# Weak Solutions of a Coupled System of Urysohn-Stieltjes Functional (Delayed) Integral Equations

Authors: A. M. A. El-Sayed; M. M. A. Al-Fadel
Journal: Complexity
 (2018)

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## Abstract

We study the existence of weak solutions for the coupled system of functional integral equations of Urysohn-Stieltjes type in the reflexive Banach spaceE. As an application, the coupled system of Hammerstien-Stieltjes functional integral equations is also studied.


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## Body

## 1. Introduction and Preliminaries

Consider the Urysohn-Stieltjes integral equation:(1)xt=pt+∫01ft,s,xsdsgt,s,t∈I=0,1,where g:I×I→R is nondecreasing in the second argument (see [1]) and the symbol ds indicates the integration with respect to s. Equations of type (1) and some of their generalizations were considered in the paper (see [2]). We remark here that when E=R, these types of equations have been studied by Banaś (see [1–5]) and also by some other authors, for example (see [6–10] and for coupled systems [11]). For the solutions in a reflexive Banach space (see [12, 13]).In this paper, letψit≤t be continuous functions on 0,1.We generalize these results to study the existence of weak solutions x,y∈CI,E×CI,E for the coupled system of Urysohn-Stieltjes functional (delayed) integral equations:
(2)xt=a1t+∫01f1t,s,yψ1sdsg1t,s,t∈I,yt=a2t+∫01f2t,s,xψ2sdsg2t,s,t∈I,in the reflexive Banach space E under the weak-weak continuity assumption imposed on fi:I×I×E→E,i=1,2.As an application, we study the existence of weak solutionsx,y∈CI,E for the coupled system of Hammerstien-Stieltjes functional integral equations:
(3)xt=a1t+∫01k1t,sh1s,yψ1sdsg1t,s,t∈I,yt=a2t+∫01k2t,sh2s,xψ2sdsg2t,s,t∈I.For the definition, background, and properties of the Stieltjes integral, we refer to Banaś [1]. However, the coupled system of integral equations has been studied, recently, by some authors (see [14, 15]).Throughout this paper, otherwise stated,E denotes a reflexive Banach space with norm ⋅ and dual E∗. Denote by CI,E the Banach space of strongly continuous functions x:I→E with sup-norm.LetX=CI,E×CI,E=ut=xt,yt:x∈CI,E,y∈CI,E,t∈I be a Banach space with the norm defined as
(4)∥x,y∥X=∥x∥CI,E+∥y∥CI,E,∀x,y∈X.Now, we shall present some auxiliary results that will be needed in this work. LetE be a Banach space (need not be reflexive) and let x:a,b→E; then
(1)
x. is said to be weakly continuous (measurable) on a,b if for every ϕ∈E∗,ϕx. is continuous (measurable) on a,b.
(2)
A functionh:E→E is said to be weakly sequentially continuous if h maps weakly convergent sequences in E to weakly convergent sequences in E.Ifx is weakly continuous on I, then x is strongly measurable and hence weakly measurable (see [16, 17]). It is evident that in reflexive Banach spaces, if x is weakly continuous function on a,b, then x is weakly Riemann integrable (see [17]). Since the space of all weakly Riemann-Stieltjes integrable functions is not complete, we will restrict our attention to the existence of weak solutions of the coupled system (2) in the space CI,E×CI,E.Definition 1.
Letf:I×E→E. Then ft,u is said to be weakly-weakly continuous at t0,u0; if given ϵ>0,ϕ∈E∗; there exists δ>0 and a weakly open set U containing u0 such that
(5)ϕft,u−ft0,u0<ϵ,whenever
(6)t−t0<δ,u∈U.Now, we have the following fixed point theorem, due to O’Regan, in the Banach space (see [18]).Theorem 1.
LetE be a Banach space, let Q be a nonempty, bounded, closed, and convex subset of CI,E, and let F:Q→Q be weakly sequentially continuous and assume that FQt is relatively weakly compact in E for each t∈I. Then F has a fixed point in set Q.Recall [19] that a subset of a reflexive Banach space is weakly compact if and only if it is closed in the weak topology and bounded in the norm topology. Thus, putting in mind that TQt is a bounded subset of E, then the condition TQt is weakly relatively compact and is automatically satisfied. Accordingly, we immediately have the following theorem.Theorem 2.
LetE be a reflexive Banach space with Q a nonempty, closed, convex, and equicontinuous subset of CI,E. Assume that T:Q→Q is weakly sequentially continuous. Then T has a fixed point in Q.Proposition 1.
In the reflexive Banach space, the subset is weakly relatively compact if and only if it is bounded in the norm topology.Proposition 2.
LetE be a normed space with y∈E and y≠0. Then there exists a φ∈E∗ with φ=1 and y=φy.


