Hi! I am currently working with a linear PDE on the form(adsbygoogle = window.adsbygoogle || []).push({});

[itex]\frac{\partial f}{\partial t} + A(v^2 - v_r^2)\frac{\partial f}{\partial \phi} + B\cos(\phi)\frac{\partial f}{\partial v} = 0[/itex].

[itex]A[/itex] and [itex]B[/itex] are constants. I wish to find a clever coordinate substitution that simplifies, or maybe even solves the problem. So far I have tried with an action-angle approach that reduces the dimensionality of the problem. I first assume a general substitution [itex]\lbrace\phi, v\rbrace \rightarrow \lbrace r(\phi, v), s(\phi, v)\rbrace[/itex], and choose

[itex]r(\phi, v) = A\left(\frac{1}{3}v^3 - v_r^2 v\right) - B\sin(\phi)[/itex].

This choice then cancels all the [itex]\partial/\partial r[/itex] terms, and one is left with

[itex]\frac{\partial f}{\partial t} + \left[A(v^2 - v_r^2)\frac{\partial s}{\partial\phi} + B\cos(\phi)\frac{\partial s}{\partial v}\right]\frac{\partial f}{\partial s} = 0[/itex].

Now I am left with the choice of [itex]s(\phi, v)[/itex]. Is there maybe a way to choose [itex]s[/itex] such that the expression in front of [itex]\partial f/\partial s[/itex] becomes constant? Am I even on the right track, or are there much more clever ways to solve the problem?

Thanks in advance,

Simon

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# Clever coordinate substitution for linear PDE

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