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Positive solutions and convergence of Mann iterative schemes for a fourth order neutral delay difference equation

Advances in Difference Equations20152015:178

https://doi.org/10.1186/s13662-015-0513-8

Received: 26 January 2015

Accepted: 19 May 2015

Published: 12 June 2015

Abstract

The existence of uncountably many positive solutions and convergence of Mann iterative schemes for a fourth order neutral delay difference equation are proved. Seven examples are included.

Keywords

fourth order neutral delay difference equationpositive solutionsMann iterative methodsBanach fixed point theorem

MSC

39A10

1 Introduction and preliminaries

This paper is concerned with the following fourth order neutral delay difference equation
$$ \begin{aligned}[b] &\Delta\bigl(a_{n}\Delta^{3}(x_{n}+b_{n}x_{n-\tau}) \bigr)+\Delta h(n,x_{h_{1n}},x_{h_{2n}},\ldots,x_{h_{kn}}) \\ &\quad {}+f(n,x_{f_{1n}},x_{f_{2n}},\ldots,x_{f_{kn}})=c_{n}, \quad \forall n\ge n_{0}, \end{aligned} $$
(1.1)
where \(\tau,k,n_{0}\in\Bbb{N}\), \(\{a_{n}\}_{n\in\Bbb{N}_{n_{0}}}\subset\Bbb{R}\setminus\{0\}\), \(\{b_{n}\}_{n\in\Bbb{N}_{n_{0}}},\{c_{n}\}_{n\in\Bbb{N}_{n_{0}}}\subset\Bbb{R}\), \(h,f\in C(\Bbb{N}_{n_{0}}\times\Bbb{R}^{k},\Bbb{R})\), \(\{h_{ln}\}_{n\in\Bbb{N}_{n_{0}}}\), \(\{f_{ln}\}_{n\in\Bbb{N}_{n_{0}}}\subseteq\Bbb{N}\) and
$$\lim_{n\to\infty}h_{ln}=\lim_{n\to\infty}f_{ln}=+ \infty,\quad l\in\{1,2,\ldots,k\}. $$
Over the past several decades, a lot of researchers paid much attention to the problems of oscillation, nonoscillation, asymptotic behavior and existence of solutions for some second and third order difference equations, see, for example, [114] and the references cited therein. In particular, the researchers [58, 12] used fixed point theorems to study the existence of bounded nonoscillatory solutions and positive solutions for the following second and third order nonlinear neutral delay difference equations
$$\begin{aligned}& \Delta^{3}x_{n}+f(n,x_{n},x_{n-\tau})=0, \quad \forall n\geq n_{0}, \\& \Delta^{2}(x_{n}+b_{n}x_{n-\tau})+ \Delta h(n,x_{h_{1n}},x_{h_{2n}},\ldots,x_{h_{kn}}) +f(n,x_{f_{1n}},x_{f_{2n}},\ldots,x_{f_{kn}})=c_{n}, \quad \forall n\ge n_{0}, \\& \Delta\bigl(a_{n}\Delta(x_{n}+b_{n}x_{n-\tau}) \bigr)+ \Delta h(n,x_{h_{1n}},x_{h_{2n}},\ldots,x_{h_{kn}}) \\& \quad{} +f(n,x_{f_{1n}},x_{f_{2n}},\ldots,x_{f_{kn}})=c_{n}, \quad \forall n\ge n_{0}, \\& \Delta\bigl(a_{n}\Delta^{2}(x_{n}+p_{n}x_{n-\tau}) \bigr) +f(n,x_{n-d_{1n}},x_{n-d_{2n}},\ldots,x_{n-d_{ln}})=g_{n}, \quad \forall n\ge n_{0} \end{aligned}$$
and
$$\Delta^{3}(x_{n}+b_{n}x_{n-\tau})+ \Delta h(n,x_{h_{1n}},x_{h_{2n}},\ldots,x_{h_{kn}})+f (n,x_{f_{1n}},x_{f_{2n}},\ldots,x_{f_{kn}})=c_{n}, \quad \forall n\ge n_{0}. $$

The main purpose of this paper is to utilize the Banach fixed point theorem and some new techniques to establish the existence of uncountably many positive solutions of Eq. (1.1). Not only do we construct a few Mann iterative algorithms for approximating these positive solutions, but we also prove convergence and the error estimates of the Mann iterative algorithms relative to these positive solutions. Moreover, seven nontrivial examples are given to illustrate our results.

Throughout this paper, we assume that Δ is the forward difference operator defined by \(\Delta x_{n}=x_{n+1}-x_{n}\), \(\Bbb{R}=(-\infty, +\infty)\), \(\Bbb{R}^{+}=[0, +\infty)\), \(\Bbb{N}_{0}\) and \(\Bbb{N}\) denote the sets of all nonnegative integers and positive integers, respectively,
$$\begin{aligned}& \Bbb{N}_{t}=\{n:n\in\Bbb{N} \text{ with } n\ge t\},\quad \forall t\in \Bbb{N}, \\& \beta=\min\bigl\{ n_{0}-\tau,\inf\{h_{ln},f_{ln}:1 \le l\le k,n\in\Bbb{N}_{n_{0}}\}\bigr\} \in\Bbb{N}, \\& H_{n}=\max\bigl\{ h^{2}_{ln}:l\in\{1,2,\ldots,k\} \bigr\} ,\qquad F_{n}=\max\bigl\{ f^{2}_{ln}:l\in\{1,2, \ldots,k\}\bigr\} ,\quad \forall n\in\Bbb{N}_{n_{0}}, \end{aligned}$$
\(l_{\beta}^{\infty}\) represents the Banach space of all real sequences \(x=\{x_{n}\}_{n\in\Bbb{N}_{\beta}}\) in \(\Bbb{N}_{\beta}\) with norm
$$\Vert x\Vert =\sup_{n\in\Bbb{N}_{\beta}}\biggl\vert \frac{x_{n}}{n^{2}} \biggr\vert < +\infty\quad \text{for each } x=\{x_{n}\}_{n\in\Bbb{N}_{\beta}} \in l_{\beta}^{\infty} $$
and
$$A(N,M)=\biggl\{ x=\{x_{n}\}_{n\in\Bbb{N}_{\beta}}\in l_{\beta}^{\infty} : N\le\frac{x_{n}}{n^{2}}\le M, n\in\Bbb{N}_{\beta}\biggr\} \quad \text{for any } M>N>0. $$
It is clear that \(A(N,M)\) is a closed and convex subset of \(l_{\beta}^{\infty}\). By a solution of Eq. (1.1), we mean a sequence \(\{x_{n}\}_{n\in\Bbb{N}_{\beta}}\) with a positive integer \(T\ge n_{0}+\tau+\beta\) such that Eq. (1.1) holds for all \(n\ge T\).

Lemma 1.1

Let \(\{p_{t}\}_{t\in\Bbb{N}}\) be a nonnegative sequence and \(n,\tau\in\Bbb{N}\). Then
$$\begin{aligned}& \sum_{v=n}^{\infty}\sum _{u=v}^{\infty}\sum_{t=u}^{\infty}p_{t} \le\sum_{t=n}^{\infty}t^{2}p_{t}; \end{aligned}$$
(1.2)
$$\begin{aligned}& \sum_{v=n}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty} \sum_{t=s}^{\infty}p_{t}\le\sum _{t=n}^{\infty}t^{3}p_{t}; \end{aligned}$$
(1.3)
$$\begin{aligned}& \sum_{i=1}^{\infty}\sum _{u=n+i\tau}^{\infty}\sum_{s=u}^{\infty} \sum_{t=s}^{\infty}p_{t} \le \frac{1}{\tau}\sum_{t=n+\tau}^{\infty}t^{3}p_{t}; \end{aligned}$$
(1.4)
$$\begin{aligned}& \sum_{i=1}^{\infty}\sum _{v=n+i\tau}^{\infty}\sum_{u=v}^{\infty} \sum_{s=u}^{\infty}\sum _{t=s}^{\infty}p_{t} \le\frac{1}{\tau}\sum _{t=n+\tau}^{\infty}t^{4}p_{t}. \end{aligned}$$
(1.5)

Proof

Note that
$$\begin{aligned} \sum_{v=n}^{\infty}\sum _{u=v}^{\infty}\sum_{t=u}^{\infty}p_{t} &=\sum_{v=n}^{\infty}\Biggl(\sum _{u=v}^{\infty}\sum_{t=u}^{\infty }p_{t} \Biggr) \\ &=\sum_{v=n}^{\infty}\Biggl(\sum _{t=v}^{\infty}p_{t}+\sum _{t=v+1}^{\infty}p_{t} +\sum _{t=v+2}^{\infty}p_{t}+\cdots\Biggr) \\ &=\sum_{v=n}^{\infty}\sum _{t=v}^{\infty}(t-v+1)p_{t}\le\sum _{v=n}^{\infty }\sum_{t=v}^{\infty}tp_{t} =\sum_{t=n}^{\infty}(t-n+1)tp_{t} \\ &\le\sum_{t=n}^{\infty}t^{2}p_{t} \end{aligned}$$
and
$$\begin{aligned} \sum_{i=1}^{\infty}\sum _{u=n+i\tau}^{\infty}\sum_{s=u}^{\infty} \sum_{t=s}^{\infty}p_{t} &=\sum _{i=1}^{\infty}\sum_{u=n+i\tau}^{\infty} \Biggl(\sum_{s=u}^{\infty }\sum _{t=s}^{\infty}p_{t}\Biggr) \\ &=\sum_{i=1}^{\infty}\sum _{u=n+i\tau}^{\infty}\sum_{t=u}^{\infty}(t-u+1)p_{t} \le\sum_{i=1}^{\infty}\sum _{u=n+i\tau}^{\infty}\sum_{t=u}^{\infty}tp_{t} \\ &=\sum_{i=1}^{\infty}\Biggl(\sum _{t=n+i\tau}^{\infty}tp_{t}+\sum _{t=n+1+i\tau}^{\infty}tp_{t} +\sum _{t=n+2+i\tau}^{\infty}tp_{t}+\cdots\Biggr) \\ &=\sum_{i=1}^{\infty}\sum _{t=n+i\tau}^{\infty}(t-n-i\tau+1)tp_{t} \le\sum _{i=1}^{\infty}\sum_{t=n+i\tau}^{\infty}t^{2}p_{t} \\ &=\sum_{t=n+\tau}^{\infty}t^{2}p_{t}+ \sum_{t=n+2\tau}^{\infty}t^{2}p_{t} +\sum_{t=n+3\tau}^{\infty}t^{2}p_{t}+ \cdots \\ &\le\sum_{t=n+\tau}^{\infty}\biggl( \frac{t-n-\tau}{\tau}+1\biggr)t^{2}p_{t} =\sum _{t=n+\tau}^{\infty}\frac{t-n}{\tau}t^{2}p_{t} \\ &\le\frac{1}{\tau}\sum_{t=n+\tau}^{\infty}t^{3}p_{t}, \end{aligned}$$
which imply (1.2) and (1.4), respectively. It follows from (1.2) and (1.4) that
$$\begin{aligned} \sum_{v=n}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty} \sum_{t=s}^{\infty}p_{t} &=\sum _{v=n}^{\infty}\Biggl(\sum _{u=v}^{\infty}\sum_{s=u}^{\infty} \sum_{t=s}^{\infty}p_{t}\Biggr) \le \sum_{v=n}^{\infty}\sum _{t=v}^{\infty}t^{2}p_{t}=\sum _{t=n}^{\infty }(t-n+1)t^{2}p_{t} \\ &\le\sum_{t=n}^{\infty}t^{3}p_{t} \end{aligned}$$
and
$$\begin{aligned} \sum_{i=1}^{\infty}\sum _{v=n+i\tau}^{\infty}\sum_{u=v}^{\infty} \sum_{s=u}^{\infty}\sum _{t=s}^{\infty}p_{t} &=\sum _{i=1}^{\infty}\sum_{v=n+i\tau}^{\infty} \sum_{u=v}^{\infty} \Biggl(\sum _{s=u}^{\infty}\sum_{t=s}^{\infty}p_{t} \Biggr) \\ &=\sum_{i=1}^{\infty}\sum _{v=n+i\tau}^{\infty}\sum_{u=v}^{\infty} \sum_{t=u}^{\infty}(t-u-1)p_{t} \\ &\le\sum_{i=1}^{\infty}\sum _{v=n+i\tau}^{\infty}\sum_{u=v}^{\infty} \sum_{t=u}^{\infty}tp_{t} \\ &\le\frac{1}{\tau}\sum_{t=n+\tau}^{\infty}t^{4}p_{t}, \end{aligned}$$
which yields (1.3) and (1.5), respectively. This completes the proof. □

2 Uncountably many positive solutions and Mann iterative sequences

In this section, we discuss the existence of uncountably many positive solutions of Eq. (1.1) and prove convergence and the error estimates of the Mann iterative algorithms with respect to these positive solutions by using the Banach fixed point theorem.

Theorem 2.1

Assume that there exist two constants M and N with \(M>N>0\) and four nonnegative sequences \(\{P_{n}\}_{n\in\Bbb{N}_{n_{0}}}\), \(\{Q_{n}\}_{n\in\Bbb{N}_{n_{0}}}\), \(\{R_{n}\}_{n\in\Bbb{N}_{n_{0}}}\) and \(\{W_{n}\}_{n\in\Bbb{N}_{n_{0}}}\) satisfying
$$\begin{aligned}& \begin{aligned}[b] &\bigl\vert f(n,u_{1},u_{2}, \ldots,u_{k})-f(n,\bar{u}_{1},\bar{u}_{2},\ldots, \bar{u}_{k})\bigr\vert \le P_{n}\max\bigl\{ \vert u_{l}-\bar{u}_{l}\vert :1\le l\le k\bigr\} , \\ &\bigl\vert h(n,u_{1},u_{2},\ldots,u_{k})-h(n, \bar{u}_{1},\bar {u}_{2},\ldots,\bar{u}_{k})\bigr\vert \le R_{n}\max\bigl\{ \vert u_{l}- \bar{u}_{l}\vert :1\le l\le k\bigr\} , \\ &\quad \forall(n,u_{l},\bar{u}_{l})\in\Bbb{N}_{n_{0}}\times\bigl(\Bbb{R}^{+}\setminus\{0\}\bigr)^{2}, 1\le l\le k ; \end{aligned} \end{aligned}$$
(2.1)
$$\begin{aligned}& \begin{aligned}[b] &\bigl\vert f(n,u_{1},u_{2}, \ldots,u_{k})\bigr\vert \le Q_{n}\quad \textit{and}\quad \bigl\vert h(n,u_{1},u_{2},\ldots,u_{k})\bigr\vert \le W_{n}, \\ &\quad \forall(n,u_{l})\in\Bbb{N}_{n_{0}}\times\bigl(\Bbb{R}^{+} \setminus\{0\}\bigr), 1\le l\le k ; \end{aligned} \end{aligned}$$
(2.2)
$$\begin{aligned}& \lim_{n\to\infty}\frac{1}{n^{2}}\sum _{i=1}^{\infty}\sum_{v=n+i\tau}^{\infty}\sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \max \{R_{s}H_{s},W_{s}\}=0; \end{aligned}$$
(2.3)
$$\begin{aligned}& \lim_{n\to \infty}\frac{1}{n^{2}}\sum _{i=1}^{\infty}\sum_{v=n+i\tau}^{\infty}\sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\sum_{t=s}^{\infty}\frac{1}{\vert a_{s}\vert } \max\bigl\{ P_{t}F_{t},Q_{t}, \vert c_{t}\vert \bigr\} =0; \end{aligned}$$
(2.4)
$$\begin{aligned}& b_{n}=-1\quad \textit{eventually.} \end{aligned}$$
(2.5)
Then
  1. (a)
    for any \(L\in(N, M)\), there exist \(\theta\in(0, 1)\) and \(T\ge n_{0}+\tau+\beta\) such that for each \(x_{0}=\{x_{0n}\}_{n\in \Bbb{N}_{\beta}}\in A(N,M)\), the Mann iterative sequence \(\{x_{m}\}_{m\in\Bbb{N}_{0}}=\{\{x_{mn}\}_{n\in\Bbb{N}_{\beta}}\}_{m\in\Bbb{N}_{0}}\) generated by the scheme:
    $$\begin{aligned} x_{m+1n}= \textstyle\begin{cases} (1-\alpha_{m})x_{mn}+\alpha_{m}\{n^{2}L\\ \quad{} -\sum_{i=1}^{\infty}\sum_{v=n+i\tau}^{\infty}\sum_{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{a_{s}} [h(s,x_{mh_{1s}},x_{mh_{2s}},\ldots,x_{mh_{ks}})\\ \quad {}-\sum_{t=s}^{\infty}(f(t,x_{mf_{1t}},x_{mf_{2t}},\ldots ,x_{mf_{kt}})-c_{t})]\} ,\quad m\ge0, n\ge T,\\ (1-\alpha_{m})\frac{n^{2}}{T^{2}}x_{mT}+\alpha_{m}\frac{n^{2}}{T^{2}}\{T^{2}L\\ \quad {}-\sum_{i=1}^{\infty}\sum_{v=T+i\tau}^{\infty}\sum_{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{a_{s}} [h(s,x_{mh_{1s}},x_{mh_{2s}},\ldots,x_{mh_{ks}})\\ \quad {}-\sum_{t=s}^{\infty}(f(t,x_{mf_{1t}},x_{mf_{2t}},\ldots ,x_{mf_{kt}})-c_{t})]\} ,\quad m\ge0, \beta\le n< T \end{cases}\displaystyle \displaystyle \end{aligned}$$
    (2.6)
    converges to a positive solution \(w=\{w_{n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\) of Eq. (1.1) with
    $$ \lim_{n\to\infty}\frac{w_{n}}{n^{2}}=L\in(N, M) $$
    (2.7)
    and has the following error estimate:
    $$ \Vert x_{m+1}-w\Vert \le e^{-(1-\theta)\sum_{i=0}^{m}\alpha_{i}}\Vert x_{0}-w\Vert ,\quad \forall m\in\Bbb{N}_{0}, $$
    (2.8)
    where \(\{\alpha_{m}\}_{m\in\Bbb{N}_{0}}\) is an arbitrary sequence in \([0, 1]\) with
    $$ \sum^{\infty}_{m=0}\alpha_{m} =+ \infty; $$
    (2.9)
     
  2. (b)

    Equation (1.1) possesses uncountably many positive solutions in \(A(N,M)\).

     

