Lyapunov stability – zero real part eigenvalues
The stability of equilibria of a nonlinear system x’=f(x) usually can be determined by using the corresponding linearized system x’=Ax. If the eigenvalues of A are all on the left open plane, the nonlinear system is (locally) asymptotically stable; if at least one eigenvalues of A is on the right open plane, the nonlinear system is (locally) unstable. But if some eigenvalues of A are on the l.o.p, and the others are on the imaginary axis, the stability can not be determined (at least directly) using the linearized system. I want to highlight two things:
(1) Why if some of eigenvalues are of zero real parts, the stability can not be determined?
The intuitive explanation is based on perturbation. The nonlinear system is different from the linearized system around the equilibrium, but still very similar when very close to the equilibrium. So if the eigenvalues at the equilibrium is negative/positive, we can expect the eigenvalues of any linearized system at a point near the equilibrium is also negative/positive, of course they would vary a little bit. But if the eigenvalue at the equilibrium is zero, then at another point close to the equilibrium, the eigenvalues can be positive or negative! In this sense, we can’t determine the stability in this case.
(2) How to tackle the zero-real-part eigenvalues?
Use Center manifold theory. Why called center manifold? because for linear systems when all eigenvalues are on the imaginary axis, the equilibrium is called center!
Edit： some notes about invariant manifold. <Hassan> page 304
(a) Manifold: x is the state vector in n-dimensional space. The solutions of the following vector equation is a manifold:
If g(x) is a k-dimensional vector, we can expect the manifold is with (n-k) dimensions. Why? consider the DOF.
(b) Invariant Manifold: A manifold is invariant for x’=f(x), if x(0) in the manifold, the trajectory will never come out. So an invariant manifold of a nonlinear system is nothing but a union of a bunch of complete trajectories. (by complete we mean not part of a trajectory).