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Use the Unscaled Papkovich–Neuber representation with $e_k=e^{ikx-ky}$ and harmonic potentialsTo see how these coefficients arise, start with $\Phi =(0,A{e}_k)$ and $\chi =B{e}_k$. The representation givesAt $y=0$, the prescribed tangential velocity requires $ikB=U$, while zero normal velocity requires $A=-kB=iU$. Hence the decaying half-space flow isThe pressure follows from $2\mu {\partial }_y{\Phi }_y$. These harmonic potentials verify the Stokes equation and incompressibility; the polynomial factors do not prevent decay as $y\to \infty$.

For the Newtonian fluid stress tensor, ${\sigma }_{xy}=\mu (u_y+v_x)$ and ${\sigma }_{yy}=-p+2\mu v_y$. On the boundary $u_y=-2kU{e}^{ikx}$, $v_x=0$ and $v_y=-ikU{e}^{ikx}$. ThereforeTake real parts for a physical sinusoidal flow. Together with the normal-velocity result supplied in the question, these formulas give the Fourier traction map for a viscous half-space; for signed wave numbers its resistance coefficient is $2\mu |k|$.

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.

A layer of half-thickness $h_0$, dynamic viscosity $\mu$ and density $\rho$, between semi-infinite fluids of viscosity $\lambda \mu$ and densities $\rho \mp \Delta \rho$, relaxes a symmetric thickness Fourier mode. In the plug-like extensional viscosity regime, mass conservation and $4\mu (h{u}_x{)}_x=h\Delta \rho g{h}_x-{\sigma }_{xz}^{+}$ together with the Fourier traction map for a viscous half-space give the displayed rate, where $\kappa =|k|h_0$. Inner longitudinal extension dominates for $\lambda \ll \kappa$; outer tangential resistance dominates for $\kappa \ll \lambda$ while the plug approximation remains valid.