basins-of-attractiondynamical systemsordinary differential equationsstability-in-odesstability-theory
Consider the system $$x'=(\epsilon x+2y)(z+1)$$
$$y'=(\epsilon y-x)(z+1)$$
$$z'=-z^3$$
(a) Show that the origin is not asymptotically stable when $\epsilon=0.$
(b) Show that when $\epsilon <0,$ the basin of attraction of the origin contains the region $z>-1.$
I did (a) showing that at least one eigenvalue of the jacobian matrix had not negative real part.
And I'm stuck in (b), I don't understand what do I have to prove?
Can someone help please?
This is usually called Fatou's theorem, I think. Good references for it are
The basic idea is this. Suppose that $f\colon \mathbb{P}^1(\mathbb{C})\to \mathbb{P}^1(\mathbb{C})$ is a rational map of degree $d\geq 2$ and that $z_0$ is an attracting fixed point, say with $f'(z_0) = \lambda$ with $0<|\lambda|<1$. Assume for contradiction there is no critical point in the immediate basin of attraction of $z_0$. Because $z_0$ is attracting, there is some small ball $U_0$ around it which lies in the basin of attraction of $z_0$, say with $f(U_0)\Subset U_0$. If $U_0$ does not contain a critical value, then there is some inverse branch $f^{-1}\colon U_0\to U_1$, with $U_0\Subset U_1$. Similarly, if $U_1$ does not contain a critical value, there is an inverse branch $f^{-1}\colon U_1\to U_2$, where $U_1\Subset U_2$. Continuing in this fashion, we can construct inverse branches $f^{-n}\colon U_0\to U_n$ with $U_0\Subset U_n$. Moreover, the $U_n$ don't meet the Julia set of $f$ by construction (they are contained in the basin of attraction of $z_0$). Thus by Montel's theorem there must be a subsequence of the $f^{-n}$ converging on $U_0$. This isn't possible, though, since $(f^{-n})'(z_0) = \lambda^{-n}\to\infty$. Contradiction.
Let $F(x)=\left|f(x)\right|$. I claim that $F(x)<\left|x\right|$ for all $x\in\left(-\sqrt{2},\sqrt{2}\right)$. This is quite easy to see from a graph:
It can be proved using the factorization $$F(x) = \left|x-x^3\right| = \left|x\right|\,\left|1-x^2\right|.$$ Now, the claim is trivial for $0 < x \leq 1$ since then $0<1-x^2<1$ so $$F(x) = \left|x\right|\,\left|1-x^2\right| < |x|.$$ If $1<x<\sqrt{2}$, then $1<x^2<2$ so that $0<x^2-1<1$ and, again, $\left|1-x^2\right|<1$.
With this lemma out of the way, your problem is easy. Any seed $x_1\in\left(0,\sqrt{2}\right)$ leads to a decreasing sequence, that is bounded below by zero, under iteration of $F$. Thus, there is a limit; that limit must be zero, since zero is the only fixed point of $F$ in $[0,\sqrt{2})$. Any seed in $\left(-\sqrt{2},0\right)$ leads to a positive first iterate to which the previous analysis applies. These results extend to $f$ since the absolute value of an orbit of $f$ is exactly an orbit of $F$. Finally, the basin is no larger than $\left(-\sqrt{2},\sqrt{2}\right)$, since those endpoints form an orbit of period 2 for $f$.
Best Answer
Hints: Instead of considering $x^2+2y^2+z^2$ it is in fact easier to consider $z$ separately from $(x,y)$.
1) Integrate the equation for $z$ and show that $z(t)$ goes to zero as $t\rightarrow +\infty$.
2) Show that the function: $L(t)=x(t)^2+2y(t)^2$ verifies the ode: $\dot{L}(t) =-2|\epsilon| L(t) \; (z(t)+1).$
3) Integrate this ode for $L$ and show that $L(t)\rightarrow 0$ as $t\rightarrow +\infty$.
(Along the way you should also conclude that neither $L$, nor $z$ can go to infinity)