@f[
G_S:= (D \mathbf{x}_S)^T \ D \mathbf{x}_S
@f]
-denote the corresponding first fundamental form, where $D
+denotes the corresponding first fundamental form, where $D
\mathbf{x}_S=\left(\frac{\partial x_{S,i}(\hat{\mathbf x})}{\partial \hat x_j}\right)_{ij}$ is the
derivative (Jacobian) of the mapping.
In the following, $S$ will be either the entire surface $\Gamma$ or,
@f]
The surface Laplacian (also called the Laplace-Beltrami operator) is then
defined as $\Delta_S:= \nabla_S \cdot \nabla_S$.
+Note that an alternate way to define the surface gradient on smooth surfaces $\Gamma$ is
+@f[
+\nabla_S v := \nabla \tilde v - \mathbf n (\mathbf n \nabla \tilde v),
+@f]
+where $\tilde v$ is a "smooth" extension of $v$ in a tubular neighborhood of $\Gamma$ and
+$\mathbf n$ is the normal of $\Gamma$.
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
<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
+ side, we have to compute the surface Laplacian 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
+ 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
=
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