I got the following dynamical system in 1-d, $$\dot x = (x+\mu)(\mu+2x-x^2)$$
Im asked to find dependence of the the stationary points of the system and its stability with respect the $\mu$ parameter. Then draw the bifurcation diagram and find at which $\mu$ the bifurcation occurs.
Before showing my attempt of solution I have to say sorry, but Im really lost and I havent found any reference to study this properly. Here we go,
Attempt of partial solution
For the stationary(fixed points), we have that $x=-\mu$ or $x=1\pm 1\sqrt{1+\mu}$. So, now we have to analyze stability for the cases $\mu >-1$ and $\mu <-1$. We will check the sign of $\frac{d((x+\mu)(\mu+2x-x^2))}{dx} = 3\mu + 4x – 2 \mu x – 3 x^2$.
Case 0: $\mu = -1$ theres only one stationary point which is x=1.
Case 1: $\mu >-1$,
1.1)For $x=-\mu$, $\frac{d((x+\mu)(\mu+2x-x^2))}{dx} = 3\mu-4\mu+2\mu^{2}-3\mu^{2}<0$. So, $x=-\mu$ is stable.
1.2)For $x= 1+1\sqrt{1+\mu}$, $\frac{d((x+\mu)(\mu+2x-x^2))}{dx}=-2 (\mu + 1)( \sqrt{\mu+ 1} + 1)<0$
1.3) For $x = 1-1\sqrt{1+\mu}$, $\frac{d((x+\mu)(\mu+2x-x^2))}{dx}=2 (1 + \mu) (-1 + \sqrt{1 + \mu})>0$ which implies it is unstable.
Case 2:$\mu <-1$
The way to work it is analogous to what we did in C.1.
Now, I have no idea how to draw the bifurcation diagram. Its only one diagram or its 1 for each case?, what exactly means that 'a bifurcation occurs' and how do I classify these points into pitchfork, saddle node and transcritical?
Thanks so much for your help, I really appreciate it. <3
Best Answer
bifucation graph
Note that the behavior of the vector field at large values of $x$ never changes in quality, it always points towards the origin. As the roots, except at the triple and crossing point, are simple, you always have that the middle root is unstable and the outer roots are stable.
Let's look again at the $x$-derivative of the right side. By the product rule, it is $$ 1(μ+2x-x^2)+(x+μ)(2-2x)=μ+2x-x^2+2(μ+(1-μ)x-x^2)=3μ+2(2-μ)x-3x^2 $$
For $x=-\mu$, this reduces to $μ-2μ-μ^2=-μ(1+μ)$, so the stability changes at the triple point $μ=-1$ and again at the crossing point $μ=0$.
For $μ>-1$ and the quadratic roots, the derivative expression reduces to $$ 1(μ+2x-x^2)+2(x+μ)(1-x)=2(μ+(1-μ)x-x^2)=-2(1+μ)x. $$ and thus for $x=1\pm\sqrt{1+μ}$ this is negative as long as both roots stay positive and changes to positive as the lower root becomes negative for $μ>0$.