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\subsubsection*{* Non-examinable material}

Laplace's equation:
\[ \nabla^2 \phi = 0 \]
Want to solve on $E$ with boundary conditions.

Two questions:
\begin{itemize}
  \item Does a solution exist?
  \item If so, is it unique?
\end{itemize}

Dirichlet used arguments like this:
\begin{align*}
  \int_E \nabla \phi \nabla phi \dd V
  &= \int_E \nabla \cdot (\phi \nabla \phi) -
  \nabla^2 \phi \dd V \\
  &= \int_E \nabla \cdot (\phi \nabla \phi) \dd V
  &\text{if $\nabla^2 \phi = 0$} \\
  &= \int_{\partial E} \phi \pfrac{\phi}{n} \dd S
  &\text{if $\phi = 0$ on surface}
\end{align*}
$\nabla \phi \cdot \nabla \phi \ge 0$, so $\nabla
\phi$ constant, so $\phi$ constant so $\phi = 0$
on $E$.

It is natural to think of
\[ \int \nabla \phi \cdot \nabla \phi \dd V \]
as an energy and Dirichlet showed that the minima
corresponds to a solution of Laplace's equation.

\textbf{Problems:}
\begin{itemize}
  \item  What are the conditions on
    $\phi$ and on the surface which actually permit us
    to use the Divergence Theorem.
  \item Does the energy $\int \nabla \phi \cdot
    \nabla \phi \dd V$ have a minimum?
  \item If given on the boundary, does there
    exist a $\phi$ with $\phi = f$ on the
    boundary and $\int \nabla \phi \cdot \nabla
    \phi < \infty$.
\end{itemize}

** This is the end of the non-examinable comments.

\subsubsection*{Examinable content again}

Interested in Laplace's equation. Start by
deciding what a boundary is.

\begin{flashcard}[closure-defn]
\begin{definition*}[Closure]
  \glsnoundefn{closure}{closure}{closures}
  \glssymboldefn{Cl}{$Cl$}{$Cl$}
  \cloze{If $E$ is a set in $(X, d)$ a metric space, then
  the closure of $E$ is
  \[ \Cl E = \{x \in E : \exists e_n \in E, e_n
  \dto x\} \]}
\end{definition*}
\end{flashcard}

\vspace{-1em}
Properties:
\begin{enumerate}[(1)]
  \item \glsref[closure]{Closure} is closed. Suppose $z_n \in \Cl E$
    and $z_n \dto z$. Then $\exists e_n \in E$
    such that $d(e_n, z_n) < \frac{1}{n}$, so by
    triangle inequality $e_n \to z$ so $z \in \Cl
    E$.
  \item \glsref[closure]{Closure} of $E$ is the smallest closed set
    containing $E$.
\end{enumerate}

\begin{flashcard}[interior-defn]
\begin{definition*}[Interior]
  \glsnoundefn{interior}{interior}{interiors}
  \glssymboldefn{Int}{$Int$}{$Int$}
  \cloze{If $E$ is a set in $(X, d)$ a metric space, then
  the interior of $E$ is
  \[ \Int E = \{x \in E : \exists \delta > 0, B(x,
  \delta) \subseteq E\} \]}
\end{definition*}
\end{flashcard}

\vspace{-1em}
The \gls{interior} of $E$ is the largest open set
contained in $E$ (proof left to reader).

\glsnoundefn{bound}{boundary}{boundaries}
\glssymboldefn{bound}{$\partial$}{$\partial$}
$\partial E$ is the boundary of $E$ and is defined
by $\partial E = \Cl E \setminus \Int E$.

\begin{remark*}
  The \gls{bound} can be a bit odd. For example,
  \[ E = \{x \in \RR^n : \|x\| = 1\} \]
  has the property that $E = \bound E$.

  However, for simple things, the notion of the
  \gls{bound} of an open set is what we would
  expect.
\end{remark*}

\begin{remark*}
  The boundary $\partial E = \Cl E \cap (\Int
  E)^c$ is closed.
\end{remark*}

\begin{remark*}
  If we work in $\RR^n$ and $E$ is bounded then
  $\partial E$ is bounded (and closed) hence
  compact.
\end{remark*}

\begin{theorem*}
  Let $\Omega \subseteq \RR^n$ is open, non-empty
  and bounded. Let $f : \Omega \to \RR$ be
  continuous.
  Let $\phi : \Cl \Omega \to \RR$ be continuous,
  and twice differentiable on $\Omega$ with
  $\nabla^2 \phi = 0$ on $\Omega$.

