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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2013 (2013), No. 23, pp. 1--19.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2013 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2013/23\hfil Darboux integrability]
{Darboux integrability and rational reversibility in
cubic systems with two invariant \\ straight lines}

\author[Dumitru Cozma\hfil EJDE-2013/23\hfilneg]
{Dumitru Cozma}  % in alphabetical order

\address{Dumitru Cozma \newline
Tiraspol State University \\
5 Gh. Iablocichin str. \\
Chi\c{s}in\u{a}u, MD-2069,  Moldova}
\email{dcozma@gmail.com}

\thanks{Submitted December 1, 2011. Published January 27, 2013.}
\subjclass[2000]{34C05}
\keywords{Cubic differential systems; center problem;
 invariant straight lines; \hfill\break\indent
Darboux integrability; rational reversibility}

\begin{abstract}
 We find conditions for a singular point $O(0,0)$ of a center or a focus type
 to be a center, in a cubic differential system with two distinct invariant
 straight lines. The presence of a center at $O(0,0)$ is proved by
 using the method of Darboux integrability and the rational reversibility.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks


\section{Introduction and statement of results}


 A cubic system with a singular point with pure imaginary
eigenvalues $(\lambda_{1,2} = \pm i$, $i^2 = -1)$ by a
nondegenerate transformation of variables and time rescaling can
be brought to the form
\begin{equation}\label{eq1}
\begin{gathered}
 \dot x=y+ax^2+cxy+fy^2+kx^3+mx^2y+pxy^2+ry^3=P(x,y), \\
 \dot y=-(x+gx^2+dxy+by^2+sx^3+qx^2y+nxy^2+ly^3)=Q(x,y),
\end{gathered}
\end{equation}
where the variables $x, y$ and coefficients $a, b,\dots, s$ in
\eqref{eq1} are assumed to be real. Then the origin $O(0,0)$ is a
singular point of a center or a focus type for \eqref{eq1}. The
problem arises of distinguishing between a center and a focus,
i.e. of finding the coefficient conditions under which $O(0,0)$
is, for example, a center. These conditions are called the
conditions for a center existence or the center conditions and the
problem - the problem of the center.

The derivation of necessary conditions for a center existence
often involves extensive use of computer algebra (see, for
example, \cite{LLR}, \cite{LP1}), in many cases making very heavy
demands on the available algorithms and hardware. The necessary
conditions are shown to be sufficient by a variety of methods. A
number of techniques, of progressively wider application, have
been developed.

A theorem of Poincar\'e  in \cite{P1} says that a singular point
$O(0,0)$ is a center for \eqref{eq1} if and only if the system has
a nonconstant analytic first integral $F$ in the neighborhood of
$O(0,0)$. It is known \cite{ALS} that the origin is a center for
system \eqref{eq1} if and only if the system has in the
neighborhood of $O(0,0)$ an analytic integrating factor of the
form
$$
 \mu(x,y)=1+\sum_{k=1}^{\infty}\mu_k(x,y),
$$
where $\mu_k$ are homogeneous polynomials of degree $k$.

There exists a formal power series $F(x,y)=\sum F_j(x,y)$ such
that the rate of change of $F(x,y)$ along trajectories of
\eqref{eq1} is a linear combination of polynomials
$\{(x^2+y^2)^j\}_{j=2}^{\infty}:\,\,
dF/dt=\sum_{j=2}^{\infty}L_{j-1}(x^2+y^2)^j.$ Quantities
$L_j,\;j=\overline{1,\infty}$ are polynomials with respect to the
coefficients of system \eqref{eq1} called to be the Lyapunov
quantities \cite{Lia}. The origin $O(0,0)$ is a center for
\eqref{eq1} if and only if $L_j=0, \; j=\overline{1,\infty}.$ The
set of these conditions, which are polynomial equations in the
coefficients of the system \eqref{eq1}, is denumerable
\cite{ColDan} and hence by Hilbert's basis theorem, it is
sufficient that a finite number of them be satisfied.

A singular point $O(0,0)$ is a center for \eqref{eq1} if the
equations of \eqref{eq1} are invariant under reflection in a line
through the origin and reversion of time, called time-reversible
systems. The classical condition is that the system is invariant
under one or other of the transformations $(x, y, t) \to (-x,
y,-t)$ or $(x, y, t) \to (x,-y,-t)$. The first corresponds to
reflection in the $y$-axis and the second to reflection in the
$x$-axis.

\.Zo\l\c{a}dek \cite{Z3} mentioned three general mechanisms
for producing centers: searching for (1) a Darboux first integral
or (2) a Darboux--Schwarz--Christoffel first integral or by (3)
generating centers by rational reversibility, and he claimed that
these are sufficient for producing all cases of real polynomial
differential systems with centers. This conjecture is still open,
even for cubic systems \eqref{eq1}.

The time-reversibility in two-dimensional autonomous systems was
studied in \cite{Rom1} and the relation between time-reversibility
and the center-focus problem was discussed in \cite{TY}.

The problem of the center was solved for quadratic systems and for
cubic symmetric systems. The problem of finding a finite number of
necessary and sufficient conditions for the center in the cubic
case (for cubic system \eqref{eq1}) is still open. It was possible
to find the conditions for the center only in some particular
cases (see, for example, \cite{Sad2, CS1, CS2, CS3, C1, C2, C3,
Ko1, LLR, Sad1, Sad3, SC1, SC2, Z1}).

The problem of the center for cubic differential systems
\eqref{eq1} with invariant straight lines (real or complex) was
considered in \cite{CS1}, \cite{CS2}, \cite{CS3}, \cite{C1},
\cite{Ko1}, \cite{SC1}, \cite{SC2}. In these papers, the problem
of the center was completely solved for cubic systems with at
least three invariant straight lines. The main results of these
works is that every center in the cubic system \eqref{eq1} with at
least three invariant straight lines comes from a Darboux
integrability or from a rational reversibility.

The goal of this paper is to obtain center conditions for a cubic
differential system \eqref{eq1} with two distinct invariant
straight lines by using the method of Darboux integrability and
rational reversibility. Our main result is the following one.


\begin{theorem}\label{mainthm}
 The origin is a center for a cubic differential system \eqref{eq1},
 with at least two invariant straight lines, if one of the conditions
{\rm (i)--(xiv), (1)-- (26)} hold.
\end{theorem}

The paper is organized as follows. In Section 2 we summarize the
results obtained for cubic differential systems with at least
three invariant straight lines and centers. In Section 3 we find
four series of conditions for the existence of two distinct
invariant straight lines. In Section 4 we study the Darboux
integrability in cubic systems with two distinct invariant
straight lines and obtain the center conditions (i)--(xiv).
In Section 5 we describe the algorithm to transform a cubic
system \eqref{eq1} to one which is symmetric in a line by means of
a rational transformation. In Section 6 for cubic system
\eqref{eq1} with at least two invariant straight lines we obtain
conditions (1)--(26)  for the system to be rationally
reversible. In the last section, we prove the main theorem.


\section{Cubic systems with at least three
invariant straight lines}

 We shall study the problem of the center for cubic
system \eqref{eq1} assuming that \eqref{eq1} has invariant
straight lines.

\begin{definition}\label{def1} \rm
An algebraic invariant curve (or an algebraic particular integral)
of \eqref{eq1} is the solution set in $\mathbb{C}^2$ of an
equation $f(x,y)=0$, where $f$ is a polynomial in $x,y$ with
complex coefficients such that
$$
\frac{df}{dt}=\dot f=\frac{\partial f}{\partial x}P+
\frac{\partial f}{\partial y}Q=fK,
$$
for some polynomial in $x,y$, $K=K(x,y)$ with complex
coefficients, called the cofactor of the invariant algebraic curve
$f(x,y)=0$.
\end{definition}

By the above definition, a straight line
\begin{equation}\label{eq2}
L\equiv C+Ax+By=0,\quad A, B, C \in \mathbb{C},\quad (A, B)\neq (0,0),
\end{equation}
is an invariant straight line for \eqref{eq1} if and only if there
exists a polynomial $K(x,y)$ such that the following identity
holds
\begin{equation}\label{eq3}
A\cdot P(x,y)+B\cdot Q(x,y)\equiv (C+Ax+By)\cdot K(x,y).
\end{equation}

If the cubic system \eqref{eq1} has complex invariant straight
lines then obviously they occur in complex conjugated pairs
$$L\equiv C+Ax+By=0\quad\mbox{and}\quad
\overline{L}\equiv\overline{C}+\overline{A}x+\overline{B}y=0. $$

According to \cite{CS1} the cubic system \eqref{eq1} cannot have
more than four nonhomogeneous invariant straight lines, i.e.
invariant straight lines of the form
\begin{equation}\label{eq4}
1+Ax+By=0,\quad (A,B)\neq (0,0).
\end{equation}
As homogeneous straight lines $Ax + By = 0$, this system can have
only the lines $x \pm iy = 0, i^2 = -1$.

