Existence and uniqueness results for implicit differential equations with generalized fractional derivative

This article concerns the existence of solutions for implicit fractional differential systems with generalized fractional derivative (or Katugampola-Caputo derivative). The Banach and Krasnoselskii fixed point theorem are used to obtain the desired results. An example is presented to illustrate the theory.


Introduction
Recently, the theory of fractional differential equations (FDEs) has gained considerable popularity and importance, and as a result, several research papers and monographs have been published in this field (see, for example, [6,12] and the references therein).In this paper, we consider the following implicit differential equation with generalized fractional derivative { ( ρ c D α 0 + x where ρ c D α 0 + is the Caputo generalized fractional derivative.Let α ∈ R, 0 < λ < 1, ρ > 0 and f : J × X × X → X, g : C([0, T ], X) → X are given continuous functions.It is seen that system (1.1) is equivalent to the following nonlinear integral equation(see [2,3] for more details).
In passing, we note that the application of nonlinear condition x(0) + g(x) = x 0 in physical problems yeilds better effect than the initial condition x(0) = x 0 [4].These days generalization of the derivatives of both Riemann-Liouville and Caputo types are introduced and shown the effect of utilizing it in equations of mathematical physics or related to probability.This was done using the definition of generalized fractional derivatives given by Katugampola [7].The author initiated a new fractional integral, which generalizes the Riemann-Liouville and the Hadamard integrals into a single form.Later, Katugampola [8] introduced a new fractional derivative, which generalizes the two derivatives in question.Motivated by the papers [8,11], we apply Katugampola-Caputo derivative for implicit fractional differential equations.The rest of this paper is organized as follows: In Section 2, we recall some useful preliminaries and results concerning the generalized fractional derivative.In Section 3, we discuss the existence and uniqueness of solutions for the problem (1.1).

Preliminaries
In this section, we introduce notations, definitions and preliminary facts that are used throughout this article.Let C(J, X) be the Banach space of continuous function x(t) with x(t) ∈ X for t ∈ J and ∥x∥ C(J,X) = sup t∈J ∥x(t)∥.
Definition 2.2.The Hadamard fractional integral and derivative are given by and respectively, for t > a ≥ 0 and Re(α) > 0.
Now we give the definitions of the generalized fractional operators introduced in [8].
Definition 2.3.The generalized left-sided fractional integral ρ I α a + f of order α ∈ C (Re(α)) is defined by for t > a, if the integral exists.The generalized fractional derivative, corresponding to the generalized fractional integral (2.3), is defined for 0 ≤ a < t, by if the integral exists.

International Scientific Publications and Consulting Services
Lemma 2.1.(Krasnoselkii's fixed point theorem) Let K be a closed convex and nonempty subset of a Banach space X.Let T and S, be two operators such that • T x + Sy ∈ K for any x, y ∈ K; • T is compact and continuous; • S is contraction mapping.
Then there exists z 1 ∈ K such that z We are ready to present our results.We adopt some ideas from [5,9].

Existence and uniqueness results
Let us list some assumptions to prove our existence results.
for any u, v, u, v ∈ X and t ∈ J.
(A4) there exist l, p, q ∈ C(J, X) with l * = sup t∈J l(t) < 1 such that Our first result is based on Banachs fixed point theorem.
Proof.Consider the operator P : C(J, X) → C(J, X). where It is clear that the fixed points of P are solution of (1.1).Let x, y ∈ C(J, X) and t ∈ J, then we have International Scientific Publications and Consulting Services and By replacing (3.8) in the inequality (3.7), we get Thus, From (3.5), it follows that P has a unique fixed point which is solution of problem (1.1).
Our second result is based on Krasnoselskii fixed point theorem.
(1−q * ) (l * + p * r), and consider B r = {x ∈ C(J, X) : |x| ≤ r}.Let A and B the two operators defined on B r by respectively.Note that x, y ∈ B r then Ax + By ∈ B r .Indeed it is easy to check the inequality and by (A4) International Scientific Publications and Consulting Services Therefore, ) , ≤ r.
Thus, Ax + By ∈ B r .
By (A3), it is also clear that B is a contraction mapping.Produced from continuity of x, the operator (Ax)(t) is continuous in accordance with (A1).Also we observe that ) .
Then A is uniformly bounded on B r .Now let's prove that (Ax)(t) is equicontinuous.Let t 1 ,t 2 ∈ J, t 2 ≤ t 1 and x ∈ B r .Using the fact f is bounded on the compact set J × B r (these sup (t,x)∈J×B r ∥K x (t)∥ := C 0 < ∞).
We will get ) .
The second integral in the right-hand side of the last inequality has the value 1 ρα (t ρ 2 − t ρ 1 ) α .For the first integral, consider the three cases α < 0, α = 0 and α > 1 separately.In the case α = 1, the integral has the value zero.For α < 1, we have (t

International Scientific Publications and Consulting Services
Combining these results, we have which is autonomous of x and head for zero as t 1 − t 2 → 0 consequently.So, A(B r ) is relatively compact.By the Arezela-Ascoli theorem, A is compact.We now conclude the results of the theorem based on the Krasnoselskii fixed point theorems.Thus, the problem (1.1) has at least one fixed point on J.