# Fixed Points of Subadditive Maps with an Application to a System of Volterra–Fredholm Type Integrodifferential Equations

Authors: Hüseyin Işık; Bahman Moeini; Hassen Aydi; Nabil Mlaiki
Journal: Mathematical Problems in Engineering
 (2019)

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## Abstract

In this paper, some fixed-point theorems are established for strongly subadditive maps onCΩ,ϒ (where CΩ,ϒ denotes the space of ϒ-valued continuous functions on a compact Hausdorff space Ω and ϒ is a unital Banach algebra). Finally, the result is applied to prove the existence and uniqueness of a solution for a system of nonlinear integrodifferential equations.


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## Body

## 1. Introduction

Volterra–Fredholm integro differential equations [1–10] appear in a number of physical models, and an important question is whether these equations can support periodic solutions. This question has been studied extensively by a number of authors; in fact, they investigated the existence of solutions for a class of these kinds of equations using Schauder’s fixed-point theorem and Green’s function. The distance between two zeros of nontrivial solutions to integrodifferential equations was estimated by Domoshnitsky [11]. Ezzinbi and Ndambomve [12] considered the control system governed by some partial functional integrodifferential equations with finite delay in Banach spaces. On the other hand, the fixed-point theory is one of the most rapidly growing topics of nonlinear functional analysis. It is a vast and interdisciplinary subject whose study belongs to several mathematical domains such as classical analysis, functional analysis, operator theory, topology, and algebraic topology. This topic has grown very rapidly perhaps due to its interesting applications in various fields within and outside mathematics such as integral equations, initial and boundary value problems for ordinary and partial differential equations, and others. In our work, using a tool of fixed-point theory, we will ensure the existence and uniqueness of a solution for a system of nonlinear integrodifferential equations.In [13], some properties of subadditive separating maps have been investigated. A fixed-point theorem was also given, and it was used to study the existence and uniqueness of a solution of nonlinear integral type equations. Consider Ω and Λ as compact Hausdorff spaces. Take ϒ as a unital Banach algebra. Denote by CΩ,ϒ (resp. CΛ,ϒ) the set of ϒ-valued continuous functions on Ω (resp. Λ). The operator Δ:CΩ,ϒ⟶CΛ,ϒ is said to be separating if ζξ⋅φξ=0 implies that Δζϱ⋅Δφϱ=0 for all ζ,φ∈CΩ,ϒ and ξ∈Ω,ϱ∈Λ. Recently, an attractive work on separating maps between different spaces of functions (as well as, operator algebras) is considered (see [14–17] and references therein). The existence, uniqueness, continuation, and other properties of solutions for nonlinear integral and integrodifferential equations are studied in [18–25].The purpose of this work is to prove some new fixed-point theorems for strongly subadditive maps and to ensure the existence and uniqueness of solutions for the following system of Volterra–Fredholm type integrodifferential equations:(1)x′t=ζt+∫0tkt,s,xsds+∫0bht,s,xsds,t∈J=0,b,x0=x0,where ζ:J⟶Ω, k,h:J×J×Ω⟶Ω are continuous and x0∈Δ (Δ is a Banach space with norm ⋅). Note that a solution of problem (1) is as follows:(2)xt=x0+∫0tζsds+∫0t∫0sks,ϱ,xϱdϱ+∫0bhs,ϱ,xϱdϱds,t∈J=0,b.


