# On Positive Solutions and Mann Iterative Schemes of a Third Order Difference Equation

Authors: Zeqing Liu; Heng Wu; Shin Min Kang; Young Chel Kwun
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

The existence of uncountably many positive solutions and convergence of the Mann iterative
schemes for a third order nonlinear neutral delay difference equation are proved. Six examples are given to illustrate the results presented in this paper.


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## Body

## 1. Introduction and Preliminaries

Recently, many researchers studied the oscillation, nonoscillation, and existence of solutions for linear and nonlinear second and third order difference equations and systems see, for example, [1–23] and the references cited therein. By means of the Reccati transformation techniques, Saker [18] discussed the third order difference equation
(1)Δ3xn+pnxn+1=0,∀n≥n0,
and presented some sufficient conditions which ensure that all solutions are to be oscillatory or tend to zero. Utilizing the Schauder fixed point theorem, Yan and Liu [22] proved the existence of a bounded nonoscillatory solution for the third order difference equation
(2)Δ3xn+f(n,xn,xn-τ)=0,∀n≥n0.
Agarwal [2] established the oscillatory and asymptotic properties for the third order nonlinear difference equation
(3)Δ3xn+qnf(xn+1)=0,∀n≥1.
Andruch-Sobiło and Migda [4] studied the third order linear difference equation of neutral type
(4)Δ3(xn-pnxσn)±qnxτn=0,∀n≥n0,
and obtained sufficient conditions which ensure that all solutions of the equation are oscillatory. Grace and Hamedani [6] discussed the difference equation
(5)Δ3(xn-xn-τ)±qn|xn-σ|3sgn⁡xn-σ=0,∀n≥0,
and gave some new criteria for the oscillation of all solutions and all bounded solutions.Our goal is to discuss solvability and convergence of the Mann iterative schemes for the following third order nonlinear neutral delay difference equation:(6)Δ3(xn+bnxn-τ)+Δh(n,xh1n,xh2n,…,xhkn)+f(n,xf1n,xf2n,…,xfkn)=cn,∀n≥n0,
whereτ, k, n0∈ℕ, {bn}n∈ℕn0, {cn}n∈ℕn0⊂ℝ, h, f∈C(ℕn0×ℝk,ℝ), {hln}n∈ℕn0, {fln}n∈ℕn0⊆ℕ, and
(7)lim⁡n→∞hln=lim⁡n→∞fln=+∞,l∈{1,2,…,k}.
By employing the Banach fixed point theorem and some new techniques, we establish the existence of uncountably many positive solutions of (6), conceive a few Mann iterative schemes for approximating these positive solutions, and prove their convergence and the error estimates. Six nontrivial examples are included.Throughout this paper, we assume thatΔis the forward difference operator defined byΔxn=xn+1-xn, ℝ=(-∞,+∞),ℝ+=[0,+∞)ℕ0and ℕ denote the sets of 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⁡{hln2:l∈{1,2,…,k}},∀n∈ℕn0,Fn=max⁡{fln2:l∈{1,2,…,k}},∀n∈ℕn0,
and lβ∞represents the Banach space of all real sequences onℕβwith norm
(9)∥x∥=sup⁡n∈ℕβ|xnn2|<+∞foreachx={xn}n∈ℕβ∈lβ∞,A(N,M)={x={xn}n∈ℕβ∈lβ∞:N≤xnn2≤M,n∈ℕβ}{.}n∈ℕβ∈lβ∞:NN≤xnn2≤M,n∈ℕβforanyM>N>0.
It is easy to see thatA(N,M)is a closed and convex subset oflβ∞. By a solution of (6), we mean a sequence{xn}n∈ℕβwith a positive integerT≥n0+τ+βsuch that (6) holds for alln≥T.Lemma 1.
Let{pt}t∈ℕbe a nonnegative sequence andτ∈ℕ.(i)
Iflim⁡n→∞(1/n2)∑t=n+τ∞‍t2pt=0, thenlim⁡n→∞(1/n2)∑i=1∞‍∑s=n+iτ∞‍∑t=s∞‍pt=0.(ii)
Iflim⁡n→∞(1/n2)∑t=n+τ∞‍t3pt=0, thenlim⁡n→∞(1/n2)∑i=1∞‍∑u=n+iτ∞‍∑s=u∞‍∑t=s∞‍pt=0.Proof.
Note that(10)0≤1n2∑i=1∞‍∑s=n+iτ∞‍∑t=s∞‍pt=1n2∑i=1∞‍(∑t=n+iτ∞‍pt+∑t=n+1+iτ∞‍pt+∑t=n+2+iτ∞‍pt+⋯)=1n2∑i=1∞‍∑t=n+iτ∞‍(1+t-n-iτ)pt≤1n2∑i=1∞‍∑t=n+iτ∞‍tpt=1n2(∑t=n+τ∞‍tpt+∑t=n+2τ∞‍tpt+∑t=n+3τ∞‍tpt+⋯)≤1n2∑t=n+τ∞‍(1+t-n-ττ)tpt=1n2∑t=n+τ∞‍t-nτtpt≤1n2τ∑t=n+τ∞‍t2pt⟶0asn⟶∞
that is,
(11)lim⁡n→∞1n2∑i=1∞‍∑s=n+iτ∞‍∑t=s∞‍pt=0.
As in the proof of (10), we infer that
(12)0≤1n2∑i=1∞‍∑u=n+iτ∞‍∑s=u∞‍∑t=s∞‍pt=1n2∑i=1∞‍∑u=n+iτ∞‍∑t=u∞‍(1+t-u)pt≤1n2∑i=1∞‍∑u=n+iτ∞‍∑t=u∞‍tpt≤1n2τ∑t=n+τ∞‍t3pt⟶0asn⟶∞,
which implies that
(13)lim⁡n→∞1n2∑i=1∞‍∑u=n+iτ∞‍∑s=u∞‍∑t=s∞‍pt=0.
This completes the proof.


## 2. Uncountably Many Positive Solutions and Mann Iterative Schemes

In this section, using the Banach fixed point theorem and Mann iterative schemes, we establish the existence of uncountably many positive solutions of (6), prove convergence of the Mann iterative schemes relative to these positive solutions, and compute the error estimates between the Mann iterative schemes and the positive solutions.Theorem 2.
Assume that there exist two constantsMandNwithM>N>0and four nonnegative sequences{Pn}n∈ℕn0, {Qn}n∈ℕn0, {Rn}n∈ℕn0and {Wn}n∈ℕn0 satisfying
(14)|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;(15)|f(n,u1,u2,…,uk)|≤Qn,|h(n,u1,u2,…,uk)|≤Wn,(n,u1,u2,…,21)∀(n,ul)∈ℕn0×(ℝ+∖{0}),1≤l≤k;(16)lim⁡n→∞1n2∑i=1∞‍∑u=n+iτ∞‍∑s=u∞‍max⁡{RsHs,Ws}=0;(17)lim⁡n→∞1n2∑i=1∞‍∑u=n+iτ∞‍∑s=u∞‍∑t=s∞‍max⁡{PtFt,Qt,|ct|}=0;(18)bn=-1eventually.
