Interpretation of Phase Portrait

Question

I have the following system $x'=f(X)$ of ODES: \begin{align} x_1'=& -4x_1^3(x_2-2)^2 \\ x_2'=& 2x_1^4(2-x_2) \end{align}

Solving for equilibria: I got $1$ at $(0, 2)$. I plotted this and I am curious about the strange behavior around the line $y=2$ (it looks like tiny oscillations). There is definitely saddle behaviour everywhere else. What does this mean:

Is there in fact only one equilibrium?

Answer 1

Solving for the critical points, we find the only CP is $(0,2)$.

However, look at what happens to the system at those points individually:

• At $x_1 = 0$, what happens to the system, regardless of $x_2$?
• At $x_2 = 2$, what happens to the system, regardless of $x_1$?

This is what you are seeing in the strange looking phase portrait as the vertical line $x_1 = 0$ or the horizontal line $x_2 = 0$ make the system zero, regardless of the other value (that is, for all other values, the system is $x'_1 = x'_2 = 0$).

We can see these behaviors in the strange looking phase portrait, but you need to pick out the observations above. We have: