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Edit File: lwt132.tex

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\begin{document}

\title{\textbf{Oscillation of Second-Order Sublinear Neutral Delay Difference
Equations}}
\author{\textbf{Wan-Tong Li}\thanks{%
Supported by the NNSF of China (10171040), the NSF of Gansu Province of
China (ZS011-A25-007-Z), the Foundation for University Key Teacher by
Ministry of Education of China, and the Teaching and Research Award Program
for Outstanding Young Teachers in Higher Education Institutions of Ministry
of Education of China.} \\
%EndAName
Department of Mathematics, Lanzhou University\\
Lanzhou, Gansu 730000, People's Republic of China. \\
E-mail: wtli@lzu.edu.cn \and \textbf{S. H. Saker} \\
%EndAName
Mathematics Department, Faculty of Science\\
Mansoura University, Mansoura, 35516, Egypt\\
E--mail: shsaker@mum.mans.edu.eg\\
Faculty of Mathematics and Computer Science\\
Adam Mickiewicz University\\
Matejki 48/49, 60--769 Poznan, Poland\\
E-mail: shsaker@amu.edu.pl}
\maketitle

\begin{abstract}
We present new oscillation criteria for the second-order sublinear neutral
delay difference equation 
\[
\Delta \left( a_n\Delta \left( x_n+p_nx_{n-\tau }\right) \right)
+q_nx_{n-\sigma }^{\gamma} =0, 
\]
where $0<{\gamma} <1$ is a quotient of odd positive integers and $%
\sum\limits_{n=0}^\infty \frac 1{a_n}=\infty .$

\textbf{Key Words and Phrases}: Oscillation, delay neutral difference
equations

\textbf{2000 AMS Subject Classifications:} 39A10
\end{abstract}

\section{Introduction}

In this note we shall consider the second-order sublinear neutral delay
difference equation 
\begin{equation}
\Delta \left( a_n\Delta \left( x_n+p_nx_{n-\tau }\right) \right)
+q_nx_{n-\sigma }^{\gamma}=0,\text{\ }n=0,1,2...  \label{1.1}
\end{equation}
where $0<{\gamma }<1$ is a quotient of odd positive integers$,$ $\Delta $
denotes the forward difference operator $\Delta x_n=x_{n+1}-x_n$ and $\Delta
^2x_n=\Delta (\Delta x_n)$ for any sequence $\{x_n\}$ of real numbers, $\tau
,$ $\sigma $ are fixed nonnegative integers, $\{a_n\},\{p_n\}$ and $\{q_n\}$
are sequences of real numbers such that 
\begin{equation}
a_n>0,\sum\limits_{n=n_0}^\infty \frac 1{a_n}=\infty ,\text{ }0\leq p_n<1%
\text{ \ for \ all \ }n\geq 0\text{ \ and \ }q_n\geq 0\,  \label{1.2}
\end{equation}
and $q_n$ is not identically zero for large $n$.

By a solution of (\ref{1.1}) we mean a nontrivial sequence $\{x_n\}$ which
is defined for $n\geq -N,$ where $N=\max \{\tau ,\sigma \},$ and satisfies
equation (\ref{1.1}) for $n=0,1,2...$ . Clearly if

\begin{equation}
x_n=A_n\text{ \ \ for }n=-N,...,-1,0  \label{1.3}
\end{equation}
are given, then Eq.(\ref{1.1}) has a unique solution satisfying the initial
conditions (\ref{1.3}). A solution $\{x_n\}$ of (\ref{1.1}) is said to be
oscillatory if for every $n_1>0$ there exists an $n\geq n_1$ such that $%
x_nx_{n+1}\leq 0,$ otherwise it is nonoscillatory. Equation (\ref{1.1}) is
said to be oscillatory if all its solutions are oscillatory.

Recently, there has been an increasing interest in the study oscillation and
asymptotic behavior of solutions of second order neutral delay difference
equations, for example see \cite[2-18]{1} and the references therein. To the
best of our knowledge, nothing is known regarding the qualitative behavior
of solutions of Eq.(\ref{1.1}) in the sublinear case. Therefore our aim in
this paper is to give several oscillation criteria of Eq.(\ref{1.1}) when (%
\ref{1.2}) holds.

\section{Main Results}

We will assume throughout this paper that $\Delta a_n\geq 0$.

