\documentclass[reqno]{amsart}
\usepackage{hyperref}
\usepackage{graphicx}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2014 (2014), No. 40, 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 2014 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2014/40\hfil Sharp bounds of the number of zeros]
{Sharp bounds of the number of zeros of Abelian integrals with parameters}

\author[X. Sun, J. Yang \hfil EJDE-2014/40\hfilneg]
{Xianbo Sun, Junmin Yang}  % in alphabetical order

\address{Xianbo Sun \newline
Department of Applied Mathematics, Guangxi University of Finance and Economics,
Nanning, 530003 Guangxi, China}
\email{xianbo01@126.com Tel +86 15977781786}

\address{Junmin Yang \newline
College of Mathematics and Information Science, Hebei Normal University \newline
Shijiazhuang, 050024 Hebei, China}
\email{jmyzhw@sina.com}

\thanks{Submitted November 11, 2013. Published February 5, 2014.}
\subjclass[2000]{34C05, 34C07, 34C08}
\keywords{Limit cycle;  Li\'enard system; Chebyshev system; bifurcation;
\hfill\break\indent heteroclinc loop}

\begin{abstract}
 In this article,  we study four Abelian integrals over compact level curves
 of four sixth-degree hyper-elliptic Hamiltonians with  parameters.
 We prove that the sharp bound of the number of zeros for each Abelian
 integral  is 2. The proofs rely mainly  on the Chebyshev criterion for Abelian
 integrals and  asymptotic expansions of
 Abelian integrals.
\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 main result}

The  second part of  Hilbert's 16th problem and its weak version are
two open problems in the qualitative theory of planar differential
equations. The first one asks for the maximal number of limit cycles
and their distribution for the following planar polynomial
differential equation of degree $n$,
 \begin{equation}\label{h1}
  \dot{x}=P_n(x,y), \ \ \dot{y}=Q_n(x,y).
 \end{equation}
A special form of \eqref{h1} is
\begin{equation}\label{eq11}
\begin{array}{l}
\dot{x}={H}_y + \varepsilon p(x,y,\delta),\quad
\dot{y}=-{H}_x + \varepsilon q(x,y,\delta),
\end{array}
\end{equation}
where $H(x,y)$, $p(x,y)$,  $q(x,y)$ are polynomials of $x$ and $y$,
and their degrees satisfy $\max\{\deg p, \deg q\}=n$ and $\deg(H)=n+1$, and
$\varepsilon$ is a positive and sufficiently small parameter. 
The unperturbed form of \eqref{eq11} is
\begin{equation}\label{eq11a}
\dot{x}={H}_y,\quad \dot{y}=-{H}_x.
\end{equation}
The Hamiltonian function $H(x,y)$  defines at least  one family of
closed curves  $L_h$ which form a period annulus of \eqref{eq11a}
denoted by $\{L_h\}$, where $h$ is  energy parameter on  an open
interval $J$.  Corresponding to system \eqref{eq11}, the following 
integral is called Abelian
integral or first order Melnikov function,
\begin{equation}\label{AA}
 A_n(h)= \oint_{L_h} q(x,y)dx-p(x,y)dy, \quad h \in J,
\end{equation}
which plays an important role in studying  the limit cycles of
\eqref{eq11} (see the Poincar\'e-Pontryagin Theorem \cite{cl}),
and finding the upper bound of the maximal number of zeros of $A_n(h)$ 
is the weak version of the second part of Hilbert's 16th problem 
(usually called weak Hilbert's 16th problem).
 Its research advances and the  recent popular and efficient
methods for  special forms of \eqref{eq11} can be found in the
survey works \cite{li2012,li2003}.

Since both problems are difficult, mathematicians try to study
special and simpler forms of \eqref{h1} and \eqref{eq11}. 
 Smale 13th problem restricts Hilbert's 16th problem to the 
Li\'enard system
$$
 \dot{x}=y-f(x), \quad \dot{y}=-x.
$$
To study the number of zeros of $A_n(h)$, many mathematicians
concentrate on a simpler form of \eqref{eq11}  as follows
\begin{equation}\label{l0}
    \dot{x}=y, \quad \dot{y}=g(x)+\varepsilon f(x)y,
\end{equation}
which is called Li\'enard system of type $(m,n)$ if
 $g(x)$ and $f(x)$ are polynomials of degree respectively $m$ and
 $n$.


 A comprehensive study has been made
in \cite{fdcr} for the cases $m +n\leq  4$, except for 
$(m, n)=(1,3)$. In all these cases, it has proven that at most one limit cycle
can appear and for  $(m, n)=(1, 3)$ the same result has been
conjectured (see \cite{ccnl}). For type $(3, 2)$, there are several
cases according to the portraits of the unperturbed system.
Dumortier and Li  \cite{ld1,ld2,ld3,ld4}  have made a complete study
on these cases   and obtained different sharp upper bounds of  the
number of zeros of Abelian integrals for different cases. 
Li, Marde\v{s}i\'c and Roussarie \cite{ld5}  investigated  some  Li\'enard
systems of type $(3, 2)$ with  symmetry  and  also obtained the
sharp bound.   Wang and Xiao  \cite{wang1,wphd}
investigated some Li\'enard system of type $(4,3)$ and proved that $4$ is
the least upper bound and 3 is the maximum lower bound of the number
of the zeros for the  corresponding Abelian integral. Some other
cases of type $(4,3)$ are investigated in \cite{rajaac,ssh2013}, and
the least upper bound and the maximal lower one are obtained.  The
results of the maximum lower bound for other systems of type $(4,3)$
can be found in \cite{yhrjmaa,yhijbc,yhcma}.


