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FURTHER NUMERICAL TESTS OF ENSEMBLE EDDY

VISCOSITY METHODS

NAN JIANG AND WILLIAM LAYTON

Abstract. This supplementary material complements the report with similar

title.

1. Introduction

The problem of computing ensembles, uj , pj , of solutions of the Navier-Stokesequations (NSE):

uj,t + uj · ∇uj − ν4uj +∇pj = fj(x, t), in Ω, j = 1, ..., J(1.1)∇ · uj = 0, and uj(x, 0) = u0j (x), in Ω and uj = 0, on ∂Ω.

was studied in a paper with similar title. This supplement presents further tests.The method uses two new ensemble eddy viscosity (EEV) type turbulence modelswith turbulent viscosity parametrizations

EEV1: νT = µ14x|u′|, andEEV2: νT = µ2|u′|24t.

These are based on direct parameterization of the energy in the turbulent fluctu-ations, 12 |u

′|2 and a redefinition of the LES lengthscale from (the usual) l = 4xto

l = distance a fluctuating eddy travels in one time step = |u ′|4t .

1.1. Methods. The euclidean length of a vector and Frobenius norm of an arrayis | · |. The symmetric part of the velocity gradient tensor is denoted ∇s. Theensemble mean , fluctuation u′j , its magnitude |u′| and the induced kineticenergy density k′ are

mean: :=1

J

J∑j=1

uj , fluctuation: u′j := uj− ,

|u′|2 :=J∑

j=1

|u′j |2 and energy density: k′ =1

2|u′|2(x, t).

Date: June 2013.

2000 Mathematics Subject Classification. Primary 65M12; Secondary 65J08 .

Key words and phrases. NSE, ensemble calculation, UQ.The research herein was partially supported by NSF grant DMS 1216465 and Air Force grant

FA 9550-12-1-0191.

1

2 NAN JIANG AND WILLIAM LAYTON

Suppress the secondary spacial discretization and let superscripts denote the timestepnumber. Thus, for example, n, u′nj denote respectively approximations to

1

J

J∑j=1

uj(·, tn) and uj(·, tn)− n, tn := n4t.

We have the method: for j = 1, ..., J, ∇ · un+1j = 0, and

un+1j − unj∆t

+ n ·∇un+1j + (unj− n) · ∇unj(1.2)

+∇pn+1j − ν∆un+1j −∇ · (νT (l

n, k′n)∇sun+1j ) = fn+1j .

The ensemble eddy viscosity parameterization is the coefficient νT (·). Briefly, theKolmogorov-Prandtl relation gives

νT (·)=Const.l√k′

l = mixing length of fluctuations,

k′ = kinetic energy in fluctuations.

Often extensive (and optimistic) modelling steps are needed to generate repre-sentations of these two quantities, e.g., [34], [29]. Algorithm (1.2) allows directcalculation of both:

k′ =1

2|u′|2 and l =

{either 4x,or |u′|4t .

This gives

EEV1: νT = µ14x|u′|, andEEV2: νT = µ2|u′|24t.

2. Supplementary Material: Further Numerical Tests of EnsembleEddy Viscosity Methods

In our report, the following tests are presented.Test 1 was for flow between offset cylinders driven by a rotating body force

(Re = 800).Test 2 compared EEV1 and EEV2 for the same geometry atRe = 800, 1200, 2400

and constant timestep 4t = 0.025.Test 3 (results given in the supplementary materials) repeated these two tests

but reinitialized the perturbations at t = 1, 2, 3, · · · . The conclusions regardingstability were not altered by reinitialization.

Test 4 was an accuracy test with a smooth, known exact solution. In Test 4both EEV1 and EEV2 produced 2 significant digits of accuracy with 4x = 0.1, anacceptable result.

Test 5 is a flow in a channel with 2 outlets and a constriction from [4, 18, 23].Both EEV1 and EEV2 gave the correct general outlines of the flow (compared toothers published results) and differences in the smaller details of the flow.

FURTHER NUMERICAL TESTS OF ENSEMBLE EDDY VISCOSITY METHODS 3

We performed further tests of EEV1 and EEV2 and variants that, while inter-esting, confirmed or were consistent with the preliminary results presented in ourreport. We present some of these explorations in this supplementary section.

