Electronic Journal of Qualitative Theory of Differential Equations 2010, No. 50, 1-13; http://www.math.u-szeged.hu/ejqtde/

Floquet Boundary Value Problem of Fractional Functional Differential Equations^{∗}

Yong Zhou^{†}

Department of Mathematics, Xiangtan University, Xiangtan, Hunan 411105, P. R. China E-mail: yzhou@xtu.edu.cn

Yuansheng Tian

Department of Mathematics, Xiangnan University Chenzhou, Hunan, 423000, P. R. China Email: tys73@163.com

Yun-Yun He

Department of Mathematics, Xiangtan University, Xiangtan, Hunan 411105, P. R. China

Abstract: In this paper, we prove the existence of positive solutions for Floquet boundary value problem concerning fractional functional differential equations with bounded delay. The results are obtained by using two fixed point theorems on appropriate cones.

Keywords: Positive solutions, Boundary value problems, Fractional functional differential equations.

1. Introduction

Fractional differential equations have gained considerable importance due to their application in various sciences, such as physics, chemistry, mechanics, engineering, etc. For details, see [1-4] and references therein. Naturally, such equations need to be solved. Recently, there are some papers focused on initial value problem of fractional functional differential equations[5-12], and boundary value problems of fractional ordinary differential equations [13-20]. But the results dealing with the boundary value problems of fractional functional differential equations are relatively scarce.

* Research supported by National Natural Science Foundation of P. R. China (10971173).

† Corresponding author.

In this paper, we consider the existence of positive solutions for the following fractional functional differential equation

CD^{α}x(t) = f(t, x_{t}), t ∈ [0, T],  (1)

with the boundary condition

Ax_{0} − x_{T} = ϕ,  (2)

where α, A, T are real numbers with 0 < α ≤ 1, A > 1 and T > 0. ^{C}D^{α} denote Caputo’s fractional derivative. f : [0, T]×C[−r, 0] → R is a given function satisfying some assumptions that will be specified later, and ϕ ∈ C[−r, 0], where 0 ≤ r < T. As usual, C[−r, 0] is the space of continuous functions on [−r, 0], equipped with ∥ϕ∥ = max_{−r≤θ≤0} |ϕ(θ)|. For any t ∈ [0, T] and x ∈ C[−r, T], the function x_{t} is defined by x_{t}(θ) = x(t + θ), − r ≤ θ ≤ 0.

The boundary value problem (1) − (2) belongs to a class of problems knows as “Floquet problems” which arise from physics (see [21]). The existence of positive solutions of the first order functional differential equations concerned with this problem was discussed by Mavridis and Tsamatos in [22].

In this paper, we firstly deduced the problem (1)−(2) to an equivalent operator equation. Next, using two fixed-point theorems, we get that the equivalent operator has (at least) a fixed point, it means that the boundary value problem (1) − (2) has (at least) one positive solution, which is upper and lower bounded by specific real numbers .

2. Preliminaries

In this section, we introduce definitions and preliminary facts which are used throughout this paper.

Let Ω be a finite or infinite interval of the real axis R = (−∞, ∞). We denote by L^{p}Ω (1 ≤ p ≤ ∞) the set of those Lebesgue measurable functions f on Ω for which ∥f∥_{L^{p}Ω} < ∞, where

∥f∥_{L^{p}Ω} = ( ∫_{Ω} |f(t)|^{p}dt)^{\frac{1}{p}} (1 ≤ p < ∞)

and

∥f∥_{L^{∞}Ω} = ess sup_{x∈Ω}|f(t)|.

Definition 2.1. [2,3] The fractional integral of order α with the lower limit t_{0} for a function f is defined as

I^{α}f(t) = \frac{1}{Γ(α)} ∫_{t_{0}}^{t} \frac{f(s)}{(t − s)^{1−α}} ds, t > t_{0}, α > 0,  (3)

provided the right side is point-wise defined on [t_{0}, ∞), where Γ is the gamma function.

Definition 2.2. [2,3] Riemann-Liouville derivative of order α with the lower limit t_{0} for a function f : [t_{0}, ∞) → R can be written as

LD^{α}f(t) = \frac{1}{Γ(n − α)} \frac{d^{n}}{dt^{n}} ∫_{t_{0}}^{t} \frac{f(s)}{(t − s)^{α+1−n}} ds, t > t_{0}, n − 1 < q < n.

