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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.

Multiply the narrow-gap centrifugal stability equation by the complex conjugate mode and use integration by parts with zero endpoint radial velocities. This gives the displayed real Rayleigh quotient. Its denominator is positive. Positive Rayleigh discriminant throughout the gap gives positive ${\omega }^2$ for every nonzero mode, excluding exponential axisymmetric growth; nonnegativity permits marginal cases. In the alternative convention $\Phi =-{\kappa }_{\mathrm{ep}}^2$, the numerator is $-k^2\int \Phi |u{|}^{2}dr$, so the stable sign is $\Phi <0$. For rigid rotation, ${\kappa }_{\mathrm{ep}}^2=4{\Omega }^2$ gives ${\omega }_n^2=4{\Omega }^2{k}^2/[k^2+(n\pi /d{)}^2]$.