Nonlinearity 11 (1998) 1521–1537. Printed in the UK
PII: S0951-7715(98)88484-3
Linear estimate for the number of zeros of Abelian
integrals with cubic Hamiltonians
Emil Horozov†§ and Iliya D Iliev‡k
† University of Sofia, Faculty of Mathematics and Informatics, 5, J. Bourchier blvd, 1126 Sofia,
Bulgaria
‡ Institute of Mathematics, Bulgarian Academy of Sciences, PO Box 373, 1090 Sofia, Bulgaria
Received 8 October 1997
Recommended by L Bunimovich
Abstract. An explicit upper
bound Z(3, n) 6 5n+15 is derived for the number of the zeros of
R
the integral h → I (h) = H =h g(x, y) dx −f (x, y) dy of degree n polynomials f, g, on the open
interval 6 for which the cubic curve {H = h} contains an oval. The
RR proof exploits the properties
of the Picard–Fuchs system satisfied by the four basic integrals H <h x i y j dx dy, i, j = 0, 1,
generating the module of complete Abelian integrals I (h) (over the ring of polynomials in h).
AMS classification scheme numbers: 34C05, 58F14
1. Introduction
A challenging problem which has arisen in attempts to solve the well known Hilbert’s 16th
problem (second part) is the following one. Let H (x, y), (x, y) ∈ R2 be a polynomial of
degree m (we call it Hamiltonian) and let f (x, y), g(x, y) be two polynomials of degrees,
which do not exceed n. Let 6 = (h1 , h2 ) ⊂ R1 be a maximal interval of existence of
a continuous family of ovals (closed connected components of the real algebraic curve
H (x, y) = h, h ∈ 6, that are free of critical points). We will denote these ovals by δ(h).
Define the complete Abelian integral
Z
[g(x, y) dx − f (x, y) dy],
h ∈ 6.
(1.1)
I (h) =
δ(h)
Then the infinitesimal 16th Hilbert problem (see [21]) is to give an estimate Z(m, n) of the
number of zeros of the Abelian integral I (h), h ∈ 6 only in terms of the degrees of the
polynomials H, f, g. In such a form the problem was explicitly formulated by Arnol’d [1]
but also can be traced back to the work of Ilyashenko [9].
We briefly recall the connection between the infinitesimal Hilbert problem and the
number of limit cycles, which is the main concern of Hilbert’s 16th problem. If we consider
the perturbed Hamiltonian system
ẋ = Hy + εf,
ẏ = −Hx + εg
(1.2)
§ E-mail: [email protected]
k E-mail: [email protected]
c 1998 IOP Publishing Ltd and LMS Publishing Ltd
0951-7715/98/061521+17$19.50 1521
1522
E Horozov and I D Iliev
where ε is a small parameter, then the zeros of the displacement function
Pε (h) − h = εI (h) + O(ε 2 )
(1.3)
give the limit cycles of (1.2). The first coefficient I (h) in the above expansion in ε is
exactly the Abelian integral (1.1). Obviously, the zeros of (1.1) give information about the
zeros of Pε (h)−h and thus about the limit cycles of (1.2). For a few nongeneric low-degree
Hamiltonians, the number of zeros of I (h) has been estimated in [13, 15–18, 4].
The first general result in the direction of solving the infinitesimal Hilbert problem was
achieved by Varchenko [20] and Khovanskii [12] who proved independently the existence
of the upper estimate Z(m, n). Their proofs give no indication of how this number depends
on the degree m of the Hamiltonian H and the maximal degree n of the perturbation
(f, g). Recently, in a series of papers, Ilyashenko, Yakovenkov and Novikov [10, 11, 14, 21]
suggested a constructive procedure which allows us to estimate the number of zeros of the
complete Abelian integral I (h) in terms of the Hamiltonian H and the degree n of the
perturbation. Their best result is that for any Hamiltonian H from a certain open set of
‘good’ Hamiltonians, there exists a constant c(H ) < ∞ such that the number of real isolated
zeros of I (h) in 6 does not exceed exp(c(H )n).
In this paper we return to the original question about the explicit estimate of the number
of zeros of I (h) only in terms of the degrees m and n. We give an upper bound for the
numbers Z(3, n), i.e. we estimate the number of zeros of I (h) only in terms of the degree
of the perturbation independently of the Hamiltonian H with a fixed degree, deg H = 3.
More precisely, our main result which we prove here is the following.
Theorem 1.1. For any cubic Hamiltonian H , the number of isolated zeros of I (h) in 6 does
not exceed 5n + 15, where n = max{deg f, deg g}.
Apparently, this upper bound is not the optimal one. For example, Z(3, 1) = 0, whilst
the sharp estimate one can expect for the quadratic case is Z(3, 2) = 2. For the open
set of generic cubic Hamiltonians with one centre and three saddles and any quadratic
perturbation, we proved [7] that (1.2) has no more than two limit cycles at all. For the
particular Hamiltonian H = y 2 + x 3 − x, Petrov [15] has given the exact estimate that the
number of zeros of I (h) does not exceed n − 1. Recently, Gavrilov [4] found that for the
Hamiltonian triangle H = 12 (y 2 + x 2 ) − 13 x 3 + xy 2 , the exact bound is [ 23 (n − 1)] ([p]
denotes the entire part of p). In all of these cases, the maximal number of zeros of I (h)
is less than the dimension of the corresponding linear space of integrals, i.e. I (h) obeys
the Chebyshev property. What we prove in section 2 is that the dimension of the linear
space of integrals I (h) in the generic case is [ 13 (4n + 1)]. In the light of these results, one
is tempted to conjecture that Z(3, n) = [ 13 (4n + 1)] − 1, i.e. that for generic H the space
of functions (1.1) also is Chebyshev in 6 (cf [3]). In section 5 we present a family of
generic cubic Hamiltonians with a cubic perturbation (f, g) for which the integral (1.1) has
five zeros whereas the conjecture would give three. Thus, for n > 3 the space of integrals
I (h) could be Chebyshev with accuracy at least 2. Our construction is obtained by first
perturbing a suitable family of Hamiltonian systems in the direction of non-Hamiltonian
ones and after that, in the direction of the Hamiltonian stratum. The latter perturbation
produces the additional two zeros. This example suggests that for generic Hamiltonians of
arbitrary degree m > 3, the Chebyshev property for n > 3 would hold with accuracy given
by the true dimension of the Hamiltonian space, which is 12 (m2 + 3m − 14) (that is the
dimension of the orbit space, under the action of the group of affine coordinate changes,
of the space of all degree m Hamiltonians). Thus, a general conjecture could be stated as
Linear estimate for the number of zeros of Abelian integrals
1523
follows
Z(m, n) = (the dimension of the space of Hamiltonians)
+(the dimension of the space of integrals) − 1.
