I'm thinking about an equivalent formulation of IMCF proposed by Gerhard Huisken and Tom Ilmanen in their 2001 work THE INVERSE MEAN CURVATURE FLOW AND THE RIEMANNIAN PENROSE INEQUALITY:
Let $M$ be a smooth Riemannian manifold of dimension $n\geq 2$ with metric $g=(g_{ij})$. A classical solution of the inverse mean curvature flow is a smooth family $x:N\times[0,T]\to M$ of hypersurfaces $N_t:=x(N,t)$ satisfying the parabolic evolution equation
$$\frac{\partial x}{\partial t}=\frac{\nu}{H},\qquad x\in N_t,\qquad 0\leq t\leq T,\tag{$*$}$$
where $H$, assumed to be positive, is the mean curvature of $N_t$ at the point $x$, $\nu$ is the outward unit normal, and $\frac{\partial x}{\partial t}$ denotes the normal velocity field along the surface $N_t$.Level-Set Description. Because it is desirable that $H$ remain positive (or at leas nonnegative), we impose a unidirectional ansatz that compels this. This is accomplished via a level-set formulation, inspired by Evans-Spruck, Chen-Giga-Goto. Assume that the flow is given by the level sets of a function $u:M\to\mathbb{R}$ via
$$E_t:=\{x:u(x)<t\},\qquad N_t:=\partial E_t.$$
Wherever $u$ is smooth with $\nabla u\neq 0$, ($*$) is equivalent to
$$\mathrm{div}_M(\frac{\nabla u}{|\nabla u|})=|\nabla u|,\tag{$**$}$$
where the left side gives the mean curvature of $\{u=t\}$ and the right side gives the inverse speed.
Let $N$ be a smooth hypersurface in the Riemannian manifold $M$ with metric $g$. The authors said that $\nabla$ is written for the connection on $M$. In this regard, I think it is more appropriate to rewrite ($**$) as
$$\mathrm{div}_M(\frac{\mathrm{grad\ } u}{|\mathrm{grad\ } u|})=|\mathrm{grad\ } u|.\tag{$***$}$$
Now I'd like to know how ($***$) can be derived. I've already known from introductory Riemannian geometry that
$$\nu=\frac{\mathrm{grad\ } u}{|\mathrm{grad\ } u|}$$
and
$$H=\mathrm{div}_M(\frac{\mathrm{grad\ } u}{|\mathrm{grad\ } u|}).$$
Then what should I do with $\frac{\partial x}{\partial t}$ in order to reach ($***$)? Also, both sides of ($***$) are real-valued functions. They don't seem to match ($*$), whose both sides are vector fields. Is ($***$) really an equivalent formulation of ($*$)? Thank you.
Best Answer
If you fix a reference point $p \in N$, then $x(t) = x(p, t)$ is a curve satisfying $x(t) \in N_t,$ i.e. $u(x(t)) = t.$ Differentiating this with the chain rule yields $$1 = Du(\frac{\partial x}{\partial t}) = g(\mathrm{grad\ } u, \frac{\partial x}{\partial t}) = \|\mathrm{grad\ } u\| g(\nu, \frac{\partial x}{\partial t}).$$
Since ($*$) tells us $\frac{\partial x}{\partial t}$ is in the direction $\nu$, this can be rewritten as $$\frac{\partial x}{\partial t} = \frac \nu {\|\mathrm{grad\ } u\|};$$ so we can write ($*$) as $$\frac \nu {\|\mathrm{grad\ } u\|} = \frac \nu H$$ or simply $\|\mathrm{grad\ } u\| = H.$ From here you just need your formula for $H$ in terms of $u$ and you're done.
You can reverse this argument to show equations ($*$) and ($***$) are equivalent so long as you assume the flow is parametrized orthogonally. This assumption is what allows the scalar equation (which you can think about as being in terms of speed) to "strengthen" to the vector field equation (in terms of velocity): you already know both sides are colinear, so you just need to compare their magnitudes.