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Perturbations of non-Hamiltonian reversible quadratic systems with cubic orbits
Nonlinear Analysis 64 (2006) 2332 – 2351 www.elsevier.com/locate/na
Perturbations of non-Hamiltonian reversible quadratic systems with cubic orbits夡 Yulin Zhao Department of Mathematics, Sun Yat-sen University, Guangzhou, 510275, PR China Received 8 April 2005; accepted 15 August 2005
Abstract This paper is concerned with degree n polynomial perturbations of a class of planar non-Hamiltonian reversible quadratic integrable system whose almost all orbits are cubics. We give an estimate of the number of limit cycles for such a system. If the first-order Melnikov function (Abelian integral) M1 (h) is not identically zero, then the perturbed system has at most 5 for n = 3 and 3n − 7 for n 4 limit cycles on the finite plane. If M1 (h) is identically zero but the second Melnikov function is not, then an upper bound for the number of limit cycles on the finite plane is 11 for n = 3 and 6n − 13 for n 4, respectively. In the case when the perturbation is quadratic (n = 2), there exists a neighborhood U of the initial non-Hamiltonian polynomial system in the space of all quadratic vector fields such that any system in U has at most two limit cycles on the finite plane. The bound for n = 2 is exact. 䉷 2005 Elsevier Ltd. All rights reserved. MSC: 34C07; 34C08; 37G15; 34M50 Keywords: The kth order Melnikov function; Limit cycles
1. Introduction Recall that the quadratic integrable system are divided into four cases [23]: QH 3 (Hamiltonian), (reversible), Q4 (codimension four) and QLV (generalized Lotka–Volterra). In this paper, we 3 H whose almost study degree n polynomial perturbations of quadratic reversible system QR \Q 3 3 all orbits are cubic, where the phrase “almost all” means “all except at most a finite number of”. QR 3
夡 Supported by NSF of China (No. 10571184), Guangdong Natural Science Foundation (No. 04009794), SRF for ROCS, SEM, and the Foundation of Sun Yat-sen University Advanced Research Center. E-mail addresses: [email protected], [email protected].
0362-546X/$ - see front matter 䉷 2005 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2005.08.016
Y. Zhao / Nonlinear Analysis 64 (2006) 2332 – 2351
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Using the results from [12,24], the reversible non-Hamiltonian quadratic systems we consider have the following normal form: x˙ = xy =
Hy , x −4
3 Hx y˙ = (A + 2) − 2(A + 1)x + Ax 2 + y 2 = − −4 , 2 x
where H (x, y) is a first integral of (1.1), defined by 1 2 A+2 y + Ax 2 − (A + 1)x + H (x, y) = x −3 2 3
(1.1)
(1.2)
with integrating factor x −4 . Using the normal form (1.1), we list the types of critical points appearing in (1.1) as follows: (1) (2) (3) (4) (5)
a center, a saddle and two nodes: A ∈ (−∞, −2); a center and a degenerate critical point: A = −2; two centers: A ∈ (−2, 0); a center: A = 0; and a center and a saddle: A ∈ (0, +∞).
The purpose of the present paper is to investigate the bifurcations of limit cycles in polynomial perturbations of system (1.1) with a center and a saddle, i.e., A ∈ (0, +∞). Consider the corresponding perturbed system x˙ = xy + f (x, y, ), y˙ = (A + 2) − 2(A + 1)x + Ax 2 + 23 y 2 + g(x, y, ),
A ∈ (0, +∞).
(1.3)
Here f (x, y, ) and g(x, y, ) are polynomials in x, y with coefficients depending analytically on the small parameter , f (x, y, ) =
∞ k=1
k−1 fk (x, y),
g(x, y, ) =
∞
k−1 gk (x, y),
(1.4)
k=1
max{deg fk (x, y), deg gk (x, y)} n, k = 0, 1, 2, . . . , +∞. Let h ⊂ {(x, y)|H (x, y) = h} be the closed orbit of system (1.1) around the center, h ∈ (hc , hs ). A key point in the investigation of the number of limit cycles in (1.3) is to reduce the original problem to counting the number of zeros of the displacement function d(h, ) = M1 (h) + 2 M2 (h) + · · · + k Mk (h) + · · ·
(1.5)
for sufficiently small ||. It is well known that the number of zeros of the first non-disappearing kth order Melnikov function Mk (h) in (1.5) determines the upper bound of the number of limit cycles in (1.3) emerging from the periodic orbits of system (1.1). The main results of this paper are the following: Theorem 1.1. Assume n4. Let K ⊂ R2 be a fixed compact domain and A ∈ (0, +∞). If M1 (h) ≡ / 0, then system (1.3) has at most 3n − 7 limit cycles in K for small enough. If M1 (h) ≡ / 0, then (1.3) has at most 6n − 13 limit cycles in K. 0 but M2 (h) ≡
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For n = 3, system (1.3) has at most 5 (resp. 11) limit cycles if M1 (h) ≡ / 0 (resp. M1 (h) ≡ / 0). 0, M2 (h) ≡ Theorem 1.2. Let K ⊂ R2 be a fixed compact domain as in Theorem 1.1 and A ∈ (0, +∞). There is a neighborhood U of (1.1) in the space of all quadratic vector fields, such that any system in U has at most two limit cycles in K. This bound is exact. In the proof of Theorem 1.1, we will show that M1 (h) has at most 3n − 7 zeros on the interval (0, 1/6); see Theorem 5.4 below. This improves the upper bound given in [24] for the total number of zeros of Abelian integral M1 (h). The result from Theorem 1.2 has been obtained in [22] in different ways. We also note that there has been a substantial amount of work devoted to estimating the number of either limit cycles or zeros of Melnikov functions; see for instance [1,4,6,10,11,14–18,21,25–27] and reference therein. The rest of this paper is organized as follows: in the next section, system (1.3) is reduced to another normal form by a Poincaré map. This new system is a small non-polynomial perturbation of a quadratic Hamiltonian vector field. The polynomial perturbations of such quadratic Hamiltonian vector fields have been studied in [5,7]. We obtain the cohomology decomposition formula of the corresponding non-polynomial differential 1-form in Section 3. Using them allows us to derive an appropriate formula of M1 (h) and M2 (h) in Section 4. Finally, in Section 5 we estimate the number of both zeros of Mk (h), k = 1, 2 and limit cycles of (1.3) by following the arguments in [5,7]. 2. Normal form In order to study limit cycles in (1.3), we consider a change of variables. The main result of this section is the following: Proposition 2.1. The perturbed system (1.3) with A ∈ (0, +∞) can be reduced to the following normal form: x˙ = y(1 + 2ax) +
∞
k−1 pk (x, y),
k=1
y˙ = −x + x 2 − ay 2 + ε
∞
k−1 qk (x, y),
(2.1)
k=1
where a ∈ (− 21 , 0), pk (x, y) =
m n
Pkmj y j
m=0 j =0
(1 + 2ax)
, m−3
qk (x, y) =
n m+1
Qkmj y j
m=0 j =0
(1 + 2ax)m−2
(2.2)
and Pkmj , Qkmj are real constants. A first integral of system (2.1)0 is (x, y) = 1 x 2 + 1 y 2 − 1 x 3 + axy 2 = h. H 2 2 3
(2.3)
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The period annulus h is defined on the interval (0, 1/6). 0 and 1/6 correspond to the center (x, y) (0, 0) and the homoclinic loop through the saddle (1, 0), respectively. The Hamiltonian H has three critical values, defined by 1 + 3a 2a + 1 1 1 − ,± (0, 0) = 0, H (1, 0) = , H = (2.4) = hl . H 2a 4a 3 24a 3 6 Proof. Let fk (x, y) =
n
Fkij x i y j ,
gk (x, y) =
i+j =0
n
Gkij x i y j ,
i+j =0
where Fkij , Gkij are real constants. When we make the Poincaré transformation 1 y , u = , dt = −z d, x x system (1.3) is changed to ⎛ ⎞⎞ ⎛ n ∞ j F u dz kij ⎠⎠ , k−1 ⎝ = uz + ⎝ zi+j −3 d z=
k=1
(2.5)
i+j =0
du 1 = − A + 2(A + 1)z − (A + 2)z2 − u2 d 2 ⎛ ⎞⎞ ⎛ ∞ n j j +1 + F u −G u kij kij ⎠⎠ . + ⎝ k−1 ⎝ zi+j −2 i+j =0
k=1
The substitution 1 a=− , A+2
v=
1 (z − 1) 2a
transforms the above system into the form ⎛ ⎞⎞ ⎛ ∞ n Fkij uj dv 1 1 ⎝ k−1 ⎝ ⎠⎠ , = u(1 + 2av) + i+j −3 d 2a 2a (1 + 2av) k=1 i+j =0 ⎛ ⎞⎞ ⎛ ∞ n j j +1 −Gkij u + Fkij u du 1 ⎠⎠ . k−1 ⎝ = −4av + 4av 2 − u2 + ⎝ 2 d (1 + 2av)i+j −2 k=1
i+j =0
Finally, we obtain (2.1) by using the scaling √ u u¯ = √ , = 2, v → x, u¯ → y, 2 2a
m = i + j.
Remark 2.2. Proposition 2.1 yields that (1) To study the number of limit cycles of system (1.3), it is sufficient to study (2.1) . H (2) (1.3) is a polynomial perturbed system of non-Hamiltonian vector field (1.1) ∈ QR 3 \Q3 , H but (2.1) is a non-polynomial perturbed system of (2.1)0 ∈ Q3 . This idea first appeared in the paper [2].
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Y. Zhao / Nonlinear Analysis 64 (2006) 2332 – 2351
System (1.3) can also be transformed into (2.1) for A = −2, A ∈ R. If A = −2, then system (1.3) is changed to ⎞⎞ ⎛ ⎛ n ∞ j Fkij u ⎠⎠ , z˙ = uz + ⎝ k−1 ⎝ zi+j −3 i+j =0
k=1
⎛ ⎞⎞ ⎛ n ∞ j j +1 −Gkij u + Fkij u 1 ⎠⎠ . u˙ = 2 − 2z − u2 + ⎝ k−1 ⎝ 2 zi+j −2
(2.6)
i+j =0
k=1
System (2.6)0 has a first integral of the form z(u2 /2 − 2 + z). Convection 2.3. From now on we always suppose a ∈ (−1/2, 0) unless the opposite is claimed. 3. The relative cohomology decomposition of non-polynomial differential 1-forms In this section we describe the decomposition of non-polynomial differential 1-form related to cubic Hamiltonian (2.3) and system (2.1) . Below, ∗ denotes a real constant for simplicity. Let = q(x, y) dx − p(x, y) dy = −
m n
∗y j
m=0 j =0
(1 + 2ax)m−3
n m+1
∗y j
m=0 j =0
(1 + 2ax)m−2
dx
dy.
(3.1)
The substitution X = 1 + 2ax
(3.2)
transforms Hamiltonian (2.3) into the form (X, y) = H0 (X, y) + hl = h, H
H0 (X, y) = X( 21 y 2 + BX 2 + CX + D).
(3.3)
Here hl is defined as (2.4), B =−
1 , 24a 3
C=
1 (a + 1), 8a 3
D=−
1 (1 + 2a). 8a 3
We use the following standard notations: ij = X i y j dX, Since
i = 0, ±1, ±2, . . . , j = 0, 1, 2, . . . .
(3.4)
j +1 yj (m − 3)y j +1 y 1 + d dy = dx , j +1 X m−3 X m−2 X m−3
the differential 1-form is represented as the form ⎞ ⎛ m n m+1 n j +1 y = ∗−(m−2),j + d ⎝ ∗ m−3 ⎠ . X m=0 j =0
m=0 j =0
(3.5)
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In the sequel, let [s] be the entire part of s and R[X, y] (resp. R[H0 ]) the ring of polynomials in two variables X, y (resp. H0 ) over R. For simplicity, we say deg (X, y) = l if l is an upper bound (usually a lowest upper bound) for the degree of (X, y) ∈ R[X, y]. To obtain the relative cohomology decomposition of the differential 1-form , first of all we prove the following lemmas: Lemma 3.1. If j is odd, j 3, then ij can be expressed as the form ij =
i+j −1
∗k1 + d(X i+1 j (X, y)) + (X i j (X, y)) dH0 ,
k=i
i = 0, ±1, ±2, . . . .
