% vim: tw=50
% 24/10/2022 10AM

\subsection{Oscillation energy}

A vibrating string has kinetic energy due to its
motion (for example particle $\half mv^2$)
\begin{flashcard}[KE-of-string]
\prompt{Kinetic energy of a vibrating string?}
\[ KE = \cloze{\half \mu \int_0^L \left( \pfrac{y}{t}
\right)^2 \dd x} \]
\end{flashcard}
and potential energy due to stretching $\Delta x$
\begin{align*}
    PE
    &= T \Delta x \\
    &= T \int_0^L \left( \sqrt{1 +
    \left( \pfrac{y}{x} \right)^2} - 1
    \right) \dd x \\
    &\approx \half T \int_0^L \left( \pfrac{y}{x}
    \right)^2 \dd x
    &&\text{for $\left| \pfrac{y}{x} \right| \ll 1$}
\end{align*}
\begin{hiddenflashcard}[PE-of-string]
\prompt{Potential energy of a vibrating string?}
\[ PE = \cloze{\half T \int_0^L \left( \pfrac{y}{x}
\right)^2 \dd x} \]
\end{hiddenflashcard}
The total summed energy becomes ($c^2 =
\frac{T}{\mu}$)
\begin{flashcard}[total-energy-of-string]
\prompt{Total energy of a vibrating string?}
\[ E = \cloze{\half \mu \int_0^L \left[ \left(
\pfrac{y}{t} \right)^2 + c^2 \left( \pfrac{y}{x}
\right)^2 \right] \dd x \tag{3.13}} \]
\end{flashcard}
Substitute (3.10) and use orthogonality (1.1)
\begin{align*}
    E
    &= \half \mu \sum_{n = 1}^\infty \int_0^L
    \left[ \left( -\frac{n\pi c}{L} C_n \sin \frac{n
    \pi ct}{L} + \frac{n\pi c}{2} D_n \cos \frac{n\pi
    ct}{L} \right)^2 \sin^2 \frac{n\pi x}{L}
    \right. \\
    &\hspace{5em} \left. + c^2
    \left( C_n \cos \frac{n\pi ct}{L} + D_n \sin
    \frac{n\pi ct}{L} \right)^2 \frac{n^2\pi^2}{L}
    \cos^2 \frac{n\pi x}{L} \right] \dd x \\
    &= \frac{1}{4} \mu \sum_{n = 1}^\infty
    \frac{n^2\pi^2 c^2}{L} \left( C_n^2 + D_n^2
    \right) \tag{3.14} \\
    &= \sum_{\text{normal modes}} [\text{energy in
    $n$-th mode}]
\end{align*}
This is constant, so energy is conserved in time
(no dissipation).

\subsection{Wave reflection and transmission}

Recall travelling wave solution (3.12). A simple
\emph{harmonic} travelling wave is
\[ y = \Re[A e^{-\omega(t - x/c)}] = |A|
\cos \left( \omega \left( t - \frac{x}{c} \right)
+ \phi \right) \]
where the \emph{phase} is $\phi = \arg A$ and
wavelength is $\frac{2\pi c}{\omega}$. \myskip
Consider a density discontinuity on a string at $x
= 0$, with
\[ \mu = \begin{cases}
    \mu_- & \text{for $x < 0$ hence $c_- =
    \sqrt{\frac{t}{\mu_-}}$} \\
    \mu_+ & \text{for $x > 0$ hence $c_+ =
    \sqrt{\frac{T}{\mu_+}}$}
\end{cases} \]
assuming constant tension. Incident wave on
junction
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/d2f372c4537e11ed.png}
\end{center}
Boundary (or junction) conditions at $x = c$:
\begin{itemize}
    \item String does not break, i.e. $y$ is
        continuous for all $t$.
        \[ \implies A + B = D \tag{$*$} \]
    \item Forces balance $T \left. \pfrac{y}{x}
        \right|_{x = 0_-} = T \left. \pfrac{y}{x}
        \right|_{x = 0_+}$ i.e. $\pfrac{y}{x}$ is
        continuous for all $t$
        \[ \implies -\frac{i\omega A}{c_-} +
        \frac{i\omega B}{c_-} = -\frac{i\omega
        D}{c_+} \tag{\dag} \]
\end{itemize}
\begin{hiddenflashcard}[boundary-conditions-for-discontinuous-density-string]
\prompt{Boundary conditions for a string with
discontinuous density?}
\begin{itemize}
    \item \cloze{$y$ continuous for all $t$ (string does
        not break).}
    \item \cloze{$\pfrac{y}{x}$ continuous for all $t$
        (forces balance).}
\end{itemize}
\end{hiddenflashcard}
\[ (*) - \frac{c_-}{i\omega} (\dag) \implies 2A =
D + D \frac{c_-}{c_+} = \frac{D}{c_+} (c_+ + c_-) \]
So given $A$, we have the solution:
\[ D = \frac{2c_+}{c_- + c_+} A \qquad B =
\frac{c_+ - c_-}{c_+ + c_-} A \tag{3.16} \]
where $D$ is the transmitted amplitude and $B$ is
the reflected amplitude. In general, different
phase shift $\phi$ is possible. \myskip
Limiting cases:
\begin{enumerate}[(1)]
    \item Continuity $c_- = c_+ \implies D = A, B
        = 0$.
    \item Dirichlet boundary conditions
        $\frac{\mu_+}{\mu_-} \to \infty$ (fixed
        end $y = 0$ at $x = 0$) then
        $\frac{c_+}{c_-} \to 0 \implies D = 0, B =
        -A$ i.e. total reflection with opposite
        phase ($\phi = \pi$)
    \item Neumann boundary conditions
        $\frac{\mu_+}{\mu_-} \to 0$ (free end of
        string - very light string $x > 0$) then
        $\frac{c_+}{c_-} \to \infty \implies D =
        2A, B = A$ (boundary condition $\left.
        \pfrac{y}{x} \right|_{x = 0}$). Total
        reflection with same phase ($\phi = 0$).
\end{enumerate}

