Consider the function $$x'=rx-\frac{x}{1+x^2}$$ I'm asked to find the bifurcation points and classify if they are saddle node, trans-critical, pitchfork, etc.
One method we have been using in class is to look at the Taylor series and cut off an appropriate number of terms to get it in the normal form for one of the various types of bifurcations. So I did this on this problem and got,
$$rx - (x-x^3+O(x^5))$$ $$rx-x+x^3$$ $$x(r-1)+x^3$$ Now if we let $\sigma=r-1$ we get $x\sigma +x^3$ which is exactly the normal form for a subcritical pitchfork bifurcation. My problem is when I graph the function as $r$ varies it looks more like a saddle node bifurcation because fixed points are being created and destroyed. For this problem when $r \geq 1$ it looks like we have exactly one fixed point, then when $0<r<1$ there are 3 fixed points and then when $r \leq 0$ we have one fixed point again. I'm unsure how to classify this equation if anyone could correct me if my thinking or work is wrong. Any help is appreciated, thanks in advance.
$0$ is always a fixed point and $r$ has critical values $0$ and $1$. If $r$ increases, then, as it passes $0$, two nonzero fixed points are created, and they’re both unstable. As $r$ passes $1$, they’re both destroyed and $0$ becomes unstable.
You should get a bifurcation diagram that looks something like
This appears to be a subcritical pitchfork bifurcation.
Doing the Taylor expansion of $f(x, r) = rx-\dfrac{x}{1+x^2}$ near $x = 0$,
$$f(x, r) \approx \underbrace{ (r - 1) x + x^3}_{\text{Normal}} + O(x^5)$$
This is the Normal form of the subcritical pitchfork bifurcation (see Section $2.1.3$) at $x_* = 0, r_* = 1$.