From 95cf011990cfd4d70763febaed677da169d332b4 Mon Sep 17 00:00:00 2001
From: bangerth <bangerth@0785d39b-7218-0410-832d-ea1e28bc413d>
Date: Sun, 2 Jan 2011 22:58:23 +0000
Subject: [PATCH] Discuss the issues with the exact solution and the remainder
 of the testcase.

git-svn-id: https://svn.dealii.org/trunk@23096 0785d39b-7218-0410-832d-ea1e28bc413d
---
 deal.II/examples/step-38/doc/intro.dox | 176 ++++++++++++++++++++++++-
 1 file changed, 170 insertions(+), 6 deletions(-)

diff --git a/deal.II/examples/step-38/doc/intro.dox b/deal.II/examples/step-38/doc/intro.dox
index 332ee93fd5..bef01c5927 100644
--- a/deal.II/examples/step-38/doc/intro.dox
+++ b/deal.II/examples/step-38/doc/intro.dox
@@ -127,13 +127,177 @@ 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.
+<ul>
+<li>
+  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. To compute the right hand
+  side, we have to compute the second <i>tangential</i> derivatives of the
+  solution function. There are (at least) two ways to do that. The first one
+  is to project away the normal derivative, i.e. to compute
+  @f[
+    -\Delta_\Gamma u
+    =
+    -
+    \left[\nabla - \mathbf n (\mathbf n \cdot \nabla)\right]
+    \cdot
+    \left[\nabla - \mathbf n (\mathbf n \cdot \nabla)\right]
+    u.
+  @f]
+  Since we are on the unit circle, $\mathbf n=\mathbf x$. Furthermore, $\nabla
+  u = \left(\begin{array}{c}-2x_2 \\ -2x_1\end{array}\right)$.
+  Consequently, we have the following identities:
+  @f{align*}
+    \left[\nabla - \mathbf n (\mathbf n \cdot \nabla)\right] u
+    &=
+    \left(\begin{array}{c}-2x_2 \\ -2x_1\end{array}\right)
+    +
+    4x_1x_2
+    \left(\begin{array}{c}x_1 \\ x_2\end{array}\right)
+    =
+    \left(\begin{array}{c}-2x_2(1-2x_1^2) \\ -2x_1(1-2x_2^2)\end{array}\right)
+    \\
+    \nabla \cdot \left[\nabla - \mathbf n (\mathbf n \cdot \nabla)\right] u
+    &=
+    16x_1x_2
+    \\
+    (\mathbf n \cdot \nabla)
+    \left[\nabla - \mathbf n (\mathbf n \cdot \nabla)\right] u
+    &=
+    \left(\begin{array}{c}
+      12x_1^2x_2 -2x_2\\
+      12x_1x_2^2 -2x_1
+    \end{array}\right)
+    \\
+    \left[\mathbf n (\mathbf n \cdot \nabla)\right] \cdot
+    \left[\nabla - \mathbf n (\mathbf n \cdot \nabla)\right] u
+    &=
+    12x_1^3x_2 -4x_1x_2 + 12x_1x_2^3
+    \\
+    \Delta_\Gamma u
+    &=
+    16x_1x_2 - (12x_1^3x_2 -4x_1x_2 + 12x_1x_2^3)
+    =
+    20x_1x_2 - 12 x_1x_2 (x_1^2+x_2^2).
+  @f}
+  In the last equation, we can note that since we only ever evaluate this
+  right hand side on the unit circle, $x_1^2+x_2^2=1$, yielding the final
+  value $-\Delta_\Gamma u = -8 x_1x_2$.
 
