FR26 · Elastic instability

Elastic Instability: Symmetry Breaking in a Four-Roll Mill

How can a slow polymer flow with no inertia at all lose its symmetry and start to oscillate? Through elastic stress alone. At the central stagnation point the flow stretches polymers into a thin, strongly stressed strand, and that stress pushes back on the flow. Above a Weissenberg number of about 5 a tiny tilt of the strand grows instead of fading, so the four rolls lose their symmetry. Near 10 the flow keeps oscillating and mixes fluid between the rolls, still with no inertia at all.

Four rolls stir a polymer solution with no inertia. Past a critical Weissenberg number, stretched polymers at the center break symmetry and the flow oscillates.

Stress asymmetry $49.4$

Flow at the probe $-0.0282$

Mixing index of the tracers $0.646$

Smallest stretch eigenvalue $0.326$

Turn up the polymer memory

Polymer relaxation time against the flow time

10

\mathrm{Wi}=\lambda/\tau_f

Artificial stress diffusion

\nu_p=0.001

The smallest stretch eigenvalue is below zero, so this grid does not resolve the stress strand. A finer grid is under Grid and time step.

Drag the crosshair to move the velocity probe. Above a Weissenberg number of about 4.9 the bright central strand tips over and the flow oscillates.

Grid and time step

Grid points per sideHalve the time step

Changing either one resamples the current stress onto the new grid and continues the run. A 128 by 128 grid costs about five times as much per step and runs more slowly.

Model, units and sources

$-\Delta\mathbf u+\nabla p=\beta\,\nabla\cdot\mathbf S+\mathbf f,\quad \overset{\nabla}{\mathbf S}=-\frac{\mathbf S-\mathbf I}{\mathrm{Wi}}+\nu_p\Delta\mathbf S,\quad \beta\,\mathrm{Wi}=\tfrac12$

The box is periodic, $[-\pi,\pi]^2$ in the model length unit. Time is in flow times: the inverse strain rate at the center of the solvent-only flow. The solver is Fourier pseudo-spectral with the Hou–Li filter and an integrating-factor third-order Runge–Kutta step. The dye is 4096 passive particles moved with the computed velocity; they do not act on the flow.

Course chapter: how the stagnation point loses symmetry

Thomases & Shelley (2007) · Thomases & Shelley (2009) · Thomases, Shelley & Thiffeault (2011)

Model assumptions

The model is the two-dimensional periodic Stokes–Oldroyd-B four-roll mill of Thomases and Shelley: a fixed body force drives four counter-rotating cells around an extensional stagnation point at the center, inertia is absent, and the polymer-to-solvent viscosity ratio is one half. The polymer stress carries a small artificial diffusion, as in the published runs; it smooths the thin stress sheets and changes long-time dynamics. The Lab runs on a 64 by 64 grid, coarser than the published 256 by 256 runs. Its symmetry-breaking threshold and its Weissenberg 10 trajectory agree with our 128 by 128 runs, but the smallest stretch eigenvalue dips below zero while the strand forms, so that stage is under-resolved. The curved-channel and swirling-flow experiments on elastic turbulence are not modelled. Every field is a simulation; no measured data are shown.

Flow at the probe

Distance from symmetry

Where quadrant I’s dye came from

More numbers and how they are computed

Model time: $0$
Polymer coupling: $0$
Artificial stress diffusion: $0$
Grid and time step: $0$
Central stagnation point: $0$
Largest stretch, trace of the conformation tensor: $0$
Smallest stretch eigenvalue since this run started: $0$
Relative residual of the Stokes energy identity: $0$
Stress spectrum in the outer third of wavenumbers: $0$
Advective Courant number: $0$
Dye released at: $0$
Model time per wall-clock second: $0$

The asymmetry is the root-mean-square difference between the stress and its average over the symmetries of the roll force; it is zero for a symmetric state. The mixing index is one for dye spread evenly over the four quadrants and zero for dye still in its starting quadrant. A negative smallest eigenvalue means the stretch field is not physical there: the grid is too coarse for the stress strand. Every curve is computed by this simulation; none is measured data or a published curve.

Physics tutorial

Symmetry lost without inertia

BackgroundFour counter-rotating rolls stir a liquid so slowly that inertia plays no part. A plain liquid then follows the force exactly: the flow has the same symmetry as the stirring and never changes. Add long polymers and the same slow stirring can produce a flow that leans to one side and keeps oscillating.

