FEM: Domain Decomposition and Homogenization for Maxwell s Equations of Large Scale Problems

Transcription

1 FEM: and Homogenization for Maxwell s Equations of Large Scale Problems Karl Hollaus Vienna University of Technology, Austria Institute for Analysis and Scientific Computing February 13, 2012

2 Outline 1 Introduction Poisson Equation Nitsche-type mortar method Hybride Nitsche-type mortar method Numerical Examples Maxwell s Equations Steady State Hybride Nitsche-type mortar method Novel hybride Nitsche-type mortar method Numerical Examples 2 Homogenization Introduction Electrostatic Problem Two-scale approach Eddy Current Problem Two-scale approach

3 Introduction Motivation can be used for: Moving parts: Electrical machines, etc. Complicated structure: Printed circuit boards, human body, Parallel computing Etc.

4 Introduction Problem Ω... Domain Ω... Surface of Ω Ω i... Subdomain Ω = Ω 1 Ω 2 Ω i... Surface of Ω i Γ... Interface Γ = Ω 1 Ω 2

5 Introduction Problem Aim: Non-matching finite element grids and continuous solutions! 2D: 3D:

6 Introduction Electrostatics Consider the corresponding Maxwell s equations curl E = 0 div D = ρ, the material relation D = ɛ E and appropriate boundary conditions. Then, an electric scalar potential V can be introduced as E = V and the Poisson equation V = ρ ɛ is obtained.

7 Poisson Equation Standard problem Classical formulation: (λ(x) u(x)) = f u = u D λ u n = g in Ω on Γ D on Γ N

8 Poisson Equation Standard problem Classical formulation: (λ(x) u(x)) = f in Ω u = u D λ u n = g on Γ D on Γ N Weak form: Multiplication of PDE by a test function v and integration over Ω yields (λ (u))v dω = fv dω, Ω Ω and integration by parts leads to λ u v dω λgvn dγ = fv dω. Γ Ω Ω

9 Poisson Equation Standard problem Classical formulation: (λ(x) u(x)) = f in Ω u = u D λ u n = 0 on Γ D on Γ N Weak form: The boundary conditions together with v = 0 on Γ D leads to the weak form for the finite element method: Find u h V D := {u h V h : u h = u D on Γ D }, such that λ u h v h dω = fv h dω v h V 0 Ω with V 0 := {v h V h : v h = 0 on Γ D }, where V h is a finite element subspace of H 1 (Ω). Ω

10 Poisson Equation Equivalent transmission problem Classical formulation: where u i = u Ωi. u i = f in Ω i, i = {1, 2} u i = 0 on Ω, i = {1, 2} u 1 u 2 = 0 on Γ u 1 n 1 + u 2 n 2 = 0 on Γ,

11 Poisson Equation Equivalent transmission problem Classical formulation: where u i = u Ωi. Definitions: u i = f in Ω i, i = {1, 2} u i = 0 on Ω, i = {1, 2} u 1 u 2 = 0 on Γ u 1 n 1 + u 2 n 2 = 0 on Γ, Jump at the interface: [[u]] := u 1 n 1 + u 2 n 2 Mean value: {u} := 1 2 (u 1 + u 2 )

12 Poisson Equation Nitsche-type mortar method Classical formulation: where u i = u Ωi. u i = f in Ω i, i = {1, 2} u i = 0 on Ω, i = {1, 2} u 1 u 2 = 0 on Γ u 1 n 1 + u 2 n 2 = 0 on Γ, Idea of Nitsche-type mortar method for domain decomposition: 2 u v dω Ω i { u} [[v]] ds = Ω i fv dω Ω i i=1

13 Poisson Equation Nitsche-type mortar method Weak form: Find u h V D := {u h V h : u h = 0 on Ω}, such that 2 u h v h dω { u h } [[v h ]] ds i=1 Ω i Γ { v h } [[u h ]] ds + αp2 [[u h ]] [[v h ]] ds = fv h dω v h V 0 Γ h Γ Ω i with V 0 := {v h V h : v h = 0 on Ω}, where V h is a finite element subspace of H 1 (Ω 1 ) H 1 (Ω 2 ). α... Stabilization factor p... Polynomial order h... Mesh size