-- 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.
-  
+  A somewhat simpler version, at least for the current case of a curve in
+  two-dimensional space, is to note that we can map the interval $t \in
+  [0,\pi]$ onto the domain $\Omega$ using the transformation
+  $\mathbf x(t)= \left(\begin{array}{c} \cos t \\ \sin t \end{array}\right)$.
+  At position $\mathbf x=\mathbf x(t)$, the value of the solution is then
+  $u(\mathbf x(t)) = -2\cos t \sin t$.
+  Taking into account that the transformation is length preserving, i.e. a
+  segment of length $dt$ is mapped onto a piece of curve of exactly the same
+  length, the tangential Laplacian then satisfies
+  @f{align*}
+    \Delta_\Gamma u 
+    &= \frac{d^2}{dt^2}(-2\cos t \sin t)
+    = -2 \frac{d}{dt}(-sin^2 t + \cos^2 t)
+    = -2 (-2 \sin t \cos t - 2 \cos t \sin t)
+    \\
+    &= 8 \sin t \cos t,
+  @f}
+  which is of course the same result as we had above.
+
+<li>
+  In 3d, the domain is again half of the surface of the unit ball, i.e. a half
+  sphere or dome. We choose $u(\mathbf x)=-2\sin(\pi x_1)\cos(\pi x_2)e^z$ as
+  the solution. We can compute the right hand side of the
+  equation, $f=-\Delta_\Gamma u$, in the same way as above, yielding an
+  awkward and lengthy expression. You can find the full expression in the
+  source code, where we use the fact that
+  @f{align*}
+    \Delta_\Gamma 
+    &=
+    \left[\nabla - \mathbf n (\mathbf n \cdot \nabla)\right]
+    \cdot
+    \left[\nabla - \mathbf n (\mathbf n \cdot \nabla)\right]
+    \\
+    &=
+    \left[\mathbf I - \mathbf n \otimes \mathbf n\right]\nabla
+    \cdot
+    \left[\mathbf I - \mathbf n \otimes \mathbf n\right]\nabla
+    \\
+    &=
+    \text{trace}\;
+    \left\{
+      (\left[\mathbf I - \mathbf n \otimes \mathbf n\right]\nabla)
+      \otimes
+      (\left[\mathbf I - \mathbf n \otimes \mathbf n\right]\nabla)
+    \right\}
+    \\
+    &=
+    \text{trace}\;
+    \left\{
+      \nabla^2
+      +
+      (\left[\mathbf n \otimes \mathbf n\right]\nabla)
+      \otimes
+      (\left[\mathbf n \otimes \mathbf n\right]\nabla)
+      -
+      \nabla \otimes (\left[\mathbf n \otimes \mathbf n\right]\nabla)
+      -
+      \left[\mathbf n \otimes \mathbf n\right]\nabla^2
+    \right\}.
+  @f}
+  Applied to the solution, we then get using the abbreviation $H=\nabla^2 u$
+  for the Hessian with derivatives in all three spatial directions:
+  @f{align*}
+    \Delta_\Gamma u
+    &=
+    \text{trace}\;
+    \left\{
+      H
+      +
+      (\left[\mathbf n \otimes \mathbf n\right]\nabla)
+      \otimes
+      (\left[\mathbf n \otimes \mathbf n\right]\nabla) u
+      -
+      \nabla \otimes (\left[\mathbf n \otimes \mathbf n\right]\nabla) u
+      -
+      \left[\mathbf n \otimes \mathbf n\right] H
+    \right\}
+    \\
+    &=
+    \text{trace}\; H
+    +
+    \text{trace}\;
+    \left\{
+      (\left[\mathbf n \otimes \mathbf n\right]\nabla)
+      \otimes
+      (\left[\mathbf n \otimes \mathbf n\right]\nabla) u
+      -
+      \nabla \otimes (\left[\mathbf n \otimes \mathbf n\right]\nabla) u
+    \right\}
+    -
+    \mathbf n^T H \mathbf n.
+  @f}
+  A lengthier computation shows that if we take into account that $\mathbf n =
+  \mathbf x$, then the middle term can be simplified in
+  such a way that we obtain
+  @f{align*}
+    \Delta_\Gamma u
+    &=
+    \text{trace}\; H
+    +
+    (2-\text{spacedim}-1) \mathbf n \cdot \nabla u
+    -
+    \mathbf n^T H \mathbf n.
+  @f}
+</ul> 
+
+In the program, we will also compute the $H^1$ seminorm error of the
+solution. Since the solution function and its numerical approximation are only
+defined on the manifold, the obvious definition of this error functional is
+$| e |_{H^1} = \left( \int_\Omega | \left[\mathbf n \otimes \mathbf
+n\right]\nabla (u-u_h) |^2 \right)^{1/2}$. This requires us to provide the
+<i>tangential</i> gradient $\left[\mathbf n \otimes \mathbf
+n\right]\nabla u$ to the function VectorTools::integrate_difference, which we
+will do by implementing the function <code>Solution::gradient</code> in the
+program below.
 
 
 <h3>Implementation</h3>
-- 
2.39.5