Divergence in Definition of Laplace-Beltrami Operator

Question

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.

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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}

Answer 1

The surface gradient operator is defined as follows

$$\eqalign{ & \mathop \nabla \limits^s = \left( {{\bf{I}} - {\bf{n}} \otimes {\bf{n}}} \right).\nabla \cr & = {\bf{I}}.\nabla - \left( {{\bf{n}} \otimes {\bf{n}}} \right).\nabla \cr & = \nabla - \left( {{\bf{n}}.\nabla } \right){\bf{n}} \cr}\tag{1}$$

• $\bf{n}$ is the unit normal vector
• $\bf{I}$ is the second order identity tenor
• $\otimes$ is the tensor product
• $.$ is the scalar product

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

$$\mathop \nabla \limits^s .\mathop \nabla \limits^s u = \nabla .\mathop \nabla \limits^s u\tag{5}$$