= Solution
Let $\zeta$ denote free-surface height and $h=\zeta-b$ the water depth. The inviscid rotating <shallow water equations> with constant <Coriolis parameter> $f$ are
$$
D_Hu-fv=-g\zeta_x,\qquad D_Hv+fu=-g\zeta_y,
\qquad h_t+(hu)_x+(hv)_y=0,
\qquad D_H=\partial_t+u\partial_x+v\partial_y.
$$
It is $\zeta$, not $h$, that supplies the horizontal pressure gradient. For a fixed bottom, mass conservation equivalently gives $D_Hh=-h(u_x+v_y)$.
Write $\xi=v_x-u_y$ for relative <vorticity>. Differentiate the second momentum equation in $x$ and the first in $y$, then subtract. The pressure terms cancel. Expanding the differentiated advection terms gives
$$
D_H\xi+(\xi+f)(u_x+v_y)=0.
$$
Consequently the <absolute vorticity> $q_a=\xi+f$ satisfies $D_Hq_a=-q_a\nabla_H\cdot\mathbf u_H$. Combining this with mass conservation yields exact <shallow-water potential vorticity> conservation:
$$
\boxed{D_H(q_a/h)=\frac{hD_Hq_a-q_aD_Hh}{h^2}=0.}
$$
A sloping bottom changes the depth and hence the stretching, but contributes no curl to the gravitational pressure force.
For the <quasi-geostrophic approximation>, choose constant reference depth $H_0$, speed $U$, and horizontal length $L$. Require <Rossby number> $\operatorname{Ro}=U/(|f|L)\ll1$, slow evolution on the advective time $L/U$, smooth horizontal scales, and $L/L_D=O(1)$ or smaller, where $L_D=\sqrt{gH_0}/|f|$ is the <Rossby deformation radius>. Write $\zeta=\zeta_0+\eta$, $b=b_0+\delta b$, and $h=H_0+\eta-\delta b$. The leading momentum balance is <geostrophic balance>:
$$
u_g=-\psi_y,\qquad v_g=\psi_x,\qquad \psi=g\eta/f.
$$
It gives $\eta/H_0=O(\operatorname{Ro}(L/L_D)^2)$, while we require $\delta b/H_0=O(\operatorname{Ro})$ and slopes on the chosen horizontal scale. Relative <vorticity> divided by $f$ is also $O(\operatorname{Ro})$. Thus
$$
\frac{q_a}{h}=\frac f{H_0}+\frac1{H_0}
\left(\nabla_H^2\psi-\frac f{H_0}\eta+\frac f{H_0}\delta b\right)
+O\left(\operatorname{Ro}^2\frac{|f|}{H_0}\right).
$$
The derivative of the constant first term vanishes. In advecting the first-order anomaly, the <ageostrophic flow> contributes only at the next order. Therefore the <shallow-water quasi-geostrophic potential vorticity> equation is
$$
\boxed{q=\nabla_H^2\psi-L_D^{-2}\psi+\frac f{H_0}\delta b,
\qquad q_t+J(\psi,q)=0,\qquad J(\psi,q)=\psi_xq_y-\psi_yq_x.}
$$
This retains the balanced vortical branch, including <Rossby waves> driven here by the topographic <potential-vorticity gradient>. It filters the freely oscillating inertia–gravity branches of the full <shallow water equations>, and does not retain independent boundary <Kelvin wave> dynamics. Small <Rossby number> alone is insufficient for the reduction: fast waves must be absent at leading order, as in well-prepared balanced initial data.
In the channel take $H_0=h_{00}$ and measure the bottom anomaly from mid-channel: $\delta b=\varepsilon(y-L/2)$. Define $\beta=\varepsilon f/H_0$. The resting <potential vorticity> is $\beta(y-L/2)$, and the linear <Rossby wave> equation is
$$
\partial_t(\nabla_H^2-L_D^{-2})\psi+\beta\psi_x=0.
$$
For nonzero zonal <wavenumber> $k$, wall conditions $\psi=0$ select
$$
\boxed{\psi=\operatorname{Re}\{A\sin(n\pi y/L)e^{i(kx-\omega t)}\},
\qquad \omega=-\frac{\beta k}{k^2+(n\pi/L)^2+L_D^{-2}},\quad n=1,2,\ldots.}
$$
The $k=0$ modes are stationary. Write $K^2=k^2+(n\pi/L)^2+L_D^{-2}$ and take real $A$ for the following sketch. The total <potential vorticity> is $q=\beta(y-L/2)-K^2\psi$. For a contour whose undisturbed position is $y_0$, its small meridional displacement is
$$
\delta y=\frac{K^2}{\beta}\psi(x,y_0,t)+O(A^2).
$$
Thus <potential-vorticity conservation> says a northward displacement in a positive background gradient carries a negative anomaly, $q'=-\beta\delta y$. <Potential-vorticity inversion> turns this into positive $\psi$ and anticyclonic motion around the bulge. At a positive cosine crest, $v=\psi_x$ is northward on its western flank and southward on its eastern flank. Advection raises the contour on the west and lowers it on the east, so the crest moves west. Algebraically, $\delta y_t=v$ and $\delta y=(K^2/\beta)\psi$ give $\omega/k=-\beta/K^2$, the same <phase velocity> as above. For $\beta<0$ all propagation directions reverse.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/iii/paper-81-pv-contour.png]
{title=Potential-vorticity contour displacement and westward Rossby-wave propagation for a positive background gradient}
For a square basin use a nonzero-frequency <normal mode> $\psi=\operatorname{Re}\{\Psi(x,y)e^{-i\omega t}\}$, with $\Psi=0$ on all walls. Its equation is
$$
(\nabla_H^2-L_D^{-2})\Psi+\frac{i\beta}{\omega}\Psi_x=0.
$$
Set $\alpha=\beta/(2\omega)$ and $\widehat\Psi=e^{i\alpha x}\Psi$. Substitution cancels the first derivative and gives
$$
\nabla_H^2\widehat\Psi+(\alpha^2-L_D^{-2})\widehat\Psi=0,
\qquad \widehat\Psi=0\text{ on the walls}.
$$
The square's Dirichlet <Laplacian eigenfunctions> are $\widehat\Psi=A\sin(m\pi x/L)\sin(n\pi y/L)$, with $m,n$ positive integers. Consequently the <square-basin topographic Rossby mode> has
$$
\boxed{\omega=\pm\frac{\beta}{2\sqrt{\pi^2(m^2+n^2)/L^2+L_D^{-2}}},
\qquad \Psi=Ae^{-i\beta x/(2\omega)}\sin(m\pi x/L)\sin(n\pi y/L).}
$$
For $\beta>0$, the positive-frequency branch therefore tends to $\beta L/[2\pi\sqrt{m^2+n^2}]$ as $L/L_D\to0$, as required. The negative-frequency solution is its complex-conjugate representation for a real disturbance; the physical frequency magnitude is $|\beta|/[2\sqrt{\pi^2(m^2+n^2)/L^2+L_D^{-2}}]$.
Back to article page
