Existence of Positive Solutions for a Functional Fractional Boundary Value Problem

and Applied Analysis 3 2. Preliminaries Firstly, we recall some definitions of fractional calculus, which can be found in 11–14 . Definition 2.1. The Riemann-Liouville fractional derivative of order α > 0 of a continuous function f : 0,∞ → R is given by Df t 1 Γ n − α ( d dt )n ∫ t 0 f s t − s α−n 1 ds, 2.1 where n α 1 and α denotes the integer part of number α, provided that the right side is pointwise defined on 0,∞ . Definition 2.2. The Riemann-Liouville fractional integral of order α of a function f : 0,∞ → R is defined as If t 1 Γ α ∫ t 0 f s t − s 1−α ds, 2.2 provided that the integral exists. The following lemma is crucial in finding an integral representation of the boundary value problem 1.3 – 1.5 . Lemma 2.3 see 4 . Suppose that u ∈ C 0, 1 ∩ L 0, 1 with a fractional derivative of order α > 0. Then IDu t u t c1tα−1 c2tα−2 · · · cntα−n, 2.3 for some ci ∈ R, i 1, 2, . . . , n, where n α 1. From Lemma 2.3, we now give an integral representation of the solution of the linearized problem. Lemma 2.4. If h ∈ C 0, 1 , then the boundary value problem Du t h t 0, 0 < t < 1, m − 1 < ρ ≤ m, 2.4 u 0 u′ 0 · · · u m−2 0 0, u m−2 1 0 2.5 has a unique solution u t ∫1 0 G t, s h s ds, 2.6 4 Abstract and Applied Analysis where G t, s 1 Γ ( ρ ) ⎧ ⎨ ⎩ tρ−1 1 − s ρ−m 1 − t − s ρ−1, s < t, tρ−1 1 − s ρ−m , t ≤ s. 2.7 Proof. Wemay apply Lemma 2.3 to reduce BVP 2.4 , 2.5 to an equivalent integral equation u t c1tρ−1 c2tρ−2 · · · cmtρ−m − ∫ t 0 t − s ρ−1 Γ ( ρ ) h s ds. 2.8 By the boundary condition 2.5 , we easily obtain that c2 c3 · · · cm 0, c1 1 Γ ( ρ ) ∫1 0 1 − s ρ−m h s ds. 2.9 Hence, the unique solution of BVP 2.4 , 2.5 is u t ∫1 0 1 Γ ( ρ ) tρ−1 1 − s ρ−m h s ds − ∫ t 0 t − s ρ−1 Γ ( ρ ) h s ds ∫1 0 G t, s h s ds. 2.10 The proof is complete. Lemma 2.5. G t, s has the following properties. i 0 ≤ G t, s ≤ B s , t, s ∈ 0, 1 , where B s 1 − s ρ−m 1 − 1 − s ρ−1 Γ ( ρ ) ; 2.11 ii G t, s ≥ tρ−1/ m − 2 B s , for 0 ≤ t, s ≤ 1. Abstract and Applied Analysis 5 Proof. It is easy to check that i holds. Next, we prove that ii holds. If t > s, thenand Applied Analysis 5 Proof. It is easy to check that i holds. Next, we prove that ii holds. If t > s, then G t, s B s tρ−1 1 − s ρ−m 1 − t − s ρ−1 1 − s ρ−m 1 − 1 − s ρ−1 tm−2 t − ts ρ−m 1 − t − s ρ−1 t − ts ρ−m 1 1 − s ρ−m 1 − 1 − s ρ−1 ≥ t m−2 t − ts ρ−m 1 − t − s m−2 t − ts ρ−m 1 1 − s ρ−m 1 − 1 − s ρ−1 tρ−m 1 1 − s ρ−m 1 [ tm−2 − t − s m−2 ] 1 − s ρ−m 1 [ 1 − 1 − s m−2 ] tρ−m 1 [ tm−3 tm−4 t − s · · · t − s m−3 ] 1 1 − s · · · 1 − s m−3 ≥ t ρ−m 1tm−3 1 1 − s · · · 1 − s m−3 ≥ t ρ−m 1tm−3 m − 2 tρ−2 m − 2 ≥ tρ−1 m − 2 . 2.12


Introduction
Fractional differential equations can describe many phenomena in various fields of science and engineering such as physics, mechanics, chemistry, control, and engineering.Due to their considerable importance and application, significant progress has been made in there are a great number of excellent works about ordinary and partial differential equations involving fractional derivatives; see, for instance, 1-15 .
As pointed out in 16 , boundary value problems associated with functional differential equations have arisen from problems of physics and variational problems of control theory appeared early in the literature; see 17,18 .Since then many authors see, e.g., 19-23 investigated the existence of solutions for boundary value problems concerning functional differential equations.Recently an increasing interest in studying the existence of solutions for boundary value problems of fractional-order functional differential equations is observed; see for example, 24-26 .For τ > 0, we denote by C τ the Banach space of all continuous functions ψ : −τ, 0 → R endowed with the sup-norm ψ −τ,0 : sup ψ s : s ∈ −τ, 0 .

