# A Nonlinear Integral Equation Related to Infectious Diseases

Authors: Mohamed Jleli; Bessem Samet
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this paper, a nonlinear integral equation related to infectious diseases is investigated. Namely, we first study the existence and uniqueness of solutions and provide numerical algorithms that converge to the unique solution. Next, we study the lower and upper subsolutions, as well as the data dependence of the solution.


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## Body

## 1. Introduction

We consider the nonlinear integral equation(1)xt=∏i=1nfit,xt+∫t−τitgis,xsds,t∈ℝ,where n≥2 is an integer and τi>0, i=1,2,⋯,n. In the case n=1 and f1≡0, (1) reduces to
(2)xt=∫t−τ1tg1s,xsdst∈ℝ.The integral equation (2) models the spread of certain infectious diseases with periodic contact rate that varies seasonally (see [1]). Several results related to certain mathematical aspects of (2) have been obtained by many authors (see, e.g., [1–9] and the references therein). In particular, in [3], using the Picard operator technique, the integral equation (2) was investigated regarding the existence and uniqueness of solutions and periodic solutions, lower and upper subsolutions, the data dependence, and the differentiability of solutions with respect to a parameter.In this paper, we are concerned with the integral equation (1). We first investigate the existence and uniqueness of solutions and provide numerical algorithms that converge to the unique solution. Next, we study the lower and upper subsolutions, as well as the data dependence of the solution.The next section is devoted to the main results of this paper. Namely, in Subsection2.1, we fix some notations that will be used throughout this paper. In Subsection 2.2, we provide some lemmas that will be used in the proofs of our main results. In Subsection 2.3, the existence and uniqueness of solutions and periodic solutions are derived using the Banach contraction principle. Moreover, an iterative algorithm based on Picard iteration for approximating the unique solution is provided. In Subsection 2.4, a Prešic′-type iterative algorithm that converges to the unique solution is provided. Lower and upper subsolutions type results are obtained in Subsection 2.5. Finally, in Subsection 2.6, the data dependence of solutions is studied.


## 2. Results

We first fix some notations.


### 2.1. Notations

LetI=α,β and J=m,M, where 0<α<β and 0<m<M. Let
(3)Cℝ×I,J=f:ℝ×I⟶J,fiscontinuous,X=Cℝ,I=f:ℝ⟶I,fiscontinuous.The functional spaceX is equipped with the norm ∥·∥X, where
(4)xX=supt∈ℝxt,x∈X.Notice thatX,·X is a Banach space.


### 2.2. Preliminaries

The following lemma will be useful later. It can be easily proved by induction.Lemma 1.
Letan and bn be two real sequences. Then, for all n≥2,
(5)a1a2⋯an−b1b2⋯bn≤a2a3⋯ana1−b1+b1a3a4⋯ana2−b2+b1b2a4a5⋯ana3−b3+⋯+anb1b2⋯bn−2an−1−bn−1+b1b2⋯bn−1an−bn.We recall the following result due to Prešic′ [10].Lemma 2.
LetX,d be a complete metric space, k a positive integer and φ:Xk⟶X a mapping satisfying the following condition:
(6)dφx1,x2,⋯,xk,φx2,x3,⋯,xk+1≤q1dx1,x2+q2dx2,x3+⋯+qkdxk,xk+1,for all x1,⋯,xk+1∈X, where q1,q2,⋯,qk are nonnegative constants such that q1+q2+⋯+qk<1. Then,
(i)
There exists a uniquex∗∈X such that(7)x∗=φx∗,x∗,⋯,x∗.(ii)
For allx1,x2,⋯,xk∈X, the sequence xp⊂X defined by(8)xp+k=φxp,xp+1,⋯,xp+k−1,p≥1is convergent to x∗.For more details about the above result, we refer to [11–15].


