1. Introduction
Certainly, the Lyapunov direct method has been, for more than 100 years, the
efficient tool for the study of stability properties of ordinary,
functional, partial differential and difference equations. Nevertheless, the
application of this method to problems of stability in differential and
difference equations with delay has encountered serious difficulties if the
delay is unbounded or if the equation has unbounded terms ([
1,
2,
3,
4,
5,
6,
7,
8 9,
10,
11,
12,
13,
14,
15,
16]). Recently, Burton, Furumochi, Zhang, Raffoul, Islam, Yankson and
others have noticed that some of these difficulties vanish or might be
overcome by means of fixed point theory (see [
17,
18,
19,
20,
21,
22,
23,
24,
25,
26,
27,
28,
29,
30,
31,
32]). The fixed
point theory does not only solve the problem on stability but has a
significant advantage over Lyapunov’s direct method. The conditions of the
former are often averages but those of the latter are usually pointwise (see
[
1]).
In this paper, we consider the following mixed type neutral difference
equation
\begin{equation}
\Delta x\left( t\right) +a\left( t\right) \Delta x\left( \tau \left(
t\right) \right) +\sum_{i=1}^{k}b_{i}\left( t\right) x\left( \sigma
_{i}\left( t\right) \right) +\sum_{j=1}^{l}c_{j}\left( t\right) x\left( \tau
_{j}\left( t\right) \right) =0, \label{1}
\end{equation}
(1)
with an assumed initial condition
\begin{equation}
x(t)=\psi (t)\text{ for }t\in \lbrack m\left( t_{0}\right) ,t_{0}]\cap
\mathbb{Z}, \label{2}
\end{equation}
(2)
where \(\psi :[m\left( t_{0}\right) ,t_{0}]\cap \mathbb{Z}\rightarrow \mathbb{\mathbb{R}}\) is a bounded sequence and for \(t_{0}\geq 0\)
\begin{equation*}
m\left( t_{0}\right) =\inf \{\sigma _{i}(s):s\geq t_{0},\ i=1,…k\}.
\end{equation*}
Here \(\Delta \) denotes the forward difference operator \(\Delta
x(t)=x(t+1)-x(t)\) for any sequence \(\left\{ x\left( t\right) ,\ t\in \mathbb{
Z}^{+}\right\} \). For more details on the calculus of difference equations,
we refer the reader to [
11] and [
24]. Throughout this paper, we
assume that \(a\), \(b_{i}\) and \(c_{j}\) are bounded sequences, and \(\tau \), \(
\sigma _{i}\) and \(\tau _{j}\) are non-negative sequences such that
\begin{eqnarray*}
\tau \left( t\right) &\rightarrow &\infty \text{ as }t\rightarrow \infty ,\
\tau \left( t\right) \geq t,\ t\geq t_{0}, \\
\sigma _{i}\left( t\right) &\rightarrow &\infty \text{ as }t\rightarrow
\infty ,\ i=1,…,k,\ \sigma _{i}\left( t\right) \leq t,\ t\geq t_{0}, \\
\tau _{j}\left( t\right) &\rightarrow &\infty \text{ as }t\rightarrow \infty
,\ j=1,…,l,\ \tau _{j}\left( t\right) \geq t,\ t\geq t_{0}.
\end{eqnarray*}
Equation (1) can be viewed as a discrete analogue of the mixed type
neutral differential equation;
\begin{equation}
x^{\prime }\left( t\right) +a\left( t\right) x^{\prime }\left( \tau \left(
t\right) \right) +\sum_{i=1}^{k}b_{i}\left( t\right) x\left( \sigma
_{i}\left( t\right) \right) +\sum_{j=1}^{l}c_{j}\left( t\right) x\left( \tau
_{j}\left( t\right) \right) =0. \label{66}
\end{equation}
(3)
In [
25], Bicer investigated (3) and obtained the asymptotic
behavior of solutions. Our purpose here is to show the asymptotic behavior
of solutions for (1). An asymptotic stability theorem with a necessary
and sufficient condition is proved by using the contraction mapping theorem.
For details on contraction mapping principle we refer the reader to [
33]
. An example is given to illustrate our main results.
2. Main results
Theorem 1.
Let \(a\), \(b_{i}\) and \(c_{j}\) non positive sequences. Assume
that the following inequality has a nonnegative solution
\begin{equation*}
-a\left( t\right) \lambda \left( \tau \left( t\right) \right)
\prod_{u=t}^{\tau \left( t\right) -1}\left( 1-\lambda \left( u\right)
\right) -\sum_{i=1}^{k}b_{i}\left( t\right) \prod_{u=t}^{\sigma _{i}\left(
t\right) -1}\left( 1-\lambda \left( u\right) \right) \notag -\sum_{j=1}^{l}c_{j}\left( t\right) \prod_{u=t}^{\tau _{j}\left( t\right)
-1}\left( 1-\lambda \left( u\right) \right) \leq\lambda \left( t\right) ,\ t\geq t_{0}, \label{3}
\end{equation*}
with \(\lambda \left( t\right) < 1\). Then, (1) has a positive solution.
