I am analyzing a program which transforms vector fields by action of diffeomorphisms and feedbacks.

Here are the operation that I don't understand (it's Mathematica code)

dyn=u F + v G;
newdyn = Inverse[T1].(dyn/.{x->tmp[[1]],y->tmp[[2]],z->tmp[[3]]});

$F,G$ are two (smooth) vector fields on $\mathbb{R}^3\ni (x,y,z)=q$ and we have a control system of the form $$ \dot q = u F(q) + v G(q) $$ where $u,v$ are scalar control functions. The components of $F,G$ are polynomials in $q=(x,y,z)^\intercal$ (they are obtained by Taylor expansions).

$$\varphi(Q) = \overbrace{(F,G,-[F,G])_{\mid x=y=z=0}^\intercal}^{A}\, \, Q= q.$$

So $\varphi$ is a linear transformation and $\varphi^{-1}(q) = A^{-1} q$ and $\frac{\partial \varphi}{\partial Q}(Q) = A$.

Then they perform the pushforward of $F,G$ by $\varphi$ so that $$ \dot Q = A^{-1} \dot q = A^{-1}\left(u\, F(AQ)+v\, G(AQ)\right). $$

Also, what could be the use of such transformation $\varphi$?

On many examples, I observe that $\varphi$ removes the constant terms in the components $F_2$ and $G_1$.

Edit: Source of the Program : Aloui thesis




  • 1
    $\begingroup$ This would make more sense to me if you defined $A=[F,G,-[F,G]]$ (without transposition). In this case, $A^{-1}F\bigg|_{Q=0}=\begin{pmatrix}1\\0\\0\end{pmatrix}$ and $A^{-1}G\bigg|_{Q=0}=\begin{pmatrix}0\\1\\0\end{pmatrix}$ so, in the neighbourhood of $0$, you system would turn to $$\begin{cases}\dot{x}=u\\\dot{y}=v\\\dot{z}=0\end{cases}.$$ $\endgroup$ – Dmitry Nov 17 '17 at 11:08
  • $\begingroup$ What is the source of the problem? $\endgroup$ – Dmitry Nov 17 '17 at 11:11
  • $\begingroup$ This paper : sci-hub.cc/https://doi.org/10.1007/BF02269424 and the source file in the question. $\endgroup$ – Smilia Nov 17 '17 at 12:46
  • 1
    $\begingroup$ I think you are right, there is no transposition and your answer makes a lot of sense $\endgroup$ – Smilia Nov 17 '17 at 12:55

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