 Solution

For zero liquid flux the reduced gradients are ${\Gamma }_X=-4J/(H\Gamma )$ and $H_X=6J/(H^2\Gamma )$. Their ratio givesIntegrate $dX/dH=H^2\Gamma /(6J)$ to obtain the requested implicit height profile: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 tosoThis zero-liquid-flux surfactant film grows to the right from a formal dry edge, with $H\sim (18JX{)}^{1/3}$ there. Its dimensionless velocity isThe 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 $U_s=H^2{H}_X/6=J/\Gamma$.

The flux curve obeys $dJ/d\delta =(\sqrt{3}/4)\sqrt{\delta }(1-\delta )$. It increases throughout the interval and reaches its formal maximum $J_{\max }=1/(5\sqrt{3})$ 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.

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.