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I have 2d dynamical non-linear system $\dot{x} = f(x)$ with $x\in \mathbb{R}^{2}$ with exactly two stable attractors $x_{1} = ( a\quad b)^{T}$ and $x_{2} = (c\quad d)^{T}$ and one saddle point (One eigenvalue of the linearized system is always positive, and the other one can be positive or negative.) Moreover, I can show that the system never goes to infinity and has no limit cycle, so that it basically reaches one of the attractor.

Is there a way to compute the unstable manifold (or another trick), to know whether the point $(a\quad d)^{T}$ will end-up in $x_{1}$ or in $x_{2}$ ? I can of course do the simulations, but my aim is to find conditions on $f$ such that this is the case.

Many thanks!

citronrose
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  • Are you asking a general question about dynamical systems, or do you have a specific dynamical system in mind? – John Barber Jun 24 '16 at 01:10
  • @John: It's a general question. – citronrose Jun 24 '16 at 05:09
  • @John: I only know $f$ is smooth and has three zeros. – citronrose Jun 24 '16 at 05:21
  • Do you actually want to calculate the separatrix for the two basins of attraction? You can solve your problem if you have some potential $\varphi$ for $f$, i.e., $f = \nabla\varphi$. – Tobias Dec 06 '16 at 09:23
  • Calculating the separatrix would be great, but for me it is sufficient to know on which side of the separatrix is the point $(a\ d)^{T}$. Using a potential would require that $\frac{\partial f_{x}}{dy}=\frac{\partial f_{y}}{dx}$, right? – citronrose Dec 06 '16 at 09:57
  • Yes, the field must be circulation free for that purpose. The location of the separatrix is at the location of the ridge of the potential. The ridge can be calculated as enveloppe of a family of curves $x\mapsto \varphi(x,y)$ with $y$ as parameter (or the other way around $y\mapsto \varphi(x,y)$ with $x$ as parameter). – Tobias Dec 06 '16 at 11:41

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