(I have posed this question over at math.se but since there were no answers I hope it's okay to post here.)

When dealing with (nonlinear) dynamical systems, one often deals with state space representation, i.e. systems of the form

$$\dot{x}=f(x),\quad x(t)\in\mathbb{R}^n.$$ Let $x^*$ be a solution of this system, then the variational equation reads

$$\dot{x}= \underbrace{\frac{\partial f}{\partial x}(x^*)}_{=:A}\cdot x\quad (1)$$

or with $D:=\frac{d}{dt}$ in operator form


where $I_n$ is the $n$-th unit matrix. The Elements of $P(D)$ in general are meromorphic functions in $D$, so we are dealing with Ore polynomial matrices with the shifting rule

$$Da=\dot{a}+aD$$ for all meromorphic functions $a$. Here is what I am very confused about: How can be assumed $\dot{x}=Dx$ when writing equation (1) in the form (2), and not $\dot{x}=Dx-xD$ (according to the shifting rule)? And why can't i factor out $D$ to the right instead, i.e. $\dot{x}=xD$ (which by the shifting rule would result in something different)?

Also, what I have encountered a lot, is implicit equations of the type $F(x,\dot{x})=0$ where the operator notation is acquired by $$0=\underbrace{\left(\frac{\partial F}{\partial \dot{x}}D+\frac{\partial F}{\partial x}\right)}_{=:P(D)}x.\quad (3)$$

Here too, I am wondering why this is not say $$0=\underbrace{\left(D\frac{\partial F}{\partial \dot{x}}+\frac{\partial F}{\partial x}\right)}_{=:P(D)}x\quad (4)$$

instead. What am I missing here?

EDIT: By $\dot{x}$ i mean the time derivative of $x$, so $\dot{x}=\frac{\partial x}{\partial t}.$

  • $\begingroup$ Please explain your notation. What is the difference between $\dot{x}$ and $Dx$? $\endgroup$ – user539887 May 17 '18 at 22:07
  • $\begingroup$ I have added what I mean by $\dot{x}.$ $\endgroup$ – emma May 17 '18 at 22:09

What you're mixing up here are the two meanings of symbols like "$x$". In $\frac{dx}{dt} = f(x(t))$, $x$ simply refers to a function $\mathbb{R}\to\mathbb{R}^n$. The commutator relation $[D,a] = \dot{a}$ on the other hand is a relationship for differential and multiplication operators, i.e. $a$ is not the function $a$ itself, it is shorthand for the multiplication operator $C^\infty \to C^\infty, f\mapsto a\cdot f$. Both the multiplication operators and the differential operators are endomorphisms of $C^\infty$ (or a similar space) and obey this commutator relationship.

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  • $\begingroup$ So, I think I see the point: $D$ is just the differentiation of $f$, and $[D,a]=\dot{a}$ is a long-winded way of writing the product rule for the derivatives. $\endgroup$ – user539887 May 17 '18 at 22:29
  • $\begingroup$ You're both correct, but I don't see how this explains my questions, e.g. that equation (3) holds instead of equation (4). $\endgroup$ – emma May 18 '18 at 5:19

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