\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2002(2002), No. 34, pp. 1--7.\newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.swt.edu  (login: ftp)}
\thanks{\copyright 2002 Southwest Texas State University.}
\vspace{1cm}}

\begin{document}

\title[\hfilneg EJDE--2001/34\hfil Centering conditions]
{Centering conditions for planar septic systems}

\author[Evgenii P. Volokitin\hfil EJDE--2001/34\hfilneg]
{Evgenii P. Volokitin}

\address{Sobolev Institute of Mathematics, Novosibirsk, 630090, Russia}
\email{volok@math.nsc.ru}


\date{}
\thanks{Submitted August 22, 2002. Published April 3, 2003.}
\thanks{Partially supported by Grant 02-01-00194 from the Russian
Foundation for Basic Research}
\subjclass[2000]{34C05, 34C25}
\keywords{centering conditions, isochronicity, commutativity}


\begin{abstract}
 We find centering conditions for the following $O$-symmetric system
 of degree 7:
 \begin{gather*}
 \dot x=y+ x (H_2 (x,y)+H_6 (x,y)),\\
 \dot y=-x+ y (H_2 (x,y)+H_6 (x,y)),
 \end{gather*}
 where $H_2 (x,y)$ and $H_6 (x,y)$ are homogeneous polynomials
 of degrees 2 and 6, respectively.
 In some cases, we can find commuting systems and first integrals for the
 original system. We also study the geometry of the central region.
\end{abstract}

\maketitle

\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\numberwithin{equation}{section}

\section{introduction}

Consider the planar autonomous system of ordinary differential
equations
\begin{equation} \label{1}
\begin{gathered}
\dot x = y + x R_{n-1} (x,y),\\
\dot y = - x +y R_{n-1} (x,y),\\
\end{gathered}
\end{equation}
where $R_{n-1}(x,y)$ is a polynomial in $x$ and $y$, of degree $n-1$,
and $R_{n-1}(0,0)=0$. This system has only one singular point
at $O(0,0)$ which is the center of the linear part of the system.
The orbits of this system move around the origin with constant
angular speed,  and the origin is so a uniformly isochronous
singular point.

Such systems have been studied in many papers; see
\cite{1}--\cite{5} and references therein. The following problem
was stated as Problem 19.1. in \cite{2}:
\begin{quote}
Identify systems \eqref{1} of odd degree
which are $O$-symmetric (not necessarily quasi homogeneous) having
$O$ as a (uniformly isochronous) center.
\end{quote}
In this article we solve this problem for some systems of degree
7. In particular, we find necessary and sufficient conditions for
system \eqref{1} with
\begin{equation} \label{2}
\begin{aligned}
R_{n-1} (x,y)=& a_0 x^2 + a_1 x y + a_2 y^2 + c_0 x^6 + c_1 x^5 y + c_2 x^4 y^2 \\
&+c_3 x^3 y^3 + c_4 x^2 y^4 + c_5 x y^5 + c_6 y^6, 
\end{aligned}
\end{equation}
where $a_0, a_1, a_2, c_0, c_1, c_2, c_3, c_4, c_5, c_6$ are real numbers.

The plan for this paper is as follows: In Section 2, we present
centering conditions. In Section 3, we investigate some properties
of systems in the presence of a center. In particular, we discuss
the question of existence of a polynomial commuting system. When
this system exists, we give a first integral of the original
system. We also study the geometry of the central region.