## 2. Preliminaries

In the sequel, denote byB=C0,b,Ω the Banach space of all continuous functions from J into Ω endowed with the norm(3)x=sups∈Jxs.LetΩ and Λ be two compact Hausdorff spaces. Let ϒ be a unital Banach algebra with a unit element e0. For ζ∈CΩ,ϒ, denote by cozζ the cozero set of ζ, i.e., cozζ=ξ∈X:ζξ≠0. ClU (resp. CU) denotes the topological closure (resp. the compliment) of the set U. For a separating map Δ:CΩ,ϒ⟶CΛ,ϒ and ϱ∈Λ, ϱ^∘Δ:CΩ,ϒ⟶ϒ and ϱ^∘Δf=Δζϱ, for all ζ∈CΩ,ϒ.Definition 1 (see [26]).
(a)
The operatorΔ:CΩ,ϒ⟶CΛ,ϒ is said to be subadditive, if(4)Δζ+φ≤Δζ+Δφ,for all ζ,φ∈CΩ,ϒ.(b)
GivenM>0 and a subadditive map Δ, if for any ζ∈CΩ,ϒ there is ε>0 so that Δζ−δφ≤MΔζ−φ, for all φ∈CΩ,ϒ verifying ζ−φ<ε, then Δ is said to be strongly subadditive.Example 1.
Given a continuous mapφ:Λ⟶Λ, take σ∈CΛ,ℂ. The operator Δ:CΩ,ℂ⟶CΛ,ℂ, defined by ζ⟶σ⋅ζ∘φ, is separating and strongly subadditive. Note that Δ it is not additive.Definition 2 (see [26]).
The operatorΔ:CΩ,ϒ⟶CΛ,ϒ is said to be pointwise subadditive if(5)Δζ+φτ≤Δζτ+Δφτ,for all ζ,φ∈CΩ,ϒ and τ∈Λ.Definition 3 (see [26]).
LetΔ:CΩ,ϒ⟶CΛ,ϒ be a separating map and U be an open subset of Ω. Such U is said to be a vanishing set for τ^∘δ if for each ζ∈CΩ,ϒ, cozζ⊆U implies that τ^Δζ=0. The support of τ^∘Δ is given as follows:(6)suppτ^∘Δ=ΩV⊆Ω:Vis a vanishing set forτ^∘Δ.
Fore∈ϒ, define 1e:Ω⟶ϒ by 1eξ=e for each ξ∈Ω. Here, 1e∈CΩ,ϒ. Suppose that Δ1e0τ≠0 for every τ∈Λ. Due to the fact that 1e0⋅0=0 and Δ is separating, we have Δ0=0. The use of 1e is to extend the scalar-valued case to vector-valued case.Remark 1 (decomposition of the identity).
IfK=ℝ or ℂ, then for each finite cover Uii=1n of open subsets of Ω, there exists a continuous decomposition of identity, eii=1n⊆CΩ,ϒ, subordinate to the Ui. That is, ∑i=1nei=1 and cozei⊆Ui, for any i=1,2,…,n [27].
Note thatΔ1e0 is not equal to zero everywhere, so ΔCΩ,ϒ separates Λ. Also, each compact Hausdorff space is a Tychonoff space. The following remark is based on Theorem 1.1 of [28].Remark 2.
LetΔ:CΩ,ϒ⟶CΛ,ϒ be a pointwise subadditive separating map. Then, for each τ∈Λ, supp τ^∘Δ is a singleton.Theorem 1 (see [13]).
LetΔ:CΩ,ϒ⟶CΩ,ϒ be a strongly subadditive map. Suppose that the following assumptions hold:(A1)
For anyζ∈CΩ,ϒ, there is ε>0 such that Δζ−ζ<ε(A2)
There isc>0 such that 0<cM<1 and Δζ≤cζ, for any ζ∈CΩ,ϒ, where M follows using the strong subadditivity of Δ
Then,Δ possesses a unique fixed point ζ0 and limn⟶∞Δnζ=ζ0, for any ζ∈CΩ,ϒ.
ExamplesA.1. and A.2. given in Appendix support Theorem 1.


