A Multigrid Method for Two Dimensional Maxwell Interface Problems

Transcription

1 A Multigrid Method for Two Dimensional Maxwell Interface Problems Susanne C. Brenner Department of Mathematics and Center for Computation & Technology Louisiana State University USA JSA 2013

2 Outline A Maxwell Interface Problem in 2D Reduction to Scalar Problems P 1 Finite Element Methods A Multigrid Method Concluding Remarks Collaborators J. Cui, J. Gedicke, Z. Nan and L.-Y. Sung

3 References S.C.B., J. Cui, Z.Nan and L.-Y. Sung Hodge decomposition for divergence-free vector fields and twodimensional Maxwell s equations Math. Comp., 2012 S.C.B., J. Gedicke and L.-Y. Sung An adaptive P 1 finite element method for two-dimensional Maxwell s equations J. Sci. Comput., 2013 Z. Nan Applications of Helmholtz/Hodge Decomposition to Finite Element Methods for Two-Dimensional Maxwell s Equations Ph. D. Thesis, 2013

4 A Maxwell Interface Problem in 2D

5 A Maxwell Interface Problem in 2D An Elliptic Weak Formulation Find u H 0 (curl;ω) H(div 0 ;Ω;) such that (µ 1 u, v)+α(u, v) =(f, v) for all v H 0 (curl;ω) H(div 0 ;Ω;). u = electric field Ω=bounded polygonal domain in R 2 f [L 2 (Ω)] 2 (, ) =inner product of L 2 (Ω) µ = permeability (µ >0, µ and µ 1 are bounded) = permittivity ( >0, and 1 are bounded) α = constant

6 A Maxwell Interface Problem in 2D An Elliptic Weak Formulation Find u H 0 (curl;ω) H(div 0 ;Ω;) such that (µ 1 u, v)+α(u, v) =(f, v) for all v H 0 (curl;ω) H(div 0 ;Ω;). v = v1 v 2 v = v 2 x 1 v 1 x 2 v = v 1 x 1 + v 2 x 2 H(curl;Ω)={v [L 2 (Ω)] 2 : v L 2 (Ω)} H 0 (curl;ω)={v H(curl;Ω): tangential component of v = 0 on Ω} H(div 0 ;Ω;) ={v [L 2 (Ω)] 2 : (v) =0}

7 A Maxwell Interface Problem in 2D An Elliptic Weak Formulation Find u H 0 (curl;ω) H(div 0 ;Ω;) such that (µ 1 u, v)+α(u, v) =(f, v) for all v H 0 (curl;ω) H(div 0 ;Ω;). There exists a sequence (Maxwell eigenvalues) 0 α 1 >α 2 >... decreasing to such that this problem is uniquely solvable if α is different from any of the α k s. We assume that α is not a Maxwell eigenvalue.

8 A Maxwell Interface Problem in 2D Strong Form for the Weak Problem Find u H 0 (curl;ω) H(div 0 ;Ω;) such that (µ 1 u)+αu = Q( 1 f) n u = 0 in Ω on Ω where Q is the orthogonal projection operator from [L 2 (Ω)] 2 to H(div 0 ;Ω;) with respect to the inner product (, ) given by (v, w) = (v w)dx Ω

9 A Maxwell Interface Problem in 2D Strong Form for the Weak Problem Find u H 0 (curl;ω) H(div 0 ;Ω;) such that (µ 1 u)+αu = Q( 1 f) n u = 0 in Ω on Ω where Q is the orthogonal projection operator from [L 2 (Ω)] 2 to H(div 0 ;Ω;) with respect to the inner product (, ) given by (v, w) = (v w)dx Ω The appearance of Q is due to the fact that [C c (Ω)] 2 is not a subspace of H 0 (curl;ω) H(div 0 ;Ω;).

10 A Maxwell Interface Problem in 2D Strong Form for the Weak Problem Find u H 0 (curl;ω) H(div 0 ;Ω;) such that (µ 1 u)+αu = Q( 1 f) n u = 0 in Ω on Ω where Q is the orthogonal projection operator from [L 2 (Ω)] 2 to H(div 0 ;Ω;) with respect to the inner product (, ) given by (v, w) = (v w)dx Ω The appearance of Q is due to the fact that [C c (Ω)] 2 is not a subspace of H 0 (curl;ω) H(div 0 ;Ω;). If f = 0, then 1 f belongs to H(div 0 ;Ω;) and hence Q( 1 f)=( 1 f)=f

11 A Maxwell Interface Problem in 2D A Comment Suppose = µ = 1. In the case where f = 0 and α = 0, a common (non-elliptic) weak formulation for Maxwell s equations is to find u H 0 (curl;ω) such that ( u, v)+α(u, v) =(f, v) for all v H 0 (curl;ω), which implies 0 =(f, ζ) ζ H 1 0(Ω) =( u, ( ζ)) + α(u, ζ) ( ζ H 0 (curl;ω)) = α(u, ζ) ζ H 1 0(Ω) and hence u H(div 0 ;Ω). This weak problem can be solved by the H(curl)-conforming edge element methods of Nédélec.

12 A Maxwell Interface Problem in 2D A Comment Suppose = µ = 1. In the case where f = 0 and α = 0, a common (non-elliptic) weak formulation for Maxwell s equations is to find u H 0 (curl;ω) such that for all v H 0 (curl;ω). ( u, v)+α(u, v) =(f, v) There is a good reason to avoid using the space H 0 (curl;ω) H(div 0 ;Ω) because a finite element subspace of H(curl;Ω) H(div;Ω) is necessarily also a subspace of [H 1 (Ω)] 2 and it is also known that H 1 -conforming finite element methods for the Maxwell equations converge in general to wrong solutions if Ω is nonconvex.

