 Solution

Put $z-z_0=ϵ\bar{z}$, with physical transverse thickness $h=\delta H$. Matching the preceding quadratic meniscus gives $\delta ={ϵ}^2$ when the inner amplitude is normalized as $H\sim {\bar{z}}^2/\sqrt{2}$. After multiplying the transformed film equation by $ϵ$, its two capillary/gravity coefficients are ${\delta }^3\hat{C}{ϵ}^{-3}$ and ${\delta }^2\hat{C}$. Setting the curvature coefficient to one therefore givesThus the inner equation is exactly the stated $ϵH_t+v_0{H}_{\bar{z}}+[H^3(H_{\bar{z}\bar{z}\bar{z}}+ϵ)/3{]}_{\bar{z}}=0$. The leading matching conditions are $H\sim {\bar{z}}^2/\sqrt{2}$ as $\bar{z}\to -\infty$ and a bounded film, with vanishing derivatives, as $\bar{z}\to +\infty$. They connect the static bath-side meniscus to a uniform entrained film.

For positive wall speed varying on the outer time scale, neglect the order-$ϵ$ terms at leading order and setThe parabolic matching coefficient is unchanged because $v_0^{2/3}/v_0^{2/3}=1$. Both remaining differential terms scale as $v_0^{4/3}$, reducing the equation to $3g_{\zeta }+(g^3{g}_{\zeta \zeta \zeta }{)}_{\zeta }=0$. The supplied universal solution therefore yields the quasisteady drag-out matching lawAbove the turn-around region, $H$ and its outer derivatives are order one, while ${\delta }^3\hat{C}={ϵ}^3$ and ${\delta }^2\hat{C}=ϵ$. Hence the leading outer equation and inflow condition areThe boundary value comes from matching, not from setting the film to zero at the wall contact point.