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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2018 (2018), No. 159, pp. 1--15.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2018 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2018/159\hfil Second-order Emden-Fowler differential equations]
{Oscillation of second-order Emden-Fowler neutral delay differential equations}

\author[Y. Wu, Y. Yu, J. Xiao \hfil EJDE-2018/159\hfilneg]
{Yingzhu Wu, Yuanhong Yu, Jinsen Xiao}

\address{Yingzhu Wu \newline
Department of Mathematics,
 Guangdong University of Petrochemical Technology,
Maoming 525000, China}
\email{yingzhu1978@163.com}

\address{Yuanhong Yu (corresponding author)\newline
Academy of Mathematics and Systems Science,
Chinese Academy of Sciences, Beijing 100190, China}
\email{yu84845366@126.com}

\address{Jinsen Xiao \newline
School of Sciences,
Guangdong University of Petrochemical Technology,
Maoming 525000, China}
\email{jinsenxiao@yahoo.com}

\thanks{Submitted August 6, 2017. Published September 4, 2018.}
\subjclass[2010]{34C10, 34K11}
\keywords{Emden-Fowler equation; half-linear differential equation;
\hfill\break\indent  Riccati method; oscillation criterion}

\begin{abstract}
 In this article, we establish new oscillation criteria for the second-order
 Emden-Fowler neutral delay differential equation
 $$
 \Big(r(t)|z'(t)|^{\alpha-1}z'(t)\Big)'+q(t)|x(\sigma(t))|^{\beta-1}
 x\big(\sigma(t)\big)=0,
 $$
 where $z(t)=x(t)+p(t)x(\tau(t)), \alpha>0$ and $\beta>0$.
 Our results improve some well-known results which were published recently
 in the literature. Some illustrative examples are also provided to show
 the significance of our 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 this article, we consider the second-order Emden-Fowler neutral delay 
differential equation
\begin{equation}
\Big(r(t)|z'(t)|^{\alpha-1}z'(t)\Big)'
 +q(t)|x\big(\sigma(t)\big)|^{\beta-1}x\big(\sigma(t)\big)=0, \label{1.1}
\end{equation}
where $z(t)=x(t)+p(t)x(\tau(t))$, $t\geq t_0>0$, $\alpha>0$, and $\beta>0$.
Here we use the following assumptions:
\begin{itemize}
\item[(A1)] $r,\sigma\in C^{1}([t_0,\infty),(0,\infty))$, $r(t)>0$, $r'(t)\geq0$
$\sigma(t)\leq t$, $\sigma'(t)>0$, and $\lim_{t\to\infty}\sigma(t)=\infty$; 

\item[(A2)] $p,q,\tau\in C([t_0,\infty),R)$, $0\leq p(t)<1$,
$q(t)\geq0$, $\tau(t)\leq t$, and $\lim_{t\to\infty}\tau(t)=\infty$.

\end{itemize}

A function $x(t)\in C^{1}([T_{x},\infty),R)$, $T_{x}\geq t_0$  is called
 a solution of \eqref{1.1} if it satisfies the property
 $r(t)|z'(t)|^{\alpha-1}z'(t)\in C^{1}([T_{x},\infty),R)$  and   
\eqref{1.1} on $[T_{x},\infty)$. In this article, we only
consider the nontrivial solutions of \eqref{1.1}, which
ensure  $\sup\{|x(t)|:t\geq T\}>0$  for the condition $T\geq T_{x}$. 
A solution of \eqref{1.1} is said to be oscillatory if it has an arbitrarily
 large zero point on $[T_{x},\infty)$; otherwise, it is called 
nonoscillatory. Equation \eqref{1.1} is said to be oscillatory if all 
its solutions are oscillatory.

Recently, there have been a large number of papers that devoted to the 
oscillation of the neutral differential equations. We refer the readers to the 
articles \cite{Agarwal1,Agarwal2,Baculikova1,Baculikova2,Candan,Dong,Erbe,
Grammatikopoulos,Hasanbulli,
Han,Li1,Li2,Qin,Liu,Rogovchenko,Li3,Sun,Tiryaki,Wang,Zeng}.

Candan \cite{Candan} studied the oscillation for second-order neutral 
differential equations with distributed deviating arguments
\begin{equation}
\Big(r(t)|z'(t)|^{\alpha-1}z'(t)\Big)'
+\int^{d}_{c}f\big(t,x\big(\sigma(t,\xi)\big)\big)d\xi=0, \label{1.2}
\end{equation}
where $z(t)=x(t)+\int^{b}_{a}p(t,\xi)x(\tau(t,\xi))d\xi$,
$|f(t,u)|\geq q(t,\xi)|u^{\alpha}|$, and $\alpha>0$.

In \cite{Candan} the following results are presented, with the notation 
$Q(t)=\int^{d}_{c}[1-p(\sigma(t,\xi))]^{\alpha}
p(t,\xi)d\xi$,
$\bar{Q}(t)=\int^{\infty}_{t}Q(s)ds$, 
$\bar{R}(t)=\frac{\alpha\sigma_1'(t)}{r^{1/\alpha}
(\sigma_1(t))}$,
and $\sigma_1(t)=\sigma(t,\alpha)$.

\begin{theorem}[{\cite[Theorem 2.1]{Candan}}] \label{theorem1}
 Assume that
\begin{gather}
\int^{\infty}_{t_0}\frac{1}{r^{1/\alpha}(t)}\,dt=\infty, \label{1.3} \\
\int^{\infty}_{t_0}Q(t)\,dt=\infty,
\end{gather}
then \eqref{1.2} is oscillatory.
\end{theorem}

\begin{theorem}[{\cite[Theorem 2.3]{Candan}}]
 Assume that \eqref{1.3} holds and
\begin{equation}
\int^{\infty}_{t_0}Q(t)\,dt<\infty.
\end{equation}
If
\begin{equation}
\liminf_{t\to\infty}\frac{1}{\bar{Q}(t)}
\int^{\infty}_{t}\bar{Q}^{\frac{\alpha+1}{\alpha}}(s)\bar{R}(s)ds>\frac{\alpha}
{(\alpha+1)^{\frac{\alpha+1}{\alpha}}},\label{1.6}
\end{equation}
then \eqref{1.2} is oscillatory.\label{theorem2}
\end{theorem}

In 2011, Li et al.\ \cite{Li1} studied the oscillatory behavior of the 
 second order Emden-Fowler delay differential equation of the neutral type
\begin{equation}
(r(t)(x(t)+p(t)x(t-\tau))')'+q(t)x^{\beta}
\big(\sigma(t)\big)=0, \label{1.7}
\end{equation}
where $\tau\geq 0,\beta\geq 1$, and $r(t)$ satisfies
\begin{equation}
\int^{\infty}_{t_0}\frac{1}{r(t)}dt<\infty,\label{1.8}
\end{equation}
and they presented the following result.

\begin{theorem}[{\cite[Theorem 2.1]{Li1}}] \label{theorem3}
 Suppose \eqref{1.8} holds. If there exists a function
 $\rho\in C^{1}([t_0,\infty),R)$, $\rho(t)\geq t$, $\rho'(t)>0$, 
$\sigma(t)\leq \rho(t)-\tau$ such that for all sufficiently large $t_1$
and any $M>0$ and $L>0$, it holds
\begin{gather}
\int^{\infty}\Big[q(t)(1-p\big(\sigma(t)\big))^{\beta}R^{\beta}
\big(\sigma(t)\big)-\frac{\beta M^{1-\beta}\sigma'(t)R^{\beta-1}
\big(\sigma(t)\big)}{r\big(\sigma(t)\big)\int^{t}_{t_1}
\frac{\sigma'(s)}{r(\sigma(s))}\,ds}\Big]dt=\infty, \label{1.9} \\
\int^{\infty}\big[q(t)(\frac{1}{1+p(\rho(t))})^{\beta}
\delta^{\beta}(t)-\frac{\beta \rho'(t)}{L^{\beta-1}\delta(t)r(\rho(t))}\big]dt
=\infty,\label{1.10}
\end{gather}
where $R(t)=\int^{t}_{t_0}r^{-1}(s)ds$ and 
$\delta(t)=\int^{t}_{\rho(t)}r^{-1}(s)ds$,
then \eqref{1.7} is oscillatory.
\end{theorem}


In 2016, Agarwal et al.\ \cite{Agarwal1} considered the oscillation criteria 
for second order half-linear neutral delay differential equation
\begin{equation}
(r(t)[(x(t)+p(t)x(\tau(t)))']^{\alpha})'
+q(t)x^{\alpha}\big(\sigma(t)\big)=0, t\geq t_0, \label{1.11}
\end{equation}
where $\alpha\geq 1$ is a quotient of odd positive integers. 
A new oscillation criterion is given as follow.

