Machine-translated from Chinese.
准备工作
对完备流形$(M,g),$ 紧集$E,$ $\dim_\mathcal{H}E=p.$
那么$\,\forall\,p'>p,$ $\,\forall\,\varepsilon,\delta>0,$
$\,\exists\,(x_i,r_i<\delta),$
有限并$\bigcup \overline{B}_{r_i}(x_i)\supset E,$
使得$\sum \omega(p')r_i^{p'}<\varepsilon.$
令$\eta\in C^\infty(\mathbb{R}),$ 使得
$$
\eta=
\left\{\begin{aligned}
&1,&&t<1\\
&\in[0,1],&&t\in [1,2]\\
&0,&&t>2.
\end{aligned}\right.
$$
我们给出接下来的定义与性质:
$$
\eta_i(x)=\eta\left(\frac{d(x,x_i)}{r_i}\right)\in \mathscr{D}(M),\quad \eta_i|_{B^g_{r_i}(x_i)}=1,\quad \eta_i|_{B^g_{2r_i}(x_i)^c}=0.
$$
计算发现它的梯度满足如下性质:
$$
|\nabla_g\eta_i|=|\eta'|\left|\frac{\nabla_g d(x,x_i)}{r_i}\right|<\frac{C}{r_i}.
$$
$$
\int_M |\nabla_g\eta_i|^{n-p'}dV_g\le \frac{C}{r_i^{n-p'} }\mathrm{vol}(B_{2r_i}^g(x_i))\le Cr_i^{p'}.
$$
接下来, 定义
$$
\phi=\max\{\eta_i\}\in C^p(M)\cap W^{1,n-p'}(M),
$$
那么由先前的条件与结论,
$$
\int_M|\nabla \phi|^{n-p'}\le C\varepsilon,
$$
$$
\,\exists\,U\subset B_{2r_i}^g(E), \quad \phi|_U\equiv 1,\quad \phi|_{\left(B_{2r_i}^g(E)\right)^c}=0.
$$
简单起见, 定义$r=d(\partial U,E),$ 那么$\phi|_{B_r^g(E)}\equiv 1.$
因此随着$\varepsilon,\delta$改变,
我们可以说明$\,\exists\,\phi_k,\delta_k,r_k,$ 使得:
$r_k<\delta_k\rightarrow 0,$
$\operatorname{supp}\phi_k\subset B_{\delta_k}^g(E),$
$\phi|_{B_{r_k}^g(E)}=1,$
$\phi_k\in C^0(M)\cap W^{1,n-p'}(M),$
$\int_M|\nabla_g\phi_k|^{n-p'}dV_g\rightarrow 0.$
由逼近可设$\phi_k\in \mathscr{D}(M).$
应用
设$\Omega$为$M$中的有界区域, $\partial\Omega=\Gamma_1\cup \Gamma_2,$
其中$\Gamma_1$维数小于$n-1$且为紧集, $\Gamma_2$是$n-1$维的,
且$\mathcal{H}^{n-1}(\Gamma_2)<+\infty.$
我们来证明对于$M$上光滑向量场$X,$
$$
\int_\Omega \operatorname{div}X=\int_{\Gamma_2} X\cdot n.
$$
任取$p'\in (p,n-1),$ 令$E=\Gamma_1,$ $\,\exists\,\phi_k$如前所述,
通过将坏点挖去, 我们有:
$$
\int_M\operatorname{div}(1-\phi_k)X=\int_{\Gamma_2}(1-\phi_k)X\cdot n.
$$
而由$\phi_k$性质与Hausdorff维数性质, 我们知道,
$$
\int_\Omega \nabla \phi_k \cdot X\rightarrow 0,\quad \int_\Omega \phi_k \operatorname{div}X\rightarrow 0,\quad \int_\Omega \phi_kX\cdot n\rightarrow 0.
$$
这就得到了前面的结论.
