# Positive Solutions and Mann Iterative Algorithms for a Second-Order Nonlinear Difference Equation

Authors: Zeqing Liu; Xin Li; Shin Min Kang; Young Chel Kwun
Journal: Journal of Function Spaces
 (2016)

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## Abstract

This paper deals with the second-order nonlinear neutral delay difference equationΔ(anh(xn-τ1n,xn-τ2n,…,xn-τmn)Δ(xn-qnxn-τ0))+f(n,xn-σ1n,xn-σ2n,…,xn-σkn)=bn, n≥n0. Using the Banach fixed point theorem, Mann iterative method with errors, and some new techniques, we prove the existence of uncountably many positive solutions and the convergence of the sequences generated by the Mann iterative method with errors relative to these solutions for the above equation. Six examples are included. Our results extend and improve essentially the known results in this field.


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## Body

## 1. Introduction and Preliminaries

Consider the second-order nonlinear neutral delay difference equation with forcing term:(1)Δanhxn-τ1n,xn-τ2n,…,xn-τmnΔxn-qnxn-τ0+fn,xn-σ1n,xn-σ2n,…,xn-σkn=bn,n≥n0,where n0∈N0, m,k,τ0∈N, ⋃i∈{1,2,…,m}{τin}n∈Nn0⊂Z, ⋃j∈{1,2,…,k}{σjn}n∈Nn0⊂Z with(2)limn→∞⁡n-τin=limn→∞⁡n-σjn=+∞,i∈1,2,…,m,j∈1,2,…,k,{an}n∈Nn0, {bn}n∈Nn0, and {qn}n∈Nn0 are real sequences with an≠0 for each n∈Nn0, f∈C(Nn0×Rk,R), and h∈C(Rm,R).In the recent ten years, there have been a lot of activities concerning the oscillation, nonoscillation, asymptotic behavior, and existence of solutions, nonoscillatory solutions, and positive solutions for various types of difference equations, which are special cases of (1); see, for example, [1–21] and the references cited therein. In 2007, Cheng [3] used the Banach fixed point theorem to investigate the second-order neutral delay linear difference equation with positive and negative coefficients: (3)Δ2xn+pxn-m+pnxn-k-qnxn-l=0,n≥n0,and got a sufficient condition for the existence of a nonoscillatory solution of (3). In 2004, Malaguti et al. [15] discussed the second-order nonlinear difference equation (4)ΔrkΔuk+qkguk+1=0,k≥0,and obtained sufficient and necessary conditions for the existence of bounded solutions of (4). In 2007, by using the Krasnoselskii fixed point theorem and Schauder fixed point theorem, Rath et al. [16] investigated the existence of solutions of the second-order nonlinear neutral delay difference equation (5)ΔrnΔxn-pnxn-m+qnGxn-k=0,n≥n0. In 2003, Agarwal et al. [1] used a nonlinear alternative of Leray-Schauder type to obtain the existence of a nonoscillatory solution for the following second-order neutral delay difference equation: (6)ΔanΔyn+pyn-τ+Fn+1,yn+1-σ=0,n≥1.In 2009, using the Banach fixed point theorem, Liu et al. [12] researched the global solvability for the second-order nonlinear neutral delay difference equation (7)ΔanΔxn+bxn-τ+fn,xn-d1n,xn-d2n,…,xn-ddn=cn,n≥n0.However, the study of the existence of bounded positive solutions for a lot of second-order neutral delay difference equations appears to be insufficient. In particular, the uncountably many bounded positive solutions for these second-order neutral delay difference equations obtain less concern.The purpose of this paper is to study the solvability and convergence of Mann iterative approximations of (1). By means of the Banach fixed point theorem, Mann iterative method with errors, and a few new techniques, we obtain the existence of uncountably many bounded positive solutions of (1) and prove that the sequences generated by the Mann iterative method with errors converge to these bounded positive solutions. Our results generalize and improve all results in [3, 12]. Six examples are constructed in order to illustrate the superiority and applications of the results presented in this paper.Throughout this paper, we assume thatR=(-∞,+∞), N0={0}∪N, N and Z denote the sets of all positive integers and integers, respectively, and Δ stands for the forward difference operator; that is, Δxn=xn+1-xn. Let (8)β=inf⁡n-τ0,n-τin,n-σjn:n∈Nn0,1≤i≤k,1≤j≤m,Zβ=n:n∈Zwithn≥β,Nt=n:n∈N0withn≥t,∀t∈N0. Assume that lβ∞ represents the Banach space of all bounded real sequences x={xn}n∈Zβ with the norm x=supn∈Zβxn and (9)AN,M=x=xn∈lβ∞:N≤xn≤M,n∈Zβ,∀M>N>0.It is easy to see that A(N,M) is a bounded closed and convex subset of the Banach space lβ∞.By a solution of (1), we mean a real sequence {xn}n∈Zβ with a positive integer D≥n0+τ0+β such that (1) holds for all n≥D.Lemma 1 (see [22]).
Let{αn}n≥0, {βn}n≥0, {γn}n≥0, and {tn}n≥0 be four nonnegative real sequences satisfying the inequality (10)αn+1≤1-tnαn+tnβn+γn,∀n≥0,where {tn}n≥0⊆[0,1], ∑n=0∞tn=+∞, limn→∞βn=0, and ∑n=0∞γn<+∞. Then, limn→∞αn=0.Lemma 2.
Letτ∈N, n0∈N0 be two constants and let {cn}n∈Nn0+1, {kn}n∈Nn0 be two nonnegative sequences. Then, (11)∑j=0∞∑i=n0+1+jτ∞ci∑s=n0i-1ks<+∞⟺∑i=n0+1∞ici∑s=n0i-1ks<+∞.Proof.
For eacht∈R, [t] denotes the largest integer not exceeding t. Notice that (12)∑j=0∞∑i=n0+1+jτ∞ci∑s=n0i-1ks=∑i=n0+1∞ci∑s=n0i-1ks+∑i=n0+1+τ∞ci∑s=n0i-1ks+∑i=n0+1+2τ∞ci∑s=n0i-1ks+⋯=∑i=n0+1∞1+i-n0-1τci∑s=n0i-1ks,limi→∞⁡1+i-n0-1/τi/τ=1.Thus, (11) follows from (12). This completes the proof.


## 2. Main Results

According to the different ranges of the sequenceqnn∈Nn0, now we use the Banach fixed point theorem to study the existence of uncountably many bounded positive solutions and convergence of the Mann iterative approximations of (1).Theorem 3.
Assume that there exist constantsM, N, q, Lh, and mh and two nonnegative sequences {Pn}n∈Nn0 and {Wn}n∈Nn0 satisfying (13)0<N<1-qM;(14)0≤qn≤q<1,eventually;(15)hu1,u2,…,um-hu¯1,u¯2,…,u¯m≤Lhmax⁡ui-u¯i:1≤i≤m,∀ui,u¯i∈N,M2,1≤i≤m;(16)hu1,u2,…,um≥mh>0,∀ui∈N,M,1≤i≤m;(17)fn,u1,u2,…,uk-fn,u¯1,u¯2,…,u¯k≤Pnmax⁡ui-u¯i:1≤i≤m,∀n,ui,u¯i∈Nn0×N,M2,1≤i≤m;(18)fn,u1,u2,…,uk≤Wn,∀n,ui∈Nn0×N,M,1≤i≤m;(19)∑i=n0+1∞1ai∑s=n0i-1max⁡Ws,Ps,bs<+∞.Then, one has the following.
