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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 128, pp. 1--14.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/128\hfil Oscillation criteria]
{Oscillation criteria for even-order nonlinear neutral difference
equations with continuous variables}

\author[S. Wu, P. S. Calay, Z. Hou \hfil EJDE-2015/128\hfilneg]
{Shuhui Wu, Pargat Singh Calay, Zhanyuan Hou}

\address{Shuhui Wu (corresponding author) \newline
School of Science, Zhejiang University of Science and Technology,
318 Liuhe Road, Hangzhou 310023, China}
\email{s.wu@zust.edu.cn, shuhui75@hotmail.com}

\address{Pargat Singh Calay \newline
Faculty of Computing, London Metropolitan University,
166--220 Holloway Road, London N7 8DB, UK}
\email{p.calay@londment.ac.uk}


\address{Zhanyuan Hou \newline
Faculty of Computing, London Metropolitan University,
166--220 Holloway Road, London N7 8DB, UK}
\email{z.how@londonmet.ac.uk}

\thanks{Submitted November 2, 2014. Published May 7, 2015.}
\subjclass[2010]{39A10, 39A21}
\keywords{Neutral difference equations; oscillation; continuous variable}

\begin{abstract}
 In this article, we study the oscillatory behavior of solutions to
 even-order nonlinear neutral difference equations of the form
 $$
 \Delta^m_\tau( x(t)-px(t-r))+f(t,x(g(t)))=0.
 $$
 Using an integral transformation, the Riccati transformation, and iteration,
 we obtain sufficient conditions for all solutions to be oscillatory.
 Examples are also given to illustrate the obtained criteria.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{example}[theorem]{Example}
\allowdisplaybreaks


\section{Introduction}\label{intro5c}

Difference equations have attracted a great deal of attention of researchers
in mathematical, biological, physical sciences, and economy. 
This is specially due to the applications in various problems of biology, 
physics, economy, and so on.
The topics studied for oscillation of the solutions have been investigated 
intensively and the references \cite{Agarwalbook2}--\cite{xzhouwzhang} 
are just a few examples.

In this article, we study even-order nonlinear neutral difference equations 
with continuous variable of the form
\begin{equation} \label{e2n1}
  \Delta^{m}_\tau( x(t)-px(t-r))+f(t,x(g(t)))=0,
  \end{equation}
where $m$ is an even integer $m\geq 4$,  $p\geq 0$, $\tau$ and $r$ are positive 
constants, $\Delta_\tau x(t)=x(t+\tau)-x(t)$, $0<g(t)<t$,
 $g\in C^1([t_0,\infty), \mathbb{R}^+)$, $g'(t)>0$, and
 $f \in C([t_0,\infty)\times \mathbb{R}, \mathbb{R})$.
 Throughout this article we assume that
 \begin{gather}
g(t+\tau)\geq g(t)+\tau \quad \text{for }  t\geq t_0, \label{em2} \\
 \label{em4}
{f(t,\;u)}/{u}\ge q(t)>0 \quad \text{for $u\neq 0$ and 
$q\in C(\mathbb{R}, \mathbb{R}^+)$}.
\end{gather}

Let $t_0^*=\min\{g(t_0),\, t_0-r\}$ and $I_0=[t_0^*,\, t_0]$.
A function $x$ is called  the \emph{solution} of \eqref{em4} with
$x(t)=\varphi(t)$ for
 $t\in I_0$ and  $\varphi\in {C(I_0, \mathbb{R})}$ if it satisfies \eqref{em4}
 for $t\ge t_0$.

A solution $x$ is called oscillatory if it has arbitrarily large zeros;
otherwise it is non-oscillatory.
$x$ is eventually positive if there exists $t_1\geq t_0$, such that
$x(t)>0$ for all $t\geq t_1$. Eventually negative definite is defined similarly.

This article is organized as follows. 
The main results is stated in section 2 and leave the proofs for section 5.
Examples will be presented in section 3 to demonstrate the 
application of the obtained results. 
In section 4, some lemmas will be given to be used in
the proofs of the main results.

\section{Statement of main results}

The assumptions on $g$ guarantee the existence 
and differentiability of its inverse $g^{-1}$. Let
\begin{equation}\label{e2n3}
 \bar{q}(t)=\alpha \min_ {t\leq s\leq t+m\tau
}\{q(s)\}\Big(\min_ {g(t)\le s\le g(t)+m\tau}\{(g^{-1}(s))'\}\Big)^m,
\end{equation}
 where $0<\alpha<1$. The function $\bar{q}$ will play an important role 
in the oscillatory criteria for \eqref{em4}.
Throughout this article,  we use the symbol $\lceil a \rceil$ to denote 
the smallest integer not less than $a$.
 
\begin{theorem}\label{th2n1}
 Assume that for some $t_1 \geq t_0$,
\begin{equation} \label{2nt1}
 \sum_{i=0}^{\infty}\bar{q}(t_1+i\tau) = \infty\,.
\end{equation}
 Then for every solution $x(t)$ of
\eqref{e2n1}, is either oscillatory, or for any $T \geq t_0$ there exists a 
$t_2 >T$ such that $|x(t_2)| \leq p|x(t_2-r)|$.
\end{theorem}

\begin{theorem} \label{th2n2}
In addition to \eqref{2nt1}, we assume that $0<p<1$ and there is a positive 
integer $k_0$ and a $t_1 \geq t_0$ satisfying 
$m_1(n)=\lceil (g(t_1+n\tau)-t_1+k_0r)/\tau \rceil \leq n$
for all large enough $n$. 
Moreover, assume that there is a sequence 
$\{n_{k}\} \to \infty$  such that 
\begin{equation} \label{2nt21}
\sum_{i=m_1(n_k)}^{n_k} \bar{q} (t_1+i\tau)\geq \frac{p^{k_0}(1-p)}{1-p^{k_0}}
\end{equation}
for all $k$ large enough. Then for every solution $x(t)$ of \eqref{e2n1}, 
either $x(t)$  or $x(t)-px(t-r)$ or both are oscillatory.
\end{theorem}

\begin{corollary} \label{co2n21}
 In addition to \eqref{2nt1},
we assume that $0<p<1$ and there is a positive integer
$k_0$ and a $t_1 \geq t_0$ satisfying
$m_1(n)=\lceil (g(t_1+n\tau)-t_1+k_0r)/\tau \rceil \leq n$
for all sufficiently large $n$. Moreover, we assume that there is a sequence
$\{n_{k}\}\to \infty$ such that
\begin{equation} \label{e2nt2c14}
\sum_{i=m_1(n_k)}^{n_k}(i-m_1+1)\bar{q}(t_1+i\tau)\geq
\frac{p^{k_0}(1-p)}{1-p^{k_0}}
\end{equation}
for all $k$ large enough. Then for every solution $x(t)$ of
\eqref{e2n1}, either $x(t)$ or $x(t)-px(t-r)$ or both are oscillatory.
\end{corollary}

\begin{remark} \label{rmk2.6}  \rm
Note that  the requirement \eqref{e2nt2c14} for $\bar{q}(t)$ is weaker
 than \eqref{2nt21} since  $(i-m_1+1) \geq 1$ holds in \eqref{e2nt2c14}.
\end{remark}

