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\subsection{3D Cylindrical Polar Coordinates}

Here
\[ \nabla^2 = \frac{1}{r} \pfrac{}{r} \left( r
\pfrac{\phi}{r} \right) + \frac{1}{r^2}
\pfrac[2]{\phi}{\theta} + \pfrac[2]{\phi}{z} = 0
\tag{5.9} \]
Substitute $\phi(r, \theta, z) =
R(r)H(\theta)Z(z)$ to find
\[ H'' = -\mu H, \quad Z'' = \lambda Z \]
\[ r(rR')' + (\lambda r^2 - \mu)R = 0 \]
\begin{itemize}
    \item Polar (as before) $\mu_m = m^2$,
        \[ H_m = \cos m\theta \text{ and } \sin
        m\theta \]
    \item Radial (Bessel's equation (3.26))
        \[ r(rR')' + (\lambda r^2 - m^2)R = 0 \]
        with solutions $R = J_m(kr)$ and $Y_m =
        (kr)$. Setting boundary conditions $R = 0$
        at $r = 1$ means
        \[ J_m(ka) \implies k = \frac{j_{mn}}{a} \]
        where $j_{mn}$ is the $n$-th zero (see
        (3.32)). Radial eigenfunction
        \[ R_{mn} = J_m \left( \frac{j_{mn}}{a} r
        \right) \tag{3.10} \]
        (eliminate $Y_m$ since $Y_m \to -\infty$
        as $r \to 0$)
    \item $Z$ equation: $Z'' = k^2 Z$ implies $Z =
        e^{-kz}$ and $z = e^{kz}$ (usually
        eliminate $e^{kz}$ with $z \to 0$ as $z
        \to \infty$.)
\end{itemize}
So general solution is
\[ \phi(r, \theta, z) = \sum_{m = 0}^\infty
\sum_{n = 1}^\infty (a_{mn} \cos m\theta + b_{mn}
\sin m\theta) \times J_m(\frac{j_{mn}}{a}r)
e^{-j_{mn} z/a} \tag{5.11} \]
Exercise: Describe steady-state heat flow in a
semi-infinite  circular wire with boundary
conditions $\phi = 0$ at $r = a$, $\phi = T_0$ at
$z = 0$ and $\phi \to 0$ as $z \to \infty$ (see
section 3.9 and 5.9). Show that the solution is
\[ \phi(r, \theta, z) = \sum_{k = 1}^\infty
\frac{2T_c}{j_{0n} J_1(j_{0n})} J_0 \left(
\frac{j_{0n}}{a} r \right) e^{-j_{0n} z/a} \]

\subsection{3D Spherical Polar Coordinates}

Recall that
\[ x = r\sin\theta \cos\theta \]
\[ y = r\sin\theta \sin\theta \]
\[ z = r\cos\theta \]
and $\dd V = r^2 \sin \theta \dd r \dd \theta \dd
\phi$, $0 \le r < \infty$, $0 \le \theta \le \pi$,
$0 \le \phi \le 2\pi$.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/b750d3965a9911ed.png}
\end{center}
Laplace's equation (5.1) becomes
\[ \frac{1}{r^2} \pfrac{}{r} \left( r^2
\pfrac{\Phi}{r} \right) + \frac{1}{r^2\sin\theta}
\pfrac{}{\theta} \left( \sin\theta
\pfrac{\Phi}{\theta} \right) +
\frac{1}{r^2\sin\theta} \pfrac[2]{\Phi}{\phi} = 0
\tag{5.12} \]

