PlanetMath: Liouville's theorem

Proof. Let

be defined as above then

$\displaystyle V(t_0+h)$	$\displaystyle =$	$\displaystyle \int_{\Phi_{t_0+h}(D)}dy$
	$\displaystyle =$	$\displaystyle \int_{\Phi_h(\Phi_{t_0}(D))}dy$
	$\displaystyle =$	$\displaystyle \int_{\Phi_{t_0}(D)} \operatorname{det}\left(\frac{\partial\Phi_h}{\partial x}(x)\right) dx.$

We claim that, for $x\in \Phi_{t_0}(D)$ ,

$\displaystyle \frac{\partial\Phi_t}{\partial x}(x) = I + t\frac{\partial f}{\partial x}(x) + o(t)$

as $t\to 0$ .

In fact,

$\displaystyle \Phi_t(x) = x + \int_{0}^t f(\Phi_s(x))ds,$

and by the Leibniz integral rule

$\displaystyle \frac{\partial \Phi_t}{\partial x}(x) = I + \int_{0}^t \frac{\partial}{\partial x}f(\Phi_s(x)) ds,$

so that

$\displaystyle \frac{\partial}{\partial t} \frac{\partial \Phi_t}{\partial x}(x) = \frac{\partial}{\partial x}f(\Phi_t(x))$

and evaluating at

we get

$\displaystyle {\frac{\partial}{\partial t} \frac{\partial \Phi_t}{\partial x}(x... ... = \frac{\partial}{\partial x}f(\Phi_0(x)) = \frac{\partial f}{\partial x} (x).$

Our claim follows from this and from the definition of derivative.

Hence

$\displaystyle \operatorname{det}\left(\frac{\partial\Phi_t}{\partial x}(x)\right)$	$\displaystyle =$	$\displaystyle \operatorname{det}\left(I + t\frac{\partial f}{\partial x}(x)\right) + o(t)$
	$\displaystyle =$	$\displaystyle \prod_{i=1}^n(1 + \frac{\partial f_i}{\partial x_i}(x)) + o(t)$
	$\displaystyle =$	$\displaystyle 1+t\sum_{i=1}^n\frac{\partial f_i}{\partial x_i}(x) +o(t)$
	$\displaystyle =$	$\displaystyle 1 + t\operatorname{div}\, f(x) + o(t)$

as $t\to0$ . It follows that

$\displaystyle V(t_0+h) = \int_{\Phi_{t_0}(D)} 1 + h\operatorname{div}\, f(x) + o(h) dx$

and

$\displaystyle \dot{V}(t_0)$	$\displaystyle =$	$\displaystyle \lim_{h\to 0}\frac{V(t_0+h)-V(t_0)}{h}$
	$\displaystyle =$	$\displaystyle \frac{\int_{\Phi_{t_0}(D)} 1 + h\operatorname{div}\, f(x) + o(h) dx -V(t_0)}{h}$
	$\displaystyle =$	$\displaystyle \frac{V(t_0) + h\int_{\Phi_{t_0}(D)}\operatorname{div}\, f(x)dx + o(h) -V(t_0)}{h}$
	$\displaystyle =$	$\displaystyle \int_{\Phi_{t_0}(D)}\operatorname{div}\, f(x)dx + \lim_{h\to 0}\frac{o(h)}{h}$
	$\displaystyle =$	$\displaystyle \int_{\Phi_{t_0}(D)}\operatorname{div}\, f(x)dx.$

$\qedsymbol$

References