Positive solutions and iterative approximations of a third order nonlinear neutral delay difference equation

Abstract

This paper deals with the third order nonlinear neutral delay difference equation with a forced term

$$\begin{aligned}& \Delta^{2} \bigl(a(n)\Delta \bigl(x(n)+c(n)x(n-\tau) \bigr) \bigr)+f \bigl(n,x \bigl(n-b_{1}(n) \bigr),x \bigl(n-b_{2}(n) \bigr), \ldots,x \bigl(n-b_{k}(n) \bigr) \bigr) \\& \quad =d(n),\quad n\geq n_{0}. \end{aligned}$$

Using the Banach fixed point theorem, we prove the existence of uncountably many bounded positive solutions for the equation, suggest some Mann iterative schemes and obtain the error estimates between these bounded positive solutions and the sequences generated by the iterative schemes. Five nontrivial examples are also included.

1 Introduction and preliminaries

In the past thirty years there has been much research activity concerning the oscillation, nonoscillation, asymptotic behavior and existence of solutions, nonoscillatory solutions and bounded positive solutions for various kinds of neutral delay difference equations, see, for example, [1–25] and the references therein. Jinfa [7] studied the existence of a bounded nonoscillatory solution for the second order neutral delay difference equation with positive and negative coefficients

$$\begin{aligned}& \Delta^{2} \bigl(x(n)+px(n-m) \bigr)+p(n)x(n-k)-q(n)x(n-l)=0,\quad n \geq n_{0} \end{aligned}$$

(1.1)

under the condition $p\neq-1$. Migda and Migda [17] got the asymptotic behavior of the second order neutral difference equation

$$\begin{aligned}& \Delta^{2} \bigl(x(n)+px(n-k) \bigr)+f \bigl(n,x(n) \bigr)=0, \quad n \geq1. \end{aligned}$$

(1.2)

Meng and Yan [16] discussed sufficient and necessary conditions of the existence of bounded nonoscillatory solutions for the second order nonlinear neutral difference equation

$$\begin{aligned}& \Delta^{2} \bigl(x(n)-px(n-r) \bigr)=\sum _{i=1}^{m}q_{i}f_{i} \bigl(x(n- \sigma_{i}) \bigr), \quad n\geq n_{0}. \end{aligned}$$

(1.3)

El-Morshedy [5] obtained the oscillation of the second order neutral difference equation with positive and negative coefficients

$$\begin{aligned}& \Delta^{2} \bigl(x(n)\pm a(n)x(n-\tau) \bigr)+p(n)x(n-\delta)-q(n)x(n- \sigma)=0, \quad n\geq0. \end{aligned}$$

(1.4)

Tripathy [22] studied the second order nonlinear neutral delay difference equation

$$\begin{aligned}& \Delta^{2} \bigl(x(n)+ p(n)x(n-m) \bigr)+q(n)G \bigl(x(n-k) \bigr)=0, \quad n \geq n_{0} \end{aligned}$$

(1.5)

and deduced sufficient conditions under which every solution of Eq. (1.5) oscillates. Rath et al. [18] investigated the second order neutral delay difference equation

$$\begin{aligned}& \Delta \bigl(r(n)\Delta \bigl(x(n)-p(n)x(n-m) \bigr) \bigr)+q(n)G \bigl(x(n-k) \bigr)=f(n), \quad n\geq n_{0} \end{aligned}$$

(1.6)

and found necessary conditions for every solution of Eq. (1.6) to oscillate or to tend to zero as $n\to\infty$. Liu, Xu and Kang [12] considered the solvability for the second order nonlinear neutral delay difference equation

$$\begin{aligned}& \Delta \bigl(a(n)\Delta \bigl(x(n)+bx(n-\tau) \bigr) \bigr)+f \bigl(n,x \bigl(n-d_{1}(n) \bigr),x \bigl(n-d_{2}(n) \bigr),\ldots,x \bigl(n-d_{k}(n) \bigr) \bigr) \\& \quad =c(n),\quad n\geq n_{0} \end{aligned}$$

(1.7)

and provided the global existence of uncountably many bounded nonoscillatory solutions for Eq. (1.7) relative to all $b\in\Bbb {R}$. Saker [19] studied the third order difference equation

$$\begin{aligned}& \Delta^{3}x(n)+p(n)x(n+1)=0, \quad n\geq n_{0} \end{aligned}$$

(1.8)

and established a few sufficient conditions for all solutions to be oscillatory or tend to zero. Yan and Liu [23] provided the existence of a bounded nonoscillatory solution for the third order difference equation

$$\begin{aligned}& \Delta^{3}x(n)+f \bigl(n,x(n),x(n-r) \bigr)=0, \quad n\geq n_{0} \end{aligned}$$

(1.9)

and got a necessary and sufficient condition for Eq. (1.9) to have a bounded nonoscillatory solution $\{x(n)\}_{n\geq n_{0}}$ with $\lim_{n\rightarrow\infty}x(n)=d$. Andruch-Sobilo and Migda [2] investigated the third order linear difference equation of neutral type

$$\begin{aligned}& \Delta^{3} \bigl(x(n)-p(n)x \bigl(\sigma(n) \bigr) \bigr)\pm q(n)x \bigl( \tau(n) \bigr)=0,\quad n\geq n_{0} \end{aligned}$$

(1.10)

and proved sufficient conditions which ensure that all solutions of Eq. (1.10) are oscillatory.

The purpose of this paper is to study the below third order nonlinear neutral delay difference equation with a forced term

$$\begin{aligned}& \Delta^{2} \bigl(a(n)\Delta \bigl(x(n)+c(n)x(n-\tau) \bigr) \bigr) +f \bigl(n,x \bigl(n-b_{1}(n) \bigr),x \bigl(n-b_{2}(n) \bigr), \ldots,x \bigl(n-b_{k}(n) \bigr) \bigr) \\& \quad =d(n),\quad n\geq n_{0}, \end{aligned}$$

(1.11)

where $\tau,k\in\Bbb {N}$, $n_{0}\in\Bbb {N}_{0}=\{0\}\cup\Bbb {N}$, $\{a(n)\}_{n\in\Bbb {N}_{n_{0}}}$, $\{c(n)\}_{n\in\Bbb {N}_{n_{0}}}$ and $\{d(n)\}_{n\in\Bbb {N}_{n_{0}}}$ are real sequences with $a(n)\ne0$ for $n\in\Bbb {N}_{n_{0}}$, $\bigcup_{i=1}^{k}\{b_{i}(n)\}_{n\in\Bbb {N}_{n_{0}}}\subseteq\Bbb {Z}$ with $\lim_{n\to \infty}(n-b_{i}(n))=+\infty$, $1\leq i\leq k$ and $f:\Bbb {N}_{n_{0}}\times\Bbb {R}^{k}\rightarrow\Bbb {R}$ is a mapping. Using the Banach fixed point theorem, we prove several existence results of uncountably many bounded positive solutions for Eq. (1.11), suggest a few Mann iterative methods for these bounded positive solutions and discuss the error estimates between these bounded positive solutions and the iterative sequences generated by the Mann iterative methods. To illustrate our results, five examples are also constructed.

Throughout this paper, we assume that Δ denotes the forward difference operator defined by $\Delta x(n)=x(n+1)-x(n)$, $\Delta^{2}x(n)=\Delta(\Delta x(n))$, $\Delta^{3}x(n)=\Delta(\Delta^{2} x(n))$, $\Bbb {R}=(-\infty,+\infty)$, $\Bbb {R}^{+}=[0,+\infty)$, $\Bbb {Z}$ and $\Bbb {N}$ stand for the sets of all integers and positive integers, respectively,

$$\begin{aligned}& \Bbb {N}_{n_{0}}=\{n:n\in\Bbb {N}_{0} \text{ with } n\ge n_{0}\},\qquad \Bbb {Z}_{\beta}=\{n:n\in\Bbb {Z}\text{ with } n\geq \beta \}, \\& \beta=\min \bigl\{ n_{0}-\tau, \inf \bigl\{ n-b_{i}(n): 1 \leq i \leq k,n\in\Bbb {N}_{n_{0}} \bigr\} \bigr\} , \end{aligned}$$

$l_{\beta}^{\infty}$ stands for the Banach space of all bounded sequences on $\Bbb {Z}_{\beta}$ with norm

$$\Vert x\Vert =\sup_{n\in\Bbb {Z}_{\beta}} \bigl\vert x(n) \bigr\vert \quad \text{for }x= \bigl\{ x(n) \bigr\} _{n\in\Bbb {Z}_{\beta}}\in l_{\beta}^{\infty}$$

and

$$A(N,M)= \bigl\{ x= \bigl\{ x(n) \bigr\} _{n\in\Bbb {Z}_{\beta}}\in l_{\beta}^{\infty}: N\leq x(n)\leq M, n\in\Bbb {Z}_{\beta}\bigr\} \quad \text{for }M>N>0. $$

By a solution of Eq. (1.11), we mean a sequence $\{x(n)\}_{n\in \Bbb {Z}_{\beta}}$ with a positive integer $n_{1}\geq n_{0}+\tau+\vert \beta \vert$ such that Eq. (1.11) holds for all $n\geq n_{1}$.

Lemma 1.1

([8])

Let $\tau\in\Bbb {N}$, $n_{0}\in \Bbb {N}_{0}$ and $B:\Bbb {N}_{n_{0}}\to\Bbb {R}^{+}$ be a mapping. Then

$$\sum_{i=0}^{\infty}\sum _{s=n_{0}+i\tau}^{\infty}\sum_{t=s}^{\infty }B(t)< + \infty \quad \Longleftrightarrow\quad \sum_{s=n_{0}}^{\infty} \sum_{t=s}^{\infty}sB(t)< +\infty. $$

2 Existence of uncountably many bounded positive solutions and Mann iterative schemes

Now we use the Banach fixed point theorem to show the existence of uncountably many bounded positive solutions for Eq. (1.11), construct Mann iterative schemes and discuss the error estimates between the bounded positive solutions and the sequences generated by the Mann iterative schemes.

Theorem 2.1

Assume that there exist two positive constants M and N with $M>N$ and two nonnegative sequences $\{P(n)\}_{n\in\Bbb {N}_{n_{0}}}$ and $\{Q(n)\}_{n\in\Bbb {N}_{n_{0}}}$ satisfying

$$\begin{aligned}& c(n)=1, \quad \textit{eventually}; \end{aligned}$$

(2.1)

$$\begin{aligned}& \bigl\vert f(n,u_{1},u_{2},\ldots,u_{k})-f(n,v_{1},v_{2}, \ldots,v_{k}) \bigr\vert \\& \quad \leq P(n)\max \bigl\{ \vert u_{i}-v_{i}\vert : 1 \leq i \leq k \bigr\} ,\quad n\in\Bbb {N}_{n_{0}}, u_{i}, v_{i} \in[N,M], 1\leq i\leq k; \end{aligned}$$

(2.2)

$$\begin{aligned}& \bigl\vert f(n,u_{1},u_{2},\ldots,u_{k}) \bigr\vert \leq Q(n),\quad n\in\Bbb {N}_{n_{0}}, u_{i} \in[N,M], 1 \leq i \leq k; \end{aligned}$$

(2.3)

$$\begin{aligned}& \sum_{t=n_{0}}^{\infty} \frac{1}{\vert a(t)\vert }\sum _{s=t}^{\infty}(s-t+1)\max \bigl\{ P(s),Q(s), \bigl\vert d(s) \bigr\vert \bigr\} < +\infty. \end{aligned}$$

(2.4)

Then

• (a) for any $L\in(N,M)$, there exist $\theta\in(0,1)$ and $n_{1}\geq n_{0}+\tau+\vert \beta \vert$ such that for each $x_{0}=\{x_{0}(n)\}_{n\in\Bbb {Z}_{\beta}}\in A(N,M)$, the Mann iterative sequence $\{x_{m}\}_{m\in\Bbb {N}_{0}}=\{x_{m}(n)\}_{(n,m)\in\Bbb {Z}_{\beta}\times\Bbb {N}_{0}}$ generated by the scheme:

  $$\begin{aligned}& x_{m+1}(n) \\& \quad =\textstyle\begin{cases} (1-\alpha_{m})x_{m}(n)+ \alpha_{m} \{L+\sum_{i=1}^{\infty} \sum_{t=n+(2i-1)\tau}^{n+2i\tau-1}\frac{1}{a(t)} \sum_{s=t}^{\infty}(s-t+1) \\ \quad {}\times [f (s,x_{m} (s-b_{1}(s) ),x_{m} (s-b_{2}(s) ),\ldots,x_{m} (s-b_{k}(s) ) )-d(s) ] \}, \\ \quad n\ge n_{1}, m\ge0, \\ (1-\alpha_{m})x_{m}(n_{1})+\alpha_{m} \{L+\sum_{i=1}^{\infty}\sum_{t=n_{1}+(2i-1)\tau}^{n_{1}+2i\tau-1} \frac{1}{a(t)}\sum_{s=t}^{\infty}(s-t+1) \\ \quad {}\times [f (s,x_{m} (s-b_{1}(s) ),x_{m} (s-b_{2}(s) ),\ldots,x_{m} (s-b_{k}(s) ) )-d(s) ] \}, \\ \quad \beta\le n< n_{1}, m\ge0 \end{cases}\displaystyle \end{aligned}$$

  (2.5)

  converges to a bounded positive solution $x\in A(N,M)$ of Eq. (1.11) and has the following error estimate:

  $$\begin{aligned}& \Vert x_{m+1}-x\Vert \le e^{-(1- \theta)\sum_{k=0}^{m}\alpha_{k}}\Vert x_{0}-x \Vert ,\quad m\in\Bbb {N}_{0}, \end{aligned}$$

  (2.6)

  where $\{\alpha_{m}\}_{m\in\Bbb {N}_{0}}$ is an arbitrary sequence in $[0,1]$ such that

  $$\begin{aligned}& \sum^{\infty}_{m=0}\alpha_{m} =+ \infty; \end{aligned}$$

  (2.7)

• (b) Eq. (1.11) possesses uncountably many bounded positive solutions in $A(N,M)$.

