PROBLEM:Let $\psi(x)$ be any $\mathbf{C}^2$ function defined on all of the three-dimensional space that vanishes outside some sphere. Prove that: $$\psi(\mathbf{0})=-\int_{D}\frac{1}{|\mathbf{x}|}\Delta\psi(\mathbf{x})\frac{d\mathbf{x}}{4\pi}$$ Note the integration is taken over the region where $\psi(\mathbf{x})$ is not $0.$

How can we show this? Can anyone help me please?

My attempt: So I do not understand why this was put on hold? I just needed alittle hint, to help me solve it. I seemed to have attempted the problem all day yesterday and I managed to solve it. It was actually nothing more than the application of Green's second identity in PDE's.

If we first define a region $D$ such that $\text{supp} (\psi)\subset D$, then let $D_{\epsilon} = D\setminus B(\mathbf{0}, \epsilon),$ where we can consider some $\epsilon$ small enough such that $\partial B(\mathbf{0},\epsilon)\cap \partial D =\emptyset$. We now apply Green's Second Identity, where $g(\mathbf{x}) = \frac{1}{|\mathbf{x}|},$

Green's Second Identity: $$\iiint_D (u\Delta v - v \Delta u )dV=\iint_{\partial D}\bigg(u\frac{\partial v}{\partial n}-v\frac{\partial u}{\partial n}\bigg)dS$$

$$\int_{D_{\epsilon}} (g\Delta \psi - \psi \Delta g )d\mathbf{x}=\int_{\partial D_{\epsilon}}\bigg(g\frac{\partial \psi}{\partial n}-\psi\frac{\partial g}{\partial n}\bigg)dS$$ From here it is not hard to make further simplifications of the ABOVE INTEGRAL:

$$\int_{D_{\epsilon}} g\Delta \psi d\mathbf{x}=\int_{\partial D_{\epsilon}}\bigg(g\frac{\partial \psi}{\partial n}-\psi\frac{\partial g}{\partial n}\bigg)dS$$ $$=-\frac{1}{\epsilon}\int_{r=\epsilon}\frac{\partial \psi}{\partial r}dS-\frac{1}{\epsilon^2}\int_{r=\epsilon}\psi dS$$ $$=-4\pi\frac{\partial \overline{\psi}}{\partial r}-4\pi \overline{\psi}$$ $$\rightarrow -4\pi \psi(\mathbf{0})$$ $$\therefore \psi(\mathbf{0})=-\int_{D}\frac{1}{|\mathbf{x}|}\Delta\psi(\mathbf{x})\frac{d\mathbf{x}}{4\pi}$$

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    $\begingroup$ Can anyone explain why this was put on hold? $\endgroup$ – Aurora Borealis Jun 9 '18 at 3:15

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