# Alternative Proof of Liouville's Theorem of Linear Flow

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.

• Can you find a linear equation satisfied by $\det\Psi_t$? That's all there is to it and your post is almost there. Sep 16 '19 at 21:07
• I think it is perhaps $\det(\Psi_{t})=\det(e^{A(t)t})=\exp(Tr(A(t))t$? Sep 16 '19 at 21:12
• 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. Sep 16 '19 at 23:08
• @JohnB Okay, let me think about it and I will edit my post. Thank you! 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 $$Tr(D(X))=div(X)=0$$.
• I am sorry that I don't see why $(a)$ implies your conclusion. How could I use $\det(\Psi_{t})$ in your argument? Sep 16 '19 at 23:30
• $Jac(\phi_t)=det(d\phi_t)$ Sep 16 '19 at 23:31
• But then how do you connect $\det(d\Psi_{t})$ with $\det(\Psi_{t})$? I am sorry if this question is dumb.... Sep 16 '19 at 23:44
• Here $d\phi_t=\Phi_t$ it is the solution of the differential equation with $A(t)=DX$ where $DX$ is the differential of $X$. Sep 16 '19 at 23:47