 Solution

For unforced incompressible flow, multiply the vorticity equation by ${\omega }_i$. The antisymmetric part of the velocity gradient tensor drops out of ${\omega }_i{\omega }_j{\partial }_j{u}_i$, leavingStatistical homogeneity removes mean divergences. Integration by parts, together with $\nabla \cdot \mathbf{\omega }=0$, givesThe last equality holds for the averages, not pointwise. It follows from the usual curl/divergence identity after the divergence terms have averaged to zero. Vortex stretching produces enstrophy, while viscous diffusion destroys it.

Let $T=\ell /u$, $ϵ\sim u^3/\ell$, and use the Kolmogorov microscales $\eta =({\nu }^3/ϵ{)}^{1/4}$ and ${\tau }_{\eta }=(\nu /ϵ{)}^{1/2}$. Since $⟨{\omega }^2⟩=ϵ/\nu$, a slowly changing mean enstrophy has a time derivative of order $ϵ/(\nu T)$. The usual small-scale estimate givesThe relative imbalance is consequently ${\tau }_{\eta }/T\sim (\nu /u\ell {)}^{1/2}={\mathrm{Re}}^{-1/2}$, yieldingThe estimate assumes an established high-Reynolds-number cascade and integral-time evolution of the mean, not a rapid initial transient. In a forced flow the exact balance includes $⟨\mathbf{\omega }\cdot \nabla \times \mathbf{f}⟩$; forcing smooth on the integral scale of turbulence is smaller than the leading small-scale terms.