Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What does smooth dynamical system mean? It is the title of a paper I am supposed to read in non linear systems.

share|cite|improve this question
Can you provide a reference to the specific paper? The Wikipedia article on dynamical systems states that smooth can mean different things in different contexts. – Daniel Geisler Apr 1 '14 at 12:21
Smooth usually means that the function is continuously differentiable. – Kyle Kanos Apr 1 '14 at 12:45

A dynamical system is a system that evolves by a rule over time. As opposed to modeling a system with a PDE, the system is modeled by some type of iterated function, $f^t(x)$. Smooth implies that the function being iterated is differentiable, so we are not talking about an iterated system like cellular automata.

share|cite|improve this answer

Many dynamical systems involve recursive sequences of simple functions, $f(f(f(...f(x))))$ which are iterated n times to lead to some discrete iteration sequence (a Picard sequence) with often tractable properties. It is possible however, to analytically continue the discrete iteration index n to a continuous non integer one, fractional, infinitesimal, or negative, through the miracle of functional conjugation and to produce continuous orbits, $f_t(x)$, whose velocity and continuous phase-space trajectories may be monitored.

The theory that makes this possible is the celebrated Schröder's equation (1870) which allows easy calculation of functional square (and higher) roots, and renormalization trajectories, since it turns out that the original Gell-Mann-Low Renormalization Group equation (1954) is, in fact, this very equation of Schröder, RG.

Typically, you use it, e.g., to extend the Beverton-Holt map $f(x)=x/(2−x)$ to its continuous iterate $$f_t(x)=\frac{x}{2^t + x(1-2^t)} ~,$$ where f(x) is the above for t=1. The resulting evolution surface is thus smooth.

share|cite|improve this answer

Your Answer


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.