The solution of time harmonic wave equations using complete families of elementary solutions

Transcription

1 The solution of time harmonic wave equations using complete families of elementary solutions Peter Monk Department of Mathematical Sciences University of Delaware Newark, DE USA Joint research with Tomi Huttunen and Teemu Luostari, Kuopio, Finland. Research supported in part by a grant from AFOSR

2 Linear time harmonic wave equations Goal: find numerical solutions of common time harmonic wave equations. Examples include: Helmholtz Equation: u + κ 2 n 2 u = 0.

3 Linear time harmonic wave equations Goal: find numerical solutions of common time harmonic wave equations. Examples include: Helmholtz Equation: Maxwell s equations: u + κ 2 n 2 u = 0. µ 1 u κ 2 ɛu = 0

4 Linear time harmonic wave equations Goal: find numerical solutions of common time harmonic wave equations. Examples include: Helmholtz Equation: Maxwell s equations: Navier s equation: u + κ 2 n 2 u = 0. µ 1 u κ 2 ɛu = 0 (λ + 2µ) ( u) µ ( u) + κ 2 ρu = 0 together with appropriate boundary conditions.

5 Examples of solutions Suppose the coefficients are constant: Helmholtz Equation: u + κ 2 n 2 u = 0, u = exp(iκnx d), d = 1.

6 Examples of solutions Suppose the coefficients are constant: Helmholtz Equation: u + κ 2 n 2 u = 0, u = exp(iκnx d), d = 1. Maxwell s equations: µ 1 u κ 2 ɛu = 0, u = p exp(iκ µɛx d), d = 1, p d.

7 Examples of solutions Suppose the coefficients are constant: Helmholtz Equation: u + κ 2 n 2 u = 0, u = exp(iκnx d), d = 1. Maxwell s equations: µ 1 u κ 2 ɛu = 0, u = p exp(iκ µɛx d), d = 1, p d. Navier s equation: (λ + 2µ) ( u) µ ( u) + κ 2 ρu = 0 u = p exp(iκ s x d) + q exp(iκ p x d) with p d, q d, κ p = ω ρ/(λ + 2µ), κ s = ω ρ/µ

8 Helmholtz equation Example: find numerical solutions of the Helmholtz equation: u + κ 2 n 2 u = 0.

9 Helmholtz Examples [Huttunen] 20 θ = x (mm) y (mm)

10 Model Acoustic Problem Given a bounded domain Ω R N, N = 2, 3, find u such that u + κ 2 n 2 u = 0 in Ω u iκu n = g on Γ := Ω. Assume the wave number κ is real. Important parameter: the wavelength (of a plane wave solution) is 2π/(κn), so large κ corresponds to short wavelength.

11 Standard Finite Elements Weak form: find u h S h H 1 (Ω) such that u h ξ κ 2 n 2 u h ξ dv iκ u h ξ da = Ω for all ξ S h. This gives rise to a matrix problem (S κ 2 M iκb) u = g Γ Ω gξ da The matrix is complex symmetric (not Hermitian) with eigenvalues having positive and negative real parts if k is large so The linear system is hard to solve efficiently. and the matrix has to be large...

12 High frequency: many DoF The discretization must become finer at high wave number κ Suppose n is constant. Melenk and Sauter 1 show, under several non-restrictive assumptions, that for a conforming hp finite element approximation of our model boundary value problem, a quasi-optimal error estimate holds with constant independent of κ if h p = δ and p = 1 + C ln(κ) nκ 1 Vienna University: ASC Report No. 31/2009, Wave-Number Explicit Convergence Analysis for Galerkin Discretizations of the Helmholtz Equation, J.M. Melenk and S. Sauter

13 High frequency: many DoF The discretization must become finer at high wave number κ Suppose n is constant. Melenk and Sauter 1 show, under several non-restrictive assumptions, that for a conforming hp finite element approximation of our model boundary value problem, a quasi-optimal error estimate holds with constant independent of κ if Note h p = δ and p = 1 + C ln(κ) nκ #DoF per wavelength = (2π/(κn))(p/h) = 2π/δ Best case: we need at least O(κ N ) degrees of freedom. 1 Vienna University: ASC Report No. 31/2009, Wave-Number Explicit Convergence Analysis for Galerkin Discretizations of the Helmholtz Equation, J.M. Melenk and S. Sauter

