Anderson acceleration for time-dependent problems...Royal Military Academy Belgium Problem setting...

RobHaelterman

Dept. Math.Royal Military

Academy

Belgium

Problemsetting

Coupledproblem

Andersonacceleration

Results

ImprovementsI

ImprovementsII

ImprovementsIII

Questions

Anderson acceleration for time-dependent problems

RobHaelterman


Academy

Belgium

Problemsetting

Coupledproblem


Results

ImprovementsI

ImprovementsII

ImprovementsIII

Questions

PROBLEM SETTING I: 1D FLOW

FIGURE : One-dimensional flow in a tube.

incompressibleinviscidgravity neglectedconservative form

RobHaelterman


Academy

Belgium

Problemsetting

Coupledproblem


Results

ImprovementsI

ImprovementsII

ImprovementsIII

Questions


(∂g

∂t+

)∂gu

∂x= 0 (1a)

∂gu

∂t+∂gu2

∂x+

1ρ

(∂gp̃

∂x− p̃

∂g

∂x

)= 0 (1b)

Inlet BC: uin(t) = uref +uref10 sin2 (πt)

Constant pressure at outlet.

RobHaelterman


Academy

Belgium

Problemsetting

Coupledproblem


Results

ImprovementsI

ImprovementsII

ImprovementsIII

Questions


Discrete variables: g→ g, u→ u, p̃→ p̃Equidistant meshCentral discretization for all space derivatives in flowequations except convective term that uses upwinddiscretisation.Backward (implicit) Euler time-discretisation

→ Root finding problem: F (p̃t+1,ut+1) = 0→ Anderson

RobHaelterman


Academy

Belgium

Problemsetting

Coupledproblem


Results

ImprovementsI

ImprovementsII

ImprovementsIII

Questions


RESULT

Solving with Anderson didn’t go well at all, compared tofsolve.

RobHaelterman


Academy

Belgium

Problemsetting

Coupledproblem


Results

ImprovementsI

ImprovementsII

ImprovementsIII

Questions

PROBLEM SETTING II: FLEXIBLE TUBE

Flexible tube with Hookean law g = g(p)Inertia neglectedp = p̃/ρ = kinematic pressure.

∂p

∂t+ u

∂p

∂x+ c2 ∂u

∂x= 0 (2a)

∂gu

∂t+∂gu2

∂x+∂gp

∂x− p

∂g

∂x= 0, (2b)

Non-reflecting BC at outlet:∂uout∂t

=1c∂pout∂t

. (3)

where the wave speed c is defined by

c2 =gdgdp

. (4)

RobHaelterman


Academy

Belgium

Problemsetting

Coupledproblem


Results

ImprovementsI

ImprovementsII

ImprovementsIII

Questions

PROBLEM SETTING II: FLEXIBLE TUBE

g = go

(po − 2c2

mk

p− 2c2mk

)2

(5)

Moens Korteweg Wave-speed: c2mk = Eh

2ρro.

Young modulus EReference radius r

Thickness of wall hNon-dimensionalize (and discretize):

κ =

√Eh

2ρro−Po

2Uo

(stiffness) and τ = Uo∆tcoL (time-step).

Fourier analysis: smaller κ and τn is more unstable forfixed point iteration.

RobHaelterman


Academy

Belgium

Problemsetting

Coupledproblem


Results

ImprovementsI

ImprovementsII

ImprovementsIII

Questions

COUPLED PROBLEM

THE COUPLED EQUATIONS{F (pt+1,gt+1;pt ,gt) = 0S(pt+1,gt+1;pt ,gt) = 0

(6)

{F (gt+1) = pt+1S(pt+1) = gt+1

(7)

{F (g) = pS(p) = g

(8)

RobHaelterman


Academy

Belgium

Problemsetting

Coupledproblem


Results

ImprovementsI

ImprovementsII

ImprovementsIII

Questions

COUPLED PROBLEM


(6)

{F (gt+1) = pt+1S(pt+1) = gt+1

(7)

{F (g) = pS(p) = g

(8)

RobHaelterman


Academy

Belgium

Problemsetting

Coupledproblem


Results

ImprovementsI

ImprovementsII

ImprovementsIII

Questions

COUPLED PROBLEM


(6)

{F (gt+1) = pt+1S(pt+1) = gt+1

(7)

{F (g) = pS(p) = g

(8)

RobHaelterman


Academy

Belgium

Problemsetting

Coupledproblem


Results

ImprovementsI

ImprovementsII

ImprovementsIII

Questions

COUPLED PROBLEM

HYPOTHESES

1 Solving each sub-problem is computationally heavy.2 No access to Jacobian of each sub-problem.3 Only access to F (g) and S(p).