## 3. Main Results

Consider the following statement:H Let y0=ζ. Define in CΩ,ϒ a sequence yn approximately to Δyn−1, i.e., yn satisfies(7)yn−Δyn−1≤σn,where σn is a sequence of positive numbers, for which limn⟶∞σn=0.Theorem 2.
LetΔ:CΩ,ϒ⟶CΩ,ϒ be a strongly subadditive map. Suppose that the following assumptions hold:(B1)
For eachζ∈CΩ,ϒ, there is ε>0 such that Δnζ−yn−1<ε, for all n≥1, where yn is defined in H(B2)
There isc>0 such that 0<cM<1 and Δζ≤cζ, for each ζ∈CΩ,ϒ, where M comes from strong subadditivity of Δ
Then,Δ admits a unique fixed point ζ0 in CΩ,ϒ. Moreover, for each ζ∈CΩ,ϒ, limn⟶∞Δnζ=ζ0.Proof.
It is sufficient to taken=1; then, by Theorem 1, Δ has a unique fixed point ζ0 in CΩ,ϒ. Also, for each ζ∈CΩ,ϒ, limn⟶∞Δnζ=ζ0.
Now, a natural question arises. If in Theorem 2, one replaces the sequenceΔnζ by yn, then under what conditions we have the case that(8)limn⟶∞yn=ζ0?
In the following, we propose a positive answer to this question.Theorem 3.
Consider the assumptions of Theorem 2 and letζ0 be the unique fixed point of Δ. Then, limn⟶∞yn=ζ0, where yn is defined by H.Proof.
Lety0=ζ. By strong subadditivity of Δ and conditions (B1) and (B2), we have(9)Δm+1y0−ym=Δm+1ζ−ym≤ΔΔmζ−Hym−1+Δym−1−ym≤MΔΔmζ−ym−1+σm≤cMΔmζ−ym−1+σm≤cMcMΔm−1ζ−ym−2+σm−1+σm⋮≤cMcM⋯cMΔζ−y0+σ1⋯+σm−1+σm,where M,c>0, M comes from strong subadditivity of Δ, and 0<cM<1. We get that(10)Δm+1ζ−ym≤∑i=1mcMm−iσi.
Thus,(11)ym−ζ0≤ym−Δm+1ζ+Δm+1ζ−ζ0≤∑i=1mcMm−iσi+Δm+1ζ−ζ0.
Now, letσ>0. Since limn⟶∞σn=0, there is N∈ℕ so that for m≥N, σm≤σ. Thus,(12)∑i=1mcMm−iσi=∑i=1NcMm−iσi+∑i=N+1mcMm−iσi≤cMm−N∑i=1NcMN−iσi+σ∑i=N+1mcMm−i.
Therefore,(13)limm⟶∞∑i=1mcMm−iσi≤σcMN+11−cM.
Sinceσ>0 is arbitrary, limm⟶∞∑i=1mcMm−iσi=0. Also, by Theorem 2, limm⟶∞Δm+1ζ−ζ0=0. Thus, limm⟶∞ym=ζ0.Theorem 4.
SupposeΔ:CΩ,ϒ⟶CΩ,ϒ is a map such that all the conditions of Theorem 1 are satisfied for ΔN (ΔN is Nth iteration of Δ for some positive integer N). Then, Δ admits a unique fixed point.Proof.
By Theorem 1,ΔN:CΩ,ϒ⟶CΩ,ϒ has a unique fixed point ζ∈CΩ,ϒ. However,(14)ΔNΔζ=ΔN+1ζ=ΔΔNζ=Δζ,so Δζ is also fixed point of ΔN. Since the fixed point of ΔN is unique, one writes that Δζ=ζ. Also, if Δφ=φ, then ΔNφ=φ. By uniqueness, we obtain that ζ=φ.
Now, we present an extension of Theorem 1.
LetS denote the class of functions α:ℝ+⟶0,1 such that kαtn⟶1⟹tn⟶0 and kαt≤1 for every t∈ℝ+ (where k>0).Theorem 5.
LetΔ:CΩ,ϒ⟶CΩ,ϒ be a strongly subadditive map. Suppose that the following assumptions hold:(C1)
For allζ∈CΩ,ϒ, there is ε>0 so that Δn+1ζ−Δnζ<ε, for each n∈ℕ(C2)
There isα∈S so that 0<Mαζ<1 and for each ζ∈CΩ,ϒ, Δζ≤αζζ, where M comes from strong subadditivity of Δ
Then,Δ admits a unique fixed point ζ0 and limn⟶∞Δnζ=ζ0, for each ζ∈CΩ,ϒ.Proof.
Fixζ∈CΩ,ϒ and let ζn=Δnζ for n=1,2,3,…. The proof is divided in two steps.Step 1. We claim thatlimn⟶∞ζn+1−ζn=0.By conditions (C1) and (C2) and strong subadditivity ofΔ, the sequence ζn+1−ζn is monotonously decreasing and is bounded below. Hence, limn⟶∞ζn+1−ζn=r≥0. Suppose that r≠0. Using the strong subadditivity of H and condition (C2), we have(15)ζn+2−ζn+1ζn+1−ζn=Δn+2ζ−Δn+1ζΔn+1ζ−Δnζ≤MΔΔn+1ζ−ΔnζΔn+1ζ−Δnζ≤MαΔn+1ζ−ΔnζΔn+1ζ−ΔnζΔn+1ζ−Δnζ=MαΔn+1ζ−Δnζ=Mαζn+1−ζn,n=1,2,3,….Takingn⟶∞, we get 1≤limn⟶∞Mαζn+1−ζn<1 (by (C2)). Since α∈S, this implies that r=0, that is, a contradiction, and so step 1 is completed.Step 2. We shall prove thatζn is a Cauchy sequence.Suppose thatlimsupm,n⟶∞ζm−ζn>0. By triangle inequality, strong subadditivity of Δ, and condition (C2), we have(16)ζm−ζn≤ζm−ζm+1+ζm+1−ζn+1+ζn+1−ζn≤ζm−ζm+1+Mαζm−ζnζm−ζn+ζn+1−ζn≤1−Mαζm−ζn−1ζm−ζm+1+ζn+1−ζn.
Under the assumption thatlimsupm,n⟶∞ζm−ζn>0, Step 1 now implies that(17)limsupm,n⟶∞1−Mαζm−ζn−1=+∞.
Therefore,(18)lim supm,n⟶∞Mαζm−ζn=1.
However, sinceα∈S, this implies that limsupm,n⟶∞ζm−ζn=0, which is again a contradiction. Therefore, Step 2 is established.
Now, letζ∈CΩ,ϒ. Since ζn=Δnζ is a Cauchy sequence in CΩ,ϒ, which is complete, we have limn⟶∞Δnζ=ζ0∈CΩ,ϒ. The continuity of Δ yields that Δζ0=ζ0. Now, we shall show that ζ0 is the unique fixed point. Let φ∈CΩ,ϒ be another fixed point of Δ. So, ζ0=limn⟶∞Δnφ=φ, and this completes the proof.


