1 Introduction

Consider differential systems of the form

$$\begin{aligned} \left\{ \begin{array}{ll} x'=-y+\sum _{k=2}^mP_k(x, y),\\ y'=x+\sum _{k=2}^m Q_k(x, y), \end{array} \right. \end{aligned}$$
(1)

where \(P_k\) and \(Q_k\) are homogeneous polynomials in x and y of degree k . The problem of determining necessary and sufficient conditions on \(P_k\) and \(Q_k\) for system (1) to have a center at the origin is known as the center-focus problem. Studying this problem has a long history. The first works are due to Poincar\(\acute{e}\) [14] and Dulac [8], and continued by Lyapunov [9] and many others. Unfortunately, the center-focus problem has been solved only for quadratic systems and some special systems [1, 2, 6, 7, 13, 15, 16] and references therein. Up to now, very little is known about the center conditions for polynomial differential systems with arbitrary degree \(m\,(m>2)\).

A center of (1) is called a Weak Center if the Poincar\(\acute{e}\)-Liapunov first integral can be written as \( H =\frac{1}{2}(x^2+y^2)(1+ h.o.t..).\) By literature [10,11,12], we know that a center of polynomial differential system (1) is a weak center if and only if it can be written as

$$\begin{aligned} \left\{ \begin{array}{@{}ll} x'=-y(1+\varLambda )+x\varOmega ,\\ y'=x(1+\varLambda )+y\varOmega , \end{array} \right. \end{aligned}$$
(2)

where \(\varLambda =\varLambda (x, y)\) and \(\varOmega =\varOmega (x,y)\) are polynomials of degree at most \(m-1\) such that \(\varLambda (0, 0) =\varOmega (0, 0) =0\). The weak centers contain the uniform isochronous centers and the holomorphic isochronous centers [10], they also contain the class of centers studied by Alwash and Lloyd [5], but they do not coincide with all classes of isochronous centers [10]. The class of differential systems (2) is called the \(\varLambda -\varOmega \) System.

In [11], the authors put forward such conjectures:

Conjecture 1

The polynomial differential system of degree \(m+1\)

$$\begin{aligned} \left\{ \begin{array}{ll} x'=-y(1+\mu (a_2x-a_1y))+x((a_1x+a_2y)+\varPhi _{m}),\\ y'=x(1+\mu (a_2x-a_1y))+y((a_1x+a_2y)+\varPhi _{m}), \end{array} \right. \end{aligned}$$
(3)

where \( (\mu +m-1)(a_1^2+a_2^2)\ne 0\) and \(\varPhi _{m}=\varPhi _{m}(x,y)\) is a homogenous polynomial of degree m, has a weak center at the origin if and only if system (3) after a linear change of variables \((x, y)\rightarrow (X, Y)\) is invariant under the transformations \((X, Y, t)\rightarrow (-X, Y, -t).\)

Conjecture 2

The polynomial differential system of degree \(m+1\)

$$\begin{aligned} \left\{ \begin{array}{ll} x'=-y(1+a_1x+a_2y)+x\varPhi _{m},\\ y'=x(1+a_1x+a_2y)+y\varPhi _{m} \end{array} \right. \end{aligned}$$
(4)

has a weak center at the origin if and only if system (4) after a linear change of variables \((x, y)\rightarrow (X, Y)\) is invariant under the transformations \((X, Y, t)\rightarrow (-X, Y, -t).\)

They have used Poincar\(\acute{e}\)-Liapunov first integral and Reeb inverse integrating factor to prove Conjecture 1 and Conjecture 2 hold for \(m =1,2,...,5.\) They remarked that the only difficulty for proving conjectures 1 and 2 for the \(\varLambda -\varOmega \)-systems of degree m with \(m >5\) is the huge number of computations for obtaining the conditions that characterize the centers.

In this paper, we use a method different from Llibre et al. [11] and a simple transformation, without huge number of computation, to prove that under several restrictive conditions, the \(\varLambda -\varOmega \) system

$$\begin{aligned} \left\{ \begin{array}{ll} x'=-y(1+\mu _1(a_2x-a_1y))+x(\mu _2(a_1x+a_2y)+{P}_2(x,y)+{P}_m(x,y)),\\ y'=x(1+\mu _1(a_2x-a_1y))+y(\mu _2(a_1x+a_2y)+{P}_2(x,y)+{P}_m(x,y)) \end{array} \right. \end{aligned}$$
(5)

has a weak center at the origin if and only if

$$\begin{aligned} \int _0^{2\pi }\sin ^{2i}\theta \,P_{2}(\cos \theta ,\sin \theta )\,\mathrm{d}\theta =0\,(i=0,1) \end{aligned}$$
(6)

and

$$\begin{aligned} \int _0^{2\pi }\sin ^j\theta \,P_{m}(\cos \theta ,\sin \theta )\,\mathrm{d}\theta =0\,(j=0,1,2,...,m). \end{aligned}$$
(7)

Here and in the following \({P}_2(x,y), {P}_m(x,y) \) are, respectively, homogenous polynomials of degree 2 and m. As corollaries, we also show that for arbitrary m, the Conjecture 1 under several restrictive conditions and the Conjecture 2 are valid.

