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Under the plug-like extensional viscosity approximation, incompressibility gives $w_z=-u_x$. Symmetry about $z=0$ fixes $w=-z{u}_x$. Remove the reference hydrostatic stress of the middle fluid, and write $S$ for its leading vertical normal stress. Vertical traction balance at the upper surface givesThe middle fluid's horizontal normal stress exceeds $S$ by $4\mu u_x$: the horizontal strain is $u_x$ and the vertical strain is $-u_x$. HenceAt a slightly inclined upper interface, tangential traction continuity gives, to the order retained in the small-outer-viscosity extensional approximation,The slope term is needed even though the interface is nearly horizontal: dropping it would miss the derivative of $h$ in the extensional resultant. Terms involving the outer deviatoric normal-stress difference times the slope are smaller in this approximation. Integrating horizontal force balance over the symmetric layer, or over its upper half, now givesSubstitution yields the extensional-flow equationThe full layer thickness is $2h$, so its conservation of mass is $(2h{)}_t+(2hu{)}_x=0$, orThis explains the factor $4\mu$ in the extensional stress rather than the shear viscosity coefficient alone.

For a Fourier mode of wave number $k>0$, set $h=h_0+\eta e^{ikx-st}$ and $u=\hat{u}{e}^{ikx-st}$. Neglect ${\sigma }_{zz}^{+}$ as stipulated. The Fourier traction map for a viscous half-space, with outer dynamic viscosity $\lambda \mu$, gives ${\sigma }_{xz}^{+}=-2\lambda \mu k\hat{u}{e}^{ikx-st}$. The two linear equations areEliminating $\hat{u}$ proves the positive decay rate for relaxation of a symmetric viscous layer:When $\lambda \ll \kappa$, inner longitudinal extension dominates the viscous dissipation, and $s\sim \Delta \rho g{h}_0/(4\mu )$. When $\kappa \ll \lambda$ while the plug approximation remains valid, tangential motion of the semi-infinite outer fluids supplies the main resistance, and $s\sim \Delta \rho g{h}_0\kappa /(2\lambda \mu )$. The PDF's second inequality is $\kappa \ll \lambda$; the TeX converts it incorrectly to $\kappa \ll 1$. At much larger outer viscosity the plug assumption fails, as the next calculation shows.