\documentclass[reqno]{amsart}
\usepackage{graphics}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2004(2004), No. 36, pp. 1--14.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2004 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}

\title[\hfilneg EJDE-2004/36\hfil Chaotic orbits of a pendulum]
{Chaotic orbits of a pendulum with variable length}

\author[M. Furi, M. Martelli, M. O'Neill, \& C. Staples\hfil EJDE-2004/36\hfilneg]
{Massimo Furi, Mario Martelli, \\
 Mike O'Neill, \&  Carolyn Staples} % in alphabetical order

\address{Massimo Furi \hfill\break
Dipartimento di Matematica Applicata ``Giovanni Sansone''\\
Universit\`a degli Studi di Firenze\\
Via S. Marta 3\\
50139 Firenze, Italy}
\email{furi@dma.unifi.it}

\address{Mario Martelli\hfill\break
Department of Mathematics\\
Claremont McKenna College \\
Claremont, CA, 91711, USA}
\email{mmartelli@mckenna.edu}

\address{Mike O'Neill\hfill\break
Department of Mathematics\\
Claremont McKenna College \\
Claremont, CA, 91711, USA}
\email{moneill@mckenna.edu}

\address{Carolyn Staples \hfill\break
Department of Mathematics\\
Claremont McKenna College \\
Claremont, CA, 91711, USA}
\email{cstaples@mckenna.edu}

\date{}
\thanks{Submitted November 25, 2003. Published March 14, 2004.}
\subjclass{34C28}
\keywords{Pendulum, orbit, chaotic, separatrix}

\begin{abstract}
 The main purpose of this investigation is to show that a
 pendulum, whose pivot oscillates vertically in a periodic fashion,
 has uncountably many chaotic orbits. The attribute \emph{chaotic} 
 is given according to the criterion we now describe. First, we
 associate to any orbit a finite or infinite sequence as follows. 
 We write 1 or $-1$ every time the pendulum crosses the position 
 of unstable equilibrium with positive (counterclockwise) or negative
 (clockwise) velocity, respectively. We write 0 whenever we find a 
 pair of consecutive zero's of the velocity separated only by a 
 crossing of the stable equilibrium, and with the understanding that 
 different pairs cannot share a common time of zero velocity. 
 Finally, the symbol $\omega$, that is used only as the ending
 symbol of a finite sequence, indicates that the orbit tends 
 asymptotically to the position of unstable equilibrium. Every 
 infinite sequence of the three symbols $\{1,-1,0\}$ represents a real 
 number of the interval $[0,1]$ written in base 3 when $-1$ is replaced 
 with 2. An orbit is considered chaotic whenever the associated 
 sequence of the three symbols $\{1,2,0\}$ is an irrational number 
 of $[0,1]$. Our main goal is to show that there are uncountably 
 many orbits of this type.
\end{abstract}

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


\section{Introduction} \label{introduction}

This introduction, although a bit technical in a couple of paragraphs, is
largely descriptive. Its aim is to present the organization of the paper and
to discuss its goal, strategies and results. It is written with the intent of
generating enough interest and curiosity to entice our readers to follow us
through the entire journey, including its most technical parts. First we
present the main goal of the paper and the ideas needed to better understand
its meaning and importance. Then we explain the organization of the paper and
the strategies we use to prove the results. Finally, we talk about some
results closely related to our effort and previously obtained by other
authors.

The main purpose of our investigation is to show, with a rather simple
argument, that a pendulum, whose pivot oscillates vertically in a periodic
fashion, has uncountably many chaotic orbits that start, with zero velocity,
from positions sufficiently close to the unstable
equilibrium.

Our readers are certainly aware that there are many different definitions
of chaos. Hence, we need to explain in what sense we say that an orbit
is chaotic. To any orbit of the pendulum we associate a finite or infinite
sequence as follows. We write $1$ or $-1$ every time the position of unstable
equilibrium is crossed with positive (counterclockwise) or negative
(clockwise) velocity, respectively. We write $0$ whenever we encounter a pair
of consecutive zero's of the velocity separated only by a crossing of the
stable equilibrium, and with the understanding that different pairs cannot
share a common time of zero velocity. \emph{Oscillation} is the name we shall
use for each pair of this type. Finally, the symbol $\omega$, that is used
only as the ending symbol of a finite sequence, indicates that the orbit
tends asymptotically to the position of unstable equilibrium. Every infinite
sequence of the three symbols $\{1,-1,0\}$ is a real number of the interval
$[0,1]$ written in base $3$ when $-1$ is replaced with $2$. An orbit of the
pendulum will be considered chaotic whenever the associated sequence of the
three symbols $\{1,2,0\}$ is an irrational number of $[0,1]$.

In Section 4 we shall prove that given any infinite sequence $S$ with
entries taken exclusively from the three symbols $1,-1,0$, or any finite
sequence $S$ ending with $\omega$ and with all remaining entries taken from
the three symbols $1,-1,0$, we can find infinitely many orbits of the
pendulum to which the sequence $S$ is associated according to the rules just
described. For example, let us suppose that the sequence is
$S=\{1,1,-1,0,0,1,-1,\dots\}$. Then, we can find infinitely many orbits that
start with two counterclockwise crossings followed by one clockwise crossing,
two oscillations, one counterclockwise and one clockwise crossing, etc. Since
the irrational numbers of $[0,1]$ are uncountable we obtain that the pendulum
has uncountably many chaotic orbits.


\begin{figure}
\includegraphics{pendulum.eps}
\label{escapeup1}
\caption{The pivot C of the pendulum
moves vertically and periodically with period $\frac{2\pi}{\mu}$.}
\end{figure}

The technical preparatory lemmas and theorems are presented in Section 3, and
the main result is proved in Section 4. Although some parts of Section 3, and
in particular Theorems \ref{theorempassing} and \ref{notover}, may look
intimidating, they are based on a very simple idea that is explained below. A
reader who feels comfortable with the idea can glance through Section 2,
where notations and definitions are introduced, skip Section 3, and go
directly to Section 4, where the main result is presented and some relevant
consequences are derived.

