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\def\gua{H.~J.~SUSSMANN}
\def\guo{HAMILTON AND WEIERSTRASS}

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\begin{center}
{\Huge{\bf HAMILTON'S EQUATIONS\\
$\,$\\
\vspace{-5mm}
AND WEIERSTRASS'\\
$\,$\\
\vspace{-4mm}
SIDE CONDITION}}\\
$\,\,$\\
{\Large{\it H\'ector J. Sussmann}}\\
$\,\,$\\
{\Large November 15, 2000}
\end{center}

\bigbreak
\begin{center}
{\Large\bf CONTENTS}
\end{center}

\medbreak

\thispagestyle{empty}
\noindent 1.\ THE TWO FORMS OH HAMILTON'S SYSTEM OF EQUATIONS\hfill 1 

\noindent \ 1.1\ The Legendre condition \hfill 2

\noindent \ 1.2\ Hamilton's missed opportunity \hfill 3

\noindent \ 1.3\ The ``control Hamiltonian'' versus Hamilton's Hamiltonian \hfill 6

\medbreak
\noindent 2.\ WEIERSTRASS MISSES ANOTHER OPPORTUNITY \hfill 7

\noindent \ 2.1\ The Weierstrass side condition \hfill 8

\noindent \ 2.2\ A Hamiltonian rewriting of the Weierstrass condition \hfill 11

\noindent \ 2.3\ What Weierstrass missed \hfill 12

\noindent \ 2.4\ A toy problem\hfill 13

\bigbreak


\ssection{THE TWO FORMS OH HAMILTON'S SYSTEM OF EQUATIONS}

\noindent We will now discuss an interesting story of ``missed 
opportunities,'' by showing how Hamilton in the 1830s, Weierstrass in
the 1870s, and Carath\'eodory\footnote{This handout just deals with Hamilton
and Weierstrass. The discussion of
Carath\'eodory's attempt to rewrite the Weierstrass condition in Hamiltonian
form will be given next week.} in the 1930s, had all the knowledge that
was needed to make a dramatic discovery and get close to what we now
call the ``maximum principle'' of optimal control, but failed to make it
for a fairly trivial reason. Specifically, it was merely the fact that
they chose one particular formalism rather than another that on the
surface looked nearly identical, that led them astray and prevented
them from making the discovery.

We will argue that

\medbreak
\begin{center}
\fbox{\fbox{\begin{minipage}{10.3cm}
{\large{\em The evolution of the calculus of variations reached a
critical fork in the road when Hamilton's equations were introduced.
The way the equations were actually written was only one
of two possible formulations, and may not have been the best one.
If Hamilton's equations had been written in a slightly
different way---using a Hamiltonian that is a function of all three
sets of variables (positions, velocities and momenta) rather than a
function of the positions and momenta only---then some important
discoveries that were not made until the 1950s would almost certainly
have been made much sooner. Weierstrass and Carath\'eodory were misled 
by the fact that only the ``wrong'' form of the equations was available 
to them.}}
\end{minipage}}}
\end{center}

\medbreak\noindent Before we get to Hamilton, Weierstrass, and 
Carath\'edory, it will be cobvenient to analyze the second-order
necessary condition due to Legendre.

\ssubsection{The Legendre condition}  Recall that the Euler-Lagrange system
is given by 
{\protect\huge
\begin{equation}\label{eulagfirst}
\fbox{$\frac{d\,\,}{dt}\Big(\frac{\partial L}
{\partial\dot x}\Big)=\frac{\partial L}{\partial x}$}
\end{equation}}
or
{\protect\Large
\begin{equation}\label{eulagvectoru} 
\fbox{$
\frac{d\,\,}{dt}\!\Biggl[\frac{\partial L}{\partial
u}\Big(\xi_*(t),\dot\xi_*(t),t\Big)\Biggr]\!\!= \frac{\partial L}{\partial
x}\Big(\xi_*(t),\dot\xi_*(t),t\Big)\,,$}
\end{equation}}
or
{\protect\Large
\begin{equation}\label{eulagvectoru-vec} 
\fbox{$
\frac{d\,\,}{dt}\!\Biggl[\frac{\partial L}{\partial
u^i}\Big(\xi_*(t),\dot\xi_*(t),t\Big)\Biggr]\!\!= \frac{\partial L}{\partial
x^i}\Big(\xi_*(t),\dot\xi_*(t),t\Big)\,,\quad i=1,\ldots,n\,.$}
\end{equation}}
This necssary condition for an optimum 
can be derived, as we shall see later, by looking at
the ``first variation'' $\delta I$ of the cost functional $I$, and requiring 
that $\delta I=0$. So in fact (\ref{eulagfirst}) is a 
necessary condition for the {\em stationarity} of $I$. 

The next natural step was to look at the second variation of $I$, 
and this was done in 1786 by {\bem Adrien-Marie 
Legendre} (1752-1833), who found an 
additional necessary  condition  for a minimum.
His condition, derived for the scalar case, is
\begin{equation}\label{legendre}
\fbox{\large{$
\frac{\partial^2 L}{\partial\dot x^2\,}\Big(\xi_*(t),\dot\xi_*(t),t\Big)\ge
0\,\,\,\,($}\small{${\rm i.e.,}$}\large{$\,\,\,
\frac{\partial^2 L}{\partial u^2}\Big(\xi_*(t),\dot\xi_*(t),t\Big)\ge 0)\,.$}}
\end{equation}
With an appropriate reinterpretation, Legendre's condition
(\ref{legendre}) is also necessary 
in the vector case: all we have to do is read 
(\ref{legendre}) as asserting that {\bem the Hessian matrix 
$\Bigl\{\frac{\partial^2 L\,\,\,\,}{\partial u^i\partial u^j}\Big(\xi_*(t),
\dot\xi_*(t),t\Big)\Bigr\}_{1\le i,j\le n}$ has to be nonnegative definite}, 
that is, that 
\begin{equation}\label{legendre-vec}
\fbox{\large{$
\sum_{i,j=1}^n\alpha^i\alpha^j\frac{\partial^2 L\,\,\,\,}{\partial u^i\partial u^j}\Big(\xi_*(t),\dot\xi_*(t),t\Big)\!\ge\!0\,\,
{\rm for\,\,all}\,\,\alpha\!=\!(\alpha^1,\ldots,\alpha^n)\!\in\!\rr^n.$}}
\end{equation} 



\ssubsection{Hamilton's missed opportunity} The work of 
{\bem William Rowan Hamilton} (1805-1865), published in 1834 and 1835,
produced a far-reaching reformulation of the Euler-Lagrange
equations. But we will see that this reformulation could have been
even more powerful if the new equations had been written in a
``slightly different'' form. To explain this, we will have to write
and analyze in detail both versions, the one that Hamilton used, and
the one that he could have used but did not, so that we can then 
compare them and show that what looks like a small difference
has in fact enormous implications. 

