Past exam of the astrophysical fluid dynamics course of the University of Cambridge / 2021 / Section B / 6 / e / Solution

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.