Electronic Journal of Qualitative Theory of Differential Equations 2010, No. 30, 1-14; http://www.math.u-szeged.hu/ejqtde/

Existence of positive solutions for boundary value problems

of fractional functional differential equations

Chuanzhi Bai^{∗}

Department of Mathematics, Huaiyin Normal University, Huaian, Jiangsu 223300, P R China

Abstract

This paper deals with the existence of positive solutions for a boundary value problem involving a nonlinear functional differential equation of fractional order α given by D^{α}u(t) + f(t, u_{t}) = 0, t ∈ (0, 1), 2 < α ≤ 3, u′(0) = 0, u′(1) = bu′(η), u_{0} = ϕ. Our results are based on the nonlinear alternative of Leray-Schauder type and Krasnosel’skii fixed point theorem. Keywords: Fractional derivative; Boundary value problem; Functional differential equation; Positive solution; Fixed point theorem

1. Introduction Recently, an increasing interest in studying the existence of solutions for boundary value problems

of fractional order functional differential equations has been observed [1-3]. Fractional differential equations describe many phenomena in various fields of science and engineering such as physics, mechanics, chemistry, control, engineering, etc. In view of the importance and applications of fractional differential equations, a significant progress has been made in this direction by exploring several aspects of these equations, see, for instance, [4-21].

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]}. If u : [−τ, 1] → R, then for any t ∈ [0, 1], we denote by u_{t} the element of C_{τ} defined by

u_{t}(θ) = u(t + θ), for θ ∈ [−τ, 0]. In this paper we investigate a fractional order functional differential equation of the form

D^{α}u(t) + f(t, u_{t}) = 0, t ∈ (0, 1), 2 < α ≤ 3, (1.1) where D^{α} is the standard Riemann-Liouville fractional order derivative, f(t, u_{t}) : [0, 1] × C_{τ} → R is a

EJQTDE, 2010 No. 30, p. 1 continuous function, associated with the boundary condition

∗E-mail address: czbai8@sohu.com

u′(0) = 0, u′(1) = bu′(η), (1.2) and the initial condition

u_{0} = ϕ, (1.3) where 0 < η < 1, 1 < b < \frac{1}{η^{α−2}} , and ϕ is an element of the space

C_{τ}^{+}(0) := {ψ ∈ C_{τ} : ψ(s) ≥ 0, s ∈ [−τ, 0], ψ(0) = 0}. To the best of the authors knowledge, no one has studied the existence of positive solutions for

problem (1.1)-(1.3). The aim of this paper is to fill the gap in the relevant literatures. The key tools in finding our main results are the nonlinear alternative of Leray-Schauder type and Krasnosel’skii fixed point theorem.

2. Preliminaries First of all, we recall some definitions of fractional calculus [16-19].

Definition 2.1. The Riemann-Liouville fractional derivative of order α > 0 of a continuous function f : (0, ∞) → R is given by

D^{α}f(t) = \frac{1}{Γ(n − α)} (\frac{d}{dt})^{n} ∫_{0}^{t} \frac{f(s)}{(t − s)^{α−n+1}} ds,

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

I^{α}f(t) = \frac{1}{Γ(α)} ∫_{0}^{t} \frac{f(s)}{(t − s)^{1−α}} ds,

provided that the integral exists.

The following lemma is crucial in finding an integral representation of the boundary value problem (1.1)-(1.3). Lemma 2.1 [11]. Suppose that u ∈ C(0, 1) ∩ L(0, 1) with a fractional derivative of order α > 0. Then

I^{α}D^{α}u(t) = u(t) + c_{1}t^{α−1} + c_{2}t^{α−2} + ⋯ + c_{n}t^{α−n}. for some c_{i} ∈ ℝ, i = 1, 2, ..., n.