Proof.
Let \(\lambda _{0}\) be a nonnegative solution of (\ref{3}). Set
\begin{equation*}
\lambda _{n}\left( t\right) =\left\{
\begin{array}{l}
\lambda _{n-1}\left( t\right) ,\text{ \ \ if }m\left( t_{0}\right) \leq
t\leq t_{0}, \\
-a\left( t\right) \lambda _{n-1}\left( \tau \left( t\right) \right)
\prod\limits_{u=t}^{\tau \left( t\right) -1}\left( 1-\lambda _{n-1}\left(
u\right) \right) -\sum\limits_{i=1}^{k}b_{i}\left( t\right)
\prod\limits_{u=t}^{\sigma _{i}\left( t\right) -1}\left( 1-\lambda
_{n-1}\left( u\right) \right) \\
-\sum\limits_{j=1}^{l}c_{j}\left( t\right) \prod\limits_{u=t}^{\tau
_{j}\left( t\right) -1}\left( 1-\lambda _{n-1}\left( u\right) \right) ,\text{
}t\geq t_{0},
\end{array}
\right.
\end{equation*}
for \(n=1,2,…\). Then, by (\ref{3}), we get
\begin{equation*}
\lambda _{0}\left( t\right) \geq -a\left( t\right) \lambda _{0}\left( \tau
\left( t\right) \right) \prod_{u=t}^{\tau \left( t\right) -1}\left(
1-\lambda _{0}\left( u\right) \right) -\sum_{i=1}^{k}b_{i}\left( t\right)
\prod_{u=t}^{\sigma _{i}\left( t\right) -1}\left( 1-\lambda _{0}\left(
u\right) \right) -\sum_{j=1}^{l}c_{j}\left( t\right) \prod_{u=t}^{\tau _{j}\left( t\right)
-1}\left( 1-\lambda _{0}\left( u\right) \right) =\lambda _{1}\left( t\right) .
\end{equation*}
Then, we obtain \(\lambda _{0}(t)\geq \lambda _{1}(t)\geq …\geq \lambda
_{n}(t)\geq 0\). So, there exists a pointwise limit \(\lambda
(t)=\lim\limits_{n\rightarrow \infty }\lambda _{n}(t)\). So, from the
Lebesgue convergence theorem, we obtain
\begin{equation*}
\lambda \left( t\right) =-a\left( t\right) \lambda \left( \tau \left(
t\right) \right) \prod_{u=t}^{\tau \left( t\right) -1}\left( 1-\lambda
\left( u\right) \right) -\sum_{i=1}^{k}b_{i}\left( t\right)
\prod_{u=t}^{\sigma _{i}\left( t\right) -1}\left( 1-\lambda \left( u\right)
\right) -\sum_{j=1}^{l}c_{j}\left( t\right) \prod_{u=t}^{\tau _{j}\left( t\right)
-1}\left( 1-\lambda \left( u\right) \right) .
\end{equation*}
Hence,
\begin{equation*}
x\left( t\right) =\left\{
\begin{array}{l}
\lambda \left( t\right) ,\text{ \ \ if }m\left( t_{0}\right) \leq t\leq
t_{0}, \\
\lambda \left( t_{0}\right) \prod\limits_{u=t_{0}}^{t-1}\left( 1-\lambda
\left( u\right) \right) ,\ t\geq t_{0},
\end{array}
\right.
\end{equation*}
is a positive solution of (1).
Theorem 2.
Let \(a\), \(b_{i}\) and \(c_{j}\) be non positive sequences and let \(
\Delta a(t)>0\), \(a\left( t_{0}\right) \neq -\infty \). If
\begin{equation*}
\sum_{u=t_{0}}^{\infty }\sum_{j=1}^{l}c_{j}\left( u\right) =-\infty ,
\end{equation*}
and \(x\) is a eventually positive solution of (1), then \(
x(t)\rightarrow \infty \) as \(t\rightarrow \infty \).
Proof.
Assume that \(x(t)>0\) for \(t\geq T_{1}\). Choose \(T\geq T_{1}\) such that \(
T_{1}\leq \inf \{\sigma _{i}(s):s\geq T,\ i=1,…,k\}\). Then \(\Delta
x(t)+a(t)\Delta x(\tau (t))\geq 0\), for \(t\geq T\),
\begin{equation*}
\Delta x\left( t\right) +a\left( t\right) \Delta x\left( \tau \left(
t\right) \right) =-\sum_{i=1}^{k}b_{i}\left( t\right) x\left( \sigma
_{i}\left( t\right) \right) -\sum_{j=1}^{l}c_{j}\left( t\right) x\left( \tau
_{j}\left( t\right) \right) ,
\end{equation*}
and
\begin{equation*}
\Delta \left[ a(t)x\left( \tau \left( t\right) \right) \right] =a\left(
t\right) \Delta x\left( \tau \left( t\right) \right) +\Delta a\left(
t\right) x\left( \tau \left( t+1\right) \right) ,
\end{equation*}
that is
\begin{equation*}
\Delta \left[ x\left( t\right) +a(t)x\left( \tau \left( t\right) \right)
\right] -\Delta a\left( t\right) x\left( \tau \left( t+1\right) \right) \geq
-\sum_{j=1}^{l}c_{j}\left( t\right) x\left( \tau _{j}\left( t\right) \right)
.
