Third Order Melnikov Functions of a Cubic Center under Cubic Perturbations

Abstract

In this paper, cubic perturbations of the integral system ${(1+x)}^{2}dH$ where $H=(x^2+y^2)/2$ are considered. Some useful formulae are deduced that can be used to compute the first three Melnikov functions associated with the perturbed system. By employing the properties of the ETC system and the expressions of the Melnikov functions, the existence of exactly six limit cycles is given. Note that there are many cases for the existence of third-order Melnikov functions, and some existence conditions are very complicated—the corresponding Melnikov functions are not presented.

1. Introduction

In the study of planar systems of differential equations, one of the most challenging problems is the second part of Hilbert’s 16th problem. Consider a planar system of differential equations in the following form

$$\begin{array}{c}\hfill \dot{x}=P(x,y),\dot{y}=Q(x,y),\end{array}$$

(1)

where P and Q are polynomials of degree n. Then, the second part of Hilbert’s 16th problem asks for the upper bound for the number of limit cycles in planar polynomial systems of degree n and their relative positions (see [1,2], for example). Due to its difficulty, a weakened version, now known as the weakened 16th Hilbert problem is posed by Arnold [3] that asks for the upper bound for the number of limit cycles of the perturbed system.

$$\begin{array}{c}\hfill \dot{x}=\frac{\partial H}{\partial y}+ϵP(x,y),\dot{y}=-\frac{\partial H}{\partial y}+ϵQ(x,y),\end{array}$$

(2)

where $\left|ϵ\right|\ll 1$ and $H(x,y)$ is a Hamiltonian function. The problem has been researched for $n=2$ by several research groups independently [4,5,6], and to answer this weakened version, various methods have been developed, among them a popular method is based on the Melnikov functions. By computing the higher-order Melnikov functions based on the algorithm of [7,8], researchers have studied the number of limit cycles bifurcated from the above perturbed system (see [9,10,11,12,13,14,15,16,17,18,19,20]). The results in [12] showed that quasi-homogeneous polynomial Hamiltonian systems have a bound on the number of limit cycle bifurcations from the period annulus at any order of Melnikov functions. Asheghi and Nabavi [16] discussed the following perturbed system

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \left\{\begin{array}{c}\dot{x}=-y(1-x^2)+ϵ{\displaystyle \sum _{i+j=0}^2}{a}_{ij}{x}^i{y}^j,\hfill \\ \dot{y}=x(1-x^2)-ϵ{\displaystyle \sum _{i+j=0}^2}{b}_{ij}{x}^i{y}^j.\hfill \end{array}\right.\end{array}\end{array}$$

(3)

They studied the limit cycles by using Melnikov functions up to order three and proved that there are six or seven limit cycles of system (3).

Yang and Yu [17] considered the following system with $n=2$

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \left\{\begin{array}{c}\dot{x}=y{(1+x)}^2-ϵ{\displaystyle \sum _{i+j=0}^n}{a}_{ij}{x}^i{y}^j,\hfill \\ \dot{y}=-x{(1+x)}^2+ϵ{\displaystyle \sum _{i+j=0}^n}{b}_{ij}{x}^i{y}^j,\hfill \end{array}\right.\end{array}\end{array}$$

(4)

using the Melnikov function of any order and proved that the upper bound for the number of limit cycles is three.

Following the work of [17], Liu [18] studied the number of limit cycles bifurcated from the origin of the following perturbed system

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \left\{\begin{array}{c}\dot{x}=-y{(1+x)}^3+ϵ{\displaystyle \sum _{i+j=0}^3}{a}_{ij}{x}^i{y}^j,\hfill \\ \dot{y}=x{(1+x)}^3-ϵ{\displaystyle \sum _{i+j=0}^3}{b}_{ij}{x}^i{y}^j.\hfill \end{array}\right.\end{array}\end{array}$$

(5)

The main results showed that the first four Melnikov functions associated with the perturbed system can yield five limit cycles.

For system (4), it is easy to see when $ϵ=0$; there exists a family of periodic orbits ${\Gamma }_h:H(x,y)=\frac{1}{2}(x^2+y^2)=h$ for $0<h<1/2$ surrounding the origin. Then, the higher order Melnikov functions can be defined in the following way: fixing a transversal segment to the flow in (4) and using the energy level h to parameterize it, the corresponding displacement function is

$$d(h,ϵ)=ϵM_1\left(h\right)+{ϵ}^2{M}_2\left(h\right)+{ϵ}^3{M}_3\left(h\right)+\cdots ,h\in \left(0,\frac{1}{2}\right),$$

(6)

where $M_k\left(h\right)$ is called the kth-order Melnikov function of system (4). For example, the first-order Melnikov function of system (4) takes the form

$$M_1\left(h\right)=\underset{H=h}{\oint }\frac{Q(x,y)}{{(1+x)}^2}dx+\frac{P(x,y)}{{(1+x)}^2}dy.$$

(7)

If $M_1\left(h\right)\not\equiv 0$, every simple zero $h_0\in \left(0,\frac{1}{2}\right)$ of $M_1\left(h\right)$ corresponds to a limit cycle of the perturbed system (4) near ${\Gamma }_{h_0}$. If $M_1\left(h\right)\equiv 0$, then the simple zero of $M_2\left(h\right)$ will become very important in investigating the number of limit cycles of the perturbed system. Similarly, we may use $M_3\left(h\right)$, $M_4\left(h\right)$,⋯ to study the number of limit cycles [9,10,13,15,17].

In this paper, we also consider system (4), but with $n=3.$ Thus, the system becomes a planar cubic system. We will study the upper bound of limit cycles of the system by using the higher-order Melnikov functions. Although, our system seems similar to the system in [17,18], it needs more-difficult 1-form decompositions than theirs.

The organization of the paper is as follows: in Section 2, some Lemmas and formulae are presented as preliminaries; in Section 3, the first three Melnikov functions are computed, and the corresponding bifurcated limit cycles are given; in Section 4, a brief discussion is presented.

2. Preliminaries

Let

$$\omega =\frac{Q(x,y)}{{(1+x)}^2}dx+\frac{P(x,y)}{{(1+x)}^2}dy,$$

then the Pfaffian form of system (4) is $dH=ϵ\omega$.

The following Lemmas and Remarks shall be used to prove the main Theorems. The algorithm of calculating $M_k\left(h\right)$ is shown by Francoise [7] and Iliev [8].

Lemma 1 ([8,17]).  

Assume ${\Gamma }_h$ is the period annulus defined by $H(x,y)=h$, the polynomial function H(x,y) and the 1-form ω satisfy $\underset{H=h}{\oint }\omega =0$ if and only if there are two analytic functions $q(x,y)$ and $S(x,y)$ in a neighborhood of ${\Gamma }_h$ such that $\omega =qdH+dS.$

Lemma 2 ([7,17]). 

Let $\omega ={\overline{q}}_{0}dH+d{\overline{Q}}_0+N_0$, then $M_1\left(h\right)=\underset{H=h}{\oint }\omega =\underset{H=h}{\oint }{N}_0$;

(1)

If $M_1\left(h\right)\equiv 0$, then $N_0={\tilde{q}}_{0}dH+d{\tilde{Q}}_0.$ Let $q_0={\overline{q}}_0+{\tilde{q}}_0,$ $Q_0={\overline{Q}}_0+{\tilde{Q}}_0;$ we have $\omega =q_{0}dH+d{Q}_0$, and $q_0\omega$ can be decomposed into $q_0\omega ={\overline{q}}_{1}dH+d{\overline{Q}}_1+N_1$; then $M_2\left(h\right)=\underset{H=h}{\oint }{q}_0\omega =\underset{H=h}{\oint }{N}_1$;

(2)

If $M_2\left(h\right)\equiv 0$, then $N_1={\tilde{q}}_{1}dH+d{\tilde{Q}}_1.$ Let $q_1={\overline{q}}_1+{\tilde{q}}_1,$ $Q_1={\overline{Q}}_1+{\tilde{Q}}_1;$ we have $q_0\omega =q_{1}dH+d{Q}_1$, and $q_1\omega$ can be decomposed into $q_1\omega ={\overline{q}}_{2}dH+d{\overline{Q}}_2+N_2$; then $M_3\left(h\right)=\underset{H=h}{\oint }{q}_1\omega =\underset{H=h}{\oint }{N}_2$;

By this way, if $M_1\left(h\right)=M_2\left(h\right)=\cdots =M_{i-1}\left(h\right)\equiv 0$, then $q_j\omega =q_{j+1}dH+d{Q}_{j+1},j\le i-2$ and we assume $q_{-1}=1$. We have $M_i\left(h\right)=\underset{H=h}{\oint }{q}_{i-2}\omega =\underset{H=h}{\oint }{N}_{i-1}$, where $q_{i-2}\omega ={\overline{q}}_{i-1}dH+d{\overline{Q}}_{i-1}+N_{i-1}.$

Remark 1.

The authors in [17] have noted that $q_i\omega =q_{i+1}dH+d{Q}_{i+1},$ $d{Q}_i$ is not used in the subsequent calculative process when $i>0.$ Thus, the specific form of $d{Q}_i$ for $i\ge 1$ will not be shown in the following sections.

In the next lemma, an algorithm to decompose $q_i\omega$ is given to simplify the expression of $M_k\left(h\right)$.