13 A Maxwell Interface Problem in 2D A Comment M. Costabel A coercive bilinear form for Maxwell s equations J. Math. Anal. Appl., 1991

14 A Maxwell Interface Problem in 2D A Comment M. Costabel A coercive bilinear form for Maxwell s equations J. Math. Anal. Appl., 1991 Nevertheless it is still desirable to solve the Maxwell equations using an elliptic formulation. M. Costabel and M. Dauge Weighted regularization of Maxwell equations in polyhedral domains Numer. Math., 2002

15 A Maxwell Interface Problem in 2D An Interface Problem Find u H 0 (curl;ω) H(div 0 ;Ω;) such that (µ 1 u, v)+α(u, v) =(f, v) for all v H 0 (curl;ω) H(div 0 ;Ω;). This is an interface problem when µ and are piecewise constant functions. nts µ 1, 1 Ω 1 s µ 2, 2 Ω 2 µ 1, 1 Ω 1 µ 2, 2 µ 3, 3 µ 3, 3 µ 4, 4 Ω 2 Ω 3 Ω 3 Ω 4

16 A Maxwell Interface Problem in 2D An Interface Problem Find u H 0 (curl;ω) H(div 0 ;Ω;) such that (µ 1 u, v)+α(u, v) =(f, v) for all v H 0 (curl;ω) H(div 0 ;Ω;). This is an interface problem when µ and are piecewise constant functions. M. Costabel, M. Dauge and S. Nicaise Singularities of Maxwell interface problems Math. Model. Numer. Anal., 1999

17 A Maxwell Interface Problem in 2D An Interface Problem Find u H 0 (curl;ω) H(div 0 ;Ω;) such that (µ 1 u, v)+α(u, v) =(f, v) for all v H 0 (curl;ω) H(div 0 ;Ω;). This is an interface problem when µ and are piecewise constant functions. s µ 2, 2 µ 1, 1 Ω 2 Ω 1 µ 3, 3 µ 4, 4 Ω 3 Ω 4 1 = = = 1 4 = 10 4 Regularity u [H t (Ω j )] 2 for t <

18 A Maxwell Interface Problem in 2D An Interface Problem Find u H 0 (curl;ω) H(div 0 ;Ω;) such that (µ 1 u, v)+α(u, v) =(f, v) for all v H 0 (curl;ω) H(div 0 ;Ω;). This is an interface problem when µ and are piecewise constant functions. nts µ 1, 1 Ω 1 µ 2, 2 µ 3, 3 Ω 2 Ω 3 1 = 1/350 2 = 1 3 = 1/350 Regularity u [H t (Ω j )] 2 for t <

19 Reduction to Scalar Problems

20 Reduction to Scalar Problems Helmholtz/Hodge Decomposition Every v H(div 0 ;Ω;) has a unique representation v = 1 φ + m j=1 c j ψ j, where φ H 1 (Ω), (φ, 1) =0, ψ 1,...,ψ m are -harmonic functions, and m + 1 equals the number of components of Ω.

21 Reduction to Scalar Problems Helmholtz/Hodge Decomposition Every v H(div 0 ;Ω;) has a unique representation v = 1 φ + m j=1 c j ψ j, where φ H 1 (Ω), (φ, 1) =0, ψ 1,...,ψ m are -harmonic functions, and m + 1 equals the number of components of Ω. m = 0 m = 1 m = 2 Γ 0 = outer boundary Γ j = inner boundary (1 j m)

22 Reduction to Scalar Problems Helmholtz/Hodge Decomposition ψ 1,...,ψ m are determined by the scalar elliptic problems ( ψ j, v) =0 v H 1 0(Ω) ψ j Γ0 = 0 ψ j Γk = δ jk = 1 j = k 0 j = k for 1 k m These are Dirichlet boundary value problems for a second order scalar elliptic equation if is smooth. (They are interface problems if is a piecewise constant function.)

23 Reduction to Scalar Problems Helmholtz/Hodge Decomposition ψ 1,...,ψ m are determined by the scalar elliptic problems ( ψ j, v) =0 v H 1 0(Ω) ψ j Γ0 = 0 ψ j Γk = δ jk = 1 j = k 0 j = k for 1 k m These are Dirichlet boundary value problems for a second order scalar elliptic equation if is smooth. (They are interface problems if is a piecewise constant function.) Note that ψ k H 0 (curl;ω) H(div 0 ;Ω;) for 1 k m.

24 Reduction to Scalar Problems Helmholtz/Hodge Decomposition Every v H(div 0 ;Ω;) has a unique representation v = 1 φ + m j=1 c j ψ j, where φ H 1 (Ω), (φ, 1) =0, ψ 1,...,ψ m are -harmonic functions, and m + 1 equals the number of components of Ω. C. Amrouche, C. Bernardi, M. Dauge and V. Girault Vector potentials in three-dimensional non-smooth domains Math. Methods Appl. Sci., 1998 V. Girault and P.-A. Raviart Finite Element Methods for Navier-Stokes Equations Springer-Verlag, 1986

25 Reduction to Scalar Problems Helmholtz/Hodge Decomposition In particular, since u H 0 (curl;ω) H(div 0 ;Ω;), we have u = 1 φ + m j=1 c j ψ j where φ H 1 (Ω), (φ, 1) =0, ψ 1,...,ψ m are -harmonic functions, and m + 1 equals the number of components of Ω.

26 Reduction to Scalar Problems Helmholtz/Hodge Decomposition In particular, since u H 0 (curl;ω) H(div 0 ;Ω;), we have u = 1 φ + m j=1 c j ψ j where φ H 1 (Ω), (φ, 1) =0, ψ 1,...,ψ m are -harmonic functions, and m + 1 equals the number of components of Ω. Idea Instead of computing u directly, we will compute ψ 1,...,ψ m, c 1,...,c m, and φ that appear in the Helmholtz/Hodge decomposition of u.

27 Reduction to Scalar Problems Helmholtz/Hodge Decomposition In particular, since u H 0 (curl;ω) H(div 0 ;Ω;), we have u = 1 φ + m j=1 c j ψ j where φ H 1 (Ω), (φ, 1) =0, ψ 1,...,ψ m are -harmonic functions, and m + 1 equals the number of components of Ω. Idea Instead of computing u directly, we will compute ψ 1,...,ψ m, c 1,...,c m, and φ that appear in the Helmholtz/Hodge decomposition of u. We need to find the problems that will determine the coefficients c 1,...,c m and the function φ in terms of the given f.

