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

\AtBeginDocument{{\noindent\small 
\emph{Electronic Journal of Differential Equations}, 
Vol. 2011 (2011), No. 15, pp. 1--12.\newline 
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu
or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2011 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2011/15\hfil Cubic and quartic planar differential systems]
{Cubic and quartic planar differential systems with exact
algebraic limit cycles}

\author[A. Bendjeddou, R. Cheurfa\hfil EJDE-2011/15\hfilneg]
{Ahmed Bendjeddou, Rachid Cheurfa}  % in alphabetical order

\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}

\thanks{Submitted April 28, 2010. Published January 26, 2011.}
\subjclass[2000]{34C05, 34A34, 34C25}
\keywords{Polynomial system; invariant curve; algebraic curve;
\hfill\break\indent limit cycle; Hilbert 16th problem}

\begin{abstract}
 We construct  cubic and quartic polynomial planar differential
 systems with exact limit cycles that are ovals of algebraic real
 curves of degree four. The result obtained for the cubic case
 generalizes a proposition of \cite{l1}. For the quartic case, we deduce
 for the first time a class of systems with four algebraic limit
 cycles and another for which nested configurations of limit
 cycles occur.
\end{abstract}

\maketitle 
\numberwithin{equation}{section}
\numberwithin{figure}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{example}[theorem]{Example}
%\allowdisplaybreaks

\section{Introduction}

In this article, we consider the autonomous planar polynomial
system of ordinary differential equations
\begin{equation}
\begin{gathered}
\dot {x}=\frac{dx}{dt}=P(x,y), \\
\dot {y}=\frac{dx}{dt}=Q(x,y),
\end{gathered}\label{e1}
\end{equation}
where $P$ and $Q$ are two polynomials of $\mathbb{R}[x,y
]$ with no common factor, the derivatives are performed
with respect to the time variable $t$. By definition, the degree
of the system \eqref{e1} is $n=\max (\deg (P),\deg (Q))$.

In the qualitative theory of planar dynamical systems see
\cite{d1,z1}, one of the most important topics is related to the
second part of the unsolved Hilbert 16th problem: what is the
maximum number $H(n)$ of limit cycles that system \eqref{e1} can
have for a given degree $n$ and what are their disposition in the
$x,y$? There is a huge literature about limit cycles, most of them
deal essentially with their detection, their number and their
stability and rare are papers concerned by giving them explicitly.
In the last years however, some papers on planar systems with one
or more exact non trivial limit cycles of higher degrees were
written (see for instance \cite{a1,b1,l1} and references therein).
We recall that in the phase plane, a limit cycle of system
\eqref{e1} is a closed periodic orbit in the set of all its
periodic orbits. A limit cycle is stable if all other solutions
approach the limit cycle either from its interior or from exterior
asymptotically as $t\to +\infty $ and when the limit cycle is
unique and stable it dominates the global behavior of the system.

An algebraic limit cycle is a non-singular compact component (or
an oval) of a real algebraic curve. Also, according to Harnack
theorem, the maximum number of ovals that an algebraic real curve
of degree $n$, can have is at most $\frac{1}{2}(n-1)(n-2)+1$ and
when this bound is reached, the corresponding curve is called an
$M$-curve. It is strongly expected \cite{d1,g2,s1} that a
polynomial planar system of degree $n$ has at most $\frac{1}{2}
(n-1)(n-2)+1$ algebraic limit cycles and generally we look for
them as non-singular compact components of invariant algebraic
curves.

For $U\in \mathbb{R}[x,y]$, the algebraic curve $U=0$ is
called an invariant curve of the polynomial system \eqref{e1},
if for some polynomial $K\in $\ $\mathbb{R}[x,y]$ called the cofactor of
the algebraic curve, we have
\begin{equation}
P\frac{\partial U}{dx}+Q\frac{\partial U}{dy}=KU.  \label{e2}
\end{equation}
Simple analysis of equation \eqref{e2} shows that the degree
of the cofactor is at most $n-1$ and that the curve $U=0$ is
formed by trajectories of the system \eqref{e1}. Also, if the curve
$U=0$ is non-singular, the equilibrium points of the system are
contained either in its non-bounded components or are located on
the curve $K=0$.

This paper is concerned by systems with maximum number of
algebraic  limit cycles, we show that this is possible for $n=3$
and $n=4$ by giving their exact analytic expressions. More
precisely, we prove that the cubic class admits two algebraic
limit cycles of degree four, by the way a result obtained in
cite{l1} becomes a particular case. Concerning the quartic case,
we present a class of systems with four exact limit cycles and an
other one with two nested limit cycles, this classes are new in
the literature. This work is organized as follows:

In the second section, we construct a class of cubic systems
admitting two algebraic limit cycles of degree four analytically
given. This generalizes the example in \cite{l1}.

