#### Abstract

We obtain in this article a solution of sequential differential equation involving the Hadamard fractional derivative and focusing the orders in the intervals and . Firstly, we obtain the solution of the linear equations using variation of parameter technique, and next we investigate the existence theorems of the corresponding nonlinear types using some fixed-point theorems. Finally, some examples are given to explain the theorems.

#### 1. Introduction

In recent years, there has been a great development in the study of fractional differential equations. This advancement is ranging from the theoretical analysis of the subject to analytical and numerical techniques (see [1–3] and the references cited therein). Among the theoretic approach, the existence theory of solutions for fractional differential models has gained attentions of many authors. Most of them have focused on using Riemann-Liouville and Caputo derivatives in representing the underlying fractional differential equation (see [4–10]). Another kind of fractional derivative is Hadamard type which was introduced in 1892 [11]. This derivative differs from aforementioned derivatives in the sense that the kernel of the integral in the definition of Hadamard derivative contains logarithmic function of arbitrary exponent. A detailed description of Hadamard fractional derivative and integral can be found in [12, 13]. Recently, the existence and uniqueness of solution for fractional differential equations in Hadamard sense were introduced in many faces by several authors ([6, 7] and references therein). We add in this article a new idea concerning the sequential definition of Hadamard fractional operator with constant coefficients of order less than three. For , the Hadamard fractional differential equation can be transformed to classical Euler-Cauchy second-order nonhomogeneous differential equation that can be solved by variation of parameter technique.

More precisely, we consider the nonlinear Hadamard fractional differential equations given bywhere , , and is a given continuous function for each . Here, is the Hadamard fractional derivative, and , , and are constants.

#### 2. Linear Fractional Differential Equations

We introduce some basic ideas of fractional calculus that may be used in the sequel of this article [12].

*Definition 1. *The Hadamard fractional integral of order is defined as provided that the integral exists.

*Definition 2. *The Hadamard derivative of fractional order is defined as where denotes the integer part of the real number

Let denote the Banach space of all real valued continuous functions defined on , and denotes the Banach space of all real valued functions such that

Lemma 3 (see [12]). *Let , , and ; then,The following result is obvious by Lemma 3.*

Lemma 4. *Let , and The fractional differential equationhas a solution given by*

Lemma 5. *Let , and . The fractional differential equationhas a solution given by*

*Proof. *Applying the fractional integral operator to (9), we havefor some constants , , and . The initial condition implies that Taking the first derivative of (11), it follows thatThe condition implies that Multiplying (12) by the integrating factor given by we get Integrating (14), and using again , we conclude that This finishes the proof.

*Remark 6. *For , the solution is still valid, since This trivial case will be negligible hereafter.

Lemma 7. *Let , , and The linear system has a solution *

*Proof. *Applying the fractional integral operator to (16), we havefor some constants , , , , , and . The initial condition implies that Taking the first derivative of (18), and using Leibniz’s rule, it follows thatThe condition implies that Multiplying by and then the second derivative would imply For the last initial condition , we get Now, multiplying (20) by , it follows that which is a second-order Cauchy-Euler differential equation that has a general solution , where and are the complementary and particular solution of (14), respectively. To find the complementary solution of (21), consider the solutions and for the homogeneous equation where and are the distinct real roots of the characteristic equationThe complementary solution of the homogeneous equation (22) is for some constants and . These two constants can be evaluated by the initial conditions given in (16), which both imply that The only solution for these algebraic equations is , since The Wronskian for the solutions and is since Applying the variation of parameter technique we can get the particular solution Therefore, the general solution is

If , we consider and Then, the complementary and the particular solutions are where the Wronskian in this case is given byThe evaluation of the constants and in the complementary solution (28) leads towhich imply Therefore, the general solution becomes again The two cases together constitute the required solution. This finishes the proof.

*Remark 8. *If , then (23) has two complex conjugate roots that will not be considered in this article. On the other hand, if , the solution is , which is a trivial case and hereafter will be negligible.

*Remark 9. *Any of initial conditions should not be used because it would imply ill-posed systems.

#### 3. Existence Theorems

We establish sufficient conditions for existence of solutions to problems (1)–(3) using different types of fixed-point theorems.

In view of Lemmas 4, 5, and 7, we transform the initial value problems (1)–(3), respectively, into operator equations asand for , we define the respective cases , and , as where If the operator , , has a fixed point, the corresponding problem in systems (1)–(3) has this fixed point as a solution.

Lemma 10. *The operator , , is completely continuous.*

*Proof. *Obviously, the continuity of the operator follows from the continuity of the function Let be a bounded proper subset of ; then for any , and , there exists a positive constant such that . Accordingly, for any , and , it follows that and for , we have Consequently, for any , the operator is bounded on Furthermore, if , such that , then for Let , and then For , let , and then Now, let , and thenAs , , for This implies that is equicontinuous on In consequence, it follows by the Arzela-Ascoli theorem that the operator is completely continuous. This finishes the proof.

We need the following Schauder’s fixed point theorem ([14]).

Theorem 11. *If is a closed, bounded, convex subset of a Banach space and the mapping is completely continuous, then has a fixed point in .*

Accordingly, if we define a closed, bounded, convex subset of on which s, as defined by (31)-(32), are completely continuous, then problems (1)–(3) have the respective solution.

Theorem 12. *Assume thatthen, each problem of (1)–(3) has a solution.*

*Proof. *For any , the given condition implies that , whenever , for a fixed number Therefore, define a subset asHence, is a closed, bounded, and convex subset of If , then , for any On the other hand, the operator is completely continuous by Lemma 10; then by Schauder’s fixed-point Theorem 11, each problem of (1)–(3) has a solution. This finishes the proof.

Our next result deals with the uniqueness of solution for problems (1)–(3) and is based on the Banach contraction theorem. The following assumptions are essential for proving the next theorem.(H1)For any , , ,(H2)Let , such that

Here , if , or , if , where

Theorem 13. *Assume that (H1) and (H2) are satisfied. Then there exists a unique solution for each of the problems (1)–(3) on *

*Proof. *Let , for We show that , where This follows by the following estimates: for any . Moreover, for , , and for each , we obtain As , is a contraction. Thus, the conclusion of the theorem follows by the contraction mapping principle. This completes the proof.

We close the existence theorems by applying Leray-Schauder degree theorem [14]. Therefore, we need the following assumption.(H3)For , , and ,where , satisfies where is defined as in (H2).

Theorem 14. *Assume that (H3) is satisfied. Then every problem of (1)–(3) has a solution.*

*Proof. *Define a ball , , for some positive real number which will be determined later. We show that satisfies for any , and ]; here denotes the boundary set of Define the homotopy Then, by Lemma 10, is completely continuous. Let denote the identity operator; then the homotopy invariance and normalization properties of topological degrees imply that since . By the nonzero property of the Leray-Schauder degree, for at least one . To find , we assume that for some and for all Then, using (H3), we have Therefore,The value of is sufficient for applicability of Leray-Schauder degree theorem. This completes the proof.

*Example 1. *Consider the following nonlinear fractional differential equation:Here , , and Simple calculations lead to , and Therefore, all previous existence theorems ensure the existence and uniqueness of the solution in for problem (53).

*Example 2. *Consider the following nonlinear fractional differential equation:Here , , , and Simple calculations lead to We apply any of the following:(1)Theorem 12, with (2)Theorem 13, with , and (3)Theorem 14, with , and Problem (55) has a solution in . The uniqueness of this solution follows from Theorem 13.

#### Conflicts of Interest

The author declares that there are no conflicts of interest regarding the publication of this paper.