28 Reduction to Scalar Problems A Useful Result Let η, ψ H 1 (Ω) such that ψ is a constant on each component of the boundary of Ω. Then we have ( η, ψ) =0

29 Reduction to Scalar Problems A Useful Result Let η, ψ H 1 (Ω) such that ψ is a constant on each component of the boundary of Ω. Then we have ( η, ψ) =0 Proof ( η, ψ) = = = = Ω Ω j=0 ( η) ( ψ) dx ( η) (nψ) ds m (ψ Γj ) ( η) n ds Γ j m (ψ dη Γj ) Γ j ds ds = 0 j=0

30 Reduction to Scalar Problems Problem for c 1,...,c m Recall u H 0 (curl;ω) H(div 0 ;Ω;) is defined by (µ 1 u, v)+α(u, v) =(f, v) for all v H 0 (curl;ω) H(div 0 ;Ω;).

31 Reduction to Scalar Problems Problem for c 1,...,c m Recall u H 0 (curl;ω) H(div 0 ;Ω;) is defined by (µ 1 u, v)+α(u, v) =(f, v) for all v H 0 (curl;ω) H(div 0 ;Ω;). Let v = ψ k H 0 (curl;ω) H(div 0 ;Ω;).

32 Reduction to Scalar Problems Problem for c 1,...,c m Recall u H 0 (curl;ω) H(div 0 ;Ω;) is defined by (µ 1 u, v)+α(u, v) =(f, v) for all v H 0 (curl;ω) H(div 0 ;Ω;). Let v = ψ k H 0 (curl;ω) H(div 0 ;Ω;). (µ 1 u, ψ k )+α(u, ψ k )=(f, ψ k ) =0

33 Reduction to Scalar Problems Problem for c 1,...,c m Recall u H 0 (curl;ω) H(div 0 ;Ω;) is defined by (µ 1 u, v)+α(u, v) =(f, v) for all v H 0 (curl;ω) H(div 0 ;Ω;). Let v = ψ k H 0 (curl;ω) H(div 0 ;Ω;). (µ 1 u, ψ k )+α(u, ψ k )=(f, ψ k ) =0 α( φ + u = 1 φ + m c j ψ j, ψ k )=(f, ψ k ) j=1 m j=1 c j ψ j

34 Reduction to Scalar Problems Problem for c 1,...,c m Recall u H 0 (curl;ω) H(div 0 ;Ω;) is defined by (µ 1 u, v)+α(u, v) =(f, v) for all v H 0 (curl;ω) H(div 0 ;Ω;). Let v = ψ k H 0 (curl;ω) H(div 0 ;Ω;). (µ 1 u, ψ k )+α(u, ψ k )=(f, ψ k ) =0 α( φ + α m j=1 u = 1 φ + c j ψ j, ψ k )=(f, ψ k ) m ( ψ j, ψ k )c j =(f, ψ k ) j=1 ( φ, ψ k )=0 m j=1 c j ψ j

35 Reduction to Scalar Problems Problem for c 1,...,c m For 1 k m, m ( ψ j, ψ k )c j =(1/α)(f, ψ k ) j=1

36 Reduction to Scalar Problems Problem for c 1,...,c m For 1 k m, m ( ψ j, ψ k )c j =(1/α)(f, ψ k ) j=1 Observe that when m 1, Ω is not simply connected and 0 is a Maxwell eigenvalue. Hence α = 0 when m 1 since we assume α is not a Maxwell eigenvalue.

37 Reduction to Scalar Problems Problem for c 1,...,c m For 1 k m, m ( ψ j, ψ k )c j =(1/α)(f, ψ k ) j=1 Observe that when m 1, Ω is not simply connected and 0 is a Maxwell eigenvalue. Hence α = 0 when m 1 since we assume α is not a Maxwell eigenvalue. The bilinear form (ρ, ζ) ( ρ, ζ) is SPD on the space spanned by the -harmonic functions because ψ 1,...,ψ m vanish on the outer boundary Γ 0 of Ω.

38 Reduction to Scalar Problems Problem for c 1,...,c m For 1 k m, m ( ψ j, ψ k )c j =(1/α)(f, ψ k ) j=1 Observe that when m 1, Ω is not simply connected and 0 is a Maxwell eigenvalue. Hence α = 0 when m 1 since we assume α is not a Maxwell eigenvalue. The bilinear form (ρ, ζ) ( ρ, ζ) is SPD on the space spanned by the -harmonic functions because ψ 1,...,ψ m vanish on the outer boundary Γ 0 of Ω. The coefficients c 1,...,c m in the Helmholtz/Hodge decomposition of u are determined by an m m SPD system defined in terms of the ψ j s (depending only on Ω) and f (given).

39 Reduction to Scalar Problems Problem for φ In order to find the problem that determines the function φ in the Helmholtz/Hodge decomposition u = 1 φ + m j=1 c j ψ j we first need to find the problem that determines the function ξ = µ 1 u

40 Reduction to Scalar Problems Problem for ξ = µ 1 u Recall (µ 1 u)+αu = Q( 1 f) Q = orthogonal projection from [L 2 (Ω)] 2 onto H(div 0 ;Ω;) with respect to the weighted inner product (, ) )

41 Reduction to Scalar Problems Problem for ξ = µ 1 u Recall Hence (µ 1 u)+αu = Q( 1 f) ξ + αu = Q( 1 f) In particular we have ξ H 1 (Ω).

42 Reduction to Scalar Problems Problem for ξ = µ 1 u Recall Hence (µ 1 u)+αu = Q( 1 f) ξ + αu = Q( 1 f) Let ζ H 1 (Ω) be arbitrary. ( ξ, 1 ζ)+α(u, 1 ζ) =(Q( 1 f), 1 ζ)

43 Reduction to Scalar Problems Problem for ξ = µ 1 u Recall Hence (µ 1 u)+αu = Q( 1 f) ξ + αu = Q( 1 f) Let ζ H 1 (Ω) be arbitrary. ( ξ, 1 ζ)+α(u, 1 ζ) =(Q( 1 f), 1 ζ) ( ξ, 1 ζ)+α(u, ζ) =(Q( 1 f), 1 ζ)

44 Reduction to Scalar Problems Problem for ξ = µ 1 u Recall Hence (µ 1 u)+αu = Q( 1 f) ξ + αu = Q( 1 f) Let ζ H 1 (Ω) be arbitrary. ( ξ, 1 ζ)+α(u, 1 ζ) =(Q( 1 f), 1 ζ) ( ξ, 1 ζ)+α(u, ζ) =(Q( 1 f), 1 ζ) ( ξ, 1 ζ)+α( u,ζ)=( 1 f, Q( 1 ζ)) u H 0 (curl;ω)

