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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2011 (2011), No. 134, pp. 1--15.\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-2011/134\hfil Longitudinal librations]
{Longitudinal librations of a satellite}

\author[M. Furi, M. Martelli, A. Vignoli\hfil EJDE-2011/134\hfilneg]
{Massimo Furi, Mario Martelli, Alfonso Vignoli}  % in alphabetical order

\address{Massimo Furi \newline
Dipartimento di Matematica Applicata\\
Universit\`a degli Studi di Firenze\\
Via S.\ Marta 3 \\
50139 Firenze, Italy}
\email{massimo.furi@unifi.it}

\address{Mario Martelli \newline
Department of Mathematics\\
Claremont Graduate University\\
710 N College Ave\\
Claremont, CA 91711, USA}
\email{mario.martelli@cgu.edu}

\address{Alfonso Vignoli \newline
Dipartimento di Matematica\\
Universit\`a di Roma Tor Vergata\\
Via della Ricerca Scientifica 1\\
00133, Roma, Italy}
\email{vignoli@mat.uniroma2.it}

\thanks{Submitted April 18, 2011. Published October 14, 2011.}
\subjclass[2000]{34C28, 70F15}
\keywords{Ordinary differential equations;
   Chaotic behavior of a satellite; librations}

\begin{abstract}
 Furi, Martelli and Landsberg gave a theoretical
 explanation of the chaotic longitudinal librations of Hyperion,
 a satellite of Saturn. The analysis was made under the simplifying
 assumption that the spin axis remains perpendicular to the orbit
 plane. Here, under the same assumption, we investigate the behavior
 of the longitudinal librations of any  satellite. Also we show that
 they are possibly chaotic depending on two parameters: a constant
 $k$ related to the principal moments of  inertia of the satellite,
 and the eccentricity $e$ of its orbit.
 We prove that the plane $k$-$e$ contains an open region $\Omega$
 with the property that the longitudinal librations of any satellite
 are possibly chaotic if the point $(k,e)$ belongs to this region.
 Since Hyperion's point is inside $\Omega$, the results of this paper
 are more general than those obtained previously.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{definition}[theorem]{Definition}


\section{Introduction} \label{introduction}

The differential equation that governs the motion of a satellite, when its
spin axis remains perpendicular to the orbit plane, takes the form
\begin{equation}
\label{fullpendulum}
\ddot x=\Big(\frac{a}{r(t)}\Big)^3\Big(\frac{16c^2 e\sin\theta(t)}{a^4(1-e^2)}-
3\frac{B-A}{C}
\sin x \Big),
\end{equation}
which, as shown in \cite{FLM2}, is equivalent to the one in
\cite{GP,WPM}. The variable $x$ is equal to twice the angle
$\varphi$ between the longest axis of the satellite and the
planet-satellite center line, and the symbol $\ddot x$ denotes the
second derivative of $x$ with respect to the time $t$. The
constants $a$ and $e$ are, respectively, the \emph{semimajor axis}
and the \emph{eccentricity} of the elliptical orbit described by
the center of mass of the satellite. The function $r(t)$ denotes
the distance between the satellite and the planet. The constant
$c$ is the \emph{areolar velocity} of the satellite, that is the
instantaneous area swept by the segment joining the centers of
mass of the planet and of the satellite. The function $\theta(t)$
indicates the \emph{polar angle}. Measured counterclockwise and
expressed in radians, it provides the angle between the major
diameter of the elliptical orbit oriented towards the
\emph{periapsis} (i.e., the point when the satellite is closest to
the planet) and the planet-satellite center line (see figure
\ref{fig:satellite}). The constants $A$, $B$, and $C$, with
$0< A \le B \le C$, are the principal \emph{moments of inertia} of the
satellite, with $C$ being the moment about the spin axis.

Equation \eqref{fullpendulum} was studied in \cite{FLM2} in the
case when all the constants involved are referred to Hyperion, one
of the many satellites of the planet Saturn. Theoretical results
and numerical estimates were used to explain  why the longitudinal
librations of this satellite are chaotic.

In this paper we investigate the behavior of the longitudinal
librations of any satellite with motion governed by
\eqref{fullpendulum}, and we show that these librations are
possibly chaotic depending on two parameters: the ratio
$k=(B-A)/C$ and the eccentricity $e$ of the orbit of the
satellite. Both constants are assumed to be positive and less than
$1$. By means of theoretical considerations we show that the plane
$k$-$e$ contains an open region $\Omega$ (see figure
\ref{fig:triangle}) with the property that the longitudinal
librations of any satellite (with spin axis perpendicular to the
orbit plane) are possibly chaotic provided that the corresponding
pair of parameters $(k,e)$ drops in this region. After having
determined, numerically, the boundary of $\Omega$, we realized
that the pair $(k,e)$ associated with Hyperion belongs to
$\Omega$, as expected. Thus, the results in this paper extend
those in \cite{FLM2}.


Some historical remarks are in order.  Wisdom,  Peale, and
Mignard \cite{WPM} investigated the irregular oscillations of
Hyperion's longest axis with respect to the planet-satellite
center line. The conclusions of the three scientists were derived
from:
\begin{itemize}
\item[(1)]
Hyperion's images transmitted by Voyager 2 \cite {SB};
\item[(2)]
a mostly numerical analysis of a differential equation
modeling a planetary motion and proposed by P.\ Goldreich and S.\ Peale
\cite{GP}.
\end{itemize}
Martelli and VignoliThe modified the equation proposed by Danby
\cite{D} to study the longitudinal librations of the moon.

In an interesting paper Wisdom, Peale, and Mignard \cite{WPM}
provided numerical evidence that Hyperion's longitudinal
librations are chaotic. Denoting by $\varphi$ the angle between
Hyperion's longest axis and the planet-satellite center line, they
plotted the pairs $(\varphi,\dot\varphi)$ when Hyperion crosses
the \emph{periapsis}, obtaining a large cloud of points which
dominates the $1/2$ and $2$ spin-orbit states.

We point out that, besides  Wisdom,  Peale, and Mignard
\cite{WPM}, other authors (see, for example,
\cite{BB,BZ1,BZ2,FMOS,FLM,H,HML,M,NT1,NT2,W,WI,ZM}) have analyzed
problems similar to the one investigated here.

