Laplace-Beltrami of the mean curvature For a surface $S$ defined in 3D space, denote its mean curvature as $H$, and the Laplace-Beltrami operator as $\Delta_S$. I know that there is a result for the Laplace-Beltrami of coordinate functions:
$$ \Delta_S \vec{r} = 2 H \vec{n}. $$
My question is that: how to calculate the Laplace-Beltrami of the mean curvature?
$$  \Delta_S H = ? $$
If there isn't a simple expression, is it possible to expand it like this way:
$$  \Delta_S H = a_1 H + a_2 H^2 + a_3 K + ...$$
where $a_i$ are some constant numbers?
 A: Here is a bit more detail on the answer to the following question, which is my interpretation of what the OP is asking:
Suppose that one knows the induced metric $g$ on a surface $S\subset\mathbb{R}^3$.  What equations involving the metric does the mean curvature function $H$ satisfy?
There have to be some nontrivial equations, since one can't freely specify the mean curvature function $H$ once the induced metric is fixed.  At the very least, $H$ has to satisfy the inequality $H^2-K\ge0$, where $K$ is the Gauss curvature of $g$, since $H^2-K = \tfrac14(\kappa_1{-}\kappa_2)^2$.  This simple inequality is not sufficient though, since, as was classically observed, if a surface $S\subset \mathbb{R}^3$ has $H=0$ (i.e., is minimal), then the (singular) metric $(-K)\,g$ is the pullback of the metric on $S^2$ via the Gauss map $\nu:S\to S^2$ of the surface $S$ and hence must have Gauss curvature $+1$, which turns out to be an equation of order $4$ on the metric $g$.
Setting aside the trivial all-umbilic case where $H^2-K=0$, one can assume that $H^2-K = r^2 > 0$ for some function $r$ on $S$.  In this case, it turns out that pursuing the structure equations, one derives a further necessary inequality of the form $F_g(H,\nabla H,\nabla^2H)\ge 0$, i.e., a polynomial inequality that is second-order in $H$, but whose coefficients involve 4 derivatives of the metric $g$.
In some cases, when  $F_g(H,\nabla H,\nabla^2H)\equiv 0$, one can actually show that there is an entire circle of isometric immersions of $(S,g)$ into $\mathbb{R}^3$ with mean curvature $H$.  Such data $(S,g,H)$ is said to constitute a Bonnet surface, after the work of Ossian Bonnet in the 19th century.
Assuming that $H$ satisfies the strict inequality $F_g(H,\nabla H,\nabla^2H)> 0$, it then turns out that there is a pair of equations
$$
M_g(H,\nabla H,\nabla^2H,\nabla^3H) = N_g(H,\nabla H,\nabla^2H,\nabla^3H) = 0,
$$
which are third-order in $H$ and fifth-order in $g$ that $H$ must satisfy in order to be the mean curvature of an isometric immersion of $(S,g)$ into $\mathbb{R}^3$  When $H$ satisfies these equations, there are at most two non-congruent isometric immersions of $(S,g)$ into $\mathbb{R}^3$ up to isometry.  (When two exist, this pair of immersions is nowadays said to be a Bonnet pair.)
In particular, it follows from the structure equations that there is no universal identity of the form $\Delta_g H = E_g(H,\nabla H)$ where $E_g$ is a (possibly nonlinear) operator constructed from the metric $g$ and its derivatives (and hence, for example, might contain $K$ and some of its derivatives).
