As we know, $\omega(x) = \sum_{i=1}^n a_i(x) dx_i = 0, x \in \mathbb{R}^n$, is Pfaffian differential form (inexact/exact differential equations). If $d\omega(x) = 0$, then it is exact differential equation; otherwise it's inexact differential equation.
For inexact differential equations, maybe there exists an integrating factor such that $\mu(x) \sum_{i=1}^n a_i(x) dx_i = df(x)$.
For $n\ge3$, there is Frobenius' theorem: $\omega(x)$ has integrating factor $\iff$ $d\omega \wedge \omega = 0$.
Especially, for 3d, $Pdx + Qdy + R dz =0$, the necessary and sufficient conditions that it can has integrating factor is $$P\left(\frac{\partial Q}{\partial z} – \frac{\partial R}{\partial y}\right) + Q\left(\frac{\partial R}{\partial x} – \frac{\partial P}{\partial z}\right) + R\left(\frac{\partial P}{\partial y} – \frac{\partial Q}{\partial x}\right) = 0$$
My questions:
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Why is there no easy criterion to determine $P(x,y) dx + Q(x,y)dy$ has integrating factor?
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Does inexact differential equation in 2d always have integrating factor, even though it's hard to find in general? Prove or give an example to disprove
Best Answer
Yes, for any $1$-form $\omega$ so that $\omega$ is nowhere $0$, you can always find an integrating factor (at least locally).
One can see this by thinking about coordinates $(x^1,x^2)$ in which the corresponding vector field becomes $\partial/\partial x^1$. But, staying just with differential forms, we know that $d\omega = \eta\wedge\omega$ for some $1$-form $\eta$, and a little bit of calculation shows that we can always change $\eta$ to $\eta'=\eta+\lambda\omega$ so that this form will be closed/exact. Then you can solve $\eta' = -d\log f$ and then $$d(f\omega) = fd\omega + df\wedge\omega = f\eta'\wedge\omega+df\wedge\omega = (f\eta'+df)\wedge\omega = 0.$$