### 2.3. Existence and Uniqueness Result

Problem (1) is investigated under the following conditions:(C1)fi,gi∈Cℝ×I,J, i=1,2,⋯,n.(C2) For alli=1,2,⋯,n, there exists a constant Lfi>0 such that for all t∈ℝ,
(9)fit,u−fit,v≤Lfiu−v,u,v∈I.(C3) For alli=1,2,⋯,n, there exists a constant Lgi>0 such that for all t∈ℝ,
(10)git,u−git,v≤Lgiu−v,u,v∈I.(C4)Mn−1∏i=1n−1τi+1∑k=1nLfk+Lgkτk<1.(C5)α/mn≤∏i=1nτi+1≤β/Mn.We have the following existence and uniqueness result.Theorem 3.
Under conditions (C1)–(C5), problem (1) admits one and only one solution x∗∈X. Moreover, for all x0∈X, the sequence xp⊂X defined by
(11)xp+1t=∏i=1nfit,xpt+∫t−τitgis,xpsds,t∈ℝconverges uniformly to x∗.Proof.
Let us define the operatorT:X⟶Cℝ,ℝ by
(12)Txt=∏i=1nTixt,x∈X,t∈ℝ,where
(13)Tixt=fit,xt+∫t−τitgis,xsds,i=1,2,⋯,n.By (C1), for all i=1,2,⋯,n and t∈ℝ, one has
(14)Tixt≤M+∫t−τitMds=τi+1M,which yields
(15)Txt≤Mn∏i=1nτi+1.Then, using (C5), one deduces that
(16)Txt≤β,t∈ℝ.Similarly, by (C1), one has
(17)Tixt≥m+∫t−τitmds=τi+1m,which yields
(18)Txt≥mn∏i=1nτi+1.Hence, using (C5), one obtains
(19)Txt≥α,t∈ℝ.Therefore, it follows from (16) and (19) that
(20)TX⊂X.Moreover, the set of solutions to the integral equation (1) coincides with the set of fixed points of the operator T. Next, by Lemma 1, for all x,y∈X and t∈ℝ, one has
(21)Txt−Tyt=∏i=1nTixt−∏i=1nTiyt≤T2xtT3xt⋯TnxtT1xt−T1yt+T3xtT4xt⋯TnxtT1ytT2xt−T2yt+T4xtT5xt⋯TnxtT1ytT2ytT3xt−T3yt+⋯+T1ytT2yt⋯Tn−1ytTnxt−Tnyt.On the other hand, by (C2) and (C3), for all i=1,2,⋯,n, one has
(22)Tixt−Tiyt≤fit,xt−fit,yt+∫t−τitgis,xs−gis,ysds≤Lfixt−yt+Lgi∫t−τitxs−ysds≤Lfi+Lgiτix−yX.Therefore, using (14), (21), and (22), one obtains
(23)∣Txt−Tyt∣≤Mn−1∏i=1n−1τi+1∑k=1nLfk+Lgkτkx−yX,which yields
(24)∥Tx−Ty∥X≤Mn−1∏i=1n−1τi+1∑k=1nLfk+Lgkτk∥x−y∥X,x,y∈X.Finally, using (C4), (20) and (24), the conclusion of the theorem follows from the Banach contraction principle.Now, we consider problem (1) under the additional condition:(C6) There existsω>0 such that for all i=1,2,⋯,n,
(25)fit+ω,u=fit,u,git+ω,u=git,u,t∈ℝ,u∈I.Theorem 4.
Under conditions (C1)–(C6), problem (1) admits one and only one ω-periodic solution x∗∈X. Moreover, for any ω-periodic function x0∈X, the sequence xp defined by (11) converges uniformly to x∗.Proof.
LetT:X⟶X be the operator defined by (12). Notice that from the proof of Theorem 3, we know that under conditions (C1)–(C5), one has TX⊂X. Let V be the closed subset of X (with respect to the norm ·X) defined by
(26)V=x∈X:xt+ω=xt,t∈ℝ.For allx∈V and t∈ℝ, using (C6), one obtains
(27)Txt+ω=∏i=1nfit+ω,xt+ω+∫t+ω−τit+ωgis,xsds=∏i=1nfit,xt+∫t−τitgiσ+ω,xσ+ωdσ=∏i=1nfit,xt+∫t−τitgiσ,xσdσ=Txt.Hence, one hasTV⊂V. On the other hand, since V⊂X, it follows from (24) that
(28)∥Tx−Ty∥X≤Mn−1∏i=1n−1τi+1∑k=1nLfk+Lgkτk∥x−y∥X,x,y∈V.Then, the conclusion of the theorem follows from the Banach contraction principle.


