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Consider two functions $\alpha,\beta: \mathbb{R}^2 \to \mathbb{R}$, where $\alpha$ is given and we look for solutions $\beta$ such that

$$*(d\alpha \wedge d\beta) = \lambda \beta$$

for some $\lambda \in \mathbb{R}$ and where $d$ is the exterior derivative, $*$ is the Hodge star operator, which turns the 2-form $d\alpha \wedge d\beta$ into a 0-form.

Formulated differently, I am looking for eigensolutions of the operator

$$D_{\alpha}(\cdot) := *(d\alpha \wedge d(\cdot))$$

on the space of 0-forms (possibly under some restriction for the domain).

Is this a well posed problem?

If so, what are methods to find solutions in this language of exterior calculus?

Can someone give an example for a simple, but non-trivial $\alpha$?

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  • $\begingroup$ Sorry, since $\alpha$ and $\beta$ map into $\mathbb{R}^3$, then how do you take the exterior derivative? Aren't they $0$-forms on $\mathbb{R}$? Do you want $\alpha,\beta:\mathbb{R}^3 \mapsto \mathbb{R}$? $\endgroup$
    – k3thomps
    Commented May 29, 2014 at 18:49
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    $\begingroup$ On a domain where $\alpha$ is a constant, this is, of course, the zero operator. On a domain $D$ on which $\alpha:D\to\mathbb{R}$ can be taken to be the first entry of a coordinate chart $(x,y)$ with $\ast(\mathrm{d}x\wedge\mathrm{d}y)=1$ and image a coordinate box (which is locally always possible on neighborhoods of points where $\mathrm{d}\alpha\not=0$), your operator is simply $D_\alpha(\beta) = \beta_y$, and this is always locally solvable. The eigenvalue problem has solutions $\beta = f(x)\ e^{\lambda y}$ where $f$ is arbitrary. $\endgroup$
    – Robert Bryant
    Commented May 29, 2014 at 19:36

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No, the problem as stated is not well posed. You need initial data.

We can rewrite the above equation in coordinates to get $$ \partial_x \alpha \partial_y \beta - \partial_y \alpha \partial_x \beta = \lambda \beta $$ This is a first order PDE and so one can use method of characteristics to solve this. I know this isn't in the language of exterior calculus like you want, but it shows that solutions can be found.

To see that it is not well posed take $\alpha(x,y) = x$ then we have $$\partial_y \beta = \lambda \beta$$ which has the solution $D(x) e^{\lambda y}$ (as pointed out by Robert Bryant in the comments above) where $D(x)$ is any function. This destroys uniqueness for you. You need some sort of initial data to rectify this.

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  • $\begingroup$ Thanks! Is there a coordinate free language of dealing with such operators or would one always go back to local coordinates for their analysis? $\endgroup$
    – madison54
    Commented May 29, 2014 at 20:10
  • $\begingroup$ To be quite honest, I don't know. Personally I don't see what the advantage to doing anything in a coordinate free manner is useful. Aesthetically it might be nice, but I don't see what it buys you. Now, I may just be mathematically immature and just because I don't know why it is useful doesn't mean it isn't. $\endgroup$
    – k3thomps
    Commented May 30, 2014 at 13:11

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