## 4. An Application to Nonlinear Integrodifferential Equations

Consider the following assumptions:(i)
For allx,y∈C0,b,Ω,(19)x0+∫0tζsds+∫0t∫0sks,ς,x+yςdς+∫0bhs,ς,x+yςdςds≤x0+∫0tζsds+∫0t∫0sks,ς,xςdς+∫0bhs,ς,xςdςds+x0+∫0tζsds+∫0t∫0sks,ς,yςdς+∫0bhs,ς,yςdςdsfor allt∈0,b.(ii)
LetM>0. For each x∈C0,b,Ω, there is ε>0 such that(20)∫0t∫0sks,ς,xς−ks,ς,yςdς+∫0bhs,ς,xς−hs,ς,yςdςds≤Mx0+∫0tζsds+∫0t∫0sks,ς,x−yςdς+∫0bhs,ς,x−yςdςds.for eachy∈C0,b,Ω satisfying x−y<ε and for each t∈0,b.(iii)
For eachx∈C0,b,Ω, there is ε>0 such that(21)x0+∫0tζsds+∫0t∫0sks,ς,xςdς+∫0bhs,ς,xςdςds−xt<ε.(iv)
For eachx∈C0,b,Ω, there is L>0 such that(22)x0+∫0tζsds+∫0t∫0sks,ς,xςdς+∫0bhs,ς,xςdςds≤Lx.Theorem 6.
LetΩ be a Banach space. Suppose that the assumptions i−iv hold. Then, the initial value problem (1) has a unique solution in C0,b,Ω, for which 0<LM<1.Proof.
We shall prove that the operatorG:C0,b,Ω⟶C0,b,Ω, given as(23)Gxt=x0+∫0tζsds+∫0t∫0sks,ς,xςdς+∫0bhs,ς,xςdςds,t∈0,b,admits a unique fixed point, which is a solution of (1). For this, by condition i, we have(24)Gx+y=x0+∫0tζsds+∫0t∫0sks,ς,x+yςdς+∫0bhs,ς,x+yςdςds=sup0≤t≤bx0+∫0tζsds+∫0t∫0sks,ς,x+yςdς+∫0bhs,ς,x+yςdςds≤sup0≤t≤bx0+∫0tζsds+∫0t∫0sks,ς,xςdς+∫0bhs,ς,xςdςds+sup0≤t≤bx0+∫0tζsds+∫0t∫0sks,ς,yςdς+∫0bhs,ς,yςdςds=Gx+Gy,for all x,y∈C0,b,Ω. Hence, G is subadditive. Let M>0. By condition ii, for each x∈C0,b,Ω, there is ϵ>0 such that(25)Gx−Gy=∫0t∫0sks,ς,xςdς+∫0bhs,ς,xςdςds−∫0t∫0sks,ς,yςdς+∫0bhs,ς,yςdςds=sup0≤t≤b∫0t∫0sks,ς,xςdς+∫0bhs,ς,xςdςds−∫0t∫0sks,ς,yςdς+∫0bhs,ς,yςdςds=sup0≤t≤b∫0t∫0sks,ς,xς−ks,ς,yςdς+∫0bhs,ς,xς−hs,ς,yςdςds≤Msup0≤t≤bx0+∫0tζsds+∫0t∫0sks,ς,x−yςdς+∫0bhs,ς,x−yςdςds=MGx−y,for each y∈C0,b,Ω satisfying x−y<ε, so G is strongly subadditive. By condition iii, for each x∈C0,b,Ω, there is ε>0 so that Gx−x<ε. Also, by condition iv, we conclude that for each x∈C0,b,Ω, there is L>0 such that Gx≤Lx, for which 0<LM<1. Therefore, all conditions of Theorem 1 hold, and so (1) has a unique solution.
Now, an example is provided in order to make use of our result.Example 2.
