somewhere near the beginning.

Linear centers of reversible systems are centers

Filed under: Mathematics — Alex @ 10:08 am 2/27/2007

We found a really clever way yesterday of answering the first problem on our quiz: show that a given differential equation has a periodic solution, regardless of the value of its parameter. In the end, it boils down to some clever algebra tricks and the fact that the system is time-reversible, so if you find a center in its linearized approximation, then you’ve found a center for the system itself. The problem is, I believe the last assertion is true, because several books state it (without proof), but we didn’t learn it in class, so I’d like to understand/prove it if I’m going to use it. Since the system is reversible, it seems like all you have to show is that if you start close enough to the center on the positive side of the x-axis, you’ll reach the negative side of the x-axis in finite time.

And I can’t figure out why this should be the case. I’ve had two ideas: one to show that when you get arbitrarily close to the center point, the trajectories of the nonlinearized and linearized systems are so close that if the linear one crosses the axis, so must the nonlinear one. But the best estimate I can get of the distance between the trajectories is via Gronwall’s inequality, which gives an exponentially increasing term in time; not good enough. The other idea is to kind of use the trajectories of the linear system as trapping boundaries on the nonlinear trajectories, but this is very sketchy.

It’s incredible that none of the books I’ve read prove this. In fact, most of the ones I’ve seen didn’t even mention reversibility, and these are generally hard-hitting books. My de facto supplementary reference, by Chicone, has this assertion as an exercise near the very front of the book. I reread the sections before the exercise, and still have no idea what if anything they contribute towards solving it. I suppose if I can’t prove this, I’ll just have to reference it as a ‘known fact’.

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