To describe the idea let us model the motion of a pendulum, when
the pivot oscillates vertically in a periodic manner, with the
initial value problem
\begin{equation}\label{standard0}
\begin{gathered}
\ddot x(t)+(1+r \sin\mu t) \sin x(t)=0 \\
x(0)=\theta_0,\quad  \dot x(0)=0,
\end{gathered}
\end{equation}
with $(r,\mu)\in(0,1)\times(0,1]$. For simplicity, we assume that $\theta_0\in
(-\pi,-\frac{\pi}{2})$ is given and $\mu=1$. Denote by $x(\theta_0,t)$ the
corresponding solution of (\ref{standard0}). Then, given $n\in N$,
$x(\theta_0,t)$ will go over the top if the downward vertical position is
reached for the first time when $t=(2n+1)\pi$, and will not go over the top
if $t=2n\pi$. The statement \emph{going over the top} means that
$\dot x(\theta_0,t)>0$ as long as $x(\theta_0,t)\leq \pi$, while \emph{not
going over the top} means that there exists $t_0>2n\pi$ such that $\dot
x(\theta_0,t)>0$ for $t\in(0,t_0)$, $\dot x(\theta_0,t_0)=0$ and
$x(\theta_0,t_0)<\pi$. Continuity with respect to initial conditions shows
that the odd-even crossings just mentioned are sufficient but not necessary
for going or not going over the top.

The reason why the odd-even crossings of the downward vertical position make
such a big difference is the same for both cases. To better understand it, we
need to remember that the function $u(t)=2\arcsin(\tanh t)$ is a solution of
the differential equation
%
\[
\ddot u(t)+\sin u(t)=0
\]
and has the property
\[
\lim_{t\to\mp\infty}u(t)=\mp\pi.
\]
%
The function $u(t)$ is called \emph{separatrix}. For $t\neq 0$ we have
$\ddot u(t) u(t)<0$. In other words the position and the acceleration of the
separatrix have opposite sign.The total energy (kinetic and potential) of the
separatrix is
%
\[
E(t)=\frac{1}{2}\dot u^2(t)+1-\cos u(t).
\]
%
Since $\dot E(t)=0$, we have $E(t)=2$ for every $t\in\mathbb{R}$. We now observe that
the gain or loss of energy which is crucial for determining whether
the solution $x(\theta_0,t)$ goes over or does not go over the top is taking
place \emph{$\pi$ units of time before and after reaching the bottom
position}. For example, let us consider the case when the first crossing is
taking place at $t=(2n+1)\pi$ with $n\geq 1$. At $t=2n\pi$ the solution is
more negative than $2\arcsin(\tanh -\pi)<-2.9688$. In the time interval
$[2n\pi,(2n+1)\pi]$ it gains energy over the separatrix since
$(1+r\sin(t))>1$. Then, in the time interval $[(2n+1)\pi,(2n+2)\pi]$ the
negative acceleration is not as strong as the one acting on the separatrix,
since $1+r\sin t<1$. Hence, at $t=(2n+2)\pi$ we have
$x(\theta_0,(2n+2)\pi)>2\arcsin(\tanh\pi)>2.9688$ and the solution has enough
energy to go over the top. The situation when $x(\theta_0,2n\pi)=0$ is just
the opposite. We provide a technical proof of this very simple idea in
Section 3. In Section 4 we show how this idea, in combination with continuity
with respect to initial conditions, produces the infinitely many orbits to
which a given sequence can be associated.

Several authors have worked on related problems during the last 20 years
(see, for example,\cite{H},\cite{HML},\cite{PZ}). Many of them have been
interested in proving the existence of chaotic orbits  for the pendulum
  (\cite{HML},\cite{W}). In some cases
numerical evidence has been proposed as the main argument \cite{H}, while in
others \cite{W} chaos in the sense of Smale \cite{S} has been proved using the
Melnikov \cite{M} method. Finally, some authors have proved the existence of
chaotic orbits for planar systems (\cite{PZ},\cite{PZ2}), while others
have obtained the presence of specific type of chaotic orbits
(\cite{H},\cite{HML}) for systems similar to ours.

To the best of our knowledge the odd-even crossing idea is new. However, our
investigation has been largely motivated by the paper of Hastings and McLeod
\cite{HML}. We end this introduction by mentioning that the presence of a
small friction term in (\ref{standard0}) can easily be incorporated into the
proofs of the results established in Sections $3$ and $4$.

\section{Notation and Definitions}
\label{models}

We mentioned in the Introduction that the motion of a pendulum, with the pivot
oscillating vertically in a periodic manner, can be modeled by the second
order ordinary differential equation
\begin{equation}
\label{pendulum1}
\ddot x(t)+(1+r \sin\mu t) \sin x(t)=0.
\end{equation}
The function $x(t)$ is an angle and it measures the displacement of the
pendulum's arm from the downward vertical position. It is taken positive when
measured counterclockwise.


