 Solution

At the crest the two depth branches meet at $h=h_c=h_1{F}^{1/3}$. For a smooth nondegenerate maximum of the bed, $D^{\ast }-D$ is proportional to $(x-x_c{)}^2$, while expansion of the depth relation gives $D^{\ast }-D\sim 3(h-h_c{)}^2/(2h_c)$. A differentiable flow through the crest therefore switches branches rather than following the cusp obtained by staying on the same branch. Subcritical upstream flow becomes supercritical downstream; supercritical upstream flow becomes subcritical downstream.

When $F<1$ and the bed returns to zero, put $y=h_{\mathrm{down}}/h_1$. The head equation becomes $2y^3-(2+F)y^2+F=0$, orThe root $y=1$ is the original subcritical branch. The other positive root is smaller and lies on the supercritical branch, givingThis is the smooth transcritical, shock-free solution. A downstream hydraulic jump would require further downstream conditions and energy loss. A degenerate flat crest can also require a branch-selection condition; the standard critical-control interpretation uses a nondegenerate smooth crest. For $F=1$, $D^{\ast }=0$, so a positive hump is incompatible with the prescribed upstream critical head. If $F=0$, the head equation is the still-water relation $h+D=h_1$; a crest of height $h_1$ dries out, so the wet transcritical-flow interpretation requires $F>0$.