I would like to obtain the following equation : $\text{div}\left( \frac{\nabla u}{\sqrt{|\nabla u|^2 + \epsilon^2}} \right) =\frac{1}{\sqrt{\epsilon^2 + |\nabla u|^2}} \left( g^{ij} - \frac{\nabla^i u \nabla^j u}{\epsilon^2 + |\nabla u|^2} \right) \nabla^2_{ij}u$ with $g$ being the metric on a manifold $M$.
I have tried using the different properties of the divergence but there is always a term missing... Is this equation just basically saying that $div(A) = Tr (\nabla A)$ ?
Thanks!
Actually if you calculate that at the center of a normal coordinates, writing $X = \sum_i X^i \frac{\partial}{\partial x^i}$, then
$$\text{div} X = \sum_i \partial_i X^i.$$
So using $|\nabla u|^2 = \sum_j u_j^2$,
$$\begin{split} \text{div} \left( \frac{\nabla u}{\sqrt{|\nabla u|^2 + \epsilon^2}}\right) &= \sum_i \left( \frac{ u_i}{\sqrt{|\nabla u|^2 + \epsilon^2}}\right)_i \\ &= \frac{1}{\sqrt{|\nabla u|^2 + \epsilon^2}}\sum_i \left( u_{ii} - \sum_j \frac{u_{ij} u_i u_j}{|\nabla u|^2 + \epsilon^2}\right) \end{split}$$
and that is what you want as (under normal coordinates)
$$\sum_i u_{ii} = \sum_{i,j} g^{ij} u_{ij},\ u_i = \nabla^i u,\ u_{ij}= \nabla^2_{ij} u.$$