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Abstract and Applied Analysis

Volume 2012 (2012), Article ID 303545, 11 pages

http://dx.doi.org/10.1155/2012/303545

## Positive Solutions of Nonlinear Fractional Differential Equations with Integral Boundary Value Conditions

Departamento de Matemáticas, Universidad de Las Palmas de Gran Canaria, Campus de Tafira Baja, 35017 Las Palmas de Gran Canaria, Spain

Received 28 February 2012; Accepted 7 September 2012

Academic Editor: Yong H. Wu

Copyright © 2012 J. Caballero 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 of the following nonlinear fractional differential equation with integral boundary value conditions, , , where , and is the Caputo fractional derivative and is a continuous function. Our analysis relies on a fixed point theorem in partially ordered sets. Moreover, we compare our results with others that appear in the literature.

#### 1. Introduction

Many papers and books on fractional differential equations have appeared recently (see, for example, [1–22]). The interest of the study of fractional-order differential equations lies in the fact that fractional-order models are more accurate than integer-order models, that is, there are more degrees of freedom in the fractional-order models.

Integral boundary conditions have various applications in chemical engineering, thermo-elasticity, population dynamics, and so forth. For a detailed description of the integral boundary conditions, we refer the reader to some recent papers (see, [23–30]) and the references therein. Recently, Cabada and Wang in [31] investigated the existence of positive solutions for the fractional boundary value problem where , , is the Caputo fractional derivative and is a continuous function.

The main tool used in [31] is the well-known Guo-Krasnoselskii fixed point theorem and the question of uniqueness of solutions is not treated. We consider our paper as an alternative answer to the results of [31]. The fixed point theorem in partially ordered sets is the main tool used in our results. The existence of fixed points in partially ordered sets has been considered recently (see, e.g. [32–34]).

#### 2. Preliminaries and Basic Facts

For the convenience of the reader, we present in this section some notations and lemmas which will be used in the proofs of our results. For details, see [35, 36].

*Definition 2.1. *The Caputo derivative of fractional order of a function is defined by
where and denotes the integer part of .

*Definition 2.2. *The Riemman-Liouville fractional integral of order of a function is defined by
provided that such integral exists.

*Definition 2.3. *The Riemman-Liouville fractional derivative of order of a function is given by
where , provided that the right hand side is pointwise defined on .

Lemma 2.4. *Let then the fractional differential equation
**
has
**
as unique solution. *

Lemma 2.5. *Let then
*

In [31], the authors obtain the Green's function associated with Problem (1.1). More precisely, they proved the following result.

Theorem 2.6 (see [31]). *Let and . Suppose that then the unique solution of
**
is , where
*

In [31], the following lemma is proved.

Lemma 2.7. *Let be the Green's function associated to Problem (2.7), which has the expression (2.8). Then: *(i)* for all if and only if . *(ii)* for all and . *(iii)*For and is a continuous function on . *

In the sequel, we present the fixed point theorem which we will be use later. This result appears in [32].

Theorem 2.8 (see [32]). *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
**
where is a continuous and nondecreasing function, is positive on and .**Assume that either is continuous or is such that
**Besides, if
**
then has a unique fixed point. *

*Remark 2.9. *Notice that the condition is superfluous in Theorem 2 of [32].

*Remark 2.10. *If we look at the proof of Theorem 2.2 in [32] we notice that the condition about the continuity of is redundant.

In fact, from the authors generate the sequence and if we put it is proved that

Consequently, is a nonnegative decreasing sequence of real numbers and hence posseses a limit .

Taking limit when in the last inequality, we obtain
and, therefore,

Suppose that , since is a decreasing sequence for all , and, since is a nondecreasing function, we have for all .

As is positive on , for all and, therefore,

This contradicts to (2.14). Consequently, .

The rest of the proof works well and the condition about the continuity of is not used.

Theorem 2.11. *Theorem 2.8 is valid without the assumption is continuous. *

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 [33] it is proved that satisfies condition (2.10) of Theorem 2.8. Moreover, for , as the function , satisfies condition (2.11) of Theorem 2.8.

#### 3. Main Result

Our starting point in this section is to present the class of function which we will use later. By we will denote the class of those functions which are nondecreasing and such that if then the following conditions are satisfied: (a) and is nondecreasing. (b). (c) is positive on .

Examples of such functions are and .

In what follows, we formulate our main result.