\begin{corollary}\label{co2n22}
 In addition to \eqref{2nt1},
we assume that $0<p<1$ and that there is a positive integer
$k_0$ and a $t_1 \geq t_0$ satisfying
$m_1(n)=\lceil (g(t_1+n\tau)-t_1+k_0r)/\tau \rceil \leq
n$ for all sufficiently large $n$.\, Moreover, assume that there is a sequence
$\{n_{k}\}\to \infty$, and an  integer $l$, $1 \leq l < m$ such that
\begin{equation} \label{e2nt2c21}
\frac{1}{l!}\sum_{i=m_1(n_k)}^{n_k} (i-m_1+1)(i-m_1+2)\dots(i-m_1+l)
\bar{q}(t_1+i\tau)
\geq \frac{p^{k_0}(1-p)}{1-p^{k_0}}
\end{equation}
for all $k$ large enough. Then for every solution $x(t)$ of
\eqref{e2n1}, either  $x(t)$  or $x(t)-px(t-r)$ or both are oscillatory.
\end{corollary}

\begin{remark} \rm
Note that \eqref{e2nt2c21} coincides with \eqref{e2nt2c14} for $l=1$.
For $l >1$, 
\[
\frac{(i-m_1+1)(i-m_1+2)\dots (i-m_1+l)}{l!} \geq i-m_1+1 \geq 1\,.
\]
Thus, \eqref{e2nt2c21} is weaker than \eqref{e2nt2c14} and \eqref{2nt21}.
\end{remark}

\begin{theorem}  \label{th2n3}
In addition to \eqref{2nt1}, we assume that $p=1$ and that there is a
 positive integer $k_0$ and a $t_1 \geq t_0$ satisfying $m_1(n)=\lceil
(g(t_1+n\tau)-t_1+k_0r)/\tau \rceil \leq n$ for all sufficiently large $n$. 
Moreover assume that there is a sequence 
$\{n_{k}\}\to \infty$ such that
\begin{equation} \label{e2nt61}
\sum_{i=m_1(n_k)}^{n(k)}\bar{q} (t_1+i\tau)\geq \frac{1}{k_0}
\end{equation}
for $k$ large enough. Then for every solution $x(t)$ of  \eqref{e2n1}, 
either $x(t)$ or $x(t)-x(t-r)$ or both are oscillatory.
\end{theorem}

\begin{corollary} \label{co2n31}
 In addition to \eqref{2nt1}, we assume that $p=1$ and that there is 
a positive integer $k_0$  and a $t_1 \geq t_0$ satisfying $m_1(n_{k})=\lceil
(g(t_1+n\tau)-t_1+k_0r)/\tau \rceil \leq n$ for all sufficiently large $n$.
Moreover assume that there is a sequence $\{n_{k}\} \to \infty$
such that
\begin{equation} \label{e2nt6c1}
\sum_{i=m_1(n_k)}^{n_k}(i-m_1+1)\bar{q} (t_1+i\tau)\geq \frac{1}{k_0}
\end{equation}
for $k$ large enough. Then for every solution $x(t)$ of \eqref{e2n1}, 
either $x(t)$ or $x(t)-x(t-r)$ or both are oscillatory.
\end{corollary}

\begin{corollary} \label{co2n32}
 In addition to \eqref{2nt1}, we
assume that $p=1$ and that there is a positive integer $k_0$
and a $t_1 \geq t_0$ satisfying $m_1(n_{k})=\lceil
(g(t_1+n\tau)-t_1+k_0r)/\tau \rceil \leq n$ for all sufficiently large $n$. 
Moreover we assume that there is a sequence $\{n_{k}\} \to \infty$ 
and an integer $l$, $ 1 \leq l \leq m-1$ such that
\begin{equation} \label{e2nt6c2}
\frac{1}{l!}\sum_{i=m_1(n_k)}^{n_k}(i-m_1+1)(i-m_1+2)\dots (i-m_1+l)\bar{q}
(t_1+i\tau)\geq \frac{1}{k_0}
\end{equation}
for $k$ large enough. Then, for every solution $x(t)$ of \eqref{e2n1}, 
either $x(t)$  or $x(t)-x(t-r)$ or both are oscillatory.
\end{corollary}

 \begin{theorem}\label{th2n4}
 In addition to \eqref{2nt1}, we
assume that $p>1$ and that there is a positive integer $k_0$
and a $t_1 \geq t_0$ satisfying $m_1(n_{k})=\lceil
(g(t_1+n\tau)-t_1+k_0r)/\tau \rceil \leq n$ for all large enough $n$. 
Moreover assume that there is a sequence $\{n_{k}\}\to \infty$  such that 
\begin{equation} \label{e2nt41}
\sum_{i=m_1(n_k)}^{n_k}\bar{q}
(t_1+i\tau)\geq \frac{p^{k_0}(1-p)}{1-p^{k_0}}
\end{equation}
holds for all $k$ large enough. Then, for every bounded solution $x(t)$
 of \eqref{e2n1}, either $x(t)$ or $x(t)-px(t-r)$ or both are
oscillatory.
\end{theorem}

\begin{corollary} \label{co2n41}
In addition to \eqref{2nt1}, we assume that $p>1$ and that there is a positive 
integer $k_0$ and a $t_1 \geq t_0$ satisfying
$m_1(n)=\lceil (g(t_1+n\tau)-t_1+k_0r)/\tau \rceil \leq
n$ for all large enough $n$. Moreover assume that there is a sequence
$\{n_{k}\}\to \infty$ such that
\begin{equation} \label{e2nt4c11}
\sum_{i=m_1(n_k)}^{n_k}(i-m_1+1)\bar{q}(t_1+i\tau)\geq
\frac{p^{k_0}(1-p)}{1-p^{k_0}}
\end{equation}
holds for all $k$ large enough. Then for every bounded solution $x(t)$ of
\eqref{e2n1}, either $x(t)$ or $x(t)-px(t-r)$ or both are oscillatory.
\end{corollary}

\begin{corollary}   \label{co2n42}
In addition to \eqref{2nt1},
we assume that $p>1$ and that there is a positive integer
$k_0$ and a $t_1 \geq t_0$ satisfying
$m_1(n)=\lceil (g(t_1+n\tau)-t_1+k_0r)/\tau \rceil \leq
n$ for all sufficiently large $n$. Moreover assume that there is a sequence
$\{n_{k}\}\to \infty$  and an integer $l$, $1  \leq l <m$, such that
\begin{equation} \label{e2nt4c41}
\frac{1}{l!} \sum_{i=m_1(n_k)}^{n_k}
(i-m_1+1)(i-m_1+2)\dots(i-m_1+l)\,\bar{q}(t_1+i\tau)\geq
\frac{p^{k_0}(1-p)}{1-p^{k_0}}
\end{equation}
holds for all $k$ large enough. Then, for every bounded solution $x(t)$ of equation
\eqref{e2n1}, either $x(t)$ or $x(t)-px(t-r)$ or both are oscillatory.
\end{corollary}

