Rob Haelterman Dept. Math. Royal Military Academy Belgium Problem Anderson acceleration for time-dependent problems setting Coupled problem Anderson acceleration Results Improvements I Improvements II Improvements III Questions
P ROBLEM SETTING I: 1D FLOW Rob Haelterman Dept. Math. Royal Military Academy Belgium Problem setting Coupled problem Anderson F IGURE : One-dimensional flow in a tube. acceleration Results Improvements I incompressible Improvements inviscid II Improvements gravity neglected III conservative form Questions
P ROBLEM SETTING I: 1D FLOW Rob Haelterman Dept. Math. Royal Military � ∂ gu Academy � ∂ g Belgium ∂ t + ∂ x = 0 (1a) Problem � ∂ g ˜ � ∂ t + ∂ gu 2 ∂ gu + 1 p ∂ g p setting ∂ x − ˜ = 0 (1b) Coupled ∂ x ρ ∂ x problem Anderson acceleration 10 sin 2 ( π t ) Inlet BC: u in ( t ) = u ref + u ref Results Improvements Constant pressure at outlet. I Improvements II Improvements III Questions
P ROBLEM SETTING I: 1D FLOW Rob Haelterman Dept. Math. Discrete variables: g → g , u → u , ˜ p → ˜ p Royal Military Academy Equidistant mesh Belgium Central discretization for all space derivatives in flow Problem equations except convective term that uses upwind setting discretisation. Coupled problem Backward (implicit) Euler time-discretisation Anderson acceleration Results → Root finding problem: F (˜ p t + 1 , u t + 1 ) = 0 Improvements I → Anderson Improvements II Improvements III Questions
P ROBLEM SETTING I: 1D FLOW Rob Haelterman Dept. Math. Royal Military Academy Belgium R ESULT Problem setting Solving with Anderson didn’t go well at all, compared to Coupled fsolve . problem Anderson acceleration Results Improvements I Improvements II Improvements III Questions
P ROBLEM SETTING II: FLEXIBLE TUBE Flexible tube with Hookean law g = g ( p ) Rob Inertia neglected Haelterman p = ˜ p /ρ = kinematic pressure. Dept. Math. Royal Military Academy Belgium ∂ p ∂ t + u ∂ p ∂ x + c 2 ∂ u ∂ x = 0 (2a) Problem setting ∂ t + ∂ gu 2 ∂ gu + ∂ gp ∂ x − p ∂ g ∂ x = 0 , (2b) Coupled ∂ x problem Anderson acceleration Non-reflecting BC at outlet: Results ∂ u out = 1 ∂ p out Improvements ∂ t . (3) I ∂ t c Improvements where the wave speed c is defined by II c 2 = g Improvements . (4) III d g Questions d p
P ROBLEM SETTING II: FLEXIBLE TUBE Rob Haelterman � � 2 p o − 2 c 2 mk Dept. Math. g = g o (5) p − 2 c 2 Royal Military Academy mk Belgium Moens Korteweg Wave-speed: c 2 Eh mk = 2 ρ r o . Problem Young modulus E setting Coupled Reference radius r problem Anderson Thickness of wall h acceleration Non-dimensionalize (and discretize): Results � 2 ρ ro − Po Eh Improvements (stiffness) and τ = U o ∆ t 2 I κ = c o L (time-step). U o Improvements Fourier analysis: smaller κ and τ n is more unstable for II fixed point iteration. Improvements III Questions