\section{The model}
\label{equation}

According to \cite{GP,WPM} the motion of a tri-axial satellite
describing an elliptical orbit around a planet, with spin axis
perpendicular to the orbit plane, can be modelled by the
second-order nonlinear differential equation
\begin{equation}
\label{wpm}
 \ddot \varphi +\ddot \theta=-
\frac{3(B-A)}{2C}\big(\frac{a}{r(t)}\big)^3
\sin(2\varphi),
\end{equation}
where the quantities involved are as in Section \ref{introduction}.
Thus, $\theta+\varphi$ is the angle between the satellite's longest
axis and the longest diameter of the elliptical orbit
(see figure~\ref{fig:satellite}).

\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.7\textwidth]{fig1} 
\end{center}
 \caption{The planet-satellite system}
 \label{fig:satellite}
\end{figure}

Since the planet lies at one focus of the orbit, its distance $r$
from the satellite is expressed by the following function of the
polar angle~$\theta$:
\begin{equation}\label{radius}
r(\theta)= \frac{p}{1+e\cos\theta},
\end{equation}
where $p=a(1-e^2)$ is the \emph{parameter} of the elliptical orbit.
The angle $\theta$ and the time $t$ are related by Kepler's second
law of planetary motion and by the initial condition $\theta(0)=0$,
obtained by selecting $t=0$ when the planet is at the periapsis.
Hence, $\theta (t)$ is the solution of the initial value problem
\begin{equation} \label{areavelocity}
\begin{gathered}
r^2(\theta) \dot \theta  =  2c,\\
\theta(0)  =  0.
\end{gathered}
\end{equation}
Differentiating \eqref{radius} and \eqref{areavelocity}, and
recalling the equality $p=a(1-e^2)$, we obtain
\begin{equation*}
\label{thetaacceleration}
 \ddot\theta(t)=-\frac{8c^2}{r^3(t)}\frac{e\sin\theta(t)}{a(1-e^2)}.
\end{equation*}
Hence, \eqref{wpm} takes the form
\begin{equation*} \label{fullequation}
\ddot\varphi=\big(\frac{a}{r(t)}\big)^3
\big(\frac{8c^2 e\sin\theta(t)}{a^4(1-e^2)}-
\frac{3(B-A)}{2C}\sin 2\varphi\big).
\end{equation*}
Setting $x=2\varphi$, we obtain the differential equation
\begin{equation*}
\label{fullpendulumequation}
\ddot x=\big(\frac{a}{r(t)}\big)^3
\big(\frac{16c^2 e\sin\theta(t)}{a^4(1-e^2)}-
\frac{3(B-A)}{C}\sin x\big).
\end{equation*}

We set $a=1$ and normalize the time so that the period of
revolution of the satellite is $2\pi$. Consequently, $t$ and
$\theta(t)$ coincide when $t=n\pi$, $n\in \mathbb{Z}$, and the
difference $\eta(t) := t - \theta(t)$ is a $2\pi$-periodic odd
function. Moreover, the area swept during a full turn by the
segment joining the satellite with the planet is $\pi
ab=\pi\sqrt{1-e^2}$. It follows that
\[
c=\frac12 \sqrt{1-e^2}.
\]
Moreover, since $p=a(1-e^2)$ and $r^2(\theta) \dot \theta = 2c$,
 with the above setting we obtain the differential equation
\begin{equation}
\label{nonautonomous}
 \ddot x=\big(\frac{1+e\cos\theta(t)}{1-e^2}\big)^3
\big(4e\sin\theta(t)-3k\sin x\big),
\end{equation}
and the initial value problem \eqref{areavelocity}, which defines
the function $\theta(t)$, becomes
\begin{equation}\label{teta1}
\begin{gathered}
\dot \theta  =  \frac{(1+e\cos\theta)^{2}}{(1-e^2)^{3/2}},\\
\theta(0)  =  0.
\end{gathered}
\end{equation}

Notice that  \eqref{nonautonomous} can be regarded as governing
the motion of a \emph{point mass} constrained on a vertical circle
and acted on by two periodic forces of the same period:
an oscillating gravitational force, and a forcing term with zero
mean acting, alternately, clockwise and counterclockwise.
To emphasize this interpretation, we observe that
\eqref{nonautonomous} can be rewritten in the form
\begin{equation*}
\label{equivalent0}
\ddot x + b(t)\sin x = a(t)\sin(t+\eta(t)),
\end{equation*}
where the functions $a(t)$ and $b(t)$ are strictly positive, even,
and $2\pi$-periodic. Moreover $\eta(t)$, which can be considered
as the \emph{phase} of a forcing term with oscillating \emph{amplitude}
$a(t)$, is odd and $2\pi$-periodic.

\section{Preliminary Results} \label{preliminary}

In this section we begin our investigation of equation
\eqref{nonautonomous}. We already mentioned that the dependent
variable $x$ can be regarded as the location (measured in radians)
of a point mass on $S^1$. The lowest position of this point
corresponds to $x = 2k\pi$, $k \in \mathbb{Z}$, and will be called
\emph{South Pole}. The \emph{North Pole} is identified with
$x=(2k+1)\pi$, $k \in \mathbb{Z}$.

To make the statements and proofs of Sections \ref{crossing}  and
\ref{chaos} more concise and transparent we now present some
terminology, lemmata, propositions, and appropriate remarks.

\begin{remark}\label{equilibrium}\rm
Regard \eqref{nonautonomous} as the motion equation of a point $x$ of unitary mass, constrained on $S^1$, and acted on by the force
\[
f(t,x) = \Big(\frac{1+e\cos\theta(t)}{1-e^2}\Big)^3
\big(4e\sin\theta(t)- 3k\sin x\big).
\]
Assume that $4e<3k$. Then $f(t,x)$ pushes towards the South Pole
(for any $t$) whenever $x$ lies in each one of the two opposite arcs
\begin{gather*}
A_W =  (-\pi+\arcsin (4e/3k),-\arcsin (4e/3k)),\\
A_E =  (\arcsin (4e/3k), \pi-\arcsin (4e/3k)),
\end{gather*}
centered at $-\pi/2$ and $\pi/2$ and defined by $-4e-3k \sin x>0$
and $4e-3k \sin x<0$, respectively.
\end{remark}

For the rest of this article,  we assume that the two positive
parameters $k$ and $e$ satisfy the inequality $4e<3k$ ensuring
the existence of the opposite arcs defined in Remark \ref{equilibrium}.
We call $A_W$ and $A_E$ the \emph{western arc} and \emph{eastern arc},
with $A_W=(\alpha_W,\beta_W)$ and $A_E=(\beta_E,\alpha_E)$.
Moreover, the arc containing the North Pole with extremes $\alpha_W$
and $\alpha_E$ will be called the \emph{northern arc} and denoted by $A_{N}$, while the arc containing the South Pole with extremes $\beta_W$ and $\beta_E$ will be called the \emph{southern arc}
and denoted by $A_{S}$. Accordingly, the points
\emph{West, East, North Pole} and
\emph{South Pole} are the centers of the four arcs $A_W,
A_E, A_{N}$ and $A_{S}$.

