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

Abstract

We perturb the vector field 𝑥̇=−𝑦𝐶(𝑥,𝑦), 𝑦̇=𝑥𝐶(𝑥,𝑦) with a polynomial perturbation of degree 𝑛, where 𝐶(𝑥,𝑦)=(1−𝑦²)^𝑚, and study the number of limit cycles bifurcating from the period annulus surrounding the origin.

1. 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 form ̇,̇,𝑥=−𝑦𝐶𝑥,𝑦+𝜀𝑃𝑥,𝑦𝑦=𝑥𝐶𝑥,𝑦+𝜀𝑄𝑥,𝑦 (1)

where 𝑃, 𝑄, and 𝐶 are real polynomials, 𝐶(0,0)≠0, and 𝜀 is a small real parameter. It is well known that the number of zeros of the Abelian integral 𝑀𝜌=_𝛾𝜌𝑄𝑥,𝑦𝑑𝑥−𝑃𝑥,𝑦𝑑𝑦𝐶,𝑥,𝑦 (2)

where 𝛾_𝜌={(𝑥,𝑦)∶𝑥²+𝑦²=𝜌²}, controls the number of limit cycles of (1) that bifurcate from the periodic orbits of the unperturbed system (1) with 𝜀=0; see [2].

The problem of finding lower and upper bounds for the number of zeros of 𝑀(𝜌), when 𝑃 and 𝑄 are arbitrary polynomials of a given degree, say 𝑛, and 𝐶 is a particular polynomial, has been faced in several recent papers. In the case of perturbing the linear center by arbitrary polynomials 𝑃 and 𝑄 of degree 𝑛, that is, considering 𝑥̇=−𝑦+𝜀𝑝(𝑥,𝑦), 𝑦̇=𝑥+𝜀𝑞(𝑥,𝑦), there are at most [(𝑛−1)/2] limit cycles up to first order in 𝜀, see [3], where [⋅] denotes the integer part function. Also it is known that perturbing the quadratic center 𝑥̇=−𝑦(1+𝑥), 𝑦̇=𝑥(1+𝑥) inside the polynomial systems of degree 𝑛 we can obtain at most 𝑛 limit cycles up to first order in 𝜀 (see [4]). The authors of [5] studied the perturbation of the cubic center 𝑥̇=−𝑦(1+𝑥)(2+𝑥), 𝑦̇=𝑥(1+𝑥)(2+𝑥) inside the polynomial differential systems of degree 𝑛, and they obtained that 2𝑛+2−(−1)^𝑛 is an upped bound for the number of limit cycles up to first order in 𝜀. In [6], the authors studied the perturbations of 𝑥̇=−𝑦(𝑎+𝑥)(𝑏+𝑦), 𝑦̇=𝑥(𝑎+𝑥)(𝑏+𝑦), and they obtained that 3[(𝑛−1)/2]+4 if 𝑎≠𝑏 and, respectively, 2[(𝑛−1)/2]+2 if 𝑎=𝑏, up to first order in 𝜀, are upper bounds for the number of the limit cycles. In [7] the authors studied the maximum number 𝜎 of limit cycles which can bifurcate from the periodic orbits of the quartic center 𝑥̇=−𝑦𝑓(𝑥,𝑦), 𝑦̇=𝑥𝑓(𝑥,𝑦) with 𝑓(𝑥,𝑦)=(𝑥+𝑎)(𝑦+𝑏)(𝑥+𝑐) and 𝑎𝑏𝑐≠0 by perturbing it inside the class of polynomial vector fields of degree 𝑛. They proved that 4[(𝑛−1)/2]+4≤𝜎≤5[(𝑛−1)/2]+14. In [8] the authors studied the bifurcation of limit cycles of the system 𝑥̇=−𝑦(𝑥²−𝑎²)(𝑦²−𝑏²)+𝜀𝑃(𝑥,𝑦), 𝑦̇=𝑥(𝑥²−𝑎²)(𝑦²−𝑏²)+𝜀𝑄(𝑥,𝑦) for 𝜀 sufficiently small, where 𝑎,𝑏∈𝑅−0 and 𝑃, 𝑄 are polynomials of degree 𝑛. They obtained that up to first order in 𝜀 the upper bound for the number of limit cycles that bifurcate from the period annulus of the quintic center given by 𝜀=0 is (3/2)(𝑛+sin²(𝑛𝜋/2))+1 if 𝑎≠𝑏 or 𝑛−1 if 𝑎=𝑏. More results can be found in [9, 10].

