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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2018 (2018), No. 83, pp. 1--7.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2018 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2018/83\hfil Normal forms and hyperbolic algebraic limit cycles]
{Normal forms and hyperbolic algebraic limit cycles for a class
of polynomial \\ differential systems}

\author[J. Llibre, C. Valls \hfil EJDE-2018/83\hfilneg]
{Jaume Llibre, Claudia Valls}

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

\address{Claudia Valls \newline
Departamento de Matem\'atica,
Instituto Superior T\'ecnico,
Universidade de Lisboa, 1049-001 Lisboa, Portugal}
\email{cvalls@math.ist.utl.pt}

\thanks{Submitted March 3, 2017. Published March 31, 2018.}
\subjclass[2010]{34C05, 34A34, 34C14}
\keywords{Limit cycles; polynomial vector fields; algebraic limit cycles}

\begin{abstract}
 We study the normal forms of polynomial systems having a set of
 invariant algebraic curves with singular points. We provide
 sufficient conditions for the existence of hyperbolic algebraic
 limit cycles.
\end{abstract}

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

\def\N{\mathbb{N}}
\def\R{\mathbb{R}}
\def\Z{\mathbb{Z}}
\def\C{\mathbb{C}}
\newcommand{\X}{{\mathcal X}}
\newcommand{\Y}{{\mathcal Y}}

\section{Introduction}

 The algebraic theory of integrability is a classical tool and is
 related with the first part of the Hilbert's 16th problem. This kind
 of integrability is usually called Darboux integrability, and it
 provides a link between the integrability of a polynomial
 differential system and its number of invariant algebraic curves.
 In this article we are interested in polynomial differential
 systems, integrable or not, having a given set of invariant algebraic 
 curves more concretely, we study the normal forms of planar vector 
 fields having a given set of invariant algebraic curves. 
 That is, we are interested in some sense in a kind of inverse theory of 
 the Darboux theory of integrability.

We deal with the following (planar) polynomial differential system
of degree $m$:
\begin{equation}\label{eq:1}
\dot x =P(x,y), \quad \dot y=Q(x,y),
\end{equation}
where $P,Q \in \C_m[x,y]$, being $\C_m[x,y]$ the set of complex
polynomials such that $\max \{\deg  P, \deg  Q\}=m$. 
The dot denotes derivative with respect to a real or complex
independent variable.

Let $F(x,y) \in \C[x,y]$ (being $\C[x,y]$ the ring of polynomials in
$x$ and $y$). The algebraic curve $F(x,y)=0$ of $\C^2$ is called an
\emph{invariant algebraic curve} of system \eqref{eq:1} if
\begin{equation}\label{eq:2}
P F_x + Q F_y = K F
\end{equation}
for some complex polynomial $K(x,y)$ which is called a
\emph{cofactor} of $F=0$. We denote by $F_x$ and $F_y$ the
derivatives of $F$ with respect to $x$ and $y$, respectively. For
simplicity we shall talk about the curve $F=0$ by only saying the
curve $F$, see for details \cite{CLS}.

Cofactors of $F$ have degree at most $m-1$. Let
$F=\prod_{l=1}^\ell F_i^{n_i}$ be  the irreducible decomposition of
$F$. Then $F$ is an invariant algebraic curve with a cofactor $K$ of
system \eqref{eq:1} if and only if $F_i$ is an invariant algebraic
curve of system \eqref{eq:1} with cofactor $K_i$. Moreover
$K=\sum_{i=1}^\ell n_i K_i$. For a proof see \cite{ChL}.

Our first main result works for complex polynomial differential
systems.

\begin{theorem}\label{main.1}
For $i=1,\ldots,p$,  let $F_i=0$ be an irreducible invariant algebraic
curves of a complex polynomial differential system, and set
$r=\sum_{i=1}^p \deg  F_i$. We assume that $F_i$'s satisfy the
following assumptions:
\begin{itemize}
\item [(i)] There are no points at which two curves $F_i$ and $F_j$
satisfy either $F_i=F_j=F_{iy}=0$ and $F_i=F_j=F_{jy}=0$, or
$F_i=F_j=F_{ix}=0$ and $F_i=F_j=F_{jx}=0$. Note that when $p=1$ then
we have no condition.

