I’m doing an exercise for a lecture on dynamical systems. We are asked to classify all bifurcations of the dynamical system $f_c = c·\sin$ for real $c > 0$.
We are given that bifurcations of $f_c$ can only occur at parameters $c$ where there’s a neutral fixed point, i.e. a point $p ∈ ℝ$ such that $|{f_c}’ (p)| = 1$.
It is easy to prove that there are no bifurcations for $c < 1$. So let $c ≥ 1$.
Now, for any $p ∈ ℝ$, we have
- ${f_c}’(p) = 1$ if and only if $p ∈ \arccos \frac{1}{c} + 2πℤ$, and
- ${f_c}’(p) = -1$ if and only if $p ∈ \arccos \frac{-1}{c} + 2πℤ$.
Therefore, since $\sin^2 \arccos x = 1 - x^2$, neutral fixed points $p$ of $f_c$ satisfy $$p^2 = f_c(p)^2 = c^2·\sin^2 (\arccos \frac{\pm 1}{c}) = c^2(1 - \frac{1}{c^2}) = c^2 - 1.$$
How can I proceed to find which one of these points $c^2 - 1$ are actually fixed points of $f_c$, i.e. satisfy $c·\sin (c^2 - 1) = c^2 - 1$ and how can I conclude anything about the bifurcations that may or may not occur at these parameters and fixed points?
A fixed point $p$ has $c \sin(p) = p$, and it is neutral if $c \cos(p) = \pm 1$. Thus $$\dfrac{1}{c} = \dfrac{\sin(p)}{p} = \pm \cos(p)$$ Here's a plot of $\sin(p)/p$ (red), $\cos(p)$ (blue) and $-\cos(p)$ (green) for $p \ge 0$: the graph is symmetric.
As you can see (and its easy to prove), for each integer $n$ there is one solution in $[n\pi, (n+1/2)\pi]$ and one in $[(n+1/2)\pi, (n+1)\pi]$. In this case you want $c > 1$, so $\sin(p)/p > 0$: this occurs for $n = 0, 2, 4, \ldots$.