From a24c1f567400d30f30672f7b322fffc210c92644 Mon Sep 17 00:00:00 2001 From: bangerth Date: Wed, 6 Sep 2006 23:26:33 +0000 Subject: [PATCH] One final note git-svn-id: https://svn.dealii.org/trunk@13848 0785d39b-7218-0410-832d-ea1e28bc413d --- deal.II/examples/step-23/doc/intro.dox | 6 +++++- 1 file changed, 5 insertions(+), 1 deletion(-) diff --git a/deal.II/examples/step-23/doc/intro.dox b/deal.II/examples/step-23/doc/intro.dox index 7dda1b6864..bfb31d8df4 100644 --- a/deal.II/examples/step-23/doc/intro.dox +++ b/deal.II/examples/step-23/doc/intro.dox @@ -294,7 +294,11 @@ It is worth noting that if we choose the same mesh on each time step (as we will in fact do in the program below), then we have the same shape functions on time step $n$ and $n-1$, i.e. $\phi^n_i=\phi_i^{n-1}=\phi_i$. Consequently, we get -$M^n=M^{n,n-1}=M$ and $A^n=A^{n,n-1}=A$. +$M^n=M^{n,n-1}=M$ and $A^n=A^{n,n-1}=A$. On the other hand, if we had +used different shape functions, then we would have to compute +integrals that contain shape functions defined on two meshes. This is a +somewhat messy process that we omit here, but that is treated in some +detail in @ref step_22 "step-22". As a final note, we remember that we have posed boundary values for $u$, but not $v$. In practice that means that we have to apply -- 2.39.5