= 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 identity>
$$
\int|u_r'|^2\,dr+\frac{k^2}{\omega^2}\int\Phi|u_r|^2\,dr
=-k^2\int|u_r|^2\,dr.
$$
Rearrange to obtain a real <Rayleigh quotient>:
$$
\boxed{\omega^2=-\frac{k^2\int\Phi(r)|u_r|^2\,dr}
{\int|u_r'|^2\,dr+k^2\int|u_r|^2\,dr}.}
$$
The denominator is strictly positive. Consequently, with the paper's convention,
$$
\boxed{\Phi(r)<0\ \text{throughout the gap guarantees oscillatory normal-mode stability}.}
$$
The weak inequality $\Phi\le0$ 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:
$$
\boxed{\frac d{dr}(r^2\Omega)^2>0.}
$$
This is the <Rayleigh centrifugal stability criterion>. In the suggested <Time-independent Schrödinger equation> analogy, a growing mode has $\omega^2<0$; if $\Phi\le0$, 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> $\kappa_{\rm ep}^2=4\Omega^2$ and $\Phi=-4\Omega^2$. In a gap of width $d=R_>-R_<$, the radial modes $u_r\propto\sin[n\pi(r-R_<)/d]$ have
$$
\omega_n^2=\frac{4\Omega^2k^2}{k^2+(n\pi/d)^2}>0,
$$
as 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.
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