Saddle-node bifurcation

Remarks

Under these conditions, the equilibria near $(0, 0)$ form a curve $\mu = m(x)$ with $m(0) = m'(0) = 0$ and $m''(0) = -f_{xx}(0, 0) / f_\mu(0, 0) \ne 0$. So for $\mu$ on one side of $0$ there are two equilibria near $0$, one stable and one unstable, and on the other side there are none.

Example: $f(x, \mu) = \mu - x^2$. Then $f(0, 0) = 0$, $f_x = -2x$ vanishes at $0$, $f_{xx} = -2 \ne 0$ and $f_\mu = 1 \ne 0$. The equilibria are $\pm\sqrt{\mu}$ for $\mu > 0$, and there are none for $\mu < 0$.

Non-example: $f(x, \mu) = \mu x - x^2$. Here $f_\mu(0, 0) = 0$, so clause 4 fails; the two equilibria $0$ and $\mu$ exist for all $\mu$ and only cross, which is a transcritical bifurcation.

For a system in $\R^n$, clause 2 is replaced by a condition on the Jacobian matrix at the equilibrium. It must have a simple eigenvalue $0$ and no other eigenvalue on the imaginary axis. The conditions 3 and 4 are then imposed on the equation reduced to a centre manifold.