## 2. Main Results

In this section, we present our main result by proving the existence of weak solutions for the coupled system of Urysohn-Stieltjes integral equation (2) in the reflexive Banach space. Let us first state the following assumptions:Assumption 1.
ai∈CI,E,i=1,2.Assumption 2.
ψi:I→I is a continuous function such that ψit≤t.Assumption 3.
fi:I×I×E→E,i=1,2 satisfy the following conditions:
(1)
fi…,s,xψis are continuous functions on I for every x∈E.
(2)
fit,… are weakly-weakly continuous functions, ∀t∈I.
(3)
There exist two continuous functionsmit,mi:I×I→I and two positive constants bi, such that fit,s,x≤mit,s+bix,i=1,2.Assumption 4.
The functionst→git,1 and t→git,0 are continuous on I.Assumption 5.
For allt1,t2∈I such that t1<t2, the functions s→git2,s−git1,s are nondecreasing on I.Assumption 6.
gi0,s=0, for any s∈I.Remark 1.
Observe that Assumptions5 and 6 imply that the function s→gt,s is nondecreasing on the interval I, for any fixed t∈I (Remark 1 in [5]). Indeed, putting t2=t,t1=0 in Assumption 5 and keeping in mind Assumption 6, we obtain the desired conclusion. From this observation, it follows immediately that, for every t∈I, the function s→gt,s is of bounded variation on I.Definition 2.
By a weak solution for the coupled system (2), we mean the pair of functions x,y∈CI,E×CI,E such that
(7)φxt=φa1t+∫01φf1t,s,yψ1sdsg1t,s,t∈I,φyt=φa2t+∫01φf2t,s,xψ2sdsg2t,s,t∈I,
for allφ∈E∗.Now, let(8)μ=maxsuptgit,1+suptgit,0,t∈I,i=1,2,M=maxmit,s:t,s∈I.Then we have the following theorem.Theorem 3.
Under the Assumptions1–6, the coupled system of Urysohn-Stieltjes integral equation (2) has at least one weak solution x,y∈CI,E×CI,E.Proof 1.
Define an operatorA by
(9)Ax,y=A1y,A2y,where
(10)A1yt=a1t+∫01f1t,s,yψ1sdsg1t,s,t∈I,A2xt=a2t+∫01f2t,s,xψ2sdsg2t,s,t∈I.
For everyxi∈CI,E, fi…,s,xψis is continuous on I, and fit,…,xψi. are weakly continuous on I; then φfit,…,xψi. are continuous for every φ∈E∗. Hence, in view of bounded variational of gi it follows, fit,s,xψis is weakly Riemann-Stieltjes integrable on I with respect to s→git,s. Thus, Ai make sense.
Now, we can prove thatA1:CI,E→CI,E and for ϵ>0, t1,t2∈I, t1<t2, and t2−t1<ϵ (without loss of generality, assume that A1yt2−A1yt1≠0), and there exists φ∈E∗, such that
(11)A1yt2−A1yt1≤φa1t2−a1t1+∫01φf1t2,s,yψ1sdsg1t2,s−∫01φf1t1,s,yψ1sdsg1t1,s≤a1t2−a1t1+∫01φf1t2,s,yψ1sdsg1t2,s−∫01φf1t1,s,yψ1sdsg1t2,s+∫01φf1t1,s,yψ1sdsg1t2,s−∫01φf1t1,s,yψ1sdsg1t1,s≤a1t2−a1t1+∫01φf1t2,s,yψ1s−f1t1,s,yψ1sds⋁z=0sg1t2,z+∫01φf1t1,s,yψ1sds⋁z=0sg1t2,z−g1t1,z≤a1t2−a1t1+f1t2,s,y−f1t1,s,y∫01dsg1t2,s+∫01f1t1,s,ydsg1t2,s−g1t1,s≤a1t2−a1t1+f1t2,s,y−f1t1,s,yg1t2,1−g1t2,0+∫01m1t,sdsg1t2,s−g1t1,s+∫01b1ydsg1t2,s−g1t1,s≤a1t2−a1t1+f1t2,s,y−f1t1,s,yg1t2,1−g1t2,0+M∫01dsgt2,s−gt1,s+b1r1∫01dsg1t2,s−g1t1,s≤a1t2−a1t1+f1t2,s,y−f1t1,s,yg1t2,1−g1t2,0+M+b1r1gt2,1−gt1,1−gt2,0−gt1,0≤a1t2−a1t1+f1t2,s,y−f1t1,s,yg1t2,1−g1t2,0+M+b1r1g1t2,1−g1t1,1+g1t2,0−g1t1,0.