Section three is devoted to an effective construction of  classes
of quartic systems with one, two and four exact algebraic limit
cycles. To obtain the result, we  use a theorem by Christoffer
\cite{c2}.

\section{Cubic Systems}

Consider the class of cubic systems
\begin{equation}
\begin{gathered}
\begin{aligned}
\dot {x} &=P(x,y)\\
&=a_{00}+a_{10}x+a_{01}y+a_{20}x^2+a_{11}xy \\
&\quad +a_{02}y^2+a_{30}x^3+a_{21}x^2y+a_{12}xy^2+a_{03}y^3,
\end{aligned}\\
\begin{aligned}
\dot {y}&=Q(x,y)\\
&=b_{00}+b_{10}x+b_{01}y+b_{20}x^2+b_{11}xy \\
&\quad +b_{02}y^2+b_{30}x^3+b_{21}x^2y+b_{12}xy^2+b_{03}y^3.
\end{aligned}
\end{gathered} \label{e3}
\end{equation}
For a sub-class of \eqref{e3}, we prove the existence of two algebraic
limit cycles, moreover these limit cycles are explicitly given.

We introduce the quartic curve $U=0$ where
\begin{equation}
U(x,y)=y^2+\delta (x^2-\alpha ^2)(x^2-\beta ^2),
\label{e4}
\end{equation}
and $\alpha ,\beta ,\gamma $ and $\delta$ are real
constants such that $\delta >0$, $0<\alpha <\beta $.
To formulate and prove the main result of this section,
we need to establish some auxiliary lemmas.

\begin{lemma}  \label{lem1}
The curve $U=0$ is a non-singular quartic formed by two ovals.
\end{lemma}

\begin{proof}
Let $\delta >0,\alpha ,\beta $ such that $\alpha <\beta $. We
observe that the curve $U=0$ exits only when
$x\in [-\beta ,-\alpha ]\cup [\alpha ,\beta ]$ and is formed
by the union of the two arcs $(C_1)$ an $(C_2)$, where
\begin{gather*}
(C_1):\{ (x,y):x\in [-\beta ,-\alpha ]\cup [
\alpha ,\beta ]\wedge y=y_1(x)\},
\\
(C_2):\{ (x,y):x\in [-\beta ,-\alpha ]\cup [
\alpha ,\beta ]\wedge y=y_2(x)\} ,
\\
y_1(x)=-\sqrt{\delta }\sqrt{(x^2-\alpha ^2)(x^2-\beta
^2)}, \quad y_2(x)=-y_1(x).
\end{gather*}
 Since $y_2(x)-y_1(x)=2\sqrt{\delta }\sqrt{(x^2-\alpha
^2)(x^2-\beta ^2)}\geq 0$, the second arc is always above
the first one with the common points $(-\beta ,0),(-\alpha ,0),(\alpha ,0),(\beta ,0)$
as points of meeting, hence $U=0$ is composed of the two closed curves
\begin{gather*}
\{ (x,y):x\in [-\beta ,-\alpha ]\wedge y=y_2(x)\}
\cup \{ (x,y):x\in [-\beta ,-\alpha ]\wedge
y=y_1(x)\} ,\\
\{ (x,y):x\in [\alpha ,\beta ]\wedge
y=y_1(x)\} \cup \{ (x,y):x\in [\alpha ,\beta
]\wedge y=y_2(x)\} .
\end{gather*}
By construction  of the curve $U=0$, the problem
of smoothness can occur only at the points of meeting $(-\beta
,0),(-\alpha ,0),(\alpha ,0),(\beta ,0)$ of $(C_1)$ and
$(C_2)$, but this is not the case since at this points the
tangent to the curve is just parallel to the $y$-axis according
to the simple fact that
\[
\frac{dy}{dx}=\pm x\frac{\sqrt{\delta }}{\sqrt{(\alpha
^2-x^2) (\beta ^2-x^2) }}(-2x^2+\alpha ^2+\beta ^2).
\]
\end{proof}

\begin{lemma} \label{lem2}
The most general cubic planar polynomial system admitting the curve
 $U=0$ as invariant curve is the system:
\begin{equation}
\begin{gathered}
\dot {x}=-\frac{1}{2}((\alpha ^2+\beta ^2)
a_{21}+\frac{1}{\delta }b_{21})
y+a_{11}xy+a_{21}x^2y+a_{12}xy^2,
\\
\begin{aligned}
\dot {y}&=2\delta \alpha ^2\beta ^2a_{11}-\frac{1}{2}( \delta
(\alpha ^2-\beta ^2) ^2a_{21}+(\alpha ^2+\beta
^2) b_{21}) x \\
&\quad +2\delta \alpha ^2\beta ^2a_{12}y-\delta (\alpha
^2+\beta ^2) a_{11}x^2+2a_{11}y^2 \\
&\quad +b_{21}x^3-\delta (\alpha ^2+\beta ^2)
a_{12}x^2y+2a_{21}xy^2+2a_{12}y^3.
\end{aligned}
\end{gathered}\label{e5}
\end{equation}
\end{lemma}

