A reentrant polyhedral edge has an H1 harmonic singularity

In polar coordinates the cross-sectional Laplacian is ${\mathrm{\partial }}_{rr}+r^{-1}{\mathrm{\partial }}_r+r^{-2}{\mathrm{\partial }}_{\theta \theta }$. For $u=r^{\alpha }\sin (\alpha \theta )$, the three terms sum to $[\alpha (\alpha -1)+\alpha -{\alpha }^2]r^{\alpha -2}\sin (\alpha \theta )=0$; the $z$ derivative is zero. Also $u(r,0,z)=u(r,\omega ,z)=0$ because $\alpha \omega =\pi$. Thus it is harmonic with zero side trace in the local geometry [F1].

The squared transverse gradient is $\mathrm{\mid }{\mathrm{\nabla }}_{r,\theta }u{\mathrm{\mid }}^2={\alpha }^2{r}^{2\alpha -2}$, so its local integral is a constant times ${\int }_0^{r_0}{r}^{2\alpha -1}dr<\mathrm{\infty }$ for $\alpha >0$; $u$ itself is also square integrable. On the other hand, $u_{rr}=\alpha (\alpha -1)r^{\alpha -2}\sin (\alpha \theta )$ is nonzero on a set of angles of positive measure, and its squared integral contains $c{\int }_0^{r_0}{r}^{2\alpha -3}dr=\mathrm{\infty }$ because $\alpha <1$. Hence $u\in H^1$ locally but $u\mathrm{\notin }{H}^2$ near the edge.