14 Poisson Equation Nitsche-type mortar method Weak form: Find u h V D := {u h V h : u h = 0 on Ω}, such that 2 u h v h dω { u h } [[v h ]] ds i=1 Ω i Γ { v h } [[u h ]] ds + αp2 [[u h ]] [[v h ]] ds = fv h dω v h V 0 Γ h Γ Ω i with V 0 := {v h V h : v h = 0 on Ω}, where V h is a finite element subspace of H 1 (Ω 1 ) H 1 (Ω 2 ). The formulation is consistent, stable and bounded w. r. t. a mesh dependant energy norm 1. 1 Becker, R., P. Hansbo and R. Stenberg, A finite element method for domain decomposition with non-matching grids, 2003

15 Poisson Equation Nitsche-type mortar method Algebraic equation system: Surface integrals introduce a coupling across the interface Γ: Ω 1 {}}{ Ω 2 {}}{ u 1 u 2 = f 1 f 2

16 Poisson Equation Nitsche-type mortar method Algebraic equation system: Surface integrals introduce a coupling across the interface Γ: Ω 1 {}}{ Ω 2 {}}{ u 1 u 2 = f 1 f 2 The corresponding entries are almost all zero. But the integrals of non-zero entries can be difficult to compute!

17 Poisson Equation Nitsche-type mortar method Coupling terms in the weak form: ϕ 1 ϕ 2 n 2 dγ Γ Fig.: Partly adjacent finite elements T i with basis functions ϕ i The determination of the intersection Γ of the two elements T 1 and T 2 can be expensive. Practically, a remeshing is required.

18 Poisson Equation Hybride Nitsche-type mortar method Definition: Additional independent variable λ, which is the restriction u 1 = u 2 on Γ Function space M := {µ L 2 ( Ω 1 Ω 2 ) : µ = 0 on Ω}

19 Poisson Equation Hybride Nitsche-type mortar method Weak form 2 : Find (u h, λ h ) V D := {(u h, λ h ) V h M h : u h = 0 on Ω}, such that { 2 u h u h v h dω i=1 Ω i Ω i n (v v h h µ h ) dγ Ω i n (u h λ h ) dγ+ αp 2 } (u h λ h )(v h µ h ) dγ = fv h dω (v h, µ h ) V 0 h Ω i Ω with (u h, λ h ) V 0 := {(v h, µ h ) V h M h : v h = 0 on Ω}, where V h is a finite element subspace of H 1 (Ω 1 ) H 1 (Ω 2 ) and M h is a subspace of L 2 (Γ). 2 Egger, H., A class of hybrid mortar finite element methods for interface problems with non-matching meshes, preprint AICES , Aachen Institute for Advanced Study in Computational Engineering Science, 2009.

20 Poisson Equation Hybride Nitsche-type mortar method Matrix of the hybrid Nitsche-type mortar method: 0 0 u 1 u 2 λ = f 1 f 2 0 Now, coupling of u 1 with u 2 exists only via the interface variable λ!

21 Poisson Equation Poisson Equation We propose a B-spline basis for M h 3 : M h = S = { λ R λ [ti,t i+1 ] P n, λ C n 1 ([0, 1]) } 3 K. Hollaus, D. Feldengut, J. Schöberl, M. Wabro, D. Omeragic: Nitsche-type Mortaring for Maxwell s Equations, PIERS Proceedings, , July 5-8, Cambridge, USA 2010.

22 Poisson Equation Poisson Equation We propose a B-spline basis for M h 3 : M h = S = { λ R λ [ti,t i+1 ] P n, λ C n 1 ([0, 1]) } Gaussian quadrature benefits from smoothness of the splines. DeBoor recursion B i,r (x) = x t i B i,r 1 (x) + t i+r x B i+1,r 1 (x) t i+r 1 t i t i+r t i+1 3 K. Hollaus, D. Feldengut, J. Schöberl, M. Wabro, D. Omeragic: Nitsche-type Mortaring for Maxwell s Equations, PIERS Proceedings, , July 5-8, Cambridge, USA 2010.

