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

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, let ψi t ≤ t be continuous functions on 0, 1 We generalize these results to study the existence of weak solutions x, y ∈ C I, E × C I, E for the coupled system of Urysohn-Stieltjes functional (delayed) integral equations:


Introduction and Preliminaries
Consider the Urysohn-Stieltjes integral equation: x t = p t + 1 0 f t, s, x s d s g t, s , t ∈ I = 0, 1 , 1 where g I × I → R is nondecreasing in the second argument (see [1]) and the symbol d s 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][2][3][4][5]) and also by some other authors, for example (see [6][7][8][9][10] and for coupled systems [11]).For the solutions in a reflexive Banach space (see [12,13]).In this paper, let ψ i t ≤ t be continuous functions on 0, 1 We generalize these results to study the existence of weak solutions x, y ∈ C I, E × C I, E for the coupled system of Urysohn-Stieltjes functional (delayed) integral equations: x t = a 1 t + 1 0 f 1 t, s, y ψ 1 s d s g 1 t, s , t ∈ I, y t = a 2 t + 1 0 f 2 t, s, x ψ 2 s d s g 2 t, s , t ∈ I, 2 in the reflexive Banach space E under the weak-weak continuity assumption imposed on f i I × I × E → E, i = 1, 2 As an application, we study the existence of weak solutions x, y ∈ C I, E 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, E denotes a reflexive Banach space with norm ⋅ and dual E * .Denote by C I, E the Banach space of strongly continuous functions x I → E with sup-norm.
Let X = C I, E × C I, E = u t = x t , y t : x ∈ C I, E , y ∈ C I, E , t ∈ I be a Banach space with the norm defined as Now, we shall present some auxiliary results that will be needed in this work.Let E 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 * , ϕ x is continuous (measurable) on a, b .
(2) A function h E → E is said to be weakly sequentially continuous if h maps weakly convergent sequences in E to weakly convergent sequences in E.
If x 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 C I, E × C I, E .Definition 1.Let f I × E → E. Then f t, u is said to be weakly-weakly continuous at t 0 , u 0 ; if given ϵ > 0, ϕ ∈ E * ; there exists δ > 0 and a weakly open set U containing u 0 such that Now, we have the following fixed point theorem, due to O'Regan, in the Banach space (see [18]).
Theorem 1.Let E be a Banach space, let Q be a nonempty, bounded, closed, and convex subset of C I, E , and let F Q → Q be weakly sequentially continuous and assume that FQ t 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 TQ t is a bounded subset of E, then the condition TQ t is weakly relatively compact and is automatically satisfied.Accordingly, we immediately have the following theorem.
Theorem 2. Let E be a reflexive Banach space with Q a nonempty, closed, convex, and equicontinuous subset of C I, 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. Let E be a normed space with y ∈ E and y ≠ 0.
Then there exists a φ ∈ E * with φ = 1 and y = φ y .

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 6. g i 0, s = 0, for any s ∈ I.
Remark 1. Observe that Assumptions 5 and 6 imply that the function s → g t, s is nondecreasing on the interval I, for any fixed t ∈ I (Remark 1 in [5]).Indeed, putting t 2 = t, t 1 = 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 → g t, 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 ∈ C I, E × C I, E such that Then we have the following theorem.where For every x i ∈ C I, E , f i … , s, x ψ i s is continuous on I, and f i t, … , x ψ i are weakly continuous on I; then φ f i t, … , x ψ i are continuous for every φ ∈ E * .Hence, in view of bounded variational of g i it follows, f i t, s, x ψ i s is weakly Riemann-Stieltjes integrable on I with respect to s → g i t, s .Thus, A i make sense.Now, we can prove that A 1 C I, E → C I, E and for ϵ > 0, t 1 , t 2 ∈ I, t 1 < t 2 , and t 2 − t 1 < ϵ (without loss of generality, assume that A 1 y t 2 − A 1 y t 1 ≠ 0), and there exists φ ∈ E * , such that Similarly, we can show that Then from the continuity of the functions, g i is required in Assumption 4; we deduce that A maps X into X Define the sets Q 1 and Q 2 by and Now, define the closed, convex, bounded, equicontinuous set Q by Now, let y ∈ Q 1 and x ∈ Q 2 ; without loss of generality, we may assume A 1 y ≠ 0, A 2 x t ≠ 0, t ∈ I.By Proposition 2, we have Similarly, we can prove that Therefore, for any u ∈ Q, , respectively.Applying the Lebesguedominated convergence theorem, then we get

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Thus, A is weakly sequentially continuous on Q.

Complexity
Since all conditions of Theorem 1 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.

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 h i I × E → E and k i I × I → R + assume that h i , k i satisfy the following assumptions: (1) h i s, x ψ i s are weakly-weakly continuous functions.
(2) There exist continuous functions m * i t and constants b i > 0 such that (
Being new for the existence of weak solutions of (3), we have the following theorem.x, y ∈ X Proof 2. Let f i t, s, x s = k i t, s h i s, x ψ i s 27 Then from Assumption 7, we find that the assumptions of Theorem 3 are satisfied and the result follows.
For example, consider the functions g i I × I → R defined by the formula g 1 t, s = tln t + s t , for t ∈ 0, 1 , s ∈ I, 0, for t = 0, s ∈ I, g 2 t, s = t t + s − 1 , t ∈ I

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It can be easily seen that the functions g 1 t, s and g 2 t, 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: x t = a 1 t + Therefore, the coupled system (29) has at least one weak solution u = x, y ∈ X, if the functions a i , ψ i , and f i satisfy Assumptions 1-3.

Theorem 3 .Proof 1 .
Under the Assumptions 1-6, the coupled system of Urysohn-Stieltjes integral equation (2) has at least one weak solution x, y ∈ C I, E × C I, E .Define an operator A by A x, y = A 1 y, A 2 y , 9

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 nonempty, uniformly bounded, and strongly equicontinuous subset of X, by the uniform boundedness of AQ; according to Proposition 1, AQ is relatively weakly compact.… and f 2 t, s, … are weakly continuous.Then f 1 t, s, y n ψ 1 s and f 2 t, s, x n ψ 2 s converge weakly to f 1 t, s, y ψ 1 s and f 2 t, s, x ψ 2 s , respectively.Furthermore, ∀φ ∈ E * φ f 1 t, s, y n ψ 1 s and φ f 2 t, s, x n ψ 2 s converge strongly to φ f 1 t, s, y ψ 1 s and φ f 2 t, s, x ψ 2 s