In a sense, the issue at stake will seem rather trivial, just a matter
of rewriting the Euler-Lagrange system in a different formalism.
However, sometimes formalisms can make a tremendous difference.  To
understand what happened and what could have happened but did not, let
us first see if we can make sense of the two necessary conditions for a
minimum that have been presented so far, namely, the Euler-Lagrange
equation (\ref{eulagfirst}) and the Legendre condition
(\ref{legendre}), and look for a single simple property that would
imply both conditions.

The Legendre condition is clearly the second-order
necessary condition for a minimum of a function, namely,
the function $\rr^n\ni u\mapsto L(\xi_*(t),u,t)$.
But (\ref{eulagfirst}) does not look at all
like the first-order condition for a minimum of that same function.
It is natural to ask whether there might be a way to relate the two
conditions. Is it possible that both could be expressed as necessary
conditions for a minimum of one and the same function?  The answer is
``yes,'' and understanding how this is done leads straight to optimal
control theory, the maximum principle, and far-reaching
generalizations of the classical theory.  But before we get there let
us tell the story of how Hamilton almost got there himself but missed,
and Weierstrass got even closer but missed as well.

Let us look at another way of writing (\ref{eulagfirst}).
Suppose a curve $t\mapsto{}\xi_*(t)$ is a solution of 
(\ref{eulagfirst}).  Define
a function $(x,u,p,t)\mapsto H(x,u,p,t)$ of three vector variables 
$x,u,p$ in $\rr^n$, and of $t\in\rr$, by letting
\begin{equation}\label{defofh}
\fbox{\large{$H(x,u,p,t)= \langle p,u\rangle-L(x,u,t)\,.$}}
\end{equation}
Then define the ``momentum'' $p$ by 
\begin{equation}\label{defofp}
\pi(t)=\frac{\partial L}{\partial u}\Big(\xi_*(t),\dot\xi_*(t),t\Big)\,.
\end{equation}
It is then clear that
$$
\frac{\partial H}{\partial p}=u\,,
$$
so along our curve $\xi_*$:
\begin{equation}\label{hamileqa}
\frac{d\xi_*}{dt\,\,\,}(t)=\frac{\partial H}{\partial p\,}
\Big(\xi_*(t),\dot\xi_*(t),\pi(t),t\Big)\,.
\end{equation}
Also,
$$
\frac{\partial H}{\partial x}=-\frac{\partial L}{\partial x}\,.
$$
Therefore (\ref{eulagvectoru}),  with $\pi(t)$ defined by (\ref{defofp}), 
says that
\begin{equation}\label{hamileqb}
\frac{d\pi}{dt}(t)=-\frac{\partial H}{\partial x}\Big(\xi_*(t),\dot
\xi_*(t),\pi(t),t\Big)\,. 
\end{equation}
Finally,
\begin{equation}\label{kangur}
\frac{\partial H}{\partial u}=p-\frac{\partial L}{\partial u}\,,
\end{equation}
so (\ref{defofp}) says:
\begin{equation}\label{hamileqc}
\frac{\partial H}{\partial u}\Big(\xi_*(t),\dot\xi_*(t),\pi(t),t\Big)=0\,.
\end{equation}
So we have shown that

\bigbreak
\begin{center}
\fbox{\fbox{
\begin{minipage}{11cm}
\protect\large The system of equations (\ref{hamileqa}), (\ref{hamileqb}), 
(\ref{hamileqc}), usually written more concisely as
\begin{equation}\label{hamileqsys}
\fbox{\Large{$
\frac{dx}{dt}=\frac{\partial H}{\partial p}\,,
\,\,\,\,\,\,\,\,\,\frac{dp}{dt}=-\frac{\partial H}{\partial x}\,,
\,\,\,\,\,\,\,\,\,\frac{\partial H}{\partial u}=0\,,$}}
\end{equation}
is {\bem exactly equivalent} to (\ref{eulagfirst}), provided that $H$ is
defined as in (\ref{defofh}).    
\end{minipage}}}\end{center}

\bigbreak

More precisely,

\begin{theorem}\label{aaaaaa}
Assume $n$ is a positive integer, 
$\Omega$ is an open subset of $\rr^n$, $a$, $b$ are real numbers such
that $a<b$, $L$ is a real-valued function on
$\Omega\times\rr^n\times[a,b]$, and $\bar x$, $\hat x$ are given
points in $\Omega$.  Assume that $L$ is a function of class $C^1$. Let
$\xi_*:[a,b]\mapsto\Omega$ be a curve of class $C^1$. Then the
Euler-Lagrange equation (\ref{eulagvectoru}) holds for all
$t\in[a,b]$ if and only if the ``control Hamilton equations''
(\ref{hamileqa}), (\ref{hamileqb}), (\ref{hamileqc}) are satisfied
for some choice of the function $t\mapsto\pi(t)$.  Moreover, in 
that case the momentum function $\pi$ is necessarily given by 
(\ref{defofp}).\proofend
\end{theorem}




\def\classical{(\ref{hamileqsysham})-(\ref{hamileqsyshamnewhham})-(\ref{pimpl})}

\medbreak\noindent We will call the function $H$ the {\bf control
Hamiltonian,} and refer to 
(\ref{hamileqsys}) as the {\bem control Hamiltonian form} of the
Euler-Lagrange equations.  In our view, Formula (\ref{defofh}) is the
definition that Hamilton should have given for the Hamiltonian, and
Equations (\ref{hamileqsys}) are {\em ``Hamilton's equations as he should
have written them.''}

