The Positive Properties of Green’s Function for Fractional Differential Equations and Its Applications

and Applied Analysis 3 for some C 1 , C 2 , . . . , C n ∈ R. Consequently, the general solution of (10) is u (t) = − 1 Γ (α) ∫ t 0 (t − s) α−1 h (s) ds + C 1 t α−1 + C 2 t α−2 + ⋅ ⋅ ⋅ + C n t α−n . (14) By u(0) = u󸀠(0) = ⋅ ⋅ ⋅ = u(n−2)(0) = 0, this is C 2 = C 3 = ⋅ ⋅ ⋅ = C n = 0. On the other hand, u(1) = βu(η) combining with u (1) = − 1 Γ (α) ∫ 1 0 (1 − s) α−1 h (s) ds + C 1 , u (η) = − 1 Γ (α) ∫ η 0 (η − s) α−1 h (s) ds + C 1 η α−1 (15)


Introduction
Fractional differential equations have been of great interest recently.This is due to the intensive development of the theory of fractional calculus itself as well as its applications.Apart from diverse areas of mathematics, fractional differential equations arise in rheology, dynamical processes in selfsimilar and porous structures, fluid flows, electrical networks, viscoelasticity, chemical physics, and many other branches of science.For details, see [1][2][3][4][5][6][7][8][9][10].
It should be noted that most of the papers and books on fractional calculus are devoted to the solvability of linear initial fractional differential equations on terms of special functions.Recently, there are some papers dealing with the existence and multiplicity of solution to the nonlinear fractional differential equations boundary value problems, see [11][12][13][14][15][16][17].
In this paper, we give some existence of positive solutions for singular boundary value problems by means of Schauder fixed-point theorem for the case:  * = 0,  * ≥ 0,  * ≤ 0,  * < 0 <  * .
The paper is organized as follows.In Section 2, we state some known results and give a property of Green's function.In Section 3, using Schauder fixed-point theorem, the existence of positive solutions to singular problems are obtained.

Background Materials
For the convenience of the reader, we present here the necessary definitions from fractional calculus theory.
Let us define Then,
Proof.Let  = ([0, 1], ‖ ⋅ ‖), and Ω is a closed convex set defined as here  = [0, 1] is the Banach space of continuous functions defined on [0, 1] with the norm and  >  > 0 are positive constants to be given below.Now, we define an operator  : Ω →  by Then, (28) is equivalent to the fixed-point problem Let  be the positive constant satisfying (H 3 ) and Then, we have  >  > 0. Now, we prove (Ω) ⊂ Ω.
In fact, for each  ∈ Ω and for all  ∈ (0, 1), by (H 1 ) and (H 3 ) On the other hand, by conditions (H 2 ) and (H 3 ), we have In conclusion, (Ω) ⊂ Ω.Finally, it is standard that  : Ω → Ω is a continuous and completely continuous operator.By a direct application of Schauder's fixed-point theorem, (28) has at least one positive solution () ∈ [0, 1], the proof is finished.

(H *
3 ) There exists a positive constant  > 0 such that and here Then, (28) has at least one positive solution.
From now on, let us define Example 9. Suppose that the nonlinearity in (28) is where  ≻ 0, 0 <  < 1 and If  * = 0, then (28) has at least one positive solution.
Proof.We will apply Corollary 8. To this end, we take then (H 1 ) and (H 2 ) are satisfied since () < +∞, and the existence condition (H * 3 ) becomes for some  > 0. Since 0 <  < 1, we can choose  > 0 large enough such that (46) is satisfied, and the proof is finished.
Proof.We will apply Corollary 11.To this end, we take (), ℎ(), and () as the same in the proof of Example 10, then (H 2 ) is satisfied, and the existence condition (H 4 ) becomes for some  > 0. So, (28) has at least one positive solution for The function () possesses a maximum at then  2 = ( 0 ).We have the desired results (i) and (ii).
(H 5 ) There exist two positive constants  >  > 0 such that here Then, (28) has at least one positive solution.
Proof.We will apply Corollary 14.Take (), () as the same in the proof of Example 9.Then, (H 2 ) is satisfied, and the existence condition (H 5 ) is satisfied if we take  >  > 0 with and () < +∞.If we fix  =  2 /  , then the first inequality holds if  satisfies or equivalently The function () possesses a minimum at Taking  =  0 , then the first inequality in (63) holds if  * ≥ ( 0 ), which is just condition (62).The second inequality holds directly from the choice of , so it remains to prove that  =  2 /  >  0 .This is easily verified through elementary computations.
We have the desired results.
Proof.We will apply Theorem 7. To this end, we take (), ℎ(), and () as the same in the proof of Example Let  = 1/( −  * ), then Then, we have Let   () = 0, then we have In fact, by (103), we have that is, .
(106) Also we have with ,  ≻ 0, but the notation becomes cumbersome.Here we consider the nonlinearity (47) only for the simplicity.