# Unbounded Positive Solutions and Mann Iterative Schemes of a Second-Order Nonlinear Neutral Delay Difference Equation

Authors: Zeqing Liu; Xiaochuan Hou; Jeong Sheok Ume; Shin Min Kang
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

This paper is concerned with solvability of the second-order nonlinear neutral delay difference equationΔ2(xn+anxn-τ)+Δh(n,xh1n,xh2n,…,xhkn)+f(n,xf1n,xf2n,…,xfkn)=bn,∀n≥n0. Utilizing the Banach fixed point theorem and some new techniques, we show the existence of uncountably many unbounded positive solutions for the difference equation, suggest several Mann-type iterative schemes with errors, and discuss the error estimates between the unbounded positive solutions and the sequences generated by the Mann iterative schemes. Four nontrivial examples are given to illustrate the results presented in this paper.


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## Body

## 1. Introduction and Preliminaries

Recently, the oscillation, nonoscillation, asymptotic behavior, and existence of solutions of different classes of linear and nonlinear second-order difference equations have been studied by many authors; see, for example, [1–26] and the references cited therein. Using the Banach fixed point theorem, Jinfa [5] discussed the existence of a bounded nonoscillatory solution for the second-order neutral delay difference equation with positive and negative coefficients:
(1)Δ2(xn+pxn-m)+pnxn-k-qnxn-l=0,∀n≥n0
under the condition p≠-1. Luo and Bainov [13] and M. Migda and J. Migda [16] considered the asymptotic behaviors of nonoscillatory solutions for the second-order neutral difference equation with maxima:
(2)Δ2(xn+pnxn-k)+qnmax{xs:n-l≤s≤n}=0,∀n≥1
and the second-order neutral difference equation:
(3)Δ2(xn+pxn-k)+f(n,xn)=0,∀n≥1,
respectively. Meng and Yan [15] studied the existence of bounded nonoscillatory solutions for the second-order nonlinear nonautonomous neutral delay difference equation:
(4)Δ2(xn-pxn-τ)=∑n=1m‍qixn-σi+f(n,xn-η1n,…,xn-ηln),∀n≥n0.
Applying the cone compression and expansion theorem in Fréchet spaces, Tian and Ge [21] established the existence of multiple positive solutions of the second-order discrete equation on the half-line:
(5)Δ2xn-1-pΔxn-1-qxn-1+f(n,xn)=0,∀n≥1
with certain boundary value conditions. But to the best of our knowledge, results on multiplicity of unbounded solutions for neutral delay difference equations are very scarce in the literature. Nothing has been done with the existence of uncountably many unbounded positive solutions for (1)~(5) and any other second-order neutral delay difference equations:Inspired and motivated by the results in [1–26], in this paper we introduce and study the second-order nonlinear neutral delay difference equation:
(6)Δ2(xn+anxn-τ)+Δh(n,xh1n,xh2n,…,xhkn)+f(n,xf1n,xf2n,…,xfkn)=bn,∀n≥n0,
where τ,k,n0∈ℕ, {an}n∈ℕn0,{bn}n∈ℕn0⊂ℝ, h,f∈C(ℕn0×ℝk,ℝ), {hln}n∈ℕn0,{fln}n∈ℕn0⊆ℕ, and
(7)limn→∞hln=limn→∞fln=+∞,l∈{1,2,…,k}.
By means of the Banach fixed point theorem and some new techniques, we establish sufficient conditions for the existence of uncountably many unbounded positive solutions of (6), suggest a few Mann iterative schemes with errors for approximating these unbounded positive solutions, and prove their convergence and the error estimates. The results obtained in this paper extend the result in [5]. Four nontrivial examples are interested in the text to illustrate the importance of our results.Throughout this paper, we assume thatΔ is the forward difference operator defined by Δxn=xn+1-xn, ℝ=(-∞,+∞), ℝ+=[0,+∞), ℤ, ℕ0, and ℕ denote the sets of all integers, nonnegative integers, and positive integers, respectively,
(8)ℕt={n:n∈ℕwithn≥t},∀t∈ℕ,β=min{n0-τ,inf{hln,fln:1≤l≤k,n∈ℕn0}}∈ℕ,Hn=max{hln:l∈{1,2,…,k}},Fn=max{fln:l∈{1,2,…,k}},∀n∈ℕn0,lβ∞ represents the Banach space of all real sequences on ℕβ with norm
(9)∥x∥=supn∈ℕβ|xnn|<+∞foreachx={xn}n∈ℕβ∈lβ∞,A(N,M)={x={xn}n∈ℕβ∈lβ∞:N≤xnn≤M,n∈ℕβ}foranyM>N>0.
It is clear that A(N,M) is a closed and convex subset of   lβ∞. By a solution of (6), we mean a sequence {xn}n∈ℕβ with a positive integer T≥n0+τ+β such that (6) holds for all n≥T.The following lemmas play important roles in this paper.Lemma 1 (see [27]).