But few of these algebraic curves have a multiple factor. In [11], the authors took 𝐶(𝑥,𝑦)=(1−𝑦)^𝑚 and proved that an upper bound for the number of zeros of 𝑀(𝜌) on (0,1) is 𝑛+𝑚−1 and that this bound is reached when 𝑚=1. The approach of [11] is mainly based on the explicit computation of 𝑀(𝜌). In [12], the authors obtained the maximum of zeros of 𝑀(𝜌), taking into account their multiplicities, is [(𝑚+𝑛)/2]−1 when 𝑛<𝑚−1 and 𝑛 when 𝑛≥𝑚−1. In [12], the authors improve the upper bound given in [11] and provide the optimal upper bound for the zeros of 𝑀(𝜌). In this paper, motivated by [12] we take 𝐶(𝑥,𝑦)=(1−𝑦²)^𝑚 and obtain the following theorem.

Theorem 1. Consider (1) with 𝐶(𝑥,𝑦)=(1−𝑦²)^𝑚. Let 𝑀𝜌=_𝛾𝜌𝑄𝑥,𝑦𝑑𝑥−𝑃𝑥,𝑦𝑑𝑦1−𝑦^2𝑚, (3)
where 𝛾_𝜌={(𝑥,𝑦)∶𝑥²+𝑦²=𝜌²}, 𝜌∈(0,1), and 𝑃 and 𝑄 are polynomials of degree 𝑛. Then the maximum number of zeros of 𝑀(𝜌), taking into account their multiplicities, is 𝑚+[(𝑛−1)/2]−1 when 𝑛<2𝑚−1 and 2[(𝑛+1)/2]−1 when 𝑛≥2𝑚−1. Moreover, when 𝑚=1,2 and 𝑛=2, the corresponding maximum number, which is 1, can be reached by taking suitable 𝑃 and 𝑄.

2. Preliminary Results

To study the property of 𝑀(𝜌), we need to make some preliminaries. First we introduce a function of the form 𝐼_𝑗𝜌=₀^2𝜋11−𝜌²sin²𝜃^𝑗𝑑𝜃, (4)

where 𝜌∈[0,1) and 𝑗 is an integer.

This section contains some preliminary computations to express the Abelian integral 𝑀(𝜌) given in (3) in terms of polynomials.

Lemma 2. Let 𝑅_𝑙(𝑥) be a polynomial of degree 𝑙 in 𝑥. Then for 𝜌∈[0,1) and 𝑚≥0 it holds that ₀^2𝜋𝑅_𝑙𝜌²sin²𝜃1−𝜌²sin²𝜃^𝑚𝑑𝜃=^𝑙_𝑗=0𝛼_𝑚−𝑗𝐼_𝑚−𝑗𝜌, (5)
for some 𝛼_𝑗∈𝑅.

Proof. Note that, for 𝑗≥0, ₀^2𝜋𝜌^2𝑗sin^2𝑗𝜃1−𝜌²sin²𝜃^𝑚=𝑑𝜃_02𝜋1−1−𝜌²sin²𝜃^𝑗1−𝜌²sin²𝜃^𝑚=𝑑𝜃_𝑗𝑘=0𝑗𝑘(−1)^𝑘₀^2𝜋1−𝜌²sin²𝜃^𝑘1−𝜌²sin²𝜃^𝑚=𝑑𝜃_𝑗𝑘=0𝑗𝑘(−1)^𝑘𝐼_𝑚−𝑘𝜌. (6)
The result follows by applying the above formula to each term of 𝑅_𝑙𝜌²sin²𝜃=^𝑙_𝑗=0𝑟_𝑗𝜌²sin²𝜃^𝑗. (7)

Lemma 3. Let 𝐼_𝑗 be the functions introduced in (4). Then, for 1≠𝑗∈𝑍, 𝜌²−1𝐼_𝑗𝜌=3−2𝑗2𝑗−22−𝜌²𝐼_𝑗−1𝜌+𝑗−2𝐼𝑗−1^𝑗−2𝜌. (8)

