 Solution

For the Equatorial Kelvin wave, set $\hat{V}=0$. The zonal momentum and mass equations give ${\omega }^2=k^2{c}^2$. Meridional trapping selects the eastward branchfor whichThe westward algebraic branch would grow away from the equator and is rejected.

For $\hat{V}\ne 0$, let $D={\omega }^2-k^2{c}^2$. The zonal momentum and mass equations giveSubstitution in meridional geostrophic balance and use of the stated Hermite differential equation gives the trapped Equatorial Rossby-wave dispersion relationFor $n=1$, define $L_e=(c/\beta {)}^{1/2}$, $Y=y/L_e$, and takeThen

Without imposing meridional geostrophic balance, the equatorial shallow-water modes satisfy the Matsuno cubicA 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 $|k|$.