#### Abstract

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

#### 1. Introduction and Preliminaries

Consider the Urysohn-Stieltjes integral equation: where is nondecreasing in the second argument (see [1]) and the symbol indicates the integration with respect to . Equations of type (1) and some of their generalizations were considered in the paper (see [2]). We remark here that when , 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, let be continuous functions on We generalize these results to study the existence of weak solutions for the coupled system of Urysohn-Stieltjes functional (delayed) integral equations: in the reflexive Banach space under the weak-weak continuity assumption imposed on

As an application, we study the existence of weak solutions for the coupled system of Hammerstien-Stieltjes functional integral equations:

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, denotes a reflexive Banach space with norm and dual . Denote by the Banach space of strongly continuous functions with sup-norm.

Let be a Banach space with the norm defined as

Now, we shall present some auxiliary results that will be needed in this work. Let be a Banach space (need not be reflexive) and let ; then (1) is said to be weakly continuous (measurable) on if for every is continuous (measurable) on .(2)A function is said to be weakly sequentially continuous if maps weakly convergent sequences in to weakly convergent sequences in .

If is weakly continuous on , then is strongly measurable and hence weakly measurable (see [16, 17]). It is evident that in reflexive Banach spaces, if is weakly continuous function on , then 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 .

*Definition 1. *Let . Then is said to be weakly-weakly continuous at ; if given ; there exists and a weakly open set containing such that
whenever

Now, we have the following fixed point theorem, due to O’Regan, in the Banach space (see [18]).

Theorem 1. *Let be a Banach space, let be a nonempty, bounded, closed, and convex subset of , and let be weakly sequentially continuous and assume that is relatively weakly compact in for each . Then has a fixed point in set .*

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 is a bounded subset of , then the condition is weakly relatively compact and is automatically satisfied. Accordingly, we immediately have the following theorem.

Theorem 2. *Let be a reflexive Banach space with a nonempty, closed, convex, and equicontinuous subset of . Assume that is weakly sequentially continuous. Then has a fixed point in .*

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. *Let be a normed space with and . Then there exists a with and .*

#### 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. *

*Assumption 2. * is a continuous function such that

*Assumption 3. * satisfy the following conditions:
(1) are continuous functions on for every (2) are weakly-weakly continuous functions, (3)There exist two continuous functions and two positive constants , such that .

*Assumption 4. *The functions and are continuous on

*Assumption 5. *For all such that , the functions are nondecreasing on .

*Assumption 6. *, for any .

*Remark 1. *Observe that Assumptions 5 and 6 imply that the function is nondecreasing on the interval , for any fixed (Remark 1 in [5]). Indeed, putting in Assumption 5 and keeping in mind Assumption 6, we obtain the desired conclusion. From this observation, it follows immediately that, for every , the function is of bounded variation on .

*Definition 2. *By a weak solution for the coupled system (2), we mean the pair of functions such that
for all .

Now, let

Then we have the following theorem.

Theorem 3. *Under the Assumptions 1–6, the coupled system of Urysohn-Stieltjes integral equation (2) has at least one weak solution .*

*Proof 1. *Define an operator by
where
For every , is continuous on and are weakly continuous on ; then are continuous for every . Hence, in view of bounded variational of it follows, is weakly Riemann-Stieltjes integrable on with respect to . Thus, make sense.

Now, we can prove that and for , , , and (without loss of generality, assume that ), and there exists , such that
Hence,
Similarly, we can show that
Now,
and
Then from the continuity of the functions, is required in Assumption 4; we deduce that maps into

Define the sets and by
and
Now, define the closed, convex, bounded, equicontinuous set by
Now, let and ; without loss of generality, we may assume . By Proposition 2, we have
Then
Similarly, we can prove that
Therefore, for any ,
that is, Thus,

Note that is a nonempty, uniformly bounded, and strongly equicontinuous subset of by the uniform boundedness of ; according to Proposition 1, is relatively weakly compact.

It remains to prove that is weakly sequentially continuous.

Let and be sequences in weakly convergent to and , respectively (), since and are weakly continuous. Then and converge weakly to and , respectively. Furthermore, and converge strongly to and , respectively. Applying the Lebesgue-dominated convergence theorem, then we get
and
Thus, is weakly sequentially continuous on .

Since all conditions of Theorem 1 are satisfied, then the operator has at least one fixed point 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. *Let and assume that satisfy the following assumptions:
(1) are weakly-weakly continuous functions.(2)There exist continuous functions and constants such that
for Moreover, we put .(3) is a continuous function such that , where is a positive constant.

*Definition 3. *By a weak solution for the coupled system (3), we mean the pair of functions such that
for all .

Being new for the existence of weak solutions of (3), we have the following theorem.

Theorem 4: *Let Assumptions 1, 2, 4, 5, and 7 be satisfied. Then the coupled system of Hammerstien-Stieltjes functional integral equation (3) has at least one weak solution *

*Proof 2. *Let

Then from Assumption 7, we find that the assumptions of Theorem 3 are satisfied and the result follows.

For example, consider the functions defined by the formula

It can be easily seen that the functions and 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:

Therefore, the coupled system (29) has at least one weak solution if the functions , and satisfy Assumptions 1–3.

#### Data Availability

There is no data availability.

#### Conflicts of Interest

The authors declare that there is no conflict of interest regarding the publication of this paper.