Define

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\omega }_{ij}^k=& \frac{x^i{y}^j}{{(1+x)}^k}dx,{\delta }_{ij}^k=\frac{x^i{y}^j}{{(1+x)}^k}dy,{\Omega }_{ij}^k=\frac{x^i{y}^{j}ln(1+x)}{{(1+x)}^k}dx,\hfill \\ \hfill {\Delta }_{ij}^k=& \frac{x^i{y}^{j}ln(1+x)}{{(1+x)}^k}dy,{\Omega }_{ij2}^k=\frac{x^i{y}^j{ln}^2(1+x)}{{(1+x)}^k}dx,\hfill \\ \hfill {\Delta }_{ij2}^k=& \frac{x^i{y}^j{ln}^2(1+x)}{{(1+x)}^k}dy,J_k\left(h\right)=\underset{H=h}{\oint }{\delta }_{00}^k,{\overline{J}}_k\left(h\right)=\underset{H=h}{\oint }{\Delta }_{00}^k,k\ge 0,\hfill \end{array}\end{array}$$

(8)

where $J_k\left(h\right)$ and ${\overline{J}}_k\left(h\right)$ are called the generators of $M_i\left(h\right).$

Therefore, $\omega$ also can be rewritten as

$$\omega ={\displaystyle \sum _{i+j=0}^3}(a_{ij}{\delta }_{ij}^2+b_{ij}{\omega }_{ij}^2).$$

(9)

Lemma 3.

(i)

For $k=2$, there exist 1-forms ${\omega }_{ij}$ and ${\delta }_{ij}$ as follows:

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\delta }_{01}^2=& \frac{\mathit{dH}}{{\left(1+x\right)}^2}-\frac{x\mathit{dx}}{{\left(1+x\right)}^2},{\delta }_{10}^2={\delta }_{00}^1-{\delta }_{00}^2,{\delta }_{02}^2=2H{\delta }_{00}^2-\mathit{dy}+2{\delta }_{00}^1-{\delta }_{00}^2,\hfill \\ \hfill {\delta }_{11}^2=& \frac{x\mathit{dH}}{{\left(1+x\right)}^2}-\frac{x^2\mathit{dx}}{{\left(1+x\right)}^2},{\delta }_{20}^2={\delta }_{00}^2-2{\delta }_{00}^1+\mathit{dy},{\omega }_{00}^2=\frac{\mathit{dx}}{{\left(1+x\right)}^2},\hfill \\ \hfill {\omega }_{10}^2=& \frac{x\mathit{dx}}{{\left(1+x\right)}^2},{\omega }_{20}^2=\frac{x^2\mathit{dx}}{{\left(1+x\right)}^2},{\omega }_{01}^2=\frac{y\mathit{dx}}{{\left(1+x\right)}^2}-\frac{\mathit{dy}}{1+x}+{\delta }_{00}^1,\hfill \\ \hfill {\omega }_{02}^2=& 2\frac{\mathit{dH}}{1+x}-2d\left(\frac{H}{1+x}\right)-\frac{x^2\mathit{dx}}{{\left(1+x\right)}^2},\hfill \\ \hfill {\omega }_{11}^2=& \frac{y\mathit{dH}}{{\left(1+x\right)}^2}-2H{\delta }_{00}^2+{\delta }_{00}^2-2{\delta }_{00}^1+\mathit{dy},\hfill \\ \hfill {\omega }_{11}^1=& d\left(-yln\left(1+x\right)\right)+{\omega }_{01}^0+{\Delta }_{00}^0,{\omega }_{03}^2=d\left(-\frac{y^3}{1+x}\right)+\frac{3y\mathit{dH}}{1+x}-3{\omega }_{11}^1,\hfill \\ \hfill {\omega }_{12}^2=& d\left(2H\left(\frac{1}{1+x}+ln\left(1+x\right)\right)\right)-2\left(\frac{1}{1+x}+ln\left(1+x\right)\right)\mathit{dH}-\frac{x^3\mathit{dx}}{{\left(1+x\right)}^2},\hfill \\ \hfill {\delta }_{03}^2=& \frac{y^2\mathit{dH}}{{\left(1+x\right)}^2}-{\omega }_{12}^2,{\omega }_{21}^2={\omega }_{11}^1-{\omega }_{11}^2,{\delta }_{12}^2=\frac{xy\mathit{dH}}{{\left(1+x\right)}^2}-{\omega }_{11}^1+{\omega }_{11}^2,\hfill \\ \hfill {\delta }_{30}^2=& 2H({\delta }_{00}^1-{\delta }_{00}^2)-{\delta }_{12}^2,{\omega }_{30}^2=\frac{x^3\mathit{dx}}{{\left(1+x\right)}^2},{\delta }_{21}^2=\frac{x^2\mathit{dH}}{{\left(1+x\right)}^2}-\frac{x^3\mathit{dx}}{{\left(1+x\right)}^2}.\hfill \end{array}\end{array}$$

(10)

(ii)

If j is even,

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\omega }_{04}^k=& \frac{{(2H-x^2)}^2}{{(1+x)}^k}dx=\frac{4H^2-4H{x}^2+x^4}{{(1+x)}^k}dx\hfill \\ \hfill =& 4H^{2}d\left(\frac{-1}{(k-1){(1+x)}^{k-1}}\right)-4Hd\left(\int \frac{x^2}{{(1+x)}^k}dx\right)+\frac{x^4}{{(1+x)}^k}dx\hfill \end{array}\end{array}$$

(11)

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill =& d\left(-\frac{4H^2}{(k-1){(1+x)}^{k-1}}\right)+\frac{8H}{(k-1){(1+x)}^{k-1}}dH-d\left(4H\int \frac{x^2}{{(1+x)}^k}dx\right)\hfill \\ & +4\int \frac{x^2}{{(1+x)}^k}dH+\frac{x^4}{{(1+x)}^k}dx\hfill \\ & =\left(\frac{8H}{(k-1){(1+x)}^{k-1}}+4\int \frac{x^2}{{(1+x)}^k}dx\right)dH+d{Q}_{04}^k(x,H),\hfill \\ \hfill {\omega }_{06}^k=& \left(\frac{24H^2}{(k-1){(1+x)}^{k-1}}+24H\int \frac{x^2}{{(1+x)}^k}dx-6\int \frac{x^4}{{(1+x)}^k}dx\right)dH+d{Q}_{06}^k(x,H),\hfill \\ \hfill {\omega }_{02}^k=& \frac{2}{(k-1){(1+x)}^{k-1}}dH+d{Q}_{02}^k(x,H),\hfill \\ \hfill {\omega }_{08}^k=& \left(\frac{64H^3}{(k-1){(1+x)}^{k-1}}+96H^2\int \frac{x^2}{{(1+x)}^k}dx-48H\int \frac{x^4}{{(1+x)}^k}dx\right.\hfill \\ & \left.+8\int \frac{x^6}{{(1+x)}^k}dx\right)dH+d{Q}_{08}^k(x,H),\hfill \end{array}\end{array}$$

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\delta }_{04}^k=& \frac{{(2H-x^2)}^2}{{(1+x)}^k}dy={\left(2H-1\right)}^2{\delta }_{00}^k+\left(8H-4\right){\delta }_{00}^{k-1}+\left(6-4H\right){\delta }_{00}^{k-2}-4{\delta }_{00}^{k-3}+{\delta }_{00}^{k-4},\hfill \\ \hfill {\delta }_{06}^k=& 8H^3{\delta }_{00}^k-12H^2\left({\delta }_{00}^k-2{\delta }_{00}^{k-1}+{\delta }_{00}^{k-2}\right)+6H\left({\delta }_{00}^k-4{\delta }_{00}^{k-1}+6{\delta }_{00}^{k-2}-4{\delta }_{00}^{k-3}+{\delta }_{00}^{k-4}\right)\hfill \\ & -{\delta }_{00}^k+6{\delta }_{00}^{k-1}-15{\delta }_{00}^{k-2}+20{\delta }_{00}^{k-3}-15{\delta }_{00}^{k-4}+6{\delta }_{00}^{k-5}-{\delta }_{00}^{k-6},\hfill \\ \hfill {\delta }_{02}^k=& 2H{\delta }_{00}^k-{\delta }_{00}^k-{\delta }_{00}^{k-2}+2{\delta }_{00}^{k-1},\hfill \\ \hfill {\Omega }_{02}^k=& \frac{(2H-x^2)ln(1+x)}{{(1+x)}^k}dx=2Hd\int \frac{ln(1+x)}{{(1+x)}^k}dx-\frac{x^{2}ln(1+x)}{{(1+x)}^k}dx\hfill \\ & =-2\int \frac{ln(1+x)}{{(1+x)}^k}dxdH+d{\overline{Q}}_{02}^k,\hfill \\ \hfill {\Omega }_{04}^k=& \left(-8H\int \frac{ln(1+x)}{{(1+x)}^k}dx+4\int \frac{x^{2}ln(1+x)}{{(1+x)}^k}dx\right)dH+d{\overline{Q}}_{04}^k,\hfill \\ \hfill {\Delta }_{02}^k=& 2H{\Delta }_{00}^k-{\Delta }_{00}^k-{\Delta }_{00}^{k-2}+2{\Delta }_{00}^{k-1},\hfill \\ \hfill {\Delta }_{04}^k=& {\left(2H-1\right)}^2{\Delta }_{00}^k+\left(8H-4\right){\Delta }_{00}^{k-1}+\left(6-4H\right){\Delta }_{00}^{k-2}-4{\Delta }_{00}^{k-3}+{\Delta }_{00}^{k-4},\hfill \end{array}\end{array}$$

If j is odd,

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\omega }_{0j}^k=& d\left(\frac{-y^j}{(k-1){(1+x)}^{k-1}}\right)+\frac{j}{k-1}{\delta }_{0,j-1}^{k-1}=d{S}_{0j}^k(x,y)+\frac{j}{k-1}{\delta }_{0,j-1}^{k-1},\hfill \\ \hfill {\delta }_{0j}^k=& \frac{y^{j-1}}{{(1+x)}^k}dH-{\omega }_{0,j-1}^{k-1}+{\omega }_{0,j-1}^k,\hfill \\ \hfill {\Omega }_{0j}^k=& d\left(y^j\int \frac{ln(1+x)}{{(1+x)}^k}dx\right)+\frac{j}{k-1}{\Delta }_{0,j-1}^{k-1}+\frac{j}{{(k-1)}^2}{\delta }_{0,j-1}^{k-1}\hfill \\ & =d{\overline{S}}_{0j}^k(x,y)+\frac{j}{k-1}{\Delta }_{0,j-1}^{k-1}+\frac{j}{{(k-1)}^2}{\delta }_{0,j-1}^{k-1},\hfill \\ \hfill {\Delta }_{0j}^k=& \frac{y^{j-1}ln(1+x)}{{(1+x)}^k}dH-{\Omega }_{0,j-1}^{k-1}+{\Omega }_{0,j-1}^k.\hfill \end{array}\end{array}$$

Proof. 

