= Solution
The centrifugal criterion concerns axisymmetric disturbances, so set $m=0$. The azimuthal equation gives
$$
v'=-\frac{2\Omega+r\Omega'}{\sigma}u'.
$$
Eliminating $w'$ with incompressibility and then $p'$ with the axial equation yields
$$
\boxed{
\frac d{dr}\left[\frac1r\frac d{dr}(ru')\right]
-k^2\left(1+\frac{\Phi(r)}{\sigma^2}\right)u'=0},
$$
where the <Rayleigh discriminant> is
$$
\boxed{
\Phi(r)=2\Omega(2\Omega+r\Omega')
=\frac1{r^3}\frac d{dr}(r^4\Omega^2)}.
$$
Impermeability at the two solid walls gives $u'(r_i)=u'(r_o)=0$.
Multiply the equation by $r\overline{u'}$, integrate between the walls, and use <integration by parts>. The boundary terms vanish and one obtains
$$
\sigma^2
=-\frac{k^2\displaystyle\int_{r_i}^{r_o}
r\Phi|u'|^2\,dr}
{\displaystyle\int_{r_i}^{r_o}
\frac{|(ru')'|^2}{r}\,dr
+k^2\displaystyle\int_{r_i}^{r_o}r|u'|^2\,dr}.
$$
The denominator is positive. Therefore $\Phi\geq0$ 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> is
$$
\boxed{
\frac d{dr}(r^2\Omega)^2\geq0
\quad\text{for centrifugal stability}}.
$$
An outward decrease of squared specific angular momentum permits an axisymmetric centrifugal instability.
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