From 4653c8e02ead6aebc143f97dd393110b72d6c363 Mon Sep 17 00:00:00 2001 From: David Wells Date: Sat, 25 Jun 2016 19:27:34 -0400 Subject: [PATCH] Fix some matrix coefficients in the step-25 intro. The above formula has the correct sign on N. Additionally, the Laplace matrix here is missing the coefficients. --- examples/step-25/doc/intro.dox | 7 ++++--- 1 file changed, 4 insertions(+), 3 deletions(-) diff --git a/examples/step-25/doc/intro.dox b/examples/step-25/doc/intro.dox index 03ae995ab2..f1a5761816 100644 --- a/examples/step-25/doc/intro.dox +++ b/examples/step-25/doc/intro.dox @@ -176,9 +176,10 @@ i.e. $N_{ij}(f,g) = \left( \cos\left[ \theta f + (1-\theta) g\right]\varphi_i, What solvers can we use for the first equation? Let's look at the matrix we have to invert: @f[ - (M+k^2\theta^2(A-N))_{ij} = - \int_\Omega (1+k^2\theta^2 \cos \alpha) - \varphi_i\varphi_j \; dx+\int_\Omega \nabla\varphi_i\nabla\varphi_j \; dx, + (M+k^2\theta^2(A + N))_{ij} = + \int_\Omega (1+k^2\theta^2 \cos \alpha) + \varphi_i\varphi_j \; dx + + k^2 \theta^2 \int_\Omega \nabla\varphi_i\nabla\varphi_j \; dx, @f] for some $\alpha$ that depends on the present and previous solution. First, note that the matrix is symmetric. In addition, if the time step $k$ is small -- 2.39.5