Proof

First of all we show (a). Let $L\in(N,M)$. It follows from (2.1) and (2.4) that there exist $\theta\in (0,1)$ and $n_{1}\geq n_{0}+\tau+\vert \beta \vert$ satisfying

$$\begin{aligned}& c(n)=1, \quad n\ge n_{1}; \end{aligned}$$

(2.8)

$$\begin{aligned}& \theta=\sum_{t=n_{1}}^{\infty}\frac{1}{\vert a(t)\vert }\sum _{s=t}^{\infty }(s-t+1)P(s); \end{aligned}$$

(2.9)

$$\begin{aligned}& \sum_{t=n_{1}}^{\infty}\frac{1}{\vert a(t)\vert }\sum _{s=t}^{\infty }(s-t+1) \bigl(Q(s)+ \bigl\vert d(s) \bigr\vert \bigr)\leq \min\{M-L ,L-N\}. \end{aligned}$$

(2.10)

Define a mapping $T_{L}: A(N,M)\rightarrow l_{\beta}^{\infty}$ by

$$\begin{aligned}& T_{L}x(n)=\textstyle\begin{cases} L+\sum_{i=1}^{\infty} \sum_{t=n+(2i-1)\tau}^{n+2i\tau-1}\frac {1}{a(t)}\sum_{s=t}^{\infty}(s-t+1) \\ \quad {}\times [f (s,x (s-b_{1}(s) ),x (s-b_{2}(s) ),\ldots,x (s-b_{k}(s) ) )-d(s) ], \\ \quad n\geq n_{1}, \\ T_{L}x(n_{1}), \quad \beta\leq n< n_{1} \end{cases}\displaystyle \end{aligned}$$

(2.11)

for each $x=\{x(n)\}_{n\in\Bbb {Z}_{\beta}}\in A(N,M)$. In view of (2.2), (2.3) and (2.9)~(2.11), we conclude that for $x=\{x(n)\}_{n\in\Bbb {Z}_{\beta}}$, $y=\{y(n)\}_{n\in\Bbb {Z}_{\beta}}\in A(N,M)$ and $n\geq n_{1}$

$$\begin{aligned}& \bigl\vert T_{L}x(n)-T_{L}y(n) \bigr\vert \\& \quad = \Biggl\vert \sum_{i=1}^{\infty}\sum _{t=n+(2i-1)\tau}^{n+2i\tau-1}\frac {1}{a(t)}\sum _{s=t}^{\infty}(s-t+1) \bigl[f \bigl(s,x \bigl(s-b_{1}(s) \bigr),x \bigl(s-b_{2}(s) \bigr),\ldots,x \bigl(s-b_{k}(s) \bigr) \bigr) \\& \qquad {} -f \bigl(s,y \bigl(s-b_{1}(s) \bigr),y \bigl(s-b_{2}(s) \bigr), \ldots,y \bigl(s-b_{k}(s) \bigr) \bigr) \bigr] \Biggr\vert \\& \quad \leq \sum_{i=1}^{\infty}\sum _{t=n+(2i-1)\tau}^{n+2i\tau-1}\frac {1}{\vert a(t)\vert }\sum _{s=t}^{\infty} (s-t+1)P(s)\max \bigl\{ \bigl\vert x \bigl(s-b_{j}(s) \bigr)-y \bigl(s-b_{j}(s) \bigr) \bigr\vert : 1\leq j\leq k \bigr\} \\& \quad \leq\sum_{i=1}^{\infty}\sum _{t=n+(2i-1)\tau}^{n+2i\tau-1}\frac {1}{\vert a(t)\vert }\sum _{s=t}^{\infty}(s-t+1)P(s)\Vert x-y\Vert \\& \quad \leq\sum_{t=n}^{\infty}\frac{1}{\vert a(t)\vert } \sum_{s=t}^{\infty }(s-t+1)P(s)\Vert x-y\Vert \\& \quad \leq\theta \Vert x-y\Vert \end{aligned}$$

and

$$\begin{aligned}& \bigl\vert T_{L}x(n)-L \bigr\vert \\& \quad = \Biggl\vert \sum_{i=1}^{\infty}\sum _{t=n+(2i-1)\tau}^{n+2i\tau-1}\frac {1}{a(t)}\sum _{s=t}^{\infty}(s-t+1) \\& \qquad {}\times \bigl[f \bigl(s,x \bigl(s-b_{1}(s) \bigr),x \bigl(s-b_{2}(s) \bigr),\ldots,x \bigl(s-b_{k}(s) \bigr) \bigr)-d(s) \bigr] \Biggr\vert \\& \quad \leq\sum_{i=1}^{\infty}\sum _{t=n+(2i-1)\tau}^{n+2i\tau-1}\frac {1}{\vert a(t)\vert }\sum _{s=t}^{\infty}(s-t+1) \\& \qquad {}\times \bigl[ \bigl\vert f \bigl(s,x \bigl(s-b_{1}(s) \bigr),x \bigl(s-b_{2}(s) \bigr),\ldots,x \bigl(s-b_{k}(s) \bigr) \bigr) \bigr\vert + \bigl\vert d(s) \bigr\vert \bigr] \\& \quad \leq\sum_{t=n}^{\infty}\frac{1}{\vert a(t)\vert } \sum_{s=t}^{\infty }(s-t+1) \bigl(Q(s)+ \bigl\vert d(s) \bigr\vert \bigr) \\& \quad \leq\min\{M-L ,L-N\}, \end{aligned}$$

which imply that

$$\begin{aligned}& \Vert T_{L}x-T_{L}y\Vert \leq\theta \Vert x-y\Vert , \quad x,y\in A(N,M)\quad \text{and}\quad T_{L} \bigl(A(N,M) \bigr) \subseteq A(N,M). \end{aligned}$$

(2.12)

Thus (2.12) ensures that $T_{L}$ is a contraction mapping on the closed subset $A(N ,M)$ and it has a unique fixed point $x=\{x(n)\}_{n\in\Bbb {Z}_{\beta}}\in A(N,M)$. It follows that for $n\geq n_{1}+\tau$

$$\begin{aligned}& x(n)=L+\sum_{i=1}^{\infty}\sum _{t=n+(2i-1)\tau}^{n+2i\tau-1}\frac {1}{a(t)}\sum _{s=t}^{\infty}(s-t+1) \\& \hphantom{x(n)=} {}\times \bigl[f \bigl(s,x \bigl(s-b_{1}(s) \bigr),x \bigl(s-b_{2}(s) \bigr),\ldots,x \bigl(s-b_{k}(s) \bigr) \bigr)-d(s) \bigr], \\& x(n-\tau)=L+\sum_{i=1}^{\infty}\sum _{t=n+2(i-1)\tau}^{n+(2i-1)\tau -1}\frac{1}{a(t)}\sum _{s=t}^{\infty}(s-t+1) \\& \hphantom{x(n-\tau)=} {}\times \bigl[f \bigl(s,x \bigl(s-b_{1}(s) \bigr),x \bigl(s-b_{2}(s) \bigr),\ldots,x \bigl(s-b_{k}(s) \bigr) \bigr)-d(s) \bigr], \end{aligned}$$

which yield that for $n\geq n_{1}+\tau$

$$\begin{aligned} x(n)+x(n-\tau) =&2L+\sum_{t=n}^{\infty} \frac{1}{a(t)}\sum_{s=t}^{\infty}(s-t+1) \\ &{}\times \bigl[f \bigl(s,x \bigl(s-b_{1}(s) \bigr),x \bigl(s-b_{2}(s) \bigr),\ldots,x \bigl(s-b_{k}(s) \bigr) \bigr)-d(s) \bigr], \end{aligned}$$

which gives that for $n\geq n_{1}+\tau$

$$\begin{aligned} \Delta \bigl(x(n)+x(n-\tau) \bigr) =&-\frac{1}{a(n)}\sum _{s=n}^{\infty}(s-n+1) \\ &{}\times \bigl[f \bigl(s,x \bigl(s-b_{1}(s) \bigr),x \bigl(s-b_{2}(s) \bigr),\ldots,x \bigl(s-b_{k}(s) \bigr) \bigr)-d(s) \bigr], \end{aligned}$$

which implies that for $n\geq n_{1}+\tau$

$$\Delta \bigl[a(n)\Delta \bigl(x(n)+x(n-\tau) \bigr) \bigr]=\sum _{s=n}^{\infty} \bigl[f \bigl(s,x \bigl(s-b_{1}(s) \bigr),x \bigl(s-b_{2}(s) \bigr),\ldots,x \bigl(s-b_{k}(s) \bigr) \bigr)-d(s) \bigr] $$

and

$$\Delta^{2} \bigl[a(n)\Delta \bigl(x(n)+x(n-\tau) \bigr) \bigr]=-f \bigl(n,x \bigl(n-b_{1}(n) \bigr),x \bigl(n-b_{2}(n) \bigr), \ldots,x \bigl(n-b_{k}(n) \bigr) \bigr)+d(n), $$

which together with (2.8) means that $x=\{x(n)\}_{n\in\Bbb {Z}_{\beta}}$ is a bounded positive solution of Eq. (1.11) in $A(N,M)$.

It follows from (2.5), (2.8), (2.9), (2.11) and (2.12) that for any $m\ge0$ and $n\ge n_{1}$,

$$\begin{aligned}& \bigl\vert x_{m+1}(n)-x(n) \bigr\vert \\& \quad = \Biggl\vert (1-\alpha_{m})x_{m}(n)+ \alpha_{m} \Biggl\{ L+\sum_{i=1}^{\infty} \sum_{t=n+(2i-1)\tau}^{n+2i\tau-1} \frac{1}{a(t)}\sum _{s=t}^{\infty}(s-t+1) \\& \qquad {} \times \bigl[f \bigl(s,x_{m} \bigl(s-b_{1}(s) \bigr),x_{m} \bigl(s-b_{2}(s) \bigr),\ldots,x_{m} \bigl(s-b_{k}(s) \bigr) \bigr)-d(s) \bigr] \Biggr\} -x(n) \Biggr\vert \\& \quad \le(1-\alpha_{m}) \bigl\vert x_{m}(n)-x(n) \bigr\vert + \alpha_{m} \bigl\vert T_{L}x_{m}(n)-T_{L}x(n) \bigr\vert \\& \quad \le(1-\alpha_{m})\Vert x_{m}-x\Vert + \alpha_{m}\theta \Vert x_{m}-x\Vert \\& \quad = \bigl(1-(1-\theta)\alpha_{m} \bigr)\Vert x_{m}-x \Vert \\& \quad \le e^{-(1-\theta)\alpha_{m}}\Vert x_{m}-x\Vert , \end{aligned}$$

which gives that

$$\Vert x_{m+1}-x\Vert \le e^{-(1-\theta)\alpha_{m}}\Vert x_{m}-x \Vert \le e^{-(1- \theta)\sum_{k=0}^{m}\alpha_{k}}\Vert x_{0}-x\Vert ,\quad m\in\Bbb {N}_{0}. $$

That is, (2.6) holds. It follows from (2.6) and (2.7) that $\lim_{m\to\infty}x_{m}=x$.

