# Notes on the history of Liouville’s theorem

Jordan Bell
April 10, 2015

## 1 Introduction

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

 $\|A\|=\sup\{\|Av\|:v\in\mathbb{R}^{n},\|v\|\leq 1\},\qquad A\in\mathscr{B}(% \mathbb{R}^{n}),$

$\mathscr{B}(\mathbb{R}^{n})$ is a Banach space.

## 2 Autonomous differential equations

###### Lemma 1.

If $A\in\mathscr{B}(\mathbb{R}^{n})$, then

 $\det(I+\epsilon A+o(\epsilon))=1+\epsilon\textrm{tr}A+o(\epsilon)$

as $\epsilon\to 0$.

###### Proof.

Let $\lambda_{1},\ldots,\lambda_{n}$ be the eigenvalues of $A$, repeated according to algebraic multiplicity. For $\epsilon>0$, the eigenvalues of $I+\epsilon A+o(\epsilon)$ repeated according to algebraic multiplicity are

 $1+\epsilon\lambda_{1}+o(\epsilon),\ldots,1+\epsilon\lambda_{n}+o(\epsilon),$

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

 $\det(I+\epsilon A+o(\epsilon))=\prod_{k=1}^{n}(1+\epsilon\lambda_{k}+o(% \epsilon)),$

as $\epsilon\to 0$. But

 $\prod_{k=1}^{n}(1+\epsilon\lambda_{k}+o(\epsilon))=1+\epsilon\sum_{k=1}^{n}% \lambda_{k}+o(\epsilon)=1+\epsilon\textrm{tr}A+o(\epsilon)$

as $\epsilon\to 0$. ∎

###### Theorem 2.

If $A\in\mathscr{B}(\mathbb{R}^{n})$, then

 $\det e^{A}=e^{\textrm{tr}A}$
###### Proof.

We have

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

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

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

Then, using Lemma 1,

 $\displaystyle\det e^{A}$ $\displaystyle=$ $\displaystyle\lim_{m\to\infty}\left(\det\left(I+\frac{A}{m}\right)\right)^{m}$ $\displaystyle=$ $\displaystyle\lim_{m\to\infty}\left(1+\frac{1}{m}\textrm{tr}A+o\left(\frac{1}{% m}\right)\right)^{m}$ $\displaystyle=$ $\displaystyle e^{\textrm{tr}A}.$

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

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

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

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

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

 $\lambda(e^{tA}U)=\int_{e^{tA}U}dy=\int_{U}|\det(De^{tA})(x)|dx=\int_{U}|\det(e% ^{tA})|dx=e^{t\textrm{tr}A}\lambda(U).$

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

## 3 Nonautonomous differential equations

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

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

is a vector space. For each $t\in I$ we define $B_{t}:X\to\mathbb{R}^{n}$ by $B_{t}(x)=x(t)$. It is apparent that for each $t\in I$ the map $B_{t}$ is linear. For each $x_{0}\in\mathbb{R}^{n}$, 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 $\phi_{1},\ldots,\phi_{n}\in X$, and for each $t\in I$ let $\Phi(t)\in\mathscr{B}(\mathbb{R}^{n})$ be defined by $\Phi(t)e_{i}=\phi_{i}(t)$. Then

 $\dot{\Phi}(t)e_{i}=\frac{d}{dt}(\Phi(t)e_{i})=\dot{\phi_{i}}(t)=A(t)\phi_{i}(t% ),\qquad t\in I,$

and hence

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

The Wronskian $W=W(\phi_{1},\ldots\phi_{n})$ of the ordered set $\phi_{1},\ldots,\phi_{n}$ is the function that assigns to each $t\in I$ the oriented volume of the parallelepiped spanned by $\phi_{1}(t),\ldots,\phi_{n}(t)$. That is,

 $W(t)=\det\Phi(t),\qquad t\in I.$
###### Theorem 3.

Suppose that $I$ is an open interval and that $A\in C(I,\mathscr{B}(\mathbb{R}^{n}))$. If $\phi_{1},\ldots,\phi_{n}\in X$, then the Wronskian $W=W(\phi_{1},\ldots,\phi_{n})$ satisfies

 $\dot{W}(t)=(\textrm{tr}A(t))W(t),\qquad t\in I.$
###### Proof.

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

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

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

 $\displaystyle W(t+\Delta)$ $\displaystyle=$ $\displaystyle\det\Phi(t+\Delta)$ $\displaystyle=$ $\displaystyle\det\Phi(t)\det(I+A(t)\Delta+o(\Delta))$ $\displaystyle=$ $\displaystyle\det\Phi(t)(1+\textrm{tr}A(t)\Delta+o(\Delta))$ $\displaystyle=$ $\displaystyle\det\Phi(t)+\det\Phi(t)\textrm{tr}A(t)\Delta+o\left(\det\Phi(t)% \Delta\right)$ $\displaystyle=$ $\displaystyle\det\Phi(t)+\det\Phi(t)\textrm{tr}A(t)\Delta+o(\Delta)$

as $\Delta\to 0$, i.e.,

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

which gives us

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

One checks that for any $t_{0}\in I$,

 $t\mapsto W(t_{0})\exp\left(\int_{t_{0}}^{t}\textrm{tr}A(\tau)d\tau\right),% \qquad t\in I$

