$\mathbf n$ is the normal of $\Gamma$.
Since $\Delta_S = \nabla_S \cdot \nabla_S$, we deduce
@f[
-\Delta_S v = \Delta \tilde v - \mathbf n^T \ D \tilde v \ \mathbf n - (\nabla \tilde v)\cdot \mathbf n (\nabla \cdot \mathbf n).
+\Delta_S v = \Delta \tilde v - \mathbf n^T \ D \tilde v \ \mathbf n - (\nabla \tilde v)\cdot \mathbf n (\nabla \cdot \mathbf n - \mathbf n \ D \mathbf n \ \mathbf n ).
@f]
+Worth mentioning, the term $\nabla \cdot \mathbf n - \mathbf n \ D \mathbf n \ \mathbf n$ appearing in the above expression is the total curvature of the surface (sum of principal curvatures).
As usual, we are only interested in weak solutions for which we can use $C^0$
finite elements (rather than requiring $C^1$ continuity as for strong
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 = \Delta u - \mathbf n^T \ D u \ \mathbf n - (\nabla u)\cdot \mathbf n (\nabla \cdot \mathbf n).
+ -\Delta_\Gamma u = \Delta u - \mathbf n^T \ D u \ \mathbf n - (\nabla u)\cdot \mathbf n \kappa,
@f]
- Since we are on the unit circle, $\mathbf n=\mathbf x$ so that
+ where $\kappa$ is the total curvature of $\Gamma$.
+ Since we are on the unit circle, $\mathbf n=\mathbf x$ and $\kappa = 1$ so that
@f[
-\Delta_\Gamma u = -8 x_1x_2.
@f]
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 the method above, yielding an
+ equation, $f=-\Delta_\Gamma u$, in the same way as the method above (in that case $\kappa = 2$), yielding an
awkward and lengthy expression. You can find the full expression in the
source code.
</li>
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