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\subsubsection*{Harmonic Functions}

Solutions to the Laplace equation
\[ \nabla^2 \psi = 0 \]
are called \emph{harmonic functions}.

\begin{claim*}[The mean value property]
    If $\psi$ is harmonic in a region $V$ that
    includes the ball with boundary
    \[ S_r : |\bf{x} - \bf{a}| = R \]
    then
    \[ \psi(\bf{a}) = \ol{\psi}(R) :=
    \frac{1}{4\pi R^2} \int{S_R} \psi(\bf{x}) \dd
    S \]
    i.e. the value in the middle of the sphere is
    equal to the average over the boundary of the
    ball.
\end{claim*}

\begin{proof}
    In spherical polar coordinates
    \[ \dd S = r^2 \sin \theta \dd \theta \dd \phi \]
    \[ \ol{\psi}(r) = \frac{1}{4\pi} \int \dd \phi
    \int \dd \theta \sin\theta \psi(x, \theta,
    \phi) \]
    \begin{align*}
        \dfrac{\ol{\psi}}{r}(R)
        &= \frac{1}{4\pi} \int \dd \phi \int \dd
        \theta \sin \theta \pfrac{\psi}{r} (R,
        \theta, \phi) \\
        &= \frac{1}{4\pi R^2} \int_{S_R}
        \pfrac{\psi}{r} \dd S
        \implies \dfrac{\ol{\psi}}{r} (R) =
        \frac{1}{4\pi R^2} \int_{S_R} \nabla \psi
        \cdot \bf{\dd S} \\
        &= \frac{1}{4 \pi R^2} \int_{\text{Ball}}
        \nabla^2 \psi \dd V \\
        &= 0
    \end{align*}
    by the divergence theorem. But $\ol{\psi}(R)
    \to \psi(\bf{a})$ as $R \to 0$ hence
    $\ol{\psi}(R) = \psi{a}$ for all $R$.
\end{proof}

\begin{claim*}
    A harmonic function can have neither a maximum
    nor a minimum in the interior of $V$. The max
    / min lie on $\partial V$.
\end{claim*}

\begin{proof}
    If $\exists$ a local maximum at $\bf{a}$ then
    $\exists$ $\eps$ such that $\psi(\bf{x}) <
    \psi(\bf{a})$ for all $|\bf{x} - \bf{a}| <
    \eps$. But this contradicts that $\ol{\psi}(R)
    = \psi(\bf{a})$ for $0 < R < \eps$.
\end{proof}

\begin{note*}
    Saddle points are allowed. The Hessian is
    \[ \frac{\partial^2 \psi}{\partial x^i
    \partial x^j} \]
    has eigenvalues $\lambda_i$, but $\nabla^2
    \psi = 0 \implies \sum_i \lambda_i = 0$ so
    $\lambda_i$ must be both positive and
    negative. (This has a loophole when all
    $\lambda_i = 0$ which is closed by our
    previous proof).
\end{note*}

\subsubsection*{Integral Solutions}

We want to solve the Poisson equation
\[ \nabla^2 \psi = -\rho(\bf{x}) \]
for a fixed $\rho(\bf{x})$. Consider
\[ \psi(\bf{x}) = \frac{\lambda}{4 \pi r} \]
for $\lambda$ fixed. Previously we showed that
this solves $\nabla^2 \psi = 0$, at least if $r
\neq 0$. By what happens at $r = 0$? Something
must be going on because
\begin{align*}
    \int_V \nabla^2 \psi \dd V
    &= \int_S \nabla \psi \cdot \bf{\dd S} \\
    &= -\lambda
\end{align*}
We can't have $\nabla^2 \psi = 0$ everywhere!
Instead, $\psi = \frac{\lambda}{4\pi r}$ must
actually solve the Poisson equation for some
source $\rho(\bf{x})$. But we know $\rho(\bf{x}) =
0$ for all $\bf{x} \neq 0$. And we must have
\[ \int \rho(\bf{x}) \dd V = \lambda \]
The source is the 3D Dirac delta function:
\[ \rho(\bf{x}) = \lambda \delta^3(\bf{x}) \]
Here $\delta^3(\bf{x})$ is an infinite spike at
the origin, such that
\[ \int_V f(\bf{x}) \delta^3(\bf{x}) \dd V =
f(\bf{x} = 0) \]
In particular
\[ \int_V \delta^3(\bf{x}) \dd V = 1 \]
So, we've learned that $\psi = \frac{\lambda}{4\pi
r}$ does \emph{not} solve the Laplace equation,
but
\[ \nabla^2 \psi = -\lambda \delta^3(\bf{x})
\implies \psi(\bf{x}) = \frac{\lambda}{4\pi r} \]

\begin{claim*}
    $\nabla^2 \psi = -\rho$ has the integral
    solution
    \[ \psi(\bf{x}) = \frac{1}{4\pi} \int_{V'}
    \frac{\rho(\bf{x}')}{|\bf{x} - \bf{x}'|} \dd
    V' \]
    ($V'$ should include any region with
    $\rho(\bf{x}') = 0$)
\end{claim*}

\begin{proof}
    Intuition is that you sum over
    ``$\frac{1}{r}$'' solutions, weighted by
    $\rho(\bf{x}')$ for each $\bf{x}'$.
    \[ \nabla^2 \psi(\bf{x}) = \frac{1}{4\pi}
    \int_{V'} \rho(\bf{x}') \nabla^2 \left(
    \frac{1}{|\bf{x} - \bf{x}'|} \right) \dd V' \]
    (here $\nabla$ differentiates $\bf{x}$ and
    treats $\bf{x}'$ as constant) but
    \[ \nabla^2 \left( \frac{1}{|\bf{x} -
    \bf{x}'|} \right) = -4\pi \delta^3(\bf{x} -
    \bf{x}') \]
    (this is our previous result that $\nabla^2
    \frac{1}{r} = -4\pi \delta^3(\bf{x})$ but with
    the origin shifted to $\bf{x}'$)
    \begin{align*}
        \implies \nabla^2 \psi
        &= -\int_{V'} \rho(\bf{x}')
        \delta^3(\bf{x} - \bf{x}') \dd V' \\
        &= -\rho(\bf{x})
    \end{align*}
\end{proof}

\myskip
This powerful technique is known as the
\emph{Green's function} approach.