Stationary points and their behavior of : $x' = y-x^3, y' = -x + y^3$ via Linearisation

Question

Find the stationary points of the following system and discuss their behavior using Linearisation

$$x'=y-x^3,y'=-x+y^3$$

My solution and where I'm stuck :

To find the stationary points, we'll solve the system :

$$\begin{cases} y-x^3= 0\\ -x + y^3 =0 \end{cases}$$

which yields the following stationary points :

$$A=(0,0)$$ $$B=(-1,-1)$$ $$C=(1,1)$$

To discuss their behavior, I'll start off by calculating the Jacobian of the non-linear system :

$$J(x,y) = \begin{bmatrix} -3x^2 & 1 \\ -1 & 3y^2 \end{bmatrix}$$

Now, we'll calculate the eigenvalues of the Jacobian at each of the stationary points to make a case about their behavior and their character :

$$J(0,0) = \begin{bmatrix} 0 & 1 \\ -1 & 0\end{bmatrix}$$

so, we get $\det(J(0,0)-λI)=λ^2 +1=0\Leftrightarrow λ = \pm i$

Now, this is exactly where I'm stuck. I know that generally, when we have purely imaginary eigenvalues, it means we have a center, but in a case of an almost linear system, the topological equality for centers does not hold between it and it's linearised system. That means, that I need a different condition to determine whether the point $A=(0,0)$ is a center or that if it is a focus/spiral for the almost-linear/non-linear system.

I have been stuck on exercises that have purely imaginary eigenvalues for some days now, since I cannot find any examples related on our book or on the internet that will help me understand what I have to do in such cases (I'm Greek, so maybe I'm searching it in internet/global textbooks using wrong terminology).

I think my question is straight forward as in previous questions, some people wondered what I was asking.

I am asking exactly the following : What do I need to do to in order to discuss whether a stationary point that leads to purely imaginary eigenvalues on the linearised system/Jacobian, is either a center or a focus ?

Please, I would really appreciate anyone that can tell me what I have to do in such cases since I've found myself stuck on numerous exercises, not knowing what I have to do and not being able to find anything myself via studying, searching or trying.

Answer 1

Sometimes we can obtain everything in a straightforward manner, in particular without using Lyapunov constants.

In your case note that $$xx'+yy'=-x^4+y^4.$$ Along the line $x=0$ we have $$(x^2+y^2)'=2(xx'+yy')>0$$ outside the origin and so the origin is unstable.

Some people would describe this solution as using Lyapunov functions (namely $x^2+y^2$) although this is not a standard nomenclature.