\begin{remark}  \rm
Note that 
\[
\frac{(i-m_1+1)(i-m_1+2)\dots (i-m_1+l-1)}{(l-1)!} \geq i-m_1+1 \geq 1
\]
holds. Thus, \eqref{e2nt4c41} is weaker that \eqref{e2nt4c11} and 
\eqref{e2nt41}.
\end{remark}

\begin{corollary}\label{co2n43}
 In addition to \eqref{2nt1}, we assume that $p>1$ and that there is a 
positive integer $k_0$ and a $t_1 \geq t_0$ satisfying
$m_1(n)=\lceil (g(t_1+n\tau)-t_1+k_0r)/\tau \rceil \leq
n$ for all sufficiently large $n$. Moreover assume that there is a sequence
$\{n_{k}\}\to \infty$ such that
\begin{equation} \label{e2nt4c42}
\frac{1}{(n-1)!}\sum_{i=m_1(n_k)}^{n_k} \frac{(i-m_1+n-1)!}{(i-m_1)!}\,\bar{q}(t_1+i\tau)\geq
\frac{p^{k_0}(1-p)}{1-p^{k_0}}
\end{equation}
holds for all $k$ large enough. Then, for every bounded solution $x(t)$ of
\eqref{e2n1}, either $x(t)$ or $x(t)-px(t-r)$ or both are oscillatory.
\end{corollary}


\section{Examples} \label{2nexa}

Three illustrating examples are given here to demonstrate the applications 
of the obtained oscillatory criteria.

\begin{example}  \label{ex2n1} \rm
Consider the linear difference equation
\begin{equation}
\Delta^{2n}_{\tau}
(x(t)-px(t-r))+\frac{1}{t}x (t-\frac{\sigma}{1+\beta t} )
=0 \label{exa2n1}
\end{equation}
for $t >0$, where $n$ is a positive integer,
 $p \geq 0$, $\beta \geq 0$, the constants $r$, $\tau$ \and $\sigma$ are positive. 
Viewing \eqref{exa2n1} as \eqref{e2n1},  we have $q(t)=1/t$ and 
$g(t)=t-\sigma/(1+\beta t)$. Then,  according to \eqref{e2n3},  
$\bar{q}(t)={\alpha}/(t+2n\tau)$ for $\beta=0$ and
\[
\bar{q}(t)=\frac{\alpha}{t+2n\tau}\Big(1-\frac{\sigma\beta }{(1+\beta
t)^2+\sigma \beta }\Big)^{2n} 
\] 
for $\beta >0$. Since $\bar{q}_{2n}(t)\geq {\alpha'}/{(t+2n\tau)}$ for some 
$\alpha'>0$ and all $t\geq 0$, $\bar{q}_{2n}$ satisfies \eqref{2nt1} with $t_1=0$. 
By Theorem~\ref{th2n1}, for every solution $x(t)$ of \eqref{exa2n1},
 either $x(t)$ is oscillatory
or for any $T \geq t_0$ there exists a $t_2 > T$ such that $|x(t_2)|< p|x(t_2-r)|$.
 In particular, when
$p=0$, every solution of \eqref{exa2n1} is oscillatory. 
\end{example}

\begin{example} \label{examp2.} \label{ex2n2}
Consider the difference equation
\begin{equation}
\Delta^{2n}_{\pi}
(x(t)-px(t-\pi))+8x(t-\pi)+\frac{8\sigma}{1+t^2}x^3(t-\pi) =0,
\label{exa2n2}
\end{equation}
where $\sigma \geq 0$ is a constant. Regarding  \eqref{exa2n2} as
\eqref{e2n1}, we have $\tau=\pi$, $r=\pi$, $g(t)=t-\pi$ 
and $q(t)=8$. Then, for some $\alpha \in(0,1)$, 
$\bar{q}_{2n}=8\alpha$  by \eqref{e2n3} so \eqref{2nt1} is
satisfied. For $p=1$, $k_0=1$ and $t_1=t$,  we have
$m_l=l$ and
\[\sum
_{s=m_l}^{l}(s+1-m_l)\bar{q}_{2n}({t_1}+s\tau)=8\alpha>1=\frac{1}{k_0}
\]
if $\alpha>1/8$. Also we  have
\[
\sum_{s=m_l}^{l}(s+1-m_l)\bar{q}_{2n}({t_1}+s\tau)=8\alpha>p
=\frac{(1-p)p^{k_0}}{1-p^{k_0}}
\]
if $p\in (0,1)\cup(1,8)\; \text{and}\; \alpha>p/8$. According to Theorems
\ref{th2n2} and \ref{th2n3}, for every solution $x(t)$ of \eqref{exa2n2}, 
either $x(t)$  or $x(t)-px(t-r)$  is oscillatory
 if $0 < p\leq 1$. Furthermore, by Theorem \ref{th2n4}, 
for every bounded solution $x(t)$ of \eqref{exa2n2}, either
 $x(t)$ or $x(t)-px(t-r)$ is oscillatory if $1<p<8$.
\end{example}

\begin{example}  \label{ex2n3}\rm
 Consider the difference equation
\begin{equation}
\Delta^{2n}_{\tau} (x(t)-x(t-r))+2^{2n+1}x(t-3)=0, \label{exa2n3}
\end{equation}
where $\tau$ and $ r$ are positive odd integers.
 Viewing  \eqref{exa2n3} as \eqref{e2n1}, we have $ g(t)=t-3$
and $q(t)=2^{2n+1}$. Then, for some $\alpha \in(0,1)$,
$\bar{q}_{2n}=\alpha 2^{2n+1}$  by \eqref{e2n3} so \eqref{2nt1} is
satisfied. For $p=1$, $k_0=1$  and $t_1=t$,  we have
$m_l=l$ and
\[
\sum _{s=m_l}^{l}(s+1-m_l)\bar{q}_{2n}({t_1}+s\tau)
=\alpha 2^{2n+1}>1=\frac{1}{k_0}
\]
if $ \alpha > 2^{-(2n+1)}$. According to Theorem \ref{th2n3},
for every solution $x(t)$ of \eqref{exa2n3}, either $x(t)$ or $x(t)-x(t-r)$ 
is oscillatory.
\end{example}

\section{Related lemmas}

In this section, we  present the lemmas which will be needed in the proofs 
of the main results. The following lemma can be found in 
\cite[page 31]{Agarwalbook2}. 

\begin{lemma} \label{le41}
Let $u(k)$ be  defined on $N(a)$, where $a \in N$, and $ u(k)>0$  with
$\Delta^{m} u(k)$ of constant sign on $ N(a) $ for any
positive integer $m$ and not identically zero.  Then, there exists
an integer $h$, $0\leq h \leq m$,  with   $m+h$ odd for
$\Delta^{m} u(k) \leq 0$ or $m+h$ even for 
$\Delta^{m} u(k) \geq 0$ such that
\begin{itemize}
\item[(i)]
 $ h \leq m-1$ implies $(-1)^{h+i}\Delta^{i}u(k) >0$
 for all $k \in N(a)$, $h\leq i \leq m-1$,