Next we show (b). Let $L_{1},L_{2}\in(N,M)$ with $L_{1}\neq L_{2}$. Similarly we infer that for each $z\in\{1,2\}$, there exist constants $\theta_{z}\in(0,1)$ and $n_{z}\geq n_{0}+\tau+\vert \beta \vert$ and a mapping $T_{L_{z}}$ satisfying (2.9)~(2.11), where θ, L and $n_{1}$ are replaced by $\theta_{z}$, $L_{z}$ and $n_{z}$, respectively, and the contraction mapping $T_{L_{z}}$ has a unique fixed point $x^{z}=\{x^{z}(n)\}_{n\in\Bbb {Z}_{\beta}}\in A(N,M)$, which is a bounded positive solution of Eq. (1.11) in $A(N,M)$, that is,

$$\begin{aligned} x^{z}(n) =&L_{z}+\sum_{i=1}^{\infty} \sum_{t=n+(2i-1)\tau}^{n+2i\tau-1}\frac {1}{a(t)}\sum _{s=t}^{\infty}(s-t+1) \\ &{} \times \bigl[f \bigl(s,x^{z} \bigl(s-b_{1}(s) \bigr),x^{z} \bigl(s-b_{2}(s) \bigr),\ldots,x^{z} \bigl(s-b_{k}(s) \bigr) \bigr)-d(s) \bigr], \quad n\ge n_{z}, \end{aligned}$$

which together with (2.2) and (2.9) yields that

$$\begin{aligned}& \bigl\vert x^{1}(n)-x^{2}(n) \bigr\vert \\& \quad = \Biggl\vert L_{1}-L_{2}+\sum _{i=1}^{\infty}\sum_{t=n+(2i-1)\tau}^{n+2i\tau -1} \frac{1}{a(t)}\sum_{s=t}^{\infty}(s-t+1) \\& \qquad {}\times \bigl[f \bigl(s,x^{1} \bigl(s-b_{1}(s) \bigr),x^{1} \bigl(s-b_{2}(s) \bigr),\ldots,x^{1} \bigl(s-b_{k}(s) \bigr) \bigr) \\& \qquad {} -f \bigl(s,x^{2} \bigl(s-b_{1}(s) \bigr),x^{2} \bigl(s-b_{2}(s) \bigr),\ldots,x^{2} \bigl(s-b_{k}(s) \bigr) \bigr) \bigr] \Biggr\vert \\& \quad \geq \vert L_{1}-L_{2}\vert -\sum _{i=1}^{\infty}\sum_{t=n+(2i-1)\tau}^{n+2i\tau -1} \frac{1}{\vert a(t)\vert }\sum_{s=t}^{\infty}(s-t+1) \\& \qquad {}\times \bigl\vert f \bigl(s,x^{1} \bigl(s-b_{1}(s) \bigr),x^{1} \bigl(s-b_{2}(s) \bigr),\ldots,x^{1} \bigl(s-b_{k}(s) \bigr) \bigr) \\& \qquad {} -f \bigl(s,x^{2} \bigl(s-b_{1}(s) \bigr),x^{2} \bigl(s-b_{2}(s) \bigr),\ldots,x^{2} \bigl(s-b_{k}(s) \bigr) \bigr) \bigr\vert \\& \quad \geq \vert L_{1}-L_{2}\vert -\sum _{i=1}^{\infty}\sum_{t=n+(2i-1)\tau}^{n+2i\tau -1} \frac{1}{\vert a(t)\vert }\sum_{s=t}^{\infty} (s-t+1)P(s) \\& \qquad {}\times \max \bigl\{ \bigl\vert x^{1} \bigl(s-b_{j}(s) \bigr)-x^{2} \bigl(s-b_{j}(s) \bigr) \bigr\vert : 1\leq j \leq k \bigr\} \\& \quad \geq \vert L_{1}-L_{2}\vert -\sum _{i=1}^{\infty}\sum_{t=n+(2i-1)\tau}^{n+2i\tau -1} \frac{1}{\vert a(t)\vert }\sum_{s=t}^{\infty} (s-t+1)P(s) \bigl\Vert x^{1}-x^{2} \bigr\Vert \\& \quad \geq \vert L_{1}-L_{2}\vert -\sum _{t=n}^{\infty}\frac{1}{\vert a(t)\vert }\sum _{s=t}^{\infty }(s-t+1)P(s) \bigl\Vert x^{1}-x^{2} \bigr\Vert \\& \quad \geq \vert L_{1}-L_{2}\vert -\max\{ \theta_{1}, \theta_{2}\} \bigl\Vert x^{1}-x^{2} \bigr\Vert ,\quad n\geq\max\{n_{1},n_{2}\}, \end{aligned}$$

which means that

$$\bigl\Vert x^{1}-x^{2} \bigr\Vert \geq \frac{\vert L_{1}-L_{2}\vert }{1+\max\{\theta_{1},\theta_{2}\}}>0, $$

that is, $x^{1}\neq x^{2}$. Thus the set of bounded positive solutions of Eq. (1.11) in $A(N,M)$ is uncountable. This completes the proof. □

Theorem 2.2

Assume that there exist two positive constants M and N with $M>N$ and two nonnegative sequences $\{P(n)\}_{n\in\Bbb {N}_{n_{0}}}$ and $\{Q(n)\}_{n\in\Bbb {N}_{n_{0}}}$ satisfying (2.2), (2.3) and

$$\begin{aligned}& c(n)=-1,\quad \textit{eventually}; \end{aligned}$$

(2.13)

$$\begin{aligned}& \sum_{t=n_{0}}^{\infty}\frac{t}{\vert a(t)\vert }\sum _{s=t}^{\infty}(s-t+1)\max \bigl\{ P(s),Q(s), \bigl\vert d(s) \bigr\vert \bigr\} < +\infty. \end{aligned}$$

(2.14)

Then

• (a) for any $L\in(N,M)$, there exist $\theta\in(0,1)$ and $n_{1}\geq n_{0}+\tau+\vert \beta \vert$ such that for each $x_{0}=\{x_{0}(n)\}_{n\in\Bbb {Z}_{\beta}}\in A(N,M)$, the Mann iterative sequence $\{x_{m}\}_{m\in\Bbb {N}_{0}}=\{x_{m}(n)\}_{(n,m)\in\Bbb {Z}_{\beta}\times\Bbb {N}_{0}}$ generated by the scheme:

  $$\begin{aligned}& x_{m+1}(n) \\& \quad = \textstyle\begin{cases} (1-\alpha_{m})x_{m}(n)+ \alpha_{m} \{L-\sum_{i=1}^{\infty} \sum_{t=n+i\tau }^{\infty}\frac{1}{a(t)} \sum_{s=t}^{\infty}(s-t+1) \\ \quad {}\times [f (s,x_{m} (s-b_{1}(s) ),x_{m} (s-b_{2}(s) ),\ldots,x_{m} (s-b_{k}(s) ) )-d(s) ] \}, \\ \quad n \ge n_{1}, m\ge0, \\ (1-\alpha_{m})x_{m}(n_{1})+\alpha_{m} \{L-\sum_{i=1}^{\infty}\sum_{t=n_{1}+i\tau}^{\infty}\frac{1}{a(t)} \sum_{s=t}^{\infty}(s-t+1) \\ \quad {}\times [f (s,x_{m} (s-b_{1}(s) ),x_{m} (s-b_{2}(s) ),\ldots,x_{m} (s-b_{k}(s) ) )-d(s) ] \}, \\ \quad \beta\le n< n_{1}, m\ge0 \end{cases}\displaystyle \end{aligned}$$

  (2.15)

  converges to a bounded positive solution $x \in A(N,M)$ of Eq. (1.11) and satisfies the error estimate (2.6), where $\{\alpha_{m}\}_{m\in\Bbb {N}_{0}}$ is an arbitrary sequence in $[0,1]$ satisfying (2.7);

• (b) Eq. (1.11) possesses uncountably many bounded positive solutions in $A(N,M)$.

Proof

First of all we show (a). Let $L\in(N,M)$. It follows from (2.13), (2.14) and Lemma 1.1 that there exist $\theta\in(0,1)$ and $n_{1}\geq n_{0}+\tau+\vert \beta \vert$ satisfying

$$\begin{aligned}& c(n)=-1,\quad n\ge n_{1}; \end{aligned}$$

(2.16)

$$\begin{aligned}& \theta=\sum_{i=1}^{\infty}\sum _{t=n_{1}+i\tau}^{\infty}\frac {1}{\vert a(t)\vert }\sum _{s=t}^{\infty}(s-t+1)P(s); \end{aligned}$$

(2.17)

$$\begin{aligned}& \sum_{i=1}^{\infty}\sum _{t=n_{1}+i\tau}^{\infty}\frac{1}{\vert a(t)\vert }\sum _{s=t}^{\infty}(s-t+1) \bigl(Q(s)+ \bigl\vert d(s) \bigr\vert \bigr)\leq \min\{M-L ,L-N\}. \end{aligned}$$

(2.18)

Define a mapping $T_{L}: A(N,M)\rightarrow l_{\beta}^{\infty}$ by

$$\begin{aligned}& T_{L}x(n)=\textstyle\begin{cases} L-\sum_{i=1}^{\infty} \sum_{t=n+i\tau}^{\infty}\frac{1}{a(t)}\sum_{s=t}^{\infty}(s-t+1) \\ \quad {}\times [f (s,x (s-b_{1}(s) ),x (s-b_{2}(s) ),\ldots,x (s-b_{k}(s) ) )-d(s) ], \\ \quad n\geq n_{1}, \\ T_{L}x(n_{1}),\quad \beta\leq n< n_{1} \end{cases}\displaystyle \end{aligned}$$

(2.19)

for each $x= \{x(n) \}_{n\in\Bbb {Z}_{\beta}}\in A(N,M)$. On account of (2.2), (2.3) and (2.17)~(2.19), we derive that for each $x= \{x(n) \}_{n\in\Bbb {Z}_{\beta}},y= \{y(n) \}_{n\in\Bbb {Z}_{\beta}}\in A(N,M)$ and $n\geq n_{1}$

$$\begin{aligned}& \bigl\vert T_{L}x(n)-T_{L}y(n) \bigr\vert \\& \quad = \Biggl\vert -\sum_{i=1}^{\infty}\sum _{t=n+i\tau}^{\infty}\frac{1}{a(t)}\sum _{s=t}^{\infty}(s-t+1) \bigl[f \bigl(s,x \bigl(s-b_{1}(s) \bigr),x \bigl(s-b_{2}(s) \bigr),\ldots,x \bigl(s-b_{k}(s) \bigr) \bigr) \\& \qquad {} -f \bigl(s,y \bigl(s-b_{1}(s) \bigr),y \bigl(s-b_{2}(s) \bigr), \ldots,y \bigl(s-b_{k}(s) \bigr) \bigr) \bigr] \Biggr\vert \\& \quad \leq\sum_{i=1}^{\infty}\sum _{t=n+i\tau}^{\infty}\frac{1}{\vert a(t)\vert }\sum _{s=t}^{\infty} (s-t+1)P(s)\max \bigl\{ \bigl\vert x \bigl(s-b_{j}(s) \bigr)-y \bigl(s-b_{j}(s) \bigr) \bigr\vert : 1\leq j\leq k \bigr\} \\& \quad \leq\sum_{i=1}^{\infty}\sum _{t=n_{1}+i\tau}^{\infty}\frac{1}{\vert a(t)\vert }\sum _{s=t}^{\infty}(s-t+1)P(s)\Vert x-y\Vert =\theta \Vert x-y\Vert , \end{aligned}$$

and

$$\begin{aligned}& \bigl\vert T_{L}x(n)-L \bigr\vert \\& \quad = \Biggl\vert -\sum_{i=1}^{\infty}\sum _{t=n+i\tau}^{\infty}\frac{1}{a(t)}\sum _{s=t}^{\infty}(s-t+1) \\& \qquad {}\times \bigl[f \bigl(s,x \bigl(s-b_{1}(s) \bigr),x \bigl(s-b_{2}(s) \bigr),\ldots,x \bigl(s-b_{k}(s) \bigr) \bigr)-d(s) \bigr] \Biggr\vert \\& \quad \leq\sum_{i=1}^{\infty}\sum _{t=n+i\tau}^{\infty}\frac{1}{\vert a(t)\vert }\sum _{s=t}^{\infty}(s-t+1) \\& \qquad {}\times \bigl[ \bigl\vert f \bigl(s,x \bigl(s-b_{1}(s) \bigr),x \bigl(s-b_{2}(s) \bigr),\ldots,x \bigl(s-b_{k}(s) \bigr) \bigr) \bigr\vert + \bigl\vert d(s) \bigr\vert \bigr] \\& \quad \leq\sum_{i=1}^{\infty}\sum _{t=n_{1}+i\tau}^{\infty}\frac{1}{\vert a(t)\vert }\sum _{s=t}^{\infty}(s-t+1) \bigl(Q(s)+ \bigl\vert d(s) \bigr\vert \bigr) \leq\min\{M-L ,L-N\}, \end{aligned}$$

which yield (2.12). Consequently $T_{L}$ is a contraction mapping on the closed subset $A(N,M)$ and it has a unique fixed point $x=\{x(n)\}_{n\in\Bbb {Z}_{\beta}}\in A(N,M)$. It follows that for $n\geq n_{1}+\tau$