All results will be stated for the case $\mu=1$. At the end of Section
\ref{main result} we will make some observations regarding the cases $\mu<1$
and $\mu>1$. We shall indicate why the results proved in Sections
\ref{technical results} and \ref{main result} continue to be valid in the
case $\mu<1$. We can prove that the same conclusion holds when
$\mu\in(1,\mu_0]$, where
\[
\mu_0=\frac{\pi}{\log(\sqrt{2}+1)-\log(\sqrt{2}-1)}.
\]
The result is false for large values of $\mu$ (see \cite{A}) although we are
unable to specify what the values of $\mu$ might be. In Section $3$ and $4$
we usually deal with the Initial Value Problem
%
\begin{equation}\label{standard}
\begin{gathered}
\ddot x(t)+(1+r \sin t) \sin x(t)=0 \\
x(0)=\theta_0,\quad \dot x(0)=0,
\end{gathered}
\end{equation}
with $r\in(0,1)$ and $\theta_0\in[-\pi,0)$. In the case when $r=0$,
(\ref{standard}) reduces to the well-known mathematical model of the
motion of a simple pendulum
%
\begin{equation} \label{standardpendulum}
\begin{gathered}
\ddot x(t)+\sin x(t)=0\\
x(0)=\theta_0,\quad \dot x(0)=0.
\end{gathered}
\end{equation}
%
The solution of (\ref{standardpendulum}) reaches the downward vertical
position at a time $T$ given by the elliptic integral
%
\begin{equation}
\label{timetodown}
T =\int_{0}^{\frac{\pi}{2}} \frac{d\phi}{\sqrt{1-k^2\rm sin^2\phi}}\,,
\end{equation}
%
where $k=\sin\frac{\theta_0}{2}$. The time $T=\pi$ is of special interest to
us. In this case, we shall follow a standard notation used by other authors,
and replace $\theta_0$ by $\alpha$. Hence the initial value problem will be
%
\begin{equation}
\label{standardpendulum2}
\begin{gathered}
\ddot q(t)+\sin q(t)=0 \\
q(0)=\alpha,\quad \dot q(0)=0.
\end{gathered}
\end{equation}
%
An easy numerical estimate shows that the angle $\alpha$ such that
$q(\alpha,\pi)=0$ for the first time, satisfies the inequality
$-2.78824<\alpha<-2.78823$. We mentioned in the previous section that the
separatrix of the equation of a simple pendulum is the function
%
\begin{equation}
\label{pendulum2}
x(t) = 2\arcsin(\tanh t)
\end{equation}
%
The derivative is $\dot x(t)=2 \mathop{\rm sech} t$ and its value for $t=0$ is $2$.

A solution of (\ref{standard}) will be denoted by $x(\theta_0,t)$. When the
initial velocity is $a\ne 0$, it will be incorporated in the notation and the
solution will be denoted by $x(\theta_0,a,t)$. The energy of $x(\theta_0,t)$
is the function
%
\begin{equation}
\label{energy}
E(t) = \frac{(\dot x (\theta_0,t))^2}{2}+1-\cos x(\theta_0,t)
\end{equation}
and its derivative is
\begin{equation}
\label{derivativeofenergy}
\dot E(t) = -r \dot x(\theta_0,t) \sin t \sin x(\theta_0,t).
\end{equation}
%
The first term of (\ref{energy}) is the \emph{kinetic energy}. The
term $1-\cos x(\theta_0,t)$ is the potential\, energy, that sometimes will
be called \emph{height of the solution}, since it represents how far up is
$x(\theta_0,t)$ with respect to the downward vertical position.

Observe that the differential equation
%
\begin{equation}\label{standard1}
\ddot x(t)+(1+r \sin t) \sin x(t)=0
\end{equation}
%
has two equilibrium solution: one stable and one unstable. The stable one
is obtained with the choice of $0$ initial position and $0$ initial
velocity. The unstable one has the same initial velocity, but the initial
position is changed to $\pm\pi$. Sometimes we shall call \emph{bottom}
and \emph{top} the stable and unstable positions, respectively.

We explained in the previous section in what sense the orbits of a pendulum
are considered chaotic. We add here a comment on how the symbol $0$ is
associated to a pair of $0$'s of the velocity separated only by a crossing of
the stable equilibrium. Two different situations may arise. The symbols
immediately before and after a string of $k$ consecutive $0's$, $k=1,2,\dots$,
may have the same or opposite sign. The velocity is $2k$ times equal to $0$
in the first case, and $2k+1$ times in the second case. The number of
oscillations will be equal to $k$ in both cases.

We are now ready for the technical details.


%%%%%
\section{Over the top or not}
\label{technical results}

This section is divided into two parts. After the preliminary Lemma
\ref{lemma1} that is used in both parts, we prove, in Lemma \ref{lemma2} and
Theorem \ref{theorempassing}, that a solution of (\ref{standard}) such that
$x(\theta_0,(2n+1)\pi)=0,n\geq 1,$ for the first time, will go over the top
before its velocity changes sign. Then, in Theorem \ref{notover}, we prove
that when $x(\theta_0,2n\pi)=0, n\geq 2,$ for the first time, the solution
will not go over the top before its velocity changes sign.

\begin{lemma} \label{lemma1}
Assume that $f:[a,b]\to\mathbb{R}$ is increasing and continuous. Then, for every
$n\in\mathbb{N}$ such that $[0,2n\pi]\subseteq[a,b]$ we have
$\big|\int_{0}^{2n\pi}\cos tf(t) \,dt\big|\leq f(b)-f(a)$.
\end{lemma}

\begin{proof}
We shall prove this result with the additional assumption that f is $C^1$.
Integration by parts gives
\begin{equation}
\label{inequality1}
\int_{0}^{2n\pi}\cos t f(t) \,dt = -\int_{0}^{2n\pi} \sin t\dot f(t)\,dt.
\end{equation}
Since
\begin{equation}
\label{inequality2}
\big|\int_{0}^{2n\pi}\sin t \dot f(t)\,dt\big| \leq \int_{0}^{2n\pi}\dot f(t)
\,dt =f(2n\pi)- f(0)\leq f(b)-f(a),
\end{equation}
the result follows.
\end{proof}

Note that Lemma \ref{inequality1} can be easily adjusted to the case when
$f$ is decreasing.

