We investigate the existence and uniqueness of positive solutions for the following singular fractional three-point boundary value problem

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., [

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

In [

In [

In [

Very recently, in [

Motivated by [

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

The Riemann-Liouville fractional integral of order

The Riemann-Liouville fractional derivative of order

The following two lemmas can be found in [

Assume that

Then the fractional differential equation

Assume that

By using Lemma

Let

Then the boundary value problem

In [

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

By

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

Let

In our considerations, we will work in the Banach space

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

Our starting point of this section is the following lemma.

Suppose that

We divide the proof into three cases.

Since

Then, we get

This proves the continuity of

Without loss of generality, we can take

In fact,

In fact, as

In fact, as

Now, following the same lines that in the proof of Case 1, we can demonstrate the continuity of

This finishes the proof.

Suppose that

If

Since

Taking into account that

Notice that the function

In fact, for

In the case,

Suppose that

By definition of

Suppose that

By definition of

By commodity, we denote by

Let

Then Problem (

Consider the cone:

It is easily checked that

Now, for

Moreover, in view of the nonnegative character of

In what follows, we check that assumptions in Theorem

Firstly, we will prove that

In fact, by (

On the other hand, for

Now, we will prove that

In contrary case, we can find

Taking into account that the nonnegative solution

This and (

Therefore,

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.

A function

Suppose that

Recall that a function

Let

We take

Since

In what follows, we will prove that the function

The function

(a) Since

Moreover, the function

Since

In order to prove that

In fact, in contrary case

Now, we suppose that

Then, we can find

Since

As

Therefore,

This contradicts the fact that

Therefore,

This proves that

Therefore,

(b) Since

By Remark

Now, we present an example which illustrates our result.

Consider the following singular fractional boundary value problem

Moreover, in this case

Notice that

Now, we check that

It is clear that

Moreover, by Lemma

Finally, Theorem

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