Dynamical Aspects of Initial/Boundary Value Problems for Ordinary Differential EquationsView this Special Issue
Bifurcation of Limit Cycles by Perturbing a Piecewise Linear Hamiltonian System
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 , Kunze , di Bernardo et al. , 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 ), 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).
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  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.  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 , 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.
2. Preliminary Lemmas
In this section, we will derive expressions of in (11). First, we have the following.
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.
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.
Proof. Since along the curve in (14) becomes
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 .
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).
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.