What is the constant in the rate of exponential convergence for mean curvature flow? Given a domain $\Omega \subset \mathbb{R}^d$ which is convex and smooth and $|
\Omega|=1$, it is well known that the metric converges exponentially fast to that of the sphere under volume preserving MCF. I would like to know the following:
Question: Does the rate of convergence depend on the domain $\Omega$? If so, in what way? If I know in particular that 
$\frac{d}{dt} \|\kappa - \bar \kappa\|_{L^2(\partial \Omega)}^2 \leq -C \|\kappa - \bar \kappa\|_{L^2(\partial \Omega)}^2$ `, can I say that $C$ is independent of $\Omega$? If not, can I see in which explicit way it does depend on $\Omega$? I suppose this is equivalent to asking if there is a particular $\Omega$ so that the convergence is slowest. 
Thanks.
 A: This is answered in a paper of N. Sesum
See here  for Mathscinet review.
Basically since knowing something about the rate only matters at large scales you can assume the flow is near a sphere (by Huisken's result).  Then the rate just depends on spectral properties of the Laplacian on the sphere.  In particular, the rate is at worst the gap between (if I recall correctly) the first non-trivial eigenvalue and the second non-trivial eigenvalue. This is sharp as one can deform the sphere a small amount in the normal direction by an amount given by the eigenfunction associated to the second non-trivial eigenvalue. 
A: Here is an explicit example, for the  Ricci flow, with which one can compare
general computations. For the MCF in dimension 1, an analogue is the
Galaktionov--Angenent oval (a.k.a. paper clip). The King-Rosenau (a.k.a.
sausage model) solution to the backward Ricci flow on $S^{2}$ can be viewed as
the two-point compactification of $g_{\tau}(x)=\frac{\operatorname{s}%
(\tau)(dx^{2}+d\theta^{2})}{\operatorname{c}(x)+\operatorname{c}(\tau)}$,
$\tau\in(0,\infty)$, on the cylinder $\mathbb{R\times(R}/4\pi\mathbb{Z})$,
where $\operatorname{s}=\sinh$ and $\operatorname{c}=\cosh$. Its area is
$\operatorname{A}_{\tau}=8\pi\tau$ and its scalar curvature is $R_{g_{\tau}
}(x)=\frac{\operatorname{c}(\tau)\operatorname{c}(x)+1}{\operatorname{s}
(\tau)\left(  \operatorname{c}(x)+\operatorname{c}(\tau)\right)  }$, with
average $r_{g_{\tau}}=\tau^{-1}$; so it satisfies $\frac{\partial}
{\partial\tau}g_{\tau}=R_{g_{\tau}}g_{\tau}$. Using a table of integrals of
rational functions of hyperbolic functions, I think we obtain $\frac{1}{4\pi
}\int(R_{g_{\tau}}-r_{g_{\tau}})^{2}d\mu_{g_{\tau}}=\frac{\operatorname{c}
(\tau)}{\operatorname{s}(\tau)}+\frac{\tau}{\operatorname{s}^{2}(\tau)}
-\frac{2}{\tau}=\frac{2}{45}\tau^{3}+\operatorname{O}(\tau^{5})$ as
$\tau\rightarrow0^{+}$. Given $a>0$, $h_{\tau}=ag\left(  a^{-1}\tau\right)  $
is also a solution. Then $\frac{1}{4\pi}\int(R_{h_{\tau}}-r_{h_{\tau}}%
)^{2}d\mu_{h_{\tau}}=\frac{a^{-1}}{4\pi}\int(R_{g_{a^{-1}\tau}}-r_{g_{a^{-1}
\tau}})^{2}d\mu_{g_{a^{-1}\tau}}=a^{-4}\frac{2}{45}\tau^{3}+\operatorname{O}
(\tau^{5})$. For the corresponding solutions to the normalized backward Ricci
flow, dilation by $a$ corresponds to time translation by $\ln a$. Let $\bar
{g}_{\tau}=\tau^{-1}g_{\tau}$, so that $\overline{\operatorname{A}}_{\tau
}=8\pi$ and $\bar{r}_{\tau}=1$; let $\bar{\tau}=\ln\tau$. Then we have the
eternal solution $\frac{\partial}{\partial\bar{\tau}}\bar{g}_{\bar{\tau}
}=(R_{\bar{g}_{\bar{\tau}}}-1)\bar{g}_{\bar{\tau}}$ and
$$
\frac{1}{4\pi}\int(R_{\bar{g}_{\bar{\tau}}}-1)^{2}d\mu_{\bar{g}_{\bar{\tau}}
}=\frac{e^{\bar{\tau}}\operatorname{c}(e^{\bar{\tau}})}{\operatorname{s}
(e^{\bar{\tau}})}+\frac{e^{2\bar{\tau}}}{\operatorname{s}^{2}(e^{\bar{\tau}}
)}-2=\frac{2}{45}e^{4\bar{\tau}}+\operatorname{O}(e^{6\bar{\tau}}
)\quad\;\text{as }\bar{\tau}\rightarrow-\infty.
$$
