Exponentially decay of volume preserving mean curvature flow

Question

As in The volume preserving mean curvature flow, consider the volume preserving mean curvature $$ \partial_tF(x,t)=(h(t)-H(x,t))\cdot\nu(x,t) ~~~x\in U, t\ge 0 \\ F(\cdot, 0 )=F_0 \\ h(t)=\frac{\int_{M_t}Hd\mu}{\int_{M_t}d\mu} $$ $F_0:R^n\supset U\rightarrow F_0(U)\subset M_0\subset R^{n+1}$ is local represent of $M_0$ . $H=g^{ij}h_{ij}$ , $h_{ij}$ is the second fundamental form. Then, there is $$ \sup_{M_t}|H-h|\rightarrow 0 \text{ as } t\rightarrow \infty $$ Then, what is the mean of red line in picture below, I mean that what decays exponentially ? And why ?

Answer 1

They are talking about the rate of change of the surface under the flow, which is given by $$ \left| \frac{\partial}{\partial t} F \right| = |H-h| $$ recalling that $F(x,t)$ describes the mapping of the manifold from coordinates in $\mathbb{R}^n$ to points on the surface in $\mathbb{R}^{n+1}$.

The equation for the flow (which describes how $F$ changes over time) defines a rate of change of the surfaces (i.e. how different is $M_{t}$ from $M_{t-\epsilon}$). Thus, the LHS of the equation just above is the "velocity of the surfaces", as the authors put it. (I guess I would call it the speed).

Hence, the quantity that is decaying exponentially is that rate. So, intuitively, as the flow goes on, the effect that it has on the surface becomes "weaker". If you could look at the surface visually (as you can in $\mathbb{R}^3$), as the flow begins, it will quickly "smooth out" all the local details, but then have less and less of a noticeable effect over time. For very high times, it will have essentially converged (to a sphere in the convex case in the paper) and you not be able to notice any change in $F$ at all.

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Now, why does it decay exponentially? (or how do they show it?)

The authors define $A=[h]_{ij}$ to be the second fundamental form. Then, $|A|^2=g^{ij}g^{kl}h_{ik}h_{jl}$. They then show: $$ |A|^2 - \frac{1}{n}H^2 \leq c \exp(-\delta t) $$ for constants $c,\delta\in\mathbb{R}$. In words, the difference between the "norm" of the second fundamental form and the mean curvature vanishes exponentially quickly in time. It's not obvious (to me anyway) why that implies the surface stops changing exponentially quickly as well, but earlier the authors show that: $$ |A|^2 - \frac{1}{n}H^2 = \frac{1}{n} \sum\limits_{i<j} (\kappa_i - \kappa_j)^2 $$ where $\kappa_i$ satisfy $h_{ij}=\kappa_i \delta_{ij}$ in some coordinate system (i.e. they are the principal curvatures).

This makes it a little easier to see what they mean: basically all the differences between the principal curvatures shrink to a constant exponentially fast.

(If anyone sees something else in the paper I missed, let me know)