I need some reference for the proof of the following theorem attributed to Liouville:

Theorem. Let $f(x):\Omega\longrightarrow \mathbb R^n$ be a $C^2$ function where $\Omega$ is an open subset of $\mathbb R^n$ and assume that

$$ \textrm{div}\, f=\sum_{i=1}^n\frac{\partial f_i}{\partial x_i}=0. $$

If $\varphi$ is the flow of the differential equation $y'=f(y)$ and we consider the homeomorphism $\pi_t:\Omega\longrightarrow\Omega$, such that $$ x\,\longmapsto\, \pi_t(x):=\varphi(x,t), $$ then the map $\pi_t$ preserves the Lebesgue measure of every measurable subset of $\Omega$. To be more precise, for every $\mu$-measurable subset $D\subseteq \Omega$ we have that $\mu(D)=\mu\big(\pi_t(D)\big)$.

The above theorem is very famous and its simpler form applied to Hamiltonian systems is often cited in texts of mechanics. However I need a proof of the general statement.

Thanks in advance


Idea of the proof.

Let $D_t=\pi_t(D)$. It suffices to show that $$ \frac{d}{dt}m(D_t)=\int_{D_t}\nabla\cdot f\,dx. $$ The Theorem of Change of Variables says that $$ m(D_t)=\int_{D_t} 1\,dx=\int_{D}\det\left(\frac{\partial\pi_t}{\partial x}\right)\,dx. $$ But $$ \frac{\partial\pi_t}{\partial x}=I+\frac{\partial f}{\partial x}t+{\mathcal O}(t^2). $$ This is due to the fact that the solution of $$ x'=f(x),\quad x(0)=x_0, $$ after small time $t$ looks like $x(t)=x_0+tf(x_0)+{\mathcal O}(t^2)$.

Using standard properties of the determinant one can show that $$ \det (I+tA)=1+t\,\mathrm{Tr}\,A+{\mathcal O}(t^2). $$ and hence $$ \frac{d}{dt}\det\left(\frac{\partial\pi_t}{\partial x}\right)=\mathrm{Tr}\,\frac{\partial f}{\partial x}=\nabla\cdot f. $$

Note. For a complete proof see Arnold, Mathematical Methods in Classical Mechanics.

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  • $\begingroup$ Is this reasoning also true for a function $f(t,x):\mathbb R\times\mathbb R^n\longrightarrow\mathbb R^n$ such that $\sum_{i=1}^n\frac{\partial f_i}{\partial x_i}=0$? I've tried to modify your proof ad hoc, and it seems that there aren't problems. $\endgroup$ – Dubious Feb 27 '14 at 17:16
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    $\begingroup$ Yes, it works without any problems. $\endgroup$ – Yiorgos S. Smyrlis Feb 27 '14 at 19:29

I am a little confused with the statement you have given, since the ODE $y'=f(y,t)$ does not generate the flow (i.e., a one-parameter group of transformations on the state space). We need an autonomous system to generate a flow: $y'=f(y)$. In this case it can be proved that $$ \frac{dV_t}{dt}=\int_{D_t}\nabla\cdot f\,dx, $$ where $D_t$ is the set where the set of initial conditions $D_0$ was mapped by the flow, $D_t=\phi(t,D_0)$, and $V_t$ is the measure of this set. A proof is given, e.g., in Verhult's Nonlinear differential equations and dynamical systems (Lemma 2.4 in the first edition). Liouville's theorem is a simple corollary of this fact.

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  • $\begingroup$ You are right, I need an autonomous system. But starting with a non-autonomous system I can always bring back to the autonomous case. $\endgroup$ – Dubious Feb 23 '14 at 7:48
  • $\begingroup$ This is correct, but the expression for divergence will change, and therefore your original statement will be still incorrect. $\endgroup$ – Artem Feb 23 '14 at 13:36

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