 Solution

The QG thermodynamic relation makes vertical velocity proportional to the material derivative of ${\psi }_z$. At a flat rigid boundary, no normal flow therefore givesor ${\hat{\psi }}^{\prime }(0)=0$ for a nonzero-frequency mode. Let $K^2=k^2+l^2$ and take the vertical structure $\hat{\psi }={\psi }_0\cos (mz)$, which satisfies this condition. Linearizing $q_t+\beta {\psi }_x=0$ gives the Baroclinic Rossby wave dispersion relationFor long horizontal waves, $K{R}_d\ll 1$, so $\omega \simeq -\beta k{R}_d^2$: the vertical-mode deformation term controls the response. For short waves, $K{R}_d\gg 1$, so $\omega \simeq -\beta k/K^2$, the Barotropic Rossby wave form. In both limits the zonal phase propagation is westward relative to the mean flow.