\subsection{Wave equation in 2D plane polars}

The 2D wave equation for $u(r, \theta, t)$ becomes
\[ \boxed{\frac{1}{c^2} \pfrac[2]{u}{t} = \nabla^2
u} \tag{3.17} \]
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/7d6a7fc6538011ed.png}
\end{center}
with boundary conditions at $r = 1$ on a unit disc
(drum)
\[ u(1, \theta, t) = 0 ~ \forall t \tag{3.18} \]
(fixed rim) and initial conditions for $t = 0$
\[ u(r, \theta, 0) = \phi(r, \theta), \quad
\pfrac{u}{t} (r, \theta, 0) = \psi(r, \theta)
\tag{3.19} \]

\subsubsection*{Temporal separation}

Substitute
\[ u(r, \theta, t) = T(t) V(r, \theta) \tag{3.20} \]
into (3.17) to get
\[ \ddot{T} + \lambda c^2 T = 0 \tag{3.21} \]
\[ \nabla^2 V + \lambda V = 0 \tag{3.22} \]
which in polars is
\[ \pfrac[2]{V}{r} + \frac{1}{r} \pfrac{V}{r} +
\frac{1}{r^2} \pfrac[2]{V}{\theta} + \lambda V = 0 \]

\subsubsection*{Spatial separation}

Now try
\[ V(r, \theta) = R(r) H(\theta) \]
in (3.22):
\[ H'' + \mu H = 0 \tag{3.23} \]
\[ r^2 R'' + rR' + (\lambda r^2 - \mu)R = 0
\tag{3.24} \]
where $\lambda, \mu$ are separation constants.
\myskip
Polar solution: Configuration implies periodic
boundary conditions
\[ H(0)=  H(2\pi) \]
with $\mu > 0$, so the eigenvalue $\mu = m^2$ ($m$
integer) with solution
\[ H_m(\theta) = A_m \cos m\theta + B_m \sin
m\theta \tag{3.25} \]
Radial equation: divide (3.24) by $r$ to bring it
into Sturm Liouville form (2.7) with $\mu = m^2$
\[ \dfrac{}{r} (rR') - \frac{m^2}{r}R = -\lambda r
R \quad (0 \le r \le 1) \tag{3.26} \]
where $p(r) = r$, $q(r) = \frac{m^2}{r}$ and
weight $\omega(r) = r$, with self-adjoint boundary
conditions with $R(1) = 0$ and bounded at $R(0)$,
since $p(0) = 0$ a regular singular point.

\subsubsection*{Bessel's equation}

Substitute $z \equiv \sqrt{\lambda r}$
in (3.26) to find
\[ \boxed{z^2 \dfrac[2]{R}{z} + z \dfrac{R}{z} + (z^2 -
m^2) R = 0} \tag{3.27} \]
which is Bessel's equation
\begin{flashcard}[bessels-equation]
\prompt{Bessel's equation?}
\[ \cloze{(zR')' + \left( z - \frac{m^2}{z}
\right) R} = \cloze{0} \]
\end{flashcard}
Frobenius solution: substitute power series
\[ R = z^p \sum_{n = 0}^\infty a_n z^n \]
to obtain
\[ \sum_n a_n [(n + p)(n + p - 1) z^{n + p} + (n +
p) z^{n + p} + z^{n + p + 2} - m^2 z^{n + p}] = 0 \]
Equate powers of $z$: considering coefficient of
$z^p$,
\[ p^2 - m^2 = 0 \implies p = m, -m \]