2 Several Lemmas

In polar coordinates, the system (1) becomes

$$\begin{aligned} \frac{\mathrm{d}r}{\mathrm{d}\theta } =\frac{\sum _{k=2}^{m}R_{k}(\theta )r^k}{1+\sum _{k=2}^m \varTheta _{k}(\theta )r^{k-1}}, \end{aligned}$$

where

$$\begin{aligned}&R_{k}(\theta ) =\cos \theta P_k(\cos \theta ,\sin \theta )+\sin \theta Q_k(\cos \theta , \sin \theta ), \\&\quad \varTheta _{k}(\theta ) =\cos \theta Q_k(\cos \theta , \sin \theta )-\sin \theta P_k(\cos \theta ,\sin \theta ). \end{aligned}$$

Lemma 2.1

[4] If there exists a trigonometric polynomial \(u(\theta )\) such that

$$\begin{aligned} R_k(\theta ) =u'(\theta )\sum r_{k j}u^j(\theta ),\, \varTheta _k(\theta ) =\sum \theta _{k j}u^j(\theta )\,\,(k=2,3,\ldots ,m), \end{aligned}$$
(8)

where \(r_{k j} ,\, \theta _{k j}\) are real numbers, then the origin point of (1) is a center and this center is called Composition Center.

The condition (8) is called Composition Condition [3, 4]. In [3], it was shown that the composition condition is a sufficient but not necessary for the origin to be a center of (1).

Lemma 2.2

[17] If

$$\begin{aligned} P_m(x,y)=\sum _{i+j=m}p_{ij}x^iy^j,\, p_{ij}\in R, \end{aligned}$$

which satisfies (7), then

$$\begin{aligned} P_m(\cos \theta ,\sin \theta )=\cos \theta \sum _{i=1}^{m}\lambda _i\sin ^{i-1}\theta \end{aligned}$$

and

$$\begin{aligned} P_m(x,y)=x\phi (x^2+y^2,y),\end{aligned}$$
(9)

where \(\lambda _i (i=1,2,\ldots ,m)\) are real numbers.

Proof

By (7) we get \(\int _0^{2\pi } P_m(\cos \theta ,\sin \theta )\mathrm{d}\theta =0,\) so \(P_m=\sum _{k=1}^m (a_k\cos k\theta +b_k\sin k\theta ),\) where

$$\begin{aligned} a_k=\frac{1}{\pi }\int _0^{2\pi } P_m(\cos \theta ,\sin \theta )\cos k\theta \, \mathrm{d}\theta ,\, b_k=\frac{1}{\pi }\int _0^{2\pi } P_m(\cos \theta ,\sin \theta )\sin k\theta d\theta . \end{aligned}$$

Applying (7) we obtain

$$\begin{aligned} b_{2i-1}=0,\, a_{2i}=0,\,( i=1,2,3,\ldots ) \end{aligned}$$

Therefore,

$$\begin{aligned} P_m(\cos \theta ,\sin \theta )&=a_1\cos \theta +b_2\sin 2\theta +a_3\cos 3\theta \\&\quad +b_4\sin 4\theta +\cdots +a_{2k-1}\cos (2k-1)\theta +b_{2k}\sin 2k\theta +\ldots \end{aligned}$$

As

$$\begin{aligned}&\sin 2k\theta =\cos \theta \sin \theta \left( \sum _{j=1}^{k}(-1)^{j-1}C_{2k}^{2j-1}\cos ^{2k-2j}\theta \sin ^{2(j-1)}\theta \right) , \\&\quad \cos (2k+1)\theta =\cos \theta \left( \sum _{j=0}^k(-1)^jC_{2k+1}^{2j}\cos ^{2k-2j}\theta \sin ^{2j}\theta \right) , \end{aligned}$$

thus,

$$\begin{aligned} P_m(\cos \theta ,\sin \theta )=\cos \theta \sum _{i=1}^{m}\lambda _i\sin ^{i-1}\theta =\cos \theta \sum _{i=1}^{m}\lambda _i\sin ^{i-1}\theta (\cos ^2\theta +\sin ^2\theta )^{\frac{m-i}{2}}, \end{aligned}$$

Thus, the conclusion of the present lemma holds.

