 Solution

Use the thin-annulus approximation stated in the source, so $u_r^{\prime }+ik{u}_z=0$. LetHere ${\kappa }_{\mathrm{ep}}^2$ is the conventional Rayleigh discriminant, also the squared local radial epicyclic frequency when positive. For $k\ne 0$ and $\omega \ne 0$, the azimuthal and axial mode equations giveThe thin-gap incompressible flow condition then gives $\pi =i\omega u_r^{\prime }/k^2$. Substitute these into the radial equation:ThusTo put this into exactly the form printed in the paper, defineThe narrow-gap centrifugal stability equation is consequentlyThe two rigid cylinders are impermeable, so the no-penetration boundary condition givesThere is no inviscid no-slip condition on the tangential or axial components.

The sign deserves emphasis: the source's printed plus sign makes its $\Phi$ the negative of the conventional Rayleigh discriminant. Using the conventional positive discriminant instead requires a minus sign in front of that potential term. These are equivalent conventions; replacing one without adjusting the other would falsely predict instability of solid-body rotation. The $k=0$ and marginal $\omega =0$ cases, excluded by these eliminations, must be treated separately or by limits.