= Solution
With no $y$ dependence, integrate the dimensionless equation once. The upstream condition fixes the constant:
$$
h^3-h^3h_x=1,
\qquad
\boxed{h_x=1-h^{-3}}.
$$
Write $h=1+\eta$ upstream. Linearization gives $\eta_x=3\eta$, so
$$
\boxed{h\sim1+Ae^{3x}},
\qquad k=3.
$$
For $A>0$, the thickness increases monotonically. At large $h$,
$$
\frac{dx}{dh}=\frac1{1-h^{-3}}
=1+h^{-3}+O(h^{-6}).
$$
Integration gives $x-x_0=h- frac12h^{-2}+O(h^{-5})$, and inversion yields
$$
\boxed{h=x-x_0+O(x^{-2})}.
$$
In dimensional variables, $dh_{\rm dim}/dx_{\rm dim}\to\alpha$. The increasing thickness therefore cancels the plane's downward slope, so the free surface becomes asymptotically horizontal. The profile represents the upslope edge of a deep viscous pool or pond held back by an obstruction.
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