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 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