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I looked around the Web for a proof of what I have heard called "the Divergence theorem". However, it seems this name implies three-dimensionality, as all I could find was a series of proofs like this which work in $\mathbb{R}^3$. What I would like a proof of is:

Divergence Theorem

If $\Omega\subseteq\mathbb{R}^n$ is a bounded $\mathcal{C}^1$ domain and $\underline w\in\mathbb{C}^1(\overline\Omega)$ is a field, then:

$$\int\limits_\Omega\operatorname{div}\underline wdV=\int\limits_{\partial\Omega}\underline w\cdot\nu d\sigma,$$

$\nu$ being the outward unit normal and $d\sigma$ the surface measure.

While we are at it, it would be nice to have an idea of where this surface measure comes from. I have been told a $\mathcal{C}^k$ hypersurface is a hypersurface locally described as the locus of zeros of a function $F\in\mathcal{C}^k(\mathbb{R}^n,\mathbb{R})$ with everywhere nonzero gradient, and that thus locally a derivative of $F$, say the $j$th, is always nonzero so by the implicit function theorem we can represent $x_j=\phi(x_1,\dotsc,x_{j-1},x_{j+1},\dotsc,x_n)$ and have the hypersurface as the locus where $x_j=\phi(\text{the other coordinates})$, and that with that the surface measure is $\sqrt{1+|\nabla\phi(x)|^2}dx_1\dotsc dx_{n-1}$. How is this expression found?

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There are two generalizations that imply your result. One is that of Green's identities, and the other is the generalized Stokes' theorem. I would advise that you look for a proof of the first of Green's identities.

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  • $\begingroup$ Actually, we did Green's identities in class as a corollary of this result :). Generalized Stokes theorem... let me see. $\endgroup$ – MickG Oct 9 '15 at 18:48
  • $\begingroup$ I think you can prove it as purely a consequence of integration by parts $\endgroup$ – Omnomnomnom Oct 9 '15 at 18:50
  • $\begingroup$ Another corollary of it in our class treatment :). I mean in higher dimensions we deduced ibp from this result. $\endgroup$ – MickG Oct 9 '15 at 18:52
  • $\begingroup$ I mean that you could prove Green's as a consequence of one-dimensional integration by parts. I could be wrong, though. $\endgroup$ – Omnomnomnom Oct 9 '15 at 18:54
  • $\begingroup$ Oh. I can't quite see how. What are you thinking? $\endgroup$ – MickG Oct 9 '15 at 18:56
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In the end, I decided to look for a reference, and I found:

enter image description here

On pp. 123-132, the book proves integration by parts and deduces the Divergence theorem from it. Since it looks very similar to the proof I saw for the 2- and 3-dimensional case in Analysis two (the only difference is having more variables and hence more $dx$'s and more derivatives), I think I will sooner or later adapt that proof into one written over here to have it available.

Update

One way to deduce it from other results is using Stokes' theorem (the one with the exterior derivatives, not the one with the integral of the curl). Said theorem states:

$$\int_U\mathrm{d}\omega=\int_{\partial U}\omega.$$

Let us find a form such that:

$$\mathrm{d}\omega=\nabla\cdot F\mathrm{d}V_{n+1},$$

where $F$ is a field on $\mathbb{R}^{n+1}$ and $\mathrm{d}V_{n+1}$ is the canonical volume form on $\mathbb{R}^{n+1}$. It is easily seen that this gives:

$$\omega=\sum_i(-1)^{i-1}F_i\ast(\mathrm{d}x_i),$$

where $\ast(\mathrm{d}x_i)$ is $\mathrm{d}V$ with $\mathrm{d}x_i$ removed. So the LHS is easy. The RHS, on the other hand, needs a pullback via a chart. The hypersurface is $\mathcal{C}^k$, so we have a $\mathcal{C}^k$ chart $G(x)=(x,\phi(x))$. I didn't calculate the pullback in general, but I did a calculation for $n+1=3$. In that case, we have:

$$\omega=F_1\mathrm{d}x_2\wedge\mathrm{d}x_3-F_2\mathrm{d}x_1\wedge\mathrm{d}x_3+F_3\mathrm{d}x_1\wedge\mathrm{d}x_3.$$

