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
-
(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)
-
(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
-
(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);
-
(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
-
(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);
-
(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
-
(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);
-
(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
-
(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);
-
(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
-
(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);
-
(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
-
(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);
-
(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. □