(i) Here, we use partial integration and the Laurent expansion method to decompose ${\omega }_{ij}^2$ and ${\delta }_{ij}^2$, and some decompositions have been proved in [17].

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\omega }_{11}^1=& \frac{xy}{1+x}dx=yd(x-ln(1+x))=d\left(y(x-ln(1+x))\right)-(x-ln(1+x))dy\hfill \\ & =d(-yln(1+x))+{\omega }_{01}^0+{\Delta }_{00}^0,\hfill \\ \hfill {\omega }_{03}^2=& y^{3}d\left(-\frac{1}{1+x}\right)=d\left(-\frac{y^3}{1+x}\right)+\frac{3y^2}{1+x}dy=d\left(-\frac{y^3}{1+x}\right)+\frac{3y}{2(1+x)}d{y}^2\hfill \\ & =d\left(-\frac{y^3}{1+x}\right)+\frac{3y}{1+x}dH-3{\omega }_{11}^1,\hfill \\ \hfill {\omega }_{12}^2=& \frac{x(2H-x^2)}{{(1+x)}^2}dx=2Hd\left(\frac{1}{1+x}+ln(1+x)\right)-\frac{x^3}{{(1+x)}^2}dx\hfill \\ \hfill =& d\left(2H\left(\frac{1}{1+x}+ln\left(1+x\right)\right)\right)-2\left(\frac{1}{1+x}+ln\left(1+x\right)\right)\mathit{dH}-\frac{x^3\mathit{dx}}{{\left(1+x\right)}^2},\hfill \\ \hfill {\delta }_{03}^2=& \frac{y^2}{2{(1+x)}^2}d{y}^2=\frac{y^2}{{(1+x)}^2}dH-{\omega }_{12}^2,\hfill \\ \hfill {\delta }_{30}^2=& \frac{x(2H-y^2)}{{(1+x)}^2}dy=2H({\delta }_{00}^1-{\delta }_{00}^2)-{\delta }_{12}^2.\hfill \end{array}\end{array}$$

(12)

The Formula (10) is obtained. The proof of (ii) is omitted. □

Lemma 4.

The generators $J_k\left(h\right)$ shown in (8) can be obtained through Maple software

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill J_1\left(h\right)=& -2\pi \left(1-\frac{1}{\sqrt{1-2h}}\right),J_2\left(h\right)=\frac{4h\pi }{{\left(1-2h\right)}^{3/2}},J_3\left(h\right)=\frac{6h\pi }{{\left(1-2h\right)}^{5/2}},\hfill \\ \hfill J_4\left(h\right)=& \frac{4h\pi \left(h+2\right)}{{\left(1-2h\right)}^{7/2}},J_5\left(h\right)=\frac{5h\pi \left(3h+2\right)}{{\left(1-2h\right)}^{9/2}},J_6\left(h\right)=\frac{6h\pi \left(h^2+6h+2\right)}{{\left(1-2h\right)}^{11/2}},\hfill \\ \hfill J_7\left(h\right)=& \frac{7h\pi \left(5h^2+10h+2\right)}{{\left(1-2h\right)}^{13/2}},J_8\left(h\right)=\frac{2h\pi \left(5h^3+60h^2+60h+8\right)}{{\left(1-2h\right)}^{15/2}},\hfill \\ \hfill J_{01}\left(h\right)=& \underset{H=h}{\oint }{\omega }_{01}^0=2\pi h.\hfill \end{array}\end{array}$$

(13)

Define

$$\begin{array}{c}\hfill F_k\left(r\right)=\underset{0}{\overset{2\pi }{\int }}\frac{{cos}^k\theta ln(1+rcos\theta )}{1+rcos\theta }d\theta ,G_k\left(r\right)={\int }_0^{2\pi }{cos}^k\theta ln(1+rcos\theta )d\theta ,\end{array}$$

(14)

then

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill F_0\left(r\right)=& \frac{2\pi }{\sqrt{1-r^2}}ln\left(\frac{2(1-r^2)}{1+\sqrt{1-r^2}}\right),\hfill \\ \hfill F_2\left(r\right)=& \frac{2\pi \left(1-\sqrt{1-r^2}\right)}{r^2}-\frac{2\pi }{r^2}ln\frac{1+\sqrt{1-r^2}}{2}+\frac{2\pi }{r^2\sqrt{1-r^2}}ln\left(\frac{2(1-r^2)}{1+\sqrt{1-r^2}}\right),\hfill \\ \hfill G_0\left(r\right)=& 2\pi ln\frac{1+\sqrt{1-r^2}}{2},G_1\left(r\right)=\frac{2\pi \left(1-\sqrt{1-r^2}\right)}{r}.\hfill \end{array}\end{array}$$

(15)

The proof of the expressions $F_0\left(r\right)$, $F_2\left(r\right)$, $G_0\left(r\right)$ and $G_1\left(r\right)$ can be seen in [10].

Lemma 5.

The following expressions can be deduced by using the above formulae

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \underset{H=h}{\oint }{\Delta }_{002}^0=& -2r^2(F_0\left(r\right)-F_2\left(r\right)),{\overline{J}}_0\left(r\right)=-r{G}_1\left(r\right),{\overline{J}}_1\left(r\right)=-r{F}_1\left(r\right)=F_0\left(r\right)-G_0\left(r\right),\hfill \\ \hfill {\overline{J}}_2\left(r\right)=& r{F}_0^{{}^{\prime }}\left(r\right)+J_2\left(r\right),{\overline{J}}_3\left(r\right)=\frac{1}{2}{J}_3\left(r\right)+\frac{r}{2}{(F_0\left(r\right)+{\overline{J}}_2\left(r\right))}^{{}^{\prime }},{\overline{J}}_{10}\left(r\right)=-r^2{G}_2\left(r\right),\hfill \end{array}\end{array}$$

(16)

where

$$\begin{array}{c}\hfill G_2\left(r\right)=\pi \left(ln\frac{1+\sqrt{1-r^2}}{2}+\frac{\sqrt{1-r^2}-1}{r^2}+\frac{1}{2}\right),r=\sqrt{2h}.\end{array}$$

(17)

Proof. 

Since

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\Omega }_{01}^1=& \frac{yln(1+x)}{1+x}dx=yd\left(\frac{1}{2}{ln}^2(1+x)\right)=d\left(\frac{1}{2}y{ln}^2(1+x)\right)-\frac{1}{2}{ln}^2(1+x)dy\hfill \\ & =d\left(\frac{1}{2}y{ln}^2(1+x)\right)-\frac{1}{2}{\Delta }_{002}^0,\hfill \end{array}\end{array}$$

(18)

therefore

$$\begin{array}{c}\hfill \underset{H=h}{\oint }{\Omega }_{01}^1=-\frac{1}{2}\underset{H=h}{\oint }{\Delta }_{002}^0.\end{array}$$

(19)

Let $r=\sqrt{2h},$ $x=rcos\theta$ and $y=rsin\theta$, then

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \underset{H=h}{\oint }{\Omega }_{01}^1=& r^2\underset{0}{\overset{2\pi }{\int }}\frac{{sin}^2\theta ln(1+rcos\theta )}{1+rcos\theta }d\theta =r^2\underset{0}{\overset{2\pi }{\int }}\frac{(1-{cos}^2\theta )ln(1+rcos\theta )}{1+rcos\theta }d\theta \hfill \\ & =r^2(F_0\left(r\right)-F_2\left(r\right)),\hfill \end{array}\end{array}$$

(20)

which implies $\underset{H=h}{\oint }{\Delta }_{002}^0=-2r^2(F_0\left(r\right)-F_2\left(r\right)).$ We also have

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\overline{J}}_0\left(r\right)=& \underset{H=h}{\oint }ln(1+x)dy=-r{\int }_0^{2\pi }cos\theta ln(1+rcos\theta )d\theta =-r{G}_1\left(r\right).\hfill \\ \hfill {\overline{J}}_1\left(r\right)=& \underset{H=h}{\oint }\frac{ln(1+x)}{1+x}dy=-r{\int }_0^{2\pi }\frac{cos\theta ln(1+rcos\theta )}{1+rcos\theta }d\theta =-r{F}_1\left(r\right)\hfill \\ \hfill =& -r{\int }_0^{2\pi }ln(1+rcos\theta )\left(\frac{1}{r}-\frac{1}{r(1+rcos\theta )}\right)d\theta =F_0\left(r\right)-G_0\left(r\right).\hfill \end{array}\end{array}$$

(21)

By

$$\begin{array}{c}\hfill F_0^{{}^{\prime }}\left(r\right)={\int }_0^{2\pi }-\frac{cos\theta \left(ln\left(1+rcos\theta \right)-1\right)}{{\left(1+rcos\theta \right)}^2}d\theta \end{array}$$