 \section{Results}

\begin{theorem} \label{thm1}
The origin is a center of (\ref{2}) if and only if
one of the following two conditions is satisfied:
\begin{itemize}
\item[(i)] $a_0=a_1=a_2=0$,  $5c_{0} + c_{2} + c_{4} + 5c_{6}=0$;

\item[(ii)]
\begin{gather*}
a_0+a_2=0,\\
5c_0 + c_2 + c_4 + 5c_6 = 0,\\
a_1(15c_0 + c_2 - c_4 - 15c_6) + 2a_0(5c_1 + 3c_3 + 5c_5) = 0,\\
(a_1^2 - 4a_0^2)(3c_0 - c_2 - c_4 + 3c_6) + 8a_0 a_1(c_1 - c_5)= 0,\\
a_1(a_1^2-12a_0^2)(c_0 - c_2 + c_4 - c_6) + 2a_0(3a_1^2
-4a_0^2)(c_1-c_3+c_5) =0.
\end{gather*}
\end{itemize}
\end{theorem}

Before proving this theorem, we consider the instance of (\ref{2})
in which $a_0=a_2=0$, that is
\begin{equation} \label{4}
\begin{gathered}
\dot x=y+x(a_1 x y + c_0 x^6 + c_1 x^5 y + c_2 x^4 y^2 +
c_3 x^3 y^3 + c_4 x^2 y^4 + c_5 x y^5 + c_6 y^6),\\
\dot y=-x+y(a_1 x y + c_0 x^6 + c_1 x^5 y + c_2 x^4 y^2 +
c_3 x^3 y^3 + c_4 x^2 y^4 + c_5 x y^5 + c_6 y^6).
\end{gathered}
\end{equation}


\begin{lemma} \label{lm1}
The origin is a center of (\ref{4}) if and only if
one of the following two conditions is satisfied:
\begin{gather}\label{5i}
 a_1=0, \quad  5c_{0} + c_{2} + c_{4} + 5c_{6}=0;\\
 c_0=c_2=c_4=c_6=0. \label{5ii}
\end{gather}
\end{lemma}

\begin{proof}
We used the software package {\it Mathematica} to find the first
six Poincar\'{e}-Lyapunov constants of (\ref{4}) (see more details
about our method in \cite{6}). Up to a positive scalar factor they
are
\begin{align*}
l_1 &=0, l_2 = 0,\\
l_3 &= 5c_0 + c_2 + c_4 + 5c_6,\\
l_4 &= -a_1(5c_0 + 3c_2 + 5c_4 + 35c_6),\\
l_5 &= a_1^2(-101c_0 - 17c_2 - 9c_4 + 19c_6),\\
l_6 &= 15a_1^3(621 c_0 + 367c_2 + 565c_4 + 3367c_6)\\
    &\quad - 56(31c_1 + 20c_3 + 49c_5)(5c_0 + c_2 + c_4 + 5c_6).
\end{align*}
Necessity of conditions \eqref{5i}--\eqref{5ii} result from
solving the simultaneous equations $l_3=l_4=l_5=l_6=0$.

In the case \eqref{5i}, the sufficiency part
of the lemma follows from the fact
that (\ref{4}) is a quasi homogeneous system of degree 7
whose coefficients satisfy the equation $5c_0+c_2+c_4+5c_6=0$
representing a necessary and sufficient centering condition \cite{1}.

In the case \eqref{5ii}, system (\ref{4}) is reversible
and  its trajectories are symmetric with respect to both coordinate axes.

It is well known that if the linear part of a reversible system
has a center at the origin,  then the origin is also a center of
the system itself. Thus, under conditions \eqref{5i}-\eqref{5ii},
the origin is a center of system (\ref{4}). This completes the
proof of the lemma. \end{proof}

