Chaos, Solitons and Fractals 23 (2005) 1787–1794 www.elsevier.com/locate/chaos
On the cyclicity of a 2-polycycle for quadratic systems
q
Maoan Han *, Cheng Yang Department of Mathematics, Shanghai Jiao Tong University, 200030, Shanghai, PR China Accepted 1 July 2004 Communicated by Prof. Ji-Huan He
Abstract It has been already known that the maximum number of limit cycles near a homoclinic loop of a quadratic Hamiltonian system under quadratic perturbations is two. However, the problem of finding the maximum number of limit cycles in the 2-polycycle case is still open. This paper addresses the problem in some detail and solves it partially. Ó 2004 Elsevier Ltd. All rights reserved.
1. Introduction Consider a quadratic system of the form x_ ¼ H y þ eP ðx; yÞ;
y_ ¼ H x þ eQðx; yÞ;
where e > 0 is a small parameter, H is cubic in (x, y), and X X P ðx; yÞ ¼ aij xi y i ; Qðx; yÞ ¼ bij xi y j : iþj62
ð1:1Þ
ð1:2Þ
iþj62
An interesting problem for (1.1) is to find the maximum number of limit cycles on the plane for e > 0 small and aij, bij bounded. The problem is closely related to the Hilbert 16th problem and is still open. Assume that for e = 0 (1.1) has a family of periodic orbits given by Lh : H ðx; yÞ ¼ h;
h2J
with J an interval. The integral I MðhÞ ¼ Q dx P dy Lh
is called the first order Melnikov function of (1.1).
q
Research supported by the National Natural Science Foundation of China (10371072). Corresponding author. E-mail address:
[email protected] (M. Han).
*
0960-0779/$ - see front matter Ó 2004 Elsevier Ltd. All rights reserved. doi:10.1016/j.chaos.2004.07.008
ð1:3Þ
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Let the family of {Lh, h 2 J} be bounded. Then its boundary is a homoclinic or heteroclinic loop, denoted by L0. If L0 is homoclinic to a hyperbolic saddle, then it has been proved in [7,8] that for e > 0 small and aij, bij fixed, (1.1) has at most two limit cycles near L0. The same conclusion was proved in [4,5] for e > 0 small and aij, bij varying in a compact set. Moreover, two limit cycles can appear in an arbitrary neighborhood of L0. Thus the cyclicity of L0 is 2 for the system (1.1). The cyclicity of the period annulus Lh is found to be 2 [1]. Now suppose L0 is a heteroclinic loop with two hyperbolic saddles. It is also called a 2-polycycle. As was mentioned in [10, Remark 6.5], the problem of finding the maximum number of limit cycles near L0 is open. In this paper we will study this problem and solve it under a condition. To state our main results, we introduce P x þ Qy ¼ A0 þ A1 x þ A2 y;
ð1:4Þ
where A0 ¼ a10 þ b01 ;
A1 ¼ 2a20 þ b11 ;
A2 ¼ a11 þ 2b02 :
ð1:5Þ
Theorem 1.1. Suppose (1.1) has a heteroclinic loop L0 with two hyperbolic saddles for e = 0. Then for any given constants K > d > 0 there exist a constant e0 = e0(K, d) > 0 and a neighborhood U = U(K, d) of L0 such that (1.1) has at most two limit cycles in U if jaij j 6 K;
jbij j 6 K for i þ j 6 2;
and
jA0 j þ jA1 j P d; 0 < e 6 e0 :
Based on the above theorem it is reasonable to give the following. Conjucture. For e > 0 small and aij, bij bounded (1.1) has at most two limit cycles near the heteroclinic loop L0. It is obvious that the conjecture is true if and only if the constant e0 in Theorem 1.1 satisfies inf e0 ðK; dÞ ¼ e0 ðKÞ > 0: d
Before stating next theorem we recall the definition of the order of a homoclinic loop. Consider a planar C1 system of the form x_ ¼ f0 ðx; yÞ þ f ðx; y; aÞ;
y_ ¼ g0 ðx; yÞ þ gðx; y; aÞ;
ð1:6Þ
m
where a 2 R is a vector parameter, f(x, y, 0) = g(x, y, 0) = 0. Suppose for a = 0 (1.6) has a homoclinic loop L with a hyperbolic saddle S. From [9,11], there exist constants C1, C2, C3, . . . such that if C1 = = Ck1 = 0, Ck 5 0, k P 1, then for all jaj small (1.6) has at most k limit cycles near L, and k limit cycles can appear for suitable f and g. In this case the homoclinic loop L is said to have order k. From [4,5] we know I of0 og0 of0 og0 C1 ¼ þ ðSÞ; C2 ¼ þ C 1 dt: ox oy ox oy L Then for quadratic system (1.1) we have Theorem 1.2. Let for e > 0 small (1.1) have a homoclinic loop L near L0. Then (i) it has order at most two unless (1.1) is Hamiltonian. (ii) The system (1.1) has no limit cycles near L0 if the homoclinic loop L has order 2. (iii) It can have two limit cycles near L0 by perturbating L. Theorem 1.3. Let for e > 0 small (1.1) have a heteroclinic loop L near L0. Then (i) (1.1) has no limit cycle on the plane, (ii) If the loop L is isolated, then it generates at most two limit cycles under further perturbations. In the next section we prove Theorems 1.1–1.3.
