We set
We now state the main result of this paper.
Theorem 3.1 Let . Assume that one of the following conditions is satisfied:
-
(i)
(Sublinear case) and .
-
(ii)
(Superlinear case) and .
If , then problem (1)-(2) admits at least one positive solution.
Theorem 3.2 Let . Assume that one of the following conditions is satisfied:
-
(i)
(Sublinear case) and .
-
(ii)
(Superlinear case) and .
If , then problem (1)-(2) admits a solution which is positive on an interval .
Proof of Theorem 3.1 Let be the Banach space of all continuous real-valued functions on endowed with the usual supremum norm .
We define the operator as
where is defined by (4).
It is clear from Lemma 2.4 that the fixed points of the operator T coincide with the solutions of problem (1)-(2).
We now define the cone
where λ is given by (6).
First, we show that .
It follows from the continuity and the non-negativity of the functions G and f on their domains of definition that if , then and for all .
For a fixed and for all , the fact that satisfies property (A) leads to the following inequalities:
Hence, .
We now show that is completely continuous.
In view of the continuity of the functions G and f, the operator is continuous.
Let be bounded, that is, there exists a positive constant such that for all . Define
Then for all , we have
for all . That is, the set is bounded.
For each and such that , we have
Clearly, the right-hand side of the above inequalities tends to 0 as and therefore the set is equicontinuous. It follows from the Arzela-Ascoli theorem that the operator is completely continuous.
We now consider the two cases.
-
(i)
Sublinear case ( and ).
Since , there exists such that for all , where satisfies
(7)
We take such that , then we have the following inequalities:
Let . Hence, we have , .
Since is a continuous function on , we can define the function:
It is clear that is non-decreasing on and since , we have (see [19])
Therefore, there exists such that for all , where satisfies
(8)
Define and let such that . Then
Hence, we have , .
Thus, by the first part of the Guo-Krasnosel’skii fixed point theorem, we conclude that (1)-(2) has at least one positive solution.
-
(ii)
Superlinear case ( and ).
Let be given as in (8).
Since , there exists a constant such that for . Take such that . Then we have
If we let , we see that for .
Now, since , there exists such that for all , where is as in (7).
Define , where . Then and imply that
and so we obtain
This shows that for . We conclude by the second part of the Guo-Krasnosel’skii fixed point theorem that (1)-(2) has at least one positive solution . □
Remark 3.1
To prove Theorem 3.2, we use the cone
where b and λ are defined in Lemma 2.6 for the case where , and in Lemma 2.7 for the case where . We skip the rest of the proof as it is similar to the proof of Theorem 3.1.
Example 3.1
Consider the fractional boundary value problem:
(9)
which is problem (1)-(2) with , , and .
First, we note that is not a solution of (9).
Clearly, and , and we also have .
We take
and consider the cone .
By the first part of Theorem 3.1, we conclude that the boundary value problem (9) has a positive solution in the cone P.