 Solution

Use plane wave perturbations proportional to $e^{i(k_{x}x+k_{z}z-\omega t)}$, with amplitudes $\hat{u},\hat{w},\hat{\pi },\hat{\varrho }$. The linear equations becomeTo eliminate pressure without dividing by $k_x$, multiply the vertical momentum equation by $k_x$ and subtract $k_z$ times the horizontal momentum equation:Use incompressible flow and the equation for mass density to obtain ${\omega }^2{\rho }_0(k_x^2+k_z^2)\hat{w}=-g{\rho }_0^{\prime }{k}_x^2\hat{w}$. Hence, for a nontrivial internal gravity wave,For stable density stratification, ${\rho }_0^{\prime }<0$ and $N=\sqrt{-g{\rho }_0^{\prime }/{\rho }_0}$ is the buoyancy frequency. Then $0\le {\omega }^2\le N^2$ and the normal modes oscillate rather than grow. For ${\rho }_0^{\prime }>0$, a mode with $k_x\ne 0$ has ${\omega }^2<0$ and grows exponentially: denser fluid lies above lighter fluid. Uniform density, ${\rho }_0^{\prime }=0$, is marginal for this restoring mechanism.

The case $k_x=0$, $k_z\ne 0$, has zero frequency in the displayed relation. Incompressible flow then requires $\hat{w}=0$, so it is a degenerate non-propagating case rather than a finite-frequency vertical oscillation. These equations concern the incompressible model of part (b); its $N^2$ does not contain the adiabatic compression correction needed in part (a).