The first−and maybe the most important−property of Riemann-Liouville fractional derivative is that Riemann-Liouville fractional differentiation operator is a left inverse to the Riemann-Liouville fractional integration operator of the same order α. Lemma 2.1.[2] Let f(t) ∈ L^{1}[t_{0}, ∞). Then

LD^{α}(I^{α}f(t)) = f(t), t > t_{0} and 0 < α < 1

Definition 2.3. [2] Caputo’s derivative of order α for a function f : [t_{0}, ∞) → R can be written as

CD^{α}f(t) = ^{L}D^{α}(f(t) − ∑_{k=0}^{n−1} \frac{(t − t_{0})^{k}}{k!} f^{(k)}(t_{0})), t > t_{0}, n − 1 < α < n.

One can show that if f(t) ∈ C^{n}[t_{0}, ∞), then

CD^{α}f(t) = \frac{1}{Γ(n − α)} ∫_{t_{0}}^{t} \frac{f^{(n)}(s)}{(t − s)^{α+1−n}} ds.

Obviously, Caputo’s derivative of a constant is equal to zero. Definition 2.4. Let X be a real Banach space. A cone in X is a nonempty, closed set P ⊂ X such that

(i) λu + μv ∈ P for all u, v ∈ P and all λ, μ ≥ 0, (ii) u, −u ∈ P implies u = 0.

Let P be a cone in a Banach space X. Then, for any b > 0, we denote by P_{b} the set P_{b} = {x ∈ P : ∥x∥ < b},

and by ∂P_{b} the boundary of P_{b} in P, i.e, the set ∂P_{b} = {x ∈ P : ∥x∥ = b}. In order to prove our results, and since we are looking for positive solutions, we will use

the following two lemmas, which are applications of the fixed point theory in a cone. Their proofs can be found in [23,24]. Lemma 2.2. Let g : \overline{P}_{b} → P be a completely continuous map such that g(x) ≠ λx for all x ∈ ∂P_{b} and λ ≥ 1, then g has a fixed point in P_{b}. Lemma 2.3. Let E = (E, ∥ · ∥) be a Banach space, P ⊂ E be a cone, and ∥ · ∥ be increasing (strictly) with respect to P. Also, σ, τ are positive constants with σ ≠ τ, suppose g : \overline{P}_{max\{σ,τ\}} → P is a completely continuous map and assume the conditions

(i) g(x) ≠ λx for every x ∈ ∂P_{σ} and λ ≥ 1, (ii) ∥g(x)∥ ≥ ∥x∥ for x ∈ ∂P_{τ} , hold, then g has at least a fixed point x with min{σ, τ} ≤ ∥x∥ ≤ max{σ, τ}.

3. Main results

Let the intervals I := [0, T] and J := [−r, 0] and set C(J ∪ I) be endowed with the ordering x ≤ y if x(t) ≤ y(t) for all t ∈ (J ∪ I), and the maximum norm, ∥x∥_{J∪I} = max_{−r≤t≤T} |x(t)|. Define the cone P ⊂ C(J ∪ I) by P = {x ∈ C(J ∪ I) | x(t) > 0} and set P_{l} = {x ∈ P | ∥x∥_{J∪I} ≤ l, l > 0} and C^{+}(J) = {x ∈ C(J)| x(t) ≥ 0, t ∈ J}.

The following assumptions are adopted throughout this section. (H_{1}) for any φ ∈ C(J), f(t, φ) is measurable with respect to t on I, (H_{2}) for any given l > 0, x ∈ P_{l}, there exist α_{1} ∈ (0, α) and a function m_{l}(t) ∈ L^{\frac{1}{α_{1}}} I such that

|f(t, x_{t})| ≤ m_{l}(t), t ∈ I, (H_{3}) f(t, φ) is continuous with respect to φ on C(J), (H_{4}) f : I × C^{+}(J) → R^{+}, ϕ(0) > 0 and ϕ(t) > −\frac{ϕ(0)}{A−1}, for t ∈ J, (H_{5}) there exists a ρ > 0 such that

ρ > \frac{ϕ(0)}{A − 1} + \frac{∥ϕ∥}{A} + \frac{AM_{ρ}}{(A − 1)Γ(α)(1 + β)^{1−α_{1}}} T^{(1+β)(1−α_{1})},

where M_{ρ} = ∥m_{ρ}(t)∥_{L^{\frac{1}{α_{1}}} I}.