Given $x_0, x_1 \in \mathbb{R}$, $x_0 \neq x_1$, we denote by
$\overline{x_0x_1}$ the closed interval
\[
[\min\{x_0,x_1\},\, \max\{x_0,x_1\}].
\]
We need the following physically meaningful result
(see e.g.\ \cite{FLM2}).
For completeness' sake we include its elementary proof.

\begin{lemma}\label{arriva e non arriva}
Let $f: \mathbb{R}\sp 2 \to \mathbb{R}$,
$f_{-}: \mathbb{R} \to \mathbb{R}$ and $f_{+}: \mathbb{R} \to
\mathbb{R}$ be continuous and such that
\[
f_{-}(x) \leq f(t,x) \leq f_{+}(x)
\]
for all $(t,x) \in \mathbb{R}\sp 2$. Let $x(t)$, $t \geq t_0$,
be any solution of the initial value problem
\begin{gather*} %\label{fund1}
 \ddot x = f(t,x),  \\
x(t_0) = x_0, \\
\dot x(t_0) = v_0.
\end{gather*}
Define the real functions $g_{-}$ and $g_{+}$ by
\begin{equation*}%\label{fund11}
g_{-}(x) = \frac {v_0\sp 2}2 + \int_{x_0}\sp x f_{-}(s)\,ds, \quad
g_{+}(x) = \frac {v_0\sp 2}2 + \int_{x_0}\sp x f_{+}(s)\,ds.
\end{equation*}
Given $x_1 \neq x_0$, denote by $u_{-}$ and $u_{+}$ the numbers
\[
u_{-} = \min \big\{g_{-}(x): x \in \overline{x_0x_1} \big\}, \quad
u_{+} = \min \big\{g_{+}(x): x \in \overline{x_0x_1} \big\}.
\]
The following four assertions hold.
\begin{itemize}
\item
If $x_1 > x_0$, $v_0 > 0$, and $u_{-} > 0$,
then $\dot x(t) \geq \sqrt{2u_{-}}$ for all $t \geq t_0$
such that $x_0 \leq x(t) \leq x_1$. In particular, $x(t)$ reaches
$x_1$ without stopping.

\item
If $x_1 < x_0$, $v_0 < 0$, and $u_{+} > 0$,
then $\dot x(t) \leq -\sqrt{2u_{+}}$ for all $t \geq t_0$
such that $x_0 \leq x(t) \leq x_1$. In particular, $x(t)$
reaches $x_1$ without stopping.

\item If $x_1 > x_0$, $v_0 > 0$, and $u_{+} < 0$, then $x(t)$
does not reach $x_1$ before it stops.

\item If $x_1 < x_0$, $v_0 < 0$, and $u_{-} < 0$, then $x(t)$
does not reach $x_1$ before it stops.
\end{itemize}
\end{lemma}

\begin{proof}
Observe that
\begin{equation} \label{energia cinetica}
\frac{\dot x\sp 2(t)}2 = \frac{v_0\sp 2}2 + \int_{t_0}\sp {t}
f(\tau,x(\tau))\dot x(\tau)\, d\tau,
\quad \forall\, t \geq t_0,
\end{equation}
as can be easily verified by differentiating both members
of \eqref{energia cinetica} and
noticing that they coincide when $t = t_0$.

We have to examine four cases, and in each one of them the initial
velocity $v_0 = \dot x(t_0)$ is different from $0$.
Therefore, it makes sense
to consider the maximal interval containing $t_0$ and contained
in the relatively open subset
$\{t \geq t_0: \dot x(t) \neq 0\}$ of $[t_0,+\infty)$.
This nonempty interval will be denoted by $[t_0, t_*)$,
with $t_0 < t_* \leq +\infty$.

Consider the first case, namely $x_1 > x_0$, $v_0 > 0$,
and $u_{-} > 0$. {}From equation \eqref{energia cinetica}
and from the inequality $f_{-}(x) \leq f(t,x)$, we obtain
\begin{equation*}
\frac{\dot x\sp 2(t)}2
\geq \frac{v_0\sp 2}2 + \int_{t_0}\sp {t} f_{-}(x(\tau))
 \dot x(\tau)\, d\tau
= \frac{v_0\sp 2}2 + \int_{x_0}\sp {x(t)} f_{-}(s)\,ds
= g_{-}(x(t))
\end{equation*}
for every $t \in [t_0, t_*)$.
Since $\dot x(t_0) > 0$, we obtain $\dot x(t) \geq \sqrt{2u_{-}}$
for all $t \in [t_0, t_*)$ such that $x(t) \leq x_1$.
Hence, the first of the four assertions follows.

The second case can be analyzed in a similar manner by taking into
account that $\dot x(\tau)$ is negative for all $\tau \in [t_0, t]
\subset [t_0, t_*)$.

Consider the third case: $x_1 > x_0$, $v_0 > 0$,
and $u_{+} < 0$. From equation \eqref{energia cinetica}
and the inequality $f(t,x) \leq f_{+}(x)$, we obtain
\begin{equation*}
\frac{\dot x\sp 2(t)}2 \leq \frac{v_0\sp 2}2 + \int_{x_0}\sp {x(t)} f_{+}(s)\,ds
 = g_{+}(x(t)), \quad \forall\, t \in [t_0,t_*).
\end{equation*}
Since $u_{+} < 0$, there exists a point $\bar x$ in the open
interval $(x_0,x_1)$ such that $g_{+}(\bar x) < 0$.
Thus, because of the above inequality, $x(t)$ cannot cross the
point $\bar x$ before it stops.