\begin{proof}
Considering system \eqref{e3}, we perform an Euclidean division of the
polynomial $P(x,y) \frac{\partial U}{dx}(x,y) +Q( x,y)
\frac{\partial U}{dy}(x,y) $ over the polynomial $U(x,y) $ with
respect to the $y$ variable. The curve $U=0$ is invariant for this
system if and only if the remainder vanishes identically. We are lead to a
linear system of sixteen equations of the twenty unknowns $a_{ij}$
and $b_{ij}$, $i+j=0,1,2,3$. Using Maple, we obtain
\begin{gather*}
a_{00}=a_{10}=a_{20}=a_{02}=a_{30}=a_{03}=b_{11}=0;
\\
a_{01}=-\frac{1}{2}((\alpha ^2+\beta ^2) a_{21}+\frac{
1}{\delta }b_{21}); \quad
b_{00}=2\delta \alpha ^2\beta ^2a_{11};
\\
b_{10}=-\frac{1}{2}(\delta (\alpha ^2-\beta ^2)
^2a_{21}+(\alpha ^2+\beta ^2) b_{21}) ;\quad
b_{01}=2\delta \alpha ^2\beta ^2a_{12};
\\
b_{20}=-\delta (\alpha ^2+\beta ^2)a_{11};\quad
b_{02}=2a_{11}; \quad b_{12}=b_{30}; \quad b_{21}=b_{03}=2a_{12}.
\end{gather*}
After the substitution of these solution into system \eqref{e3} and
rewriting it in the standard form, we obtain the system \eqref{e5}. Then
straightforward computations show that
\[
P(x,y) \frac{\partial U}{dx}(x,y) +Q(x,y) \frac{\partial
U}{dy}(x,y) =(4a_{12}y^2+( 4a_{11}+4xa_{21}) y) U(x,y),
\]
which implies that the curve $U=0$ is an invariant curve and the
associated cofactor is $K(x,y) =4a_{12}y^2+4( a_{11}+a_{21}x)y$.
\end{proof}

Since we are interested by limit cycles rather than by invariant curves, we
are constrained to impose additional conditions in order that system
\eqref{e5}
admits the ovals of the curve $U=0$ as periodic solutions. If we put
$a_{21}=a_{11}=0$, this system reduces to
\begin{equation}
\begin{gathered}
\dot {x}=-\frac{1}{2\delta }b_{21}y+a_{12}xy^2, \\
\begin{aligned}
\dot {y}&=-\frac{1}{2}(\alpha ^2+\beta ^2)
b_{21}x+2\delta \alpha ^2\beta ^2a_{12}y \\
&\quad +b_{21}x^3-\delta (\alpha ^2+\beta ^2)
a_{12}x^2y+2a_{12}y^3.
\end{aligned}
\end{gathered}\label{e6}
\end{equation}

\begin{lemma} \label{lem3}
Suppose that $a_{12}\neq 0,b_{21}\neq 0$ and
\begin{equation}
\big(\frac{b_{21}}{a_{12}}\big) ^2>\tfrac{8}{27}\delta ^3( (\alpha
^2+\beta ^2) (\alpha ^4-\tfrac{5}{2}\alpha ^2\beta
^2+\beta ^4)-(\alpha ^4-\beta ^2\alpha ^2+\beta ^4)
^{3/2}) . \label{e7}
\end{equation}
Then the ovals of the curve $U=0$ are periodic solutions of
\eqref{e6}.
\end{lemma}

\begin{proof}
The singular points at finite distance of system \eqref{e6} are the
origin $(0,0) $, the points $(\pm \frac{\sqrt{2}}{2}\sqrt{\alpha
^2+\beta ^2},0) $, and for $y\neq 0$, the others that can
probably lie on the curve $U=0$ are the points
$(x_0,\frac{1}{2\delta } \frac{b_{21}}{a_{12}x_0}) $ where
$x_0$ is solution of the equation $4\delta
^3a_{12}^2(x^2-\beta ^2) (x^2-\alpha ^2)
x^2+b_{21}^2=0$. Thanks to condition \eqref{e7}, this equation has
no real solutions. Consequently, these ovals are periodic
solutions, one oval surrounds the point
$(-\frac{\sqrt{2}}{2}\sqrt{ \alpha ^2+\beta ^2},0) $, the
other surrounds the point $( \frac{\sqrt{2}}{2}\sqrt{\alpha
^2+\beta ^2},0) $.
\end{proof}

