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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 ?

Dylan
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Moaaz Sherif
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    The number $4\pi$ is the surface of the unit sphere. Usually that's where it comes from. (Not a bad question, BTW. +1) – Giuseppe Negro Nov 02 '17 at 18:36
  • See for example this derivation: https://math.stackexchange.com/questions/2076653/greens-function-of-laplace-operator/2076654 (for the case $\beta = 0$, but the origin of $-1/4\pi$ remains the same). If you do the derivation for a general dimension $D$ then the prefactor turns out to be $-\frac{1}{S_{D-1}}$ where $S_D$ is the surface area of the unit $D$-sphere. – Winther Nov 02 '17 at 18:43

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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!

Mark Viola
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  • does this way holds if we assume that the sphere radius would tend to infinity (or any other value) ,instead of zero ?, i mean that epsilon tends to infinity .... – Moaaz Sherif Nov 02 '17 at 20:00
  • Yes, indeed. I've edited to show the equality is true regardless of the size to the sphere over which we integrate. In fact, it is easy to show that we need not choose the integration region to be a sphere. Any sufficiently smooth region suffices – Mark Viola Nov 02 '17 at 20:02
  • my final question.. how did we know that if we integrate over sphere we can know the integration constant c .. i mean we usually need a initial or boundary condition to determine the value of c how could we determine c without a boundary condition ? – Moaaz Sherif Nov 02 '17 at 20:04
  • One can reformulate the problem without appealing to the Dirac Delta. In that case, we have the PDE $(\nabla +\beta^2) G=0$ for $r\ne 0$ and the boundary conditions that the waves are outwardly propagating and $\left.\left (r^2\frac{\partial G}{\partial r})\right)\right|_{r= 0}=-1/4\pi$ – Mark Viola Nov 02 '17 at 20:12
  • can we use the Gauss theorem of divergence to reach the same result as your proof ? – Moaaz Sherif Nov 02 '17 at 20:28
  • It was used to convert the volume integral of $\nabla^2 G$ to the surface integral of $\hat n\cdot \nabla G$. – Mark Viola Nov 02 '17 at 20:35
  • And you're welcome. It was my pleasure. – Mark Viola Nov 02 '17 at 20:35
  • sorry so much sir .. but i have a question does this method is applicable to all linear second order PDEs ? the method of using volume integration to determine the constants... if so why don't textbooks use to solve the wave equation in its various applications ? – Moaaz Sherif Nov 09 '17 at 05:36