Tent map invariant density Is there a formula for the invariant density for the tent map $f_t$ (for $\sqrt{2}\leq t\leq 2$)?
 $$ f_t = \begin{cases} tx, & 0\leq x<1/2 \\ t-tx, & 1/2 \leq x\leq 1. \end{cases} $$
By invariant density I mean the density of absolutely continuous (wrt Lebesgue) ergodic invariant (wrt $f_t$) measure. Or, equivalently, the fixed point of Perron-Frobenius operator.
 A: There is no general formula. The problem is that the set $I_t := [f_t^2 (1/2), f_t(1/2)]$ is invariant and attracting. Hence, any invariant density is supported by $I_t$, and we only need to look at the restricted dynamics. By conjugation, it is isomorphic to a map :
$$T_{a,b} : \left\{ \begin{array}{lll} [0,1] & \to & [0,1] \\ x & \mapsto & \left\{ \begin{array}{lll} b+\frac{1-b}{a}x & \text{ if } & x \in [0,a] \\ \frac{1-x}{1-a} & \text{ if } & x \in (a,1] \end{array}\right. \end{array}\right.,$$
where $a \in (0,1)$ and $b \in (0,1-a)$. We should have $a = 1-t^{-1}$ and $b=2-t$, but that is beside the point.
The problem with this map is that it is piecewise expanding, but it does not fit the framework e.g. of expanding maps of the circle (there would be a discontinuity at $0$). General theory (going back to Lasota-Yorke) says that the density of the invariant absolutely continuous measure has bounded variation. This can be proved by letting the Perron-Frobenius operator act on the space of functions with bounded variation.
If the map is Markov (in this case, if $1/2$ is preperiodic), then there is a finite number of discontinuities ; since the map $T_{a,b}$ is piecewise affine, finding the invariant density becomes a problem of linear algebra (equivalent to finding the normalized main eigen-covector of a stochastic matrix). Otherwise, discontinuities are countable with no hope of finding a closed formula.
I think this article gives the most comprehensive answer that you can hope for.
A: I thought that the question is considered to be forgotten but as people continue to post answers I have to answer my own question.
As D.Thomine mentioned, there is a general theory for Perron-Frobenius operators of piecewise-expanding (PE) interval maps. It gives existence of unique invariant density from the space of functions with bounded variation.
Then there is a general thing from functional analysis saying that each function with bounded variation can be expressed as a sum of continuous part and a saltus part i.e. sum of Heaviside functions with coefficients. I first saw it in the artcle by V. Baladi on susceptibility function for PE maps.
There are three problems now 


*

*to split saltus part from the continuous part

*to determine the coefficients Heaviside functions

*to determine the place where Heaviside function have discontinuities


It seems that if one takes a characteristic function of the invariant interval and directly applies PF operator and looks at jumps at different points then he sees that the only jump is at $f_t(1/2)$. Then it is easy to see that if the invariant density has a jump at some point, it has also jumps (with decaying size) on the forward orbit of the same point.
The advantage of this is that it works also in nonlinear setup.
However it is completely unclear to me how to calculate continuous part in the general nonlinear case (PE unimodal map with nonlinear branches).
There is also an article of 2008 by Pawel Gora that gives explicit formulae for the kind of maps (piecewise-LINEAR maps). It does not extend to the case when branches are nonlinear however.
The paper D. Thomine mentinoed is also very interesting and does in a sense similar thing to constructing the saltus part inductively as described above. 