Then one has the following.(a)
For anyL∈(N,M), there existθ∈(0,1)andT≥n0+τ+βsuch that, for eachx0={x0n}n∈ℕβ∈A(N,M), the Mann iterative sequence{xm}m∈ℕ0={{xmn}n∈ℕβ}m∈ℕ0 generated by the scheme
(19)xm+1n={(1-αm)xmn+αm{h(s,xmh1s,xmh2s,…,xmhks)]n2L+∑i=1∞‍∑u=n+iτ∞‍∑s=u∞‍[h(s,xmh1s,xmh2s,…,xmhks)-∑t=s∞‍(f(t,xmf1t,xmf2t,…,xmfkt)-cth(s,xmh1s,xmh2s,…,xmhks))]},n≥T,m≥0,(1-αm)xmT+αm{h(s,xmh1s,xmh2s,…,xmhks)T2L+∑i=1∞‍∑u=T+iτ∞‍∑s=u∞‍[h(s,xmh1s,xmh2s,…,xmhks)-∑t=s∞‍(f(t,xmf1t,xmf2t,…,xmfkt)-cth(s,xmh1s,xmh2s,…,xmhks))]},β≤n<T,m≥0,converges to a positive solutionz={zn}n∈ℕβ∈A(N,M)of (6) with lim⁡n→∞zn=+∞ and has the following error estimate:
(20)∥xm+1-z∥≤e-(1-θ)∑i=0mαi∥xm-z∥,∀m∈ℕ0,
where {αm}m∈ℕ0 is an arbitrary sequence in[0,1]such that
(21)∑m=0∞‍αm=+∞.(b)
Equation (6) possesses uncountably many positive solutions inA(N,M).Proof.
Firstly, we show that (a) holds. PutL∈(N,M). It follows from (16)~(18) that there existθ∈(0,1)andT≥n0+τ+βsatisfying
(22)θ=1T2∑i=1∞‍∑u=T+iτ∞‍∑s=u∞‍(RsHs+∑t=s∞‍PtFt);(23)1T2∑i=1∞‍∑u=T+iτ∞‍∑s=u∞‍(Ws+∑t=s∞‍(Qt+|ct|))<min⁡{M-L,L-N};(24)bn=-1,∀n≥T.
Define a mapping SL:A(N,M)→lβ∞ by
(25)SLxn={n2L+∑i=1∞‍∑u=n+iτ∞‍∑s=u∞‍{h(s,xh1s,xh2s,…,xhks)-∑t=s∞‍[f(t,xf1t,xf2t,…,xfkt)-ct]},n≥T,SLxT,β≤n<T,
for eachx={xn}n∈ℕβ∈A(N,M). In light of (14), (15), (22), (23), and (25), we obtain that for eachx={xn}n∈ℕβ,y={yn}n∈ℕβ∈A(N,M)(26)|SLxnn2-SLynn2|≤1n2∑i=1∞‍∑u=n+iτ∞‍∑s=u∞‍[|h(s,xh1s,xh2s,…,xhks)-h(s,yh1s,yh2s,…,yhks)|+∑t=s∞‍|f(t,xf1t,xf2t,…,xfkt)-f(t,yf1t,yf2t,…,yfkt)|]≤1n2∑i=1∞‍∑u=n+iτ∞‍∑s=u∞‍[Rsmax⁡{|xhls-yhls|:1≤l≤k}+∑t=s∞‍Ptmax⁡{|xflt-yflt|:1≤l≤k}]≤∥x-y∥n2∑i=1∞‍∑u=n+iτ∞‍∑s=u∞‍[∑t=s∞‍Rsmax⁡{hls2:1≤l≤k}+∑t=s∞‍Ptmax⁡{flt2:1≤l≤k}]≤∥x-y∥T2∑i=1∞‍∑u=T+iτ∞‍∑s=u∞‍(RsHs+∑t=s∞‍PtFt)=θ∥x-y∥,|SLxnn2-L|=|1n2∑i=1∞‍∑u=n+iτ∞‍∑s=u∞‍{∑t=s∞‍h(s,xh1s,xh2s,…,xhks)-∑t=s∞‍[f(t,xf1t,xf2t,…,xfkt)-ct]}|≤1n2∑i=1∞‍∑u=n+iτ∞‍∑s=u∞‍{∑t=s∞‍|h(s,xh1s,xh2s,…,xhks)|∑t=s∞‍+∑t=s∞‍[|f(t,xf1t,xf2t,…,xfkt)|+|ct||f(t,xf1t,xf2t,…,xfkt)|]∑t=s∞‍}≤1T2∑i=1∞‍∑u=T+iτ∞‍∑s=u∞‍[Ws+∑t=s∞‍(Qt+|ct|)]<min⁡{M-L,L-N},
which yield that
(27)SL(A(N,M))⊆A(N,M),∥SLx-SLy∥≤θ∥x-y∥,∀x,y∈A(N,M),
which implies thatSLis a contraction inA(N,M). The Banach fixed point theorem and (27) ensure thatSLhas a unique fixed pointz={zn}n∈ℕβ∈A(N,M); that is,
(28)zn=n2L+∑i=1∞‍∑u=n+iτ∞‍∑s=u∞‍{∑t=s∞‍h(s,zh1s,zh2s,…,zhks)-∑t=s∞‍[f(t,zf1t,zf2t,…,zfkt)-ct]},-∑t=s∞‍[f(t,zf1t,zf2t,…,zfkt)-ct]}∀n≥T,zn-τ=(n-τ)2L+∑i=1∞‍∑u=n+(i-1)τ∞‍∑s=u∞‍{h(s,zh1s,zh2s,…,zhks)-∑t=s∞‍[h(s,zh1s,zh2s,…,zhks)f(t,zf1t,zf2t,…,zfkt)-cth(s,zh1s,zh2s,…,zhks)]},∀n≥T+τ,
which mean that
(29)zn-zn-τ=(2nτ-τ2)L-∑u=n∞‍∑s=u∞‍{∑t=s∞‍h(s,zh1s,zh2s,…,zhks)-∑t=s∞‍[f(t,zf1t,zf2t,…,zfkt)-ct]},-∑t=s∞‍[f(t,zf1t,zf2t,…,zfkt)-ct]}∀n≥T+τ,
which yields that
(30)Δ(zn-zn-τ)=2τL+∑s=n∞‍{∑s=n∞‍h(s,zh1s,zh2s,…,zhks)-∑t=s∞‍[f(t,zf1t,zf2t,…,zfkt)-ct]},-∑t=s∞‍[f(t,zf1t,zf2t,…,zfkt)-ct]}∀n≥T+τ,Δ2(zn-zn-τ)=-h(n,zh1n,zh2n,…,zhkn)+∑t=n∞‍[f(t,zf1t,zf2t,…,zfkt)-ct],∀n≥T+τ,
which gives that
(31)Δ3(zn-zn-τ)=-Δh(n,zh1n,zh2n,…,zhkn)-f(t,zf1n,zf2n,…,zfkn)+cn,∀n≥T+τ,
which together with (24) implies thatz={zn}n∈ℕβis a positive solution of (6) inA(N,M). Note that
(32)N≤znn2≤M,∀n∈ℕβ,
which guarantees thatlim⁡n→∞zn=+∞. It follows from (19), (22), (24), (25), and (27) that for anym∈ℕ0andn≥T(33)|xm+1nn2-znn2|=1n2|∑i=1∞‍(1-αm)xmn+αm{xmfkt)-ct)]n2L+∑i=1∞‍∑u=n+iτ∞‍∑s=u∞‍[h(s,xmh1s,xmh2s,…,xmhks)-∑t=s∞‍(f(t,xmf1t,xmf2t,…,xmfkt)-ct)]}-zn∑i=1∞‍|≤(1-αm)|xmn-zn|n2+αm|SLxmn-SLzn|n2≤(1-αm)∥xm-z∥+θαm∥xm-z∥≤[1-(1-θ)αm]∥xm-z∥,∀m∈ℕ0,n≥T,
which implies that
(34)∥xm+1-z∥≤e-(1-θ)∑i=0mαi∥xm-z∥,∀m∈ℕ0.
