Range of stability for iterative map

Question

Using linear stability analysis, I would like to compute the range of stability of the fixed points and the $2$-cycles of the following iterative map: $x_n = x_{n-1}^{2} - 3\mu$.

Setting $x = x^{2} - 3\mu$, I calculated the fixed points: $x_1 = \dfrac{\sqrt{1+12\mu}}{2}$ and $x_2 = \dfrac{\sqrt{1-12\mu}}{2}$.
And the two cycles are the fixed points and two other additional points: $x_3 = \dfrac{-1-\sqrt{3(-1+4\mu)}}{2}$ and $x_4 = \dfrac{-1+\sqrt{3(-1+4\mu)}}{2}$.

Now I am struggling to find the range of stability of the fixed points and $2$-cycles.
What I know is we need to compute the slope of the map at the points. So if we differentiate $x^{2} - 3\mu$ and calculate the derivative at the fixed point $x_1 = \dfrac{\sqrt{1+12\mu}}{2}$, we get $2(x_1) = 1+\sqrt{1+12\mu}$. And similarly for $x_2$, $2(x_2) = 1-\sqrt{1+12\mu}$. Then we set the absolute values < 1. So we get $|1+\sqrt{1+12\mu}| < 1$ and $|1-\sqrt{1+12\mu}| < 1$. Now here is the problem, I can only find the range of stability for $x_2$ but not for $x_1$. Here is my (hands) work:
$$\begin{align} |1-\sqrt{1+12\mu}| < 1\\ \implies-1<1-\sqrt{1+12\mu} < 1 \\\implies-2<-\sqrt{1+12\mu} < 0\\ \implies-\frac{1}{3} < \mu < \frac{3}{13}\end{align}$$

But now I try to find the range of stability of $x_1$:
$$\begin{align}|1+\sqrt{1+12\mu}| < 1 \\ \implies -1<1+\sqrt{1+12\mu} < 1 \\ \implies -2<\sqrt{1+12\mu} < 0 \end{align}$$ and there is no solution for this inequality.

My questions are:
1. How can we compute the range of stability for $x_1$?
2. What can we conclude for the range of stability for $x_1$? Am I correct to say that its range of stability is zero?
3. Are there any other methods to compute range of stability of fixed points and n-cycles?

Many thanks in advance.

Answer 1

It is always slightly surprising when the graph of the function $u$ defining a dynamical system $x_{n+1}=u(x_n)$ one is interested in, is not even mentioned.

In the present case, the graph $y=u(x)$ of the function $u:x\mapsto x^2-3\mu$ meets the diagonal $y=x$ at $x=x^*$ with $x^*=\frac12(1+\sqrt{1+12\mu})$ and at another point $x^{**}$ which will not interest us. (Your choice of the notations $x_1$ and $x_2$ for these fixed points is rather unfortunate since $x_1$ and $x_2$ already denote members of the sequence.)

Since the derivative at $x^*$ is $u'(x^*)\gt1$, $x^*$ is repulsive in the sense that, for every $x$ close enough to $x^*$, $x\gt x^*$ implies $u(x)\gt x$, while $x\lt x^*$ implies $u(x)\lt x$. Furthermore, $u'(x)\gt1$ uniformly on $x\geqslant x^*$ hence for every starting point $x_0\gt x^*$, the sequence $(x_n)$ is increasing and $x_n\to+\infty$.

The situation for $x_0\lt x^*$, even when $x_0$ is close to $x^*$, is more complicated. Truly enough, the first few terms of the sequence $(x_n)$ are decreasing but what happens afterwards is not determined by the local behaviour of $u$ at $x^*$.

Thus, if "range of stability" (a phrase I never met before, except with an altogether different meaning) means "set of initial conditions $x_0$ such that the sequence $(x_n)$ converges", I am afraid that an awful more lot of work is needed to determine it. Note for example that, even when $x_0$ is close to $x^*$ with $x_0\lt x^*$, it may well happen that, for some well chosen values of $x_0$, $x_n=x^*$ for every $n$ large enough (or even that $x_n=x^{**}$ for every $n$ large enough).