Consider the expression:

$$\lim\limits_{\delta \to 0} \iiint_{V-\delta} \nabla \cdot \mathbf{F}(x,y,z)\ dV \tag1$$

where $\delta$ is a small volume inside volume $V$

Now can we apply the divergence theorem and write $(1)$ as:

$$\iint_{\partial V} \mathbf{F}(x,y,z) \cdot \hat{n}\ dS +\lim\limits_{\delta \to 0} \iint_{\partial \delta} \mathbf{F}(x,y,z) \cdot \hat{n}\ dS$$

  • $\begingroup$ What does the notation $$\int_{\partial \delta}$$ even mean? What is $\delta$ here? Is it some small neighborhood around a singularity of $F$? $\endgroup$ – user7530 Jun 20 at 7:54
  • $\begingroup$ Yes, you can do it, provided the field $\mathbf{F}$ satisfies appropriate conditions and the domain $\delta$ has a sufficiently regular boundary in order to apply an appropriate version of the divergence theorem: this method is used classically (i.e. before the introduction of the theory of distributions) in order to define integrals of fields with non integrable singularities. You can see an example in this Math.SE Q&A $\endgroup$ – Daniele Tampieri Jun 20 at 7:55
  • $\begingroup$ Yes. $\delta$ is a small volume around a singularity of $\mathbf{F}$ and $\partial \delta$ is its boundary $\endgroup$ – Joe Jun 20 at 7:57

The divergence (Gauss-Green) theorem can be used to define the improper integral of the divergence of (weakly) singular vector fields $\mathbf{F}$ with isolated singular points $\mathbf{p}_o=(x_0,y_o,z_0)\in V$. Customarily, the definition goes as follows $$ \begin{split} \int\limits_{V}\nabla\cdot\mathbf{F}(x,y,z)\,\mathrm{d}V& \triangleq \lim_{R\to 0} \Bigg[\,\int\limits_{V\setminus B(\mathbf{p}_o,R)} \nabla\cdot\mathbf{F}(x,y,z)\, \mathrm{d}V - \int\limits_{\partial B(\mathbf{p}_o,R)} \mathbf{F}(x,y,z) \cdot \hat{n}\ dS\Bigg]\\ \\ &\triangleq \int\limits_{\partial V} \mathbf{F}(x,y,z) \cdot \hat{n}\ dS, \end{split}\label{1}\tag{1} $$ where, for the small volume $\delta$, a small ball $B(\mathbf{p}_o,R)$ with radius $R>0$ centered on the singular point of $\mathbf{F}$ is customarily chosen. The definition is clearly consistent if and only if the limits of the two integrals in formula \eqref{1} exist and are finite.

An example.
The most famous example of use of \eqref{1} as a definition is perhaps the calculation of the integral of the divergence of the following field: $$ \begin{split} \mathbf{F}(x,y,z)&=\nabla{\bigg[\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2 }\,\bigg]^{-1}}\\ &=\nabla\frac{1}{|\;\mathbf{p}-\mathbf{p}_o|} \end{split} $$ where $\mathbf{p}=(x,y,z)\in V$. This vector field is, apart from a multiplicative constant, the gradient of the fundamental solution of the laplacian: therefore, the integral of the divergence of this vector field is zero in every domain $V\subset\Bbb R^3\setminus\mathbf{p}_o$ since $\nabla\cdot\mathbf{F}$ is zero in such domains. However, applying \eqref{1} we have $$ \begin{split} \int\limits_{V}\nabla\cdot\mathbf{F}(\mathbf{p})\,\mathrm{d}V&= -\lim_{R\to 0} \int\limits_{ \partial B(\mathbf{p},R)} \nabla\frac{1}{|\;\mathbf{p}-\mathbf{p}_o|} \cdot\hat{n}\, \mathrm{d}S \\ &= -\lim_{R\to 0} \int\limits_{ \partial B(\mathbf{p},R)} \frac{ \partial }{\partial \hat{n}} \frac{1}{|\;\mathbf{p}-\mathbf{p}_o|} \mathrm{d}S\\ &=-\lim_{R\to 0} \int\limits_{ \partial B(\mathbf{p},R)} \frac{ \partial }{\partial r} \frac{1}{r}\mathrm{d}S\\ &=\lim_{R\to 0} \frac{1}{R^2} \int\limits_{ \partial B(\mathbf{p},R)} \mathrm{d}S = 4\pi, \end{split} $$ and thus we can also define the flux of $\mathbf{F}$ throug $\partial V$.