That is, (20) holds. Thus Lemma 1, (20), and (21) guarantee that lim⁡m→∞xm=z.
Next we show that (b) holds. LetL1, L2∈(N,M)andL1≠L2. As in the proof of (a), we deduce similarly that, for eachc∈{1,2}, there exist constantsθc∈(0,1)andTc≥n0+τ+βand a mappingSLcsatisfying (22)~(27), whereθ, L, andTare replaced byθc, Lc, andTc, respectively, and the mappingSLchas a fixed pointzc={znc}n∈ℕβ∈A(N,M), which is a positive solution of (6) inA(N,M)withlim⁡n→∞znc=+∞. It follows that
(35)znc=n2Lc+∑i=1∞‍∑u=n+iτ∞‍∑s=u∞‍{∑t=s∞‍h(s,zh1sc,zh2sc,…,zhksc)-∑t=s∞‍[f(t,zf1tc,zf2tc,…,zfktc)-ct]},-∑t=s∞‍[f(t,zf1tc,zf2tc,…,zfktc)-ct]},∀n≥Tc,
which together with (14) and (20) means that forn≥max⁡⁡{T1,T2}(36)|zn1n2-zn2n2|≥|L1-L2|-1n2∑i=1∞‍∑u=n+iτ∞‍∑s=u∞‍|h(s,zh1s1,zh2s1,…,zhks1)-h(s,zh1s2,zh2s2,…,zhks2)|+∑t=s∞‍|f(t,zf1t1,zf2t1,…,zfkt1)-f(t,zf1t2,zf2t2,…,zfkt2)|≥|L1-L2|-1n2∑i=1∞‍∑u=n+iτ∞‍∑s=u∞‍[∑t=s∞‍Rsmax⁡{|zhls1-zhls2|:1≤l≤k}+∑t=s∞‍Ptmax⁡{|zflt1-zflt2|:1≤l≤k}]≥|L1-L2|-∥z1-z2∥n2∑i=1∞‍∑u=n+iτ∞‍∑s=u∞‍(RsHs+∑t=s∞‍PtFt)≥|L1-L2|-∥z1-z2∥max⁡{T12,T22}∑i=1∞‍∑u=max⁡{T1,T2}+iτ∞‍∑s=u∞‍(RsHs+∑t=s∞‍PtFt)≥|L1-L2|-max⁡{θ1,θ2}∥z1-z2∥,
which yields that
(37)∥z1-z2∥≥|L1-L2|1+max⁡{θ1,θ2}>0;
that is,z1≠z2.This completes the proof.Theorem 3.
Assume that there exist two constantsMandNwithM>N>0and four nonnegative sequences{Pn}n∈ℕn0, {Qn}n∈ℕn0, {Rn}n∈ℕn0, and{Wn}n∈ℕn0satisfying (14), (15), and
(38)lim⁡n→∞1n2∑u=n∞‍∑s=u∞‍max⁡⁡{RsHs,Ws}=0;(39)lim⁡n→∞1n2∑u=n∞‍∑s=u∞‍∑t=s∞‍max⁡⁡{PtFt,Qt,|ct|}=0;(40)bn=1eventually.
Then one has the following.(a)
For anyL∈(N,M), there existθ∈(0,1)andT≥n0+τ+βsuch that, for eachx0={x0n}n∈ℕβ∈A(N,M), the Mann iterative sequence{xm}m∈ℕ0={{xmn}n∈ℕβ}m∈ℕ0generated by the scheme
(41)xm+1n={(1-αm)xmn+αm{h(s,xmh1s,xmh2s,…,xmhks)]n2L-∑i=1∞‍∑u=n+(2i-1)τn+2iτ-1‍∑s=u∞‍[h(s,xmh1s,xmh2s,…,xmhks)-∑t=s∞‍(f(t,xmf1t,xmf2t,…,xmfkt)-cth(s,xmh1s,xmh2s,…,xmhks))]},n≥T,m≥0,(1-αm)xmT+αm{h(s,xmh1s,xmh2s,…,xmhks)T2L-∑i=1∞‍∑u=T+(2i-1)τT+2iτ-1‍∑s=u∞‍[h(s,xmh1s,xmh2s,…,xmhks)-∑t=s∞‍(f(t,xmf1t,xmf2t,…,xmfkt)-cth(s,xmh1s,xmh2s,…,xmhks))]},β≤n<T,m≥0,converges to a positive solutionz={zn}n∈ℕβ∈A(N,M)of (6) withlim⁡n→∞zn=+∞and has the error estimate (20), where{αm}m∈ℕ0is an arbitrary sequence in[0,1]satisfying (21).(b)
Equation (6) possesses uncountably many positive solutions inA(N,M).Proof.
LetL∈(N,M). It follows from (38)~(40) that there existθ∈(0,1)andT≥n0+τ+βsatisfying
(42)θ=1T2∑u=T∞‍∑s=u∞‍(RsHs+∑t=s∞‍PtFt);(43)1T2∑u=T∞‍∑s=u∞‍(Ws+∑t=s∞‍(Qt+|ct|))<min⁡{M-L,L-N};(44)bn=1,∀n≥T.
Define a mappingSL:A(N,M)→lβ∞by
(45)SLxn={n2L-∑i=1∞‍∑u=n+(2i-1)τn+2iτ-1‍∑s=u∞‍{h(s,xh1s,xh2s,…,xhks)-∑t=s∞‍[f(t,xf1t,xf2t,…,xfkt)-ct]},n≥T,SLxT,β≤n<T,
for eachx={xn}n∈ℕβ∈A(N,M). Using (14), (15), (42), (43), and (45), we get that for eachx={xn}n∈ℕβ,y={yn}n∈ℕβ∈A(N,M)andn≥T(46)|SLxnn2-SLynn2|≤1n2∑i=1∞‍∑u=n+(2i-1)τn+2iτ-1‍∑s=u∞‍[|h(s,xh1s,xh2s,…,xhks)-h(s,yh1s,yh2s,…,yhks)|+∑t=s∞‍|f(t,xf1t,xf2t,…,xfkt)-f(t,yf1t,yf2t,…,yfkt)|]≤∥x-y∥n2∑i=1∞‍∑u=n+(2i-1)τn+2iτ-1‍∑s=u∞‍(RsHs+∑t=s∞‍PtFt)≤∥x-y∥T2∑u=T∞‍∑s=u∞‍(RsHs+∑t=s∞‍PtFt)=θ∥x-y∥,|SLxnn2-L|≤1n2∑i=1∞‍∑u=n+(2i-1)τn+2iτ-1‍∑s=u∞‍{|h(s,xh1s,xh2s,…,xhks)|+∑t=s∞‍[|f(t,xf1t,xf2t,…,xfkt)|+|ct||f(t,xf1t,xf2t,…,xfkt)|]}≤1T2∑u=T∞‍∑s=u∞‍(Ws+∑t=s∞‍[Ws+∑t=s∞‍(Qt+|ct|)])<min⁡{M-L,L-N},
which imply (27). The rest of the proof is similar to the proof of Theorem 2 and is omitted. This completes the proof.Theorem 4.