45 Reduction to Scalar Problems Problem for ξ = µ 1 u Recall Hence (µ 1 u)+αu = Q( 1 f) ξ + αu = Q( 1 f) Let ζ H 1 (Ω) be arbitrary. ( ξ, 1 ζ)+α(u, 1 ζ) =(Q( 1 f), 1 ζ) ( ξ, 1 ζ)+α(u, ζ) =(Q( 1 f), 1 ζ) ( ξ, 1 ζ)+α( u,ζ)=( 1 f, Q( 1 ζ)) ( ξ, 1 ζ)+α(µξ, ζ) =(f, 1 ζ) 1 ζ H(div 0 ;Ω;)

46 Reduction to Scalar Problems Problem for ξ = µ 1 u ξ = µ 1 u H 1 (Ω) satisfies ( ) ( ξ, 1 ζ)+α(µξ, ζ) =(f, 1 ζ) ζ H 1 (Ω)

47 Reduction to Scalar Problems Problem for ξ = µ 1 u ξ = µ 1 u H 1 (Ω) satisfies ( ) ( ξ, 1 ζ)+α(µξ, ζ) =(f, 1 ζ) ζ H 1 (Ω) Since ( ξ, 1 ζ) =( ξ, 1 ζ), this is a second order scalar elliptic problem with the homogeneous Neumann boundary condition, where the right-hand side is defined in terms of f (given). (This is an interface problem if is a piecewise constant function.)

48 Reduction to Scalar Problems Problem for ξ = µ 1 u ξ = µ 1 u H 1 (Ω) satisfies ( ) ( ξ, 1 ζ)+α(µξ, ζ) =(f, 1 ζ) ζ H 1 (Ω) Since ( ξ, 1 ζ) =( ξ, 1 ζ), this is a second order scalar elliptic problem with the homogeneous Neumann boundary condition, where the right-hand side is defined in terms of f (given). For α = 0, the problem ( ) is well-posed under the assumption that α is not a Maxwell eigenvalue and it implies (µξ, 1) =0. (consistent with Ω u dx = 0 for u H 0 (curl;ω))

49 Reduction to Scalar Problems Problem for ξ = µ 1 u ξ = µ 1 u H 1 (Ω) satisfies ( ) ( ξ, 1 ζ)+α(µξ, ζ) =(f, 1 ζ) ζ H 1 (Ω) Since ( ξ, 1 ζ) =( ξ, 1 ζ), this is a second order scalar elliptic problem with the homogeneous Neumann boundary condition, where the right-hand side is defined in terms of f (given). For α = 0, the problem ( ) is well-posed under the assumption that α is not a Maxwell eigenvalue and it implies (µξ, 1) =0. For α = 0 and a simply connected Ω, this is a singular but consistent problem and ξ is determined by ( ) and the constraint (µξ, 1) =0.

50 Reduction to Scalar Problems Problem for φ We can now derive the problem that determines the function φ H 1 (Ω) in the Helmholtz/Hodge decomposition u = 1 φ + m c j ψ j j=1

51 Reduction to Scalar Problems Problem for φ We can now derive the problem that determines the function φ H 1 (Ω) in the Helmholtz/Hodge decomposition u = 1 φ + Let ζ H 1 (Ω) be arbitrary. m j=1 c j ψ j

52 Reduction to Scalar Problems Problem for φ We can now derive the problem that determines the function φ H 1 (Ω) in the Helmholtz/Hodge decomposition u = 1 φ + Let ζ H 1 (Ω) be arbitrary. ( 1 φ, ζ) =(u m j=1 c j ψ j m c j ψ j, ζ) j=1

53 Reduction to Scalar Problems Problem for φ We can now derive the problem that determines the function φ H 1 (Ω) in the Helmholtz/Hodge decomposition u = 1 φ + Let ζ H 1 (Ω) be arbitrary. ( 1 φ, ζ) =(u m j=1 m j=1 =(u, ζ) c j ψ j c j ψ j, ζ) ( ψ j, ζ) =0

54 Reduction to Scalar Problems Problem for φ We can now derive the problem that determines the function φ H 1 (Ω) in the Helmholtz/Hodge decomposition u = 1 φ + Let ζ H 1 (Ω) be arbitrary. ( 1 φ, ζ) =(u m j=1 m j=1 =(u, ζ) =( u,ζ) c j ψ j c j ψ j, ζ) ( ψ j, ζ) =0 u H 0 (curl;ω)

55 Reduction to Scalar Problems Problem for φ We can now derive the problem that determines the function φ H 1 (Ω) in the Helmholtz/Hodge decomposition u = 1 φ + Let ζ H 1 (Ω) be arbitrary. ( 1 φ, ζ) =(u m j=1 m j=1 =(u, ζ) =( u,ζ) =(µξ, ζ) c j ψ j c j ψ j, ζ) ( ψ j, ζ) =0 u H 0 (curl;ω) ξ = µ 1 u

56 Reduction to Scalar Problems Problem for φ φ H 1 (Ω) satisfies ( φ, 1 ζ) =(µξ, ζ) ζ H 1 (Ω) where ξ = µ 1 u.

57 Reduction to Scalar Problems Problem for φ φ H 1 (Ω) satisfies ( φ, 1 ζ) =(µξ, ζ) ζ H 1 (Ω) where ξ = µ 1 u. This is a singular second order scalar elliptic problem with the homogeneous Neumann boundary condition. (It is an interface problem if is a piecewise constant function). It is well-posed because of the compatibility condition (µξ, 1) =0 and then φ is uniquely determined by the constraint (φ, 1) =0.