In the following we denote: \(\bar{P}=\int _0^{\theta }P(\cos \tau ,\sin \tau )\mathrm{d}\tau ,\,\overline{\sin ^k\theta P}=\int _0^{\theta }sin^k\tau P(\cos \tau ,\sin \tau )\mathrm{d}\tau .\) \(\square \)

Lemma 2.3

Suppose that

$$\begin{aligned}&\frac{\mathrm{d}\varGamma _0}{\mathrm{d}\theta }=e_0P_m;\,\frac{\mathrm{d}\varGamma _1}{\mathrm{d}\theta }=e_1\sin \theta P_m; \end{aligned}$$
(10)
$$\begin{aligned}&\quad \frac{\mathrm{d}\varGamma _k}{\mathrm{d}\theta }=\sum _{j=1}^{[\frac{k}{2}]}\sum _{i=0}^{k-2j}jA_{3+i}C_{3+i}^j\nonumber \\&\quad \sum _{i_1+i_2+\cdots +i_j=k-2j-i}\delta _{i_1}\delta _{i_2}...\delta _{i_{j}-1}\varGamma _{i_j}\sin ^{k-2-i_j-i}\theta +\eta _k\sin ^k\theta P_m, \nonumber \\&\quad (k=2,3,4,\ldots ,m) \end{aligned}$$
(11)

where

$$\begin{aligned}&A_{3+i}=p_{11} d_i\sin ^{i+1}\theta \cos \theta ; \\&\quad \delta _k=\frac{p_{11}}{k+2}\left( d_k+\sum _{j=1}^{[\frac{k}{2}]}\sum _{i=0}^{k-2j}C_{3+i}^jd_i\sum _{i_1+i_2+\cdots +i_j=k-2j-i}\delta _{i_1}\delta _{i_2}\ldots \delta _{i_j}\right) ; \end{aligned}$$

\(d_i,e_i,\eta _i, p_{11}\) are real numbers, \(P_m=P_m(\cos \theta ,\sin \theta ).\) Then

$$\begin{aligned}&\varGamma _0=\gamma _0^0\bar{P}_m,\,\gamma _0^0=e_0,\,\varGamma _1=\gamma _1^1\overline{\sin \theta P_m},\,\gamma _1^1=e_1,\end{aligned}$$
(12)
$$\begin{aligned}&\quad \varGamma _k=\sum _{j=0}^{k-2}\gamma _k^j\sin ^{k-j}\theta \overline{\sin ^j\theta P_m}+\gamma _k^k\overline{\sin ^k\theta P_m},\,(k=2,3,4,...,m)\end{aligned}$$
(13)
$$\begin{aligned}&\quad \gamma _k^j=C_{k+1-j}^1\delta _{k-2-j}\gamma _j^j,\,(j=0,1,2,\ldots ,k-2),\, \gamma _{k}^{k-1}=0,\,\gamma _k^k=\eta _k-\sum _{j=0}^{k-2}\gamma _k^j.\nonumber \\ \end{aligned}$$
(14)

Proof

Obviously, solving (10), we get (12). By (11), we obtain

$$\begin{aligned} \varGamma _2'=A_3C_3^1\varGamma _0+\eta _2\sin ^2\theta P_m=C_3^1d_0\lambda \cos \theta \sin \theta \gamma _0^0\bar{P}_m+\eta _2\sin ^2\theta P_m, \end{aligned}$$

solving this equation we have

$$\begin{aligned} \varGamma _2=\gamma _2^0\sin ^2\theta \bar{P}_m+\gamma _2^2\overline{\sin ^2\theta P_m}, \end{aligned}$$

where

$$\begin{aligned} \gamma _2^0=C_3^1\delta _0\gamma _0^0,\,\gamma _2^1=0,\,\gamma _2^2=\eta _2-\gamma _2^0. \end{aligned}$$

This means that the conclusion of the present lemma is valid when \(k=2.\) Now suppose that the conclusion of the present lemma is valid for \(k\le l-1\), next we will show that for \(k=l\) the conclusion is correct, too. In fact, using (11) and inductive hypothesis, we get

$$\begin{aligned}&\varGamma _k'=\sum _{j=1}^{[\frac{k}{2}]}\sum _{i=0}^{k-2j}jp_{11} d_i C_{3+i}^j\sum _{i_1+i_2+\cdots +i_j=k-2j-i}\delta _{i_1}\delta _{i_2}\ldots \delta _{i_{j}-1} \sin ^{k-1-i_j}\theta \cos \theta \\&\quad \left( \sum _{l=0}^{i_j-2}\gamma _{i_j}^l\sin ^{i_j-l}\theta \overline{\sin ^l\theta P_m}+\gamma _{i_j}^{i_j}\overline{\sin ^{i_j}\theta P_m}\right) +\eta _k\sin ^k\theta P_m, \end{aligned}$$

solving this equation we get

$$\begin{aligned} \varGamma _k=\sum _{l=0}^{k-2}\gamma _k^l\sin ^{k-l}\theta \overline{\sin ^l\theta P_m}+\gamma _k^k\overline{\sin ^k\theta P_m}, \end{aligned}$$

