basins-of-attractiondynamical systems
I want to find the basin of attraction of a fixed point.
For example, I have $f(x)=\frac 1{x+1}$, whose fixed points are $\frac{-1\pm \sqrt{5}}{2}$. Now, I must create a neighborhood around the $x$ point that would consist of all points around it that would attract to it. How can I figure out the radius of a neighborhood?
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 $f$.
As pointed out by Did, for non-attracting fixed points $x_0$ the set of points $x$ with $\lim_{t\to \infty} x(t)=x_0$ does not have to be open.
For an attractive fixed point, if the dynamical system is continuous then the attraction basin is indeed open (Note: For discontinuous dynamics this is clearly no true; for a counterexample, just pick any set $S$ containing a neighbourhood of $x_0$ and send it to $x_0$ and send the rest to some point $y_0$ not in $S$; the basin of attraction is now $S$).
To see this for continuous dynamics, argue as follows:
Start with $x_1$ in the attraction basin. We will find a neighborhood of it also in the attraction basin.
By definition of attractive fixed point, there exists open $U$ around $x_0$ such that for any $x\in U$ $\lim_{t\to \infty} x(t)=x_0$. The fact that $x_1$ is attracted of $x_0$ means that $\lim_{t\to \infty} x_1(t)=x_0$. In particular, for some large $T$ we have $x_1(T)\in U$. Since the dynamics is continuous, the preimage of $U$ under $\phi_T$ (lets call it $V$) is open, and contains $x_1$.
I claim that all of $V$ is also in the basin of attraction -- which is obvious because any $x\in V$ ends up in $U$ after time $T$ and then attracts to $x_0$; in formulas $\lim_{t\to \infty} x(t)=\lim_{(s=t-T)\to \infty} x(T)(s) =x_0$.
So there you have it: any $x_1$ in the basin has an open neighbourhood $V$ around it which is also entirely in the basin.
Best Answer
For the endpoints $a,b$ of the immediate basin of attraction of an attracting fixed point, the possibilities are:
$\pm \infty$.
A singularity (i.e. where $f$ is undefined).
A repelling fixed point.
$(a,b)$ is a $2$-cycle, i.e. $f(a) = b$ and $f(b) = a$.
Take the closest of these possibiities on each side of your fixed point.