Linear estimate of the number of zeros of Abelian integrals for ...

SCIENCE IN CHINA (Series A)

Vol. 45 No. 8

August 2002

Linear estimate of the number of zeros of Abelian integrals for quadratic centers having almost all their orbits formed by cubics


 ) , LI Weigu (  & ZHANG Zhifen (   )

ZHAO Yulin (

1

 

)2 , LI Chengzhi (

2

)2

1. Department of Mathematics, Zhongshan University, Guangzhou 510275, China; 2. School of Mathematical Sciences, Peking University, Beijing 100871, China Correspondence should be addressed to Zhao Yulin (email: [email protected]) Received March 13, 2001; revised July 2, 2001 Abstract We study the number of zeros of Abelian integrals for the quadratic centers having almost all their orbits formed by cubics, when we perturb such systems inside the class of all polynomial systems of degree n. Keywords:

1

Abelian integrals, quadratic integrable system.

Introduction and statement of the main results We consider the polynomial perturbations of the planar polynomial integrable system ( x˙ = X(x, y) + εP (x, y), y˙ = Y (x, y) + εQ(x, y),

(1.1)ε

where ε is a small parameter, max{deg X, deg Y } = m, max{deg P, deg Q} = n. We assume that for ε = 0, system (1.1)ε has a first integral H(x, y) with integrating factor M (x, y). This means ∂H/∂x −∂H/∂y , Y (x, y) = . X(x, y) = M M Let Σ ⊂ R be the maximal connected intervals of existence of a continuous family of ovals (closed connected components of the real curve H(x, y) = h, h ∈ Σ, that are free of critical points). We denote the oval by Γh . The Abelian integrals for (1.1)ε is defined as I I(h) = (M (x, y)Q(x, y))dx − (M (x, y)P (x, y))dy,

h ∈ Σ.

(1.2)

Γh

The problem of finding an upper bound for the number of isolated zeros of Abelian integrals I(h) is called the weakened 16th Hilbert problem, posed by Arnold[1] . This problem is closely related to the problem of determining an upper bound for the number of limit cycles of perturbed integrable system (1.1)ε . The relationship between both problems comes from the following two facts: (i) If I(h∗ ) = 0 and I 0 (h∗ ) 6= 0, then there exists a hyperbolic limit cycle Lh∗ of system (1.1)ε such that Lh∗ → Γh∗ as ε → 0; and conversely, if there exists a hyperbolic limit cycle Lh∗ of system (1.1)ε such that Lh∗ → Γh∗ as ε → 0, then I(h∗ ) = 0[2] . (ii) The total number of isolated zeros of I(h) (taking into account their multiplicity) is an

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upper bound for the number of limit cycles of system (1.1)ε , which tends to some periodic orbits Γh of system (1.1)0 when ε → 0. Up to now most of the results on the weakened 16th Hilbert problem is concerned with the perturbation of Hamiltonian system, i.e. the case M ≡ 1, deg H = m + 1. For example, Khovansky[3] and Varchenko[4] proved independently that the upper bound of the number of zeros of I(h) is a finite number N (n, m) which depends only on the degree of polynomial H(x, y), P (x, y) and Q(x, y) , but an explicit expression for N (n, m) is unknown. Many authors have contributed to estimating or giving an upper bound for the number N (n, m), usually they fix H(x, y) and take arbitrary polynomials P (x, y) and Q(x, y) with degree n fixed or not (see refs.[5—18]). Some results concerned the cyclicity of homoclinic loop (cf. refs. [19, 20]) and the reference therein). For the integrable but non-Hamiltonian system, since M (x, y) is not a constant any more, also the functions M (x, y)P (x, y), M (x, y)Q(x, y) and H(x, y) are in general not polynomials any more, the study of Abelian integrals in this case is much more difficult than in the Hamiltonian case. As far as we know there are few papers studying the non-Hamiltonian centers (see refs. [21—27]). In our opinion, it should start with integrable systems that all their orbits are algebraic curves of low degree. We have obtained some results in this direction. In ref. [26], we have studied the quadratic centers whose almost all orbits are conics. In the sequel, the phrase “almost all” means “all except at most a finite number of ”. We proved that such a center must be an isochronous one or a linear center after removing the linear factor. We got also for these centers the least upper bound of the number of isolated zeros of Abelian integral under polynomial perturbations of degree n, which are linearly dependent on n. In this paper, we study the number of isolated zeros of Abelian integrals for the planar quadratic centers having almost all their orbits formed by cubics under small polynomial perturbations. Using the results from ref. [23] and taking a complex coordinate z = x + iy, the list of quadratic centers at (0, 0) having almost all their orbits is cubic looks as follows: i) The Hamiltonian system QH 3 : z 2, z˙ = −iz − z 2 + 2|z|2 + (b + ic)¯   b c 1 − 1 x3 + cx2 y − (1 + b)xy 2 − y 3 . H(x, y) = (x2 + y 2 ) + 2 3 3

