Notes on the history of Liouville’s theorem

1 Introduction

We denote by $ℬ({ℝ}^n)$ the set of all linear maps ${ℝ}^n\to {ℝ}^n$. We take it as known that with the operator norm

$$\parallel A\parallel =sup\{\parallel Av\parallel :v\in {ℝ}^n,\parallel v\parallel \le 1\},A\in ℬ({ℝ}^n),$$

$ℬ({ℝ}^n)$ is a Banach space.

2 Autonomous differential equations

Lemma 1.

If $A\in ℬ({ℝ}^n)$, then

$$det(I+ϵA+o(ϵ))=1+ϵ\text{tr}A+o(ϵ)$$

as $ϵ\to 0$.

Proof.

Let ${\lambda }_1,\mathrm{\dots },{\lambda }_n$ be the eigenvalues of $A$, repeated according to algebraic multiplicity. For $ϵ>0$, the eigenvalues of $I+ϵA+o(ϵ)$ repeated according to algebraic multiplicity are

$$1+ϵ{\lambda }_1+o(ϵ),\mathrm{\dots },1+ϵ{\lambda }_n+o(ϵ),$$

as $ϵ\to 0$. The determinant of a linear map ${ℝ}^n\to {ℝ}^n$ is the product of its eigenvalues according to algebraic multiplicity, so

$$det(I+ϵA+o(ϵ))=\prod _{k=1}^n(1+ϵ{\lambda }_k+o(ϵ)),$$

as $ϵ\to 0$. But

$$\prod _{k=1}^n(1+ϵ{\lambda }_k+o(ϵ))=1+ϵ\sum _{k=1}^n{\lambda }_k+o(ϵ)=1+ϵ\text{tr}A+o(ϵ)$$

as $ϵ\to 0$. ∎

Theorem 2.

If $A\in ℬ({ℝ}^n)$, then

$$det{e}^A=e^{\text{tr}A}$$

Proof.

We have

$$e^A=\underset{m\to \mathrm{\infty }}{lim}{\left(I+\frac{A}{m}\right)}^m.$$

As $det:ℬ({ℝ}^n)\to ℝ$ is continuous, we have

$$det{e}^A=\underset{m\to \mathrm{\infty }}{lim}det{\left(I+\frac{A}{m}\right)}^m.$$

Then, using Lemma 1,

$det{e}^A$ $=$ $\underset{m\to \mathrm{\infty }}{lim}{\left(det\left(I+{\displaystyle \frac{A}{m}}\right)\right)}^m$

$=$ $\underset{m\to \mathrm{\infty }}{lim}{\left(1+{\displaystyle \frac{1}{m}}\text{tr}A+o\left({\displaystyle \frac{1}{m}}\right)\right)}^m$

$=$ $e^{\text{tr}A}.$

∎

If $A\in ℬ({ℝ}^n)$, then the flow of the vector field $A$ is

$$(t,x)\mapsto e^{tA}x.$$

For each $t$ we have $e^{tA}\in ℬ({ℝ}^n)$, and by Theorem 2 we have

$$det(e^{tA})=e^{\text{tr}(tA)}=e^{t\text{tr}A}.$$

Let $\lambda$ be Lebesgue measure on ${ℝ}^n$. If $U$ is an open subset of ${ℝ}^n$, then

$$\lambda (e^{tA}U)={\int }_{e^{tA}U}𝑑y={\int }_U|det(D{e}^{tA})(x)|𝑑x={\int }_U|det(e^{tA})|𝑑x=e^{t\text{tr}A}\lambda (U).$$

Therefore, $\lambda$ is an invariant measure for the flow if and only if $\text{tr}A=0$, namely, if and only if $A$ is skew-symmetric.