### 2.4. Prešic′-Type Approximation of the Unique Solution

Let us consider the integral equation (1) under conditions (C1)–(C5). Notice that by Theorem 3, (1) admits one and only one solution x∗∈X.Theorem 5.
Under conditions (C1)–(C5), for any x1,x2,⋯,xn∈X, the sequence xp defined by
(29)xp+nt=f1t,xpt+∫t−τ1tg1s,xpsdsf2t,xp+1t+∫t−τ2tg2s,xp+1sds⋯fnt,xp+n−1t+∫t−τntg2s,xp+n−1sds,p≥1,t∈ℝconverges uniformly to x∗.Proof.
Consider the functionφ:Xn⟶X defined by
(30)φx1,x2,⋯,xnt=∏i=1nfit,xit+∫t−τitg1s,xisds,t∈ℝ,that is,
(31)φx1,x2,⋯,xnt=∏i=1nTixit,where for all i=1,2,⋯,n, the operator Ti is defined by (13). Notice that from the considered assumptions, one has φXn⊂X, so φ is well-defined. On the other hand, using Lemma 1, for all x1,x2,⋯,xn,xn+1∈X and t∈ℝ, on has
(32)φx1,x2,⋯,xnt−φx2,x3,⋯,xn+1t=∏i=1nTixit−∏i=1nTixi+1t≤T2x2t⋯TnxntT1x1t−T1x2t+T1x2tT3x3t⋯TnxntT2x2t−T2x3t+⋯+T1x2t⋯Tn−1xntTnxnt−Tnxn+1t.Next, using (14), it holds that
(33)φx1,x2,⋯,xnt−φx2,x3,⋯,xn+1t≤Mn−1∏i=1n−1τi+1∑k=1nTkxkt−Tkxk+1t.On the other hand, under the considered assumptions, for allk=1,2,⋯,n, one has
(34)Tkxkt−Tkxk+1t≤Lfk+τkLgkxk−xk+1X.Hence, one deduces that(35)φx1,x2,⋯,xn−φx2,x3,⋯,xn+1X≤Mn−1∏i=1n−1τi+1∑k=1nLfk+τkLgkxk−xk+1X.Finally, using (C4) and Lemma 2, the desired result follows.


