Let $\dot{x}=u(x)$ a dynamical system ($x\in\Gamma$) with solution $x(t)=\Phi^t_u(y)$ and $\mu$ a $\Phi^t_u$-invariant measure on $\sigma_\Gamma$. I want to show that the smooth density $\rho=d\mu/dV$ satisfy the continuity equation $$ \partial_t\rho(t,x(t))+\nabla_x\cdot(\rho u(x(t)))=0,\qquad(1) $$ when $dV$ is the Lebesgue measure on $\Gamma$.
What i have try..
An invariant measure is such that $d\mu=d\mu\circ\Phi^t_u$, i.e.
$$
\mu(A)=\mu(\Phi^t_u(A)),
$$
so differentiating this one with respect to time, (1) must follow.
$$
0=\frac{d}{dt}\int\!\chi_{\Phi^t_u(A)}(x(t))\rho(t,x(t))dV(x(t)),
$$
$$
0=\frac{d}{dt}\int\!\chi_A(\Phi^{-t}_u(x))\rho(t,x(t))dV(x(t)).
$$
Using $y=\Phi^{-t}_u(x)$ as variable change and the fact that the measure is invariant, we find
$$
0=\int_A\frac{d}{dt}\Big[\rho(t,x(t))\det{\frac{\partial x(t)}{\partial y}}\Big]dV(y),
$$
where $\det{\frac{\partial x(t)}{\partial y}}$ is the determinant of the jacobian of the trasformation. When the total derivative act on the latter we obtain the result using the fact that
$$
\frac{d}{dt}\det{\frac{\partial x(t)}{\partial y}}=det\frac{\partial x(t)}{\partial y}\nabla_x\cdot\dot{x}=det\frac{\partial x(t)}{\partial y}\nabla_x\cdot u.
$$
So my ipotesi is that time derivative bypass the determinant acting on its argument. Is this correct? Is the result correlated with the fact that
$$
\det(1+\epsilon A)=1+\epsilon\text{Tr}A+o(\epsilon)?
$$