Mean curvature equals to the Laplacian of the position vector? My question is based on the manuscript titled On Mean Curvature Diffusion in Nonlinear Image Filtering (link: https://escholarship.org/content/qt0736r63m/qt0736r63m_noSplash_78bb00066d49daa019a4de37e1324b85.pdf?t=ptt37d). Specifically,  1) In page 5, before equation (15), it is said that $-2\mathrm{H}{\bf n} = \nabla^2 {\bf x}$. To be honest, I am a bit confused about the validity of this relation (although it is an exercise of another reference book cited in the paper). As ${\bf x} = (x_1,x_2,x_3)$, shouldn't the Laplcian of ${\bf x}$ be the zero vector since $\nabla {\bf x} = (1,1,1)$ ? I admit that I do not have a good background in differential geometry. So any elaboration of this is greatly appreciated.
2) In page 6, before equation (18), it is stated that "applying the same limit in (8) results in $2\mathrm{H}$ approaching the Laplcian $\nabla^2 I$". May I know why this is the case? Since $g(x) = x_3 - I(x_1,x_2)$, I obtain $$\frac{\partial g}{\partial t} = \frac{\partial x_3}{\partial t} - I_{x_1}\frac{\partial x_1}{\partial t} -  I_{x_2}\frac{\partial x_2}{\partial t} \approx  \frac{\partial x_3}{\partial t} $$under the limit $\nabla I → {\bf 0}$, which is not what we want... Thanks for your help
 A: This is indeed a correct equation regarding the mean curvature. The key point here is that $\mathbf{x}$ is the position vector for points on the surface. Think of $\mathbf{x}$ as representing a smooth map embedding the surface into Euclidean space. You should be able to derive the equation
$$H\hat{\mathbf{n}} = g^{ij}\nabla_i\nabla_j\mathbf{x},$$
where $g_{ij}$ is the metric tensor and $\nabla$ is the covariant derivative ,using the Weingarten equations.
Remember that $\mathbf{x}$ is a vector,
so
$$\Delta\mathbf{x} = (\Delta x_1, \Delta x_2, \Delta I).$$
Perhaps this will help see how the limit will work out.
Edit: The issue with your argument is that you're just not considering the right quantity. It's the mean curvature $H$ that is approaching $\Delta I$. Of course, under this flow, $2H=\partial_t g$, but that won't guarantee that the limiting behavior of $H$ will pop out of a direct computation with $g$. Your computation looks ok, but it just means you'll have
$$\partial_t I = c\Delta I.$$