14 Other methods We will now look at a DG method using non-standard shape functions called the Ultra Weak Variational Formulation 2 (UWVF) For an h version analysis of plane wave DG methods in general (but not UWVF) see [ 3 ] below. For other methods using special basis functions see for example Trefftz methods, PUFEM 4, Discontinuous Enrichment 5, Integral equations 6,... 2 O. Cessenat and B. Després, Application of the ultra-weak variational formulation of elliptic PDEs to the 2-dimensional Helmholtz problem. SIAM J. Numer. Anal., 35:255 99, C. Gittelson, R. Hiptmair, and I. Perugia. Plane wave discontinuous Galerkin methods. ESAIM: Mathematical Modeling and Numerical Analysis, 43: , J.M. Melenk. On generalized finite element methods. PhD thesis, University of Maryland, College Park, MD, M. Amara, R. Djellouli, and C. Farhat. Convergence analysis of a discontinuous Galerkin method with plane waves and Lagrange multipliers for the solution of Helmholtz problems. SIAM J. Numer. Anal., 47: , F. Ecevit and F. Reitich. Analysis of multiple scattering iterations for high-frequency scattering problems. I: the two-dimensional case. Numer. Math., 114: , 2009.

15 Time harmonic symmetric hyperbolic system 7 Given a domain Ω R N, N = 2, 3, and temporal frequency ω, we seek u = u(x) C m (so U(x, t) = R(u(x) exp( iωt))) such that N iωau + A j u xj + Bu = 0 in Ω j=1 where all the matrices are real symmetric. Also A(x) is positive definite, B(x) positive semidefinite and we assume that the A j are constant. Boundary condition: (D N)u = g on Γ := Ω where D = N j=1 A jn j and n is the unit outward normal to Ω. N is chosen to enforce the desired boundary condition. 7 G.Gabard. Discontinuous Galerkin methods with plane waves for time-harmonic problems. J. Comput. Phys., 225(2): , 2007.

16 Example: Helmholtz s equation in R 2 We seek to approximate the function p that satisfies p + (ω/c) 2 n(x)p = 0 in Ω and ( ) c p c iω n + σp = Q p iω n σp + g on Γ Introduce v such that i(ω/c)v = p. Define k = ω/c. Then if u = (p, v 1, v 2 ) T, we see that u satisfies the hyperbolic system with B = 0 and A 1 = A = n(x)/c /c /c, A 2 =, ,

17 Boundary conditions for the 2D example The boundary condition matrix D = n 1 A 1 + n 2 A 2 0 n 1 n 2 D = n 1 0 0, n Let L ± = 1 2σ (±σ, n 1, n 2 ). Note that L ± u = (±σp n 1 v 1 n 2 v 2 )/ 2σ corresponds to an impedance boundary condition. We set D + = (L + ) T L + and D = (L ) T L so that D = D + + D On the boundary Γ we can choose N = D 2(D + QL (L + ) T ), Q 1.

18 The Mesh Approximate the domain Ω by a tetrahedral finite element mesh consisting of N h tetrahedra K m, m = 1,, N h of maximum diameter h. N iωau + A j u xj + Bu = 0 j=1 in Ω

19 Derivation of the UWVF Let K denote an element in the mesh with faces f K and outward normal n K. Let u K = u K. Multiply the equation by a smooth function φ K on K and integrate by parts: iω + K f K Au K φ K dv f K N j=1 D K u K φ K da = 0 ( A j u φ K x j + Bu φ K ) dv Here the overbar represents complex conjugation, and we use D K to denote D on K.

20 Derivation continued Splitting the flux D K = D K,+ + D K, and using the boundary condition and symmetry of the matrices we get N u K iωaφ K A j φ K x j + Bφ K dv K + K f K K Γ K j=1 ( (D K,+ u K ) φ K da + f ((D K + N)u K + g) φ K da = 0. f ) (D K, u K ) φ K da

21 Non standard basis Given Y K X K (L 2 ( K )) m we choose φ K to be a solution of the adjoint equation with impedance boundary condition iωaφ K N A j φ K x j + Bφ K = 0 on K j=1 We obtain (D K,+ u K ) φ K da + (D K, u K ) φ K da K K K Γ f K K ((D K + N)u K + g) φ K da = 0. f

22 We can write D K,± = ±L K,± (L K,± ) T and N = D K 2(D K, + QL K, (L K,+ ) T ). Using also L K, = L K,+ we get: (L K,+ ) T u K (L K,+ ) T φ K da K + ((L K,+ ) T u K ) (L K, ) T φ K da K f K K f Q((L K,+ ) T u K + g) (L K, ) T φ K da = 0. Γ K Set Y K = (L K,+ ) T φ K. Define the unknown impedance flux and operator F K : X K X K by X K = (L K,+ ) T u K and F K (Y K ) = (L K, ) T φ K.

