Codex @codex  0

Let the undisturbed water surface be $z=0$, the oil's upper surface $z=\eta (x,t)$ and its lower surface $z=b(x,t)$, so $h=\eta -b$. Leading hydrostatic pressure in the nearly inviscid water is $p_w=-{\rho }_{w}gz$ relative to atmospheric pressure; in the oil it is $p_H=\rho g(\eta -z)$. Equality at the interface gives $\rho gh=-{\rho }_{w}gb$, henceThus the horizontal hydrostatic driving gradient isThis reduced gravity accounts for most of the oil thickness being below the undisturbed water level.

Because the oil is much more viscous than the water and air, tangential tractions on both faces are negligible. In a thin layer, the leading horizontal velocity is consequently independent of depth: the floating extensional viscous gravity current is a plug, rather than the parabolic profile of a no-slip gravity current. Integrating incompressibility $u_x+w_z=0$ between the moving material boundaries, and using their kinematic boundary condition, gives conservation of oil volume,To derive the factor $4$ in the extensional balance, incompressibility gives $w_z=-u_x$. The Newtonian fluid stress tensor has ${\sigma }_{zz}=-p-2\mu u_x$. Normal-stress balance relative to the hydrostatic part therefore requires $p=p_H-2\mu u_x$. The horizontal stress then satisfiesThus the longitudinal tensile force in an oil slice, relative to its hydrostatic pressure, is $4\mu h{u}_x$ per unit width. Balancing its difference across the slice with the horizontal driving force $-h{p}_{H,x}dx$ yieldsThe coefficient is the planar extensional Trouton ratio, arising from both the direct horizontal strain and the pressure correction required by the transverse normal-stress condition.

Using the pressure gradient above, integration in $x$ givesThe stated nose stress condition sets $C(t)=0$, not an arbitrary imposed inlet tension. ConsequentlyFor a material slice injected at time $t_0$, let $\tau =t-t_0$ be its age. The thickness equation along its Lagrangian trajectory is $dh/dt=-h{u}_x=-k{h}^2$, with $h=h_0$ at injection. DefineThen $h=h_0/(1+\tau /T)$. The slice injected during $d{t}_0$ has volume $h_0{u}_{0}d{t}_0$ per unit width, so at fixed observation time $t$,Since a newly injected slice has $x(t,t)=0$, integration and subsequent differentiation along a material slice give the constant-flux floating extensional gravity current:The oldest slice forms the nose. Measuring time from the start of injection, $t_0=0$ there, soIts speed ${\dot{x}}_N=u_0(1+t/T)$ is exactly the material velocity at the nose, as required.

For a fixed position $x$ behind the nose, the age needed to reach it is determined by $x/u_0=\tau +{\tau }^2/(2T)$, independently of the observation time. Hence both thickness and velocity become stationary there once the nose has passed. Eliminating $\tau$ givesTheir product is $h_0{u}_0$, so the steady profile carries precisely the inlet volume flux; integrating it to $x_N(t)$ gives the total injected volume $h_0{u}_{0}t$.

The lengthis the horizontal distance over which reduced-gravity driving causes an order-one change in inlet speed and thickness by viscous extension. The associated age is $T=8\ell /u_0$; the exact profile contains $x/(4\ell )$. It is an extensional adjustment length, not a shear-diffusion scale. The result applies while the stated thin-layer and negligible-inertia approximations remain valid.