# Existence and Iterative Algorithms of Positive Solutions for a Higher Order Nonlinear Neutral Delay Differential Equation

Authors: Zeqing Liu; Ling Guan; Sunhong Lee; Shin Min Kang
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

This paper is concerned with the higher order nonlinear neutral delay differential equation[a(t)(x(t)+b(t)x(t-τ))(m)](n-m)+[h(t,x(h1(t)),…,x(hl(t)))](i)+f(t,x(f1(t)),…,x(fl(t)))=g(t), for all t≥t0. Using the Banach fixed point theorem, we establish the existence results of uncountably many positive solutions for the equation, construct Mann iterative sequences for approximating these positive solutions, and discuss error estimates between the approximate solutions and the positive solutions. Nine examples are included to dwell upon the importance and advantages of our results.


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## Body

## 1. Introduction and Preliminaries

In recent years, the existence problems of nonoscillatory solutions for neutral delay differential equations of first, second, third, and higher order have been studied intensively by using fixed point theorems; see, for example, [1–12] and the references therein.Using the Banach, Schauder, and Krasnoselskii fixed point theorems, Zhang et al. [9] and Liu et al. [7] considered the existence of nonoscillatory solutions for the following first order neutral delay differential equations:
(1)[x(t)+P(t)x(t-τ)]′+Q1(t)x(t-τ1)-Q2(t)x(t-τ2)=0,∀t≥t0,[x(t)+c(t)x(t-τ)]′+h(t)f(x(t-σ1),x(t-σ2),…,x(t-σk))=g(t),∀t≥t0,
where P∈C([t0,+∞),ℝ∖{±1}) and c∈C([t0,+∞),ℝ). Making use of the Banach and Krasnoselskii fixed point theorems, Kulenović and Hadžiomerspahić [2] and Zhou [10] studied the existence of a nonoscillatory solution for the following second order neutral differential equations:
(2)[x(t)+cx(t-τ)]′′+Q1(t)x(t-σ1)-Q2(t)x(t-σ2)=0,∀t≥t0,[r(t)(x(t)+P(t)x(t-τ))′]′+∑i=1mQi(t)fi(x(t-σi))=0,∀t≥t0,
where c∈ℝ∖{±1} and P∈C([t0,∞),ℝ). Zhou and Zhang [11], Zhou et al. [12], and Liu et al. [4], respectively, investigated the existence of nonoscillatory solutions for the following higher order neutral delay differential equations:
(3)[x(t)+cx(t-τ)](n)+(-1)n+1[P(t)x(t-σ)-Q(t)x(t-δ)]=0,∀t≥t0,[x(t)+P(t)x(t-τ)](n)+∑i=1mQi(t)fi(x(t-σi))=g(t),∀t≥t0,[x(t)+ax(t-τ)](n)+(-1)n+1f(t,x(t-σ1),x(t-σ2),…,x(t-σk))=g(t),∀t≥t0,
where c∈ℝ∖{±1}, P∈C([t0,∞),ℝ) and a∈ℝ∖{-1}. Candan [1] proved the existence of a bounded nonoscillatory solution for the higher order nonlinear neutral differential equation:
(4)[r(t)(x(t)+P(t)x(t-τ))(n-1)]′+(-1)n[Q1(t)g1(x(t-σ1))-Q2(t)g2(x(t-σ2))-f(t)]=0,∀t≥t0,
where P∈C([t0,∞),ℝ∖{±1}).Motivated by the results in [1–12], in this paper we consider the following higher order nonlinear neutral delay differential equation:
(5)[a(t)(x(t)+b(t)x(t-τ))(m)](n-m)=+[h(t,x(h1(t)),…,x(hl(t)))](i)=+f(t,x(f1(t)),…,x(fl(t)))=g(t),∀t≥t0,
where m,n∈ℕ and i∈ℕ0 with i≤n-m-1, τ>0, a∈C([t0,+∞),ℝ∖{0}), b,g,fj,hj∈C([t0,+∞),ℝ), h∈Ci([t0,+∞)×ℝl,ℝ) and f∈C([t0,+∞)×ℝl,ℝ) with
(6)limt→+∞hj(t)=limt→+∞fj(t)=+∞,j∈{1,2,…,l}.
It is clear that (5) includes (1)–(4) as special cases. Utilizing the Banach fixed point theorem, we prove several existence results of uncountably many positive solutions for (5), construct a few Mann iterative schemes, and discuss error estimates between the sequences generated by the Mann iterative schemes and the positive solutions. Nine examples are given to show that the results presented in this paper extend substantially the existing ones in [1, 2, 4, 5, 8, 9, 11].Throughout this paper, we assume thatℝ=(-∞,+∞),ℝ+=[0,+∞), ℕ denotes the set of all positive integers, ℕ0=ℕ∪{0},
(7)Hj=1(m-1)!(n-m-j-1)!,j∈{0,i},γ=min{t0-τ,inft≥t0hj(t),inft≥t0fj(t):j∈{1,2,…,l}},CB([γ,+∞),ℝ) stands for the Banach space of all continuous and bounded functions in [γ,+∞) with norm ∥x∥=supt≥γ|x(t)|, and for any M>N>0(8)Ω1(N,M)={x∈CB([γ,+∞),ℝ):N≤x(t)≤M,∀t≥γ},Ω2(N,M)={x∈CB([γ,+∞),ℝ):Nb(t+τ)≤x(t)≤Mb(t+τ),∀t≥T;Nb(T+τ)ai≤x(t)≤Mb(T+τ),∀t∈[γ,T)},Ω3(N,M)={x∈CB([γ,+∞),ℝ):-Nb(t+τ)≤x(t)≤-Mb(t+τ),∀t≥T;-Nb(T+τ)aii≤x(t)≤-Mb(T+τ),∀t∈[γ,T)}.
It is easy to check that Ω1(N,M), Ω2(N,M) and Ω3(N,M) are closed subsets of CB([γ,+∞),ℝ).By a solution of (5), we mean a function x∈C([γ,+∞),ℝ) for some T>1+|t0|+τ+|γ|, such that a(t)(x(t)+b(t)x(t-τ))(m) are n-m times continuously differentiable in [T,+∞) and such that (5) is satisfied for t≥T.Lemma 1.
Letτ>0,c≥0, F∈C([c,+∞)3,ℝ+) and G∈C([c,+∞)2,ℝ+). Then(a)
∫c+∞∫r+∞∫u+∞rF(s,u,r)dsdudr<+∞⇔∑j=0∞∫c+jτ+∞∫r+∞∫u+∞F(s,u,r)dsdudr<+∞;(b)
∫c+∞∫u+∞uG(s,u)dsdu<+∞⇔∑j=0∞∫c+jτ+∞∫u+∞G(s,u)dsdu<+∞;(c)
if∫c+∞∫r+∞∫u+∞rF(s,u,r)dsdudr<+∞, then
(9)∑j=1∞∫t+jτ+∞∫r+∞∫u+∞F(s,u,r)dsdudr≤1τ∫t+τ+∞∫r+∞∫u+∞rF(s,u,r)dsdudr<+∞,∀t≥c;(d)
if∫c+∞∫u+∞uG(s,u)dsdu<+∞, then
(10)∑j=1∞∫t+jτ+∞∫u+∞G(s,u)dsdu≤1τ∫t+τ+∞∫u+∞uG(s,u)dsdu<+∞,∀t≥c.Proof.
Let[t] denote the largest integral number not exceeding t∈ℝ. Note that
(11)limr→+∞[(r-c)/τ]+1r=1τ,(12)c+nτ≤r<c+(n+1)τ⟺n≤r-cτ<n+1,∀n∈ℕ0.
Clearly (12) means that
(13)∑j=0∞∫c+jτ+∞‍∫r+∞‍∫u+∞‍F(s,u,r)dsdudr=∫c+∞‍∫r+∞‍∫u+∞‍F(s,u,r)dsdudr+∫c+τ+∞‍∫r+∞‍∫u+∞‍F(s,u,r)dsdudr+∫c+2τ+∞‍∫r+∞‍∫u+∞‍F(s,u,r)dsdudr+∫c+3τ+∞‍∫r+∞‍∫u+∞‍F(s,u,r)dsdudr+⋯=∫cc+τ‍∫r+∞‍∫u+∞‍F(s,u,r)dsdudr+2∫c+τc+2τ‍∫r+∞‍∫u+∞‍F(s,u,r)dsdudr+3∫c+2τc+3τ‍∫r+∞‍∫u+∞‍F(s,u,r)dsdudr+4∫c+3τc+4τ‍∫r+∞‍∫u+∞‍F(s,u,r)dsdudr+⋯=∑n=0∞∫c+nτc+(n+1)τ‍∫r+∞‍∫u+∞‍(n+1)F(s,u,r)dsdudr=∑n=0∞∫c+nτc+(n+1)τ‍∫r+∞‍∫u+∞‍([r-cτ]+1)×F(s,u,r)dsdudr=∫c+∞‍∫r+∞‍∫u+∞‍([r-cτ]+1)F(s,u,r)dsdudr.
Thus (a) follows from (11) and (13).
Assume that∫c+∞‍∫r+∞‍∫u+∞‍rF(s,u,r)dsdudr<+∞. As in the proof of (a), we infer that
(14)∑j=1∞‍∫t+jτ+∞‍∫r+∞‍∫u+∞‍F(s,u,r)dsdudr=∫t+τ+∞‍∫r+∞‍∫u+∞‍[r-tτ]F(s,u,r)dsdudr≤1τ∫t+τ+∞‍∫r+∞‍∫u+∞‍rF(s,u,r)dsdudr≤1τ∫c+∞‍∫r+∞‍∫u+∞‍rF(s,u,r)dsdudr<+∞,∀t≥c,
that is, (c) holds.
Similar to the proofs of (a) and (c), we conclude that (b) and (d) hold. This completes the proof.


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

Now we show the existence of uncountably many positive solutions for (5) and discuss the convergence of the Mann iterative sequences to these positive solutions.Theorem 2.
Assume that there exist three constantsM, N, and b0 and four functions P,Q,R,W∈C([t0,+∞),ℝ+) satisfying
(15)0<N<M,b0<M-N2M,|b(t)|≤b0eventually;(16)|f(t,u1,…,ul)-f(t,u-1,…,u-l)|≤P(t)max{|uj-u-j|:1≤j≤l},|h(t,u1,…,ul)-h(t,u-1,…,u-l)|≤R(t)max{|uj-u-j|:1≤j≤l},∀(t,u1,…,ul,u-1,…,u-l)∈[t0,+∞)×[N,M]2l;(17)|f(t,u1,…,ul)|≤Q(t),|h(t,u1,…,ul)|≤W(t),∀(t,u1,…,ul)∈[t0,+∞)×[N,M]l;(18)∫t0+∞‍∫u+∞‍|u|m-1|a(u)|[|s|n-m-1max{P(s),Q(s),|g(s)|}avagagga+|s|n-m-i-1max{R(s),W(s)}]dsdu<+∞.
Then
(
a
) for any L∈(b0M+N,(1-b0)M), there exist θ∈(0,1) and T>1+|t0|+τ+|γ| such that for each x0∈Ω1(N,M), the Mann iterative sequence {xk}k∈ℕ0 generated by the following scheme
(19)xk+1(t)={(1-αk)xk(t)+αk{∫t+∞‍∫u+∞‍(s-u)n-m-1(u-t)m-1a(u)L-b(t)xk(t-τ)+(-1)nH0×∫t+∞‍∫u+∞‍(s-u)n-m-1(u-t)m-1a(u)×[(xk(fl(s)))g(s)-f((fl(s))(xk(fl(s)))s,xk(f1(s)),…,xk(fl(s)))]dsdu+(-1)n-i-1Hi×∫t+∞‍∫u+∞‍(s-u)n-m-i-1(u-t)m-1a(u)×h((xk(hl(s)))xk(h1(s)),…,xk(hl(s)))dsdu∫t+∞‍(s-u)n-m-1(u-t)m-1a(u)},t≥T,k∈ℕ0,(1-αk)xk(T)+αk{(s-u)n-m-i-1(u-T)m-1a(u)L-b(T)xk(T-τ)+(-1)nH0×∫T+∞‍∫u+∞‍(s-u)n-m-1(u-T)m-1a(u)×[(xk(fl(s)))g(s)-f(s,xk(f1(s)),…,xk(fl(s)))]dsdu+(-1)n-i-1Hi×∫T+∞‍∫u+∞‍(s-u)n-m-i-1(u-T)m-1a(u)×h((xk(hl(s)))s,xk(h1(s)),…,xk(hl(s)))dsdu∫T+∞‍(s-u)n-m-i-1(u-T)m-1a(u)},t0≤t<T,k∈ℕ0
converges to a positive solution x∈Ω1(N,M) of (5) and has the following error estimate:
(20)∥xk+1-x∥≤e-(1-θ)∑p=0kαp∥x0-x∥,∀k∈ℕ0,
where {αk}k∈ℕ0 is an arbitrary sequence in [0,1] such that
(21)∑k=0∞αk=+∞;
(
b
) Equation (5) has uncountably many positive solutions in Ω1(N,M).Proof.