Final notes

  • All the above development is done under the hypothesis $\delta=B(\mathbf{p},R)$: however, formula \eqref{1} is valid for more general classes of small volumes $\delta$, provided that the singularity of $\mathbf{F}$ is sufficiently "weak".
  • Let's precise the exact meaning of the locution "weak singularity" in the context of fields with a single isolated singularity. Being the field $\mathbf{F}$ singular, we can say that, near the singular point $\mathbf{p}_o\in V$, $$ |\mathbf{F}(\mathbf{p})|\le K{|\;\mathbf{p}-\mathbf{p}_o|^{-\alpha(|\mathbf{p}-\mathbf{p}_o)|}}\label{2}\tag{2} $$ where $\alpha:\Bbb R_+\to \Bbb R_+$ is a non negative function ($\alpha\ge0$). Then we have that $$ \lim_{R\to 0}\bigg|\int\limits_{ \partial B(\mathbf{p},R)}\mathbf{F}(\mathbf{p})\cdot\hat{n}\, \mathrm{d}S \Bigg|<\infty\iff \lim_{R\to 0}R^{-\alpha(R)+2}<\infty\iff \lim_{R\to 0} \alpha(R)\le 2 $$ Be it noted that this condition is stronger than the simple condition of local integrability of the field $\mathbf{F}$: this one implies that $$ \lim_{R\to 0} \alpha(R)<3 $$ in estimate \eqref{2}, thus there are locally integrable vector fields for which formula \eqref{1} is not applicable.
  • As I stated clearly at the beginning of my answer, formula \eqref{1} is not really a divergence (Gauss-Green) theorem for singular vector fields: it is a definition which uses the standard theorem applied to non-singular regions of $F$ (by eventually cutting small volume pieces $\delta$) to extend its range of applicability to a class of singular vector fields. Therefore you will not find it stated as a theorem: however, in books on partial differential equations which do not use the theory of distributions, \eqref{1} is silently used in the proof of Green's formula. See for example Tikhonov and Samarskii [1], chapter IV, §2.1, pp. 316-318.

[1] A. N. Tikhonov and A. A. Samarskii (1990) [1963], "Equations of mathematical physics", New York: Dover Publications, pp. XVI+765 ISBN 0-486-66422-8, MR0165209, Zbl 0111.29008.

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    $\begingroup$ I think there should be a minus in the middle term of equation (1) $\endgroup$ – Joe Jun 21 at 9:28
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    $\begingroup$ @Joe I think you're right. I missed the orientation of $\hat{n}$ on $B(\mathbf{p}_o,R)$: corrected. $\endgroup$ – Daniele Tampieri Jun 21 at 9:40
  • $\begingroup$ By sufficiently weak singularity of $\mathbf{F}$, do you mean the limit of integral (with respect to $V$) of divergence of $\mathbf{F}$ exist? $\endgroup$ – Joe Jun 22 at 4:28
  • $\begingroup$ No. I mean that $$\lim_{R\to 0} \bigg|\int\limits_{ \partial B(\mathbf{p},R)} \mathbf{F}\cdot\hat{n}\, \mathrm{d}S \bigg|<\infty $$ which is a weaker condition, as you can see from the example from potential theory. $\endgroup$ – Daniele Tampieri Jun 22 at 7:36
  • $\begingroup$ I see... The inequality $$\lim_{R\to 0} \bigg|\int\limits_{ \partial B(\mathbf{p},R)} \mathbf{F}\cdot\hat{n}\, \mathrm{d}S \bigg|<\infty$$ and the statement $$\lim_{R\to 0} \int\limits_{ \partial B(\mathbf{p},R)} \mathbf{F}\cdot\hat{n}\, \mathrm{d}S$$ exist mean the same thing. (Am I correct?) $\endgroup$ – Joe Jun 22 at 7:50

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