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In elementary calculus texts, Green's theorem is proved for regions enclosed by piecewise smooth, simple closed curves (and by extension, finite unions of such regions), including regions that are not simply connected.

Can Green's theorem be further generalized? In particular, are there regions on which Green's theorem definitely does not hold?

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    $\begingroup$ I'll concede that the phrasing of the question is a little naive, and from context I would guess that the author is not a mathematical researcher, but nevertheless a question like "how far can you push Green's theorem using state of the art geometric measure theory" probably is a good MO question. Then again it might require a little more care to formulate that question well. $\endgroup$ Apr 26, 2020 at 16:42
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    $\begingroup$ Thank you both for the advice. $\endgroup$
    – carl
    Apr 26, 2020 at 18:48
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    $\begingroup$ I have attempted to rewrite the question and have nominated it for reopening. $\endgroup$ Apr 26, 2020 at 23:08
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    $\begingroup$ @YCor: The latest title edit is ungrammatical. $\endgroup$ Apr 27, 2020 at 19:08
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    $\begingroup$ @R..GitHubSTOPHELPINGICE your last comment is a bit of an overstatement — “On which regions can’t Green’s theorem be applied?” is perfectly grammatical (to my native BrE-speaker ear). But I agree the original phrasing with “can … not” better, emphasising the “not” more clearly. $\endgroup$ Apr 27, 2020 at 22:28

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I think this is an interesting and sort of deep question, so I'm going to answer it in part with the hope that my answer attracts even better answers.

I'll start with my first thought: surely there's no hope of formulating Green's theorem for an unbounded region, say the region $y > 0$. But then I thought about it for a moment, and observed that if you consider a smooth vector field $F(v)$ on the plane such that $F(v) \to 0$ rapidly as $v \to \infty$ then we can extend $F$ to the sphere by stereographic projection; this sends $y > 0$ to a hemisphere and the boundary curve $y = 0$ to the bounding great circle, and you can apply Stokes' theorem to this situation. Unwinding the calculations, this would give you a version of "Green's theorem" even for unbounded regions, albeit one that applies only to a certain class of vector field.

Then I thought about regions whose boundary is pathological, like the interior of the Koch snowflake. Here the boundary has infinite length, so surely there is no real hope of even defining the "boundary side" of Green's theorem. But then I noted that the Koch snowflake - like many pathological plane curves - has a very nice polygonal approximation, and it didn't sound insane that the boundary side could be defined as a limit of integrals over these approximations (again, maybe not for all vector fields). Sure enough, this has been worked out, and there is indeed a version of Green's theorem for fractal boundaries:

There are other crazy things to try, like removing a non-measurable set from the plane or something. But Green's theorem (and its parent, the fundamental theorem of calculus) is based on a very resilient idea, something like "when you sum differences, things cancel". So in the spirit of the principle, "The fastest way to find something is to assert that it doesn't exist on the internet", I'll make a bold conjecture: Green's theorem can be generalized to any subset of the plane.

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  • $\begingroup$ Thank you. Your response has been very helpful. $\endgroup$
    – carl
    Apr 26, 2020 at 18:54
  • $\begingroup$ To extend the first idea: the entire plane? Or the plane except a single point? $\endgroup$ Apr 27, 2020 at 13:25
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    $\begingroup$ "The best way to get the right answer on the Internet is not to ask a question; it's to post the wrong answer."Ward Cunningham's law. $\endgroup$
    – Pål GD
    Apr 27, 2020 at 18:42
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    $\begingroup$ "I'll start with my first thought: surely there's no hope of formulating Green's theorem for an unbounded region" Don't tell people solving external acoustic problems that. The entire foundation of their boundary element numerical methods is Green's functions on unbounded regions of $\mathbb{R}^3$. $\endgroup$
    – alephzero
    Apr 27, 2020 at 20:40
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    $\begingroup$ It can be generalized even beyond subsets. A subset of the plane may be viewed as a function $f : \mathbb{R}^2 \to {0,1}$, but we can also generalize the range to be $\mathbb{R}$, to allow fractional points to be in that subset. We can then define the integral over the "generalized subset" $\int_f g$ = $\int f g$. In this case Greene's theorem becomes the same as integration by parts. Now we can generalize $f$ further from a function to a Schwartz distribution, and perhaps beyond. $\endgroup$
    – Jules
    Sep 25, 2020 at 13:50
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As mentioned elsewhere on this site, Sauvigny's book Partial Differential Equations provides a proof of Green's theorem (or the more general Stokes's theorem) for oriented open sets in manifolds, as long as the boundary has capacity zero, and the differential form that you integrate has compact support. (Sauvigny just assumes that the open set is bounded in Euclidean space, but the same proof works without that hypothesis, as long as the differential form is compactly supported, $C^1$ in the interior and $C^0$ up to the boundary). The precise definition of capacity is complicated, so you would need to read the book to get that, but it includes reasonable things like corners and cone points. He also indicates the problems that arise with capacity nonzero, and (if I remember correctly) there are always problems. Of course, there is a problem with making sense of the integration if you allow objects that are too wildly behaved.

