1
$\begingroup$

Picture below is from 260th page of G. Huisken's Flow by mean curvature of convex surfaces into spheres.

First, I think it should be $$\widetilde h_{ij}(x,\widetilde t)=\psi(\widetilde t) h_{ij}(x,\widetilde t).$$

Second, I don't know how to differentiate (14). Consider locally, then $$ \int_{\widetilde U_t} d\widetilde\mu=\int_{\widetilde F(U,t)}\sqrt {\widetilde g(x,t)} dS=\int_{\psi(t) F(U,t)} \psi(t)\sqrt{g(x,t)} dS $$ Then I am not sure how to proceed.

enter image description here

$\endgroup$
0
$\begingroup$

Just one way of all, and there should be some way is more easy. In fact , the area of $M_t$ is $$ \int_U \sqrt {\widetilde g} dx =\int _U\psi\sqrt g dx $$ Then $$ 0=\partial_t \int _U\psi\sqrt g dx =\psi '\int\sqrt g+ \psi\int -\frac{1}{n}H^2\sqrt g $$ So $$ \frac{\psi'}{\psi}=\frac{1}{n}\frac{\int H^2\sqrt g}{\int\sqrt g} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.