EJQTDE, 2010 No. 30, p. 2 Lemma 2.2. Let 0 < η < 1 and b ≠ \frac{1}{η^{α−2}} . If g ∈ C[0, 1], then the boundary value problem

D^{α}u(t) + g(t) = 0, 0 < t < 1, (2.1) u(0) = u′(0) = 0, u′(1) = bu′(η), (2.2)

has a unique solution

u(t) = ∫_{0}^{1} G(t, s)g(s)ds + \frac{bt^{α−1}}{(α − 1)(1 − bη^{α−2})} ∫_{0}^{1} H(η, s)g(s)ds, (2.3)

where

G(t, s) = \frac{1}{Γ(α)} \{ \begin{matrix}t^{α−1}(1 − s)^{α−2} − (t − s)^{α−1}, & s ≤ t, \\ t^{α−1}(1 − s)^{α−2}, & t ≤ s, \end{matrix} (2.4)

and

H(t, s) := \frac{∂G(t, s)}{∂t} = \frac{α − 1}{Γ(α)} \{ \begin{matrix}t^{α−2}(1 − s)^{α−2} − (t − s)^{α−2}, & s ≤ t, \\ t^{α−2}(1 − s)^{α−2}, & t ≤ s. \end{matrix} (2.5)

Proof. By Lemma 2.1, the solution of (2.1) can be written as

u(t) = c_{1}t^{α−1} + c_{2}t^{α−2} + c_{3}t^{α−3} − ∫_{0}^{t} \frac{(t − s)^{α−1}}{Γ(α)} g(s)ds.

Using the boundary conditions (2.2), we find that c_{2} = c_{3} = 0, and

c_{1} = \frac{1}{(1 − bη^{α−2})Γ(α)} [∫_{0}^{1}(1 − s)^{α−2}g(s)ds − ∫_{0}^{η} b(η − s)^{α−2}g(s)ds] .

Hence, the unique solution of BVP (2.1), (2.2) is

u(t) = − ∫_{0}^{t} \frac{(t − s)^{α−1}}{Γ(α)} g(s)ds + (\frac{t^{α−1}}{Γ(α)} + \frac{bη^{α−2}t^{α−1}}{(1 − bη^{α−2})Γ(α)}) ∫_{0}^{1}(1 − s)^{α−2}g(s)ds

−\frac{bt^{α−1}}{(1 − bη^{α−2})Γ(α)} ∫_{0}^{η}(η − s)^{α−2}g(s)ds

= \frac{1}{Γ(α)} ∫_{0}^{t}(t^{α−1}(1 − s)^{α−2} − (t − s)^{α−1})g(s)ds + \frac{1}{Γ(α)} ∫_{t}^{1} t^{α−1}(1 − s)^{α−2}g(s)ds

+\frac{bη^{α−2}t^{α−1}}{(1 − bη^{α−2})Γ(α)} ∫_{0}^{1}(1 − s)^{α−2}g(s)ds − \frac{bt^{α−1}}{(1 − bη^{α−2})Γ(α)} ∫_{0}^{η}(η − s)^{α−2}g(s)ds

= ∫_{0}^{1} G(t, s)h(s)ds + \frac{bt^{α−1}}{(α − 1)(1 − bη^{α−2})} ∫_{0}^{1} H(η, s)g(s)ds.

The proof is complete.

To establish the existence of solutions for (1.1)-(1.3), we need the following known results. Theorem 2.3 (Nonlinear alternative of Leray-Schauder [22]). Let E be a Banach space with C ⊂ E

EJQTDE, 2010 No. 30, p. 3 closed and convex. Assume that U is a relatively open subset of C with 0 ∈ U and T : \overline{U} → C is completely continuous. Then either

(i) T has a fixed point in \overline{U}, or

(ii) there exists u ∈ ∂U and γ ∈ (0, 1) with u = γT u. Theorem 2.4 (Krasnosel’skii [23]). Let E be a Banach space and let K be a cone in E. Assume that Ω_{1} and Ω_{2} are open subsets of E, with 0 ∈ Ω_{1} ⊂ \overline{Ω}_{1} ⊂ Ω_{2}, and let A : K ∩ (\overline{Ω}_{2} \ Ω_{1}) → K be a completely continuous operator such that either

∥Au∥ ≤ ∥u∥, u ∈ K ∩ ∂Ω_{1}, and ∥Au∥ ≥ ∥u∥, u ∈ K ∩ ∂Ω_{2}, or

∥Au∥ ≥ ∥u∥, u ∈ K ∩ ∂Ω_{1}, and ∥Au∥ ≤ ∥u∥, u ∈ K ∩ ∂Ω_{2},

Then A has a fixed point in K ∩ (\overline{Ω}_{2} \ Ω_{1}). Lemma 2.5. G(t, s) has the following properties.

