Let $\dot x=f(x)$ (where $x\in\mathbb R^n$) an ODE and let $\bar x$ an hyperbolic equilibrium point, i.e. the real part of the eigenvalues of $\nabla f(\bar x)$ are non-zero. Let $$\dot y_t=\nabla f(\bar x)y_t,$$
the linearized equation near to $\bar x(t)=\bar x$.
As far as I understand, Hartman-Grobman theorem say that there is an homeomorphism between orbits of the ODE near $\bar x$ and the orbits of the linearized system near $0$ that preserves the sense of orbits.
Q1) Does this implies that stability (resp. unstability) of $0$ in the linearized system is equivalent to the stability (resp. unstability) of $\bar x$ in the non-linear system ?
Q2) What other connection between linearized and non linearized system arise from this theorem ?
Best Answer
The Hartman-Grobman theorem guarantees that, essentially, if we have a hyperbolic fixed point (Jacobian at fixed point has eigenvalues with non-zero real part), then the non-linear system locally looks like the linearized system near the origin. In this case, we can deduce stability results from the linear system. If the fixed point is not hyperbolic, then we can't use Hartman-Grobman anymore. However, all hope is not lost; there are still things that we can say. In this case, however, we must resort to using center manifold theory, which is a bit more complicated. See e.g. Perko.