[Math] Finding a Hopf Bifucation with eigenvalues

Question

I am trying to show that the following 2D system has a Hopf bifurcation at $\lambda=0$:
\begin{align}
x' =& y + \lambda x \\
y' =& -x + \lambda y – x^2y
\end{align}
I know that I could easily plot the system with a CAS but I wish to analytical methods. So, I took the Jacobian:
\begin{equation}
J = \begin{pmatrix} \lambda&1\\-1-2xy&\lambda-x^2\end{pmatrix}
\end{equation}
My book says I should look at the eigenvalues of the Jacobian and find where the real part of the eigenvalue switches from $-$ to $+$. This would correspond to where the
system changes stability. So I took the $\det(J)$:
\begin{align}
\det(J) =& -\lambda x^2 + 2xy + \lambda^2 + 1 = 0
\end{align}
I am stuck here with algebra and am not quite sure how to find out where the eigenvalues switch from negative real part to positive real part. I would like to use the
quadratic formula but the $2xy$ term throws me off.

How do I proceed? Thanks for all the help!

Answer 1

Suppose that $(\bar{x},\bar{y})$ is an equilibrium of the system. A criterion for the equilibrium $(\bar{x},\bar{y})$ to undergo a Hopf bifurcation at $\lambda=0$ is that the eigenvalues of the Jacobian evaluated at the equilibrium are purely imaginary if $\lambda=0$.

So, first you need to solve for the equilibria and then start testing which of these equilibria could undergo the Hopf bifurcation by evaluating the jacobian at each of them.

Setting $\dot{x}=\dot{y}=0$ and solving, it is easy to show that $(0,0)$ is an equilibrium for any value of $\lambda$ and that if $\lambda\geq 0$, there are two additional equilibria

$$\left(\pm\sqrt{\frac{\lambda^2+1}{\lambda}},\mp\lambda\sqrt{\frac{\lambda^2+1}{\lambda}}\right).$$

We know that Hopf bifurcations do not remove/add equilibria, they just change the stability of a single equilibrium. So, since the second and third equilibria do not exist if $\lambda<0$ and the bifurcation occurs when $\lambda=0$, we do not have to consider these when looking for the bifurcation.

So plugging in $\bar{x}=\bar{y}=0$ into the Jacobian, it is rather quick to show that the eigenvalues of the $J$--that is, the complex numbers $\alpha_1$ and $\alpha_2$ such that $\det(\alpha_1 I-J)=\det(\alpha_2 I-J)=0$ (not to be confused with your parameter $\lambda$)--are purely imaginary if $\lambda=0$.

Did that help?

EDIT: Technically, I should've said that the criterion is that "the real part of the eigenvalues of the Jacobian evaluated at the equilibrium switches sign" instead of "the eigenvalues are purely imaginary".