Abstract and Applied Analysis
If u : −τ, 1 → R, then for any t ∈ 0, 1 , we denote by u t the element of C τ defined by In this paper we investigate a fractional-order functional differential equation of the form and the initial condition where φ is an element of the space To the best of the authors knowledge, no one has studied the existence of positive solutions for problem 1.3 -1.5 .The aim of this paper is to fill the gap in the relevant literatures.The key tool in finding our main results is the following well-known fixed point theorem due to Krasnoselski 27 .Theorem 1.1.Let B be a Banach space and let K be a cone in B. Assume that Ω 1 and Abstract and Applied Analysis 3

Preliminaries
Firstly, we recall some definitions of fractional calculus, which can be found in 11-14 .
Definition 2.1.The Riemann-Liouville fractional derivative of order α > 0 of a continuous function f : 0, ∞ → R is given by where n α 1 and α denotes the integer part of number α, provided that the right side is pointwise defined on 0, ∞ .Definition 2.2.The Riemann-Liouville fractional integral of order α of a function f : 0, ∞ → R is defined as provided that the integral exists.
The following lemma is crucial in finding an integral representation of the boundary value problem 1.3 -1.5 .
From Lemma 2.3, we now give an integral representation of the solution of the linearized problem.

Lemma 2.4. If h ∈ C 0, 1 , then the boundary value problem
has a unique solution where

2.7
Proof.We may apply Lemma 2.3 to reduce BVP 2.4 , 2.5 to an equivalent integral equation By the boundary condition 2.5 , we easily obtain that

2.9
Hence, the unique solution of BVP 2.4 , 2.5 is

2.10
The proof is complete.
Lemma 2.5.G t, s has the following properties.
Proof.It is easy to check that i holds.Next, we prove that ii holds.If t > s, then

2.13
The proof is complete.

Main Result
In the sequel we will denote by C 0 0, 1 the space of all continuous functions x : 0, 1 → R with x 0 0. This is a Banach space when it is furnished with the usual sup-norm

3.1
For each φ ∈ C τ 0 and x ∈ C 0 0, 1 , we define and observe that x t ; φ ∈ C τ .By a solution of the boundary value problem 1.3 -1.5 , we mean a function u ∈ C 0 0, 1 such that D ρ u exists on 0, 1 and u satisfies boundary condition 1.4 , and for a certain φ, the relation holds for all t ∈ 0, 1 .By Lemma 2.4 we know that a function u is a solution of the boundary value problem 1.3 -1.5 if and only if it satisfies We set Define the cone P ⊂ C 0 0, 1 by where 0 < τ < 1.By Lemma 2.4, the boundary value problem 1.3 -1.5 is equivalent to the integral equation

3.7
In this paper, we assume that 0 < τ < 1, φ ∈ C τ 0 , and we make use of the following assumptions.
Lemma 3.2.If 0 < τ < 1 and u ∈ P , then we have where Proof.From the definition of u t s; φ , for t ≥ τ, we have u t s; φ u t s , s ∈ −τ, 0 .

3.12
Thus, we get for u ∈ P that

3.13
We are now in a position to present and prove our main result.
Proof.If u ∈ P with u 0 M, then from 3.8 , 3.14 , and Lemma 2.4 i , we get for any t ∈ 0, 1 that

3.16
Now if we set Without loss of generality, we suppose that m < M. For u ∈ P with u 0 m, we have from Lemma 3.2 and 3.9 that

3.18
Now if we set Hence by the first part of Theorem 1.1, T φ has a fixed point u ∈ P ∩ Ω 1 \ Ω 2 , and accordingly, u is a solution of 1.3 -1.5 .
Having in mind the proof of Theorem 3.3, one can easily conclude the following results.

3.23
From H 4 , we will consider two cases in the following.

3.24
We choose a positive constant

3.25
For u ∈ P with u 0 M 1 , we have

3.29
If u ∈ P with u 0 M 2 , then from 3.8 , 3.14 , and Lemma 2.4 i , we get for any t ∈ 0, 1 that

3.30
Set M 3 max{M 1 , M 2 , 2m}.Then in either case we may put and φ −1/2,0 ≤ 4, which implies that condition 3.14 holds with M 4. Finally, we observe that lim x → ∞ η x /x ∞, and therefore condition 3.20 is satisfied.Then all assumptions of Theorem 3.4 hold.Thus, with Theorem 3.4, problem 1.3 -1.5 has at least one positive solution.