58 Reduction to Scalar Problems A Numerical Procedure

59 Reduction to Scalar Problems A Numerical Procedure In the case where m 1, i.e., when Ω is not simply connected, solve the following scalar elliptic problems numerically to find approximations ψj of ψ j for 1 j m. This can be carried out once Ω is given. ( ψ j, v) =0 v H 1 0(Ω) ψ j Γ0 = 0 ψ j Γk = δ jk = 1 j = k 0 j = k for 1 k m

60 Reduction to Scalar Problems A Numerical Procedure In the case where m 1, i.e., when Ω is not simply connected, solve the following scalar elliptic problems numerically to find approximations ψj of ψ j for 1 j m. This can be carried out once Ω is given. Recall m ( ψ j, ψ k )c j = 1 α (f,ψ k) j=1 Therefore we can replace ψ j and ψ k by ψ j and ψ k and solve the m m SPD system m ( ψj, ψk)c j = 1 α (f,ψ k) j=1 to obtain the approximations c j of c j (1 j m).

61 Reduction to Scalar Problems A Numerical Procedure Solve the following problem numerically to find an approximation ξ H 1 (Ω) of ξ = µ 1 u. ( ξ, 1 ζ)+α(µξ, ζ) =(f, 1 ζ) ζ H 1 (Ω) (together with the constraint (µξ, 1) =0 if Ω is simply connected and α = 0)

62 Reduction to Scalar Problems A Numerical Procedure Solve the following problem numerically to find an approximation ξ H 1 (Ω) of ξ = µ 1 u. ( ξ, 1 ζ)+α(µξ, ζ) =(f, 1 ζ) ζ H 1 (Ω) (together with the constraint (µξ, 1) =0 if Ω is simply connected and α = 0) If µ = = 1, Ω is convex, ξ is smooth and ξ is obtained by the P 1 finite element method, then ξ ξ L2 (Ω) = O(h 2 ) by a standard duality argument.

63 Reduction to Scalar Problems A Numerical Procedure Solve the following problem numerically to find an approximation ξ H 1 (Ω) of ξ = µ 1 u. ( ξ, 1 ζ)+α(µξ, ζ) =(f, 1 ζ) ζ H 1 (Ω) (together with the constraint (µξ, 1) =0 if Ω is simply connected and α = 0) Solve the following problem numerically to find an approximation φ H 1 (Ω) of φ. ( φ, 1 ζ) =(µξ,ζ) ζ H 1 (Ω) (φ, 1) =0 ( φ, 1 ζ) =(µξ, ζ)

64 Reduction to Scalar Problems A Numerical Procedure The approximation u of the solution u of Maxwell s equations is given by m u = 1 φ + c j ψj. j=1 Solve m scalar Dirichlet problems to compute ψj. Solve an m m SPD system to compute c j. Solve a scalar Neumann problem to compute ξ (which approximates ξ = µ 1 u). Solve a scalar Neumann problem to compute φ.

65 Reduction to Scalar Problems A Numerical Procedure The approximation u of the solution u of Maxwell s equations is given by m u = 1 φ + c j ψj. j=1 Solve m scalar Dirichlet problems to compute ψj. Solve an m m SPD system to compute c j. Solve a scalar Neumann problem to compute ξ (which approximates ξ = µ 1 u). Solve a scalar Neumann problem to compute φ. These scalar elliptic problems can be solved by many numerical methods.

66 P 1 Finite Element Methods

67 P 1 Finite Element Methods A Standard P 1 Finite Element Method We consider homogeneous media with µ = = 1. The scalar elliptic problems are solved by the P 1 finite element method on uniform grids consisting of isosceles rightangled triangles with diameter 2h. a piecewise linear approximation ξ h of ξ = u. a piecewise constant approximation u h of u given by u h = φ h + m c j,h ψ j,h j=1

68 P 1 Finite Element Methods A Standard P 1 Finite Element Method We consider homogeneous media with µ = = 1. The scalar elliptic problems are solved by the P 1 finite element method on uniform grids consisting of isosceles rightangled triangles with diameter 2h. a piecewise linear approximation ξ h of ξ = u. a piecewise constant approximation u h of u given by u h = φ h + m j=1 A standard error analysis shows that c j,h ψ j,h u ξ h L2 (Ω) + u u h L2 (Ω) Ch β f L2 (Ω) where β = min 1, (π/ω) and ω is the maximum interior angle of Ω.

69 P 1 Finite Element Methods An Example for the Unit Square Ω=(0, 1) (0, 1) α = 1 N = number of DOFs h 2 sin πx2 Exact Solution u = sin πx 1 We compare the performance of the P 1 finite element method and the performance of the lowest order edge element method (from Nédélec s first family).

70 P 1 Finite Element Methods An Example for the Unit Square Ω=(0, 1) (0, 1) α = 1 N = number of DOFs h 2 sin πx2 Exact Solution u = sin πx 1 The errors for the approximations of u and u by the edge element method are both O(h) =O(N 1/2 ). For the P 1 finite element method, the error for the approximation of u is O(h) =O(N 1/2 ), while the error for the approximation of ξ = u is O(h 2 )=O(N 1 ).

71 P 1 Finite Element Methods An Example for the Unit Square 10 0 u u l, l, u u l Nd,!(u ul Nd ) u u uniform l l uniform Nd u u l uniform 1 1 1/2 1!(u u l Nd ) uniform N

72 P 1 Finite Element Methods An Example for an L-Shaped Domain Ω=( 1, 1) 2 \ [0, 1] 2 Exact solution u = r 2/3 cos 2 3 θ π φ(x) 3 (r,θ)=polar coordinates at the reentrant corner φ(x) =(1 x 2 1 )2 (1 x 2 2 )2

73 P 1 Finite Element Methods An Example for an L-Shaped Domain Ω=( 1, 1) 2 \ [0, 1] 2 Exact solution u = r 2/3 cos 2 3 θ π φ(x) 3 (r,θ)=polar coordinates at the reentrant corner φ(x) =(1 x 2 1 )2 (1 x 2 2 )2 u has the correct singularity at the reentrant corner. ω =(3π)/2 and β = 2/3

74 P 1 Finite Element Methods An Example for an L-Shaped Domain h u ξ h L2 f L2 Order h u u h L2 f L2 Order α = 1 1/8 6.77E /8 1.06E / E / E / E / E / E / E / E / E / E / E The order of convergence for u h is approaching β = 2/3. The order of convergence for ξ h is 4/3 because ξ = u H 5/3 (Ω) instead of just in H 1 (Ω) (the case under the assumption that f [L 2 (Ω)] 2 ).

