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Take an isothermal hydrostatic reference atmosphere, ${\rho }_0(z)={\rho }_s{e}^{-z/H}$, and linearize about rest. The horizontal acceleration and pressure-gradient force retain the equatorial Coriolis parameter $f\simeq \beta y$, with $\beta =2\Omega /a$. The traditional approximation retains only its vertical component. Quadratic perturbation advection is neglected. The vertical acceleration is balanced by perturbation buoyancy and pressure gradient after removing the reference hydrostatic balance.

The continuity constraint is the anelastic approximation, $\nabla \cdot ({\rho }_0\mathbf{u})=0$, giving $u_x+v_y+w_z-w/H=0$. It retains stratification-driven expansion while filtering acoustic density fluctuations. It is therefore not the full compressible continuity equation. Adiabatic displacement in a stably stratified background gives $b_t+N^{2}w=0$, with $N$ the Brunt–Väisälä frequency; upward displacement develops restoring negative buoyancy.

Pressure convention matters. With the entropy-buoyancy equation just described, a consistent reduced-pressure form writes the pressure acceleration as $-\nabla \Phi$, where $\Phi =p^{\prime }/{\rho }_0$. Its vertical term is $-{\partial }_z(p^{\prime }/{\rho }_0)$, not simply $-{\rho }_0^{-1}{p}_z^{\prime }$. The original source explicitly specifies variable ${\rho }_0(z)$ but prints the latter. Those differ by $p^{\prime }/({\rho }_{0}H)$, so the two cannot silently be equated. Parts b and c display the consequence and then solve the explicitly corrected reduced-pressure model that yields the requested wave equation. This is a mathematical inconsistency in the printed set, not missing source material.

Work with the explicitly corrected reduced-pressure system in b. Write $d=1/(2H)$ and use the complex wave amplitude, taking real parts at the end. The vertical factor satisfies $({\partial }_z^2-H^{-1}{\partial }_z)w=-(m^2+d^2)w$, so the wave equation givesFor a Kelvin wave, continuity, zonal momentum and the buoyancy equation then determine all polarizations:In particular $u,w,b,\Phi$ have amplitude proportional to $e^{z/(2H)}$, whereas pressure $p^{\prime }$ has amplitude proportional to $e^{-z/(2H)}$. Directly substituting these expressions into vertical momentum gives ${\Phi }_z=-i\omega (m^2+d^2)w/k^2=i(\omega -N^2/\omega )w$, verifying both the sign and the dispersion relation.

Meridional momentum is $\beta yu+{\Phi }_y=0$. Since $\Phi =cu$, it integrates to a Gaussian $\exp [-\beta y^2/(2c)]$. ThusThe eastward intrinsic branch $c>0$ is selected by decay away from the equator; choosing $c<0$ would make the meridional amplitude grow without bound. Although the squared dispersion relation has both frequency signs, a meridionally trapped Kelvin wave has this one phase-propagation direction.

For short vertical wavelengths, $m^2\gg k^2+d^2$, so ${\omega }^2\simeq N^2{k}^2/m^2$, $c\simeq N/|m|$ and $L_D^2\simeq N/(\beta |m|)$. The wave is a slowly oscillating, hydrostatic internal gravity wave with strong vertical variation and relatively narrow trapping. Here $u\simeq -(m/k)w$ and $p^{\prime }/{\rho }_0\simeq -(\omega m/k^2)w$; the buoyancy expression above remains valid.

For long vertical wavelengths, $|m|\ll \sqrt{k^2+d^2}$, the exact limiting speed is $c=N/\sqrt{k^2+d^2}$. To obtain the stated height-controlled limit also requires long horizontal wavelength, $|k|H\ll 1$. With both $|m|H\ll 1$ and $|k|H\ll 1$,The finite density scale height prevents the hydrostatic $1/|m|$ phase speed from diverging as $m\to 0$. The polarizations tend to $u\simeq -iw/(2Hk)$, $\Phi \simeq -iNw/k$ and $b\simeq -iNw/(2Hk)$. Long vertical wavelength alone at fixed large $kH$ would not justify $c=2NH$. In the opposite horizontally short limit the frequency approaches $N$ instead. The growth $e^{z/(2H)}$ compensates decreasing reference density in wave-energy density; a radiating wave train is a local modal description, not a finite-total-energy eigenfunction over an infinite vertical domain.