# Differental equation system and Lyapunov stability

1. Jan 14, 2014

### prehisto

1. The problem statement, all variables and given/known data
Example:
x'=y-x^3
y'=-x-y^3

2. Relevant equations

3. The attempt at a solution

Linear system
x'=y
y'=-x
Is stable because Det(P-$\lambda$E)=$\lambda$2+1
$\lambda$1,2=+-i

So if Im not mistaken,than Ishould use Lyapunov stability,because the linear system is stable and I cant say anything about original system.
( I dont know why i cant tell anything about the original system, I just now it like "algorithm")

So The Lyapunov function in general looks like V=ax^2+by^2
So V'=2axx'+2byy'
I substitute x' and y' from original system:

V'=2axy-2ax^4-2byx-2by^4
So my book says that xy is not relevent and in order to get rid of them
2a-2b=0 -> a=1 and b=1
So V=x^2+y^2

Now I have the function which will allow me to determine stabilty.
V'=2xx'+2yy' Again I do the same - take x' and y' from original system
V'=2xy-2x^4-2yx-2y^4=-2(x^4-y^4)
Can i say the function is asympt.stable because V'=-V ?

In which cases i have to use Lyapunov stability,linearization is not enough ( or its not so easy to determine) ?
And do I have to determine a and b constants allways or I can just use V=x^2+y^2?

2. Jan 14, 2014

### pasmith

The origin is not asymptotically stable, since trajectories of the linear system are circles centered on the origin. This is due to the fact that the eigenvalues of the linear system have zero real part. If there is an eigenvalue with zero real part then the linearization doesn't tell you everything about the stability of the fixed point in the full system.

Or you could substitute $a = b = 1$ in the expression for $V'$ you already have.

No, because $-V = -(x^2 + y^2) \neq -2(x^4 + y^4) = V'$. But it is the case that the origin is asymptotically stable because $V' < 0$ everywhere except at the origin.