\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2018 (2018), No. 184, pp. 1--23.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2018 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2018/184\hfil 
 Center problem for  $\Lambda$-$\Omega$ differential systems]
{Center problem for generalized $\Lambda$-$\Omega$ \\ differential systems}

\author[J. Llibre, R. Ram\'irez, V. Ram\'irez \hfil EJDE-2018/184\hfilneg]
{Jaume Llibre, Rafael Ram\'irez, Valent\'in Ram\'irez}

\address{Jaume Llibre \newline
Departament de Matem\`atiques, 
Universitat Aut\`onoma de Barcelona,
08193 Bellaterra, Barcelona,
Catalonia, Spain}
\email{jllibre@mat.uab.cat}

\address{Rafael Ram\'irez \newline
Departament d'Enginyeria Inform\`{a}tica i Matem\`{a}tiques,
Universitat Rovira i Virgili, Avinguda dels Pa\"{\i}sos Catalans 26,
43007 Tarragona, Catalonia, Spain}
\email{rafaelorlando.ramirez@urv.cat}

\address{Valent\'in Ram\'irez \newline
Departament de Matem\`atiques,
Universitat Aut\`onoma de Barcelona,
08193 Bellaterra, Barcelona,
Catalonia, Spain}
\email{valentin.ramirez@e-campus.uab.cat}

\thanks{Submitted July 9, 2018. Published November 14, 2018.}
\subjclass[2010]{34C05, 34C07}
\keywords{Linear type center; Darboux first integral; weak center;
\hfill\break\indent Poincar\'e-Liapunov theorem; Reeb integrating factor}

\begin{abstract}
 $\Lambda$-$\Omega$ differential systems are the real  planar
 polynomial differential equations of degree $m$ of the form
 \[
 \dot{x}=-y(1+\Lambda)+x\Omega,\quad \dot{y}=x(1+\Lambda)+y\Omega,
 \]
 where $\Lambda=\Lambda(x,y)$ and $\Omega=\Omega(x,y)$ are
 polynomials of degree at most  $m-1$ such that $\Lambda(0,0)=\Omega(0,0)=0$.
 A planar vector field with linear type center can be written as a
 $\Lambda$-$\Omega$ system if and only if the Poincar\'e-Liapunov
 first integral is of the form $F=\frac{1}{2}(x^2+y^2)(1+O(x,y))$.
 The main objective of this article is to study the center problem for
 $\Lambda$-$\Omega$ systems of degree $m$ with
 $\Lambda=\mu(a_2x-a_1y)$, and
 $\Omega=a_1x+a_2y+\sum_{j=2}^{m-1}\Omega_j$, where
 $\mu,a_1,a_2$ are constants and  $\Omega_j= \Omega_j(x,y)$ is
 a homogenous polynomial of degree $j$, for $j=2,\dots,m-1$.
 We prove the following results. Assuming that $m=2,3,4,5$ and
 \[
 (\mu+(m-2))(a^2_1+a^2_2)\ne 0 \quad \text{and}\quad
 \sum_{j=2}^{m-2}\Omega_j\ne 0
 \]
 the  $\Lambda$-$\Omega$ system has a weak center at the origin
 if and only if these systems after a linear change of variables
 $(x,y)\to (X,Y)$ are invariant under the transformations
 $(X,Y,t)\to (-X,Y,-t)$.
 If $(\mu+(m-2))(a^2_1+a^2_2)=0$  and $\sum_{j=1}^{m-2}\Omega_j=0$
 then the origin is a  weak center. We observe that the main difficulty
 in proving this  result for $m>6$ is related to the huge computations.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{conjecture}[theorem]{Conjecture}
\allowdisplaybreaks

\section{Introduction}

Let $ \mathcal{X}=P\frac{\partial}{\partial x}+Q\frac{\partial}{\partial y}$ 
be the real planar polynomial vector field associated to the real
planar polynomial differential system
\begin{equation}\label{1}
\dot{x}=P(x,y),\quad \dot{y}=Q(x,y),
\end{equation}
where the dot denotes derivative with respect to an independent
variables here called the time $t$, and $P$ and $Q$ are real coprime
polynomials in $\mathbb{R}[x,y]$.  We say that the polynomial
differential system \eqref{1} has {\it degree}
$m=\max\{\deg{P},\,\deg{Q}\}$.

In what follows we assume that the  origin $O:=(0,0)$ is a singular
or equilibrium point, i.e. $P(0,0)=Q(0,0)=0$.

The equilibrium point $O$ is a {\it center} if there exists an open
neighborhood $U$ of $O$ where all the orbits contained in
$U\setminus \{O\}$ are periodic.

We shall work with the polynomial differential systems of degree $m$
such that
\begin{equation}\label{3}
\dot{x}=- y+X,\quad \dot{y}= x+Y,
\end{equation}
where  $X=X(x,y)$ and $Y=Y(x,y)$ are polynomials starting at least
with quadratic terms in the neighborhood of the origin, so
 $m=\max\{\deg{X},\,\deg{Y}\}\geq 2$. The {\it
center-focus problem} asks about conditions on the coefficients of
$X$ and $Y$ under which the origin of system \eqref{3}  is a center.
To know centers help for studying the limit cycles which can
bifurcate from the periodic orbits of the centers when we perturb
them, see for instance \cite{LLM}.

If a system \eqref{3} has a local first integral at the origin of
the form
\begin{equation*}\label{R0ZXY1}
F=\frac{1}{2}(x^2+y^2)\Phi(x,y),
\end{equation*}
where $\Phi=\Phi(x,y)$ is an analytic function such that
$\Phi(0,0)=1$, then the origin of system \eqref{3} is a center
called a {\it weak center.} The weak center contain the uniform
isochronous centers and the holomorphic isochronous centers (for a
prof of these results see \cite{LRV}), but they do not coincide with
the all class of isochronous centers (see \cite[Remark 19]{LRV}).

In this paper we shall study  the particular case of differential
systems \eqref{3} of the form
\begin{equation}\label{ZXY1}
\dot{x}=-y(1+\Lambda)+x\Omega,\quad \dot{y}=x(1+\Lambda)+y\Omega,
\end{equation}
where $\Lambda=\Lambda(x,y)$ and $\Omega=\Omega(x,y)$ are
polynomials such $m=\max\{\deg{\Lambda },\,\deg{\Omega}\}+1$.


 By applying the inverse approach in ordinary differential
equations see \cite{LR0} the following theorem is proved and shows
the importance of system  \eqref{ZXY1} in the theory of ordinary
differential equations (see \cite[Theorem 15]{LRV}).

\begin{theorem}\label{ValF1}
The polynomial differential system  \eqref{3}  has a weak center at
the origin if and only if it can be written as \eqref{ZXY1} with
\begin{gather*}
\Lambda=\sum_{j=2}^{m}\Big(\frac{j+1}{2}\Upsilon_{j-1}
+\frac{j}{2}g_1\Upsilon_{j-2}
+\dots+\frac{3}{2}g_{j-2}\Upsilon_1+g_{j-1}
\Big), \\
\Omega=\frac{1}{2}\sum_{j=2}^{m}\Big(\{\Upsilon_{j-1},H_2\}
+g_1\{\Upsilon_{j-2},H_2\} +\dots+g_{j-2}\{\Upsilon_{1},H_2\}\Big),
\end{gather*}
where $g_j$ and  $\Upsilon_j$ are homogenous polynomials of degree
$j$ for $j\geq 1$ and has a first integral of the form
\begin{equation*}
H=H_2\Phi=H_2(1+\mu_1\Upsilon_1+\dots+\mu_{m-1}\Upsilon_{m-1}),
\end{equation*}
where $H_2=(x^2+y^2)/2$, and $\mu_j=\mu_j(x,y)$ is a convenient
analytic function in the neighborhood of the origin for
$j=1,\dots,m-1$.
\end{theorem}

\section{Statement of main results}

In this section we  give the statements  of our main results which
will be proved in sections  \ref{ORaf3} and \ref{ORaf4}, also we
state some conjectures.

\begin{conjecture}\label{Conj2}
The polynomial differential system of degree $m$
\begin{equation}\label{centerR0}
\begin{gathered}
\dot{x}=-y(1+\mu(a_2x-a_1y))+x(a_1x+a_2y
 +\sum_{j=2}^{m-1}\Omega_j(x,y)), \\
\dot{y}=x(1+\mu(a_2x-a_1y))+y(a_1x+a_2y+\sum_{j=2}^{m-1}\Omega_j(x,y)),
\end{gathered}
\end{equation}
 under the assumptions
$(\mu+(m-2))(a^2_1+a^2_2)\ne 0$ and 
$\sum_{j=2}^{m-2}\Omega_j\ne 0 $, where
$\Omega_j=\Omega_j(x,y)$ is a homogenous polynomial of degree $j$
for $j=2,\dots,m-1$, has a weak center at the origin if and only if
system \eqref{centerR0} after a linear change of variables
$(x,y)\to (X,Y)$ is invariant under the transformations
$(X,Y,t)\to (-X,Y,-t)$.  Moreover differential system
\eqref{centerR0} in the variables $X,Y$ becomes
\begin{gather*} %\label{alfa1}
\dot{X}=-Y(1+\mu\,Y)+X^2\Theta(X^2,Y)=-Y(1+\mu\,Y)+X\{H_2,\Phi\}, \\
\dot{Y}=X(1+\mu\,Y)+XY\Theta(X^2,Y)=X(1+\mu\,Y)+Y\{H_2,\Phi\},
\end{gather*}
where $\Theta(X^2,Y)$ is a polynomial of degree $m-2$, and $\Phi$ is
a polynomial of degree $m-1$ such that
$\{H_2,\Phi\}=X\Theta(X^2,Y)$.
\end{conjecture}

The case when  $ (\mu+(m-2))(a^2_1+a^2_2)= 0$  
and $\sum_{j=2}^{m-2}\Omega_j= 0 $ was study in \cite{LRV1}.

\begin{theorem}\label{th2}
Conjecture \ref{Conj2} holds for $m=2,3$ and for $m=4$ with $\mu=0$.
\end{theorem}

The proof of Theorem \ref{th2} for $\mu=0$ and  $m=2$  goes back to
Loud \cite{Loud}. 
The proof of Theorem \ref{th2} for $\mu=0$ and
$m=3$ was done by Collins \cite{Collins}. 
The proof of Theorem \ref{th2} for $\mu=0$ and $m=4$  goes back to 
\cite{Algaba, Algaba1, chava}. However, in the proof of this last result 
there are some mistakes. The phase portraits of these systems are classified in
\cite{Artes, Itikawa, Itikawa2}. The proof that these centers are
 weak centers has been done in Theorem \ref{ValF1}.