In order to gain our results, firstly, we must reformulate our boundary value problem (1) − (2) into an abstract operator equation. This is done in the following lemma. Lemma 3.1. Assume that (H_{1}) − (H_{3}) hold. Then a function x ∈ P_{l} is a solution of the boundary value problem (1) − (2) if and only if x(t) = F x(t), t ∈ J ∪ I, where F : P_{l} → C(J ∪ I) is given by the formula

F x(t) = \{ \begin{matrix}\frac{ϕ(0)}{A−1} + \frac{1}{(A−1)Γ(α)} ∫_{0}^{T}(T − s)^{α−1}f(s, x_{s})ds + \frac{1}{Γ(α)} ∫_{0}^{t}(t − s)^{α−1}f(s, x_{s})ds, t ∈ I, \\ \frac{ϕ(0)}{A(A−1)} + \frac{1}{A(A−1)Γ(α)} ∫_{0}^{T}(T − s)^{α−1}f(s, x_{s})ds \\ +\frac{1}{AΓ(α)} ∫_{0}^{T +t}(T + t − s)^{α−1}f(s, x_{s})ds + \frac{ϕ(t)}{A}, t ∈ J. \end{matrix}

Proof. Firstly, it is easy to obtain that f(t, x_{t}) is Lebesgue measurable in I according to conditions (H_{1}) and (H_{3}). The direct calculation gives that (t−s)^{α−1} ∈ L^{\frac{1}{1−α_{1}}} [0, t], for t ∈ I. In light of Hölder inequality and the condition (H_{2}), we obtain that (t − s)^{α−1}f(s, x_{s}) is Lebesgue integrable with respect to s ∈ [0, t] for all t ∈ I, and

∫_{0}^{t} |(t − s)^{α−1}f(s, x_{s})|ds ≤ ∥(t − s)^{α−1}∥_{L^{\frac{1}{1−α_{1}}} [0,t]}∥m_{l}(t)∥_{L^{\frac{1}{α_{1}}} I}.  (4)

Hence, F x exists. From the formula of F x, we have

F x(0 ) = F x(0^{+}), x ∈ P_{l}.

So it is clear that F x is a continuous function for every x ∈ P_{l}. It is to say that F : P_{l} → C(J ∪ I). Moreover, from (1), we have

I^{α} ^{C}D^{α}x(t) = I^{α}f(t, x_{t}), t ∈ I,

i.e.,

x(t) = x(0) + \frac{1}{Γ(α)} ∫_{0}^{t}(t − s)^{α−1}f(s, x_{s})ds, t ∈ I.  (5)

Since r < T, if θ ∈ J, then T + θ ∈ I. Thus from (2) and (6) we get

Ax(θ) − (x(0) + \frac{1}{Γ(α)} ∫_{0}^{T +θ}(T + θ − s)^{α−1}f(s, x_{s})ds) = ϕ(θ).  (6)

Therefore, for θ = 0, we get

x(0) = \frac{ϕ(0)}{A − 1} + \frac{1}{(A − 1)Γ(α)} ∫_{0}^{T}(T − s)^{α−1}f(s, x_{s})ds.  (7)

Using (6) and (8) we have

x(t) = \frac{ϕ(0)}{A − 1} + \frac{1}{(A − 1)Γ(α)} ∫_{0}^{T}(T − s)^{α−1}f(s, x_{s})ds + \frac{1}{Γ(α)} ∫_{0}^{t}(t − s)^{α−1}f(s, x_{s})ds, t ∈ I.

Also, by (7) and (8), for t ∈ J we get

x(t) = \frac{ϕ(0)}{A(A − 1)} + \frac{1}{A(A − 1)Γ(α)} ∫_{0}^{T}(T − s)^{α−1}f(s, x_{s})ds

+\frac{1}{AΓ(α)} ∫_{0}^{T +t}(T + t − s)^{α−1}f(s, x_{s})ds + \frac{ϕ(t)}{A}.

So,

x(t) = F x(t), t ∈ J ∪ I.

On the other hand, if x ∈ P_{l} is such that x(t) = F x(t), t ∈ J ∪ I, then, by Definition 2.3 and Lemma 2.1, for every t ∈ I we have

CD^{α}x(t) = ^{L}D^{α}F x(t) = f(t, x_{t}).

Also for any θ ∈ J, it is clear that

Ax_{0}(θ) − x_{T} (θ) = Ax(θ) − x(T + θ)
= ϕ(θ).

The proof is complete. Lemma 3.2. Assume that (H_{1}) − (H_{4}) hold. Then F x(t) > 0, for t ∈ J ∪ I, x ∈ P_{l}, and F : P_{l} → P is completely continuous operator. Proof. From Lemma 3.1, we get F : P_{l} → C(J ∪ I), and by (H_{4}), we easily obtain F x(t) > 0 for x ∈ P_{l}. Also, it is clear that F : P_{l} → P is continuous according to condition (H_{3}).

