Delta function in 3D

**isatis55** · November 8th 2012, 02:48 AM

Hi,I need to prove the following relation $\nabla^{2} = −4π\delta (R)$ where $R=(x-y), x,y \in R^3$ and $\nabla^{2}$ is taken over x_{i} and $\delta(R)$ is the 3D delta function (=1 iff x=y)Thanks!Isatis

**chiro** · November 8th 2012, 08:44 PM

Hey isatis55.

Can you please define specifically the definition of the 3D delta function given <x,y,z> in R^3 where you map <x,y,z> -> w?

**zzephod** · November 8th 2012, 10:18 PM

Originally Posted by chiro

Hey isatis55.

Can you please define specifically the definition of the 3D delta function given <x,y,z> in R^3 where you map <x,y,z> -> w?

3D delta functional with the usual abuse of notation, for any function on $\mathbb{R}^3$ continuous at $\bf{0}$ :

$\delta(f) = \int_{\mathbb{R}^3} f({\bf{x}}) \delta({\bf{x}}) d{\bf{x}}=f({\bf{0})$

(the first and last terms above are the definition, the middle term is a common suggestive abuse of notation.)

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