as at least one positive solution for and sufficiently small.

Proof. We consider the system of ordinary fractional differential equations

(6)

with the coupled integral boundary conditions

(7)

with and.

The above problem (6)-(7) has the solution

(8)

where is defined in (J1). By assumption (J1) we obtain and for all.

We define the functions and, by

where is a solution of (S)-(BC). Then (S)-(BC) can be equivalently written as

(9)

with the boundary conditions

(10)

Using the Green’s functions, from Lemma 1, a pair is a solution of problem (9)-(10) if and only if is a solution for the nonlinear integral equations

(11)

where and, are given in (8).

We consider the Banach space with the supremum norm, the space with the norm, and we define the set

We also define the operators and by

for all, and.

For sufficiently small and, by (J3), we deduce

Then, by using Lemma 3, we obtain, for all and. By Lemma 4, for all, we have

and

Therefore.

Using standard arguments, we deduce that S is completely continuous. By Theorem 1, we conclude that S has a fixed point, which represents a solution for problem (9)-(10). This shows that our problem (S)-(BC) has a positive solution with for sufficiently small and.

In what follows, we present sufficient conditions for the nonexistence of positive solutions of (S)-(BC).

Theorem 3. Assume that assumptions (J1), (J2) and (J4) hold. Then problem (S)-(BC) has no positive solution for and sufficiently large.

Proof. We suppose that is a positive solution of (S)-(BC). Then with, is a solution for problem (9)-(10), where is the solution of problem (6)-(7) (given by (8)). By (J2) there exists such that, and then, , ,. Now by using Lemma 3, we have, for all, and by Lemma 5 we obtain and.

Using now (8), we deduce that and. Therefore, we obtain and.

We now consider. By using (J4), for R defined above, we conclude that there exists such that, for all. We consider and sufficiently large such that and. By (J2), (9), (10) and the above inequalities, we deduce that and.

Now by using Lemma 4 and the above considerations, we have

Therefore, we obtain, which is a contradiction, because. Then, for and sufficiently large, our problem (S)-(BC) has no positive solution.

4. An Example

We consider, for all, , , , for all, then and. We also consider the functions, , , for all, with. We have.

Therefore, we consider the system of fractional differential equations

(S0)

with the boundary conditions

(BC0)

Then we obtain

We also deduce

, for all. For the functions, , we obtain

Then we deduce that assumptions (J1), (J2) and (J4) are satisfied. In addition, by using the above functions, , we obtain, ,

, , and then. We choose and if we select, then we conclude that, for all. For example, if and, then the above conditions for f and g are satisfied. So,

assumption (J3) is also satisfied. By Theorems 2 and 3 we deduce that problem (S0)-(BC0) has at least one positive solution for sufficiently small and, and no positive solution for sufficiently large and.

Acknowledgements

The work of R. Luca and A. Tudorache was supported by a grant of the Romanian National Authority for Scientific Research, CNCS-UEFISCDI, project number PN-II-ID-PCE-2011-3-0557.

Cite this paper

JohnnyHenderson,RodicaLuca,AlexandruTudorache, (2015) Positive Solutions for Systems of Coupled Fractional Boundary Value Problems. Open Journal of Applied Sciences,05,600-608. doi: 10.4236/ojapps.2015.510059

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