# Positive Solutions and Mann Iterations of a Fourth Order Nonlinear Neutral Delay Differential Equation

Authors: Zeqing Liu; Jingjing Zhu; Jeong Sheok Ume; Shin Min Kang
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

This paper deals with a fourth order nonlinear neutral delay differential equation. By using the Banach fixed point theorem, we establish the existence of uncountably many bounded positive solutions for the equation, construct several Mann iterative sequences with mixed errors for approximating these positive solutions, and discuss some error estimates between the approximate solutions and these positive solutions. Seven nontrivial examples are given.


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## Body

## 1. Introduction and Preliminaries

The oscillation, nonoscillation, and existence of solutions for various kinds of second order and third order neutral delay differential equations have been extensively studied over the last decades; for example, see [1–12]. Elbert [2] and Huang [3] established a few oscillation and nonoscillation criteria for the second order linear differential equation
(1)x′′(t)+q(t)x(t)=0,t≥0,
where q∈C([0,+∞),R+). Tang and Liu [9] studied the existence of bounded oscillation for the second order linear delay differential equation of unstable type
(2)x′′(t)=p(t)x(t-τ),t≥t0,
where τ>0, p∈C([t0,+∞),R+), and p(t)≢0 on any interval of length τ. Using the Banach fixed point theorem, Kulenović and Hadžiomerspahić [4] deduced the existence of a nonoscillatory solution for the second order linear neutral delay differential equation with positive and negative coefficients
(3)(x(t)+cx(t-τ))′′+Q1(t)x(t-σ1)-Q2(t)x(t-σ2)=0,t≥t0,
where c∈R∖{-1,1}, τ>0, σ1,  σ2∈[0,+∞), Q1,Q2∈C([t0,+∞),R+). Lin [5] suggested a few sufficient conditions for oscillation and nonoscillation of the second order nonlinear neutral differential equation
(4)(x(t)+p(t)x(t-τ))′′+q(t)f(x(t-σ))=0,t≥t0,
where τ>0, σ≥0, p∈C([t0,+∞),R+), q∈C(R,R+), f∈C(R,R), f is nondecreasing, and xf(x)>0, x≠0. Qin et al. [8] and Yang et al. [11] developed several oscillation criteria for the second order differential equation
(5)(r(t)(x(t)+p(t)x(t-τ))′)′+q(t)f(x(t-δ))=0,t≥t0,
where τ and δ are nonnegative constants, r,p,q∈C([t0,+∞),R), and f∈C(R,R). By utilizing Krasnoselskii’s fixed point theorem, Zhou [12] discussed the existence of nonoscillatory solutions of the second order nonlinear neutral differential equation
(6)(r(t)(x(t)+p(t)x(t-τ))′)′+∑i=1m‍Qi(t)fi(x(t-σi))=0,t≥t0,
where m≥1 is an integer, τ>0, σi≥0, r,p,Qi∈C([t0,+∞),R), and fi∈C(R,R) for i∈{1,2,…,m}. Yu and Wang [10] studied the existence of a nonoscillatory solution for the second order nonlinear neutral delay differential equations with positive and negative coefficients
(7)(r(t)(x(t)+P(t)x(t-τ))′)′+Q1(t)f(x(t-σ1))-Q2(t)g(x(t-σ2))=0,t≥t0,
where τ>0, σ1,σ2∈[0,+∞), P,Q1,Q2,r∈C([t0,+∞),R), f,g∈C(R,R). Liu and Kang [7] investigated the existence of nonoscillatory solutions of the second order nonlinear neutral delay differential equation
(8)(a(t)(x(t)+b(t)x(t-τ))′)′+(h(t,x(h1(t)),x(h2(t)),…,x(hk(t))))′+f(t,x(f1(t)),x(f2(t)),…,x(fk(t)))=g(t),hhhhhhhhhhhhhhhhhhhhhhhhhhhxxhhhhhhht≥t0,
where τ>0, a,b,g∈C([t0,+∞),R) with a(t)>0 for t≥t0, h∈C1([t0,+∞)×Rk,R), f∈C([t0,+∞)×Rk,R), hl∈C1([t0,+∞),R), and fl∈C([t0,+∞),R) with
(9)lim⁡t→+∞hl(t)=lim⁡t→+∞fl(t)=+∞,l=1,…,k.
Kang et al. [13] discussed the existence of nonoscillatory solutions of the third order nonlinear neutral delay differential equation
(10)(α(t)(β(t)(x(t)+p(t)x(t-τ))′)′)′+f(t,x(σ1(t)),x(σ2(t)),…,x(σn(t)))=0,hhhhhhhhhhhhhhhhxihhhhhhhhhhhhhhit≥t0,
where n≥1 is an integer, τ>0, α,β∈C([t0,+∞),R+∖{0}), p∈C([t0,+∞),R), and f∈C([t0,+∞)×Rn,R).Motivated by the papers mentioned above, in this paper, we investigate the following fourth order nonlinear neutral delay differential equation:(11)(α(t)(β(t)(γ(t)(x(t)+p(t)x(t-τ))′)′)′)′+(h(t,x(h1(t)),…,x(hk(t))))′+f(t,x(f1(t)),x(f2(t)),…,x(fk(t)))=g(t),hhhhhhhhhhhhhhhhhhhxihhhhhhhhhhhhhhht≥t0,
where k∈N, τ>0, α,β,γ∈C([t0,+∞),R∖{0}), p,g∈C([t0,+∞),R), h∈C1([t0,+∞)×Rk,R), f∈C([t0,+∞)×Rk,R), hl∈C1([t0,+∞),R), and fl∈C([t0,+∞),R) with
(12)lim⁡t→+∞hl(t)=lim⁡t→+∞fl(t)=+∞,l=1,2,…,k.
Utilizing the contraction mapping principle, we show the existence of uncountably many bounded positive solutions for (11), construct a few Mann type iterative schemes with mixed errors for these positive solutions, and discuss error estimates between the approximate solutions and the bounded positive solutions. Seven nontrivial examples are considered to illustrate our results.Throughout this paper, we assume thatR=(-∞,+∞), N denotes the set of positive integers, N0={0}∪N, and
(13)a=min⁡{t0-τ,inf⁡{hl(t),fl(t):t∈[t0,+∞),1≤l≤k}}.
By a solution of (11), we mean a function x∈C([a,+∞),R) for some T≥t0+|a|+τ, such that α(t)(β(t)(γ(t)(x(t)+p(t)x(t-τ))′)′)′ is continuously differentiable in [T,+∞) and (11) is satisfied for t≥T.LetCB([a,+∞),R) denote the Banach space of all continuous and bounded functions on [a,+∞) with the norm ∥x∥=sup⁡t≥a|x(t)| for each x∈CB([a,+∞),R) and
(14)A(N,M)={x∈CB([a,+∞),R):N≤x(t)≤M,t≥a}hhhhhhhhhhhhhhhhxihhhhhhhhhhhhhhforM>N>0.
It is easy to see that A(N,M) is a bounded closed and convex subset of CB([a,+∞),R).The following lemma plays an important role in this paper.Lemma 1 (see [6]).
Let{αn}n∈N0,  {βn}n∈N0,  {γn}n∈N0, and {tn}n∈N0 be four nonnegative real sequences satisfying the inequality
(15)αn+1≤(1-tn)αn+tnβn+γn,n∈N0,
where {tn}n∈N0⊂[0,1], ∑n=0∞‍tn=+∞, lim⁡n→∞βn=0, and ∑n=0∞‍γn<+∞. Then lim⁡n→∞αn=0.


## 2. Uncountably Many Bounded Positive Solutions and Iterative Approximations

Now we study the solvability of (11).Theorem 2.
Assume that there exist constantsM,N,c1,c2, and T0≥t0 and functions Q1,Q2,R1, and R2∈C([t0,+∞),R+) satisfying
(16)|f(t,u1,u2,…,uk)-f(t,u¯1,u¯2,…,u¯k)|≤Q1(t)max⁡{|ul-u¯l|:1≤l≤k},|h(t,u1,u2,…,uk)-h(t,u¯1,u¯2,…,u¯k)|≤Q2(t)max⁡{|ul-u¯l|:1≤l≤k}fort∈[t0,+∞),ul,u¯l∈[N,M],1≤l≤k;(17)|f(t,u1,u2,…,uk)|≤R1(t),|h(t,u1,u2,…,uk)|≤R2(t),fort∈[t0,+∞),ul∈[N,M],hhhhhhhhhhhhhhhhiii1≤l≤k;(18)∫t0+∞‍∫w+∞‍∫v+∞‍1|α(u)β(v)γ(w)|×max⁡⁡{Q2(u),R2(u)}dudvdw<+∞,∫t0+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1|α(u)β(v)γ(w)|×max⁡{Q1(s),R1(s),|g(s)|}dsdudvdw<+∞;(19)min⁡⁡{c1,c2}≥0,c1+c2<1,0<N<(1-c1-c2)M;(20)-c2≤p(t)≤c1,t≥T0.
