Rossby wave

Remarks

Indeed $\Delta \psi = -(k^2 + l^2) \psi$, so the equation becomes $\big( -\omega (k^2 + l^2) - \beta k \big) A \sin(k x + l y - \omega t) = 0$. The zonal phase speed $\omega / k = -\beta / (k^2 + l^2)$ is negative, so the crests move westward. On a sphere of radius $a$ at latitude $\varphi_0$, $\beta = 2 \abs{\Omega} \cos \varphi_0 / a$. In a uniform eastward flow $U$ the relation becomes $\omega = U k - \beta k / (k^2 + l^2)$, and the waves are stationary when $U = \beta / (k^2 + l^2)$.

Example: with $\beta = 1.6 \times 10^{-11}$ m$^{-1}$s$^{-1}$ and a wavelength of $6000$ km in $x$ with $l = 0$, $k \approx 1.05 \times 10^{-6}$ m$^{-1}$, and the phase speed is about $-15$ m/s.

Non-example: on an $f$-plane, where $\beta = 0$, the relation gives $\omega = 0$; there is no propagating Rossby wave.