\begin{conjecture}\label{Conj3}
Assume that the  polynomial differential system of degree $m-1$
\begin{gather*}% \label{centerR}
\dot{x}=-y(1+\mu(a_2x-a_1y))+x(a_1x+a_2y+\sum_{j=2}^{m-2}\Omega_j(x,y)),\\
\dot{y}=x(1+\mu(a_2x-a_1y))+y(a_1x+a_2y+\sum_{j=2}^{m-2}\Omega_j(x,y)),
\end{gather*}
where  $a_1a_2\ne 0$, and $\Omega_j=\Omega_j(x,y)$ is a
homogenous polynomial of degree $j$  for $j=2,\dots,m-2$, after a
linear change of variables $(x,y)\to (X,Y)$ it is
invariant under the transformations $(X,Y,t)\to (-X,Y,-t)$. 
Then the  polynomial differential system of degree $m$
\begin{gather*} % \label{RcenterR}
\dot{x}=-y(1+\mu(a_2x-a_1y))+x(a_1x+a_2y+\sum_{j=2}^{m-1}\Omega_j(x,y)), \\
\dot{y}=x(1+\mu(a_2x-a_1y))+y(a_1x+a_2y+\sum_{j=2}^{m-1}\Omega_j(x,y)),
\end{gather*}
has a weak center at the origin if and only if the system
\begin{equation}\label{RcenterR1}
\begin{gathered}
\dot{x}=-y(1+\mu(a_2x-a_1y))+x(a_1x+a_2y+\Omega_{m-1}(x,y)), \\
\dot{y}=x(1+\mu(a_2x-a_1y))+y(a_1x+a_2y+\Omega_{m-1}(x,y)),
\end{gathered}
\end{equation}
under the assumption $ (\mu+(m-2))(a^2_1+a^2_2)\ne 0$ and after a
linear change of variables $(x,y)\to (X,Y)$ it is
invariant under the transformations $(X,Y,t)\to (-X,Y,-t)$.
\end{conjecture}

The existence of the weak center of \eqref{RcenterR1} was solve in \cite{LRV1}.

\begin{theorem}\label{th3}
Conjecture \ref{Conj3} holds for $m=3,4,5,6$.
\end{theorem}

We note that when system \eqref{centerR0} with $\mu=0$ has a center
at the origin this center is a uniform isochronous center, i.e.\ if
we write these systems in polar coordinates $(r,\theta)$ we obtain
that $\dot{\theta}$ is constant. Clearly if $\mu=0$ then the weak
centers are uniform isochronous centers.
Also note that Conjecture \ref{Conj3} is a particular case of Conjecture
\ref{Conj2}.


\section{Preliminary results}

In the proofs of Theorems \ref{th2} and \ref{th3}, 
the following results and notation, which we can find in \cite{LRV},
plays a very important role.
As usual the {\it Poisson bracket} of the functions $f(x,y)$ and
$g(x,y)$ is defined as
\[
\{f,g\}:=\frac{\partial f}{\partial x}\frac{\partial g}{\partial
y}-\frac{\partial f}{\partial y}\frac{\partial g}{\partial x}.
\]
The following result is a simple consequence of the Liapunov result
given in \cite[Theorem 1, page 276]{Liapunov}.

 \begin{corollary}\label{Chetaevvv1}
Let $U=U(x,y)$ be a homogenous polynomial of degree $m$.   The linear
partial differential equation $\{H_2,V\}=U$,
 has a unique homogenous polynomial  solution $V$
of degree $m$ if $m$ is odd; and if $V$ is a homogenous polynomial
solution when $m$ is even then any other homogenous polynomial
solution is  of the form $V+c(x^2+y^2)^{m/2}$ with
$c\in\mathbb{R}$. Moreover, for $m$ even these solutions exist if
and only if
$\int_0^{2\pi}U(x,y)\big|_{x=\cos{t},\,y=\sin{t}}dt=0$.
\end{corollary}

\begin{proposition}[{see \cite[Proposition 6]{LRV}}] \label{TTT}
The  relation
\begin{equation*}\label{a55}
\int_0^{2\pi}\{H_2,\Psi\}\big|_{x=\cos{t},\,y=\sin{t}}dt=0
\end{equation*}
holds for an arbitrary $C^1$ function $\Psi=\Psi(x,y)$ defined in the
interval $[0,2\pi]$.
\end{proposition}

\begin{proposition}[{\cite[Proposition 24]{LRV}}] \label{important}
Consider the polynomial differential system \eqref{1} of degree $m$
which satisfies
\begin{equation*} %\label{xx1}
\int_0^{2\pi}\Big(\frac{\partial
P}{\partial\,x}+\frac{\partial
Q}{\partial\,y}\Big)\Big|_{x=\cos{t},\,y=\sin{t}} dt=0.
\end{equation*}
Then there exist polynomials $F=F(x,y)$ and $G=G(x,y)$ of degree
$m+1$ and $m-1$ respectively such that system \eqref{1} can be
written as
\begin{equation*} %\label{xx2}
\dot{x}=P=\{F,x\}+(1+G)\{H_2,x\},\quad
\dot{y}=Q=\{F,y\}+(1+G)\{H_2,y\},
\end{equation*}
with $G(0,0)=0$.
\end{proposition}

We need the following definitions and notion.

  A function $V=V (x, y)$ is an {\it inverse integrating factor} of
system \eqref{1} in an open subset $U\subset \mathbb{R}^2$ if 
$V\in C^1 (U), V \not\equiv 0$ in $U$ and 
$ \frac{\partial }{\partial x}\big(\frac{P}{V}\big) 
+\frac{\partial }{\partial y}\big(\frac{Q}{V}\big)=0 $.

\begin{theorem}[[Reeb 's criterion, \cite{Reeb}] \label{Lia33}
 The analytic differential system
\[
\dot{x}=-y+\sum_{j=2}^\infty,\quad
 X_j,\quad\dot{y}=x+\sum_{j=2}^\infty\, Y_j 
\]
 has a center at the origin if and only if there is a local nonzero analytic inverse
integrating factor of the form $V=1+\text{higher order terms}$
  in a neighborhood of the origin.
\end{theorem}

An analytic inverse integrating factor of the form $V=1+h.o.t$.  in a
neighborhood of the origin is called a \emph{Reeb inverse integrating
factor}. The analytic function
\begin{equation*}\label{a1}
H=\sum_{j=2}^\infty\,H_j(x,y)=\frac{1}{2}(x^2+y^2)
+\sum_{j=3}^\infty\,H_j(x,y),
\end{equation*}
where  $H_j$ is homogenous polynomials of degree $j>1$, is called
the {\it Poincar\'e-Liapunov local first integral} if $H$ is
constant on the solutions of \eqref{3}.

\begin{theorem}[{see \cite[Theorem 13 and Remark 14 ]{LRV}]\label{RRLia}}]
Consider the polynomial vector field 
$\mathcal{X}=(-y+\sum_{j=2}^m\, X_j)\frac{\partial }{\partial
x}+(x+\sum_{j=2}^m\, Y_j)\frac{\partial }{\partial y}$.
Then this vector field has a Poincar\'e-Liapunov local first
integral $H$ if and only if it has a Reeb inverse integrating factor
$V$. Moreover,  the differential system associated to the vector
field $\mathcal{X}$ for which $H=(x^2+y^2)/2+$h.o.t.  is a local first
integral can be written as
\begin{equation}\label{inverse1}
\begin{aligned}
\dot{x}=&V\{H,x\} \\
=&\{H_{m+1},x\}+(1+g_1)\{H_{m},x\}+\dots
+ (1+g_1+\dots+g_{m-1})\{H_2,x\}, \\
\dot{y}=&V\{H,y\} \\
=&\{H_{m+1},y\}+(1+g_1)\{H_{m},y\}+\dots +
(1+g_1+\dots+g_{m-1})\{H_2,y\},
\end{aligned}
\end{equation}
and   $V$ and $H$  are such that
\begin{equation}\label{RRR1}
\begin{gathered}
V=1+\sum_{j=1}^{\infty}g_j, \\
\begin{aligned}
H=&\frac{1}{2}(x^2+y^2)+\sum_{j=2}^{\infty}H_j
=\tau_{1}H_{m+1}+\tau_{2}H_m+\dots+\tau_mH_2 \\
=&\int_\gamma\Big(\frac{dH_{m+1}}{V}+\frac{(1+g_1)dH_{m}}{V}+\dots
+\frac{(1+g_1+\dots+g_{m-1})dH_2}{V}\Big),
\end{aligned}
\end{gathered}
\end{equation}
where $\gamma$ is an oriented curve (see \cite{stenberg}),
$\tau_j=\tau_j(x,y)$ is a convenient analytic
function in the neighborhood of the origin such that
$\tau_j(0,0)=1$, and $g_j=g_j(x,y)$  is an arbitrary homogenous
polynomial of degree $j$ which we choose in such a way that $V$ is
the inverse Reeb integrating factor which satisfies the first order
partial differential equation
\begin{equation}\label{RRR11}
\{H_{m+1},\frac{1}{V}\}+\{H_{m},\frac{1+g_1}{V}\}+\dots +
\{H_2,\frac{1+g_1+\dots+g_{m-1}}{V}\}=0.
\end{equation}
\end{theorem}

\begin{remark}[{see \cite[formula (44) and the proof of Theorem 13]{LLVR}}] 
\label{rem11} \rm
From \eqref{RRR11} and \eqref{RRR1} the following infinite number of equations
must hold
\begin{equation}\label{aqu}
\begin{gathered}
\{H_{m+1},g_1\}+\{H_{m},g_2\}+\dots +
\{H_3,g_{m-1}\}+\{H_2,g_{m}\}=0, \\
\begin{aligned}
&\{H_{m+1},g^2_1-g_2\}+\{H_{m},g_1g_2-g_3\}+\dots \\
&+\{H_3,g_1g_{m-1}-g_m\}+\{H_2,g_1g_{m}+g_{m+1}\}=0,
\end{aligned} \\
\dots\,.
\end{gathered}
\end{equation}
Consequently
\begin{equation}\label{aqu00}
\begin{gathered}
\int_0^{2\pi}\left(\{H_{m+1},g_1\}+\{H_{m},g_2\}+\dots
+ \{H_3,g_{m-1}\}\right)\big|_{x=\cos(t),\, y=\sin(t)}dt=0, \\
\begin{aligned}
&\int_0^{2\pi}\big(\{H_{m+1},g^2_1-g_2\}+\{H_{m},g_1g_2-g_3\}+\dots\\
&+\{H_3,g_1g_{m-1}-g_m\}\big)\big|_{x=\cos(t),\, y=\sin(t)}dt=0,
\end{aligned}\\
\dots\,.
\end{gathered}
\end{equation}
Conditions \eqref{aqu} and \eqref{aqu00} are equivalent to the following
relations.
\begin{equation}\label{euro}
\begin{gathered}
\{H_{m+j+1},g_1\}+\{H_{m+j},g_2\}+\dots +
\{H_3,g_{m+j-1}\}+\{H_2,g_{m+j}\}=0, \\
\begin{aligned}
&\int_0^{2\pi}\Big(\{H_{m+j+1},g_1\}+\{H_{m+j},g_2\}+\dots \\
&+ \{H_3,g_{m+j-1}\}\Big)\Big|_{x=\cos(t),y=\sin(t)}dt=0, 
\end{aligned}
\end{gathered}
\end{equation}
for $j\geq 0$.  Theorem \ref{RRLia} can be applied to  determine  the
Poincar\'e-Liapunov first integral,  Reeb inverse integrating
factor and Liapunov constants for the case when the polynomial
differential system is given (see \cite[section 8]{LRV}. Indeed,
given a polynomial vector field $\mathcal{X}$ of degree $m$ with a linear
type center at the origin of coordinates, using  \eqref{inverse1} we
determine its first integral $H$ and its Reeb inverse integrating
factor. Thus, if in \eqref{3} $X=\sum_{j=2}^mX_j$ and
$Y=\sum_{j=2}^mY_j$ with $X_j$ and $Y_j$ homogenous
polynomials of degree $j$, from \eqref{inverse1} and from the proof
of Theorem \ref{RRLia} equating the terms of the same degree we get
\begin{gather*} %\label{Mir30}
\{H_{j+1},x\}+g_1\{H_j,x\}+\dots + g_{j-1}\{H_2,x\}=X_j  \\
\{H_{j+1},y\}+g_1\{H_j,y\}+\dots + g_{j-1}\{H_2,y\}=Y_j,  \\
\{H_{k+1},x\}+g_1\{H_{k},x\}+\dots + g_{k-1}\{H_2,x\}=0  \\
\{H_{k+1},y\}+g_1\{H_{k},y\}+\dots + g_{k-1}\{H_2,y\}=0,
\end{gather*}
for $j=2,\dots,m$, and $k>m$.   Then the compatibility condition of
these equations are
\begin{equation}\label{Mir3}
\begin{gathered}
\{H_j,g_1\}+\dots
+\{H_2,g_{j-1}\}=\frac{\partial\,X_j}{\partial
x}+\frac{\partial\,Y_j}{\partial y}\quad \text{for }
j=2,\dots,m,  \\
\{H_k,g_1\}+\dots+\{H_2,g_{k-1}\}=0\quad\text{for } k>m,
\end{gathered}
\end{equation}
for $k>1$.

