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In the article you can find at http://www.cc.gatech.edu/~turk/my_papers/schange.pdf, precisely at page 2 of the .pdf, there is a functional E which is said to be a measure of the aggregate squared curvature of the surface f(x,y) over the region of interest. I don't know why this functional represent this. Thanks.

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Nice handle, my friend. It reminds me of Martian names in SciFi books... –  Georges Elencwajg Sep 23 '13 at 11:21

1 Answer 1

The second fundamental form (a matrix which represents extrinsic curvatures) of the graph of $f$ is given by $A = (1 + |\nabla f|)^{-1/2} \nabla^2 f$ where $\nabla^2 f$ is the matrix of second partial derivatives. Thus the functional is

$$ E(f) = \int_\Omega (1+|\nabla f|) |A|^2, $$

so up to the factor of $(1 + |\nabla f|)$ it is just the $L^2$ norm of the second fundamental form. In particular it is zero only when the graph of $f$ is a plane, and will be large when the graph of $f$ is highly curved; so it is a (slightly warped due to the $\nabla f$ term) measurement of the total extrinsic curvature of the surface.

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Thank you. We use the square of the curvature because the curvature can be positive or negative? –  fghtygibjsdfds Sep 25 '13 at 14:34
    
@fghtygibjsdfds: something like that... there are plenty of other positive measurements of deviation from flatness (e.g. just add the absolute values of the matrix entries instead of their squares), but taking the squared norm is the most common thing to do in this kind of situation. It certainly leads to nicer gradient flow equations - I'm not sure whether it makes a big difference (in either quality of outcome or performance) in your application or not. –  Anthony Carapetis Sep 25 '13 at 15:20

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