Suppose $U$ is a domain in $\mathbb R^n$ containing $0$, and let $\Phi$ be the fundamental solution to the Laplace PDE $\Delta u=0$. What is the value of $\int_{\partial U} \dfrac{\partial \Phi(y)}{\partial \nu}dA(y)$? Here $\nu$ is the outward normal at $y\in\partial U$ and $dA$ is the surface area element.
By Green formula, this should be $\int_U \Delta \Phi(y)dy$ (here $dy$ is the usual volume element in $\mathbb R^n$). But $\Phi$ is not in $C^2(U)$, in particular it diverges at $0$. So I was wondering if this holds still. Since $\Phi$ satisfied $\Delta \Phi(y)=0$ for $y\neq0$, I want to say the answer is 0, but I am not sure.
Best Answer
Greens identity (according to Wikipedia) states that if $\Phi(x,y) = -1/4\pi\|x-y\|$ (the fundamental solution) and $\psi$ twice continuously differentiable in $U$ that $$ \int_U \Phi(x,y) \Delta \psi(x) dV_x - \psi(y) = \int_{\partial U}\left( \Phi(x,y)\frac{\partial \psi}{\partial y}(x)-\psi(x)\frac{\partial \Phi(x,y)}{\partial y} \right)dA(x) $$ Now taking $\psi=1$ which is $C^2$ and harmonic (i.e. $\Delta \psi = 0$) we deduce $$1 = \int_{\partial U} \frac{\partial \Phi(x,y)}{\partial y}dA(x)$$
You have to be aware that if $\Psi$ is the fundamental solution to the Laplace equation then, by definition, $\Delta \Psi(x) = \delta(x)$, i.e. the Dirac delta. This is a distribution defined by $\delta[f] := f(0)$ for $f$ a suitable test function. So if you look at $\int_U \Delta \Phi(y)dy$ you actually look at $\delta$ being applied to the function $\psi=1$ on $U$ (which only works if $U$ is bounded by the way). And naturally $\delta[\psi] = \psi(0) =1$.