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Mapped Tent Pitching Method for Hyperbolic Conservation Laws - PowerPoint PPT Presentation

Mapped Tent Pitching Method for Hyperbolic Conservation Laws Christoph Wintersteiger Institute for Analysis and Scientific Computing, TU Wien Jay Gopalakrishnan Portland State University, USA Joachim Sch oberl Institute for Analysis and


  1. Inverse transformation G − 1 ( ˆ U ) ˆ U ≡ G (ˆ u ) := ˆ u − f (ˆ u ) ∇ ϕ (3) Convection equation b ∈ R 2 Flux function: f ( u ) := bu , ˆ U ˆ U = (1 − b · ∇ ϕ )ˆ ⇔ u = ˆ u 1 − b · ∇ ϕ 1 solvable if |∇ ϕ | < | b | 11

  2. Inverse transformation G − 1 ( ˆ U ) ˆ U ≡ G (ˆ u ) := ˆ u − f (ˆ u ) ∇ ϕ (3) Convection equation b ∈ R 2 Flux function: f ( u ) := bu , ˆ U ˆ U = (1 − b · ∇ ϕ )ˆ ⇔ u = ˆ u 1 − b · ∇ ϕ 1 solvable if |∇ ϕ | < | b | (known CFL-condition) 11

  3. Inverse transformation G − 1 ( ˆ U ) ˆ U ≡ G (ˆ u ) := ˆ u − f (ˆ u ) ∇ ϕ (3) Convection equation b ∈ R 2 Flux function: f ( u ) := bu , ˆ U ˆ U = (1 − b · ∇ ϕ )ˆ ⇔ u = ˆ u 1 − b · ∇ ϕ 1 solvable if |∇ ϕ | < | b | (known CFL-condition) Theorem If there holds |∇ ϕ | < 1 c , then (3) has a unique solution ˆ u . c . . . maximal speed 11

  4. Inverse transformation G − 1 ( ˆ U ) Find ( ρ, m , E ) : Ω × (0 , T ] → R × R N × R s.t.     ρ m    1  ∂ t  + div ρ m ⊗ m + pI  = 0 m   m E ρ ( E + p ) 12

  5. Inverse transformation G − 1 ( ˆ U ) Find ( ρ, m , E ) : Ω × (0 , T ] → R × R N × R s.t.     ρ m    1  ∂ t  + div ρ m ⊗ m + pI  = 0 m   m E ρ ( E + p ) ˆ U ≡ G (ˆ u ) := ˆ u − f (ˆ u ) ∇ ϕ (3) 12

  6. Inverse transformation G − 1 ( ˆ U ) Find ( ρ, m , E ) : Ω × (0 , T ] → R × R N × R s.t.     ρ m    1  ∂ t  + div ρ m ⊗ m + pI  = 0 m   m E ρ ( E + p ) ˆ U ≡ G (ˆ u ) := ˆ u − f (ˆ u ) ∇ ϕ (3) m , ˆ u = (ˆ ˆ ρ, ˆ E ) U = ( ˆ ˆ R , ˆ M , ˆ F ) m , ˆ E ) = ˆ G − 1 ( ˆ R , ˆ M , ˆ (ˆ ρ, ˆ F ) 12

  7. Inverse transformation G − 1 ( ˆ U ) Find ( ρ, m , E ) : Ω × (0 , T ] → R × R N × R s.t.     ρ m    1  ∂ t  + div ρ m ⊗ m + pI  = 0 m   m E ρ ( E + p ) ˆ R 2 ˆ ρ = a 1 − 2 d |∇ ϕ | 2 a 3 m = ˆ ρ M + 2 ( ˆ ˆ d a 3 ∇ ϕ ) ˆ R E = ˆ ρ F + 2 a 3 ˆ ( ˆ ρ ∇ ϕ · ˆ m ) ˆ d ˆ R where a 2 a 1 = ˆ R − ˆ a 2 = 2 ˆ F ˆ R − | ˆ M | 2 , M · ∇ ϕ, a 3 = . � 1 − 4( d +1) a 2 |∇ ϕ | 2 a 2 a 1 + d 2 12