RobHaelterman


Academy

Belgium

Problemsetting

Coupledproblem


Results

ImprovementsI

ImprovementsII

ImprovementsIII

Questions

COUPLED PROBLEM

COUPLED PROBLEM {F (g) = pS(p) = g

ROOT FINDING PROBLEM

F (S(p))︸︷︷︸H(p)

−p

︸︷︷︸K (p)

= 0 (9)

RobHaelterman


Academy

Belgium

Problemsetting

Coupledproblem


Results

ImprovementsI

ImprovementsII

ImprovementsIII

Questions

ANDERSON ACCELERATION


F (S(p))︸︷︷︸H(p)

−p

︸︷︷︸K (p)

= 0

AA “TYPE II” FOR K (p) = 01 Startup:

1 Take an initial value po.2 Compute p1 = (1− ω)po + ωH(po).3 Set s = 1.

2 Loop until sufficiently converged:1 Compute K (ps).2 Construct the approximate inverse Jacobian M̂ ′

s.3 Quasi-Newton step: ps+1 = ps − M̂ ′

sK (ps).4 Set s = s + 1.

RobHaelterman


Academy

Belgium

Problemsetting

Coupledproblem


Results

ImprovementsI

ImprovementsII

ImprovementsIII

Questions

ANDERSON ACCELERATION


F (S(p))︸︷︷︸H(p)

−p

︸︷︷︸K (p)

= 0

AA: JACOBIAN

M̂ ′o = −1ω

I (de facto) , (10a)

M̂ ′s = Ws((Vs)T Vs)

−1(Vs)T − I for s > 0, (10b)

where δKi = K (pi+1)− K (pi ) (i = 0, . . . , s − 1),Vs = [δKs−1 δKs−2 . . . δK0]δHi = H(pi+1)− H(pi ) (i = 0, . . . , s − 1),Ws = [δHs−1 δHs−2 . . . δH0]

(11)

RobHaelterman


Academy

Belgium

Problemsetting

Coupledproblem


Results

ImprovementsI

ImprovementsII

ImprovementsIII

Questions

RESULTS

n = 1000, 10 time-steps.

κ τ Broyden Type II Anderson Type II103 10−4 9 (8.7) 8 (6.9)102 10−4 div 19 (15.8)10 10−3 div 22 (16.5)10 10−4 div 58 (51.8)

RobHaelterman


Academy

Belgium

Problemsetting

Coupledproblem


Results

ImprovementsI

ImprovementsII

ImprovementsIII

Questions

IMPROVEMENTS: MGM

RANK s UPDATE

M̂ ′s+1 = −I + Ws((Vs)T Vs)

−1(Vs)T︸︷︷︸

Rank s

RANK 1 UPDATE

M̂ ′s+1 = M̂ ′s +(δps − M̂ ′sδKs)

〈ds, δKs〉dT

s︸︷︷︸vT

= M̂ ′s + uvT

ds = (I − Ls(Ls)T )δKs ⊥ δKj (j < s)

where columns of Ls form orthonormal base forR(Vs) = span{δKs−1, δKs−2, . . . , δKo}.

v⊥δKs−1 ⇒ well-studied by Martinez et al.

RobHaelterman


Academy

Belgium

Problemsetting

Coupledproblem


Results

ImprovementsI

ImprovementsII

ImprovementsIII

Questions

IMPROVEMENTS: MGM

RANK s UPDATE

M̂ ′s+1 = −I + Ws ((Vs)T Vs)

−1(Vs)T

M̂ ′s+1 = −I + (Ws − Vs − (−I)Vs)((Vs)T Vs)

−1(Vs)T

M̂ ′s+1 = A + (Ws − Vs − A Vs)((Vs)T Vs)

−1(Vs)T

will respect multi-secant condition

M̂ ′s+1Vs = [δps−1 δps−2 . . . δp0] = Ws − Vs

Take A as Jacobian from coarser grid.

RobHaelterman


Academy

Belgium

Problemsetting

Coupledproblem


Results

ImprovementsI

ImprovementsII

ImprovementsIII

Questions

IMPROVEMENTS: MGM

nc = nf/3→

κ τ Anderson II Anderson II MGM103 10−4 8 (6.9) 5(4.7)F + 8(7.2)C

≈ 7.67 (7.1)102 10−4 19 (15.8) 7(8.1)F + 20(20.8)C

≈ 13.67 (15.03)10 10−3 22 (16.5) 8(7.9)F + 21(20.5)C

≈ 15 (14.73)10 10−4 58 (51.8) 36(35.7)F + 54(52.0)C

≈ 54 (53.03)

nc < nf ⇒ cheaper but also more unstable: far highernumber needed on coarse grid.