  Then there can exist at most one such $\phi$
  with $\phi = f$ on $\bound \Omega$.
\end{theorem*}

\vspace{-1em}
Suffices to prove:

\begin{lemma*}
  If $\Omega \subseteq \RR^n$ is open,
  non-empty and bounded, with:
  \begin{itemize}
    \item $\phi = \Cl \Omega \to \RR$ continuous
    \item $\phi$ twice differentiable on $\Omega$
    \item $\phi$ continuous on $\Cl\Omega$
    \item $\phi = 0$ on $\partial \Omega$
  \end{itemize}
  then $\phi = 0$.
\end{lemma*}

\begin{proof}[Proof of Theorem from Lemma (easy)]
  Let $\phi_1$ and $\phi_2$ satisfy the conditions
  of the theorem. Set $\phi = \phi_1 - \phi_2$.
  Then $\phi$ satisfies the conditions of the
  lemma, so $\phi = 0$ on $\Cl E$, so $\phi_1 =
  \phi_2$ on $\Cl E$.
\end{proof}

\begin{proof}
  \textbf{Key step:} If $\Omega$ is open, non-empty
  and bounded with $\phi$ continuous on $\Cl \Omega$
  and $\phi$ twice differentiable on $\Omega$ with
  $\nabla^2 \phi > 0$, then it can have no interior
  minimum, i.e. the minimum (know exists by
  compactness) must be attained on $\partial
  \Omega$.

  Suppose $\bf{y} \in \Omega$. Then we can find
  $\delta > 0$ such that
  \[ \prod [y_j - \delta, y_j + \delta] \subseteq
  \Omega \]
  Since $\nabla^2 \phi > 0$ at $\bf{y}$ there must
  exist a $k$ such that
  $\pfrac[2]{\phi}{x_k}{\bf{y}} > 0$ so $\exists 0
  < \eta < \delta$ such that $\pfrac[2]{\phi}{x_k}
  > 0$ on $\prod [y_j - \eta, y_j + \eta]$. Look
  at
  \[ g(t) = \phi(y_1, y_2, \ldots, y_{k - 1}, y_k
  + t, y_{k + 1}, \ldots, y_m) \]
  $g''(0) > 0$ so no minimum at $t = 0$ so $\phi$
  has no minimum at $\bf{y}$ \contradiction.

  \textbf{Next step:} Conditions as before, but
  replace $\nabla^2 \phi > 0$ by $\nabla^2 \phi =
  0$ on $\Omega$. Then the global minimum for
  $\phi$ is on the boundary.

  Set $\phi_N(\bf{x}) = \phi(\bf{x}) + \frac{1}{N}
  (x_1^2 + x_2^2 + \cdots + x_m^2)$. Then
  \[ \nabla^2 \phi_N(\bf{x}) = \nabla^2 \phi +
  \frac{1}{N} (2 + 2 + 2 \cdots + 2) =
  \frac{2m}{N} > 0 .\]
  By previous result, there exists $\bf{x}_N \in
  \bound \Omega$ such that $\phi(\bf{x}_N) \ge
  \phi(\bf{z})$ for all $\bf{z} \in \Cl\Omega$.
  There exists $N(j) \to \infty$ and $\bf{x}^* \in
  \bound \Omega$ (compactness) such that
  $\bf{x}_{N(j)} \to \bf{x}^*$. By continuity,
  $\phi(\bf{x}_{N(j)}) \to \phi(\bf{x}^*)$.
  \[ \phi(\bf{x}_{N(j)}) + \frac{2m}{N(j)} \ge
  \phi(z) + \frac{2m}{N(j)} \ge \phi(\bf{z})
  \qquad \forall z \in \Cl \Omega \]
  so
  \[ \phi(\bf{x}^*) \ge \phi(\bf{z}) \qquad
  \forall \bf{z} \in \Cl \Omega .\]

  We have shown $\nabla^2 = 0$ on $\Omega$ implies
  that we have a maximum on the \gls{bound}. But
  $\nabla^2 = \phi$ then $\nabla^2(-\phi) = 0$ so
  $-\phi$ has maximum on \gls{bound} so if $\phi =
  0$ on $\bound \Omega$ we have $\phi = 0$ on $\Cl
  \Omega$.
\end{proof}