$x(x-1)^2+\alpha$ bifurcation diagram

Question

I have to find all the equilibrium points of $$\dot{x}=x(x-1)^2+\alpha$$ and sketch the corresponding bifurcation diagram, but I don't see how to start, since the roots of this polynomial don't have 'nice' expressions. Do you know how to deal with this kind of systems?

Answer 1

Let $$ f_\alpha(x)=x(x-1)^2+\alpha. $$ Then $f_\alpha'(x)=3x^2-4x+1$ has two roots $x=1,x=\frac13$. In order to find a bifurcation value of $\alpha$, let either $f_\alpha(1)=0$ or $f_\alpha(\frac13)=0$. So $\alpha=0$ or $\alpha=-\frac{8}{27}$. Therefore the DE $$ \dot{x}=f_\alpha(x)$$ has two bifurcation values $\alpha=0$, $\alpha=-\frac{8}{27}$. The bifurcation diagram is the following

Answer 2

Hint: Using the graph of the RHS function (e.g. see https://www.desmos.com/calculator/ia9fmlfbo9) one can argue qualitatively for which values of $a$ how many equilibrium points there are and what their behaviours are.