Liouville’s theorem

Let

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

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

$$V(t)={\int }_{{\mathrm{\Phi }}_t(D)}𝑑x$$

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

Theorem 1 (Liouville’s theorem).

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

$$\dot{V}(t)={\int }_{{\mathrm{\Phi }}_t(D)}\mathrm{div}f(x)𝑑x$$

Proof.

Let $V(t)$ be defined as above then

$V(t_0+h)$ $=$ ${\displaystyle {\int }_{{\mathrm{\Phi }}_{t_0+h}(D)}}𝑑y$

$=$ ${\displaystyle {\int }_{{\mathrm{\Phi }}_h({\mathrm{\Phi }}_{t_0}(D))}}𝑑y$

$=$ ${\displaystyle {\int }_{{\mathrm{\Phi }}_{t_0}(D)}}\det \left({\displaystyle \frac{\partial {\mathrm{\Phi }}_h}{\partial x}}(x)\right)𝑑x.$

We claim that, for $x\in {\mathrm{\Phi }}_{t_0}(D)$,

$$\frac{\partial {\mathrm{\Phi }}_t}{\partial x}(x)=I+t\frac{\partial f}{\partial x}(x)+o(t)$$

as $t\to 0$.

In fact,

$${\mathrm{\Phi }}_t(x)=x+{\int }_0^{t}f({\mathrm{\Phi }}_s(x))𝑑s,$$

and by the Leibniz integral rule

$$\frac{\partial {\mathrm{\Phi }}_t}{\partial x}(x)=I+{\int }_0^t\frac{\partial }{\partial x}f({\mathrm{\Phi }}_s(x))𝑑s,$$

so that

$$\frac{\partial }{\partial t}\frac{\partial {\mathrm{\Phi }}_t}{\partial x}(x)=\frac{\partial }{\partial x}f({\mathrm{\Phi }}_t(x))$$

and evaluating at $t=0$ we get

$${\frac{\partial }{\partial t}\frac{\partial {\mathrm{\Phi }}_t}{\partial x}(x)|}_{t=0}=\frac{\partial }{\partial x}f({\mathrm{\Phi }}_0(x))=\frac{\partial f}{\partial x}(x).$$

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

Hence

$\det \left({\displaystyle \frac{\partial {\mathrm{\Phi }}_t}{\partial x}}(x)\right)$ $=$ $\det \left(I+t{\displaystyle \frac{\partial f}{\partial x}}(x)\right)+o(t)$

$=$ ${\displaystyle \prod _{i=1}^n}(1+{\displaystyle \frac{\partial f_i}{\partial x_i}}(x))+o(t)$

$=$ $1+t{\displaystyle \sum _{i=1}^n}{\displaystyle \frac{\partial f_i}{\partial x_i}}(x)+o(t)$

$=$ $1+t\mathrm{div}f(x)+o(t)$

as $t\to 0$. It follows that

$$V(t_0+h)={\int }_{{\mathrm{\Phi }}_{t_0}(D)}1+h\mathrm{div}f(x)+o(h)dx$$

and

$\dot{V}(t_0)$ $=$ $\underset{h\to 0}{lim}{\displaystyle \frac{V(t_0+h)-V(t_0)}{h}}$

$=$ ${\displaystyle \frac{{\int }_{{\mathrm{\Phi }}_{t_0}(D)}1+h\mathrm{div}f(x)+o(h)dx-V(t_0)}{h}}$

$=$ ${\displaystyle \frac{V(t_0)+h{\int }_{{\mathrm{\Phi }}_{t_0}(D)}\mathrm{div}f(x)𝑑x+o(h)-V(t_0)}{h}}$

$=$ ${\displaystyle {\int }_{{\mathrm{\Phi }}_{t_0}(D)}}\mathrm{div}f(x)𝑑x+\underset{h\to 0}{lim}{\displaystyle \frac{o(h)}{h}}$

$=$ ${\displaystyle {\int }_{{\mathrm{\Phi }}_{t_0}(D)}}\mathrm{div}f(x)𝑑x.$

∎

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. ∎

Title	Liouville’s theorem
Canonical name	LiouvillesTheorem
Date of creation	2013-03-22 15:14:55
Last modified on	2013-03-22 15:14:55
Owner	Koro (127)
Last modified by	Koro (127)
Numerical id	20
Author	Koro (127)
Entry type	Theorem
Classification	msc 34A34