# Derivation of the volume preserving mean curvature flow

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Picture above is from Huisken, Gerhard, The volume preserving mean curvature flow, J. Reine Angew. Math. 382, 35-48 (1987). ZBL0621.53007..

I want to get the (2) from (1) . First , choice a positive function $\varphi(t)$ such that the volume enclosed by surface $\widetilde M_t$ given by $$\widetilde F(x,t) = \varphi(t)F(x,t) ~~~~~~~~~~~~~~~~~~~~~(3)$$ is equal to the volume enclosed by $M_0$. So I have $$\frac{1}{n+1}\int _U \widetilde F(x,t)\cdot \widetilde\nu(x,t) \sqrt{\widetilde g(x,t)} dx =\frac{1}{n+1}\int_U F(x,0)\cdot \nu(x,0) \sqrt{g(x,0)} dx ~~~~~~~~~~~~~~(4)$$ And $$\widetilde g_{ij}=\varphi^2 g_{ij} ~~~~~~~~~~~~~~~\widetilde h_{ij}=\varphi h_{ij} \\ ~~~~~~\widetilde H=\varphi^{-1} H ~~~~~~~~~~~~~~\sqrt{\widetilde g}= \varphi^n\sqrt{g}~~~~ \\ ~~~~~~~\widetilde\nu(x,t) =\nu(x,t) ~~~~~~~~~~~~~~\partial_t \nu(x,t)=\nabla H(x,t) \\ \partial_t\sqrt g =-H^2 \sqrt g~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$ Differentiate (4), I have $$0=\int_U (n+1)\varphi^n\varphi' F\cdot\nu\sqrt g - \varphi^{n+1} H \sqrt g +\varphi^{n+1}F\cdot \nabla H \sqrt g -\varphi ^{n+1} F\cdot\nu H^2 \sqrt g dx$$ so I have $$\frac{\varphi ' }{\varphi} =\frac{1}{n+1} \frac{\int_U (H-F\cdot \nabla H +F\cdot \nu H^2 )\sqrt g}{\int_U F\cdot \nu \sqrt g }$$ define $h(t)$ as $$h(t)\triangleq\frac{\int_U (H-F\cdot \nabla H +F\cdot \nu H^2 )\sqrt g}{\int_U F\cdot \nu \sqrt g }$$ introduce a new time
$$\widetilde t (t) = \int _0^t \varphi^2(\tau) d\tau$$ so I have $$\frac{d\widetilde t}{dt}=\varphi^2(t) ~~~~~~~~~~~ \frac{d t}{d\widetilde t}=\varphi^{-2}(t(\widetilde t))$$ So , I have $$\partial_{\widetilde t} \widetilde F =\partial _t \widetilde F \varphi^{-2} =\varphi^{-2}[\partial_t\varphi F+\varphi \partial_t F] =\varphi^{-1}Fh-\varphi H\nu =\widetilde F\widetilde h-\widetilde H \widetilde \nu$$ So , I have the volume preserving mean curvature flow $$\partial_t F =Fh-H\nu$$ In fact, it really be preserving volume. Because $$\partial_t \int _U F\cdot \nu \sqrt g dx =0$$ But it is not same with (2) in above picture. How to get the (2) ? I think the form of (2) is better than mine.

PS: After long time, I can't deal it. Thanks for any help.