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

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Homework Help Overview

The discussion revolves around the stability of a nonlinear system described by a set of differential equations. The original poster seeks to demonstrate that the zero solution is nominally stable by finding a change of variable that transforms the system into a linear one.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the linearization of the system and the eigenvalues of the Jacobian matrix at the equilibrium point (0,0). There are questions about the necessity of transformations and the implications of different equilibrium points. The original poster expresses confusion regarding the term 'nominally stable' compared to 'locally stable'.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and questioning the assumptions related to stability and linearization. Some guidance has been offered regarding the linearization process, but there is no explicit consensus on the approach to take.

Contextual Notes

The problem specifies the need to show nonlinear stability and to find a transformation to a linear system, which raises questions about the feasibility of such a transformation given the presence of multiple equilibrium points.

alejandrito29
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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...
 
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What does the linearized matrix look like?
 
No, it is pretty obvious just looking at this system what the eigenvalues are and they are both negative.
 
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).
 
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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
 
alejandrito29 said:
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

alejandrito29 said:
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?:rolleyes:
 
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