Question:
Determine the transcritical bifurcation for $x_{n+1}=\alpha x_{n}\left ( 1-x_{n} \right )^{2}$
I have determined the fixed point to be $x^{\ast}=0$ and $x^{\ast}=1$
Also, for values $\alpha$ such that the region $D \in \left [ 0,1 \right ]$ constitute a trapping region, the values of alpha is the set of values in the interval $\alpha \in \left [ 1,\infty \right ]$.
The stability of the fixed point can be determined via the criterion for map stability:
$\left | \frac{\mathrm{d} f}{\mathrm{d} x} \right |_{x^{\ast}} >1 $
or
$\left | \frac{\mathrm{d} f}{\mathrm{d} x} \right |_{x^{\ast}} <1 $
The former indicating the fixed point to be unstable and the latter, stable fixed point.
In this question,
$\left | \frac{\mathrm{d} f}{\mathrm{d} x} \right |_{x^{\ast}=0} =\left | \alpha \right |$ is stable for $\left | \alpha \right | <1$ and unstable for $\left | \alpha \right | > 1$
$\left | \frac{\mathrm{d} f}{\mathrm{d} x} \right |_{x^{\ast}=1} =\left | \alpha \right |=0 $. Less than 1 so stable fixed point.
Any help is appreciated.
Thanks in advance.