On 2/12/19 6:15 PM, SebG wrote: > > sorry about not getting back to this for while. I have addressed your second > point and created an example that is much easier to understand. Now the > geometry is the unit cube. The vector field is A = [0 , 0 , y] and then > curl(A) = [1 , 0 , 0]. > > I have attached three screenshots which compare the analytic result > (left-hand > side) to the numerical approximation (right-hand side). The approximation of > the field, A, seems fine except for a strange wiggle of the magnitude on the > finest mesh, see A_magnitude.png. From looking at the graphical output > in**curlA_glyphs.png, the approximation of the curl of A does not work well. > > Regarding your third point, I have created the following convergence table: > > cycle cells dofs field curl > error rate error rate > 0 1 54 0.1866 - 1.4410 - > 1 8 300 0.1967 0.95 3.0428 0.47 > 2 64 1944 0.1576 1.25 4.8742 0.62 > 3 512 13872 0.1429 1.10 8.8342 0.55 > > Regarding the field everything looks good. For the curl, the order seems to > be > one half although the error increases. This is a bit surprising for me > because > the overall error increases with the number of dofs. Apart from this, I am > asking myself if the errors should not be much smaller because the field is > linear and its curl is constant. Hence, these functions should be an element > of the approximating finite element space. If I am not wrong, a single > element > should be sufficient for an approximation in this case. With FE_NedelecSZ the > results are similar.
Seb -- as you may have noticed, none of the main developers work on curl problems with the Nedelec elements, and so it is difficult to get definitive answers about questions. If you're still chasing answers, let's first focus on the potential (which I think the table above labels as "field error". I don't know how the error rate is computed, but the numbers for the rates can't be correct: I think they show a reduction *factor*, not a rate. The rate is the exponent in error = C h^r where from one line to the next, h is reduced by a factor of 2. So for the last two rows, for example, you need to have 0.1429 = (1/2)^r 0.1576 or r = -log(0.1429/0.1576)/log(2) = 0.14. So what I see here is that the convergence of the potential (field) is already not correct unless you happen to have a very non-smooth right hand side of the equation you are trying to solve. When you later compute the error of the curl of the potential, one typically expects that the convergence rate is one lower for the derivative than the one for the field itself. This would suggest that you would get a rate of -0.86 -- which turns out to be almost exactly what you have in your table above when one computes the rate with the formula above. Best W. -- ------------------------------------------------------------------------ Wolfgang Bangerth email: [email protected] 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 [email protected]. For more options, visit https://groups.google.com/d/optout.