For the reader’s convenience we describe the main steps in our work. In section 2 we
present our Abelian integral I (h) (in fact, in the case we consider it is complete elliptic
integral) as a polynomial envelope [10, 21] of certain fixed integrals. For generic cubic
Hamiltonians we show that the envelope has the form
I (h) = α(h)X(h) + β(h)Y (h) + γ (h)K(h) + δ(h)M(h)
RR
i j
where M = I00 , X = I10 , Y = I01 , K = I11 and Iij (h) =
H <h x y dx dy for
h ∈ 6 = (0, 16 ). See section 2 for the rest of the (nongeneric) Hamiltonians. Our next step
is to derive a four-dimensional Picard–Fuchs system (Ah + B)J 0 = CJ for the vector
of basic integrals J = col(X, Y, K, M). One of our main observations is that there exists
a linear combination Z of these integrals such that the right-hand side (C − A)J 0 of the
corresponding Picard–Fuchs system for the derivative J 0 depends only on Z 0 and M 0 (the
fact that the matrix C − A has rank two follows from the existence of two forms having
covariant derivatives without residues, see section 3 for more details). This implies that
the ratio Z 0 /M 0 satisfies a Riccati equation. Using these results, in section 4 we prove our
main theorem 1.1. This is done by first considering I 0 (h) which has the same number of
zeros and then eliminating the integrals X 0 and Y 0 . Reducing the initial problem to counting
the zeros of a rational envelope of only Z 0 , M 0 , we can use the related Riccati equation to
obtain the estimate.
2. Polynomial envelopes and their dimensions
In what follows we shall derive a formula for the Abelian integral I (h) given by (1.1). We
will represent it as a polynomial envelope of four basic integrals. Before that we recall the
normal form from [7] for all cubic Hamiltonians having a centre (i.e. those which only
make sense to consider here).
Lemma 2.1 ([7]). Any cubic Hamiltonian H (x, y) having a critical point of a centre type at
the origin can be put via affine change of variables into a normal form
H (x, y) = 12 (x 2 + y 2 ) − 13 x 3 + axy 2 + 13 by 3
where the parameters a, b are taken from the set
 = {− 12 6 a 6 1, 0 6 b 6 (1 − a)(1 + 2a)1/2 }.
(2.1)
The closed ovals around the centre at the origin are defined for Hamiltonian values h ∈
6 = (0, 16 ). The generic Hamiltonians are presented by the internal points of  and the
nongeneric ones are presented by the points on its boundary (see figure 1).
Indeed H being an odd-order polynomial must have a saddle as another critical point.
Placing it at (1, 0), we obtain the required normal form. Provided H has more saddles, we
place at (1, 0) a saddle from the connection surrounding the centre at the origin. Hereafter
we shall use only the above normal form, without explicitly mentioning it.
Denote by ωij the differential forms ωij = x i+1 y j dy, i, j = 0, 1, . . .. Let h ∈ 6 and
denote by δ(h) the oval of the curve 0h given by the equation H (x, y) = h (itR surrounds
the centre at the origin). Then introduce the integrals Iij (h) = (i + 1)−1 δ(h) ωij =
RR
1
i j
H <h x y dx dy for h ∈ (0, 6 ). Denote M = I00 , X = I10 , Y = I01 , K = I11
1524
E Horozov and I D Iliev
Figure 1. The domain  in the plane (a, b). Its boundary ∂ consists of the line AC containing
the loop cases and the arc CA containing the segment cases. The bifurcation points on ∂ are
A(− 12 , 0), B(0, 0), C(1, 0) and D( 12 , √1 ). The pictures around  are attached respectively to
2
the point A, the open segment AB, the point B etc. The bifurcation curves inside  are 02 :
{b2 + 8a 2 + 4a = 0}, presenting Hamiltonians with a cusp, and 0∞ : {4a 3 − b2 = 0}, presenting
Hamiltonians with three finite critical points.
(all these integrals have a mechanical meaning, see [7]). Finally, introduce the vector
J = col(X, Y, K, M) and put
c = 4a 3 − b2 .
(2.2)
Our main result in this section is the following lemma.
Lemma 2.2. Assume that bc =
6 0. Then for any polynomial P (x, y) of degree n − 1, the
integral
ZZ
P (x, y) dx dy,
h ∈ (0, 16 )
I (h) =
H <h
can be expressed as
I (h) =
X
αij k hk Iij (h)
ij k
where αij k are constants and the summation is taken along 0 6 i, j 6 1, i + j + 3k 6 n − 1.
Thus,
I (h) = α(h)X(h) + β(h)Y (h) + γ (h)K(h) + δ(h)M(h)
where α, β, γ , δ are polynomials of degrees deg α = deg β =
[ 13 (n − 3)] = q, deg δ = [ 13 (n − 1)] = r, respectively.
[ 13 (n
(2.3)
− 2)] = p, deg γ =
Linear estimate for the number of zeros of Abelian integrals
1525
Proof. We intend to first prove that for any d > 3 the integrals Iij , i + j = d can be
expressed as linear combinations of Ikl , k + l = d − 1 and hIkl , k + l = d − 3. For this, we
integrate the one-forms x k y l H (x, y) dx and x Rk y l h dx, k + l = d − 2, kR> 0, l > 0, along
the oval δ(h) ⊂ {H = h} and get equations δ(h) x k y l H (x, y) dx = h δ(h) x k y l dx which
imply
ZZ
x k y l−1 (lH + yHy ) dx dy = lhIk,l−1 ,
k + l = d − 2.