Firstly, we prove that (a) holds. SetL∈(b0M+N,(1-b0)M). From (15) and (18), we know that there exist θ∈(0,1) and T>1+|t0|+τ+|γ| satisfying
(22)|b(t)|≤b0,∀t≥T;(23)θ=b0+∫T+∞‍∫u+∞‍um-1|a(u)|[Hisn-m-i-1R(s)H0sn-m-1P(s)+Hisn-m-i-1R(s)]dsdu;(24)∫T+∞‍∫u+∞‍um-1|a(u)|[H0sn-m-1(|g(s)|+Q(s))+Hisn-m-i-1W(s)]dsdu<min{(1-b0)M-L,L-b0M-N}.
Define a mapping SL:Ω1(N,M)→CB([γ,+∞),ℝ) by
(25)SLx(t)={L-b(t)x(t-τ)+(-1)nH0×∫t+∞‍∫u+∞‍(s-u)n-m-1(u-t)m-1a(u)×[g(s)-f(((fl(s)))s,x(f1(s)),…,x(fl(s)))]dsdu+(-1)n-i-1Hi×∫t+∞‍∫u+∞‍(s-u)n-m-i-1(u-t)m-1a(u)×h(s,x(h1(s)),…,x(hl(s)))dsdu,t≥T,x∈Ω1(N,M),SLx(T),γ≤t<T,x∈Ω1(N,M).
It is obvious that SLx is continuous for each x∈Ω1(N,M). By means of (16), (22), (23), and (25), we deduce that for any x,y∈Ω1(N,M) and t≥T(26)|SLx(t)-SLy(t)|≤|b(t)||x(t-τ)-y(t-τ)|+H0∫t+∞‍∫u+∞‍(s-u)n-m-1(u-t)m-1|a(u)|×|f(s,x(f1(s)),…,x(fl(s)))-f(s,y(f1(s)),…,y(fl(s)))|dsdu+Hi∫t+∞‍∫u+∞‍(s-u)n-m-i-1(u-t)m-1|a(u)|×|h(s,x(h1(s)),…,x(hl(s)))-h(s,y(h1(s)),…,y(hl(s)))|dsdu≤b0∥x-y∥+∥x-y∥×∫T+∞‍∫u+∞‍um-1|a(u)|[H0sn-m-1P(s)+Hisn-m-i-1R(s)]dsdu=θ∥x-y∥,
which yields that
(27)∥SLx-SLy∥≤θ∥x-y∥,∀x,y∈Ω1(N,M).
On the basis of (17), (22), (24), and (25), we acquire that for any x∈Ω1(N,M) and t≥T(28)SLx(t)≤L+|b(t)|x(t-τ)+H0∫t+∞‍∫u+∞‍sn-m-1um-1|a(u)|×[|f(s,x(f1(s)),…,x(fl(s)))||g(s)|+|f(s,x(f1(s)),…,x(fl(s)))|]dsdu+Hi∫t+∞‍∫u+∞‍sn-m-i-1um-1|a(u)|×|h(s,x(h1(s)),…,x(hl(s)))|dsdu≤L+b0M+∫T+∞‍∫u+∞‍um-1|a(u)|×[H0sn-m-1(|g(s)|+Q(s))+Hisn-m-i-1W(s)]dsdu<L+b0M+min{(1-b0)M-L,L-b0M-N}≤M,SLx(t)≥L-|b(t)|x(t-τ)-H0∫t+∞‍∫u+∞‍sn-m-1um-1|a(u)|×[|f(s,x(f1(s)),…,x(fl(s)))||g(s)|+|f(s,x(f1(s)),…,x(fl(s)))|]dsdu-Hi∫t+∞‍∫u+∞‍sn-m-i-1um-1|a(u)|×|h(s,x(h1(s)),…,x(hl(s)))|dsdu≥L-b0M-∫T+∞‍∫u+∞‍um-1|a(u)|×[H0sn-m-1(|g(s)|+Q(s))+Hisn-m-i-1W(s)]dsdu>L-b0M-min{(1-b0)M-L,L-b0M-N}≥N,
which guarantee that SL(Ω1(N,M))⊆Ω1(N,M). Consequently, (27) gives that SL is a contraction mapping in Ω1(N,M) and it has a unique fixed point x∈Ω1(N,M). It is easy to see that x∈Ω1(N,M) is a positive solution of (5).
It follows from (19), (25), and (27) that
(29)|xk+1(t)-x(t)|=|∫t+∞‍∫u+∞‍(s-u)n-m-1(u-t)m-1a(u)(1-αk)xk(t)+αk{∫t+∞‍∫u+∞‍(s-u)n-m-1(u-t)m-1a(u)L-b(t)xk(t-τ)+(-1)nH0×∫t+∞‍∫u+∞‍(s-u)n-m-1(u-t)m-1a(u)×[(s,xk(f1(s)),…,xk(fl(s)))g(s)-f(s,xk(f1(s)),…,xk(fl(s)))]dsdu+(-1)n-i-1Hi×∫t+∞‍∫u+∞‍(s-u)n-m-i-1(u-t)m-1a(u)×h(s,xk(h1(s)),…,xk(hl(s)))dsdu∫u+∞‍(s-u)n-m-1(u-t)m-1a(u)}-x(t)∫t+∞‍∫u+∞‍(s-u)n-m-1(u-t)m-1a(u)|≤(1-αk)|xk(t)-x(t)|+αk|SLxk(t)-SLx(t)|≤(1-αk)|xk(t)-x(t)|+αkθ|xk(t)-x(t)|=(1-(1-θ)αk)|xk(t)-x(t)|≤e-(1-θ)αk∥xk-x∥≤e-(1-θ)∑p=0kαp∥x0-x∥,∀k∈ℕ0,t≥T,
which yields that
(30)∥xk+1-x∥≤e-(1-θ)∑p=0kαp∥x0-x∥,∀k∈ℕ0.
That is, (20) holds. Thus (20) and (21) ensure that limk→∞xk=x.
Secondly, we show that (b) holds. LetL1,L2∈(b0M+N,(1-b0)M) with L1≠L2. In light of (15) and (18), we know that for each p∈{1,2}, there exist θp∈(0,1), Tp and T* with Tp>1+|t0|+τ+|γ| and T*>max{T1,T2} satisfying (22)–(24) and
(31)∫T*+∞‍∫u+∞‍um-1|a(u)|[H0sn-m-1P(s)+Hisn-m-i-1R(s)]dsdu<|L1-L2|,
where θ and T are replaced by θp and Tp, respectively. Let the mapping SLp be defined by (25) with L and T replaced by Lp and Tp, respectively. As in the proof of (a), we deduce easily that the mapping SLp possesses a unique fixed point zp∈Ω1(N,M), that is, zp is a positive solution of (5) in Ω1(N,M). In order to prove (b), we need only to show that z1≠z2. In fact, (25) means that for each t≥T* and p∈{1,2}(32)zp(t)=Lp-b(t)zp(t-τ)+(-1)nH0×∫t+∞‍∫u+∞‍(s-u)n-m-1(u-t)m-1a(u)×[g(s)-f(s,zp(f1(s)),…,zp(fl(s)))]dsdu+(-1)n-i-1Hi∫t+∞‍∫u+∞‍(s-u)n-m-i-1(u-t)m-1a(u)×h(s,zp(h1(s)),…,zp(hl(s)))dsdu.
It follows from (16), (22), (31), and (32) that for each t≥T*(33)|z1(t)-z2(t)|≥|L1-L2|-|b(t)||z1(t-τ)-z2(t-τ)|-H0∫t+∞‍∫u+∞‍(s-u)n-m-1(u-t)m-1|a(u)|×|f(s,z1(f1(s)),…,z1(fl(s)))-f(s,z2(f1(s)),…,z2(fl(s)))|dsdu-Hi∫t+∞‍∫u+∞‍(s-u)n-m-i-1(u-t)m-1|a(u)|×|h(s,z1(h1(s)),…,z1(hl(s)))-h(s,z2(h1(s)),…,z2(hl(s)))|dsdu≥|L1-L2|-b0∥z1-z2∥-∥z1-z2∥∫T*+∞‍∫u+∞‍um-1|a(u)|×[H0sn-m-1P(s)+Hisn-m-i-1R(s)]dsdu≥|L1-L2|-(b0+|L1-L2|)∥z1-z2∥,
which implies that
(34)∥z1-z2∥≥|L1-L2|1+b0+|L1-L2|>0,
that is, z1≠z2. This completes the proof.Theorem 3.
Assume that there exist three constantsM, N, and b0 and four functions P,Q,R,W∈C([t0,+∞),ℝ+) satisfying (16)–(18) and
(35)0<N<M,b0<M-NM,0≤b(t)≤b0eventually.
Then(a)
for anyL∈(b0M+N,M), there exist θ∈(0,1) and T>1+|t0|+τ+|γ| such that for each x0∈Ω1(N,M), the Mann iterative sequence {xk}k∈ℕ0 generated by (19) converges to a positive solution x∈Ω1(N,M) of (5) and has the error estimate (20), where {αk}k∈ℕ0 is an arbitrary sequence in [0,1] satisfying (21);(b)
Equation (5) has uncountably many positive solutions in Ω1(N,M).Proof.
LetL∈(b0M+N,M). Equations (18) and (36) ensure that there exist θ∈(0,1) and T>1+|t0|+τ+|γ| satisfying (23),
(36)0≤b(t)≤b0,∀t≥T;(37)∫T+∞‍∫u+∞‍um-1|a(u)|[H0sn-m-1(|g(s)|+Q(s))+Hisn-m-i-1W(s)]dsdu<min{M-L,L-b0M-N}.
Define a mapping SL:Ω1(N,M)→CB([γ,+∞),ℝ) by (25). Obviously, SLx is continuous for every x∈Ω1(N,M). Using (16), (23), (25), and (36), we conclude that for any x,y∈Ω1(N,M) and t≥T(38)|SLx(t)-SLy(t)|≤b(t)|x(t-τ)-y(t-τ)|+H0∫t+∞‍∫u+∞‍(s-u)n-m-1(u-t)m-1|a(u)|×|f(s,x(f1(s)),…,x(fl(s)))-f(s,y(f1(s)),…,y(fl(s)))|dsdu+Hi∫t+∞‍∫u+∞‍(s-u)n-m-i-1(u-t)m-1|a(u)|×|h(s,x(h1(s)),…,x(hl(s)))-h(s,y(h1(s)),…,y(hl(s)))|dsdu≤b0∥x-y∥+∥x-y∥×∫T+∞‍∫u+∞‍um-1|a(u)|[H0sn-m-1P(s)+Hisn-m-i-1R(s)]dsdu=θ∥x-y∥,
which implies that (27) holds. In light of (17), (25), (36), and (37), we know that for any x∈Ω1(N,M) and t≥T(39)SLx(t)≤L+H0∫t+∞‍∫u+∞‍sn-m-1um-1|a(u)|×[(s,x(f1(s)),…,x(fl(s)))|g(s)|+|(x(fl(s)))f((fl(s))s,x(f1(s)),…,x(fl(s)))|]dsdu+Hi∫t+∞‍∫u+∞‍sn-m-i-1um-1|a(u)|×|h(s,x(h1(s)),…,x(hl(s)))|dsdu≤L+∫T+∞‍∫u+∞‍um-1|a(u)|[H0sn-m-1(|g(s)|+Q(s))+Hisn-m-i-1W(s)]dsdu<L+min{M-L,L-b0M-N}≤M,SLx(t)≥L-|b(t)|x(t-τ)-H0∫t+∞‍∫u+∞‍sn-m-1um-1|a(u)|×[|g(s)|+|(x(fl(s)))f(s,x(f1(s)),…,x(fl(s)))|]dsdu-Hi∫t+∞‍∫u+∞‍sn-m-i-1um-1|a(u)|×|h(s,x(h1(s)),…,x(hl(s)))|dsdu≥L-b0M-∫T+∞‍∫u+∞‍um-1|a(u)|×[H0sn-m-1(|g(s)|+Q(s))+Hisn-m-i-1W(s)]dsdu>L-b0M-min{M-L,L-b0M-N}≥N,
which mean that SL(Ω1(N,M))⊆Ω1(N,M). Equation (27) guarantees that SL is a contraction mapping in Ω1(N,M) and it possesses a unique fixed point x∈Ω1(N,M). As in the proof of Theorem 2, we infer that x∈Ω1(N,M) is a positive solution of (5). 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 three constantsM, N, and b0 and four functions P,Q,R,W∈C([t0,+∞),ℝ+) satisfying (16)–(18) and
(40)0<N<M,b0<M-NM,-b0≤b(t)≤0eventually.
Then(a)
for anyL∈(N,(1-b0)M), there exist θ∈(0,1) and T>1+|t0|+τ+|γ| such that for each x0∈Ω1(N,M), the Mann iterative sequence {xk}k∈ℕ0 generated by (19) converges to a positive solution x∈Ω1(N,M) of (5) and has the error estimate (20), where {αk}k∈ℕ0 is an arbitrary sequence in [0,1] satisfying (21);(b)
Equation (5) has uncountably many positive solutions in Ω1(N,M).Proof.
SetL∈(N,(1-b0)M). It follows from (18) and (40) that there exist θ∈(0,1) and T>1+|t0|+τ+|γ| satisfying (23),
(41)-b0≤b(t)≤0,∀t≥T;(42)∫T+∞‍∫u+∞‍um-1|a(u)|[H0sn-m-1(|g(s)|+Q(s))+Hisn-m-i-1W(s)]dsdu<min{L-N,(1-b0)M-L}.