(3.6)
If (i, j ) = (0, 2), (−1, 0) and (i, j ) = (i, 0), i = −1, then we have 02 = d(2H0 ln X − 23 BX 3 − CX 2 − 2DX) − 2 ln X dH0 , i+1 X −1,0 = d ln X, i0 = d . i+1
(3.7)
If j is even and i + j 3, i 2, j 2, then ij is decomposed into the form ⎞ ⎛ k =−1 ∗X k+1 + ij (H0 ) ln X + X i+1 j (X, y)⎠ ij = d ⎝ i k i+j
+ (ij ln X + X i j (X, y)) dH0
(3.8)
with (a) ij (H0 ) = ij ≡ 0 for i + j − 2 or (i, j ) = (1, 2); (b) deg ij (H0 ) = 0, ij = 0 for i + j = −1, 0, 1; and (c) deg ij (H0 ) = 1 for i 0 and i + j = 2, 3, where ij is a real constant, deg j (X, y) = j , deg j (X, y) = j − 2, and j (X, y) = j (X, y) =
[(j −2)/2]
2m
m=0
km =0
[(j −2)/2]
2m
m=0
km =0
∗X km y j −2m , ∗X km y j −2m−2 .
(3.9)
Proof. It follows from (3.3) that dH0 = ( 21 y 2 + 3BX 2 + 2CX + D) dX + Xy dy
(3.10)
which implies X i y j −2 dH0 =
1 2
ij + 3Bi+2,j −2 + 2Ci+1,j −2
+ Di,j −2 + X i+1 y j −1 dy,
j 2.
(3.11)
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Y. Zhao / Nonlinear Analysis 64 (2006) 2332 – 2351
Since X i+1 y j −1 dy = d
X i+1 y j j
−
i+1 ij , j
we obtain from (3.11) that for j = 2(i + 1), 2j − 3Bi+2,j −2 − 2Ci+1,j −2 − Di,j −2 ij = j − 2(i + 1) i+1 j X y i j −2 dH0 . +X y −d j
(3.12)
If j is odd and j 3, then j = 2(i + 1) for ∀i ∈ {0, ±1, ±2, . . .}. It follows from (3.12) by induction that ij =
2
∗i+s1 ,j −2 + d(∗X i+1 y j ) + ∗X i y j −2 dH0
s1 =0
=
2 2
⎛ ∗i+s1 +s2 ,j −4 + d ⎝∗X i+1 y j +
s1 =0 s2 =0
⎛
+ ⎝∗Xi y j −2 +
2
2
⎞ ∗X i+s1 +1 y j −2 ⎠
s1 =0
⎞ ∗X i+s1 y j −4 ⎠ dH0
s1 =0
= ··· 2
=
∗i+s1 +s2 +···+s(j −1)/2 ,1
s1 ,s2 ,...,s(j −1)/2=0
⎛
+ d ⎝X i+1 y j +
2
∗X i+s1 +1 y j −2 + · · ·
s1 =0 2
+
⎞
X i+s1 +s2 +···+s(j −3)/2 +1 y 3 ⎠
s1 ,s2 ,...,s(j −3)/2 =0
⎛
+ ⎝Xi y j −2 +
2
X i+s1 y j −4 + · · ·
s1 =0
+
2
⎞
X i+s1 +s2 +...+s(j −3)/2 y ⎠ dH0 ,
s1 ,s2 ,···,s(j −3)/2 =0
which yields (3.6) for odd j. If j is even and i + j 3, i 3, j 0, then j = 2(i + 1) if and only if (i, j ) = (0, 2), (−1, 0). Since (3.3) shows that y 2 = 2H0 X −1 − 2BX 2 − 2CX − 2D, we immediately obtain the representation for 02 . The expression for −1,0 and i,0 , i = −1 follows by direct computations.
Y. Zhao / Nonlinear Analysis 64 (2006) 2332 – 2351
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Substituting (i, j ) = (1, 2) into (3.12), one obtains the decomposition of 12 . If j = 2(i + 1), i.e. (i, j ) = (−1, 0), (0, 2), then using (3.12) again, one obtains ⎞ ⎛ (j −4)/2 2m ij = ∗k,2 + ∗02 + d ⎝X i+1 ∗X km y j −2m ⎠ i k i+j −2,k=0
⎛
(j −4)/2
2m
m=0
km =0
+ Xi ⎝
⎞
m=0
km =0
∗X km y j −2m−2 ⎠ dH0 ,
where the term 02 disappears if either i + j 1 or i 1. Representation (3.8) follows from (3.7), (3.12)|j =2 and the above decomposition. Lemma 3.2. Assume i 3. Then the differential 1-form −i,1 can be decomposed into the form −i,1 =
1 H0i−1
( −i,1 (H0 )01 + −i,1 (H0 )11 + −i,1 (H0 )21
+ dU−i,1 (X, y) + X −1 V−i,1 (X, y) dH0 ),
(3.13)
where −i,1 (H0 ), −i,1 (H0 ), −i,1 (H0 ) are polynomials of H0 with deg −i,1 (H0 )=[(2i −3)/3], deg −i,1 (H0 ) = deg −i,1 (H0 ) = [(2i − 5)/3], U−i,1 (x, y) and V−i,1 (X, y) are polynomials of degrees 2i − 1 and 2i − 3, respectively. If i=2, then −2,1 is decomposed into (3.13) with deg −2,1 (H0 )=deg −2,1 (H0 )=0, −2,1 (H0 ) ≡ 0, deg U−2,1 (x, y) = 3, deg V−2,1 (X, y) = 1. For i = 1, we have −1,1 =
1 (−2D01 − 5C11 − 8B21 − d(Xy 3 ) + 3y dH0 ). H0
(3.14)
Proof. Using (3.3) again, we obtain H0 −i,1 = 21 −i+1,3 + B−i+3,1 + C−i+2,1 + D−i+1,1 . It follows from (3.12) that 6 −i+1,3 = − 3B−i+3,1 − 2C−i+2,1 − D−i+1,1 2i − 1 −i+2 3 y X −i+1 +X −d y dH0 . 3
(3.15)
(3.16)
Substituting (3.16) into (3.15), we have −i,1 =
1 (2(i − 2)D−i+1,1 + (2i − 7)C−i+2,1 + (2i − 10)B−i+3,1 (2i − 1)H0 − d(X −i+2 y 3 ) + 3X −i+1 y dH0 ). (3.17)
Taking i = 1, 2 into (3.17), we obtain decomposition (3.14) and the representation of −2,1 , respectively. In what follows we prove (3.13) by induction. Noting that the term −i+3,1 (resp. −i+1,1 ) in (3.17) disappears if i =5 (resp. i =2), it follows from (3.14) and (3.17) that −k,1 , k =3, 4, 5, can
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Y. Zhao / Nonlinear Analysis 64 (2006) 2332 – 2351
be decomposed into form (3.13). Suppose that decomposition (3.13) holds for 3 k i −1, i 6. Then using (3.17) again, i−1 1 −i+2 3 −i+1 −i,1 = ∗−k,1 − d(X y ) + 3X y dH0 H0 k=i−3 i−1 1 1 ( −k,1 (H0 )01 + −k,1 (H0 )11 + −k,1 (H0 )21 ∗ = k−1 H0 H 0 k=i−3 + dU−k,1 (X, y) +X−1 U−k,1 (X, y) dH0 ) −d(X =
−i+2 3
y ) + 3X
1 H0i−1
i−1
−i+1
y dH0
∗( −k,1 (H0 )H i−k−1 01 + −k,1 (H0 )H0i−k−1 11
k=i−3
+ −k,1 (H0 )H0i−k−1 21 + H0i−k−1 dU−k,1 (X, y) + H0i−k−1 X −1 V−k,1 (X, y) dH0 )
−H0i−2 d(X −i+2 y 3 ) + 3H0i−2 X −i+1 y dH0 which shows that −i,1 can be expressed as (3.13) with −i,1 (H0 ) =
i−1
∗ −k,1 (H0 )H i−k−1 ,
k=i−3
−i,1 (H0 ) =
i−1
i−1
−i,1 (H0 ) =
∗ −k,1 (H0 )H i−k−1 ,
k=i−3
∗ −k,1 (H0 )H i−k−1 ,
k=i−3
U−i,1 (X, y) =
i−1
∗H i−k−1 U−k,1 (X, y) + H0i−2 X −i+2 y 3 ,
k=i−3
V−i,1 (X, y) = X
−1
i−1
(∗XH i−k−2 U−k,1 (X, y) + ∗H0i−k−1 V−k,1 (X, y))
k=i−3
+ ∗ H0i−3 X −i+3 y 3
+ 3H0i−2 X −i+2 y
.
Therefore, deg −i,1 (H0 )
max
i−3 k i−1
i−k−1+
2k − 3 3
=
2i − 3 . 3
Y. Zhao / Nonlinear Analysis 64 (2006) 2332 – 2351
2341
Using the same arguments as above, we obtain the upper bounds for deg −i,1 (H0 ) and deg −i,1 (H0 ), respectively. Since H0 (X, y), defined in (3.3), is a polynomial of degree 3, and H0i−2 X −i+2 y 3 = X i−2 ( 21 y 2 + BX 2 + CX + D)i−2 X −i+2 y 3 = ( 21 y 2 + BX 2 + CX + D)i−2 y 3 , which implies deg U−i,1 (H0 ) max
max
i−3 k i−1
{3(i − k − 1) + deg U−k,1 (X, y)}, 2(i − 2) + 3
= 2i − 1. Similarly, one obtains deg V−i,1 (X, y) 2i − 3.
Lemma 3.3. Suppose j is odd and j 3, i 2, −i + j 3; then, −i,j can be expressed in the form −i,j =
1 ( −i,j (H0 )01 + −i,j (H0 )11 + −i,j (H0 )21 H i−1 + dU−i,j (X, y) + X −1 V−i,j (X, y) dH0 ),
(3.18)
where U−i,j (X, y) and V−i,j (X, y) are polynomials of degrees 2i + j − 2 and 2i + j − 4, respectively, and (a) deg −i,j (H0 ) = [(2i + j − 4)/3], deg −i,j (H0 ) = deg −i,j (H0 ) = [(2i + j − 6)/3] for j = i − 1, i + 2; (b) deg −i,i−1 (H0 ) = deg −i,i−1 (H0 ) = i − 2, deg −i,i−1 (H0 ) = i − 3; and (c) deg −i,i+2 (H0 ) = deg −i,i+2 (H0 ) = i − 1, deg −i,i+2 (H0 ) = i − 2. Proof. Representation (3.18) follows from (3.6) and Lemma 3.2.
The main result of this section is the following. Proposition 3.4. Assume n4. Then any differential 1-form , defined in (3.5), can be decomposed into =
1 H0n−3
( (H0 )01 + (H0 )11 + (H0 )21 + d( (H0 ) ln X + (X, y))
+ ((H0 ) ln X + X −1 (X, y)) dH0 ),
(3.19)
where deg (H0 )=deg (H0 )=deg (H0 )=deg (H0 )=n−3, deg (H0 )=n−2, deg (X, y)= 3n−5, deg (X, y)=3n−7, (H0 ) and (H0 ) have a zero at H0 =0 with multiplicity n−3, n−4, respectively. If n = 1, 2, then can be expressed as = (H0 )01 + (H0 )11 + (H0 )21 + d( (H0 ) ln X + (X, y)) + ((H0 ) ln X + X −1 (X, y)) dH0 ,
(3.20)
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Y. Zhao / Nonlinear Analysis 64 (2006) 2332 – 2351
where (H0 ) = (H0 ) = (H0 ) ≡ 0 for n = 1 and deg (H0 ) = 1, deg (H0 ) = deg (H0 ) = 0 for n = 2, respectively, deg (H0 ) = deg (H0 ) = 0, deg (x, y) = 4, deg (X, y) = 2, (0) = 0. If n = 3, then can be represented as =
1 ( (H0 )01 + (H0 )11 + (H0 )21 + d( (H0 ) ln X + (X, y)) H0 + ((H0 ) ln X + X −1 (X, y)) dH0 ),
(3.21)
with deg (H0 ) = deg (H0 ) = deg (H0 ) = deg (H0 ) = 1, deg (H0 ) = 2, deg (X, y) = 7, deg (X, y) = 5, (0) = 0. Proof. Decomposition (3.20) follows from (3.5)–(3.8). From (3.6) we obtain that −1,3 =
1
∗k1 + d(3 (X, y)) + X −1 3 (X, y) dH0 .
