% vim: tw=50
% 02/03/2023 10AM

\noindent
eg Small oscillations of a gass bubble $a = a_0 +
\eta(t)$, $|\eta| \ll a_0$. Linearise (1)
$\implies$ $\rho a_0 \ddot{\eta} = p(a) -
p_\infty$. Assume $p_\infty$ is constant. Gas in
bubble obeys $pV^\gamma = \text{const}$ (**
adiabatic variation in an ideal gas $\gamma =
\frac{C_\rho}{C_V} = 1$ for air **)

\myskip
Then $\frac{\delta p}{p} = -\delta \frac{\delta
V}{V} = -3\gamma \frac{\delta a}{a}$ because $V
\propto a^3$.
\[ \implies \rho a_0 \ddot{\eta} = p_0 (-3\gamma)
\frac{\eta}{a_0} \]
$\implies$ SHM with $\omega = \left( \frac{3\gamma
p_0}{\rho a_0^2} \right)^{\half} \approx 2 \times
10^4 \mathsf{s^{-1}}$ for a $1\mathsf{mm}$ bubble.

\newpage
\section{Geophysical Flows}

\subsection{Water Waves}

Particularly successful application of potential
flow. Consider
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/63627868b8e311ed.png}
\end{center}

\subsubsection{Governing Equations}

Assume water is inviscid, and motion starts from
rest. Hence, flow is and remains irrotational.
\[ \implies \bf{u} = \nabla \phi \quad \text{and}
\quad \boxed{\nabla^2 \phi = 0 \text{ in $-h < z <
\zeta(x, y, t)$}} \tag{1} \]
Kinematic boundary conditions
\begin{itemize}
    \item At the rigid bottom
        \[ \boxed{\pfrac{\phi}{z} = 0 \text{ at }
        z = -h} \tag{2} \]
    \item At the air-water interface, often called
        the free surface as it is free to move,
        \[ \boxed{\pfrac{\zeta}{t} + u \pfrac{\zeta}{x} +
        v \pfrac{\zeta}{y} = w \text{ at } z =
        \zeta} \tag{3} \]
        (See section 1.4)
\end{itemize}
Dynamic boundary condition
\[ \boxed{p_{\text{water}} = p_{\text{air}} \text{
at } z = \zeta(x, y, t)} \]
As $p_{\text{air}} \ll p_{\text{water}}$, assume
that pressure variations in the air are $\ll$
those in the water. Hence take $p_{\text{air}} =
\text{const} = p_0$. Bernoulli for potential flow
\[\rho \pfrac{\phi}{t} + \half \rho u^2 + p + \chi
= f(t) \]
independent of $\bf{x}$ ($\chi = \rho gz$). Apply
to the surface and use $p = p_0 = \text{const}$.
\[ \implies \boxed{\rho \pfrac{\phi}{t} + \half
\rho |\nabla \phi|^2 + \rho g \zeta = f(t) \quad
\text{ at } \quad z = \zeta(x, y, t)} \tag{4} \]
Equations (1) - (4) give the full nonlinear
problem. (1) is linear but on an unknown domain;
(3) and (4) are complicated nonlinear boundary
conditions to be applied at an unknown position
$\zeta$.

\subsubsection{Linear Water Waves}

For small-amplitudes ($\zeta \ll h$ and
$\pfrac{\zeta}{x}, \pfrac{\zeta}{y} \ll 1$,
waveheight $\ll$ fluid depth and wavelenth) can
linearise the prolem (about a state of rest):
\begin{itemize}
    \item Ignore the terms quadratic in
        disturbance quantities, for example $u
        \pfrac{\zeta}{x}$, $\rho u^2$.
    \item Use Taylor series to expand boundary
        conditions at $z = \zeta$ in terms of
        information at $z = 0$. For example
        \[ \left. \pfrac{\phi}{z} \right|_{z =
        \zeta} = \left. \pfrac{\phi}{z} \right|_{z
        = 0} + \zeta \left. \pfrac[2]{\phi}{z}
        \right|_{z = 0} + \cdots \]
        (on the RHS we ignore everything other than the
        $\left. \pfrac{\phi}{z} \right|_{z = 0}$
        term, as the others are quadratic or
        higher).
\end{itemize}
The linearise problem is
\begin{align*}
    \nabla^2
    &= 0
    &&\text{in $-h < z < 0$}
    \tag{1} \\
    \pfrac{\phi}{z}
    &= 0
    &&\text{at $z = -h$}
    \tag{2} \\
    \pfrac{\zeta}{t}
    &= \pfrac{\phi}{z}
    &&\text{at $z = 0$} 
    \tag{3} \\
    \rho \pfrac{\phi}{t} + \rho g \zeta
    &= f(t)
    &&\text{independent of $x, y$ at $z = 0$}
    \tag{4}
\end{align*}
Using IB Michaelmas Methods, we can
\begin{itemize}
    \item either look for separable solutions
        $\phi(x, y, z, t) = \Phi(z) X(x) Y(y)
        T(t)$ and find
        \[ X'' = -k^2 X \quad Y'' = -l^2 T \quad
        T'' = -\omega^2 T \]
        or, on an infinite domain, take Fourier
        transforms with respect to $x, y, t$.
\end{itemize}
In 2D, end up looking for solution $\phi =
e^{i(kx - \omega t)} \Phi(z)$, $\zeta =
\zeta_0 e^{i(kx - \omega t)}$ (with real
part understood). (1) $\nabla^2 \phi = 0$
implies $\Phi'' - k^2\Phi = 0$ ($*$).
(Could write down $\Phi = e^{\pm kz}$, but
bottom boundary condition $\Phi'(-h) = 0$
suggests cosh and sinh better with origin
shifted to $z = -h$, ie:)
\[ (*) \implies \phi = A \cosh(k (z + h))
+ B \sinh(k(z + h)) \]
and boundary condition (2) implies $B =
0$. In (4), LHS $\propto e^{ikx} \implies
f(t) = 0$ and LHS $= 0$. So (3) and (4)
\[ \begin{cases}
    -i\omega \zeta_0 = Ak\sinh kh \\
    -i\omega A \cosh kh + g\zeta_0 = 0
\end{cases} \]
Linear, homogeneous equations, with
non-zero solution iff
\[ \left|
\begin{matrix}
    -k\sinh kh & -i\omega \\
    -i\omega \cosh kh & g
\end{matrix}
\right| = 0 \]
\[ \implies \omega^2 \cosh kh = gk \sinh
kh \]
Solutions must satisfy the \emph{dispersion
relationship}
\[ \boxed{\omega^2 = gk\tanh kh} \]
Wave crest moves with the \emph{phase speed}
\[ \boxed{c = \frac{\omega}{k}} \]
Long waves propagate faster.
\begin{center}
    \includegraphics[width=0.3\linewidth]
    {images/7fbf07ceb8e811ed.png}
\end{center}
Unlike light and sound, water waves are
\emph{dispersion}: waves of different frequencies
at different speeds and thus disperse as they
propagate away from a local disturbance (splash,
storm in the Atlantic).