With $K=\rho g/(3\mu)$ and formal edge conditions $h(\pm L,t)=0$, the <lubrication theory> equation is $h_t=K(h^3h_x)_x$. Its positive separated long-time profile has
$$
h(x,t)=\left(\frac{L^2}{Kt}\right)^{1/3}H(x/L),\qquad
(H^3H')'=-\frac H3.
$$
Writing $C=B(4/5,1/2)$ in terms of the <beta function>, the centre height is $h_0^3=10\mu L^2/(\rho gC^2t)$. The <edge region of a draining viscous film> determines how the singular outer profile matches the flow over the physical edge.

Draining viscous film on a finite horizontal plate

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.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 329 / 3 / c / Solution

The outer <similarity solution> approaches a plate edge as $h/h_0\sim[(4C/5)(1-x/L)]^{1/4}$, where $C=B(4/5,1/2)$. Its slope becomes order one at
$$
\ell\sim L(h_0/L)^{4/3}.
$$
This region lies outside <lubrication theory> because its vertical and horizontal scales are comparable. The outer zero-thickness boundary condition is consistent at leading order since the matching height is $O(\ell)\ll h_0$, while the actual edge flow requires an inner solution. A finite outward flux survives even though the outer thickness vanishes, because the slope diverges.

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 Incoming links: Edge region of a draining viscous film