Define a mapping SL:Ω1(N,M)→CB([γ,+∞),ℝ) by (25). Distinctly, SLx is continuous for each x∈Ω1(N,M). In terms of (16), (23), (25), and (41), we reason that for any x,y∈Ω1(N,M) and t≥T(43)|SLx(t)-SLy(t)|≤|b(t)||x(t-τ)-y(t-τ)|+H0∫t+∞‍∫u+∞‍(s-u)n-m-1(u-t)m-1|a(u)|×|f(s,x(f1(s)),…,x(fl(s)))-f(s,y(f1(s)),…,y(fl(s)))|dsdu+Hi∫t+∞‍∫u+∞‍(s-u)n-m-i-1(u-t)m-1|a(u)|×|h(s,x(h1(s)),…,x(hl(s)))-h(s,y(h1(s)),…,y(hl(s)))|dsdu≤b0∥x-y∥+∥x-y∥×∫T+∞‍∫u+∞‍um-1|a(u)|[H0sn-m-1P(s)+Hisn-m-i-1R(s)]dsdu=θ∥x-y∥,
which means that (27) holds. Owing to (17), (25), (41), and (42), we earn that for any x∈Ω1(N,M) and t≥T(44)SLx(t)≤L+|b(t)|x(t-τ)+H0∫t+∞‍∫u+∞‍sn-m-1um-1|a(u)|×[|g(s)|+|f(((x(fl(s))))s,x(f1(s)),…,x(fl(s)))|]dsdu+Hi∫t+∞‍∫u+∞‍sn-m-i-1um-1|a(u)|×|h(s,x(h1(s)),…,x(hl(s)))|dsdu≤L+b0M+∫T+∞‍∫u+∞‍um-1|a(u)|×[H0sn-m-1(|g(s)|+Q(s))+Hisn-m-i-1W(s)]dsdu<L+b0M+min{L-N,(1-b0)M-L}≤M,SLx(t)≥L-H0∫t+∞‍∫u+∞‍sn-m-1um-1|a(u)|×[|g(s)|+|f((x(fl(s)))s,x(f1(s)),…,x(fl(s)))|]dsdu-Hi∫t+∞‍∫u+∞‍sn-m-i-1um-1|a(u)|×|h(s,x(h1(s)),…,x(hl(s)))|dsdu≥L-∫T+∞‍∫u+∞‍um-1|a(u)|×[H0sn-m-1(|g(s)|+Q(s))+Hisn-m-i-1W(s)]dsdu>L-min{L-N,(1-b0)M-L}≥N,
which yield that SL(Ω1(N,M))⊆Ω1(N,M). Thus (27) ensures that SL is a contraction mapping in Ω1(N,M) and it owns a unique fixed point x∈Ω1(N,M). As in the proof of Theorem 2, we infer that x∈Ω1(N,M) is a positive solution of (5). The rest of the proof is parallel to that of Theorem 2, and hence is elided. This completes the proof.Theorem 5.
Assume that there exist three constantsM, N, and b0 and four functions P,Q,R,W∈C([t0,+∞),ℝ+) satisfying (18) and
(45)M>N>0,b0>MM-N,b(t)≥b0eventually;(46)|f(t,u1,…,ul)-f(t,u-1,…,u-l)|≤P(t)max{|uj-u-j|:1≤j≤l},|h(t,u1,…,ul)-h(t,u-1,…,u-l)|≤R(t)max{|uj-u-j|:1≤j≤l},∀(t,u1,…,ul,u-1,…,u-l)∈[t0,+∞)×[0,Mb0]2l;(47)|f(t,u1,…,ul)|≤Q(t),|h(t,u1,…,ul)|≤W(t),∀(t,u1,…,ul)∈[t0,+∞)×[0,Mb0]l.
Then
(
a
) for any L∈(N+M/b0,M), there exist θ∈(0,1) and T>1+|t0|+τ+|γ| such that for each x0∈Ω2(N,M), the Mann iterative sequence {xk}k∈ℕ0 generated by the following scheme
(48)xk+1(t)={(1-αk)xk(t)+αkb(t+τ)×{∫u+∞‍(s-u)n-m-1(u-t-τ)m-1a(u)L-xk(t+τ)+(-1)nH0×∫t+τ+∞‍∫u+∞‍(s-u)n-m-1(u-t-τ)m-1a(u)×[g(s)-f(s,xk(f1(s)),…,xk(fl(s)))]dsdu+(-1)n-i-1Hi×∫t+τ+∞‍∫u+∞‍(s-u)n-m-i-1(u-t-τ)m-1a(u)×h((xk(hl(s)))s,xk(h1(s)),…,xk(hl(s)))dsdu∫u+∞‍(s-u)n-m-1(u-t-τ)m-1a(u)},t≥T,k∈ℕ0,(1-αk)xk(T)+αkb(t+τ)×{∫u+∞‍(s-u)n-m-1(u-t-τ)m-1a(u)L-xk(T+τ)+(-1)nH0×∫T+τ+∞‍∫u+∞‍(s-u)n-m-1(u-T-τ)m-1a(u)×[g(s)-f(s,xk(f1(s)),…,xk(fl(s)))]dsdu+(-1)n-i-1Hi×∫T+τ+∞‍∫u+∞‍(s-u)n-m-i-1(u-T-τ)m-1a(u)×h((xk(hl(s)))s,xk(h1(s)),…,xk(hl(s)))dsdu∫u+∞‍(s-u)n-m-1(u-t-τ)m-1a(u)},γ≤t<T,k∈ℕ0
converges to a positive solution x∈Ω2(N,M) of (5) and has the error estimate (20), where {αk}k∈ℕ0 is an arbitrary sequence in [0,1] with (21);
(
b
) Equation (5) has uncountably many positive solutions in Ω2(N,M).Proof.
First of all, we show that (a) holds. SetL∈(N+M/b0,M). It follows from (18) and (45) that there exist θ∈(0,1) and T>1+|t0|+τ+|γ| such that
(49)b(t)≥b0,∀t≥T;(50)θ=1b0+1b0∫T+∞‍∫u+∞‍um-1|a(u)|×[H0sn-m-1P(s)+Hisn-m-i-1R(s)]dsdu;(51)∫T+∞‍∫u+∞‍um-1|a(u)|[H0sn-m-1(|g(s)|+Q(s))+Hisn-m-i-1W(s)]dsdu<min{M-L,L-Mb0-N}.
Define a mapping SL:Ω2(N,M)→CB([γ,+∞),ℝ) by
(52)SLx(t)={Lb(t+τ)-x(t+τ)b(t+τ)+(-1)nH0b(t+τ)×∫t+τ+∞‍∫u+∞‍(s-u)n-m-1(u-t-τ)m-1a(u)×[g(s)-f(s,x(f1(s)),…,x(fl(s)))]dsdu+(-1)n-i-1Hib(t+τ)×∫t+τ+∞‍∫u+∞‍(s-u)n-m-i-1(u-t-τ)m-1a(u)×h(s,x(h1(s)),…,x(hl(s)))dsdu,t≥T,x∈Ω2(N,M),SLx(T),γ≤t<T,x∈Ω2(N,M).
In light of (46), (49), (50), and (52), we conclude that for x,y∈Ω2(N,M) and t≥T(53)|SLx(t)-SLy(t)|≤1b(t+τ)|x(t+τ)-y(t+τ)|+H0b(t+τ)×∫t+τ+∞‍∫u+∞‍sn-m-1um-1|a(u)|×|f(s,x(f1(s)),…,x(fl(s)))-f(s,y(f1(s)),…,y(fl(s)))|dsdu+Hib(t+τ)×∫t+τ+∞‍∫u+∞‍sn-m-i-1um-1a(u)×|h(s,x(h1(s)),…,x(hl(s)))-h(s,y(h1(s)),…,y(hl(s)))|dsdu≤1b0∥x-y∥+1b0∥x-y∥×∫T+∞‍∫u+∞‍um-1|a(u)|[H0sn-m-1P(s)+Hisn-m-i-1R(s)]dsdu=θ∥x-y∥,
which yields that
(54)∥SLx-SLy∥≤θ∥x-y∥,∀x,y∈Ω2(N,M).
In view of (47), (49), (51), and (52), we obtain that for any x∈Ω2(N,M) and t≥T(55)SLx(t)≤1b(t+τ)×{∫u+∞‍sn-m-i-1um-1|a(u)|L-x(t+τ)+H0∫t+τ+∞‍∫u+∞‍sn-m-1um-1|a(u)|×[|g(s)|+|f(s,x(f1(s)),…,x(fl(s)))|]dsdu+Hi∫t+τ+∞‍∫u+∞‍sn-m-i-1um-1|a(u)|×|h(s,x(h1(s)),…,x(hl(s)))|dsdu∫t+τ+∞‍}≤1b(t+τ)×{∫T+∞‍∫u+∞‍um-1|a(u)|L-Nb(t+τ)+∫T+∞‍∫u+∞‍um-1|a(u)|×[H0sn-m-1(|g(s)|+Q(s))+Hisn-m-i-1W(s)]dsdu∫T+∞‍∫u+∞‍um-1|a(u)|}<1b(t+τ)×(L-Nb(t+τ)+min{M-L,L-Mb0-N})≤Mb(t+τ),SLx(t)≥1b(t+τ)×{sn-m-i-1um-1|a(u)|L-x(t+τ)-H0∫t+τ+∞‍∫u+∞‍sn-m-1um-1|a(u)|×[|g(s)|+|f(s,x(f1(s)),…,x(fl(s)))|]dsdu-Hi∫t+τ+∞‍∫u+∞‍sn-m-i-1um-1|a(u)|×|h(s,x(h1(s)),…,x(hl(s)))|dsdu∫t+τ+∞‍}≥1b(t+τ)×{∫u+∞‍um-1|a(u)|L-Mb(t+τ)-∫T+∞‍∫u+∞‍um-1|a(u)|×[H0sn-m-1(|g(s)|+Q(s))+Hisn-m-i-1W(s)]dsdu∫T+∞‍}>1b(t+τ)×(L-Mb(t+τ)-min{M-L,L-Mb0-N})≥Nb(t+τ),
which imply that SL(Ω2(N,M))⊆Ω2(N,M). It follows from (50) and (54) that SL is a contraction mapping in Ω2(N,M) and it has a unique fixed point x∈Ω2(N,M). It is clear that x∈Ω2(N,M) is a positive solution of (5).
Note that (48), (52), and (54) undertake that
(56)|xk+1(t)-x(t)|=|(1-αk)xk(t)+αkb(t+τ)×{∫u+∞‍(s-u)n-m-1(u-t-τ)m-1a(u)L-xk(t+τ)+(-1)nH0×∫t+τ+∞‍∫u+∞‍(s-u)n-m-1(u-t-τ)m-1a(u)×[g(s)-f(s,xk(f1(s)),…,sxk(fl(s)))]dsdu+(-1)n-i-1Hi×∫t+τ+∞‍∫u+∞‍(s-u)n-m-i-1(u-t-τ)m-1a(u)×h(s,xk(h1(s)),…,xk(hl(s)))dsdu∫u+∞‍(s-u)n-m-1(u-t-τ)m-1a(u)}-x(t)αkb(t+τ)|≤(1-αk)|xk(t)-x(t)|+αk|SLxk(t)-SLx(t)|≤(1-αk)|xk(t)-x(t)|+αkθ|xk(t)-x(t)|=(1-(1-θ)αk)|xk(t)-x(t)|≤e-(1-θ)αk∥xk-x∥≤e-(1-θ)∑p=0kαp∥x0-x∥,∀k∈ℕ0,t≥T,
which indicates that (20) holds. Thus (20) and (21) assure that limk→∞xk=x.
Next we prove that (b) holds. LetL1,L2∈(N+M/b0,M) with L1≠L2. As in the proof of (a) we infer that for each p∈{1,2} there exist θp∈(0,1), Tp>1+|t0|+τ+|γ| and SLp satisfying (49)–(52), where L, θ, T, and SL are replaced by Lp, θp, Tp, and SLp, respectively, and SLp has a unique fixed point zp∈Ω2(N,M), which is a positive solution of (5) in Ω2(N,M). It follows that for each t≥Tp and p∈{1,2}(57)zp(t)=Lpb(t+τ)-zp(t+τ)b(t+τ)+(-1)nH0b(t+τ)×∫t+τ+∞‍∫u+∞‍(s-u)n-m-1(u-t-τ)m-1a(u)×[g(s)-f(s,zp(f1(s)),…,zp(fl(s)))]dsdu+(-1)n-i-1Hib(t+τ)×∫t+τ+∞‍∫u+∞‍(s-u)n-m-i-1(u-t-τ)m-1a(u)×h(s,zp(h1(s)),…,zp(hl(s)))dsdu.
On behalf of proving (b), we need only to show that z1≠z2. Notice that (18) guarantees that there exits T3>max{T1,T2} satisfying
(58)∫T3+∞‍∫u+∞‍um-1|a(u)|[H0sn-m-1P(s)+Hisn-m-i-1R(s)]dsdu<|L1-L2|1+2∥z1-z2∥.
