Topic: Stability of Autonomous Non-linear ODEs
I'm wondering whether having a hyperbolic critical point that's not asymptotically linearly stable (ALS) in the linearisation of a system implies that the critical point is not asymptotically stable (AS) in the full non-linear system...
I know that in general, not ALS doesn't imply not AS, but it seems that the Hartman-Grobman theorem should make it true for hyperbolic critical points.
I'll lay out an example, just to make it a bit clearer.
Say we have the system:
\begin{align} \dot{x} &= -6y + 2xy - 8\\ \dot{y} &= y^2 - x^2 \end{align}
With critical points $(-1, -1)$ and $(4, 4)$. The Jacobian is:
$$ DF(x, y) = \begin{bmatrix} 2y & 2x - 6\\ -2x & 2y \end{bmatrix} $$
Now, for $(4, 4)$ the linearisation is:
$$ DF(4, 4) = \begin{bmatrix} 8 & 2\\ -8 & 8 \end{bmatrix} $$
Which has trace $\tau = 16$ and determinant $\delta = 80$. The eigenvalues are $8 \pm 4i$.
Hence, $(4, 4)$ is hyperbolic in the linearisation and describes a spiral source. By the Hartman-Grobman theorem, we can conclude that the critical point also "looks like" a spiral source in the full non-linear system. My question is, in general, is it safe to conclude that the critical point is not asymptotically stable, i.e. it is not the case that solutions in some neighbourhood tend to the critical point as time goes to infinity?
In this case, plotting the thing with Mathematica reveals that it is properly unstable at $(4, 4)$ (sorry about the lack of axes).
As nonlinearism says, the answer is yes by the Hartman-Grobman theorem.
(Got my assignment back, with tacit approval for the idea, i.e. no marks taken off)