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 $\alpha$?



Eigenvalue problem of an operator involving the exterior derivative of differential forms
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.