Due to (46), (51), (57), and (58), we conclude that for each t≥T3(59)|z1(t)-z2(t)+z1(t+τ)b(t+τ)-z2(t+τ)b(t+τ)|≥1b(t+τ)×(∫T3+∞‍|L1-L2|-H0∫T3+∞‍∫u+∞‍sn-m-1um-1|a(u)|×|f((s,z1(f1(s)),…,z1(fl(s)))-f(s,z2(f1(s)),…,z2(fl(s))))|dsdu-Hi∫T3+∞‍∫u+∞‍sn-m-i-1um-1|a(u)|×|h(s,z1(f1(s)),…,z1(fl(s)))-h(s,z2(f1(s)),…,z2(fl(s)))|dsdu∫u+∞‍sn-m-1um-1|a(u)|)≥1b(t+τ)×(∫T3+∞‍|L1-L2|-∥z1-z2∥×∫T3+∞‍∫u+∞‍um-1|a(u)|[H0sn-m-1P(s)+Hisn-m-i-1R(s)]dsdu∫T3+∞‍)>1b(t+τ)(|L1-L2|-∥z1-z2∥|L1-L2|1+2∥z1-z2∥)>|L1-L2|2b(t+τ)>0,
which yields that z1≠z2. This completes the proof.Theorem 6.
Assume that there exist three constantsM, N, and b0 and four functions P,Q,R,W∈C([t0,+∞),ℝ+) satisfying (18), (46), (47), and
(60)0<N<M,MM-N<b0,b(t)≤-b0eventually.
Then
(
a
) for any L∈(N,(1-1/b0)M), there exist θ∈(0,1) and T>1+|t0|+τ+|γ| such that for each x0∈Ω3(N,M), the Mann iterative sequence {xk}k∈ℕ0 generated by the following scheme
(61)xk+1(t)={(1-αk)xk(t)+αkb(t+τ)×{∫u+∞‍(s-u)n-m-1(u-t-τ)m-1a(u)-L-xk(t+τ)+(-1)nH0×∫t+τ+∞‍∫u+∞‍(s-u)n-m-1(u-t-τ)m-1a(u)×[g(s)-f(s,xk(f1(s)),…,xk(fl(s)))]dsdu+(-1)n-i-1Hi×∫t+τ+∞‍∫u+∞‍(s-u)n-m-i-1(u-t-τ)m-1a(u)×h(s,xk(h1(s)),…,xk(hl(s)))dsdu∫u+∞‍(s-u)n-m-1(u-t-τ)m-1a(u)},t≥T,k∈ℕ0,(1-αk)xk(T)+αkb(t+τ)×{∫u+∞‍(s-u)n-m-1(u-T-τ)m-1a(u)-L-xk(T+τ)+(-1)nH0×∫T+τ+∞‍∫u+∞‍(s-u)n-m-1(u-T-τ)m-1a(u)×[g(s)-f(s,xk(f1(s)),…,xk(fl(s)))]dsdu+(-1)n-i-1Hi×∫T+τ+∞‍∫u+∞‍(s-u)n-m-i-1(u-T-τ)m-1a(u)×h(s,xk(h1(s)),…,xk(hl(s)))dsdu∫u+∞‍(s-u)n-m-1(u-T-τ)m-1a(u)},γ≤t<T,k∈ℕ0
converges to a positive solution x∈Ω3(N,M) of (5) and has the error estimate (20), where {αk}k∈ℕ0 is an arbitrary sequence in [0,1] satisfying (21);
(
b
) Equation (5) has uncountably many positive solutions in Ω3(N,M).Proof.
PutL∈(N,(1-1/b0)M). It follows from (18) and (60) that there exist θ∈(0,1) and T>1+|t0|+τ+|γ| satisfying (50) and
(62)b(t)≤-b0,∀t≥T;∫T+∞‍∫u+∞‍um-1|a(u)|[H0sn-m-1(|g(s)|+Q(s))+Hisn-m-i-1W(s)]dsdu<min{M(1-1b0)-L,L-N}.
Define a mapping SL:Ω3(N,M)→CB([γ,+∞),ℝ) by
(63)SLx(t)={-Lb(t+τ)-x(t+τ)b(t+τ)+(-1)nH0b(t+τ)×∫t+τ+∞‍∫u+∞‍(s-u)n-m-1(u-t-τ)m-1a(u)×[x(fl(s))g(s)-f(s,x(f1(s)),…,x(fl(s)))]dsdu+(-1)n-i-1Hib(t+τ)×∫t+τ+∞‍∫u+∞‍(s-u)n-m-i-1(u-t-τ)m-1a(u)×h(s,x(h1(s)),…,x(hl(s)))dsdu,t≥T,x∈Ω3(N,M),SLx(T),γ≤t<T,x∈Ω3(N,M).
By virtue of (47), (62), and (63), we know that for any x∈Ω3(N,M) and t≥T(64)SLx(t)≤1b(t+τ)×(∫t+τ+∞‍∫u+∞‍(s-u)n-m-1(u-t-τ)m-1|a(u)|-L-x(t+τ)-H0×∫t+τ+∞‍∫u+∞‍(s-u)n-m-1(u-t-τ)m-1|a(u)|aaaaaaaaaaa×[|g(s)|+|f(s,x(f1(s)),…,aaaaaaaaaaax(fl(s)))|]dsdu-Hi∫t+τ+∞‍∫u+∞‍(s-u)n-m-i-1(u-t-τ)m-1|a(u)|×|h(s,x(h1(s)),…,x(hl(s)))|dsdu∫t+τ+∞‍∫u+∞‍(s-u)n-m-1(u-t-τ)m-1|a(u)|)≤1b(t+τ)×(-L+Mb(t+τ)-∫T+∞‍∫u+∞‍um-1|a(u)|×[H0sn-m-1(|g(s)|+Q(s))aaaaaaaaaaa+Hisn-m-i-1W(s)]dsdu∫u+∞‍um-1|a(u)|)<1b(t+τ)(-L+Mb(t+τ)-min{M(1-1b0)-L,L-N})≤-Mb(t+τ),SLx(t)≥1b(t+τ)×(∫t+τ+∞‍∫u+∞‍(s-u)n-m-1(u-t-τ)m-1|a(u)|-L-x(t+τ)+H0×∫t+τ+∞‍∫u+∞‍(s-u)n-m-1(u-t-τ)m-1|a(u)|aaaaaaaaaaa×[|g(s)|+|f(s,x(f1(s)),…,aaaaaaaaaaax(fl(s)))|]dsdu+Hi∫t+τ+∞‍∫u+∞‍(s-u)n-m-i-1(u-t-τ)m-1|a(u)|×|h(s,x(h1(s)),…,x(hl(s)))|dsdu∫t+τ+∞‍∫u+∞‍(s-u)n-m-1(u-t-τ)m-1|a(u)|)≥1b(t+τ)×(-L+Nb(t+τ)+∫T+∞‍∫u+∞‍um-1|a(u)|×[H0sn-m-1(|g(s)|+Q(s))aaaaaaaaaaa+Hisn-m-i-1W(s)]dsdu∫u+∞‍um-1|a(u)|)>1b(t+τ)(-L+Nb(t+τ)+min{M(1-1b0)-L,L-N})≥-Nb(t+τ),
which imply that SL(Ω3(N,M))⊆Ω3(N,M). The rest of the proof is identical with the proof of Theorem 5 and hence is omitted. This completes the proof.Theorem 7.
Letm≥2. Assume that there exist two constants M, N with M>N>0 and four functions P,Q,R,W∈C([t0,+∞),ℝ+) satisfying (16)–(18) and
(65)b(t)=1eventually.
Then
(
a
) for any L∈(N,M), there exist θ∈(0,1) and T>1+|t0|+τ+|γ| such that for each x0∈Ω1(N,M), the Mann iterative sequence {xk}k∈ℕ0 generated by the following scheme
(66)xk+1(t)={(1-αk)xk(t)+αk×{∑j=1∞∫t+(2j-1)τt+2jτ‍∫r+∞‍∫u+∞‍(s-u)n-m-1(u-r)m-2a(u)L+(-1)n(m-1)H0×∑j=1∞∫t+(2j-1)τt+2jτ‍∫r+∞‍∫u+∞(((s-u)n-m-1(u-r)m-2)‍×(a(u))-1)×[g(s)-f(s,xk(f1(s)),…,xk(fl(s)))]dsdudr+(-1)n-i-1(m-1)Hi×∑j=1∞∫t+(2j-1)τt+2jτ‍∫r+∞‍∫u+∞‍(((s-u)n-m-i-1(u-r)m-2)×(a(u))-1)×h(s,xk(h1(s)),…,xk(hl(s)))dsdudr∑j=1∞∫t+(2j-1)τt+2jτ‍∫r+∞‍∫u+∞‍(s-u)n-m-1(u-r)m-2a(u)},t≥T,k∈ℕ0,(1-αk)xk(T)+αk×{∑j=1∞∫T+(2j-1)τT+2jτ‍∫r+∞‍∫u+∞‍(s-u)n-m-1(u-r)m-2a(u)L+(-1)n(m-1)H0×∑j=1∞∫T+(2j-1)τT+2jτ‍∫r+∞‍∫u+∞‍(((s-u)n-m-1(u-r)m-2)×(a(u))-1)×[g(s)-f(s,xk(f1(s)),…,xk(fl(s)))]dsdudr+(-1)n-i-1(m-1)Hi×∑j=1∞∫T+(2j-1)τT+2jτ‍∫r+∞‍∫u+∞‍(((s-u)n-m-i-1(u-r)m-2)×(a(u))-1)×h(s,xk(h1(s)),…,xk(hl(s)))dsdudr∑j=1∞∫T+(2j-1)τT+2jτ‍∫r+∞‍∫u+∞‍(s-u)n-m-1(u-r)m-2a(u)},γ≤t<T,k∈ℕ0
converges to a positive solution x∈Ω1(N,M) of (5) and has the error estimate (20), where {αk}k∈ℕ0 is an arbitrary sequence in [0,1] with (21);
(
b
) Equation (5) has uncountably many positive solutions in Ω1(N,M).Proof.
LetL∈(N,M). It follows from (18) and (65) that there exist θ∈(0,1) and T>1+|t0|+τ+|γ| satisfying
(67)b(t)=1,∀t≥T;(68)θ=∫T+∞‍∫u+∞‍um-1|a(u)|[H0sn-m-1P(s)+Hisn-m-i-1R(s)]dsdu;(69)∫T+∞‍∫u+∞‍um-1|a(u)|[H0sn-m-1(|g(s)|+Q(s))+Hisn-m-i-1W(s)]dsdu<min{M-L,L-N}.
Define a mapping SL:Ω1(N,M)→CB([γ,+∞),ℝ) by
(70)SLx(t)={L+(-1)n(m-1)H0×∑j=1∞∫t+(2j-1)τt+2jτ‍∫r+∞‍∫u+∞‍(((s-u)n-m-1(u-r)m-2)×(a(u))-1)×[g(s)-f(s,x(f1(s)),…,x(fl(s)))]dsdudr+(-1)n-i-1(m-1)Hi×∑j=1∞∫t+(2j-1)τt+2jτ‍∫r+∞‍∫u+∞‍(((s-u)n-m-i-1(u-r)m-2)×(a(u))-1)×h(s,x(h1(s)),…,x(hl(s)))dsdudr,t≥T,x∈Ω1(N,M),SLx(T),γ≤t<T,x∈Ω1(N,M).
With a view to (16), (68), and (70), we derive that for any x,y∈Ω1(N,M) and t≥T(71)|SLx(t)-SLy(t)|≤(m-1)H0×∑j=1∞∫t+(2j-1)τt+2jτ‍∫r+∞‍∫u+∞‍(s-u)n-m-1(u-r)m-2|a(u)|×|f(s,x(f1(s)),…,x(fl(s)))-f(s,y(f1(s)),…,y(fl(s)))|dsdudr+(m-1)Hi×∑j=1∞∫t+(2j-1)τt+2jτ‍∫r+∞‍∫u+∞‍(s-u)n-m-i-1(u-r)m-2|a(u)|×|h(s,x(h1(s)),…,x(hl(s)))-h(s,y(h1(s)),…,y(hl(s)))|dsdudr≤(m-1)H0∥x-y∥×∫t+∞‍∫r+∞‍∫u+∞‍(s-u)n-m-1(u-r)m-2|a(u)|P(s)dsdudr+(m-1)Hi∥x-y∥×∫t+∞‍∫r+∞‍∫u+∞‍(s-u)n-m-i-1(u-r)m-2|a(u)|R(s)dsdudr=H0∥x-y∥×∫t+∞‍∫u+∞‍(s-u)n-m-1(u-t)m-1|a(u)|P(s)dsdu+Hi∥x-y∥×∫t+∞‍∫u+∞‍(s-u)n-m-i-1(u-t)m-1|a(u)|R(s)dsdu≤∥x-y∥×∫T+∞‍∫u+∞‍um-1|a(u)|[H0sn-m-1P(s)+Hisn-m-i-1R(s)]dsdu=θ∥x-y∥,
which gives (27). By virtue of (17), (69), and (70), we deduce that for any x∈Ω1(N,M) and t≥T(72)SLx(t)≤L+(m-1)H0×∫t+∞‍∫r+∞‍∫u+∞‍(s-u)n-m-1(u-r)m-2|a(u)|×[|g(s)|+|f(s,x(f1(s)),…,x(fl(s)))|]dsdudr+(m-1)Hi×∫t+∞‍∫r+∞‍∫u+∞‍(s-u)n-m-i-1(u-r)m-2|a(u)|×|h(s,x(h1(s)),…,x(hl(s)))|dsdudr≤L+(m-1)H0×∫t+∞‍∫r+∞‍∫u+∞‍(s-u)n-m-1(u-r)m-2|a(u)|×[|g(s)|+Q(s)]dsdudr+(m-1)Hi×∫t+∞‍∫r+∞‍∫u+∞‍(s-u)n-m-i-1(u-r)m-2|a(u)|W(s)dsdudr=L+H0∫t+∞‍∫u+∞‍(s-u)n-m-1(u-t)m-1|a(u)|×[|g(s)|+Q(s)]dsdu+Hi∫t+∞‍∫u+∞‍(s-u)n-m-i-1(u-t)m-1|a(u)|×W(s)dsdu≤L+∫T+∞‍∫u+∞‍um-1|a(u)|×[H0sn-m-1(|g(s)|+Q(s))+Hisn-m-i-1W(s)]dsdu<L+min{M-L,L-N}≤M,SLx(t)≥L-(m-1)H0×∫t+∞‍∫r+∞‍∫u+∞‍(s-u)n-m-1(u-r)m-2|a(u)|×[(x(fl(s)))|g(s)|+|f(x(fl(s))s,x(f1(s)),…,x(fl(s)))|]dsdudr-(m-1)Hi×∫t+∞‍∫r+∞‍∫u+∞‍(s-u)n-m-i-1(u-r)m-2|a(u)|×|h(s,x(h1(s)),…,x(hl(s)))|dsdudr≥L-∫T+∞‍∫u+∞‍um-1|a(u)|×[H0sn-m-1(|g(s)|+Q(s))+Hisn-m-i-1W(s)]dsdu>L-min{M-L,L-N}≥N,
which mean that SL(Ω1(N,M))⊆Ω1(N,M). Coupled with (27) and (68), we get that SL is a contraction mapping in Ω1(N,M) and it possesses a unique fixed point x∈Ω1(N,M). Clearly, x∈Ω1(N,M) is a positive solution of (5).