Proof. Note that 𝐼_𝑗−1𝜌=₀^2𝜋1−𝜌²sin²𝜃1−𝜌²sin²𝜃^𝑗𝑑𝜃=𝐼_𝑗𝜌−𝜌²₀^2𝜋sin²𝜃1−𝜌²sin²𝜃^𝑗𝑑𝜃. (9)
Using integration by parts in the last integral, we obtain 𝜌²₀^2𝜋sin²𝜃1−𝜌²sin²𝜃^𝑗𝑑𝜃=𝜌²₀^2𝜋sin𝜃1−𝜌²sin²𝜃^𝑗𝑑()=−cos𝜃−𝜌²sin𝜃cos𝜃1−𝜌²sin²𝜃^𝑗₀^2𝜋+₀^2𝜋𝜌²cos²𝜃1−𝜌²sin²𝜃^𝑗+2𝑗𝜌⁴sin²𝜃cos²𝜃1−𝜌²sin²𝜃^𝑗−1×1−𝜌²sin²𝜃^2𝑗−1=𝑑𝜃_02𝜋𝜌²cos²𝜃1−𝜌²sin²𝜃^𝑗𝑑𝜃+2𝑗₀^2𝜋𝜌⁴sin²𝜃cos²𝜃1−𝜌²sin²𝜃^𝑗+1𝑑𝜃=𝜌²𝐼_𝑗+₀^2𝜋1−𝜌²sin²𝜃−11−𝜌²sin²𝜃^𝑗𝑑𝜃+2𝑗₀^2𝜋𝜌⁴sin²𝜃1−sin²𝜃1−𝜌²sin²𝜃^𝑗+1𝑑𝜃. (10)
Note that ₀^2𝜋𝜌⁴sin²𝜃1−sin²𝜃1−𝜌²sin²𝜃^𝑗+1𝑑𝜃=𝜌²₀^2𝜋𝜌²sin²𝜃1−𝜌²sin²𝜃^𝑗+1−𝑑𝜃_02𝜋𝜌⁴sin⁴𝜃1−𝜌²sin²𝜃^𝑗+1𝑑𝜃=𝜌²₀^2𝜋1−1−𝜌²sin²𝜃1−𝜌²sin²𝜃^𝑗+1−𝑑𝜃_02𝜋1−1−𝜌²sin²𝜃²1−𝜌²sin²𝜃^𝑗+1𝑑𝜃=𝜌²𝐼_𝑗+1−𝐼_𝑗−𝐼_𝑗+1−2𝐼_𝑗+𝐼_𝑗−1. (11)
Then 𝜌²₀^2𝜋sin²𝜃1−𝜌²sin²𝜃^𝑗𝑑𝜃=𝜌²𝐼_𝑗+𝐼_𝑗−1−𝐼_𝑗+2𝑗𝜌²𝐼_𝑗+1−𝐼_𝑗−𝐼_𝑗+1+2𝐼_𝑗−𝐼_𝑗−1. (12)
Substituting the formula above into (9), we find 𝐼_𝑗−1𝜌=1−2𝑗2−𝜌²𝐼_𝑗𝜌+2𝑗−1𝐼_𝑗−1𝜌+2𝑗1−𝜌²𝐼_𝑗+1𝜌. (13)
Replacing 𝑗 by 𝑗−1, we can obtain the conclusion.

Lemma 4. The functions in (4) satisfy 𝐼_𝑗𝜌=11−𝜌^2𝑗−1/2𝐼_1−𝑗𝜌,𝑗∈𝑍. (14)
Moreover, 𝐼_𝑗𝜌=𝑅_(−𝑗)𝜌²𝐼,𝑗≤0,_𝑗𝜌=𝑅_𝑗−1𝜌²1−𝜌^2𝑗−1/2,𝑗≥1, (15)
where 𝑅_𝑙 denotes a polynomial of (exact) degree 𝑙.

