Abstract and Applied Analysis

Volume 2014 (2014), Article ID 419514, 15 pages

http://dx.doi.org/10.1155/2014/419514

Research Article

Solvability for a Fractional Order Three-Point Boundary Value System at Resonance

Zigen Ouyang and Hongliang Liu

School of Mathematics and Physics, University of South China, Hengyang 421001, China

Received 31 March 2014; Revised 15 May 2014; Accepted 16 May 2014; Published 11 June 2014

Academic Editor: Chuangxia Huang

Copyright © 2014 Zigen Ouyang and Hongliang Liu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

A class of fractional order three-point boundary value system with resonance is investigated in this paper. Using some techniques of inequalities, a completely new method is incorporated. We transform the problem into an integral equation with a pair of undetermined parameters. The topological degree theory is applied to determine the particular value of the parameters so that the system has a solution.

1. Introduction

In this paper, we consider the following fractional differential system:

where is standard Riemann-Liouville fractional derivative of order , and is a nonlinear two-dimension continuous vector function.

In the last few decades, many authors have focused on the dynamics of differential equations [1–7]; most of them have investigated fractional differential equations which have been applied in many fields such as physics, mechanics, chemistry, and engineering; see [8–13]. In particular, the positive solutions of the boundary value problem have attracted many authors’ attention [14–25].

Recently, the existence of solutions of three-point boundary value problem

where is standard Riemann-Liouville fractional derivative of order has been studied by many authors under the case that . They obtained some nice results by using some fixed point theorems; see [26–28].

In [29], Ahmad and Nieto considered the existence results for following three-point boundary value problem for a coupled system of nonlinear fractional differential equations given by

where ,  ,  ,  ,  ,  ,  ,   is standard Riemann-Liouville fractional derivative and are given continuous functions. An existence result was proved in their paper by applying the Schauder fixed point theorem.

However, few authors have investigated fractional differential boundary value problems with resonance [1, 2, 30–32].

In this paper, we establish some sufficient conditions for the existence of the boundary value system (1) by using intermediate value theorems. To present the main results, we assume that satisfies the following.

, . Suppose that there exist nonnegative functions ,     , with , , for any such that

 where   , , and . For any real numbers and , the functions satisfy

 Furthermore, assume that

 We have the following results.

Theorem 1. Assume that holds. If

where

then (1) has at least one solution in .

Also, we consider the following special case of as follows.

(), . Suppose that there exist nonnegative functions ,     , with , , , for any such that

 where   , , and . The functions satisfy

 Furthermore, assume that (7) and (8) hold.

From Theorem 1, we have the following corollary.

Corollary 2. Assume that and (9) hold; then (1) has at least one solution in .

2. Some Lemmas

In this section, we first introduce some definitions and preliminary facts and some lemmas which will be used in this paper.

Definition 3 (see [21]). The fractional integral of order of a function is given by

provided that the right integral converges.

Definition 4 (see [21]). The standard Riemann-Liouville fractional derivative of order of a continuous function is given by

where , provided that the right integral converges.

Lemma 5 (see [21]). Assume that with a fractional derivative of order that belongs to .Then

for some , , where is the smallest integer greater than or equal to .

The following lemma is a fixed point theorem in a particular Banach space:

equipped with the norm

It is easy to show that if , then .

Lemma 6 (see [33]). Let be a Banach space with closed and convex. Assume that is a relatively open subset of with and is completely continuous. Then either

(i) has a fixed point in , or(ii)there exist an and with .

To use this lemma to prove our main result, we need transfer (1) into an integral operator.

Lemma 7 (see [34]). Problem (1) is equivalent to the following integral equation:

where

Lemma 8 (see [34]). For any , is continuous, and for any .

Let

where

Then (1) is equivalent to the following integral equation:

The new Green’s function has the following properties.

Lemma 9 (see [34]). is continuous for and

Furthermore, for .