Let{αn}n∈ℕ0,{βn}n∈ℕ0,{γn}n∈ℕ0, and {tn}n∈ℕ0 be four nonnegative sequences satisfying the inequality
(10)αn+1≤(1-tn)αn+tnβn+γn,∀n∈ℕ0,
where {tn}n∈ℕ0⊂[0,1], ∑n=0∞‍tn=+∞, limn→∞βn=0, and ∑n=0∞‍γn<+∞. Then limn→∞αn=0.Lemma 2 (see [11]).
Letτ,n0∈ℕ and {bn}n∈ℕn0 be a nonnegative sequence. Then(a)
∑i=0∞‍∑s=n0+iτ∞‍bs<+∞⇔∑s=n0∞‍sbs<+∞;(b)
∑i=0∞‍∑s=n0+iτ∞‍∑t=s∞‍bt<+∞⇔∑s=n0∞‍∑t=s∞‍sbt<+∞.


## 2. Existence of Uncountably Many Unbounded Positive Solutions

Using the Banach fixed point theorem and the Mann iterative schemes with errors, we next discuss the existence of uncountably many unbounded positive solutions of (6), prove that the Mann iterative schemes with errors converge to these unbounded positive solutions, and compute the error estimates between the Mann iterative schemes with errors and the unbounded positive solutions.Theorem 3.
Assume that there exist two constantsM and N with M>N>0 and four nonnegative sequences {Pn}n∈ℕn0,{Qn}n∈ℕn0,{Rn}n∈ℕn0, and {Wn}n∈ℕn0 satisfying
(11)|f(n,u1,u2,…,uk)-f(n,u-1,u-2,…,u-k)|≤Pnmax{|ul-u-l|:1≤l≤k},|h(n,u1,u2,…,uk)-h(n,u-1,u-2,…,u-k)|≤Rnmax{|ul-u-l|:1≤l≤k},∀(n,ul,u-l)∈ℕn0×(ℝ+∖{0})2,1≤l≤k,(12)|f(n,u1,u2,…,uk)|≤Qn,|h(n,u1,u2,…,uk)|≤Wn,∀(n,ul)∈ℕn0×(ℝ+∖{0}),1≤l≤k,(13)limn→∞1n∑i=0∞‍∑s=n+iτ∞max{RsHs,Ws}=0,(14)limn→∞1n∑i=0∞∑s=n+iτ∞‍∑t=s∞max{PtFt,Qt,|bt|}=0,(15)an=-1eventually.
Then
(a) for anyL∈(N,M), there exist θ∈(0,1) and T≥n0+τ+β such that for each x0={x0n}n∈ℕβ∈A(N,M), the Mann iterative sequence with errors {xm}m∈ℕ0={xmn}(m,n)∈ℕ0×ℕβ generated by the scheme
(16)xm+1n={(1-αm-βm)xmn+αm{∑t=s∞(f(t,xmf1t,xmf2t,…,nL+αm-∑i=1∞∑s=n+iτ∞[∑t=s∞(f(t,xmf1t,xmf2t,…,xmfkt)-bt)h(s,xmh1s,xmh2s,…,xmhks)+αm-∑t=s∞(f(t,xmf1t,xmf2t,…,+αmxmfkt)-bt)∑t=s∞(f(t,xmf1t,xmf2t,…,xmfkt)]}+βmγmn,n≥T,m≥0,(1-αm-βm)xmT+αm{∑t=s∞(f(t,xmf1t,xmf2t,…,xmfkt)TL+αm-∑i=1∞∑s=T+iτ∞[∑t=s∞(f(t,xmf1t,xmf2t,…,xmfkt)-bt)h(s,xmh1s,xmh2s,…,xmhks)∑i=1∞∑s=T+iτ∞2003-∑t=s∞(f(t,xmf1t,xmf2t,…,+αmxmfkt)-bt)∑t=s∞(f(t,xmf1t,xmf2t,…,xmfkt)]}+βmγmT,β≤n<T,m≥0
converges to an unbounded positive solution x∈A(N,M) of (6) and has the following error estimate:
(17)∥xm+1-x∥≤[1-(1-θ)αm]∥xm-x∥+2Mβm,∀m∈ℕ0,
where {γm}m∈ℕ0={γmn}(m,n)∈ℕ0×ℕβ is an arbitrary sequence in A(N,M) and {αm}m∈ℕ0 and {βm}m∈ℕ0 are any sequences in [0,1] such that
(18)∑m=0∞αm=+∞,∑m=0∞βm<+∞orthereexistsasequence{ξm}m∈ℕ0⊂[0,+∞)satisfyingβm=ξmαm,m∈ℕ0,limm→∞ξm=0;(b)
equation (6) possesses uncountably many unbounded positive solutions in A(N,M).Proof.
First of all we show that (a) holds. SetL∈(N,M). It follows from (13), (14), and (15) that there exist θ∈(0,1) and T≥n0+τ+β satisfying
(19)θ=1T∑i=1∞∑s=T+iτ∞(RsHs+∑t=s∞‍PtFt),(20)1T∑i=1∞‍∑s=T+iτ∞‍(Ws+∑t=s∞‍Qt+∑t=s∞‍|bt|)<min{M-L,L-N},(21)an=-1,∀n≥T.
