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\centerline{\Large \bf Dirichlet's Principle}
\bigskip
By 1840 it was known that if $S\subset \R$ is a closed and bounded set
and $f:S \to \R$ is a continuous function, then there are points $p$
and $q$ in $S$ where $f$ has its maximum and minimum value.
\smallskip
Mathematicians and physicists were considering more complicated
functions, such as, on a smooth surface $S $ in $\R^3$ finding the
shortest path in the surface joining the two points $p$ and $q$. If
we write the curve as $\vec{\gamma}(t)=(x(t),\, y(t),\,z(t))\subset S$
where $\vec{\gamma(0)}=p$ and $\vec{\gamma}(1)=q$, then the length of
the curve is
$$
J(\vec{\gamma})=\int_0^1 \abs{\vec{\gamma}\,'(t)}\,dt.
$$
To find the curve minimizing the distance we need to look at all
curves in the surface and find the curve minimizing $J$. Thus we seek
functions $x(t)$, $y(t)$, and $z(t)$. If the surface is smooth, is
there always a minimizing curve? If so, is it smooth?
\medskip
Historically, the first interesting problem of this sort was to study
a bead, starting from rest, sliding down a curve under the influence of
gravity. In particular, given the points $P$ and $Q$, find the
equation of curve $y=f(x)$ from $P$ to $Q$ so the particle arrives at
$Q$ in the least time. This is called the {\it Brachistochrone
Problem}. The solution is interesting -- and not at all obvious.
Look it up.
\medskip
One problem for a function $u(x,y)$ in several variables arose in a
number of applications. Let $\Omega\subset\R^2$ be a bounded region
with a smooth boundary $\partial\Omega$. and let $f(x,y)$ be a smooth
function defined on the boundary, $\partial\Omega$. We seek a
function $u(x,y)$ that minimizes the ``energy"
\begin{equation} \label{Dir}
J(v)=\iint_\Omega [v_x^2 + v_y^2]\,dx\,dy =\iint_\Omega\abs{\nabla
v}^2\,dx\,dy
\end{equation}
among all functions $v(x,y)$ that agree with $f$ on the boundary:
$v(x,y)=f(x,y)$ for $(x,\,y)\in \partial\Omega$. In 1851, for his
proof of what we call the Riemann Mapping Theorem, Riemann was seeking a
minimizer since this minimizer it would be a solution of the Laplace
equation:
\begin{equation}\label{laplace}
u_{xx} + u_{yy}= 0 \ \ \text{ in } \ \Omega\quad\text{with } \ u=f \
\text{ on } \partial\Omega.
\end{equation}
It is easy to show this. Since we want to minimize something, the
idea is to use that at a minimum of a real-valued function
$\varphi(t)$ its first derivative is zero.
\medskip
The computation we will use to seek a minimum of the (smooth) function
$J(v)$ closely follows that used for a real-valued function $f(X)$ of
several variables, $X=(x_1, x_2, \ldots,x_n)$. We recall this. Say
$f$ has a local minimum at an interior point $X_0$ of the set where $f$ is
defined. For any vector $Z\in\R^n$ let
$$
\varphi(t) = f(X_0 + tZ).
$$
Observe that $f(X_0 + tZ) \ge f(X_0)=\varphi(0)$, that is,
$\varphi(0) \le \varphi(t)$ for all $t$ near zero. Thus
$\varphi$ has a local min at $t=0$ so its derivative at $0$ is zero:
$\varphi'(0)=0$ (this is the directional derivative of $f$ at $X_0$ in
the direction of the vector $Z$). But by the chain rule,
$$
0=\varphi'(0)=\nabla f(X_0)\cdot Z.
$$
Because $Z$ ia an arbitrary vector, this says that $\nabla f$ is
orthogonal to all vectors so it must be zero, that is, $\nabla
(f(X_0)=0$.
\medskip
We use the same procedure to find the minima of $J(v)$ in equation
\eqref{Dir}. Say a function $u$ satisfying the boundary condition
minimizes $J$. Let $h(x,y)$ be any smooth function that is zero on
the boundary $\partial\Omega$. Then for any real $t$ the function
$u(x,y) + th(x,y)$ also satisfies the boundary conditions. Thus the
function
$$
\varphi(t) := J(u+th)=\iint_\Omega \big[\abs{\nabla u}^2
+2t\nabla u\cdot\nabla h + t^2\abs{\nabla h}^2\big]\,dx\,dy
$$
has a minimum at $t=0$. Therefore $\varphi'(0)=0$. That is,
\begin{equation}\label{variation}
\iint_\Omega \nabla u\cdot\nabla h\,dx\,dy = 0
\end{equation}
for {\it any} smooth function $h$ that is zero on the boundary.
