Separated Boundary Value Problems of Sequential Caputo and Hadamard Fractional Differential Equations

1 Intelligent and Nonlinear Dynamic Innovations Research Center, Department of Mathematics, Faculty of Applied Science, King Mongkut’s University of Technology North Bangkok, Bangkok 10800,Thailand 2Division of Sciences and Liberal Arts, Mahidol University Kanchanaburi Campus, Kanchanaburi 71150, Thailand 3Department of Mathematics, University of Ioannina, 451 10 Ioannina, Greece 4Nonlinear Analysis and Applied Mathematics (NAAM) Research Group, Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia

Sequential fractional differential equations are also found to be of much interest [16,17].In fact, the concept of sequential fractional derivative is closely related to the nonsequential Riemann-Liouville derivatives, for details, see [3].For some recent results on boundary value problems for sequential fractional differential equations; see [18][19][20][21][22] and references cited therein.
It can be observed that the sequential Caputo-Hadamard and Hadamard-Caputo fractional differential equations in (1) and ( 2

Preliminaries
In this section, we introduce some notations and definitions of fractional calculus [4,5] and present preliminary results needed in our proofs later.
Definition 1 (see [5]).For an at least -times differentiable function  : [, ∞) → R, the Caputo derivative of fractional order  is defined as where [] denotes the integer part of the real number .
Definition 2 (see [5]).The Riemann-Liouville fractional integral of order  of a function  : [, ∞) → R is defined as provided the integral exists.
Definition 4 (see [5]).The Hadamard fractional integral of order  is defined as provided the integral exists.
In order to define the solution of the boundary value problem (1), we consider the linear variant where  ∈ ([, ], R).
Then, the unique solution of the separated boundary value problem of sequential Caputo and Hadamard fractional differential equation ( 12) is given by the integral equation Proof.Taking the Riemann-Liouville fractional integral of order  to the first equation of ( 12), we get Again taking the Hadamard fractional integral of order  to the above equation, we obtain Substituting  =  in ( 15)-( 16) and applying the first boundary condition of (12), it follows that For  =  in equations ( 15)-( 16) and using the second boundary condition of (12), it yields Solving the linear system of ( 17) and (18) for finding two constants  1 ,  2 , we get and Substituting constants  1 and  2 in ( 16), we get the integral equation ( 14).The converse follows by direct computation.
The proof is completed.
In the same way, we can prove the following lemma, which concerns a linear variant of problem (2): where  ∈ ([, ], R).

Lemma 8. Let
Then, the unique solution of the separated boundary value problem of sequential Caputo and Hadamard fractional differential equation ( 21) is given by the integral equation

Main Results
We set some abbreviate notations for sequential Riemann-Liouville and Hadamard fractional integrals of a function with two variables as and where  ∈ {, }.Also we use this one for a single Riemann-Liouville and Hadamard fractional integrals of orders  and , respectively.In this section, we will use fixed point theorems to prove the existence and uniqueness of solution for problems (1) and (2).To accomplish our purpose, we define the Banach space C = ([, ], R), of all continuous functions on [, ] to R endowed with the norm ‖‖ = sup{|()|,  ∈ [, ]}.In addition, we define the operator K : C → C by where Ω ̸ = 0 is defined by ( 13) and   () = (, ()).Note that the separated boundary value problem (1) has solutions if and only if  = K has fixed points.
For computational convenience we put To prove the existence theorems of problem (2), we define the operator A : C → C by Now, we prove the existence and uniqueness result for problem (1).For problem (2)  Our second existence result is based on Krasnoselskii's fixed point theorem.
Theorem 11 ((Krasnoselskii's fixed point theorem) [24]).Let  be a closed, bounded, convex, and nonempty subset of a Banach space .Let ,  be operators such that Note that the ball   is a closed, bounded, and convex subset of the Banach space C. Now, we will show that K 1  + K 2  ∈   for satisfying condition (a) of Theorem 11.Setting ,  ∈   , then we have which implies that the set K 1   is uniformly bounded.Now we are going to prove that K 1   is equicontinuous.For  1 ,  2 ∈ [, ] such that  1 <  2 and for  ∈   , we have which is independent of  and also tends to zero as  1 →  2 .Hence the set K 1   is equicontinuous.Therefore the set K 1   is relatively compact.By applying the Arzelá-Ascoli theorem, the operator K 1 is compact on   .Therefore the operators K 1 and K 2 satisfy the assumptions of Theorem As  1 →  2 , the right-hand side of the above inequality tends to zero independently of  ∈   .Hence, by applying the Arzelá-Ascoli theorem, the operator K : C → C is completely continuous.The result will be followed from the Leray-Schauder nonlinear alternative if we prove the boundedness of the set of the solutions to equation  = ]K for ] ∈ (0, 1).Let  be a solution of the operator equation  = K.Then, for  ∈ [, ], by directly computation, we have It is easy to see that the operator K :  → C is continuous and completely continuous.From the choice of , there is no  ∈  such that  = ]K for some ] ∈ (0, 1).Therefore, by the nonlinear alternative of Leray-Schauder type (Theorem 15), we deduce that the operator K has a fixed point  ∈  which is a solution of problem (1).The proof is completed.