Define a mapping SL:A(N,M)→lβ∞ by
(22)SLxn={nL-∑i=1∞‍∑s=n+iτ∞‍{∑t=s∞‍h(s,xh1s,xh2s,…,xhks)-∑t=s∞‍[f(t,xf1t,xf2t,…,xfkt)-bt]},n≥T,SLxT,β≤n<T
for each x={xn}n∈ℕβ∈A(N,M). In view of (11), (12), (19), (20), and (22), we deduce that for each x={xn}n∈ℕβ,y={yn}n∈ℕβ∈A(N,M) and for all n≥T(23)|SLxnn-SLynn|≤1n∑i=1∞‍∑s=n+iτ∞‍[∑t=s∞‍|h(s,xh1s,xh2s,…,xhks)≤1n∑i=1∞‍∑s=n+iτ∞‍-h(s,yh1s,yh2s,…,yhks)|≤1n∑i=1∞‍∑s=n+iτ∞‍+∑t=s∞‍|f(t,xf1t,xf2t,…,xfkt)≤1n∑i=1∞‍∑s=n+iτ∞‍-f(t,yf1t,yf2t,…,yfkt)|∑t=s∞‍]≤1n∑i=1∞‍∑s=n+iτ∞‍[∑t=s∞‍Rsmax{|xhls-yhls|:1≤l≤k}×∑i=1∞‍∑s=n+iτ∞‍+∑t=s∞‍Ptmax{|xflt-yflt|:1≤l≤k}]≤∥x-y∥n∑i=1∞‍∑s=n+iτ∞‍[∑t=s∞‍Rsmax{hls:1≤l≤k}∑i=1∞‍∑s=n+iτ∞‍+∑t=s∞‍Ptmax{flt:1≤l≤k}]≤∥x-y∥T∑i=1∞‍∑s=T+iτ∞‍(RsHs+∑t=s∞‍PtFt)=θ∥x-y∥,|SLxnn-L|=|1n∑i=1∞‍∑s=n+iτ∞‍{∑t=s∞‍h(s,xh1s,xh2s,…,xhks)∑s=n+iτ∞‍∑s=n+iτ∞‍-∑t=s∞‍[f(t,xf1t,xf2t,…,xfkt)-bt]}|≤1n∑i=1∞‍∑s=n+iτ∞‍{∑t=s∞‍|h(s,xh1s,xh2s,…,xhks)|∑s=n+iτ∞‍∑s=n+iτ∞‍+∑t=s∞‍[|f(t,xf1t,xf2t,…,xfkt)|+|bt|]}≤1T∑i=1∞‍∑s=T+iτ∞‍[Ws+∑t=s∞‍(Qt+|bt|)]<min{M-L,L-N},
which yield that
(24)SL(A(N,M))⊆A(N,M),∥SLx-SLy∥≤θ∥x-y∥∀x,y∈A(N,M),
which means that SL is a contraction in A(N,M). It follows from the Banach fixed point theorem that SL has a unique fixed point x={xn}n∈ℕβ∈A(N,M), that is,
(25)xn=nL-∑i=1∞‍∑s=n+iτ∞‍{∑t=s∞‍h(s,xh1s,xh2s,…,xhks)-∑i=1∞‍∑s=n+iτ∞‍-∑t=s∞‍[f(t,xf1t,xf2t,…,xfkt)-bt]},∀n≥T,xn-τ=(n-τ)L-∑i=1∞‍∑s=n+(i-1)τ∞‍{∑t=s∞‍[f(t,xf1t,xf2t,…,xfkt)-bt]h(s,xh1s,xh2s,…,xhks)-∑i=1∞‍∑s=n+(i-1)τ∞‍-∑t=s∞‍[f(t,xf1t,xf2t,…,xfkt)-bt]},∀n≥T+τ,
which imply that
(26)xn-xn-τ=τL+∑s=n∞‍{∑t=s∞‍h(s,xh1s,xh2s,…,xhks)τL+∑s=n∞‍-∑t=s∞‍[f(t,xf1t,xf2t,…,xfkt)-bt]},∀n≥T+τ,
which yields that
(27)Δ(xn-xn-τ)=-h(n,xh1n,xh2n,…,xhkn)+∑t=n∞‍[f(t,xf1t,xf2t,…,xfkt)-bt],∀n≥T+τ,Δ2(xn-xn-τ)=-Δh(n,xh1n,xh2n,…,xhkn)-f(n,xf1n,xf2n,…,xfkn)+bn,∀n≥T+τ,
which together with (21) gives that x={xn}n∈ℕβ is an unbounded positive solution of (6) in A(N,M). It follows from (16), (19), (21), (22), and (24) that for any m∈ℕ0 and n≥T(28)|xm+1nn-xnn|=1n|∑s=n+iτ∞‍(1-αm-βm)xmn+αm{∑t=s∞‍nL+αm-∑i=1∞‍∑s=n+iτ∞[∑t=s∞‍h(s,xmh1s,xmh2s,…,xmhks)+αm-∑i=1∞‍∑s=n+iτ∞-∑t=s∞‍(f(t,xmf1t,xmf2t,…,+αm-∑i=1∞‍∑s=n+iτ∞-xmfkt)-bt)∑t=s∞‍nL]}+βmγmn-xn∑i=1∞‍|≤(1-αm-βm)|xmn-xn|n+αm|SLxmn-SLxn|n+βm|γmn-xn|n≤(1-αm-βm)∥xm-x∥+θαm∥xm-x∥+2Mβm≤[1-(1-θ)αm]∥xm-x∥+2Mβm,
which implies that
(29)∥xm+1-x∥≤[1-(1-θ)αm]∥xm-x∥+2Mβm,∀m∈ℕ0.