(i) 0 ≤ G(t, s) ≤ h(s), t, s ∈ [0, 1], where

h(s) = \frac{(1 − s)^{α−2} − (1 − s)^{α−1}}{Γ(α)} ;

(ii) G(t, s) ≥ t^{α−1}h(s), for 0 ≤ t, s ≤ 1. Proof. It is easy to check that (i) holds. Next, we prove (ii) holds. If t ≥ s, then

\frac{G(t, s)}{h(s)} = \frac{t^{α−1}(1 − s)^{α−2} − (t − s)^{α−1}}{(1 − s)^{α−2} − (1 − s)^{α−1}}

= \frac{t(t − ts)^{α−2} − (t − s)(t − s)^{α−2}}{(1 − s)^{α−2} − (1 − s)^{α−1}}

≥ \frac{t(t − ts)^{α−2} − (t − s)(t − ts)^{α−2}}{(1 − s)^{α−2} − (1 − s)^{α−1}}

= \frac{(t − ts)^{α−2}}{(1 − s)^{α−2}} = t^{α−2} ≥ t^{α−1}.

If t ≤ s, then

\frac{G(t, s)}{h(s)} = \frac{t^{α−1}(1 − s)^{α−2}}{(1 − s)^{α−2} − (1 − s)^{α−1}} = \frac{t^{α−1}}{s} ≥ t^{α−1}.

The proof is complete.

Lemma 2.6. Let g ∈ C^{+}[0, 1] := \{u ∈ C[0, 1], u(t) ≥ 0, t ∈ [0, 1]\}. If 1 < b < \frac{1}{η^{α−2}} , then the unique solution u(t) of BVP (2.1), (2.2) is positive, and satisfying

min_{t∈[τ,1]} u(t) ≥ τ^{α−1}∥u∥,

where 0 < τ < 1. Proof. Let g ∈ C^{+}[0, 1], we have from (2.3)-(2.5) that u(t) ≥ 0. By Lemma 2.3, we get

∥u∥ = max_{0≤t≤1} |u(t)| = ∫_{0}^{1} G(t, s)g(s)ds + \frac{bt^{α−1}}{(α − 1)(1 − bη^{α−2})} ∫_{0}^{1} H(η, s)g(s)ds

≤ ∫_{0}^{1} h(s)g(s)ds + \frac{bt^{α−1}}{(α − 1)(1 − bη^{α−2})} ∫_{0}^{1} H(η, s)g(s)ds. (2.6)

On the other hand, from Lemma 2.3 and (2.6), we obtain for each t ∈ [0, 1] that

u(t) = ∫_{0}^{1} G(t, s)g(s)ds + \frac{bt^{α−1}}{(α − 1)(1 − bη^{α−2})} ∫_{0}^{1} H(η, s)g(s)ds

≥ t^{α−1} ∫_{0}^{1} h(s)g(s)ds + \frac{bt^{α−1}}{(α − 1)(1 − bη^{α−2})} ∫_{0}^{1} H(η, s)g(s)ds

≥ t^{α−1} [∫_{0}^{1} h(s)g(s)ds + \frac{bt^{α−1}}{(α − 1)(1 − bη^{α−2})} ∫_{0}^{1} H(η, s)g(s)ds]

≥ t^{α−1}∥u∥.

Thus,

min_{t∈[τ,1]} u(t) ≥ τ^{α−1}∥u∥, 0 < τ < 1.

3. Main results In the sequel we shall 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

∥u∥ = max_{t∈[0,1]} |u(t)|. For each ϕ ∈ C_{τ}^{+}(0) and x ∈ C_{0}[0, 1] we define

x_{t}(s; ϕ) := \{ \begin{matrix}x(t + s), & t + s ≥ 0, \\ ϕ(t + s), & t + s ≤ 0, s ∈ [−τ, 0], \end{matrix}

and observe that x_{t}(s; ϕ) ∈ C_{τ}.

By a solution of the boundary value problem (1.1)-(1.3) we mean a function u ∈ C_{0}[0, 1] such that D^{α}u exists on [0, 1] and u satisfies boundary condition (1.2) and for a certain ϕ the relation

D^{α}u(t) + f(t, u_{t}(·; ϕ)) = 0

holds for all t ∈ [0, 1].

