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Smoothness of distance function

June 11, 2013 – 5:53 pm

Suppose ${M}$ is a complete Riemannian manifold. ${p\in M}$ define ${\rho(x)=d(x,p)}$ . Obviously ${\rho(x)}$ is continuous, what can we say about the smoothness of ${\rho}$ .

(1) ${\rho(x)}$ is not ${C^1}$ near ${p}$ ;

(2) If ${M}$ is compact, then ${\rho(x)}$ is not ${C^1}$ in ${M\backslash \{p\}}$

It seems that ${\rho(x)}$ is not so smooth, let us consider ${\rho^2(x)}$

(3) ${\rho^2(x)}$ is smooth at neighborhood ${U}$ of ${p}$ and ${D^2\rho^2}$ is positive definite in ${U}$ .

(4) If ${M}$ is simply connected complete manifold with ${Sec_M\leq 0}$ , then ${\rho^2}$ is ${C^\infty}$ on whole ${M}$ and ${D^2\rho^2}$ is positive definite on ${M}$ .

By Sun's World | Posted in Geometry, Riemannian Geo | Comments (0)

Derivate of distance function and inner normal of boundary of a domain

October 11, 2012 – 3:27 pm

$\mathbf{Problem:}$ Let $\Omega\subset \mathbb{R}^n$ has non-empty boundary $\partial \Omega\in C^2$ . Let $\nu(y)$ denote the unit inner normal to $\partial \Omega$ at $y$ . WLOG assume $\partial \Omega$ is given by $x_n=\phi(x')$ where $x'=(x_1,x_2,\cdots,x_{n-1})$ and $\psi\in C^2(\mathcal{N}\cap{x_n=0})$ and $D\phi(y'_0)=0$ , where $\mathcal{N}$ is a neighborhood of $y_0$ . Also assume $\nu(y_0)$ is the $x_n$ coordinate axis.

The unit inner normal vector $\nu(y)$ at a point $y=(y',\psi(y'))\in \mathcal{N}\cap \partial \Omega$ is given by

$\displaystyle \nu_i(y)=\frac{-D_i\psi(y')}{\sqrt{1+|D\psi(y')|^2}}$ $i=1,2,\cdots,n-1.$ $\quad \nu_n(y)=\displaystyle \frac{1}{\sqrt{1+|D\psi(y')|^2}}\quad (1)$

Since $\partial \Omega\in C^2$ , there exists a neighborhood of $\partial \Omega$ , say $\chi_\epsilon=\{x\in \overline{\Omega}|d(x)<\epsilon\}$ such that for $\forall\, x\in\chi_\epsilon$ , there will exist a unique point $y=y(x)\in\partial \Omega$ such that $|x-y|=d(x)$ . Then $x,y$ satisfy

$x=y+\mathbf{\nu}(y)d\quad (2)$

Prove that for each point $x\in \chi_\epsilon$ , we have $Dd(x)=\nu(y(x))$ .

$\mathbf{Proof:}$ Since $|\nu(y)|=1$ , (2) means $d=(x-y)\cdot\nu(y)$ . Then $y=y(x)$ is $C^1$ (see GT’s book p355).

$\displaystyle d=\sum\limits_{j=1}^n(x_j-y_j)\nu_j(y)$ , $y_n=\phi(y')$ and $\nu(y)=\nu(y',y_n)$ is given by (1)

$\displaystyle D_kd(x)=\nu_k(y)+\sum\limits_{j,\,l=1}^nx_jD_l\nu_j D_k y_l-\sum\limits_{j=1}^nD_ky_j\nu_j(y)-\sum\limits_{j,\,l=1}^ny_jD_l\nu_j(y)D_ky_l$

$\displaystyle =\nu_k(y)+\sum\limits_{j,\,l=1}^n(x_j-y_j)D_l\nu_j D_k y_l-\sum\limits_{j=1}^nD_ky_j\nu_j(y)$

$\displaystyle =\nu_k(y)+d\sum\limits_{j,\,l=1}^n\nu_jD_l\nu_j D_k y_l-\sum\limits_{j=1}^nD_ky_j\cdot\nu_j(y)$

Actually the last two terms are equal to 0. In fact.

$\displaystyle \sum\limits_{j,\,l=1}^n\nu_jD_l\nu_j D_k y_l=D_k|\nu(y)|^2=0$ because $|v|=1$

Using (1)

$\displaystyle \sum\limits_{j=1}^nD_ky_j\cdot\nu_j(y)=\sum\limits_{i=1}^{n-1}\frac{-D_i\psi(y')D_ky_i}{\sqrt{1+|D\psi(y')|^2}}+\frac{\sum_j D_j\psi(y')D_ky_j}{\sqrt{1+|D\psi(y')|^2}}=0$

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$ GT’s book p355. Chapter 14. Boundary curvatures and Distance Function.

