= Solution
Use the thin-annulus approximation stated in the source, so $u_r'+iku_z=0$. Let
$$
A(r)=2\Omega+r\Omega'=\frac1r(r^2\Omega)',\qquad
\kappa_{\rm ep}^2=2\Omega A=\frac1{r^3}(r^4\Omega^2)'.
$$
Here $\kappa_{\rm ep}^2$ is the conventional <Rayleigh discriminant>, also the squared local <radial epicyclic frequency> when positive. For $k\ne0$ and $\omega\ne0$, the azimuthal and axial mode equations give
$$
u_\phi=-\frac{iA}{\omega}u_r,\qquad u_z=\frac k\omega\pi.
$$
The thin-gap <incompressible flow> condition then gives $\pi=i\omega u_r'/k^2$. Substitute these into the radial equation:
$$
-i\omega u_r+\frac{2i\Omega A}{\omega}u_r=-\pi',\qquad
\frac{i\omega}{k^2}u_r''=\frac{i(\omega^2-\kappa_{\rm ep}^2)}\omega u_r.
$$
Thus
$$
u_r''=k^2\left(1-\frac{\kappa_{\rm ep}^2}{\omega^2}\right)u_r.
$$
To put this into exactly the form printed in the paper, define
$$
\boxed{\Phi(r)=-\kappa_{\rm ep}^2
=-\frac{2\Omega}{r}(r^2\Omega)'
=-\frac1{r^3}(r^4\Omega^2)'.}
$$
The <narrow-gap centrifugal stability equation> is consequently
$$
\boxed{-u_r''+\frac{k^2}{\omega^2}\Phi(r)u_r=-k^2u_r.}
$$
The two rigid cylinders are impermeable, so the <no-penetration boundary condition> gives
$$
\boxed{u_r(R_<)=u_r(R_>)=0.}
$$
There 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.
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