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


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.

  • 1
    $\begingroup$ The general approach exists for sure and the solution is called Lyapunov coefficients. On MSE you can refer for Lyapunov coefficients to this answer. While the problem in this question is different from yours, you still can try to compute first and second Lyapunov coefficients using that answer. $\endgroup$ – Evgeny Nov 16 '17 at 11:00
  • $\begingroup$ @Evgeny I'm afraid we haven't been taught about Lyapunov coefficients, which I am 100% sure about as I've been at every lecture. There should be another way. I am currently considering of the standard linearisation conditions involving polar transformations. $\endgroup$ – Rebellos Nov 16 '17 at 11:25

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.

  • $\begingroup$ No: as I said, $(x^2+y^2)'>0$ along that line so the type of stability is quite obvious right? $\endgroup$ – John B Nov 16 '17 at 23:53

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