An Implicit Hybrid Delay Functional Integral Equation: Existence of Integrable Solutions and Continuous Dependence

: In this work, we are discussing the solvability of an implicit hybrid delay nonlinear functional integral equation. We prove the existence of integrable solutions by using the well known technique of measure of noncompactnes. Next, we give the sufﬁcient conditions of the uniqueness of the solution and continuous dependence of the solution on the delay function and on some functions. Finally, we present some examples to illustrate our results.


Introduction
The study of implicit differential and integral equations has received much attention over the last 30 years or so. For instance, Nieto et al. [1] have studied IFDE via the Liouville-Caputo Derivative. The integrable solutions of IFDEs has been studied in [2]. Moreover, IFDEs have recently been studied by several researchers; Dhage and Lakshmikantham [3] have proposed and studied hybrid differential equations. Zhao et al. [4] have worked at hybrid fractional differential equations and expanded Dhage's approach to fractional order. A fractional hybrid two-point boundary value problem had been studied by Sun et al. [5]. The technique of measure of noncompactness is found to be a fruitful one to obtain the existence results for a variety of differential and integral equations, for example, see [6][7][8][9][10][11][12][13][14].
Srivastava et al. [15] have studied the existence of monotonic integrable a.e. solution of nonlinear hybrid implicit functional differential inclusions of arbitrary fractional orders by using the measure of noncompactness technique.
Here, we investigate the existence of integrable solutions of an implicit hybrid delay functional integral equation g(t, x(t)) = f 1 t, x(t) − h(t, x(t)) g(t, x(t)) , s, x(s) − h(s, x(s)) g(s, x(s)) ds , t ∈ [0, 1]. (1) where ϕ : [0, 1] → [0, 1], ϕ(t) ≤ t is nondecreasing continuous function. The main tool of our study is the technique of measure noncompactness. Furthermore, we studied the continuous dependence on the delay function ϕ and on the two functions f 1 and f 2 . Our article is organized as follows: In Section 2 we introduce some preliminaries. Existence results are presented in Section 3. Section 4 contains the continuous dependence of the unique solution on the delay function ϕ and of the two functions f 1 and f 2 . Section 5 presents two examples to verify our theorems. Lastly, conclusions are stated.

Preliminaries
We present here some definitions and basic auxiliary results that will be needed to achieve our aim.
Let L 1 = L 1 (I) be the class of Lebesgue integrable functions on the interval Now, let (E, . ) denote an arbitrary Banach space with zero element θ and X a nonempty bounded subset of E. Moreover, denote by B r = B(θ, r) the closed ball in E centered at θ and with the radius r.
The measure of weak noncompactness defined by De Blasi [16] is given by The function β(X) possesses several useful properties that may be found in De Blasi's paper [16]. The convenient formula for the function β(X) in L 1 was given by Appell and De Pascale [17] as follows: where the symbol meas D stands for Lebesgue measure of the set D.
Next, we shall also use the notion of the Hausdorff measure of noncompactness χ [6] defined by In the case when the set X is compact in measure, the Hausdorff and De Blasi measures of noncompactness will be identical. Namely, we have the following [16].
Theorem 1. Let X be an arbitrary nonempty bounded subset of L 1 . If X is compact in measure, then β(X) = χ(X).
Now, we will recall the fixed point theorem from Banaś [18].
Theorem 2. Let Q be a nonempty, bounded, closed, and convex subset of E, and let T : Q → Q be a continuous transformation, which is a contraction with respect to the Hausdorff measure of noncompactness χ, that is, there exists a constant α ∈ [0, 1] such that χ(TX) ≤ αχ(X) for any nonempty subset X of Q. Then, T has at least one fixed point in the set Q.
We present some criterion for compactness in measure in the next section; the complete description of compactness in measure was given in Banaś [6], but the following sufficient condition will be more convenient for our purposes [6]. Theorem 3. Let X be a bounded subset of L 1 . Assume that there is a family of measurable subsets (Ω c ) 0≤c≤b−a of the interval (a, b) such that meas Ω c = c. If for every c ∈ [0, b − a], and for every x ∈ X, then, the set X is compact in measure.

