= Solution
Let $H=h_{00}+\zeta$ and write the horizontal <shallow water equations> on the constant <f-plane> as
$$
\frac{D\mathbf u_H}{Dt}+f\widehat{\mathbf z}\times\mathbf u_H=-g\nabla_H\zeta,\qquad H_t+\nabla_H\cdot(H\mathbf u_H)=0.
$$
Keeping only terms linear in the disturbance about rest gives the <linearized shallow water equations>
$$
u_t-fv=-g\zeta_x,\qquad v_t+fu=-g\zeta_y,\qquad \zeta_t+h_{00}(u_x+v_y)=0.
$$
Define the horizontal <divergence> $\delta=u_x+v_y$ and relative <vorticity> $q=v_x-u_y$. Taking respectively the <divergence> and vertical <curl> of the momentum equation yields
$$
\boxed{\delta_t=fq-g\nabla_H^2\zeta,\qquad q_t=-f\delta.}
$$
Together with $\zeta_t=-h_{00}\delta$, this proves
$$
\boxed{\partial_t\left(\frac q{h_{00}}-\frac{f\zeta}{h_{00}^2}\right)=0.}
$$
The nonlinear <shallow-water potential vorticity> is $Q=(f+q)/H$. The nonlinear vertical <vorticity equation> gives $D(f+q)/Dt=-(f+q)\delta$, while <mass conservation> gives $DH/Dt=-H\delta$; their ratio is therefore materially conserved. Expanding about rest,
$$
Q=\frac f{h_{00}}+\frac q{h_{00}}-\frac{f\zeta}{h_{00}^2}+O(\text{disturbance}^2).
$$
Advection of its constant background contributes nothing at first order, explaining why the displayed linear anomaly is conserved at each fixed position.
A further time derivative eliminates $q$ and $\zeta$ from the <divergence> equation:
$$
\delta_{tt}+f^2\delta-c_0^2\nabla_H^2\delta=0,\qquad c_0=\sqrt{gh_{00}}.
$$
For a <plane wave> with horizontal <wavevector> $(k,l)$, write $K=\sqrt{k^2+l^2}$. More directly, substituting the <plane wave> into the three linear equations gives a coefficient <matrix> with <determinant> proportional to $\omega[\omega^2-f^2-c_0^2K^2]$. Thus the full <linear rotating shallow-water dispersion relation> has one balanced branch and two wave branches:
$$
\boxed{\omega=0\quad\hbox{or}\quad\omega=\pm\sqrt{f^2+c_0^2K^2}.}
$$
The latter are <inertia-gravity waves>. The positive branch starts at $f$ with zero slope and tends to the straight line $c_0K$; the negative branch is its reflection. The requested sketch, including the stationary branch, is
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2001/iii/paper-48-dispersion.png]
{title=Rotating shallow-water dispersion and positive-branch phase and group speeds}
For the positive-frequency branch, the magnitudes of the <phase velocity> and <group velocity> are
$$
c_p=\frac\omega K=\sqrt{c_0^2+\frac{f^2}{K^2}},\qquad c_g=\frac{d\omega}{dK}=\frac{c_0^2K}{\omega},\qquad c_pc_g=c_0^2.
$$
The <group velocity> vector is $c_0^2(k,l)/\omega$; the negative branch reverses its direction. Near $\omega=f$, namely $KR_D\ll1$ with the <Rossby deformation radius> $R_D=c_0/f$, $c_p\sim f/K$ is large while $c_g\sim c_0^2K/f$ is small. The $K=0$ limit is an <inertial oscillation>, for which a spatial phase speed is not defined. Far above $f$, both speeds approach $c_0$, the <shallow water> gravity-wave speed, with $c_p>c_0$ and $c_g<c_0$ at finite $K$.
On the zero-frequency branch, <geostrophic balance> gives
$$
u=-\frac gf\zeta_y,\qquad v=\frac gf\zeta_x,\qquad \boxed{q=\frac gf\nabla_H^2\zeta,\qquad\delta=0.}
$$
For a nonconstant <plane wave>, $\widehat q=-(gK^2/f)\widehat\zeta$: relative <vorticity> and elevation have opposite signs. They can both be nonzero and spatially structured, whereas the horizontal <divergence> vanishes identically. A constant elevation is also a zero-frequency solution, with zero velocity and zero relative <vorticity>.
For the initially resting ridge, set $\zeta_0(x)=\epsilon h_{00}e^{-x^2/L^2}$ and assume an unbounded domain, uniform in $y$. Initial <potential vorticity> fixes
$$
q-\frac f{h_{00}}\zeta=-\frac f{h_{00}}\zeta_0.
$$
In the balanced remainder, insert $q_s=(g/f)\zeta_s''$ to obtain
$$
\boxed{(1-R_D^2\partial_x^2)\zeta_s=\zeta_0,\qquad u_s=0,\qquad v_s=\frac gf\zeta_s'.}
$$
The decaying <Green's function> of this <modified Helmholtz equation> gives the <Geostrophic adjustment of a Gaussian height ridge> explicitly:
$$
\zeta_s(x)=\frac1{2R_D}\int_{-\infty}^{\infty}e^{-|x-x'|/R_D}\zeta_0(x')\,dx'.
$$
The initial outward pressure force creates a cross-ridge current; the <Coriolis acceleration> turns that current and builds the along-ridge <geostrophic flow>. The remainder is radiated <inertia-gravity waves>, not a dissipative disappearance of energy.
For precision, use the <Fourier transform> $\widehat\zeta(k)=\int\zeta(x)e^{-ikx}\,dx$. Initial rest implies $\zeta_t(x,0)=0$, and the conserved anomaly gives
$$
\widehat\zeta_{tt}+(f^2+c_0^2k^2)\widehat\zeta=f^2\widehat\zeta_0.
$$
Consequently
$$
\widehat\zeta_0=\epsilon h_{00}L\sqrt\pi\,e^{-k^2L^2/4},\qquad
\widehat\zeta_s=\frac{\widehat\zeta_0}{1+R_D^2k^2},\qquad
\widehat\zeta(t)=\widehat\zeta_s+(\widehat\zeta_0-\widehat\zeta_s)\cos(\sqrt{f^2+c_0^2k^2}\,t).
$$
These formulas satisfy both initial conditions. They also show that the waves carry no linear <potential vorticity> anomaly: $q=(f/h_{00})(\zeta-\zeta_0)$ and $\delta=-\zeta_t/h_{00}$ reconstruct the remaining fields, with $u$ and $v$ obtained from their derivatives. On the infinite line, dispersive wave packets propagate away and decay locally, leaving the balanced ridge and its oppositely directed flanking currents. <Conservation of energy> still holds globally. In a reflecting finite basin, standing waves may persist, so a permanent pointwise relaxation is not guaranteed without dissipation.
If $L\gg R_D$, most of the initial elevation remains balanced: $\zeta_s\simeq\zeta_0$ with small smoothing corrections. If $L\ll R_D$, the balanced ridge spreads to width $R_D$, and away from the original narrow core
$$
\zeta_s(x)\simeq\frac{\epsilon h_{00}L\sqrt\pi}{2R_D}e^{-|x|/R_D}.
$$
Thus the final central height is much smaller than the initial one and the wave component is substantial. The integral of the balanced elevation equals the initial integral, since the <Green's function> has unit integral; the oscillatory component has zero net added volume.
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