Existence and Uniqueness of Positive Solution for a Boundary Value Problem of Fractional Order

We are concerned with the existence and uniqueness of positive solutions for the following nonlinear fractional boundary value problem: D α 0 (cid:2) u (cid:3) t (cid:4) (cid:2) f (cid:3) t,u (cid:3) t (cid:4)(cid:4) (cid:5) 0, 0 ≤ t ≤ 1, 3 < α ≤ 4, u (cid:3) 0 (cid:4) (cid:5) u (cid:3) (cid:3) 0 (cid:4) (cid:5) u (cid:3)(cid:3) (cid:3) 0 (cid:4) (cid:5) u (cid:3)(cid:3) (cid:3) 1 (cid:4) (cid:5) 0, where D α 0 (cid:2) denotes the standard Riemann-Liouville fractional derivative. Our analysis relies on a ﬁxed point theorem in partially ordered sets. Some examples are also given to illustrate the results.


Introduction
Differential equations of fractional order occur more frequently in different research and engineering areas such as physics, chemistry, economics, and control of dynamical. Indeed, we can find numerous applications in viscoelasticity, electrochemistry control, porous media, and electromagnetism. see, e.g., 1-7 . For an extensive collection of results about this type of equations we refer the reader to the monograph by Kilbas  On the other hand, some basic theory for the initial value problems of fractional differential equations involving the Riemann-Liouville differential operator has been discussed by Lakshmikantham Since boundary values are nonzero, the Riemann-Liouville fractional derivative D α 0 is not suitable and the author used the Caputo fractional derivative c D α 0 . Motivated by these works, in this paper we discuss the existence and uniqueness of positive solutions for the following nonlinear boundary value problem of fractional order: This problem was studied in 21 , where the authors use lower and upper solution method and the Schauder fixed point theorem which cannot ensure the uniqueness of the solution. The practical relevance of 3 < α ≤ 4 appears in problems related with other areas as physics and economics which can be modeled by these fractional boundary values problems. Particularly, these problems appear in the Hamiltonian formulation for the lagrangians depending on fractional derivatives of coordinates when the systems are nonconservative see, e.g., 7 .
Our main interest in this paper is to give an alternative answer to the main results of the paper 21 .
The main tool used in our study is a fixed point theorem in partially ordered sets which gives us uniqueness of the solution.

Preliminaries and Previous Results
For the convenience of the reader, we present here some definitions, lemmas and basic results that will be used in the proofs of our theorems.
provided that the right-hand side is pointwise defined on 0, ∞ and where Γ α denotes the gamma function.
Definition 2.2. The Riemann-Liouville fractional derivative of order α > 0 of a function f : 0, ∞ → R is given by where n α 1 and α denotes the integer part of α. The following two lemmas can be found in 17, 22 . Lemma 2.3. Let α > 0 and u ∈ C 0, 1 ∩ L 1 0, 1 . Then fractional differential equation for some c i ∈ R (i 1, 2, . . . n) and n α 1 as unique solution.
for some c i ∈ R i 1, . . . , n and n α 1.
Using Lemma 2.4, in 21 the following result is proved.

Lemma 2.5.
Given f ∈ C 0, 1 and f t ≥ 0, the unique nonnegative solution for

4 Abstract and Applied Analysis
In the sequel, we present the fixed-point theorems which we will use later. These results appear in 23 .
Theorem 2.6. Let X, ≤ be a partially ordered set and suppose that there exists a metric d in X such that X, d is a complete 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 ∀n ∈ N.

2.9
Let T : X → X be a nondecreasing mapping such that where ψ : 0, ∞ → 0, ∞ is a continuous and nondecreasing function such that ψ is positive in Moreover, if X, ≤ satisfies the following condition: for x, y ∈ X there exists z ∈ X which is comparable to x and y, 2.11 which appears in 24 , the following result is proved 23 . In our considerations we will work in the Banach space C 0, 1 {x : 0, 1 → R, 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 In 24 it is proved that C 0, 1 , ≤ with the classical metric given by satisfies condition 2.9 of Theorem 2.6. Moreover, for x, y ∈ C 0, 1 , as the function max x, y ∈ C 0, 1 , C 0, 1 , ≤ satisfies condition 2.11 . Finally, by F we denote the class of functions ψ : 0, ∞ → 0, ∞ continuous, nondecreasing, positive in 0, ∞ and ψ 0 0. By J we denote the class of functions ϕ : 0, ∞ → 0, ∞ continuous, nondecreasing, satisfying that I − ϕ ∈ F, where I denotes the identity mapping on 0, ∞ . Abstract and Applied Analysis 5

Main Result
The main result of the paper is the following. H2 There exists t 0 ∈ 0, 1 such that f t 0 , 0 > 0.
H3 There exists 0 < λ ≤ α − 2 Γ α 1 /2 such that, for x, y ∈ 0, ∞ with y ≥ x and t ∈ 0, 1 , Before the proof of Theorem 3.1, we will need some properties of Green's function appearing in Lemma 2.5. Proof. The continuity of G is easily checked. In order to prove the nonnegativness of G t, s , In the case of 0 ≤ s ≤ t ≤ 1 with t / 0, we have

3.4
Taking into account that the function g α t α with α > 0 and t ∈ 0, 1 is decreasing we have

3.5
The last inequality and 3.3 give us G t, s ≥ 0 with t / 0. Finally, notice that G 0, s 0, and this finishes the proof.
we deduce that ϕ t 1 0 G t, s ds is strictly increasing and, consequently,

3.9
In the sequel, we give the proof of Theorem 3.1.

Proof of Theorem 3.1. Consider the cone
Obviously, P is a closed set of C 0, 1 , and, thus, P is a complete metric space with the distance given by d u, v sup t∈ 0,1 {|u t − v t |}. P can be equipped with a partial order defined by x, y ∈ P, x ≤ y ⇐⇒ x t ≤ y t , for t ∈ 0, 1 .