In what follows we shall denote by $u(\beta,\gamma,t)$ the solution of the
initial value problem
%
\begin{equation}\label{usolution1}
\begin{gathered}
\ddot u(t)+\sin u(t)=0 \\
u(0)=\beta,\quad \dot u(0)=\gamma.
\end{gathered}
\end{equation}
%
We may simply write $u(\beta,t)$ when $\gamma=0$.

Let $\theta_1$ be such that $u(\theta_2,a,\pi)=0$, where $a=\dot
u(\theta_1,\pi)$ and $\theta_2=u(\theta_1,\pi)+a\pi$. Notice that $\theta_1$
is selected so that in a time interval of $3\pi$ we reach the downward
vertical position by first following $u(\theta_1,t)$ for $t\in [0,\pi]$, then
advancing with constant speed $a=\dot u(\theta_1,\pi)$ for $t\in [\pi,2\pi]$
and, finally, following $u(\theta_2,a,t)$ for $t\in[2\pi,3\pi]$.

\begin{lemma} \label{lemma2}
Let $r\in(0,1)$ be given. Denote by $x(\theta_0,t)$ the solution of the
initial value problem (\ref{standard}) with $\theta_0\in (-\pi,0)$.
Assume that $n\geq 1$ is such that $x(\theta_0,(2n+1)\pi)=0$ for the first
time. Let $\beta=x(\theta_0,2n\pi)$. Then $\beta\leq\theta_2$, where
$\theta_2 $ was defined above.
\end{lemma}

\begin{proof}
The proof is divided into three parts. First we establish that $a<\dot
x(\theta_0,2n\pi)$ implies that $x(\theta_0,2n\pi)\leq\theta_2$. Then
we show that $\theta_1<\theta_0$ is not an acceptable alternative since
it would require $a<\dot x(\theta_0,2n\pi)$ and
$\theta_2<x(\theta_0,2n\pi)$. Finally we show that the only other option
$\theta_0\leq\theta_1$ always implies $x(\theta_0,2n\pi)\leq\theta_2$.

For the first part assume that $a<\dot x(\theta_0,2n\pi)$, where
$a$ was defined above. We want to show that $x(\theta_0,2n\pi)\leq\theta_2$.
To see why this is true let $b=\dot x(\theta_0,2n\pi)$ and
$\theta_3=x(\theta_0,2n\pi)$. The inequality $\theta_2<\theta_3$ would imply
that $u(\theta_3,b,t)$ would reach the downward vertical position in a time
$T_1<\pi$. In fact, the integral giving the time $\pi$ for $u(\theta_2,a,t)$
to reach the downward vertical position is strictly larger than the integral
that provides the time $T_1$ needed by $u(\theta_3,b,t)$ to reach the same
position. At this point the conclusion for the first part is derived from the
inequality $u(\theta_3,b,t)\leq x(\theta_3,b,t)=x(\theta_0,2n\pi+t)$ for every
$t\in(0,\pi)$, that can be easily established using an energy argument. In
fact, since $\dddot u(\theta_3,b,0)<\dddot x(\theta_0,2n\pi)$ and the
energy of $x(\theta_0,2n\pi+t)$ is larger than the energy of
$u(\theta_3,b,t)$ for every $t\in(0,\pi]$ we obtain that $\dot
u(\theta_3,b,t)<\dot x(\theta_0,2n\pi+t)$ for every $t\in(0,\pi]$.

For the second part let us assume that $\theta_1<\theta_0$. It is not hard to
show that $x(\theta_0,2n\pi)<-\frac{\pi}{2}$. As a consequence of this
we obtain that $\ddot u(\theta_1,t)<\ddot x(\theta_0,(2n-2)\pi+t)$ for every
$t\in (0,\pi)$. It follows that $a<\dot x(\theta_0,(2n-1)\pi)$ and
$u(\theta_1,\pi)<x(\theta_0,(2n-1)\pi)$. Consequently,
$\theta_2<x(\theta_0,2n\pi)$ and $a<\dot x(\theta_0,2n\pi)$, in contradiction
to the first part of the proof.

In the third part it remains to show that the only alternative left, namely
$\theta_0\leq\theta_1$, implies $x(\theta_0,2n\pi)\leq\theta_2$. Using an
energy argument we can show that $u(\theta_1,\pi)\leq x(\theta_0,(2n-1)\pi)$
would imply $a=\dot u(\theta_1,\pi)<\dot x(\theta_0,(2n-1)\pi)$. Consequently,
we would obtain the same unacceptable conclusion already seen in the second
part of the proof. Hence, we must have
$x(\theta_0,(2n-1)\pi)<u(\theta_1,\pi)$. Now let us look at $\dot
x(\theta_0,2n\pi)$. On the one hand, we know that if this velocity is
strictly larger than $a$ then we must have $x(\theta_0,2n\pi)\leq \theta_2$.
On the other hand, if $\dot x(\theta_0,2n\pi)\leq a$, then from the inequality
$x(\theta_0,(2n-1)\pi)<u(\theta_1,\pi)$, we easily derive
$x(\theta_0,2n\pi)\leq \theta_2$.
\end{proof}

Theorem \ref{theorempassing} can be labeled as the \emph{over the top}
theorem. Notice that, as a consequence of continuity with respect to initial
conditions, the result stated in it continues to be valid for all solutions
with initial conditions sufficiently close to the ones included in the
theorem.

\begin{theorem} \label{theorempassing}
Let $r>0$ be given and let $\phi \in (-\pi,0)$ be such that
$1+\cos\phi<0.1r$. The following two statements hold.
\begin{itemize}
\item[i.]
There exists a positive integer $N\geq 1$ such that for every $n
\geq N$ there is at least one initial position $\theta_0\in(-\pi,\phi)$
such that the unique solution of the initial value problem (\ref{standard})
reaches the downward vertical position for the first time when
$t=(2n+1)\pi$.
\item[ii.]
There exists $t_2>(2n+1)\pi$ such that
$x(\theta_0,t_2)=\pi$ and $ \dot x(\theta_0,t)>0$ for every
   $ t\in (0,t_2].$
\end{itemize}
\end{theorem}

\begin{proof}
The first part of Theorem \ref{theorempassing} is an easy consequence
of the continuity with respect to initial conditions, since, as we
approach $-\pi$, the time needed to reach the downward vertical position goes
to infinity.