\begin{theorem}[{\cite[Theorem 2.2]{Agarwal1}}] \label{theorem4}
Assume that
\begin{equation}
\pi(t_0)<\infty,\quad \text{where }
\pi(t)=\int^{\infty}_{t}r^{-1/\alpha}(s)ds. \label{1.12}
\end{equation}
If there exist the functions $\rho,\delta\in C^{1}([t_0,\infty),(0,\infty))$ 
such that
\begin{gather}
\limsup_{t\to\infty}\int^{t}_{t_0}\Big[\rho(s)q(s)(1-p(\sigma(s))
)^{\alpha}-\frac{(\rho_{+}'(s))^{\alpha+1}
r(\sigma(s))}{(\alpha+1)^{\alpha+1}(\rho(s)\sigma'(s))^{\alpha}}]ds
=\infty, \label{1.13} \\
\limsup_{t\to\infty}\int^{t}_{t_0}[\psi(s)
-\frac{\delta(s)r(s)(\varphi_{+}(s))^{\alpha+1}}{(\alpha+1)^{\alpha+1}}]ds
=\infty,\label{1.14}
\end{gather}
where
\begin{gather*}
\psi(t)=\delta(t)[q(t)(1-p(\sigma(s))\frac{\pi(\tau\big(\sigma(t)\big))}
{\pi\big(\sigma(t)\big)})^{\alpha}+\frac{1-\alpha}{r^{1/\alpha}(t)
\pi^{\alpha+1}(t)}], \\
p(t)<\frac{\pi(t)}{\pi(\tau(t))},\quad 
\varphi(t)=\frac{\delta'(t)}{\delta(t)}+\frac{1+\alpha}{r^{1/\alpha}(t)\pi(t)},
\end{gather*}
$\rho_{+}'(t)=\max\{0,\rho'(t)\}$, and $\varphi_{+}(t)=\max\{0,\varphi(t)\}$,
then \eqref{1.11} is oscillatory. 
\end{theorem}

We see that the neutral delay Emden-Fowler equation \eqref{1.7} and neutral
 delay half-linear equation \eqref{1.11} are not mutually inclusive 
each other. However, equations \eqref{1.7} and \eqref{1.11} are included 
in the \eqref{1.1}. Therefor, it will be of great interest to find some 
oscillation criteria for the neutral differential equation \eqref{1.1}.

Our aim in this article is to establish some new sufficient conditions for 
the oscillation of \eqref{1.1}, by using  generalized Riccati inequalities.
  To the best of our knowledge, very little is known regarding the oscillation 
criterion of \eqref{1.1}. The relevance of our theorems becomes clear in the
carefully selected examples.

The rest of article is organized as follows. In Section 2, we state and 
prove our main results. In Section 3, we show several examples.


\section{Main Results}

The following inequalities contain the variable $t$, in which we assume 
that they hold for the sufficiently large $t$, if there is no other statement.

\begin{theorem}  \label{theorem5}
Assume that
\begin{gather}
\int^{\infty}_{t_0}(\frac{1}{r(t)})^{1/\alpha}dt=\infty, \label{2.1} \\
\int^{\infty}_{t_0}[1-p\big(\sigma(t)\big)]^{\beta}q(t)dt=\infty\,.\label{2.2}
\end{gather}
Then \eqref{1.1} is oscillatory. 
\end{theorem}

\begin{proof} 
Let $x(t)$ be a nonoscillatory solution of \eqref{1.1}. We assume without 
loss of generality that $x(t)$ is eventually positive, that is, there exists 
a $t_0\geq 0$ such that $x(t)>0$ for $t\geq t_0$ and thus there exists a 
$t_1\geq t_0$ such that $x(\tau(t))>0$, and $x(\sigma(t))>0$ for $t\geq t_1$.
 If $x(t)$ is an eventually negative solution, it can be proved by the similar
 manner. From \eqref{1.1}, we have
\begin{equation}
\Big(r(t)|z'(t)|^{\alpha-1}z'(t)\Big)'\leq -q(t)x^{\beta}\big(\sigma(t)\big)
\leq 0.\label{2.3}
\end{equation}
Hence, $r(t)|z'(t)|^{\alpha-1}z'(t)$ is decreasing. Thus, we have two 
possible cases for $z'(t)$.
\smallskip

\noindent\textbf{Case I.}
 $z'(t)<0$ for $t\geq t_1$. Using the decreasing property of
 $r(t)|z'(t)|^{\alpha-1}z'(t)$, we obtain
\begin{equation}
r(t)|z'(t)|^{\alpha-1}z'(t)\leq r(t_2)|z'(t_2)|^{\alpha-1}z'(t_2), \quad
t\geq t_2\geq t_1.\label{2.4}
\end{equation}
Dividing both sides of \eqref{2.4} by $r(t)$, integrating from $t_2$ to $t$
and using \eqref{2.1}, we have
$$
z(t)\leq z(t_2)-r^{1/\alpha}(t_2)|z'(t_2)|\int^{t}_{t_2}r^{-1/\alpha}(s)
ds\to-\infty, \ \text{as}\ t\to\infty,
$$
which contradicts positivity of $z(t)$.
\smallskip

\noindent\textbf{Case II.}
 $z'(t)>0$ for $t>t_1$. Since $z(t)>x(t)$ and $z(t)$ is increasing, we have
$$
z(t)=x(t)+p(t)x(\tau(t))\leq x(t)+p(t)z(\tau(t))\leq x(t)+p(t)z(t).
$$
Thus, 
$$
(1-p(t))z(t)\leq x(t), t\geq t_2^{*}\geq t_1
$$
or
\begin{equation}
[1-p\big(\sigma(t)\big)]^{\beta}z^{\beta}\big(\sigma(t)\big)
\leq x^{\beta}\big(\sigma(t)\big), t\geq t_{3}\geq t_2^{*}.\label{2.5}
\end{equation}
Substituting \eqref{2.5} into \eqref{2.3}, we have
\begin{equation}
(r(t)(z'(t))^{\alpha})'\leq -q(t)[1-p\big(\sigma(t)\big)]^{\beta}z^{\beta}
\big(\sigma(t)\big).\label{2.6}
\end{equation}
On the other hand, since $r(t)(z'(t))^{\alpha}$ is decreasing, we have
$$
r(t)(z'(t))^{\alpha}\leq r(\sigma(t))(z'\big(\sigma(t)\big))^{\alpha}
$$
or
\begin{equation}
\Big(\frac{r(t)}{r\big(\sigma(t)\big)}\Big)^{1/\alpha}
\leq \frac{z'\big(\sigma(t)\big)}{z'(t)}.\label{2.7}
\end{equation}
Set the function
\begin{equation}
w(t):=\frac{r(t)(z'(t))^{\alpha}}{z^{\beta}\big(\sigma(t)\big)},
\quad t\geq t_{3}.\label{2.8}
\end{equation}
It is obvious that $w(t)>0$. Taking the derivative of $w(t)$, using 
\eqref{2.6}, \eqref{2.7} and \eqref{2.8}, we have
\begin{equation}
\begin{aligned}
w'(t)
&=\frac{(r(t)(z'(t))^{\alpha})'}{z^{\beta}\big(\sigma(t)\big)}
-\frac{\beta r(t)(z'(t))^{\alpha}z'\big(\sigma(t)\big)\sigma'(t)}{z^{\beta+1}
\big(\sigma(t)\big)}\\
&\leq -q(t)[1-p\big(\sigma(t)\big)]^{\beta}
-\frac{\beta \sigma'(t)(r^{1/\alpha}(t)z'(t))^{\alpha+1}}{r^{1/\alpha}
\big(\sigma(t)\big)z^{\beta+1}\big(\sigma(t)\big)}.
\end{aligned} \label{2.9}
\end{equation}
In view of the positivity of $z(t)$ and $z'(t)$, we obtain
\begin{equation}
w'(t)+q(t)[1-p(\sigma(t))]^{\beta} \leq 0.\label{2.10}
\end{equation}
Integrating both sides of \eqref{2.10} from $t_{3}$ to $t$ and using \eqref{2.2}, 
we obtain
$$
w(t)\leq w(t_{3})-\int^{t}_{t_{3}}q(s)[1-p(\sigma(s))]^{\beta} \,ds\to-\infty, 
 \quad \text{as } t\to\infty.
$$
which contradicts the fact $w(t)>0$. The proof is complete.
\end{proof}

Note that Theorem \ref{theorem5} is an improvement of 
\cite[Theorem 1]{Grammatikopoulos}.