For the type $(5,4)$,   many works concentrate on the following  Li\'enard 
systems  with  symmetry
\begin{equation}\label{LL}
    \dot{x}=y, \quad \dot{y}=\eta x (x^2-a)(x^2-b)
+\varepsilon (\alpha+\beta x^2+\gamma x^4)y,
\end{equation}
where $\eta=\pm1$, $\alpha$, $\beta$ and $\gamma$ are real bounded number.
Assume the portraits of system \eqref{LL}$_{\varepsilon=0}$ 
has at least one periodic annulus,  there are 12 cases  according 
to the value of $a$, $b$ and $\eta$, see Figure \ref{fig1}.

\begin{figure}[ht]
\begin{center}
\begin{tabular}{@{}cccc@{}} % {@{}c@{}c@{}c@{}c@{}}
 \includegraphics[width=0.23 \textwidth]{fig1} % b1.eps
&\includegraphics[width=0.23 \textwidth]{fig2} % b2.eps
&\includegraphics[width=0.23 \textwidth]{fig3} % b3.eps
&\includegraphics[width=0.23 \textwidth]{fig4} % b4.eps
\\
 \parbox{29mm}{1: $\eta=-1$, $a>0$, $b>0$, $a\neq b$}
&\parbox{29mm}{2:  $\eta=-1$, $ab\neq0$, $a=b$}
&\parbox{29mm}{3:  $\eta=-1$, $ab=0$, ${\rm sgn}(a)+{\rm sgn}(b)=1$}
&\parbox{29mm}{4:  $\eta=-1$, $ab<0$}
\\[12pt]
 \includegraphics[width=0.23 \textwidth]{fig5} % b5.eps
&\includegraphics[width=0.23 \textwidth]{fig6} % b6.eps
&\includegraphics[width=0.23 \textwidth]{fig7} % b7.eps
&\includegraphics[width=0.23 \textwidth]{fig8} % f1.eps
\\
 \parbox{29mm}{5:  $\eta=-1$, $ab=0$, ${\rm sgn}(a)+{\rm sgn}(b)=-1$} 
&\parbox{29mm}{6:  $\eta=-1$, $a^2+b^2=0$}
&\parbox{29mm}{7:  $\eta=-1$, $a<0$, $b<0$}
&\parbox{29mm}{8:  $\eta=1$, $0<\frac{b}{a}<\frac{1}{3}$ or $\frac{b}{a}>{3}$}
\\[12pt]
 \includegraphics[width=0.23 \textwidth]{fig9} % f2.eps
&\includegraphics[width=0.23 \textwidth]{fig10} % f3.eps
&\includegraphics[width=0.23 \textwidth]{fig11} % f5.eps
&\includegraphics[width=0.23 \textwidth]{fig12} % f6.eps
\\
 \parbox{29mm}{9: $\eta=1$, $\frac{b}{a}=\frac{1}{3}$ or $\frac{b}{a}=3$}
&\parbox{29mm}{10: $\eta=1$, $\frac{1}{3}<\frac{b}{a}<{1}$ or $1<\frac{b}{a}<{3}$}
&\parbox{29mm}{11: $\eta=1$, $ab=0$, ${\rm sgn}(a)+{\rm sgn}(b)=1$}
&\parbox{29mm}{12: $\eta=1$, $ab<0$}
\end{tabular}
\end{center}
\caption{Twelve cases of \eqref{LL} each having at least one annulus 
surrounding a center} \label{fig1}
\end{figure}


For case 1, Zhang et al.~\cite{z1} proved that system  \eqref{LL}
with $a=1/2$, $b=2$ has at most $3$ zeros of the corresponding Abelian
integral. 
For case 2, Asheghi and Zangeneh  studied
\eqref{LL} with $a=b=1$  and   proved that the least upper bound
for the number of zeros of the related Abelian integral inside the
eye-figure loop is 2 in \cite{a1} and both inside and outside the
eye-figure loop is 4 in \cite{a2}. 
For case 3,   Asheghi and
Zangeneh \cite{a3}  studied \eqref{LL}  by taking $a=0$, $b=1$ and
proved that the corresponded Abelian integral  has at most 2 zeros
inside the double cuspidal loops.
For case 3,  Zhao \cite{zhaonon2014} studied system \eqref{LL} 
with $a=0$ and $b=1$ and obtained that $2$ is the sharp bound of the 
number of zeros of Abelian integral associated on the the two bounded 
period annuluses.  
For case 8, Xu and Li \cite{xl} proved that  system  \eqref{LL}  
has at least $5$ limit cycles
bifurcated from $3$ annuluses of the system
\eqref{LL})$_{\varepsilon=0}$ with $a=1/4$, $b=1$. 
For case 9, Sun \cite{sunade} proved there are at most $4$ zeros 
for the corresponding Abelian integral. Later, Zhao \cite{zhaodouble}
 proved the sharp bound of number of zeros for the corresponding Abelian 
integral is 2. 
For case 10, Qi and Zhao \cite{qz} proved that system
\eqref{LL} with $a=\frac{21-\sqrt{41}}{20}$ and
$b=\frac{21+\sqrt{41}}{20}$ has at most $2$ limit cycles bifurcated
from each annulus.

In this article, we  study   the cases 5, 6, 7 and  12  with some
parameters.  Without loss of generality we fix $\gamma=1$ in all cases.
For case 12  we take $a=1$, $b=-\lambda$ without loss of generality. 
For convenience we assume $\lambda\geq1$, then system  \eqref{LL}  becomes
\begin{equation}
\dot{x}=y, \quad \dot{y}=\ x (x^2-1)(x^2+\lambda)
+\varepsilon (\alpha+\beta x^2+ x^4)y, \label{eA}
\end{equation}
with the Hamiltonian function
 \begin{equation}\label{HL}
  \widetilde{ H}(x,y)=\frac{y^2}{2}+\frac{\lambda}{2} {x}^2
- \frac{\lambda-1}{4}x^4 -\frac{1}{6} {x}^6.
 \end{equation}
The level sets (i.e. $\widetilde{H}(x, y) = h$) of Hamiltonian function
\eqref{HL} are sketched in Figure \ref{fig2}.
$\widetilde{H}(x, y)=h$ defines one family of ovals which correspond to a
period annulus  of system \eqref{eA}$_{\varepsilon=0}$ denoted by
 $\{\Gamma_h\}$.   $H(x,y)=\frac{3\lambda+1}{12}$ defines a 2-polycycles
 $\Gamma^*=\{(x,y)|H(x,y)=\frac{3\lambda+1}{12}\}$ which consists of
two heteroclinic  orbits. $\Gamma_{0}$ is an elementary center.
 The Abelian integral on $\Gamma_h$  is
\begin{equation}\label{Ml}
I(h,\delta)=\oint_{\Gamma_h}(\alpha+\beta x^2+  x^4)y dx
\equiv \alpha I_0(h)+\beta I_1(h)+ I_2(h),
\end{equation}
for $h\in (0,(3\lambda+1)/12)$, where $\delta=(\alpha,\beta,1)$,
$I_i(h)=\oint_{\Gamma_h} x^{2i} y dx$, $i=0,1,2$.
\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.5 \textwidth]{fig13} % a.eps
\end{center}
\caption{The level set of  $\widetilde{H}(x,y)$}
\label{fig2}
\end{figure}