(a) For anyc∈(N,(1-q)M), there exist θ∈(0,1) and D≥1+n0+τ0+β such that, for each x0={x0,n}n∈Zβ∈A(N,M), the Mann iterative sequence with errors {xη}η∈N0={xη,n}(η,n)∈N0×Zβ generated by the scheme (20)
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,converges to a bounded positive solution x∈A(N,M) of (1) and has the following error estimate: (21)xη+1-x≤1-1-θαηxη-x+2Mβη,η≥0,where {γη}η∈N0={γη,n}(η,n)∈N0×Zβ is an arbitrary sequence in A(N,M) and {αη}η∈N0 and {βη}η∈N0 are any sequences in [0,1] such that (22)∑η=0∞αη=+∞,∑η=0∞βη<+∞orthereexistsasequenceξηη≥0⊂0,+∞satisfyingβη=ξηαη,∀η∈N0,limη→∞⁡ξη=0.
(b) Equation (1) has uncountably many bounded positive solutions in A(N,M).Proof.
First, we show that (a) holds. Letc∈(N,(1-q)M). It follows from (14) and (19) that there exist θ∈(0,1) and D≥1+n0+τ0+β sufficiently large such that (23)0≤qn≤q<1,∀n≥D,θ=q+1mh2∑i=D∞1ai∑s=n0i-1mhPs+LhWs+bs,1mh∑i=D∞1ai∑s=n0i-1Ws+bs<min⁡c-N,1-qM-c. Define an operator T:A(N,M)→lβ∞ by (24)Txn=c+qnxn-τ0+∑i=n∞1aihxi-τ1i,xi-τ2i,…,xi-τmi∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σks-bs,n≥D,TxD,β≤n<D,for any x={xn}n∈Zβ∈A(N,M). On account of (15)~(18) and (23)~(24), we get that, for every x={xn}n∈Zβ,y={yn}n∈Zβ∈A(N,M), and n≥D, (25)Txn=c+qnxn-τ0+∑i=n∞1aihxi-τ1i,xi-τ2i,…,xi-τmi∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σks-bs≥c-1mh∑i=n∞1ai∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σks+bs≥c-1mh∑i=n∞1ai∑s=n0i-1Ws+bs≥c-min⁡c-N,1-qM-c≥N,Txn=c+qnxn-τ0+∑i=n∞1aihxi-τ1i,xi-τ2i,…,xi-τmi∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σks-bs≤c+qM+1mh∑i=n∞1ai∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σks+bs≤c+qM+1mh∑i=n∞1ai∑s=n0i-1Ws+bs≤c+qM+min⁡c-N,1-qM-c≤M,Txn-Tyn=qnxn-τ0-yn-τ0+∑i=n∞1aihxi-τ1i,xi-τ2i,…,xi-τmi∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σks-bs-∑i=n∞1aihyi-τ1i,yi-τ2i,…,yi-τmi∑s=n0i-1fs,ys-σ1s,ys-σ2s,…,ys-σks-bs≤qx-y+∑i=n∞1ai∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σkshxi-τ1i,xi-τ2i,…,xi-τmi-fs,ys-σ1s,ys-σ2s,…,ys-σkshxi-τ1i,xi-τ2i,…,xi-τmi+fs,ys-σ1s,ys-σ2s,…,ys-σkshxi-τ1i,xi-τ2i,…,xi-τmi-fs,ys-σ1s,ys-σ2s,…,ys-σkshyi-τ1i,yi-τ2i,…,yi-τmi+bs1hxi-τ1i,xi-τ2i,…,xi-τmi-1hyi-τ1i,yi-τ2i,…,yi-τmi≤qx-y+∑i=n∞1ai∑s=n0i-1Psmhmax⁡xs-σjs-ys-σjs:1≤j≤k+Lhmh2fs,ys-σ1s,ys-σ2s,…,ys-σks+bsmax⁡xi-τji-yi-τji:1≤j≤m≤qx-y+∑i=n∞1ai∑s=n0i-1Psmhx-y+LhWs+bsmh2x-y≤q+1mh2∑i=D∞1ai∑s=n0i-1mhPs+LhWs+bsx-y=θx-y, which yield that (26)TAN,M⊆AN,M,Tx-Ty≤θx-y,∀x,y∈AN,M;that is, T is a contraction operator in A(N,M) and T has a unique fixed point x={xn}n∈Zβ∈A(N,M), which means that for all n≥D(27)xn=c+qnxn-τ0+∑i=n∞1aihxi-τ1i,xi-τ2i,…,xi-τmi∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σks-bs,which yields that for all n≥D(28)Δxn-qnxn-τ0=-1anhxn-τ1n,xn-τ2n,…,xn-τmn∑s=n0n-1fs,xs-σ1s,xs-σ2s,…,xs-σks-bs,Δanhxn-τ1n,xn-τ2n,…,xn-τmnΔxn-qnxn-τ0=-fn,xn-σ1n,xn-σ2n,…,xn-σkn+bn,which implies that x={xn}n∈Zβ is a bounded positive solution of (1) in A(N,M).
By (20), (24), and (26), we infer that for any η≥0 and n≥D(29)xη+1,n-xn=1-αη-βηxη,n+αηc+qnxη,n-τ0+∑i=n∞1aihxη,i-τ1i,xη,i-τ2i,…,xη,i-τmi∑s=n0i-1fs,xη,s-σ1s,xη,s-σ2s,…,xη,s-σks-bs+βηγη,n-xn≤1-αη-βηxη,n-xn+αηTxη,n-Txn+βηγη,n-xn≤1-αη-βηxη-x+αηθxη-x+2Mβη≤1-1-θαηxη-x+2Mβη,which implies that (30)xη+1-x≤1-1-θαηxη-x+2Mβη,η≥0.That is, (21) holds. Consequently, Lemma 1 and (21)~(22) imply that limη→∞xη=x.
Next, we show that (b) holds. Letc1, c2 denote two arbitrary constants in (N,(1-q)M) with c1≠c2. As in the proof of (a), we infer similarly that for each l∈{1,2} there exist constants θl∈(0,1), Dl>1+n0+τ0+β, and an operator Tl:A(N,M)→A(N,M) satisfying (23), (24), and (26), where θl, Dl, and Tl take the place of θ, D, and T, respectively. Moreover, the contraction operator Tl has a unique fixed point xl={xnl}n∈Zβ∈A(N,M), which is also a bounded positive solution of (1); that is, (31)xnl=cl+qnxn-τ0l+∑i=n∞1aihxi-τ1il,xi-τ2il,…,xi-τmil∑s=n0i-1fs,xs-σ1sl,xs-σ2sl,…,xs-σksl-bs,∀n≥Dl,which together with (15)~(18), (23), (24), and (26) gives that for each n≥max⁡{D1,D2}(32)xn1-xn2=c1+qnxn-τ01+∑i=n∞1aihxi-τ1i1,xi-τ2i1,…,xi-τmi1∑s=n0i-1fs,xs-σ1s1,xs-σ2s1,…,xs-σks1-bs-c2-qnxn-τ02-∑i=n∞1aihxi-τ1i2,xi-τ2i2,…,xi-τmi2∑s=n0i-1fs,xs-σ1s2,xs-σ2s2,…,xs-σks2-bs≥c1-c2-qnxn-τ01-xn-τ02-∑i=n∞1ai∑s=n0i-1fs,xs-σ1s1,xs-σ2s1,…,xs-σks1hxi-τ1i1,xi-τ2i1,…,xi-τmi1-fs,xs-σ1s2,xs-σ2s2,…,xs-σks2hxi-τ1i2,xi-τ2i2,…,xi-τmi2+bs1hxi-τ1i1,xi-τ2i1,…,xi-τmi1-1hxi-τ1i2,xi-τ2i2,…,xi-τmi2≥c1-c2-qx1-x2-∑i=n∞1ai∑s=n0i-1fs,xs-σ1s1,xs-σ2s1,…,xs-σks1-fs,xs-σ1s2,xs-σ2s2,…,xs-σks2hxi-τ1i1,xi-τ2i1,…,xi-τmi1+fs,xs-σ1s2,xs-σ2s2,…,xs-σks2hxi-τ1i1,xi-τ2i1,…,xi-τmi1-fs,xs-σ1s2,xs-σ2s2,…,xs-σks2hxi-τ1i2,xi-τ2i2,…,xi-τmi2+bsmh2hxi-τ1i1,xi-τ2i1,…,xi-τmi1-hxi-τ1i2,xi-τ2i2,…,xi-τmi2≥c1-c2-qx1-x2-∑i=n∞1ai∑s=n0i-1Psmhmax⁡xs-σls1-xs-σls2:1≤l≤k+LhWs+bsmh2max⁡xi-τli1-xi-τli2:1≤l≤m≥c1-c2-qx1-x2-x1-x2mh2∑i=n∞1ai∑s=n0i-1mhPs+LhWs+bs≥c1-c2-q+1mh2∑i=max⁡D1,D2∞1ai∑s=n0i-1mhPs+LhWs+bsx1-x2≥c1-c2-max⁡θ1,θ2x1-x2, which yields that (33)x1-x2≥c1-c21+max⁡θ1,θ2>0; that is, x1≠x2. This completes the proof.Theorem 4.
