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.
Am Mittwoch, 23. Mai 2018 16:11:44 UTC+2 schrieb Daniel Arndt: > > >> > I tried to implement the static heat equation in cylindrical >> coordinates on a >> > 2d-plane, while being radially symmetric at x = 0. That leads me to the >> > following weak formulation of the problem (while neglecting the factor >> 2\pi): >> > >> > \int\limits_0^R \nabla U\nabla\phi r dr dz = \int\limits_0^R f\phi r dr >> dz >> > >> > Compared to the cartesian version >> > >> > \int\limits_0^Y \nabla U\nabla\phi dx dy = \int\limits_0^Y f\phi dx dy >> > >> > the onliest difference is the factor r in the equation. >> > In case you are trying to solve the problem in cylindrical coordinates, > you also have to take into account > that the laplacian looks different with respect to these coordinates. > > Best, > Daniel > -- 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 [email protected]. For more options, visit https://groups.google.com/d/optout.