Then,(a)
for anyL∈(N+c1M,(1-c2)M), there existθ∈(0,1)andT>t0+|a|+τ+|T0|such that for eachx0∈A(N,M), the Mann iterative sequence{xm}m∈N0with mixed errors generated by the scheme(21)xm+1(t)={(1-αm-βm)xm(t)+αm×{(s,xm(f1(s)),…,xm(fk(s)))L-p(t)xm(t-τ)1α(u)β(v)γ(w)+∫t+∞‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×h(u,xm(h1(u)),…,xm(hk(u)))dudvdw-∫t+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,xm(f1(s)),…,xm(fk(s)))f(s,xm(f1(s)),…,xm(fk(s)))-g(s)]1α(u)β(v)γ(w)dsdudvdw}+βmγm(t),t≥T,m∈N0,(1-αm-βm)xm(T)+αm×{[f(s,xm(f1(s)),…,xm(fk(s)))-g(s)]L-p(T)xm(T-τ)1α(u)β(v)γ(w)+∫T+∞‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×h((hk(u))u,xm(h1(u)),…,xm(hk(u)))dudvdw-∫T+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,xm(f1(s)),…,xm(fk(s)))(s,xm(f1(s)),…,xm(fk(s)))-g(s)]1α(u)β(v)γ(w)dsdudvdw}+βmγm(T),a≤t<T,m∈N0converges to a bounded positive solutionx∈A(N,M)of (11) and has the following error estimate:(22)∥xm+1-x∥≤(1-(1-θ)αm)∥xm-x∥+2Mβm,m∈N0,where{γm}m∈N0is an arbitrary sequence inA(N,M)and{αm}m∈N0and{βm}m∈N0are any sequences in[0,1]such that(23)∑m=0∞‍αm=+∞,(24)∑m=0∞βm<+∞orthereexistsasequence{ξm}m∈N0⊆[0,+∞)satisfyingβm=ξmαm,m∈N0,lim⁡m→∞ξm=0;(b)
equation (11) possesses uncountably many bounded positive solutions inA(N,M).Proof.
Firstly, we show that (a) holds. LetL∈(N+c1M,(1-c2)M). It follows from (18) and (19) that there exist θ∈(0,1) and T>t0+|a|+τ+|T0| satisfying
(25)θ=c1+c2+∫T+∞‍∫w+∞‍∫v+∞‍Q2(u)|α(u)β(v)γ(w)|dudvdw+∫T+∞‍∫w+∞‍∫v+∞‍∫u+∞‍Q1(s)|α(u)β(v)γ(w)|dsdudvdw,(26)∫T+∞‍∫w+∞‍∫v+∞‍R2(u)|α(u)β(v)γ(w)|dudvdw+∫T+∞‍∫w+∞‍∫v+∞‍∫u+∞‍R1(s)+|g(s)||α(u)β(v)γ(w)|dsdudvdw<min⁡{(1-c2)M-L,L-N-c1M}.
Define a mapping SL:A(N,M)→CB([a,+∞),R) by
(27)SLx(t)={L-p(t)x(t-τ)+∫t+∞‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×h(u,x(h1(u)),…,x(hk(u)))dudvdw-∫t+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,x(f1(s)),…,x(fk(s)))(s,x(f1(s)),…,x(fk(s)))-g(s)]dsdudvdw,t≥T,x∈A(N,M),SLx(T),a≤t<T,x∈A(N,M).
Obviously, SLx is continuous for each x∈A(N,M). Combining (16), (17), (19), (20), and (25)–(27), we derive that for x,y∈A(N,M) and t≥T(28)|SLx(t)-SLy(t)|≤|p(t)||x(t-τ)-y(t-τ)|+∫t+∞‍∫w+∞‍∫v+∞‍1|α(u)β(v)γ(w)|×|h(u,x(h1(u)),…,x(hk(u)))-h(u,y(h1(u)),…,y(hk(u)))|dudvdw+∫t+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1|α(u)β(v)γ(w)|×|f(s,x(f1(s)),…,x(fk(s)))-f(s,y(f1(s)),…,y(fk(s)))|dsdudvdw≤(c1+c2)∥x-y∥+∥x-y∥×∫t+∞‍∫w+∞‍∫v+∞‍Q2(u)|α(u)β(v)γ(w)|dudvdw+∥x-y∥×∫t+∞‍∫w+∞‍∫v+∞‍∫u+∞‍Q1(s)|α(u)β(v)γ(w)|dsdudvdw≤[c1+c2+∫T+∞‍∫w+∞‍∫v+∞‍Q2(u)|α(u)β(v)γ(w)|dudvdw+∫T+∞‍∫w+∞‍∫v+∞‍∫u+∞‍Q1(s)|α(u)β(v)γ(w)|dsdudvdw]×∥x-y∥=θ∥x-y∥,SLx(t)=L-p(t)x(t-τ)+∫t+∞‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×h(u,x(h1(u)),…,x(hk(u)))dudvdw-∫t+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,x(f1(s)),…,x(fk(s)))-g(s)]dsdudvdw≤L+c2M+∫t+∞‍∫w+∞‍∫v+∞‍1|α(u)β(v)γ(w)|×|h(u,x(h1(u)),…,x(hk(u)))|dudvdw+∫t+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1|α(u)β(v)γ(w)|×[|f(s,x(f1(s)),…,x(fk(s)))|+|g(s)|]dsdudvdw≤L+c2M+∫T+∞‍∫w+∞‍∫v+∞‍R2(u)|α(u)β(v)γ(w)|dudvdw+∫T+∞‍∫w+∞‍∫v+∞‍∫u+∞‍R1(s)+|g(s)||α(u)β(v)γ(w)|dsdudvdw<L+c2M+min⁡⁡{(1-c2)M-L,L-N-c1M}≤M,SLx(t)≥L-c1M-∫t+∞‍∫w+∞‍∫v+∞‍1|α(u)β(v)γ(w)|×|h(u,x(h1(u)),…,x(hk(u)))|dudvdw-∫t+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1|α(u)β(v)γ(w)|×|[f(s,x(f1(s)),…,x(fk(s)))-g(s)]|dsdudvdw≥L-c1M-∫T+∞‍∫w+∞‍∫v+∞‍R2(u)|α(u)β(v)γ(w)|dudvdw-∫T+∞‍∫w+∞‍∫v+∞‍∫u+∞‍R1(s)+|g(s)||α(u)β(v)γ(w)|dsdudvdw>L-c1M-min⁡⁡{(1-c2)M-L,L-N-c1M}≥N,
which mean that
(29)∥SLx-SLy∥≤θ∥x-y∥,x,y∈A(N,M),SL(A(N,M))⊆A(N,M).
That is, SL is a contraction mapping in A(N,M) and has a unique fixed point x∈A(N,M), which is a bounded positive solution of (11). By virtue of (21), (27), and (29), we get that for any m∈N0 and t≥T(30)|xm+1(t)-x(t)|=|(s,xm(f1(s)),…,xm(fk(s)))(1-αm-βm)xm(t)+αm(s,xm(f1(s)),…,xm(fk(s)))×{(s,xm(f1(s)),…,xm(fk(s)))L-p(t)xm(t-τ)+∫t+∞‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×h(u,xm(h1(u)),…,xm(hk(u)))dudvdw-∫t+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,xm(f1(s)),…,xm(fk(s)))(s,xm(f1(s)),…,xm(fk(s)))-g(s)]dsdudvdw}+βmγm(t)-x(t)|≤(1-αm-βm)|xm(t)-x(t)|+αm|SLxm(t)-SLx(t)|+βm|γm(t)-x(t)|≤(1-αm-βm)|xm(t)-x(t)|+αmθ|xm(t)-x(t)|+2Mβm≤(1-(1-θ)αm)|xm(t)-x(t)|+2Mβm≤(1-(1-θ)αm)∥xm-x∥+2Mβm,
which yielded that
(31)∥xm+1-x∥≤(1-(1-θ)αm)∥xm-x∥+2Mβm,hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhm∈N0.
That is, (22) holds. Thus Lemma 1, (23), and (24) ensure that lim⁡m→∞xm=x.
Next we show that (b) holds. LetL1,L2∈(N+c1M,(1-c2)M) with L1≠L2. As in the proof of (a), we conclude that for each j∈{1,2} there exist θj∈(0,1),  Tj>t0+|a|+τ+|T0|, and SLj:A(N,M)→A(N,M) satisfying (25)–(27), where θ,T,SL are replaced by θj,Tj, and SLj, respectively, and the contraction mapping SLj has a unique fixed point zj∈A(N,M), which is also a bounded positive solution of (11). In order to prove (b), we need only to show that z1≠z2. Put T*=max⁡{T1,T2}. Note that for t≥T* and j∈{1,2}(32)zj(t)=SLzj(t)=Lj-p(t)zj(t-τ)+∫t+∞‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×h(u,zj(h1(u)),…,zj(hk(u)))dudvdw-∫t+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,zj(f1(s)),…,zj(fk(s)))-g(s)]dsdudvdw,
which together with (16) and (25) implies that for t≥T*(33)|z1(t)-z2(t)|=|-f(s,z2(f1(s)),…,z2(fk(s)))]dsdudvdwL1-L2-p(t)(z1(t-τ)-z2(t-τ))+∫t+∞‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×[h(u,z1(h1(u)),…,z1(hk(u)))-h(u,z2(h1(u)),…,z2(hk(u)))]dudvdw-∫t+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,z1(f1(s)),…,z1(fk(s)))-f(s,z2(f1(s)),…,z2(fk(s)))]dsdudvdw|≥|L1-L2|-(c1+c2)∥z1-z2∥-∥z1-z2∥×∫T*+∞‍∫w+∞‍∫v+∞‍Q2(u)|α(u)β(v)γ(w)|dudvdw-∥z1-z2∥×∫T*+∞‍∫w+∞‍∫v+∞‍∫u+∞‍Q1(s)|α(u)β(v)γ(w)|dsdudvdw≥|L1-L2|-∥z1-z2∥max⁡{θ1,θ2},
which yields that
(34)∥z1-z2∥≥|L1-L2|1+max⁡{θ1,θ2}>0.
That is, z1≠z2. This completes the proof.Theorem 3.
Assume that there exist constantsM,N,c1,c2, and T0≥t0 and functions Q1,Q2,R1,  and R2∈C([t0,+∞),R+) satisfying (16)–(18) and
(35)0<(1-c2)N<(1-c1)M,0≤c2≤p(t)≤c1<1,t≥T0.
Then(a)
for anyL∈(N+c1M,M+c2N), there existθ∈(0,1)andT>t0+|a|+τ+|T0|such that, for eachx0∈A(N,M), the Mann iterative sequence{xm}m∈N0generated by the scheme (21) with (23) and (24) converges to a bounded positive solutionx∈A(N,M)of (11) and has the error estimate (22);(b)
equation (11) has uncountably many bounded positive solutions inA(N,M).Proof.
