I know how to compute the discrete Laplacian of a graph and of a mesh (the Laplace-Beltrami operator). Is there an analogous definition for the computation of the Laplacian of a parametric curve ?

For instance the Laplacian matrix of a graph is defined as $L=D−A$ where $D$ is the diagonal degree matrix and $A$ is the adjacency matrix.

In my case the parametric curve is defined as :

$ \phi(t) : \mathbb{R} \to \mathbb{R} ^2 $

$ \phi(t) = (x(t), y(t)) $

  • $\begingroup$ If the curve has no intersection point, you can of course just take the Laplace-Beltrami on the curve -- which at every point will essentially just be the second derivative of the function in tangential direction. $\endgroup$ Jan 28 '15 at 3:44
  • $\begingroup$ Thank you for the comment, sure I can evaluate the derivative using the chain rule. $ \frac {\partial \phi(t)}{\partial x}= \frac {\partial \phi(t)}{\partial t} \frac {1} {\frac {\partial x}{\partial t} } $ For meshes laplace operators take into account the geometry of a surface (e.g. the angles at the nodes) and they are not based on finite difference method. So my question is there an analogous definition for the laplacian (for curves) that is not based on finite difference method ? $\endgroup$ Jan 28 '15 at 9:19
  • $\begingroup$ I don't think I understand what you mean by your question. My fundamental problem with your question is that you mix discrete things like the graph Laplacian (a matrix operator) with continuous things like the Laplace-Beltrami operator (a differential operator). Can you elaborate? $\endgroup$ Jan 29 '15 at 3:40
  • $\begingroup$ Thank you for your comment, I think that the mention of the Laplace-Beltrami operator is confusing ... The discrete Laplace operator is an analog of the continuous Laplace operator, defined so that it has meaning on a graph or a discrete grid. For a graph it's called the Laplacian matrix. I have a discrete curve defined by a sequence of 2D points I want to define a laplace operators that take into account the geometry and curvature of the curve. For an example a Laplacian on a mesh is a (n,n) matrix, where n is the number of vertices and there are several forms of laplacian. $\endgroup$ Jan 29 '15 at 9:50
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    $\begingroup$ Well, then the way I read your comment is that you want to consider your curve defined by $n$ points to be some sort of generalization of the graph that connects these points in a linear way (i.e., a chain). If you'd like to define some generalized graph Laplacian on this chain graph, then you have to say what properties it is supposed to have that goes beyond the normal graph Laplacian? $\endgroup$ Jan 29 '15 at 23:44

There is already a very clear analog of the Laplacian when restricted to a curve: the second derivative along the curve. It's not clear to me why and how you believe that the shape of your curve should have any explicit effect on the analogue of the Laplacian you seek. The second derivative along the curve implicitly accounts for the curve's shape since the values (and therefore their derivatives) presumably depend on the curve's shape. Of course this also relies on choosing the correct metric and coordinates (global or curve-local) for the derivatives. Whether the curve is defined continuously or discretely seems to make no difference to the concept, it only impacts the method used to compute this derivative.

I suggest you also look at this very similar question over on Math.Stackexchange which essentially just derives a finite difference approximation of the second derivative based on discrete points along a curve. In the case where the points lie on some underlying continuous curve and you take the limit as the edge lengths go to zero this is just equal to the second derivative along the curve.

If you still don't agree that this is the logical analogue to the Laplacian on a discrete curve you will need to more explicitly state precisely how you expect the curvature to play a role other than it's implicit influence on the values at the points.


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