C OUPLED PROBLEM Rob Haelterman T HE COUPLED EQUATIONS Dept. Math. Royal Military � F ( p t + 1 , g t + 1 ; p t , g t ) = 0 Academy (6) Belgium S ( p t + 1 , g t + 1 ; p t , g t ) = 0 Problem setting � F ( g t + 1 ) = p t + 1 Coupled (7) problem S ( p t + 1 ) = g t + 1 Anderson acceleration Results � F ( g ) = p Improvements (8) I S ( p ) = g Improvements II Improvements III Questions
C OUPLED PROBLEM Rob Haelterman T HE COUPLED EQUATIONS Dept. Math. Royal Military � F ( p t + 1 , g t + 1 ; p t , g t ) = 0 Academy (6) Belgium S ( p t + 1 , g t + 1 ; p t , g t ) = 0 Problem setting � F ( g t + 1 ) = p t + 1 Coupled (7) problem S ( p t + 1 ) = g t + 1 Anderson acceleration Results � F ( g ) = p Improvements (8) I S ( p ) = g Improvements II Improvements III Questions
C OUPLED PROBLEM Rob Haelterman T HE COUPLED EQUATIONS Dept. Math. Royal Military � F ( p t + 1 , g t + 1 ; p t , g t ) = 0 Academy (6) Belgium S ( p t + 1 , g t + 1 ; p t , g t ) = 0 Problem setting � F ( g t + 1 ) = p t + 1 Coupled (7) problem S ( p t + 1 ) = g t + 1 Anderson acceleration Results � F ( g ) = p Improvements (8) I S ( p ) = g Improvements II Improvements III Questions
C OUPLED PROBLEM Rob Haelterman Dept. Math. Royal Military Academy Belgium H YPOTHESES 1 Solving each sub-problem is computationally heavy. Problem setting 2 No access to Jacobian of each sub-problem. Coupled problem 3 Only access to F ( g ) and S ( p ) . Anderson acceleration Results Improvements I Improvements II Improvements III Questions
C OUPLED PROBLEM Rob Haelterman Dept. Math. C OUPLED PROBLEM Royal Military � F ( g ) Academy = p Belgium S ( p ) = g Problem setting R OOT FINDING PROBLEM Coupled problem Anderson F ( S ( p )) − p = 0 (9) acceleration � �� � Results H ( p ) � �� � Improvements I K ( p ) Improvements II Improvements III Questions
A NDERSON A CCELERATION R OOT FINDING PROBLEM Rob Haelterman F ( S ( p )) − p = 0 � �� � Dept. Math. Royal Military H ( p ) Academy � �� � K ( p ) Belgium AA “T YPE II” FOR K ( p ) = 0 Problem setting 1 Startup: Coupled problem 1 Take an initial value p o . Anderson 2 Compute p 1 = ( 1 − ω ) p o + ω H ( p o ) . acceleration 3 Set s = 1. Results 2 Loop until sufficiently converged: Improvements I 1 Compute K ( p s ) . Improvements 2 Construct the approximate inverse Jacobian ˆ M ′ s . II 3 Quasi-Newton step: p s + 1 = p s − ˆ M ′ s K ( p s ) . Improvements III 4 Set s = s + 1. Questions
A NDERSON A CCELERATION R OOT FINDING PROBLEM Rob Haelterman F ( S ( p )) − p = 0 � �� � Dept. Math. H ( p ) Royal Military Academy � �� � K ( p ) Belgium AA: J ACOBIAN Problem setting Coupled − 1 problem ˆ M ′ = ω I (de facto) , (10a) o Anderson acceleration W s (( V s ) T V s ) − 1 ( V s ) T − I for s > 0 , ˆ M ′ = (10b) s Results Improvements where I δ K i = K ( p i + 1 ) − K ( p i ) ( i = 0 , . . . , s − 1 ) , V s = [ δ K s − 1 δ K s − 2 . . . δ K 0 ] Improvements (11) II δ H i = H ( p i + 1 ) − H ( p i ) ( i = 0 , . . . , s − 1 ) , W s = [ δ H s − 1 δ H s − 2 . . . δ H 0 ] Improvements III Questions