(3.22)
k=−1
Use of (3.22), (3.8), (3.14) and (3.20) yields expression (3.21). Lemma 3.3 and (3.8) imply the following decomposition, provided m 4: ⎞ ⎛ m+1 m j +1 y ⎠ ∗−(m−2),j + d ⎝ X m−3 j =0
=
j =0
1
(−(m−2) (H0 )01 H0m−3 + −(m−2) (H0 )11 + −(m−2) (H0 )21 + d(−(m−2) (H0 ) ln X
+ −(m−2) (X, y)) + (−(m−2) (H0 ) ln X + X −1 −(m−2) (X, y)) dH0 ),
where deg −(m−2) (H0 ) = m − 3, deg −(m−2) (H0 ) = deg −(m−2) (H0 ) = m − 3 + ((−1)m − 1)/2, deg −(m−2) (H0 ) = m − 2, deg −(m−2) (H0 ) = m − 3, deg −(m−2) (X, y) = 3m − 5, deg −(n+2) (X, y)=3m−7, −(m−2) (H0 ) and −(m−2) (H0 ) have a zero at H0 =0 with multiplicity m − 3 and m − 4, respectively. By induction, one obtains (3.19) from the above decomposition, (3.20) and (3.21). 4. Abelian integrals and the second Melnikov functions In this section, we will derive, by Francoise’s recursive procedure [3], an appropriate formula for Mk (h), k = 1, 2. Before that we rewrite system (2.1) as a Pfaff equation − 1 − 2 2 − · · · − k k − · · · = 0, dH
(4.1)
where k = qk (x, y) dx − pk (x, y) dy and pk (x, y), qk (x, y) are defined in (2.2). Then M1 (h) = 1 . h
(4.2)
(4.3)
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(x, y) = h} are closed orbits of system (4.1) . As in [5,7], we introduce the Here h ⊂ {(x, y)|H following basic 1-forms: 01 = y dx,
11 = xy dx,
21 = x 2 y dx,
−1,1 =
y dx 1 + 2ax
and the corresponding integrals i1 . Ji (h) = h
(4.4)
(4.5)
Comparing (4.4) with (3.4), we have −1,1 = 2a −1,1 , 01 = 2a 01 , 11 = 2a( 01 + 2a 11 ), 21 = 2a( 01 + 4a 11 + 4a 2 21 ).
(4.6)
This yields the following. Proposition 4.1. Assume n4. Then M1 (h) can be expressed in the form M1 (h) =
1 (h − hl )n−3
( 1 (h)J0 (h) + 1 (h)J1 (h) + 1 (h)J2 (h)),
(4.7)
where deg 1 (h) = deg 1 (h) = deg 1 (h) = n − 3. If n = 1, 2, then M1 (h) = 1 (h)J0 (h) + 1 (h)J1 (h) + deg 1 (h)J2 (h),
(4.8)
where deg 1 (h) = deg 1 (h) = deg 1 (h) = 0. If n = 3, then M1 (h) =
1 ( 1 (h)J0 (h) + 1 (h)J1 (h) + 1 (h)J2 (h)), h − hl
(4.9)
where deg 1 (h) = deg 1 (h) = deg 1 (h) = 1. Proof. Assume that 1 is decomposed into (3.19) and n 4. It follows from Propositions 3.4 and (4.6) that M1 (h) can be expressed as (4.7) with 1 (h) = 2a( (h − hl ) + (h − hl ) + (h − hl )),
1 (h) = 4a 2 ( (h − hl ) + 2 (h − hl )), 1 (h) = 8a 3 (h − hl )
(4.10)
which yields the assertion for n4. Repeating the same arguments, we obtain (4.8) and (4.9). To obtain M2 (h) in an appropriate form, we next recall in brief Francoise’s recursive procedure [3] for a calculation of the kth order Melnikov function Mk (h), related system (4.1) . Lemma 4.2 (Francoise’s recursion formula [3]). Assume M1 (h) = M2 (h) = · · · = Mk−1 (h) ≡ 0 in (1.5), k 2. Then k , (4.11) Mk (h) = h
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Y. Zhao / Nonlinear Analysis 64 (2006) 2332 – 2351
where 1 = 1 ,
m = m +
ri j ,
2 m k
(4.12)
i+j =m
and the functions ri (x, y, H ), 1 i k − 1 are determined successively from the representations i = dRi (x, y, H ) + ri (x, y, H ) dH .
(4.13)
The proof of this lemma can be found in many papers; see for instance [3,7,13], etc. To apply Francoise’s recursion formula to determine M2 (h) for system (4.1) , first of all we have to obtain the representation 1 = 1 , defined in (4.13). Lemma 4.3. Assume that differential 1-form 1 is decomposed into the form (3.19) for n 4, (3.20) for n = 1, 2 and (3.21) for n = 3, respectively. Then M1 (h) ≡ 0 if and only if (H0 ) = (H0 ) = (H0 ) ≡ 0. Proof. Proposition 4.1 implies that M1 (h) ≡ 0 if and only if 1 (h)J0 (h) + 1 (h)J1 (h) + 2 (h)J2 (h) ≡ 0.
(4.14)
We say that is a polynomial 1-form of degree m, if = −f(x, y) dy + g (x, y) dx,
f(x, y), g (x, y) ∈ R[x, y]
(4.15)
and max{deg f(x, y), deg g (x, y)} = m. Let I be the real vector space of Abelian integral of the form (x, y). I(h) = (4.16) h
Then an Abelian integral I(h) belongs to I if and only if I(h) can be expressed as a linear combination of J0 (h), J1 (h), J2 (h) with polynomial coefficients in h [5]. It follows from [5, Proposition 8] that the R[h] module I is freely generated by the Abelian integrals J0 (h), J1 (h), J2 (h), which implies that (4.14) holds if and only if 1 (h) = 1 (h) = 1 (h) ≡ 0 (or see the proof of Proposition 3 in [7]). Using (4.10) again, we know that 1 (h) = 1 (h) = 1 (h) ≡ 0 if and only if (H0 ) = (H0 ) = (H0 ) ≡ 0. This proves the assertions. From the above discussions, we obtain the following. Proposition 4.4. Assume M1 (h) ≡ 0. Then the second-order Melnikov function for (4.1) and (2.3) can be expressed in the form M2 (h) = M2 (h) =
1 (h − hl )3
( 2 (h)J0 (h) + 2 (h)J1 (h) + 2 (h)J2 (h)),
1 (h − hl )2n−5
n = 3,
( 2 (h)J0 (h) + 2 (h)J1 (h) + 2 (h)J2 (h)),
n4,
(4.17) (4.18)
where deg 2 (h)=deg 2 (h)=deg 2 (h)=3 for n=3 and deg 2 (h)=deg 2 (h)=deg 2 (h)=2n−5 for n 4, respectively.
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Proof. Assume that 1 has been expressed in the form (3.19) for n 4. By Lemma 4.3, we have (H0 )= (H0 )= (H0 ) ≡ 0. Therefore, the 1-form 1 can be decomposed into the form (4.13)i=1 with r1 =
1 H0n−2
( 1 (H0 ) ln X + (−n + 3) (X, y) + X −1 H (X, y)),
where 1 (H0 ) = (−n + 3) (H0 ) + H (H0 )). From Lemma 4.3, we obtain (the calculations below are performed modulo 1-forms dR + r dH0 ): r1 (X, y, H0 )
(d( (H0 ) ln X + (X, y)) + ((H0 ) ln X + X −1 (X, y)) dH0 ) H0n−3 r1 (X, y, H0 ) 1 = (H0 ) dX + d (X, y) X H0n−3 1 (H0 ) 1 −1 = 2n−5 dX − (X, y)H0 (H0 ) X H0 (X, y) X X H0 +X−1 H0 (X, y) d ,
r1 1 =
where we use the following identities:
(X, y) d (X, y) = d( 21 2 (X, y)), 1 (H0 ) (H0 )
ln X dX = d( 21 1 (H0 ) (H0 )(ln X)2 ) X − 21 ( 1 (H0 ) (H0 )) (ln X)2 dH0 ,
1 (H0 ) ln X d = d( (X, y) 1 (H0 ) ln X) − (X, y) 1 (H0 ) ln X dH0 1 − (X, y) 1 (H0 ) dX. X It follows from (3.3) that X−1 H0 (X, y), (H0 ), X−1 (X, y)H0 (H0 ) are polynomials in X, y of degrees 3n − 5, 3n − 6 and 6n − 12, respectively. Since Proposition 3.4 asserts (0) = 0, we conclude that (H0 )/X is a polynomial in X, y of degree 3n − 7. Therefore, r1 1 can be rewritten in the form r1 1 =
1 H02n−5
2 ,
(4.19)
where 2 is a polynomial 1-form of degree 6n−12 (see (4.15) for the definition). Abelian integrals I(h), defined by (4.16), can be expressed as I(h) = (h)J0 (h) +
(h)J1 (h) + (H )J2 (h),
(4.20)
with deg (h) = [(m − 1)/3], deg
(h) = [(m − 2)/3], deg (H ) = [(m − 3)/3] (see [9, Remark 2.3]). This yields 1 r1 1 = ( 2 (h)J0 (h) +
2 (h)J1 (h) + 2 (h)J2 (h)), (4.21) (h − hl )2n−5 h
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Y. Zhao / Nonlinear Analysis 64 (2006) 2332 – 2351
where deg 2 (h) = deg
(h) = deg (H ) = 2n − 5. On the other hand, by the same arguments as in Proposition 4.1 we obtain that 1 2 = (¯ 2 (h)J0 (h) + ¯ 2 (h)J1 (h) + ¯ 2 (H )J2 (h)), (4.22) (h − hl )n−3 h ¯ with deg ¯ 2 (h) = deg (h) = deg ¯ (H ) = n − 3. Application of Lemma 4.2 allow us to obtain M2 (h) = 2 + r1 1 . 2 = h
h
h
Expression (4.18) follows from the above discussions. Repeating the same arguments, we obtain (4.17). Define the space of integrals n M1 = M1 (h)|M1 (h) =
1 1 , h ∈ 0, 6 h
,
Then we have the following proposition: Proposition 4.5. We have y n M1 = span J0 (h), J1 (h), J2 (h), J−k (h) = dx, k h (1 + 2ax) k = 1, 2, . . . , n − 2, ,
(4.23)
which yields dim Mn1 n + 1. Proof. Assertion (4.23) follows from (3.5) and (3.6).
In the paper [8], the authors proved that the linear space Mn1 has a dimension n + 1 if A = −2 in (1.2). According to this conclusion and Proposition 4.5, the following conjecture is reasonable: Conjecture 4.6. J−n+2 (h), J−n+3 (h), . . . , J0 (h), J1 (h), J2 (h) are linearly independent, so dim Mn1 = n + 1. If the above conjecture holds, we could obtain Conjecture 4.7. Let n M2 = M2 (h)|M2 (h) =
1 2 , M1 (h) ≡ 0, h ∈ 0, 6 h
.
Then dim Mn2 = 2n − 1. 5. Bounds of the number of limit cycles In this section, we will repeat the arguments of [5] and [7] to obtain the main results of this paper.
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Let J (h) = 1 (h)J0 (h) + 1 (h)J1 (h) + 1 (h)J2 (h).
(5.1)
Then we have Lemma 5.1. The following expression holds: J (h) = r(h)J0 (h) + s(h)J1 (h) + u(h)Z (h),
(5.2)
where Z(h) = (3a − 1)J1 (h) − 4aJ 2 (h), r(h), s(h), u(h) are real polynomials with deg r(h) = deg s(h) = deg u(h) = 0 for n = 1, 2, deg r(h) = deg s(h) = deg u(h) = 1 for n = 3, and deg r(h) = deg s(h) = deg u(h) = n − 3 for n 4, respectively. Proof. It follows from Remark 3.4 in [9] that J0 (h), J1 (h), J2 (h) satisfy the following Picard–Fuchs equation: −6ah −1 a+1 J0 0 −12a 2 h + a + 1 2a 2 − a − 1 J1 0 6h − 1 12ah − 2a J2 −4a 0 0 J0 = a(a − 1) −12a 2 0 J1 −(a + 1) −3a + 7 16a J2 which yields (5.2).
