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This is a question about the derivative of Jacobian of the tangent map of a smooth flow. I think it might have an intrinsic formula, but I don't have a clue what it might be.

Let $M$ be an $n$-dimensional Riemannian manifold (say $C^\infty$, orientable, etc if it can save us from some trouble), $X:M\to\mathfrak{X}(M)$ be a smooth vector field (say nowhere vanishing), $\phi_t:M\to M$ be the flow induced by $X$. Now let's consider the tangent map $D_x\phi_t:T_xM\to T_{\phi_tx}M$.

Since each tangent space has an inner product structure induced by the metric, we can define the Jacobian of $\phi_t$ at $x$ as $J(t,x)=\det(D_x\phi_t)$.

Now we want to take the derivative with respect to $t$: $J_t(t,x)$. Is there any formula for this derivative? In fact a formula for $J_t(0,x)$ will be good enough.

We can choose a local orthonormal frame around $x$ and then write $D_x\phi_t$ as a time-dependent matrix, say $A(t,x)=(a_{jk}(t,x))=({\bf a}_1(t,x),\cdots,{\bf a}_n(t,x))$. Then $J(t,x)=\det A(t,x)$ and $\displaystyle J_t(t,x)=\sum_{1\le j\le n}\det({\bf a}_1(t,x),\cdots,{\bf \dot{a}}_i(t,x),\cdots,{\bf a}_n(t,x))$. Then letting $t\to0$, we get that ${\bf a}_j(0,x)={\bf e}_j$ (since $\phi_0=\text{Id}$) and $\displaystyle J_t(0,x)=\sum_{1\le j\le n}\det({\bf e}_1,\cdots,{\bf \dot{a}}_j(0,x),\cdots,{\bf e}_n)=\sum_{1\le j\le n}\dot{a}_{jj}(0,x)$.

I feel like it should be a fundamental fact in Ordinary Differential Equations. But I don't know where to start. Thank you!


A warmup: consider $M=\mathbb{R}$ and the equation $\dot{x}=f(x)$. Let $\phi(t,x)=\phi_t(x)$ be the solution with initial condition $\phi(0,x)=x$. Then $J(t,x)=\partial_x \phi(t,x)$ and $J_t(t,x)=\partial_t\partial_x \phi(t,x)=\partial_x\partial_t\phi(t,x)=\partial_x f(x(t))$. In particular $J_t(0,x)=f'(x)$.

Then let $M=\mathbb{R}^2$ and the equation $\dot{x}_i=f_i({\bf x})$. Let $\vec{\phi}(t,{\bf x})=\vec{\phi}_t({\bf x})$ be the solution with initial condition $\vec{\phi}(0,{\bf x})={\bf x}$. Then $J(t,x)=\partial_1\phi_1\cdot\partial_2\phi_2-\partial_2\phi_1\cdot\partial_1\phi_2$. Taking derivative w.r.t. $t$, $\dot{\phi}_i(t,x)=f_i({\bf x}(t))$ and hence $J_t(0,x)=\sum_j \partial_jf_j(x)=\mathrm{div}(f_1,f_2)$.

So in general $J_t(0,x)=(\mathrm{div}X)(x)$.

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  • $\begingroup$ @nonlinearism Yes! I should start with $\mathbb{R}^n$. No, maybe with $\mathbb{R}$. $\endgroup$
    – Pengfei
    Feb 12, 2013 at 16:53
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    $\begingroup$ Starting with $\mathbb{R}^n$ is probably more illustrative of the actual issues -- the problem with $\mathbb{R}$ is that everything commutes and so the ODE is easy to solve as $D_x\phi_t = e^{\int_0^t J(x(s),s)\\,ds}$, whereas in $\mathbb{R}^n$ this doesn't necessary work, because $J(x(t),t)$ may not commute with $J(x(s),s)$ for $s\neq t$. $\endgroup$ Feb 12, 2013 at 17:54
  • $\begingroup$ Here, of course, by "problem with $\mathbb{R}$" I mean "the reason that $\mathbb{R}$ is too nice to be as informative as we want..." $\endgroup$ Feb 12, 2013 at 17:55
  • $\begingroup$ Oh I see! Suppose $\phi(t,x)$ solves the ODE $\dot{x}=F(x)$ with the initial $\phi(0,x)=x$. That is, $\dot{\phi}(t,x)=F(\phi(t,x))$. Taking differential w.r.t. $x$, we get a matrix equation: $D_x\dot{\phi}(t,x)=D_xF(\phi(t,x))\cdot D_x\phi(t,x)$. Then check that $\dot{J}(t,x)=\frac{d}{d t}(\det D_x\phi(t,x))=\mathrm{tr}(D_xF(\phi(t,x)))\cdot \det D_x\phi(t,x)=\mathrm{div}F(\phi(t,x))\cdot J(t,x)$. So the Jacobian satisfies $J(t,x)=e^{\int_0^t \mathrm{div}F(\phi(s,x))dt}$. In particular $\dot{J}(0,x)=\mathrm{div}F(x)$. $\endgroup$
    – Pengfei
    Feb 12, 2013 at 21:08
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    $\begingroup$ @Pengfei You can (and should) post an answer to your own question. I promise an upvote for your effort. $\endgroup$
    – user53153
    Feb 20, 2013 at 21:59

1 Answer 1

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Following the first and third commends above, this question turns out to be a simple exercise in ODE. Let $F:\mathbb{R}^n\to\mathbb{R}^n$ be a smooth vector field and consider the ODE $\dot{x}=F(x)$ ---- $(\ast)$

For simplicity we assume $F$ is bounded and hence there exists a smooth flow $\phi:\mathbb{R}\times\mathbb{R}^n\to\mathbb{R}^n$, $(t,x)\mapsto \phi_t(x)$. For each fixed $t$, $\phi_t$ is a diffeomorphism and let $J(t,x)$ be the Jacobian of $\phi_t$ at $x$, that is, $J(t,x)=\det(D_x\phi_t)$.

To find an expression for $J(t,\cdot)$, we first rewrite $(\ast)$ as

  • $\dot{\phi}(t,x)=F(\phi(t,x))$.

Then taking differential of $x$, the left side gives $D_x(\dot{\phi}(t,x))=\frac{d}{dt}D_x\phi(t,x)$, while the right side gives $D_x(F\circ \phi(t,x))=(D_xF)(\phi(t,x))\cdot D_x\phi(t,x)$. So we get a differential matrix equation:

  • $\frac{d}{dt}D_x\phi(t,x)=(D_xF)(\phi(t,x))\cdot D_x\phi(t,x)$.

Using the same proof of Liouville's Theorem (here), we calculate

  • $\dot{J}(t,x)=\frac{d}{dt}\det D_x\phi(t,x)=\mathrm{Tr}(D_xF)(\phi(t,x))\cdot J(t,x)=\mathrm{div}F(\phi(t,x))\cdot J(t,x)$.

Letting $t=0$, we see $J(0,x)=1$ since $\phi_0=Id$ and hence $\dot{J}(0,x)=\mathrm{div}F(x)$.

More generally by taking integration, we have $J(t,x)=e^{\int_0^t\mathrm{div}F(\phi(s,x))ds}$.

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