H <h
Writing the left-hand side in terms of Iij , we obtain
− 13 lIk+3,l−1 + (l + 2)aIk+1,l+1 + 13 (l + 3)bIk,l+2
= lhIk,l−1 − 12 lIk+2,l−1 − 12 (l + 2)Ik,l+1 ,
k l
k + l = d − 2.
(2.4)
k l
Similarly, integrating the forms x y H (x, y)dy and x y hdy yields
− 13 (k + 3)Ik+2,l + (k + 1)aIk,l+2 + 13 kbIk−1,l+3
= khIk−1,l − 12 (k + 2)Ik+1,l − 12 kIk−1,l+2 ,
k + l = d − 2.
Elementary manipulations reduce system (2.4), (2.5) to
3k
k + 2d + 4
k
−Ik+2,l + aIk,l+2 =
hIk−1,l −
Ik+1,l −
Ik−1,l+2 ,
d +2
2(d + 2)
2(d + 2)
k + l = d − 2,
3l
l
l + 2d + 4
hIk,l−1 −
Ik+2,l−1 −
Ik,l+1 ,
2aIk+1,l+1 + bIk,l+2 =
d +2
2(d + 2)
2(d + 2)
k + l = d − 2.
(2.5)
(2.6)
(2.7)
Next, we use (2.6) with l = 0, 1, 2, . . . , d − 2 and (2.7) with l = d − 3, d − 2, respectively
to obtain a linear algebraic system for I = col(Id,0 , Id−1,1 , . . . , I0,d ) of the form
T
∗
(2.8)
AI = B,
A=
0 A3
where T is upper triangular matrix of order d − 2 which diagonal elements are equal to −1
and
!
−1 0 a
A3 = 2a b 0 .
0 2a b
Since det A3 = c 6= 0 and B contains only the integrals we claimed above, the first step of
the proof is complete.
For k = l = 0 equations (2.6) and (2.7) become
−I2,0 + aI0,2 = −X,
2aK + bI0,2 = −Y.
(2.9)
Provided b 6= 0 this yields that the integrals Iij , i + j = 2 are expressed as linear
combinations of X, Y and K.
Now, for n = 1, 2 the assertion of lemma 2.2 is obvious; for n = 3, it follows from the
argument just stated, and for n > 4 it follows by induction. Lemma 2.2 is proved.
In fact, in the generic case the dimension of the space of functions in (2.3) is [ 4n+1
], see
3
the end of this section. Moreover, given any polynomial envelope (2.3), then integrating
along the oval the polynomial one-form of degree n, defined by
ω = −[α(H )xy + 12 β(H )y 2 + 12 γ (H )xy 2 + δ(H )y] dx
yields (2.3). Hence, the linear spaces in (1.1) and (2.3) have the same dimension.
1526
E Horozov and I D Iliev
Remark 2.3. The nongeneric cubic Hamiltonians are presented by the points (a, b) from
the boundary of , see (2.1). Assume that c 6= 0 in (2.2). Then the following cases appear.
(1) a ∈ (− 12 , 1), b = (1 − a)(1 + 2a)1/2 (segment cases). The line bx + (1 − a)y = 0
is a symmetry axis for the Hamiltonian. Then the integrals X, Y are linearly dependent,
bX = (a − 1)Y . So in the formulation of lemma 2.2 one can replace (2.3) with
I (h) = α(h)X(h) + γ (h)K(h) + δ(h)M(h)
(2.10)
where α, γ and δ have the same degrees as in (2.3). The dimension of the space of integrals
I (h) in this case is n.
(2) b = 0, a 6= 1 (loop cases). The symmetry axis is y = 0 now. Hence Y and K do
vanish and we can take as the basic integrals M, X and L = I2,0 . The proof stated above
still holds, the only difference is that from (2.9) we can express Iij , i + j = 2 as linear
combinations of X and L. Thus (2.3) becomes
I (h) = α(h)X(h) + γ (h)L(h) + δ(h)M(h)
(2.11)
where α, γ and δ have the same degrees as in (2.3). The dimension of the space of integrals
I (h) is n as well.
(3) a = 1, b = 0 (the Hamiltonian triangle). Then X, Y and K do vanish and we can
take as the basic integrals M and L. From (2.9), Iij , i + j = 2 are proportional to L and
(2.3) becomes
I (h) = γ (h)L(h) + δ(h)M(h)
(2.12)
where γ and δ have the same degrees as in (2.3). The dimension of the space of integrals
] in this case.
I (h) is [ 2n+1
3
Remark 2.4. The proof of lemma 2.2 does not work if c = 0. This is the case when H
has a singular point escaped to infinity (these Hamiltonians are not ‘nice’, in the sense of
[9, 14]). In this situation, we denote Z(h) = aX − bY . Then (2.3) has to be replaced by
I (h) = α(h)X(h) + β(h)Z(h) + δ(h)M(h)
(2.13)
], deg δ = [ n−1
]. This would set up an upper bound of [ 3n
]
where deg α = deg β = [ n−2
2
2
2
(which will be used hereafter) for the dimension of the integral in (2.13). We believe that
the true dimension for this case should be lower; one can expect the exact value would also
], see the conjecture below. As this value will not improve the main result of the
be [ 4n+1
3
paper, we did not attempt to obtain it.
Two particular cases
√ appear.
(1) a = 12 , b = 1/ 2 (the parabolic segment). Then X and Y being dependent, reduce
(2.13) to
I (h) = α(h)X(h) + δ(h)M(h)
(2.14)
where α and δ have the same degrees as in (2.13). The dimension is again n.
(2) a = b = 0 (the Bogdanov–Takens loop). Then (2.14) holds, which is well known
result due to Petrov [22].
To be more concrete, in the case where c = 0, we proceed as follows. Using the last
three equations in system (2.8) (with d replaced by d + 1) we find that the right-hand sides
satisfy an equation
d −1
3d + 4
3d + 5
d −2
2
aI3,d−3 −
bI2,d−2 + 4a +
a I1,d−1 −
bI0,d
d +3
2(d + 3)
d +3
2(d + 3)
6(d − 2)
3(d − 1)
ahI1,d−3 −
bhI0,d−2 .
