 Solution

For the Keplerian shearing sheet velocity $\mathbf{U}=-(3/2)\Omega x{\mathbf{e}}_y$, its material acceleration vanishes because $\mathbf{U}\cdot \nabla \mathbf{U}=0$. The $x$ component of the Coriolis acceleration is $-3{\Omega }^{2}x$, which is cancelled by $-\nabla {\Phi }_t=+3{\Omega }^{2}x{\mathbf{e}}_x$ with the signs in the stated equation; constant $P_0$ supplies no force. The flow is also incompressible.

Let $\mathbf{u}=\mathbf{U}+{\mathbf{u}}^{\prime }$ and $h^{\prime }=P^{\prime }/\rho$. Axisymmetry removes $\mathbf{U}\cdot \nabla {\mathbf{u}}^{\prime }$, while ${\mathbf{u}}^{\prime }\cdot \nabla \mathbf{U}=-(3/2)\Omega u_x^{\prime }{\mathbf{e}}_y$. Keeping first-order terms givesDifferentiate the momentum equations in time and use incompressibility to eliminate $h^{\prime }$. The radial velocity obeysFor a plane wave proportional to $e^{i(k_{x}x+k_{z}z-\omega t)}$, this gives the inertial wave dispersion relation

When $k_x=0$ and $k_z\ne 0$, incompressibility forces $u_z^{\prime }=0$, and the vertical momentum equation then forces $h^{\prime }=0$. The remaining motion has $\omega =\pm \Omega$ and satisfies $u_y^{\prime }=\mp i{u}_x^{\prime }/2$: each horizontal layer executes an epicyclic motion, with the phase varying vertically but no pressure or vertical-velocity perturbation.