Abstract

We are concerned with a type of impulsive fractional differential equations attached with integral boundary conditions and get the existence of at least one positive solution via global bifurcation techniques.

1. Introduction

Fractional differential equations have been extensively studied in recent years (see, for instance, [1–7] and their references). In addition, since Rabinowitz established unilateral global bifurcation theorems, there have been many researches in global bifurcation theory and it has been applied to obtain the existence and multiplicity for solutions of differential equations (see, for instance, [8–16] and their references). However, the previous researches seldom involve both global bifurcation techniques and fractional differential equations. In [16], the following problem was studied. where the fractional difference was of Riemann-Liouville type. Under suitable conditions, the existence of at least one positive solution and one negative solution was got. We would like to conduct further research on the above problem. For instance, in practical applications, may rely not only on but also on , which will give rise to additional difficulties for the study. Moreover, what if the boundary value conditions are nonlocal rather than local? Can we add impulsive terms into the system? As a reply to above questions, we will tackle the following impulsive Caputo fractional differential equations attached with integral boundary value conditions in this paper.where satisfies when and ; ; ; ; We set throughout this paper.

Through global bifurcation techniques, we get the existence of at least one positive solution of (2) (see Theorem 14). Moreover, for the sake of convenience of use, we give a corollary of Theorem 14 (see Theorem 15).

The rest of this paper is organized as follows. In Section 2, we will present some preliminary knowledge and some conditions for (2) that we need. We will prove some properties of several functions in Section 3 as a preparation of Section 4. In Section 4, we will present our main results and prove them. In Section 5, we present an example as an application of the main results. Finally, an appendix is given to prove a formula which will be used in the proof of the main results.

2. Preliminary

Firstly, we introduce several spaces that this paper needs.

In this paper, we set as well as

Secondly, we introduce some knowledge of fractional integral and fractional derivative.

Definition 1 (see [3]). The fractional integral of order for a function is defined as

Definition 2 (see [3]). The Riemann-Liouville derivative of fractional order for a function is defined as For , it is well known that holds almost everywhere on . What is more, holds on if .

Definition 3 (see [3]). The Caputo derivative of fractional order for a function is defined as Clearly, (8) is equivalent to Moreover, for and , it is well known that However, (10) does not necessarily hold if does not belong to . Since and considered in (2) are not continuous at , we need some modification on Definition 3. So we generalize Definition 3 in the following way, which will be applied to (2).

Definition 4. The Caputo derivative of fractional order for a function is defined as where

With the above definition, we have a conclusion similar to (10), which will be represented as (13).

Lemma 5. For and , if for , there must be

Proof. Suppose thatthat is We know that Then Since , we know that . Then by (17) we know that So (13) turns to be that is Choosing suitable , we finish the proof.

If we set where , and are constant real numbers, then we will have the following conclusion.

Lemma 6. One has

Proof. It is easy to see that for and then it is clear that

Lemma 7. For , is a solution of

if and only if

Proof. If is a solution of (24), by Lemma 5 we know that Since , , , and , we know that Then that is, satisfies (25a) and (25b).
If (25a) and (25b) hold, by Lemma 6 we know that Since , we know that are continuous on as well as and , so (29) turns to be All the other conditions of (24) can be verified by direct computations and we omit it here.

Thirdly, we introduce some knowledge of order cone.

Definition 8. Let be a real Banach space and let be a subset of . Then is called an order cone if(i) is closed, nonempty, and ;(ii);(iii) and . On this basis, is denoted by , while means that and . Moreover, is called to be solid if ; that is, has interior points. means that is an interior point of .

In this paper, we set Then is a Banach space endowed with the norm

We set the cone Clearly is a solid cone in . By the way, for any , it is obvious that and , which will be useful lately in this paper.

Fourthly, we introduce two lemmas which are very important in this paper.

Lemma 9 (see [17, Corollary  15.12]). We set If (H1) and (H2) are satisfied, then is a bifurcation point of and contains an unbounded solution component which passes through .
If additionally (H3) is satisfied, then and always imply that and .

The conditions that Lemma 9 needs are stated as below.The operators are compact on the real Banach space . is positive: that is, when . is linear and as . Moreover, has an order cone with .The spectral radius of is positive. is strongly positive: that is, for .

Lemma 10 (see [18, Theorem  19.3]). Let be a Banach space and let be a solid cone. is linear, compact, and strongly positive. Then we have the following:(a), is a simple eigenvalue with an eigenvector and there is no other eigenvalue with positive eigenvector.(b)For , the equation has no solution in .(c)Let be a linear operator. If on , then , while if for .

At last, we present here some conditions that we need in this paper.At least one of the following two conditions is satisfied:(1) does not identically vanish on any subinterval of .(2).There exist and such that for all uniformly. What is more, , .There exists a function such that for all . In addition,

3. Property of and

In this paper, we set

Lemma 11. is positive and compact.