Let β = \frac{α−1}{1−α_{1}} ∈ (−1, 0). For every t ∈ I, we have

|F x(t)| ≤ \frac{ϕ(0)}{A − 1} + \frac{M_{l}}{(A − 1)Γ(α)}∥(T − s)^{α−1}∥_{L^{\frac{1}{1−α_{1}}} I} + \frac{M_{l}}{Γ(α)}∥(t − s)^{α−1}∥_{L^{\frac{1}{1−α_{1}}} [0,t]}
≤ \frac{ϕ(0)}{A − 1} + \frac{M_{l}}{(A − 1)Γ(α)(1 + β)^{1−α_{1}}} T^{(1+β)(1−α_{1})} + \frac{M_{l}}{Γ(α)(1 + β)^{1−α_{1}}} T^{(1+β)(1−α_{1})}.

Also, for every t ∈ J, we have

|F x(t)| ≤ \frac{∥ϕ∥}{A − 1} + \frac{M_{l}}{A(A − 1)Γ(α)}∥(T − s)^{α−1}∥_{L^{\frac{1}{1−α_{1}}} I} + \frac{M_{l}}{AΓ(α)}∥(T + t − s)^{α−1}∥_{L^{\frac{1}{1−α_{1}}} [0,T +t]}
≤ \frac{∥ϕ∥}{A − 1} + \frac{M_{l}}{A(A − 1)Γ(α)(1 + β)^{1−α_{1}}} T^{(1+β)(1−α_{1})} + \frac{M_{l}}{AΓ(α)(1 + β)^{1−α_{1}}} T^{(1+β)(1−α_{1})}.

Hence, LP_{l} is bounded.

Now, we will prove that LP_{l} is equicontinuous.

In the following, we divide the proof into three cases. Case 1. x ∈ P_{l}, 0 ≤ t_{1} < t_{2} ≤ T,

|F x(t_{2}) − F x(t_{1})|

≤ |\frac{1}{Γ(α)} ∫_{0}^{t_{2}}(t_{2} − s)^{α−1}f(s, x_{s})ds − \frac{1}{Γ(α)} ∫_{0}^{t_{1}}(t_{1} − s)^{α−1}f(s, x_{s})ds|

≤ \frac{1}{Γ(α)} ∫_{0}^{t_{1}} |(t_{2} − s)^{α−1} − (t_{1} − s)^{α−1})|f(s, x_{s})ds + \frac{1}{Γ(α)} ∫_{t_{1}}^{t_{2}} (t_{2} − s)^{α−1}f(s, x_{s})ds

≤ \frac{M_{l}}{Γ(α)}( ∫_{0}^{t_{1}}[(t_{1} − s)^{α−1} − (t_{2} − s)^{α−1})]^{\frac{1}{1−α_{1}}} ds)^{1−α_{1}} + \frac{M_{l}}{Γ(α)}( ∫_{t_{1}}^{t_{2}}[(t_{2} − s)^{α−1}]^{\frac{1}{1−α_{1}}} ds)^{1−α_{1}}

≤ \frac{M_{l}}{Γ(α)}( ∫_{0}^{t_{1}}(t_{1} − s)^{β} − (t_{2} − s)^{β}ds)^{1−α_{1}} + \frac{M_{l}}{Γ(α)}( ∫_{t_{1}}^{t_{2}}(t_{2} − s)^{β}ds)^{1−α_{1}}

≤ \frac{M_{l}}{Γ(α)(1 + β)^{1−α_{1}}} (t_{1}^{1+β} − t_{2}^{1+β} + (t_{2} − t_{1})^{1+β})^{1−α_{1}}

+\frac{M_{l}}{Γ(α)(1 + β)^{1−α_{1}}} (t_{2} − t_{1})^{(1+β)(1−α_{1})}

≤ \frac{2M_{l}}{Γ(α)(1 + β)^{1−α_{1}}} (t_{2} − t_{1})^{(1+β)(1−α_{1})}.

Case 2. x ∈ P_{l}, −r ≤ t_{1} < t_{2} ≤ 0,

|F x(t_{2}) − F x(t_{1})|

≤ |\frac{ϕ(t_{2})}{A} − \frac{ϕ(t_{1})}{A}|

+\frac{1}{AΓ(α)}| ∫_{0}^{T +t_{2}}(T + t_{2} − s)^{α−1}f(s, x_{s})ds − ∫_{0}^{T +t_{1}}(T + t_{1} − s)^{α−1}f(s, x_{s})ds|

≤ \frac{1}{A}|ϕ(t_{2}) − ϕ(t_{1})| + \frac{M_{l}}{AΓ(α)}( ∫_{0}^{T +t_{1}}[(T + t_{1} − s)^{α−1} − (T + t_{2} − s)^{α−1})]^{\frac{1}{1−α_{1}}} ds)^{1−α_{1}}