If \eqref{Mir3} holds then by considering that $H_n$ for $n>1$ are
homogenous polynomials of degree $n$.  Then applying  Euler's Theorem
for homogenous polynomials we obtain  the homogenous polynomial
$H_n$ as follows
\begin{equation}\label{Mir03}
\begin{gathered}
H_{j+1}=-\frac{1}{j+1}\left(yX_j-xX_j+jg_1H_j+\dots+2g_{j-1}H_2\right),  \\
H_{k+1}=-\frac{1}{k+1}\left(kg_1H_k+\dots+2g_{k-1}H_2\right),
\end{gathered}
\end{equation}
for $j=2,\dots,m$, and $k>m$.
\end{remark}
We need  the following results.


 Let
\begin{equation}\label{letra}
x=\kappa_1X-\kappa_2Y,\quad y=\kappa_2X+\kappa_1Y,
\end{equation}
 be a non-degenerated linear transformation, i.e.\
$\kappa^2_1+\kappa^2_2\ne 0$.   Then the differential system
\eqref{ZXY1} becomes
\begin{equation}\label{Rcomp1}
\begin{gathered}
\dot{X}=-Y\big(1+\tilde{\Lambda}(X,Y)\big)+X\tilde{\Omega}(X,Y), \\
\dot{Y}=X\big(1+\tilde{\Lambda}(X,Y)\big)+Y\tilde{\Omega}(X,Y),
\end{gathered}
\end{equation}
where
$\tilde{\Lambda}(X,Y)=\Lambda(\kappa_1X-\kappa_2Y,\kappa_2X+\kappa_1Y)$
and
$\tilde{\Omega}(X,Y)=\Omega(\kappa_1X-\kappa_2Y,\kappa_2X+\kappa_1Y)$.
Here we say that system \eqref{3} is {\it reversible with respect to
a straight line $l$} through the origin if it is invariant with
respect to reversion about $l$ and a reversion of time $t$ (see for
instance \cite{Conti}). In particular Poincar\'e's Theorem is
applied for the case when \eqref{3} is invariant under the
transformations $(x,y,t)\to (-x,y,-t)$, or
$(x,y,t)\to (x,-y,-t)$.

In the proof of the results which we give later on we need the
Poincare's Theorem (see \cite[p.122]{Stepanov}).

\begin{theorem} \label{Poincare}
The origin of system \eqref{3} is a center  if the system is
reversible.
\end{theorem}

Since a rotation with respect to the origin of coordinates is a
particular transformation of type \eqref{letra}, when a center of
system \eqref{ZXY1} is invariant with respect to a straight line it
is not restrictive to assume that such straight line is the  x-axis.
So the center of system \eqref{ZXY1} will be invariant by the
transformation $(X,Y,t)\to (-X,Y,-t)$ or
$(X,Y,t)\to (X,-Y,-t)$.  Without loss of the generality
we shall study only the first case, i.e.\  we shall suppose that the
$\Lambda$-$\Omega$ system is invariant with respect to the
transformation $(X,Y,t)\to (-X,Y,-t)$.
The following proposition is easy to prove (see \cite{Malkin}).

\begin{proposition}\label{rever}
Differential system \eqref{Rcomp1} is invariant under the
transformation $(X,Y,t)\to(-X,Y,-t)$ if and only if it
can be written as
\begin{equation}\label{Rcomp2}
\begin{gathered}
\dot{X}=-Y\left(1+\Theta_1(X^2,Y)\right)+X^2\Theta_2(X^2,Y), \\
\dot{Y}=X\left(1+\Theta_1(X^2,Y)\right)+XY\Theta_2(X^2,Y).
\end{gathered}
\end{equation}
 \end{proposition}

 The following result was proved  in \cite[Corollary 15]{LRV1}.

\begin{corollary}
 Polynomial differential system \eqref{Rcomp2}  can be written as
\begin{equation}\label{Rcomp22}
\begin{gathered}
\dot{X}=-Y\left(1+\Theta_1(X^2,Y)\right)+X\{H_2,\,\Phi\}, \\
\dot{Y}=X\left(1+\Theta_1(X^2,Y)\right)+Y\{H_2,\,\Phi\},
\end{gathered}
\end{equation}
where $\Phi=\Phi(x,y)$ is a polynomial of degree at most $m-1$ and
such that $\{H_2,\,\Phi\}=X\Theta_2(X^2,Y)$.
 \end{corollary}

\begin{corollary}
Any weak centers of the type
\begin{equation}\label{RR}
\begin{gathered}
\dot{x}=-y\left(1+\Lambda\right)+x\{H_2,\,\Phi\}=p, \\
\dot{y}=x\left(1+\Lambda\right)+y\{H_2,\,\Phi\}=q,
\end{gathered}
\end{equation}
satisfies that the integral of the divergence on the unit circle is
zero. Moreover differential system \eqref{Rcomp22} can be written as
\begin{equation}\label{RR11}
\begin{gathered}
\dot{x}=\{\Phi,x\}+(1+G)\{H_2,x\}:=p, \\
\dot{y}=\{\Phi,y\}+(1+G)\{H_2,y\}:=q,
\end{gathered}
\end{equation}
where $G=G(x,y)$ is a polynomial of degree $m-1$.
\end{corollary}

\begin{proof}
Indeed from the relations
\begin{align*}
\frac{\partial p}{\partial x}+\frac{\partial q}{\partial y}
&= 2\{H_2,\,\Phi\}+x\frac{\partial \{H_2,\,\Phi\}}{\partial x}
 +y\frac{\partial\{H_2,\,\Phi\}}{\partial y}+\{H_2,\Lambda\} \\
&= \{H_2,2\Phi+x\frac{\partial \Phi}{\partial x}
 +y\frac{\partial \Phi}{\partial y}+\Lambda\},
\end{align*}
and by Proposition \ref{TTT} we obtain 
\[
\int_0^{2\pi}\Big(\frac{\partial p}{\partial
x}+\frac{\partial q}{\partial y}\Big)
\Big|_{x=\cos(t),\ ,x=\sin(t)} dt=0.
\]
Consequently  from Proposition \ref{important} we get that
\eqref{RR} becomes \eqref{RR11}.  Thus the proof is complete.
\end{proof}

\section{Proof of Theorem \ref{th2}}\label{ORaf3}

The proof of Theorem \ref{th2} for $m=2$ and $m=3$    follows from
the proof of \cite[Theorem 7]{LRV1}.  For $m=4$  we prove
Theorem \ref{th3} in the following propositions.

\begin{proposition}\label{rever11}
The fourth polynomial differential system
\begin{equation}\label{ROever3}
\begin{aligned}
\dot{x}&=-y+x\Big(a_1x+a_2y+a_3x^2+a_4y^2 \\
&\quad +a_5xy +a_6x^3+a_7y^3+a_8x^2y+a_9xy^2\Big):=P, \\
\dot{y}&= x+y\Big(a_1x+a_2y+a_3x^2+a_4y^2 \\
&\quad +a_5xy +a_6x^3 +a_7y^3+a_8x^2y+a_9xy^2\Big):=Q,
\end{aligned}
\end{equation}
where $a^2_1+a^2_2+a^2_3+a^2_4+a^2_5\ne 0$  has a weak center at the
origin if and only if after a linear change of variables
$(x,y)\to (X,Y)$ it is invariant under the
transformations $(X,Y,t)\to (-X,Y,-t)$ or
$(X,Y,t)\to(X,-Y,-t)$.  Moreover,
\begin{itemize}
\item[(i)] if $a^2_1+a^2_2\ne 0$, then system \eqref{ROever3} has a weak center
at the origin if and only if
\begin{equation}\label{TT1b}
\begin{gathered}
a_3+a_4=0,\quad a_5a_1a_2+(a^2_2-a^2_1)a_4=0, \\
a_1^3a_7-a_1^2a_2a_9+a_1a_2^2a_8-a_2^3a_6=0, \\
3a_1a_2^2a_7-3a_1^2a_2a_6+(a_1^3-2a_1a_2^2)a_8+(2a_1^2a_2-a_2^3)a_9=0.
\end{gathered}
\end{equation}
Consequently
\begin{itemize}
\item[(a)]
\begin{equation}\label{RORever3}
\begin{gathered}
a_3+a_4=0,\quad a_5+\frac{(a^2_2-a^2_1)}{a_1a_2}a_4=0, \\
a_6+\frac{1}{2a^3_2}\left(a_7(a^3_1-3a^2_2a_1)
+a_9(a^3_2-a^2_1a_2)\right)=0, \\
a_8+\frac{1}{2a^2_2a_1}\left(a_7(3a^3_1
-3a_1a^2_2)+a_9(a^3_2-3a^2_1a_2)  \right)=0.
\end{gathered}
\end{equation}
when $a_1a_2\ne 0$,
\item[(b)] $a_2=a_3=a_4=a_7=a_8=0$, when $a_1\ne 0$,
\item[(c)] $a_1=a_3=a_4=a_6=a_9=0$,  when $a_2\ne 0$.
\end{itemize}

\item[(ii)] If $a_1=a_2=0$ and $a_4a_5\ne 0$ then
system \eqref{ROever3} has a weak center at the origin if and only
if
\begin{gather*} %\label{RORever33}
a_3+a_4=0, \\
\lambda a_5+(1-\lambda^2)a_4=0, \\
\lambda^3a_7-\lambda^2a_9+\lambda\,a_8-a_6=0, \\
3\lambda^2a_7+3\lambda a_6+(\lambda^3-
2\lambda^2)a_8+(2\lambda^2-1)a_9=0,
\end{gather*}
where $\lambda=\frac{a_5+\sqrt{4a^2_4+a^2_5}}{2a_4}$. 
Moreover the weak center in this case after a linear change of variables
$(x,y)\to (X,Y)$ it is invariant under the
transformations $(X,Y,t)\to (-X,Y,-t)$.

