\documentclass[reqno]{amsart}
\usepackage{hyperref}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 71, pp. 1--7.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2017 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2017/71\hfil Coexistence of algebraic and non-algebraic limit cycles]
{Coexistence of algebraic and non-algebraic limit cycles for
 quintic polynomial differential systems}

\author[A. Bendjeddou, R. Cheurfa \hfil EJDE-2017/71\hfilneg]
{Ahmed Bendjeddou, Rachid Cheurfa}

\address{Ahmed Bendjeddou \newline
D\'epartement de Math\'ematiques, Facult\'e des Sciences,
Universit\'e de S\'etif, 19000 S\'etif, Alg\'erie}
\email{Bendjeddou@univ-setif.dz}

\address{Rachid Cheurfa \newline
D\'epartement de Math\'ematiques, Facult\'e des Sciences,
Universit\'e de S\'etif, 19000 S\'etif, Alg\'erie}
\email{rcheurfa@yahoo.fr}


\dedicatory{Communicated by Mokhtar Kirane}

\thanks{Submitted January 2, 2017. Published March 14, 2017.}
\subjclass[2010]{34C29, 34C25}
\keywords{Non-algebraic limit cycle; invariant curve; Poincar\'e return-map;
\hfill\break\indent  first integral; Riccati equation}

\begin{abstract}
 In the work by Gin\'e and  Grau \cite{11}, a planar differential
 system of degree nine admitting a nested configuration formed by
 an algebraic and a non-algebraic limit cycles explicitly given was presented.
 As an improvement, we obtain by a new method a similar result for a family of
 quintic polynomial differential systems.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\numberwithin{figure}{section}
\newtheorem{theorem}{Theorem}[section]
\allowdisplaybreaks

\section{Introduction}

 In the qualitative theory of autonomous and planar differential systems,
the study of limit cycles is very attractive because of their relation
with the applications to other areas of sciences; see for instance \cite{9,17}.
Nevertheless, most of researchers on that domain focus their
attention on the number, stability and location in the phase plane of the
limit cycles for the system of degree $n=\max \{ \deg P_n,\deg Q_n\}$,
\begin{equation}
\begin{gathered}
\dot{x}=\frac{dx}{dt}=P_n( x,y) , \\
\dot{y}=\frac{dy}{dt}=Q_n( x,y) ,
\end{gathered} \label{e1}
\end{equation}
where $P_n( x,y) $ and $Q_n( x,y) $ are coprime
polynomials of $\mathbb{R}[ x,y] $. We recall that a limit cycle
of system \eqref{e1} is an isolated periodic orbit in the set of its
periodic orbits and it is said to be algebraic if it is contained in the
zero set of an invariant algebraic curve of the system. We recall that an
algebraic curve defined by $U( x,y) =0$ is an invariant curve for
\eqref{e1} if there exists a polynomial $K( x,y) $ (called the
cofactor) such that
\begin{center}
\begin{equation}
P_n( x,y) \frac{\partial U}{\partial x}+Q_n( x,y)
\frac{\partial U}{\partial x}=K( x,y) U( x,y) .
\label{e2}
\end{equation}
\end{center}

Another interesting and also a natural problem is to express analytically
the limit cycles. Until recently, the only limit cycles known in an explicit
way were algebraic (see for instance \cite{4,5,12, 14} and references therein).
It is surprising that exact algebraic limit cycles
where obtained by Abdelkadder \cite{1} and  Bendjeddou and Cheurfa
\cite{4} for a class of Li\'enard equation.

Limit cycles of planar polynomial differential systems are not in general
algebraic. For instance, the limit cycle appearing in the van der Pol
equation is non-algebraic as it is proved by Odani \cite{15}. In the
chronological order the first examples of systems were explicit
non-algebraic limit cycles appeared are those of  Gasull \cite{10} and by
Al-Dossary \cite{2} for $n=5$, Bendjeddou and al.\ \cite{3} for $n=7$ and by
Benterki and Llibre \cite{6} for $n=3$. Another class of quintic systems with
homogeneous nonlinearity has been studied via averaging theory by
Benterki and  Llibre \cite{7} . The first result for the coexistence of
algebraic and non-algebraic limit cycles goes back to  Gin\'e and
Grau \cite{11} for $n=9$. These last authors transform their system into a
Ricatti equation which is itself transformed into a variable coefficients
second order linear differential equation using the classic linearization
method. From the principal result of an earlier work (see details from page
5 of their paper) they obtain a first integral and by the way the explicit
equations of the possible limit cycles.

