Positive solutions for boundary value problem of nonlinear fractional differential equation

In this paper, we investigate the existence of three positiv e solutions for the nonlinear fractional boundary value problem D �+ u(t) + a(t) f (t, u(t), u 00 (t)) = 0, 0 < t < 1, 3 < � � 4, u(0) = u 0 (0) = u 00 (0) = u 00 (1) = 0, where D �+ is the standard Riemann-Liouville fractional derivative. The method involves applications of a new fixed-point theorem due to Bai and Ge. The interesting point l ies in the fact that the nonlinear term is allowed to depend on the second order derivative u 00 .


Introduction
Many papers and books on fractional calculus and fractional differential equations have appeared recently, see for example [1][2][3][7][8][9][10][11][12].Very recently, El-Shahed [5] used the Krasnoselskii's fixedpoint theorem on cone expansion and compression to show the existence and non-existence of positive solutions of nonlinear fractional boundary value problem : where D α 0+ is the standard Riemann-Liouville fractional derivative.Kaufmann and Mboumi [6] studied the existence and multiplicity of positive solutions of nonlinear fractional boundary value problem : Motivated by the above works, in this paper we study the existence of three positive solutions for the following nonlinear fractional boundary value problem : ) by using a new fixed-point theorem due to Bai and Ge [4].Here, the interesting point lies in the fact that the nonlinear term f is allowed to depend on the second order derivative u .To the best of the authors knowledge, no one has studied the existence of positive solutions for nonlinear fractional boundary value problems (1.1)-(1.2).
Throughout this paper, we assume that the following conditions hold.
The rest of this paper is organized as follows: In section 2, we present some preliminaries and lemmas.Section 3 is devoted to prove the existence of three positive solutions for BVP (1.1) and (1.2).

Preliminaries
For the convenience of the reader, we present some definitions from the cone theory on ordered Banach spaces.
Definition 2.1.The map ψ is said to be a nonnegative continuous concave functional on a cone P of a real Banach space E provided that ψ : P → [0, ∞) is continuous and Similarly, we say the map φ is a nonnegative continuous convex functional on a cone P of a real Banach space E provided that φ : P → [0, ∞) is continuous and Definition 2.2.Let r > a > 0, L > 0 be given and ψ be a nonnegative continuous concave functional and γ, β be nonnegative continuous convex functionals on the cone P. Define convex sets: EJQTDE, 2008 No. 24, p. 2 Suppose that the nonnegative continuous convex functionals γ, β on the cone P satisfy (A 1 ) there exists M > 0 such that x ≤ M max{γ(x), β(x)}, for x ∈ P; (A 2 ) P(γ, r; β, L) ∅, for any r > 0, L > 0.
Lemma 2.1.[4] Let P be a cone in a real Banach space E and constants.Assume that γ, β are nonnegative continuous convex functionals on P such that (A 1 ) and (A 2 ) are satisfied.ψ is a nonnegative continuous concave functional on P such that ψ(x) ≤ γ(x) for all x ∈ P(γ, r 2 ; β, L 2 ) and let T : P(γ, r 2 ; β, L 2 ) → P(γ, r 2 ; β, L 2 ) be a completely continuous operator.
The above fixed-point theorem is fundamental in the proof of our main result.
Next, we give some definitions from the fractional calculus.
The following lemma is crucial in finding an integral representation of the boundary value problem (1), (2).
Lemma 2.2.[3] Suppose that u ∈ C(0, 1) ∩ L(0, 1) with a fractional derivative of order α > 0. Then From Lemmas 2.2, we now give an integral representation of the solution of the linearized problem.
then the boundary value problem ) has a unique solution where Proof.From Lemma 2.2, we get By (2.2), there are c 2 = c 3 = c 4 = 0, and Hence, the unique solution of BVP (2.1), (2.2) is The proof is complete.
Lemma 2.4.G(t, s) has the following properties. where Proof.It is easy to check that (i) holds.Next, we prove (ii) holds.If t ≥ s, then The proof is complete.

Main results
Then we have the following lemma.
By Lemma 3.1, X is a Banach space when it is endowed with the norm u = u 0 .
It is easy to know that We define the operator T by Thus, T : X → X.By Lemma 2. Define the cone P ⊂ X by where 0 < ω < 1 as in (H 2 ).
We are now in a position to present and prove our main result.
Theorem 3.2.Assume that (H 1 ) and (H 2 ) hold.Suppose there exist constants then BVP (1.1)-(1.2) has at least three positive solutions u 1 , u 2 , and u 3 such that Proof.By (H 1 ), (H 2 ), Lemma 2.4 and (3.3), for u ∈ P, we have T u(t) ≥ 0, ∀t ∈ [0, 1], and Thus, T (P) ⊂ P.Moreover, it is easy to check by the Arzela-Ascoli theorem that the operator T is completely continuous.We now show that all the conditions of Lemma 2.1 are satisfied.
Finally, we give an example to illustrate the effectiveness of our result.