To prove the second part we first show that when the solution arrives at the
bottom position its energy is at least $2+0.8r$. Lastly, we prove
that with this energy the solution will go over the top.

An easy computation shows that
%
\begin{equation} \label{velocity0}
\begin{aligned}
\frac{\dot x^2(\theta_0,(2n+1)\pi)}{2}
&=2-\delta+2r\int_{0}^{2n\pi}\cos t\sin^2\frac{x(\theta_0,t)}{2}\,dt \\
&\quad +2r\int_{2n\pi}^{(2n+1)\pi}\cos t
\sin^2\frac{x(\theta_0,t)}{2}\,dt,
\end{aligned}
\end{equation}
where $0<\delta = 1+\cos \theta_0<0.1r$.
Using the estimate provided by Lemma \ref{lemma2}
we find that
\[
x(\theta_0,2n\pi)\leq -2.9688.
\]
Hence, from Lemma \ref{lemma1}, we derive
%
\begin{equation}\label{velocity3}
2r\big|\int_{0}^{2n\pi}\cos t \sin^2\frac{x(t)}{2}\,dt\big|\leq 0.014884r.
\end{equation}
%
We now need to estimate the last integral of (\ref{velocity0}). To
accomplish this task we split the integral into two parts: the first from
$2n\pi$ to $2n\pi+\frac{\pi}{2}$ and the second from $2n\pi+\frac{\pi}{2}$
to $(2n+1)\pi$.

The first part of the integral is positive. We obtain a lower estimate of
its value using the function $u(\theta_3,a_3,t)$ where
$\theta_3=2\rm arc\sin (\tanh(-\pi))$ and $a_3=\frac{2}{\cosh(-\pi)}$. The
given position and velocity are selected so that $u(\theta_3,a_3,\pi)=0$ and
\[
0\leq \cos t \sin^{2}
\frac{u(\theta_3,a_3,t)}{2}\leq\cos t \sin^{2}
\frac{x(\theta_0,2n\pi+t)}{2}\,,
\]
for $t\in[0,\frac{\pi}{2}]$. The second part of the integral is negative and
we provide a lower estimate of its value using the solution of the initial
value problem
%
\begin{equation}\label{usolution2}
\begin{gathered}
\ddot v(t)+2\sin v(t)=0\\
v(0)=\theta_4,\quad  \dot v(0)=a_4,
\end{gathered}
\end{equation}
%
where $\theta_4=2\rm arc\sin (\tanh(-\sqrt{2}\,\pi))$ and
$a_4=\frac{2\sqrt{2}}{\cosh(-\sqrt{2}\,\pi)}$. The initial position and
velocity are selected so that
$v(\theta_4,a_4,\pi)=0$ and
\[
\cos t \sin^{2}
\frac{v(\theta_4,a_4,t)}{2}\leq\cos t \sin^{2}
\frac{x(\theta_0,2n\pi+t)}{2}\leq 0\,,
\]
for $t\in[\frac{\pi}{2},\pi]$.

The first estimate provides a positive value exceeding $1.938527$ and the
second estimate provides a negative value not smaller than $-0.829164$.
Putting together all estimates and assuming the worst possible
situation we have
\[
0.8r\leq (-0.1-0.014884+1.938527-0.829164)r.
\]

As $x(\theta_0,t)$ moves past the downward vertical position, it travels
faster than the separatrix and at $t=(2n+2)\pi$ we have
$x(\theta_0,(2n+2)\pi)>2\arcsin\tanh\pi>2.9688$. Hence, the energy needed to
go over the top does not exceed $r (1+\cos 2.9688) <0.014892r$. Recall that
at the bottom position the energy surplus was at least $0.8r$ and observe
that in the interval $[(2n+1)\pi,(2n+2)\pi]$ the solution is losing less
kinetic energy than the separatrix. Hence, the solution will make it over the
top.
\end{proof}

Theorem \ref{notover} addresses the case when the solution reaches the
downward vertical position for $t=2n\pi$ with $n\geq2$. The technical details
are similar to the ones introduced in the proofs of Lemma \ref{lemma2} and
Theorem \ref{theorempassing}. The estimates are obtained using different
functions, but the basic ideas and strategy are the same. Therefore, we
will simply mention the results without including the technical details.

\begin{figure}
\includegraphics{over5.eps}
\label{escapeup3}
\caption{The solution with $r=0.001$ that reaches the position $\theta=0$
when $t=5\pi$ crosses over the unstable equilibrium before $8\pi$}
%\end{center}
\end{figure}

\begin{theorem} \label{notover}
Let $r>0$ be given and let $\phi \in (-\pi,0)$ be such that
$1+cos\phi\leq 0.1r$. The following two statements hold.
\begin{itemize}
\item[i.]
There exists a positive integer $N\geq 1$ such that for every $n
\geq N$ there is at least one initial position $\theta_0\in(-\pi,\phi)$
such that the solution of the initial value problem (\ref{standard})
reaches the downward vertical position for the first time when $t=2n\pi$.
\item[ii.] There exists $t_3>2n\pi$ such that $\dot x(\theta_0,t)>0$ for
$t\in(0,t_3)$, $\dot x(\theta_0,t_3)=0$ and
$ x(\theta_0,t_3)<\pi$.
\end{itemize}
\end{theorem}

\begin{proof}
The first part of Theorem \ref{notover} is an easy consequence
of continuity with respect to initial conditions combined with the fact
that as we approach $-\pi$ the time needed to reach the downward vertical
position goes to infinity.

