The number of zeros of Abelian integrals for a perturbation of a hyperelliptic Hamiltonian system with a nilpotent center and a cuspidal loop

In this paper we consider the number of isolated zeros of Abelian integrals associated to the perturbed system ẋ = y, ẏ = −x3(x− 1)2 + ε(α + βx + γx3)y, where ε > 0 is small and α, β, γ ∈ R. The unperturbed system has a cuspidal loop and a nilpotent center. It is proved that three is the upper bound for the number of isolated zeros of Abelian integrals, and there exists some α, β and γ such that the Abelian integrals could have three zeros which means three limit cycles could bifurcate from the nilpotent center and period annulus. The proof is based on a Chebyshev criterion for Abelian integrals, asymptotic behaviors of Abelian integrals and some techniques from polynomial algebra.


Introduction
Consider the following polynomial Liénard equations of type (m, n), i.e. ẋ = y, ẏ = −g(x) − ε f (x)y, where ε > 0 is small, f (x) and g(x) are polynomials of degree m and n, respectively.For ε = 0 the above system reduces to ẋ = y, ẏ = −g(x), which is a Hamiltonian system with the Hamiltonian function where G is a polynomial in x of degree n + 1.The level curves are rational for n = 0, 1, elliptic for n = 2, 3 and hyper-elliptic for n ≥ 4. Suppose the unperturbed system (1.2) has a family of Figure 1.1:The level curves of H(x, y) = h.
periodic orbits Γ h defined by H(x, y) = h.Then for the associated perturbed system (1.1) there exists an Abelian integral or so-called first-order Melnikov function of the following form: According to the Poincaré-Pontryagin-Andronov theorem, it is known that the total number of isolated zeros (counting their multiplicities) of the Abelian integral I(h, δ) is an upper bound for the number of limit cycles bifurcated from the periodic annulus of the unperturbed system (1.2).The second part of Hilbert's 16th problem for system (1.1), asks for an upper bound for the number of limit cycles in terms of m and n and their relative distributions.Some of the previous works that focus on hyper-elliptic case are as follows: Asheghi et al. in [1] studied the Chebyshev's property of a 3-dimensional vector space of Abelian integrals by integrating the 1-form (α 0 + α 1 x + α 2 x 2 )ydx over the compact level curves of a hyper-elliptic Hamiltonian of degree 7. Wang in [8] investigated the number of different phase portraits of hyper-elliptic Hamiltonian system of degree five and obtained 40 different phase portraits.Kazemi et al. in [5] studied the zeros of Abelian integrals obtained by integrating 1-form (a + bx + cx 3 + x 4 )ydx over the compact level curves of (1.1)| ε=0 with g(x) = −x(x − 1 2 )(x − 1) 3 and proved that the upper bound of the number of isolated zeros of Abelian integral is three.Wang in [7] studied the zeros of Abelian integrals obtained by integrating 1-form (α + βx + γx 2 )ydx over the compact level curves of the hyper-elliptic Hamiltonian of degree five H(x, y) = y 2 2 + 1 4 x 4 − 1 5 x 5 and proved that the upper bound of the number of isolated zeros of Abelian integral is two.
A limit cycle is an isolated periodic orbit in the set of periodic orbits.The Abelian integral I(h, δ) is a suitable tool for studying limit cycles of system (1.4).We recall that a limit cycle of system (1.4) corresponds to an isolated zero of the Abelian integral I(h, δ).
The rest of the paper is presented in two sections.In Section 2, we show that the Abelian integral (1.5) has the Chebyshev property with accuracy one for h ∈ (0, 1 60 ).Hence, by the criterion introduced in [6] we get that the upper bound for the number of isolated zeros of I(h, δ) in any compact subinterval of (0, 1 60 ) is three.In Section 3, we calculate the asymptotic expansions of Abelian integral I(h, δ) near the nilpotent center and the cuspidal loop, using that we get that there exists some α, β, γ such that the Abelian integral I(h, δ) can have three isolated zeros in (0, 1  6 ), which means that the system (1.4) can have three limit cycles.

Bifurcation of limit cycles from the period annulus
In this section we study the maximum number of limit cycles which bifurcate from the period annulus of system (1.3) for h ∈ (0, 1 60 ).We use an algebraic criterion given in [6] to study the related Abelian integral I(h, δ) of system (1.4).But first we give the following definitions.
