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The supplied positive free-space kernel satisfies $-\Delta H=\delta$, so use the same sign convention for the Dirichlet Green function. For an interior source $Q=(\xi ,\eta ,\zeta )$, requireAway from $Q$, the function is harmonic, and $G-H$ is regular near $Q$. The method of images reflects the source in each coordinate plane, reversing its sign on each reflection. ThereforeOn either face, the terms cancel in pairs. Only the original source lies in the domain, so the other three terms are harmonic there; the required singularity and decay also follow. This is the Dirichlet Green function for a three-dimensional quarter-space. Using $\Delta G=\delta$ instead would change the sign of the kernel and the representation formula together, as in sign conventions for a Dirichlet Green function.

For the face $z=0$ of $x>0,z>0$, set $r_{\mp }^2=(x\mp \xi {)}^2+(y-\eta {)}^2+z^2$, with source boundary coordinates $\xi >0$, $\eta \in \mathbb{R}$. The Poisson kernel for data on that face and zero data on $x=0$ is the displayed positive difference. It follows by differentiating the Dirichlet Green function for a three-dimensional quarter-space in the inward source direction. Integrating it against bounded continuous face data gives the bounded harmonic solution away from compatibility issues at the edge. Without a boundedness or suitable infinity condition, uniqueness fails because the harmonic function $xz$ vanishes on both faces.

Its integral over this one face is $2\arctan (x/z)/\pi$, the harmonic measure of that face. To calculate it, integration in $\eta$ first gives $z\{[(x-\xi {)}^2+z^2{]}^{-1}-[(x+\xi {)}^2+z^2{]}^{-1}\}/\pi$; the remaining half-line integral is an elementary arctangent difference. This mass lies between zero and one: the other face carries the remaining boundary influence.