\begin{proof}[Proof of Theorem \ref{thm1}]
In the previous lemma we considered a particular case.
Now, we consider the general system (\ref{2}).
The first Poincar\'{e}-Lyapunov constant $l_1$ of (\ref{2}) is
$$
l_1=2(a_0 + a_2).
$$
If
$a_0+a_2=0$ then the change of variables
$x\mapsto x \cos \vartheta + y \sin \vartheta$,
$y\mapsto -x \sin \vartheta + y \cos \vartheta$,
with $\vartheta$ defined from the condition
\begin{equation}\label{6}
a_0 \cos^2 \vartheta + a_1 \sin \vartheta \cos \vartheta
-a_0 \sin^2 \vartheta=0,
\end{equation}
reduces (\ref{2}) to a system of the form (\ref{4}):
\begin{gather*}
\dot x=y+x(a_{1}' x y + c_{0}' x^6 + c_{1}' x^5 y + c_{2}' x^4 y^2 +
c_{3}' x^3 y^3 + c_{4}' x^2 y^4 + c_{5}' x y^5 + c_{6}' y^6),\\
\dot y=-x+y(a_{1}' x y + c_{0}' x^6 + c_{1}' x^5 y + c_{2}' x^4 y^2 +
c_{3}' x^3 y^3 + c_{4}' x^2 y^4 + c_{5}' x y^5 + c_{6}' y^6)
\end{gather*}
whose coefficients are expressible in terms of the coefficients of
(\ref{2}).
In particular, we have
\begin{equation} \label{7}
\begin{aligned}
&a_{1}'=a_1 \cos^2\vartheta - 4 a_0 \cos\vartheta \sin\vartheta
- a_1 \sin^2\vartheta,\\
&c_{0}' =    (2d_0 - d_2 + 2d_4 - d_6)/32,\\
&c_{2}' =  (6d_0 - d_2 - 10 d_4 + 15d_6)/32,\\
&c_{4}' =    (6d_0+d_2-10d_4-15d_6)/32,\\
&c_{6}' = (2d_0+d_2+2d_4+d_6)/32,
\end{aligned}
\end{equation}
where
\begin{align*}
&d_0 = 5c_0 + c_2 + c_4 + 5c_6,\\
&d_2 = (15c_0 + c_2 - c_4 - 15c_6)\cos2\vartheta + (5c_1 + 3c_3 + 5c_5)\sin2\vartheta,\\
&d_4 = (3c_0 - c_2 - c_4 + 3c_6)\cos4\vartheta + 2(c_1 - c_5)\sin4\vartheta,\\
&d_6 = (c_0 - c_2 + c_4 - c_6)\cos6\vartheta + (c_1 - c_3 + c_5)\sin6\vartheta,
\end{align*}
and  $\vartheta$ is defined in (\ref{6}).

Using Lemma \ref{lm1},  we see that the origin is
a center of system (\ref{2}) if and only if one of the following two
conditions is satisfied:
\begin{gather} \label{8i}
a_0+a_2=0, a_{1}'=0, \quad
5c_{0}' + c_{2}' + c_{4}' + 5c_{6}'=0;\\
 a_0+a_2=0, \quad
c_{0}'=c_{2}'=c_{4}'=c_{6}'=0. \label{8ii}
\end{gather}
In the case \eqref{8i},  $a_1'=0$ and (\ref{6}) amount
to $a_0=a_1=0$. Next, (\ref{7}) yields
$5c_{0}' + c_{2}' + c_{4}' + 5c_{6}'=d_0=5c_{0} + c_{2} + c_{4} + 5c_{6}$.
This proves the theorem in the case (\mbox{i}).

Next, $c_{0}'=c_{2}'=c_{4}'=c_{6}'=0$ amounts to
$d_0=d_2=d_4=d_6=0$. Using (\ref{6}), we eliminate $\vartheta$
from the last equations, thus arriving at the conditions of the
case \eqref{8ii} expressed in terms of the coefficients of the
original system  (\ref{2}). These conditions coincide with those
in the case (ii) of Theorem \ref{thm1} and the proof is complete.
\end{proof}

\section{Properties of systems with center}

Consider system (\ref{4}) with coefficients that satisfy the
centering conditions \eqref{5i}--\eqref{5ii}. For a planar
polynomial system, the presence of an isochronous center is well
known to be equivalent to the existence of a transverse analytic
system commuting with it \cite{7}.

