Tag Archives: distance function

Smoothness of distance function

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}.

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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.

Possion equation for unbounded function

\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

\displaystyle \Delta(d^\beta)=d^{\beta-2}[\beta(\beta-1)+\beta d\Delta d]\rightarrow -\infty uniformly as x\to \partial \Omega

So there exists a neighborhood \Gamma' of \partial \Omega such that \Gamma'\subset \Gamma and \exists C'=C'(C,\beta,\Omega)

\Delta(C'd^\beta-v)\leq 0 in \Gamma' and C'd^\beta-v\geq 0 on \partial \Gamma'

By the maximum principle, v(x)\leq C'd^\beta(x) in \Gamma'.

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

\mathbf{Remark:} Gilbarg, Trudinger. Chapter 4, exercise 4.6. p71

Also see, J.H. Michael. A general theory for linear elliptic partial differential equations. 1977

Gary M. Lieberman. Elliptic equations with strongly singular lower order terms. 2008