For case 7, we take $a=-\lambda_1, b=-\lambda_2$, where $\lambda_1, \lambda_2>0$, 
then system \eqref{LL}  becomes
\begin{equation}
\dot{x}=y, \quad \dot{y}= -x (x^2+\lambda_1)(x^2+\lambda_2)
+\varepsilon (\alpha+\beta x^2+ x^4)y. \label{eB}
\end{equation}
For case 6, we take $a=0, b=-\lambda_3$,  where $\lambda_3>0$,
then system \eqref{LL}  becomes
\begin{equation}
\dot{x}=y, \quad \dot{y}= - x^3 (x^2+\lambda_3)
+\varepsilon (\alpha+\beta x^2+ x^4)y. \label{eC}
\end{equation}
For case 5, we take $a=b=0$,   then system \eqref{LL}  becomes
\begin{equation}
\dot{x}=y, \quad \dot{y}= - x^5+\varepsilon (\alpha+\beta x^2+ x^4)y. \label{eD}
\end{equation}
The corresponding Abelian integrals of systems
\eqref{eB}, \eqref{eC}, \eqref{eD} are, respectively,
\begin{gather*}
I^b(h,\delta)=\oint_{\Gamma_h^b}(\alpha+\beta x^2+  x^4)y dx
 \equiv \alpha I_0^b(h)+\beta I_1^b(h)+ I_2^b(h),\\
I^c(h,\delta)=\oint_{\Gamma_h^c}(\alpha+\beta x^2+  x^4)y dx
 \equiv \alpha I_0^c(h)+\beta I_1^c(h)+ I_2^c(h), \\
I^d(h,\delta)=\oint_{\Gamma_h^d}(\alpha+\beta x^2+  x^4)y dx
 \equiv \alpha I_0^d(h)+\beta I_1^d(h)+ I_2^d(h),
\end{gather*}
where $I(h)$, $I^b(h)$ and $I^c(h)$ have parameters $\lambda$,
$\lambda_1$, $\lambda_2$, $\lambda_3$. Using some algebraic  method,
some polynomial techniques and expansions of Abelian integrals, the
following results are obtained.

\begin{theorem} \label{thmA} 
For all $\alpha$ and $\beta$, each of
$I(h,\delta)$, $I^b(h,\delta)$, $I^c(h,\delta)$  and $I^d(h,\delta)$
has at most $2$ zeros, counting the multiplicity. 
Taking $0<\alpha \ll -\beta \ll 1$, two zeros of each Abelian integral
   appear in some small intervals near $h=0$. Therefore, 2 is  the sharp  bound.
\end{theorem}

By the Poincar\'e-Pontryagin theorem and Theorem \ref{thmA},  each of system 
\eqref{eA}, \eqref{eB}, \eqref{eC}, \eqref{eD} has at most
 2 limit cycles bifurcated from the corresponding period annulus, 
and there exist some $(\alpha,\beta)$ and $0<\varepsilon\ll 1$
such that  each system has  2 limit cycles bifurcated from the 
 corresponding period annulus.
The rest of the article is organized as follows: in section 2 we
will   introduce some definitions and the new criteria  which are
used to determine the number of zeros of the Abelian integrals. In
sections 3 and 4,  we will prove the main results.

\section{Preliminary lemmas and definitions}

The method we will introduce proposes some criterion functions 
defined directly by Hamiltonian and integrands of Abelian
integrals, through which the problem whether the basis
of the vector space generated by Abelian integrals is a
Chebyshev system could be reduced to the problem
whether the family of criterion functions form a Chebyshev
system, since the latter can be tackled by checking the nonvanishing
properties of its Wronskians. For this paper to be
self-contained, we list some related definitions and criterions.
For more details, \cite{mp,MG} is referred.


\begin{definition} \label{def2.1} \rm
 Suppose $f_0, f_1, f_2, \dots , f_{n-1}$ are analytic functions
on an real open interval $J$.

(i) The family of polynomials $\{f_0, f_1, f_2, \dots , f_{n-1}\}$  
 is called Chebyshev system (T-system for short) provided that any
nontrivial linear combination
$$
k_0f_0(x) + k_1f_1(x)+ \dots  + k_{n-1}f_{n-1}(x)
$$
has at most $n-1$ isolated zeros on $J$.

(ii) An ordered set of $n$  functions $\{f_0, f_1, f_2, \dots , f_{n-1}\}$ is
called complete Chebyshev system (CT-system for
short) provided any nontrivial linear combination
$k_0f_0(x) + k_1f_1(x)+ \dots  + k_{i-1}f_{i-1}(x)$ has at most $i-1$ zeros for
all $i = 1,2,\dots ,n$,  moreover it is called extended complete
Chebyshev system (ECT-system for short) if the
multiplicities of zeros taken into account.