It is proved in \cite{5} that in the case \eqref{5i}
system (\ref{4}) commutes with a polynomial system of the form
\begin{equation} \label{9}
\begin{gathered}
\dot x=x+x Q(x,y),\\
\dot y=y+y Q(x,y),
\end{gathered}
\end{equation}
where
\[
 Q(x,y)=q_0 x^6 + q_1 x^5 y + q_2 x^4 y^2 +
 q_3 x^3 y^3 + q_4 x^2 y^4 + q_5 x y^5 + q_6 y^6
\]
which is a homogeneous polynomial of degree 6
satisfying
\begin{equation} \label{10}
y Q_x(x,y) - x Q_y(x,y) = 6 P_6(x,y)
\end{equation}
with
\[
P_6(x,y)=c_0 x^6 + c_1 x^5 y + c_2 x^4 y^2 +
c_3 x^3 y^3 + c_4 x^2 y^4 + c_5 x y^5 + c_6 y^6\,.
\]
To satisfy this equation,  we can take
\begin{equation}\label{11}
\begin{gathered}
q_0=0, \quad q_1=-6c_0, \quad q_2=-3c_1, q_3=-2(5c_0+c_2), \\
q_4=-3(c_1+c_3/2),\quad q_5=6c_6, \quad q_6=-(c_1+c_3/2+c_5).
\end{gathered}
\end{equation}
Note that if we add $c(x^2+y^2)^3$ to the polynomial $Q(x,y)$ with
the coefficients (\ref{11}) then the resultant system of the form (\ref{9})
commutes with (\ref{4}).

Following \cite{8}, we say that a function $C: R^2 \to R$ and the curve
$C=0$ are invariants for a system $\dot x=p(x,y)$, $\dot y=q(x,y)$
if there is a polynomial $L$ such that $\dot C=C L$, where
$\dot C=C_x p+C_y q$. The polynomial $L$ is called the cofactor of $C$.

Note that the functions
$$
C_1=x^2+y^2, \quad  C_2=1+Q(x,y)
$$
are invariants for (\ref{4}). This enables us to find the  first Darboux
integral
\begin{equation}\label{12}
H(x,y)=\frac{(x^2+y^2)^3}{1+Q(x,y)}
\end{equation}
for this system.
The Darboux method is presented, for instance, in \cite{8,9}.

By \cite{2},  the center of (\ref{4}) is of type $B^k$, and
the boundary of the center domain is the union of $k$ open unbounded
trajectories ($1\leq k\leq 6$).
In the case under study,  we can describe this boundary explicitly and
indicate the possible values of $k$ more precisely.

Passing to the polar coordinates $x=\varrho \cos \varphi$,
$y=\varrho \sin \varphi$ in (\ref{12}),
we can show that in the case \eqref{5i},
the boundary of the central region is defined by
the equation
$$
\varrho=\frac{1}{(c_0-Q(\cos \varphi, \sin \varphi))^{1/6}}
$$
with $c_0=\max_{[0,2\pi]} Q(\cos \varphi, \sin \varphi)$.

The central region is a curvilinear $k$-polygon whose  vertices
are points at infinity in the intersection
of the equator of the Poincar\'{e} sphere
with the rays
$x=r \cos\varphi_i, y=r \sin\varphi_i, r>0$,
where the values of $\varphi_i$ are determined from the conditions
$$
Q(\cos \varphi_i, \sin \varphi_i)=c_0,
0\leq \varphi_i < 2\pi.
$$
Note that the values $\varphi_i$ are solutions of the equation
$P_6(\cos \varphi, \sin \varphi)=0$.
Indeed, from (\ref{10}) we deduce:
\begin{align*}
0&=\frac{d}{d \varphi} Q(\cos \varphi, \sin \varphi)|_{\varphi=\varphi_i}\\
&=-Q_x(\cos \varphi_i, \sin \varphi_i) \sin \varphi_i
+Q_y(\cos \varphi_i, \sin \varphi_i)\cos \varphi_i\\
&=6P_6(\cos \varphi_i, \sin \varphi_i).
\end{align*}

The trigonometric polynomial $Q(\cos \varphi, \sin \varphi)$
of degree 6 satisfies the condition
$Q(\cos(\varphi+\pi), \sin(\varphi+\pi))=Q(\cos \varphi, \sin \varphi)$
and takes its every value,  $c_0$ inclusively,
on the interval $[0,2\pi)$ an even number of times.
Thus, in the case \eqref{5i} the central region is symmetric
about the origin and its boundary is the union of an even number of
unbounded trajectories. Therefore, the center is of type $B^k$,
where $k=2, 4, 6$. Moreover, a ``generic" system
has a center of type $B^2$. For the center to be
of type $B^4$ or $B^6$,  the trigonometric polynomial
$Q(\cos \varphi, \sin \varphi)$ must take its greatest value
$c_0$ on the interval $[0, 2\pi)$ more
than twice. This requires extra restrictions on the
coefficients of the system.

