Recently I am reading the mean curvature flow from the lecture notes of Carlo Mantegazza where I found that

Under mean curvature flow given by$$\begin{cases}{\partial\over \partial t}\varphi(p,t)=H(p,t)\nu (p,t),\\\varphi(p,0)=\varphi_0(p)\end{cases}$$ among all the velocity functions with fixed $L^2(\mu)$ norm equal to $(\int _MH^2d\mu)^{1/2}$,the one such that the Area of hypersurface decreases most rapidly.

Actually I can not understand the argument that why it decreases the area most rapidly and also why we must select the $L^2$ norm?


Assume that there is a family of immersions $\varphi_t : M^n \to \mathbb R^{n+1}$ so that

$$\frac{\partial }{\partial t} \varphi_t = f_t\ \nu_t,$$

where $f_t$ are smooth functions on $M$. Let $A_t$ denote the area given by $\varphi_t$. Then one has

$$\frac{\partial}{\partial t} A_t = -\int_M f_t H d\mu = -\langle f_t, H\rangle_t$$

Note that $\langle f , H\rangle_t \le \|f_t\|_2 \cdot \| H\|_2$ and equality holds if and only if $f = CH$ for some $C>0$. Thus if we require that $\|f\|_2 = \|H\|_2$ (that is, $C=1$), then the area will be decreasing most raidly when $f = H$, that is the case for the mean curvature flow.

  • $\begingroup$ Yes I understood the final argument.But can you clarify me how $\frac{\partial}{\partial t} A_t = -\int_M f_t H d\mu $ holds? $\endgroup$ – debabrata chakraborty Jun 11 '15 at 6:33
  • 1
    $\begingroup$ @debabratachakraborty: this is the first variation of area formula - you should be able to find a derivation easily by searching the web. $\endgroup$ – Anthony Carapetis Jun 11 '15 at 6:42
  • $\begingroup$ yes.i got it.thanks $\endgroup$ – debabrata chakraborty Jun 11 '15 at 7:09

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