# Liouville’s theorem

Let

 $\dot{x}=f(x)$ (1)

be a autonomous  ordinary differential equation  in $\mathbb{R}^{n}$ defined by a smooth vector field $f\colon\mathbb{R}^{n}\to\mathbb{R}^{n}$ and the Jacobian of $f$ is denoted $\frac{\partial f}{\partial x}$. Also let $\Phi_{t}(x)$ be the flow (http://planetmath.org/Flow2) associated with (1). Let

 $V(t)=\int_{\Phi_{t}(D)}dx$

be the volume of the image of $D$ under this flow after a time $t$.

###### Theorem 1 (Liouville’s theorem).

If $D\subseteq\mathbb{R}^{n}$ is a bounded measurable domain. Then

 $\dot{V}(t)=\int_{\Phi_{t}(D)}\operatorname{div}\,f(x)dx$
###### Proof.

Let $V(t)$ 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 $x\in\Phi_{t_{0}}(D)$,

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

as $t\to 0$.

In fact,

 $\Phi_{t}(x)=x+\int_{0}^{t}f(\Phi_{s}(x))ds,$

and by the Leibniz integral rule

 $\frac{\partial\Phi_{t}}{\partial x}(x)=I+\int_{0}^{t}\frac{\partial}{\partial x% }f(\Phi_{s}(x))ds,$

so that

 $\frac{\partial}{\partial t}\frac{\partial\Phi_{t}}{\partial x}(x)=\frac{% \partial}{\partial x}f(\Phi_{t}(x))$

and evaluating at $t=0$ we get

 ${\frac{\partial}{\partial t}\frac{\partial\Phi_{t}}{\partial x}(x)}\Big{|}_{t=% 0}=\frac{\partial}{\partial x}f(\Phi_{0}(x))=\frac{\partial f}{\partial x}(x).$

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 $t\to 0$. It follows that

 $V(t_{0}+h)=\int_{\Phi_{t_{0}}(D)}1+h\operatorname{div}\,f(x)+o(h)dx$

and

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

###### Corollary 1.

The flow of an Hamiltonian system (http://planetmath.org/HamiltonianEquations) preserves volume.

###### Proof.

It follows directly since the vector field of an Hamiltonian system has divergence equal to zero. Hence $\dot{V}=0$ implies that the volume is constant. ∎

## References

Title Liouville’s theorem LiouvillesTheorem 2013-03-22 15:14:55 2013-03-22 15:14:55 Koro (127) Koro (127) 20 Koro (127) Theorem msc 34A34