3.11
Using a similar argument to that in 24 , it can be proved that P, ≤ satisfies condition 2.9 of Theorem 2.6. Moreover, as for x, y ∈ P the function max x, y ∈ P , P, ≤ satisfies condition 2.11 .
Abstract and Applied Analysis 7 Now, we consider the operator T defined on P and given by G t, s f s, u s ds, for u ∈ P.

3.12
By H1 and Lemma 3.2, T applies P into itself.
In the sequel we check that T satisfies the assumptions of Theorem 2.6. Firstly, we prove that T is a nondecreasing operator. In fact, by H1 , for u, v ∈ P with u ≥ v and t ∈ 0, 1 , we have Now, we prove that T satisfies the contractive condition appearing in Theorem 2.6. In fact, for u, v ∈ P and u ≥ v, taking into account assumption H3 , we get

3.14
As ψ ∈ J and, thus, ψ is nondecreasing and by Lemma 3.3, from the last inequality we obtain

3.15
Using the fact that λ ≤ 2/ α − 2 Γ α 1 assumption H3 , we have Put ϕ x x − ψ x , As ψ ∈ J, this means that ϕ ∈ F. The last inequality gives us This proves that T satisfies the contractive condition of Theorem 2.6. Finally, as G t, s ≥ 0 Lemma 3.2 and f ≥ 0 assumption H1 , we have where 0 denotes the zero function.

Abstract and Applied Analysis
Now, Theorem 2.6 shows that problem 1.3 has at least one nonnegative solution. As P, ≤ satisfies condition 2.11 , we obtain the uniqueness of the solution.
In what follows, we will prove that this solution is positive this means that x t > 0, for t ∈ 0, 1 .
Finally, we will prove that the zero function is not the solution for problem 1.3 . In fact, in contrary case, the zero function is a fixed point of T and, thus, we have The nonnegative character of the functions G and f and the last expression give us G t, s · f s, 0 0 a.e. s , for t ∈ 0, 1 .

3.20
This and the fact that G t, s / 0 a.e. s for any t ∈ 0, 1 because G t, s is given by a polynomial implies f s, 0 0 a.e. s .

3.21
Taking into account assumption H2 , f t 0 , 0 > 0 for certain t 0 ∈ 0, 1 . By the continuity of f we can find a set A ⊂ 0, 1 with t 0 ∈ A and μ A > 0, where μ is the Lebesgue measure, such that f t, 0 > 0 for t ∈ A. This contradicts 3.21 . This proves that the zero function is not the solution for problem 1.3 . Now, we will prove that the solution x is positive.
In the contrary case, we find 0 < t * < 1 such that x t * 0. As the solution x is a fixed point of the operator T , this means that Since x ∈ P and, thus, x ≥ 0 and by the fact that f is nondecreasing in the second variable and G t, s ≥ 0, we can get and this inequality implies Using a similar reasoning to the one above used we obtain a contradiction. Therefore, x t > 0, for t ∈ 0, 1 . This finishes the proof.
Remark 3.4. In Theorem 3.1, condition H2 seems to be a strong condition in order to obtain a positive solution for problem 1.3 , but when there is uniqueness of solution one will see that this condition is a very adjusted one. More precisely, under the assumption that problem 1.3 has a unique nonnegative solution x t one has f t 0 , 0 > 0 for certain t 0 ∈ 0, 1 if and only if x t is a positive solution.

3.25
In fact, if f t 0 , 0 > 0 for certain t 0 ∈ 0, 1 the argument used in the proof of Theorem 3.1 give us that x t is a positive solution.
For the other implication, suppose that f t, 0 0 for any t ∈ 0, 1 . Under this assumption, our problem 1.3 admits as solutions the function x t and the zero function and this contradicts the hypothesis about uniqueness of solution to problem 1.3 . Therefore, f t 0 , 0 > 0 for certain t 0 ∈ 0, 1 .

Remark 3.5.
Notice that the assumptions in Theorem 3.1 are invariant by additive perturbations. More precisely, if f t, 0 0 for any t ∈ 0, 1 and f satisfies H1 and H3 of Theorem 3.1, then g t, u a t f t, u , with a : 0, 1 → 0, ∞ a nondecreasing continuous function with a t 0 / 0 for certain t 0 ∈ 0, 1 , satisfies H1 , H2 , and H3 of Theorem 3.1 and the following nonlinear boundary value problem of fractional order: has a unique positive solution by Theorem 3.1 .
In the sequel we present an example where the results can be applied.
Besides, for u ≥ v and t ∈ 0, 1 , we have

3.28
A straightforward calculation gives us that ψ x ln 1 x satisfies that ψ ∈ J.