$$\begin{aligned}& x(n)=L-\sum_{i=1}^{\infty}\sum _{t=n+i\tau}^{\infty}\frac{1}{a(t)}\sum _{s=t}^{\infty}(s-t+1) \\& \hphantom{x(n)=} {}\times \bigl[f \bigl(s,x \bigl(s-b_{1}(s) \bigr),x \bigl(s-b_{2}(s) \bigr),\ldots,x \bigl(s-b_{k}(s) \bigr) \bigr)-d(s) \bigr], \\& x(n-\tau) \\& \quad =L-\sum_{i=1}^{\infty}\sum _{t=n+(i-1)\tau}^{\infty}\frac {1}{a(t)}\sum _{s=t}^{\infty}(s-t+1) \\& \qquad {}\times \bigl[f \bigl(s,x \bigl(s-b_{1}(s) \bigr),x \bigl(s-b_{2}(s) \bigr),\ldots,x \bigl(s-b_{k}(s) \bigr) \bigr)-d(s) \bigr], \end{aligned}$$

which guarantee that for $n\geq n_{1}+\tau$

$$\begin{aligned}& \Delta \bigl(x(n)-x(n-\tau) \bigr) \\& \quad =-\frac{1}{a(n)}\sum_{s=n}^{\infty}(s-n+1) \bigl[f \bigl(s,x \bigl(s-b_{1}(s) \bigr),x \bigl(s-b_{2}(s) \bigr),\ldots,x \bigl(s-b_{k}(s) \bigr) \bigr)-d(s) \bigr], \\& \Delta \bigl(a(n)\Delta \bigl(x(n)-x(n-\tau) \bigr) \bigr) \\& \quad =\sum _{s=n}^{\infty } \bigl[f \bigl(s,x \bigl(s-b_{1}(s) \bigr),x \bigl(s-b_{2}(s) \bigr),\ldots,x \bigl(s-b_{k}(s) \bigr) \bigr)-d(s) \bigr] \end{aligned}$$

and

$$\begin{aligned}& \Delta^{2} \bigl(a(n)\Delta \bigl(x(n)-x(n-\tau) \bigr) \bigr) \\& \quad =-f \bigl(n,x \bigl(n-b_{1}(n) \bigr),x \bigl(n-b_{2}(n) \bigr), \ldots,x \bigl(n-b_{k}(n) \bigr) \bigr)+d(n), \end{aligned}$$

which together with (2.16) implies that $x=\{x(n)\}_{n\in\Bbb {Z}_{\beta}}\in A(N,M)$ is a bounded positive solution of Eq. (1.11).

It follows from (2.12), (2.15), (2.17) and (2.19) that for any $m\ge0$ and $n\ge n_{1}$,

$$\begin{aligned}& \bigl\vert x_{m+1}(n)-x(n) \bigr\vert \\& \quad = \Biggl\vert (1-\alpha_{m})x_{m}(n)+ \alpha_{m} \Biggl\{ L-\sum_{i=1}^{\infty} \sum_{t=n+i\tau}^{\infty} \frac{1}{a(t)}\sum _{s=t}^{\infty}(s-t+1) \\& \qquad {} \times \bigl[f \bigl(s,x_{m} \bigl(s-b_{1}(s) \bigr),x_{m} \bigl(s-b_{2}(s) \bigr),\ldots,x_{m} \bigl(s-b_{k}(s) \bigr) \bigr)-d(s) \bigr] \Biggr\} -x(n) \Biggr\vert \\& \quad \le(1-\alpha_{m}) \bigl\vert x_{m}(n)-x(n) \bigr\vert + \alpha_{m} \bigl\vert T_{L}x_{m}(n)-T_{L}x(n) \bigr\vert \\& \quad \le(1-\alpha_{m})\Vert x_{m}-x\Vert + \alpha_{m}\theta \Vert x_{m}-x\Vert \\& \quad = \bigl(1-(1-\theta)\alpha_{m} \bigr)\Vert x_{m}-x \Vert \\& \quad \le e^{-(1-\theta)\alpha_{m}}\Vert x_{m}-x\Vert , \end{aligned}$$

which gives (2.6). Thus (2.6) and (2.7) guarantee that $\lim_{m\to\infty}x_{m}=x$.

Next we show (b). Let $L_{1},L_{2}\in(N,M)$ and $L_{1}\neq L_{2}$. Analogously we deduce that for each $z\in\{1,2\}$, there exist constants $\theta_{z}\in(0,1)$ and $n_{z}\geq n_{0}+\tau+\vert \beta \vert$ and a mapping $T_{L_{z}}$ satisfying (2.17)~(2.19), where θ, L and $n_{1}$ are replaced by $\theta_{z}$, $L_{z}$ and $n_{z}$, respectively, and the contraction mapping $T_{L_{z}}$ has a unique fixed point $x^{z}=\{x^{z}(n)\}_{n\in\Bbb {Z}_{\beta}}\in A(N,M)$, which is a bounded positive solution of Eq. (1.11) in $A(N,M)$, that is,

$$\begin{aligned} x^{z}(n) =&L_{z}-\sum_{i=1}^{\infty} \sum_{t=n+i\tau}^{\infty}\frac {1}{a(t)}\sum _{s=t}^{\infty}(s-t+1) \\ &{} \times \bigl[f \bigl(s,x^{z} \bigl(s-b_{1}(s) \bigr),x^{z} \bigl(s-b_{2}(s) \bigr),\ldots,x^{z} \bigl(s-b_{k}(s) \bigr) \bigr)-d(s) \bigr],\quad n\ge n_{z}, \end{aligned}$$

which together with (2.2) and (2.17) gives that

$$\begin{aligned}& \bigl\vert x^{1}(n)-x^{2}(n) \bigr\vert \\& \quad = \Biggl\vert L_{1}-L_{2}-\sum _{i=1}^{\infty}\sum_{t=n+i\tau}^{\infty} \frac {1}{a(t)}\sum_{s=t}^{\infty}(s-t+1) \\& \qquad {}\times \bigl[f \bigl(s,x^{1} \bigl(s-b_{1}(s) \bigr),x^{1} \bigl(s-b_{2}(s) \bigr),\ldots,x^{1} \bigl(s-b_{k}(s) \bigr) \bigr) \\& \qquad {} -f \bigl(s,x^{2} \bigl(s-b_{1}(s) \bigr),x^{2} \bigl(s-b_{2}(s) \bigr),\ldots,x^{2} \bigl(s-b_{k}(s) \bigr) \bigr) \bigr] \Biggr\vert \\& \quad \geq \vert L_{1}-L_{2}\vert -\sum _{i=1}^{\infty}\sum_{t=n+i\tau}^{\infty} \frac {1}{\vert a(t)\vert }\sum_{s=t}^{\infty}(s-t+1) \\& \qquad {}\times \bigl\vert f \bigl(s,x^{1} \bigl(s-b_{1}(s) \bigr),x^{1} \bigl(s-b_{2}(s) \bigr),\ldots,x^{1} \bigl(s-b_{k}(s) \bigr) \bigr) \\& \qquad {} -f \bigl(s,x^{2} \bigl(s-b_{1}(s) \bigr),x^{2} \bigl(s-b_{2}(s) \bigr),\ldots,x^{2} \bigl(s-b_{k}(s) \bigr) \bigr) \bigr\vert \\& \quad \geq \vert L_{1}-L_{2}\vert -\sum _{i=1}^{\infty}\sum_{t=n+i\tau}^{\infty} \frac{1}{\vert a(t)\vert } \sum_{s=t}^{\infty}(s-t+1)P(s) \\& \qquad {}\times \max \bigl\{ \bigl\vert x^{1} \bigl(s-b_{j}(s) \bigr)-x^{2} \bigl(s-b_{j}(s) \bigr) \bigr\vert : 1\leq j \leq k \bigr\} \\& \quad \geq \vert L_{1}-L_{2}\vert -\sum _{i=1}^{\infty}\sum_{t=n+i\tau}^{\infty} \frac{1}{\vert a(t)\vert } \sum_{s=t}^{\infty}(s-t+1)P(s) \bigl\Vert x^{1}-x^{2} \bigr\Vert \\& \quad \geq \vert L_{1}-L_{2}\vert -\max\{ \theta_{1}, \theta_{2}\} \bigl\Vert x^{1}-x^{2} \bigr\Vert ,\quad n\geq\max\{n_{1},n_{2}\}, \end{aligned}$$

which yields that

$$\bigl\Vert x^{1}-x^{2} \bigr\Vert \geq \frac{\vert L_{1}-L_{2}\vert }{1+\max\{\theta_{1},\theta_{2}\}}>0, $$

that is, $x^{1}\neq x^{2}$. Hence Eq. (1.11) possesses uncountably many bounded positive solutions in $A(N,M)$. This completes the proof. □

Theorem 2.3

Assume that there exist positive constants M and N, nonnegative constants $c_{1}$ and $c_{2}$ and nonnegative sequences $\{P(n)\}_{n\in\Bbb {N}_{n_{0}}}$ and $\{Q(n)\}_{n\in\Bbb {N}_{n_{0}}}$ satisfying $(1-c_{1}-c_{2})M>N$, (2.2)~(2.4) and

$$\begin{aligned}& -c_{1}\leq c(n)\leq c_{2},\quad \textit{eventually}. \end{aligned}$$

(2.20)

Then

• (a) for any $L\in (c_{2}M+N,(1-c_{1})M )$, there exist $\theta\in(0,1)$ and $n_{1}\geq n_{0}+\tau+\vert \beta \vert$ such that for each $x_{0}=\{x_{0}(n)\}_{n\in\Bbb {Z}_{\beta}}\in A(N,M)$, the Mann iterative sequence $\{x_{m}\}_{m\in\Bbb {N}_{0}}=\{x_{m}(n)\}_{(n,m)\in \Bbb {Z}_{\beta}\times\Bbb {N}_{0}}$ generated by the scheme:

  $$\begin{aligned}& x_{m+1}(n) \\& \quad =\textstyle\begin{cases} (1-\alpha_{m})x_{m}(n)+ \alpha_{m} \{L-c(n)x(n-\tau) \\ \quad {}+\sum_{t=n}^{\infty } \frac{1}{a(t)}\sum_{s=t}^{\infty}(s-t+1) \\ \quad {}\times [f (s,x_{m} (s-b_{1}(s) ),x_{m} (s-b_{2}(s) ),\ldots,x_{m} (s-b_{k}(s) ) )-d(s) ] \}, \\ \quad n \ge n_{1}, m\ge0, \\ (1-\alpha_{m})x_{m}(n_{1})+\alpha_{m} \{L-c(n_{1})x(n_{1}-\tau) \\ \quad {}+\sum_{t=n_{1}}^{\infty} \frac{1}{a(t)}\sum_{s=t}^{\infty}(s-t+1) \\ \quad {}\times [f (s,x_{m} (s-b_{1}(s) ),x_{m} (s-b_{2}(s) ),\ldots,x_{m} (s-b_{k}(s) ) )-d(s) ] \}, \\ \quad \beta\le n< n_{1}, m\ge0 \end{cases}\displaystyle \end{aligned}$$

  (2.21)

  converges to a bounded positive solution $x\in A(N,M)$ of Eq. (1.11) and satisfies the error estimate (2.6), where $\{\alpha_{m}\}_{m\in\Bbb {N}_{0}}$ is an arbitrary sequence in $[0,1]$ satisfying (2.7);

• (b) Eq. (1.11) possesses uncountably many bounded positive solutions in $A(N,M)$.

