Why do we need to find the multiple roots? (bifurcation curve)

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Consider the system $$ \dot{x}=x+ay-y^3,\quad \dot{y}=b-2y+x. $$ The task is to give the bifurcation curve for the equilibria.

First of all, equilibria are determined by $$ x+ay-y^3=0\quad b-2y+x=0. $$

This gives $x=2y-b$. Hence $$ y^3-(a+2)y+b=0. $$

Now, it is said, we need to determine the multiple roots of the last equation. A multiple root of the last equation is $3y^2=a+2$, so we get the bifurcation curve $$ b=\pm 2\left(\frac{a+2}{3}\right)^{3/2}. $$

There are two things I do not understand:

(1) Why do we need to find the multiple roots?

(2) How do we find them?

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Why do we need to find the multiple roots?

In general, the roots will depend on the parameters. Depending on the value of the parameters, a multiple root might "split" into different roots, which would be a bifurcation.

Take a look at the equation $$y^3-(a+2)y+b=0$$

with $a = -2$ and $b=0$ it simplifies to $$y^3=0$$ which has $0$ as its only root. Now let's say $a \neq -2$ but still $b=0$, you can see that $y_0=0$ remains a root of the equation $$y\,(y^2-(a+2))=0$$

The other two roots depend on the parameter $a$ now $$ \begin{align} y^2-(a+2)&=0\\ y^2&=a+2\\ y_{1,2}&=\pm \sqrt{a+2}\\ \end{align} $$

For $a\lt-2$, there are no real solutions, thus $y_0=0$ is the only valid equilibrium, but for $a\gt-2$ there are 3 possible values $y_0=0$ and $y_{1,2}=\pm \sqrt{a+2}$

Coming from $a=-\infty$ going towards $a=+\infty$, you will only have 1 equilibrium $y_0$ up to $a=-2$ after which you have 3 equilibria $y_{0,1,2}$

How do we find them?

If you take a look at the nature of the roots of a cubic equation of the form $\color{red}{a}x^3+\color{red}{b}x^2+cx+d=0$ (the wikipedia coefficients are red, to prevent confusion with yours) you will find that the value of its discriminant $\Delta$ is as follows

$\Delta =18\color{red}{ab}cd-4\color{red}{b}^3d+\color{red}{b}^2c^2-4\color{red}{a}c^3-27\color{red}{a}^2d^2$

  • If $\Delta \gt 0$, then the equation has three distinct real roots.
  • If $\Delta = 0$, then the equation has a multiple root and all its roots are real.
  • If $\Delta \lt 0$, then the equation has one real root and two nonreal complex conjugate roots.

The number of real roots and their multiple-ness depends on $\Delta$, which in turn depends on the parameters.

In your case $\color{red}{b} = 0$ and $\color{red}{a} = 1$ which simplifies $\Delta$ to

$$\Delta =-4c^3-27d^2$$

finally, with $c=-(a+2)$ and $d=b$ you have

$$\Delta =4a^3+24a^2+48a-27b^2+32$$

A contour plot in wolfram alpha looks like this:

contour plot

Now I'm not an expert at using that software and can't tell what the colors mean exactly, but the simple case of $a = -2$ and $b=0$ that I explained above is right in the center. You should tweak the plot (or use a different software) to display the areas where $\Delta$ is $\lt0, \gt0, =0$ which in turn correspond to how many equilibria you have in the system.

So the answer would be not to find the roots, but only how their count is influenced by the parameters.