# skew-symmetric form

 Register Blogs Members List Search Today's Posts Mark Forums Read

 May 10, 2002, 09:50 skew-symmetric form #1 Annie Guest   Posts: n/a Sponsored Links 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!

 May 16, 2002, 08:38 Re: skew-symmetric form #2 Nicola Guest   Posts: n/a 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

 Thread Tools Display Modes Linear Mode

 Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is OffTrackbacks are On Pingbacks are On Refbacks are On Forum Rules

 Similar Threads Thread Thread Starter Forum Replies Last Post mnabi Main CFD Forum 0 March 5, 2011 17:58 siw Main CFD Forum 0 August 17, 2010 03:18 Demonwolf Main CFD Forum 2 October 29, 2009 20:53 Christoph FLUENT 0 November 3, 2005 06:50 santosh FLUENT 1 August 16, 2005 14:38