Proof

In the first place we show that (a) holds. Set \(L\in(N, M)\). It follows from (2.3)-(2.5) that there exist \(\theta\in(0, 1)\) and \(T\ge n_{0}+\tau+\beta\) satisfying
$$\begin{aligned}& \theta=\frac{1}{T^{2}}\sum_{i=1}^{\infty} \sum_{v=T+i\tau}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl(R_{s}H_{s}+\sum _{t=s}^{\infty}P_{t}F_{t}\Biggr); \end{aligned}$$
(2.10)
$$\begin{aligned}& \frac{1}{T^{2}}\sum_{i=1}^{\infty}\sum _{v=T+i\tau}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{\vert a_{s}\vert }\Biggl(W_{s}+\sum_{t=s}^{\infty} \bigl(Q_{t}+\vert c_{t}\vert \bigr)\Biggr)< \min\{M-L, L-N\}; \end{aligned}$$
(2.11)
$$\begin{aligned}& b_{n}=-1,\quad \forall n\ge T. \end{aligned}$$
(2.12)
Define a mapping \(S_{L}: A(N,M)\to l_{\beta}^{\infty}\) by
$$ S_{L}x_{n}= \textstyle\begin{cases} n^{2}L-\sum_{i=1}^{\infty}\sum_{v=n+i\tau}^{\infty}\sum_{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{a_{s}} \{h(s,x_{h_{1s}},x_{h_{2s}},\ldots,x_{h_{ks}})\\ \quad {}-\sum_{t=s}^{\infty}[f(t,x_{f_{1t}},x_{f_{2t}}, \ldots,x_{f_{kt}})-c_{t}]\},\quad n\ge T,\\ \frac{n^{2}}{T^{2}}S_{L}x_{T},\quad \beta\le n< T \end{cases} $$
(2.13)
for each \(x=\{x_{n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\). By virtue of (2.1), (2.2), (2.10), (2.11) and (2.13), we gain that for each \(x=\{x_{n}\}_{n\in\Bbb{N}_{\beta}},y=\{y_{n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\)
$$\begin{aligned} &\biggl\vert \frac{S_{L}x_{n}}{n^{2}}-\frac{S_{L}y_{n}}{n^{2}}\biggr\vert \\ &\quad \le\frac{1}{n^{2}}\sum_{i=1}^{\infty}\sum _{v=n+i\tau}^{\infty}\sum_{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl( \bigl\vert h(s,x_{h_{1s}},x_{h_{2s}},\ldots,x_{h_{ks}}) -h(s,y_{h_{1s}},y_{h_{2s}},\ldots,y_{h_{ks}})\bigr\vert \\ &\qquad {}+\sum_{t=s}^{\infty}\bigl\vert f(t,x_{f_{1t}},x_{f_{2t}},\ldots ,x_{f_{kt}}) -f(t,y_{f_{1t}},y_{f_{2t}},\ldots,y_{f_{kt}})\bigr\vert \Biggr) \\ &\quad \le\frac{1}{n^{2}}\sum_{i=1}^{\infty}\sum _{v=n+i\tau}^{\infty}\sum_{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl(R_{s}\max\bigl\{ \vert x_{h_{ls}}-y_{h_{ls}} \vert :1\le l\le k\bigr\} \\ &\qquad {}+\sum_{t=s}^{\infty}P_{t} \max\bigl\{ \vert x_{f_{lt}}-y_{f_{lt}}\vert :1\le l\le k\bigr\} \Biggr) \\ &\quad \le\frac{\Vert x-y\Vert }{n^{2}}\sum_{i=1}^{\infty}\sum _{v=n+i\tau}^{\infty}\sum_{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl(R_{s}\max\bigl\{ h^{2}_{ls}:1\le l\le k \bigr\} \\ &\qquad {}+\sum_{t=s}^{\infty}P_{t}\max \bigl\{ f^{2}_{lt}:1\le l\le k\bigr\} \Biggr) \\ &\quad \le\frac{\Vert x-y\Vert }{T^{2}}\sum_{i=1}^{\infty}\sum _{v=T+i\tau}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl(R_{s}H_{s}+\sum _{t=s}^{\infty}P_{t}F_{t}\Biggr) \\ &\quad =\theta \Vert x-y\Vert ,\quad \forall n\ge T, \\ &\biggl\vert \frac{S_{L}x_{n}}{n^{2}}-\frac{S_{L}y_{n}}{n^{2}}\biggr\vert =\biggl\vert \frac {S_{L}x_{T}}{T^{2}}-\frac{S_{L}y_{T}}{T^{2}}\biggr\vert \le\theta \Vert x-y\Vert , \quad \beta\le n< T, \\ &\biggl\vert \frac{S_{L}x_{n}}{n^{2}}-L\biggr\vert \\ &\quad =\Biggl\vert \frac{1}{n^{2}}\sum_{i=1}^{\infty} \sum_{v=n+i\tau}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{a_{s}} \Biggl(h(s,x_{h_{1s}},x_{h_{2s}}, \ldots,x_{h_{ks}}) \\ &\qquad {}-\sum_{t=s}^{\infty} \bigl[f(t,x_{f_{1t}},x_{f_{2t}},\ldots ,x_{f_{kt}})-c_{t} \bigr]\Biggr)\Biggr\vert \\ &\quad \le\frac{1}{n^{2}}\sum_{i=1}^{\infty}\sum _{v=n+i\tau}^{\infty}\sum_{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl( \bigl\vert h(s,x_{h_{1s}},x_{h_{2s}},\ldots,x_{h_{ks}}) \bigr\vert \\ &\qquad {}+\sum_{t=s}^{\infty} \bigl[\bigl\vert f(t,x_{f_{1t}},x_{f_{2t}},\ldots,x_{f_{kt}}) \bigr\vert +\vert c_{t}\vert \bigr]\Biggr) \\ &\quad \le\frac{1}{T^{2}}\sum_{i=1}^{\infty}\sum _{v=T+i\tau}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl(W_{s}+\sum_{t=s}^{\infty} \bigl(Q_{t}+\vert c_{t}\vert \bigr)\Biggr) \\ &\quad < \min\{M-L, L-N\},\quad \forall n\ge T, \\ &\biggl\vert \frac{S_{L}x_{n}}{n^{2}}-L\biggr\vert =\biggl\vert \frac{S_{L}x_{T}}{T^{2}}-L \biggr\vert < \min\{M-L, L-N\},\quad \beta\le n< T, \end{aligned}$$
which yield that
$$ S_{L}\bigl(A(N,M)\bigr)\subseteq A(N,M),\qquad \Vert S_{L}x-S_{L}y \Vert \le\theta \Vert x-y\Vert ,\quad \forall x,y\in A(N,M), $$
(2.14)
which means that \(S_{L}\) is a contraction in \(A(N,M)\). Utilizing the Banach fixed point theorem, we conclude that \(S_{L}\) has a unique fixed point \(w=\{w_{n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\), that is,
$$\begin{aligned} w_{n}=S_{L}w_{n} ={}&n^{2}L-\sum _{i=1}^{\infty}\sum_{v=n+i\tau}^{\infty}\sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\frac{1}{a_{s}} \Biggl\{ h(s,w_{h_{1s}},w_{h_{2s}},\ldots,w_{h_{ks}}) \\ &{}-\sum_{t=s}^{\infty}\bigl[f(t,w_{f_{1t}},w_{f_{2t}}, \ldots ,w_{f_{kt}})-c_{t}\bigr]\Biggr\} ,\quad \forall n\geq T \end{aligned}$$
(2.15)
and
$$ w_{n}=S_{L}w_{n}=\frac{n^{2}}{T^{2}}S_{L}w_{T}= \frac{n^{2}}{T^{2}}w_{T},\quad \beta\le n< T. $$
(2.16)
It is obvious that (2.15) yields that
$$\begin{aligned}& \begin{aligned} w_{n-\tau} ={}&(n-\tau)^{2}L-\sum _{i=1}^{\infty}\sum_{v=n+(i-1)\tau}^{\infty}\sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\frac{1}{a_{s}} \Biggl\{ h(s,w_{h_{1s}},w_{h_{2s}},\ldots,w_{h_{ks}}) \\ &{}-\sum_{t=s}^{\infty}\bigl[f(t,w_{f_{1t}},w_{f_{2t}}, \ldots,w_{f_{kt}})-c_{t}\bigr]\Biggr\} ,\quad \forall n\ge T+ \tau, \end{aligned} \\& \begin{aligned} w_{n}-w_{n-\tau} ={}&\bigl(2n\tau- \tau^{2}\bigr) L+\sum_{v=n}^{\infty} \sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\frac {1}{a_{s}}\Biggl\{ h(s,w_{h_{1s}},w_{h_{2s}},\ldots,w_{h_{ks}}) \\ &{}-\sum_{t=s}^{\infty}\bigl[f(t,w_{f_{1t}},w_{f_{2t}}, \ldots ,w_{f_{kt}})-c_{t}\bigr]\Biggr\} ,\quad \forall n\ge T+ \tau, \end{aligned} \end{aligned}$$
which implies that
$$\begin{aligned}& \begin{aligned} \Delta(w_{n}-w_{n-\tau})={}&2\tau L-\sum _{u=n}^{\infty}\sum_{s=u}^{\infty}\frac{1}{a_{s}}\Biggl\{ h(s,w_{h_{1s}},w_{h_{2s}}, \ldots,w_{h_{ks}}) \\ &{}-\sum_{t=s}^{\infty}\bigl[f(t,w_{f_{1t}},w_{f_{2t}}, \ldots ,w_{f_{kt}})-c_{t}\bigr]\Biggr\} ,\quad \forall n\ge T+ \tau, \end{aligned} \\& \begin{aligned} \Delta^{2}(w_{n}-w_{n-\tau})={}& \sum_{s=n}^{\infty}\frac{1}{a_{s}}\Biggl\{ h(s,w_{h_{1s}},w_{h_{2s}},\ldots,w_{h_{ks}}) \\ &{}-\sum_{t=s}^{\infty}\bigl[f(t,w_{f_{1t}},w_{f_{2t}}, \ldots ,w_{f_{kt}})-c_{t}\bigr]\Biggr\} ,\quad \forall n\ge T+ \tau, \end{aligned} \\& \begin{aligned} a_{n}\Delta^{3}(w_{n}-w_{n-\tau})={}&{-}h(n,w_{h_{1n}},w_{h_{2n}}, \ldots ,w_{h_{kn}}) \\ &{}+\sum_{t=n}^{\infty}\bigl[f(t,w_{f_{1t}},w_{f_{2t}}, \ldots ,w_{f_{kt}})-c_{t}\bigr],\quad \forall n\ge T+\tau \end{aligned} \end{aligned}$$
and
$$\begin{aligned} \Delta\bigl(a_{n}\Delta^{3}(w_{n}-w_{n-\tau}) \bigr) ={}&{-}\Delta h(n,w_{h_{1n}},w_{h_{2n}},\ldots,w_{h_{kn}}) \\ &{}-f(n,w_{f_{1n}},w_{f_{2n}},\ldots,w_{f_{kn}})+c_{n}, \quad \forall n\ge T+\tau, \end{aligned} $$
which together with (2.12) means that \(w=\{w_{n}\}_{n\in\Bbb{N}_{\beta}}\) is a positive solution of Eq. (1.1) in \(A(N,M)\). It follows from (2.2)-(2.4) and (2.15) that
$$\begin{aligned} &\biggl\vert \frac{w_{n}}{n^{2}}-L\biggr\vert \\ &\quad =\Biggl\vert \frac{1}{n^{2}}\sum_{i=1}^{\infty} \sum_{v=n+i\tau}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{a_{s}} \Biggl(h(s,w_{h_{1s}},w_{h_{2s}}, \ldots,w_{h_{ks}}) \\ &\qquad {}-\sum_{t=s}^{\infty} \bigl[f(t,w_{f_{1t}},w_{f_{2t}},\ldots ,w_{f_{kt}})-c_{t} \bigr]\Biggr)\Biggr\vert \\ &\quad \le\frac{1}{n^{2}}\sum_{i=1}^{\infty} \sum_{v=n+i\tau}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl(\bigl\vert h(s,w_{h_{1s}},w_{h_{2s}}, \ldots,w_{h_{ks}}) \bigr\vert \\ &\qquad {}+\sum_{t=s}^{\infty} \bigl[\bigl\vert f(t,w_{f_{1t}},w_{f_{2t}},\ldots,w_{f_{kt}}) \bigr\vert +\vert c_{t}\vert \bigr]\Biggr) \\ &\quad \le\frac{1}{n^{2}}\sum_{i=1}^{\infty} \sum_{v=n+i\tau}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl(W_{s}+\sum_{t=s}^{\infty} \bigl(Q_{t}+\vert c_{t}\vert \bigr)\Biggr) \\ &\quad \to0\quad \text{as } n\to\infty, \end{aligned}$$
that is, (2.7) holds. It follows from (2.6), (2.10), (2.12) and (2.14)-(2.16) that
$$\begin{aligned} &\frac{\vert x_{m+1n}-w_{n}\vert }{n^{2}} \\ &\quad =\frac{1}{n^{2}}\Biggl\vert (1-\alpha_{m})x_{mn}+ \alpha_{m}\Biggl\{ n^{2}L -\sum_{i=1}^{\infty} \sum_{v=n+i\tau}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{a_{s}} \Biggl[h(s,x_{mh_{1s}},x_{mh_{2s}}, \ldots,x_{mh_{ks}}) \\ &\qquad {}-\sum_{t=s}^{\infty}\bigl[f(t,x_{mf_{1t}},x_{mf_{2t}}, \ldots ,x_{mf_{kt}})-c_{t}\bigr]\Biggr]\Biggr\} -w_{n} \Biggr\vert \\ &\quad \le(1-\alpha_{m})\frac{\vert x_{mn}-w_{n}\vert }{n^{2}}+\alpha_{m} \frac {\vert S_{L}x_{mn}-S_{L}w_{n}\vert }{n^{2}} \\ &\quad \le(1-\alpha_{m})\Vert x_{m}-w\Vert +\theta \alpha_{m}\Vert x_{m}-w\Vert \\ &\quad =\bigl[1-(1-\theta)\alpha_{m}\bigr]\Vert x_{m}-w\Vert \\ &\quad \le e^{-(1-\theta)\alpha_{m}}\Vert x_{m}-w\Vert ,\quad \forall m\in\Bbb{N}_{0}, n\ge T \end{aligned}$$
and
$$\begin{aligned} &\frac{\vert x_{m+1n}-w_{n}\vert }{n^{2}} \\ &\quad =\frac{1}{n^{2}}\Biggl\vert (1-\alpha_{m})\frac{n^{2}}{T^{2}}x_{mT}\\ &\qquad {}+ \alpha_{m}\frac {n^{2}}{T^{2}}\Biggl\{ T^{2}L -\sum _{i=1}^{\infty}\sum_{v=T+i\tau}^{\infty}\sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\frac{1}{a_{s}} \Biggl[h(s,x_{mh_{1s}},x_{mh_{2s}}, \ldots, x_{mh_{ks}}) \\ &\qquad {}-\sum_{t=s}^{\infty}\bigl[f(t,x_{mf_{1t}},x_{mf_{2t}}, \ldots ,x_{mf_{kt}})-c_{t}\bigr]\Biggr]\Biggr\} -w_{n} \Biggr\vert \\ &\quad \le(1-\alpha_{m})\frac{\vert x_{mT}-w_{T}\vert }{T^{2}}+\alpha_{m} \frac {\vert S_{L}x_{mT}-S_{L}w_{T}\vert }{T^{2}} \\ &\quad \le\bigl[1-(1-\theta)\alpha_{m}\bigr]\Vert x_{m}-w\Vert \\ &\quad \le e^{-(1-\theta)\alpha_{m}}\Vert x_{m}-w\Vert ,\quad \forall m\in\Bbb{N}_{0}, \beta\le n< T, \end{aligned}$$
which imply that
$$\Vert x_{m+1}-w\Vert \le e^{-(1-\theta)\alpha_{m}}\Vert x_{m}-w \Vert \le e^{-(1-\theta)\sum_{i=0}^{m}\alpha_{i}}\Vert x_{0}-w\Vert ,\quad \forall m\in\Bbb{N}_{0}, $$
that is, (2.8) holds. Thus (2.8) and (2.9) guarantee that \(\lim_{m\to\infty}x_{m}=w\).
In the next place we show that (b) holds. Let \(L_{1},L_{2}\in(N, M)\) and \(L_{1}\neq L_{2}\). As in the proof of (a), we deduce similarly that for each \(c\in\{1,2\}\), there exist constants \(\theta_{c}\in (0, 1)\), \(T_{c}\ge n_{0}+\tau+\beta\) and a mapping \(S_{L_{c}}\) satisfying (2.10)-(2.14), where θ, L and T are replaced by \(\theta_{c}\), \(L_{c}\) and \(T_{c}\), respectively, and the mapping \(S_{L_{c}}\) has a fixed point \(z^{c}=\{z^{c}_{n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\), which is a positive solution of Eq. (1.1) in \(A(N,M)\), that is,
$$\begin{aligned} z^{c}_{n}={}&n^{2}L_{c}-\sum _{i=1}^{\infty}\sum_{v=n+i\tau}^{\infty}\sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\frac{1}{a_{s}} \Biggl\{ h \bigl(s,z^{c}_{h_{1s}},z^{c}_{h_{2s}}, \ldots,z^{c}_{h_{ks}}\bigr) \\ &{}-\sum_{t=s}^{\infty}\bigl[f \bigl(t,z^{c}_{f_{1t}},z^{c}_{f_{2t}},\ldots ,z^{c}_{f_{kt}}\bigr)-c_{t}\bigr]\Biggr\} ,\quad \forall n \ge T_{c}, \end{aligned}$$
which together with (2.1), (2.10) and (2.12) implies that
$$\begin{aligned} &\biggl\vert \frac{z^{1}_{n}}{n^{2}}-\frac{z^{2}_{n}}{n^{2}}\biggr\vert \\ &\quad \ge \vert L_{1}-L_{2}\vert -\frac{1}{n^{2}}\sum _{i=1}^{\infty}\sum_{v=n+i\tau}^{\infty}\sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl(\bigl\vert h \bigl(s,z^{1}_{h_{1s}},z^{1}_{h_{2s}}, \ldots,z^{1}_{h_{ks}}\bigr) \\ &\qquad {}-h\bigl(s,z^{2}_{h_{1s}},z^{2}_{h_{2s}}, \ldots,z^{2}_{h_{ks}}\bigr)\bigr\vert +\sum_{t=s}^{\infty}\bigl\vert f \bigl(t,z^{1}_{f_{1t}},z^{1}_{f_{2t}},\ldots ,z^{1}_{f_{kt}}\bigr) -f\bigl(t,z^{2}_{f_{1t}},z^{2}_{f_{2t}}, \ldots,z^{2}_{f_{kt}}\bigr)\bigr\vert \Biggr) \\ &\quad \ge \vert L_{1}-L_{2}\vert -\frac{1}{n^{2}}\sum _{i={1}}^{\infty}\sum _{v=n+i\tau}^{\infty}\sum_{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl(R_{s}\max\bigl\{ \bigl\vert z^{1}_{h_{ls}}-z^{2}_{h_{ls}} \bigr\vert : 1\leq l\leq k\bigr\} \\ &\qquad {}+\sum_{t=s}^{\infty}P_{t}\max\bigl\{ \bigl\vert z^{1}_{f_{lt}}-z^{2}_{f_{lt}} \bigr\vert :1\leq l\leq k\bigr\} \Biggr) \\ &\quad \ge \vert L_{1}-L_{2}\vert -\frac{\Vert z^{1}-z^{2}\Vert }{n^{2}}\sum _{i={1}}^{\infty}\sum _{v=n+i\tau}^{\infty}\sum_{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl(R_{s}H_{s}+\sum_{t=s}^{\infty}P_{t}F_{t} \Biggr) \\ &\quad \ge \vert L_{1}-L_{2}\vert -\frac{\Vert z^{1}-z^{2}\Vert }{\max\{T_{1}^{2},T_{2}^{2}\}} \sum _{i={1}}^{\infty}\sum _{v=\max\{T_{1},T_{2}\}+i\tau}^{\infty}\sum_{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl(R_{s}H_{s}+\sum_{t=s}^{\infty}P_{t}F_{t} \Biggr) \\ &\quad \ge \vert L_{1}-L_{2}\vert -\max\{\theta_{1}, \theta_{2}\}\bigl\Vert z^{1}-z^{2}\bigr\Vert ,\quad \forall n\ge\max\{T_{1},T_{2}\}, \end{aligned}$$
which yields that
$$\bigl\Vert z^{1}-z^{2}\bigr\Vert \ge \frac{\vert L_{1}-L_{2}\vert }{1+\max\{\theta_{1},\theta_{2}\}}>0, $$
that is, \(z^{1}\neq z^{2}\). This completes the proof. □

Theorem 2.2

Assume that there exist two constants M and N with \(M>N>0\) and four nonnegative sequences \(\{P_{n}\}_{n\in\Bbb{N}_{n_{0}}}\), \(\{Q_{n}\}_{n\in\Bbb{N}_{n_{0}}}\), \(\{R_{n}\}_{n\in\Bbb{N}_{n_{0}}}\) and \(\{W_{n}\}_{n\in\Bbb{N}_{n_{0}}}\) satisfying (2.1), (2.2),
$$\begin{aligned}& \lim_{n\to \infty}\frac{1}{n^{2}}\sum _{v=n}^{\infty}\sum_{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{\vert a_{s}\vert }\max \{R_{s}H_{s},W_{s}\}=0; \end{aligned}$$
(2.17)
$$\begin{aligned}& \lim_{n\to \infty}\frac{1}{n^{2}}\sum _{v=n}^{\infty}\sum_{u=v}^{\infty}\sum_{s=u}^{\infty}\sum _{t=s}^{\infty}\frac{1}{\vert a_{s}\vert } \max\bigl\{ P_{t}F_{t},Q_{t},\vert c_{t} \vert \bigr\} =0; \end{aligned}$$
(2.18)
$$\begin{aligned}& b_{n}=1\quad \textit{eventually}. \end{aligned}$$
(2.19)
Then
  1. (a)
    for any \(L\in(N, M)\), there exist \(\theta\in(0, 1)\) and \(T\ge n_{0}+\tau+\beta\) such that for each \(x_{0}=\{x_{0n}\}_{n\in \Bbb{N}_{\beta}}\in A(N,M)\), the Mann iterative sequence \(\{x_{m}\}_{m\in\Bbb{N}_{0}}=\{\{x_{mn}\}_{n\in\Bbb{N}_{\beta}}\}_{m\in\Bbb{N}_{0}}\) generated by the scheme:
    $$\begin{aligned} x_{m+1n}= \textstyle\begin{cases} (1-\alpha_{m})x_{mn}+\alpha_{m}\{n^{2}L\\ \quad {}+\sum_{i=1}^{\infty}\sum_{v=n+(2i-1)\tau}^{n+2i\tau-1}\sum_{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{a_{s}} [h(s,x_{mh_{1s}},x_{mh_{2s}},\ldots,x_{mh_{ks}})\\ \quad {}-\sum_{t=s}^{\infty}(f(t,x_{mf_{1t}},x_{mf_{2t}},\ldots ,x_{mf_{kt}})-c_{t})]\},\quad m\ge0, n\ge T,\\ (1-\alpha_{m})\frac{n^{2}}{T^{2}}x_{mT}+\alpha_{m}\frac{n^{2}}{T^{2}}\{T^{2}L\\ \quad {}+\sum_{i=1}^{\infty}\sum_{v=T+(2i-1)\tau}^{T+2i\tau-1}\sum_{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{a_{s}} [h(s,x_{mh_{1s}},x_{mh_{2s}},\ldots,x_{mh_{ks}})\\ \quad {}-\sum_{t=s}^{\infty}(f(t,x_{mf_{1t}},x_{mf_{2t}},\ldots ,x_{mf_{kt}})-c_{t})]\},\quad m\ge0, \beta\le n< T \end{cases}\displaystyle \displaystyle \end{aligned}$$
    (2.20)
    converges to a positive solution \(w=\{w_{n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\) of Eq. (1.1) with (2.7) and has the error estimate (2.8), where \(\{\alpha_{m}\}_{m\in\Bbb{N}_{0}}\) is an arbitrary sequence in \([0, 1]\) satisfying (2.9);
     
  2. (b)

    Equation (1.1) possesses uncountably many positive solutions in \(A(N,M)\).