From \eqref{eq3} it results that \eqref{eq4} is an invariant
straight line of \eqref{eq1} if and only if $A$ and $B$ are the
solutions of the system
\begin{equation}\label{eq5}
\begin{gathered}
 F_1(A,B) \equiv AB^2 - fAB + bB^2 + rA - lB = 0,\\
 F_2(A,B) \equiv A^2B + aA^2 - gAB - kA + sB = 0,\\
 F_3(A,B) \equiv B^3 - 2A^2B + fA^2 + (c - b)AB - dB^2 - pA + nB =0,\\
 F_4(A,B) \equiv A^3 - 2AB^2 - cA^2 + (d - a)AB + gB^2 + mA - qB = 0.
\end{gathered}
\end{equation}
The cofactor of \eqref{eq4} is
\begin{align*}
K(x, y) &= -Bx + Ay + (aA - gB + AB)x^2+ (cA - dB + B^2 -A^2)xy\\
&\quad + (fA - bB - AB)y^2.
\end{align*}
The problem of the center for cubic systems with at least three
invariant straight lines was completely solved. The main results
of these works are summarized in the following three theorems.

\begin{theorem}[\cite{CS1,CS2}] \label{thm1}
Let the cubic differential system have four invariant straight lines
(real, real and complex, complex). Then any singular point with pure
imaginary eigenvalues of this system is a center if and only if the
first two Liapunov quantities vanish ($L_1 = L_2 = 0$).
\end{theorem}

\begin{theorem}[\cite{CS3,C1,SC1,SC2}] \label{thm2}
Let the cubic differential system have exactly three invariant straight
lines (real, real and complex). Then any singular point with pure
imaginary eigenvalues of this system is a center if and only if
the first seven Liapunov quantities vanish
($L_j = 0$, $j=1,\dots,7$).
\end{theorem}

\begin{theorem} \label{thm3}
Every center in the cubic differential system \eqref{eq1} with:
\begin{itemize}
\item[(1)] four invariant straight lines comes from a Darboux first
integral or a Darboux integrating factor;

\item[(2)] three invariant straight lines comes from a Darboux integrating
factor or a rational reversibility.
\end{itemize}
\end{theorem}




\section{Cubic systems with two invariant straight lines}


 Let the cubic system \eqref{eq1} have two distinct
invariant straight lines $L_1$ and $L_2$ real or complex. If
$L_1,\;L_2$ are complex and $L_2\neq \overline{L_1}$, then the
straight lines $ \overline{L_1}, \overline{L_2} $ conjugate with
$L_1,\;L_2$ will be also invariant for \eqref{eq1} (the
coefficients in \eqref{eq1} are real). In this case the system
\eqref{eq1} has four distinct invariant straight lines and the
problem of the center is solved by Theorem \ref{thm1}. If $L_1$ is complex
and $L_2$ is real, then the problem of the center is solved by
Theorem \ref{thm2}.

In this section, we shall consider cubic systems \eqref{eq1} with
two distinct invariant straight lines, where $L_1,\, L_2$ are real
or $L_1,\, L_2$ are complex ($L_2= \overline{L_1}$). It is easy to
see that for the relative positions of two distinct invariant
straight lines three cases can occur:
\begin{itemize}
\item[(1)] two parallel invariant straight lines;

\item[(2)] two homogeneous invariant straight lines;

\item[(3)] two nonhomogeneous and nonparallel invariant straight lines.
\end{itemize}

\subsection{Two parallel invariant straight lines}

 Let the cubic system \eqref{eq1} have two parallel
invariant straight lines $L_1,\,L_2$, then by a rotation of axes
we can make them parallel to the axis of ordinates ($Oy $). Note
that by a rotation of axes of coordinates the linear part of
\eqref{eq1} preserves the form.

Assume $L_1 $ and $L_2 $ are complex, then $L_2= \overline{L_1}$.
From $L_1 ||\overline{L_1}$, it follows that $L_1 $ looks as $
1+A(x+By)=0, $ where $A$ is a complex number and $B$ is real. In
this case, via a rotation of axes about the origin, it is also
possible to make the straight lines $L_1 $ and $L_2 $ to be
parallel to the axis $Oy.$

In order that the cubic system \eqref{eq1} had two invariant
straight lines $L_1, L_2 $ parallel to the axis $Oy $, it is
necessary and sufficient that the following coefficient conditions
to be satisfied
\begin{equation}\label{eq6}
a=f=k=p=r=0,\quad m(c^2-4m)\neq 0.
\end{equation}
In this case the invariant straight lines $L_1$ and $L_2$ are
\begin{equation}\label{eq7}
 L_{1,2}\equiv 1+\frac{c\pm
\sqrt{c^2-4m}}{2}x=0.
\end{equation}


\subsection{Two homogeneous invariant straight lines}

 For homogeneous invariant straight lines, it is easily
verified that the cubic system \eqref{eq1} can have only the lines
$x\pm iy=0$, $i^2=-1$. These lines are invariant if and only if the
following conditions hold
\begin{equation}\label{eq8}
g = b + c,\quad f = a + d,\quad  q = p + l-k,\quad s = m+n - r.
\end{equation}


\subsection{Two nonhomogeneous and nonparallel invariant
straight lines}


 Let the cubic system \eqref{eq1} have two nonhomogeneous
and nonparallel invariant straight lines $L_1,\, L_2$ intersecting
at a point $(x_0,y_0)$. The intersection point $(x_0,y_0)$ is a
singular point for \eqref{eq1} and has real coordinates. By
rotating the system of coordinates ($x\to
x\cos\varphi-y\sin\varphi, y\to x\sin\varphi+y\cos\varphi)$ and
rescaling the axes of coordinates $(x\to \alpha x,\,y\to \alpha
y)$, we obtain $ L_1\cap L_2=(0,1)$. In this case the invariant
straight lines can be written as
\begin{equation}\label{eq9}
 L_j\equiv 1+ A_jx-y=0,\quad A_j\in \mathbb{C},\quad j=1,2;\; A_1-A_2\neq 0.
\end{equation}

As the point $(0, 1)$ is a singular point for \eqref{eq1}, then
$P(0, 1) = Q(0, 1) = 0$. These equalities yield
$r = -f - 1, l = -b$. Substituting $B = -1, r = -f - 1$ and
 $l = -b$ in \eqref{eq5}
we find that the straight lines \eqref{eq9} are invariant for
\eqref{eq1} if and only if $A_1$ and $A_2$ are the solutions of
the system
\begin{equation}\label{eq10}
\begin{gathered}
 F_2(A_1) \equiv (a - 1)A_1^2 + (g - k)A_1 - s = 0,\\
 F_3(A_1) \equiv (f + 2)A_1^2 + (b - c - p)A_1 - d - n - 1 = 0,\\
 F_4(A_1) \equiv A_1^3 - cA_1^2 + (a - d + m - 2)A_1 + g + q = 0,\\
 F_2(A_2) \equiv (a - 1)A_2^2 + (g - k)A_2 - s = 0,\\
 F_3(A_2) \equiv (f + 2)A_2^2 + (b - c - p)A_2 - d - n - 1 = 0,\\
 F_4(A_2) \equiv A_2^3 - cA_2^2 + (a - d + m - 2)A_2 + g + q = 0.
\end{gathered}
\end{equation}
It is easy to see from \eqref{eq10} that the system \eqref{eq1}
can have two distinct invariant straight lines of the form
\eqref{eq9} if and only if the following coefficient conditions
are satisfied
\begin{equation}\label{eq11}
\begin{gathered}
 k = (a - 1)(A_1 + A_2) + g,\quad l = - b,\quad  r =  - f - 1, \\
 m = - A_1^2 - A_1A_2 - A_2^2 + c(A_1 + A_2) - a + d + 2,\\
 n = -(f + 2)A_1A_2 - (d + 1),\quad  s = (1 - a)A_1A_2,\\
 p = (f + 2)(A_1 + A_2) + b - c,\quad q = (A_1 + A_2 - c)A_1A_2-g.
\end{gathered}
\end{equation}

\begin{theorem} \label{thm4}
The cubic differential system \eqref{eq1} has at least two
distinct invariant straight lines if and only if one of the sets
of conditions \eqref{eq6}, \eqref{eq8} and \eqref{eq11} holds.
\end{theorem}



\section{Darboux integrability in cubic systems
with two invariant straight lines}


 Let the cubic system \eqref{eq1} have sufficiently many
invariant algebraic curves $f_j(x,y)=0,\; j=1,\dots,q$ with
cofactors $K_j(x,y)$. Then in most cases a first integral (an
integrating factor) can be constructed in the Darboux form
\cite{Dar1}
\begin{equation}\label{eq12}
f_1^{\alpha_1}f_2^{\alpha_2}\cdots f_q^{\alpha_q}
\end{equation}
and we say that the cubic system \eqref{eq1} is Darboux
integrable. The function \eqref{eq12}, with $\alpha_i\in
\mathbb{C}$ not all zero, is a first integral (an integrating
factor) for \eqref{eq1} if and only if
$$
\sum_{i=1}^q\alpha_iK_i \equiv 0\quad
\Bigl(\sum_{i=1}^q\alpha_iK_i \equiv \frac{\partial Q}{\partial
y}- \frac{\partial P}{\partial x}\Bigr).
$$
The method of Darboux turns out to be very useful and elegant one
to prove integrability for some classes of systems depending on
parameters.

In this section we shall find center conditions for cubic system
\eqref{eq1} with two invariant straight lines by constructing an
integrating factor of the Darboux form
\begin{equation}\label{eq13}
\mu=L_1^{\alpha_1}L_2^{\alpha_2},
\end{equation}
where $L_j=0$, $j=1,2$ are invariant straight lines for \eqref{eq1}
with cofactor $K_j(x,y)$ and $\alpha_j\in \mathbb{C}$. The cubic
system \eqref{eq1} will have an integrating factor of the form
\eqref{eq13} if and only if the numbers $\alpha_j$ satisfy the
following identity
\begin{equation}\label{eq14}
\alpha_1K_1(x,y)+\alpha_2K_2(x,y)\equiv \frac{\partial Q}{\partial
y}- \frac{\partial P}{\partial x}.
\end{equation}


\subsection{Centers of system \eqref{eq1} with two
parallel invariant straight lines and Darboux integrability}

\begin{lemma} \label{lem1}
The following set of conditions is sufficient condition for the
origin to be a center for system \eqref{eq1}:
\begin{itemize}
\item[(i)]
$a=d=f=k=l=p=q=r=0$.
\end{itemize}
\end{lemma}

\begin{proof}
 Let the conditions \eqref{eq6} hold, then the cubic
system \eqref{eq1} has two invariant straight lines of the form
\eqref{eq7} with cofactors $K_{1,2}(x,y)=[y(c + 2mx \pm
\sqrt{c^2-4m}\,)]/2$. Taking into account the cofactors, the
identity \eqref{eq14} yields $d = q = l = 0$ and
$$
 \alpha_{1,2}=[(n-m)\sqrt{c^2-4m} \pm
(2bm-cn)]/(m\sqrt{c^2-4m}\,),
$$
we obtain the center conditions (i).
\end{proof}

\subsection{Centers of system \eqref{eq1} with two
homogeneous invariant straight lines and Darboux integrability}

\begin{lemma} \label{lem2}
The following three sets of conditions are sufficient conditions
for the origin to be a center for system \eqref{eq1}:
\begin{itemize}
\item[(ii)]
$ c =  - 2b$, $d =  - 2a$, $f =  - a$, $g =  - b$,
$n =  2r- m$, $p = - l$,  $q =  - k$, $s = r$;