Since f : [0, 1] × C_{τ} → R is a continuous function, set f(t, u_{t}(·; ϕ)) := g(t) in Lemma 2.2, we have by Lemma 2.2 that a function u is a solution of the boundary value problem (1.1)-(1.3) if and only if it satisfies

u(t) = ∫_{0}^{1} G(t, s)f(s, u_{s}(·; ϕ))ds + \frac{bt^{α−1}}{(α − 1)(1 − bη^{α−2})} ∫_{0}^{1} H(η, s)f(s, u_{s}(·; ϕ))ds, t ∈ [0, 1].

(3.1)

We set

C_{0}^{+}[0, 1] = {u ∈ C_{0}[0, 1] : u(t) ≥ 0, t ∈ [0, 1]}. Define the cone P ⊂ C_{0}[0, 1] by

P = {y ∈ C_{0}^{+}[0, 1] : min_{τ≤t≤1} y(t) ≥ τ^{α−1}∥y∥} ,

where 0 < τ < 1.

For u ∈ P, we define the operator T_{ϕ} as follows:

(T_{ϕ}u)(t) = ∫_{0}^{1} G(t, s)f(s, u_{s}(·; ϕ))ds + \frac{bt^{α−1}}{(α − 1)(1 − bη^{α−2})} ∫_{0}^{1} H(η, s)f(s, u_{s}(·; ϕ))ds, t ∈ [0, 1].

(3.2)

It is easy to know that fixed points of T_{ϕ} are solutions of the BVP (1.1)-(1.3).

In this paper, we assume that 0 < τ < 1, ϕ ∈ C_{τ}^{+}(0), and we make use of the following assumption: (H_{1}) f : [0, 1] × C_{τ}^{+}(0) → [0, +∞) is a continuous function.

Lemma 3.1. Let (H_{1}) holds. Then T_{ϕ} : P → P is completely continuous. Proof. By (H_{1}), we have (T_{ϕ}u)(t) ≥ 0, for u ∈ P and t ∈ [0, 1]. It follows from (3.2) and Lemma 2.5 that

∥T_{ϕ}u∥ = max_{0≤t≤1} |(T_{ϕ}u)(t)| ≤ ∫_{0}^{1} h(s)f(s, u_{s}(·; ϕ))ds + \frac{b}{(α − 1)(1 − bη^{α−2})} ∫_{0}^{1} H(η, s)f(s, u_{s}(·; ϕ))ds.

In view of Lemma 2.5, we have

(T_{ϕ}u)(t) ≥ ∫_{0}^{1} t^{α−1}h(s)f(s, u_{s}(·; ϕ))ds + \frac{bt^{α−1}}{(α − 1)(1 − bη^{α−2})} ∫_{0}^{1} H(η, s)f(s, u_{s}(·; ϕ))ds

≥ t^{α−1} [∫_{0}^{1} h(s)f(s, u_{s}(·; ϕ))ds + \frac{bt^{α−1}}{(α − 1)(1 − bη^{α−2})} ∫_{0}^{1} H(η, s)f(s, u_{s}(·; ϕ))ds]

≥ t^{α−1}∥T_{ϕ}u∥, t ∈ [0, 1].

So,

min_{τ≤t≤1}(T_{ϕ}u)(t) ≥ τ^{α−1}∥T_{ϕ}u∥,

which shows that T_{ϕ}P ⊂ P. Moreover, similar to the proof of Lemma 3.2 in [10], it is easy to check that T_{ϕ} : P → P is completely continuous. Lemma 3.2. If 0 < τ < 1 and u ∈ P, then we have

∥u_{t}(·; ϕ)∥_{[−τ,0]} ≥ τ^{α−1}∥u∥, t ∈ [τ, 1].

Proof. From the definition of u_{t}(s; ϕ), for t ≥ τ, we have

u_{t}(s; ϕ) = u(t + s), s ∈ [−τ, 0]. Thus, we get for u ∈ P that

∥u_{t}(·; ϕ)∥_{[−τ,0]} = max_{s∈[−τ,0]} u(t + s) ≥ u(t) ≥ t^{α−1}∥u∥ ≥ τ^{α−1}∥u∥, t ≥ τ.