\end{equation*}
From this, we can write
\begin{equation*}
\Delta \left[ x\left( t\right) +a(t)x\left( \tau \left( t\right) \right)
\right] \geq -\sum_{j=1}^{l}c_{j}\left( t\right) x\left( \tau _{j}\left(
t\right) \right) ,
\end{equation*}
so
\begin{equation*}
\Delta \left[ x\left( t\right) +a(t)x\left( \tau \left( t\right) \right)
\right] \geq -x\left( T\right) \sum_{j=1}^{l}c_{j}\left( t\right) ,
\end{equation*}
which implies
\begin{equation*}
x\left( t\right) +a(t)x\left( \tau \left( t\right) \right) \geq
a(t_{0})x\left( \tau \left( t_{0}\right) \right) -x\left( T\right)
\sum_{u=t_{0}}^{t-1}\sum_{j=1}^{l}c_{j}\left( u\right) .
\end{equation*}
So, we get
\begin{equation*}
x\left( t\right) \geq a(t_{0})x\left( \tau \left( t_{0}\right) \right)
-x\left( T\right) \sum_{u=t_{0}}^{t-1}\sum_{j=1}^{l}c_{j}\left( u\right) .
\end{equation*}
Then \(x(t)\rightarrow \infty \) as \(t\rightarrow \infty \).
Theorem 3.
Let \(a(t)>0\), \(b_{i}\) and \(c_{j}\) be nonnegative sequences and
let \(\Delta a(t)< 0\), \(a\left( t_{0}\right) \neq \infty \). If
\begin{equation*}
\sum_{u=t_{0}}^{\infty }\sum_{j=1}^{l}c_{j}\left( u\right) =\infty ,
\end{equation*}
and \(x\) is a eventually positive solution of (1), then \(
x(t)\rightarrow 0\) as \(t\rightarrow \infty \).
Proof.
For \(t\geq T_{1}\), since \(x(t)>0\) we Choose \(T\geq T_{1}\) such that \(
T_{1}\leq \inf \{\sigma _{i}(s):s\geq T,\ i=1,…,k\}\). Then \(\Delta
x(t)+a(t)\Delta x(\tau (t))\leq 0\), for \(t\geq T\), and
\begin{equation*}
\Delta x\left( t\right) +a\left( t\right) \Delta x\left( \tau \left(
t\right) \right) \leq -\sum_{j=1}^{l}c_{j}\left( t\right) x\left( \tau
_{j}\left( t\right) \right) ,
\end{equation*}
that is
\begin{equation*}
\Delta \left[ x\left( t\right) +a(t)x\left( \tau \left( t\right) \right)
\right] -\Delta a\left( t\right) x\left( \tau \left( t+1\right) \right) \leq
-\sum_{j=1}^{l}c_{j}\left( t\right) x\left( \tau _{j}\left( t\right) \right)
.
\end{equation*}
From this, we can write
\begin{equation*}
\Delta \left[ x\left( t\right) +a(t)x\left( \tau \left( t\right) \right)
\right] \leq -\sum_{j=1}^{l}c_{j}\left( t\right) x\left( \tau _{j}\left(
t\right) \right) ,
\end{equation*}
so
\begin{equation*}
\Delta \left[ x\left( t\right) +a(t)x\left( \tau \left( t\right) \right)
\right] \leq -x\left( T\right) \sum_{j=1}^{l}c_{j}\left( t\right) ,
\end{equation*}
which implies
\begin{equation*}
x\left( t\right) +a(t)x\left( \tau \left( t\right) \right) \leq
a(t_{0})x\left( \tau \left( t_{0}\right) \right) -x\left( T\right)
\sum_{u=t_{0}}^{t-1}\sum_{j=1}^{l}c_{j}\left( u\right) .
\end{equation*}
So, we get
\begin{equation*}
x\left( t\right) \leq a(t_{0})x\left( \tau \left( t_{0}\right) \right)
-x\left( T\right) \sum_{u=t_{0}}^{t-1}\sum_{j=1}^{l}c_{j}\left( u\right) .
\end{equation*}
Since \(x(t)>0\), we get a contradiction.Then \(x(t)\rightarrow 0\) as \(
t\rightarrow \infty \).
Now, we investigate the asymptotic behavior of solutions of (1), free
of the sign of the coefficients. During the process of inverting (1),
an summation by parts will have to performed on the term involving \(\Delta
x(\tau (t))\).