\item [(ii)] The highest order terms of $F_i$ have no repeated
factors.

\item [(iii)] If two curves intersect at a point in the
finite plane, they are transversal at this point.

\item [(iv)] No more than two curves $F_i=0$ meet
at any point in the finite plane.

\item [(v)] No two curves having a common factor in the
highest order terms.
\end{itemize}
Then any polynomial vector field $\X$ of degree $m$ tangent to all
$F_i=0$ satisfies one of the following statements:
\begin{itemize}
\item [(a)] If $r < m+1$ then
\[
\X=\Big(\prod_{i=1}^p F_i \Big)\Y + \sum_{i=1}^p h_i
\Big(\prod_{j=1, j \ne i}^p F_j \Big) \X_{F_i},
\]
where $\chi_{F_i} = (-F_{iy },F_{ix})$ is a Hamiltonian vector
field, the $h_i$ are polynomials of degree no more than $m-r+1$, and
$\Y$ is a polynomial of degree no more than $m-r$.

\item [(b)] If $r=m+1$ then
\[
\X=\sum_{i=1}^p \alpha_i \Big(\prod_{j=1, j \ne i}^p F_j \Big)
\X_{F_i}, \quad \alpha_i \in \C.
\]
\item [(c)] If $r > m+1$ then $\X=0$.
\end{itemize}
\end{theorem}

Theorem \ref{main.1} is proved in section \ref{sec.2}. 
When condition (i) in Theorem \ref{main.1} is replaced by the 
more restrictive condition: ``there are no points at which $F_i$ 
and its first derivatives all vanish, i.e. the curve $F_i=0$ 
does not have singular points'' was proved in Theorem 1 of \cite{CLPZ}. 
Note that our condition (i) allows that the curves $F_i=0$ have 
singular points, but a singular point of a curve $F_i=0$ cannot also 
be a singular point of the curve $F_j=0$ if $j\ne i$.

Theorem 1 of \cite{CLPZ} goes back to Christopher \cite{C0}, and it
was stated in several papers without a proof such as in 
\cite{C0, KC}, and used in other papers, see \cite{CFL, LPR}.
 Zholadek in \cite{Z1} stated a similar result with an analytical 
proof while in \cite{CLPZ} the proof is algebraic. In all the cases they work with
the assumption that the invariant algebraic curves $F_i=0$ do not
have singular points. The vector field of statement (b) is Darboux
integrable as it was proved in \cite{KC}.

Since the next two results are about limit cycles,  the
polynomial differential systems  and the independent
variables under consideration are real.

\begin{theorem}\label{main.2}
Let $F(x,y)=0$ be the unique irreducible invariant algebraic curve
of degree $n$ of a real polynomial vector field $\X$ of degree $m$.
Then $\X$ can be written as
\begin{equation*}%\label{eq:4}
\X = (\lambda_3 F -\lambda_1 F_y, \lambda_2 F +\lambda_1 F_x)
\end{equation*}
where $\lambda_\nu=\lambda_\nu(x,y)$ for $\nu=1,2,3$ are
polynomials. Assume that the following conditions hold.
\begin{itemize}
\item[(I)] The intersection of the periodic orbits of $F=0$ with
the algebraic curve $\lambda_1=0$ is empty.

\item[(II)] If $\gamma$ is an isolated periodic solution of $\X$ which
does not intersect the curve $\lambda_1=0$ then
\[
\int_\gamma \frac{\lambda_3 \, d y -\lambda_2 \, d x}{\lambda_1} =
\int \int_\Gamma \Big(\Big(\frac{\lambda_3}{\lambda_1}\Big)_x +
\Big(\frac{\lambda_2}{\lambda_1}\Big)_y \Big) \, d x \, d y \ne 0,
\]
where $\Gamma$ is the bounded region limited by $\gamma$.

