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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.

For a Keplerian shearing sheet, the shear rate is $S=3\Omega /2$ and the shearing-sheet tidal potential isA uniform steady solution isThe Coriolis acceleration of this linear shear flow balances the radial tidal acceleration. The only nonzero background component of the viscous stress tensor that matters is $T_{xy}=-\nu ({\Sigma }_0){\Sigma }_{0}S$, which is spatially constant. Hence $\nabla \cdot \mathbf{T}=0$: a local uniform patch has no stress gradient or torque divergence to drive an accretion flow.