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If Laplace's equation has a solution, then it is
unique.

We now give an example of Zaremba which shows that
(at least in the general form we have given) there
need not be a solution.

We take $E = \{\bf{x} \in \RR^2 : 0 < \|x\| \le
1\}$.
\begin{center}
  \includegraphics[width=0.2\linewidth]{images/2198d82968fc4bb3.png}
\end{center}
Observe $\Int E = \{\bf{x} : 0 < \|x\| \le 1\}$,
$\bound E = \{x : \|x = 1\|\} \cup \{\bf{0}\}$.

We show that there does not exist $\phi$ such
that $\phi$ is continuous on $\Cl E = \{\bf{x} :
\|x\| \le 1\}$ which is twice differentiable on
$\Int E = \{\bf{x} : 0 < \|\bf{x}\| < 1\}$ with
$\nabla^2 \phi = 0$ on $\Int E$ satisfying
$\phi(x) = 0$ on $\|\bf{x}\| = 1$ with
$\phi(\bf{0}) = 1$.

Suppose such a $\phi$ exists. Let $R_\theta$ be
a rotation through $\theta$ about $\bf{0}$. Then
$\phi_\theta(\bf{x}) = \phi(R_\theta \bf{x})$
also solve the problem, so by uniqueness we must
have $\phi_\theta = \phi$ for all $\theta$. So
$\phi$ is radially symmetric: $\phi(x, y) =
f(r)$, $r = \sqrt{x^2 + y^2}$.

Recall from \courseref[Vector Calculus]{VC} the
formula for $\nabla^2$ in polar coordinates. So we
seek to solve
\[ \frac{1}{r} \dfrac{}{r} (rf'(r)) = 0 \]
subject to $f(0) = 1$, $f(1) = 0$. So:
\begin{align*}
  \dfrac{}{r} (rf'(r))
  &= 0 \\
  \implies rf'(r)
  &= A \\
  \implies f'(r)
  &= \frac{A}{r} \\
  \implies f(r)
  &= A \log r + B
\end{align*}
But we want $f(r) \to 1$ as $r \to \infty$, so $A
= 0$. So $f(1) = B$, contradiction.

This completes our discussion of Laplace's
equation. One can show that the equation always
has a solution in 2D, provided that the boundary
is a Jordan curve, but the proof of this result is
very hard. The result is similar to the Riemann
mapping theorem. For higher dimensions, there are
no particularly nice theorems about existence of
solutions.

\newpage
\section{Brouwer's Fixed Point Theorem}

\begin{lemma*}
  If $f : [0, 1] \to [0, 1]$ is continuous then
  $\exists x \in [0, 1]$ such that $f(x) = x$.
\end{lemma*}

\begin{proof}
  Set $g(t) = f(t) - t$. Then $g : [0, 1] \to \RR$
  is continuous. $g(0) = f(0) \ge 0$, $g(1) = f(1)
  - 1 \le 0$. So by IVT, there exists $x$ such
  that $g(x) = 0$.
\end{proof}

\begin{flashcard}[brouwer]
\begin{theorem*}[Brouwer]
  \label{brouwer}
  \cloze{If $\ol{D} = \{\bf{x} \in \RR^2 : \|\bf{x}\| \le
  1\}$ and $f: \ol{D} \to \ol{D}$ is continuous,
  then there exists $\bf{y}$ such that $f(\bf{y})
  = \bf{y}$.}
\end{theorem*}
\end{flashcard}

\fcscrap{
\begin{remark*}
  \phantom{}
  \begin{enumerate}[(a)]
    \item This reusult works in $\RR^n$. Most of
      our proof will work with obvious
      modifications in $\RR^n$. One bit,
      ``Sperner's Lemma'' requires work (but not
      enromous changes).
    \item If $E \subseteq \RR^2$ and $T : E \to
      \ol{D}$ is a homeomorphism (i.e. $T$
      bijective, $T$ continuous and $T^{-1}$
      continuous) then Brouwer continues to work.
      Suppose $g : E \to E$ is continuous. Then $T
      \circ g \circ T^{-1} : \ol{D} \to \ol{D}$ is
      continuous, so has a fixed point $\bf{z}$,
      satisfying $TgT^{-1}(z) = z$, so $gT^{-1}(z)
      = T^{-1}(z)$.
  \end{enumerate}
\end{remark*}
}

The proof goes through many steps
\[ A \iff B \iff C \iff D \]
so to understand the proof we need to understand
the strategy (ie $A$, $B$, \ldots\ and the tactics
$\iff$).

We start with the ``no retraction'' theorem.

\begin{theorem*}[No Retraction Theorem]
  \label{no_retrac_thm}
  There does not exist a $g : \ol{D} \to \bound
  D$ continuous with $g(\bf{x}) = \bf{x} ~\forall
  \bf{x} \in \bound D$.
\end{theorem*}

We will start by showing \nameref{no_retrac_thm}
$\iff$ \nameref{brouwer}.

\begin{proof}
  Suppose \nameref{no_retrac_thm} is false. Then
  there exists $g : \ol{D} \to \ol{D}$ continuous
  with $g(\bf{x}) = \bf{x}$ for all $\bf{x} \in
  \bound D$. Now let $R$ be a rotation about the
  origin through angle $\pi$. Then $Rg : \ol{D}
  \to \bound D \subseteq \ol{D}$ has no fixed
  point. So \nameref{brouwer} would have to be
  false.

  If the \nameref{no_retrac_thm} is true then we
  now want to show that \nameref{brouwer} holds.
  Suppose $f : \ol{D} \to \ol{D}$ is continuous
  without a fixed point. Then we define $g(\bf{x})$ to
  be the point of intersection of the ray $f(\bf{x})$
  to $\bf{x}$ with $\bound D$. Notice $g$ is
  continuous, and fixed $\bound D$, so $g$ is a
  retraction mapping.
\end{proof}