We are now in a position to present and prove our main results. Theorem 3.3. Let (H_{1}) holds. Suppose that the following conditions are satisfied:

(H_{2}) there exist a continuous function a : [0, 1] → [0, +∞) and a continuous, nondecreasing function F : [0, +∞) → [0, +∞) such that

f(t, ψ) ≤ a(t)F (∥ψ∥_{[−τ,0]}), (t, ψ) ∈ [0, 1] × C_{τ}^{+}(0), (H_{3}) there exists r > ∥ϕ∥_{[−τ,0]}, with

\frac{r}{F (r)} > ∫_{0}^{1} h(s)a(s)ds + \frac{b}{(α − 1)(1 − bη^{α−2})} ∫_{0}^{1} H(η, s)a(s)ds. (3.3)

then BVP (1.1)-(1.3) has at least one positive solution. Proof. We shall apply Theorem 2.3 (the nonlinear alternative of Leray-Schauder type) to prove that T_{ϕ} has at least one positive solution.

Let U = {u ∈ P : ∥u∥ < r}, where r is as in (H_{3}). Assume that there exist u ∈ P and λ ∈ (0, 1) such that u = λT_{ϕ}u, we claim that ∥u∥ ≠ r. In fact, if ∥u∥ = r, we have

u(t) = λ ∫_{0}^{1} G(t, s)f(s, u_{s}(·; ϕ))ds + \frac{λbt^{α−1}}{(α − 1)(1 − bη^{α−2})} ∫_{0}^{1} H(η, s)f(s, u_{s}(·; ϕ))ds

EJQTDE, 2010 No. 30, p. 7

≤ ∫_{0}^{1} h(s)a(s)F (∥u_{s}(·; ϕ)∥_{[−τ,0]})ds

+\frac{bt^{α−1}}{(α − 1)(1 − bη^{α−2})} ∫_{0}^{1} H(η, s)a(s)F (∥u_{s}(·; ϕ)∥_{[−τ,0]})ds. (3.4)

By the definition of u_{s}(·; ϕ), we easily obtain that

∥u_{s}(·; ϕ)∥_{[−τ,0]} ≤ max{∥u∥, ∥ϕ∥_{[−τ,0]}} = max{r, ∥ϕ∥_{[−τ,0]}} = r. (3.5) Thus by (3.4), (3.5) and the nondecreasing of F, we get that

r = ∥u∥ ≤ ∫_{0}^{1} h(s)a(s)F (r)ds + \frac{b}{(α − 1)(1 − bη^{α−2})} ∫_{0}^{1} H(η, s)a(s)F (r)ds.

Consequently

\frac{r}{F (r)} ≤ ∫_{0}^{1} h(s)a(s)ds + +\frac{b}{(α − 1)(1 − bη^{α−2})} ∫_{0}^{1} H(η, s)a(s)ds,

which contradict (3.3). Hence u ∉ ∂U. By Theorem 2.3, T_{ϕ} has a fixed point u ∈ \overline{U}. Therefore, BVP (1.1)-(1.3) has at least one positive solution. Theorem 3.4. Let (H_{1}) holds. Suppose that the following conditions are satisfied:

(H_{4}) There exist constant r_{2} > ∥ϕ∥_{[−τ,0]}, as well as continuous function p ∈ C[0, 1] and nondecreasing continuous functions L : R^{+} → R^{+} such that

f(t, ψ) ≤ p(t)L(∥ψ∥_{[−τ,0]}), (t, ψ) ∈ [0, 1] × C_{τ}^{+}(0), ∥ψ∥_{[−τ,0]} ≤ r_{2}. (3.6) (H_{5}) There exist functions ω : [0, 1] → [0, τ], continuous c : [0, 1] → R^{+}, and nondecreasing J :

R^{+} → R^{+} such that

f(t, ψ) ≥ c(t)J(ψ(−ω(t))), (t, ψ) ∈ [0, 1] × C_{τ}^{+}(0), (3.7) If r_{2} ( as in (H_{4})) satisfying

\frac{L(r_{2})}{r_{2}} (∫_{0}^{1} h(s)p(s)ds + \frac{b}{(α − 1)(1 − bη^{α−2})} ∫_{0}^{1} H(η, s)p(s)ds) ≤ 1. (3.8)

