[Émilie.du]

Hi everyone,
I need help with this problem, please.
Consider on \( \mathbb{R}^3 \) the differential form \( \omega=ydx\wedge dy \). Determine all 1-forms \( \eta=pdx+qdy \) satisfying \( d\eta=\omega \).

My solution: It is easy to prove that \( \partial{\eta}=\partial{p}\wedge{dx}+\partial{q}\wedge{dy}=\left (\dfrac{\partial{q}}{\partial{x}}-\dfrac{\partial{p}}{\partial{y}}  \right )dx\wedge dy \). Then \( \dfrac{\partial{q}}{\partial{x}}-\dfrac{\partial{p}}{\partial{y}}=y \) 

[Luis Fuentes]

Hi

Hi everyone,
I need help with this problem, please.
Consider on \( \mathbb{R}^3 \) the differential form \( \omega=ydx\wedge dy \). Determine all 1-forms \( \eta=pdx+qdy \) satisfying \( d\eta=\omega \).

My solution: It is easy to prove that \( \partial{\eta}=\partial{p}\wedge{dx}+\partial{q}\wedge{dy}=\left (\dfrac{\partial{q}}{\partial{x}}-\dfrac{\partial{p}}{\partial{y}}  \right )dx\wedge dy \). Then \( \dfrac{\partial{q}}{\partial{x}}-\dfrac{\partial{p}}{\partial{y}}=y \) 

Any pair of funcion satisfying \( \dfrac{\partial{q}}{\partial{x}}-\dfrac{\partial{p}}{\partial{y}}=y \) are solutions.

Yo can fix \( p \), and integrate to compute \( q \).

Best regards.

[Émilie.du]

Any pair of funcion satisfying \( \dfrac{\partial{q}}{\partial{x}}-\dfrac{\partial{p}}{\partial{y}}=y \) are solutions.

Yo can fix \( p \), and integrate to compute \( q \).

Best regards.

Thanks for replying, but I do not understand what you mean. If we integrated with respect to \( x \) we obtains:
\( p,q:\mathbb{R^3} \rightarrow \mathbb{R} \)
\( q=\displaystyle \int \dfrac{\partial{p}}{\partial{y}}dx+xy \Rightarrow \eta=pdx+qdy=pdx+\left (\int \dfrac{\partial{p}}{\partial{y}}dx+xy  \right )dy \)
Sorry, I'm a little lost.

Regards.

[Luis Fuentes]

Hi

Thanks for replying, but I do not understand what you mean. If we integrated with respect to \( x \) we obtains:
\( p,q:\mathbb{R^3} \rightarrow \mathbb{R} \)
\( q=\displaystyle \int \dfrac{\partial{p}}{\partial{y}}dx+xy \Rightarrow \eta=pdx+qdy=pdx+\left (\int \dfrac{\partial{p}}{\partial{y}}dx+xy  \right )dy \)
Sorry, I'm a little lost.

What I mean is that you cannot give a more explicit solution.

P.S. We forgot a detail. We are working in \( \mathbb{R^3} \) so:

\( d\eta=p_ydy\wedge dx+p_zdz\wedge d_x+q_xdx\wedge dy+q_z\wedge dy \)

Thus we also have the conditions:

\( \dfrac{\partial p}{\partial z}=\dfrac{\partial q}{\partial z}=0 \),

that is \( p \) and \( q \) does not depend on \( z \).

But now, thing works like I explained in my first answer.

[Émilie.du]

Thanks you