     

Proof

Set \(L\in(N, M)\). It follows from (2.17)-(2.19) that there exist \(\theta\in(0, 1)\) and \(T\ge n_{0}+\tau+\beta\) satisfying
$$\begin{aligned}& \theta=\frac{1}{T^{2}}\sum_{v=T}^{\infty}\sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl(R_{s}H_{s}+ \sum_{t=s}^{\infty}P_{t}F_{t} \Biggr); \end{aligned}$$
(2.21)
$$\begin{aligned}& \frac{1}{T^{2}}\sum_{v=T}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty}\frac {1}{\vert a_{s}\vert }\Biggl(W_{s}+\sum_{t=s}^{\infty} \bigl(Q_{t}+\vert c_{t}\vert \bigr)\Biggr)< \min\{M-L, L-N\}; \end{aligned}$$
(2.22)
$$\begin{aligned}& b_{n}=1,\quad \forall n\ge T. \end{aligned}$$
(2.23)
Define a mapping \(S_{L}: A(N,M)\to l_{\beta}^{\infty}\) by
$$ S_{L}x_{n}= \textstyle\begin{cases} n^{2}L+\sum_{i=1}^{\infty}\sum_{v=n+(2i-1)\tau}^{n+2i\tau-1}\sum_{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{a_{s}} \{h(s,x_{h_{1s}},x_{h_{2s}},\ldots,x_{h_{ks}})\\ \quad {}-\sum_{t=s}^{\infty}[f(t,x_{f_{1t}},x_{f_{2t}},\ldots ,x_{f_{kt}})-c_{t}]\},\quad n\ge T,\\ \frac{n^{2}}{T^{2}}S_{L}x_{T},\quad \beta\le n< T \end{cases} $$
(2.24)
for each \(x=\{x_{n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\). Using (2.1), (2.2), (2.21)-(2.24), we obtain that for each \(x=\{x_{n}\}_{n\in\Bbb{N}_{\beta}}\), \(y=\{y_{n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\)
$$\begin{aligned}& \biggl\vert \frac{S_{L}x_{n}}{n^{2}}-\frac{S_{L}y_{n}}{n^{2}}\biggr\vert \\& \quad \le\frac{1}{n^{2}}\sum_{i=1}^{\infty}\sum _{v=n+(2i-1)\tau}^{n+2i\tau -1}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl(\bigl\vert h(s,x_{h_{1s}},x_{h_{2s}}, \ldots,x_{h_{ks}}) -h(s,y_{h_{1s}},y_{h_{2s}}, \ldots,y_{h_{ks}})\bigr\vert \\& \qquad {}+\sum_{t=s}^{\infty}\bigl\vert f(t,x_{f_{1t}},x_{f_{2t}},\ldots ,x_{f_{kt}}) -f(t,y_{f_{1t}},y_{f_{2t}},\ldots,y_{f_{kt}})\bigr\vert \Biggr) \\& \quad \le\frac{\Vert x-y\Vert }{n^{2}}\sum_{i=1}^{\infty}\sum _{v=n+(2i-1)\tau }^{n+2i\tau-1}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl(R_{s}H_{s}+\sum _{t=s}^{\infty}P_{t}F_{t}\Biggr) \\& \quad \le\frac{\Vert x-y\Vert }{T^{2}}\sum_{v=T}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{\vert a_{s}\vert }\Biggl(R_{s}H_{s}+\sum _{t=s}^{\infty} P_{t}F_{t}\Biggr) \\& \quad =\theta \Vert x-y\Vert ,\quad \forall n\ge T, \\& \biggl\vert \frac{S_{L}x_{n}}{n^{2}}-\frac{S_{L}y_{n}}{n^{2}}\biggr\vert =\biggl\vert \frac{S_{L}x_{T}}{T^{2}}-\frac{S_{L}y_{T}}{T^{2}}\biggr\vert \le\theta \Vert x-y\Vert ,\quad \beta\le n< T, \\& \begin{aligned} \biggl\vert \frac{S_{L}x_{n}}{n^{2}}-L\biggr\vert \le{}&\frac{1}{n^{2}} \sum_{i=1}^{\infty}\sum _{v=n+(2i-1)\tau}^{n+2i\tau -1}\sum_{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl( \bigl\vert h(s,x_{h_{1s}},x_{h_{2s}},\ldots,x_{h_{ks}}) \bigr\vert \\ & {}+\sum_{t=s}^{\infty}\bigl[\bigl\vert f (t,x_{f_{1t}},x_{f_{2t}},\ldots,x_{f_{kt}})\bigr\vert + \vert c_{t}\vert \bigr]\Biggr) \\ \le{}&\frac{1}{T^{2}}\sum_{v=T}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl(W_{s}+\sum_{t=s}^{\infty} \bigl(Q_{t}+\vert c_{t}\vert \bigr)\Biggr) \\ < {}&\min\{M-L, L-N\},\quad \forall n\ge T \end{aligned} \end{aligned}$$
and
$$\biggl\vert \frac{S_{L}x_{n}}{n^{2}}-L\biggr\vert =\biggl\vert \frac{S_{L}x_{T}}{T^{2}}-L \biggr\vert < \min\{M-L, L-N\},\quad \beta\le n< T, $$
which mean (2.14). Consequently, (2.14) gives that \(S_{L}\) is a contraction in \(A(N,M)\) and has a unique fixed point \(w=\{w_{n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\), that is,
$$\begin{aligned} w_{n}=S_{L}w_{n} =n{}&^{2}L+\sum _{i=1}^{\infty}\sum_{v=n+(2i-1)\tau}^{n+2i\tau-1} \sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\frac{1}{a_{s}} \Biggl\{ h(s,w_{h_{1s}},w_{h_{2s}},\ldots,w_{h_{ks}}) \\ &{}-\sum_{t=s}^{\infty}\bigl[f(t,w_{f_{1t}},w_{f_{2t}}, \ldots ,w_{f_{kt}})-c_{t}\bigr]\Biggr\} ,\quad \forall n\ge T \end{aligned}$$
(2.25)
and (2.16) holds. It follows from (2.25) that
$$\begin{aligned} \Delta(w_{n}+w_{n-\tau})={}&(4n+2-2\tau)L-\sum _{u=n}^{\infty}\sum_{s=u}^{\infty} \frac{1}{a_{s}} \Biggl\{ h(s,w_{h_{1s}},w_{h_{2s}}, \ldots,w_{h_{ks}}) \\ &{}-\sum_{t=s}^{\infty}\bigl[f(t,w_{f_{1t}},w_{f_{2t}}, \ldots ,w_{f_{kt}})-c_{t}\bigr]\Biggr\} ,\quad \forall n\ge T+ \tau, \\ \Delta^{2}(w_{n}+w_{n-\tau})={}&4L+\sum _{s=n}^{\infty}\frac{1}{a_{s}} \Biggl\{ h(s,w_{h_{1s}},w_{h_{2s}},\ldots,w_{h_{ks}}) \\ &{}-\sum_{t=s}^{\infty}\bigl[f(t,w_{f_{1t}},w_{f_{2t}}, \ldots ,w_{f_{kt}})-c_{t}\bigr]\Biggr\} ,\quad \forall n\ge T+ \tau, \\ a_{n}\Delta^{3}(w_{n}+w_{n-\tau})={}&{-}h(n,w_{h_{1n}},w_{h_{2n}}, \ldots ,w_{h_{kn}}) \\ &{}+\sum_{t=n}^{\infty}\bigl[f(t,w_{f_{1t}},w_{f_{2t}}, \ldots ,w_{f_{kt}})-c_{t}\bigr],\quad \forall n\ge T+\tau \end{aligned}$$
and
$$\begin{aligned} \Delta\bigl(a_{n}\Delta^{3}(w_{n}+w_{n-\tau}) \bigr)={}&{-}\Delta h (n,w_{h_{1n}},w_{h_{2n}},\ldots,w_{h_{kn}}) \\ &{}-\bigl[f(n,w_{f_{1n}},w_{f_{2n}},\ldots,w_{f_{kn}})-c_{n} \bigr],\quad \forall n\ge T+\tau, \end{aligned}$$
that is, \(w=\{w_{n}\}_{n\in\Bbb{N}_{\beta}}\) is a positive solution of Eq. (1.1) in \(A(N,M)\). In terms of (2.2), (2.17), (2.18) and (2.25), we infer that
$$\begin{aligned} \biggl\vert \frac{w_{n}}{n^{2}}-L\biggr\vert =&\frac{1}{n^{2}}\Biggl\vert \sum_{i=1}^{\infty}\sum _{v=n+(2i-1)\tau}^{n+2i\tau -1}\sum_{u=v}^{\infty} \sum_{s=u}^{\infty}\frac{1}{a_{s}} \Biggl\{ h(s,w_{h_{1s}},w_{h_{2s}},\ldots,w_{h_{ks}}) \\ &{}-\sum_{t=s}^{\infty}\bigl[f(t,w_{f_{1t}},w_{f_{2t}}, \ldots ,w_{f_{kt}})-c_{t}\bigr]\Biggr\} \Biggr\vert \\ \le&\frac{1}{n^{2}}\sum_{i=1}^{\infty}\sum _{v=n+(2i-1)\tau}^{n+2i\tau -1}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty} \frac{1}{\vert a_{s}\vert } \Biggl(\bigl\vert h(s,w_{h_{1s}},w_{h_{2s}}, \ldots,w_{h_{ks}})\bigr\vert \\ &{}+\sum_{t=s}^{\infty}\bigl[\bigl\vert f(t,w_{f_{1t}},w_{f_{2t}},\ldots ,w_{f_{kt}})\bigr\vert + \vert c_{t}\vert \bigr]\Biggr) \\ \le&\frac{1}{n^{2}}\sum_{i=1}^{\infty}\sum _{v=n+(2i-1)\tau}^{n+2i\tau -1}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty} \frac{1}{\vert a_{s}\vert } \Biggl(W_{s}+\sum_{t=s}^{\infty} \bigl(Q_{t}+\vert c_{t}\vert \bigr)\Biggr) \\ \le&\frac{1}{n^{2}}\sum_{v=n}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty} \frac{1}{\vert a_{s}\vert } \Biggl(W_{s}+\sum_{t=s}^{\infty} \bigl(Q_{t}+\vert c_{t}\vert \bigr)\Biggr) \\ \to&0\quad \text{as } n\to\infty, \end{aligned}$$
that is, (2.7) holds. Linking (2.14), (2.16), (2.20), (2.21) and (2.25), we infer that
$$\begin{aligned} &\frac{\vert x_{m+1n}-w_{n}\vert }{n^{2}} \\ &\quad =\frac{1}{n^{2}}\Biggl\vert (1-\alpha_{m})x_{mn}+ \alpha_{m}\Biggl\{ n^{2}L\\ &\qquad {}+\sum_{i=1}^{\infty} \sum_{v=n+(2i-1)\tau}^{n+2i\tau-1} \sum _{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{a_{s}} \Biggl[h(s,x_{mh_{1s}},x_{mh_{2s}}, \ldots,x_{mh_{ks}}) \\ &\qquad {}-\sum_{t=s}^{\infty}\bigl[f(t,x_{mf_{1t}},x_{mf_{2t}}, \ldots ,x_{mf_{kt}})-c_{t}\bigr]\Biggr]\Biggr\} -w_{n} \Biggr\vert \\ &\quad \le(1-\alpha_{m})\frac{\vert x_{mn}-w_{n}\vert }{n^{2}}+\alpha_{m} \frac {\vert S_{L}x_{mn}-S_{L}w_{n}\vert }{n^{2}} \\ &\quad \le(1-\alpha_{m})\Vert x_{m}-w\Vert +\theta \alpha_{m}\Vert x_{m}-w\Vert \\ &\quad =\bigl[1-(1-\theta)\alpha_{m}\bigr]\Vert x_{m}-w\Vert \\ &\quad \le e^{-(1-\theta)\alpha_{m}}\Vert x_{m}-w\Vert ,\quad \forall m\in\Bbb{N}_{0}, n\ge T \end{aligned}$$
and
$$\begin{aligned} &\frac{\vert x_{m+1n}-w_{n}\vert }{n^{2}} \\ &\quad =\frac{1}{n^{2}}\Biggl\vert (1-\alpha_{m})\frac{n^{2}}{T^{2}}x_{mT}\\ &\qquad {}+ \alpha_{m}\frac {n^{2}}{T^{2}}\Biggl\{ T^{2}L +\sum _{i=1}^{\infty}\sum_{v=T+(2i-1)\tau}^{T+2i\tau-1} \sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\frac{1}{a_{s}} \Biggl[h(s,x_{mh_{1s}},x_{mh_{2s}}, \ldots,x_{mh_{ks}}) \\ &\qquad {}-\sum_{t=s}^{\infty}\bigl[f(t,x_{mf_{1t}},x_{mf_{2t}}, \ldots ,x_{mf_{kt}})-c_{t}\bigr]\Biggr]\Biggr\} -w_{n} \Biggr\vert \\ &\quad \le(1-\alpha_{m})\frac{\vert x_{mT}-w_{T}\vert }{T^{2}}+\alpha_{m} \frac {\vert S_{L}x_{mT}-S_{L}w_{T}\vert }{T^{2}} \\ &\quad \le\bigl[1-(1-\theta)\alpha_{m}\bigr]\Vert x_{m}-w\Vert \\ &\quad \le e^{-(1-\theta)\alpha_{m}}\Vert x_{m}-w\Vert ,\quad \forall m\in\Bbb{N}_{0}, \beta\le n< T, \end{aligned}$$
which imply that
$$\Vert x_{m+1}-w\Vert \le e^{-(1-\theta)\alpha_{m}}\Vert x_{m}-w \Vert \le e^{-(1-\theta)\sum_{i=0}^{m}\alpha_{i}}\Vert x_{0}-w\Vert ,\quad \forall m\in\Bbb{N}_{0}, $$
that is, (2.8) holds. It follows from (2.8) and (2.9) that \(\lim_{m\to\infty}x_{m}=w\).
Next we show that (b) holds. Let \(L_{1},L_{2}\in(N, M)\) and \(L_{1}\neq L_{2}\). Similar to the proof of (a), we get that for each \(c\in \{1,2\}\), there exist constants \(\theta_{c}\in(0, 1)\), \(T_{c}\ge n_{0}+\tau+\beta\) and a mapping \(S_{L_{c}}\) satisfying (2.21)-(2.24), where θ, L and T are replaced by \(\theta_{c}\), \(L_{c}\) and \(T_{c}\), respectively, and the mapping \(S_{L_{c}}\) has a fixed point \(z^{c}=\{z^{c}_{n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\), which is a positive solution of Eq. (1.1) in \(A(N,M)\), that is,
$$\begin{aligned} z^{c}_{n}={}&n^{2}L_{c}-\sum _{i=1}^{\infty}\sum_{v=n+(2i-1)\tau}^{n+2i\tau } \sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\frac{1}{a_{s}} \Biggl\{ h \bigl(s,z^{c}_{h_{1s}},z^{c}_{h_{2s}}, \ldots,z^{c}_{h_{ks}}\bigr) \\ &{}-\sum_{t=s}^{\infty}\bigl[f \bigl(t,z^{c}_{f_{1t}},z^{c}_{f_{2t}},\ldots ,z^{c}_{f_{kt}}\bigr)-c_{t}\bigr]\Biggr\} ,\quad \forall n \ge T_{c}, \end{aligned}$$
which together with (2.1), (2.10) and (2.23) implies that
$$\begin{aligned} &\biggl\vert \frac{z^{1}_{n}}{n^{2}}-\frac{z^{2}_{n}}{n^{2}}\biggr\vert \\ &\quad \ge \vert L_{1}-L_{2}\vert -\frac{1}{n^{2}}\sum _{i=1}^{\infty}\sum_{v=n+(2i-1)\tau }^{n+2i\tau} \sum_{u=v}^{\infty}\sum _{s=u}^{\infty} \frac{1}{\vert a_{s}\vert } \Biggl(\bigl\vert h \bigl(s,z^{1}_{h_{1s}},z^{1}_{h_{2s}}, \ldots,z^{1}_{h_{ks}}\bigr)\\ &\qquad {} -h\bigl(s,z^{2}_{h_{1s}},z^{2}_{h_{2s}}, \ldots,z^{2}_{h_{ks}}\bigr)\bigr\vert +\sum_{t=s}^{\infty}\bigl\vert f \bigl(t,z^{1}_{f_{1t}},z^{1}_{f_{2t}},\ldots ,z^{1}_{f_{kt}}\bigr) -f\bigl(t,z^{2}_{f_{1t}},z^{2}_{f_{2t}}, \ldots,z^{2}_{f_{kt}}\bigr)\bigr\vert \Biggr) \\ &\quad \ge \vert L_{1}-L_{2}\vert -\frac{1}{n^{2}}\sum _{i={1}}^{\infty}\sum _{v=n+(2i-1)\tau }^{n+2i\tau}\sum_{u=v}^{\infty}\sum_{s=u}^{\infty} \frac{1}{\vert a_{s}\vert } \Biggl(R_{s}\max\bigl\{ \bigl\vert z^{1}_{h_{ls}}-z^{2}_{h_{ls}} \bigr\vert : 1\leq l\leq k\bigr\} \\ &\qquad {}+\sum_{t=s}^{\infty}P_{t}\max\bigl\{ \bigl\vert z^{1}_{f_{lt}}-z^{2}_{f_{lt}} \bigr\vert :1\leq l\leq k\bigr\} \Biggr) \\ &\quad \ge \vert L_{1}-L_{2}\vert -\frac{\Vert z^{1}-z^{2}\Vert }{n^{2}}\sum _{i={1}}^{\infty}\sum _{v=n+(2i-1)\tau}^{n+2i\tau} \sum_{u=v}^{\infty} \sum_{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl(R_{s}H_{s}+\sum_{t=s}^{\infty}P_{t}F_{t} \Biggr) \\ &\quad \ge \vert L_{1}-L_{2}\vert -\frac{\Vert z^{1}-z^{2}\Vert }{\max\{T_{1}^{2},T_{2}^{2}\}}\sum _{v=\max\{T_{1},T_{2}\}}^{\infty} \sum _{u=v}^{\infty}\sum_{s=u}^{\infty} \frac{1}{\vert a_{s}\vert } \Biggl(R_{s}H_{s}+\sum _{t=s}^{\infty}P_{t}F_{t}\Biggr) \\ &\quad \ge \vert L_{1}-L_{2}\vert -\max\{\theta_{1}, \theta_{2}\}\bigl\Vert z^{1}-z^{2}\bigr\Vert ,\quad \forall n\ge\max\{T_{1},T_{2}\}, \end{aligned}$$
which yields that
$$\bigl\Vert z^{1}-z^{2}\bigr\Vert \ge \frac{\vert L_{1}-L_{2}\vert }{1+\max\{\theta_{1},\theta_{2}\}}>0, $$
that is, \(z^{1}\neq z^{2}\). This completes the proof. □

Theorem 2.3

Assume that there exist three constants b, M and N with \((1-b)M>N>0\) and four nonnegative sequences \(\{P_{n}\}_{n\in\Bbb{N}_{n_{0}}}\), \(\{Q_{n}\}_{n\in\Bbb{N}_{n_{0}}}\), \(\{R_{n}\}_{n\in\Bbb{N}_{n_{0}}}\) and \(\{W_{n}\}_{n\in\Bbb{N}_{n_{0}}}\) satisfying (2.1), (2.2), (2.17), (2.18) and
$$ 0\le b_{n}\le b< 1\quad \textit{eventually}. $$
(2.26)
Then
  1. (a)
    for any \(L\in(bM+N, M)\), there exist \(\theta\in (0, 1)\) and \(T\ge n_{0}+\tau+\beta\) such that for each \(x_{0}=\{x_{0n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\), the Mann iterative sequence \(\{x_{m}\}_{m\in\Bbb{N}_{0}}=\{\{x_{mn}\}_{n\in\Bbb{N}_{\beta}}\}_{m\in\Bbb{N}_{0}}\) generated by the scheme:
    $$\begin{aligned} x_{m+1n}= \textstyle\begin{cases} (1-\alpha_{m})x_{mn}+\alpha_{m}\{n^{2}L-b_{n}x_{mn-\tau}\\ \quad {}+\sum_{v=n}^{\infty}\sum_{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{a_{s}} [h(s,x_{mh_{1s}},x_{mh_{2s}},\ldots,x_{mh_{ks}})\\ \quad {}-\sum_{t=s}^{\infty}(f(t,x_{mf_{1t}},x_{mf_{2t}},\ldots ,x_{mf_{kt}})-c_{t})]\},\quad m\ge0, n\ge T,\\ (1-\alpha_{m})\frac{n^{2}}{T^{2}}x_{mT}+\alpha_{m}\frac{n^{2}}{T^{2}}\{ T^{2}L-b_{T}x_{mT-\tau}\\ \quad {}+\sum_{s=T}^{\infty}\frac{1}{a_{s}}[h (s,x_{mh_{1s}},x_{mh_{2s}},\ldots,x_{mh_{ks}})\\ \quad {}-\sum_{t=s}^{\infty}(f(t,x_{mf_{1t}},x_{mf_{2t}},\ldots ,x_{mf_{kt}})-c_{t})]\},\quad m\ge0, \beta\leq n< T \end{cases}\displaystyle \displaystyle \end{aligned}$$
    (2.27)
    converges to a positive solution \(w=\{w_{n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\) of Eq. (1.1) with
    $$ \lim_{n\to\infty}\frac{w_{n}+b_{n}w_{n-\tau}}{n^{2}}=L $$
    (2.28)
    and has the error estimate (2.8), where \(\{\alpha_{m}\}_{m\in\Bbb{N}_{0}}\) is an arbitrary sequence in \([0, 1]\) satisfying (2.9);
     
  2. (b)

    Equation (1.1) possesses uncountably many positive solutions in \(A(N,M)\).