in which

$$\begin{aligned}&\gamma _k^l=\frac{p_{11}}{k-l}\sum _{j=1}^{[\frac{k-l-2}{2}]}\sum _{i=0}^{k-l-2-2j}jC_{3+i}^jd_i\sum _{i_1+i_2+\cdots +i_j=k-l-2-2j-i}\delta _{i_1}\delta _{i_2}\ldots \delta _{i_j-1}\gamma _{i_j}^l \\&\quad =\frac{p_{11}}{k-l}\sum _{j=1}^{[\frac{k-l-2}{2}]}\sum _{i=0}^{k-l-2-2j}jC_{3+i}^jd_i\sum _{i_1+i_2+\cdots +i_j=k-l-2-2j-i}\delta _{i_1}\delta _{i_2}\ldots \delta _{i_j-1}\delta _{i_j-2-l}C_{i_j+1-l}^1\gamma _{l}^l \\&\quad =C_{k+1-l}^1\frac{p_{11}}{k-l}(d_{k-2-l}+\sum _{j=1}^{[\frac{k-2-l}{2}]}\sum _{i=0}^{k-2-l-2j}C_{3+i}^jd_i\sum _{i_1+i_2+\cdots +i_j=k-2-l-2j-i}\delta _{i_1}\delta _{i_2}\ldots \delta _{i_j})\gamma _{l}^l \end{aligned}$$

      \(=C_{k+1-l}^1\delta _{k-2-l}\gamma _l^l,\,(l=0,1,2,\ldots ,k-2)\)

\( \qquad \gamma _k^{k-1}=0,\,\gamma _k^k=\eta _k-\sum _{i=0}^{k-2}\gamma _k^i.\)

Thus, by induction, the conclusion of the present lemma is valid. \(\square \)

3 Main Results

Consider the differential system (5).

If \(a_1^2+a_2^2\ne 0,\) taking the linear changes of variables

$$\begin{aligned} X=a_1x+a_2y,\, Y=-a_2x+a_1y, \end{aligned}$$

the system (5) becomes

$$\begin{aligned} \left\{ \begin{array}{ll} X'=-Y(1-\mu _1 Y)+X(\mu _2 X+\tilde{P}_{2}(X,Y)+\tilde{P}_{m}(X,Y)),\\ Y'=X(1-\mu _1 Y)+Y( \mu _2 X+\tilde{P}_{2}(X,Y)+\tilde{P}_{m}(X,Y)). \end{array} \right. \end{aligned}$$
(15)

Case 1 If \(\mu _2\ne 0\), applying the transformation: \(X=\frac{1}{\mu _2}x,\,Y=\frac{1}{\mu _2}Y,\) the system (15) becomes

$$\begin{aligned} \left\{ \begin{array}{ll} x'=-y(1-{\mu } y)+y(x+{P}_{2}(x,y)+{P}_{m}(x,y)),\\ y'=x(1-{\mu } y)+y(x+{P}_{2}(x,y)+{P}_{m}(x,y)), \end{array} \right. \end{aligned}$$
(16)

where \({\mu }=\frac{\mu _1}{\mu _2},\) \(\,P_k(x,y)=\sum _{i+j=k}p_{ij}x^iy^j,\,p_{ij}\in R,\,(k=2,m).\)

Obviously, if \({P}_{2}=x\psi (x^2+y^2,y),\,{P}_{m}=x\phi (x^2+y^2,y),\) then the \(\varLambda -\varOmega \) system (5) after a linear change of variables \((x, y)\rightarrow (X, Y)\) is invariant under the transformations \((X, Y, t)\rightarrow (-X, Y, -t).\) By Lemma 2.2, to find the necessary and sufficient conditions for (5) to have a weak center, only need to seek the conditions under which the identities (6) and (7) are held.

Theorem 3.1

If \(\mu =1\), \(m\ge 5\) and \(\prod _{i=2}^m\gamma _i^i p_{11}\ne 0,\) then the origin point of (16) is a center if and only if (6) and (7) are held. Moreover, this center is a composition center and weak center. Where \(\gamma _k^k (k=2,3,4,\ldots ,m)\) are the same as they in Lemma 2.3.

