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For a given dynamical system, the normal form of a transcritical bifurcation is given by $$ g_{\alpha}(x) = (1+\alpha)x-x^2. $$ To determine the fixed points, we have $g_{\alpha}(x) = x$, which implies $x_1 = 0$, $x_2 = \alpha$ are both fixed points.

To analyse the stability we use the Linearised Stability Theorem which states: a fixed point $x^*$ is stable if $|g'_{\alpha}(x^*)|<1$. Applying this $$|g'_{\alpha}(x=0)|<1 \iff |1+\alpha|<1 \iff -2<\alpha<0.$$ But according to my lecturer, a transcritial bifurcation with fixed point $x=0$ is stable for $\alpha<0$.

HarryT
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1 Answers1

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The normal form for a transcritical bifurcation is $g_\alpha=\alpha x -x^2$.

We linearize around $x=0$,

$L(x) = g_\alpha(0) + g'_\alpha(0)(x-0)=(\alpha-2x)|_{x=0}(x-0) = \alpha x$

This is stable iff the eigenvalues of $L$ are negative. Since $L$ is a scalar function the eigenvalues are just $\alpha$. So if $\alpha<0$ we have stability at the fixed point $x=0$.

The issue was that you were using the criteria for stability of a discrete dynamical system.

Jürgen Sukumaran
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