Estimating a dynamical system's behavior without using Liapunov theorem

Question

Assume that we have the following dynamical system $$x'=(\epsilon x+2y)(1+z)$$ $$y'=(-x+\epsilon y)(1+z)$$ $$z'=-z^3$$ Then how can I show that any solution that started from the region $z>-1$ tends to the only equilibrium of this system, origin? Since I'm trying to verify the Liapunov theorem for this case I want not to use Liapunov theorem to show this. My trial was letting $m=\frac{y}{x}$ so that $$x=Ce^{-{\epsilon\over\sqrt{2}}\arctan(\sqrt{2}m)}\frac{1}{\sqrt{2m^2+1}}$$ and $$y=mx$$ and used the fact that the solution keeps sprialing aroung the origin so $m$must be $\infty$ at some moment so that $x$ becomes zero. But this proof has a lot of gaps and is not so persuasive. How can I show it rigorously, or at least without serious gaps?

Answer 1

Let $A(u,v)=(\epsilon u+2v,-u+\epsilon v)$, then $(x',y')=A(x,y)\cdot(1+z)$ and $z'=-z^3$ hence $(x,y)$ stays on the trajectories of the system $(u',v')=A(u,v)$, only, with a time change and possibly turning backwards. The (trivial) phase diagram of $z'=-z^3$ shows that, for every starting point, $z\to0$ hence, after a time, $\frac12\leqslant z+1\leqslant2$ and there is no more turning point.

The eigenvalues of the $(u,v)$ linear system are $\epsilon\pm\mathrm i\sqrt2$ hence:

• If $\epsilon\lt0$, then $(x,y,z)\to(0,0,0)$
• If $\epsilon=0$, then $z\to0$ and $(x,y)$ circles around $(0,0)$ on the ellipse $x^2+2y^2=x_0^2+2y_0^2$ clockwise
• If $\epsilon\gt0$ and $(x_0,y_0)=(0,0)$, then $(x,y,z)\to(0,0,0)$ since $(x,y)=(0,0)$ uniformly and $z\to0$
• If $\epsilon\gt0$ and $(x_0,y_0)\ne(0,0)$, then $z\to0$ and $(x,y)$ explodes in the sense that $\|x\|\to\infty$ and $\|y\|\to\infty$