Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
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 2 '11 at 3:02
Cartan's book Elementary theory of analytic functions ... gives a very readable answer to your question. –  t.b. Dec 2 '11 at 3:14

1 Answer 1

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

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.