\begin{lemma}  \label{lemma1}
Assume that $x(t)$ is an eventually positive solution of \eqref{1.1}, 
and $w(t)$ is  defined by \eqref{2.8}. Then 
\begin{equation}
w'(t)\leq -q(t)(1-p\big(\sigma(t)\big))^{\beta}
-\frac{\xi K\sigma'(t)}{r^{1/\xi}(\theta(t))}
w^{\frac{\xi+1}{\xi}}(t),\label{2.11}
\end{equation}
where $\xi=\min\{\alpha,\beta\}$
and
$$
K=
\begin{cases} 1,& \alpha=\beta\\
{\rm const}>0, & \alpha\neq\beta,
\end{cases}
\quad 
\theta(t)=\begin{cases}
t,& \alpha>\beta\\
\sigma(t),& \alpha\leq\beta.
\end{cases}
$$
\end{lemma}

\begin{proof} 
Proceeding as in the proof of Theorem \ref{theorem5}, we obtain \eqref{2.9}; 
that is
\begin{equation}
\begin{aligned}
w'(t)&\leq -q(t)[1-p\big(\sigma(t)\big)]^{\beta}
 -\frac{\beta\sigma'(t)[r^{1/\alpha}(t)z'(t)]^{\alpha+1}}{r^{1/\alpha}
 (\sigma(t))z^{\beta}(\sigma(t))} \\
&\leq -q(t)[1-p(\sigma(t))]^{\beta}-\frac{\beta\sigma'(t)}{r^{1/\alpha}
 \big(\sigma(t)\big)}[z\big(\sigma(t)\big)]^{\frac{\beta-\alpha}{\alpha}}
w^{\frac{\alpha+1}{\alpha}}(t).
\end{aligned} \label{2.12}
\end{equation}

If $\beta\geq \alpha$, in view of $z\big(\sigma(t)\big)$ being increasing, 
then there exist constants $K_1>0$ and $t_4\geq t_{3}$ such that
$[z(\sigma(t))]^{\frac{\beta-\alpha}{\alpha}}\geq K_1$ for $t\geq t_4$.
Thus, \eqref{2.12} gives
\begin{equation}
w'(t)\leq -q(t)[1-p\big(\sigma(t)\big)]^{\beta}
-\frac{\alpha K_1\sigma'(t)}{r^{1/\alpha}\big(\sigma(t)\big)}
w^{\frac{\alpha+1}{\alpha}}(t). \label{2.13}
\end{equation}
It is easy to check that $K_1=1$ for $\alpha=\beta$.

Next, if $\alpha> \beta$, since $(r(t)(z'(t))^{\alpha})'\leq 0$~and $r'(t)\geq 0$,
 we obtain $z''(t)\leq0$, which implies that $z'(t)$
is decreasing and $[z'(t)]^{\frac{\beta-\alpha}{\beta}}$ is increasing. 
Then there exist constant $K_2>0$ and $t_5\geq t_4$ such that
$[z'(t)]^{\frac{\beta-\alpha}{\beta}}\geq K_2$ for $t\geq t_5$.
Hence, by \eqref{2.12} it has
\begin{equation}
\begin{aligned}
w'(t)&\leq -q(t)[1-p(\sigma(t))]^{\beta}
 -\frac{\beta\sigma'(t)}{r^{1/\beta}(t)}[z'(t)
 ]^{\frac{\beta-\alpha}{\beta}}w^{\frac{\beta+1}{\beta}}(t)
 \\
&\leq -q(t)[1-p(\sigma(t))]^{\beta}
 -\frac{\beta K_2\sigma'(t)}{r^{1/\beta}(t)}
 w^{\frac{\beta+1}{\beta}}(t),t\geq t_5.
\end{aligned}\label{2.14}
\end{equation}
Combining \eqref{2.13} and \eqref{2.14}, we have that inequality \eqref{2.11} 
holds for all $\alpha>0$ and $\beta>0$.
\end{proof}

We now consider the case when \eqref{2.2} does not hold. 
We use the following notation for simplicity:
\begin{equation}
Q(t)=\int^{\infty}_{t}q(s)[1-p(\sigma(s))]^{\beta}ds, \quad
A(t)=\frac{\xi K\sigma'(t)}{r^{1/\xi}(\theta(t))}. \label{ast}
\end{equation}
Define a sequence of functions $\{y_{n}(t)\}^{\infty}_{n=0}$ by
$$
y_0(t)=Q(t),\quad t\geq t_0
$$
and
\begin{equation}
y_{n}(t)=\int^{\infty}_{t}A(s)y_{n-1}^{\frac{\xi+1}{\xi}}(s)ds+y_0(t),\quad
t\geq t_0,\; n=1,2,3,\dots\,. \label{2.15}
\end{equation}
By induction we see that  $y_0\leq y_{n+1}(t),t\geq t_0,n=1,2,3,\dots $

\begin{lemma} \label{lemma2}
Assume that $x(t)$ is an eventually positive solution of \eqref{1.1}. 
Then $y_{n}(t)\leq w(t)$, where $w(t)$ and $y_{n}(t)$ are defined by 
\eqref{2.8} and \eqref{2.15}, respectively.
Also, there exits a positive function $y(t)$ on $[T,\infty)$, such that
$\lim_{n\to\infty}y_{n}(t)=y(t)$ for $t\geq T\geq t_0$ and
\begin{equation}
y(t)=\int^{\infty}_{t}A(s)y^{\frac{\xi+1}{\xi}}(s)ds+y_0(s),\quad
t\geq T.\label{2.16}
\end{equation}
\end{lemma}

\begin{proof} 
Proceeding as in the proof of Lemma \ref{lemma1}, we have  inequality 
\eqref{2.11} or
\begin{equation}
w'(t)\leq -q(t)[1-p(\sigma(t))]^{\beta}-A(t)w^{\frac{\xi+1}{\xi}}(t). \label{2.17}
\end{equation}
Integrating both sides of \eqref{2.17} from $t$ to $t'$, we obtain
\begin{equation}
w(t')-w(t)+\int^{t'}_{t}q(s)[1-p(\sigma(s))]^{\beta}ds
-\int^{t'}_{t}w^{\frac{\xi+1}{\xi}}(s)A(s)ds\leq 0. \label{2.18}
\end{equation}
Then it is clear that
\begin{equation}
w(t')-w(t)+\int^{t'}_{t}w^{\frac{\xi+1}{\xi}}(s)A(s)ds\leq 0. \label{2.19}
\end{equation}
It follows  that
\begin{equation}
\int^{\infty}_{t}w^{\frac{\xi+1}{\xi}}(s)A(s)ds< \infty, \quad t\geq T. \label{2.20}
\end{equation}

Otherwise, $w(t')\leq w(t)-\int^{t'}_{t}w^{\frac{\xi+1}{\xi}}(s)A(s)ds\to -\infty$ 
as $t'\to \infty$, which contradicts to the fact that $w(t)>0$. 
Since $w(t)$ is positive and decreasing $\lim_{t\to\infty}w(t)=l\geq 0$.
By  \eqref{2.20}, we have $l=0$. Thus, from \eqref{2.18}, we have
$$
w(t)\geq Q(t)+\int^{\infty}_{t}w^{\frac{\xi+1}{\xi}}(s)A(s)ds
=y_0(t)+\int^{\infty}_{t}w^{\frac{\xi+1}{\xi}}(s)A(s)ds,
$$
i.e.
\begin{equation}
w(t)\geq Q(t)=y_0(t).\label{2.21}
\end{equation}
Moreover, by induction we can also see that
$w(t)\geq y_{n}(t)$ for $t\geq t_0$, $n=1,2,3\dots$ Thus, since the 
sequence $\{y_{n}(t)\}^{\infty}_{n=0}$ monotone increasing and bounded above,
it converges to $y(t)$. Letting $n\to \infty$ in \eqref{2.15} and 
using Lebesgue's monotone convergence theorem, we obtain \eqref{2.16}.
\end{proof}

The following theorem provides a new oscillation criterion of \eqref{1.1}
 with respect to that the condition \eqref{2.2} of Theorem \ref{theorem5} 
does not hold.


\begin{theorem} \label{theorem6}
Assume that \eqref{2.1} holds and \eqref{2.2} is not valid. If
\begin{equation}
\liminf_{t\to\infty}\frac {1}{Q(t)}\int^{\infty}_{t}Q^{\frac{\xi+1}{\xi}}(s)A(s)ds
>\frac{\xi}{(\xi+1)^{\frac{\xi+1}{\xi}}},\label{2.22}
\end{equation}
where $\xi,Q(t)$ and $A(t)$ are defined by \eqref{2.11} and \eqref{ast},
then \eqref{1.1} is oscillatory.
\end{theorem}

\begin{proof} 
Let $x(t)$ be a nonoscillatory solution of \eqref{1.1}. Proceeding as in the 
proof of Lemma \ref{lemma1} and Lemma \ref{lemma2}, we obtain \eqref{2.21} and  have
\begin{equation}
\frac{w(t)}{Q(t)}\geq 1+\frac{1}{Q(t)}\int^{\infty}_{t}A(s)Q^{\frac{\xi+1}{\xi}}(s)
\Big(\frac{w(s)}{Q(s)}\Big)^{\frac{\xi+1}{\xi}}ds,\quad t\geq T.\label{2.23}
\end{equation}
Let $\lambda=\inf_{t\geq T}\frac{w(t)}{Q(t)}$, then obviously $\lambda\geq 1$.

