Divergence Formal Definition Meaning of $\lim_{R_{(x,y,z)} \to 0}$ of Region R

Question

A formal definition of divergence for 3 dimensions is the following,

$$\lim_{R_{(x,y,z)} \to 0}\frac{1}{|R_{(x,y,z)}|}\iint_SF\cdot\hat{n}D_\sum$$

My question is and the part I do not understand is R going to 0. What is the role of the volume (region) going to 0? What is it's purpose and reason for existing and how does it relate to the flux integral?

From my point of view, the region R seems to be an arbitrary concept that does not add any value to the formula and makes it more complex than it needs to be. I hope someone can help me understand it.

Answer 1

This definition is motivated by Gauss's theorem: $$\iiint_R \text{div}(\mathbf F)\,dV = \iint_S \mathbf F \cdot \mathbf n\,dA$$ Where $S = \partial R$ is the surface of the region $R$ and $\mathbf n$ is the unit outward normal to $S$.

Intuitively, if we think of $\mathbf F$ as being the steady-state flow of some ethereal incompressible fluid, the value of the divergence at a point $p, \text{div}(\mathbf F)\vert_p$ is the amount of fluid that is emerging at $p$ (negative if the fluid is disappearing at $p$). The theorem says that the amount of fluid emerging over the entire region $R$ is equal to the amount of fluid flowing through the surface of $R$.

If $\text{div}(\mathbf F)$ is continuous, then for each $p$ and any $\epsilon > 0$, there is a neighborhood $R_\epsilon$ of $p$ such that if $q \in R_\epsilon$, then $|\text{div}(\mathbf F)\vert_q - \text{div}(\mathbf F)\vert_p| < \epsilon$ and therefore $$(\text{div}(\mathbf F)\vert_p -\epsilon)\text{Vol}(R_\epsilon)\le \iiint_{R_\epsilon} \text{div}(\mathbf F)\,dV \le (\text{div}(\mathbf F)\vert_p +\epsilon)\text{Vol}(R_\epsilon)$$ or $$\left| \text{div}(\mathbf F)\vert_p - \dfrac 1{\text{Vol}(R_\epsilon)}\iiint_{R_\epsilon} \text{div}(\mathbf F)\,dV \right |\le \epsilon$$ Plugging in Gauss's Law: $$\left| \text{div}(\mathbf F)\vert_p - \dfrac 1{\text{Vol}(R_\epsilon)}\iint_{\partial R_\epsilon} \mathbf F \cdot \mathbf n\,dA \right |\le \epsilon$$

More generally, the same holds for any neighborhoods of $p$ smaller than $R_\epsilon$.

So the idea here is to turn this around and use it to define $\text{div}(\mathbf F)\vert_p$ in the first place:

$\text{div}(\mathbf F)\vert_p = L$ if for every $\epsilon > 0$, there is a neighborhood $R_\epsilon$ of $p$ such that for all neighborhood regions $R$ of $p$ with $R \subseteq R_\epsilon$, $$\left| L - \dfrac 1{\text{Vol}(R)}\iint_{\partial R} \mathbf F \cdot \mathbf n\,dA \right |\le \epsilon$$

or written in a limit notation: $$\text{div}(\mathbf F)\vert_p = \lim_{R \to p} \dfrac 1{\text{Vol}(R)}\iint_{\partial R}\mathbf F \cdot \mathbf n\,dA$$

The integral $\iint_{\partial R}\mathbf F \cdot \mathbf n\,dA$ depends not only on the behavior of $\mathbf F$ at $p$, but on its behavior over the entire region. By letting $R \to p$, one removes the impact of $\mathbf F$'s behavior at other points, leaving only the behavior at $p$.

This is entirely analogous to the definition of the derivative itself, where the limit is taken to turn the average rate of change over an interval into the rate of change of the function at a single point.