Skip to main content
20 events
when toggle format what by license comment
Dec 1, 2013 at 8:25 vote accept Justin Dong
Dec 1, 2013 at 5:07 comment added Jesse Chan Ah, I didn't notice that. Thanks for the catch. However, $x^2, y^2, xy, x, y, 1$ is still missing two basis functions before spanning $P_2$, no? The actual construction of quadratic basis functions on triangles involves vertex blending functions, which seemed a bit long to write out.
Nov 29, 2013 at 15:59 comment added Jed Brown @JLC It would be unconventional (at best) to use a biquadratic space on a triangular mesh. Those spaces are generally used on quadrilateral meshes.
Nov 29, 2013 at 10:00 history edited boyfarrell
Added a tag.
Nov 29, 2013 at 8:07 answer added chris timeline score: 5
Nov 28, 2013 at 13:42 history tweeted twitter.com/#!/StackSciComp/status/406055609407406080
Nov 26, 2013 at 20:48 comment added Justin Dong ...diffusion problem (I know my original post is advection diffusion but I just went back to diffusion to test this). And in my definition of the basis functions, I would need $p=4$ to get zero error for $x(x-1)y(y-1)$, which I do...
Nov 26, 2013 at 20:46 comment added Justin Dong Hmm, this is weird. Using those additional basis functions, I don’t obtain zero error for $x(x-1)y(y-1)$, but I see why the error should be zero. For $p=2$, the basis functions $1$, $x$, $y$, $xy$, $x^{2}$, $y^{2}$, $x^{2}y^{2}$, $x^{2}y$, $xy^{2}$ should give zero error. This is concerning since it seems to suggest there was an error in implementing my formulation. The part that confuses me is that when I use the basis functions as described by Riviere (i.e. degree $p$ implies sum of exponents of a basis function must equal $p$), I get all of the convergence rates correctly for the linear...
Nov 26, 2013 at 17:05 comment added Jesse Chan That's probably because your space doesn't span 2D quadratic polynomials. You should also be including $x^2y, xy^2, x^2y^2$, and so on.
Nov 26, 2013 at 16:52 comment added Justin Dong Also, if the manufactured solution is $u(x,y) = x^{3} + y^{3}$, then for $p=1$ the convergence rates are fine. For $p=2$ they are suboptimal (2 and 1 in the $L^{2}$ and energy norms, resp.) but for $p=3$ the errors are approximately zero.
Nov 26, 2013 at 16:46 history edited Justin Dong CC BY-SA 3.0
added 29 characters in body
Nov 26, 2013 at 16:41 comment added Justin Dong My basis functions for $p=2$ (considered on a reference triangle) are $x^{2}$, $y^{2}$, $xy$, $x$, $y$ and $1$. I guess I’m not seeing how these basis functions could exactly capture the solution. In any case, for $p=2$ I don’t capture the solution exactly in my code. Convergence rates are $2$ and $1$ in the $L^{2}$ and energy norms, resp.
Nov 26, 2013 at 16:37 comment added Jesse Chan The solution is quadratic in each direction.
Nov 26, 2013 at 16:34 comment added Justin Dong @chris Isn’t the manufactured solution quartic and requires $p=4$ to capture the solution exactly? For $p=4$, the solution is captured with error on the order of $10^{-14}$, which is close enough to machine precision I think.
Nov 26, 2013 at 15:49 comment added chris your manufactured solution is quadratic, so you should capture the solution exactly for $p=2$ (zero discretization error) on any grid. Do you? If you don't, switch off the terms one by one to find the problem. If you still can't find the problem, use a linear manufactured solution, and eventually a constant. Once fixed, use a manufactured solution with sines to check convergence rates for any $p$.
Nov 26, 2013 at 14:22 history edited Justin Dong CC BY-SA 3.0
added 31 characters in body
Nov 26, 2013 at 13:23 comment added Christian Waluga How do you solve your problem?
Nov 25, 2013 at 22:47 history edited Justin Dong CC BY-SA 3.0
added 37 characters in body
Nov 25, 2013 at 22:36 history edited Justin Dong CC BY-SA 3.0
added 9 characters in body
Nov 25, 2013 at 21:58 history asked Justin Dong CC BY-SA 3.0