Boundary value problems for Caputo fractional differential equations with nonlocal and fractional integral boundary conditions

In this paper, we study the existence and uniqueness of solutions for fractional differential equations with nonlocal and fractional integral boundary conditions. New existence and uniqueness results are established using the Banach contraction principle. Other existence results are obtained using O’Regan fixed point theorem and Burton and Kirk fixed point. In addition, an example is given to demonstrate the application of our main results.


Introduction
Fractional differential equations have gained much importance and attention due to the fact that they have been proved to be valuable tools in the modeling of many phenomena in engineering and sciences such as physics, mechanics, economics, biology, etc. (see, for example, [16,19,23,24] and references therein). Actually, the concepts of fractional derivatives are not the only generalization of the ordinary derivatives, but also it has been found that they can efficiently and properly describe the behavior of many physical systems (real-life phenomena) more accurately than integer-order derivatives.
Motivated by the above papers, in this paper, we establish various existence and uniqueness results of solutions for a boundary value problem of nonlinear fractional differential equations of order α ∈ (2, 3] with nonlocal and fractional integral boundary conditions given by: where c D κ 0 + is the Caputo fractional derivative of order κ ∈ {α, β, β 1 }, such that 2 < α ≤ 3, and 0 < β, β 1 ≤ 1, and f : [0, 1]×R 2 −→ R and g : C([0, 1], R) −→ R are continuous functions, I σ i 0 + is the Riemann-Liouville fractional integral of order σ i > 0, i = 1, 2, η 1 , η 2 , a, b are suitably chosen real constants, such that: The paper is organized as follows. In Sect. 2, we collect some background material and basic results about fractional calculus that will be used later. In Sect. 3, we establish the existence of solutions for the boundary value problem (1) using Banach contraction principle, O'Regan fixed point theorem, and Burton and Kirk fixed point. In the fourth section, we give an example to demonstrate the application of our main results. Finally, we add a conclusion to the paper in the last section.

Preliminaries
In this section, we will recall some notions and definition which we will need in the sequel (details can be found, e.g., in [16,23,24]).

Lemma 2.8 The fractional integral I
Proof Let u ∈ L 1 ([0, 1], R), and then: Let us now introduce the space , R) equipped with the norm: Clearly, (E, · E ) is a Banach space [26].

Main results
For the existence of solutions for the problem (1), we need the following lemma: Lemma 3.1 For a given h ∈ C(J, R), the unique solution of the linear fractional boundary value problem: supplemented with boundary conditions is given by: where and Δ is defined by (2).
Proof Applying the Riemann-Liouville fractional integral operator of order α to both sides of (3) and using Lemma 2.6, we have: Applying the boundary conditions (4) in (7), we find that: Using the boundary conditions of (4) in (7), we find a system of equations: Solving the system of equations (8) and (9) together with notations (6), we find that: Substituting the values of c 0 , c 1 , c 2 in (7), we obtain the solution (5). This completes the proof.
For computational convenience, we introduce the notations: Define the integral operator T : E −→ E by: Next, we introduce two operators T 1,2 : E −→ E as follows: and Clearly: Remark 3.2 In our work, we assume that the restriction Δ = 0 is satisfied, because in the case when Δ = 0, the problem (1) will be under the resonant conditions, and hence, the previously mentioned theorems are not applicable, and then, the problem can be managed using the method of Mawhin continuation theorem.
Proof Let u be a solution of (1). Then, using the same method as used in Lemma 3.1, we can prove that: Conversely, u satisfies: and denotes the right-hand side of the equation by v(t). Then, by Lemma 2.4, we obtain: is a solution of the fractional differential Eq. (1). Also, it is easy to verify by direct computation that v satisfies conditions (4). This completes the proof.
To simplify the computations in the main results, we set some notations: In the sequel, we need the following assumptions.
for all u 1 , u 2 , v 1 , v 2 ∈ R; g ∈ L 1 (J, R + ) and t ∈ J. (H3) There exist a positive constant ω < 1 and a continuous function φ : such that: for (H7) There exists non-negative function p ∈ L 1 (J, R + ), such that: for all u, v ∈ R and t ∈ J .
Our first results is based on the Banach contraction principle.
holds, then the problem (1) has a unique solution on J.
Proof Transform the problem (1) into a fixed point problem. Clearly, the fixed points of the operator T defined by (10) are solutions of the problem (1). Let u, v ∈ E and t ∈ J , and from the definition of T , we can write: By (H2), we can find that: According to the Lemmas 2.7, 2.8, and (H 3), we have: Thus: Observe that: using similar techniques as to get (19), it yields: Combining (19) and (20), we get: Therefore, we have: In view of the given condition γ 1 = (M 1 + M 2 ) g L 1 + (N 1 + N 2 )ω) < 1, it follows that the mapping T is a contraction. Hence, by the Banach fixed point theorem, T has a unique fixed point which is a unique solution of Problem (1). This completes the proof.

