solution to wave equation

Question

I have this differential equation from magnetic vector potential analysis
$$\nabla^2 G + \beta^2 G = \delta(r) $$ and here is its solution according to the textbook $$G = - \frac{e^{-j\beta r}}{4\pi r}$$ what i really don't understand is where the constant $ -1/4\pi $ came from ?

Answer 1

Note that with $G=C\frac{e^{-j\beta r}}{r}$, heuristically (or in distribution) we have for any $R>0$

$$\begin{align} \int_{|\vec r|\le R}\left(\nabla^2 G+\beta^2G\right)\,dV&= \oint_{|\vec r|=R}\nabla G\cdot \hat n\,\,dS+\beta^2\,\int_{|\vec r|\le R} G\,dV\\\\ &=\int_0^{2\pi}\int_0^\pi C \left(-j\beta\, \frac{e^{-j\beta R}}{\epsilon}-\frac{e^{-j\beta R}}{R^2}\right)\,R^2\,\sin(\theta)\,d\theta\,d\phi\\\\ &+\beta^2 \,\int_0^R\int_0^{2\pi}\int_0^\pi C\,\frac{e^{-j\beta r}}{r}\,r^2\,\sin(\theta)\,d\theta\,d\phi\\\\ &=-4\pi C\tag 1 \end{align}$$

And in distribution we have $$\int_{|\vec r|\le R}\,\delta(r)\,dV=1 \tag 2$$

Equating $(1)$ and $(2)$ and solving for $C$ yields

$$C=-\frac{1}{4\pi}$$

as was to be shown!