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Abstract and Applied Analysis

Volume 2013 (2013), Article ID 575390, 19 pages

http://dx.doi.org/10.1155/2013/575390

## Bifurcation of Limit Cycles by Perturbing a Piecewise Linear Hamiltonian System

Department of Mathematics, Shanghai Normal University, Shanghai 200234, China

Received 5 October 2012; Accepted 21 November 2012

Academic Editor: Yonghui Xia

Copyright © 2013 Yanqin Xiong and Maoan Han. 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

This paper concerns limit cycle bifurcations by perturbing a piecewise linear Hamiltonian system. We first obtain all phase portraits of the unperturbed system having at least one family of periodic orbits. By using the first-order Melnikov function of the piecewise near-Hamiltonian system, we investigate the maximal number of limit cycles that bifurcate from a global center up to first order of .

#### 1. Introduction and Main Results

Recently, piecewise smooth dynamical systems have been well concerned, especially in the scientific problems and engineering applications. For example, see the works of Filippov [1], Kunze [2], di Bernardo et al. [3], and the references therein. Because of the variety of the nonsmoothness, there can appear many complicated phenomena in piecewise smooth dynamical systems such as stability (see [4, 5]), chaos (see [6]), and limit cycle bifurcation (see [7–10]). Here, we are more concerned with bifurcation of limit cycles in a perturbed piecewise linear Hamiltonian system: where is a sufficiently small real parameter, with , , , and real numbers satisfying , and with compact. Then system (1) has two subsystems which are called the right subsystem and the left subsystem, respectively. For , systems (5a) and (5b) are Hamiltonian with the Hamiltonian functions, respectively,

Note that the phase portrait of the linear system with has possibly the following four different phase portraits on the plane (see Figure 1).

Then, one can find that system (1) can have 13 different phase portraits (see Figure 2) when at least one family of periodic orbits appears.

We remark that in Figure 2, GC: global center, Ho: homoclinic, He: heteroclinic, : center in the region , : center in the region , : saddle in the region , : saddle in the region , : curvilinear or straightline in the region , : curvilinear or straightline in the region .

It is easy to obtain the following Table 1 which shows conditions for each possible phase portrait appearing above. Also, cases (3), (5), (7), (9), and (13) in Figure 2 are equivalent to cases (2), (6), (8), (10), and (12), respectively, by making the transformation together with time rescaling .

The authors Liu and Han [7] studied system (1) in a subcase of the case (1) of Figure 2 by taking . By using the first order Melnikov function, they proved that the maximal number of limit cycles on Poincaré bifurcations is up to first-order in . The authors Liang et al. [8] considered system (1) in the case (5) of Figure 2 by taking , ,, and . By using the same method, they gave lower bounds of the maximal number of limit cycles in Hopf, and Homoclinic bifurcations, and derived an upper bound of the maximal number of limit cycles bifurcating from the periodic annulus between the center and the Homoclinic loop up to the first-order in . Clearly, the maximal number of limit cycles in the case (7) or (8) of Figure 2 is on Poincaré, Hopf and Homoclinic bifurcations up to first-order in , by using the first order Melnikov function.

This paper focuses on studying the limit cycle bifurcations of system (1) in the case (1) of Figure 2 by using the first order Melnikov function. That is, system (1) satisfies Clearly, system (1) satisfying (9) has a family of periodic orbits such that the limit of as is the origin. The intersection points of the closed curve with the positive -axis and the negative -axis are denoted by and , respectively. Let Then, from Liu and Han [7], the first-order Melnikove function corresponding to system (1) is Let denote the maximal number of zeros of for and the cyclicity of system (1) at the origin. Then, we can obtain the following.

Theorem 1. *Let (9) be satisfied. For any given , one has Table 2. *

This paper is organized as follows. In Section 2, we will provide some preliminary lemmas, which will be used to prove the main results. In Section 3, we present the proof of Theorem 1.

#### 2. Preliminary Lemmas

In this section, we will derive expressions of in (11). First, we have the following.