2. Proof of the main results We first give some preliminary lemmas. Suppose for e = 0 (1.1) has a family of periodic orbits {Lh, h 2 J} with a 2polycycle L0 = L1 [ L2 as the boundary. From [12], one of the connections L1 and L2 lies on a straight line. Without loss of generality, we assume that L1 lies on the y-axis with end points (i.e. saddles) (0,±1). It implies that H has the form
M. Han, C. Yang / Chaos, Solitons and Fractals 23 (2005) 1787–1794
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A H ðx; yÞ ¼ x y 2 1 x2 þ Bxy þ Cx : 3 In this case we have Hy = x[2y + Bx]. Set Y ¼ y þ B2 x. Then the system x_ ¼ H y ;
y_ ¼ H x
becomes ~ 2 2Cx; Y_ ¼ 1 Y 2 þ Ax
x_ ¼ 2xY ;
~ ¼ A þ 3 B2 . For definiteness, suppose the family {Lh, h 2 J} is in the right half plane. If C 5 0, by letting where A 4 X = jCjx we further have ~ A Y_ ¼ 1 Y 2 þ 2 X 2 2X : C
X_ ¼ 2XY ;
Hence, we have the following results. Lemma 2.1. Assume for e = 0 the quadratic system (1.1) has a 2-polycycle L0. Then (1.1) (e = 0) can be transformed into y_ ¼ H x ;
x_ ¼ H y ; where
or
or
ð2:1Þ
A H ¼ x y 2 1 x2 þ x ; 3
3 A< ; 4
ð2:2Þ
A H ¼ x y 2 1 x2 x ; 3
A < 0;
ð2:3Þ
A H ¼ x y 2 1 x2 ; 3
A < 0:
ð2:4Þ
We can write (2.2)–(2.4) uniformly as A H ¼ x y 2 1 x2 þ lx ; 3
l ¼ 0; 1:
The 2-polycycle L0 is given by H(x, y) = 0, x P 0, jyj 6 1 or L1 : x ¼ 0;
jyj 6 1; A L2 : y 2 ¼ 1 þ x2 lx; 3
0 6 x 6 x ðlÞ;
ð2:5Þ
where x* > 0 is the least positive zero of A3 x2 lx þ 1 ¼ 0. It is easy to see that 8 1; l ¼ 1; A ¼ 0 > > qffiffiffiffiffiffiffiffiffiffiffiffi
> > > 3 4A > l ¼ 1; A 6¼ 0 > < 2A 1 1 3 ; q ffiffiffiffiffiffiffiffiffiffiffiffi
x ¼ 3 4A > > > 2A 1 þ 1 3 ; l ¼ 1; A < 0 > > q ffiffiffiffi > > : 3; l ¼ 0; A < 0: A
For the phase portraits of (2.1) near L0, see Figs. 1 and 2. Let Z Q dx P dy; i ¼ 1; 2: Mi ¼ Li
ð2:6Þ
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M. Han, C. Yang / Chaos, Solitons and Fractals 23 (2005) 1787–1794
Fig. 1. (l = 1).
Fig. 2. (l = 1, 0).