Assume that there exist constantsM, N, q, Lh, and mh and two nonnegative sequences {Pn}n∈Nn0 and {Wn}n∈Nn0 satisfying (15)~(19): (34)0<N<1+qM;(35)-1<q≤qn≤0,eventually.Then, one has the following.
(a) For anyc∈(N-qM,M), there exist θ∈(0,1) and D≥1+n0+τ0+β such that, for each x0={x0,n}n∈Zβ∈A(N,M), the Mann iterative sequence with errors {xη}η∈N0={xη,n}(η,n)∈N0×Zβ generated by (20) converges to a bounded positive solution x∈A(N,M) of (1) and has the error estimate (21), where {γη}η∈N0={γη,n}(η,n)∈N0×Zβ is an arbitrary sequence in A(N,M) and {αη}η∈N0 and {βη}η∈N0 are any sequences in [0,1] satisfying (22).
(b) Equation (1) has uncountably many bounded positive solutions in A(N,M).Proof.
First, we show that (a) holds. Letc∈(N-qM,M). It follows from (19) and (35) that there exist θ∈(0,1) and D≥1+n0+τ0+β sufficiently large such that (36)-1<q≤qn≤0,∀n≥D,θ=-q+1mh2∑i=D∞1ai∑s=n0i-1mhPs+LhWs+bs,1mh∑i=D∞1ai∑s=n0i-1Ws+bs<min⁡c+qM-N,M-c.Let the operator T:A(N,M)→lβ∞ be defined by (24). By virtue of (15)~(18), (24), and (36), we obtain that, for every x={xn}n∈Zβ, y={yn}n∈Zβ∈A(N,M), and n≥D, (37)Txn=c+qnxn-τ0+∑i=n∞1aihxi-τ1i,xi-τ2i,…,xi-τmi∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σks-bs≥c+qM-1mh∑i=n∞1ai∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σks+bs≥c+qM-1mh∑i=n∞1ai∑s=n0i-1Ws+bs≥c+qM-min⁡c+qM-N,M-c≥N,Txn=c+qnxn-τ0+∑i=n∞1aihxi-τ1i,xi-τ2i,…,xi-τmi∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σks-bs≤c+1mh∑i=n∞1ai∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σks+bs≤c+1mh∑i=n∞1ai∑s=n0i-1Ws+bs≤c+min⁡c+qM-N,M-c≤M.The rest of the proof is similar to that of Theorem 3 and is omitted. This completes the proof.Theorem 5.
Assume that there exist constantsM, N, q_, q¯, Lh, and mh and two nonnegative sequences {Pn}n∈Nn0 and {Wn}n∈Nn0 satisfying (15)~(19): (38)0<q¯N<q_-1M;(39)1<q_≤qn≤q¯,eventually.Then, one has the following.
(a) For anyc∈(q¯N,(q_-1)M), there exist θ∈(0,1) and D≥1+n0+τ0+β such that, for each x0={x0,n}n∈Zβ∈A(N,M), the Mann iterative sequence with errors {xη}η∈N0={xη,n}(η,n)∈N0×Zβ generated by the scheme (40)
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s
-
σ
k
s
-
b
s
+
β
η
γ
η
,
D
,
β
≤
n
<
D
,
η
≥
0
,converges to a bounded positive solution x∈A(N,M) of (1) and has the error estimate (21), where {γη}η∈N0={γη,n}(η,n)∈N0×Zβ is an arbitrary sequence in A(N,M) and {αη}η∈N0 and {βη}η∈N0 are any sequences in [0,1] satisfying (22).
(b) Equation (1) has uncountably many bounded positive solutions in A(N,M).Proof.
First, we show that (a) holds. Letc∈(q¯N,(q_-1)M). It follows from (19) and (39) that there exist θ∈(0,1) and D≥1+n0+τ0+β sufficiently large such that (41)1<q_≤qn≤q¯,∀n≥D;θ=1q_+1q_mh2∑i=D∞1ai∑s=n0i-1mhPs+LhWs+bs;1mh∑i=D∞1ai∑s=n0i-1Ws+bs<min⁡cq¯-Nq_,q_M-c-M.Define an operator T:A(N,M)→lβ∞ as follows: (42)Txn=cqn+τ0+xn+τ0qn+τ0-1qn+τ0∑i=n+τ0∞1aihxi-τ1i,xi-τ2i,…,xi-τmi∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σks-bs,n≥D,TxD,β≤n<D,for any x={xn}n∈Zβ∈A(N,M). In view of (15)~(18) and (41)~(42), we get that, for every x={xn}n∈Zβ, y={yn}n∈Zβ∈A(N,M), and n≥D, (43)Txn=cqn+τ0+xn+τ0qn+τ0-1qn+τ0∑i=n+τ0∞1aihxi-τ1i,xi-τ2i,…,xi-τmi∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σks-bs≥cq¯-1q_mh∑i=n+τ0∞1ai∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σks+bs≥cq¯-1q_mh∑i=n+τ0∞1ai∑s=n0i-1Ws+bs≥cq¯-1q_min⁡cq¯-Nq_,q_M-c-M≥N,Txn=cqn+τ0+xn+τ0qn+τ0-1qn+τ0∑i=n+τ0∞1aihxi-τ1i,xi-τ2i,…,xi-τmi∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σks-bs≤cq_+Mq_+1q_mh∑i=n+τ0∞1ai∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σks+bs≤cq_+Mq_+1q_mh∑i=n+τ0∞1ai∑s=n0i-1Ws+bs≤cq_+Mq_+1q_min⁡cq¯-Nq_,q_M-c-M≤M,Txn-Tyn=xn+τ0-yn+τ0qn+τ0-1qn+τ0∑i=n+τ0∞1aihxi-τ1i,xi-τ2i,…,xi-τmi∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σks-bs+1qn+τ0∑i=n+τ0∞1aihyi-τ1i,yi-τ2i,…,yi-τmi∑s=n0i-1fs,ys-σ1s,ys-σ2s,…,ys-σks-bs≤1q_x-y+1q_∑i=n+τ0∞1ai∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σkshxi-τ1i,xi-τ2i,…,xi-τmi-fs,ys-σ1s,ys-σ2s,…,ys-σkshxi-τ1i,xi-τ2i,…,xi-τmi+fs,ys-σ1s,ys-σ2s,…,ys-σkshxi-τ1i,xi-τ2i,…,xi-τmi-fs,ys-σ1s,ys-σ2s,…,ys-σkshyi-τ1i,yi-τ2i,…,yi-τmi+bs1hxi-τ1i,xi-τ2i,…,xi-τmi-1hyi-τ1i,yi-τ2i,…,yi-τmi≤1q_x-y+1q_∑i=n+τ0∞1ai∑s=n0i-1Psmhmax⁡xs-σjs-ys-σjs:1≤j≤k+LhWs+bsmh2max⁡xs-τjs-ys-τjs:1≤j≤m≤1q_+1q_mh2∑i=n∞1ai∑s=n0i-1mhPs+LhWs+bsx-y≤θx-y,which yield that (26) holds; that is, T is a contraction operator in A(N,M) and T has a unique fixed point x={xn}n∈Zβ∈A(N,M), which gives that for all n≥D(44)xn=cqn+τ0+xn+τ0qn+τ0-1qn+τ0∑i=n+τ0∞1aihxi-τ1i,xi-τ2i,…,xi-τmi∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σks-bs,which implies that for all n≥D(45)Δxn-qnxn-τ0=-1anhxn-τ1n,xn-τ2n,…,xn-τmn∑s=n0n-1fs,xs-σ1s,xs-σ2s,…,xs-σks-bs,Δanhxn-τ1n,xn-τ2n,…,xn-τmnΔxn-qnxn-τ0=-fn,xn-σ1n,xn-σ2n,…,xn-σkn+bn,which means that x={xn}n∈Zβ is a bounded positive solution of (1) in A(N,M).