\item[(iii)]
$ a = d = f = 0$, $g = b + c$, $k = l$, $m = (2br + cn -cr)/(2b)$,
 $p = q=[l(b + c)]/b$, $s = (2bn + cn - cr)/(2b)$;

\item[(iv)]
$ c = (bd)/a$,  $f = a + d$, $g = [b(a + d)]/a$, $p = q=[l(a + d)]/a$,
$k = l$, $m = (2ar + dn - dr)/(2a)$, $s = (2an + dn - dr)/(2a)$.
\end{itemize}
\end{lemma}

\begin{proof}  Assume the conditions \eqref{eq8} are satisfied, then
the cubic system \eqref{eq1} has two homogeneous invariant
straight lines $x\pm iy=0$ with cofactors
\begin{gather*}
\begin{aligned}
 K_1(x,y) &= - i + (a - ib - ic)x - (b + ia + id)y+
(k - im - in + ir)x^2\\
&\quad + (r-n - il - ip)xy  - (l+ir)y^2,
\end{aligned}\\
K_2=\overline{K_1}.
\end{gather*}
In this case the system \eqref{eq1} will have an integrating
factor of the form \eqref{eq13} if and only if the identity
\eqref{eq14} holds. Substituting in this identity the expressions
of the cofactors and identifying the coefficients of
$x^0, x, y, x^2, xy$ and $y^2$, we obtain that $\alpha_2 = \alpha_1$ and
$\alpha_1$ obey the following system of algebraic equations:
\begin{equation}\label{eq15}
\begin{gathered}
 (r-n)\alpha_1 + m - n=0,\quad  2a(\alpha_1 + 1) - d=0,\quad
 2b(\alpha_1 + 1) - c=0,\\
  2k(\alpha_1+2) - l - p=0,\quad (\alpha_1 + 2)(k - l)=0.
\end{gathered}
\end{equation}

Let $\alpha_1 =- 2$, then from \eqref{eq15} we obtain the
conditions (ii). Assume $\alpha_1 \neq - 2$, then $k=l$. If $a=0$,
then $b\neq 0$, $\alpha_1=(c - 2b)/(2b)$ and from \eqref{eq15} we
get the conditions (iii). If $a\neq 0$, then
$\alpha_1= (d -2a)/(2a)$ and \eqref{eq15} implies the conditions (iv).

In each of the cases (ii)--(iv), the system \eqref{eq1} has an
integrating factor of the form \eqref{eq13} and therefore the
origin is a center for \eqref{eq1}.
\end{proof}



\subsection{Centers of system \eqref{eq1} with two nonhomogeneous and nonparallel
invariant straight lines and Darboux integrability}

 Let the coefficient conditions \eqref{eq11} hold. Denote
$\lambda=a-1$, $\gamma=f+2$ and consider the following two cases:

\subsubsection{$\lambda = 0$} In this case $a=1$ and
\eqref{eq11} yields the following conditions
\begin{equation}\label{eq16}
\begin{gathered}
a = 1,\quad k = g,\quad l =  - b,\quad
  q=[(d+n+1)(c\gamma+b-c-p)-g\gamma^2]/\gamma^2,\\
 m=[(\gamma(d+1)+c^2)(\gamma-1)-(b-p)(c(\gamma-2)+b-p)-n\gamma]/\gamma^2,\\
 r =  1-\gamma,\quad s = 0,\quad \gamma[(b-c-p)^2 + 4\gamma(d+n+1)] \neq  0
\end{gathered}
\end{equation}
for the existence of two distinct invariant straight lines of the
form \eqref{eq9} where $A_j$, $j=1,2$ are the solutions of the
equation
\begin{equation}\label{eq17}
\gamma A^2 + (b - c - p)A - d - n - 1 = 0.
\end{equation}

\begin{lemma}
The following five sets of conditions are sufficient conditions
for the origin to be a center for system \eqref{eq1}:
\begin{itemize}
\item[(v)]
$a = \gamma=1$, $d = -2$,  $k = -q=g$, $p=-l =  b$, $m =  - n$,
$r = s = 0$;

\item[(vi)]
$ a = n=1$, $b=l=s=0$, $d = -2$,
$k = -q=g$, $p=c(\gamma-1)$,
$f=\gamma-2$, $m=-1$,  $r=1-\gamma$;

\item[(vii)]
$a = n=1$, $d = -2$, $f=\gamma-2$, $k = -q=g$,
$l = - b$, $r=1-\gamma$, $s=0$,
$c = [2b(\gamma - 2)]/\gamma$, $m = -(4b^2\gamma - 4b^2 +
\gamma^2)/\gamma^2$, $p=b(4-3\gamma)/\gamma$;


\item[(viii)]
$a = 1$, $d = -1$,  $f=(-3)/2$, $k =g=q=s=0$, $l = - b$,
$m=-2n$, $p=(2b-c)/2$, $r=1/2$;


\item[(ix)]
$a = 1$, $k = g$, $l =  - b$,
$q=[(d+n+1)(c\gamma+b-c-p)-g\gamma^2]/\gamma^2$,
$m=[(\gamma(d+1)+c^2)(\gamma-1)-(b-p)(c(\gamma-2)+b-p)-n\gamma]/\gamma^2$,
$f=\gamma-2$, $r =  1-\gamma$,  $s = 0$, $p = b(1-d) + (c-2b)\gamma - c$,
$g=[b((d+\gamma)^2-\gamma^2+(n+1)(d+2\gamma))]/[(d+2)\gamma^2]$,
$n=[b(d+2\gamma)(c\gamma-2bd-2b\gamma)+d\gamma(d+1)(\gamma-1)]/[\gamma(d+2\gamma)]
$.
\end{itemize}
\end{lemma}

\begin{proof}
 Indeed, if the conditions \eqref{eq16} hold, then the
cubic system \eqref{eq1} has two invariant straight lines of the
form $L_{1,2}\equiv 1+A_{1,2}x-y=0$ with cofactors
$$
K_{1,2}(x,y)= x+ A_{1,2}y + gx^2 + (1 + d - A_{1,2}^2
+ cA_{1,2})xy + ((\gamma- 1)A_{1,2} + b)y^2,
$$
 where $A_1$, $A_2$ are the roots of the equation
\eqref{eq17}:
$$
A_{1,2}=(p-b+c\pm
\sqrt{(b-c-p)^2+4\gamma(d+n+1)}\;)/(2\gamma).
$$

In this case system \eqref{eq1} will have an integrating
factor of the form \eqref{eq13} if and only if the identity
\eqref{eq14} holds. Substituting in \eqref{eq14} the expressions
of the cofactors and identifying the coefficients of $x, y, x^2,
xy$ and $y^2$, we obtain that
$$
\alpha_1 = d - 2-\alpha_2,\quad
\alpha_2 = [(d-2)A_1 -2b + c]/(A_1 - A_2)
$$
and
\begin{equation}\label{eq18}
\begin{gathered}
 p = b(1-d) + (c-2b)\gamma - c,\; g(d+2)\gamma^2-b(d+n+1)(d+2\gamma)=0,\\
 (d^2 - 4b^2 + 2bc + d - 2n)\gamma^2 + d\gamma(bc - 6b^2 - d - n-1) - 2b^2d^2=0.
\end{gathered}
\end{equation}

Let $d=-2$. If $\gamma=1$, then from \eqref{eq18} we obtain the
conditions (v); if $\gamma\neq 1$ and $b=0$ -- the conditions (vi);
if $b(\gamma-1)\neq 0$ and $n=1$ -- the conditions (vii).

Assume $d\neq -2$. If $d+2\gamma=0$, then we get the conditions
(viii) and if $d+2\gamma\neq 0$ -- the conditions (ix). In each of
the cases (v)--(ix), the system \eqref{eq1} has an integrating
factor of the form \eqref{eq13} and therefore the origin is a
center for \eqref{eq1}.
\end{proof}


\subsubsection{$\lambda \neq  0$} In this case
\eqref{eq11} yields the following conditions
\begin{equation}\label{eq19}
\begin{gathered}
 p=[(b-c)\lambda+(k-g)\gamma]/\lambda,\quad
 q=[\lambda(cs-g\lambda)+s(g-k)]/\lambda^2,\\
 l =  - b,\quad  m=[(d-\lambda+1)\lambda^2+\lambda(c(k-g)-s)-(k-g)^2]/\lambda^2,\\
 r = 1-\gamma,\quad n=[s\gamma-(1+d)\lambda]/\lambda,\quad
(g-k)^2+4s\lambda\neq 0
\end{gathered}
\end{equation}
for the existence of two distinct invariant straight lines of the
form \eqref{eq9} where $A_j,\; j=1,2$ are the solutions of the
equation
\begin{equation}\label{eq20}
\lambda A^2 + (g - k)A - s = 0.
\end{equation}

\begin{lemma} \label{lem4}
The following five sets of conditions are sufficient conditions
for the origin to be a center for system \eqref{eq1}:
\begin{itemize}
\item[(x)]
$a = \lambda+1$, $b = - (2c\lambda + g)/(2\lambda)$,
$d=2\lambda-1$, $f=(-3)/2$,
$k = c\lambda + g$,  $l = (2c\lambda + g)/(2\lambda)$,
$ m =(\lambda^2 - s)/\lambda$, $q = - g$,
$ n = (s-4\lambda^2 )/(2\lambda)$,  $p =  - (3c\lambda +g)/(2\lambda)$,
$r = 1/2$;

\item[(xi)]
$ a = \lambda+1$, $d = -2$, $f = \lambda - 1$, $k = c\lambda + g$,
$l=-b = c-g$, $q =  - g$,
$m = -\lambda - 1 - s\lambda^{-1}$, $n = s + 1 + s\lambda^{-1}$,
$p = c(\lambda - 1) + g$,  $r = - \lambda$;

