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Old   May 10, 2002, 09:50
Default skew-symmetric form
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Annie
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Dear all,

I found there are two versions of the so-called skew-symmetric form for the convective term in the incompressible flow, i.e.,

1) u \cdot \nabla u = 1/2 ( u \cdot \nabla u ) + 1/2 \nabla (u \cdot u);

2) u \cdot \nabla u = 1/2 ( u \cdot \nabla u ) + 1/2( (\p u_i u_j)/(\p x_j));

The second version is easy to understand but i think the first one never holds, though it is so widely used. Any one can give some explanations ?

Thank you so much!
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Old   May 16, 2002, 08:38
Default Re: skew-symmetric form
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Nicola
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Hi Annie,

this is a simple problem of differential calculus. If i have not made a mistake in understanding the formula, the following should be the answer to your question. The thesis is: u \cdot \nabla u = \nabla (u \cdot u) that is U_j d/dX_j (U_i) = d/dX_i (U_j U_j) (using the summation convention for repeated indices). We have simply: U_j d/dX_j (U_i) = d/dX_j (U_j U_i)- U_i d/dX_j (U_j) and the last term is zero because it is the divergence of the velocity vector. I hope this have cleared your doubts. Best regards,

Nicola
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