75 P 1 Finite Element Methods An Example for an L-Shaped Domain h u ξ h L2 f L2 Order h u u h L2 f L2 Order α = 0 1/8 1.12E /8 1.35E / E / E / E / E / E / E / E / E / E / E h u ξ h L2 f L2 Order h u u h L2 f L2 Order α = 1 1/8 3.57E /8 3.19E / E / E / E / E / E / E / E / E / E / E

76 P 1 Finite Element Methods An Example for an L-Shaped Domain u u l, ξ ξ l, u u l Nd, (u ul Nd ) u u l uniform ξ ξ uniform l Nd u u uniform l Nd (u u ) uniform l 1 1/3 2/ N

77 P 1 Finite Element Methods An Example for a Doubly Connected Domain Ω=(0, 4) 2 \ [1, 3] 2 Right-hand side 1 + x1 f = x 2 if x 1 < x 2 and 3 < x 1 < 4, otherwise.

78 P 1 Finite Element Methods An Example for a Doubly Connected Domain Ω=(0, 4) 2 \ [1, 3] 2 Right-hand side 1 + x1 f = x 2 if x 1 < x 2 and 3 < x 1 < 4, otherwise. The piecewise constant u h is given by ω =(3π)/2 and β = 2/3 u h = φ h + c h ψ h

79 P 1 Finite Element Methods An Example for a Doubly Connected Domain h u ξ h L2 f L2 Order c h Order u u h L2 f L2 Order α = 1 1/4 1.03E E /8 4.04E E / E E / E E / E E The order of convergence for u h is β = 2/3. The order of convergence for c h is β = 7/6. The order of convergence for ξ h is higher than β because f is piecewise smooth.

80 P 1 Finite Element Methods An Example for a Doubly Connected Domain h u ξ h L2 f L2 Order c h Order u u h L2 f L2 Order α = 1 1/4 1.03E E /8 4.04E E / E E / E E / E E h u ξ h L2 f L2 Order c h Order u u h L2 f L2 Order α = 1 1/4 1.72E E /8 5.28E E / E E / E E / E E

81 P 1 Finite Element Methods An Adaptive P 1 Finite Element Method

82 P 1 Finite Element Methods An Adaptive P 1 Finite Element Method The adaptive algorithm uses a residual-based error estimator η R for u u h L2 (Ω). η 2 R = T T h h 2 Tξ h 2 L 2 (T) + e E h h e [[ φ h ]] t e 2 L 2 (e) + m h e [[ ψ j,h ]] n e 2 L 2 (e) j=1 e E i h E h = the set of the edges of T h E i h = the set of the edge of T h interior to Ω [[ ]] = jump across the edge t e = unit tangent of the edge e n e = unit normal of e

83 P 1 Finite Element Methods An Adaptive P 1 Finite Element Method The adaptive algorithm uses a residual-based error estimator η R for u u h L2 (Ω). η 2 R = T T h h 2 Tξ h 2 L 2 (T) + e E h h e [[ φ h ]] t e 2 L 2 (e) + m h e [[ ψ j,h ]] n e 2 L 2 (e) j=1 e E i h It follows from a standard argument that u u h L2 (Ω) C η R + ξ ξ h L2 (Ω) h.o.t.

84 P 1 Finite Element Methods An Adaptive P 1 Finite Element Method Adaptive Loop Solve Estimate Mark Refine Solve: We apply the P 1 finite element methods based on Helmholtz/Hodge Decomposition. Estimate: We evaluate a local error indicator based on η R. Mark: We mark the triangles using the bulk criterion of Dörfler. Refine: We refine the marked triangles followed by a closure algorithm that removes the hanging nodes.

85 P 1 Finite Element Methods An Example for an L-Shaped Domain Ω=( 1, 1) 2 \ [0, 1] 2 α = 1 Exact solution u = r 2/3 cos 2 3 θ π φ(x) 3 (r,θ)=polar coordinates at the reentrant corner φ(x) =(1 x 2 1 )2 (1 x 2 2 )2

86 P 1 Finite Element Methods An Example for an L-Shaped Domain 10 1 u u l, ξ ξ l, u u l Nd, (u ul Nd ) u u l for η R adaptive ξ ξ l for η R adaptive 1 1 1/2 u u l Nd adaptive (u u l Nd ) adaptive N

87 P 1 Finite Element Methods An Example for a Doubly Connected Domain Ω=(0, 4) 2 \ [1, 3] 2 Right-hand side 1 + x1 f = x 2 if x 1 < x 2 and 3 < x 1 < 4, otherwise.

88 P 1 Finite Element Methods An Example for a Doubly Connected Domain /3 η R, u u l, ξ ξ l / / u u l uniform η R adaptive 1 u u l for η R adaptive 1 ξ ξ for η adaptive l R ξ ξ l uniform N

89 A Multigrid Method

90 A Multigrid Method A Maxwell Interface Problem We consider heterogeneous media where µ and are piecewise constant functions. The weak problem of finding u H 0 (curl;ω) H(div 0 ;Ω;) such that (µ 1 u, v)+α(u, v) =(f, v) for all v H 0 (curl;ω) H(div 0 ;Ω;) becomes an interface problem. For simplicity we assume Ω is simply connected so that u = 1 φ where φ H 1 (Ω) satisfies (φ, 1) =0.

91 A Multigrid Method A Maxwell Interface Problem Helmholtz/Hodge Decomposition u = 1 φ where φ H 1 (Ω) and (φ, 1) =0 Problem for φ ( φ, 1 ζ) =(µξ, ζ) ζ H 1 (Ω) Problem for ξ = µ 1 u ( ξ, 1 ζ)+α(µξ, ζ) =(f, 1 ζ) ζ H 1 (Ω)

92 A Multigrid Method A Maxwell Interface Problem Helmholtz/Hodge Decomposition u = 1 φ where φ H 1 (Ω) and (φ, 1) =0 Problem for φ ( φ, 1 ζ) =(µξ, ζ) ζ H 1 (Ω) Problem for ξ = µ 1 u ( ξ, 1 ζ)+α(µξ, ζ) =(f, 1 ζ) ζ H 1 (Ω) These are interface problems for the Laplace operator.

