I have come across an interesting property of a dynamical system, being transformed by a map, but i haven't been able to figure out why this is happening (for quite some time now actually). Any help is greatly appreciated. Here goes then:

Let M be a n-D manifold and $\dot x=F(x)u_1, F\in \mathbb{R}^{n\times m}, x \in \mathbb{R}^{n}, u_1 \in \mathbb{R}^{m}$ be a control system evolving on M (F is the system matrix i.e. state transition function, and $u_1$ is the input of the system. For all practical purposes $u_1$ is an m-vector from an input space $\mathbb{R}^{m}$). Now let $x=\Psi (y)$ be a coordinate change on M and $u_2=M(y)u_1$ a transformation of the input $u_1$ of the first system. By applying these maps on the system, you get the new equations $\dot y=F(y)u_2$. As you may notice, F is the same in both systems. The problem is why is this happening i.e. for what systems and transformations does this property hold?

A little more elaboration

It is useful to investigate the maps more closely. In the general case one has

$\dot x=D\Psi \dot y$
$\dot x= F(x)u_1$

$\dot y=D\Psi ^{-1} F(x)u_1$, (1)

where $D\Psi$ is the Jacobian matrix of $\Psi$. In our case it actually turns out that:

$\dot y=F(y)M(y)u_1$. (2)

You can then consider that $u_2=M(y)u_1$ and get the final system,

$\dot y=F(y)u_2$,

that is, the same system. By (1),(2) you get,

$D\Psi ^{-1} F(x)u_1=F(y)M(y)u_1 \Rightarrow (D\Psi ^{-1} F(x)-F(y)M(y))u_1=0$.

Since this holds for every $u_1$, you have the condition,

$F(\Psi (y))=D\Psi F(y)M(y)$

So, what does this condition imply? What systems F and maps $\Psi$ hold this property (of system invariance)? I should note that F is nonlinear and a case study where this actually happens is the kinematic model of a unicycle robot i.e. this. Any ideas?

  • $\begingroup$ Are you sure your elaboration is correct? In particular, it seems to me that in your very first line of your elaboration, you have not applied the chain rule properly. If you 'fix' that, you'll get a condition which is much more reasonable. $\endgroup$ Dec 18, 2011 at 14:15
  • $\begingroup$ Jacque, if you're referring to $\dot x=D\Psi \dot y$, then this is correct. You can calculate it using basic analysis. $\endgroup$
    – Jorge
    Dec 18, 2011 at 17:32
  • $\begingroup$ Ok, I forgot that notation, I guess. Chain rule gives me $\dot{x} = \dot{y} \cdot D\Psi y$ for $x=\Psi y$. So there is a $y$ embedded in your $D\Psi$ which you elide (and some explicit computation confirm) that threw me off. Then the formula reads just like the change of variables formula on manifolds, no? $\endgroup$ Dec 19, 2011 at 13:49
  • $\begingroup$ Ok let me just be a little more formal on the math. The vectors x,y∈Rn (column vectors), thus differentiating gives $\dot x=DΨ \dot y$ ($D\Psi \in \mathbb{R}^{n\times n}$- a square matrix). It is indeed the change of variables formula for the derivative. $\endgroup$
    – Jorge
    Dec 19, 2011 at 16:04
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    $\begingroup$ I understand that (2) is a particular phenomenon arising in a special case Jorge is dealing with. $\endgroup$ Feb 13, 2015 at 0:37

2 Answers 2


Let me reply taking M to be the identity (indeed M is somewhat cosmetic to the discussion).

The identity $F\circ Ψ=DΨ\circ F$ is what one considers for example in the Grobman-Hartman theorem, passing from a dynamics to its linearization say at a fixed point. The possibilities for F are endless; $F$ would then be a topological conjugacy, perhaps locally, although maybe not very regular, more precisely at most Hölder continuous in general. Moreover, $F$ need not satisfy any invariance properties, which if I understand correctly is your main concern.


This is not a surprising fact for driftless systems to have symmetries. In our previous work we considered only nonlinear systems with drift and classified the symmetries accordingly. In case of a driftless system \dot x=F(x)u you can interpret $F$ (that's what it is) as a distribution of vector fields $f_1, \dots, f_m$ (column vectors of the matrix) and whenever such distribution is involutive ($[f_i, f_j]=\lambda_1(x)f_1+\cdots+\lambda_m(x)f_m$ then necessarily the system admits nontrivial symmetries (this is due to the Frobenius Theorem). That's the case you have here with the unicycle with $f_1=(\cos \theta, \sin \theta, 0), f_2=(0,0, 1)$. I can write in more extensive way but I hope you would get the idea from there.


  • $\begingroup$ OOPS!!!! The third variable in your system is \theta so there is no involutivity here!!! Hmmm let's think again. Sorry for the previous posting. Issa $\endgroup$
    – user32618
    Mar 29, 2013 at 2:42
  • $\begingroup$ Part of my initial answer (except for your system being involutive) holds true; it is indeed more general than just the involutivity. It is a well-known fact that vector fields often have symmetries. If you have a driftless system $\dot x=F(x)u$ or a linear in input system $ \dot x=f_1(x)u_1+\cdots+f_m(x)u_m$ you can consider the associated dynamical systems: $\Xi_i: \dot x=f_i(x), 1\leq i \leq m$ and the associated single input systems $\Sigma_i : \dot x=f_i(x)u_i, 1\leq i \leq m$. Denote by $G_i$ (resp. $\bar{G}_i$ the group of symmetries of $\Xi_i$ (resp. $\Sigma_i$). Obviously $G_i \subset $\endgroup$
    – Issa
    Mar 29, 2013 at 14:47
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    $\begingroup$ You should not add a 'new' answer every time you think of something more to say. You should just edit your original answer (and correct its mistakes). Having your answer scattered into several parts like this will cause confusion when someone puts in another answer and it gets mixed in with yours. (Also, there is no guarantee that the different parts will stay in the same order, in case someone votes for the last one and not the first one.) $\endgroup$ Mar 29, 2013 at 17:51
  • $\begingroup$ @Robert Bryant: I think that users are allowed to edit their posts only if they have high enough reputation. But maybe it applies only to questions and answers can be edited always. $\endgroup$ Apr 3, 2013 at 17:39
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    $\begingroup$ I've just taken advantage of the new capabilities of MO 2.0 to merge some multiple answers into comments. Robert's point stands, and for everyone else, remember you now have the ability to flag for moderator attention requesting conversion into a comment. $\endgroup$ Jun 25, 2013 at 15:32

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