Proof. It is easy to check that equality (14) is true for 𝑗=0,1. Now we prove that it is true for any 𝑗≥2 by induction.
Suppose that it is true for 𝑗 and 𝑗+1; that is, 1−𝜌^2𝑗−1/2𝐼_𝑗𝜌=𝐼_1−𝑗𝜌,1−𝜌^2𝑗+1/2𝐼_𝑗+1𝜌=𝐼_−𝑗𝜌. (16)
We need to prove that 1−𝜌^2𝑗+3/2𝐼_𝑗+2𝜌=𝐼_−(𝑗+1)𝜌. (17)
By Lemma 3, we have 1−𝜌²𝐼_𝑗+2𝜌=−2𝑗+1𝜌2𝑗+2²−2𝐼_𝑗+1𝜌−𝑗𝐼𝑗+1^𝑗𝜌. (18)
Hence, 1−𝜌^2𝑗+3/2𝐼_𝑗+2𝜌=−2𝑗+1𝜌2𝑗+2²−21−𝜌^2𝑗+1/2𝐼_𝑗+1𝜌−𝑗𝑗+11−𝜌^2𝑗+1/2𝐼_𝑗𝜌. (19)
Then it follows from assumption (16) that 1−𝜌^2𝑗+3/2𝐼_𝑗+2𝜌=−2𝑗+1𝜌2𝑗+2²−2𝐼_−𝑗𝜌−𝑗𝑗+11−𝜌²𝐼_1−𝑗𝜌. (20)
By Lemma 3 again we have 1−𝜌²𝐼_1−𝑗𝜌=−2𝑗+1𝜌2𝑗²−2𝐼_−𝑗𝜌−𝑗+1𝑗𝐼_−(𝑗+1)𝜌. (21)
Substituting it into (20) we obtain 1−𝜌^2𝑗+3/2𝐼_𝑗+2𝜌=−2𝑗+1𝜌2𝑗+2²−2𝐼_−𝑗𝜌−𝑗−𝑗+12𝑗+1𝜌2𝑗²−2𝐼_−𝑗𝜌−𝑗+1𝑗𝐼_−(𝑗+1)𝜌=𝐼_−(𝑗+1)𝜌, (22)
which gives (17). Hence equality (14) holds for 𝑗≥0.
If 𝑗≤−1, then 𝑗̃=1−𝑗≥2. By applying (14) for 𝑗̃≥2, it is easy to see that (14) holds for all 𝑗≤−1.
The first formula in (15) for the case of 𝑗≤0 follows directly from (4). The second one follows from the first one together with (14). This completes the proof.

Some explicit expressions of 𝐼_𝑗(𝜌) are 𝐼₁𝜌=2𝜋1−𝜌^21/2,𝐼₂𝜌=2−𝜌²𝜋1−𝜌^23/2,𝐼₃𝜌=3𝜌⁴−8𝜌²+8𝜋41−𝜌^25/2,𝐼₀𝜌𝐼=2𝜋,₋₁𝜌=2−𝜌²𝐼𝜋,₋₂𝜌=3𝜌⁴−8𝜌²+8𝜋4. (23)

Lemma 5. For any nonnegative integer numbers 𝑝, 𝑞, and 𝑚, one has ₀^2𝜋sin^2𝑝+1𝜃cos^𝑞𝜃1−𝜌²sin²𝜃^𝑚𝑑𝜃=0,_02𝜋sin^𝑝𝜃cos^2𝑞+1𝜃1−𝜌²sin²𝜃^𝑚𝑑𝜃=0. (24)(25)

Proof. Since the integrand is an odd function of 𝜃, (24) follows. Further ₀^2𝜋sin^𝑝𝜃cos^2𝑞+1𝜃1−𝜌²sin²𝜃^𝑗𝑑𝜃=₀^2𝜋sin^𝑝𝜃1−sin²𝜃^𝑞1−𝜌²sin²𝜃^𝑗𝑑sin𝜃=0. (26)
Then (25) follows, and the proof is ended.

Lemma 6. Let 𝑀(𝜌) be the Abelian integral given in (3). Then, there exist polynomials 𝑅_𝑙 of degree 𝑙, 𝑙=0,1,…,[(𝑛+1)/2], such that, for 𝜌∈[0,1), 𝑀𝜌=^{[(𝑛+1)/2]+1}_𝑗=1𝑅_{[(𝑛+1)/2]+1−𝑗}𝜌²𝐼_{𝑚+1−𝑗}𝜌. (27)