     

Proof

Put \(L\in(bM+N, M)\). It follows from (2.17), (2.18) and (2.26) that there exist \(\theta\in(0, 1)\) and \(T\ge n_{0}+\tau+\beta\) satisfying
$$\begin{aligned}& \theta=b+\frac{1}{T^{2}}\sum_{v=T}^{\infty}\sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl(R_{s}H_{s}+ \sum_{t=s}^{\infty}P_{t}F_{t} \Biggr); \end{aligned}$$
(2.29)
$$\begin{aligned}& \frac{1}{T^{2}}\sum_{v=T}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty}\frac {1}{\vert a_{s}\vert }\Biggl(W_{s}+\sum_{t=s}^{\infty} \bigl(Q_{t}+\vert c_{t}\vert \bigr)\Biggr)< \min \{M-L,L-bM-N\}; \end{aligned}$$
(2.30)
$$\begin{aligned}& 0\le b_{n}\le b< 1,\quad \forall n\ge T. \end{aligned}$$
(2.31)
Define a mapping \(S_{L}: A(N,M)\to l_{\beta}^{\infty}\) by
$$ S_{L}x_{n}= \textstyle\begin{cases} n^{2}L-b_{n}x_{n-\tau}+\sum_{v=n}^{\infty}\sum_{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{a_{s}} \{h(s,x_{h_{1s}},x_{h_{2s}},\ldots,x_{h_{ks}})\\ \quad {}-\sum_{t=s}^{\infty}[f(t,x_{f_{1t}},x_{f_{2t}}, \ldots,x_{f_{kt}})-c_{t}]\},\quad n\ge T,\\ \frac{n^{2}}{T^{2}}S_{L}x_{T},\quad \beta\le n< T \end{cases} $$
(2.32)
for each \(x=\{x_{n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\). According to (2.1), (2.2) and (2.29)-(2.32), we obtain that for each \(x=\{x_{n}\}_{n\in\Bbb{N}_{\beta}}\), \(y=\{y_{n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\)
$$\begin{aligned}& \biggl\vert \frac{S_{L}x_{n}}{n^{2}}-\frac{S_{L}y_{n}}{n^{2}}\biggr\vert \\& \quad \le b_{n}\cdot\frac{(n-\tau)^{2}}{n^{2}}\biggl\vert \frac{x_{n-\tau}-y_{n-\tau }}{(n-\tau)^{2}} \biggr\vert \\& \qquad {}+\frac{1}{n^{2}}\sum_{v=n}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl(\bigl\vert h(s,x_{h_{1s}},x_{h_{2s}}, \ldots,x_{h_{ks}}) -h(s,y_{h_{1s}},y_{h_{2s}}, \ldots,y_{h_{ks}})\bigr\vert \\& \qquad {}+\sum_{t=s}^{\infty}\bigl\vert f(t,x_{f_{1t}},x_{f_{2t}},\ldots ,x_{f_{kt}}) -f(t,y_{f_{1t}},y_{f_{2t}},\ldots,y_{f_{kt}})\bigr\vert \Biggr) \\& \quad \le\Biggl[b+\frac{1}{T^{2}}\sum_{v=T}^{\infty}\sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl(R_{s}H_{s}+ \sum_{t=s}^{\infty}P_{t}F_{t} \Biggr)\Biggr]\Vert x-y\Vert \\& \quad =\theta \Vert x-y\Vert ,\quad \forall n\ge T, \\& \biggl\vert \frac{S_{L}x_{n}}{n^{2}}-\frac{S_{L}y_{n}}{n^{2}}\biggr\vert = \frac {n^{2}}{T^{2}}\biggl\vert \frac{S_{L}x_{T}}{n^{2}}-\frac{S_{L}y_{T}}{n^{2}}\biggr\vert \le\theta \Vert x-y\Vert ,\quad \beta\le n< T, \\& \begin{aligned} \frac{S_{L}x_{n}}{n^{2}} \le{}& L+\frac{1}{n^{2}}\sum _{v=n}^{\infty}\sum_{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl( \bigl\vert h(s,x_{h_{1s}},x_{h_{2s}},\ldots,x_{h_{ks}}) \bigr\vert \\ &{}+\sum_{t=s}^{\infty}\bigl[\bigl\vert f(t,x_{f_{1t}},x_{f_{2t}},\ldots ,x_{f_{kt}})\bigr\vert + \vert c_{t}\vert \bigr]\Biggr) \\ \le {}&L+\frac{1}{T^{2}}\sum_{v=T}^{\infty}\sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl(W_{s}+ \sum_{t=s}^{\infty}\bigl(Q_{t}+\vert c_{t}\vert \bigr)\Biggr) \\ < {}&L+\min\{M-L,L-bM-N\} \\ \le{}& M,\quad \forall n\ge T, \end{aligned} \\& \frac{S_{L}x_{n}}{n^{2}}=\frac{n^{2}}{T^{2}}\cdot\frac{S_{L}x_{T}}{n^{2}}\le M,\quad \beta \le n< T, \\& \begin{aligned} \frac{S_{L}x_{n}}{n^{2}} ={}&L-b_{n}\frac{x_{n-\tau}}{(n-\tau)^{2}}\cdot \frac{(n-\tau)^{2}}{n^{2}} +\frac{1}{n^{2}}\sum_{v=n}^{\infty}\sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\frac{1}{a_{s}} \Biggl(h(s,x_{h_{1s}},x_{h_{2s}}, \ldots,x_{h_{ks}}) \\ &{}-\sum_{t=s}^{\infty}\bigl[f(t,x_{f_{1t}},x_{f_{2t}}, \ldots,x_{f_{kt}})-c_{t}\bigr]\Biggr) \\ \ge {}&L-bM-\frac{1}{n^{2}}\sum_{v=n}^{\infty}\sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl(\bigl\vert h(s,x_{h_{1s}},x_{h_{2s}},\ldots,x_{h_{ks}})\bigr\vert \\ &{}+\sum_{t=s}^{\infty}\bigl[\bigl\vert f(t,x_{f_{1t}},x_{f_{2t}},\ldots ,x_{f_{kt}})\bigr\vert + \vert c_{t}\vert \bigr]\Biggr) \\ \ge {}&L-bM-\frac{1}{T^{2}}\sum_{v=T}^{\infty}\sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl(W_{s}+ \sum_{t=s}^{\infty}\bigl(Q_{t}+\vert c_{t}\vert \bigr)\Biggr) \\ >{}&L-bM-\min\{M-L,L-bM-N\} \\ \ge{}& N,\quad \forall n\ge T \end{aligned} \end{aligned}$$
and
$$\frac{S_{L}x_{n}}{n^{2}}=\frac{n^{2}}{T^{2}}\cdot\frac{S_{L}x_{T}}{n^{2}}\ge N,\quad \beta \le n< T, $$
which give (2.14), in turns, which implies that \(S_{L}\) is a contraction in \(A(N,M)\) and possesses a unique fixed point \(w=\{w_{n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\), that is,
$$\begin{aligned} w_{n}=S_{L}w_{n}={}& n^{2}L-b_{n}w_{n-\tau}+ \sum_{v=n}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty} \frac{1}{a_{s}} \Biggl\{ h(s,w_{h_{1s}},w_{h_{2s}}, \ldots,w_{h_{ks}}) \\ &{}-\sum_{t=s}^{\infty}\bigl[f(t,w_{f_{1t}},w_{f_{2t}}, \ldots ,w_{f_{kt}})-c_{t}\bigr]\Biggr\} ,\quad \forall n\ge T \end{aligned}$$
(2.33)
and (2.16) is satisfied. It is easy to verify that (2.33) yields that
$$\begin{aligned}& \begin{aligned} \Delta(w_{n}+b_{n}w_{n-\tau})={}&(2n+1)L-\sum _{u=n}^{\infty}\sum _{s=u}^{\infty }\frac{1}{a_{s}} \Biggl\{ h(s,w_{h_{1s}},w_{h_{2s}},\ldots,w_{h_{ks}}) \\ &{}-\sum_{t=s}^{\infty}\bigl[f(t,w_{f_{1t}},w_{f_{2t}}, \ldots ,w_{f_{kt}})-c_{t}\bigr]\Biggr\} ,\quad \forall n\ge T+\tau, \end{aligned} \\& \begin{aligned} \Delta^{2}(w_{n}+b_{n}w_{n-\tau})={}&2L+ \sum_{s=n}^{\infty}\frac{1}{a_{s}} \Biggl\{ h(s,w_{h_{1s}},w_{h_{2s}},\ldots,w_{h_{ks}}) \\ &{}-\sum_{t=s}^{\infty}\bigl[f(t,w_{f_{1t}},w_{f_{2t}}, \ldots ,w_{f_{kt}})-c_{t}\bigr]\Biggr\} ,\quad \forall n\ge T+\tau, \end{aligned} \\& \begin{aligned} a_{n}\Delta^{3}(w_{n}+b_{n}w_{n-\tau})={}&{-}h(n,w_{h_{1n}},w_{h_{2n}}, \ldots ,w_{h_{kn}}) \\ &{}+\sum_{t=n}^{\infty}\bigl[f(t,w_{f_{1t}},w_{f_{2t}}, \ldots ,w_{f_{kt}})-c_{t}\bigr],\quad \forall n\ge T+\tau \end{aligned} \end{aligned}$$
and
$$\begin{aligned}\Delta\bigl(a_{n}\Delta^{3}(w_{n}+b_{n}w_{n-\tau}) \bigr)={}&{-}\Delta h (n,w_{h_{1n}},w_{h_{2n}},\ldots,w_{h_{kn}}) \\ &{}-f(n,w_{f_{1n}},w_{f_{2n}},\ldots,w_{f_{kn}})+c_{n},\quad \forall n\ge T+\tau, \end{aligned} $$
that is, \(w=\{w_{n}\}_{n\in\Bbb{N}_{\beta}}\) is a positive solution of Eq. (1.1) in \(A(N,M)\). Making use of (2.17), (2.18) and (2.33), we infer that
$$\begin{aligned} \biggl\vert \frac{w_{n}+b_{n}w_{n-\tau}}{n^{2}}-L\biggr\vert \le&\frac{1}{n^{2}} \sum_{v=n}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty} \frac{1}{\vert a_{s}\vert } \Biggl(\bigl\vert h(s,w_{h_{1s}},w_{h_{2s}}, \ldots,w_{h_{ks}})\bigr\vert \\ &{}+\sum_{t=s}^{\infty}\bigl[\bigl\vert f(t,w_{f_{1t}},w_{f_{2t}},\ldots ,w_{f_{kt}})\bigr\vert + \vert c_{t}\vert \bigr]\Biggr) \\ \le&\frac{1}{n^{2}}\sum_{v=n}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty} \frac{1}{\vert a_{s}\vert } \Biggl(W_{s}+\sum_{t=s}^{\infty} \bigl(Q_{t}+\vert c_{t}\vert \bigr)\Biggr) \\ \to&0\quad \text{as } n\to\infty, \end{aligned}$$
which gives (2.28). In light of (2.14), (2.16), (2.27), (2.29) and (2.33), we deduce that
$$\begin{aligned} \frac{\vert x_{m+1n}-w_{n}\vert }{n^{2}} ={}&\frac{1}{n^{2}}\Biggl\vert (1- \alpha_{m})x_{mn}+\alpha_{m}\Biggl\{ n^{2}L-b_{n}x_{mn-\tau} \\ &{}+\sum_{v=n}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty} \frac{1}{a_{s}} \Biggl[h(s,x_{mh_{1s}},x_{mh_{2s}}, \ldots,x_{mh_{ks}}) \\ &{}-\sum_{t=s}^{\infty}\bigl[f(t,x_{mf_{1t}},x_{mf_{2t}}, \ldots ,x_{mf_{kt}})-c_{t}\bigr]\Biggr]\Biggr\} -w_{n} \Biggr\vert \\ \le{}&(1-\alpha_{m})\frac{\vert x_{mn}-w_{n}\vert }{n^{2}}+\alpha_{m} \frac {\vert S_{L}x_{mn}-S_{L}w_{n}\vert }{n^{2}} \\ \le{}&(1-\alpha_{m})\Vert x_{m}-w\Vert +\theta \alpha_{m}\Vert x_{m}-w\Vert =\bigl[1-(1-\theta)\alpha_{m}\bigr]\Vert x_{m}-w\Vert \\ \le{}& e^{-(1-\theta)\alpha_{m}}\Vert x_{m}-w\Vert ,\quad \forall m\in\Bbb{N}_{0}, n\ge T \end{aligned}$$
and
$$\begin{aligned} \frac{\vert x_{m+1n}-w_{n}\vert }{n^{2}} ={}&\frac{1}{n^{2}}\Biggl\vert (1- \alpha_{m})\frac{n^{2}}{T^{2}}x_{mT}+\alpha_{m} \frac {n^{2}}{T^{2}}\Biggl\{ T^{2}L-b_{T}x_{mT-\tau} \\ &{}+\sum_{v=T}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty} \frac{1}{a_{s}} \Biggl[h(s,x_{mh_{1s}},x_{mh_{2s}}, \ldots,x_{mh_{ks}}) \\ &{}-\sum_{t=s}^{\infty}\bigl[f(t,x_{mf_{1t}},x_{mf_{2t}}, \ldots ,x_{mf_{kt}})-c_{t}\bigr]\Biggr]\Biggr\} -w_{n} \Biggr\vert \\ \le{}&(1-\alpha_{m})\frac{\vert x_{mT}-w_{T}\vert }{T^{2}}+\alpha_{m} \frac {\vert S_{L}x_{mT}-S_{L}w_{T}\vert }{T^{2}} \\ \le{}&\bigl[1-(1-\theta)\alpha_{m}\bigr]\Vert x_{m}-w\Vert \\ \le{}& e^{-(1-\theta)\alpha_{m}}\Vert x_{m}-w\Vert ,\quad \forall m\in\Bbb{N}_{0}, \beta\le n< T, \end{aligned}$$
which imply that
$$\Vert x_{m+1}-w\Vert \le e^{-(1-\theta)\alpha_{m}}\Vert x_{m}-w \Vert \le e^{-(1-\theta)\sum_{i=0}^{m}\alpha_{i}}\Vert x_{0}-w\Vert ,\quad \forall m\in\Bbb{N}_{0}, $$
that is, (2.8) holds. It follows from (2.8) and (2.9) that \(\lim_{m\to\infty}x_{m}=w\).
Next we show that (b) holds. Let \(L_{1},L_{2}\in(bM+N, M)\) and \(L_{1}\neq L_{2}\). Similar to the proof of (a), we get that for each \(c\in\{1,2\}\) there exist constants \(\theta_{c}\in(0, 1)\), \(T_{c}\ge n_{0}+\tau+\beta\) and a mapping \(S_{L_{c}}\) satisfying (2.29)-(2.32), where θ, L and T are replaced by \(\theta_{c}\), \(L_{c}\) and \(T_{c}\), respectively, and the mapping \(S_{L_{c}}\) has a fixed point \(z^{c}=\{z^{c}_{n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\), which is a positive solution of Eq. (1.1) in \(A(N,M)\), that is,
$$\begin{aligned} z^{c}_{n}={}&n^{2}L_{c}-b_{n}z^{c}_{n-\tau}+ \sum_{v=n}^{\infty}\sum _{u=v}^{\infty }\sum_{s=u}^{\infty} \frac{1}{a_{s}} \Biggl\{ h\bigl(s,z^{c}_{h_{1s}},z^{c}_{h_{2s}}, \ldots,z^{c}_{h_{ks}}\bigr) \\ &{}-\sum_{t=s}^{\infty}\bigl[f \bigl(t,z^{c}_{f_{1t}},z^{c}_{f_{2t}},\ldots ,z^{c}_{f_{kt}}\bigr)-c_{t}\bigr]\Biggr\} ,\quad \forall n \ge T_{c}, \end{aligned}$$
which together with (2.1), (2.29) and (2.31) means that
$$\begin{aligned} &\biggl\vert \frac{z^{1}_{n}}{n^{2}}-\frac{z^{2}_{n}}{n^{2}}\biggr\vert \\ &\quad \ge \vert L_{1}-L_{2}\vert -b_{n} \frac{\vert z^{1}_{n}(n-\tau)-z^{2}_{n}(n-\tau)\vert }{(n-\tau)^{2}}\cdot \frac{(n-\tau)^{2}}{n^{2}} \\ &\qquad {}-\frac{1}{n^{2}}\sum_{v=n}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty} \frac{1}{\vert a_{s}\vert } \Biggl(\bigl\vert h\bigl(s,z^{1}_{h_{1s}},z^{1}_{h_{2s}}, \ldots,z^{1}_{h_{ks}}\bigr) -h\bigl(s,z^{2}_{h_{1s}},z^{2}_{h_{2s}}, \ldots,z^{2}_{h_{ks}}\bigr)\bigr\vert \\ &\qquad {}+\sum_{t=s}^{\infty}\bigl\vert f \bigl(t,z^{1}_{f_{1t}},z^{1}_{f_{2t}},\ldots ,z^{1}_{f_{kt}}\bigr) -f\bigl(t,z^{2}_{f_{1t}},z^{2}_{f_{2t}}, \ldots,z^{2}_{f_{kt}}\bigr)\bigr\vert \Biggr) \\ &\quad \ge \vert L_{1}-L_{2}\vert -b\bigl\Vert z^{1}-z^{2}\bigr\Vert -\frac{1}{n^{2}}\sum _{v=n}^{\infty }\sum_{u=v}^{\infty} \sum_{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl(R_{s}\max\bigl\{ \bigl\vert z^{1}_{h_{ls}}-z^{2}_{h_{ls}} \bigr\vert : 1\leq l\leq k\bigr\} \\ &\qquad {}+\sum_{t=s}^{\infty}P_{t}\max\bigl\{ \bigl\vert z^{1}_{f_{lt}}-z^{2}_{f_{lt}} \bigr\vert :1\leq l\leq k\bigr\} \Biggr) \\ &\quad \ge \vert L_{1}-L_{2}\vert -b\bigl\Vert z^{1}-z^{2}\bigr\Vert -\frac{\Vert z^{1}-z^{2}\Vert }{n^{2}}\sum _{v=n}^{\infty}\sum_{u=v}^{\infty} \sum_{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl(R_{s}H_{s}+\sum_{t=s}^{\infty}P_{t}F_{t} \Biggr) \\ &\quad \ge \vert L_{1}-L_{2}\vert -b\bigl\Vert z^{1}-z^{2}\bigr\Vert -\frac{\Vert z^{1}-z^{2}\Vert }{\max\{ T_{1}^{2},T_{2}^{2}\}} \sum _{v=\max\{T_{1},T_{2}\}}^{\infty}\sum_{u=v}^{\infty} \sum_{s=u}^{\infty }\frac{1}{\vert a_{s}\vert } \Biggl(R_{s}H_{s}+\sum_{t=s}^{\infty}P_{t}F_{t} \Biggr) \\ &\quad \ge \vert L_{1}-L_{2}\vert -\max\{\theta_{1}, \theta_{2}\}\bigl\Vert z^{1}-z^{2}\bigr\Vert ,\quad \forall n\ge\max\{T_{1},T_{2}\}, \end{aligned}$$
which yields that
$$\bigl\Vert z^{1}-z^{2}\bigr\Vert \ge \frac{\vert L_{1}-L_{2}\vert }{1+\max\{\theta_{1},\theta_{2}\}}>0, $$
that is, \(z^{1}\neq z^{2}\). This completes the proof. □

Theorem 2.4

Assume that there exist constants b, M and N with \((1+b)M>N>0\) and four nonnegative sequences \(\{P_{n}\}_{n\in\Bbb{N}_{n_{0}}}\), \(\{Q_{n}\}_{n\in\Bbb{N}_{n_{0}}}\), \(\{R_{n}\}_{n\in\Bbb{N}_{n_{0}}}\), \(\{W_{n}\}_{n\in\Bbb{N}_{n_{0}}}\) satisfying (2.1), (2.2), (2.17), (2.18) and
$$ -1< b\le b_{n}\le0\quad \textit{eventually}. $$
(2.34)
Then
  1. (a)

    for any \(L\in(N, (1+b)M)\), there exist \(\theta\in (0, 1)\) and \(T\ge n_{0}+\tau+\beta\) such that for each \(x_{0}=\{x_{0n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\), the Mann iterative sequence \(\{x_{m}\}_{m\in\Bbb{N}_{0}}=\{\{x_{mn}\}_{n\in\Bbb{N}_{\beta}}\}_{m\in\Bbb{N}_{0}}\) generated by (2.27) converges to a positive solution \(w=\{w_{n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\) of Eq. (1.1) with (2.28) and has the error estimate (2.8), where \(\{\alpha_{m}\}_{m\in\Bbb{N}_{0}}\) is an arbitrary sequence in \([0, 1]\) satisfying (2.9);

     
  2. (b)

    Equation (1.1) possesses uncountably many positive solutions in \(A(N,M)\).