Definition 2.1.The base functions {I i (h, δ), i = 0, 2, . . ., n − 1} in the Abelian integral I(h, δ) is said to be a Chebyshev system with accuracy k, if the number of zeros of any nontrivial linear combination . Consider a Hamiltonian function with the following special form which is analytic in some open subset of the plane and has a local minimum at the origin.We fix that H(0, 0) = 0, then (0, 0) is the center critical point of the associated vector field.So, there exists a period annulus P foliated by the set of ovals Γ h ⊂ {H(x, y) = h} surrounding the origin.The period annulus can be parametrized by the energy levels h ∈ (0, h 0 ) for some h 0 ∈ (0, +∞].In the sequel, we denote the projection of P on the x-axis by (x , x r ).It is easy to verify that, under the above assumptions, xA (x) > 0 for any x ∈ (x , x r ) \ {0} and B(0) > 0. Thus by implicit function theorem, there exists a smooth unique analytic function z(x) with x < z(x) < 0 such that A(x) = A(z(x)) for 0 < x < x r .Theorem A in [6] is as follows.
The efficiency of Theorem 2.3 comes from the fact that finding an upper bound for the number of zeros of Abelian integrals I(h, δ) follows just from some pure algebraic computations.
In the sequel, we will apply Theorem 2.3 to show that the Abelian integral (1.5) has Chebyshev property with accuracy one in the interval (0, 1 60 ).Using the notation in Theorem 2.3 we have and s = 1, n = 3.The period annulus inside Γ 1 6 is foliated by the level curves Figure 2.1: The involution of x and z(x) defined by A(x) = A(z(x)).
whose images on the x-axis is an open interval (x l , 1), where with negative half x-axis, which satisfies It is easy to check that xA (x) > 0 for all x ∈ (x l , 1) \ {0}.Therefore, there exists an analytic function z(x) with x l < z(x) < 0 such that A(x) = A(z(x)) as 0 < x < 1, see Fig. 2.1.
To apply Theorem 2.3, we notice that I k (h) = Γ h x k ydx, hence m = 1, n = 3 and s = 1, so that the hypothesis (c) (s > m(n + k − 2)) of Theorem 2.3 is not fulfilled (note that, as we shall see, in our case k = 1).However it is possible to overcome this problem using the following result (see Lemma 4.1 in [2]), and obtain new Abelian integrals for which the corresponding s is large enough to verify the inequality.Lemma 2.4.Let γ h be an oval inside the level curve {A(x) + B(x)y 2 = h} and we consider a function F such that F/A is analytic at x = 0.Then, for any k ∈ N, Here we have to promote the power s to three such that the condition s > n − 1 holds.On the oval Γ h we have Now we apply Lemma 2.4 with k = 3 and F(x) = 2x i A(x) to the first integral above to get x−1) 3 , and By (2.2) we obtain Again we apply Lemma 2.4 with k = 5 and F(x) = 2(x i + G i (x))A(x) to the first integral above to get 6 , and gi = 1000 + 1100 i + 100 i 2 x i+6 − 6000 + 7000 i + 680 i 2 x i+5 From (2.3) we obtain where } is a Chebyshev system with accuracy one on (0, 1 60 ) if and only if {I 0 , I 1 , I 3 } is a Chebyshev system with accuracy one on the same interval.As s = 3, n = 3 and the condition s > n − 1 is satisfied, now, we can apply Theorem 2.3 to study the Chebyshev property of { Ĩ0 , Ĩ1 , Ĩ3 } in the interval (0, 1 60 ).For i = 0, 1, 3, let We know that for ) is equivalent to 1 60 (x − z)q(x, z) = 0, where q(x, z) = 10 x 5 − 24 x 4 + 10 zx 4 + 15 x 3 − 24 zx 3 + 10 z 2 x 3 + 15 zx 2 − 24 z 2 x 2 + 10 x 2 z 3 + 15 z 2 x − 24 xz 3 So, A(x) = A(z(x)) is equivalent to q(x, z(x)) = 0. We need to prove that {m 0 , m 1 , m 3 } satisfy hypothesis (i)-(iii) of Theorem 2.3 with k = 1.To do this we prove the following lemma.