What Hamilton actually wrote was (in our notation, not his)
\begin{equation}\label{hamileqsysham}
\frac{dx}{dt}=\frac{\partial\hham}{\partial p}\,,
\quad\,\,\,\,\frac{dp}{dt}=-\frac{\partial\hham}{\partial x}\,,
\end{equation}
where $(x,p,t)\mapsto\hham(x,p,t)$ is a function of $x,p$ and $t$ alone, 
defined by the formula 
\begin{equation}\label{hamileqsyshamnewhham}
\hham(x,p,t)=\langle p,\dot x\rangle-L(x,\dot x,t)\,,
\end{equation}
which resembles (\ref{defofh}) but is not at all the same. The 
difference is that in Hamilton's definition $\dot x$ is supposed to be
treated {\sl not} as an independent variable, but as a function 
of  $x,p,t$,   defined implicitly by the equation
\begin{equation}\label{pimpl}
p=\frac{\partial L}{\partial\dot x}(x,\dot x,t)\,.
\end{equation}
It is easy to prove the following:

\medbreak
\begin{center}
\fbox{\fbox{\begin{minipage}{10.5cm}
{\bf If the map $(x,\dot x,t)\mapsto{}(x,p,t)$ defined
by (\ref{pimpl}) can be inverted,} {\em i.e., if we can ``solve (\ref{pimpl})
for $\dot x$ as a function of $x,p,t$,''  then (\ref{hamileqsysham})
is equivalent to (\ref{hamileqsys}).}   
\end{minipage}}}\end{center}

More precisely,

\begin{theorem} Assume $n$ is a positive integer, 
$\Omega$ is an open subset of $\rr^n$, $a$, $b$ are real numbers such
that $a<b$, $L$ is a real-valued function on
$\Omega\times\rr^n\times[a,b]$, and $\bar x$, $\hat x$ are given
points in $\Omega$.  Assume that $L$ is a function of class $C^1$. Let
$\xi_*:[a,b]\mapsto\Omega$ be a curve of class $C^1$. 
Let $\pi$ be the function given by (\ref{defofp}).
Assume that for some $\delta>0$  there exists a map
$$
U_\delta\ni(x,p,t)\mapsto v(x,p,t)\,,
$$
of class $C^1$, defined on the set
$$
U_\delta=\{(x,p,t):a\le t\le b\,,\,\|x-\xi_*(t)\|<\delta\,,\,
\|p-\pi(t)\|<\delta\}\,,
$$
such that 
\begin{equation}\label{gugug}
p=\frac{\partial L}{\partial u}(x,u,t)\Longleftrightarrow
u=v(x,p,t)\quad
{\rm whenever}\quad (x,p,t)\in U_\delta\,,\,\,u\in\rr^n\,.
\end{equation}
Define 
\begin{equation}\label{hamileqsyshamnewhham-bababa}
\hham(x,p,t)=\langle p,v(x,p,t)\rangle-L(x,v(x,p,t),t)\,.
\end{equation}
Then the Euler-Lagrange equation (\ref{eulagvectoru}) holds for all
$t\in[a,b]$ if and only if the ``Hamilton equations''
\begin{equation}\label{hamileqsysham-bababa}
\frac{dx}{dt}=\frac{\partial\hham}{\partial p}\,,
\quad\,\,\,\,\frac{dp}{dt}=-\frac{\partial\hham}{\partial x}\,,
\end{equation}
are satisfied along $(\xi_*,\pi)$, that is, if and only if
\begin{eqnarray*}\label{hamexpanded}
\dot\xi_*(t)&=&\phantom{-}\frac{\partial\hham}{\partial p}(\xi_*(t),
\pi(t),t)\,,\\
\dot\pi(t)\,\,&=&-\frac{\partial\hham}{\partial x}(\xi_*(t),\pi(t),t)\,,
\end{eqnarray*}
for all $t\in[a,b]$.\proofend
\end{theorem}
\begin{proof} 
If $(x,p,t)\in U_\delta$, then it is clear that
$$
\hham(x,p,t)=H(x,v(x,p,t),p,t)\,.
$$
So the chain rule tells us that
\begin{equation}
\frac{\partial\hham}{\partial x}=\frac{\partial H}{\partial x}
+\frac{\partial H}{\partial u}\cdot \frac{\partial v}{\partial x}\,,
\end{equation}
that is, 
\begin{equation}
\frac{\partial\hham}{\partial x}(x,p,t)=\frac{\partial H}{\partial x}(x,
v(x,p,t),p,t)+\frac{\partial H}{\partial u}(x,v(x,p,t),p,t)\cdot
\frac{\partial v}{\partial x}(x,p,t)\,.
\end{equation}
Now, (\ref{kangur}) implies
\begin{equation}\label{kangur-bbb}
\frac{\partial H}{\partial u}(x,v(x,p,t),p,t)=
p-\frac{\partial L}{\partial u}(x,v(x,p,t),p,t)=p-p=0\,.
\end{equation}
Therefore 
\begin{equation}\label{kangur-aaa}
\frac{\partial\hham}{\partial x}(x,p,t)=\frac{\partial H}{\partial x}(x,
v(x,p,t),p,t)\,,
\end{equation}
and, similarly,
\begin{equation}\label{kangur-aaa-aa}
\frac{\partial\hham}{\partial p}(x,p,t)=\frac{\partial H}{\partial p}(x,
v(x,p,t),p,t)\,.
\end{equation}
If $\xi_*$ is a solution of Euler-Lagrange and $\pi$ is defined by
(\ref{defofp}), then (\ref{defofp}) and (\ref{gugug}) imply
$\dot\xi_*(t)=v(\xi_*(t),\pi(t),t)$, and Theorem \ref{aaaaaa} 
tells us that 
\begin{eqnarray*}
\dot\xi_*(t)&=&\phantom{-}\frac{\partial H}{\partial p}(\xi_*(t),
\dot\xi_*(t),\pi(t),t)\,,\\
\dot\pi(t)&=&-\frac{\partial H}{\partial x}(\xi_*(t),\dot\xi_*(t),\pi(t),
t)\,,
\end{eqnarray*}
so 
\begin{eqnarray*}
\dot\xi_*(t)&=&\phantom{-}\frac{\partial H}{\partial p}(\xi_*(t),
\dot\xi_*(t),\pi(t),t)\\
&=&\phantom{-}\frac{\partial H}{\partial p}(\xi_*(t),
v(\xi_*(t),\pi(t),t),\pi(t),t)\\
&=&\phantom{-}\frac{\partial\hham}{\partial p}(\xi_*(t),\pi(t),t)\,,\\
\dot\pi(t)&=&-\frac{\partial H}{\partial x}(\xi_*(t),
\dot\xi_*(t),\pi(t),t)\\
&=&-\frac{\partial H}{\partial x}(\xi_*(t),
v(\xi_*(t),\pi(t),t),\pi(t),t)\\
&=&-\frac{\partial\hham}{\partial x}(\xi_*(t),\pi(t),t)\,,
\end{eqnarray*}
showing that (\ref{hamileqsysham-bababa}) is satisfied.
The converse is also easily proved.%\proofend
\end{proof}


