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Green's theorem is usually stated as follows:

Let $U \subseteq \mathbb{R}^2$ be an open bounded set. Suppose its boundary $\partial U$ is the range of a closed, simple, piecewise $C^1$, positively oriented curve $\phi: [0,1] \to \mathbb{R}^2$ with $\phi(t) = (x(t),y(t))$. Let $f,g: \overline{U} \to \mathbb{R}$ be continuous with continuous, bounded partial derivatives in $U$. Then $$ \int_U \left(\frac{\partial f}{\partial x}(x,y) - \frac{\partial g}{\partial y}(x,y)\right) dx dy = \int_{[0,1]} f(\phi(t))y'(t) dt + \int_{[0,1]} g(\phi(t))x'(t) dt. $$

Is there a complete rigorous proof of this theorem somewhere? Most texts (Rudin, Munkres, Spivak) proves the generalized Stokes' theorem first, and shows Green's theorem as a corollary. However, the "piecewise" criteria is never addressed, since the version of Stokes' proven in the texts above do not work for manifolds with corners. Hence the version of Green's theorem proven is always with full $C^1$ assumption.

Because Green's theorem is only on the plane, I'm wondering if there is an easy way to obtain the "piecewise" generalization from just the $C^1$ assumption. Spivak mentions

Green's theorem is true for a square ... can be proved by approximating the square ... by manifolds with boundary.

Is this easy to do in the context of just $\mathbb{R}^2$?

Wakaka
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    One way is the prove Stokes' theorem for triangles (which can be proved using parameterized rectangles, where one side has length $0$). This implies Stokes' theorem for regions that can be chopped up into triangles with smooth sides. You of course, put a vertext at each point where the curve is not $C^1$. – Deane Jun 19 '22 at 18:17

2 Answers2

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It is easy to do by approximation -- just replace each corner of the square by an arc of a small circle, and show that the errors in both the area integral and the line integral approach zero as the radius of the circle approaches zero.

Alternatively, you can prove directly that Stokes's theorem applies to manifolds with corners. I do this in my Introduction to Smooth Manifolds (2nd ed.), Theorem 16.25. See also this MathOverflow question and its answers for more references.

Community
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Jack Lee
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  • Thanks for the book reference, I will definitely look at that someday! About the approximation, would it be as easy for generic piecewise $C^1$ curves though? I'm not sure how I would describe the arc to connect lines that are not "straight"... – Wakaka Dec 24 '16 at 17:40
  • Just do it in local coordinates near the vertex. It shouldn't be hard to find an arc of a hyperbola that matches up with your curves and their tangent vectors at two points. – Jack Lee Dec 24 '16 at 21:11
  • What if the the directions of one-sided limits of $\frac {d\phi} {dt}$ at a point are opposite? In this case $\overline U$ will not be a manifold with corners. – Zhang Yuhan Dec 03 '20 at 15:30
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    @Zhang: That's right. In that case you have to use an approximation argument. But you can still replace the vertex by an arc of a very small circle to get a $C^1$ boundary curve. Once you choose points on the two edges near the corner, it's just a question of finding a circle that passes through those two points and has specific tangent lines there. – Jack Lee Dec 03 '20 at 15:40
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    @Zhang: Sorry, I just realized that you need a little more flexibility than a circle provides. But an ellipse will do. – Jack Lee Dec 03 '20 at 16:34
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In "A First Course in Real Analysis" by Murray H. Protter and Charles B. Jr. Morrey Green's theorem is proved in paragraph 16.4 prior to proving Stokes' theorem. They prove Green's Theorem for so-called regular regions, which are regions whose boundary is given by piecewise-differentiable curves (along with some other conditions).

Sha Vuklia
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