By Roussarie’s Theorem [19,20], we have either J (h) ≈ |(h − 16 )m/2 ln(h − 16 )|,
where m is even
(5.3)
or J (h) ≈ |(h − 16 )(m−1)/2 |,
where m is odd,
(5.4)
holds, and any function of the form J¯ (h) = (holomorphic function) ln(h − 16 ) + holomorphic function, sufficiently close to J (h) can have at most m zeros in {|h − 1/6| < d1 } ∩ D, where d1 > 0 is a small enough constant and D = C\[1/6, ∞). Definition 5.2 (Gavrilov [5], Gavrilov and Iliev [7]). We shall say J (h), h ∈ D, has a zero of multiplicity m at 1/6, provided that either (5.3) or (5.4) holds. Proposition 5.3. If J (h) has a zero of multiplicity m at 1/6, then it has at most 3n − 7 − m zeros in D. Proof. This proof literally repeats the arguments in [5] and [7]. Indeed, we just change the context in [5,7] concerning the degrees of polynomials and some symbols and notations. We present the proof here for the reader’s convenience.
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We know from [5, Proposition 10] that 0 is a first kind 1-form and hence J0 (h) does not disappear. Proposition 12 of [5] asserts that J (h) is a holomorphic function in D. So we can define the analytic function F (h) = J (h)/J0 (h) in D. In what follows we will estimate the number of zeros of F (h) by the argument principle. Denote by F + (h) (resp. F − (h)) the analytic continuation of F (h) on [1/6, +∞) along the path contained in the half-plane I mh > 0 (resp. I mh < 0). The Picard–Lefschetz formula implies I mF ± (h) = ±
, 0 ) W1 ,2 ( , |J0 (h)|2
where 1 , 2 are the cycles disappearing at h = 0 and h = 1/6, respectively, and 1 1 0 . )1 + 1 (H )2 , W , ( )0 + 1 (H = 1 (H , ) = det 0 1 2 2 2 0 It follows from Lemma 5.1 that W1 ,2 ( 1 , 0 ) + u(h)W1 ,2 ( Z , 0 ), , 0 ) = s(h)W1 ,2 ( where Z = (3a − 1) 1 − 4a 2 . By Proposition 10 in [5], the function W1 ,2 (Z , 0 ) is single valued in h on C and has no poles. For |h| ≈ ∞ the asymptotic estimate |J0 (h)| ≈ |h|−1/3 ,
|Z (h)| ≈ |h|1/3
implies that W1 ,2 ( Z , 0 ) is bounded in h. It follows that it is a non-zero constant. Further, the reciprocity law for meromorphic differentials of the first and third kind 0 and 1 implies P+ √ 0 , W1 ,2 ( 1 , 0 ) = 2 −1Res P+ 1 P−
where the path of integration from P− to P+ in the integral above is contained in h cut along P the loops 1 (h) and 2 (h). Since Res P+ 1 is a purely imaginary constant in h and P−+ 0 is imaginary too, we obtain that on the interval [1/6, ∞) holds P+ √ 0 + , 0 ) = s(h)2 −1 u(h), W1 ,2 ( P−
where s(h), q (h) are polynomials with deg s(h) = deg q (h) = n − 3. By Lemma 3 of [5], W1 ,2 ( , 0 ) has at most 2n − 5 zeros in the complex domain C\(−∞, hl ). Noting hl < 1/6 and |W1 ,2 ( , 0 )| < ≈ |h − 16 |[m/2] , the imaginary part of F (h) has at most 2n − 5 − [m/2] zeros on the interval (1/6, ∞). Let d∞ be large enough constant and ds be small enough constant. Denote by D the set obtained from D ∩ {|h| < d∞ } by removing a disc of radius d1 around h = 1/6. We shall evaluate the increment of argument of F (h) along the boundary D which has a positive orientation. Suppose that m is even. Then I mF (h) has at most 2n − 5 − m/2 zeros on the interval (1/6, ∞). Along the circle {|h| = d∞ }, we have F (h) < ≈ |h|n−3 .
Y. Zhao / Nonlinear Analysis 64 (2006) 2332 – 2351
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Finally, if h ≈ 1/6, then |F (h)| ≈ |h − 1/6|m/2 . This yields that the increment of the argument F (h) along the boundary D is equal to or less than 2(1 + n − 3 + 2n − 5 − m/2 − m/2) = 2(3n − 7 − m) and hence F (h) can have at most 3n − 7 − m, zeros in D . If m is odd, then we can obtain the assertion by the same arguments as in the proof of Theorem 5 in [5]. We omit the details here. Theorem 5.4. Assume n 4. Then M1 (h) and M2 (h) have at most 3n − 7 and 6n − 13 zeros on the interval (0, 1/6), respectively. If n = 1, 2, then M1 (h) has at most two zeros in (0, 1/6); If n = 3, then M1 (h) and M2 (h) have at most 5 and 11 zeros in (0, 1/6), respectively. Proof. Since J (0) = 0, it follows from Proposition 5.3 that J (h) has at most 3n − 7 zeros in (0, 1/6) for n 4. Therefore, M1 (h), defined in (4.7), has at most 3n − 7 zeros in (0, 1/6). Using the same arguments as in Proposition 5.3, we obtain that 2 (h)J0 (h)+ 2 (h)J1 (h)+ 2 (h) has at most 6n − 13 zeros in (0, 1/6) for n 4. The same is true, of course, for M2 (h). Repeating the same arguments as above, we can obtain the upper bounds of the number of zeros of M1 (h), n = 1, 2, 3 and M2 (h), n = 3, respectively. Proof of Theorem 1.1. To study the number of limit cycles in (4.1) , we divide the limit cycles into three types. As → 0, a limit cycle in (4.1) tends either to (a) the center (0, 0), or to (b) periodic orbits of (4.1)0 , or to (c) the homoclinic loop 1/6 . Let M1 (h) ≡ / 0. M1 (h) is analytic at h = 0 [20]. If M1 (h) = ∗hk + o(hk ), then the number of limit cycles of type (a) is less than or equal to k. Since hl < 1/6, we know that 1/(h − hl ) is analytic at h = 1/6, which implies that M1 (h) can be expressed as form (5.3) or (5.4) modulo a real constant factor. If J (h) (and hence M1 (h)) has a zero of multiplicity m at h = 1/6, then the number of limit cycles of type (c) is less than or equal to m. Finally, the number of limit cycles of type (b) is less than or equal to 3n − 7 − m − k by using Proposition 5.3. Therefore, system (4.1) (and hence (1.3)) has at most 3n − 7 − m − k + k + m = 3n − 7 limit cycles in the compact domain K ⊂ R for small enough. If M1 (h) ≡ 0 but M2 (h) ≡ / 0, then we obtain the second assertion of Theorem 1.1 by repeating the same arguments as above. Proof of Theorem 1.2. The exact upper bound for the number of limit cycles produced by the periodic annulus of system (1.1) under quadratic perturbations is equal to the maximum number of zeros of the related Abelian integrals as follows [12]: 1 (h) = M
h
x −4 ( x +
+ x −1 )y dx,
where ,
, are real constants. Therefore, we consider the following system instead of (1.3): x˙ = xy,
y˙ = (A + 2) − 2(A + 1)x + Ax 2 + 23 y 2 + ( x +
+ x −1 )y.
(5.5)
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Using the same arguments as in Proposition 2.1, system (5.5) can be reduced to x˙ = y(1 + 2ax),
y˙ = −x + x 2 − ay 2 + ( + x + x 2 )y,
(5.6)
where , are real constants. The related Melnikov function is M1 (h)= J0 (h)+ J1 (h)+ J2 (h). Repeating the arguments as in the proof of Theorem 1.1 or using Theorem 3 of [5] (the case n=3), one obtains that system (5.5) has at most two limit cycles in K and the bound is exact. It is obvious that the same is true for (1.3) with n = 2. This proves the theorem. References [1] Shui Nee Chow, Chengzhi Li, Yingfei Yi, The cyclicity of period annuli of degenerate quadratic Hamiltonian systems with elliptic segment loops, Ergodic Theory Dynamics Systems 22 (2) (2002) 349–374. [2] F. Dumortier, C. Li, Z. Zhang, Unfolding of a quadratic integrable system with two centers and two unbounded heteroclinic loops, J. Differential Equations 139 (1997) 146–193. [3] J.P. Francoise, Successive derivative of first return map, application to the study of quadratic vector fields, Ergodic Theory Dynamics Systems 1 (1996) 87–96. [4] L. Gavrilov, Petrov modules and zeros of Abelian integrals, Bull. Sci. Math. 122 (1998) 571–584. [5] L. Gavrilov, Abelian integrals related to Morse polynomials and perturbations of plane Hamiltonian vector fields, Ann. Inst. Fourier (Grenoble) 49 (1999) 611–652. [6] L. Gavrilov, The infinitesimal 16th Hilbert problem in the quadratic case, Invent. Math. 143 (3) (2001) 449–497. [7] L. Gavrilov, I.D. Iliev, Second-order analysis in polynomially perturbed reversible quadratic Hamiltonian systems, Ergodic Theory Dynamics Systems 20 (2000) 1671–1686. [8] L. Gavrilov, I.D. Iliev, Two-dimensional Fuchsian systems and the Chebyshev property, J. Differential Equations 191 (1) (2003) 105–120. [9] E. Horozov, I.D. Iliev, Linear estimate of the number of zeros of Abelian integrals with cubic Hamiltonians, Nonlinearity 11 (1998) 1521–1537. [10] I.D. Iliev, The cyclicity of the period annulus of the quadratic Hamiltonian triangle, J. Differential Equations 128 (1996) 309–326. [11] I.D. Iliev, Inhomogeneous Fuchs equations and the limit cycles in a class of near-integrable quadratic systems, Proc. Roy. Soc. Edinburgh 127A (1997) 1207–1217. [12] I.D. Iliev, Perturbations of quadratic centers, Bull. Sci. Math. 122 (1998) 107–161. [13] I.D. Iliev, On the limit cycles available from polynomial perturbations of the Bogdanov–Takens Hamiltonian, Israel J. Math. 115 (2000) 269–284. [14] I.D. Iliev, C. Li, J. Yu, Bifurcations of limit cycles from quadratic non-Hamiltonian systems with two centres and two unbounded heteroclinic loops, Nonlinearity 18 (2005) 305–330. [15] J. Li, Hilbert’s 16th problem and bifurcations of planar polynomial vector fields, Internat. J. Bifurcation Chaos 13 (1) (2003) 47–106. [16] Weigu Li, Yulin Zhao, Chengzhi Li, Zhifen Zhang, Abelian integrals for quadratic centres having almost all their orbits formed by quartics, Nonlinearity 15 (3) (2002) 863–885. [17] D. Novikov, S. Yakovenko, Redundant Picard–Fuchs system for abelian integrals, J. Differential Equations 177 (2) (2001) 267–306. [18] G.S. Petrov, On the nonoscillation of elliptic integrals, Funktsional. Anal. i Prilozhen 31 (4) (1997) 47–51; translation in Funct. Anal. Appl. 31(4) (1998) (262–265) (in Russian). [19] R. Roussarie, On the number of limit cycles which appear by perturbation of separatrix loop of planar vector fields, Bol. Soc. Brasil. Mat. 17 (2) (1986) 67–101. [20] R. Roussarie, Bifurcation of planar vector fields and Hilbert’s sixteenth problem, Progress in Mathematics, vol. 164, Birkhäauser Verlag, Basel, 1998. [21] S. Yakovenko, Bounded decomposition in Brieskorn lattice and Pfaffian Picard–Fuchs systems for Abelian integrals, Bull. Sci. Math. 126 (2002) 535–554. [22] J. Yu, C. Li, Bifurcation of a class of planar non-Hamiltonian integrable systems with one center and one-homoclinic loop, J. Math. Anal. Appl. 269 (1) (2002) 227–243. [23] H. Zoladek, Quadratic systems with centers and their perturbations, J. Differential Equations 109 (1994) 223–273. [24] Y. Zhao, W. Li, C. Li, Z. Zhang, Linear estimate of the number of zeros of Abelian integrals for quadratic centers having almost all their orbits formed by cubics, Sci. China (Ser. A) 45 (2002) 964–974.
