Theorem 3.1.

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.

Proof.

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

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

Besides, for

As the function is nondecreasing then, for ,

and from last inequality we get

Put . Obviously, is continuous, nondecreasing, positive in , and .

Thus, for

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])

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 [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

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,

It is trivial that

Case 2.

For and , we have

Now, and then

Hence, taking into account the last inequality and (3.16), we obtain .

Case 3.

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.

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

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 .