Assume that there exist three constantsb, M, andNwith(1-b)M>N>0and four nonnegative sequences{Pn}n∈ℕn0, {Qn}n∈ℕn0, {Rn}n∈ℕn0and{Wn}n∈ℕn0satisfying (14), (15), (38), (39) and
(47)0≤bn≤b<1eventually.
Then one has the following.(a)
For anyL∈(bM+N,M), there existθ∈(0,1)andT≥n0+τ+βsuch that, for anyx0={x0n}n∈ℕβ∈A(N,M), the Mann iterative sequence{xm}m∈ℕ0={{xmn}n∈ℕβ}m∈ℕ0 generated by the scheme
(48)xm+1n={(1-αm)xmn+αm{h(s,xmh1s,xmh2s,…,xmhks)]n2L-bnxmn-τ-∑u=n∞‍∑s=u∞‍[h(s,xmh1s,xmh2s,…,xmhks)-∑t=s∞‍(f(t,xmf1t,xmf2t,…,xmfkt)-cth(s,xmh1s,xmh2s,…,xmhks))]},n≥T,m≥0,(1-αm)xmT+αm{h(s,xmh1s,xmh2s,…,xmhks)T2L-bTxmT-τ-∑u=T∞‍∑s=u∞‍[h(s,xmh1s,xmh2s,…,xmhks)-∑t=s∞‍(f(t,xmf1t,xmf2t,…,xmfkt)-cth(s,xmh1s,xmh2s,…,xmhks))]},β≤n<T,m≥0,converges to a positive solutionz={zn}n∈ℕβ∈A(N,M)of (6) withlim⁡n→∞zn=+∞and has the error estimate (20), where{αm}m∈ℕ0is an arbitrary sequence in[0,1]satisfying (21).(b)
Equation (6) possesses uncountably many positive solutions inA(N,M).Proof.
PutL∈(bM+N,M). It follows from (38), (39), and (47) that there existθ∈(0,1)andT≥n0+τ+βsatisfying
(49)θ=b+1T2∑u=T∞‍∑s=u∞‍(RsHs+∑t=s∞‍PtFt);1T2∑u=T∞‍∑s=u∞‍(Ws+∑t=s∞‍(Qt+|ct|))<min⁡{M-L,L-bM-N};0≤bn≤b<1,∀n≥T.
Define a mappingSL:A(N,M)→lβ∞by
(50)SLxn={n2L-bnxn-τ-∑u=n∞‍∑s=u∞‍{∑t=s∞‍h(s,xh1s,xh2s,…,xhks)-∑t=s∞‍[f(t,xf1t,xf2t,…,xfkt)-ct]},n≥T,SLxT,β≤n<T,
for eachx={xn}n∈ℕβ∈A(N,M). In view of (14), (15), and (49) and (50), we obtain that for eachx={xn}n∈ℕβ,y={yn}n∈ℕβ∈A(N,M)andn≥T(51)|SLxnn2-SLynn2|≤bn|xn-τ-yn-τn2|+1n2∑u=n∞‍∑s=u∞‍[|h(s,xh1s,xh2s,…,xhks)-h(s,yh1s,yh2s,…,yhks)|+∑t=s∞‍|f(t,xf1t,xf2t,…,xfkt)-f(t,yf1t,yf2t,…,yfkt)|]≤bn|xn-τ-yn-τ(n-τ)2|(n-τ)2n2+1n2∑u=n∞‍∑s=u∞‍[∑t=s∞‍Rsmax⁡{|xhls-yhls|:1≤l≤k}+∑t=s∞‍Ptmax⁡{|xflt-yflt|:1≤l≤k}]≤b∥x-y∥+∥x-y∥n2∑u=n∞‍∑s=u∞‍[∑t=s∞‍PtRsmax⁡{hls2:1≤l≤k}+∑t=s∞‍Ptmax⁡{flt2:1≤l≤k}]≤[b+1T2∑u=T∞‍∑s=u∞‍(RsHs+∑t=s∞‍PtFt)]∥x-y∥=θ∥x-y∥,SLxnn2≤L+1n2∑u=n∞‍∑s=u∞‍{∑t=s∞‍|h(s,xh1s,xh2s,…,xhks)|+∑t=s∞‍[|f(t,xf1t,xf2t,…,xfkt)|+|ct|]}≤L+1T2∑u=T∞‍∑s=u∞‍(Ws+∑t=s∞‍(Qt+|ct|))<L+min⁡{M-L,L-bM-N}≤M,SLxnn2≥L-bM-1n2∑u=n∞‍∑s=u∞‍{∑t=s∞‍|h(s,xh1s,xh2s,…,xhks)|+∑t=s∞‍[|f(t,xf1t,xf2t,…,xfkt)|+|ct|]}≥L-bM-1T2∑u=T∞‍∑s=u∞‍[Ws+∑t=s∞‍(Qt+|ct|)]>L-bM-min⁡{M-L,L-bM-N}≥N,
which imply (27). The rest of the proof is similar to that of Theorem 2 and is omitted. This completes the proof.Theorem 5.
Assume that there exist constantsb, M, andNwith(1+b)M>N>0and four nonnegative sequences{Pn}n∈ℕn0, {Qn}n∈ℕn0, {Rn}n∈ℕn0, and {Wn}n∈ℕn0satisfying (14), (15), (38), (39), and
(52)-1<b≤bn≤0eventually.
Then one has the following.(a)
For anyL∈(N,(1+b)M), there existθ∈(0,1)andT≥n0+τ+βsuch that, for anyx0={x0n}n∈ℕβ∈A(N,M), the Mann iterative sequence{xm}m∈ℕ0={{xmn}n∈ℕβ}m∈ℕ0generated by (48) converges to a positive solutionz={zn}n∈ℕβ∈A(N,M)of (6) withlim⁡n→∞zn=+∞and has the error estimate (20), where{αm}m∈ℕ0is an arbitrary sequence in[0,1]satisfying (21).(b)
Equation (6) possesses uncountably many positive solutions inA(N,M).Proof.
PutL∈(N,(1+b)M). It follows from (38), (39), and (52) that there existθ∈(0,1)andT≥n0+τ+βsatisfying
(53)θ=-b+1T2∑u=T∞‍∑s=u∞‍(RsHs+∑t=s∞‍PtFt);(54)1T2∑u=T∞‍∑s=u∞‍(Ws+∑t=s∞‍(Qt+|ct|))<min⁡{(1+b)M-L,L-N};(55)-1<b≤bn≤0,∀n≥T.