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[25] Y. Zhao, Z. Zhang, Linear estimate of the number of zeros of Abelian integrals for a kind of quadratic Hamiltonians, J. Differential Equations 155 (1999) 73–88. [26] Y. Zhao, S. Zhu, Perturbations of the non-generic Hamiltonian vector fields with hyperbolic segment, Bull. Sci. Math 125 (2) (2001) 109–138. [27] Y. Zhao, S. Zhu, Higher order Melnikov function for a quartic Hamiltonian with cuspidal loop, Discrete Contin. Dynamic Syst. 8 (2002) 995–1018.
Perturbations of non-Hamiltonian reversible quadratic systems with cubic orbits夡 Yulin Zhao Department of Mathematics, Sun Yat-sen University, Guangzhou, 510275, PR China Received 8 April 2005; accepted 15 August 2005
Abstract This paper is concerned with degree n polynomial perturbations of a class of planar non-Hamiltonian reversible quadratic integrable system whose almost all orbits are cubics. We give an estimate of the number of limit cycles for such a system. If the first-order Melnikov function (Abelian integral) M1 (h) is not identically zero, then the perturbed system has at most 5 for n = 3 and 3n − 7 for n 4 limit cycles on the finite plane. If M1 (h) is identically zero but the second Melnikov function is not, then an upper bound for the number of limit cycles on the finite plane is 11 for n = 3 and 6n − 13 for n 4, respectively. In the case when the perturbation is quadratic (n = 2), there exists a neighborhood U of the initial non-Hamiltonian polynomial system in the space of all quadratic vector fields such that any system in U has at most two limit cycles on the finite plane. The bound for n = 2 is exact. 䉷 2005 Elsevier Ltd. All rights reserved. MSC: 34C07; 34C08; 37G15; 34M50 Keywords: The kth order Melnikov function; Limit cycles
1. Introduction Recall that the quadratic integrable system are divided into four cases [23]: QH 3 (Hamiltonian), (reversible), Q4 (codimension four) and QLV (generalized Lotka–Volterra). In this paper, we 3 H whose almost study degree n polynomial perturbations of quadratic reversible system QR \Q 3 3 all orbits are cubic, where the phrase “almost all” means “all except at most a finite number of”. QR 3
夡 Supported by NSF of China (No. 10571184), Guangdong Natural Science Foundation (No. 04009794), SRF for ROCS, SEM, and the Foundation of Sun Yat-sen University Advanced Research Center. E-mail addresses: [email protected], [email protected].
0362-546X/$ - see front matter 䉷 2005 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2005.08.016
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Using the results from [12,24], the reversible non-Hamiltonian quadratic systems we consider have the following normal form: x˙ = xy =
Hy , x −4
3 Hx y˙ = (A + 2) − 2(A + 1)x + Ax 2 + y 2 = − −4 , 2 x
where H (x, y) is a first integral of (1.1), defined by 1 2 A+2 y + Ax 2 − (A + 1)x + H (x, y) = x −3 2 3
(1.1)
(1.2)
with integrating factor x −4 . Using the normal form (1.1), we list the types of critical points appearing in (1.1) as follows: (1) (2) (3) (4) (5)
a center, a saddle and two nodes: A ∈ (−∞, −2); a center and a degenerate critical point: A = −2; two centers: A ∈ (−2, 0); a center: A = 0; and a center and a saddle: A ∈ (0, +∞).
The purpose of the present paper is to investigate the bifurcations of limit cycles in polynomial perturbations of system (1.1) with a center and a saddle, i.e., A ∈ (0, +∞). Consider the corresponding perturbed system x˙ = xy + f (x, y, ), y˙ = (A + 2) − 2(A + 1)x + Ax 2 + 23 y 2 + g(x, y, ),
A ∈ (0, +∞).
(1.3)
Here f (x, y, ) and g(x, y, ) are polynomials in x, y with coefficients depending analytically on the small parameter , f (x, y, ) =
∞ k=1
k−1 fk (x, y),
g(x, y, ) =
∞
k−1 gk (x, y),
(1.4)
k=1
max{deg fk (x, y), deg gk (x, y)} n, k = 0, 1, 2, . . . , +∞. Let h ⊂ {(x, y)|H (x, y) = h} be the closed orbit of system (1.1) around the center, h ∈ (hc , hs ). A key point in the investigation of the number of limit cycles in (1.3) is to reduce the original problem to counting the number of zeros of the displacement function d(h, ) = M1 (h) + 2 M2 (h) + · · · + k Mk (h) + · · ·
(1.5)
for sufficiently small ||. It is well known that the number of zeros of the first non-disappearing kth order Melnikov function Mk (h) in (1.5) determines the upper bound of the number of limit cycles in (1.3) emerging from the periodic orbits of system (1.1). The main results of this paper are the following: Theorem 1.1. Assume n4. Let K ⊂ R2 be a fixed compact domain and A ∈ (0, +∞). If M1 (h) ≡ / 0, then system (1.3) has at most 3n − 7 limit cycles in K for small enough. If M1 (h) ≡ / 0, then (1.3) has at most 6n − 13 limit cycles in K. 0 but M2 (h) ≡
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Y. Zhao / Nonlinear Analysis 64 (2006) 2332 – 2351
For n = 3, system (1.3) has at most 5 (resp. 11) limit cycles if M1 (h) ≡ / 0 (resp. M1 (h) ≡ / 0). 0, M2 (h) ≡ Theorem 1.2. Let K ⊂ R2 be a fixed compact domain as in Theorem 1.1 and A ∈ (0, +∞). There is a neighborhood U of (1.1) in the space of all quadratic vector fields, such that any system in U has at most two limit cycles in K. This bound is exact. In the proof of Theorem 1.1, we will show that M1 (h) has at most 3n − 7 zeros on the interval (0, 1/6); see Theorem 5.4 below. This improves the upper bound given in [24] for the total number of zeros of Abelian integral M1 (h). The result from Theorem 1.2 has been obtained in [22] in different ways. We also note that there has been a substantial amount of work devoted to estimating the number of either limit cycles or zeros of Melnikov functions; see for instance [1,4,6,10,11,14–18,21,25–27] and reference therein. The rest of this paper is organized as follows: in the next section, system (1.3) is reduced to another normal form by a Poincaré map. This new system is a small non-polynomial perturbation of a quadratic Hamiltonian vector field. The polynomial perturbations of such quadratic Hamiltonian vector fields have been studied in [5,7]. We obtain the cohomology decomposition formula of the corresponding non-polynomial differential 1-form in Section 3. Using them allows us to derive an appropriate formula of M1 (h) and M2 (h) in Section 4. Finally, in Section 5 we estimate the number of both zeros of Mk (h), k = 1, 2 and limit cycles of (1.3) by following the arguments in [5,7]. 2. Normal form In order to study limit cycles in (1.3), we consider a change of variables. The main result of this section is the following: Proposition 2.1. The perturbed system (1.3) with A ∈ (0, +∞) can be reduced to the following normal form: x˙ = y(1 + 2ax) +
∞
k−1 pk (x, y),
k=1
y˙ = −x + x 2 − ay 2 + ε
∞
k−1 qk (x, y),
(2.1)
k=1
where a ∈ (− 21 , 0), pk (x, y) =
m n
Pkmj y j
m=0 j =0
(1 + 2ax)
, m−3
qk (x, y) =
n m+1
Qkmj y j
m=0 j =0
(1 + 2ax)m−2
(2.2)
and Pkmj , Qkmj are real constants. A first integral of system (2.1)0 is (x, y) = 1 x 2 + 1 y 2 − 1 x 3 + axy 2 = h. H 2 2 3
(2.3)
Y. Zhao / Nonlinear Analysis 64 (2006) 2332 – 2351
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The period annulus h is defined on the interval (0, 1/6). 0 and 1/6 correspond to the center (x, y) (0, 0) and the homoclinic loop through the saddle (1, 0), respectively. The Hamiltonian H has three critical values, defined by 1 + 3a 2a + 1 1 1 − ,± (0, 0) = 0, H (1, 0) = , H = (2.4) = hl . H 2a 4a 3 24a 3 6 Proof. Let fk (x, y) =
n
Fkij x i y j ,
gk (x, y) =
i+j =0
n
Gkij x i y j ,
i+j =0
where Fkij , Gkij are real constants. When we make the Poincaré transformation 1 y , u = , dt = −z d, x x system (1.3) is changed to ⎛ ⎞⎞ ⎛ n ∞ j F u dz kij ⎠⎠ , k−1 ⎝ = uz + ⎝ zi+j −3 d z=
k=1
(2.5)
i+j =0
du 1 = − A + 2(A + 1)z − (A + 2)z2 − u2 d 2 ⎛ ⎞⎞ ⎛ ∞ n j j +1 + F u −G u kij kij ⎠⎠ . + ⎝ k−1 ⎝ zi+j −2 i+j =0
k=1
The substitution 1 a=− , A+2
v=
1 (z − 1) 2a
transforms the above system into the form ⎛ ⎞⎞ ⎛ ∞ n Fkij uj dv 1 1 ⎝ k−1 ⎝ ⎠⎠ , = u(1 + 2av) + i+j −3 d 2a 2a (1 + 2av) k=1 i+j =0 ⎛ ⎞⎞ ⎛ ∞ n j j +1 −Gkij u + Fkij u du 1 ⎠⎠ . k−1 ⎝ = −4av + 4av 2 − u2 + ⎝ 2 d (1 + 2av)i+j −2 k=1
i+j =0
Finally, we obtain (2.1) by using the scaling √ u u¯ = √ , = 2, v → x, u¯ → y, 2 2a
m = i + j.
Remark 2.2. Proposition 2.1 yields that (1) To study the number of limit cycles of system (1.3), it is sufficient to study (2.1) . H (2) (1.3) is a polynomial perturbed system of non-Hamiltonian vector field (1.1) ∈ QR 3 \Q3 , H but (2.1) is a non-polynomial perturbed system of (2.1)0 ∈ Q3 . This idea first appeared in the paper [2].
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Y. Zhao / Nonlinear Analysis 64 (2006) 2332 – 2351
System (1.3) can also be transformed into (2.1) for A = −2, A ∈ R. If A = −2, then system (1.3) is changed to ⎞⎞ ⎛ ⎛ n ∞ j Fkij u ⎠⎠ , z˙ = uz + ⎝ k−1 ⎝ zi+j −3 i+j =0
k=1
⎛ ⎞⎞ ⎛ n ∞ j j +1 −Gkij u + Fkij u 1 ⎠⎠ . u˙ = 2 − 2z − u2 + ⎝ k−1 ⎝ 2 zi+j −2
(2.6)
i+j =0
k=1
System (2.6)0 has a first integral of the form z(u2 /2 − 2 + z). Convection 2.3. From now on we always suppose a ∈ (−1/2, 0) unless the opposite is claimed. 3. The relative cohomology decomposition of non-polynomial differential 1-forms In this section we describe the decomposition of non-polynomial differential 1-form related to cubic Hamiltonian (2.3) and system (2.1) . Below, ∗ denotes a real constant for simplicity. Let = q(x, y) dx − p(x, y) dy = −
m n
∗y j
m=0 j =0
(1 + 2ax)m−3
n m+1
∗y j
m=0 j =0
(1 + 2ax)m−2
dx
dy.
(3.1)
The substitution X = 1 + 2ax
(3.2)
transforms Hamiltonian (2.3) into the form (X, y) = H0 (X, y) + hl = h, H
H0 (X, y) = X( 21 y 2 + BX 2 + CX + D).
(3.3)
Here hl is defined as (2.4), B =−
1 , 24a 3
C=
1 (a + 1), 8a 3
D=−
1 (1 + 2a). 8a 3
We use the following standard notations: ij = X i y j dX, Since
i = 0, ±1, ±2, . . . , j = 0, 1, 2, . . . .
(3.4)
j +1 yj (m − 3)y j +1 y 1 + d dy = dx , j +1 X m−3 X m−2 X m−3
the differential 1-form is represented as the form ⎞ ⎛ m n m+1 n j +1 y = ∗−(m−2),j + d ⎝ ∗ m−3 ⎠ . X m=0 j =0
m=0 j =0
(3.5)
Y. Zhao / Nonlinear Analysis 64 (2006) 2332 – 2351
2337
In the sequel, let [s] be the entire part of s and R[X, y] (resp. R[H0 ]) the ring of polynomials in two variables X, y (resp. H0 ) over R. For simplicity, we say deg (X, y) = l if l is an upper bound (usually a lowest upper bound) for the degree of (X, y) ∈ R[X, y]. To obtain the relative cohomology decomposition of the differential 1-form , first of all we prove the following lemmas: Lemma 3.1. If j is odd, j 3, then ij can be expressed as the form ij =
i+j −1
∗k1 + d(X i+1 j (X, y)) + (X i j (X, y)) dH0 ,
k=i
i = 0, ±1, ±2, . . . .
