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I found the following definition in a book (S. Osher, R. Fedkiw, "Level Set Methods and Dynamic Implicit Surfaces", p. 140):

[the context is reconstruction of surfaces from unorganized point sets]

Let $S$ denote a set of points in $\mathbb{R}^3$, and define $d(x) = \mathrm{dist}(x, S)$ the shortest distance between $x$ and any of the points in $S$. Consider the following energy function:

$$ E(\Gamma) = \left[ \int_\Gamma d^p(x)\ ds \right]^\frac{1}{p} \quad, \qquad 1 \le p \le \infty $$ in which $\Gamma$ is a surface and $ds$ is the surface area.

The author now gives an expression for the gradient flow of this energy functional:

$$ \frac{d\Gamma}{dt} = -\left[ \int_\Gamma d^p(x)\ ds \right]^{\frac{1}{p} - 1} d^{p-1}(x) \left[ \nabla d(x) \cdot N + \frac{1}{p}d(x)\kappa \right] N $$

In this, $N$ is the surface normal and $\kappa$ the mean curvature. Also, $d(x) \kappa$ is called the surface tension.

Now, I realize that I may be in a bit over my head here, but I really would like to know how this expression was derived... unfortunately, the book is somewhat terse on the basics.

My take on this is that the underlying equation is

$$ \frac{d\Gamma}{dt} = -\nabla_{\!F}\ E(\Gamma)$$

in which $F$ is the space of deformations, and $\nabla_{\!F}\ E(\Gamma)$ is the gradient vector field of $E$. It seems the derivation above somehow applies the chain rule to this equation...

I have some more information, but I honestly don't know how much of this is standard stuff. I would be grateful if someone could give me some hints as to how to arrive at the result given.

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    In the book, is it said what $N$ and $\kappa$ are?2011-09-06
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    @Willie: They are the surface normal and mean curvature, respectively. I edited accordingly...2011-09-07
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    Note that one can find a pdf version of the text online quite easily.2014-08-01
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    Hi, I have just posted an answer to your interesting question. I apologize for the rather long text, but the topic is quite complex and requires a careful in-depth clarification. I hope that my answer could be useful to you.2014-08-05

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