To prove the second part we first show that when the solution arrives at the
downward vertical position its energy does not exceed $2-0.8r$. After,
we prove that the solution will not go over the top.

An easy computation shows that
%
\begin{equation}\label{velocity2}
\begin{aligned}
\frac{\dot x^2(2n\pi)}{2}
&=2-\delta+2r\int_{0}^{(2n-1)\pi}\cos t\sin^2\frac{x (t)}{2}\,d t\\
&\quad +2r\int_{(2n-1)\pi}^{2n\pi} \cos t \sin^2\frac{x (t)}{2}\,dt,
\end{aligned}
\end{equation}
%
where $0<\delta = 1+\cos\theta_0\leq0.1r$. With a strategy similar to the one
used in Lemma \ref{lemma2} we obtain that
$x(\theta_0,(2n-1)\pi)\leq -2.65314$. Hence, by Lemma \ref{lemma1} we have
%
\begin{equation}
\label{velocity4}
2r\big|\int_{0}^{(2n-1)\pi}\cos t \sin^2\frac{x(t)}{2}\,dt\big|\leq 0.11694r.
\end{equation}

We now estimate the last integral of (\ref{velocity2}). In the interval
$[(2n-1)\pi,(2n-1)\pi+\frac{\pi}{2}]$ the integral is more negative than
$-1.6308916r$ and in the interval $[(2n-1)\pi+\frac{\pi}{2},2n\pi]$ is
bounded above by $0.5680262r$.
Putting all estimates together we
obtain that the energy of the solution at the downward vertical position does
not exceed
$2-0.8r$.

With this loss of energy the solution will not make it
over the top. The proof of this last step is divided into two parts. In the
first we consider those solutions such that
\[
\dot x(\theta_0,2n\pi)\leq\sqrt{2(1-\cos\alpha)},
\]
where $\alpha$ was defined and numerically estimated in Section \ref{models}.
In the second we consider those solutions whose velocity at the bottom is
larger than $\sqrt{2(1-\cos\alpha)}$.

All solutions of the first group will come to a rest point at a time
$t<\pi$ and before reaching the top position. Here is why. The solution
\[
u(0,\sqrt{2(1-\cos\alpha)},t)
\]
reaches zero velocity at $t=\pi$,
$u(0,\sqrt{2(1-\cos\alpha)},\pi)<\pi$, and in the interval
$(2n\pi,(2n+1)\pi)$ we have
\[
\ddot x(\theta_0,t)< \ddot u(0,\sqrt{2(1-\cos\alpha)},t)<0.
\]

For the solutions of the second group we use the estimate on the energy
at the bottom to derive that $r\leq0.077$. We now use the solution
of the initial value problem
%
\begin{equation}\label{goingup2}
\begin{gathered}
\ddot w(t)+ (1.077)\sin w(t)=0 \\
w(0)=0,\quad \dot w(0)=\sqrt{2(1-\cos\alpha)}.
\end{gathered}
\end{equation}
%
It can be easily verified that
\[
w(0,\sqrt{2(1-\cos\alpha)},\pi)\leq x(\theta_0,(2n+1)\pi)\leq 2\arcsin\tanh
\pi.
\]
Hence, for any choice of $r\in(0,0.077)$, a solution of (\ref{standard}) that
reaches the bottom position at a time $t=2n\pi$ will be at least as high as
the solution of (\ref{goingup2}) at $t=(2n+1)\pi$. An easy numerical
estimate shows that $1-\cos w(0,\sqrt{2(1-\cos\alpha)},\pi)\geq 2-0.2$.
Consequently, the largest excess of energy $x(\theta_0,t)$ can gain to reach
the top after $t=(2n+1)\pi$ is at most $0.2r$. Since at the bottom there was
already a loss of energy of at least $0.8r$ and in the following time interval
$[2n\pi,(2n+1)\pi]$, the solution $x(\theta_0,t)$ is losing more energy than
the separatrix, it will not have enough energy to reach the top before its
velocity changes sign.
\end{proof}

\begin{figure}
\includegraphics{overnot.eps}
\label{escapeup4}
\caption{The solution with $r=0.001$ that reaches the downward vertical
position  when $t=4\pi$ does not cross over the unstable equilibrium
before $t=5\pi$}
%\end{center}
\end{figure}

We now add an important remark to the results established previously.

\begin{remark} \label{remark one} \rm
Systems of the form
   \begin{equation} \label{different times}
\begin{gathered}
\ddot z(t)+(1+r\sin t)\sin z(t) = 0\\
z(t_0)=\alpha_0,\quad  \dot z(t_0) = 0.
\end{gathered}
\end{equation}
%
are sometimes considered and will be needed in the proof of our main result.
Given $r>0$ and $t_0$, we can determine $\alpha_0 \in(-\pi,0)$
sufficiently close to $-\pi$ so that all conclusions reached before
are still valid. In a similar manner we can handle cases in which
$\alpha_0=-\pi$ and small velocities are assumed in either direction. Both
situations will arise in the proof of the next theorem.
\end{remark}

\section{Chaotic Orbits} \label{main result}

We are now ready to state and prove the main result of this paper concerning
orbits of a pendulum with an oscillating pivot. We introduce a preliminary
definition.

\begin{definition} \label{rotation oscillation}\rm
The symbols $1,-1$ denote a crossing of the unstable equilibrium in a
counterclockwise or clockwise direction, respectively. The symbol $0$
denotes two times of zero velocity separated only by a crossing of the
position of stable equilibrium. The symbol $\omega$ indicates that an orbit
tends asymptotically to the position of unstable equilibrium.
\end{definition}

Since an orbit may tend asymptotically to the top position either in a
counterclockwise or clockwise manner, it would be more precise to use
$+\omega$ in one case, and $-\omega$ in the other case. However, this
distinction does not really add an important information on the orbit.
Hence, we have decided not to use it.