Proof In polar coordinates, the system (16) can be written as

$$\begin{aligned} \frac{dr}{d\theta }=\frac{r^2\cos \theta +P_{2}r^3+P_{m}r^{m+1}}{1- r\sin \theta }, \end{aligned}$$

where \(P_{2}=P_{2}(\cos \theta ,\sin \theta ),P_{m}=P_{m}(\cos \theta ,\sin \theta ).\)

Taking \(\rho =\frac{r}{e^{r\,\sin \theta }},\) the above equation becomes

$$\begin{aligned} \frac{\mathrm{d}\rho }{\mathrm{d}\theta }=\rho ^3e^{2r\sin \theta }P_{2}+\rho ^{m+1}e^{mr\sin \theta }P_{m} \end{aligned}$$
(17)

Applying the Lagrange-B\(\ddot{u}\)rman formula [1], we have

$$\begin{aligned} e^{m r\sin \theta }=1+m\sum _{n=1}^{\infty }\frac{(m+n)^{n-1}}{n!}\rho ^n\sin ^n\theta . \end{aligned}$$

Thus, the Eq. (17) can be written as

$$\begin{aligned} \frac{d\rho }{d\theta }=\sum _{k=3}^{\infty }A_k\rho ^k+\sum _{k=m+1}^\infty B_k\rho ^k, \end{aligned}$$
(18)

where

$$\begin{aligned}&A_k=d_{k-3}\sin ^{k-3}\theta \,P_2,\,B_k=e_{k-m-1}\sin ^{k-m-1}\theta \,P_m, \\&\quad d_0=e_0=1,\, d_k=\frac{2(k+2)^{k-1}}{k!},\,e_k=\frac{m(k+m)^{k-1}}{k!}, \, (k=1,2,\ldots ). \end{aligned}$$

Therefore, the system (16) has a center at (0, 0) if and only if all the solutions \(\rho (\theta )\) of Eq. (18) near \(\rho =0\) are periodic [1].

Let \( \rho (\theta , c)\) be the solution of (18) such that \( \rho (0,c)=c \,(0<c\ll 1).\) We write

$$\begin{aligned} \rho (\theta ,c)=c\sum _{n=0}^{\infty }a_n(\theta )c^n, \end{aligned}$$

where \(a_0(0)=1\) and \(a_n(0)=0\) for \(n\ge 1\). The origin of (18) is a center if and only if \(\rho (\theta +2\pi ,c)=\rho (\theta ,c)\), i.e., \(a_0(2\pi )=1,\,a_n(2\pi )=0\,(n=1,2,3, ...)\) [5].

Substituting \(\rho (\theta ,c)\) into (18) we obtain

$$\begin{aligned} c\sum _{i=0}^{\infty }a_i'(\theta )c^n&=\sum _{k=3}^{\infty }A_k(c\sum _{i=0}^{\infty }a_i(\theta )c^i)^{k}\nonumber \\&\quad + \sum _{k=m+1}^\infty B_k(c\sum _{i=0}^{\infty }a_i(\theta )c^i)^{k}. \end{aligned}$$
(19)

Equating the corresponding coefficients of \(c^n\) of (19) yields

$$\begin{aligned} a_0'(\theta )=0, a_0(0)=1;\,a_1'(\theta )=0, a_1(0)=0; \,a_2'(\theta )=A_3 a_0^3, a_2(0)=0, \end{aligned}$$

so

$$\begin{aligned} a_0(\theta )= 1,\,a_1(\theta )\equiv 0, a_{2}(\theta )=\bar{A}_{3}. \end{aligned}$$

Rewriting

$$\begin{aligned} \rho =c(1+c^{2}h),\, h=\sum _{i=0}^{\infty }h_i(\theta )c^i,\,h_i(0)=0,\, (i=0,1,2\ldots ). \end{aligned}$$

Substituting it into (18) we get

$$\begin{aligned} \sum _{k=0}^{\infty }h_k'(\theta )c^k=\sum _{k=3}^{\infty }A_kc^{k-3} \sum _{i=0}^kC_k^ih^ic^{2i}+\sum _{k=m+1}^\infty B_k c^{k-3}\sum _{i=0}^kC_k^ih^ic^{2i}. \end{aligned}$$
(20)

Equating the corresponding coefficients of \(c^k\) of the Eq. (20), we obtain

$$\begin{aligned} h_0=\bar{A}_3,\,h_1=\bar{A}_4,\,h_2=\bar{A}_5+\frac{3}{2}\bar{A}_3^2. \end{aligned}$$

As \(d_k\ne 0(k=0,1,2)\), from \(h_k(2\pi )=0 \,(k=0,1,2)\) follow that

$$\begin{aligned} \int _0^{2\pi }\sin ^k\theta \,P_{2}(\cos \theta ,\sin \theta )d\theta =0,\,(k=0,1,2) \end{aligned}$$

i.e., the condition (6) is a necessary condition for \(\rho =0\) to be a center. By Lemma 2.2 which implies that

$$\begin{aligned} P_{2}=p_{11}\cos \theta \sin \theta ,\,\bar{P}_{2}=\frac{p_{11}}{2}\sin ^2\theta . \end{aligned}$$
(21)