(1.3)

ii) The Hamiltonian triangle: z˙ = −iz + z¯2 ,   1 2 1 2 H(x, y) = (1 − 2x) y − (x + 1) . 2 6

(1.4)

iii) The reversible system: z 2 , b 6= −1. z˙ = −iz + (2b + 1)z 2 + 2|z|2 + b¯    1 1 − 3b 2 3−b 1 b−1 X X + H(x, y) = X −3 y 2 + + 2 , 2 8(b + 1)2 b + 1 b+1 3b + 3 where X = 1 + 2(b + 1)x and M (x, y) = X −4 .

(1.5)

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iv) The generic Lotka-Volterra system: 1 z˙ = −iz + (1 − ci)z 2 + ci¯ z 2, c = ± √ , 3 √ !−3 √ 3 1 y H(x, y) = − (1 + 2x)(1 + x ± 3y) 1 + x ± , M (x, y) = 2 3

(1.6)± √

3 y 1+x± 3

!−4

.

In ref. [7], Horozov and Iliev proved that the number of zeros of Abelian integrals for the Hamiltonian center (1.3) and (1.4) does not exceed 5n + 15. In the present paper, we give an upper bound of the number of zeros of I(h) for (1.5) and (1.6)± . More precisely, our main result is as follows: Theorem 1.1.

For the system (1.5) and (1.6)± , the number of isolated zeros of I(h) in Σ,

taking into account their multiplicities, does not exceed 7n, where n = max{deg P (x, y), deg Q(x, y)}. In view of the above theorem and the results in ref. [7], it is natural to get Theorem 1.2. For the planar quadratic centers having almost all their orbits formed by cubic algebraic curves, the number of isolated zeros of the Abelian integrals I(h) in Σ associated with (1.1)ε does not exceed 7n + 15. Remark. Since the quadratic system (1.1)0 has at most two centers, the ovals Γh are defined at most in two connected components. For simplicity, in this paper we always denote Σ as a simple connected component. The rest of this paper is organized as follows: In sec. 2, we give another normal form for (1.5) , (1.6)± , and denote the Abelian integral as a linear combination of several basic integrals with polynomial coefficients. In sec. 3, the Picard-Fuchs equation will be derived, which implies a Riccati equation we need. In sec. 4, the initial problem is reduced to counting the number of isolated zeros of certain Abelian integral which is expressed as a linear combination of only two basic integrals. Using the above results, we get the main theorem of this paper.

2

The algebraic structure of the related Abelian integrals In what follows we are going to express the Abelian integrals (1.2), related to (1.5) and

(1.6)± , as a linear combination of several basic integrals. Returning back to real coordinates, we give another normal form of the perturbed system (1.1)ε corresponding to (1.5) and (1.6)± . More precisely, we have Lemma 2.1.

