Definition of divergence

We see in certain places that $$\operatorname{div} \mathbf{F}|_p = \lim_{V \to \{p\}} \iint_{\partial V} \frac{\mathbf{F} \cdot \mathbf{\hat n}}{|V|} \: dS.$$

But what is $V$? Neighborhoods of $p$? Balls around $p$? How can you take a limit that is not as one number approaches another? What is the actual rigorous definition of this limit, and furthermore, divergence?

I am already familiar with the $\nabla \cdot F$ definition.

• $V$ is a volume, which could be a ball or any other continuous shape, since all that matters is that it can be deformed to a single point $p$. – David G. Stork Sep 17 '18 at 22:14
• How do you rigorously define this limit (you clearly can't do epsilon-delta right?)? Define "deformed?" Edit: I'll take it to be it is homotopic to a point. – Jeffery Opoku-Mensah Sep 17 '18 at 22:15
• Yes... homotopic. – David G. Stork Sep 17 '18 at 22:19
• @DavidG.Stork If you take e.g. a cube instead of a ball you have a "continuous shape" but it is impossible to define the outward unit normal vector field, so you don't know how to compute the flux. You do need some degree of regularity: see my answer below. – giobrach Sep 18 '18 at 0:09

Let's put it this way. Suppose you have defined the divergence as a differential operator: that is, for a general function $\mathbf f : A \to \mathbb R^n$, where $A$ is a subset of $\mathbb R^n$, and for $\mathbf x_0 \in A \setminus \partial A$ at which $\mathbf f$ is differentiable, let the divergence of $\mathbf f$ at $\mathbf x_0$ be defined as the real number $$\operatorname{div} \mathbf f (\mathbf x_0) := \operatorname{tr} J\mathbf f(\mathbf x_0) = \sum_{i=1}^n \frac{\partial f_i}{\partial x_i}(\mathbf x_0),$$ where $J\mathbf f(\mathbf x_0)$ is the Jacobian matrix of $\mathbf f$ at $\mathbf x_0$ and $\operatorname{tr}$ indicates the trace operator.

Then the following theorem holds:

Theorem. Let $\Omega$ be an open subset of $\mathbb R^n$, and let $\mathbf f : \Omega \to \mathbb R^n$ be of class $C^1$. Suppose furthermore that $\mathbf x_0 \in \Omega$, and $\{A_k\}_{k \in \mathbb N}$ is a sequence of subsets of $\Omega$ such that

1. For all $k$, $A_k$ is a regular open set (see below);
2. For all $k$, $A_k$ contains the point $\mathbf x_0$;
3. For all $\varepsilon>0$ there is an index $k \in \mathbb N$ such that $\operatorname{diam} A_k < \varepsilon$ — or, equivalently, $$\lim_{k\to\infty} \operatorname{diam}A_k = 0.$$

Then, if $\hat{\mathbf n}_k : \partial A_k \to \mathbb R^n$ is the function associating, to each point of $\partial A_k$, the unit normal vector pointing outward w.r.t. $A_k$, $$\operatorname{div} \mathbf f(\mathbf x_0) = \lim_{k\to \infty} \frac{1}{\operatorname{vol}_n A_k } \int_{\partial A_k} \mathbf f \cdot \hat{\mathbf n}_k \mathop{}\!da.$$

By $\operatorname{diam}A_k$ we mean the diameter of the set $A_k$, i.e. the greatest possible distance between two points in the closure of $A_k$: $$\operatorname{diam} A_k := \sup_{\mathbf x, \mathbf y \in A_k} |\mathbf x-\mathbf y|;$$ also by $\operatorname{vol}_n$ we mean the $n$-dimensional standard measure associated with $\mathbb R^n$. For $n = 3$ we're dealing with the usual notion volume.

By regular open set we mean that $A_k$ should have a "nice" border, i.e. that $\partial A_k$ should be a $(n-1)$-dimensional compact manifold of certain regularity (otherwise we will have problems defining the outward unit normal function $\hat{\mathbf n}_k$). More precisely, for an open set $U \subseteq \mathbb R^n$ to be regular, we require that it be bounded, and that for all $\mathbf x_0 \in \partial U$ there exist a neighborhood $I$ of $\mathbf x_0$, a neighborhood $J$ of the origin of $\mathbb R^n$, and a bijection $\mathbf r : J \to I$ of class at least $C^1$ such that

1. $\mathbf r^{-1} : I \to J$ is of the same regularity as $\mathbf r$;
2. $I \cap U = \mathbf r(J \cap \mathbb R^n_+)$;
3. $I \cap \partial U = \mathbf r(J \cap \partial \mathbb R^n_+)$;
4. $I \setminus \overline{U} = \mathbf r(J \setminus \overline{\mathbb R^n_+})$.

Of course, by $\mathbb R^n_+$ we mean the (open) region of $\mathbb R^n$ where $x^n > 0$ (or for that matter any other one of the coordinates in place of $x^n$); this is also called the $n$-dimensional upper half-space.

There is also a useful characterization of these regular open sets:

Proposition. A bounded open set $U \subset \mathbb R^n$ is regular if and only if there exists a function $g : \mathbb R^n \to \mathbb R$ that is at least of class $C^1$, that verifies $$\begin{split} U &= \{\mathbf x \in \mathbb R^n\ |\ g(\mathbf x) < 0 \}, \\ \partial U &= \{ \mathbf x \in \mathbb R^n\ |\ g(\mathbf x) = 0\}, \\ \mathbb R^n \setminus \overline U &= \{\mathbf x \in \mathbb R^n\ |\ g(\mathbf x) > 0\}, \end{split}$$ and is such that $\nabla g(\mathbf x) \neq \mathbf 0$ for all $\mathbf x \in \partial U$.

Of course, that famous theorem which states that contour lines of regular functions can be locally parametrized by maps of the same regularity prevents the border $\partial U$ from having any "sharp edges" or "corners" or other similar singular portions. That is why smooth balls are usually employed as the prototypical choice of $A_k$.

As an exercise: try to rewrite the statement of the theorem above by replacing the (discrete) sequence $\{A_k\}_{k\in\mathbb N}$ with the (continuous) family of open balls $\{B_{\varepsilon}(\mathbf x_0)\}_{\varepsilon > 0}$. (Spoiler: it becomes much simpler)

• Update: I have added a proof of the Theorem in the special case of balls, that does not use the divergence theorem, here (it was too long to be posted here) – synmath.blogspot.com/2018/09/… – giobrach Sep 21 '18 at 23:02
• I would be very pleased if you update the link showing the proof or give some references to check this theorem! I am very interested! – user55268 Apr 4 at 21:15
• @MdW Here is the new link: deltapegasi.blogspot.com/2018/09/… – giobrach Apr 5 at 16:40