On the other hand, from \eqref{2.22} we know that there exists a constant $C>0$ 
such that
\begin{equation}
\liminf_{t\to\infty}\frac {1}{Q(t)}\int^{\infty}_{t}Q^{\frac{\xi+1}{\xi}}(s)A(s)ds
>C>\frac{\xi}{(\xi+1)^{\frac{\xi+1}{\xi}}}.\label{2.24}
\end{equation}
Then, from \eqref{2.23} and \eqref{2.24}, we see that
\begin{equation}
\lambda\geq 1+\lambda^{\frac{\xi+1}{\xi}}C.\label{2.25}
\end{equation}
Using the inequality
$$
Bu-Au^{\frac{\xi+1}{\xi}}\leq\frac{\xi^{\xi}}{(\xi+1)^{\xi+1}}
\frac{B^{\xi+1}}{A^{\xi}},
$$
where $A>0$, $B\geq 0$ and $\xi>0$. We get
\begin{equation}
\lambda-C\lambda^{\frac{\xi+1}{\xi}}
\leq \frac{\xi^{\xi}}{(\xi+1)^{\xi+1}}\frac{1}{C^{\xi}}.\label{2.26}
\end{equation}
Combining \eqref{2.24} and \eqref{2.26}, we see that
$$
\lambda<1+C\lambda^{\frac{\xi+1}{\xi}},
$$
which contradicts with \eqref{2.25}. The proof is complete.
\end{proof}

Theorem \ref{theorem6} improves  Theorem \ref{theorem2} and the corresponding
 result in \cite{Dong}. 
In the following, we establish new oscillation criteria of \eqref{1.1} 
with respect to that the condition \eqref{2.1} of Theorem \ref{theorem5} is invalid.


\begin{theorem}  \label{theorem7}
Assume that \eqref{1.12} holds. If there exists a function  
$\rho$ in the space $C^{1}([T_0,\infty),(0,\infty))$ such that for all
 sufficiently large  $T$ and any $K>0$, $M>0$, it holds
\begin{gather}
\limsup_{t\to\infty}\int^{t}_{T}\Big[\rho(s)q(s)(1-p(\sigma(s)))^{\beta}
-\frac{(\rho_{+}'(s))^{\xi+1}r(\theta(s))}{(\xi+1)^{\xi+1}(K\rho(s)\sigma'(s))^{\xi}}
\Big]ds=\infty, \label{2.27} \\
\limsup_{t\to\infty}\int^{t}_{T}[\pi^{\eta}(s)q(s)(1-p(\sigma(s))
 \frac{\pi(\tau(\sigma(s)))}{\pi(\sigma(s))})^{\beta}
 -\frac{\mu}{\pi(s)r^{1/\alpha}(s)}]ds=\infty,\label{2.28}
\end{gather}
where  $p(t)<\frac{\pi(t)}{\pi(\tau(t))},\xi=\min\{\alpha,\beta\}$,
$\eta=\max\{\alpha,\beta\}$,
$$
\theta(t)=
\begin{cases}
t,& \alpha>\beta,\\
\sigma(t),& \alpha\leq\beta,
\end{cases}
$$ 
$\rho_{+}'(t)=\max\{0,\rho'(t)\}$, and
$\mu=(\frac{\eta}{\eta+1})^{\eta+1}(\frac{\eta}{M})^{\eta}$
(when  $\alpha=\beta$, $K=1,M=\alpha$),
then \eqref{1.1} is oscillatory.
\end{theorem}

\begin{proof}
 Let $x(t)$ be a nonoscillatory solution of \eqref{1.1}. Without loss of
 generality, we assume that there exists a $t_1\geq t_0>0$ such that
 $x(t)>0,x(\tau(t))>0$ and $x(\sigma(t))>0$ for $t\geq t_1$. 
Hence, $z(t)>0, t\geq t_1$. On the other hand, from  \eqref{1.1} we see that
\begin{equation}
\Big(r(t)|z'(t)|^{\alpha-1}z'(t)\Big)'\leq0,\quad t\geq t_1,\label{2.29}
\end{equation}
which implies that $r(t)|z'(t)|^{\alpha-1}z'(t)$ is decreasing. Hence, $z'(t)$ 
does not eventually change signs, that is, there exists a $t_2\geq t_1$
such that either $z'(t)>0$ or $z'(t)<0$ for all $t\geq t_2$.
\smallskip

\noindent\textbf{Case I.}  $z'(t)>0$ for $t\geq t_2$. It follows from the 
definition of $z(t)$ that
\begin{equation}
x(t)=z(t)-p(t)x(\tau(t))\geq z(t)-p(t)z(\tau(t))\geq (1-p(t))z(t).\label{2.30}
\end{equation}
It follows from equations \eqref{1.1} and \eqref{2.30} that
\begin{equation}
(r(t)(z'(t))^{\alpha})'+q(t)(1-p\big(\sigma(t)\big))^{\beta}z^{\beta}
\big(\sigma(t)\big)\leq 0,t\geq t_{3}\geq t_2.\label{2.31}
\end{equation}
Define a function $u(t)$ by
\begin{equation}
u(t):=\rho(t)\frac{r(t)(z'(t))^{\alpha}}{z^{\beta}\big(\sigma(t)\big)},\quad
t\geq t_{3}.\label{2.32}
\end{equation}
Then, $u(t)>0, t\geq t_{3}$. Taking differentiation on both sides of \eqref{2.32}, 
we have
\begin{equation}
u'(t)\leq-\rho(t)q(t)(1-p(\sigma(t)))^{\beta}
+\frac{\rho'(t)}{\rho(t)}u(t)-\frac{\rho(t)r(t)(z'(t))^{\alpha}\beta\sigma'(t)
z'(\sigma(t))}{z^{\beta+1}\big(\sigma(t)\big)}.\label{2.33}
\end{equation}
For this inequality,  if $\alpha\leq \beta$, in view of  
$r^{1/\alpha}(t)z'(t)\leq r^{1/\alpha}(\sigma(t))z'(\sigma(t))$, we see that
$$
u'(t)\leq-\rho(t)q(t)(1-p(\sigma(t)))^{\beta}+\frac{\rho'(t)}{\rho(t)}u(t)
-\frac{\beta\sigma'(t)}{(\rho(t)r(\sigma(t)))^{1/\alpha}}
[z(\sigma(t))]^{\frac{\beta-\alpha}{\alpha}}u^{\frac{\alpha+1}{\alpha}}(t).
$$
Because $z(\sigma(t))$ is increasing,  there exists constants $K_1>0$ and 
$t_4\geq t_{3}$ such that
$[z(\sigma(t))]^{\frac{\beta-\alpha}{\alpha}}\geq K_1,\ t\geq t_4$.
Thus, the above inequality gives
\begin{equation}
u'(t)\leq-\rho(t)q(t)(1-p(\sigma(t)))^{\beta}+\frac{\rho'(t)}{\rho(t)}u(t)
-\frac{\alpha K_1\sigma'(t)}{(\rho(t)r\big(\sigma(t)\big))^{1/\alpha}}
u^{\frac{\alpha+1}{\alpha}}(t).\label{2.34}
\end{equation}
Obviously, if $\alpha=\beta$, then $K_1=1$.

