= Solution
For the <Equatorial Kelvin wave>, set $\widehat V=0$. The zonal momentum and mass equations give $\omega^2=k^2c^2$. Meridional trapping selects the eastward branch
$$
\boxed{\omega=kc},
$$
for which
$$
\boxed{
\widehat U=U_0e^{-\beta y^2/(2c)},
\qquad
\widehat\Phi=cU_0e^{-\beta y^2/(2c)}}.
$$
The westward algebraic branch would grow away from the equator and is rejected.
For $\widehat V\ne0$, let $D=\omega^2-k^2c^2$. The zonal momentum and mass equations give
$$
\widehat U=\frac{i(\omega\beta y\widehat V-kc^2\widehat V_y)}D,
\qquad
\widehat\Phi=\frac{ic^2(k\beta y\widehat V-\omega\widehat V_y)}D.
$$
Substitution in meridional geostrophic balance and use of the stated <Hermite differential equation> gives the trapped <Equatorial Rossby-wave dispersion relation>
$$
\boxed{\omega=-\frac{kc}{2n+1}},
\qquad n=1,2,\ldots.
$$
For $n=1$, define $L_e=(c/\beta)^{1/2}$, $Y=y/L_e$, and take
$$
\widehat V=2V_0Y e^{-Y^2/2}.
$$
Then
$$
\boxed{
\widehat U=\frac{3iV_0}{4kL_e}(3-2Y^2)e^{-Y^2/2}},
$$
$$
\boxed{
\widehat\Phi=-\frac{3icV_0}{4kL_e}(1+2Y^2)e^{-Y^2/2}}.
$$
Without imposing meridional geostrophic balance, the equatorial shallow-water modes satisfy the Matsuno cubic
$$
\boxed{
\omega^3-[k^2c^2+(2n+1)\beta c]\omega
-\beta kc^2=0}.
$$
A dispersion diagram therefore contains high-frequency eastward and westward inertia--gravity branches as well as westward Rossby branches, plus the separate straight Kelvin branch $\omega=kc$. The Rossby curves approach $-kc/(2n+1)$ only in the long-wave limit and bend toward zero like $-\beta/k$ at short wavelength. The geostrophic model retains the Kelvin and long-wave Rossby lines but filters the inertia--gravity modes and misses Rossby-wave dispersion at larger $|k|$.
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