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\def\gua{H.~J.~SUSSMANN AND J.~C.~WILLEMS}
\def\guo{THE BRACHISTOCHRONE PROBLEM AND MODERN CONTROL THEORY}

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\newcommand{\hdleft}{{\rm H. J. Sussmann}}
\newcommand{\hdright}{{\rm \hfil Hamilton and Weierstrass}}
\newcommand{\acknowhector}{The writing of this paper began while the first
author was Johann Bernoulli Visiting Professor at the University of
Groningen, in May, June and July 1999, and continued while he was back
at Rutgers University, partially supported by the National Science Foundation
under NSF Grant DMS$\,$98-03411. It  was concluded while he was
Weston Visiting Professor at the Weizmann Institute of Science 
in Rehovot, Israel, from October 2000 to January 2001.  He
is grateful to the University of Groningen, the Weizmann Institute, 
and NSF, for their help and support.}

\begin{center}

{\Huge {\bf The Brachistochrone Problem\\

\vspace{0.3cm}
and Modern Control Theory}}

\end{center}

\vspace{0.9cm}

\noindent{\huge{\it H\'ector J. Sussmann\Ft{\acknowhector}
$\quad\,\,\,$Jan C. Willems}}
\vspace{0.2cm}

\noindent\mbox{\begin{minipage}{6.6cm}
{\large Department of Mathematics\\
Rutgers University\\
Hill center, Busch Campus\\
Piscataway, NJ 08854, USA

\vspace{0.07cm}

{\it sussmann@math.rutgers.edu}}

\vspace{0.07cm}

http://www.math.rutgers.edu/$\,\widetilde{\,\,}\,$sussmann
\end{minipage}}
\mbox{\begin{minipage}{6.4cm}
{\large Department of Mathematics\\
University of Groningen\\
P.O.~Box 800, 9700 AV Groningen\\
The Netherlands

\vspace{0.07cm}
{\it J.C.Willems@math.rug.nl}}

\vspace{0.07cm} 

http://www.math.rug.nl/$\,\widetilde{\,\,}\,$willems

\end{minipage}}

\vspace{0.9cm}

\begin{center}
{\mymagtwoit Dedicated to Velimir Jurdjevic on his 60th birthday}
\end{center}

\vspace{0.7cm}


\section{Introduction}

\noindent The purpose of this paper is to show that {\em modern control
theory}, both in the form of the ``classical'' ideas developed in the 1950s
and 1960s, and in that of later, more recent methods such as the
``nonsmooth,'' ``very nonsmooth'' and ``differential-geometric''
approaches, provides the best and mathematically most natural setting to do
justice to Johann Bernoulli's famous 1696 ``brachistochrone problem.''
(For the classical theory, especially the smooth version of the Pontryagin
Maximum Principle, see, e.g., Pontryagin {\it et al.}  \cite{pontrybook},
Lee and Markus \cite{leemarkus}, Berkovitz \cite{berkovitz}; for the
nonsmooth approach, and the version of the Maximum Principle for locally
Lipschitz vector fields, cf., e.g., Clarke \cite{clarkeminimal,
clarkebook}, Clarke {\em et al.}  \cite{clarkebooknew}; for very nonsmooth
versions of the Maximum Principle, see Sussmann
\cite{suss1,suss2,suss5,suss6,suss7,suss9,suss10,suss11}; for the
differential-geometric approach, cf., e.g., Jurdjevic \cite{jurdjevic},
Isidori \cite{isidori}, Nijmeijer and van der Schaft \cite{nijmeijer},
Jakubczyk and Respondek \cite{jakures}, Sussmann \cite{susscofree,suss6}.)

We will make our case in favor of modern control theory in two main ways.
{\em First}, we will look at four approaches to the brachistochrone
problem, presenting them in chronological order, and comparing them. {\em
Second}, we will look at several ``variations on the theme of the
brachistochrone,'' that is, at several problems closely related to the one of
Johann Bernoulli.

The first of our two lines of inquiry will be pursued in sections
\ref{standardfor}, \ref{spurious}, \ref{opti-ap}, and \ref{brachprinc},
devoted, respectively, to

\medbreak
\begin{itemize}
\item[1.]  Johann Bernoulli's own solution based on an
analogy with geometrical optics, 
\item[2.] the solution based on the classical
calculus of variations, 
\item[3.] the optimal control method,
\end{itemize}
and, finally, 
\begin{itemize}
\item[4.] the differential-geometric approach.  
\end{itemize}

\medbreak\noindent We will show that {\em each of the three
transitions from one of the methods in the above list to the next one leads to
real progress for the brachistochrone problem, by making it possible to
derive stronger conclusions about it than those that were obtainable by
means of previously existing techniques.} 

Specifically, we will point out that

\begin{itemize}
\item[1.a] the differential equation derived by Johann
Bernoulli has spurious solutions (i.e., solutions
other than the cycloids), as was first noticed by Taylor in 1715,
\item[but]
\item[1.b] the application of the Euler-Lagrange condition of the calculus
of variations eliminates these solutions and leaves only the cycloids,
\end{itemize}
and that
\begin{itemize}
\item[2.a] with the calculus of variations approach, the existence
of solutions is a delicate problem, due to the non-coercivity 
of the Lagrangian, 
\item[but]
\item[2.b] the optimal control control method renders this question
trivial, reducing the proof of existence to a straightforward
application of the Ascoli-Arzel\`a theorem,
\end{itemize}
and, moreover, 
\begin{itemize}
\item[3.a] the usual calculus of variations formulations require
that one postulate that the optimal curve is the graph of a function
$x\mapsto y(x)$,
\item[but]
\item[3.b]  with the optimal control control method this postulate
becomes a provable conclusion.
\end{itemize}

\medbreak\noindent These observations show that the first two of the 
three transitions brought
improvements to the analysis of Johann Bernoulli's problem.  But the core
of our argument is \S \ref{brachprinc}, where we look at the third
transition and the advent of differential-geometric control theory.
We will argue that, 

\medbreak\noindent on the one hand
\begin{itemize}
\item[4.a] previous methods require that we assume extra knowledge of physics
(in the form of the law of conservation of energy) in order even 
to be able to translate Johann's problem into mathematics, 
\item[and]
\item[5.a] with those methods an anomalous situation appears to arise
when one looks for a state space formulation of the problem and 
asks for its ``Hamilton principal function'' $V$ (that is, the value 
function regarded as a function of both the initial and terminal states),
since $V$ ought to be a function of an even number $\nu$ of variables,
but for the brachistochrone problem the best choice of $\nu$ seems
to be {\em five};
\end{itemize}
but on the other hand
\begin{itemize}
\item[4.b] in the differential-geometric framework one does not
need to bring in conservation of energy in order
to {\em formulate} the problem; instead, one writes
the equations as a control system in $\rr^4$, and then computes
the  accessibility distribution, which 
turns out to be three-dimensional at each point; this shows that 
a nontrivial conserved quantity exists locally, and direct computation
shows that it exists globally and is equal to the energy,
\item[and]
\item[5.b] the anomaly described in  5.a disappears, and the 
principal function $V$ turns out to be  defined on $\rr^4\times\rr^4$
(although $V(q,q')<+\infty$ if and only if $q$ and $q'$ belong to the same
leaf of the foliation of $\rr^4$ by the level 
submanifolds of the energy function); the method 
provides a full explanation for the apparent anomaly, by showing that
the full optimal synthesis requires impulse controls that lead to 
instantaneous jumps.
\end{itemize}

\medbreak\noindent The discussion in \S \ref{brachprinc} will also show
that the differential-geometric approach brings into the problem
interesting connections with other mathematical ideas. To begin with, the
analysis leads naturally to a minimum time optimal control problem with a
four-dimensional state space and an unbounded scalar control entering
linearly into the dynamics. Moreover, this problem is quite degenerate,
because its accessibility distribution happens to be three-dimensional, as
pointed out above.  Since the problem is affine in the control but not
linear, complete controllability on each leaf is in principle
problematic. The accessibility Lie algebra turns out to be
the ``diamond Lie algebra,'' a well known four-dimensional solvable Lie
algebra that occurs, for example, in the study of the quantized harmonic
oscillator.  To establish controllability on each leaf one needs additional
arguments, and it turns out that the issue can be settled by computing the
Jurdjevic-Kupka Lie saturate (cf. Bonnard {\it et al.}
\cite{bonnard,bonnard-b}, Jurdjevic-Kupka \cite{jukup-a,jukup-b}, Jurdjevic
\cite{jurdjevic}) of the family of system vector fields.  The minimum time
problem then makes sense in principle for every pair of points belonging to
the same leaf.  But, since the control is unbounded, it is not at all
obvious that an optimal control joining $A$ to $B$ exists for every pair
$(A,B)$ of points belonging to a given leaf $L$.  An application of the
Maximum Principle yields a vector field $Z$ tangent to the leaves such that
all optimal arcs are integral curves of $Z$. Since, in general, two
arbitrarily chosen points $A$, $B$ of a leaf $L$ do not lie on the same
integral curve of $Z$, it follows that for general pairs $(A,B)\in L\times
L$ an optimal control joining $A$ to $B$ does not exist.  On the other
hand, when a control system is completely controllable and a suitable
properness condition is satisfied\Ft{See Definition \ref{proper-cond}
for a precise statement of the properness condition, and Proposition
\ref{proper-brach} for the proof that the condition is satisfied
for the brachistochrone problem.}, then most
mathematicians would agree that optimal controls ought to exist, and if
they do not then this can only happen
because we have not made a good choice of our
space $\mathcal{U}$ of admissible controls, so $\mathcal{U}$ has to be
enlarged by adding suitably defined ``generalized controls.'' (For a
trivial example, consider the problem of moving from $A$ to $B$ in minimum
time with the dynamics $\dot x=u$, $u$ arbitrary, so in particular no bound
is prescribed for the velocity $u$. Clearly, the ``minimum time,'' is $0$,
and to attain it one needs to allow instantaneous jumps, which means
that the class of ordinary  controls has to be enlarged to allow
for Delta functions.  We will show in \S \ref{brachprinc} that something
like this happens for the brachistochrone.) In our case, the
true complete optimal synthesis of the problem turns out to require impulse
controls, a fact known to be common in problems with ``cheap'' unbounded
controls, and studied in other (linear quadratic) contexts by Jurdjevic and
Kupka, cf. Jurdjevic-Kupka \cite{jukup-c}, Kupka \cite{kupka-deg}.

Our second group of arguments in favor of modern control is presented in \S
\ref{fourvar}, and consists of ``five modern variations on the theme of the
brachistochrone.''  Following a suggestion made by M. Hestenes in 1956, we
``speculate on what type of brachistochrone problem the Bernoulli brothers
might have formulated had they lived in modern times'' (Hestenes
\cite{hestenespap}, p.~64), and we extend Hestenes' idea by speculating
not only on ``what type of $\ldots$ problem 
the $\ldots$ brothers might have formulated'' but also on 
{\em what type of answer they might have liked for such problems.}

We deal with five questions. The first one
is Hestenes' own ``variation of the brachistochrone problem,''
which turns out to be a problem solvable by optimal control but not by
classical calculus of variations techniques. 

The second ``variation'' is the ``reflected brachistochrone,'' a very
natural extension of Johann's problem, in which the state space is the
whole plane rather than a half plane. This problem is no longer solvable by
means of traditional smooth or nonsmooth optimal control methods---because
it is a minimum time optimal control problem with a non-Lipschitzian
right-hand side---but turns out to lend itself to the ``very nonsmooth''
approach, which yields a solution with a simple geometric interpretation
that, in our view, the Bernoulli brothers would have liked.

The third ``variation'' is the {\em brachistochrone problem with friction},
which corresponds, mathematically, to ``unfolding'' the original Johann
Bernoulli question by embedding it in a one-parameter family of problems,
the parameter being, of course, the friction coefficient $\rho$. These
problems turns out to be nondegenerate, in the sense that, for example,
their accessibility distribution has rank four at each point. Consequently,
their time-optimal controls behave qualitatively like the time-optimal
controls for generic control-affine systems in $\rr^4$ with scalar
unbounded controls. In particular, through every point there pass a
one-parameter family of optimal curves, since the optimal control is given
in feedback form as a function of the state and of an additional scalar
parameter that stays constant along each trajectory.  The analysis
clarifies the way in which the original problem of Johann Bernoulli
is degenerate: the accessibility Lie algebra is eight-dimensional
for $\rho\neq 0$ but four-dimensional for $\rho=0$; the 
accessibility distribution has rank four everywhere
if $\rho\neq 0$ but has rank three everywhere if $\rho=0$; the 
optimal ``feedback'' control is a function of the state and an extra 
parameter, but this parameter only occurs in a term that
is multiplied by $\rho$, so that when $\rho=0$ the optimal
control is a function of the state only. 

The fourth ``variation'' deals with the derivation of Snell's
law of refraction from general necessary conditions for an optimum.
Since Snell's law plays a crucial role in Johann's solution of the
brachistochrone problem, we speculate that a natural question
for him to have asked ``if he had lived in modern times'' is whether
the general theories about minimizing curves that were created
in the three centuries after  Johann's work actually cover
the Snell law case. Our answer is negative {\em if only the classical
calculus of variations and pre-1990 smooth and nonsmooth optimal
control are considered}, but becomes positive if we add
to the list our recent very nonsmooth methods 

To conclude the paper, we take the liberty of including a fifth
``variation'' that we find amusing but does not really have much to do
with control theory.  In this ``variation''---which is a digression on
a topic of great importance and related to geometry, to the
brachistochrone, and to Johann Bernoulli---we ``speculate'' that
Johann Bernoulli would have found it very interesting to
know\Ft{Although he probably would not have thought of asking the
question.}  that with his work on the brachistochrone he had been
extremely close to solving one of the most important open problems in
Geometry, namely, that of the independence of Euclid's fifth postulate
{}from the other ones. (The problem had intrigued mathematicians since
ancient times---cf., e.g., Bonola \cite{bonola}---and was very much in
the forefront of research in geometry in Johann's times, as shown by
the publication of Saccheri's book \cite{saccheri-euc} in 1733. A
definitive solution was found by Beltrami in 1868, cf. Beltrami
\cite{beltrami} and Stillwell \cite{stillwell-hyp}.)

\begin{remarksec}
{\rm An earlier, related version of the argument 
presented here appeared in \cite{sussjan}. That paper also dealt with
the evolution from the brachistochrone to the calculus of variations
and optimal control, but focused on a different set of issues, 
especially on the search for the correct Hamiltonian formulation of the 
necessary condition for an optimum and how opportunities were missed
before the ``control Hamiltonian'' was finally discovered in the 
1950s.}\proofend
\end{remarksec}

\section{Johann Bernoulli and the brachistochrone problem}\label{chalen}

\noindent In the June 1696 issue of the journal 
{\em Acta Eruditorum}, Johann Bernoulli---a Swiss-born 
professor of mathematics
at the University of Groningen, in the \mbox{Netherlands}\Ft{
The Bernoulli family was originally from Antwerp in Flanders, where they
lived until 1583.  
Like many other Flemish Protestants, they fled Flanders to escape religious
oppression by the Spanish rulers. They spent some time in Frankfurt, 
and finally settled in Basel, Switzerland, early in the seventeenth
century.}---challenged mathematicians to 
solve the ``problema novum'' of finding the curve that---following Leibniz'
suggestion---he named the ``brachistochrone,''  
{}from the Greek $\beta\rho\acute{\alpha}\chi\iota\sigma\tau
o\varsigma$: shortest, and $\chi\rho\acute{o}\nu o\varsigma$: time\Ft{
So the brachistochrone is the solution of a
{\bem minimum time} problem, i.e., of a genuine {\bem optimal
control} problem in the modern sense of the term!}.
The journal's May 1697 issue contains articles on this problem by six
of the most renowned mathematicians of the time: Johann Bernoulli,
Johann's elder brother Jakob, Leibniz, Newton, the marquis de
l'H\^{o}pital, and Tschirnhaus.  

Leibniz called Johann Bernoulli's question a ``splendid'' problem.
The fact that he and five other thinkers of such unparalleled
distinction chose to become actively engaged in the search for its
solution shows that at the time of its formulation the problem was
perceived by the leaders of the mathematical community as a major
challenge, lying at the very frontiers of their research.

It would not have been easy at that point to predict the future impact 
on science of Johann Bernoulli's challenge. Yet, the six famous authors of the
1697 papers must have had their own reasons to sense that an exceptionally
promising line of inquiry was being opened, and the developments of the 300
years that followed have lent ample support to their intuition.

