From f85688d9f07f38e94ffcaebb0f1490e237fea0c3 Mon Sep 17 00:00:00 2001
From: bangerth <bangerth@0785d39b-7218-0410-832d-ea1e28bc413d>
Date: Sat, 1 Jan 2011 03:27:36 +0000
Subject: [PATCH] A bit more.

git-svn-id: https://svn.dealii.org/trunk@23095 0785d39b-7218-0410-832d-ea1e28bc413d
---
 deal.II/examples/step-38/doc/intro.dox | 33 ++++++++++++++++++--------
 deal.II/examples/step-38/step-38.cc    | 25 +++++++++----------
 2 files changed, 34 insertions(+), 24 deletions(-)

diff --git a/deal.II/examples/step-38/doc/intro.dox b/deal.II/examples/step-38/doc/intro.dox
index 0d4cb87c1e..332ee93fd5 100644
--- a/deal.II/examples/step-38/doc/intro.dox
+++ b/deal.II/examples/step-38/doc/intro.dox
@@ -115,18 +115,31 @@ and
 [Bonito, Nochetto, and Pauletti, SIAM J. Numer. Anal. 48(5), 2010].
 
 
-<h3>The numerical approximation</h3>
+
+<h3>Testcase</h3>
+
+In general when you want to test numerically the accuracy and/or order of
+convergence of an algorithm you need to provide an exact solution. The usual
+trick is to pick a function that we want to be the solution, then apply the
+differential operator to it that defines a forcing term for the right hand
+side. This is what we do in this example. In the current case, the form of the
+domain is obviously also essential.
+
+We produce one test case for a 2d problem and another one for 3d:
+
+- In 2d, let's choose as domain a half circle. On this domain, we choose the
+  function $u(\mathbf x)=-2x_1x_2$ as the solution.
+
+- In 3d, the domain is again half of the surface of the unit sphere but this
+  time without the disk that closes it. We choose $u(\mathbf x)=-2\sin(\pi
+  x_1)\cos(\pi x_2)e^z$ as the solution.
+  
+
 
 <h3>Implementation</h3>
 
-<h3>Testcase</h3>
-In general when you want to test numerically the accuracy and/or 
-order of convergence of an algorithm you need to provide an exact
-solution. 
-The usual trick is to pick a function that we want to be the solution,
-then apply the differential operator to it that defines a forcing term 
-for the right hand side.
-This is what we do in this example.
-We produce one test case for a 2d problem and another one for the 3d one.
 
 Mapping objects
+
+
+
diff --git a/deal.II/examples/step-38/step-38.cc b/deal.II/examples/step-38/step-38.cc
index 67a10713ec..f7442bde8a 100644
--- a/deal.II/examples/step-38/step-38.cc
+++ b/deal.II/examples/step-38/step-38.cc
@@ -45,19 +45,17 @@
 
 using namespace dealii;
 
-				 /**
-				    @sect3{The <code>LaplaceBeltramiProblem</code> class template}
-
-				    This class is extremely similar to the
-				    <code>LaplaceProblem</code> class as in 
-				    example 4.
-				    One difference is that now some members
-				    will be defined with two template parameters
-				    one for the dimension of the mesh @p dim,
-				    and the other for the dimension of
-				    the embedding space @p spacedim.
-				    Now <code>MappingQ</code> appears.
-				 */
+				 // @sect3{The <code>LaplaceBeltramiProblem</code> class template}
+
+				 // This class is extremely similar to the
+				 // <code>LaplaceProblem</code> class as in 
+				 // example 4.
+				 // One difference is that now some members
+				 // will be defined with two template parameters
+				 // one for the dimension of the mesh @p dim,
+				 // and the other for the dimension of
+				 // the embedding space @p spacedim.
+				 // Now <code>MappingQ</code> appears.
 template <int spacedim>
 class LaplaceBeltramiProblem 
 {
@@ -225,7 +223,6 @@ LaplaceBeltramiProblem<spacedim>::LaplaceBeltramiProblem (const unsigned degree)
 template <int spacedim>
 void LaplaceBeltramiProblem<spacedim>::make_grid_and_dofs ()
 {
-  
   Triangulation<spacedim> volume_mesh;
   GridGenerator::half_hyper_ball(volume_mesh);
   
-- 
2.39.5