Corresponding to system (1.5) and (1.6)± , the perturbed system (1.1)ε can

be reduced to the following normal form   x˙ = xy + εf (x, y),

(2.1)ε  y˙ = 3C + 2Bx + Ax2 + 3 y 2 + εg(x, y), 2 where f (x, y) and g(x, y) are polynomial of x, y with max{deg f (x, y), deg g(x, y)} = n. A first integral of system (2.1)0 is given by   1 2 −3 2 H(x, y) = x (2.2) y + Ax + Bx + C = h 2 with the integrating factor M (x, y) = x−4 . i) B = −A − 1, C = (A + 2)/3, if (2.1)ε comes from (1.5);

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ii) A = −9/8, B = 3/4, C = −1/8, if (2.1)ε comes from (1.6)± . √

We get this lemma by linear transformations. For example, taking X = 1 + x + √ 3y/3, Y = −x/2 + 3y/2, we obtain (2.1)ε from (1.6)+ . Proof. Denote Ii,j (h) =

and

I

i j

M (x, y)x y dx = Γh

I

xi−4 y j dx, i = −1, 0, 1, · · · , j = 0, 1, 2, · · · , Γh

Ji (h) = Ii,1 (h), i = −1, 0, 1, · · · , where Γh is the compact component of H = h, defined in (2.2). Since H(x, −y) = H(x, y), we obviously have Ii,2k = 0 for k = 0, 1, 2, · · ·. Our main result in this section is the following Suppose C 6= 0, n > 4 in (2.1)ε , then the Abelian integral I(h) can be

Proposition 2.1. expressed as I(h) =

1 hn−3

J(h), J(h) = α(h)J−1 (h) + β(h)J0 (h) + γ(h)J1 (h),

(2.3)

where α(h), β(h) and γ(h) are polynomials of h with degα(h) 6 n − 3, degβ(h) 6 n − 3, degγ(h) 6 n − 3. Moreover, I(h) = α(h)J−1 +β(h)J0 with degα(h) = degβ(h) = 0 for n = 1; I(h) = α(h)J−1 + β(h)J0 + γ(h)J1 with degα(h) =degβ(h) = degγ(h) = 0 for n = 2; and I(h) = (1/h)(α(h)J−1 + β(h)J0 + γ(h)J1 ) with degα(h) = 1, degβ(h) = degγ(h) = 1 for n = 3. Proof. By integration by parts, we get I I 1 i−4 xi−4 dy j+1 = − Ii−1,j+1 , i 6 0. M (x, y)xi y j dy = j + 1 j +1 Γh Γh

Therefore, in the rest of this paper, we express the Abelian integral (1.2) as the form I I I(h) = xi−4 g(x, y)dx, i > 1. Γh

(2.4)

Γh

where degg(x, y) = n. In what follows we split the proof into three steps. 1) The Abelian integral (2.4) can be denoted as the form I(h) =

n−1 X

c i Ji .

(2.5)

i=−1

It follows from (2.2) that

∂y 3 (2.6) − x−4 y 2 − Ax−2 − 2Bx−3 − 3Cx−4 = 0. ∂x 2 Multiplying (2.6) by xi y j−2 and integrating it over Γh , we get 2j Ii,j = − (AIi+2,j−2 + 2BIi+1,j−2 + 3CIi,j−2 ). (2.7) 2i + 3j − 6 The condition 2i + 3j − 6 = 0 holds if and only if (i, j) = (3, 0) or (0, 2). Noting I30 = I02 = 0, the equality (2.7) implies the result of this step. 2) The Abelian integral Ji (h), i > 4, can be denoted as the form 1 Ji (h) = i−2 (αi,−1 (h)J−1 (h) + βi,0 (h)J0 (h) + γi,1 (h)J1 (h)), (2.8) h x−3 y

