So, you would suggest simply adding the prefactors, after I then can keep
my problem in 2d?
Concerning the point with the laplacian:
When starting with the laplacian in cylinder coordinates
\int\frac{1}{r}\partial_r\left(r\partial_r U\right)\phi r dr dz =
\int\partial_r\left(r\partial_r U\right)\phi dr dz = -\int\left(r\partial_r
U\right)\partial_r\phi dr dz
which is exactly the same as if I calculate the laplacian in cartesian
coordinates, except the factor r. Is that correct?
Am Mittwoch, 23. Mai 2018 16:35:09 UTC+2 schrieb Wolfgang Bangerth:
>
> On 05/23/2018 10:29 PM, 'Maxi Miller' via deal.II User Group wrote:
> > Yes, I know, but at least the angle-dependent part can be neglected in
> my
> > case. Nevertheless, would switching the triangulation to a cylinder
> already
> > include that change? If yes, then I think the easiest way for me is to
> simply
> > switch the triangulation.
>
> You'd have to deal with a 3d problem, though you can now pose it in
> Cartesian
> coordinates.
>
> But of course solving a 3d problem is MUCH more expensive than solving in
> 2d.
> That's particularly annoying if you don't NEED to solve in 3d because you
> know
> that your solution doesn't depend on the angle :-)
>
> Best
> W.
>
> --
> ------------------------------------------------------------------------
> Wolfgang Bangerth email: [email protected]
> <javascript:>
> www: http://www.math.colostate.edu/~bangerth/
>
>
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