A Study of Caputo-Hadamard-Type Fractional Differential Equations with Nonlocal Boundary Conditions

Abstract

Existence and uniqueness results of positive solutions to nonlinear boundary value problems for Caputo-Hadamard fractional differential equations by using some fixed point theorems are presented. The related Green’s function for the boundary value problem is given, and some useful properties of Green’s function are obtained. Example is presented to illustrate the main results.

1. Introduction

In this paper, we study the existence and uniqueness of positive solutions for the -point boundary value problems for Caputo-Hadamard fractional differential equations of the formwhere is the Caputo-Hadamard fractional derivative of order , , , for all , and

The derivative is a kind of fractional derivatives attributed to Hadamard in 1892 [1]; this fractional derivative differs from the Riemann-Liouville and Caputo fractional derivatives in the sense that the kernel of the integral contains a logarithmic function of arbitrary exponent. The Riemann-Liouville and Hadamard derivative have their own disadvantages as well, one of which is the fact that the derivative of a constant is not equal to zero in general. The subject of Hadamard-type fractional differential equations has received much attention by many researchers. Some new results on the existence of solutions for a fractional boundary value problem involving Hadamard-type fractional differential inclusions and integral boundary conditions can be found in [2].

In [3], Tariboon et al. studied the existence and uniqueness of solutions to the boundary value problems consisting of a fractional differential equation of Riemann-Liouville type, subject to the Hadamard fractional integral equations. Thiramanus et al. [4] investigated the existence and uniqueness of solutions for a Hadamard fractional differential equations with nonlocal fractional integral boundary conditions. Ahmad and Ntouyas [5, 6] studied the existence results for a boundary value problem of nonlinear fractional hybrid differential inclusions of Hadamard type with Dirichlet boundary condition and for a coupled system of Hadamard fractional differential equations and Hadamard-type integral boundary conditions, respectively.

Jarad et al. [7] modified the Hadamard fractional derivative into a more suitable one having physically interpretable initial conditions similar to the ones in the Caputo setting.

Basic definitions and properties of fractional calculus and Hadamard-type fractional calculus can be found in [8–10].

The idea of this paper is to demonstrate sufficient conditions on existence and uniqueness of positive solutions to nonlinear boundary value problems (1) for modified Hadamard fractional (Caputo-Hadamard) differential equations, by using Banach’s fixed point theorem, Leray-Schauder nonlinear alternative theorem for single valued maps, Krasnoselskii’s fixed point theorem, and some properties of the Green function.

2. Preliminaries

In this section, we introduce some notations and definitions of Hadamard-type fractional calculus.

Definition 1 (see [11, 12]). The Hadamard derivative of fractional order for a function is defined aswhere denotes the integer part of the real number and

Definition 2 (see [11, 12]). The Hadamard fractional integral of order of a function , is defined as

Definition 3 (see [1, 7]). Let , and ThenHere and In particular, if , then

Theorem 4 (see [1, 7]). Let and Then and exist everywhere on and
(a) if , (b) if ,In particular,

Lemma 5 (see [1, 7]). Let , and
If or , then

Lemma 6 (see [1, 7]). Let or and , and then

Lemma 7 (see [1, 7]). For , and , the unique solution of the problemis given by where

Proof. In view of Lemma 6, the solution of the Hadamard differential equation (12) can be written as In view of the boundary conditions, we conclude that Substituting the values of , we obtain

Lemma 8. The functions , defined by (15) satisfy,,, where, where

Proof. It is clear that (i) holds. So we prove that (ii) is true.
(ii) In view of the expression for , it follows that for all .
If , we haveIf , we haveThusTherefore, (iii) If , If , we have (iv) Since , then is nonincreasing in , so

3. Existence Results

Let us denote by the Banach space of all continuous functions from to endowed with the norm and let be the cone

Through this paper, we assume that the function satisfies the following conditions of Carathéodory type:

: (i) is Lebesgue measurable with respect to on ,

(ii) is continuous with respect to on

By Lemma 6, we obtain an operator as It should be noticed that problem (1) has solutions if and only if the operator has fixed points.

The first existence and uniqueness result is based on the Banach contraction principle.

Theorem 9. Assume that the condition holds and there exists a real-valued function such thatThen problem (1) has a unique solution provided , where

Proof. We set and choose
Now, we show that , where For any , we have It follows that For and for each , we have Hence it follows that , where Therefore is a contraction. Hence by the contraction mapping principle, problem (1) has a uniqueness solution.

Theorem 10 (nonlinear alternative for single valued maps [13]). Let be a Banach space, a closed, convex subset of , an open subset of , and Suppose that is a continuous, compact (i.e., is a relatively compact subset of ) map. Then either(i) has a fixed point in or(ii)There is (the boundary of in ) and with

Theorem 11. Assume that and the following conditions hold:
There exist two nonnegative real-valued functions such that There exists a constant , such thatThen the boundary value problem (1) has at least one solution on

Proof. First, we show that the operator is continuous.
For any with Thus, by condition (ii) of , we have So, we can conclude that On the other hand,Hence, This means that is continuous.
Now, we show that maps bounded sets into bounded sets in . It suffices to show that, for any , there exists a positive constant such that, for each , we have By (33), for each , we have which implies that Further, we let with and , where is a bounded set of , and then we find thatHence, for , we have This implies that maps bounded sets into equicontinuous sets of .
Thus, by the Arzelà-Ascoli theorem, the operator is completely continuous.
Next, we consider the set and show that the set is bounded, let , and then . For any , we haveThus, for any , so that set is bounded. Thus, by the conclusion of Theorem 10, the operator has at least one fixed point, which implies that problem (1) has at least one solution.

Theorem 12 (Krasnoselskii fixed point theorem [14]). Let be a Banach space and is a cone in . Assume that and are open subsets of with and . Let be completely continuous operator. In addition suppose that either(i) and or(ii) and holds. Then has a fixed point in .
To state the last result of this section, we set

Theorem 13. Suppose that there exist two positive constants and :(i), for (ii), for Then (1) has at least a positive solution.

Proof. Let From the proof of Theorem 11, we know that the operator defined by (28) is completely continuous on .
For any , it follows that that is,
On the other hand, for any , it follows thatthat is,
In view of Theorem 12, has a fixed point in , which is a positive solution to (1).

4. Examples

In this section, we exemplify our theoretical results obtained in Section 3.

Example 1. Consider the following BVP for Hadamard fractional differential equation:Here , , , , andUsing the given data, we find that , , , , and Hence, we obtain Therefore, by Theorem 9, problem (45) has a unique solution on

Example 2. Consider the following BVP for Hadamard fractional differential equation:Here , , , . Let with , Here,
It is easy to verify that
Then, by Theorem 11, problem (48) has at least one solution on

Example 3. Consider the following BVP for Hadamard fractional differential equation:where , , , , , , . Let It is easy to verify that Choosing , , we have Hence all conditions of Theorem 13 are satisfied; then problem (50) has at least one positive solution such that

Competing Interests

The author declares no competing interests.