=
(2.15)
d +3
d +3
Linear estimate for the number of zeros of Abelian integrals
1527
Then system (2.8) in which the last equation is replaced by (2.15) has a determinant
det A = 6(−1)d−2 a 3 (a + 1)(2d + 3)/(d + 3) and hence for any d > 3 the elements of
I are determined through Ikl , k + l = d − 1, hIkl , k + l = d − 3, and the function
(2d − 4)ahI1,d−3 − (d − 1)bhI0,d−2 (except for the Bogdanov–Takens Hamiltonian). For
1
bI2,0 + (4a 2 + 2a)K − 11
bI0,2 = − 35 bhM. Together with (2.9)
d = 2, (2.15) becomes − 10
10
this yields that Ikl , k + l = 2 are expressed through X, Z and hM if a 6= 0. Then (2.13)
follows by induction. The Bogdanov–Takens case can be treated similarly.
According to the above present results, the following conjecture is reasonable.
Conjecture 2.5. The true dimension of the linear space of integrals I (h) is as follows.
], for the Hamiltonian triangle (three axes of symmetry of the Hamiltonian vector
(i) [ 2n+1
3
field dH = 0),
(ii) n, for all remaining Hamiltonians from the boundary of  (having just one axis of
symmetry),
], for all Hamiltonians from the interior of  (no axis of symmetry in the field
(iii) [ 4n+1
3
dH = 0).
In practice, it remains to verify the above conjecture only for Hamiltonians corresponding
to the curve c = 0, a ∈ (0, 12 ). For all other cases, the assertions in the conjecture could be
proved by using the calculations above in combination with the results from [3], as Gavrilov
kindly explained to us. For instance, in the same manner the statement in (i) was proved
in [4]. The proof of conjecture 2.5 for c = 0 seems to require other ideas. Since in what
follows we will only use the fact that the corresponding dimensions do not exceed the above
bounds (which we have already established), we shall not give a detailed proof to all the
remaining cases as well.
3. Picard–Fuchs and Riccati equations
As mentioned in the introduction, one of the main ingredients of our construction is the
existence of two differential forms of the second kind. The explanation of this fact is the
following. The three differential forms x 2 dy, xy dy, x 2 y dy (here it is convenient to denote
them by ω1 , ω2 and ω3 , respectively) have poles at the three points P1 , P2 , P3 at infinity of
the algebraic curves H (x, y) = h. The residues of ωk are linear in h, hence their covariant
residues. A well known theorem in the theory of
derivatives ω10 , ω20 , ω30 have constant
P
Riemann surfaces [6] states that j ResPj ωk0 = 0, k = 1, 2, 3. Using this, we see that there
exists a nonzero linear combination ω of ω1 , ω2 and ω3 such that ResPj ω0 = 0, j = 1, 2, 3.
Hence ω0 has no residues, i.e. it is a form of the second kind. Then the corresponding
integrals of ω0 and of the holomorphic form ω00 , where ω0 = x dy, satisfy, over any fixed
cycle of the curve 0h , a system of differential equations.
In what follows we compute the form ω. In order to find its explicit expression one can
compute the residues of ω10 , ω20 and ω30 and find an appropriate linear combination with zero
residue. We prefer to use another method which also gives us the Picard–Fuchs and Riccati
equations required. The first step is to find the Picard–Fuchs system for X, Y , K, M.
1528
E Horozov and I D Iliev
Lemma 3.1. The integrals X, Y, K, M satisfy a Picard–Fuchs system
bX0 − (a + 1)Y 0 − 2a(a + 1)K 0 − 6bhM 0 + 4bM = 0,
(6ch + a + 1)Y 0 + [4a(a + 1)2 − c]K 0 − 6cY + b(a + 1)M = 0,
b(6h − 1)X 0 + (12a 2 h + a)Y 0 + (2a 2 + 3a + 1)K 0 − 6bX − 12a 2 Y + bM = 0,
(3.1)
[24a 2 (a + 1)h + a]Y 0 + [(6h − 1)c + (a + 1)(2a + 1)2 ]K 0 + b(a + 1)X
+[c − 28a 2 (a + 1)]Y − 8cK + abM = 0.
Rewritten as a first-order matrix equation for J = col(X, Y, K, M), this system reads
(Ah + B)J 0 = CJ
where
(3.2)


0
0
0 −6b
6c
0
0 
 0
A=
,
0
0
6b
12a 2
0 24a 2 (a + 1) 6c
0


b −(a + 1)
−2a(a + 1)
0
0
a+1
4a(a + 1)2 − c
 0
B=
,
−b
a
(a + 1)(2a + 1)
0
0
a
(a + 1)(2a + 1)2 − c 0


0
0
0
−4b
0
6c
0 −b(a + 1) 

C=
.
0
−b
6b
12a 2
2
−ab
−b(a + 1) 28a (a + 1) − c 8c
Proof. In fact, system (3.1) has already been derived in [7, lemma 3.3]. Indeed, we
assume for a moment that c 6= 0 and rewrite the system (3.5) from [7] as a matrix equation
(A0 h + B0 )J 0 = C0 J . Then multiplying from the left this equation with the matrix


c
0
0 0
c
0 0 
0
c−1 

1 0
0
2a 2
2
0 4a (a + 1) 0 −1
we get (3.2). The case c = 0 follows by continuity. The lemma is proved.
We observe that the matrix C − A contains in the upper-left corner a zero submatrix of
order 3, while its last row is {−b(a + 1), b2 + 4a 2 , 2c, −ab}. Taking Z = b(a + 1)X − (b2 +
4a 2 )Y −2cK, this means that the function Z 0 will together with M 0 satisfy a two-dimensional
Picard–Fuchs system.
To be more concrete, we consider first the generic case, with c 6= 0. Let us denote by
D(h) = h(h − 16 )(h − h0 )(h − h1 ) the critical polynomial corresponding to H (recall that
the critical levels h0 and h1 may be complex-conjugated; otherwise 16 < h0 6 h1 ). Below
we denote by Pk , Qk , Rk , Sk , Uk and Vk polynomials in h of degree k.
