I have to draw the phase portrait of this ODE:
$$\ddot{x} = -x (x – 1) (x + 2)$$
Since this ODE is an autonomous equation I know that it can be represented as:
\begin{align}
\dot{x} &= y\\
\dot{y} &= -x (x – 1) (x + 2), (x, y) \in \mathbb{R}^2
\end{align}
For this specific system, the nullclines are given by:
$$\dot{x}=0 \rightarrow y = 0$$
$$\dot{y}=0 \rightarrow x=0, x=1, x=-2$$
We can see that $\dot{y}$ changes sign at each of the equilibrium points. Specifically, it is positive for $x<-2$, negative for $-2<x<0$, positive for $0<x<1$, and negative for $x>1$. This means that in the $x$–$y$ phase plane, $y$ will increase when $x<-2$ or $0<x<1$, and $y$ will decrease when $-2<x<0$ or $x>1$. I don't know how to proceed from here, what should I do?
Best Answer
Find the critical points where $x' = y' = 0$ simultaneously
$$(x, y) = (-2, 0), (0,0),(1,0)$$
Find the Jacobian of your matrix
$$\mbox{jac}(x,y) = \begin{pmatrix} 0 & 1 \\ -3 x^2-2 x+2 & 0 \\ \end{pmatrix}$$
Evaluate the eigenvalues at each critical point of the Jacobian
At $(x,y) = (-2,0)$, $\lambda_{1,2}, = i \sqrt{6},-i \sqrt{6}$, trajectories are ellipses.
At $(x,y) = (0,0)$, $\lambda_{1,2}, =-\sqrt{2},\sqrt{2}$, trajectories are a saddle point.
At $(x,y) = (1,0)$, $\lambda_{1,2}, = i \sqrt{3},-i \sqrt{3}$, trajectories are ellipses.
That is enough to give a rough idea of what the phase portrait looks like.
Here are some notes to draw the phase portrait by hand and these notes for stability.
Using this online tool, we get the phase portrait (do it by hand using the above)
Here is the same thing using Mathematica
Here is a much more detailed phase portrait using Mathematica