I have a differential equation: $y'=ay-y^2$ and I have to find the critical bifurcation values. My attempt: I first found the equilibrium values and saw that $y'= 0$ for $y= 0$ or $y = a$. So then I tried to find out equilibrium values of values $a<0$, and saw that I have two equilibrium points (one being $y=0$) then as I go to $a = 0$, I see that I have only one equilibrium point ($y=0$) and when I go to $a >0$, I again have two equilibrium values. My question: Is $a$ the critical bifurcation value? I am confused because for $a =0$ we only have 1 equilibrium value but for any other $a$, we have two such values. Shouldn't the equilibrium values diminish beyond a certain point for that point to be called critical bifurcation value?
Critical Bifurcation Values of a differential equation
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ordinary-differential-equations
bifurcation
1 Answers
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We say that a bifurcation occurs if when a small change to the parameter value cause a sudden change in the behaviour of the system. For the given differential equation, the parameter is $a$; so the critical bifurcation value is the value of $a$ such that if you just change $a$ to $a+\varepsilon$ for any $\varepsilon>0$, the behaviour of the system changes completely.
In this case, the critical bifurcation value is $a=0$. For $a=0$, we only have one equilibrium point $y^*=0$, which is semistable. For $a\neq 0$, we have two equilibrium points $y^*=0$ and $y^*=a$, but their stability depends on the sign of $a$.