= Solution
Under the plug-like <extensional viscosity> approximation, <incompressibility> gives $w_z=-u_x$. Symmetry about $z=0$ fixes $w=-zu_x$. 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 gives
$$
S=-\Delta\rho gh+\sigma_{zz}^+.
$$
The middle fluid's horizontal normal stress exceeds $S$ by $4\mu u_x$: the horizontal strain is $u_x$ and the vertical strain is $-u_x$. Hence
$$
\sigma_{xx}^{\mathrm{middle}}=S+4\mu u_x.
$$
At a slightly inclined upper interface, tangential traction continuity gives, to the order retained in the small-outer-viscosity extensional approximation,
$$
\sigma_{xz}^{\mathrm{middle}}(h)=\sigma_{xz}^++4\mu h_xu_x.
$$
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 gives
$$
0=h\partial_x(S+4\mu u_x)+\sigma_{xz}^{\mathrm{middle}}(h).
$$
Substitution yields the \b[extensional-flow equation]
$$
\boxed{4\mu\partial_x(hu_x)=h\partial_x(\Delta\rho gh-\sigma_{zz}^+)-\sigma_{xz}^+.}
$$
The full layer thickness is $2h$, so its <conservation of mass> is $(2h)_t+(2hu)_x=0$, or
$$
\boxed{h_t+(hu)_x=0.}
$$
This 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 $u=\widehat u e^{ikx-st}$. Neglect $\sigma_{zz}^+$ as stipulated. The <Fourier traction map for a viscous half-space>, with outer <dynamic viscosity> $\lambda\mu$, gives $\sigma_{xz}^+=-2\lambda\mu k\widehat u e^{ikx-st}$. The two linear equations are
$$
-4\mu h_0k^2\widehat u=ik\Delta\rho gh_0\eta+2\lambda\mu k\widehat u,
\qquad -s\eta+ikh_0\widehat u=0.
$$
Eliminating $\widehat u$ proves the \b[positive decay rate] for <relaxation of a symmetric viscous layer>:
$$
\boxed{s=\frac{\Delta\rho gh_0}{2\mu}\frac\kappa{\lambda+2\kappa},\qquad\kappa=kh_0.}
$$
When $\lambda\ll\kappa$, inner longitudinal extension dominates the <viscous dissipation>, and $s\sim\Delta\rho gh_0/(4\mu)$. When $\kappa\ll\lambda$ while the plug approximation remains valid, tangential motion of the semi-infinite outer fluids supplies the main resistance, and $s\sim\Delta\rho gh_0\kappa/(2\lambda\mu)$. The PDF's second inequality is $\kappa\ll\lambda$; the TeX converts it incorrectly to $\kappa\ll1$. At much larger outer viscosity the plug assumption fails, as the next calculation shows.
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