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

$\mathbf{Problem:}$ 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.