(22)

we have

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\overline{J}}_2\left(r\right)=& \underset{H=h}{\oint }\frac{ln(1+x)}{{(1+x)}^2}dy=-r{\int }_0^{2\pi }\frac{cos\theta ln(1+rcos\theta )}{{(1+rcos\theta )}^2}d\theta \hfill \\ & =r\left(F_0^{{}^{\prime }}\left(r\right)-{\int }_0^{2\pi }\frac{cos\theta }{{(1+rcos\theta )}^2}d\theta \right)=r{F}_0^{{}^{\prime }}\left(r\right)+J_2\left(r\right).\hfill \end{array}\end{array}$$

(23)

Because

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill F_{02}\left(r\right):=& {\int }_0^{2\pi }\frac{ln(1+rcos\theta )}{{(1+rcos\theta )}^2}d\theta \hfill \\ \hfill =& {\int }_0^{2\pi }ln(1+rcos\theta )\left(\frac{1}{1+rcos\theta }-\frac{rcos\theta }{{(1+rcos\theta )}^2}\right)d\theta =F_0\left(r\right)+{\overline{J}}_2\left(r\right),\hfill \end{array}\end{array}$$

(24)

on the other hand, we have

$$\begin{array}{c}\hfill \begin{array}{c}\hfill F_{02}^{{}^{\prime }}\left(r\right)={\int }_0^{2\pi }\frac{cos\theta \left(1-2ln\left(1+rcos\theta \right)\right)}{{\left(1+rcos\theta \right)}^3}d\theta ,\end{array}\end{array}$$

(25)

therefore,

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\overline{J}}_3\left(r\right)=& -r{\int }_0^{2\pi }\frac{cos\theta ln(1+rcos\theta )}{{(1+rcos\theta )}^3}d\theta =-r\left(\frac{1}{2}{\int }_0^{2\pi }\frac{cos\theta }{{(1+rcos\theta )}^3}-\frac{1}{2}{F}_{02}^{{}^{\prime }}\left(r\right)\right)\hfill \\ & =\frac{1}{2}{J}_3\left(r\right)+\frac{r}{2}{(F_0\left(r\right)+{\overline{J}}_2\left(r\right))}^{{}^{\prime }}.\hfill \\ \hfill \underset{H=h}{\oint }{\Delta }_{10}^0=& \underset{H=h}{\oint }xln(1+x)dy=-r^2{\int }_0^{2\pi }{cos}^2\theta ln(1+rcos\theta )d\theta =-r^2{G}_2\left(r\right),\hfill \\ \hfill G_2^{{}^{\prime }}\left(r\right)=& {\int }_0^{2\pi }\frac{{cos}^3\theta }{1+rcos\theta }d\theta =-\frac{1}{r^3}\underset{H=h}{\oint }\frac{x^2}{1+x}dy\hfill \\ & =\frac{2}{r^3}{\int }_{-r}^r\frac{x^3}{(1+x)\sqrt{r^2-x^2}}dx=\frac{\pi \left(r^4+r^2+2\sqrt{1-r^2}-2\right)}{r^3\left(r^2-1\right)},\hfill \\ \hfill G_2\left(r\right)=& \pi \left(ln\frac{1+\sqrt{1-r^2}}{2}+\frac{\sqrt{1-r^2}-1}{r^2}+\frac{1}{2}\right)\mathrm{for}{G}_2\left(0\right)=0.\hfill \end{array}\end{array}$$

(26)

□

The Chebyshev criterion will be used in the following (see [16,21,22] for example).

Let $f_0$, $f_1$, …,$f_{n-1}$ be analytic functions on an open interval L of R. An ordered set $(f_0,f_1,...,f_{n-1})$ is an extended complete Chebyshev system (in short, ECT system) on L if, for all $i=1,2,...,n-1,$ any nontrivial linear

$${\lambda }_0{f}_0\left(x\right)+{\lambda }_1{f}_1\left(x\right)+...+{\lambda }_i{f}_i\left(x\right)$$

(27)

has at most $i-1$ isolated zeros on L counted with multiplicities.

Lemma 6 ([21]). 

$(f_0,f_1,...,f_{n-1})$ is an ECT system on L if, and only if, for each $i=1,2,...,n,$

$$\begin{array}{c}\hfill W_{i-1}\left(x\right)=W[f_0\left(x\right),f_1\left(x\right),\cdots ,f_{i-1}\left(x\right)]\ne 0\end{array}$$

(28)

for all $x\in L.$

Remark 2.

If $(f_0,f_1,...,f_{n-1})$ is an ECT system on L, then for each $i=1,2,...,n-1$ there exists a linear combination (30) with exactly $i-1$ simple zeros on L (see, for instance, Remark 3.7 in [22]).

3. The Calculation of M_k(h)

3.1. The Melnikov Function of First-Order M₁(h)

Lemma 7.

The 1-form ω of (9) can be expressed

$$\omega ={\overline{q}}_{0}dH+d{\overline{Q}}_0+N_0,$$

(29)

where

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\overline{q}}_0=& {\overline{q}}_0(x,y)=\frac{a_{03}{y}^2+\left(b_{11}-b_{21}\right)y}{{\left(1+x\right)}^2}+\frac{\left(a_{12}-a_{30}+3b_{03}\right)y}{1+x}+\frac{a_{01}-a_{11}+a_{21}}{{\left(1+x\right)}^2}\hfill \\ & +\frac{a_{11}-2a_{21}+2a_{03}+2b_{02}-2b_{12}}{1+x}+a_{21}+2\left(a_{03}-b_{12}\right)ln\left(1+x\right),\hfill \end{array}\end{array}$$

(30)

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill d{\overline{Q}}_0=& d{\overline{Q}}_0(x,y,H)=-2\left(\frac{a_{03}+b_{02}-b_{12}}{1+x}+\left(a_{03}-b_{12}\right)ln\left(1+x\right)\right)\mathit{dH}+\hfill \\ & \left(\frac{b_{03}{y}^3-\left(b_{12}-a_{03}-b_{02}\right)y^2+b_{01}y}{{\left(1+x\right)}^2}+\frac{\left(b_{12}-a_{03}\right)y^2+\left(a_{12}-a_{30}+3b_{03}-b_{21}\right)y}{1+x}\right)\mathit{dx}\hfill \\ & -\frac{\left(\left(a_{21}-b_{30}\right)x^3+\left(a_{11}-b_{20}\right)x^2+\left(a_{01}-b_{10}\right)x-b_{00}\right)dx}{{\left(1+x\right)}^2}\hfill \\ & +\left(\left(a_{12}-a_{30}+3b_{03}-b_{21}\right)ln(1+x)+a_{20}-a_{02}-a_{30}+a_{12}\right.\hfill \\ & \left.-\frac{b_{01}}{1+x}+b_{11}-b_{21}-\frac{3b_{03}{y}^2}{1+x}\right)\mathit{dy},\hfill \\ \hfill N_0=& A_0{\omega }_{01}^0+\left(H{A}_{21}+A_{20}\right){\delta }_{00}^2+\left(H{A}_{11}+A_{10}\right){\delta }_{00}^1+A_0{\Delta }_{00}^0,\mathrm{with}\hfill \\ \hfill A_0=& a_{30}-a_{12}+b_{21}-3b_{03},A_{21}=2a_{02}-2a_{12}-2b_{11}+2b_{21},\hfill \\ \hfill A_{20}=& a_{00}-a_{10}+a_{20}-a_{02}-a_{30}+a_{12}+b_{11}-b_{21},\hfill \\ \hfill A_{11}=& 2a_{30},A_{10}=2a_{02}+a_{10}-2a_{12}-2a_{20}+2a_{30}+b_{01}-2b_{11}+2b_{21}.\hfill \end{array}\end{array}$$

(31)

Theorem 1.

$M_1\left(h\right)$ has at most three zeros, i.e., system (4) has at most three limit cycles by the first-order Melnikov function, and the upper bound for the number of limit cycles is reached.

Proof. 

Let $z=\sqrt{1-2h},$ $z\in (0,1);$ by Lemmas 2, 4 and 5,

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill M_1\left(h\right)& =\underset{H=h}{\oint }{N}_0=A_0{\overline{J}}_1\left(h\right)+\left(h{A}_{21}+A_{20}\right)J_2\left(h\right)+\left(h{A}_{11}+A_{10}\right)J_1\left(h\right)+A_0{J}_0\left(h\right)\hfill \\ & =M_1\left(\frac{1-z^2}{2}\right)=\frac{\pi \left(1-z\right)}{z^3}\left(n_4{z}^4+n_3{z}^3+n_2{z}^2+n_0\left(z+1\right)\right),\hfill \\ \hfill n_0=& 2\left(a_0-a_{10}+a_{20}-a_{30}\right),\hfill \\ \hfill n_2=& 2\left(a_2+a_{10}-a_{12}-2a_{20}+3a_{30}+b_1-b_{11}+b_{21}\right),\hfill \\ \hfill n_3=& \left(-2a_2+3a_{12}-a_{30}+3b_3+2b_{11}-3b_{21}\right),n_4=\left(-a_{12}-a_{30}-3b_3+b_{21}\right).\hfill \end{array}\end{array}$$

(32)

It is easy to verify that $(z+1,z^2,z^3,z^4)$ is an ECT system for $z>0$ by Lemma 6 and since $\frac{\partial (n_0,n_2,n_3,n_4)}{\partial (a_{10},a_{30},b_{11},b_{21})}=-8\ne 0$. Therefore, $M_1\left(h\right)$ has exactly three zeros when taking appropriate coefficients by Remark 2. □

