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Put $v=0$ and use amplitudes proportional to $e^{i(kx+mz-\omega t)}$. Zonal momentum giveswhile incompressibility, buoyancy evolution, and hydrostatic balance giveHenceMeridional geostrophic balance requiressoDecay as $|y|\to \infty$, together with $\beta >0$ and $k>0$, requires $\omega >0$. ThereforeThe negative-frequency root makes the Gaussian exponent positive and the solution diverge away from the equator. The acceptable branch is the eastward Equatorial Kelvin wave.

Solved by gpt-5.6-sol high.

Every forced propagating mode must have the imposed zonal frequencyFor $C>0$, only the Equatorial Kelvin wave has the correct sign. Upward group velocity in the fluid $z>0$ selectsWith $K(y)=\exp [-\beta y^2/(2C)]$, its pressure field can be writtenHere $v_K=0$, $u_K={\varphi }_K/C$, and the complex vertical-velocity amplitude isThe lower boundary is matched only ifafter choosing the phase of $a_K$ to make the prescribed cosine amplitude real.

For $C<0$, the upward-radiating Equatorial Rossby waves haveChoose each pair $({\hat{v}}_n,{\hat{\varphi }}_n)$ as in part c for this $m_n$. The propagating sum isThe other fields follow mode by mode from part c and ${\hat{w}}_n=-kC{m}_n{\hat{\varphi }}_n/N^2$. Thus the boundary matching condition is

For $C>0$, the space of upward-radiating wave profiles is only the one-dimensional Gaussian Kelvin profile, so a generic $W(y)$ cannot be matched by propagating waves. Its Kelvin projection radiates upward; the remaining forcing produces a balanced, vertically evanescent response trapped near the lower boundary. This is the equatorial analogue of the fact that quasi-geostrophic Rossby waves have westward rather than eastward phase propagation.

Solved by gpt-5.6-sol high.