Derivation of the volume preserving mean curvature flow

Question

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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: this question has asked in https://math.stackexchange.com/questions/2193510/derivation-of-the-volume-preserving-mean-curvature-flow , but not any answer or hint, so I ask it here.

Answer 1

I do not think the flows are equivalent.

Suppose I have two flows $F : M \times [0, T) \to \mathbb{R}^{n+1}$ and $G : M \times [0, T) \to \mathbb{R}^{n+1}$. Let's call $F$ and $G$ equivalent if $F(M, t) = G(M, t)$ for every $t$. In other words if the image hypersurfaces are the same at each time $t$. This is equivalent to $$ G(x, t) = F(\varphi(x, t), t) $$ where $\varphi_t(\cdot) = \varphi(\cdot, t) : M \to M$ is a diffeomorphism for each $t$.

This is also equivalent to $$ \langle \partial_t G, \nu_G \rangle|_{(x, t)} = \langle \partial_t F, \nu_F \rangle|_{(\varphi_t(x), t)} $$ where $\nu_F$ is the unit normal to $F$ and $\nu_G$ is the unit normal to $G$.

Clearly (1) and (2) are not equivalent in this sense since the normal component of the speed is different.

Let's call $F$ and $G$ equivalent up to scale if $$ G(M, t) = \lambda(t) F(M, t) $$ for some smooth, positive function $\lambda$ of $t$.

Note: We really should include a translation here like in this answer but I'll let you work out that part. For the purpose of this answer, let's just suppose that both flows collapse to the origin which can easily be arranged by a time-independent translation if we also suppose that both flows collapse to a point (not necessarily the same point).

Equivalent formulations of equivalence up to scale: $$ G(x, t) = \lambda(t) F(\varphi_t(x), t) $$ and $$ \tag{eq}\label{eq} \langle \partial_t G, \nu_G \rangle|_{(x, t)} = \left[\partial_t \lambda \langle F, \nu_F \rangle + \lambda \langle \partial_t F, \nu_F \rangle\right]|_{(\varphi_t(x), t)} $$

Now you have a little work to do. Let's say $F$ is the MCF. Then we want $$ \lambda(t) = \left(\frac{\operatorname{Enclosed Vol}(F(M, 0))}{\operatorname{Enclosed Vol}(F(M, t))}\right)^{1/(n+1)} $$ to keep the enclosed volume fixed.

Then you can try to work out $\lambda$ and you should see it's not equivalent. But by inspection you can see that neither $\lambda$ nor equation \eqref{eq} will produce $h$, the average mean curvature anywhere.

In the curve case, equation (2) in A comparison theorem for the isoperimetric profile under curve shortening flow gives the flow equivalent to the curve shortening flow up to scale (after also a change in time parameter).

Answer 2

In the ansatz (3), you seem to rescale the solution of the mean curvature flow from the origin. If you start with a small sphere far away, this sphere is going to shrink under the mean curvature flow, and your ansatz will rescale it from $0$, with the effect that the round sphere will be parallelly translated to infinity eventually, whereas Huisken's flow will keep it constant.

On the other hand, to get (2), you compensate the volume change effected by the mean curvature flow evenly on the whole area of $M$.