Consider the systems $$ \begin{cases}\dot{x}=-y-2x-4x^3\\\dot{y}=-2y-x\end{cases}\text{ and }\begin{cases}\dot{x}=-x-7y\\\dot{y}=-2y+10x\end{cases}. $$ Decide whether they are topologically equivalent or not.

I am not sure how to decide that.

Both systems have equilibrium $(0,0)$. If I linearize the first system at $(0,0)$ then for the linearized system, the equilibrium $(0,0)$ is a stable one. For the second system, $(0,0)$ is a stable equilibrium, too.

That's all I can say by now. Maybe there is a theorem, corollar, lemma etc. that only need these information to say something about topological equivalence. Or you just have an idea. Thanks!

  • $\begingroup$ Except that one is a stable focus while the other is a stable node! $\endgroup$ – Did Jan 10 '16 at 23:51
  • $\begingroup$ @Did For the linearized first system, $(0,0)$ is a stable node (all orbits tend to it tangentially as $t\to +\infty$), for the second system, $(0,0)$ is a stable focus (all orbits tend to it as a spiral as $t\to +\infty$). - I am sorry, but I am not sure what that tells me about topol. equivalence. $\endgroup$ – Rhjg Jan 11 '16 at 9:41
  • $\begingroup$ Well, it seems you have two distinct problems on your hands, the first one being to check whether the nonlinear system $\dot{x}=-y-2x-4x^3$, $\dot{y}=-2y-x$ and its linearized version $\dot{x}=-y-2x$, $\dot{y}=-2y-x$, are topologically equivalent or not, the second one being to check whether linear systems corresponding to a stable focus and to a stable node respectively, for example $\dot x=-y$, $\dot y=x$, and $\dot x=-x$, $\dot y=-y$, are topologically equivalent. Any idea about these two problems? $\endgroup$ – Did Jan 11 '16 at 10:24

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