3.2. The Melnikov Function of Second-Order M₂(h)

$M_1\left(h\right)\equiv 0$ implies that $n_0=n_2=n_3=n_4=0$; solving them, we have

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill a_{00}=& -a_{30}+a_{20}-b_{01},a_{10}=-2a_{30}+2a_{20}-b_{01},a_{12}=-a_{30}+b_{21}-3b_{03},\hfill \\ \hfill b_{11}=& 2a_{30}+3b_{03}+a_{02}.\hfill \end{array}\end{array}$$

(33)

Let (33) hold; we compute $M_2\left(h\right)$. First, it has

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill N_0=& a_{30}\left(2H{\delta }_{00}^1-2H{\delta }_{00}^2+2{\Delta }_{00}^0-2{\delta }_{00}^1+{\delta }_{00}^2+2{\omega }_{01}\right)\hfill \\ \hfill =& a_{30}\left(\frac{\left(2{\left(1+x\right)}^{2}ln\left(1+x\right)+x^3+x{y}^2-2x-1\right)\mathit{dy}}{{\left(1+x\right)}^2}+2y\mathit{dx}\right)\hfill \\ \hfill =& a_{30}\left(\frac{x{y}^{2}dy}{{(1+x)}^2}+2ln(1+x)+\frac{\left(x^2-x-1\right)\mathit{dy}}{1+x}+2y\mathit{dx}\right)\hfill \\ \hfill =& a_{30}\left(\frac{xy\mathit{dH}}{{\left(1+x\right)}^2}+d\left(2ln(1+x)+\frac{x^2-x-1}{1+x}\right)y\right),\hfill \end{array}\end{array}$$

(34)

which implies that

$$\begin{array}{c}\hfill \begin{array}{c}\hfill {\tilde{q}}_0=\frac{a_{30}yx}{{\left(1+x\right)}^2},d{\tilde{Q}}_0=d\left(2ln(1+x)+\frac{x^2-x-1}{1+x}\right)y\end{array}\end{array}$$

(35)

by Lemma 2. Then

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill q_0={\overline{q}}_0+& {\tilde{q}}_0=f_0\left(x\right)+g_0(x,y),\hfill \\ \hfill f_0\left(x\right)=& \frac{a_{01}-a_{11}+a_{21}}{{\left(1+x\right)}^2}+\frac{a_{11}-2a_{21}+2a_{03}+2b_{02}-2b_{12}}{1+x}+a_{21}+\hfill \\ & 2ln\left(1+x\right)\left(a_{03}-b_{12}\right),\hfill \\ \hfill g_0(x,y)=& \frac{a_{03}{y}^2+\left(2a_{30}+3b_{03}+a_{02}-b_{21}\right)y}{{\left(1+x\right)}^2}+\frac{\left(b_{21}-2a_{30}\right)y}{1+x}+\frac{a_{30}yx}{{\left(1+x\right)}^2},\hfill \\ \hfill d{Q}_0=& d{\overline{Q}}_0+d{\tilde{Q}}_0={\Omega }_1\left(x\right)+{\Omega }_2(x,y)+h\left(x\right)dH,\hfill \\ \hfill {\Omega }_1\left(x\right)=& -\frac{\left(\left(a_{21}-b_{30}\right)x^3+\left(a_{11}-b_{20}\right)x^2+\left(a_{01}-b_{10}\right)x-b_{00}\right)\mathit{dx}}{{\left(1+x\right)}^2},\hfill \\ \hfill {\Omega }_2(x,y)=& \left(\frac{b_{03}{y}^3}{{\left(1+x\right)}^2}-\frac{\left(\left(a_{03}-b_{12}\right)x-b_{02}\right)y^2}{{\left(1+x\right)}^2}+\frac{\left(x^2{a}_{30}+2x{a}_{30}+b_{01}\right)y}{{\left(1+x\right)}^2}\right)\mathit{dx}\hfill \\ & +\left(\frac{x^2{a}_{30}+\left(a_{20}-a_{30}\right)x-3b_{03}{y}^2+a_{20}-a_{30}-b_{01}}{1+x}\right)\mathit{dy},\hfill \\ \hfill h\left(x\right)=& -\frac{2(a_{03}+b_{02}-b_{12})}{1+x}-2\left(a_{03}-b_{12}\right)ln\left(1+x\right),\mathrm{and}\omega =q_{0}dH+d{Q}_0.\hfill \end{array}\end{array}$$

(36)

Lemma 8.

The 1-form $q_0\omega$ can be decomposed into

$$\begin{array}{c}\hfill \begin{array}{c}\hfill q_0\omega ={\overline{q}}_{1}dH+d{\overline{Q}}_1+N_1,\end{array}\end{array}$$

(37)

where

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\overline{q}}_1=& q_{0}h\left(x\right)+q_0^2+q_{10},\hfill \\ \hfill q_{10}=& 2{\left(a_{03}-b_{12}\right)}^2{ln}^2(1+x)+\left(2{\beta }_{03}^2-2{\alpha }_{02}^1+4{\alpha }_{04}^3+\frac{4\left(a_{03}-b_{12}\right)\left(a_{03}+b_{02}-b_{12}\right)}{1+x}\right)\hfill \\ & \times ln(1+x)+\frac{2{\alpha }_{02}^4+3{\beta }_{01}^3}{3{\left(1+x\right)}^3}+\frac{{\alpha }_{02}^3+{\beta }_{01}^2}{{\left(1+x\right)}^2}+\frac{2{\alpha }_{02}^2+{\beta }_{01}^1}{1+x}+\frac{4{\alpha }_{04}^4\left(-3x^2+2H-3x-1\right)}{3{\left(1+x\right)}^3}\hfill \\ & -\frac{{\beta }_{03}^3\left(2x^2-y^2+3x+1\right)}{{\left(1+x\right)}^3}+\frac{y\left(3{\alpha }_{03}^2+{\beta }_{02}^1\right)}{1+x}+\frac{2{\alpha }_{04}^3\left(2H+3+4x\right)}{{\left(1+x\right)}^2}\hfill \\ & +\frac{{\beta }_{03}^2\left(y^2+2x+2\right)}{{\left(1+x\right)}^2}+\frac{2\left(a_{03}-b_{12}\right)\left(3y{b}_{03}+2a_{03}+2b_{02}-2b_{12}\right)}{1+x},\hfill \\ \hfill N_1=& \left(4\left(\frac{5{\alpha }_{05}^4}{3}+{\beta }_{04}^3\right)H^2+2\left({\beta }_{02}^3-\frac{10{\alpha }_{05}^4}{3}-2{\beta }_{04}^3+{\alpha }_{03}^4\right)H-{\beta }_{02}^3+\frac{5}{3}{\alpha }_{05}^4+{\beta }_{04}^3-{\alpha }_{03}^4\right.\hfill \\ & \left.+\frac{1}{3}{\alpha }_{01}^4+{\beta }_{00}^3\right){\delta }_{00}^3+\left(\left(2{\beta }_{02}^2+\frac{40{\alpha }_{05}^4}{3}+8{\beta }_{04}^3+3{\alpha }_{03}^3\right)H+2{\beta }_{02}^3-{\beta }_{02}^2-\frac{20{\alpha }_{05}^4}{3}\right.\hfill \\ & \left.-4{\beta }_{04}^3+2{\alpha }_{03}^4-\frac{3}{2}{\alpha }_{03}^3+\frac{1}{2}{\alpha }_{01}^3+{\beta }_{00}^2\right){\delta }_{00}^2+\left(2\left(a_{03}-b_{12}\right)\left(b_{01}-a_{30}\right)\right.\hfill \\ & \left.-4\left(\frac{5{\alpha }_{05}^4}{3}+{\beta }_{04}^3\right)H-{\beta }_{02}^3+2{\beta }_{02}^2+10{\alpha }_{05}^4+6{\beta }_{04}^3+{\alpha }_{01}^2-{\alpha }_{03}^4+3{\alpha }_{03}^3+{\beta }_{00}^1\right){\delta }_{00}^1\hfill \\ & -\left(2\left(a_{03}-b_{12}\right)\left(a_{30}+3b_{03}\right)+\frac{5}{3}{\alpha }_{05}^4+{\beta }_{04}^3+{\beta }_{10}^0-{\alpha }_{01}^0+{\beta }_{02}^1+3{\alpha }_{03}^2\right){\omega }_{01}^0\hfill \\ & +\left(2\left(a_{03}-b_{12}\right)\left(a_{20}-3b_{03}-3a_{30}\right)-{\alpha }_{01}^1-{\beta }_{02}^1-3{\alpha }_{03}^2\right){\Delta }_{00}^0.\hfill \end{array}\end{array}$$

(38)

Proof. 

Following (36), we have

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill q_0\omega =& q_0(q_{0}dH+d{Q}_0)=q_0^{2}dH+q_{0}d{Q}_0=q_0^{2}dH+(f_0\left(x\right)+g_0(x,y))({\Omega }_1\left(x\right)\hfill \\ & +{\Omega }_2(x,y)+h\left(x\right)dH)\hfill \\ \hfill =& (q_0^2+q_{0}h\left(x\right))dH+f_0\left(x\right){\Omega }_1\left(x\right)+g_0(x,y){\Omega }_1\left(x\right)+q_0(x,y){\Omega }_2(x,y)\hfill \\ \hfill =& (q_0^2+q_{0}h\left(x\right))dH+f_0\left(x\right){\Omega }_1\left(x\right)+{\displaystyle \sum _{k=0}^4}{\displaystyle \sum _{j=1}^{k+1}}{\alpha }_{0j}^k{\omega }_{0j}^k+{\displaystyle \sum _{k=0}^3}{\displaystyle \sum _{j=0}^{k+1}}{\beta }_{0j}^k{\delta }_{0j}^k\hfill \\ & +{\displaystyle \sum _{j=1}^3}{\overline{\alpha }}_{0j}^2{\Omega }_{0j}^2+{\overline{\alpha }}_{02}^1{\Omega }_{02}^1+{\overline{\alpha }}_{01}^0{\Omega }_{01}^0\hfill \\ & +{\beta }_{10}^0{\delta }_{10}^0+{\overline{\beta }}_{02}^1{\Delta }_{02}^1+{\overline{\beta }}_{00}^1{\Delta }_{00}^1+{\overline{\beta }}_{00}^0{\Delta }_{00}^0+{\overline{\beta }}_{10}^0{\Delta }_{10}^0,\hfill \end{array}\end{array}$$

(39)

and the expressions of ${\alpha }_{ij}^k$ and ${\beta }_{ij}^k$ are omitted.

