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% 07/03/2023 10AM

\begin{note*}[Non-examinable]
    Can show (Part II Waves) that a wave packet (a
    group of nearly monochromatic waves) moves with
    the \emph{group velocity} $c_g =
    \dfrac{q}{k}$.
\end{note*}

\subsubsection*{Limiting Cases}

\begin{enumerate}[(1)]
    \item Deep-water limit. $kh \gg 1$, $\tanh kh
        \simeq 1$. So $w = \sqrt{gk}$, $c =
        \sqrt{g / k}$ and $c_g = \half c$
        independent of $h$. For example Atlantic
        storm produces swell waves with period
        $15\mathsf{s}$, $\omega = \frac{2\pi}{15}
        = 0.4\mathsf{s^{-1}}$, $k^{-1} =
        \frac{5}{\omega^2} = 60\mathsf{m} \ll
        \text{ocean depth} = 4\mathsf{km}$
        (average). $c = \frac{\omega}{k} =
        25\mathsf{ms^{-1}} = 2000 \mathsf{km /
        day}$ so arrive before the storm.
    \item Shallow water limit: $kh \ll 1$, $\tanh
        kh \simeq kh$. So $\omega = \sqrt{gh} k$,
        $c = \sqrt{gh}$, $c_y = c$ independent of
        $k$. For example flood waves on a river,
        $h = 2\mathsf{m}$ $\implies$ $c =
        \sqrt{gh} = 4 \mathsf{ms^{-1}} =
        16\mathsf{kmh^{-1}}$. Or tsunami, $h =
        4\mathsf{km}$ (average) $\implies$ $c =
        \sqrt{gh} = 200 \mathsf{ms^{-1}}$ roughly
        the speed of a commercial plane.
        Wavelength $\lambda \simeq 500
        \mathsf{km}$. (Boxing Day 2004).
\end{enumerate}

\subsubsection*{Velocities}

$\zeta = \zeta_0 e^{i(kx - \omega t)}$, $\phi =
-\frac{i\omega \zeta_0}{k} \frac{\cosh k(z +
h)}{\sinh kh} e^{i(kx - \omega t)}$.
\[ \bf{u} = \nabla \phi = (\cosh k(z + h),
-i\sinh k(z + h)) \frac{\zeta_0 w}{\sinh kh}
e^{i(kx - \omega t)} \]
(real part understood). $u_x$ and $\zeta$ are in
phase.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/d0902784bcd311ed.png}
\end{center}

\subsubsection*{Deep water limit ($kh \gg 1$)}

$\cosh, \sinh \simeq e^{+}$. Hence $\bf{u} = (1,
-i) e^{k\zeta} \zeta_0 \omega e^{i(kx - \omega
t)}$ circular motion, smaller at depth.

\subsubsection*{Shallow water limit ($kh \ll 1$)}

$\bf{u} \simeq \left( \frac{1}{kh}, -i \frac{z +
h}{h} \right) \zeta_0 \omega e^{i(kx - \omega t)}$
($\frac{1}{kh}$ large, so horizontal velocity
$\gg$ vertical velocity).

\subsubsection{Standing Waves in a Container}

Consider water in a deep rectangular box $0 < x <
a$, $0 < y < b$, $z < 0$. Look for linearised
waves with surface at $z = \zeta(x, y, t)$, $|z|
\ll a, b$, $|\nabla \zeta| \ll 1$.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/0119c5eabcd411ed.png}
\end{center}
Want to solve $\nabla^2 \phi = 0$ in $z < 0$,
$\pfrac{\phi}{n} = 0$ on side boundaries. $\nabla
\phi \to 0$ as $z \to -\infty$.
Also want on $z = 0$,
$\pfrac{\zeta}{t} = \pfrac{\phi}{z}$ and
$\rho \pfrac{\phi}{t} + \rho g \zeta$ independent of $x, y$.
Try separation of variables $\phi(x, y, z, t) = A
\cos \frac{m\pi x}{a} \cos \frac{n\pi y}{b} e^{kx}
e^{-i\omega t}$.
\myskip
Where Laplace's equation $-\frac{m^2\pi^2}{a^2} -
\frac{n^2\pi^2}{b^2} + k^2 = 0$. Kinematic
boundary condition
\[ \pfrac{\zeta}{t} = \pfrac{\phi}{z} \implies
\zeta = \frac{iAk}{\omega} \cos \frac{m\pi x}{a}
\cos \frac{n\pi y}{b} e^{-i\omega t} \]
dynamic boundary condition
\[ \pfrac{\phi}{t} + g\zeta \implies -i\omega A +
g \frac{iAk}{\omega} = 0 \implies \omega^2 = gk \]
as before for deep-water. Only difference is that
$k$ is quantised by the side walls:
\[ k_{mn} = \left( \frac{m^2\pi^2}{a^2} +
\frac{n^2\pi^2}{b^2} \right)^{\half} \]
These are standing waves / normal modes. $\zeta =
f_{mn}(x, y) \cos(\omega_{mn} t)$.

\subsubsection*{The Rayleigh Instability}

Consider an upside down container
\begin{center}
    \includegraphics[width=0.3\linewidth]
    {images/44edad70bcd611ed.png}
\end{center}
Obvious instability! Just let $g \to -g \implies
\omega^2 = -gk$ $\implies$ $\omega = \pm
i\sqrt{gk}$, $e^{-i\omega t} = e^{\pm \sqrt{gk}
t}$.