93 A Multigrid Method Laplace Interface Problems S. Nicaise Polygonal Interface Problems Peter Lang, 1993 S. Nicaise and A. Sändig General Interface Problems-I, II Math. Methods. Appl. Sci., 1994

94 A Multigrid Method The Laplace Interface Problem for φ Find φ H 1 (Ω) such that (φ, 1) =0 and ( φ, 1 ζ) =(µξ, ζ) ζ H 1 (Ω) Strong Form φ j = j µ j ξ j in Ω j φ n = 0 on Ω nts µ 1, 1 Ω 1 φ = φ + on interface µ 2, 2 µ 3, 3 1 φ n = 1 + φ + n on interface Ω 2 Ω 3 φ j = φ Ωj ξ j = ξ Ωj

95 A Multigrid Method The Laplace Interface Problem for φ Singularity at the Cross Point φ = r λ Θ(θ) (Θ = Θ j in Ω j ) (3π)/2 0 1 Θ 2 (θ)dθ = 1

96 A Multigrid Method The Laplace Interface Problem for φ A Sturm-Liouville Problem for Θ(θ) Θ j = λ 2 Θ j 1 j 3 at θ = π/2 Θ 1 = 0 at θ = π Θ 1 =Θ 2 at θ = 3π/2 Θ 2 =Θ 3 at θ = 2π Θ 3 = Θ 1 = 1 2 Θ Θ 2 = 1 3 Θ 3 ts µ 1, 1 Ω 1 µ 2, 2 µ 3, 3 Ω 2 Ω 3

97 A Multigrid Method The Laplace Interface Problem for φ A Sturm-Liouville Problem for Θ(θ) Θ j = λ 2 Θ j 1 j 3 at θ = π/2 Θ 1 = 0 at θ = π Θ 1 =Θ 2 at θ = 3π/2 Θ 2 =Θ 3 at θ = 2π Θ 3 = Θ 1 = 1 2 Θ Θ 2 = 1 3 Θ 3 ts µ 1, 1 Ω 1 µ 2, 2 µ 3, 3 Ω 2 Ω 3 λ = = 3 = 1/350 2 = 1

98 A Multigrid Method The Laplace Interface Problem for φ Singularity at the Cross Point φ = r λ Θ(θ) (Θ = Θ j in Ω j ) Regularity of u u = 1 φ On each subdomain Ω j, for any t <λ. u belongs to [H t (Ω j )] 2

99 A Multigrid Method The Laplace Interface Problem for φ Singularity at the Cross Point φ = r λ Θ(θ) (Θ = Θ j in Ω j ) Regularity of u u = 1 φ On each subdomain Ω j, u belongs to [H t (Ω j )] 2 for any t <λ. For the case where 1 = 3 = 1/350 and 2 = 1, u belongs to [H t (Ω j )] 2 for any t <

100 A Multigrid Method P 1 Finite Element Method The strong singularity of u around the cross point renders the standard P 1 finite element method ineffective.

101 A Multigrid Method P 1 Finite Element Method The strong singularity of u around the cross point renders the standard P 1 finite element method ineffective. There are many ways to use the information on the singularity to improve the convergence rate (singular function method, dual singular function method, etc.).

102 A Multigrid Method P 1 Finite Element Method The strong singularity of u around the cross point renders the standard P 1 finite element method ineffective. There are many ways to use the information on the singularity to improve the convergence rate (singular function method, dual singular function method, etc.). We will use a multigrid approach that is based on the singular function representations for the solutions of the Laplace interface problems and extraction formulas for the stress intensity factors. S.C.B. and L.-Y. Sung Multigrid methods for the computation of singular solutions and stress intensity factor III: interface singularities Comp. Methods. Appl. Mech. Engrg., 2003

103 A Multigrid Method A Multigrid Method for the Laplace Interface Problem Find ζ H 1 (Ω) such that (ζ,1) =0 and ( ζ, 1 v) =(g, v) v H 1 (Ω) where g L 2 (Ω). (This is the interface problem for φ when g = µξ.)

104 A Multigrid Method A Multigrid Method for the Laplace Interface Problem Find ζ H 1 (Ω) such that (ζ,1) =0 and ( ζ, 1 v) =(g, v) v H 1 (Ω) where g L 2 (Ω). Singular Function Representation ζ = κψ+ζ R ζ R Ωj H 2 (Ω j ) Ψ(r,θ)=r λ Θ(θ)ρ(r) λ and Θ are determined by the Sturm- Liouville problem and ρ is a smooth cutoff function that equals 1 near r = 0. nts µ 1, 1 Ω 1 µ 2, 2 µ 3, 3 Ω 2 Ω 3

105 A Multigrid Method A Multigrid Method for the Laplace Interface Problem Find ζ H 1 (Ω) such that (ζ,1) =0 and ( ζ, 1 v) =(g, v) v H 1 (Ω) where g L 2 (Ω). Singular Function Representation where κ = 1 2λ Ω [gψ + 1 ζ Ψ ]dx Ψ (r,θ)=r λ Θ(θ)ρ(r) is the dual singular function. ζ = κψ+ζ R nts µ 1, 1 Ω 1 µ 2, 2 µ 3, 3 Ω 2 Ω 3

106 A Multigrid Method A Multigrid Method for the Laplace Interface Problem Let T 0, T 1,... be triangulations of Ω obtained by uniform subdivisions and V k be the corresponding P 1 finite element space. On each level we look for an approximation of ζ in the form of ζ k = κ k Ψ+w k where w k V k.

107 A Multigrid Method A Multigrid Method for the Laplace Interface Problem On level-0 we set κ 0 = 0 and compute w 0 V 0 such that (w 0, 1) =0 and ( w 0, 1 v) =(g, v) v V 0 ( ζ, 1 v) =(g, v) Then we set κ 1 = 1 2λ Ω [gψ + 1 w 0 Ψ ]dx κ = 1 2λ Ω [gψ + 1 ζ Ψ ]dx

108 A Multigrid Method A Multigrid Method for the Laplace Interface Problem On level-k for k 1, we have an approximation κ k for the stress intensity factor from level-(k 1) and we compute w k V k such that ( (κ k Ψ+w k ), 1 v) =(g, v) v V k or equivalently ( ζ, 1 v) =(g, v) ( w k, 1 v) =(g, v) κ k ( Ψ, 1 v) v V k Then we set κ k+1 = 1 2λ Ω [gψ + 1 (κ k Ψ+w k ) Ψ ]dx

