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Let's consider the autonomous differential equation $x' = f(x) $ with phase space in $\mathbb{R}^n$ and flow function $\phi(t,x)$. Let $K$ be an invariant set in the phase space.

According to the version of Liouville's theorem I am currently trying to come to terms with, it holds for any $A \subset K$ that $$ \frac{d}{dt} m(\phi(t,A)) = \iint_{\phi(t,A)} \text{div} f(x) \text{d}x,$$ where $m$ is the $n$-dimensional Lebesgue measure.

The proof I am trying to understand uses Liouville's formula to show for any $h \in \mathbb{R}$ that $$m(\phi(t+h,A)) = \iint_{\phi(t,A)} \exp \left( \int_0^h (\text{div} f)(\phi(s,x)) \text{d}s \right) \text{d}x,$$

which seems straightforward. It follows quite obviously that $$ \frac{m(\phi(t+h,A)) - m(\phi(t,A))}{h} = \frac{1}{h} \iint_{\phi(t,A)} \left[ \exp \left( \int_0^h (\text{div} f)(\phi(s,x)) \text{d}s \right) - 1 \right] \text{d}x.$$

We know that $$f(x) = \frac{\partial}{\partial t} \phi(t,x) |_{t=0}~~~~ \text{and }~~ \text{div} f(x) = \text{trace}\frac{\partial f}{\partial x}(x),$$ and from Liouville's formula we get $$ \exp \left( \int_0^h (\text{div} f)(\phi(s,x)) \text{d}s \right) = \text{det} \left( \frac{\partial \phi}{\partial x} (h,x) \right). $$ My question is, how do we get from here to $$\lim_{h \to 0} \frac{m(\phi(t+h,A)) - m(\phi(t,A))}{h} = \iint_{\phi(t,A)} \text{div} f(x) \text{d}x ~~\text{?}$$

Marcos Gonzalez
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1 Answers1

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If you are comfortable moving the limit under the integral sign (which can be done with mild hypotheses on $f$), then you have

$$\lim_{h\to 0} \frac{m(\phi(t+h,A))-m(\phi(t,A))}{h} = \iint_{\phi(t,A)}\lim_{h\to 0}\frac{1}{h}\left[\exp\left(\int_0^h(\text{div} f)(\phi(s,x))ds\right)-1\right]dx$$

Considering the limit definition of the derivative, the limit under the integral sign is exactly

$$\frac{d}{dt}\exp\left(\int_0^{t} (\text{div}f)(\phi(s,x))ds\right)\bigg\vert_{t=0} $$

But we can calculate this derivative using the chain rule and fundamental theorem of calculus as

$$\begin{align}\frac{d}{dt}\exp\left(\int_0^{t} (\text{div}f)(\phi(s,x))ds\right)\bigg\vert_{t=0} &=(\text{div}f)(\phi(t,x))\exp\left(\int_0^{t} (\text{div}f)(\phi(s,x))ds\right)\bigg\vert_{t=0} \\ &= (\text{div} f)(\phi(0,x))\exp\left(\int_0^{0} (\text{div}f)(\phi(s,x))ds\right)\\ &=(\text{div} f)(x)\end{align}$$

In the last line, we used that a flow $\phi(x,t)$ is the identity for $t=0$.

podiki
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