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Hi,
In this question After we expand $\dot{x}$ around $x^*$ = 0 , we get $\dot{x}$ = (1-ab)x + ($\frac{1}{2}$a$b^2$)$x^2$ + O($x^3$).
How do we jump from this to knowing that "Hence a transcritcal bifurcation occurs when ab=1"?
Is it just that: since we assume x to be small around $x^*$=0, wee can render all other terms neglegable except for the first, and therefore ab=1 is approximately the point around which you have a change of sign for $\dot{x}$?
Thanks in advance and best regards, Chen
Suppose in general we are given a system $\dot{x} = f(x)$. Recall that if $f'(x^*) > 0$ in a 1D system, then the fixed point is stable, while if $f'(x^*) < 0$, it is unstable. Here, from the approximation made by Strogatz, $$f'(x) = 1-ab + ab^2x + O(x^2).$$ To show that $ab = 1$ is the condition for a transcritical bifurcation to occur, we must show that one of the fixed points is stable while the other is unstable in both scenarios when $ab < 1$ and $ab > 1$ (this is the definition of a transcritical bifurcation -- an interchange of stabilities occurs). Given the other fixed point $x^*_2 \approx \frac{2(ab-1)}{ab^2}$, we plug into the equation above.
Suppose $ab < 1$. Then $f'(0) = 1 - ab > 0$, meaning this is a stable fixed point. However, $$f'(x_2^*) = 1 - ab + 2(ab -1) = 1-ab -2(1-ab) = - (1-ab) < 0.$$
An analogous analysis can be performed to show that one of the fixed points is stable and the other is non-stable for the case $ab > 1$.
Therefore, since an interchange of stabilities occurs for the fixed points, we conclude that $ab =1$ gives the equation of the bifurcation curve for the system.