In this work, we obtain by a more intuitive and understandable method a
similar result for a class of systems of degree $n=5$. We show that our
system admits an invariant algebraic curve, corresponding of course to a
particular solution of the Ricatti equation obtained when the suited
transformations are performed on the system, so the first integral can be
easily obtained. The limit cycles are also exactly given and form
a nested configuration, the inner one is algebraic, while the outer is
non-algebraic.

\section{Main result}

As a main result, we shall prove the following theorem.

\begin{theorem} \label{thm2.1}
The quintic two-parameters system
\begin{equation}
\begin{gathered}
\dot{x}=P_5( x,y) , \\
\dot{y}=Q_5( x,y) ,
\end{gathered} \label{e3}
\end{equation}
where
\begin{align*}
P_5( x,y) &=  x+x( x^2+y^2-1) (ax^2-4bxy+ay^2)  \\
&\quad+ ( x^2+y^2) ( -2x+2y+x^{3}+xy^2) ,
\end{align*}
and
\begin{align*}
Q_5( x,y) &=  y+y( x^2+y^2-1) (ax^2-4bxy+ay^2)  \\
&\quad+ ( x^2+y^2) ( -2x-2y+y^{3}+x^2y) ,
\end{align*}
in which $a\in \mathbb{R}_{+}^{\ast }$ and
$b\in \mathbb{R}^{\ast} $ possesses exactly two limit cycles: the circle
$( \gamma_1) : $ $x^2+y^2-1=0$ surrounding a transcendental and stable
limit cycle $( \gamma _{\ast }) $ explicitly given in polar
coordinates $( r,\theta ) $, by the equation
\begin{equation}
r( \theta ;r_{\ast })
=\sqrt{\frac{\exp ( a\theta +b\cos
2\theta ) ( \frac{r_{\ast }^2}{( r_{\ast }^2-1)
e^{b}}+f( \theta ) ) }{-1+\exp ( a\theta +b\cos
2\theta ) ( \frac{r_{\ast }^2}{( r_{\ast }^2-1)
e^{b}}+f( \theta ) ) }},  \label{e4}
\end{equation}
with
\[
f( \theta ) =\int_0^{\theta }\exp (-as-b\cos 2s) ds, \quad
r_{\ast }=\sqrt{\frac{f( 2\pi )
e^{b+2\pi a}}{( f( 2\pi ) +1) e^{2\pi a}-1}},
\]
 if the following conditions are fulfilled:
\begin{gather}
4b^2-a^2<0, \label{e5} \\
f( 2\pi ) \neq \frac{1+e^{-2\pi a}}{1-e^{b}},\quad
f(2\pi ) \neq e^{-b-2\pi a}\Big( \frac{1+( e^{2\pi a}-1)
r_{\ast }^2\pm \sqrt{2}e^{\pi a}\sqrt{r_{\ast }}}{1-r_{\ast }^2}\Big) .
\label{e6}
\end{gather}
 Moreover, $( \gamma _{\ast }) $ defines an unstable
limit cycle when $b+\pi a=0$.
\end{theorem}

\begin{proof}
Firstly, we have $yP_5( x,y) -xQ_5( x,y) =2(x^2+y^2) ^2$,
thus the origin is the unique critical point at
finite distance. Moreover it is not difficult to see that the circle
$(\gamma _1) :x^2+y^2-1=0$ is an invariant curve, the associated
cofactor being
\begin{equation*}
K( x,y) =2( x^2+y^2) P_2( x,y) ,
\end{equation*}
where $P_2( x,y) =( a+1) x^2-4bxy+(a+1) y^2-1$.