By Sun's World | Posted in Elliptic PDE, PDE, Real Analysis | Comments (0)

Possion equation for unbounded function

August 26, 2012 – 7:27 pm

$\mathbf{Thm:}$ Let $\Omega\subset \mathbb{R}^n$ is a bounded $C^2$ domain. Fix $\beta\in(0,1)$ , $f$ is a function in $\Omega$ such that $\displaystyle \sup\limits_{x\in B}d^{2-\beta}(x)|f(x)|\leq N<\infty$ , where $d(x)=dist(x,\partial \Omega)$ is the distance function. Then

$\displaystyle \begin{cases} \Delta u=f\quad \text{ in } \Omega\\u=0\quad \text{ on }\partial \Omega\end{cases}$

has a unique solution $u\in C^2(\Omega)\cap C^0(\overline{\Omega})$ satisfying

$\displaystyle \sup\limits_{x\in B}d^{-\beta}(x)|u(x)|\leq CN$

where the constant $C$ depends only on $\beta$ and $\Omega$ .

Firstly, recall some basic properties of distance function. By lemma 14.16 in GT’s book, if $\Omega$ is $C^2$ and bounded, there is a neighborhood of $\partial \Omega$ in $\Omega$ , say $\Gamma$ , such that $d(x)\in C^2(\Gamma)$ . And $\Delta d$ is bounded in $\Gamma$ .

$\mathbf{Proof:}$ $\Omega$ is $C^2$ means there exists $\delta$ such that $d(x)$ is $C^2$ in $B_\delta(x_\alpha)$ for any $x_\alpha\in\partial \Omega$ .

Suppose $\eta\in C^\infty_0(B_\delta(0))$ with $\eta(0)=\frac {1}{\beta(1-\beta)}$ . Denote $\eta_\alpha(y)=\eta(y-x_\alpha)$ .

Fix $x_\alpha\in \partial \Omega$ ,

$\Delta(d^\beta\eta_\alpha)=\eta_\alpha\Delta (d^\beta)+2\nabla (d^\beta)\cdot\nabla\eta_\alpha+d^\beta\Delta\eta_\alpha$

$\displaystyle =\left[\beta(\beta-1)d^{\beta-2}|\nabla d|^2+\beta d^{\beta-1}\Delta d\right]\eta_{\alpha}+2\beta d^{\beta-1}\nabla d\cdot\nabla\eta_\alpha+d^\beta\Delta\eta_\alpha$

$=\displaystyle -d^{\beta-2}\left[\beta(1-\beta)\eta_\alpha-\beta d\Delta d-2\beta d\nabla d\cdot\nabla\eta_\alpha-d^2\Delta\eta_\alpha\right]$
$\leq -\frac 12d^{\beta-2}$ $\quad$ if $0<d<r$ small enough.

$\Delta(d^\beta\eta_\alpha)\leq -\frac 12d^{\beta-2}\quad when \quad x\in B_r(x_\alpha)$

$\Delta(d^\beta\eta_\alpha)\leq C=C(\beta,\Omega)\quad when \quad x\in \Omega\backslash B_r(x_\alpha)$

Since $\partial \Omega$ is compact, there exists finitely many $x_1,x_2,\cdots,x_m$ such that $\displaystyle \bigcup\limits_{i=1}^m B_{r}(x_i)$ covers $\partial \Omega$ . Let $w$ is solution of $\Delta v=-mC$ in $\Omega$ and $v=0$ on $\partial \Omega$ .

Define $w=\sum\limits_{i=1}^md^\beta\eta_i +v$ , then $w=0$ on $\partial \Omega$ and $\Delta w\leq -\frac 12d^{\beta-2}$ in $\Omega$ . So

$\displaystyle \Delta(2Nw\pm u)\leq 0$ in $\Omega$ and $2Nw\pm u=0$ on $\partial \Omega$

Consequently, by the maximum principle,

$|u(x)|\leq 2Nw=d^\beta 2N( \sum\limits_{i=1}^m\eta_i )+2Nv$

Since $v$ has an upper bound only depends on the geometry of $\Omega$ and $m, C$ , we only need to prove $v(x)\leq C'd^\beta(x)$ when $x$ is near the boundary, where $C'=C'(\beta,\Omega)$ . Note that