Is the Gauss Divergence theorem applicable when the Divergence of a vector field is a dirac delta function?

Question

(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)?

Answer 1

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 $\vec r\ne0$.

Answer 2

Let $\Omega \subset \mathbb{R}^n$ be a bounded region and $\mathbf{F}$ be a distribution-valued vector field which is $C^\infty$ on $\partial\Omega$. Then the product $(\nabla\cdot\mathbf{F})\chi_\Omega,$ where $\chi_\Omega$ is the indicator function of $\Omega,$ is defined as a distribution. Since this product has compact support (as $\Omega$ is bounded) the action of it on $\mathbf{1},$ the function that has value $1$ everywhere, is defined: $\langle (\nabla\cdot\mathbf{F})\chi_\Omega, \mathbf{1} \rangle.$ If $\mathbf{F}$ is smooth, this is exactly $\int_\Omega (\nabla\cdot\mathbf{F}) \, dV,$ where $V$ is the volume measure of $\mathbb{R}^n.$

Now, by the product law, $ (\nabla\cdot\mathbf{F})\chi_\Omega = \nabla\cdot(\mathbf{F}\,\chi_\Omega) - \mathbf{F}\cdot\nabla\chi_\Omega , $ so $$ \langle (\nabla\cdot\mathbf{F})\chi_\Omega, \mathbf{1} \rangle = \langle \nabla\cdot(\mathbf{F}\,\chi_\Omega), \mathbf{1} \rangle - \langle \mathbf{F}\cdot\nabla\chi_\Omega, \mathbf{1} \rangle . $$ Here the first term on the right hand side vanishes since, by definition of derivatives of distributions, the derivative is moved to the test function, where here is $\mathbf{1},$ and $\nabla\mathbf{1}=0.$ Thus we have $$ \langle (\nabla\cdot\mathbf{F})\chi_\Omega, \mathbf{1} \rangle = - \langle \mathbf{F}\cdot\nabla\chi_\Omega, \mathbf{1} \rangle . $$ But $\nabla\chi_\Omega$ only has support on $\partial\Omega$ since $\chi_\Omega$ is constant on both the interior and the exterior of $\Omega$ so the gradient vanishes there. Also, the gradient points in the direction in which the function grows fastest, which is into $\Omega,$ i.e. in the negative direction of the normal. We can therefore identify the right hand side as $\oint_{\partial\Omega} \mathbf{F} \cdot \mathbf{n} \, dS,$ where $\mathbf{n}$ is the outwards pointing normal and $S$ is the surface measure.

Thus, writing $\int_\Omega (\nabla\cdot\mathbf{F}) \, dV$ for $\langle (\nabla\cdot\mathbf{F})\chi_\Omega, \mathbf{1} \rangle$ and $\oint_{\partial\Omega} \mathbf{F}\cdot\mathbf{n}\,dS$ for $\langle \mathbf{F}\cdot(-\nabla\chi_\Omega), \mathbf{1} \rangle$ we have $$ \int_\Omega (\nabla\cdot\mathbf{F}) \, dV = \oint_{\partial\Omega} \mathbf{F}\cdot\mathbf{n}\,dS $$ also for distribution-valued vector fields under some assumptions.