I am working on this problem:

(a) Consider a time dependent ODE $\dot{x}=A(t)x$. Let $\Psi_{t}$ by the time dependent linear flow with $\Psi_{0}=Id$. Prove by writing down the determinant directly that $$\dfrac{d}{dt}\Big|_{t=0}\det(\Psi_{t})=Tr(A(0)).$$

(b)Use part $(a)$ to generate an alternative proof of the following:

Let $\varphi_{t}$ be a flow generated by a vector field $X$. Then $\varphi_{t}$ is volume preserving if and only if $div(X)=0$ everywhere on $\mathbb{R}^{n}.$

I have shown part $(a)$ as following:

Recall that $\Psi_{t}$ is the solution of the ODE $\dot{x}=A(t)x,$ and thus it satisfies $$\dfrac{d}{dt}\Psi_{t}=A(t)\Psi_{t}.$$

Now, by Jacobi's Formula, we have \begin{align*} \dfrac{d}{dt}\det(\Psi_{t})&=Tr\Big(adj(\Psi_{t})\cdot\dfrac{d\Psi_{t}}{dt}\Big)\\ &=Tr\Big(adj(\Psi_{t})\cdot A(t)\cdot\Psi_{t}\Big)\ \text{by above recall}. \end{align*}

Then, at $t=0$, we have \begin{align*} \dfrac{d}{dt}\Big|_{t=0}\det(\Psi_{t})&=Tr\Big(adj(\Psi_{0})\cdot\Psi_{0}\cdot A(0)\Big)\\ &=Tr\Big(adj(Id)\cdot Id\cdot A(0)\Big)\\ &=Tr\Big(Id\cdot Id\cdot A(0)\Big)\\ &=Tr\Big(A(0)\Big), \ \text{as desried}. \end{align*}

However, I don't see the connection between $(a)$ and Liouville's Theorem.

The proof of Liouville's Theorem I learnt can be sketched as following:

(1) Let $m_{0}$ denote the Lebesgue measure on $\mathbb{R}^{n}$, and let $m_{t}$ denote $m_{0}$ transported forward by $\varphi_{t}$, i.e. $$m_{t}(V)=m_{0}(\varphi_{-t}(V)).$$ Then, we show that $m_{t}=m_{0}$ if and only if $\dfrac{d}{dt}\Big|_{t=0}m_{t}(V)=0$ for all Borel $V.$

(2) The rate of flow of measure across a cross-section $\Sigma$ to the flow is given by $$\int_{\Sigma}X\cdot \textbf{n}ds,$$ where $\textbf{n}$ is the unit normal vector to $\Sigma$ in the direction of the flow, and $ds$ is surface area on $\Sigma$.

(3) Let $V\subset\mathbb{R}^{n}$ be a bounded region with smooth $\partial V$. Then by Gauss-Green Theorem, we have $$\dfrac{d}{dt}\Big|_{t=0}m_{t}(V)=\int_{\partial V}X\cdot\textbf{n}ds=\int_{V}div(X),$$ which concludes the proof.

However, I did not see how to apply $(a)$ to this proof since we never used $\det(\Psi_{t})$ in the original proof.

  • $\begingroup$ Can you find a linear equation satisfied by $\det\Psi_t$? That's all there is to it and your post is almost there. $\endgroup$
    – John B
    Sep 16 '19 at 21:07
  • $\begingroup$ I think it is perhaps $\det(\Psi_{t})=\det(e^{A(t)t})=\exp(Tr(A(t))t$? $\endgroup$ Sep 16 '19 at 21:12
  • $\begingroup$ Not like that. Try to show that it is a solution of $x'=({\rm Tr} A(t)) x$. As I wrote: your post is almost there. $\endgroup$
    – John B
    Sep 16 '19 at 23:08
  • $\begingroup$ @JohnB Okay, let me think about it and I will edit my post. Thank you! $\endgroup$ Sep 16 '19 at 23:15

Let $\phi_t$ be the flow of $X$ and $\Omega$ the standard volume form of $\mathbb{R}^n$. We have $\phi_t^*\Omega=Jac(\phi_t)\Omega$, we deduce that $\phi_t$ preserves $\Omega$ if and only if ${d\over{dt}}_{t=0}Jac(\phi_t)=0$. Remark that $Jac(\phi_t)$ is the solution of the differential equation:

${d\over{dt}}\Phi_t=DX(t)\Phi_t$, where $DX$ is the differential of $X$, so $a)$ implies that $\phi_t$ preserves $\Omega$ if and only if $Tr(D(X))=div(X)=0$.

  • $\begingroup$ I am sorry that I don't see why $(a)$ implies your conclusion. How could I use $\det(\Psi_{t})$ in your argument? $\endgroup$ Sep 16 '19 at 23:30
  • $\begingroup$ $Jac(\phi_t)=det(d\phi_t)$ $\endgroup$ Sep 16 '19 at 23:31
  • $\begingroup$ But then how do you connect $\det(d\Psi_{t})$ with $\det(\Psi_{t})$? I am sorry if this question is dumb.... $\endgroup$ Sep 16 '19 at 23:44
  • $\begingroup$ Here $d\phi_t=\Phi_t$ it is the solution of the differential equation with $A(t)=DX$ where $DX$ is the differential of $X$. $\endgroup$ Sep 16 '19 at 23:47
  • $\begingroup$ Oh! I got you! Thank you so much! $\endgroup$ Sep 16 '19 at 23:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.