We recall that our goal is to show that the ovals of the curve $U=0$
are if fact limit cycles. The main result of this section is as follows.

\begin{theorem} \label{thm4}
If the constants $\alpha ,\beta ,\gamma $ and $\delta$ are such
that $\delta >0$, $0<\alpha <\beta $ and the constants
$a_{12}\neq 0$ and $b_{21}$ satisfy the inequality \eqref{e7}, the
system \eqref{e6} admits as limit cycles the two ovals of the algebraic
curve $U=0$.
\end{theorem}

\begin{proof}
From the preceding lemmas, the cofactor is $K(x,y)=4a_{12}y^2$,
the two ovals of the algebraic curve $U=0$ are periodic orbits
of  \eqref{e6}.

 If we denote by $T_{i}$, $i=1,2$, their corresponding periods, then
since $a_{12}\neq 0$,
\[
\int_0^{T_{i}}K(x(t),y(t))dt=4a_{12}\int _0^{T_{i}}y^2(t)dt\neq 0.
\]
By \cite[Theorem 1]{g1} we conclude. The hyperbolicity of the two
limit cycles depends on the sign of $a_{12}$.
\end{proof}

\subsection{Nature of the singular points}

It is useful to discuss the nature of the singular points of the
system \eqref{e6} in order to draw its phase portrait near this
points. We just outline the situation for the interesting cases:
\begin{itemize}
\item For $b_{21}\neq \frac{1}{3}$, $4( \frac{b_{21}}{a_{12}} )
^2(3b_{21}-1) <\frac{\delta ^3(\alpha ^2-\beta ^2) ^4}{(\alpha
^2+\beta ^2) }$ and $a_{12}<0$, the points $(\pm
\frac{\sqrt{2}}{2}\sqrt{\alpha ^2+\beta ^2},0) $ are stable foci;

\item For $b_{21}\neq \frac{1}{3}$, $4( \frac{b_{21}}{a_{12}} )
^2(3b_{21}-1) <\frac{\delta ^3(\alpha ^2-\beta ^2) ^4}{(\alpha
^2+\beta ^2) }$ and $a_{12}>0$, the points $(\pm
\frac{\sqrt{2}}{2}\sqrt{\alpha ^2+\beta ^2},0) $ are unstable
foci.
\end{itemize}
 Finally, the origin is always a saddle and as the line $y=0$ is an
invariant curve, no other orbit can cross the $y$-axis.


\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.7\textwidth]{fig21}
\end{center}
\caption{Limit cycles and the null-clines}
\label{fig21}
\end{figure}


\subsection{Example}

This example shows that  \cite[proposition 19]{l1} is a particular
result of Theorem \ref{thm4}. In fact if we take $\delta =\frac{1}{2}$,
$a_{12}=2$, $b_{21}=-20$, $\alpha ^2\beta ^2=\frac{1}{2}$ and
$\alpha ^2+\beta ^2=2$, the system \eqref{e5} reads
\begin{equation}
\begin{gathered}
\dot {x}=20y+2xy^2, \\
\dot {y}=20x+y-20x^3-2x^2y+4y^3,
\end{gathered} \label{e8}
\end{equation}
which is nothing but system \cite[(50)]{h1}. It is easy to see
that Theorem \ref{thm4} applies, and that the two ovals of the curve
$y^2+\frac{1}{2} x^4-x^2+\frac{1}{4}=0$ are the corresponding
algebraic limit cycles enclosing the unstable foci $(\pm 1,0)$
(see the disposition of the limit cycles in figure \ref{fig21},
where the vertical isocline $\frac{dx}{dt}=0$ is composed of the
straight line $y=0$ and the hyperbola $y=-10/x$, while the
horizontal isocline $y=0$ is the cubic
$20x+y-20x^3-2x^2y+4y^3=0$). Inside the limit cycles, all
solutions recede from the foci and spiral clockwise approaching
the limit cycles asymptotically as $t\to +\infty $.

\section{Quartic System}

In this section, we give a feasible construction of quartic systems
admitting exact hyperbolic algebraic limit cycles of degree four.
In fact, since nested configurations of ovals can occur for an
algebraic curve of degree equal or greater than $4$, this allow
us to present for the first time a class of quartic systems with
nested configuration of algebraic limit
cycles and also an other class with four exact algebraic limit cycles.
Moreover this limit cycles are explicitly given. For that,
we consider the well known planar system
\begin{equation}
\begin{gathered}
\dot {x}=P(x,y)=af(x,y) -D(x,y) f_{y}(x,y) , \\
\dot {y}=Q(x,y)=bf(x,y) +D(x,y) f_{x}(x,y) ,
\end{gathered}\label{e9}
\end{equation}
 where $f\in \mathbb{R}[x,y]$ is a polynomial of degree
$m$ and $D(x,y) =ux+vy$ $+w$. This planar system belongs to a more
general class of systems intervening in the inverse approach of
dynamical systems. Christoffer \cite{c2} has proved that if the
line $D(x,y)=0$ lies outside all non-singular compact components
(ovals) of the algebraic curve $f=0$ and the constants $a$ and $b$
are chosen such that $au+bv\neq 0$, then this system admits all
the bounded components of the curve $f=0$ as hyperbolic limit
cycles. Furthermore, the vector field $(3.1)$ has no other limit
cycles.

