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`Journal of Function SpacesVolume 2018, Article ID 6974046, 8 pageshttps://doi.org/10.1155/2018/6974046`
Research Article

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

1Intelligent 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

Received 24 July 2018; Accepted 31 October 2018; Published 15 November 2018

Guest Editor: Lishan Liu

#### Abstract

In this paper, we discuss the existence and uniqueness of solutions for new classes of separated boundary value problems of Caputo-Hadamard and Hadamard-Caputo sequential fractional differential equations by using standard fixed point theorems. We demonstrate the application of the obtained results with the aid of examples.

#### 1. Introduction

Fractional differential equations have been of increasing importance for the past decades due to their diverse applications in science and engineering such as biophysics, bioengineering, virology, control theory, signal and image processing, blood flow phenomena, etc.; see [16]. Many interesting results of the existence of solutions of various classes of fractional differential equations have been obtained; see [715] and the references therein.

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 [1822] and references cited therein.

In this paper, we discuss existence and uniqueness of solutions for two sequential Caputo-Hadamard and Hadamard-Caputo fractional differential equations subject to separated boundary conditions asandwhere and are the Caputo and Hadamard fractional derivatives of orders and , respectively, , is a continuous function, and , .

It can be observed that the sequential Caputo-Hadamard and Hadamard-Caputo fractional differential equations in (1) and (2) are different type when and , since and for , respectively.

The rest of the paper is arranged as follows. In Section 2, we establish basic results that lay the foundation for defining a fixed point problem equivalent to the given problems (1) and (2). The main results, based on Banach’s contraction mapping principle, Krasnoselskii’s fixed point theorem, and nonlinear alternative of Leray-Schauder type, are obtained in Section 3. Illustrating examples are discussed in Section 4.

#### 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 , 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 is defined as provided the integral exists.

Definition 3 (see [5]). For an at least -times differentiable function , the Caputo-type Hadamard derivative of fractional order is defined as where , .

Definition 4 (see [5]). The Hadamard fractional integral of order is defined as provided the integral exists.

Lemma 5 (see [5]). For , the general solution of the fractional differential equation is given by where , .

In view of Lemma 5, it follows that for some ,   ().

Lemma 6 (see [23]). Let or   and , where . Then, we have where , .

In order to define the solution of the boundary value problem (1), we consider the linear variantwhere

Lemma 7. LetThen, 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 getAgain taking the Hadamard fractional integral of order to the above equation, we obtainSubstituting in (15)-(16) and applying the first boundary condition of (12), it follows thatFor in equations (15)-(16) and using the second boundary condition of (12), it yieldsSolving the linear system of (17) and (18) for finding two constants , we getandSubstituting constants and 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

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

#### 3. 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 , of all continuous functions on to endowed with the norm In addition, we define the operator bywhere is defined by (13) and Note that the separated boundary value problem (1) has solutions if and only if has fixed points.

For computational convenience we putTo prove the existence theorems of problem (2), we define the operator by

Now, we prove the existence and uniqueness result for problem (1). For problem (2) the proof is similar and omitted.

Theorem 9. Suppose that
there exists a function , such thatIf , where , then the separated boundary value problem (1) has a unique solution on .

Proof. Firstly, we define a ball as , where the constant satisfies where . Next, we will show that For any and using the triangle inequality , we have which implies that Let , then which yields that Since , we deduce that the operator is a contraction. By Banach contraction mapping principle the operator has a unique fixed point, which leads that problem (1) has a unique solution on .

Theorem 10. Let in Theorem 9 holds. If , where then the separated boundary value problem (2) has a unique solution on .

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 (a) where ;(b) is compact and continuous;(c) is a contraction mapping. Then there exists such that .

Theorem 12. Let be a continuous function satisfying in Theorem 9. In addition, assume that
() , and .
Ifthen the separated boundary value problem (1) has at least one solution on .

