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Let $\Omega\subseteq \mathbb{R}^2$ be an open set and $\omega=\omega^1dx_1+\omega^2dx_2$ an $1$-form on $\Omega$ and

$$L=\omega^2\frac{\partial}{\partial x_1}-\omega_1\frac{\partial}{\partial x_2}\in\mathfrak{X}(\Omega)$$

How to prove that exist $f\in C^{\infty}(\Omega)$, such that $f>0$ and $d(f\omega)=0$, iff there exists an $u\in C^{\infty}(\Omega)$, such that $$Lu=\frac{\partial\omega^2}{\partial x_1}-\frac{\partial\omega^1}{\partial x_2}$$

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$ curl(graf)= 0 $ for example perhaps if $ w^{1}= \frac{\partial f}{\partial x} $ and $ w^{2}= \frac{\partial f}{\partial y} $ –  Jose Garcia Jul 4 '13 at 15:38
    
@JoseGarcia I edited the question, it is $\mathbb{R}^2$ not $\mathbb{R}^N$. –  user34870 Jul 4 '13 at 18:21
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1 Answer

up vote 2 down vote accepted

The standard result is that if $d(f\omega) = 0$, with $f\ne 0$, then $d\omega\wedge\omega = 0$. The converse follows (locally, at least) from the Frobenius Theorem on differential systems/integral submanifolds.

First I'm going to assume $N=2$. Since $-(Lu)\, dx_1\wedge dx_2 = \omega\wedge du$ by definition, your criterion $-(Lu)\,dx_1\wedge dx_2=d\omega$ can be rephrased as $$\omega\wedge du = d\omega\,.$$ But $$d(f\omega) = f\,d\omega + df\wedge\omega = 0 \iff d\omega = -\frac{df}f\wedge\omega \iff d\omega = \omega\wedge \frac{df}f\,,$$ so taking $du = df/f$, i.e., $u=\log f$ does it. Conversely, given the $u$, we take $f=e^u$ and $d(f\omega)=0$.

Now, in general, with $N>2$, we add a "$\wedge dx_3\wedge\dots\wedge dx_N$" to the above computations. It is certainly still true that if $d(f\omega)=0$, then there exists $u$. But the converse doesn't seem to work: With the hypothesis on $Lu$, we'll only get $d(f\omega)=0 \pmod{\langle dx_3,\dots,dx_N\rangle}$. Counterexample forthcoming?

Edit: Ah, I just saw you amended it to be $N=2$, so I won't construct a counterexample for $N=3$ :)

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By which definition $-(Lu)\, dx_1\wedge dx_2 = \omega\wedge du$? Thanks! –  Eden Harder Jan 1 at 6:56
    
@EdenHarder: Write out $du=\dfrac{\partial u}{\partial x_1}dx_1+\dfrac{\partial u}{\partial x_2}dx_2$ and use the standard properties of $\wedge$. –  Ted Shifrin Jan 1 at 14:02
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