### 2.5. Lower and Upper Subsolutions

We consider problem (1) under conditions (C1)–(C5). We recall that by Theorem 3, problem (1) admits one and only one solution x∗∈X. We suppose also thatFor alli=1,2,⋯,n and t∈ℝ, the functions
(36)fit,·:I⟶Jandgit,·:I⟶Jare nondecreasing.Theorem 6.
Suppose that conditions (C1)–(C5) and (C6′) are satisfied. If x∈Cℝ,I satisfies
(37)xt≤∏i=1nfit,xt+∫t−τitgis,xsds,t∈ℝ,then
(38)xt≤x∗t,t∈ℝ.Proof.
LetT:X⟶X be the operator defined by (12). Then, (39) is equivalent to
(39)xt≤Txt,t∈ℝ.We shall prove thatT is a nondecreasing operator, that is,
(40)u,v∈X,ut≤vt,t∈ℝ⟹Tut≤Tvt,t∈ℝ.Letu,v∈X be such that
(41)ut≤vt,t∈ℝ.By (C6′), for all i=1,2,⋯,n and t∈ℝ, one obtains
(42)0≤fit,ut+∫t−τitgis,usds≤fit,vt+∫t−τitgis,vsds,which yields
(43)∏i=1nfit,ut+∫t−τitgis,usds≤∏i=1nfit,vt+∫t−τitgis,vsds,that is,
(44)Tut≤Tvt.This proves (40). Next, by (39), it holds that
(45)xt≤Txt≤T2xt≤⋯≤Tpxt,for all nonnegative integer p and t∈ℝ, where
(46)T0xt=xtandTp+1xt=TTpxt.Hence, it holds that(47)xt≤xpt,t∈ℝ,wherexp is the sequence defined by (11) with x0=x. On the other hand, by Theorem 3, one has
(48)limp→∞xpt=x∗t,t∈ℝ.Therefore, passing to the limit asp⟶∞ in (47), (38) follows.(A1) Forj=1,2,3 and i=1,2,⋯,n, let fij,gij∈Cℝ×I,J. We suppose that(A2) For alli=1,2,⋯,n and j=1,2,3, there exists a constant Lfij>0 such that for all t∈ℝ,
(49)fijt,u−fijt,v≤Lfiju−v,u,v∈I.For alli=1,2,⋯,n and j=1,2,3, there exists a constant Lgij>0 such that for all t∈ℝ,
(50)gijt,u−gijt,v≤Lgiju−v,u,v∈I.(A3)Mn−1∏i=1n−1τi+1∑k=1nLfkj+Lgkjτk<1, j=1,2,3.(A4)α/mn≤∏i=1nτi+1≤β/Mn.(A5) For alli=1,2,⋯,n and t∈ℝ, the functions
(51)fi2t,·:I⟶Jandgi2t,·:I⟶J,are nondecreasing.(A6) For alli=1,2,⋯,n, t∈ℝ and u∈I,
(52)fi1t,u≤fi2t,u≤fi3t,uandgi1t,u≤gi2t,u≤gi3t,u.Notice that by (A1)–(A4), it follows from Theorem 3 that for all j=1,2,3, the integral equation
(53)xt=∏i=1nfijt,xt+∫t−τitgijs,xsds,t∈ℝ,admits one and only one solution xj∈X. Moreover, for all j=1,2,3 and x0j∈X, the sequence xpj⊂X defined by
(54)xp+1jt=∏i=1nfijt,xpjt+∫t−τitgijs,xpjsds,t∈ℝconverges uniformly to xj.Theorem 7.
Under conditions (A1)–(A6), one has
(55)x1t≤x2t≤x3t,t∈ℝ.Proof.
For allj=1,2,3, let Tj:X⟶X be the operator defined by
(56)Tjxt=∏i=1nfijt,xt+∫t−τitgijs,xsds,x∈X,t∈ℝ.From condition (A5), the operator T2 is nondecreasing, that is,
(57)u,v∈X,ut≤vt,t∈ℝ⟹T2ut≤T2vt,t∈ℝ.Moreover, by (A6), one has
(58)T1ut≤T2ut≤T3ut,u∈X,t∈ℝ.Letx01,x02,x03∈X be such that
(59)x01t≤x02t≤x03t,t∈ℝ.Hence, by (57), one obtains
(60)T2x01t≤T2x02t≤T2x03t,t∈ℝ.On the other hand, by (58), one has
(61)T1x01t≤T2x01t,t∈ℝ,(62)T2x03t≤T3x03t,t∈ℝ.Therefore, using (60), (61), and (62), one deduces that
(63)T1x01t≤T2x02t≤T3x03t,t∈ℝ,that is,
(64)x11t≤x12t≤x13t,t∈ℝ.Repeating the same argument, by induction, one deduces that for all nonnegative integerp and t∈ℝ,
(65)xp1t≤xp2t≤xp3t,where xpj, j=1,2,3, is the sequence defined by (54). Finally, passing to the limit as p⟶∞ in (65), the desired result follows.


### 2.6. Data Dependence of Solutions

Suppose that conditions (C1)–(C5) are satisfied. Then, by Theorem 3, the integral equation admits one and only one solution x∗∈X. Consider now the perturbed problem
(66)yt=∏i=1nFit,yt+∫t−τitGis,ysds,t∈ℝ,where Fi,Gi∈Cℝ×I,J, i=1,2,⋯,n. Suppose that y∗∈X is a solution to the integral equation (66).We have the following data dependence result.Theorem 8.
Suppose that for alli=1,2,⋯,n, there exist σi,ηi>0 such that
(67)fit,u−Fit,u≤σi,git,u−Git,u≤ηi,t∈ℝ,u∈I.Then,(68)x∗−y∗X≤Mn−1∏i=1n−1τi+1∑k=1nσk+τkηk1−Mn−1∏i=1n−1τi+1∑k=1nLfk+Lgkτk.Proof.
Let(69)Syt=∏i=1nSiyt,t∈ℝ,where
(70)Siyt=Fit,yt+∫t−τitGis,ysds,i=1,2,⋯,n.Then, for allt∈ℝ, one has
(71)x∗t−y∗t=Tx∗t−Sy∗t=∏i=1nTix∗t−∏i=1nSiy∗t,where the operator T is defined by (12). Next, by Lemma 1 and (14), one obtains
(72)∣x∗t−y∗t∣≤T2x∗t⋯Tnx∗t∣T1x∗t−S1y∗t∣+S1y∗tT3x∗t⋯Tnx∗t∣T2x∗t−S2y∗t∣+⋯+S1y∗t⋯Sn−1y∗t∣Tnx∗t−Sny∗t∣≤Mn−1∏i=1n−1τi+1∑k=1nTkx∗t−Sky∗t.On the other hand, using (C2) and (C3) and (67), for all k=1,2,⋯,n, one has
(73)Tkx∗t−Sky∗t≤fkt,x∗t−Fkt,y∗t+∫t−τktgks,x∗s−Gks,y∗sds≤fkt,x∗t−fkt,y∗t+fkt,y∗t−Fkt,y∗t+∫t−τktgks,x∗s−gks,y∗sds+∫t−τktgks,y∗s−Gks,y∗sds≤Lfkx∗−y∗X+σk+Lgkτkx∗−y∗X+τkηk=Lfk+Lgkτkx∗−y∗X+σk+τkηk.Hence, by (72), it holds that
(74)x∗−y∗X≤Mn−1∏i=1n−1τi+1∑k=1nLfk+Lgkτkx∗−y∗X+σk+τkηk,which yields
(75)1−Mn−1∏i=1n−1τi+1∑k=1nLfk+Lgkτkx∗−y∗X≤Mn−1∏i=1n−1τi+1∑k=1nσk+τkηk.Finally, by (C4), the desired result follows.


