> > > Well, both are the solution -- one is a function of x, the other is a > vector > of nodal values, but they ultimately represent the same object. U_i is the > vector you get out of the linear solver. >
I'm sorry to make you confused I should have been more specific. But, my question was that in usual tutorial program of deal.II, the vector solution contains values for U_i not u_h(x). Am I right...? > > Not easily. To see the problem, think about using quadratic finite > elements in > 1d. Let's say you have only one cell [0,1], so the shape functions are > defined > at x=0, x=1/2, and x=1. Assume you have nodal values 1,0,0, then your > solution > vector is > > u_h(x) = sum_i U_i \varphi_i(x) = 1 * \varphi_0(x) > = 2(1-x)(1/2-x) > = 1 - 3x + 2x^2 > > Note that while the minimal value of U_i is zero, the function u_h(x) has > a > mimimum at u_h'(x)=-3+4x=0, i.e., at x=3/4 where its value is > u_h(3/4) = 1 - 9/4 + 2*9/16 = -1/8 > > In other words, even though your nodal values are 1,0,0, the minimum of > this > function is -1/8 < 0 ! > > In general, the location and the value of the minimum depend in complex > ways > on the nodal values and can not easily be predicted. The best you can do > is to > use a quadrature formula with sufficiently many points and sample the > solution > at these points to get upper bounds for the minimum, and lower bounds for > the > maximum of the solution function u_h(x). > Thank you very much for your detailed explanation. I can understand it. Kyusik. -- 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.