Test 1: Stability of no EEV vs. EEV2 for flow between offset circles.Recall that the domain is

Ω = {(x, y) : x2 + y2 ≤ r21 and (x− c1)2 + (y − c2)2 ≥ r22}

a disk with a smaller, off-center obstacle inside with no-slip boundary conditionson both circles. Let r1 = 1, r2 = 0.1, c = (c1, c2) = (

12 , 0). The flow is driven by:

f(x, y, t) = (−4y ∗ (1− x2 − y2), 4x ∗ (1− x2 − y2))T .

The mesh has n = 40 mesh points around the outer circle and m = 10 mesh pointsaround the immersed circle, and extended to Ω as a Delaunay mesh. We begin withRe = 800 and then increase Re.

Generation of the initial conditions. Perturbations of u0j , j = 1, 2, and u00

(with � ≡ 0, ‘no perturbation’), are generated by solving the steady Stokes problemon the same geometry with � = 10−3 and

f1(x, y, t) = f(x, y, t) + � ∗ (sin(3πx)sin(3πy), cos(3πx)cos(3πy))T ,f2(x, y, t) = f(x, y, t)− � ∗ (sin(3πx)sin(3πy), cos(3πx)cos(3πy))T ,

The Navier-Stokes equations are then solved with these initial conditions, givingu1, u2, uave = (u1 + u2)/2 and u0 (initial condition u

00 -‘no perturbation’).

Quantities plotted. The report gave plots of volume averaged statistics. Herewe supplement these with streamlines and contour plots of vorticity over 0 ≤ t ≤ 10. The timestep is adapted as in the report. First we plot total energy dissipationrates of EEV2 and noEV and power input rates of both.

Next plotted is EEV2 and noEV velocity vectors at ν = 1/800. Note that at thisReynolds number the flow already begins to have interesting features and that thenoEV velocities begin to exhibit radial oscillations.

This pattern is confirmed by the contour plots of vorticity. The noEV vorticitycontours reveal small scale oscillations clearly.

Next the Reynolds number was increased to Re = 1200 and 2400. We observedthat EEV2 remained stable with 4t = 0.05 at both Re = 1200 and 2400. WithNoEV, adapting the timestep ensured stability but forced a very small timestepand execution time so long that the method failed to reach the final time. TheEEV2 solutions are given below.

Test 2: Stability of EEV1 vs. EEV2. Test 1 was repeated comparingEEV1 and EEV2 for Re = 800 and constant timestep. We take ∆t = 0.025, Re =800, µ2 = 1, µ1∆x = 0.2. The streamlines very clearly show that by t = 10.0,the EEV1 approximation returns to the Stokes flow initial condition while EEV2continues its evolution.

The over diffusion in EEV1 vs. EEV2 is also clear from the kinetic energy plot.Test #3: Re-initialization: We repeat Test 2 but reinitialize the perturbation

at t = 1, 2, 3, · · ·, restarting with initial data being the average produced up to thatpoint and the same perturbations as at t = 0, as in Test #1. We use the sameconstant timestep ∆t = 0.025, Re = 800, µ2 = 1, µ1∆x = 0.2 . The plots show thatEEV1 results improved somewhat but still EEV1 over-diffuses while EEV2 doesnot.


Dissipation

0 1 2 3 4 5 6 7 8 9 100

1

2

3

4

5

6

7

8

9

epsilon1= 0.001, epsilon2= −0.001, Re=800

time

Dis

sip

atio

n

EV2−u

1

EV2−u2

NoEV−u1

NoEV−u2

Power Input

0 1 2 3 4 5 6 7 8 9 100

1

2

3

4

5

6

7

8

9

epsilon1= 0.001, epsilon2= −0.001, Re=800

time

Po

we

r In

pu

t

EV2−u

1

EV2−u2

NoEV−u1

NoEV−u2

Figure 1. EV2 vs. NoEV: Dissipation & Power Input, ν = 1/800

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t=0 t=0.5

t=5.0 t=10.0

Figure 2. EV2: Velocity, ν = 1/800

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t=0 t=0.540771

t=5.09414 t=10.0

Figure 3. NoEV: Velocity, ν = 1/800

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t=0 t=0.5IsoValue122.656233.046343.437453.827564.217674.608784.998895.3891005.781116.171226.56

