This question is concerned with how to prove Rayleigh's stability criterion for a rotating inviscid fluid. I can follow the details of the proof up to the final line, but I cannot see immediately how the conclusion is deduced.
This stability criterion considers a purely azimuthal flow ${\bf u} = V(r) \,{\bf e}_\theta = r\,\Omega\, {\bf e}_\theta$, where $\Omega$ is the angular velocity of the flow and ${\bf e}_r,\; {\bf e}_\theta$ are unit vectors in a cylindrical coordinate system. The criterion states that a necessary and sufficient for condition for the stability of this flow to axisymmetric disturbances is that $$\frac{d}{dr} \left( \Omega r^2 \right)^2 \geq 0 $$ throughout the flow. (See for example `Elementary Fluid Dynamics', D. J. Acheson, pg 318). This result can be demonstrated through energy arguments that involve displacing small elements of fluid and arguing for conservation of angular momentum. However, I am instead interested in a rigorous proof of this result based directly on the perturbation equations of linear stability. To do this we introduce infinitesimal perturbations of the form ${u}_r = \hat{u}_r \, e^{st + inz}$ (with similar expressions for the other variables), where $s$ is in general complex and $n$ real. Note that the stability of the flow is determined by the sign of the real part of $s$. When $\Re(s) > 0$ the disturbance will grow (i.e. the flow is unstable) while the disturbance will decay (the flow is stable) when $\Re(s) < 0$. We eventually end up with the equation (see Acheson pg 382) $$ s^2 \int_{r_1}^{r_2} \left\{ r \mid \hat{u}_r' \mid^2 + r\left(n^2 + \frac{1}{r^2}\right) \mid \hat{u}_r \mid^2 \right\} \, dr = - \int_{r_1}^{r_2} \frac{n^2}{r^2}(r^2 V^2)' \mid \hat{u}_r \mid^2 \, dr $$ Since the integral on the lhs is positive, we can thus deduce : \begin{eqnarray*} (r^2 V^2)' &= \frac{d}{dr} \left( \Omega r^2 \right)^2 > 0 \text{ everywhere } \Rightarrow s^2 < 0,\\ (r^2 V^2)' &= \frac{d}{dr} \left( \Omega r^2 \right)^2 < 0 \text{ everywhere } \Rightarrow s^2 > 0. \\ (r^2 V^2)' & \text {changes sign } \Rightarrow s^2 \text{ can have either sign } \end{eqnarray*}
So far so good! The question is how do we for example then deduce stability (to axisymmetric disturbances) when $\frac{d}{dr} \left( \Omega r^2 \right)^2 > 0 $? In this case we have that $s^2 < 0$ which means that $s$ must be purely imaginary. This means that the real part of $s$ is zero which is surely 'neutral stability' (a disturbance which neither grows nor decays), rather than stability ? Furthermore when $\frac{d}{dr} \left( \Omega r^2 \right)^2 < 0 $ we have that $s^2 > 0$ which means that $s$ must be real. However here $s$ can be either positive (instability) or negative (stability) - since we haven't proven it is positive, how can we claim the flow is unstable?
$\Re(s)$) to denote that. Your notation makes it confusing when you later write $s^2$ (which one might take to mean that $R=2$). $\endgroup$