Proof. In polar coordinates, 𝑥=𝜌cos𝜃 and 𝑦=𝜌sin𝜃, the integral 𝑀(𝜌) writes as 𝑀𝜌=−₀^2𝜋𝑄×𝜌cos𝜃,𝜌sin𝜃𝜌sin𝜃+𝑃𝜌cos𝜃,𝜌sin𝜃𝜌cos𝜃1−𝜌²sin²𝜃^𝑚−1𝑑𝜃. (28)
Note that ₀^2𝜋sin^𝑝𝜃cos^2𝑞𝜃1−𝜌²sin²𝜃^𝑚𝑑𝜃=₀^2𝜋sin^𝑝𝜃1−sin²𝜃^𝑞1−𝜌²sin²𝜃^𝑚𝑑𝜃. (29)
Then by Lemma 5, we have 𝑀𝜌=^𝑛_𝑗=1𝑅_𝑗(cos𝜃,sin𝜃)𝜌^𝑗1−𝜌²sin²𝜃^𝑚=𝑑𝜃_{[(𝑛+1)/2]𝑗=1}𝑆_𝑗sin²𝜃𝜌^2𝑗1−𝜌²sin²𝜃^𝑚𝑑𝜃, (30)
where 𝑅_𝑗(𝑥,𝑦) denotes a homogeneous polynomial of degree 𝑗 in (𝑥,𝑦), and 𝑆_𝑗(𝑥) denotes a polynomial of degree 𝑗 in 𝑥 having the form 𝑆_𝑗sin²𝜃=^𝑗_𝑙=0𝑠_2𝑗,2𝑙sin²𝜃^𝑙. (31)
By the above formula, we get ^[(𝑛+1)/2]_𝑗=1𝑆_𝑗sin²𝜃𝜌^2𝑗=𝑠_2,2sin²𝜃𝜌²+𝑠_4,4sin⁴𝜃𝜌⁴+⋯+𝑠_{2[(𝑛+1)/2],2[(𝑛+1)/2]}sin^{2[(𝑛+1)/2]}𝜃𝜌^{2[(𝑛+1)/2]}+𝜌²𝑠_2,0+𝑠_4,2sin²𝜃𝜌²+⋯+𝑠_{2[(𝑛+1)/2],2[(𝑛+1)/2]−2}×sin^{2[(𝑛+1)/2]−2}𝜃𝜌^{2[(𝑛+1)/2]−2}+⋯+𝜌^{2[(𝑛+1)/2]−2}𝑠_{2[(𝑛+1)/2]−2,0}+𝑠_{2[(𝑛+1)/2],2}sin²𝜃𝜌²+𝜌^{2[(𝑛+1)/2]}𝑠_{2[(𝑛+1)/2],0}=𝑇_[(𝑛+1)/2]𝜌²sin²𝜃+𝜌²𝑇_{[(𝑛+1)/2]−1}𝜌²sin²𝜃+𝜌⁴𝑇_{[(𝑛+1)/2]−2}𝜌²sin²𝜃+⋯+𝜌^{2[(𝑛+1)/2]}𝑇₀𝜌²sin²𝜃=^[(𝑛+1)/2]_𝑗=0𝜌^2𝑗𝑇_{[(𝑛+1)/2]−𝑗}𝜌²sin²𝜃, (32)
where 𝑇_𝑖 is a polynomial of degree 𝑖 in 𝜌²sin²𝜃, 𝑖=0,1,…,[(𝑛+1)/2].
Hence by (30) and Lemma 2, 𝑀𝜌=^[(𝑛+1)/2]_𝑗=1𝑆_𝑗sin²𝜃𝜌^2𝑗1−𝜌²sin²𝜃^𝑚=𝑑𝜃_{[(𝑛+1)/2]𝑗=0}𝜌^2𝑗₀^2𝜋𝑇_{[(𝑛+1)/2]−𝑗}𝜌²sin²𝜃1−𝜌²sin²𝜃^𝑚=𝑑𝜃_{[(𝑛+1)/2]𝑗=0}𝜌^2𝑗𝛼_𝑚𝐼_𝑚𝜌+𝛼_𝑚−1𝐼_𝑚−1𝜌+⋯+𝛼_{𝑚−[(𝑛+1)/2]+𝑗}𝐼_{𝑚−[(𝑛+1)/2]+𝑗}𝜌=^{[(𝑛+1)/2]+1}_𝑗=1𝑅_{([(𝑛+1)/2]+1−𝑗)}𝜌²𝐼_{𝑚+1−𝑗}𝜌. (33)
This ends the proof.

Assume 𝛿=1−𝜌², 𝜌∈[0,1). By Lemma 4, we have 𝐼^∗_𝑗(𝛿)=𝑅^∗_−𝑗𝛿²𝐼,𝑗≤0,_∗𝑗(𝛿)=𝑅^∗_𝑗−1𝛿²𝛿^2𝑗−1,𝑗≥1, (34)

where 𝑅^∗_−𝑗𝛿²=𝑅_−𝑗1−𝛿²,𝐼^∗_𝑗(𝛿)=𝐼_𝑗1−𝛿², (35)

where 𝑅^∗_𝑙 denotes a polynomial of degree 𝑙.