IsoValue25.010947.520770.030592.5403115.05137.56160.07182.579205.089227.599250.109

t=5.0 t=10.0IsoValue24.412546.383768.354990.3261112.297134.269156.24178.211200.182222.153244.125

IsoValue27.233451.743576.2535100.764125.274149.784174.294198.804223.314247.824272.334

Figure 4. EV2: Contours of Vorticity, ν = 1/800

[34] P. Sagaut, Large eddy simulation for Incompressible flows, Springer, Berlin, 2001.

[35] D.J. Stensrud, Parameterization schemes: Keys to numerical weather prediction models,

Cambridge U. Press, 2009.[36] G.I. Taylor, On decay of vortices in a viscous fluid, Phil. Mag., 46 (1923), 671-674.

[37] Z. Toth and E. Kalney, Ensemble forecasting at NMC: The generation of perturbations,

Bull. Amer. Meteor. Soc., 74 (1993), 2317-2330.

(All authors) Dept. of Mathematics, Univ. of Pittsburgh, Pittsburgh, PA 15260, USAE-mail address: [email protected], [email protected]

URL: www.math.pitt.edu/~wjl


t=0 t=0.540771IsoValue-1226.56-1103.9-981.248-858.592-735.936-613.28-490.624-367.968-245.312-245.312-122.6560122.656245.312367.968490.624613.28735.936858.592981.248

IsoValue-189.511-170.56-151.609-132.658-113.707-94.7556-75.8045-56.8534-37.9022-37.9022-18.9511018.951137.902256.853475.804594.7556113.707132.658151.609

t=5.09414 t=10.0IsoValue-218.375-196.537-174.7-152.862-131.025-109.187-87.3498-65.5124-43.6749-43.6749-21.8375021.837543.674965.512487.3498109.187131.025152.862174.7

IsoValue-144.006-129.605-115.204-100.804-86.4034-72.0028-57.6022-43.2017-28.8011-28.8011-14.4006014.400628.801143.201757.602272.002886.4034100.804115.204

Figure 5. NoEV: Contours of Vorticity, ν = 1/800

t=0 t=0.45

t=5.2 t=10.0

Figure 6. EEV2: Velocity, ν = 1/1200


t=0 t=0.45IsoValue183.984349.57515.155680.741846.3261011.911177.51343.081508.671674.251839.84

IsoValue10.886120.683630.481140.278650.076159.873669.671179.468689.266199.0636108.861

t=5.2 t=10.0IsoValue20.454138.862957.271675.680394.0891112.498130.907149.315167.724186.133204.541

IsoValue22.448942.65362.85783.0611103.265123.469143.673163.877184.081204.285224.489

Figure 7. EEV2: Contours of Vorticity, ν = 1/1200

t=0 t=0.523633

t=5.01582 t=10.0

Figure 8. EEV2: Velocity, ν = 1/2400


t=0 t=0.523633IsoValue367.968699.1391030.311361.481692.652023.822354.992686.173017.343348.513679.68

IsoValue13.969326.541739.114151.686464.258876.831289.4036101.976114.548127.121139.693

t=5.01582 t=10.0IsoValue11.258621.391431.524141.656851.789661.922372.055182.187892.3206102.453112.586

IsoValue17.067132.427647.78863.148478.508893.8692109.23124.59139.95155.311170.671

Figure 9. EEV2: Contours of Vorticity, ν = 1/2400

t=0 t=0.5

t=5.0 t=10.0

Figure 10. EV1 gives Stokes flow: Velocity, ν = 1/800


t=0 t=0.5

t=5.0 t=10.0

Figure 11. EV2: Velocity, ν = 1/800

t=0 t=0.5IsoValue122.656233.046343.437453.827564.217674.608784.998895.3891005.781116.171226.56

IsoValue8.7285916.584324.4432.295840.151548.007255.86363.718771.574479.430287.2859

t=5.0 t=10.0IsoValue0.7643431.452252.140162.828073.515984.203894.891795.57976.267616.955527.64343