Lemma 1.
A sequence \(x\) is a solution of (1)–(2) if and
only if
\begin{eqnarray}
x\left( t\right) & =&\left( x\left( t_{0}\right) +a\left( t_{0}-1\right)
x\left( \tau \left( t_{0}\right) \right) \right) \prod_{s=t_{0}}^{t-1}\left(
1-B(s)\right) -a(t-1)x(\tau \left( t\right) ) \notag \\
& &+\sum_{r=t_{0}}^{t-1}\prod_{s=r+1}^{t-1}\left( 1-B(s)\right) h\left(
r\right) x\left( \tau \left( r\right) \right)
-\sum_{r=t_{0}}^{t-1}\prod_{s=r+1}^{t-1}\left( 1-B(s)\right) B\left(
r\right) x\left( r+1\right) \notag \\
&& -\sum_{r=t_{0}}^{t-1}\prod_{s=r+1}^{t-1}\left( 1-B(s)\right)
\sum_{i=1}^{k}b_{i}\left( r\right) x\left( \sigma _{i}\left( r\right) \right) -\sum_{r=t_{0}}^{t-1}\prod_{s=r+1}^{t-1}\left( 1-B(s)\right)
\sum_{j=1}^{l}c_{j}\left( r\right) x\left( \tau _{j}\left( r\right) \right) ,
\label{5}
\end{eqnarray}
(4)
for \(t\geq t_{0}\), where
\begin{equation*}
B\left( t\right) =\sum_{i=1}^{k}b_{i}\left( t\right)
+\sum_{j=1}^{l}c_{j}\left( t\right) ,\ 0 < B\left( t\right) < 1,
\end{equation*}
and
\begin{equation}
h\left( t\right) =a\left( t\right) -a\left( t-1\right) \left( 1-B(t)\right) .
\label{h1}
\end{equation}
(5)
Proof.
Since
\begin{eqnarray*}
x\left( \tau _{j}\left( t\right) \right) =x\left( t+1\right)
+\sum_{u=t+1}^{\tau _{j}\left( t\right) -1}\Delta x\left( u\right) , \text{ and }
x\left( \sigma _{i}\left( t\right) \right) =x\left( t+1\right)
+\sum_{u=t+1}^{\sigma _{i}(t)-1}\Delta x\left( u\right).
\end{eqnarray*}
We can rewrite (1) as
\begin{align}
\Delta x\left( t\right) & =-a\left( t\right) \Delta x\left( \tau \left(
t\right) \right) -\sum_{i=1}^{k}b_{i}\left( t\right) \sum_{u=t+1}^{\sigma
_{i}(t)-1}\Delta x\left( u\right) -\sum_{j=1}^{l}c_{j}\left( t\right) \sum_{u=t+1}^{\tau _{j}(t)-1}\Delta
x\left( u\right) -B\left( t\right) x\left( t+1\right) . \label{6}
\end{align}
(6)
Multiplying both sides of (6) with \(\prod\limits_{s=t_{0}}^{t}\left(
1-B(s\right) )^{-1}\), by summing from \(t_{0}\) to \(t-1\), we obtain
\begin{align*}
& \underset{r=t_{0}}{\overset{t-1}{\sum }}\Delta \left[ \prod
\limits_{s=t_{0}}^{r-1}\left( 1-B(s\right) )^{-1}x\left( r\right) \right] =-\sum_{r=t_{0}}^{t-1}\prod_{s=t_{0}}^{r}\left( 1-B(s\right) )^{-1}a\left(
r\right) \Delta x\left( \tau \left( r\right) \right)
-\sum_{r=t_{0}}^{t-1}\prod_{s=t_{0}}^{r}\left( 1-B(s\right) )^{-1}B\left(
r\right) x\left( r+1\right) \\
&\,\,\, -\sum_{r=t_{0}}^{t-1}\prod_{s=t_{0}}^{r}\left( 1-B(s\right)
)^{-1}\sum_{i=1}^{k}b_{i}\left( t\right) \sum_{u=t+1}^{\sigma
_{i}(r)-1}\Delta x\left( u\right) -\sum_{r=t_{0}}^{t-1}\prod_{s=t_{0}}^{r}\left( 1-B(s\right)
)^{-1}\sum_{j=1}^{l}c_{j}\left( t\right) \sum_{u=t+1}^{\tau _{j}(r)}\Delta
x\left( u\right) .
