Positive Solutions for a System of Riemann–Liouville Type Fractional-Order Integral Boundary Value Problems

: In this paper, we use the ﬁxed-point index to establish positive solutions for a system of Riemann–Liouville type fractional-order integral boundary value problems. Some appropriate concave and convex functions are used to characterize coupling behaviors of our nonlinearities.

On the other hand, from Lemma 3 (iv), we obtain . Furthermore, we have This completes the proof. where .
Proof.We use Lemma 4 (ii)-(iii).Indeed, we have and This completes the proof.
To obtain our main results, we consider the following auxiliary problem where f satisfies the condition: From Lemma 2, (3) is equivalent to the following Hammerstein type integral equation From this, we obtain an integral equation Lemma 7. (i) If ϕ * is a positive solution for (4), then ϕ * + w is a positive solution for (5); (ii) If ϕ * * is a positive solution for (5) and ϕ * * (t) ≥ w(t), t ∈ [0, 1], then ϕ * * − w is a positive solution for (4).

Proof. Note that
Therefore, we have Thus, ϕ * + w is a positive solution for (5).
On the other hand, we have and thus This implies that ϕ * * − w is a positive solution for (4).This completes the proof.
From Lemma 7, we only study solutions of (5), which are greater than w.For this we define an operator A : P → P as follows: and we turn to study the fixed points of A, which also are required to be greater than w.If there exists ϕ * ∈ P such that Aϕ * = ϕ * and ϕ * (t) ≥ w(t), t ∈ [0, 1], then this, together with Lemmas 2 and 6, implies that ϕ * ∈ P 0 , and 1) 1) and 1) As a result, there exists ϕ * ∈ P such that Aϕ * = ϕ * with ϕ * ≥ Θ Q , and then ϕ * − w is a positive solution for (4).Now, we begin to study (1).Let 1) can be transformed into the following system of fractional-order integral boundary value problems where f , g satisfy the condition: From Lemma 2, ( 6) is equivalent to the following system of Hammerstein type integral equations In what follows, we establish an appropriate operator equation for problem (6).Note that E 2 = E × E is also a Banach space with norm (ϕ, φ) = ϕ + φ , and P 2 = P × P a cone on E 2 .Let and where then (ϕ * − w 1 , φ * − w 2 ) is a positive solution for (6).Let 1) and 1) Then, if ϕ * ≥ Θ Q f and φ * ≥ Θ Q g , we obtain that (8) holds true.
Lemma 8 (see [22]).Let E be a real Banach space and P a cone on E. Suppose that Ω ⊂ E is a bounded open set and that A : Ω ∩ P → P is a continuous compact operator.If there exists ω 0 ∈ P\{0} such that ω − Aω = λω 0 , ∀λ ≥ 0, ω ∈ ∂Ω ∩ P, then i(A, Ω ∩ P, P) = 0, where i denotes the fixed point index on P.

Main Results
Let L k = H(1) + . Now, we list the selection of assumptions for our nonlinearities.
Step 1.We shall prove that where B ρ = {ϕ ∈ E : ϕ < ρ}, ρ > 0. Suppose the contrary.Then, there exist ), and from (H5), we have and Therefore, we have This contradicts (10), and thus (9) holds.Hence, Lemma 9 implies that Step 2. We claim that there exist sufficiently large where i (i = 1, 2) are fixed elements in P 0 .Assume the contrary.Then, there exist This implies that Note that i ∈ P 0 (i = 1, 2), and from Lemma 6 we have From (H4) (ii), we have uniformly for t ∈ [0, 1], and there exist c 1 > 0 and c 2 > 0 such that From these inequalities, we have and Consequently, we have 1) 1) ds.Then, we have From (H4) (iii), there exists c 4 > 0 such that Hence, we obtain and then Multiply by ζ(t) on both sides of the above, integrate over [0, 1] and use Lemma 5 to obtain Solving this inequality, we obtain .
Note that ϕ 2 ∈ P 0 , we have and Multiply by ζ(t) on both sides of (13), integrate over [0, 1] and use Lemma 5 to obtain Note that φ 2 , w 2 ∈ P 0 and By the concavity of ξ 1 , we have and thus, Therefore, we have < +∞.
Note that ξ 1 is a strictly increasing function, and there exits N φ 2 > 0 such that From ( 11) and ( 15), we have Therefore, there exits (ϕ is a positive solution for (1).Thus, (1) has at least one positive solution.This completes the proof.
Step 1.We shall verify where i (i = 3, 4) are given elements in P. Assume the contrary.Suppose there exist This implies that From these inequalities, we have and then On the other hand, from (H6), we have These two inequalities imply that This contradicts (17).Hence, Lemma 8 implies that Step 2. We claim that there exist sufficiently large Then, from (20), we have Note that from (H7) (iii), there exists c 6 > 0 such that Therefore, from (H7) (i) we have and thus Multiply by ζ(t) on both sides of the above, integrate over [0, 1] and use Lemma 5 to obtain Solving this inequality, we obtain .
Note that ϕ 4 ∈ P 0 and we have .