From: sebastianGl31415 Date: Thu, 28 Jan 2016 16:42:23 +0000 (+0100) Subject: Update intro.dox X-Git-Tag: v8.4.0-rc2~45^2 X-Git-Url: https://gitweb.dealii.org/cgi-bin/gitweb.cgi?a=commitdiff_plain;h=refs%2Fpull%2F2121%2Fhead;p=dealii.git Update intro.dox I fixed language mistakes. --- 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