RobHaelterman


Academy

Belgium

Problemsetting

Coupledproblem


Results

ImprovementsI

ImprovementsII

ImprovementsIII

Questions

IMPROVEMENTS: JACOBIAN FROM PREVIOUS

TIME-STEP

RANK s UPDATE

M̂ ′s+1 = A + (Ws − Vs − A Vs)((Vs)T Vs)

−1(Vs)T

Take A as Jacobian from previous time-step.

RobHaelterman


Academy

Belgium

Problemsetting

Coupledproblem


Results

ImprovementsI

ImprovementsII

ImprovementsIII

Questions


TIME-STEP

κ τ Anderson II + MGM + Jac103 10−4 8 (6.9) 7.67 (7.1) 8 (3.6)102 10−4 19 (15.8) 13.67 (15.03) 19 (5.9)10 10−3 22 (16.5) 15 (14.73) 22 (7.0)10 10−4 58 (51.8) 54 (53.03) 58 (21.5)

Take MGM Jacobian for first time-step only ?

RobHaelterman


Academy

Belgium

Problemsetting

Coupledproblem


Results

ImprovementsI

ImprovementsII

ImprovementsIII

Questions

IMPROVEMENTS: INPUT-OUTPUT VECTORS

RANK s UPDATE

M̂ ′s+1 =

−I + [Ws|Wprev ](([Vs|Vprev ])T [Vs|Vprev ])

−1([Vs|Vprev ])T

Will still respect secant condition(s) at current time-step, butalso those of previous time-steps.

RobHaelterman


Academy

Belgium

Problemsetting

Coupledproblem


Results

ImprovementsI

ImprovementsII

ImprovementsIII

Questions


TIME-STEP

κ τ And. II + Jac + I/O (5) + I/O (10)103 10−4 8 (6.9) 8 (3.6) 8 (4.1) 8 (4.0)102 10−4 19 (15.8) 19 (5.9) 19 (6.9) 19 (5.3)10 10−3 22 (16.5) 22 (7.0) 22 (8.2) 22 (6.5)10 10−4 58 (51.8) 58 (21.5) 58 (21.4) 58 (16.0)

How many extra I/O ?

RobHaelterman


Academy

Belgium

Problemsetting

Coupledproblem


Results

ImprovementsI

ImprovementsII

ImprovementsIII

Questions

OTHER APPLICATIONS

FIGURE : Three-dimensional flow in a tube.

RobHaelterman


Academy

Belgium

Problemsetting

Coupledproblem


Results

ImprovementsI

ImprovementsII

ImprovementsIII

Questions

OTHER APPLICATIONS

FIGURE : Flapping tail.

RobHaelterman


Academy

Belgium

Problemsetting

Coupledproblem


Results

ImprovementsI

ImprovementsII

ImprovementsIII

Questions

Questions ?

RobHaelterman


Academy

Belgium

Problemsetting

Coupledproblem


Results

ImprovementsI

ImprovementsII

ImprovementsIII

Questions

REFERENCES

Equivalence of QN-LS and BQN-LS for affine problemsR Haelterman, B Lauwens, H Bruyninckx, J PetitJournal of Computational and Applied Mathematics 278, 48-51On the similarities between the quasi-Newton least squares method andGMResR Haelterman, et al.Journal of Computational and Applied Mathematics 273, 25-28On the non-singularity of the quasi-Newton-least squares methodR Haelterman, J Petit, B Lauwens, H Bruyninckx, J VierendeelsJournal of Computational and Applied Mathematics 257, 129-131Bubble simulations with an interface tracking technique based on apartitioned fluid-structure interaction algorithmJ Degroote, P Bruggeman, R Haelterman, J VierendeelsJournal of computational and applied mathematics 234 (7), 2303-2310Performance of partitioned procedures in fluid-structure interactionJ Degroote, R Haelterman, S Annerel, P Bruggeman, J VierendeelsComputers & structures 88 (7), 446-457On the similarities between the quasi-Newton inverse least squares methodand GMResR Haelterman, J Degroote, D Van Heule, J VierendeelsSIAM Journal on Numerical Analysis 47 (6), 4660-4679

RobHaelterman


Academy

Belgium

Problemsetting

Coupledproblem


Results

ImprovementsI

ImprovementsII

ImprovementsIII

Questions

REFERENCES

The quasi-newton least squares method: A new and fast secant methodanalyzed for linear systemsR Haelterman, J Degroote, D Van Heule, J VierendeelsSIAM Journal on numerical analysis 47 (3), 2347-2368

Coupling techniques for partitioned fluid-structure interaction simulationswith black-box solversJ Degroote, R Haelterman, S Annerel, J Vierendeels10th MpCCI User Forum, 82-91

Stability of a coupling technique for partitioned solvers in FSI applicationsJ Degroote, P Bruggeman, R Haelterman, J VierendeelsComputers & Structures 86 (23), 2224-2234

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Anderson acceleration for time-dependent problems...Royal Military Academy Belgium Problem setting...

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