From 91ede93b5447d5adf303a3656e43ce12c6cd2dcd Mon Sep 17 00:00:00 2001 From: Wolfgang Bangerth Date: Thu, 19 Dec 2019 14:33:35 -0700 Subject: [PATCH] Update results.dox --- examples/step-71/doc/results.dox | 25 ++++++++++++++++++------- 1 file changed, 18 insertions(+), 7 deletions(-) diff --git a/examples/step-71/doc/results.dox b/examples/step-71/doc/results.dox index 13ba2a879c..d89d08f861 100644 --- a/examples/step-71/doc/results.dox +++ b/examples/step-71/doc/results.dox @@ -4,15 +4,19 @@ We run the program with a right hand side that will produce the solution $u = \sin(\pi x) \sin(\pi y)$ and with clamped boundary conditions in the domain $\Omega = (0,1)^2$. We test this setup using $Q_2$, $Q_3$, and $Q_4$ elements, which one can -change `fe\_degree` in `main()`. With mesh +change via the `fe_degree` variable in the `main()` function. With mesh refinement, the $L_2$ convergence rates, $H_1$-seminorm convergence and $H_2$-seminorm convergence of $u$ should then be around 2, 2, 1 for $Q_2$ , 4, 3, 2 for -$Q_3$, and 5, 4, 3 for $Q_4$ separately. -We use different penalties $\eta = 1$, $2$, and $p(p+1)$ where $p$ -is the degree of polynomials, -and compare convergence rates of numerical solutions computed by these -penalties. +$Q_3$, and 5, 4, 3 for $Q_4$, respectively. + +From the papers by Brenner et al., it is not immediately clear what +the penalty parameter $\eta$ should be. Educated guesses, comparing +to the discontinuous Galerkin formulations for the Laplace equation, +suggest that $\eta = 1$, $2$, and $p(p+1)$ would all be reasonable, +where $p$ is the degree of polynomials. This is easy to change +in the code from its current default. Below we show results for +all of these.

Test results on Q2 with \eta = p(p+1)

@@ -124,7 +128,7 @@ to round-off. But the $L_2$ error is pretty small in that case. Although $L_2$ norm and $H_1$-seminorm convergence rates of $u$ follow the theoretical expectations, $H_2$-seminorm does not converge. -Comparing results from $\eta = 1$ and $\eta = p(p+1)$, +Comparing results from $\eta = 1$ and $\eta = p(p+1)$, it is clear that $\eta = p(p+1)$ is a better penalty.

Test results on Q2 with \eta = 2

@@ -153,3 +157,10 @@ In the case, all convergence rates of $u$ follow the theoretical expectations. But compared to the results from $\eta = p(p+1)$, it does not show a good convergence on $L_2$ errors. + + +

Conclusions for the choice of the penalty parameter + +The conclusions for which of the "reasonable" choices one should use for the penalty parameter +is that $\eta=p(p+1)$ yields the expected results. It is, consequently, what the code +uses as currently written. -- 2.39.5