\end{align*}
By dividing both sides of the above expression by \(\prod
\limits_{s=t_{0}}^{t-1}\left( 1-B(s\right) )^{-1}\) we get
\begin{eqnarray}
x\left( t\right) & =&x\left( t_{0}\right) \prod_{s=t_{0}}^{t-1}\left(
1-B(s)\right) -\sum_{r=t_{0}}^{t-1}\prod_{s=r+1}^{t-1}\left( 1-B(s\right)
)a\left( r\right) \Delta x\left( \tau \left( r\right) \right) -\sum_{r=t_{0}}^{t-1}\prod_{s=r+1}^{t-1}\left( 1-B(s)\right) B\left(
r\right) x\left( r+1\right) \notag \\
&& -\sum_{r=t_{0}}^{t-1}\prod_{s=r+1}^{t-1}\left( 1-B(s\right)
)\sum_{i=1}^{k}b_{i}\left( r\right) \sum_{u=t+1}^{\sigma _{i}(r)-1}\Delta
x\left( u\right) -\sum_{r=t_{0}}^{t-1}\prod_{s=r+1}^{t-1}\left( 1-B(s\right)
)\sum_{j=1}^{l}c_{j}\left( r\right) \sum_{u=t+1}^{\tau _{j}(r)-1}\Delta
x\left( u\right) . \label{7}
\end{eqnarray}
(7)
By performing an summation by parts, we get
\begin{align}
& \sum_{r=t_{0}}^{t-1}\prod_{s=r+1}^{t-1}\left( 1-B(s\right) )a\left(
r\right) \Delta x\left( \tau \left( r\right) \right) \notag \\
& =a\left( t-1\right) x\left( \tau \left( t\right) \right) -a\left(
t_{0}-1\right) x\left( \tau \left( t_{0}\right) \right)
\prod_{s=t_{0}}^{t-1}\left( 1-B(s\right) )-\sum_{r=t_{0}}^{t-1}\Delta \left[ \prod_{s=r}^{t-1}\left( 1-B(s\right)
)a\left( r-1\right) \right] x\left( \tau \left( r\right) \right) . \label{8}
\end{align}
(8)
But,
\begin{align*}
\sum_{r=t_{0}}^{t-1}\Delta \left[ \prod_{s=r}^{t-1}\left( 1-B(s\right)
)a\left( r-1\right) \right] x\left( \tau \left( r\right) \right) =\sum_{r=t_{0}}^{t-1}\prod_{s=r+1}^{t-1}\left( 1-B(s\right) )\left[
a(r)-a\left( r-1\right) \left( 1-B(r)\right) \right] x\left( \tau \left(
r\right) \right) \\
& =\sum_{r=t_{0}}^{t-1}\prod_{s=r+1}^{t-1}\left( 1-B(s\right) )h\left(
r\right) x\left( \tau \left( r\right) \right) ,
\end{align*}
where \(h\) is given by (5). We obtain (4) by replacing (8)
into (7). Since each step is reversible, the converse follows easily.
This completes the proof.
Theorem 4.
Assume that \(0< B\left( t\right) < 1\) and the following
conditions hold
\begin{equation}
\prod\limits_{s=t_{0}}^{t-1}\left( 1-B(s)\right) \rightarrow 0\text{ as }
t\rightarrow \infty , \label{9}
\end{equation}
(9)
and
\begin{align}
& \left\vert a(t-1)\right\vert
+\sum_{r=t_{0}}^{t-1}\prod_{s=r+1}^{t-1}\left( 1-B(s)\right) \left\vert
h\left( r\right) \right\vert +\sum_{r=t_{0}}^{t-1}\prod_{s=r+1}^{t-1}\left(
1-B(s)\right) \left\vert B\left( r\right) \right\vert \notag \\
& +\sum_{r=t_{0}}^{t-1}\prod_{s=r+1}^{t-1}\left( 1-B(s)\right)
\sum_{i=1}^{k}\left\vert b_{i}\left( r\right) \right\vert +\sum_{r=t_{0}}^{t-1}\prod_{s=r+1}^{t-1}\left( 1-B(s)\right)
\sum_{j=1}^{l}\left\vert c_{j}\left( r\right) \right\vert\leq\beta< 1. \label{10}
\end{align}
(10)
Then for each initial condition (2), every solution of (1)
converges to zero.
Proof.
Let \(x\in C([m\left( t_{0}\right) ,\infty )\cap \mathbb{Z})\) is the space of
all bounded sequences and
\(
M=\{x\in C([m\left( t_{0}\right) ,\infty )\cap \mathbb{Z}):x(t)\rightarrow 0
\text{ as }t\rightarrow \infty \},
\)
be a closed subspace. Then \((M,\left\Vert .\right\Vert )\) is a Banach space
with the norm
\(
\left\Vert x\right\Vert =\sup_{t\geq m\left( t_{0}\right) }\left\vert
x\left( t\right) \right\vert .
