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Journal of Function Spaces
Volume 2019, Article ID 5923490, 6 pages
https://doi.org/10.1155/2019/5923490
Research Article

The Unique Positive Solution for Singular Hadamard Fractional Boundary Value Problems

1School of Mathematics, Qufu Normal University, Qufu, Shandong, 273165, China
2Department of Mathematics, Jining University, Qufu, Shandong, 273155, China

Correspondence should be addressed to Jinxiu Mao; moc.361@2891uixnijoam

Received 21 February 2019; Accepted 31 May 2019; Published 20 June 2019

Guest Editor: Pedro Garrancho

Copyright © 2019 Jinxiu Mao et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

In this paper, we investigate singular Hadamard fractional boundary value problems. The existence and uniqueness of the exact iterative solution are established only by using an iterative algorithm. The iterative sequences have been proved to converge uniformly to the exact solution, and estimation of the approximation error and the convergence rate have also been derived.

This paper is dedicated to our advisors

1. Introduction

Fractional differential operators play an important role in describing phenomenons in many fields such as physics, chemistry, control, and electromagnetism [19]. They have many applications in fractional differential equations and fractional integral equations. In [10], authors investigate a class of fractional integral equations arising from a symmetric transition model where and are the left and right Riemann-Liouville fractional integrals of order , respectively, and is the gradient of at .

The fractional order diffusion equation where and , contains fractional differential operators which can describe different diffusion process.

In the past ten years, fractional differential equations have been considered in many papers (see [1122]). Most of the works on the topic have been based on Riemann-Liouville type and Caputo type fractional differential equations. By means of fixed point theorems and variational methods, authors obtain at least one or multiple positive solutions for boundary value problems of fractional differential equations. Very recently, more studies have been carried out on the boundary value problems of nonlinear Hadamard fractional differential equations. An important characteristic of Hadamard fractional derivative is that it contains logarithmic function of arbitrary exponent. However, there are few results about this topic (see [2328]).

By using the Krasnoselskii-Zabreiko fixed point theorem, Yang [26] obtained at least one positive solution for the boundary value problem where , . was the Hadamard fractional derivative of order . , .

In [28], the authors studied the Hadamard fractional differential equation with boundary conditions where denoted Hadamard fractional derivative of order and denoted Hadamard fractional integral of order , . By employing the complete continuity of the associated integral operator and the monotone iterative method, the authors obtained twin positive solutions and the unique positive solution.

Most of the above works required the associated integral operators to be completely continuous because fixed point theorems could be applied. Furthermore, the uniqueness of positive solutions was rarely investigated while the existence and multiplicity of positive solutions were investigated widely.

Inspired by the above results, in this work, we study the existence and uniqueness of positive solutions for the following boundary value problem:where denotes Hadamard fractional derivative of order ; is continuous.

In this work, only by using the monotone iterative technique, we aim to establish the unique positive solution for problem (6). The main contributions of this work are as follows: (a) the nonlinear term can be singular at and ; (b) we do not need the continuity and complete continuity of the associated integral operator; (c) we get the unique positive solution.

Throughout this work, we assume that the following conditions hold without further mention.

is continuous.

: For , is nondecreasing in and there exists a constant such that, for ,

It is easy to verify that if , then

2. Preliminaries

The way to attack this new problem follows a scheme similar to that used in [21], with the necessary adaptations that Hadamard fractional derivative contains logarithmic function of arbitrary exponent.

In this section, we present some basic concepts and conclusions needed in the proof of our main results.

Definition 1. The Hadamard fractional integral of order of a function is given by

Definition 2. The Hadamard fractional derivative of order is defined by where , .
Specifically,

Lemma 3. Suppose that , Then the Hadamard type fractional differential equationhas a unique solution where

Proof. As argued in [9], the solution of the Hadamard differential equation in (10) can be written as the equivalent integral equationFrom , we have Thus (47) reduces toUsing we obtainSubstituting (15) into (14), we obtain

Take and for , . We can prove that have the following properties.

Lemma 4 (see [26]). For , Green’s function satisfies the following properties:
(i) is continuous on and ,
(ii) ,
(iii) ,
(iv)
From , and (ii) we get

In this paper, we will work in the Banach space with the norm

Define a set as follows.

there are constants such that Evidently Therefore, is not empty.

3. The Main Results

Theorem 5. Assume hold. And Then problem (6) has at least one positive solution

Proof. Define the operator byWe can see easily the equivalence between is a solution of (6) and is a fixed point of .
Claim 1. The operator is nondecreasing.
In fact, for , it is obvious that , For any and , from (17), and where and are positive constants satisfying Thus, it follows that there are constants such that, for , Therefore, for any , , is the operator From (19), it is easy to see that is nondecreasing with respect to . Hence, Claim 1 holds.
Claim 2. There exist a nondecreasing sequence and a nonincreasing sequence and there exists such thatuniformly on
First, there exist two constants with since Take and to be fixed numbers satisfyingObviously,
We construct two iterative sequences as follows:ThenIn fact, from (26), we have and Furthermore, From and nondecreasing, (28) holds. Let ; then It follows from (7) that And, for any natural number , Thus, for any natural number and , we havewhich implies that is a cauchy sequence in So there exists such thatand from (32)From , we have Combining with nondecreasing on , Let , which implies is a positive solution of problem (6).

Theorem 6. Assume hold. Then, we have the following:
(i)Problem (6) has unique positive solution in and there exist constants with such that (ii)For any initial value , there exists a sequence that uniformly converges to the unique positive solution , and we have the error estimationwhere is a constant with and determined by ; represents the same order infinitesimal of

Proof. Let be defined in (26) and (27).
(i) It follows from Theorem 5 that problem (6) has a positive solution , which implies that there exists constants and with such that Let be another positive solution of problem (6). Then there exist constants and with such that Let defined in (25) be small enough so that and defined in (25) be large enough so that Then Note that and is nondecreasing; we have Letting , we obtain that Hence, the positive solution of problem (6) is unique.
(ii) For any , there exist constants and with such that Similar to (i), take and to be defined by (25) satisfying and Then Let Note that is nondecreasing, Letting , it follows from (33) and (34) that uniformly on .
At the same time, (38) follows from (32).

Remark 7. We just investigate a simple form of boundary value problems for Hadamard differential equations. We can easily apply the monotone iterative technique to multipoint or multistrip boundary value problems.

Remark 8. Suppose that are nonnegative continuous functions on , which may be unbounded at the end points of . is the set of functions which satisfy the conditions and . Then we have the following conclusions:
(1) , where .
(2) If and , then .
(3) If , then .
(4) If , then .
The above four facts can be verified directly. This indicates that there are many kinds of functions which satisfy the conditions and .

4. An Example

Consider the following boundary value problem:where , ,

Analysis 1. First, and so holds.
For any , we take and have Then holds.
Obviously where . Hence all conditions of Theorem 5 are satisfied, and consequently we have the following corollary.

Corollary 9. Problem (47) has unique positive solution . For any initial value , the successive iterative sequence generated by uniformly converges to the unique positive solution on . We have the error estimation where is a constant with and determined by the initial value And there are constants with such that

Data Availability

Data sharing is not applicable to this article as no datasets were generated or analysed during the current study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Acknowledgments

This paper is supported by the Natural Science Foundation of China (11571197) and the Science Foundation of Qufu Normal University of China (XJ201112).

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