Let $D \subset \mathbb{R}^3, \textbf{x} \in \textrm{ and }\partial D, \textbf{x$_o$} \in D$, where $\partial D$ denotes the boundary of $D$. Suppose that there exists a Green's Function for the operator $\Delta$ that satisfies the following properties.

  1. $\Delta G(\textbf{x}, \textbf{x$_o$}) = 0$ in $D$ where $\textbf{x} \neq \textbf{x$_o$}$
  2. $\dfrac{\partial G}{\partial n} = 0 \; \forall \; \textbf{x} \in \partial D$
  3. $G - \dfrac{1}{4\pi|\textbf{x} - \textbf{x$_o$}|}$ is finite

My question is, would it be fair to assume that $\Delta G$ is like a "delta function"? If so, does that imply $$\displaystyle \int_{D}{\Delta G(\textbf{x}, \textbf{x$_o$}) \phi(\textbf{x}) \; d\textbf{x}} = \phi(\textbf{x$_o$})$$


Yes, you can even calculate it rigorously, that $\Delta G(x,x_o)$ acts like a delta function. From property 3, you get that $G \approx \frac{1}{C(n)V(B_1(0))}\frac{1}{|x-x_0|}$. Splitting the integral $$ \int_D \Delta G(x,x_0) \phi(x)dx $$ over "critical" (close to $x_0$, say $B_{\epsilon}(x_0)$) and "non-critical" away from $x_0$ (e.g. D \ $B_{\epsilon}(x_0)$ ) and integrating by parts will give you the result, as you can computer derivatives of $\partial_{x_i}G(x,x_0)$ explicitly and $\phi$ is usually sufficently smooth and/or bounded.
If you look for the full proof, I suggest "Partial Differential Equations" by L.C. Evans. It can be found in the first chapter and is usually part of every university library.

  • $\begingroup$ That's how I thought I should proceed, but I didn't try the integration by parts one. Thank you! $\endgroup$ – Simon Feb 1 '17 at 23:52
  • $\begingroup$ In all fairness, I should warn you. I looked up the proof and its quite long. The most important "trick" is that you see that the derivative of G is associated to the unit normal of a ball and the factor in front is the "volume of the surface" of a ball. $\endgroup$ – F. Conrad Feb 2 '17 at 13:00

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