The last case can be studied in a similar manner using the fact
that now $\dot x(\tau) < 0$, $\forall\, \tau \in [t_0, t]$.
\end{proof}

\begin{remark}\label{stime}\rm
The equation \eqref{nonautonomous} is of the type
$\ddot x = f(t,x)$, with
\begin{equation*}
\label{inequality1}
f_{-}(x) \leq f(t,x) \leq f_{+}(x),
\end{equation*}
for all $(t,x) \in \mathbb{R}\sp 2$, where
\begin{equation}\label{comparison}
\begin{gathered}
f_{-}(x)  =  \frac{-4e-3k\sin x}{(1+e\operatorname{sign}(-4e-3k\sin x))^3},
\\
f_{+}(x)  = \frac{4e-3k\sin x}{(1-e\operatorname{sign}(4e-3k \sin x))^3}.
\end{gathered}
\end{equation}
\end{remark}

For the rest of this article, we shall assume that $f_{-}(x)$
and $f_{+}(x)$ are as in \eqref{comparison}.
Observe that in this case,
\begin{equation*}\label{fundequality}
f_{-}(-x)=-f_{+}(x), \quad \forall x \in \mathbb{R}.
\end{equation*}
Moreover, under the assumption $4e < 3k$, which implies the
existence of the arcs $A_W = (\alpha_{_W}, \beta_{_W})$ and
$A_E = (\beta_{_E}, \alpha_{_E})$, we have
$\alpha_{_W} = -\alpha_{_E}$ and $\beta_{_W} = -\beta_{_E}$.
Consequently,
\begin{equation*}
\int_{\alpha_{_W}}^{\beta_{_W}}f_{-}(x)dx =
\int_{\alpha_{_E}}^{\beta_{_E}}f_{+}(x)dx\,, \quad
\int_{\beta_{_W}}^{0}f_{-}(x)dx =
\int_{\beta_{_E}}^{0}f_{+}(x)dx\,.
\end{equation*}

The following function of $k$ and $e$, defined on the open triangle
\begin{equation*}%\label{triangle}
T = \big\{(k,e) \in \mathbb{R}^2: 0 < 4e < 3k < 3\big\},
\end{equation*}
plays an important role in the sequel:
\begin{equation}\label{defineH}
h(k,e) := \int_{\alpha_{_W}}^{0}f_{-}(x)dx =
\int_{\alpha_{_E}}^{0}f_{+}(x)dx.
\end{equation}
Notice that
$h(k,e) = h_{+}(k,e) + h_{-}(k,e)$, with
\begin{gather*}
h_{+}(k,e) := \int_{\alpha_{_W}}^{\beta{_W}}f_{-}(x)dx =
\int_{\alpha_{_E}}^{\beta{_E}}f_{+}(x)dx,\\
h_{-}(k,e) := \int_{\beta_{_W}}^{0}f_{-}(x)dx =
\int_{\beta_{_E}}^{0}f_{+}(x)dx\,.
\end{gather*}
Moreover, $h_{+}(k,e) > 0$ and $h_{-}(k,e) < 0$
for all $(k,e)$ in the triangle $T$, since $f_{-}(x)$
is positive for $x \in (\alpha_{_W}, \beta_{_W})$ and
negative for $x \in (\beta_{_W}, 0)$. Actually, elementary
computations show that
\[
h_{+}(k,e) = \frac{-4e(\pi-2\arcsin(4e/3k))
+6k\sqrt{1-(4e/3k)^2}}{(1+e)^3}
\]
and
\[
h_{-}(k,e) = \frac{-4e\arcsin(4e/3k)
+3k(1-\sqrt{1-(4e/3k)^2})}{(1-e)^3}.
\]


\begin{proposition}\label{falling-down}
Let $h(k,e)>0$, and assume that when $t=t_0$ the point mass crosses
one of the positions $\alpha_W$ or $\alpha_E$ and it travels towards
the South Pole.
Then it reaches it at some $t_1>t_0$ and $\dot x(t)\ne 0$ for
every $t\in [t_0,t_1]$.
\end{proposition}

\begin{proof}
We consider first the case when the point mass crosses $\alpha_W$
with positive velocity. With the notation of Lemma
\ref{arriva e non arriva}, let $x_0 = \alpha_W$ be the highest
point of the west
arc and let $x_1 = 0$ be the South Pole. Consider the function
$g_{-} : [x_0,x_1] \to \mathbb{R}$ defined by
\[
g_{-}(x) = \frac{v_0^2}{2} + \int_{x_0}\sp x f_{-}(s)\,ds,
\]
where $v_0 := \dot x(t_0) > 0$ and $f_{-}(s)$ is as in
Remark \ref{stime}. As pointed out before, $f_{-}(s)$
is positive in the west arc $(x_0, \beta_W)$ and negative
between $\beta_W$ and $x_1$ (the South Pole).
Therefore, the function $g_{-}(x)$ is increasing up to $\beta_W$
and then decreasing. Moreover, at the extremes of the
interval $[x_0,x_1]$ we have
\[
g_{-}(x_0) = \frac{v_0^2}{2} > 0 \quad \text{and} \quad
g_{-}(x_1) = \frac{v_0^2}{2} + \int_{x_0}\sp{x_1} f_{-}(s)\,ds
 = \frac{v_0^2}{2} + h(k,e) > 0.
\]
Thus, the number
\[
u_{-} = \min \big\{g_{-}(x): x \in \overline{x_0x_1} \big\}
\]
is positive.
Consequently, the first assertion of Lemma \ref{arriva e non arriva}
shows that $x(t)$ reaches the South Pole with positive velocity
without stopping.

The situation in which the point mass crosses $\alpha_E$
clockwise can be analyzed as in the previous case, using the
equality
\[
h(k,e) = \int_{\alpha_{_E}}^{0}f_{+}(x)dx
\]
and the second assertion of Lemma \ref{arriva e non arriva}.
\end{proof}

\section{over or not} \label{crossing}

The two results of this section deal with \emph{going over the top}
(see Theorem \ref{overnotover}) and \emph{not going
over the top} (see Theorem \ref{overnotover2}). We establish results
similar to the ones proved in \cite{FLM} for the satellite Hyperion.
 The procedure, however, is quite different since we are dealing with
any satellite having its motion around a planet described by
\eqref{nonautonomous}. Recall that we are assuming $4e < 3k$,
which is verified for Hyperion.

As a first change, in equation \eqref{nonautonomous} we substitute
the independent variable $t$ with $\theta$, which can be regarded
as well as an independent variable since the correspondence
$t \mapsto \theta(t)$ is invertible. The reason of this substitution
is that \eqref{nonautonomous} contains the function $\theta(t)$
which depends on the satellite under consideration, and our
numerical approach requires that we decide when $f(t) = \sin\theta(t)$
is equal to $+1$ or to $-1$. By using $\theta$ as the independent
variable we avoid the complication of redetermining the values
of $t$ such that $\theta(t) = \pm \pi/2$ each time we change the
parameters $k$ and $e$.

