From 6ec72b1544154a934c3082f9a27ea924153da458 Mon Sep 17 00:00:00 2001 From: bangerth Date: Sat, 20 Sep 2008 03:11:58 +0000 Subject: [PATCH] Document smaller time step. git-svn-id: https://svn.dealii.org/trunk@16870 0785d39b-7218-0410-832d-ea1e28bc413d --- deal.II/examples/step-31/doc/results.dox | 29 ++++++++++-------------- 1 file changed, 12 insertions(+), 17 deletions(-) diff --git a/deal.II/examples/step-31/doc/results.dox b/deal.II/examples/step-31/doc/results.dox index 64e1418aa4..83542ab4e6 100644 --- a/deal.II/examples/step-31/doc/results.dox +++ b/deal.II/examples/step-31/doc/results.dox @@ -218,33 +218,28 @@ can take.
Conclusions
-Concluding, $\beta=0.03$ appears to be a good choice for the -stabilization parameter in 2d, and $\beta=0.05$ in 3d. In a dimension -independent way, we can model this as $\beta=0.015d$. As we have seen -in the sections above, in 2d -$k=\frac 14 \frac 1{q_T}\frac{h_K}{\|\mathbf{u}\|_{L^\infty(K)}}$ -is an appropriate time step, where $q_T$ is the polynomial degree of -the temperature shape functions (in the program, this corresponds to -the variable temperature_degree). To reconcile this with -the findings in 3d for the same $\beta$, we could write this as -$k=\frac 1{2\sqrt{2}\sqrt{d}} \frac -1{q_T}\frac{h_K}{\|\mathbf{u}\|_{L^\infty(K)}}$ -but this doesn't take into account that we also have to increase -$\beta$ in 3d. The final form that takes all these factors in reads as -follows: +Concluding, from the simple computations above, $\beta=0.03$ appears to be a +good choice for the stabilization parameter in 2d, and $\beta=0.05$ in 3d. In +a dimension independent way, we can model this as $\beta=0.015d$. If one does +longer computations (several thousand time steps) on finer meshes, one +realizes that that's not quite small enough and that for stability one will +have to reduce the above values a bit more (by about a factor of $\frac 78$). + +As a consequence, a formula that reconciles 2d, 3d, and variable polynomial +degree and takes all factors in account reads as follows: @f{eqnarray*} k = - \frac 1{2\sqrt{2}} \frac 1{\sqrt{d}} + \frac 1{2 \cdot 1.6} \frac 1{\sqrt{d}} \frac 2d \frac 1{q_T} \frac{h_K}{\|\mathbf{u}\|_{L^\infty(K)}} = - \frac 1{d\sqrt{2}\sqrt{d}} + \frac 1{1.6 d\sqrt{d}} \frac 1{q_T} \frac{h_K}{\|\mathbf{u}\|_{L^\infty(K)}}. @f} In the first form (in the center of the equation), $\frac -1{2\sqrt{2}}$ is a universal constant, $\frac 1{\sqrt{d}}$ +1{2 \cdot 1.6}$ is a universal constant, $\frac 1{\sqrt{d}}$ is the factor that accounts for the difference between cell diameter and grid point separation, $\frac 2d$ accounts for the increase in $\beta$ with space dimension, -- 2.39.5