By means of the fixed-point theorem in the cone of strict-set-contraction operators, we consider the existence of a nonlinear multi-point boundary value problem of fractional integro-differential equation in a Banach space. In addition, an example to illustrate the main results is given.

1. Introduction

The purpose of this paper is to establish the existence results of positive solution to nonlinear fractional boundary value problem in a Banach space , where is the zero element of , and ,  ,   is Riemann-Liouville fractional derivative, and where ,  , and denotes the set of all nonnegative numbers.

Fractional differential equations have gained importance due to their numerous applications in many fields of science and engineering including fluid flow, rheology, diffusive transport akin to diffusion, electrical networks, and probability. For details see [13] and the references therein. In recent years, there are some papers dealing with the existence of the solutions of initial value problems or linear boundary value problems for fractional differential equations by means of techniques of nonlinear analysis (fixed-point theorems, Leray-Schauder theory, lower and upper solutions method, and so forth), see for example, [423].

In [8], by means of the fixed-point theorem for the mixed monotone operator, the authors considers unique existence of positive to singular boundary value problems for fractional differential equation where is Riemann-Liouville fractional derivative of order .

In [11], El-Shahed and Nieto study the existence of nontrivial solutions for a multi-point boundary value problem for fractional differential equations where ,  , and   is Riemann-Liouville fractional derivative. Under certain growth conditions on the nonlinearity, several sufficient conditions for the existence of nontrivial solution are obtained by using Leray-Schauder nonlinear alternative. And then, Goodrich [24] was concerned with a partial extension of the problem (1.3) and the authors derived the Green function for the problem (1.5) and showed that it satisfies certain properties.

By the contraction mapping principle and the Krasnoselskii’s fixed-point theorem, Zhou and Chu [13] discussed the existence and uniqueness results for the following fractional differential equation with multi-point boundary conditions: where is the Caputo’s fractional derivative, , and are real numbers, , and .

In [20], Stanĕk has discussed the existence of positive solutions for the singular fractional boundary value problem

However, to the best of the author’s knowledge, a few papers can be found in the literature dealing with the existence of solutions to boundary value problems of fractional differential equations in Banach spaces. In [25], Salem investigated the existence of Pseudo solutions for the following nonlinear -point boundary value problem of fractional type in a reflexive Banach space , where is the Pseudo fractional differential operator of order ,  .

In [26], by the monotone iterative technique and mönch fixed-point theorem, Lv et al. investigated the existence of solution to the following Cauchy problems for differential equation with fractional order in a real Banach space where is the Caputo’s derivative order, .

By means of Darbo’s fixed-point theorem, Su [27] has established the existence result of solutions to the following boundary value problem of fractional differential equation on unbounded domain in a Banach space .   is the Riemann-Liouville fractional derivative.

Motivated by the above mentioned papers [8, 13, 24, 25, 27, 28] but taking a quite different method from that in [2629]. By using fixed-point theorem for strict-set-contraction operators and introducing a new cone , we obtain the existence of at least two positive solutions for the BVP (1.1) under certain conditions on the nonlinear term in Banach spaces. Our results are different from those of [8, 13, 24, 25, 28, 30]. Note that the nonlinear term depends on and its derivatives .

2. Preliminaries and Lemmas

Let the real Banach space with norm be partially ordered by a cone of ; that is, if and only if ; and is said to be normal if there exists a positive constant such that implies , where the smallest is called the normal constant of . For details on cone theory, see [31].

The basic space used in this paper is . For any , evidently, is a Banach space with norm , and for is a cone of the Banach space .

Definition 2.1 (see [31]). Let be a bounded set in a real Banach space , and , all the diameters of . Clearly, .   is called the Kuratovski measure of noncompactness.
We use ,   to denote the Kuratowski noncompactness measure of bounded sets in the spaces , respectively.

Definition 2.2 (see [31]). Let be real Banach spaces, .   is a continuous and bounded operator. If there exists a constant , such that , then is called a -set contraction operator. When ,   is called a strict-set-contraction operator.

Lemma 2.3 (see [31]). If is bounded and equicontinuous, then is continuous on and where .

Definition 2.4 (see [2, 3]). The left-sided Riemann-Liouville fractional integral of order of a function is given by

Definition 2.5 (see [2, 3]). The fractional derivative of order of a function is given by where denotes the integer part of number , provided that the right side is pointwise defined on .

Lemma 2.6 (see [2, 3]). Let . Then the fractional differential equation has a unique solution ; here .

Lemma 2.7 (see [2, 3]). Let . Then the following equality holds for , for some ,  ; here is the smallest integer greater than or equal to .

