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Consider the system

$\frac{dx}{dt}$=$2-(b+1)x+ax^{2}y$

$\frac{dy}{dt}$=$bx-ax^{2}y$

For what values of a and b does a Hopf Bifurcation occure, given that a and b are positive, and in the region x,y ≥0

I found that the system had only one fixed point (2,$\frac{b}{2a}$) and computed the jacobian at that point. Should I be looking for eigenvalues here and figuring out for what values of a, b different fixed points occur?

Clayton
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Robbie Meaney
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1 Answers1

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You should find that the Jacobian of this system has two complex eigenvalues (which are complex conjugates of each other) at the fixed point which you have correctly found. A Hopf bifurcation occurs when a fixed point loses stability. That means it happens when the real part of the eigenvalues goes from negative (stable) to positive (unstable). Since you have the eigenvalues as a function of $a$ and $b$, you can calculate a relationship between $a$ and $b$ when that real part is zero, which is where the Hopf bifurcation occurs.

John Barber
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  • So would it be possible to find set values for a and b or would it suffice to say a Hopf Bifurcation occurs when b=4a+1 because that's when the real part equals to zero. – Robbie Meaney Aug 08 '18 at 16:57
  • I think the latter, although I don't get that the condition is $b = 4a+1$. I get $1 + 4 a + b = 2 a^2 b$. – John Barber Aug 08 '18 at 19:24