[대학원-전자기학]문제풀이 델타함수와 발산

Problem Prove that $$\nabla^2 (\frac{1}{r}) = -4\pi \delta(\vec{x})$$ where $r=|\vec{x}|$
Answer

$$\begin{aligned} \nabla^2 \frac{1}{r} = \nabla \cdot (\nabla \frac{1}{r}) = \nabla \cdot (-\frac{1}{r^2}\hat{r}) = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \left( \frac{-1}{r^2}\right) \right) \nonumber\end{aligned}$$
Hence, the intergral of $\nabla^2 \frac{1}{r} = 0$
$$\int_v \nabla^2 \frac{1}{r} dv = 0$$
However, when we use the divergence theorem, we get different result.
$$\begin{aligned} \int_v \nabla^2 \frac{1}{r} dv &= \oint_s -\frac{1}{r^2}\hat{r} d\vec{a} \nonumber \\ &=\oint_s - \frac{1}{r^2} r^2\sin\theta d\theta d\phi \nonumber \\ &= -4\pi \nonumber\end{aligned}$$
This problem arise in the origin. Hence, we introduce delta function to solve this problem.
$$\nabla^2 (\frac{1}{r}) = -4\pi \delta(\vec{x})$$
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