Differental equation system and Lyapunov stability

[prehisto]

Homework Statement

Example:
x'=y-x^3
y'=-x-y^3

Homework Equations

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 I am not mistaken,than Ishould use Lyapunov stability,because the linear system is stable and I can't say anything about original system.
( I don't know why i can't 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 relevant 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?

 

[pasmith]

Homework Statement

Example:
x'=y-x^3
y'=-x-y^3

Homework Equations

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

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.

So if I am not mistaken,than Ishould use Lyapunov stability,because the linear system is stable and I can't say anything about original system.
( I don't know why i can't 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 relevant 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

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

V'=2xy-2x^4-2yx-2y^4=-2(x^4-y^4)
Can i say the function is asympt.stable because V'=-V ?

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.