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.

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,

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
.