 We introduce the real algebraic curve $f=0$ given analytically for
$\gamma \neq 0$ by
\begin{equation}
f(x,y)=y^4+x^4+\alpha y^2+\beta x^2+\gamma =0.
\label{e10}
\end{equation}
 The main result of this section is as follows.

\begin{theorem} \label{thm5}
Let the planar differential system
\begin{equation}
\begin{gathered}
\begin{aligned}
\dot {x}&=a\gamma -2w\alpha y+a\beta x^2-2u\alpha xy+(
a-2v) \alpha y^2\\
&\quad - 4wy^3+ax^4-4uxy^3+(a-4v) y^4,
\end{aligned} \\
\begin{aligned}
\dot {y}&=b\gamma +2w\beta x+(b+2u) \beta x^2+2v\beta
xy+b\alpha y^2\\
&\quad + 4wx^3+(b+4u) x^4+4vx^3y+by^4,
\end{aligned}
\end{gathered} \label{e11}
\end{equation}
where the real constants $\alpha ,\beta ,\gamma \neq
0,a,b,u,v$ and $w$ are such that $au+bv\neq 0$ and the line
$ux+vy$ $+w=0$ do not intersect the algebraic curve \eqref{e10}. If we
assume additionally the conditions:
\begin{gather}
(\alpha ,\beta ,\gamma ) \in \mathbb{R}^3\setminus \mathbb{R}
^{+}\times \mathbb{R}^{+}\times \mathbb{R}^{\ast +} \label{e12}
\\
\alpha ^2+\beta ^2-4\gamma >0;  \label{e13}
\\
\alpha ^2-4\gamma \neq 0;  \label{e14}
\\
\beta ^2-4\gamma \neq 0;  \label{e15}
\end{gather}
this system possesses exactly:
\begin{itemize}
\item Four limit cycles each located strictly in one of the
four quarter of the plane and having the four points $(\pm
\frac{\sqrt{2}}{2 }\sqrt{-\beta },\pm
\frac{\sqrt{2}}{2}\sqrt{-\alpha }) $ as centers if $\alpha
<0$, $\beta <0$ and $\max \{ \alpha ^2,\beta ^2\}<4\gamma$;

\item a nested configuration of two limit cycles with the origin as
center for $\alpha <0$, $\beta <0$ and
$0<4\gamma <\min \{ \alpha ^2,\beta ^2\}$;

\item two limit cycles symmetric with respect to the
$y$-axis and centered at \\
$(\pm \frac{\sqrt{2}}{2}\sqrt{-\beta },0) $
if  $\alpha >0$, $\beta <0$ and $\alpha ^2<4\gamma <\beta ^2$;

\item two limit cycles symmetric with respect to the
$x$-axis and centered at \\
$(0,\pm \frac{\sqrt{2}}{2}\sqrt{-\alpha }) $
for $\alpha <0$, $\beta >0$ and $\beta ^2<4\gamma <\alpha ^2$;

\item one limit cycle centered at the origin if $\alpha \leq 0$,
$\beta \leq 0$ and $\gamma <0$.

\item Moreover, these limit cycles are hyperbolic and
analytically given as the ovals of the curve \eqref{e10}.
\end{itemize}
\end{theorem}

To prove this theorem, we need some lemmas.

\begin{lemma} \label{lem6}
The curve \eqref{e10} is composed only by ovals.
\end{lemma}

