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Given nonlinear dynamical system, if one is asked to show that this system is topologically conjugate, is it asking that the flow of nonlinear system and the flow of linearization of the nonlinear system have a homeomorphism between them?

So, let $x'=f(x)$, $x\in R^n$ be given. then should I show that the flow of this system, say $\phi_t (x)$, and the flow, say $\psi_t (x)$, of the linearized system $x'=Ax+g(x)$, where $A$ is a matrix and $g$ is remaining nonlinear term, have a homeomorphism $h(x)$ such that $h(\phi_t)=\psi_t(h(x))$?(EDIT: this part is solved)

In addition, I understand the homeomorphism of systems shows that the systems have same dynamical structure, like sink, source, etc. Am I right? and then, is it possible that the system and its linearized system can not be homeomorphic? Because, the linearized system is an approximation of the system. I think at least the local behavior should be same, so they must be homeomorphic(at leat locally). Thank you.

rekt
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  • If you're speaking about hyperbolic steady states (all equlibria those Jacobi matrix doesn't have eigenvalues on imaginary line), then it's true that local structure of non-linear equilibrium state is the same as local structure of it's linearization. Things get really different when you have non-hyperbolic steady states.One of the classical examples is system like $\dot{x} = y - x (x^2+y^2 - 1)$, $\dot{y} = - x - y (x^2 + y^2 -1)$ : for linearization origin is a center, but non-linear system doesn't have center-like behaviour of trajectories in any small neighbourhood of origin. – Evgeny May 20 '15 at 06:36
  • @Evgeny Thank you so much. that's very interesting :D – rekt May 20 '15 at 08:39

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