# Existence results of positive solutions for boundary value problems of fractional differential equations

Guoqing Chai

Author Affiliations

College of Mathematics and Statistics, Hubei Normal University, Hubei, 435002, P.R. China

Boundary Value Problems 2013, 2013:109  doi:10.1186/1687-2770-2013-109

 Received: 15 October 2012 Accepted: 15 April 2013 Published: 29 April 2013

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

### Abstract

In this paper, we are concerned with the following fractional equation:

with the boundary value conditions

where is the standard Caputo derivative with and δ, γ are constants with , . By applying a new fixed point theorem on cone and Krasnoselskii’s fixed point theorem, some existence results of positive solution are obtained.

MSC: 34A08, 34B15, 34B18.

##### Keywords:
fractional differential equations; existence results; fixed point theorem; positive solution

### 1 Introduction

In this paper, we are concerned with the existence of positive solutions for the fractional equation

(1.1)

with the boundary value conditions

(1.2)

where is the standard Caputo derivative with and δ, γ are constants with , .

Differential equations of fractional order have recently proved to be valuable tools in the modeling of many phenomena in various fields of science and engineering. Indeed, we can find numerous applications in viscoelasticity, electrochemistry, control, porous media, electromagnetism, etc. (see [1-5]). There has been a significant development in the study of fractional differential equations and inclusions in recent years, see the monographs of Podlubny [5], Kilbas et al.[6], Lakshmikantham et al.[7], Samko et al.[8], Diethelm [9], and the survey by Agarwal et al.[10]. For some recent contributions on fractional differential equations, see [9-25] and the references therein.

On the other hand, it is well known that the fourth-order boundary value problem describes the deformations of an elastic beam in equilibrium state. Owing to its importance in physics, the existence of solutions to this problem has been studied by many authors; see, for example, [26-30] and references therein. Recently, there have been a few papers dealing with the existence of solutions for fractional equations of order .

In [14], Xu et al. discussed the problem

where and is nonnegative, is the Riemann-Liouville fractional derivative of order α. The existence results of positive solutions are obtained by applying the Leray-Schauder nonlinear alternative theorem.

In [15], Liang and Zhang studied the following nonlinear fractional boundary value problem:

where , is nondecreasing relative to u, is the Riemann-Liouville fractional derivative of order α. By means of the lower and upper solution method and fixed point theorems, some results on the existence of positive solutions were obtained.

In [16], Agarwal and Ahmad studied the solvability of the following anti-periodic boundary value problem for a nonlinear fractional differential equation:

where . The existence results were obtained by the nonlinear alternative theorem.

Inspired by above work, the author will be concerned with the boundary value problem (BVP for short in the sequel) (1.1)-(1.2). To the best of our knowledge, no contribution exists concerning the existence of solutions for BVP (1.1)-(1.2). In the present paper, by applying a new fixed point theorem on cone and Krasnoselskii’s fixed point theorem, some existence results of positive solution for BVP (1.1)-(1.2) are obtained. It is worth to point out that the results in this paper are also new even for relative to the corresponding literature with regard to the fourth-order boundary value problem. In addition, the conditions imposed in this paper are easily verified.

The organization of this paper is as follows. In Section 2, we present some necessary definitions and preliminary results that will be used to prove our main results. In Section 3, we put forward and prove our main results. Finally, we give two examples to demonstrate our main results.

### 2 Preliminaries

In this section, we introduce some preliminary facts which are useful throughout this paper.

Let ℕ be the set of positive integers, ℝ be the set of real numbers, , and . Let . Denote by the Banach space endowed with the norm , where for .

Definition 2.1[6]

The Riemann-Liouville fractional integral of order of a function is given by

Definition 2.2[6]

The Riemann-Liouville fractional derivative of order of a function is given by

where , denotes the integer part of α.

Definition 2.3[6]

The Caputo fractional derivative of order of a function y on is defined via the above Riemann-Liouville derivatives by

Lemma 2.1[6]

Letand. Then

holds on.

Lemma 2.2[17]

Letwith, . Ifand, then

holds on.

For convenience, we first list some hypotheses which will be used throughout this paper.

() .

() , , .

For , consider the following BVP:

(2.1)

(2.2)

We have the following lemma, which is important in this paper.

Lemma 2.3Let () hold. Thenis a solution of BVP (2.1)-(2.2) iffhas the expression as follows:

(2.3)

where

(2.4)

and

(2.5)

(2.6)

Proof Let be a solution of (2.1)-(2.2). Then by Lemma 2.2, we have

(2.7)

and so

Thus, by the boundary value condition (2.2), we can obtain

(2.8)

(2.9)

(2.10)

(2.11)

From (2.11), we have

(2.12)

Substituting (2.12) into (2.10), we get

(2.13)

So, by (2.13), (2.12), and (2.9), we have

(2.14)

Thus, from (2.8), we have

(2.15)

Hence, from (2.7) together with (2.12)-(2.15), it follows that

(2.16)

Noticing that

by Definition 2.1, we have

Conversely, if u has the expression (2.3), then from the fact that , we can easily verify that

(2.17)

(2.18)

(2.19)

hold for , and u satisfies the boundary condition (2.2).