+\frac{M_{l}}{AΓ(α)}( ∫_{T +t_{1}}^{T +t_{2}}(T + t_{2} − s)^{β}ds)^{1−α_{1}}

≤ \frac{1}{A}|ϕ(t_{2}) − ϕ(t_{1})| + \frac{M_{l}}{AΓ(α)(1 + β)^{1−α_{1}}} [(T + t_{1})^{1+β} − (T + t_{2})^{1+β} + (t_{2} − t_{1})^{1+β}]^{1−α_{1}}

+\frac{M_{l}}{AΓ(α)(1 + β)^{1−α_{1}}} (t_{2} − t_{1})^{(1+β)(1−α_{1})}

≤ \frac{1}{A}|ϕ(t_{2}) − ϕ(t_{1})| + \frac{2M_{l}}{AΓ(α)(1 + β)^{1−α_{1}}} (t_{2} − t_{1})^{(1+β)(1−α_{1})}.

Case 3. x ∈ P_{l}, −r ≤ t_{1} < 0 ≤ t_{2} ≤ T,

|F x(t_{2}) − F x(t_{1})|

≤ |F x(t_{2}) − F x(0)| + |F x(0) − F x(t_{1})|

≤ \frac{1}{Γ(α)} ∫_{0}^{t_{2}}(t_{2} − s)^{α−1}f(s, x_{s})ds + \frac{|ϕ(0) − ϕ(t_{1})|}{A}

\frac{1}{AΓ(α)}| ∫_{0}^{T}(T − s)^{α−1}f(s, x_{s})ds − ∫_{0}^{T +t_{1}}(T + t_{1} − s)^{α−1}f(s, x_{s})ds|

≤ \frac{M_{l}}{Γ(α)(1 + β)^{1−α_{1}}} t_{2}^{(1+β)(1−α_{1})} + \frac{|ϕ(0) − ϕ(t_{1})|}{A} + \frac{2M_{l}}{AΓ(α)(1 + β)^{1−α_{1}}} (−t_{1})^{(1+β)(1−α_{1})}.

According to the continuity of ϕ and t^{(1+β)(1−α_{1})}, we can easily obtain LP_{l} is equicontinuous. Then F : P_{l} → P is a completely continuous operator by Arzela-Ascoli theorem. The proof is complete. Theorem 3.1. Assume that the conditions (H_{1}) − (H_{5}) hold. Then the boundary value problem (1) − (2) has at least one positive solution x, such that

\frac{ϕ(0)}{A − 1} ≤ ∥x∥_{J∪I} < ρ.

Proof. From Lemma 3.2, F : \overline{P}_{ρ} → P is a completely continuous operator.

Furthermore, we will show that λx ≠ F x for every λ ≥ 1 and x ∈ ∂P_{ρ}. Otherwise, let x ∈ ∂P_{ρ} and λ ≥ 1 such that λx = F x. Then for every t ∈ I, we have

|x(t)| ≤ λ|x(t)|
= \frac{ϕ(0)}{A − 1} + \frac{1}{(A − 1)Γ(α)} ∫_{0}^{T}(T − s)^{α−1}f(s, x_{s})ds + \frac{1}{Γ(α)} ∫_{0}^{t}(t − s)^{α−1}f(s, x_{s})ds
≤ \frac{ϕ(0)}{A − 1} + \frac{M_{ρ}}{(A − 1)Γ(α)}( ∫_{0}^{T}(T − s)^{β}ds)^{1−α_{1}} + \frac{M_{ρ}}{Γ(α)}( ∫_{0}^{t}(t − s)^{β}ds)^{1−α_{1}}
≤ \frac{ϕ(0)}{A − 1} + \frac{M_{ρ}}{(A − 1)Γ(α)} \frac{1}{(1 + β)^{1−α_{1}}} T^{(1+β)(1−α_{1})} + \frac{M_{ρ}}{Γ(α)} \frac{1}{(1 + β)^{1−α_{1}}} T^{(1+β)(1−α_{1})}
≤ \frac{ϕ(0)}{A − 1} + \frac{AM_{ρ}}{(A − 1)Γ(α)(1 + β)^{1−α_{1}}} T^{(1+β)(1−α_{1})}.

