Let $\Sigma\subset \Bbb{R}^n$ be some compact hypersurface without boundary in $\Bbb{R}^n$,fixed $v = -H$ where $H$ is mean curvature vector, assume we have a family of hypersurface $$\Sigma_{s} = \{x+sv(x)\mid x\in \Sigma\}$$
Prove the differentiation under the integral sign formula
$$\frac{d}{dt}\int_{\Sigma_t} u = \int_{\Sigma_t} u_t -\int _{\Sigma_t}u|H|^2$$
where $H$ is the mean curvature vector.
My attempt using the differentiation under the integral sign formula in calculus we have $$\frac{d}{dt}\int_{\Sigma_t}u\ dV_g= \int_{\Sigma_t}u_t \ d V_g + \int_{\partial\Sigma_t} u \langle v,n\rangle\ dS$$
Correct where $v$ is the velocity direction of the change of $\Sigma _t$ and $n$ is the out normal direction of $\Sigma_s$.
However, as $\Sigma_t$ does not has boundary, the second term on the right hand side is zero?Sorry for the silly question, I can't see where goes wrong with my reasoning?
Best Answer
The problem of the reasoning has been stated in the comment.
Let's see how to deduce it correctly (by the way, the similar argument has already given in Colding & Minicozzi's minimal surface book), I just mimic it.
Consider the embedding $F:\Sigma \times (-\epsilon, \epsilon) \to \Bbb{R}^{N}$ that maps $x\mapsto x- H(x)t\ $ with image $F(\Sigma , t) = \Sigma_t$, therefore the manifold $\Sigma _t$ isometric isomorphic to $(\Sigma,(F(\cdot,t))^*(\bar{g}))$ where $F(\cdot, t): \Sigma\to \Bbb{R}^N$ and $\bar{g}$ is standard Euclidean metric. therefore the new metric on $\Sigma$ is $$g_{i,j}(t) = \bar{g}(dF(\partial_i),dF(\partial_j))$$
therefore we have under the local coordinate $$\int_{\Sigma_t} u(x,t) \ dV_{t} = \int_{\Sigma} u \sqrt{\det{g_{ij}(t)} \det{g^{ij}(0)}}\sqrt{\det(g_{ij}(0))} dx^1\wedge...\wedge dx^n$$
denote $\nu(t) = \sqrt{\det{g_{ij}(t)} \det{g^{ij}(0)}}$Now use the fact in the book referenced above page 7, we have
$$\frac{d}{dt}\nu(t) = \text{div}_{\Sigma}(F_t) = \text{div}_{\Sigma}(-H)$$
therefore $$\frac{d}{dt}\int_{\Sigma_t}u = \frac{d}{dt}\int_{\Sigma} u\nu dV_0 = \int_{\Sigma} u_t \nu + \frac{d}{dt}\nu u dV_0 = \int_{\Sigma_t} u\ dV_t - \int_{\Sigma}|H|^2 u$$
For the global case, just use the partition of unity which finish the proof.