Proof

Let $L\in (c_{2}M+N,(1-c_{1})M )$. It follows from (2.4) that there exist $\theta\in(0,1)$ and $n_{1}\geq n_{0}+\tau+\vert \beta \vert$ satisfying

$$\begin{aligned}& -c_{1}\leq c(n)\leq c_{2},\quad n\ge n_{1}; \end{aligned}$$

(2.22)

$$\begin{aligned}& \theta=c_{1}+c_{2}+\sum_{t=n_{1}}^{\infty} \frac{1}{\vert a(t)\vert }\sum_{s=t}^{\infty }(s-t+1)P(s); \end{aligned}$$

(2.23)

$$\begin{aligned}& \sum_{t=n_{1}}^{\infty}\frac{1}{\vert a(t)\vert }\sum _{s=t}^{\infty }(s-t+1) \bigl(Q(s)+ \bigl\vert d(s) \bigr\vert \bigr)\leq \min \bigl\{ M(1-c_{1})-L ,L-c_{2}M-N \bigr\} . \end{aligned}$$

(2.24)

Define a mapping $T_{L}: A(N,M)\rightarrow l_{\beta}^{\infty}$ by

$$\begin{aligned}& T_{L}x(n)= \textstyle\begin{cases} L-c(n)x(n-\tau)+\sum_{t=n}^{\infty}\frac{1}{a(t)}\sum_{s=t}^{\infty }(s-t+1) \\ \quad \times [f (s,x (s-b_{1}(s) ),x (s-b_{2}(s) ),\ldots,x (s-b_{k}(s) ) )-d(s) ], \\ \quad n\geq n_{1}, \\ T_{L}x(n_{1}),\quad \beta\leq n< n_{1} \end{cases}\displaystyle \end{aligned}$$

(2.25)

for $x= \{x(n) \}_{n\in\Bbb {Z}_{\beta}}\in A(N,M)$. Using (2.2), (2.3) and (2.22)~(2.25), we derive that for each $x= \{x(n) \}_{n\in\Bbb {Z}_{\beta}}$, $y= \{y(n) \}_{n\in\Bbb {Z}_{\beta}}\in A(N,M)$ and $n\geq n_{1}$

$$\begin{aligned}& \bigl\vert T_{L}x(n)-T_{L}y(n) \bigr\vert \\& \quad = \Biggl\vert -c(n) \bigl(x(n-\tau)-y(n-\tau) \bigr) \\& \qquad {} +\sum_{t=n}^{\infty} \frac{1}{a(t)}\sum_{s=t}^{\infty }(s-t+1) \bigl[f \bigl(s,x \bigl(s-b_{1}(s) \bigr),x \bigl(s-b_{2}(s) \bigr),\ldots,x \bigl(s-b_{k}(s) \bigr) \bigr) \\& \qquad {} -f \bigl(s,y \bigl(s-b_{1}(s) \bigr),y \bigl(s-b_{2}(s) \bigr), \ldots,y \bigl(s-b_{k}(s) \bigr) \bigr) \bigr] \Biggr\vert \\& \quad \leq \bigl\vert c(n) \bigr\vert \bigl\vert x(n-\tau)-y(n-\tau) \bigr\vert \\& \qquad {} +\sum_{t=n}^{\infty}\frac{1}{\vert a(t)\vert } \sum_{s=t}^{\infty}(s-t+1) P(s)\max \bigl\{ \bigl\vert x \bigl(s-b_{j}(s) \bigr)-y \bigl(s-b_{j}(s) \bigr) \bigr\vert : 1\leq j\leq k \bigr\} \\& \quad \leq \Biggl[c_{1}+c_{2}+\sum _{t=n}^{\infty} \frac{1}{\vert a(t)\vert }\sum _{s=t}^{\infty }(s-t+1)P(s) \Biggr]\Vert x-y\Vert \\& \quad \leq\theta \Vert x-y\Vert , \\& T_{L}x(n)=L-c(n)x(n-\tau)+\sum_{t=n}^{\infty} \frac{1}{a(t)}\sum_{s=t}^{\infty}(s-t+1) \\& \hphantom{T_{L}x(n)=}{} \times \bigl[f \bigl(s,x \bigl(s-b_{1}(s) \bigr),x \bigl(s-b_{2}(s) \bigr),\ldots,x \bigl(s-b_{k}(s) \bigr) \bigr)-d(s) \bigr] \\& \hphantom{T_{L}x(n)}\leq L+c_{1}M+\sum_{t=n_{1}}^{\infty} \frac{1}{\vert a(t)\vert }\sum_{s=t}^{\infty }(s-t+1) \bigl(Q(s)+ \bigl\vert d(s) \bigr\vert \bigr) \\& \hphantom{T_{L}x(n)}\leq L+c_{1}M+\min \bigl\{ M(1-c_{1})-L ,L-c_{2}M-N \bigr\} \\& \hphantom{T_{L}x(n)}\leq M \end{aligned}$$

and

$$\begin{aligned} T_{L}x(n) =&L-c(n)x(n-\tau)+\sum_{t=n}^{\infty} \frac{1}{a(t)}\sum_{s=t}^{\infty}(s-t+1) \\ &{} \times \bigl[f \bigl(s,x \bigl(s-b_{1}(s) \bigr),x \bigl(s-b_{2}(s) \bigr),\ldots,x \bigl(s-b_{k}(s) \bigr) \bigr)-d(s) \bigr] \\ \geq& L-c_{2}M-\sum_{t=n_{1}}^{\infty} \frac{1}{\vert a(t)\vert }\sum_{s=t}^{\infty }(s-t+1) \bigl(Q(s)+ \bigl\vert d(s) \bigr\vert \bigr) \\ \geq& L-c_{2}M-\min \bigl\{ M(1-c_{1})-L ,L-c_{2}M-N \bigr\} \\ \geq& N, \end{aligned}$$

which imply (2.12). Hence $T_{L}$ is a contraction mapping on the closed subset $A(N,M)$ and it has a unique fixed point $x=\{x(n)\}_{n\in\Bbb {Z}_{\beta}}\in A(N,M)$. That is,

$$\begin{aligned} x(n) =&L-c(n)x(n-\tau)+\sum_{t=n}^{\infty} \frac{1}{a(t)}\sum_{s=t}^{\infty}(s-t+1) \\ &{} \times \bigl[f \bigl(s,x \bigl(s-b_{1}(s) \bigr),x \bigl(s-b_{2}(s) \bigr),\ldots,x \bigl(s-b_{k}(s) \bigr) \bigr)-d(s) \bigr],\quad n\geq n_{1}, \end{aligned}$$

which gives that for $n\geq n_{1}+\tau$

$$\begin{aligned}& \Delta \bigl(x(n)+c(n)x(n-\tau) \bigr) \\& \quad =-\frac{1}{a(n)}\sum_{s=n}^{\infty}(s-n+1) \bigl[f \bigl(s,x \bigl(s-b_{1}(s) \bigr),x \bigl(s-b_{2}(s) \bigr),\ldots,x \bigl(s-b_{k}(s) \bigr) \bigr)-d(s) \bigr], \end{aligned}$$

which implies that for $n\geq n_{1}+\tau$

$$\begin{aligned}& \Delta \bigl(a(n)\Delta \bigl(x(n)+c(n)x(n-\tau) \bigr) \bigr) \\& \quad =\sum_{s=n}^{\infty} \bigl[f \bigl(s,x \bigl(s-b_{1}(s) \bigr),x \bigl(s-b_{2}(s) \bigr),\ldots,x \bigl(s-b_{k}(s) \bigr) \bigr)-d(s) \bigr] \end{aligned}$$

and

$$\begin{aligned}& \Delta^{2} \bigl(a(n)\Delta \bigl(x(n)+c(n)x(n-\tau) \bigr) \bigr) \\& \quad =-f \bigl(n,x \bigl(n-b_{1}(n) \bigr),x \bigl(n-b_{2}(n) \bigr), \ldots,x \bigl(n-b_{k}(n) \bigr) \bigr)+d(n), \end{aligned}$$

which means that $x=\{x(n)\}_{n\in\Bbb {Z}_{\beta}}$ is a bounded positive solution of Eq. (1.11) in $A(N,M)$.

By means of (2.12), (2.21), (2.23) and (2.25), we conclude that for any $m\ge0$ and $n\ge n_{1}$

$$\begin{aligned}& \bigl\vert x_{m+1}(n)-x(n) \bigr\vert \\& \quad = \Biggl\vert (1-\alpha_{m})x_{m}(n)+ \alpha_{m} \Biggl\{ L-c(n)x(n-\tau)+\sum_{t=n}^{\infty} \frac{1}{a(t)}\sum_{s=t}^{\infty}(s-t+1) \\& \qquad {} \times \bigl[f \bigl(s,x_{m} \bigl(s-b_{1}(s) \bigr),x_{m} \bigl(s-b_{2}(s) \bigr),\ldots,x_{m} \bigl(s-b_{k}(s) \bigr) \bigr)-d(s) \bigr] \Biggr\} -x(n) \Biggr\vert \\& \quad \le(1-\alpha_{m}) \bigl\vert x_{m}(n)-x(n) \bigr\vert + \alpha_{m} \bigl\vert T_{L}x_{m}(n)-T_{L}x(n) \bigr\vert \\& \quad \le(1-\alpha_{m})\Vert x_{m}-x\Vert + \alpha_{m}\theta \Vert x_{m}-x\Vert \\& \quad = \bigl(1-(1-\theta)\alpha_{m} \bigr)\Vert x_{m}-x \Vert \\& \quad \le e^{-(1-\theta)\alpha_{m}}\Vert x_{m}-x\Vert , \end{aligned}$$

which implies (2.6). Thus (2.6) and (2.7) ensure that $\lim_{m\to\infty}x_{m}=x$.

Let $L_{1},L_{2}\in (c_{2}M+N,(1-c_{1})M )$ and $L_{1}\neq L_{2}$. Homoplastically we conclude that for each $z\in\{1,2\}$, there exist constants $\theta_{z}\in(0,1)$ and $n_{z}\geq n_{0}+\tau+\vert \beta \vert$ and a mapping $T_{L_{z}}$ satisfying (2.22)~(2.25), where θ, L and $n_{1}$ are replaced by $\theta_{z}$, $L_{z}$ and $n_{z}$, respectively, and the contraction mapping $T_{L_{z}}$ has a unique fixed point $x^{z}=\{x^{z}(n)\}_{n\in\Bbb {Z}_{\beta}}$, which is a bounded positive solution of Eq. (1.11) in $A(N,M)$, that is,

$$\begin{aligned} x^{z}(n) =&L_{z}-c(n)x^{z}(n-\tau)+\sum _{t=n}^{\infty}\frac{1}{a(t)}\sum _{s=t}^{\infty}(s-t+1) \\ &{} \times \bigl[f \bigl(s,x^{z} \bigl(s-b_{1}(s) \bigr),x^{z} \bigl(s-b_{2}(s) \bigr),\ldots,x^{z} \bigl(s-b_{k}(s) \bigr) \bigr)-d(s) \bigr], \quad n\ge n_{z}, \end{aligned}$$

which together with (2.2), (2.22) and (2.23) yield that

$$\begin{aligned}& \bigl\vert x^{1}(n)-x^{2}(n) \bigr\vert \\& \quad = \Biggl\vert L_{1}-L_{2}-c(n) \bigl(x^{1}(n- \tau)-x^{2}(n-\tau) \bigr) \\& \qquad {} +\sum_{t=n}^{\infty} \frac{1}{a(t)}\sum_{s=t}^{\infty}(s-t+1) \bigl[f \bigl(s,x^{1} \bigl(s-b_{1}(s) \bigr),x^{1} \bigl(s-b_{2}(s) \bigr),\ldots,x^{1} \bigl(s-b_{k}(s) \bigr) \bigr) \\& \qquad {} -f \bigl(s,x^{2} \bigl(s-b_{1}(s) \bigr),x^{2} \bigl(s-b_{2}(s) \bigr),\ldots,x^{2} \bigl(s-b_{k}(s) \bigr) \bigr) \bigr] \Biggr\vert \\& \quad \geq \vert L_{1}-L_{2}\vert - \bigl\vert c(n) \bigr\vert \bigl\vert x^{1}(n-\tau)-x^{2}(n-\tau) \bigr\vert \\& \qquad {} -\sum_{t=n}^{\infty}\frac{1}{\vert a(t)\vert } \sum_{s=t}^{\infty}(s-t+1) \bigl\vert f \bigl(s,x^{1} \bigl(s-b_{1}(s) \bigr),x^{1} \bigl(s-b_{2}(s) \bigr),\ldots,x^{1} \bigl(s-b_{k}(s) \bigr) \bigr) \\& \qquad {} -f \bigl(s,x^{2} \bigl(s-b_{1}(s) \bigr),x^{2} \bigl(s-b_{2}(s) \bigr),\ldots,x^{2} \bigl(s-b_{k}(s) \bigr) \bigr) \bigr\vert \\& \quad \geq \vert L_{1}-L_{2}\vert -(c_{1}+c_{2}) \bigl\Vert x^{1}-x^{2} \bigr\Vert \\& \qquad {} -\sum_{t=n}^{\infty}\frac{1}{\vert a(t)\vert } \sum_{s=t}^{\infty}(s-t+1) P(s)\max \bigl\{ \bigl\vert x^{1} \bigl(s-b_{j}(s) \bigr)-x^{2} \bigl(s-b_{j}(s) \bigr) \bigr\vert : 1\leq j\leq k \bigr\} \\& \quad \geq \vert L_{1}-L_{2}\vert - \Biggl[c_{1}+c_{2}+ \sum_{t=\max\{n_{1},n_{2}\}}^{\infty}\frac{1}{\vert a(t)\vert } \sum _{s=t}^{\infty}(s-t+1)P(s) \Biggr] \bigl\Vert x^{1}-x^{2} \bigr\Vert \\& \quad \geq \vert L_{1}-L_{2}\vert -\max\{ \theta_{1}, \theta_{2}\} \bigl\Vert x^{1}-x^{2} \bigr\Vert ,\quad n\geq\max\{n_{1},n_{2}\}, \end{aligned}$$

which means that

$$\bigl\Vert x^{1}-x^{2} \bigr\Vert \geq \frac{\vert L_{1}-L_{2}\vert }{1+\max\{\theta_{1},\theta_{2}\}}>0, $$

that is, $x^{1}\neq x^{2}$. This completes the proof. □

Theorem 2.4

Assume that there exist four constants M, N, $c_{1}$ and $c_{2}$ and two nonnegative sequences $\{P(n)\}_{n\in \Bbb {N}_{n_{0}}}$ and $\{Q(n)\}_{n\in\Bbb {N}_{n_{0}}}$ satisfying $M>N$, $c_{2} (c_{1}^{2}-c_{2})M>c_{1} (c_{2}^{2}-c_{1})N>0$, (2.2)~(2.4) and

$$\begin{aligned}& 1< c_{1}\leq c(n)\leq c_{2},\quad \textit{eventually}. \end{aligned}$$

