- Research Article
- Open Access

# Existence and Uniqueness of Positive and Nondecreasing Solutions for a Class of Singular Fractional Boundary Value Problems

- J Caballero Mena
^{1}, - J Harjani
^{1}and - K Sadarangani
^{1}Email author

**2009**:421310

https://doi.org/10.1155/2009/421310

© J. Caballero Mena et al. 2009

**Received:**24 April 2009**Accepted:**14 June 2009**Published:**19 July 2009

## Abstract

We establish the existence and uniqueness of a positive and nondecreasing solution to a singular boundary value problem of a class of nonlinear fractional differential equation. Our analysis relies on a fixed point theorem in partially ordered sets.

## Keywords

- Continuous Function
- Unique Solution
- Partial Order
- Fractional Derivative
- Fixed Point Theorem

## 1. Introduction

Many papers and books on fractional differential equations have appeared recently. Most of them are devoted to the solvability of the linear fractional equation in terms of a special function (see, e.g., [1, 2]) and to problems of analyticity in the complex domain [3]. Moreover, Delbosco and Rodino [4] considered the existence of a solution for the nonlinear fractional differential equation , where and , is a given continuous function in . They obtained results for solutions by using the Schauder fixed point theorem and the Banach contraction principle. Recently, Zhang [5] considered the existence of positive solution for equation , where and is a given continuous function by using the sub- and super-solution methods.

In this paper, we discuss the existence and uniqueness of a positive and nondecreasing solution to boundary-value problem of the nonlinear fractional differential equation

where , is the Caputo's differentiation and with (i.e., is singular at ).

Note that this problem was considered in [6] where the authors proved the existence of one positive solution for (1.1) by using Krasnoselskii's fixed point theorem and nonlinear alternative of Leray-Schauder type in a cone and assuming certain hypotheses on the function . In [6] the uniqueness of the solution is not treated.

In this paper we will prove the existence and uniqueness of a positive and nondecreasing solution for the problem (1.1) by using a fixed point theorem in partially ordered sets.

Existence of fixed point in partially ordered sets has been considered recently in [7–12]. This work is inspired in the papers [6, 8].

For existence theorems for fractional differential equation and applications, we refer to the survey [13]. Concerning the definitions and basic properties we refer the reader to [14].

Recently, some existence results for fractional boundary value problem have appeared in the literature (see, e.g., [15–17]).

## 2. Preliminaries and Previous Results

For the convenience of the reader, we present here some notations and lemmas that will be used in the proofs of our main results.

Definition 2.1.

provided that the right-hand side is pointwise defined on .

Definition 2.2.

where , provided that the right-hand side is pointwise defined on .

The following lemmas appear in [14].

Lemma 2.3.

where ,

Lemma 2.4.

is valid when , , .

The following lemmas appear in [6].

Lemma 2.5.

Remark 2.6.

Note that for and (see [6]).

Lemma 2.7.

is continuous on [0,1], where is the Green function defined in Lemma 2.5.

Now, we present some results about the fixed point theorems which we will use later. These results appear in [8].

Theorem 2.8.

where is continuous and nondecreasing function such that is positive in , and . If there exists with then has a fixed point.

If we consider that satisfies the following condition:

then we have the following theorem [8].

Theorem 2.9.

Adding condition (2.10) to the hypotheses of Theorem 2.8 one obtains uniqueness of the fixed point of .

In our considerations, we will work in the Banach space with the standard norm .

Note that this space can be equipped with a partial order given by

In [10] it is proved that with the classic metric given by

satisfies condition (2) of Theorem 2.8. Moreover, for , as the function is continuous in , satisfies condition (2.10).

## 3. Main Result

Theorem 3.1.

Then one's problem (1.1) has an unique nonnegative solution.

Proof.

Note that, as is a closed set of , is a complete metric space.

Now, for we define the operator by

Hence, .

In what follows we check that hypotheses in Theorems 2.8 and 2.9 are satisfied.

Firstly, the operator is nondecreasing since, by hypothesis, for

Put . Obviously, is continuous, nondecreasing, positive in , and .

Thus, for

Finally, take into account that for the zero function, , by Theorem 2.8 our problem (1.1) has at least one nonnegative solution. Moreover, this solution is unique since satisfies condition (2.10) (see comments at the beginning of this section) and Theorem 2.9.

Remark 3.2.

In [6, lemma 3.2] it is proved that is completely continuous and Schauder fixed point theorem gives us the existence of a solution to our problem (1.1).

In the sequel we present an example which illustrates Theorem 3.1.

Example 3.3.

Note that .

Theorem 3.1 give us that our fractional differential (3.10) has an unique nonnegative solution.

This example give us uniqueness of the solution for the fractional differential equation appearing in [6] in the particular case and

Remark 3.4.

where is continuous and satisfies

(a) and nondecreasing;

(b) ;

(c) is positive in ;

(d) .

Examples of such functions are and .

Remark 3.5.

Note that the Green function is strictly increasing in the first variable in the interval . In fact, for fixed we have the following cases

Case 1.

Case 2.

Hence, taking into account the last inequality and (3.16), we obtain .

Case 3.

and, as for , it can be deduced that and consequently, .

This completes the proof.

Remark 3.5 gives us the following theorem which is a better result than that [6, Theorem 3.3] because the solution of our problem (1.1) is positive in and strictly increasing.

Theorem 3.6.

Under assumptions of Theorem 3.1, our problem (1.1) has a unique nonnegative and strictly increasing solution.

Proof.

Taking into account Remark 3.4 and the fact that , we get .

Now, if we suppose that then and as, we deduce that a.e.

On the other hand, if a.e. then

and this contradicts that a.e.

Thus, for with . Finally, as we have that for .

## Declarations

### Acknowledgment

This research was partially supported by "Ministerio de Educación y Ciencia" Project MTM 2007/65706.

## Authors’ Affiliations

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