# Boundary behavior of Greens functions on smooth bounded (planar) domains

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 $$y\in\mathring\Omega$$.

• 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. Apr 10 '19 at 15:37
• Let’s say $C^\infty$. Apr 10 '19 at 15:37
• Also in the title you mention "planar" domains while in the text consider $R^n$. Apr 10 '19 at 15:38
• 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. Apr 10 '19 at 15:40
• In addition to Alexandre Eremenko's answer: the normal derivative of the Green function (in a sufficiently smooth domain) is the Poisson kernel. 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 $$|x| then $$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.$$