Lemma 7. Let 𝑀(𝜌) be the Abelian integral given in (3). Then for 𝛿∈(0,1]𝑀𝜌=𝑅^∗_{)]𝑚+[(𝑛−1/2}𝛿²𝛿^2𝑚−1𝑀𝜌=𝑅,𝑛<2𝑚−1,_{∗2)][(𝑛+1/2}(𝛿)𝛿(),𝑀𝜌=𝑅,𝑚=1𝑛≥2𝑚−1_{∗)]𝑚+[(𝑛+1/2−1}𝛿²𝛿^2𝑚−1+𝑅^∗_{)][(𝑛+1/2−𝑚}𝛿²,𝑚≠1(𝑛≥2𝑚−1). (36)

Proof. We have the following cases.
Case  1 (𝑚+1−([(𝑛+1)/2]+1)>0). In this case we have [(𝑛+1)/2]<𝑚, or 𝑛<2𝑚−1.
By Lemmas 4 and 6, we have 𝑀𝜌=𝑅_[(𝑛+1)/2]𝜌²𝐼_𝑚𝜌+𝑅_{[(𝑛+1)/2]−1}𝜌²𝐼_𝑚−1𝜌+⋯+𝑅₀𝜌²𝐼_{𝑚−[(𝑛+1)/2]}𝜌. (37)
Let 𝑀(𝜌)=𝑀^∗(𝛿). Then 𝑀^∗(𝛿)=𝑅^∗_)][(𝑛+1/2𝛿²𝐼^∗_𝑚(𝛿)+𝑅^∗_{)][(𝑛+1/2−1}𝛿²𝐼^∗_𝑚−1(𝛿)+⋯+𝑅^∗₀𝛿²𝐼^∗_{)]𝑚−[(𝑛+1/2}(𝛿)=𝑅^∗_)][(𝑛+1/2𝛿²𝑅^∗_𝑚−1𝛿²𝛿^2𝑚−1+𝑅^∗_{)][(𝑛+1/2−1}𝛿²𝑅^∗_𝑚−2𝛿²𝛿^2𝑚−3+⋯+𝑅^∗₀𝛿²𝑅^∗_{)]𝑚−[(𝑛+1/2−1}𝛿²𝛿^{2𝑚−2[(𝑛+1)/2]−1}=𝑅^∗_{()𝑚+[𝑛−1/2]}𝛿²𝛿^2𝑚−1, (38)
where 𝑅^∗_{)][(𝑛+1/2+𝑚−1}𝛿²=𝑅^∗_)][(𝑛+1/2𝛿²𝑅^∗_𝑚−1𝛿²+𝛿²𝑅^∗_{)][(𝑛+1/2−1}𝛿²𝑅^∗_𝑚−2𝛿²+⋯+𝛿^{(2[(𝑛+1)/2])}𝑅^∗₀𝛿²𝑅^∗_{)]𝑚−[(𝑛+1/2−1}𝛿², (39)
which is a polynomial of degree [(𝑛+1)/2]+𝑚−1   (=[(𝑛−1)/2]+𝑚) in 𝛿².
Hence, 𝑀𝜌=𝑀^∗(𝛿)=𝑅^∗_{)][(𝑛−1/2+𝑚}𝛿²𝛿^2𝑚−1. (40)
Case  2 (𝑚+1−([(𝑛+1)/2]+1)≤0). In this case we have [(𝑛+1)/2]≥𝑚 or 𝑛≥2𝑚−1.

(i) When 𝑚=1, we have 𝑀𝜌=^{[(𝑛+1)/2]+1}_𝑗=1𝑅_{[(𝑛+1)/2]+1−𝑗}𝜌²𝐼_2−𝑗𝜌=𝑅_{[(𝑛+1)/2])}𝜌²𝐼₁𝜌+𝑅_{[(𝑛+1)/2]−1}𝜌²𝐼₀𝜌+𝑅_{[(𝑛+1)/2]−2}𝜌²𝐼₋₁𝜌+⋯+𝑅₀𝜌²𝐼_{1−[(𝑛+1)/2]}𝜌. (41)

