# Harmonic functions vanishing on the boundary and distance function asymptotics

Let $$\Omega \subset \mathbb R^N$$ be a $$C^2$$ domain. Let $$u$$ be a function such that $$u \in W^{2,2}(\Omega)$$ and $$u = \Delta u = 0$$ on $$\partial \Omega$$. Is it true that $$c \le \frac{u}{[\mathrm{dist}(x, \partial \Omega)]^2} \le C$$ or some other similar estimate holds? Can we obtain similar results if $$\Omega$$ is less regular?

## 1 Answer

$$\DeclareMathOperator{\dist}{dist}$$ $$\newcommand{\bR}{\mathbb{R}}$$ $$\newcommand{\pa}{\partial}$$ Suppose that $$N=2$$ and $$\Omega$$ is is the unit disk. Choose $$u= -1+ar^4+br^5\in C^2(\overline{\Omega}).$$ Then $$u=0$$ along $$\pa \Omega$$ implies $$a+b=1$$. Next $$\Delta u=\frac{1}{r}\pa_r\big(\; r \pa_r u\;\big)= \frac{1}{r}\pa_r(4ar^4+5br^5)=16ar^2+25br^3.$$ The equality $$\Delta u=0$$ along $$\pa \Omega$$ implies $$16a+25b=0$$. Since $$a=1-b$$ we deduce $$16-16b+25b=0\implies b=-\frac{16}{9},\;\;a=\frac{25}{9}.$$

Note that $$\pa_r u=\frac{1}{9}\big(\; 100 r^3-80 r^4\;\big).$$

Along the boundary we have $$\pa_ru=\frac{20}{9}$$ which implies that

$$u(x)\sim \frac{20}{9}\dist\big(x,\Omega\big)\;\;\mbox{near \pa\Omega}.$$

• Is there any way to recover the square under additional assumptions? – Lao May 7 at 20:16
• For example, what if we require also $\nabla u = 0$ on $\partial \Omega$? – Lao May 7 at 22:23
• You also need $\frac{\partial^2 u}{\partial \nu^2}\neq 0$. – Liviu Nicolaescu May 8 at 0:35
• Why do you need that? – Lao May 8 at 7:01
• Think Taylor expansion in normal direction. – Liviu Nicolaescu May 8 at 9:27