When is this function Lyanpunov?

I went to my prof for help on this and he couldn't figure it out.

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.

Re: When is this function Lyanpunov?

I'll tell you what we normally did in our ODE course.

We looked at systems of the form X'=AX+K (*)

(where X' is the vector (x',y') , A is 2x2 matrix, X is the vector (x,y) and K vector).

K is called the perturbation, or small disturbance, vector.

In your case, A is the matrix (-1,1;0,-1)

Meaning, A is the linear part of the system. The other terms go into the vector K.

One would hope that A is diagonalizable, so you could write A=EDE^-1, multiply the (*) equation from the left with E^-1, and you'd get X1' = DX1 + (E^-1)K

where X1 = (E^-1) X

You would then define a Lyapnunov function similar to what you've written, but in terms of the new variables (x1,y1)=X1. (i.e V=x1^2 + w * y1^2)

(To avoid confusion with the k in your original post, I've changed it to w)

You could in this fashion, find that domain (if it exists) where the system is asymptotically stable near the origin.

BUT, in your case A is not diagonalizable, in fact it's in a Jordan form, with -1 as an eigenvalue.

In our course we have not dealt with Jordan forms. But you could try it (I will definitely try it tomorrow out of curiosity ).

You'd write A=EJE^-1 and hope for the best :-)

Re: When is this function Lyanpunov?

Thanks for the reply, zokomoko. 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.