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.
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 . Moreover, Delbosco and Rodino  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  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  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  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.
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.
The Riemman-Liouville fractional integral of order of a function is given by
provided that the right-hand side is pointwise defined on .
The Caputo fractional derivative of order of a continuous function is given by
where , provided that the right-hand side is pointwise defined on .
The following lemmas appear in .
Let , . Then
is valid when , , .
The following lemmas appear in .
Given and , the unique solution of
is given by
Note that for and (see ).
Let , and is a continuous function with . Suppose that is a continuous function on . Then the function defined by
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 .
Let be a partially ordered set and suppose that there exists a metric in such that is a complete metric space. Assume that satisfies the following condition: if is a non decreasing sequence in such that then . Let be a nondecreasing mapping such that
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 .
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  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
Let , , is continuous and , is a continuous function on . Assume that there exists such that for with and
Then one's problem (1.1) has an unique nonnegative solution.
Consider the cone
Note that, as is a closed set of , is a complete metric space.
Now, for we define the operator by
By Lemma 2.7, . Moreover, taking into account Remark 2.6 and as for by hypothesis, we get
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
As the function is nondecreasing then, for ,
and from last inequality we get
Put . Obviously, is continuous, nondecreasing, positive in , and .
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.
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.
Consider the fractional differential equation (this example is inspired in )
In this case, for . Note that is continuous in and . Moreover, for and we have
because is nondecreasing on , and
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  in the particular case and
Note that our Theorem 3.1 works if the condition (3.1) is changed by, for with and
where is continuous and satisfies
(a) and nondecreasing;
(c) is positive in ;
Examples of such functions are and .
Note that the Green function is strictly increasing in the first variable in the interval . In fact, for fixed we have the following cases
For and as, in this case,
It is trivial that
For and , we have
Now, and then
Hence, taking into account the last inequality and (3.16), we obtain .
For and , we have
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.
Under assumptions of Theorem 3.1, our problem (1.1) has a unique nonnegative and strictly increasing solution.
By Theorem 3.1 we obtain that the problem (1.1) has an unique solution with . Now, we will prove that this solution is a strictly increasing function. Let us take with , then
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
Now, as , then for there exists such that for with we get . Observe that , consequently,
and this contradicts that a.e.
Thus, for with . Finally, as we have that for .
This research was partially supported by "Ministerio de Educación y Ciencia" Project MTM 2007/65706.
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