Pitchfork bifurcation

Remarks

Under these conditions, besides $x = 0$, the equilibria near $(0, 0)$ form a curve $\mu = m(x)$ with $m$ even, $m(0) = 0$ and $m''(0) = -f_{xxx}(0, 0) / (3 f_{x\mu}(0, 0))$. So on one side of $\mu = 0$ there is a symmetric pair of new equilibria, and the stability of $0$ changes at $\mu = 0$. The bifurcation is supercritical when the new equilibria are stable and subcritical when they are unstable.

Example: $f(x, \mu) = \mu x - x^3$. It is odd in $x$, $f_x(0, 0) = 0$, $f_{x\mu} = 1$ and $f_{xxx} = -6$. The new equilibria are $\pm\sqrt{\mu}$ for $\mu > 0$, and they are stable since $f_x(\pm\sqrt{\mu}, \mu) = -2\mu < 0$. The equation $x' = \mu x + x^3$ gives the subcritical case.

Non-example: $f(x, \mu) = \mu x - x^2$ is not odd in $x$, so the symmetry hypothesis fails; it gives a transcritical bifurcation. Without the symmetry, a small perturbation such as $\mu x - x^3 + \varepsilon$ breaks the pitchfork into a saddle-node and a separate branch.