Thus, 𝑀^∗(𝛿)=𝑅^∗_)][(𝑛+1/2𝛿²𝑅^∗₀𝛿²𝛿+𝑅^∗_{)][(𝑛+1/2−1}𝛿²⋅2𝜋+𝑅^∗_{)][(𝑛+1/2−2}𝛿²𝑅^∗₁𝛿²+⋯+𝑅^∗₀𝛿²𝑅^∗_{)][(𝑛+1/2−1}𝛿²=𝑅^∗_{()2[𝑛+1/2]}(𝛿)𝛿, (42)

where 𝑅^∗_{2)][(𝑛+1/2}(𝛿)=𝑅^∗_)][(𝑛+1/2𝛿²𝑅^∗₀𝛿²+2𝜋𝛿𝑅^∗_{)][(𝑛+1/2−1}𝛿²+𝛿𝑅^∗_{)][(𝑛+1/2−2}𝛿²𝑅^∗₁𝛿²+⋯+𝛿𝑅^∗₀𝛿²𝑅^∗_{)][(𝑛+1/2−1}𝛿². (43)

Then 𝑀^∗(𝛿)=𝑅^∗_{()2[𝑛+1/2]}(𝛿)𝛿. (44)

(ii) When 𝑚≠1, by Lemmas 4 and 6, we have 𝑀𝜌=𝑅_{[(𝑛+1)/2]+𝑚−1}𝜌²1−𝜌^2𝑚−1/2+^{[(𝑛+1)/2]−𝑚}_𝑗=0𝑅_{[(𝑛+1)/2]−𝑚−𝑗}𝜌²𝐼_−𝑗𝜌=𝑅_{[(𝑛+1)/2]+𝑚−1}𝜌²1−𝜌^2𝑚−1/2+𝑅_{[(𝑛+1)/2]−𝑚}𝜌²𝐼₀𝜌+𝑅_{[(𝑛+1)/2]−𝑚−1}𝜌²𝐼₋₁𝜌+⋯+𝑅₀𝜌²𝐼_{−([(𝑛+1)/2]−𝑚)}𝜌. (45)

Thus, 𝑀^∗(𝛿)=𝑅^∗_{)][(𝑛+1/2+𝑚−1}𝛿²𝛿^2𝑚−1+𝑅^∗_{)][(𝑛+1/2−𝑚}𝛿²𝑅^∗₀𝛿²+𝑅^∗_{)][(𝑛+1/2−𝑚−1}𝛿²𝑅^∗₁𝛿²+⋅⋅⋅+𝑅^∗₀𝛿²𝑅^∗_{)][(𝑛+1/2−𝑚}𝛿²=𝑅^∗_{)][(𝑛+1/2+𝑚−1}𝛿²𝛿^2𝑚−1+𝑅^∗_{)][(𝑛+1/2−𝑚}𝛿². (46)

This ends the proof.

The following lemma can be found in [5].

Lemma 8. Consider the family of functions 𝐹(𝑥)=𝐴_𝑖(𝑥)+𝐵_𝑗(𝑥)(𝑎−𝑥)^𝛼, (47)
defined on (−∞,𝑎), where 𝐴_𝑖 and 𝐵_𝑗 are polynomials of degrees 𝑖 and 𝑗, respectively, and 𝛼∉𝑍. Then each nontrivial function of the form has at most 𝑖+𝑗+1 real zeros, taking into account their multiplicities. Moreover, there exist polynomials 𝐴_𝑖 and 𝐵_𝑗 such that the corresponding function has exactly this number of zeros on (−∞,𝑎).

3. Proof of Theorem 1

Proof. By Lemma 7, when 𝑛<2𝑚−1 we have 𝑀𝜌=𝑅^∗_{)]𝑚+[(𝑛−1/2}𝛿²𝛿^2𝑚−1. (48)
As the numerator of the above expression is a polynomial in 𝛿² of degree 𝑚+[(𝑛−1)/2], the maximum number of zeros of 𝑀(𝜌) in [0,1) is 𝑚+[(𝑛−1)/2].
When 𝑚=1 (𝑛≥2𝑚−1), then 𝑀𝜌=𝑅^∗_{()2[𝑛+1/2]}(𝛿)𝛿. (49)
As before, the maximum number of zeros of 𝑀(𝜌) in [0,1) is 2[(𝑛+1)/2].
When 𝑚≠1(𝑛≥2𝑚−1), we have 𝑀𝜌=𝑀𝛿²=𝑅^∗_{)]𝑚+[(𝑛+1/2−1}𝛿²𝛿^2𝑚−1+𝑅^∗_{)][(𝑛+1/2−𝑚}𝛿². (50)
Then, using Lemma 8, 𝑀(𝛿²) has at most (𝑚+[(𝑛+1)/2]−1)+([(𝑛+1)/2]−𝑚)+1=2[(𝑛+1)/2] zeros in 𝛿²∈(0,1].
Finally, for all 𝑛 and 𝑚 we know that 𝑀(0)=𝑀(1)=0. Then the maximum number of zeros of 𝑀(𝜌), taking into account their multiplicities, is 𝑚+[(𝑛−1)/2]−1 when 𝑛<2𝑚−1 and 2[(𝑛+1)/2]−1 when 𝑛≥2𝑚−1. The proof is completed.

