Anderson acceleration for time-dependent problems setting Coupled - - PowerPoint PPT Presentation

Rob Haelterman

• Dept. Math.

Royal Military Academy Belgium Problem setting Coupled problem Anderson acceleration Results Improvements I Improvements II Improvements III Questions

Anderson acceleration for time-dependent problems

Rob Haelterman

• Dept. Math.

Royal Military Academy Belgium Problem setting Coupled problem Anderson acceleration Results Improvements I Improvements II Improvements III Questions

PROBLEM SETTING I: 1D FLOW

FIGURE : One-dimensional flow in a tube.

incompressible inviscid gravity neglected conservative form

Rob Haelterman

• Dept. Math.

Royal Military Academy Belgium Problem setting Coupled problem Anderson acceleration Results Improvements I Improvements II Improvements III Questions

PROBLEM SETTING I: 1D FLOW

∂g ∂t + ∂gu ∂x = 0 (1a) ∂gu ∂t + ∂gu2 ∂x + 1 ρ ∂g˜ p ∂x − ˜ p ∂g ∂x

• = 0

(1b) Inlet BC: uin(t) = uref + uref

10 sin2 (πt)

Constant pressure at outlet.

Rob Haelterman

• Dept. Math.

Royal Military Academy Belgium Problem setting Coupled problem Anderson acceleration Results Improvements I Improvements II Improvements III Questions

PROBLEM SETTING I: 1D FLOW

Discrete variables: g → g, u → u, ˜ p → ˜ p Equidistant mesh Central discretization for all space derivatives in flow equations except convective term that uses upwind discretisation. Backward (implicit) Euler time-discretisation → Root finding problem: F(˜ pt+1, ut+1) = 0 → Anderson

Rob Haelterman

• Dept. Math.

Royal Military Academy Belgium Problem setting Coupled problem Anderson acceleration Results Improvements I Improvements II Improvements III Questions

PROBLEM SETTING I: 1D FLOW

RESULT Solving with Anderson didn’t go well at all, compared to fsolve.

Rob Haelterman

• Dept. Math.

Royal Military Academy Belgium Problem setting Coupled problem Anderson acceleration Results Improvements I Improvements II Improvements III Questions

PROBLEM SETTING II: FLEXIBLE TUBE

Flexible tube with Hookean law g = g(p) Inertia neglected p = ˜ p/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 = 1 c ∂pout ∂t . (3) where the wave speed c is defined by c2 = g

dg dp

. (4)

Rob Haelterman

• Dept. Math.

Royal Military Academy Belgium Problem setting Coupled problem Anderson acceleration Results Improvements I Improvements II Improvements III Questions

PROBLEM SETTING II: FLEXIBLE TUBE

g = go

• po − 2c2

mk

p − 2c2

mk

2 (5) Moens Korteweg Wave-speed: c2

mk = Eh 2ρro .

Young modulus E Reference radius r Thickness of wall h Non-dimensionalize (and discretize): κ =

• Eh

2ρro − Po 2

Uo

(stiffness) and τ = Uo∆t

coL (time-step).

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

Rob Haelterman

• Dept. Math.

Royal Military Academy Belgium Problem setting Coupled problem Anderson acceleration Results Improvements I Improvements II Improvements III Questions

COUPLED PROBLEM

THE COUPLED EQUATIONS F(pt+1, gt+1; pt, gt) = 0 S(pt+1, gt+1; pt, gt) = 0 (6) F(gt+1) = pt+1 S(pt+1) = gt+1 (7) F(g) = p S(p) = g (8)

Rob Haelterman

• Dept. Math.

Royal Military Academy Belgium Problem setting Coupled problem Anderson acceleration Results Improvements I Improvements II Improvements III Questions

COUPLED PROBLEM

THE COUPLED EQUATIONS F(pt+1, gt+1; pt, gt) = 0 S(pt+1, gt+1; pt, gt) = 0 (6) F(gt+1) = pt+1 S(pt+1) = gt+1 (7) F(g) = p S(p) = g (8)

Rob Haelterman

• Dept. Math.

Royal Military Academy Belgium Problem setting Coupled problem Anderson acceleration Results Improvements I Improvements II Improvements III Questions

COUPLED PROBLEM

THE COUPLED EQUATIONS F(pt+1, gt+1; pt, gt) = 0 S(pt+1, gt+1; pt, gt) = 0 (6) F(gt+1) = pt+1 S(pt+1) = gt+1 (7) F(g) = p S(p) = g (8)

Rob Haelterman

• Dept. Math.

Royal Military Academy Belgium Problem setting Coupled problem Anderson acceleration Results Improvements I Improvements II Improvements III 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).

Rob Haelterman

• Dept. Math.

Royal Military Academy Belgium Problem setting Coupled problem Anderson acceleration Results Improvements I Improvements II Improvements III Questions

COUPLED PROBLEM

COUPLED PROBLEM F(g) = p S(p) = g ROOT FINDING PROBLEM F(S(p))

• H(p)

−p

• K(p)

= 0 (9)

Rob Haelterman

• Dept. Math.

Royal Military Academy Belgium Problem setting Coupled problem Anderson acceleration Results Improvements I Improvements II Improvements III Questions

ANDERSON ACCELERATION

ROOT FINDING PROBLEM F(S(p))

• H(p)

−p

• K(p)

= 0 AA “TYPE II” FOR K(p) = 0

1 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 ˆ

M′

s.

3 Quasi-Newton step: ps+1 = ps − ˆ

M′

sK(ps).

4 Set s = s + 1.

Rob Haelterman

• Dept. Math.