And there exists a constant r_{1} > 0 (r_{1} < r_{2}) satisfying

\frac{J(τ^{α−1}r_{1})}{r_{1}} ∫_{τ}^{1} τ^{α−1}h(s)c(s)ds ≥ 1. (3.9)

Then BVP (1.1)-(1.3) has a positive solution. Proof. If u ∈ P with ∥u∥ = r_{2}, then from (3.6), (3.5), and Lemma 2.5, we get for any t ∈ [0, 1] that

(T_{ϕ}u)(t) = ∫_{0}^{1} G(t, s)f(s, u_{s}(·; ϕ))ds + \frac{bt^{α−1}}{(α − 1)(1 − bη^{α−2})} ∫_{0}^{1} H(η, s)f(s, u_{s}(·; ϕ))ds

≤ ∫_{0}^{1} h(s)f(s, u_{s}(·; ϕ))ds + \frac{b}{(α − 1)(1 − bη^{α−2})} ∫_{0}^{1} H(η, s)f(s, u_{s}(·; ϕ))ds

≤ ∫_{0}^{1} h(s)p(s)L(∥u_{s}(·; ϕ)∥_{[−τ,0]})ds

+\frac{b}{(α − 1)(1 − bη^{α−2})} ∫_{0}^{1} H(η, s)p(s)L(∥u_{s}(·; ϕ)∥_{[−τ,0]})ds

≤ L(r_{2}) (∫_{0}^{1} h(s)p(s)ds + \frac{b}{(α − 1)(1 − bη^{α−2})} ∫_{0}^{1} H(η, s)p(s)ds)

≤ r_{2}. (3.10)

Now if we set

Ω_{1} = {u ∈ C[0, 1] : ∥u∥ < r_{2}}, then (3.10) shows that ∥T_{ϕ}u∥ ≤ ∥u∥ for u ∈ P ∩ ∂Ω_{1}.

For u ∈ P with ∥u∥ = r_{1}, we have from (3.7), (3.9), and Lemmas 2.5 and 3.2 that (T_{ϕ}u)(t) ≥ ∫_{0}^{1} G(t, s)f(s, u_{s}(·; ϕ))ds

≥ ∫_{τ}^{1} G(t, s)f(s, u_{s}(·; ϕ))ds ≥ ∫_{τ}^{1} τ^{α−1}h(s)f(s, u_{s}(·; ϕ))ds

≥ ∫_{τ}^{1} τ^{α−1}h(s)c(s)J(u_{s}(−ω(s); ϕ))ds

≥ J(τ^{α−1}∥u∥) ∫_{τ}^{1} τ^{α−1}h(s)c(s)ds

= J(τ^{α−1}r_{1}) ∫_{τ}^{1} τ^{α−1}h(s)c(s)ds ≥ r_{1}. (3.11)

Now if we set

Ω_{2} = {u ∈ C[0, 1] : ∥u∥ < r_{1}}, then (3.11) shows that ∥T_{ϕ}u∥ ≥ ∥u∥ for u ∈ P ∩ ∂Ω_{2}.

Hence by the second part of Theorem 2.4, T_{ϕ} has a fixed point u ∈ P ∩ (\overline{Ω}_{1} \ Ω_{2}), and accordingly, u is a solution of BVP (1.1)-(1.3).

Set

M = (\frac{1}{(α − 1)Γ(α + 1)} + \frac{bη^{α−2}(1 − η)}{(α − 1)Γ(α)(1 − bη^{α−2})})^{−1} , (3.12)

m = \frac{(α − 1)Γ(α + 1)}{τ^{2(α−1)}[α(1 − τ)^{α−1} − (α − 1)(1 − τ)^{α}]}. (3.13)

Corollary 3.5. Let (H_{1}) holds. If there exist two positive constants r_{2} > r_{1} > 0 (r_{2} > ∥ϕ∥_{[−τ,0]}) such that

(H_{6}) f(t, ψ) ≤ Mr_{2}, (t, ψ) ∈ [0, 1] × C_{τ}^{+}(0), ∥ψ∥_{[−τ,0]} ≤ r_{2}. (3.14) (H_{7}) f(t, ψ) ≥ mr_{1}, (t, ψ) ∈ [0, 1]× ∈ C_{τ}^{+}(0), ∥ψ∥_{[−τ,0]} ≤ r_{1}. (3.15)

