no closed orbits and Liapunov functions

What are some techniques to finding/picking Liapunov functions?

What does "by considering straight lines connecting fixed points show that there are no closed orbits" mean? If fixed points are saddle point, unstable node, stable node, how are there straight lines connecting the fixed points?

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For searching purposes: "Liapunov" is also spelled "Lyapunov". – J. M. Oct 16 '11 at 13:06
I presume you are looking at a specific ODE problem? The quoted text doesn't allow you to rule out periodic orbits unless you have some extra information. – Sam Lisi Oct 17 '11 at 19:42
The word "saddle" refers to the function's behavior in the image, so there's no issue with drawing a line in the domain. – anon Oct 17 '11 at 20:01

1 Answer

Unfortunately, there are many ad hoc methods, but nothing systematic I am aware of.

Are you looking at the flow of a vector field on a surface? If so, you can use trajectories connecting fixed points to cut the surface in pieces. By uniqueness of solutions, you can't cross these trajectories. If you are lucky, you end up simplifying the problem.

Another trick is to construct a region with piecewise smooth boundary. If the corners are fixed points, and the flow is always transverse to the boundary, then you can say something about the global structure of the flow. I think the question is asking you to construct a polygon by connecting the fixed points with straight lines. If the vector field always points outwards along the polygon, then you cannot have any periodic orbit that goes through the region bounded by the polygon.

This argument can be seen as a construction of a Lyapunov function. Let P be the polygon we constructed before. Then, define p to be a defining function for P, and extend it to the whole surface by making it 0 inside and 1 outside a small tubular neighbourhood of P.

Another way of finding a Lyapunov function is to exploit some global structure of the problem, but I don't think this is what you are interested in.

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