Laplace-Beltrami on a Curve

Is there a way to write out Laplace-Beltrami operator explicitly for a sufficiently smooth plane curve given by implicit equation $\,s(x,y)=0\,$?

If I knew knew the parametrization of the curve I could have computed metric tensor $\,g_{ij}\,$ and use well-known expression $$\Delta_{LB} u = \left\lvert g\right\rvert^{-1/2}\partial_i\big(\left\lvert g\right\rvert \,g^{ij}\,\partial_j u \big),$$ where $g^{ij}$ is the dual metric, and $\,\left\lvert g\right\rvert\,$ is the determinant of $\,g_{ij}$.

However, I am looking for an explicit formula for $\,\Delta_{LB}u\,$ which could be useful without knowing parametrization of the curve.

Let $\gamma$ be an arclength parametrization of the curve, so that $\Delta_{\rm LB} u = \frac{d^2}{dt^2}(u \circ \gamma)$. From the definition of the curve we have $s \circ \gamma = 0$, so differentiating this twice we get the equations $$\langle \nabla s , \dot \gamma \rangle = 0 \; \text{and} \; \nabla^2 s(\dot \gamma, \dot \gamma) + \langle \nabla s, \ddot \gamma\rangle = 0.$$

Since $\gamma$ is unit-speed we know $\langle \dot \gamma, \ddot \gamma \rangle = 0$ and thus $\ddot \gamma$ and $\nabla s$ are parallel; so putting these together we get

$$\ddot \gamma = \frac{\langle\ddot \gamma, \nabla s\rangle \nabla s }{|\nabla s|^2 } = \frac{-\nabla^2 s(\dot\gamma,\dot\gamma)}{|\nabla s|^2}\nabla s.$$

Since $|\dot\gamma|=1$ we can assume without loss of generality that $\dot \gamma = \frac{J \nabla s}{|\nabla s|}$, where $J$ is either a clockwise or counterclockwise rotation by $\pi/2$. (It turns out not to matter which.) We can now write $\dot \gamma$ and $\ddot \gamma$ in terms of the derivatives of $s$, so the Laplacian is

$$\frac{d}{dt}( \nabla u \cdot \dot \gamma ) = \nabla^2 u( \dot \gamma, \dot \gamma) + \nabla u \cdot \ddot \gamma = \frac{\nabla^2 u ( J \nabla s, J \nabla s ) }{|\nabla s|^2} - \frac{\nabla^2 s(J \nabla s,J \nabla s)}{|\nabla s|^4}\langle\nabla s,\nabla u\rangle.$$

If we define the operator $Q(f) = \frac{\nabla^2 f ( J \nabla s, J \nabla s ) }{|\nabla s|^2}$ (which is simply the second derivative of $f$ in the direction tangent to $\gamma$) then this is easier to understand:

$$\Delta_{\rm LB} u = Q(u) - Q(s)\frac{\langle \nabla s, \nabla u \rangle}{|\nabla s|^2}.$$

The first term is the second derivative along the line tangent to the level curve of $s$ and the second term accounts for the deviation of the level curve from this straight line.

• If $\nabla s = \left[ \begin{smallmatrix} s_x \\ x_y \end{smallmatrix}\right]$, then $J \nabla s = \left[ \begin{smallmatrix} - s_y & x_x \end{smallmatrix}\right]$, is that correct (assuming $J$ rotates counterclockwise)? – Vlad Jun 1 '15 at 5:33
• @Vlad: If $\nabla s = (s_x, s_y)$ then $J \nabla s = (-s_y, s_x)$. – Anthony Carapetis Jun 1 '15 at 5:39
• I assume $f = f( \, \cdot\, , \cdot \, )$ depends on two variables. What do you mean then by $f ( J \nabla s, J \nabla s )$? Are there four variables involved? – Vlad Jun 1 '15 at 5:41
• @Vlad: When I write $\nabla^2 f (v,w)$, I mean the second derivative/Hessian of $f$ acting on the vectors $v,w$. You could write it as $v^T H w = \sum_{i,j} v_i H_{ij} w_j$ where $H_{ij} = \partial^2 f / \partial x_i \partial x_j$ is the Hessian matrix of $f$. – Anthony Carapetis Jun 1 '15 at 5:43