23 The UWVF Summarizing, we see that X K X K (L 2 ( K )) m satisfies X K Y K da + X K F K (Y K ) da K K f K K f Q(X K + g) F K (Y K ) da = 0. K Γ for all Y K X K (L 2 ( K )) m and all tetraheda K in the mesh. This is the UWVF of Cessenat and Després before discretization Note that for the acoustic case m = 1, X K = L 2 ( K ), and for electromagnetism m = 3, while X K = L 2 t ( K ). Cessenat and Després prove existence of a unique solution for acoustics and electromagnetism.

24 Useful result Lemma (Isometry Lemma) For all Y K X K, F K (Y K ) L 2 ( K ) YK L 2 ( K ) with equality if B = 0. To see this: let Y K = (L K,+ ) T φ K. Then using the adjoint equation and integration by parts D K φ K φ K da = 2 φ K Bφ K dv. K Splitting D K = D K,+ + D K, and using the definition of F K completes the demonstration. K

25 The discrete UWVF Let Vp K denote a finite dimensional space of p := p(k ) solutions of the adjoint problem on an element K. Let { } Xp K = v X K : v = (L K,+ ) T φ K for some φ K Vp K The discrete problem is to seek X K h X K p such that K X K h YK da + K Γ K f K K f X K h Q(X K h + g) F K (Y K ) da = 0 for all Y K X K p and all elements K. F K (Y K ) da

26 Model Acoustic Problem Given a bounded domain Ω R N, N = 2, 3, find u such that Assume κ is real. u + κ 2 n 2 u = 0 in Ω u iκu n = g on Γ := Ω.

27 2D acoustic plane wave basis Let n K = n K assumed piecewise constant. Using p directions d K m = (cos(θ m ), sin(θ m )) T, θ m = 2πm/p we set { } = span exp(ikn K x d K m), 1 m p V K p A function Y K X K p can then be written Y K = 1 ( ) 1 p + I ym K exp(ikn K x d K 2 ik ν m). K m=1 Then F K (Y K ) is easy to compute since F K (Y K ) = 1 ( ) 1 p I ym K exp(ikn K x d K 2σ ik ν m). K m=1 X K h is expanded in the same way as YK.

28 Some comments on plane wave bases On straight tets/triangles, the UWVF integrals can be computed in a closed form, therefore assembly is rapid and we avoid the need for complicated quadratures (c.f. PUFEM). Ill-conditioning may occur when elements are small or at low frequencies since exp(ikx d) = 1 + ikx d +... Fourier-Bessel functions offer an attractive alternative.

29 Properties of the acoustic UWVF [Cessenat/Després, 2D] If p = 2µ + 1 X X h L 2 (Γ) Chµ 1/2 u C µ+1 (Ω) [Buffa-Monk, 2D] 8 Under the above conditions (with u h reconstructed from X h ), and if Ω is convex u u h L 2 (Ω) C(κ)hµ 1 u C µ+1 (Ω) 8 A. Buffa and P. Monk. Error estimates for the Ultra Weak Variational Formulation of the Helmholtz equation. ESAIM: Mathematical Modeling and Numerical Analysis, 42:925 40, P. Monk and D.Q. Wang. A least squares method for the Helmholtz equation. Comput. Meth. Appl. Mech. Eng., 175:121 36, 1999

30 Properties of the acoustic UWVF [Cessenat/Després, 2D] If p = 2µ + 1 X X h L 2 (Γ) Chµ 1/2 u C µ+1 (Ω) [Buffa-Monk, 2D] 8 Under the above conditions (with u h reconstructed from X h ), and if Ω is convex u u h L 2 (Ω) C(κ)hµ 1 u C µ+1 (Ω) Proof: We relate the UWVF back to an upwind DG method. Then usual DG results gives an estimate in DG norms. Using duality we can then estimate the L 2 norm, 9 loosing a half power of h along the way. 8 A. Buffa and P. Monk. Error estimates for the Ultra Weak Variational Formulation of the Helmholtz equation. ESAIM: Mathematical Modeling and Numerical Analysis, 42:925 40, P. Monk and D.Q. Wang. A least squares method for the Helmholtz equation. Comput. Meth. Appl. Mech. Eng., 175:121 36, 1999