(iii) The continuous Wronskian of  $\{f_0, f_1, f_2, \dots , f_{n-1}\}$  
at $x \in R$ is
\begin{align*}
W[f_0, f_1, f_2, \dots , f_{k-1}]
&=\det(f_i^j)_{0\leq i,j\leq k-1}\\
&=  \begin{vmatrix}
 f_0(x) & f_1(x) & \dots  & f_{k-1} \\
 f_0'(x) & f_1'(x) & \dots  & f_{k-1}'(x) \\
 \dots  & \dots  & \dots  & \dots  \\
 f_0^{(k-1)}(x) & f_1^{(k-1)}(x) & \dots  & f_{k-1}^{(k-1)}(x)
 \end{vmatrix},
\end{align*}
where $f'(x)$ is the first order derivative of $f(x)$ and $f^{(i)}(x)$ is
the $i$th order derivative of $f(x)$, $i\geq 2$.
\end{definition}

The above definitions imply that the function tuple $\{f_0, f_1, \dots ,
f_{k-1}\}$ is an ECT-system on $J$, therefore it is a CT-system on
$J$, and then a T-system on $J$, however the inverse implications
are  not true at all.

Recall that the authors of \cite{MG} studied the number of isolated
zero of Abelian integrals using a  purely algebraic criteria which is
developed from the idea introduced in \cite{lizhang}. Let $H(x, y) =
A(x) + \frac{1}{2}y^2$ be an analytic function in some open subset
of the plane which has a local minimum at $(0,0)$. Then there
exists a punctured neighborhood $P$ of the origin foliated by ovals
$L_h:{H(x, y) =h}$ which correspond to the clockwise closed orbits
of $(\ref{eq11a})$. The set of ovals $L_h$ inside the period
annulus, is parameterized by the energy levels $h \in(0, h_1)=J$ for
some $h_1 \in (0,+\infty]$. The projection of $P$ on the $x$-axis is
an interval $(x_l, x_r)$ with $x_l <0 < x_r$. Under the above
assumptions it is easy to verify that $x A'(x) > 0$ for all
$x\in(x_l, x_r)\setminus \{0\}$, $A(x)$ has a zero of even
multiplicity at $x = 0$ and there exists an analytic involution
$z(x)$ such that
$$
A(x) = A(z(x))
$$
for all $x\in(x_l, x_r)$. It is obvious that  
 $z(x)=-x$  if $A(x)$ is a even function.

For the number of isolated zeros of nontrivial linear combination 
of some  integrals of special form, the algebraic criterion in
 \cite[Theorem B]{MG} can be stated as follows:

\begin{lemma} \label{lem2.1} 
 Assume that the function $f_i(x)$ is
analytic on the interval $(x_l,x_r)$ for $i = 0, 1, \dots ,n-1$, and
consider
$$
A_{i}(h)=\int_{L_h} f_i(x)y^{2s-1}dx, \quad i=0, 1,\dots , n-1,
$$
where for each $h \in(0,h_0)$, $L_h$ is the oval surrounding the
origin inside an level curve $\{A(x)+\frac{1}{2}y^2=h\}$. We
define
$$
l_i(x):=\frac{f_i(x)}{A'(x)}- \frac{f_i(z(x))}{A'(z(x))}.
$$
Then,
 $\{A_0, A_1,\dots , A_{n-1}\}$   is an ECT-system  on 
$(0,h_1)$ if $\{l_0, l_1,\dots , l_{n-1}\}$
 is a CT-system
on $(x_l,0)$ or $(0, x_r )$ and $s > n-2$. And $\{l_0, l_1,\dots ,
l_{n-1}\}$   is an ECT-system on $(x_0,x_r)$ or $(x_l,x_0)$ if and
only if the continuous Wronskian of $\{l_0,l_1,\dots ,l_{k-1}\}$
does not vanish for $ \forall x \in (0, x_r)$ or for all 
$z\in (x_l,0)$ and $k = 1,\dots ,n$.
\end{lemma}

Usually $s$ is not big enough,  Lemma \ref{lem2.1} can not be applied  directly. 
To overcome this problem the next result (see \cite[Lemma 4.1]{MG}) 
is useful to increase the power of $y$ in $A_i(h)$.

\begin{lemma} \label{lem2.2} 
 Let $L_h$ be an oval inside the level curve ${A(x) + \frac{1}{2}(x)y^2 = h}$ 
and consider a function $F(x)$ satisfying
$\frac{F(x)}{A'(x)}$ is analytic at $x = 0$. Then, for any $k \in N$,
$$
\oint_{L_h} F(x)y^{k-2} dx = \oint_{L_h}G(x)y^k dx
$$
where $G(x) = \frac{1}{k}(\frac{F}{A'})'(x)$.
\end{lemma}

\section{Proof of main result}

For briefness we prove only case 12, other cases can be proved similarly. 
In what follows, we proved that the following generating elements 
of $I(h,\delta)$,
$$
I_i(h)=\int_{\Gamma_h}x^{2i}ydx,\quad i=0, 1, 2
$$
have the Chebyshev property for  
$h\in (0,\frac{3\lambda+1}{12})$.

 By Lemma \ref{lem2.1},
 $A(x)=\widetilde{H}(x,0)=-\frac{3}{2} {x}^2+{x}^4-\frac{1}{6} {x}^6$
and $s = 1$, $n = 3$  for system \eqref{eA}.
 The period annulus is foliated by the ovals $\Gamma_h$, and
the projection of the period annulus on the  plan is an open interval $(-1, 1)$. 
Noting that $xA'(x) > 0$ for
all $x \in (-1, 1)\setminus \{0\}$, therefore there
exists an analytic involution $z(x)$  such that
 $$
A(x)=A(z(x)).
$$
Our goal is to prove that the vector space generated by
Abelian integral $I_i(h)$ has the Chebyshev property for
 $x\in (0, 1)$ by Lemma \ref{lem2.1}. 
However, for $s = 1$ and $n = 3$ it does not satisfy
the hypothesis $s> n - 2$ in Lemma \ref{lem2.1}. Thus  the power $s$ of $y$
in the integrand of $I_i(h)$ should be increased   such that the
condition $s > n -2$ holds.