Lemma 3.2.
(i) The following relations hold in the generic case, provided c 6= 0:
DX00 = hU1 Z 0 + hU2 M 0
DY 00 = hV1 Z 0 + hV2 M 0
DZ 00 = hP2 Z 0 + hQ2 M 0
DM 00 = R2 Z 0 + hS2 M 0 .
(3.3)
(3.4)
Linear estimate for the number of zeros of Abelian integrals
1529
(ii) The vector I = col(X, Y, Z, M) satisfies a matrix equation of the form
(A1 h + B1 )I 0 = I
where B1 is some constant matrix and

4c 0
 0 4c
A1 = (4c)−1 
0 0
0 0
(3.5)
0
0
3c
0

c − 2a 2 (a + 1)
b(a + 1)

.
3
b[c
+
a 2 (a + 1)2 ]
2
6c
Proof. The proof follows by computation. We use the following constant matrices,




1 0
2
0
2c
0
0
0
0
2c
0
0
 0 1 −a − 1 0 

S=
T =
,
,
0 0
1
0
b(a + 1) c − 4a 2 (a + 1) −1 0
0
0
0 2c
0 0
−a
1
to transform (3.2) into an appropriate form. Take I = col(X, Y, Z, M), then 2cJ = T I
and (3.5) directly follows. We next differentiate (3.2) and then multiply the result by S.
One obtains
(SAT h + SBT )I 00 = S(C − A)T I 0 ⇔ (A2 h + B2 )I 00 = C2 I 0 .
The calculation yields

0 −2b
4b
8a 2
0 
 −2b(a + 1) 2c − 4a 2 (a + 1) 0
A2 = 6c 
,
2b
4a 2
0
0
0
b(a + 1)
c + 4a 2 (a + 1) −1


0 0
0
0
0
0 
0 0
C2 = 
.
0 0
0
−2bc
0 0 −2c
0
Also note that the first, second and fourth rows of the matrix SB are linearly dependent
and moreover, its last column is zero. Hence, the same remains true for B2 . Using these
facts, we see that the statement in (i) follows from Cramer’s formula.

Corollary 3.3. Assume that c 6= 0. Then in the generic case the ratio w(h) = Z 0 (h)/M 0 (h)
satisfies the Riccati equation
D(h)w 0 (h) = −R2 (h)w2 (h) + h[P2 (h) − S2 (h)]w(h) + hQ2 (h).
(3.6)
In order to investigate the rest of the cases we have to derive for them the analogues of
(3.3), (3.4), (3.6). Their treatment is simpler. We have listed the corresponding equations
in the following two remarks.
Remark 3.4.
(1) In the segment cases (except for the parabolic segment) we obtain instead of (3.3),
(3.4) and (3.6) the equations
DX00 = hU0 Z 0 + hV1 M 0
DZ 00 = hP1 Z 0 + hQ1 M 0
(3.7)
DM 00 = R1 Z 0 + hS1 M 0
D(h)w 0 (h) = −R1 (h)w2 (h) + h[P1 (h) − S1 (h)]w(h) + hQ1 (h),
(3.8)
1530
E Horozov and I D Iliev
with a critical polynomial D(h) = h(h − 16 )(h − h1 ) where 16 < h1 . In this case the integrals
X and Y are dependent which reduces the dimension of system (3.2) by 1, so we can take
I = col(X, Z, M), Z = b(a 3 + 1)X + c(a − 1)K. The argument is the same as above.
(2) In the loop cases (except for the Bogdanov–Takens loop) we can derive a Picard–
Fuchs system for integrals J = col(X, L, M). For this purpose, we could use system (3.1).
As the integrals Y and K vanish for b = 0, we can write Y (b, h) = bYb (h) + O(b2 ),
K(b, h) = bKb (h) + O(b2 ). We put these values in (3.1), divide the system by b and then
take a limit b → 0. Operating with the equations of the system thus obtained, we are able
to extract from it a three-dimensional system that the integrals X, Yb + 2aKb and M satisfy.
Next, using (2.9) we find
Yb (h) + 2aKb (h) = ∂b [Y (b, h) + 2aK(b, h)]|b=0 = −I02 (h) = (X(h) − L(h))/a.
Replacing, we come to the following system
−X 0 + (a + 1)L0 − 6ahM 0 + 4aM = 0,
(−12a 2 h + a + 1)X 0 + (2a 2 − a − 1)L0 + 12a 2 X + a(1 − a)M = 0, (3.9)
(6h − 1)X 0 + (12ah − 2a)L0 + (3a − 7)X − 16aL + (a + 1)M = 0.
Denoting Z = (3a − 1)X − 4aL, I = col(X, Z, M) and using this system, just as before
we find equations (3.7) and (3.8). Let us note that in the considered case the third critical
value h1 = (3a + 1)/24a 3 lies for a ∈ (− 12 , − 13 ) in the interval 6. Also note that in the
particular case when a = − 13 , the system for X, Z, M becomes
8
−Z 0 − 4hM 0 + M = 0,
3
4
4hX0 + Z 0 − 4X + M = 0,
3
4
0
(1 − 6h)Z + 8Z + M = 0
3
and hence contains an invariant subsystem of size 2. This yields a Riccati equation directly
for w(h) = Z(h)/M(h),
h(6h − 1)w 0 (h) = 2w2 (h) + (4h + 1)w(h) + 43 h.
This particular case has been considered (partly) by Petrov [15], see also [18] (indeed,
taking appropriate coordinates, one can rewrite the Hamiltonian as H = xy + x 3 + y 3 ).
(3) Since X = 0 for the Hamiltonian triangle, (3.9) reduces to a two-dimensional system
which immediately yields a Riccati equation for w(h) = L(h)/M(h):
h(6h − 1)w 0 (h) = − 83 w2 (h) + (4h + 1)w(h) − h.
