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Abstract and Applied Analysis
Volume 2012 (2012), Article ID 803417, 18 pages
http://dx.doi.org/10.1155/2012/803417
Research Article

Existence and Uniqueness of Positive Solutions for a Singular Fractional Three-Point Boundary Value Problem

Department of Mathematics, University of Las Palmas de Gran Canaria, Campus de Tafira Baja, 35017 Las Palmas de Gran Canaria, Spain

Received 12 June 2012; Accepted 29 July 2012

Academic Editor: Juan J. Trujillo

Copyright © 2012 I. J. Cabrera 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

We investigate the existence and uniqueness of positive solutions for the following singular fractional three-point boundary value problem , where , is the standard Riemann-Liouville derivative and with (i.e., is singular at ). Our analysis relies on a fixed point theorem in partially ordered metric spaces.

1. Introduction

Fractional differential equations have been of great interest recently. It is caused both by the intensive development of the theory of fractional calculus itself and by the applications (see, e.g., [15]).

Recently, many papers have appeared dealing with the existence of solutions of nonlinear fractional boundary value problems.

In [6], the authors studied the existence and multiplicity of positive solutions for the boundary value problem: where and is continuous, by using some fixed point theorem on cones.

In [7], the authors considered the following nonlinear fractional boundary value problem: where and is continuous. They obtained their results by using lower and upper solution method and fixed point theorems.

In [8] the authors investigated the existence and uniqueness of positive and nondecreasing solutions for a class of singular fractional boundary value problems by using a fixed point theorem in partially ordered metric spaces.

Very recently, in [9] the authors studied the existence of solutions of the following three-point boundary value problem: where , , and is continuous.

Motivated by [8, 9], in this paper we discuss the existence and uniqueness of positive solutions for Problem (1.3) assuming that is such that (i.e., is singular at ). Our main tool is a fixed point theorem in partially ordered metric spaces which appears in [10].

2. Preliminaries and Basic Facts

For the convenience of the reader, we present some notations and lemmas which will be used in the proof of our results.

Definition 2.1 (see [5]). The Riemann-Liouville fractional integral of order of a function is given by provided that the right-hand side is pointwise defined on and where denotes the classical gamma function.

Definition 2.2 (see [2]). The Riemann-Liouville fractional derivative of order of a function is defined as where and denotes the integer part of .

The following two lemmas can be found in [2] and they are crucial in finding an integral representation of the boundary value problem (1.3).

Lemma 2.3 (see [2]). Assume that and .
Then the fractional differential equation has where and , as unique solution.

Lemma 2.4 (see [2]). Assume that with fractional derivative of order that belongs to . Then for some and .

By using Lemma 2.4, in [9] the authors proved the following result.

Lemma 2.5 (see [9]). Let and and .
Then the boundary value problem where , has as unique solution where

Remark 2.6. In [9] it is proved that is a continuous function on , , and

In the sequel, we present the fixed point theorem which we will use later. Previously, we present the following class of functions.

By we denote the class of functions satisfying the following condition: Examples of functions belonging to are with and .

The fixed point theorem which we will use later appears in [10].

Theorem 2.7 (see [10]). Let be a partially ordered set and suppose that there exists a metric in such that is a complete metric space. Let be a nondecreasing mapping such that there exists an element with . Suppose that there exists such that Assume that either is continuous or is such that Besides, if then has a unique fixed point.

In our considerations, we will work in the Banach space with the classical metric given by .

Notice that this space can be equipped with a partial order given by In [11] it is proved that satisfies condition (2.12) of Theorem 2.7. Moreover, for , as the function , satisfies condition (2.13).

3. Main Result

Our starting point of this section is the following lemma.

Lemma 3.1. Suppose that , , and   is a continuous function with . If is a continuous function on then the function defined by is continuous on , where is the Green’s function appearing in Lemma 2.5.

Proof. We divide the proof into three cases.
Case 1 (). It is clear that .
Since is a continuous function on , we can find a constant such that for any .
Then, we get If in the integral we use the change of variables then we have This and (3.2) give us and letting , we see that .
This proves the continuity of at .
Case 2(). We take and we have to prove that .
Without loss of generality, we can take (the same argument works for ).
In fact, where In the sequel, we will prove that when .
In fact, as , then Moreover, and, as we have that the sequence converges pointwise to the zero function and is bounded by a function belonging to , then by Lebesgue’s dominated convergence theorem Now, we will prove that when .
In fact, as and letting and, taking into account that , from the last expression we get Finally, from (3.5), (3.10), and (3.12) we get This proves the continuity of at .
Case3 (). It is easily checked that .
Now, following the same lines that in the proof of Case 1, we can demonstrate the continuity of at .
This finishes the proof.

Lemma 3.2. Suppose that , , , and is a continuous function with .
If is a continuous function on then the function defined by is continuous on , where is the function appearing in Lemma 2.5.

