Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 77 / 2

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.

Applying the stated long-wave expression formally in the intended high-viscosity regime, the numerator is dominated by $2\lambda {\kappa }^2/3$ and the displayed denominator by $\lambda$. It givesThe mechanism is inner transverse shear: the outer fluids almost immobilize the interfaces tangentially, so the middle layer has a pressure-driven parabolic velocity instead of the plug-like extensional velocity of part (b). Indeed, with nearly stationary faces at $z=\pm h$, its integrated flux is $-2h^3{p}_x/(3\mu )$. With $p_x=\Delta \rho g{h}_x$, conservation of mass gives precisely the displayed decay rate. The leading viscous dissipation is then in the middle layer, even though its neighbours have much greater dynamic viscosity.

There is a nonuniformity in taking this conclusion literally for every $\lambda \gg {\kappa }^{-1}$. A long-wave remainder written for fixed $\lambda$ cannot be assumed uniform as $\lambda$ diverges. Here this matters physically because sufficiently viscous outer fluids also resist normal interface motion. The distinction can be resolved directly, without an unspecified remainder, from the Stokes equation. For a symmetric thickness mode, the middle vertical velocity has the formIt is odd in $z$, and the middle horizontal velocity is even. At $z=h_0$, the exact outer traction map is ${\sigma }_{xz}^{+}=-2\lambda \mu ku$ and ${\sigma }_{zz}^{+}=-2\lambda \mu kw$. Tangential continuity and the normal hydrostatic jump therefore giveThe latter uses $p=\mu (w_{zzz}-k^2{w}_z)/k^2$, obtained from horizontal force balance. The kinematic condition is $w(h_0)=-s\eta$. Writing $S=\sinh \kappa$ and $C=\cosh \kappa$, tangential continuity fixesEliminating $A,B$ in the remaining two equations gives the exact symmetric-layer rateFor $\kappa \ll 1$ and $\lambda \kappa \gg 1$, this proves the uniform high-viscosity limit of viscous-layer relaxationThus the intended inner-shear answer is valid in the distinguished range ${\kappa }^{-1}\ll \lambda \ll {\kappa }^{-3}$. If $\lambda {\kappa }^3\gg 1$, the main resistance is instead outer normal motion andBoth ranges satisfy the printed $\lambda \gg {\kappa }^{-1}\gg 1$. For example $\lambda ={\kappa }^{-4}$ lies in the latter range and has $s\sim [\Delta \rho g{h}_0/(2\mu )]{\kappa }^3$, rather than a rate proportional to ${\kappa }^2$. The exact denominator contains a ${\lambda }^2{\kappa }^3$ contribution hidden by a fixed-parameter remainder. The inner-shear limit is the intended result, but an additional upper restriction on $\lambda$ is needed to make it a uniform conclusion.