\item[(iii)] if $a_1=a_2=a_3=a_4=a_5=0$, then the origin is a weak
center.
\end{itemize}
\end{proposition}

\begin{proof}
{\it Sufficiency:} First of all we observe that the polynomial
differential system \eqref{ROever3}  after the linear change of
variables  \eqref{letra} would be invariant under the transformation
$(X,Y,t)\to(-X,Y,-t)$  if and only if
\begin{equation}\label{Rfil1}
\begin{gathered}
\kappa_2a_1-\kappa_1a_2=0, \\
\kappa^2_1a_3+\kappa^2_2a_4+\kappa_1\kappa_2a_5=0, \\
\kappa^2_2a_3+\kappa^2_1a_4-\kappa_1\kappa_2a_5=0, \\
\kappa_1^3a_7-\kappa_1^2\kappa_2a_9+\kappa_1\kappa_2^2a_8-\kappa_2^3a_6=0, \\
3\kappa_1\kappa_2^2a_7-3\kappa_1^2\kappa_2a_6+(\kappa_1^3
-2\kappa_1\kappa_2^2)a_8+(2\kappa_1^2\kappa_2-\kappa_1\,\kappa_2^3)a_9=0.
\end{gathered}
\end{equation}
We suppose that \eqref{Rfil1} holds, and consequently the origin of the new 
system is a  center. When $a^2_1+a^2_2\ne 0$, after the change 
$x=a_1X-a_2Y$, $y=a_2X+a_1Y$, we get that the system has the form of 
system \eqref{Rcomp2} with $m=4$, here $\kappa_1=a_1$ and $\kappa_2=a_2$
and consequently this system is invariant under the change
$(X,Y,t)\to(-X,Y,-t)$ i.e. it is reversible. Thus in
view of the Poincar\'e Theorem we get that the origin is a center.
Hence system \eqref{ROever3} under conditions \eqref{TT1} has a weak
center at the origin.  Thus  the sufficiency under assumption (i) is
proved.

When $\kappa_1\kappa_2\ne 0$  then by solving \eqref{Rfil1} with
respect to $\kappa_1$ and $\kappa_2$, and if we denote by
$\kappa_1=a_1$ and $\kappa_2=a_2$ we obtain \eqref{RORever3}. For
the case when $\kappa_2=0$ and $k_1\ne 0$, then from \eqref{Rfil1}
it follows that
\begin{equation}\label{fil100}
a_2=a_3=a_4=a_7=a_8=0.
\end{equation}
If \eqref{fil100} holds then system \eqref{ROever3} becomes
\begin{gather*} %\label{algaba1}
\dot{x}=-y+x^2\big(a_1+a_5y +a_6x^2
+a_9y^2\big), \\
\dot{y}=x+yx\big(a_1+a_5y +a_6x^2 +a_9y^2\big),
\end{gather*}
which is invariant under the change
$(x,y,t)\to(-x,y,-t)$.  If  $\kappa_1=0$ and $k_2\ne 0$
then from \eqref{Rfil1} it follows that
\begin{equation}\label{Rfil1100}
a_1=a_3=a_4=a_6=a_9=0.
\end{equation}
If \eqref{Rfil1100} holds then \eqref{ROever3} becomes
\begin{gather*} %\label{algaba2}
\dot{x}=-y+xy\big(a_2+a_5x +a_7y^2
+a_8x^2\big), \\
\dot{y}=x+y^2\big(a_2+a_5x +a_7y^2 +a_8x^2\big),
\end{gather*}
which is invariant under the change $(x,y,t)\to(x,-y,-t)$.


When $a_1=a_2=0$ and $a_4a_5\ne 0$, then  by taking
\[
\kappa_1=\cos{\theta}:=\frac{\lambda}{\sqrt{1+\lambda^2}},\quad
\kappa_2=\sin \theta:=\frac{1}{\sqrt{1+\lambda^2}},
\]
where $\lambda$ is a solution of the equation
$\lambda^2-\frac{a_5}{a_4}\lambda-1=0$.
 After the rotation $ x=\cos \theta\,X-\sin \theta\,Y$,
$y=\sin\theta\,X+\cos \theta\, Y$ then in
view of \eqref{Rfil1} we get that \eqref{ROever3} becomes
\begin{gather*} %\label{algaba3}
\dot{X}=-Y+\frac{1+\lambda^2}{2\lambda}X^2\Big(-2a_4Y
+\frac{(a_9-3\lambda\,a_7)}{\sqrt{1+\lambda^2}}Y^2+\frac{\lambda^3a_7
-\lambda^2\,a_9-2\lambda\,a_7}{\sqrt{1+\lambda^2}}  X^2\Big), \\
\dot{Y}=X+\frac{1+\lambda^2}{2\lambda}XY\Big(-2a_4Y
+\frac{(a_9-3\lambda\,a_7)}{\sqrt{1+\lambda^2}}Y^2+\frac{\lambda^3a_7
-\lambda^2\,a_9-2\lambda\,a_7}{\sqrt{1+\lambda^2}} X^2\Big).
\end{gather*}
Thus this system is invariant under the change
$(X,Y,t)\to(-X,Y,-t)$, i.e.\ it is reversible. thus in
view of the Poincar\'e Theorem we get that the origin is a center.
Therefore the sufficiency is proved and (ii) holds.

If $a_1=a_2=a_3=a_4=a_5=0$, then system \eqref{ROever3} becomes
\begin{gather*}
\dot{x}=-y+x\Big(a_6x^3+a_9xy^2+a_7y^3+a_8x^2y\Big)=-y+x\Omega_3, \\
\dot{y}=x+y\Big(a_6x^3 +a_9xy^2+a_7y^3+a_8x^2y\Big)=x+y\Omega_3,
\end{gather*}
By considering that $\int_0^{2\pi}\Omega_3(\cos(t),\sin(t))dt=0$,
then in view  of \cite[Corollary  4]{LRV1} we get that the origin 
is a weak center which in general is not reversible.
 Thus the sufficiency of the proposition follows.


  {\it Necessity in case (i)} We shall study only the case (a). 
The case (b) and (c)  can be studied in analogous form. Therefore we assume 
that $a_1a_2\ne 0$.  Now we suppose that the origin
is a center of  \eqref{ROever3} and we prove that \eqref{RORever3}
holds. Indeed, from Theorem \ref{RRLia} it follows that differential
system \eqref{ROever3} can be written as
\begin{equation}\label{TTram1000}
\begin{gathered}
\begin{aligned}
P&=\{H_5,\,x\}+(1+g_1)\{H_4,x\}+(1+g_1+g_2)\{H_3,x\} \\
&\quad +(1+g_1+g_2+g_3)\{H_2,x\} \\
&=-y+x\Big(a_1x+a_2y+a_4y^2+a_3x^2+a_5xy +a_6x^3\\
&\quad +a_7y^3+a_8x^2y+a_9xy^2\Big) , 
\end{aligned}\\
\begin{aligned}
Q&=\{H_5,\,y\}+(1+g_1)\{H_4,y\}+(1+g_1+g_2)\{H_3,y\}\\
&\quad +(1+g_1+g_2+g_3)\{H_2,y\}, \\
&=x+y\Big(a_1x+a_2y+a_4y^2+a_3x^2+a_5xy +a_6x^3\\
&\quad +a_7y^3+a_8x^2y+a_9xy^2\Big)
\end{aligned}
\end{gathered}
\end{equation}
In view of Corollary \ref{Chetaevvv1} and assisted by an algebraic
computer we can obtain the solutions of \eqref{TTram1000}, i.e.\ the
homogenous polynomials $H_5,H_3,g_1,g_3$ of degree odd are  unique
and the homogenous polynomials  $H_4,\,g_2$ of degree even are
obtained modulo an  arbitrary polynomial of the form $c(x^2 +y^2)^k$
where $k=1,2$.   Indeed taking the homogenous part of these equations
of degree two we obtain
\begin{gather*}
\{H_3,x\}+g_1\{H_2,x\}= x(a_1x+a_2y), \\
\{H_3,y\}+g_1\{H_2,y\}= y(a_1x+a_2y).
\end{gather*}
The solutions of these equations are
\[
g_1=3(a_1y-a_2x),\quad  H_3=2H_2(a_2x-a_1y).
\]
The homogenous part of \eqref{TTram1000} of degree 3 is
\begin{equation}\label{mad}
\begin{gathered}
\{H_4,x\}+g_1\{H_3,x\}+g_2\{H_2,x\}=x(a_4y^2+a_3x^2+a_5xy)=x\Omega_2, \\
\{H_4,x\}+g_1\{H_3,x\}+g_2\{H_2,x\}=y(a_4y^2+a_3x^2+a_5xy)=y\Omega_2.
\end{gathered}
\end{equation}
The compatibility condition  of these two last equations   becomes
of $\{H_3,g_1\}+\{H_2,g_2\}=4\Omega_2$, and by considering that
 $\{H_3,g_1\}=\{H_2,-3(a_2x-a_1y)^2\}$ since
\[
\{H_2,g_2-3(a_2x-a_1y)^2\}=4\Omega_2.
\]
Hence,  in view of proposition \ref{TTT},  we obtain
\[
\int_0^{2\pi}\Omega_2(\cos(t),\sin(t))dt=2\pi (a_3+a_4)=0.
\]
 So $a_3+a_4=0$.  Therefore $g_2=3(a_2x-a_1y)^2 -a_4xy-2a_5x^2+c_1H_2$, 
where $c_1$ is a constant. Then from system \eqref{mad} by considering that
 $H_4$ is a homogenous polynomial of degree four we obtain the solution
\begin{align*}
H_4
=&-\frac{1}{4}\left(3g_1H_3+2g_2H_2\right)+c_1H^2_2 \\
=&H_2\left(3\left((a^2_2-a^2)x^2-a_1a_2xy\right)+a_5x^2+2a_4xy\right)+c_1H^2_2
\end{align*}
Inserting these previous solutions
$g_1,\,H_3,\,g_2$ and $H_4$ into the partial differential equations
\begin{equation}\label{mad1}
\begin{gathered}
\begin{aligned}
&\{H_5,\,x\}+g_1\{H_4,x\}+g_2\{H_3,x\}+g_3\{H_2,x\} \\
&=x(a_6x^3 +a_7y^3+a_8x^2y+a_9xy^2)=x\Omega_3:=X_4,
\end{aligned} \\
\begin{aligned}
&\{H_5,y\}+g_1\{H_4,y\}+g_2\{H_3,y\}+g_3\{H_2,y\} \\
&=y(a_6x^3 +a_7y^3+a_8x^2y+a_9xy^2)=y\Omega_3:=Y_4,
\end{aligned}
\end{gathered}
\end{equation}
we get that these differential equations have a unique solution.
Indeed, in this case the compatibility condition is
\begin{equation}\label{mad2}
\{H_4,g_1\}+\{H_3,g_2\}+\{H_2,g_3\}=5\Omega_3,
\end{equation}
because $\frac{\partial\,X_4}{\partial x}+\frac{\partial\,Y_4}{\partial x}=5\Omega_3$, 
and $\Omega_3$ is a homogenous polynomial of degree 3. Consequently there exists a
unique solution $g_3$ of \eqref{mad2} such that
\begin{align*}
g_3:=&\Big(-6a_2a_1^2-a_2^3+\frac{11}{3}a_2a_5-\frac{5}{3}a_1a_4
 -\frac{10}{3}a_7-\frac{5}{3}a_8\Big)x^3 \\
&+\Big((2a_1^3-a_1a_2^2)\mu^2+(8a_1^3-2a_1a_2^2-2a_1a_5-a_2a_4-4a_1c_1)\mu \\
&+6a_1^3+3a_1a_2^2-2a_1a_5+9a_2a_4+5a_6-4a_1c_1\Big)x^2y \\
&+ \Big( -a_2a_1^2\mu^2+(a_1a_4+4a_2c_1+a_1a_4)\mu-9a_2a_1^2+4c_1a_2
 -9a_1a_4-5a_7\Big)xy^2 \\
&+ \Big( \frac{5}{3}a_1^3\mu^2+\frac{1}{3}(22a_1^3-5a_1a_5-5a_2a_4-4a_1c_1)\mu \\
&+\frac{1}{3}(21a_1^3+5a_1a_5+5a_2a_4+5a_9+10a_6-12a_1c_1 )\Big)y^3,
\end{align*}
Thus the homogenous polynomial $H_5$ can be computed as 
\[
H_5=-\frac{1}{5}\left(4g_1H_4+3g_2H_3+2g_3H_2 \right),
\]
using \eqref{mad1}.

