= Solution
For the horizontal film, part (a) gives the <lubrication theory> equation
$$
h_t=K(h^3h_x)_x,\qquad K=\frac{\rho g}{3\mu}.
$$
The length scale is fixed at $L$, so balancing $h/t$ against $Kh^4/L^2$ gives $h\sim(L^2/(Kt))^{1/3}$. The <draining viscous film on a finite horizontal plate> therefore has the long-time <similarity solution>
$$
\boxed{h(x,t)=\left(\frac{L^2}{Kt}\right)^{1/3}H(\eta),\qquad
\eta=\frac xL,\qquad h\propto t^{-1/3}.}
$$
An additive shift of the time origin can represent the initial transient without changing the long-time law. The positive profile is even, with $H'(0)=0$ and $H(\pm1)=0$. Substitution yields
$$
(H^3H')'=-\frac H3.
$$
To integrate, put $q=H^3H'$ and use $q'=q\,dq/dH\,/H^3$. Then
$$
q\frac{dq}{dH}=-\frac{H^4}{3},\qquad
q^2=\frac2{15}(H_0^5-H^5),\qquad H_0=H(0).
$$
For $0<\eta<1$ the decreasing branch has $q<0$, so inversion gives
$$
\boxed{\eta(H)=\int_H^{H_0}F(s)\,ds,\qquad
F(s)=\sqrt{\frac{15}{2}}\frac{s^3}{\sqrt{H_0^5-s^5}}.}
$$
The negative half of the profile follows by symmetry.
At the edge, $\eta(0)=1$. Set $u=(s/H_0)^5$ in the integral to obtain
$$
1=\frac15\sqrt{\frac{15}{2}}H_0^{3/2}
\int_0^1u^{-1/5}(1-u)^{-1/2}\,du
=\frac15\sqrt{\frac{15}{2}}H_0^{3/2}C,
\qquad C=B(4/5,1/2).
$$
Here $B$ is the <beta function>. Thus $H_0^3=10/(3C^2)$, and the centre thickness is
$$
\boxed{h_0(t)=h(0,t)=\left(\frac{10\mu L^2}{\rho gC^2t}\right)^{1/3}.}
$$
For $H\ll H_0$, the implicit integral remaining between $0$ and $H$ gives
$$
1-\eta\sim\sqrt{\frac{15}{2}}\frac{H^4}{4H_0^{5/2}},\qquad
\left(\frac H{H_0}\right)^4\sim\frac{4C}{5}(1-\eta).
$$
Consequently the <edge region of a draining viscous film> has
$$
\boxed{k=\frac14,\qquad
h\sim h_0\left[\frac{4C}{5}\frac{L-x}{L}\right]^{1/4}.}
$$
Its slope satisfies $|h_x|\sim(h_0/L)(1-x/L)^{-3/4}$ up to a constant factor, so it becomes order one at
$$
\boxed{\ell=L-x\sim L\left(\frac{h_0}{L}\right)^{4/3}.}
$$
The local height is then also $O(\ell)$, violating the shallow geometry required by <lubrication theory>. A full local flow is needed to turn the fluid over the edge.
The imposed zero thickness is nevertheless a consistent leading outer boundary condition: $\ell/h_0\sim(h_0/L)^{1/3}\ll1$, so the unresolved edge height is small compared with the film's bulk height. It should not be interpreted as an exact pointwise prediction inside the edge region. Moreover, $H^3H'\to-\sqrt{2/15}\,H_0^{5/2}$ at the right edge, so the outward <volume flux> remains finite even as the outer thickness tends to zero. The singular slope is what allows this outer solution to describe drainage.
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