= Solution
Put $z-z_0=\epsilon\bar z$, with physical transverse thickness $h=\delta H$. Matching the preceding quadratic <meniscus> gives $\delta=\epsilon^2$ when the inner amplitude is normalized as $H\sim\bar z^2/\sqrt2$. After multiplying the transformed film equation by $\epsilon$, its two capillary/gravity coefficients are $\delta^3\widehat C\epsilon^{-3}$ and $\delta^2\widehat C$. Setting the <curvature> coefficient to one therefore gives
$$
\boxed{\epsilon=\widehat C^{-1/3},\qquad\delta=\widehat C^{-2/3},\qquad\delta^2\widehat C=\epsilon.}
$$
Thus the inner equation is exactly the stated $\epsilon H_t+v_0H_{\bar z}+[H^3(H_{\bar z\bar z\bar z}+\epsilon)/3]_{\bar z}=0$. The leading matching conditions are $H\sim\bar z^2/\sqrt2$ 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-$\epsilon$ terms at leading order and set
$$
H=v_0^{2/3}g(\zeta),\qquad\zeta=\bar z/v_0^{1/3}.
$$
The 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 $3g_\zeta+(g^3g_{\zeta\zeta\zeta})_\zeta=0$. The supplied universal solution therefore yields the <quasisteady drag-out matching law>
$$
\boxed{H\longrightarrow C_0v_0^{2/3},\qquad C_0=0.948,\quad\bar z\to+\infty.}
$$
Above the turn-around region, $H$ and its outer derivatives are order one, while $\delta^3\widehat C=\epsilon^3$ and $\delta^2\widehat C=\epsilon$. Hence the leading outer equation and inflow condition are
$$
\boxed{H_t+v_0(t)H_z=0,\qquad H(z_0^+,t)=C_0[v_0(t)]^{2/3}.}
$$
The boundary value comes from matching, not from setting the film to zero at the wall contact point.
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