 Hence  partial differential system \eqref{mad1}
has a solution if and only if $a_3+a_4=0$.   On the other hand from
\eqref{aqu} for $m=4$ and assuming that $a_1a_2\ne 0$ and denoting
\begin{gather*} % \label{trian}
\lambda_1:=a_5-\frac{(a^2_1-a^2_2)a_4}{a_1a_2}, \\
\lambda_2:=a_6-\frac{1}{2a^3_2}\left(a_7(a^3_1-3a^2_2a_1)
+a_9(a^3_2-a^2_1a_2)\right), \\
\lambda_3:=a_8-\frac{1}{2a^2_2a_1}\left(a_7(3a^3_1
-3a_1a^2_2)+a_9(a^3_2-3a^2_1a_2)  \right). \\
\end{gather*}
From Remak \ref{rem11} with $m=4$ we obtain
\begin{align*}%\label{Raj1}
I_1:=&\int_0^{2\pi}\left(\{H_5,\,g_1\}+\{H_4,\,g_2\}
+\{H_3,\,g_3\}\right)\big|_{x=\cos(t),y=\sin(t)}dt \\
=&(3/2)\pi \left(2a_1a_2\lambda_1+2a_2\lambda_2-2a_1\lambda_3\right)=0.
\end{align*}
Under this condition the first  differential equation of 
\eqref{aqu}with $m=4$ becomes
\[
\{H_5,g_1\}+\{H_4,g_2\}+\{H_3,g_3\}+\{H_2,g_4\}=0.
\]
It has a solution $g_4$ which in view of Corollary \ref{Chetaevvv1}
can be obtained as follows
\[
g_4=G_4(x,y)+8c_1x(2a_4y+2a_5x)H_2+4c_2H^2_2,
\]
where $G_4=G_4(x,y)$ is a convenient homogenous polynomial of degree
four, $c_2$ is a constant. Using formula \eqref{Mir03} with $k=1$
$X_5=Y_5=0$ we obtain  the homogenous polynomial $H_6$ as follows
\[
H_6=-\frac{5}{6}g_1H_5-\frac{4}{6}g_2H_4-\frac{3}{6}g_3H_3-\frac{2}{6}g_4H_2.
\]
By considering that the integral of the homogenous polynomial of
degree 5,
\[
\int_0^{2\pi}\left(\{H_6,\,g_1\}+\{H_5,\,g_2\}
+\{H_4,\,g_3\}+\{H_3,\,g_4\}\right)\Big|_{x=\cos(t),y=\sin(t)}dt\equiv 0,
\] 
then we obtain that there is a unique solution for
the homogenous polynomial $g_5$ of degree 5  of the equation
\[
\{H_6,g_1\}+\{H_5,g_2\}+\{H_4,g_3\}+\{H_3,g_4\}+\{H_2,g_5\}=0,
\]
which comes from the first equation of \eqref{euro} with $m=4$ and
$j=1$.


Using formula \eqref{Mir03} with $k=2$  $X_6=Y_6=0$  we obtain  the
homogenous polynomial 
\[
H_7=-\frac{6}{7}g_1H_5-\frac{5}{7}g_2H_5-\frac{4}{7}g_3H_4
-\frac{3}{7}g_4H_3-\frac{2}{7}g_5H_2
\]
and inserting it into the next integral of the homogenous polynomials of degree 6
 we obtain
\begin{equation}\label{Raj2}
\begin{aligned}
I_2:=&\int_0^{2\pi}\left(\{H_7,\,g_1\}
+\{H_6,g_2\}+\{H_4,g_3\}+\{H_3,g_4\}\right)\big|_{x=\cos(t),y=\sin(t)}dt \\
=&\pi\left(\nu_2\lambda_1\lambda_2+\nu_4\lambda_1+\nu_5\lambda_2
 +\nu_6\lambda_3\right).
\end{aligned}
\end{equation}
where
\begin{gather*}
\nu_4=-\frac{2\big(4(a_1a_2)^3+16a_1a^5_2+2a_2^4a_4+(5a_1a^2_2-a^3_1)a_7
 +(a^2_1a_2-a^3_2)a_9\big)}{a_2^2}, \\
\nu_2=-4a_2,\quad
\nu_5=\frac{-24a_1^3-88a_1a^3_2-8a^2_2a_4}{a_1},\quad
\nu_6=-8a_2(a^2_1+3a^2_2)
\end{gather*}
By solving $I_1=0$ and $I_2=0$ and assuming
that $a_1(4a^2_2+\lambda_1)+2a_2a_4\ne 0$, we obtain
\begin{equation}\label{bob}
\begin{gathered}
\lambda_2=\frac{a_1\lambda_1\left(
-4a_1a^5_2-2a^4_2a_4+(a^3_1-5a_1a^2_2)a_7+(a^3_2-a^2_1a_2)a_9\right)}
{2a^3_2(a_1(4a^2_2+\lambda_1)+2a_2a_4)}, \\
\lambda_3=\frac{\lambda_1(-4a_1a^5_2+2a_1a^3_2\lambda_1-2^4_2a_4
 +(3a^3_1-15a_1a^2_2)a_7+(3a^3_2-3a^2_1a_2)a_9)}{2a^3_2(a_1(4a^2_2
 +\lambda_1)+2a_2a_4)}.
\end{gathered}
\end{equation}
By continuing this process, the following relation must hold
\begin{equation}\label{Raj2b}
\begin{aligned}
I_3:=&\int_0^{2\pi}\left(\{H_9,\,g_1\}
+\{H_8,\,g_2\}+\{H_7,\,g_3\}+\dots+\{H_3,\,g_7\}\right)
\big|_{x=\cos(t),y=\sin(t)}dt \\
=&p(\lambda_1,\lambda_2,\lambda_3)=0,
\end{aligned}
\end{equation}
where $p$ is a convenient polynomial of degree five in the variables
$\lambda_1,\lambda_2,\lambda_3$.  Inserting into  $I_3$ the values of
$\lambda_2$ and $\lambda_3$ from \eqref{bob}  we get that the
following relations must hold
\begin{equation}\label{bob1}
\tilde{p}=p(\lambda_1,\lambda_2,\lambda_3)
\big|=\lambda_1\left(e_4\lambda^4_1+e_3\lambda^3_1+e_2\lambda^2_2
+e_1\lambda_1+e_0\right)=0,
\end{equation}
where
\begin{equation}\label{bob2}
\begin{gathered}
e_4=6550\pi a^4_2a^4_1, \\
 e_3=41280\pi a^4_2a^4_1c_1+r^{(3)}_{0}, \\
e_2=\left(99840\pi a^4_2a^4_1\pi\right)c^2_1+r^{(2)}_1, \\
e_1=\left(10a_2a_1(79872a^3_1a^5_2+3993a^2_1a^4_2a_4) \right)c^2_1+r^{(1)}_1, \\
e_0=\pi (20a_1a_2+10a_4)\left(79872a^3_1a^7_2+39936a^2_1a^6_2a_4\right)c^2_1
+r^{(0)}_1,
\end{gathered}
\end{equation}
where $r^{(k)}_j$ is a convenient polynomial of degree $j$ in the
variable $c_1$ for $k=0,1,2,3,$.   Now we show that the polynomial
$\tilde{p}$ has only one real root. Indeed from  the results given
in \cite{Lu} we get that a quartic polynomial with  real
coefficients $e_4x^4+e_3x^3+e_2x^2+e_1x+e_0$ with $e_4\ne 0$ has
four complex roots if
\begin{equation}\label{kak}
\begin{gathered}
D_2=3e_3^2-8e_2e_4\leq 0, \\
\begin{aligned}
D_4&=256e_4^3e^3_0-27e^2_4e^4_1-192e^2_4e_1e^2_0e_3-27e^4_3e^2_0
 -6e_4e_3^2e_0e^2_3+e^2_2e^2_1e^2_3 \\
&\quad -4e_4e^3_2e^2_1+18e_2e_3^3e_1e_0+144e_4e_2e_0^2e_3^2-80e_4e_2^2e_0e_3e_1
 +18e_4e_2e_1^3e_3 \\
&\quad -4e_2^3e_0e_3^2-4e^3_3e^3_1+16e_4e_2^4e_0-128e_4^2e_2^2e_0^2
 +144e_4^2e_2e_0e_1^2>0.
\end{aligned}
\end{gathered}
\end{equation}
After some computations we can prove that for the  $e_j$'s given in
\eqref{bob2} for $j=0,1,2,3,4$ obtain
\begin{gather*}
D_2=\big(-119500800\pi^2a^8_1a^8_2\big)c^2_1+q^{(2)}_1 , \\
D_4=\Big(3584286725689459049392896000000\pi^6a^{21}_1a^{27}_2
 (2a_1a_2+a_4)^3\Big)c^9_1+q^{(4)}_8,
\end{gather*}
where $q^{(k)}_j$is a convenient polynomial of degree $j$ in the
variable $c_1$, for $k=2,4$. Taking the arbitrary constant $c_1$
big enough and such that $a_1a_2(2a_1a_2+a_4)c_1>0$  we obtain that
the polynomial $\tilde{p}$ has the unique real root $\lambda_1=0$,
and consequently $\lambda_2=\lambda_3=0$.