     

Proof

Put \(L\in(N, (1+b)M)\). It follows from (2.17), (2.18) and (2.34) that there exist \(\theta\in(0, 1)\) and \(T\ge n_{0}+\tau+\beta\) satisfying
$$\begin{aligned}& \theta=-b+\frac{1}{T^{2}}\sum_{v=T}^{\infty}\sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl(R_{s}H_{s}+ \sum_{t=s}^{\infty}P_{t}F_{t} \Biggr); \end{aligned}$$
(2.35)
$$\begin{aligned}& \frac{1}{T^{2}}\sum_{v=T}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty}\frac {1}{\vert a_{s}\vert }\Biggl(W_{s}+\sum_{t=s}^{\infty} \bigl(Q_{t}+\vert c_{t}\vert \bigr)\Biggr)< \min\bigl\{ (1+b)M-L, L-N\bigr\} ; \end{aligned}$$
(2.36)
$$\begin{aligned}& -1< b\le b_{n}\le0,\quad \forall n\geq T. \end{aligned}$$
(2.37)
Define a mapping \(S_{L}: A(N,M)\to l_{\beta}^{\infty}\) by (2.32). By virtue of (2.2), (2.32), (2.36) and (2.37), we easily verify that
$$\begin{aligned}& \begin{aligned} \frac{S_{L}x_{n}}{n^{2}} \le{}& L-b_{n}\frac{x_{n-\tau}}{(n-\tau)^{2}}\cdot \frac{(n-\tau)^{2}}{n^{2}}+\frac{1}{n^{2}} \sum_{v=n}^{\infty}\sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl(\bigl\vert h(s,x_{h_{1s}},x_{h_{2s}},\ldots,x_{h_{ks}})\bigr\vert \\ &{}+\sum_{t=s}^{\infty}\bigl[\bigl\vert f(t,x_{f_{1t}},x_{f_{2t}},\ldots ,x_{f_{kt}})\bigr\vert + \vert c_{t}\vert \bigr]\Biggr) \\ \le{}& L-bM+\frac{1}{T^{2}}\sum_{v=T}^{\infty}\sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl(W_{s}+ \sum_{t=s}^{\infty}\bigl(Q_{t}+\vert c_{t}\vert \bigr)\Biggr) \\ < {}&L-bM+\min\bigl\{ (1+b)M-L, L-N\bigr\} \\ \le{}& M,\quad \forall n\geq T, \end{aligned} \\& \frac{S_{L}x_{n}}{n^{2}}=\frac{n^{2}}{T^{2}}\cdot\frac{S_{L}x_{T}}{n^{2}}\le M,\quad \beta \le n< T,\\& \begin{aligned} \frac{S_{L}x_{n}}{n^{2}} \ge {}&L-\frac{1}{n^{2}}\sum _{v=T}^{\infty}\sum_{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl( \bigl\vert h(s,x_{h_{1s}},x_{h_{2s}},\ldots,x_{h_{ks}}) \bigr\vert \\ &{} +\sum_{t=s}^{\infty}\bigl[\bigl\vert f(t,x_{f_{1t}},x_{f_{2t}},\ldots ,x_{f_{kt}})\bigr\vert +\vert c_{t}\vert \bigr]\Biggr) \\ \ge{}& L-\frac{1}{T^{2}}\sum_{v=T}^{\infty}\sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl(W_{s}+ \sum_{t=s}^{\infty}\bigl(Q_{t}+\vert c_{t}\vert \bigr)\Biggr) \\ >{}&L-\min\bigl\{ (1+b)M-L, L-N\bigr\} \\ \ge{}& N,\quad \forall n\geq T \end{aligned} \end{aligned}$$
and
$$\begin{aligned} \frac{S_{L}x_{n}}{n^{2}}=\frac{n^{2}}{T^{2}}\cdot\frac{S_{L}x_{T}}{n^{2}}\ge N,\quad \beta \le n< T, \end{aligned}$$
which yield that \(S_{L}(A(N,M))\subseteq A(N,M)\). The rest of the proof is similar to that of Theorem 2.3 and is omitted. This completes the proof. □

Theorem 2.5

Assume that there exist constants q, \(b_{*}\), \(b^{*}\), M and N and four nonnegative sequences \(\{P_{n}\}_{n\in\Bbb{N}_{n_{0}}}\), \(\{Q_{n}\}_{n\in\Bbb{N}_{n_{0}}}\), \(\{R_{n}\}_{n\in\Bbb{N}_{n_{0}}}\), \(\{W_{n}\}_{n\in\Bbb{N}_{n_{0}}}\) satisfying (2.1), (2.2), (2.17), (2.18) and
$$\begin{aligned}& q^{2}b^{*}< 1< b_{*}q,\qquad b^{*}(Mq+N)< \frac{M}{q}+\frac{N}{qb^{*}}, \end{aligned}$$
(2.38)
$$\begin{aligned}& 1< b_{*}\le b_{n}\le b^{*},\quad \textit{eventually}. \end{aligned}$$
(2.39)
Then
  1. (a)
    for any \(L\in(b^{*}(Mq+N), \frac{M}{q}+\frac{N}{qb^{*}})\), there exist \(\theta\in(0, 1)\) and \(T\ge n_{0}+\tau+\beta\) such that for each \(x_{0}=\{x_{0n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\), the Mann iterative sequence \(\{x_{m}\}_{m\in\Bbb{N}_{0}}=\{\{x_{mn}\}_{n\in\Bbb{N}_{\beta}}\}_{m\in\Bbb{N}_{0}}\) generated by the scheme:
    $$\begin{aligned} x_{m+1n}= \textstyle\begin{cases} (1-\alpha_{m})x_{mn}+\alpha_{m}\{\frac{(n+\tau)^{2}L}{b_{n+\tau}}-\frac {x_{mn+\tau}}{b_{n+\tau}}\\ \quad {}+\frac{1}{b_{n+\tau}}\sum_{v=n+\tau}^{\infty}\sum_{u=v}^{\infty }\sum_{s=u}^{\infty}\frac{1}{a_{s}} [h(s,x_{mh_{1s}},x_{mh_{2s}},\ldots,x_{mh_{ks}})\\ \quad {}-\sum_{t=s}^{\infty}(f(t,x_{mf_{1t}},x_{mf_{2t}},\ldots ,x_{mf_{kt}})-c_{t})]\},\quad m\ge0, n\ge T,\\ (1-\alpha_{m})\frac{n^{2}}{T^{2}}x_{mT}+\alpha_{m}\frac{n^{2}}{T^{2}}\{\frac {(T+\tau)^{2}L}{b_{T+\tau}}-\frac{x_{m{T+\tau}}}{b_{T+\tau}}\\ \quad {}+\sum_{v=T+\tau}^{\infty}\sum_{u=v}^{\infty}\sum_{s=u}^{\infty }\frac{1}{a_{s}} [h(s,x_{mh_{1s}},x_{mh_{2s}},\ldots,x_{mh_{ks}})\\ \quad {}-\sum_{t=s}^{\infty}(f(t,x_{mf_{1t}},x_{mf_{2t}},\ldots ,x_{mf_{kt}})-c_{t})]\},\quad m\ge0, \beta\leq n< T \end{cases}\displaystyle \displaystyle \end{aligned}$$
    (2.40)
    converges to a positive solution \(w=\{w_{n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\) of Eq. (1.1) with (2.28) and has the error estimate (2.8), where \(\{\alpha_{m}\}_{m\in\Bbb{N}_{0}}\) is an arbitrary sequence in \([0, 1]\) satisfying (2.9);
     
  2. (b)

    Equation (1.1) possesses uncountably many positive solutions in \(A(N,M)\).

     

Proof

Let \(L\in(b^{*}(Mq+N), \frac{M}{q}+\frac{N}{qb^{*}})\). It follows from (2.17), (2.18), (2.38) and (2.39) that there exist \(\theta\in(0, 1)\) and \(T\ge n_{0}+\tau+\beta\) satisfying
$$\begin{aligned}& \theta=q+\frac{1}{b_{*}T^{2}}\sum_{v=T}^{\infty} \sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl(R_{s}H_{s}+ \sum_{t=s}^{\infty}P_{t}F_{t} \Biggr); \end{aligned}$$
(2.41)
$$\begin{aligned}& \frac{1}{b_{*}T^{2}}\sum_{v=T}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty} \frac{1}{\vert a_{s}\vert }\Biggl(W_{s}+\sum_{t=s}^{\infty} \bigl(Q_{t}+\vert c_{t}\vert \bigr)\Biggr) \\& \quad < \min\biggl\{ M-qL+\frac{N}{b^{*}}, \frac {L}{b^{*}}-Mq-N\biggr\} ; \end{aligned}$$
(2.42)
$$\begin{aligned}& \biggl(1+\frac{\tau}{n}\biggr)^{2}< b_{*}q,\quad 1< b_{*} \le b_{n}\le b^{*},\forall n\ge T. \end{aligned}$$
(2.43)
Define a mapping \(S_{L}: A(N,M)\to l_{\beta}^{\infty}\) by
$$ S_{L}x_{n}= \textstyle\begin{cases} \frac{(n+\tau)^{2}L}{b_{n+\tau}}-\frac{x_{n+\tau}}{b_{n+\tau}}+\frac {1}{b_{n+\tau}} \sum_{v=n+\tau}^{\infty}\sum_{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{a_{s}} \{h(s,x_{h_{1s}},x_{h_{2s}},\ldots,x_{h_{ks}})\\ \quad {}-\sum_{t=s}^{\infty}[f(t,x_{f_{1t}},x_{f_{2t}},\ldots ,x_{f_{kt}})-c_{t}]\},\quad n\ge T,\\ \frac{n^{2}}{T^{2}}S_{L}x_{T},\quad \beta\le n< T \end{cases} $$
(2.44)
for each \(x=\{x_{n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\). On account of (2.1), (2.2) and (2.41)-(2.44), we ensure that for each \(x=\{x_{n}\}_{n\in\Bbb{N}_{\beta}}, y=\{y_{n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\)
$$\begin{aligned}& \begin{aligned} &\biggl\vert \frac{S_{L}x_{n}}{n^{2}}-\frac{S_{L}y_{n}}{n^{2}}\biggr\vert \\ &\quad \le\frac{1}{b_{n+\tau}}\cdot\frac{(n+\tau)^{2}}{n^{2}}\cdot\frac{ \vert x_{n+\tau}-y_{n+\tau} \vert }{(n+\tau)^{2}} \\ &\qquad {}+\frac{1}{b_{n+\tau}n^{2}}\sum_{v=n+\tau}^{\infty}\sum _{u=v}^{\infty }\sum_{s=u}^{\infty} \frac{1}{\vert a_{s}\vert } \Biggl(\bigl\vert h(s,x_{h_{1s}},x_{h_{2s}}, \ldots,x_{h_{ks}}) -h(s,y_{h_{1s}},y_{h_{2s}}, \ldots,y_{h_{ks}})\bigr\vert \\ &\qquad {}+\sum_{t=s}^{\infty}\bigl\vert f(t,x_{f_{1t}},x_{f_{2t}},\ldots ,x_{f_{kt}}) -f(t,y_{f_{1t}},y_{f_{2t}},\ldots,y_{f_{kt}})\bigr\vert \Biggr) \\ &\quad \le\frac{1}{b_{*}}\biggl(1+\frac{\tau}{T}\biggr)^{2}\Vert x-y\Vert \\ &\qquad {}+\frac{1}{b_{*}T^{2}}\sum_{v=T}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty} \frac{1}{\vert a_{s}\vert } \Biggl(R_{s}\max\bigl\{ \vert x_{h_{ls}}-y_{h_{ls}} \vert :1\le l\le k\bigr\} \\ &\qquad {} +\sum_{t=s}^{\infty}P_{t} \max\bigl\{ \vert x_{f_{lt}}-y_{f_{lt}}\vert :1\le l\le k\bigr\} \Biggr) \\ &\quad \le\Biggl[q+\frac{1}{b_{*}T^{2}}\sum_{v=T}^{\infty} \sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl(R_{s}H_{s}+ \sum_{t=s}^{\infty}P_{t}F_{t} \Biggr)\Biggr]\Vert x-y\Vert \\ &\quad =\theta \Vert x-y\Vert ,\quad \forall n\ge T, \end{aligned} \\& \biggl\vert \frac{S_{L}x_{n}}{n^{2}}-\frac{S_{L}y_{n}}{n^{2}}\biggr\vert = \frac{n^{2}}{T^{2}} \biggl\vert \frac{S_{L}x_{T}}{n^{2}}-\frac{S_{L}y_{T}}{n^{2}}\biggr\vert \le\theta \Vert x-y\Vert ,\quad \beta\le n< T, \\& \begin{aligned} \frac{S_{L}x_{n}}{n^{2}} ={}&\biggl(1+\frac{\tau}{n}\biggr)^{2} \frac{L}{b_{n+\tau}}-\frac{1}{b_{n+\tau }}\biggl(1+\frac{\tau}{n} \biggr)^{2}\frac{x_{n+\tau}}{(n+\tau)^{2}} \\ &{}+\frac{1}{b_{n+\tau}n^{2}}\sum_{v=n+\tau}^{\infty}\sum _{u=v}^{\infty }\sum_{s=u}^{\infty} \frac{1}{a_{s}} \Biggl(h(s,x_{h_{1s}},x_{h_{2s}}, \ldots,x_{h_{ks}}) \\ &{}-\sum_{t=s}^{\infty}\bigl[f(t,x_{f_{1t}},x_{f_{2t}}, \ldots ,x_{f_{kt}})-c_{t}\bigr]\Biggr) \\ \le{}&\biggl(1+\frac{\tau}{n}\biggr)^{2}\frac{L}{b_{*}}- \frac{N}{b^{*}} +\frac{1}{b_{*}n^{2}}\sum_{v=n+\tau}^{\infty} \sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl\{ \bigl\vert h(s,x_{h_{1s}},x_{h_{2s}},\ldots,x_{h_{ks}})\bigr\vert \\ &{}+\sum_{t=s}^{\infty}\bigl[\bigl\vert f(t,x_{f_{1t}},x_{f_{2t}},\ldots ,x_{f_{kt}})\bigr\vert + \vert c_{t}\vert \bigr]\Biggr\} \\ \le{}& qL-\frac{N}{b^{*}}+\frac{1}{b_{*}T^{2}}\sum_{v=T}^{\infty} \sum_{u=v}^{\infty}\sum _{s=u}^{\infty} \frac{1}{\vert a_{s}\vert }\Biggl(W_{s}+ \sum_{t=s}^{\infty} \bigl(Q_{t}+ \vert c_{t}\vert \bigr)\Biggr) \\ < {}&qL-\frac{N}{b^{*}}+\min\biggl\{ M-qL+\frac{N}{b^{*}}, \frac {L}{b^{*}}-Mq-N\biggr\} \\ \le {}&M,\quad \forall n\ge T, \end{aligned} \\& \frac{S_{L}x_{n}}{n^{2}} =\frac{n^{2}}{T^{2}}\cdot\frac{S_{L}x_{T}}{n^{2}}\le M,\quad \beta\le n< T, \\& \begin{aligned} \frac{S_{L}x_{n}}{n^{2}} \ge{}&\frac{L}{b^{*}}-\frac{M}{b_{*}}\biggl(1+ \frac{\tau}{n}\biggr)^{2} -\frac{1}{b_{*}n^{2}}\sum _{v=n+\tau}^{\infty}\sum_{u=v}^{\infty} \sum_{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl( \bigl\vert h(s,x_{h_{1s}},x_{h_{2s}},\ldots,x_{h_{ks}}) \bigr\vert \\ &{}+\sum_{t=s}^{\infty}\bigl[\bigl\vert f(t,x_{f_{1t}},x_{f_{2t}},\ldots ,x_{f_{kt}})\bigr\vert + \vert c_{t}\vert \bigr]\Biggr) \\ \ge{}&\frac{L}{b^{*}}-Mq-\frac{1}{b_{*}T^{2}}\sum_{v=T}^{\infty} \sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl(W_{s}+ \sum_{t=s}^{\infty}\bigl(Q_{t}+\vert c_{t}\vert \bigr)\Biggr) \\ >{}&\frac{L}{b^{*}}-Mq-\min\biggl\{ M-qL+\frac{N}{b^{*}}, \frac {L}{b^{*}}-Mq-N\biggr\} \\ \ge {}&N,\quad \forall n\ge T \end{aligned} \end{aligned}$$
and
$$\begin{aligned} \frac{S_{L}x_{n}}{n^{2}} =\frac{n^{2}}{T^{2}}\cdot\frac{S_{L}x_{T}}{n^{2}}\ge N,\quad \beta\le n< T, \end{aligned}$$
which mean (2.14). It follows from the Banach fixed point theorem that the contraction mapping \(S_{L}\) possesses a unique fixed point \(w=\{w_{n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\), that is,
$$\begin{aligned} w_{n}=S_{L}w_{n}={}& \frac{(n+\tau)^{2}}{b_{n+\tau}}L- \frac{w_{n+\tau}}{b_{n+\tau}}+\frac {1}{b_{n+\tau}} \sum_{v=n+\tau}^{\infty} \sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\frac{1}{a_{s}} \Biggl\{ h(s,w_{h_{1s}},w_{h_{2s}},\ldots,w_{h_{ks}}) \\ &{}-\sum_{t=s}^{\infty}\bigl[f(t,w_{f_{1t}},w_{f_{2t}}, \ldots ,w_{f_{kt}})-c_{t}\bigr]\Biggr\} ,\quad \forall n\ge T \end{aligned}$$
(2.45)
and (2.16) is satisfied. It is easy to verify that (2.45) yields that
$$\begin{aligned}& \begin{aligned}[b] w_{n}+b_{n}w_{n-\tau}={}&n^{2}L+\sum _{v=n}^{\infty}\sum_{u=v}^{\infty} \sum_{s=u}^{\infty}\frac{1}{a_{s}} \Biggl\{ h(s,w_{h_{1s}},w_{h_{2s}},\ldots,w_{h_{ks}}) \\ &{}-\sum_{t=s}^{\infty}\bigl[f(t,w_{f_{1t}},w_{f_{2t}}, \ldots ,w_{f_{kt}})-c_{t}\bigr]\Biggr\} ,\quad \forall n\ge T+\tau, \end{aligned} \\& \begin{aligned} \Delta(w_{n}+b_{n}w_{n-\tau})={}&(2n+1)L-\sum _{u=n}^{\infty}\sum _{s=u}^{\infty }\frac{1}{a_{s}} \Biggl\{ h(s,w_{h_{1s}},w_{h_{2s}},\ldots,w_{h_{ks}}) \\ &{}-\sum_{t=s}^{\infty}\bigl[f(t,w_{f_{1t}},w_{f_{2t}}, \ldots ,w_{f_{kt}})-c_{t}\bigr]\Biggr\} ,\quad \forall n\ge T+\tau, \end{aligned} \\& \begin{aligned} \Delta^{2}(w_{n}+b_{n}w_{n-\tau})={}&2L+ \sum_{s=n}^{\infty}\frac{1}{a_{s}} \Biggl\{ h(s,w_{h_{1s}},w_{h_{2s}},\ldots,w_{h_{ks}}) \\ &{}-\sum_{t=s}^{\infty}\bigl[f(t,w_{f_{1t}},w_{f_{2t}}, \ldots ,w_{f_{kt}})-c_{t}\bigr]\Biggr\} ,\quad \forall n\ge T+\tau, \end{aligned} \\& \begin{aligned} a_{n}\Delta^{3}(w_{n}+b_{n}w_{n-\tau}) ={}&{-}h(n,w_{h_{1n}},w_{h_{2n}},\ldots,w_{h_{kn}}) \\ &{}+\sum_{t=n}^{\infty}\bigl[f(t,w_{f_{1t}},w_{f_{2t}}, \ldots ,w_{f_{kt}})-c_{t}\bigr],\quad \forall n\ge T+\tau \end{aligned} \end{aligned}$$
(2.46)
and
$$\begin{aligned} \Delta\bigl(a_{n}\Delta^{3}(w_{n}+b_{n}w_{n-\tau}) \bigr) ={}&{-}\Delta h(n,w_{h_{1n}},w_{h_{2n}},\ldots,w_{h_{kn}}) \\ &{}-f(n,w_{f_{1n}},w_{f_{2n}},\ldots,w_{f_{kn}})+c_{n},\quad \forall n\ge T+\tau, \end{aligned}$$
that is, \(w=\{w_{n}\}_{n\in\Bbb{N}_{\beta}}\) is a positive solution of Eq. (1.1) in \(A(N,M)\). Making use of (2.17), (2.18) and (2.46), we infer that
$$\begin{aligned} \biggl\vert \frac{w_{n}+b_{n}w_{n-\tau}}{n^{2}}-L\biggr\vert \le&\frac{1}{n^{2}} \sum_{v=n}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty} \frac{1}{\vert a_{s}\vert } \Biggl(\bigl\vert h(s,w_{h_{1s}},w_{h_{2s}}, \ldots,w_{h_{ks}})\bigr\vert \\ &{}+\sum_{t=s}^{\infty}\bigl[\bigl\vert f(t,w_{f_{1t}},w_{f_{2t}},\ldots ,w_{f_{kt}})\bigr\vert + \vert c_{t}\vert \bigr]\Biggr) \\ \le&\frac{1}{n^{2}}\sum_{v=n}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty} \frac{1}{\vert a_{s}\vert } \Biggl(W_{s}+\sum_{t=s}^{\infty} \bigl(Q_{t}+\vert c_{t}\vert \bigr)\Biggr) \\ \to&0\quad \text{as } n\to\infty, \end{aligned}$$
which gives (2.28). In terms of (2.14), (2.16), (2.40), (2.44) and (2.45), we deduce that
$$\begin{aligned} \frac{\vert x_{m+1n}-w_{n}\vert }{n^{2}} ={}&\frac{1}{n^{2}}\Biggl\vert (1- \alpha_{m})x_{mn}+\alpha_{m}\Biggl\{ \frac{(n+\tau )^{2}L}{b_{n+\tau}}-\frac{x_{n+\tau}}{b_{n+\tau}} \\ &{}+\frac{1}{b_{n+\tau}}\sum_{v=n+\tau}^{\infty}\sum _{u=v}^{\infty }\sum_{s=u}^{\infty} \frac{1}{a_{s}} \Biggl[h(s,x_{mh_{1s}},x_{mh_{2s}}, \ldots,x_{mh_{ks}}) \\ &{}-\sum_{t=s}^{\infty}\bigl[f(t,x_{mf_{1t}},x_{mf_{2t}}, \ldots ,x_{mf_{kt}})-c_{t}\bigr]\Biggr]\Biggr\} -w_{n} \Biggr\vert \\ \le{}&(1-\alpha_{m})\frac{\vert x_{mn}-w_{n}\vert }{n^{2}}+\alpha_{m} \frac {\vert S_{L}x_{mn}-S_{L}w_{n}\vert }{n^{2}} \\ \le{}&(1-\alpha_{m})\Vert x_{m}-w\Vert +\theta \alpha_{m}\Vert x_{m}-w\Vert \\ ={}&\bigl[1-(1-\theta)\alpha_{m}\bigr]\Vert x_{m}-w\Vert \\ \le{}& e^{-(1-\theta)\alpha_{m}}\Vert x_{m}-w\Vert ,\quad \forall m\in\Bbb{N}_{0}, n\ge T \end{aligned}$$
and
$$\begin{aligned} \frac{\vert x_{m+1n}-w_{n}\vert }{n^{2}} ={}&\frac{1}{n^{2}}\Biggl\vert (1- \alpha_{m})\frac{n^{2}}{T^{2}}x_{mT}+\alpha_{m} \frac {n^{2}}{T^{2}}\Biggl\{ \frac{(T+\tau)^{2}L}{b_{T+\tau}}-\frac{x_{m{T+\tau}}}{b_{T+\tau}} \\ &{}+\sum_{v=T+\tau}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty } \frac{1}{a_{s}} \Biggl[h(s,x_{mh_{1s}},x_{mh_{2s}}, \ldots,x_{mh_{ks}}) \\ &{}-\sum_{t=s}^{\infty}\bigl[f(t,x_{mf_{1t}},x_{mf_{2t}}, \ldots ,x_{mf_{kt}})-c_{t}\bigr]\Biggr]\Biggr\} -w_{n} \Biggr\vert \\ \le{}&(1-\alpha_{m})\frac{\vert x_{mT}-w_{T}\vert }{T^{2}}+\alpha_{m} \frac {\vert S_{L}x_{mT}-S_{L}w_{T}\vert }{T^{2}} \\ \le{}&\bigl[1-(1-\theta)\alpha_{m}\bigr]\Vert x_{m}-w\Vert \\ \le{}& e^{-(1-\theta)\alpha_{m}}\Vert x_{m}-w\Vert ,\quad \forall m\in\Bbb{N}_{0}, \beta\le n< T, \end{aligned}$$
which imply that
$$\Vert x_{m+1}-w\Vert \le e^{-(1-\theta)\alpha_{m}}\Vert x_{m}-w \Vert \le e^{-(1-\theta)\sum_{i=0}^{m}\alpha_{i}}\Vert x_{0}-w\Vert ,\quad \forall m\in\Bbb{N}_{0}, $$
that is, (2.8) holds. It follows from (2.8) and (2.9) that \(\lim_{m\to\infty}x_{m}=w\).
Next we show that (b) holds. Let \(L_{1},L_{2}\in(b^{*}(Mq+N), \frac{M}{q}+\frac{N}{qb^{*}})\) and \(L_{1}\neq L_{2}\). Similar to the proof of (a), we get that for each \(c\in\{1,2\}\) there exist constants \(\theta_{c}\in(0, 1)\), \(T_{c}\ge n_{0}+\tau+\beta\) and a mapping \(S_{L_{c}}\) satisfying (2.41)-(2.44), where θ, L and T are replaced by \(\theta_{c}\), \(L_{c}\) and \(T_{c}\), respectively, and the mapping \(S_{L_{c}}\) has a fixed point \(z^{c}=\{z^{c}_{n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\), which is a positive solution of Eq. (1.1) in \(A(N,M)\), that is,
$$\begin{aligned} z^{c}_{n}={}&\frac{(n+\tau)^{2}}{b_{n+\tau}}L_{c}- \frac{z^{c}_{n+\tau }}{b_{n+\tau}} +\frac{1}{b_{n+\tau}}\sum_{v=n}^{\infty} \sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\frac{1}{a_{s}} \Biggl\{ h \bigl(s,z^{c}_{h_{1s}},z^{c}_{h_{2s}}, \ldots,z^{c}_{h_{ks}}\bigr) \\ &{}-\sum_{t=s}^{\infty}\bigl[f \bigl(t,z^{c}_{f_{1t}},z^{c}_{f_{2t}},\ldots ,z^{c}_{f_{kt}}\bigr)-c_{t}\bigr]\Biggr\} ,\quad \forall n \ge T_{c}, \end{aligned}$$
which together with (2.1), (2.41) and (2.43) means that
$$\begin{aligned} &\biggl\vert \frac{z^{1}_{n}}{n^{2}}-\frac{z^{2}_{n}}{n^{2}}\biggr\vert \\ &\quad \ge\frac{1}{b_{n+\tau}}\biggl(1+\frac{\tau}{n}\biggr)^{2}\vert L_{1}-L_{2}\vert -\frac{1}{b_{n+\tau}}\cdot\frac{(n+\tau)^{2}}{n^{2}} \cdot\frac {\vert z^{1}_{n}(n+\tau)-z^{2}_{n}(n+\tau)\vert }{(n+\tau)^{2}} \\ &\qquad {}-\frac{1}{b_{n+\tau}n^{2}}\sum_{v=n+\tau}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty} \frac{1}{\vert a_{s}\vert } \Biggl(\bigl\vert h\bigl(s,z^{1}_{h_{1s}},z^{1}_{h_{2s}}, \ldots,z^{1}_{h_{ks}}\bigr) -h\bigl(s,z^{2}_{h_{1s}},z^{2}_{h_{2s}}, \ldots,z^{2}_{h_{ks}}\bigr)\bigr\vert \\ &\qquad {}+\sum_{t=s}^{\infty}\bigl\vert f \bigl(t,z^{1}_{f_{1t}},z^{1}_{f_{2t}},\ldots ,z^{1}_{f_{kt}}\bigr) -f\bigl(t,z^{2}_{f_{1t}},z^{2}_{f_{2t}}, \ldots,z^{2}_{f_{kt}}\bigr)\bigr\vert \Biggr) \\ &\quad \ge\frac{\vert L_{1}-L_{2}\vert }{b^{*}}-\frac{1}{b_{*}}\biggl(1+\frac{\tau}{n} \biggr)^{2} \bigl\Vert z^{1}-z^{2}\bigr\Vert \\ &\qquad {} - \frac{1}{b_{*}n^{2}}\sum_{v=n+\tau}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty} \frac{1}{\vert a_{s}\vert } \Biggl(R_{s}\max\bigl\{ \bigl\vert z^{1}_{h_{ls}}-z^{2}_{h_{ls}}\bigr\vert : 1 \leq l\leq k\bigr\} \\ &\qquad {}+\sum_{t=s}^{\infty}P_{t}\max\bigl\{ \bigl\vert z^{1}_{f_{lt}}-z^{2}_{f_{lt}} \bigr\vert :1\leq l\leq k\bigr\} \Biggr) \\ &\quad \ge\frac{\vert L_{1}-L_{2}\vert }{b^{*}}-q\bigl\Vert z^{1}-z^{2}\bigr\Vert -\frac{\Vert z^{1}-z^{2}\Vert }{b_{*}\max\{T_{1}^{2},T_{2}^{2}\}} \sum_{v=\max\{T_{1},T_{2}\}}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty } \frac{1}{\vert a_{s}\vert } \Biggl(R_{s}H_{s}+\sum _{t=s}^{\infty}P_{t}F_{t}\Biggr) \\ &\quad \ge\frac{\vert L_{1}-L_{2}\vert }{b^{*}}-\max\{\theta_{1},\theta_{2}\}\bigl\Vert z^{1}-z^{2}\bigr\Vert ,\quad \forall n\geq\max \{T_{1},T_{2}\}, \end{aligned}$$
which yields that
$$\bigl\Vert z^{1}-z^{2}\bigr\Vert \ge \frac{\vert L_{1}-L_{2}\vert }{b^{*}(1+\max\{\theta_{1},\theta_{2}\})}>0, $$
that is, \(z^{1}\neq z^{2}\). This completes the proof. □