For the rest of this article, we denote by $x'$ and $x''$ the
first and second derivatives of $x$ with respect to $\theta$.
By abuse of terminology, those will be called \emph{velocity}
and \emph{acceleration} of the point mass as if $\theta$ were the time.
Recall that $\dot x$ and $\ddot x$ denote the first and second
derivatives of $x$ with respect to $t$.

Observe that
\begin{equation*}
\frac{d}{d\theta}=\frac{1}{g(\theta)}\frac{d}{dt},
\end{equation*}
where, according to \eqref{teta1},
\begin{equation*}%\label{teta-t}
g(\theta)=\frac{d\theta}{dt}=\frac{(1+e\cos \theta)^2}{(1-e^2)^{3/2}}.
\end{equation*}
Consequently,
\begin{equation*} %\label{derivativewithteta}
x''=\frac{\ddot x}{g^{2}(\theta)}-\frac{g'(\theta)}{g(\theta)}x'.
\end{equation*}
Since (see \eqref{nonautonomous})
\begin{equation*} %\label{eguaglianzabase}
\ddot x= \Big( \frac{1+e\cos\theta(t)}{1-e^2}\Big)^3
\big(4e\sin\theta(t)-3k\sin x\big),
\end{equation*}
we obtain
\begin{equation}
\label{funzionediteta}
x''=\frac{1}{1+e\cos\theta}(2e(x'+2)\sin\theta-3k\sin x).
\end{equation}

\begin{remark}\label{equivariance2}\rm
Notice that whenever $x(\theta)$ is
a solution of \eqref{funzionediteta}, every function of the form
\begin{equation}\label{additional4}
z_{k}(\theta):=x(\theta+2k\pi)
\end{equation}
is also a solution for every $k\in \mathbb{Z}$. Moreover,
additional solutions are provided by
the function
$w(\theta):=-x(-\theta)$ (\emph{equivariance}) and by its
translates according to \eqref{additional4}.
\end{remark}

The following definition will be used repeatedly.

\begin{definition}\label{immediately}\rm
Given two events regarding the point mass $x(\theta)$, the
expression \emph{the second event happens immediately after
the first one} means that the last event occurs for the first
time after the previous one and, between the two events,
the derivative $x'(\theta)$ never vanishes.
\end{definition}

Notice that, with the new independent variable $\theta$,
 Proposition \ref{falling-down} can be reformulated as follows.

\begin{proposition}\label{falling-down with theta}
Let $h(k,e)>0$, and assume that when $\theta=\theta_0$ the point
mass crosses one of the positions $\alpha_W$ or $\alpha_E$ and
it travels downward.
Then it reaches the South Pole immediately after $\theta_0$.
\end{proposition}


We now introduce symbols to be used in this section.
In the definition of any of them, one of the following two events
is always considered: the point mass $x(\theta)$ crosses $S$
(the South Pole) when the variable $\theta$ is either
$\theta_{S} = \pi/2$ or $\theta_{S} = -\pi/2$.
The crossing can be counterclockwise or clockwise and must
happen or not happen immediately before or after one of the
following events:
\begin{itemize}
\item[(1)]
leaving the northern arc $A_{N}$,
\item[(2)]
entering $A_{N}$,
\item[(3)]
crossing $A_{N}$ \emph{strictly} (meaning that $x'(\theta)$ never
vanishes during the crossing).
\end{itemize}
The plus sign is used when $\theta_{S}=\pi/2$, while the minus
sign is used when $\theta_{S}=-\pi/2$. The letter $P$ stands for
\emph{past} and it denotes that the event under consideration
takes place before the crossing of $S$. The letter $F$ stands for
\emph{future} and it denotes that the event happens after crossing
$S$. The letter $v$ always stands for the absolute value of
$x'(\theta_{S})$. An arrow located over the letter $v$ indicates
the supremum of $v$ and an arrow located below $v$ indicates the
infimum of $v$. The orientation of the arrow is important, since
it indicates the direction of the point mass when it crosses the
South Pole. For example, $\leftarrow$ means that (for $\theta =
\theta_s = \pm \pi/2$) we have $x(\theta_s)=0$ and
$x'(\theta_s) < 0$; that is, the motion is clockwise.

Clearly, there are $2^4=16$ possibilities. However, only $8$ of
them are of interest to us. Four of them regard the
counterclockwise motion; the other ones the clockwise motion. We
give their description below.

Recall first that in all cases, even if not specified, the
crossing  of $S$ happens when the variable $\theta$ is either
$\theta_{S} = \pi/2$ or $\theta_{S} = -\pi/2$ according to the
plus or minus sign, and the crossing is counterclockwise or
clockwise according to the direction of the arrow.

The symbols
\[
\underrightarrow{v}_{p}^{+} \quad \text{and} \quad
\underleftarrow{v}_{p}^{-}
\]
denote the \emph{infimum} of $v=|x'(\theta_{S})|$ such that the
point mass crosses $S$ immediately after leaving the northern arc
$A_{N}$.

Similarly,
\[
\underleftarrow{v}_{{F}}^{+} \quad \text{and} \quad
\underrightarrow{v}_{{F}}^{-}
\]
indicate the \emph{infimum} of $|x'(\theta_{S})|$ sufficient for
the point mass to arrive at the northern arc immediately after
crossing $S$. Consequently, with a smaller $v$ the point mass
cannot reach $A_{N}$ immediately after leaving $S$.

Likewise
\[
\overleftarrow{v}_{p}^{+} \quad \text{and} \quad
\overrightarrow{v}_{p}^{-}
\]
denote the \emph{supremum} of $|x'(\theta_{S})|$ such that the
point mass crosses $S$ immediately after coming from $A_{N}$,
which, however, should not have been crossed \emph{strictly}.

Finally,
\[
\overrightarrow{v}_{{F}}^{+} \quad \text{and} \quad
\overleftarrow{v}_{{F}}^{-}
\]
stand for the \emph{supremum} of $|x'(\theta_{S})|$ such that
the point mass does not cross \emph{strictly} the northern arc
$A_{N}$ immediately after leaving $S$.

