On the existence of solutions of a set-valued functional integral equation of Volterra–Stieltjes type and some applications

This paper is concerned with the existence of continuous solutions of a set-valued functional integral equation of Volterra–Stieltjes type. The continuous dependence of the solution on the set of selections of the set-valued function will be proven. As an application, we study the existence of solutions to an initial-value problem of arbitrary fractional-order differential inclusion.


Introduction
Consider the set-valued functional integral equation of Volterra-Stieltjes type First, we establish some notation. We will denote by I = [0, T] a fixed interval, where T > 0 is arbitrarily fixed and by C(I) = C[0, T] the Banach space consisting of all continuous functions acting from the interval I into R with the standard norm Define the Banach space X = C(I) × C(I) with the norm (x, y) X = x C + y C . Definition 2.1 Let F be a set-valued map defined on a Banach space E, f is called a selection of F if f (x) ∈ F(x), for every x ∈ E and we denote by the set of all selections of F (for the properties of the selection of F see [1][2][3]).

Definition 2.2 ([4])
A set-valued map F from I × E to family of all nonempty closed subsets of E is called Lipschitzian if there exists k > 0 such that, for all t ∈ I and all x 1 , x 2 ∈ E, we have h F(t, x 1 ), F(s, x 2 ) ≤ k |t -s| + |x 1x 2 | , (2.1) where h(A, B) is the Hausdorff distance between the two subsets A, B ∈ I × E.
(For properties of the Hausdorff distance see [5].) The following theorem [5,Sect. 9, Chap. 1, Th. 1] assumes the existence of a Lipschitzian selection. In what follows, we discuss a few auxiliary facts concerning functions of bounded variation (cf. [7]). To this end assumes that x is a real function defined on a fixed interval [a, b]. By the symbol b a x we will denote the variation of the function x on the interval [a, b]. In the case when b a x is finite we say that x is of bounded variation on [a, b]. In the case of a function u(t, s) =: [a, b] × [c, d] → R we can consider the variation q t=p u(t, s) of the function t → u(t, s) (i.e., the variation of the function u(t, s) with respect to the variable t) on the interval [p, q] ⊂ [a, b]. Similarly, we define the quantity q s=p u(t, s). We will not discuss the properties of the variation of functions of bounded variation, we refer to [7] for the mentioned properties. Furthermore, assume that x and φ are two real functions defined on the interval [a, b]. Then, under some extra conditions (cf. [7]), we can define the Stieltjes integral (more precisely, the Riemann-Stieltjes integral) of the function x with respect to the function φ on the interval [a, b] In such a case, we say that x is Stieltjes integrable on the interval [a, b] with respect to φ.
In the relevant literature, we may encounter a lot of conditions guaranteeing the Stieltjes integrability [7][8][9]. One of the most frequently exploited condition requires that x is continuous and φ is of bounded variation on [a, b].
Next, we recall a few properties of the Stieltjes integral which will be used in our considerations (cf. [7]).

Lemma 2.5 Let x 1 and x 2 be Stieltjes integrable functions on the interval
In the sequel, we will also consider the Stieltjes integrals of the form b a x(s) d s g(t, s), where g : [a, b] × [a, b] → R and the symbol d s indicates the integration with respect to the variable s. The details concerning the integral of such a type will be given later.

Existence of at least one continuous solution
Consider now the set-valued integral equation (1.1) under the following assumptions.
(i) p : is a Lipschitzian set-valued map with a nonempty compact convex subset of 2 R + . (iii) ϕ : I → I is continuous function.
(iv) f 2 : I × R → R is continuous and there exist two constants a and b such that (v) The function g i is continuous on the triangle i , for i = 1, 2, where (vi) The function s → g i (t, s) is of bounded variation on [0, t] for each t ∈ I (i = 1, 2). (vii) For any > 0 there exists δ > 0 such that, for all t 1 ; t 2 ∈ I such that t 1 < t 2 and t 2t 1 ≤ δ, the following inequality holds: It is clear that, from Theorem 2.3 and assumption (ii), the set of Lipschitz selection of F 1 is non-empty. So, the solution of the single-valued integral equation where f 1 ∈ S F 1 , is a solution of inclusion (1.1).
It must be noted that f 1 satisfies the Lipschitz selection Obviously, we will assume that g i satisfies assumptions (v)-(viii). For our purposes, we only need the following lemmas.
Further, let us observe that based on Lemma 3.3 we infer that there exists a finite positive constant K i , such that where T > 0 is arbitrarily fixed and i = 1, 2.
We now introduce some functions that will be useful in our further studies: In our considerations, we will examine the double Stieltjes integral of the form , 2) and the symbol d y indicates the integration with respect to the variable y (similarly, we define the symbol d s ). Now, let then the nonlinear functional integral equation (3.1) can be written in the form     Proof Define the set Q r by where r = p * +f * 1 K 1 1-kK 1 + aK 2 1-bK 2 with kK 1 < 1, bK 2 < 1. It is clear that the set Q r is nonempty, bounded, closed and convex. Let A be any operator defined by where for u = (x, y) ∈ Q r , and from Remark 3.5 we have Then Then From the above estimate we derive the following inequality: Hence, AQ r ⊂ Q r and the class {Au}, u ∈ Q r is uniformly bounded. Now, for u = (x, y) ∈ Q r , for all > 0, δ > 0 and for each t 1 , t 2 ∈ [0, T], t 1 < t 2 , such that Further, for the operator A and u ∈ Q r we have Then This means that the class of functions Au is equi-continuous on Q r . Then by the Arzela-Ascoli theorem [11] the operator A is compact. It remains to prove the continuity of A : Q r → Q r . Let u n = (x n , y n ) is a sequence in Q r with x n → x, and y n → x and since f 1 (t, y(t)) and f 2 (t, x(t)) is continuous in C[0, T] × R then f 1 (t, y n (t)) and f 2 (t, x n (t)) converge to f 1 (t, y(t)) and f 2 (t, x(t)), thus f 2 (t, x n (ϕ(t))) converges to f 2 (t, x(ϕ(t))) (see assumption (ii)). Using assumption (iii) and applying Lebesgue dominated convergence theorem, we get