That is, (17) holds. Thus, Lemma 1 and (17) and (18) guarantee that limm→∞xm=x.
Next we show that (b) holds. LetL1,L2∈(N,M) and L1≠L2. As in the proof of (a), we deduce similarly that for each c∈{1,2} there exist constants θc∈(0,1), Tc≥n0+τ+β, and a mapping SLc satisfying (19)~(24), where θ,L, and T are replaced by θc,Lc and Tc, respectively, and the mapping SLc has a fixed point zc={znc}n∈ℕβ∈A(N,M), which is an unbounded positive solution of (6) in A(N,M), that is,
(30)znc=nLc-∑i=1∞‍∑s=n+iτ∞‍{∑t=s∞‍h(s,zh1sc,zh2sc,…,zhksc)×-∑i=1∞‍∑s=n+iτ∞‍-∑t=s∞‍[f(t,zf1tc,zf2tc,…,zfktc)-bt]},∀n≥Tc,
which together with (11) and (17) implies that for n≥max{T1,T2}(31)|zn1n-zn2n|≥|L1-L2|-1n∑i=1∞∑s=n+iτ∞{∑t=s∞|f(t,zf1t1,zf2t1,…,zfkt1)|h(s,zh1s1,zh2s1,…,zhks1)-∑i=1∞‍∑s=n+iτ∞‍-h(s,zh1s2,zh2s2,…,zhks2)|-∑i=1∞‍∑s=n+iτ∞‍+∑t=s∞|f(t,zf1t1,zf2t1,…,zfkt1)-∑s=n+iτ∞‍-f(t,zf1t2,zf2t2,…,zfkt2)|∑t=s∞|f(t,zf1t1,zf2t1,…,zfkt1)}≥|L1-L2|-1n∑i=1∞∑s=n+iτ∞[∑t=s∞Ptmax{|zflt1-zflt2|:1≤l≤k}Rsmax{|zhls1-zhls2|:1≤l≤k}-∑i=1∞‍∑s=n+iτ∞‍+∑t=s∞Ptmax{|zflt1-zflt2|:1≤l≤k}]≥|L1-L2|-∥z1-z2∥n∑i=1∞∑s=n+iτ∞(RsHs+∑t=s∞‍PtFt)≥|L1-L2|-∥z1-z2∥max{T1,T2}×∑i=1∞∑s=max{T1,T2}+iτ∞(RsHs+∑t=s∞‍PtFt)≥|L1-L2|-max{θ1,θ2}∥z1-z2∥,
which yields that
(32)∥z1-z2∥≥|L1-L2|1+max{θ1,θ2}>0,
that is, z1≠z2. This completes the proof.Theorem 4.
Assume that there exist two constantsM and N with M>N>0 and four nonnegative sequences {Pn}n∈ℕn0, {Qn}n∈ℕn0, {Rn}n∈ℕn0, and {Wn}n∈ℕn0 satisfying (11), (12),
(33)limn→∞1n∑s=n∞‍max{RsHs,Ws}=0,(34)limn→∞1n∑s=n∞‍∑t=s∞‍max{PtFt,Qt,|bt|}=0,(35)an=1eventually.
Then
(a) for anyL∈(N,M), there exist θ∈(0,1) and T≥n0+τ+β such that for each x0={x0n}n∈ℕβ∈A(N,M), the Mann iterative sequence with errors {xm}m∈ℕ0={xmn}(m,n)∈ℕ0×ℕβ generated by the scheme
(36)xm+1n={(1-αm-βm)xmn+αm{∑t=s∞(f(t,xmf1t,xmf2t,…,xmfkt)-bt)nL+αm+∑i=1∞∑s=n+(2i-1)τn+2iτ-1[∑t=s∞(f(t,xmf1t,xmf2t,…,xmfkt)-bt)h(s,xmh1s,xmh2s,…,xmhks)+αm-∑t=s∞(f(t,xmf1t,xmf2t,…,+αmxmfkt)-bt)∑t=s∞(f(t,xmf1t,xmf2t,…,]}+βmγmn,n≥T,m≥0,(1-αm-βm)xmT+αm{∑s=T+(2i-1)τT+2iτ-1TL+αm+∑i=1∞∑s=T+(2i-1)τT+2iτ-1[∑t=s∞(f(t,xmf1t,xmf2t,…,xmfkt)-bt)h(s,xmh1s,xmh2s,…,xmhks)+αm-∑t=s∞(f(t,xmf1t,xmf2t,…,+αmxmfkt)-bt)∑t=s∞(f(t,xmf1t,xmf2t,…,]}+βmγmT,β≤n<T,m≥0
converges to an unbounded positive solution x∈A(N,M) of (6) and has the error estimate (17), where {γm}m∈ℕ0={γmn}(m,n)∈ℕ0×ℕβ is an arbitrary sequence in A(N,M), and {αm}m∈ℕ0 and {βm}m∈ℕ0 are any sequences in [0,1] satisfying (18);
(b) equation (6) possesses uncountably many unbounded positive solutions in A(N,M).Proof.