\item[(xii)]
$a = \lambda + 1$, $b=l=0$, $c = [g(2\lambda - d - 2)]/(2\lambda)$,
$k = [g(2\lambda - d)]/2$,
$f=\gamma-2$, $p=[(b-c)\lambda+(k-g)\gamma]/\lambda$,
$ n=[s\gamma-(1+d)\lambda]/\lambda$,
$m=[(d-\lambda+1)\lambda^2+\lambda(c(k-g)-s)-(k-g)^2]/\lambda^2$,
$r = 1-\gamma$,
$q=-g$, $s = [\lambda(d^2 - 2d\lambda + 3d - 4\lambda + 2)]/(d +
2\gamma - 2\lambda)$;


\item[(xiii)]
$a=\lambda+1$, $f=-2$, $d=2\lambda$, $n =  - (2\lambda + 1)$,
$b=[(c\lambda+g-k)(c\lambda+2(g-k))-2\lambda^2(\lambda+1)]
/[2\lambda(c\lambda+g-k)]$,
$ p=(b-c)$, $q=[\lambda(cs-g\lambda)+s(g-k)]/\lambda^2$,
$r=1$,
$l =  - b$, $m=[\lambda^2(\lambda+1)+\lambda(c(k-g)-s)-(k-g)^2]/\lambda^2$,
$s=-\lambda^2(2\lambda^2(\lambda+1)+(k+g)(k-g-c\lambda))/(c\lambda+g-k)^2$;

\item[(xiv)]
$a = \lambda + 1$, $f=\gamma-2$, $r = 1-\gamma$, $l =  - b$,
$b = [\gamma(c\lambda + g - k)]/[\lambda(d + 2(\gamma -
\lambda))]$,
$n=[s\gamma-(1+d)\lambda]/\lambda$,
$p=[(b-c)\lambda+(k-g)\gamma]/\lambda$,
$q=[\lambda(cs-g\lambda)+s(g-k)]/\lambda^2$,
$m=[(d-\lambda+1)\lambda^2+\lambda(c(k-g)-s)-(k-g)^2]/\lambda^2$,
$s=[\lambda^2((2b-c-2g)\lambda+3k-g+dg)]/(c\lambda+g-k)$,
$2(d+2)\lambda^3+((c-b)^2-b^2-d^2-3d-2s-2)\lambda^2
 +((3c-2b)(g-k)+(d+2\gamma)s)\lambda +2(g-k)^2=0$.
\end{itemize}
\end{lemma}

\begin{proof}
 Indeed, if the conditions \eqref{eq19} hold, then the
cubic system \eqref{eq1} has two invariant straight lines of the
form $L_{1,2}\equiv 1+A_{1,2}x-y=0$ with cofactors
$K_{1,2}(x,y)= x+ A_{1,2}y + (g+\lambda A_{1,2})x^2 + (1 + d + cA_{1,2} -
A_{1,2}^2 )xy + (b + (\gamma-1)A_{1,2})y^2$, where $A_1$, $A_2$
are the roots of the equation \eqref{eq20}:
$$
A_{1,2}=(k-g\pm \sqrt{(g-k)^2+4\lambda
s}\,)/(2\lambda).
$$
In this case the system \eqref{eq1} will have an integrating
factor of the form \eqref{eq13} if and only if the identity
\eqref{eq14} holds. Substituting in this identity the expressions
of the cofactors and identifying the coefficients of
 $x, y, x^2, xy$ and $y^2$, we obtain that
$$
\alpha_1 = d - 2(\lambda+1) - \alpha_2 ,\; \alpha_2 =
[c-2b +(d-2-2\lambda)A_1]/(A_1 - A_2)
$$
and
\begin{equation}\label{eq21}
\begin{gathered}
 b=[(c+2g)\lambda^3+(g-dg-3k)\lambda^2+(c\lambda+g-k)s]/(2\lambda^3),\\
 2b\lambda^2+((c-2b)\gamma-bd)\lambda+(g-k)\gamma=0,\\
\begin{aligned}
& 2(d+2)\lambda^3+((c-b)^2-b^2-d^2-3d-2s-2)\lambda^2 \\
& +((3c-2b)(g-k)+(d+2\gamma)s)\lambda +2(g-k)^2=0.
\end{aligned}
\end{gathered}
\end{equation}

Let $k=c\lambda+g$. If $d=2(\lambda-\gamma)$, then from
\eqref{eq21} we obtain the conditions (x) and (xi); if
$d\neq 2(\lambda-\gamma)$, then \eqref{eq21} implies the
conditions (xii).

Let $k\neq c\lambda+g$. If $d=2(\lambda-\gamma)$, then from
\eqref{eq21} we get the conditions (xiii) and  if
$d\neq 2(\lambda-\gamma)$, then \eqref{eq21} yields the conditions
(xiv).

In each of the cases (x)--(xiv), the system \eqref{eq1} has an
integrating factor of the form \eqref{eq13} and therefore the
origin is a center for \eqref{eq1}.
\end{proof}

Taking into account Lemmas \ref{lem1}--\ref{lem4},
for cubic differential system \eqref{eq1} with two distinct invariant
straight lines (real or complex conjugated), it was proved the following
theorem.

\begin{theorem} \label{thm5}
The  differential system \eqref{eq1} with two distinct
invariant straight lines  has a Darboux integrating factor of the
form \eqref{eq13} if and only if one of the sets of conditions
{\rm (i)--(xiv)} is satisfied.
\end{theorem}




\section{Rational transformation in cubic systems}

 It is well known from Poincar\'e \cite{P1} that if a
differential system with a singular point $O(0,0)$ of a center or
a focus type is invariant by the reflection with respect, for
example, to the axis $X=0$ and reversion of time then $O(0,0)$ is
a center for \eqref{eq1} ($X=0$ is called the axis of symmetry).
It is clear that \eqref{eq1} has a center at $O(0,0)$ if there
exists a diffeomorphism
\begin{equation}\label{eq21a}
\Phi:U\to V,\quad \Phi=\{X= \varphi(x,y),\, Y=\psi(x,y)\},\quad
\Phi(0,0)=(0,0),
\end{equation}
which brings system \eqref{eq1} to a system with the axis of
symmetry. In particular, if $\varphi(x,y)$ and $\psi(x,y)$ are
rational functions in \eqref{eq21a}, then we say that \eqref{eq1}
is rationally reversible (\cite{Z4}).

In \cite{LP1} is described an algorithm based on application of
Gr\"oebner bases in the search for a bilinear transformation,
which is invertible in a neighbourhood of the origin and transform
a given system to one which is symmetric in a line. This algorithm
is applied to find center conditions for some cubic systems.

In this section we shall consider a general mechanism to produce
center by rational reversibility. We seek a transformation of the
form
\begin{equation}\label{eq22}
x=\frac{a_1X+b_1Y}{a_3X+b_3Y-1},\quad
y=\frac{a_2X+b_2Y}{a_3X+b_3Y-1}
\end{equation}
with $a_1b_2-b_1a_2\neq 0$ and $a_j, b_j\in \mathbb{R}$,
$j=1,2,3$. The condition $a_1b_2-b_1a_2\neq 0$ guarantees that
\eqref{eq22} is invertible in a neighborhood of $O(0,0)$ and the
singular point is mapped to $X=Y=0$. Applying the transformation
\eqref{eq22} to \eqref{eq1} we obtain a system of the form
$$
\dot X=\frac{P(X,Y)}{R(X,Y)},\quad \dot Y=\frac{Q(X,Y)}{R(X,Y)},
$$
whose orbits in some neighborhood of $O(0,0)$ are the same as
those of the system
\begin{equation}\label{eq23}
\dot X=\sum_{i+j=0}^4U_{ij}X^iY^j\equiv P(X,Y),\quad
\dot Y=\sum_{i+j=0}^4V_{ij}X^iY^j\equiv Q(X,Y),
\end{equation}
where $U_{ij}, V_{ij}$ are polynomials in the coefficients of the
original system and the parameters $a_1, a_2, a_3, b_1, b_2, b_3$
of the transformation.

The requirement is to show that $a_1, a_2, a_3, b_1, b_2, b_3$ can
be chosen so that the system \eqref{eq23} is symmetric in the
$Y$-axis; i.e. the transformation \eqref{eq22} brings in some
neighborhood of $O(0,0)$ the system \eqref{eq1} to one equivalent
with a polynomial system
\begin{equation}\label{eq24}
\frac{dX}{dt}=Y+M(X^2,Y),\quad   \frac{dY}{dt}=-X(1+N(X^2,Y)).
\end{equation}
The obtained system has an axis of symmetry $X=0$ and therefore
$O(0,0)$ is a center for \eqref{eq1}. The system \eqref{eq24} is
equivalent to the system \eqref{eq23} if  the following conditions
are satisfied:
$$
 U_{31}\equiv V_{22}=0,\quad U_{13}\equiv V_{04}=0,\quad
 U_{10}\equiv V_{01}=0,\quad V_{40}= 0,\quad U_{00}= 0,\quad V_{00}= 0
$$
and
\begin{gather}
 V_{04}\equiv a_3[sb_1^4 + ((k + q)b_1^2 + (m +
n)b_1b_2 + (l + p)b_2^2)b_1b_2 + rb_2^4]=0, \nonumber\\
\begin{aligned}
 V_{22}&\equiv a_3[ma_1^2b_2^2 + ca_1b_2^2a_3+ (2p -
3k -q)a_1a_2b_2^2 + da_2b_1^2a_3 \\
&\quad +  (3l + p - 2q)a_2^2b_1b_2 + (3(r +s)
 - 2(m +n))a_2^2b_2^2 + na_2^2b_1^2\\
&\quad +  (2b + c - 2g)a_2b_1b_2a_3 +
(2f - 2a- d)a_2b_2^2a_3 + a_3^2]=0,
\end{aligned} \nonumber\\
\begin{aligned}
 U_{30}&\equiv 2aa_1^2b_2a_3 + [(m - s)a_1 + (p -
q)a_2 +2(c - g)a_3]a_1a_2b_2\\
&\quad + ka_1^3b_2 +  a_2^3(lb_1 - nb_2 +
rb_2)+ 2a_2^2a_3(bb_1 - db_2 + fb_2)=0,
\end{aligned} \nonumber
\\
\begin{aligned}
U_{12}&\equiv (qa_2 + 2ga_3)b_1^3 + [2(a+d)a_3 + (m+ 2n - 3s)a_2]b_1^2b_2
 + \big[(3l-3k \\
&\quad+ 2p - 2q)a_2  + 2(b+c)a_3\big]b_1b_2^2 + [pa_1 - (2m + n - 3r)a_2
+ 2fa_3]b_2^3=0,
\end{aligned} \nonumber\\
\begin{aligned}
 V_{03}&\equiv (ka_2 - ga_3)b_1^3 + [(m-s)a_2 -
(a+d)a_3]b_1^2b_2\\
&\quad +[(p - q)a_2 - (b+c)a_3]b_1b_2^2 +(la_1 - (n-r)a_2 - fa_3)b_2^3=0,
\end{aligned} \nonumber\\
\begin{aligned}
 V_{21}&\equiv qa_1^3b_2 + (m + 2n - 3s)a_1^2a_2b_2 + (d - a)a_1^2a_3b_2 \\
&\quad + (3l - 3k + 2p - 2q)a_1a_2^2b_2 +
[pb_1 + (3r - 2m - n)b_2]a_2^3 \\
&\quad +  (2b - g)a_1a_2a_3b_2 + [(f - 2a)b_2 -(b-c)b_1]a_2^2a_3=0,
\end{aligned} \nonumber\\
 V_{02}\equiv aa_2b_1^2  + (c - g)b_1b_2a_2 + (ba_1 -da_2 + fa_2)b_2^2-a_3=0,
\nonumber\\
 V_{20}\equiv ga_1^3 + (a+d)a_1^2a_2 +(b+c)a_1a_2^2 + fa_2^3 + 2a_3=0,
\nonumber\\
 U_{11}\equiv  [db_1+(2b + c - 2g)b_2]a_2b_1 + [ca_1+(2f-2a - d)a_2]b_2^2+3a_3=0,
\nonumber\\
 U_{01}\equiv b_1^2 + b_2^2 - 1=0,\quad
 U_{10}\equiv a_1b_1 + a_2b_2=0,\quad V_{10}\equiv a_1^2 + a_2^2 - 1=0.
\label{eq25}
\end{gather}

