Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

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

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

share|improve this question
    
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. –  Jacques Carette Dec 18 '11 at 14:15
    
Jacque, if you're referring to $\dot x=D\Psi \dot y$, then this is correct. You can calculate it using basic analysis. –  Jorge Dec 18 '11 at 17:32
    
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? –  Jacques Carette Dec 19 '11 at 13:49
    
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. –  Jorge Dec 19 '11 at 16:04
    
How do you obtain (2) from (1)? I don't see why there's (2) –  Shuchang Sep 21 '13 at 1:18

1 Answer 1

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.

Issa

share|improve this answer
    
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 –  user32618 Mar 29 '13 at 2:42
    
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 –  Issa Mar 29 '13 at 14:47
2  
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.) –  Robert Bryant Mar 29 '13 at 17:51
    
@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. –  Vít Tuček Apr 3 '13 at 17:39
3  
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. –  Scott Morrison Jun 25 '13 at 15:32

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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