# Determining stability of equilibrium points for a non linear system

Given the system:

$$\left\{ \begin{array}{} \dot x=-x^3y^2 \\ \dot y = -2x^2y^3 \end{array} \right.$$

I need to find the equilibrium points and to determine whether the system is stable around them. I'v found $(0,0)$ to be a stable equilibrium point, using the Lyapunov function $V(x)=x^2+y^2$.

The rest of the equilibrium points are $(x_0,0), (0,y_0) \; , \; x_0,y_0 \in \Bbb R$. I'm having trouble with determining wether they are stable or not. Linearization is not useful in this case, and I couldn't find any Lyapunov function.

• Just asking... why linearization is not useful? – Mateus Rocha Jul 5 '18 at 6:36
• The equilibiruim points are not hyperbolic, meaning we can't use the linearization theorem. – user401516 Jul 5 '18 at 6:50

Your system is $$\frac{d}{dt} \begin{pmatrix} x \\ y \end{pmatrix} = x^2 y^2 \begin{pmatrix} -x \\ -2y \end{pmatrix} ,$$ which only differs from the linear system $$\frac{d}{dt} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -x \\ -2y \end{pmatrix}$$ by the nonnegative factor $x^2 y^2$.
This means that except for the $x$ and $y$ axes (where $x^2 y^2=0$), the trajectories for your system will be the same as for the linear system, up to a reparametrization of time.
Using this, you should be able to see what happens if you perturb an equilibrium $(x_0,0)$ or $(0,y_0)$ a little (so that you move off the axis).
• Do you mean that for each starting point, the solution will convert to the point $(0,0)$, meaning the system is unstable for every other equilibrium point? – user401516 Jul 5 '18 at 7:20
• Yes, it's rigorous. Given a trajectory $(x(s),y(s))$ of one system, one can let $s=s(t)$ and derive an ODE that $s$ has to satisfy in order for $(x(s(t)),y(s(t)))$ to be a solution of the other system. The solution of that ODE gives the reparametrization. (Intuitively, it's quite clear that the phase portrait only depends on the direction of the vector field; the magnitude is only important if you want to know how fast you are moving along the trajectory at any given point – exactly the information which is suppressed in the phase portrait.) – Hans Lundmark Jul 5 '18 at 7:38