\item[(ii)] $h \geq 1$  implies $\Delta^{i} u(k)>0$ for all 
$k \in N(a)$, $1 \leq i \leq h-1$.
\end{itemize}
\end{lemma}

By applying the above result to the difference with continuous variables,  
we have the following lemma.

\begin{lemma} \label{le42}  
Let $y(t)$ be defined on $[t_0, +\infty)$ where $t_0 \in \mathbb{R}$,  and
$y(t)>0$  with  $\Delta^{m}_{\tau} y(t)$ of constant sign on
$[t_0, +\infty)$ for any positive integer $m$ and not
identically zero. Then, there exists an integer $h, \; 0\leq h
\leq m, \text{ with } m+h$ odd for $\Delta^{m}_{\tau} y(t)
\leq 0 \;\text{ or }\; m+h \;\text{ even for }\; \Delta^{m}_{\tau}
y(t) \geq 0$ such that
\begin{itemize}
\item[(i)]  $ h \leq m-1 $ implies $(-1)^{h+i}\Delta^{i}_{\tau} y(t)>0$
 for all $t \in [t_0, \infty)$, $h\leq i \leq m-1$,

\item[(ii)] $h \geq 1$ implies $\Delta^{i}_{\tau} y(t)>0$  for all
$t \in [t_0, +\infty)$, $1 \leq i \leq h-1$.
\end{itemize}
\end{lemma}

\begin{proof} 
Let $t_1$ be any constant real number in $[t_0, +\infty)$. 
For this fixed $t_1$, by the assumption, we have $y(t_1+k\tau)$ defined 
for any $k \in \{0,1, \dots\}$, and $y(t_1+k\tau)>0$  with
$\Delta^{m}_{\tau} y(t_1+k\tau)$ of constant sign for any 
$k \in \{0, 1, \dots\}$ and for any positive integer $m$ and not
identically zero. Thus, by
Lemma~\ref{le41}, the conclusion holds with the replacement of 
$t \text{ by } t_1+k\tau \text{ for all } k \in N$. Since 
$t_1 \in [t_0, \infty)$ is arbitrary, we can see that the conclusion holds
for $t \in [t_0, +\infty)$.
\end{proof}

\begin{lemma}[{\cite[page 289]{Agarwalbook2}}]  \label{le43} 
Let $y(t)$ be an $m$ times differentiable function on $\mathbb{R}_{+}$ of constant
sign satisfying  $y^{(m)}(t) \not\equiv 0$  and 
$y^{(m)}(t)y(t) \leq 0$  on $[t_1, \infty)$. Then the following statements hold.
\begin{itemize}
\item[(i)] There exists a $t_{2} \geq t_1$ such that the functions
$y^{(j)}(t)$, $j=1, 2, \ldots, m-1,$ are of constant sign on
$[t_{2}, \infty)$.

\item[(ii)] There exists an integer $k <m $ which is odd (even) when 
$m$ is even (odd), such that
\begin{gather*}
y(t)y^{(j)}(t)>0 \quad \text{for } j=0, 1, \ldots, k,\; t \geq t_{2},\\
(-1)^{m+j+1}y(t)y^{(j)}(t)>0 \quad \text{for } j=k+1, \ldots, m,
\; t \geq t_{2}.
\end{gather*}
\end{itemize}
\end{lemma}

\begin{lemma}[{\cite[page 289]{Erbebook}}]\label{le44}
 Assume that $y(t),\, y'(t),  \ldots, y^{(m-1)}(t)$ are absolutely continuous
 and of constant sign on the interval $(t_0, \infty)$, and assume 
$y^{(m)}(t)y(t) \geq 0$.
Then either $y^{(k)}(t) y(t) \geq 0$, $k=0, 1, \ldots, m$ or
there exists an integer $l, 0\leq l \leq m-2$, which is even
 (odd) when $m$ is even (odd), such that
\begin{gather*}
y^{(k)}(t)y(t) \geq 0, \quad \text{for } k=0, 1, \ldots, l,\\
(-1)^{m+k} y^{(k)}(t)y(t) \geq 0, \quad \text{for } k=l+1, \ldots, m.
\end{gather*}
\end{lemma}

\begin{lemma} \label{le2n1} 
Assume that $x(t)$ is an eventually
positive (negative) solution of  \eqref{e2n1} such that $y(t)=x(t)-px(t-r)>0$
$(<0)$ eventually. 
Then $\Delta_{\tau}y(t)>0\, (<0)$  and
$\Delta_{\tau}^{m-1} y(t)>0\, (<0)$ hold eventually.
\end{lemma}

\begin{proof} Suppose $x(t)>0$ and $y(t)>0 $ hold eventually. Due to
$g(t)<t$,  $g'(t)>0$ and \eqref{em2},  there exists a $t_1>t_0$ such that 
$x(g(t))>0$  for all $t \geq t_1$. Further,
\eqref{e2n1} becomes
\[
\Delta_{\tau}^{m}y(t)+f(t, x(g(t)))=0.
\]
According to \eqref{em4},   $f(t, x(g(t))) \geq q(t)x(g(t))>0 $ for  
$t \geq t_1$ hold. Therefore,
\begin{equation} \label{e2nl1}
\Delta^{m}_{\tau}y(t) \leq -q(t)x(g(t)) < 0
\end{equation}
for all large enough $t$,  namely,  $\Delta^m_{\tau}y(t) < 0 $
eventually.
By Lemma~\ref{le42}, $h$ could be odd  with $1 \leq h \leq m-1$.
 For all cases,  we could obtain $\Delta_{\tau}y(t)>0$  and
 $\Delta_{\tau}^{m-1}y(t)>0$ eventually. If $x(t)<0 \text{ and } y(t)<0$ hold 
eventually, then \eqref{e2nl1} becomes $\Delta^{m}_{\tau}y(t) \geq -q(t)x(g(t))>0$. 
Applying Lemma \ref{le42} to $-y(t)$, we obtain 
$\Delta_{\tau}y(t)<0$ and $\Delta_{\tau}^{m-1}y(t)<0$.
\end{proof}

\begin{lemma} \label{le2n2}
 Let the hypothesis of Lemma~\ref{le2n1} be satisfied. Moreover, 
let $\bar{q}(t)$ be defined by \eqref{e2n3}.
 Set
\[
u(t)=\int_{t}^{t+\tau}\,dt_1\int_{t_1}^{t_1+\tau}\,dt_{2}\dots\int_{t_{m-2}}^
{t_{m-2}+\tau}\,dt_{m-1}
\int_{t_{m-1}}^{t_{m-1}+\tau}y(\theta)\,d\theta.
\]
Then $u$ satisfies $u^{(m)}(t)=\Delta^{m}_{\tau}y(t) < 0$ 
$(>0)$, $u(t)>0$ $(<0)$,  $u'(t)>0$ $(<0)$,  $ u^{(m-1)}(t)>0$ 
$(<0)$, $\Delta_{\tau}^{m-1} u(t) >0$  $(<0)$,
and
\[
\Delta^{m}_{\tau} u(t) +\bar{q}(t)\, u(g(t)-kr)\sum_{i=0}^{k}p^{i} \leq 0\, (\geq 0)
\]
for each fixed number $k$ and for all large enough $t$.
\end{lemma}

