(This question is motivated by electrostatics)
In electrostatics one comes across discontinuous vector fields all the time, for example, the vector field $ \vec E=\frac{\vec r}{r^3} $ which has a missing point discontinuity at $\vec r=0$, whose divergence $\nabla\cdot \vec E$ is the dirac delta distribution (to be precise, $\vec \nabla \cdot \vec E=4\pi\delta(\vec r)$, using test functions and so on). It is not clear to me how one can assert that the volume integral of this dirac delta distribution, $\vec \nabla \cdot \vec E$, is equal to the surface integral of the field, which is discontinuous.
Yet, physics textbooks still apply the Gauss Divergence theorem, to say that $\iiint_{V}\nabla\cdot\vec E d\tau=\iint_{S}\vec E\cdot\hat ndS$, where V contains the singularity $\vec r =0$, but don't explain why one is allowed to do this, when the vector field is clearly discontinuous in the domain in which the integral is being done. Alas, the integral $\iint_{S}\vec E \cdot \hat n dS= \iint_S\frac{\hat r \cdot \hat n dS}{r^2}$ comes out to be $4\pi$, but my question still holds.
Now, I am aware that the result for this case comes out to be exactly the same, $4\pi$, through different calculations, but is this a coincicende? More precicely, given a vector field $\vec E$ which is defined , continuous and differentiable everywhere except at $\vec r=0$, whose divergence $\vec \nabla \cdot \vec E$ comes out to be the dirac delta distribution, can we apply gauss divergence theorem in such a case (case where we include the origin)?
Best Answer
For any $\phi\in C_C^\infty$, the distribution $\nabla \cdot \vec E$, for $\vec E=\frac{\vec r}{r^3}$ satisfies
$$\begin{align} \langle \nabla \cdot \vec E,\phi\rangle &=-\sum_{i=1}^3 \langle E_i, \partial_i \phi \rangle\\\\ &=-\int_{\mathbb{R^3}}\vec E\cdot \nabla \phi\, d^3r\tag1\\\\ &=-\lim_{\varepsilon\to 0}\int_{\mathbb{R^3}\setminus \{\vec r||\vec r|<\varepsilon\}}\vec E\cdot \nabla \phi\, d^3r\tag2\\\\ &=\lim_{\varepsilon\to 0}\int_0^{2\pi}\int_0^\pi \phi(\vec r)\,\sin(\theta)\,d\theta\,d\phi \tag3\\\\ &=4\pi \phi(0) \end{align}$$
Therefore, in distribution $\nabla \cdot \vec E=4\pi \delta(\vec r)$.
NOTES:
In going from $(1)$ to $(2)$, we exclude the ball $|\vec r|<\varepsilon$ from the integral over $\mathbb{R^3}$ and take the limit as $\varepsilon\to 0$.
In going from $(2)$ to $(3)$, we integrate by parts using $\nabla \cdot (\phi \vec E)=\vec E\cdot \nabla \phi+\phi\nabla \cdot \vec E$, the Divergence Theorem, which applies to the region $\mathbb{R^3}\setminus \{\vec r||\vec r|<\varepsilon\}$, and exploited the fact that $\nabla \cdot \vec E=0$ for $\vec r\ne0$.