文章最后更新于 2022-09-17 22:53:07
Preparation
For complete manifold $(M,g),$ compact set $E,$ $\dim_\mathcal{H}E=p.$
Then $\,\forall\,p'>p,$ $\,\forall\,\varepsilon,\delta>0,$
$\,\exists\,(x_i,r_i<\delta),$
Limited Union$\bigcup \overline{B}_{r_i}(x_i)\supset E,$
Make$\sum \omega(p')r_i^{p'}<\varepsilon.$
Let $\eta\in C^\infty(\mathbb{R}),$ make
$$
\eta=
\left\{\begin{aligned}
&1,&&t<1\\
&\in[0,1],&&t\in [1,2]\\
&0,&&t>2.
\end{aligned}\right.
$$
We give the following definitions and properties:
$$
\eta_i(x)=\eta\left(\frac{d(x,x_i)}{r_i}\right)\in \mathscr{D}(M),\quad \eta_i|_{B^g_{r_i}(x_i)}=1,\quad \eta_i|_{B^g_{2r_i}(x_i)^c}=0.
$$
Calculation found that its gradient satisfies the following properties:
$$
|\nabla_g\eta_i|=|\eta'|\left|\frac{\nabla_g d(x,x_i)}{r_i}\right|<\frac{C}{r_i}.
$$
$$
\int_M |\nabla_g\eta_i|^{n-p'}dV_g\le \frac{C}{r_i^{n-p'} }\mathrm{vol}(B_{2r_i}^g(x_i))\le Cr_i^{p'}.
$$
Next, define
$$
\phi=\max\{\eta_i\}\in C^p(M)\cap W^{1,n-p'}(M),
$$
Then from the previous conditions and conclusions,
$$
\int_M|\nabla \phi|^{n-p'}\le C\varepsilon,
$$
$$
\,\exists\,U\subset B_{2r_i}^g(E), \quad \phi|_U\equiv 1,\quad \phi|_{\left(B_{2r_i}^g(E)\right)^c}=0.
$$
For simplicity, define $r=d(\partial U,E),$ then $\phi|_{B_r^g(E)}\equiv 1.$
Therefore as $\varepsilon,\delta$ changes,
We can show $\,\exists\,\phi_k,\delta_k,r_k,$ such that:
$r_k<\delta_k\rightarrow 0,$
$\operatorname{supp}\phi_k\subset B_{\delta_k}^g(E),$
$\phi|_{B_{r_k}^g(E)}=1,$
$\phi_k\in C^0(M)\cap W^{1,n-p'}(M),$
$\int_M|\nabla_g\phi_k|^{n-p'}dV_g\rightarrow 0.$
It can be set by approximation $\phi_k\in \mathscr{D}(M).$
Application
Let $\Omega$ be the bounded area in $M$, $\partial\Omega=\Gamma_1\cup \Gamma_2,$
Among them, the dimension of $\Gamma_1$ is smaller than $n-1$ and is a compact set, $\Gamma_2$ is of dimension $n-1$,
And $\mathcal{H}^{n-1}(\Gamma_2)<+\infty.$
Let us prove that for $M$ the smooth vector field $X,$
$$
\int_\Omega \operatorname{div}X=\int_{\Gamma_2} X\cdot n.
$$
Take $p'\in (p,n-1),$ arbitrarily and let $E=\Gamma_1,$ $\,\exists\,\phi_k$ be as mentioned above,
By digging out the bad pixels, we have:
$$
\int_M\operatorname{div}(1-\phi_k)X=\int_{\Gamma_2}(1-\phi_k)X\cdot n.
$$
And from the $\phi_k$ property and the Hausdorff dimension property, we know that,
$$
\int_\Omega \nabla \phi_k \cdot X\rightarrow 0,\quad \int_\Omega \phi_k \operatorname{div}X\rightarrow 0,\quad \int_\Omega \phi_kX\cdot n\rightarrow 0.
$$
This leads to the previous conclusion.
The article was last updated on 2022-09-17 22:53:07
