]> https://gitweb.dealii.org/ - dealii-svn.git/commitdiff
Fix typo.
authorbangerth <bangerth@0785d39b-7218-0410-832d-ea1e28bc413d>
Mon, 21 May 2012 09:52:44 +0000 (09:52 +0000)
committerbangerth <bangerth@0785d39b-7218-0410-832d-ea1e28bc413d>
Mon, 21 May 2012 09:52:44 +0000 (09:52 +0000)
git-svn-id: https://svn.dealii.org/trunk@25522 0785d39b-7218-0410-832d-ea1e28bc413d

deal.II/examples/step-15/doc/intro.dox

index 5578a5b06ff3b63fb0de14fbc3133fc73aec6e65..d731db4eb7ba0378abdf50f269691a70001c70df 100644 (file)
@@ -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.

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