  8. Conservation ˆ ˆ K i K i t t ν i ˆ U ( x , 1) Φ ν i − 1 ˆ U ( x , 0) x x 13

  9. Conservation ˆ ˆ K i K i t t ν i ˆ U ( x , 1) Φ ν i − 1 ˆ U ( x , 0) x x Parametrizations γ i − 1 : x �→ ( x , τ i − 1 ( x )) , γ i : x �→ ( x , τ i ( x )) and space-time unit normal vectors ν i − 1 , ν i . 13

  10. Conservation ˆ ˆ K i K i t t ν i ˆ U ( x , 1) Φ ν i − 1 ˆ U ( x , 0) x x Parametrizations γ i − 1 : x �→ ( x , τ i − 1 ( x )) , γ i : x �→ ( x , τ i ( x )) and space-time unit normal vectors ν i − 1 , ν i . 13

  11. Conservation ˆ ˆ K i K i t t ν i ˆ U ( x , 1) Φ ν i − 1 ˆ U ( x , 0) x x Parametrizations γ i − 1 : x �→ ( x , τ i − 1 ( x )) , γ i : x �→ ( x , τ i ( x )) and space-time unit normal vectors ν i − 1 , ν i . Conservation After time step ˆ U ( x , 0) → ˆ U ( x , 1) there holds � � � � � � f ( u ) f ( u ) · ν i d s = · ν i − 1 d s . u u γ i γ i − 1 13

  12. Conservation Conservation After time step ˆ U ( x , 0) → ˆ U ( x , 1) there holds � � � � � � f ( u ) f ( u ) · ν i d s = · ν i − 1 d s . u u γ i γ i − 1 14

  13. Conservation Conservation After time step ˆ U ( x , 0) → ˆ U ( x , 1) there holds � � � � � � f ( u ) f ( u ) · ν i d s = · ν i − 1 d s . u u γ i γ i − 1 � � � � −∇ τ i ( x ) −∇ τ i − 1 ( x ) ν i ≈ , ν i − 1 ≈ 1 1 14

  14. Conservation Conservation After time step ˆ U ( x , 0) → ˆ U ( x , 1) there holds � � � � � � f ( u ) f ( u ) · ν i d s = · ν i − 1 d s . u u γ i γ i − 1 � � � � −∇ τ i ( x ) 0 ν i ≈ , ν i − 1 = 1 1 ν i ν i − 1 14

  15. Conservation Conservation After time step ˆ U ( x , 0) → ˆ U ( x , 1) there holds � � � � f ( u ) · ν i d s = u d s . u γ i γ i − 1 � � � � −∇ τ i ( x ) 0 ν i ≈ , ν i − 1 = 1 1 ν i ν i − 1 14

  16. Conservation Conservation After time step ˆ U ( x , 0) → ˆ U ( x , 1) there holds � � � � � � f ( u ) f ( u ) · ν i d s = · ν i − 1 d s . u u γ i γ i − 1 � � � � 0 −∇ τ i − 1 ( x ) ν i = , ν i − 1 ≈ 1 1 ν i ν i − 1 14

  17. Conservation Conservation After time step ˆ U ( x , 0) → ˆ U ( x , 1) there holds � � � � f ( u ) u d s = · ν i − 1 d s . u γ i γ i − 1 � � � � 0 −∇ τ i − 1 ( x ) ν i = , ν i − 1 ≈ 1 1 ν i ν i − 1 14

  18. Tent pitching algorithm in 1D t Advancing front τ ∇ ϕ = 0 x ∇ ϕ = 0 ϕ ( x , ˆ t ) := (1 − ˆ t ) τ i − 1 ( x ) + ˆ t τ i ( x ) 15

  19. Tent pitching algorithm in 1D t Advancing front τ ∇ ϕ = 0 x ∇ ϕ = 0 ϕ ( x , ˆ t ) := (1 − ˆ t ) τ i − 1 ( x ) + ˆ t τ i ( x ) ˆ U = ˆ u − f (ˆ u ) ∇ ϕ 15