The statement \emph{solution starting from} followed by the indication of a
position angle will be used to denote the solution of initial value problems
like (\ref{pendulum20}) below. We may also say that the solution
\emph{corresponds to} followed by the indication of the position angle. When
the initial velocity is not mentioned it will be assumed equal to $0$. The
statement \emph{a sequence corresponds to a solution} means that a
sequence of symbols is associated to the solution according to the rules
stated in Definition \ref{rotation oscillation}.

\begin{theorem} \label{porbits}
Let $r\in(0,1)$ be given. Select any infinite sequence of entries from the
symbols $1,-1,0$ or any finite sequence of entries from the same symbols and
ending with $\omega$. Then there are infinitely many initial conditions
$(\theta_0,0)$ such that the given sequence of symbols corresponds to the
solution of the initial value problem
\begin{equation} \label{pendulum20}
\begin{gathered}
\ddot x(t)+(1+r \sin t)\sin x(t)=0 \\
x(0)=\theta_0, \quad \dot x(0)=0.
\end{gathered}
\end{equation}
\end{theorem}

\begin{proof}
The procedure to follow in the case of a finite sequence will be evident
from the proof we present when the sequence is infinite. To obtain the
desired result we will produce a family of nested intervals
$I_n=[a_n,b_n]$, such that all orbits of (\ref{pendulum20}) with initial
position $\theta_0 \in I_n$ will complete the first $n$ steps of the
sequence, except when $\theta_0$ is equal to either one of the
border points. These two initial positions will produce solutions satisfying
only the first $n-1$ steps of the sequence and terminating with $\omega$.
Since $\cap I_n=I_{\infty}\neq\emptyset$ we obtain the desired orbit by
selecting $\theta_0 \in I_{\infty}$.

To better understand how the sequence of intervals can be constructed let us
keep in mind that the set of initial conditions with corresponding orbits
satisfying the first $k$ entries of the infinite sequence is an open set. This
is a direct consequence of the continuity with respect to initial conditions.
Let us also keep in mind that to any solution we can associate a sequence,
although different solutions need not have different sequences. For example,
the sequence $\{0,0,\dots,0,\dots\}$ corresponds to all solutions that will
indefinitely oscillate around the position of stable equilibrium.

Let us assume that the sequence starts with
$1$. The cases when the sequence starts with $-1$ or $0$ are handled
similarly.

Given $r \in(0,1)$ we select $N$ large enough so that for all $n\geq
N$ we can determine $\theta_0$ so that the solution of the initial value
problem
%
\begin{equation}\label{pendulum22}
\begin{gathered}
\ddot x(t)+(1+r\sin t)\sin x(t)=0\\
x(0)=\theta_0, \quad \dot x(0)=0
\end{gathered}
\end{equation}
%
reaches the downward vertical position at time $t=(2n+1)\pi$. Hence,
the solution will go over the top. Take the largest interval of the form
$[a_1,b_1]$ where $a_1<\theta_0<b_1$ and $[a_1,b_1]$ is such that for all
$\theta\in (a_1,b_1)$ the corresponding solution will go over the top at
least once, while for $\theta=a_1$ or $\theta=b_1$ the corresponding solution
will not go over the top but will tend asymptotically to it as
$t\rightarrow +\infty$. Consequently, the sequence corresponding to these two
orbits will be simply denoted by $S=\{\omega\}$. Set $I_1=[a_1,b_1]$. Observe
that this first interval can be selected in infinitely many different
ways. In fact, for every  $n\geq N$, we can find an open interval
$(a_1,b_1)$ such that every solution with initial position in $(a_1,b_1)$
will go over the top at least once before its velocity changes sign.

We now indicate how to construct $I_2$. We shall assume that the second entry
of the sequence is $0$. The cases with second entry equal to $\pm 1$ are
handled similarly. $I_2$ will be constructed so that it is contained in
$I_1$ and all its points, except the border points, provide solutions having
$\{1,0\}$ in the first two entries of the corresponding sequence. The
solutions corresponding to the border points will have $1$ in the first
entry, and $\omega$ in the second entry. First we select in $I_1$ an initial
condition $\theta_1$ so that the velocity of the corresponding solution over
the unstable equilibrium will be very small and the downward vertical position
will be reached at time $t=2k\pi$, with $2k\pi>>t_0$ and $t_0$ the time when
the solution is over the top. This can obviously be accomplished, since as
the initial condition in $I_1$ approaches $b_1$ (or $a_1$) the corresponding
solution will arrive at the top with progressively smaller velocity. Hence,
we can also consider an initial condition smaller than $\theta_1$ and larger
than $a_1$ so that the corresponding solution will reach the downward
vertical position at a time that is an odd multiple of $\pi$. The two initial
conditions will be separated by one generating a solution that after going
over the top will tend asymptotically to the unstable equilibrium. From this
discussion the interested reader can understand how the choice of
$\theta_1$ can be made so that the border points of the interval
$I_2$ as selected below are contained in $(a_1,b_1)$. Moreover, we can also
satisfy the requirement imposed by the magnitude of $r$ and mentioned
in the statements of Theorems \ref{theorempassing} and \ref{notover} of
starting close enough to the unstable equilibrium to insure the validity of
all inequalities previously established.