Then BVP (1.1)-(1.3) has a positive solution. Proof. In (H_{4}), let p(t) ≡ M and L(x) ≡ x, then by (3.12) and (3.14), we have that (3.8) holds. In fact, we have

L(r_{2}) (∫_{0}^{1} h(s)p(s)ds + \frac{b}{(α − 1)(1 − bη^{α−2})} ∫_{0}^{1} H(η, s)p(s)ds)

= r_{2}M (∫_{0}^{1} h(s)ds + \frac{b}{(α − 1)(1 − bη^{α−2})} ∫_{0}^{1} H(η, s)ds)

= r_{2}M (\frac{1}{(α − 1)Γ(α + 1)} + \frac{bη^{α−2}(1 − η)}{(α − 1)Γ(α)(1 − bη^{α−2})})

= r_{2}.

Moreover, let c(t) ≡ m and J(x) ≡ x in (H_{5}), then by (3.13) and (3.15), we have that (3.9) holds. In fact, we have

J(τ^{α−1}r_{1}) ∫_{τ}^{1} τ^{α−1}h(s)c(s)ds = τ^{2(α−1)}r_{1}m ∫_{τ}^{1} h(s)ds

= τ^{2(α−1)}r_{1}m\frac{α(1 − τ)^{α−1} − (α − 1)(1 − τ)^{α}}{(α − 1)Γ(α + 1)} = r_{1}.

So all conditions of Theorem 3.4 are satisfied. By Theorem 3.4, BVP (1.1)-(1.3) has at least one positive solution.

Having in mind the proof of Theorem 3.4, one can easily conclude the following results. Theorem 3.6. Let (H_{1}), (H_{4}), (H_{5}) and (3.8) hold. If the function η satisfies the condition

lim sup_{x→0+} \frac{J(x)}{x} > \frac{1}{τ^{2(α−1)} ∫_{τ}^{1} h(s)c(s)ds}. (3.16)

Then BVP (1.1)-(1.3) has a positive solution.

EJQTDE, 2010 No. 30, p. 10 4. Examples

To illustrate our results we present the following examples. Example 4.1. Consider the boundary value problem of fractional order functional differential equations

D^{2.5}u(t) + \frac{1}{2}(1 + t^{2})u^{2} (t − \frac{1}{3}) + 3 = 0, 0 < t < 1, (4.1)

u′(0) = 0, u′(1) = bu′(η), (4.2) u(t) = ϕ(t), t ∈ [−τ, 0], (4.3)

where b = 1.2, η = \frac{2}{3}, τ = \frac{1}{2}, and ϕ ∈ C_{τ}^{+}(0) with ∥ϕ∥_{[−τ,0]} < \frac{3}{2}.

Let f(t, ψ) = \frac{1}{2}(1+t^{2})ψ^{2}(−\frac{1}{3})+3, (t, ψ) ∈ [0, 1]×C_{τ}^{+}, and α = 2.5. Obviously, 1 < b < \frac{1}{η^{α−2}} = 1.2247. By simple calculation, we obtain that

M = (\frac{1}{(α − 1)Γ(α + 1)} + \frac{bη^{α−2}(1 − η)}{(α − 1)Γ(α)(1 − bη^{α−2})})^{−1} = 3.0864,

m = \frac{(α − 1)Γ(α + 1)}{τ^{2(α−1)}[α(1 − τ)^{α−1} − (α − 1)(1 − τ)^{α}]} = 64.489.