Proof. Since is continuous on , there exists a constant such that for any .
Taking into account that we have and, consequently, the function is continuous at any point .

Remark 3.3. Notice that the function appearing in Lemma 2.5 which is defined as is continuous function on and, moreover, .
In fact, for it is clear that .
In the case, , we have This proves the nonnegative character of the function on .

Lemma 3.4. Suppose that . Then where is the function appearing in Lemma 2.5.

Proof. By definition of , we have As we saw in Case 1 of Lemma 3.1. and, therefore, Now, using elemental calculus it is easily seen that the function is increasing on the interval and, therefore,

Lemma 3.5. Suppose that then where is the function appearing in Lemma 2.5.

Proof. By definition of , we have By a similar argument that the one used in the Case 1 of Lemma 3.1, we have This finishes the proof.

By commodity, we denote by the constant given by Moreover, we introduce the following class of functions which will be used in the main result of the paper. By we denote the class of functions satisfying the following: (i)is nondecreasing. (ii) for any . (iii), where is the class of functions introduced in Remark 2.6.

Theorem 3.6. Let , , , , and is a continuous function with and such that is a continuous function on . Assume that there exists such that for with and where .
Then Problem (1.3) has a unique positive solution (this means that for ).

Proof. Consider the cone: Since is a closed set of , is a complete metric space with the distance given by , for .
It is easily checked that satisfies conditions (2.12) and (2.13) of Theorem 2.7.
Now, for we define the operator by By Lemmas 3.1 and 3.2, for we have .
Moreover, in view of the nonnegative character of , , and , we have that for .
In what follows, we check that assumptions in Theorem 2.7 are satisfied.
Firstly, we will prove that is nondecreasing.
In fact, by (3.29), for we have This proves that is a nondecreasing operator.
On the other hand, for and , we have Since is nondecreasing, the last inequality implies Now, from Lemmas 3.4 and 3.5 it follows: Since , from the last inequality we obtain and, since , Since this inequality is obviously satisfied for , we have Finally, since the zero function satisfies , Theorem 2.7 says us that the operator has a unique fixed point in , or, equivalently, Problem (1.3) has a unique nonnegative solution in .
Now, we will prove that is a positive solution.
In contrary case, we can find such that .
Taking into account that the nonnegative solution of Problem (1.3) is a fixed point of the operator, we have and, particularly, Since both summands in the right hand are nonnegative (see Remarks 2.6 and 3.3) we have Given the nonnegative character of , , and , we have Taking into account that , this means that for we can find such that for we have . Notice that and , where is the Lebesgue measure on .
This and (3.42) give us that and this is a contradiction since and are rational functions in the variable .
Therefore, for .
This finishes the proof.

In order to present an example which illustrates our results, we need to prove some properties about the hyperbolic tangent function.

Previously, we recalled some definitions.

Definition 3.7. A function is said to be subadditive if it satisfies An example of subadditive function is the square root function, that is, .

Remark 3.8. Suppose that is subadditive and then In fact, sincethis inequality implies that
Recall that a function is concave if for any and .

Lemma 3.9. Let be a concave function with . Then is subadditive.

Proof. We take .
Since is concave and , we get Adding these inequalities, we have This proves the lemma.

In what follows, we will prove that the function belongs to the class previously defined.

Lemma 3.10. The function defined as satisfies: (a). (b) is subadditive.

Proof. (a) Since for , is nondecreasing.
Moreover, the function has as derivative and, consequently, is strictly nondecreasing on .
Since , we have for or, equivalently, for .
In order to prove that , notice that if then the sequence is a bounded sequence.
In fact, in contrary case and we have which contradicts the fact that .
Now, we suppose that and .
Then, we can find such that for each there exists with .
Since is a bounded sequence (because ) we can find a subsequence of , which we will denote of the same way, such that .
As , it follows that and, as the unique solution of the equation on is , we deduce that .
Therefore, and this implies that there exists such that for .
This contradicts the fact that for any .
Therefore, .
This proves that .
Therefore, .
(b) Since and this means that is a concave function with and, by Lemma 3.9, is subadditive.

Remark 3.11. By Remark 3.8 and by (b) of Lemma 3.9, for with

Now, we present an example which illustrates our result.

Example 3.12. Consider the following singular fractional boundary value problem In this case, , , and .
Moreover, in this case for .
Notice that is continuous in and .
Now, we check that satisfies assumptions appearing in Theorem 3.6.
It is clear that is a continuous function on .
Moreover, by Lemma 3.10 and Remark 3.11, for and we have where is a function belonging to (see, Lemma 3.10).
Finally, Theorem 3.6 says us that Problem (3.59) has a unique positive solution for Or, equivalently, for .

Acknowledgment

This research was partially supported by “Universidad de Las Palmas de Gran Canaria”, Project ULPGC 2010–006.

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