\begin{proof}
Suppose $x(t)>0$ and $y(t)>0$ hold eventually.  According to the
definition of $u(t)$  and \eqref{e2nl1}, we can see that $u(t)>0$, 
$u^{(m)}(t)= \Delta^{m}_{\tau} y(t) < 0 $  and
\begin{equation} \label{e2nl4}
\Delta^{m}_{\tau}y(t)+q(t) x(g(t)) \leq 0
\end{equation}
for sufficiently large $t$.
Taking into account the definition of $y(t)$,  we have
\[
\Delta^{m}_{\tau}y(t)+q(t)(y(g(t))+px(g(t)-r)) \leq 0.
\]
By repeating the above process $k$ times, we deduce
\[
\Delta^{m}_{\tau}y(t)+q(t)\sum_{i=0}^{k}p^{i} y(g(t)-ir)+q(t)
p^{k+1}x(g(t)-(k+1)r) \leq 0.
\]
Therefore, since $q(t)p^{k+1}x(g(t)-(k+1)r) \geq 0$, it follows that
\[
\Delta^{m}_{\tau}y(t)+q(t)\sum_{i=0}^{k}p^{i} \,y(g(t)-ir)\leq 0.
\]
Furthermore,
\begin{equation} \label{e2nl5}
u^{(m)}(t)+q(t)\sum_{i=0}^{k}p^{i}\,y(g(t)-ir) \leq 0.
\end{equation}
Then, for large enough $t$,  the assumptions on $g$ and $q$
give
\begin{align*}
&\int_{t}^{t+\tau}  ds_1 \int_{s_1}^{s_1+\tau}
\,ds_{m-2} \dots
\int_{s_{m-2}}^{s_{m-2}+\tau}\,ds_{m-1}\int_{s_{m-1}}^{s_{m-1}+\tau}
y(g(\theta)-ir)q(\theta)\,d\theta\\
&\geq \min_{t \leq l \leq t+m\tau}\{q(l)\}\int_{t}^{t+\tau}\, ds_1 \int_{s_1}^{s_1+\tau}
   \,ds_{m-2} \dots\\
&\quad\times \int_{s_{m-2}}^{s_{m-2}+\tau}   \,ds_{m-1}\int_{s_{m-1}}^{s_{m-1}+\tau}
y(g(\theta)-ir)\,d\theta \\
&\geq \min_{t \leq l \leq t+m\tau} \{q(l)\}
\int_{g(t)}^{g(t+\tau)} (g^{-1}(s_1))'ds_1 
\int_{s_1}^{g(g^{-1}(s_1)+\tau)}(g^{-1}(s_{2}))'\,ds_{2} \dots \\
&\quad\times \int_{s_{m-2}}^{g(g^{-1}(s_{m-2})+\tau)}(g^{-1}(s_{m-1}))'ds_{m-1}
\int_{s_{m-1}}^{g(g^{-1}(s_{m-1})+\tau)}
y(\theta-ir)(g^{-1}(\theta))'d\theta\\
&\geq \min_{t \leq l \leq t+m\tau} \{q(l)\}
\Big(\min_{g(t)\leq s \leq g(t)+m\tau}(g^{-1}(s))'\Big)^{m}
\int_{g(t)}^{g(t)+\tau} ds_1 \int_{s_1}^{s_1+\tau}
ds_{2} \dots \\
&\quad\times \int_{s_{m-2}}^{s_{m-2}+\tau}ds_{m-1}
\int_{s_{m-1}}^{s_{m-1}+\tau}
y(\theta-ir)d\theta\\
&\geq \min_{t \leq l \leq t+m\tau}\{q(l)\} \Big(\min_{g(t)\leq s
\leq g(t)+m\tau}(g^{-1}(s))'\min)^{m} u(g(t)-ir)\\
&\geq \bar{q}(t)u(g(t)-ir).
\end{align*}
Thus, integration on both sides of  \eqref{e2nl5} gives
\begin{equation}\label{e2nl6}
\Delta^{m}_{\tau}u(t)+\bar{q}(t)\sum_{i=0}^{k} p^{i}  u(g(t)-ir) \leq 0.
\end{equation}
According to the definition of $u(t)$, the equality
\[
u'(t)=\int_{t}^{t+\tau}\,dt_{2}\int_{t_{2}}^{t_{2}+\tau}\,dt_{3}
\dots\int_{t_{m-2}}^{t_{m-2}+\tau}\,dt_{m-1}
\int_{t_{m-1}}^{t_{m-1}+\tau}\Delta_{\tau}y(\theta)\,d\theta
\]
 holds. Then it follows from Lemma~\ref{le2n1} that $u'(t)>0$. 
Similarly, we have
 \[
u^{(m-1)}(t)=\int_{t}^{t+\tau} \Delta_{\tau}^{m-1} y(\theta) d \theta
\]
 so $u^{(m-1)}(t)>0$ from Lemma~\ref{le2n1}. Hence,
 \[ 
\Delta^{m-1}_{\tau} u(t)=\int_{t}^{t+\tau}dt_1 \int_{t_1}^{t_1+\tau}dt_{2}
\dots  \int_{t_{m-2}}^{t_{m-2}+\tau}u^{(m-1)}(\theta)d\theta>0.
\]
 Further, \eqref{e2nl6} implies
\[
\Delta^{m}_{\tau} u(t)+\bar{q}(t) u(g(t)-kr) \sum_{i=0}^{k} p^{i} \leq 0
\]
for each fixed natural number $k$ and for all large enough $t$. 
If $x(t)<0$ and $y(t)<0$ hold eventually, then $u(t)<0$, 
$u^{(m)}(t)=\Delta_{\tau}^{m}y(t)>0$  and 
$\Delta_{\tau}^{m}y(t)+q(t)x(g(t)) \geq 0$ for large enough $t$.
 Moreover, \eqref{e2nl5} becomes
\[
u^{(m)}(t)+q(t)\sum_{i=0}^{k}p^{i}y(g(t)-ir) \geq 0
\]
and \eqref{e2nl6} becomes
\[
\Delta_{\tau}^{m}u(t)+\bar{q}(t)\sum_{i=0}^{k}p^{i}u(g(t)-ir) \geq 0.
\]
That $u'(t)<0$  and $u^{(m-1)}(t) <0$ follow from 
$\Delta_{\tau}y(t)<0$  and $\Delta_{\tau}^{m-1}y(t)<0$. 
Then $\Delta_{\tau}^{m-1}u(t)<0$ follows from the integration of
 $u^{(m-1)}(t)$. Since $u(t)$ is decreasing, each $u(g(t)-ir)$ can be 
replaced by $u(g(t)-kr)$ in the above inequality.
\end{proof}