The solution starting from $\theta_1$ will come to a rest on the
right-hand-side before reaching the top a second time. Since the set of
solutions with this property is open, we consider the largest interval in
$I_1$ of the form $(c_2,d_2)$ with $c_2<\theta_1<d_2$ and such that for all
initial conditions of this interval the corresponding solution will come to a
rest before reaching the top a second time. The solutions corresponding to
$\theta=c_2$ or $\theta = d_2$ will go over the top once and then will tend
asymptotically to the unstable equilibrium. Hence, the sequence corresponding
to both will be $S=\{1,\omega\}$. Among the initial conditions of
$(c_2,d_2)$ there will be some with corresponding solution coming down to the
downward vertical position at a time that is even multiple of $\pi$ and
others with corresponding solutions coming down at a time that is odd multiple
of $\pi$. These will be separated by initial conditions with corresponding
solutions that after going over the top and coming to a rest point on the
right-hand-side, will tend asymptotically to the unstable equilibrium as they
move up on the left-hand-side. Pick an initial condition in $(c_2,d_2)$ so
that the corresponding solution comes down again at a time that is an even
multiple of $\pi$ and consider the largest open interval in $(c_2,d_2)$
containing this initial condition and such that for all $\theta$ in this open
interval the corresponding solution will have a rest point on the
left-hand-side separated from the one on the right-hand-side by a single
crossing of the position of stable equilibrium. The closure of this interval
is $I_2=[a_2,b_2]$. Clearly, for all $\theta\in(a_2,b_2)$ the corresponding
solution will be represented by a sequence having $\{1,0\}$ in the first and
second position. For $\theta=b_2$ or $\theta=a_2$ the corresponding solution
will be represented by the sequence $\{1,\omega\}$. An induction argument can
now be used to conclude the proof.
\end{proof}

\begin{remark} \rm
We have not included the presence of a friction term\, $k\dot x(t)$ in our
analysis. However, it is not difficult to see that the principles on
which the proofs are based will continue to be valid if a small friction term
is added. While it is hard to establish the magnitude of the constant $k$,
we can say that given $r>0$ there exists $k_0$ such that for all
$k<k_0$ the inclusion of a friction term with constant $k>0$ will not affect
the validity of the results we have established.
\end{remark}

\begin{remark} \rm
We now examine the case when $\mu\neq1$ and still $\mu>0$. The
separatrix of the problem
%
\begin{equation}
\label{pendulum34}
\ddot u(t)+c\sin u(t)=0
\end{equation}
%
is given by
\[
u(t)=2\arcsin(\tanh(\sqrt{c}t)).
\]
At the downward vertical position its velocity is $2\sqrt{c}$.

The results proved before remain unchanged if $\mu<1$. In fact, with a
suitable change of variable we can rewrite equation \eqref{pendulum1}
in the form
%
\begin{equation}
\label{pendulum24}
\begin{array}{{ccc}}
\ddot \theta(t)+c(1+r \sin t) \sin\theta(t)=0
\end{array}
\end{equation}
%
where $c=\frac{1}{\mu^2}$, and we see that all inequalities remain true due to
the fact that the system must start from a more negative position to reach
the downward vertical position at the required time. Hence, for every
$(r,\mu) \in(0,1)\times(0,1]$ Theorem \ref{porbits} holds true.

The case $\mu>1$ is more complicated. The approach we used before is still
valid for $\mu\in[1,\mu_0)$, where
\[
\mu_0=\frac{\pi}{\log(\sqrt{2}+1)-\log(\sqrt{2}-1)}\,.
\]
For all these values of $\mu$ one can show, using exactly the same approach
outlined in Lemmas \ref{lemma1} and \ref{lemma2}, that the estimates of
energy gain or loss are the same as previously determined. We simply have to
multiply them by $\frac{1}{\mu^2}$. More precisely, we can prove that the
kinetic energy of a solution that reaches the bottom position at an odd
multiple of $\pi$ is at least $\frac{2+ 0.8r}{\mu^2}$. For example, for
$\mu=\mu_0$ and with $1+\cos \theta_0<0.1r $ the energy surplus at the
downward vertical position is at least $\frac{.8r}{\mu_0^2}$ and the same
reasoning used in the proof of Theorem \ref{theorempassing} shows that the
solution will go over the top. Similarly, when the bottom position is reached
at a time that is an even multiple of $\pi$ and $\mu\leq\mu_0$, the kinetic
energy cannot exceed $\frac{2- 0.8r}{\mu^2}$ and the solution will not make
it over the top. The only caveat is that the multiples of $\pi$ may need to
have $n$ very large, but this is obviously not a problem.

For $\mu_0<\mu$, the situation is more complex, particularly when
$2\mu_0<\mu$. Numerical experiments suggest that the result is still true
when $\mu$ is not too large. One has to be careful in selecting the time
needed to come down to the position of stable equilibrium. The reader would
certainly remember that an appropriate choice was also included in Theorems
\ref{theorempassing} and \ref{notover}. Hence, this is nothing new. The
results on the stability of the inverted pendulum (see \cite{A}, Chapter 5)
show the existence of large $\mu$ values for which it is
hard to establish what the behavior of the system might be at least for
certain choices of the initial position and velocity. Hence, from this point
of view, some additional work needs to be done.
\end{remark}

There are also some interesting questions we have not been able
to answer. One of the most puzzling is the amount of energy an orbit can
accumulate, given $r$ and $\mu$. We have done some experiments with
$\mu=1$ and we have observed that the energy fluctuates between specific
values. In each case we have started with $0$ initial velocity. The energy
never grows too large or becomes too small. Although this behavior makes
sense, we have not been able to prove it, let alone establish what an upper
and lower bound for the energy must be.

We sincerely hope that some of these questions, and others that are not
mentioned here, will rouse the curiosity of some interested readers, who will
further explore the intricacies of these simple, yet fascinating systems.


%%%%%
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\end{document}