Lemma 10 (see [34]). is nonincreasing with respect to for any . In particular, for any , , and for . That is, , where

Let

Then , and (23) gives

Let

Then , and . From (27), (28) can be rewritten as

with . Now the integral equation (27) is equivalent to (29). It can be seen from (29) that the solution of (29) is dependent on the value . Now, instead of (29), we replace with a real vector and consider

For any , let

equipped with the norm . Define an operator in as follows:

Using a similar method of Lemmas 3.5 and  3.6 in [34], we obtain that is completely continuous in , and (30) has at least a solution for any given real constant vector ; the solution is dependent on the given vector . We note the solution .

3. The Proof of Theorem 1

From Lemma 10, for any real vector , the integral equation (30) has at least a solution . Therefore, to show that problem (1) has a solution, it remains to show that there exists a , such that , or .

In what follows, we will use the method of topological degree to prove our main result.

Let be an open subset of the plane with the boundary being a simple closed curve; is a continuous mapping from to . Let . Denote by a variable point on the boundary . As traverses the boundary, assume that its image traces out a closed curve that does not pass through the point . As in complex analysis, we can define the winding number of this curve with respect to , by measuring the total change of the argument of the vector joining and the variable point . For two-dimensional space, this number is equivalent to the topological degree of the mapping at .

We introduce a proposition from [8] as follows.

Proposition 11. If the degree of a continuous mapping with respect to a point is nonzero, then the equation has a solution .

From Section 2, for any parameters , , , there exists a solution of (30). At the point , we denote , . It is obvious that the parameters , depend on the parameters , , so we define a map as follows:

Therefore, if we can find a domain with its boundary as a closed curve , so that its image contains the point in it, then it is implied by Proposition 11 that there exists a point in such that . Thus, the function is a solution of (29), where

We now proceed to find such . For convenience, we take a curve , where , , , , and , , , and are a part of line. The image . We want to show that the point is inside the closed curve as the parameters , , and are large enough. In fact, we will prove that the line () lies in the left (right) side of the -axis, and the line () lies above (under) the -axis as , , and are large enough.

Let

From (9) and (), and , and , , we may take , , and large enough satisfying

Then, the points , , , and can be expressed as follows:

Now the proof of Theorem 1 is reduced as the following lemmas.

Lemma 12. Suppose that and (9) hold. Then, for large enough, lies in the third quadrant.

Proof. From (30), we have

By the definition of , in (33), we may rewrite (39) as follows:

Let . From (40), we have

Now we will show that as . We only show that and the proof of is similar. Assume on the contrary that . Thus, there exists a sequence , such that .

Recall that

Now we claim that it is impossible to have

as is sufficiently large. Indeed, assume that (43) is true. Then, by the first equation of (42), we have

for all . Therefore, we obtain

for . We define some sets as follows:

We have assumed in (7) that

It is easy to show from (42), , and our assumption that the set is not empty, and . We have the following:

Using conditions (6) and (8), we have from (45) that there exists a constant such that

for and any large enough. From the second formula of (42), (45)–(49), one gets

which implies that

It contradicts (48).

Thus, for any large enough, there exists some , such that

Now we define

Then, is not empty.

We can further divide the set into two sets and , and divide the set into two sets and as follows:

It is easy to know that , and , .

We claim that the set is not empty for large enough. Otherwise, the function is bounded from above. In fact, assume that is unbounded from above for large enough; then we have from that there exist a sequence and a subsequence of such that

Using a similar method of (51), we can derive a contradiction. Therefore, is bounded from above. From (42), is bounded from above, which implies that as . If (where is defined in (46)), then , which contradicts our assumption. Thus, . Using a similar method of getting (51) also gives a contradiction. Therefore, is not empty.

Similarly as getting (51) again, we conclude that the function is bounded above by a constant for and . From the condition , if  (or ) and (or ), then (or is also bounded from above by a constant for . Therefore, from the definition of , , there exists a constant , independent of and such that

Let

From the definitions of and , we have

Since is not empty, it follows that as . Recall from (9) and (35) that . Therefore, we can choose large enough so that

for , where

Now, for later use, for any integral in a domain

we define a subset as

Thus, the integral in (62) can be rewritten as

From , (42), and the definitions of , and , , for , we have

which yields from (56) and the definition in (63) that

Further, one gets from (56) that

which gives

That is,

If , then we have from (69) that

which contradicts (59).

If , using a similar method of (69), we can estimate as

Substituting this into (69), we obtain

which finally contradicts (59). Therefore, our result is proved.