\subsubsection*{Axisymmetric case (no $\phi$
dependence)}

Seek separable $\Phi(r, \theta) = R(r)H(\theta)$.
\[ (\sin\theta H')' + \lambda \sin\theta H = 0 \]
\[ (r^2 R')' = \lambda R = 0 \tag{5.13} \]
Polar (Legendre's) equation: Substitute $x =
\cos\theta$ with
\[ \dfrac{x}{\theta} = -\sin\theta \implies
\dfrac{H}{\theta} = -\sin\theta \dfrac{H}{x} \]
\[ -\cancel{\sin\theta} \dfrac{}{x} \left[ -\sin^2
\theta \dfrac{H}{x} \right] +
\lambda\cancel{\sin\theta} H = 0 \]
\[ \dfrac{}{x} \left( (1 - x^2) \dfrac{H}{x}
\right) + \lambda H = 0 \]
which is Legendre's equation (2.21) with
eigenvalues $\lambda_l = l(l + 1)$ and
eigenfunctions (2.23)
\[ H_l(\theta) = P_l(x) = P_l(\cos\theta)
\tag{5.14} \]
(see section 2.5)
\begin{itemize}
    \item Radial equation:
        \[ (r^2 R')' - l(l + 1)R = 0 \]
        Seek solutions $R = \alpha r^\beta$.
        \[ \beta(\beta + 1) - l(l + 1) = 0
        \implies \left( \beta + \half \right)^2 =
        \left( l + \half \right)^2 \]
        with two solutions $\beta = l$ and $\beta
        = -l - 1$.
        \[ R_l = r^l \text{ and } r^{-l - 1} \]
\end{itemize}
\begin{flashcard}[general-axisymmetric-laplacian-solution]
General axisymmetric solution\prompt{ of
Laplacian in spherical polar coordinates}:
\[ \cloze{\Phi(r, \theta)} = \cloze{\sum_{l = 0}^\infty (a_l r^l
+ b_l r^{-l - 1}) P_l(\cos\theta) \tag{5.15}} \]
\end{flashcard}
where $a_l$, $b_l$ determined by boundary
conditions, usually at fixed $r = r_0$. Use
orthogonality conditions for $P_l$'s, see (2.24).
\myskip
Unit sphere solution: Solve $\nabla^2 \Phi = 0$
for $r < 1$ given axisymmetric boundary conditions
at $r = 1$, $\Phi(1, \theta) = f(\theta)$.
Regularity implies that $b_l = 0$, so we have
\[ f(\theta) = \sum_{l = 0}^\infty a_l
P_l(\cos\theta) \]
or with $f(\theta) = F(\cos\theta) = F(x)$,
\[ F(x) = \sum_{l = 0}^\infty a_l P_l(x) \]
so by (2.25) so
\[ a_l = \frac{2l + 1}{2} \int_{-1}^1 F(x) P_l(x)
\dd x \]
Exercise: Show $f(\theta) = \sin^2\theta$ yields
solution
\[ \Phi(r, \theta) = \frac{2}{3} (1 -
P_2(\cos\theta) r^2) \]

\subsubsection*{Generating function for $P_l(x)$
(2.23a)}

Consider a charge on $z$-axis at $z = 1$,
$\bf{r}_0 = (0, 0, 1)$ then the potential $P$
becomes
\begin{align*}
    \Phi(\bf{r})
    &= \frac{1}{|\bf{r} - \bf{r}_0|} \\
    &= \frac{1}{(x^2 + y^2 + (z - 1)^2)^\half} \\
    &= \frac{1}{(r^2 \sin^2\theta + r^2 \cos^2
    \theta - 2r\cos\theta + 1)^\half} \\
    &= \frac{1}{\sqrt{r^2 - 2r\cos\theta + 1}} \\
    &= \frac{1}{\sqrt{r^2 - 2r \ol{x} + 1}}
\end{align*}
($\ol{x} = \cos\theta$) \\
Exercise: verify $\Phi = \frac{1}{|\bf{r} -
\bf{r}_0|}$ satisfies $\nabla^2 \Phi = 0$ whenever
$\bf{r} \neq \bf{r}_0$. \\
We can represent any
axisymmetric solution (5.12) as a sum (5.15) (with
$b_n = 0$) for $r < 1$:
\[ \frac{1}{\sqrt{r^2 - 2rx + 1}} = \sum_{l =
0}^\infty a_l P_l(x) r^l \]
with norm at $x = 1$, $P_l(1) = 1$, we get
\[ \frac{1}{1 - r} = \sum_{l = 0}^\infty a_l r^l \]
so $a_l = 1$ (for a geometric series). Thus
generating function for $P_l(x)$ is
\[ \frac{1}{\sqrt{1 - 2xr + r^2}} = \sum_{l =
0}^\infty P_l(x) r^l \tag{5.16} \]
Expand LHS with binomial theorem to find $P_l(x)$
(coefficient of the $r^l$ term) Use to obtain norm
condition (2.24). (Example sheet 2, Q5)
\begin{example*}[Electric multipole]
    \phantom{}
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/4cffc1885a9d11ed.png}
    \end{center}
\end{example*}