\begin{proof}
We recall that the curve \eqref{e10} is non-singular if and only if
the following system has no solutions:
\begin{gather*}
\frac{\partial f}{\partial x}(x,y)=0, \\
\frac{\partial f}{\partial y}(x,y)=0, \\
f(x,y)=0.
\end{gather*}
 Taking into account the symmetries of this curve, we
are just lead to examine the following possible critical points of
the function $f(x,y)$: $(0,0),( 0,\frac{\sqrt{2}}{2}\sqrt{-\alpha
}) $ for 
$\alpha \leq 0,( \frac{1}{2}\sqrt{2}\sqrt{-\beta },0)$ for
$\beta \leq 0$ and the last one is
$(\frac{\sqrt{2}}{2}\sqrt{-\beta },\frac{
\sqrt{2}}{2}\sqrt{-\alpha }) $ for
$\alpha \leq 0$ and $\beta \leq 0$. But from the assumptions of
Theorem \ref{thm5},
\begin{gather*}
f(0,0)=-\gamma \neq 0, \\
f(0,\frac{\sqrt{2}}{2}\sqrt{-\alpha })=-\frac{1}{4}(\alpha
^2-4\gamma ) \neq 0, \\
f(\frac{\sqrt{2}}{2}\sqrt{-\beta },0)=-\frac{1}{4}(\beta
^2-4\gamma ) \neq 0,\\
f(\frac{\sqrt{2}}{2}\sqrt{-\beta
},\frac{\sqrt{2}}{2}\sqrt{-\alpha })=- \frac{1}{4}(\alpha
^2+\beta ^2-4\gamma ) <0,
\end{gather*}
 so this curve is non-singular.

 Since $y=\pm \frac{\sqrt{2}}{2}\sqrt{-\alpha \pm \sqrt{
-4x^4-4\beta x^2+\alpha ^2-4\gamma }}$ cannot approach $\pm
\infty $, the curve remains at finite distance from the origin, it
is then bounded and consequently composed just by ovals. We can
also deduce that this curve is bounded if we put \eqref{e12} on the
form:
\[
(y^2+\frac{\alpha }{2})^2+(x^2+\frac{\beta
}{2})^2=\frac{1}{4} (\alpha ^2+\beta ^2-4\gamma ),
\]
 which shows by the same that conditions \eqref{e12} and \eqref{e13}
are necessary and sufficient for the curve to be non-empty and
not formed by a finite set of points. This complete the proof.
\end{proof}

The following lemma enumerates the number of ovals.

\begin{lemma} \label{lem7}
The curve \eqref{e10} is composed of:
\begin{itemize}
\item Four ovals strictly located each strictly in one of
the four quarters of the plane and having the four points
$( \pm \frac{\sqrt{2}}{ 2}\sqrt{-\beta },\pm
\frac{\sqrt{2}}{2}\sqrt{-\alpha }) $ as centers if $\alpha
<0$, $\beta <0$ and $\max \{ \alpha ^2,\beta ^2\} <4\gamma $;

\item a nested configuration of two ovals centered at the
origin for $\alpha <0$, $\beta <0$ and $0<4\gamma <\min \{
\alpha ^2,\beta ^2\}$;

\item two ovals symmetric with respect to the $y$-axis with
the points $(\pm \frac{\sqrt{2}}{2}\sqrt{-\beta },0) $ as centers
if $\alpha >0$, $\beta <0$ and $\alpha ^2<4\gamma <\beta ^2$;

\item  two ovals symmetric with respect to the $x$-axis with
the points $(0,\pm \frac{\sqrt{2}}{2}\sqrt{-\alpha }) $ as centers
for $\alpha <0$, $\beta >0$ and $\beta ^2<4\gamma <\alpha ^2$;

\item one oval centered at the origin if $\alpha \leq
0$, $\beta \leq 0$ and $\gamma <0$.
\end{itemize}
\end{lemma}

\begin{proof}
The symmetries of the curve $f=0$ allow us to avoid a cumbersome
proof. Let $N_1$ (resp. $N_2$) be the number of intersecting
points of this curve with the $x$-axis (resp. the $y$-axis). To
compute $N_1$ and $N_2$, we consider respectively the
equations
\begin{gather}
x^4+\beta x^2+\gamma =0 \label{E-1},\\
y^4+\alpha y^2+\gamma =0 \label{E-2},
\end{gather}
and let $\Delta (x)=-4x^4-4\beta x^2+\alpha ^2-4\gamma $.

\textbf{Case 1:} $\alpha ^2-4\gamma <0$ and $\beta
^2-4\gamma <0$. Equations \eqref{E-1}\ and \eqref{E-2} have no
solutions so $N_1=0$ and $N_2=0$. Restricted to $x>0$ and
$y>0$, the curve $f=0$ is composed of an oval centered at the
point $(\frac{\sqrt{2}}{2}\sqrt{ -\beta
},\frac{\sqrt{2}}{2}\sqrt{-\alpha }) $ and formed by the union of
the arcs
\[
x\mapsto y(x)=\frac{\sqrt{2}}{2}\sqrt{-\alpha +\sqrt{\Delta
(x)}}, \quad
x\mapsto y(x)=\frac{\sqrt{2}}{2}\sqrt{-\alpha
-\sqrt{\Delta (x)}},
\]
 and $x \in [x_1,x_2]$, with
\[
x_1=\frac{\sqrt{2}}{2}\sqrt{-\beta -\sqrt{\alpha ^2+\beta
^2-4\gamma } },\quad
x_2=\frac{\sqrt{2}}{2}\sqrt{-\beta
+\sqrt{\alpha ^2+\beta ^2-4\gamma }}.
\]
 The same statement holds in the remaining quarters of the plan.