Define a mappingSL:A(N,M)→lβ∞by (50). By virtue of (15), (50), (53), and (55), we infer that for allx={xn}n∈ℕβ, y={yn}n∈ℕβ∈A(N,M) andn≥T(56)|SLxnn2-SLynn2|≤bn|xn-τ-yn-τn2|+1n2∑u=n∞‍∑s=u∞‍[|h(s,xh1s,xh2s,…,xhks)-h(s,yh1s,yh2s,…,yhks)|+∑t=s∞‍|f(t,xf1t,xf2t,…,xfkt)-f(t,yf1t,yf2t,…,yfkt)|]≤[-b+1T2∑u=T∞‍∑s=u∞‍(RsHs+∑t=s∞‍PtFt)]∥x-y∥=θ∥x-y∥,SLxnn2≤L-bM+1n2∑u=n∞‍∑s=u∞‍{∑t=s∞‍|h(s,xh1s,xh2s,…,xhks)|+∑t=s∞‍[|f(t,xf1t,xf2t,…,xfkt)|+|ct|]}≤L-bM+1T2∑u=T∞‍∑s=u∞‍(Ws+∑t=s∞‍(Qt+|ct|))<L-bM+min⁡{(1+b)M-L,L-N}≤M,SLxnn2≥L-1n2∑u=n∞‍∑s=u∞‍{∑s=u∞‍|h(s,xh1s,xh2s,…,xhks)|+∑t=s∞‍[|f(t,xf1t,xf2t,…,xfkt)|+|ct|]}≥L-1T2∑u=T∞‍∑s=u∞‍[Ws+∑t=s∞‍(Qt+|ct|)]>L-min⁡{(1+b)M-L,L-N}≥N.
That is, (27) holds. The rest of the proof is similar to that of Theorem 2 and is omitted. This completes the proof.Theorem 6.
Assume that there exist constantsb, M, and N with(1-1/b)M>N>0and four nonnegative sequences{Pn}n∈ℕn0, {Qn}n∈ℕn0, {Rn}n∈ℕn0, and {Wn}n∈ℕn0satisfying (14), (15), (38),  (39), and
(57)bn≥b>1eventually.
Then one has the following.(a)
For anyL∈((1/b)M+N,M), there existθ∈(0,1)andT≥n0+τ+βsuch that, for any x0={x0n}n∈ℕβ∈A(N,M), the Mann iterative sequence {xm}m∈ℕ0={{xmn}n∈ℕβ}m∈ℕ0 generated by the scheme
(58)xm+1n={(1-αm)xmn+αm{h(s,xmh1s,xmh2s,…,xmhks)]n2L-xmn+τbn+τ-1bn+τ×∑u=n+τ∞‍∑s=u∞‍[h(s,xmh1s,xmh2s,…,xmhks)-∑t=s∞‍(f(t,xmf1t,xmf2t,…,xmfkt)-cth(s,xmh1s,xmh2s,…,xmhks))]xmn+τbn+τ},n≥T,m≥0,(1-αm)xmT+αm{h(s,xmh1s,xmh2s,…,xmhks)T2L-xmT+τbT+τ-∑u=T+τ∞‍∑s=u∞‍[h(s,xmh1s,xmh2s,…,xmhks)-∑t=s∞‍(f(t,xmf1t,xmf2t,…,xmfkt)-cth(s,xmh1s,xmh2s,…,xmhks))]xmT+τbT+τ},β≤n<T,m≥0,converges to a positive solutionz={zn}n∈ℕβ∈A(N,M)of (6) withlim⁡n→∞zn=+∞and has the error estimate (20), where{αm}m∈ℕ0is an arbitrary sequence in[0,1]satisfying (21).(b)
Equation (6) possesses uncountably many positive solutions inA(N,M).Proof.
PutL∈((1/b)M+N,M). It follows from (38), (39), and (57) that there existθ∈(0,1)andT≥n0+τ+βsatisfying
(59)θ=1b[(1+τT)2+1T2∑u=T∞‍∑s=u∞‍(RsHs+∑t=s∞‍PtFt)];(60)1bT2∑u=T∞‍∑s=u∞‍(Ws+∑t=s∞‍(Qt+|ct|))<min⁡{M-L,L-1bM-N};(61)bn≥b>1,∀n≥T.
Define a mappingSL:A(N,M)→lβ∞by
(62)SLxn={n2L-xn+τbn+τ-1bn+τ×∑u=n+τ∞‍∑s=u∞‍{h(s,xh1s,xh2s,…,xhks)-∑t=s∞‍[(f(t,xf1t,xf2t,…,xfkt)-cth(s,xmh1s,xmh2s,…,xmhks))]},n≥T,SLxT,β≤n<T,
for eachx={xn}n∈ℕβ∈A(N,M). In view of (14), (15), and (59)~(62), we obtain that for eachx={xn}n∈ℕβ, y={yn}n∈ℕβ∈A(N,M)andn≥T(63)|SLxnn2-SLynn2|≤1bn+τ|xn+τ-yn+τn2|+1bn+τn2∑u=n+τ∞‍∑s=u∞‍[|h(s,xh1s,xh2s,…,xhks)-h(s,yh1s,yh2s,…,yhks)|+∑t=s∞‍|f(t,xf1t,xf2t,…,xfkt)-f(t,yf1t,yf2t,…,yfkt)|]≤1bn+τ|xn+τ-yn+τ(n+τ)2|(n+τ)2n2+1bn+τn2∑u=n∞‍∑s=u∞‍[Rsmax⁡{|xhls-yhls|:1≤l≤k}+∑t=s∞‍Ptmax⁡{|xflt-yflt|:1≤l≤k}]≤1b[(1+τT)2+1T2∑u=T∞‍∑s=u∞‍(RsHs+∑t=s∞‍PtFt)]∥x-y∥=θ∥x-y∥,SLxnn2≤L+1bn2∑u=n+τ∞‍∑s=u∞‍{|h(s,xh1s,xh2s,…,xhks)|+∑t=s∞‍[|f(t,xf1t,xf2t,…,xfkt)|+|ct||f(t,xf1t,xf2t,…,xfkt)|]}≤L+1bT2∑u=T∞‍∑s=u∞‍(Ws+∑t=s∞‍(Qt+|ct|))<L+min⁡{M-L,L-1bM-N}≤M,SLxnn2≥L-1bM-1bn2∑u=n+τ∞‍∑s=u∞‍{∑t=s∞‍|h(s,xh1s,xh2s,…,xhks)|+∑t=s∞‍[|f(t,xf1t,xf2t,…,xfkt)|+|ct|]}≥L-1bM-1bT2∑u=T∞‍∑s=u∞‍(Ws+∑t=s∞‍(Qt+|ct|))>L-1bM-min⁡{M-L,L-1bM-N}≥N,
which imply (27). The rest of the proof is similar to that of Theorem 2 and is omitted. This completes the proof.Theorem 7.
Assume that there exist constantsb, M, andNwith(1+1/b)M>N>0and four nonnegative sequences{Pn}n∈ℕn0, {Qn}n∈ℕn0, {Rn}n∈ℕn0, and {Wn}n∈ℕn0 satisfying (14), (15), (38), (39) and
(64)bn≤b<-1eventually.
