Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 333 / 4 / ii

For an Equatorial Kelvin wave, set $\hat{v}=0$. The zonal and meridional momentum equations givesoHydrostatic balance and buoyancy evolution give $\hat{\sigma }=im\hat{\varphi }$ and $\hat{w}=-\omega m\hat{\varphi }/N^2$. Continuity then givesBecause the mode must decay as $|y|\to \infty$, one needs $k/\omega >0$. With the stipulated $\omega >0$, the acceptable branch is thereforeThe branch $\omega =-Nk/|m|$ would require $k<0$ when $\omega >0$, making the Gaussian exponent positive and the mode unbounded; it is not an equatorially trapped solution.

For real $k,m,\omega$,Upward group propagation has $m<0$, while a Kelvin wave has $k>0$, so $\overline{u^{\prime }{w}^{\prime }}>0$. Dissipation makes this eastward momentum flux decrease with height; therefore $-{\partial }_z\overline{u^{\prime }{w}^{\prime }}>0$ and the wave exerts an eastward force on the mean flow.