= Solution
With zero surfactant flux and positive concentration, the nondiffusive surface <velocity> is zero. The reduced equations become $\Gamma_X=6Q/H^2$, $H_X=-12Q/H^3$. Integrating, with the prescribed initial height and concentration, gives
$$
\boxed{H^4=H_0^4-48QX,\qquad
\Gamma=1+\frac{H_0^2-H^2}{4}.}
$$
At the far end, the imposed concentration drop fixes $H(1)^2=H_0^2+4\delta$, hence
$$
\boxed{Q=-\frac{\delta(H_0^2+2\delta)}6,\qquad
H(X)=\big[H_0^4+8\delta(H_0^2+2\delta)X\big]^{1/4}.}
$$
This <steady surfactant film with zero surface flux> rises to the right while its concentration decreases. The liquid flux is \b[negative], despite the rightward Marangoni traction: the adverse hydrostatic gradient is strong enough to immobilize the surface and drive the interior leftwards.
In dimensionless height $Y=y/h_*$, the <velocity> profile is
$$
\boxed{U(Y)=\tfrac12H_XY(Y-H),\qquad 0\le Y\le H.}
$$
It vanishes at both boundaries and is negative in the interior when $\delta>0$. Its integral is $Q=-H^3H_X/12$. For $\delta=0$, the nondegenerate film is flat and stationary. If its initial thickness is zero, the formal $X^{1/4}$ edge has an unbounded slope, so the lubrication approximation applies away from that edge rather than at the exact dry point.
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