BIFURCATIONS OF LIMIT CYCLES FROM CUBIC HAMILTONIAN SYSTEMS WITH A CENTER AND A HOMOCLINIC SADDLE-LOOP

It is proved in this paper that the maximum number of limit cycles of system { dx dt = y, dy dt = kx − (k + 1)x2 + x3 + (α + βx + γx2)y is equal to two in the finite plane, where k > 11+ √ 33 4 , 0 < | | 1, |α| + |β| + |γ| = 0. This is partial answer to the seventh question in [2], posed by Arnold.


Consider the Abelian integral
where H(x, y), X(x, y) and Y (x, y) are real polynomial of x and y, Γ h is the compact component of H(x, y) = h, Σ is the maximal interval of existence of Γ h .Finding the lowest upper bound for the number of zeros of I(h) is called the weakend Hilber-16th problem [1], which is closed related to determining the number of limit cycles of perturbed system where 0 < | | 1.
In particular, suppose H(x, y) = 1 2 y 2 + U (x) = h, (1.3) where U (x) is a real polynomial of x with degree n.In this case, finding the number of zeros of I(h) is one of the ten problems in [2].When n = 3, this problem was investigated by many authors (e.g.[7], [8], [10], [11]).When n = 4, some results were given by [5], [12], [13], [16], [17], but this case is far from complete solving.In this paper, we study the case n = 4 and the Hamiltonian vector field dH = 0 possesses one center and one homoclinic saddle-loop, which has the following normal form where k > 2.
The system (1.4) has the first integral and the phase portrait is shown in Figure 1 } corresponds the saddle point (0, 0) and homoclinic loop.The critical point (k, 0) is a saddle.
x i y dx, i = 0, 1, 2, (1.6) I(h) = αI 0 (h) + βI 1 (h) + γI 2 (h), (1.7) where the ovals Γ h , h ∈ ( −2k+1 12 , 0), has negative (clockwise) orientation coinciding with the orientation of the vector field (1.4), α, β and γ are arbitrary constants.The central result of this paper is the following theorem: Theorem 1.1.The maximum number of limit cycles of the , either I(h) vanishes identically or its lowest upper bound of the number of zeros is equal to two, which is partial answer to the seventh problem in [2].
The paper is organized as follows: In section 2, Picard-Fuchs equation satisfied by I 0 (h), I 1 (h) and I 2 (h) is derived and the expansions of I(h) near its endpoints are given, the latter results reveal the connection between the Abelian integrals I(h) and the limit cycles of system (1.8) which tend to the center (1, 0) or homoclinic loop of (1.4) as → 0. In section 3, instead of estimating the number of zeros of I(h), we will prove that I (h) has at most two zeros, i.e., I(h) has at most two inflection points in ( −2k+1 12 , 0), which implies the lowest upper bound of the number of zeros of I(h) does not exceed three in the same interval.Using the fact ω(h) = I 1 (h) 12 , 0)} can be defined.It is readily seen that the intersection points of line α + βω + γν = 0 with Ω in ων-plane correspond the zeros of I (h), which shows that the convexity of Ω determinates the number of the zeros of I (h).
In section 4, we make precise connection between the intersection points of L : α + βP + γQ = 0 with the centroid curve Ω = {(P, I1(h) } on one hand and the zeros of Abelian integral I(h) on the other hand.Finally, the main results of this paper are proved in section 5. Some techniques in section 4 and section 5 are borrowed from [4].
Remark.Unfortunately, the techniques we use in this present paper do not fit for the case of 2 < k < 11+ √ 33 4 . Therefore, throughout this paper, we suppose k > 11+ √ 33 4 > 4 unless the opposite is claimed.Some computation in this paper is done by the computer program "Mathematica".

Picard-Fuchs equation and the asymptotic expansions of I(h) near its endpoints
In this section we shall derive Picard-Fuchs equation satisfied by I i (h) and describe the behaviours of I(h) near h = 0 and h = −2k+1 12 .
Repeating the same arguments, we obtain the third equation.The lemma has been proved.

iv)
Proof: The results i) and ii) follows from Green's formula.
Proof: (i) It follows from Theorem C of [14].
Proof: Differentiating both sides of (2.1) yields where Eliminating I 0 from the first two equations of (3.3), we get (3.2).
Lemma 3.3.The integral I 0 , I 1 satisfy the following equation where
2) The multiplicity of zero of I(h) is at most three.If h = h 0 is the zero of multiplicity 3, then h = h 0 is an unique zero of I(h).
3) If h = h 0 is the zero of multiplicity two of I(h), then another zero h = h 1 (if there exists) must be simple.
Obviously, h = h 0 satisfies I(h 0 ) = I (h 0 ) = 0, I (h 0 ) = 0. Without loss of generality, suppose I (h 0 ) > 0, i.e., h = h 0 is minimal point of I(h).Suppose h 1 > h 0 .Then there must exist two inflection points between −2k+1 12 and h 1 .Hence, it follows from step 1) that h = h 1 must be simple zero of I(h).In the case of h 1 < h 0 , we can get the result by the same arguments as above.
Summing up above discussion, we get the theorem.

The geometry of the centriod curve
Definition 4.1.In P Q-plane, the curve is called centroid curve.
It has been proved in Corollary 3.7 that P (h) < 0. Therefore, P can be taken as a new parameter and denote Ω as where h(P ) is the inverse function of P = P (h).
The importance of concept of the centroid curve lies in the fact that its geometry contains the complete information of I(h) although the definition of Ω depends only on H(x, y) = h.
Using same arguments as [4], we have Theorem 4.2.
Proof: (i) Part i) of the statement follows from (2.19).
(iii) The condition I (h 0 ) = 0 when I(h 0 ) = I (h 0 ) = 0 is equivalent to This and (4.3) imply the result.ii) The coefficient c 0 is zero if and only if L passes through (P (0), Q(0)).
iii) The coefficient c 0 = c 1 = 0 is equivalent to L = L s , where c 0 , c 1 is defined as (2.31).

Proof of main theorem
Theorem 4.2-4.4reduce the proof of Theorem 1.1 to showing that each line L : α + βP + γQ = 0 intersects the centriod curve Ω in at most two points, which implies Ω is a strictly concave curve.
As a sequence of Theorem 3.9 and Theorem 4.2, the following assertion holds: Lemma 5.1.If the line L does not pass through (1, 1) or (P (0), Q(0)), then L intersects Ω in at most three points (counted with their multiplicities).

Lemma 5.2. Each line L intersects the centriod curve Ω in at most two points (counted with their multiplicities).
Proof: We split the proof in several steps.1) Each line L, passing through (1, 1) or (P (0), Q(0)), intersects Ω in at most two points (counted with their multiplicities).
For L = L c , L s or L cs , we have proved the conclusion in Proposition 4.5 and Proposition 4.6.Suppose now that L is a line through Proof of Theorem 1.1:For a given perturbation (1.8) , if β = γ = 0, then either the divergenve in (1.8) vanishes identically or it is nowhere zero.In the first case, (1.8) is a Hamiltonian system and in second one no limit cycle can appear in (1.8) .Suppose |β| + |γ| = 0, which means that the line L is defined.By Lemma 5.2, Theorem 2.3, Theorem 2.6 and Theorem 4.2-4.4,the theorem follows.

dP h=0 = dQ dh dh dP h=0 = lim h→0 I 2 ICorollary 4 . 7 .
0 − I 0 I 2 I 1 I 0 − I 0 I 1 the equation of L s is (4.5).The analysis we have done shows that The centriod curve Ω is entirely placed in the triangle formed by L s , L c and L cs , seeFigure 4.1.