From Lemma 3, we can calculate the decomposed expressions of ${\omega }_{0j}^k$ and ${\delta }_{0j}^k$ for $k\ge 2.$ Together with

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\omega }_{01}^1=& d(yln(1+x))-{\Delta }_{00}^0,{\omega }_{02}^1=d(2Hln(1+x))-2ln\left(1+x\right)\mathit{dH}-\frac{x^2}{1+x}dx,\hfill \\ \hfill {\Omega }_{03}^2=& d\left(-\frac{1+ln(1+x)}{1+x}{y}^3\right)+3y^2\frac{1+ln(1+x)}{1+x}dy\hfill \\ & =d\left(-\frac{1+ln(1+x)}{1+x}{y}^3\right)+3({\delta }_{02}^1+{\Delta }_{02}^1),\hfill \\ \hfill {\Omega }_{02}^2=& \frac{2(1+ln(1+x\left)\right)}{1+x}\mathit{dH}+d{\overline{Q}}_{02}^2(x,H),\hfill \\ \hfill {\Omega }_{02}^1=& -{\left(ln\left(1+x\right)\right)}^2\mathit{dH}+d{\overline{Q}}_{02}^1(x,H),{\Omega }_{01}^1=d\left(\frac{1}{2}y{ln}^2(1+x)\right)-\frac{1}{2}{\Delta }_{002}^0,\hfill \\ \hfill {\Omega }_{01}^2=& d\left(-\frac{1+ln(1+x)}{1+x}y\right)+\frac{1+ln(1+x)}{1+x}dy={\delta }_{00}^1+{\Delta }_{00}^1+d{\overline{S}}_{01}^2(x,y),\hfill \\ \hfill {\Omega }_{01}^0=& dy(1+x)(ln(1+x)-1)-(1+x)(ln(1+x)-1)dy\hfill \\ & =-{\Delta }_{00}^0-{\Delta }_{10}^0-{\omega }_{01}^0+d{\overline{S}}_{01}^0(x,y),\hfill \\ \hfill {\Delta }_{02}^1=& \frac{y^{2}ln(1+x)}{1+x}dy=\frac{yln(1+x)}{2(1+x)}d{y}^2=\frac{yln(1+x)}{1+x}dH-\frac{xyln(1+x)}{1+x}dx\hfill \\ \hfill =& \frac{yln(1+x)}{1+x}dH-yd(\left(1+x\right)\left(ln\left(1+x\right)-1\right)-\frac{1}{2}{ln}^2\left(1+x\right))\hfill \\ \hfill =& \frac{yln(1+x)}{1+x}dH+d{\overline{S}}_{02}^1(x,y)+{\Delta }_{00}^0+{\Delta }_{10}^0+{\omega }_{01}^0-\frac{1}{2}{\Delta }_{002}^0,\hfill \\ \hfill {\delta }_{02}^1=& \frac{y}{2(1+x)}d{y}^2=\frac{y}{1+x}dH-{\omega }_{11}^1=\frac{y}{1+x}dH+d\left(yln\left(1+x\right)\right)-{\omega }_{01}^0-{\Delta }_{00}^0.\hfill \end{array}\end{array}$$

(40)

By (39) and (40) and the expressions of ${\omega }_{0j}^k$ and ${\delta }_{0j}^k,$ (37) is given. □

Theorem 2.

Let $M_1\left(h\right)\equiv 0$; $M_2\left(h\right)$ has at most four zeros, then system (4) has at most four limit cycles by the second-order Melnikov function, and the maximum number can be attained.

Proof. 

Let $z=\sqrt{1-2h}$, $z\in (0,1)$. Together with Lemmas 2, 4 and 5,

$$\begin{array}{c}\hfill \begin{array}{c}\hfill M_2\left(\frac{1-z^2}{2}\right)=\frac{\pi (z-1)}{z^5}(l_6{z}^6+l_5{z}^5+l_4{z}^4+l_3{z}^2(z+1)+l_1(z+1)).\end{array}\end{array}$$

(41)

The function $(z+1,z^2(z+1),z^4,z^5,z^6)$ is an ECT system for $z>0$ by Lemma 6. Furthermore, when we take $b_{03}=a_{21}=b_{30}=b_{10}=a_{03}=a_{01}=b_{02}=b_{00}=b_{20}=a_{20}=0$, then $\frac{\partial (l_1,l_3,l_4,l_5,l_6)}{\partial (a_{02},a_{30},a_{11},b_{12},b_{01})}=4a_{11}{b}_{21}(\left(5a_{30}-3b_{21}\right)a_{11}^2+b_{12}\left(4a_{02}+21a_{30}-2b_{01}-7b_{21}\right)a_{11}+3b_{12}^2\left(-2b_{01}+3a_{30}+a_{02}-b_{21}\right))\ne 0$; it follows from Remark 2 that $M_2\left(h\right)$ has exactly four zeros by choosing appropriate coefficients $h\in (0,\frac{1}{2})$. □

3.3. The Third-Order Melnikov Function

Let $M_2\left(h\right)\equiv 0$, i.e., $l_1=l_3=l_4=l_5=l_6=0$ of Theorem 2. By using the Maple software, there are 28 cases. Some cases are short; however, some cases are very long, and we cannot continue to compute the higher-order Melnikov functions. Here, we only show five cases.

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill Case& \left(1\right)a_{02}=3b_{21},a_{20}=2b_{21},a_{30}=b_{01}=b_{21},b_{02}=-a_{03}-\frac{1}{2}{a}_{11}+a_{21},\hfill \\ \hfill b_{03}=& -b_{21},b_{12}=a_{03};\hfill \\ \hfill Case& \left(2\right)a_{01}=a_{21}+\frac{4}{3}{a}_{03}-2b_{02},a_{02}=-3b_{03},a_{11}=2a_{21}+\frac{4}{3}{a}_{03}-2b_{02},\hfill \\ \hfill a_{20}=& -6b_{03},a_{30}=b_{21}=0,b_{01}=-6b_{03},b_{12}=\frac{5}{3}{a}_{03};\hfill \\ \hfill Case& \left(3\right)a_{01}=a_{21}+\frac{2}{3}{b}_{12}-2b_{02},a_{03}=\frac{2}{3}{b}_{12},a_{11}=2a_{21}+\frac{2}{3}{b}_{12}-2b_{02},\hfill \\ \hfill a_{20}=& b_{01},a_{02}=a_{30}=b_{03}=b_{21}=0;\hfill \\ \hfill Case& \left(4\right)a_{01}=a_{21}-2b_{02},a_{02}=3b_{21},a_{03}=b_{12},a_{11}=2(a_{21}-b_{02}),\hfill \\ \hfill a_{20}=& a_{30}=b_{21},b_{03}=-b_{21};\hfill \\ \hfill Case& \left(5\right)a_{02}=a_{20},a_{03}=b_{12}=0,b_{21}=a_{30}=-b_{03},\hfill \\ \hfill b_{00}=& -\frac{\left(a_{20}-b_{01}+b_{03}\right)\left(2a_{01}-a_{11}+2b_{02}\right)}{a_{20}+3b_{03}},\hfill \\ \hfill b_{10}=& -\frac{2a_1{a}_{20}+2a_{01}{b}_{03}-2a_{11}{b}_{01}-a_{11}{b}_{03}+4a_{20}{b}_{02}-2a_{20}{b}_{20}-2b_{01}{b}_2+2b_{01}{b}_{20}}{2(a_{20}+3b_{03})}\hfill \\ & -\frac{5b_{02}{b}_{03}-3b_{03}{b}_{20}}{2(a_{20}+3b_{03})},\hfill \\ \hfill b_{30}=& -\frac{a_{11}{b}_{03}-a_{20}{b}_{02}-a_{20}{b}_{20}-2b_{02}{b}_{03}-4b_{03}{b}_{20}}{2(a_{20}+3b_{03})}.\hfill \end{array}\end{array}$$

(42)

Assume that $M_1\left(h\right)=M_2\left(h\right)\equiv 0,$ then

$$q_1\omega ={\overline{q}}_{2}dH+d{\overline{Q}}_2+N_2.$$

(43)

Lemma 9.

