= 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 $-3\Omega^2x$, which is cancelled by $-\nabla\Phi_t=+3\Omega^2x\mathbf e_x$ with the signs in the stated equation; constant $P_0$ supplies no force. The flow is also incompressible.
Let $\mathbf u=\mathbf U+\mathbf u'$ and $h'=P'/\rho$. Axisymmetry removes $\mathbf U\cdot\nabla\mathbf u'$, while $\mathbf u'\cdot\nabla\mathbf U=-(3/2)\Omega u_x'\mathbf e_y$. Keeping first-order terms gives
$$
\boxed{\partial_tu_x'=-\partial_xh'+2\Omega u_y'},
\qquad
\boxed{\partial_tu_y'=-\frac12\Omega u_x'},
$$
$$
\boxed{\partial_tu_z'=-\partial_zh'},
\qquad
\boxed{\partial_xu_x'+\partial_zu_z'=0}.
$$
Differentiate the momentum equations in time and use incompressibility to eliminate $h'$. The radial velocity obeys
$$
\boxed{\partial_t^2(\nabla^2u_x')
+\Omega^2\partial_z^2u_x'=0}.
$$
For a plane wave proportional to $e^{i(k_xx+k_zz-\omega t)}$, this gives the <inertial wave> dispersion relation
$$
\boxed{\omega^2=\Omega^2
\frac{k_z^2}{k_x^2+k_z^2}}.
$$
When $k_x=0$ and $k_z\ne0$, incompressibility forces $u_z'=0$, and the vertical momentum equation then forces $h'=0$. The remaining motion has $\omega=\pm\Omega$ and satisfies $u_y'=\mp i u_x'/2$: each horizontal layer executes an <epicyclic motion>, with the phase varying vertically but no pressure or vertical-velocity perturbation.
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