Hence,(12)A1yt2−A1yt1≤a1t2−a1t1+f1t2,s,y−f1t1,s,yg1t2,1−g1t2,0+M+b1r1g1t2,1−g1t1,1+g1t2,0−g1t1,0.
Similarly, we can show that(13)A2xt2−A2xt1≤a2t2−a2t1+f2t2,s,y−f2t1,s,yg2t2,1−g2t2,0+M+b2r2g2t2,1−g2t1,1+g2t2,0−g2t1,0.
Now,(14)Ax,yt2−Ax,yt1=A1yt2,A2xt2−A1yt1,A2xt1=A1yt2−A1yt1,A2xt2−A2xt1,and
(15)Ax,yt2−Ax,yt1≤A1yt2−A1yt1+A2xt2−A2xt1≤a1t2−a1t1+f1t2,s,y−f1t1,s,yg1t2,1−g1t2,0+M+b1r1g1t2,1−g1t1,1+g1t2,0−g1t1,0+a2t2−a2t1+f2t2,s,y−f2t1,s,yg2t2,1−g2t2,0+M+b2r2g2t2,1−g2t1,1+g2t2,0−g2t1,0.
Then from the continuity of the functions,gi is required in Assumption 4; we deduce that A maps X into X.
Define the setsQ1 and Q2 by
(16)Q1=y∈CI,E:y≤r1,r1=∥a1∥+Mμ1−b1μ,and
(17)Q2=x∈CI,E:x≤r2,r2=∥a2∥+Mμ1−b2μ.
Now, define the closed, convex, bounded, equicontinuous setQ by
(18)Q=u=x,y∈X:u≤r1+r2,ut2−ut1≤a1t2−a1t1+f1t2,s,y−f1t1,s,yg1t2,1−g1t2,0+M+b1r1g1t2,1−g1t1,1+g1t2,0−g1t1,0+a2t2−a2t1+f2t2,s,y−f2t1,s,yg2t2,1−g2t2,0+M+b2r2g2t2,1−g2t1,1+g2t2,0−g2t1,0.
Now, lety∈Q1 and x∈Q2; without loss of generality, we may assume A1y≠0,A2xt≠0,t∈I. By Proposition 2, we have
(19)A1yt=φA1yt≤φa1t+φ∫01f1t,s,yψ1sdsg1t,s≤a1+∫01∣φf1t,s,yψ1s∣ds⋁z=0sg1t,z≤a1+∫01∥f1t,s,yψ1s∥ds⋁z=0sg1t,z≤a1+∫01m1t,sdsg1t,s+∫01b1ydsg1t,s≤a1+M∫01dsg1t,s+b1r1∫01dsg1t,s≤a1+Mg1t,1−g1t,0+b1r1g1t,1−g1t,0≤∥a1∥+M+b1r1supt∈Ig1t,1+supt∈Ig1t,0≤a1+M+b1r1μ.
Then(20)A1yt≤a1+M+b1r1μ=r1.
Similarly, we can prove that(21)A2xt≤a2+M+b2r2μ=r2.
Therefore, for anyu∈Q,
(22)Aut=Ax,yt=A1yt,A2xt≤A1yt+A2xt≤a1+M+b1r1μ+a2+M+b2r2μ=r1+r2,that is, ∀u∈Q⇒Au∈Q⇒AQ⊂Q. Thus, A:Q→Q.
Note thatQ is a nonempty, uniformly bounded, and strongly equicontinuous subset of X, by the uniform boundedness of AQ; according to Proposition 1, AQ is relatively weakly compact.
It remains to prove thatA is weakly sequentially continuous.
Letynt and xnt be sequences in CI,E weakly convergent to yt and xt, respectively (∀t∈I), since f1t,s,… and f2t,s,… are weakly continuous. Then f1t,s,ynψ1s and f2t,s,xnψ2s converge weakly to f1t,s,yψ1s and f2t,s,xψ2s, respectively. Furthermore, ∀φ∈E∗φf1t,s,ynψ1s and φf2t,s,xnψ2s converge strongly to φf1t,s,yψ1s and φf2t,s,xψ2s, respectively. Applying the Lebesgue-dominated convergence theorem, then we get
(23)φ∫01f1t,s,ynψ1sdsg1t,s=∫01φf1t,s,ynψ1sdsg1t,s⟶∫01φf1t,s,yψ1sdsg1t,s,∀φ∈E∗,t∈I,and
(24)φ∫01f2t,s,xnψ2sdsg2t,s=∫01φf2t,s,xnψ2sdsg2t,s⟶∫01φf2t,s,xψ2sdsg2t,s,∀φ∈E∗,t∈I.
Thus,A is weakly sequentially continuous on Q.
Since all conditions of Theorem1 are satisfied, then the operator A has at least one fixed point x,y=u∈Q and the coupled system of Urysohn-Stieltjes integral equation (2) has at least one weak solution.


