From: sebastianGl31415 <glane@tu-berlin.de>
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=9a2f1521a1de8506f3c11d525a88dc3ee75e1fc8;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