## 2.1. Notations

LetI=α,β and J=m,M, where 0<α<β and 0<m<M. Let
(3)Cℝ×I,J=f:ℝ×I⟶J,fiscontinuous,X=Cℝ,I=f:ℝ⟶I,fiscontinuous.The functional spaceX is equipped with the norm ∥·∥X, where
(4)xX=supt∈ℝxt,x∈X.Notice thatX,·X is a Banach space.


## 2.2. Preliminaries

The following lemma will be useful later. It can be easily proved by induction.Lemma 1.
Letan and bn be two real sequences. Then, for all n≥2,
(5)a1a2⋯an−b1b2⋯bn≤a2a3⋯ana1−b1+b1a3a4⋯ana2−b2+b1b2a4a5⋯ana3−b3+⋯+anb1b2⋯bn−2an−1−bn−1+b1b2⋯bn−1an−bn.We recall the following result due to Prešic′ [10].Lemma 2.
LetX,d be a complete metric space, k a positive integer and φ:Xk⟶X a mapping satisfying the following condition:
(6)dφx1,x2,⋯,xk,φx2,x3,⋯,xk+1≤q1dx1,x2+q2dx2,x3+⋯+qkdxk,xk+1,for all x1,⋯,xk+1∈X, where q1,q2,⋯,qk are nonnegative constants such that q1+q2+⋯+qk<1. Then,
(i)
There exists a uniquex∗∈X such that(7)x∗=φx∗,x∗,⋯,x∗.(ii)
For allx1,x2,⋯,xk∈X, the sequence xp⊂X defined by(8)xp+k=φxp,xp+1,⋯,xp+k−1,p≥1is convergent to x∗.For more details about the above result, we refer to [11–15].


