Uniaxial alpha dynamo threshold (${\gamma }^2=(q^2+m^2{)}^2/(m^2+\delta q^2)$)

For $\mathbf{\alpha }=\mathrm{diag}({\alpha }_0,{\alpha }_0,{\alpha }_1)$, set $\gamma ={\alpha }_0/\eta$, $\delta ={\alpha }_1/{\alpha }_0$, and $q^2=k^2+l^2$. A steady nonzero Fourier mode obeys the displayed condition when the denominator is nonzero. At fixed $q>0$ and $0<\delta <1$, minimizing over $m^2\ge 0$ gives $m_{\ast }^2=(1-2\delta )q^2$ and ${\gamma }_{\min }^2=4(1-\delta )q^2$ for $\delta <1/2$; for $\delta \ge 1/2$ it gives $m_{\ast }^2=0$ and ${\gamma }_{\min }^2=q^2/\delta$. This is a boundary-constrained threshold of an anisotropic alpha-squared dynamo.