If $\alpha> \beta$, since $(r(t)(z'(t))^{\alpha})'\leq 0$~and $r'(t)\geq 0$, 
we obtain $z''(t)\leq0$, which implies that $z'(t)$  is decreasing and
 $[z'(t)]^{\frac{\beta-\alpha}{\beta}}$ is increasing. Then there exist 
constants $K_2>0$, $t_5\geq t_4$ such that 
$[z'(t)]^{\frac{\beta-\alpha}{\beta}}\geq K_2$, $t\geq t_5$.
Thus,  inequality \eqref{2.33} becomes
\begin{equation}
\begin{aligned}
u'(t)
&\leq-\rho(t)q(t)(1-p(\sigma(t)))^{\beta}+\frac{\rho'(t)}{\rho(t)}u(t) \\
&\quad -\frac{\beta\sigma'(t)}{(\rho(t)r(t))^{1/\beta}}
[z'(t)]^{\frac{\beta-\alpha}{\beta}}u^{\frac{\beta+1}{\beta}}(t)
 \\
&\leq -\rho(t)q(t)(1-p(\sigma(t)))^{\beta}+\frac{\rho'(t)}{\rho(t)}u(t) \\
&\quad -\frac{\beta K_2\sigma'(t)}{(\rho(t)r(t))^{1/\beta}}
 u^{\frac{\beta+1}{\beta}}(t),\quad t\geq t_5.
\end{aligned}\label{2.35}
\end{equation}
Combining \eqref{2.34} and \eqref{2.35}, we obtain for any $\alpha>0$ 
and $\beta>0$ that
\begin{equation}
u'(t)\leq-\rho(t)q(t)(1-p(\sigma(t)))^{\beta}+\frac{\rho'(t)}{\rho(t)}u(t)
-\frac{\xi K\sigma'(t)}{(\rho(t)r(\theta(t)))^{1/\xi}}
u^{\frac{\xi+1}{\xi}}(t), \label{2.36}
\end{equation}
for $t\geq t_5$, where $\xi=\min\{\alpha,\beta\}$,
and
$$
K=\begin{cases}
1,& \alpha=\beta\\
K>0,& \alpha\neq\beta,
\end{cases}
\quad \theta(t)=
\begin{cases}
t,& \alpha>\beta\\
\sigma(t),& \alpha\leq\beta.
\end{cases}
$$

Let $y=u(t),D=\frac{\rho'(t)}{\rho(t)}$, and 
$C=\frac{\xi K\sigma'(t)}{(\rho(t)r(\theta(t)))^{1/\xi}}$.
By \eqref{2.36} and the inequality
\begin{equation}
Dy-Cy^{\frac{\xi+1}{\xi}}\leq\frac{\xi^{\xi}}{(\xi+1)^{\xi+1}}
\frac{D_{+}^{\xi+1}}{C^{\xi}},\label{2.37}
\end{equation}
where $C>0,y\geq0$, and $D_{+}=\max\{0,D\}$, we obtain
\begin{equation}
u'(t)\leq -\rho(t)q(t)(1-p(\sigma(t)))^{\beta}
+\frac{(\rho_{+}'(t))^{\xi+1}r(\theta(t))}{(\xi+1)^{\xi+1}(K\rho(t)
\sigma'(t))^{\xi}}.\label{2.38}
\end{equation}
Integrating both sides of \eqref{2.38} from $T>t_5$ to $t$, we obtain
\begin{equation}
u(t)\leq u(T)-\int^{t}_{T}\Big[\rho(s)q(s)(1-p(\sigma(s)))^{\beta}
-\frac{(\rho_{+}'(s))^{\xi+1}r(\theta(s))}{(\xi+1)^{\xi+1}(K\rho(s)
\sigma'(s))^{\xi}}\Big]ds.\label{2.39}
\end{equation}
Letting $t\to \infty$ in the above inequality, we obtain a contradiction 
with \eqref{2.27}.
\smallskip

\noindent\textbf{Case II.}  $z'(t)<0$ for $t>t_2$. By \eqref{2.29} we have
\begin{equation}
(r(t)(-z'(t))^{\alpha})'\geq 0,\quad t\geq t_2.\label{2.40}
\end{equation}
Then, $r^{1/\alpha}(t)(-z'(t))$ is an increasing function and thus
\begin{equation}
z'(s)\leq \Big(\frac{r(t)}{r(s)}\Big)^{1/\alpha}z'(t),\quad 
s\geq t\geq t_2.\label{2.41}
\end{equation}
Integrating the above inequality from $t$ to $l$, we obtain
$$
z(l)\leq z(t)+r^{1/\alpha}(t)z'(t)\int^{l}_{t}r^{-1/\alpha}(s)ds,l\geq t\geq t_2.
$$
Letting $t\to\infty$, we then have
\begin{equation}
z(t)\geq \pi(t)r^{1/\alpha}(t)(-z'(t)), t\geq t_2.\label{2.42}
\end{equation}
It follows that
\begin{equation}
z^{\alpha}(t)\geq \pi^{\alpha}(t)r(t)(-z'(t))^{\alpha}, \quad
t\geq T_1\geq t_2.\label{2.43}
\end{equation}
If $\alpha\geq \beta$, then $z^{\alpha-\beta}(t)$ is a decreasing function
 and thus there exists a constant $l_1>0$ such that $z^{\alpha-\beta}(t)\leq l_1$ 
and $t\geq T_1$.

Define  a function  $V(t)$ by
\begin{equation}
V(t):=\frac{r(t)(-z'(t))^{\alpha}}{z^{\beta}(t)},\quad t\geq T_1.\label{2.44}
\end{equation}
Hence, $V(t)>0$, $t\geq T_1$  and we have
\begin{equation}
l_1\geq z^{\alpha-\beta}(t)\geq \pi^{\alpha}(t)V(t), \quad \alpha\geq \beta.\label{2.45}
\end{equation}

On the other hand,  from \eqref{2.42} it follows that
\begin{equation}
z^{\beta}(t)\geq \pi^{\beta}(t)\Big(r^{1/\alpha}(t)(-z'(t))\Big)^{\beta-\alpha+\alpha}.
\label{2.46}
\end{equation}
Note that $\big(r^{1/\alpha}(t)(-z'(t))\big)^{\beta-\alpha}$ is an increasing 
function for $\beta>\alpha$. Then there exists a constant $l_2>0$ such that
\begin{equation}
l_2\geq \Big(r^{1/\alpha}(t)(-z'(t))\Big)^{\alpha-\beta}\geq \pi^{\beta}(t)V(t),\quad
 \beta>\alpha.\label{2.47}
\end{equation}
Combining \eqref{2.45} and \eqref{2.47}, we have
\begin{equation}
0<\pi^{\eta}(t)V(t)\leq l,\label{2.48}
\end{equation}
where $\eta=\max\{\alpha,\beta\}$ and $l=\max\{l_1,l_2\}$.

We further observe that  \eqref{2.42} gives $(\frac{z(t)}{\pi(t)})'\geq 0$ 
for $t\geq t_2$.
Then $\frac{z(t)}{\pi(t)}$ is an increasing function and thus
$$
x(t)=z(t)-p(t)x(\tau(t))\geq z(t)-p(t)z(\tau(t))
\geq \Big(1-p(t)\frac{\pi(\tau(t))}{\pi(t)}\Big)z(t).
$$
Note that $z'(t)<0$. Hence we find
\begin{equation}
x^{\beta}(\sigma(t))\geq \Big(1-p(\sigma(t))\frac{\pi(\tau(t))}{\pi(\sigma(t))}
\Big)^{\beta}z^{\beta}(t).\label{2.49}
\end{equation}
Combining \eqref{1.1} and \eqref{2.49}, we obtain
\begin{equation}
(r(t)(-z'(t))^{\alpha})'-Q(t)z^{\beta}(t)\geq 0, t\geq T_1\geq t_2,\label{2.50}
\end{equation}
where
\begin{equation}
Q(t)= q(t)\Big(1-p(\sigma(t))\frac{\pi(\tau(\sigma(t)))}{\pi(\sigma(t))}\Big)^{\beta}.
\label{2.51}
\end{equation}
Differentiating on both sides of \eqref{2.44}, using \eqref{2.50}, we obtain
\begin{equation}
V'(t)\geq Q(t)+\frac{\beta r(t)(-z'(t))^{\alpha+1}}{z^{\beta+1}(t)},
\quad t\geq T_1.\label{2.52}
\end{equation}

For this inequality, if $\alpha\geq\beta$, because 
$[z'(t)]^{\frac{\beta-\alpha}{\alpha}}$ is an increasing function, there exist 
constants $M_1>0$, $T_2\geq T_1$, such that 
$[z'(t)]^{\frac{\beta-\alpha}{\alpha}}\geq M_1,t\geq T_2$. 
From\eqref{2.52}, we obtain
\begin{equation}
V'(t)\geq Q(t)+\frac{\beta}{r^{1/\alpha}(t)}[z(t)]^{\frac{\beta-\alpha}{\alpha}}
V^{\frac{\alpha+1}{\alpha}}(t)\geq Q(t)
+\frac{\beta M_1}{r^{1/\alpha}(t)}V^{\frac{\alpha+1}{\alpha}}(t),
\quad t\geq T_2.\label{2.53}
\end{equation}
Note that if $\alpha=\beta$, then $ M_1=1$.