(2.26)

Then

• (a) for any $L\in (\frac{c_{2}}{c_{1}}M+c_{2}N ,\frac{c_{1}}{c_{2}}N+c_{1}M )$, there exist $\theta\in(0,1)$ and $n_{1}\geq n_{0}+\tau+\vert \beta \vert$ such that for each $x_{0}=\{x_{0}(n)\}_{n\in\Bbb {Z}_{\beta}}\in A(N,M)$, the Mann iterative sequence $\{x_{m}\}_{m\in\Bbb {N}_{0}}=\{x_{m}(n)\}_{(n,m)\in\Bbb {Z}_{\beta}\times\Bbb {N}_{0}}$ generated by the scheme:

  $$\begin{aligned}& x_{m+1}(n)=\textstyle\begin{cases} (1-\alpha_{m})x_{m}(n)+ \alpha_{m} \{\frac{L}{c(n+\tau)}-\frac{x(n+\tau)}{c(n+\tau)} \\ \quad {}+\frac{1}{c(n+\tau)}\sum_{t=n+\tau}^{\infty} \frac{1}{a(t)}\sum_{s=t}^{\infty}(s-t+1) \\ \quad {}\times [f (s,x_{m} (s-b_{1}(s) ),x_{m} (s-b_{2}(s) ),\ldots, x_{m} (s-b_{k}(s) ) ) \\ \quad {}-d(s) ] \}, \\ \quad n \ge n_{1}, m\ge0, \\ (1-\alpha_{m})x_{m}(n_{1})+\alpha_{m} \{\frac{L}{c(n_{1}+\tau)}-\frac {x(n_{1}+\tau)}{c(n_{1}+\tau)} \\ \quad {}+\frac{1}{c(n_{1}+\tau)}\sum_{t=n_{1}+\tau}^{\infty} \frac{1}{a(t)}\sum_{s=t}^{\infty}(s-t+1) \\ \quad {}\times [f (s,x_{m} (s-b_{1}(s) ),x_{m} (s-b_{2}(s) ),\ldots,x_{m} (s-b_{k}(s) ) ) \\ \quad {}-d(s) ] \}, \\ \quad \beta\le n< n_{1}, m\ge0 \end{cases}\displaystyle \end{aligned}$$

  (2.27)

  converges to a bounded positive solution $x\in A(N,M)$ of Eq. (1.11) and satisfies the error estimate (2.6), where $\{\alpha_{m}\}_{m\in\Bbb {N}_{0}}$ is an arbitrary sequence in $[0,1]$ satisfying (2.7);

• (b) Eq. (1.11) possesses uncountably many bounded positive solutions in $A(N,M)$.

Proof

Let $L\in (\frac{c_{2}}{c_{1}}M+c_{2}N ,\frac{c_{1}}{c_{2}}N+c_{1}M )$. Note that (2.4) and (2.26) imply that there exist $\theta\in(0,1)$ and $n_{1}\geq n_{0}+\tau+\vert \beta \vert$ satisfying

$$\begin{aligned}& c_{1}\leq c(n)\leq c_{2},\quad n\ge n_{1}; \end{aligned}$$

(2.28)

$$\begin{aligned}& \theta=\frac{1}{c_{1}}+\frac{1}{c_{1}}\sum_{t=n_{1}}^{\infty} \frac {1}{\vert a(t)\vert }\sum_{s=t}^{\infty}(s-t+1)P(s); \end{aligned}$$

(2.29)

$$\begin{aligned}& \sum_{t=n_{1}}^{\infty}\frac{1}{\vert a(t)\vert }\sum _{s=t}^{\infty }(s-t+1) \bigl(Q(s)+ \bigl\vert d(s) \bigr\vert \bigr) \\& \quad \leq \min \biggl\{ c_{1}M+\frac{c_{1}N}{c_{2}}-L , \frac{c_{1}L}{c_{2}}-M-c_{1}N \biggr\} . \end{aligned}$$

(2.30)

Define a mapping $T_{L}: A(N,M)\rightarrow l_{\beta}^{\infty}$ by

$$\begin{aligned}& T_{L}x(n)=\textstyle\begin{cases} \frac{L}{c(n+\tau)}-\frac{x(n+\tau)}{c(n+\tau)}+ \frac{1}{c(n+\tau)} \sum_{t=n+\tau}^{\infty} \frac{1}{a(t)}\sum_{s=t}^{\infty}(s-t+1) \\ \quad {}\times [f (s,x (s-b_{1}(s) ),x (s-b_{2}(s) ),\ldots,x (s-b_{k}(s) ) )-d(s) ], \\ \quad n\geq n_{1}, \\ T_{L}x(n_{1}), \quad \beta\leq n< n_{1} \end{cases}\displaystyle \end{aligned}$$

(2.31)

for each $x= \{x(n) \}_{n\in\Bbb {Z}_{\beta}}\in A(N,M)$. It follows from (2.2), (2.3) and (2.28)~(2.31) that for each $x= \{x(n) \}_{n\in\Bbb {Z}_{\beta}}$, $y= \{y(n) \}_{n\in\Bbb {Z}_{\beta}}\in A(N,M)$ and $n\geq n_{1}$

$$\begin{aligned}& \bigl\vert T_{L}x(n)-T_{L}y(n) \bigr\vert \\& \quad = \Biggl\vert -\frac{x(n+\tau)-y(n+\tau)}{c(n+\tau)} \\& \qquad {} +\frac{1}{c(n+\tau)} \sum_{t=n+\tau}^{\infty} \frac{1}{a(t)}\sum_{s=t}^{\infty }(s-t+1) \bigl[f \bigl(s,x \bigl(s-b_{1}(s) \bigr),x \bigl(s-b_{2}(s) \bigr),\ldots,x \bigl(s-b_{k}(s) \bigr) \bigr) \\& \qquad {} -f \bigl(s,y \bigl(s-b_{1}(s) \bigr),y \bigl(s-b_{2}(s) \bigr), \ldots,y \bigl(s-b_{k}(s) \bigr) \bigr) \bigr] \Biggr\vert \\& \quad \leq\frac{\vert x(n+\tau)-y(n+\tau)\vert }{c(n+\tau)} \\& \qquad {} +\frac{1}{c(n+\tau)} \sum_{t=n+\tau}^{\infty} \frac{1}{\vert a(t)\vert }\sum_{s=t}^{\infty}(s-t+1) P(s) \\& \qquad {}\times \max \bigl\{ \bigl\vert x \bigl(s-b_{j}(s) \bigr)-y \bigl(s-b_{j}(s) \bigr) \bigr\vert : 1\leq j\leq k \bigr\} \\& \quad \leq\frac{1}{c_{1}}\Vert x-y\Vert +\frac{1}{c_{1}}\sum _{t=n_{1}}^{\infty}\frac{1}{\vert a(t)\vert } \sum _{s=t}^{\infty}(s-t+1)P(s)\Vert x-y\Vert \\& \quad =\theta \Vert x-y\Vert , \\& T_{L}x(n)=\frac{L}{c(n+\tau)}-\frac{x(n+\tau)}{c(n+\tau)}+ \frac {1}{c(n+\tau)} \sum_{t=n+\tau}^{\infty} \frac{1}{a(t)}\sum _{s=t}^{\infty}(s-t+1) \\& \hphantom{T_{L}x(n)=}{} \times \bigl[f \bigl(s,x \bigl(s-b_{1}(s) \bigr),x \bigl(s-b_{2}(s) \bigr),\ldots,x \bigl(s-b_{k}(s) \bigr) \bigr)-d(s) \bigr] \\& \hphantom{T_{L}x(n)}\leq\frac{L}{c_{1}}-\frac{N}{c_{2}}+\frac{1}{c_{1}}\sum _{t=n_{1}}^{\infty}\frac {1}{\vert a(t)\vert } \sum _{s=t}^{\infty}(s-t+1) \bigl(Q(s)+ \bigl\vert d(s) \bigr\vert \bigr) \\& \hphantom{T_{L}x(n)}\leq\frac{L}{c_{1}}-\frac{N}{c_{2}}+\frac{1}{c_{1}}\min \biggl\{ c_{1}M+\frac{c_{1}N}{c_{2}}-L ,\frac{c_{1}L}{c_{2}}-M-c_{1}N \biggr\} \\& \hphantom{T_{L}x(n)}\leq M \end{aligned}$$

and

$$\begin{aligned} T_{L}x(n) =&\frac{L}{c(n+\tau)}-\frac{x(n+\tau)}{c(n+\tau)}+ \frac {1}{c(n+\tau)} \sum_{t=n+\tau}^{\infty} \frac{1}{a(t)}\sum _{s=t}^{\infty}(s-t+1) \\ &{} \times \bigl[f \bigl(s,x \bigl(s-b_{1}(s) \bigr),x \bigl(s-b_{2}(s) \bigr),\ldots,x \bigl(s-b_{k}(s) \bigr) \bigr)-d(s) \bigr] \\ \geq&\frac{L}{c_{2}}-\frac{M}{c_{1}}-\frac{1}{c_{1}}\sum _{t=n_{1}}^{\infty}\frac {1}{\vert a(t)\vert } \sum _{s=t}^{\infty}(s-t+1) \bigl(Q(s)+ \bigl\vert d(s) \bigr\vert \bigr) \\ \geq&\frac{L}{c_{2}}-\frac{M}{c_{1}}-\frac{1}{c_{1}}\min \biggl\{ c_{1}M+\frac{c_{1}N}{c_{2}}-L ,\frac{c_{1}L}{c_{2}}-M-c_{1}N \biggr\} \\ \geq& N, \end{aligned}$$

which yield (2.12), that is, $T_{L}$ is a contraction mapping on the closed subset $A(N,M)$ and it has a unique fixed point $x=\{x(n)\}_{n\in\Bbb {Z}_{\beta}}\in A(N,M)$, which is a bounded positive solution of Eq. (1.11).

It follows from (2.12), (2.27), (2.29) and (2.31) that for any $m\ge0$ and $n\ge n_{1}$,

$$\begin{aligned}& \bigl\vert x_{m+1}(n)-x(n) \bigr\vert \\& \quad = \Biggl\vert (1-\alpha_{m})x_{m}(n)+ \alpha_{m} \Biggl\{ \frac{L}{c(n+\tau)}-\frac {x(n+\tau)}{c(n+\tau)} + \frac{1}{c(n+\tau)}\sum_{t=n+\tau}^{\infty} \frac{1}{a(t)}\sum_{s=t}^{\infty}(s-t+1) \\& \qquad {} \times \bigl[f \bigl(s,x_{m} \bigl(s-b_{1}(s) \bigr),x_{m} \bigl(s-b_{2}(s) \bigr),\ldots,x_{m} \bigl(s-b_{k}(s) \bigr) \bigr)-d(s) \bigr] \Biggr\} -x(n) \Biggr\vert \\& \quad \le(1-\alpha_{m}) \bigl\vert x_{m}(n)-x(n) \bigr\vert + \alpha_{m} \bigl\vert T_{L}x_{m}(n)-T_{L}x(n) \bigr\vert \\& \quad \le(1-\alpha_{m})\Vert x_{m}-x\Vert + \alpha_{m}\theta \Vert x_{m}-x\Vert \\& \quad = \bigl(1-(1-\theta)\alpha_{m} \bigr)\Vert x_{m}-x \Vert \\& \quad \le e^{-(1-\theta)\alpha_{m}}\Vert x_{m}-x\Vert , \end{aligned}$$

which gives (2.6). Thus (2.6) and (2.7) guarantee that $\lim_{m\to\infty}x_{m}=x$.

