## Friday, 6 May 2016

### 2014 exam question 4 (v)

A student asked me about the model answer, which is very short and perhaps not so obviously correct.
'there exists $$x$$ in $$A$$ such that the $$\omega-$$limit set of $$x$$ is not contained in the stable manifold of the singularity. Hence $$A$$ contains a periodic orbit.' .
The $$\omega$$-limit set of a point $$x$$ are the points to which $$\phi^t(x)$$  accumulates, ie points $$y$$ so that there exists an increasing sequence $$t_n$$  such that $$lim_{n\to\infty} \phi^{t_n}(x)=y$$.
If an equilibrium $$y$$ has a stable manifold, then the $$\omega$$-limit set of every point on this manifold equals $$y$$ (as the point $$x$$ converges to $$y$$  under the flow). However, if we have a saddle point, there exists also an unstable manifold. If an initial point $$x$$ does not lie on the stable manifold of an equilibrium, then by definition it does not converge to the equilibrium. It is a more subtle question whether it could accumulate to the equilibrium. I did not make this exam question, but the model answer if perhaps a bit too brief. It could namely be that a for point $$x$$  that does not lie on the stable manifold of an equilibrium $$y$$, we still have $$y\in\omega(x)$$. For instance, there could be a heteroclinic or homoclinic cycle (consisting of equilibria and connecting orbits) to which $$x$$ accumulates, and that contains the saddle equilibrium $$y$$. In this exam question, there is only one equilibrium, so this implies that we only could have a homoclinic cycle (connecting orbit to one saddle), but this homoclinic cycle to a saddle would imply the existence of an equilibrium inside the area enclosed by the homoclinic loop (by PB arguments similar to the conclusion about the existence of an equilibrium inside the area enclosed by a periodic orbit) and as there is only one equilibrium in the system under consideration, this cannot be the case here. So there is no homoclinic cycle, there is only one equilibrium in $$A$$ and orbits cannot leave $$A$$ but also do not converge to the equilibrium. Then by PB they need to accumulate to a periodic solution (in $$A$$).

#### 1 comment:

1. I am sure this helped the student to answer the question correctly. keep on helping others and spreading positivity among people. Be optimistic!