Dirichlet Green function for a three-dimensional quarter-space ($G(P,Q)={\sum }_{i,j=0}^1(-1{)}^{i+j}/(4\pi |P-Q_{ij}|)$)

For $D=\{x>0,z>0\}\subset {\mathbb{R}}^3$, take the Dirichlet Green function convention $-{\Delta }_{P}G(P,Q)={\delta }_Q(P)$. Reflect $Q=(\xi ,\eta ,\zeta )$ to $Q_{ij}=((-1{)}^i\xi ,\eta ,(-1{)}^j\zeta )$. The displayed image sum vanishes on both faces, has exactly one positive unit source inside $D$, and decays at infinity. Pairwise cancellation proves the boundary conditions, while the three image terms are harmonic inside $D$. Its inward source-normal derivative gives the Poisson kernel of a three-dimensional quarter-space. The source sign must be kept consistent with Green's second identity.

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.