\ssubsection{The ``control Hamiltonian'' versus Hamilton's Hamiltonian}
It should be clear from the above discussion that the Hamiltonian
reformulation of the Euler-Lagrange equations in terms of the ``control
Hamiltonian'' is at least as natural as the classical version
\classical, and perhaps even
simpler.  Moreover, the control formulation has at least one obvious
advantage, namely,

\medbreak 

\begin{center}
\fbox{\fbox{\begin{minipage}{10.5cm}
{\bf (A1)} {\em the control version of the Hamilton equations is
{\bem exactly equivalent} to the Euler-Lagrange system under
completely general conditions, whereas the classical version
\classical{} only makes sense when the transformation (\ref{pimpl})
can be inverted, at least locally, to solve for $\dot x$ as a function
of $x,p,t$.}
\end{minipage}}}\end{center}

\medbreak\noindent 
We now show that (A1) is not the only advantage of the control view 
over the classical one. To see this, we must take another look at Legendre's
condition (\ref{legendre}).   Since $H(x,u,p,t)$ is equal to 
$-L(x,u,t)$ plus a linear function of $u$, (\ref{legendre}) is completely
equivalent to 
\begin{equation}\label{legendreham}
\frac{\partial^2 H}{\partial u^2}\Big(\xi_*(t),\dot\xi_*(t),\pi(t),
t\Big)\le 0\,,
\,\,\,\,\,\,\,\,{\rm i.e.,\,\,\,\,\,\,\,\,}\frac{\partial^2 H}{\partial
\dot x^2\,}\Big(\xi_*(t),\dot\xi_*(t),\pi(t),t\Big)\le 0\,.
\end{equation}
Now let us write (\ref{legendreham}) alongside the third
equation of  
(\ref{hamileqsys}):
\begin{equation}\label{legendrehamtog}
\fbox{\Large {$
\frac{\partial H}{\partial u}=0\,\,\,\,\,\,\,\,$}\small{${\rm and}
$}\Large{$\,\,\,\,\,\,\,\,\frac{\partial^2 H}{\partial u^2}\le 0$}}\,\,\,,
\end{equation}
and stare at the result for a few seconds.

These equations unmistakably suggest something!   They show that 

\medbreak
\begin{center}
\fbox{\begin{minipage}{10.5cm}
\noindent{\bem the Euler-Lagrange and Legendre conditions have to do 
with critical points---with  respect to the velocity variable 
$u$---of a function $H$.  Precisely, 
Euler-Lagrange says that the gradient  $\nabla_uH$ of 
this function has to vanish, and then Legendre adds to this the extra 
requirement that the second 
$u$-derivative of $H$---i.e., the Hessian matrix 
$\{\frac{\partial^2 H}{\partial u_i\partial u_j}\}_{1\le i,j\le n}$---must 
be nonpositive definite.}
\end{minipage}}\end{center}

\medbreak\noindent The message should be clear:  {\bf what must
be happening here is that $H$ has a maximum as a function of $u$.}

We state this formally as a  conjecture.

\medbreak
\begin{center}
\fbox{\fbox{\begin{minipage}{10.5cm}
\noindent{\bem CONJECTURE M:}\ {\em Besides 
(\ref{hamileqsys}) (or the equivalent form (\ref{eulagfirst})), 
an additional necessary condition for optimality 
of an arc $\xi_*$ should be that the function
$\rr^n\ni u\mapsto H(\xi_*(t),u,\pi(t),t)\in\rr$ have a
maximum at $\dot\xi_*(t)$ for each t. That is, ``the velocity 
that is actually used must maximize the Hamiltonian---for
fixed time, position and momentum---among all possible values
of the velocity vector.''} 
\end{minipage}}}\end{center}

\medbreak \noindent We contend  that

\bi
\item[1.] Conjecture M is an extremely natural consequence of 
rewriting Hamilton's equations ``as Hamilton should have done it,'' in terms 
of we are calling the ``control Hamiltonian.''
\item[2.] It is reasonable to guess that, if Hamilton had actually written 
``Hamilton's equations'' in this alternative way, then he
himself, or some other 19th century mathematician, would have written
(\ref{legendrehamtog}) and been led by it to the conjecture. 
\item[3.] On the other hand, it is only by using the Hamiltonian of 
(\ref{defofh}), as opposed to Hamilton's own form of the Hamiltonian, that one
can see that  the Euler-Lagrange and Legendre condition really are 
first- and second-order conditions about critical points 
{\bf of the same function.}   This function
cannot be $L$ itself, because the first order conditions do {\it not} say
that $\frac{\partial L}{\partial u}=0$. Nor can it be Hamilton's
Hamiltonian $\hham$, which isn't even a function of $u$. {\em Only the use
of the ``control'' Hamiltonian leads naturally to Conjecture M.}
\ei

\medbreak\noindent 
It will turn out that 


\medbreak
\begin{center}
\begin{minipage}{10.5cm}
\noindent{\bf Conjecture M is true. Moreover, once its truth is known
then vast generalizations are possible.}
\end{minipage}\end{center}

\bigbreak

\ssection{WEIERSTRASS MISSES ANOTHER OPPORTUNITY}

\noindent The contributions of {\bem Karl Weierstrass} (1815-1897) to 
the calculus of variations are contained in the notes of his three
series of lectures on the topic, given in 1875, 1879 and 1882, and not
formally published as books or journal articles.

