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