 Solution

For the preceding sign convention, Green's second identity gives the harmonic representationprovided the far-boundary term vanishes or the bounded-solution representation applies. The face $x=0$ contributes nothing because $g_1=0$. On $\zeta =0$, the outward normal is $-e_{\zeta }$, so the remaining kernel is $-{\partial }_{n_Q}G={\partial }_{\zeta }G$. Differentiating the image sum givesThus the requested constants areThe Poisson kernel of a three-dimensional quarter-space is the difference of the upper-half-space kernel and its reflection in $x=0$. For bounded, suitably continuous data it produces the bounded harmonic solution, taking the prescribed traces away from the edge. Continuous data at the common edge must be compatible. Without any growth restriction at infinity, boundary values alone do not give uniqueness: a multiple of the harmonic function $xz$ can be added while leaving both face values zero.