Also, for every t ∈ J, we have

|x(t)| ≤ λ|x(t)|
≤ \frac{ϕ(0)}{A(A − 1)} + \frac{∥ϕ∥}{A} + \frac{M_{ρ}}{A(A − 1)Γ(α)}( ∫_{0}^{T}(T − s)^{β}ds)^{1−α_{1}}

+\frac{M_{ρ}}{AΓ(α)}( ∫_{0}^{T +t}(T + t − s)^{β}ds)^{1−α_{1}}

≤ \frac{ϕ(0)}{A(A − 1)} + \frac{∥ϕ∥}{A} + \frac{M_{ρ}}{A(A − 1)Γ(α)} \frac{1}{(1 + β)^{1−α_{1}}} T^{(1+β)(1−α_{1})}

+\frac{M_{ρ}}{AΓ(α)(1 + β)^{1−α_{1}}} T^{(1+β)(1−α_{1})}

≤ \frac{ϕ(0)}{A(A − 1)} + \frac{∥ϕ∥}{A} + \frac{M_{ρ}}{(A − 1)Γ(α)(1 + β)^{1−α_{1}}} T^{(1+β)(1−α_{1})}.

Consequently, for every t ∈ J ∪ I, it holds

∥x∥_{J∪I} < ρ,

which contradicts with x ∈ ∂P_{ρ}.

So applying Lemma 2.2, we can obtain that L has at least a fixed point, what means that the boundary value problem (1) − (2) has at least one positive solution x, such that

∥x∥_{J∪I} < ρ.

Then, taking into account the formula of L and the fact A > 1, we easily conclude that

x(t) ≥ \frac{ϕ(0)}{A − 1}, t ∈ I,

which implies that

∥x∥_{J∪I} ≥ \frac{ϕ(0)}{A − 1}.

Observe that \frac{ϕ(0)}{A−1} < ρ. Therefore, we finally have

\frac{ϕ(0)}{A − 1} ≤ ∥x∥_{J∪I} < ρ.

The proof is complete.

In order to gain our second result, we need the following assumption: (H_{6}) There exists an interval E ⊆ I, and functions u : E → [0, r], continuous v : E → [0, +∞) with sup{v(t) : t ∈ E} > 0 and nonincreasing w : [0, +∞) → [0, +∞) such that

f(t, y) ≥ v(t)w(y(−u(t))), (t, y) ∈ E × C^{+}(J).

Let

μ := \frac{1}{A(A − 1)} ∫_{E} v(s)ds, Λ := \frac{ϕ(0)}{A(A − 1)} + \frac{ϕ(−r)}{A}.

Theorem 3.2. Suppose that (H_{1}) − (H_{6}) hold, also suppose that there exists τ > 0, such that

τ ≤ \frac{T^{α−1}}{Γ(α)} μw(τ).  (8)

Then the boundary value problem (1) − (2) has at least one positive solution x, such that

d ≤ ∥x∥_{J∪I} ≤ max{τ, ρ},

where

d = \{ \begin{matrix}ρ, if τ > ρ, \\ max\{τ, Λ\}, if τ < ρ, \end{matrix}

and ρ is the constant involved in (H_{5}) and τ ≠ ρ. Proof. From Lemma 3.2, F : \overline{P}_{max\{τ,ρ\}} → P is a completely continuous map.

As we did in Theorem 3.1, we can prove that F x ≠ λx for every λ ≥ 1 and x ∈ ∂P_{ρ}. Now we will prove that ∥F x∥_{J∪I} ≥ ∥x∥_{J∪I} for every x ∈ ∂P_{τ} and t ∈ J ∪ I. By (H_{4}) and

(H_{6}), we have

F x(−r) = \frac{ϕ(0)}{A(A − 1)} + \frac{1}{A(A − 1)Γ(α)} ∫_{0}^{T}(T − s)^{α−1}f(s, x_{s})ds

+\frac{1}{AΓ(α)} ∫_{0}^{T −r}(T − r − s)^{α−1}f(s, x_{s})ds + \frac{ϕ(−r)}{A}

≥ \frac{1}{A(A − 1)Γ(α)} ∫_{0}^{T}(T − s)^{α−1}f(s, x_{s})ds

≥ \frac{1}{A(A − 1)Γ(α)} ∫_{E} (T − s)^{α−1}v(s)w(x(s − u(s)))ds
≥ \frac{1}{A(A − 1)Γ(α)}T^{α−1} ∫_{E} v(s)w(x(s − u(s)))ds
≥ \frac{T^{α−1}}{Γ(α)} μw(τ)
≥ τ.

Therefore, for every x ∈ ∂P_{τ}, we have ∥F x∥_{J∪I} ≥ ∥x∥_{J∪I} = τ.

Applying Lemma 2.3, we get that L has a fixed point, which means that the boundary value problem (1) − (2) has at least one positive solution x, such that

min{τ, ρ} ≤ ∥x∥_{J∪I} ≤ max{τ, ρ}.