We now set
\begin{equation}\label{first}
\begin{gathered}
\delta_{\to}^{+}=\underrightarrow{v}_{p}^{+}
-\overrightarrow{v}_{{F}} ^{+}, \quad
\delta_{\to}^{-}=\underrightarrow{v}_{{F}}^{-}
-\overrightarrow{v}_{p}^{-},
\\
\delta_{\leftarrow}^{+}=\underleftarrow{v}_{{F}}^{+}-\overleftarrow{v}_{p}
^{+},\quad
 \delta_{\leftarrow }^{-}=\underleftarrow{v}_{p}^{-}
-\overleftarrow{v}_{{F}}^{-}.
\end{gathered}
\end{equation}
The equivariance mentioned in Remark \ref{equivariance2} implies
that
\begin{equation}\label{second}
\begin{gathered}
\underrightarrow{v}_{P}^{+}=\underrightarrow{v}_{F}^{-},\quad
\overrightarrow{v}_{F}^{+}=\overrightarrow{v}_{P}^{-},
\\
\underleftarrow{v}_{{F}}^{+}=\underleftarrow{v}_{P}^{-},\quad
\overleftarrow{v}_{P}^{+}=\overleftarrow{v}_{{F}}^{-}.
\end{gathered}
\end{equation}
From \eqref{first} and \eqref{second} we obtain
\begin{equation}\label{third}
\begin{gathered}
\delta_{\to}^{+}=\delta_{\to}^{-},\\
\delta_{\leftarrow}^{+}=\delta_{\leftarrow}^{-}.
\end{gathered}
\end{equation}
To emphasize the dependence of \eqref{third} on the ratio $k$ and
the eccentricity $e$, we shall write
\[
\delta_{\to}(k,e),\quad \delta_{\leftarrow}(k,e).
\]
Moreover, we set
\begin{equation}\label{delta}
 \delta(k,e) = \min\{\delta_{\to}(k,e),\; \delta_{\leftarrow}(k,e)\}.
 \end{equation}

Given a pair $(k,e)$, to estimate the four numbers of \eqref{second}
we use a program, written by the first author, that simulates the
motion of the point mass on the circle $S^1$ satisfying the
differential equation \eqref{funzionediteta}.
On this circle the marks $\alpha_W$, $\beta_W$, $\beta_E$, $\alpha_E$
are indicated.

For example, to compute $\underleftarrow{v}_{{F}}^{+}$ we use the
shooting method starting from the South Pole $S$ when
$\theta=\pi/2$ and the parameter $v=x'(\pi/2)$ is negative; that
is, the motion is clockwise. The corresponding solution is
observed in the \emph{future} (i.e., for $\theta>\pi/2$).
Equivalently, because of the equality
$\underleftarrow{v}_{F}^{+}=\underleftarrow{v}_{P}^{-}$, one can
evaluate the same number by starting from $S$ when $\theta=-\pi/2$
and $v=x'(-\pi/2) < 0$. The motion is again clockwise, but in this
case the point mass must be observed in the \emph{past} (so that
the point appears to be moving counterclockwise).

The curves $h(k,e)=0$ and $\delta(k,e)=0$ intersect at two points
(see figure \ref{fig:triangle}) that have been evaluated to be
\[
P_l=(0.179\pm10^{-3},0.088\pm10^{-3}), \quad P_r=(0.753\pm 10^{-3},0.279\pm 10^{-3}).
\]
For $P_l$ the four velocities are
\begin{gather*}
\underrightarrow{v}_{P}^{+}=\underrightarrow{v}_{{F}}^{-}
= 1.689 \pm 10^{-3},\quad
\overrightarrow{v}_{F}^{+}=\overrightarrow{v}_{P}^{-}
= 1.161 \pm 10^{-3},\\
\underleftarrow{v}_{{F}}^{+}=\underleftarrow{v}_{P}^{-}
 = 1.444 \pm 10^{-3},\quad
\overleftarrow{v}_{P}^{+}=\overleftarrow{v}_{{F}}^{-}
= 1.444 \pm 10^{-3}.
\end{gather*}
For $P_r$ we obtain
\begin{gather*}
\underrightarrow{v}_{P}^{+}=\underrightarrow{v}_{{F}}^{-}
= 4.337 \pm 10^{-3},\quad
\overrightarrow{v}_{F}^{+}=\overrightarrow{v}_{P}^{-}
= 1.526 \pm 10^{-3},\\
\underleftarrow{v}_{{F}}^{+}=\underleftarrow{v}_{P}^{-}
= 2.970 \pm 10^{-3},\quad
\overleftarrow{v}_{P}^{+}=\overleftarrow{v}_{{F}}^{-}
= 1.970 \pm 10^{-3}.
\end{gather*}
For Hyperion $k=0.26$, $e=0.11$, and we obtain
\begin{gather*}
\underrightarrow{v}_{P}^{+}=\underrightarrow{v}_{{F}}^{-}
= 2.177 \pm 10^{-3},\quad
\overrightarrow{v}_{F}^{+}=\overrightarrow{v}_{P}^{-}
= 1.308 \pm 10^{-3},\\
\underleftarrow{v}_{{F}}^{+}=\underleftarrow{v}_{P}^{-}
= 1.787 \pm 10^{-3},\quad
\overleftarrow{v}_{P}^{+}=\overleftarrow{v}_{{F}}^{-}
= 1.729 \pm 10^{-3}.
\end{gather*}
It follows that
$\delta(0.26,0.11) = 0.058 \pm 2\cdot 10^{-3}$.

The following theorem deals with \emph{going over the top}.

\begin{theorem}\label{overnotover}
Assume $ \delta(k,e)>0 $ and consider a
solution $x(\theta)$ of the differential equation
\eqref{funzionediteta} which, for some $\theta^0 \in \mathbb{R}$,
satisfies the condition $x(\theta^0)\in A_{N}$. Suppose that, for
some $k \in \mathbb{Z}$ and immediately after leaving $A_{N}$,
the point mass $x(\theta)$ reaches counterclockwise the South Pole
when $\theta=\theta_{S}=\pi/2 + 2k\pi$ or clockwise when
$\theta=\theta_{S}=-\pi/2 + 2k\pi$.
Then the point mass crosses $A_{N}$ immediately after
$\theta_{S}$.
\end{theorem}

\begin{proof}
First of all notice that the equation \eqref{funzionediteta} is
$2\pi$ periodic. Thus, if $x(\theta)$ is a solution, so is
$y(\theta):=x(\theta+ 2k\pi)$ for any $k\in \mathbb{Z}$.
Therefore, we can set $\theta_{S}=\pi/2$ in the first case,
and $\theta_{S}=-\pi/2$ in the second one.

Let us assume that point mass crosses the arc $A_W$ reaching the
South Pole for $\theta_{S}=\pi/2$. Then, from the condition
$\delta_{\to}(k,e)>0$ we derive that
$\underrightarrow{v}_{p}^{+}>\overrightarrow{v}_{{F}}^{+}$. In
other words, the speed of the point mass at the South Pole is
sufficient for \emph{going over the top}.