23 Poisson Equation Poisson Equation Numerical Example in 2D: f = x in the circle, else f = 0 Rectangle: a = 2, b = 1 Circle: r = 0.3, at: x = 0.7, y = 0.5 FE-order: 5 Spline-order: 6, no. of splines: 6 Solution u of the Poisson equation

24 Poisson Equation Poisson Equation Numerical Example in 2D: f = x in the circle, else f = 0 Rectangle: a = 2, b = 1 Circle: r = 0.3, at: x = 0.7, y = 0.5 FE-order: 5 Spline-order: 6, no. of splines: 6 ( u) x of the Poisson equation

25 Poisson Equation Poisson Equation Numerical Example in 3D: f = xz in the cylinder, else f = 0 Cuboid: a = 3, b = 2, c = 1 Cylinder: r = 0.4, at: x = 1.0, y = 1.0 FE-order: 5 Spline-order: 6, No. splines: 6 Solution u of the Poisson equation

26 Poisson Equation Poisson Equation Numerical Example in 3D: f = xz in the cylinder, else f = 0 Cuboid: a = 3, b = 2, c = 1 Cylinder: r = 0.4, at: x = 1.0, y = 1.0 FE-order: 5 Spline-order: 6, No. splines: 6 ( u) x of the Poisson equation

27 Maxwell s Equations Complex Representation The time harmonic case, steady state: Consider the partial differential equations curlh = J + jωd in Ω curle = jωb div B = 0 div D = ρ, boundary conditions E n = 0 H n = K B n = b on Γ E on Γ H on Γ B and material relations J = σe B = µh D = ɛh.

28 Maxwell s Equations Standard Problem Magnetic vector potential A: A magnetic vector potential A can be introduced as B = curl A. Taking acount of the above relations, the physical quantaties can be written as E = iωa, H = µ 1 curl A and after known manpulations curl µ 1 curl A + κa = J in Ω with the complex conductivity κ = jωσ ω 2 ɛ is obtained.

29 Maxwell s Equations Standard Problem Classical formulation: curl µ 1 curl A + κa = J 0 in Ω where Ω = Ω H + Ω B + Ω E. µ 1 curl A n = K on Ω H A n = α on Ω B A n = 0 on Ω E,

30 Maxwell s Equations Standard Problem Weak form: Considering A n = 0 on Ω lead to the finite element approximation: Find A h V D := {A h V h : A h n = α h on Ω}, such that µ 1 curl A h curl v h dω + jω κa h v h dω = J 0 v h dω Ω Ω Ω 0 for all v h V 0 := {v h V h : v h n = 0 on Ω}, where V h is a finite element subspace of H(curl, Ω).

31 Maxwell s Equations Equivalent transmission problem Classical formulation: curl µ 1 curl A i + κa i = J 0 in Ω i, i = {1, 2} A i n = 0 on Ω, i = {1, 2} A 1 n 1 + A 2 n 2 = 0 on Γ µ 1 curl A 1 n 1 + µ 1 curl A 2 n 2 = 0 on Γ, where A i = A Ωi.

32 Maxwell s Equations Hybride Nitsche-type mortar method Weak form: Considering homogeneous boundary conditions on Ω and introducing symmetry, stabilization and hybridization yields the hybrid formulation: Find (A h, λ) V B := {A h V h : A h n = 0 on Γ B }, such that { 2 µ 1 {curl A curl v +κa v}+ Ω i µ 1 curl A [(v µ) n] Ω i i=1 + µ Ωi 1 curl v [(A λ) n]+ αp2 µh for all (v, µ). [(A λ) n] [(v µ) n] Ω i = J 0 v Ω }

33 Maxwell s Equations Novel hybride Nitsche-type mortar method Novel hybride Nitsche-type mortar method 4 : Considering symmetry, stability and hybridization also for φ leads to { 2 {µ 1 1 curl A curl v +κav}+ curl A [(v ψ µ) n]+ i=1 Ω i Ω i µ 1 curl v [(A φ λ) n]+ Ω i µ α 1 p 2 1 [(A φ λ) n][(v ψ µ) n] κa n (ψ ψ Γ ) h Ω i µ Ω i κv n (φ φ Γ )+ Ωi α 2p 2 } { 2 } κ(φ φ Γ )(ψ ψ Γ ) = J 0 v J n ψ h Ω i Ω i Ω i i=1 4 K. Hollaus, D. Feldengut, J. Schöberl, M. Wabro, D. Omeragic: Nitsche-type Mortaring for Maxwell s Equations, PIERS Proceedings, , July 5-8, Cambridge, USA 2010.