## 3. Hammerstein-Stieltjes Coupled System

This section, as an application, deals with the existence of weak continuous solution for the coupled system of Hammerstein-Stieltjes functional integral equation (3). Consider the following assumption:Assumption 7.
Lethi:I×E→E and ki:I×I→R+ assume that hi,ki satisfy the following assumptions:
(1)
his,xψis are weakly-weakly continuous functions.
(2)
There exist continuous functionsmi∗t and constants bi>0 such that
(25)hit,x≤mi∗t+bix,for t,s∈I,x∈E. Moreover, we put M∗=maxmi∗t:t∈I,M∗>0.
(3)
kit,s is a continuous function such that K=suptkit,s, where K is a positive constant.Definition 3.
By a weak solution for the coupled system (3), we mean the pair of functions x,y∈CI,E×CI,E such that
(26)φxt=φa1t+∫01k1t,sφh1s,yψ1sdsg1t,s,t∈I,φyt=φa2t+∫01k2t,sφh2s,xψ2sdsg2t,s,t∈I,for all φ∈E∗.Being new for the existence of weak solutions of (3), we have the following theorem.Theorem 4:
Let Assumptions1, 2, 4, 5, and 7 be satisfied. Then the coupled system of Hammerstien-Stieltjes functional integral equation (3) has at least one weak solution x,y∈X.Proof 2.
Let(27)fit,s,xs=kit,shis,xψis.Then from Assumption7, we find that the assumptions of Theorem 3 are satisfied and the result follows.For example, consider the functionsgi:I×I→R defined by the formula
(28)g1t,s=tlnt+st,fort∈0,1,s∈I,0,fort=0,s∈I,g2t,s=tt+s−1,t∈I.It can be easily seen that the functionsg1t,s and g2t,s satisfy assumptions (iv)-(vi) given in Theorem given in Theorem 3. In this case, the coupled system of Urysohn-Stieltjes integral equation (2) has the following form:
(29)xt=a1t+∫01tt+sf1t,s,yψ1sds,t∈I,(30)yt=a2t+∫01tf2t,s,xψisds,t∈I.Therefore, the coupled system (29) has at least one weak solution u=x,y∈X, if the functions ai, ψi, and fi satisfy Assumptions 1–3.



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