In Case (1), there exists (43), where

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill q_1=& {\overline{q}}_1+{\tilde{q}}_1=f_1\left(x\right)+g_1(x,y),\hfill \\ \hfill f_1\left(x\right)=& \frac{{\left(a_{01}-a_{11}+a_{21}\right)}^2}{{\left(1+x\right)}^4}+\frac{2a_{03}\left(a_{01}-a_{11}+a_{21}+b_0-b_{10}+b_{20}-b_{30}\right)}{3{(1+x)}^3}\hfill \\ & +\frac{-2\left(a_{01}-a_{11}+a_{21}\right)\left(4a_{03}-a_{11}+2a_{21}\right)}{3{(1+x)}^3}\hfill \\ & -\frac{-2a_{03}^2+\left(3a_{01}-a_{11}-a_{21}-3b_{10}+6b_{20}-9b_{30}\right)a_{03}-6\left(a_{01}-a_{11}+a_{21}\right)a_{21}}{3{(1+x)}^2}\hfill \\ & +\frac{2a_{03}\left(-a_{21}+4a_{03}-a_{11}+3b_{20}-9b_{30}\right)}{3(1+x)}+a_{21}^2+2a_{03}\left(a_{21}-b_{30}\right)M,\hfill \end{array}\end{array}$$

(44)

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill g_1(x,y)=& \frac{y}{{(1+x)}^4}\left(a_{03}^2{y}^3+\frac{2}{3}{a}_{03}{b}_{21}\left(5x+2\right)y^2+\left(2\left(a_{01}-a_{11}+a_{21}\right)a_{03}+b_{21}^2\right)y\right.\hfill \\ & \left.-2\left(a_{01}-a_{11}+a_{21}\right)b_{21}\right)\hfill \\ & -\frac{y\left(\left(10a_{03}^2-a_{03}{a}_{11}+2a_{03}{a}_{21}+6b_{21}^2\right)y-3b_{21}\left(4a_{01}+2a_{03}-5a_{11}+6a_{21}\right)\right)}{3{\left(1+x\right)}^3}\hfill \\ & +\frac{y\left(2b_{21}\left(a_{03}+a_{21}\right)x+\left(2a_{03}{a}_{21}+b_{21}^2\right)y-b_{21}\left(2a_{03}-a_{11}+2a_{21}\right)\right)}{{\left(1+x\right)}^2},\hfill \\ \hfill {\overline{q}}_2=& q_1{q}_0+q_{1}h\left(x\right)+q_{20},\hfill \\ \hfill N_2=& \left(4b_{21}{a}_{03}\left(a_{21}-b_{30}\right)+{\xi }_{01}^0-{\eta }_{02}^1-3{\xi }_{03}^2-\frac{5}{3}{\xi }_{05}^4-{\eta }_{04}^3\right){\omega }_{01}^0+\left(4\left(a_{21}-b_{30}\right)a_{03}{b}_{21}\right.\hfill \\ & \left.-{\xi }_{01}^1-{\eta }_{02}^1-3{\xi }_{03}^2\right){\Delta }_{00}^0+\left({\eta }_{10}^0-\frac{7}{5}{\xi }_{07}^6-{\eta }_{06}^5\right){\delta }_{10}^0+\left(\frac{1}{5}{\xi }_{01}^6\right.\hfill \\ & +\frac{1}{5}{\left(2H-1\right)}^3\left(7{\xi }_{07}^6+5{\eta }_{06}^5\right)+{\left(2H-1\right)}^2\left({\xi }_{05}^6+{\eta }_{04}^5\right)\hfill \\ & \left.+\frac{1}{5}\left(5{\eta }_{02}^5+3{\xi }_{03}^6\right)\left(2H-1\right)\right){\delta }_{00}^5+\left(\frac{1}{4}{\xi }_{01}^5+{\eta }_{00}^4+\frac{1}{20}{\left(2H-1\right)}^2\left(20{\eta }_{04}^4\right.\right.\hfill \\ & \left.+120{\eta }_{06}^5+25{\xi }_{05}^5+168{\xi }_{07}^6\right)+\frac{1}{4}\left(4{\eta }_{02}^4+16{\eta }_{04}^5+3{\xi }_{03}^5+16{\xi }_{05}^6\right)\left(2H-1\right)\hfill \\ & \left.+\frac{6}{5}{\xi }_{03}^6+2{\eta }_{02}^5\right){\delta }_{00}^4+\left(3{\xi }_{03}^3-{\xi }_{03}^4-5{\xi }_{05}^5+{\xi }_{01}^2+{\xi }_{05}^6+{\eta }_{04}^5\right.\hfill \\ & -4{\eta }_{04}^4+2{\eta }_{02}^2-{\eta }_{02}^3+{\eta }_{00}^1+\frac{3}{5}\left(7{\xi }_{07}^6+5{\eta }_{06}^5\right)\left(2H-5\right)\hfill \\ & \left.-\frac{2}{3}\left(5{\xi }_{05}^4+3{\eta }_{04}^3\right)\left(2H-3\right)\right){\delta }_{00}^1+\left(\frac{1}{2}{\xi }_{01}^3-\frac{3}{4}{\xi }_{03}^5-{\eta }_{02}^4+{\eta }_{00}^2+2{\xi }_{03}^4-4{\xi }_{05}^6\right.\hfill \\ & -4{\eta }_{04}^5+2{\eta }_{02}^3+\frac{1}{6}\left(6{\eta }_{02}^2+24{\eta }_{04}^3+9{\xi }_{03}^3+40{\xi }_{05}^4\right)\left(2H-1\right)\hfill \\ & \left.-\frac{1}{2}\left(4{\eta }_{04}^4+5{\xi }_{05}^5\right)\left(2H-3\right)-\frac{4}{5}\left(7{\xi }_{07}^6+5{\eta }_{06}^5\right)\left(6H-5\right)\right){\delta }_{00}^2\hfill \\ & +\left(\frac{1}{3}{\xi }_{01}^4-\frac{3}{5}{\xi }_{03}^6-{\eta }_{02}^5+{\eta }_{00}^3+\frac{3}{2}{\xi }_{03}^5+2{\eta }_{02}^4\right.\hfill \\ & +\left({\eta }_{02}^3+4{\eta }_{04}^4+{\xi }_{03}^4+5{\xi }_{05}^5\right)\left(2H-1\right)+\frac{1}{3}{\left(2H-1\right)}^2\left(5{\xi }_{05}^4+3{\eta }_{04}^3\right)\hfill \\ & \left.-2\left(2H-3\right)\left({\xi }_{05}^6+{\eta }_{04}^5\right)-\frac{3}{5}\left(2H-1\right)\left(2H-5\right)\left(7{\xi }_{07}^6+5{\eta }_{06}^5\right)\right){\delta }_{00}^3.\hfill \end{array}\end{array}$$

(45)

Proof. 

Under Case (1), there exist $M_1\left(h\right)=M_2\left(h\right)\equiv 0,$ Together with Lemma 2, $N_1$ can be written as $N_1={\tilde{q}}_{1}dH+d{\tilde{Q}}_1$ by $\underset{H=h}{\oint }{N}_1=0.$ By Case (1), we have

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill N_1=& \frac{1}{3}{b}_{21}\left(2\left(2H-1\right)\left(4H{a}_{03}+3a_{01}-2a_{03}-3a_{11}+3a_{21}\right){\delta }_{00}^3+\left(8H{a}_{03}\right.\right.\hfill \\ & \left.+9a_{01}-4a_{03}-9a_{11}+9a_{21}\right){\delta }_{00}^2\left.-2a_{03}\left(8H{\delta }_{00}^1+3{\Delta }_{00}^0-3{\delta }_{00}^1+8{\omega }_{01}^0\right)\right)\hfill \\ \hfill =& \left(\frac{4a_{03}{b}_{21}{y}^4}{3{\left(1+x\right)}^3}-\frac{b_{21}\left(4a_{03}{x}^2\left(x+2\right)-2\left(3a_{01}+a_{03}-3a_{11}+3a_{21}\right)x\right)}{3{\left(1+x\right)}^2}\right.\hfill \\ & \left.-\frac{b_{21}\left(-3a_{01}-6a_{03}+3a_{11}-3a_{21}\right)}{3{\left(1+x\right)}^2}-\frac{2b_{21}\left(2x{a}_{03}-a_{01}+2a_{03}+a_{11}-a_{21}\right)y^2}{{\left(1+x\right)}^3}\right)\mathit{dy}\hfill \\ & -\frac{2}{3}{a}_{03}{b}_{21}\left(3ln(1+x)\mathit{dy}+8y\mathit{dx}\right)\hfill \end{array}\end{array}$$

(46)

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill =& \widetilde{q_1}dH+d{\tilde{Q}}_1,\mathrm{where}\hfill \\ \hfill {\tilde{q}}_1=& -\frac{2b_{21}y\left(3a_{03}{x}^2-2y^2{a}_{03}+9x{a}_{03}-3a_{01}+6a_{03}+3a_{11}-3a_{21}\right)}{3{\left(1+x\right)}^3},\hfill \\ \hfill d{\tilde{Q}}_1=& d\left(\frac{2b_{21}{a}_{03}{y}^3\left(1+2x\right)}{3{\left(1+x\right)}^2}+b_{21}\left(\frac{2\left(a_{03}+a_{01}-a_{11}+a_{21}\right)}{1+x}-\frac{\left(a_{01}-a_{11}+a_{21}\right)}{{\left(1+x\right)}^2}\right.\right.\hfill \\ & \left.\left.-\frac{2}{3}{b}_{21}{a}_{03}\left(2x+3M\right)\right)y\right).\hfill \end{array}\end{array}$$

(47)

By Lemma 2, $q_1={\overline{q}}_1+{\tilde{q}}_1:=f_1\left(x\right)+g_1(x,y).$ Then, we have $q_1\omega =q_1{q}_{0}dH+q_{1}d{Q}_0,$ where

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill q_{1}d{Q}_0=& q_1({\Omega }_1\left(x\right)+{\Omega }_1(x,y)+h\left(x\right)dH)=q_{1}h\left(x\right)dH+f_1\left(x\right){\Omega }_1\left(x\right)\hfill \\ & +g_1(x,y){\Omega }_1\left(x\right)+q_1{\Omega }_2(x,y)\hfill \\ \hfill =& q_{1}h\left(x\right)dH+f_1\left(x\right){\Omega }_1\left(x\right)+{\displaystyle \sum _{k=j-1}^6}{\displaystyle \sum _{j=1}^7}{\xi }_{0j}^k{\omega }_{0j}^k+{\displaystyle \sum _{k=j-1}^5}{\displaystyle \sum _{j=2}^6}{\eta }_{0j}^k{\delta }_{0j}^k\hfill \\ & +{\displaystyle \sum _{k=0}^4}({\eta }_{00}^k{\delta }_{00}^k+{\eta }_{01}^k{\delta }_{01}^k)+{\eta }_{10}^0{\delta }_{10}^0+{\Xi }_{01}^0{\Omega }_{01}^0+{\Xi }_{02}^2{\Omega }_{02}^2+{\Xi }_{32}^2{\Omega }_{03}^2\hfill \\ & +{\overline{\eta }}_{02}^1{\Delta }_{02}^1+{\overline{\eta }}_{10}^0{\Delta }_{10}^0.\hfill \end{array}\end{array}$$

(48)

□

Theorem 3.

