From 50a3cf43aaf8f12a4a76450088134b216c964302 Mon Sep 17 00:00:00 2001 From: bonito Date: Tue, 4 Jan 2011 15:26:23 +0000 Subject: [PATCH] Modified the intro, the code needs to be fixed accordingly git-svn-id: https://svn.dealii.org/trunk@23120 0785d39b-7218-0410-832d-ea1e28bc413d --- deal.II/examples/step-38/doc/intro.dox | 141 ++----------------------- 1 file changed, 11 insertions(+), 130 deletions(-) diff --git a/deal.II/examples/step-38/doc/intro.dox b/deal.II/examples/step-38/doc/intro.dox index abfb18ec63..d7065e99d5 100644 --- a/deal.II/examples/step-38/doc/intro.dox +++ b/deal.II/examples/step-38/doc/intro.dox @@ -147,56 +147,14 @@ We produce one test case for a 2d problem and another one for 3d: solution function. There are (at least) two ways to do that. The first one is to project away the normal derivative as described above using the natural extension of $u(\mathbf x)$ (still denoted by $u$) over $\mathbb R^d$, 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] - where, of course, $\nabla - \mathbf n (\mathbf n \cdot \nabla)=\nabla_\Gamma$. - 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 $-\Delta_\Gamma u = -8 x_1x_2$. + -\Delta_\Gamma u = \Delta u - \mathbf n^T D u \mathbf n - (\nabla u)\cdot \mathbf n (\nabla \cdot \mathbf n). + @f] + Since we are on the unit circle, $\mathbf n=\mathbf x$ so that + @f[ + -\Delta_\Gamma u = -8 x_1x_2$. + @f] - A somewhat simpler version, at least for the current case of a curve in + A somewhat simpler way, 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 $\mathbf x(t)= \left(\begin{array}{c} \cos t \\ \sin t \end{array}\right)$. @@ -216,92 +174,15 @@ We produce one test case for a 2d problem and another one for 3d: &= 8 x_1x_2, @f} which is of course the same result as we had above. - +
  • 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 + equation, $f=-\Delta_\Gamma u$, in the same way as the method 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} + source code. +
    • In the program, we will also compute the $H^1$ seminorm error of the -- 2.39.5