Remark 3.5. The case when c = 0 can be treated similarly. We put c = 0 in (3.1) and
then eliminate K 0 from the obtained system which yields
2bX 0 − (2a + 1)Y 0 − 12bhM 0 + 9bM = 0,
4ab(6h − 1)X 0 + (48a 3 h + 4a 2 − 2a − 1)Y 0 − 24abX − 48a 3 Y + (2a − 1)bM = 0,
[96a 3 (a + 1)h − 4a − 1]Y 0 + 4ab(a + 1)X − 112a 3 (a + 1)Y − (4a + 1)bM = 0.
Taking Z = aX −bY , I = col(X, Z, M) we get just as above (3.7) (where the first equation
is replaced by DX00 = hU1 Z 0 + hV1 M 0 ) and then (3.8). It remains to consider the following
two particular cases.
(1) The parabolic segment. A Riccati equation
w2 (h) + (3h + 1)w(h) − 12 h
h(6h − 1)w 0 (h) = − 15
4
(3.10)
Linear estimate for the number of zeros of Abelian integrals
1531
obtains for w(h) = X(h)/M(h) in this case.
(2) The Bogdanov–Takens Hamiltonian. Using (3.9) with a = 0, we easily find the well
known equation for w(h) = X(h)/M(h):
h(6h − 1)w0 (h) = − 76 w2 (h) + (2h + 1)w(h) − h.
4. Proof of the linear estimate in the generic case
In this section we prove the main theorem for generic Hamiltonians (here, genericity
means that the Hamiltonian parameters (a, b) correspond to internal point of , satisfying
c = 4a 3 − b2 6= 0). The proofs in the rest of the cases (see remarks 3.4 and 3.5) follow
the same line of argument. Moreover, being technically simpler, the estimates for them are
even better. The details for these cases are sketched in the next section.
As far as I (0) = 0, the function I (h) has no more zeros in 6 than I 0 (h). From (2.3)
and (3.5) we obtain the following expression of I1 = I 0 :
I1 (h) = α1 (h)X0 (h) + β1 (h)Y 0 (h) + γ1 (h)Z 0 (h) + δ1 (h)M 0 (h)
(4.1)
where deg α1 = deg β1 = p, deg γ1 = q, deg δ1 = r. Our strategy is to reduce the problem
of estimating the zeros of I1 (h) to a problem of estimating the zeros of certain Abelian
integral, expressed as a linear combination of only Z 0 , M 0 with rational coefficients (cf [8]).
This will be done in two steps by consecutively eliminating X0 and Y 0 . Denote by N the
set of the zeros of α1 (h) in 6.
Lemma 4.1. In 6 \ N we have the representation, provided c 6= 0,
I1 (h) 0
I2 (h)
= 2
α1 (h)
α1 (h)D(h)
(4.2)
where
I2 (h) = β3 (h)Y 0 (h) + γ3 (h)Z 0 (h) + δ3 (h)M 0 (h)
and deg β3 = 2p + 2, deg γ3 = p + q + 3, deg δ3 = p + r + 3, β3 (0) = δ3 (0) = 0.
Proof. By (4.1), (3.3) and (3.4), we have in 6 \ N
(I1 /α1 )0 = (1/α1 )[α1 X 00 + β1 Y 00 + γ1 Z 00 + δ1 M 00 ] + (β1 /α1 )0 Y 0 + (γ1 /α1 )0 Z 0 + (δ1 /α1 )0 M 0
= 1/(α1 D)[γ2 Z 0 + δ2 M 0 ] + (β1 /α1 )0 Y 0 + (γ1 /α1 )0 Z 0 + (δ1 /α1 )0 M 0
= 1/(α12 D)[β3 Y 0 + γ3 Z 0 + δ3 M 0 ]
It is easy to see that deg γ2 = q + 3, deg δ2 = r + 3, deg β3 = 2p + 2, deg γ3 = p + q + 3,
deg δ3 = p + r + 3 and that β3 (0) = δ3 (0) = 0.
Therefore we need to estimate the zeros of I2 (h). Indeed, denote by λ and by µ
respectively the number of the zeros in 6 of I1 (h) and I2 (h). Then we have the following
estimate.
Lemma 4.2. The number λ of zeros of I1 (h) in 6 does not exceed µ + p + 1.
Proof. Denote by ξ0 , ξ1 , . . . , ξk+1 , k > 0, the elements of the set N ∪ {0} ∪ { 16 }, where
0 = ξ0 < ξ1 < · · · < ξk+1 = 16 . Let us choose around each point ξj , j = 0, . . . , k + 1 a
small interval ωj = [ξj − ε0 , ξj + ε0 ] where the functions I1 (h), I2 (h) and α1 (h) do not have
zeros different from ξj . Consider the interval j = (ξj + ε0 , ξj +1 − ε0 ), j = 0, . . . , k. If
1532
E Horozov and I D Iliev
we denote the number of the zeros of I1 (h) in j by λj and correspondingly the number of
the zeros of I2 (h) by µj , then obviously λj 6 µj + 1. In the intervals ωj , 0 6 j 6 k + 1,
the only zeros possessed by I1 (h) and I2 (h) could eventually be at ξj . Let us denote the
multiplicity of the zero of I1 (h) and I2 (h) at ξj , 1 6 j 6 k, by lj > 0 and by mj > 0
respectively. We will show that lj 6 mj . Assume that the multiplicity of the zero of α1 (h)
at ξj is aj . Then we see that in the right-hand side of (4.2) the power of the first term in the
Laurent series around h = ξj is mj − 2aj . At the same time, in the left-hand side it is either
lj − aj − 1, provided lj 6= aj , or at least zero otherwise. This gives lj 6 mj − aj + 1 6 mj
for j = 1, . . . , k. Now the number of the zeros of I1 (h) in (0, 16 ) is estimated by
λ = λ0 +
k
k
X
X
(lj + λj ) 6 µ0 + 1 +
(mj + µj + 1) = µ + k + 1 6 µ + p + 1,
j =1
j =1
using the fact that k 6 p. This proves the statement of the lemma.
In the same manner as in lemma 4.1 we can omit Y 0 .
