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.
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
(11)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.
The Riemman-Liouville fractional integral of order
of a function
is given by
(21)provided that the right-hand side is pointwise defined on
.
Definition 2.2.
The Caputo fractional derivative of order
of a continuous function
is given by
(22)where
, provided that the right-hand side is pointwise defined on
.
The following lemmas appear in [14].
Lemma 2.3.
Let
,
. Then
(23)where
, 
Lemma 2.4.
The relation
(24)is valid when
,
,
.
The following lemmas appear in [6].
Lemma 2.5.
Given
and
, the unique solution of
(25)is given by
(26)where
(27)Remark 2.6.
Note that
for
and
(see [6]).
Lemma 2.7.
Let
,
and
is a continuous function with
. Suppose that
is a continuous function on
. Then the function defined by
(28)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.
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
(29)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:
(210)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
(211)In [10] it is proved that
with the classic metric given by
(212)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.
Let
,
,
is continuous and
,
is a continuous function on
. Assume that there exists
such that for
with
and 
(31)Then one's problem (1.1) has an unique nonnegative solution.
Proof.
Consider the cone
(32)Note that, as
is a closed set of
,
is a complete metric space.
Now, for
we define the operator
by
(33)By Lemma 2.7,
. Moreover, taking into account Remark 2.6 and as
for
by hypothesis, we get
(34)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 
(35)Besides, for 
(36)As the function
is nondecreasing then, for
,
(37)and from last inequality we get
(38)Put
. Obviously,
is continuous, nondecreasing, positive in
,
and
.
Thus, for 
(39)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.
Consider the fractional differential equation (this example is inspired in [6])
(310)In this case,
for
. Note that
is continuous in
and
. Moreover, for
and
we have
(311)because
is nondecreasing on
, and
(312)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.
Note that our Theorem 3.1 works if the condition (3.1) is changed by, for
with
and 
(313)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.
For
and
as, in this case,
(314)It is trivial that
(315)Case 2.
For
and
, we have
(316)Now,
and
then
(317)Hence, taking into account the last inequality and (3.16), we obtain
.
Case 3.
For
and
, we have
(318)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.
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
(319)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
(320)Now, as
, then for
there exists
such that for
with
we get
. Observe that
, consequently,
(321)and this contradicts that
a.e.
Thus,
for
with
. Finally, as
we have that
for
.
Acknowledgment
This research was partially supported by "Ministerio de Educación y Ciencia" Project MTM 2007/65706.
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