\)
Define the operator \(\phi :M\rightarrow M\) by
\begin{equation}
\left( \phi x\right) \left( t\right) =\left\{
\begin{array}{l}
\psi \left( t\right) ,\text{ \ \ if }m\left( t_{0}\right) \leq t\leq t_{0,}
\\
\left( x\left( t_{0}\right) +a\left( t_{0}-1\right) x\left( \tau \left(
t_{0}\right) \right) \right) \prod\limits_{s=t_{0}}^{t-1}\left(
1-B(s)\right) -a(t-1)x(\tau \left( t\right) ) \\
+\sum\limits_{r=t_{0}}^{t-1}\prod\limits_{s=r+1}^{t-1}\left( 1-B(s)\right)
h\left( r\right) x\left( \tau \left( r\right) \right)
-\sum\limits_{r=t_{0}}^{t-1}\prod\limits_{s=r+1}^{t-1}\left( 1-B(s)\right)
B\left( r\right) x\left( r+1\right) \\
-\sum\limits_{r=t_{0}}^{t-1}\prod\limits_{s=r+1}^{t-1}\left( 1-B(s)\right)
\sum\limits_{i=1}^{k}b_{i}\left( r\right) x\left( \sigma _{i}\left( r\right)
\right)-\sum\limits_{r=t_{0}}^{t-1}\prod\limits_{s=r+1}^{t-1}\left( 1-B(s)\right)
\sum\limits_{j=1}^{l}c_{j}\left( r\right) x\left( \tau _{j}\left( r\right)
\right) ,\text{\ }t\geq t_{0}.
\end{array}
\right. \label{11}
\end{equation}
(11)
It is clear that for \(x\in M\), \(\phi x\) is bounded. Now, we will show that \(
\phi \) is a contraction. Let \(x\) and \(y\) be two bounded sequences on \(
[m\left( t_{0}\right) ,\infty )\cap \mathbb{Z}\) and satisfying same initial
condition (2). Then for \(t\geq t_{0}\), we get
\begin{align*}
& \left\vert \left( \phi x\right) \left( t\right) -\left( \phi y\right)
\left( t\right) \right\vert \leq \left\vert a(t-1)\right\vert \left\vert x\left( \tau \left( t\right)
\right) -y\left( \tau \left( t\right) \right) \right\vert +\sum_{r=t_{0}}^{t-1}\prod_{s=r+1}^{t-1}\left( 1-B(s)\right) \left\vert
h\left( r\right) \right\vert \left\vert x\left( \tau \left( r\right) \right)
-y\left( \tau \left( r\right) \right) \right\vert \\
& +\sum_{r=t_{0}}^{t-1}\prod_{s=r+1}^{t-1}\left( 1-B(s)\right) \left\vert
x\left( r+1\right) -y\left( r+1\right) \right\vert \left\vert B\left(
r\right) \right\vert +\sum_{r=t_{0}}^{t-1}\prod_{s=r+1}^{t-1}\left( 1-B(s)\right)
\sum_{i=1}^{k}\left\vert b_{i}\left( r\right) \right\vert \left\vert x\left(
\sigma _{i}\left( r\right) \right) -y\left( \sigma _{i}\left( r\right)
\right) \right\vert \\
& +\sum_{r=t_{0}}^{t-1}\prod_{s=r+1}^{t-1}\left( 1-B(s)\right)
\sum_{j=1}^{l}\left\vert c_{j}\left( r\right) \right\vert \left\vert x\left(
\tau _{j}\left( r\right) \right) -y\left( \tau _{j}\left( r\right) \right)
\right\vert \leq \beta \left\Vert x-y\right\Vert .
\end{align*}
Thus, the operator \(\phi \) has a unique fixed point in \(M\), which solves (
1). Now, we will show that, \(\left( \phi x\right) \left( t\right)
\rightarrow 0\) as \(t\rightarrow \infty \). Actually, for \(x\in M\), we have
\begin{eqnarray}
\left\vert \left( \phi x\right) \left( t\right) \right\vert &\leq& \left\vert \left( x\left( t_{0}\right) +a\left( t_{0}-1\right)
x\left( \tau \left( t_{0}\right) \right) \right) \right\vert
\prod_{s=t_{0}}^{t-1}\left( 1-B(s)\right) +\left\vert a(t-1)\right\vert
\left\vert x\left( \tau \left( t\right) \right) \right\vert \notag \\
&& +\sum_{r=t_{0}}^{t-1}\prod_{s=r+1}^{t-1}\left( 1-B(s)\right) \left\vert
h\left( r\right) \right\vert \left\vert x\left( \tau \left( r\right) \right)
\right\vert +\sum_{r=t_{0}}^{t-1}\prod_{s=r+1}^{t-1}\left( 1-B(s)\right) \left\vert
x\left( r+1\right) \right\vert \left\vert B\left( r\right) \right\vert
\notag \\
&& +\sum_{r=t_{0}}^{t-1}\prod_{s=r+1}^{t-1}\left( 1-B(s)\right)
\sum_{i=1}^{k}\left\vert b_{i}\left( r\right) \right\vert \left\vert x\left(
\sigma _{i}\left( r\right) \right) \right\vert +\sum_{r=t_{0}}^{t-1}\prod_{s=r+1}^{t-1}\left( 1-B(s)\right)
\sum_{j=1}^{l}\left\vert c_{j}\left( r\right) \right\vert \left\vert x\left(
\tau _{j}\left( r\right) \right) \right\vert .\notag\\&& \label{12}
\end{eqnarray}
(12)
Note that by (9),
\begin{equation*}
\left\vert \left( x\left( t_{0}\right) +a\left( t_{0}-1\right) x\left( \tau
\left( t_{0}\right) \right) \right) \right\vert \prod_{s=t_{0}}^{t-1}\left(
1-B(s)\right) \rightarrow 0\text{ as }t\rightarrow \infty .