(3.6)
If (i, j ) = (0, 2), (−1, 0) and (i, j ) = (i, 0), i = −1, then we have 02 = d(2H0 ln X − 23 BX 3 − CX 2 − 2DX) − 2 ln X dH0 , i+1 X −1,0 = d ln X, i0 = d . i+1
(3.7)
If j is even and i + j 3, i 2, j 2, then ij is decomposed into the form ⎞ ⎛ k =−1 ∗X k+1 + ij (H0 ) ln X + X i+1 j (X, y)⎠ ij = d ⎝ i k i+j
+ (ij ln X + X i j (X, y)) dH0
(3.8)
with (a) ij (H0 ) = ij ≡ 0 for i + j − 2 or (i, j ) = (1, 2); (b) deg ij (H0 ) = 0, ij = 0 for i + j = −1, 0, 1; and (c) deg ij (H0 ) = 1 for i 0 and i + j = 2, 3, where ij is a real constant, deg j (X, y) = j , deg j (X, y) = j − 2, and j (X, y) = j (X, y) =
[(j −2)/2]
2m
m=0
km =0
[(j −2)/2]
2m
m=0
km =0
∗X km y j −2m , ∗X km y j −2m−2 .
(3.9)
Proof. It follows from (3.3) that dH0 = ( 21 y 2 + 3BX 2 + 2CX + D) dX + Xy dy
(3.10)
which implies X i y j −2 dH0 =
1 2
ij + 3Bi+2,j −2 + 2Ci+1,j −2
+ Di,j −2 + X i+1 y j −1 dy,
j 2.
(3.11)
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Y. Zhao / Nonlinear Analysis 64 (2006) 2332 – 2351
Since X i+1 y j −1 dy = d
X i+1 y j j
−
i+1 ij , j
we obtain from (3.11) that for j = 2(i + 1), 2j − 3Bi+2,j −2 − 2Ci+1,j −2 − Di,j −2 ij = j − 2(i + 1) i+1 j X y i j −2 dH0 . +X y −d j
(3.12)
If j is odd and j 3, then j = 2(i + 1) for ∀i ∈ {0, ±1, ±2, . . .}. It follows from (3.12) by induction that ij =
2
∗i+s1 ,j −2 + d(∗X i+1 y j ) + ∗X i y j −2 dH0
s1 =0
=
2 2
⎛ ∗i+s1 +s2 ,j −4 + d ⎝∗X i+1 y j +
s1 =0 s2 =0
⎛
+ ⎝∗Xi y j −2 +
2
2
⎞ ∗X i+s1 +1 y j −2 ⎠
s1 =0
⎞ ∗X i+s1 y j −4 ⎠ dH0
s1 =0
= ··· 2
=
∗i+s1 +s2 +···+s(j −1)/2 ,1
s1 ,s2 ,...,s(j −1)/2=0
⎛
+ d ⎝X i+1 y j +
2
∗X i+s1 +1 y j −2 + · · ·
s1 =0 2
+
⎞
X i+s1 +s2 +···+s(j −3)/2 +1 y 3 ⎠
s1 ,s2 ,...,s(j −3)/2 =0
⎛
+ ⎝Xi y j −2 +
2
X i+s1 y j −4 + · · ·
s1 =0
+
2
⎞
X i+s1 +s2 +...+s(j −3)/2 y ⎠ dH0 ,
s1 ,s2 ,···,s(j −3)/2 =0
which yields (3.6) for odd j. If j is even and i + j 3, i 3, j 0, then j = 2(i + 1) if and only if (i, j ) = (0, 2), (−1, 0). Since (3.3) shows that y 2 = 2H0 X −1 − 2BX 2 − 2CX − 2D, we immediately obtain the representation for 02 . The expression for −1,0 and i,0 , i = −1 follows by direct computations.
Y. Zhao / Nonlinear Analysis 64 (2006) 2332 – 2351
2339
Substituting (i, j ) = (1, 2) into (3.12), one obtains the decomposition of 12 . If j = 2(i + 1), i.e. (i, j ) = (−1, 0), (0, 2), then using (3.12) again, one obtains ⎞ ⎛ (j −4)/2 2m ij = ∗k,2 + ∗02 + d ⎝X i+1 ∗X km y j −2m ⎠ i k i+j −2,k=0
⎛
(j −4)/2
2m
m=0
km =0
+ Xi ⎝
⎞
m=0
km =0
∗X km y j −2m−2 ⎠ dH0 ,
where the term 02 disappears if either i + j 1 or i 1. Representation (3.8) follows from (3.7), (3.12)|j =2 and the above decomposition. Lemma 3.2. Assume i 3. Then the differential 1-form −i,1 can be decomposed into the form −i,1 =
1 H0i−1
( −i,1 (H0 )01 + −i,1 (H0 )11 + −i,1 (H0 )21
+ dU−i,1 (X, y) + X −1 V−i,1 (X, y) dH0 ),
(3.13)
where −i,1 (H0 ), −i,1 (H0 ), −i,1 (H0 ) are polynomials of H0 with deg −i,1 (H0 )=[(2i −3)/3], deg −i,1 (H0 ) = deg −i,1 (H0 ) = [(2i − 5)/3], U−i,1 (x, y) and V−i,1 (X, y) are polynomials of degrees 2i − 1 and 2i − 3, respectively. If i=2, then −2,1 is decomposed into (3.13) with deg −2,1 (H0 )=deg −2,1 (H0 )=0, −2,1 (H0 ) ≡ 0, deg U−2,1 (x, y) = 3, deg V−2,1 (X, y) = 1. For i = 1, we have −1,1 =
1 (−2D01 − 5C11 − 8B21 − d(Xy 3 ) + 3y dH0 ). H0
(3.14)
Proof. Using (3.3) again, we obtain H0 −i,1 = 21 −i+1,3 + B−i+3,1 + C−i+2,1 + D−i+1,1 . It follows from (3.12) that 6 −i+1,3 = − 3B−i+3,1 − 2C−i+2,1 − D−i+1,1 2i − 1 −i+2 3 y X −i+1 +X −d y dH0 . 3
(3.15)
(3.16)
Substituting (3.16) into (3.15), we have −i,1 =
1 (2(i − 2)D−i+1,1 + (2i − 7)C−i+2,1 + (2i − 10)B−i+3,1 (2i − 1)H0 − d(X −i+2 y 3 ) + 3X −i+1 y dH0 ). (3.17)
Taking i = 1, 2 into (3.17), we obtain decomposition (3.14) and the representation of −2,1 , respectively. In what follows we prove (3.13) by induction. Noting that the term −i+3,1 (resp. −i+1,1 ) in (3.17) disappears if i =5 (resp. i =2), it follows from (3.14) and (3.17) that −k,1 , k =3, 4, 5, can
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Y. Zhao / Nonlinear Analysis 64 (2006) 2332 – 2351
be decomposed into form (3.13). Suppose that decomposition (3.13) holds for 3 k i −1, i 6. Then using (3.17) again, i−1 1 −i+2 3 −i+1 −i,1 = ∗−k,1 − d(X y ) + 3X y dH0 H0 k=i−3 i−1 1 1 ( −k,1 (H0 )01 + −k,1 (H0 )11 + −k,1 (H0 )21 ∗ = k−1 H0 H 0 k=i−3 + dU−k,1 (X, y) +X−1 U−k,1 (X, y) dH0 ) −d(X =
−i+2 3
y ) + 3X
1 H0i−1
i−1
−i+1
y dH0
∗( −k,1 (H0 )H i−k−1 01 + −k,1 (H0 )H0i−k−1 11
k=i−3
+ −k,1 (H0 )H0i−k−1 21 + H0i−k−1 dU−k,1 (X, y) + H0i−k−1 X −1 V−k,1 (X, y) dH0 )
−H0i−2 d(X −i+2 y 3 ) + 3H0i−2 X −i+1 y dH0 which shows that −i,1 can be expressed as (3.13) with −i,1 (H0 ) =
i−1
∗ −k,1 (H0 )H i−k−1 ,
k=i−3
−i,1 (H0 ) =
i−1
i−1
−i,1 (H0 ) =
∗ −k,1 (H0 )H i−k−1 ,
k=i−3
∗ −k,1 (H0 )H i−k−1 ,
k=i−3
U−i,1 (X, y) =
i−1
∗H i−k−1 U−k,1 (X, y) + H0i−2 X −i+2 y 3 ,
k=i−3
V−i,1 (X, y) = X
−1
i−1
(∗XH i−k−2 U−k,1 (X, y) + ∗H0i−k−1 V−k,1 (X, y))
k=i−3
+ ∗ H0i−3 X −i+3 y 3
+ 3H0i−2 X −i+2 y
.
Therefore, deg −i,1 (H0 )
max
i−3 k i−1
i−k−1+
2k − 3 3
=
2i − 3 . 3
Y. Zhao / Nonlinear Analysis 64 (2006) 2332 – 2351
2341
Using the same arguments as above, we obtain the upper bounds for deg −i,1 (H0 ) and deg −i,1 (H0 ), respectively. Since H0 (X, y), defined in (3.3), is a polynomial of degree 3, and H0i−2 X −i+2 y 3 = X i−2 ( 21 y 2 + BX 2 + CX + D)i−2 X −i+2 y 3 = ( 21 y 2 + BX 2 + CX + D)i−2 y 3 , which implies deg U−i,1 (H0 ) max
max
i−3 k i−1
{3(i − k − 1) + deg U−k,1 (X, y)}, 2(i − 2) + 3
= 2i − 1. Similarly, one obtains deg V−i,1 (X, y) 2i − 3.
Lemma 3.3. Suppose j is odd and j 3, i 2, −i + j 3; then, −i,j can be expressed in the form −i,j =
1 ( −i,j (H0 )01 + −i,j (H0 )11 + −i,j (H0 )21 H i−1 + dU−i,j (X, y) + X −1 V−i,j (X, y) dH0 ),
(3.18)
where U−i,j (X, y) and V−i,j (X, y) are polynomials of degrees 2i + j − 2 and 2i + j − 4, respectively, and (a) deg −i,j (H0 ) = [(2i + j − 4)/3], deg −i,j (H0 ) = deg −i,j (H0 ) = [(2i + j − 6)/3] for j = i − 1, i + 2; (b) deg −i,i−1 (H0 ) = deg −i,i−1 (H0 ) = i − 2, deg −i,i−1 (H0 ) = i − 3; and (c) deg −i,i+2 (H0 ) = deg −i,i+2 (H0 ) = i − 1, deg −i,i+2 (H0 ) = i − 2. Proof. Representation (3.18) follows from (3.6) and Lemma 3.2.
The main result of this section is the following. Proposition 3.4. Assume n4. Then any differential 1-form , defined in (3.5), can be decomposed into =
1 H0n−3
( (H0 )01 + (H0 )11 + (H0 )21 + d( (H0 ) ln X + (X, y))
+ ((H0 ) ln X + X −1 (X, y)) dH0 ),
(3.19)
where deg (H0 )=deg (H0 )=deg (H0 )=deg (H0 )=n−3, deg (H0 )=n−2, deg (X, y)= 3n−5, deg (X, y)=3n−7, (H0 ) and (H0 ) have a zero at H0 =0 with multiplicity n−3, n−4, respectively. If n = 1, 2, then can be expressed as = (H0 )01 + (H0 )11 + (H0 )21 + d( (H0 ) ln X + (X, y)) + ((H0 ) ln X + X −1 (X, y)) dH0 ,
(3.20)
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Y. Zhao / Nonlinear Analysis 64 (2006) 2332 – 2351
where (H0 ) = (H0 ) = (H0 ) ≡ 0 for n = 1 and deg (H0 ) = 1, deg (H0 ) = deg (H0 ) = 0 for n = 2, respectively, deg (H0 ) = deg (H0 ) = 0, deg (x, y) = 4, deg (X, y) = 2, (0) = 0. If n = 3, then can be represented as =
1 ( (H0 )01 + (H0 )11 + (H0 )21 + d( (H0 ) ln X + (X, y)) H0 + ((H0 ) ln X + X −1 (X, y)) dH0 ),
(3.21)
with deg (H0 ) = deg (H0 ) = deg (H0 ) = deg (H0 ) = 1, deg (H0 ) = 2, deg (X, y) = 7, deg (X, y) = 5, (0) = 0. Proof. Decomposition (3.20) follows from (3.5)–(3.8). From (3.6) we obtain that −1,3 =
1
∗k1 + d(3 (X, y)) + X −1 3 (X, y) dH0 .
