Uniqueness of positive solutions for boundary value problems of singular fractional differential equations

Abstract

In this article, we study the existence and uniqueness of a positive solution for the singular nonlinear fractional differential equation boundary value problem where 3 < α ≤ 4 is a real number, is the Riemann–Liouville fractional derivative and f : (0, 1] × [0, +∞) → [0, +∞) is continuous, (i.e. f is singular at t = 0). Our analysis relies on a fixed point theorem in partially ordered sets. As an application, an example is presented to illustrate the main results.

Keywords:

• singular fractional differential equation
• boundary value problem
• positive solution
• fractional Green's function
• fixed point

AMS Subject Classifications::

• 34A08
• 34B16

1. Introduction

Fractional differential equations have been of great interest recently. It is caused both by the intensive development of the theory of fractional calculus itself and by the applications, see 1–5. It should be noted that most papers and books on fractional calculus are devoted to the solvability of linear fractional differential equations in terms of special functions.

Recently, there are some papers dealing with the existence of solutions (or positive solutions) of nonlinear initial fractional differential equation by the use of techniques of nonlinear analysis (fixed-point theorems, Leray–Schauder theory, Adomian decomposition method, etc.), see 6–11.

Delbosco and Rodino 6 considered the existence of a solution for the nonlinear fractional differential equation where 0 < t < 1, and f : [0, a] × ℝ → ℝ is a given function, continuous in (0, a) × ℝ → ℝ. They obtained results for solutions by using the Schauder fixed theorem and the Banach contraction principle.

Zhang 10 studied the existence of positive solution for equation where 0 < t < 1, and f : [0, 1] × [0, +∞) → [0, +∞) is a given continuous function, by using the upper and lower solution methods.

In fact, there are the same requirements for boundary conditions, see 12–23. However, there are few papers which have considered the singular boundary value problems of fractional differential equations, see 17–23.

Xu et al. 22 considered the existence of positive solutions for the following problem: where 3 < α ≤ 4 is a real number and is the Riemann–Liouville fractional differentiation. Using Leray–Schauder nonlinear alternative and a fixed point theorem on cones, they obtained some new existence criteria for singular problem when the right side function f(t, u) was singular at u = 0 (i.e. ), where f ∈ C([0, 1] × (0, +∞), [0, +∞)).

Existence of fixed point in partially ordered sets has been considered recently in 23–29. Caballero et al. 23 discussed the existence and uniqueness of a positive and nondecreasing solution to boundary value problem of the nonlinear fractional differential equation where 2 < α ≤ 3 is a real number and is Caputo's fractional derivative, and f : (0, 1] × [0, +∞) → [0, +∞) is continuous, (i.e. f is singular at t = 0).

From the above-mentioned works, we investigate the existence and uniqueness of a positive solution for the singular nonlinear fractional differential equation boundary value problem (1) (2) where 3 < α ≤ 4 is a real number, is Riemann–Liouville fractional derivative and f : (0, 1] × [0, +∞) → [0, +∞) is continuous, (i.e. f is singular at t = 0). In this article, by using a fixed point theorem in partially ordered sets, existence and uniqueness of a positive solution for this boundary value problem is given.

2. Preliminaries

For the convenience of the reader, we give some background materials from fractional calculus theory to facilitate analysis of problem (1.1) and (1.2). These materials can be found in the recent literature, see 22,24,30.

Definition 2.1 30

The Riemann–Liouville fractional derivative of order α > 0 of a continuous function f : (0, +∞) → ℝ is given by where n = [α] + 1, [α] denotes the integer part of number α, provided that the right side is pointwise defined on (0, +∞).

Definition 2.2 30

The Riemann–Liouville fractional integral of order α > 0 of a function f : (0, +∞) → ℝ is given by provided that the right side is pointwise defined on (0, +∞).

From the definition of the Riemann–Liouville derivative, we can obtain the following statement.

Lemma 2.1 22

Let α > 0. If we assume u ∈ C(0, 1) ∩ L(0, 1), then the fractional differential equation has u(t) = c₁t^α−1 + c₂t^α−2 + ··· + c_Nt^α−N, c_i ∈ ℝ, i = 1, 2, … , N, as unique solutions, where N is the smallest integer greater than or equal to α.

Lemma 2.2 22

Assume that u ∈ C(0, 1) ∩ L(0, 1) with a fractional derivative of order α > 0 that belongs to C(0, 1) ∩ L(0, 1). Then where N is the smallest integer greater than or equal to α.

In the following, we present the Green function of fractional differential equation boundary value problem.

Lemma 2.3 22

Let h ∈ C[0, 1] and 3 < α ≤ 4. The unique solution of problem (3) (4) is where (5) Here G(t, s) is called the Green function of boundary value problem (2.1) and (2.2).

The following properties of the Green function play important roles in this article.

Lemma 2.4 22

The function G(t, s) defined by (2.3) satisfies the following conditions:

1.	G(t, s) = G(1 − s, 1 − t), for t, s ∈ (0, 1),
2.	(α − 2)tα−2(1 − t)2s2(1 − s)α−2 ≤ Γ(α)G(t, s) ≤ M0s2(1 − s)α−2, for t, s ∈ (0, 1),
3.	G(t, s) > 0, for t, s ∈ (0, 1),
4.	(α − 2)s2(1 − s)α−2tα−2(1 − t)2 ≤ Γ(α)G(t, s) ≤ M0tα−2(1 − t)2, for t, s ∈ (0, 1),

here M₀ = max{α − 1, (α − 2)²}.

The following two lemmas are fundamental in the proofs of our main results.

