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Given the equations: \begin{align} \frac{dx}{dt}&=1-(b-1)x+ax^{2}y\\ \frac{dy}{dt}&= bx-ax^{2}y \end{align} a) Fix $a$ and vary $b$. Show that a Hopf Bifurcation occurs at $b=a+1$.

So I got two fixed points $(0,0)$ and $(1,\frac{b}{a})$ I computed the Jacobian at $(1,\frac{b}{a})$ and from there was able to characterise the fixed points depending on the values of a and b. Finally I set the real part of the equation $\frac{b-a-1}{2}=0$ and solved to get $b=a+1$.

My question is if this is correct and why does it only work at the second fixed point and not the first? Is it because $(0,0)$ is not an isolated fixed point? How could I use this to discuss the dependence of the period of the limit cycle close to this $b=a+1$ Hopf Bifurcation?

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  • $\begingroup$ I don't think $(0,0)$ is an equilibrium point since $\dot{x}=1$ (fixed point is rather used in the context $f(x)=x$). $\endgroup$ – WalterJ Aug 11 '18 at 13:04
  • $\begingroup$ I thought so too except if you get $\endgroup$ – Robbie Meaney Aug 11 '18 at 13:09
  • $\begingroup$ x(b-axy)=0 from $\frac{dy}{dt}$=0 gives x=0 as a solution, and plugging that into y=$\frac{(b-1)x-1}{ax^{2}}$ gives 0 also $\endgroup$ – Robbie Meaney Aug 11 '18 at 13:11

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