  20. Tent pitching algorithm in 1D t Advancing front τ ∇ ϕ = 0 x ∇ ϕ = 0 ϕ ( x , ˆ t ) := (1 − ˆ t ) τ i − 1 ( x ) + ˆ t τ i ( x ) ∇ ϕ =0 ˆ ˆ U = ˆ u − f (ˆ u ) ∇ ϕ = ⇒ U = ˆ u 15

  21. The wave equation Find ψ : Ω × (0 , T ] → R s.t. ∂ tt ψ − div( ∇ ψ ) = 0 in Ω × (0 , T ] . 16

  22. The wave equation Find ψ : Ω × (0 , T ] → R s.t. ∂ tt ψ − div( ∇ ψ ) = 0 in Ω × (0 , T ] . With � � � � −∇ ψ q := , µ ∂ t ψ 16

  23. The wave equation Find ψ : Ω × (0 , T ] → R s.t. ∂ tt ψ − div( ∇ ψ ) = 0 in Ω × (0 , T ] . With � � � � −∇ ψ q := , µ ∂ t ψ we obtain � � � � q I µ + div = 0 . ∂ t q ⊤ µ 16

  24. The wave equation Find ψ : Ω × (0 , T ] → R s.t. ∂ tt ψ − div( ∇ ψ ) = 0 in Ω × (0 , T ] . With � � � � −∇ ψ q := , µ ∂ t ψ we obtain � � � � q I µ + div = 0 . ∂ t q ⊤ µ Mapping to space-time cylinder leads to �� � � � � � � �� q ˆ I ˆ µ I ˆ µ ∂ ˆ − ∇ ϕ + div δ = 0 . t q ⊤ q ⊤ ˆ ˆ ˆ µ 16

  25. The wave equation Find ψ : Ω × (0 , T ] → R s.t. ∂ tt ψ − div( ∇ ψ ) = 0 in Ω × (0 , T ] . With � � � � q −∇ ψ := , µ ∂ t ψ we obtain � � � � q I µ ∂ t + div = 0 . q ⊤ µ Mapping to space-time cylinder leads to �� � � �� � � I −∇ ϕ q ˆ I δ ˆ µ ∂ ˆ + div = 0 . t −∇ ϕ ⊤ q ⊤ 1 µ ˆ δ ˆ 16

  26. The wave equation �� � � �� � � I −∇ ϕ q ˆ I δ ˆ µ ∂ ˆ + div = 0 . (4) t −∇ ϕ ⊤ q ⊤ 1 ˆ δ ˆ µ 17

  27. The wave equation �� � � �� � � I −∇ ϕ q ˆ I δ ˆ µ ∂ ˆ + div = 0 . (4) t −∇ ϕ ⊤ q ⊤ 1 ˆ δ ˆ µ Space-discretization by DG leads to time-dependent mass matrix ∂ ˆ t M ˆ u + A ˆ u = 0 . 17

  28. The wave equation �� � � �� � � I −∇ ϕ q ˆ I δ ˆ µ ∂ ˆ + div = 0 . (4) t −∇ ϕ ⊤ q ⊤ 1 ˆ δ ˆ µ Space-discretization by DG leads to time-dependent mass matrix ∂ ˆ t M ˆ u + A ˆ u = 0 . Introduce a new variable y = M ˆ u and discretize transformed system t y + AM − 1 y = 0 ∂ ˆ by a Runge-Kutta method. 17

  29. The wave equation Find ψ : Ω × (0 , T ] → R s.t. ∂ tt ψ − div( ∇ ψ ) = 0 in Ω × (0 , T ] . With � � � � q −∇ ψ := , µ ∂ t ψ we obtain � � � � q I µ ∂ t + div = 0 . q ⊤ µ √ Domain Ω = [0 , π ] 2 , T = 2 π , √ 1 ψ ( x , t ) = √ cos( x 1 ) cos( x 2 ) sin( 2 t ) 2 18

