Thin-film equations with insoluble surfactant ($h_t+q_x=0,C_t+(C{u}_s{)}_x=0$)

For a no-slip planar liquid film with an insoluble surfactant, lubrication theory gives $q=-h^3{p}_x/(3\mu )+h^2{\gamma }_x/(2\mu )$ and surface velocity $u_s=-h^2{p}_x/(2\mu )+h{\gamma }_x/\mu$. Mass conservation and conservation of insoluble surfactant on a moving interface give $h_t+q_x=0$ and $C_t+(C{u}_s{)}_x=0$. With $\gamma ={\gamma }_0-AC$ and small slopes, $p_x=\rho g{h}_x-(\gamma h_{xx}{)}_x$. The different coefficients in $q$ and $u_s$ are essential: Marangoni stress and hydrostatic return flow transport fluid and surfactant differently.

In the same gravity–Marangoni reduction, zero liquid flux gives $\Gamma =1+(H_0^2-H^2)/3$ and $6JX=(1+H_0^2/3)(H^3-H_0^3)/3-(H^5-H_0^5)/15$. The lower part of the liquid flows backwards while the surface and upper third flow forwards. For zero initial height and concentration drop $\delta$, $J={\delta }^{3/2}(1-3\delta /5)/(2\sqrt{3})$ is largest formally at complete depletion. Since surface velocity is $J/\Gamma$, that endpoint is singular and requires physical regularization by diffusion, capillarity or an edge model. The algebraic maximum is not by itself a regular zero-concentration outflow.

In the steady nondiffusive gravity–Marangoni model with capillarity neglected, zero surfactant flux and positive concentration make the surface velocity zero. With $Q=-H^2{\Gamma }_X/2-H^3{H}_X/3$ and $J=-H\Gamma {\Gamma }_X-H^2\Gamma H_X/2$, integration gives $H^4=H_0^4-48QX$ and $\Gamma =1+(H_0^2-H^2)/4$. A concentration drop $\delta$ over unit length therefore selects $Q=-\delta (H_0^2+2\delta )/6$. The liquid flows opposite to the imposed Marangoni traction, with velocity proportional to $H_{X}Y(Y-H)/2$ and an immobilized surface. This reduction is singular at a dry endpoint; its small-slope assumption does not resolve that edge.

A localized amount $M=\int Cdx$ of insoluble surfactant spreads over a planar film of depth scale $h_0$. Neglect gravity, capillarity and diffusion. Then $C\sim M/\ell$ and Marangoni stress drives $u_s\sim AM{h}_0/(\mu {\ell }^2)$, giving $\ell \sim (AM{h}_{0}t/\mu {)}^{1/3}$. The thin-film equations with insoluble surfactant admit a similarity solution in which the film height and concentration depend on $x/\ell$, conserving total surfactant and fluid volume relative to the undisturbed layer.

When the smoothing width exceeds the capillary length, gravity dominates capillarity. Balancing hydrostatic flux $\rho g{h}_0^4/(\mu \Delta )$ against the Marangoni stress flux gives $\Delta \sim \rho g{h}_0^2{x}_N^2/(AM)\propto t^{2/3}$. The condition is ${\Delta }^2\gg {\gamma }_0/(\rho g)$ in physical units. The smoothing region catches up with the spreading pool at $x_N\sim AM/(\rho g{h}_0^2)$ and $t\sim \mu (AM{)}^2/[(\rho g{)}^3{h}_0^7]$, after which the sharp-front outer similarity solution ceases to apply.

After gravity smoothing of a Marangoni front reaches the whole spreading pool, the film approaches uniform depth while surface transport continues. A small height gradient creates a hydrostatic pressure return flow that almost cancels the net Marangoni stress flux. Setting $q\simeq 0$ in the thin-film equations with insoluble surfactant gives $h_x\simeq -3A{C}_x/(2\rho g{h}_0)$ and fractional depth variation $O(AM/(\rho g{h}_0^2{x}_N))\ll 1$. The depressed region under the surfactant therefore becomes progressively shallower.

The height jump in linear similarity profiles for surfactant spreading is smoothed by surface tension. If $C_x\sim M/x_N^2$ through a region of width $\Delta$, the Marangoni stress flux is $AM{h}_0^2/(\mu x_N^2)$ and the capillary flux is ${\gamma }_0{h}_0^4/(\mu {\Delta }^3)$. Their balance gives $\Delta \sim ({\gamma }_0{h}_0^2{x}_N^2/(AM){)}^{1/3}\propto t^{2/9}$. This estimate assumes gravity is negligible and the smoothing region remains narrow compared with the pool.

For symmetric finite-mass Marangoni spreading on a liquid film, the leading outer profiles are $h=2h_0|x|/x_N$ and $C=(M/x_N)(1-|x|/x_N)$ inside $|x|<x_N$, where $x_N=(6AM{h}_{0}t/\mu {)}^{1/3}$. Outside, $h=h_0$ and $C=0$. The similarity solution follows from $H{\Gamma }^{\prime }=-\eta /3$, $\eta H^{\prime }=H$, $2\int \Gamma d\eta =1$ and $\int (H-1)d\eta =0$. The film reaches height $2h_0$ immediately behind the front, leaving an outer height jump that gravity or surface tension must smooth. These are ideal leading outer profiles, not a description of the microscopic central dry region or the smoothed edge.