LetL∈(N+c1M,M+c2N). It follows from (18) and (35) that there exist θ∈(0,1) and T>t0+|a|+τ+|T0| satisfying
(36)θ=c1+∫T+∞‍∫w+∞‍∫v+∞‍Q2(u)|α(u)β(v)γ(w)|dudvdw+∫T+∞‍∫w+∞‍∫v+∞‍∫u+∞‍Q1(s)|α(u)β(v)γ(w)|dsdudvdw,(37)∫T+∞‍∫w+∞‍∫v+∞‍R2(u)|α(u)β(v)γ(w)|dudvdw+∫T+∞‍∫w+∞‍∫v+∞‍∫u+∞‍R1(s)+|g(s)||α(u)β(v)γ(w)|dsdudvdw<min⁡{M+c2N-L,L-N-c1M}.
Define a mapping SL:A(N,M)→CB([a,+∞),R) by (27). Clearly SLx is continuous for each x∈A(N,M). On account of (16), (17), (27), (35), (36), and (37), we infer that for x,y∈A(N,M) and t≥T(38)|SLx(t)-SLy(t)|≤|p(t)||x(t-τ)-y(t-τ)|+∫t+∞‍∫w+∞‍∫v+∞‍1|α(u)β(v)γ(w)|×|h(u,x(h1(u)),…,x(hk(u)))-h(u,y(h1(u)),…,y(hk(u)))|dudvdw+∫t+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1|α(u)β(v)γ(w)|×|f(s,x(f1(s)),…,x(fk(s)))-f(s,y(f1(s)),…,y(fk(s)))|dsdudvdw≤c1∥x-y∥+∥x-y∥×∫t+∞‍∫w+∞‍∫v+∞‍Q2(u)|α(u)β(v)γ(w)|dudvdw+∥x-y∥×∫t+∞‍∫w+∞‍∫v+∞‍∫u+∞‍Q1(s)|α(u)β(v)γ(w)|dsdudvdw≤(c1+∫T+∞‍∫w+∞‍∫v+∞‍Q2(u)|α(u)β(v)γ(w)|dudvdw+∫T+∞‍∫w+∞‍∫v+∞‍∫u+∞‍Q1(s)|α(u)β(v)γ(w)|dsdudvdw)×∥x-y∥=θ∥x-y∥,SLx(t)=L-p(t)x(t-τ)+∫t+∞‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×h(u,x(h1(u)),…,x(hk(u)))dudvdw-∫t+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,x(f1(s)),…,x(fk(s)))-g(s)]dsdudvdw≤L-c2N+∫t+∞‍∫w+∞‍∫v+∞‍1|α(u)β(v)γ(w)|×|h(u,x(h1(u)),…,x(hk(u)))|dudvdw+∫t+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1|α(u)β(v)γ(w)|×[|f(s,x(f1(s)),…,x(fk(s)))|+|g(s)|]dsdudvdw≤L-c2N+∫T+∞‍∫w+∞‍∫v+∞‍R2(u)|α(u)β(v)γ(w)|dudvdw+∫T+∞‍∫w+∞‍∫v+∞‍∫u+∞‍R1(s)+|g(s)||α(u)β(v)γ(w)|dsdudvdw<L-c2N+min⁡{M+c2N-L,L-N-c1M}≤M,SLx(t)≥L-c1M-∫t+∞‍∫w+∞‍∫v+∞‍1|α(u)β(v)γ(w)|×|h(u,x(h1(u)),…,x(hk(u)))|dudvdw-∫t+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1|α(u)β(v)γ(w)|×|[f(s,x(f1(s)),…,x(fk(s)))-g(s)]|dsdudvdw≥L-c1M-∫T+∞‍∫w+∞‍∫v+∞‍R2(u)|α(u)β(v)γ(w)|dudvdw-∫T+∞‍∫w+∞‍∫v+∞‍∫u+∞‍R1(s)+|g(s)||α(u)β(v)γ(w)|dsdudvdw>L-c1M-min⁡⁡{M+c2N-L,L-N-c1M}≥N,
which imply that (29) holds. The rest of the proof is similar to that of Theorem 2 and is omitted. This completes the proof.Theorem 4.
Assume that there exist constantsM,N,c1,c2 and T0≥t0 and functions Q1,Q2,R1 and R2∈C([t0,+∞),R+) satisfying (17), (18), and
(39)0<(1-c2)N<(1-c1)M;-1<-c1≤p(t)≤-c2≤0,t≥T0.
Then (a)
for anyL∈((1-c2)N,(1-c1)M),
there existθ∈(0,1)andT>t0+|a|+τ+|T0|such that, for eachx0∈A(N,M), the Mann iterative sequence{xm}m∈N0generated by the scheme (21) with (23) and (24) converges to a bounded positive solutionx∈A(N,M)of (11) and has the error estimate (22);(b)
equation (11) has uncountably many bounded positive solutions inA(N,M).Proof.
LetL∈((1-c2)N,(1-c1)M). Equations (18) and (39) guarantee that there exist θ∈(0,1) and T>t0+|a|+τ+|T0| satisfying (36) and
(40)∫T+∞‍∫w+∞‍∫v+∞‍R2(u)|α(u)β(v)γ(w)|dudvdw+∫T+∞‍∫w+∞‍∫v+∞‍∫u+∞‍R1(s)+|g(s)||α(u)β(v)γ(w)|dsdudvdw<min⁡{(1-c1)M-L,(c2-1)N+L}.
Let the mapping SL:A(N,M)→CB([a,+∞),R) be defined by (27). Using (17), (27), (39), and (40), we deduce that for any x∈A(N,M) and t≥T(41)SLx(t)=L-p(t)x(t-τ)+∫t+∞‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×h(u,x(h1(u)),…,x(hk(u)))dudvdw-∫t+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,x(f1(s)),…,x(fk(s)))-g(s)]dsdudvdw≤L+c1M+∫t+∞‍∫w+∞‍∫v+∞‍1|α(u)β(v)γ(w)|×|h(u,x(h1(u)),…,x(hk(u)))|dudvdw+∫t+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1|α(u)β(v)γ(w)|×[|f(s,x(f1(s)),…,x(fk(s)))||f(s,x(f1(s)),…,x(fk(s)))|+|g(s)|]dsdudvdw≤L+c1M+∫T+∞‍∫w+∞‍∫v+∞‍R2(u)|α(u)β(v)γ(w)|dudvdw+∫T+∞‍∫w+∞‍∫v+∞‍∫u+∞‍R1(s)+|g(s)||α(u)β(v)γ(w)|dsdudvdw<L+c1M+min⁡{(1-c1)M-L,(c2-1)N+L}≤M,SLx(t)≥L+c2N-∫T+∞‍∫w+∞‍∫v+∞‍R2(u)|α(u)β(v)γ(w)|dudvdw-∫T+∞‍∫w+∞‍∫v+∞‍∫u+∞‍R1(s)+|g(s)||α(u)β(v)γ(w)|dsdudvdw>L+c2N-min⁡{(1-c1)M-L,(c2-1)N+L}≥N,
which mean that (29) holds. That is, SL is a contraction mapping and possesses a unique fixed point x∈A(N,M), which is a bounded positive solution of (11). 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 constantsM,N, and T0≥t0 and functions Q1,Q2,R1,  and R2∈C([t0,+∞),R+) satisfying (16)–(18) and
(42)0<N<M,p(t)=1,t≥T0.
Then(a)
for anyL∈(N,M), there existθ∈(0,1)andT>t0+|a|+τ+|T0|such that, for eachx0∈A(N,M), the Mann iterative sequence{xm}m∈N0generated by the scheme(43)xm+1(t)={(1-αm-βm)xm(t)+αm×{L+∑i=1∞‍∫t+(2i-1)τt+2iτ‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×h(u,xm(h1(u)),…,xm(hk(u)))dudvdw-∑i=1∞‍∫t+(2i-1)τt+2iτ‍∫w+∞‍∫v+∞‍∫u+∞1α(u)β(v)γ(w)‍×[f(s,xm(f1(s)),…,xm(fk(s)))f(s,xm(f1(s)),…,xm(fk(s)))-g(s)]dsdudvdw∑i=1∞‍∫t+(2i-1)τt+2iτ‍∫t+(2i-1)τt+2iτ‍∫w+∞‍}+βmγm(t),t≥T,m∈N0,(1-αm-βm)xm(T)+αm×{L+∑i=1∞‍∫T+(2i-1)τT+2iτ‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×h(u,xm(h1(u)),…,xm(hk(u)))dudvdw-∑i=1∞‍∫T+(2i-1)τT+2iτ‍∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,xm(f1(s)),…,xm(fk(s)))(s,xm(f1(s)),…,xm(fk(s)))-g(s)]dsdudvdw∑i=1∞‍∫T+(2i-1)τT+2iτ‍}+βmγm(T),a≤t<T,m∈N0converges to a bounded positive solutionx∈A(N,M)of (11) and has the error estimate (22), where{γm}m∈N0is an arbitrary sequence inA(N,M)and{αm}m∈N0and{βm}m∈N0are any sequences in[0,1]satisfying (23) and (24);(b)
equation (11) possesses uncountably many bounded positive solutions inA(N,M).Proof.
First of all, we show that (a) holds. LetL∈(N,M). It follows from (18) that there exist θ∈(0,1) and T>t0+|a|+τ+|T0| satisfying
(44)θ=∫T+∞‍∫w+∞‍∫v+∞‍Q2(u)|α(u)β(v)γ(w)|dudvdw+∫T+∞‍∫w+∞‍∫v+∞‍∫u+∞‍Q1(s)|α(u)β(v)γ(w)|dsdudvdw,(45)∫T+∞‍∫w+∞‍∫v+∞‍R2(u)|α(u)β(v)γ(w)|dudvdw+∫T+∞‍∫w+∞‍∫v+∞‍∫u+∞‍R1(s)+|g(s)||α(u)β(v)γ(w)|dsdudvdw<min⁡{M-L,L-N}.
