Find the values of $\epsilon$ at the two pair bifurcation points of the polynomial: $\phi(x,\epsilon):=(x-2)^2(x-3)+\epsilon^2=0$.

Question

The question is from Murdock's textbook on Perturbation theory.

The full question is the following:

Again using the graph of $y=(x-2)^2(x-3)$, sketch the bifurcation diagram of $(x-2)^2(x-3)+\epsilon^2=0$ for all real $\epsilon$. Notice that this diagram is symmetrical about the $x$ axis. Find the values of $\epsilon$ at the two pair bifurcation points.

Is there an algebraic method to solve this problem or do I need to look at the bifurcation diagram? and then what should I look for in that case?

Thanks.

Answer 1

Algebraically, we can compute the discriminant $\Delta$ of the cubic polynomial $P(x) = (x - 2)^2 (x - 3) + \epsilon^2$. There are three roots when $\Delta > 0$ and one root when $\Delta < 0$. The sign of $\Delta$ changes at the values of $\epsilon$ corresponding to pair bifurcation points.

Or, since varying $\epsilon$ is the same as shifting the graph of $P(x)$ vertically, we can conclude that a pair bifurcation occurs when the sign of a local extremum of $P(x)$ changes.