Next we shall study the compatibility of \eqref{eq25} assuming
that the cubic system \eqref{eq1} has two distinct invariant
straight lines (real or complex conjugated). If \eqref{eq25} is
compatible, then the cubic system \eqref{eq1} with two distinct
invariant straight lines is rationally reversible and a singular
point $O(0,0)$ is a center.


\section{Rationally reversible cubic systems with
at least two  invariant straight lines}
In this section we shall find conditions on the
coefficients, for cubic system \eqref{eq1} with two distinct
invariant straight lines, which allow us to transform the system
to the system \eqref{eq24}, symmetric in a line, by means of the
rational transformations \eqref{eq22}.

 It is easy to verify that the
equations $ U_{01}=0,\; V_{10}=0 $ of \eqref{eq25} admit the
following parametrization
\begin{gather*}
 a_1 = (2u)/(u^2 + 1),\quad a_2 = (u^2 - 1)/(u^2 + 1),\\
 b_1 = (2v)/(v^2 + 1),\quad b_2 = (v^2 - 1)/(v^2 + 1),
\end{gather*}
where $u$ and $v$ are some real parameters. In this case
$U_{10}\equiv j_1j_2=0$, where $j_1=uv+u-v+1$, $j_2=uv-u+v+1$.

Next assume  $j_1=0$, then $v = (1 + u)/(1-u)$ and
$ U_{10}\equiv 0 $. The case $j_2=0$ is equivalent with $j_1=0$ if we take into
consideration that $j_2(u,v)=j_1(-u,-v)$.


\subsection{Centers of system \eqref{eq1} with two
parallel invariant straight lines and reversibility}


Consider the system of algebraic equations \eqref{eq25} and let
the conditions \eqref{eq6} hold.

\subsubsection{$a_3=0$} In this case $V_{04}\equiv 0$ and
$ V_{22}\equiv 0$.
 If $u=0;\, u=-1$ or $u(u+1)\neq 0$, then from the
equations of \eqref{eq25} we obtain, respectively, the following
three sets of conditions for the existence of a center:
\begin{itemize}
\item[(1)] $a=d=f=k=l=p=q=r=0$;

\item[(2)] $a=b=c=f=g=k=l=p=q=r=0$;

\item[(3)] $ a=f=k=p=r=0$, $l = [4mu(u^6 - 7u^4 + 7u^2 - 1)]/(u^2 + 1)^4$,
$b= [c(6u^2-u^4 - 1)]/(u^2 + 1)^2$,
$s = [m(u^4 - 6u^2 + 1)^2]/(u^2 + 1)^4$,
$g = -b$, $q = -3l$, $d = [2cu(10u^2 - 3u^4 - 3)]/[(u^2 + 1)^2(u^2 - 1)]$,
$n = [-2m(u^8 - 20u^6 + 54u^4 - 20u^2 + 1)]/(u^2 + 1)^4$.
\end{itemize}

\subsubsection{$a_3\neq 0$} In this case from the equation
$V_{02}=0$ of \eqref{eq25} we have
$$
a_3=[2u((g-c)u^4-2du^3+2(2b+c-g)u^2+2du+g-c)]/(u^2+1)^3.
$$
If $u=0$, then  \eqref{eq25} yields the center conditions which
are contained in  (1).

If $u=-1$, then from the equations of \eqref{eq25} we obtain the
following conditions for the existence of a center
\begin{itemize}
\item[(4)]
$ a=f=k=l=p=r=0$, $c=-3b$, $g=-2b$, $m=2b^2$, $q=-bd$.
\end{itemize}
Assume $u(u+1)\neq 0$, then from the equations
$\{U_{11}=0,\,V_{04}=0,\, U_{12}=0,\, V_{03}=0,\, V_{21}=0,\, U_{30}=0\}$ of
\eqref{eq25} we express, $g, l, s, m, q, n$, respectively and
$$
V_{22}\equiv V_{20}\equiv (3b + c)(3u^2 - 1)(u^2 -
3)u - d(u^4 - 10u^2 + 1)(u^2-1)=0.
$$


If $(3u^2 - 1)(u^2 - 3)=0$, then we obtain the following two sets
of conditions for the existence of a center:
\begin{itemize}
\item[(5)]
$ a=d=f=k=p=r=0$, $g = (c - 7b)/5$, $n = [3c(2b - c)]/20$,
$l =[\sqrt{3}(8b^2 + 2bc - 3c^2)]/100$, $m = [2(2c^2 - 2b^2 - 3bc)]/25$,
$q = [3\sqrt{3}(3c^2 - 8b^2 - 2bc)]/100$,
 $s = [3(16b^2 - 6bc - c^2)]/100$;

\item[(6)]
$a=d=f=k=p=r=0$, $g = (c - 7b)/5$, $n = [3c(2b - c)]/20$,
$l =[\sqrt{3}(-8b^2 - 2bc + 3c^2)]/100$, $m = [2(2c^2 - 2b^2 - 3bc)]/25$,
$q = [3\sqrt{3}(-3c^2 + 8b^2 + 2bc)]/100$,
$s = [3(16b^2 - 6bc -c^2)]/100$.
\end{itemize}


If $(3u^2 - 1)(u^2 - 3)\neq 0$, then we get the following
conditions for the existence of a center
\begin{itemize}
\item[(7)]
$a=f=k=p=r=0$,  $m=[h(du^2-4bu-8cu-d)]/(100u^2)$,
$g=[4(b+2c)(u^5+u)+d(1-19u^2+19u^4-u^6)-8(4b+3c)u^3]/[10u(u^2-1)^2]$,
$l=[h(u^2-1)(d(u^6-19u^4+19u^2-1)-(19b+3c)(u^5+u)
+6(7b-c)u^3)]/[25u(u^2+1)^4]$,
$ h=d+4bu-2cu-du^2$,
\begin{align*}
n&=[h(d(1-9u^2+230u^4-230u^6+9u^8-u^{10})+3(3b+c)(u+u^9)\\
&\quad + 16(c-12b)(u^7+u^3)+2(279b+13c)u^5)]/[50u^2(u^2+1)^4],\\
q&=[h(d(1-21u^2+458u^4-458u^6+21u^8-u^{10})+12(2b-c)(u+u^9)\\
&\quad+ 40(c-9b)(u^3+u^7)+8(144b+13c)u^5)]/[50u(u^2-1)(u^2+1)^4],\\
s&=[2hu(d(4u^7-76u^5+76u^3-4u)+5(b-c)(u^8+1)\\
&\quad+ 8(c-7b)(u^6+u^2)+2(99b+13c)u^4)]/[25(u^2-1)^2(u^2+1)^4],
\end{align*}
$c = [d(u^4 - 10u^2 + 1)(u^2-1)]/[( 3u^2 - 1)(u^2 - 3)u] - 3b$.
\end{itemize}

\begin{remark} \label{rmk1} \rm
In each of the cases (3) and (7) the system \eqref{eq1} has
four invariant straight lines. Thus, in conditions (3) besides the
invariant straight lines \eqref{eq7}, the system \eqref{eq1} has
two more invariant straight lines $ L_{3,4}=(c \pm
\sqrt{c^2-4m})[(u^4 - 6u^2 + 1)x + 4u(1-u^2)y] + 2(u^2 + 1)^2$;
 in conditions (7): $L_3= [4bu(3u^2-1)(u^2-3)-2d(u^2 + 2u - 1)(u^2 - 2u
- 1)(u^2-1)](u^2y+2ux-y)- (3u^2 - 1)(u^2 + 1)^2(u^2 - 3)$,
$L_4= (3u^2-1)(u^2-3)[4bu(u^2-1)(u^2x-2uy-x)-2u(u^2+1)^2]-d(u^2-1)(u^8x
- 12u^6x + 32u^5y + 38u^4x - 32u^3y - 12u^2x + x)$.
\end{remark}

\subsection{Centers of system \eqref{eq1} with two
homogeneous invariant straight lines and reversibility}


Consider the system of algebraic equations \eqref{eq25} and let
the conditions \eqref{eq8} hold.

