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We can also look in cylindrical polar coordinates
and seek solutions of the form
\[ \psi = \psi(r) \]
($r^2 = x^2 + y^2$). Now we have
\[ \nabla^2 \psi = \dfrac[2]{\psi}{r} +
\frac{1}{r} \dfrac{\psi}{r} = \frac{1}{r}
\dfrac{}{r} \left( r \dfrac{\psi}{r} \right) = 0 \]
\[ \implies \psi = A \log r + B \]

\begin{note*}
    In $\RR^n$, with $n \ge 3$, $\nabla^2 \psi =
    0$ has the symmetric solution
    \[ \psi = \frac{A}{r^{n - 2}} + B \]
\end{note*}

\noindent
We can get more solutions by differentiating, for
example if $\psi = \frac{1}{r}$ is a solution in
$\RR^3$, then so too
\[ \tilde{\psi}(\bf{x}) = \bf{d} \cdot \nabla
\left( \frac{1}{r} \right) = -\frac{\bf{d} \cdot
\bf{x}}{r^3} \]
with $\bf{d}$ a constant vector. (This is a
potential for a \emph{dipole} in
electromagnetism).

\subsubsection*{Boundary Conditions}

Often boundary conditions are important. For
example: Solve
\[ \nabla^2 \psi = \begin{cases}
    -\rho_0 & r \le R \\
    0 & r > R
\end{cases} \]
with $\rho_0$ constant. (for example
gravitational potential for a planet with
constant density). To get a unique solution, we
require
\begin{itemize}
    \item $\psi(r = 0)$ non-singular
    \item $\psi(r) \to 0)$ as $r \to \infty$
    \item $\psi$ and $\psi'$ continuous.
\end{itemize}
Can check that if $\psi(r) = r^p$ then
\[ \nabla^2 = p(p + 1)r^{p - 2} \]
(in $\RR^3$). With spherical symmetry, we have
\[ \psi(r) = \frac{A}{r} + B - \frac{1}{6} \rho_0
r^2 \qquad r \le R \]
$\psi(r = 0)$ non-singular $\implies A = 0$. For
$r > R$,
\[ \psi(r) = \frac{c}{r} + D \qquad r > R \]
$\psi(r \to \infty) \to 0 \implies D = 0$. Then we
patch these together at $r = R$:
\[ \psi(r = R) = B - \frac{1}{6} \rho_0 R^2 =
\frac{C}{R} \]
\[ \psi'(r = R) = -\frac{1}{3} \rho_0 R =
-\frac{c}{R^2} \]
\[ \implies \psi(r) = \begin{cases}
    \frac{1}{6} \rho_0 (3R^2 - r^2) & r \le R \\
    \frac{1}{3} \rho_0 \frac{R^3}{r} & r > R
\end{cases} \]
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/76b61b3e9a4111ec.png}
\end{center}

\subsubsection*{General Results}

If we solve $\nabla^2 \psi = -\rho$ in some region
$V \subset \RR^3$, then there are two common
boundary conditions on $\partial V$.
\begin{itemize}
    \item Dirichlet (D): Fix $\psi(\bf{x}) =
        f(\bf{x})$ on $\partial V$.
    \item Neumann (N): Fix $\bf{n} \cdot \nabla
        \psi(\bf{x}) = g(\bf{x})$ on $\partial V$,
        with $\bf{n}$ the outward pointing normal.
\end{itemize}

\begin{notation*}
    Sometimes this is written as
    \[ \dfrac{\psi}{\bf{n}} \equiv \bf{n} \cdot
    \nabla \psi \]
    or even
    \[ \dfrac{\psi}{n} \equiv \bf{n} \cdot \nabla
    \psi \]
\end{notation*}

\begin{claim*}
    There is a unique solution to the Poisson
    equation on $V$ with either D or N boundary
    conditions specified on each $\partial V$.
\end{claim*}

\begin{note*}
    Unique up to constant for N.
\end{note*}

\begin{proof}
    Let $\psi_1$ and $\psi_2$ satisfy Poisson and
    \[ \psi = \psi_1 - \psi_2 \]
    Then $\nabla^2 \phi = 0$ with $\psi = 0$ or
    $\bf{n} \cdot \nabla \psi = 0$ on each
    $\partial V$.
    \begin{align*}
        \int_v \nabla \cdot (\psi \nabla \psi) \dd
        V
        &= \int_V (\nabla \psi \cdot \nabla \psi +
        \psi \nabla^2 \psi) \dd V \\
        &= \int_V |\nabla \psi|^2 \dd V \\
        &= \int_{\partial V} \psi \nabla \psi
        \cdot \bf{\dd s}
        &&\text{by Divergence Theorem} \\
        &= \int_{\partial V} \psi(\bf{n} \cdot
        \nabla \psi) \dd s \\
        &= 0
        &&\text{by one of the boundary conditions}
        \\
        \implies |\nabla \psi|^2
        &= 0
        &&\text{in $V$} \\
        \implies \nabla \psi
        &= 0
        &&\text{in $V$} \\
        \implies \psi
        &\text{is constant}
    \end{align*}
    If Dirichlet $\implies \psi = 0$ on $\partial
    V \implies \psi = 0$ everywhere.
\end{proof}

\begin{note*}
    Strictly for bounded $V$, but we can work
    harder and extend to, for example $\RR^3$.
\end{note*}

\begin{note*}
    \begin{itemize}
        \item If we can find, for example an
            isotropic solution, then this is
            \emph{the} solution.
        \item Sometimes there may be \emph{no}
            solution.
            \begin{example*}
                Solve $\nabla^2 \psi = \rho(x)$ with $\bf{n}
                \cdot \nabla \psi = g(\bf{x})$ on $\partial
                V$. Then
                \[ \int_V \nabla^2 \psi \dd V = \int_{\partial
                V} \nabla \psi \cdot \bf{\dd s} \]
                so a solution can exist only if
                \[ \int_V \rho \dd v = \int_{\partial V}
                g(\bf{x}) \dd s \]
            \end{example*}
        \item The proof uses Green's first
            identity:
            \begin{align*}
                \int_V \psi \nabla^2 \psi \dd V
                &= -\int \nabla \phi \cdot \nabla
                \psi \dd V + \int_S \phi \nabla
                \psi \cdot \bf{\dd s}
            \end{align*}
            (with $\phi = \psi$) This follows
            from the divergence theorem. Or by
            anti-symmetry
            \[ \int_V (\phi \nabla^2 \psi - \psi
            \nabla^2 \phi) \dd V = \int_S (\phi
            \nabla \psi - \psi \nabla \phi) \cdot
            \bf{\dd s} \]
            This is Green's second identity.
    \end{itemize}
\end{note*}