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For axisymmetric inviscid modes between impermeable cylinders in a thin annulus, let $A=2\Omega +r{\Omega }^{\prime }$ and ${\kappa }_{\mathrm{ep}}^2=2\Omega A=r^{-3}(r^4{\Omega }^2{)}^{\prime }$. The linear azimuthal equation gives $u_{\varphi }=-iA{u}_r/\omega$ and the axial equation gives $u_z=k\pi /\omega$, where $\pi$ is the kinematic pressure perturbation. Thin-gap incompressibility implies $\pi =i\omega u_r^{\prime }/k^2$. Eliminating it from the radial equation gives the title equation with $u(R_{<})=u(R_{>})=0$. If a plus sign is used for the potential, its coefficient must be defined as $\Phi =-{\kappa }_{\mathrm{ep}}^2$, the negative of the usual Rayleigh discriminant.

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.

Let $u_r$ be a nonzero admissible mode, multiply the equation of part (d) by its complex conjugate and integrate between the cylinders. The endpoint conditions remove the boundary term in integration by parts, giving the Rayleigh stability integral identityRearrange to obtain a real Rayleigh quotient:The denominator is strictly positive. Consequently, with the paper's convention,The weak inequality $\Phi \le 0$ excludes exponentially growing axisymmetric modes, but permits marginal cases and possible algebraic growth when the restoring effect vanishes. Strict negativity avoids that degeneracy. Equivalently, the conventional Rayleigh discriminant is positive, or the squared specific angular momentum increases outward:This is the Rayleigh centrifugal stability criterion. In the suggested Time-independent Schrödinger equation analogy, a growing mode has ${\omega }^2<0$; if $\Phi \le 0$, its effective potential $k^2\Phi /{\omega }^2$ is nonnegative. The nonnegative kinetic term and that potential cannot have the required negative eigenvalue $-k^2$, which gives the same exclusion.

As a sign check, for solid-body rotation ${\kappa }_{\mathrm{ep}}^2=4{\Omega }^2$ and $\Phi =-4{\Omega }^2$. In a gap of width $d=R_{>}-R_{<}$, the radial modes $u_r\propto \sin [n\pi (r-R_{<})/d]$ haveas expected for inertial waves. The argument establishes axisymmetric inviscid stability within the stated thin-gap model, rather than a claim about all perturbations of a viscous flow.

The pressure-square term in the Cowling energy principle for a rotating barotropic star is nonnegative, and ${\mathcal{N}}^2>0$ makes its stratification term nonnegative. Write $j=R^2\Omega$ for specific angular momentum. ThenFor a regular star reaching the rotation axis, $j(0)=0$. The given $dj/dR>0$ therefore makes $j>0$ for $R>0$, and consequently ${\kappa }^2>0$. Equivalently, with the usual nonnegative angular-velocity convention the sign follows directly. All three energy terms are nonnegative, soThis is stability within the Cowling approximation used throughout. The orientation-independent rotational condition is $d(j^2)/dR\ge 0$, the Rayleigh discriminant criterion. If a fluid region excludes the axis and negative angular velocity is allowed, $dj/dR>0$ alone needs the additional sign of $j$; the squared-angular-momentum criterion avoids that ambiguity.

The centrifugal criterion concerns axisymmetric disturbances, so set $m=0$. The azimuthal equation givesEliminating $w^{\prime }$ with incompressibility and then $p^{\prime }$ with the axial equation yieldswhere the Rayleigh discriminant isImpermeability at the two solid walls gives $u^{\prime }(r_i)=u^{\prime }(r_o)=0$.

Multiply the equation by $r\overline{u^{\prime }}$, integrate between the walls, and use integration by parts. The boundary terms vanish and one obtainsThe denominator is positive. Therefore $\Phi \ge 0$ throughout the annulus excludes positive real ${\sigma }^2$ and gives centrifugal stability. Since $r^4{\Omega }^2=(r^2\Omega {)}^2$ is the square of the specific angular momentum, Rayleigh's circulation criterion isAn outward decrease of squared specific angular momentum permits an axisymmetric centrifugal instability.

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 ${\omega }_n^2=4{\Omega }^2{k}^2/[k^2+(n\pi /d{)}^2]$.