Similarly, we can show that . Thus, the point lies in the third quadrant. The proof is completed.

Lemma 13. Suppose that and (9) hold. Then, for large enough, lies in the second quadrant.

Proof. It suffices to show that and .

First, we claim that . On the contrary, we assume that there exists a sequence such that . By a similar method in Lemma 12, we know it is impossible to have

as is sufficiently large.

Now, for large , we define

Then, is not empty.

As in Lemma 12, we can further divide the set into two sets and and divide the set into two sets and as follows:

Then ,   and ,  .

Using a similar method as in the proof of Lemma 12, we can show that the set is not empty. Furthermore, the function is bounded below by a constant for and . If (or and (or , then (or is also bounded below by a constant for . From the definition of , and the condition , there exists a constant , independent of and such that

Let

From the definitions of and , we have

and it follows that as . Therefore, we can choose large enough so that

for , where

Notice that , . From , (42), and the definitions of , and , , for , we have

Thus,

which follows from (76) and the definition of (63) that

Thus,

which implies that

If , from (85), we have

which contradicts (79).

If . Using a similar method of (85), we have

Substituting it into (85), we obtain

which also contradicts (79). Therefore, .

Now, we show that .

On the contrary, assume that there exists a vector sequence such that and

Similarly as before, it is impossible to have

as is sufficiently large.

Now for large , we define

Then, is not empty.

We can further divide the set into two sets and and divide the set into two sets and as follows:

It is easy to know that , and , .

Using a similar method of the proof of Lemma 12, we obtain that the set is not empty. Furthermore, there exists a constant , independent of and such that

Let

From the definitions of and , we have

Since is not empty, it follows that as . Therefore, we can choose large enough so that

for , where

From and (42), we have

which follows from (93) and the definition in (63) that

Thus,

which implies that

Therefore, we have

If , then we have from (76) that

which contradicts (102).

If . Using a similar method to that in Lemma 12, we have

which also contradicts (96). Thus, the point lies in the second quadrant. The proof is completed.

Lemma 14. Suppose that and (9) hold. Then, for large enough, the line lies in the left of -axis.

Proof. For any point in , it suffices to show that as for any .

On the contrary, we assume that there exists a vector sequence satisfying and a point such that as . We define some sets , , and , , , and some numbers , as in Lemma 13. Using a similar method of the proof of Lemma 13, we have

It follows from (93)–(96) that

Notice that ; from (35), one gets

which implies that

It is easy to show that

for , which contradicts (102), and

for , which contradicts (96) also. Thus, the line lies in the left of -axis. The proof is completed.

Similar to the proof of Lemma 12, we can show that the image point of the point lies in the first quadrant. From (37), we have . Using a similar method of Lemma 13, we can show that the image point of the point lies in the fourth quadrant.

Using the conditions (36) and (37), similar to Lemma 14, we can show that the image line of the line lies above the -axis, lies in the right of the -axis, and lies under the -axis. Therefore, we have the following lemmas.

Lemma 15. Suppose that and (9) hold. For large enough, lies above the -axis, lies in the second quadrant, and lies in the first quadrant.

Lemma 16. Suppose that and (9) hold. For large enough, lies in the right of the -axis and lies in the fourth quadrant.

Lemma 17. Suppose that and (9) hold. For large enough, lies below the -axis.

Proof of Theorem 1. From Lemmas 12–17, when , , and are large enough and satisfy (36) and (37), then the image of the curve will contain the zero in it. From Proposition 11, it follows that there exists a vector such that the solution of (30) satisfies , which implies that the integral equation (27) has a solution . From (26), it follows that (19) has a solution. Therefore, the problem (1) has at least one solution. The proof is completed.

4. Examples

Example 1. Consider the following boundary value system:

where

It is obvious that

where

It is easy to check that the function satisfies (8)–(12). Notice that

Consider

which satisfies (9). Therefore, all conditions of Theorem 1 hold and thus the problem (111) has at least a solution.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The work was supported by the NSF of Hunan province (no. 13JJ3074), Major project of department of education of Hunan province (no. 13A088), the Scientific Research Foundation of Hengyang city (no. J1), and the Construct Program in USC.

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