= Solution
Let
$$
\delta=(ah_0)^{1/2},
\qquad
\xi=\frac{x-L}{\delta},
\qquad
H=\frac h{h_0},
\qquad
V=\frac{\mu U}{\gamma}\left(\frac a{h_0}\right)^{1/2}.
$$
Flux conservation gives $u=U/H$. Substitution into the extensional-force equation yields the third-order equation
$$
\boxed{HH_{\xi\xi\xi}
-4V(H^{-1}H_\xi)_\xi=0}.
$$
Since
$$
HH_{\xi\xi\xi}
=\left(HH_{\xi\xi}-\frac12H_\xi^2\right)_\xi,
$$
one integration, using $H\to1$ and its derivatives tending to zero on the flat-film side, gives
$$
HH_{\xi\xi}-\frac12H_\xi^2
=4V\frac{H_\xi}{H}.
$$
Put $q(H)=H_\xi$. After division by $q$, this becomes
$$
H\frac{dq}{dH}-\frac12q=\frac{4V}{H}.
$$
The <integrating factor> $H^{-1/2}$ gives
$$
\frac d{dH}(qH^{-1/2})=4VH^{-5/2}.
$$
A second integration and $q(1)=0$ therefore give
$$
\boxed{H^{-1/2}H_\xi
=\frac{8V}{3}(1-H^{-3/2})}.
$$
On the bubble side, matching to its cylindrical shape gives $H\sim\xi^2/2$, so $H^{-1/2}H_\xi\to\sqrt2$. Taking $H\to\infty$ in the integrated equation yields $8V/3=\sqrt2$. Hence
$$
\boxed{
U=\frac{3\gamma}{8\mu}
\left(\frac{2h_0}{a}\right)^{1/2}}.
$$
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