3 Nonautonomous differential equations

Suppose that $I$ is an open interval and $A\in C(I,ℬ({ℝ}^n))$. The set $X$ of all functions $x:I\to {ℝ}^n$ that satisfy the differential equation

$$\dot{x}(t)=A(t)x(t)$$

is a vector space. For each $t\in I$ we define $B_t:X\to {ℝ}^n$ by $B_t(x)=x(t)$. It is apparent that for each $t\in I$ the map $B_t$ is linear. For each $x_0\in {ℝ}^n$, by the existence and uniqueness theorem for ordinary differential equations there is a unique $x\in X$ for which $B_0(x)=x_0$, hence for each $t\in I$ we get that $B_t$ is a bijection, and hence a linear isomorphism.

Suppose that ${\varphi }_1,\mathrm{\dots },{\varphi }_n\in X$, and for each $t\in I$ let $\mathrm{\Phi }(t)\in ℬ({ℝ}^n)$ be defined by $\mathrm{\Phi }(t)e_i={\varphi }_i(t)$. Then

$$\dot{\mathrm{\Phi }}(t)e_i=\frac{d}{dt}(\mathrm{\Phi }(t)e_i)=\dot{{\varphi }_i}(t)=A(t){\varphi }_i(t),t\in I,$$

and hence

$$\dot{\mathrm{\Phi }}(t)=A(t)\mathrm{\Phi }(t),t\in I.$$ (1)

The Wronskian $W=W({\varphi }_1,\mathrm{\dots }{\varphi }_n)$ of the ordered set ${\varphi }_1,\mathrm{\dots },{\varphi }_n$ is the function that assigns to each $t\in I$ the oriented volume of the parallelepiped spanned by ${\varphi }_1(t),\mathrm{\dots },{\varphi }_n(t)$. That is,

$$W(t)=det\mathrm{\Phi }(t),t\in I.$$

Theorem 3.

Suppose that $I$ is an open interval and that $A\in C(I,ℬ({ℝ}^n))$. If ${\varphi }_1,\mathrm{\dots },{\varphi }_n\in X$, then the Wronskian $W=W({\varphi }_1,\mathrm{\dots },{\varphi }_n)$ satisfies

$$\dot{W}(t)=(\text{tr}A(t))W(t),t\in I.$$

Proof.

By (1), for each $t\in I$ we have

$\mathrm{\Phi }(t+\mathrm{\Delta })$ $=$ $\mathrm{\Phi }(t)+\dot{\mathrm{\Phi }}(t)\mathrm{\Delta }+o(\mathrm{\Delta })$

$=$ $\mathrm{\Phi }(t)+A(t)\mathrm{\Phi }(t)\mathrm{\Delta }+o(\mathrm{\Delta })$

$=$ $\mathrm{\Phi }(t)+A(t)\mathrm{\Phi }(t)\mathrm{\Delta }+o(\mathrm{\Phi }(t)\mathrm{\Delta })$

$=$ $\mathrm{\Phi }(t)(I+A(t)\mathrm{\Delta }+o(\mathrm{\Delta }))$

$=$ $\mathrm{\Phi }(t)(I+A(t)\mathrm{\Delta }+o(\mathrm{\Delta }))$

as $\mathrm{\Delta }\to 0$. Using Lemma 1 we get

$W(t+\mathrm{\Delta })$ $=$ $det\mathrm{\Phi }(t+\mathrm{\Delta })$

$=$ $det\mathrm{\Phi }(t)det(I+A(t)\mathrm{\Delta }+o(\mathrm{\Delta }))$

$=$ $det\mathrm{\Phi }(t)(1+\text{tr}A(t)\mathrm{\Delta }+o(\mathrm{\Delta }))$

$=$ $det\mathrm{\Phi }(t)+det\mathrm{\Phi }(t)\text{tr}A(t)\mathrm{\Delta }+o\left(det\mathrm{\Phi }(t)\mathrm{\Delta }\right)$