Let us now assume that the point mass crosses the arc $A_E$
reaching the South Pole for $\theta_{S}=-\pi/2$. Then from the
condition $\delta_{\leftarrow}(k,e)>0$ we derive
$\underleftarrow{v}_{p}^{-}>\overleftarrow{v}_{{F}}^{-}$. Hence,
the speed of arrival of the point mass at $S$ is larger than the
maximum speed such that $A_{N}$ is not crossed. Consequently,
$A_{N}$ is crossed.
\end{proof}

Our next result deals with the case of \emph{not going over the top}.

\begin{theorem}\label{overnotover2}
Suppose $ \delta(k,e)>0 $ and let $x(\theta)$
be a solution of \eqref{funzionediteta} that, for some
$\theta_0\in \mathbb{R}$, verifies the following conditions:
\begin{itemize}
\item
$x(\theta_0)\in A_{N}$;
\item
$x'(\theta_0)=0$;
 \item
for some $k \in \mathbb{Z}$ and immediately after $\theta_0$, the
point mass $x(\theta)$ reaches the South Pole when
$\theta=\theta_{S}=-\pi/2 + 2k\pi$ in the counterclockwise
case or when $\theta=\theta_{S}=\pi/2 + 2k\pi$ in the
clockwise case.
\end{itemize}
Then the point mass will not cross the northern arc immediately
after $\theta_{S}$.
\end{theorem}

\begin{proof}
The proof repeats almost verbatim the one of Theorem
\ref{overnotover}. In the counterclockwise case, the condition
$\delta_{\to}(k,e)>0$ implies that
$\underrightarrow{v}_{{F}}^{-}>\overrightarrow{v}_{p} ^{-}$.
Therefore, the point mass cannot reach $A_{N}$   immediately after
leaving $S$. A similar conclusion is obtained in the clockwise
case, since the condition $\delta_{\leftarrow}(k,e)>0$ implies
$\underleftarrow{v}_{{F}}^{+}>\overleftarrow{v}_{p} ^{+}$.
\end{proof}


\section{chaotic behavior}\label{chaos}

This section is similar to the corresponding portion of \cite{FLM2}.
There are however, some important differences, since the independent
variable $t$ has been replaced by the independent variable $\theta$.
Moreover, throughout this section, we assume that the pair $(k,e)$
belongs to the open triangle $T$ given by $ 0 < 4e < 3k < 3$, so
that the four arcs $A_W$, $A_E$, $A_{N}$, $A_{S}$ are well defined,
together with the function $h(k,e)$.
In particular, we are interested in the following open subset of $T$
(see figure \ref{fig:triangle}):
\[
\Omega = \big\{(k,e) \in T: h(k,e)>0,\, \delta(k,e)>0\big\}.
\]


\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.7\textwidth]{fig2}
\end{center}
\caption{The region $\Omega$ is inside the triangle $T$.
It is bounded to the right and to the left by the curve
$\delta(k,e)=0$ (represented as a sequence of dots, obtained
numerically), and above by the curve $h(k,e)=0$
(see equation \eqref{defineH}). The dot in $\Omega$,
namely the point with coordinates $(0.26,0.11)$, corresponds
to Hyperion}
\label{fig:triangle}
\end{figure}

Let $\theta_{N} \in \mathbb{R}$ and $x_{N} \in A_{N}$ be fixed.
Given any $v \in \mathbb{R}$, let $x_v(\theta)$ denote the
solution of the initial value problem
\begin{equation}\label{autonomous3bis}
 \begin{gathered}
 x''=\frac{1}{1+e\cos\theta}\big(2e(x'+2)\sin\theta
-3k\sin x\big)\\
x(\theta_{N})=x_{N},\quad x'(\theta_{N})=v.
\end{gathered}
\end{equation}


\begin{definition}\label{def1}\rm
Whenever, for $\theta > \theta_{N}$, $x_v(\theta)$ crosses $A_W$
with counterclockwise velocity or $A_E$ with clockwise velocity,
we say that the crossing is a \emph{significant event}.
In the first case we label the crossing with the number $1$;
and in the second case, with $-1$.
\end{definition}


\begin{definition}\rm
Denote by $\Sigma$ the set of all sequences in $\{-1,1\}$. Given $\sigma \in \Sigma$ and $n\in\mathbb{N}$, we say that $x_v(\theta)$
\emph{$n$-fulfills $\sigma$} if after $\theta_{N}$ it has at least
$n$ significant events and the list of labels associated to them
coincides with the first $n$ elements of $\sigma$, with their order
preserved.
We say that $x_v(\theta)$ \emph{fulfills} $\sigma \in \Sigma$ if
$n$-fulfills $\sigma$ for every~$n\in \mathbb{N}$.
\end{definition}

Let $\sigma \in \Sigma$ and $n\in \mathbb{N}$ be given.
\begin{itemize}
\item
We denote by $I_n\sp\sigma \subset \mathbb{R}$ the set of
$v \in \mathbb{R}$ such that the solution $x_v(\theta)$
of \eqref{autonomous3bis} $n$-fulfills $\sigma$.
\item
Given $v\in I_n\sp\sigma$, denote by $\theta_n\sp\sigma(v)$
the value of $\theta$ such that $x_v(\theta)$ crosses the
South Pole immediately after $n$-significant events.
\end{itemize}

Notice that continuity with respect to initial conditions implies
that any $I_n\sp\sigma$ is open. Moreover, as a consequence of
Proposition \ref{falling-down with theta}, the function
$\theta_n\sp\sigma: I_n\sp\sigma \to \mathbb{R}$ is well defined
and continuous.

We are now ready to state and prove the main result of this paper.

\begin{theorem}\label{porbits}
Suppose $(k,e) \in \Omega$. Then, given any
$\sigma \in \Sigma$, there exists $v \in \mathbb{R}$ such that the
solution $x_v(\theta)$ of \eqref{autonomous3bis} fulfills
$\sigma$.
\end{theorem}

\begin{proof}
We shall determine a sequence $\{J_n \colon n\in \mathbb{N}\}$ of nonempty
 bounded open intervals such that the following two conditions are
satisfied:
\begin{itemize}%\label{crucial}
\item[($a_n$)]
$J_n \subset I_n\sp\sigma$;
\item[($b_n$)]
the closure of $J_{n+1}$ is contained in $J_n$.
\end{itemize}
The two properties just mentioned imply that
\begin{equation}
\label{intersection}
J_{\infty}=\cap_{n \ge 1}J_n \ne \emptyset,
\end{equation}
and a solution of \eqref{autonomous3bis} with initial velocity $v\in
J_{\infty}$ fulfills $\sigma$.


