Suppose we are observing chaotic continuos maps, the Perron-Frobenius operator $P$ satisfies:
$P\phi_{n}(t) = \frac{d}{dt} \int_{f^{-1}([a,t])} \phi(x)dx$
I don't understand how for the shift map, $S(x) = 2x$ mod $1$, in $I = [0,1]$:
$S^{-1}([0,t]) = [0,t/2] \cup [1/2,1/2 + t/2]$
In the logistic map with bifurcation parameter $k=4$, $f(x) = 4x(1-x)$ in $I=[0,1]$:
$f^{-1}([0,t]) = [0,1/2-1/2\sqrt{1-t}] \cup [1/2 + 1/2\sqrt{1-t}, 1]$
How are they getting these values, they are stated to be trivial but how does one compute them? I'm probably missing something very obvious so sorry for that.
Well, in both cases you know that it is made of two strictly monotone pieces (branches). Then when you take inverse of a particular point you just take preimages with respect to each of the branches (because both has range equal to the whole segment). When you take the inverse of a segment, you do the same noting that inverse of the strictly monotone function is strictly monotone and under strictly monotone function $\phi$ you have $\phi([a,b]) = [\phi(a),\phi(b)]$.
P.S. I have never seen the definition of PF operator in this form.