Choosing r_{1} = \frac{1}{22}, r_{2} = \frac{3}{2} > ∥ϕ∥_{[−τ,0]}, we have

f(t, ψ) = \frac{1}{2}(1 + t^{2})ψ^{2} (−\frac{1}{3}) + 3 ≤ ∥ψ∥_{[−τ,0]} + 3 ≤ 4.5

< 4.6026 = 3.0864 · 1.5 = Mr_{2}, (t, ψ) ∈ [0, 1] × C_{τ}^{+} with ∥ψ∥_{[−τ,0]} ≤ \frac{3}{2},

f(t, ψ) = \frac{1}{2}(1 + t^{2})ψ^{2} (−\frac{1}{3}) + 3 ≥ 3 > 2.9313

= 64.489 · \frac{1}{22} = mr_{1}, (t, ψ) ∈ [0, 1] × C_{τ}^{+} with ∥ψ∥_{[−τ,0]} ≤ \frac{1}{22},

Hence all conditions of Corollary 3.5 are satisfied. By Corollary 3.5, BVP (4.1)-(4.3) has at least one positive solution u such that \frac{1}{22} ≤ ∥u∥ ≤ \frac{3}{2}. Example 4.2. Consider the boundary value problem of fractional order functional differential equations

D^{2.2}u(t) + \sqrt{t} sin^{2} t · u^{4} (t − \frac{1}{4}) = 0, 0 < t < 1, (4.4)

u′(0) = 0, u′(1) = bu′(η), (4.5) u(t) = ϕ(t), t ∈ [−τ, 0], (4.6) where b = 1.1, η = 0.2, τ = \frac{1}{3} and ϕ ∈ C_{τ}^{+}(0) with ∥ϕ∥_{[−τ,0]} < \frac{3}{4}.

Let f(t, ψ) = \sqrt{t} sin^{2} t · u^{4} (t − \frac{1}{4}), (t, ψ) ∈ [0, 1] × C_{τ}^{+}, and α = 2.2.

Obviously, 1 < b < \frac{1}{η^{α−2}} = 1.3797, and

f(t, ψ) = \sqrt{t} sin^{2} t · ψ^{4} (−\frac{1}{4}) ≤ a(t)F (∥ψ∥_{[−τ,0]}),

where a(t) = \sqrt{t} and F (x) = x^{4}, which implies that (H_{2}) holds.

With the aid of computation we have that

∫_{0}^{1} h(s)a(s)ds + \frac{b}{(α − 1)(1 − bη^{α−2})} ∫_{0}^{1} H(η, s)a(s)ds

= \frac{1}{Γ(2.2)} ∫_{0}^{1}((1 − s)^{0.2} − (1 − s)^{1.2}))s^{0.5}ds

+\frac{1.1}{1.2(1 − 1.1 · 0.2^{0.2})} [∫_{0}^{0.2} \frac{1.2}{Γ(2.2)}(0.2^{0.2} · (1 − s)^{0.2} − (0.2 − s)^{0.2})s^{0.5}ds

+ ∫_{0.2}^{1} \frac{1.2}{Γ(2.2)}0.2^{0.2} · (1 − s)^{0.2}s^{0.5}ds]

≤ \frac{1}{Γ(2.2)} ∫_{0}^{1}(1 − s)^{0.2}s^{0.5}ds − \frac{1}{Γ(2.2)} ∫_{0}^{1}(1 − s)^{1.2}s^{0.5}ds

+\frac{1.1}{1.2(1 − 1.1 · 0.2^{0.2})} · \frac{1.2}{Γ(2.2)}0.2^{0.2} ∫_{0}^{1}(1 − s)^{0.2}s^{0.5}ds

= \frac{1}{Γ(2.2)}B(1.5, 1.2) − \frac{1}{Γ(2.2)}B(1.5, 2.2) + \frac{1.1 · 0.2^{0.2}}{(1 − 1.1 · 0.2^{0.2})Γ(2.2)}B(1.5, 1.2)

= 2.1458,

where B(·, ·) is a Beta function. Choosing r = \frac{3}{4}, then r > ∥ϕ∥_{[−τ,0]}, and

\frac{r}{F (r)} = \frac{1}{r^{3}} = 2.3704

> 2.1458 = ∫_{0}^{1} h(s)a(s)ds + \frac{b}{(α − 1)(1 − bη^{α−2})} ∫_{0}^{1} H(η, s)a(s)ds,

that is, condition (H_{3}) holds. Thus all conditions of Theorem 3.3 are satisfied. By Theorem 3.3, BVP (4.4)-(4.6) has at least one positive solution.

Acknowledgements

The author would like to thank the referees for their valuable suggestions and comments. Project supported by the National Natural Science Foundation of China (10771212) and the Natural Science Foundation of Jiangsu Education Office (06KJB110010).

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(Received January 21, 2010)