## 2.3. Existence and Uniqueness Result

Problem (1) is investigated under the following conditions:(C1)fi,gi∈Cℝ×I,J, i=1,2,⋯,n.(C2) For alli=1,2,⋯,n, there exists a constant Lfi>0 such that for all t∈ℝ,
(9)fit,u−fit,v≤Lfiu−v,u,v∈I.(C3) For alli=1,2,⋯,n, there exists a constant Lgi>0 such that for all t∈ℝ,
(10)git,u−git,v≤Lgiu−v,u,v∈I.(C4)Mn−1∏i=1n−1τi+1∑k=1nLfk+Lgkτk<1.(C5)α/mn≤∏i=1nτi+1≤β/Mn.We have the following existence and uniqueness result.Theorem 3.
Under conditions (C1)–(C5), problem (1) admits one and only one solution x∗∈X. Moreover, for all x0∈X, the sequence xp⊂X defined by
(11)xp+1t=∏i=1nfit,xpt+∫t−τitgis,xpsds,t∈ℝconverges uniformly to x∗.Proof.
Let us define the operatorT:X⟶Cℝ,ℝ by
(12)Txt=∏i=1nTixt,x∈X,t∈ℝ,where
(13)Tixt=fit,xt+∫t−τitgis,xsds,i=1,2,⋯,n.By (C1), for all i=1,2,⋯,n and t∈ℝ, one has
(14)Tixt≤M+∫t−τitMds=τi+1M,which yields
(15)Txt≤Mn∏i=1nτi+1.Then, using (C5), one deduces that
(16)Txt≤β,t∈ℝ.Similarly, by (C1), one has
(17)Tixt≥m+∫t−τitmds=τi+1m,which yields
(18)Txt≥mn∏i=1nτi+1.Hence, using (C5), one obtains
(19)Txt≥α,t∈ℝ.Therefore, it follows from (16) and (19) that
(20)TX⊂X.Moreover, the set of solutions to the integral equation (1) coincides with the set of fixed points of the operator T. Next, by Lemma 1, for all x,y∈X and t∈ℝ, one has
(21)Txt−Tyt=∏i=1nTixt−∏i=1nTiyt≤T2xtT3xt⋯TnxtT1xt−T1yt+T3xtT4xt⋯TnxtT1ytT2xt−T2yt+T4xtT5xt⋯TnxtT1ytT2ytT3xt−T3yt+⋯+T1ytT2yt⋯Tn−1ytTnxt−Tnyt.On the other hand, by (C2) and (C3), for all i=1,2,⋯,n, one has
(22)Tixt−Tiyt≤fit,xt−fit,yt+∫t−τitgis,xs−gis,ysds≤Lfixt−yt+Lgi∫t−τitxs−ysds≤Lfi+Lgiτix−yX.Therefore, using (14), (21), and (22), one obtains
(23)∣Txt−Tyt∣≤Mn−1∏i=1n−1τi+1∑k=1nLfk+Lgkτkx−yX,which yields
(24)∥Tx−Ty∥X≤Mn−1∏i=1n−1τi+1∑k=1nLfk+Lgkτk∥x−y∥X,x,y∈X.Finally, using (C4), (20) and (24), the conclusion of the theorem follows from the Banach contraction principle.Now, we consider problem (1) under the additional condition:(C6) There existsω>0 such that for all i=1,2,⋯,n,
(25)fit+ω,u=fit,u,git+ω,u=git,u,t∈ℝ,u∈I.Theorem 4.
Under conditions (C1)–(C6), problem (1) admits one and only one ω-periodic solution x∗∈X. Moreover, for any ω-periodic function x0∈X, the sequence xp defined by (11) converges uniformly to x∗.Proof.
LetT:X⟶X be the operator defined by (12). Notice that from the proof of Theorem 3, we know that under conditions (C1)–(C5), one has TX⊂X. Let V be the closed subset of X (with respect to the norm ·X) defined by
(26)V=x∈X:xt+ω=xt,t∈ℝ.For allx∈V and t∈ℝ, using (C6), one obtains
(27)Txt+ω=∏i=1nfit+ω,xt+ω+∫t+ω−τit+ωgis,xsds=∏i=1nfit,xt+∫t−τitgiσ+ω,xσ+ωdσ=∏i=1nfit,xt+∫t−τitgiσ,xσdσ=Txt.Hence, one hasTV⊂V. On the other hand, since V⊂X, it follows from (24) that
(28)∥Tx−Ty∥X≤Mn−1∏i=1n−1τi+1∑k=1nLfk+Lgkτk∥x−y∥X,x,y∈V.Then, the conclusion of the theorem follows from the Banach contraction principle.


## 2.4. Prešic′-Type Approximation of the Unique Solution

Let us consider the integral equation (1) under conditions (C1)–(C5). Notice that by Theorem 3, (1) admits one and only one solution x∗∈X.Theorem 5.
Under conditions (C1)–(C5), for any x1,x2,⋯,xn∈X, the sequence xp defined by
(29)xp+nt=f1t,xpt+∫t−τ1tg1s,xpsdsf2t,xp+1t+∫t−τ2tg2s,xp+1sds⋯fnt,xp+n−1t+∫t−τntg2s,xp+n−1sds,p≥1,t∈ℝconverges uniformly to x∗.Proof.
Consider the functionφ:Xn⟶X defined by
(30)φx1,x2,⋯,xnt=∏i=1nfit,xit+∫t−τitg1s,xisds,t∈ℝ,that is,
(31)φx1,x2,⋯,xnt=∏i=1nTixit,where for all i=1,2,⋯,n, the operator Ti is defined by (13). Notice that from the considered assumptions, one has φXn⊂X, so φ is well-defined. On the other hand, using Lemma 1, for all x1,x2,⋯,xn,xn+1∈X and t∈ℝ, on has
(32)φx1,x2,⋯,xnt−φx2,x3,⋯,xn+1t=∏i=1nTixit−∏i=1nTixi+1t≤T2x2t⋯TnxntT1x1t−T1x2t+T1x2tT3x3t⋯TnxntT2x2t−T2x3t+⋯+T1x2t⋯Tn−1xntTnxnt−Tnxn+1t.Next, using (14), it holds that
(33)φx1,x2,⋯,xnt−φx2,x3,⋯,xn+1t≤Mn−1∏i=1n−1τi+1∑k=1nTkxkt−Tkxk+1t.On the other hand, under the considered assumptions, for allk=1,2,⋯,n, one has
(34)Tkxkt−Tkxk+1t≤Lfk+τkLgkxk−xk+1X.Hence, one deduces that(35)φx1,x2,⋯,xn−φx2,x3,⋯,xn+1X≤Mn−1∏i=1n−1τi+1∑k=1nLfk+τkLgkxk−xk+1X.Finally, using (C4) and Lemma 2, the desired result follows.


