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March 3rd, 2012, 08:05 AM   #1
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When is this Lyapunov?

I went to my prof for help on this and he couldn't figure it out, although he assured me he has figured it out in the past.

I am trying to determine the values of k for which V(x,y)=x^2+ky^2 is Lyapunov for the following first order system:

x'=-x+y-x^2-y^2+(x)(y^2) ; y'=-y+xy-y^2-(x^2)(y)

I'd also like to like to be able to say something about the domain of stability for the case k=1. So far I've intuitively determined that there is a neighborhood around (0,0) for which the Lyanpunov's(k=1) derivative is always positive, thus unstable, thus there is no domain of stability for the case k=1. I'm having some trouble showing it in a rigorous manner, though.

Any help would be greatly appreciated.
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March 6th, 2012, 04:56 PM   #2
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Re: When is this Lyapunov?

I have actually figured this one out. I made a mistake for the k=1 case. I think I misplaced a negative or something. It turns out that when k=1, the derivate of V(x,y) can be factored as -2(x+y+1)(x^2-xy+y^2) and since (x^2-xy+y^2) is always positive in any deleted neighborhood of the origin, we need for x+y+1 to be greater than 0. This gives y>-x-1. So we have asymptotic stability at the origin and the domain of stability could be a ball of radius sqrt(2)/2 centered at the origin. Sqrt(2)/2 is the infimum of the distances from (0,0) to the line y=-x-1 and it is in the region where y>-x-1.

As for finding the values of k for which V(x,y)=x^2+ky^2 is Lyapunov for the system in question, here is what I did:
As (x,y) gets close to the origin, the only terms of the gradient derivative that aren't negligible are the 2 degree terms, meaning V'(x,y)~-2x^2+2xy-2ky^2. So I'm saying that in a small enough neighborhood of the origin, all of the terms of 3rd and 4th degree are small. Okay, so then we need -2x^2+2xy-2ky^2<0 or equivalently, k>(x/y)-(x/y)^2. Then I substitute cot(a)=(x/y) and maximimize the right side giving k > 1/4. Sorry, I am skipping some steps. If something doesn't make sense let me know.
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