Of course $( \gamma _1) $ defines a periodic solution of system
\eqref{e3}, since it do not pass through the origin. To see whether or not
$( \gamma _1) $ is in fact a limit cycle, we can proceed as
follow: Let $T$ denotes be the period of $( \gamma _1) $, we
consider the integral $I( \gamma _1 ) $, where
\begin{equation}
I( \gamma _1) =\int_0^{T}Div( x( t),y( t) ) dt.  \label{e7}
\end{equation}
We know from \cite{12} that can be computed via
\begin{equation}
I( \gamma _1) =\int_0^{T}K( x( t),y( t) ) dt.  \label{e8}
\end{equation}
 From \eqref{e5}, we have $4b^2-a^2<0$, so the curve $P_2( x,y) =0$
do not cross $( \gamma _1) $. But $P_2( 0,0) <0$, hence $K( x,y) <0$ inside
$(\gamma _1) /\{ ( 0,0)\} $, so $I(\gamma _1) <0$. Consequently
$( \gamma _1) $ defines a stable algebraic limit cycle for system \eqref{e3}.
The search for the non-algebraic limit cycle, requires the
integration of the system. In polar coordinates, this system becomes
\begin{equation}
\begin{aligned}
\dot{r}&=( -2b\sin 2\theta +a+1) r^{5}+( 2b\sin 2\theta -a-2) r^{3}+r, \\
\dot{\theta }&=-2r^2.
\end{aligned}\label{e9}
\end{equation}
Since $\dot{\theta }$ is negative for all $t$, the orbits $( r( t) ,\theta ( t) ) $
of system \eqref{e8} have the opposite orientation with respect to those
 $( x(t) ,y( t) ) $ of system \eqref{e3}. Taking $\theta $
as an independent variable, we obtain the equation
\begin{equation}
\frac{dr}{d\theta }=-\frac{1}{2}( -2b\sin 2\theta +a+1) r^{3}-
\frac{1}{2}( 2b\sin 2\theta -a-2) r-\frac{1}{2r}.
\label{e10}
\end{equation}
Via the change of variables $\rho =r^2$, this equation is transformed into
the Riccati equation
\begin{equation}
\frac{d\rho }{d\theta }=( 2b\sin 2\theta -a-1) \rho ^2+(
-2b\sin 2\theta +a+2) \rho -1.  \label{e11}
\end{equation}
Fortunately, this equation is integrable, since it possesses the
particular solution $\rho =1$ corresponding of course to the limit cycle
$( \gamma _1) $. The general solution of this equation is
\begin{equation*}
\rho ( \theta ) =\frac{\exp ( a\theta +b\cos 2\theta )
( k+f( \theta ) ) }{-1+\exp ( a\theta +b\cos
2\theta ) ( k+f( \theta ) ) },
\end{equation*}
 with $f( \theta ) =\int_0^{\theta }\exp (-as-b\cos 2s) ds$.
Consequently, the general solution of  \eqref{e10} is
\begin{equation}
r( \theta ;k) =\sqrt{\frac{\exp ( a\theta +b\cos 2\theta
) ( k+f( \theta ) ) }{-1+\exp ( a\theta
+b\cos 2\theta ) ( k+f( \theta ) ) }}\,,
\label{e12}
\end{equation}
as given in the theorem.

By passing to Cartesian coordinates, we deduce the first integral is
\begin{equation}
\begin{aligned}
F( x,y) &=  \Big(\tfrac{x^2+y^2}{x^2+y^2-1}-\exp (
a\arctan \frac{y}{x}+b\cos ( 2\arctan \frac{y}{x}) )\\
&\quad\times \int_0^{\arctan \frac{y}{x}}\exp ( -b\cos 2s-as) ds\Big)\\
&\quad \div \exp ( a\arctan \frac{y}{x}+b\cos ( 2\arctan \frac{y}{x}) ).
\end{aligned}
\label{e13}
\end{equation}
 The trajectories of system \eqref{e3} are the level curves
$F( x,y) =k$, $k$ $\in \mathbb{R}$ and since these curves are
obviously all non-algebraic (if we exclude of course the curve
$(\gamma _1) $ corresponding to $k\to +\infty $), thus any
other limit cycle, if exists, should also be non-algebraic.

