Using Hartman-Grobman to determine stability of ODE

Question

Take the system of ODES:

$\dot x=(\epsilon x+2y)(x+1)$

$\dot y=(-x+\epsilon y)(x+1)$

• Linearise this system and find the eigenvalues of its Jacobian at the origion

Answer: Eigenvalues $\lambda_{\pm}=\epsilon \pm \sqrt 2i$

• For which values of $\epsilon$ is the origin is a hyperbolic equilibrium point for the system above. Apply Hartman-Grobman theorem to determine the stability of the origin depending on the value of $\epsilon$.

Attempt: I know the first part is $\epsilon\neq \pm\sqrt 2i$ but I'm not sure what to write about the Hartman-Grobman application to determine its stability? I know it must be simple because its only worth one mark.

Any help would be great!

Answer 1

By Hartman-Grobman your system is behaving like the linearized system around the origin and the stability is the same.

So the linearized system has the eigenvalues $\lambda_{\pm}=\epsilon \pm \sqrt{2} i.$ We look at the real part of the eigenvalues to determine stability. If $\text{Re} (\lambda)\leq0$ for all $\lambda$, then your equilibrium is stable and additionally asymptotically stable if the eigenvalues have a strictly negative real part. If there is a eigenvalue such that $\text{Re} (\lambda)>0$ then it is unstable.

Notice $\text{Re}(\lambda_\pm)=\text{Re}(\epsilon)$.

Answer 2

With the given eigenvalues, $$\lambda_{\pm}=\epsilon \pm \sqrt 2i$$ the linearized system is spiraling out for positive values of $\epsilon $ and spiraling in for negative values of $\epsilon $ and the origin is a center for $\epsilon=0 $

Thus for the linearized system we have stability for $\epsilon \le 0 $

The Hartman-Grobman theorem applies to hyperbolic systems and we do not have a hyperbolic system in this problem.