\item[(III)] the polynomial $(\lambda_3 \lambda_{1x}+\lambda_2
\lambda_{1y})|_{ \lambda_1=0}$ is not zero in $\R^2 \setminus
\{F=0\}$.
\end{itemize}
Then all periodic orbits of the invariant algebraic curve $F=0$ are
hyperbolic limit cycles of $\X$. Furthermore $\X$ has no other limit
cycles.
\end{theorem}

The statement of Theorem \ref{main.2} coincides with the statement
of \cite[Theorem 4]{LRS}, but Theorem 4 has the additional
assumption that the algebraic curve $F=0$ cannot have singular
points, i.e., there are no points at which $F=0$ and all its first
derivatives all vanish. So our Theorem \ref{main.2} improves 
\cite[Theorem 4]{LRS} because it allows that $F=0$ has singular points.
But the proof of Theorem \ref{main.2} is exactly the proof of
\cite[Theorem 4 ]{LRS}, the difference is that such a proof uses
the next Lemma \ref{L1} that it works when the polynomial
differential system has a unique invariant algebraic curve with
singular points. Other papers related with algebraic limit cycles
are \cite{BC, BL, XZ}.

Now we extend Theorem \ref{main.2} to the case with two algebraic curves.

\begin{theorem}\label{main.3}
Let $F_\nu(x,y)=0$ for $\nu=1,2$ be the two unique irreducible
invariant algebraic curves of the polynomial vector field
\begin{equation*}%\label{eq:6}
\X = (\lambda_4 F_1 F_2-r_1 F_{1y} -r_2 F_{2y}, \lambda_3 F_1 F_2
+r_1 F_{1x} +r_2 F_{2x}),
\end{equation*}
of degree $n$, where $r_1=\lambda_1 F_2$, $r_2 =\lambda_2 F_1$ and
$\lambda_j=\lambda_j(x,y)$ for $j=1,2,3,4$ are polynomials. Assume
that the curves $F_\nu(x,y)=0$ satisfy condition {\rm (i)} in
Theorem \ref{main.1} and that the following conditions hold:
\begin{itemize}
\item[(I)] For $\nu,\mu=1,2$, the intersection of the periodic orbits
of $F_\nu=0$ with the algebraic curve $r_\mu=0$ is empty when $\nu
\ne \mu$.

\item[(II)] If $\gamma$ is an isolated periodic solution of $\X$ which
does not intersect the curve $r_\nu=0$ for $\nu=1,2$, then
\[
\begin{split}
I_1 &= \int_\gamma \Big(\frac{\lambda_4 \, d y -\lambda_3 \, d x}{\lambda_1} -
\frac{\lambda_2}{\lambda_1} d (\log |F_2|) \Big) \ne 0, \\
I_2 &= \int_\gamma \Big(\frac{\lambda_4 \, d y -\lambda_3 \, d x}{\lambda_2} -
\frac{\lambda_1}{\lambda_2} d (\log |F_1|) \Big) \ne 0.
\end{split}
\]

\item[(III)] The two polynomials
\[
\begin{split}
& (\lambda_4 r_{1x} + \lambda_3 r_{1y} + \lambda_2 \{F_2,\lambda_1\})_{|r_1=0}, \\
& (\lambda_4 r_{2x} + \lambda_3 r_{2y} + \lambda_1 \{F_1,\lambda_2\})_{|r_2=0},
\end{split}
\]
are not zero in $\R^2 \setminus
\{F_1 F_2=0\}$, where $\{f,g\}=f_x g_y -f_y g_x$.
\end{itemize}
Then all periodic orbits of $F_\nu=0$ for $\nu=1,2$ are hyperbolic
limit cycles of $\X$. Furthermore $\X$ has no other limit cycles.
\end{theorem}

Theorem \ref{main.3} when we additionally assume that the invariant
algebraic curves have no singular points was proved in Theorem 6 of
\cite{LRS}. In fact, our proof is exactly the same than in Theorem 6
of \cite{LRS}, the difference is that we use the next Lemma
\ref{L2}, allowing the two invariant curves to have singular points,
but such singular points are not shared by both curves.

\section{Proof of Theorem \ref{main.1}}\label{sec.2}

We will use the following
well-known Hilbert's Nullstellensatz (see for instance, \cite{Fu}).

\begin{theorem}\label{Nuns}
Set $A,B_i\in \mathbb C[x,y]$ for $i=1,\cdots,r$. If $A$ vanishes in
$\mathbb C^2$ whenever the polynomials $B_i$ vanish simultaneously,
then there exist polynomials $M_i\in\mathbb C[x,y]$ and a
nonnegative integer $n$ such that $A^n=\sum\limits_{i=1}
\limits^{r}M_iB_i$. If all $B_i$ have no common zero, then there
exist polynomial $M_i$ such that $\sum\limits_{i=1}
\limits^{r}M_iB_i=1$.
\end{theorem}