\textbf{Theorem 2.1. }Assume that (\ref{1.2})\ holds. Furthermore, assume
that there exists a positive sequence $\mathit{\{}\rho _n\}$\ such that for\
every $\alpha \geq 1$

\begin{equation}
\lim_{n\rightarrow \infty }\sup \sum\limits_{l=0}^n\left[ \rho _lQ_l-\frac{%
a_{l-\sigma }\left( \alpha (l+1-\sigma )\right) ^{1-{\gamma }}(\Delta \rho
_l)^2}{4{\gamma }\rho _l}\right] =\infty .  \label{2.1}
\end{equation}
where $Q_n=q_n(1-p_{n-\sigma })^{\gamma}.$\ Then every solution of Eq.(\ref
{1.1}) oscillates\textit{.}

\textbf{Proof.} Assume for the sake of contradiction that Eq.(\ref{1.1}) has
a nonoscillatory solution $\{x_n\},$ we may assume without loss of
generality that $x_{n-N}>0$ for $n\geq n_0>0.$ (the case when $x_n<0$ is
similar and hence is omitted). Set 
\begin{equation}
z_n=x_n+p_nx_{n-\tau }  \label{2.2}
\end{equation}
By, assumption (\ref{1.2}$)$ we have $z_n>0$ for $n\geq n_0$ and from (\ref
{1.1}) it follows that 
\[
\Delta (a_n\Delta z_n)=-q_nx_{n-\sigma }^\gamma \leq 0,\,\ \ \ \ n\geq n_0 
\]
and so $\{a_n\Delta z_n\}$ is an eventually nonincreasing sequence. We first
show that $\Delta z_n\geq 0$ \ for $n\geq n_0.\,$In fact, if there exists an
integer $n_1\geq n_0$ such that $a_{n_1}\Delta z_{n_1}=c<0,$ then $a_n\Delta
z_n\leq c$ for $n\geq n_1$, that is 
\[
\Delta z_n\leq \frac c{a_n}, 
\]
and, hence 
\[
z_n\leq z_{n_1}+c\sum\limits_{i=n_1}^{n-1}\frac 1{a_i}\rightarrow -\infty 
\text{ \ as }n\rightarrow \infty , 
\]
which contradicts the fact that $z_n>0$ for $n\geq n_0.$ Also we claim that $%
\Delta ^2z_n\leq 0.$ If not there exists $n_1\geq n_0$ such that $\Delta
^2z_n>0$ for $n\geq n_1$ and this implies that $\Delta z_{n+1}>\Delta z_n$,
so that since $\Delta a_n\geq 0,$ $a_{n+1}\left( \Delta z_{n+1}\right)
>a_{n+1}\left( \Delta z_n\right) \geq a_n\left( \Delta z_n\right) $ and this
contradicts the fact that $\{a_n\left( \Delta z_n\right) \}$ is
nonincreasing sequence, then $\Delta ^2z_n\leq 0$ and therefore we have 
\begin{equation}
z_n>0,\text{ \ \ \ \ \ }\Delta z_n\geq 0\,\ \ \ \text{and}\,\ \ \Delta
^2z_n\leq 0\text{ \ for }n\geq n_0,  \label{2.4}
\end{equation}
and then from (\ref{2.2}) and (\ref{2.4}) we have $x_n\geq (1-p_n)z_n$ and
this implies that for $n\geq n_1=n_0+\sigma $ 
\[
x_{n-\sigma }\geq (1-p_{n-\sigma })z_{n-\sigma }. 
\]
>From Eq.(\ref{1.1}) and the last inequality, we have 
\begin{equation}
\Delta (a_n\Delta z_n)+Q_nz_{n-\sigma }^\gamma \leq 0,\,\ \ n\geq n_1.
\label{2.3}
\end{equation}
Define the function 
\begin{equation}
