= 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 $F_u\ne0$. 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,
$$
F=16u^2-(30x+24y)u+9y^2-40xy
$$
has two branches $u=(15x+12y\pm5\sqrt{9x^2+40xy})/16$. On them $F_u=\pm10\sqrt{9x^2+40xy}$, so the graph <projection> folds on the nonzero boundary points of $9x^2+40xy\ge0$. 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.
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