Main Results
Now, let I = [0, 1] and consider the following: (H 1 ) (i) f 1 : I × R × R → R is a Carathéodory function which is measurable in t ∈ I, ∀u, v ∈ R × R and continuous in u, v ∈ R × R, ∀t ∈ I.
(ii) There exists a measurable and bounded function m 1 : I → I and nonnegative constant b 1 such that (iii) f 1 is nondecreasing on the set I × R × R with respect to all the three variables, i.e., for almost all (t 1 , t 2 ) ∈ I 2 such that t 1 ≤ t 2 and for all u 1 ≤ u 2 and v 1 ≤ v 2 (H 2 ) f 2 : I × I × R → R is a Carathéodory function, and a continuous function m 2 : I × I → R, and nonnegative constant b 2 such that such that Moreover, f 2 is nondecreasing on the set I × R × R with respect to all the three variables. (H 3 ) ϕ : I → I, ϕ(t) ≤ t is nondecreasing. function. (H 4 ) g : I × R → R \ {0} and h : I × R → R satisfy the following: (i) They are nondecreasing on the set I × R with respect to both variables, i.e., for almost all (t 1 , t 2 ) ∈ I 2 such that t 1 ≤ t 2 and for all (ii) They are measurable in t ∈ I for every x ∈ R and continuous in x ∈ R for every t ∈ I, and there exist two integrable functions a i ∈ L 1 (I) and two positive constants l i , (i = 1, 2.) such that |h(t, x)| ≤ |a 1 (t)| + l 1 |x| and |g(t, x)| ≤ |a 2 (t)| + l 2 |x|. Let then the integral Equation (1) can be reduced to where x satisfies the Equation.
Thus, we have proved the following result.

Proof. Define the set
Consider the integral Equation (6) and define the operator Let y ∈ Q ρ , then Hence the operator maps the ball B ρ into itself where Now, Q ρ contains all positive and nondecreasing functions a.e. on I. obviously the set Q ρ is nonempty, bounded and convex. To prove that Q ρ is closed we have {x m } ⊂ Q ρ , which converges strongly to x. Then {x m } converges in measure to x and we deduce the existence of a subsequence {x k } of {x m } which converges to x a.e. on I (see [19]). Therefore, x is nondecreasing a.e. on I which means that x ∈ Q ρ . Hence the set Q ρ is compact in measure(see Lemma 2 in [7], p. 63).
Using (H 1 )-(H 3 ), then maps Q ρ into itself, is continuous on Q ρ , and transforms a nondecreasing a.e. and positive function into a function with same type (see [7]).
To show that the operator is a contraction with respect to the weak noncompactness measure β. Let us start by fixing > 0 and X ⊂ Q ρ . Furthermore, if we select a measurable subset D ⊂ I as such meas D ≤ , then for any x ∈ X using the same assumptions and argument as in [6,7], we obtain with β is the De Blasi measure of weak noncompactness. The set X is compact in measure, so Hausdorff and De Blasi measures of noncompactness will be identical [16], then where χ is the Hausdorff measure of noncompactness. Since b 1 + b 1 b 2 < 1, it follows, from fixed point theorem [18], that is a contraction with regard to the measure of noncompactness χ and has at least one fixed point in Q ρ which show that Equation (6) has at least one positive a.e. nondecreasing solution y ∈ L 1 .

Solvability of Equation (4)
In this section, the existence of a.e. nondecreasing solutions x ∈ L 1 for the Equation (7) will be studied x(t)), t ∈ I Theorem 5. Let the assumptions (H 2 ), (H 4 ) be satisfied. Let the assumptions of Theorem 4 be satisfied. Assume that l 1 + M l 2 < 1. Then there is at least one a.e. nondecreasing solution x ∈ L 1 to (7).

Proof. Interpret the set in the form
Let x ∈ L 1 and M = sup t∈I |y(t)|, then by assumptions (H 2 )-(H 4 ), we find that Then for t ∈ I, we have Hence A maps B r into itself where Allowing Q r to be a subset of B r containing all functions that are nonnegative and a.e. nondecreasing on I, we may conclude that Q r is nonempty, closed, convex, bounded, and compact in measure ( [6], p. 780). Now Q r is a bounded subset of L 1 that contains all positive and nondecreasing a.e. functions on I, then Q r is compact in measure (see Lemma 2 in [7], p. 63). As a result of assumption (H 4 ), A maps Q r into itself, is continuous on Q r , and turns a nondecreasing a.e. and positive function into a function of the same type (see [7]). Thus, A is shown to be a contraction with regard to the weak noncompactness measure β. Let us start by fixing > 0 and X ⊂ Q r . Furthermore, if we select a measurable subset D ⊂ I as such meas D ≤ , then for any x ∈ X using the same assumptions and argument as in [6,7], we obtain Then we find β(Ax(t)) ≤ (l 1 + M l 2 ) β(x(t)).

