Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 352 (Non-Newtonian Fluid Mechanics)

Let $z\in [0,h]$ measure distance through the thin Bingham plastic layer, with a stationary wall at $z=0$ and large-scale velocity $u_b$ at $z=h$. To leading lubrication order the shear stress ${\tau }_b$ is uniform across the layer. If $|{\tau }_b|\le {\tau }_c$, the material is unyielded and the no-slip wall makes the entire layer stationary. After yield,Integrating across the depth gives the Bingham sliding lawEquivalently, for nonzero sliding,

Write ${\tau }_d=\rho gH\sin \theta$. For unidirectional velocity $\mathbf{u}=(u(y,z),0,0)$, the nontrivial force balances areFor the power-law fluid,At the free surface $z=H$, $p=p_{\mathrm{atm}}$ and ${\tau }_{xz}=0$. At the bed $z=0$, continuity of tangential velocity and traction gives the local sliding conditionbecause the stated inequalities keep the downhill bed yielded. The velocity and horizontal shear traction are continuous across $y=0$, and the flow tends to uniform states as $y\to \pm \infty$.

When horizontal shear dominates, $u=U(y)+O(ϵ)$ andIntegrating downhill momentum through $0<z<H$ givesThe Bingham sliding law then yields the ODEDefineThenThe speed decreases monotonically across the material transition, with a smooth margin layer centered at $y=0$.

In either half-plane, let $U_{\infty }$ denote its far-field value. Multiplying the ODE by $U^{\prime }$ and integrating from the far field givesSince the profile decreases,Continuity of $U^{\prime }$ at $y=0$ gives

For $n=1$, let $\ell =A^{-1}=\sqrt{hHK/\mu }$ and $\Delta U=U_{-}-U_{+}$. ThenThe Newtonian disturbance therefore has exponential tails.

For $n<1$ and $y>0$, put $W=U-U_{+}$. Integration givesThe corresponding expression on $y<0$ follows by reflection about $(0,U_0)$. A shear thinning power-law material has algebraic rather than exponential margin tails. Measurements of how rapidly an ice-stream speed approaches its far-field values could therefore distinguish an approximately Newtonian rheology from power-law shear thinning and estimate $n$.

If $\delta$ is the cross-stream transition width and $\Delta U$ its speed change, dominant horizontal shear requiresBalancing the two terms in the reduced ODE gives the scaleThe approximation is strongest in the margin where $U^{\prime }$ is large, for a thin basal layer $h\ll H$ and parameters making this lateral scale short compared with the scale on which vertical shear changes the plug velocity. It fails sufficiently far into either uniform region because $U^{\prime }\to 0$ while the $O(ϵ)$ vertical shear needed to transmit the driving stress remains. It can also fail in narrow neighborhoods where neglected free-surface, bed-transition, or three-dimensional effects vary on the same scale.

For $n>1$, the exponent $m=2/(n+1)$ in $U^{\prime }=-A|U-U_{\infty }{|}^m$ is below one. Its integral reaches $U=U_{\infty }$ at a finite distance, after which the constant far-field solution can be attached. The ideal power law therefore predicts a compactly supported transition with finite-width edges rather than the algebraic tails found for $n<1$. Near those edges horizontal shear vanishes and the neglected vertical or regularizing physics becomes important, so the sharp termination should not be interpreted literally.

A plain material derivative of the conformation tensor is not an objective time derivative: an observer undergoing a time-dependent rigid rotation would infer a different constitutive response, and even rigid-body rotation could appear to change polymer deformation. The upper-convected derivative subtracts deformation and rotation carried by the velocity gradient and is frame indifferent.

The Oldroyd-B model is often inadequate because its Hookean dumbbells are infinitely extensible. It predicts constant shear viscosity rather than shear thinning, zero second normal-stress difference, and an unbounded extensional viscosity at a finite extension rate. Real polymer chains have finite extensibility and commonly exhibit shear thinning, bounded extensional stress, multiple relaxation times, and nonlinear solvent or concentration effects.

Let $a=\lambda \dot{\gamma }$. For simple shear, the steady conformation equation isIts nonzero components areSince $\mathbf{\tau }={\eta }_s\dot{\mathbf{\gamma }}+({\eta }_p/\lambda )(F\mathbf{C}-\mathbf{I})$,Thus the FENE-P model has

The trace isSubstituting it into $F=(L^2-3)/(L^2-T)$ givesFor $\beta \ll 1$, $F=1+\beta +O({\beta }^2)$. For $\beta \gg 1$, $F\sim {\beta }^{1/3}$. The effective shear viscosity is thereforeThe FENE-P curve decreases from ${\eta }_s+{\eta }_p$ toward the solvent plateau ${\eta }_s$, displaying shear thinning. Oldroyd-B has $F=1$ and remains at the constant value ${\eta }_s+{\eta }_p$.

