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`Abstract and Applied AnalysisVolume 2014, Article ID 419514, 15 pageshttp://dx.doi.org/10.1155/2014/419514`
Research Article

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

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

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 [17]; most of them have investigated fractional differential equations which have been applied in many fields such as physics, mechanics, chemistry, and engineering; see [813]. In particular, the positive solutions of the boundary value problem have attracted many authors’ attention [1425].

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 [2628].

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, 3032].

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