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Consider Duffing's equation

$\ddot x + \delta \dot x + \alpha x + \beta x^3 = \gamma \cos{\omega t},$

where $\delta, \alpha, \beta, \gamma$ and $\omega$ are real parameters, $t$ represents time and $\dot x := dx/dt$.

Since there is an explicit dependence on time, this is classified as a non-autonomous system; however (following Guckenheimer and Holmes) the system can be rewritten as an autonomous system

$\dot u = v$,
$\dot v = \gamma \cos{\omega \theta} - \delta v - \alpha u - \beta u^3$,
$\dot \theta = 1$,

with $(u,v,\theta) \in \mathbb{R}^2 \times S^1$. My questions:

  1. Are there examples of systems where the above procedure doesn't work?
  2. If so, what are the implications?

Please suggest edits if the question is to broad - I'm still a novice in this area!

Best regards, \T

trolle3000
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1 Answers1

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Given a non-autonomous system $x'=f(x,t)$, you can introduce new vector function $u(t)=(x(t),t)$ which satisfies the autonomous system $u'=g(u)$ with $g(u)=(f(u),1)$. So the answer is yes, you can always turn a system into autonomous.

The implication is that the dimension of the system goes up. And while autonomous systems are often easier to understand by analysis of their equilibria, we do not get a free lunch here: the new system $u'=g(u)$ has no equilibria ($g$ never vanishes).

40 votes
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    So, basically: 1) It's always possible to do for ODE's, 2) the dimension of the phase space increases by one, and 3) equilibria disappear - that's all? – trolle3000 Jul 12 '13 at 12:34
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    @trolle All I can think of. A simple, general transformation like this can't magically make the problem much easier, or much harder. – 40 votes Jul 12 '13 at 14:09
  • that makes sense! – trolle3000 Jul 12 '13 at 14:57
  • https://math.stackexchange.com/questions/4221034/transfer-function-of-a-linear-non-autonomous-system – ayr Aug 10 '21 at 05:50