Lemma 2.8 (see [31]). Let be a cone in a Banach space . Assume that are open subsets of with ,  . If is a strict-set-contraction operator such that either: (i),  , and ,  , or (ii),  , and ,  ,then has a fixed point in .

3. Main Results

For convenience, we list some following assumptions. There exist and such that , for any ,   is uniformly continuous on and there exist nonnegative constants ,  , with such that where ,  ,  .

Lemma 3.1. Given and hold. Then the unique solution of is where

Proof. Deduced from Lemma 2.7, we have for some . Consequently, the general solution of (3.4) is By boundary value conditions , there is , and Therefore, the solution of problem (3.4) is The proof is complete.

Moreover, there is one paper [8] in which the following statement has been shown.

Lemma 3.2. The function defined in (3.7) satisfying the following properties: (1) is continuous on , and for all ;(2)there exists a positive function such that , where and where is the solution of For our purpose, one assumes that and ,  , where ,   are the constants in (2) of Lemma 3.2.

Remark 3.3. We note that if holds, then the function defined in (3.4) is satisfying the following properties: (i)   is continuous on , and ,   for all ,  ,   where ;(ii) for all ,  , where Indeed, it is obvious from (1) of Lemma 3.2 and (3.6) that

Lemma 3.4. Let ,  . Then the problem (1.1) can be transformed into the following modified problem: where ,  ,  . Moreover, if is a solutions of problem (3.16), then the function is a solution of (1.1).

The proof follows by routine calculations.

To obtain a positive solution, we construct a cone by where ,  .


Lemma 3.5. Assume that (H1)–(H3) hold. Then is a strict-set-contraction operator.

Proof. Let . Then, it follows from Remark 3.3. that here, by , we know that and .
From (3.6) and (3.18), we obtain which implies that ;  that is,  .
Next, we prove that is continuous on . Let , and . Hence is a bounded subset of . Thus, there exists such that and . It is clear that According to the properties of ,  for all  , there exists such that for ,  for all  .
Therefore,  for all  , for any and , we get This implies that is continuous on .
By the properties of continuous of , it is easy to see that is equicontinuous on .
Finally, we are going to show that is a strict-set-contraction operator. Let be bounded. Then by condition (H1), Lemma 3.1 implies that . It follows from (3.18) that which implies Obviously, Using a similar method as in the proof of Theorem   in [31], we have Therefore, it follows from (3.26)–(3.29) that Noticing that (3.3), we obtain that is a strict-set-contraction operator. The proof is complete.

Theorem 3.6. Let cone be normal and conditions hold. In addition, assume that the following conditions are satisfied. There exist ,   and such that There exist ,  , and such that There exists a such that where . Then problem (1.1) has at least two positive solutions.

Proof. Consider condition (H4), there exists an , such that Therefore, Take Then for ,  , we have, by (3.18), Hence, and consequently, Similarly, by condition (H5), there exists , such that where is given in (2) of Lemma 3.2. Therefore, Choose Then for ,  ,  , we have which implies that is,
On the other hand, according to (ii) of Remark 3.3 and (3.18), we get By condition (H1), for ,  ,  , we have Therefore,
Applying Lemma 2.7 to (3.39), (3.45), and (3.48) yields that has a fixed-point , and a fixed-point . Noticing (3.48), we get and . This and Lemma 3.4 complete the proof.

Theorem 3.7. Let cone be normal and conditions hold. In addition, assume that the following condition is satisfied: Then problem (1.1) has at least one positive solution.

Proof. By (H4), we can choose . As in the proof of Theorem 3.6, it is easy to see that (3.39) holds. On the other hand, considering (3.49), there exists such that where satisfies Choose . Then for ,  ,  , it follows from (3.46) that and consequently, Since , applying Lemma 2.7 to (3.39) and (3.53) yield that has a fixed-point . This and Lemma 3.4 complete the proof.

4. An Example

Consider the following system of scalar differential equations of fractional order

Conclusion. The problem (4.1) has at least two positive solutions.

Proof. Let with the norm , and . Then is a normal cone in with normal constant , and system (4.1) can be regarded as a boundary value problem of the form (1.1). In this situation, , in which Observing the inequality , we get, by (4.2), Hence (H1) is satisfied for and
Now, we check condition (H2). Obviously, , for any , and is uniformly continuous on . Let , where and , in which For any and bounded subsets , from (4.5) and by the diagonal method, we have It follows from (4.6) that that is, condition (H2) holds for .
On the other hand, take . Then , which implies that condition (H3) holds. By (4.2), we have