Lemma 2. *Suppose system (1) satisfies (9). Then,* * in (11) can be written as
* * where
* *with
* * in (11) can be expressed as
* *where
* *with
*

*Proof. *We only prove (i) since (ii) can be verified in a similar way. By (11), we obtain
which follows that by Green formula and (3)
where
Then, by Green formula again
where
By (3), (4), and the above formulas, we have
Combining (20)–(25) gives (13) and (14). Thus, the proof is ended.

Then, using Lemma 2 and (6) we can obtain the following three lemmas.

Lemma 3. *
If , then in (11) has form
**
where
** If , then we have
**
where
*

*Proof. *Note that along . Then, inserting it into (14) follows that
Let . Then we have and the above integral can be carried into
Thus, using (30) and the above equation we can write (13) as
where
which gives (i) by (15). Thus, (i) holds and we can prove (ii) in the same way by (16)–(18). This ends the proof.

Lemma 4. *Let system (5a) satisfy (3) and (4). Then * * If has the expression
* *where
* * If can be written as
* *or
* *where
* *each is a nonzero constant and are polynomials of degree , respectively. *

*Proof. *Since along the curve in (14) becomes
where
Let . Then, we have and the above equation becomes

For , make the transformation . Then, we have by (41)
Substituting the above formula into (37), together with (13), gives that
where
Thus, by (15) and the above discussion we know that (i) holds.

For , (41) can be represented as
where
Recall that
Then, by (46) and the above equation we obtain that
It follows that
where
which is a polynomial of degree in . For convenience, introduce
Then, combining (49) and (51) gives that
Further, by using the formula
we have that
It follows that
where
which is a polynomial of degree in . Let
Then, we have that by (55) and the above
Substituting the above equation into (52), one can find that
where for odd, for even, and
which is a polynomial of degree in . Combining (37), (45), and (59) gives that
which implies (35), together with (13) and (15).

Note that
Then, we have
Inserting the above formula into (35), we can obtain (36). Hence, the proof is finished.

Similar to Lemma 4, we can obtain the following lemma about .

Lemma 5. *Let system (5b) satisfy (3) and (4). Then **
If , in (11) has the expression
**
where
** If in (11) has the form
**
or
**
where
**
each is nonzero constant and are polynomials of degree , respectively. *

#### 3. Proof of Theorem 1

In this section, we will prove the main results. Obviously, under (9) there are the following 9 subcases:(1), (2),(3), (4), (5), (6), (7), (8), (9).

We only give the proof of Subcases 1, 2, 3, 4, 5, and 6. And the Subcases 7, 8, and 9 can be verified, similar to Subcases 3, 4, and 5, respectively.

*Subcase 1. *. From (12) and Lemma 3, one can obtain that
which implies that has at most isolated positive zeros for . To show that this bound can be reached, take . Then, by (27) and (29), (69) has the form
where
Hence, using (70) we can take as free parameters to produce simple positive zeros of near , which gives limit cycles correspondingly near the origin. Thus, in this case. This ends the proof.

*Subcase 2. * Similar to the above and using (32) and (64), in (12) has the expression of the form
where . Further, taking , then, by (34) and (65), in (72) becomes
where
Thus, from (72) and (73), we can discuss similar to Subcase 1. This finishes the proof.

*Subcase 3. * By Lemmas 3, and 4 and (12), we can have that
Let us prove that has at most zeros on the open interval . For the purpose, let . Then, for , in (75) has the expression
where
To prove has at most zeros, it suffices to prove has at most zeros for . By Rolles theorem we need only to prove that has at most zeros for . From (76), we can have that
which shows that has at most zeros for . Thus, has at most zeros for . To prove zeros can appear, we only need to prove that in (75) can appear zeros for small. Let , and . Then in (75) can be expressed as by (29) and (34)
where
Thus, by changing the sign of , in turn such that
we can find simply positive zeros with . For , we can discuss in a similar way. Thus, this bound can be reached and . The proof is finished.

*Subcase 4. * From (12) and Lemmas 3 and 4, we get that
where
which is a polynomial of degree in . Let . Then , and (82) becomes
where
One can see that , where
Denote by the number of zeros of the function in the interval taking into account their multiplicities. Note that
and they are even functions in . Therefore,
Then, from [8], we can obtain that
which implies that . Now, we verify .

Make the transformation . Then in (35) becomes
which follows that as small
Note that
Inserting the above formula into (91) gives that
where