Locally exact form $P\;dx+Q\;dy$ , and the property $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$

Question

This is a very known result, but I don't have some proof. Someone known or has some proof of it?

Let be $\omega = P\;dx + Q\;dy$ be a $C^1$ differential form on a domain $D$. If $$\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} ,$$ then $\omega$ is locally exact.

This is the Poincaré-lemma in its most basic form. Since this is closely related to your other question I strongly recommend that you do some reading on differential forms and the wedge product (written $\wedge$) — t.b., Dec 02 '11 at 03:02
Cartan's book Elementary theory of analytic functions ... gives a very readable answer to your question. — t.b., Dec 02 '11 at 03:14

score 2 · Answer 1 · edited Apr 13 '17 at 12:19

If you have a curl-free field $W = (W_1, W_2, W_3)$ in a neighborhood of the origin, it is the gradient of a function $f$ given by $$ f(x,y,z) = \int_0^1 \; \left( \; x W_1(tx, ty,tz) + y W_2(tx, ty,tz) + z W_3(tx, ty,tz) \; \right) dt.$$

In your case, take $W_3 = 0$ and drop the dependence on $z$ from $f, \; W_1$ and $W_2.$ Note how this is set up so that $f=0$ at the origin.

There is more information at Anti-curl operator

Locally exact form $P\;dx+Q\;dy$ , and the property $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$

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