From (27), (66), and (70), we gain that
(73)|xk+1(t)-x(t)|=|×h(s,xk(h1(s)),…,xk(hl(s)))dsdudr∑j=1∞∫t+(2j-1)τt+2jτ‍∫r+∞‍∫u+∞‍(s-u)n-m-1(u-r)m-2a(u)}(1-αk)xk(t)+αk{∑j=1∞∫t+(2j-1)τt+2jτ‍∫r+∞‍∫u+∞‍(s-u)n-m-1(u-r)m-2a(u)L+(-1)n(m-1)H0×∑j=1∞∫t+(2j-1)τt+2jτ‍∫r+∞‍∫u+∞‍(s-u)n-m-1(u-r)m-2a(u)×[g(s)-f(s,xk(f1(s)),…,xk(fl(s)))]dsdudr+(-1)n-i-1(m-1)Hi×∑j=1∞∫t+(2j-1)τt+2jτ‍∫r+∞‍∫u+∞‍(s-u)n-m-i-1(u-r)m-2a(u)×h(s,xk(h1(s)),…,xk(hl(s)))dsdudr∑j=1∞∫t+(2j-1)τt+2jτ‍∫r+∞‍∫u+∞‍(s-u)n-m-1(u-r)m-2a(u)}-x(t)|≤(1-αk)|xk(t)-x(t)|+αk|SLxk(t)-SLx(t)|≤(1-αk)|xk(t)-x(t)|+αkθ|xk(t)-x(t)|=(1-(1-θ)αk)|xk(t)-x(t)|≤e-(1-θ)αk∥xk-x∥≤e-(1-θ)∑p=0kαp∥x0-x∥,∀k∈ℕ0,t≥T,
which yields (20). It follows from (20) and (21) that limk→∞xk=x.
Now we prove that (b) holds. LetL1,L2∈(N,M) and L1≠L2. As in the proof of (a), we conclude that for each p∈{1,2}, there exist θp∈(0,1), Tp>1+|t0|+τ+|γ| and SLp:Ω1(N,M)→Ω1(N,M) satisfying (69)–(77), where L, θ, T, and SL are replaced by Lp, θp, Tp, and SLp, respectively, and SLp has a unique fixed point zp∈Ω1(N,M), which is a positive solution of (5) in Ω1(N,M), that is,
(74)zp(t)=Lk+(-1)n(m-1)H0×∑j=1∞∫t+(2j-1)τt+2jτ‍∫r+∞‍∫u+∞‍(s-u)n-m-1(u-r)m-2a(u)×[g(s)-f(s,zp(f1(s)),…,zp(fl(s)))]dsdudr+(-1)n-i-1(m-1)Hi×∑j=1∞∫t+(2j-1)τt+2jτ‍∫r+∞‍∫u+∞‍(s-u)n-m-i-1(u-r)m-2a(u)×h(s,zp(h1(s)),…,zp(hl(s)))dsdudr,∀t≥Tp,p∈{1,2}.
For purpose of proving (b), we just need to show that z1≠z2. It follows from (16), (27), (68), and (74) that
(75)|z1(t)-z2(t)|≥|L1-L2|-(m-1)H0∥z1-z2∥×∫t+∞‍∫r+∞‍∫u+∞(((s-u)n-m-1(u-r)m-2)‍×(|a(u)|)-1)P(s)dsdudr-(m-1)Hi∥z1-z2∥×∫t+∞‍∫r+∞‍∫u+∞‍(((s-u)n-m-i-1(u-r)m-2)×(|a(u)|)-1)R(s)dsdudr≥|L1-L2|-∥z1-z2∥×∫max{T1,T2}+∞‍∫u+∞‍um-1|a(u)|×[H0sn-m-1P(s)+Hisn-m-i-1R(s)]dsdu>|L1-L2|-max{θ1,θ2}∥z1-z2∥,∀t≥max{T1,T2},
which yields that
(76)∥z1-z2∥≥|L1-L2|1+max{θ1,θ2}>0,
that is, z1≠z2. This completes the proof.Theorem 8.
Letm=1. Assume that there exist two constants M, N with M>N>0 and four functions P,Q,R,W∈C([t0,+∞),ℝ+) satisfying (16), (17), (65), and
(77)∫t0+∞‍∫u+∞‍1|a(u)|×[|s|n-2max{P(s),Q(s),|g(s)|}+|s|n-i-2max{R(s),W(s)}]dsdu<+∞.
Then
(
a
) for any L∈(N,M), there exist θ∈(0,1) and T>1+|t0|+τ+|γ| such that for each x0∈Ω1(N,M), the Mann iterative sequence {xk}k∈ℕ0 generated by the following scheme
(78)xk+1(t)={(1-αk)xk(t)+αk{∑j=1∞∫t+(2j-1)τt+2jτ‍∫u+∞‍(s-u)n-2a(u)L+(-1)n(n-2)!×∑j=1∞∫t+(2j-1)τt+2jτ‍∫u+∞‍(s-u)n-2a(u)×[g(s)-f(s,xk(f1(s)),…,xk(fl(s)))]dsdu+(-1)n-i-1(n-i-2)!×∑j=1∞∫t+(2j-1)τt+2jτ‍∫u+∞‍(s-u)n-i-2a(u)×h(s,xk(h1(s)),…,xk(hl(s)))dsdu∑j=1∞∫t+(2j-1)τt+2jτ‍∫u+∞‍(s-u)n-2a(u)},t≥T,k∈ℕ0,(1-αk)xk(T)+αk{∑j=1∞∫T+(2j-1)τT+2jτ‍∫u+∞‍(s-u)n-2a(u)L+(-1)n(n-2)!×∑j=1∞∫T+(2j-1)τT+2jτ‍∫u+∞‍(s-u)n-2a(u)×[g(s)-f(s,xk(f1(s)),…,xk(fl(s)))]dsdu+(-1)n-i-1(n-i-2)!×∑j=1∞∫T+(2j-1)τT+2jτ‍∫u+∞‍(s-u)n-i-2a(u)×h(s,xk(h1(s)),…,xk(hl(s)))dsdu(-1)n(n-2)!∑j=1∞∫T+(2j-1)τT+2jτ‍∫u+∞‍(s-u)n-2a(u)},γ≤t<T,k∈ℕ0
converges to a positive solution x∈Ω1(N,M) of (5) and has the error estimate (20), where {αk}k∈ℕ0 is an arbitrary sequence in [0,1] with (21);
(
b
) Equation (5) has uncountably many positive solutions in Ω1(N,M).Proof.
LetL∈(N,M). It follows from (65) and (77) that there exist θ∈(0,1) and T>1+|t0|+τ+|γ| satisfying (67),
(79)θ=∫T+∞‍∫u+∞‍1|a(u)|∫T+∞‍∫u+∞‍×[sn-2(n-2)!P(s)+sn-i-2(n-i-2)!R(s)]dsdu,(80)∫T+∞‍∫u+∞‍1|a(u)|[sn-2(n-2)!(|g(s)|+Q(s))∫T+∞‍∫u+∞‍1|a(u)|+sn-i-2(n-i-2)!W(s)]dsdu<min{M-L,L-N}.
Define a mapping SL:Ω1(N,M)→CB([γ,+∞),ℝ) by
(81)SLx(t)={L+(-1)n(n-2)!∑j=1∞∫t+(2j-1)τt+2jτ‍∫u+∞‍(s-u)n-2a(u)×[g(s)-f(s,x(f1(s)),…,x(fl(s)))]dsdu+(-1)n-i-1(n-i-2)!∑j=1∞∫t+(2j-1)τt+2jτ‍∫u+∞‍(s-u)n-i-2a(u)×h(s,x(h1(s)),…,x(hl(s)))dsdu,t≥T,x∈Ω1(N,M),SLx(T),γ≤t<T,x∈Ω1(N,M).
By virtue of (16), (79), and (81), we derive that for any x,y∈Ω1(N,M) and t≥T(82)|SLx(t)-SLy(t)|≤1(n-2)!∑j=1∞∫t+(2j-1)τt+2jτ‍∫u+∞‍(s-u)n-2|a(u)|×|f(s,x(f1(s)),…,x(fl(s)))-f(s,y(f1(s)),…,y(fl(s)))|dsdu+1(n-i-2)!∑j=1∞∫t+(2j-1)τt+2jτ‍∫u+∞‍(s-u)n-i-2|a(u)|×|h(s,x(h1(s)),…,x(hl(s)))-h(s,y(h1(s)),…,y(hl(s)))|dsdu≤1(n-2)!∥x-y∥∫T+∞‍∫u+∞‍(s-u)n-2|a(u)|P(s)dsdu+1(n-i-2)!∥x-y∥×∫T+∞‍∫u+∞‍(s-u)n-i-2|a(u)|R(s)dsdu≤∥x-y∥∫T+∞‍∫u+∞‍1|a(u)|[sn-2(n-2)!P(s)∥x-y∥∫T+∞‍∫u+∞‍1|a(u)|+sn-i-2(n-i-2)!R(s)]dsdu=θ∥x-y∥,
which gives (27). It follows from (17), (80), and (81) that for any x∈Ω1(N,M) and t≥T(83)SLx(t)≤L+1(n-2)!∑j=1∞∫t+(2j-1)τt+2jτ‍∫u+∞‍(s-u)n-2|a(u)|×[|g(s)|+|f(s,x(f1(s)),…,x(fl(s)))|]dsdu+1(n-i-2)!∑j=1∞∫t+(2j-1)τt+2jτ‍∫u+∞‍(s-u)n-i-2|a(u)|×|h(s,x(h1(s)),…,x(hl(s)))|dsdu≤L+∫T+∞‍∫u+∞‍1|a(u)|×[sn-2(n-2)!(|g(s)|+Q(s))+sn-i-2(n-i-2)!W(s)]dsdu<L+min{M-L,L-N}≤M,SLx(t)≥L-1(n-2)!∑j=1∞∫t+(2j-1)τt+2jτ‍∫u+∞‍(s-u)n-2|a(u)|×[|g(s)|+|f(s,x(f1(s)),…,x(fl(s)))|]dsdu-1(n-i-2)!∑j=1∞∫t+(2j-1)τt+2jτ‍∫u+∞‍(s-u)n-i-2|a(u)|×|h(s,x(h1(s)),…,x(hl(s)))|dsdu≥L-∫T+∞‍∫u+∞‍1|a(u)|×[sn-2(n-2)!(|g(s)|+Q(s))+sn-i-2(n-i-2)!W(s)]dsdu>L-min{M-L,L-N}≥N,
which mean that SL(Ω1(N,M))⊆Ω1(N,M). Combined with (27) and (79), we know that SL is a contraction mapping in Ω1(N,M) and it possesses a unique fixed point x∈Ω1(N,M). Obviously, x∈Ω1(N,M) is a positive solution of (5).
In light of (27), (78), and (81), we gain that
(84)|xk+1(t)-x(t)|=|∑j=1∞∫t+(2j-1)τt+2jτ‍∫u+∞‍(s-u)n-2a(u)(1-αk)xk(t)+αk{L+(-1)n(n-2)!∑j=1∞∫t+(2j-1)τt+2jτ‍∫u+∞‍(s-u)n-2a(u)×[g(s)-f(s,xk(f1(s)),…,xk(fl(s)))]dsdu+(-1)n-i-1(n-i-2)!∑j=1∞∫t+(2j-1)τt+2jτ‍∫u+∞‍(s-u)n-i-2a(u)×h(s,xk(h1(s)),…,xk(hl(s)))dsdu∑j=1∞∫t+(2j-1)τt+2jτ‍∫u+∞‍(s-u)n-2a(u)}-x(t)|≤(1-αk)|xk(t)-x(t)|+αk|SLxk(t)-SLx(t)|≤(1-αk)|xk(t)-x(t)|+αkθ|xk(t)-x(t)|=(1-(1-θ)αk)|xk(t)-x(t)|≤e-(1-θ)αk∥xk-x∥≤e-(1-θ)∑p=0kαp∥x0-x∥,∀k∈ℕ0,t≥T,
which yields (20). It follows from (20) and (21) that limk→∞xk=x.