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There is a fun reverse definition that is used for so called "currents" in geometric measure theory, objects for which then in Green's theorem always ends up trivially being true. But then using the resulting theory, one can then show that Green's theorem is always true in a more proper sense, whenever the two integrals in it are defined, even if only in a very weak measure theoretic way.

Let $\Omega \subset \mathbb{R}^2$ be a $\mathcal{H}^2$-measureable set ($\mathcal{H}^2$ is the 2-dimensional Hausdorff measure).¹ Then we can define the corresponding linear operator

$$ \begin{array}{rccl} [\Omega] : & C_c^\infty(\mathbb{R}^2) &\to &\mathbb{R}\\ & f & \mapsto & \displaystyle\int_\Omega f dx \end{array} $$ This is what is called a $2$-current, i.e. an element of the topological dual $\mathcal{D}_2 := C_c^\infty(\mathbb{R}^2)'$. The theory of currents is quite similar to those of distributions, but a bit more geometrical. In particular, for any $T \in \mathcal{D}_2$ we define its boundary by $$ \begin{array}{rcl} \partial T: & C_c^\infty(\mathbb{R}^2;\mathbb{R}^2) & \to & \mathbb{R}\\ & F &\mapsto & T(\operatorname{curl}F) \end{array} $$ The resulting object is what is called a $1$-current.² As one would expect, those correspond to integral along suitably generalized curves. Using this definition, Green's theorem is always automatically true, since $\partial[\Omega](F) = [\Omega](\operatorname{curl} F)$ by definition.

One can of course then reformulate the question to make it interesting again. Let $\Gamma$ be a $\mathcal{H}^1$-measurable set of locally finite measure and $\tau: \Gamma \to \mathbb{R}^2$ some $\mathcal{H}^1$-measurable "unit tangent" orienting that set. Then we can define similarly $$ \begin{array}{rccl} [\Gamma] : & C_c^\infty(\mathbb{R}^2;\mathbb{R}^2) &\to &\mathbb{R}\\ & F &\mapsto& \displaystyle\int_\Gamma F \cdot \tau d \mathcal{H}^1 \end{array} $$ which is likely the weakest way to give some sense to the other integral in Green's theorem. Now the question is, for which $\Omega$ does this commute, i.e. $\partial [\Omega] = [\partial \Omega]$? Here the topological boundary $\partial \Omega$ is easy enough to define, but it turns out that the key-question here is what is the tangential vector $\tau$? The resulting notion is that of rectifiability. Roughly the condition for $\tau$ to be "the" tangential vector of $\Gamma$ at $x$ is that for any double-cone in direction $\tau$, most of $B_\epsilon(x)\cap \Gamma$ lies in that cone (the details are technical). If such a $\tau$ exists $\mathcal{H}^1$-almost everywhere, then $\Gamma$ is called rectifiable.

Now there are some further minor details involving the orientation of $\tau$ and possible multiplicity, but fundamentally it turns out that whenever the topological boundary $\partial \Omega$ is rectifiable, then there is a matching $\tau$ such that $\partial$ and $[\,]$ commute, i.e. Green's theorem holds.

The proper citation for all of this is Federer's "Geometric measure theory", but as it is one of those books, I'd recommend picking up Morgan's "Geometric measure theory: A beginner's guide" first.

¹Asking if Green's theorem holds for non-measurable sets should probably only be done by Zen Buddhists.

²As you can see, the number refers to the "dimensionality" of the object. To be more precise one should actually use differential forms of the dimension 2 in place of $f$ (resp. dimension 1 in place of $F$). This, and using the exterior derivative instead of $\operatorname{curl}$, also give the right generalisation to higher dimensions.

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