From 5a809562ce3d5b8bd16ed89d29560867467a790e Mon Sep 17 00:00:00 2001 From: bangerth Date: Fri, 23 Mar 2007 19:34:45 +0000 Subject: [PATCH] Add a missing factor c_0^2. git-svn-id: https://svn.dealii.org/trunk@14598 0785d39b-7218-0410-832d-ea1e28bc413d --- deal.II/examples/step-24/doc/intro.dox | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/deal.II/examples/step-24/doc/intro.dox b/deal.II/examples/step-24/doc/intro.dox index e785e1d0a9..8d95a41cb5 100644 --- a/deal.II/examples/step-24/doc/intro.dox +++ b/deal.II/examples/step-24/doc/intro.dox @@ -73,13 +73,13 @@ on the right hand side can be rewritten as an initial value problem as follows: @f{eqnarray*} \Delta \bar{p}- \frac{1}{c_0^2} \frac{\partial^2 \bar{p}}{\partial t^2} & = & 0 \\ -\bar{p}(0,\mathbf r) &=&\lambda a(\mathbf r) = b(\mathbf r) \\ +\bar{p}(0,\mathbf r) &=& c_0^2 \lambda a(\mathbf r) = b(\mathbf r) \\ \frac{\partial\bar{p}(0,\mathbf r)}{\partial t} &=& 0. @f} (A derivation of this transformation into an initial value problem is given at the end of this introduction as an appendix.) -In the inverse problem, it is the initial condition $b(\mathbf r) = \lambda a(\mathbf r)$ that +In the inverse problem, it is the initial condition $b(\mathbf r) = c_0^2 \lambda a(\mathbf r)$ that one would like to recover, since it is a map of absorption strengths for microwave energy, and therefore presumably an indicator to discern healthy from diseased tissue. -- 2.39.5