Finding the critical value of bifurcation

Question

I want to find the critical value/bifurcation value, $a_c$, I'm not sure if I'm approaching this the correct way.

$$ \dot{x} = x(1-x) - \frac{1}{4}(a+1)^2\left(\frac{x}{x+a}\right)$$

A previous example I did broke the function up into two separate pieces and then took the derivative to find $x$. Used the $x$ back into the original function to find a critical.

So I think how it should be done is as follows:

$$ d/dx (x -x^2) = -d/dx(\frac{1}{4}(a+1)^2(\frac{x}{x+a}))$$ $$ 1 -2x = -\frac{1}{4}(a+1)^2(a)$$ $$ x = \frac{1}{8}(a+1)^2a -1 $$

Then I would plug this x value into $\dot{x} = 0$ to solve for a-critical. Is this the correct approach? Is there an easier way to do this?

Answer 1

The equilibrium points $x^* \in \lbrace{0, \tfrac{1}{2}(1-a)}\rbrace$ are obtained by solving $\dot x = f(x) = 0$ with respect to $x$, where $$ f(x) = \frac{-x}{x+a}\left(\frac{1-a}{2} - x\right)^2 . $$ The local stability of the equilibrium points can be investigated by examining the sign of the derivative $$ f'(x) = \frac{2 x^2 + 3 a x - a\frac{1-a}{2}}{(x + a)^2} \left(\frac{1 - a}{2} - x\right) . $$ The derivative takes the following particular values $f'(0) = -\frac{1}{4a}(1-a)^2$ and $f'\big(\tfrac{1}{2}(1-a)\big) = 0$ at the equilibrium points $x^*$. Hence, $x^*=0$ is either unstable or asymptotically stable. The equilibrium $x^*=\tfrac{1}{2}(1-a)$ is hyperbolic, which requires to examine the variations of $f'$ in its vicinity.

For $a<0$, $x^* = 0$ is unstable.

• Case $a<-1$. The equilibrium $x^* = \tfrac{1}{2}(1-a)$ is semi-stable (asymptotically left-stable).
• Case $a=-1$. The equilibrium $x^* = \tfrac{1}{2}(1-a)$ is asymptotically stable.
• Case $-1< a< 0$. The equilibrium $x^* = \tfrac{1}{2}(1-a)$ is semi-stable (asymptotically right-stable).

For $a=0$, there is only one equilibrium $x^* = \tfrac{1}{2}(1-a)$ which is semi-stable (asymptotically right-stable).

For $a>0$, $x^* = 0$ is asymptotically stable.

• Case $0< a< 1$. The function $f$ is decreasing to the right of the equilibrium point $x^* = \tfrac{1}{2}(1-a)$ ($f$ has a local maximum). Hence, the equilibrium is semi-stable (asymptotically right-stable).
• Case $a = 1$. Both equilibria merge together, and they are asymptotically stable.
• Case $a > 1$. The function $f$ is decreasing to the left of the equilibrium point $x^* = \tfrac{1}{2}(1-a)$ ($f$ has a local minimum). Hence, the equilibrium is semi-stable (asymptotically left-stable).

If the study is restricted to positive $a$, a bifurcation occurs at $a=1=a_c$.