\end{equation*}
Moreover, since \(x(t)\rightarrow 0\) as \(t\rightarrow \infty \), for each \(
\varepsilon >0\), there exists \(T_{1}>t_{0}\) such that \(u\geq T_{1}\) implies
that \(\left\vert x(\tau (u))\right\vert < \frac{\varepsilon }{2}\). Thus, for \(
t\geq T_{1}\), the third term \(I_{3}\) in (12) satisfies
\begin{align*}
I_{3}& =\sum_{r=t_{0}}^{t-1}\prod_{s=r+1}^{t-1}\left( 1-B(s)\right)
\left\vert h\left( r\right) \right\vert \left\vert x\left( \tau \left(
r\right) \right) \right\vert \\&\leq \sum_{r=t_{0}}^{T_{1}-1}\prod_{s=r+1}^{t-1}\left( 1-B(s)\right)
\left\vert h\left( r\right) \right\vert \left\vert x\left( \tau \left(
r\right) \right) \right\vert +\sum_{r=T_{1}}^{t-1}\sum_{r=t_{0}}^{t-1}\prod_{s=r+1}^{t-1}\left(
1-B(s)\right) \left\vert h\left( r\right) \right\vert \left\vert x\left(
\tau \left( r\right) \right) \right\vert\\& \leq \sum_{r=t_{0}}^{T_{1}-1}\sum_{r=t_{0}}^{t-1}\prod_{s=r+1}^{t-1}\left(
1-B(s)\right) \left\vert h\left( r\right) \right\vert \left\vert x\left(
\tau \left( r\right) \right) \right\vert +\frac{\varepsilon }{2}\sum_{r=T_{1}}^{t-1}\sum_{r=t_{0}}^{t-1}
\prod_{s=r+1}^{t-1}\left( 1-B(s)\right) \left\vert h\left( r\right)
\right\vert \\& \leq \frac{\varepsilon }{2}+\beta \frac{\varepsilon }{2}\\&\leq \varepsilon .
\end{align*}
Thus \(I_{3}\rightarrow 0\) as \(t\rightarrow \infty \). By a similar technique,
we can prove that the rest of terms in (12) tend zero as \(t\rightarrow
\infty \). Therefore \((\phi x)(t)\rightarrow 0\) as \(t\rightarrow \infty \).
This completes the proof.
Theorem 5.
Suppose that \(0< B\left( t\right) < 1\). If all solutions of (1) converge to zero, then (9) holds.
Proof.
Suppose that (9) does not holds. That is,
\begin{equation}
\lim_{t\rightarrow \infty }\prod\limits_{s=t_{0}}^{t-1}\left( 1-B(s)\right)
=\delta \neq 0. \label{13}
\end{equation}
(13)
So, from (13), we can write \(\delta \neq 0\). Then, there exists a
sequence \(\left\{ t_{n}\right\} \) approaching \(\infty \), such that
\begin{equation*}
\prod\limits_{s=t_{0}}^{t_{n}-1}\left( 1-B(s)\right) \rightarrow \delta
\text{ as }n\rightarrow \infty .
\end{equation*}
For \(x(t_{0})\neq 0\), let \(x\) be a solution. Then,
\begin{align}
& \lim_{n\rightarrow \infty }\left\vert \left( x\left( t_{0}\right) +a\left(
t_{0}-1\right) x\left( \tau \left( t_{0}\right) \right) \right) \right\vert
\prod_{s=t_{0}}^{t_{n}-1}\left( 1-B(s)\right)=\left\vert \left( x\left( t_{0}\right) +a\left( t_{0}-1\right) x\left(
\tau \left( t_{0}\right) \right) \right) \right\vert \delta \neq 0.