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where αi,−1 (h), βi,0 (h) and γi,1 (h) are polynomials of h with degαi,−1 (h) 6 [(2i − 7)/3], degβi,0 (h) 6 [(2i − 7)/3], degγi,1 (h) 6 [(2i − 5)/3]. [s] denotes the integer part of s. For i = 2, 3, we have 1 1 J2 (h) = (−8CJ−1 − 5BJ0 − 2AJ1 ), J3 = (−2CJ0 − BJ1 ). (2.9) h h To prove (2.8) and (2.9), we first derive some equalities. Rewrite (2.2) in the form 1 2 y + Ax2 + Bx + C = hx3 , (2.10) 2 which yields 1 hIi,j = Ii−3,j+2 + AIi−1,j + BIi−2,j + CIi−3,j . (2.11) 2 It follows from (2.7) that 2(j + 2) (AIi−1,j + 2BIi−2,j + 3CIi−3,j ). Ii−3,j+2 = − (2.12) 2i + 3j − 6 Substituting (2.12) into (2.11), we obtain h(2i + 3j − 6)Ii,j = 2(i + j − 4)AIi−1,j + (2i + j − 10)BIi−2,j + 2(i − 6)CIi−3,j ,

(2.13)

which implies 1 {2(i − 3)AJi−1 + (2i − 9)BJi−2 + 2(i − 6)CJi−3 }. (2.14) h(2i − 3) Taking i = 2, 3 in (2.14) respectively, we get (2.9). In the following we prove (2.8) by induction. Using (2.14), it is easy to prove by direct computation that (2.8) holds for i = 4, 5, 6. Suppose for k 6 i − 1, Jk (h) can be denoted as (2.8) with degαk,−1 (h) 6 [(2k − 7)/3], degβk,0 (h) 6 [(2k − 7)/3], degγk,1 (h) 6 [(2k − 5)/3]. Then for k = i, it follows from (2.14) that Ji (h) can be expressed as (2.8) with 1 {2A(i − 3)αi−1,−1 + (2i − 9)Bhαi−2,−1 + 2(i − 6)Ch2 αi−3,−1 }, αi,−1 (h) = 2i − 3 1 {2A(i − 3)βi−1,0 + (2i − 9)Bhβi−2,0 + 2(i − 6)Ch2 βi−3,0 }, βi,0 (h) = 2i − 3 1 {2A(i − 3)γi−1,1 + (2i − 9)Bhγi−2,1 + 2(i − 6)Ch2 γi−3,1 }. γi,1 (h) = 2i − 3 Therefore,         2i − 9 2i − 11 2i − 13 2i − 7 , 1+ , 2+ = . degαi,−1 (h) 6 max 3 3 3 3 Similarly, we get degβi,0 (h) 6 [(2i − 7)/3], degγi,1 (h) 6 [(2i − 5)/3]. 3) Finally, we get the main result of this proposition from (2.5), (2.8) and (2.9). Proposition 2.2. Suppose n > 4 in (2.1)ε and A = −2, B = 1, C = 0 (corresponding to the nongeneric case of (1.5)), then the Abelian integrals (1.2) can be denoted as 1 (2.15) I(h) = n−2 J(h), J(h) = α(h)J−1 (h) + β(h)J0 (h), h where α(h) and β(h) are polynomials of h with degα(h) 6 n − 2, degβ(h) 6 n − 2. Moreover, I(h) = α(h)J−1 + β(h)J0 with degα(h) = degβ(h) = 0 for n = 1; I(h) = (1/h)(α(h)J−1 (h) + β(h)J0 (h)) with degα(h) = 1, degβ(h) = 1 for n = 2; and I(h) = (1/h2 ) (α(h)J0 (h) + β(h)J1 (h)) with degα(h) = 2, degβ(h) = 2 for n = 3. Proof. Using the same arguments as in the proof of Proposition 2.1, we get 1 1 1 J1 = (7J−1 − 8J0 ), J2 = 2 (28J−1 − (5h + 32)J0 ), Ji = i−1 (αi,−1 (h)J−1 + βi,0 (h)J0 ), h h h where degαi,−1 (h) 6 [(i − 3)/2], degβi,0 (h) 6 [(i − 2)/3], i > 3. The expression (2.15) follows from (2.5). Ji =