In what follows if we have a polynomial $A$ we will denote its
degree by $a$. We will also denote by $F^c$ the homogeneous part of
degree $c$ of the polynomial $F$. We shall need several auxiliary
results. The first one is proved in \cite{CLPZ}.

\begin{lemma}\label{L0}
If $F^f$ has no repeated factors then $(F_x,F_y)=1$.
\end{lemma}

Now we consider the case in which system \eqref{eq:1} has a given
invariant algebraic curve. The next result improves Lemma 6 of
\cite{CLPZ} showing that the result of \cite{CLPZ} also holds for
invariant algebraic curves with singular points satisfying
conditions {\rm (i)} of Theorem \ref{main.1}.

\begin{lemma}\label{L1}
Assume that the polynomial system \eqref{eq:1} of degree $m$ has an
invariant algebraic curve $F=0$ of degree $f$.
\begin{itemize}
\item[(a)] If $(F_x,F_y)=1$, then system \eqref{eq:1} has the following
normal form
\begin{equation}\label{eq:9}
\dot x = A F - D F_y, \quad \dot y =B F + D F_x,
\end{equation}
where $A,B$ and $D$ are suitable polynomials.

\item[(b)] If $F$ satisfies condition {\rm (ii)} of Theorem \ref{main.1},
then system \eqref{eq:1} has the normal form \eqref{eq:9} with $a, b
\le m -f$ and $d \le m-f+1$. Moreover, if the highest order term
$F^f$ of $F$ does not have the factors $x$ and $y$, then $a \le
p-f$, $b \le q-f$ and $d \le \min\{p,q\}-f+1$.
\end{itemize}
\end{lemma}

We recall that under affine changes of coordinates system
\eqref{eq:9} preserves its form and the degrees of the polynomials.
Indeed, using the affine change $u=a_1 x +b_1 y + c_1$ and $v=a_2 x
+b_2 y + c_2$ with $a_1 b_2 -a_2 b_1 \ne 0$, system \eqref{eq:9}
becomes
\[
\dot u = (a_1 A + b_1 B)F-(a_1 b_2 -a_2 b_1) D F_v, \quad
\dot v = (a_2 A +b_2 B)F +(a_1 b_2 -a_2 b_1) D F_u.
\]

\begin{proof}
We recall that $F$ satisfies \eqref{eq:2} for some polynomial $K$
called the cofactor.

First note that if $F=0$ has no singular points then it follows from
Lemma 6 in \cite{CLPZ} and there is nothing to prove. By condition
(ii) of Theorem \ref{main.1} all the singular points of $F=0$ are
finite (since they are equilibrium points of the polynomial
differential system and its number is $\le m^2$ (see Theorem 4 of
\cite{Ts})), there exists a linear polynomial $L$ such that $F$,
$F_x$, $F_y$ and $L+K$ do not vanish simultaneously, and $L+K$ is
not a constant.

By Theorem \ref{Nuns} (Hilbert's Nullstellensatz) we obtain that
there exist polynomials $E,G,T$ and $H$ such that
\begin{equation}\label{eq:10}
E F_x +T F_y +G F +H (L+K)=1
\end{equation}

From \eqref{eq:2} and \eqref{eq:10} we obtain
\[
K(1-H (L+K))=(K E +G P)F_x + (KT +GQ)F_y.
\]
Substituting $K$ into \eqref{eq:2} we obtain
\begin{align*}
& [(1-H (L+K))P - F(KE+GP)]F_x \\
&= - [(1-H(L+K))Q - F (KT+FQ)]F_y.
 \end{align*}
Since $(F_x,F_y)=1$, there exists a polynomial $D$ such that
\begin{gather*}
(1-H (L+K))P - F(KE+GP)  =-D F_y, \\
 (1-H(L+K))Q - F (KT+FQ) =D F_x.
 \end{gather*}
By scaling the time variable we can write system \eqref{eq:1} as
\begin{equation}\label{eq:1.bis}
\dot x =(1-H (L+K))P, \quad \dot y=(1-H (L+K))Q
\end{equation}
Note that $1-H (L+K)$ is not identically zero, otherwise 
$L+K=1/H$ must be constant, and this is not possible by the choice of $L$.
Hence system \eqref{eq:1.bis} has the form \eqref{eq:9} with
\[
A= KE +GP, \quad B=KT+FQ.
\]
This proves statement (a).