Lemma 2.2. Suppose (2.2) and (2.3) or (2.4) hold. Then 2 M 1 ¼ a02 2a00 ; 3 2 M 2 ¼ N 0 A0 þ N 1 A1 þ a02 þ 2a00 ; 3 where N0 and N1 are positive constants depending on A. Proof. By (2.6) and GreenÕs formula, we have Z Z Z Z ðP x þ Qy Þ dx dy ¼ M1 þ M2 ¼ Int L0
ðA0 þ A1 x þ A2 yÞ dx dy ¼ N 0 A0 þ N 1 A1 ; Int L0
where N0 ¼
Z Z
N1 ¼
dx dy > 0; Int L0
Z Z x dx dy > 0: Int L0
By (2.5) and (2.6), we have Z 1 2 ða00 þ a11 y þ a02 y 2 Þ dy ¼ a02 2a00 : M1 ¼ 3 1 Hence, the conclusion follows.
h
i
Let Si = (0, (1) ) + O(e) = (xi, yi) denote the saddle point of (1.1) near (0, (1)i). Also, let f ¼ H y þ eP ;
g ¼ H x þ eQ:
M. Han, C. Yang / Chaos, Solitons and Fractals 23 (2005) 1787–1794
Denote by ki1 (>0) and ki2 (<0) the eigenvalues of the matrix ri = ki2/ki1, i = 1, 2. We have
oðf ;gÞ ðS i Þ. oðx;yÞ
1791
Then the hyperbolicity ratio of (1.1) at Si is
Lemma 2.3. The product of r1 and r2 has the expansion below for e small r ¼ r1 r2 ¼ 1 þ r0 e þ Oðe2 Þ;
r0 ¼ A0 :
Proof. Let oðf ; gÞ ðS i Þ ¼ Bi ðeÞ ¼ Bi ð0Þ þ OðeÞ; oðx; yÞ where Bi ð0Þ ¼
2ð1Þi 2l
0 2ð1Þi
! ;
i ¼ 1; 2:
Then the eigenpolynomial of Bi is k2 ui k þ vi ¼ 0; with ui ¼ trBi ¼ eðP x þ Qy ÞðS i Þ ¼ e½A0 þ ð1Þi A2 þ Oðe2 Þ ¼ eu0i þ Oðe2 Þ; vi ¼ det Bi ¼ 4 þ ev0i þ Oðe2 Þ;
v0i 2 R:
It is easy to obtain 1 ki1 ¼ 2 þ ð2u0i v0i Þe þ Oðe2 Þ; 4 1 ki2 ¼ 2 þ ð2u0i þ v0i Þe þ Oðe2 Þ: 4 Therefore, ri ¼
ki2 2 14 ð2u0i þ v0i Þe þ Oðe2 Þ 1 1 0 0 2 0 0 2 ¼ 1 ¼ þ v Þe þ Oðe Þ 1 v Þe þ Oðe Þ ð2u ð2u i i i i ki1 2 þ 14 ð2u0i v0i Þe þ Oðe2 Þ 8 8
1 ¼ 1 u0i e þ Oðe2 Þ; 2
i ¼ 1; 2:
Thus, we have 1 r ¼ r1 r2 ¼ 1 ðu01 þ u02 Þe þ Oðe2 Þ ¼ 1 A0 e þ Oðe2 Þ: 2 The proof is finished.
h
The following lemma is a corollary of Theorem 1.1 [6]. Lemma 2.4. Let a = (aij, bij) denote a vector parameter and a0 a given value of a. If (M1 + M2)ja=a0 5 0, then (1.1) has no limit cycles near L0 for e + ja a0j small. If (M1 + M2)ja=a0 = 0, r0ja=a0 5 0, then (1.1) has at most two limit cycles near L0 for e + ja a0j small. Proof of Theorem 1.1. We now prove Theorem 1.1 by using Lemmas 2.1–2.4. First by Lemma 2.1. We can suppose one of (2.2)–(2.4) holds. If the conclusion is not true, then there exist sequences ek ! 0, ak ¼ ðakij ; bkij Þ satisfying jakij j 6 K;
jbkij j 6 K;
ðjA0 j þ jA1 jÞja¼ak P d;
such that for (e, a) = (ek, ak), k ! 1, (1.1) has three limit cycles Lkj ! L0 ; j ¼ 1; 2; 3. We can suppose limk ! 1ak = a0, for some a0 ¼ ða0ij ; b0ij Þ. Obviously, we have ja0ij j 6 K;
jb0ij j 6 K;
ðjA0 j þ jA1 jÞja¼a0 P d:
ð2:7Þ
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On the other hand, it follows from Lemma 2.4 that ðM 1 þ M 2 Þja¼a0 ¼ 0;
r0 ja¼a0 ¼ 0:
Hence, (A0, A1)a=a0 = (0, 0) from Lemmas 2.2 and 2.3. This contradicts to (2.7). The proof is completed.