Theorem 2.6

Assume that there exist constants \(b_{*}\), \(b^{*}\), M and N with \(N\frac{1+b_{*}}{1+b^{*}}>M>N>0\) and four nonnegative sequences \(\{P_{n}\}_{n\in\Bbb{N}_{n_{0}}}\), \(\{Q_{n}\}_{n\in\Bbb{N}_{n_{0}}}\), \(\{R_{n}\}_{n\in\Bbb{N}_{n_{0}}}\) and \(\{W_{n}\}_{n\in\Bbb{N}_{n_{0}}}\) satisfying (2.1), (2.2), (2.17), (2.18) and
$$ b_{*}\le b_{n}\le b^{*}< -1 \quad \textit{eventually.} $$
(2.47)
Then
  1. (a)

    for any \(L\in(N(1+b_{*}), M(1+b^{*}))\), there exist \(\theta\in(0, 1)\) and \(T\ge n_{0}+\tau+\beta\) such that for each \(x_{0}=\{x_{0n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\), the Mann iterative sequence \(\{x_{m}\}_{m\in\Bbb{N}_{0}}=\{\{x_{mn}\}_{n\in\Bbb{N}_{\beta}}\}_{m\in\Bbb{N}_{0}}\) generated by (2.40) converges to a positive solution \(w=\{w_{n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\) of Eq. (1.1) with (2.28) and has the error estimate (2.8), where \(\{\alpha_{m}\}_{m\in\Bbb{N}_{0}}\) is an arbitrary sequence in \([0, 1]\) satisfying (2.9);

     
  2. (b)

    Equation (1.1) possesses uncountably many positive solutions in \(A(N,M)\).

     

Proof

Put \(L\in(N(1+b_{*}), M(1+b^{*}))\). Observe that
$$\begin{aligned} \lim_{n\to\infty}\biggl[N\biggl(1+b_{*}\biggl(1+ \frac{\tau}{n} \biggr)^{-2}\biggr)\biggr] &=N(1+b_{*})< L< M\bigl(1+b^{*} \bigr) \\ &=\lim_{n\to\infty}\biggl[M\biggl(1+b^{*}\biggl(1-\frac{\tau}{n} \biggr)^{-2}\biggr)\biggr] \\ &=\lim_{n\to\infty}\biggl[M\biggl(1+b^{*}\biggl(1+\frac{\tau}{n} \biggr)^{-2}\biggr)\biggr], \end{aligned}$$
which implies that there exists \(T_{0}\in\Bbb{N}_{n_{0}+\tau+\beta}\) satisfying
$$\begin{aligned} L&\in\biggl(N\biggl(1+b_{*}\biggl(1+\frac{\tau}{n}\biggr)^{-2} \biggr), M \biggl(1+b^{*}\biggl(1-\frac{\tau}{n}\biggr)^{-2}\biggr) \biggr) \\ &\subset\bigl(N(1+b_{*}), M\bigl(1+b^{*}\bigr)\bigr) \\ &\subset\biggl(N(1+b_{*}), M\biggl(1+b^{*}\biggl(1+\frac{\tau}{n} \biggr)^{-2}\biggr)\biggr),\quad \forall n\in\Bbb{N}_{T_{0}}. \end{aligned}$$
(2.48)
It follows from (2.17), (2.18) and (2.47) that there exist \(\theta\in(0, 1)\) and \(T\geq T_{0}\) satisfying
$$\begin{aligned}& \theta=-\frac{1}{b^{*}}\Biggl[\biggl(1+\frac{\tau}{T} \biggr)^{2}+\frac {1}{T^{2}}\sum_{v=T}^{\infty} \sum_{u=v}^{\infty} \sum _{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl(R_{s}H_{s}+ \sum_{t=s}^{\infty}P_{t}F_{t} \Biggr)\Biggr]; \end{aligned}$$
(2.49)
$$\begin{aligned}& -\frac{1}{b^{*}T^{2}}\sum_{v=T}^{\infty} \sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl(W_{s}+ \sum_{t=s}^{\infty}\bigl(Q_{t}+\vert c_{t}\vert \bigr)\Biggr) \\& \quad < \min\biggl\{ M+\biggl(1+\frac{\tau}{T}\biggr)^{2} \frac{M-L}{b^{*}}, \biggl(1+\frac{\tau}{T}\biggr)^{2} \frac{L-N}{b_{*}}-N\biggr\} ; \end{aligned}$$
(2.50)
$$\begin{aligned}& b_{n}\le b< -1,\quad \forall n\ge T. \end{aligned}$$
(2.51)
Define a mapping \(S_{L}: A(N,M)\to l_{\beta}^{\infty}\) by (2.44). Making use of (2.1), (2.2), (2.44) and (2.48)-(2.51), we conclude that for each \(x=\{x_{n}\}_{n\in\Bbb{N}_{\beta}},y=\{y_{n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\)
$$\begin{aligned}& \begin{aligned} &\biggl\vert \frac{S_{L}x_{n}}{n^{2}}-\frac{S_{L}y_{n}}{n^{2}}\biggr\vert \\ &\quad \le-\frac{1}{b_{n+\tau}}\cdot\frac{(n+\tau)^{2}}{n^{2}}\biggl\vert \frac {x_{n+\tau}-y_{n+\tau}}{(n+\tau)^{2}} \biggr\vert \\ &\qquad {}-\frac{1}{b_{n+\tau}n^{2}}\sum_{v=n+\tau}^{\infty}\sum _{u=v}^{\infty }\sum_{s=u}^{\infty} \frac{1}{\vert a_{s}\vert } \Biggl(\bigl\vert h(s,x_{h_{1s}},x_{h_{2s}}, \ldots,x_{h_{ks}}) -h(s,y_{h_{1s}},y_{h_{2s}}, \ldots,y_{h_{ks}})\bigr\vert \\ &\qquad {}+\sum_{t=s}^{\infty}\bigl\vert f(t,x_{f_{1t}},x_{f_{2t}},\ldots ,x_{f_{kt}}) -f(t,y_{f_{1t}},y_{f_{2t}},\ldots,y_{f_{kt}})\bigr\vert \Biggr) \\ &\quad \le-\frac{1}{b^{*}}\biggl(1+\frac{\tau}{T}\biggr)^{2}\Vert x-y\Vert \\ &\qquad {}-\frac{1}{b^{*}T^{2}}\sum_{v=T}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty} \frac{1}{\vert a_{s}\vert } \Biggl(R_{s}\max\bigl\{ \vert x_{h_{ls}}-y_{h_{ls}} \vert :1\le l\le k\bigr\} \\ &\qquad {} +\sum_{t=s}^{\infty}P_{t} \max\bigl\{ \vert x_{f_{lt}}-y_{f_{lt}}\vert :1\le l\le k\bigr\} \Biggr) \\ &\quad \le-\frac{1}{b^{*}}\Biggl[\biggl(1+\frac{\tau}{T}\biggr)^{2}+ \frac{1}{T^{2}}\sum_{v=T}^{\infty}\sum _{u=v}^{\infty} \sum _{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl(R_{s}H_{s}+ \sum_{t=s}^{\infty}P_{t}F_{t} \Biggr)\Biggr]\Vert x-y\Vert \\ &\quad =\theta \Vert x-y\Vert ,\quad \forall n\ge T, \end{aligned} \\& \biggl\vert \frac{S_{L}x_{n}}{n^{2}}-\frac{S_{L}y_{n}}{n^{2}}\biggr\vert = \biggl\vert \frac{n^{2}}{T^{2}}\cdot\frac{S_{L}x_{T}}{n^{2}}-\frac{n^{2}}{T^{2}}\cdot \frac {S_{L}y_{T}}{n^{2}}\biggr\vert \le\theta \Vert x-y\Vert ,\quad \beta\le n< T, \\& \begin{aligned} \frac{S_{L}x_{n}}{n^{2}} ={}&\biggl(1+\frac{\tau}{n}\biggr)^{2} \frac{L}{b_{n+\tau}}-\frac{1}{b_{n+\tau }}\biggl(1+\frac{\tau}{n} \biggr)^{2} \frac{x_{n+\tau}}{(n+\tau)^{2}} \\ &{}+\frac{1}{b_{n+\tau}n^{2}}\sum_{v=n+\tau}^{\infty}\sum _{u=v}^{\infty }\sum_{s=u}^{\infty} \frac{1}{a_{s}} \Biggl\{ h(s,x_{h_{1s}},x_{h_{2s}}, \ldots,x_{h_{ks}}) \\ &{}-\sum_{t=s}^{\infty}\bigl[f(t,x_{f_{1t}},x_{f_{2t}}, \ldots,x_{f_{kt}})-c_{t}\bigr]\Biggr\} \\ \le{}&\biggl(1+\frac{\tau}{n}\biggr)^{2}\frac{L}{b^{*}}- \biggl(1+\frac{\tau }{n}\biggr)^{2}\frac{M}{b^{*}} \\ &{}- \frac{1}{b^{*}n^{2}}\sum_{v=n+\tau}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty} \frac{1}{\vert a_{s}\vert } \Biggl(\bigl\vert h(s,x_{h_{1s}},x_{h_{2s}}, \ldots,x_{h_{ks}})\bigr\vert \\ &{}+\sum_{t=s}^{\infty}\bigl[\bigl\vert f(t,x_{f_{1t}},x_{f_{2t}},\ldots ,x_{f_{kt}})\bigr\vert + \vert c_{t}\vert \bigr]\Biggr) \\ \le{}&\biggl(1+\frac{\tau}{T}\biggr)^{2}\frac{L-M}{b^{*}} - \frac{1}{b^{*}T^{2}}\sum_{v=T}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty} \frac{1}{\vert a_{s}\vert }\Biggl(W_{s}+\sum_{t=s}^{\infty} \bigl(Q_{t}+\vert c_{t}\vert \bigr)\Biggr) \\ < {}&\biggl(1+\frac{\tau}{T}\biggr)^{2}\frac{L-M}{b^{*}} \\ &{}+\min \biggl\{ M+\biggl(1+\frac{\tau}{T}\biggr)^{2}\frac{M-L}{b^{*}}, \biggl(1+\frac{\tau}{T}\biggr)^{2}\frac{L-N}{b_{*}}-N\biggr\} \\ \le {}&M,\quad \forall n\ge T, \end{aligned} \\& \frac{S_{L}x_{n}}{n^{2}}=\frac{n^{2}}{T^{2}}\cdot\frac{S_{L}x_{T}}{n^{2}}\le M,\quad \beta \le n< T, \\& \begin{aligned} \frac{S_{L}x_{n}}{n^{2}} \ge{}&\biggl(1+\frac{\tau}{n}\biggr)^{2} \frac{L}{b_{*}}-\biggl(1+\frac{\tau }{n}\biggr)^{2} \frac{N}{b_{*}} \\ &{}+\frac{1}{b^{*}n^{2}}\sum_{v=n+\tau}^{\infty} \sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl(\bigl\vert h(s,x_{h_{1s}},x_{h_{2s}},\ldots,x_{h_{ks}})\bigr\vert \\ &{}+\sum_{t=s}^{\infty}\bigl[\bigl\vert f(t,x_{f_{1t}},x_{f_{2t}},\ldots ,x_{f_{kt}})\bigr\vert + \vert c_{t}\vert \bigr]\Biggr) \\ \ge{}&\biggl(1+\frac{\tau}{n}\biggr)^{2}\frac{L-N}{b_{*}} + \frac{1}{b^{*}T^{2}}\sum_{v=T}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty} \frac{1}{\vert a_{s}\vert } \Biggl(W_{s}+\sum_{t=s}^{\infty} \bigl(Q_{t}+\vert c_{t}\vert \bigr)\Biggr) \\ >{}&\biggl(1+\frac{\tau}{T}\biggr)^{2}\frac{L-N}{b_{*}}\\ &{}-\min\biggl\{ M+ \biggl(1+\frac{\tau}{T}\biggr)^{2}\frac{M-L}{b^{*}}, \biggl(1+ \frac{\tau}{T}\biggr)^{2}\frac{L-N}{b_{*}}-N\biggr\} \\ \ge{}& N,\quad \forall n\ge T \end{aligned} \end{aligned}$$
and
$$\begin{aligned} \frac{S_{L}x_{n}}{n^{2}}=\frac{n^{2}}{T^{2}}\cdot\frac{S_{L}x_{T}}{n^{2}}\ge N,\quad \beta \le n< T, \end{aligned}$$
which yield (2.14). The rest of the proof is similar to that of Theorem 2.5 and is omitted. This completes the proof. □

Theorem 2.7

Assume that there exist constants b, M and N with \((1-2b)M>N>0\) and four nonnegative sequences \(\{P_{n}\}_{n\in\Bbb{N}_{n_{0}}}\), \(\{Q_{n}\}_{n\in\Bbb{N}_{n_{0}}}\), \(\{R_{n}\}_{n\in\Bbb{N}_{n_{0}}}\), \(\{W_{n}\}_{n\in\Bbb{N}_{n_{0}}}\) satisfying (2.1), (2.2), (2.17), (2.18) and
$$ \vert b_{n}\vert \le b< \frac{1}{2}\quad \textit{eventually}. $$
(2.52)
Then
  1. (a)

    for any \(L\in(N+bM, (1-b)M)\), there exist \(\theta\in (0, 1)\) and \(T\ge n_{0}+\tau+\beta\) such that for any \(x_{0}=\{x_{0n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\), the Mann iterative sequence \(\{x_{m}\}_{m\in\Bbb{N}_{0}}=\{\{x_{mn}\}_{n\in\Bbb{N}_{\beta}}\}_{m\in\Bbb{N}_{0}}\) generated by (2.27) converges to a positive solution \(w=\{w_{n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\) of Eq. (1.1) with (2.28) and has the error estimate (2.8), where \(\{\alpha_{m}\}_{m\in\Bbb{N}_{0}}\) is an arbitrary sequence in \([0, 1]\) satisfying (2.9);

     
  2. (b)

    Equation (1.1) possesses uncountably many positive solutions in \(A(N,M)\).

