ODE existence of specific solutions

Question

I have difficulties with this question :

Given the ODE named (1) : $$x'=y+\sin (x^2y)$$ $$y'=x+\sin(xy^2)$$

and the :

Definition. A Petal is a solution $(x(t),y(t))$ that verifies $\displaystyle \lim _{t \to \pm \infty} (x(t),y(t)) =(0,0)$.

How can I show that the exists at most two distinct petals ?

First I remarked that the question has sense since :

$$\|(x',y')\| \leqslant \|(x,y)\| + C$$ where $C$ is a wisely chosen constant. It follows that the solution $(x(t),y(t))$ is defined on $\mathbb{R}$.

Next, the linearization in $(0,0)$ shows that $(0,0)$ is an hyperbolic point.

But how can I say more ?

Thank you

Answer 1

It looks to me that the non-linear perturbation is not strong enough to alter the behaviour of the linearized dynamical system:

$\hspace2in$

so there are no petals at all: solutions cannot cross each other, and if we rotate our reference system $45^\circ$ counter-clockwise, any solution stays in the same quadrant all the time.

Answer 2

What is important is that you've found that origin is a hyperbolic saddle point. There are exactly 2 trajectories in $\omega$-limit set (incoming separatrices) and 2 trajectories in $\alpha$-limit set (outgoing separatrices) (if we don't count the origin itself, but it's not needed here) — this is a consequence of Grobman-Hartman theorem. So this pretty answers your question: if there is a petal for this saddle, then it must contain one of the incoming and one of the outgoing separatrices. And since we have only 2 of each kind, we can't have more than 2 petals for hyperbolic planar saddle.

If I'm not confusing here, complex systems of form $\dot{z} = z^m, \; m>1,$ can be adjusted to have as many petals as needed.