POSITIVE SOLUTIONS FOR BOUNDARY VALUE PROBLEM OF SINGULAR FRACTIONAL FUNCTIONAL DIFFERENTIAL EQUATION

In this paper, we consider a class of singular fractional order functional differential equations with delay. By means of a fixed point theorem on cones, some sufficient conditions for the existence of at least one or two positive solutions for the boundary value problem are established. We also give examples to illustrate the applicability of our main results. AMS Subject Classification: 26A33, 34B18


Introduction
Fractional differential equations have been of great interest recently.It is caused both by the intensive development of the theory of fractional calculus itself and by the applications (see [1], [2], [3], [4], [5]).
In [6], [7], [8], [9], [10], they studied the existence and multiplicity of positive solutions to fractional differential equations, and obtained some results.However, to the best of the author knowledge, there are few articles studying the functional differential equations of fractional order.In [11], the authors studied the existence of positive solution for boundary value problem of fourthorder FDE.Motivated by the work above, this paper investigates the existence of positive solutions for singular fractional order functional differential equation with boundary conditions where D α 0 + is the standard Riemann-Liouville fractional derivative of order α; φ(t) 2 is a constant.
(A 2 ) r(t) is a nonnegative measurable function defined on [0, 1], and satisfies We would mention that r(t) is allowed to be zero on some subset of E and has singularity at points t = 0 and t = 1.Suppose that u(t) is a solution of BVP(1), then Suppose that x(t) is the solution of BVP(1) with f ≡ 0 , then Let x(t) = u(t) − x(t), then we have from ( 3)and(4) that We define For 0 < t < 1, x ∈ K, we obtain from ( 5) and Lemma 3 that (T x)(t) > 0, and Then we have the following lemma.
Lemma 4. The operator T : K → K is completely continuous.
Proof.We can obtain the continuity of T from the continuity of f .In face, suppose x (n) , x ∈ K and x (n) − x → 0 as n → ∞, then we get Thus, for t ∈ [−τ, 1] we have from (5) and Lemma 3 that which implies the boundedness of T (Ω).Furthermore, we have for 0 By means of the Arzela-Ascoli theorem, T : K → K is completely continuous.Lemma 5. ( [12]) Let X be a Banach space, and let K ⊂ X be a cone in X. Assume that Ω 1 , Ω 2 are open subsets of X with 0 ∈ Ω 1 ⊂ Ω 1 ⊂ Ω 2 , and let T : K → K be a completely continuous operator such that, either

Main Results
For convenience, we introduce the following notations.
In the next, we let 16 is a constant.Theorem 6. Assume that one of the following conditions is satisfied:
which implies that x s ∈ C * for s ∈ E and Note that when s ∈ E we have xs = 0. Thus, we obtain which leads to T x ≥ x , ∀x ∈ K ∂Ω ρ 2 .According to the first part of Lemma 5, it follows that T has a fixed point . Furthermore, by a similar argument as (6), we have x s ∈ C * , x s C ≥ γ 0 x = γ 0 ρ 1 , s ∈ E. Thus, we have an analogous result to (7) On the other hand, since Thus, we have By the second part of Lemma 5, it follows that T has a fixed point t) ∈ K and x(t) ≥ 0 , we conclude that u(t) is a positive solution of BVP(1).This completes the proof.Theorem 7. If the following conditions are satisfied: Then BVP (1) Hence, we obtain an analogous inequality: Hence, we obtain an analogous inequality: which implies that T x ≤ x , ∀x ∈ K ∂Ω p .According to Lemma 5, it follows that T has two fixed points This completes the proof.
Similarly, we have the following result.