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I'm trying to prove the following without the divergence theorem(for soon to be obvious reasons):

Let $F: \mathbb{R^3} \to \mathbb{R^3}$ be a differentiable function. In $\mathbb{R}^3$, let $V$ be a closed and path-connected subset. Then there exists a $p \in V$ such that

$$\operatorname{div} F|_{p} = \frac{1}{|V|}\iint_S F \cdot \hat{\mathbf{n}} \, \text{d}S = \color{red} {f_0}$$ Here, the divergence is defined as

$$\operatorname{div} F|_{p} = \lim_{V \to \{p\}} \frac{1}{|V|}\iint_{\partial V} F \cdot \hat{\mathbf{n}} \, \text{d}S $$

The following is my proof. Is it correct? Can it be simplified?


If there are two points $a,b$ such that $\operatorname{div} F|_a < f_0$ and $\operatorname{div} F|_b > f_0$, let $\gamma$ be a path from $a$ to $b$. Then

$$\gamma : [0,1] \to \mathbb{R}^3 \implies \operatorname{div} F|_\gamma : [0,1] \to \mathbb{R} $$ Moreover, $\operatorname{div}F|_{\gamma(0)} < f_0$ and $\operatorname{div}F|_{\gamma(1)} > f_0$, so by the Intermediate Value Theorem, there is some $c \in [0,1]$ such that $\operatorname{div} F|_\gamma(c) = f_0$.

Otherwise, for every point in $V$, without loss of generality, the divergence is $\geq f_0$. For ease, define for volume $Q$, $$\operatorname{flux}(Q) = \iint_{\partial Q} F \cdot \hat{\mathbf{n}} \, \text{d}S$$ $$f(Q) = \frac{\operatorname{flux}(Q)}{|Q|}$$ Let a plane partition a volume $V$ into two. The resulting sets are called chunks of $V$.

Partition $V = V_0$ into 2 chunks. Then pick one of the sets and call it $V_1$. Partition $V_1$ into 2 chunks as well. Repeat by partitioning into smaller chunks and picking a chunk. Partition in such a manner that the $x$, $y$, and $z$ diameters are cut in half every three steps.

By Cantor's intersection theorem $\bigcap V_n$ is non-empty, containing a single point $p$. And by the definition of divergence, the sequence $\{f(V_n)\}$ must converge.

If $f(V_n)$ is always $f_0$, then $\operatorname{div} F|_p = f_0$.

Otherwise let $V_k$ be the first $V_i$ such that $\operatorname{flux}(V_i) \neq f_0 |V_i|$. Partition $V_{k-1}$ into $V_{k}$ and $V_{k-1} \setminus V_{k}$. Note

$$ \begin{align} \operatorname{flux}(V_{k}) + \operatorname{flux}(V_{k-1} \setminus V_{k}) &= \operatorname{flux}(V_{k-1}) \\ f(V_{k})|V_{k}| + f(V_{k-1} \setminus V_{k})|V_{k-1} \setminus V_{k}| &= f_0|V_{k-1}| \\ f(V_{k})|V_{k}| + f(V_{k-1} \setminus V_{k})|V_{k-1} \setminus V_{k}| &= f_0|V_{k}| + f_0|V_{k-1} \setminus V_{k}| \end{align} $$

which implies, as $f(V_{k}) \neq f_0$, that one of $f(V_{k}), f(V_{k-1} \setminus V_{k})$ is less than $f_0$.

So pick $V_{k}$ or $V_{k-1} \setminus V_{k}$, call it $W_{1}$, such that $f(W_{1})$ is less than $f_0$(the other chunk is is greater than $f_0$). Now, given $W_n$ with $f(W_n) < 0$, we can partition $W_{n}$ into two chunks such that the $f$-values are as follows. $$ \begin{align} (<f_0) &\longrightarrow (<f_0) \text{ and } (<f_0) \\ (<f_0) &\longrightarrow (<f_0) \text{ and } (=f_0) \\ (<f_0) &\longrightarrow (<f_0) \text{ and } (>f_0) \end{align} $$ In either case, we can pick $W_{n+1}$ with $f(W_{n+1})<f_0$. Finally, we have a new sequence $\{f(W_n)\}$ bounded above by $f_0$, and by Cantor's we can set $p = \bigcap W_n$. There are no points with divergence less than $f_0$, thus $\{f(W_n)\}$ must converge to $f_0$, i.e, $\operatorname{div} F|_p$ must be $f_0$.

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  • $\begingroup$ If anything, does it work on the 1-dimensional analogue? $\endgroup$ – Jeffery Opoku-Mensah May 19 '18 at 12:49

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