Applying (21) we get

$$\begin{aligned}&\int _{0}^{2\pi }\sin ^k\theta \,P_{2} d\theta =0,(k=0,1,2,\ldots ),\nonumber \\&\quad \bar{A}_k=\frac{p_{11}}{k-1}d_{k-3}\sin ^{k-1}\theta ,\, \bar{A}_k(2\pi )=0\,\,(k=3,4,\ldots ) .\end{aligned}$$
(22)
$$\begin{aligned}&\quad h_0=\delta _0\sin ^2\theta ,\,h_1=\delta _1\sin ^3\theta ,\,h_2=\delta _3\sin ^4\theta ,\end{aligned}$$
(23)
$$\begin{aligned}&\quad \delta _0=\frac{1}{2}d_0p_{11},\,\delta _1=\frac{1}{3}d_1 p_{11},\,\delta _2=\frac{1}{4}p_{11}(d_2+C_3^1d_0\delta _0). \end{aligned}$$
(24)

Equating the corresponding coefficients of \(c^{k}\) of the Eq. (20) we obtain

$$\begin{aligned} h_{k}'=A_{k+3}+\sum _{j=1}^{[\frac{k}{2}]}\sum _{i=0}^{k-2j}A_{3+i}C_{3+i}^j\sum _{i_1+i_2+\cdots +i_j=k-2j-i}h_{i_1}h_{i_2}....h_{i_j} , \,(k=3,4,\ldots ,m-3) \end{aligned}$$

solving these equations and using (22–24) we get

$$\begin{aligned}&h_k=\delta _k\sin ^{k+2}\theta ,\,(k=0,1,2,\ldots ,m-3) \end{aligned}$$
(25)
$$\begin{aligned}&\quad \delta _k=\frac{p_{11}}{k+2}(d_k+\sum _{j=1}^{[\frac{k}{2}]}\sum _{i=0}^{k-2j}C_{3+i}^jd_i\sum _{i_1+i_2+...+i_j=k-2j-i}\delta _{i_1}\delta _{i_2}\ldots \delta _{i_j}),(k=0,1,2,\ldots )\nonumber \\ \end{aligned}$$
(26)

where \(C_k^j=\frac{k!}{j!(k-j)!}\) and when \(j>k, C_k^j=0.\)

By (20) we get

$$\begin{aligned} h_{m-2+k}'=B_{m+1+k}+\sum _{j=0}^{[\frac{k}{2}]}\sum _{i=0}^{k-2j}B_{m+1+i}C_{m+1+i}^j\sum _{i_1+i_2+\cdots +i_j=k-2j-i}h_{i_1}h_{i_2}...h_{i_j}+ \nonumber \\ A_{m+k+1}+\sum _{j=1}^{[\frac{m+k-2}{2}]}\sum _{i=0}^{m-2+k-2j}A_{3+i}C_{3+i}^j\sum _{i_1+i_2+\cdots +i_j=m-2+k-2j-i}h_{i_1}h_{i_2}....h_{i_j}, \nonumber \\ (k=0,1,2,...,m)\nonumber \\ \end{aligned}$$
(27)

solving these equations we get

$$\begin{aligned}&h_{m-2+k}=\delta _{m-2+k}\sin ^{m+k}\theta +\varGamma _k,\,(k=0,1,2,\ldots ,m-1),\end{aligned}$$
(28)
$$\begin{aligned}&\quad h_{2m-2}=\delta _{2m-2}\sin ^{2m}\theta +\varGamma _{m}+\frac{1}{2}C_{m+1}^1\bar{B}_{m+1}^2, \end{aligned}$$
(29)

where

$$\begin{aligned} \varGamma _0'=B_{m+1},\,\varGamma _1'=B_{m+2}, \end{aligned}$$
$$\begin{aligned}&\varGamma _k'=\sum _{j=1}^{[\frac{k}{2}]}\sum _{i=0}^{k-2j}jA_{3+i}C_{3+i}^j\sum _{i_1+i_2+\cdots +i_j=k-2j-i}\delta _{i_1}\delta _{i_2}\ldots \delta _{i_{j}-1}\varGamma _{i_j}\sin ^{k-2-i_j-i}\theta +\eta _k\sin ^k\theta P_m, \\&\quad (k=2,3,4,\ldots ,m) \\&\quad \eta _k=e_k+\sum _{j=1}^{[\frac{k}{2}]}\sum _{i=0}^{k-2j}C_{m+1+i}^je_i\sum _{i_1+i_2+\cdots +i_j=k-2j-i}\delta _{i_1}\delta _{i_2}\ldots \delta _{i_j}, \end{aligned}$$