h
If we take (aij, bij) as constants, we can prove the following theorem. Theorem 2.1. Let P and Q be quadratic polynomials with constant coefficients aij and bij. If one of the following conditions holds: (i) ja10 + b01j + j2a20 + b11j 5 0, (ii) a11 + 2b02 = 0, then there exist e0 = e0(aij, b,ij) > 0 and a neighborhood U = U(aij, bij) such that (1.1) has at most two limit cycles in U for 0 < e 6 e0. Proof. Under condition (i) we have A0 5 0 or A1 5 0. Hence, by Lemmas 2.2 and 2.3, M1 + M2 5 0 or M1 + M2 = 0 and r0 5 0. Hence, the conclusion follows from Lemma 2.4. Under condition (ii), we have A2 = 0. By the conclusion (i) we may assume A0 = A1 = 0. Thus, Px + Qy = 0. That is, (1.1) is Hamiltonian and no limit cycles exist. This finishes the proof. h As we know, there exist functions d i ðeÞ ¼ eM i þ Oðe2 Þ;
i ¼ 1; 2
such that for e > 0 small (1.1) has a heteroclinic orbit connecting S1 and S2 near Li if and only if di(e) = 0. Also, from [6, p. 255] there exists a function dðeÞ ¼ eðM 1 þ M 2 Þ þ OðeÞ such that for e > 0 small (1.1) has a homoclinic orbit near L0 if and only if d(e) = 0 and di 5 0. Thus by Lemma 2.2 and the implicit function theorem, we have the following lemma. Lemma 2.5. There exist functions u1 ðeÞ ¼ 3a00 þ OðeÞ;
u2 ðeÞ ¼
N0 2 a02þ2a00 N 1 þ OðeÞ; A0 3 N1
uðeÞ ¼
N0 A0 þ OðeÞ; N1
such that for e > 0 small (1.1) has (i) a heteroclinic orbit near L if and only if a02 = u1(e), (ii) a heteroclinic orbit near L2 if and only if A1 = u2(e), (iii) a homoclinic orbit near L0 if and only if A1 = u(e) 5 u2(e).
Proof of Theorem 1.2. Let C i ðeÞ ¼ ðP x þ Qy ÞðS i Þ ¼ A0 þ ð1Þi A2 þ OðeÞ;
i ¼ 1; 2:
By Lemma 2.5, (1.1) has a homoclinic loop L* near L0 if and only if A1 = u(e) 5 u2(e). In this case, for definiteness, we can suppose L* is homoclinic to S1. Then a unique function q1(e) = A2 + O(e) exists such that for e > 0 small C 1 ðeÞ P 0 () A0 P q1 ðeÞ:
ð2:8Þ
It is obvious that L* is of order 1 if A0 5 q1(e). For A0 = q1(e), we have C 2 ðeÞ ¼ 2A2 þ OðeÞ: Thus a unique function q2(e) = O(e) exists such that for e > 0 small, C 2 ðeÞ P 0 () A2 P q2 ðeÞ:
ð2:9Þ
We claim that L* is of order 2 if A = q1(e), A2 5 q2 (e) and is non-isolated if A0 = q1(e), A2 = q2(e). In fact, we can write I I I IðA2 ; eÞ ¼ ðP x þ Qy Þ dt ¼ ðP x þ Qy Þ dt þ ðP x þ Qy Þ dt; ð2:10Þ L
L1
L2
M. Han, C. Yang / Chaos, Solitons and Fractals 23 (2005) 1787–1794
1793
where L1 is the intersection of L* with a neighborhood of S2, and L2 ¼ L L1 . Since C1(e) = 0, it is easy to prove by using the normal form near S1 that the second term in the right hand side of (2.10) approaches a finite limit as e goes to zero. Let e0 > 0 be a small constant. If A2 P e0, then jC 2 j P 12 e0 for e > 0 sufficiently small, and hence, I ðP x þ Qy Þ dt ! þ1; as e ! 0: C2 L1
Therefore, by (2.10) we have I C2 ðP x þ Qy Þ dt > 0 for jA2 j P e0 and e > 0 small:
ð2:11Þ
L
Now let jA2j 6 e0. From the above discussion, the function ðA1 ; A0 ; A2 Þ ¼ ðu ðeÞ; q1 ðeÞ; q2 ðeÞÞ is a unique solution of the equations d = C1 = C2 = 0, where q1 ðeÞ ¼ q1 ðeÞjA2 ¼q2 ðeÞ ;