But x is a positive solution of the boundary value problem(1) − (2), this means that x = F x and it is easy to see that x(−r) = F x(−r) ≥ Λ, which implies that

∥x∥_{J∪I} ≥ Λ.

Moreover, it is clear that Λ ≤ ρ. Hence

d ≤ ∥x∥_{J∪I} ≤ max{τ, ρ}.

The proof is complete.

Now, we give the following assumption (H_{6})′, which is similar to assumption (H_{6}), when the function w is nondecreasing. (H_{6})′ There exists an interval E ⊆ I, and functions u : E → [0, r], continuous v : E → [0, +∞) with sup{v(t) : t ∈ E} > 0 and nondecreasing w : [0, +∞) → [0, +∞) such that

f(t, y) ≥ v(t)w(y(−u(t))), (t, y) ∈ E × C^{+}(J).

Theorem 3.3. Suppose that (H_{1}) − (H_{5}), (H_{6})′ hold, and there exists τ > 0 such that

τ ≤ \frac{T^{α−1}}{Γ(α)} μw(0).  (9)

Then the boundary value problem (1) − (2) has at least one positive solution x, such that

d ≤ ∥x∥_{J∪I} ≤ max{τ, ρ},

where d is defined in Theorem 3.2 , ρ is the constant involved in (H_{5}) and τ ≠ ρ. Proof. F : \overline{P}_{max\{τ,ρ\}} → P is a completely continuous map.

As we did in Theorem 3.1, we can prove that F x ≠ λx for every λ ≥ 1 and x ∈ ∂P_{ρ}.

Now we will prove that ∥F x∥_{J∪I} ≥ ∥x∥_{J∪I} for every x ∈ ∂P_{τ} .

As in Theorem 3.2, using (H_{4}) and (H_{6})′, we obtain

F x(−r) ≥ \frac{1}{A(A − 1)Γ(α)} ∫_{0}^{T}(T − s)^{α−1}f(s, x_{s})ds
≥ \frac{1}{A(A − 1)Γ(α)}T^{α−1} ∫_{E} v(s)w(x(s − u(s)))ds
≥ \frac{T^{α−1}}{Γ(α)} μw(0)
≥ τ.

Therefore, for every x ∈ ∂P_{τ}, we have ∥F x∥_{J∪I} ≥ ∥x∥_{J∪I} = τ.

Applying Lemma 2.3, we get that L has at least a fixed point, which means that the boundary value problem (1) − (2) has at least one positive solution x, such that

min{τ, ρ} ≤ ∥x∥_{J∪I} ≤ max{τ, ρ}.

Then

∥x∥_{J∪I} ≥ Λ, Λ ≤ ρ.

So d ≤ ∥x∥_{J∪I} ≤ max{τ, ρ}. The proof is complete. Theorem 3.4. Suppose that (H_{1})−(H_{6}) (respectively (H_{1})−(H_{5}), (H_{6})′) hold, additionally, there exist τ > 0 such that (9) (respectively (10)) holds. Then if ρ < τ, the boundary value problem (1) − (2) has at least two positive solutions x_{1}, x_{2}, such that

\frac{ϕ(0)}{A − 1} ≤ ∥x_{1}∥_{J∪I} < ρ < ∥x_{2}∥_{J∪I} < τ.

Example 3.1. Consider the boundary value problem

D^{\frac{2}{3}} x(t) = (sin t)\sqrt{x_{t}}, t ∈ I := [0, 1],  (10)
5x_{0} − x_{1} = \frac{1}{2},  (11)

where f(t, x_{t}) = (sin t)\sqrt{x_{t}}, for t ∈ I, α = \frac{2}{3}, A = 5 and ϕ(t) = \frac{1}{2}. For any x ∈ C[−\frac{1}{2}, 1], The x_{t} defined by x_{t}(θ) = x(t + θ), −\frac{1}{2} ≤ θ ≤ 0.

For any given l > 0, x ∈ P_{l}, choose m_{l}(t) = \sqrt{l}t and α_{1} = \frac{1}{2} such that m_{l}(t) ∈ L^{\frac{1}{α_{1}}} I. So the assumption (H_{2}) holds and M_{l} = ∥m_{l}(t)∥_{L^{2}I} = \sqrt{\frac{l}{3}}.

Now observe that assumptions (H_{1}), (H_{3}) − (H_{4}) hold. For ρ = 1.27,

\frac{ϕ(0)}{A − 1} + \frac{∥ϕ∥}{A} + \frac{AM_{ρ}}{(A − 1)Γ(α)(1 + β)^{1−α_{1}}} T^{(1+β)(1−α_{1})} = \frac{9}{40} + \frac{5\sqrt{ρ}}{4Γ(\frac{2}{3})} < ρ.