Finally we study the case when $2a_1a_2+a_4$.  By repeating the
process of the previous case we finally obtain that from the
equations $I_j=0$ for $j=1,2,3$ we obtain
\begin{gather*}
\lambda_3=\frac{3a_2}{a_1}\lambda_2, \\
\begin{aligned}
0&=\lambda_2\Big(174a^3_2\lambda_2+a_2(87a^2_1-29a^2_2)a_9
 +a_2(261a^2_2-87a^2_1)a_7  \\
&\quad +a^3_2a_1(605a^2_2-995a^2_1)+704a_1a^3_2c_1\Big).
\end{aligned}
\end{gather*}
By choosing the arbitrary constant properly, we can obtain that the
unique solution of $I_j=0$ for $j=1,2,3$ is
$\lambda_1=\lambda_2=\lambda_3=0$.  Thus the origin is a weak center
in this particular case.  
Thus the necessity of the proposition is proved.
\end{proof}

We observe that Proposition \ref{rever11} provides the necessary and
sufficient conditions for the existence of  quartic uniform
isochronous centers. We observe that this problem was study in
\cite{chava,Algaba,Algaba1}, but in these papers there are some
mistakes. For more details see the appendix.

Proposition \ref{rever11} can be generalized as follows and the
proof is similar.

\begin{proposition}\label{rever111}
The fourth polynomial differential system
\begin{equation}\label{cloro}
\begin{aligned}
\dot{x}&=-y(1+\mu(a_2x-a_1y))+x\Big(a_1x+a_3x^2+a_2y+a_4y^2 \\
&\quad +a_5xy +a_6x^3+a_7y^3+a_8x^2y+a_9xy^2\Big), \\
\dot{y}&=x(1+\mu(a_2x-a_1y))+y\Big(a_1x+a_2y+a_3x^2+a_4y^2 \\
&\quad +a_5xy +a_6x^3 +a_7y^3+a_8x^2y+a_9xy^2\Big),
\end{aligned}
\end{equation}
where $(\mu+m-2)(a^2_1+a^2_2)+a^2_3+a^2_4+a^2_5\ne 0$  has a weak
center at the origin if and only if after a linear change of
variables $(x,y)\to (X,Y)$ it is invariant under the
transformations $(X,Y,t)\to (-X,Y,-t)$ or
$(X,Y,t)\to(X,-Y,-t)$.  Moreover,
\begin{itemize}
\item[(i)] if $a^2_1+a^2_2\ne 0$, then system \eqref{cloro} has a weak center
at the origin if and only if
\begin{equation}\label{TT1}
\begin{gathered}
a_3+a_4=0,\quad a_5a_1a_2+(a^2_2-a^2_1)a_4=0, \\
a_1^3a_7-a_1^2a_2a_9+a_1a_2^2a_8-a_2^3a_6=0, \\
3a_1a_2^2a_7-3a_1^2a_2a_6+(a_1^3-2a_1a_2^2)a_8+(2a_1^2a_2-a_2^3)a_9=0.
\end{gathered}
\end{equation}
Consequently
\begin{itemize}
\item[(a)]
\begin{gather*} %\label{cloro1}
a_3+a_4=0,\quad a_5+\frac{(a^2_2-a^2_1)}{a_1a_2}a_4=0, \\
a_6+\frac{1}{2a^3_2}\left(a_7(a^3_1-3a^2_2a_1)
+a_9(a^3_2-a^2_1a_2)\right)=0, \\
a_8+\frac{1}{2a^2_2a_1}\left(a_7(3a^3_1
-3a_1a^2_2)+a_9(a^3_2-3a^2_1a_2)  \right)=0.
\end{gather*}
when $a_1a_2\ne 0$,

\item[(b)] $a_2=a_3=a_4=a_7=a_8=0$, when $a_1\ne 0$,

\item[(c)] $a_1=a_3=a_4=a_6=a_9=0$,  when $a_2\ne 0$.
\end{itemize}

\item[(ii)] If $a_1=a_2=0$ and $a_4a_5\ne 0$ then 
 system \eqref{cloro} has a weak center
at the origin if and only if
\begin{gather*} %\label{RORever330}
a_3+a_4=0, \\
\lambda a_5+(1-\lambda^2)a_4=0, \\
\lambda^3a_7-\lambda^2a_9+\lambda a_8-a_6=0, \\
3\lambda^2a_7+3\lambda a_6+\left(\lambda^3-
2\lambda^2\right)a_8+\left(2\lambda^2-1)\right)a_9=0,
\end{gather*}
where $\lambda=\frac{a_5+\sqrt{4a^2_4+a^2_5}}{2a_4}$.  Moreover the
weak center in this case after a linear change of variables
$(x,y)\to (X,Y)$ it is invariant under the
transformations $(X,Y,t)\to (-X,Y,-t)$.

\item[(iii)] if $a_1=a_2=a_3=a_4=a_5=0$, then the origin is a weak
center.

\item[(iv)] $\mu+2=a_3=a_4=a_5=0$, then the origin is a weak center.
\end{itemize}
\end{proposition}

\section{Proof of Theorem \ref{th3}}\label{ORaf4}

The proof follows from the next  propositions.

\begin{proposition}\label{gat1}
A cubic polynomial differential system
\begin{equation}\label{RRever212}
\begin{gathered}
\dot{x}=-y(1+\mu(a_2x-a_1y))+x(a_1x+a_2y+a_3x^2+a_4y^2+a_5xy), \\
\dot{y}=x(1+\mu(a_2x-a_1y)) +y(a_1x+a_2y+a_3x^2+a_4y^2+a_5xy),
\end{gathered}
\end{equation}
has a weak center at the origin if and only if
\begin{equation}\label{Raqu1}
a_3+a_4=0,\quad a_1a_2a_5+(a^2_2-a^2_1)a_4=0,
\end{equation}
Moreover system \eqref{RRever212} under condition \eqref{Raqu1} and
$(\mu+1)(a^2_1+a^2_2)\ne 0$, after a linear change of variables
$(x,y)\to (X,Y)$ it is invariant under the
transformations $(X,Y,t)\to (-X,Y,-t)$.
\end{proposition}

Proposition \ref{RRever212} is proved in \cite[Proposition 23]{LRV1}.
We give the proof of Proposition \ref{rever001}. The proofs of
Propositions \ref{Rori10} and \ref{Rori110} are analogous to the
proofs of Proposition \ref{rever001}.

\begin{proposition}\label{rever001}
A polynomial differential system of degree four
\begin{equation}\label{EEever3}
\begin{gathered}
\begin{aligned}
\dot{x}&=-y(1+\mu(a_2x-a_1y))+x\Big(a_1x+a_2y+a_4\Big(y^2-x^2
 -\frac{(a^2_2-a^2_1)}{a_1a_2}xy\Big)\\
&\quad +a_6x^3+a_7y^3+a_8x^2y+a_9xy^2\Big), 
\end{aligned}\\
\begin{aligned}
\dot{y}&=x(1+\mu(a_2x-a_1y))+y\Big(a_1x+a_2y+a_4\Big(y^2-x^2
 -\frac{(a^2_2-a^2_1)}{a_1a_2}xy\Big) \\
&\quad +a_6x^3 +a_7y^3+a_8x^2y+a_9xy^2\Big),
\end{aligned}
\end{gathered}
\end{equation}
where $a_1a_2\ne 0$  has a weak center at the origin if and only if
the following  conditions hold.
\begin{equation}\label{tur}
\begin{gathered}
\lambda_1:=a_9+\frac{1}{2a_2a_1^2}\big((3a_1a_2^2-a_1^3\big)a_8
+\dots)=0,\\
\lambda_2;=a_7+\frac{1}{2a_1^3}\big((a_2^3-3a_2a_1^2)a_8+\dots\big)=0
\end{gathered}
\end{equation}

Moreover system \eqref{EEever3} under conditions \eqref{tur} and after
a linear change of variables $(x,y)\to (X,Y)$ it is
invariant under the transformations $(X,Y,t)\to (-X,Y,-t)$.
\end{proposition}

\begin{proof}
 {\it Sufficiency:}  First we observe that the differential system
 \eqref{EEever3} under the linear transformation \eqref{letra} can
 be written as \eqref{Rcomp1} with $m=4$, and
\begin{gather*}
 \Lambda=\mu(a_2x-a_1y)=0, \\
\begin{aligned}
\Omega&= a_1x+a_2y+a_4(y^2-x^2-\frac{a^2_2-a^2_1)}{a_1a_2}xy ) \\
&\quad +a_6x^3+a_7y^3+a_8xy+a_9xy^2=0.
\end{aligned}
\end{gather*}
This differential system  is invariant under the transformation
$(X,Y,t)\to(-X,Y,-t)$ if and only if
\begin{equation}\label{RRRaqu10}
\begin{gathered}
\kappa_1a_2-\kappa_2a_1=0, \\
\kappa_1(\kappa^2_1a_7+\kappa_2^2a_8)-\kappa_2(\kappa^2_2a_6+\kappa^2_1a_9)=0, \\
3\kappa_1\kappa_2(a_7\kappa_2-a_6\kappa_1)+\kappa_1(\kappa^2_1-2\kappa^2_2)a_8
+\kappa_1(2\kappa^2_1-\kappa^2_2)a_9=0,
\end{gathered}
\end{equation}
We suppose that \eqref{tur} holds and show that then the origin is a 
center of system \eqref{EEever3}. Assume that $a_1a_2\ne 0$.  Then after the
 transformation
\begin{equation}\label{letra1}
x=a_1X-a_2Y,\, y=a_2X+a_1Y,
\end{equation}
we get that this system can be written as system \eqref{Rcomp2} for $m=4$ 
and with $\kappa_1=a_1$ and $\kappa_2=a_2$, and consequently the conditions 
\eqref{RRRaqu10} becomes
\begin{gather*}
a_1(a^2_1a_7+a_2^2a_8)-a_2(a^2_2a_6+a^2_1a_9)=0, \\
3a_1a_2(a_7a_2-a_6a_1)+a_1(a^2_1-2a^2_2)a_8+a_1(2a^2_1-a^2_2)a_9=0.
\end{gather*}
By solving these two equations with respect to $a_7$ and $a_9$ we
get \eqref{tur}. Hence  \eqref{EEever3} is invariant, after the
given linear change \eqref{letra1} is invariant under the
transformation $(X,Y,t)\to(-X,Y,-t)$, i.e. it  is
reversible. Thus in view of the Poincar\'e Theorem we get that the
origin is a center of  \eqref{EEever3} if \eqref{tur} holds.

