 Solution

For fixed source $(\xi ,\eta )\in D$, define the Dirichlet Green function by $-{\Delta }_{x,y}G={\delta }_{(\xi ,\eta )}$ in $D$, $G=0$ on $\partial D$. It is harmonic away from the source and has singularity $-(2\pi {)}^{-1}\log \sqrt{(x-\xi {)}^2+(y-\eta {)}^2}$. The outward normal vector derivative is ${\partial }_{n}G=\mathbf{n}\cdot {\nabla }_{x,y}G$ at a boundary point of arc-length parameter $s$.

Green's second identity gives ${\int }_D(u\Delta G-G\Delta u)={\int }_{\partial D}(u{\partial }_{n}G-G{\partial }_{n}u)ds$. Substituting the two PDEs and the boundary data yieldsThe singular version follows by excising a small source disk and taking its radius to zero, or directly by the distributional identity.

For the upper half-plane use the opposite-sign image source at $(\xi ,-\eta )$:The image lies outside the domain, so it is harmonic inside; the true source has the required unit singularity, and the two distances agree at $y=0$, giving zero boundary value. On that boundary $\mathbf{n}=-{\mathbf{e}}_y$, soThe assumed far-field decay makes the large semicircle contribution vanish. With $f=0$, the representation becomesFinally set $s=\xi +\eta \tan \theta$, $ds=\eta {\sec }^2\theta d\theta$. The Jacobian matrix cancels the denominator, givingIf $g$ is even, the substitution $s\mapsto -s$ in the Poisson integral gives $u(-\xi ,\eta )=u(\xi ,\eta )$. At $\eta =0$ this agrees with the continuous even boundary trace.