34 Maxwell s Equations Novel hybride Nitsche-type mortar method Novel hybride Nitsche-type mortar method: We propose to descretize by u, v... H (curl) conforming finite elements of order p in Ω i φ, ψ... H 1 conforming finite elements of order p + 1 on Ω i Γ φ Γ, ψ Γ... tensor product scalar B-splines of order m and λ, µ... Nedelec-type B-splines satisfying an exact sequence at Γ

35 Maxwell s Equations Numerical Example Numerical Example: Permanent Magnet PDE for magnetostatics: curl µ 1 curl A i = f (M) in Ω i A... Magnetic vector potential M... Magnetisation: M 0 in the permanent magnet, else M = 0 Fig.: Two domains (green and blue) with the permanent magnet (red)

36 Maxwell s Equations Numerical Example Numerical Example: Permanent Magnet Fig.: Vector plot of the magnetic field intensity H

37 Maxwell s Equations Numerical Example Numerical Example: Permanent Magnet Fig.: Scalar plot of the magnetic field intensity H x

38 Maxwell s Equations Numerical Example LWD-Tool LWD... logging while drilling Detail of the tool.

39 Maxwell s Equations Numerical Example LWD-Tool Ambient soil with a borehole.

40 Maxwell s Equations Numerical Example LWD-Tool LWD-Tool to reconstruct the ambient material parameters Detail of borehole with exciting antenna (red ring) and receiver (blue ring)

41 Maxwell s Equations Numerical Example LWD-Tool Vector plot of the magnetic field intensity H

42 Maxwell s Equations Numerical Example LWD-Tool Field lines of the magnetic field intensity H

43 Maxwell s Equations Numerical Example LWD-Tool Frequency in khz Standard: Voltage in µ V NVF: Voltage in µ V i i i i i i i i 5290 Table: Numerical results for first order finite elements. Frequency in khz Standard: Voltage in µ V NVF: Voltage in µ V i i i i i i i i 5255 Table: Numerical results for second order finite elements. Spline order: 5, spline no.: n ϕ = 5, n z = 150

44 Homogenization Introduction Motivation Laminated transformer core: Large transformer Principle of a transformer Laminated transformer core

45 Homogenization Introduction Motivation Fig.: FE-Model with 100 laminates. Fig.: FE-Model with 100 laminates (Detail, lower right corner). Finite element models of laminated media lead to large equations systems. Homogenization overcomes this problem!

46 Homogenization Introduction Motivation λ i... Material parameter n... No. laminations d = d 1 + d 2... periodic structure!!! ff... Fill factor ff = d1 d 1+d 2 For example: Electrostatics Problem with laminated medium. λ=ɛ... Electric permittivity u=v... Electric scalar potential

47 Homogenization Electrostatic Problem Electrostatic Problem Boundary value problem: Ω... Domain Ω = Ω 0 Ω m Γ... Boundary Γ = Γ D Γ N (λ(x, y) u(x, y)) = 0 in Ω (1) u = u D on Γ D (2) λ u n n = α on Γ N (3)

48 Homogenization Electrostatic Problem Standard problem Classical formulation: (λ(x) u(x)) = f in Ω u = u D λ u n = 0 on Γ D on Γ N Weak form: The boundary conditions together with v = 0 on Γ D leads to the weak form for the finite element method: Find u h V D := {u h V h : u h = u D on Γ D }, such that λ u h v h dω = 0 v h V 0 Ω with V 0 := {v h V h : v h = 0 on Γ D }, where V h is a finite element subspace of H 1 (Ω).

49 Homogenization Electrostatic Problem Electrostatic Problem in 2D Numerical Example: Fig.: Numerical model. ff... Fill factor ff = d 1 d 1+d 2 ff = 0.9 λ 0 = 1 λ 1 = 1000 λ 2 = 1 u b u a = 2 No. laminates = 10

50 Homogenization Electrostatic Problem Electrostatic Problem Reference solution: Each laminate is modeled individually! Fig.: Reference solution u of one halfe of the domain.

51 Homogenization Electrostatic Problem Electrostatic Problem Reference solution: Solution u, mean value u 0 and envelope u 1 along x at y = 0. Periodic micro-shape function φ(x), one periode is shown.

