Can we just set it?

Question

Suppose that we have $G(\overline{x},\overline{y})=-\frac{1}{4 \pi} \frac{1}{||\overline{x}-\overline{y}||}$ for $\overline{x}, \overline{y} \in \mathbb{R}^3$.

I want to calculate $\Delta{G(\overline{x}, \overline{y})}$.

I have found that $\frac{\partial^2}{\partial{x_1^2}} \left( -\frac{1}{4 \pi} \frac{1}{\sqrt{(x_1-y_1)^2+(x_2-y_2)^2+(x_3-y_3)^2}}\right)=\frac{1}{4 \pi} \frac{(x_2-y_2)^2+(x_3-y_3)^2-2(x_1-y_1)^2}{((x_1-y_1)^2+(x_2-y_2)^2+(x_3-y_3)^2)^{\frac{5}{2}}}$

Making the corresponding for $x_2$ and $x_3$ we get that $\Delta{G(\overline{x}, \overline{y})}=0$ for $\overline{x} \neq \overline{y}$.

Can we just say that $\Delta{G(\overline{x}, \overline{y})}=\delta(\overline{x}-\overline{y})$ if $\overline{x}=\overline{y}$ or do we have to show something in order to set that?

Answer 1

From what you have shown above, we know that $\Delta G$ is a distribution supported at the origin. To show that $\Delta G(x,y)=\delta(x-y)$, we can use its homogeneity of $-3$, $\frac1{\|x\|}$ has homogeneity $-1$ and $\Delta$ reduces homogeneity by $2$, and the Divergence Theorem: $$ \begin{align} \int_{B(0,r)}\Delta G(x,0)\,\mathrm{d}x &=\int_{\partial B(0,r)}n\cdot\nabla G(x,0)\,\mathrm{d}\sigma\\ &=\int_{\partial B(0,r)}\frac x{\|x\|}\cdot\frac x{4\pi\|x\|^3}\,\mathrm{d}\sigma\\ &=\int_{\partial B(0,r)}\frac1{4\pi\|x\|^2}\,\mathrm{d}\sigma\\[6pt] &=1\tag{1} \end{align} $$

Answer 2

First fix $y=0$. We can recover the general case by setting $x=\bar{x}-\bar{y}$. Now $\Delta G(x,0) = \delta(x)$ by definition means that $\int_{\mathbb{R}^3}G(x)\Delta\phi(x)dx = \phi(0)$ for any smooth compactly supported function $\phi$. This is by the definition of the distribution derivative (weak derivative).

We know $\int_{\mathbb{R}^3}G(x)\Delta\phi(x)dx =\int_{B_R(0)}G(x)\Delta\phi(x)dx +\int_{B_R(0)^C}G(x)\Delta\phi(x)dx =\int_{B_R(0)}G(x)\Delta\phi(x)dx.$ This is because $\int_{B_R(0)^C}G(x)\Delta\phi(x)dx=\int_{B_R(0)^C}\Delta G(x)\phi(x)dx=0$, using integration by parts (which we can do because $G$ is differentiable away from the origin).

At this point you can evaluate $\int_{B_R(0)}G(x)\Delta\phi(x)dx$ by using some form of the divergence theorem as robjohn shows.