LetL∈(N,M). It follows from (33)~(35) that there exist θ∈(0,1) and T≥n0+τ+β satisfying
(37)θ=1T∑s=T∞‍(RsHs+∑t=s∞‍PtFt),(38)1T∑s=T∞‍(Ws+∑t=s∞‍(Qt+|bt|))<min{M-L,L-N},(39)an=1,∀n≥T.
Define a mapping SL:A(N,M)→lβ∞ by
(40)SLxn={nL+∑i=1∞‍∑s=n+(2i-1)τn+2iτ-1‍{∑t=s∞‍h(s,xh1s,xh2s,…,xhks)-∑t=s∞‍[f(t,xf1t,xf2t,…,xfkt)-bt]},n≥T,SLxT,β≤n<T
for each x={xn}n∈ℕβ∈A(N,M). Using (11), (12), (37), (38), and (40), we get that for each x={xn}n∈ℕβ,y={yn}n∈ℕβ∈A(N,M) and n≥T(41)|SLxnn-SLynn|≤1n∑i=1∞‍∑s=n+(2i-1)τn+2iτ-1‍[∑t=s∞‍|h(s,xh1s,xh2s,…,xhks)∑i=1∞‍∑s=n+(2i-1)τn+2iτ-1‍-h(s,yh1s,yh2s,…,yhks)|∑i=1∞‍∑s=n+(2i-1)τn+2iτ-1‍+∑t=s∞‍|f(t,xf1t,xf2t,…,xfkt)∑i=1∞‍∑s=n+(2i-1)τn+2iτ-1‍+∑t=s∞‍+∑t=s∞‍-f(t,yf1t,yf2t,…,yfkt)|∑t=s∞‍]≤∥x-y∥n∑i=1∞‍∑s=n+(2i-1)τn+2iτ-1‍(RsHs+∑t=s∞‍PtFt)≤∥x-y∥T∑s=T∞‍(RsHs+∑t=s∞‍PtFt)=θ∥x-y∥,|SLxnn-L|≤1n×∑i=1∞‍∑s=n+(2i-1)τn+2iτ-1‍{∑t=s∞‍|h(s,xh1s,xh2s,…,xhks)|1n∑i=1∞‍∑s=n+(2i-1)τn+2iτ-1‍+∑t=s∞‍[|f(t,xf1t,xf2t,…,xfkt)|+|bt|]}≤1T∑i=T∞‍[Ws+∑t=s∞‍(Qt+|bt|)]<min{M-L,L-N},
which imply (24). Consequently, (24) means that SL is a contraction in A(N,M) and has a unique fixed point x={xn}n∈ℕβ∈A(N,M), which is also an unbounded positive solution of (6) in A(N,M). The rest of the proof is similar to the proof of Theorem 3 and is omitted. This completes the proof.Theorem 5.
Assume that there exist three constantsa,M, and N with (1-a)M>N>0 and four nonnegative sequences {Pn}n∈ℕn0, {Qn}n∈ℕn0, {Rn}n∈ℕn0, and {Wn}n∈ℕn0 satisfying (11), (12), (33), (34) and
(42)0≤an≤a<1eventually.
Then
(a) for anyL∈(aM+N,M), there exist θ∈(0,1) and T≥n0+τ+β such that for any x0={x0n}n∈ℕβ∈A(N,M), the Mann iterative sequence with errors {xm}m∈ℕ0={xmn}(m,n)∈ℕ0×ℕβ generated by the scheme
(43)xm+1n={(1-αm-βm)xmn+αm{∑s=n∞nL-anxmn-τ+αm+∑s=n∞[∑t=s∞(f(t,xmf1t,xmf2t,…,xmfkt)-bt)h(s,xmh1s,xmh2s,…,xmhks)+αm-∑t=s∞(f(t,xmf1t,xmf2t,…,+αmxmfkt)-bt)∑t=s∞(f(t,xmf1t,xmf2t,…,xmfkt)-bt)]}+βmγmn,n≥T,m≥0,(1-αm-βm)xmT+αm{∑t=s∞(f(t,xmf1t,xmf2t,…,xmfkt)-bt)TL-aTxmT-τ+αm+∑s=T∞[∑t=s∞(f(t,xmf1t,xmf2t,…,xmfkt)-bt)h(s,xmh1s,xmh2s,…,xmhks)+αm-∑t=s∞(f(t,xmf1t,xmf2t,…,+αmxmfkt)-bt)∑t=s∞(f(t,xmf1t,xmf2t,…,xmfkt)-bt)]}+βmγmT,β≤n<T,m≥0
converges to an unbounded positive solution x∈A(N,M) of (6) and has the error estimate (17), where {γm}m∈ℕ0={γmn}(m,n)∈ℕ0×ℕβ is an arbitrary sequence in A(N,M) and {αm}m∈ℕ0 and {βm}m∈ℕ0 are any sequences in [0,1] satisfying (18);
(b) equation (6) possesses uncountably many unbounded positive solutions in A(N,M).Proof.