Define a mapping SL:A(N,M)→CB([a,+∞),R) by
(46)SLx(t)={L+∑i=1∞‍∫t+(2i-1)τt+2iτ‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×h(u,x(h1(u)),…,x(hk(u)))dudvdw-∑i=1∞‍∫t+(2i-1)τt+2iτ‍∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,x(f1(s)),…,x(fk(s)))(s,x(f1(s)),…,x(fk(s)))-g(s)]dsdudvdw,t≥T,x∈A(N,M),SLx(T),a≤t<T,x∈A(N,M).
Clearly, SLx is continuous for each x∈A(N,M). Notice that (16), (17), (42), and (44)–(46) ensure that for x,y∈A(N,M) and t≥T(47)|SLx(t)-SLy(t)|≤∑i=1∞‍∫t+(2i-1)τt+2iτ‍∫w+∞‍∫v+∞‍1|α(u)β(v)γ(w)|×|h(u,x(h1(u)),…,x(hk(u)))-h(u,y(h1(u)),…,y(hk(u)))|dudvdw+∑i=1∞‍∫t+(2i-1)τt+2iτ‍∫w+∞‍∫v+∞‍∫u+∞‍1|α(u)β(v)γ(w)|×|f(s,x(f1(s)),…,x(fk(s)))-f(s,y(f1(s)),…,y(fk(s)))|dsdudvdw≤∥x-y∥×∑i=1∞‍∫t+(2i-1)τt+2iτ‍∫w+∞‍∫v+∞‍Q2(u)|α(u)β(v)γ(w)|dudvdw+∥x-y∥×∑i=1∞‍∫t+(2i-1)τt+2iτ‍∫w+∞‍∫v+∞‍∫u+∞‍Q1(s)|α(u)β(v)γ(w)|dsdudvdw≤(∫T+∞‍∫w+∞‍∫v+∞‍Q2(u)|α(u)β(v)γ(w)|dudvdw+∫T+∞‍∫w+∞‍∫v+∞‍∫u+∞‍Q1(s)|α(u)β(v)γ(w)|dsdudvdw)×∥x-y∥=θ∥x-y∥,SLx(t)=L+∑i=1∞‍∫t+(2i-1)τt+2iτ‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×h(u,x(h1(u)),…,x(hk(u)))dudvdw-∑i=1∞‍∫t+(2i-1)τt+2iτ‍∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,x(f1(s)),…,x(fk(s)))-g(s)]dsdudvdw≤L+∫t+∞‍∫w+∞‍∫v+∞‍R2(u)|α(u)β(v)γ(w)|dudvdw+∫t+∞‍∫w+∞‍∫v+∞‍∫u+∞‍R1(s)+|g(s)||α(u)β(v)γ(w)|dsdudvdw<L+min⁡⁡{M-L,L-N}≤M,SLx(t)≥L-∫t+∞‍∫w+∞‍∫v+∞‍R2(u)|α(u)β(v)γ(w)|dudvdw-∫t+∞‍∫w+∞‍∫v+∞‍∫u+∞‍R1(s)+|g(s)||α(u)β(v)γ(w)|dsdudvdw>L-min⁡{M-L,L-N}≥N,
which imply that (29) holds. That is, SL is a contraction mapping and has a unique fixed point x∈A(N,M). It follows that (48)x(t)=L+∑i=1∞‍∫t+(2i-1)τt+2iτ‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×h(u,x(h1(u)),…,x(hk(u)))dudvdw-∑i=1∞‍∫t+(2i-1)τt+2iτ‍∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,x(f1(s)),…,x(fk(s)))(s,x(f1(s)),…,x(fk(s)))-g(s)]dsdudvdw,t≥T,(49)x(t-τ)=L+∑i=1∞‍∫t+(2i-2)τt+(2i-1)τ‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×h(u,x(h1(u)),…,x(hk(u)))dudvdw-∑i=1∞‍∫t+(2i-2)τt+(2i-1)τ‍∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,x(f1(s)),…,x(fk(s)))f(s,x(f1(s)),…,x(fk(s)))-g(s)]dsdudvdw,t≥T+τ.
Adding (48) and (49), we infer that
(50)x(t)+x(t-τ)=2L+∑i=1∞‍[∫t+(2i-1)τt+2iτ‍+∫t+(2i-2)τt+(2i-1)τ‍]×∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×h(u,x(h1(u)),…,x(hk(u)))dudvdw-∑i=1∞‍[∫t+(2i-1)τt+2iτ‍+∫t+(2i-2)τt+(2i-1)τ‍]×∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,x(f1(s)),…,x(fk(s)))-g(s)]dsdudvdw=2L+∫t+∞‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×h(u,x(h1(u)),…,x(hk(u)))dudvdw-∫t+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,x(f1(s)),…,x(fk(s)))(s,x(f1(s)),…,x(fk(s)))-g(s)]dsdudvdw,t≥T+τ,
which yields that x is a bounded positive solution of (11). By means of (29), (43), and (46), we know that for any t≥T and m∈N0(51)|xm+1(t)-x(t)|=|∫t+(2i-1)τt+2iτ‍(1-αm-βm)xm(t)+αm×{L+∑i=1∞‍∫t+(2i-1)τt+2iτ‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×h(u,xm(h1(u)),…,xm(hk(u)))dudvdw-∑i=1∞‍∫t+(2i-1)τt+2iτ‍∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,xm(f1(s)),…,xm(fk(s)))(s,xm(f1(s)),…,xm(fk(s)))-g(s)]dsdudvdw∑i=1∞‍∫t+(2i-1)τt+2iτ‍}+βmγm(t)-x(t)|≤(1-αm-βm)|xm(t)-x(t)|+αm|SLxm(t)-SLx(t)|+βm|γm(t)-x(t)|≤(1-αm-βm)|xm(t)-x(t)|+αmθ|xm(t)-x(t)|+2Mβm≤(1-(1-θ)αm)|xm(t)-x(t)|+2Mβm,
which gives (22). Thus Lemma 1, (23), and (24) ensure that lim⁡m→∞xm=x.
Now we show that (b) holds. LetL1,L2∈(N,M) with L1≠L2. As in the proof of (a), for each ∈{1,2}, we infer that there exists θj∈(0,1),  Tj>t0+|a|+τ+|T0|, and SLj:A(N,M)→A(N,M) satisfying (44)–(46), where θ,T,SL are replaced by θj,Tj,SLj, respectively, and the contraction mapping SLj possesses a unique fixed point zj∈A(N,M), and zj is a bounded positive solution of (11); that is,
(52)zj(t)=SLzj(t)=Lj+∑i=1∞‍∫t+(2i-1)τt+2iτ‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×h(u,zj(h1(u)),…,zj(hk(u)))dudvdw-∑i=1∞‍∫t+(2i-1)τt+2iτ‍∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,zj(f1(s)),…,zj(fk(s)))(s,zj(f1(s)),…,zj(fk(s)))-g(s)]dsdudvdw,t≥Tj.
Put T*=max⁡{T1,T2}. Using (16), (44), and (52), we conclude that for t>T*(53)|z1(t)-z2(t)|=|f(s,z2(f1(s)),…,z2(fk(s)))L1-L2+∑i=1∞‍∫t+(2i-1)τt+2iτ‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×[h(u,z1(h1(u)),…,z1(hk(u)))-h(u,z2(h1(u)),…,z2(hk(u)))]dudvdw-∑i=1∞‍∫t+(2i-1)τt+2iτ‍∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,z1(f1(s)),…,z1(fk(s)))-f(s,z2(f1(s)),…,z2(fk(s)))]dsdudvdw|≥|L1-L2|-∥z1-z2∥×∑i=1∞‍∫t+(2i-1)τt+2iτ‍∫w+∞‍∫v+∞‍Q2(u)|α(u)β(v)γ(w)|dudvdw-∥z1-z2∥×∑i=1∞‍∫t+(2i-1)τt+2iτ‍∫w+∞‍∫v+∞‍∫u+∞‍Q1(s)|α(u)β(v)γ(w)|dsdudvdw≥|L1-L2|-∥z1-z2∥×(∫T*+∞‍∫w+∞‍∫v+∞‍Q2(u)|α(u)β(v)γ(w)|dudvdw+∫T*+∞‍∫w+∞‍∫v+∞‍∫u+∞‍Q1(s)|α(u)β(v)γ(w)|dsdudvdw)≥|L1-L2|-∥z1-z2∥max⁡{θ1,θ2},
which yields that
(54)∥z1-z2∥≥|L1-L2|1+max⁡⁡{θ1,θ2}>0;
that is, z1≠z2. This completes the proof.Theorem 6.
Assume that there exist constantsM,N and T0≥t0 and functions Q1,Q2,R1 and R2∈C([t0,+∞),R+) satisfying (16), (17), and
(55)0<N<M,p(t)=-1,t≥T0,(56)∑i=1∞‍∫t0+iτ+∞‍∫w+∞‍∫v+∞‍max⁡{Q2(u),R2(u)}|α(u)β(v)γ(w)|dudvdw<+∞;∑i=1∞‍∫t0+iτ+∞‍∫w+∞‍∫v+∞‍∫u+∞‍max⁡{Q1(s),R1(s),|g(s)|}|α(u)β(v)γ(w)|<+∞.