For uniaxial extension,The diagonal steady conformation tensor iswhere physical solutions require $F>2a$. The extensional viscosity isThe implicit closure is

As $\dot{ϵ}\to 0$, $a\to 0$, $F\to 1$, andThe Trouton ratio is therefore three. This is also the small-rate Oldroyd-B result because finite extensibility is irrelevant while polymer deformation remains small.

For FENE-P at $a\gg 1$, write $F=2a+\delta$. The trace constraint approaches the finite maximum $T\to L^2$, so $C_{zz}=1/\delta \to L^2$ and $\delta \to L^{-2}$ to leading order. Henceup to corrections that vanish as $a^{-1}$. Its extensional viscosity rises from the Newtonian plateau and saturates at a finite-extensibility plateau.

For Oldroyd-B, $F=1$ andIt diverges as $\lambda \dot{ϵ}\uparrow 1/2$ and has no physical steady homogeneous branch beyond that point. This extensional catastrophe is removed by finite chain extensibility.

Define the viscous numberThe pressure relation givesUsing $\eta \hat{\mu }{\varphi }^2|\dot{\gamma }|/({\varphi }_m-\varphi {)}^2=\hat{\mu }{p}_s$, the shear stress at imposed pressure isThe material therefore behaves as a pressure-dependent Bingham plastic with yield stress $\hat{\mu }{p}_s$, while its steady concentration dilates as shear rate increases.

After a step in shear rate, the particles must rearrange and undergo shear-induced dilation or compaction. Because the sample is saturated, that volume change requires pore fluid to migrate through the packing, so pore pressure and effective particle pressure relax over a finite poro-viscous time. Measuring the transient stress can therefore constrain the permeability, and with an independently known permeability can constrain the suspension's compressibility or dilatancy law.

LetAxial momentum balance gives shear-stress magnitude $|{\tau }_{rz}|=Gr/2$. The wall stress must exceed the pressure-dependent yield stress, so motion requiresWhen this holds, define the plug radiusThe no-slip solid velocity isIn the sheared annulus,so

The volume flux carried with the particle skeleton isBy Darcy law, the fluid moves relative to the solids at $kG/\eta$. The total mixture-volume flux is thereforeThis is explicit because $\varphi (r)$ is given above. For example, ifthen

At fixed $G=\Delta P/L$, increasing $p_s$ increases the plug radius and lowers both the particle velocity and porosity. The total flux decreases from a nearly particle-free Poiseuille value toward a nonzero Darcy seepage flux through a jammed packing. Particles stop when $a=R$, namely at

Put $\delta =1-p_s/p_s^{\ast }=1-a/R$. As $\delta \downarrow 0$, almost the whole cross-section is a plug at concentration ${\varphi }_m$ and speed $G(R-a{)}^2/(4\eta )$. The particle flux is thereforeThe jammed seepage flux is approximatelyIt appreciably changes the total flux when $Q_p\lesssim Q_{\mathrm{seep}}^{\ast }$, orBecause $k\ll R^2$, seepage matters principally in a narrow pressure interval just below jamming.

At fixed $Q_T$, the pressure gradient is no longer fixed. As $p_s\to 0$, the constitutive law gives a dilute particle fraction and hence $Q_p\to 0$. Raising $p_s$ first adds particles and increases their flux. At high $p_s$, the packing becomes dense and nearly jammed; maintaining $Q_T$ then requires a large pressure gradient that sends most of the liquid through the packing by Darcy seepage, while $Q_p$ falls back toward zero. Thus $Q_p(p_s)$ has an interior maximum, and every particle flux below that maximum occurs at two solid pressures.

On the low-$p_s$ branch the suspension is dilute, shear is distributed broadly, particle motion carries much of the total volume, and the required $\Delta P$ is relatively small. On the high-$p_s$ branch the suspension contains a large nearly jammed plug, particles move slowly, seepage carries a substantial fraction of $Q_T$, and a much larger $\Delta P$ is required. Equal particle flux therefore does not imply equal concentration, flow structure, or pumping cost.