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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?

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$\DeclareMathOperator{\dist}{dist}$ $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\pa}{\partial}$ 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

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

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

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