$=$ $det\mathrm{\Phi }(t)+det\mathrm{\Phi }(t)\text{tr}A(t)\mathrm{\Delta }+o(\mathrm{\Delta })$

as $\mathrm{\Delta }\to 0$, i.e.,

$$W(t+\mathrm{\Delta })=W(t)+W(t)\text{tr}A(t)\mathrm{\Delta }+o(\mathrm{\Delta }),$$

which gives us

$$\dot{W}(t)=(\text{tr}A(t))W(t).$$

∎

One checks that for any $t_0\in I$,

$$t\mapsto W(t_0)\exp \left({\int }_{t_0}^t\text{tr}A(\tau )𝑑\tau \right),t\in I$$

is a solution of the differential equation in Theorem 3. For each $t\in I$ we have that $v\mapsto (\text{tr}A(t))v$ is linear, and in particular is locally Lipschitz, so by the existence and uniqueness theorem it follows that

$$W(t)=W(t_0)\exp \left({\int }_{t_0}^t\text{tr}A(\tau )𝑑\tau \right),t\in I.$$

4 Jacobi’s formula

Let $\mathrm{\Omega }$ be the volume form on ${ℝ}^n$, and let $A\in ℬ({ℝ}^n)$. One checks that

$$\mathrm{\Omega }(x_1,\mathrm{\dots },x_n)detA=\mathrm{\Omega }(A{x}_1,\mathrm{\dots },A{x}_n),x_1,\mathrm{\dots },x_n\in {ℝ}^n.$$ (2)

If $\omega$ is an $(n-1)$-form on ${ℝ}^n$, then there is a unique $x_{\omega }\in {ℝ}^n$ such that for all $x_1,\mathrm{\dots },x_{n-1}\in {ℝ}^n$,

$$\omega (x_1,\mathrm{\dots },x_{n-1})=\mathrm{\Omega }(x_{\omega },x_1,\mathrm{\dots },x_{n-1}).$$

Thus, if $x_0\in {ℝ}^n$ and we define an $(n-1)$-form $\omega$ by

$$\omega (x_1,\mathrm{\dots },x_{n-1})=\mathrm{\Omega }(x_0,A{x}_1,\mathrm{\dots },A{x}_n),$$

then there is some $x_{\omega }$ with which

$$\mathrm{\Omega }(x_0,A{x}_1,\mathrm{\dots },A{x}_n)=\mathrm{\Omega }(x_{\omega },x_1,\mathrm{\dots },x_{n-1}).$$

We define $\mathrm{adj}A\in ℬ({ℝ}^n)$ by $(\mathrm{adj}A)(x_0)=x_{\omega }$. Thus, for $x_1,\mathrm{\dots },x_n\in {ℝ}^n$, we have

$$\mathrm{\Omega }(x_1,A{x}_2,\mathrm{\dots },A{x}_n)=\mathrm{\Omega }((\mathrm{adj}A)x_1,x_2,\mathrm{\dots },x_n).$$ (3)

Hence

$$\mathrm{\Omega }(A{x}_1,A{x}_2,\mathrm{\dots },A{x}_n)=\mathrm{\Omega }((\mathrm{adj}A)A{x}_1,x_2,\mathrm{\dots },x_n),$$

therefore

$$\mathrm{\Omega }((\mathrm{adj}A)A{x}_1,x_2,\mathrm{\dots },x_n)=\mathrm{\Omega }(x_1,\mathrm{\dots },x_n)detA,$$

and because this holds for all $x_1,\mathrm{\dots },x_n\in {ℝ}^n$, it follows that

$$(\mathrm{adj}A)A=(detA)I.$$

Furthermore, if $A\in ℬ({ℝ}^n)$, then one checks that for all $x_1,\mathrm{\dots },x_n\in {ℝ}^n$,

$$\mathrm{\Omega }(x_1,\mathrm{\dots },x_n)\text{tr}A=\sum _{i=1}^n\mathrm{\Omega }(x_1,\mathrm{\dots },A{x}_i,\mathrm{\dots },x_n).$$ (4)

If $I$ is an open interval and $A\in C^1(I,ℬ({ℝ}^n))$, we have, using (2) for the first equality, (3) for the third equality, and (4) for the fourth equality,

${\displaystyle \frac{d}{dt}}\left(\mathrm{\Omega }(e_1,\mathrm{\dots },e_n)detA(t)\right)$ $=$ ${\displaystyle \frac{d}{dt}}\mathrm{\Omega }(A(t)e_1,\mathrm{\dots },A(t)e_n)$