{\it Necessity:} Now we suppose that the origin is a center of
  \eqref{EEever3} and we prove that \eqref{tur}  holds.
Indeed, from Theorem \ref{RRLia} it follows that differential system
\eqref{EEever3} can be written as
\begin{gather*}
\begin{aligned}
&\{H_5,\,x\}+(1+g_1)\{H_4,x\}+(1+g_1+g_2)\{H_3,x\}+(1+g_1+g_2+g_3)\{H_2,x\} \\
&=-y+x(a_1x+a_2y+a_4\Big(y^2-x^2-\frac{(a^2_2-a^2_1)}{a_1a_2}xy\Big)
+a_6x^3 +a_7y^3+a_8x^2y+a_9xy^2)   , 
\end{aligned} \\
\begin{aligned}
&\{H_5,\,y\}+(1+g_1)\{H_4,y\}+(1+g_1+g_2)\{H_3,y\}+(1+g_1+g_2+g_3)\{H_2,y\}, \\
&=x+y(a_1x+a_2y+a_4\Big(y^2-x^2-\frac{(a^2_2-a^2_1)}{a_1a_2}xy\Big)
+a_6x^3 +a_7y^3+a_8x^2y+a_9xy^2).
\end{aligned}
\end{gather*}
Hence
\begin{equation}\label{podem1}
\begin{gathered}
\{H_3,x\}+g_1\{H_2,x\}=-y\mu(a_1y-a_2x)+x(a_1x+a_2y)=X_2, \\
\{H_3,y\}+g_2\{H_2,y\}=x\mu(a_1y-a_2x)+y(a_1x+a_2y)=Y_2, \\
\{H_4,x\}+g_1\{H_3,x\}+g_2\{H_2,x\}
 =a_4x\Big(y^2-x^2-\frac{(a^2_2-a^2_1)}{a_1a_2}xy\Big) 
 =x\Omega_2=X_3,  
\\
\{H_4,y\}+g_1\{H_3,y\}+g_2\{H_2,y\}=a_4y
 \Big(y^2-x^2-\frac{(a^2_2-a^2_1)}{a_1a_2}xy\Big) \\
=y\Omega_2=Y_3, 
 \\
\begin{aligned}
&\{H_5,\,x\}+g_1\{H_4,x\}+g_2\{H_3,x\}+g_3\{H_2,x\}  \\
&=x\left(a_6x^3+a_7y^3+a_8x^2y+a_9xy^2\right):=x\Omega_3=X_4, \\
&\{H_5,\,y\}+g_1\{H_4,y\}+g_2\{H_3,y\}+g_3\{H_2,y\} \\
=&y\left(a_6x^3+a_7y^3+a_8x^2y+a_9xy^2\right):=y\Omega_3=Y_4,
\end{aligned} \\
\end{gathered}
\end{equation}
The two first equations of \eqref{podem1} are compatible if and only
if $g_1$ satisfies
\[
\{H_2,g_1\}=-(\mu-3)(a_1x+a_2y)=\frac{\partial X_2}{\partial
x}+\frac{\partial Y_2}{\partial y}.
\]
Thus $g_1=-(\mu-3)(a_1y-a_2x)$, and consequently from the first part of
\eqref{podem1} we obtain that  $H_3=-(x^2+y^2)\left(a_1y-a_2x\right) $

From the third and fourth equations of \eqref{podem1}  we get that
these equations are compatible if and only if
\[
\{H_3,g_1\}+\{H_2,g_2\}
=3a_4\Big(y^2-x^2-\frac{(a^2_2-a^2_1)}{a_1a_2}xy\Big)
=\frac{\partial X_3}{\partial x}+\frac{\partial Y_3}{\partial y},
\]
and  in view of  Proposition \ref{TTT} we get that this equation has 
the polynomial solution $g_2$ if and only if
\[
\int_{0}^{2\pi}\Big(\{H_3,g_1\}
+3a_4\Big(y^2-x^2-\frac{(a^2_2-a^2_1)}{a_1a_2}xy\Big)\Big)
\Big|_{x=\cos(t),\,y=\sin(t)}dt= 0, \\
\]
which  holds identically. Thus we obtain the homogenous polynomial
\begin{align*}
g_2&=\Big((a^2_1+2a^2_2)(\mu-3a_1a_2)+\frac{(a^2_1-a^2_2)a_4}{a_1a_2}
\Big)x^2-2(a_1a_2(\mu-3)-2a_4)xy \\
&\quad +\Big((
a^2_2+2a^2_1)(\mu-3)+\frac{(a^2_2-a^2_1)a_4}{a_1a_2}\Big)y^2+c_1H_2,
\end{align*}
where $c_1$ is an arbitrary constant. From \eqref{Mir03} with $j=3$
we obtain the homogenous polynomial
\begin{align*}
H_4&=-\frac{1}{4}\left(3g_1H_3+2g_2H_2\right)\\
& =c_1H^2_2+  \frac{H_2}{2}\Big(\mu+\frac{1}{a_1a_2}(a_4-3a_1a_2)((a_1-a_2)x \\
&\quad +(a_1+a_2)y)((a_1-a_2)y+(a_1+a_2)x)\Big).
\end{align*}
From \eqref{Mir3} with $j=4$ we compute
\[
\{H_4,g_1\}+\{H_3,g_2\}+\{H_2,g_3\}
=4\Omega_3
=\frac{\partial X_4}{\partial x}+\frac{\partial Y_4}{\partial y}.
\]
This last equation  has a unique homogenous polynomial solution
$g_3$, which we insert in the expression for
 $H_5$ (see \eqref{Mir03} when $j=4$)
 and we obtain
\[
H_5=-4g_1H_4/5-3g_2H_3/5-2g_3H_2/5.
\]

Hence the homogenous polynomials $H_5,H_3,g_1,g_3$ are determined
and $H_4,\,g_2$ are obtained  with and arbitrary term
of the type $c_k(x^2 +y^2)^k$ where $k=1,2$, respectively. 
On the other hand from \eqref{aqu00} with $m=4$ and assuming that
$a_1a_2\ne 0$  we get
\begin{align*} %\label{MRaj1}
I_1:=&\int_0^{2\pi}\left(\{H_5,\,g_1\}+\{H_4,\,g_2\}
+\{H_3,\,g_3\}\right)\big|_{x=\cos(t),y=\sin(t)}dt \\
&=3\pi\left(a_2\lambda_1-3a_1\lambda_2\right)=0
\end{align*}
under this condition the partial differential equation (coming from
\eqref{Mir3} with $k=5$)
\[
\{H_5,g_1\}+\{H_4,g_2\}+\{H_3,g_3\}+\{H_2,g_4\}=0,
\]
has a homogenous polynomial solution $g_4$ which in view of
Corollary \ref{Chetaevvv1} can be obtained with arbitrary term of
the type $c(x^2+y^2)^2$.


The homogenous polynomial $H_6$ can be determined as follows (see
\eqref{Mir03} when k=5)
\[
H_6=-\frac{5}{6}g_1H_5-\frac{4}{6}g_2H_4-\frac{3}{6}g_3H_3-\frac{2}{6}g_4H_2.
\]
Since the integral of the homogenous polynomial of degree 5,
\[
\int_0^{2\pi}\left(\{H_6,g_1\}+\{H_5,g_2\}
+\{H_4,g_3\}+\{H_3,g_4\}\right)\big|_{x=\cos(t),y=\sin(t)}dt
\]
is zero,  we obtain that there is a unique homogenous polynomial
$g_5$ of degree 5  solution  of the equation
\[
\{H_6,g_1\}+\{H_5,g_2\}+\{H_4,g_3\}+\{H_3,g_4\}+\{H_2,g_5\}=0.
\]
Calculating the homogenous polynomial of degree 7 (see \eqref{Mir03} when $k=6$)
 we obtain
\[
H_7=-\frac{6}{7}g_1H_5-\frac{5}{7}g_2H_5-\frac{4}{7}g_3H_4
-\frac{3}{7}g_4H_3-\frac{2}{7}g_5H_2,
\]
and inserting it into the integral of the homogenous polynomial of
degree 6,
\begin{align*} %\label{MRaj2}
I_2:=&\int_0^{2\pi}\left(\{H_7,g_1\}+\{H_6,g_2\}+\{H_4,g_3\}+\{H_3,g_4\}\right)
\big|_{x=\cos(t),y=\sin(t)}dt \\
=&\pi(\mu-3)\left( \nu_1\lambda_1+\nu_2\lambda_2\right)=0,
\end{align*}
where
\begin{gather*}
\nu_1=-\frac{\pi}{42}\Big(\left( 4203a^3_2+108ca_2\right)a_1
-3255a^3_1a_2+(157a^2_1-489a^2_2)a_4\Big), \\
\nu_2=-\frac{\pi}{42}\Big((1401a^3_2+36ca_2)a_1-2601a^3_1a_2+(147a^2_1-163a^2_2)a_4\Big).
\end{gather*}
By solving the linear system $ I_1=0,\quad I_2=0 $ with respect to
$\lambda_1$ and $\lambda_2$, and by considering that the
determinant of the matrix of this system  is
$\Delta=\frac{2\pi^2a^2_1}{7}(71a_4-1137a_1a_2)$.  Assuming that
$\Delta\ne 0$ we deduce that $\lambda_1=\lambda_2=0$.

 The case when $71a_4-1137a_1a_2=0$ can be analyzed in analogous form. 

By solving $I_j=p_j(\lambda_1,\lambda_2)$ for $j=1,2$ we obtain
that $\lambda_2=\lambda_2(\lambda_1)$. Inserting these
expressions into $I_3=0$  we get that
$\lambda_1\left(e_4\lambda^4_1+e_3\lambda^3_1+e_2\lambda^2_1+e_1\lambda_1
+e_0\right)=0$,
where
\begin{gather*}
e_4=166446510550a^2_2a^2_1\pi, \\
e3=1048994191680a^2_1a^2_2\pi\,c^2_1+r^{(3)}_0, \\
e_2=2537102231040a^2_2a^2_1\pi c^2_1+r^{(2)}_1, \\
e_1=182814295971840a^2_1a^4_2\pi c^2_1+r^{(4)}_1, \\
e_0=329323217673216a^2_1a^6_2c^2_1+r^{(0)}_1,
\end{gather*}
where $r^{(n)}_j$ are convenient polynomials of degree $j$ in the
variable $c_1$.  By applying the result given in \cite{Lu} with $D_2$
and $D_4$ given in \eqref{kak} and choosing the arbitrary constant
$c_1$ conveniently 
we deduce that the unique real solution of $I_3 = 0$ is
$\lambda_1=0$. Consequently $\lambda_2=\lambda_3=0$.  In
short the proof complete.
\end{proof}

The following two propositions can be proved in analogous way of the
proof of Proposition \ref{rever001}. 