Royal Military Academy Belgium Problem setting Coupled problem Anderson acceleration Results Improvements I Improvements II Improvements III Questions

ANDERSON ACCELERATION

ROOT FINDING PROBLEM F(S(p))

• H(p)

−p

• K(p)

= 0 AA: JACOBIAN ˆ M′

• =

− 1 ωI (de facto) , (10a) ˆ M′

s

= Ws((Vs)TVs)−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)

Rob Haelterman

• Dept. Math.

Royal Military Academy Belgium Problem setting Coupled problem Anderson acceleration Results Improvements I Improvements II Improvements III Questions

RESULTS

n = 1000, 10 time-steps. κ τ Broyden Type II Anderson Type II 103 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)

Rob Haelterman

• Dept. Math.

Royal Military Academy Belgium Problem setting Coupled problem Anderson acceleration Results Improvements I Improvements II Improvements III Questions

IMPROVEMENTS: MGM

RANK s UPDATE ˆ M′

s+1 = −I + Ws((Vs)TVs)−1(Vs)T

• Rank s

RANK 1 UPDATE ˆ M′

s+1 = ˆ

M′

s + (δps − ˆ

M′

sδKs)

ds, δKs dT

s

• vT

= ˆ M′

s + uvT

ds = (I − Ls(Ls)T)δKs ⊥ δKj (j < s) where columns of Ls form orthonormal base for R(Vs) = span{δKs−1, δKs−2, . . . , δKo}. v⊥δKs−1 ⇒ well-studied by Martinez et al.

Rob Haelterman

• Dept. Math.

Royal Military Academy Belgium Problem setting Coupled problem Anderson acceleration Results Improvements I Improvements II Improvements III Questions

IMPROVEMENTS: MGM

RANK s UPDATE ˆ M′

s+1

= −I + Ws ((Vs)TVs)−1(Vs)T ˆ M′

s+1

= −I + (Ws − Vs − (−I)Vs)((Vs)TVs)−1(Vs)T ˆ M′

s+1

= A + (Ws − Vs − A Vs)((Vs)TVs)−1(Vs)T will respect multi-secant condition ˆ M′

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

Take A as Jacobian from coarser grid.

Rob Haelterman

• Dept. Math.

Royal Military Academy Belgium Problem setting Coupled problem Anderson acceleration Results Improvements I Improvements II Improvements III Questions

IMPROVEMENTS: MGM

nc = nf/3 → κ τ Anderson II Anderson II MGM 103 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 higher number needed on coarse grid.

Rob Haelterman

• Dept. Math.

Royal Military Academy Belgium Problem setting Coupled problem Anderson acceleration Results Improvements I Improvements II Improvements III Questions

IMPROVEMENTS: JACOBIAN FROM PREVIOUS

TIME-STEP

RANK s UPDATE ˆ M′

s+1

= A + (Ws − Vs − A Vs)((Vs)TVs)−1(Vs)T Take A as Jacobian from previous time-step.

Rob Haelterman

• Dept. Math.

Royal Military Academy Belgium Problem setting Coupled problem Anderson acceleration Results Improvements I Improvements II Improvements III Questions

IMPROVEMENTS: JACOBIAN FROM PREVIOUS

TIME-STEP

κ τ Anderson II + MGM + Jac 103 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 ?

Rob Haelterman

• Dept. Math.

Royal Military Academy Belgium Problem setting Coupled problem Anderson acceleration Results Improvements I Improvements II Improvements III Questions

IMPROVEMENTS: INPUT-OUTPUT VECTORS

RANK s UPDATE ˆ M′

s+1 =

−I + [Ws|Wprev](([Vs|Vprev])T[Vs|Vprev])−1([Vs|Vprev])T Will still respect secant condition(s) at current time-step, but also those of previous time-steps.

Rob Haelterman

• Dept. Math.

Royal Military Academy Belgium Problem setting Coupled problem Anderson acceleration Results Improvements I Improvements II Improvements III Questions

IMPROVEMENTS: JACOBIAN FROM PREVIOUS

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 ?

Rob Haelterman

• Dept. Math.

Royal Military Academy Belgium Problem setting Coupled problem Anderson acceleration Results Improvements I Improvements II Improvements III Questions

OTHER APPLICATIONS

FIGURE : Three-dimensional flow in a tube.

Rob Haelterman

• Dept. Math.

Royal Military Academy Belgium Problem setting Coupled problem Anderson acceleration Results Improvements I Improvements II Improvements III Questions

OTHER APPLICATIONS

FIGURE : Flapping tail.

Rob Haelterman

• Dept. Math.

Royal Military Academy Belgium Problem setting Coupled problem Anderson acceleration Results Improvements I Improvements II Improvements III Questions

Questions ?

Rob Haelterman

• Dept. Math.

Royal Military Academy Belgium Problem setting Coupled problem Anderson acceleration Results Improvements I Improvements II Improvements III Questions

REFERENCES

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

Rob Haelterman

• Dept. Math.

Royal Military Academy Belgium Problem setting Coupled problem Anderson acceleration Results Improvements I Improvements II Improvements III Questions

REFERENCES

The quasi-newton least squares method: A new and fast secant method analyzed for linear systems R Haelterman, J Degroote, D Van Heule, J Vierendeels SIAM Journal on numerical analysis 47 (3), 2347-2368 Coupling techniques for partitioned fluid-structure interaction simulations with black-box solvers J Degroote, R Haelterman, S Annerel, J Vierendeels 10th MpCCI User Forum, 82-91 Stability of a coupling technique for partitioned solvers in FSI applications J Degroote, P Bruggeman, R Haelterman, J Vierendeels Computers & Structures 86 (23), 2224-2234