Stability of Nonlinear System: Can the Zero Solution be Nominally Stable?

[alejandrito29]

i need show that at the following system the zero solution is nominally stable, using some change of variable that transforme in a linear system

$$\frac{dx}{dt}=-x + \beta (x^2+ y^2)$$

$$\frac{dy}{dt}=-2y + \gamma x y$$

i tried with the eigenvalues of the Jacobian matrix at (0,0), but one of them is positive , then the system is unstable...

 

[voko]

What does the linearized matrix look like?

 

[HallsofIvy]

No, it is pretty obvious just looking at this system what the eigenvalues are and they are both negative.

 

[epenguin]

What point are you linearising about?

OK now I see you say only (0, 0) is required. Doesn't seem to me you need any transformation for that point.

If I am not mistaken there is the possibility of three 'equilibrium' points - the other two may be more interesting than (0, 0).

 

[alejandrito29]

the problem says:
"show that the zero solution is nonlinear stable. For this, find the change of variable that transforms this system in a linear system"...

i don't understand

 

[epenguin]

i need show that at the following system the zero solution is nominally stable, using some change of variable that transforme in a linear system

the problem says:
"show that the zero solution is nonlinear stable. For this, find the change of variable that transforms this system in a linear system"...

i don't understand

Can anyone tell me what 'nominally stable' means? I know what 'locally stable' is, which would usually be the question.

To transform the whole system in which there are in general three different stationary points into a linear one would seem on the face of it impossible, isn't it?