Existence of Positive Solutions to a Boundary Value Problem for a Delayed Nonlinear Fractional Differential System

Abstract

Though boundary value problems for fractional differential equations have been extensively studied, most of the studies focus on scalar equations and the fractional order between 1 and 2. On the other hand, delay is natural in practical systems. However, not much has been done for fractional differential equations with delays. Therefore, in this paper, we consider a boundary value problem of a general delayed nonlinear fractional system. With the help of some fixed point theorems and the properties of the Green function, we establish several sets of sufficient conditions on the existence of positive solutions. The obtained results extend and include some existing ones and are illustrated with some examples for their feasibility.

1. Introduction

In the past decades, fractional differential equations have been intensively studied. This is due to the rapid development of the theory of fractional differential equations itself and the applications of such construction in various sciences such as physics, mechanics, chemistry, and engineering [1, 2]. For the basic theory of fractional differential equations, we refer the readers to [3–7].

Recently, many researchers have devoted their attention to studying the existence of (positive) solutions of boundary value problems for differential equations with fractional order [8–23]. We mention that the fractional order involved is generally in with the exception that in [12, 23] and in [8, 17]. Though there have been extensive study on systems of fractional differential equations, not much has been done for boundary value problems for systems of fractional differential equations [18–20].

On the other hand, we know that delay arises naturally in practical systems due to the transmission of signal or the mechanical transmission. Though theory of ordinary differential equations with delays is mature, not much has been done for fractional differential equations with delays [24–31].

As a result, in this paper, we consider the following nonlinear system of fractional order differential equations with delays,

(1.1)

where is the standard Riemann-Liouville fractional derivative of order   for some integer , for , for , and is a nonlinear function from to . The purpose is to establish sufficient conditions on the existence of positive solutions to (1.1) by using some fixed point theorems and some properties of the Green function. By a positive solution to (1.1) we mean a mapping with positive components on such that (1.1) is satisfied. Obviously, (1.1) includes the usual system of fractional differential equations when for all and . Therefore, the obtained results generalize and include some existing ones.

The remaining part of this paper is organized as follows. In Section 2, we introduce some basics of fractional derivative and the fixed point theorems which will be used in Section 3 to establish the existence of positive solutions. To conclude the paper, the feasibility of some of the results is illustrated with concrete examples in Section 4.

2. Preliminaries

We first introduce some basic definitions of fractional derivative for the readers' convenience.

Definition 2.1 (see [3, 32]).

The fractional integral of order   of a function is defined as

(2.1)

provided that the integral exists on , where is the Gamma function.

Note that has the semigroup property, that is,

(2.2)

Definition 2.2 (see [3, 32]).

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

(2.3)

provided that the right-hand side is pointwise defined on , where .

It is well known that if then . Furthermore, if and then for .

The following results on fractional integral and fractional derivative will be needed in establishing our main results.

Lemma 2.3 (see [10]).

Let . Then solutions to the fractional equation can be written as

(2.4)

where , .

Lemma 2.4 (see [10]).

Let . Then

(2.5)

for some , .

Now, we cite the fixed point theorems to be used in Section 3.

Lemma 2.5 (the Banach contraction mapping theorem [33]).

Let be a complete metric space and let be a contraction mapping. Then has a unique fixed point.

Lemma 2.6 (see [16, 34]).

Let be a closed and convex subset of a Banach space . Assume that is a relatively open subset of with and is completely continuous. Then at least one of the following two properties holds:

(i) has a fixed point in ;

(ii)there exists and with .

Lemma 2.7 (the Krasnosel'skii fixed point theorem [33, 35]).

Let be a cone in a Banach space . Assume that and are open subsets of with and . Suppose that is a completely continuous operator such that either

(i) for and for

or

(ii) for and for .

Then has a fixed point in .

3. Existence of Positive Solutions

Throughout this paper, we let . Then is a Banach space, where

(3.1)

In this section, we always assume that .

Lemma 3.1.

System (1.1) is equivalent to the following system of integral equations:

(3.2)

where

(3.3)

Proof.

It is easy to see that if satisfies (3.2) then it also satisfies (3.2). So, assume that is a solution to (1.1). Integrating both sides of the first equation of (1.1) of order with respect to gives us

(3.4)

for , . It follows that

(3.5)

for , . This, combined with the boundary conditions in (1.1), yields

(3.6)

Similarly, one can obtain

(3.7)

(3.8)

  . Then it follows from (3.8) and the boundary condition that

(3.9)

Therefore, for ,

(3.10)

This completes the proof.

The following two results give some properties of the Green functions .

Lemma 3.2.

For is continuous on and for .

Proof.

Obviously, is continuous on . It remains to show that for . It is easy to see that for . We only need to show that for . For , let

(3.11)

(3.12)

Then

(3.13)

Note that and for . It follows that and hence for .