w_n=\rho _n\frac{a_n\Delta z_n}{z_{n-\sigma }^\gamma }.  \label{2.5}
\end{equation}
Then $w_n>0$ and 
\begin{equation}
\Delta w_n=a_{n+1}\Delta z_{n+1}\Delta \left[ \frac{\rho _n}{z_{n-\sigma
}^\gamma }\right] +\frac{\rho _n\Delta (a_n\Delta z_n)}{z_{n-\sigma }^\gamma 
}.  \label{2.6}
\end{equation}
>From the fact that 
\[
\Delta z_n\geq 0\,\ \ \ \text{and}\,\ \ \ \Delta (a_n\Delta z_n)\leq 0\text{
\ for }n\geq n_1, 
\]
we have 
\begin{equation}
a_{n-\sigma }\Delta z_{n-\sigma }\geq a_{n+1}\Delta z_{n+1}\ \text{and \ \ \ 
}z_{n+1-\sigma }\geq z_{n-\sigma }.  \label{2.7}
\end{equation}
Then, from Eq.(\ref{1.1}), (\ref{2.5}) and (\ref{2.6}), we get 
\begin{equation}
\Delta w_n=-\rho _nQ_n+\frac{\Delta \rho _n}{\rho _{n+1}}w_{n+1}-\frac{\rho
_na_{n+1}\Delta z_{n+1}\Delta z_{n-\sigma }^\gamma }{z_{n+1-\sigma }^\gamma
z_{n-\sigma }^\gamma }.  \label{2.8}
\end{equation}
>From (\ref{2.7}) and (\ref{2.8}), we have 
\begin{equation}
\Delta w_n\leq -\rho _nQ_n+\frac{\Delta \rho _n}{\rho _{n+1}}w_{n+1}-\frac{%
\rho _na_{n+1}\Delta z_{n+1}\Delta z_{n-\sigma }^\gamma }{\left(
z_{n+1-\sigma }^\gamma \right) ^2}.  \label{2.9}
\end{equation}
Now, by using the inequality (cf. \cite[p. 39]{5}) 
\[
x^\gamma -y^\gamma \geq \gamma x^{\gamma -1}(x-y)\text{ for all }x\geq y>0%
\text{ and }0<\gamma \leq 1, 
\]
we find that 
\begin{eqnarray}
\Delta z_{n-\sigma }^\gamma &=&z_{n+1-\sigma }^\gamma -z_{n-\sigma }^\gamma
\geq {\gamma }\left( z_{n+1-\sigma }\right) ^{{\gamma }-1}(z_{n+1-\sigma
}-z_{n-\sigma })  \label{2.10} \\
&=&{\gamma }\left( z_{n+1-\sigma }\right) ^{{\gamma }-1}\left( \Delta
z_{n-\sigma }\right) .  \nonumber
\end{eqnarray}
Substitute from (\ref{2.10}) in (\ref{2.9}), we have 
\begin{equation}
\Delta w_n\leq -\rho _nQ_n+\frac{\Delta \rho _n}{\rho _{n+1}}w_{n+1}-{\gamma 
}\left( z_{n+1-\sigma }\right) ^{{\gamma }-1}\rho _na_{n+1}\frac{\Delta
z_{n+1}\left( \Delta z_{n-\sigma }\right) }{\left( z_{n+1-\sigma }^\gamma
\right) ^2}.  \label{2.11}
\end{equation}
Again, from (\ref{2.7}) in (\ref{2.11}), we obtain 
\begin{equation}
\Delta w_n\leq -\rho _nQ_n+\frac{\Delta \rho _n}{\rho _{n+1}}w_{n+1}-{\gamma 
}\left( z_{n+1-\sigma }\right) ^{{\gamma }-1}\rho _n\frac{\left(
a_{n+1}\right) ^2}{\left( a_{n-\sigma }\right) }\frac{\left( \Delta
z_{n+1}\right) ^2}{\left( z_{n+1-\sigma }^\gamma \right) ^2},  \label{2.12}
\end{equation}
and hence, 
\begin{equation}
\Delta w_n\leq -\rho _nQ_n+\frac{\Delta \rho _n}{\rho _{n+1}}w_{n+1}-{\gamma 
}\rho _n\frac{\left( a_{n+1}\right) ^2}{\left( a_{n-\sigma }\right) \left(
z_{n+1-\sigma }\right) ^{1-{\gamma }}}\frac{\left( \Delta z_{n+1}\right) ^2}{%
