hi Roy,
thanx for the explaination. But how did you solve ((u
* grad)u, v)_Omega? It's a square term. I heard there
are some other methods, streamline, least square FEM
...  I would like to hear your comments.

pan 



--- Roy Stogner <[EMAIL PROTECTED]> wrote:

> 
> On Thu, 24 Jan 2008, li pan wrote:
> 
> > I've worked with Newton type flow equation. To
> make it
> > sure, I would like to know the exact expression of
> > equation in ex13. Can you tell me?
> 
> The system of equations with variables (u,p) is :
> (partial u)/(partial t) = - (u * grad)u - div(sigma)
> div(u) = 0
> 
> Where sigma is the stress tensor (normalized to have
> unit viscosity)
> sigma = ((grad(u) + transpose(grad(u)))/2 - pI)
> 
> Then the weak form we use in ex13 and ex18, with
> test functions (v,q)
> is:
> ((partial u)/(partial t), v)_Omega = - ((u * grad)u,
> v)_Omega
>    + (sigma, grad v)_Omega + (sigma * n, v)_dOmega
> (div(u), q) = 0
> 
> In ex13 we use Dirichlet boundaries everywhere, so v
> = 0 on the
> boundary and we drop the dOmega term.  Otherwise,
> you'd substitute
> into that term the natural boundary condition:
> sigma * n = 0
> 
> which is actually what David wanted in the first
> place.  ;-)
> 
> You know, we probably ought to have something like
> this in the
> comments heading examples 13 and 18.  "The
> Navier-Stokes equations" is
> definitive enough, but the fact that we integrate
> all of sigma
> (including the pressure term) by parts isn't set in
> stone.
> ---
> Roy
> 
>
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