This question is taken from Glendinning's textbook "Stability, instability and chaos": Given the map $x_{n+1} = \mu - x_n^2$, determine the type of bifurcation which occurs at $\mu = -\frac{1}{4}$ and $\mu = \frac{3}{4}$.
I'm used to answering such questions in the continuous case, but am unsure how to proceed in the discrete case. So far I've done the following:
The fixed points of the map occur at $x^{\ast}$ which satisfied $x^{\ast} = \mu - (x^{\ast})^2$. Solving the quadratic gives that fixed points occur at $x^{\ast} = -\frac{1}{2} \pm \sqrt{\frac{1}{4} + \mu}$.
So we can see that for $\mu < -\frac{1}{4}$, we have $0$ fixed points, for $\mu = -\frac{1}{4}$ we have $1$ fixed point at $x^{\ast} = -\frac{1}{2}$, and for $\mu > -\frac{1}{4}$ we have $2$ fixed points.
So then the bifurcation diagram, without instability looks like the following (just plotting $x$ against $\mu$):
So it seems fairly obvious that there is a saddle node bifurcation occurring at $\mu = -\frac{1}{4}$ since we go from $0$ fixed points to two fixed points.
My questions are the following:
How do I figure out the stability/instability of the fixed points for the bifurcation diagram. Usually in the continuous case I deal with a problem of the form $\dot{x} = f(x)$, and to find stability, I just plot $\dot{x}$ against $x$ and make inferences from that. In this case should I be plotting $x_{n+1}$ against $x_n$, or should I be plotting the quadratic $(x^{\ast})^2 + x^{\ast} - \mu = 0$?
What happens at $\mu = \frac{3}{4}$. My bifurcation diagram makes it seem like nothing changes at that point?
Thanks
A fixed point $x$ of the discrete system $x_{n+1} = f(x_n)$ is classified as locally stable (attracting) if $ |f'(x)|<1$, locally unstable (repelling) if $|f'(x)|>1$, and indeterminant (possibly transitioning from one type to the other) if $|f'(x)|=1$.