PutL∈(aM+N,M). It follows from (33), (34), and (42) that there exist θ∈(0,1) and T≥n0+τ+β satisfying
(44)θ=a+1T∑s=T∞‍(RsHs+∑t=s∞‍PtFt),1T∑s=T∞‍[Ws+∑t=s∞‍(Qt+|bt|)]<min{M-L,L-aM-N},0≤an≤a<1,∀n≥T.
Define a mapping SL:A(N,M)→lβ∞ by
(45)SLxn={nL-anxn-τ+∑s=n∞{∑s=n∞h(s,xh1s,xh2s,…,xhks)-∑t=s∞[f(t,xf1t,xf2t,…,xfkt)-bt]},n≥T,SLxT,β≤n<T,
for each x={xn}n∈ℕβ∈A(N,M). In view of (11), (12), and (44) and (45), we obtain that for each x={xn}n∈ℕβ,y={yn}n∈ℕβ∈A(N,M) and n≥T,
(46)|SLxnn-SLynn|≤an|xn-τ-yn-τn|+1n∑s=n∞[∑s=T∞‍|h(s,xh1s,xh2s,…,xhks)‍+1n-h(s,yh1s,yh2s,…,yhks)|+1n+∑t=s∞|f(t,xf1t,xf2t,…,xfkt)+1n-f(t,yf1t,yf2t,…,yfkt)|∑s=T∞‍]≤[a+1T∑s=T∞‍(RsHs+∑t=s∞‍PtFt)]∥x-y∥=θ∥x-y∥,SLxnn≤L+1n∑s=n∞{∑t=s∞‍|h(s,xh1s,xh2s,…,xhks)|+∑t=s∞‍[|f(t,xf1t,xf2t,…,xfkt)|+|bt|]}≤L+1T∑s=T∞‍[Ws+∑t=s∞‍(Qt+|bt|)]<L+min{M-L,L-aM-N}≤M,SLxnn≥L-aM-1n∑s=n∞{∑t=s∞‍|h(s,xh1s,xh2s,…,xhks)|‍+∑t=s∞‍[|f(t,xf1t,xf2t,…,xfkt)|+|bt|]}≥L-aM-1T∑s=T∞‍[Ws+∑t=s∞‍(Qt+|bt|)]>L-aM-min{M-L,L-aM-N}≥N,
which give (24), in turn, which implies that SL is a contraction in A(N,M) and possesses a unique fixed point x={xn}n∈ℕβ∈A(N,M), which is an unbounded positive solution of (6) in A(N,M). The rest of the proof is similar to that of Theorem 3 and is omitted. This completes the proof.Theorem 6.
Assume that there exist constantsa,M, and N with (1+a)M>N>0 and four nonnegative sequences {Pn}n∈ℕn0, {Qn}n∈ℕn0, {Rn}n∈ℕn0, and {Wn}n∈ℕn0 satisfying (11), (12), (33), (34), and
(47)-1<a≤an≤0eventually.
Then
(a) for anyL∈(N,(1+a)M), there exist θ∈(0,1) and T≥n0+τ+β such that for any x0={x0n}n∈ℕβ∈A(N,M) and the Mann iterative sequence with errors {xm}m∈ℕ0={xmn}(m,n)∈ℕ0×ℕβ generated by (43) converges to an unbounded positive solution x∈A(N,M) of (6) and has the error estimate (17), where {γm}m∈ℕ0={γmn}(m,n)∈ℕ0×ℕβ is an arbitrary sequence in A(N,M), {αm}m∈ℕ0 and {βm}m∈ℕ0 are any sequences in [0,1] satisfying (18);
(b) equation (6) possesses uncountably many unbounded positive solutions in A(N,M).Proof.
PutL∈(N,(1+a)M). It follows from (33), (34), and (47) that there exist θ∈(0,1) and T≥n0+τ+β satisfying
(48)θ=-a+1T∑s=T∞‍(RsHs+∑t=s∞‍PtFt),(49)1T∑s=T∞‍[Ws+∑t=s∞‍(Qt+|bt|)]<min{(1+a)M-L,L-N},(50)-1<a≤an≤0,∀n≥T.
Define a mapping SL:A(N,M)→lβ∞ by (45). By virtue of (12), (45), (48), and (50), we easily verify that
(51)SLxnn≤L-aM+1n∑s=n∞‍{∑s=n∞‍|h(s,xh1s,xh2s,…,xhks)|+∑t=s∞‍[|f(t,xf1t,xf2t,…,xfkt)|+|bt|]}≤L-aM+1T∑s=T∞‍[Ws+∑t=s∞‍(Qt+|bt|)]<L+min{(1+a)M-L,L-N}≤M,SLxnn≥L-1n∑s=Tn-1‍{∑t=s∞‍|h(s,xh1s,xh2s,…,xhks)|+∑t=s∞‍[|f(t,xf1t,xf2t,…,xfkt)|+|bt|]}≥L-1T∑s=T∞‍[Ws+∑t=s∞‍(Qt+|bt|)]>L-min{(1+a)M-L,L-N}≥N,
which yield that SL(A(N,M))⊆A(N,M). The rest of the proof is similar to that of Theorem 5 and is omitted. This completes the proof.Remark 7.
