Are there any methods for finding an (infinite, absolutely continuous with respect to lebesgue measure) invariant measure $d\mu=f(u,v)dudv$ for something like the following?

$$ T:Q\to Q, \ Q=\{u^2+v^2\leq1, u\geq0,v\geq0\} $$ $$ T(u,v)= \left\{ \begin{array}{cc} \left(\frac{|u^2+(v-1/2)^2-1/4|}{u^2+v^2},\frac{|(u-1/2)^2+v^2-1/4|}{u^2+v^2}\right)&(u-1)^2+(v-1)^2\leq1,\\ \frac{1}{2}\left(\frac{|u^2+(v-1/2)^2-1/4|}{(u-1/2)^2+(v-1/2)^2},\frac{|(u-1/2)^2+v^2-1/4|}{(u-1/2)^2+(v-1/2)^2}\right)&(u-1)^2+(v-1)^2\geq1\\ \end{array} \right. $$

$T$ is continuous, piece-wise rational, and 4,5, or 6-to-1 (the domain is broken into seven regions, four of which get mapped onto all of $Q$ and three of which get mapped onto three of the regions).

Anyway, the details of my map aren't important. What I'm looking for is references or ideas for constructing an invariant measure. Thanks for any help.

  • $\begingroup$ I don't know if you can make it work in your setting, but there is a result for one dimensional maps in researchgate.net/publication/… I'd be interested in some sense of the details of your map. They might be instructive. $\endgroup$ – user24142 Jul 23 '15 at 1:09
  • 1
    $\begingroup$ Do you need to actually find the measure (if yes, analytically or numerically?) or to prove its existence? $\endgroup$ – demitau Jul 23 '15 at 8:40

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