),
\end{equation}
where $v(t)$ is defined in \eqref{e36}. Then every solution of
 \eqref{e1} is oscillatory when $\psi(x(t))\equiv1$.
\end{theorem}

\begin{proof} 
For the sake of contradiction, assume that \eqref{e1} has a non-oscillatory 
solution. Without loss of generality, we may assume that \eqref{e1} 
has an eventually positive  $x(t)>0 $ for all $t\geq t_0 $. Define
\[
u(t)=v(t)r(t)\Big(\frac{|x'(t)|^{\alpha-1}x'(t)}{x(t)}-h(t)\Big).
\]
The rest of the proof is similar to Theorem \ref{thm3}, and it is omitted here.
\end{proof}

\subsection*{Acknowledgments}
The author express a sincere gratitude to Professor Julio G. Dix 
and to the anonymous referees for their useful comments and suggestions.


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