Now we prove that (b) holds. LetL1,L2∈(N,M) and L1≠L2. As in the proof of (a), we conclude that for each p∈{1,2}, there exist θp∈(0,1), Tp>1+|t0|+τ+|γ| and SLp:Ω1(N,M)→Ω1(N,M) satisfying (67) and (79)–(81), where L, θ, T, and SL are replaced by Lp, θp, Tp, and SLp, respectively, and SLp has a unique fixed point zp∈Ω1(N,M), which is a positive solution of (5) in Ω1(N,M), that is,
(85)zp(t)=Lp+(-1)n(n-2)!×∑j=1∞∫t+(2j-1)τt+2jτ‍∫u+∞‍(s-u)n-2a(u)×[g(s)-f(s,zp(f1(s)),…,zp(fl(s)))]dsdu+(-1)n-i-1(n-i-2)!∑j=1∞∫t+(2j-1)τt+2jτ‍∫u+∞‍(s-u)n-i-2a(u)×h(s,zp(h1(s)),…,zp(hl(s)))dsdu,∀t≥Tp,p∈{1,2}.
In order to prove (b), we just need to show that z1≠z2. In view of (16), (27), (79), and (85), we get that
(86)|z1(t)-z2(t)|≥|L1-L2|-1(n-2)!∑j=1∞∫t+(2j-1)τt+2jτ‍∫u+∞‍(s-u)n-2|a(u)|×|f(s,z2(f1(s)),…,z2(fl(s)))-f(s,z1(f1(s)),…,z1(fl(s)))|dsdu-1(n-i-2)!∑j=1∞∫t+(2j-1)τt+2jτ‍∫u+∞‍(s-u)n-i-2|a(u)|×|h(s,z1(h1(s)),…,z1(hl(s)))-h(s,z2(h1(s)),…,z2(hl(s)))|dsdu≥|L1-L2|-∥z1-z2∥(n-2)!∫t+∞‍∫u+∞‍sn-2|a(u)|P(s)dsdu-∥z1-z2∥(n-i-2)!∫t+∞‍∫u+∞‍sn-i-2|a(u)|R(s)dsdu≥|L1-L2|-∥z1-z2∥×∫max{T1,T2}+∞‍∫u+∞‍1|a(u)|×[sn-2(n-2)!P(s)+sn-i-2(n-i-2)!R(s)]dsdu>|L1-L2|-max{θ1,θ2}∥z1-z2∥,∀t≥max{T1,T2},
which implies that
(87)∥z1-z2∥≥|L1-L2|1+max{θ1,θ2}>0,
that is, z1≠z2. This completes the proof.Theorem 9.
Letm≥2. Assume that there exist two constants M, N with M>N>0 and four functions P,Q,R,W∈C([t0,+∞),ℝ+) satisfying (16), (17),
(88)∫t0+∞‍∫r+∞‍∫u+∞‍|r||u|m|a(u)|×[|s|n-m-i-1max{R(s),W(s)}|s|n-m-1max{P(s),Q(s),|g(s)|}∫t0+∞‍∫r+∞‍∫u+∞‍+|s|n-m-i-1max{R(s),W(s)}]dsdudr<+∞,(89)b(t)=-1eventually.
Then
(
a
) for any L∈(N,M), there exist θ∈(0,1) and T>1+|t0|+τ+|γ| such that for each x0∈Ω1(N,M), the Mann iterative sequence {xk}k∈ℕ0 generated by the following scheme
(90)xk+1(t)={(1-αk)xk(t)+αk{∑j=1∞∫t+jτ+∞‍∫r+∞‍∫u+∞‍(s-u)n-m-1(u-r)m-2a(u)L+(-1)n-1(m-1)H0×∑j=1∞∫t+jτ+∞‍∫r+∞‍∫u+∞‍(s-u)n-m-1(u-r)m-2a(u)×[g(s)-f(s,xk(f1(s)),…,xk(fl(s)))]dsdudr+(-1)n-i(m-1)Hi×∑j=1∞∫t+jτ+∞‍∫r+∞‍∫u+∞‍(s-u)n-m-i-1(u-r)m-2a(u)×h(s,xk(h1(s)),…,xk(hl(s)))dsdudr∑j=1∞∫t+jτ+∞‍∫r+∞‍∫u+∞‍(s-u)n-m-1(u-r)m-2a(u)∑j=1∞∫t+jτ+∞‍∫r+∞‍∫u+∞‍(s-u)n-m-1(u-r)m-2a(u)},t≥T,k∈ℕ0,(1-αk)xk(T)+αk{∑j=1∞∫t+jτ+∞‍∫r+∞‍∫u+∞‍(s-u)n-m-1(u-r)m-2a(u)L+(-1)n-1(m-1)H0×∑j=1∞∫T+jτ+∞‍∫r+∞‍∫u+∞‍(s-u)n-m-1(u-r)m-2a(u)×[g(s)-f(s,xk(f1(s)),…,xk(fl(s)))]dsdudr+(-1)n-i(m-1)Hi×∑j=1∞∫T+jτ+∞‍∫r+∞‍∫u+∞‍(s-u)n-m-i-1(u-r)m-2a(u)×h(s,xk(h1(s)),…,xk(hl(s)))dsdudr∑j=1∞∫T+jτ+∞‍∫r+∞‍∫u+∞‍(s-u)n-m-1(u-r)m-2a(u)},γ≤t<T,k∈ℕ0
converges to a positive solution x∈Ω1(N,M) of (5) and has the error estimate (20), where {αk}k∈ℕ0 is an arbitrary sequence in [0,1] with (21);
(
b
) Equation (5) has uncountably many positive solutions in Ω1(N,M).Proof.
SetL∈(N,M). In view of (88) and (89), there exist θ∈(0,1) and T>1+|t0|+τ+|γ| such that
(91)b(t)=-1,∀t≥T;(92)θ=m-1τ∫T+∞‍∫r+∞‍∫u+∞‍rum-2|a(u)|×[H0sn-m-1P(s)+Hisn-m-i-1R(s)]dsdudr;(93)m-1τ∫T+∞‍∫r+∞‍∫u+∞‍rum-2|a(u)|×[H0sn-m-1(|g(s)|+Q(s))+Hisn-m-i-1W(s)]dsdudr<min{M-L,L-N}.
Define a mapping SL:Ω1(N,M)→CB([γ,+∞),ℝ) by
(94)SLx(t)={L+(-1)n-1(m-1)H0×∑j=1∞∫t+jτ+∞‍∫r+∞‍∫u+∞‍(s-u)n-m-1(u-r)m-2a(u)×[g(s)-f(s,x(f1(s)),…,x(fl(s)))]dsdudr+(-1)n-i(m-1)Hi×∑j=1∞‍∫t+jτ+∞‍∫r+∞‍∫u+∞‍(s-u)n-m-i-1(u-r)m-2a(u)×h(s,x(h1(s)),…,x(hl(s)))dsdudr,t≥T,x∈Ω1(N,M),SLx(T),γ≤t<T,x∈Ω1(N,M).
By virtue of (16), (92), (94), and Lemma 1, we acquire that for any x,y∈Ω1(N,M) and t≥T(95)|SLx(t)-SLy(t)|≤(m-1)H0∑j=1∞∫t+jτ+∞‍∫r+∞‍∫u+∞‍(s-u)n-m-1(u-r)m-2|a(u)|×|f(s,x(f1(s)),…,x(fl(s)))-f(s,y(f1(s)),…,y(fl(s)))|dsdudr+(m-1)Hi∑j=1∞∫t+jτ+∞‍∫r+∞‍∫u+∞‍(s-u)n-m-i-1(u-r)m-2|a(u)|×|h(s,x(h1(s)),…,x(hl(s)))-h(s,y(h1(s)),…,y(hl(s)))|dsdudr≤(m-1)H0τ∥x-y∥×∫T+∞‍∫r+∞‍∫u+∞‍(s-u)n-m-1(u-r)m-2r|a(u)|P(s)dsdudr+(m-1)Hiτ∥x-y∥×∫T+∞‍∫r+∞‍∫u+∞‍(s-u)n-m-i-1(u-r)m-2r|a(u)|R(s)dsdudr≤m-1τ∫T+∞‍∫r+∞‍∫u+∞‍rum-2|a(u)|×[H0sn-m-1P(s)+Hisn-m-i-1R(s)]dsdudr=θ∥x-y∥,
which yields that (27) holds. From (17), (94), (98), and Lemma 1, we obtain that for any x∈Ω1(N,M) and t≥T(96)|SLx(t)-L|≤(m-1)H0∑j=1∞∫t+jτ+∞‍∫r+∞‍∫u+∞‍(s-u)n-m-1(u-r)m-2|a(u)|×[|g(s)|+|f(s,x(f1(s)),…,x(fl(s)))|]dsdudr+(m-1)Hi∑j=1∞∫t+jτ+∞‍∫r+∞‍∫u+∞‍(s-u)n-m-i-1(u-r)m-2|a(u)|×|h(s,x(h1(s)),…,x(hl(s)))|dsdudr≤(m-1)H0τ∫T+∞‍∫r+∞‍∫u+∞‍rsn-m-1um-2|a(u)|&≤(m-1)H0τ∫T+∞‍∫r+∞‍×[|g(s)|+Q(s)]dsdudr+(m-1)Hiτ∫T+∞‍∫r+∞‍∫u+∞‍rsn-m-i-1um-2|a(u)|+(m-1)Hiτ∫T+∞‍∫r+∞‍∫u+∞‍×W(s)dsdudr=m-1τ∫T+∞‍∫r+∞‍∫u+∞‍rum-2|a(u)|×[H0sn-m-1(|g(s)|+Q(s))+Hisn-m-i-1W(s)]dsdudr<min{M-L,L-N},
which means that SL(Ω1(N,M))⊆Ω1(N,M). It follows from (27) and (92) that SL is a contraction mapping and it has a unique fixed point x∈Ω1(N,M). It is clear that x∈Ω1(N,M) is a positive solution of (5).
On the basis of (27), (90), and (94), we deduce that
(97)|xk+1(t)-x(t)|=|∑j=1∞∫t+jτ+∞‍∫r+∞‍∫u+∞‍(s-u)n-m-1(u-r)m-2a(u)(1-αk)xk(t)+αk{∑j=1∞∫t+jτ+∞‍∫r+∞‍∫u+∞‍(s-u)n-m-1(u-r)m-2a(u)L+(-1)n-1(m-1)H0×∑j=1∞∫t+jτ+∞‍∫r+∞‍∫u+∞‍(s-u)n-m-1(u-r)m-2a(u)×[g(s)-f(s,xk(f1(s)),…,xk(fl(s)))]dsdudr+(-1)n-i(m-1)Hi×∑j=1∞∫t+jτ+∞‍∫r+∞‍∫u+∞‍(s-u)n-m-i-1(u-r)m-2a(u)×h(s,xk(h1(s)),…,xk(hl(s)))dsdudr∑j=1∞∫t+jτ+∞‍∫r+∞‍∫u+∞‍(s-u)n-m-i-1(u-r)m-2a(u)∑j=1∞‍∫t+jτ+∞‍∫r+∞‍∫u+∞‍}-x(t)∑j=1∞∫t+jτ+∞‍∫r+∞‍∫u+∞‍(s-u)n-m-i-1(u-r)m-2a(u)|≤(1-αk)|xk(t)-x(t)|+αk|SLxk(t)-SLx(t)|≤(1-αk)|xk(t)-x(t)|+αkθ|xk(t)-x(t)|=(1-(1-θ)αk)|xk(t)-x(t)|≤e-(1-θ)αk∥xk-x∥≤e-(1-θ)∑p=0kαp∥x0-x∥,∀k∈ℕ0,t≥T,
which signifies that (20) holds. It follows from (20) and (21) and that limk→∞xk=x.
Now we show that (b) holds. LetL1,L2∈(N,M) and L1≠L2. As in the proof of (a), we conclude that for each p∈{1,2}, there exist θp∈(0,1), Tp>1+|t0|+τ+|γ| and SLp:Ω1(N,M)→Ω1(N,M) satisfying (91)–(94), where L, θ, T, and SL are replaced by Lp, θp, Tp, and SLp, respectively, and SLp has a unique fixed point zp∈Ω1(N,M), which is a positive solution of (5) in Ω1(N,M), that is,
(98)zp(t)=Lp+(-1)n-1(m-1)H0×∑j=1∞∫t+jτ+∞‍∫r+∞‍∫u+∞‍(s-u)n-m-1(u-r)m-2a(u)×[g(s)-f(s,zp(f1(s)),…,zp(fl(s)))]dsdudr+(-1)n-i(m-1)Hi×∑j=1∞∫t+jτ+∞‍∫r+∞‍∫u+∞‍(s-u)n-m-i-1(u-r)m-2a(u)×h(s,zp(h1(s)),…,zp(hl(s)))dsdudr,∀t≥Tp,p∈{1,2}.
