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\noindent
Assume the initial conditions are such that $u, v$
independent of $z$ $\implies$ remain so.

\subsubsection*{Mass conservation}

Consider a cylindrical volume
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/e402c236c25011ed.png}
\end{center}
\[ \dfrac{}{t} \int_A \rho h \dd A +
\int_{\partial A} \rho h \bf{u}_H \cdot \bf{n} \dd
s = 0 \]
\[ \implies \int_A \pfrac{h}{t} + \nabla_H \cdot
(h\bf{u}_H) =\dd A = 0 \]
\[ \implies \pfrac{h}{t} + \pfrac{}{x} (hu) +
\pfrac{}{y} (hv) = 0 \]
or
\[ \boxed{\Dfrac{h}{t} = -h \left( \pfrac{u}{x} +
\pfrac{v}{y} \right)} \]

\subsubsection*{Geostrophic balance}

If $R_0 \equiv \frac{U}{\Omega L} \ll 1$
($\implies \bf{u} \cdot \nabla \bf{u} \ll
2\bf{\Omega} \times \bf{u}$) and steady
($\pfrac{}{t} = 0$) then (1) and (2)
\begin{align*}
    \implies -fv
    &= -\frac{1}{\rho} \pfrac{p}{x}
    = -g \pfrac{h}{x} \\
    +fu
    &= -\frac{1}{\rho} \pfrac{p}{y}
    = -g \pfrac{h}{y}
\end{align*}
$\implies$ flow given by 2D stream-function $\psi
= -\frac{p}{\rho f} = -\frac{gh}{f}$, i.e.
pressure / height contours are streamlines. Hence
winds blow parallel to isobars not perpendicular!
$\rho(f\bf{e}_z) \times \bf{u} = -\nabla_H p$ is
called \emph{geostrophic balance}. Note, if given
$h(x, y)$ can calculate steady flow, but no
equation for $h$ or $\pfrac{h}{t}$.

\subsubsection*{Potential Vorticity}

Recall
\begin{align*}
    \Dfrac{}{t} (\zeta + f)
    &= -(\zeta + f) \left( \pfrac{u}{x} +
    \pfrac{v}{y} \right) \tag{v} \\
    \Dfrac{}{t} h
    &= -h \left( \pfrac{u}{x} + \pfrac{v}{y}
    \right) \tag{m}
\end{align*}
\[ \implies \boxed{\Dfrac{}{t} \left( \frac{\zeta
+ f}{h} = 0 \right)} \tag{$\rho v$} \]
$\boxed{\frac{\zeta + f}{h}}$ is called the
\emph{potential vorticity}. PV is conserved for a
material cylinder; the total vorticity $\zeta + f$
is $\propto$ the height of cylinder because
vertical stretching $\to$ smaller radius $\to$
spins faster. PV is useful because $\nabla \times
(\text{momentum})$ eliminates the largest terms
$\nabla p$ and $2\rho \bf{\Omega} \times \bf{u}$
of nearly geostrophic balance and reveals how the
smaller terms give the evolution of $h$ with time.

\subsubsection*{Linearised Evolution}

Suppose $h = h_0 + \eta(x, y, t)$, $|\eta| \ll
h_0$, $|\nabla \eta| \ll \frac{H}{L} \ll 1$.
Linearised equations are
\begin{align*}
    \pfrac{u}{t} - fv
    &= -g \pfrac{\eta}{x}
    \tag{1} \\
    \pfrac{v}{t} + fu
    &= -g \pfrac{\eta}{y}
    \tag{2} \\
    \pfrac{\eta}{t} + h_0 \left( \pfrac{u}{x} +
    \pfrac{v}{y} \right) \tag{m}
\end{align*}
and
\[ \pfrac{}{t} \left( \frac{f}{h_0} \frac{1( +
\zeta / f)}{(1 + \eta / h_0)} \right) = 0 \implies
\frac{\zeta}{f} - \frac{\eta}{h_0}(+ O(z)) \]
is constant, i.e. $\zeta h_0 - \eta f = $ initial
value $\zeta_0 h_0 - \eta_0 f$. (pv). Now $h_0
\left( \pfrac{(1)}{x} + \pfrac{(2)}{y} \right) -
\pfrac{(m)}{t}$
\[ \implies -\pfrac[2]{\eta}{t} + gh_0 \nabla^2
\eta - h_0 f\zeta = 0 .\]
Comine with (pv) $\implies$
\[ \boxed{\pfrac[2]{\eta}{t} - c^2 \nabla^2 \eta +
f^2 \eta = f^2\eta_0 - h_0 f\zeta_0} \]
$c = \sqrt{gh_0} =
\text{shallow-water wave speed}$.


\begin{example*}
    $\eta_0 = \eps \sgn(x)$, $\bf{u}_0 = \bf{0}
    \implies \zeta_0 = 0$.
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/86787102c25411ed.png}
    \end{center}
    Without rotation ($f = $) just have
    $\pfrac[2]{\eta}{t} = c^2 \pfrac[2]{\eta}{x}$
    with d'Alembert solution
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/7c5a2f98c25511ed.png}
    \end{center}
    With rotation, the final steady state solves
    \[ \pfrac[2]{\eta_\infty}{x} - \frac{f^2}{c^2}
    \eta_\infty = -\frac{f^2}{c^2} \eps \sgn(x) \]
    \[ \implies \eta_\infty = \eps \sgn(x) (1 -
    e^{-|x| f / c}) \]
    \[ \implies u_\infty = -\frac{g}{f}
    \pfrac{\eta_\infty}{y} = 0 \]
    \[ v_\infty = \frac{g}{f}
    \pfrac{\eta_\infty}{x} = \eps \frac{g}{c}
    e^{-|x| f / c} \]
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/03519d7ec25611ed.png}
    \end{center}
    Completely different!
\end{example*}