Again, from (2.16) and Lemma 2.1, we have that , . In addition, noting that , it is easy to see that from (2.19). □

For the forthcoming analysis, we need to introduce some new notations.

Let , and . Denote

It is easy to verify that , and noting that , , .

We also need the following lemma, which will play an important role in obtaining our main results in Section 3.

Lemma 2.4Under the assumption (), Green’s functionhas the following properties:

(1) Gis continuous on;

(2) , ; , , ;

(3) , ; , , ;

, , ; , , , .

Proof (1) Observing the expression of Green’s function given by (2.4)-(2.6), the conclusion (1) of Lemma 2.4 is obvious.

(2) We first show that , , .

In fact, if , then by (2.5) we have

(2.20)

Owing to the fact that and , we have that . Thus, we immediately obtain that for from (2.20) together with the condition , .

Similarly, we can deduce that

(2.21)

for , and so for .

To summarize, for all with .

Now, since for , and for with , it follows that for all .

(3) The proof is divided into four steps.

Step 1. We show that for .

(i) If , then by (2.5) and the assumption that , , and , we have

(2.22)

(ii) If , then by an argument similar to (2.22), we have

Summing up the above analysis (i)-(ii), we obtain

Step 2. We show that for and .

In fact, if and , then by (2.5) combined with the assumption that , , , we have

(2.23)

where , because .

If with , then by an argument similar to (2.23), we have

(2.24)

So, by (2.23)-(2.24), we have

for , and .

Step 3. Now, we show that

(i) If , then by (2.20) and keeping in mind that , , , it follows that

(2.25)

(ii) If , then by an argument similar to (2.25), from (2.21), we have

Summing up the above analysis (i)-(ii), and noting Step 2 of the proof as before, it follows that

Step 4. It remains to show that

(i) If , then by (2.20) and the fact that , we know that the relations

hold for , with , where , and

Similarly, we can obtain that

The proof is complete. □

Now, we introduce a cone as follows:

It is easy to check that the above set P is a cone in the space , which will be used in the sequel.

We define an operator T on P as follows:

(2.26)

Obviously, under the assumption ()-(), the operator T is well defined. Moreover,

(2.27)

where , , , , and , are given by (2.20)-(2.21), respectively.

A function is a positive solution of BVP (1.1)-(1.2) if , , , and u satisfies BVP (1.1)-(1.2).

By Lemma 2.3, it is easy to know that a function is a positive solution of BVP (1.1)-(1.2) iff is a nonzero fixed point of T. So, we can focus on seeking the existence of a nonzero fixed point of T in P.

Finally, for the remainder of this section, we give the following two theorems, which are fundamental in the proof of our main results.

Let X be a Banach space, and let be a cone. Suppose that the functions α, β satisfy the following condition:

(D) are continuous convex functionals satisfying , for , ; for , and for with , where is a constant.

Lemma 2.5[31]

Assume that, , Lare constants with, , and

Set. Suppose thatis a completely continuous operator satisfying

() , ; , ;

() , ;

() there is asuch thatandfor alland.

ThenThas at least one fixed point in.

Lemma 2.6[32]

Assume that, are two open subsets ofXwith, and letbe a completely continuous operator such that either

(i) , ; , , or

(ii) , ; , .

ThenThas a fixed point in.

### 3 Main results

We first prove the following lemma to obtain our main results.

Lemma 3.1Suppose that ()-() hold. Then the operatorTdefined by (2.26) mapsPintoP, andTis completely continuous.

Proof It is well known that the norms and are equivalent on . So, we can consider that the Banach space is equipped with the norm in the following proof.

For any , in view of the conclusion (1)-(2) of Lemma 2.4 and the hypotheses ()-(), it is easy to see that , , , and , observing (2.26)-(2.27). Moreover, the conclusion (3) of Lemma 2.4 implies that

(3.1)

and

(3.2)

for .

From (3.1)-(3.2), it follows that , , . Thus,

Similarly, we can obtain

Now, we show that the operator T is compact on P.

In fact, let U be an arbitrary bounded set in P. Then there exists a positive number L such that for all , and so such that , for all .

In terms of Lemma 2.4, it follows from (2.26)-(2.27) that

(3.3)

(3.4)

Because the functions and are integrable on I, the formulae (3.3)-(3.4) yield that , , where . So, . That is, TU is uniformly bounded.

On the other hand, for any with , by setting , the formula (2.26) implies that

(3.5)

According to (2.4)-(2.5) and by applying the mean value theorem, we have

and so

(3.6)

where .

Similarly, there is another constant such that

(3.7)

Again, because the function is integrable on I, the absolute continuity of integral of on ensures that there exists a constant such that

(3.8)

So, (3.5) together with (3.6)-(3.8) implies that there exists a constant N such that the inequality

holds for any and with . That is, the set TU is equicontinuous.

Similarly, we can deduce that the set is also equicontinuous in terms of (2.27).

