Boundary Value Problems for Hybrid Caputo Fractional Differential Equations

In this paper, we discuss the existence of solutions for a hybrid boundary value problem of Caputo fractional differential equations. The main tool used in our study is associated with the technique of measures of noncompactness. As an application, we give an example to illustrate our results.

The area of differential equations devoted to quadratic perturbations of nonlinear differential equations (called hybrid differential equations) has been considered more important and served as special cases of dynamical systems.This class of equations involves the fractional derivative of an unknown function hybrid with the nonlinearity depending on it.Some recent results on hybrid differential equations can be found in a series of papers [20][21][22][23][24][25][26][27][28][29][30][31].
Dhage and Lakshmikantham [23] discussed the existence and uniqueness theorems of the solution to the ordinary first-order hybrid differential equation with perturbation of first type where f ∈ C(J × R, R \ {0}) and g ∈ C(J × R, R), where J = [t 0 , t 0 + a] is a bounded interval in R for some t 0 and a ∈ R with a > 0. Zhao et al. [32] studied existence and uniqueness results for the following hybrid differential equations involving Riemann-Liouville differential operators: x(0) = 0, where 0 A fixed point theorem in Banach algebras is the main tool used in this work.Hilal and Kajouni [25] extended the results to the following boundary value problem for fractional hybrid differential equations involving Caputo's derivative: where 0 ) and a, b, c are real constants with a + b = 0.They proved the existence of the boundary fractional hybrid differential equations under mixed Lipschitz and Carathéodory conditions.Some fundamental fractional differential inequalities are also established, which are utilized to prove the existence of extremal solutions.Motivated by some recent studies on hybrid fractional differential equations, we consider the following boundary value problem of nonlinear fractional hybrid differential equations: where 0 < α ≤ 1, a, b, c are real constants such that a 3) is reduced to the problem in Equation (1).
The main tool of our study is a generalization of Darbo's fixed point theorem for the product of two operators associated to measures of noncompactness.We can use a numerical method to solve the problem in Equation (3) (see [16]).

Preliminaries
First, we present the necessary definitions and lemmas from fractional calculus theory.
In the sequel, we recall some definitions and basic facts about measures of noncompactness.Assume that (E, • ) is a real Banach space with zero element 0. By B(x, r), we denote the closed ball in E centered at x with the radius r.By B r , we denote the ball B(0, r).If X is non-empty subset of E, then X and ConvX denote the closure and the closed convex closure of X, respectively.When X is a bounded subset, diamX denotes the diameter of X and X the quantity given by X = sup{ x : x ∈ X}.Further, by M E , we denote the family of the nonempty and bounded subsets of E and by N E its subfamily consisting of the relatively compact subsets.
We accept the following definition of measure of noncompactness.

Definition 3 ([33]
).A mapping σ : M E −→ [0, ∞) is said to be a measure of noncompactness if it satisfies the following conditions: (1) The An important theorem about fixed point theorem in the context of measures of noncompactness is the following Darbo's fixed point theorem.
Theorem 1 ([34]).Let Ω be a nonempty, bounded, closed and convex subset of a Banach space E and let T : Ω → Ω be a continuous mapping.Suppose that there exists k ∈ [0, 1) such that σ(T X) ≤ kσ(X), for any non-empty subset X of Ω, where σ is a measure of noncompactness in E.Then, T has a fixed point in Ω.Now, we introduce the following class A of functions ϕ : (0, ∞) → (1, ∞) satisfying the following condition: For any sequence (t n ) ⊂ (0, ∞), A generalization of Theorem 1, which is very useful for our study, is the following theorem (see [35]).

Theorem 2.
Let Ω be a nonempty, bounded, closed and convex subset of a Banach space E and let T : Ω → Ω be a continuous mapping.Suppose that there exist ϕ ∈ A and k ∈ [0, 1) such that, for any nonempty subset X of Ω with σ(T X) > 0, for any non-empty subset X of Ω, where σ is a measure of noncompactness in E.Then, T has a fixed point in Ω.
Next, we assume that the space E has structure of Banach algebra.By xy, we denote the product of two elements x, y ∈ X and by XY we denote the set defined by XY = {xy : x ∈ X, y ∈ Y}.Definition 4. Let E be a Banach algebra.A measure of noncompactness σ in E is said to satisfy condition (m) if it satisfies the following condition: This definition appears in [36].As is known, the family of all real valued and continuous functions defined on interval I is a Banach space with the standard norm Notice that (C(I), • ) is a Banach algebra, where the multiplication is defined as the usual product of real functions.
In our considerations, we use a measure of noncompactness defined in [33].To o recall the definitions of that measure, let us fix a set X ∈ C(I).For x ∈ X and for a given ε > 0, denote by ω(x, ε) the modulus of continuity of x, i.e., and The authors of [33] showed that function ω 0 is a measure of noncompactness in space C(I).Lemma 3. The measure of noncompactness ω 0 on C(I) satisfies condition (m).
Proof.Let X, Y be a fixed subset of M C(I) , ε > 0 and t, s ∈ I with |t − s| ≤ ε.Then, for x ∈ X and y ∈ Y, we have Thus, and Therefore, we get This completes the proof.
Lemma 4. For any y ∈ C(I), the unique solution of the hybrid fractional differential equation, with boundary conditions is given by where Here, G(t, s) is called the Green function of the boundary value problem in Equations (4) and (5).
Proof.We may apply Lemma 2 to reduce Equation ( 4) to an equivalent integral equation Consequently, the general solution of Equation ( 4) is Applying the boundary conditions in Equation (5) to Equation ( 8), we find that Therefore, we have Substituting the value of c 0 into Equation ( 9), we get Equation (6).
that can be written as where G is defined by Equation ( 7).The proof is complete.

