Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 78 / 2

Choose $x=0$ at the contact line, with $x>0$ pointing into the mountains and the plate moving in the positive $x$ direction at speed $U>0$. Let $z=0$ be the plate. For viscous mountain spreading over an underthrust plate, the lubrication approximation assumes slow horizontal variation, hydrostatic pressure $p=\rho g(h-z)$, and $\mu u_{zz}=p_x=\rho g{h}_x\equiv G$. Impose no slip at the plate and zero tangential traction at the free surface. Without sediment,The horizontal mass flux is $\rho q$. For a steady mountain range with no mountain material entering across its contact line, $q=0$. Therefore $h_x=3\mu U/(\rho g{h}^2)$, and the vanishing height at the bare contact gives the mountain profileThe large slope extremely close to $x=0$ is an inner region where the lubrication approximation ceases to be quantitatively valid.

Now let sediment occupy $0<z<d$ and mountains $d<z<h$, with $H=h-d$ the mountain-material thickness and $M={\mu }_s/\mu$. Equal densities make $G=\rho g{h}_x$ the same in both layers. Continuity of velocity and shear stress at $z=d$, no slip at $z=0$ and zero shear at $z=h$ give the shear stress $G(z-h)$ throughout. Thus the two velocity profiles areTheir volume fluxes areMultiplying by $\rho$ gives the separate mass fluxes. The coupled conservation of mass equations areThis is the sediment-lubricated viscous mountain model. It includes shear-driven transport by the plate and pressure-driven gravitational spreading.

In steady state the mountain-material flux is zero, while the upstream uniform sediment layer has $q_s=U{d}_{\infty }$. For $H>0$ this givesEliminating $G$ proves the algebraic thickness relation:At the contact $H\to 0^{+}$, the last coefficient must vanish. The nonzero physical solution is consequentlyThe condition comes from the limiting zero flux per unit mountain thickness, not merely the automatic vanishing of $q_m$ at $H=0$. The surface slope anywhere under the mountains isLower sediment viscosity flattens the near-contact surface, while its contact height remains $3d_{\infty }$. Far inside, where $H\gg d/M$, the mountains recover the bare-layer slope scaling. The formulas determine the mountain profile by the algebraic relation plus this first-order equation. For example, put $\delta =d/d_{\infty }$ and $w=H/d$; the physical root satisfies $2M(\delta -1)w^2+3(\delta -2)w+\delta -3=0$, taking $w\ge 0$ for $1<\delta \le 3$. This selects the branch smoothly connected to the contact.

In front of the contact, $x<0$, there is only sediment, so $h=d$ andDefine $\ell =\rho g{d}_{\infty }^3/(3{\mu }_{s}U)$ and $P(\delta )={\delta }^3/3+{\delta }^2/2+\delta +\ln (\delta -1)$. Integrating from the contact yields the upstream basin profile of a sediment-lubricated mountain:Equivalently, $d(x)=d_{\infty }{P}^{-1}(P(3)+x/\ell )$ on $1<\delta \le 3$. The inverse is uniquely defined because $P^{\prime }(\delta )={\delta }^3/(\delta -1)>0$; an elementary explicit inverse is unnecessary. This profile tends to $d_{\infty }$ as $x\to -\infty$, approaches $3d_{\infty }$ at the contact and has the same surface slope there as the mountain branch. Its far-field excess decays exponentially on length scale $\ell$, which increases as the sediment viscosity decreases.