Hopf Bifurcation - Planar System

Question

Consider the planar system, $$x'=x(\mu-2x)-xy, \ \ \ y'=y(x-1)+y^2 \ \ \ (\mu\in\mathbb{R}).$$

It can be shown that $(x^*,y^*)=\left(\mu-1, \ 2-\mu\right)$ is a fixed point of the above system. But what is the bifurcation phenomena of this fixed point? The stability matrix associated with this fixed point is $$A=\begin{pmatrix} 2-2\mu & 1-\mu \\ 2-\mu & 2-\mu \end{pmatrix}.$$ The eigenvalues of $A$ can be expressed as $$\lambda=\frac{(4-3\mu)\pm\sqrt{5\mu^2-12\mu+8}}{2}.$$ My initial thought was that $\mu=\frac{4}{3}$ was a Hopf bifurcation, as $\Re(\lambda)=0$. But from my understanding, $\lambda$ must be purely imaginary when $\mu=\frac{4}{3}$ for this to be a Hopf bifurcation, and this is not the case.

Perhaps a bifurcation point does not exist?

Answer 1

Hint.

As the roots for $5\mu^2-12\mu + 8 = 0$ are complex, no existence of Hopf bifurcation.

Follows the plot for the eigenvalues

$$ \frac{1}{2} \left(4-3\mu\pm\sqrt{5 \mu ^2-12 \mu +8}\right) $$

Answer 2

Hint, you have more equilibrium points, namely $(0,0), (\frac{1}{2}\mu,0), (0,1)$. What happens with $(\mu-1,2-\mu)$ when $\mu=1$ or $\mu=2$?

Answer 3

The $\text{tr}(A)=0$ and $\text{det}(A)<0$ when $\mu=\frac{4}{3}$. Then, your eigenvalues cannot be purely imaginary. Therefore, Hopf bifurcation cannot occur at $(x^*,y^*)=\left(\mu-1, \ 2-\mu\right)$.