\left( z_{n+1-\sigma }^\gamma \right) ^2}.  \label{2.13}
\end{equation}
>From (\ref{2.5}) and (\ref{2.13}), we find 
\begin{equation}
\Delta w_n\leq -\rho _nQ_n+\frac{\Delta \rho _n}{\rho _{n+1}}w_{n+1}-\frac{{%
\gamma }\rho _n}{\left( a_{n-\sigma }\right) \left( \rho _{n+1}\right) ^2}%
w_{n+1}^2\frac 1{\left( z_{n+1-\sigma }\right) ^{1-{\gamma }}}.  \label{2.14}
\end{equation}
>From (\ref{2.4}), we conclude that 
\[
z_n\leq z_{n_0}+\Delta z_{n_0}(n-n_0),\text{ \ \ \ \ \ \ }n\geq n_0, 
\]
and consequently there exists a $n_1\geq n_0$ and appropriate constant $%
\alpha \geq 1$ such that 
\[
z_n\leq \alpha n\text{ \ \ for }n\geq n_1, 
\]
and this implies that 
\[
z_{n+1-\sigma }\leq \alpha (n+1-\sigma )\text{ \ \ for }n\geq n_2=n_1+\sigma
-1, 
\]
and, hence 
\[
\frac 1{\left( z_{n+1-\sigma }\right) ^{1-{\gamma }}}\geq \frac 1{\left(
\alpha (n+1-\sigma )\right) ^{1-{\gamma }}}. 
\]
Substitute from the last inequality in (\ref{2.14}), we find 
\begin{eqnarray}
\Delta w_n &\leq &-\rho _nQ_n+\frac{\Delta \rho _n}{\rho _{n+1}}w_{n+1}-%
\frac{{\gamma }\rho _n}{\left( \rho _{n+1}\right) ^2a_{n-\sigma }\left(
\alpha (n+1-\sigma )\right) ^{1-{\gamma }}}w_{n+1}^2  \label{2.15} \\
&=&-\rho _nQ_n+\frac{a_{n-\sigma }\left( \alpha (n+1-\sigma )\right) ^{1-{%
\gamma }}(\Delta \rho _n)^2}{4{\gamma }\rho _n}  \nonumber \\
&&-\left[ \frac{\sqrt{{\gamma }\rho _n}}{\rho _{n+1}\sqrt{\left( \alpha
(n+1-\sigma )\right) ^{1-{\gamma }}a_{n-\sigma }}}w_{n+1}-\frac{\sqrt{\left(
\alpha (n+1-\sigma )\right) ^{1-{\gamma }}a_{n-\sigma }}\Delta \rho _n}{2%
\sqrt{{\gamma }\rho _n}}\right] ^2  \nonumber \\
&<&-\left[ \rho _nQ_n-\frac{\left( \alpha (n+1-\sigma )\right) ^{1-{\gamma }%
}a_{n-\sigma }(\Delta \rho _n)^2}{4{\gamma }\rho _n}\right] .  \nonumber
\end{eqnarray}
Then, we have 
\begin{equation}
\Delta w_n<-\left[ \rho _nQ_n-\frac{\left( a_{n-\sigma }\right) \left(
\alpha (n+1-\sigma )\right) ^{1-{\gamma }}(\Delta \rho _n)^2}{4{\gamma }\rho
_n}\right] .  \label{2.151}
\end{equation}
Summing (\ref{2.151}) from $n_2$ to $n$, we obtain 
\[
-w_{n_2}<w_{n+1}-w_{n_2}<-\sum\limits_{l=n_2}^n\left[ \rho _lQ_l-\frac{%
\left( a_{l-\sigma }\right) \left( \alpha (l+1-\sigma )\right) ^{1-{\gamma }%
}(\Delta \rho _l)^2}{4{\gamma }\rho _l}\right] , 
\]
which yields 
\[
\sum\limits_{l=n_2}^n\left[ \rho _lQ_l-\frac{a_{l-\sigma }\left( \alpha
(l+1-\sigma )\right) ^{1-{\gamma }}(\Delta \rho _l)^2}{4{\gamma }\rho _l}%
\right] <c_1, 
\]
for all large n, which is contrary to (\ref{2.1}). The proof is complete.