Theorems3~6 extend and improve Theorem 1 in [5].


## 3. Examples

In this section we suggest four examples to explain the results presented in Section2. Note that Theorem 1 in [5] is useless for all these examples.Example 8.
Consider the second-order nonlinear neutral delay difference equation:(52)Δ2(xn-xn-τ)+Δ(sin2xn-2n4)+1(n6+n3+2)(1+|xn2|3)=n2-3n+1n7+n4+1,n≥3,
where τ∈ℕ is fixed. Let n0=3, k=1,  β=min{3-τ,1}, M and N two positive constants with M>N and
(53)an=-1,bn=n2-3n+1n7+n4+1,f(n,u)=1(n6+n3+2)(1+|u|3),h(n,u)=sin2un4,Fn=f1n=n2,Hn=h1n=n-2,Pn=3M2(1+N3)2n6,Qn=1n6,Rn=2n4,Wn=1n4,∀(n,u)∈ℕn0×ℝ.
It is easy to see that (11), (12), and (15) are satisfied. Note that
(54)∑s=n∞‍smax{RsHs,Ws}=∑s=n∞‍smax{2(s-2)s4,1s4}=∑s=n∞‍2(s-2)s3<+∞,∀n∈ℕn0,(55)∑s=n∞‍∑t=s∞‍smax{PtFt,Qt,|bt|}=∑s=n∞‍∑t=s∞‍smax{3M2(1+N3)2t4,1t6,|t2-3t+1|t7+t4+1}≤max{1,3M2(1+N3)2}∑s=n∞‍∑t=s∞‍st4≤max{1,3M2(1+N3)2}∑s=n∞‍∑t=s∞‍1t3=max{1,3M2(1+N3)2}∑t=n∞‍t-n+1t3≤max{1,3M2(1+N3)2}∑t=n∞‍1t2<+∞,∀n∈ℕn0,
which together with Lemma 2 yield that (13) and (14) hold. It follows from Theorem 3 that (52) possesses uncountably many unbounded positive solutions in A(N,M). On the other hand, for any L∈(N,M), there exist θ∈(0,1) and T≥n0+τ+β such that for each x0={x0n}n∈ℕβ∈A(N,M), the Mann iterative sequence with errors {xm}m∈ℕ0={xmn}(m,n)∈ℕ0×ℕβ generated by (16) converges to an unbounded positive solution x∈A(N,M) of (52) and has the error estimate (17), where {γm}m∈ℕ0={γmn}(m,n)∈ℕ0×ℕβ is an arbitrary sequence in A(N,M) and {αm}m∈ℕ0 and {βm}m∈ℕ0 are any sequences in [0,1] satisfying (18).Example 9.
Consider the second-order nonlinear neutral delay difference equation:(56)Δ2(xn+xn-τ)+Δ(sin2x3n3+1n3(n2+2)(1+x2n2-34))+(-1)nn3(xn2-n-1+x(n+1)(n+2))(n11+n5+1)(1+xn2-n-12+x(n+1)(n+2)2)=n2-lnnn6+n5+1,n≥5,
where τ∈ℕ is fixed. Let n0=5, k=2,  β=5-τ, M and N two positive constants with M>N and
(57)an=1,bn=n2-lnnn6+n5+1,f(n,u,v)=(-1)nn3(u+v)(n11+n5+1)(1+u2+v2),h(n,u,v)=sin2vn3(n2+2)(1+u4),f1n=n2-n-1,Fn=f2n=(n+1)(n+2),h1n=2n2-3,Hn=h2n=3n3+1,Pn=Qn=4n8,Rn=Wn=10n5,∀(n,u,v)∈ℕn0×ℝ2.
It is clear that (11), (12), and (35) are fulfilled. Note that
(58)limn→∞1n∑s=n∞‍max{RsHs,Ws}=limn→∞1n∑s=n∞‍max{10(3s3+1)s5,10s5}=0,1n∑s=n∞‍∑t=s∞‍max{PtFt,Qt,|bt|}=1n∑s=n∞‍∑t=s∞‍max{4(t+1)(t+2)t8,4t8,t2-lntt6+t5+1}≤4n∑s=n∞‍∑t=s∞‍1t4≤4n∑t=n∞‍1t3→0asn→∞,
which yields that
(59)limn→∞1n∑s=n∞∑t=s∞max{PtFt,Qt,|bt|}=0.
