= Solution
With no mean flow and constant isotropic coefficients, the <mean-field dynamo> equation is
$$
\partial_t\overline{\mathbf B}=\alpha\nabla\times\overline{\mathbf B}+\eta_T\nabla^2\overline{\mathbf B},\qquad \nabla\cdot\overline{\mathbf B}=0,\qquad \eta_T=\eta+\beta>0.
$$
If the symbol $\beta$ already denotes the total diffusivity in the chosen mean-field convention, replace $\eta_T$ by $\beta$. Let $\nabla\times\mathbf B_0=\kappa\mathbf B_0$. This <Beltrami field> is <solenoidal> for $\kappa\ne0$, and taking its <curl> again gives $\nabla^2\mathbf B_0=-\kappa^2\mathbf B_0$. Thus
$$
\boxed{\overline{\mathbf B}(\mathbf x,t)=e^{st}\mathbf B_0(\mathbf x),\qquad s=\alpha\kappa-\eta_T\kappa^2.}
$$
For example, $\mathbf B_0=(\cos kz,-\sin kz,0)$ has $\kappa=k$, whereas reversing the sine sign gives $\kappa=-k$. Choose the helicity sign to agree with $\alpha$. The <homogeneous alpha-squared dynamo growth criterion> is then
$$
\boxed{0<|\kappa|<\frac{|\alpha|}{\eta_T}.}
$$
On an unbounded or suitably periodic domain with freely selectable <wavenumber>, this gives exponentially growing modes for every $\alpha\ne0$. Maximizing their <growth rate> gives $\kappa_* =\alpha/(2\eta_T)$ and $s_{\max}=\alpha^2/(4\eta_T)$. In a specified bounded domain, the permitted modes and boundary conditions impose a threshold; the optimal wavelength must also remain large compared with the turbulent scale for <mean-field electrodynamics> to apply.
\b[A nonzero alpha effect is necessary in this constant-coefficient model.] If $\alpha=0$, the mean equation is pure diffusion and the claimed growing modes do not exist. Positivity of $\beta$ alone does not imply growth.
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