# Developing a divergence expression

1. Jul 20, 2010

### selfstudier

Hi all,

I'm trying to fill in some mathematical steps from a derivation i'm reading in a research paper.

I want to see that the following expression

$$\text{Div}\left(\nabla u\frac{ \phi ' \left(|\nabla u|\right)}{|\nabla u|}\right)$$

is equivalent to

$$(\frac{\phi ' \left(|\nabla u|\right)}{|\nabla u|} u_{\text{TT}} + \phi '' \left(|\nabla u|\right) u_{\text{NN}}$$

Here $$u \equiv u(x,y) , \phi \equiv \phi (x)$$
$$N = \frac{\nabla u}{|\nabla u|}$$ and
$$T$$ is a unit vector perpendicular to it (so N and T form a local coordinate system attached to each point (x,y))

$$u_{NN}$$ and $$u_{TT}$$ represent the second derivative of u with respect to N and T respectively:
$$u_{NN} = N^t \nabla^2 u N$$
and similarly for T

(hope any unexplained symbols and unmentioned assumptions are clear; it's very tiresome to write so much LaTeX!)

I can see that applying the "product rule" for divergence to the first expression gives
$$\nabla\left(\frac{\phi ' \left(|\nabla u|\right)}{|\nabla u|}\right) \cdot \nabla u + \frac{\phi ' \left(|\nabla u|\right)}{|\nabla u|} \Delta u$$
and I know that
$$\Delta u = u_{xx} + u_{yy} = u_{NN} + u_{TT}$$
due to the rotation invariance property of the Laplacian.

Is that the way to proceed, to show the equivalence of these two expressions (would seem there's some tedious work ahead with the first term obtained from the product rule)? Or is there a property/properties that I'm missing which could make it easier? I would like to be able to derive the 2nd expression from the 1st one, without knowing the 2nd one beforehand.

Thank you!
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