\textbf{Case 2:} These conditions $\alpha <0$, $\beta <0$ and 
$0<4\gamma <\min \{ \alpha ^2,\beta ^2\}$
imply that the equations
\eqref{E-1} and \eqref{E-2} admit four solutions each, so
$N_1=N_2=4$. Let us denote increasingly by
$-x_2<-x_1<x_1<x_2$ and $-y_2<-y_1<y_1<y_2$ these
solutions. In this case, the curve $f=0$ is composed by two nested
ovals centered both at the origin, the outer one passes through
the points $(x_2,0) ,(0,y_2) ,( -x_2,0) $ and $(0,-y_2) $,
and the inner one passes through the points
$(x_1,0) ,(0,y_1),(-x_1,0) $ and $(0,-y_2) $. For $y>0$,
the union of the two arcs
\[
x\mapsto y(x)=\frac{\sqrt{2}}{2}\sqrt{-\alpha +\sqrt{\Delta
(x)}}, \quad
x\mapsto y(x)=\frac{\sqrt{2}}{2}\sqrt{-\alpha
-\sqrt{\Delta (x)}}
\]
forms the two halves located above the $x$-axis of the two nested
ovals, while the union of the arcs
\[
x\mapsto y(x)=-\frac{\sqrt{2}}{2}\sqrt{-\alpha +\sqrt{\Delta (x)}},\quad
x\mapsto y(x)=-\frac{\sqrt{2}}{2}\sqrt{-\alpha -\sqrt{\Delta (x)}}
\]
composes the remaining parts of these ovals located
below the $x$-axis.

\textbf{Case 3:}  $\alpha >0,\beta <0$ and $\alpha ^2<4\gamma
<\beta ^2 $.  Equation \eqref{E-1} possesses four solutions
$-x_2<-x_1<x_1<x_2 $ and \eqref{E-2} no solutions, so
$N_1=4$ and $N_2=0$. The curve $f=0$ is composed by two ovals
symmetric with respect to the $y$-axis, the first one is centered
at $(-\frac{\sqrt{2}}{2}\sqrt{ -\beta },0) $ and passes through
the points $(-x_2,0)$ and $(-x_1,0)$ and the second centered
at $(\frac{\sqrt{2}}{2}\sqrt{ -\beta },0) $ passes through the
points $(x_1,0)$ and $(x_2,0)$.

\textbf{Case 4:}
$\alpha <0$, $\beta >0$ and $\beta ^2<4\gamma <\alpha ^2$:
The arguments are similar to those in the third case. The curve
$f=0$ is composed by two ovals where the first one passes through
the points $(0,-y_2)$ and $(0,-y_1)$ and the second through
the points $(0,y_1)$ and $(0,y_2)$.

\textbf{Case 5:} $\alpha \leq 0$, $\beta \leq 0$ and
$\gamma <0$. Equations \eqref{E-1} and \eqref{E-2} have two
solutions each denoted by $-x_1,x_1$ and $-y_1,y_1$ so $N_1=N_2=2$
and the curve $f=0$ admits the origin as center and is formed by
a single oval passing through the points $(x_1,0),(0,y_1),(-x_1,0)$ and
$(0,-y_1)$.
\end{proof}

We can now give the proof of the main result of this section.

\begin{proof}[Proof of Theorem \ref{thm5}]
If we choose $f(x,y)=y^4+x^4+\alpha
y^2+\beta x^2+\gamma $ in system $(9)$, we obtain system \eqref{e11}
and of course the curve $f=0$ is an invariant curve of the later,
we can see that
\[
K(x,y)=a\frac{\partial f}{\partial x}+b\frac{\partial f}{\partial
y} =4ax^3+2a\beta x+4by^3+2b\alpha y
\]
is the associated cofactor.

 From Lemma \ref{lem6}, no equilibrium point of the system \eqref{e11}
can lie on the curve $f=0$, so the ovals of this curve are
periodic orbits. Since we have assumed that the line $ux+vy$
$+w=0$ do not intersect the algebraic curve $f=0$, from the
theorem by Christoffer \cite{c2} we deduce that these ovals are
the only hyperbolic limit cycles of system \eqref{e11}. Their
number is discussed in  Lemma \ref{lem7} and this completes the proof.
\end{proof}

