I'm looking for more insight into geomatric operators on a 2d (parametric) curve. I know the laplace beltrami operator is given by $$\Delta_S u = |g|^{-1/2} \sum_{i,j}^{}\partial_i(|g|g^{ij} \partial_j u),$$

and from wikipedia : "in the usual (orthonormal) Cartesian coordinates $x^i$ on the Euclidean space, the metric is reduced to the Kronecker delta, giving $|g|=1.$ Consequently, in this case, the operator reduces to the ordinary Laplacian.

I have also seen from other posts here that the laplace beltrami operator on a 2d curve simplifies to the second derivative with respect to arc length for a curve parametrized by arc length.

I'm looking for help in piecing this all together. How would the laplace beltrami operator simplify for a 2d curve given by $(x(r), y(r))$ where $r$ is angle not arc length? I'm looking to set up a problem for a pde (say the advection-diffusion equation) set up on a parametric curve.

Even insight into how the derivation of the laplace beltrami operator $\to$ the derivative with respect to arc length occurs would help!


1 Answer 1


If you have a curve $\gamma$ parametrized by a variable $s,$ then this $s$ also forms a 1-dimensional coordinate system for the curve. In this coordinate, the metric is the $1\times1$ matrix with entry $g_{ss} = g(\partial_s, \partial_s).$ Since we're defining $g$ to be the induced metric from $\mathbb R^2,$ we have $g(\partial_s, \partial_s) = \gamma'(s) \cdot \gamma'(s) = |\gamma'(s)|^2.$

From this we can easily calculate the determinant and inverse as $|g| = |\gamma'|^2$ and $g^{ss} = |\gamma'|^{-2}.$

Thus the Laplace-Beltrami formula becomes $$\Delta u = \frac{1}{\sqrt {|g|}} \partial_s \left(\sqrt{|g|} g^{ss} \partial_s f \right) =\frac1{|\gamma'|} \partial_s\left( \frac 1 {|\gamma'|} \partial_sf\right).$$

When your parametrization is by arc length, you have $|\gamma '|=1$ and thus this just reduces to the second derivative.

  • $\begingroup$ This is so helpful, thank you! Would you be able to provide a bit more insight as to how g_ss becomes this dot product? $\endgroup$
    – lrs417
    Oct 5, 2020 at 14:16
  • $\begingroup$ @lrs417: The metric on the curve is defined to be the metric induced by the immersion $\gamma: I \to \mathbb R^2,$ meaning that by definition we have $g(X,X) = d\gamma(X)\cdot d\gamma(X).$ Since $d\gamma(\partial_s) = \gamma'(s),$ the formula I wrote in my answer follows. (If you're not using the induced metric, the Laplace-Beltrami operator will depend on the metric you choose.) $\endgroup$ Oct 6, 2020 at 0:35

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