\(\delta _k\) is represented by (26). By Lemma 2.3 we have

$$\begin{aligned}&\varGamma _0=\gamma _0^0\bar{P}_m,\, \varGamma _1=\gamma _1^1\overline{\sin \theta P_m},\gamma _0^0=e_0=1,\,\gamma _1^1=e_1=m, \nonumber \\&\quad \varGamma _k=\sum _{j=0}^{k-2}\gamma _k^j\sin ^{k-j}\theta \overline{\sin ^j\theta P_m}+\gamma _k^k\overline{\sin ^k\theta P_m},\,(k=2,3,4,\ldots ,m) \nonumber \\&\quad \gamma _k^j=C_{k+1-j}^1\delta _{k-2-j}\gamma _j^j,\,(j=0,1,2,\ldots ,k-2),\, \gamma _{k}^{k-1}=0,\,\gamma _k^k=\eta _k-\sum _{j=0}^{k-2}\gamma _k^j.\nonumber \\ \end{aligned}$$
(30)

Using (28) and (29), from \(h_{m-2+k}(2\pi )=0\) implies that \(\varGamma _k(2\pi )=0 \,(k=0,1,2,\ldots ,m).\) From \(\varGamma _i(2\pi )=0,\,(i=0,1)\) we get

$$\begin{aligned} \int _0^{2\pi }\sin ^i\theta P_m(\cos \theta ,\sin \theta )\mathrm{d}\theta =0,\,(i=0,1). \end{aligned}$$
(31)

Applying (30) and (31) and \(\gamma _2^2\ne 0\), from \(\varGamma _2(2\pi )=0\) follows that

$$\begin{aligned} \int _0^{2\pi }\sin ^2\theta P_m(\cos \theta ,\sin \theta )d\theta =0. \end{aligned}$$
(32)

Similar, using (30) and \(\gamma _k^k\ne 0\, (k=0,1,2,\ldots ,m),\) we get

$$\begin{aligned} \int _0^{2\pi }\sin ^k\theta P_m(\cos \theta ,\sin \theta )\mathrm{d}\theta =0, \,(k=0,1,2,\ldots ,m). \end{aligned}$$
(33)

In summary, under the assumptions of the present theorem, the (6) and (7) are the necessary conditions for \(\rho =0\) to be a center of (18). Therefore, the necessity has been proved. On the other hand, by Lemma 2.2, if the conditions (6) and (7) are satisfied, then \(P_2=p_{11}\cos \theta \sin \theta ,\,P_m=\cos \theta \sum _{i=1}^{m}\lambda _i\sin ^{i-1}\theta \), by Lemma 2.1 we see that \(\rho =0\) is a center and composition center of Eq. (18), this means that the sufficiency is proved. By Lemma 2.2 this center is a weak center.

The proof of the present theorem is finished.

Corollary 3.2

If \(\mu =1\) and \(p_{11}=0\), then the origin point of system (16) is a center, if and only if, \(p_{20}=p_{02}=0\) and (7) is satisfied.

Proof

From the proof process of the Theorem 3.1 we see that if \(\rho =0\) is a center of Eq. (18) then \(P_2=p_{11}\cos \theta \sin \theta =0\) and (6) is valid. So, \(\delta _k=0\, (k=0,1,2,...)\) and \(\gamma _k^k=\eta _k=e_k=\frac{m(k+m)^{k-1}}{k!}\ne 0.\) Therefore, the present corollary is directly deduced from Theorem 3.1. \(\square \)

Remark 1

By Corollary 3.2, when \(\mu =1\), the Conjecture 1 is correct for arbitrary m.

When \(\mu \ne 1\), the system (16) in polar coordinates becomes

$$\begin{aligned} \frac{dr}{d\theta }=\frac{r^2\cos \theta +P_{2}r^3+P_{m}r^{m+1}}{1- r\mu \sin \theta }. \end{aligned}$$
(34)

Taking \(\rho =\frac{r}{(1+(1-\mu )r\sin \theta )^{\frac{1}{1-\mu }}},\) the Eq. (34) becomes

$$\begin{aligned} \frac{d\rho }{d\theta }=\sum _{k=3}^{\infty }\tilde{A}_k\rho ^k+\sum _{k=m+1}^{\infty }\tilde{B}_k\rho ^k, \end{aligned}$$
(35)

where

$$\begin{aligned}&\tilde{A}_k=\tilde{d}_{k-3}sin^{k-3}\theta \, P_2,\,\tilde{B}_k=\tilde{e}_{k-m-1}\sin ^{k-m-1}\theta \,P_m, \\&\quad \tilde{d}_k=\frac{1+\mu }{k!}\prod _{i=0}^{k-2}(k+2\mu -i(1-\mu )),\,(k=2,3,\ldots ),\,\tilde{d}_0=1,\,\tilde{d}_1=1+\mu , \\&\quad \tilde{e}_k=\frac{m-1+\mu }{k!}\prod _{i=0}^{k-2}(k+m-2+2\mu -i(1-\mu )),\\&\quad (k=2,3,\ldots ),\tilde{e}_0=1,\,\tilde{e}_1=m-1+\mu . \end{aligned}$$

Similar to Theorem 3.1 we can get the following result.