Today, it is clear that the events of 1696-97 represented a critical
turning point in the history of mathematics, and it is widely agreed that
with them an important new field, later to be called the ``calculus of
variations,'' was born. 

\begin{remarksec}
{\rm The words ``calculus of variations'' were first
used by Euler in a 1760 paper, cf. \cite{euler}. Euler had received, in
1755, a letter from a 19-year-old mathematician from Turin, called Ludovico
de la Grange Tournier, where a new method was proposed for the study of
variational problems, by using what we would now call ``variations''
(cf. Goldstine \cite{goldstine}, p.~110.).  This was quite different from
the approach Euler had been using until then, and presented in his 1744
book entitled \textit{Methodus inveniendi maximi minimeve proprietate
gaudentes sive solutio problematis isoperimetrici latissimo sensu accepti},
that is ``A method for discovering curved lines having a maximum or minimum
property or the solution of the isoperimetric problem taken in its widest
sense.''  Euler's earlier method was based on a time-discretization
followed by a passage to the limit\Ft{Euler himself talked about ``letting
$h$ be infinitesimal.''}, as the size $h$ of the time intervals goes to
$0$.  Euler was so enthusiastic about the new idea of the unknown youngster
{}from Turin that he dropped his own method, adopted that of Lagrange
instead, and renamed the subject the {\it calculus of variations}.  In the
summary to his first paper using variations, Euler says ``Even though the
author of this [Euler] had meditated a long time and had revealed to
friends his desire yet the glory of first discovery was reserved to the
very penetrating geometer of Turin LA GRANGE, who having used analysis
alone, has clearly attained the same solution which the author had deduced
by geometrical considerations.''}\proofend
\end{remarksec}

The newborn subject was destined to become a major area of
mathematics. Moreover, its implications would soon extend to physics, whose
very foundations would be shown in the 18th century---by Euler, Maupertuis,
Lagrange, and others---to depend on principles formulated in the language
of the calculus of variations.

So it is not surprising that the story of the brachistochrone is told in
many books, and that the importance of this curve is stressed 
in most histories of mathematics.  The brachistochrone has 
received honors not commonly bestowed on mathematical curves.  For example,
it is depicted in a stained-glass window of the Academy building
(the main venue of the university) in Groningen.

Moreover, {\em the brachistochrone is probably the only mathematical
curve that has been deemed worthy of a monument in its honor.} In
1994, on the occasion of the 375th anniversary of its foundation, the
University of Groningen decided to honor in various ways the most
famous former members of its faculty.  In the case of Johann
Bernoulli, who had been a professor from 1695 to 1705, a decision was
made to honor him by erecting a monument to the brachistochrone, one
of the most significant discoveries---and one of which he was
particularly proud---that he made during his Groningen period.  The
unveiling of the brachistochrone monument, located in the Zernike
complex of the University, took place in 1996, coinciding with  the
300th anniversary of the publication of Johann's challenge in {\em
Acta Eruditorum}.

\subsection{Johann Bernoulli's challenge}\label{subsec-thechal}
The June 1696 challenge took the form of an ``Invitation to all
Mathematicians to solve a new problem'':
     
\bqsl If in a vertical plane two points $A$ and $B$ are given, then it
is required to specify the orbit $AMB$ of the movable point $M$, along
which it, starting from $A$, and under the influence of its own
weight, arrives at $B$ in the shortest possible time. So that those
who are keen of such matters will be tempted to solve this problem, is
it good to know that it is not, as it may seem, purely speculative and
without practical use. Rather it even appears, and this may be hard to
believe, that it is very useful also for other branches of science
than mechanics.  In order to avoid a hasty conclusion, it should be
remarked that the straight line is certainly the line of shortest
distance between $A$ and $B$, but it is not the one which is traveled
in the shortest time. However, the curve $AMB$---which I shall
disclose 
if by the end of this year nobody else has found it---is very
well known among geometers. 
\eqsl 

\noindent Later, Johann Bernoulli changed his original plan to disclose 
the solution by the end of 1696 and, following a suggestion by Leibniz,
extended the deadline until Easter 1697.

Responses came from the best mathematical minds of the time.  In addition 
to \mbox{Johann's} own solution, there was one by Leibniz, who had communicated
it to Johann in a letter dated June 16, 1696; another one by 
Jakob; one by Tschirnhaus; one by l'H\^{o}pital and,
finally, one by Newton. Newton's solution was presented to the Royal Society on
February 24, 1697 and published, anonymously and without proof, in the
{\em Philosophical Transactions}.  (The identity of the anonymous
author was clear to Johann Bernoulli, since he thought that you can know
{\em ex ungue leonem}, i.e., that you can know ``the lion from its claws.'')  
The May 1697 issue of {\em
Acta Eruditorum} contains articles with the solutions
by Johann \cite{bernoulli-a} and Jakob \cite{bernoulli-b},
as well as contributions by Tschirnhaus
and l'H\^{o}pital, a short
note by Leibniz, remarking that he would not reproduce his own
solution, since it was similar to that of the Bernoullis, and Newton's
anonymous paper, reprinted from the {\em Philosophical Transactions}.

\subsection{The cycloid's double role as the brachistochrone and the 
tautochrone}\label{sec-tauto}
As is apparent from Johann Bernoulli's announcement, he was under the
impression that the problem was new. However, Leibniz knew better: Galileo
in his book on the {\em Two New Sciences} in 1638 formulated the
brachistochrone problem and even suggested that the solution was
a circle.  (Galileo had showed---correctly---that an arc of a circle
always did better than a straight line---except, of course, when $B$
lies on the same vertical line as $A$---but he then wrongly concluded from
this that the circle is optimal.)

Johann Bernoulli considered the fact that Galileo had been wrong on two
counts, in thinking that the catenary was a parabola, and that the
brachistochrone was a circle, as definitive evidence of the superiority of
differential calculus (or the {\em Nova Methodus}, as he called this new
set of techniques,  following Leibniz).

He was thrilled by his discovery that the
brachistochrone was a cycloid. This curve had been introduced by Galileo
who gave it its name: {\em related to the circle}. Huygens had discovered
a remarkable fact about the cycloid: {\em it is the only curve with
the property that, when a body oscillates by 
falling under its own weight while being guided by this
curve, then the period of the oscillations is
independent of the initial point where the body is released}.\Ft{Contrary to
what Galileo thought, the circle has this property only approximately: the
period of oscillation of a pendulum is a function of its amplitude.}
Therefore Huygens called this curve, the cycloid, the {\em tautochrone}
(from $\tau\alpha\acute{\upsilon}\tau \acute{o}\varsigma$: equal,
and $\chi\rho\acute{o}\nu o\varsigma$: time).

Johann Bernoulli was amazed and somewhat puzzled, it
seems, by the fact that the same curve had these two remarkable properties
related to the time traveled on it by a body falling under its own weight:


\bqsl
Before I end I must voice once more the admiration that I feel for the
unexpected identity of Huygens' tautochrone and my brachistochrone.
I consider it especially remarkable that that this coincidence can take
place only under the hypothesis of Galilei [HJS \&{} JCW: 
that is, ``that the velocities
of falling bodies are to each other as the square roots of the altitudes
traversed''], so that we even obtain from this
a proof of its correctness. Nature always tends to act in the simplest way,
and so here lets one curve serve two different functions, while under any
other hypothesis we would need two curves, one for tautochrone
oscillations, the other for the most rapid fall. If, for example, the
velocities were as the altitudes, then both curves 
would be algebraic, the one a circle, the other one a straight line.
\eqsl

\ssubsection{Why was the brachistochrone so important in 1696-97?}
\label{subsec-specu}

Perhaps the crucial fact about Johann Bernoulli's question is that
he is telling us very clearly that {\bem the solution is 
a curve that everybody knows}, and yet {\em there appears to be no way 
to guess which curve it is by means of some 
intuitive geometric argument.} 

This situation should be compared with that arising in another famous
and much older question about curve minimization, the {\bem
isoperimetric problem}. In this problem, we are asked to find the
simple closed rectifiable curve that encloses the largest possible
area among all simple closed rectifiable curves of a given length, and
the answer also turns out to be a curve that is certainly ``very well
known among geometers,'' namely, a circle. But in this case it is
fairly easy to guess directly that the answer is a circle, even though
it is a delicate matter to give a rigorous proof.  For example, it
suffices to observe that the question is rotationally symmetric, so
the answer ought to be a rotationally symmetric curve, and the only
such curve is the circle.  Although this argument does not amount to a
completely rigorous proof\Ft{The argument assumes the existence of a
solution, as well as its uniqueness up to translations. Indeed, it
might conceivably happen that the solution is not unique, and in that
case the only conclusion one can draw from rotational symmetry is that
every rotation will map a solution $S$ to a solution $S'$, which may be
different from $S$.}, it is undoubtedly rather convincing, and makes
it quite obvious that the solution must be a circle.

Therefore, if Johann Bernoulli had challenged ``the best mathematical
minds'' of his time to solve the isoperimetric problem, it would not
have made sense for him to say, as he did in his brachistochrone
challenge, that ``the curve---which I shall disclose if by the end of
this year nobody else has found it---is very well known among
geometers,'' since everybo,y would have guessed immediately that the
mysterious curve was the circle.

In the case of the brachistochrone, {\bem there is no way to
guess}\Ft{Even now, 300 years later, we are not aware of any simple
geometric argument that might lead to the cycloid without using
Calculus.}.  The problem was posed and solved at a time when the
Calculus was being invented, and cannot be solved without using
Calculus.  So it provides an impressive illustration of the power of
the Calculus to solve problems that cannot be solved in any other way,
and to discover truths about the world that can be stated and
understood in ``well known'' terms\Ft{This is an important point.
Compare the brachistochrone, for example, with Newton's 1685 work on
the shape of a body with minimal drag. In this case, the solution also
requires the Calculus, but once we have solved the problem all we find
is a curve defined by some strange formula.  This curve was not ``well
known to geometers,'' and would not have had any particular
significance to mathematicians, who would only retain from going
through Newton's solution the fact that the solution is ``some
curve.''  In the case of the brachistochrone, one does not need the
Calculus to explain what the solution is, since cycloids were already
well understood curves, but one does need it to discover that the
solution is a cycloid.}, but can only be arrived at by using this new
mathematical tool.

\section{The standard formulation and Johann Bernoulli's 
solution}\label{standardfor}

\noindent We now discuss the ``standard formulation'' of the brachistochrone
problem in modern mathematical language, as it is usually found in
modern books, and Johann Bernoulli's ingenious solution based 
on an analogy with geometrical optics.  

In our view, the standard formulation is questionable. We will explain
our objections in \S \ref{brachprinc} below, where we will make a case for
a different approach, based on differential-geometric control. But
first it is necessary to present the classical point of view in
detail.

The key fact is that the ``movable point $M$'' is supposed to move ``under
the influence of its own weight,'' which presumably means ``under the
influence of its own weight {\em and nothing else other than whatever
force is needed to keep $M$ on the given curve}.'' This means, in
particular, that {\em energy is conserved}. 

\ssubsection{Formulation of the brachistochrone problem}\label{stanformu}
We choose $x$ and $y$ axes in the plane with the
$y$ axis pointing downwards, use $(a,\alpha)$ and $(b,\beta)$ to
denote, respectively the coordinates of the endpoints $A$ and
$B$ of our curve, and fix a number $E\in\rr$, thought of as the total
energy of the particle. The condition characterizing the feasible paths
\begin{equation}\label{thepath}
[0,T]\ni t\mapsto\xi(t)=(x(t),y(t))\in\rr^2
\end{equation}
will then be---using $m$, $g$ to denote, respectively,
the particle's mass and the gravitational acceleration---that the 
sum of the kinetic energy 
$\frac{m}{2}(\dot x(t)^2+\dot y(t)^2)$ and the potential energy 
$-gmy(t)$ is equal to $E$. We can certainly choose units so that
$m=1$. The condition then becomes
$$
\frac{1}{2}(\dot x(t)^2+\dot y(t)^2)-gy(t)=E\,.
$$
So we formally define 
an $(A,B,E)$-{\em feasible trajectory} (or $(A,B,E)$-{\em feasible path})
to be a map (\ref{thepath}), defined on some interval $[0,T]$,  such that

\medbreak
\bi
\item[(i)] $\xi(0) = A$, $\xi(T)=B$, 
\item[(ii)] $\xi$ is absolutely continuous,
\item[(iii)] $\frac{1}{2}(\dot x(t)^2 + \dot y(t)^2) = E+gy(t)$
for almost all $t\in [0,T]$.
\ei

\medbreak\noindent Notice that Condition (iii) implies in particular that 
the derivatives $\dot x$ and $\dot y$ are bounded, so a feasible path
is in fact Lipschitz continuous. 

Condition (i) states that the path $\xi$ must start at $A$ and end at
$B$. The number $E+g\alpha$ is the initial kinetic energy of the body.
(Often, only the case $E+g\alpha=0$ is considered, corresponding to
the body being dropped at time $0$ with zero initial speed.)
Condition (ii) is of course a modern technicality that would have made
no sense to Johann Bernoulli, who was far from imagining that there
could exist curves that cannot be recovered by integrating their
derivative.  Condition (iii) reflects conservation of energy.  (The
law that the kinetic energy of a body which has fallen from a height
$h$ is increased by an amount proportional to $h$ was due to Galileo,
and was well known in Bernoulli's time.)

\medskip

An $(A,B,E)$-feasible path $\xi:[0,T]\rightarrow \rr^2$
is said to be {\em optimal} if there exists no $(A,B,E)$-feasible path 
$\tilde\xi:[0,\tilde T]\rightarrow \rr^2$ for which 
$\tilde T<T$. An $(A,B,E)$-{\em brachistochrone} is either (a)~an
optimal $(A,B,E)$-feasible path, or (b)~the geometric locus in $\rr^2$ of
such a path, that is, a subset $\mathcal{B}$ of $\rr^2$ such that
$$
\mathcal{B}= \{(x,y) \in \rr^2:\mbox{ there exists } t\in [0,T^\ast] \mbox{,  such
that } (x,y) = \xi^\ast(t)\}\,,
$$
for some optimal $(A,B,E)$-feasible path $\xi^\ast:[0,T^\ast]\mapsto{}\rr^2$. 
A {\em brachistochrone} is an object which is an
$(A,B,E)$-brachistochrone for some $A,B,E$. 

Obviously, a feasible path\Ft{From now on, as long as $A$, $B$ and $E$ 
are fixed, we just talk about ``feasible paths'' and 
``brachistochrones,'' without explicitly specifying  $A$, $B$ and $E$.}
(\ref{thepath})  must be entirely contained in the closed half-plane
\begin{equation}
H_+(\eta)=\{(x,y):y\ge-\eta\}\,,\,\,\,\,\,\,\,{\rm  where\,\,\,\,\,\,}
\eta=\frac{E}{g}\,,
\end{equation}
since, for $y(t)$ to be $<-\eta$,  the kinetic energy would have to be
negative, which is of course impossible.

It is clear that {\sl the solution
cannot always be a straight line}, as 
Johann Bernoulli rightly points out in his ``Invitation.''
For example, consider the extreme  case when
$E=0$ and $a=\alpha=\beta=0\neq b$. (That is, when the endpoints of
the curve are at the same height but do not coincide, and the initial
kinetic energy vanishes.) We will show that in
this situation there is a 
feasible path that goes from $A$ to $B$ in finite time. Since 
the straight-line segment $S$ from $A$ to $B$ 
is horizontal, (iii) implies that the speed of motion along $S$ vanishes,
and then it will follow that $S$ cannot be an optimal path, because the motion
along $S$ takes infinite time. To construct a feasible path from $A$ to
$B$ we follow a circle $C$ with center on the $x$ axis.  The circle can
be parametrized by $x$, taking values in the interval $[0,b]$. The speed
$v$ of motion along $C$ will be proportional to $\sqrt{y}$, and hence
bounded away from $0$ as long as $x$ is bounded away from $0$ and $b_1$. Near
$A$, we can also use $y$ as a parameter, and then $y\downarrow 0$ as
$x\downarrow 0$, and $v$ behaves like $\frac{dy}{dt}$ and also like
$\sqrt{y}$, so $\frac{dt}{dy}\sim\frac{1}{\sqrt{y}}$, showing that $t$
stays finite as $x$ and $y$ approach $0$, since the function
$y\mapsto{}\frac{1}{\sqrt{y}}$ is integrable near $y=0$.
A similar argument shows that $t$
stays finite near $B$. So $C$ joins $A$ to $B$ in finite time. \proofend

\subsection{Johann Bernoulli's solution}\label{subsec-johann-sol} It 
turns out that
the brachistochrone is a cycloid.  When $E=0$, it is the 
curve described by a point $P$ in a circle that rolls without slipping on
on the $x$ axis, in such a way that $P$
passes through $A$ and then through $B$, without hitting 
the $x$ axis in between.  (It is easy to see that this defines the cycloid
uniquely.)