Now if $\alpha<\beta$, $[r^{1/\alpha}(t)(-z'(t))]^{\frac{\beta-\alpha}{\beta}}$  
is an increasing function and there exist constants
$M_2>0$ and $T>T_2$, such that 
$[r^{1/\alpha}(t)(-z'(t))]^{\frac{\beta-\alpha}{\beta}}>M_2,t\geq T$. 
By \eqref{2.52}, we have
\begin{equation}
\begin{aligned}
V'(t)
&\geq Q(t)+\frac{\beta}{r^{1/\alpha}(t)}\big[r^{1/\alpha}(t)(-z'(t))
\big]^{\frac{\beta-\alpha}{\beta}}V^{\frac{\beta+1}{\beta}}(t) \\
&\geq Q(t) +\frac{\beta M_2}{r^{1/\alpha}(t)}V^{\frac{\beta+1}{\beta}}(t),
\quad t\geq T.
\end{aligned}  \label{2.54}
\end{equation}
Combining \eqref{2.53} and \eqref{2.54}, we  obtain
\begin{equation}
V'(t)\geq Q(t)+\frac{ M}{r^{1/\alpha}(t)}V^{\frac{\eta+1}{\eta}}(t),
\quad t\geq T,\label{2.55}
\end{equation}
where $\eta=\max\{\alpha,\beta\}$, and
$M= \begin{cases}
\alpha,& \alpha=\beta\\
K>0,& \alpha\neq\beta.
\end{cases}$

Multiplying both sides of \eqref{2.55} by $\pi^{\eta}(t)$ and integrating  
from $T$ to $t$, yields
\begin{equation}
\begin{aligned}
\int^{t}_{T}\pi^{\eta}(s)Q(s)ds  
&\leq \int^{t}_{T}\pi^{\eta-1}(s)r^{-1/\alpha}(s)
 [\eta V(s)-M\pi(s)V^{\frac{\eta+1}{\eta}}(s)]ds \\
&\quad +\pi^{\eta}(t)V(t)-\pi^{\eta}(T)V(T).
\end{aligned}\label{2.56}
\end{equation}
Let $y=V(s)$, $D=\eta$ and $C=M\pi(s)$. Again by the inequality \eqref{2.37},
 we have
\begin{equation}
\int^{t}_{T}\pi^{\eta}(s)Q(s)ds
\leq \int^{t}_{T}\frac{\mu}{\pi(s)r^{1/\alpha}(s)}ds
+\pi^{\eta}(t)V(t)-\pi^{\eta}(T)V(T).\label{2.57}
\end{equation}
Combining \eqref{2.57}, \eqref{2.51}, and \eqref{2.48}, we have
\begin{equation}
\int^{t}_{T}[\pi^{\eta }(s)q(s)(1-p(\sigma(s))
\frac{\pi(\tau(\sigma(s)))}{\pi(\sigma(s))})^{\beta} 
-\frac{\mu}{\pi(s)r^{1/\alpha}(s)}]ds\leq l,\label{2.58}
\end{equation}
where $\mu=(\frac{\eta}{\eta+1})^{\eta+1}(\frac{\eta}{M})^{\eta}$,
which contradicts condition \eqref{2.28}. The proof is complete.
\end{proof}


Setting $\alpha=\beta$ in \eqref{1.1}, by Theorem \ref{theorem3} we immediately have
the following result.

\begin{corollary} \label{corollary1}
Suppose that $\alpha=\beta$ and \eqref{1.12} holds. If there exists a function
 $\rho$ in the space $C^{1}([t_0,\infty),(0,\infty))$ such that for all 
sufficiently large  $T,T\geq t_0$, it holds that
\begin{gather}
\limsup_{t\to\infty}\int^{t}_{T}[\rho(s)q(s)(1-p(\sigma(s)))^{\alpha}
-\frac{(\rho_{+}'(s))^{\alpha+1}r(\sigma(s))}{(\alpha+1)^{\alpha+1}
(\rho(s)\sigma'(s))^{\alpha}}]ds=\infty, \label{2.59} \\
\limsup_{t\to\infty}\int^{t}_{T}[\pi^{\alpha}(s)q(s)(1-p(\sigma(s))
\frac{\pi(\tau(\sigma(s)))}{\pi(\sigma(s))})^{\alpha}
-\frac{\varepsilon}{\pi(s)r^{1/\alpha}(s)}]ds=\infty,\label{2.60}
\end{gather}
where $p(t)<\frac{\pi(t)}{\pi(\tau(t))}$, 
$\varepsilon=(\frac{\alpha}{\alpha+1})^{\alpha+1}$ and 
$\rho_{+}'(t)=\max\{0,\rho'(t)\}$, then \eqref{1.1} is  oscillatory.
\end{corollary}


Corollary \ref{corollary1} holds for any $\alpha>0$ while Theorem \ref{theorem4} 
holds for $\alpha\geq1$, which is a quotient of odd positive integers. 
On the other hand, condition \eqref{2.60} is more general than condition \eqref{1.14} 
of Theorem \ref{theorem4}. We shall illustrate this in Example \ref{example3},
 given in next section.

Note that in \eqref{1.1}, if $\alpha=1$ and $\beta>1$, then \eqref{1.1} 
is super-linear and Theorem \ref{theorem7} has the following corollary.

\begin{corollary} \label{corollary2}
Suppose \eqref{1.8} holds. If there exists a function 
$\rho$ in the space $C^{1}([t_0,\infty),(0,\infty))$, 
and the constants $K>0$ and $M>0$, 
such that for all sufficiently large $T\geq t_0$, it holds
\begin{gather}
\limsup_{t\to\infty}\int^{t}_{T}\Big[\rho(s)q(s)(1-p(\sigma(s)))^{\beta}
-\frac{(\rho_{+}'(s))^{2}r(\sigma(s))}{4(K\rho(s)\sigma'(s))}\Big]ds
=\infty , \label{2.61} \\
\limsup_{t\to\infty}\int^{t}_{T}[\pi^{\beta}(s)q(s)(1-p(\sigma(s))
 \frac{\pi(\tau(\sigma(s)))}{\pi(\sigma(s))})^{\beta}
 -\frac{\mu_1}{\pi(s)r(s)}]ds=\infty,\label{2.62}
\end{gather}
where 
$p(t)<\frac{\pi(t)}{\pi(\tau(t))}$,
$\rho_{+}'(t)=\max\{0,\rho'(t)\}$,
$\mu_1=\big(\frac{\beta}{\beta+1}\big)^{\beta+1}(\frac{\beta}{M})^{\beta}$, and \\
$\pi(t)=\int^{\infty}_{t}\frac{1}{r(s)}ds$,
then \eqref{1.7} is oscillatory.
\end{corollary}

Note that in equation \eqref{1.1}, if $\alpha=1$ and $0<\beta<1$, 
then \eqref{1.1} is sub-linear and Theorem \ref{theorem7} has the 
following corollary.

\begin{corollary} \label{corollary3}
Suppose \eqref{1.8} holds. If there exists a function 
$\rho$ in the space $C^{1}([t_0,\infty),(0,\infty))$, and the constants 
$K>0$ and $M>0$, 
such that for all sufficiently large $T\geq t_0$, we have
\begin{gather} \label{2.63}
\limsup_{t\to\infty}\int^{t}_{T}\Big[\rho(s)q(s)(1-p(\sigma(s)))^{\beta}
-\frac{(\rho_{+}'(s))^{\beta+1}r(s)}{(\beta+1)^{\beta+1}(K\rho(s)
\sigma'(s))^{\beta}}\Big]ds=\infty\,, \\
\limsup_{t\to\infty}\int^{t}_{T}\Big[\pi(s)q(s)(1-p(\sigma(s))
\frac{\pi(\tau(\sigma(s)))}{\pi(\sigma(s))})^{\beta}
-\frac{\mu_2}{\pi(s)r(s)}\Big]ds=\infty,\label{2.64}
\end{gather}
where $\mu_2=1/(4M)$, then \eqref{1.7} is oscillatory.
\end{corollary}


\section{Examples}

In this section, we provide some examples to illustrate our results.

\begin{example} \label{example1} \rm
 Consider the  neutral delay differential equation
\begin{equation}
\Big(|z'(t)|^{\alpha-1}z'(t)\Big)'+e^{\mu\beta t}(t)|x(\mu t)|^{\beta-1}x(\mu t)=0,
 \label{3.1}
\end{equation}
where $z(t)=x(t)+(1-e^{-t})x(t-1)$,  $\alpha>0$, $\beta>0$, and  
$0<\mu<1$.
\end{example}

Comparing  \eqref{3.1} with \eqref{1.1}, we see that 
$r(t)=1$, $q(t)=e^{\mu\beta t}$, $\sigma(t)=\mu t$, and $p(t)=1-e^{-t}$,
then $q(t)[1-p\big(\sigma(t)\big)]^{\beta}=1$.
Clearly one can see that conditions of Theorem \ref{theorem5} are satisfied. 
Hence, \eqref{3.1} is oscillatory.

