darker vortices it is directed clockwise. The cross-section shown in the figure is roughly $4000 \times 4000\, km^2$, where the typical scale of a convective cell is $1000\, km$.\
\[fig:v0\]](f1.eps){width="47.00000%"}

One can attempt to bypass the non-linear problem by adopting a much smaller “turbulent” value of the Reynolds number [@HP92]. For this “ad hoc” value of the Reynolds number the average dissipation would be the same as in the high Reynolds number active turbulence limit. Linearizing equation (\[eq:adim2\]) (with $n=1$ and $Re_1 = Re$), we obtain $$\label{eq:lind2}
\frac{\partial \mathbf{b}_{\perp} }{\partial t} 
=  c_\mathcal{A} \frac{\partial  \mathbf{u}_{\perp}}{\partial z}
+ \frac{1}{Re} \mathbf{ \nabla}^2_{\perp}  \mathbf{b}_{\perp}.$$ Taking into account that the forcing velocities are dominated by components at the injection scale $\ell_{c}$ (see eq. (\[eq:bc\])), the relation $\mathbf{\nabla}^2_{\perp}  \varphi = - \left( 2 \pi / \ell_{c} \right)^2 
\varphi$, where $\ell_{c} = \ell / k_{c}$ with the average wavenumber $k_{c} \sim 3.4$, is approximately valid. Integrating then eq. (\[eq:lind2\]) over $z$ and dividing by the length $L$, we obtain for $\mathbf{b}_{\perp}$ averaged along $z$: $$\label{eq:eqdiff}
\frac{\partial \mathbf{b}_{\perp} }{\partial t} 
=  \frac{c_\mathcal{A}}{L} 
\left[ \mathbf{u}^L \left( x, y \right) - \mathbf{u}^0 \left( x, y \right) \right]
- \frac{\left( 2 \pi \right)^2}{\ell_{c}^2 Re}  \mathbf{b}_{\perp}.$$ Indicating with $\mathbf{u}_{ph} = \mathbf{u}^L 