Fold boundary of a characteristic graph

A smooth characteristic surface in three-dimensional $(x,y,u)$ space need not remain the graph of a differentiable function after projection to the $(x,y)$ plane. Locally, if its implicit equation is $F(x,y,u)=0$, the implicit function theorem produces a smooth graph wherever $F_u\ne 0$. At a point where $F_u=0$, the projection can fold and two graph branches can meet. A real radical formula may continue to that boundary even though its partial derivatives diverge there. Thus the closed set on which the radical is real can be larger than the open domain of a classical solution.

For example,has two branches $u=(15x+12y\pm 5\sqrt{9x^2+40xy})/16$. On them $F_u=\pm 10\sqrt{9x^2+40xy}$, so the graph projection folds on the nonzero boundary points of $9x^2+40xy\ge 0$. Its gradient in the base variables does not vanish at those points, and differentiation of the radical gives the transverse derivative singularity. The characteristic surface itself can be smooth there: its graph coordinate is the part which fails. Boundary values may still be defined continuously, but they do not create an open classical-solution neighborhood.