Lemma 4.3. If c 6= 0, then the function I2 (h) has the representation
I3 (h)
I2 (h) 0
,
I3 (h) = γ5 (h)Z 0 (h) + δ5 (h)M 0 (h)
= 2
β3 (h)
β3 (h)D(h)
(4.3)
where deg γ5 = 3p + q + 8, deg δ5 = 3p + r + 8 and γ5 (0) = δ5 (0) = δ50 (0) = 0.
Proof. We have, as above,
(I2 /β3 )0 = (1/β3 )[β3 Y 00 + γ3 Z 00 + δ3 M 00 ] + (γ3 /β3 )0 Z 0 + (δ3 /β3 )0 M 0
= 1/(β3 D)[γ4 Z 0 + δ4 M 0 ] + (γ3 /β3 )0 Z 0 + (δ3 /β3 )0 M 0
= 1/(β32 D)[γ5 Z 0 + δ5 M 0 ].
It is easily checked that deg γ4 = p + q + 6, deg δ4 = p + r + 6, deg γ5 = 3p + q + 8,
deg δ5 = 3p + r + 8 and that γ5 , δ5 have the needed behaviour at the origin.
Denote by ν the number of the zeros in 6 of the function I3 (h). Repeating the argument
of lemma 4.2 we obtain the following.
Lemma 4.4. The number µ of the zeros of I2 (h) in 6 does not exceed ν + 2p + 2. Hence,
λ 6 ν + 3p + 3.
Proof. The only difference with the proof previously stated is that the condition β3 (0) = 0
sets an upper bound deg β3 − 1 = 2p + 1 for the number of different zeros of β3 in 6. Now we are ready to give a proof of the main theorem in the generic case.
Theorem 4.5. Let H be generic. Then the number λ of the zeros of I (h) in 6 does not
exceed 5n + 15.
Proof. Obviously our main concern will be to estimate the number ν of the zeros of
I3 . We are going to use the Riccati equation (3.6). Denote by W = W (h) the ratio
I3 /M 0 = γ5 w + δ5 . As M 0 has no zeros and poles, W has in 6 just ν zeros. A simple
computation using (3.6) shows that W satisfies the Riccati equation
γ5 (h)D(h)W 0 (h) = −R2 (h)W 2 (h) + hγ6 (h)W (h) + h3 γ7 (h)
(4.4)
Linear estimate for the number of zeros of Abelian integrals
1533
where deg γ6 = 3p + q + 10, deg γ7 = 6p + q + r + 16. Now denote by 1j , j = 1, . . . , k,
k 6 deg γ5 the open intervals into which 6 is split by the consecutive zeros of γ5 . Consider
the following autonomous system (cf [5, 18])
ḣ = γ5 (h)D(h),
Ẇ = −R2 (h)W 2 + hγ6 (h)W + h3 γ7 (h)
(4.5)
which is equivalent to (4.4). Take an interval 1j . Denote by λj the number of zeros of γ7
in 1j . Then between any two zeros of W in 1j there is a zero of γ7 (h) (that is, there exists
a phase curve of (4.5) for which the h-axis is a horizontal tangent). Hence the number of
the zeros of W in 1j is at most λj + 1. We have
k
X
(λj + 1) 6 deg γ7 + deg γ5 .
j =1
From this we obtain
X
ν6
(λj + 1) 6 (6p + q + r + 16) + (3p + q + 8).
(4.6)
The above bound holds under the restriction that (4.5) has no critical points of type (h∗ , 0)
where h∗ ∈ 6. However, it remains true in the general case as well (clearly this estimate
is stable under a slight modification of the coefficients in (4.5)).
Finally, we obtain by (4.6) and by lemma 4.4 that λ 6 12p + 2q + r + 27 6 5n + 15.
Theorem 4.5 is proved.
5. End of the proof of the main theorem. Concluding remarks
In order to complete the proof of theorem 1.1 we need to find estimates for the degenerate
Hamiltonians. As mentioned in the introduction, for some of them the linear estimate is
known. Up to now, all the known results have been established by using Petrov’s method.
For some of the cases, this method has allowed us to obtain the optimal bounds. However,
it works provided all the critical levels of the Hamiltonian are real, which is not generally
true. In proposition 5.1 we list the results which have been previously obtained and in
proposition 5.2 we add to them the cases that have not been considered. Their analysis is
simpler, due to the fact that we need to eliminate here only X0 and in some of the cases
even this is not needed. To formulate the propositions we introduce the following notation.
Given a degenerate Hamiltonian H , put Z(H, n) = (number of zeros of I (h) = IH (h) in
6).
Proposition 5.1.
(i) (Petrov [15]) For the Bogdanov–Takens loop (a = b = 0):
Z(H, n) 6 n − 1.
(ii) (Gavrilov [4]) For the Hamiltonian triangle (a = 1, b = 0):
2n − 2
.
Z(H, n) 6
3
(iii) (Stoev [19]) For the elliptic segment case a ∈ (− 12 , 12 ), b = (1 − a)(1 + 2a)1/2 :
4n − 3
.
Z(H, n) 6
3
1534
E Horozov and I D Iliev
To obtain bounds for the rest of the degenerate cases, we follow the line of argument
presented in the previous section.
Proposition 5.2.
(i) For the loop cases (b = 0, a 6= 0, 1), the estimate holds
7n + 13
Z(H, n) 6
.
3
(ii) For the hyperbolic segment case a ∈ ( 12 , 1), b = (1 − a)(1 + 2a)1/2 :
7n + 7
Z(H, n) 6
.
3
(iii) For the parabolic segment case (a = 12 , b =
3n − 2
Z(H, n) 6
.
2
√1 ):
2
(iv) For Hamiltonians with three finite critical points, a ∈ (0, 12 ), c = 4a 3 − b2 = 0:
7n + 4
Z(H, n) 6
.
2
Proof. The proof follows easily by the argument used in the proof of theorem 4.5, taking
into account remarks 3.4 and 3.5.
The bounds from propositions 5.1 and 5.2 together with theorem 4.5 give our main
result.
Theorem 5.3. The following linear estimate is satisfied for the cubic case m = 3:
Z(3, n) 6 5n + 15.