Next we consider the resultant r 2 (x) between p 2 (x, z) and q(x, z) with respect to z, and obtain r 2 (x) = x 16 (x − 1) 28 k 2 (x), where k 1 (x) is a polynomial in x of degree 128.Applying Sturm's theorem, we get that k 2 (x) = 0 in (0, 1), which implies that W[m 0 , m 1 ](x) = 0 for x ∈ (0, 1).Finally, we compute the resultant r 3 (x) between p 3 (x, z) and q(x, z) to get where k 3 (x) is a polynomial in x of degree 198.By applying Sturm's theorem, we get that k 3 (x) has only one root in (0, 1).Using the algorithm of real root isolation to k 3 (x) and using function realroot with accuracy 1 1000 in Maple 15, we get that the root is located in the closed subinterval [ 57 64 , 913  1024 ].On the other hand, we get the resultant r 3 (z) between p 3 (x, z) and q(x, z) with respect to x as follows r 3 (z) = z 34 (z − 1) 44 k 3 (z), where k 3 (z) is a polynomial in z of degree 198.By applying Sturm's theorem, we get that k 3 (z) has only one root in (x l , 0).Again, using algorithm of real root isolation to k 3 (z) and using function realroot with accuracy 1 1000 in Maple 15, we can prove that the root is in the closed subinterval [− 223  512 , − 445 1024 ].Therefore, there exists a unique x * ∈ (0, 1), with 57  64 ≤ x * ≤ 913 1024 , so that W[m 0 , m 1 , m 3 ](x * ) = 0. We will now show that x * is simple root.Let us denote W[m 0 , m 1 , m 3 ](x) by W 0 (x, z(x)) and calculate its derivative, that is where p 4 (x, z) is a polynomial in (x, z).The resultant with respect to z between q(x, z) and where k 4 (x) is a polynomial of degree 220 in x.By applying Sturm's theorem, we find that k 4 (x) has no zeros in [ 57 64 , 913 1024 ].Therefore, W[m 0 , m 1 , m 3 ](x) has exactly one simple root in the interval (0, 1).This ends the proof.So far we have proved the following.
Theorem 2.6.The collection {I 0 (h), I 1 (h), I 3 (h)} is a Chebyshev system with accuracy one on the interval 0, 1 60 .Hence, if the Abelian integral I(h, δ) is not identical to zero, then for all values of parameters (α, β, γ) it has at most three zeros, counting multiplicities, in any compact subinterval of 0, 1 60 , and the number of limit cycles bifurcating from the periodic annulus is at most three.

Asymptotic expansions of Abelian integral I(h, δ)
In this section we study the asymptotic expansion of Abelian integral I(h, δ) at h = 0 and h = 1 6 , respectively.Using these asymptotic expansions we prove the following theorem.
Proof.i) We follow the idea given in [4] and [7] and adapt our notations according to [7].To obtain the asymptotic expansion of Abelian integral I(h, δ) as h → 0 + , we compute I(h, δ) near the nilpotent center (0, 0).Let us denote the intersection points of the oval Γ h with the negative and positive half x-axis by x l (h) and x r (h), respectively.We know that A (x) = Applying the implicit function theorem to ψ(x, u) = 0 at (x, u) = (0, 0), we know that there exists a unique analytic function x = ϕ(u) and a small positive number 0 < ρ 1 such that ψ(ϕ(u), u) = 0 for |u| < ρ.It can be checked that ϕ(u) has the following expression: where the first five coefficients are as follows and s, then we have Using the change of variable s 4 = τ, we obtain where β(a, b) is the following Beta-function for a > 0, b > 0, It is obvious that E k = 0 when k is odd.From the relation between Beta-function and Gammafunction, we compute the three elliptic integrals E k for k = 0, 2, 4 as follows where Γ(a) = +∞ 0 x a−1 e −x dx is the Gamma-function and Γ( 3 4 ) ≈ 1.225416702.Substituting (3.6) into (3.4)we get where When c 0 (δ) = c 2 (δ) = 0, we get a unique solution hence by Theorem 1.3 in [4], we get that, if γ = 0, then the Abelian integral (1.5) can have at least two zeros for some (α, β) near (α * , β * ).Therefore the perturbed system (1.4) could have at least two limit cycles near the origin, for ε sufficiently small.
For ε = 0 system (3.9) has the Hamiltonian This system has a cusp critical point at (0, 0) and a cuspidal loop Γ 0 .There are two families of periodic orbits of (3.9) near Γ 0 given by Γ ± h : H(x, y) = h, 0 < ±h 1.Then the two corresponding Abelian integrals are given by It was proved in [3] that  Thus, by Theorem 3.2 of [3] we know that there exists some (α, β) close to (ᾱ, β) for which the Abelian integral (1.5) could have 3 zeros near h = 1 6 .This means that the perturbed system (1.4) could have 3 limit cycles near cuspidal loop, for ε sufficiently small.
There are two different distributions of 3 limit cycles near cuspidal loop: (1, 2) and (2, 1), where (i, j) denotes that i limit cycles are outside the loop while j limit cycles inside the loop.