\begin{proposition} \label{Rori10}
A  polynomial differential system of degree five
\begin{equation}\label{Rever33}
\begin{gathered}
\begin{aligned}
\dot{x}
&=-y(1+\mu(a_2x-a_1y)) \\
&\quad +x\Big(a_1x+a_2y++a_4(y^2-x^2+\frac{(a^2_2-a^2_1)}{a_1a_2}xy)+a_6x^3 \\
&\quad +\frac{1}{2a^3_1}((3a_2a^2_1-a^3_2)a_6+(a^2_2a_1-a^3_1)a_8))y^3+a_8x^2y \\
&\quad +\frac{1}{2a_2a^2_1}(3(a^2_1a_2-a^3_2)a_6+(3a_1a^2_2-a^3_1)a_8))xy^2 \\
&\quad +a_{10}x^4+a_{11}x^3y+a_{12}x^2y^2+a_{13}xy^3+a_{14}y^4\Big),
\end{aligned} \\
\begin{aligned}
\dot{y}&=x(1+\mu(a_2x-a_1y)) \\
&\quad +y\Big(a_1x+a_2y+a_4(y^2-x^2+\frac{(a^2_2-a^2_1)}{a_1a_2}xy)+a_6x^3 \\
&\quad +\frac{1}{2a^3_1} \big((3a_2a^2_1-a^3_2)a_6+(a^2_2a_1-a^3_1)a_8\big)
 \Big)y^3+a_8x^2y \\
&\quad +\frac{1}{2a_2a^2_1}\Big(\big(3(a^2_1a_2-a^3_2)a_6
 +(3a_1a^2_2-a^3_1)a_8\big)\Big)xy^2 \\
&\quad +a_{10}x^4+a_{11}x^3y+a_{12}x^2y^2+a_{13}xy^3+a_{14}y^4,
\end{aligned}
\end{gathered}
\end{equation}
where $a_1a_2\ne 0$  has a weak center at the origin if and only if
the following conditions  hold
\begin{gather*}
a_{12}+ 3(a_{10}+a_{14})=0, \\
2a^3_1a^3_2a_{13}-\left(a^6_1+7(a^2_1a^4_2-a^4_1a^2_2\right)a_{10}
-\left(a^5_1a_2-4a^3_1a^3_2+a_1a^5_2\right)a_{11}=0, \\
2a^2_1a^2_2a_{14}-\left(a^4_1-4a^2_1a^2_2+a^4_2\right)a_{10}
-(a^3_1a_2-a_1a^3_2)a_{11}=0.
\end{gather*}
Moreover system \eqref{Rever33} under these conditions  and after a
linear change of variables $(x,y)\to (X,Y)$ it is
invariant under the transformations $(X,Y,t)\to (-X,Y,-t)$.
\end{proposition}


\begin{proposition}\label{Rori110}
A  polynomial differential system of degree six,
\begin{equation} \label{RRRever33}
\begin{gathered}  
\begin{aligned}
\dot{x}
&=-y(1+\mu(a_2x-a_1y))+x\Big(a_1x+a_2y+a_4(y^2-x^2 \\
&\quad  +\frac{(a^2_2-a^2_1)}{a_1a_2}xy) +a_6x^3 
 +\frac{1}{2a^3_1}((3a_2a^2_1-a^3_2)a_6\\
&\quad +(a^2_2a_1-a^3_1)a_8))y^3+a_8x^2y  \\
&\quad +\frac{1}{2a_2a^2_1}(3(a^2_1a_2-a^3_2)a_6+(3a_1a^2_2-a^3_1)a_8)
)xy^2 \\
&\quad +a_{10}x^4+a_{11}x^3y+a_{12}x^2y^2+a_{13}xy^3+a_{14}y^4\Big), 
\end{aligned} \\
\begin{aligned}
\dot{y}
&=x(1+\mu(a_2x-a_1y))+y\Big(a_1x+a_2y++a_4(y^2-x^2 \\
&\quad +\frac{(a^2_2-a^2_1)}{a_1a_2}xy) +a_6x^3 
+\frac{1}{2a^3_1}((3a_2a^2_1-a^3_2)a_6+(a^2_2a_1-a^3_1)a_8)
)y^3 \\
&\quad +a_8x^2y +\frac{1}{2a_2a^2_1}(3(a^2_1a_2-a^3_2)a_6+(3a_1a^2_2-a^3_1)a_8)
)xy^2 \\
&\quad +a_{10}x^4+a_{11}x^3y+a_{12}x^2y^2+a_{13}xy^3+a_{14}y^4 \Big),
\end{aligned} 
\end{gathered}
\end{equation}
where $a_1a_2\ne 0$  has a weak center at the origin if and only if
the following conditions hold
\begin{gather*}
\begin{aligned} %\label{RRRRSever444}
\lambda_1
&=a_{15}+\frac{1}{8a^2_1a^5_2}\Big(\left(2a^5_1a^2_2-4a^3_1a^4_2
 +2a_1a^6_2\right)a_{19} \\
&\quad -\left(3a^7_1 -15a^5_1a^2_2+25a^3_1a^4_2-5a_1a^6_2\right)a_{16} \\
&\quad +\left(a^7_2+11a^4_1a^3_2-9a^2_1a^5_2-3a^6_1a_2\right)a_{20}\Big)=0, 
\end{aligned} \\
\begin{aligned}
\lambda_2&=a_{17} -\frac{1}{8a_1^3a_2^4
}\Big((15a_1^7-55a_1^5a_2^2+45a^3_1a^4_2-5a_1a^6_2)a_{16}\\
&\quad +(10a^5_1a^2_2-12a^3_1a^4_2+2a_1a^6_2)a_{19} \\
&\quad +(-15a^6_1a_2+35a^4_1a^3_2-13a^2_1a^5_2+a^7_2)a_{20}\Big)=0,
\end{aligned} \\
\begin{aligned}
\lambda_3&=  a_{18}+\frac{1}{2a^2_1a^3_2 }
 \Big(-(5a^5_1-10a^3_1a^2_2+5a_1a^4_2)a_{16}  \\
&\quad -(4a^3_1a^2_2-2a_1a^4_2)a_{19}-(6a^2_1a_2^3-5a^4_1a_2-a^5_2)a_{20}\Big)=0.
\end{aligned}
\end{gather*}
Moreover system \eqref{RRRever33} under these conditions  and after
a linear change of variables $(x,y)\to (X,Y)$ is
invariant under the transformations $(X,Y,t)\to (-X,Y,-t)$.
\end{proposition}

\begin{remark} \rm
A weak center in general is not invariant  with respect to a
straight line.  Indeed, the cubic $\Lambda$-$\Omega$ system with a
weak center at the origin \cite{Zh}
\begin{equation}\label{zh}
\begin{gathered}
\dot{x}=-y\Big(1+y+\frac{y^2}{2}\Big)+\frac{x}{2}(x-y-y^2), \\
 \dot{y}= x\Big(1+y+\frac{y^2}{2}\Big)+\frac{y}{2}(x-y-y^2),
\end{gathered}
\end{equation}
is not invariant with respect to the straight line.
\end{remark}


\section{Appendix}

The classification of the isochronous centers of Proposition
\ref{rever11} for system \eqref{ROever3} has been previously studied
in  \cite{chava,Algaba1}. But in both papers there are some mistakes.
More precisely, in \cite{chava} they write system \eqref{ROever3} in
in polar coordinates as 
\begin{equation}\label{Artu1100}
\dot r=P_2(\varphi)r^2+P_3(\varphi)r^3+P_4(\varphi)r^4,\quad \dot
\varphi=1,
\end{equation}
where
\begin{gather*}
P_2(\varphi)=R_1\cos \varphi+ r_1 \sin\varphi,\\
P_3(\varphi)=R_2\cos 2\varphi +r_2\sin 2\varphi,\\
P_4(\varphi)=R_3\cos 3\varphi +r_3\sin 3\varphi +R_4\cos
\varphi+r_4\sin \varphi.
\end{gather*}
In \cite{chava} they forgot to write the term $r_1
\sin\varphi$. The relations between the parameters of
\eqref{ROever3} and the parameters of system \eqref{Artu1100} are
\begin{gather*}
R_1 = a_1,\quad r_1=a_2,\quad R_2 = (a_3-a_4)/2,\quad r_2 = a_5/2, \\
R_0 = (a_3+a_4)/2,\quad R_3 = (a_6-a_9)/4,\quad r_3 = (a_8-a_7)/4, \\
R_4 = (3a_6+a_9)/4,\quad  r_4 = (3a_7+a_8)/4.
\end{gather*}

In \cite{Algaba1} they write system \eqref{ROever3} in complex
notation as 
\begin{equation}\label{Artu100}
\dot{z}=iz+z\Big( Az+\bar{A}\bar{z}
+Bz^2+2(b_1+b_3)z\bar{z}+\bar{B}\bar{z}^2+Cz^3
+Dz^2\bar{z}+\bar{D}\bar{z}z^2+\bar{C}\bar{z}^3
\Big), 
\end{equation}
being  $z=x+iy$, $\bar{z}=x-iy$,
$A=(a_1-ia_2)/2$, $B=(b_1+b_3-ib_2)/4,\,C=(d_1-id_2)/8$ and
$D=(d_3-id_4)/8$ where $a_1,a_2,b_1,b_2,b_3,d_1,d_2,d_3,d_4$ are
real constants. 
The relations between the parameters of system
\eqref{ROever3} and the parameters of system \eqref{Artu100} are
\begin{gather*} % \label{Artu1}
a_1=a_1,\quad a_2=a_2, \quad a_3=5 (b_1+b_3)/2,\\\
a_4=3(b_1+b_3)/2,\quad a_5=b_2, \quad a_6=(d_3+d_1)/4,\\
 a_7=(d_4-d_2)/4,\quad a_8=(d_4+3d_2)/4,\quad a_9=(d_3-3d_1)/4.
\end{gather*}

The following sets of conditions are equivalent
\begin{itemize}
\item $r_1=r_4=R_0=R_4=0$ and $r_3\ne 0$ for system \eqref{Artu1100},

\item $a_2=b_1+b_3=d_3=d_4=0$ and $b_2\ne 0$ for system \eqref{Artu100},

\item $a_2=3a_7+a_8=3a_6+a_9=a_3+a_4=0$ and $a_5\ne 0$ for system \eqref{ROever3}.
\end{itemize}

In \cite{chava,Algaba1} they claim that system
\eqref{ROever3} under the previous conditions has a center, but this
is incorrect because such a system has a week focus because
their Liapunov constants are not all zero. 
Thus its first non-zero Liapunov constant is $\pi a_1^2 a_3 /2$. 
For more details on Liapunov constants see  \cite[chapter 5]{DLA}.

\subsection*{Acknowledgments}
J. Llibre was y supported by the Ministerio de
Econom\'ia, Industria y Competitividad, Agencia Estatal de
Investigaci\'on grant MTM2016-77278-P (FEDER),by  the Ag\`encia de
Gesti\'o d'Ajuts Universitaris i de Recerca grant 2017 SGR 1617, and
by the European project Dynamics-H2020-MSCA-RISE-2017-777911.
R. Ram\'irez was supported by the Spanish Ministry of
Education through projects TIN2014-57364-C2-1-R, TSI2007-65406-C03-01
``AEGIS''.

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\end{document}