\label{14}
\end{align}
(14)
From Lemma 1, \(x(t_{n})\) satisfies (4). On the other hand,
we know that
\begin{eqnarray}
&&\lim_{n\rightarrow \infty }\left[ \sum_{r=t_{0}}^{t_{n}-1}
\prod_{s=r+1}^{t_{n}-1}\left( 1-B(s)\right) \left\vert h\left( r\right)
\right\vert \left\vert x\left( \tau \left( r\right) \right) \right\vert
-\left\vert a(t_{n}-1)\right\vert \left\vert x\left( \tau \left(
t_{n}\right) \right) \right\vert \right. \notag \\
&&+\sum_{r=t_{0}}^{t_{n}-1}\prod\limits_{s=r+1}^{t_{n}-1}\left(
1-B(s)\right) \left\vert x\left( r+1\right) \right\vert \left\vert B\left(
r\right) \right\vert +\sum_{r=t_{0}}^{t_{n}-1}\prod\limits_{s=r+1}^{t_{n}-1}\left(
1-B(s)\right) \sum_{i=1}^{k}\left\vert b_{i}\left( r\right) \right\vert
\left\vert x\left( \sigma _{i}\left( r\right) \right) \right\vert \notag \\
&&\left. +\sum_{r=t_{0}}^{t_{n}-1}\prod_{s=r+1}^{t_{n}-1}\left(
1-B(s)\right) \sum_{j=1}^{l}\left\vert c_{j}\left( r\right) \right\vert
\left\vert x\left( \tau _{j}\left( r\right) \right) \right\vert \right]
\begin{array}{c}
=
\end{array}
0. \label{15}
\end{eqnarray}
(15)
Since all solutions tend zero, from (4), (14) and (15), we
get
\begin{equation*}
\lim_{n\rightarrow \infty }x\left( t_{n}\right) =\left\vert \left( x\left(
t_{0}\right) +a\left( t_{0}-1\right) x\left( \tau \left( t_{0}\right)
\right) \right) \right\vert \delta \neq 0,
\end{equation*}
which contradicts all solutions of (1) converge to zero. The proof is
completed.
We end the paper with the following example.
Example 1.
consider the mixed type neutral difference equation
\begin{equation}
\Delta x\left( t\right) +a\left( t\right) \Delta x\left( \tau \left(
t\right) \right) +b_{1}\left( t\right) x\left( \sigma _{1}\left( t\right)
\right) +c_{1}\left( t\right) x\left( \tau _{1}\left( t\right) \right) =0,
\label{4.1}
\end{equation}
(16)
with an assumed initial condition
\begin{equation*}
x(t)=\psi (t)\text{ for }t\in \lbrack m\left( t_{0}\right) ,t_{0}]\cap
\mathbb{Z},
\end{equation*}
where \(t_{0}=0\), \(m\left( t_{0}\right) =-2\), \(\psi (t)=t/3\), \(a\left(
t\right) =\dfrac{1}{3^{t+2}}\), \(b_{1}\left( t\right) =1-\dfrac{1}{2^{t}}\), \(
c_{1}\left( t\right) =\dfrac{1}{2^{t+1}}\), \(\tau \left( t\right) =3t/2\), \(
\sigma _{1}\left( t\right) =t/2-2\), \(\tau _{1}\left( t\right) =5t/2\). We have
\begin{equation*}
B\left( t\right) =1-\dfrac{1}{2^{t+1}},\ \prod\limits_{s=0}^{t-1}\left(
1-B(s)\right) =\prod\limits_{s=0}^{t-1}\dfrac{1}{2^{s+1}}\rightarrow 0\text{
as }t\rightarrow \infty ,
\end{equation*}
and
\begin{align*}
& \left\vert a(t-1)\right\vert
+\sum_{r=t_{0}}^{t-1}\prod_{s=r+1}^{t-1}\left( 1-B(s)\right) \left\vert
h\left( r\right) \right\vert +\sum_{r=t_{0}}^{t-1}\prod_{s=r+1}^{t-1}\left(
1-B(s)\right) \left\vert B\left( r\right) \right\vert \\
& +\sum_{r=t_{0}}^{t-1}\prod_{s=r+1}^{t-1}\left( 1-B(s)\right) \left\vert
b_{1}\left( r\right) \right\vert +\sum_{r=0}^{t-1}\prod_{s=r+1}^{t-1}\left(
1-B(s)\right) \left\vert c_{1}\left( r\right) \right\vert \\
& =\dfrac{1}{3^{t+1}}+\sum_{r=0}^{t-1}\prod_{s=r+1}^{t-1}\dfrac{1}{2^{s+1}}
\left\vert \frac{1}{3^{r+2}}-\frac{1}{3^{r+1}\times 2^{r+1}}\right\vert
+\sum_{r=0}^{t-1}\prod_{s=r+1}^{t-1}\dfrac{1}{2^{s+1}}\left( 1-\dfrac{1}{
2^{r+1}}\right) \\
& +\sum_{r=0}^{t-1}\prod_{s=r+1}^{t-1}\dfrac{1}{2^{s+1}}\left( 1-\dfrac{1}{
2^{r}}\right) +\sum_{r=0}^{t-1}\prod_{s=r+1}^{t-1}\dfrac{1}{2^{s+1}}\times
\dfrac{1}{2^{r+1}} \simeq 0.722< 1.
\end{align*}
Thus all the conditions of Theorem 4 are satisfied and every
solution of (16) converges to zero.
3. Concluding remarks
In this article, a neutral mixed type difference equation is considered. The
asymptotic behavior of solutions is obtained with a necessary and sufficient
condition by using fixed point theorems. The results are supported with a
suitable illustrative example.
Author Contributions
All authors contributed equally to the writing of this paper. All authors read and approved the
final manuscript.
Competing Interests
The author(s) do not have any competing interests in the manuscript.