Let $M_1\left(h\right)=M_2\left(h\right)\equiv 0$ and Case (1) hold. $M_3\left(h\right)$ has exactly six simple zeros; moreover, system (4) has exactly six limit cycles by the third-order Melnikov function.

Proof. 

Let $z=\sqrt{1-2h}$, $z\in (0,1)$. By Lemmas 2–5, it yields

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill M_3\left(h\right)& =\underset{H=h}{\oint }{N}_2=\frac{b_{21}\pi }{24z^9}(z-1)(k_{10}{z}^{10}+k_9{z}^9+k_8{z}^8+k_7{z}^6(z+1)+k_5{z}^4(z+1)\hfill \\ & +k_3{z}^2(z+1)+k_1(z+1)),\hfill \end{array}\end{array}$$

(49)

It is easy to verify that $(z+1,z^2(z+1),z^4(z+1),z^6(z+1),z^8,z^9,z^{10})$ is an ECT system according to Lemma 6. Let $a_{11}=a_{21}=0;$ one can obtain that

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & \frac{\partial (k_1,k_3,k_5,k_7,k_8,k_9,k_{10})}{\partial (a_{01},a_{03},b_{00},b_{10},b_{20},b_{21},b_{30})}=-513684799488b_{21}{a}_{01}\left(160\left(a_{03}-3b_{30}\right)a_{01}^4\right.\hfill \\ & -4a_{03}\left(28a_{03}+40b_{10}+64b_{20}-247b_{30}\right)a_{01}^3+a_{03}^2\left(218a_{03}-16b_{10}+328b_{20}-777b_{30}\right)a_{01}^2\hfill \\ & \left.-2a_{03}^3\left(14a_{03}+16b_0-27b_{10}+97b_{20}-167b_{30}\right)a_{01}+56a_{03}^4\left(b_{00}-b_{10}+b_{20}-b_{30}\right)\right)\ne 0\hfill \end{array}\end{array}$$

(50)

by Remark 2, one can choose enough coefficients such that $M_3\left(h\right)$ has exactly six simple zeros. □

By taking similar steps, we consider Cases (2)–(5).

Theorem 4.

Let $M_1\left(h\right)=M_2\left(h\right)\equiv 0$, then

(i)

in Case (2), $M_3\left(h\right)$ has at most six zeros;

(ii)

in Case (3), $M_3\left(h\right)$ has exactly four simple zeros;

(iii)

in Case (4), $M_3\left(h\right)$ has exactly five simple zeros;

(iv)

in Case (5), $M_3\left(h\right)$ has exactly six simple zeros.

Proof. 

Let $z=\sqrt{1-2h}$. (i) In Case (2),

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill M_3\left(h\right)& =M_3\left(\frac{1-z^2}{2}\right)=\underset{H=h}{\oint }{N}_2\hfill \\ & =-\frac{a_{03}{b}_{03}\pi }{18z^7}[32a_{03}{z}^9-4\left(46a_{03}+45a_{21}-45b_{30}\right)z^8+3\left(262a_{03}+360a_{21}\right.\hfill \\ & \left.+15b_2+60b_{20}-480b_{30}\right)z^7-3\left(386a_{03}+444a_{21}+105b_{02}-40b_{10}\right.\hfill \\ & \left.+380b_{20}-1164b_{30}\right)z^6-3\left(2a_{03}-72a_{21}-15b_{00}-99b_{02}+155b_{10}\right.\hfill \\ & \left.-367b_{20}+651b_{30}\right)z^4\left(z+1\right)+27\left(4a_{03}-17b_{00}-6b_{02}+25b_{10}-33b_{20}\right.\hfill \\ & \left.+41b_{30}\right)z^2\left(z+1\right)+270\left(b_{00}-b_{10}+b_{20}-b_{30}\right)\left(z+1\right)].\hfill \end{array}\end{array}$$

(51)

Here, we can prove that the function $(z+1,z^2(z+1),z^4(z+1),z^6,z^7,z^8,z^9)$ is an ECT system. However, the corresponding seven coefficients are not independent, and $M_3\left(h\right)$ has at most six zeros.

(ii) In Case (3), by Lemmas 2–5

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill M_3\left(h\right)& =M_3\left(\frac{1-z^2}{2}\right)=\underset{H=h}{\oint }{N}_2\hfill \\ & =\frac{b_{12}{b}_{01}\pi (z-1)}{27z^7}[\left(108a_{21}+54b_{12}-108b_{30}\right)z^7+\left(-180a_{21}-36b_{02}\right.\hfill \\ & \left.-74b_{12}-108b_{20}+396b_{30}\right)z^6+\left(36a_{21}+45b_{02}-54b_{10}+b_{12}+144b_{20}\right.\hfill \\ & \left.-270b_{30}\right)z^4\left(z+1\right)-9\left(7b_{00}+3b_{02}-11b_{10}-b_{12}+15b_{20}-19b_{30}\right)\hfill \\ & z^2\left(z+1\right)+45\left(z+1\right)\left(b_0-b_{10}+b_{20}-b_{30}\right)].\hfill \end{array}\end{array}$$

(52)

Since the function $(z+1,z^2(z+1),z^4(z+1),z^6,z^7)$ is an ECT system and the corresponding five coefficients are free parameters, by Lemma 6 and Remark 2, $M_3\left(h\right)$ has exactly four simple zeros.

(iii) In Case (4),

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill M_3\left(h\right)=& M_3\left(\frac{1-z^2}{2}\right)=\frac{\pi (z-1)}{12z^7}[K_8{z}^8+K_7{z}^7+K_6{z}^6+K_5{z}^4\left(z+1\right)\hfill \\ & +K_3{z}^2\left(z+1\right)+K_1\left(z+1\right)].\hfill \end{array}\end{array}$$

(53)

One can verify that the function $(z+1,z^2(z+1),z^4(z+1),z^6,z^7,z^8)$ is an ECT system, and when $b_{02}=0$, then we have

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & \frac{\partial (K_1,K_3,K_5,K_6,K_7,K_8)}{\partial (a_{21},b_{30},b_{00},b_{10},b_{20},b_{12})}=-29859840b_{01}{b}_{12}^3{b}_{21}^2\left(5b_{01}-2b_{21}\right)\left(36a_{21}^2{b}_{21}^2\left(b_{00}\right.\right.\hfill \\ & \left.-b_{10}+b_{20}-b_{30}\right)-b_{12}\left(\left(8b_{00}-11b_{10}-21b_{12}+20b_{20}-29b_{30}\right)a_{21}\right.\hfill \\ & \left.-21b_{12}\left(4b_{00}-b_{10}\right)\right)b_{21}^2+3b_{01}{b}_{12}\left(\left(32b_{00}-36b_{10}-29b_{12}+48b_{20}-60b_{30}\right)a_{21}\right.\hfill \\ & \left.\left.-b_{12}\left(96b_{00}-12b_{10}-8b_{20}-b_{30}\right)\right)b_{21}+12b_{01}^2{b}_{12}^2\left(a_{21}+8b_{20}-17b_{30}\right)\right)\ne 0.\hfill \end{array}\end{array}$$

(54)

By Lemma 6 and Remark 2, $M_3\left(h\right)$ has exactly five simple zeros.

(iv) In Case (5),

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill M_3\left(h\right)=& M_3\left(\frac{1-z^2}{2}\right)\hfill \\ & =-\frac{\pi (z-1)}{4(a_{20}+3b_{03})z^9}[c_{10}{z}^{10}+c_9{z}^9+c_8{z}^8+c_7{z}^6(z+1)+c_5{z}^4(z+1)+\hfill \\ & c_3{z}^2(z+1)+c_1(z+1)],\hfill \end{array}\end{array}$$

(55)

By Lemma 6, we know that $(z+1,z^2(z+1),z^4(z+1),z^6(z+1),z^8,z^9,z^{10})$ is an ECT system. Furthermore, we have $\frac{\partial (c_1,c_3,c_5,c_7,c_8,c_9,c_{10})}{\partial (a_{11},a_{01},a_{20},b_{01},b_{20},b_{02},b_{03})}\ne 0;$ then $M_3\left(h\right)$ has exactly six simple zeros by Remark 2. □

4. Conclusions

Based on the previous results, by deducing some new formulae in Lemmas 3 and 5, the first three Melnikov functions of system (4) with $n=3$ are considered. There are many third-order Melnikov functions; since some cases are difficult to compute, we just list five cases. We obtain exactly six limit cycles. It is difficult to obtain the conditions of the existence of the fourth-order Melnikov functions; we do not continue to consider the fourth-order Melnikov function.