It follows from (26), (40), and (42) that for any η≥0 and n≥D(46)xη+1,n-xn=1-αη-βηxη,n+αηcqn+τ0+xη,n+τ0qn+τ0-1qn+τ0∑i=n+τ0∞1aihxη,i-τ1i,xη,i-τ2i,…,xη,i-τmi∑s=n0i-1fs,xη,s-σ1s,xη,s-σ2s,…,xη,s-σks-bs+βηγη,n-xn≤1-αη-βηxη,n-xn+αηTxη,n-Txn+βηγη,n-xn≤1-αη-βηxη-x+αηθxη-x+2Mβη≤1-1-θαηxη-x+2Mβη,which implies that (47)xη+1-x≤1-1-θαηxη-x+2Mβη,η≥0.That is, (21) holds. Consequently, Lemma 1 and (21)~(22) imply that limη→∞xη=x.
Next, we show that (b) holds. Letc1,c2∈(q¯N,(q_-1)M) with c1≠c2. As in the proof of (a), we conclude similarly that for each l∈{1,2} there exist constants θl∈(0,1), Dl>1+n0+τ0+β, and an operator Tl:A(N,M)→A(N,M) satisfying (26) and (41)~(42), where θ, D, and T are replaced by θl, Dl, and Tl, respectively, and the contraction operator Tl has a unique fixed point xl={xnl}n∈Zβ∈A(N,M), which is also a bounded positive solution of (1); that is, (48)xnl=clqn+τ0+xn+τ0lqn+τ0-1qn+τ0∑i=n+τ0∞1aihxi-τ1il,xi-τ2il,…,xi-τmil∑s=n0i-1fs,xs-σ1sl,xs-σ2sl,…,xs-σksl-bs,∀n≥Dl,which together with (15)~(18), (26), and (41)~(42) guarantees that for each n≥max⁡D1,D2(49)x1-x2=c1qn+τ0+xn+τ01qn+τ0-1qn+τ0∑i=n+τ0∞1aihxi-τ1i1,xi-τ2i1,…,xi-τmi1∑s=n0i-1fs,xs-σ1s1,xs-σ2s1,…,xs-σks1-bs-c2qn+τ0-xn+τ02qn+τ0+1qn+τ0∑i=n+τ0∞1aihxi-τ1i2,xi-τ2i2,…,xi-τmi2∑s=n0i-1fs,xs-σ1s2,xs-σ2s2,…,xs-σks2-bs≥c1-c2qn+τ0-xn+τ01-xn+τ02qn+τ0-1qn+τ0∑i=n+τ0∞1ai∑s=n0i-1fs,xs-σ1s1,xs-σ2s1,…,xs-σks1hxi-τ1i1,xi-τ2i1,…,xi-τmi1-fs,xs-σ1s2,xs-σ2s2,…,xs-σks2hxi-τ1i2,xi-τ2i2,…,xi-τmi2+bs1hxi-τ1i1,xi-τ2i1,…,xi-τmi1-1hxi-τ1i2,xi-τ2i2,…,xi-τmi2≥c1-c2q¯-x1-x2q_-1q_∑i=n+τ0∞1ai∑s=n0i-1fs,xs-σ1s1,xs-σ2s1,…,xs-σks1-fs,xs-σ1s2,xs-σ2s2,…,xs-σks2hxi-τ1i1,xi-τ2i1,…,xi-τmi1+fs,xs-σ1s2,xs-σ2s2,…,xs-σks2hxi-τ1i1,xi-τ2i1,…,xi-τmi1-fs,xs-σ1s2,xs-σ2s2,…,xs-σks2hxi-τ1i2,xi-τ2i2,…,xi-τmi2+bsmh2hxi-τ1i1,xi-τ2i1,…,xi-τmi1-hxi-τ1i2,xi-τ2i2,…,xi-τmi2≥c1-c2q¯-x1-x2q_-1q_∑i=n∞1ai∑s=n0i-1Psmhmax⁡xs-σls1-xs-σls2:1≤l≤k+LhWs+bsmh2max⁡xi-τli1-xi-τli2:1≤l≤m≥c1-c2q¯-1q_+1mh2q_∑i=max⁡D1,D2∞1ai∑s=n0i-1mhPs+LhWs+bsx1-x2≥c1-c2q¯-max⁡θ1,θ2x1-x2,which yields that (50)x1-x2≥c1-c2q¯1+max⁡θ1,θ2>0;that is, x1≠x2. This completes the proof.Theorem 6.
Assume that there exist constantsM, N, q_, q¯, Lh, and mh and two nonnegative sequences {Pn}n∈Nn0 and {Wn}n∈Nn0 satisfying (15)~(19): (51)0<q_q¯N<q_+q¯2M;(52)-q¯2<q_≤qn≤q¯<-1,eventually.Then, one has the following.
(a) For anyc∈(q¯M,q_N-q_M/q¯), there exist θ∈(0,1) and D≥1+n0+τ0+β such that, for each x0={x0,n}n∈Zβ∈A(N,M), the Mann iterative sequence with errors {xη}η∈N0={xη,n}(η,n)∈N0×Zβ generated by (40) converges to a bounded positive solution x∈A(N,M) of (1) and has the error estimate (21), where {γη}η∈N0={γη,n}(η,n)∈N0×Zβ is an arbitrary sequence in A(N,M) and {αη}η∈N0 and {βη}η∈N0 are any sequences in [0,1] satisfying (22).
(b) Equation (1) has uncountably many bounded positive solutions in A(N,M).Proof.
First, we show that (a) holds. Letc∈(q¯M,q_N-q_M/q¯). It follows from (19) and (52) that there exist θ∈(0,1) and D≥1+n0+τ0+β satisfying (53)θ=-1q¯-1q¯mh2∑i=D∞1ai∑s=n0i-1mhPs+LhWs+bs,–q¯2<q_≤qn≤q¯<-1,∀n≥D,1mh∑i=D∞1ai∑s=n0i-1Ws+bs<min⁡q¯N-q¯cq_-M,c-q¯M.Define an operator T:A(N,M)→lβ∞ by (42). In light of (15)~(19), (42), and (53), we get that, for every x={xn}n∈Zβ∈A(N,M), y={yn}n∈Zβ∈A(N,M), and n≥D+1, (54)Txn=cqn+τ0+xn+τ0qn+τ0-1qn+τ0∑i=n+τ0∞1aihxi-τ1i,xi-τ2i,…,xi-τmi∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σks-bs≥cq_+Mq¯+1q¯∑i=n+τ0∞1aimh∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σks+bs≥cq_+Mq¯+1q¯mh∑i=n+τ0∞1ai∑s=n0i-1Ws+bs≥cq_+Mq¯+1q¯min⁡q¯N-q¯cq_-M,c-q¯M≥N,Txn=cqn+τ0+xn+τ0qn+τ0-1qn+τ0∑i=n+τ0∞1aihxi-τ1i,xi-τ2i,…,xi-τmi∑s=Di-1fs,xs-σ1s,xs-σ2s,…,xs-σks-bs≤cq¯-1q¯∑i=n+τ0∞1aimh∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σks+bs≤cq¯-1q¯mh∑i=n+τ0∞1ai∑s=n0i-1Ws+bs≤cq¯-1q¯min⁡q¯N-q¯cq_-M,c-q¯M≤M.The rest of the proof is similar to that of Theorem 5 and is omitted. This completes the proof.Theorem 7.
