Poincaré Bifurcations of Two Classes of Polynomial Systems
Abstract
Using bifurcation methods and the Abelian integral, we investigate the number of the limit cycles that bifurcate from the period annulus of the singular point when we perturb the planar ordinary differential equations of the form , with an arbitrary polynomial vector field, where C(x, y) = 1 − x3 or C(x, y) = 1 − x4.
1. Introduction and Main Results
In the qualitative theory of real planar differential systems, one of the main problems is to determine the existence and number of the limit cycles of the polynomial differential system. In the general case, this is a very difficult task. Therefore, the researchers consider the weak Hilbert 16th problem. In addition, the existence of invariant algebraic curves in polynomial systems may influence the number of limit cycles. For example, the planar quadratic systems with one invariant line or conic curve or cubic curve can have at most one limit cycle [1–4]. In [3], the authors proved that the cubic systems with four invariant lines have at most one limit cycle. In [5, 6], the authors proved that a real polynomial system of degree m with irreducible invariant algebraic curves has at most 1 + (m − 1)(m − 2)/2 limit cycles if m is even and (m − 1)(m − 2)/2 limit cycles if m is odd.
For the arbitrary polynomials P(x, y), Q(x, y) of given degree n, the number of limit cycles of (2) depends on the different choices of C(x, y). At present, several works have figured out this problem for the particular choices of C(x, y). In [8], the authors studied the system (1) with C(x, y) = 1 + x and proved that the number of limit cycles that bifurcate from the period orbits is at most n. The authors in [9, 10] studied the number of limit cycles which bifurcate from (1) when ɛ = 0 with C(x, y) = 1 + Bx + Ax2 and C(x, y) = 1 + ax + by + cx(x2 + y2), respectively. The authors in [11] studied the number of limit cycles of system (1) with C(x, y) = y2 + Ax2 + Bx + C. In [12], the authors studied the number of limit cycles of system (1) with C(x, y) = (x + a)(x + b) and obtain that the system can have at most 3[(n − 1)/2] + 2 limit cycles if a ≠ b and 2[(n − 1)/2] + 1 if a = b, respectively. In [13], the authors studied the case the curves C(x, y) = 0 are three lines, two of them parallel and one perpendicular, and [14, 15] studied the case the curves are k (k > 3) lines, and any two of them are parallel or perpendicular directions. The authors in [16] studied the case the curves are consistent by k nonzero points. The authors in [17] considered system (1) with C(x, y) = 1 + x4 and proved that 3[(n + 1)/2] − 2 limit cycles can at most bifurcate from the periodic orbits of the unperturbed system. In [18], the authors proved that the system (1) with C(x, y) = (1 − x) m has at most n + m − 1 limit cycles.
The aim of this paper is to investigate the upper bound of the number of limit cycles bifurcate from the periodic annulus of the center of the unperturbed system (1) (ɛ = 0) with the perturbed polynomials P(x, y), Q(x, y) of given degree n, and C(x, y) = 1 − x3 or C(x, y) = 1 − x4.
Theorem 1. If C(x, y) = 1 − x3, the lower bound of the maximum number of limit cycles bifurcating from the period orbits of system (5) with ɛ = 0 is 4[(n + 1)/2] − 2.
Theorem 2. If C(x, y) = 1 − x4, the upper bound of the maximum number of limit cycles bifurcating from the period orbits of system (5) with ɛ = 0 is 3[(n + 1)/2] − 2.
Our primary purpose is to calculate the concrete expression of Φ(h); then we can obtain the number of limit cycles of the perturbed system (5) by determining the isolated real zeros of Abelian integral Φ(h). In Sections 2 and 3, we prove these two theorems with the different methods, respectively.
2. The proof of Theorem 1
Firstly, we have the following obvious result.
Lemma 3. If j is odd, Φk,j = 0 (k ≥ 0).
Lemma 4. Abelian integral Φ(h) has an expansion in the form
Proof. Firstly, we have
Using (12) and (13), (17) can be written as follows:
-
if j < l,
() -
if j = l + 3u − 1 (u = 1,2, 3, …),
() -
if j = l + 3u (u = 0,1, 2, ⋯),
() -
if j = l + 3u − 2 (u = 1,2, 3, …),
()
For matrices Jj, Hj and Il, we have Lemmas 5 and 6, respectively.
Lemma 5. For j ≥ 1, det(Jj) ≠ 0 and det(Hj) ≠ 0. That is, rank(Jj) = j + 1, rank(Hj) = j.
Proof. According to Lemma 4 and (24), we can obtain that
Define (j + 1)×(j + 1) matrix
We add entries of the i + 1th (0 < i ≤ j − k + 1) column which times −(2(j + 2 − i) − 1)/2(j + 2 − i) to the ith column and obtain
where . We can write as follows:
Summarizing above results, we have
therefore, rank(Jj) = j + 1.
According to Lemma 4 and (25), we have
By the above proof procedure, we can obtain rank(Hj−1) = j − 1 in a similar way. That is rank(Hj) = j. The proof is completed.
Lemma 6. For l ≥ 1, rank(Il) = 4l − 1.
Proof. Firstly, if l = 1,
If l = 2, we have
For l ≥ 3, let , , , ; we have
By Lemma 6, it is obvious that p0, p1, p2, …, pl−2, pl−1, pl, … , p4l−4, p4l−3, p4l−1 are independent. Now, we have the following lemma.