In order to prove (b), it is sufficient to show that z1≠z2. Note that (16), (92), (98), and Lemma 1 lead to
(99)|z1(t)-z2(t)|≥|L1-L2|-(m-1)H0τ∥z1-z2∥×∫t+τ+∞‍∫r+∞‍∫u+∞‍rsn-m-1um-2|a(u)|P(s)dsdudr-(m-1)Hiτ∥z1-z2∥×∫t+τ+∞‍∫r+∞‍∫u+∞‍rsn-m-i-1um-2|a(u)|R(s)dsdudr≥|L1-L2|-(m-1)∥z1-z2∥τ×∫max{T1,T2}+∞‍∫r+∞‍∫u+∞‍rum-2|a(u)|×[H0sn-m-1P(s)+Hisn-m-i-1R(s)]dsdudr>|L1-L2|-max{θ1,θ2}∥z1-z2∥,∀t≥max{T1,T2},
which means that
(100)∥z1-z2∥≥|L1-L2|1+max{θ1,θ2}>0,
that is, z1≠z2. This completes the proof.Theorem 10.
Letm=1. Assume that there exist two constants M, N with M>N>0 and four functions P,Q,R,W∈C([t0,+∞),ℝ+) satisfying (16), (17), (89), and
(101)∫t0+∞‍∫u+∞‍|u||a(u)|[|s|n-2max{P(s),Q(s),|g(s)|}+|s|n-i-2max{R(s),W(s)}]dsdu<+∞.
Then
(
a
) for any L∈(N,M), there exist θ∈(0,1) and T>1+|t0|+τ+|γ| such that for each x0∈Ω1(N,M), the Mann iterative sequence {xk}k∈ℕ0 generated by the following scheme
(102)xk+1(t)={(1-αk)xk(t)+αk{L+(-1)n-1(n-2)!∑j=1∞∫t+jτ+∞‍∫u+∞‍(s-u)n-2a(u)×[g(s)-f(s,xk(f1(s)),…,xk(fl(s)))]dsdu+(-1)n-i(n-i-2)!∑j=1∞∫t+jτ+∞‍∫u+∞‍(s-u)n-i-2a(u)×h(s,xk(h1(s)),…,xk(hl(s)))dsdu(-1)n-1(n-2)!∑j=1∞∫t+jτ+∞‍∫u+∞‍(s-u)n-2a(u)},t≥T,k∈ℕ0,(1-αk)xk(T)+αk{L+(-1)n-1(n-2)!∑j=1∞∫T+jτ+∞‍∫u+∞‍(s-u)n-2a(u)×[g(s)-f(s,xk(f1(s)),…,xk(fl(s)))]dsdu+(-1)n-i(n-i-2)!∑j=1∞∫T+jτ+∞‍∫u+∞‍(s-u)n-i-2a(u)×h(s,xk(h1(s)),…,xk(hl(s)))dsdu(-1)n-1(n-2)!∑j=1∞∫T+jτ+∞‍∫u+∞‍(s-u)n-2a(u)},γ≤t<T,k∈ℕ0,
converges to a positive solution x∈Ω1(N,M) of (5) and has the error estimate (20), where {αk}k∈ℕ0 is an arbitrary sequence in [0,1] with (21);
(
b
) Equation (5) has uncountably many positive solutions in Ω1(N,M).Proof.
SetL∈(N,M). Due to (101), there exist θ∈(0,1) and T>1+|t0|+τ+|γ| satisfying (91),
(103)θ=1τ∫T+∞‍∫u+∞‍u|a(u)|×[sn-2(n-2)!P(s)+sn-i-2(n-i-2)!R(s)]dsdu,(104)1τ∫T+∞‍∫u+∞‍u|a(u)|×[sn-2(n-2)!(|g(s)|+Q(s))+sn-i-2(n-i-2)!W(s)]dsdu<min{M-L,L-N}.
Define a mapping SL:Ω1(N,M)→CB([γ,+∞),ℝ) by
(105)SLx(t)={L+(-1)n-1(n-2)!∑j=1∞∫t+jτ+∞‍∫u+∞‍(s-u)n-2a(u)×[g(s)-f(s,x(f1(s)),…,x(fl(s)))]dsdu+(-1)n-i(n-i-2)!∑j=1∞∫t+jτ+∞‍∫u+∞‍(s-u)n-i-2a(u)×h(s,x(h1(s)),…,x(hl(s)))dsdu,t≥T,x∈Ω1(N,M),SLx(T),γ≤t<T,x∈Ω1(N,M).
In view of (16), (103), (105), and Lemma 1, we achieve that for any x,y∈Ω1(N,M) and t≥T(106)|SLx(t)-SLy(t)|≤1(n-2)!∑j=1∞∫t+jτ+∞‍∫u+∞‍(s-u)n-2|a(u)|×|f(s,x(f1(s)),…,x(fl(s)))-f(s,y(f1(s)),…,y(fl(s)))|dsdu+1(n-i-2)!∑j=1∞∫t+jτ+∞‍∫u+∞‍(s-u)n-i-2|a(u)|×|h(s,x(h1(s)),…,x(hl(s)))-h(s,y(h1(s)),…,y(hl(s)))|dsdu≤∥x-y∥τ(n-2)!∫t+τ+∞‍∫u+∞‍usn-2|a(u)|P(s)dsdu+∥x-y∥τ(n-i-2)!∫t+τ+∞‍∫u+∞‍usn-i-2|a(u)|R(s)dsdu≤∥x-y∥τ∫T+∞‍∫u+∞‍u|a(u)|×[sn-2(n-2)!P(s)+sn-i-2(n-i-2)!R(s)]dsdu=θ∥x-y∥,
which means that (27) holds. It follows from (17), (104), (105), and Lemma 1 that for any x∈Ω1(N,M) and t≥T(107)|SLx(t)-L|≤1(n-2)!∑j=1∞∫t+jτ+∞‍∫u+∞‍(s-u)n-2|a(u)|×[|g(s)|+|f(s,x(f1(s)),…,x(fl(s)))|]dsdu+1(n-i-2)!∑j=1∞∫t+jτ+∞‍∫u+∞‍(s-u)n-i-2|a(u)|×|h(s,x(h1(s)),…,x(hl(s)))|dsdu≤1τ(n-2)!∫t+τ+∞‍∫u+∞‍usn-2|a(u)|(|g(s)|+Q(s))dsdu+1τ(n-i-2)!∫t+τ+∞‍∫u+∞‍usn-i-2|a(u)|W(s)dsdu≤1τ∫T+∞‍∫u+∞‍u|a(u)|×[sn-2(n-2)!(|g(s)|+Q(s))+sn-i-2(n-i-2)!W(s)]dsdu<min{M-L,L-N},
which means that SL(Ω1(N,M))⊆Ω1(N,M). Coupled with (27), we know that SL is a contraction mapping and it has a unique fixed point x∈Ω1(N,M). It follows that x∈Ω1(N,M) is a positive solution of (5).
In view of (27), (102), and (105), we deduce that
(108)|xk+1(t)-x(t)|=|∑j=1∞∫t+jτ+∞‍∫u+∞‍(s-u)n-2a(u)(1-αk)xk(t)+αk{∑j=1∞∫t+jτ+∞‍∫u+∞‍(s-u)n-2a(u)L+(-1)n-1(n-2)!∑j=1∞∫t+jτ+∞‍∫u+∞‍(s-u)n-2a(u)×[g(s)-f(s,xk(f1(s)),…,xk(fl(s)))]dsdu+(-1)n-i(n-i-2)!∑j=1∞∫t+jτ+∞‍∫u+∞‍(s-u)n-i-2a(u)×h(s,xk(h1(s)),…,xk(hl(s)))dsdu(-1)n-1(n-2)!∑j=1∞∫t+jτ+∞‍∫u+∞‍(s-u)n-2a(u)}-x(t)|≤(1-αk)|xk(t)-x(t)|+αk|SLxk(t)-SLx(t)|≤(1-αk)|xk(t)-x(t)|+αkθ|xk(t)-x(t)|=(1-(1-θ)αk)|xk(t)-x(t)|≤e-(1-θ)αk∥xk-x∥≤e-(1-θ)∑p=0kαp∥x0-x∥,∀k∈ℕ0,t≥T,
which signifies that (20) holds. It follows from (20) and (21) that limk→∞xk=x.
Now we show that (b) holds. LetL1,L2∈(N,M) and L1≠L2. As in the proof of (a), we conclude that for each p∈{1,2}, there exist θp∈(0,1), Tp>1+|t0|+τ+|γ| and SLp:Ω1(N,M)→Ω1(N,M) satisfying (91) and (103)–(105), where L, θ, T, and SL are replaced by Lp, θp, Tp and SLp, respectively, and SLp has a unique fixed point zp∈Ω1(N,M), which is a positive solution of (5) in Ω1(N,M). It follows that for any t≥Tp and p∈{1,2}(109)zp(t)=Lp+(-1)n-1(n-2)!∑j=1∞∫t+jτ+∞‍∫u+∞‍(s-u)n-2a(u)×[g(s)-f(s,zp(f1(s)),…,zp(fl(s)))]dsdu+(-1)n-i(n-i-2)!∑j=1∞∫t+jτ+∞‍∫u+∞‍(s-u)n-i-2a(u)×h(s,zp(h1(s)),…,zp(hl(s)))dsdu.
In order to prove (b), we just need to show that z1≠z2. Notice that (16), (103), (109), and Lemma 1 ensure that
(110)|z1(t)-z2(t)|≥|L1-L2|-1(n-2)!∑j=1∞∫t+jτ+∞‍∫u+∞‍(s-u)n-2|a(u)|×|f(s,z2(f1(s)),…,z2(fl(s)))-f(s,z1(f1(s)),…,z1(fl(s)))|dsdu-1(n-i-2)!∑j=1∞∫t+jτ+∞‍∫u+∞‍(s-u)n-i-2|a(u)|×|h(s,z1(h1(s)),…,z1(hl(s)))-h(s,z2(h1(s)),…,z2(hl(s)))|dsdu≥|L1-L2|-∥z1-z2∥τ(n-2)!∫t+τ+∞‍∫u+∞‍usn-2|a(u)|P(s)dsdu-∥z1-z2∥τ(n-i-2)!∫t+τ+∞‍∫u+∞‍usn-i-2|a(u)|R(s)dsdu≥|L1-L2|-∥z1-z2∥τ∫max{T1,T2}+∞‍∫u+∞‍u|a(u)|×[sn-2(n-2)!P(s)+sn-i-2(n-i-2)!R(s)]dsdu>|L1-L2|-max{θ1,θ2}∥z1-z2∥,
which yields that
(111)∥z1-z2∥≥|L1-L2|1+max{θ1,θ2}>0,
that is, z1≠z2. This completes the proof.


## 3. Remark and Examples

Remark 11.
Theorems2–10 extend, improve, and unifies Theorems 1–4 in [1], the theorem in [2], Theorems 2.1–2.4 in [4], Theorems 2.1–2.5 in [5, 8], Theorems 1–3 in [9], and Theorems 1–4 in [11], respectively. The examples below prove that Theorems 2–10 extend substantially the corresponding results in [1, 2, 4, 5, 8, 9, 11]. Note that none of the known results can be applied to these examples.Example 12.
Consider the higher order nonlinear neutral delay differential equation(112)[(tm+1+1)(x(t)+sin(2t2)-cos(t5-1)7+2sin(8t3+2t-1)x(t-τ))(m)](n-m)+(t2x(t-3)x2(t-4)tn-m-i+3+t2+1)(i)+t3x3(t2-t)-x4(t-1)tn-m+4+t+2=tln(1+t2)-cos2(t2-t+1)t2n-m+3+1,∀t≥2,
where τ>0 and i≤n-m-1. Let l=2, t0=2, γ=min{2-τ,-2}, M=10, N=1, b0=2/5 and
(113)h1(t)=t-3,h2(t)=t-4,f1(t)=t2-t,f2(t)=t-1,a(t)=tm+1+1,b(t)=sin(2t2)-cos(t5-1)7+2sin(8t3+2t-1),h(t,u,v)=t2uv2tn-m-i+3+t2+1,f(t,u,v)=t3u3-v4tn-m+4+t+2,g(t)=tln(1+t2)-cos2(t2-t+1)t2n-m+3+1,P(t)=M2(3t3+4M)tn-m+4+t+2,Q(t)=M3(t3+M)tn-m+4+t+2,R(t)=3M2t2tn-m-i+3+t2+1,W(t)=M3t2tn-m-i+3+t2+1,∀(t,u,v)∈[t0,+∞)×[N,M]2.
It is easy to verify that the conditions of Theorem 2 are satisfied. Thus Theorem 2 ensures that (112) has uncountably many positive solutions in Ω1(1,10), and for any L∈(5,6), there exist θ∈(0,1) and T>1+|t0|+τ+|γ| such that the Mann iterative sequence {xk}k∈ℕ0 generated by (19) and (21) converges to a positive solution x∈Ω1(1,10) of (112) and has the error estimate (20).Example 13.