ð2:12Þ
u ðeÞ ¼ uðeÞjA2 ¼q2 ðeÞ;A0 ¼q ðeÞ : 1
On the other hand, the equations A1 = A0 = A2 = 0 implies Px + Qy = 0, and hence d = C1 = C2 = 0. This shows that (A1, A0, A2) = (0, 0, 0) is also a solution of i the equations d = C1 = C2 = 0. Thus, by the uniqueness of solutions we obtain u ¼ 0;
q1 ¼ 0;
q2 ¼ 0;
ð2:13Þ
and (A1, A0A2) = (u, q1, q2) implies (A1, A0A2) = (0, 0, 0). That is, if (A1, A0 A2) = (u, q1, q2) then Px + Qy = 0 and therefore L* is non-isolated. It suffices to prove that if (A1, A0) = (u, q1) and A2 5 q2 then I C 2 IðA2 ; eÞ ¼ C 2 ðP x þ Qy Þ dt > 0 for 0 < jA2 j 6 e0 and e > 0 small: ð2:14Þ L
First, from (2.12), (2.13) and (2.9) we have q1 ðeÞ ¼ A2 ð1 þ OðeÞÞ; N0 uðeÞ ¼ A0 ð1 þ OðeÞÞ þ OðeA2 Þ; N1 C 2 ðeÞ ¼ 2A2 ð1 þ OðeÞÞ: Thus, inserting (2.15) into (2.10) we have for A1 = u(e), A0 = q1(e) I I C2 IðA2 ; eÞ ¼ A2 ð1 þ OðeÞÞ ðg0 ðeÞ þ g1 ðeÞx þ g2 ðeÞyÞ dt ¼ 2 L
ð2:15Þ
Gðe; x; yÞ dt; L
where g0 ð0Þ ¼ 1;
g1 ð0Þ ¼
N0 ; N1
g2 ð0Þ ¼ 1;
and Gðe; x; yÞ ¼ g0 þ g1 x þ g2 y ¼ A12 ðP x þ Qy Þ satisfies, Gðe; S 1 Þ 0;
Gðe; S 2 Þ ¼ 2 þ OðeÞ > 0:
Similar to (2.11) we have I Gðe; x; yÞ dt ! þ1 as e ! 0: L
Hence, (2.14) follows and the claim is proved. Therefore the conclusion (i) is obtained. For the second conclusion, let A1 = u(e), A0 = q1(e) and A2 5 0. By the third equality in (2.15), (2.11) and (2.14), the loop L* is stable (unstable) if A2 < 0 (A2 > 0) If (1.1) has a limit cycle, denoted by C1 near inside L*, then similar to (2.11) and (2.14) we can prove I ðP x þ Qy Þ dt > 0 for A2 6¼ 0 and e > 0 small: A2 C1
Thus C* and L* have the same stability, a contradiction. Then, the conclusion (ii) follows.
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M. Han, C. Yang / Chaos, Solitons and Fractals 23 (2005) 1787–1794
For (iii) let A2 P e0. Then by (2.11), L* is unstable. Fix A2 and e and vary A0 such that A0 < q1 and 0 < jA0 q1j A2. Then by (2.8), L* has changed its stability from unstable into stable, and hence an unstable limit cycle, say C, inside L* has appeared. Finally, change A1 such that 0 < uðeÞ A1 jA0 q1 ð) dðeÞ < 0Þ: Then L* has broken and a stable limit cycle surrounding C has appeared. Then (iii) follows.
h
Proof of Theorem 1.3. The conclusion (ii) is a direct corollary of [2,3] For (i), by Lemma 2.5, suppose a02 = u1(e), A1 ¼ u2 ðeÞ ¼ NN 01 A0 þ ðeÞ. By [12] L1 is, a straight line, and if (1.1) has a limit cycle C, it is unique and simple. We can suppose L = L1 [ L2 is stable. Then change A1 such that 0 < u2(e) A1 1()d2 < 0). Hence L2 has broken and a stable limit cycle C 0 has appeared. In the same time, the simple limit cycle C remains to exist. Hence, (1.1) has an invariant line and two limit cycles in the same time. This shows a contradiction. h
Acknowledgments The authors would like to thank the referees for their helpful comments and suggestions which improved the presentation of this work.
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