Lemma 7. For l ≥ 1, one can write pj as follows:
Proof. According to Lemma 4, if j = l + 3u − 1 (u = 1,2, 3, …),
If j = l + 3u − 2 (u = 1,2, 3, …), in a similar way, we can prove that the second formula holds.
If j = l + 3u (u = 0,1, 2, …), we have
Now, we prove Theorem 1.
Proof. For h ∈ (0,1), Abelian integral Φ(h) has an expansion of the following form:
Let p0 = p1 = p2 = ⋯ = p4l−1 = p4l−3 = 0 and p4u−1 = 1; then, by Lemma 7, p4l−2 = 0. Thus (44) becomes Φ(h) = h4l + o(h4l), and Φ(h) > 0 if h ∈ (0,1). Furthermore, we take p0 = p1 = p2 = ⋯ = p4l−5 = p4l−4 = 0, then p4l−2 = 0 still holds. Choosing proper p4l−3 ∈ (−1,0) such that Φ(h) = p4l−3h4l−2 + h4l + o(h4l) < 0, by Descartes’ rule of signs, (44) has a root h1 on interval (0,1).
Let p0 = p1 = p2 = ⋯ = p4l−6 = p4l−5 = 0, and choose proper p4l−4 (p4l−4 ∈ (0,1)) so that Φ(h) = p4l−4h4l−3 + p4l−3h4l−2 + h4l + o(h4l) > 0; then (44) has the second root h2 on interval (0, 1). In a similar way, we take proper pi (i = 4l − 6,4l − 7, …, 2,1, 0) in turn such that pipi−1 < 0 and |pi | ∈(0, 1). According to Descartes’ rule of signs, we can obtain 4l − 4 zeros h3, h4, …, h4l−3, h4l−2 on interval (0,1).
Applying the Poincaré-Pontryagin theorem, the system (5) with C(x, y) = 1 − x3 can have at least 4[(n + 1)/2] − 2 limit cycles for suitable ak,j and bk,j (0 ≤ k + j ≤ n). The proof of Theorem 1 is completed.
3. The proof of Theorem 2
In this section, we will prove Theorem 2. At first, all the primary computations to express the Abelian integral Φ(h) and some concerned lemmas are presented.
Lemma 8. Let ω1 = 1, ω2 = −1, ω3 = i, ω4 = −i; then
where s = 1, 2, 3, 4. And
Proof. We use the residue theorem to compute the Ψs (s = 1,2, 3,4).
When , we have
Let eiθ = z; then cos θ = (z2 + 1)/2z, dθ = dz/iz. The previous formula becomes
Lemma 9. If j is odd, the integrands in and are odd functions with respect to the variable θ; therefore, and .
Define
Lemma 10. If j is even, then
Lemma 11. (i) If k is odd, .
(ii) If k is even, then
Proof. When k > 1,
We use the residue theorem to compute the integrals M1 and M2. Denote eiθ = z; thus, cos θ = (z2 + 1)/2z, dθ = dz/iz. We have
In a similar way, we can prove the following lemma.
Lemma 12. (i) If k is odd, .
(ii) If k is even, then
Using Lemmas 11 and 12, it is easy to see that, if k is even, Φk,0 = 0. Therefore, by Lemma 10 and the definition of Φk,j, if n is even, Σk+j=nΦk,j = 0.
Lemma 13. Consider a function of the form
To prove Lemma 13, we need the following lemma and a known principle, the Derivation-division algorithm.
Lemma 14. For any n ≥ 0, m ≥ 1 and the real constants α,
The previous lemma has been proved in [14] and [19]. Now, we will prove Lemma 13.
Proof. Differentiating F(x) in formula (68) n0 + 1 times, . According to Lemma 14 and dividing the expression , which does not vanish in U = [0,1), let a = −1/2 − (n0 + 1); we can obtain
Differentiating F1(x) in formula (71) n + 1 times and applying Lemma 14 again and dividing the expression (1 − x) a−(n+1)/(1 + x) a+(n+1), which does not vanish in interval U = [0,1), we have
Now, we prove Theorem 2.
Proof. From Lemma 9, we have
By Lemmas 11 and Lemma 12, the previous formula becomes
According to Lemma 10, we have that, if j is even,
if j is odd,
According to (75) and (76), we can obtain that, if k is odd, j is even,
From (77) and (78), we have, if k is odd, j is even,
Let m = [(n + 1)/2] and
According to (79), the Abelian integral Φ(h) of system (5) with C(x, y) = 1 − x4 has the following form:
According to Lemma 13, taking into account their multiplicities, the maximum number of real zeros of (81) in interval U = [0,1) is 3m − 1. From (73), (75), and (76), we know that Φ(0) = 0. Hence, Φ(h) has at least 3m − 2 real zeros in the open interval (0,1).
Applying the Poincaré-Pontryagin theorem, the upper bound of number of limit cycles for the system (5) with C(x, y) = 1 − x4 is 3m − 2. That is, the maximum number of limit cycles bifurcating from the period orbits of system (5) with ɛ = 0 is 3[(n + 1)/2] − 2. The proof of Theorem 2 is completed.
Acknowledgments
The second author is partially supported by the National Natural Science Foundation of China, Grants no. 11171309 and no. 11172269, and the Zhejiang Provincial Natural Science Foundation of China Grant no. Y6110195.