\begin{lemma} \label{lem3.1}  
For $i=0,1,2$, we have
$$
2h I_i(h)=\int_{\Gamma_h}{f}_i(x)y^3dx,
$$
where $f_i(x)=  \frac {{x}^{2 i} \widetilde{f}_i(x)}{18(x-1)^2(x+1)^2
(x^2+\lambda)^2}$ with
\begin{align*}
\widetilde{f}_i(x)
&=  20{x}^8+39{x}^6\lambda+21{\lambda}^2{x}^4-39{\lambda}^
{2}{x}^2+24{\lambda}^2+4 i{x}^8-10 i{x}^6+6 i{x}^4\\
&\quad +12i{\lambda}^2+10 i{x}^6\lambda -28 i{x}^4\lambda
+18i\lambda{x}^2+6 i{\lambda}^2{x}^4-18 i{\lambda}^2{x}^2\\
&\quad +21{x}^4- 70{x}^4\lambda+39\lambda{x}^2-39{x}^6.
\end{align*}
\end{lemma}

\begin{proof}
 It is clear that on every periodic orbits
$\Gamma_h:\{\widetilde{H}(x,y)=h\}$, $\frac{2A(x)+y^2}{2h}=1$ holds.
Therefore,
 \begin{equation}\label{51}
 I_i(h)=\frac{1}{2h}\int_{\Gamma_h}(2A(x)+ y^2)x^{2i}ydx
=\frac{1}{2h}\int_{\Gamma_h}2x^{2i}A(x)ydx
  + \frac{1}{2h}\int_{\Gamma_h}x^{2i}y^3dx,
 \end{equation}
for $i=0,1,2$.
Noting that the functions $\frac{2x^{2i}A(x)}{A'(x)}$ are analytic
on $x=1$, by  Lemma \ref{lem2.2},   we have
\begin{equation}\label{32}
\int_{\Gamma_h}2x^{2i}A(x)ydx=\int_{\Gamma_h}G_i(x)y^3dx,
\end{equation}
where
$$
G_i(x)= \frac{ x^{2i} g_i(x)}{( x-1)^2 (x+1)^2 (x^2+\lambda)^2}
$$ 
with
\begin{align*}
g_i(x)& = 2{x}^8+3{x}^6\lambda+3{\lambda}^2{x}^4-3{\lambda}^2{
x}^2+6{\lambda}^2+4 i{x}^8-10 i{x}^6+6 i{x}^4\\
&\quad +12 i{\lambda}^2+10 i{x}^6\lambda  -28 i{x}^4\lambda+18 i\lambda{x}^2
 +6 i{\lambda}^2{x}^4\\
&\quad -18 i{\lambda}^2{x}^2+3{x}^4+2{x}^4\lambda+3\lambda{x}^2-3{x}^6.
\end{align*}
Combine  \eqref{51} and \eqref{32},  so  Lemma \ref{lem3.1} is proved.
\end{proof}

 Let 
$$
\widetilde{I}_i(h)=\int_{\Gamma_h}{f}_i(x)y^3dx.
$$
Then $\{I_0, I_1,I_2\}$ is an  ECT-system   on $(0,\frac{3\lambda+1}{12})$ if
and only if $\{\widetilde{I}_0, \widetilde{I}_1,\widetilde{I}_2\}$
is as well. Since $s = 2$, $n = 3$ and the condition $s > n-2$
holds, lemma \ref{lem2.1} can be used to study if 
$\{\widetilde{I}_0, \widetilde{I}_1,\widetilde{I}_2\}$  
is an ECT-system on $(0,\frac{3\lambda+1}{12})$. Thus, setting the
criteria functions
\begin{equation}\label{lx}
l_i(x)=(\frac{f_i}{A'})(x)-(\frac{f_i}{A'})(z(x)), \ 0 < x < 1,  i= 0, 1, 2,
\end{equation}
where $z(x)$ is the analytic
involution $z(x)$ defined by $A(x)=A(z)$.  By  symmetry of system \eqref{eA}, it is obvious $z(x)=-x$.

Inserting $z(x)=-x$ in \eqref{lx} gives
$$
l_i(x)=- {\frac {( {x}^{2 i}+ (-x) ^{2 i})\widetilde{l_i}(x)}{18 (x-1) ^3
 (x+1)^3 ({x}^2+\lambda)^3x}}
$$ 
with
\begin{align*}
\widetilde{l_i}(x)
&= 20{x}^8+21{\lambda}^2{x}^4-39{
\lambda}^2{x}^2+24{\lambda}^2+4 i{x}^8-10 i{x}^6+6 i{x}^4\\
&\quad +12 i{\lambda}^2+39{x}^6\lambda+10 i{x}^6\lambda
 -28 i{x}^4\lambda+18 i\lambda{x}^2+6 i{\lambda}^2{x}^4\\
&\quad -18 i{\lambda}^2{x}^2+21{x}^4-70{x}^4\lambda+39\lambda{x}^{2}-39{x}^6.
\end{align*}

Next, we  check that the ordered set of criterion
functions $\{l_1(x), l_2(x), l_0(x)\}$ is an ECT-system for 
$x \in(0, 1)$ by verifying the non-vanishing property of continuous
Wronskians $W[l_1]$, $W[l_1, l_2]$, $W[ l_1, l_2, l_0]$. 

\begin{lemma} \label{lem3.2}
The function tuple $\{l_1(x), l_2(x), l_0(x)\}$ is an ECT-system for $x \in (0, 1)$.
\end{lemma}

