Bifurcation Help - Fixed and Varied Constants

Question

Question

The dynamic system is two dimensional. The variables $\alpha$, $\beta$, $A$ and $B$ are real constants. However both $A$ and $B$ are nonzero. $$f(x,y) = A(y-x^2)$$

$$g(x,y) = B(x-\alpha)(y-\beta x+x^2)$$

(a) Show that Hopf bifurcation cannot exist in this system.

(b) Show this dynamic system can have all three other types of local bifurcations as $\alpha$ varies and $\beta$ is fixed and vice versa.

To show that the Hopf bifurcation cannot exist, I have founded all the fixed points and the corresponding eigenvalues to which none have imaginary parts. From this I draw the conclusion that a Hopf bifurcation is impossible. However, I am not really sure if this is actually the case.

I am completely stuck on (b). I have founded fixed points before, I do not see how I can transform my system into any of the bifurcations.

Results:

Let $f = g = 0$. Solving the equations and labelling the new function $h$ gives

$$h(x) = x(x-\alpha)(2x-\beta)$$

Taking the roots of this function gives the fixed points $x^* = 0$, $\alpha$ and $\beta/2$.

Answer 1

Hint.

Plotting the zero locus for $f(x,y)$ in blue and $g(x,y)$ in red we can observe the variety of possible intersections. The vertical red line depends on $\alpha$ and red parabola depends on $\beta$. The blue and red parabolas are tangent only when $\beta = 0$ otherwise always intersect twice. There is a unique intersection point when $\alpha=\beta=0$

Answer 2

To show that the Hopf bifurcation cannot exist, solve the following system:

$$ \begin{align} f(x,y)&=0,\\ g(x,y)&=0,\\ \text{tr}(J(x,y))&=0, \end{align} $$ for $(x,y,A)$ and with this solution evaluate the determinant of Jacobian Matrix $J(x,y)$. Then, the same for $(x,y,B)$, $(x,y,\alpha)$ and $(x,y,\beta)$. In all cases, the determinant is negative, therefore the Hopf bifurcation cannot exist in your system.