\section{Proofs of the main results}

\begin{proof}[Proof of Theorem \ref{th2n1}]
 Let $x(t)$ be a solution of  \eqref{e2n1}  satisfying  $x(t)>0$ and  
$x(t)-px(t-r)>0$  for all large $t$.  Let $y(t)$ be as in Lemma~\ref{le2n1} 
and $u(t)$ be as in Lemma \ref{le2n2}. Furthermore, for any positive integer 
$k$, we have
\[ 
\Delta^{m}_{\tau}u(t)+\bar{q}(t)u(g(t)-kr) \sum_{i=0}^{k}p^{i} \leq 0,
\]
where $u(g(t)-kr)>0$. Define the Riccati transformation by
\[
v(t)=\frac{\Delta^{m-1}_{\tau}u(t)}{u(g(t)-kr)}.
\]
Notice that $v(t)>0$. Moreover we deduce
\begin{align*}
\Delta_{\tau} v(t)
&=  v(t+\tau)-v(t)\\
&= \frac{\Delta_{\tau}^{m-1}u(t+\tau)}{u(g(t+\tau)-kr)}
 -\frac{\Delta^{m-1}_{\tau}u(t)}{u(g(t)-kr)}\\
&= \frac{u(g(t)-kr)\Delta^{m-1}_{\tau}u(t+\tau)
 -u(g(t+\tau)-kr)\Delta^{m-1}_{\tau}u(t)}{u(g(t+\tau)-kr)u(g(t)-kr)}\\
&= \frac{u(g(t)-kr)\Delta^{m-1}_{\tau}u(t+\tau)
 +u(g(t+\tau)-kr)(\Delta^{m}_{\tau}u(t)-
\Delta^{m-1}_{\tau}u(t+\tau))}{u(g(t+\tau)-kr)u(g(t)-kr)}\\
&\leq \frac{\Delta^{m}_{\tau}u(t)}{u(g(t)-kr)}
 -\frac{\Delta^{m-1}_{\tau}u(t+\tau)
\Delta_{\tau}u(g(t)-kr)}{u(g(t)-kr)u(g(t+\tau)-kr)} \\
&\leq  -\bar{q}(t) \sum_{i=0}^{k}p^{i}-v(t+\tau)
 \frac{\Delta_{\tau}u(g(t)-kr)}{u(g(t)-kr)}\\
&\leq  -\bar{q}(t) \sum_{i=0}^{k}p^{i}.
\end{align*}
Therefore, there exists a $t_1>t_0$ such that
\begin{equation} \label{2nt2}
\Delta_{\tau}v(t_1+j\tau)+\bar{q}(t_1+j\tau)\sum_{i=0}^{k}p^{i} \leq 0.
\end{equation}
Summing both sides of \eqref{2nt2} from $0$ to $n$, we have
\[
v(t_1+(n+1)\tau)-v(t_1)+\sum_{i=0}^{k}p^{i}
\sum_{j=0}^{n}\bar{q}(t_1+j\tau) \leq 0.
\] 
Thus
\[
\sum_{i=0}^{k}p^{i}\sum_{j=0}^{n}\bar{q}(t_1+j\tau)<v(t_1)< \infty,
\]
which leads to a contradiction to \eqref{2nt1}. 
If $x(t)$ is a solution of \eqref{e2n1} satisfying $x(t)<0$ and $y(t)<0$ eventually,
 from Lemmas \ref{le2n1} and \ref{le2n2}, the above argument about $v(t)$ 
is still valid and also leads to a contradiction. 
Therefore, the conclusion of the theorem holds.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{th2n2}]
According to Theorem \ref{th2n1}, if \eqref{2nt1} holds, we have that every 
solution $x(t)$ of \eqref{e2n1} is either oscillatory or for any $T \geq t_0$, 
there exists one $t_2>T$ such that $|x(t_2)| \leq p|x(t_2-r)|$.

Assume that \eqref{e2n1} has an eventually positive solution $x(t)$
 such that $y(t)=x(t)-px(t-\tau)$ is not oscillatory. 
Then from Theorem~\ref{th2n1}, we deduce that $y(t) < 0$ for all large enough $t$.
Let $z(t)=-y(t)$. Therefore, $z(t)>0$ and
\[
\Delta^{m}_{\tau} z(t)- f(t, x(g(t)))=0.
\]
Moreover,
\[
\Delta^{m}_{\tau} z(t) \geq q(t)x(g(t)) > 0
\]
so
\begin{equation} \label{2nt22}
\Delta^{m}_{\tau} z(t)-q(t)x(g(t)) \geq 0.
\end{equation}

For $z(t)$, according to Lemma \ref{le42}, $h$ is even. 
So $\Delta_{\tau}^{i} z(t)>0$ for all even number 
$i$ with $2 \leq i \leq m-2$,
 and $|\Delta_{\tau}^{j} z(t)|>0$ for all odd number  
$j$ with $1 \leq j \leq m-1$.

 We show that $\Delta_{\tau}z(t)<0$. Indeed, if $\Delta_{\tau}z(t)>0$, 
then, since $\Delta^{2}_{\tau}z(t)>0$, we may assume
$\Delta_{\tau}z(t_1+k\tau) >l>0$ for a large enough $t_1$ and
all $k \in N$. Then
\[
\sum_{i=0}^{d} \Delta_{\tau} z(t_1+i\tau)=z(t_1+(d+1)\tau)-z(t_1) \geq (d+1)l.
\]
Let $d \to \infty$, then $z(t_1+(d+1)\tau) \to +\infty$.   We
have $\lim_{t \to \infty} x(t)=0$ by repeating
$x(t) <px(t-r)$ for $0<p<1$. Thus,
by the definition of $z(t)$, we have 
$\lim_{t \to \infty} z(t)=0$ which contradicts 
$z(t_1+d\tau) \to +\infty$ as $d \to \infty$. Thus, $\Delta_{\tau} z(t)<0$.

So, according to Lemma~\ref{le42} again, $h=0$. Thus,  
$\Delta_{\tau}^{i} z(t)>0$ for all even number 
$i$ with $2 \leq i \leq m-2$,  and $\Delta_{\tau}^{j} z(t)<0$ for all odd 
number  $j$  with $1 \leq j \leq m-1$.

Notice $x(t)=(x(t+r)+z(t+r))/p$. Hence, from \eqref{2nt22}, it follows that
\[
\Delta_{\tau}^{m}z(t)-\frac{q(t)}{p}  z(g(t)+r)-\frac{q(t)}{p}x(g(t)+r) \geq 0,
\]
and further
\[
\Delta_{\tau}^{m}z(t)-q(t)  \sum_{i=1}^{k}\frac{1}{p^{i}}\,z(g(t)+ir)-
\frac{q(t)}{p^{k}}x(g(t)+kr) \geq 0.
\] 
So,
\begin{equation} \label{e2nt23}
\Delta_{\tau}^{m}z(t)-q(t)\sum_{i=1}^{k}\frac{1}{p^{i}}\,z(g(t)+ir)>0
\end{equation}
since $x(g(t)+kr)>0$. Let
\[
u(t)=\int_0^{\tau}ds_1\int_{s_1}^{s_1+\tau}ds_{2}\dots 
\int_{s_{m-2}}^{s_{m-2}+\tau}ds_{m-1}
\int_{t+s_{m-1}}^{t+s_{m-1}+\tau}z(\theta)d\theta.
\]
Then we have $u^{(m)}(t) > 0\, \text{ and } u(t)>0$.
Since
\[
u'(t)=\int_0^{\tau}ds_1\int_{s_1}^{s_1+\tau}ds_{2}\dots 
\int_{s_{m-2}}^{s_{m-2}+\tau} \Delta_{\tau}z(t+s_{m-1})ds_{m-1},
\]
then $\Delta_{\tau}z(t) <0$ implies $u'(t)<0$.
Moreover,  $ u^{(i)}(t)>0$ for all even number 
$i$  with $2 \leq i \leq m-2$,  and $u^{(j)}(t)<0$ for all odd number  
$j$ with $1 \leq j \leq m-1$.

