I would like to know, is there any dynamical system where the eigenvalues of a fixed point fail to explain the nature of the fixed point(whether they are stable or unstable).
Hi Suresh
Consider the Rossler system in 3D:
xdot=-y-z, ydot=x+y, zdot=1+xz-2z.
There is one non-hyperbolic critical point at (1,-1,1) and the eigenvalues are {0, -i, i}.
Best wishes
Stephen
Hi !
http://www.scholarpedia.org/article/Equilibrium#Hyperbolic_Equilibria
"The equilibrium is said to be hyperbolic if all eigenvalues of the Jacobian matrix have non-zero real parts."
"local phase portrait of a hyperbolic equilibrium of a nonlinear system is equivalent to that of its linearization. This statement has a mathematically precise form known as the Hartman-Grobman Theorem."
In other words, eigenvalues may fail to detect properties of fixed points, which are non-hyperbolic (i.e. their Jacobian matrices have at least one eigenvalue whose real part is zero).
One of the simpliest examples is:
x' = f(x) = -x^3
In this case we have a fixed point x=0 with df/dx(0) = 0. Nevertheless, the fixed point is stable. Therefore, linearization does not predict properties of the system properly.
If you want to find out more, I suggest "Nonlinear dynamics and chaos" by Strogatz for further reading.
Have a good day !
Marek
Hello Suresh,
It seems to me that you can answer my question here in my researchgate. Could you please take a look.
The stability of those systems (with a non-hyperbolic equilibrium) that do not have an eigenvalue with a positive real part cannot be determined from the eigenvalues and requires a nonlinear analysis.
If some or all eigenvalues of the jacobian matrix are zero then we can not say anything about the stability of the system. for example:-the system x'=[0, 1: 0, 0]x is unstable, and the system x'=[-1, 0,0; 0,0,0;0,0,0]x is stable.
In a numerical calculation of the Lyapunov Exponents from the data (not from the equations of the dynamical system), how can one be sure that the Real part of the Eigenvalue is zero? There is always an error, right? One could find a positive Real part (depending, probably on the chosen procedure and on its parameters) and think that the fixed point is unstable.
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