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Proposition 2.3. Suppose A = −9/8, B = 3/4, C = −1/8 in (2.1)ε (corresponding to (1.6)± ), then the Abelian integrals (1.2) can be denoted as 1 I(h) = n−3 J(h), J(h) = α(h)J−1 + β(h)J0 (h), (2.16) h where degα(h) 6 n − 3, degβ(h) 6 n − 3, n > 4. Moreover, I(h) = α(h)J−1 + β(h)J0 with degα(h) = degβ(h) = 0 for n = 1, 2; and I(h) = (1/h)(α(h)J−1 + β(h)J0 ) with degα(h) = 1, degβ(h) 6 1 for n = 3. Proof. In this case, the first integral (2.2) of (2.1)0 can be reduced to  2 ! 1 2 9 1 −3 y − H(x, y) = x = h. x− 2 8 3 2 Denote by (xi (h), 0), i = 1, 2, the intersection points of Γh and x-axis, hence (9/4)x−3 i (xi −1/3) = −2h, h ∈ Σ = (−1/2, 0). Therefore, s  2 Z x2 (h) 9 1 −4 3 J1 (h) − J0 (h) = 2 x (x − 1) 2hx + dx x− 4 3 x1 (h)  s Z x2 (h) 8h 1 −4 x (x − 1) x − + 1dx = 3 −3 3 9x (x − 13 )2 x1 (h) s  2 ! Z 4 x2 (h) 8h 9 −3 1 x = − x−
2 + 1d 3 x1 (h) 4 3 9x−3 x − 13 s s   2 2 4 9 −3 9 1 1 = − x 2h + x−3 x − x− 3 4 3 4 3  s s   2 2  x2 (h) 9 9 1 1  x−3 x − +2h ln  2h + x−3 x − +  4 3 4 3 x1 (h)

= 0,

which means J0 (h) ≡ J1 (h). Using this result, we get (2.16) from (2.3).

3

Picard-Fuchs equations and Riccati equations

In this section, we shall derive the Picard-Fuchs equation of Ji (h), i = −1, 0, 1, which implies the Riccati equation we need. Lemma 3.1. Suppose C 6= 0 in (2.1)ε , then the vector V = col(J−1 (h), J0 (h), J1 (h)) satisfies the following Picard-Fuchs equation (Dh + S)V 0 = U V ,

(3.1)

where 

 D= 

−6C 2

2BC

2(2AC − B 2 )

0

−6C

2B

0

0

−3 

 U = 





2B(B 2 − 3AC) 2A(B 2 − 2AC) 0

   , S =  2(2AC − B 2 )   B  −8C 2 BC 2AC − B 2  . 0 −6C B  0 0 −2

−2AB 2A



 0  , 0

970

Proof.

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Vol. 45

We can get (3.1) by a standard way. It follows from (2.10) that x3 ∂y = , ∂h y

(3.2)

which implies Ji0 (h) = Therefore,

I

xi−1 dx. y

Γh

(3.3)

xi−4 y 2 dx y Γ Ih xi−4 (2hx3 − 2Ax2 − 2Bx − 2C) = dx y Γh

Ji (h) =

I

0 0 0 = 2hJi0 − 2AJi−1 − 2BJi−2 − 2CJi−3 .

Differentiating (2.10) with respect to x, we obtain ∂y y + 2Ax + B = 3hx2 . ∂x Using (3.3), (3.5) and integration by parts, we have I I 1 1 i−3 Ji (h) = ydx =− xi−3 dy i − 3 Γh i − 3 Γh I xi−3 (3hx2 − 2Ax − B) 1 dx = − i − 3 Γh y 1 0 0 = − (3hJi0 − 2AJi−1 − BJi−2 ), i−3 i.e. 0 0 (i − 3)Ji = −3hJi0 + 2AJi−1 + BJi−2 .

(3.4)

(3.5)

(3.6)

It follows from (3.4) and (3.6) that 0 0 J1 = 2hJ10 − 2AJ00 − 2BJ−1 − 2CJ−2 ,

J0 =

2hJ00

−

0 2AJ−1

−

0 2BJ−2

−

0 2CJ−3 ,

0 0 0 −4J−1 = −3hJ−1 + 2AJ−2 + BJ−3 .

(3.7) (3.8) (3.9)

0 0 , J−2 from (3.7)—(3.9), we get the first equation of (3.1). Eliminating J−3 Taking i = 0 in (3.6), we obtain 0 0 −3J0 = −3hJ00 + 2AJ−1 + BJ−2 .

