Let $M$ be a smooth manifold and let $v$ be a tangent vector field on $M$. Consider a system of ordinary differential equations $$ \dot x = v(x), $$ in local coordinates $x = (x_1, \ldots, x_n)$. Consider a hypersurphase $\Lambda \subset M$ transversal to the vector field $v$ (i.e., for any $x \in \Lambda : v(x) \notin T_x \Lambda$). For any $y \in \Lambda$ let $\gamma_y (t)$ - the solution of the main equation with initial conditions $\gamma_y (0) = y$. We assume that after some time $\gamma_y(t)$ returns to $\Lambda$. Let $\tau > 0$ be the time moment of the first return: $\tau = \{t > 0 \mid \gamma_y (t) \in \Lambda\}$. Then by definition the point $z = \gamma_y (\tau)$ is the image of $y$ w.r.t. the Poincare map $P$.
I want to prove the following problem.
$\bf{Problem}$: Suppose that the system $\dot x = v(x)$ has an invariant measure with the smooth density $\rho \ge 0$. Then the corresponding Poincare map also has an invariant measure with a smooth nonnegative density.
$\bf{My}$ $\bf{solution}$: Consider the cylinder $Z_h = I \times H$, where $I \subset \Lambda$. Let $\mu$ be an invariant measure on $M$. Measure of a cylinder is $$ \int\limits_{Z_h} \hat \mu = \int\limits_{I} \hat \mu \int\limits_{H} \hat \mu, \quad \hat \mu = \rho(x) {d} x. $$ Measure of $I$ is: $\mu (I) = \int\limits_{I} \rho^*(y) {d}y$, where $\rho^*$ density of some measure on $\Lambda$. Let differentiate upper equation by $h \in H$, we get: $$ \frac{{d}}{{d} h} \int\limits_{Z_h} \hat \mu = \int\limits_{I} \rho^*(y) {d} y \cdot \big ( \frac{{d}}{{d} h} \int\limits_{H} \hat \mu \big ). $$ Derrivative $\frac{{d}}{{d} h} \int\limits_{H} \hat \mu$ is equal to 1. Consider new cylinder $\tilde Z_h = g^{\tau} (Z_h)$, where $g^t$ - phase flow of main system. We get another equality: $$ \frac{{d}}{{d} h} \int\limits_{g^{\tau} (Z_h)} \hat \mu = \int\limits_{g^{\tau}(I)} \rho^*(y) {d} y \cdot \big ( \frac{{d}}{{d} h} \int\limits_{g^{\tau} (H)} \hat \mu \big ). $$ Because $\mu$ - invariant measure, then $\frac{{d}}{{d} h} \int\limits_{Z_h} \hat \mu = \frac{{d}}{{d} h} \int\limits_{g^{\tau} (Z_h)} \hat \mu$. From this we get the following equation: $$ \int\limits_{I} \rho^*(y) {d} y = \int\limits_{g^{\tau}(I)} \rho^*(y) {d} y $$ Is it true that $g^{\tau} (I) = I$? Are there any other ways to prove the problem?