\textbf{Remark 2.1.} Note that from Theorem 2.1, we can obtain different
conditions for oscillation of all solutions of Eq.(\ref{1.1}) when (\ref{1.2}%
) holds by different choices of $\{\rho _n\}.$ Let $\rho _n=n^{{\lambda}} ,$ 
$n\geq n_0$ and ${\lambda} >1$ is a constant. By Theorem 2.2 we have the
following result.

\textbf{Corollary 2.1.} Assume that all the assumption of Theorem 2.1 hold,
except the condition (\ref{2.1}) is replaced by 
\begin{equation}
\lim_{n\rightarrow \infty }\sup \sum\limits_{s=n_{0}}^{n}\left[ s^{\lambda
}Q_{s}-\frac{a_{s-\sigma }\left( \alpha (s+1-\sigma )\right) ^{1-{\gamma }%
}\left( (s+1)^{\lambda }-s^{\lambda }\right) ^{2}}{4{\gamma }s^{\lambda }}%
\right] =\infty \mathit{,}  \label{2.16}
\end{equation}
where $Q_{n}=q_{n}(1-p_{n-\sigma })^{\gamma }.$ Then every solution of Eq.(%
\ref{1.1}) oscillates.

\textbf{Remark 2.2. }When ${\gamma} =1$, Eq.(\ref{1.1}) reduces to the
linear delay difference equation

\begin{equation}
\Delta (a_n(\Delta x_n+p_nx_{n-\tau }))+q_nx_{n-\sigma }=0,\text{ \ \ \ \ }%
n=0,1,2...  \label{2.17}
\end{equation}
and the condition (\ref{2.1}) in Theorem 2.1 reduces to

\begin{equation}
\lim_{n\rightarrow \infty }\sup \sum\limits_{l=0}^n\left[ \rho
_lq_l(1-p_{l-\sigma })-\frac{\left( a_{l-\sigma }\right) (\Delta \rho _l)^2}{%
4\rho _l}\right] =\infty .  \label{2.18}
\end{equation}
Then Theorem 2.1 and Theorem 1 in \cite{9} are the same in linear case. Also
when $p_n=0$ and ${\gamma }=1$ Theorem 2.1 and Corollary 1 in \cite{10} are
the same.

\textbf{Theorem 2.2.} Assume that (\ref{1.2}) holds, and let $\{\rho
_{n}\}_{n=0}^{\infty }$\ be a positive sequence$.$\ Furthermore, we assume
that there exists a double sequence $\{H_{m,n}:m\geq n\geq 0\}$\ such that
(i) $H_{m,m}=0$\ for $m\geq 0,$ (ii) $\,H_{m,n}>0\,$for $m>n>0$, (iii) $%
\Delta _{2}H_{m,n}=H_{m,n+1}-H_{m,n}.$ If 
\begin{equation}
\lim_{m\rightarrow \infty }\sup \frac{1}{H_{m,0}}\sum\limits_{n=n_{0}}^{m-1}%
\left[ H_{m,n}\rho _{n}Q_{n}-\frac{\rho _{n+1}^{2}}{4\stackrel{-}{\rho }_{n}}%
\left( h_{m,n}\sqrt{H_{m,n}}-\frac{\Delta \rho _{n}}{\rho _{n+1}}%
H_{m,n}\right) ^{2}\right] \mathit{=\infty ,}  \label{2.19}
\end{equation}
where 
\[
h_{m,n}=\frac{-\Delta _{2}H_{m,n}}{\sqrt{H_{m,n}}},\text{ }\stackrel{-}{\rho 
}_{n}={\gamma }\rho _{n}/\left( (\alpha (n+1-\sigma ))^{1-{\gamma }}\left(
a_{n-\sigma }\right) \right) . 
\]
Then every solution of Eq.(\ref{1.1}) oscillates.