We now describe how to define $J_1, J_2$ and $J_3$ for any
$\sigma \in \Sigma$. An induction procedure can be used to define
$J_n$ for all $n \in N$ so that \eqref{intersection} holds.

Without loss of generality we can assume that the first element of
$\sigma$ is $1$. Notice that a point mass with a large and
counterclockwise initial speed will first cross $A_W$.
Thus, the open set $I_1\sp\sigma$ is nonempty.
Analogously, if the initial speed is large enough and clockwise,
the point mass will first cross $A_E$.
Hence, $I_1\sp\sigma$ is bounded below. Therefore, it
contains a bounded interval $J_1 = (\omega_1,v_1)$ such that
$\omega_1 \notin I_1\sp\sigma$.
Continuity with respect to initial conditions implies that the unique
solution of \eqref{autonomous3bis} with initial velocity $\omega_1$
cannot have  $-1$ as the first significant event.
Thus, this solution will always remain in $A_{N}$ oscillating
indefinitely.
Hence, continuity with respect to initial conditions implies
\begin{equation}
\label{limitveloc1}
\lim_{v \to \omega_{1\sp +}} \theta_1\sp\sigma (v)=+\infty.
\end{equation}

Suppose now that the second element of $\sigma$ is different from the
previous one; i.e., it is $-1$. The continuity of
$ \theta_1\sp\sigma$, equation
\eqref{limitveloc1} and Theorem \ref{overnotover} imply the existence
of an initial velocity $v_2 \in J_1$ such that the corresponding
solution of \eqref{autonomous3bis} crosses $A_W$ a second time without
stopping.
Now observe that continuity with respect to initial conditions implies that
any solution of \eqref{autonomous3bis} with initial velocity $v \in J_1$
close to $\omega_1$ will stop inside $A_{N}$ for some value $\theta_{N} \ge 0$
before the first significant event (recall that the solution of
\eqref{autonomous3bis} with $v = \omega_1$ oscillates
indefinitely inside $A_{N}$). Thus, equation \eqref{limitveloc1} and
Theorem \ref{overnotover2} imply the existence of an initial
$w_2 \in (\omega_1,v_2)$ such that the corresponding solution of
\eqref{autonomous3bis} does not reach the northern arc $A_{N}$ immediately
after the first significant event. Therefore, continuity with respect to
initial conditions implies the existence of a solution of
\eqref{autonomous3bis} with initial velocity $u_2 \in (w_2,v_2)$ which
enters $A_{N}$ (immediately after the first significant
event) and then goes back crossing $A_E$ with clockwise velocity.
Consequently, its second significant event is $-1$. This shows that the open
set $I_2\sp\sigma \cap (u_2,v_2)$ is nonempty. Moreover, since this set is
clearly strictly contained in $(u_2,v_2)$, it contains an interval
$J_2 = (u_2,\omega_2)$ with $\omega_2 \notin I_2\sp\sigma$.
Continuity with respect to initial conditions implies that the solution of
\eqref{autonomous3bis} with initial velocity $\omega_2$ cannot have $1$ or
$-1$ as the second significant event.
Thus, this solution, after the first significant event, will enter
$A_{N}$ and remain there oscillating indefinitely.

At this point, the situation is as follows:

\begin{itemize}
\item[($a_2$)]
the interval $J_2 = (u_2,\omega_2)$ is contained in $I_2\sp\sigma$;
\item[($b_2$)]
the closure of $J_2$ is contained in $J_1$;
\item[($c_2$)]
$\omega_2$ is the initial velocity of a solution of \eqref{autonomous3bis}
that after the first significant event enters $A_{N}$ and remains there
oscillating indefinitely.
\end{itemize}


Continuity with respect to initial conditions implies that
\begin{equation}\label{limitveloc2}
\lim_{v \to \omega_{2\sp -}} \theta_2\sp\sigma(v)=+\infty.
\end{equation}


Let us assume that the third element of $\sigma$ is the same as the previous
one; i. e., it is again $-1$. Observe that, because of
\eqref{limitveloc2} and condition ($c_2$), we can find in $J_2$ an
initial velocity of a solution
of \eqref{autonomous3bis} that satisfies the assumptions of Theorem
\ref{overnotover} for some $\theta^0$ after the first significant
event and before the second (which, we point out, is labelled $-1$).
Similarly, in $J_2$ we can find an initial velocity of a solution that
satisfies the assumptions of Theorem \ref{overnotover2} for some
$\theta_0$ between the first and the second significant events.
Therefore, with a procedure analogous to the one described for the
construction of $J_2$, we can select an open interval $J_3$ with the
following properties:

\begin{itemize}
\item[($a_3$)]
all velocities of the interval $J_3$ $3$-fulfill the
sequence $\sigma$;
\item[($b_3$)]
the closure of $J_3$ is contained in $J_2$;
\item[($c_3$)]
one of the extremes of $J_3$ is the initial velocity of a solution of
\eqref{autonomous3bis} that, after the second significant event, enters
$A_{N}$ and remains there oscillating indefinitely.
\end{itemize}


An induction argument can now be used to complete the proof.
\end{proof}

Figure \ref{fig:quadrilateral} below contains a quadrilateral
in which, according to our numerical estimates, the assumptions
of Theorem \ref{porbits} are satisfied.

\begin{figure}[ht]
\begin{center} 
\includegraphics[width=0.7\textwidth]{fig3}
\end{center}
 \caption{Quadrilateral in which
   the assumptions of Theorem \ref{porbits} are satisfied}
 \label{fig:quadrilateral}
\end{figure}

The vertices are $(0.15,0.01)$, $(0.85,0.01)$, $(0.75,0.27)$,
$(0.19,0.09)$. Thus, a point $(k,e)$ belongs to the quadrilateral
if and only if
\begin{gather*}
(0.85-0.15)(e-0.01)-(0.01-0.01)(k-0.15) \geq 0,\\
(0.75-0.85)(e-0.01)-(0.27-0.01)(k-0.85) \geq 0,\\
(0.19-0.75)(e-0.27)-(0.09-0.27)(k-0.75) \geq 0,\\
(0.15-0.19)(e-0.09)-(0.01-0.09)(k-0.19) \geq 0.
\end{gather*}
Notice that the above conditions are satisfied by the point
$(k,e)$ corresponding to the satellite Hyperion.

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