R ESULTS Rob Haelterman Dept. Math. Royal Military n = 1000, 10 time-steps. Academy Belgium κ τ Broyden Type II Anderson Type II Problem 10 3 10 − 4 9 (8.7) 8 (6.9) setting 10 2 10 − 4 Coupled div 19 (15.8) problem 10 − 3 10 div 22 (16.5) Anderson 10 − 4 acceleration 10 div 58 (51.8) Results Improvements I Improvements II Improvements III Questions
I MPROVEMENTS : MGM R ANK s UPDATE Rob Haelterman ˆ s + 1 = − I + W s (( V s ) T V s ) − 1 ( V s ) T M ′ Dept. Math. � �� � Royal Military Rank s Academy Belgium R ANK 1 UPDATE Problem s + ( δ p s − ˆ M ′ s δ K s ) setting ˆ s + 1 = ˆ = ˆ d T s + uv T M ′ M ′ M ′ s Coupled � d s , δ K s � ���� problem v T Anderson acceleration d s = ( I − L s ( L s ) T ) δ K s ⊥ δ K j ( j < s ) Results Improvements where columns of L s form orthonormal base for I R ( V s ) = span { δ K s − 1 , δ K s − 2 , . . . , δ K o } . Improvements II Improvements v ⊥ δ K s − 1 ⇒ well-studied by Martinez et al. III Questions
I MPROVEMENTS : MGM Rob Haelterman R ANK s UPDATE Dept. Math. Royal Military Academy ˆ (( V s ) T V s ) − 1 ( V s ) T M ′ = − I + W s s + 1 Belgium ˆ − I + ( W s − V s − ( − I ) V s )(( V s ) T V s ) − 1 ( V s ) T M ′ = s + 1 Problem ˆ A + ( W s − V s − A V s )(( V s ) T V s ) − 1 ( V s ) T M ′ setting = s + 1 Coupled problem will respect multi-secant condition Anderson acceleration ˆ M ′ s + 1 V s = [ δ p s − 1 δ p s − 2 . . . δ p 0 ] = W s − V s Results Improvements I Improvements Take A as Jacobian from coarser grid. II Improvements III Questions
I MPROVEMENTS : MGM Rob n c = n f / 3 → Haelterman Dept. Math. κ τ Anderson II Anderson II MGM Royal Military Academy 10 3 10 − 4 8 (6.9) 5(4.7)F + 8(7.2)C Belgium ≈ 7.67 (7.1) Problem 10 2 10 − 4 19 (15.8) 7(8.1)F + 20(20.8)C setting ≈ 13.67 (15.03) Coupled problem 10 − 3 10 22 (16.5) 8(7.9)F + 21(20.5)C Anderson ≈ 15 (14.73) acceleration 10 − 4 10 58 (51.8) 36(35.7)F + 54(52.0)C Results Improvements ≈ 54 (53.03) I Improvements n c < n f ⇒ cheaper but also more unstable: far higher II Improvements number needed on coarse grid. III Questions
I MPROVEMENTS : J ACOBIAN FROM PREVIOUS TIME - STEP Rob Haelterman Dept. Math. Royal Military Academy Belgium R ANK s UPDATE Problem setting ˆ A + ( W s − V s − A V s )(( V s ) T V s ) − 1 ( V s ) T M ′ = Coupled s + 1 problem Anderson Take A as Jacobian from previous time-step. acceleration Results Improvements I Improvements II Improvements III Questions
I MPROVEMENTS : J ACOBIAN FROM PREVIOUS TIME - STEP Rob Haelterman Dept. Math. Royal Military Academy κ τ Anderson II + MGM + Jac Belgium 10 3 10 − 4 8 (6.9) 7.67 (7.1) 8 (3.6) Problem 10 2 10 − 4 19 (15.8) 13.67 (15.03) 19 (5.9) setting 10 − 3 Coupled 10 22 (16.5) 15 (14.73) 22 (7.0) problem 10 − 4 10 58 (51.8) 54 (53.03) 58 (21.5) Anderson acceleration Results Take MGM Jacobian for first time-step only ? Improvements I Improvements II Improvements III Questions
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