= Solution
For zero liquid flux the reduced gradients are $\Gamma_X=-4J/(H\Gamma)$ and $H_X=6J/(H^2\Gamma)$. Their ratio gives
$$
\boxed{\Gamma=B-\frac{H^2}{3},\qquad B=1+\frac{H_0^2}{3}.}
$$
Integrate $dX/dH=H^2\Gamma/(6J)$ to obtain the requested implicit height profile:
$$
\boxed{X=\frac1{6J}\left[\frac B3(H^3-H_0^3)-\frac1{15}(H^5-H_0^5)\right].}
$$
The endpoint relation is $H_1^2=H_0^2+3\delta$, and the same expression at $X=1$ fixes $J$. For $H_0=0$ it simplifies to
$$
\Gamma=1-H^2/3,\qquad
6JX=H^3/3-H^5/15,\qquad H_1=\sqrt{3\delta},
$$
so
$$
\boxed{J=\frac{\delta^{3/2}(1-3\delta/5)}{2\sqrt3}.}
$$
This <zero-liquid-flux surfactant film> grows to the right from a formal dry edge, with $H\sim(18JX)^{1/3}$ there. Its dimensionless <velocity> is
$$
\boxed{U(Y)=\tfrac12H_XY(Y-2H/3).}
$$
The lower two thirds flow left, the upper third flows right, and the integrated liquid flux is zero. Surface advection carries the positive surfactant flux, with $U_s=H^2H_X/6=J/\Gamma$.
The flux curve obeys $dJ/d\delta=(\sqrt3/4)\sqrt\delta(1-\delta)$. It increases throughout the interval and reaches its formal maximum \b[$J_{\max}=1/(5\sqrt3)$ at $\delta=1$]. Larger depletion strengthens the Marangoni driving, so increasing transport is plausible. But finite flux at a zero-concentration endpoint is a singular prediction: $U_s=J/\Gamma$ and the height gradient diverge there. Neglected diffusion, capillarity or an endpoint region must regularize the physical limit, and can alter its maximum. The dry initial edge is likewise outside a uniform small-slope approximation.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/iii/paper-73-film-profiles.png]
{title=Zero-surfactant-flux and zero-liquid-flux film shapes with velocity profiles, and the reduced surfactant-flux curve versus depletion}
{height=620}
The profiles and <velocity> arrows illustrate the opposing interior-flow directions and the formal endpoint maximum; neither sketch treats the singular edges as resolved lubrication regions.
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