\ssubsection{The Weierstrass side condition} Weierstrass considered the 
problem of minimizing an integral $I(\xi)$ of
the form 
$$
I(\xi)=\int_a^bL(\xi(s),\dot\xi(s))\,ds\,,
$$
for Lagrangians $L$ such that

\medbreak
\begin{center}
\begin{minipage}{12cm}
\bi
\item[{\bem (W)}] {\em $L(x,\dot x)$ is positively homogeneous
with respect to the velocity $\dot x$ (that is, $L(x,\alpha\,\dot x)=
\alpha L(x,\dot x)$ for all $x,\dot x$ and all $\alpha\ge 0$)
and does not depend on time.}\proofend
\ei
\end{minipage}\end{center}

\medbreak\noindent As will become clear from Remark \ref{snott}, we
have a good reason for  
using  $s$ rather than $t$ as the ``time'' variable in the
expression for $I$.

\begin{remark}{\rm \label{snott} In a sense, one can always assume (W) 
``without loss of generality.''  Indeed, suppose we are interested in
a completely arbitrary problem $\mathcal{P}(L,a,b,\bar x,\hat x)$
of the standard form considered earlier, namely,
\begin{equation}\label{standard}
\left\{\begin{array}{rcl}
{\rm given\,\,}&\,\,& L,\,a,\,b,\,\bar x,\,\hat x\,,\\
\,\,&\,\,&\,\,\\
{\rm minimize\,\,}&\,\,& I=\int_a^bL(\xi(t),\dot\xi(t),t)dt\,,\\
\,\,&\,\,&\,\,\\
{\rm subject\,\,to\,\,}&\,\,&\xi(a)=\bar x\,\,\,{\rm and}\,\,\xi(b)=\hat x\,.
\end{array}\right.
\end{equation} 
Then we can always define
a new function $\Lambda(x,t,u,\tau)=\tau L(x,u/\tau,t)$, and think of
$t$ as a new $x$ variable, say $x^0$, and of $\tau$ as\,,
$\frac{dx^0}{ds\phantom{^0}}$, were $s$ is a new time variable, or 
``pseudotime,''
not to be confused with the true time variable $t$.  Then, using
$x_{new}$ for $(x,x^0)$---so that $x_{new}=(x,t)$---we have
\begin{eqnarray*}
\int L(x,\dot x,t)dt&=&\int L(x,\frac{dx}{dt},t)dt\\
&=&\int L\Biggl(x,\Big(\frac{dt}{ds}\Big)^{-1}\frac{dx}{ds},t\Biggr)
\frac{dt}{ds}\,ds\\
&=&\int\Lambda\Biggl(x_{new},\frac{dx_{new}}{ds\phantom{_{new}}}\Biggr)\,ds\,,
\end{eqnarray*}
so the minimization problem for $\Lambda$ is equivalent to
(\ref{standard}). Moreover, if $\alpha\!\!>\!\!0$, then
$$
\Lambda(x,t,\alpha\,u,\alpha\,\tau)\!=\!\alpha\,\Lambda(x,t,u,\tau)\,,
$$
i.~e.,
$$
\Lambda(x_{new},\alpha\,\dot{x}_{new})\!=\!\alpha\,\Lambda(x_{new},
\dot{x}_{new})\,,
$$
where we are now using the dot to denote differentiation
with respect to $s$ rather than $t$.  So the new Lagrangian $\Lambda$ is
positively homogeneous with respect to the new velocity variable
$\dot{x}_{new}$ and does not depend on the new time $s$. \proofend }
\end{remark}

So when Weierstrass imposed Condition (W) on his Lagrangians, he was doing
so ``without loss of generality.''  However, ``without loss of generality''
is a dangerous phrase, and does not at all entail ``without loss of
insight.''  We shall argue below that this restriction, in conjunction with
the dominant view that Hamilton's equations had to be written in the form
(\ref{hamileqsysham}), may have served to conceal from Weierstrass the true
meaning and the far-reaching implications of the new condition he
discovered.

Weierstrass introduced what is now known as

\medbreak
\begin{center}
\fbox{\fbox{\begin{minipage}{11cm}
\protect\Large
\null\medbreak
\begin{center}
THE WEIERSTRASS EXCESS FUNCTION,
\end{center}
\begin{equation}\label{excess1}
\mathcal{E}(x,\bar u,u)=L(x,u)-\frac{\partial L}{\partial u}(x,\bar u)
\cdot u\,.
\end{equation}
\end{minipage}}}\end{center}

\medbreak\noindent This function depends on three sets of independent 
variables, namely, $x$, $\bar u$ and $u$.  That is, 
{\bem $\mathcal{E}$ is a function of position and two velocity vectors.}

He then proved  his {\em side condition}:

\bigbreak
\begin{center}
\begin{minipage}{11cm}
\bi
\item[{\large{\bem (SC)}}] {\large{\em For a curve $s\mapsto{}\xi_*(s)$ to
be a solution of the 
minimization problem, the function $\mathcal{ E}$ has to be $\ge 0$ when
evaluated for $x=\xi_*(s)$, $\bar u=\dot\xi_*(s)$, and a completely arbitrary 
$u$.}}
\ei 
\end{minipage}\end{center}

\bigbreak\noindent 
Notice that Lagrangians with Property (W) satisfy the identity
\begin{equation}\label{excess-hom}
L(x,u)=\frac{\partial L}{\partial u}(x,u)\cdot u\,.
\end{equation}
Therefore Weierstrass could equally well have written his excess function as 