\subsubsection{$a_3=0$} In this case $V_{04}\equiv
V_{22}\equiv 0$ and we have the following possibilities:

 If $ u=-1$ or $u(u+1)\neq 0$, then from the equations of
\eqref{eq25} we obtain respectively the following two sets of
conditions for the existence of a center:
\begin{itemize}
\item[(8)] $b=c=g=k=l=p=q=0$, $f=a+d$, $r=m+n-s$;

\item[(9)]
$ b = [a(1 - u^2)]/(2u)$, $c = [d(1 - u^2)]/(2u)$,
$g = [(a+d)(1-u^2)]/(2u)$,
$n = [(q-3k)(u^4-6u^2+1)+4m(u^3-u)]/[4u(u^2 - 1)]$,
$l = k$,
$f = a + d,\; r=[(q-k)(u^4-6u^2+1)+4m(u^3-u)]/[4u(u^2-1)]$,
$s=[k(6u^2-u^4-1)+2mu(u^2-1)]/[2u(u^2-1)]$, $p=q$.
\end{itemize}
If $u=0$, then  \eqref{eq25} yields the symmetric set of
conditions to (8).


\subsubsection{$a_3\neq 0$} In this case from the equation
$V_{02}=0$ of \eqref{eq25} we find
$$
a_3=[a(u^2-1)+2bu]/(u^2 + 1).
$$
If  $u=-1$, then \eqref{eq25} yields the following conditions for
the existence of a center
\begin{itemize}
\item[(10)]
$ c =  - 3b$, $f = a + d$, $g =  - 2b$,
$k =  - 2ab$, $l = b(a + d)$,
$m = 2b^2$, $p =  - 2b(a + d)$, $q = b(a - d)$, $r = 0$,
$s = 2b^2 + n$.
\end{itemize}
In the case $u=0$, we get the symmetric to (10) set of conditions
for the existence of a center.

Let  $u(u+1)\neq 0$, then from the equations of \eqref{eq25} we
obtain  the following conditions for the existence of a center
\begin{itemize}
\item[(11)]
$c=[(3a+d)(1-u^2)-6bu]/(2u)$, $g=[(3a+d)(1-u^2)-4bu]/(2u)$,
\begin{gather*}
 l=[a(3a+d)(u^2-1)+2(3ab+bd+k)u]/(2u),\\
\begin{aligned}
m&=[r(u^2+1)^4+2(au^2-a+2bu)((5a+2d)(u^6-1)\\
&\quad + (11a-2d)(u^2-u^4)+b(10u^5-12u^3+10u))]/(u^2+1)^4,
\end{aligned}\\
\begin{aligned}
s&=[n(u^2+1)^4+2(au^2-a+2bu)((5a+2d)(u^6-1)\\
&\quad + (11a-2d)(u^2-u^4)+b(10u^5-12u^3+10u))]/(u^2+1)^4,
\end{aligned}\\
 f=a+d,\quad q=[2pu+(3a+d)(au^2-a+2bu)]/(2u),\\
\begin{aligned}
 r&=[2(5ab+bd+k)(u^{11}-u)+2(4b^2-9a^2-3ad)(u^{10}+u^2)\\
&\quad + a(3a+d)(u^{12}+1)+2(3k-5bd-33ab)(u^9-u^3)\\
&\quad + (61a^2-ad-64b^2)(u^8+u^4)+4(45ab-3bd+k)(u^7-u^5)\\
&\quad + 4(28b^2-23a^2+3ad)u^6]/[4u^2(u^2+1)^4],
\end{aligned}\\
\begin{aligned}
n&=[2(k-10ab-2bd)(u^9+u)+8(10ab+k)(u^7+u^3)\\
&\quad + 2(14a^2+ad-12b^2)(u^8-u^2)+4(10b^2+ad-8a^2)(u^6-u^4)\\
&\quad + 4(3k-14ab+2bd)u^5+2a(2a+d)(1-u^{10})]/[(u^2+1)^4(u^2-1)],
\end{aligned}\\
\begin{aligned}
p&=[(12ab+2bd+k)(u^9+u)+(12b^2-5ad-19a^2)(u^8-u^2)\\
&\quad + 4(k-16ab-2bd)(u^7+u^3)+2(21a^2-3ad-26b^2)(u^6-u^4)\\
&\quad + a(3a+d)(u^{10}-1)+2(52ab-10bd+3k)u^5]/[u(u^2+1)^4].
\end{aligned}
\end{gather*}
\end{itemize}

\begin{remark} \label{rmk2}\rm
In each of the cases (10) and (11) the system \eqref{eq1} has three
invariant straight lines. Thus, in conditions (10) besides the
invariant straight lines $x\pm iy=0$, the system \eqref{eq1} has
one more invariant straight line $ L_3=1 - 2bx$; in conditions
(11): $L_3= (au^2-a+2bu)[4ux+2(u^2-1)y]-(u^2+1)^2=0$.
\end{remark}



\subsection{Centers of system \eqref{eq1} with two nonhomogeneous and nonparallel
 invariant straight lines and reversibility}

Consider the system of algebraic equations \eqref{eq25} and let
the conditions \eqref{eq11} hold.

\subsubsection{$a_3=0$} In this case $V_{04}\equiv
V_{22}\equiv 0$ and we have the following possibilities:

If $u=0$ or $ u=-1$, then from the equations of \eqref{eq25} we
obtain, respectively, the following three sets of conditions for
the existence of a center:
\begin{itemize}
\item[(12)]
$ a=b=d=f=k=l=p=q=0$, $c=2g$, $m = g^2 + 1$,
$n=1$, $r=s=-1$,  $A_1 = g - A_2$, $A_2^2 - gA_2 - 1=0$;

\item[(13)]
$ b=c=g=k=l=p=q=0$, $r = - (f + 1)$, $s=(a-1)(d-a-m+2)$,
$n=(f+2)(d-a-m+2)-d-1$, $A_1=-A_2$, $A_2^2+a+m-d-2=0$;

\item[(14)]
$b=c=g=l=p=q=s=0$, $f=-2$, $n =  - (d + 1)$,
$r=1$, $k=(a-1)A_2$, $A_1=0$, $A_2^2+a-d+m-2=0$.
\end{itemize}


If $u(u+1)\neq 0$, then from the equations $\{V_{02}=0,\,
V_{20}=0,\, U_{11}=0\}$ of \eqref{eq25} we express, $b, g$ and
$d$, respectively. Then $V_{03}\equiv f_1f_2f_3=0$, where
\begin{gather*}
 f_1=2uA_1+1-u^2,\; f_2=2uA_2+1-u^2,\\
f_3=a(u^4+1)+2(A_1+A_2-c)(u^3-u)+2(2f-a+4)u^2.
\end{gather*}

Let $f_1=0$, then $A_1=(u^2-1))/(2u)$ and
$V_{03}\equiv V_{21}\equiv 0$. Express $A_2$ from $U_{30}=0$ and denote
$z=u^4-6u^2+1$, then $U_{12}\equiv h_1c+h_2=0$, where
$h_1=2(fz+(u^2+1)^2-8au^2)(u^2-1)u$ and
$h_2=f(u^2+1)^4-32a^2u^4+f^2(u^2-1)^2z-4a(f-2)(u^2+1)^2u^2$.

If $h_1=0$, then $U_{12}=0$ yields $f=-1$ and $a=1$. In this case
we get the following conditions for the existence of a center
\begin{itemize}
\item[(15)]
$a=1$, $b= [u^6-15u^4+15u^2-1-2cuz]/[2u(u^2 + 1)^2]$, $f=-1$,
$d = [4(u^2-1)z -2cu(3z+8u^2)]/[(u^2+1)^2(u^2-1)]$, $g = k = l =-b$,
$p=q=b$, $m = (bz)/[2u(1-u^2)]$, $r=s = 0$, $n=-m$, $z=u^4-6u^2+1$,
$A_1 = (u^2 - 1)/(2u)$, $A_2 = (2cu - u^2 + 1)/(2u)$.
\end{itemize}

If $h_1\neq 0$, then express $c$ form $U_{12}=0$ and obtain the
following conditions for the existence of a center
\begin{itemize}
\item[(16)]
$g=[(a(1-u^2)+2cu)z+8f(1-u^2)u^2]/[2u(1+u^2)^2]$,
$s = (1 - a)A_1A_2$,
$d=[2(a-f)(u^2-1)z-2cu(3z+8u^2)]/[(u^2-1)(u^2+1)^2]$,
\begin{align*}
c&=-[f(u^2+1)^4-32a^2u^4+f^2(u^2-1)^2z-4a(f-2)(u^2+1)^2u^2]\\
&\quad\div [2(fz+(u^2+1)^2-8au^2)(u^2-1)u],
\end{align*}
$b=-g+(a+f)(1-u^2)/(2u)$,
$k = (a - 1)(A_1 + A_2) + g$, $q = (A_1 + A_2 - c)A_1A_2 -g$,
$r =  - f - 1$, $n = A_1A_2( - f - 2) - (d + 1)$,
$p = (f + 2)(A_1 + A_2) + b -c$, $l = - b$,
$m = - A_1^2 - A_1A_2 - A_2^2 + c(A_1 + A_2) - a + d + 2$,
$z=u^4-6u^2+1$,
$A_1 = (u^2 - 1)/(2u)$, $A_2=2(a-1)u/(u^2-1)+(2cu+fu^2-f)/(2u)$.
\end{itemize}

The case $f_2=0$ can be reduced to $f_1 = 0$ if we replace $A_2$
by $A_1$.

Assume $f_1f_2\neq 0$ and $f_3=0$. We express $A_1$ from $f_3=0$,
$a$ from $V_{21}=0$ and calculate the resultant of the equations
$\{U_{30}=0,U_{12}=0\}$ by $A_2$. We find that
$\operatorname{Res}(U_{30},U_{12},A_2)\equiv
64g_1^2g_2^2(u^2+1)^{12}(u^2-1)^4u^2$, where
$g_1=2uz(u^2-1)c+f(u^8+1)-8(f-1)(u^6+u^2)+2(23f+24)u^4$,
$g_2=(u^2+ 1)^2f + 8u^2$.