Assume that there exist constantsM, N, Lh, and mh with 0<N<M and two nonnegative sequences {Pn}n∈Nn0 and {Wn}n∈Nn0 satisfying (15)~(19) and (55)qn=-1,eventually.Then, one has the following.
(a) For anyc∈(N,M), there exist θ∈(0,1) and D≥1+n0+τ0+β such that, for each x0={x0,n}n∈Zβ∈A(N,M), the Mann iterative sequence with errors {xη}η∈N0={xη,n}(η,n)∈N0×Zβ generated by the scheme (56)
x
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+
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=
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-
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=
1
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=
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1
a
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-
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=
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f
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,converges to a bounded positive solution x∈A(N,M) of (1) and has the error estimate (21), where {γη}η∈N0={γη,n}(η,n)∈N0×Zβ is an arbitrary sequence in A(N,M) and {αη}η∈N0 and {βη}η∈N0 are any sequences in [0,1] satisfying (22).
(b) Equation (1) has uncountably many bounded positive solutions in A(N,M).Proof.
First, we show that (a) holds. Letc∈(N,M). It follows from (19) that there exist θ∈(0,1) and D≥1+n0+τ0+β sufficiently large such that (57)θ=1mh2∑i=D∞1ai∑s=n0i-1mhPs+LhWs+bs;1mh∑i=D∞1ai∑s=n0i-1Ws+bs<minc-N,M-c.Define an operator T:A(N,M)→lβ∞ as follows: (58)Txn=c+∑j=1∞∑i=n+2j-1τ0n+2jτ0-11aihxi-τ1i,xi-τ2i,…,xi-τmi∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σks-bs,n≥D,TxD,β≤n<D,for each x={xn}n∈Zβ∈A(N,M). In view of (15)~(18) and (57)~(58), we conclude that, for every x={xn}n∈Zβ∈A(N,M), y={yn}n∈Zβ∈A(N,M), and n≥D, (59)Txn=c+∑j=1∞∑i=n+2j-1τ0n+2jτ0-11aihxi-τ1i,xi-τ2i,…,xi-τmi∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σks-bs≥c-∑i=n+τ0∞1aimh∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σks+bs≥c-1mh∑i=n+τ0∞1ai∑s=n0i-1Ws+bs≥c-min⁡c-N,M-c≥N,Txn=c+∑j=1∞∑i=n+2j-1τ0n+2jτ0-11aihxi-τ1i,xi-τ2i,…,xi-τmi∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σks-bs≤c+∑i=n+τ0∞1aimh∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σks+bs≤c+1mh∑i=n+τ0∞1ai∑s=n0i-1Ws+bs≤c+min⁡c-N,M-c≤M,Txn-Tyn=∑j=1∞∑i=n+2j-1τ0n+2jτ0-11aihxi-τ1i,xi-τ2i,…,xi-τmi∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σks-bs-∑j=1∞∑i=n+2j-1τ0n+2jτ0-11aihyi-τ1i,yi-τ2i,…,yi-τmi∑s=n0i-1fs,ys-σ1s,ys-σ2s,…,ys-σks-bs≤∑j=1∞∑i=n+2j-1τ0n+2jτ0-11ai∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σkshxi-τ1i,xi-τ2i,…,xi-τmi-fs,ys-σ1s,ys-σ2s,…,ys-σkshxi-τ1i,xi-τ2i,…,xi-τmi+fs,ys-σ1s,ys-σ2s,…,ys-σkshxi-τ1i,xi-τ2i,…,xi-τmi-fs,ys-σ1s,ys-σ2s,…,ys-σkshyi-τ1i,yi-τ2i,…,yi-τmi+bs1hxi-τ1i,xi-τ2i,…,xi-τmi-1hyi-τ1i,yi-τ2i,…,yi-τmi≤∑j=1∞∑i=n+2j-1τ0n+2jτ0-11ai∑s=n0i-1Psmhmax⁡xs-σjs-ys-σjs:1≤j≤k+1mh2fs,ys-σ1s,ys-σ2s,…,ys-σks+bshxi-τ1i,xi-τ2i,…,xi-τmi-hyi-τ1i,yi-τ2i,…,yi-τmi≤∑j=1∞∑i=n+2j-1τ0n+2jτ0-11ai∑s=n0i-1Psmhx-y+LhWs+bsmh2max⁡xi-τji-yi-τji:1≤j≤m≤1mh2∑i=n+τ0∞1ai∑s=n0i-1mhPs+LhWs+bsx-y≤θx-y,which yield (26); that is, T is a contraction operator in A(N,M) and T has a unique fixed point x={xn}n∈Zβ∈A(N,M), which means that for all n≥D(60)xn=c+∑j=1∞∑i=n+2j-1τ0n+2jτ0-11aihxi-τ1i,xi-τ2i,…,xi-τmi∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σks-bs,which gives that for all n≥D(61)xn+xn-τ0=2c+∑j=1∞∑i=n+2j-1τ0n+2jτ0-11aihxi-τ1i,xi-τ2i,…,xi-τmi∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σks-bs=2c+∑i=n∞1aihxi-τ1i,xi-τ2i,…,xi-τmi∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σks-bs,Δxn+xn-τ0=-1anhxn-τ1n,xn-τ2n,…,xn-τmn∑s=n0n-1fs,xs-σ1s,xs-σ2s,…,xs-σks-bs,which implies that for all n≥D(62)Δanhxn-τ1n,xn-τ2n,…,xn-τmnΔxn+xn-τ0=-fn,xn-σ1n,xn-σ2n,…,xn-σkn+bn,which means that x={xn}n∈Zβ is a bounded positive solution of (1) in A(N,M).
Using (20), (26), and (58), we deduce that for any η≥0 and n≥D(63)xη+1,n-xn=1-αη-βηxη,n+αηc+∑j=1∞∑i=n+2j-1τ0n+2jτ0-11aihxη,i-τ1i,,xη,i-τ2i,…,xη,i-τmi∑s=n0i-1fs,xη,s-σ1s,xη,s-σ2s,…,xη,s-σks-bs+βηγη,n-xn≤1-αη-βηxη,n-xn+αηTxη,n-Txn+βηγη,n-xn≤1-αη-βηxη-x+αηθxη-x+2Mβη≤1-1-θαηxη-x+2Mβη,which implies that (64)xη+1-x≤1-1-θαηxη-x+2Mβη,η≥0.That is, (21) holds. Consequently, Lemma 1 and (21)~(22) ensure that limη→∞xη=x.