Lemma 2.5 24

Let (X, ≤) be a partially ordered set and suppose that there exists a metric space. Assume that X satisfies the following condition: if {x_n} is a nondecreasing sequence in X such that x_n → x then x_n ≤ x for all n ∈ ℕ. Let T : X → X be a nondecreasing mapping such that (6) where ϕ : [0, +∞) → [0, +∞) is continuous and nondecreasing function such that ϕ is positive in (0, +∞) and ϕ(0) = 0. If there exists x₀ ∈ X with x₀ ≤ T(x₀), then T has a fixed point.

If we consider that (X, ≤) satisfies the following condition: (7) then we have the following lemma in 24.

Lemma 2.6 24

Adding condition (2.5) to the hypotheses of Lemma 2.5, one obtains uniqueness of the fixed point of T.

For convenience, we set q(t) = t^α−2(1 − t)², k(s) = s²(1 − s)^α−2, then

3. Main results

In this section, we establish the existence and uniqueness of a positive solution for boundary value problem (1.1) and (1.2).

Theorem 3.1

Let 0 < σ < 1, 3 < α ≤ 4, F : (0, 1] → ℝ is continuous and . Suppose that t^σF(t) is continuous function on [0, 1]. Then the function is continuous on [0, 1].

Proof

By the continuity of t^σF(t) and It is easy to check that H(0) = 0. The proof is divided into three cases:

Case 1 t₀ = 0 ∀t ∈ (0, 1].

Since t^σF(t) is continuous in [0, 1], there exists a constant M > 0, such that , for t ∈ [0, 1]. Hence, where B(·, ·) denotes the beta function.

Case 2 t₀ ∈ (0, 1) ∀t ∈ (t₀, 1].

Case 3 t₀ ∈ (0, 1] ∀t ∈ [0, t₀) The proof is similar to that of Case 2, so we omit it.

From the above, for ∀ε > 0, t, t₀ ∈ [0, 1], there exists δ > 0 such that |t − t₀| < δ, we have |H(t) − H(t₀)| < ε. Thus, the function is continuous on [0, 1].

The proof is complete.

Let Banach space E = C[0, 1] be endowed with the norm ‖u‖ = max_0≤t≤1|u(t)|. Note that this space can be equipped with a partial order given by (8) In 27, it is proved that (E, ≤) with the classic metric given by (9) satisfies condition (2.4) of Lemma 2.5. Moreover, for x, y ∈ E, as the function max{x, y} is continuous in [0, 1], (E, ≤) satisfies condition (2.5).

Define the cone P ⊂ E by Note that, as P is a closed subset of E, P is a complete metric space.

Suppose that u is a solution of boundary value problem (1.1) and (1.2). Then We define an operator A : P → E as follows By Theorem 3.1, Au ∈ C[0, 1]. Moreover, taking into account Lemma 2.4 and as t^σf(t, u) ≥ 0 for (t, u) ∈ [0, 1] × [0, +∞) by hypothesis, we get Hence, A(P) ⊂ P.

Theorem 3.2

Let 0 < σ < 1, 3 < α ≤ 4, f : (0, 1] × [0, +∞) → [0, +∞) is continuous and , t^σf(t, u) is continuous function on [0, 1] × [0, +∞). Assume that there exists 0 < λ ≤ (1 − σ)Γ(α − σ + 1)/(2Γ(3 − σ)) such that for u, v ∈ [0, +∞) with u ≥ v and t ∈ [0, 1], where φ : [0, +∞) → [0, +∞) is continuous and nondecreasing, ϕ(u) = u − φ(u) satisfies

a.	ϕ : [0, +∞) → [0, +∞) and nondecreasing,
b.	ϕ(0) = 0,
c.	ϕ is positive in (0, +∞).

Then the boundary value problem (1.1) and (1.2) has an unique positive solution.

Proof

Firstly, the operator A is nondecreasing since, by hypothesis, for u ≥ v, Besides, for u ≥ v, As the function φ(u) is nondecreasing then, for u ≥ v, and from the last inequality we get Assume that ϕ(u) = u − φ(u) and ϕ : [0, +∞) → [0, +∞) is continuous, nondecreasing, positive in (0, +∞) and ϕ(0) = 0. Then, for u ≥ v, d(Au, Av) ≤ d(u, v) − ϕ(d(u, v)). Finally, take into account that for the zero function, A0 ≥ 0, by Lemma 2.5, the boundary value problem (1.1) and (1.2) has at least one positive solution. Moreover, this solution is unique since (P, ≤) satisfies condition (2.5) and Lemma 2.6. This completes the proof.

In the following, we present an example which illustrates Theorem 3.2.

Example 3.1

Consider the following singular nonlinear fractional differential equation boundary value problem (10) (11) In this case, for (t, u) ∈ (0, 1] × [0, +∞). Note that f is continuous in (0, 1] × [0, +∞) and . Let σ = 1/2, φ(u − v) = ln(u − v + 1). Moreover, for u ≥ v and t ∈ [0, 1], we have because g(x) = ln(2 + x) is nondecreasing on [0, +∞), and Note that By Theorem 3.2, the boundary value problem (3.3) and (3.4) has one positive solution.

Acknowledgements

The authors sincerely thank the referees for their constructive suggestions which improved the content of this article. This research is supported by the Natural Science Foundation of China (11071143, 60904024, 11026112), China Postdoctoral Science Foundation funded project (200902564), the Shandong Provincial Natural Science Foundation (ZR2010AL002, ZR2009AL003, Y2008A28), University of Jinan Research Funds for Doctors (XBS0843) and Innovation Funds for Graduate Students of University of Jinan (YCX09014).