Then(a)
for anyL∈(N,M), there existθ∈(0,1)andT>t0+|a|+τ+|T0|such that, for eachx0∈A(N,M), the Mann iterative sequence{xm}m∈N0generated by the scheme(57)xm+1(t)={(1-αm-βm)xm(t)+αm×{L-∑i=1∞‍∫t+iτ+∞‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×h(u,xm(h1(u)),…,xm(hk(u)))dudvdw+∑i=1∞‍∫t+iτ+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,xm(f1(s)),…,xm(fk(s)))(s,xm(f1(s)),…,xm(fk(s)))-g(s)]dsdudvdw∫v+∞∑i=1∞‍‍}+βmγm(t),t≥T,m∈N0,(1-αm-βm)xm(T)+αm×{L-∑i=1∞‍∫T+iτ+∞‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×h(u,xm(h1(u)),…,xm(hk(u)))dudvdw+∑i=1∞‍∫T+iτ+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,xm(f1(s)),…,xm(fk(s)))(s,xm(f1(s)),…,xm(fk(s)))-g(s)]dsdudvdw∑i=1∞‍∫T+iτ+∞‍∫T+iτ+∞‍}+βmγm(T),a≤t<T,m∈N0converges to a bounded positive solutionx∈A(N,M)of (11) and has the error estimate (22), where{γm}m∈N0is an arbitrary sequence inA(N,M), and{αm}m∈N0and{βm}m∈N0are any sequences in[0,1]satisfying (23) and (24);(b)
equation (11) possesses uncountably many bounded positive solutions inA(N,M).Proof.
Firstly, we show that (a) holds. LetL∈(N,M). It follows from (56) that there exist θ∈(0,1) and T>t0+|a|+τ+|T0| satisfying
(58)θ=∑i=1∞‍∫T+iτ+∞‍∫w+∞‍∫v+∞‍Q2(u)|α(u)β(v)γ(w)|dudvdw+∑i=1∞‍∫T+iτ+∞‍∫w+∞‍∫v+∞‍∫u+∞‍Q1(s)|α(u)β(v)γ(w)|dsdudvdw,(59)∑i=1∞‍∫T+iτ+∞‍∫w+∞‍∫v+∞‍R2(u)|α(u)β(v)γ(w)|dudvdw+∑i=1∞‍∫T+iτ+∞‍∫w+∞‍∫v+∞‍∫u+∞‍R1(s)+|g(s)||α(u)β(v)γ(w)|dsdudvdw<min⁡{M-L,L-N}.
Define a mapping SL:A(N,M)→CB([a,+∞),R) by
(60)SLx(t)={L-∑i=1∞‍∫t+iτ+∞‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×h(u,x(h1(u)),…,x(hk(u)))dudvdw+∑i=1∞‍∫t+iτ+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,x(f1(s)),…,x(fk(s)))(s,x(f1(s)),…,x(fk(s)))-g(s)]dsdudvdw,t≥T,x∈A(N,M),SLx(T),a≤t<T,x∈A(N,M).
Clearly, SLx is continuous for each x∈A(N,M). In light of (16), (17), and (58)–(60), we conclude that for x,y∈A(N,M) and t≥T(61)|SLx(t)-SLy(t)|≤∑i=1∞‍∫t+iτ+∞‍∫w+∞‍∫v+∞‍1|α(u)β(v)γ(w)|×|h(u,x(h1(u)),…,x(hk(u)))-h(u,y(h1(u)),…,y(hk(u)))|dudvdw+∑i=1∞‍∫t+iτ+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1|α(u)β(v)γ(w)|×|f(s,x(f1(s)),…,x(fk(s)))-f(s,y(f1(s)),…,y(fk(s)))|dsdudvdw≤∥x-y∥×∑i=1∞‍∫t+iτ+∞‍∫w+∞‍∫v+∞‍Q2(u)|α(u)β(v)γ(w)|dudvdw+∥x-y∥×∑i=1∞‍∫t+iτ+∞‍∫w+∞‍∫v+∞‍∫u+∞‍Q1(s)|α(u)β(v)γ(w)|dsdudvdw≤(∑i=1∞‍∫T+iτ+∞‍∫w+∞‍∫v+∞‍Q2(u)|α(u)β(v)γ(w)|dudvdw+∑i=1∞‍∫T+iτ+∞‍∫w+∞‍∫v+∞‍∫u+∞‍Q1(s)|α(u)β(v)γ(w)|dsdudvdw)×∥x-y∥=θ∥x-y∥,SLx(t)=L-∑i=1∞‍∫t+iτ+∞‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×h(u,x(h1(u)),…,x(hk(u)))dudvdw+∑i=1∞‍∫t+iτ+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,x(f1(s)),…,x(fk(s)))-g(s)]dsdudvdw≤L+∑i=1∞‍∫t+iτ+∞‍∫w+∞‍∫v+∞‍R2(u)|α(u)β(v)γ(w)|dudvdw+∑i=1∞‍∫t+iτ+∞‍∫w+∞‍∫v+∞‍∫u+∞‍R1(s)+|g(s)||α(u)β(v)γ(w)|dsdudvdw<L+min⁡{M-L,L-N}≤M,SLx(t)≥L-∑i=1∞‍∫t+iτ+∞‍∫w+∞‍∫v+∞‍R2(u)|α(u)β(v)γ(w)|dudvdw-∑i=1∞‍∫t+iτ+∞‍∫w+∞‍∫v+∞‍∫u+∞‍R1(s)+|g(s)||α(u)β(v)γ(w)|dsdudvdw>L-min⁡{M-L,L-N}≥N,
which yield that (29) holds. That is, SL is a contraction mapping in A(N,M) and has a unique fixed point x∈A(N,M); that is,
(62)x(t)=L-∑i=1∞‍∫t+iτ+∞‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×h(u,x(h1(u)),…,x(hk(u)))dudvdw+∑i=1∞‍∫t+iτ+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,x(f1(s)),…,x(fk(s)))(s,x(f1(s)),…,x(fk(s)))-g(s)]dsdudvdw,t≥T,(63)x(t-τ)=L-∑i=1∞‍∫t+(i-1)τ+∞‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×h(u,x(h1(u)),…,x(hk(u)))dudvdw+∑i=1∞‍∫t+(i-1)τ+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,x(f1(s)),…,x(fk(s)))(s,x(f1(s)),…,x(fk(s)))-g(s)]dsdudvdw,t≥T+τ.
By virtue of (62) and (63), we get that
(64)x(t)-x(t-τ)=∫t+∞‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×h(u,x(h1(u)),…,x(hk(u)))dudvdw-∫t+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,x(f1(s)),…,x(fk(s)))(s,x(f1(s)),…,x(fk(s)))-g(s)]dsdudvdw,t≥T+τ,
which yields that
(65)(x(t)-x(t-τ))′=-∫t+∞‍∫v+∞‍1α(u)β(v)γ(t)×h(u,x(h1(u)),…,x(hk(u)))dudv+∫t+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(t)×[f(s,x(f1(s)),…,x(fk(s)))(s,x(f1(s)),…,x(fk(s)))-g(s)]dsdudv,t≥T+τ,(γ(t)(x(t)-x(t-τ))′)′=∫t+∞‍1α(u)β(t)×h(u,x(h1(u)),…,x(hk(u)))du-∫t+∞‍∫u+∞‍1α(u)β(t)×[f(s,x(f1(s)),…,x(fk(s)))(s,x(f1(s)),…,x(fk(s)))-g(s)]dsdu,t≥T+τ,
which ensures that
(66)[β(t)(γ(t)(x(t)-x(t-τ))′)′]′=-1α(t)h(u,x(h1(u)),…,x(hk(u)))+∫t+∞‍1α(t)[f(s,x(f1(s)),…,x(fk(s)))-g(s)]ds,xxxxxxxxxxxxxxxxxxxxxxixxxxxxxxmxxxxxxxit≥T+τ,[α(t)[β(t)(γ(t)(x(t)-x(t-τ))′)′]′]′=-[h(t,x(h1(t)),…,x(hk(t)))]′-[f(t,x(f1(t)),…,x(fk(t)))-g(t)],xxxxxxxxxxxxxmixxxmxxxxxxxxxt≥T+τ,
which gives that x is a bounded positive solution of (11). By means of (29), (57), (58), and (60), we deduce that, for any m∈N0 and t≥T,
(67)|xm+1(t)-x(t)|=|∑i=1∞‍∫t+iτ+∞‍(1-αm-βm)xm(t)+αm×{L-∑i=1∞‍∫t+iτ+∞‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×h(u,xm(h1(u)),…,xm(hk(u)))dudvdw+∑i=1∞‍∫t+iτ+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,xm(f1(s)),…,xm(fk(s)))(s,xm(f1(s)),…,xm(fk(s)))-g(s)]dsdudvdw∑i=1∞‍∫t+iτ+∞‍∫w+∞‍∫v+∞‍}+βmγm(t)-x(t)∑i=1∞‍∫t+iτ+∞‍|≤(1-αm-βm)|xm(t)-x(t)|+αm|SLxm(t)-SLx(t)|+βm|γm(t)-x(t)|≤(1-αm-βm)|xm(t)-x(t)|+αmθ|xm(t)-x(t)|+2Mβm≤(1-(1-θ)αm)|xm(t)-x(t)|+2Mβm,
which implies (22). Thus Lemma 1, (23), and (24) ensure that lim⁡m→∞xm=x.
Next we show that (b) holds. LetL1,L2∈(N,M) with L1≠L2. As in the proof of (a), for each j∈{1,2}, we infer that there exists θj∈(0,1),Tj>t0+|a|+τ+|T0|, and SLj:A(N,M)→A(N,M) satisfying (58)–(60), where θ,T,SL are replaced by θj,Tj,SLj, respectively, and the contraction mapping SLj possesses a unique fixed point zj∈A(N,M), and zj is a bounded positive solution of (11); that is,
(68)zj(t)=Lj-∑i=1∞‍∫t+iτ+∞‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×h(u,zj(h1(u)),…,zj(hk(u)))dudvdw+∑i=1∞‍∫t+iτ+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,zj(f1(s)),…,zj(fk(s)))(s,zj(f1(s)),…,zj(fk(s)))-g(s)]dsdudvdw,t≥Tj.
