 Solution

Define $A(H)=(H-1)/H^3$ for positive thickness. Its derivative is $A^{\prime }(H)=(3-2H)/H^4$, so its maximum is attained at $H=3/2$ and equals $4/27$. At $\theta =0$, the equation requires $A(H)=\alpha$, proving the necessary condition $\alpha \le 4/27$.

It is also sufficient for a continuous finite periodic branch. On $0<H\le 3/2$, $A$ increases monotonically from $-\infty$ to $4/27$, so every value $\alpha \cos \theta$ in the required range has exactly one root on this branch. It joins $H=1$ where $\cos \theta =0$. The second large-thickness root on the ascending side does not extend to a finite periodic branch through those points.

Thus the limiting maximum thickness occurs at the ascending point of the critical film:Subcritical films satisfy a strict thickness bound there. Equivalently the largest carried flux and critical absolute thickness areThe mechanism is a transport capacity limit: wall transport grows only as $Uh$, while opposing gravitational drainage grows as $h^3$. At the critical point, adding thickness no longer increases the upward flux. Indeed $\partial Q/\partial h=U(3/H-2)$ on the steady branch and vanishes at $H=3/2$. Excess liquid cannot be maintained as this globally wall-entrained film and must redistribute or drain.

The algebra establishes existence and the limiting branch of the steady lubrication model. A full temporal free-surface stability result requires its time-dependent equation and, when relevant, pressure and capillary corrections; the sign of $\partial Q/\partial h$ alone is not a proof of growth of perturbations.