IsoValue0.7643431.452252.140162.828073.515984.203894.891795.57976.267616.955527.64343

Figure 12. EV1 gives Stokes flow: Contours of Vorticity, ν = 1/800


t=0 t=0.5IsoValue122.656233.046343.437453.827564.217674.608784.998895.3891005.781116.171226.56

IsoValue15.49529.440543.38657.331671.277185.222699.1681113.114127.059141.005154.95

t=5.0 t=10.0IsoValue25.540748.527371.513994.5005117.487140.474163.46186.447209.434232.42255.407

IsoValue24.159345.902667.64689.3893111.133132.876154.619176.363198.106219.849241.593

Figure 13. EV2: Contours of Vorticity, ν = 1/800

0 1 2 3 4 5 6 7 8 9 100

5

10

15epsilon1= 0.001, epsilon2= −0.001, dt=0.025, Re=800

time

Ener

gy

EV1

EV2

Figure 14. Energy: EV1 Vs EV2, ∆t = 0.025, ν = 1/800


EV1

0 1 2 3 4 5 6 7 8 9 100

0.5

1

1.5

2

2.5

3

3.5

4epsilon1= 0.001, epsilon2= −0.001, dt= 0.025, Re= 800

time

An

gula

r M

om

entu

m

u1

u2

Average

no perturbation

EV2

0 1 2 3 4 5 6 7 8 9 100

1

2

3

4

5

6


time

Angula

r M

om

entu

m

u1

u2

Average

no perturbation

Figure 15. Reinitialize: Angular Momentum, ν = 1/800


EV1

0 1 2 3 4 5 6 7 8 9 100

1

2

3

4

5


time

En

erg

y

u1

u2

Average

no perturbation

EV2

0 1 2 3 4 5 6 7 8 9 100

1

2

3

4

5

6

7


time

En

erg

y

u1

u2

Average

no perturbation

Figure 16. Reinitialize: Energy, ν = 1/800


EV1

0 1 2 3 4 5 6 7 8 9 100

200

400

600

800

1000

1200

1400

1600


time

En

str

op

hy

u1

u2

Average

no perturbation

EV2

0 1 2 3 4 5 6 7 8 9 100

200

400

600

800

1000

1200

1400

1600

1800


time

En

str

op

hy

u1

u2

Average

no perturbation

Figure 17. Reinitialize: Enstrophy, ν = 1/800


t=0 t=1

t=5.0 t=10.0

Figure 18. Reinitialize EV1: Stokes flow velocity, ν = 1/800

t=0 t=1

t=5.0 t=10.0

Figure 19. Reinitialize: EV2: Velocity, ν = 1/800


t=0 t=1IsoValue122.656233.046343.437453.827564.217674.608784.998895.3891005.781116.171226.56

IsoValue1.541392.928644.31595.703157.09048.477659.8649111.252212.639414.026715.4139

t=5.0 t=10.0IsoValue0.7643431.452252.140162.828073.515984.203894.891795.57976.267616.955527.64343

IsoValue0.7643431.452252.140162.828073.515984.203894.891795.57976.267616.955527.64343

Figure 20. Reinitialize: EV1: Contours of Vorticity, ν = 1/800

t=0 t=1IsoValue122.656233.046343.437453.827564.217674.608784.998895.3891005.781116.171226.56

IsoValue15.931630.2744.608458.946973.285387.6237101.962116.301130.639144.977159.316

t=5.0 t=10.0IsoValue19.593237.22754.860972.494790.1286107.762125.396143.03160.664178.298195.932

IsoValue26.322950.013473.70497.3946121.085144.776168.466192.157215.847239.538263.229

Figure 21. Reinitialize: EV2: Contours of Vorticity, ν = 1/800


0 1 2 3 4 5 6 7 8 9 100

1

2

3

4

5

6

7

8epsilon1= 0.001, epsilon2= −0.001, dt=0.025, Re=800

time

Energ

y

EV1

EV2

Figure 22. Reinitialize: Energy: EV1 Vs EV2, ∆t = 0.025, ν = 1/800

Introduction - University of Pittsburgh · events from PDF ensemble, via REA method, on Senegal...

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