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".

Answer 2

Step 1: As jkn stated, the first step is to find the equilibrium such that $\dot x=\dot y=0$. It is easy to obtain the equilibrium is $(0, 0)$.

Step 2: Compute the eigenvalue around the equilibrium $(0, 0)$. $$J= \left(\begin{array}{cc} \lambda &1\\ -1& \lambda\\ \end{array}\right)$$ Thus the characteristic equation is $$\det (\tilde\lambda I-J)=0$$

The aobve equation admits two eigenvalue $\lambda\pm i$. It is called the imaginary $\omega=1$.

It is obvious that $\lambda:=0$ such that the dynamical system has a pair of pure imaginary roots $\pm i$, moverover $$\frac{d\tilde\lambda}{d\lambda}|_{\lambda=0}=\frac{d}{d\lambda}|_{\lambda=0}(\lambda\pm i)=1>0$$ There the dynamical system exhibits a Hopf bifurcation at $\lambda=0$.

Step 3, Compute the Hopf bifurcation. Consider $$(J-\tilde\lambda I)\left(\begin{array}{cc} q_1\\ q_2\\ \end{array}\right)=0$$ we get the eigenvector $$\left(\begin{array}{cc} q_1\\ q_2\\ \end{array}\right)=\left(\begin{array}{cc} i\\ 1\\ \end{array}\right)$$. On the other hand, consider$$(J^T-\overline{\tilde\lambda} I)\left(\begin{array}{cc} p_1\\ p_2\\ \end{array}\right)=0$$ we get the eigenvector $$\left(\begin{array}{cc} p_1\\ p_2\\ \end{array}\right)=\left(\begin{array}{cc} i\\ 1\\ \end{array}\right)$$ It is always possible to normalize $\mathbf p$ with respect to $\mathbf p$: $$<\mathbf p, \mathbf q>=1, \mbox{ with } <\mathbf p, \mathbf q>=\bar p_1q_1+\bar p_2 q_2 $$ Thus $$\left(\begin{array}{cc} p_1\\ p_2\\ \end{array}\right)=\frac{1}{2}\left(\begin{array}{cc} i\\ 1\\ \end{array}\right)$$

We denote the nonlinear term $$\mathbf F:=\left(\begin{array}{cc} F_1\\ F_2\\ \end{array}\right) =\left(\begin{array}{cc} 0\\ -x^2y\\ \end{array}\right)$$ Introduce the complex variable $$z:=<\left(\begin{array}{cc} p_1\\ p_2\\ \end{array}\right), \left(\begin{array}{cc} x\\ y\\ \end{array}\right)>$$ Then the dynamical system becomes to $$\dot z=(\lambda+i)z+g(z, \bar z,\lambda) \mbox{ with }g(z, \bar z,\lambda) =<\left(\begin{array}{cc} p_1\\ p_2\\ \end{array}\right), \left(\begin{array}{cc} F_1\\ F_2\\ \end{array}\right)>$$ Some direct computations show $$g(z, \bar z,\lambda)=\frac{1}{2}(z^3-z^2\bar z-z\bar z^2+\bar z^3):=\frac{g_{30}}{6}z^3+\frac{g_{21}}{2}z^2\bar z+\frac{g_{12}}{2}z\bar z^2+\frac{g_{03}}{6}\bar z^3$$

We recall that the first Lyapunov coefficient is $$l_1=\frac{1}{2\omega^2}\mbox{Re} (ig_{20}g_{11}+\omega g_{21})$$

Since $\omega=1$, $g_{20}=g_{11}=0$, $g_{21}=-1$, we obtain $$l_1=-\frac{1}{2}<0$$ Therefore $\lambda=0$ is a super crirical Hopf bifurcation.