Theorem 3.3

If \(\mu \ne 1,\) \(m\ge 5\) and \(\prod _{i=1}^m\tilde{\gamma }_i^ip_{11}\ne 0.\) Then the origin point of (16) is a center if and only if (6) and (7) are held. Moreover, this center is a composition center and weak center.

Where \(\tilde{\gamma }_k^k (k=1,2,3,\ldots ,m)\) are obtained by replacing \(d_k\) and \(e_k\) with \(\tilde{d}_k\) and \(\tilde{e}_k\), respectively, in Theorem 3.1.

Corollary 3.4

If \(\mu \ne 1\) and \(p_{11}=0\) and \(\prod _{i=1}^m\tilde{e}_i\ne 0,\) then the origin point (0,0) is a center of system (16) if and only if \(p_{20}=p_{02}=0\) and (7) is valid.

Remark 2

By Corollary 3.4, if \(\mu \ne 1\), the Conjecture 1 is valid for arbitrary m when \(\prod _{i=1}^m\tilde{e}_i\ne 0.\)

Case 2 \(\mu _2=0,\,\mu _1\ne 0.\)

Consider the differential system

$$\begin{aligned} \left\{ \begin{array}{ll} x'=-y(1-{\mu _1} y)+y({P}_{2}(x,y)+{P}_{m}(x,y)),\\ y'=x(1-{\mu _1} y)+y({P}_{2}(x,y)+{P}_{m}(x,y)). \end{array} \right. \end{aligned}$$
(36)

In polar coordinates, this system becomes

$$\begin{aligned} \frac{dr}{d\theta }=\sum _{k=3}^{\infty }\hat{A}_kr^k+\sum _{k=m+1}^{\infty }\hat{B}_kr^k, \end{aligned}$$
(37)

where

$$\begin{aligned}&\hat{A}_k=\hat{d}_{k-3}\sin ^{k-3}\theta \, P_2,\,\hat{B}_k=\hat{e}_{k-m-1}\sin ^{k-m-1}\theta \, P_m, \\&\quad \hat{d}_k=\mu _1^k,\,\hat{e}_k=\mu _1^k,\,(k=0,1,2,\ldots ) \end{aligned}$$

Similar to Theorem 3.1 we get the following conclusion.

Theorem 3.5

If \(\mu _2=0,\mu _1\ne 0,\) \(m\ge 5\) and \(\prod _{i=2}^m\hat{\gamma }_i^ip_{11}\ne 0 \), then the origin point of (16) is a center if and only if (6) and (7) are held. Moreover, this center is a composition center and weak center. Where \(\hat{\gamma }_k^k (k=2,3,4,\ldots ,m)\) are obtained by replacing \(d_k\) and \(e_k\) with \(\hat{d}_k\) and \(\hat{e}_k\), respectively, in Theorem 3.1.

Corollary 3.6

If \(\mu _2=0,\mu _1\ne 0\) and \(p_{11}=0\), then the origin point is a center of system (16) if and only if \(p_{20}=p_{02}=0\) and (7) is valid.

Remark 3

By Corollary 3.6 and [11], the Conjecture 2 is correct for arbitrary m.

Case 3 \(\mu _2=\mu _1=0\) or \(a_1^2+a_2^2=0.\)

Consider system

$$\begin{aligned} \left\{ \begin{array}{ll} x'=-y+y({P}_{2}(x,y)+{P}_{m}(x,y)),\\ y'=x+y({P}_{2}(x,y)+{P}_{m}(x,y)). \end{array} \right. \end{aligned}$$
(38)

By [18] we get the following result.

Theorem 3.7

\((\dot{1}).\) If \(p_{20}^2+p_{11}^2\ne 0\) and \(m=2k\), then (0, 0) is a center of (38) if and only if

$$\begin{aligned}&\int _0^{2\pi }P_2(\cos \theta ,\sin \theta )\mathrm{d}\theta =0, \\&\quad \int _0^{2\pi }P_{2k}(\cos \theta ,\sin \theta )\left( \int _0^{\theta }P_2(\cos \tau ,\sin \tau )\mathrm{d}\tau \right) ^i\mathrm{d}\theta ,\,(i=0,1,2,\ldots ,k). \end{aligned}$$

            \((\dot{2}).\) If \( p_{20}^2+p_{11}^2= 0,\) then (0, 0) is a center of (38) if and only if \(p_{02}=0\) and \(\int _0^{2\pi }P_m(\cos \theta ,\sin \theta )d\theta =0.\)

Summarizing, the center-focus problem of system (5) has been solved except for a few cases. In other cases, to get the center conditions is very difficult due to the large amount of calculation. We will discuss it in detail in the next article.