 Solution

Green's second identity, with the outward normal derivative, isSet $u=\Phi$ and $v=G(\cdot ;{\mathbf{r}}^{\prime })$, interpreting the logarithmic singularity by a small excised disc or distribution theory. The volume integral is $\Phi ({\mathbf{r}}^{\prime })$ because $\Delta \Phi =0$ and $\Delta G=\delta$. On the boundary $G=0$ and $\Phi =f$, leavingFor the exterior problem below, apply the identity first in a large truncated region. For bounded harmonic functions the extra circle at infinity contributes zero: $G$ tends to a constant, its radial derivative is $O(r^{-2})$, and a bounded exterior harmonic function also has radial derivative $O(r^{-2})$. This is the condition under which the same boundary representation applies there.