Integrating \eqref{e2nt23} and from the proof of Lemma~\ref{le2n2} 
replacing  $y(t)$ by  $z(t)$,  we have
\[
\Delta_{\tau}^{m}u(t)-\bar{q}(t)\sum_{i=1}^{k}\frac{1}{p^{i}} u(g(t)+ir)>0,
\]
which leads to
\[
\Delta_{\tau}^{m}u(t)-\bar{q}(t) u(g(t)+kr)\sum_{i=1}^{k}\frac{1}{p^{i}}>0.
\]
Due to $\sum_{i=1}^{k}1/p^{i}=(1-p^{k})/(p^{k}(1-p))$, we deduce that
\[
\Delta_{\tau}^{m}u(t) \geq  \frac{1-p^{k}}{p^{k}(1-p)} 
 \bar{q}(t)\, u(g(t)+kr) > 0.
\]
Replacing  $k$ by $k_0$ and $t$ by $t_1+i\tau$ in the above
inequalities  yield
\[
\Delta_{\tau}^{m}u(t_1+i\tau) \geq \frac{1-p^{k_0}}
{p^{k_0}(1-p)}\, \bar{q}(t_1+i\tau) u(g(t_1+i\tau)+k_0r).
\] 
Summing up both sides of the above inequality for $i$ from $s$ to $n$ 
and since $u'(t)<0$, we have
\[
\Delta_{\tau}^{m-1}u(t_1+(n+1)\tau)-\Delta_{\tau}^{m-1}u(t_1+s\tau) \geq
\frac{1-p^{k_0}}
{p^{k_0}(1-p)}  u(g(t_1+n\tau)+k_0r)\sum_{i=s}^{n}\bar{q}(t_1+i\tau),
\]
 which implies
\begin{equation} \label{2nt23}
-\Delta_{\tau}^{m-1}u(t_1+s\tau) > \frac{1-p^{k_0}}
{p^{k_0}(1-p)} u(g(t_1+n\tau)+k_0r)\sum_{i=s}^{n}\bar{q}(t_1+i\tau)
\end{equation}
due to $\Delta^{m-1}_{\tau} u(t) <0$.
For the above inequality, we will reduce the order of 
 $\Delta_{\tau}^{j}u(t_1+s\tau)$ by rewriting it as
 $\Delta_{\tau}^{j-1} u(t+(s+1)\tau)-\Delta_{\tau}^{j-1}u(t+s\tau)$  for 
$j=1, 2, \dots , n-1$. Taking into account the fact that all even terms 
are positive and all odd terms are negative,
 we will write off all the negative terms from the left hand side of this inequality. 
It yields 
\[
u(t_1+ s\tau) > \frac{1-p^{k_0}}{p^{k_0}(1-p)}u(g(t_1+n\tau)+k_0r)
\sum_{i=s}^{n}\bar{q}(t_1+i\tau).
\]
 Since $g(t_1+n\tau)+k_0r \leq t_1+m_1\tau \text{ and } u$ is decreasing,
by taking $s=m_1$, we obtain
\[
u(t_1+m_1\tau) > u(t_1+m_1\tau)\frac{1-p^{k_0}}{p^{k_0}(1-p)}
\sum_{i=m_1}^{n}\bar{q}(t_1+i\tau),
\]
i.e.,
\[
\sum_{i=m_1}^{n}\bar{q}(t_1+i\tau)<\frac{p^{k_0}(1-p)}{1-p^{k_0}}.
\]
This inequality contradicts \eqref{2nt21}.  If $x(t)$ is an eventually 
negative solution such that $y(t)$ is not oscillatory, then $y(t)>0$ 
holds eventually. The above reasoning with an obvious minor modification 
also leads to a contradiction. Therefore, for every solution $x(t)$, 
either $x(t)$ or $y(t)$ is oscillatory.
\end{proof}

\begin{proof}[Proof of Corollary \ref{co2n21}]
Without loss of generality, we suppose \eqref{e2n1} has an eventually 
positive solution $x(t)$ such that $y(t)=x(t)-px(t-r)$ is not oscillatory.  
The proof is the same as that of  Theorem \ref{th2n2} up to  \eqref{2nt23}.
By the same technique we reduce the order of the difference on the left 
hand side of this inequality down to the second order and it yields
\[
\Delta_{\tau}^{2} u(t_1+s \tau) > \frac{1-p^{k_0}}
{p^{k_0}(1-p)} u(g(t_1+n\tau)+k_0r)\sum_{i=s}^{n}\bar{q}(t_1+i\tau).
\]
Summing  the above inequality for $s$ from $m_1$ to $n$, we have
\[
\Delta_{\tau}u(t_1+(n+1)\tau)-\Delta_{\tau}u(t_1+m_1\tau)
> \frac{1-p^{k_0}}
{p^{k_0}(1-p)} u(g(t_1+n\tau)+k_0r)\sum_{s=m_1}^{n}\sum_{i=s}^{n}\bar{q}(t_1+i\tau).
\]
Due to $\Delta_{\tau}u(t)<0$, it follows from the above inequality that
\[
-\Delta_{\tau}u(t_1+m_1\tau)
> \frac{1-p^{k_0}}
{p^{k_0}(1-p)} u(g(t_1+n\tau)+k_0r)\sum_{s=m_1}^{n}\sum_{i=s}^{n}\bar{q}(t_1+i\tau),
\]
so
\[
u(t_1+m_1\tau)
> \frac{1-p^{k_0}}
{p^{k_0}(1-p)} u(g(t_1+n\tau)+k_0r)\sum_{s=m_1}^{n}
\sum_{i=s}^{n}\bar{q}(t_1+i\tau).
\]
According to $g(t_1+n\tau)+k_0r \leq t_1+m_1\tau$ \,and $u$
is decreasing,  it follows that
\[
u(t_1+m_1\tau) > \frac{1-p^{k_0}}{p^{k_0}(1-p)}u(t_1+m_1\tau)
\sum_{i=m_1}^{n}(i-m_1+1)\bar{q}(t_1+i\tau),
\]
i.e.,
\[
\sum_{i=m_1}^{n}(i-m_1+1)\bar{q}(t_1+i\tau)<
\frac{p^{k_0}(1-p)}{1-p^{k_0}}.
\]
This inequality contradicts \eqref{e2nt2c14}.
Thus  conclusion holds.
\end{proof}