Then one has the following.(a)
For anyL∈(-(1+1/b)M,-N), there existθ∈(0,1)andT≥n0+τ+βsuch that, for anyx0={x0n}n∈ℕβ∈A(N,M), the Mann iterative sequence{xm}m∈ℕ0={{xmn}n∈ℕβ}m∈ℕ0generated by the scheme
(65)xm+1n={(1-αm)xmn+αm{h(s,xmh1s,xmh2s,…,xmhks)]-n2L-xmn+τbn+τ-1bn+τ×∑u=n+τ∞‍∑s=u∞‍[h(s,xmh1s,xmh2s,…,xmhks)-∑t=s∞‍(f(t,xmf1t,xmf2t,…,xmfkt)-cth(s,xmh1s,xmh2s,…,xmhks))]xmn+τbn+τ},n≥T,m≥0,(1-αm)xmT+αm{h(s,xmh1s,xmh2s,…,xmhks)-T2L-xmT+τbT+τ-∑u=T+τ∞‍∑s=u∞‍[h(s,xmh1s,xmh2s,…,xmhks)-∑t=s∞‍(f(t,xmf1t,xmf2t,…,xmfkt)-cth(s,xmh1s,xmh2s,…,xmhks))]xmn+τbn+τ},β≤n<T,m≥0,converges to a positive solutionz={zn}n∈ℕβ∈A(N,M) of (6) with lim⁡n→∞zn=+∞ and has the error estimate (20), where {αm}m∈ℕ0 is an arbitrary sequence in[0,1]satisfying (21).(b)
Equation (6) possesses uncountably many positive solutions inA(N,M).Proof.
PutL∈(-(1+1/b)M,-N). It follows from (38), (39), and (64) that there existθ∈(0,1)andT≥n0+τ+βsatisfying
(66)θ=-1b[(1+τT)2+1T2∑u=T∞‍∑s=u∞‍(RsHs+∑t=s∞‍PtFt)];(67)-1bT2∑u=T∞‍∑s=u∞‍(Ws+∑t=s∞‍(Qt+|ct|))<min⁡{(1+1b)M-L,L-1bM-N};(68)bn≥b>1,∀n≥T.
Define a mappingSL:A(N,M)→lβ∞by
(69)SLxn={-n2L-xn+τbn+τ-1bn+τ×∑u=n+τ∞‍∑s=u∞‍{h(s,xh1s,xh2s,…,xhks)-∑t=s∞‍[f(t,xf1t,xf2t,…,xfkt)-ct]},n≥T,SLxT,β≤n<T,
for eachx={xn}n∈ℕβ∈A(N,M). Making use of (15), (66), (68), and (69), we conclude that
(70)|SLxnn2-SLynn2|≤-1bn+τ|xn+τ-yn+τn2|-1bn+τn2∑u=n+τ∞‍∑s=u∞‍[|h(s,xh1s,xh2s,…,xhks)-h(s,yh1s,yh2s,…,yhks)|+∑t=s∞‍|f(t,xf1t,xf2t,…,xfkt)-f(t,yf1t,yf2t,…,yfkt)|]≤-1bn+τ|xn+τ-yn+τ(n+τ)2|(n+τ)2n2-1bn+τn2∑u=n∞‍∑s=u∞‍[Rsmax⁡{|xhls-yhls|:1≤l≤k}+∑t=s∞‍Ptmax⁡{|xflt-yflt|:1≤l≤k}]≤-1b[(1+τT)2+1T2∑u=T∞‍∑s=u∞‍(RsHs+∑t=s∞‍PtFt)]∥x-y∥=θ∥x-y∥,SLxnn2≤-L-Mb-1bn2×∑u=n+τ∞‍∑s=u∞‍{∑t=s∞‍|h(s,xh1s,xh2s,…,xhks)|+∑t=s∞‍[|f(t,xf1t,xf2t,…,xfkt)|+|ct|]}≤-L-Mb-1bT2∑u=T∞‍∑s=u∞‍(Ws+∑t=s∞‍(Qt+|ct|))<-L-Mb+min⁡{(1+1b)M+L,-L-N}≤M,SLxnn2≥-L+1bn2∑u=n+τ∞‍∑s=u∞‍{∑t=s∞‍|h(s,xh1s,xh2s,…,xhks)|+∑t=s∞‍[|f(t,xf1t,xf2t,…,xfkt)|+|ct|]}≥-L+1bT2∑u=T∞‍∑s=u∞‍[Ws+∑t=s∞‍(Qt+|ct|)]>-L-min⁡{(1+1b)M+L,-L-N}≥N,
which yield (27). The rest of the proof is similar to that of Theorem 2 and is omitted. This completes the proof.


## 3. Examples

In this section, we suggest six examples to explain the results presented in Section2.Example 1.
Consider the third order nonlinear neutral delay difference equation(71)Δ3(xn-xn-τ)+Δ(sin2xn-3n7)+1(n9+2n5+1)(1+xn22)=n2-2nn8+n3+1,∀n≥4,
whereτ∈ℕ is fixed. Letn0=4, k=1, and β=min⁡{4-τ,1}, and let MandNbe two positive constants withM>Nand
(72)bn=-1,cn=n2-2nn8+n3+1,f(n,u)=1(n9+2n5+1)(1+u2),h(n,u)=sin2un7,f1n=n2,Fn=n4,h1n=n-3,Hn=(n-3)2,Pn=2Mn9,Qn=1n9,Rn=2n7,Wn=1n7,∀(n,u)∈ℕn0×ℝ.
It is easy to see that (14), (15), and (18) are satisfied. Note that
(73)1n2∑t=n+τ∞‍t2max⁡{RtHt,Wt}=1n2∑t=n+τ∞‍t2max⁡{2(t-3)2t7,1t7}=∑t=n+τ∞‍2(t-3)2+1t5≤2n2∑t=n+τ∞‍1t3⟶0asn⟶∞,1n2∑t=n+τ∞‍t3max⁡{PtFt,Qt,|ct|}=1n2∑t=n+τ∞‍t3max⁡{2Mt5,1t9,|t2-2t|t8+t3+1}≤max⁡⁡{1,2M}n2∑t=n+τ∞‍1t2⟶0asn⟶∞,
which together with Lemma 1 yield that (16) and (17) hold. It follows from Theorem 2 that (71) possesses uncountably many positive solutions inA(N,M). On the other hand, for anyL∈(N,M), there existθ∈(0,1)andT≥n0+τ+βsuch that, for eachx0={x0n}n∈ℕβ∈A(N,M), the Mann iterative sequence{xm}m∈ℕ0={{xmn}n∈ℕβ}m∈ℕ0generated by (19) converges to a positive solutionz={zn}n∈ℕβ∈A(N,M)of (71) withlim⁡n→∞zn=+∞and has the error estimate (20), where{αm}m∈ℕ0is an arbitrary sequence in[0,1]satisfying (21).Example 2.
Consider the third order nonlinear neutral delay difference equation(74)Δ3(xn+xn-τ)+Δ(sin2x3n3+1n3(n6+2)(1+x2n2-34))+(-1)nn3(xn2-n-1+x(n+1)(n+2))(n13+n5+1)(1+xn2-n-12+x(n+1)(n+2)2)=n2-ln⁡nn6+n5+1,∀n≥5,
whereτ∈ℕ is fixed. Letn0=5, k=2, and β=5-τ, and let MandNbe two positive constants withM>Nand
(75)bn=1,cn=n2-ln⁡nn6+n5+1,f(n,u,v)=(-1)nn3(u+v)(n13+n5+1)(1+u2+v2),h(n,u,v)=sin2vn3(n6+2)(1+u4),f1n=n2-n-1,f2n=(n+1)(n+2),Fn=(n+1)2(n+2)2,h1n=2n2-3,h2n=3n3+1,Hn=(3n3+1)2,Pn=Qn=4n10,Rn=Wn=10n9,∀(n,u,v)∈ℕn0×ℝ2.