\medbreak\begin{center}
\fbox{\fbox{
\begin{minipage}{12cm}
\protect\large
\begin{equation}\label{excess2}
\mathcal{E}(x,\bar u,u)=L(x,u)-\frac{\partial L}{\partial u}(x,\bar u)\cdot u
-\Bigl(L(x,\bar u)-\frac{\partial L}{\partial u}(x,\bar u)\cdot\bar u\Bigr)\,,
\end{equation}
$\,$
\end{minipage}}}\end{center}
and he could even have generalized this formula to the time-dependent
case, by writing

\medbreak\begin{center}
\fbox{\fbox{
\begin{minipage}{12cm}
\protect\large
\begin{equation}\label{excess2-aaa}
\mathcal{E}(x,\bar u,u,t)\!\!=\!\!L(x,u,t)\!-\!\!\frac{\partial L}{\partial u}
(x,\bar u,t)\!\cdot\!u\!-\!\!\Bigl(L(x,\bar u,t)\!-\!\!\frac{\partial L}
{\partial u}(x,\bar u,t)\!\cdot\!\bar u\Bigr).
\end{equation}
$\,$
\end{minipage}}}\end{center}


\begin{remark}\label{interp-excess}
{\rm Formulae (\ref{excess2}) and (\ref{excess2-aaa}) 
have a simple geometric interpetation. 
Assume, for simplicity, that $L$ does not depend on $t$.
For a given point $x$, let $G_x$ be the  graph of 
$$
y=L(x,u)\equaldef L_x(u)
$$ 
as a function of $u$. Fix a value $\bar u$ of $u$, and let 
$$
\bar y=L(x,\bar u)\,,
$$
so the point 
$$
\bar P=(\bar u,\bar y)
$$
lies on $G_x$. Let $\Gamma_{x,\bar u}$ be the tangent hyperplane to $G_x$
at $\bar P$.  Then the equation of $\Gamma_{x,\bar u}$ is 
$$
y=\bar y+\frac{\partial L}{\partial u}(x,\bar u)\cdot(u-\bar u)\,,
$$
that is 
$$
y=L(x,\bar u)+\frac{\partial L}{\partial u}(x,\bar u)\cdot(u-\bar u)\,.
$$

\noindent Given any $u$, the ordinates $y_{graph}(u)$, $y_{plane}(u)$ 
of the points of $G_x$, $\Gamma_{x,\bar u}$,
that lie on the line $\{u\}\times\rr$ are given by 
$$
y_{graph}(u)=L(x,u)\,,\quad y_{plane}(u)=
L(x,\bar u)+\frac{\partial L}{\partial u}(x,\bar u)\cdot(u-\bar u)\,.
$$
Therefore
$$
\mathcal{E}(x,\bar u,u)=y_{graph}(u)-y_{plane}(u)\,.
$$
In other words (cf. Figure \ref{excess-fig}): 

\medbreak
\begin{center}
\begin{minipage}{11cm}
{\large{\bem The number $\mathcal{E}(x,\bar u,u)$ tells 
us by how much the actual value of the function $y=L_x(u)$ 
at a particular $u$ is larger than the value at $u$  of the linearization
(i.e., tangent approximation) of this function at the point
$(\bar u,L(x,\bar u))$.}}
\end{minipage}
\end{center}

\bigskip
\vbox{
\[
\epsfbox{uxcess.eps}
\]
\figurecap{The Weierstrass side condition\label{excess-fig}}
}

\newpage

Therefore the Weierstrass side condition says that 

\medbreak
\begin{center}
\begin{minipage}{11cm}
\bi
\item[{\large{\bem (SC)}}] {\large{\em 
If a curve $t\mapsto{}\xi_*(t)$ is a solution of the 
minimization problem, then for every $t$, if we draw the tangent 
hyperplane to the graph of the function $u\mapsto L(\xi_*(t),u,t)$ 
at the point of this graph corresponding to $u=\dot\xi_*(t)$, then the graph
must lie entirely above the hyperplane.}}\proofend  
\ei 
\end{minipage}\end{center}}
\end{remark}

\ssubsection{A Hamiltonian rewriting of the Weierstrass condition}
Using
$$
\pi(t)=\frac{\partial L}{\partial u}(\xi_*(t),\dot\xi_*(t),t)\,,
$$ 
as in (\ref{defofp}), we see that 
\begin{eqnarray}
\label{pig}&\,&\mathcal{E}(\xi_*(t),\dot\xi_*(t),u,t)\\
&\,&\qquad=\Big(L(\xi_*(t),u,t)-\langle \pi(t),u\rangle\Bigr)
-\Bigl(L(\xi_*(t),\dot\xi_*(t),t)-\langle\pi(t),\dot\xi_*(t)\rangle
\Bigr)\,,\nonumber
\end{eqnarray}
which the reader will immediately recognize as saying that
\begin{equation}\label{indep}
\mathcal{E}(\xi_*(t),\dot\xi_*(t),u,t)=
H(\xi_*(t),\dot\xi_*(t),\pi(t),t)-H(\xi_*(t),u,\pi(t),t)\,,
\end{equation}
where $H$ is our ``control  Hamiltonian.''    So Weierstrass' condition,
expressed in terms of the control Hamiltonian, simply says that

\medbreak
\begin{center}
\begin{minipage}{11.4cm}
\bi
\item[{\large{\bem (MAX)}}] {\large{\em  Along an optimal curve $t\mapsto{}
\xi_*(t)$, if we define 
$\pi(t)$ via (\ref{defofp}), then for every $t$ the value $u=\dot\xi_*(t)$
must maximize the (control) Hamiltonian $H(\xi_*(t),u,\pi(t),t)$ as a function
of $u$.}}
\ei
\end{minipage}\end{center}

Moreover, (MAX) can be considerably simplified.  Indeed, 

\medbreak
\begin{center}
{\bf\em The requirement that  $\pi(t)$ be defined via (\ref{defofp}) is now 
redundant.}
\end{center}

