Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 73 / 3 / Solution

Take $z$ positive upward, and write $b=\tilde{\rho }g$. For a straight steady viscous buoyant conduit, the inner and outer axial equations areRegularity gives $w_i^{\prime }(0)=0$, the wall has $w_o(R)=0$, and velocity and tangential traction are continuous at $r=a_0$: $w_i=w_o$, $\lambda \mu w_i^{\prime }=\mu w_o^{\prime }$. With no interfacial tension specified, normal traction is continuous; for this straight steady flow that sets $P_i=P_o$ at the interface. Thus the modified gradients are common in this case.

When their value is zero, integration givesConsequentlyThe large center-to-interface speed ratio explains why the inner parabolic component dominates at small viscosity ratio.

For a slowly varying radius, set $B=b-P_{i,z}$. The local inner solution has $w_i=w_a+B(a^2-r^2)/(4\lambda \mu )$, soThe shear-driven outer interface speed has scale $a^{2}B\ln (R/a)/\mu$. The pressure-driven return flow needed for zero total flux adds a velocity of order $q/R^2$ and a shear-driven contribution of order $a^{2}B/\mu$. Relative to the inner parabolic velocity these terms are $O(\lambda \ln (R/a))$ and $O((a/R{)}^2)$, which are small under the stated assumptions. Axial viscous derivatives are also smaller because the plume is slender. ThereforeFor clarity, the outer pressure-gradient estimate must include both return flux and transmitted interface shear. Their flux scales are $R^4{P}_{o,z}/\mu$, $a^2{R}^{2}B/\mu$ and $q$. Zero total flux hence impliesThis justifies the negligible outer pressure gradient. We subsequently choose its negligible pressure reference as zero, as in the reduced model.

Near the moving interface the outer radial source flow is $u_r=a{a}_t/r$: the interface's axial advection is smaller because its axial velocity is small compared with the inner core speed. Its radial strain is ${\partial }_r{u}_r=-a{a}_t/r^2$. Continuity of normal traction at $r=a$, neglecting the much smaller inner normal viscous stress, givesWith $A=a^2/a_0^2$, this is $P_i=\mu A_t/A$ after dropping $P_o$. Conservation of plume volume is $\pi (a^2{)}_t+q_z=0$.

Let the axial and velocity scales beSince $\mu U_0/(b{\ell }^2)=1$, substitution gives the conduit equationThe buoyant flux and the viscous normal-stress pressure are both retained at this scaling.

Linearizing about unit area gives $a_T+2a_Z-a_{TZZ}=0$. ThusWave crests propagate upward. The group velocity is $2(1-k^2)/(1+k^2{)}^2$: it is downward for $|k|>1$, so the phase direction is not the direction of every wave packet.

For a travelling wave $f(\zeta )$, integrating once and using its uniform far field givesDivide by $f^2$, multiply by $f^{\prime }/f$ and integrate again. A convenient potential, defined up to an additive constant, isAt the positive crest $f=\alpha f_0$, the derivative is zero. Subtracting the far-field potential yieldsThis is the solitary-wave amplitude-speed relation for the conduit equation. A nontrivial elevation wave has $\alpha >1$ and $c>2f_0$; as $\alpha \downarrow 1$, the speed tends to the long-wave speed $2f_0$. Positive area is required throughout the derivation.