In the case \eqref{5ii} system (\ref{4}) takes the form
\begin{gather*}
\dot x = y+x(a_1 x y + c_1 x^5 y + c_3 x^3 y^3 + c_5 x y^5),\\
\dot y = -x+y(a_1 x y + c_1 x^5 y + c_3 x^3 y^3 + c_5 x y^5).
\end{gather*}
If $a_1=0$ then we arrive at the case \eqref{5i}.
If $a_1\neq 0$ then we may assume that $a_1=1$. The general case is
reduced to this by the change of variables
$x \mapsto x/\sqrt{a}, y \mapsto y/\sqrt{a}$ for $a>0$ or
$x \mapsto y/\sqrt{-a}, y \mapsto x/\sqrt{-a}, t \mapsto -t$ for $a<0.$

According to \cite{5}, the system
\begin{equation} \label{13}
\begin{gathered}
\dot x = y+x(x y + c_1 x^5 y + c_3 x^3 y^3 + c_5 x y^5)\equiv y+ x P(x,y),\\
\dot y = -x+y(x y + c_1 x^5 y + c_3 x^3 y^3 + c_5 x y^5)\equiv -x+ y P(x,y)
\end{gathered}
\end{equation}
commutes with an analytic system of the form
\begin{equation} \label{14}
\dot x = x Q(x,y),\quad
\dot y= y Q(x,y),
\end{equation}
where the function $Q(x,y)$ meets the equation
\begin{equation}  \label{15}
\begin{aligned}
&x(Q_y(x,y)+P_x(x,y)Q(x,y)-P(x,y)Q_x(x,y))\\
&+y(-Q_x(x,y)+P_y(x,y)Q(x,y)-P(x,y)Q_y(x,y))=0.
\end{aligned}
\end{equation}
Suppose that the function $Q(x,y)$ is a polynomial
in the variables $x$ and $y$ which has
degree $N$ in $y$: $Q(x,y)=Q_0(x) + Q_1(x)y+\ldots+Q_N(x)y^N$.
After insertion of $Q(x,y)$ in (\ref{15}),  the left-hand side becomes
a polynomial of degree at most $N+5$ and the coefficient of $y^{N+5}$ is
$$
c_5 x((6-N)Q_N(x)-x Q_N'(x)).
$$
Therefore, we must have $x Q_N'(x)=(6-N) Q_N(x)$ or $c_5 =0$.
If $c_5\neq0$,  this yields $N\leq6$.
The same bound of $N$ can  be shown to be true
also in the case when $c_5=0$.
Likewise,  we can show that the degree of $Q(x,y)$ in $x$ is at most 6.

Substituting the polynomial $Q(x,y)=\sum_{i,j=0}^6 q_{ij}x^i y^j$
in (\ref{15}) yields a system of linear equations
in the coefficients $q_{ij}$.
We derived this system and investigated its properties by
using the software package {\it Mathematica}.
In this way we found out that the necessary solvability condition is
\begin{equation}\label{16}
c_3^2-4c_1c_5=0.
\end{equation}
This condition is also sufficient.