Assuming the minimizer $u$ is smooth,
$$
\nabla u\cdot\nabla h = \nabla\cdot (h\nabla u) -h\Delta u,
$$
where $\Delta u = \nabla\cdot\nabla u= u_{xx}+u_{yy}$ is the Laplacian.
Thus integrating by parts (the Divergence Theorem), equation
\eqref{variation} implies that
\begin{equation}\label{variation2}
\iint_\Omega (\Delta u)h\,dx\,dy = 0
\end{equation}
for all $h$ that are zero on $\partial\Omega$. This implies that
$\Delta u=0$ throughout $\Omega$ (Proof: say $\Delta u >0$ in a small disk
$Q\subset\Omega$. Pick any $h$ that is positive on this disk and zero
outside it. But then for this $h$ we have
$$
\iint_\Omega (\Delta u)h\,dx\,dy =\iint_Q (\Delta u)h\,dx\,dy > 0.
$$
contradicting equation \eqref{variation2}. Thus, finding a minimizer
of \eqref{Dir} gives a solution of the Laplace equation \eqref{laplace}
with the desired boundary values.
\medskip
Riemann's innovation was using the existence of a minimizer of
\eqref{Dir} to prove the existence of a solution of the boundary value
problem \eqref{laplace}. He call this {\it Dirichlet's Principle}.
Since in \eqref{Dir} $J(v)$ is bounded below (by zero), it is clear
that $J$ has an infimum among all functions $v$ satisfying the boundary
condition. It is not at all clear that there is a twice differentiable
function $u$ that actually minimizes $J$. To illustrate the difficulty
Weierstrass gave an explicit example of a related problem
$$
\text{Minimize } J(v):=\int_{-1}^1 x^2v'\,^2(x)\,dx
$$
for all $v$ with $v(-1)=-1$ and $v(1)=1$. Following his
reasoning, we show that $J$ have an infimum but does not have a
minimum. He considered the sequence of functions
$$
v_n(x)=\begin{cases}-1& \text{ if $-1\le x\le -1/n$},\\
nx& \text{ if $-1/n \le x \le 1/n$}.\\
1& \text{ if $1/n\le x\le 1$} \end{cases}.
$$
By an easy calculation has $J(v_n)=2/(3n)\to 0$. This shows that
the $\inf J(v)=0$ (if you prefer a smooth sequence of functions you
can use $v_n(x):=\frac{\tanh nx}{\tanh n}$). However, if there is a
$v$ with $J(v)=0$ then $v'=0$, so $v$ must be a constant-- and that
can't satisfy the boundary conditions. Thus this $J$ has an {\it inf}
but not a {\it min}.
\medskip
Mathematicians generally believed the idea behind Riemann's proof of
the existence of a solution to \eqref{laplace} -- but there certainly
was a gap in the proof. It took about 50 years to develop the ideas
such as compactness needed to understand the situation adequately.
\bigskip
{\sc Toy Example: } Here is a toy (but not obvious) example where the
idea behind Dirichlet's Principle works immediately. Say you are
seeking a solution $(x,y)$ of the two equations
\begin{equation}\label{grad}
\begin{split}
2x(x^2+y^2) + y -1 &=0\\
2y(x^2+y^2) +2y^3 +x+2&=0
\end{split}
\end{equation}
Idea: find a function $f(x,y)$ that has a local minimum somewhere and
with the property that equations \eqref{grad} are the equations
$f_x=0$ and $f_y=0$, so they will be satisfied at this local minimum.
Consider the function
$$
f(x,y)=(x^2+y^2)^2 + y^4 + 2xy -2x -+4y -3
$$
Computing $f_x$ and $f_y$, except for a factor of $2$, these are exactly the
equations \eqref{grad} we wanted to solve. Thus, if we can show that
$f$ has a local minimum somewhere, then at least one solution exists,
namely $(x_0,y_0)$.
With this problem in mind, in Homework Set 2 Problem 4 you found a
number $R$ so that if $x^2+y^2 \ge R^2$ then $f(x,y)\ge 1$. Since the
disk $x^2+y^2 \le R^2$ is compact, there is at least one point
$(x_0,y_0)$ in this disk where $f$ attains its minimum. Because
$f(0,0) = -3<1$, this point is not on the boundary of the disk so it
is an interior point. Thus, at this point, the gradient of $f$ is
zero, that is, $f_x(x_0,y_0)=0$ and $f_y(x_0,y_0)=0$.
\bigskip
Last revised \today
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