Put T*=max⁡{T1,T2}. Using (16), (58), and (68), we conclude that for t>T*(69)|z1(t)-z2(t)|=|(s,z2(f1(s)),…,z2(fk(s)))L1-L2-∑i=1∞‍∫t+iτ+∞‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×[h(u,z1(h1(u)),…,z1(hk(u)))-h(u,z2(h1(u)),…,z2(hk(u)))]dudvdw+∑i=1∞‍∫t+iτ+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,z1(f1(s)),…,z1(fk(s)))-f(s,z2(f1(s)),…,z2(fk(s)))]dsdudvdw|≥|L1-L2|-∥z1-z2∥×∑i=1∞‍∫t+iτ+∞‍∫w+∞‍∫v+∞‍Q2(u)|α(u)β(v)γ(w)|dudvdw-∥z1-z2∥×∑i=1∞‍∫t+iτ+∞‍∫w+∞‍∫v+∞‍∫u+∞‍Q1(s)|α(u)β(v)γ(w)|dsdudvdw≥|L1-L2|-∥z1-z2∥×(∑i=1∞‍∫max⁡⁡{T1,T2}+iτ+∞‍∫w+∞‍∫v+∞‍Q2(u)|α(u)β(v)γ(w)|dudvdw+∑i=1∞‍∫max⁡⁡{T1,T2}+iτ+∞‍∫w+∞‍∫v+∞‍∫u+∞‍Q1(s)|α(u)β(v)γ(w)|dsdudvdw)≥|L1-L2|-∥z1-z2∥max⁡{θ1,θ2}, which yields that
(70)∥z1-z2∥≥|L1-L2|1+max⁡{θ1,θ2}>0;
that is, z1≠z2. This completes the proof.Theorem 7.
Assume that there exist constantsM,N,c1,c2 and T0≥t0 and functions Q1,Q2,R1 and R2∈C([t0,+∞),R+) satisfying (16)–(18) and
(71)1<c2<c1<c22,0<c12-c2c1N<c22-c1c2M;(72)c2≤p(t)≤c1,t≥T0.
Then(a)
for anyL∈((c1/c2)M+c1N,(c2/c1)N+c2M), there existθ∈(0,1)andT>t0+|a|+τ+|T0|such that, for eachx0∈A(N,M), the Mann iterative sequence{xm}m∈N0with mixed errors generated by the scheme(73)xm+1(t)={(1-αm-βm)xm(t)+αm×{Lp(t+τ)-xm(t+τ)p(t+τ)+1p(t+τ)×∫t+τ+∞‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×h(u,xm(h1(u)),…,xm(hk(u)))dudvdw-1p(t+τ)×∫t+τ+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,xm(f1(s)),…,xm(fk(s)))(s,xm(f1(s)),…,xm(fk(s)))-g(s)]dsdudvdwLp(t+τ)}+βmγm(t),xxxxxxxxxxxxxxxxxxt≥T,m∈N0,(1-αm-βm)xm(T)+αm×{Lp(T+τ)-xm(T+τ)p(T+τ)+1p(T+τ)×∫T+τ+∞‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×h(u,xm(h1(u)),…,xm(hk(u)))dudvdw-1p(T+τ)∫T+τ+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,xm(f1(s)),…,xm(fk(s)))(s,xm(f1(s)),…,xm(fk(s)))-g(s)]dsdudvdw{Lp(T+τ)-xm(T+τ)p(T+τ)(s,xm(f1(s)),…,xm(fk(s)))}+βmγm(T),xxxxxxxxxxxxxa≤t<T,m∈N0with (23) and (24) converges to a bounded positive solutionx∈A(N,M)of (11) and has the error estimate (22);(b)
equation (11) possesses uncountably many bounded positive solutions inA(N,M).Proof.
In the first place, we prove that (a) holds. LetL∈((c1/c2)M+c1N,(c2/c1)N+c2M). It follows from (18) and (71) that there exist θ∈(0,1) and T>t0+|a|+τ+|T0| satisfying
(74)θ=1c2(1+∫T+∞‍∫w+∞‍∫v+∞‍Q2(u)|α(u)β(v)γ(w)|dudvdw+∫T+∞‍∫w+∞‍∫v+∞‍∫u+∞‍Q1(s)|α(u)β(v)γ(w)|dsdudvdw),(75)∫T+∞‍∫w+∞‍∫v+∞‍R2(u)|α(u)β(v)γ(w)|dudvdw+∫T+∞‍∫w+∞‍∫v+∞‍∫u+∞‍R1(s)+|g(s)||α(u)β(v)γ(w)|dsdudvdw<min⁡{c2M-L+c2c1N,-c2N+c2c1L-M}.
Define a mapping SL:A(N,M)→CB([a,+∞),R) by
(76)SLx(t)={Lp(t+τ)-x(t+τ)p(t+τ)+1p(t+τ)×∫t+τ+∞‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×h(u,x(h1(u)),…,x(hk(u)))dudvdw-1p(t+τ)×∫t+τ+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,x(f1(s)),…,x(fk(s)))(s,x(f1(s)),…,x(fk(s)))-g(s)]dsdudvdw,t≥T,x∈A(N,M),SLx(T),a≤t<T,x∈A(N,M).
Obviously, SLx is continuous for each x∈A(N,M). In view of (16), (17), (71), (72), and (74)–(76), we conclude that for x,y∈A(N,M) and t≥T(77)|SLx(t)-SLy(t)|≤1|p(t+τ)||x(t+τ)-y(t+τ)|+1|p(t+τ)|×∫t+τ+∞‍∫w+∞‍∫v+∞‍1|α(u)β(v)γ(w)|×|h(u,x(h1(u)),…,x(hk(u)))-h(u,y(h1(u)),…,y(hk(u)))|dudvdw+1|p(t+τ)|∫t+t+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1|α(u)β(v)γ(w)|×|f(s,x(f1(s)),…,x(fk(s)))-f(s,y(f1(s)),…,y(fk(s)))|dsdudvdw≤1c2∥x-y∥×(1+∫T+∞‍∫w+∞‍∫v+∞‍Q2(u)|α(u)β(v)γ(w)|dudvdw+∫T+∞‍∫w+∞‍∫v+∞‍∫u+∞‍Q1(s)|α(u)β(v)γ(w)|dsdudvdw)=θ∥x-y∥,SLx(t)=Lp(t+τ)-x(t+τ)p(t+τ)+1p(t+τ)∫t+τ+∞‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×h(u,x(h1(u)),…,x(hk(u)))dudvdw-1p(t+τ)∫t+τ+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,x(f1(s)),…,x(fk(s)))(s,x(f1(s)),…,x(fk(s)))-g(s)]dsdudvdw≤Lc2-Nc1+1c2×∫T+∞‍∫w+∞‍∫v+∞‍R2(u)|α(u)β(v)γ(w)|dudvdw+1c2∫T+∞‍∫w+∞‍∫v+∞‍∫u+∞‍R1(s)+|g(s)||α(u)β(v)γ(w)|dsdudvdw<Lc2-Nc1+1c2×min⁡⁡{c2M-L+c2c1N,-c2N+c2c1L-M}≤M,SLx(t)≥Lc1-Mc2-1c2×∫T+∞‍∫w+∞‍∫v+∞‍R2(u)|α(u)β(v)γ(w)|dudvdw-1c2∫T+∞‍∫w+∞‍∫v+∞‍∫u+∞‍R1(s)+|g(s)||α(u)β(v)γ(w)|dsdudvdw>Lc1-Mc2-1c2×min⁡⁡{c2M-L+c2c1N,-c2N+c2c1L-M}≥N,
which imply that (29) holds. That is, SL is a contraction mapping in A(N,M) and has a unique fixed point x∈A(N,M), which is a bounded positive solution of (11). By means of (29), (73), and (76), we obtain that for any m∈N0 and t≥T(78)|xm+1(t)-x(t)|=|Lp(t+τ)(1-αm-βm)xm(t)+αm×{Lp(t+τ)-x(t+τ)p(t+τ)+1p(t+τ)×∫t+τ+∞‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×h(u,xm(h1(u)),…,xm(hk(u)))dudvdw-1p(t+τ)∫t+τ+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,xm(f1(s)),…,xm(fk(s)))(s,xm(f1(s)),…,xm(fk(s)))-g(s)]dsdudvdwLp(t+τ)}+βmγm(t)-x(t)|≤(1-αm-βm)|xm(t)-x(t)|+αm|SLxm(t)-SLx(t)|+βm|γm(t)-x(t)|≤(1-αm-βm)|xm(t)-x(t)|+αmθ|xm(t)-x(t)|+2Mβm≤(1-(1-θ)αm)|xm(t)-x(t)|+2Mβm≤(1-(1-θ)αm)∥xm-x∥+2Mβm,
which gives (22). Thus Lemma 1, (23), and (24) mean that lim⁡m→∞xm=x.