     

Proof

Put \(L\in(N+bM, (1-b)M)\). It follows from (2.17), (2.18) and (2.52) that there exist \(\theta\in(0, 1)\) and \(T\ge n_{0}+\tau+\beta\) satisfying (2.29),
$$\begin{aligned}& \frac{1}{T^{2}}\sum_{v=T}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty}\frac {1}{\vert a_{s}\vert }\Biggl(W_{s}+\sum_{t=s}^{\infty} \bigl(Q_{t}+\vert c_{t}\vert \bigr)\Biggr)< \min\bigl\{ (1-b)M-L, L-bM-N\bigr\} ; \end{aligned}$$
(2.53)
$$\begin{aligned}& \vert b_{n}\vert \le b,\quad \forall n\ge T. \end{aligned}$$
(2.54)
Define a mapping \(S_{L}: A(N,M)\to l_{\beta}^{\infty}\) by (2.32). By virtue of (2.2), (2.32), (2.53) and (2.54), we easily verify that for each \(x=\{x_{n}\}_{n\in\Bbb{N}_{\beta}},y=\{y_{n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\)
$$\begin{aligned}& \begin{aligned} \frac{S_{L}x_{n}}{n^{2}} \le{}& L-\vert b_{n}\vert \biggl(1- \frac{\tau}{n}\biggr)^{2}\frac{x_{n-\tau}}{(n-\tau)^{2}} \\ &{}+\frac{1}{n^{2}}\sum_{v=n}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl(\bigl\vert h(s,x_{h_{1s}},x_{h_{2s}}, \ldots,x_{h_{ks}}) \bigr\vert \\ &{}+\sum_{t=s}^{\infty} \bigl[\bigl\vert f(t,x_{f_{1t}},x_{f_{2t}},\ldots,x_{f_{kt}}) \bigr\vert +\vert c_{t}\vert \bigr]\Biggr) \\ \le{}& L+bM+\frac{1}{T^{2}}\sum_{v=T}^{\infty}\sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl(W_{s}+ \sum_{t=s}^{\infty}\bigl(Q_{t}+\vert c_{t}\vert \bigr)\Biggr) \\ < {}&L+bM+\min\bigl\{ (1-b)M-L,L-bM-N\bigr\} \\ \le{}& M,\quad \forall n\ge T, \end{aligned} \\& \frac{S_{L}x_{n}}{n^{2}}=\frac{n^{2}}{T^{2}}\cdot\frac{S_{L}x_{T}}{n^{2}}\le M,\quad \beta \le n< T, \\& \begin{aligned} \frac{S_{L}x_{n}}{n^{2}} \ge{}& L-\vert b_{n}\vert \biggl(1- \frac{\tau}{n}\biggr)^{2}\frac{x_{n-\tau}}{(n-\tau )^{2}} \\ &{}-\frac{1}{n^{2}}\sum_{v=n}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl(\bigl\vert h(s,x_{h_{1s}},x_{h_{2s}}, \ldots,x_{h_{ks}})\bigr\vert \\ & {} +\sum_{t=s}^{\infty} \bigl[\bigl\vert f(t,x_{f_{1t}},x_{f_{2t}},\ldots ,x_{f_{kt}})\bigr\vert +\vert c_{t}\vert \bigr]\Biggr) \\ \ge{}& L-bM-\frac{1}{T^{2}}\sum_{v=T}^{\infty}\sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \Biggl(W_{s}+ \sum_{t=s}^{\infty}\bigl(Q_{t}+\vert c_{t}\vert \bigr)\Biggr) \\ >{}&L-bM-\min\bigl\{ (1-b)M-L,L-bM-N\bigr\} \\ \ge{}& N,\quad \forall n\ge T \end{aligned} \end{aligned}$$
and
$$\begin{aligned} \frac{S_{L}x_{n}}{n^{2}}=\frac{n^{2}}{T^{2}}\cdot\frac{S_{L}x_{T}}{n^{2}}\ge N,\quad \beta \le n< T, \end{aligned}$$
which yield that \(S_{L}(A(N,M))\subseteq A(N,M)\). The rest of the proof is similar to that of Theorem 2.3 and is omitted. This completes the proof. □

3 Examples

In this section, we suggest seven examples to explain the results presented in Section 2.

Example 3.1

Consider the fourth order neutral delay difference equation
$$\begin{aligned} &\Delta\bigl(\bigl(n^{2}-n+1\bigr)\Delta^{3}(x_{n}-x_{n-\tau}) \bigr)+\Delta\biggl(\frac {\sin^{2}(x_{n-3}-nx_{n^{2}-1})}{n^{18}+3n^{6}-4n^{3}+1}\biggr) \\ &\quad {}+\frac{3n-\sqrt{n}}{(n^{15}+2n^{5}-n+1) (1+x_{n^{2}}^{2}+x^{2}_{n-2})} =\frac{(-1)^{n}\ln^{2}n}{n^{11}+2n^{5}-n^{4}+1},\quad \forall n\ge4, \end{aligned}$$
(3.1)
where \(\tau\in\Bbb{N}\) is fixed. Let \(n_{0}=4, k=2, \beta=\min\{4-\tau,1\}=1\in\Bbb{N}\), M and N be two positive constants with \(M>N\) and
$$\begin{aligned}& a_{n} =n^{2}-n+1,\qquad b_{n}=-1,\qquad c_{n}=\frac{(-1)^{n}\ln ^{2}n}{n^{11}+2n^{5}-n^{4}+1},\qquad f_{1n}=n^{2}, \\& f_{2n} =n-2,\qquad F_{n}=n^{4},\qquad h_{1n}=n-3,\qquad h_{2n}=n^{2}-1,\qquad H_{n}=\bigl(n^{2}-1 \bigr)^{2}, \\& f(n ,u,v)=\frac{3n-\sqrt{n}}{(n^{15}+2n^{5}-n+1)(1+u^{2}+v^{2})},\qquad h(n,u,v)=\frac{\sin^{2}(u-nv)}{n^{18}+3n^{6}-4n^{3}+1}, \\& P_{n} = Q_{n}=\frac{20}{n^{14}},\qquad R_{n}= \frac{4}{n^{17}},\qquad W_{n}=\frac{1}{n^{18}},\quad \forall(n,u,v)\in\Bbb{N}_{n_{0}}\times\Bbb{R}^{2}. \end{aligned}$$
It is easy to see that (2.1), (2.2) and (2.5) are satisfied. Note that Lemma 1.1 means that
$$\begin{aligned} &\frac{1}{n^{2}}\sum_{i=1}^{\infty}\sum_{v=n+i\tau}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \max\{R_{s}H_{s},W_{s}\} \\ &\quad \le\frac{1}{n^{2}\tau}\sum_{s=n+\tau}^{\infty} \frac{s^{3}}{s^{2}-s+1}\max \biggl\{ \frac{4(s^{2}-1)^{2}}{s^{17}},\frac{1}{s^{18}}\biggr\} \le\frac{4}{n^{2}\tau}\sum_{s=n+\tau}^{\infty} \frac{1}{s^{10}}\to 0\quad \text{as } n\to\infty \end{aligned}$$
and
$$\begin{aligned} &\frac{1}{n^{2}}\sum_{i=1}^{\infty}\sum_{v=n+i\tau}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty}\sum_{t=s}^{\infty}\frac{1}{\vert a_{s}\vert } \max\bigl\{ P_{t}F_{t},Q_{t},\vert c_{t} \vert \bigr\} \\ &\quad =\frac{1}{n^{2}}\sum_{i=1}^{\infty}\sum _{v=n+i\tau}^{\infty}\sum_{u=v}^{\infty}\sum_{s=u}^{\infty}\sum _{t=s}^{\infty}\frac{1}{s^{2}-s+1}\max \biggl\{ \frac{20}{t^{10}},\frac{20}{t^{14}},\frac{\ln ^{2}t}{t^{11}+2t^{5}-t^{4}+1}\biggr\} \\ &\quad \le\frac{20}{n^{2}}\sum_{i=1}^{\infty}\sum _{v=n+i\tau}^{\infty}\sum_{u=v}^{\infty}\sum_{s=u}^{\infty}\sum _{t=s}^{\infty}\frac{1}{t^{8}} \\ &\quad \le\frac{20}{n^{2}\tau}\sum_{t=n+\tau}^{\infty}\frac{1}{t^{6}}\to 0\quad \text{as } n\to\infty, \end{aligned}$$
which give that
$$\lim_{n\to\infty}\frac{1}{n^{2}}\sum_{i=1}^{\infty}\sum_{v=n+i\tau}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{\vert a_{s}\vert } \max\{R_{s}H_{s},W_{s}\}=0 $$
and
$$\lim_{n\to \infty}\frac{1}{n^{2}}\sum _{i=1}^{\infty}\sum_{v=n+i\tau}^{\infty}\sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\sum_{t=s}^{\infty}\frac{1}{\vert a_{s}\vert } \max\bigl\{ P_{t}F_{t},Q_{t}, \vert c_{t}\vert \bigr\} =0. $$
That is, (2.3) and (2.4) hold. Consequently Theorem 2.1 implies that Eq. (3.1) possesses uncountably many positive solutions in \(A(N,M)\). Moreover, for each \(L\in(N, M)\), there exist \(\theta\in(0, 1)\) and \(T\ge n_{0}+\tau+\beta\) such that for each \(x_{0}=\{x_{0n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\), the Mann iterative sequence \(\{x_{m}\}_{m\in\Bbb{N}_{0}}=\{\{x_{mn}\}_{n\in\Bbb{N}_{\beta}}\}_{m\in\Bbb{N}_{0}}\) generated by (2.6) converges to a positive solution \(w=\{w_{n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\) of Eq. (3.1) with (2.7) and (2.8), where \(\{\alpha_{m}\}_{m\in\Bbb{N}_{0}}\) is an arbitrary sequence in \([0, 1]\) satisfying (2.9).

Example 3.2

Consider the fourth order neutral delay difference equation
$$\begin{aligned} &\Delta\bigl((-1)^{n}n^{2}\Delta^{3}(x_{n}+x_{n-\tau}) \bigr) +\Delta\biggl(\frac{\cos ^{2}(n^{14}x_{n-4}-2)}{(n^{34}+28n^{22}-1)(1+x^{4}_{2n-3})}\biggr) \\ &\qquad{} +\frac {(n^{20}-n^{13}+(-1)^{n})(x_{n^{2}-16}+x_{n^{2}-20})}{(n^{36}+10n^{28}-\sqrt {n})(1+x^{2}_{n^{2}-16}+x^{2}_{n^{2}-20})} \\ &\quad =\frac{(-1)^{n}n^{3}+4n^{2}-\sqrt{\ln n}}{n^{19}+20n^{15}-n^{4}+1},\quad n\ge5, \end{aligned}$$
(3.2)
where \(\tau\in\Bbb{N}\) is fixed. Let \(n_{0}=5, k=2, \beta=5-\tau\in\Bbb{N}\), M and N be two positive constants with \(M>N\) and
$$\begin{aligned}& a_{n} =(-1)^{n}n^{2},\qquad b_{n}=1,\qquad c_{n}=\frac{(-1)^{n}n^{3}+4n^{2}-\sqrt{\ln n}}{n^{19}+20n^{15}-n^{4}+1},\qquad f_{1n}=n^{2}-16, \\& f_{2n} =n^{2}-20,\qquad F_{n}=\bigl(n^{2}-16 \bigr)^{2},\qquad h_{1n}=2n-3,\\& h_{2n}=n-4,\qquad H_{n}=(2n-3)^{2}, \\& f(n ,u,v)=\frac{(n^{20}-n^{13}+(-1)^{n})(u+v)}{(n^{36}+10n^{28}-\sqrt {n})(1+u^{2}+v^{2})},\\& h(n,u,v)=\frac{\cos^{2}(n^{14}v-2)}{(n^{34}+28n^{22}-1)(1+u^{4})}, \\& P_{n} =Q_{n}=\frac{4}{n^{12}},\qquad R_{n}= W_{n}=\frac{10}{n^{13}},\quad \forall(n,u,v)\in\Bbb{N}_{n_{0}}\times \Bbb{R}^{2}. \end{aligned}$$
It is clear that (2.1), (2.2) and (2.19) are fulfilled. Note that Lemma 1.1 ensures that
$$\begin{aligned} &\frac{1}{n^{2}}\sum_{v=n}^{\infty}\sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\frac {1}{\vert a_{s}\vert }\max\{R_{s}H_{s},W_{s} \} \\ &\quad \le\frac{1}{n^{2}}\sum_{s=n}^{\infty}\frac{s^{2}}{\vert (-1)^{s}s^{2}\vert }\max\biggl\{ \frac{10(2s-3)^{2}}{s^{13}},\frac{10}{s^{13}}\biggr\} \\ &\quad \leq\frac{40}{n^{2}}\sum_{s=n}^{\infty}\frac{1}{s^{11}}\to 0\quad \text{as } n\to\infty \end{aligned}$$
and
$$\begin{aligned} &\frac{1}{n^{2}}\sum_{v=n}^{\infty}\sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\sum_{t=s}^{\infty}\frac{1}{\vert a_{s}\vert } \max\bigl\{ P_{t}F_{t},Q_{t}, \vert c_{t}\vert \bigr\} \\ &\quad =\frac{1}{n^{2}}\sum_{v=n}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty}\sum_{t=s}^{\infty}\frac{1}{\vert (-1)^{s}s^{2}\vert }\max \biggl\{ \frac{4\vert t^{2}-16\vert ^{2}}{t^{12}},\frac{4}{t^{12}}, \biggl\vert \frac{(-1)^{t}t^{3}+4t^{2}-\sqrt{\ln t}}{t^{19}+20t^{15}-t^{4}+1} \biggr\vert \biggr\} \\ &\quad \leq\frac{4}{n^{2}}\sum_{v=n}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty}\sum_{t=s}^{\infty}\frac{1}{t^{8}} \\ &\quad \leq\frac{4}{n^{2}}\sum_{t=n}^{\infty}\frac{1}{t^{5}}\to 0\quad \text{as } n\to\infty, \end{aligned}$$
which mean that
$$\lim_{n\to \infty}\frac{1}{n^{2}}\sum _{v=n}^{\infty}\sum_{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{\vert a_{s}\vert }\max \{R_{s}H_{s},W_{s}\}=0 $$
and
$$\lim_{n\to \infty}\frac{1}{n^{2}}\sum _{v=n}^{\infty}\sum_{u=v}^{\infty}\sum_{s=u}^{\infty}\sum _{t=s}^{\infty}\frac{1}{\vert a_{s}\vert } \max\bigl\{ P_{t}F_{t},Q_{t},\vert c_{t} \vert \bigr\} =0. $$
That is, (2.17) and (2.18) hold. Consequently Theorem 2.2 implies that Eq. (3.2) possesses uncountably many positive solutions in \(A(N,M)\). Moreover, for each \(L\in(N, M)\), there exist \(\theta\in(0, 1)\) and \(T\ge n_{0}+\tau+\beta\) such that for each \(x_{0}=\{x_{0n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\), the Mann iterative sequence \(\{x_{m}\}_{m\in\Bbb{N}_{0}}=\{\{x_{mn}\}_{n\in\Bbb{N}_{\beta}}\}_{m\in\Bbb{N}_{0}}\) generated by (2.20) converges to a positive solution \(w=\{w_{n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\) of Eq. (3.2) with (2.7) and (2.8), where \(\{\alpha_{m}\}_{m\in\Bbb{N}_{0}}\) is an arbitrary sequence in \([0, 1]\) satisfying (2.9).

Example 3.3

Consider the fourth order neutral delay difference equation
$$\begin{aligned} &\Delta\biggl(\sqrt{n^{5}+1}\Delta^{3}\biggl(x_{n}+ \frac {3n^{3}-2}{4n^{3}+3}x_{n-\tau}\biggr)\biggr) \\ &\qquad {}+\Delta\biggl(\frac{\sin(n^{8}\vert x_{n-1}\vert -\sqrt{n})}{n^{24}+n^{4}-\sqrt{n}+1} -\frac{n^{5}-(-1)^{n}n+1}{(n^{19}+6n^{8}-n^{2}+1)2^{\vert x_{2n-1}\vert }}\biggr) \\ &\qquad {}+\frac{(-1)^{n}n^{9}-3n^{4}+2n^{2}+1}{(n^{17}+n^{5}+1)(1+x^{2}_{2n-4})} -\frac{n^{15}\sin ^{5}(3n^{8}-1)+n^{3}-1}{(n^{25}+4n^{24}+n^{7}-1)(1+x^{2}_{n-3})} \\ &\quad =\frac{(-1)^{n}n^{21}-n^{7}+2n^{3}-1}{n^{28}+8n^{14}-2n^{7}+1}, \quad \forall n\ge7, \end{aligned}$$
(3.3)
where \(\tau\in\Bbb{N}\) is fixed. Let \(n_{0}=7, k=2, b=\frac{3}{4},\beta=\min\{7-\tau,5\}\in\Bbb{N}\), M and N be two positive constants with \(M>4N\) and
$$\begin{aligned}& a_{n} =\sqrt{n^{5}+1},\qquad b_{n}= \frac{3n^{3}-2}{4n^{3}+3}, \qquad c_{n}=\frac{(-1)^{n}n^{21}-n^{7}+2n^{3}-1}{n^{28}+8n^{14}-2n^{7}+1}, \qquad f_{1n}=2n-4, \\& f_{2n} =n-3,\qquad F_{n}=(2n-4)^{2},\qquad h_{1n}=n-1,\qquad h_{2n}=2n-1,\qquad H_{n}=(2n-1)^{2}, \\& f(n ,u,v)=\frac{(-1)^{n}n^{9}-3n^{4}+2n^{2}+1}{(n^{17}+n^{5}+1)(1+u^{2})} -\frac{n^{15}\sin^{5}(3n^{8}-1)+n^{3}-1}{(n^{25}+4n^{24}+n^{7}-1)(1+v^{2})}, \\& h(n ,u,v)=\frac{\sin(n^{8}\vert u\vert -\sqrt{n})}{n^{24}+n^{4}-\sqrt{n}+1} -\frac{n^{5}-(-1)^{n}n+1}{(n^{19}+6n^{8}-n^{2}+1)2^{\vert v\vert }}, \\& P_{n} =Q_{n}=\frac{3}{n^{8}},\qquad R_{n}=W_{n}= \frac{2}{n^{10}},\quad \forall (n,u,v)\in\Bbb{N}_{n_{0}}\times\Bbb{R}^{2}. \end{aligned}$$
It is not difficult to verify that (2.1), (2.2) and (2.26) are fulfilled. Note that Lemma 1.1 implies that
$$\begin{aligned} &\frac{1}{n^{2}}\sum_{v=n}^{\infty}\sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\frac {1}{\vert a_{s}\vert }\max\{R_{s}H_{s},W_{s} \} \\ &\quad \le\frac{1}{n^{2}}\sum_{s=n}^{\infty}\frac{s^{2}}{\sqrt{s^{5}+1}} \max\biggl\{ \frac{2\vert 2s-1\vert ^{2}}{s^{10}},\frac{2}{s^{10}}\biggr\} \\ &\quad \leq\frac{8}{n^{2}}\sum_{s=n}^{\infty} \frac{1}{s^{6}}\to 0\quad \text{as } n\to\infty \end{aligned}$$
and
$$\begin{aligned} &\frac{1}{n^{2}}\sum_{v=n}^{\infty}\sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\sum_{t=s}^{\infty}\frac{1}{\vert a_{s}\vert }\max\bigl\{ P_{t}F_{t},Q_{t}, \vert c_{t}\vert \bigr\} \\ &\quad =\frac{1}{n^{2}}\sum_{v=n}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty}\sum_{t=s}^{\infty}\frac{1}{\vert \sqrt{s^{5}+1}\vert }\max \biggl\{ \frac{3\vert 2t-4\vert ^{2}}{t^{8}},\frac{3}{t^{8}},\frac{ \vert (-1)^{t}t^{21}-t^{7}+2t^{3}-1\vert }{t^{28}+8t^{14}-2t^{7}+1}\biggr\} \\ &\quad \leq\frac{12}{n^{2}}\sum_{v=n}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty}\sum_{t=s}^{\infty}\frac{1}{t^{6}} \\ &\quad \leq\frac{12}{n^{2}}\sum_{t=n}^{\infty} \frac{1}{t^{3}}\to 0\quad \text{as } n\to\infty, \end{aligned}$$
which mean that
$$\lim_{n\to \infty}\frac{1}{n^{2}}\sum _{v=n}^{\infty}\sum_{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{\vert a_{s}\vert }\max \{R_{s}H_{s},W_{s}\}=0 $$
and
$$\lim_{n\to \infty}\frac{1}{n^{2}}\sum _{v=n}^{\infty}\sum_{u=v}^{\infty}\sum_{s=u}^{\infty}\sum _{t=s}^{\infty}\frac{1}{\vert a_{s}\vert } \max\bigl\{ P_{t}F_{t},Q_{t},\vert c_{t} \vert \bigr\} =0. $$
That is, (2.17) and (2.18) hold. Consequently Theorem 2.3 implies that Eq. (3.3) possesses uncountably many positive solutions in \(A(N,M)\). Moreover, for each \(L\in(bM+N, M)\), there exist \(\theta\in(0, 1)\) and \(T\ge n_{0}+\tau+\beta\) such that for each \(x_{0}=\{x_{0n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\), the Mann iterative sequence \(\{x_{m}\}_{m\in\Bbb{N}_{0}}=\{\{x_{mn}\}_{n\in\Bbb{N}_{\beta}}\}_{m\in\Bbb{N}_{0}}\) generated by (2.27) converges to a positive solution \(w=\{w_{n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\) of Eq. (3.3) with (2.28) and (2.7), where \(\{\alpha_{m}\}_{m\in\Bbb{N}_{0}}\) is an arbitrary sequence in \([0, 1]\) satisfying (2.9).