The proof of statement (b) is the same as the proof of statement (b)
of \cite[Lemma 6]{CLPZ}.
\end{proof}

The following lemma improves \cite[Lemma 7]{CLPZ} showing that it
also holds for invariant algebraic curves with singular points
satisfying condition  (i) of Theorem \ref{main.1}.

\begin{lemma}\label{L2}
Assume that $F=0$ and $G=0$ are different irreducible invariant
algebraic curves of system \eqref{eq:1} of degree $m$ that satisfy
the conditions {\rm (i)} and {\rm (iii)} of Theorem \ref{main.1}.
\begin{itemize}
\item[(a)] If $(F_x,F_y)=1$ and $(G_x,G_y)=1$, then system \eqref{eq:1}
has the normal form
\begin{equation}\label{eq:12}
\dot x = A F G -E F_y G -N C G_y, \quad \dot y = BFG+E F_x G +N F G_x
\end{equation}
\item[(b)] If $F$ and $G$ satisfy additionally conditions {\rm (ii)}
and {\rm (v)}, then system \eqref{eq:1} has the normal form
\eqref{eq:12} with $a,b \le m-f-g$ and $e,n \le m-f-g+1$.
\end{itemize}
\end{lemma}

\begin{proof}
Since $(F,G)=1$, the curves $F$ and $G$ have finitely many
intersection points. By assumption (i), we can assume that both
systems $F=G=G_y=0$ and $F=G=F_y=0$ have no solutions (the case when
$F=G=G_x=0$ and $F=G=F_x=0$ can be proved in a similar way). The
rest of the proof follows the same steps than the proof of 
\cite[Lemma 7]{CLPZ}.
\end{proof}

The next lemma follows as in \cite[Lemma 8]{CLPZ}.

\begin{lemma}\label{L3}
Let $F_i=0$ for $i=1,\ldots,p$ be different irreducible invariant
algebraic curves of system \eqref{eq:1} with $\deg F_i=f_i$. Assume
that $F_i$ satisfy conditions (i), (iii) and (iv) of Theorem
\ref{main.1}. Then the following statements hold.
\begin{itemize}
\item[(a)] If $(F_{ix},F_{iy})=1$ for $i=1,\ldots,p$ then system
\eqref{eq:1} has the normal form
\begin{equation}\label{eq:25}
\dot x = \Big(B - \sum_{i=1}^p \frac{A_i F_{iy}}{F_i} \Big) \prod_{i=1}^p
F_i, \quad
\dot y = \Big(C + \sum_{i=1}^p \frac{A_i F_{ix}}{F_i} \Big) \prod_{i=1}^p
F_i, \quad
\end{equation}
where $B,C$ and $A$ are suitable polynomials.

\item[(b)] If $F_i$ satisfy additionally conditions (ii) and (v) of
Theorem \ref{main.1}, then system \eqref{eq:1} has the normal form
\eqref{eq:25} with $b,c \le m- \sum_{i=1}^p f_i$ and $a_i \le m
-\sum_{i=1}^p f_i +1$.
\end{itemize}
\end{lemma}

\begin{proof}
By Lemma \ref{L3} we obtain statement (a) of Theorem \ref{main.1}.
From the degrees of the polynomials $A_i$, $B$ and $C$ in
statement (b) of Lemma \ref{L3} we obtain statement (b) of Theorem
\ref{main.1}.

By statement (a) of Lemma \ref{L3}, we can rewrite system
\eqref{eq:1} into the form \eqref{eq:25}. Finally, by statement (b)
of Lemma \ref{L3} we obtain $B=0$, $C=0$ and $A_i=0$. This completes
the proof of statement (c) of Theorem \ref{main.1}.
\end{proof}


\subsection*{Acknowledgements}
J. Llibre was partially supported by a FEDER-MINECO grant
MTM2016-77278-P, a MINECO grant MTM2013-40998-P, and an AGAUR grant
number 2014SGR-568. C. Valls was partially supported by
FCT/Portugal through the project UID/MAT/04459/2013.


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\end{document}