Next we show that (b) holds. LetL1,L2∈((c1/c2)M+c1N,(c2/c1)N+c2M) with L1≠L2. As in the proof of (a), we conclude that, for each ∈{1,2}, there exist θj∈(0,1),Tj>t0+|a|+τ+|T0| and SLj:A(N,M)→A(N,M) satisfying (74)–(76), where θ,T,SL are replaced by θj,Tj, and SLj, respectively, and the contraction mapping SLj has a unique fixed point zj∈A(N,M), which is also a bounded positive solution of (11). In order to prove (b), we need only to show that z1≠z2. Put T*=max⁡{T1,T2}. Note that for t≥T* and j∈{1,2}(79)zj(t)=Ljp(t+τ)-zj(t+τ)p(t+τ)+1p(t+τ)×∫t+τ+∞‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×h(u,zj(h1(u)),…,zj(hk(u)))dudvdw-1p(t+τ)×∫t+τ+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,zj(f1(s)),…,zj(fk(s)))(s,zj(f1(s)),…,zj(fk(s)))-g(s)]dsdudvdw,
which together with (16) and (74) implies that for t≥T*(80)|z1(t)-z2(t)|=|L1-L2p(t+τ)-z1(t+τ)-z2(t+τ)p(t+τ)+1p(t+τ)×∫t+τ+∞‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×[h(u,z1(h1(u)),…,z1(hk(u)))-h(u,z2(h1(u)),…,z2(hk(u)))]dudvdw-1p(t+τ)×∫t+τ+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,z1(f1(s)),…,z1(fk(s)))-f(s,z2(f1(s)),…,z2(fk(s)))]dsdudvdwz1(t+τ)-z2(t+τ)p(t+τ)|≥|L1-L2|c1-∥z1-z2∥c2-∥z1-z2∥c2×∫T*+∞‍∫w+∞‍∫v+∞‍Q2(u)|α(u)β(v)γ(w)|dudvdw-∥z1-z2∥c2×∫T*+∞‍∫w+∞‍∫v+∞‍∫u+∞‍Q1(s)|α(u)β(v)γ(w)|dsdudvdw≥|L1-L2|c1-∥z1-z2∥max⁡{θ1,θ2},
which yields that
(81)∥z1-z2∥≥|L1-L2|c1(1+max⁡{θ1,θ2})>0;
that is, z1≠z2. This completes the proof.Theorem 8.
Assume that there exist constantsM,N,c1,c2 and T0≥t0 and functions Q1,Q2,R1 and R2∈C([t0,+∞),R+) satisfying (16)–(18) and
(82)0<(c1-1)N<(c2-1)M;-∞<-c1≤p(t)≤-c2<-1,t≥T0.
Then (a)
for anyL∈((c1-1)N,(c2-1)M), there existθ∈(0,1)andT>t0+|a|+τ+|T0|such that, for eachx0∈A(N,M), the Mann iterative sequence{xm}m∈N0with mixed errors generated by the scheme(83)xm+1(t)={(1-αm-βm)xm(t)+αm×{-Lp(t+τ)-xm(t+τ)p(t+τ)+1p(t+τ)×∫t+τ+∞‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×h(u,xm(h1(u)),…,xm(hk(u)))dudvdw-1p(t+τ)∫t+τ+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,xm(f1(s)),…,xm(fk(s)))(s,xm(f1(s)),…,xm(fk(s)))-g(s)]dsdudvdw-Lp(t+τ)}+βmγm(t),xxxxxxxxxxxxxxxt≥T,m∈N0,(1-αm-βm)xm(T)+αm×{-Lp(T+τ)-xm(T+τ)p(T+τ)+1p(T+τ)×∫T+τ+∞‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×h(u,xm(h1(u)),…,xm(hk(u)))dudvdw-1p(T+τ)×∫T+τ+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,xm(f1(s)),…,xm(fk(s)))(s,xm(f1(s)),…,xm(fk(s)))-g(s)]dsdudvdw-Lp(T+τ)}+βmγm(T),xxxxxxxxxxxxa≤t<T,m∈N0with (23) and (24) converges to a bounded positive solutionx∈A(N,M)of (11) and has the error estimate (22);(b)
equation (11) possesses uncountably many bounded positive solutions inA(N,M).Proof.
First of all, we prove that (a) holds. LetL∈((c1-1)N,(c2-1)M). It follows from (18) and (82) that there exist θ∈(0,1) and T>t0+|a|+τ+|T0| satisfying (74) and
(84)∫T+∞‍∫w+∞‍∫v+∞‍R2(u)|α(u)β(v)γ(w)|dudvdw+∫T+∞‍∫w+∞‍∫v+∞‍∫u+∞‍R1(s)+|g(s)||α(u)β(v)γ(w)|dsdudvdw<min⁡{(c2-1)M-L,(c2c1-c2)N+c2c1L}.
Define a mapping SL:A(N,M)→CB([a,+∞),R) by
(85)SLx(t)={-Lp(t+τ)-x(t+τ)p(t+τ)+1p(t+τ)×∫t+τ+∞‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×h(u,x(h1(u)),…,x(hk(u)))dudvdw-1p(t+τ)×∫t+τ+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,x(f1(s)),…,x(fk(s)))(s,x(f1(s)),…,x(fk(s)))-g(s)]dsdudvdw,t≥T,x∈A(N,M),SLx(T),a≤t<T,x∈A(N,M).
Obviously SLx is continuous for each x∈A(N,M). On account of (17), (82), (84), and (85), we get that for any x∈A(N,M) and t≥T(86)SLx(t)=-Lp(t+τ)-x(t+τ)p(t+τ)+1p(t+τ)×∫t+τ+∞‍∫w+∞‍∫v+∞‍1α(u)β(v)γ(w)×h(u,x(h1(u)),…,x(hk(u)))dudvdw-1p(t+τ)×∫t+τ+∞‍∫w+∞‍∫v+∞‍∫u+∞‍1α(u)β(v)γ(w)×[f(s,x(f1(s)),…,x(fk(s)))-g(s)]dsdudvdw≤Lc2+Mc2+1c2×∫T+∞‍∫w+∞‍∫v+∞‍R2(u)|α(u)β(v)γ(w)|dudvdw+1c2×∫T+∞‍∫w+∞‍∫v+∞‍∫u+∞‍R1(s)+|g(s)||α(u)β(v)γ(w)|dsdudvdw<Lc2+Mc2+1c2×min⁡{(c2-1)M-L,(c2c1-c2)N+c2c1L}≤M,(87)SLx(t)≥Lc1+Nc1-1c2×∫T+∞‍∫w+∞‍∫v+∞‍R2(u)|α(u)β(v)γ(w)|dudvdw-1c2×∫T+∞‍∫w+∞‍∫v+∞‍∫u+∞‍R1(s)+|g(s)||α(u)β(v)γ(w)|dsdudvdw>Lc1+Nc1-1c2×min⁡⁡{(c2-1)M-L,(c2c1-c2)N+c2c1L}≥N,
which imply (29). That is, SL is a contraction mapping in A(N,M) and has a unique fixed point x∈A(N,M), which is a bounded positive solution of (11). The rest of the proof is similar to that of Theorem 7 and is omitted. This completes the proof.


## 3. Examples

Now we construct seven examples as applications of the results presented in Section2.Example 9.
Consider the following fourth order nonlinear neutral delay differential equation:(88)((-t3(x(t)+3t2cos⁡t-(1/2)1+10t2x(t-τ))′)′)′(t2+1)×((-t3(x(t)+3t2cos⁡t-1/21+10t2x(t-τ))′)′2t4+1ln⁡⁡(t+1)×(-t3(x(t)+3t2cos⁡t-1/21+10t2x(t-τ))′)′)′)′+(tsin2(tx3(t-1))(t+3)2(t+1)3)′+cos⁡⁡(t)x3(t2)t3=sin⁡(2-t2)+13t4,t≥3,
where τ>0, t0=T0=3, k=1, N=2, M=20, c1=3/10, c2=3/5, a=min⁡{3-τ,2}, and
(89)α(t)=t2+1,β(t)=2t4+1ln⁡(t+1),γ(t)=-t3,p(t)=3t2cos⁡t-1/21+10t2,g(t)=sin(2-t2)+13t4,h1(t)=t-1,f1(t)=t2,f(t,u)=u3cos⁡(t)t3,h(t,u)=tsin2(tu3)(t+3)2(t+1)3,Q1(t)=3M2t3,Q2(t)=6t2M2(t+3)2(t+1)3,R1(t)=M3t3,R2(t)=t(t+3)2(t+1)3,(t,u)∈[t0,+∞)×R.
It is easy to verify that (16)–(20) are satisfied. It follows from Theorem 2 that (11) possesses uncountably many bounded positive solutions in A(N,M). On the other hand, for any L∈(N+c1M,(1-c2)M), there exist θ∈(0,1) and T>t0+|a|+τ+|T0| such that the Mann iterative sequence {xm}m∈N0 with mixed errors generated by (21) converges to a bounded positive solution x∈A(N,M) of (11) and has the error estimate (22), 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 (23) and (24).Example 10.
Consider the following fourth order nonlinear neutral delay differential equation:(90)((t4(-t7+1(x(t)+t33t3+2x(t-τ))′)′)t6ln⁡⁡(1+t2)1+t×(t4(-t7+1(x(t)+t33t3+2x(t-τ))′)′)′)′+(x2(t-2)+x2(t+1)-x(t+1)1+t3)′+2x2(t-1)+x2(t-t)2t4+1=1-tt4+cos⁡2t,t≥10,
where τ>0, t0=T0=10, k=2, N=5, M=21, c1=1/3, c2=1/4, a=min⁡{10-τ,11}, and
(91)α(t)=t6ln⁡(1+t2)1+t,β(t)=t4,γ(t)=-t7+1,p(t)=t33t3+2,g(t)=1-tt4+cos⁡2t,h1(t)=t-2,h2(t)=t+1,f1(t)=t-1,f2(t)=t-t,f(t,u,v)=2u2+v22t4+1,h(t,u,v)=u2+v2-v1+t3,Q1(t)=6M2t4+1,Q2(t)=4M+11+t3,R1(t)=3M22t4+1,R2(t)=2M2+M1+t3,(t,u,v)∈[t0,+∞)×R2.
It is easy to verify that (16)–(18) and (35) are satisfied. It follows from Theorem 3 that (11) possesses uncountably many bounded positive solutions in A(N,M). On the other hand, for any L∈(N+c1M,M+c2N), there exist θ∈(0,1) and T>t0+|a|+τ+|T0| such that the Mann iterative sequence {xm}m∈N0 with mixed errors generated by (21) converges to a bounded positive solution x∈A(N,M) of (11) and has the error estimate (22), 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 (23) and (24).Example 11.
