Regularity of Neumann eigenfunctions at vertices of polygons

Question

Given a bounded polygonal domain $D$ in $\mathbb{R}^2$, the Neumann eigenfunctions have continuous version on $\overline{D}$. The eigenfunctions also have critical points at vertices of $D$ (I have been taught the simple proof by Prof. Mateusz Kwaśnicki at this page).

However, the modulus of continuity of the eigenfunctions at vertices seem not obvious. For example, when $D$ is an equilateral triangle and $\phi$ is a Neumann eigenfunction on $D$, can we describe the modulus of continuity of $\phi$ at a vertex $p$ ? It holds that $|(\nabla \phi)(p)|=0$. Can we expect that $|\phi(x)-\phi(p)|=o(|x-p|^\alpha)$ as $x \to p$ for some $\alpha>1$.

If so, what is the optimal exponent?

Answer 1

Let me develop a little my comment. Consider a corner of the domain, with tip at the origin and one edge horizontal. Denote $\alpha$ its angle of aperture. The leading singularity of a solution of $\Delta u=f$ with a smooth $f$, and Neumann condition, is $A\Re(z^\kappa)$. With $z=e^{i\theta}$, its normal derivative is $r^{-1}\partial_\theta$. Whence the condition $\sin(\kappa\alpha)=0$, where we must take the smallest root $\kappa$, in absence of better information: $$\kappa=\frac\pi\alpha.$$ For $\alpha=\frac\pi3$, this gives $\kappa=3$, which is not so bad as the eigenfunction is $C^{3-\epsilon}$. But the larger $\alpha$, the less regular are the eigenfunctions.

Edit. When $f=\lambda u$, which is in $H^1(\Omega)$, it has enough "smoothness" that the above applies.