The vorticity transport equation is obtained from the (dimensionless) Navier-Stokes momentum equation
$$\frac{\partial \mathbf{u}}{\partial t} + \mathbf{u} \cdot \nabla\mathbf{u}= -\nabla p \, + \, \frac{1}{Re} \nabla^2 \mathbf{u}.$$
Taking the curl of both sides and applying some vector identities we get
$$\frac{\partial \boldsymbol\omega}{\partial t} + \mathbf{u} \cdot \nabla\boldsymbol\omega= \boldsymbol\omega \cdot \nabla \mathbf{u} - \boldsymbol\omega \, (\nabla \cdot \mathbf{u}) \, + \, \frac{1}{Re} \nabla^2 \boldsymbol\omega.$$
The first term on the RHS (vortex-line stretching) vanishes in 2D flow and the second on the RHS vanishes for incompressible flow since $\nabla \cdot \mathbf{u} = 0,$ resulting in
$$\frac{\partial \boldsymbol\omega}{\partial t} + \mathbf{u} \cdot \nabla\boldsymbol\omega= \frac{1}{Re} \nabla^2 \boldsymbol\omega.$$
You have already derived the equation
$$\tag{*}-\nabla^2 \mathbf{u} = \nabla \times \boldsymbol\omega,$$
using the definition of vorticity, $ \boldsymbol\omega = \nabla \times \mathbf{u},$ and the incompressibility condition, $ \nabla \cdot \mathbf{u} = 0.$ These are independent of the momentum equation and vorticity transport equations.
You are not going to derive (*) from the vorticity transport equation. It is just a kinematic relationship that incorporates the incompressibility condition. The incompressibility condition expresses conservation of mass. The vorticity transport equation expresses conservation of angular momentum.
Best Answer
Re-write the main given equation in index notation (following the Einstein summation convention)
$$ D_t \Omega_{ij} + \Omega_{ik}\mathcal{D}_{kj} + \mathcal{D}_{ik}\Omega_{kj} = \nu\triangle \Omega_{ij} \tag{1}$$
Small $\omega$ is defined by $$ \Omega_{ik}h^k = \frac12 \epsilon_{ijk}\omega_j h^k \tag{2}$$ which is the cross product definition. The $\epsilon_{ijk}$ is the Levi-Civita symbol (or fully antisymmetric tensor with $\epsilon_{123} = 1$).
Plugging in (2) (which implies that $\Omega_{ij} = \frac12 \epsilon_{ikj}\omega_k$) into (1) we have that
$$ \epsilon_{ilj} D_t\omega_l + \epsilon_{ilk}\mathcal{D}_{kj}\omega_l + \mathcal{D}_{ik}\epsilon_{klj}\omega_l = \nu \epsilon_{ilj}\triangle \omega_l \tag{3}$$
Next we use the property of the Levi-Civita tensor, $$ \epsilon_{jik}\epsilon_{jlk} = 2 \delta_{jl} \tag{4}$$ which means that multiplying (3) by $\epsilon_{imj}$ gives $$ 2D_t\omega_m + \left(\epsilon_{ilk}\epsilon_{imj}\mathcal{D}_{kj} + \epsilon_{klj}\epsilon_{imj}\mathcal{D}_{ik}\right) \omega_l = \nu \triangle \omega_m \tag{5}$$ The antisymmetry properties of the Levi-Civita tensor, as well as the symmetry of the tensor $\mathcal{D}$ can be used to show that $$ \epsilon_{ilk}\epsilon_{imj}\mathcal{D}_{kj} = \epsilon_{klj}\epsilon_{imj}\mathcal{D}_{ik} $$
So by another property of the Levi-Civita tensor, $$ \epsilon_{ijk}\epsilon_{imn} = \delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km} \tag{6}$$ we conclude that (5) is equivalent to $$ D_t\omega_m + \omega_m (\delta_{kj}\mathcal{D}_{kj} - \mathcal{D}_{jm}) = \nu \triangle \Omega_m ~.$$
Which shows that you in fact omitted one necessary condition for your equation to hold, which is that $\mathcal{D}$, in addition to being symmetric, is also trace-free.
If you have learned about differential forms, one should treat $\Omega$ as a differential two form on $\mathbb{R}^3$ and $\omega$ as a differential one form on $\mathbb{R}^3$ related by the Hodge star operator $\Omega = *\omega$. From this point of view the equation you want (the one for $\omega$) is merely obtained by taking the Hodge dual of the equation you are given (the one for $\Omega$) plus a little bit of multilinear algebra.