\medbreak\noindent To see this, observe that,
if $H(\xi_*(t),u,\pi(t),t)$, regarded as a function of $u$, 
has a maximum at $u=\dot\xi_*(t)$, then
$\frac{\partial H}{\partial u}(\xi_*(t),\dot\xi_*(t),\pi(t),t)$ has to vanish,
so  $\pi(t)$ must be given by (\ref{defofp}). Moreover, the vanishing of   
$\frac{\partial H}{\partial u}(\xi_*(t),\dot\xi_*(t),\pi(t),t)$ is also one of 
the conditions  of  (\ref{hamileqsys}). So we can just drop
(\ref{defofp}) altogether, and state (\ref{hamileqsys}) and (MAX) together:

\medbreak
\begin{center}
\fbox{\fbox{\begin{minipage}{12cm}
\begin{minipage}{1.6cm}
\null$\,$\\
\noindent{\large{\bem (NCO)}}\\
\vspace{4cm}
\end{minipage}
\begin{minipage}{10.35cm}
{\large{\em  If a curve $t\mapsto{}\xi_*(t)$  is a solution of the  
minimization problem (\ref{standard}), then there has to exist a function
$t\mapsto{}\pi(t)$ 
such that the following three conditions hold for all $t$:}
\begin{eqnarray}
\quad\quad\dot\xi_*(t)&=&\phantom{-}\frac{\partial H}{\partial p}(\xi_*(t),
\dot{\xi}_*(t),\pi(t),t)\,,\label{nco1}\\
\quad\quad\dot\pi(t)&=&-\frac{\partial H}{\partial x}(\xi_*(t),
\dot\xi_*(t),\pi(t),t)\,,\label{nco2}\\
\quad\quad
H(\xi_*(t),\dot{\xi}_*(t),\pi(t),t)&=&\max_{u}
H(\xi_*(t),u,\pi(t),t)\,.\label{nco3} 
\end{eqnarray}}
\end{minipage}\end{minipage}}}\end{center}

\medskip\noindent As a version of the necessary conditions for
optimality, (NCO) encapsulates in one single statement the combined
power of the Euler-Lagrange necessary conditions and the Weierstrass
side condition as well, of course, as the Legendre condition, which
obviously follows from (NCO).  Notice the elegance and economy of
language achieved by this unified statement: there is no need to bring
in an extra entity called the ``excess function.'' Nor does one need
to include a formula specifying how $\pi(t)$ is defined, since
(\ref{nco3}) does this automatically.  So the addition of the new
Weierstrass condition to the three equations of (\ref{hamileqsys})
results in a new set of three, rather than four, conditions, a set
much ``simpler than the sum of its parts.''  Notice moreover that
(\ref{nco3}) is exactly Conjecture M.  So we can surmise at this point that
(NCO), as stated, could probably have been discovered soon after the
work of Hamilton---since it is strongly suggested by
(\ref{legendrehamtog})---and almost certainly by Weierstrass, if only
Hamilton's equations had been written in the form (\ref{defofh}),
(\ref{hamileqsys}).

So we can now add two new items to our list of advantages of the ``control
formulation'' of Hamilton's equations over the classical one:

\medbreak\noindent{\bf (A2)}\ {\em Using the control Hamiltonian, it would
have been an obvious 
next step to write Legendre's condition in ``Hamiltonian form,''
as in (\ref{legendrehamtog}), and this would have led immediately 
to the formulation of Conjecture M, a proof of which would then have 
been found soon after.}

\medbreak\noindent{\bf (A3)}\ {\em In terms of the control Hamiltonian, 
the Weierstrass side condition
has a much simpler statement, not requiring the introduction of an 
``excess function,'' and can be combined with the Hamilton equations
into an elegant unified formulation (NCO) of the necessary conditions for 
optimality.}

\ssubsection{What Weierstrass missed}
This is by no means the end of our story.  There is
much more 
to the new formulation (NCO) than just elegance and simplicity.  If you
compare (NCO) with all the other necessary conditions that we had written
earlier, a remarkable new fact becomes apparent.  
Quite amazingly, {\bem the derivatives with respect to the $u$ variable are
completely gone.}  
All the earlier equations involved $u$-derivatives of
$L$ or of $H$, and even if we use the classical version
(\ref{hamileqsysham}) of Hamilton's equations---which involves no functions
of $u$ and therefore no $u$-derivatives---the fact remains that in order to
get to (\ref{hamileqsysham}) we first have to solve (\ref{pimpl}), which
does involve a $u$-derivative.

Now, if our necessary conditions for optimality can be stated without any
references to $u$-derivatives, it is reasonable to guess that the
$u$-derivatives of $L$ are not needed.  Also, the minimization that occurs
in (\ref{nco3}) makes sense over any subset $U$ of $\rr^n$, so there is
no longer any reason to insist that the range of values of $u$ be the whole
space.   This leads us to 

\medbreak
\begin{center}
\begin{minipage}{12cm}
\noindent{\bem CONJECTURE  M2:}\ {\em  (NCO)  should still be a 
necessary condition for
optimality even for problems where the derivative $\dot x$ is restricted 
to belong to some subset $U$ of $\rr^n$, and $L(x,u,t)$ is not required 
to be differentiable or even continuous with respect to $u$.} \proofend
\end{minipage}\end{center}

\bigskip\noindent
Naturally, once $u$ is restricted to a subset $U$ of $\rr^n$, and
$L(x,u,t)$ is not required to be differentiable with respect to $u$,
the maximization condition (\ref{nco3}) no longer implies
(\ref{defofp}).  So we see that {\bf\em (\ref{nco3}) is more
fundamental than (\ref{defofp})}. 