\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.7\textwidth]{fig31}
\end{center}
\caption{Four limit cycles for system \eqref{e16}} 
\label{fig31}
\end{figure}


\begin{example} \label{exa8} \rm
Let $a=b=1$, $u=w=0$, $v=1$. The system
\begin{equation}
\begin{gathered}
\dot {x}=\frac{5}{2}-3x^2+2y^2+x^4-3y^4, \\
\dot {y}=\frac{5}{2}-3x^2-6xy-2y^2+x^4+4x^3y+y^4,
\end{gathered}\label{e16}
\end{equation}
 admits exactly four hyperbolic limit cycles  that
are the ovals of the algebraic curve
$y^4+x^4-2y^2-3x^2+\frac{5}{2}=0$, each oval encloses one of
the four equilibrium points approximately given by ($x=\pm 1.29$,
$y=\pm 0.89$). See Figure \ref{fig31}.
\end{example}

\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.7\textwidth]{fig32}
\end{center}
\caption{A nested configutation of two limit cycles for system \eqref{e17}} 
\label{fig32}
\end{figure}

\begin{example} \label{exa9} \rm
Let $a=b=1$, $u=0$, $v=1$, $w=-3$:
 the system
\begin{equation}
\begin{gathered}
\dot {x}=1-36y-6x^2+6y^2+12y^3+x^4-3y^4, \\
\dot {y}=1+36x-6x^2-12xy-6y^2-12x^3+x^4+4x^3y+y^4,
\end{gathered} \label{e17}
\end{equation}
 admits two hyperbolic limit cycles that are the ovals
of the algebraic curve $y^4+x^4-6y^2-6x^2+1=0$. The inner
oval contains the equilibrium points
($x=-2.78\times 10^{-2}$, $y=2.78\times 10^{-2}$), and the set
\begin{gather*}
(x=1.62,y=-0.21),\quad (x=-1.87,y=-1.56), \quad
(x=-1.83,y=-0.21), \\
(x=-2.15,y=2.15),\quad (x=0.15,y=-1.65),\quad
( x=1.56,y=-1.56)
\end{gather*}
of equilibrium points lie between the two ovals. See Figure
\ref{fig32}.
\end{example}

\begin{remark} \label{rmk10}\rm
The class of bounded non-singular quartic curves defined
 by \eqref{e10} is the
simplest one with all possible configurations of ovals and regular
transforms of the plane or slight deformation directly operated on its
equation do not alter its topology. We can illustrate this by
the next example.
\end{remark}

\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.7\textwidth]{fig33}
\end{center}
\caption{A  nested configuration of limit cycles for system \eqref{e18}} 
\label{fig33}
\end{figure}

\begin{example} \label{exa11} \rm
 The system
\begin{equation}
\begin{gathered}
\begin{aligned}
\dot {x}&=-\frac{3}{2}+40x-5y-66x^2+30xy-12y^2-88x^3-252x^2y \\
&\quad -66xy^2-21y^3+40x^4-24x^3y-144x^2y^2-68xy^3-13y^4,
\end{aligned}\\
\begin{aligned}
\dot {y}&=-5-8x-62y-84x^2+108xy-30y^2+176x^3-72x^2y \\
&\quad +132xy^2-6y^3-64x^4+32x^3y-48x^2y^2+24xy^3,
\end{aligned}
\end{gathered}\label{e18}
\end{equation}
admits the nested asymmetric configuration of ovals composing the
curve given by the full term equation:
\begin{align*}
&17y^4-24xy^3+48x^2y^2+24x^3y+17x^4+32y^3-48xy^2\\
&+24x^2y-4x^3-6y^2-24xy-24x^2-16y+8x-4=0
\end{align*}
 as the only hyperbolic limit cycles. This result is
derived from the preceding example when we perform the regular
change of variables $x\to 2x+y$, $y\to x-2y-1$ on system
\eqref{e17}. See Figure \ref{fig33}.
\end{example}

\subsection*{Conclusion and perspectives}

The elementary method used in this paper seems to be fruitful to
investigate more general planar dynamical systems in order to
obtain explicitly some or all their limit cycles at least when it
is question of the algebraic ones. In the spirit of the inverse
approach to dynamical systems, we look for them as the ovals of
suitably chosen invariant algebraic curves. The following
questions can be raised:

 - Does system \eqref{e6} admits additional limit cycles? Can one
transforms it into another cubic system with two algebraic limit
cycles of degree greater than four?

 - Are there quartic systems with three algebraic limit cycles?

 - Is it right that the number $\frac{(n-2)(n-1)}{2}+1$
is the upper bound of algebraic limit cycles for system \eqref{e1}
in general and for system for which this bound is reached (as for
the cubic and quartic classes studied in this paper), do we have
$H(n)-[\frac{(n-2)(n-1)}{2}+1 ]=0$?

 - Finally, we know (see \cite{c1}) that nested configurations of
algebraic limit cycles are not possible in quadratic systems, we
claim that they are not possible in cubic systems to.

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\end{document}