\begin{proof}[Proof of Corollary \ref{co2n22}]
 The proof is the same as that of  Theorem~\ref{th2n2} up to  \eqref{2nt23}. 
We reduce the order of the difference at the left hand of this inequality 
down to the $l$th order as we did in the proof of Theorem~\ref{th2n2}.
Since $ 1 \leq l < m$, $l$ is odd, we obtain
\[
-\Delta_{\tau}^{l} u(t_1+s \tau) > \frac{1-p^{k_0}}
{p^{k_0}(1-p)} u(g(t_1+n\tau)+k_0r)\sum_{i=s}^{n}\bar{q}(t_1+i\tau),
\]
and if $l$ is even,
\[
\Delta_{\tau}^{l} u(t_1+s \tau) > \frac{1-p^{k_0}}
{p^{k_0}(1-p)} u(g(t_1+n\tau)+k_0r)\sum_{i=s}^{n}\bar{q}(t_1+i\tau).
\]
We can reach the same conclusion for the above two cases. 
Thus, we only give the details of the proof when $l$ is odd.
Summing up the above inequality for $s$ from $m_{l}$ to
$n$, we have
\begin{align*}
&-\Delta_{\tau}^{l-1}u(t_1+(n+1)\tau)   +\Delta_{\tau}^{l-1}u(t_1+m_{l}\tau)\\
& >\frac{1-p^{k_0}}{p^{k_0}(1-p)} u(g(t_1+n\tau)+k_0r)
 \sum_{s=m_{l}}^{n}\sum_{i=s}^{n}\bar{q}(t_1+i\tau).
\end{align*}
Since $\Delta_{\tau}^{l-1}u(t)>0$, the above inequality implies
\[
\Delta_{\tau}^{l-1}u(t_1+m_{l}\tau)
> \frac{1-p^{k_0}} {p^{k_0}(1-p)} u(g(t_1+n\tau)+k_0r)
\sum_{s=m_{l}}^{n}\sum_{i=s}^{n}\bar{q}(t_1+i\tau).
\]
By repeating the above procedure, we obtain
\[
u(t_1+m_1\tau)
> \frac{1-p^{k_0}}
{p^{k_0}(1-p)} u(g(t_1+n\tau)+k_0r)\sum_{m_{2}=m_1}^{n}\sum_{m_{3}=m_{2}}^{n}
\dots \sum_{s=m_{l}}^{n}\sum_{i=s}^{n}\bar{q}(t_1+i\tau).
\]
Because $g(t_1+n\tau)+k_0r \leq t_1+m_1\tau$ and $u$
is decreasing,  we have
\begin{align*}
1 &> \frac{1-p^{k_0}}
{p^{k_0}(1-p)}\sum_{m_{2}=m_1}^{n}\sum_{m_{3}=m_{2}}^{n}\dots
\sum_{s=m_{l}}^{n}\sum_{i=s}^{n}\bar{q}(t_1+i\tau)\\
&= \frac{1-p^{k_0}}
{p^{k_0}(1-p)}
\Bigl(\sum_{m_{2}=m_1}^{n}\sum_{m_{3}=m_{2}}^{n}\dots \\
&\quad\times \sum_{i=m_{l-1}}^{n}
\frac{1}{2!}(i-m_{l-1}+1)(i-m_{l-1}+2)\bar{q}(t_1+i\tau)\Big)\\
& \dots   \\
&=\frac{1-p^{k_0}} {p^{k_0}(1-p)} \Big(
\sum_{i=m_1}^{n}\bar{q}(t_1+i\tau)\sum_{m_{2}=m_1}^{i}\frac{1}{(l-1)!}
(i-m_{2}+1)(i-m_{2}+2)\\
&\quad\times \dots   (i-m_{2}+(l-1))\Big)
\\
&=\frac{1-p^{k_0}} {p^{k_0}(1-p)}
\Bigl( \sum_{i=m_1}^{n}\frac{1}{l!}
(i-m_1+1)(i-m_1+2)\times  \dots(i-m_1+l)\bar{q}(t_1+i\tau) \Big),
\end{align*}
i.e.,
\[
\frac{1}{l!}\sum_{i=m_1}^{n}(i-m_1+1)\dots(i-m_1+l)\bar{q}(t_1+i\tau)<
\frac{p^{k_0}(1-p)}{1-p^{k_0}}.
\]
This inequality contradicts \eqref{e2nt2c21}. Thus the conclusion holds.
\end{proof}


\begin{proof}[Proof of Theorem \ref{th2n3}] 
The proof is similar to that of Theorem~\ref{th2n2}. However, 
the proof of the feature of $z(t)$ is different from that of 
Theorem~\ref{th2n2} due to $p=1$.
 We, hence, just give the proof about the feature of $z(t)$.
 For $z(t)$, by Lemma~\ref{le42}, we notice $h$ could be even with 
$2 \leq h \leq m-2$. So $\Delta_{\tau}^{i} z(t)>0$ for all even number 
$i$ with $2 \leq i \leq m-2$, and $|\Delta_{\tau}^{j} z(t)|>0$ for all odd 
number  $j$ with $1 \leq j \leq m-1$.

 If $\Delta_{\tau}z(t)>0$, from the proof of Theorem~\ref{th2n2} we have  
$z(t_1+d\tau) \to +\infty \text{ as } d \to \infty$  for some $ t_1 \geq t_0$.
Since $p=1$, from
$0< x(t) <x(t-r)$, we know that $x(t)$ is bounded on $[t_0, \infty)$. Thus,
 $z(t)$ is bounded on $[t_0, \infty)$. This contradicts 
$z(t_1+d\tau) \to +\infty$  as $d \to \infty$. Thus, $\Delta_{\tau} z(t)<0$. 
So, according to Lemma~\ref{le42} again, $h=0$. Thus, 
 $\Delta_{\tau}^{i} z(t)>0$ for all even number 
$i$ with $2 \leq i \leq m-2$ and $\Delta_{\tau}^{j} z(t)<0$ for all odd number 
 $j$ with $1 \leq j \leq m-1$.

The rest of the proof is as in Theorem~\ref{th2n2},
replacing  $p^{k_0}(1-p)/(1-p^{k_0})$  by  $1/k_0$.
\end{proof}

The proofs of the following corollaries are very similar to those
 of Corollaries \ref{co2n21}-\ref{co2n22} except minor changes.
 Thus, we  omit them.

\begin{proof}[Proof of Theorem \ref{th2n4}]
Suppose that $x(t)$ is a bounded
eventually positive solution of \eqref{2nt1}. 
The proof of Theorem~\ref{th2n2} is then still valid for Theorem~\ref{th2n4}
subject to a few obvious minor changes. Therefore, we omit the proof 
of the results following equation \eqref{e2n1} with $p>1$.
\end{proof}

\subsection*{Acknowledgements}
S. Wu acknowledges support from the NNSF of China via Grant 11171306
and the interdisciplinary research funding from Zhejiang University 
of Science and Technology via Grant 2012JC09Y.


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