Integrating by parts source terms and incompressiblity

[Félix Bunel]

Hi,

I have two simple question :

The first one is :

Is it possible to integrate source terms to make them easier to calculate ?

Basically, I can write this : the function xi is my test function, and sigma is just an independant function.

But will I be able to calculate it in my right hand side ?

The expression of sigma is complicated and it is a derivativ of a numerical function solved in another part of the code. So if I have to calculate the gradient of sigma I would have to use second order derivativs of this function which is not good I believe.

The second one is : do test function also have to respect the incompressibility equation ?

I have two variables that verifies an incompressibility equation :

Then, during my calculation of the weak part, I use a test function written

And I have termes in and that appears in my weak form. Can I write to simplify my weak formulation ?

Thanks a lot.

[Wolfgang Bangerth]

Felix,

> The first one is :
> Is it possible to integrate source terms to make them easier to calculate ?

Yes. But you have to pay attention to the fact that you get boundary
integrals. You also have to think about whether your source term is
continuously differentiable and in which sense you integrate by parts (e.g.,
cell-by-cell or for the whole domain).

> The second one is : do test function also have to respect the
> incompressibility equation ?

You mean in an equation like the Stokes equations? No, you are enforcing
incompressibility through the equation
div u = 0
but you have no corresponding equation for the test functions.

> I have two variables that verifies an incompressibility equation :

> Then, during my calculation of the weak part, I use a test function written

> And I have termes in and that appears in my weak form. Can I write to simplify
> my weak formulation ?

Nope :-)

Best
W.

--
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Wolfgang Bangerth email: bang...@colostate.edu
www: http://www.math.colostate.edu/~bangerth/