Next, we show that (b) holds. Letc1,c2∈(N,M) with c1≠c2. As in the proof of (a), we deduce similarly that for each l∈{1,2} there exist constants θl∈(0,1), Dl>1+n0+τ0+β, and an operator Tl:A(N,M)→A(N,M) satisfying (26) and (57)~(58), where θ, D, and T are replaced by θl, Dl, and Tl, respectively, and the contraction operator Tl has a unique fixed point xl={xnl}n∈Zβ∈A(N,M), which is also a bounded positive solution of (1); that is, (65)xnl=cl+∑j=1∞∑i=n+2j-1τ0n+2jτ0-11aihxi-τ1il,xi-τ2il,…,xi-τmil∑s=n0i-1fs,xs-σ1sl,xs-σ2sl,…,xs-σksl-bs,∀n≥Dl,which together with (15)~(18), (26), and (57)~(58) means that for each n≥max{D1,D2}(66)xn1-xn2=c1+∑j=1∞∑i=n+2j-1τ0n+2jτ0-11aihxi-τ1i1,xi-τ2i1,…,xi-τmi1∑s=n0i-1fs,xs-σ1s1,xs-σ2s1,…,xs-σks1-bs-c2-∑j=1∞∑i=n+2j-1τ0n+2jτ0-11aihxi-τ1i2,xi-τ2i2,…,xi-τmi2∑s=n0i-1fs,xs-σ1s2,xs-σ2s2,…,xs-σks2-bs≥c1-c2-∑j=1∞∑i=n+2j-1τ0n+2jτ0-11ai∑s=n0i-1fs,xs-σ1s1,xs-σ2s1,…,xs-σks1hxi-τ1i1,xi-τ2i1,…,xi-τmi1-fs,xs-σ1s2,xs-σ2s2,…,xs-σks2hxi-τ1i1,xi-τ2i1,…,xi-τmi1+fs,xs-σ1s2,xs-σ2s2,…,xs-σks2hxi-τ1i1,xi-τ2i1,…,xi-τmi1-fs,xs-σ1s2,xs-σ2s2,…,xs-σks2hxi-τ1i2,xi-τ2i2,…,xi-τmi2+bs1hxi-τ1i1,xi-τ2i1,…,xi-τmi1-1hxi-τ1i2,xi-τ2i2,…,xi-τmi2≥c1-c2-∑j=1∞∑i=n+2j-1τ0n+2jτ0-11ai∑s=n0i-1Psmhmax⁡xs-σjs1-xs-σjs2:1≤j≤k+Lhmh2fs,xs-σ1s2,xs-σ2s2,…,xs-σks2+bsmax⁡xi-τji1-xi-τji2:1≤j≤m≥c1-c2-1mh2∑i=max⁡D1,D2∞1ai∑s=n0i-1mhPs+LhWs+bsx1-x2≥c1-c2-max⁡θ1,θ2x1-x2,which yields that (67)x1-x2≥c1-c21+max⁡θ1,θ2>0,which ensures that x1≠x2. This completes the proof.Theorem 8.
Assume that there exist constantsM, N, Lh, and mh with 0<N<M and two nonnegative sequences {Pn}n∈Nn0 and {Wn}n∈Nn0 satisfying (15)~(18): (68)qn=1,eventually,(69)∑i=n0+1∞iai∑s=n0i-1max⁡Ws,Ps,bs<+∞.Then, one has the following.
(a) For anyc∈(N,M), there exist θ∈(0,1) and D≥1+n0+τ0+β such that, for each x0={x0,n}n∈Zβ∈A(N,M), the Mann iterative sequence with errors {xη}η∈N0={xη,n}(η,n)∈N0×Zβ generated by the scheme (70)
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,converges to a bounded positive solution x∈A(N,M) of (1) and has the error estimate (21), where {γη}η∈N0={γη,n}(η,n)∈N0×Zβ is an arbitrary sequence in A(N,M) and {αη}η∈N0 and {βη}η∈N0 are any sequences in [0,1] satisfying (22).
(b) Equation (1) has uncountably many bounded positive solutions in A(N,M).Proof.
First, we show that (a) holds. Letc∈(N,M). It follows from Lemma 2 and (69) that (71)∑j=0∞∑i=n0+1+jτ0∞1ai∑s=n0i-1max⁡Ws,Ps,bs<+∞,which guarantees that there exist θ∈(0,1) and D≥1+n0+τ0+β sufficiently large such that (72)θ=1mh2∑j=1∞∑i=D+jτ0∞1ai∑s=n0i-1mhPs+LhWs+bs;1mh∑j=1∞∑i=D+jτ0∞1ai∑s=n0i-1Ws+bs<min⁡c-N,M-c. Define an operator T:A(N,M)→lβ∞ by (73)Txn=c+∑j=1∞∑i=n+jτ0∞1aihxi-τ1i,xi-τ2i,…,xi-τmi∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σks-bs,n≥D,TxD,β≤n<D,for each x={xn}n∈Zβ∈A(N,M). By virtue of (15)~(18) and (72)~(73), we get that, for every x={xn}n∈Zβ∈A(N,M), y={yn}n∈Zβ∈A(N,M), and n≥D, (74)Txn=c+∑j=1∞∑i=n+jτ0∞1aihxi-τ1i,xi-τ2i,…,xi-τmi∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σks-bs≥c-∑j=1∞∑i=n+jτ0∞1aimh∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σks+bs≥c-1mh∑j=1∞∑i=D+jτ0∞1ai∑s=n0i-1Ws+bs≥c-min⁡c-N,M-c≥N,Txn=c+∑j=1∞∑i=n+jτ0∞1aihxi-τ1i,xi-τ2i,…,xi-τmi∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σks-bs≤c+∑j=1∞∑i=n+jτ0∞1aimh∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σks+bs≤c+1mh∑j=1∞∑i=D+jτ0∞1ai∑s=n0i-1Ws+bs≤c+min⁡c-N,M-c≤M,Txn-Tyn=∑j=1∞∑i=n+jτ0∞1aihxi-τ1i,xi-τ2i,…,xi-τmi∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σks-bs-∑j=1∞∑i=n+jτ0∞1aihyi-τ1i,yi-τ2i,…,yi-τmi∑s=n0i-1fs,ys-σ1s,ys-σ2s,…,ys-σks-bs≤∑j=1∞∑i=n+jτ0∞1ai∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σkshxi-τ1i,xi-τ2i,…,xi-τmi-fs,ys-σ1s,ys-σ2s,…,ys-σkshxi-τ1i,xi-τ2i,…,xi-τmi+fs,ys-σ1s,ys-σ2s,…,ys-σkshxi-τ1i,xi-τ2i,…,xi-τmi-fs,ys-σ1s,ys-σ2s,…,ys-σkshyi-τ1i,yi-τ2i,…,yi-τmi+bs1hxi-τ1i,xi-τ2i,…,xi-τmi-1hyi-τ1i,yi-τ2i,…,yi-τmi≤∑j=1∞∑i=n+jτ0∞1ai∑s=n0i-1Psmhmax⁡xs-σjs-ys-σjs:1≤j≤k+Lhmh2fs,ys-σ1s,ys-σ2s,…,ys-σks+bsmax⁡xi-τji-yi-τji:1≤j≤m≤1mh2∑j=1∞∑i=n+jτ0∞1ai∑s=n0i-1mhPs+LhWs+bsx-y≤θx-y,which yield (26); that is, T is a contraction operator in A(N,M) and T has a unique fixed point x={xn}n∈Zβ∈A(N,M) with (75)xn=c+∑j=1∞∑i=n+jτ0∞1aihxi-τ1i,xi-τ2i,…,xi-τmi∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σks-bs,∀n≥D,which yields that (76)xn-xn-τ0=-∑i=n∞1aihxi-τ1i,xi-τ2i,…,xi-τmi∑s=n0i-1fs,xs-σ1s,xs-σ2s,…,xs-σks-bs,∀n≥D,Δxn-xn-τ0=-1anhxn-τ1n,xn-τ2n,…,xn-τmn∑s=n0n-1fs,xs-σ1s,xs-σ2s,…,xs-σks-bs,∀n≥D,which implies that (77)Δanhxn-τ1n,xn-τ2n,…,xn-τmnΔxn-xn-τ0=-fn,xn-σ1n,xn-σ2n,…,xn-σkn+bn,∀n≥D,which means that x={xn}n∈Zβ is a bounded positive solution of (1) in A(N,M).
