I'm trying to understand the equilibrium for the below problem. This is what I have - was hoping someone can check for any glaring errors?
Consider the piecewise function $f$ defined by
$\begin{align*} f(x)&=\begin{cases} bx & \text{if } x <\frac{a}{b}\\ a & \text{if } x \geq \frac{a}{b} \end{cases} \end{align*}$
where $a,b>0$.
I'm tasked with finding the equilibrium points of $X_{t+1}=f(X_t)$.
An equilibrium $x^*$ of $f$ is a solution to $x^*=f(x^*)$. Therefore, I need to solve
$\begin{align*} \begin{cases} x= bx & \text{if } x <\frac{a}{b}\\ x = a & \text{if } x \geq \frac{a}{b} \end{cases} \end{align*}$
The top equation has solution $x=0 < \frac{a}{b}$. The bottom equation has solution $x=a$ so long as $a\geq \frac{a}{b}\iff b\geq 1$.
So the two equilibria are $x^*=0$ and $x^*=a$ when $b\geq 1$. In terms of analysing the stability (determining when the equilibrium are stable), the equilibrium $x^*$ is stable when $|f'(x^*)|<1$.
We have:
$\begin{align*} f'(x)&=\begin{cases} b & \text{if } x <\frac{a}{b}\\ 0 & \text{if } x > \frac{a}{b}\\ \text{undefined} &\text{if } x=\frac{a}{b} \end{cases} \end{align*}$
So $f'(0)=b$, so $x^*=0$ is a stable equilibrium when $b<1$. The equilibrium $x^*=a$ (when $b > 1$) is always stable. Needing to take the strict inequality for $b$ now.