Remark 2.
From the expression of G(t, s), the function G(t, s) is not continuous on I × I, however the function t → 1 0 G(t, s)ds is continuous on I, so it is bounded.
Thanks to Lemma 4, the proposed problem is equivalent to the following integral equation

Proof of Main Results
Firstly, we list some assumptions: (H1) The functions µ, ν :

R).
(H3) There exists a constant k ∈ (0, 1) such that (H5) There exists r > 0 such that Now, we are in a position to state and prove our main result in this paper.Proof.To prove this result using Theorem 2, we consider the operator T on the Banach algebra C(I) as follows We define operators F and G on the Banach algebra C(I) in the following way: and We divide the proof into five steps.
Step 1: T transforms C(I) into itself.In fact, since the product of continuous functions is a continuous function, it is sufficient to prove that F x, Gx ∈ C(I) for any x ∈ C(I).Now, from Assumptions (H1) and (H2), it follows that if x ∈ C(I) then F x ∈ C(I).Next, we prove that if x ∈ C(I) then Gx ∈ C(I).To do this, let ε > 0 be fixed; take x ∈ C(I) and t 1 , t 2 ∈ I with t 2 − t 1 ≤ ε; and assume that t 1 ≤ t 2 .Then, in view of Assumption (H4), we get From the above inequality, we conclude that |Gx(t 2 ) − Gx(t 1 )| → 0 when ε → 0. Therefore, Gx ∈ C(I).This proves that if x ∈ C(I) then T x ∈ C(I).
Step 2: An estimate of T x for x ∈ C(I).Now, let us fix x ∈ C(I), then using our assumptions for t ∈ I, we obtain By Assumption (H5), we deduce that the operator T maps the ball B r ⊂ C(I) into itself.Moreover, let us observe that, from the last estimates, we obtain Step 3: The operators F and G are continuous on the ball B r .In fact, firstly we prove that the operator F is continuous on the ball B r .To do this, fix ε > 0 and take arbitrary x, y ∈ B r such that x − y ≤ ε.Then, for t ∈ I, we have and, since (ε + 1) k − 1 → 0 when ε → 0, from the above inequality, the operator F is continuous on the ball B r .
Next, we prove that the operator G is continuous on the ball B r .To do this, we take a sequence {x n } ⊂ B r and x ∈ B r such that x n − x → 0 as n → ∞, and we have to prove that Gx n − Gx → 0 as n → ∞.Since g(t, x) is uniformly continuous on the compact I × [−r, r], we may denote Thus, the above inequality shows that the operator G is continuous in B r .Hence, we conclude that T is continuous operator on B r .
Thus, the contractive condition appearing in Theorem 1 is satisfied with ϕ(t) = t + 1, where ϕ ∈ A. By applying Theorem 2, we get that the operator T has at least one fixed point in the ball B r .Consequently, the problem in Equation ( 3) has at least one solution in B r .This completes the proof.

An Example
Consider the following fractional hybrid problem On the other hand, since the function β(x) = 1 + |x| − 1 is concave (because β (x) ≤ 0) and β(0) = 0, we infer that β is subadditive and, therefore, for any t ∈ I and x 1 , x 2 ∈ R, we have Thus, Assumption (H3) holds, with k = 1/2.Moreover, for any t ∈ I and x ∈ R, we have Hence, Assumption (H4) holds, where ψ(x) = 1 3 x.Observe that Assumption (H5) is equivalent to Thus, Assumption (H5)is satisfied for all 0 < r ≤ √ π.Thus, all assumptions of Theorem 3 are satisfied, and consequently the problem in Equation ( 14) has at least one solution in C(I).

Theorem 3 .
Assume that Assumptions (H1)-(H5) hold.Then, the problem in Equation (3) has at least one solution in the Banach algebra C(I).

1 0G
(t, s)g(s, x(ν(s)))ds + c a + b for t ∈ I.By virtue of Lemma 4, a fixed point of T gives us the desired result.