  30. The wave equation, 2+1 dimensions Figure 1: Convergence rates in two space dimensions with RK2 for various spatial polynomial degrees p of approximation, with e 2 = � q ( · , T ) − q h � 2 L 2 (Ω) + � µ ( · , T ) − µ h � 2 L 2 (Ω) . p = 1 10 − 1 p = 2 p = 3 10 − 2 p = 4 10 − 3 O ( h ) O ( h 2 ) 10 − 4 e O ( h 3 ) 10 − 5 O ( h 4 ) 10 − 6 10 − 7 10 2 10 3 ndofs 19

  31. The wave equation Instead of t M ˆ u + A ˆ u = 0 , ∂ ˆ 20

  32. The wave equation Instead of t M ˆ u + A ˆ u = 0 , ∂ ˆ consider the system t ˆ U + A ˆ u = 0 , (5a) ∂ ˆ ˆ U = M ˆ u . (5b) 20

  33. The wave equation Instead of t M ˆ u + A ˆ u = 0 , ∂ ˆ consider the system t ˆ U + A ˆ u = 0 , (5a) ∂ ˆ ˆ U = M ˆ u . (5b) i ˆ i ˆ u = � U = � u i and ˆ i ˆ i ˆ Expansion of ˆ t t U i , with 20

  34. The wave equation Instead of t M ˆ u + A ˆ u = 0 , ∂ ˆ consider the system t ˆ U + A ˆ u = 0 , (5a) ∂ ˆ ˆ U = M ˆ u . (5b) i ˆ i ˆ u = � U = � u i and ˆ i ˆ i ˆ Expansion of ˆ t t U i , with 1 ˆ U n +1 = n + 1 A ˆ u n , U n +1 − M ′ ˆ u n +1 = ˆ M ˆ u n . 20

  35. The wave equation, 2+1 dimensions Figure 2: Convergence rates in two space dimensions with 2 Taylor steps for various spatial polynomial degrees p of approximation, with e 2 = � q ( · , T ) − q h � 2 L 2 (Ω) + � µ ( · , T ) − µ h � 2 L 2 (Ω) . p = 1 10 − 1 p = 2 p = 3 10 − 2 p = 4 10 − 3 O ( h ) O ( h 2 ) 10 − 4 e O ( h 3 ) 10 − 5 O ( h 4 ) 10 − 6 10 − 7 10 2 10 3 ndofs 21

  36. The wave equation, 2+1 dimensions Figure 3: Convergence rates in two space dimensions with 4 Taylor steps for various spatial polynomial degrees p of approximation, with e 2 = � q ( · , T ) − q h � 2 L 2 (Ω) + � µ ( · , T ) − µ h � 2 L 2 (Ω) . p = 1 10 − 1 p = 2 p = 3 p = 4 10 − 3 O ( h ) O ( h 2 ) e 10 − 5 O ( h 3 ) O ( h 4 ) 10 − 7 10 − 9 10 2 10 3 ndofs 22

  37. The wave equation Find ψ : Ω × (0 , T ] → R s.t. ∂ tt ψ − div( α ∇ ψ ) = 0 in Ω × (0 , T ] . With � � � � q −∇ ψ := , µ ∂ t ψ we obtain � � � � q I µ + div = 0 ∂ t q ⊤ µ Domain Ω = [0 , π ] 3 , T = 2 π 3 , √ √ 1 ψ ( x , t ) = √ cos( x 1 ) cos( x 2 ) cos( x 3 ) sin( 3 t ) 3 23

  38. The wave equation, 3+1 dimensions Figure 4: Convergence rates in three space dimensions for various spatial polynomial degrees p of approximation and p Taylor steps, with e 2 = � q ( · , T ) − q h � 2 L 2 (Ω) + � µ ( · , T ) − µ h � 2 L 2 (Ω) . p = 2 10 − 2 p = 3 p = 4 10 − 3 O ( h 2 ) 10 − 4 O ( h 3 ) O ( h 4 ) 10 − 5 e 10 − 6 10 − 7 10 − 8 10 − 9 10 4 10 5 10 6 ndofs 24