Johann Bernoulli's ingenious derivation of the brachistochrone has been the
subject of numerous accounts, but since this event plays a crucial
role in our own tale, we will briefly tell the story again.

Johann Bernoulli based his derivation on Fermat's minimum time principle.
If we imagine for a moment that instead of dealing
with the motion of a moving body we are dealing with a light ray, Condition
(iii) above gives us a formula for the ``speed of light'' $c(x,y)$ as a 
function of the position $(x,y)$: 
$$
c(x,y)=\sqrt{2E+2gy}\,.
$$

We now change coordinates so
that $E=0$.  (This amounts to moving the $x$ axis upward to a height
$\eta$.) Then $\eta$ is now equal to $0$, so our feasible paths live
in the half plane $H_+\equaldef H_+(0)$. Also, let us rescale---or, if the
reader so 
prefers, ``change our choice of physical units''---so that $2g=1$. Then our
problem is exactly equivalent to that of determining the light
rays---i.e., the minimum-time paths---in a plane medium where the speed of
light $c$ varies continuously as a function of position according to the
formula 
$$
c(x,y)=\sqrt{y}\,.
$$

It is then reasonable to expect that, if we use a suitable
discretization of our problem, then the optimal paths of the
discretized problem will approach those of the original problem as the
discretization parameter $\delta$ goes to $0$. Johann Bernoulli
did this as follows. Let us define 
$$
y_k^\delta=k\delta\quad{\rm for}\quad\delta>0\,\,{\rm and}\,\,k=0,1,\ldots\,,
$$
and then divide the half-plane $H_+$ into horizontal strips 
$$
S_k^\delta\equaldef\{(x,y)\in\rr^2:y_k^\delta\le y\le y_{k+1}^\delta\}
\quad{\rm for}\quad k=0,1,\ldots\,,
$$ 
of height $\delta$. Johann's discretization is then obtained
by replacing $c$ by a constant $c_k^\delta$ 
in each strip $S_k^\delta$. The constants $c_k^\delta$ are defined by
$$
c_k^\delta=\sqrt{y_{k+1}^\delta}
\quad{\rm for\quad each}\quad k\,.
$$
He then computed the light rays for the original problem 
by taking the limit of the light rays for the discretized problem 
as $\delta\downarrow 0$.  

The light rays of the discretized problem
can be studied using \snell{}'s law.  Clearly, the paths will be straight-line
segments within each individual strip, and all that needs to be done is to
determine how these rays bend as they cross the boundary between two
strips. The answer is provided by the laws of optics as developed by
\snell{}, Descartes, Fermat, Leibniz and Huygens.

\snell{} had observed that, if the speeds of light 
on the two sides $H_1$, $H_2$ of a line $L$ in the plane
are different constants $v_1$, $v_2$, 
and a light ray $R$ traverses $H_1$ and then  enters $H_2$ after being
refracted at $L$, then the ratio 
$\frac{\sin \theta_1}{\sin \theta_2}$
of the sines of the incidence and refraction
angles is a constant, independent of $\theta_1$. 
(By definition, $\theta_i$ is, for $i=1,2$, 
the angle formed by the part $R_i$ of $R$ that lies in
$H_i$ with the line perpendicular to $L$.  It is clear that
the $R_i$ are straight lines.) 
Fermat subsequently showed that this is precisely what
happens when light is assumed to follow a minimum-time path. The 
law relating the incidence angles to the velocities of propagation
is due to Leibniz and Huygens, and implies the law of \snell{}. It says that
\begin{equation}\label{snellaw}
\frac{\sin \theta_1}{\sin \theta_2} = \frac{v_1}{v_2}\,\,\,\,\,\,\,
\mbox{ or, equivalently,  }\,\,\,\,\,\,
\frac{\sin \theta_1}{v_1} = \frac{\sin \theta_2}{v_2}\,,
\end{equation}

Johann Bernoulli used (\ref{snellaw}) to study the light rays of the 
discretized problem. If $R$ is such a ray, and we let $R_k$ be the part 
of $R$ that lies in the strip $S_k^\delta$, and use $\theta_k^\delta$ to denote
the angle of $R_k$ with a vertical line, then (\ref{snellaw}) 
enabled him to conclude that the quantity 
$$
\nu=\frac{\sin{\theta_k^\delta}}{\sqrt{y_{k+1}^\delta}}
$$ 
is a constant, since 
in each strip $S_k^\delta$ the speed of our light ray is 
$\sqrt{y_{k+1}^\delta}$.


Passing to the limit as $\delta\downarrow 0$, we conclude that
the sine of the angle $\theta$ between the tangent to the brachistochrone
and the vertical axis must be proportional to $\sqrt{y}$. Since 
$$
\sin\theta=\frac{dx}{\sqrt{dx^2+dy^2}}\,,
$$
we find that
$$
\frac{dx^2}{dx^2+dy^2}=\nu y\,,
$$
where $\nu$ is a constant. Then
$$
\frac{dx^2+dy^2}{dx^2}=\frac{1}{\nu y}\,,
$$ 
i.~e., 
$$
1+y'(x)^2=\frac{C}{y}\,,\,\,\,{\rm where}\,\,\,C=\frac{1}{\nu}\,. 
$$
So the curve described by expressing the
$y$-coordinate of the brachistochrone as a function of its $x$-coordinate
will satisfy the differential equation
\begin{equation}\label{diffeq}
y'(x) = \sqrt{\frac{C-y(x)}{y(x)}}\,,
\end{equation}
for some constant $C$.

The parametrized curves $\varphi\mapsto(x(\varphi),y(\varphi))$ given by 
\begin{equation}\label{eqcycloid}
x(\varphi)  =  x_0+\frac{C}{2}(\varphi - \sin \varphi) \,,\,\,\,\,\,\,\,\,\,\,
y(\varphi) =  \frac{C}{2} (1 - \cos \varphi)\,,\,\,\,\,\,\,\,\,\,\,
0\le\varphi\le 2\pi\,, 
\end{equation}
satisfy (\ref{diffeq}). It is easily seen that these
equations specify the cycloid generated by a point $P$ on a circle of 
diameter $C$ that rolls without slipping on the
horizontal axis, in such a way that $P$ is at $(x_0,0)$ when
$\varphi=0$.   Moreover, it is also easy to check that 
\bi
\item[(*)] {\em given two points $A$ and $B$ in $H_+$ there is exactly one 
curve passing through $A$ and $B$ and belonging to
the family of curves (\ref{eqcycloid}).}
\ei 
(The parameter count for (*) is obviously right: (\ref{eqcycloid})
describes a 2-parameter family of curves; requiring that the curve
pass through $A$ fixes one of the parameters, and then asking that it
pass through $B$ fixes the other.  Proving (*) requires a bit more
work, but is still quite easy.)

\section{Spurious solutions and the calculus of variations 
approach}\label{spurious}

\subsection{Remarks on Johann Bernoulli's argument}  The argument that we
have presented is the one of Johann Bernoulli, and 
Equation (\ref{diffeq}) is the one that he wrote in his paper, followed
by the statement ``from which I conclude that the {\em Brachistochrone}
is the ordinary {\em Cycloid}.''   (He actually wrote
$dy=dx\sqrt{\frac{x}{a-x}}$, but he was using $x$ for the vertical
coordinate and $y$ for the horizontal one. Cf. \cite{struik}, p.~394.) 
In present day notation, it is customary to adopt the convention that 
the symbol $\sqrt{r}$ stands for the {\em nonnegative} square root 
of $r$, but it is obvious that Johann Bernoulli did not have this in mind. 
What he meant was, clearly, what we would write today as 
\begin{equation}\label{plumin}
y'(x) = \pm\sqrt{\frac{C-y(x)}{y(x)}}\,,
\end{equation}
or, equivalently, 
\begin{equation}\label{diffeqac}
y(x)(1+y'(x)^2)={\rm constant}\,.
\end{equation}
In particular, the solution curves should be allowed to have a negative slope,
contrary to what a ``modern'' interpretation of (\ref{diffeq}) might
suggest.  But $y'$ should stay continuous, so that a switching from a 
$+$ to a $-$ solution of (\ref{plumin}) is not permitted. 

Even with the more accurate rewriting (\ref{diffeqac}),  the differential
equation derived by Johann Bernoulli also has {\bem spurious solutions},
not given by (\ref{eqcycloid}). Indeed, for any $\bar y\!>\!0$, the 
constant function $y(x)\!=\!\bar y$ is a solution, corresponding to 
$C\!\!=\!\!\bar y$.   More generally, one can take an ordinary cycloid
given by  (\ref{eqcycloid}), follow it up to $\varphi\!\!=\!\!\pi$---so that
$dy/dx\!\!=\!\!0$---then follow the constant solution $y(x)\!\!=\!\!C$ for an
arbitrary time $T$, and 
then continue with a cycloid given by (\ref{eqcycloid}).   
Such paths are, indeed, compatible with Huygens' law of refraction.  

\begin{remark}{\rm 
The existence of the spurious solutions was pointed out by Brook Taylor
(1685-1731) in 1715. 
(Taylor is known to mathematicians as the discoverer
of ``Taylor's expansion,''  the inventor of integration by parts, 
and the creator of the calculus of finite differences.  
Like many other mathematicians of his time, he was involved in a priority
dispute with Johann Bernoulli, involving Taylor's 1708 solution of the 
problem of the center of oscillation, that went unpublished until 
1714.) \proofend
}\end{remark}

\ssubsection{The calculus of variations approach} 
The spurious solutions, and all the other problems, such as the apparent
arbitrariness of the requirement that $y'$ be continuous, can be eliminated
in a number of ways. For example, one can prove directly that the spurious
trajectories are not optimal.  Or one can use, as an alternative to Johann
Bernoulli's method, the calculus of variations approach, based on writing
the Euler-Lagrange equation. This equation gives a necessary
condition for a function
$y_*$ to minimize the integral 
$$
J=\int_a^bL(y(x),y'(x),x)\,dx
$$
within the
class $\mathcal{Y}$ of all functions 
$$
[a,b]\ni x\mapsto y(x)\in\rr
$$
that satisfy some appropriate technical conditions (for example, the
requirement that $y(\cdot)$ is Lipschitz continuous)
and are such that $y(a)$ and $y(b)$ have given values $\alpha$, $\beta$.

Here, the ``Lagrangian'' 
$$
\Omega\times\rr\times[a,b]\ni(y,u,x)\mapsto L(y,u,x)\in\rr
$$
is a given function, and $\Omega$ is an open interval in $\rr$. 
Under suitable technical conditions on $L$, $y_*$, and the set $\mathcal{Y}$,
if $y_*$ is a minimizer then the condition
\begin{equation}\label{eulagfirst}
\frac{d\,\,}{dx}\Biggl(\frac{\partial L}{\partial u}
\Big(y_*(x),y_*'(x),x\Big)\Biggr)=
\frac{\partial L}{\partial y}\Big(y_*(x),y_*'(x),x\Big)
\end{equation}
must be satisfied for almost all $x$.

It is easy to see that the above result can be applied to the
brachistochrone problem, provided that we {\bem postulate}\Ft{With
optimal control, this ``postulate'' becomes a {\em
provable conclusion}, as will be shown in \S \ref{opti-ap} below.}   that 
it suffices to consider curves  in the $x,y$ plane that are graphs of
functions $x\mapsto y(x)$ defined on  the interval 
$[a,b]$.     Then Constraint (iii) can be written (taking $E=0$
and $2g=1$ as before) as  
$$
dx^2+dy^2=y\,dt^2\,,
$$
which gives
$$
dt=\frac{\sqrt{dx^2+dy^2}}{\sqrt{y}}=L(y,y')dx\,,
$$
where
\begin{equation}\label{eqforlag}
L(y,u)=\frac{\sqrt{1+u^2}}{\sqrt{y}}\,.
\end{equation}
So Johann Bernoulli's problem becomes that of minimizing the integral
$$
J=\int_{a}^{b}L(y(x),y'(x))dx\,,
$$
subject to $y(a)=\alpha$ and $y(b)=\beta$, where $L$ is given by 
(\ref{eqforlag}).

This gives the equation 
\begin{equation}\label{truediffeq}
1+y'(x)^2+2y(x)y''(x)=0\,, 
\end{equation}
which is {\em stronger} than (\ref{diffeqac}), since (\ref{diffeqac}) is
equivalent to $y'+{y'}^3+2yy'y''=0$, i.e., to $y'(1+{y'}^2+2yy'')=0$, whose
solutions are those of (\ref{truediffeq}) plus the spurious solutions found
earlier. 

It is easy to verify that the solutions of the Euler-Lagrange equation
(\ref{truediffeq}) are {\em exactly} the curves given by (\ref{eqcycloid}),
without any extra spurious solutions, showing that,
for the brachistochrone problem,  {\bem the
Euler-Lagrange method gives better results than Johann Bernoulli's 
approach.}  

We now show that {\bem optimal control is even better.}  

\section{The optimal control approach}\label{opti-ap}

\noindent Having shown that the Euler-Lagrange equation 
gives better results than Johann Bernoulli's method, we now 
look at another important step in the evolution of the theory of 
minimizing curves, namely, optimal control theory. We show that 
optimal control provides even better results in two ways:

\begin{itemize}
\item[1.] With optimal control it is no longer necessary to postulate
that the optimal curve is the graph of a function $x\mapsto y(x)$.
\item[2.] In the optimal control setting, the problem of the existence 
of an optimum becomes trivial.
\end{itemize}

\subsection{The brachistochrone as an optimal control
problem} We can formulate Johann Bernoulli's question as an optimal
control problem, in which the motion 
takes place in the $x,y$ plane, and the dynamical behavior is given by
\begin{equation}\label{braccontrol}
\dot{x}=u\sqrt{|y|}\,,\,\,\,\,\,\dot{y}=v\sqrt{|y|}\,. 
\end{equation}
Here the control is a 2-dimensional vector $(u,v)$ taking values in
the set 
\begin{equation}\label{braccontrol-xxx}
U=\{(u,v):u^2+v^2=1\}\,.
\end{equation}
(Actually, for the ``true'' brachistochrone problem
the motion is restricted to the upper half plane, so we could have
written $\sqrt{y}$ rather than $\sqrt{|y|}$. We choose to
use the more general 
expression $\sqrt{|y|}$ because later, in the study of the reflected
brachistochrone, we will want to work in the whole plane.)

Let us use $p$, $q$, $p_0$ to denote, respectively,
the momentum variables conjugate to $x$ and $y$, and the abnormal multiplier.
Then the control theory Hamiltonian 
$H(x,y,u,v,p,q,p_0,t)$  is given (using
$\sigma={\rm{sgn}}\,y$) by the formula 
$$
H(x,y,u,v,p,q,p_0,t)=(pu+qv)\sqrt{\sigma y}-p_{0}\,.
$$
The Pontryagin Maximum Principle then tells us that if a curve
$$
[0,T]\ni t\mapsto\xi(t)=(x(t),y(t))
$$ 
is optimal then there exist absolutely continuous functions
$t\mapsto p(t)$ and $t\mapsto q(t)$, and a nonnegative constant $p_0$, 
such that, if we let $u=\dot x$, $v=\dot y$, 
and write $\vec{p}$ to denote the momentum vector $(p,q)$, and 
$\|\vec{p}\|$ to denote its Euclidean norm $\sqrt{p^2+q^2}$, 
then the Hamiltonian maximization conditions 
\begin{equation}\label{eqforuv}  
u(t)=\frac{p(t)}{\|\vec{p}(t)\|}\,, \,\,\,\,\,\,v(t)=\frac{q(t)}{\|
\vec{p}(t)\|}\,, 
\end{equation}
as well as the ``adjoint system'' of differential equations
\begin{equation}\label{eqpdot}
\dot{p}(t)=0\,,\,\,\,\,\,\,\,\dot{q}(t)=-\sigma\,\frac{p(t)u(t)+q(t)v(t)}
{2\sqrt{\sigma{}y(t)}}=-\frac{\sigma\|\vec{p}\|}{2\sqrt{\sigma y(t)}}\,,
\end{equation} 
are satisfied for almost all $t$. 
(Notice that $\|\vec{p}(t)\|\neq 0$. Indeed, the Maximum Principle also 
gives the condition that $H=0$. So 
$\|\vec{p}\|=0$ would imply $p_0=0$, contradicting the nontriviality of 
the triple $(p(t),q(t),p_0)$.)

