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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?

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3 Answers 3

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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

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

enter image description here

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  • $\begingroup$ There seems to be a change in stability. Is there a bifurcation point (not a Hopf) at $\mu=1$ and $\mu=2$? $\endgroup$
    – Bell
    Commented Aug 9, 2019 at 9:24
  • $\begingroup$ @Bell As can be depicted, for $\mu \lt 1$ we have sources. For $1\lt \mu\lt 2$ we have saddle points and for $\mu > 2$ we have sinks. This regarding this equilibrium point. $\endgroup$
    – Cesareo
    Commented Aug 9, 2019 at 9:34
  • $\begingroup$ I am sorry but I am unsure of some of your terminology (e.g. sinks), but I think I understand. $\mu=1$ is NOT a bifurcation but $\mu=2$ IS a bifurcation? $\endgroup$
    – Bell
    Commented Aug 9, 2019 at 9:45
  • $\begingroup$ @Bell Hopf bifurcations occur when a periodic solution appears. Please. See the Hopf bifurcation definition in Wikipedia. Have a look also at staff.www.ltu.se/~larserik/applmath/chap9en/part7.html $\endgroup$
    – Cesareo
    Commented Aug 9, 2019 at 9:51
  • $\begingroup$ I agree that there are no HOPF bifurcations, but I believe $\mu=2$ is a bifurcation point (for $1<\mu<2$ we have an unstable hyperbolic saddle point and for $\mu>2$ we have a stable improper node... is this not indicative of a bifurcation at $\mu=2$?). I also appreciate the link, it has helped with the terminology. $\endgroup$
    – Bell
    Commented Aug 9, 2019 at 9:55
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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 $\mu=2$?

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  • $\begingroup$ We get $(0,1)$ and $(1,0)$. I'm not sure of the significance of this. Can we relate this too the results obtain for the other equilibrium points? That would be neat. $\endgroup$
    – Bell
    Commented Aug 9, 2019 at 9:43
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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 $(x^*,y^*)=\left(\mu-1, \ 2-\mu\right)$.

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