52 Homogenization Electrostatic Problem Electrostatic Problem Two-scale approach: Thus, the following approach can be made: u(x, y) = u 0 (x, y) + φ(x)u 1 (x, y) u 0... mean value u 1... envelope of the staggered part φ... periodic-micro shape function Inserting this approach into the bilinear form λ u v dω leads to Ω Ω λ (u 0 + φu 1 ) (v 0 + φv 1 ) dω = 0.

53 Homogenization Electrostatic Problem Electrostatic Problem Homogenization of the weak form: The finite element matrix would be calculated by Ω FE u 0 x u 0 y u 1 u 1 x u 1 y with φ x = ( φ) x. λ 0 λφ x λφ 0 0 λ 0 0 λφ λφ x 0 λφ 2 x λφφ x 0 λφ 0 λφφ x λφ λφ 0 0 λφ 2 T The micro shape function φ is a highly oscillating function. v 0 x v 0 y v 1 v 1 x v 1 y dω, To homogenize the weak form, the coefficients λ, λ φ, λφ, etc. are averaged over the periode d: λ = 1 d d 0 λ(x)dx = λ 1d 1 + λ 2 d 2 d

54 Homogenization Electrostatic Problem Electrostatic Problem Homogenization of the weak form: λφ x = 1 d λφ = 1 d λφ 2 x = 1 d λφ x φ = 1 d λφ 2 = 1 d d 0 d 0 d 0 d 0 d 0 λ(x)φ x (x)dx = 2 λ 1 λ 2 d λ(x)φ(x)dx = 0 λ(x)φ x (x)φ x (x)dx = 4 d (λ 1 d 1 + λ 2 d 2 ) λ(x)φ x (x)φ(x)dx = 0 λ(x)φ(x)φ(x)dx = λ 1d 1 + λ 2 d 2 3d

55 Homogenization Electrostatic Problem Electrostatic Problem Numerical experiments have shown that neglecting the derivatives in φ u 1 and φ v 1, respectively, yields a more accurate solution. Weak form: Find (u h0, u h1 ) V D := {(u h0, u h1 ) : u h0 V h, u h1 W h, u h0 = u D on Γ D }, such that Ω xu 0 y u 0 u 1 λ 0 λφ x x v 0 0 λ 0 y v 0 dω = 0 λφ x 0 λφ 2 v x 1 T for all (v h0, v h1 ) V 0 := {(v h0, v h1 ) : v h0 V h, v h1 W h, v h0 = 0 on Γ D }, where V h is a finite element subspace of H 1 (Ω) and W h a finite element subspace of L 2 (Ω m ), respectively. The micro-shape function φ is in the space of periodic and continuous functions H per (Ω m ).

56 Homogenization Electrostatic Problem Electrostatic Problem in 2D Numerical Example: Fig.: Numerical model. ff... Fill factor ff = d 1 d 1+d 2 ff = 0.9 λ 0 = 1 λ 1 = 1000 λ 2 = 1 u b u a = 2 No. laminations = 10

57 Homogenization Electrostatic Problem Electrostatic Problem Comparison of the results: Fig.: Reference solution. Fig.: Homogenized solution. The agreement between reference solution and homogenized solution is obviously excellent!

58 Homogenization Electrostatic Problem Electrostatic Problem Comparison of the results: flux density λ ( φ) x Fig.: Reference solution. Fig.: Homogenized solution.

59 Homogenization Electrostatic Problem Eddy Current Problem in 2D Boundary value problem: Ω... Domain Ω = Ω 0 Ω m Γ... Boundary Γ = Γ H Γ B µ... Magnetic permeability σ... Electric conductivity J 0... Impressed current density in a coil B... Magnetic flux density Assumptions: - Linear material properties - Time harmonic case, steady state

60 Homogenization Eddy Current Problem Eddy Current Problem Maxwell s equations for the eddy current problem: curlh = J in Ω m curle = jωb div B = 0 J = σe B = µh curlh = J 0 in Ω 0 div B = 0 B = µh H n = K on Γ H B n = b on Γ B

61 Homogenization Eddy Current Problem Standard Problem Weak form: Considering homogeneous boundary conditions on Γ leads to the finite element approximation: Find A h V B := {A h V h : A h n = α h on Γ B }, such that µ 1 curl A h curl v h dω + jω σa h v h dω = J 0 v h dω Ω Ω Ω 0 for all v h V 0 := {v h V h : v h n = 0 on Ω B }, where V h is a finite element subspace of H(curl, Ω).