It follows from (20), (26), and (73) that for any η≥0 and n≥D(78)xη+1,n-xn=1-αη-βηxη,n+αηc+∑j=1∞∑i=n+jτ0∞1aihxη,i-τ1i,,xη,i-τ2i,…,xη,i-τmi∑s=n0i-1fs,xη,s-σ1s,xη,s-σ2s,…,xη,s-σks-bs+βηγη,n-xn≤1-αη-βηxη,n-xn+αηTxη,n-Txn+βηγη,n-xn≤1-αη-βηxη-x+αηθxη-x+2Mβη≤1-1-θαηxη-x+2Mβη,which implies that (79)xη+1-x≤1-1-θαηxη-x+2Mβη,η≥0.That is, (21) holds. Thus, Lemma 1 and (21)~(22) ensure that limη→∞xη=x.
Next, we show that (b) holds. Letc1,c2∈(N,M) with c1≠c2. As in the proof of (a), we deduce that for each l∈{1,2} there exist constants θl∈(0,1), Dl>1+n0+τ0+β, and an operator Tl:A(N,M)→A(N,M) satisfying (26) and (72)~(73), where θ, D, and T are replaced by θl, Dl, and Tl, respectively, and the contraction operator Tl has a unique fixed point xl={xnl}n∈Zβ∈A(N,M), which is also a bounded positive solution of (1); that is, (80)xnl=cl+∑j=1∞∑i=n+jτ0∞1aihxi-τ1il,xi-τ2il,…,xi-τmil∑s=n0i-1fs,xs-σ1sl,xs-σ2sl,…,xs-σksl-bs,∀n≥Dl,which together with (15)~(18), (26), and (72)~(73) gives that for each n≥max{D1,D2}(81)xn1-xn2=c1+∑j=1∞∑i=n+jτ0∞1aihxi-τ1i1,xi-τ2i1,…,xi-τmi1∑s=n0i-1fs,xs-σ1s1,xs-σ2s1,…,xs-σks1-bs-c2-∑j=1∞∑i=n+jτ0∞1aihxi-τ1i2,xi-τ2i2,…,xi-τmi2∑s=n0i-1fs,xs-σ1s2,xs-σ2s2,…,xs-σks2-bs≥c1-c2-∑j=1∞∑i=n+jτ0∞1ai∑s=n0i-1fs,xs-σ1s1,xs-σ2s1,…,xs-σks1hxi-τ1i1,xi-τ2i1,…,xi-τmi1-fs,xs-σ1s2,xs-σ2s2,…,xs-σks2hxi-τ1i1,xi-τ2i1,…,xi-τmi1+fs,xs-σ1s2,xs-σ2s2,…,xs-σks2hxi-τ1i1,xi-τ2i1,…,xi-τmi1-fs,xs-σ1s2,xs-σ2s2,…,xs-σks2hxi-τ1i2,xi-τ2i2,…,xi-τmi2+bs1hxi-τ1i1,xi-τ2i1,…,xi-τmi1-1hxi-τ1i2,xi-τ2i2,…,xi-τmi2≥c1-c2-∑j=1∞∑i=n+jτ0∞1ai∑s=n0i-1Psmhmax⁡xs-σjs1-xs-σjs2:1≤j≤k+Lhmh2fs,xs-σ1s2,xs-σ2s2,…,xs-σks2+bsmax⁡xi-τji1-xi-τji2:1≤j≤m≥c1-c2-1mh2∑j=1∞∑i=max⁡D1,D2+jτ0∞1ai∑s=n0i-1mhPs+LhWs+bsx1-x2≥c1-c2-max⁡θ1,θ2x1-x2,which yields that (82)x1-x2≥c1-c21+max⁡θ1,θ2>0;that is, x1≠x2. This completes the proof.Remark 9.
Theorems3~8 extend and unify Theorem 1 in [3] and Theorems  2.1~2.7 in [12], respectively.


## 3. Examples

In what follows, we construct six examples to demonstrate that Theorems3~8 extend properly the corresponding results in [3, 12], respectively. Note that none of the known results can be applied to the examples.Example 1.
Consider the second-order nonlinear neutral delay difference equation with forcing term(83)Δ-1n-1n3-n2+1xn2+n-12+xnn+1/2Δxn-n+12n+3xn-τ0+n2x2n2-n+xn-52n4=sin⁡n-n-nn2,n≥1,where τ0∈N is fixed. Let n0=1, m=k=2, q=1/2, M=6, N=2, β=min{1-τ0,-4}, (84)an=-1n-1n3-n2+1,qn=n+12n+3,bn=sinn-n-nn2,τ1n=-n2+1,τ2n=n1-n2,σ1n=2n1-n,σ2n=5,fn,u,v=n2u+v2n4,hu,v=u2+v,Pn=n2+2Mn4,Wn=n2M+M2n4,Lh=2M+1,mh=N2+N,n∈Nn0,u,v∈R.It is easy to show that (13)~(18) are satisfied. Note that (85)∑i=n0+1∞1ai∑s=n0i-1max⁡Ws,Ps,bs=∑i=n0+1∞1i3-i2+1∑s=n0i-1max⁡s2M+M2s4,s2+2Ms4,sin⁡s-s-ss2≤2M1+M∑i=n0+1∞1i3-i2+1<+∞;that is, (19) holds. It follows from Theorem 3 that (83) has uncountably many bounded positive solutions in A(N,M). On the other hand, for each c∈(N,(1-q)M), there exist θ∈(0,1) and T≥n0+τ0+β such that the Mann iterative sequence with errors {xm}m≥0 generated by (20) converges to a bounded positive solution x∈A(N,M) of (83) and has the error estimate (21), where {γη}η≥0 is an arbitrary sequence in A(N,M) and {αη}η≥0 and {βη}η≥0 are any sequences in [0,1] satisfying (22).Example 2.
Consider the second-order nonlinear neutral delay difference equation with forcing term(86)Δ-1n-1n5xn3+6x4n-321+xn32Δxn+4n55n5+3xn-τ0+n2+1x2n+n-12-n3xn+6n7+2n5+1=-1n-1n2+5n-1n6+20n5+3n4+n+5,n≥2,where τ0∈N is fixed. Let n0=m=k=2, q=-4/5, M=8, N=1, β=2-τ0, (87)an=-1n-1n5,qn=-4n55n5+3,bn=-1n-1n2+5n-1n6+20n5+3n4+n+5,τ1n=n-n3,τ2n=-3n+3,σ1n=-2n+1,σ2n=-6,fn,u,v=n2+1u2-n3vn7+2n5+1,hu,v=v+6u21+v2,Pn=4Mn5+1n4,Wn=2M2n5+Mn4,Lh=7+19M21+N22,mh=N+6N21+M2,n∈Nn0,u,v∈R. It is clear that (15)~(18), (34), and (35) hold. Notice that (88)∑i=n0+1∞1ai∑s=n0i-1max⁡Ws,Ps,bs=∑i=n0+1∞1i5∑s=n0i-1max⁡4Ms5+1s4,M2s5+Ms4,-1s-1s2+5s-1s6+20s5+3s4+s+5≤∑i=n0+1∞1i5∑s=n0i-12M2s4≤4M2∑i=n0+1∞1i5<+∞,which implies (19). It follows from Theorem 4 that (86) has uncountably bounded positive solutions in A(N,M). On the other hand, for any c∈(N-qM,M), there exist θ∈(0,1) and T≥n0+τ0+β such that the Mann iterative sequence with errors {xm}m≥0 generated by (20) converges to a bounded positive solution x∈A(N,M) of (86) and has the error estimate (21), where {γη}η≥0 is an arbitrary sequence in A(N,M) and {αη}η≥0 and {βη}η≥0 are any sequences in [0,1] satisfying (22).Example 3.