## 2.5. Lower and Upper Subsolutions

We consider problem (1) under conditions (C1)–(C5). We recall that by Theorem 3, problem (1) admits one and only one solution x∗∈X. We suppose also thatFor alli=1,2,⋯,n and t∈ℝ, the functions
(36)fit,·:I⟶Jandgit,·:I⟶Jare nondecreasing.Theorem 6.
Suppose that conditions (C1)–(C5) and (C6′) are satisfied. If x∈Cℝ,I satisfies
(37)xt≤∏i=1nfit,xt+∫t−τitgis,xsds,t∈ℝ,then
(38)xt≤x∗t,t∈ℝ.Proof.
LetT:X⟶X be the operator defined by (12). Then, (39) is equivalent to
(39)xt≤Txt,t∈ℝ.We shall prove thatT is a nondecreasing operator, that is,
(40)u,v∈X,ut≤vt,t∈ℝ⟹Tut≤Tvt,t∈ℝ.Letu,v∈X be such that
(41)ut≤vt,t∈ℝ.By (C6′), for all i=1,2,⋯,n and t∈ℝ, one obtains
(42)0≤fit,ut+∫t−τitgis,usds≤fit,vt+∫t−τitgis,vsds,which yields
(43)∏i=1nfit,ut+∫t−τitgis,usds≤∏i=1nfit,vt+∫t−τitgis,vsds,that is,
(44)Tut≤Tvt.This proves (40). Next, by (39), it holds that
(45)xt≤Txt≤T2xt≤⋯≤Tpxt,for all nonnegative integer p and t∈ℝ, where
(46)T0xt=xtandTp+1xt=TTpxt.Hence, it holds that(47)xt≤xpt,t∈ℝ,wherexp is the sequence defined by (11) with x0=x. On the other hand, by Theorem 3, one has
(48)limp→∞xpt=x∗t,t∈ℝ.Therefore, passing to the limit asp⟶∞ in (47), (38) follows.(A1) Forj=1,2,3 and i=1,2,⋯,n, let fij,gij∈Cℝ×I,J. We suppose that(A2) For alli=1,2,⋯,n and j=1,2,3, there exists a constant Lfij>0 such that for all t∈ℝ,
(49)fijt,u−fijt,v≤Lfiju−v,u,v∈I.For alli=1,2,⋯,n and j=1,2,3, there exists a constant Lgij>0 such that for all t∈ℝ,
(50)gijt,u−gijt,v≤Lgiju−v,u,v∈I.(A3)Mn−1∏i=1n−1τi+1∑k=1nLfkj+Lgkjτk<1, j=1,2,3.(A4)α/mn≤∏i=1nτi+1≤β/Mn.(A5) For alli=1,2,⋯,n and t∈ℝ, the functions
(51)fi2t,·:I⟶Jandgi2t,·:I⟶J,are nondecreasing.(A6) For alli=1,2,⋯,n, t∈ℝ and u∈I,
(52)fi1t,u≤fi2t,u≤fi3t,uandgi1t,u≤gi2t,u≤gi3t,u.Notice that by (A1)–(A4), it follows from Theorem 3 that for all j=1,2,3, the integral equation
(53)xt=∏i=1nfijt,xt+∫t−τitgijs,xsds,t∈ℝ,admits one and only one solution xj∈X. Moreover, for all j=1,2,3 and x0j∈X, the sequence xpj⊂X defined by
(54)xp+1jt=∏i=1nfijt,xpjt+∫t−τitgijs,xpjsds,t∈ℝconverges uniformly to xj.Theorem 7.
Under conditions (A1)–(A6), one has
(55)x1t≤x2t≤x3t,t∈ℝ.Proof.
For allj=1,2,3, let Tj:X⟶X be the operator defined by
(56)Tjxt=∏i=1nfijt,xt+∫t−τitgijs,xsds,x∈X,t∈ℝ.From condition (A5), the operator T2 is nondecreasing, that is,
(57)u,v∈X,ut≤vt,t∈ℝ⟹T2ut≤T2vt,t∈ℝ.Moreover, by (A6), one has
(58)T1ut≤T2ut≤T3ut,u∈X,t∈ℝ.Letx01,x02,x03∈X be such that
(59)x01t≤x02t≤x03t,t∈ℝ.Hence, by (57), one obtains
(60)T2x01t≤T2x02t≤T2x03t,t∈ℝ.On the other hand, by (58), one has
(61)T1x01t≤T2x01t,t∈ℝ,(62)T2x03t≤T3x03t,t∈ℝ.Therefore, using (60), (61), and (62), one deduces that
(63)T1x01t≤T2x02t≤T3x03t,t∈ℝ,that is,
(64)x11t≤x12t≤x13t,t∈ℝ.Repeating the same argument, by induction, one deduces that for all nonnegative integerp and t∈ℝ,
(65)xp1t≤xp2t≤xp3t,where xpj, j=1,2,3, is the sequence defined by (54). Finally, passing to the limit as p⟶∞ in (65), the desired result follows.


