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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.

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  • $\begingroup$ 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
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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 $Tr(D(X))=div(X)=0$.

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  • $\begingroup$ 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

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