(3.22)
k=−1
Use of (3.22), (3.8), (3.14) and (3.20) yields expression (3.21). Lemma 3.3 and (3.8) imply the following decomposition, provided m 4: ⎞ ⎛ m+1 m j +1 y ⎠ ∗−(m−2),j + d ⎝ X m−3 j =0
=
j =0
1
(−(m−2) (H0 )01 H0m−3 + −(m−2) (H0 )11 + −(m−2) (H0 )21 + d(−(m−2) (H0 ) ln X
+ −(m−2) (X, y)) + (−(m−2) (H0 ) ln X + X −1 −(m−2) (X, y)) dH0 ),
where deg −(m−2) (H0 ) = m − 3, deg −(m−2) (H0 ) = deg −(m−2) (H0 ) = m − 3 + ((−1)m − 1)/2, deg −(m−2) (H0 ) = m − 2, deg −(m−2) (H0 ) = m − 3, deg −(m−2) (X, y) = 3m − 5, deg −(n+2) (X, y)=3m−7, −(m−2) (H0 ) and −(m−2) (H0 ) have a zero at H0 =0 with multiplicity m − 3 and m − 4, respectively. By induction, one obtains (3.19) from the above decomposition, (3.20) and (3.21). 4. Abelian integrals and the second Melnikov functions In this section, we will derive, by Francoise’s recursive procedure [3], an appropriate formula for Mk (h), k = 1, 2. Before that we rewrite system (2.1) as a Pfaff equation − 1 − 2 2 − · · · − k k − · · · = 0, dH
(4.1)
where k = qk (x, y) dx − pk (x, y) dy and pk (x, y), qk (x, y) are defined in (2.2). Then M1 (h) = 1 . h
(4.2)
(4.3)
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(x, y) = h} are closed orbits of system (4.1) . As in [5,7], we introduce the Here h ⊂ {(x, y)|H following basic 1-forms: 01 = y dx,
11 = xy dx,
21 = x 2 y dx,
−1,1 =
y dx 1 + 2ax
and the corresponding integrals i1 . Ji (h) = h
(4.4)
(4.5)
Comparing (4.4) with (3.4), we have −1,1 = 2a −1,1 , 01 = 2a 01 , 11 = 2a( 01 + 2a 11 ), 21 = 2a( 01 + 4a 11 + 4a 2 21 ).
(4.6)
This yields the following. Proposition 4.1. Assume n4. Then M1 (h) can be expressed in the form M1 (h) =
1 (h − hl )n−3
( 1 (h)J0 (h) + 1 (h)J1 (h) + 1 (h)J2 (h)),
(4.7)
where deg 1 (h) = deg 1 (h) = deg 1 (h) = n − 3. If n = 1, 2, then M1 (h) = 1 (h)J0 (h) + 1 (h)J1 (h) + deg 1 (h)J2 (h),
(4.8)
where deg 1 (h) = deg 1 (h) = deg 1 (h) = 0. If n = 3, then M1 (h) =
1 ( 1 (h)J0 (h) + 1 (h)J1 (h) + 1 (h)J2 (h)), h − hl
(4.9)
where deg 1 (h) = deg 1 (h) = deg 1 (h) = 1. Proof. Assume that 1 is decomposed into (3.19) and n 4. It follows from Propositions 3.4 and (4.6) that M1 (h) can be expressed as (4.7) with 1 (h) = 2a( (h − hl ) + (h − hl ) + (h − hl )),
1 (h) = 4a 2 ( (h − hl ) + 2 (h − hl )), 1 (h) = 8a 3 (h − hl )
(4.10)
which yields the assertion for n4. Repeating the same arguments, we obtain (4.8) and (4.9). To obtain M2 (h) in an appropriate form, we next recall in brief Francoise’s recursive procedure [3] for a calculation of the kth order Melnikov function Mk (h), related system (4.1) . Lemma 4.2 (Francoise’s recursion formula [3]). Assume M1 (h) = M2 (h) = · · · = Mk−1 (h) ≡ 0 in (1.5), k 2. Then k , (4.11) Mk (h) = h
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where 1 = 1 ,
m = m +
ri j ,
2 m k
(4.12)
i+j =m
and the functions ri (x, y, H ), 1 i k − 1 are determined successively from the representations i = dRi (x, y, H ) + ri (x, y, H ) dH .
(4.13)
The proof of this lemma can be found in many papers; see for instance [3,7,13], etc. To apply Francoise’s recursion formula to determine M2 (h) for system (4.1) , first of all we have to obtain the representation 1 = 1 , defined in (4.13). Lemma 4.3. Assume that differential 1-form 1 is decomposed into the form (3.19) for n 4, (3.20) for n = 1, 2 and (3.21) for n = 3, respectively. Then M1 (h) ≡ 0 if and only if (H0 ) = (H0 ) = (H0 ) ≡ 0. Proof. Proposition 4.1 implies that M1 (h) ≡ 0 if and only if 1 (h)J0 (h) + 1 (h)J1 (h) + 2 (h)J2 (h) ≡ 0.
(4.14)
We say that is a polynomial 1-form of degree m, if = −f(x, y) dy + g (x, y) dx,
f(x, y), g (x, y) ∈ R[x, y]
(4.15)
and max{deg f(x, y), deg g (x, y)} = m. Let I be the real vector space of Abelian integral of the form (x, y). I(h) = (4.16) h
Then an Abelian integral I(h) belongs to I if and only if I(h) can be expressed as a linear combination of J0 (h), J1 (h), J2 (h) with polynomial coefficients in h [5]. It follows from [5, Proposition 8] that the R[h] module I is freely generated by the Abelian integrals J0 (h), J1 (h), J2 (h), which implies that (4.14) holds if and only if 1 (h) = 1 (h) = 1 (h) ≡ 0 (or see the proof of Proposition 3 in [7]). Using (4.10) again, we know that 1 (h) = 1 (h) = 1 (h) ≡ 0 if and only if (H0 ) = (H0 ) = (H0 ) ≡ 0. This proves the assertions. From the above discussions, we obtain the following. Proposition 4.4. Assume M1 (h) ≡ 0. Then the second-order Melnikov function for (4.1) and (2.3) can be expressed in the form M2 (h) = M2 (h) =
1 (h − hl )3
( 2 (h)J0 (h) + 2 (h)J1 (h) + 2 (h)J2 (h)),
1 (h − hl )2n−5
n = 3,
( 2 (h)J0 (h) + 2 (h)J1 (h) + 2 (h)J2 (h)),
n4,
(4.17) (4.18)
where deg 2 (h)=deg 2 (h)=deg 2 (h)=3 for n=3 and deg 2 (h)=deg 2 (h)=deg 2 (h)=2n−5 for n 4, respectively.
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Proof. Assume that 1 has been expressed in the form (3.19) for n 4. By Lemma 4.3, we have (H0 )= (H0 )= (H0 ) ≡ 0. Therefore, the 1-form 1 can be decomposed into the form (4.13)i=1 with r1 =
1 H0n−2
( 1 (H0 ) ln X + (−n + 3) (X, y) + X −1 H (X, y)),
where 1 (H0 ) = (−n + 3) (H0 ) + H (H0 )). From Lemma 4.3, we obtain (the calculations below are performed modulo 1-forms dR + r dH0 ): r1 (X, y, H0 )
(d( (H0 ) ln X + (X, y)) + ((H0 ) ln X + X −1 (X, y)) dH0 ) H0n−3 r1 (X, y, H0 ) 1 = (H0 ) dX + d (X, y) X H0n−3 1 (H0 ) 1 −1 = 2n−5 dX − (X, y)H0 (H0 ) X H0 (X, y) X X H0 +X−1 H0 (X, y) d ,
r1 1 =
where we use the following identities:
(X, y) d (X, y) = d( 21 2 (X, y)), 1 (H0 ) (H0 )
ln X dX = d( 21 1 (H0 ) (H0 )(ln X)2 ) X − 21 ( 1 (H0 ) (H0 )) (ln X)2 dH0 ,
1 (H0 ) ln X d = d( (X, y) 1 (H0 ) ln X) − (X, y) 1 (H0 ) ln X dH0 1 − (X, y) 1 (H0 ) dX. X It follows from (3.3) that X−1 H0 (X, y), (H0 ), X−1 (X, y)H0 (H0 ) are polynomials in X, y of degrees 3n − 5, 3n − 6 and 6n − 12, respectively. Since Proposition 3.4 asserts (0) = 0, we conclude that (H0 )/X is a polynomial in X, y of degree 3n − 7. Therefore, r1 1 can be rewritten in the form r1 1 =
1 H02n−5
2 ,
(4.19)
where 2 is a polynomial 1-form of degree 6n−12 (see (4.15) for the definition). Abelian integrals I(h), defined by (4.16), can be expressed as I(h) = (h)J0 (h) +
(h)J1 (h) + (H )J2 (h),
(4.20)
with deg (h) = [(m − 1)/3], deg
(h) = [(m − 2)/3], deg (H ) = [(m − 3)/3] (see [9, Remark 2.3]). This yields 1 r1 1 = ( 2 (h)J0 (h) +
2 (h)J1 (h) + 2 (h)J2 (h)), (4.21) (h − hl )2n−5 h
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where deg 2 (h) = deg
(h) = deg (H ) = 2n − 5. On the other hand, by the same arguments as in Proposition 4.1 we obtain that 1 2 = (¯ 2 (h)J0 (h) + ¯ 2 (h)J1 (h) + ¯ 2 (H )J2 (h)), (4.22) (h − hl )n−3 h ¯ with deg ¯ 2 (h) = deg (h) = deg ¯ (H ) = n − 3. Application of Lemma 4.2 allow us to obtain M2 (h) = 2 + r1 1 . 2 = h
h
h
Expression (4.18) follows from the above discussions. Repeating the same arguments, we obtain (4.17). Define the space of integrals n M1 = M1 (h)|M1 (h) =
1 1 , h ∈ 0, 6 h
,
Then we have the following proposition: Proposition 4.5. We have y n M1 = span J0 (h), J1 (h), J2 (h), J−k (h) = dx, k h (1 + 2ax) k = 1, 2, . . . , n − 2, ,
(4.23)
which yields dim Mn1 n + 1. Proof. Assertion (4.23) follows from (3.5) and (3.6).
In the paper [8], the authors proved that the linear space Mn1 has a dimension n + 1 if A = −2 in (1.2). According to this conclusion and Proposition 4.5, the following conjecture is reasonable: Conjecture 4.6. J−n+2 (h), J−n+3 (h), . . . , J0 (h), J1 (h), J2 (h) are linearly independent, so dim Mn1 = n + 1. If the above conjecture holds, we could obtain Conjecture 4.7. Let n M2 = M2 (h)|M2 (h) =
1 2 , M1 (h) ≡ 0, h ∈ 0, 6 h
.
Then dim Mn2 = 2n − 1. 5. Bounds of the number of limit cycles In this section, we will repeat the arguments of [5] and [7] to obtain the main results of this paper.
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Let J (h) = 1 (h)J0 (h) + 1 (h)J1 (h) + 1 (h)J2 (h).
(5.1)
Then we have Lemma 5.1. The following expression holds: J (h) = r(h)J0 (h) + s(h)J1 (h) + u(h)Z (h),
(5.2)
where Z(h) = (3a − 1)J1 (h) − 4aJ 2 (h), r(h), s(h), u(h) are real polynomials with deg r(h) = deg s(h) = deg u(h) = 0 for n = 1, 2, deg r(h) = deg s(h) = deg u(h) = 1 for n = 3, and deg r(h) = deg s(h) = deg u(h) = n − 3 for n 4, respectively. Proof. It follows from Remark 3.4 in [9] that J0 (h), J1 (h), J2 (h) satisfy the following Picard–Fuchs equation: −6ah −1 a+1 J0 0 −12a 2 h + a + 1 2a 2 − a − 1 J1 0 6h − 1 12ah − 2a J2 −4a 0 0 J0 = a(a − 1) −12a 2 0 J1 −(a + 1) −3a + 7 16a J2 which yields (5.2).