To go a steep further, we remark that the solution such as
$r(0;r_0) =r_0>0$, corresponds to the value
$k=\frac{r_0^2}{( r_0^2-1) e^{b}}$ provided a rewriting of the general
solution of \eqref{e10} as
\begin{equation}
r( \theta ;r_0) =\sqrt{g( \theta ) },  \label{e14}
\end{equation}
 where
\begin{equation}
g( \theta ) =\frac{\exp ( a\theta +b\cos 2\theta )
\big( \frac{r_0^2}{( r_0^2-1) e^{b}}+f( \theta ) \big) }
{-1+\exp ( a\theta +b\cos 2\theta ) \big( \frac{r_0^2}{( r_0^2-1) e^{b}}
+f( \theta ) \big) }\label{e15}
\end{equation}
A periodic solution of system \eqref{e3} must satisfy the
condition
\begin{equation}
r( 2\pi ;r_0) =r_0,  \label{e16}
\end{equation}
provided two distinct values of $r_0$: $r_1=1$ and thanks to
\eqref{e6}, the well defined second value
\[
r_{\ast }=\sqrt{\frac{f( 2\pi ) e^{b+2\pi a}}{( f( 2\pi ) +1) e^{2\pi a}-1}}.
\]
Obviously, the first value of $r_0$ corresponds to the algebraic
limit cycle $( \gamma _1) $.

By inserting the second value $r_{\ast }$ of $r_0$ in
\eqref{e14}, we obtain the second candidate solution given by the statement of
the theorem through  \eqref{e4}. In the sequel, the notation
$r(\theta ,r_{\ast }) $ or $( \gamma _{\ast }) $ both refer to
this curve solution.

 To show that it is a periodic solution, we have to show that

\noindent$\bullet $ the function $\theta \to g( \theta ) $ is $2\pi $-periodic,
where in this case
\begin{equation}
g( \theta ) =\frac{\exp ( a\theta +b\cos 2\theta )
\big( \frac{e^{2\pi a}}{1-e^{2\pi a}}f( 2\pi ) +f( \theta
) \big) }{-1+\exp ( a\theta +b\cos 2\theta )
\big(\frac{e^{2\pi a}}{1-e^{2\pi a}}f( 2\pi ) +f( \theta )\big) }\,.  \label{e17}
\end{equation}


\noindent$\bullet $ $g( \theta ) >0$ for all $\theta \in [ 0,2\pi [$.

\noindent The last condition ensures that $r( \theta ,r_{\ast }) $
is well defined for all $\theta \in [ 0,2\pi [ $ and the periodic
solution do not pass through the unique equilibrium point $( 0,0)$ 
of system \eqref{e3}.
\smallskip

\noindent\textbf{Periodicity.} Let $\theta \in [ 0,2\pi [ $, then
\begin{equation*}
g( \theta +2\pi ) =\frac{\exp ( a\theta +2\pi a+b\cos
2\theta ) \big( \frac{e^{2\pi a}}{1-e^{2\pi a}}f( 2\pi )
+f( \theta +2\pi ) \big) }{-1+\exp ( a\theta +2\pi a+b\cos
2\theta ) \big( \frac{e^{2\pi a}}{1-e^{2\pi a}}f( 2\pi )
+f( \theta +2\pi ) \big) }.
\end{equation*}
However,
\begin{align*}
f( \theta +2\pi )
& =\int_0^{\theta +2\pi }\exp (-as-b\cos 2s) ds \\
& =f( 2\pi ) +\int_{2\pi }^{\theta +2\pi }\exp ( -as-b\cos 2s) ds.
\end{align*}
In the integral case $\int_{2\pi }^{\theta +2\pi }\exp ( -as-b\cos
2s) ds$, we make the change of variable $u=s-2\pi $, we obtain
\begin{align*}
f( \theta +2\pi )
& =f( 2\pi ) +\int_0^{\theta }\exp ( -a( u+2\pi ) -b\cos 2( u+2\pi ) ) \\
& =f( 2\pi ) +e^{-2\pi a}f( \theta ) .
\end{align*}
 Taking into account \eqref{e5},  after some calculations we obtain
that $g( \theta +2\pi ) =g( \theta ) $, hence $g$ is $2\pi $-periodic.
\smallskip

\noindent\textbf{Strict positivity of $g( \theta ) $ for
$\theta \in [ 0,2\pi [ $.}
Let $\phi (\theta ) =\frac{e^{2\pi a}}{1-e^{2\pi a}}f( 2\pi ) +f(
\theta ) $. Since $\frac{d\phi }{d\theta }( \theta ) =\exp
( -a\theta -b\cos 2\theta ) >0$ for all $\theta \in [ 0,2\pi [ $,
the function $\theta \to \phi ( \theta ) $ is
strictly increasing with
$\phi ( 0) =\frac{e^{2\pi a}}{1-e^{2\pi a}}f( 2\pi ) $ and
$\phi ( 2\pi ) =\frac{1}{1-e^{2\pi a}}f( 2\pi ) $.
Since $a>0$, then $\phi ( 2\pi ) <0\Longrightarrow
\phi ( \theta ) <0$, thus $\exp ( a\theta +b\cos 2\theta
) \phi ( \theta ) <0$, hence $g( \theta ) >0$
for all $\theta \in [ 0,2\pi [ $.