(3.10)

The second equation of (3.1) follows from (3.7) and (3.10). Substituting i = 1 into (3.6), we get the third equation of (3.1). The lemma is proved. Differentiating both sides of (3.1), we get (Dh + S)V 00 = (U − D)V 0 , where



 U −D = 

−2C 2

−BC

B 2 − 2AC

0

0

−B

0

0

1

(3.11) 

 , 

which contains zero submatrix of order two in the lower-left corner. Therefore, taking Z = −2C 2 J−1 − BCJ0 ,

(3.12)

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we get the following proposition from (3.11) and (3.12): Proposition 3.1. The following relation holds in the generic case, provided B = −A − 1, C = (A + 2)/3 6= 0:     00 ! a−1,0 (h) a−1,1 (h) J−1  Z0    00   , (3.13) ∆(h)  a0,1 (h)   J0  Z  =  a0,0 (h) 1 00 J1 a1,1 (h) a1,0 (h) where

  A−1 4 ∆(h) = − (A + 2) h − h{3(A + 2)2 h − A2 (A + 3)}, 3 3 a−1,0 (h) = 2h{3(A + 2)h − A(A + 1)}, 4 a−1,1 (h) = (A + 2)h2 , 3 4 a0,0 (h) = (A + 2)h{(A − 1)(A + 1)(A + 3) − 3(A + 2)2 h}, 9 2 a0,1 (h) = − (A + 1)2 h{A(A − 1)(A + 1)(A + 3) − (A + 2)(3A2 + 6A − 1)h}, 27 2 a1,0 (h) = {A(A − 1)(A + 3) − 3(A + 1)(A + 2)h}, 3 a1,1 (h) = −a0,0 (h). Since the straight line x = 0 is invariant line of system (2.1)0 , we conclude that the curve Γh does not intersect y-axis, which means x 6= 0 for x ∈ Γh and x ∈ intΓh . This yields Z Z Z x2 (h) xi−1 √ Ji (h) = xi−4 dxdy 6= 0, Ji0 (h) = 2 dx 6= 0, 2hx3 − 2Ax2 − 2Bx − 2C x1 (h) intΓh where (x1 (h), 0) and (x2 (h), 0) are the intersection points of Γh and x-axis. Hence, we can define Z 0 (h) . ω(h) = 0 J1 (h) Corollary 3.1. Suppose B = −A − 1, C = (A + 2)/3 6= 0, then the ratio ω(h) satisfies the following Riccati equation ∆(h)ω 0 = r0 (h)ω 2 + r1 (h)ω + r2 (h),

(3.14)

where r0 (h) = −a1,0 (h), r1 (h) = 2a0,0 (h), r2 (h) = a0,1 (h). Proof. Eq. (3.14) follows directly from the second and the third equations of system (3.13). Proposition 3.2. Suppose A = −2, B = 1, C = 0, then i) The Abelian integral J−1 (h), J0 (h) satisfy the following Picard-Fuchs equation ! ! ! 0 J−1 7h −3h J−1 = . (3.15) 4h(h + 1) −7 5h + 8 J00 J0 ii) The ratio v(h) = J0 /J−1 satisfies the following Riccati equation 4h(h + 1)v 0 = 3hv 2 + (−2h + 8)v − 7. Proof.

(3.16)

In the case A = −2, B = 1, C = 0, it follows from (3.4) and (3.6) that 0 0 0 0 − 2Ji−2 , (i − 3)Ji = −3hJi0 − 4Ji−1 + Ji−2 . Ji (h) = 2hJi0 + 4Ji−1

Using the same arguments as in the proof of Proposition 2.1, we get (3.15), which implies (3.16). Proposition 3.3. Suppose A = −9/8, B = 3/4, C = −1/8.