Letla (a≥2) be the map as in Example A.1. (given later in Appendix section). Take Ω=ℝ+∪0 and x∈C0,1,Ω. We consider the following initial value problem:(26)x′t=11+120052/1199∫0ttls+2xsds+∫01sxsds,t∈0,1,x0=0..
For alls,t∈0,1 and x∈C0,1,Ω, we define in (1):(27)ζt=0,kt,s,xs=11+120052/1199tls+2xsandht,s,xs=11+120052/1199sxs,and the norm x−y=sup0≤t≤1xt−yt on C0,1,Ω. Then, C0,1,Ω is a Banach space.(i)
For allx,y∈C0,1,Δ and t∈0,1, we have(28)x0+∫0tζsds+∫0t∫0sks,ς,x+yςdς+∫01hs,ς,x+yςdςds=11+120052/1199∫0t∫0sslς+2x+yςdς+∫01ςx+yςdςds≤11+120052/1199∫0t∫0sslς+2xςdς+∫01ςxςdςds+11+120052/1199∫0t∫0sslς+2yςdς+∫01ςyςdςds=x0+∫0tζsds+∫0t∫0sks,ς,xςdς+∫01hs,ς,xςdςds+x0+∫0tζsds+∫0t∫0sks,ς,xςdς+∫01hs,ς,xςdςds.(ii)
For everyt∈0,1, we have(29)∫0t∫0sks,ς,xς−ks,ς,yςdς+∫01hs,ς,xς−hs,ς,yςdςds=11+120052/1199∫0t∫0sslς+2xς−slς+2yςdς+∫01ςxς−ςyςdςds≤11+120052/1199∫0t∫0s1+12001+ς21199slς+2x−yςdς+∫01ςxς−yςdςds≤1+120032/11991+120052/1199∫0t∫0sslς+2x−yςdς+∫01ςxς−yςdςds=Mx0+∫0tζsds+∫0t∫00ks,ς,x−yςdς+∫01hs,ς,x−yςdςds,withM=1+120032/1199 and for each y∈C0,1,Ω satisfying x−y<ε.(iii)
For everyx∈C0,1,Ω, we have(30)x0+∫0tζsds+∫0t∫0sks,ς,xςdς+∫01hs,ς,xςdςds−xt=11+120052/1199∫0t∫0sslς+2xςdς+∫01ςxςdςds−xt≤11+120052/1199∫0t∫0sslς+2xςdς+∫01ςxςdςds+xt≤sup0≤t≤1∫0t∫0ssxςdς+∫01ςxςdςds+M1≤sup0≤t≤1∫0ts2M1+12M1ds+M1<2M1=ϵ.
Again byiii, we have(31)x0+∫0tζsds+∫0t∫0sks,ς,xςdς+∫01hs,ς,xςdςds≤11+120052/119956x<11+120052/1199x.
Therefore,(32)x0+∫0tζsds+∫0t∫0sks,ς,xςdς+∫01hs,ς,xςdςds≤Lx,with L=1/1+120052/1199. Also, 0<LM=1+120032/1199/1+120052/1199<1. Thus, all conditions of Theorem 6 hold, and so the initial value problem (26) possesses a unique solution in C0,1,Ω.


## 5. Conclusion

We established some fixed-point theorems for strongly subadditive maps. We applied the obtained results to ensure the existence and uniqueness of a solution for a system of nonlinear integrodifferential equations. Moreover, some related examples have been presented in support of the given results.



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