If the constant $p$ vanishes, then $\dot x\equiv 0$, so we get a vertical
line. Otherwise, $\dot x$ is always $\neq 0$, showing that we can use 
$x$ to parametrize our solution.  Since 
\begin{equation}
y'(x)=\frac{dy}{dx}=\frac{\dot y}{\dot x}=\frac{v}{u}=\frac{q}{p}\,,
\end{equation}
we have 
\begin{equation}\label{ifwemul}
1+y'(x)^2=\frac{\|\vec{p}\|^2}{p^2}
\end{equation}
and 
\begin{equation}\label{eqfory2prime}
y''(x)=\frac{1}{p}\cdot\frac{dq}{dx}
=\frac{\dot q}{p\dot x}\,. 
\end{equation}
But (\ref{braccontrol}) and (\ref{eqforuv}) imply that 
\begin{equation}
\dot x=\frac{p\sqrt{\sigma y}}{\|\vec{p}\|}\,,
\end{equation}
and then (\ref{eqpdot}) and (\ref{eqfory2prime}) yield
\begin{equation}
y''(x)=-\frac{\|\vec{p}\|^2}{2 yp^2}\,. 
\end{equation}
So 
$2yy''=-\frac{\|\vec{p}\|^2}{p^2}=-(1+{y'}^2)$, and then 
\begin{equation}
1+{y'}^2+2yy''=0\,, 
\end{equation}
which is exactly 
Equation (\ref{truediffeq}). As before, this leads
to the cycloids, with no ``spurious solutions.''  Notice that this
argument does not involve any 
discretization or any use of \snell{}'s law of refraction.

Notice also that in our control argument {\em we have not postulated
that the solution curves could be represented as graphs of functions
$y(x)$.  We have proved it!} This should be contrasted 
with the analysis based on the calculus of variations, where 
this fact had to be postulated.)

This is one example showing that, for the
brachistochrone problem, 
{\bem the optimal  control method gives better results than the
classical calculus of variations}.

\ssubsection{Optimality proofs and the existence problem} 
\label{subsec-existence} 

For a second example showing the advantages of the optimal control
method over 
the classical calculus of variations for the specific case of the
brachistochrone problem, we turn to the question of the rigorous proof of
the optimality of Bernoulli's cycloids. Clearly, no argument based solely on
necessary conditions for optimality will ever prove that a trajectory is
optimal, because it could happen, for example, that there are no optimal
trajectories, in which case the statement that ``every optimal trajectory
is a cycloid'' would be vacuously true.  If we really want
to prove the optimality of Bernoulli's cycloids, an extra step is needed.

To give a rigorous proof of the optimality of the cycloids, valid 
for all choices of  $A$ and $B$ in the closed half-plane 
$H_+\!=\!\{x,y)\!:\!y\!\ge\! 0\}$, one can proceed in two ways.

One possibility (cf. Bliss \cite{bliss}) is to use Hamilton-Jacobi theory.
Another, much simpler, approach, is to prove first 
the {\em existence} of an optimal trajectory. Indeed, once this 
existence result is known
for all $A,B\in H_+$, then it is clear that any optimal trajectory 
$\xi_*:\![0,T]\mapsto H_+$ going from $A$ to $B$ must be such that its
restriction 
$\xi_*\restr[\alpha,\beta]$  to
every subinterval $[\alpha,\beta]$ of $[0,T]$ is also optimal 
for the problem $P(\alpha,\beta)$ with endpoints $\xi_*(\alpha)$,
$\xi_*(\beta)$.    
(This is a rather obvious point, known in optimal  control theory as the
``principle of optimality,'' and noticed in the case of the
brachistochrone by Jakob Bernoulli.)  If one assumes that 
$\xi_*(\alpha)$ and  $\xi_*(\beta)$ are interior points of
$H_+$ for $0<\alpha<\beta<T$, 
then the necessary conditions for optimality give a unique candidate 
for the solution of $P(\alpha,\beta)$, namely, the unique curve in the
family given by 
(\ref{eqcycloid}) that goes through 
$\xi_*(\alpha)$ and  $\xi_*(\beta)$. 
So $\xi_*$ is such that $\xi_*\restr[\alpha,\beta]$  is given by
(\ref{eqcycloid}) whenever $0<\alpha<\beta<T$, 
and this easily implies that $\xi_*$ itself is a cycloid
given by (\ref{eqcycloid}).   


To complete the argument, one has to exclude the possibility of a solution 
of the problem with endpoints $A$, $B$ which intersects the $x$ axis
at some other point. This can be done by a simple qualitative argument,
using the following lemma.
\begin{lemma}\label{the-lem}
Suppose $L$ is a horizontal line contained in the interior of the upper
half-plane $H_+$.   Let $P$, $Q$ be distinct points of $L$. Then the
straight-line segment joining $P$ to $Q$ is strictly 
faster than any trajectory $\zeta$ 
{}from $P$ to $Q$ such that $\zeta$ lies entirely in the lower
closed half-plane determined by $L$ and at least one point
of $\zeta$ lies strictly below $L$.
\end{lemma}

\begin{proof}
Let $T_\xi$, $T_\zeta$ be the corresponding times. 
Let $\bar y$ be the $y$ coordinate 
of the points of $L$, so $\bar y>0$. Write $P=(p,\bar y)$, 
$Q=(q,\bar y)$. 
Let $\xi:[0,T]\mapsto H_+$ be the horizontal segment from $P$ to $Q$.
Then 
$$
T_\xi=\frac{|q-p|}{\sqrt{\bar y}}\,.
$$
Now let 
$[a,b]\ni t\mapsto\zeta(t)=(x(t),y(t))\in H_+$ be 
a trajectory from $P$ to $Q$ as in our statement. Then
$$
\dot x(t)\le\|\dot\zeta(t)\|=\sqrt{y(t)}\le\sqrt{\bar y}\,,
$$
and the inequality is strict for almost all $t$ in some interval
$I$ of positive measure. So 
$$
|q-p|=|x(b)-x(a)|=\Big|\int_a^b\dot x(t)\,dt\Big|
<(b-a)\sqrt{\bar y}=T_\zeta\sqrt{\bar y}\,.
$$
Therefore
$$
T_\zeta>\frac{|q-p|}{\sqrt{\bar y}}=T_\xi\,,
$$
and our proof is complete.\proofend
\end{proof}

The following is an immediate consequence of the lemma.
\begin{corollary}
Let $A$, $B$ be points of $H_+$ such that $A\neq B$, and assume that
$\xi:[a,b]\mapsto H_+$ is time optimal. Then 
$\xi(t)$ is an interior  point of $H_+$ whenever 
$a<t<b$. 
\end{corollary}
\begin{proof}
Let $t$ be such that $a<t<b$. Assume $\xi(t)$ belongs to the boundary
$\partial H_+$ of $H_+$. Then it cannot be the case that 
$\xi(s)\in\partial H_+$ for every $s\in[a,t]$, because
if this was the case then the speed of motion on the interval
$[a,t]$ would vanish, so $\xi(a)=\xi(t)$,  and then
we could obtain a trajectory with the same endpoints
as $\xi$ and strictly faster than $\xi$ by just 
replacing $\xi$ by its restriction to $[t,b]$.
A similar argument shows
that it is not true that $\xi(s)\in\partial H_+$ for all 
$s\in[t,b]$. It then follows that there exist a positive
number $\delta$, and times $s_1$, $s_2$ such that
$a\le s_1<t<s_2\le b$, having the property
that $\xi(s_1)\in L$ and $\xi(s_2)\in L$, where 
$L$ is the line $\{(x,y):y=\delta\}$. The set 
$S=\{s\in[a,b]:\xi(s)\notin L\}$ is a relatively open subset
of the interval $[a,b]$. Let $I$ be the connected component
of $S$ that contains $t$. Then $I$ is an interval, and
$I$ is relatively open in $[a,b]$. Since
$a\le s_1<t<s_2\le b$, $s_1\notin S$, and $s_2\notin S$, the
interval $I$ is in fact open in $\rr$, so 
$I=]\tau_1,\tau_2[$ where $s_1\le\tau_1<t<\tau_2\le s_2$. 
Let $\zeta$ be the restriction of $\xi$ to the compact
interval $[\tau_1,\tau_2]$. Then Lemma \ref{the-lem} implies
that $\zeta$ is not optimal, contradicting the fact that
$\xi$ is optimal. Therefore $\xi(t)\notin\partial H_+$.\proofend
\end{proof}

So everything hinges upon the existence result.  It turns out that
this is not at all trivial if the brachistochrone problem is
reformulated in terms of the classical calculus of 
variations\Ft{We thank F. H. Clarke for bringing this point
to our attention.}, because

\begin{itemize} \item[(a)] the Lagrangian $L$ given by
$L(y,u)=\frac{\sqrt{1+u^2}}{\sqrt{y}}$ has a
singularity at $y=0$,
\end{itemize}
and, even more importantly,
\begin{itemize} \item[(b] $L$ is not sufficiently coercive
for the usual existence theorems to apply.
\end{itemize}

\noindent(The precise meaning of (b) is that, as a function of 
$u$, $L$ just grows like a constant times $|u|$, whereas for the usual
existence theorems to apply 
$L$ would have to grow like a constant times $|u|^r$ for
some $r$ such that $r>1$.
More \mbox{sophisticated} existence theorems for classical calculus of
variations problems that include the brachistochrone are stated
in Cesari \cite{cesari}, \S 14.3 and \S 14.4,
but these conditions are not at all simple, and in addition they depend
on other results by Cesari that were to appear in a book that
was never published.)

In the control theory setting, however, the existence question is
easily settled by a completely trivial application of the
Ascoli-Arzel\`a theorem: given $A,B\in H_+$, let $T$ be the infimum of
the times of all trajectories going from $A$ to $B$; let $\{\xi_j\}$
be a sequence of trajectories from $A$ to $B$, defined on intervals
$[0,T_j]$, such that $T_j\downarrow T$. Then the $\xi_j$ are all
contained in a fixed compact subset of $H_+$, as can easily be shown
by applying Gronwall's inequality and using the fact that the
right-hand side of  (\ref{braccontrol})
has less than linear growth as a function
of $x$ and $y$. Then  (\ref{braccontrol}) implies that  the derivatives
$\dot\xi_j$ are uniformly bounded as well. So we can use the
Ascoli-Arzel\`a theorem to extract a subsequence
that converges uniformly to a curve $\xi:[0,T]\mapsto H_+$.  In
principle, $\xi$ is just a trajectory of the ``convexified'' system,
in which the evolution equations are those of (\ref{braccontrol}) but
the control constraint is $u^2+v^2\le 1$ rather than $u^2+v^2=1$.
However, it is easy to see that $\xi$ is a true trajectory of the
nonconvexified system because, if the inequality $u(t)^2+v(t)^2\le 1$
was strict on a set of positive measure, then it would be possible to
find an even faster trajectory from $A$ to $B$ by just reparametrizing
$\xi$.  So we see, once again, that {\em in the control setting a
question that is hard from the point of view of the calculus of
variations can become easy or even trivial.}

The key point in the above argument is that there is at least one
trajectory going from $A$ to $B$ in finite time, as can easily be
verified. (If one of the points $A$, $B$ belongs to $\partial H_+$,
and $A\neq B$, then the argument would not work if the speed of light
was $y$ rather than $\sqrt{y}$, for in that case there would be no
path joining $A$ and $B$ in finite time.)

\section{The differential-geometric connection}\label{brachprinc}

\subsection{Is the standard formulation of the brachistochrone problem
sa\-tis\-factory?} The mathematical formulation of the brachistochrone
problem, as \mbox{presented} in \S \ref{stanformu},
is not completely natural, because it takes it
for granted that we know that energy is conserved.  More precisely,
``energy conservation'' is ``assumed'' in at least two conceptually
different ways.  First, {\bf\em the fact of energy conservation is a
prerequisite for even stating the problem}.  This is clearly not
intellectually satisfactory, because it would be much better to state the
problem first, and then 
let the mathematical analysis discover energy conservation for us,
after which we could use this new fact to find the solution.
{\em Second}, {\bf\em the problem of 
``finding the minimum time curve from
$A$ to $B$'' does not make sense for general $A$ and $B$, unless the
value of the energy $E$ is specified}. The problem usually studied in
textbooks is another one, namely, that of ``finding the minimum-time
curve from $A$ to $B$ subject to an extra constraint $E=\bar E$, for a
given $\bar E$.''  If this constraint is not given, then we really
have, for each given pair $(A,B)$ of endpoints,
a one-parameter family of problems, one
for each value of $\bar E$. (Alternatively, we may want to look at the
{\em unconstrained} problem, in which we are also minimizing over all
values of $\bar E$.  But this is clearly not what Johann Bernoulli or
anyone else who studied the problem had in mind, and in addition is
easy to solve: the infimum of the times is zero, and there is no
minimizer.)

\ssubsection{Hamilton's principal function}\label{hamilprinc}
At this point, we digress to discuss an important idea of Hamilton,
namely, the ``principal'' or ``characteristic'' function.  This idea
was not truly appreciated for more than century, until optimal control
theory emerged to provide the right setting that would do it justice.
So we will only describe it here in broad outline, following the 1944
article \cite{synge} by J.~L.~Synge.  For the specific case of the
brachistochrone, we will show below that {\em the search
for Hamilton's principal function leads directly to the discovery of
the differential-geometric structure underlying the problem}.

Synge wrote that \label{synge-quot}

{\em
\begin{itemize}
\item[] To Hamilton, optics and dynamics were merely two aspects of the 
calculus of
variations. He was not interested in experiments. To him, optics was the
investigation of the mathematical properties of of curves giving stationary
values to an integral of the type 
\begin{equation}\label{syngeint}
V=\int v\Big(x,y,z,\frac{dx}{du},\frac{dy}{du},\frac{dz}{du}\Big)du\,.
\end{equation}
Hamilton was, in fact, a great contributor---probably the gratest single
contributor of all time---to the calculus of variations. 

\item[] 
Consider two points, $A'$ with coordinates $x',y',z'$ and $A$ with
coordinates $x,y,z$. Consider all possible curves connecting 
$A'$ and $A$.  To each curve corresponds a value of the above 
integral. {\rm [Note by HJS\&JCW: Hamilton was working with Lagrangians
that are homogeneous of degree one with respect to the velocities.
In that case, the choice of the interval of definition of the curves
is immaterial, since every curve can be reparametrized to be defined
on [0,1] without changing the value of the cost integral.]}
Now compare these
values. Is there one curve that gives a smaller value to the integral than
all others? It would seem that there must be, but it would be a rash
conclusion. Suffice it to say that ``in general'' there is a curve giving a
smallest value to the integral.  Then such a curve is a {\bf ray} in 
optics. In the language of the calculus of variations, it is an extremal. 

\item[] 
Now here is Hamilton's great central idea. {\bf Regard the minimum (or
\mbox{stationary}) value of the integral as a function of the six coordinates
$x',y',z',x,y,z$ of $A'$ and $A$.}  This function is Hamilton's
characteristic or principal function.  There is nothing hard to understand
about that. What is hard to follow is Hamilton's plan to develop all the
properties of the extremals from the characteristic function. In fact, it
is so hard that few mathematicians in the ensuing century have given
serious consideration to this plan. 
\end{itemize}}

\medbreak\noindent With the rise of optimal control, ``Hamilton's plan
to develop all the properties of the extremals from the characteristic
function'' has become one of the best and most widely used tools in
the modern analysis of optimal control problems, under the name of
the ``dynamic programming'' approach.  Synge's complaint that ``few
mathematicians in the ensuing century have given serious consideration
to this plan'' may have been true in 1944 but is certainly not true
today.