Proof. Firstly, we will prove that for all . With all the other properties of easily verified, we only show here that converges, while .
We know that So, Secondly, we show that is positive.
In fact, for , observing the expression of , it is easy to see that and for .
Thirdly, we will prove that is compact. To this end, we only need to show that and are uniformly bounded and equicontinuous on for any bounded subset of named . We only prove the later one since the proof of another one is easier.
Since is bounded, we can choose a constant number such that for all , which implies that and and hence . Then there exists a constant number such that for all So we have which means that is uniformly bounded on .
Next, we will prove that is equicontinuous on . We know that By the uniform continuity of on and (41), we know that, for any , there exists such that implies Moreover, when , we have that So We set ; by (42) and (44), we know that for all and satisfy .

Lemma 12. are linear, compact, and positive.

Proof. Similar to Lemma 11, we can verify that are compact and positive. Moreover, they are obviously linear operators.

Lemma 13. If (C1) holds and , then and are linear, compact, and strongly positive.

Proof. It is obvious that is a positive linear operator. So if , will have a positive linear bounded inverse operator: Since and are compact, we know that and are compact. In the next, we will only prove that is strongly positive since the proof for another one is similar.
For any given , , if we set we will know that (noticing the expression of , we can see that will turn any positive element of to be a positive element of ). By (46), we know that and hence Since , we know that for , so by and (48), we will know that On the other hand, for , by and , we will know that What is more, similar to (39), we have By (50) and (51) and the piecewise continuity of on , we know that there must exist a positive number such that Equations (49) and (52) exactly mean that .

4. Main Results and the Proof

Theorem 14. Suppose that (C1)–(C3) hold and ; then there must exist at least one positive solution of (2) if or .

Proof. The proof is divided into three parts.
In Part 1, we consider the auxiliary equation (see (53)) whose solutions of the kind will be the solutions of (2) and we transform it into a functional operator equation (see (57)).
In Part 2, we will verify that , satisfy all the conditions required to apply Lemmas 9 and 10, where will be defined in Part 1.
In Part 3, we apply Lemmas 9 and 10 to get the existence of at least one positive solution of (2).
Part 1. We consider the following problem. We call to be a solution of (53) if it satisfies (53). It is clear that any solution of (53) of the form yields a solution of (2).
Due to Lemma 7, is a solution of (53) if and only if that is, If we set then is a solution of (53) if and only if Part 2. By Lemmas 11–13, we have confirmed that are compact, is positive, and is strongly positive. Now we only need to verify that , by , there must exist a number such that and imply Then, if , there will be Moreover, we know that By (60) and (61), we will know that Equation (62) together with the boundedness of will lead to Part 3. Applying Lemmas 9 and 10, we can draw a conclusion as below.
For (57), from there emanates an unbounded continua of positive solutions , where Furthermore, and always imply that .
To verify the existence of at least one positive solution of (2), we only need to show that crosses the hyperplane in . To this end, it will be enough to show that joins to .
Suppose that satisfy . To begin with, we will show that is bounded. Defining as similar to Lemma 13, we can verify that is linear, compact, and strongly positive. Then, due to Lemma 10(b), we see that there is a contradiction for sufficiently large in the following inequality: Then, is bounded and hence .
We choose a subsequence of converging to . Without loss of generality, we relabel the subsequence to be just for convenience.
Now we set then Since is compact, we can find a subsequence of named converging to Moreover, we know that so It is obvious that and . Let in ; we know that Since is linear, compact, and strongly positive, by Lemma 10 and (71) we know that which implies that joins to .

As a corollary of Theorem 14, we have the following conclusion.

Theorem 15. Suppose that (C1)–(C3) hold. If then there must exist at least one positive solution of (2) when or .

Proof. With all the other conditions of Theorem 14 satisfied, we only need to verify that . For any and , we have For any and , we have For any and , we have For any and , we have By (74)–(77), we know that .

5. Example

Example 1. Consider the following problem. If we set then there must be at least one positive solution of when . (We can verify that by Lemma 10.)

Proof. Let , , , , , ; then turns to be (2). Now we set , , , , . We can verify that hold. Then by Theorem 15 we know that there must be at least one positive solution of when .

Appendix

At the very first, we recall that

For any and , since for all uniformly, similar to (60)-(61) we can find sufficiently large number such that where What is more, because is bounded on , we can find sufficiently large such that where So, by (A.3) and (A.6), we know that, for with , there must be On the other hand, by (A.4) and (A.7), we know that, for with , there must be By (A.9) and (A.10), we know that Equation (A.11) together with the boundedness of will lead to

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Authors’ Contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Acknowledgments

The research is supported by the National Natural Science Foundation of China [11571197 and 11471187] and STPF of University in Shandong Province [no. J17KA161].