It is clear that (14), (15), and (40) are fulfilled. Note that
(76)1n2∑u=n∞‍∑s=u∞‍max⁡{RsHs,Ws}=1n2∑u=n∞‍∑s=u∞‍max⁡{10(3s3+1)2s9,10s9}≤160n2∑u=n∞‍∑s=u∞‍1s3≤160n2∑s=n∞‍1s2⟶0asn⟶∞,
which means that
(77)lim⁡n→∞1n2∑u=n∞‍∑s=u∞‍max⁡{RsHs,Ws}=0.
Observe that
(78)1n2∑u=n∞‍∑s=u∞‍∑t=s∞‍max⁡{PtFt,Qt,|ct|}=1n2∑u=n∞‍∑s=u∞‍∑t=s∞‍max⁡{4(t+1)2(t+2)2t10,4t10,t2-ln⁡tt6+t5+1}≤196n2∑u=n∞‍∑s=u∞‍∑t=s∞‍1t4=196n2∑u=n∞‍∑t=u∞‍t-u+1t4≤196n2∑u=n∞‍∑t=u∞‍1t3≤196n2∑t=n∞‍1t2⟶0asn⟶∞,
which yields that
(79)lim⁡n→∞1n2∑u=n∞‍∑s=u∞‍∑t=s∞‍max⁡{PtFt,Qt,|ct|}=0.
Thus Theorem 3 guarantees that (74) possesses uncountably positive solutions inA(N,M). On the other hand, for anyL∈(N,M), there existθ∈(0,1)andT≥τ+n0+βsuch that the Mann iterative sequence{xm}m∈ℕ0={{xmn}n∈ℕβ}m∈ℕ0generated by (41) converges to a positive solutionz={zn}n∈ℕβ∈A(N,M)of (74) withlim⁡n→∞zn=+∞and has the error estimate (20), where{αm}m∈ℕ0is an arbitrary sequence in[0,1]satisfying (21).Example 3.
Consider the third order nonlinear neutral delay difference equation(80)Δ3(xn+1+3ln⁡n2+4ln⁡nxn-τ)+Δ((-1)nsin⁡(e-n2|x5n2-3|)n15-n+3+n2+(-1)n(n+1)/2(n12+6n10+7)e|x2n3+1|)+(-1)nn6(1+xn-32)-1(n7+2n4-1)(1+xn+42)=3(-1)nn29n10ln⁡3n,∀n≥7,
whereτ∈ℕ is fixed. Letn0=7, k=2, b=3/4, and β=min⁡{7-τ,4}, and let M and N be two positive constants withM>4Nand
(81)bn=1+3ln⁡n2+4ln⁡n,cn=3(-1)nn29n10ln⁡3n,f(n,u,v)=(-1)nn6(1+u2)-1(n7+2n4-1)(1+v2),h(n,u,v)=(-1)nsin(e-n2|u|)n15-n+3+n2+(-1)n(n+1)/2(n12+6n10+7)e|v|,f1n=n-3,f2n=n+4,Fn=(n+4)2,h1n=5n2-3,h2n=2n3+1,Hn=(2n3+1)2,Pn=Qn=3n6,Rn=Wn=2n10,∀(n,u,v)∈ℕn0×ℝ2.
It is not difficult to verify that (14), (15), and (47) are fulfilled. Note that
(82)1n2∑u=n∞‍∑s=u∞‍max⁡{RsHs,Ws}=1n2∑u=n∞‍∑s=u∞‍max⁡{2(2s3+1)2s10,2s10}≤18n2∑u=n∞‍∑s=u∞‍1s4≤18n2∑s=n∞‍1s3⟶0asn⟶∞,
which implies that
(83)lim⁡n→∞1n2∑u=n∞‍∑s=u∞‍max⁡{RsHs,Ws}=0.
Observe that
(84)1n2∑u=n∞‍∑s=u∞‍∑t=s∞‍max⁡{PtFt,Qt,|ct|}=1n2∑u=n∞‍∑s=u∞‍∑t=s∞‍max⁡{3(t+4)2t6,3t6,|3(-1)tt29t10ln⁡3t|}≤12n2∑u=n∞‍∑s=u∞‍∑t=s∞‍1t4≤12n2∑u=n∞‍∑t=u∞‍1t3≤12n2∑t=n∞‍1t2⟶0asn⟶∞,
which means that
(85)lim⁡n→∞1n2∑u=n∞‍∑s=u∞‍∑t=s∞‍max⁡{PtFt,Qt,|ct|}=0.
That is, (38) and (39) hold. Consequently Theorem 4 implies that (80) possesses uncountably many positive solutions inA(N,M).On the other hand, for anyL∈((3/4)M+N,M), there existθ∈(0,1)andT≥n0+τ+βsuch that the Mann iterative sequence{xm}m∈ℕ0={{xmn}n∈ℕβ}m∈ℕ0generated by (41) converges to a positive solutionz={zn}n∈ℕβ∈A(N,M)of (80) withlim⁡n→∞zn=+∞and has the error estimate (20), where{αm}m∈ℕ0is an arbitrary sequence in[0,1]satisfying (21).Example 4.
Consider the third order nonlinear neutral delay difference equation(86)Δ3(xn+1-5n32+6n3xn-τ)+Δ(2n2+n-1(n8+3n6+2)(1+x3n-72))+sin(n2x3n2-2)(n+14)22=(-1)nn3+5n2+4n-2n9+n8+2n5+n3+7,∀n≥9,
where τ∈ℕ is fixed. Letn0=9, k=1, b=-5/6, and β=9-τ, and let M andNbe two positive constants with M>6N and
(87)bn=1-5n32+6n3,cn=(-1)nn3+5n2+4n-2n9+n8+2n5+n3+7,f(n,u)=sin(n2u)(n+14)22,h(n,u)=2n2+n-1(n8+3n6+2)(1+u2),f1n=3n2-2,Fn=(3n2-2)2,h1n=3n-7,Hn=(3n-7)2,Pn=Qn=3n9,Rn=Wn=1n5,∀(n,u)∈ℕn0×ℝ.
Obviously, (14), (15), and (52) are satisfied. Note that
(88)1n2∑u=n∞‍∑s=u∞‍max⁡{RsHs,Ws}=1n2∑u=n∞‍∑s=u∞‍max⁡{(3s-7)2s5,1s5}≤9n2∑u=n∞‍∑s=u∞‍1s3≤9n2∑s=n∞‍1s2⟶0asn⟶∞,
which implies that
(89)lim⁡n→∞1n2∑u=n∞‍∑s=u∞‍max⁡{RsHs,Ws}=0.