\begin{example} \label{example2} \rm
 Consider the neutral differential equation
\begin{equation}
\Big(e^{-(\frac{\alpha}{2}+\frac{\alpha}{\xi})t}|z'(t)|^{\alpha-1}z'(t)\Big)'
+e^{-t}|x(t-2)|^{\beta-1}x(t-2)=0, \label{3.2}
\end{equation}
where $z(t)=x(t)+\frac{1}{2}x(t-\tau)$, $\tau>0$, $\alpha>0$, $\beta>0$, and
 $\xi=\min\{\alpha,\beta\}$.
\end{example}

Comparing the \eqref{3.2} with \eqref{1.1}, we see that 
$r(t)=e^{-(\frac{\alpha}{2}+\frac{\alpha}{\xi})t},q(t)=e^{-t}$,
then 
\begin{gather*}
Q(t)=\int^{\infty}_{t}q(s)[1-p(\sigma(s))]^{\beta}ds
=(\frac{1}{2})^{\beta}e^{-t}, \\
A(t)
=\frac{\xi K\sigma'(t)}{r^{1/\alpha}(\theta(t))}
=\xi Ke^{(\frac{1}{2}+\frac{1}{\xi})\theta(t)}.
\end{gather*}
In view of $\theta(t)= \begin{cases}
t,& \alpha>\beta\\
\sigma(t),& \alpha\leq\beta,
\end{cases}$
 we have that for $\alpha>\beta$,
 \begin{align*}
&\liminf_{t\to\infty}\frac{1}{Q(t)}\int^{\infty}_{t}[Q(s)]^{\frac{\xi+1}{\xi}}A(s)ds\\
&=\liminf_{t\to\infty}2^{\beta}e^{t}\int^{\infty}_{t}
\big[(\frac{1}{2})^{\beta}e^{-s}\big]^{\frac{\xi+1}{\xi}}\xi 
Ke^{(\frac{1}{2}+\frac{1}{\xi})s}ds\\
&=\liminf_{t\to\infty}\xi K2^{-\frac{\beta}{\xi}}e^{t}\int^{\infty}_{t}
e^{-\frac{s}{2}}ds=\infty.
\end{align*}
If $\alpha\leq \beta$ we have
\begin{align*}
&\liminf_{t\to\infty}\frac{1}{Q(t)}
 \int^{\infty}_{t}[Q(s)]^{\frac{\xi+1}{\xi}}A(s)ds\\
&=\liminf_{t\to\infty}2^{\beta}e^{t}\int^{\infty}_{t}
\big[(\frac{1}{2})^{\beta}e^{-s}\big]^{\frac{\xi+1}{\xi}}\xi Ke^{(\frac{1}{2}
 +\frac{1}{\xi})(s-2)}ds\\
&=\liminf_{t\to\infty}\xi K2^{-\frac{\beta}{\xi}}
 e^{-(\frac{\xi+1}{\xi})}e^{t}\int^{\infty}_{t}e^{-\frac{s}{2}}ds
=\infty.
\end{align*}
Clearly one can see that all conditions of Theorem \ref{theorem6} 
are satisfied, therefore, \eqref{3.2} is oscillatory.

\begin{example} \rm
 Consider the half-linear delay differential equation of neutral type
\begin{equation}
\Big(t^{6}\big[\big(x(t)+\frac{1}{6}x(\frac{t}{3})\big)'\big]^3\Big)'
+Kt^{2}x^3(\frac{t}{2})=0,t\geq 1.\label{3.3}
\end{equation}\label{example3}
\end{example}

We claim that this equation satisfies the conditions of Corollary \ref{corollary1}. 
First, in \eqref{3.3}, $\alpha=\beta=3,K>0$. If we choose $\rho(t)=1$ then
 $\rho'(t)=0$, and we have
$$
\rho(t)q(t)(1-p(\sigma(s)))^{\alpha}=\frac{125K}{216}t^{2},
$$
then condition \eqref{2.59} is satisfied.

By \eqref{1.12}, we have $\pi(t)=\frac{1}{t}$, and condition \eqref{1.12} holds.
 Notice that $\tau(t)=\frac{t}{3},\sigma(t)=\frac{t}{2}$, and thus
$\frac{\pi(\tau\big(\sigma(t)\big))}{\pi(\sigma(t))}=3$; then
$\pi^{\alpha}(t)q(t)(1-p(\sigma(t))
\frac{\pi(\tau(\sigma(t)))}{\pi(\sigma(t))})^{\alpha}=\frac{K}{8t}$,
where $\varepsilon=(\frac{3}{4})^{4},\frac{\varepsilon}{\pi(t)r^{1/\alpha}(t)}
=\frac{81}{256t}$.

If we set $K>\frac{81}{32}$, condition \eqref{2.60} is satisfied. 
By Corollary \ref{corollary1}, equation \eqref{3.3} is oscillatory for 
$K>81/32$.

Now if we use Theorem \ref{theorem4} to work through this example,
 we need to satisfy condition \eqref{1.14}. However, \eqref{1.11} 
requires the function $\psi(t)>0$, but where
\[
q(t)(1-p(\sigma(t))\frac{\pi(\tau(\sigma(t)))}{\pi(\sigma(t))})^{\alpha}
=\frac{K}{8}t^{2},\frac{1-\alpha}{r^{1/\alpha}(t)\pi^{\alpha+1}(t)}=-2t^{2},
\]
then $\psi(t)=\delta(t)t^{2}(\frac{K}{8}-2)$.
Hence, $\psi(t)>0$ holds for $K>16$. However Corollary \ref{corollary1} 
only requires $K>\frac{81}{32}$. Consequently, Corollary \ref{corollary1} 
improves Theorem \ref{theorem4}.

\begin{example} \label{example4} \rm
Consider the neutral-type equation
\begin{equation}
\Big(t^{8}|z'(t)|^{\alpha-1}z'(t)\Big)'
+t^{5}|x(\frac{t}{3})|^{\beta-1}x(\frac{t}{3})=0,t\geq1,\label{3.4}
\end{equation}
where $z(t)=x(t)+\frac{1}{4}x(\frac{t}{2}),\alpha=4,\beta=2$.
\end{example}

We now use Theorem \ref{theorem7} to show that this equation is oscillatory. 
Notice that $\pi(t)=\frac{1}{t}$ in \eqref{3.4}, then \eqref{1.12} holds.
 If we choose $\rho(t)=1$, then \eqref{2.27} is satisfied.
To verify condition \eqref{2.28}, we have
\begin{align*}
&\limsup_{t\to\infty}\int^{t}_{T}\Big[\pi^{\eta}(s)q(s)
\Big(1-p(\sigma(s))\frac{\pi(\tau(\sigma(s)))}{\pi(\sigma(s))}\Big)^{\beta}
-\frac{\mu}{\pi(s)r^{1/\alpha}(s)}\Big]ds\\
&=\limsup_{t\to\infty}\int^{t}_{T}\Big[\frac{1}{s^{4}}s^{5}(\frac{1}{2})^{2}
 -\frac{(\frac{4}{5})^{5}(\frac{4}{M})^{4}}{s}\Big]ds
=\infty.
\end{align*}
Then \eqref{2.28} holds. Hence, by Theorem \ref{theorem7}, equation \eqref{3.4}
 is oscillatory.