Consider the higher order nonlinear neutral delay differential equation(114)[(tn+1)(x(t)+3t24t2+3x(t-τ))(m)](n-m)+(t2x(tlnt)-(t+1)x(tlnt)x(2t)tn+m-i+3)(i)+(2-t2)arctant+tx3(t3+t2)x2(t2)tn+m+3+x2(t2)=1-8t3+13t5+5t6cos(t3-1)t2n+m+4,∀t≥1,
where τ>0 and i≤n-m-1. Let l=2, t0=1, γ=min{1-τ,0}, M=6, N=1, b0=3/4 and
(115)h1(t)=tlnt,h2(t)=2t,f1(t)=t3+t2,f2(t)=t2,a(t)=tn+1,b(t)=3t24t2+3,h(t,u,v)=t2u-(t+1)uvtn+m-i+3,f(t,u,v)=(2-t2)arctant+tu3v2tn+m+3+v2,g(t)=1-8t3+13t5+5t6cos(t3-1)t2n+m+4,P(t)=5M4tn+m+4+2M(2+t2)arctant+5M6t(tn+m+3+N2)2,Q(t)=(2+t2)arctant+M5ttn+m+3+N2,R(t)=t2+2M(t+1)tn+m-i+3,W(t)=Mt2+M2(t+1)tn+m-i+3,∀(t,u,v)∈[t0,+∞)×[N,M]2.
It is easy to check that the conditions of Theorem 3 are satisfied. Therefore (114) has uncountably many positive solutions in Ω1(1,6), and for any L∈(11/2,6), there exist θ∈(0,1) and T>1+|t0|+τ+|γ| such that the Mann iterative sequence {xk}k∈ℕ0 generated by (19) and (21) converges to a positive solution x∈Ω1(1,6) of (114) and has the error estimate (20).Example 14.
Consider the higher order nonlinear neutral delay differential equation(116)[(t+1)(x(t)-arctant2x(t-τ))(m)](n-m)+(t2x(t3+t)-x2(t3-1)t2n+m-i+3+x2(t3+t))(i)+tx(t-1)sin(tx(t-sint))t2n+m+3+t3+2=tt+1sin2(t2+2t+1)t2n+m+3+1,∀t≥0,
where τ>0 and i≤n-m-1. Let l=2, t0=0, γ=min{-τ,-1}, M=8, N=1/2, b0=7/8 and
(117)h1(t)=t3+t,h2(t)=t3-1,f1(t)=t-1,f2(t)=t-sint,a(t)=t+1,b(t)=-12arctant,h(t,u,v)=t2u-v2t2n+m-i+3+u2,f(t,u,v)=tusin(tv)t2n+m+3+t3+2,g(t)=tt+1sin2(t2+2t+1)t2n+m+3+1,P(t)=Mt2+tt2n+m+3+t3+2,Q(t)=Mtt2n+m+3+t3+2,R(t)=t2n+m-i+5+M2t2+2Mt2n+m-i+3+4M3(t2n+m-i+3+N2)2,W(t)=Mt2+M2t2n+m-i+3+N2,∀(t,u,v)∈[t0,+∞)×[N,M]2.
It is easy to prove that the conditions of Theorem 4 are satisfied. Hence (116) has uncountably many positive solutions in Ω1(1/2,8), and for any L∈(1/2,1), there exist θ∈(0,1) and T>1+|t0|+τ+|γ| such that the Mann iterative sequence {xk}k∈ℕ0 generated by (19) and (21) converges to a positive solution x∈Ω1(1/2,8) of (116) and has the error estimate (20).Example 15.
Consider the higher order nonlinear neutral delay differential equation(118)[(t2i+1)(x(t)+2t2+1x(t-τ))(m)](n-m)+(t+1x2(t-1)x(t2)tn+i+1+tln(1+|x(t-2)|)+4)(i)+t2x(t-12)x(t2-9)t2n+3+t|x(t-3)|+3=t2cos(2t)+arctant3tn+4+sin2(1-t3+t4)+1,∀t≥3,
where τ>0 and i≤n-m-1. Let l=3, t0=0, γ=min{3-τ,-9}, M=12, N=5, b0=2 and
(119)h1(t)=t-1,h2(t)=t2,h3(t)=t-2,f1(t)=t-12,f2(t)=t2-9,f3(t)=t-3,a(t)=t2i+1,b(t)=2t2+1,h(t,u,v,w)=t+1u2vtn+i+1+tln(1+|w|)+4,g(t)=t2cos(2t)+arctant3tn+4+sin2(1-t3+t4)+1,f(t,u,v,w)=t2uvt2n+3+t|w|+3,P(t)=Mt2(6b0+3M2t+2b0t2n+3)b02(t2n+3+3)2,Q(t)=M2t2b02(t2n+3+3),R(t)=M2t+1b02(tn+i+1+4)2×[3tn+i+1+12+Mb0t+3tln(1+Mb0)],W(t)=M3t+1b03(tn+i+1+4),∀(t,u,v,w)∈[t0,+∞)×[0,Mb0]3.
It is easy to verify that the conditions of Theorem 5 are satisfied. Hence Theorem 5 ensures that (118) has uncountably many positive solutions in Ω2(5,12), and, for any L∈(11,12), there exist θ∈(0,1) and T>1+|t0|+τ+|γ| such that the Mann iterative sequence {xk}k∈ℕ0 generated by (48) and (21) converges to a positive solution x∈Ω2(5,12) of (118) and has the error estimate (20).Example 16.
Consider the higher order nonlinear neutral delay differential equation(120)[(t3n+2tn+1+1)(x(t)-(t2+2t+4)x(t-τ))(m)](n-m)+(tx(t-3)x2(t-4)tn+10+x2(t-3))(i)+t+1x3(tlnt)-tx4(t2-t)(t+1)n+3+tx2(tlnt)=tln(1+t2)-t+3cos3(t3+1)tn+5+4t3-1-tsin5(t2-3),∀t≥1,
where τ>0, and i≤n-m-1. Let l=2, t0=1, γ=min{1-τ,-3}, M=6, N=2, b0=3 and
(121)h1(t)=t-3,h2(t)=t-4,f1(t)=tlnt,f2(t)=t2-t,a(t)=t3n+2tn+1+1,b(t)=-t2-2t-4,h(t,u,v)=tuv2tn+10+u2,f(t,u,v)=t+1u3-tv4(t+1)n+3+tu2,g(t)=tln(1+t2)-t+3cos3(t3+1)tn+5+4t3-1-tsin5(t2-3),P(t)=M2b02(t+1)2n+6×[3(t+1)n+7/2+M2b02tt+1+4Mb0t(t+1)n+3+6M3b03t2],Q(t)=M3b03(t+1)n+8(t+1+Mb0t),R(t)=3M2b02t2n+19(tn+10+M2b02),W(t)=M3b03tn+9,∀(t,u,v)∈[t0,+∞)×[0,Mb0]2.
It is easy to check that the conditions of Theorem 6 are satisfied. Thus Theorem 6 ensures that (120) has uncountably many positive solutions in Ω3(2,6), and, for any L∈(2,4), there exist θ∈(0,1) and T>1+|t0|+τ+|γ| such that the Mann iterative sequence {xk}k∈ℕ0 generated by (61) and (21) converges to a positive solution x∈Ω3(2,6) of (120) and has the error estimate (20).Example 17.
Consider the higher order nonlinear neutral delay differential equation(122)[(t2lnt)(x(t)+x(t-τ))(m)](n-m)+(x(t-2)+x(2t-1)t3n-m-i+2+x2(t-cost))(i)+x2(t-4)+x(t-1)x2(t-sin(t9+1))t2n-m+3+1=sin13(t5-t+1)tn+7/2+1,∀t≥4,
where τ>0, m≥2 and i≤n-m-1. Let l=3, t0=4, γ=min{4-τ,0}, M=100, N=1 and
(123)h1(t)=t-2,h2(t)=2t-1,h3(t)=t-cost,f1(t)=t-4,f2(t)=t-1,f3(t)=t-sin(t9+1),a(t)=t2lnt,b(t)=1,h(t,u,v,w)=u+vt3n-m-i+2+w2,f(t,u,v,w)=u2+vw2t2n-m+3+1,g(t)=sin13(t5-t+1)tn+7/2+1,P(t)=2M+3M2t2n-m+3+1,Q(t)=M2+M3t2n-m+3+1,R(t)=2t3n-m-i+2+6M2(t3n-m-i+2+N2)2,W(t)=2Mt3n-m-i+2+N2,∀(t,u,v,w)∈[t0,+∞)×[N,M]3.
It is easy to check that the conditions of Theorem 7 are satisfied. Thus (122) has uncountably many positive solutions in Ω1(1,100), and, for any L∈(1,100), there exist θ∈(0,1) and T>1+|t0|+τ+|γ| such that the Mann iterative sequence {xk}k∈ℕ0 generated by (66) and (21) converges to a positive solution x∈Ω1(1,100) of (122) and has the error estimate (20).Example 18.
Consider the higher order nonlinear neutral delay differential equation(124)[2+sin(t+t)t2(x(t)+x(t-τ))′](n-1)+(x3(t-2)-t2x4(t+1-1)t5n+3+t+1)(i)+x3(t-3)-tt3n+4+x2(t-(-1)n)=sin(t4-t2+1)tn+3+lnt,∀t≥3,
where τ>0, m=1 and i≤n-2. Let l=2, t0=3, γ=min{3-τ,0}, M=10, N=9 and
(125)h1(t)=t-2,h2(t)=t+1-1,f1(t)=t-3,f2(t)=t-(-1)n,a(t)=2+sin(t+t)t2,b(t)=1,h(t,u,v)=u3-t2v4t5n+3+t+1,f(t,u,v)=u3-tt3n+4+v2,g(t)=sin(t4-t2+1)tn+3+lnt,P(t)=M(5M3+2t+2Mt3n+4)(t3n+4+N2)2,Q(t)=M3+tt3n+4+N2,R(t)=M2(3+4Mt2)t5n+3+t+1,W(t)=M3(1+Mt2)t5n+3+t+1,∀(t,u,v)∈[t0,+∞)×[N,M]2.
It is easy to check that the conditions of Theorem 8 are satisfied. Thus Theorem 8 ensures that (124) has uncountably many positive solutions in Ω1(9,10), and, for any L∈(9,10), there exist θ∈(0,1) and T>1+|t0|+τ+|γ| such that the Mann iterative sequence {xk}k∈ℕ0 generated by (78) and (21) converges to a positive solution x∈Ω1(9,10) of (124) and has the error estimate (20).Example 19.
Consider the higher order nonlinear neutral delay differential equation(126)[tm+1ln(4+sin(t2-t))(x(t)-x(t-τ))(m)](n-m)+(ttn-m+4+x4(t-t))(i)+sin(t3-2t+t3+1)tn+m+|t-x(t-3)|=tcos5(t7-t4+1)tn+2t-cos3(t2-3),∀t≥4,
where τ>0, m≥2 and i≤n-m-1. Let l=1, t0=4, γ=min{4-τ,1}, M=7, N=5 and
(127)h1(t)=t-t,f1(t)=t-3,a(t)=tm+1ln(4+sin(t2-t)),b(t)=-1,h(t,u)=ttn-m+4+u4,f(t,u)=sint3(-2t+t3+1)tn+m+|t-u|,g(t)=tcos5(t7-t4+1)tn+2t-cos3(t2-3),P(t)=1t2n+2m,Q(t)=1tn+m,R(t)=4M3t2n-2m+7,W(t)=1tn-m+3,∀(t,u)∈[t0,+∞)×[N,M]2.
It is easy to check that the conditions of Theorem 9 are satisfied. Thus Theorem 9 ensures that (126) has uncountably many positive solutions in Ω1(5,7), and, for any L∈(5,7), there exist θ∈(0,1) and T>1+|t0|+τ+|γ| such that the Mann iterative sequence {xk}k∈ℕ0 generated by (90) and (21) converges to a positive solution x∈Ω1(5,7) of (126) and has the error estimate (20).Example 20.
Consider the higher order nonlinear neutral delay differential equation(128)[1t3(x(t)-x(t-τ))′](n-1)+(t-sin(t8-4t5-1)tn+7+|x(t-1)-x3(t-2)|)(i)+ln(1+x2(t-arctant))t2n+6+x2(t-4)=tln(t+cos(t3-1))tn+8+1,∀t≥5,
where τ>0, m=1 and i≤n-2. Let l=2, t0=5, γ=min{5-τ,1}, M=4, N=2 and
(129)h1(t)=t-1,h2(t)=t-2,f1(t)=t-arctant,f2(t)=t-4,a(t)=1t3,b(t)=-1,h(t,u,v)=t-sin(t8-4t5-1)tn+7+|u-v3|,f(t,u,v)=ln(1+u2)t2n+6+v2,g(t)=tln(t+cos(t3-1))tn+8+1,P(t)=2M(2M2+t2n+6)t4n+12,Q(t)=M2t2n+6,R(t)=2+6M2t2n+13,W(t)=2tn+6,∀(t,u,v)∈[t0,+∞)×[N,M]2.
It is easy to check that the conditions of Theorem 10 are satisfied. Thus Theorem 10 ensures that (128) has uncountably many positive solutions in Ω1(2,4), and, for any L∈(2,4), there exist θ∈(0,1) and T>1+|t0|+τ+|γ| such that the Mann iterative sequence {xk}k∈ℕ0 generated by (102) and (21) converges to a positive solution x∈Ω1(2,4) of (128) and has the error estimate (20).



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