Consider the following fourth order nonlinear neutral delay differential equation:(92)(((2t+1)3t(x(t)+1-t3+5tx(t-τ))′)′)′-t5((x(t)+1-t3+5tx(t-τ))′(3t2+2)×((2t+1)3t(x(t)+1-t3+5tx(t-τ))′)′)′)′+(tx2(2t-3)x2(et)(1+t2)(2+sin⁡(x(2t-3)x(t2))))′+t21+t7x2((t+1)2)+ln⁡⁡(t+1)et+1sin2(tx(3t+1)x3(2t))=1+t2t4+1,t≥2,
where τ>0, t0=T0=2, k=3, N=100, M=500, c1=4/5, c2=1/13, a=min⁡{2-τ,1}, and
(93)α(t)=-t5,β(t)=3t2+2,γ(t)=(2t+1)3t,p(t)=1-t3+5t,g(t)=1+t2t4+1,h1(t)=2t-3,h2(t)=t2,h3(t)=et,f1(t)=(t+1)2,f2(t)=3t+1,f3(t)=2t,f(t,u,v,w)=t2u21+t7+ln⁡(t+1)et+1sin2(tvw3),h(t,u,v,w)=tu2w2(1+t2)(2+sin(uv)),Q1(t)=2Mt21+t7+4tM3ln⁡(t+1)et+1,Q2(t)=t(2M5+12M3)1+t2,R1(t)=t2M21+t7+ln⁡(t+1)et+1,R2(t)=tM41+t2,(t,u,v,w)∈[t0,+∞)×R3.
It is easy to verify that (17), (18), and (39) are satisfied. It follows from Theorem 4 that (11) possesses uncountably many bounded positive solutions in A(N,M). On the other hand, for any L∈((1-c2)N,(1-c1)M), there exist θ∈(0,1) and T>t0+|a|+τ+|T0| such that the Mann iterative sequence {xm}m∈N0 with mixed errors generated by (21) converges to a bounded positive solution x∈A(N,M) of (11) and has the error estimate (22), 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 (23) and (24).Example 12.
Consider the following fourth order nonlinear neutral delay differential equation:(94)(2t41+t((t2+1)(-t8(x(t)+x(t-τ))′)′)′)′+(t3+1t10ln⁡⁡(t+2)x2(t-2)x(t-1))′+x2(t-3)(1+t3)(1+x2(t-4))=1+t2-sin(2t2+1)t8,t≥3,
where τ>0, t0=T0=3, k=2, N=10, M=20, a=min⁡{3-τ,-1}, and
(95)α(t)=2t41+t,β(t)=t2+1,γ(t)=-t8,p(t)=1,g(t)=1+t2-sin(2t2+1)t8,h1(t)=t-2,h2(t)=t-1,f1(t)=t-3,f2(t)=t-4,f(t,u,v)=u2(1+t3)(1+v2),h(t,u,v)=(t3+1)u2vt10ln⁡(t+2),Q1(t)=2M+4M3(1+t3)(1+N2)2,Q2(t)=3M2(1+t3)t10ln⁡(2+t),R1(t)=M2(1+t3)(1+N2),R2(t)=M3(1+t3)t10ln⁡⁡(t+2),(t,u,v)∈[t0,+∞)×R2.
It is easy to verify that (16)–(18) and (42) are satisfied. It follows from Theorem 5 that (11) possesses uncountably many bounded positive solutions in A(N,M). On the other hand, for any L∈(N,M), there exist θ∈(0,1) and T>t0+|a|+τ+|T0| such that the Mann iterative sequence {xm}m∈N0 with mixed errors generated by (43) converges to a bounded positive solution x∈A(N,M) of (11) and has the error estimate (22), 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 (23) and (24).Example 13.
Consider the following fourth order nonlinear neutral delay differential equation:(96)(x(2t-3)cos⁡2(t-1)1+x2(3t-2)-et(t5((1-t7)(x(t)-x(t-τ))′)′)′)′+((x2(t-3)x2(t-1)-2x3(t+1)1+1+x2(t-3))11+t6×(x2(t-3)x2(t-1)-2x3(t+1)1+1+x2(t-3)))′+13t4+2t(x(2t-3)cos⁡2(t-1)1+x2(3t-2)x3(t-4)(1+sinx(2t-3))+x(2t-3)cos⁡2(t-1)1+x2(3t-2))=ln⁡(1+t)1+t10,t≥4,
where τ>0, t0=T0=4, k=3, N=100, M=200, a=min⁡{4-τ,0}, and
(97)α(t)=-et,β(t)=t5,γ(t)=1-t7,(98)p(t)=-1,g(t)=ln⁡⁡(1+t)1+t10,h1(t)=t+1,h2(t)=t-3,h3(t)=t-1,f1(t)=t-4,f2(t)=2t-3,f3(t)=3t-2,f(t,u,v,w)=13t4+2t(u3(1+sinv)+vcos⁡2(t-1)1+w2),h(t,u,v,w)=11+t6(v2w2-2u31+1+v2),Q1(t)=13t4+2t(M2(6+M)+1+3M2(1+N2)2cos⁡2(t-1)),Q2(t)=11+t6×(4M3+6M2(1+1+M2)(1+1+N2)2+2M4(1+1+N2)21+N2),R1(t)=13t4+2t(2M3+M1+N2),R2(t)=11+t6(M4+2M31+1+N2),(t,u,v,w)∈[t0,+∞)×R3.
It is easy to verify that (16), (17), (55), and (56) are satisfied. It follows from Theorem 6 that (11) possesses uncountably many bounded positive solutions in A(N,M). On the other hand, for any L∈(N,M), there exist θ∈(0,1) and T>t0+|a|+τ+|T0| such that the Mann iterative sequence {xm}m∈N0 with mixed errors generated by (57) converges to a bounded positive solution x∈A(N,M) of (11) and has the error estimate (22), 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 (23) and (24).Example 14.
Consider the following fourth order nonlinear neutral delay differential equation:(99)(×(t51+t(x(t)+4+5t3+2tx(t-τ))′)′)′3+sintt2×((t51+t(x(t)+4+5t3+2tx(t-τ))′)′tln⁡3t×(t51+t(x(t)+4+5t3+2tx(t-τ))′)′)′)′+((1-t2)x3(3t-1)x(t-5)(1+t)6+t|x(t-2)|)′+t3x3(t-sint)-t5x2(t-cos⁡t)1+t11+x2(t-22)=t31+t8,xxxxxxxxxxxxxxxxxxxxixxxxxxxxxt≥20,
where τ>0, t0=T0=20, k=3, N=3, M=20, c1=3, c2=2, a=min⁡{20-τ,-2}, and
(100)α(t)=3+sintt2,β(t)=tln⁡3t,γ(t)=t51+t,p(t)=4+5t3+2t,g(t)=t31+t8,h1(t)=3t-1,h2(t)=t-5,h3(t)=t-2,f1(t)=t-sint,f2(t)=t-cos⁡t,f3(t)=t-22,f(t,u,v,w)=t3u3-t5v21+t11+w2,h(t,u,v,w)=(1-t2)u3v(1+t)6+t|w|,Q1(t)=3(t3+t14)M2+5t3M4+2(t5+t16)M+4t5M3(1+t11+N2)2,Q2(t)=4(1+t2)(1+t)6M3+5t(1+t2)M4[(1+t)6+tN]2,R1(t)=t3M3+t5M21+t11+N2,R2(t)=(1+t2)M4(1+t)6+tN,(t,u,v,w)∈[t0,+∞)×R3.
It is easy to verify that (16)–(18), (71), and (72) are satisfied. It follows from Theorem 7 that (11) possesses uncountably many bounded positive solutions in A(N,M). On the other hand, for any L∈((c1/c2)M+c1N,(c2/c1)N+c2M), there exist θ∈(0,1) and T>t0+|a|+τ+|T0| such that the Mann iterative sequence {xm}m∈N0 with mixed errors generated by (73) converges to a bounded positive solution x∈A(N,M) of (11) and has the error estimate (22), 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 (23) and (24).Example 15.
Consider the following fourth order nonlinear neutral delay differential equation:(101)((x(t)+(-3-sint)x(t-τ))′tln⁡2(1+t)1+sin2t)′)′ln⁡tt(t22+cos⁡t(tln⁡2(1+t)1+sin2t×(x(t)+(-3-sint)x(t-τ))′tln⁡2(1+t)1+sin2t)′)′)′+((1-t)x2(t-5)x2(t-7)1+t3+sin2(tx(t-3)))′+2-x2(t)-x2(t-1)x3(t-10)1+t5+ln⁡⁡(t2+3)=3+cos⁡(t2+1)t5+2t3,t≥9,
where τ>0, t0=T0=9, k=3, N=10, M=50, c1=4, c2=2, a=min⁡{9-τ,-4}, and
(102)α(t)=ln⁡tt,β(t)=t22+cos⁡t,γ(t)=tln⁡2(1+t)1+sin2t,p(t)=-3-sint,g(t)=3+cos⁡(t2+1)t5+2t3,h1(t)=t-5,h2(t)=t-7,h3(t)=t-3,f1(t)=t,f2(t)=t-1,f3(t)=t-10,h(t,u,v,w)=(1-t)u2v21+t3+sin2(tw),f(t,u,v,w)=2-u2-v2w31+t5+ln⁡(t2+3),Q1(t)=2M+5M41+t5+ln⁡(t2+3),Q2(t)=2M3(1+t)(2t3+4+tM)(1+t3)2,R1(t)=2+M2+M51+t5+ln⁡(t2+3),R2(t)=(1+t)M41+t3,(t,u,v,w)∈[t0,+∞)×R3.
It is easy to verify that (16)–(18) and (82) are satisfied. It follows from Theorem 8 that (11) possesses uncountably many bounded positive solutions in A(N,M). On the other hand, for any L∈((c1-1)N,(c2-1)M), there exist θ∈(0,1) and T>t0+|a|+τ+|T0| such that the Mann iterative sequence {xm}m∈N0 with mixed errors generated by (83) converges to a bounded positive solution x∈A(N,M) of (11) and has the error estimate (22), 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 (23) and (24).



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*Source: 105719-2014-05-29.xml*