Example that :$u\in C^\infty(U)$ not $u \in C^\infty(\overline{U})$ for elliptic equation

Question

Let $U$ be arbitrary open subset in $\Bbb{R}^n$,Let $u \in H^1(U)$ be the solution for the following equation with $f\in C^\infty(U)$:

$$\int_U \nabla u \cdot\nabla \varphi = \int_U f\varphi$$

for all $\varphi \in C^\infty_c(U)$ .

By interior regularility result for elliptic equation we know $u \in C^\infty(U)$.But It needs not to be a solution that $u\in C(\overline{U})$.

Is there some good example ?

Answer 1

Let $U$ be the circle segment (in polar coordinates) $$ U = \{ (r,\theta): \ r\in(0,1), \ \theta \in (0,\omega)\} $$ with $\omega \in (\pi,2\pi)$. Then $U$ is non-convex. Define $$ u(r,\theta) = r^{\pi/\omega} \sin(\pi\frac \theta\omega). $$ Then $u\in H^1(U)$, $-\Delta u=0$, so it satisfies the requirements with $f=0$. But $u\not\in C^1(\bar u)$, as $|\nabla u(r,\theta)|\to\infty$ for $r\searrow0$.