Notice that
(90)1n2∑u=n∞‍∑s=u∞‍∑t=s∞‍max⁡{PtFt,Qt,|ct|}=1n2∑u=n∞‍∑s=u∞‍∑t=s∞‍max⁡{|(-1)tt3+5t2+4t-2t9+t8+2t5+t3+7|3(3t2-2)2t9,3t9,|(-1)tt3+5t2+4t-2t9+t8+2t5+t3+7|3(3t2-2)2t9}≤27n2∑u=n∞‍∑s=u∞‍∑t=s∞‍1t5≤27n2∑u=n∞‍∑t=u∞‍1t4≤27n2∑t=n∞‍1t3⟶0asn⟶∞,
which gives that(91)lim⁡n→∞1n2∑u=n∞‍∑s=u∞‍∑t=s∞‍max⁡{PtFt,Qt,|ct|}=0.
That is, (38) and (39) hold. Thus Theorem 5 shows that (86) possesses uncountably many positive solutions inA(N,M).On the other hand, for anyL∈(N,(1/6)M), there existθ∈(0,1)andT≥n0+τ+βsuch that the Mann iterative sequence{xm}m∈ℕ0={{xmn}n∈ℕβ}m∈ℕ0generated by (48) converges to a positive solutionz={zn}n∈ℕβ∈A(N,M)of (86) withlim⁡n→∞zn=+∞and has the error estimate (20), where{αm}m∈ℕ0is an arbitrary sequence in[0,1]satisfying (21).Example 5.
Consider the third order nonlinear neutral delay difference equation(92)Δ3(xn+(π2+nsin1n)xn-τ)+Δ((-1)n(n+1)/2(n+4)8(n+5)3(1+cos⁡⁡(n2x2n+1)))+nsin(nxn-2)2+(n+5)16=(-1)n-1cos⁡3(n2+1)n16+ln⁡n,∀n≥3,
whereτ∈ℕ is fixed. Letn0=3, k=1, b=π/2, and β=min⁡⁡{3-τ,1}, and let M and N be two positive constants with(1-2/π)M>Nand
(93)bn=π2+nsin1n,cn=(-1)n-1cos⁡3(n2+1)n16+ln⁡n,f(n,u)=nsin(nu)2+(n+5)16,h(n,u)=(-1)n(n+1)/2(n+4)8(n+5)3(1+cos⁡⁡(n2u)),f1n=n-2,Fn=(n-2)2,h1n=2n+1,Hn=(2n+1)2,Pn=Qn=1n14,Rn=Wn=2n9,∀(n,u)∈ℕn0×ℝ.
Clearly, (14), (15), and (61) are satisfied. Note that
(94)1n2∑u=n∞‍∑s=u∞‍max⁡{RsHs,Ws}=1n2∑u=n∞‍∑s=u∞‍max⁡{2(2s+1)2s9,2s9}≤18n2∑u=n∞‍∑s=u∞‍1s7≤18n2∑s=n∞‍1s6⟶0asn⟶∞,
which means that
(95)lim⁡n→∞1n2∑u=n∞‍∑s=u∞‍max⁡{RsHs,Ws}=0,1n2∑u=n∞‍∑s=u∞‍∑t=s∞‍max⁡{PtFt,Qt,|ct|}=1n2∑u=n∞‍∑s=u∞‍∑t=s∞‍max⁡{|(-1)t-1cos⁡3(t2+1)t16+ln⁡t|(t-2)2t14,1t14,|(-1)t-1cos⁡3(t2+1)t16+ln⁡t|}≤1n2∑u=n∞‍∑s=u∞‍∑t=s∞‍1t14≤1n2∑u=n∞‍∑t=u∞‍1t13≤1n2∑t=n∞‍1t12⟶0asn⟶∞,
which implies that
(96)lim⁡n→∞1n2∑u=n∞‍∑s=u∞‍∑t=s∞‍max⁡{PtFt,Qt,|ct|}=0.
That is, (38) and (39) hold. Consequently Theorem 6 implies that (92) possesses uncountably many positive solutions inA(N,M).On the other hand, for anyL∈((2/π)M+N,M), there existθ∈(0,1)andT≥n0+τ+βsuch that the Mann iterative sequence{xm}m∈ℕ0={{xmn}n∈ℕβ}m∈ℕ0generated by (58) converges to a positive solutionz={zn}n∈ℕβ∈A(N,M)of (92) withlim⁡n→∞zn=+∞and has the error estimate (20), where{αm}m∈ℕ0is an arbitrary sequence in[0,1]satisfying (21).Example 6.
Consider the third order nonlinear neutral delay difference equation(97)Δ3(xn-2n5+9n2-1n5+3n2+2xn-τ)+Δ(cos⁡((-1)nen)(n+7)61+|xn-2|)+sin(n2xn-1)n9+3n5+2n2+1=(-1)n-1n4+4n2+n-1n11+6n3+7n+2,∀n≥6,
where τ∈ℕ is fixed. Letn0=6, k=1, b=-2, and β=min⁡{6-τ,3}, and let M and N be two positive constants with (1/2)M>Nand
(98)bn=-2n5+9n2-1n5+3n2+2,cn=(-1)n-1n4+4n2+n-1n11+6n3+7n+2,f(n,u)=sin(n2u)n9+3n5+2n2+1,h(n,u)=cos⁡((-1)nen)(n+7)61+|u|,f1n=n-1,Fn=(n-1)2,h1n=n-2,Hn=(n-2)2,Pn=Qn=1n7,Rn=Wn=1n6,∀(n,u)∈ℕn0×ℝ.
Obviously, (14), (15), and (64) are satisfied. Note that
(99)1n2∑u=n∞‍∑s=u∞‍max⁡{RsHs,Ws}=1n2∑u=n∞‍∑s=u∞‍max⁡{(s-2)2s6,1s6}≤1n2∑u=n∞‍∑s=u∞‍1s4≤1n2∑s=n∞‍1s3⟶0asn⟶∞,
which means that
(100)lim⁡n→∞1n2∑u=n∞‍∑s=u∞‍max⁡{RsHs,Ws}=0.
It is clear that
(101)1n2∑u=n∞‍∑s=u∞‍∑t=s∞‍max⁡{PtFt,Qt,|ct|}=1n2∑u=n∞‍∑s=u∞‍∑t=s∞‍max⁡{|(-1)t-1t4+4t2+t-1t11+6t3+7t+2|(t-1)2t7,1t7,|(-1)t-1t4+4t2+t-1t11+6t3+7t+2|}≤1n2∑u=n∞‍∑s=u∞‍∑t=s∞‍1t5≤1n2∑u=n∞‍∑t=u∞‍1t4≤1n2∑t=n∞‍1t3⟶0asn⟶∞,
which implies that
(102)lim⁡n→∞1n2∑u=n∞‍∑s=u∞‍∑t=s∞‍max⁡{PtFt,Qt,|ct|}=0.
That is, (38) and (39) hold. Consequently Theorem 7 implies that (97) possesses uncountably many positive solutions inA(N,M).On the other hand, for anyL∈(-M/2,-N), there existθ∈(0,1)andT≥n0+τ+βsuch that the Mann iterative sequence{xm}m∈ℕ0={{xmn}n∈ℕβ}m∈ℕ0 generated by (65) converges to a positive solutionz={zn}n∈ℕβ∈A(N,M)of (97) withlim⁡n→∞zn=+∞and has the error estimate (20), where{αm}m∈ℕ0is an arbitrary sequence in[0,1]satisfying (21).



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*Source: 470181-2014-01-28.xml*