(t), A 2 x(t) = Au(t).
Since all conditions of the Schauder fixed-point theorem [12] hold, A has a fixed point u ∈ Q r , and then the system (3.3), (3.2) has at least one continuous solution u = (x, y) ∈ Q r , x; y ∈ C[0, T].
Consequently, the functional integral equation (3.1) has at least one solution x ∈ C[0, T].

Existence of a unique solution
In this section, we study the uniqueness of the solutions x ∈ C[0, T] of the functional integral inclusion (1.1). Proof Let x 1 and x 2 be two solutions of Eq. (3.1), then Using the Lipschitz condition for f 1 , we obtain Using Lipschitz condition for f 2 , we obtain Then This proves the uniqueness of the solution of the functional integral equation (3.1). Proof Let f 1 (t, x(t)) and f * 1 (t, x(t)) be two different Lipschitzian selections of F 1 (t, x(t)) such that

Continuous dependence
then for the two corresponding solutions x f 1 (t) and x f * 1 (t) of (1.1) we have Thus from last inequality, we get This proves the continuous dependence of the solution on the set S F 1 of all Lipschitzian selections of F 1 . This completes the proof.

Volterra integral inclusion of fractional order
In this section, we will consider the fractional integral inclusion, which has the form where t ∈ I = [0, T] and α ∈ (0, 1). Moreover, Γ (α) denotes the gamma function. Let us mention that (5.1) represents the so-called nonlinear Volterra integral inclusion of fractional orders. Recently, the inclusion of such a type was intensively investigated in some papers [13][14][15][16][17][18]. Now, we show that the functional integral inclusion of fractional orders (5.1) can be treated as a particular case of the set-valued functional integral equation of Volterra-Stieltjes (1.1) studied in Sect. 3.
Indeed, we can consider the functions g i (w, z) = g i : i → R (i = 1, 2) defined by the formulas Note that the functions g 1 and g 2 satisfy assumptions (v)-(viii) in Theorem 3.6; see [10,19]. Now, we can formulate the following existence results concerning with the Volterra integral inclusion of fractional order (5.1).

Existence of the maximal and minimal solutions
In this section, we establish the existence of the maximal and minimal solutions of the nonlinear Volterra integral inclusion of fractional order (5.1). It is clear that, from Theorem 2.3 and assumption (ii) of Theorem 3.6, the set of Lipschitz selections of F 1 is non-empty. So, the solution of the nonlinear functional integral equation of fractional order where f 1 ∈ S F 1 , is a solution of inclusion (5.1).
where one of them is strict.
Suppose f 1 and f 2 are monotonic nondecreasing functions in x, then Proof Let the conclusion (6.2) be false, then there exists t 1 such that From the monotonicity of the functions f 1 and f 2 in x, we get x(t 1 ) < y(t 1 ).
This contradicts the fact that x(t 1 ) = y(t 1 ). Then

x(t) < y(t).
Now, for the existence of the continuous maximal and minimal solutions of the nonlinear functional integral inclusion (6.1) we have the following theorem. To do this let x(t) be any solution of (6.1), then x(t) = p(t) + I α f 1 t, I β f 2 t, x ϕ(t) , (6.5) and also x (t) = p(t) + I α f 1 t, I β f 2 t, x ϕ(t) , x (t) = p(t) + I α f 1 t, I β f t, x ϕ(t) + I β + I α , x (t) > p(t) + I α f 1 t, I β f 2 t, x ϕ(t) . (6.6) Applying Lemma 6.2 and (6.5) and (6.6), we get From the uniqueness of the maximal solution (see [12]), it is clear that x (t) tends to m(t) uniformly in [0, T] as → ∞.
Similarly, we can prove the existence of the minimal solution. We set and thus we prove the existence of a minimal solution.

Differential inclusion
Consider now the initial-value problem of the differential inclusion (1.2) with the initial data (1.3). Proof Let y(t) = dx(t) dt , then the inclusion (1.2) will be y(t) ∈ I α F 1 t, I 1-τ y(t) .