109 A Multigrid Method A Multigrid Method for the Laplace Interface Problem The idea is that, as we move up the levels, the effect of the singular part of the solution is lessened and we end up approximating the smooth part of the solution, which improves the rate of convergence of the P 1 finite element method. ( w k, 1 v) =(g, v) κ k ( Ψ, 1 v) =( 1 (κψ+ζ R ), v) κ k ( Ψ, 1 v) =( ζ R, 1 v)+(κ κ k )( ψ, 1 v) small

110 A Multigrid Method A Multigrid Method for the Laplace Interface Problem The idea is that, as we move up the levels, the effect of the singular part of the solution is lessened and we end up approximating the smooth part of the solution, which improves the rate of convergence of the P 1 finite element method. We can solve the k-th level problem using a fixed number of iterations of a multigrid algorithm, with the solution w k 1 V k 1 from the previous level as the initial guess. The resulting algorithm is a full multigrid algorithm where the right-hand side changes from level to level.

111 A Multigrid Method A Multigrid Method for the Laplace Interface Problem The idea is that, as we move up the levels, the effect of the singular part of the solution is lessened and we end up approximating the smooth part of the solution, which improves the rate of convergence of the P 1 finite element method. We can solve the k-th level problem using a fixed number of iterations of a multigrid algorithm, with the solution w k 1 V k 1 from the previous level as the initial guess. The resulting algorithm is a full multigrid algorithm where the right-hand side changes from level to level. Since we are computing the smooth part of the solution, the convergence of κ k Ψ+w k to κψ+ζ R is of order O(h k ), and since we are using a multigrid algorithm, the computational cost is proportional to the dimension of V k.

112 A Multigrid Method A Multigrid Method for the Laplace Interface Problem Find ξ H 1 (Ω) such that ( ξ, 1 ζ)+α(µξ, ζ) =(f, 1 ζ) ζ H 1 (Ω) where (µξ, 1) =0 if α = 0. Under the assumption that f is piecewise H 1, we can write ξ = κψ+ξ R where ξ R H 2 (Ω) and Ψ=r λ Θ(θ)ρ(r). Moreover, we have κ = 1 ( 1 f αµξ)ψ + 1 ξ Ψ dx 2λ Ω 1 2λ Ω 1 (n f)ψ ds + Γ [[ 1 n f]]ψ ds

113 A Multigrid Method An Example for an L-Shaped Domain Ω=( 1, 1) 2 \ [0, 1] 2 1 = 1/50, 2 = 1, 3 = 1/50 µ 1 = 50, µ 2 = 1, µ 3 = 50 λ = nts µ 1, 1 Ω 1 µ 2, 2 µ 3, 3 Ω 2 Ω 3 Find u H 0 (curl;ω) H(div 0 ;Ω;) such that (µ 1 u, v) =(f, v) for all v H 0 (curl;ω) H(div 0 ;Ω;). (α = 0) Exact solution u = 1 r λ Θ(θ)(1 x1 4)2 (1 x2 4)2

114 A Multigrid Method An Example for an L-Shaped Domain For this example we have φ = 1Ψ+φ R and ξ = 0Ψ+ξ R We solve the Laplace interface problem for ξ and the Laplace interface problem for φ by full multigrid algorithms that exploit the extraction formulas.

115 A Multigrid Method An Example for an L-Shaped Domain For this example we have φ = 1Ψ+φ R and ξ = 0Ψ+ξ R Multigrid solution ξ k = κ ξ k Ψ+w k (h 0 = 1/12) k ξ R w k H 1 (Ω) κ ξ k

116 A Multigrid Method An Example for an L-Shaped Domain For this example we have φ = 1Ψ+φ R and ξ = 0Ψ+ξ R Multigrid solution φ k = κ φ k Ψ+v k (h 0 = 1/12) k φ R v k H 1 (Ω) κ φ k

117 A Multigrid Method An Example for an L-Shaped Domain Ω=( 1, 1) 2 \ [0, 1] 2 1 = 1/350, 2 = 1, 3 = 1/350 µ 1 = µ 2 = µ 3 = 1 λ = nts µ 1, 1 Ω 1 µ 2, 2 µ 3, 3 Ω 2 Ω 3 Find u H 0 (curl;ω) H(div 0 ;Ω;) such that (µ 1 u, v) =(f, v) for all v H 0 (curl;ω) H(div 0 ;Ω;). (α = 0) Exact solution ξ(= µ 1 u) =Ψ

118 A Multigrid Method An Example for an L-Shaped Domain For this example we have ξ =Ψ (ξ R = 0) Multigrid solution for ξ is ξ k = κ k Ψ+w k k κ k w k H 1 (Ω) E E E E E E E 05 h 0 = 1/24

119 A Multigrid Method An Example for an L-Shaped Domain Multigrid solution for u k u k+1 u k L2 (Ω,) order h 0 = 1/24

120 Concluding Remarks We have investigated P 1 finite element methods for two dimensional Maxwell problems that are based on a reduction to scalar elliptic problems through the Helmholtz/Hodge decomposition.

121 Concluding Remarks We have investigated P 1 finite element methods for two dimensional Maxwell problems that are based on a reduction to scalar elliptic problems through the Helmholtz/Hodge decomposition. There are many off-the-shelf solvers for these scalar elliptic problems. For example, when α is a large negative number, the elliptic problem for ξ can be solved by any good numerical method for the Helmholtz equation.

122 Concluding Remarks We have investigated P 1 finite element methods for two dimensional Maxwell problems that are based on a reduction to scalar elliptic problems through the Helmholtz/Hodge decomposition. There are many off-the-shelf solvers for these scalar elliptic problems. For example, when α is a large negative number, the elliptic problem for ξ can be solved by any good numerical method for the Helmholtz equation. We have only discussed problems with the perfectly conducting boundary condition. But this approach is also applicable to problems with the impedance boundary condition.

123 Concluding Remarks In this approach the permeability µ only appears in a lower order term. Therefore we can also handle interface problems where the permittivity has a fixed sign but the sign of µ is arbitrary. s 2 > 0 µ 2 < 0 1 > 0 µ 1 > 0 3 > 0 4 > 0 µ 3 > 0 µ 4 < 0

124 Acknowledgement