\textbf{Proof.} Proceeding as in Theorem 2.1, we assume that Eq.(\ref{1.1})
has a nonoscillatory solution, say $x_{n}>0$ and $x_{n-\sigma }>0$ for all $%
n\geq n_{0}$. From the proof of Theorem 2.1 we obtain (\ref{2.15}) for all $%
n\geq n_{2}$. From (\ref{2.15}), since $\Delta \rho _{n}\leq 0,$ we have for 
$n\geq n_{2}$%
\begin{equation}
\Delta w_{n}\leq -\rho _{n}Q_{n}+\frac{\Delta \rho _{n}}{\rho _{n+1}}w_{n+1}-%
\frac{\stackrel{-}{\rho }_{n}}{\left( \rho _{n+1}\right) ^{2}}w_{n+1}^{2},
\label{2.20}
\end{equation}
or 
\begin{equation}
\rho _{n}Q_{n}\leq -\Delta w_{n}+\frac{\Delta \rho _{n}}{\rho _{n+1}}w_{n+1}-%
\frac{\stackrel{-}{\rho }_{n}}{\left( \rho _{n+1}\right) ^{2}}w_{n+1}^{2}.
\label{2.21}
\end{equation}
Therefore, we have 
\begin{equation}
\sum\limits_{n=n_{2}}^{m-1}H_{m,n}\rho _{n}Q_{n}\leq
-\sum\limits_{n=n_{2}}^{m-1}H_{m,n}\Delta
w_{n}+\sum\limits_{n=n_{2}}^{m-1}H_{m,n}\frac{\Delta \rho _{n}}{\rho _{n+1}}%
w_{n+1}-\sum\limits_{n=n_{2}}^{m-1}H_{m,n}\frac{\stackrel{-}{\rho }_{n}}{%
\left( \rho _{n+1}\right) ^{2}}w_{n+1}^{2}.  \label{2.22}
\end{equation}
which yields, after summing by parts, 
\begin{eqnarray*}
&&\sum\limits_{n=n_{2}}^{m-1}H_{m,n}\rho _{n}q_{n+1} \\
&\leq &H_{m,n_{2}}w_{n_{2}}+\sum\limits_{n=n_{2}}^{m-1}w_{n+1}\Delta
_{2}H_{m,n}+\sum\limits_{n=n_{2}}^{m-1}H_{m,n}\frac{\Delta \rho _{n}}{\rho
_{n+1}}w_{n+1}-\sum\limits_{n=n_{2}}^{m-1}H_{m,n}\frac{\stackrel{-}{\rho }%
_{n}}{\left( \rho _{n+1}\right) ^{2}}w_{n+1}^{2} \\
&=&H_{m,n_{2}}w_{n_{2}}-\sum\limits_{n=n_{2}}^{m-1}h_{m,n}\sqrt{H_{m,n}}%
w_{n+1}+\sum\limits_{n=n_{2}}^{m-1}H_{m,n}\frac{\Delta \rho _{n}}{\rho _{n+1}%
}w_{n+1}-\sum\limits_{n=n_{2}}^{m-1}H_{m,n}\frac{\stackrel{-}{\rho }_{n}}{%
\left( \rho _{n+1}\right) ^{2}}w_{n+1}^{2} \\
&=&H_{m,n_{2}}w_{n_{2}} \\
&&-\sum\limits_{n=n_{2}}^{m-1}\left[ \frac{\sqrt{H_{m,n}\stackrel{-}{\rho }%
_{n}}}{\rho _{n+1}}w_{n+1}+\frac{\rho _{n+1}}{2\sqrt{H_{m,n}\stackrel{-}{%
\rho }_{n}}}\left( h_{m,n}\sqrt{H_{m,n}}-\frac{\Delta \rho _{n}}{\rho _{n+1}}%
H_{m,n}\right) \right] ^{2} \\
&&+\frac{1}{4}\sum\limits_{n=k}^{m-1}\frac{\left( \rho _{n+1}\right) ^{2}}{%
\stackrel{-}{\rho }_{n}}\left( h_{m,n}-\frac{\Delta \rho _{n}}{\rho _{n+1}}%
\sqrt{H_{m,n}}\right) ^{2}
\end{eqnarray*}
Then, 
\[
\sum\limits_{n=n_{2}}^{m-1}\left[ H_{m,n}\rho _{n}q_{n}-\frac{\rho _{n+1}^{2}%
}{4\bar{\rho}_{n}}\left( h_{m,n}-\frac{\Delta \rho _{n}}{\rho _{n+1}}\sqrt{%
H_{m,n}}\right) ^{2}\right] <H_{m,n_{2}}w_{n_{2}}\leq H_{m,0}w_{n_{2}} 
\]
which implies that 
\[
\sum\limits_{n=0}^{m-1}\left[ H_{m,n}\rho _{n}q_{n}-\frac{\rho _{n+1}^{2}}{4%
\bar{\rho}_{n}}\left( h_{m,n}-\frac{\Delta \rho _{n}}{\rho _{n+1}}\sqrt{%
H_{m,n}}\right) ^{2}\right] <H_{m,0}\sum_{n=0}^{n_{2}-1}\rho
_{n}q_{n+1}+H_{m,0}w_{n_{2}}. 
\]
Hence 
\[
\limsup_{m\rightarrow \infty }\frac{1}{H_{m,0}}\sum\limits_{n=0}^{m-1}\left[
H_{m,n}\rho _{n}q_{n}-\frac{\rho _{n+1}^{2}}{4\bar{\rho}_{n}}\left( h_{m,n}-%
\frac{\Delta \rho _{n}}{\rho _{n+1}}\sqrt{H_{m,n}}\right) ^{2}\right]
<\infty , 
\]
which is contrary to (\ref{2.19}). The proof is complete.

