The Number of Limit Cycles of a Polynomial System on the Plane

We perturb the vector field ̇𝑥 = −𝑦𝐶(𝑥,𝑦), ̇𝑦 = 𝑥𝐶(𝑥,𝑦) with a polynomial perturbation of degree 𝑛 , where 𝐶(𝑥,𝑦) = (1 − 𝑦 2 ) 𝑚 , and study the number of limit cycles bifurcating from the period annulus surrounding the origin.


Introduction and Main Result
The main task in the qualitative theory of real plane differential systems is to determine the number of limit cycles, which is related to Hilbert's 16th problem as well as weakened Hilbert's 16th problem, posed by Arnold in [1]. Consider a planar system of the forṁ = − ( , ) + ( , ) , where , , and are real polynomials, (0, 0) ̸ = 0, and is a small real parameter. It is well known that the number of zeros of the Abelian integral where = {( , ) : 2 + 2 = 2 }, controls the number of limit cycles of (1) that bifurcate from the periodic orbits of the unperturbed system (1) with = 0; see [2].
The problem of finding lower and upper bounds for the number of zeros of ( ), when and are arbitrary polynomials of a given degree, say , and is a particular polynomial, has been faced in several recent papers. In the case of perturbing the linear center by arbitrary polynomials and of degree , that is, considerinġ= − + ( , ),̇= + ( , ), there are at most [( −1)/2] limit cycles up to first order in , see [3], where [⋅] denotes the integer part function.

Preliminary Results
To study the property of ( ), we need to make some preliminaries. First we introduce a function of the form where ∈ [0, 1) and is an integer. This section contains some preliminary computations to express the Abelian integral ( ) given in (3) in terms of polynomials.

Lemma 2. Let ( ) be a polynomial of degree in . Then for
∈ [0, 1) and ≥ 0 it holds that for some ∈ .
Proof. Note that, for ≥ 0, The result follows by applying the above formula to each term of Lemma 3. Let be the functions introduced in (4). Then, for 1 ̸ = ∈ , Proof. Note that Using integration by parts in the last integral, we obtain Abstract and Applied Analysis 3 Note that Substituting the formula above into (9), we find Replacing by − 1, we can obtain the conclusion. (4) satisfy

Lemma 4. The functions in
Moreover, where denotes a polynomial of (exact) degree .
Proof. It is easy to check that equality (14) is true for = 0, 1. Now we prove that it is true for any ≥ 2 by induction. Suppose that it is true for and + 1; that is, We need to prove that By Lemma 3, we have Hence, Then it follows from assumption (16) that By Lemma 3 again we have Substituting it into (20) we obtain which gives (17). Hence equality (14) holds for ≥ 0.
The first formula in (15) for the case of ≤ 0 follows directly from (4). The second one follows from the first one together with (14). This completes the proof. Some explicit expressions of ( ) are (23)

Lemma 5. For any nonnegative integer numbers , , and , one has
Proof. Since the integrand is an odd function of , (24) follows. Further Proof. In polar coordinates, = cos and = sin , the integral ( ) writes as ) .
Hence by (30) and Lemma 2, This ends the proof.
where * where * denotes a polynomial of degree .