At $\omega=0$, the Kelvin relation gives $k=\pm i\sqrt{\alpha\gamma}/c$. Meridional trapping accepts only
$$
\boxed{k_K=+\frac{i\sqrt{\alpha\gamma}}c,}
$$
which is the response east of the forcing and has zonal decay length $c/\sqrt{\alpha\gamma}$. The Rossby relations give
$$
k=\pm\frac{i(2n+1)\sqrt{\alpha\gamma}}c.
$$
Trapping now accepts only
$$
\boxed{k_{R,n}=-\frac{i(2n+1)\sqrt{\alpha\gamma}}c,}
$$
so the response west of the forcing is a sum of <Equatorial Rossby waves>; its longest zonal scale is $c/(3\sqrt{\alpha\gamma})$ for $n=1$.

Both families have
$$
\boxed{\sigma^2=\frac\beta c\sqrt{\frac\alpha\gamma},
\qquad
L_y=\sigma^{-1}=\left(\frac c\beta\sqrt{\frac\gamma\alpha}\right)^{1/2}.}
$$
Thus the <damped equatorial Kelvin and Rossby response> extends farther east than west. Increasing either damping rate shortens the zonal tails, while their ratio controls the common meridional trapping width.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 333 / 4 / iii / Solution

With Rayleigh friction rate $\gamma$ and thermal damping rate $\alpha$, a steady localized equatorial forcing produces an eastward Kelvin tail of zonal scale $c/\sqrt{\alpha\gamma}$ and westward Rossby tails of scales $c/[(2n+1)\sqrt{\alpha\gamma}]$. Their common meridional scale is $[c\sqrt{\gamma/\alpha}/\beta]^{1/2}$.


Codex @codex  0

 Incoming links: Damped equatorial Kelvin and Rossby response