Note that Theorem \ref{theorem4} cannot be applied to the oscillation 
of \eqref{3.4}.

\begin{example} \label{example5} \rm
Consider the super-linear Emden-Fowler equation
\begin{equation}
\Big(t^{2}\Big(x(t)+\frac{1}{8}x(\frac{t}{4})\Big)'\Big)'
+t|x(\frac{t}{5})|^{\beta}\operatorname{sgn} x(\frac{t}{5})=0,\quad
t\geq1.\label{3.5}
\end{equation}
\end{example}
In this example,  $\alpha=1$, $\beta=\frac{3}{2}>1$, and $\pi(t)=\frac{1}{t}$; 
as a result, \eqref{1.12} holds. By Letting $\rho(t)=1$, condition \eqref{2.61} 
is satisfied. On the other hand,
\begin{align*}
&\limsup_{t\to\infty}\int^{t}_{T}\Big[\pi^{\eta}(s)q(s)\Big(1-p(\sigma(s))
\frac{\pi(\tau(\sigma(s)))}{\pi(\sigma(s))}\Big)^{\beta}
 -\frac{\mu_1}{\pi(s)r(s)}\Big]ds\\
&=\limsup_{t\to\infty}\int^{t}_{T}\Big[\frac{1}{\sqrt{s}}(\frac{1}{2})^{3/2}
-(\frac{3}{5})^{5/2}(\frac{3}{2M})^{3/2}\frac{1}{s}\Big]ds
=\infty.
\end{align*}
This shows that \eqref{2.62} holds. Then by Corollary \ref{corollary2}, 
equation \eqref{3.5} is oscillatory.


\begin{example} \label{example6} \rm
Consider the sub-linear Emden-Fowler equation
\begin{equation}
\Big(t^{2}\Big(x(t)+\frac{1}{8}x(\frac{t}{4})\Big)'\Big)'
+t|x(\frac{t}{5})|^{\beta}\operatorname{sgn} x(\frac{t}{5})=0,\quad 
t\geq1,\label{3.6}
\end{equation}
where $0<\beta=1/2<1$. 
\end{example}

In this example, it is easy to find that \eqref{1.12} and \eqref{2.63}
 are satisfied. We also see that
\begin{align*}
&\limsup_{t\to\infty}\int^{t}_{T}\Big[\pi(s)q(s)\Big(1-p(\sigma(s))
\frac{\pi(\tau(\sigma(s)))}{\pi(\sigma(s))}\Big)^{\beta}
-\frac{\mu_2}{\pi(s)r(s)}\Big]ds\\
&=\limsup_{t\to\infty}\int^{t}_{T}
\Big[(\frac{1}{2})^{1/2}-\frac{1}{4M}\frac{1}{s}\Big]ds
=\infty,
\end{align*}
which shows that \eqref{2.64} is satisfied. By Corollary \ref{corollary3}, 
we can say that \eqref{3.6} is oscillatory.

However, Theorem \ref{theorem3} cannot be applied to this example because 
it requires $\beta\geq 1$.

\begin{example} \label{example7} \rm
Consider the linear neutral-type equation
\begin{equation}
\Big(t^{2}\Big(x(t)+px(\frac{t}{m})\Big)'\Big)'+qx(t)=0,\label{3.7}
\end{equation}
where $m>1$, $0\leq p<\frac{1}{m}$, $q>0$, and $\alpha=\beta=1$.
\end{example}

Observe that \eqref{1.12} and \eqref{2.27} of Theorem \ref{theorem7} are
 satisfied. Because $\pi(t)=\frac{1}{t},r(t)=t^{2},\mu=\frac{1}{4}$, 
by condition \eqref{2.28} we have
\begin{align*}
&\limsup_{t\to\infty}\int^{t}_{T}\Big[\pi(s)q(s)\Big(1-p(\sigma(s))
\frac{\pi(\tau(\sigma(s)))}{\pi(\sigma(s))}\Big)^{\beta}
 -\frac{\mu}{\pi(s)r(s)}\Big]ds\\
&=\limsup_{t\to\infty}\int^{t}_{T}\big[\frac{q}{s}(1-mp)-\frac{1}{4s}\big]ds.
\end{align*}
Hence, if $q(1-mp)>1/4$, then condition \eqref{2.28} holds. 
According to Theorem \ref{theorem7}, the neutral-type equation \eqref{3.7} 
is oscillatory. If we set $p=0$ in \eqref{3.7}, then the second-order 
Euler equation $(t^{2}x'(t))'+qx(t)=0$ is oscillatory as $q>1/4$.

We remark that Theorem \ref{theorem7} can be applied to the linear equation 
\eqref{3.7},  the half-linear equation \eqref{3.3}, the super-linear
 equation \eqref{3.5},  and the sub-linear equation \eqref{3.6}. 
This gives four types of equations  with uniform oscillation criterion and 
improves the results in the literature such as  \cite{Agarwal1,Agarwal2,Li1,Liu}.

\subsection*{Acknowledgments}
This research is supported by the National Natural Science Foundation
of China (No. 11501131), by the Guangdong Engineering Technology Research
Center of Cloud Robot (No. 2015B090903084),
by the Science and Technology Project of Guangdong Province, China,
and by the Science and Technology Project of Maoming, China.


\begin{thebibliography}{00}

\bibitem{Agarwal1} R. P. Agarwal, C. Zhang, T. Li;
\emph{Same remarks on oscillation of second order neutraldifferential equations}, 
Appl. Math. Comput., \textbf{274} (2016), 178-181.

\bibitem{Agarwal2} R. P. Agarwal, M. Bohner, T. Li, C. Zhang;
\emph{Oscillation of second-order Emden-Fowler neutral delay differential equations}, 
Annali. Di. Mathmatica, \textbf{193} (2014), 1861-1875.

\bibitem{Baculikova1} B. Baculikova, J. Dzurina;
\emph{Oscillation theorems for second order nonlinear neutral differential equations},
 Comput Math Appl., \textbf{62} (2011), 4472-4478.

\bibitem{Baculikova2} B. Baculikova, T. Li, J. Dzurina;
\emph{Oscillation theorems for second order super-linear neutral differential 
equations}, Mathematica Slovaca, \textbf{63} (2013), 123-134.

\bibitem{Candan} T. Candan;
\emph{Oscillatory behavior of second order nonlinear neutral differential 
equations with distributed deviating arguments}, Comput. Math. Appl.,
 \textbf{59} (2010), 3710-3717.

\bibitem{Dong} J. G. Dong;
\emph{Oscillation behavior of second-order nonlinear neutral differential 
equations with deviating arguments}, Comput. Math. Appl., \textbf{59} (2010),
 3710-3717.

\bibitem{Erbe} L. Erbe, T. S. Hassan, A. Peterson;
\emph{Oscillation of second order neutral delay differential equations}, 
Adv. Dynam. Syst. Appl., \textbf{3} (2008), 53-71.

\bibitem{Grammatikopoulos} M. K. Grammatikopoulos, G. Ladas, A. Meimaridou;
\emph{Oscillations of second order neutral delay differential equations}, 
Rad. Math., \textbf{1} (1985), 267-274.

\bibitem{Hasanbulli}  M. Hasanbulli, Y. V. Rogovchenko;
\emph{Oscillation of nonlinear neutral functional differential equations}, 
Dyn. Contin. Discrete. Impuls. Syst., \textbf{16} (2009), 227-233.

\bibitem{Han} Z. Han, T. Li, S. Sun, W. Chen;
\emph{Oscillation of second order quasi-linear delay differential equations}, 
J. Appl. Math. Comput., \textbf{40} (2012), 143-152.

\bibitem{Li1} T. X. Li, Z. L. Han, C. H. Zhang, S. R. Sun;
\emph{On the oscillation of second order Emden-Fowler neutral differential equations},
 J. Appl. Math. Comput., \textbf{37} (2011), 601-610.

\bibitem{Li2} T. Li, Y. V. Rogovchenko, C. Zhang;
\emph{Oscillation of second order neutral differential equations}, Funkc. Ekvacioj.,
\textbf{56} (2013), 111-120.

\bibitem{Qin} H. Qin, N. Shang, Y. Lu;
\emph{A note on oscillation criteria of second order nonlinear neutral delay 
differential equations}, Comput. Math. Appl., \textbf{56} (2008), 2987-2992.

\bibitem{Liu} H. D. Liu, F. W. Meng, P. H. Liu;
\emph{Oscillation and asymptotic analysis on a new generalized Emden-Fowler equation},
 Appl. Math. Comput., \textbf{219} (2012), 2739-2748.

\bibitem{Rogovchenko} Y. V. Rogovchenko, F. Tuncay;
\emph{Oscillation criteria for second order nonlinear differential equations
 with damping}, Nonlinear. Anal., \textbf{69} (2008), 208-221.

\bibitem{Li3} T. Li, R. P. Agarwal, M. Bohner;
\emph{Some oscillation results for second order neutral differential equations}, 
J. Indian Math. Soc., \textbf{79} (2012), 97-106.

\bibitem{Sun} S. Sun, T. Li, Z. Han, H. Li;
\emph{Oscillation theorem for second-order quasi-linear neutral functional 
differential equations}, Abstra. Appl. Anal., \textbf{2012} (2012), 1-17.

\bibitem{Tiryaki} A. Tiryaki;
\emph{Oscillation criteria for a certain second order nonlinear differential 
equations with deviating arguments}, EJQTDE, \textbf{61} (2009), 1-11.

\bibitem{Wang} X. L. Wang, F. W. Meng;
\emph{Oscillation criteria of second order quasi-linear neutral delay 
differential equations}, Math. Comput. Modelling, \textbf{46} (2007), 415-421.

\bibitem{Zeng} Y.H. Zeng, L. P. Lou, Y. H. Yu;
\emph{Oscillation for Emden-Fowler delay differential equations of neutral type},
 Acta. Math. Sci., \textbf{35A} (2015), 803-814.

\end{thebibliography}

\end{document}