\subsection{What is Hamilton's principal function for the
brachistochrone?}
Let us look at the brachistochrone problem from the modern
dynamic programming point of view. The natural questions to ask
are

\medbreak\begin{itemize}
\item[1.] {\em what is the state space for the brachistochrone problem,}
\end{itemize}
and
\begin{itemize}
\item[2.] {\em what is the problem's 
``principal function'' in Hamilton's sense?}
\end{itemize}

\medbreak\noindent More precisely, we would like to know

\medbreak\begin{itemize}
\item[3.] {\em what, exactly, is the
controlled dynamical system
\begin{equation}\label{systo}
\dot q=f(q,u)\,,\,\,\,q\in Q\,,\,\,\,u\in U
\end{equation}
that corresponds to the brachistochrone problem?} (and, in particular,
{\em what is the system's state space $Q$?})
\item[4.] {\em what is the cost functional $J$?}
\end{itemize}
and 
\begin{itemize}
\item[5.] {\em what is the value function $V$, regarded as a function
$$
Q\times Q\ni(q_i,q_t)\mapsto V(q_i,q_t)\in\rr\cup\{-\infty,+\infty\}
$$
of the initial and terminal points, as in Hamilton's definition?}
\end{itemize}

\begin{remark}\label{value-fun-inf}
{\rm Naturally, Hamilton's definition and Synge's version of it
have to be modified to adjust  them to contemporary standards
of mathematical precision. Synge's text quoted above seems to 
require the existence of an optimum for the principal function
to be well defined. We take the point of view
that {\em $V(q_i,q_t)$ is always well defined as an extended
real number---i.e., a member of $\rr\cup\{-\infty,+\infty\}$.}
Precisely,  $V(q_i,q_t)$ is the {\em infimum} of the costs of
all the trajectories going from $q_i$ to $q_t$. In particular,
$V(q_i,q_t)=+\infty$ if and only if there exist no trajectories
{}from $q_i$ to $q_t$ with cost $<+\infty$.}\proofend
\end{remark}


\medbreak\noindent We want to address these questions taking seriously the
facts that {\em energy conservation need not be known for it to be possible
to state the problem mathematically,} and that, consequently, {\em there
should not be a different brachistochrone problem for each value $E$ of the
energy, but only one brachistochrone problem.}

It is clear that $J$ should be time.  But what exactly should we
choose for $Q$, $U$, and $f$?  Can we, for example, 
take $Q=\rr^2$?
This would mean that one can specify arbitrary points $q_i$ and $q_t$
in $\rr^2$ and define $V(q_i,q_t)$ unambiguously.  But we know that
this is not so. Given $q_i$ and $q_t$, the minimization problem as
formulated in \S \ref{stanformu} only makes sense {\em for a fixed
value $E$ of the energy}, and {\em only if both $q_i$ and $q_t$ belong
to the $E$-dependent closed half-plane $H_+\Big(\frac{E}{g}\Big)$}.
Therefore there is a well defined value $V^E(q_i,q_t)$ for each value
of $E$ such that $q_i\in H_+\Big(\frac{E}{g}\Big)$ and
$q_t\in H_+\Big(\frac{E}{g}\Big)$, but these values are in general
different, and the infimum of all of them is zero.

A common way to get around this difficulty is by stipulating that we are
only interested in dropping the particle with zero initial kinetic
energy.  (One can even interpret Johann Bernoulli's statement of the
problem, quoted in \S \ref{subsec-thechal},
as suggesting this possibility.) This, however, will not do,
because it would have the following unnatural consequence: {\em if we
pick two points $q,q'$ in an optimal arc $\Gamma$ going from $q_i$ to
$q_t$, and use $\Gamma(q,q')$ to denote the piece of $\Gamma$ from $q$ to
$q'$, then $\Gamma(q,q')$ will not in general be an optimal arc from
$q$ to $q'$} (because the 
particle that started falling with zero velocity at 
$q_i$ will not in general have zero velocity at $q$).  In modern
terminology, {\em the problem with zero initial kinetic energy does
not satisfy the principle of optimality}.\Ft{Jakob Bernoulli's
solution of the brachistochrone problem makes heavy use of the fact
that a portion of a minimizer is itself a minimizer.  So Jakob cannot
possibly have been thinking  that one should only consider 
the case of zero initial kinetic energy.}   

Perhaps the state space is bigger than the plane?  In other words, it
might be the case that one can specify the position and something else at
the starting time and a similar object at the terminal time.  This 
will not do either. We cannot, for example, specify the initial and
terminal positions and velocity vectors, or even just the positions
and the lengths of the velocities.  What can be specified is $q_i$, $q_t$,
and the energy---or something yielding equivalent information,
e.g. the length of the initial velocity vector---and from this there 
is no way to derive a state space $S$ such that $V$ is defined on $S$. 

The key problem is that {\em Hamilton's principal function
must be a function of an even number of parameters,} since it is a function
of the initial and terminal states, but {\em the natural number of 
variables for the value function of the brachistochrone problem seems to
be {\bf five}, namely, two coordinates for the initial position, two 
for the terminal position, and one for the energy.}


\ssubsection{Lie brackets, integrals of motion, and the
differential-geometric perspective} A much better way to pose the
problem is to write down the equations of motion that truly correspond to
Johann Bernoulli's problem as he stated it in June 1696.  One should then 
let the mathematical analysis lead us to the discovery of energy
conservation and the simplification resulting from it.  This will lead us
naturally to a four-dimensional problem whose analysis is best carried out
{}from the differential-geometric perspective, by means of Lie-algebraic
calculations.

The true equations of motion for the brachistochrone are 
\begin{equation}\label{trueeq}
\left\{\begin{array}{rcl}
\dot{x}&=&\phantom{-}v\,,\\
\dot{y}&=&\phantom{-}w\,,\\
\dot{v}&=&-uw\,,\\
\dot{w}&=&\phantom{-}uv-g\,.\end{array}\right.
\end{equation}
Here $x$, $y$, $v$ and $w$ are the state variables, and $u$ is the control.
The values of $u$ are allowed to be arbitrary real numbers.

The variables
$x$ and $y$ are the horizontal and vertical coordinates of our moving
point, and $v$, $w$ are the components of the velocity vector.   
The requirement that the point is freely falling ``under the influence
of its own weight'' means that the force effectively acting on the point 
should be equal to a vector proportional to $[0,-g]$ plus a 
``virtual force'' that does no work, i.e., is perpendicular to the 
velocity.   Equation (\ref{trueeq}) captures these requirements, by 
introducing a virtual force vector of the form $[uw,-uv]$, where 
$u$ is an arbitrary ``control,'' taking values in $\rr$.

Using ``$^\dagger$'' to denote ``transpose,'' and writing
$q=[x,\,y,\,v,\, w]^{\dagger}$, we get a familiar equation, namely, 
\begin{equation}\label{fami}
\dot{q}=F(q)+uG(q)\,,
\end{equation}
where $F=[v,w,0,-g]^\dagger$, and 
$G=[0,0,-w,v]^\dagger$. 

This is a {\em 4-dimensional system}, with state space $Q$ equal to 
$\rr^4$, and control space $U$ equal to $\rr$.  

\ssubsection{The diamond Lie algebra}  Having expressed our
equations of motion in the form (\ref{fami}), the first thing that one
should do, from the differential-geometric perspective, is to compute the
iterated Lie brackets of $F$ and $G$, and determine the 
{\bem Lie algebra structure of our problem.}  

Precisely, we use $L$ to denote the ``accessibility Lie algebra'' of
our system, that is, the Lie algebra of vector fields generated by $F$
and $G$.  Our first task will then be to compute $L$.

It turns out that 
the four vector fields $F$, $G$, $[F,G]$ and $[F,[F,G]]$ are linearly
independent over $\rr$ (that is, there exists no nontrivial linear
combination of them with constant coefficients that vanishes 
identically), and in addition the relations 

\begin{eqnarray}
\,[G,[F,G]]&=&F\,,\label{gfgf-a}\\
\,[F,[F,[F,G]]]&=&[G,[F,[F,G]]]=0\,\label{gfgf-b}
\end{eqnarray}
hold. It then follows easily that $L$ is four-dimensional.  As an
abstract Lie algebra, $L$ is isomorphic to the {\em diamond Lie algebra} $G_4$,
a solvable Lie algebra that plays an important role in representation
theory (cf. \cite{berna}, p. 59).   This Lie algebra has a basis 
$H,P,Q,E$, with the commutation relations 
\begin{eqnarray*}
\,[H,P]&=&\phantom{-}Q\,,\\
\,[H,Q]&=&-P\,,\\
\,[P,Q]&=&\phantom{-}E\,,\\
\,[H,E]&=&\phantom{-}[P,E]=[Q,E]=0\,.
\end{eqnarray*}
(To construct an isomorphism $\Phi$ from $G_4$ to $L$, define 
$$
\Phi(P)=[F,G]\,,\,\,\Phi(Q)=F\,,\,\,\Phi(H)=G\,,\,\,
\Phi(E)=-[F,[F,G]]\,.
$$
A concrete realization of $G_4$ as an algebra
of differential operators---acting on \mbox{smooth} functions on the real 
line---can be defined by ``quantizing the harmonic oscillator,'' i.e., by 
taking 
\begin{eqnarray*}
P&=&\frac{d\,\,}{dx}\,,\\
Q&=&{\rm multiplication\,\,\,by\,\,\,}x\,, \\
E&=&{\rm identity}\,,\\
H&=&\frac{1}{2}(-P^2+Q^2)=\frac{1}{2}\Big(-\frac{d^2\,\,}{dx^2}+x^2\Big)\,,
\end{eqnarray*}
so that $H$ is the Hermite operator.   

\ssubsection{Energy conservation} We now compute the {\em accessibility
distribution} $\Lambda$ generated by $L$. Formally, $q\mapsto\Lambda(q)$
is the map that assigns to each state $q$ the linear space
$$
\Lambda(q)=\{X(q):X\in L\}\,.
$$
To find $\Lambda$ we first observe that, even though  
the vector fields
$F$, $G$, $[F,G]$ and $[F,[F,G]]$ are linearly independent over
$\rr$, a simple computation shows that 
\begin{equation}\label{eqobse}
G=\zeta_1F+\zeta_{2}[F,G]+\zeta_{3}[F,[F,G]]\,,
\end{equation}
where the $\zeta_i$ are smooth functions
of $q$, given by 
\begin{equation}
\zeta_1=-\frac{v}{g}\,,\,\,\,\,\zeta_2=-\frac{w}{g}\,,\,\,\,\,
\zeta_3=-\frac{v^2+w^2}{2g^2}\,.  
\end{equation}
\medbreak\noindent{}From these facts  one can easily show that
every iterated Lie bracket of $F$ and $G$ is a linear combination of $F$,
$[F,G]$, and $[F,[F,G]]$ with smooth coefficients.  Since $F(q)$,
$[F,G](q)$, and $[F,[F,G]](q)$ are linearly independent at each point
$q\in\rr^4$, we can conclude that 
{\em $\Lambda$ is 3-dimensional at each point}. 
This means---using, for example, Frobenius' theorem on the existence of
integral manifolds---that, at least locally, there is a nontrivial integral of
motion, i.e., a function with nonzero gradient which is constant along all
integral curves of $F$ and $G$, and then also along all solutions of
(\ref{fami}). This integral of motion can be easily computed and turns out
to be the energy $E$, given by 
\begin{equation}
E(x,y,v,w)=\frac{v^2+w^2}{2}+gy\,.
\end{equation} 
Notice that $E$ is a smooth function on $\rr^4$, with nowhere vanishing
gradient. Therefore the level hypersurfaces  of $E$---i.e.,
the sets 
$$
S_{\bar E}\equaldef\{q:E(q)=\bar E\}\,,
$$
for $\bar E\in\rr$---are smooth $3$-dimensional
manifolds. (The set $S_{\bar E}$ is given by the equation
$$
y=\frac{2\bar E-v^2-w^2}{2g}\,,
$$
which shows that each $S_{\bar E}$ is the graph
of a smooth function $(x,v,w)\mapsto y$.)

\ssubsection{Na\"{\i}ve application of the Maximum 
Principle}\label{no-info}
The fact that our 4-dimensional state space $\rr^4$
is foliated by 3-dimensional ``leaves'' $S_E$ has profound
implications for the optimization problem.  Indeed, 
{\em if we try to find the optimal trajectories by applying the Maximum
Principle, we get} {\bem nothing}. This is a consequence of more
general result, namely, that, 

\medbreak\noindent{\bf (\#)} {\em  whenever a control system satisfies a
nontrivial ``holonomic constraint'' such as $E={\rm constant}$, then the
Maximum Principle is uninformative, because {\bem every trajectory is an
extremal}}.  

\medbreak\noindent(The word ``nontrivial'' here means that $\nabla E$ 
never vanishes.) 

\begin{remark}\label{no-use}
{\rm The reason for {\bf (\#)} is as follows.  The necessary condition
given by the Maximum Principle 
for a trajectory-control pair $(\xi,\eta)$ to be optimal
is that $(\xi,\eta)$ be either a ``normal extremal'' or an ``abnormal
extremal.'' Moreover, being an abnormal extremal is a necessary
condition for the pair $(\xi,\eta)$ to be such  that
the terminal point of $\xi$ does not belong to the interior
of the reachable set from the initial point. If a nontrivial constraint
such as $E={\rm constant}$ is satisfied, then the reachable 
set from any initial point $q$ is contained in a submanifold
of the state space of positive codimension. Hence the reachable set 
has empty interior. Therefore all trajectory-control pairs
are abnormal extremals.}\proofend
\end{remark}

\ssubsection{Controllability} We now investigate the controllability
properties of our system. First of all, it is clear that two points 
$q_i$, $q_t$ cannot 
be joined by a trajectory of (\ref{trueeq}) unless they 
belong to the same leaf $S_E$. We now show that the converse is true, that
is, {\em any two points $q_i$, $q_t$ 
belonging to the same leaf of (\ref{trueeq}) can
be joined by a trajectory.}   In other words, {\em the restriction 
of (\ref{trueeq}) to each leaf is completely controllable.}

To prove our statement, we use the technique of ``Lie saturates''
introduced by Jurdjevic and Kupka. Fix $E$ and let (\ref{trueeq}.$E$)
denote the restriction of (\ref{trueeq}) to $S_E$. By construction,
the system (\ref{trueeq}.$E$) has the accessibility property.
So to prove complete controllability it suffices to establish 
complete controllability of the Lie saturate family of vector fields
$\mathcal{L}$. It is clear that $F$, $G$ and $-G$ belong
to $\mathcal{L}$. So the desired conclusion will follow if we show that
$-F\in\mathcal{L}$. But this follows easily, because the identity
$[G,[G,F]]=-F$ implies 
$$
e^{\pi G}Fe^{-\pi G}=-F\,.
$$

\subsection{The existence problem}\label{exi-ham-val}
A precise definition of Hamilton's principal function $V$ was given in
Remark \ref{value-fun-inf}. It is clear that, if $q_i$ and $q_t$
belong to $\rr^4$, then $V(q_i,q_t)<+\infty$ if and only if $q_i$ and
$q_t$ belong to the same leaf $S_E$.

It is then reasonable to ask whether an optimal control steering
$q_i$ to $q_t$ exists
whenever $V(q_i,q_t)<+\infty$. Equivalently, we want to know
if the infimum that occurs in the definition of $V$ is in fact a minimum
whenever it is finite.


The usual way to prove existence of an optimum is to pick a sequence
$\{\eta_j\}_{j\in\nn}$ of controls steering $q_i$ to $q_t$ with costs $c_j$
that converge to the infimum, look at the corresponding trajectories
$\xi_j$, and then try to produce a subsequence
$\{\eta_{j(\ell)}\}_{\ell\in\nn}$ of $\{\eta_j\}_{j\in\nn}$ that converges
in some sense to a limiting control $\eta_\infty$ and is such that the
$\xi_{j(\ell)}$ converge to a trajectory $\xi_\infty$ of $\eta_\infty$ and
the costs $c_{j(\ell)}$ converge to the cost $c_\infty$ of
$(\xi_\infty,\eta_\infty)$. 