  39. The Maxwell equations The Maxwell equations � � � � curl H ε E = ∂ t µ H − curl E 25

  40. The Maxwell equations The Maxwell equations � � � � curl H ε E = ∂ t µ H − curl E can be written as � � � � − skew H ε E + div = 0 , ∂ t µ H skew E with (skew E ) ij := ε ijk E k . 25

  41. The Maxwell equations Figure 5: Resonator, 489k curved elements, largest to smallest element: 5:1 26

  42. The Maxwell equations Figure 6: H y at t=260, 260 time slabs, 148k tents per slab, p 2 local Taylor time-steps Shared memory server, 4 E7-8867 CPUs with 16 cores each. 27

  43. The Maxwell equations Figure 6: H y at t=260, 260 time slabs, 148k tents per slab, p 2 local Taylor time-steps Shared memory server, 4 E7-8867 CPUs with 16 cores each. p=2: 29 374 980 dofs, 20 min 27

  44. The Maxwell equations Figure 6: H y at t=260, 260 time slabs, 148k tents per slab, p 2 local Taylor time-steps Shared memory server, 4 E7-8867 CPUs with 16 cores each. p=2: 29 374 980 dofs, 20 min p=3: 58 751 160 dofs, 3 h 33 min 27

  45. The Maxwell equations Figure 7: Resonator with sharp edges, 224k curved elements, largest to smallest element: 10:1 28

  46. The Maxwell equations Figure 8: H y at t=260, 260 time slabs, 66k tents per slab, p 2 local Taylor time-steps Shared memory server, 4 E7-8867 CPUs with 16 cores each. 29

  47. The Maxwell equations Figure 8: H y at t=260, 260 time slabs, 66k tents per slab, p 2 local Taylor time-steps Shared memory server, 4 E7-8867 CPUs with 16 cores each. p=2: 13 452 000 dofs, 8 min 29

  48. The Maxwell equations Figure 8: H y at t=260, 260 time slabs, 66k tents per slab, p 2 local Taylor time-steps Shared memory server, 4 E7-8867 CPUs with 16 cores each. p=2: 13 452 000 dofs, 8 min p=3: 26 904 000 dofs, 1 h 27 min 29

  49. Euler equations Find ( ρ, m , E ) : Ω × (0 , T ] → R × R N × R s.t.     ρ m    1  ∂ t  + div ρ m ⊗ m + pI  = 0 m   m E ρ ( E + p ) 30

  50. Entropy admissibility condition Entropy admissibility condition E ( u ) ∈ R . . . entropy, F ( u ) ∈ R N . . . entropy flux 31

  51. Entropy admissibility condition Entropy admissibility condition E ( u ) ∈ R . . . entropy, F ( u ) ∈ R N . . . entropy flux ⇒ ∂ t E ( u ) + div F ( u ) ≤ 0 31

  52. Entropy admissibility condition Entropy admissibility condition E ( u ) ∈ R . . . entropy, F ( u ) ∈ R N . . . entropy flux ⇒ ∂ t E ( u ) + div F ( u ) ≤ 0 The pair ( E , F ) is called the entropy pair. 31

  53. Entropy admissibility condition Entropy admissibility condition E ( u ) ∈ R . . . entropy, F ( u ) ∈ R N . . . entropy flux ⇒ ∂ t E ( u ) + div F ( u ) ≤ 0 The pair ( E , F ) is called the entropy pair. ˆ E ( w ) = E ( w ) − F ( w ) ∇ ϕ, ˆ F ( w ) = δ F ( w ) . 31

  54. Entropy admissibility condition Entropy admissibility condition E ( u ) ∈ R . . . entropy, F ( u ) ∈ R N . . . entropy flux ⇒ ∂ t E ( u ) + div F ( u ) ≤ 0 The pair ( E , F ) is called the entropy pair. ˆ E ( w ) = E ( w ) − F ( w ) ∇ ϕ, ˆ F ( w ) = δ F ( w ) . Mapped entropy admissibility condition t ˆ u ) + div ˆ ∂ ˆ E (ˆ F (ˆ u ) = δ ( ∂ t E ( u ) + div F ( u )) ◦ Φ ≤ 0 31

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