Therefore, for and the proof is complete.

Lemma 3.3.

• (i) If , then for .

• (ii) If , then for .

Proof.

• (i) Obviously, for . Now, for , we have

  

  (3.14)

where is the function defined by (3.11). It follows that for . In summary, we have proved (i).

• (ii) Again, one can easily see that for . When , we have in this case that

  

  (3.15)

which implies that for . To summarize, we have proved (ii) and this completes the proof.

Now, we are ready to present the main results.

Theorem 3.4.

Suppose that there exist functions , , 2, , such that

(3.16)

for , ,. If

(3.17)

(3.18)

then (1.1) has a unique positive solution.

Proof.

Let

(3.19)

It is easy to see that is a complete metric space. Define an operator on by

(3.20)

where and

(3.21)

Because of the continuity of and , it follows easily from Lemma 3.2 that maps into itself. To finish the proof, we only need to show that is a contraction. Indeed, for , by (3.16) we have

(3.22)

This, combined with Lemma 3.3 and (3.17) and (3.18), immediately implies that is a contraction. Therefore, the proof is complete with the help of Lemmas 3.1 and 2.5.

The following result can be proved in the same spirit as that for Theorem 3.4.

Theorem 3.5.

For , suppose that there exist nonnegative function and nonnegative constants such that and

(3.23)

for , . If

(3.24)

then (1.1) has a unique positive solution.

Theorem 3.6.

For , suppose that there exist nonnegative real-valued functions such that

(3.25)

for almost every and all . If

(3.26)

then (1.1) has at least one positive solution.

Proof.

Let and be defined by (3.19) and (3.20), respectively. We first show that is completely continuous through the following three steps.

Step 1.

Show that is continuous. Let be a sequence in such that . Then is bounded in . Since is continuous, it is uniformly continuous on any compact set. In particular, for any , there exists a positive integer such that

(3.27)

for and . Then, for , we have

(3.28)

for and . Therefore,

(3.29)

which implies that is continuous.

Step 2.

Show that maps bounded sets of into bounded sets. Let be a bounded subset of . Then is bounded. Since is continuous, there exists an such that

(3.30)

It follows that, for and ,

(3.31)

Immediately, we can easily see that is a bounded subset of .

Step 3.

Show that maps bounded sets of into equicontinuous sets. Let be a bounded subset of . Similarly as in Step 2, there exists such that

(3.32)

Then, for any and and ,

(3.33)

Now the equicontituity of on follows easily from the fact that is continuous and hence uniformly continuous on .

Now we have shown that is completely continuous. To apply Lemma 2.6, let

(3.34)

Fix and define

(3.35)

We claim that there is no such that for some . Otherwise, assume that there exist and such that . Then

(3.36)

If , then

(3.37)

Similarly, we can have if . To summarize, , a contradiction to . This proves the claim. Applying Lemma 2.6, we know that has a fixed point in , which is a positive solution to (1.1) by Lemma 3.1. Therefore, the proof is complete.

As a consequence of Theorem 3.6, we have the following.

Corollary 3.7.

If all , , are bounded, then (1.1) has at least one positive solution.

To state the last result of this section, we introduce

(3.38)

Theorem 3.8.

Suppose that there exist and positive constants with such that

(i)  for

and

(ii), for ,

where . Then (1.1) has at least a positive solution.

Proof.

Let be defined by (3.19) and . Obviously, is a cone in . From the proof of Theorem 3.6, we know that the operator defined by (3.20) is completely continuous on . For any , it follows from Lemma 3.3 and condition (ii) that

(3.39)

that is,

(3.40)

On the other hand, for any , it follows from Lemma 3.3 and condition (i) that, for ,

(3.41)

if , whereas

(3.42)

if . In summary,

(3.43)

Therefore, we have verified condition (ii) of Lemma 2.7. It follows that has a fixed point in , which is a positive solution to (1.1). This completes the proof.

4. Examples

In this section, we demonstrate the feasibility of some of the results obtained in Section 3.

Example 4.1.

Consider

(4.1)

Here

(4.2)

One can easily see that (3.16) is satisfied with

(4.3)

Moreover,

(4.4)

and hence

(4.5)

It follows from Theorem 3.4 that (4.1) has a unique positive solution on .

Example 4.2.

Consider

(4.6)

Here

(4.7)

where

(4.8)

Hence, and satisfy (3.25). Moreover, simple calculations give us

(4.9)

Then and

(4.10)

Choose , and

(4.11)

Then, for and , we have

(4.12)

for and , we have

(4.13)

By now we have verified all the assumptions of Theorem 3.8. Therefore, (4.6) has at least one positive solution satisfying .