Liouville equation

Dispense §

Consider a generic initial probability distribution $f_0$ (w.r.t. Lebesque measure). Let $f(t,x)$ be the probability density at time $t$. For every subset of the phase space $A$ should hold the following

$${\int }_{A}dx{f}_0(x)={\int }_{A_t}dxf(t,x)$$

that is, probability is conserved. From this identity, we can solve for $f(t,x)$: just make a change of variable $x={\Phi }^{t,0}(y)$ (this makes appear the Jacobian of the flux) flow of an EDO:

$${\int }_{A}dx{f}_0(x)={\int }_{A_t}dxf(t,x)={\int }_{A}dxf(t,{\Phi }^{t,0}(x))J(x,t,0)$$

(where we renamed the new variable $x$). Since this must hold for all $A$ measurable sets, the integrands must be equal a.s:

$$f_0(x)=f(t,{\Phi }^{t,0}(x))J(x,t,0)$$

We are using a simple lemma Lemma Let $g(x)$ be a locally integrable function, if for every measurable set $A$

$${\int }_{A}g(x)=0$$

then $g(x)$ is zero a.s. Proof Let’s call $A^{+}\subseteq A$ the subset of $A$ where $g(x)>0$. Since $g$ is measurable, $A^{+}$ is measurable. Then, this set must have zero measure. The same is true for $A^{-}$. So $g(x)=0$ a.s. $\square$

The equation we have found is just the local conservation of probability, the Jacobian $J$ tells us how each volume $dx$ need to change to preserve probability.

$$f_0(x)dx=f(t,{\Phi }^{t,0}(x))J(x,t,0)dx$$

One can find that this holds also for a non-zero initial time $s$. We can find a PDE that $f(t,x)$ solves by taking the derivative w.r.t.

$${\partial }_t[f(t,{\Phi }^{t,0}(x))]J+f{\partial }_{t}J=0$$

the first term is

$$({\partial }_{t}f+u\cdot \nabla f){|}_{t,{\Phi }^{t,0}(x)}$$

we already computed the time derivative of the Jacobian in the second term jacobian of the flux

$$(f\nabla \cdot uJ){|}_{t,{\Phi }^{t,0}(x)}$$

putting all togheter

$$({\partial }_{t}f+u\cdot \nabla f)J+f\text{ div}uJ=0$$

$J$ starts as $1$ (the jacobian of the flux is the identity at time $0$), and since it solves a linear edo, it can’t be zero in finite time.

THINK

Hence we can cancel $J$. The first term is a transport term of $f$, the second is the infinitesimal volume variation. With a bit of algebra we can obain the Liouville equation:

$${\partial }_{t}f+\text{ div }uf=0$$

In sostanza

L’equazione di Lioville è una legge di conservazione locale della probabilità, con la solita interpretazione del cambio di probabilità totale in $A$ è dovuto ad un flusso $uf$di probabilità attraverso il bordo dell’insieme $\partial A$.

Arnold §

Theorem (Liouville) Let be given a system of ODEs $\dot{x}=f(x)$ with $x=(x_1,\dots ,x_n)$ such that solutions are defined for all times. Then we can define the associated flow $g^t$. Let $D_0$ be a region in the phase space, call $v_0$ its volume. For all times:

$$D(t)=g^t{D}_0,v(t)=\text{volume }D(t).$$

Then

$$\frac{dv}{dt}{|}_{t=0}={\int }_{D_0}\text{div }fdx$$

Proof

$$v(t)={\int }_{D(t)}dx={\int }_{D_0}\det \frac{\partial g^{t}x}{\partial x}dx$$

we did a change of variable, so the determinat of the Jacobian appears. From the definition of flow

$$g^t(x)=x+f(x)t+o(t^2)$$

so its Jacobian is

$$\frac{\partial g^{t}x}{\partial x}=I+\frac{\partial f}{\partial x}t+o(t^2)$$

using the Determinant near identity lemma, we get

$$v(t)={\int }_{D_0}[1+t\text{div }f+o(t^2)]dx$$

applying the definition of derivative we obtain the result. $\square$

Corollary Hamiltonian systems preserves volumes. Proof Just compute the divergence:

$$\text{ div }f={\partial }_q({\partial }_{p}H)+{\partial }_p(-{\partial }_{q}H)=0$$

so by Liouville’s theorem volumes in the phase space are preserved under time evolution. $\square$