As an example, assume that $c_1=\alpha^2$, $c_5=\beta^2$,
and $c_3=2\alpha\beta$.
Then (\ref{4}) takes the form
\begin{equation} \label{17}
\begin{gathered}
\dot x=y+x^2 y (1 + (\alpha x^2 +\beta y^2)^2),\\
\dot y=-x+x y^2 (1 + (\alpha x^2 +\beta y^2)^2).
\end{gathered}
\end{equation}
Straightforward calculations show that (\ref{17}) commutes with
the polynomial system
\begin{gather*}
\dot x=x(\alpha-\beta + (\alpha x^2 +\beta y^2) +(\alpha x^2 +\beta y^2)^3),\\
\dot y=y(\alpha-\beta + (\alpha x^2 +\beta y^2) +(\alpha x^2 +\beta y^2)^3).
\end{gather*}
Likewise, we can study the case when
$c_1=\alpha^2$, $c_5=\beta^2$, $c_3=-2\alpha\beta$
and the case when
$c_1=-\alpha^2$, $c_5=-\beta^2$, and $c_3=\pm 2\alpha\beta$.

We have thus proved that system (\ref{13}) commutes with a polynomial system
of the form (\ref{14}) if and only if (\ref{16}) holds.

Recall that every uniformly isochronous $O$-symmetric quintic system
satisfying the center conditions commutes with some polynomial system
of the same degree \cite{6}.
At the same time,
an arbitrary uniformly isochronous (not necessarily
$O$-symmetric) quintic system with a center may fail to commute
with any polynomial system \cite{4}.

The functions
$$
C_1=x^2+y^2,
C_2=\alpha-\beta + (\alpha x^2 +\beta y^2) +(\alpha x^2 +\beta y^2)^3
$$
are invariants for (\ref{17}) with the respective cofactors
$$
L_1=2xy(1+(\alpha x^2+\beta y^2)^2), \quad
L_2=2xy(1+3(\alpha x^2+\beta y^2)^2).
$$
Moreover, if $\alpha\neq\beta$ then the function
$$
C_3=\exp\Big(\int_0^{\alpha x^2+\beta y^2} \frac{dt}{\alpha-\beta+t+t^3}\Big)
$$
is an invariant with the cofactor $L_3=2xy$.
We have $3L_1-L_2-2L_3=0.$ In this case the function
\begin{align*}
H(x,y)&=\frac{C_1^3}{C_2 C_3^2}\\
&=\frac{(x^2+y^2)^3}{\alpha-\beta + (\alpha x^2 +\beta y^2) +(\alpha x^2
+\beta y^2)^3}\\
&\quad \times\frac1{
\exp(2\int_0^{\alpha x^2+\beta y^2} dt/(\alpha-\beta+t+t^3)}
\end{align*}
is a first Darboux integral of (\ref{17}).

When $\alpha=\beta$, the change of variables
$x=\sqrt{r} \cos \varphi$, $y=\sqrt{r} \sin \varphi$ reduces
(\ref{17}) to the system
$$
\dot r=r^2(1+\alpha^2 r^2) \sin 2 \varphi,
\dot \varphi=-1
$$
for which the function
$$
G(x,y)=\frac{1}{r}+\sin^2 \varphi + \arctan \alpha r
$$
is a first integral. Therefore, the function
$$
H(x,y)=\frac{x^2+y^2}{1+x^2+\alpha(x^2+y^2)\arctan \alpha (x^2+y^2)}
$$
is a first integral of (\ref{17}) for $\alpha=\beta$.

As in the case \eqref{5i},  system (\ref{17}) has a center of type $B^k$
$(1\leq k\leq 6)$.
Since the system under study is reversible,
the central region is symmetric with respect to both coordinate axes.
Therefore,  the center is of type $B^2$, $B^4$, or $B^6$.
The singular points on the equator of the Poincar\'{e} sphere,
vertices of the symmetric boundary of the central region,
are saddle points. They have common separatrices only in exceptional cases.
So in the case \eqref{8i} the center is  ``generically" of type $B^2$.
Clearly, all our results about system (\ref{4}) can be
translated to system (\ref{2}).

\subsection*{Acknowledgments}
The author is grateful to the referees for their
useful remarks and suggestions. The author thanks Dr. N. Dairbekov for
his help with the translation of this paper.


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%%%!!!                                                       |
%%%!!!                                               May be here is ";"
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