Thus, Theorem 4 guarantees that (56) possesses uncountably unbounded positive solutions in A(N,M). On the other hand, for any L∈(N,M), there exist θ∈(0,1) and T≥τ+n0+β such that the Mann iterative sequence with error {xm}m∈ℕ0 generated by (36) converges to an unbounded positive solution x∈A(N,M) of (56) and has the error estimate (17), where {γm}m∈ℕ0 is an arbitrary sequence in A(N,M) and {αm}m∈ℕ0 and {βm}m∈ℕ0 are any sequences in [0,1] satisfying (18).Example 10.
Consider the second-order nonlinear neutral delay difference equation:(60)Δ2(xn+3n3-14n3+2xn-τ)+Δ(sin(ln(1+n2|x3n2-1|))n9-n-4+Δ-n2-(-1)n(n-1)/2(n7+3n5-1)2|x4n3+1|)+(-1)nn6(1+xn-22)-1(n5+1)1+|xn+4|=(-1)nn3-1n8ln3n+1,n≥7,
where τ∈ℕ is fixed. Let n0=7, k=2,  a=3/4, β=min{7-τ,5}, M and N two positive constants with M>4N and
(61)an=3n3-14n3+2,bn=(-1)nn3-1n8ln3n+1,f(n,u,v)=(-1)nn6(1+u2)-1(n5+1)1+|v|,h(n,u,v)=sin(ln(1+n2|u|))n9-n-4-n2-(-1)n(n-1)/2(n7+3n5-1)2|v|,f1n=n-2,Fn=f2n=n+4,h1n=3n2-1,Hn=h2n=4n3+1,Pn=Qn=2n4,Rn=Wn=2n5,∀(n,u,v)∈ℕn0×ℝ2.
It is not difficult to verify that (11), (12), and (42) are fulfilled. Note that
(62)limn→∞1n∑s=n∞‍max{RsHs,Ws}=limn→∞1n∑s=n∞‍max{2(4s3+1)s5,2s5}=limn→∞1n∑s=n∞‍2(4s3+1)s5=0,limn→∞1n∑s=n∞‍∑t=s∞‍max{PtFt,Qt,|bt|}=limn→∞1n∑s=n∞‍∑t=s∞‍max{2t+8t4,2t4,|(-1)tt3-1|t8ln3t+1}=limn→∞1n∑s=n∞‍∑t=s∞‍2t+8t4=0.
That is, (33) and (34) are satisfied. Consequently Theorem 5 implies that (60) possesses uncountably many unbounded positive solutions in A(N,M). On the other hand, for any L∈((3/4)M+N,M), there exist θ∈(0,1) and T≥n0+τ+β such that the Mann iterative sequence with error {xm}m∈ℕ0 generated by (43) converges to an unbounded positive solution x∈A(N,M) of (60) and has the error estimate (17), where {γm}m∈ℕ0 is an arbitrary sequence in A(N,M) and {αm}m∈ℕ0 and {βm}m∈ℕ0 are any sequences in [0,1] satisfying (18).Example 11.
Consider the second-order nonlinear neutral delay difference equation:(63)Δ2(xn+1-2n22+3n2xn-τ)+Δ(n2-1(n5+2n3-1)(1+x2n-152))+sin2(n3x5n2-2)(n+2)8=(-1)nn4+n3+3n2-1n9+n7+3n6+n4+1,n≥11,
where τ∈ℕ is fixed. Let n0=11,  k=1, a=-4/5,  β=min{11-τ,7}, M and N two positive constants with M>5N and
(64)an=1-2n22+3n2,bn=(-1)nn4+n3+3n2-1n9+n7+3n6+n4+1,f(n,u)=sin2(n3u)(n+2)8,h(n,u)=n2-1(n5+2n3-1)(1+u2),Fn=f1n=5n2-2,Hn=h1n=2n-15,Pn=2n5,Qn=1n8,Rn=2n3,Wn=1n3,∀(n,u)∈ℕn0×ℝ.
Obviously, (11), (12), and (50) are satisfied. Note that
(65)limn→∞1n∑s=n∞‍max{RsHs,Ws}=limn→∞1n∑s=n∞‍max{4s-30s3,1s3}=0,1n∑s=n∞‍∑t=s∞‍max{PtFt,Qt,|bt|}=1n∑s=n∞‍∑t=s∞‍max{2(5t2-2)t5,1t8,|(-1)tt4+t3+3t2-1|t9+t7+3t6+t4+1}≤10n∑s=n∞‍∑t=s∞‍1t3→0asn→∞,
which gives that
(66)limn→∞1n∑s=n∞‍∑t=s∞‍max{PtFt,Qt,|bt|}=0.
That is, (33) and (34) hold. Thus, Theorem 6 shows that (63) possesses uncountably many unbounded positive solutions in A(N,M). On the other hand, for any L∈(N,M/5), there exist θ∈(0,1) and T≥n0+τ+β such that the Mann iterative sequence with error {xm}m∈ℕ0 generated by (43) converges to an unbounded positive solution x∈A(N,M) of (63) and has the error estimate (17), where {γm}m∈ℕ0 is an arbitrary sequence in A(N,M) and {αm}m∈ℕ0 and {βm}m∈ℕ0 are any sequences in [0,1] satisfying (18).



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*Source: 245012-2013-02-24.xml*