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

 $W(t)=W(t_{0})\exp\left(\int_{t_{0}}^{t}\textrm{tr}A(\tau)d\tau\right),\qquad t% \in I.$

## 4 Jacobi’s formula

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

 $\Omega(x_{1},\ldots,x_{n})\det A=\Omega(Ax_{1},\ldots,Ax_{n}),\qquad x_{1},% \ldots,x_{n}\in\mathbb{R}^{n}.$ (2)

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

 $\omega(x_{1},\ldots,x_{n-1})=\Omega(x_{\omega},x_{1},\ldots,x_{n-1}).$

Thus, if $x_{0}\in\mathbb{R}^{n}$ and we define an $(n-1)$-form $\omega$ by

 $\omega(x_{1},\ldots,x_{n-1})=\Omega(x_{0},Ax_{1},\ldots,Ax_{n}),$

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

 $\Omega(x_{0},Ax_{1},\ldots,Ax_{n})=\Omega(x_{\omega},x_{1},\ldots,x_{n-1}).$

We define $\mathrm{adj}\,A\in\mathscr{B}(\mathbb{R}^{n})$ by $(\mathrm{adj}\,A)(x_{0})=x_{\omega}$. Thus, for $x_{1},\ldots,x_{n}\in\mathbb{R}^{n}$, we have

 $\Omega(x_{1},Ax_{2},\ldots,Ax_{n})=\Omega((\mathrm{adj}\,A)x_{1},x_{2},\ldots,% x_{n}).$ (3)

Hence

 $\Omega(Ax_{1},Ax_{2},\ldots,Ax_{n})=\Omega((\mathrm{adj}\,A)Ax_{1},x_{2},% \ldots,x_{n}),$

therefore

 $\Omega((\mathrm{adj}\,A)Ax_{1},x_{2},\ldots,x_{n})=\Omega(x_{1},\ldots,x_{n})% \det A,$

and because this holds for all $x_{1},\ldots,x_{n}\in\mathbb{R}^{n}$, it follows that

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

Furthermore, if $A\in\mathscr{B}(\mathbb{R}^{n})$, then one checks that for all $x_{1},\ldots,x_{n}\in\mathbb{R}^{n}$,

 $\Omega(x_{1},\ldots,x_{n})\textrm{tr}A=\sum_{i=1}^{n}\Omega(x_{1},\ldots,Ax_{i% },\ldots,x_{n}).$ (4)

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

 $\displaystyle\frac{d}{dt}\Big{(}\Omega(e_{1},\ldots,e_{n})\det A(t)\Big{)}$ $\displaystyle=$ $\displaystyle\frac{d}{dt}\Omega(A(t)e_{1},\ldots,A(t)e_{n})$ $\displaystyle=$ $\displaystyle\sum_{i=1}^{n}\Omega(A(t)e_{1},\ldots,\dot{A}(t)e_{i},\ldots,A(t)% e_{n})$ $\displaystyle=$ $\displaystyle\sum_{i=1}^{n}\Omega(e_{1},\ldots,(\mathrm{adj}\,A(t))\dot{A}(t)e% _{i},\ldots,e_{n})$ $\displaystyle=$ $\displaystyle\Omega(e_{1},\ldots,e_{n})\textrm{tr}((\mathrm{adj}\,A(t))\dot{A}% (t)),$

that is,

 $\frac{d}{dt}\det A(t)=\textrm{tr}((\mathrm{adj}\,A(t))\dot{A}(t)).$

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 $\phi$ and $U$ is a bounded open subset of $\mathbb{R}^{n}$ with piecewise smooth boundary and $f:\mathbb{R}^{n}\times\mathbb{R}\to\mathbb{R}^{n}$ is smooth, then with $U_{t}=\phi_{t}(U)$,

 $\int_{U_{t}}f(y,t)dy=\int_{U}f(\phi_{t}(x),t)\det(D\phi_{t})(x)dx;$

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

 $\displaystyle\frac{d}{dt}\int_{U_{t}}f(y,t)dy$ $\displaystyle=$ $\displaystyle\frac{d}{dt}\int_{U}f(\phi_{t}(x),t)\det(D\phi_{t})(x)dx$ $\displaystyle=$ $\displaystyle\int_{U}(Df)(\phi_{t}(x),t)\dot{\phi}_{t}(x)\det(D\phi_{t})(x)$ $\displaystyle+\frac{\partial f}{\partial t}(\phi_{t}(x),t)\det(D\phi_{t})(x)+f% (\phi_{t}(x),t)\frac{d}{dt}\det(D\phi_{t})(x)dx$ $\displaystyle=$ $\displaystyle\int_{U}\frac{Df}{Dt}(\phi_{t}(x),t)\det(D\phi_{t})(x)+f(\phi_{t}% (x),t)\frac{d}{dt}\det(D\phi_{t})(x)dx.$

Writing $J_{t}(x)=\det(D\phi_{t})(x)$, we have

 $\displaystyle\frac{d}{dt}\int_{U_{t}}f(y,t)dy$ $\displaystyle=$ $\displaystyle\int_{U}\frac{Df}{Dt}(\phi_{t}(x),t)J_{t}(x)+f(\phi_{t}(x),t)% \frac{d}{dt}J_{t}(x)dx$ $\displaystyle=$ $\displaystyle\int_{U}\frac{D(fJ)}{Dt}$

Reynolds [73, pp. 12–13, art. 14]

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

## 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]

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]

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