• The common definition of a hyperbolic point for a flow of a vector field $f$ is a fixed point in which the eigenvalues of the Jacobian matrix of $f$ all have non-zero real parts ($\mathrm{Re}(\lambda)\neq 0$).

  • Another definition is the hyperbolic point of a map $T$: there is no eigenvalues $\lambda$ of the Jacobian of $T$ in the considered point on the unit cercle ($|\lambda|\neq1)$.

QUESTION: What is the exact link between these two definitions?

Obviously if $\lambda=0+i\,\omega$, then $|e^{\lambda t}|=1$, but if $\lambda=1+0\,i$ (therefore $\mathrm{Re}(\lambda)\neq 0$), $|e^{\lambda t}|$ still equals to one.

(Very minor additional question: if $\mathrm{Re}(\lambda)=0$ for at least one eigenvalue $\lambda$, do we call the point an "elliptic point" of the vector field?)

Edit I just realized my sentence "if $\lambda=1+0\,i$, $|e^{\lambda t}|$ still equals to one" ($\forall t$) is of course wrong: I omitted that $t$ also appears in the exponential! So the link between the two definitions is probably the flow of the vector field, but this is still a bit unclear to me.

  • $\begingroup$ In both cases, this means there exist directions in which perturbations don't grow in the linearized system. $\endgroup$ – nonlinearism Aug 1 '14 at 22:12
  • $\begingroup$ It looks like the first one is given for continuous systems and the second one is given for discrete systems. But I am not sure. $\endgroup$ – obareey Aug 1 '14 at 22:15
  • $\begingroup$ @nonlinearism: OK, but why is it written in such ways? $\endgroup$ – anderstood Aug 1 '14 at 22:36
  • $\begingroup$ @obareey: That's true, but I don't see the link between the continuous and discrete system. $\endgroup$ – anderstood Aug 1 '14 at 22:37

The Hartman-Grobman theorem states that a fixed point is topologically equivalent to the linearized system around the fixed point if it is hyperbolic. Intuitively this means we can't say anything about the stability of the fixed point just by looking the first term of the Taylor expansion (Jacobian) if it is at a critical point, which is $\text{Re}(\lambda)=0$ for continuous systems and $|\lambda|=1$ for discrete systems.

They are critical points because of the solutions of the systems, which are $e^{At} x_0$ and $A^k x_0$ respectively.

  • $\begingroup$ Simple, clear, and the answer comes at the end: perfect! If ever you also have the answer to this question, you help with be very much appreciated :D. Thank you anyway! $\endgroup$ – anderstood Aug 2 '14 at 8:09

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