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i) The Abelian integrals J−1 (h), J0 (h) satisfy the following Picard-Fuchs equation ! ! ! 0 2(4h + 3) −8h − 9 J−1 J−1 = . 3h(1 + 2h) 2 4h − 3 J00 J0

Vol. 45

(3.17)

ii) The ratio u(h) = J0 /J−1 satisfies the following Riccati equation 3h(1 + 2h)u0 = (8h + 9)u2 − (4h + 9)u + 2.

(3.18)

Proof. In the proof of Proposition 2.3, we get J1 (h) ≡ J0 (h). Substituting J1 = J0 into (3.1), we obtain the equation (3.17), which implies (3.18).

4

Linear estimate for the number of zeros of I(h)

In this section, we prove the main results of this paper. To be more concrete, we only consider in Lemmas 4.1—4.3 the generic case, provided B = −A − 1, C = (A + 2)/3 6= 0, n > 4. Our strategy is to reduce the problem of estimating the number of zeros of I(h) (i.e.J(h)) to a problem of estimating the number of zeros of certain Abelian integral expressed only as a linear combination of Z 0 and J10 with rational coefficients. This will be done by eliminating J00 (h) from J 0 (h). From (2.3), (3.1) and (3.12), we get 0 + β1 (h)Z 0 + γ1 (h)J10 , J 0 (h) = α1 (h)J−1

(4.1)

where degα1 (h) 6 n − 3, degβ1 (h) 6 n − 3, degγ1 (h) 6 n − 3. In what follows let Σ = (hc , hs ). Without loss of generality, suppose that h = hc is the Hamiltonian value corresponding to the center of (2.1)0 . Lemma 4.1. Suppose ℵ is the set of zeros of α1 (h) in Σ, then in Σ\ℵ, we have  0 0 J (h) G(h) , (4.2) = α1 (h) ∆(h)α21 (h) where G(h) = β2 (h)Z 0 + γ2 (h)J10 (4.3) with degβ2 (h) 6 2n − 4, degγ2 (h) 6 2n − 4. Proof. It follows from (4.1) and (3.12) that      0  0  0 0 β1 γ1 β1 γ1 J (h) 00 = J−1 + Z0 + J10 Z 00 + J100 + α1 (h) α1 α1 α1 α1 1 (β2 (h)Z 0 + γ2 (h)J10 ), = ∆(h)α21 (h) where β2 (h) = α21 a−1,0 (h) + α1 β1 a0,0 (h) + α1 γ1 a1,0 (h) + (α1 β10 − α01 β1 )∆(h), γ2 (h) = α21 a−1,1 (h) + α1 β1 a0,1 (h) + α1 γ1 a1,1 (h) + (α1 γ10 − α01 γ1 )∆(h), which implies degβ2 (h) 6 2n − 4, deg γ2 (h) 6 2n − 4. Lemma 4.2. Suppose µ, ν, η are the numbers of isolated zeros of J 0 (h), G(h) and α1 (h) in Σ respectively, then µ 6 ν + η + 3. (4.4) Proof. Since hc must be one of three values h = (A − 1)/3, h = 0 and h = (A2 (A + 3))/ (3(A + 2)2 ), we conclude that ∆(h) has at most two zeros in Σ. Without loss of generality, in the following we suppose ∆(h) has two zeros in Σ. Denote by ξ1 , ξ2 , · · · , ξη+2 , the elements of zeros of ∆(h)α1 (h) in Σ, ξ0 = hc , ξη+3 = hs , ξj < ξj+1 , j = 0, 1, · · · , η + 2. Around each ξj , choose a small subinterval Oj = [ξj − δ, ξj + δ] such that