If $g_1=0$ or $g_2=0$, then we get the following center
conditions, respectively:
\begin{itemize}
 \item[(17)]
$ a=(8u^2-fz)/[2(u^2-1)^2]$, $n=2pu/(u^2-1)+(z-16u^2)/z$,
\begin{gather*}
c=[8(f-1)(u^6+u^2)-2(23f+24)u^4-f(u^8+1)]/[2zu(u^2-1)],\\
d=[4(3f+8)(u^6+u^2)-8(5f+12)u^4]/[z(u^2-1)^2],\quad l = -b,\\
b=[2(f+2)u]/(u^2-1),\quad g=[f(u^4+1)+2(5f+12)u^2]/[4u(1-u^2)],\\
k=[32(f+1)u^2(u^4+1)+f^2(u^2+1)^4]/[8u(u^2-1)^3],\quad r =  - (f + 1), \\
m=[qz(u^2-1)-2u(5u^4-14u^2+5)]/[2zu],\quad z=u^4-6u^2+1,\\
p=[f^2(u^2+1)^4+48u^2(u^2-1)^2+8fu^2(7u^4-10u^2+7)]/[4uz(1-u^2)],\\
s=[(f+2)((f-2)(u^8+6u^4+1)+4(f+6)(u^6+u^2))]/[4(u^2-1)^4],\\
\begin{aligned}
q&=[f^2(u^2+1)^4(u^4-14u^2+1)+32(f+3)(u^{10}+10u^6+u^2)\\
&\quad -192(3f+5)(u^8+u^4)]/[8zu(u^2-1)^3],
\end{aligned}\\
A_1=[4z(1-u^2)uA_2- 4(f+10)(u^6+u^2)-6(f-8)u^4-f(u^8+1)]/[4zu(u^2-1)],\\
\begin{aligned}
&4zu(u^2-1)^2A_2^2+(u^2-1)(8(5u^4-6u^2+5)u^2+f(u^2+1)^4)A_2\\
&+2u((f-2)(u^8+6u^4+1)+4(f+6)(u^6+u^2))=0;
\end{aligned}
\end{gather*}

\item[(18)]
$a=(8u^2)/(u^2+1)^2$, $f=-a$, $l=g=-b$, $q=3b$, $p=-3k$,
$b=[4u(u^2-1)(u^4-14u^2+1)-cz(u^2+1)^2]/(u^2+1)^4$,
$d=[32z(u^4-u^2)-2cu(3u^4-10u^2+3)(u^2+1)^2]/[(u^2+1)^4(u^2-1)]$,
$k=[4u(u^2-1)]/(u^2+1)^2,\quad  s=(-bz)/[4u(u^2-1)],\quad r=-z/(u^2+1)^2$,
$m=[4u(u^2-1)(u^4-22u^2+1)+c(u^2+1)^4]/[4u(u^2-1)(u^2+1)^2]$,
\begin{align*}
n&=[c(u^8 - 20u^6 + 54u^4 - 20u^2 + 1)(u^2 + 1)^2 -2u(u^8 - 68u^6 +246u^4\\
&\quad  - 68u^2 + 1)(u^2-1)]/[2u(u^2 +1)^4(1-u^2)],\; z=u^4-6u^2+1,
\end{align*}
$A_1=c-A_2-8(u^3-u)/(u^2+1)^2$, $4u(u^2-1)[(u^2+1)^2(A_2^2-cA_2)
+8u(u^2-1)A_2-u^4+14u^2-1]+ cz(u^2+1)^2=0$.
\end{itemize}

\subsubsection{$a_3\neq 0$} In this case the equation $V_{02}=0$ of
\eqref{eq25} yields
$$
 a_3=[a(u^6-1)+2(g-c)(u^5+u)+(3a+4d-4f)(u^2-u^4)
+ 4(2b+c-g)u^3]/(u^2 + 1)^3.
$$

If $u=0$ or $ u=-1$, then from the equations of \eqref{eq25} we
obtain, respectively, the following three sets of conditions for
the existence of a center:
\begin{itemize}
\item[(19)] $b=l=s=0$, $a=r=1$, $d=-3$, $n=-f=2$, $k=g$, $p=-c$,
$q=-2g$, $A_2^3-cA_2^2+(m+2)A_2-g=0$, $A_1^2+(A_2-c)A_1+A_2^2-cA_2+m+2=0$;

\item[(20)]
$r=s=0$, $a=1/2$, $c = b + 2g$, $d = ( - 3)/2$, $f = -1$, $k = g/2$,
$l =  - b$,  $m = g(b + g)$, $n = 1/2$,  $p =  - g$, $q =  -g$,
$A_1 = 0$, $A_2 = g$;

\item[(21)]
$c =  - 3b$, $f =-1$, $g =  - 2b$, $k =  - 2ab$, $l =  - b$,
$m=2b^2$,
$n = 1 - a$,  $p = 2b$, $q = b(a - d)$,   $s = 3a - a^2 +
ad  - d -2$,
$r = 0$, $A_1 =  - A_2 - 2b$, $A_2^2 + 2bA_2 + a - d - 2$.
\end{itemize}


If $u(u+1)\neq 0$, then we express $d$ from $V_{20}=0$ and replace
in $V_{04}=0$. Factoring we obtain $V_{04}\equiv f_1f_2f_3=0$,
where
\begin{gather*}
 f_1=A_1(u^2-1)+2u,\; f_2=A_2(u^2-1)+2u,\\
f_3=(a-1)(u^4+1)+2(A_1+A_2-c)(u^3-u)+2(3-a+2f)u^2.
\end{gather*}

Let $f_1=0$, then $A_1=(2u)/(1-u^2)$ and we find
$U_{12}\equiv g_1g_2=0$, where
\begin{gather*}
 g_1=(2a+2f+1)(u^4+1)+4(b+g)(u^3-u)-2(2a+2f-1)u^2,\\
 g_2=(2a+f)(u^4+1)+2(b-c+g)(u^3-u)+2(f-2a)u^2.
\end{gather*}

Assume $g_1=0$ and express $g$ from $g_1=0$, then $U_{12}\equiv
U_{30}\equiv 0$.  Replacing $g$ in $V_{03}=0$ and factoring we
obtain $V_{03}\equiv h_1h_2=0$, where
\begin{gather*}
 h_1=4u(1-u^2)A_2+u^4-6u^2+1,\\
 h_2=(2a-1)(u^4+1)+4(A_2-c)(u^3-u)+2(3-2a+4f)u^2.
\end{gather*}

If $h_1=0$, then $A_2 = (u^4 - 6u^2 + 1)/[4u(u^2 - 1)]$. We
express $a$ from $V_{22}\equiv V_{21}\equiv U_{11}=0$ and obtain
the following set of conditions for the existence of a center
\begin{itemize}
\item[(22)]
\begin{align*}
a&=[(f+1)(8u^6-u^8+8u^2-1)+2(b+c)(u-u^7)+2(b-7c)(u^3-u^5)\\
&\quad + 2(9-7f)u^4]/[8u^2(u^2-1)^2],
\end{align*}
$ k = (a - 1)(A_1 + A_2) + g$, $l =  - b$,
\begin{align*}
d&=[(f+1)(-u^8-1)+2(b+c)(u-u^7)+2(4f-1)(u^6+u^2)\\
&\quad + 2(7b+3c)(u^5-u^3)-2(7f+1)u^4]/[4u^2(u^2-1)^2],
\end{align*}
$r =  - (f + 1)$,
\begin{align*}
g&=[(f+1)(u^8+1)+2(b+c)(u^7-7u^5+7u^3-u)\\
&\quad -4(4f+3)(u^6+u^2)+ 2(15f-13)u^4]/[16u^3(u^2-1)],
\end{align*}
$n = -( f + 2)A_1A_2 - d - 1$,
$m = c(A_1+A_2) - A_1^2 - A_1A_2  - A_2^2  - a + d + 2$,
$s = (1 - a)A_1A_2$,
$q = A_1A_2(A_1 + A_2 - c) - g$, $p = (f + 2)(A_1 + A_2)+ b - c$,
$A_1 = (2u)/(1-u^2)$, $A_2 = (u^4 - 6u^2 + 1)/[4u(u^2 - 1)]$.
\end{itemize}

If $h_1\neq 0$, $h_2=0$, then express $A_2$ from $h_2=0$ and $a$
from $U_{11}\equiv V_{22}=0$. We obtain the following two sets of
conditions for the existence of a center
\begin{itemize}
\item[(23)]
$a=[2cu(u^2-1)-4fu^2]/(u^2-1)^2$, $k = (a - 1)(A_1 + A_2) + g$,
$l =  - b$,
$b=[(f+1)(10u^2-u^4-1)+2c(u-u^3)]/[2(u^3-u)]$,
$r =  - (f + 1)$, $d=[6(4f+5)u^2-(4f+9)(u^4+1)+4c(u-u^3)]/[2(u^2-1)^2]$,
$g=[u^4-2(4f+11)u^2+1]/[4(u^3-u)]$, $n = -( f + 2)A_1A_2 - d - 1$,
$m = c(A_1+A_2) - A_1^2 - A_1A_2  - A_2^2  - a + d + 2$,
$s = (1 - a)A_1A_2$, $q = A_1A_2(A_1 + A_2 - c) - g$,
$p = (f + 2)(A_1 + A_2)+ b - c$,
$A_1 = (2u)/(1-u^2)$, $A_2 = (u^4 - 6u^2 + 1)/[4u(u^2 - 1)]$;


\item[(24)]
$a=[2(3f+4)u^2-f(u^4+1)]/[2(u^2-1)^2]$,
$r =  - (f + 1)$,
\begin{gather*}
\begin{aligned}
b&=[2c(u-7u^3+7u^5-u^7)-(f+1)(u^8+1)+4(3f+2)(u^6+u^2)\\
&\quad - 2(19f+7)u^4]/[2u(u^2-1)(u^2+1)^2],
\end{aligned}\\
 k = (a - 1)(A_1 + A_2) + g,\\
\begin{aligned}
 d&=[28(2f+1)(u^6+u^2)-3(2f+3)(u^8+1)+4c(3u-13u^3 \\
&\quad + 13u^5-3u^7)-6(22f+9)u^4]/[2(u^4-1)^2],
\end{aligned}\\
 s = (1 - a)A_1A_2,\\
\begin{aligned}
g&=[(f+1)(u^8-28u^6-28u^2+1)+4c(u^7-7u^5+7u^3-u)\\
&\quad + 2(35f+3)u^4]/[4u(u^2+1)^2(u^2-1)],
\end{aligned}\\
 n = -( f + 2)A_1A_2 - d - 1,\quad
 m = c(A_1+A_2) - A_1^2 - A_1A_2  - A_2^2  - a + d + 2,\\ 
 l =  - b,\quad
 q = A_1A_2(A_1 + A_2 - c) - g,\quad
 p = (f + 2)(A_1 + A_2)+ b - c,\\
 A_1 = (2u)/(1-u^2),\quad  A_2 = [(f+1)(u^4-14u^2+1)]/[4u(u^2 -1)]+c.
\end{gather*}
\end{itemize}

Assume now $g_1\neq 0$ and $g_2=0$. We express $g$ from $g_2=0$,
$A_2$ from $V_{03}=0$, $c$ from $U_{30}=0$ and $a$ from
$V_{22}=0$. Then $U_{11}\equiv g_1\neq 0$.

