It is well known that for any smooth bounded (connected) domain $\Omega\subset\mathbb R^d$ with $d\ge2$, we can define a Green's function $G:\Omega\times\Omega\to\mathbb R$ in $\Omega$ which is smooth on $\mathring\Omega\times\mathring\Omega\setminus\Delta$, such that $$ G*\phi(x) = \int_\Omega\! G(x,y)\phi(y)\,\mathrm dy$$ solves the Poisson equation $-\Delta(G*\phi)=\phi$ in $\Omega$, and $G*\phi=0$ on $\partial\Omega$. Distributionally, we have that $-\Delta_x G(x,y)=\delta_y(x)$, and that $G(x,y)=0$ for $x\in\partial\Omega$, $y\in\mathring\Omega$. We know that we may write $$ G(x,y) = g(x-y) + f(x,y) $$ where $$ g(x):= \begin{cases}-\frac{1}{2\pi}\log|x| & d=2\\ \frac{\alpha_d}{|x|^{d-2}} & d\ge3\end{cases}$$ for some smooth $f:\mathring\Omega\times\mathring\Omega\to\mathbb R$ obtained by solving the Laplace equation.

What I was wondering was whether we can assert that $\nabla_x G(x,y)$ is nonvanishing on the boundary $x\in\partial\Omega$ for $y\in\mathring\Omega$.

  • $\begingroup$ In the a smooth domain, the answer is generally "yes". If you want a more precise answer, specify more precisely just how smooth your domains are. $\endgroup$ Apr 10 '19 at 15:37
  • $\begingroup$ Let’s say $C^\infty$. $\endgroup$
    – D. Wynter
    Apr 10 '19 at 15:37
  • $\begingroup$ Also in the title you mention "planar" domains while in the text consider $R^n$. $\endgroup$ Apr 10 '19 at 15:38
  • $\begingroup$ The specific case I’m using (that is, studying point vortices in bounded domains) requires $d=2$, but I’d be curious if a more general result can also be stated. $\endgroup$
    – D. Wynter
    Apr 10 '19 at 15:40
  • $\begingroup$ In addition to Alexandre Eremenko's answer: the normal derivative of the Green function (in a sufficiently smooth domain) is the Poisson kernel. $\endgroup$ Apr 10 '19 at 17:41

This follows from the so-called (Eberhard) Hopf Minimum Principle. If you have a positive (super-) harmonic function $u$ in a ball, and $u(z_0)=0$ for some boundary point $z_0$, then the normal derivative at $z_0$ is non-zero. This is a simple exercise: If the ball ix $|x|<R$ then $$m(r)=\min\{u(x):|x|=r\}$$ is concave in a certain sense (concave with respect to $r^{2-m}$ when $m>2$ and with respect to $\log r$ when $m=2$, where $m$ is the dimension of the space; this is called the Hadamard 3-circles theorem), and $m(R)=0$, so $m'(R-0)<0.$

If your boundary is sufficiently smooth, you can touch every boundary point by a ball which lies inside the domain.

(Similar arguments also show that if you can touch your boundary point by a ball from outside of the region then the normal derivative is finite at this point).


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