 To show that it is in fact a limit cycle, we consider \eqref{e15},
and introduce the Poincar\'e return map
$r_0\to \Pi (2\pi;r_0)=r( 2\pi ;r_0) =\sqrt{g( 2\pi ) }$, with the
positive $x$-axis as section. We compute
$\frac{d\Pi }{dr_0}(2\pi ;r_0) $ at the value $r_0=r_{\ast }$. We find that
\begin{align*}
&\frac{d\Pi }{dr_0}(2\pi ;r_0)\big| _{r_0=r\ast }\\
&=e^{\pi a}r_{\ast }\frac{\sqrt{( e^{2\pi a}+Ae^{b+2\pi a}-1)
r_{\ast }^2-Ae^{b+2\pi a}}}{\sqrt{( ( Ae^{b}+1) r_{\ast
}^2-Ae^{b}) }( ( e^{2\pi a}+Ae^{b+2\pi a}-1) r_{\ast
}^2-Ae^{b+2\pi a}) ^2}.
\end{align*}
 Taking into account \eqref{e6}, we deduce that
$ \frac{d\Pi}{dr_0}(2\pi ;r_0)\big|_{r_0=r\ast }\neq 1$, and finally that
$(\gamma _{\ast })$ is the expected non-algebraic limit cycle. Obviously
$(\gamma _{\ast })$ lies inside $(\gamma _1)$ when $r_{\ast }<1$.
 Since the Poincar\'e return map do not possess other fixed points,
the system \eqref{e3} admits exactly two limit cycles.
\end{proof}

\section{Example}

As an example let $a=4$, $b=1$. then  system \eqref{e3} becomes
\begin{equation}
\begin{gathered}
x^{\prime }=x+x( x^2+y^2-1) ( 4x^2-4xy+4y^2)
+( x^2+y^2) ( -2x+2y+x^{3}+xy^2) , \\
y^{\prime }=y+y( x^2+y^2-1) ( 4x^2-4xy+4y^2)
+( x^2+y^2) ( -2x-2y+y^{3}+x^2y) .%
\end{gathered}
\label{e18}
\end{equation}
Then we have $f( 2\pi ) =\int_0^{2\pi }\exp (
-4s-\cos 2s) ds\simeq 0.121\,24$ and then
\begin{equation}
r_{\ast }=\sqrt{\frac{( 0.121\,24) e^{1+8\pi }}{( (
0.121\,24) +1) e^{8\pi }-1}}\simeq 0.542\,15.  \label{e19}
\end{equation}
It is easy to verify that all conditions of Theorem \ref{thm2.1} are
satisfied. We conclude that system \eqref{e18} has two limit cycles.
Since $r_{\ast }<1$, the non-algebraic lies inside the algebraic one as shown on the
Poincar\'e disc in Figure \ref{fig1}:


\begin{figure}[ht]
\begin{center}
 \includegraphics[width=0.5\textwidth]{fig1} % example coexistence LC}
\end{center}
\caption{Limit Cycles of System \eqref{e18}}
\label{fig1}
\end{figure}

\subsection*{Conclusion}
In this work, we have extend the result
obtained in \cite{11} by reducing  the degree of the
differential system from $n=9$ to $n=5$. The method used
is intuitive. Obtaining interesting results of this kind becomes more
and more difficult for lower values of $n$.  Nevertheless it is not
forbidden to undertake the study of the following problems:
\begin{itemize}
\item  coexistence of two explicit non-algebraic limit cycles
for a quintic system;

\item  coexistence of explicit algebraic and non-algebraic
limit cycles for $n=3$;

\item obtaining a quadratic system with exact non-algebraic limit
cycle (this question is due to Benterki and Llibre \cite{6}).
\end{itemize}

\subsection*{Acknowledgments}
 The authors would like to express their gratitude
to the referee for pointing out some references to our attention and for his
valuable remarks.


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\end{document}