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.

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

Here's a hint:

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).

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    $\begingroup$ 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? $\endgroup$ – user401516 Jul 5 '18 at 7:20
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    $\begingroup$ Exactly! (For any starting point off the axes, to be precise, but I suppose that's what you meant.) $\endgroup$ – Hans Lundmark Jul 5 '18 at 7:22
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    $\begingroup$ Yes, that's what I meant. Very nice thinking, but is this claim formal enough? I'm not sure I completely understand why the trajectories of my system will be the same as the linear system up to reparametrization of time. $\endgroup$ – user401516 Jul 5 '18 at 7:26
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    $\begingroup$ 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.) $\endgroup$ – Hans Lundmark Jul 5 '18 at 7:38

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