Proof of Liouville's Theorem in Dynamics 📂Dynamics

Proof of Liouville's Theorem in Dynamics

Theorem

Consider the Euclidean space $\mathbb{R}^{n}$ and the function $f : \mathbb{R}^{n} \to \mathbb{R}^{n}$ given the following vector field represented by a differential equation. $$ \dot{x} = f(x) $$ For the flow $\phi_t ( \cdot )$ of this system and the region $D_{0} \subset \mathbb{R}^{n}$, let $D_{t} := \phi_{t} \left( D_{0} \right)$ denote the region displaced by the flow after time $t$, whose volume is represented as $V(t) \equiv V \left( D_{t} \right)$. If $\nabla \cdot f = 0$, then the following holds for all $D_{0} \subset \mathbb{R}^{n}$ and $t \in \mathbb{R}$. $$ V \left( D_{t} \right) = V \left( D_{0} \right) $$

Explanation of the Equations

$\nabla \cdot f$ represents the divergence of the vector field, describing how the vector field spreads out or converges.

$V$ denotes the volume of a given region in the vector field. The Liouville theorem is easier to understand through equations: simply put, if the divergence is $0$ everywhere, the volume of the region moved by the flow does not change.

Proof ¹

Strategy: This involves a bit of vector calculus. By representing vector functions and the axes of each vector as follows, we can directly deduce using basic tools of calculus and linear algebra. $$ f := \left( f_{1} , \cdots , f_{n} \right) \\ x := \left( x_{1} , \cdots , x_{n} \right) $$

Part 1. When $t_{0} = 0$, then $\displaystyle \left.{{ d V } \over { d t }}\right|_{t = t_{0}} = \int_{D_{0}} \nabla \cdot f dx$

Volume definition: When $\textbf{u} \in \mathbb{R}^{n}$ is transformed by the vector function $\textbf{f} : \mathbb{R}^{n} \to \mathbb{R}^{n}$ like $\textbf{f} \left( \textbf{u} \right) = \left( f_{1} (\textbf{u}) , \cdots , f_{n} (\textbf{u}) \right)$, the volume of $D$ is as follows: $$ V(D) = \int_{D} \left| {{ \partial \textbf{f} (\textbf{u}) } \over { \partial \textbf{u} }} \right| d u_{1} d u_{2} \cdots d u_{n} $$

Following the definition of volume $$ V(t) = \int_{D_{0}} \det {{ \partial \phi_{t} (x) } \over { \partial x }} dx $$ If we expand flow $\phi_{t} (x)$ in a Taylor series at $t=0$ $$ \phi_{t} (x) = x + f(x) t + O \left( t^{2} \right) $$ By taking partial derivatives with respect to $x$ and considering the identity matrix $n\times n$ $$ {{ \partial \phi_{t} (x) } \over { \partial x }} = E + {{ \partial f } \over { \partial x }} t + O \left( t^{2} \right) $$ The determinant of this matrix is $$ \begin{align*} \det {{ \partial \phi_{t} (x) } \over { \partial x }} =& \det \left[ E + {{ \partial f } \over { \partial x }} t \right] + O \left( t^{2} \right) \\ =& 1 + \text{tr} \left[ {{ \partial f } \over { \partial x }} \right] t + O \left( t^{2} \right) \\ =& 1 + \left( {{ d f_{1} } \over { d x_{1} }} + \cdots + {{ d f_{n} } \over { d x_{n} }} \right) t + O \left( t^{2} \right) \\ =& 1 + t \nabla \cdot f + \mathcal{O} \left( t^{2} \right) \end{align*} $$ Here $\text{tr}$ is the trace, the sum of the diagonal elements of the matrix. Now, taking $\int_{D_{0}} \cdot dx$ for both sides $$ \begin{align*} V(t) =& \int_{D_{0}} \det {{ \partial \phi_{t} (x) } \over { \partial x }} dx \\ =& \int_{D_{0}} 1 dx + \int_{D_{0}} t \nabla \cdot f dx + O \left( t^{2} \right) \\ =& V \left( D_{0} \right) + t \int_{D_{0}} \nabla \cdot f dx + O \left( t^{2} \right) \end{align*} $$ If we move $V \left( D_{0} \right) = V(0)$ to the left-hand side $$ V(t) - V(0) = t \int_{D_{0}} \nabla \cdot f dx + O \left( t^{2} \right) $$ And divide both sides by $(t - 0)$, $$ {{ V(t) - V(0) } \over { t-0 }} = \int_{D_{0}} \nabla \cdot f dx + O \left( t^{1} \right) $$ Since we used Taylor approximation around $t = 0$ and let $t \to 0$, then $O \left( t^{1} \right) \to 0$. $$ \left.{{ d V } \over { d t }}\right|_{t = 0} = \int_{D_{0}} \nabla \cdot f dx $$

Part 2. Extension to $t = t_{0}$

By letting $y := \phi_{t_{0}} (x)$, we proceed similarly as in Part 1 to obtain the following. $$ \left.{{ d V } \over { d t }}\right|_{t = t_{0}} = \int_{D_{0}} \nabla \cdot f dy $$ If for some constant $c \in \mathbb{R}$, we have $\nabla \cdot f = c$, $$ \left.{{ d V } \over { d t }}\right|_{t = t_{0}} = \int_{D_{0}} \nabla \cdot f dy = \int_{D_{0}} c dy = c \int_{D_{0}} dy $$ Being $y$ the transformation of $x$ by the flow for time $t_{0}$, consequently making $\int_{D_{0}} dy$ the volume at time $t_{0}$. The same holds for any $t_{0} \in \mathbb{R}$, thus $$ V’ = c V $$ The above differential equation possesses a trivial solution $V(t) = e^{ct} V (0)$.

Part 3.

From the final equation in Part 2, if $c = 0$, then $V(t) = e^{ct} V (0) = V(0)$.

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