By Roussarie’s Theorem [19,20], we have either J (h) ≈ |(h − 16 )m/2 ln(h − 16 )|,
where m is even
(5.3)
or J (h) ≈ |(h − 16 )(m−1)/2 |,
where m is odd,
(5.4)
holds, and any function of the form J¯ (h) = (holomorphic function) ln(h − 16 ) + holomorphic function, sufficiently close to J (h) can have at most m zeros in {|h − 1/6| < d1 } ∩ D, where d1 > 0 is a small enough constant and D = C\[1/6, ∞). Definition 5.2 (Gavrilov [5], Gavrilov and Iliev [7]). We shall say J (h), h ∈ D, has a zero of multiplicity m at 1/6, provided that either (5.3) or (5.4) holds. Proposition 5.3. If J (h) has a zero of multiplicity m at 1/6, then it has at most 3n − 7 − m zeros in D. Proof. This proof literally repeats the arguments in [5] and [7]. Indeed, we just change the context in [5,7] concerning the degrees of polynomials and some symbols and notations. We present the proof here for the reader’s convenience.
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We know from [5, Proposition 10] that 0 is a first kind 1-form and hence J0 (h) does not disappear. Proposition 12 of [5] asserts that J (h) is a holomorphic function in D. So we can define the analytic function F (h) = J (h)/J0 (h) in D. In what follows we will estimate the number of zeros of F (h) by the argument principle. Denote by F + (h) (resp. F − (h)) the analytic continuation of F (h) on [1/6, +∞) along the path contained in the half-plane I mh > 0 (resp. I mh < 0). The Picard–Lefschetz formula implies I mF ± (h) = ±
, 0 ) W1 ,2 ( , |J0 (h)|2
where 1 , 2 are the cycles disappearing at h = 0 and h = 1/6, respectively, and 1 1 0 . )1 + 1 (H )2 , W , ( )0 + 1 (H = 1 (H , ) = det 0 1 2 2 2 0 It follows from Lemma 5.1 that W1 ,2 ( 1 , 0 ) + u(h)W1 ,2 ( Z , 0 ), , 0 ) = s(h)W1 ,2 ( where Z = (3a − 1) 1 − 4a 2 . By Proposition 10 in [5], the function W1 ,2 (Z , 0 ) is single valued in h on C and has no poles. For |h| ≈ ∞ the asymptotic estimate |J0 (h)| ≈ |h|−1/3 ,
|Z (h)| ≈ |h|1/3
implies that W1 ,2 ( Z , 0 ) is bounded in h. It follows that it is a non-zero constant. Further, the reciprocity law for meromorphic differentials of the first and third kind 0 and 1 implies P+ √ 0 , W1 ,2 ( 1 , 0 ) = 2 −1Res P+ 1 P−
where the path of integration from P− to P+ in the integral above is contained in h cut along P the loops 1 (h) and 2 (h). Since Res P+ 1 is a purely imaginary constant in h and P−+ 0 is imaginary too, we obtain that on the interval [1/6, ∞) holds P+ √ 0 + , 0 ) = s(h)2 −1 u(h), W1 ,2 ( P−
where s(h), q (h) are polynomials with deg s(h) = deg q (h) = n − 3. By Lemma 3 of [5], W1 ,2 ( , 0 ) has at most 2n − 5 zeros in the complex domain C\(−∞, hl ). Noting hl < 1/6 and |W1 ,2 ( , 0 )| < ≈ |h − 16 |[m/2] , the imaginary part of F (h) has at most 2n − 5 − [m/2] zeros on the interval (1/6, ∞). Let d∞ be large enough constant and ds be small enough constant. Denote by D the set obtained from D ∩ {|h| < d∞ } by removing a disc of radius d1 around h = 1/6. We shall evaluate the increment of argument of F (h) along the boundary D which has a positive orientation. Suppose that m is even. Then I mF (h) has at most 2n − 5 − m/2 zeros on the interval (1/6, ∞). Along the circle {|h| = d∞ }, we have F (h) < ≈ |h|n−3 .
Y. Zhao / Nonlinear Analysis 64 (2006) 2332 – 2351
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Finally, if h ≈ 1/6, then |F (h)| ≈ |h − 1/6|m/2 . This yields that the increment of the argument F (h) along the boundary D is equal to or less than 2(1 + n − 3 + 2n − 5 − m/2 − m/2) = 2(3n − 7 − m) and hence F (h) can have at most 3n − 7 − m, zeros in D . If m is odd, then we can obtain the assertion by the same arguments as in the proof of Theorem 5 in [5]. We omit the details here. Theorem 5.4. Assume n 4. Then M1 (h) and M2 (h) have at most 3n − 7 and 6n − 13 zeros on the interval (0, 1/6), respectively. If n = 1, 2, then M1 (h) has at most two zeros in (0, 1/6); If n = 3, then M1 (h) and M2 (h) have at most 5 and 11 zeros in (0, 1/6), respectively. Proof. Since J (0) = 0, it follows from Proposition 5.3 that J (h) has at most 3n − 7 zeros in (0, 1/6) for n 4. Therefore, M1 (h), defined in (4.7), has at most 3n − 7 zeros in (0, 1/6). Using the same arguments as in Proposition 5.3, we obtain that 2 (h)J0 (h)+ 2 (h)J1 (h)+ 2 (h) has at most 6n − 13 zeros in (0, 1/6) for n 4. The same is true, of course, for M2 (h). Repeating the same arguments as above, we can obtain the upper bounds of the number of zeros of M1 (h), n = 1, 2, 3 and M2 (h), n = 3, respectively. Proof of Theorem 1.1. To study the number of limit cycles in (4.1) , we divide the limit cycles into three types. As → 0, a limit cycle in (4.1) tends either to (a) the center (0, 0), or to (b) periodic orbits of (4.1)0 , or to (c) the homoclinic loop 1/6 . Let M1 (h) ≡ / 0. M1 (h) is analytic at h = 0 [20]. If M1 (h) = ∗hk + o(hk ), then the number of limit cycles of type (a) is less than or equal to k. Since hl < 1/6, we know that 1/(h − hl ) is analytic at h = 1/6, which implies that M1 (h) can be expressed as form (5.3) or (5.4) modulo a real constant factor. If J (h) (and hence M1 (h)) has a zero of multiplicity m at h = 1/6, then the number of limit cycles of type (c) is less than or equal to m. Finally, the number of limit cycles of type (b) is less than or equal to 3n − 7 − m − k by using Proposition 5.3. Therefore, system (4.1) (and hence (1.3)) has at most 3n − 7 − m − k + k + m = 3n − 7 limit cycles in the compact domain K ⊂ R for small enough. If M1 (h) ≡ 0 but M2 (h) ≡ / 0, then we obtain the second assertion of Theorem 1.1 by repeating the same arguments as above. Proof of Theorem 1.2. The exact upper bound for the number of limit cycles produced by the periodic annulus of system (1.1) under quadratic perturbations is equal to the maximum number of zeros of the related Abelian integrals as follows [12]: 1 (h) = M
h
x −4 ( x +
+ x −1 )y dx,
where ,
, are real constants. Therefore, we consider the following system instead of (1.3): x˙ = xy,
y˙ = (A + 2) − 2(A + 1)x + Ax 2 + 23 y 2 + ( x +
+ x −1 )y.
(5.5)
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Using the same arguments as in Proposition 2.1, system (5.5) can be reduced to x˙ = y(1 + 2ax),
y˙ = −x + x 2 − ay 2 + ( + x + x 2 )y,
(5.6)
where , are real constants. The related Melnikov function is M1 (h)= J0 (h)+ J1 (h)+ J2 (h). Repeating the arguments as in the proof of Theorem 1.1 or using Theorem 3 of [5] (the case n=3), one obtains that system (5.5) has at most two limit cycles in K and the bound is exact. It is obvious that the same is true for (1.3) with n = 2. This proves the theorem. References [1] Shui Nee Chow, Chengzhi Li, Yingfei Yi, The cyclicity of period annuli of degenerate quadratic Hamiltonian systems with elliptic segment loops, Ergodic Theory Dynamics Systems 22 (2) (2002) 349–374. [2] F. Dumortier, C. Li, Z. Zhang, Unfolding of a quadratic integrable system with two centers and two unbounded heteroclinic loops, J. Differential Equations 139 (1997) 146–193. [3] J.P. Francoise, Successive derivative of first return map, application to the study of quadratic vector fields, Ergodic Theory Dynamics Systems 1 (1996) 87–96. [4] L. Gavrilov, Petrov modules and zeros of Abelian integrals, Bull. Sci. Math. 122 (1998) 571–584. [5] L. Gavrilov, Abelian integrals related to Morse polynomials and perturbations of plane Hamiltonian vector fields, Ann. Inst. Fourier (Grenoble) 49 (1999) 611–652. [6] L. Gavrilov, The infinitesimal 16th Hilbert problem in the quadratic case, Invent. Math. 143 (3) (2001) 449–497. [7] L. Gavrilov, I.D. Iliev, Second-order analysis in polynomially perturbed reversible quadratic Hamiltonian systems, Ergodic Theory Dynamics Systems 20 (2000) 1671–1686. [8] L. Gavrilov, I.D. Iliev, Two-dimensional Fuchsian systems and the Chebyshev property, J. Differential Equations 191 (1) (2003) 105–120. [9] E. Horozov, I.D. Iliev, Linear estimate of the number of zeros of Abelian integrals with cubic Hamiltonians, Nonlinearity 11 (1998) 1521–1537. [10] I.D. Iliev, The cyclicity of the period annulus of the quadratic Hamiltonian triangle, J. Differential Equations 128 (1996) 309–326. [11] I.D. Iliev, Inhomogeneous Fuchs equations and the limit cycles in a class of near-integrable quadratic systems, Proc. Roy. Soc. Edinburgh 127A (1997) 1207–1217. [12] I.D. Iliev, Perturbations of quadratic centers, Bull. Sci. Math. 122 (1998) 107–161. [13] I.D. Iliev, On the limit cycles available from polynomial perturbations of the Bogdanov–Takens Hamiltonian, Israel J. Math. 115 (2000) 269–284. [14] I.D. Iliev, C. Li, J. Yu, Bifurcations of limit cycles from quadratic non-Hamiltonian systems with two centres and two unbounded heteroclinic loops, Nonlinearity 18 (2005) 305–330. [15] J. Li, Hilbert’s 16th problem and bifurcations of planar polynomial vector fields, Internat. J. Bifurcation Chaos 13 (1) (2003) 47–106. [16] Weigu Li, Yulin Zhao, Chengzhi Li, Zhifen Zhang, Abelian integrals for quadratic centres having almost all their orbits formed by quartics, Nonlinearity 15 (3) (2002) 863–885. [17] D. Novikov, S. Yakovenko, Redundant Picard–Fuchs system for abelian integrals, J. Differential Equations 177 (2) (2001) 267–306. [18] G.S. Petrov, On the nonoscillation of elliptic integrals, Funktsional. Anal. i Prilozhen 31 (4) (1997) 47–51; translation in Funct. Anal. Appl. 31(4) (1998) (262–265) (in Russian). [19] R. Roussarie, On the number of limit cycles which appear by perturbation of separatrix loop of planar vector fields, Bol. Soc. Brasil. Mat. 17 (2) (1986) 67–101. [20] R. Roussarie, Bifurcation of planar vector fields and Hilbert’s sixteenth problem, Progress in Mathematics, vol. 164, Birkhäauser Verlag, Basel, 1998. [21] S. Yakovenko, Bounded decomposition in Brieskorn lattice and Pfaffian Picard–Fuchs systems for Abelian integrals, Bull. Sci. Math. 126 (2002) 535–554. [22] J. Yu, C. Li, Bifurcation of a class of planar non-Hamiltonian integrable systems with one center and one-homoclinic loop, J. Math. Anal. Appl. 269 (1) (2002) 227–243. [23] H. Zoladek, Quadratic systems with centers and their perturbations, J. Differential Equations 109 (1994) 223–273. [24] Y. Zhao, W. Li, C. Li, Z. Zhang, Linear estimate of the number of zeros of Abelian integrals for quadratic centers having almost all their orbits formed by cubics, Sci. China (Ser. A) 45 (2002) 964–974.
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