Stationary points of : $y' = y^3 + y$

Question

I have a question regarding stationary points and their behavior.

Exercise :

Discuss the behavior of the stationary points of the differential equation : $$y' = y^3 + y$$

Attempt :

We can easily see that the stationary points are :

$$y^3 + y = 0 \Leftrightarrow y(y^2 + 1) = \Leftrightarrow y = \begin{cases}0\\ \pm1 \end{cases}$$

So, our stationary points are at : $a=0, b=i, c = -i$ .

To find their behavior :

First of all, we have that if $y<0$ then $y'<0 \forall y\in(-\infty,0)$, which (by theory) means :

$$\lim_{t\to -\infty}y(t) = 0, \lim_{t\to\infty}y(t) = -\infty$$

Is this correct, first of all ?

Secondly, I wanted to ask, how do I handle the $imaginary$ stationary points ? How will I study their behavior since I cannot proceed (or I can't see how to proceed) with analysing the sign of $y$ ?

Answer 1

Linearize the system at the equilibrium points to obtain:

$$\Delta y' = [3y_{\text{eq}}^2+1]\Delta y$$

We obtain for $y_{\text{eq}}=0$

$$\Delta y' = [3\cdot0^2+1]\Delta y=\Delta y$$

which is unstable.

For $y_{\text{eq}}=\pm i$ we obtain $$\Delta y' = [3(i)^2+1]\Delta y= -2\Delta y.$$

This would imply that the complex equilibria are stable (whatever the meaning of this might be). Like @Hans Lundmark suggested normally you would ignore these equilibria.