Successive approximations for random coupled Hilfer fractional differential systems

In this paper, we study the global convergence of successive approximations as well as the uniqueness of the random solution of a coupled random Hilfer fractional differential system. We prove a theorem on the global convergence of successive approximations to the unique solution of our problem. In the last section, we present an illustrative example.

In [6], the authors studied the existence of random solutions for a random coupled Hilfer and Hadamard fractional differential systems in generalized Banach spaces. In this paper we study the uniformly convergence of successive approximations for the coupled random Hilfer fractional differential system: with the initial conditions: where This paper initiates the study of system (1)-(2) using successive approximations.

Preliminaries
We denote by C the Banach space of all continuous functions from I into R m with the supremum (uniform) norm · ∞ . As usual, AC(I ) denotes the space of absolutely continuous functions from I into R m . By L 1 (I ), we denote the space of Lebesgue-integrable functions v : I → R m with the norm: By C γ (I ) and C 1 γ (I ), we denote the weighted spaces of continuous functions defined by and In addition, by C:= C γ 1 × C γ 2 , we denote the product weighted space with the norm: Now, we will give some necessary definitions of fractional calculus.
Definition 2.1 [1,15] The left-sided mixed Riemann-Liouville integral of order l > 0 of a function v ∈ L 1 (I ) is defined by where (·) is the Gamma function. For all l, l 1 , l 2 > 0 and each v ∈ C, we have I l 0 v ∈ C, and ; for a.e. t ∈ I. Definition 2.2 [1,15] The Riemann-Liouville fractional derivative of order l ∈ (0, 1] of a function v ∈ L 1 (I ) is defined by: In addition, if I 1−l 0 v ∈ C 1 1−γ (I ), then Definition 2.4 [1,15] The Caputo fractional derivative of order l ∈ (0, 1] of a function v ∈ L 1 (I ) is defined by The Hilfer fractional derivative of order α and type β of w is defined as

Corollary 2.7
Let χ ∈ C γ (I ). Then, the Cauchy problem has the unique solution Let β R m be the Borel σ -algebra. A mapping ξ : → R m is said to be measurable if for any B ∈ β R m ; one has ξ −1 (B) = {w ∈ : ξ(w) ∈ B} ⊂ A. Definition 2.8 Let A × β R m be the direct product of the σ -algebras A and β R m those defined in and R m , respectively. A mapping T : × R m → R m is called jointly measurable if for any D ∈ β R m , one has A random operator is a mapping T : ; we also say that T (w) is a random operator on R m . The random operator T (w) on E is called continuous (resp. compact, totally bounded, and completely continuous) if T (w, v) is continuous (resp. compact, totally bounded, and completely continuous) in v for all w ∈ (for more details, see [14]). Definition 2.10 [8] Let P(X ) be the family of all nonempty subsets of X and D be a mapping from into

Successive approximations and uniqueness results
In this section, we will give the main result of the global convergence of approximations of the problem (1) and (2).  (1) and (2) we mean coupled measurable functions (u, v) ∈ C γ i × C γ 2 that satisfies the system (1) on I and the system (2).
There exist a constant ρ > 0 and continuous functions g i : such that g i (t, ·, ·, w) is nondecreasing for any w ∈ and each t ∈ I, and for any w ∈ and each t ∈ I, u, satisfying the integral inequalities: and with η ≤ λ ≤ 1. (3), for any w ∈ and each t ∈ I, u ∈ C γ 1 , v ∈ C γ 2 , and i = 1, 2, we get

Remark 3.2 From
Define the operators L 1 : C× → C γ 1 , and L 2 : C× → C γ 2 by Consider the operator L : C× → C : For any w ∈ , we define the successive approximations of the problem (1) and (2) as follows: Proof From (H 1 ) the successive approximations are well defined. Thus, there exist θ 1 , θ 2 > 0, such that Next, for any w ∈ , and each t 1 , t 2 ∈ I with t 1 < t 2 , we have Then, from Remark 3.2, we get Thus, In addition, we obtain that Hence So, the sequence {(u n , v n ); n ∈ N} is equi-continuous on I, for any w ∈ . Let τ := sup η ∈ [0, 1] : {(u n , v n )} converges uniformly on I η , for any w ∈ .
If τ = 1, then we have the global convergence of successive approximations. Suppose that τ < 1, then the sequence {(u n , v n )} converges uniformly on I τ . Since this sequence is equi-continuous, then it converges uniformly to a continuous function (ũ(t),ṽ(t)). If we prove that there exists λ ∈ (τ, 1], such that {(u n , v n )} converges uniformly on I λ , for any w ∈ . This will yield a contradiction. Put u(t, w) =ũ(t, w) and v(t, w) =ṽ(t, w); for each t ∈ I τ and any w ∈ . From (H 3 ) , there exists a constant ρ > 0 and a function g i : (3). In addition, there exist λ ∈ [τ, 1] and n 0 ∈ N, such that for all t ∈ I λ and any w ∈ , and n, m > n 0 , we have For each t ∈ I λ , and any w ∈ , we put ,m) (t, w), Since the sequence (V k (t, w), W k (t, w)) is non-increasing, it is convergent to a function (V (t, w), W (t, w)) for each t ∈ I λ , and any w ∈ . From the equi-continuity of {(V k (t, w), W k (t, w))} it follows that lim k→∞ V k (t, w) = V (t, w) and lim k→∞ W k (t, w) = W (t, w) uniformly on I λ . Furthermore, for each t ∈ I λ , and any w ∈ , and for n, m ≥ k, we have Thus, from (3), we get By the Lebesgue dominated convergence theorem, we get In addition, we find that Then, from (H 1 ) and (H 3 ), we get V ≡ 0 and W ≡ 0 on I λ × , which yields that lim k→∞ (V k (t, w), W k (t, w)) = (0, 0) uniformly on I λ × . Thus {(u k (t, w), v k (t, w))} ∞ k=1 is a Cauchy sequence on I λ × . Consequently  {(u k (t, w), v k (t, w))} ∞ k=1 is uniformly convergent on I λ which yields the contradiction.
Thus, {(u k (t, w), v k (t, w))} ∞ k=1 converges uniformly on I for any w ∈ to a continuous function (u * (t, w), v * (t, w)). By the Lebesgue dominated convergence theorem, we get for each t ∈ I. This yields that (u * , v * ) is a solution of the problem (1) and (2).

An example
We equip the space R * − := (−∞, 0) with the usual σ -algebra consisting of Lebesgue measurable subsets of R * − . Consider the following random coupled Hilfer fractional differential system: ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩  u(t, w), v(t, w), w)) (I 1 4 0 u)(0, w) = 2 sin w (I 1 4 0 v)(0, w) = 2 cos w, where Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article's Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article's Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/.