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
.