The case $f_2=0$ can be reduced to $f_1 = 0$ if we replace $A_2$
by $A_1$.

Assume $f_1f_2\neq 0$ and $f_3=0$. We reduce the equations of
\eqref{eq25} by $c$ from $f_3=0$. Factoring we obtain that
$V_{03}\equiv e_1e_2=0$, where
\begin{gather*}
 e_1=(a+f+1)(u^2-1)+2(b+g-A_1)u,\\
 e_2=(a+f+1)(u^2-1)+2(b+g-A_2)u.
\end{gather*}

Let $e_1=0$, then $A_1=[(a+f+1)(u^2-1)+2(b+g)u]/(2u)$. From the
equations  $V_{21}\equiv V_{22}=0$, $U_{12}=0$ and $f_3=0$ of
\eqref{eq25} we express $b$, $g$ and $A_2$, respectively.

If $a=1$, then $\{U_{11}=0,\,U_{30}=0\}$ yields $f=-2$ and we
obtain the following conditions for the existence of a center
\begin{itemize}
\item[(25)]
$a=r=1$, $f=-2$, $k = g$, $l =  - b$,
$q =A_1A_2(A_1 + A_2 - c) - g$,
$b=[z(z-4u^2-2u(u^2-1)c)]/[2u(u^2-1)(u^2+1)^2]$,  $p = b - c$,
$d = [2u(3A_2u^4 - 10A_2u^2 + 3A_2 + 8u^3 - 8u)]/[(u^2+1)^2(1-u^2)]$,
$g=[(A_2(u^2-1)+2u)z]/[(u^2+1)^2(u^2-1)]$, $z=u^4-6u^2+1$,
$m = c(A_1+A_2) - A_1^2 - A_1A_2  - A_2^2  + d + 1$,
$n = - d -1$, $s = 0$,
$A_1 = b+g$, $A_2 = [4c(u^3 - u) - u^4 + 14u^2 - 1]/[4(u^2 -1)u]$.
\end{itemize}

If $a\neq 1$, then express $c$ from $U_{11}=0$ and
$U_{30}\equiv(2a+f)(a-1)(u^8+1)-4(u^6+u^2)(2a^2+4af+f^2-2f-1)+
2u^4(6a^2+15af+2a+12f^2+9f+4)=0$. This equation admits the
following parametrization
\begin{gather*}
 a=[(u^4-6u^2+1)(w^2+16u^4)+w(u^8-8u^6+46u^4-8u^2+1)]/[w(u^4-1)^2],\\
 f=[(u^4-6u^2+1)(4u^2-2w)-2w^2]/[w(u^2+1)^2], w\neq 0.
\end{gather*}
In this case we obtain the following conditions for the existence
of a center
\begin{itemize}
\item[(26)]
$ a=[(u^4-6u^2+1)(w^2+16u^4)+w(u^8-8u^6+46u^4-8u^2+1)]/[w(u^4-1)^2]$,
$b=[2u(w-4u^2)(u^4+10u^2-2w+1)]/[(u^2+1)^2(u^2-1)w]$,
\begin{gather*}
\begin{aligned}
c&=[2u^{10}(u^4+6u^2-7w - 113) + u^8(w^2 + 88w + 552)- 2u^2(4w^2+ 7w - 1) \\
&\quad-2u^6(4w^2 + 154w + 113) + 2u^4(23w^2 + 44w + 6)+w^2]\\
&\quad \div [uw(u^2-1)(u^4- 6u^2+1)(u^2+1)^2],
\end{aligned}\\
 f=[(u^4-6u^2+1)(4u^2-2w)-2w^2]/[w(u^2+1)^2],\\
g = [(u^4(12u^2 - w + 56)+ u^2(12-10w) - w)(4u^2 -
w)]/[2(u^2 + 1)^2(u^2 - 1)uw],
\\
\begin{aligned}
 d&=[(2a+f)(u^6-1)+2(b-c+2g)(u^5+u)+(2a-5f)(u^2-u^4) \\
&\quad + 4(3b+c)u^3]/[4u^2(u^2-1)],
\end{aligned}\\
q = (A_1 + A_2 - c)A_1A_2 -g,\quad r = - (f + 1), \\
m = - A_1^2 - A_1A_2 - A_2^2 + c(A_1 + A_2) - a + d +2,\quad
s = (1 - a)A_1A_2,\\
l = - b,\;n = -(f + 2)A_1A_2 - (d + 1),\quad
p = (f + 2)(A_1 + A_2) + b - c,\\
A_1=[2u((u^2+1)^2-w)]/[(u^2-1)w],\quad
k = (a - 1)(A_1 + A_2) + g,\\
A_2=[2cwu(u^2-1)-4u^2(u^2+1)^2-8fwu^2-w^2]/[2wu(u^2-1)].
\end{gather*}
\end{itemize}

The case $e_2=0$ can be reduced to $e_1 = 0$ if we replace $A_2$
by $A_1$.

\begin{remark} \label{rmk3} \rm
In each of the cases (14), (20), (21), (24) the system \eqref{eq1} has
three invariant straight lines. Thus, in conditions (14) besides
the invariant straight lines \eqref{eq9}, the system \eqref{eq1}
has one more invariant straight line $ L_3=1+(d+1)y$; in
conditions (20): $L_3= 1+gx$; in conditions (21): $L_3=1-2bx$; in
conditions (24):
\begin{align*}
L_3&=(u^8+1)(f+1)x+4(u^7-u)(cx - fy - y +
1)-4(u^6+u^2)(4cy + 5fx + 3x)\\
&\quad +4(u^3-u^5)(7cx - 15fy - 7y - 1)+2u^4(16cy + 43fx + 19x).
\end{align*}
\end{remark}

\begin{remark} \label{rmk4} \rm
In each of the cases (12), (15), (16), (17), (18), (19), (25), (26) the
system \eqref{eq1} has four invariant straight lines. Thus, in
conditions (12) besides the invariant straight lines \eqref{eq9},
the system \eqref{eq1} has two more invariant straight lines
$L_{3,4}=2+(g\pm\sqrt{g^2+4})x+2y=0$; in conditions (15):
$L_3=(u^2-1)(bx-1)-2buy$,
$L_4= (u^6-1)x - 2(u^5+u)(cx+3y+1)+(u^4-u^2)(8cy-15x)+4u^3(3cx+5y-1)$; in
conditions (16):
$L_3=[2(a-1)ux-(fy+y+1)(u^2-1)](u^2+1)^2-(u^2-1)(fz-8au^2)$,
\begin{align*}
L_4&=(u^{10}-1)(f+1)^2x-2(u^9+u)(f+1)(fy+y+1)+
(u^2-u^8)[4a(3f+2)\\
&\quad +(3f+1)^2]x+8(u^7+u^3)(3afy + 2ay + 2a + 2f^2y +fy + 2f)\\
&\quad +2(u^6-u^4)(16a^2 + 26af  - 4a + 11f^2 - 4f - 1)x+4u^5(1 -
16a^2y \\
&\quad - 20afy + 8ay - 8a - 7f^2y + 6fy - 7f + y);
\end{align*}
in conditions (17):
$L_{3,4}=fx(u^8+1)-4(f+2)(u^7+3u^6x+3u^2x-u)-4(f+10)(u^5-u^3)-2(13f+24)u^4x\pm
((u^2+1)^2x+4u(u^2-1))\sqrt{A}-8(f+1)u(u^2+1)^2(u^2-1)y$,
$A= f^2(u^2+1)^4+48fu^2(u^2+1)^2+64(u^4+3u^2+1)u^2$; in conditions
(18): $L_{3,4}=(u^3-u)[(cx-2)(u^8+1)+4(c-2x)(u^2+1)(u^5-u)+
4(cx-6)(u^6+u^2)+6(cx+14)u^4]\pm
((u^2+1)^2x+4u(u^2-1))\sqrt{A}-(2u(u^2-1)(u^2+1)^4)y$,
$A=u(u^2-1)[c(u^2+1)(u^4-1)-4uz](cu(u^2+1)^2-u^6-9u^4+9u^2+1)$;
in conditions (19): $L_3=1-2y$, $L_4=1+(c-a_1-a_2)x-y$; in
conditions (25): $L_3=2ux+(u^2-1)(y-1)$,
\begin{align*}
L_4&=(u^8+1)x+4(cx+y+1)(u-u^7)+4(4cy-7x)(u^6+u^2)\\
&\quad +4(7cx+23y-1)(u^5-u^3)+2(67x-16cy)u^4;
\end{align*}
in conditions (26):
$L_3=2(2ux+u^2y-y)(4u^2-w)+w(u^2-1)$,
$L_4=(8u^2-w)[(4u^2-w)(u^4-6u^2+1)x+2u(u^2-1)(u^4-6u^2+2w+1)y]+
2uw(u^2-1)(u^4-6u^2+1)$.
\end{remark}

In this way we have  proved the following theorem.

\begin{theorem} \label{thm6}
The cubic differential system \eqref{eq1} with at least two
invariant straight lines is rationally reversible if and only if
one of the sets of conditions {\rm (1)-- (26)} is
satisfied.
\end{theorem}

The proof of the main result, Theorem \ref{mainthm},  follows directly 
from  Theorems \ref{thm5} and \ref{thm6}.

\subsection*{Acknowledgments}
The author wants to thank the anonymous referees for their
suggestions and contribution in improving the content of the manuscript.

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\end{document}