Abstract and Applied Analysis

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

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

Research Article

Bifurcation of Limit Cycles by Perturbing a Piecewise Linear Hamiltonian System

Yanqin Xiong and Maoan Han

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).

(a)

(b)

(c)

(d)  

Figure 1: The possible phase portraits of system (7).  

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.

Figure 2: The possible phase portraits of system (1) . (1) GC. (2) . (3) . (4) . (5) . (6) . (7) . (8) . (9) . (10) . (11) . (12) . (13) .

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 .

Table 1: Coefficient conditions for phase portraits (1)–(13).

Coefficient

conditions

(1)
(2)

(5)

(3)
(4)
(7)
(9)

(8)

(11) (),

(6)
(10)

(12) (),

(13) ()

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.

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

Take . Then, by (28), (35), and (93), we can obtain that for small

where

with being linear combination, . One can find that

where is a matrix,

with are matrix satisfying

Hence, we can obtain that from (97)

where

and are given in (99). We claim that . We only need to prove by the above formula. Using elementary transformations to by multiplying th column by , we can obtain that by (99) and (101)

where

Now we will use elementary transformations to as follows.

(1) Add the first column multiplying by to th column, .(2) Add the second column multiplying by to th column, .(3) Add the third column multiplying by to th column,  

. Add the th column multiplying by to th column,

. multiply th column by , .

Then, becomes, together with (94)

where

with . For in (104) by adding a column on the left and a row on the above, we can obtain that, together with adding th column to th column with

which implies that by (102) and (104) if . We claim that and

Now, we prove them by induction on . For we have

which means that (106) and (107) hold for . Suppose (106) and (107) hold for . That is, we have for

Then for , we have

Note that by the first equation of (109) there only exist such that

since , where is given in (109). If , then we can obtain that from (109) and (111)

Note that

which follows that, together with (112)

By the second equation in (109) and the above formula, we have

This is a contradiction with . Hence , which means that (106) holds for . Since in (110), there only exist such that

with since the last row in the second formula of (110) is linearly independent with all rows in the first formula of (110), which means that (107) holds for . In a similar way, we can prove (106) and (107) hold for . Hence, the claim holds. So, from (99) we can know that can be taken as free parameters. So we can choose these values such that

which yields that by (97) and (96) can appear positive zeros for small. We also can know that . Hence, the conclusion is proved.

Subcase 5. By (36) and (67), one can see that

where

For convenience, we denote by any polynomial of degree although its coefficients may be different when it appears in different place. Then, we claim that for any

Now, we verify this claim by induction on . For , by (118) we can obtain that

which follows that

Hence, (120) holds for . Suppose (120) holds for  . Then for , we have

which implies that (120) holds for . Thus, the claim is proved. Then, taking , we can obtain that by differentiating it

One can find that

where

where is a polynomial of degree . Since from (118), it is easy to see that has at most zeros for by Rolle theorem. As the above discussion, we only prove as small, which implies . For the purpose, take . Then using (49), (51), we can write in (12) as

For , (127) can be written as

where

which implies that can appear zeros in small by using the same method with the Subcase 4. For , we can discuss by (127) in a similar way. Hence, the conclusion holds.

Subcase 6. . We have as the above

Similar to the Subcase 5, we can prove that for any

Taking and differentiating the above twice follow that

If , then it is easy to see that (132) has the same form with (125). Hence, we can know that has at most zeros for . If , then (132) can be written as

where

where is a polynomial of degree in . By Rolle theorem, we can obtain that has at most zeros for since . Now, we only need to prove . Take . Then, by Lemmas 4, 5, one can see that for small

Similarly, we can discuss the above formula such that such that can appear zeros for small. Hence, the conclusion is proved.

Acknowledgment

The project was supported by National Natural Science Foundation of China (11271261), a grant from Ministry of Education of China (20103127110001) and FP7-PEOPLE-2012-IRSES-316338.

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