# What fraction of a charge is induced on a surface via balayage?

Consider a smooth, bounded domain $$\Omega \subset \mathbb{R}^3$$, and place a charge $$q>0$$ at a point $$z\in\mathbb{R}^3\setminus\overline\Omega$$. Via the concept of balayage, there is an induced surface charge density (i.e. signed Radon measure) $$\nu$$ supported on $$\partial\Omega$$ such that $$\int \frac{d\nu(y)}{|x-y|} = \frac{q}{|x-z|}$$ for all $$x\in\Omega$$. Moreover, the total variation of $$\nu$$ is bounded by $$q$$. I think it is known that if $$z$$ lies in a bounded component of $$\mathbb{R}^3\setminus\overline\Omega$$, then this becomes an equality i.e. the total variation of $$\nu$$ equals $$q$$. My question is what fraction of $$q$$ is the total variation of $$\nu$$ if $$\overline\Omega$$ does not fully enclose $$z$$? Is it proportional to the solid angle subtended by $$\overline\Omega$$ from the perspective of point $$z$$ (i.e. the fraction of the view of infinity from $$z$$'s perspective that is blocked by $$\overline\Omega$$)?

• This seems to be the probability that a 3-D Brownian motion started at $z$ ever hits $\overline{\Omega}$. It is not possible to express this quantity in a closed-form way for general $\Omega$. – Mateusz Kwaśnicki Jul 3 at 6:26
• $q$ is missing in your displayed formula. – Alexandre Eremenko Jul 3 at 14:40
• @AlexandreEremenko fixed, thank you – Ben Ciotti Jul 3 at 19:37
• @MateuszKwaśnicki ah I didn't realize that, thank you for the insight. I wonder if there is a shape such that the total variation of the induced surface charge is some prescribed fraction of q? – Ben Ciotti Jul 3 at 19:40

If $$\Omega = B(0, r)$$ is a ball, then the fraction is given by $$r/|z|$$. More precisely, the harmonic reduction of $$u(x) = (4\pi)^{-1} q |x-z|^{-1}$$ in $$\mathbb{R}^3 \setminus \Omega$$ is given by $$v(x) = (4\pi)^{-1} q r |z|^{-1} |x-z^*|^{-1}$$ for $$x \in \mathbb{R}^3 \setminus \overline{\Omega}$$, where $$z^* = r^2 |z|^{-2} z$$ is the image of $$z$$ under inversion with respect to the sphere $$\partial B(0, r)$$. Of course, $$v(x) = u(x)$$ in $$\Omega$$. The expression for $$v$$ follows by a standard calculation related to the Kelvin transformation in classical potential theory: one needs to verify that $$v(x) = u(x)$$ for $$x \in \partial \Omega$$ (a lengthy calculation), that $$v$$ is harmonic in the complement of $$\Omega$$, and that it decays at infinity. Since in the complement of $$\Omega$$ the function $$v$$ is simply the potential of a point mass $$q r |z|^{-1}$$ at $$z^*$$, it follows that $$v$$ is the potential of a measure with total mass $$q r |z|^{-1}$$, as desired.