Notice also that there has been a subtle change in the formulation of
the necessary conditions for optimality.  In the classical conditions
of the calculus of variations (including Weierstrass' own version of
the side condition), the covector $\pi(t)$---the ``momentum''---is defined
by (\ref{defofp}), and the necessary conditions are equations happening
along the trajectory $\xi_*$. These equations can be formulated somewhat
more simply using $\pi(t)$, but $\pi(t)$ can also be completely eliminated,
\mbox{since} one can always substitute for it its value given by
(\ref{defofp}).  The new version of the necessary conditions for an optimum
given by Conjecture M2 is of a different
form: rather than state that certain equations hold along $\xi_*$, the
conditions are now an {\em existential statement}, asserting the
{\bem existence} of a vector-valued function $t\mapsto\pi(t)$ with certain
properties.  The elimination of $\pi(t)$, by expressing $\pi(t)$ in terms of 
$\xi_*(t)$, $\dot\xi_*(t)$ and $t$, is only possible when $U=\rr^n$
and $L$ is differentiable with respect to $u$, in which case
(\ref{defofp}) follows from (\ref{nco3}).  Otherwise, $\pi(t)$ is a ``new,''
``independent'' object.        

Since Weierstrass was working under assumptions that imply 
(\ref{defofp}), he missed the chance to see that his condition
was really about $\pi(t)$, not about $\frac{\partial L}{\partial u}$
evaluated along $\xi_*$, and that, stated this way, his condition
was valid for much more general problems, for which $\pi(t)$ 
no longer equals $\frac{\partial L}{\partial u}(\xi_*(t),\dot\xi_*(t),t)$. 

\ssubsection{A toy problem} To test Conjecture M2, let us
look at a toy problem where the answer is  
reasonably easy to guess directly, and see whether using 
(NCO) gives the same result. 

\begin{example}{\rm \label{toy2} 
Let $T$ be a fixed positive number.
Suppose that we want to find a real-valued (Lipschitz) function
$t\mapsto{}x(t)$ on the interval $[0,T]$ that satisfies
\begin{equation}\label{theco}
x(0)=1\,,\,\,\,\,x(T)=1\,,\,\,{\rm and}\,\,\,|\dot x(t)|\le 1
\,\,\,{\rm for\,\,\, all\,\,\,}t\in[0,T]\,,
\end{equation}
and minimizes the integral 
$$
I=\int_0^Tx(t)^2dt
$$
among all the functions
satisfying the constraints (\ref{theco}).  This looks
exactly like a calculus of variations problem of the classical type---with
a Lagrangian $L$ given by
$$
L(x,\dot x,t)=x^2\,,
$$
except that the derivative $\dot x$ is
required to satisfy an ``inequality constraint'' $|\dot x|\le 1$.  If we
apply the classical Euler-Lagrange equations formally, we get
$$
\frac{\partial L}{\partial x}=0\,,
$$
since $L$ does not depend on $\dot x$, so 
$\frac{\partial L}{\partial\dot x}=0$,  and then
$$
\frac{\partial L}{\partial x}=\frac{d\,\,}{dt}
\Biggl(\frac{\partial L}{\partial\dot x}\Biggr)=0\,.
$$
But
$$
\frac{\partial L}{\partial x}=2x\,.
$$
So $x(t)\equiv 0$, contradicting the endpoint constraints
$x(0)=1$, $x(T)=1$.
So {\bem there is no way at all to
satisfy Euler-Lagrange together with the boundary conditions}.


On the other hand, we can apply (NCO) formally, making the sensible guess
that in this case the maximization with respect to $u$ should be made over
the set $U=[-1,1]$ of permissible values of $u$. Then (\ref{nco1}) and
(\ref{nco2}) become the equations $\dot x=u$, $\dot p=2x$. This
suggests, among other things, that $p$ as a function of $t$ should be of
class $C^1$, since $\dot p$ has to be continuous. In addition, (\ref{nco3})
says that $u=1$ when $p>0$, and $u=-1$ when $p<0$.  (When $p=0$ all values
of $u\in U$ are solutions.)  So when $p>0$ we have $\ddot p=2$, and when
$p<0$ we have $\ddot p=-2$.  This shows that $p(\cdot)$ is a convex
function of $t$ when $p(t)>0$ and a concave function when $p(t)<0$.  Now, a
continuous function on $[0,T]$ which is convex wherever it is $>0$ and
concave wherever it is $<0$ can only be of one of the 13 types $P$, $N$,
$Z$, $PZ$, $PN$, $ZP$, $ZN$, $NP$, $NZ$, $PZP$, $PZN$, $NZP$ or $NZN$,
where ``$Z$'' means ``identically zero,'' ``$P$ means ``$>0$,'' ``$N$''
means ``$<0$,'' and, for example, a symbol such as $PZN$ means ``a positive
piece followed by a zero piece and then by a negative piece.''  (This is
easy to prove. For example, $ZPZ$ is impossible because the $P$ part would
have to vanish at two points and be convex and $>0$ in between, which
clearly cannot happen.  The reader should analyze all possibilities and
verify that only our 13 cases can arise.)  Clearly, $P$ corresponds to
$u=1$, and $N$ to $u=-1$.  As for $Z$, it corresponds to $p(t)\equiv 0$,
which yields $x(t)\equiv 0$ (since $\dot p=2x$) and then $u(t)\equiv 0$
(since $\dot x=u$).  So $Z$ corresponds to $x=u=0$.  (Notice that
(\ref{nco3}) does not directly determine the value of $u$ when $p=0$, since
in this case every value of $u$ satisfies (\ref{nco3}), but by suitably
combining the three conditions of (NCO) we have still been able to find $u$
for $p=0$.)  With this in mind, it is clear that the only way to satisfy
the endpoint conditions $x(0)=1$, $x(T)=1$, is with a $PN$ or an $NP$ or an
$NZP$ curve.  Of these three possibilities, $PN$ is obviously not optimal,
so we are left with $NP$ and $NZP$, and then a direct analysis shows that
the minimum is $NP$ if $T\le 2$ and $NZP$ if $T>2$. This result can of
course be proved directly, and corresponds to the obvious intuition that to
minimize $\int x^2$ one should try to move from the initial value $x=1$ towards
$x=0$ as fast as possible, stay at $x=0$ as long as as possible, and then
move back towards $x=1$, choosing the departure time {}from $x=0$ so as to
arrive to $x=1$ exactly at time $T$. 

So {\bem (NCO) gives us the correct solution whereas the Euler-Lagrange
equation is not satisfied by this solution}. \proofend 
}\end{example}

\end{document}







