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I found a proof of Green's theorem on a pdf and tried to understand it. Becuase the proof in itself is clear to me, I'm wondering if the hypothesis are too restrictive or if it can be a "good" general proof. Let $D$ be a limited subset of $R^2$, simple with respect to the $x-axis$, that is $D=[(x,y)\in R^2: a\le y\le b, f_1(y)\le x\le f_2(y)]$, and simple with respect to the $y-axis$, that is $D=[(x,y)\in R^2: c\le x\le b, g_1(x)\le y\le g_2(x)]$, with its boundary, $\partial D$, being the union of (the supports of) a finite number of regular curves. We also say that its boundary, $\partial D$, is positively oriented if its oriented in the counter-clockwise direction and we write in this case $\partial D^+$. Let $F(x,y)=(F_1(x,y),F_2(x,y))$ be a vector field on $D$ with $F\in C^1(A)$ where $A$ is an open set containing $D$. With these hypothesis, we have $\oint_{\partial D^+}F\cdot \tau=\iint_{D}(F_{2x}-F_{1y})dxdy$ which is Green's formula.

  • I suspect you are interested in ways to generalize or "weaken" the hypotheses while still implying Green's formula. You did not identify the source of the proof you've studied, but probably the described hypotheses were chosen to make an especially clear proof possible. It might be awkward to articulate the most general hypotheses needed, but an outline of ways to state them might be of interest to you. – hardmath Apr 14 '23 at 11:27
  • I can put the link of the pdf, but it's not in English: it's in Italian. That means not everyone can read the proof and also I don't know if I can do that for policies like copyright and MathStackExchange's guidelines. –  Apr 14 '23 at 12:40

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Theorem 1.7 - Let $D$ be a bounded domain in $\mathbb{R}^2$ that's simple with respect to both axes. If $\mathbf{F} = P\mathbf{i}+Q\mathbf{j} \in C^1(D)$, then the formula holds:

$$ \iint_D (Q_x-P_y)\,\text{d}x\,\text{d}y = \int_{\partial^+D} P\,\text{d}x+Q\,\text{d}y\,. \tag{1.6} $$

Formula (1.6) holds for much more general domains than those considered so far. We limit ourselves to observing that this class includes domains $D$ which can be expressed as a union of simple domains, which intersect only along the boundary. The positive orientation of $\partial D$ is obtained by orienting the single curves that compose it in such a way that, following them, the domain is left to one's left. The typical case is that in which $\partial D$ is the union of a $\gamma$ curve and $N$ curves $\gamma_1,\gamma_2,\dots,\gamma_N$ internal to $\gamma$ as in the figure:

$\quad\quad\quad\quad\quad\quad\quad\quad$enter image description here

from which:

$$ \iint_D (Q_x-P_y)\,\text{d}x\,\text{d}y = \int_{\gamma^+\cup\gamma_1^+\cup\gamma_2^+} P\,\text{d}x+Q\,\text{d}y\,. \tag{1.8} $$

This is the simplest version of the Gauss-Green theorem in the plane that I copied from Matematica - Calcolo Infinitesimale e Algebra Lineare - Bramanti, Pagani, Salsa - Zanichelli (first edition: September 2000). It's also the only one I know and that has always been enough for me in decades of application work.


EDIT: I recommend also consulting the excellent Analisi Matematica II - Teoria ed esercizi - Canuto, Tabacco - Springer (second edition: November 2014) where they directly prove the version of the theorem referring to a G-admissible domain, i.e. of the type of figure.

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    In the pdf I used to study the proof there was the definition of a "decomposable" domain $D\in R^2$, that is it can be written as the finite union of sets such that they are all simple with respect to both the $x-axis$ and $y-axis$, their boundaries are regular and their interiors are two by two disjoint. After that, it said that after writing Green's formulas for all the simple subsets and adding them the proof was also valid in this more general case. It's clear at this point that a "G-admissible domain" is a "decomposable" domain. Thanks, I will check the proof in the book you referred to. –  Apr 14 '23 at 15:53