Timeline for Convergence of interior penalty DG methods
Current License: CC BY-SA 3.0
20 events
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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 |
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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 |
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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 |
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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 |
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Nov 25, 2013 at 22:36 | history | edited | Justin Dong | CC BY-SA 3.0 |
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Nov 25, 2013 at 21:58 | history | asked | Justin Dong | CC BY-SA 3.0 |