Let $L_{1},L_{2}\in (\frac{c_{2}}{c_{1}}M+c_{2}N ,\frac{c_{1}}{c_{2}}N+c_{1}M )$ and $L_{1}\neq L_{2}$. Similarly we deduce that for each $z\in\{1,2\}$, there exist constants $\theta_{z}\in(0,1)$ and $n_{z}\geq n_{0}+\tau+\vert \beta \vert$ and a mapping $T_{L_{z}}$ satisfying (2.28)~(2.31), where θ, L and $n_{1}$ are replaced by $\theta_{z}$, $L_{z}$ and $n_{z}$, respectively, and the contraction mapping $T_{L_{z}}$ has a unique fixed point $x^{z}=\{x^{z}(n)\}_{n\in\Bbb {Z}_{\beta}}$, which is a bounded positive solution of Eq. (1.11) in $A(N,M)$, that is,

$$\begin{aligned} x^{z}(n) =&\frac{L_{z}}{c(n+\tau)}-\frac{x^{z}(n+\tau)}{c(n+\tau)}+ \frac {1}{c(n+\tau)} \sum_{t=n+\tau}^{\infty} \frac{1}{a(t)}\sum _{s=t}^{\infty}(s-t+1) \\ &{} \times \bigl[f \bigl(s,x^{z} \bigl(s-b_{1}(s) \bigr),x^{z} \bigl(s-b_{2}(s) \bigr),\ldots,x^{z} \bigl(s-b_{k}(s) \bigr) \bigr)-d(s) \bigr], \quad n\ge n_{z}, \end{aligned}$$

which together with (2.2), (2.28) and (2.29) yields that

$$\begin{aligned}& \bigl\vert x^{1}(n)-x^{2}(n) \bigr\vert \\& \quad = \Biggl\vert \frac{L_{1}-L_{2}}{c(n+\tau)}-\frac{x^{1}(n-\tau)-x^{2}(n-\tau)}{c(n+\tau)} \\& \qquad {} +\frac{1}{c(n+\tau)}\sum_{t=n+\tau}^{\infty} \frac{1}{a(t)}\sum_{s=t}^{\infty}(s-t+1) \\& \qquad {}\times \bigl[f \bigl(s,x^{1} \bigl(s-b_{1}(s) \bigr),x^{1} \bigl(s-b_{2}(s) \bigr),\ldots,x^{1} \bigl(s-b_{k}(s) \bigr) \bigr) \\& \qquad {} -f \bigl(s,x^{2} \bigl(s-b_{1}(s) \bigr),x^{2} \bigl(s-b_{2}(s) \bigr),\ldots,x^{2} \bigl(s-b_{k}(s) \bigr) \bigr) \bigr] \Biggr\vert \\& \quad \geq\frac{\vert L_{1}-L_{2}\vert }{c(n+\tau)}-\frac{ \vert x^{1}(n+\tau)-x^{2}(n+\tau) \vert }{c(n+\tau)} \\& \qquad {} -\frac{1}{c(n+\tau)}\sum_{t=n}^{\infty} \frac{1}{\vert a(t)\vert }\sum_{s=t}^{\infty}(s-t+1) \\& \qquad {}\times \bigl\vert f \bigl(s,x^{1} \bigl(s-b_{1}(s) \bigr),x^{1} \bigl(s-b_{2}(s) \bigr),\ldots,x^{1} \bigl(s-b_{k}(s) \bigr) \bigr) \\& \qquad {} -f \bigl(s,x^{2} \bigl(s-b_{1}(s) \bigr),x^{2} \bigl(s-b_{2}(s) \bigr),\ldots,x^{2} \bigl(s-b_{k}(s) \bigr) \bigr) \bigr\vert \\& \quad \geq\frac{\vert L_{1}-L_{2}\vert }{c_{2}}-\frac{ \Vert x^{1}-x^{2} \Vert }{c_{1}} \\& \qquad {} -\frac{1}{c_{1}} \sum_{t=n}^{\infty} \frac{1}{\vert a(t)\vert }\sum_{s=t}^{\infty}(s-t+1) P(s) \max \bigl\{ \bigl\vert x^{1} \bigl(s-b_{j}(s) \bigr)-x^{2} \bigl(s-b_{j}(s) \bigr) \bigr\vert : 1\leq j \leq k \bigr\} \\& \quad \geq\frac{\vert L_{1}-L_{2}\vert }{c_{2}}-\frac{ \Vert x^{1}-x^{2} \Vert }{c_{1}}-\frac{1}{c_{1}} \sum _{t=\max\{n_{1},n_{2}\}}^{\infty}\frac{1}{\vert a(t)\vert }\sum _{s=t}^{\infty }(s-t+1)P(s) \bigl\Vert x^{1}-x^{2} \bigr\Vert \\& \quad \geq\frac{\vert L_{1}-L_{2}\vert }{c_{2}}-\max\{\theta_{1},\theta_{2}\} \bigl\Vert x^{1}-x^{2} \bigr\Vert ,\quad n\geq\max \{n_{1},n_{2}\}, \end{aligned}$$

which means that

$$\bigl\Vert x^{1}-x^{2} \bigr\Vert \geq \frac{\vert L_{1}-L_{2}\vert }{c_{2} (1+\max\{\theta_{1},\theta _{2}\} )}>0, $$

that is, $x^{1}\neq x^{2}$. This completes the proof. □

Theorem 2.5

Assume that there exist four constants M, N, $c_{1}$ and $c_{2}$ and two nonnegative sequences $\{P(n)\}_{n\in \Bbb {N}_{n_{0}}}$ and $\{Q(n)\}_{n\in\Bbb {N}_{n_{0}}}$ satisfying $(1+c_{2})M<(1+c_{1})N<0$, (2.2)~(2.4) and

$$\begin{aligned}& c_{1}\leq c(n)\leq c_{2}< -1, \quad \textit{eventually}. \end{aligned}$$

(2.32)

Then

• (a) for any $L\in ((1+c_{2})M,(1+c_{1})N )$ there exist $\theta\in(0,1)$ and $n_{1}\ge n_{0}+\tau+\vert \beta \vert$ such that for each $x_{0}=\{x_{0}(n)\}_{n\in\Bbb {Z}_{\beta}}\in A(N,M)$, the Mann iterative sequence $\{x_{m}\}_{m\in\Bbb {N}_{0}}=\{x_{m}(n)\}_{(n,m)\in\Bbb {Z}_{\beta}\times\Bbb {N}_{0}}$ generated by (2.27) converges to a bounded positive solution $x \in A(N,M)$ of Eq. (1.11) and has the error estimate (2.6), where $\{\alpha_{m}\}_{m\in\Bbb {N}_{0}}$ is an arbitrary sequence in $[0,1]$ satisfying (2.7);

• (b) Eq. (1.11) possesses uncountable bounded positive solutions.

Proof

Let $L\in ((1+c_{2})M,(1+c_{1})N )$. It follows from (2.4) and (2.32) that there exist $\theta\in(0,1)$ and $n_{1}\geq n_{0}+\tau+\vert \beta \vert$ satisfying

$$\begin{aligned}& c_{1}\leq c(n)\leq c_{2}< -1,\quad n\ge n_{1}; \end{aligned}$$

(2.33)

$$\begin{aligned}& \theta=-\frac{1}{c_{2}}-\frac{1}{c_{2}}\sum_{t=n_{1}}^{\infty} \frac {1}{\vert a(t)\vert }\sum_{s=t}^{\infty}(s-t+1)P(s); \end{aligned}$$

(2.34)

$$\begin{aligned}& \sum_{t=n_{1}}^{\infty}\frac{1}{\vert a(t)\vert }\sum _{s=t}^{\infty }(s-t+1) \bigl(Q(s)+ \bigl\vert d(s) \bigr\vert \bigr) \\& \quad \leq \min \biggl\{ L-(1+c_{2})M, \frac{c_{2}(1+c_{1})N}{c_{1}}- \frac{c_{2}L}{c_{1}} \biggr\} . \end{aligned}$$

(2.35)

Let the mapping $T_{L}: A(N,M)\rightarrow l_{\beta}^{\infty}$ be defined by (2.31). It follows from (2.2), (2.3), (2.31) and (2.33)~(2.35) that for $x=\{x(n)\}_{n\in\Bbb {Z}_{\beta}},y=\{y(n)\}_{n\in\Bbb {Z}_{\beta}}\in A(N,M)$ and $n\ge n_{1}$

$$\begin{aligned}& \bigl\vert T_{L}x(n)-T_{L}y(n) \bigr\vert \\& \quad = \Biggl\vert -\frac{x(n+\tau)-y(n+\tau)}{c(n+\tau)} \\& \qquad {} +\frac{1}{c(n+\tau)}\sum_{t=n+\tau}^{\infty} \frac{1}{a(t)}\sum_{s=t}^{\infty}(s-t+1) \bigl[f \bigl(s,x \bigl(s-b_{1}(s) \bigr),x \bigl(s-b_{2}(s) \bigr),\ldots,x \bigl(s-b_{k}(s) \bigr) \bigr) \\& \qquad {} -f \bigl(s,y \bigl(s-b_{1}(s) \bigr),y \bigl(s-b_{2}(s) \bigr), \ldots,y \bigl(s-b_{k}(s) \bigr) \bigr) \bigr] \Biggr\vert \\& \quad \leq-\frac{\Vert x-y\Vert }{c_{2}}-\frac{1}{c_{2}}\sum _{t=n+\tau}^{\infty} \frac {1}{\vert a(t)\vert }\sum _{s=t}^{\infty}(s-t+1) P(s) \\& \qquad {}\times \max \bigl\{ \bigl\vert x \bigl(s-b_{j}(s) \bigr)-y \bigl(s-b_{j}(s) \bigr) \bigr\vert : 1\leq j\leq k \bigr\} \\& \quad \leq-\frac{\Vert x-y\Vert }{c_{2}} -\frac{1}{c_{2}}\sum _{t=n_{1}}^{\infty} \frac{1}{\vert a(t)\vert } \sum _{s=t}^{\infty}(s-t+1)P(s) \Vert x-y\Vert \\& \quad =\theta \Vert x-y\Vert , \\& T_{L}x(n)=\frac{L}{c(n+\tau)}-\frac{x(n+\tau)}{c(n+\tau)} + \frac{1}{c(n+\tau)} \sum_{t=n+\tau}^{\infty} \frac{1}{a(t)}\sum _{s=t}^{\infty}(s-t+1) \\& \hphantom{T_{L}x(n)=}{} \times \bigl[f \bigl(s,x \bigl(s-b_{1}(s) \bigr),x \bigl(s-b_{2}(s) \bigr),\ldots,x \bigl(s-b_{k}(s) \bigr) \bigr)-d(s) \bigr] \\& \hphantom{T_{L}x(n)}\leq\frac{L}{c_{2}}-\frac{M}{c_{2}}-\frac{1}{c_{2}}\sum _{t=n_{1}}^{\infty}\frac {1}{\vert a(t)\vert } \sum _{s=t}^{\infty}(s-t+1) \bigl(Q(s)+ \bigl\vert d(s) \bigr\vert \bigr) \\& \hphantom{T_{L}x(n)}\leq\frac{L}{c_{2}}-\frac{M}{c_{2}}-\frac{1}{c_{2}} \min \biggl\{ L-(1+c_{2})M,\frac{c_{2}(1+c_{1})N}{c_{1}}-\frac{c_{2}L}{c_{1}} \biggr\} \\& \hphantom{T_{L}x(n)}\leq M \end{aligned}$$

and

$$\begin{aligned} T_{L}x(n) =&\frac{L}{c(n+\tau)}-\frac{x(n+\tau)}{c(n+\tau)} + \frac{1}{c(n+\tau)} \sum_{t=n+\tau}^{\infty} \frac{1}{a(t)}\sum_{s=t}^{\infty}(s-t+1) \\ &{} \times \bigl[f \bigl(s,x \bigl(s-b_{1}(s) \bigr),x \bigl(s-b_{2}(s) \bigr),\ldots,x \bigl(s-b_{k}(s) \bigr) \bigr)-d(s) \bigr] \\ \geq&\frac{L}{c_{1}}-\frac{N}{c_{1}}+\frac{1}{c_{2}}\sum _{t=n_{1}}^{\infty}\frac {1}{\vert a(t)\vert } \sum _{s=t}^{\infty}(s-t+1) \bigl(Q(s)+ \bigl\vert d(s) \bigr\vert \bigr) \\ \geq&\frac{L}{c_{1}}-\frac{N}{c_{1}}+\frac{1}{c_{2}} \min \biggl\{ L-(1+c_{2})M,\frac{c_{2}(1+c_{1})N}{c_{1}}-\frac{c_{2}L}{c_{1}} \biggr\} \\ \geq& N, \end{aligned}$$

which implies (2.12). The rest of the proof is similar to that of Theorem 2.4 and is omitted. This completes the proof. □

3 Examples

In this section, we construct five examples to illustrate our results.