This method may fail for various reasons. The simplest one
is that the $\xi_j$ may ``go off to infinity.'' That is, 
there may not exist a subsequence 
$\{\xi_{j(\ell)}\}_{\ell\in\nn}$ that stays in a fixed compact set.
In the following definition, we give a name to the property
that this particular obstruction to the existence of optimal 
controls does not occur.

\begin{definition}\label{proper-cond}
{\rm We say that an optimal control problem with dynamics (\ref{systo})
and Lagrangian cost functional with Lagrangian $L$ satisfies
the {\em properness condition} if for every pair $q_i,q_t$ of states
such that $V(q_i,q_t)<+\infty$ 
there exist a number $c\in\rr$ and a compact subset $K$ 
of $Q$ such that 
\begin{itemize}
\item[(1)] $V(q_i,q_t)<c$,
\item[(2)] whenever $\eta$ is a control steering $q_i$ to $q_t$ with cost $<c$,
and $\xi$ is the corresponding trajectory, it follows 
that $\xi$ is entirely contained on $K$.\proofend
\end{itemize}}
\end{definition}

When a problem satisfies the properness condition, it still does not follow
that optimal controls exist, because the minimizing sequences
$\{\eta_j\}_{j\in\nn}$ need not have subsequences that converge 
in any reasonable sense.  On the other hand, it is commonly agreed
that in this case optimal controls really exist if one enlarges 
the original class $\mathcal{U}$ of admissible controls
by adding to it some ``generalized controls.'' We
will show that this is indeed what happens for our brachistochrone
problem, and that in this case the generalized controls can be easily
described in concrete terms, and turn out to be none other than
``impulse controls.'' 

To begin with, we prove the following.
\begin{proposition}\label{proper-brach}
The minimum time problem for the brachistochrone system
(\ref{trueeq}) satisfies the properness condition.
\end{proposition}
\begin{proof}
Let $q_i$, $q_t$ be two points of $\rr^4$ having the same energy $E$. 
Fix an arbitrary $c\in\rr$ such that
$c>V(q_i,q_t)$. We have to find a constant $A$ such that
$\|\xi(t)\|\le A$ for all $t\in[0,\tau]$ whenever 
$\tau<c$ and 
$\xi:[0,\tau]\mapsto\rr^4$ is a trajectory of 
(\ref{trueeq}) for some control $t\mapsto \eta(t)$ such that
$\xi(0)=q_i$ and $\xi(\tau)=q_t$. 

Write $\xi(t)=[x(t),y(t),v(t),w(t)]^\dagger$. Let 
$$
\theta(t)=[v(t),w(t)]^\dagger\,,\quad\zeta(t)=[v(t),w(t)]^\dagger\,.
$$
Then (\ref{trueeq}) implies that 
$$
\frac{d\,\,}{dt}\,\|\zeta(t)\|^2=-2g\cdot w(t)\,.
$$
Therefore 
$$
\Biggl|\frac{d\,\,}{dt}\,\|\zeta(t)\|^2\Biggr|
=2g|w(t)|\le g^2+w(t)^2\le g^2+\|\zeta(t)\|^2\,,
$$
and then, if we write 
$$
\psi(t)=\Big|\,\|\zeta(t)\|^2-\|\zeta(0)\|^2\Big|\,,
$$
we find that 
$$
\psi(t)\le\int_0^t(g^2+\|\zeta(s)\|^2)\,ds\le c(g^2+\|\zeta(0)\|^2)
+\int_0^t\psi(s)\,ds\,.
$$
Then Gronwall's inequality implies that
$$
\psi(t)\le c(g^2+\|\zeta(0)\|^2)e^t\le  c(g^2+\|\zeta(0)\|^2)e^c\,.
$$
It then follows that 
$$
\|\zeta(t)\|^2\le\|\zeta(0)\|^2+c(g^2+\|\zeta(0)\|^2)e^c\,,
$$
which gives an {\em a priori} pointwise bound for $\zeta(t)$ in terms
of $q_i$ and $c$. 

To conclude the proof, we have to find an 
{\em a priori} pointwise bound for $\theta(t)$ in terms
of $q_i$ and $c$. But this is trivial, since
$\theta(t)=\theta(0)+\int_0^t\zeta(s)\,ds$. \proofend
\end{proof}

\ssubsection{Application of the Maximum Principle on
manifolds}Using the integral of motion $E$, we can regard $\rr^4$ as
foliated by the 3-dimensional level hypersurfaces of $E$, and conclude that
every trajectory is contained in one of the leaves $S_{\bar E}$.  If
$\bar{E}\in\rr$, then the Maximum Principle on manifolds (cf. 
\cite{susscofree}) can be applied to the problem restricted to
$S_{\bar{E}}$.
The conclusion turns out to be exactly the same as that for the
unrestricted problem, except that the nontriviality condition for the
momentum is slightly stronger. This extra strength is actually
fundamental for, as explained in \S \ref{no-info}, without it 
the conclusion would be completely uninformative.

If $[a,b]\ni t\mapsto\xi_*(t)\in\rr^4$ is an optimal 
trajectory for our problem, corresponding to a control
$[a,b]\ni t\mapsto\eta_*(t)\in\rr$, then the usual version
of the Maximum Principle yields the existence of a
nontrivial momentum vector field 
$$
[a,b]\ni t\mapsto\pi(t)\in\rr^4
$$
(often called an ``adjoint vector'') that satisfies the adjoint equation,
the Hamiltonian maximization condition, and the additional condition
that the maximized value of the Hamiltonian is a nonnegative 
constant\Ft{The Maximum Principle is usually stated as yielding
a pair $(\pi,\pi_0)$ consisting of a momentum field $\pi$ and 
a nonnegative ``abnormal multiplier'' $\pi_0$. 
The value of the maximized Hamiltonian is supposed to vanish,
and the nontriviality 
condition says that $(\pi(t),\pi_0)\neq(0,0)$ for all $t$.
These conditions are in terms of the Hamiltonian 
$\mathcal{H}$ given by $\mathcal{H}=H-\pi_0$, where 
$H$ is the Hamiltonian defined by (\ref{equham}).
It then follows that $\pi(t)\neq 0$---since $\pi(t)=0$ would imply
$\pi_0=0$, contradicting the nontriviality condition---
and one recovers the form of the necessary conditions stated
in the text.}.  The principle 
applied to the restriction to the leaves yields the stronger
conclusion that $\pi(t)$ cannot be orthogonal to 
the leaf $S_{\bar E}$ that contains $\xi_*$.  (In
differential-geometric terms, $\pi(t)$ is really a {\em covector} on
$S_{\bar E}$ at $\xi_*(t)$, and has to be nontrivial as such. Equivalently,
if one insists on regarding $\pi(t)$ as a vector in $\rr^4$, then it should
not be orthogonal to the tangent space to $S_{\bar E}$ at $\xi_*(t)$.)
(Naturally, one could also avoid using the Maximum Principle on manifolds by
invoking instead the fact that $S_{\bar E}$ is identified with $\rr^3$ 
in an obvious way, since it is the graph of a smooth function 
$(x,v,w)\mapsto y(x,v,w)$. This would amount to
eliminating $y$ from the equations by substituting for it its expression in
terms of the other variables, for each fixed $\bar E$.)

The Hamiltonian is the function
$$
\rr^4\times\rr\times\rr^4\ni(q,u,p)\mapsto\rr
$$
defined by 
\begin{equation}\label{equham}
H(q,u,p)=\varphi(p,q)+u\,\psi(q,p)\,,
\end{equation}
where
the functions $\varphi$, $\psi$ are defined by 
\begin{eqnarray*}
\varphi(p,q)&=&\langle p,F(q)\rangle\,,\\
\psi(q,p)&=&\langle p,G(q)\rangle\,.
\end{eqnarray*}
Since $u$ is
completely unrestricted, $H$ cannot have a maximum as a function of $u$
unless $\psi(q,p)=0$.  So 
\begin{equation}\label{equswi-bb}
\psi(\xi_*(t),\pi(t))=0 \quad{\rm for\,\,\,all}\quad t\,.
\end{equation}
Differentiation of (\ref{equswi}) yields 
\begin{equation}\label{equswi}
\rho(\xi_*(t),\pi(t))
=0 \quad{\rm for\,\,\,all}\quad t\,,
\end{equation}
where 
$$
\rho(q,p)\equaldef\langle p,[F,G](q)\rangle\,.
$$
A further differentiation---together with the fact
that $[G,[F,G]]=F$---yields 
\begin{equation}\label{fio-zz}
\theta(\xi_*(t),\pi(t))+u(t)\varphi(\xi_*(t),\pi(t))=0\,, 
\end{equation}
where 
$$
\theta(q,p)\equaldef\langle p,[F,[F,G]](q)\rangle\,.
$$
This, together with the equalities $\psi=\rho=0$, determine $u$ as a
function of $q$, i.e., provide an optimal control in the form of a 
state space feedback law  $u=u(q)$, as we now show.  

First of all, (\ref{eqobse}) implies that
\begin{equation}\label{fio-yy}
\psi=\zeta_1\varphi+\zeta_{2}\rho+\zeta_{3}\theta
\end{equation}
along our trajectory. Next, we use the crucial fact that
\begin{equation}\label{fio-x} 
\varphi(\pi(t),\xi_*(t))\neq 0\,,
\end{equation} 
which follows from the stronger version of the Maximum Principle.
(The precise argument is as follows.
The covector $\pi(t)$ must be nontrivial as a linear functional
on the tangent space at $\xi_*(t)$ of the leaf containing
$\xi_*(t)$. Since this space is spanned by the three vectors
$F(\xi_*(t))$, $[F,G](\xi_*(t))$ and $[F,[F,G]](\xi_*(t))$, 
one of the numbers 
$\varphi(\pi(t),\xi_*(t))$, $\rho(\pi(t),\xi_*(t))$,
$\theta(\pi(t),\xi_*(t))$ must be nonzero. We know that
$\rho(\pi(t),\xi_*(t))=0$.  If $\varphi(\pi(t),\xi_*(t))=0$, 
then (\ref{fio-zz}) would imply that 
$\theta(\pi(t),\xi_*(t))=0$, and we would get a contradiction.
Hence (\ref{fio-x}) holds.)

Therefore (\ref{fio-zz}) implies
\begin{equation}\label{fiu}
u(t)=-\,\frac{\theta(\xi_*(t),\pi(t))}{\varphi(\xi_*(t),\pi(t))}\,, 
\end{equation}
that is
\begin{equation}\label{fio} 
u(t)=-\frac{\langle\pi(t),[F,[F,G]](\xi_*(t))\rangle}
{\langle\pi(t),F(\xi_*(t))\rangle}\,.
\end{equation} 

Moreover, it follows from (\ref{fio-yy}), together with the 
fact that $\psi=\rho=0$ along our trajectory, that
\begin{equation}\label{fio-ww}
\zeta_1(\xi_*(t))\varphi(\pi(t),\xi_*(t))+\zeta_{3}(\xi_*(t))
\theta(\pi(t),\xi_*(t))=0\,.
\end{equation}
Then (\ref{fiu}) implies
\begin{equation}\label{fio-gg}
u(t)=\frac{\zeta_1(\xi_*(t))}{\zeta_{3}(\xi_*(t))}\,,
\end{equation}
as long as $\zeta_{3}(\xi_*(t))\neq 0$. Therefore
\begin{equation}\label{fio-ll}
u(t)=\omega(\xi_*(t))\,,
\end{equation}
where 
\begin{equation}\label{fio-ss}
\omega(q)=\omega(x,y,v,w)\equaldef\frac{2vg}{v^2+w^2}\,.  
\end{equation}

It is then clear that $u$ is a smooth function on the complement 
of the set
\begin{equation}
\Sigma=\{(x,y,v,w):v^2+w^2=0\}\,.
\end{equation}
The ``singular'' set $\Sigma$ corresponds, naturally, to the points where 
the kinetic energy vanishes, which as we know from our other analyses 
is where some singular behavior occurs.  

If we define 
\begin{equation}
Z(q)=F(q)+\omega(q)G(q)\,,
\end{equation}
we see that {\em $Z$ is a smooth vector field on $\rr^4\backslash \Sigma$
whose trajectories are the extremals in the sense of the Maximum
Principle.}

The resulting trajectories are easily computed. First, observe
that, if we use the feedback $u=\omega(q)$, the dynamic equations
become
\begin{equation}\label{trueeq-feed}
\left\{\begin{array}{rcl}
\dot{x}&=&\phantom{-}v\,,\\
\dot{y}&=&\phantom{-}w\,,\\
\dot{v}&=&-2g(v^2+w^2)^{-1}vw\,,\\
\dot{w}&=&\phantom{-}2g(v^2+w^2)^{-1}v^2-g\\
&=&\phantom{-}(v^2+w^2)^{-1}(2gv^2-g(v^2+w^2))\\
&=&\phantom{-}g(v^2+w^2)^{-1}(v^2-w^2)\,.
\end{array}\right.
\end{equation}
 



\medbreak\noindent If we introduce a complex variable $z$
defined by $z=v+iw$, then the last two equations
of (\ref{trueeq-feed}) say that
\begin{equation}\label{trueeq-feed-com}
\dot z=\frac{giz^2}{|z|^2}\,,\quad{\rm that\,\,\,is,}
\quad \dot z=\frac{giz}{\bar z}\,.
\end{equation}
The formulas of (\ref{trueeq-feed-com}) define a well known
dynamical system of the complex plane $\cc$. The right-hand side of the 
equations is a smooth vector field on $\cc\backslash\{0\}$.
The integral curves are circles passing through the origin
and having center on the real axis (plus the imaginary axis,
parametrized by $\rr\ni t\to(0,-gt)$, which is also a solution
since along it  (\ref{trueeq-feed-com}) just says that
$\dot z=-gi$). The parametric equation of the circles is
\begin{equation}\label{evol-vel}
z(t)=r(1+\sigma\,e^{\frac{igt}{r}})\,,
\end{equation}
where $\sigma\in\{+1,-1\}$ and $r$ is a positive real number.
		
Once it has been shown that the velocity vector $(v,w)$ 
evolves according to (\ref{evol-vel}), a trivial integration
shows that the trajectories described by the position vector 
$(x,y)$ are cycloids.

\ssubsection{The complete synthesis and Hamilton's 
principal function for the 4-dimensional brachistochrone 
problem} We haven't yet completely answered all our earlier 
questions and determined Hamilton's principal function $V$. 

Let $q_i$, $q_t$ be two points of $\rr^4$. It is clear that
$V(q_i,q_t)=+\infty$ unless both $q_i$ and $q_t$ belong to the same
energy hypersurface, because if the energies are different then the
points cannot be joined by a trajectory, and the infimum of the empty
set is $+\infty$.


On the other hand, if the points  $q_i$ and $q_t$ belong to the same level
hypersurface $S_{\bar E}$, then we know that they can be joined by a 
trajectory, so $V(q_i,q_t)$  will be finite.
Moreover, it is reasonable to expect that such a curve will exist,
and if it does not then we ought to be able to explain why.     

Write $q_i=(A_i,a_i)$, $q_t=(A_t,a_t)$, where 
$A_i$, $a_i$, $A_t$, $a_t$, belong to $\rr^2$.
The classical analysis carried out earlier seems to suggest that, once
$A_i$, $A_t$ and $\bar E$ are specified, 
there will be only one optimal arc joining them. So,
if we specify $A_i$ and $A_t$ but do not fix the energy, then there should be a
1-parameter family of such arcs (one arc for each value of $E$). Yet we
now seem to need a 3-parameter family, because we want to accommodate all
possible pairs $(a_i,a_t)$ (i.e., four parameters) as long as they give 
rise to the
same energy (one constraint, and $4-1=3$). Equivalently, {\bem given
$\bar E$, we ought to be able to choose the directions $\theta_i$,
$\theta_t$, of the velocity vectors $a_i$ and $a_t$ in a completely 
arbitrary way}. (Their magnitudes are
determined by the fact that the energy at $q_i$ and $q_t$ must equal
$\bar{E}$.)  But {\bem given $A_i$, $A_t$ and $\bar{E}$ there is a unique
cycloid from $A_i$ to $A_t$ corresponding to the energy $\bar{E}$}, and in
particular the directions $\bar{\theta}_i$, $\bar{\theta}_t$
at $A_i$, $A_t$ of this cycloid 
are completely determined by $A_i$, $A_t$ and $\bar{E}$.

