= Solution
Take the curl of the <Navier-Stokes equation>. The pressure term disappears, curl commutes with the Laplacian, and the incompressible vector identity
$$
\nabla\times((\mathbf u\cdot\nabla)\mathbf u)
=(\mathbf u\cdot\nabla)\boldsymbol\omega
-(\boldsymbol\omega\cdot\nabla)\mathbf u
$$
gives
$$
\frac{\partial\boldsymbol\omega}{\partial t}
+(\mathbf u\cdot\nabla)\boldsymbol\omega
=(\boldsymbol\omega\cdot\nabla)\mathbf u
+\nu\nabla^2\boldsymbol\omega.
$$
Therefore the <material derivative> form is
$$
\boxed{
\frac{D\boldsymbol\omega}{Dt}
=(\boldsymbol\omega\cdot\nabla)\mathbf u
+\nu\nabla^2\boldsymbol\omega}.
$$
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