From 9678dfa819a518e7c11de9647295d8807c6c2f55 Mon Sep 17 00:00:00 2001 From: bangerth Date: Mon, 21 May 2012 09:52:44 +0000 Subject: [PATCH] Fix typo. git-svn-id: https://svn.dealii.org/trunk@25522 0785d39b-7218-0410-832d-ea1e28bc413d --- deal.II/examples/step-15/doc/intro.dox | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/deal.II/examples/step-15/doc/intro.dox b/deal.II/examples/step-15/doc/intro.dox index 5578a5b06f..d731db4eb7 100644 --- a/deal.II/examples/step-15/doc/intro.dox +++ b/deal.II/examples/step-15/doc/intro.dox @@ -180,7 +180,7 @@ $B$ is symmetric, and so $A$ is symmetric as well. On the other hand, $B$ is also positive definite, which confers the same property onto $A$. This can be seen by noting that the vector $v_1 = \frac{\nabla u^n}{|\nabla u^n|}$ is an eigenvector of $B$ with eigenvalue -$\lamba_1=1-\frac{1}{1+|\nabla u^n|^2} > 0$ while all vectors $v_2\ldots v_d$ +$\lambda_1=1-\frac{1}{1+|\nabla u^n|^2} > 0$ while all vectors $v_2\ldots v_d$ that are perpendicular to $v_1$ and each other are eigenvectors with eigenvalue $1$. Since all eigenvalues are positive, $B$ is positive definite and so is $A$. We can thus use the CG method for solving the Newton steps. -- 2.39.5