## 2.6. Data Dependence of Solutions

Suppose that conditions (C1)–(C5) are satisfied. Then, by Theorem 3, the integral equation admits one and only one solution x∗∈X. Consider now the perturbed problem
(66)yt=∏i=1nFit,yt+∫t−τitGis,ysds,t∈ℝ,where Fi,Gi∈Cℝ×I,J, i=1,2,⋯,n. Suppose that y∗∈X is a solution to the integral equation (66).We have the following data dependence result.Theorem 8.
Suppose that for alli=1,2,⋯,n, there exist σi,ηi>0 such that
(67)fit,u−Fit,u≤σi,git,u−Git,u≤ηi,t∈ℝ,u∈I.Then,(68)x∗−y∗X≤Mn−1∏i=1n−1τi+1∑k=1nσk+τkηk1−Mn−1∏i=1n−1τi+1∑k=1nLfk+Lgkτk.Proof.
Let(69)Syt=∏i=1nSiyt,t∈ℝ,where
(70)Siyt=Fit,yt+∫t−τitGis,ysds,i=1,2,⋯,n.Then, for allt∈ℝ, one has
(71)x∗t−y∗t=Tx∗t−Sy∗t=∏i=1nTix∗t−∏i=1nSiy∗t,where the operator T is defined by (12). Next, by Lemma 1 and (14), one obtains
(72)∣x∗t−y∗t∣≤T2x∗t⋯Tnx∗t∣T1x∗t−S1y∗t∣+S1y∗tT3x∗t⋯Tnx∗t∣T2x∗t−S2y∗t∣+⋯+S1y∗t⋯Sn−1y∗t∣Tnx∗t−Sny∗t∣≤Mn−1∏i=1n−1τi+1∑k=1nTkx∗t−Sky∗t.On the other hand, using (C2) and (C3) and (67), for all k=1,2,⋯,n, one has
(73)Tkx∗t−Sky∗t≤fkt,x∗t−Fkt,y∗t+∫t−τktgks,x∗s−Gks,y∗sds≤fkt,x∗t−fkt,y∗t+fkt,y∗t−Fkt,y∗t+∫t−τktgks,x∗s−gks,y∗sds+∫t−τktgks,y∗s−Gks,y∗sds≤Lfkx∗−y∗X+σk+Lgkτkx∗−y∗X+τkηk=Lfk+Lgkτkx∗−y∗X+σk+τkηk.Hence, by (72), it holds that
(74)x∗−y∗X≤Mn−1∏i=1n−1τi+1∑k=1nLfk+Lgkτkx∗−y∗X+σk+τkηk,which yields
(75)1−Mn−1∏i=1n−1τi+1∑k=1nLfk+Lgkτkx∗−y∗X≤Mn−1∏i=1n−1τi+1∑k=1nσk+τkηk.Finally, by (C4), the desired result follows.



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