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Consider the following DPE system:

$\left\{ \begin{array}{rcl} g_x - f_y& = &1-x^2 \\ h_x - f_z &= &3x^2 \\ h_y - g_z &=& -1 \ \end{array}\right .$

This comes from trying to prove that a certain 2-form is exact (and of course they never told us about PDEs before), and maybe the point is that I am not approaching the problem in the right way, but if so, is there any reasonable way to solve that DPE system? It is to note that I attempted Schwarz theorem, which took me to the (ridiculous) point where $g_x - f_y = 0$, so $1-x^2 = 0$, which is absurd since the 2-form is defined on the whole space.

Any ideas?

For any interest, the 2-form is, of course, $\omega = (1-x^2)dx \wedge dy + 3x^2 dx \wedge dz - dy \wedge dz$ and, in order for it to be exact, i am supposing that there is some $\eta = fdx + gdy + hdz \in \Omega^1(\mathbb{R}^3)$ such that $d\eta = \omega$.

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    Yep, but... is it closed? EDIT: oh, ok. I see your point, let me give it try, thanks for the hint!2012-10-14

2 Answers 2

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Here's a way of constructing some solutions.

Let's see if we can find a solution with one of the three functions $f$, $g$, $h$ being zero; let's say $g = 0$. Note, it is impossible to find a solution with two of the three functions being zero.

With $g = 0$, the system becomes

$-f_y = 1 - x^2,\qquad h_x - f_z = 3x^2,\qquad h_y = -1.$

The first equation gives

$f(x, y, z) = -y(1-x^2) + \alpha(x, z)$

and the second equation gives

$h(x, y, z) = -y + \beta(x, z).$

Substituting these expressions into the second equation, it reduces to

$\beta_x - \alpha_z = 3x^2.$

As we're just looking for a solution, let's see if we can find one with $\alpha = 0$. With this requirement, the equation becomes

$\beta_x = 3x^2$

and so

$\beta(x, z) = x^3 + \gamma(z).$

Again, as we are looking for a solution, let's set $\gamma = 0$. Putting all the pieces together, we have

$f = -y(1-x^2),\qquad g = 0,\qquad h = x^3 - y.$

Let's check to see whether this gives us a solution or not:

\begin{align*} d(-y(1-x^2)dx + (x^3 - y)dz) &= -(1-x^2)dy\wedge dx + 3x^2dx\wedge dy -dy\wedge dz\\ &= (1-x^2)dx\wedge dy + 3x^2dx\wedge dy - dy\wedge dz. \end{align*}

So we have constructed a solution to the system.

Note, we can find many other solutions by modifying our choices. In particular, instead of setting $g = 0$, we could have set $f = 0$ or $h = 0$. Also, we could have set $\beta = 0$ instead of $\alpha = 0$, and we could have taken any non-zero function $\gamma$. Even looking at all these combinations of choices will not necessary generate all the solutions to the system.

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More a guess than an answer... The PDE system above seems to be complete, the integrability condition generated by the first two equations is already implied by the third equation. Say we order the functions and variables so that $f_z$, $f_y$ and $g_z$ are principal derivatives and the other are parametric. Let $h$ be any arbitrary function of the variables. Then we can integrate/solve the last equation for $g$. Placing the resulting $g$ in any of the two equations of the system, we get a ODE in either $y$ or $z$. We solve any one of these to get $f$. I'm curious to hear if someone more knowledgeable can tell me if this works or not.