No. 8

ABELIAN INTEGRALS FOR QUADRATIC CENTERS

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J 0 (h), G(h) and ∆(h)α1 (h) has no zero in it except ξj . Denote by µj , νj the number of zeros of J 0 (h), G(h) in the open interval oej = (ξj + δ, ξj+1 − δ) respectively. The equality (4.2) implies µj 6 νj + 1. In the interval Oj , only the point ξj could be a common zero of J 0 (h) and G(h) with multiplicity lj , mj respectively. If ξj is a zero of α1 (h) with multiplicity aj and lj 6= aj , then (4.2) shows that lj − aj − 1 = mj − 2aj if ∆(ξj ) 6= 0, or lj − aj − 1 = mj − 2aj − 1, if ∆(ξj ) = 0, ∆0 (ξj ) 6= 0, which implies lj 6 mj . By the same arguments we get lj 6 mj for the other cases. Hence, we have η+2 η+2 X X µ = µ0 + (µj + lj ) 6 ν0 + 1 + (νj + mj + 1) 6 ν + η + 3. j=1

j=1

In what follows let φ(n), ψ(n) be the number of isolated zeros of J(h), I(h) in Σ respectively. Lemma 4.3. Suppose B = −A − 1, C = (A + 2)/3 6= 0, n > 4, then φ(n) 6 7n − 7. Proof. Since J(hc ) ≡ 0, we have φ(n) 6 µ. Our main concern will be the estimate of the number ν of the zeros of G(h). Denote G(h) = β2 (h)ω + γ2 (h). W (h) = 0 J1 (h) As J10 (h) 6= 0 for h ∈ Σ, W (h) just has ν zeros. By (3.14), we know that W = W (h) satisfies the following Riccati equation ∆(h)β2 (h)W 0 (h) = R0 (h)W 2 + R1 (h)W + R2 (h),

(4.5)

R2 (h) = ∆(h)(β2 γ20 − β20 γ2 ) + r0 γ22 − r1 β2 γ2 + r2 β22 .

(4.6)

where It follows from Proposition 3.1, Corollary 3.1 and Lemma 4.1 that deg R2 (h) 6 4n − 6. If e h1 , e h2 are any two consecutive zeros of W (h) in Σ, then β2 (e hi )∆(e hi )W 0 (e hi ) = R2 (e hi ), i = e e 1, 2. Therefore, there is at least one zero of R2 (h) or ∆(h)β2 (h) in (h1 , h2 ). This gives ν 6 the number of zeros of R2 (h) + the number of zeros of ∆(h)β2 (h) + 1 6 degR2 (h) + degβ2 (h) + 2 + 1

6 (4n − 6) + (2n − 4) + 3 = 6n − 7. Noting η 6 degα1 (h) 6 n − 3, it follows from (4.4) that φ(n) 6 µ 6 7n − 7. Proof of Theorem 1.1. We have proved in Lemma 4.3 that ψ(n) = φ(n) 6 7n − 7 for the perturbed system (2.1)ε with B = −A − 1, C = (A + 2)/3 6= 0, n > 4. In what follows we consider the other cases. 1) For (2.1)ε with B = −A − 1, C = (A + 2)/3 6= 0, it is obviously that ψ(1) 6 7. Using the same arguments as in Lemma 4.3, we have ψ(2) 6 14, ψ(3) 6 21. 2) For (2.1)ε with A = −2, B = 1, C = 0, n > 4, it follows from Proposition 2.2 and Proposition 3.2 that the ratio W = W (h) = J(h)/J−1 (h) satisfies e0 (h)W 2 + R e1 (h)W + R e2 (h), 4h(h + 1)β(h)W 0 = R

e2 (h) 6 2n − 3. Using the same arguments as in the proof of Lemma 4.3, we get where deg R ψ(n) 6 3n − 4, n > 4. It is easy to check that in this case ψ(1) 6 2, ψ(2) 6 5, ψ(3) 6 8. 3) For (2.1)ε with A = −9/8, B = 3/4, C = −1/8, using the same arguments as the above, we get ψ(n) 6 3n − 6 for n > 4 and ψ(1) = ψ(2) 6 3, ψ(3) 6 6. Summing up the above discussions, we have ψ(n) 6 7n. Acknowledgements The first author wants to thank School of Mathematical Sciences of Peking University for its support and hospitality during the period when this paper was elaborated. This work was supported by the

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SCIENCE IN CHINA (Series A)

Vol. 45

National Natural Science Foundation of China (Grant No. 10101031), Guangdong Natural Science Foundations (Grant No. 001289) and Natural Science Foundation of Zhongshan University for young teachers.

References 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27.

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