Why it mattersAt the center of the four rolls is a stagnation point where the flow stretches polymers without limit. The stretched polymers form a thin strand of stress that pushes back on the flow. This is the simplest setting in which elastic stress alone breaks a symmetry, the first step toward elastic turbulence.

Start with the essentials

Focus question: How can a slow polymer flow with no inertia at all lose its symmetry and start to oscillate?
One-sentence intuition: The Weissenberg number $\mathrm{Wi}$ compares how long polymers remember a stretch with the flow time, in which the plain flow stretches the fluid at the center by a factor e. Once $\mathrm{Wi}$ passes about 4.9, the symmetric stress strand is unstable: a small tilt of the strand bends the flow so that it stretches the strand further in the same direction.

Core mathematical model

Slow flow driven by the rolls and the polymers

-\Delta\mathbf u+\nabla p=\beta\,\nabla\cdot\mathbf S+\mathbf f,\qquad \nabla\cdot\mathbf u=0

Stokes flow: viscosity balances the fixed roll force $\mathbf f$ and the polymer stress. With $\mathbf S=\mathbf I$ the flow is four symmetric cells around a central stagnation point.

How the polymers stretch and relax

\partial_t\mathbf S+\mathbf u\cdot\nabla\mathbf S-\nabla\mathbf u\,\mathbf S-\mathbf S\,\nabla\mathbf u^{\mathsf T}=-\frac{\mathbf S-\mathbf I}{\mathrm{Wi}}+\nu_p\Delta\mathbf S

The Oldroyd-B conformation tensor $\mathbf S$ is carried and stretched by the flow and relaxes back to $\mathbf I$ . The small diffusion $\nu_p$ is a numerical device the published runs also use; it is not a property of real polymers.

Stretching at the stagnation point

\frac{d S_{xx}}{dt}=2\dot\varepsilon\, S_{xx}-\frac{S_{xx}-1}{\mathrm{Wi}},\qquad \dot\varepsilon\,\mathrm{Wi}>\tfrac12\ \Rightarrow\ S_{xx}\ \text{grows exponentially}

$\dot\varepsilon$ is the strain rate at the center in flow-time units: 1 for the solvent alone, and lower once the polymers resist, $2/3$ in the limit of small $\mathrm{Wi}$ . Past $\dot\varepsilon\,\mathrm{Wi}=1/2$ stretching outruns relaxation; without diffusion the central stress then grows without bound, from $\mathrm{Wi}\approx0.9$ in Thomases and Shelley (2007). Here diffusion and the feedback on the flow keep the strand finite.

Measuring the lost symmetry

A=\left\langle\left\|\mathbf Q-\mathcal P\mathbf Q\right\|_F^2\right\rangle^{1/2},\qquad \mathbf Q=\mathbf S-\mathbf I

$\mathcal P$ averages the stress over the symmetries of the roll force. $A$ is zero for a symmetric state and grows exponentially once the symmetric strand is unstable.

Common difficulties

Instability needs inertia?

Typical misconceptionA flow this slow cannot become unstable; instabilities and turbulence need a large Reynolds number.

Better mental modelHere the Reynolds number is zero by construction. The nonlinearity that drives the instability is in the polymer equation: the flow stretches the polymers and the stretched polymers bend the flow.

The numbers are converged

Typical misconceptionThe threshold and periods shown here are the exact answer of the model.

Better mental modelThe threshold 4.86 agrees between our 64 by 64 and 128 by 128 grids and with the published 4.8 to 4.9. At the default grid the smallest eigenvalue of $\mathbf S$ briefly turns negative while the strand forms, so that stage is under-resolved. Changing $\nu_p$ changes the long-time behavior.

Run the experiment

01
Watch the oscillation
Start with Oscillation that mixes and follow the probe chart and the colored dye.
What to observe: The bright stress strand sways off center, the probe velocity swings with a long and a short period, and dye from quadrants II and IV keeps entering quadrant I, while little arrives from the opposite quadrant III.
02
Stay below the threshold
Choose Symmetric stress strand, then drag the Weissenberg slider from 4 up past 5.
What to observe: At 4 the strand stays straight and the asymmetry decays. Above about 4.9 it starts to rise instead; at 6 it climbs as a straight line on the log axis and the strand tips over after a few hundred flow times.
03
Smear the stress
Choose Too much stress diffusion and compare it with the reference oscillation.
What to observe: Ten times the published diffusion delays the symmetry breaking to about 590 flow times and leaves only a weak slow wobble.