Theorem 3.1. *Suppose that , and satisfies the following assumptions: *(i)* is continuous. *(ii)* is nondecreasing respect to the second argument for each . *(iii)* There exist and such that
* *for with and . **Then Problem (1.1) has a unique nonnegative solution. *

* Proof. *Consider the cone

Notice that, as is a closed set of , is a complete metric with the distance given by satisfying conditions (2.10) and (2.11) of Theorem 2.8.

Now, for we define the operator by
where is the Green's function defined by (2.8).

By Lemma 2.7 and assumption (i), it is clear that applies into itself.

In the sequel, we will check that the assumptions of Theorem 2.11 are satisfied.

Firstly, the operator is nondecreasing. In fact, by (ii), for we have

Besides, for and taking into account our assumptions, we can obtain

Since is nondecreasing and , we have
and, from the last inequality it follows

By Lemma 2.7(ii) and since , we can obtain

Since , satifies the properties (a), (b), and (c) mentioned at the beginning of this section, for we have

Finally, taking into account that the zero function, (where with for all ), by Theorem 2.11, Problem (1.1) has a unique nonnegative solution.

Now, we present a sufficient condition for the existence and uniqueness of a positive solution for Problem (1.1) (positive solution means a solution satisfying for ).

Theorem 3.2. *Under assumptions of Theorem 3.1 and adding the following condition
**
we obtain existence and uniqueness of a positive solution for Problem (1.1). *

* Proof. *Consider the nonnegative solution for Problem (1.1) whose existence is guaranteed by Theorem 3.1.

Notice that satisfies

In what follows, we will prove that for .

In fact, in contrary case we can find such that , and therefore,

Since and (see, Lemma 2.7) and, taking into account the nondecreasing character with respect to the second argument of the function , from the last inequality it follows

Thus, . This fact and the nonnegative character of the functions and give us

Since for , we get

By (3.10), since for certain , this means that , and taking into account the continuity of , we can find a set with and (where is the Lebesgue measure) such that for any . This contradicts to (3.15).

Therefore, .

*Remark 3.3. *In Theorem 3.2, the condition for certain seems to be a strong condition in order to obtain a positive solution for Problem (1.1), but when the solution is unique we will see that the condition is very adjusted one. In fact, under the assumption that Problem (1.1) has a unique nonnegative solution we have that

Indeed, if for any then it is easily seen that the zero function is a solution for Problem (1.1) and the uniqueness of solution gives us .

The reverse implication is obvious.

*Remark 3.4. *Notice that assumptions in Theorem 3.1 are invariant by nonnegative and continuous perturbations. More precisely, if for any and satisfies conditions (i), (ii), and (iii) of Theorem 3.1 then , where continuous and satisfies assumptions of Theorem 3.2 and, consequently, the boundary value problem
with , has a unique positive solution.

*Example 3.5. *Now, consider the following boundary value problem

In this case, . Obviously, and is continuous. Since , satisfies condition (ii) of Theorem 3.1.

Moreover, for and we have
where . It is easily seen that belongs to the class . In this case, , consequently , and for , Problem (3.18) satisfies (iii) of Theorem 3.1. Since for , Theorem 3.2 says that Problem (3.18) has a unique positive solution for , where .

#### 4. Some Remarks and Examples

In [31] the authors consider Problem (1.1).

In order to present the main result of [31] we need the following notation. Denote by and the following limits:

The main result of [31] is the following theorem.

Theorem 4.1. *Assume that one of the two following conditions is fulfilled: *(i)*(sublinear case) and , *(ii)*(superlinear case) , and there exist and for which for all . ** Then, Problem (1.1) has at least one positive solution that belongs to
**
where is the identity mapping on . *

Notice that in Example 3.5, and in this case we have

Consequently,

Therefore, for Example 3.5 cannot be treated by Theorem 4.1.

*Example 4.2. *Consider the following boundary value problem
In this case, , .

It is easily proved that satisfies condition (i) and (ii) of Theorem 3.1. In [37], it is proved that if

Using this fact, for and , we have
where . It is easily proved that .

Then, for , the function satisfies condition (iii) of Theorem 3.1. Moreover, since , Theorem 3.2 gives us the existence and uniqueness of a positive solution for Problem (4.5) when .

On the other hand, we have

Consequently, this example corresponds to the sublinear case of Theorem 4.1. Therefore, Theorem 4.1 gives us the existence of at least one positive solution for . The question of uniqueness of solutions is not treated in [31].

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