In other words, {\bem we ought to be able to choose freely the two angles
$\theta_i$, $\theta_t$, but it appears that we cannot do so.}

The solution of this apparent paradox is as follows.  Suppose we fix an
energy level $\bar E$.  Given any $A_i$, the set $C(A_i,\bar E)$ of those $a$'s
such that $(A_i,a)\in S_{\bar E}$ is either empty or a circle, except for the
``borderline'' case when $(A_i,0)\in S_{\bar E}$ and $C(A_i,\bar E)$ 
consists of a single point.  If $C(A_i,\bar E)$ is a circle, then through 
each point of
$\{A_i\}\times C(A_i,\bar E)$ there passes an integral curve of $Z$.  The set
of all points lying on such curves is 2-dimensional, and intersects the
1-dimensional set $\{A_t\}\times C(A_t,\bar E)$ at just one point. This means
that there is a unique pair $(\bar a_i,\bar a_t)$ such that 
$(A_i,\bar a_i)$ and $(A_t,\bar a_t)$ lie on the same integral curve of $Z$.  
The portion of that
curve joining $(A_i,\bar a_i)$ to $(A_t,\bar a_t)$ is, of course, our famous
cycloid.

What is then the optimal trajectory joining $(A_i,a_i)$ to $(A_t,a_t)$
for general $a_i$, $a_t$? What is the corresponding value of
$V(A_i,a_i,A_t,a_t)$?  The answer is quite simple: the ``optimal
trajectory'' is obtained by

\begin{itemize}
\item[1.] applying an {\bem impulse control}
to rotate $a_i$ to $\bar a_i$ instantaneously, 
\item[2.] then
following the cycloid from $(A_i,\bar a_i)$ to $(A_t,\bar a_t)$, 
\item[3.] finally applying another impulse control to rotate 
$\bar a_t$ to $a_t$.  
\end{itemize}
(These are not
true trajectories, of course, but it is easy to translate the above
into a statement about a minimizing sequence of true trajectories.)
The value $V(A_i,a_i,A_t,a_t)$ is therefore equal to 
$V(A_i,\bar a_i,A_t,\bar a_t)$.

\section{Five modern variations on the theme of the brachistochrone}
\label{fourvar}

\subsection{Hestenes' ``variation of the brachistochrone
problem''}\label{hestebrach}  In 1956, M. Hestenes wrote
(in \cite{hestenespap}, p.~64):

\begin{itemize}
\item[] {\em ... it is interesting to speculate on what
type of brachistochrone problem the Bernoulli brothers might have
formulated had they lived in modern times.  An interesting variation
of the brachistochrone problem can be described as follows:
Consider an idealized rocket ship moving in a vertical plane. The ship is 
to be considered as a particle acted upon by a gravitational force $g$
and a thrust of constant magnitude $F$ and variable inclination $\varphi$ ...
There are no other forces acting on this ship. The pilot steers the ship 
by controlling the inclination $\varphi$, that is, by controlling the 
direction of the thrust. The problem is to find a path of least time
under suitable initial and terminal conditions. Analytically, one
seeks to find among all functions 
$$
x=x(t)\qquad y=y(t)\qquad \varphi=\varphi(t)\qquad 0\le t\le T
$$
subject to the constraints (with unit mass)
$$
\ddot x= F \cos\varphi\qquad\ddot y= F \sin\varphi-g
$$
and with $x$, $y$, $\dot x$, $\dot y$, prescribed at $t=0$ and $t=T$, 
one which minimizes the time of flight $T$.} 

\item[] {\em It can be shown that the path of least time has the following
interesting property: there is a line $L$ with direction fixed in
space, and with position fixed relative to the ship, such that if a
particle $M$ is moving along $L$ with constant speed, the ship will
traverse a path of least time by keeping the thrust aimed at the
particle $M$.}
\end{itemize}

\medbreak\noindent Hestenes later on proceeds to solve the problem. 	
Let us present a solution based on optimal control.  The dynamical equations
can be rewritten as 
$$
\dot x= z\qquad \dot z=F \cos\varphi\qquad\dot y= w\qquad \dot w
=F \sin\varphi-g\,.
$$
If $[0,T]\ni t\mapsto (x_*(t),z_*(t),y_*(t),w_*(t))\in\rr^2$ is an 
optimal trajectory, corresponding to a control function 
$[0,T]\ni t\mapsto\varphi_*(t)\in\rr$, then the Maximum Principle
yields a nonnegative constant $\pi_0$ and
and a vector-valued function 
$$
[0,T]\ni t\mapsto\pi(t)=(\pi_x(t),\pi_z(t),\pi_y(t),\pi_w(t))\,.
$$
The Hamiltonian is given by 
$$
H(x,z,y,w,\varphi,p_x,p_z,p_y,p_w,p_0)=
p_x\,z+p_z\,F\cos\varphi+p_y\,w+p_w\,(F\sin\varphi-g)-p_0\,.
$$
The adjoint equation becomes
$$
\dot\pi_x=0\qquad\dot\pi_z=-\pi_x\qquad\dot\pi_y=0\qquad
\dot\pi_w=-\pi_y\,.
$$
Then, if we write $\omega(t)=(\pi_z(t),\pi_w(t))$, 
we have 
$$
\omega(t)=\omega_0+t\omega_1\,,
$$
where $\omega_0$ and $\omega_1$ are constant two-dimensional vectors.

The Hamiltonian maximization condition then says that
the vector
$$
{\bf u}(t)=(\cos\varphi(t),\sin\varphi(t))
$$
is given by
\begin{equation}\label{hestenescontrol}
{\bf u}(t)=\frac{\omega(t)}{\|\omega(t)\|}\,.
\end{equation}
The Hamiltonian must vanish along our trajectory.
Therefore, $\pi(t)$ can never \mbox{vanish}, for if $\pi(t)=0$ 
it would follow that $\pi_0=0$, contradicting the nontriviality
condition. Hence $\omega(t)$ cannot vanish identically. 
So $\omega$ is a linear function which is not identically equal to zero.
Therefore there is at most one value of $t$ such that
$\omega(t)=0$. It follows that the function 
${\bf u}$ of (\ref{hestenescontrol}) is well defined 
except possibly at one point $\bar t$. Moreover, if such a $\bar t$ exists,
then $\omega(t)=(t-\bar t)\omega_1$, so 
$$
{\bf u}(t)=\left\{\begin{array}{rll}
-\omega_1&{\rm if}&t<\bar t\\
\omega_1&{\rm if}&t>\bar t.\end{array}\right.
$$
The ``line $L$'' of Hestenes' ``interesting property''
is the line $L_t$ with direction $\omega_1$, with a point $\lambda_t$
having a ``position fixed relative to the ship'' given by
$$
\lambda_t=(x_*(t),y_*(t))+\omega_0\,.
$$
So in fact $L_t$ is moving along with the ship.
If a point $q_t$ on $L_t$ moves with constant velocity $\omega_1$, 
starting at $\lambda_0$ at time $0$, then the point's 
position at time $t$ is $\lambda_t+t\omega_1$. Therefore
the vector from the ship's position to $q_t$ is 
precisely $\omega(t)$. We have shown that the direction of the thrust
is $\omega(t)$, from which Hestenes' ``geometric property'' follows.


\ssubsection{The reflected  brachistochrone problem}\label{refbrach}  

\noindent In \S \ref{standardfor},
our analysis of the bra\-chis\-to\-chrone problem was
carried out in a half-plane, by looking only for optimal trajectories
that are entirely above the $x$ axis, as in Johann Bernoulli's
original problem.

Let us pursue Hestenes' idea that ``it is interesting to speculate on
what type of brachistochrone problem the Bernoulli brothers might have
formulated had they lived in modern times.''  An obvious question for
the brothers to ask would have been that of finding out what happens
if we work in the whole plane.  This is in fact a much more natural
mathematical setting for the brachistochrone problem.  We shall refer
to this new question as the {\em reflected brachistochrone problem.}
(It is because we were thinking all along of this extension that we
wrote $\sqrt{|y|}$ rather that $\sqrt{y}$ in (\ref{braccontrol})). 
For a recent study of the reflected brachistochrone problem from the
dynamic programming perspective, see Malisoff 
\cite{malisoff-a,malisoff-b}.

Solving this more general problem amounts to
finding  the light rays when the medium is {\sl the whole
plane}, and the speed of light is $\sqrt{|y|}$.  In other words,
{\em we want to find the minimum time paths for the control system
(\ref{braccontrol}) with control set $U$ given by (\ref{braccontrol-xxx})
and state space equal to the whole plane.}

Notice that this problem is ``completely controllable,'' in the sense
that 

\medbreak
\begin{itemize}
\item[(C)] {\em any two points $A$, $B$ of $\rr^2$ can be joined by a
feasible path, even if they lie on opposite sides of the $x$ axis.}
\end{itemize}

\medbreak\noindent To prove (C), observe first that the desired conclusion 
is trivial
if $A$ and $B$ both lie in the open upper half plane or in the 
open lower half plane. So it clearly suffices to prove the conclusion
if $A$ lies in the open upper half plane and $B$ belongs to 
the $x$ axis. Moreover, we can assume that $A$ lies on the same vertical
line as $B$, for otherwise we can pick $A'$ on the same vertical
as $B$ but strictly above $B$, and join $A$ to $A'$ first in finite time.
So let us suppose that $A=(a,\alpha)$ and  $B(a,0)$. Then the curve
$[0,\alpha]\ni s\mapsto(a,\alpha-s)$ certainly goes from $A$ to $B$.
Along this curve, $dy=-\sqrt{y}dt$, so
$$
dt=-\frac{dy}{\sqrt{y}}=\frac{ds}{\sqrt{\alpha-s}}\,.
$$
Therefore the total time $T$ required to move from $A$ to $B$ along
this curve is given by
$$
T=\int_0^\alpha \frac{ds}{\sqrt{\alpha-s}}=2\sqrt{\alpha}\,,
$$
so $T<+\infty$, and (C) is proved.

The crucial point here is that {\em the right-hand side of
(\ref{braccontrol}) vanishes along the $x$ axis, but this does not
prevent the existence of feasible paths crossing the $x$-axis
in finite time, because the ``speed of light function''
$$
(x,y)\mapsto c(x,y)=\sqrt{|y|}
$$
is not Lipschitz near the $x$ axis.}

\begin{remark}\label{hedid}
{\rm If the function $c$ was Lipschitz-continuous, then the well known
\mbox{uniqueness} 
theorem for  ordinary differential equations would imply that
every solution passing through a point
on the $x$ axis is a constant curve.  This would obviously 
render it impossible for feasible curves starting at a point 
not on the $x$ axis  to reach the $x$ axis in finite time. An 
important example of this situation is the case when the speed o
light is $|y|$ rather than $\sqrt{|y|}$. That case was also studied
by Johann Bernoulli, as is clearly seen in the text we quoted at the end
of \S \ref{sec-tauto}, where he says that
{\em if, for example, the
velocities were as the altitudes, then both curves} [HJS \&{} JCW:
that is, the brachistochrone and the tautochrone]
{\em would be algebraic, the one a circle, the other one a straight line.}
It turns out that the ``brachistochrones'' for the case
when the speed of light is $|y|$ are {\em half circles whose center 
lies on the $x$ axis.} Moreover, {\em these circles 
take infinite time to reach the $x$ axis}, precisely because the
function $y\mapsto|y|$ is Lipschitz.}\proofend
\end{remark}

We have seen that the complete controllability of (\ref{braccontrol-xxx})
depends essentially on the non-Lipschitzian character of its speed of light
function.  However, {\em this same non-Lipschitzian character also renders
the Maximum Principle inapplicable, in its classical and nonsmooth versions
(including the so-called ``\L{}ojasiewicz version,'' cf. Sussmann
\cite{suss1}), since all these results require a Lipschitz reference vector
field.}

Can we still use necessary conditions to  find the solution, perhaps
by invoking a more general version of the Maximum Principle
that would be applicable to this nonsmooth, non-Lipschitzian case?

Suppose, for example, that
we want to find an optimal trajectory $\xi$ going from 
$A$ to $B$, where $A$ lies in the upper half-plane and $B$ is in the lower
half-plane. Then one can easily show, first of all, that an optimal trajectory
$\xi$ exists, using the Ascoli-Arzel\`a theorem as before
(cf. \S \ref{subsec-existence}). Next, using the usual necessary
conditions for optimality, e.g. the Euler-Lagrange equation or the
classical version of the Maximum Principle, one shows that any portion of 
an optimal curve which is entirely contained in the closed upper half plane
or in the closed lower half plane is a cycloid given by (\ref{eqcycloid}),
or a reflection or such a cycloid with respect to the $x$ axis. 
Next, one sees that $\xi$ cannot traverse the $x$ axis more than once. 
(This is a very straightforward consequence of Lemma \ref{the-lem}.)

Therefore $\xi$ must consist of a cycloid
$\Gamma_1$  going from $A$ to a point $X$ 
in the $x$ axis, followed by a second cycloid $\Gamma_2$ 
going from $X$ to $B$.  It remains to find $X$.  

Naturally, $X$ is the only point on $\xi$ in a neighborhood of which
the usual smooth and nonsmooth versions of the Maximum Principle
are inapplicable. On the other hand, it is easy to produce a nonrigorous 
formal argument yielding the missing condition, as we now show.

Let us apply the Maximum Principle formally, even though the technical
conditions required for the validity of the usual versions are not verified.
The maximized value of the Hamiltonian 
is clearly equal to $\|\vec{p}\|\sqrt{|y|}-p_0$, and we know that this
number is in fact equal to zero. So 
\begin{equation}\label{ifwemul-a}
\|\vec{p}(t)\|\sqrt{|y|}=p_0\quad{\rm for\,\,\,all}\quad t\,.
\end{equation}
If we multiply (\ref{ifwemul}) by $|y(t)|$, and use
(\ref{ifwemul-a}), we find
\begin{equation}\label{ifwemul-b}
|y(t)|\Big(1+y'(x)^2\Big)=\frac{|y(t)|\cdot\|\vec{p}\|^2}{p^2}
=\frac{p_0^2}{p^2}\,.
\end{equation}
The adjoint system (\ref{eqpdot}) tells us that $p$ is constant 
on each time interval that does not contain the point $\tau$
such  that $\xi(\tau)=X$. If we make the {\it Ansatz}
that $p$ is actually constant on the whole time interval, i.e.,
that the constant values of $p$ on both sides of $\tau$ are equal,
we find that {\bem the cycloids $\Gamma_1$, $\Gamma_2$
must be such that the quantity 
\begin{equation}
C=|y(x)|(1+y'(x)^2)\,.
\end{equation}
which is constant along each $\Gamma_i$, actually 
has the same value for both curves}. This extra condition determines $X$. 

The above argument is, of course, not rigorous. It turns out, however, that
{\bem the recent ``very nonsmooth'' versions of the Maximum Principle
developed by one of us---cf. Sussmann
\cite{suss5,suss6,suss7,suss9,suss10,suss11}---apply to this problem and
provide a rigorous justification for our formal argument.}  The reason for
this is that the new results do not require Lipschitz continuity---or even
continuity---of the reference vector field.  All that is needed is that the
{\em flow maps} generated by the reference flow be differentiable in an
appropriate sense\Ft{For the specific case of the reflected
brachistochrone, classical differentiability suffices. For more general
problems, one needs other theories of ``generalized differentiation,'' as
explained in \cite{suss5,suss6,suss7,suss9,suss10,suss11}}.

Remarkably, our ``extra condition'' has a simple geometric
meaning. Recall that the cycloids of Johann Bernoulli's problem are
generated by a circle rolling along the $x$ axis. The extra condition
then says that

\begin{center}
\fbox{\begin{minipage}{11cm}
{\Large
\begin{itemize}
\item[{\bf (C)}] {\bem The rolling circles that generate the upper and lower
cycloids must have equall radii.}
\end{itemize}}
\end{minipage}}
\end{center}


\ssubsection{A modest claim}  If we are going to ``speculate''
as Hestenes did, it is natural to ask not only what questions
the Bernoulli brothers would have asked if they had lived 
in modern times, but also what kinds of answers they would have liked.

This is of course a hard question to answer, and anything we say 
will have to be highly subjective. Granted this, 
we are firmly convinced that 

\medbreak
\begin{itemize}
\item[1.] {\em The Bernoulli brothers would have liked 
the ``interesting property'' of the solution of Hestenes' problem}
(which is almost certainly the reason why Hestenes chose to 
emphasize this property)
\end{itemize}
and also that 
\begin{itemize}
\item[2.] {\em they would have found 
elegant the geometric condition of our solution of the reflected
brachistochrone problem.} 
\end{itemize}

\medbreak\noindent This seems clear to us 
but, since {\em de gustibus non est disputandum},
we shall not pursue this particular argument further.

\subsection{The brachistochrone with friction}
In the problem studied in \S \ref{brachprinc}, we assumed that there was no 
friction. This implied that energy was conserved and that 
the four-dimensional state space was foliated by three-dimensional
orbits of the system's accessibility distribution. 