So, as a consequence of the Arzelà-Ascoli theorem, we have that TU is a compact set.

Now, we come to prove the operator T is continuous on P.

Let be an arbitrary sequence in P with . Then there exists an such that

According to the uniform continuity of f on , for an arbitrary number , there is a number such that

(3.9)

for all , whenever .

Thus, in view of Lemma 2.4, from (2.26)-(2.27) and (3.9), it follows that

and

whenever . That is, T is continuous on P. □

We are now in a position to state and prove the first theorem in the article. Let constants , satisfy , .

Theorem 3.1Suppose that ()-() hold. In addition, there are two constants, withsuch thatfsatisfies the following condition:

() , for;

, for.

Then BVP (1.1)-(1.2) has at least one positive solutionusatisfyingand.

Proof We already know that is completely continuous by Lemma 3.1.

Let , for . It is easy to verify that the functions α, β satisfy the condition (D).

Choose a constant L large enough so that , where , and , . Set , , . Define the function on as , , where , .

Consider the following ancillary BVP:

(3.10)

Obviously, the function is continuous on according to the continuity of f. Thus, by an argument similar to that in Lemma 3.1, the operator given by is also completely continuous on P and maps P into P.

We will prove that T has at least one nonzero fixed point in P by applying Lemma 2.5. The approach is divided into four steps.

Step 1. We first show that

(3.11)

In fact, for any , owing to the condition , we have that , , and so .

On the other hand, applying the mean value theorem, for any , we have that for some . Therefore, we have that from the fact that , because . So,

keeping in mind that , for .

Step 2. Now, we come to verify that the conditions corresponding to () in Lemma 2.5 hold.

For any with , we have that , , , and . Thus, in view of (3.11), we have that , , . So, according to (), we have

Thus, from (2.27) and Lemma 2.4, it follows that

Thus, , noting that the assumption . That is, .

For any with , then , , , and . Thus, from (3.11), we obtain

and so

from the condition (). Therefore, in view of Lemma 2.4, we have

Thus, , noting that . That is, .

Step 3. We verify that the conditions corresponding to () in Lemma 2.5 hold.

For any , owing to the fact that , , from the meaning of M, we have immediately that

Thus,

Hence, , and so from the choice of L. Thus, .

Step 4. Finally, take with . Then, by an argument similar to that in Lemma 3.1, we can know that . Moreover, , , and , from (3.11). Thus, , . Again,

for any , .

So, the conditions corresponding to () in Lemma 2.5 hold.

Summing up the above steps 1-4 and applying Lemma 2.5, we obtain that BVP (3.10) has at least one positive solution . That is, . , and so from the fact that and by (3.13). Thus, , , , , and . Hence, , , and so u is a positive solution of BVP (1.1)-(1.2). The proof is complete. □

Now, we state another theorem in this paper. Let us begin with introducing some notations.

Let . Denote with , and with . Put , , where , , , and () are given in Lemma 2.4.

Theorem 3.2Assume that ()-() hold. If, , then BVP (1.1)-(1.2) has at least one positive solution.

Proof As described in the proof of Theorem 3.1, is completely continuous. Again, from , it follows that there exists an such that

(3.12)

holds when with , .

Take . Set . Now, we show that the following relation holds:

(3.13)

In fact, for any , we have that with . Then

Thus,

(3.14)

Thus, from (3.12), (3.14), it follows that

(3.15)

Hence, (2.26) together with (3.15) implies that

So,

(3.16)

Similarly, we can obtain

(3.17)

Therefore, noting that , from (3.16)-(3.17), we have

So, the relation (3.13) holds.

Now, from , it follows that there exists an such that

(3.18)

whenever with , .

Take . Set . We prove the following relation holds:

(3.19)

In fact, for any , we have and . Thus, , , , and , . So, by (3.18), it follows that

Therefore, from (2.26) and in view of Lemma 2.4, we have

(3.20)

So,

(3.21)

Similarly, we can obtain

(3.22)

Consequently, noting that , from (3.21)-(3.22), it follows that

So, the relation (3.19) holds.

Summing up (3.13) and (3.19), applying Lemma 2.6, the operator T has at least one fixed point . Thus u is a positive solution of BVP (1.1)-(1.2). The proof is complete. □

Example 3.1 Consider the following BVP:

(3.23)

where , , , and f is given by

where constants , are two positive numbers. Then BVP (3.23) has at least one positive solution.

In fact, assume that the notations , , , and are described in Theorem 3.1. Take , . Then the inequality

holds for , and the inequality

holds for .

So, by Theorem 3.1, BVP (3.23) has at least one positive solution.

Example 3.2 Consider the following BVP:

(3.24)

where , , , and f is given by

where constants , are two positive numbers and a constant . Then BVP (3.24) has at least one positive solution.

In fact, observing that , , the conclusion follows from Theorem 3.2.

### Competing interests

The author declares that he has no competing interests.

### Acknowledgements

The author sincerely thanks the anonymous referees for their valuable suggestions and comments which have greatly helped improve this article. Article is supported by the Natural Science Foundation of Hubei Provincial Education Department (D20102502).

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