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I am reading Majda & Bertozzi (Vorticity and Incompressible Flow). In page 12 the following equation appears:

$\frac{D \Omega}{Dt} + \Omega \mathcal{D} + \mathcal{D} \Omega = \nu \Delta \Omega$

where $\frac{D}{Dt}$ is the convective/lagrangian/material derivative. $\Omega$ and $\mathcal{D}$ are $3$ by $3$ matrices, the first antisymmetric and the second symmetric, and $\nu$ is a scalar. Using that $\Omega$ is defined by $\Omega h = \frac{1}{2} \omega \times h $, where $\omega$ is a vector function representing vorticity, I should be able to get the following vorticity equation (which apparently plays a crucial role in the rest of the book): $$ \frac{D \omega}{Dt} = \mathcal{D} \omega + \nu \Delta \omega. $$

Any idea how?

Here is a link to the book

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Are you sure you typed your first equation correct? I'd imagine the convective derivative hitting on $\Omega$ and not on $\mathcal{D}$. – Willie Wong Jul 6 '12 at 8:56
@ Wong: You are absolutely right! – chango Jul 6 '12 at 9:02
up vote 2 down vote accepted

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.

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Thanks. So there is more in this than I thought there was. I don't know anything about tensors. Can you recommend some text or notes about them. – chango Jul 6 '12 at 9:49
Huh, I'm surprised that there isn't already a question about that on MSE. Here's one possibly relevant question, and here's another with a link to a introductory reader given in the answers. – Willie Wong Jul 6 '12 at 10:03
But since you are interested in fluids now, you may consider taking a look at Aris' Vectors, Tensors, and the Basic Equations of Fluid Mechanics now published by Dover. – Willie Wong Jul 6 '12 at 10:05

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