62 Homogenization Eddy Current Problem Eddy Current Problem in 2D Numerical example: α = 1.0Vs/m ff = 0.9 µ 0 = 4π10 7 Vs/Am Fig.: Numerical example, dimensions in mm. µ = µ r µ 0 µ r1 = 1000 σ 1 = S/m f = 50Hz δ... Penetration depth δ = 1.6mm

63 Homogenization Eddy Current Problem Eddy Current Problem Reference solution: Eddy currents in laminates, 2D problem Two-scale approach: A = A 0 + φ(0, A 1 ) T + (φw) A 0... represents the mean value of the solution φ(0, A 1 ) T considers J y and w (φw) models J x at the end of the laminates.

64 Homogenization Eddy Current Problem Eddy Current Problem Homogenized weak form: Inserting the approach into the bilinear form yields: µ 1 curl ( A 0 + φ(0, A 1 ) T + (φw) ) curl ( v 0 + φ(0, v 1 ) T + (φq) ) dω Ω +jω Ω σ ( A 0 + φ(0, A 1 ) T + (φw) ) ( v 0 + φ(0, v 1 ) T + (φq) ) dω = Ω 0 J 0 v dω (4)

65 Homogenization Eddy Current Problem Eddy Current Problem Finite element stiffness matrix: Bilinearform: Ω µ 1 curla curlv dω Homogenized finite element matrix: ( ) T ( ) ( ) curla0 ν νφx curlv 0 dω, νφ x v 1 Ω FE A 1 with the reluctivity ν = µ 1. νφ 2 x

66 Homogenization Eddy Current Problem Eddy Current Problem Finite element mass matrix: Bilinearform: jω σa v dω Ω Homogenized finite element matrix: (A 0 ) T x σ 0 0 σφ x σφ 0 (v 0 ) x (A 0 ) y 0 σ σφ 0 0 σφ (v jω A 1 0 σφ σφ σφ 2 0 ) y v Ω FE w σφ w x 0 0 σφ 2 1 x σφ x φ 0 q dω x σφ 0 0 σφ x φ σφ 2 0 q x w y 0 σφ σφ σφ 2 q y

67 Homogenization Eddy Current Problem Eddy Current Problem Homogenized weak form: Find (A 0h, A 1h, w h ) V B := {(A 0h, A 1h, w h ) : A 0h U h, A 1h V h, w h W h and A 0h n = α h on Γ B }, such that A(A 0h, A 1h ; v 0h, v 1h ) + B(A 0h, A 1h, w h ; v 0h, v 1h, q h ) = f (v 0h ) (5) for all (v 0h, v 1h, q h ) V 0 := {(v 0h, v 1h, q h ) : v 0h U h, v 1h V h, q h W h and v 0h n = 0 on Γ B }, where U h is a finite element subspace of H(curl, Ω), V h a finite element subspace of L 2 (Ω m ) and W h a finite element subspace of H 1 (Ω m ), respectively. The micro-shape function φ is in the space of periodic and continuous functions H per (Ω m ). Again, a coarse finite element mesh suffices for the homogenized medium!

68 Homogenization Eddy Current Problem Eddy Current Problem in 2D Numerical example: α = 1.0Vs/m ff = 0.9 µ 0 = 4π10 7 Vs/Am Fig.: Numerical example, dimensions in mm. µ = µ r µ 0 µ r1 = 1000 σ 1 = S/m f = 50Hz δ... Penetration depth δ = 1.6mm

69 Homogenization Eddy Current Problem Eddy Current Problem Comparison of the results: Eddy current density J Reference solution. Homogenized solution.... solution in the upper left corner: d 1 = 1.8mm, 10 laminations The agreement between reference solution and homogenized solution is obviously excellent!

70 Homogenization Eddy Current Problem Eddy Current Problem Comparison of the results: Losses and computational costs Table: Eddy current losses Losses in W/m d in mm Reference solution Homogenized solution Table: Computational costs for d=0.5mm and 40 laminations Model FE NDOF Reference solution Homogenized solution FE... No. finite elements NDOF... No. degrees of freedom

71 Homogenization Eddy Current Problem Thank you for your attention!