The number of individuals in a certain population (in arbitrary real units) obeys, at discrete time intervals, the equation $$y_{n+1} = y_n(2-y_n) \hspace{2 mm}\text{for} \hspace{1 mm} n = 0,1,2,\ldots,$$
where $y_0$ is the initial population.(a) Find all "steady-state" solutions $y^*$ such that, if $y_0 = y^*,$ then $y_n = y^*$ for $n = 1,2,\ldots$.
(b) Prove that if $y_0$ is any number in $(0,1)$, then the sequence $\{y_n\}$ converges monotonically to one of the "steady-state" solutions found in (a).
My question: Is a "steady-state" solution exactly as they describe it? A solution $y^*$ such that $y_n = y^*$ for $n = 1,2,\ldots$?
Best Answer
Let $f(x) = x (2-x)$. Note that $f$ is concave, $f'(x) = 2(1-x)$ and has a maximum at $x=1$, hence $f(x) \le 1$ for all $x$. Furthermore, $f'(x) \ge 0$ for $x \le 1$ so $f$ is non decreasing on $(-\infty,1]$.
Hence if $y_n \le 1$, then $y_{n+1} =f(y_n)\le 1$ and $y_{n+1} \ge y_n$.
Hence $y_n$ is a bounded non decreasing sequence and such sequences have limits. So suppose $y_n \to y$, then by continuity $y=f(y)$ and so we must have $y \in \{0,1\}$ (since these are the only solutions of $y=f(y)$).
In fact, if $y_0=0$ then $y_n = 0$ for all $y$ and if $y_0 \in (0,1]$, then the above shows that $y_n \to 1$.