I am trying to derive an explicit formula for Laplace-Beltrami operator in global Cartesian coordinates for a special case of plane curve.
I have found this article, and I would like to match their expression (6) for LB on a curve with the standard definition in terms of metric tensor.
According to formula $(6)$ in the paper, Laplace-Beltrami operator on plane curve can be written as
\begin{align}
\Delta_{LB}\, u & = \Delta u + \kappa\,u_{n} – u_{nn}
\\ & = \tag{$\star$}
\Delta u + \kappa\,\vec{n}\cdot\nabla u – \vec{n}\cdot\nabla\left(\vec{n}\cdot\nabla u\right)
\end{align}
- $\,\vec{n}\,$ is unit normal vector,
- $\,\kappa=-\nabla\cdot\vec{n}\,$ is curvature,
- $\,u_{n} = \vec{n}\cdot\nabla u\,$ and $\,u_{nn} = \vec{n}\cdot\nabla \left(\vec{n}\cdot\nabla u\right)\,$ are first and second normal derivatives,
- $\,\nabla u\,$ and $\,\Delta u\,$ are respectively gradient and Laplacian of $\,u\,$.
I am having troubles deriving $(\star)$ or matching it with metric tensor expression for LB operator
\begin{align}\tag{$\ast$}
\Delta_{LB}\, u = \dfrac{1}{\sqrt{\left\lvert g\right\rvert}}\,\partial_i\,\Big(\sqrt{\left\lvert g\right\rvert} \,g^{ij}\,\partial_j \,u \Big)
\end{align}
I can derive $(\star)$ from the Laplace-Beltrami expression $\,\Delta_{LB}\,u = \nabla_{s}\cdot\big(\nabla_{s}\,u\big)\,$ assuming surface divergence of a vector equals to the regular divergence of its projection to the curve.
This is a BIG assumption, and I do not know how to justify it.
I will appreciate if someone could help me to justify my assumption, or to derive $(\star)$ without assumptions on (surface) divergence.
My attempt to derive $(\star)$:
let $\,\nabla_{s}\,$, and $P$ denote surface gradient and projecting operator, then
\begin{align}
\Delta_{LB}\, u & = \nabla_{s}\cdot\big(\nabla_{s}\,u\big) \stackrel{\color{red}{\huge ?}}{=} \nabla\cdot\big(\nabla_{s}\,u\big)
\\ & = \nabla\cdot\big(P\;\nabla \,u\big)
= \nabla\cdot\Big(\nabla\,u-\big(\vec{n}\cdot\nabla\,u\big)\,\vec{n}\Big)
\\ & = \Delta\,u-\left(\nabla\cdot\vec{n}\right)\left(\vec{n}\cdot\nabla u\right)-
\vec{n}\cdot\nabla\left(\vec{n}\cdot\nabla u\right)
\\ & = \Delta u + \kappa\,u_{n} – u_{nn}
\end{align}
Best Answer
The surface gradient operator is defined as follows
$$\eqalign{ & \mathop \nabla \limits^s = \left( {{\bf{I}} - {\bf{n}} \otimes {\bf{n}}} \right).\nabla \cr & = {\bf{I}}.\nabla - \left( {{\bf{n}} \otimes {\bf{n}}} \right).\nabla \cr & = \nabla - \left( {{\bf{n}}.\nabla } \right){\bf{n}} \cr}\tag{1}$$
As you can see in $(1)$ we have subtracted the normal component of the $\nabla $ from it and hence the name surface gradient.
Use $(1)$ to derive your formula. Consider the following
$$\mathop \nabla \limits^s .{\bf{F}} = \left( {\nabla - \left( {{\bf{n}}.\nabla } \right){\bf{n}}} \right).{\bf{F}} = \nabla .{\bf{F}} - \left( {{\bf{n}}.\nabla } \right){\bf{n}}.{\bf{F}} = \nabla .{\bf{F}} - {\bf{n}}.\nabla \left( {{\bf{n}}.{\bf{F}}} \right)\tag{2}$$
Now, if you put ${\bf{F}} = \mathop \nabla \limits^s u$ you can have
$$\mathop \nabla \limits^s .\mathop \nabla \limits^s u = \nabla .\mathop \nabla \limits^s u - {\bf{n}}.\nabla \left( {{\bf{n}}.\mathop \nabla \limits^s u} \right)\tag{3}$$
but
$${\bf{n}}.\mathop \nabla \limits^s u = {\bf{n}}.\left( {\nabla u - \left( {{\bf{n}}.\nabla u} \right){\bf{n}}} \right) = {\bf{n}}.\nabla u - \left( {{\bf{n}}.\nabla u} \right)\left( {{\bf{n}}.{\bf{n}}} \right) = {\bf{n}}.\nabla u - {\bf{n}}.\nabla u = 0\tag{4}$$
and hence
$$\mathop \nabla \limits^s .\mathop \nabla \limits^s u = \nabla .\mathop \nabla \limits^s u\tag{5}$$