Consider the second-order nonlinear neutral delay difference equation with forcing term(89)Δ-1nn+1/2n6+3n+23+xn2+-1nn2/31+x5n+12Δxn-5n+1+45n+1xn-τ0+nxn2+n+x3n2+n-22n5+2n=-1nn+arctan⁡nn3+4n2+2,n≥1, where τ0∈N is fixed. Let n0=1, m=k=2, q_=2, q¯=5, M=9, N=1, β=1-τ0, (90)an=-1nn+1/2n6+3n+2,qn=5n+1+45n+1,bn=-1nn+arctan⁡nn3+4n2+2,τ1n=-n2--1nn+n,τ2n=-4n-1,σ1n=-n2,σ2n=-3n2+2,fn,u,v=nu+v2n5+2n,hu,v=3+u2/31+v2,Pn=2M+nn5+2n,Wn=nM+M2n5+2n,Lh=6M3,mh=3+N2/31+M2,n∈Nn0,u,v∈R. It is easy to show that (15)~(18), (38), and (39) are satisfied. Note that (91)∑i=n0+1∞1ai∑s=n0i-1max⁡Ws,Ps,bs=∑i=n0+1∞1i6+3i+2∑s=n0i-1max⁡sM+M2s5+2s,2M+ss5+2s,-1ss+arctan⁡ss3+4s2+2≤∑i=n0+1∞13i∑s=n0i-1M+M2s2≤M+M23<+∞,which means (19). It follows from Theorem 5 that (89) has uncountably bounded positive solutions in A(N,M). On the other hand, for any c∈(q¯N,(q_-1)M), there exist θ∈(0,1) and T≥n0+τ0+β such that the Mann iterative sequence with errors {xm}m≥0 generated by (40) converges to a bounded positive solution x∈A(N,M) of (89) and has the error estimate (21), where {γη}η≥0 is an arbitrary sequence in A(N,M) and {αη}η≥0 and {βη}η≥0 are any sequences in [0,1] satisfying (22).Example 4.
Consider the second-order nonlinear neutral delay difference equation with forcing term(92)Δn32+cos⁡n2-1xn3+n-12+xn3+n-12+xn-32Δxn+n3+3n+7n3+n+1xn-τ0+n2x9n+3+n-1x2n-83n4+x2n-82=-1nn+1/2n2+nn4+n3+1,n≥1,where τ0∈N is fixed. Let n0=1, m=k=2, q¯=-3, q_=-7, M=21, N=1, β=min{1-τ0,-6}, (93)an=n32+cosn2-1,qn=-n3+3n+7n3+n+1,bn=-1nn+1/2n2+nn4+n3+1,τ1n=3,τ2n=-n3+1,σ1n=-8n-3,σ2n=-n+8,fn,u,v=n2u+n-1v3n4+v2,hu,v=v2+v2+u2,Pn=7M4n2,Wn=2M3n2,Lh=2+4M+3M2+4M32+N22,mh=N2+N2+M2,n∈Nn0,u,v∈R. It is easy to show that (15)~(18), (51), and (52) are satisfied. Note that (94)∑i=n0+1∞1ai∑s=n0i-1max⁡Ws,Ps,bs=∑i=n0+1∞1i32+cos⁡i2-1∑s=n0i-1max⁡2M3s2,7M4s2,-1ss+1/2s2+ss4+s3+1≤14M4∑i=n0+1∞1i3<+∞;that is, (19) holds. It follows from Theorem 6 that (92) has uncountably bounded positive solutions in A(N,M). On the other hand, for any c∈(q¯M,q_N-q_M/q¯), there exist θ∈(0,1) and T≥n0+τ0+β such that the Mann iterative sequence with errors {xm}m≥0 generated by (40) converges to a bounded positive solution x∈A(N,M) of (92) and has the error estimate (21), where {γη}η≥0 is an arbitrary sequence in A(N,M) and {αη}η≥0 and {βη}η≥0 are any sequences in [0,1] satisfying (22).Example 5.
Consider the second-order nonlinear neutral delay difference equation with forcing term(95)Δ-1n-1n2ln3⁡n+12xnn+1/2+xn+521+xn+52Δxn+xn-τ0+n2xn+3x2n-5+nxn+32+x2n-5n5+xn+3=n2-2n+2n5+4n3+1,n≥1, where τ0∈N is fixed. Let n0=1, m=k=2, M=6, N=2, β=min{1-τ0,-3}, (96)an=-1n-1n2ln3⁡n+1,qn=-1,bn=n2-2n+2n5+4n3+1,τ1n=-nn-12,τ2n=-5,σ1n=5-n,σ2n=-3,fn,u,v=n2uv+nu2+vn5+u,hu,v=2u+v21+v2,Pn=4M2n3,Wn=3M2n3,Lh=7M21+N22,mh=2N+N21+M2,n∈Nn0,u,v∈R. It is easy to show that (15)~(18) and (55) hold. Note that (97)∑i=n0+1∞1ai∑s=n0i-1max⁡Ws,Ps,bs=∑i=n0+1∞1i2ln3⁡i+1∑s=n0i-1max⁡3M2s3,4M2s3,s2-2s+2s5+4s3+1=∑i=n0+1∞1i2ln3⁡i+1∑s=n0i-14M2s3≤8M2∑i=n0+1∞1i2ln3⁡i+1<+∞;that is, (19) holds. It follows from Theorem 7 that (95) has uncountably bounded positive solutions in A(N,M). On the other hand, for any c∈(N,M), there exist θ∈(0,1) and T≥n0+τ0+β such that the Mann iterative sequence with errors {xm}m≥0 generated by (56) converges to a bounded positive solution x∈A(N,M) of (95) and has the error estimate (21), where {γη}η≥0 is an arbitrary sequence in A(N,M) and {αη}η≥0 and {βη}η≥0 are any sequences in [0,1] satisfying (22).Example 6.
Consider the second-order nonlinear neutral delay difference equation with forcing term(98)Δ-1nn+1/2n5x2n+12+14xn2+n+12+4xn2+n+1+5Δxn-xn-τ0+xn+2+x4n2+n-1sin2nn4+xn+24+x4n2+n-12=-1n+nn6+4n5+1,n≥1,where τ0∈N is fixed. Let n0=1, m=k=2, M=5, N=2, β=1-τ0, (99)an=-1nn+1/2n5,qn=1,bn=-1n+nn6+4n5+1,τ1n=-2n+n-1,τ2n=-n2-1,σ1n=-2,σ2n=-4n2+1,fn,u,v=u+vsin2nn4+u4+v2,hu,v=v2+14u2+4u+5,Pn=2n4+4M21+2M2n8,Wn=2Mn4,Lh=4+18M+4M2+24M34N2+4N+52,mh=14M2+4M+5,n∈Nn0,u,v∈R.It is easy to show that (15)~(18) and (68) are satisfied. Note that (100)∑i=n0+1∞iai∑s=n0i-1max⁡Ws,Ps,bs=∑i=n0+1∞1i4∑s=n0i-1max⁡2Ms4,2s4+4M21+2M2s4,-1s+ss6+4s5+1≤∑i=n0+1∞1i4∑s=n0i-12+4M21+2M2s4≤41+2M2+4M4∑i=n0+1∞1i4<+∞;that is, (69) holds. It follows from Theorem 8 that (98) has uncountably bounded positive solutions in A(N,M). On the other hand, for any b∈(N,M), there exist θ∈(0,1) and T≥n0+τ0+β such that the Mann iterative sequence with errors {xm}m≥0 generated by (70) converges to a bounded positive solution x∈A(N,M) of (98) and has the error estimate (21), where {γη}η≥0 is an arbitrary sequence in A(N,M) and {αη}η≥0 and {βη}η≥0 are any sequences in [0,1] satisfying (22).



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*Source: 8317567-2016-04-03.xml*