 Fold of a solution branch

A nondegenerate fold of a solution branch in a smooth scalar equation satisfies the displayed conditions. The implicit function theorem still gives $\lambda =\lambda (x)$ because $F_{\lambda }\ne 0$. Implicit differentiation gives ${\lambda }^{\prime }(x_{\ast })=0$ and ${\lambda }^{\prime \prime }(x_{\ast })=-F_{xx}/F_{\lambda }\ne 0$: the parameter has a genuine local extremum on the curve.

In the cubic polynomial, $F_x=0$ gives $x_{\ast }=\pm 1/\sqrt{3}$, and substituting into $F=0$ gives ${\lambda }_{\ast }=2x_{\ast }/3$. Furthermore, $F_{xx}=6x_{\ast }\ne 0$. Yet $\nabla F=(3x^2-1,1)$ never vanishes, so the full solution branch remains a regular level set everywhere. The obstruction belongs to the projection onto $\lambda$, not to the curve itself.

For systems, the cited continuation account uses a one-dimensional kernel of $F_u$ and $F_{\lambda }$ outside its image of a linear map to define a simple fold (its Definition 2.2). These conditions make $[F_u{F}_{\lambda }]$ have full rank and support a transverse correction. A nonzero second-order term along the kernel is additionally needed to describe a nondegenerate quadratic turning point; in the scalar example this is $F_{xx}\ne 0$.^[ref]