Hence, the condition (H_{5}) holds. Therefore, by applying Theorem 3.1, we can get that the boundary value problem (11)-(12) has at least one positive solution x satisfying 0.125 ≤ ∥x∥_{J∪I} ≤ 1.27.

References

[1] V. Lakshmikantham, S. Leela, J. Vasundhara Devi, Theory of Fractional Dynamic Systems. Cambridge Scientific Publishers, Cambridge, 2009.

[2] A.A. Kilbas, Hari M. Srivastava, J. Juan Trujillo, Theory and Applications of Fractional Differential Equations, Elsevier, Amsterdam, 2006.

[3] I. Podlubny, Fractional Differential Equations, Academic Press, San Diego, 1999.

[4] K.S. Miller, B. Ross, An Introduction to the Fractional Calculus and Differential Equations, John Wiley, New York, 1993.

[5] M. Benchohra, J. Henderson, S. K. Ntouyas and A. Ouahab, Existence results for fractional order functional differential equations with infinite delay, J. Math. Anal. Appl. 338(2008) 1340-1350.

[6] J. Vasundhara Devi and V. Lakshmikantham, Nonsmooth analysis and fractional differential equations, Nonlinear Anal. 70(12)(2009) 4151-4157.

[7] Thabet Maraaba, Dumitru Baleanu and Fahd Jarad, Existence and uniqueness theorem for a class of delay differential equations with left and right Caputo fractional derivatives, J. Math. Phys. 49, (2008) 083507.

[8] V. Lakshmikantham, Theory of fractional functional differential equations, Nonlinear Anal. 69(2008) 3337-3343.

[9] V. Lakshmikantham and A. S. Vatsala, Basic theory of fractional differential equations, Nonlinear Anal. 69(2008) 2677-2682.

[10] Wei Lin, Global existence theory and chaos control of fractional differential equations, J. Math. Anal. Appl. 332(2007) 709-726.

[11] Yong Zhou, Feng Jiao, Jing Li, Existence and uniqueness for p−type fractional neutral differential equations, Nonlinear Anal. 71 (2009) 2724-2733.

[12] Yong Zhou, Feng Jiao, Jing Li, Existence and uniqueness for fractional neutral differential equations with infinite delay, Nonlinear Anal. 71 (2009) 3249-3256.

[13] R. P. Agarwal, M. Benchohra, S. Hamani, A survey on existence results for boundary value problems of nonlinear fractional differential equations, Acta Appl. Math. 109 (2010) 973-1033.

[14] M. Benchohra, S. Hamani and S. K. Ntouyas, Boundary value problems for differential equations with fractional order, Surv. Math. Appl. 3(2008), 1-12.

[15] Zhangbing Bai and Haishen Lü, Positive solutions for boundary value problem of nonlinear fractional differential equation, J. Math. Anal. Appl. 311(2005) 495-505.

[16] Y. K. Chang, J. J. Nieto, Some new existence results for fractional differential inclusions with boundary conditions, Math. Comput. Modelling 49(2009) 605-609.

[17] Moustafa El-Shahed, Positive solutions for boundary value problem of nonlinear fractional differential equation, Abstract and Applied Analysis, (2007) Article ID 10368.

[18] S. Zhang, Positive solutions to singular boundary value problem for nonlinear fractional differential equation, Comput. Math. Appl. 59(3)(2009) 1300-1309.

[19] S. Zhang, Existence of solution for a boundary value problem of fractional order, Acta Mathematica Scientia, 26B(2)(2006) 220-228.

[20] C. F. Li, X. N. Luo and Yong Zhou, Existence of positive solutions of the boundary value problem for nonlinear fractional differential equations, Comput. Math. Appl. 59(3)(2009) 1363-1375.

[21] S. Kasprzyk and J. Myjak, On the existence of solutions of the Floquet problem for ordinary differential equations, Zeszyty Nauk. Uniw. Jagiello. Prace Mat. 13 (1969) 35-39.

[22] K. G. Mavridis and P. Ch. Tsamatos, Positive solutions for a Floquet functional boundary value problem, J. Math. Anal. Appl. 296 (2004) 165-182.

[23] H. Amann, Fixed point equations and nonlinear eigenvalue problems in ordered Banach spaces, SIAM Rev. 18 (1976) 620-709.

[24] K. Deimling, Nonlinear Functional Analysis, Springer-Verlag, New York, 1985.

(Received February 26, 2010)