Lemma 3.5 Denote by U an open set in a closed, convex set C of a Banach space E. Assume 0 ∈ U . Also assume that T (U ) is bounded and that F : U −→ C is given by T = T 1 + T 2 , in which T 1 : U −→ E is continuous and completely continuous and T 2 : U −→ E is a nonlinear contraction (i.e., there exists a non-negative nondecreasing function
. Then, either (c1) T has a fixed point in U , or (c2) there exists u ∈ ∂U and λ ∈ (0, 1), such that u = λT u, where U and ∂U , respectively, represent the closure and boundary of U .
and denote the maximum number by: Proof From (H6), there exists a number r 0 > 0, such that: We shall prove that T 1 and T 2 defined by (11) and (12), respectively, satisfy the conditions of Lemma 3.5. We split the proof into several steps.
Step 1 : The operator T 1 is completely continuous. It is clear that T 1 is continuous, since f is continuous. Next, we shall prove that the operator T 1 maps bounded sets into bounded sets in E. For u ∈ B r 0 and for each t ∈ J , we have: In view of (15), we obtain: By Lemmas (2.7), ( 2.8), we obtain: Thus: With the same arguments as in (22), we get: Combining (22) with (23), we get: which shows that T 1 is uniform bounded. Now, we will prove that T 1 (B r 0 ) is equi-continuous. Let t 1 , t 2 ∈ [0, 1], t 1 < t 2 , u ∈ B r 0 : Some computations give: In a like manner: .
In consequence, we obtain: as t 1 → t 2 independent of u ∈ B r 0 . Therefore, the operator T 1 is equi-continuous and uniformly bounded. Hence, by the Arzelá-Ascoli theorem, it follows that T 1 (B r 0 ) is relatively compact in E.
Step 2 : The operator T 2 : B r 0 −→ E is contractive. Let u, v ∈ and t ∈ J by (H3), we can show that: which, in view of (H3), implies that T 2 is a contraction.
Step 3: The set T 2 (B r 0 ) is bounded. For u ∈ B r 0 and for each t ∈ J , and by (H3) and (H4), we obtain: Hence, T 2 (B r 0 ) is bounded.
Step 4 : Finally, we show that the case (c2) in Lemma 3.5 does not occur. To this end, we suppose that (c2) holds. Let u ∈ ∂ B r 0 , and then, u = λT u, for some 0 < λ < 1. Therefore, we have u E = r 0 , and: With hypotheses (H3)-(H6), we can write: which implies that: which contradicts (16). Consequently, we have proved that the operators T 1 and T 2 satisfy all the conditions in Lemma 3.5. Hence, the operator T has at least one fixed point u ∈ B r 0 , which is the solution of the boundary value problem (1).
Our next result relies on the following fixed point theorem due to Burton and Kirk [8]

Theorem 3.7 Let E be a Banach space, and A, B : E −→ E be two operators, such that A is a contraction and B is completely continuous. Then, either
-the operator equation u = A(u) + B(u) has a solution, or -the set Ω = u ∈ E : λA( u λ ) + λB(u) = u is unbounded for λ ∈ (0, 1). Then, there exists z ∈ U , such that z = Az + Bz.
Then, the fractional boundary value problem (1) has at least one solution on J .
Proof We define the map T : E −→ E by: where T 1 and T 2 are defined by (11) and (12), respectively. Notice that Problem (1) is equivalent to the fixed point problem T (u) = u. We shall prove that T 1 and T 2 satisfy the conditions of Theorem 3.7. For clarity, we will divide the remain of the proof into several steps.
Step 1: T 1 is continuous. The continuity of f implies that the operator T 1 is continuous.
Step 2: The operator T 1 maps bounded sets into bounded sets in E. In a similar manner as previous proof of Theorem 3.6, we can find that: Step 3: The operator T 1 maps bounded sets into equi-continuous sets. By a discussion similar to that of Theorem 3.6, we can get: as t 2 → t 1 . This implies that: Step 4: The operator T 2 is a contraction. This was established in Step 2 of Theorem 3.6.
Step 5: A priori bounds on solutions. Now, it remains to show that the set Ω = u ∈ E : λT 2 u λ +λT 1 (u) = u is unbounded for λ ∈ (0, 1). Let λ ∈ (0, 1), and u ∈ E be a solution of the integral equation: This implies by (H3), (H4), and (H7) that, for each t ∈ J , we have: According to the Lemmas 2.7 and 2.8, we have: for all t ∈ J . In a similar manner, we obtain: Combining (27) with (28), we get: This shows that the set Ω is bounded, since (M 1 + M 2 ) p L 1 + (N 1 + N 2 )ω < 1. Hence, T has a fixed point by Theorem 3.7, and consequently, the problem (1) has a solution. This completes the proof.

Conclusion
We have established the existence and uniqueness of solutions for fractional boundary value problem under more general assumptions. Using the techniques of fixed point theorems such as Banach contraction principle, O'Regan fixed point theorem, and Burton and Kirk fixed point theorem, we establish the main results of the paper. Finally, we conclude the paper with an example to make our results clear.