\mathstrut

By choosing the sequence $\{H_{m,n}\}$ in appropriate manners, we can derive
several oscillation criteria for (1.1). For instance, let us consider the
double sequence $\{H_{m,n}\}$ defined by 
\[
H_{m,n}=(m-n)^{{\lambda }}, 
\]
or 
\[
H_{m,n}=\left( \log \left( \frac{m+1}{n+1}\right) \right) ^{{\lambda }}, 
\]
where $\ {\lambda }\geq 1,$ $m\geq n\geq 0$. Then $H_{m,m}=0$ for $m\geq 0$
and $H_{m,n}>0$ and $\Delta _{2}H_{m,n}\leq 0$ for $m>n>0.$ Hence we have
the following results.

\textbf{Corollary 2.2.} Assume that all the assumptions of Theorem 2.2 hold,
except the condition (\ref{2.19}) is replaced by 
\[
\lim_{m\rightarrow \infty }\sup \frac{1}{m^{{\lambda }}}\sum%
\limits_{n=0}^{m}\left[ (m-n)^{{\lambda }}\rho _{n}Q_{n}-\frac{\rho
_{n+1}^{2}}{4\overline{\rho }_{n}}\left( \lambda (m-n)^{\frac{\lambda -2}{2}%
}-\frac{\Delta \rho _{n}}{\rho _{n+1}}\sqrt{(m-n)^{\lambda }}\right) \right]
=\infty \mathit{.} 
\]
where $Q_{n}=q_{n}(1-p_{n-\sigma })^{\gamma }$. Then every solution of Eq.(%
\ref{1.1}) oscillates.

\textbf{Corollary 2.3.} Assume that all the assumptions of Theorem 2.2 hold,
except the condition (\ref{2.19}) is replaced by 
\[
\lim_{m\rightarrow \infty }\sup \frac{1}{\left( \log (m+1)\right) ^{\lambda }%
}\sum\limits_{n=0}^{m}\left[ \left( \log \left( \frac{m+1}{n+1}\right)
\right) ^{\lambda }\rho _{n}Q_{n}-\frac{\rho _{n+1}^{2}}{4\overline{\rho }%
_{n}}B_{m,n}\right] =\infty . 
\]
where 
\[
B_{m,n}=\left( \frac{\lambda }{n+1}\left( \ln \frac{m+1}{n+1}\right) ^{\frac{%
\lambda -2}{2}}-\frac{\Delta \rho _{n}}{\rho _{n+1}}\sqrt{\left( \ln \frac{%
m+1}{n+1}\right) ^{\lambda }}\right) ^{2} 
\]
Then every solution of Eq.(\ref{1.1}) oscillates.

\mathstrut

Another $H_{m,n}$ may be chosen as 
\[
H_{m,n}=\phi (m-n),\;m\geq n\geq 0, 
\]
\[
H_{m,n}=(m-n)^{(\lambda )}\,\ \ \lambda >2,\text{ }m\geq n\geq 0. 
\]
where $\phi :[0,\infty )\rightarrow \lbrack 0,\infty )$ is a continuously
differentiable function which satisfies $\phi (0)=0$ and $\phi (u)>0,$ $\phi
^{\prime }(u)\geq 0$ for $u>0,$ and $(m-n)^{(\lambda
)}=(m-n)(m-n+1)...(m-n+\lambda -1)$ and 
\[
\Delta _{2}(m-n)^{(\lambda )}=(m-n-1)^{(\lambda )}-(m-n)^{(\lambda
)}=-\lambda (m-n)^{(\lambda -1)} 
\]
Corresponding corollaries can also be stated.

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\end{document}