Hi so I'm trying to work out how to find the fixed points of a piecewise function and do a stability analysis. One of the past exam papers in this complex systems course I'm doing has this question:
Consider the map $f_\beta:[0,1]\to[0,1]$ defined by:
$$f_\beta(x)=\begin{cases}
\frac x\beta\:\:\,\,\text{ if }0\leq x\leq\beta \\
\frac{1-x}{1-\beta}\text{ if }\beta<x\leq 1
\end{cases}$$
where $0<\beta<1$.
Find the fixed points of the map and determine their linear stability.
My maths knowledge isn't amazing, just algebra, some calculus, linear algebra, probability etc. But not to any expert level. I have done an introductory complex systems course before which was intended to teach all the maths necessary, I understood this pretty well. But this question is really confusing me.
My initial instinct is to set $\frac{df_\beta(x)}{dx}=0$ and solve for $x$ in order to get the fixed points. But of course this is impossible because both parts of the function are linear.
However if we evaluate the function at certain points we find that $f_\beta(x)=x$. For example:
$$x^*=0=f_\beta(x^*)$$
In the answers provided for this question it says that another fixed point is $x^*=\frac1{2-\beta}$. But I don't know how it was arrived at.
Firstly I would like to know how one arrives at obtaining the fixed points without just plugging in values and testing. Also, how is it that this doesn't match up with the method of setting the derivative to $0$ and solving for $x$ when this works with polynomials? Lastly, how would one even begin to do a stability analysis when you can't even find the second derivative (I think this was the previous method I was taught for a stability analysis)?
Best Answer
To find the fixed points of $f_{\beta}$ you need to solve $f_{\beta}(c)=c.$ You should get $c=0$ and $c=\dfrac{1}{2-\beta}.$
In order to study the stability you need to get $|f'(c)|.$ Since $|f'(0)|<1$ we have that $c=0$ is stable. But $f'\left(\dfrac{1}{2-\beta}\right)=\dfrac{-1}{1-\beta}<-1$ and thus this fixed point is not stable.