\begin{proof}  
By the Definition \ref{def2.1} (iii) about the continuous Wronskian,  with the
 aid of Maple 13, we have
\begin{gather*}
 W[l_1(x)]=   \frac{- x w_1(x,\lambda)}{9(x-1)^3 (x+1)^3  (x^2+\lambda)^3}, \\
 W[l_1(x),l_2(x)] =   \frac{2x^3 w_2(x,\lambda)}{81(x-1)^5 (x+1)^5  (x^2+\lambda)^5},  \\
 W[l_1(x),l_2(x),l_0(x)] =   \frac{- 16 w_3(x,\lambda)}{243 (x-1)^7 (x+1)^7 
 (x^2+\lambda)^7},
 \end{gather*}
where 
\begin{gather*}
\begin{aligned}
w_1(x,\lambda)
& = 24{x}^8+27{\lambda}^2{x}^4-57{\lambda}^2{x}^2+36{
\lambda}^2-49{x}^6+27{x}^4+49{x}^6\lambda\\
&\quad -98{x}^4\lambda+57\lambda{x}^2,
\end{aligned} \\
\begin{aligned}
w_2(x,\lambda)
& = 672{x}^{12}-2072{x}^{10}+2072{x}^{10}\lambda-6174{x}^8
\lambda+2295{x}^8+2295{x}^8{\lambda}^2\\
&\quad -891{x}^6+6993{x}^6\lambda 
 -6993{\lambda}^2{x}^6+891{x}^6{\lambda}^3
 +8382 {\lambda}^2{x}^4-2871{\lambda}^3{x}^4\\
&\quad -2871{x}^4\lambda -3672{\lambda}^2{x}^2+3672{\lambda}^3{x}^2
  -1728{\lambda}^3,
\end{aligned}\\
\begin{aligned}
 w_3(x,\lambda)
 & = 13824{\lambda}^4+6237{x}^8+40392{\lambda}^2{x}^4+22275\,
{x}^6\lambda+6237{\lambda}^4{x}^8-22275{\lambda}^4{x}^6\\
&\quad +40392{\lambda}^4{x}^4 -37152{\lambda}^4{x}^2+4480{x}^{16}
-17248{x}^{14}+27636{x}^{12}\\
&\quad -20979{x}^{10}-71280{x}^8{ \lambda}^3+27636{x}^{12}{\lambda}^2
 -92073{x}^{10}{\lambda}^2 +17248{x}^{14}\lambda\\
&\quad -59304{x}^{12}\lambda +20979{x}^{10}{\lambda}^3+37152{\lambda}^3{x}^2
  -71280{x}^8\lambda-122265{\lambda}^2{x}^6\\
&\quad  -106128{\lambda}^3{x}^4 +92073{x}^{10}\lambda+149094{x}^8{\lambda}^2
 +122265{x}^6{\lambda}^3,
\end{aligned}
\end{gather*}
of degree $8$, $12$ and $16$, respectively.


To check if three Wronskians vanish for  $x \in (0,1)$, we only need
check if three two-variable polynomials $w_1(x,\lambda)$,
$w_2(x,\lambda)$ and $w_3(x,\lambda)$ vanish for  $x \in (0,1)$. In
order to avoid complicated symbolic computation, such as regular
chains with parameter, and real roots isolation, we introduce some
transforms.

First, let $\alpha>0$ and  introduce $x=\frac{1}{1+\alpha}$, which satisfies
 $0<x<1$. Then $w_1(x,\lambda)$ becomes 
$w_1(x,\lambda)=\frac{p_1(\alpha,\lambda)}{(1+\alpha)^8}$,
where
\begin{align*}
p_1(\alpha,\lambda)
&= 2+8\lambda+6{\lambda}^2+108{\alpha}^3+27{\alpha}^4+113\,
{\alpha}^2+54{\lambda}^2\alpha+315{\lambda}^2{\alpha}^2\\
&\quad +984{\lambda}^2{\alpha}^3+1692{\lambda}^2{\alpha}^4
 + 1674{\lambda}^2{\alpha}^5+951{\lambda}^2{\alpha}^6+288{\lambda}
^2{\alpha}^7+36{\lambda}^2{\alpha}^8\\
&\quad +48\lambda\alpha+316 \lambda{\alpha}^2+748\lambda{\alpha}^3
 +757\lambda{\alpha}^4 + 342\lambda{\alpha}^5+57\lambda{\alpha}^6+10 \alpha,
\end{align*}
which does not vanish on 
$\{(\alpha,\lambda)|\alpha>0, \lambda\geq 1\}$ since its coefficients are 
all positive.  Therefore, $W[l_1(x)]$ has not root for $x\in(0,1)$  obviously.

Second, let  $\alpha>0$, $\beta\geq 0$ and introduce
$$
x=\frac{1}{1+\alpha}, \quad \lambda=1+\beta.
$$  
Then
$w_2(x,\lambda)=-p_2(\alpha,\beta)/(1+\alpha)^{12}$, where 
$p_2(\alpha,\beta)$ is
a polynomial  with positive coefficients and has no root on
$\{(\alpha,\lambda):\alpha>0, \beta\geq 0\}$ (see Appendix A). Hence,
$W[l_1(x),l_2(x)]$ has no roots for $x\in(0,1)$.

Last,  taking  $\alpha>0$ and substituting   $x=1/(1+\alpha)$ into
$w_3(x,\lambda)$ yields
$w_3(x,\lambda)= \frac{p_3(\alpha,\lambda)}{(1+\alpha)^{16}}$.
The polynomial $p_3(\alpha,\lambda)$ has positive coefficients (see Appendix A),
 and it  has no root  on $\{(\alpha,\lambda)|\alpha>0, \lambda\geq 1\}$. Hence,
$W[l_1(x),l_2(x),l_0(x)]$ has no root for $x\in(0,1)$.
Lemma \ref{lem3.2} is proved.
\end{proof}

By Lemmas \ref{lem2.1} and \ref{lem3.2},   $\{\widetilde{I}_1(h),
\widetilde{I}_2(h),\widetilde{I}_0(h)\}$ is an ECT-system on
 $(0,(3\lambda+1)/12)$, and so $\{ {I}_1,  {I}_2, {I}_0\}$ is
as well. Therefore,  $I(h,\delta)$ has at most $2$ zeros.

\begin{remark} \rm
With the same methods and  techniques, it is not difficult to prove 
each of $I^b(h,\delta)$, $I^c(h,\delta)$ and $I^d(h,\delta)$ has at most 
$2$ zeros, we omit the proofs here for brevity.
\end{remark}

\section{Finding zeros in small intervals}

Usually, it is  difficult  to find  zeros of $A_n(h)$.  One popular method 
is to detect the expansions of $A_n(h)$ near a center, homoclinic loop 
and heteroclinic loop of system \eqref{eq11a}, see \cite{hanexp}.
 When the annulus $\{L_h\}$ of  system \eqref{eq11a} has a homoclinic loop, 
a heteroclinic loop as the outer boundary,  the expansions of $A_n(h)$ near 
these  outer boundaries  was studied in \cite{jdde2008} and the expression 
of coefficients  are also given. When  the inner boundary of $\{L_h\}$ 
is a center, the expansion of $A_n(h)$ near an elementary center 
is investigated in \cite{hanijbc2009}. and the expansion of $A_n(h)$ near 
a nilpotent  center   is investigated in \cite{yanghamca}.