Proof. Let , where a constant satisfying and We decompose the operator into two operators and on withNote that the ball is a closed, bounded, and convex subset of the Banach space .
Now, we will show that for satisfying condition (a) of Theorem 11. Setting , then we haveThis means that . To prove that is a contraction mapping, for , we haveby condition , which is a contraction, by (34). Therefore, the condition (c) of Theorem 11 is satisfied. Next we will show that the operator is compact and continuous. By using the continuity of the function on , we can conclude that the operator is continuous. For , it follows that wherewhich implies that the set is uniformly bounded. Now we are going to prove that is equicontinuous. For , such that and for , we have which is independent of and also tends to zero as . Hence the set is equicontinuous. Therefore the set is relatively compact. By applying the Arzelá-Ascoli theorem, the operator is compact on . Therefore the operators and satisfy the assumptions of Theorem 11. By the conclusion of Theorem 11, we get that the separated boundary value problem (1) has at least one solution on . This completes the proof.

Theorem 13. Assume that and are fulfilled. If , then the separated boundary value problem (2) has at least one solution on .

The above theorem can be proved by applying Krasnoselskii’s fixed point theorem to the operator defined in (28).

Remark 14. If the operators and are interchanged, then we have the existence results as follows: (i)If , then problem (1) has at least one solution on .(ii)If , then problem (2) has at least one solution on , where However, in application to existence theory, the computation of values and is easier than and , respectively.

The third existence result will be proved by applying Leray-Schauder nonlinear alternative.

Theorem 15 ((nonlinear alternative for single valued maps) [25]). Let be a Banach space, a closed, convex subset of an open subset of , and Suppose that is a continuous; compact (that is, is a relatively compact subset of ) map. Then either (i) has a fixed point in or(ii)there is a (the boundary of in ) and with

Let us state and prove the existence theorem.

Theorem 16. Suppose that
() there exist a continuous nondecreasing function and a function such that() there exists a constant such that Then the separated boundary value problem (1) has at least one solution on .

Proof. Let the operator be defined in (26). Let us prove that the operator maps bounded sets (balls) into bounded sets in . For a constant , we define a bounded ball . Then for , one has which implies thatAfter that we will show that the operator maps bounded sets into equicontinuous sets of . Let be any two points in such that . Then for , we have As , the right-hand side of the above inequality tends to zero independently of . Hence, by applying the Arzelá-Ascoli theorem, the operator 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 for Let be a solution of the operator equation . Then, for , by directly computation, we have which leads to From the assumption , there exists a positive constant such that Let us set It is easy to see that the operator is continuous and completely continuous. From the choice of , there is no such that for some . Therefore, by the nonlinear alternative of Leray-Schauder type (Theorem 15), we deduce that the operator has a fixed point which is a solution of problem (1). The proof is completed.

Theorem 17. Assume that the condition in Theorem 16 is satisfied. If a positive constant satisfyingthen the separated boundary value problem (2) has at least one solution on .

The next two special cases can be obtained by setting , and , with two constants , .

Corollary 18. Let be a continuous function satisfying , for all . Then (i)if , then the separated boundary value problem (1) has at least one solution on ;(ii)if , then the separated boundary value problem (2) has at least one solution on .

#### 4. Examples

In this section, we present some examples to illustrate our results.

Example 1. Consider the following sequential Caputo-Hadamard fractional differential equations with separated boundary conditions

Here , , , , , , , and . From given information, we find that , , and which yield .

(i) Let withIt follows that Then condition is satisfied with . Thus . Hence, by Theorem 9, problem (51) with (52) has a unique solution on .

(ii) Given byObserve that the function defined in (54) satisfies with . But the Theorem 9 can not be applied to this case because the value of . However, by the benefit of Theorem 12, we have . By the conclusion of Theorem 12, problem (51) with (54) has at least one solution on .

Example 2. Consider the following sequential Hadamard-Caputo fractional differential equations with separated boundary conditions

Here , , , , , , , and . From above information, we can find that , , and which can be computed the value of .

(i) The function is defined bySetting and , we see that the condition of Theorem 16 is satisfied with the above function . In addition, we can find that . Then there exists a constant such that satisfying inequality (50). Therefore, applying Theorem 17, problem (55) with (56) has at least one solution on .

(ii) Let byIt is easy to see that the function defined in (57) can be expressed as . Then . Using (ii) of the Corollary 18, the problem (55) with (57) has at least one solution on .

#### Data Availability

No data were used to support this study.

#### Conflicts of Interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.

#### Acknowledgments

This research was funded by King Mongkut’s University of Technology North Bangkok, Contract no. KMUTNB-60-ART-105.

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