(5.1)
In view of the above-listed results it is natural to ask how far from the exact estimate is
(5.1). First we recall that the dimensions of the polynomial envelopes in (2.3), (2.10)–(2.12),
(2.14) do not exceed [ 4n+1
]. One is tempted to conjecture that the exact estimate would
3
] − 1, as it is in some particular cases
establish the Chebyshev property, Z(3, n) 6 [ 4n+1
3
(see proposition 5.1(i), (ii) above). The examples stated below show that it is not the case.
In fact, it turns out that for n > 2 or for m > 3, n > 1 the system of functions included in
I (h) does not generally satisfy the Chebyshev’s property. As an example we can consider
the generic case for m = n = 3. Then α, β, γ and δ in (2.3) are constants. It is not hard
to construct a linear combination having at least five zeros in 6. For this, we will use the
asymptotic expansions of integrals X, Y , K, M near the critical level h = 0 corresponding
to the centre at the origin. The first few terms of the expansions were computed in [7].
Thus, denoting m0 = M 0 (0) > 0, we have as h → 0
M ∼ m0 h,
X ∼ 12 m0 (1 − a)h2 ,
bX + (1 − a)Y ∼ 56 m0 b(1 − 3a 2 + 2a 3 − b2 )h3
and
Z = (5a − 1)[bX + (1 − a)Y ] − 5(1 − 3a 2 + 2a 3 − b2 )K = O(h4 ).
We are going to show that there is a curve 0 in the Hamiltonian parameter space (a, b) ∈ 
along which Z = O(h5 ) and a unique point P on 0 at which Z = O(h6 ), see figure 2.
Linear estimate for the number of zeros of Abelian integrals
1535
Figure 2. Location of the curve 0 and the point P inside .
The terms required in the expansion of Z can be easily computed from (3.2). We take
the kth derivative of (3.2) at h = 0,
BJ (k+1) + kAJ (k) = CJ (k)
which yields a linear system for determination of X(k+1) , Y (k+1) , K (k+1) , M (k) through X (k) ,
Y (k) , K (k) , M (k−1) . In this way we obtain
Z∼
35
m b(b2
24 0
− 2a 3 + 3a 2 − 1)(b2 − 3a 2 − 6a + 1)h4
and hence 0 = {(a, b) ∈  : b2 − 3a 2 − 6a + 1 = 0}. Clearly, 0 is a curve in the space of
generic cubic Hamiltonians. Further,
Z|0 ∼ − 72 m0 b(a + 1)2 (a 2 − 4a + 1)(11a 3 + 23a 2 − 21a + 3)h5 .
Thus, Z|P ∼ m6 h6 where P is the point on 0 having
as a first coordinate
√ √
√ the root of the
equation 11a 3 + 23a 2 − 21a + 3 = 0 with (2 − 3)/ 3 < a < 2 − 3. Now, we can
find a nearby point Q on 0 such that Z|Q = m5 h5 + m6 h6 + O(h7 ) where m5 m6 < 0 and
|m5 | |m6 |. Moving Q outside 0, we further obtain Z|Q = m4 h4 + m5 h5 + m6 h6 + O(h7 )
where m4 m5 < 0 and |m4 | |m5 |. We now freeze the point Q = (a, b) thus
obtained and then choose suitable constants α, β, γ for which the linear combination
I = αM + βX + γ [bX + (1 − a)Y ] + Z will have the expansion
I (h) =
6
X
mk hk (1 + O(h)),
mk mk+1 < 0,
|mk | |mk+1 |.
k=1
Hence, the function I possesses at least five zeros in 6 although it belongs to a fourdimensional function space.
Similarly, using that K ∼ 16 m0 b(5a − 1)h3 , we see that in the segment case, for (a, b)
√
near the point ( 15 , 45√75 ), the integral I (h) = αX + γ K + δM has at least three zeros in 6.
The scheme by which we have constructed the above examples prompts the conjecture
on Z(m, n) we formulated in the introduction.
We finish the paper by another remark concerning rather the cycles in (1.2) than the zeros
of I (h). As is well known, the Abelian integral I (h) in (1.1) defines the so-called Melnikov
function of the perturbation (1.2). We have established that in the generic case the integral
]. At the same time, the codimension of
I (h) produces a linear space of dimension [ 4n+1
3
1536
E Horozov and I D Iliev
the Hamiltonian stratum in the perturbed system (1.2) is 12 n(n + 1). As a consequence,
if n > 3, then the annihilation of I (h) does not imply that the perturbation is integrable,
even in the generic case. A similar situation occurs for generic Hamiltonians of any degree
m, provided n > m (contrary to the case n < m, when the condition I (h) ≡ 0 forces the
form ω = g(x, y) dx − f (x, y) dy to be exact [9]). In this situation, it is necessary to use
higher-order Melnikov functions in order to estimate the number of the limit cycles in (1.2).
Let us consider in more detail the case when m = n = 3. Using the linear independence
of
R
X, Y , K, M, it is easy to obtain that any degree 3 form ω satisfying I (h) = δ(h) ω ≡ 0 can
be written as ω = dG + λH dx + µH dy. Here, G(x, y) is a polynomial of degree 4 and λ,
µ are constants. To calculate the second Melnikov function I2 (h), we use the Françoise’s
procedure [2]. Expressing ω as ω = −(λx + µy) dH + d[G + (λx + µy)H ], then
Z
ZZ
(λx + µy)ω =
(µGx − λGy ) dx dy
I2 (h) = −
δ(h)
H <h
= αX + βY + γ K + (δ + ηh)M
with suitable constants α, β, etc. We observe that the second Melnikov function defines
a linear space of higher dimension, compared with (1.1). The same is true in the general
case as well. One poses the problem: given n, find the order of the Melnikov function in
(1.2), which will produce an envelope of maximal dimension. This would give us a method
to determine the maximal number of limit cycles in (1.2) which can be born from periodic
trajectories of the Hamiltonian system.
Acknowledgments
The authors thank L Gavrilov for stimulating discussions and several useful comments.
Many suggestions made by the referees helped us to improve the presentation of the text.
This research was partially supported by a grant MM-523/95 from the NSF of Bulgaria.
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