Using Index Theory to show a system has no closed orbits

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SUMMARY

The discussion centers on the application of index theory to demonstrate that a nonlinear dynamical system described by the equations \(\dot{x} = x(4-y-x^{2})\) and \(\dot{y} = y(x-1)\) has no closed orbits. The analysis reveals four fixed points, with the fixed point at (1,3) being a stable spiral with an index of 1. However, the presence of a saddle at (2,0) introduces an unstable manifold that prevents the existence of closed orbits, as trajectories cannot cross this manifold. The user expresses skepticism about Strogatz's argument and suggests that visualizing the phase portrait may clarify the situation.

PREREQUISITES
  • Understanding of nonlinear dynamics and fixed points
  • Familiarity with index theory in dynamical systems
  • Knowledge of stable and unstable manifolds
  • Proficiency in using Mathematica for plotting dynamical systems
NEXT STEPS
  • Study the concept of global unstable manifolds in dynamical systems
  • Learn about homoclinic orbits and their implications in nonlinear dynamics
  • Explore the iteration method for calculating unstable manifolds
  • Practice drawing phase portraits to visualize trajectories and fixed points
USEFUL FOR

Students and researchers in nonlinear dynamics, particularly those studying index theory and phase portraits, will benefit from this discussion. It is also relevant for anyone analyzing the stability of dynamical systems.

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Homework Statement


I'm doing a course in nonlinear dynamics using Strogatz. One of the exercises in the book is "Using index theory show that the system has no closed orbits"


Homework Equations


\dot{x} = x(4-y-x^{2}) , \dot{y} = y(x-1)



The Attempt at a Solution


Turns out there are 4 fixed points and we know that the sum of the indices of the fp within the closed orbit must equal 1.
No problem, except for the fp (1,3) which is a stable spiral(hence index of 1), which could in theory have a closed orbit around it.
Strogatz argues there can't be a closed orbit around it due to the saddle at (2,0) which has an unstable manifold in the direction of the spiral and the cycle can't cross the manifold as it would intersect another trajectory.
But as I understand things at the moment, the unstable manifold is only valid in the region close to the saddle, very clearly demonstrated by homoclinic orbits where the trajectory runs away on the unstable manifold but loops around and returns on the stable manifold.

I plotted the system in mathematica(see attachment) and it's clear that there indeed are no closed orbits, but Strogatz' argument doesn't convince me that what I drew in on the plane isn't possible.
 

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I agree, as is, this isn't terribly convincing, unless Strogatz is talking about the global unstable manifold. Is this perhaps one of those rare examples where you can explicitly calculate the global unstable manifold? Sometimes this happens; the iteration method for finding the unstable manifold might converge after a finite number of steps. I haven't read Strogatz mind you, so I don't know if he covers the iteration method.

EDIT: Also, this probably wouldn't be very helpful, since then there's the matter of just making the closed orbit "smaller" so it avoids the unstable manifold.

You might be able to convince yourself by drawing a few pictures. I know, it's a bit of a cop-out, but try it. Draw a circle for your closed orbit, and draw a saddle nearby. Extend the saddle to some global unstable and stable manifold in the configuration that you think would cause a problem. Now, try to fill the phase portrait with arrows. Can you find any points for which the trajectory couldn't be continued backwards? I think you'll find that you run into problems for some points that are near the closed orbit.
 

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