Example 3.4

Consider the fourth order neutral delay difference equation
$$\begin{aligned} &\Delta\biggl((-1)^{n}\ln^{3}(n+2)\Delta^{3} \biggl(x_{n}+\frac{2-7\ln^{9}n}{3+8\ln ^{9}n}x_{n-\tau}\biggr)\biggr) +\Delta \biggl(\frac{-3n^{2}+\ln^{2}n-1}{(n^{9}+6n^{6}+1)(1+x^{4}_{3n-7})}\biggr) \\ &\quad {}+\frac{\sin^{2}(n^{12}x_{2n^{2}-1}-3n^{4}+1)}{2n^{26}+3n^{8}+1} =\frac{(-1)^{n}n^{3}+n-2}{ n^{9}+9n^{6}-3n^{3}+1},\quad \forall n\ge9, \end{aligned}$$
(3.4)
where \(\tau\in\Bbb{N}\) is fixed. Let \(n_{0}=9, k=1, b=-\frac{7}{8}, \beta=9-\tau\in\Bbb{N}\), M and N be two positive constants with \(M>8N\) and
$$\begin{aligned}& a_{n} =(-1)^{n}\ln^{3}(n+2),\qquad b_{n}=\frac{2-7\ln^{9}n}{3+8\ln^{9}n},\\& c_{n}=\frac{(-1)^{n}n^{3}+n-2}{ n^{9}+9n^{6}-3n^{3}+1},\qquad f_{1n}=2n^{2}-1, \\& F_{n} =\bigl(2n^{2}-1\bigr)^{2},\qquad f(n,u)= \frac{\sin^{2}(n^{12}u-3n^{4}+1)}{2n^{26}+3n^{8}+1}, \\& h(n,u)=\frac{-3n^{2}+\ln^{2}n-1}{(n^{9}+6n^{6}+1)(1+u^{4})}, \\& h_{1n} =3n-7, \qquad H_{n}=(3n-7)^{2},\qquad P_{n}=Q_{n}=\frac{3}{n^{11}},\\& R_{n}=W_{n}= \frac{5}{n^{7}},\quad \forall (n,u)\in\Bbb{N}_{n_{0}}\times\Bbb{R}. \end{aligned}$$
Obviously, (2.1), (2.2) and (2.34) are satisfied. Note that
$$\begin{aligned} &\frac{1}{n^{2}}\sum_{v=n}^{\infty}\sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\frac {1}{\vert a_{s}\vert }\max\{R_{s}H_{s},W_{s} \} \\ &\quad \le\frac{1}{n^{2}}\sum_{s=n}^{\infty}\frac{s^{2}}{\vert (-1)^{s}\ln^{3}(s+2)\vert }\max\biggl\{ \frac{5\vert 3s-7\vert ^{2}}{s^{7}},\frac{5}{s^{7}}\biggr\} \\ &\quad \le\frac{45}{n^{2}}\sum_{s=n}^{\infty}\frac{1}{s^{3}}\to 0\quad \text{as } n\to\infty \end{aligned}$$
and
$$\begin{aligned} &\frac{1}{n^{2}}\sum_{v=n}^{\infty}\sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\sum_{t=s}^{\infty}\frac{1}{\vert a_{s}\vert }\max\bigl\{ P_{t}F_{t},Q_{t}, \vert c_{t}\vert \bigr\} \\ &\quad =\frac{1}{n^{2}}\sum_{v=n}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty}\sum_{t=s}^{\infty}\frac{1}{\vert (-1)^{s}\ln^{3}(s+2)\vert }\max \biggl\{ \frac{3(2t^{2}-1)^{2}}{t^{11}},\frac{3}{t^{11}},\frac{ \vert (-1)^{t}t^{3}+t-2\vert }{t^{9}+9t^{6}-3t^{3}+1}\biggr\} \\ &\quad \leq\frac{12}{n^{2}}\sum_{v=n}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty}\sum_{t=s}^{\infty}\frac{1}{t^{7}} \leq\frac{12}{n^{2}}\sum_{t=n}^{\infty} \frac{1}{t^{4}}\to 0\quad \text{as } n\to\infty, \end{aligned}$$
which yield that
$$\lim_{n\to \infty}\frac{1}{n^{2}}\sum _{v=n}^{\infty}\sum_{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{\vert a_{s}\vert }\max \{R_{s}H_{s},W_{s}\}=0 $$
and
$$\lim_{n\to \infty}\frac{1}{n^{2}}\sum _{v=n}^{\infty}\sum_{u=v}^{\infty}\sum_{s=u}^{\infty}\sum _{t=s}^{\infty}\frac{1}{\vert a_{s}\vert } \max\bigl\{ P_{t}F_{t},Q_{t},\vert c_{t} \vert \bigr\} =0. $$
That is, (2.17) and (2.18) hold. Thus Theorem 2.4 shows that Eq. (3.4) possesses uncountably many positive solutions in \(A(N,M)\). Moreover, for each \(L\in(N, (1+b)M)\), there exist \(\theta\in(0, 1)\) and \(T\ge n_{0}+\tau+\beta\) such that for each \(x_{0}=\{x_{0n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\), the Mann iterative sequence \(\{x_{m}\}_{m\in\Bbb{N}_{0}}=\{\{x_{mn}\}_{n\in\Bbb{N}_{\beta}}\}_{m\in\Bbb{N}_{0}}\) generated by (2.27) converges to a positive solution \(w=\{w_{n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\) of Eq. (3.4) with (2.28) and (2.8), where \(\{\alpha_{m}\}_{m\in\Bbb{N}_{0}}\) is an arbitrary sequence in \([0, 1]\) satisfying (2.9).

Example 3.5

Consider the fourth order neutral delay difference equation
$$\begin{aligned} &\Delta\biggl(\bigl(n^{3}-n^{2}+1\bigr)\Delta^{3} \biggl(x_{n}+\biggl(3+\frac{3}{n} \biggr)x_{n-\tau} \biggr)\biggr) \\ &\quad {}+\Delta\biggl(\frac{n^{2}-3n+\arctan^{2}n}{(n^{17}+9n^{2}+1)(1+\vert \cos (n^{4}x_{2n-1}-n)\vert )}\biggr) \\ &\quad {}+\frac{n\cos(n^{3}x_{n-2})-1}{n^{18}+2n^{16}+\ln^{3}n} =\frac{(-1)^{n-1}n^{4}-2n^{3}+\sqrt{n+1}}{n^{21}+3n^{15}-2n^{11}+1},\quad \forall n\ge3, \end{aligned}$$
(3.5)
where \(\tau\in\Bbb{N}\) is fixed. Let \(n_{0}=3\), \(k=1\), \(b^{*}=4\), \(b_{*}=3\), \(q=\frac{\sqrt{2}}{3}\), \(\beta=\min\{3-\tau ,1\}=1\), \(M=300\), \(N=1\) and
$$\begin{aligned}& a_{n} =n^{3}-n^{2}+1,\qquad b_{n}=3+\frac{3}{n},\qquad c_{n}=\frac{(-1)^{n-1}n^{4}-2n^{3}+\sqrt{n+1}}{n^{21}+3n^{15}-2n^{11}+1}, \\& f_{1n} =n-2,\qquad F_{n}=(n-2)^{2},\qquad h_{1n}=2n-1,\qquad H_{n}=(2n-1)^{2}, \\& f(n ,u)=\frac{n\cos(n^{3}u)-1}{n^{18}+2n^{16}+\ln^{3}n},\qquad h(n,u)=\frac{n^{2}-3n+\arctan^{2}n}{(n^{17}+9n^{2}+1)(1+\vert \cos(n^{4}u-n)\vert )}, \\& P_{n} =Q_{n}=\frac{1}{n^{14}}, \qquad R_{n}=W_{n}= \frac{2}{n^{9}},\quad \forall (n,u)\in\Bbb{N}_{n_{0}}\times\Bbb{R}. \end{aligned}$$
Clearly, (2.1), (2.2) and (2.39) are satisfied. Note that Lemma 1.1 yields that
$$\begin{aligned} &\frac{1}{n^{2}}\sum_{v=n}^{\infty}\sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\frac {1}{\vert a_{s}\vert }\max\{R_{s}H_{s},W_{s} \} \\ &\quad \le\frac{1}{n^{2}}\sum_{s=n}^{\infty}\frac{s^{2}}{s^{3}-s^{2}+1} \max\biggl\{ \frac{2(2s-1)^{2}}{s^{9}},\frac{2}{s^{9}}\biggr\} \\ &\quad \le\frac{8}{n^{2}}\sum_{s=n}^{\infty}\frac{1}{s^{5}}\to 0\quad \text{as } n\to\infty \end{aligned}$$
and
$$\begin{aligned} &\frac{1}{n^{2}}\sum_{v=n}^{\infty}\sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\sum_{t=s}^{\infty}\frac{1}{\vert a_{s}\vert }\max\bigl\{ P_{t}F_{t},Q_{t}, \vert c_{t}\vert \bigr\} \\ &\quad =\frac{1}{n^{2}}\sum_{v=n}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty}\sum_{t=s}^{\infty}\frac{1}{s^{3}-s^{2}+1}\max \biggl\{ \frac{\vert t-2\vert ^{2}}{t^{14}},\frac{1}{t^{14}}, \frac{ \vert (-1)^{t-1}t^{4}-2t^{3}+\sqrt{t+1}\vert }{t^{21}+3t^{15}-2t^{11}+1}\biggr\} \\ &\quad \le\frac{1}{n^{2}}\sum_{v=n}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty}\sum_{t=s}^{\infty}\frac{1}{t^{12}} \\ &\quad \le\frac{1}{n^{2}}\sum_{t=n}^{\infty} \frac{1}{t^{9}}\to 0\quad \text{as } n\to\infty, \end{aligned}$$
which mean that
$$\lim_{n\to \infty}\frac{1}{n^{2}}\sum _{v=n}^{\infty}\sum_{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{\vert a_{s}\vert }\max \{R_{s}H_{s},W_{s}\}=0 $$
and
$$\lim_{n\to \infty}\frac{1}{n^{2}}\sum _{v=n}^{\infty}\sum_{u=v}^{\infty}\sum_{s=u}^{\infty}\sum _{t=s}^{\infty}\frac{1}{\vert a_{s}\vert } \max\bigl\{ P_{t}F_{t},Q_{t},\vert c_{t} \vert \bigr\} =0. $$
That is, (2.17) and (2.18) hold. Thus Theorem 2.5 shows that Eq. (3.5) possesses uncountably many positive solutions in \(A(N,M)\). Moreover, for each \(L\in(b^{*}(Mq+N), \frac{M}{q}+\frac{N}{qb^{*}})\), there exist \(\theta\in(0, 1)\) and \(T\ge n_{0}+\tau+\beta\) such that for each \(x_{0}=\{x_{0n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\), the Mann iterative sequence \(\{x_{m}\}_{m\in\Bbb{N}_{0}}=\{\{x_{mn}\}_{n\in\Bbb{N}_{\beta}}\}_{m\in\Bbb{N}_{0}}\) generated by (2.40) converges to a positive solution \(w=\{w_{n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\) of Eq. (3.5) with (2.28) and (2.8), where \(\{\alpha_{m}\}_{m\in\Bbb{N}_{0}}\) is an arbitrary sequence in \([0, 1]\) satisfying (2.9).

Example 3.6

Consider the fourth order neutral delay difference equation
$$\begin{aligned} &\Delta\biggl(n^{4}\Delta^{3}\biggl(x_{n}- \frac {2n^{12}+9n^{11}-1}{n^{12}+3n^{11}+2}x_{n-\tau}\biggr)\biggr) +\Delta\biggl( \frac{(-1)^{n}\cos(n^{30}-2\sqrt{n+1})}{(n+3)^{9}\sqrt {n\vert x_{n-2}\vert +1}}\biggr) \\ &\quad {}+\frac{n^{4}-\ln^{3}n}{n^{15}+2n^{2}+\sin(n^{3}x_{n-1})} =\frac{(-1)^{n-1}n^{4}+5\ln^{5}n-1}{n^{13}+12n^{11}+1},\quad \forall n\ge6, \end{aligned}$$
(3.6)
where \(\tau\in\Bbb{N}\) is fixed. Let \(n_{0}=6, k=1, b^{*}=-2, b_{*}=-3, \beta=\min\{6-\tau,4\}\in\Bbb{N}\), M and N be two positive constants with \(2N>M>N\) and
$$\begin{aligned}& a_{n} =n^{4},\qquad b_{n}=- \frac{2n^{12}+9n^{11}-1}{n^{12}+3n^{11}+2},\qquad c_{n}=\frac{(-1)^{n-1}n^{4}+5\ln^{5}n-1}{n^{13}+12n^{11}+1}, \\& f_{1n} =n-1,\qquad F_{n}=(n-1)^{2},\qquad h_{1n}=n-2, \qquad H_{n}=(n-2)^{2}, \\& f(n ,u)=\frac{n^{4}-\ln^{3}n}{n^{15}+2n^{2}+\sin(n^{3}u)},\qquad h(n,u)=\frac{(-1)^{n}\cos(n^{30}-2\sqrt{n+1})}{(n+3)^{9}\sqrt {n\vert u\vert +1}}, \\& P_{n} =Q_{n}=R_{n}=W_{n}= \frac{1}{n^{8}},\quad \forall(n,u)\in\Bbb{N}_{n_{0}}\times\Bbb{R}. \end{aligned}$$
Obviously, (2.1), (2.2) and (2.47) are satisfied. Note that Lemma 1.1 guarantees that
$$\begin{aligned} &\frac{1}{n^{2}}\sum_{v=n}^{\infty}\sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\frac {1}{\vert a_{s}\vert }\max\{R_{s}H_{s},W_{s} \} \\ &\quad \le\frac{1}{n^{2}}\sum_{s=n}^{\infty}\frac{s^{2}}{s^{4}}\max\biggl\{ \frac {\vert s-2\vert ^{2}}{s^{8}},\frac{1}{s^{8}}\biggr\} \\ &\quad \le\frac{1}{n^{2}}\sum_{s=n}^{\infty}\frac{1}{s^{8}}\to 0\quad \text{as } n\to\infty \end{aligned}$$
and
$$\begin{aligned} &\frac{1}{n^{2}}\sum_{v=n}^{\infty}\sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\sum_{t=s}^{\infty}\frac{1}{\vert a_{s}\vert } \max\bigl\{ P_{t}F_{t},Q_{t}, \vert c_{t}\vert \bigr\} \\ &\quad =\frac{1}{n^{2}}\sum_{v=n}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty}\sum_{t=s}^{\infty}\frac{1}{s^{4}}\max \biggl\{ \frac{(t-1)^{2}}{t^{8}},\frac{1}{t^{8}}, \frac{\vert (-1)^{t-1}t^{4}+5\ln ^{5}t-1\vert }{t^{13}+12t^{11}+1}\biggr\} \\ &\quad \le\frac{1}{n^{2}}\sum_{v=n}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty}\sum_{t=s}^{\infty}\frac{1}{t^{6}} \\ &\quad \le\frac{1}{n^{2}}\sum_{t=n}^{\infty} \frac{1}{t^{3}}\to 0\quad \text{as } n\to\infty, \end{aligned}$$
which imply that
$$\lim_{n\to \infty}\frac{1}{n^{2}}\sum _{v=n}^{\infty}\sum_{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{\vert a_{s}\vert }\max \{R_{s}H_{s},W_{s}\}=0 $$
and
$$\lim_{n\to \infty}\frac{1}{n^{2}}\sum _{v=n}^{\infty}\sum_{u=v}^{\infty}\sum_{s=u}^{\infty}\sum _{t=s}^{\infty}\frac{1}{\vert a_{s}\vert } \max\bigl\{ P_{t}F_{t},Q_{t},\vert c_{t} \vert \bigr\} =0. $$
That is, (2.17) and (2.18) hold. Thus Theorem 2.6 shows that Eq. (3.6) possesses uncountably many positive solutions in \(A(N,M)\). Moreover, for each \(L\in(N(1+b_{*}), M(1+b^{*}))\), there exist \(\theta\in(0, 1)\) and \(T\ge n_{0}+\tau+\beta\) such that for each \(x_{0}=\{x_{0n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\), the Mann iterative sequence \(\{x_{m}\}_{m\in\Bbb{N}_{0}}=\{\{x_{mn}\}_{n\in\Bbb{N}_{\beta}}\}_{m\in\Bbb{N}_{0}}\) generated by (2.40) converges to a positive solution \(w=\{w_{n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\) of Eq. (3.6) with (2.28) and (2.8), where \(\{\alpha_{m}\}_{m\in\Bbb{N}_{0}}\) is an arbitrary sequence in \([0, 1]\) satisfying (2.9).

Example 3.7

Consider the fourth order neutral delay difference equation
$$\begin{aligned} &\Delta\biggl(n\ln^{2}(n+3)\Delta^{3}\biggl(x_{n}+ \frac {2(-1)^{n}n^{8}-n+1}{5n^{8}+3n-1}x_{n-\tau}\biggr)\biggr) +\Delta\biggl( \frac{\sin x_{n-6}}{n^{12}+nx^{2}_{n^{2}-2}}\biggr) \\ &\quad {}+\frac{(-1)^{n-1}n^{3}\cos^{3}(4n^{9}-3\ln^{2}n)}{n^{15}+\ln ^{8}n+\vert nx_{3n-1}-x_{2n-3}\vert } =\frac{(-1)^{n}n^{8}-5n^{7}-4n^{3}+1}{n^{25}+30n^{16}-2n^{7}+1},\quad \forall n\ge8, \end{aligned}$$
(3.7)
where \(\tau\in\Bbb{N}\) is fixed. Let \(n_{0}=4, k=2, b=\frac{2}{5},\beta=\min\{4-\tau,2\}\in\Bbb{N}\), M and N be two positive constants with \(M>5N\) and
$$\begin{aligned}& a_{n} =n\ln^{2}(n+3), \qquad b_{n}= \frac{2(-1)^{n}n^{8}-n+1}{5n^{8}+3n-1},\qquad c_{n}=\frac {(-1)^{n}n^{8}-5n^{7}-4n^{3}+1}{n^{25}+30n^{16}-2n^{7}+1}, \\& f_{1n} =3n-1,\qquad f_{2n}=2n-3,\qquad F_{n}=(3n-1)^{2}, \\& h_{1n} =n^{2}-2,\qquad h_{2n}=n-4, \qquad H_{n}= \bigl(n^{2}-2\bigr)^{2}, \\& f(n ,u,v)=\frac{(-1)^{n-1}n^{3}\cos^{3}(4n^{9}-3\ln^{2}n)}{n^{15}+\ln ^{8}n+\vert nu-v\vert },\qquad h(n,u,v)=\frac{\sin u}{n^{12}+nv^{2}}, \\& P_{n} =Q_{n}=R_{n}=W_{n}= \frac{4}{n^{11}},\quad \forall(n,u,v)\in\Bbb{N}_{n_{0}}\times\Bbb{R}^{2}. \end{aligned}$$
It is not difficult to verify that (2.1), (2.2) and (2.52) are fulfilled. Note that Lemma 1.1 gives that
$$\begin{aligned} &\frac{1}{n^{2}}\sum_{v=n}^{\infty}\sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\frac {1}{\vert a_{s}\vert }\max\{R_{s}H_{s},W_{s} \} \\ &\quad \le\frac{1}{n^{2}}\sum_{s=n}^{\infty}\frac{s^{2}}{s\ln^{2}(s+3)} \max\biggl\{ \frac{4(s^{2}-2)^{2}}{s^{11}},\frac{4}{s^{11}}\biggr\} \\ &\quad \leq\frac{4}{n^{2}}\sum_{s=n}^{\infty} \frac{1}{s^{6}}\to 0\quad \text{as } n\to\infty \end{aligned}$$
and
$$\begin{aligned} &\frac{1}{n^{2}}\sum_{v=n}^{\infty}\sum_{u=v}^{\infty}\sum _{s=u}^{\infty}\sum_{t=s}^{\infty}\frac{1}{\vert a_{s}\vert } \max\bigl\{ P_{t}F_{t},Q_{t}, \vert c_{t}\vert \bigr\} \\ &\quad =\frac{1}{n^{2}}\sum_{v=n}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty}\sum_{t=s}^{\infty}\frac{1}{s\ln^{2}(s+3)}\max \biggl\{ \frac{4(3t-1)^{2}}{t^{11}},\frac{4}{t^{11}},\frac {\vert (-1)^{t}t^{8}-5t^{7}-4t^{3}+1\vert }{t^{25}+30t^{16}-2t^{7}+1}\biggr\} \\ &\quad \leq\frac{36}{n^{2}}\sum_{v=n}^{\infty}\sum _{u=v}^{\infty}\sum_{s=u}^{\infty}\sum_{t=s}^{\infty}\frac{1}{t^{9}} \\ &\quad \leq\frac{36}{n^{2}}\sum_{t=n}^{\infty} \frac{1}{t^{6}}\to 0\quad \text{as } n\to\infty, \end{aligned}$$
which mean that
$$\lim_{n\to \infty}\frac{1}{n^{2}}\sum _{v=n}^{\infty}\sum_{u=v}^{\infty}\sum_{s=u}^{\infty}\frac{1}{\vert a_{s}\vert }\max \{R_{s}H_{s},W_{s}\}=0 $$
and
$$\lim_{n\to \infty}\frac{1}{n^{2}}\sum _{v=n}^{\infty}\sum_{u=v}^{\infty}\sum_{s=u}^{\infty}\sum _{t=s}^{\infty}\frac{1}{\vert a_{s}\vert } \max\bigl\{ P_{t}F_{t},Q_{t},\vert c_{t} \vert \bigr\} =0. $$
That is, (2.17) and (2.18) hold. Consequently Theorem 2.7 implies that Eq. (3.7) possesses uncountably many positive solutions in \(A(N,M)\). Moreover, for each \(L\in(N+bM, (1-b)M)\), there exist \(\theta\in(0, 1)\) and \(T\ge n_{0}+\tau+\beta\) such that for each \(x_{0}=\{x_{0n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\), the Mann iterative sequence \(\{x_{m}\}_{m\in\Bbb{N}_{0}}=\{\{x_{mn}\}_{n\in\Bbb{N}_{\beta}}\}_{m\in\Bbb{N}_{0}}\) generated by (2.27) converges to a positive solution \(w=\{w_{n}\}_{n\in\Bbb{N}_{\beta}}\in A(N,M)\) of Eq. (3.7) with (2.28) and (2.8), where \(\{\alpha_{m}\}_{m\in\Bbb{N}_{0}}\) is an arbitrary sequence in \([0, 1]\) satisfying (2.9).

Declarations

Acknowledgements

This research was supported by the Science Research Foundation of Educational Department of Liaoning Province (L2012380) and the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Science, ICT & Future Planning (2013R1A1A2057665).

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
Department of Mathematics, Liaoning Normal University, Dalian, People’s Republic of China
(2)
Department of Mathematics, Changwon National University, Changwon, Korea
(3)
Department of Mathematics and RINS, Gyeongsang National University, Jinju, Korea

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