I refer to the part where the spatial residual is defined as B(w_n)(z)=... There you integrate over G(w_n), z over \partial\Omega, but I would have expected to integrate over \Omega. I have not compared that equation with the code yet, after I already had problems with that step.
Am Mittwoch, 16. August 2017 15:38:44 UTC+2 schrieb Wolfgang Bangerth: > > On 08/15/2017 01:15 AM, 'Maxi Miller' via deal.II User Group wrote: > > I try to follow the math in example 33, but am not fully able to > understand > > how the right hand side is included. According to my understanding, I > have the > > function/vector G, multiply it with the test function z, and integrate > over > > both over the whole area. After there are no derivatives included, I do > not > > have to do any reformulations. Why do I then still end up with an > integral > > over both (G\cdot z), but integrated over the border? > > Can you point out where in the code this happens? > > There is of course a boundary term that comes from integrating by parts > the > equation. But G itself should not appear in a boundary integral. > > Best > W. > > -- > ------------------------------------------------------------------------ > Wolfgang Bangerth email: bang...@colostate.edu > <javascript:> > www: http://www.math.colostate.edu/~bangerth/ > > -- The deal.II project is located at http://www.dealii.org/ For mailing list/forum options, see https://groups.google.com/d/forum/dealii?hl=en --- You received this message because you are subscribed to the Google Groups "deal.II User Group" group. To unsubscribe from this group and stop receiving emails from it, send an email to dealii+unsubscr...@googlegroups.com. For more options, visit https://groups.google.com/d/optout.
