Uniqueness of Positive Solutions for a Perturbed Fractional Differential Equation

We are concerned with the existence and uniqueness of positive solutions for the following nonlinear perturbed fractional two-point boundary value problem:D 0 u t f t, u, u ′, . . . , u n−2 g t 0, 0 < t < 1, n − 1 < α ≤ n, n ≥ 2, u 0 u′ 0 · · · u n−2 0 u n−2 1 0, where D 0 is the standard Riemann-Liouville fractional derivative. Our analysis relies on a fixed-point theorem of generalized concave operators. An example is given to illustrate the main result.


Introduction
In this paper, we are interested in the existence and uniqueness of positive solutions for the following nonlinear perturbed fractional two-point boundary value problem: where D α 0 is the standard Riemann-Liouville fractional derivative and g : 0, 1 → 0, ∞ is continuous.
Fractional differential equations arise in many fields, such as physics, mechanics, chemistry, economics, engineering, and biological sciences; see 1-15 , for example.In recent years, the study of positive solutions for fractional differential equation boundary value problems has attracted considerable attention, and fruits from research into it emerge continuously.For a small sample of such work, we refer the reader to 16-26 and the references therein.On the other hand, the uniqueness of positive solutions for nonlinear fractional differential equation boundary value problems has been studied by some authors; see [19][20][21][22]25 , for example.
When g t ≡ 0, Yang and Chen 22 investigated the existence and uniqueness of positive solutions for the problem 1.1 by means of a fixed-point theorem for u 0 concave operators.They present the following result.
Theorem 1.1 see 22 .Assume that and f is not identically vanishing; Then the problem 1.1 with g t ≡ 0 has a unique positive solution.
In this paper, we will remove the condition H 3 and improve on H 2 .And we will use a fixed-point theorem of generalized concave operators to show the existence and uniqueness of positive solutions for the problem 1.1 .Our main result is summarized in the following theorem.Theorem 1.2.Assume that H 1 holds and H 4 for any c ∈ 0, 1 and y i ≥ 0, i 1, 2, . . ., n − 1, there exists a number ϕ c ∈ c, 1 such that f t, cy 1 , cy 2 , . . ., cy n−1 ≥ ϕ c f t, y 1 , y 2 , . . ., y n−1 .

1.3
Then the problem 1.1 has a unique positive solution is called the Riemann-Liouville fractional integral of order α, where α > 0 and Γ α denotes the gamma function.
Definition 2.2 see 4, page 36-37 .For a function f x given in the interval 0, ∞ , the expression where n α 1, α denotes the integer part of number α, is called the Riemann-Liouville fractional derivative of order α.
By using the same method in 21 , the problem 1.1 can be transformed into the following boundary value problem:

2.3
Moreover, from Lemma 2.5 and Lemma 2.7 in 21 , we can easily obtain the following result. where Here G t, s is called the Green function of the problem 2.3 .Evidently, G t, s ≥ 0 for t, s ∈ 0, 1 .
The following property of the Green function plays important roles in this paper.

Journal of Function Spaces and Applications
Lemma 2.4 see 22 .The Green function G t, s in Lemma 2.3 has the following property:

2.6
In the sequel, we present some basic concepts in ordered Banach spaces for completeness and a fixed-point theorem which we will be used later.For convenience of readers, we suggest that one refers to 27, 28 for details.
Suppose that E, • is a real Banach space which is partially ordered by a cone P ⊂ E, that is, x ≤ y if and only if y − x ∈ P .If x ≤ y and x / y, then we denote x < y or y > x.By θ we denote the zero element of E. Recall that a nonempty closed convex set P ⊂ E is a cone if it satisfies i x ∈ P , λ ≥ 0 ⇒ λx ∈ P ; ii x ∈ P , −x ∈ P ⇒ x θ.
P is called normal if there exists a constant M > 0 such that, for all x, y ∈ E, θ ≤ x ≤ y implies x ≤ M y ; in this case M is called the normality constant of is called the order interval between x 1 and x 2 .For all x, y ∈ E, the notation x ∼ y means that there exist λ > 0 and μ > 0 such that λx ≤ y ≤ μx.Clearly, ∼ is an equivalence relation.Given h > θ i.e., h ≥ θ and h / θ , we denote by P h the set P h {x ∈ E | x ∼ h}.It is easy to see that P h ⊂ P is convex and λP h P h for all λ > 0.
In a recent paper 28 , Zhai et al. considered the following operator equation: x Ax x 0 .