It is then natural to consider the more general case of the
{\em  brachistochrone with friction}, and to expect that 
for this problem there will be no nontrivial conserved 
functions of the state, so the accessibility distribution
will be four-dimensional, and the system will have the accessibility 
property.

We now outline some of the results, without carrying out a complete
analysis. 

The dynamical behavior of the system is given by the equations


\begin{equation}\label{brach-fric}
\left\{\begin{array}{rcl}
\dot x&=&\phantom{-u}\xiv\,,\\
\dot y&=&\phantom{-u}\etav\,,\\
\dot\xiv&=&-u\etav-\rho\xiv\,,\\
\dot\etav&=&\phantom{-}u\xiv-\rho\etav-g\,,\end{array}\right.
\end{equation}
where, as before, $u$ is an unbounded scalar control. The parameter
$\rho$ is a real number, the friction coefficient. The
problem with friction corresponds to $\rho>0$. When 
$\rho=0$ we get the frictionless case considered earlier.


If we write 
$$
q=\left[\begin{array}{c}x\\y\\\xiv\\\etav\end{array}\right]\,,
\qquad F(q)=\left[\begin{array}{l}\phantom{-\rho}\xiv\\
\phantom{-\rho}\etav\\-\rho\xiv\\-\rho\etav-g
\end{array}\right]\,,\qquad G(q)=\left[\begin{array}{r}0\\
0\\-\etav\\\xiv\end{array}\right]\,.
$$
then (\ref{brach-fric}) takes the form
\begin{equation}\label{brach-fric-vec}
\dot q = F(q)+uG(q)\,.
\end{equation}

{}From now on, we assume that $\rho\neq 0$, since the case when
$\rho=0$ has already been discussed.
Let $L$ be the Lie algebra of vector fields
generated by $F$ and $G$. In order to analyze the structure of
$L$, it will be convenient to introduce five new vector fields
$Z$, $V$, $W$, $X$, and $Y$. 
The following eight formulas express $F$, $G$, and $[G,F]$
in terms of the usual differential operator notation for vector fields,
and define $Z$, $V$, $W$, $X$, and $Y$:

$$
\left.\begin{array}{rcrcrcrcr}
F&=&\xiv\partial_x&+&\etav\partial_y&-&\rho\xiv\partial_\xiv&-&
(\rho\etav+g)\partial_\etav\,,\\
G&=&\,&\,&\,&-&\etav\partial_\xiv&+&\xiv\partial_\etav\,,\\
\,[G,F]&=&-\etav\partial_x&+&\xiv\partial_y&-&g\partial_\xiv&\,&\,,\\
Z&=&\,&\,&\,&\,&\xiv\partial_\xiv&+&\etav\partial_\etav\,,\\
V&=&\,&\,&\,&\,&\partial_\xiv\,&\,&\,,\\
W&=&\,&\,&\,&\,&\,&\,&\partial_\etav\,,\\
X&=&\partial_x&\,&\,&\,&\,&\,&\,,\\
Y&=&\,&\,&\partial_y&\,&\,&\,&\,.\end{array}\right.
$$

Let $S$ be the linear span of the eight vector fields
$F$, $G$, $[G,F]$, $Z$, $V$, $W$, $X$, and $Y$. It then turns out
that $S\subseteq L$, because
\begin{eqnarray*}
Z&=&\sigma(F+[G,[G,F]])\,,\\
W&=&\gamma([F,Z]+F+\rho Z)\,,\\
V&=&[G,W]\,,\\
X&=&\rho V-[F,V]\,,\\
Y&=&\rho W-[F,W]\,,
\end{eqnarray*}
where
$$
\sigma=\frac{1}{\rho}\,\,\,{\rm and}\,\,\,\gamma=\frac{1}{2g}\,.
$$

On the other hand, the formulas

\begin{eqnarray*}
\,[F,[G,F]]&=&-\rho[G,F]+2gX-2\rho gV\,,\\
\,[G,[G,F]]&=&-F-\rho Z\,,\\
\,[F,Z]&=&-F-\rho Z-2gW\,,\\
\,[F,V]&=&-X+\rho V\,,\\
\,[F,W]&=&-Y+\rho W\,,\\
\,[F,X]&=&[F,Y]=0\,,\\
\,[G,Z]&=&[G,X]=[G,Y]=0\,,\\
\,[G,V]&=&-W\,,\\
\,[G,W]&=&V\,,
\end{eqnarray*}
show that $[F,S]\subseteq S$ and $[G,S]\subseteq S$. 
So $L=S$.

A straightforward calculation shows that 
{\em $F$, $G$, $[G,F]$, $Z$, $V$, $W$, $X$, and $Y$
are linearly independent over $\rr$.}
So {\em $L$ is an $8$-dimensional Lie algebra}.  

Let $\Lambda$ be the accessibility distribution of our system,
so $\Lambda(q)\equaldef\{h(q):h\in L\}$
if $q\in\rr^4$. Since the vector fields $X$, $Y$, $V$ and $W$ 
belong to $L$, it is clear that $\Lambda(q)=\rr^4$ for every $q$.
So {\em our system has the accessibility property}.

An analysis similar to the one carried out in the previous section
for the frictionless problem yields an optimal ``feedback control''
\begin{equation}\label{optifeed}
u=\frac{2g\xiv}{\xiv^2+\etav^2}+2\rho\frac{\zeta\xiv^2-\xiv\etav}
{\xiv^2+\etav^2}\,,
\end{equation}
involving an additional parameter $\zeta$, given by
$$
\zeta=\frac{\pi_Y}{\pi_X}\,.
$$
(Here $\pi_X$, $\pi_Y$ are the $X$ and $Y$ components of the momentum, 
which are easily seen to be constant along extremals.)

In other words, {\em through every point $q\in\rr^4$ there passes
a one-parameter family of optimal trajectories, depending on the parameter
$\zeta$.}

\subsection{The classical brachistochrone as a degenerate
problem} The family of brachistochrone problems with friction coefficient
$\rho$ constitutes an ``unfolding'' of the classical frictionless
problem. For $\rho\neq 0$, the situation is generic, in the sense
that the accessibility distribution is four-dimensional.

The generic behavior of the time-optimal trajectories
for such problems is precisely the one described here: the control 
is given in ``feedback'' form, depending on the state 
and an extra parameter. Formula
(\ref{optifeed}) shows that the extra parameter 
only occurs as part of a term that is multiplied by $\rho$. 
It follows that when $\rho=0$ the extra parameter disappears, and we get
a state space feedback as determined earlier.

\ssubsection{Deriving \snell{}'s law from the Maximum 
Principle}\label{snellderi}

\noindent Our third ``modern variation'' will involve going back
to ancient times, long before 1696, 
and considering Fermat's minimum time principle and \snell{}'s
law of refraction. 

We have already explained how Snell's law is intimately related to
the brachistochrone problem, since it played a crucial role
in Johann Bernoulli's solution. So, if we pursue Hestenes' line of
inquiry further, and speculate on other questions
that the Bernoulli brothers might have asked ``had they lived in modern 
times,'' it is easy to imagine them asking the following.

\begin{itemize}
\item[(Q)] {\em Given that in these modern times powerful general theories 
have been developed, going far beyond the particular examples that were
dealt with by ad hoc methods up to
the 17th and early 18th centuries, do these general theories
actually cover {\bf all} these special examples
whose study preceded and begat them?}
\end{itemize}

\medbreak \noindent As an important special case of this question,
the brothers would certainly want to know 
whether the necessary conditions of the calculus of variations
and optimal control theory apply, in particular, to the derivation
of Snell's law.

So let us consider a medium such as the one we described when
discussing \snell{}'s law, in which the speed of light is constant on
each of the two sides $H_-$, $H_+$ of a hyperplane $H$ in $\rr^n$, but
the constant values $c_-$, $c_+$, are different.  Let us ask the
obvious question, namely, whether \snell{}'s law be derived from the
Maximum Principle.

It is easy to see that there is no way to formulate this question
as a minimum time problem with a controlled dynamics 
$\dot q=f(q,u,t)$ in which $f$ is continuous with respect to $q$, since
$||f||$ has to jump as we cross $H$.

In the usual versions of the Maximum Principle, it is not required
that $f$ be differentiable, or even continuous, with respect to
the control $u$. But here the situation is different, since
we need to be able to deal with an $f$ which is {\em discontinuous
with respect to the state variable $q$.}

Once again, it turns out that {\em the recent versions the Maximum
Principle developed in Sussmann
\cite{suss5,suss6,suss7,suss9,suss10,suss11} apply to this problem as
well, and lead to Snell's law}.  The reason is that in these version all
that is needed is that the reference trajectory $\xi_*$ arise from a flow
$\{\Phi_{t',t}\}$ consisting of differentiable maps, and there is no need
for this flow to be generated by a continuous vector field.

\subsection{Was Johann Bernoulli close to discovering non-Euclidean
geometry?} 

\noindent Given a ``speed  of light function'' 
$(x,y)\mapsto c(x,y)$ on some plane region, the time along a curve is given by 
$$
dt=\frac{\sqrt{dx^2+dy^2}}{c(x,y)}\,,
$$
since $\sqrt{{\frac{dx}{dt}}^2+{\frac{dy}{dt}}^2}=c(x,y)$. This means
that, if we adopt a slightly different point of view, and think of time 
as ``length,'' then the function $c$ defines a {\em Riemannian metric}  
on the set $\Omega_c=\{(x,y):c(x,y)>0\}$, provided that $c$ is sufficiently
smooth. Let us assume that $c(x,y)>0$ whenever $y>0$, as was the case 
in our two examples. Then $c$ gives rise to a Riemannian metric 
$$
ds^2=\frac{dx^2+dy^2}{c(x,y)^2}
$$
on the upper half plane. The ``brachistochrones'' are then the {\em minimum
length} curves, that is, the {\em geodesics}. 
In the case of the ``true brachistochrone,'' when $c(x,y)=\sqrt{|y|}$, 
these geodesics are cycloids (plus vertical segments) but 
{\em maximal geodesics have finite length}. On the other hand, 
in the case when $c(x,y)=|y|$, so that the Riemannian metric on the upper 
half plane is given by 
\begin{equation}\label{geodes-circle}
ds^2=y^{-2}(dx^2+dy^2)\,,
\end{equation}
the geodesics are, as we have seen, {\em half circles of infinite length}. 

So, using modern terminology, the circles that occur as solutions of this
modified version of Johann Bernoulli's brachistochrone problem are the
geodesics of a Riemannian metric on the upper half plane given by 
(\ref{geodes-circle}). We now know that this Riemannian metric defines
the so called {\em Poincar\'e half plane}, which is the simplest
model of non-Euclidean geometry.

This raises an obvious question: {\em how close did Johann Bernoulli get to
solving the famous problem of the provability of Euclid's fifth postulate?}
What renders the question particularly interesting is that 

\begin{itemize}
\item[1.] Johann Bernoulli lived at a time when the problem of Euclid's
parallel postulate was very much at the center of mathematical research.
For example, G. Saccheri published in 1733 his book \cite{saccheri-euc},
entitled {\em Euclides ab omni naevo vindicatus} (that is, ``Euclid
vindicated from every flaw''), where he pushed very far the development of
non-Euclidean geometry, only to end up dismissing it by claiming that a
particular conclusion he had reached had to be rejected because it was
``repugnant to the nature of a straight line.''
\item[2.] The cycloids that occur in the original brachistochrone problem
have many of the incidence properties of Euclidean lines,
and therefore
come close to satisfying all Euclid's postulates other than the
fifth. But they do not yet provide a good model for non-Euclidean
geometry because they ``go to infinity'' (i.e.,  approach the $x$ axis)
with finite length.  
\item[3.] The half-circles of the Poincar\'e model have all
the desired incidence properties and also have infinite length, so
they furnish a perfect model where all the Euclidean axioms other than
the fifth are true. 
\item[4.] Johann Bernoulli did study these half circles, as we pointed
out in Remark \ref{hedid}. So in a sense 
he did discover the Poincar\'e model.
\item[5.] Johann Bernoulli was definitely interested in the question, 
since he made significant contributions to the
early stages of the development of the idea of geodesics on curved 
surfaces, by studying the curves of minimum length on a surface, which 
later\Ft{The word ``geodesic'' was coined much later by Liouville
(1809-1892).} came to be known as ``geodesics.''
\item[6.] The solution of the problem of the independence of Euclid's 
postulate is extremely simple by today's standards, even though
mathematicians took about 2200 years to find it. Actually, 
the solution is so simple that it can be explained to any undergraduate
with a moderate understanding of plane analytic geometry.   
All that is required is to take any reasonable way to compute 
``time'' along curves, rename ``time'' by calling it ``length,'' 
and then call the length-minimizing curves (that is, the 
``brachistochrones'') ``lines.'' For example, the case
when $c=|y|$ leads us to taking 
the half circles with center on the $x$ axis (plus the vertical 
half lines) and calling them ``lines.'' Once this trivial renaming 
has been carried out, it is easy to verify that all of Euclid's postulates
other than the fifth hold, but the fifth postulate fails. 
\end{itemize}

\medbreak\noindent So {\em Johann Bernoulli had essentially all the tools
he needed to solve the problem and prove that Saccheri was wrong, by
showing that Euclid's fifth postulate does not follows from the other ones.
Why didn't he do it?}

One can surmise that he lacked the understanding 
of the {\em geometric} significance of the cycloids and half circles
that he had found as solutions of his minimization problems.    He
certainly saw these curves as ``geometric,''  but failed to see that
they could serve as the ``lines'' of a different geometry. 
	
In modern terms, this is a trivial remark, namely, that one can turn
around the well known fact that the segment is the shortest path between
two points, and make it the {\em definition of the notion of
segment}, while at  the same time using a more
general prescription to measure ``length.'' But the idea
that minimization
problems can be the source of new geometries evolved very slowly over two
millennia, and took a long time to crystallize, reaching final 
form in the nineteenth century, in the era of Gauss and Riemann,
when it revolutionized mathematics and then moved on, at the beginning 
of the twentieth century,  to bring a 
a fundamental change in our understanding of the physical universe.

One can already find the idea in a rudimentary form in Fermat's least time
principle\Ft{which, as we have seen, was a direct ancestor of Johann
Bernoulli's solution of the brachistochrone problem}, which can be
interpreted as saying that light rays follow the shortest paths, with the
proviso that ``length'' must now be defined in terms of a new and more
refined geometry, in which the physical properties of space influence the
geometry, and shortest paths can be curved or bent. This interpretation is
suggested, for example, by G.~Lochak, in \cite{lochak}, p.~57--58 [our
translation]:

\bqrm
[Fermat's principle] anticipated the greatest revolution to be undergone by
geometry in the future, namely, the discovery of non-Euclidean geometries,
because it showed for the first time that the physical properties of a
medium might cause the shortest paths not to be straight lines.
\eqrm

In fact, Leibniz definitely uses the word ``shortest'' in his analysis of
the refraction problem in his 1684 \textit{Nova methodus} paper, showing
that he was at least entertaining the possibility that light rays in a
medium with variable speed of light truly were ``shortest paths'' in some
new geometry.

But it would appear that, in spite of these earlier glimpses of what would
become one of the greatest scientific ideas of all times, 
the era of Johann Bernoulli was not yet ready to take the trivial step 
of turning around the lenght minimization property of segments. 

Johann Bernoulli passionately wanted to succeed and be famous, and was
obsessed by the desire to surpass his contemporaries, especially his
brother Jacob and his son Daniel\Ft{Johann was very jealous of
Daniel's success.  He once threw Daniel out of the house for having
won a French Academy of Sciences prize for which Johannn had also been
a candidate (cf. Bell \cite{bell}, p.~134).  Probably with the help of
his loyal disciple Euler (a student of Johann in Basel and a colleague
of Daniel in Saint Petersburg), Johann plagiarized Daniel's
path-breaking work on hydrodynamics (cf. Guillen \cite{guillen}).}. It
is thus sadly ironic that he, of all people, got so close to solving
an important 2000-year old problem but missed because he failed to
take a small step, a step so simple that today it can easily be explained
to beginners.

Had Johann Bernoulli ``lived in modern times,'' he probably would have 
been torn between the curiosity to find out about non-Euclidean
geometry and the desire not to hear about the tragedy of
his missed opportunity to reach eternal fame and glory.

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\end{document}
