From 93acdcd09b50dfb732d9b5f481c23ac81412bce3 Mon Sep 17 00:00:00 2001 From: Wolfgang Bangerth Date: Mon, 13 Jan 2020 17:28:17 -0700 Subject: [PATCH] Add a few links to step-6. --- examples/step-6/doc/results.dox | 10 ++++++++-- 1 file changed, 8 insertions(+), 2 deletions(-) diff --git a/examples/step-6/doc/results.dox b/examples/step-6/doc/results.dox index 229f662a62..5ccda26cdb 100644 --- a/examples/step-6/doc/results.dox +++ b/examples/step-6/doc/results.dox @@ -493,10 +493,16 @@ solution. In general, if the coefficient $a(\mathbf x)$ is discontinuous along a line in 2d, or a plane in 3d, then the solution may have a kink, but the gradient of the solution will not go to infinity. That means, that the solution is at least -still in the space $W^{1,\infty}$. On the other hand, we know that in the most +still in the Sobolev space +$W^{1,\infty}$ (i.e., roughly speaking, in the +space of functions whose derivatives are bounded). On the other hand, +we know that in the most extreme cases -- i.e., where the domain has reentrant corners, the right hand side only satisfies $f\in H^{-1}$, or the coefficient $a$ is only in -$L^\infty$ -- all we can expect is that $u\in H^1$, a much larger space than +$L^\infty$ -- all we can expect is that $u\in H^1$ (i.e., the +Sobolev +space of functions whose derivative is square integrable), a much larger space than $W^{1,\infty}$. It is not very difficult to create cases where the solution is in a space $H^{1+s}$ where we can get $s$ to become as small as we want. Such cases are often used to test adaptive finite element -- 2.39.5