Before pulling back $\omega$, we pull back $\mathrm{d}x_i$. Since the first two coordinates of $\mathrm{d}G$ are the identity, $\mathrm{d}x_1$ and $\mathrm{d}x_2$ are pulled back to "themselves". Instead, $\mathrm{d}x_3$ is pulled back to $\mathrm{d}\phi$. Now we use the linearity and distributivity over wedges of the pullback:

\begin{align*} G^\ast\omega={}&F_1G^\ast(\mathrm{d}x_2)\wedge G^\ast(\mathrm{d}x_3)-F_2G^\ast(\mathrm{d}x_1)\wedge G^\ast(\mathrm{d}x_3)+F_3G^\ast(\mathrm{d}x_1)\wedge G^\ast(\mathrm{d}x_2)={} \\ {}={}&F_1\mathrm{d}x_2\wedge\mathrm{d}\phi-F_2\mathrm{d}x_1\wedge\mathrm{d}\phi+F_3\mathrm{d}x_1\wedge\mathrm{d}x_2={} \\ {}={}&F_1\mathrm{d}x_2\wedge(\partial_1\phi\mathrm{d}x_1+\partial_2\phi\mathrm{d}x_2)-F_2\mathrm{d}x_1\wedge(\partial_1\mathrm{d}x_1+\partial_2\mathrm{d}x_2)+F_3\mathrm{d}x_1\wedge\mathrm{d}x_2={} \\ {}\underset{\mathrm{d}V':=\mathrm{d}x_1\wedge\mathrm{d}x_2}{=}{}&(-F_1\partial_1\phi-F_2\partial_2\phi+F_3)\mathrm{d}V'={} \\ {}={}&(-\nabla\phi,1)\cdot F\mathrm{d}V'. \end{align*}

So trusting this calculation generalizes to the same formula in arbitrary dimensions, we get exactly the theorem desired.

While we are at it, let us see where the surface measure come from. The orientation on the boundary $\partial U$ is the interior product of the canonical volume form $\mathrm{d}V=\bigwedge_1^{n+1}\mathrm{d}x_i$ and the normal field $\nu=(-\nabla\phi,1)$. Let us compute this:

$$\iota_\nu\mathrm{d}V=\sum(-1)^{i-1}N_i\mathrm{d}V_{\widehat{\mathrm{d}x_i}}=\sum_{i=1}^n(-1)^i\partial_i\phi\mathrm{d}V_{\widehat{\mathrm{d}x_i}}+(-1)^n\mathrm{d}V_{\widehat{\mathrm{d}x_{n+1}}}.$$

Let us, once again, set $n=2$. Then:

$$\iota_\nu\mathrm{d}V=-\partial_1\phi\mathrm{d}x_2\wedge\mathrm{d}x_3+\partial_2\phi\mathrm{d}x_1\wedge\mathrm{d}x_3+\mathrm{d}x_1\wedge\mathrm{d}x_2.$$

Pull it back:

\begin{align*} G^\ast(\iota_\nu\mathrm{d}V)={}&-\partial_1\phi\mathrm{d}x_2\wedge\mathrm{d}\phi+\partial_2\phi\mathrm{d}x_1\wedge\mathrm{d}\phi+\mathrm{d}x_1\wedge\mathrm{d}x_2={} \\ {}={}&(\partial_1\phi)^2\mathrm{d}V+(\partial_2\phi)^2\mathrm{d}V'+\mathrm{d}V'={} \\ {}={}&(\nabla\phi\cdot\nabla\phi+1)\mathrm{d}V', \end{align*}

where $\mathrm{d}V'$ is as defined above. Oh. The square root is not present. But then again, we were supposed to normalize the normal before $\iota_\nu$-ing, so the normalizing factor makes the square root appear again. Fantastic!

Thanks for the suggestion, Omnomnomnom (is that the right number of "nom"s :)?).

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