By the results  of \cite{jdde2008,hanijbc2009,yanghamca}, the
expansions of  $I(h,\delta)$, $I^b(h,\delta)$, $I^c(h,\delta)$  
and $I^d(h,\delta)$   near the centers are as follows
\begin{gather*}
 I(h,\delta)   =b_0(\delta)h+b_1(\delta)h^2+b_2(\delta)h^3+h.o.t.,  \quad
  h\in (0,\varepsilon_1),  \\
 I^b(h,\delta) =\widetilde{b}_0(\delta)h+\widetilde{b}_1(\delta)h^2
 +\widetilde{b}_2(\delta)h^3+h.o.t.,  \quad h\in (0,\varepsilon_2), \\
 I^c(h,\delta) =\overline{b}_0(\delta)h^{\frac{3}{4}}
 +\overline{b}_1(\delta)h^{\frac{5}{4}}+\overline{b}_2(\delta)h^{\frac{7}{4}}+h.o.t.,  \ \ \ h\in(0,\varepsilon_3),  \\
 I^d(h,\delta) =\widehat{b}_0(\delta)h^{\frac{4}{6}}+\widehat{b}_1(\delta)
 h^{\frac{8}{6}}+\widehat{b}_2(\delta)h^{\frac{10}{6}}+h.o.t.,  \quad
 h\in(0,\varepsilon_4),
 \end{gather*}
where $0<\varepsilon_1, \varepsilon_2, \varepsilon_3, \varepsilon_4 \ll 1$ and
\begin{gather*}
 b_0(\delta)=2\pi \alpha, \quad
b_1(\delta)=\frac{3\pi}{4}(\lambda-1)\alpha+\pi \beta,  \\
b_2(\delta)=\frac{5\pi}{96}(21\lambda^2-42\lambda+37)\alpha
+\frac{5\pi}{4}(\lambda-1)\beta+\pi,
\\
\widetilde{b}_0(\delta)=2\pi \alpha, \quad
\widetilde{b}_1(\delta)= \frac{3\pi}{4}(\lambda_1-\lambda_2)\alpha+ \pi \beta,  \\
\widetilde{b}_2(\delta)= \frac{105\pi}{96}( \lambda_1^2+2
\lambda_1\lambda_2+ \lambda_2^2+\frac{80}{105})\alpha+
\frac{5\pi}{4}(\lambda_1+\lambda_2) \beta+ \pi,
\\
\overline{b}_0(\delta)=\frac{4 \pi^{\frac{3}{2}} \sqrt{2} \alpha}{3
\Gamma^2 (\frac{3}{4}) \sqrt[4]{\lambda_3}}, \quad
\overline{b}_1(\delta)=\frac{8\sqrt{2}
\Gamma^2(\frac{3}{4})(2\lambda_3\beta-\alpha)}{5
{\lambda_3}^{\frac{7}{4}}  \sqrt{\pi}}, 
\\
\overline{b}_2(\delta)=\frac{2\sqrt{2}\pi^{\frac{3}{2}} (15 \alpha-20
\lambda_3 \beta+24 \lambda_3^2)}{63\Gamma^2(\frac{3}{4}){\lambda_3}^{\frac{13}{4}}},
\quad
 \widehat{b}_0(\delta)=\frac {\sqrt{2}{6}^{\frac{1}{6}}{\pi }
 ^{\frac{3}{2}} \alpha} {\Gamma(\frac{5}{6})\Gamma(\frac{2}{3})}, \\
\widehat{b}_1(\delta ) =  \frac{2\sqrt{3} \pi \beta}{3},  \quad
\widehat{b}_2(\delta ) = \frac{3\times 6^{\frac{4}{3}}
\Gamma(\frac{5}{6})\Gamma(\frac{2}{3}) }{8 \sqrt{\pi}},
\end{gather*}
Taking $\alpha=\beta=0$, then 
$b_0=b_1=\widetilde{b}_0=\widetilde{b}_1=\overline{b}_0=\overline{b}_1
=\widehat{b}_0=\widehat{b}_1=0$ and
$b_2=\widetilde{b}_2=\overline{b}_2=\widehat{b}_2=\pi$.
It is easy to find that
$$
\det \frac{\partial (b_0, b_1)}{\partial (a_0, a_1)}
= \det \frac{\partial (\widetilde{b}_0, \widetilde{b}_1)}{\partial (a_0, a_1)}
= \det \frac{\partial (\overline{b}_0, \overline{b}_1)}{\partial (a_0, a_1)}
= \det \frac{\partial (\widehat{b}_0, \widehat{b}_1)}{\partial (a_0, a_1)}= 2.
$$
Let us take $\alpha\ll -\beta \ll 1$, then $b_0\ll -b_1 \ll b_2$, 
$\widetilde{b}_0\ll -\widetilde{b}_1 \ll \widetilde{b}_2$, 
$\overline{b}_0\ll -\overline{b}_1 \ll \overline{b}_2$  and
 $\widehat{b}_0\ll -\widehat{b}_1 \ll \widehat{b}_2$,
 which imply there exist $2$ zeros for each Abelian integral $I(h,\delta)$, 
$I^b(h,\delta)$, $I^c(h,\delta)$ and $I^d(h,\delta)$.


\subsection{Conclusion}
The number  of zeros of  Abelian integral for system \eqref{LL} has been 
studied for all 12 cases except for case $4$.  Up to now, the sharp bounds  
of the numbers of zeros for the corresponding Abelian integrals defined on 
all period annuluses for one case of system \eqref{LL} are obtained for 
case $5$, $6$, $7$, $12$ and case $9$, the sharp bound for other cases  
are our further research.