Example 3.1

Consider the third order nonlinear neutral delay difference equation

$$\begin{aligned}& \Delta^{2} \bigl( \bigl(n^{2}-n^{\frac{3}{2}}+2 \bigr) \Delta \bigl(x(n)+x(n-\tau) \bigr) \bigr) +\frac{(-1)^{n}[x^{2}(n^{2})+x^{4}(n^{3}-2n+1)]}{(n+1)^{3}+n\ln^{2}n+x^{2}(n^{2}-n)} \\& \quad =\frac{2n-(n+3)\sin(5n^{4}-3n+1)}{n^{4}-n^{2}+1},\quad n\geq1, \end{aligned}$$

(3.1)

where $\tau\in\Bbb {N}$ is fixed. Let $n_{0}=1$, $k=3$, $M=2$, $N=1$, $\beta=\min\{1-\tau,0\}$,

$$\begin{aligned}& a(n)=n^{2}-n^{\frac{3}{2}}+2, \qquad c(n)=1,\qquad b_{1}(n)=n-n^{2}, \\& b_{2}(n)=3n-n^{3}-1,\qquad b_{3}(n)=2n-n^{2}, \\& d(n)=\frac{2n-(n+3)\sin(5n^{4}-3n+1)}{n^{4}-n^{2}+1},\qquad f(n,u_{1},u_{2},u_{3})= \frac{(-1)^{n}(u_{1}^{2}+u_{2}^{4})}{(n+1)^{3}+n\ln^{2}n+u_{3}^{2}}, \\& P(n)=\frac{6M^{5}+4M^{3}+2M(1+2M^{2})[(n+1)^{3}+n\ln^{2}n]}{((n+1)^{3}+n\ln^{2}n+N^{2})^{2}}, \\& Q(n)=\frac{M^{4}+M^{2}}{(n+1)^{3}+n\ln^{2} n+N^{2}}, \quad (n,u_{1},u_{2},u_{3}) \in\Bbb {N}_{n_{0}} \times\Bbb {R}^{3}. \end{aligned}$$

It is clear that (2.1)~(2.4) hold. It follows from Theorem 2.1 that Eq. (3.1) possesses uncountably many bounded positive solutions in $A(N,M)$ and for each $L\in(N,M)$, there exist $\theta\in(0,1)$ and $n_{1}\geq n_{0}+\tau+\vert \beta \vert$ such that for each $x_{0}\in A(N,M)$, the Mann iterative sequence $\{x_{m}\}_{m\geq0}$ generated by (2.5) and (2.7) converges to a bounded positive solution $x\in A(N,M)$ of Eq. (3.1) and has the error estimate (2.6).

Example 3.2

Consider the third order nonlinear neutral delay difference equation

$$\begin{aligned}& \Delta^{2} \bigl( \bigl(n^{3}+n \ln^{2}n+1 \bigr) \Delta \bigl(x(n)-x(n-\tau) \bigr) \bigr) + \frac{(-1)^{n+1}x^{3} (\frac{n(n+1)}{2} )}{ n^{4}+n^{3}x^{4}(n-3)+\sqrt {n^{5}+1}x^{2}(n^{2}-2)} \\& \quad =\frac{3n^{4}-n^{2}-(-1)^{n-1}n+2}{n^{7}-n^{6}+n^{3}+1},\quad n\geq1, \end{aligned}$$

(3.2)

where $\tau\in\Bbb {N}$ is fixed. Let $n_{0}=1$, $k=3$, $M=\frac{3}{2}$, $N=\frac{1}{2}$, $\beta=\min\{1-\tau,-2\}$,

$$\begin{aligned}& a(n)=n^{3}+n\ln^{2} n+1,\qquad c(n)=-1,\qquad b_{1}(n)= \frac{-n(n-1)}{2}, \\& b_{2}(n)=3,\qquad b_{3}(n)=-n^{2}+n+2, \\& d(n)=\frac{3n^{4}-n^{2}-(-1)^{n-1}n+2}{n^{7}-n^{6}+n^{3}+1},\qquad f(n,u_{1},u_{2},u_{3})= \frac{(-1)^{n+1}u_{1}^{3}}{n^{4}+n^{3}u_{2}^{4}+\sqrt {n^{5}+1}u_{3}^{2}}, \\& P(n)=\frac{M^{2}(3n^{4}+7M^{4}n^{3}+5M^{2}\sqrt{n^{5}+1})}{(n^{4}+n^{3}N^{4}+\sqrt{n^{5}+1}N^{2})^{2}}, \\& Q(n)=\frac{M^{3}}{n^{4}+n^{3}N^{4}+\sqrt{n^{5}+1}N^{2}}, \quad (n,u_{1},u_{2},u_{3}) \in\Bbb {N}_{n_{0}} \times\Bbb {R}^{3}. \end{aligned}$$

It is easy to verify that (2.2), (2.3), (2.13) and (2.14) are fulfilled. It follows from Theorem 2.2 that Eq. (3.2) possesses uncountably many bounded positive solutions in $A(N,M)$ and for each $L\in(N,M)$, there exist $\theta\in(0,1)$ and $n_{1}\geq n_{0}+\tau+\vert \beta \vert$ such that for each $x_{0}\in A(N,M)$, the Mann iterative sequence $\{x_{m}\}_{m\geq0}$ generated by (2.7) and (2.15) converges to a bounded positive solution $x\in A(N,M)$ of Eq. (3.2) and has the error estimate (2.6).

Example 3.3

Consider the third order nonlinear neutral delay difference equation

$$\begin{aligned}& \Delta^{2} \biggl( \biggl(\frac{n^{3}}{\ln n}-n\ln n+n^{2} \biggr)\Delta \biggl(x(n)+\frac{(-1)^{n}n}{3\sqrt{n^{2}+n}+n}x(n-\tau) \biggr) \biggr) + \frac{\sqrt{n}x^{2}(3n^{2}-2)}{n^{3}+1} \\& \quad {}+\frac{ (n^{\frac{3}{2}}-\ln n+3)x(n-8)x^{3}(n^{5}-1)}{(n^{2}+n-1)^{2}}=\frac{n^{2}-3\ln n+2}{4n^{5}-3n^{3}+n^{2}+5n+1},\quad n\geq2, \end{aligned}$$

(3.3)

where $\tau\in\Bbb {N}_{n_{0}}$ is fixed. Let $n_{0}=2$, $k=3$, $M=5$, $N=2$, $c_{1}=c_{2}=\frac{1}{4}$, $\beta=\min\{2-\tau,-6\}$,

$$\begin{aligned}& a(n)=\frac{n^{3}}{\ln n}-n\ln n+n^{2}, \qquad b_{1}(n)=-3n^{2}+n+2, \\& b_{2}(n)=8,\qquad b_{3}(n)=-n^{5}+n+1, \\& c(n)=\frac{(-1)^{n}n}{3\sqrt{n^{2}+n}+n}, \qquad d(n)=\frac{n^{2}-3\ln n+2}{4n^{5}-3n^{3}+n^{2}+5n+1}, \\& f(n,u_{1},u_{2},u_{3})=\frac{\sqrt{n}u_{1}^{2}}{n^{3}+1}+ \frac{ (n^{\frac {3}{2}}-\ln n+3)u_{2}u_{3}^{3}}{(n^{2}+n-1)^{2}}, \\& P(n)=\frac{2M\sqrt{n}}{n^{3}+1}+\frac{4M^{3} (n^{\frac{3}{2}}-\ln n+3)}{(n^{2}+n-1)^{2}}, \\& Q(n)=\frac{M^{2}\sqrt{n}}{n^{3}+1}+\frac{M^{4} (n^{\frac{3}{2}}-\ln n+3)}{(n^{2}+n-1)^{2}}, \quad (n,u_{1},u_{2},u_{3}) \in\Bbb {N}_{n_{0}}\times \Bbb {R}^{3}. \end{aligned}$$

Clearly (2.2)~(2.4) and (2.20) hold. It follows from Theorem 2.3 that Eq. (3.3) possesses uncountably many bounded positive solutions in $A(N,M)$ and for every $L\in (c_{1}M+N, (1-c_{2})M )$, there exist $\theta\in(0,1)$ and $n_{1}\geq n_{0}+\tau+\vert \beta \vert$ such that for each $x_{0}\in A(N,M)$, the Mann iterative sequence $\{x_{m}\}_{m\geq0}$ generated by (2.7) and (2.21) converges to a bounded positive solution $x\in A(N,M)$ of Eq. (3.3) and has the error estimate (2.6).

Example 3.4

Consider the third order nonlinear neutral delay difference equation

$$\begin{aligned}& \Delta^{2} \biggl(\frac{(-1)^{n-1}(2n^{5}-7n^{3}+11)}{\ln^{2}(n+2)+6n}\Delta \biggl(x(n) + \frac{3n^{2}+2n+4}{n^{2}+n+1}x(n-\tau) \biggr) \biggr)+\frac {x^{2}(n^{2}+n)}{n^{3}+2n+1} \\& \quad {}+\frac{x^{3}(n-(-1)^{n})}{n^{2}\ln^{2}(n+1)}=\frac{(-1)^{n}(n-\sqrt{n}+3)}{2n^{4}+3n^{2}-1},\quad n\geq1, \end{aligned}$$

(3.4)

where $\tau\in\Bbb {N}$ is fixed. Let $n_{0}=1$, $k=2$, $M=\frac{7}{3}$, $N=\frac{3}{7}$, $c_{1}=2$, $c_{2}=3$, $\beta=\min\{1-\tau,1\}=1-\tau$,

$$\begin{aligned}& a(n)=\frac{(-1)^{n-1}(2n^{5}-7n^{3}+11)}{\ln^{2}(n+2)+6n}, \qquad c(n)=\frac{3n^{2}+2n+4}{n^{2}+n+1}, \\& b_{1}(n)=-n^{2},\qquad b_{2}(n)=(-1)^{n}, \\& d(n)=\frac{(-1)^{n}(n-\sqrt{n}+3)}{2n^{4}+3n^{2}-1},\qquad f(n,u_{1},u_{2})= \frac{u_{1}^{2}}{n^{3}+2n+1}+\frac{u_{2}^{3}}{n^{2}\ln^{2} (n+1)}, \\& P(n)=\frac{2M}{n^{3}+2n+1}+\frac{3M^{2}}{n^{2}\ln^{2} (n+1)}, \\& Q(n)=\frac{M^{2}}{n^{3}+2n+1}+ \frac{M^{3}}{n^{2}\ln^{2} (n+1)}, \quad (n,u_{1},u_{2}) \in\Bbb {N}_{n_{0}}\times\Bbb {R}^{2}. \end{aligned}$$

Obviously (2.2)~(2.4) and (2.26) hold. It follows from Theorem 2.4 that Eq. (3.4) possesses uncountably many bounded positive solutions in $A(N,M)$ and for any $L\in (\frac{c_{2}}{c_{1}}M+c_{2}N ,\frac{c_{1}}{c_{2}}N+c_{1}M )$, there exist $\theta\in(0,1)$ and $n_{1}\geq n_{0}+\tau+\vert \beta \vert$ such that for each $x_{0}\in A(N,M)$, the Mann iterative sequence $\{x_{m}\}_{m\geq0}$ generated by (2.7) and (2.27) converges to a bounded positive solution $x\in A(N,M)$ of Eq. (3.4) and has the error estimate (2.6).

Example 3.5

Consider the third order nonlinear neutral delay difference equation

$$\begin{aligned}& \Delta^{2} \biggl(n^{2} \bigl((-1)^{n}n-3 \bigr) \Delta \biggl(x(n)-\frac {5n+(-1)^{n}n}{2n+1}x(n-\tau) \biggr) \biggr) \\& \quad {} +\frac{(-1)^{n} x (\frac{n(n-1)}{2} )x(n^{2})}{n^{3}\ln^{2} n+n^{2}+x^{2} (\frac{n(n+1)(n+2)}{3} )}=\frac{n^{2}-(-1)^{n-1}n\ln (1+\sqrt{2n+1})}{n^{5}-2n^{3}+3},\quad n\geq1, \end{aligned}$$

(3.5)

where $\tau\in\Bbb {N}$ is fixed. Let $n_{0}=1$, $k=3$, $M=10$, $N=3$, $c_{1}=-3$, $c_{2}=-2$, $\beta=\min\{1-\tau,0\}$,

$$\begin{aligned}& a(n)=n^{2} \bigl((-1)^{n}n-3 \bigr),\qquad c(n)=- \frac{5n+(-1)^{n}n}{2n+1}, \\& b_{1}(n)=\frac{n(3-n)}{2},\qquad b_{2}(n)=-n^{2}+n, \\& b_{3}(n)=\frac{-n(n^{2}+3n-1)}{3},\qquad d(n)=\frac{n^{2}-(-1)^{n-1}n\ln(1+\sqrt {2n+1})}{n^{5}-2n^{3}+3}, \\& f(n,u_{1},u_{2},u_{3})=\frac{(-1)^{n}u_{1}u_{2}}{n^{3}\ln^{2} n+n^{2}+u_{3}^{2}}, \\& P(n)= \frac{4M^{3}+2M(n^{3}\ln^{2} n+n^{2})}{(n^{3}\ln^{2} n+n^{2}+N^{2})^{2}}, \\& Q(n)=\frac{M^{2}}{n^{3}\ln^{2} n+n^{2}+N^{2}},\quad (n,u_{1},u_{2},u_{3}) \in \Bbb {N}_{n_{0}}\times\Bbb {R}^{3}. \end{aligned}$$

It is easy to verify that (2.2)~(2.4) and (2.32) are fulfilled. It follows from Theorem 2.5 that Eq. (3.5) possesses uncountably many bounded positive solutions in $A(N,M)$ and for each $L\in ((1+c_{2})M,(1+c_{1})N )$, there exist $\theta\in (0,1)$ and $n_{1}\geq n_{0}+\tau+\vert \beta \vert$ such that for each $x_{0}\in A(N,M)$, the Mann iterative sequence $\{x_{m}\}_{m\geq0}$ generated by (2.7) and (2.27) converges to a bounded positive solution $x\in A(N,M)$ of Eq. (3.5) and has the error estimate (2.6).