Let $M ≔ ℝ^{n+1}$ be equipped with some Riemannian metric $g$, let $u: ℝ^n → n$ be smooth and define
- $F: M = ℝ^n × ℝ → ℝ, (x, z) ↦ z - u(x)$
- $Σ ≔ \operatorname{graph}(u) = F^{-1}(\{0\}) ⊂ M$
What is the mean curvature $H$ of $Σ$, expressed in terms of $u$? Is there a nice formula for it?
I am aware that $H$ with respect to the normal field $N ≔ \frac{∇F}{‖∇F‖}$ on $Σ$ can be written as:
$$ H = - \operatorname{div}_Σ N = - \Big( \frac{∆_M F}{‖∇F‖} - \frac{\operatorname{Hess}{F}(∇F, ∇F)}{‖∇F‖³} \Big) $$
where $∆_M$ is the Laplace-Beltrami operator on $M$ and $∇F$ denotes the gradient of $F$ but I think I remember seeing a particularly neat formula for $H$ in terms of $u$ somewhere that looked like the usual minimal-graph equation $H = \operatorname{div}(\frac{∇u}{\sqrt{1+‖∇u‖²}})$ with only one additional (lower-order) term, involving a logarithm somewhere(?)
Unfortunately, I forgot where I found that formula and when I tried to derive it myself, I ended up with quite a mess. So I'd be grateful for a hint as to what the final result needs to look like. (EDIT: To be clear, I'm not necessarily looking for the formula I think I saw, I'm looking for any relatively nice looking formula for $H$ in terms of $u$.)