From 9a2f1521a1de8506f3c11d525a88dc3ee75e1fc8 Mon Sep 17 00:00:00 2001 From: sebastianGl31415 Date: Thu, 28 Jan 2016 17:42:23 +0100 Subject: [PATCH] Update intro.dox I fixed language mistakes. --- examples/step-34/doc/intro.dox | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/examples/step-34/doc/intro.dox b/examples/step-34/doc/intro.dox index d296c1cbff..0dce094763 100644 --- a/examples/step-34/doc/intro.dox +++ b/examples/step-34/doc/intro.dox @@ -273,7 +273,7 @@ The reason why this is possible can be understood if we consider the fact that the solution of a pure Neumann problem is known up to an arbitrary constant $c$, which means that, if we set the Neumann data to be zero, then any constant $\phi = \phi_\infty$ will be a solution. -Inserting constant solution and the Neumann boundary condition in the +Inserting the constant solution and the Neumann boundary condition in the boundary integral equation, we have @f{align*} \alpha\left(\mathbf{x}\right)\phi\left(\mathbf{x}\right) @@ -285,7 +285,7 @@ boundary integral equation, we have +\int_{\Gamma}\frac{ \partial G(\mathbf{y}-\mathbf{x}) }{\partial \mathbf{n}_y} \, ds_y \right] @f} -The integral on $\Gamma_\infty$ is unity, see above, division by the constant $\phi_\infty$ gives us the explicit +The integral on $\Gamma_\infty$ is unity, see above, so division by the constant $\phi_\infty$ gives us the explicit expression above for $\alpha(\mathbf{x})$. While this example program is really only focused on the solution of the -- 2.39.5