4. Two Illustration Examples on the Maximum Number

Consider the system 𝑥̇=−𝑦1−𝑦²,̇+𝜀𝑃𝑥,𝑦𝑦=𝑥1−𝑦²+𝜀𝑄𝑥,𝑦, (51)

where 𝑃𝑥,𝑦=_{𝑖+𝑗≤2}𝑝_𝑖𝑗𝑥^𝑖𝑦^𝑗,𝑄𝑥,𝑦=_{𝑖+𝑗≤2}𝑞_𝑖𝑗𝑥^𝑖𝑦^𝑗 (52)

and 𝜀 is a small real parameter.

Let 𝑀𝜌=_𝛾𝜌𝑄𝑥,𝑦𝑑𝑥−𝑃𝑥,𝑦𝑑𝑦1−𝑦², (53)

where 𝛾_𝜌={(𝑥,𝑦)∶𝑥²+𝑦²=𝜌²}, 𝜌∈(0,1). Assuming 𝑥=𝜌cos𝜃,𝑦=𝜌sin𝜃, we have 𝑀𝜌=−₀^2𝜋𝑝₁₀𝜌²+𝑞₀₁−𝑝₁₀𝜌²sin²𝜃1−𝜌²sin²𝜃=𝑑𝜃−𝑝₁₀𝜌²−𝑞₀₁−𝑝₁₀𝐼₁+𝑞₀₁−𝑝₁₀𝐼₀, (54)

where 𝐼_𝑗𝜌=₀^2𝜋11−𝜌²sin²𝜃^𝑗𝑑𝜃,𝜌∈[0,1), (55)

Let 𝛿=1−𝜌², 𝜌∈[0,1). Then 𝑀(𝛿)𝜌=()𝑝=𝑀𝛿−1₁₀𝛿+𝑞₀₁𝛿. (56)

Obviously 𝑀(𝛿)=0 for 𝛿∈(0,1) if and only if 𝛿=−𝑞₀₁/𝑝₁₀. Thus for system (51) the function 𝑀(𝜌) can have 1 simple zero in 𝜌∈(0,1).

Now we consider the system 𝑥̇=−𝑦1−𝑦²²+𝜀𝑃,̇𝑥,𝑦𝑦=𝑥1−𝑦²²+𝜀𝑄𝑥,𝑦, (57)

where 𝑃𝑥,𝑦=_{𝑖+𝑗≤2}𝑝_𝑖𝑗𝑥^𝑖𝑦^𝑗,𝑄𝑥,𝑦=_{𝑖+𝑗≤2}𝑞_𝑖𝑗𝑥^𝑖𝑦^𝑗, (58)(59)

and 𝜀 is a small real parameter.

Let 𝑀𝜌=_𝛾𝜌𝑄𝑥,𝑦𝑑𝑥−𝑃𝑥,𝑦𝑑𝑦1−𝑦²²=−𝑝₁₀𝜌²−𝑞₀₁−𝑝₁₀𝐼₂+𝑞₀₁−𝑝₁₀𝐼₁, (60)

where 𝛾_𝜌={(𝑥,𝑦)∶𝑥²+𝑦²=𝜌²}, 𝜌∈(0,1).

As before, let 𝛿=1−𝜌², 𝜌∈[0,1). Then 𝑀(𝛿)=𝑀𝜌=1−𝛿²−𝑝₁₀𝛿²−𝑞₀₁𝛿³. (61)

It follows that 𝑀(𝛿)=0 for ∈(0,1) if and only if 𝛿=−𝑞₀₁/𝑝₁₀. Thus for system (57) the function 𝑀(𝜌) can have 1 simple zero in 𝜌∈(0,1).

Acknowledgments

The project was supported by the National Natural Science Foundation of China (11271261), a grant from the Ministry of Education of China (20103127110001), and FP7-PEOPLE-2012-IRSES-316338.