$=$ ${\displaystyle \sum _{i=1}^n}\mathrm{\Omega }(A(t)e_1,\mathrm{\dots },\dot{A}(t)e_i,\mathrm{\dots },A(t)e_n)$

$=$ ${\displaystyle \sum _{i=1}^n}\mathrm{\Omega }(e_1,\mathrm{\dots },(\mathrm{adj}A(t))\dot{A}(t)e_i,\mathrm{\dots },e_n)$

$=$ $\mathrm{\Omega }(e_1,\mathrm{\dots },e_n)\text{tr}((\mathrm{adj}A(t))\dot{A}(t)),$

that is,

$$\frac{d}{dt}detA(t)=\text{tr}((\mathrm{adj}A(t))\dot{A}(t)).$$

Kline p. 798, Jacobi [40], [41, §17], Felix Klein, 19th century, chapter V

5 Reynolds transport theorem

If $V$ is a vector field with flow $\varphi$ and $U$ is a bounded open subset of ${ℝ}^n$ with piecewise smooth boundary and $f:{ℝ}^n\times ℝ\to {ℝ}^n$ is smooth, then with $U_t={\varphi }_t(U)$,

$${\int }_{U_t}f(y,t)𝑑y={\int }_{U}f({\varphi }_t(x),t)det(D{\varphi }_t)(x)dx;$$

this presumes that $det(D{\varphi }_t)(x)>0$. Write $\frac{D}{Dt}=\frac{\partial }{\partial t}+V\cdot D$. We then have

${\displaystyle \frac{d}{dt}}{\displaystyle {\int }_{U_t}}f(y,t)𝑑y$ $=$ ${\displaystyle \frac{d}{dt}}{\displaystyle {\int }_U}f({\varphi }_t(x),t)det(D{\varphi }_t)(x)dx$

$=$ ${\displaystyle {\int }_U}(Df)({\varphi }_t(x),t){\dot{\varphi }}_t(x)det(D{\varphi }_t)(x)$

$+{\displaystyle \frac{\partial f}{\partial t}}({\varphi }_t(x),t)det(D{\varphi }_t)(x)+f({\varphi }_t(x),t){\displaystyle \frac{d}{dt}}det(D{\varphi }_t)(x)dx$

$=$ ${\displaystyle {\int }_U}{\displaystyle \frac{Df}{Dt}}({\varphi }_t(x),t)det(D{\varphi }_t)(x)+f({\varphi }_t(x),t){\displaystyle \frac{d}{dt}}det(D{\varphi }_t)(x)dx.$

Writing $J_t(x)=det(D{\varphi }_t)(x)$, we have

${\displaystyle \frac{d}{dt}}{\displaystyle {\int }_{U_t}}f(y,t)𝑑y$ $=$ ${\displaystyle {\int }_U}{\displaystyle \frac{Df}{Dt}}({\varphi }_t(x),t)J_t(x)+f({\varphi }_t(x),t){\displaystyle \frac{d}{dt}}{J}_t(x)dx$

$=$ ${\displaystyle {\int }_U}{\displaystyle \frac{D(fJ)}{Dt}}$

Reynolds [73, pp. 12–13, art. 14]

Amann and Escher [4, p. 425, Theorem 2.11]

6 Symplectic geometry

7 Geodesic flow

Invariance of a kinematic measure on the unit tangent bundle.

8 References

Jacobi [42, p. 93]

Truesdell [87, pp. 101, 105, 351]

Whittaker [90, p. 323, §148]

Hartman [35, p. 91]

Barrow-Green [6, p. 83]

Goroff [70, p. I79]

Gray [32, p. 380]

Ostrogradskii [51, pp. 122–123]

Cajori [13, vol. II, p. 101, §464]

Gibbs [30, Chapter XII]

Sklar [77, p. 130] on phase space

Boltzmann [10, pp. 274–290, 443]

Jeans [43, p. 258, §206]

Kac [45, p. 63]

Lenzen [55, p. 129]