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differentiation of the l1 norm of gradient

 
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Jul31-12, 03:13 AM   #1
 

differentiation of the l1 norm of gradient


Hi everyone, I need help with a derivation I'm working on, it is the differentiation of the norm of the gradient of function F(x,y,z):

[itex]\frac{∂}{∂F}[/itex](|∇F|[itex]^{α}[/itex])

The part of [itex]\frac{∂}{∂F}[/itex]([itex]\frac{∂F}{∂x}[/itex]) is bit confusing.

(The answer with α=1 is div([itex]\frac{∇F}{|∇F|}[/itex]), where div stands for divergence.)

Any ideas? many thanks!
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Jul31-12, 05:39 AM   #2
 
Hey wilsonnl and welcome to the forums.

It looks like what you are trying to get is the total differential of the gradient. You can express the total differential in terms of dF = dF/dx*dx + dF/dy*dy + dF/dz*dz.

The thing is you are trying to differentiate with respect to the total differential which does make sense in the context of a norm (since this differential will correspond to the total change in each orthogonal component summed up).

I'd suggest you look into whether the total differential is what you want since it can be used as an analog to see how the total change of the length of something is changing with respect to all elements.

The norm, if it's euclidean will have things in terms of squared-components (i.e |x^2 + y^2 + z^2| for the 1-norm) so you can translate that into functions of (x,y,z) and then look at the total differential dF of this function.
Jul31-12, 08:29 AM   #3
 
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Since this has nothing to do with "differential equations", I am moving it to "Calculus and Analysis".
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