 Solution

Write $A_{ij}={\partial }_j{u}_i$ and use incompressibility, $A_{ii}=0$. The displayed vector can be expressed asOn taking its divergence, ${\partial }_i{A}_{ij}=0$. Commuting the remaining partial derivatives giveswhich cancels the derivative of the second term of $D_i$. The derivatives falling on $u_k$ leaveIn homogeneous turbulence, averaging commutes with differentiation and $⟨D_i⟩$ is position independent, provided these moments exist. Consequently $⟨\mathrm{tr}(A^3)⟩=0$.

Using the strain-rate tensor and vorticity decomposition supplied in the question now gives the Betchov relation:This uses homogeneity and incompressibility; isotropy is unnecessary.