This implies
with β is the De Blasi measure of weak noncompactness. The set X is compact in measure, so Hausdorff and De Blasi measures of noncompactness will be identical [16], then where χ is the Hausdorff measure of noncompactness. Since l 1 + M l 2 < 1, it follows, from fixed point theorem [18], that A is a contraction with regard to the measure of noncompactness χ and has at least one fixed point in Q r which show that Equation (7) has at least one positive nondecreasing a.e. solution x ∈ L 1 . Now, we are in position to state an existence result for the hybrid implicit functional Equation (1). Theorem 6. Let the assumptions of Theorems 4 and 5 be satisfied. Then the implicit hybrid delay functional integral Equation (1) has at least one nondecreasing a.e. solution x ∈ L 1 which satisfies (7) where y ∈ L 1 is the nondecreasing a.e. solution of (6).

Continuous Dependence
Here, we investigate the continuous dependence of the unique solution x ∈ L 1 on the delay function ϕ and on the two functions f 1 and f 2 .
(iii) f 1 is nondecreasing on the set I × R × R with respect to all the three variables, i.e., for almost all (t 1 , t 2 ) ∈ I 2 such that t 1 ≤ t 2 and for all u 1 ≤ u 2 and v 1 ≤ v 2 Moreover, f 2 is nondecreasing on the set I × R × R with respect to all the three variables. (H * 3 ) g : I × R → R \ {0}, and h : I × R → R, are measurable in t ∈ I for every x, y ∈ R as well as meet the Lipschitz condition for all t ∈ I and u, v ∈ R. Moreover h, g are nondecreasing a.e. in the two arguments. Note Proof. Let y 1 , y 2 be solutions of Equation (6), then Taking supermum for t ∈ I, we have which implies y 1 = y 2 . Hence the solution of the problem (6) is unique.
Next, we prove the following result. Proof. Firstly, Theorem 5 proved that the functional Equation (7) has at least one solution. Now let x 1 , x 2 ∈ L 1 (I) be two solutions of (7). Then for t ∈ I, we have Then for t ∈ I, and |y(t)| < M, we obtain Hence and then the solution of (7)   Proof. Let y be the unique solution of the functional integral Equation (6) and y * is the one of the equation Then | f 2 (t, s, y(s)) − f 2 (t, s, y * (s))|ds. Now, |ϕ(t) − ϕ * (t)| ≤ δ and by Lebesgue Theorem [20], we have | f 2 (t, s, y(s))|ds Hence Therefore, y ∈ L 1 of the problem (6) depends continuously on ϕ. This completes the proof.

Continuous
(ii) The solution y ∈ L 1 of Equation (6) depends continuously on the function f * 2 (t, s, y(s))ds | ≤ δ ⇒ y − y * ≤ , t ∈ I. Theorem 11. Assume that assumptions of Theorem 7 are verified. Then the solution y ∈ L 1 of Equation (6) depends continuously on the function f 1 .
Proof. Let y be the unique solution of the functional integral Equation (6) and y * is the solution of the functional integral equation Then f 2 (t, s, y(s)) − f 2 (t, s, y * (s)) ds Taking supermum for t ∈ I, we have Hence ||y − y * || ≤ .
Hence, the solution of (6) depends continuously on the function f 1 . This completes the proof.
By the same way we can prove the following theorem.
Theorem 12. Assume that assumptions of Theorem 7 are verified. Then the solution y ∈ L 1 of the functional integral Equation (6) depends continuously on the function f 2 . (7) depends continuously on the function y, if

Definition 3. The solution of functional Equation
Theorem 13. Let the assumptions of Theorem 7 be satisfied. Then the solution of the Equation (7) depends continuously on the function y.
Then for t ∈ I, |x(t)| < N and |y * (t)| < M, we have Hence the solution of problem (7) depends continuously on the function y.

Special Cases and Examples
We can deduce the following particular cases.