2.7
They established the existence and uniqueness of positive solutions for the above equation, and they present the following interesting result.

Proof of Theorem 1.2
In this section, we apply Theorem 2.5 to study the problem 1.1 , and we obtain a new result on the existence and uniqueness of positive solutions.The method used here is new to the literature and so is the existence and uniqueness result to the fractional differential equations.
In our considerations we will work in the Banach space C 0, 1 {x : 0, 1 → R is continuous} with the standard norm x sup{|x t | : t ∈ 0, 1 }.Notice that this space can be equipped with a partial order given by x, y ∈ C 0, 1 , x ≤ y ⇐⇒ x t ≤ y t for t ∈ 0, 1 .

3.1
Set P {x ∈ C 0, 1 | x t ≥ 0, t ∈ 0, 1 }, the standard cone.It is clear that P is a normal cone in C 0, 1 and the normality constant is 1.

3.2
For any v ∈ P , we define where G t, s is given as in Lemma 2.3.Noting that I n−2 0 v s , I n−3 0 v s , . . ., I 1 0 v s , v s ≥ 0 and G t, s ≥ 0, it follows from H 1 that A : P → P .In the sequel we check that A and x 0 satisfy all assumptions of Theorem 2.5.
Firstly, we prove that A : P → P is an increasing operator.In fact, for v i ∈ P , i 1, 2 with v 1 ≤ v 2 , we know that v 1 t ≤ v 2 t , t ∈ 0, 1 , by the monotonicity of Riemann-Liouville fractional integral I δ 0 , δ > 0 and H 1 ,

3.4
That is Av 1 ≤ Av 2 .Hence, the condition D 1 in Theorem 2.5 is satisfied.
Next we show that the condition D 3 holds.From H 4 , for any γ ∈ 0, 1 and v ∈ P , we obtain

3.5
That is A γv ≥ ϕ γ Av, for all v ∈ P , γ ∈ 0, 1 .So the condition D 3 in Theorem 2.5 is satisfied.Now we show that the condition D 2 is also satisfied.On one hand, it follows from H 1 and Lemma 2.4 that Ah t 1 0 G t, s f s, I n−2 0 h s , I n−3 0 h s , . . ., I 1 0 h s , h s ds . ., I 1 0 h s , h s ds, t ∈ 0, 1 .

3.6
On the other hand, also from H 1 and Lemma 2.4, we obtain 0 h s , I n−3 0 h s , . . ., I 1 0 h s , h s ds.

3.8
Since f is continuous and f / ≡ 0, we can get 0 < λ h ≤ μ h .Consequently, λ h h t ≤ Ah t ≤ μ h h t . 3.9 Next we consider x 0 .If g t ≡ 0, then x 0 t ≡ 0; if g t / ≡ 0, let l max t∈ 0,1 g t , then l > 0. It is easy to prove that 0

3.12
Hence x 0 Ah ∈ P h .Finally, using Theorem 2.5, v Av x 0 has a unique solution v * in P h .That is, v * is a unique positive solution of the problem 2.3 in P h .So there are μ where a, g : 0, 1 → 0, ∞ are continuous with a / ≡ 0.

3.14
Hence, all the conditions of Theorem 1.2 are satisfied.An application of Theorem 1.2 implies that the problem 3.13 has a unique positive solution.
3. Some examples of ϕ t which satisfy the condition H 4 are Definition 2.1 see 4, Definition 2.1 .The integral the solution of the problem 1.1 .Evidently, u * ∈ C 0, 1 , 0, ∞ is a unique positive solution of the problem 1.1 .Remark 3.1.Let f ≡ C > 0. Then the conditions H 1 , H 4 are satisfied and the problem 2.3 has a unique solution v t 1 0 G t, s C g t ds, t ∈ 0, 1 .From Lemma 2.4, the unique solution v is a positive solution and satisfies v ∈ P h P t α−n 1 1−t .So u I n−2 0 v is a unique positive solution of the problem 1.1 .To illustrate how our main result can be used in practice we present an example.