Alpha-squared Omega dynamo threshold (${\Omega }_0^2=4{\eta }^2{\ell }^4$)

For constant coefficients and a transverse wavenumber $\ell \ne 0$, a planar mean-field dynamo retaining both alpha feedback and shear has dispersion $(\sigma +\eta p{)}^2={\alpha }^{2}p+i\alpha \Omega k$, $p=k^2+{\ell }^2$. Marginal stability gives ${\alpha }_m^2=4{\eta }^4{p}^4/(4{\eta }^2{p}^3+{\Omega }^2{k}^2)$. Differentiate with respect to $k^2$: an interior minimum solves $4{\eta }^2{p}^3+{\Omega }^2(3k^2-{\ell }^2)=0$. Its derivative with respect to $k^2$ is positive, so a unique positive minimizing $k^2$ exists only above the displayed shear threshold. Below it the minimum is at $k=0$. The sign of the marginal oscillation frequency is fixed by $2\eta p\omega =\alpha \Omega k$.