\subsection*{Appendix A}
This section shows two polynomials  with positive coefficients that have
 no root on $\{(\alpha,\lambda):\alpha>0, \lambda\geq 1\}$.
\begin{align*} 
&p_2(\alpha,\beta) \\
&=1728{\alpha}^{12}+380160{\alpha}^9+114048{\alpha}^{10}+20736\,
{\alpha}^{11}+238656{\alpha}^3+672144{\alpha}^4\\
&\quad +1220736{\alpha}^5 +1522752{\alpha}^6 +1347456{\alpha}^7+852720{
\alpha}^8+49632{\alpha}^2+5952\alpha\\
&\quad +64\beta+36{\beta}^3+20736{\beta}^3{\alpha}^{11}
 +1728{\beta}^3{\alpha}^{12}+3622848\beta{\alpha}^7
 +2395560\beta{\alpha}^8\\
&\quad +3226296{\beta}^2{\alpha}^7+2235831{\beta}^2{\alpha}^8
 +950904{\beta}^3{\alpha}^7  +692991{\beta}^3{\alpha}^8
 +1103760\beta{\alpha}^9 \\
&\quad +338472\beta{\alpha}^{10}
 +1067040{\beta}^2{\alpha}^ {9}+334800{\beta}^2{\alpha}^{10}
 +343440{\beta}^3{\alpha}^9 +110376{\beta}^3{\alpha}^{10}\\
&\quad +62208\beta{\alpha}^{11}+5184 \beta{\alpha}^{12}
  +62208{\beta}^2{\alpha}^{11}+5184{\beta}^2{\alpha}^{12}
  +11360\beta\alpha +101296\beta{\alpha}^2 \\
&\quad +496896\beta{\alpha}^3+1491744\beta{\alpha}^4+96{\beta}^2
  +7356{\beta}^2\alpha+69162{\beta}^2{\alpha}^2+349356{\beta}^2{\alpha}^3\\
&\quad  +1102515{\beta}^2{\alpha}^4  +2910624\beta{\alpha}^5 
  +3875376\beta{\alpha}^6
  +2293896{\beta}^2{\alpha}^5 +3258564{\beta}^2{\alpha}^6\\
&\quad +1638{\beta}^3\alpha  +15831{ \beta}^3{\alpha}^2 
 +82476{\beta}^3{\alpha}^3
 +271845{\beta}^3{\alpha}^4+598662{\beta}^3{\alpha}^5\\
&\quad  +905049{\beta}^3{\alpha}^6.
\end{align*}
\begin{align*}
&p_3(\alpha,\lambda)\\
&= 126+1012\lambda+1026{\lambda}^4+2988{\lambda}^3+2784{\lambda}^2
  +40236{\alpha}^3+149541{\alpha}^4+223398{\alpha}^5\\
&\quad +153657{\alpha}^6+49896{\alpha}^7+6237{\alpha}^8
  +8519{\alpha}^2+12912{\lambda}^2\alpha+123300{\lambda}^2{\alpha}^2\\
&\quad  +832788{\lambda}^2{\alpha}^3+3401511{\lambda}^2{\alpha}^4
  +8976510{\lambda}^2{\alpha}^5 +15729117{\lambda}^2{\alpha}^6\\
&\quad +18511416{\lambda}^2{\alpha}^7
 +14641209{\lambda}^2{\alpha}^8
 +2228\lambda\alpha+49054\lambda{\alpha}^2+285564\lambda{\alpha}^3\\
&\quad +1009941\lambda{\alpha}^4 +2174058\lambda{\alpha}^5
 +2773983\lambda{\alpha}^6+70\alpha 
 +484704{\lambda}^2{\alpha}^{11}\\
&\quad+40392{\lambda}^2{\alpha}^{12} +222750\lambda{\alpha}^9
  +22275\lambda{\alpha}^{10} +57553848{\lambda}^3{\alpha}^7\\
&\quad +64464741{\lambda}^3{\alpha}^8 +52252794{\lambda}^3{\alpha}^9
  +30306969{\lambda}^3{\alpha}^{10}+12249792{\lambda}^3{\alpha}^{11}\\
&\quad +3274704{\lambda}^3{\alpha}^{12} +520128{\lambda}^3{\alpha}^{13}
  +37152{\lambda}^3{\alpha}^{ 14}+2102760\lambda{\alpha}^7
  +931095\lambda{\alpha}^8\\
&\quad+ 7663590{\lambda}^2{\alpha}^9
  +2543607{\lambda}^2{\alpha}^{10} +12906{\lambda}^4\alpha
  +116181{\lambda}^4{\alpha}^2+780624{\lambda}^4{\alpha}^3\\
&\quad  +3723408{\lambda}^4{\alpha}^4
  +12731364 {\lambda}^4{\alpha}^5
 +31954230{\lambda}^4{\alpha}^6+60008256{\lambda}^4{\alpha}^7\\
&\quad  +85345326{\lambda}^4{\alpha}^8+92431746{\lambda}^4{\alpha}^9
  +76157037{\lambda}^4{\alpha}^{10}+47344608{\lambda}^4{\alpha}^{11}\\
&\quad  +21819240{\lambda}^4{\alpha}^{12}+7221312{\lambda}^4{\alpha}^{13}
 +1621728{\lambda}^{4}{\alpha}^{14}+221184{\lambda}^4{\alpha}^{15}\\
&\quad  +13824{\lambda}^4 {\alpha}^{16}+24876{\lambda}^3\alpha 
 +197154{\lambda}^3{\alpha}^2+1274868{\lambda}^3{\alpha}^3
 +5656527{\lambda}^3{\alpha}^4 \\
&\quad +17269902{\lambda}^3{\alpha}^5 +37205973{\lambda}^3{\alpha}^6.
\end{align*}


\subsection*{Acknowledgements}
 This work was supported by the National Natural Science
Foundations of China (No. 11101118, 11261013),  Natural Science
Foundation of Hebei Province (A2012205074) and  Research project
of Guangxi Universities (2013YB216).


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