31 Properties of the acoustic UWVF [Cessenat/Després, 2D] If p = 2µ + 1 X X h L 2 (Γ) Chµ 1/2 u C µ+1 (Ω) [Buffa-Monk, 2D] 8 Under the above conditions (with u h reconstructed from X h ), and if Ω is convex u u h L 2 (Ω) C(κ)hµ 1 u C µ+1 (Ω) Proof: We relate the UWVF back to an upwind DG method. Then usual DG results gives an estimate in DG norms. Using duality we can then estimate the L 2 norm, 9 loosing a half power of h along the way. Problem: we we usually fix a grid and use a p version. 8 A. Buffa and P. Monk. Error estimates for the Ultra Weak Variational Formulation of the Helmholtz equation. ESAIM: Mathematical Modeling and Numerical Analysis, 42:925 40, P. Monk and D.Q. Wang. A least squares method for the Helmholtz equation. Comput. Meth. Appl. Mech. Eng., 175:121 36, 1999

32 Much improved analysis [Hiptmair, Moiola, Perugia, 2009] 10 p-version error estimate (using κ-dependent norms): ( ) log(p) r 1/2 κ u u h L 2 (Ω) Chr 1 u r+1,κ,ω p 10 R. Hiptmair, A. Moiola, and I. Perugia, Plane wave discontinuous Galerkin methods for the 2D Helmholtz equation: analysis of the p-version

33 Much improved analysis [Hiptmair, Moiola, Perugia, 2009] 10 p-version error estimate (using κ-dependent norms): ( ) log(p) r 1/2 κ u u h L 2 (Ω) Chr 1 u r+1,κ,ω p [Cessenat, Despres] The discrete problem has the form (D C)x = b where D is Hermitian positive definite and the eigenvalues of D 1 C lie in the closure of the unit disk excluding 1. The condition number of D satisfies cond(d) C h 2p 2 10 R. Hiptmair, A. Moiola, and I. Perugia, Plane wave discontinuous Galerkin methods for the 2D Helmholtz equation: analysis of the p-version

34 Numerical results: 2D mesh refinement 11 We take u(x) = i 4 H(1) 0 (κ x x 0 ) with x 0 is outside the computational domain h = 0.5 h = T. Huttunen, P. Monk, and J.P. Kaipio. Computational aspects of the Ultra Weak Variational Formulation. J. Comput. Phys., 182:27 46, 2002.

35 Results for κ = 20, p = 15 (µ = 7) Mesh size h L 2 (Ω) Error (%) Order cond(d) Order Measured global convergence O(h 7.8 ). Cessenat and Després predict that the condition number of D will increase O(h 12 )

36 Results for κ = 40, p = 21 (µ = 10). Mesh size h L 2 (Ω) Error (%) Order cond(d) Order Measured global convergence O(h 10 ). Cessenat and Després predict that the condition number of D increases O(h 18 ).

37 Typical Electromagnetic Application [Tomi Huttunen] Example: Simulate microwave interaction with wood (part of a non-destructive testing project). A transmitting and receiving antenna are shown.

38 Maxwell s equations: Cavity problem E, H: Electric/magnetic field k : Wave-number ɛ r : Relative permittivity µ r : Relative permeability Ω: Bounded domain ikɛ r E H = 0 in Ω ikµ r H + E = 0 in Ω H n ηe T = Q(H n + ηe T ) 2ηg on Γ = Ω where η > 0 and Q 1.

39 Required continuity between elements Let E k = E Ωk and E j = E Ωj (similarly for H) then if E, H is to be a solution of the Maxwell s equations we need E k n K k = E j n K j and H k n K k = H j n K j on Σ j,k In the Ultra Weak Variational Formulation (UWVF) this is achieved by demanding that impedance data agree on the interfaces, so on Σ j,k E k n K k + η(h k n K k ) n K k = E j n K j + η(h j n K j ) n K j E k n K k η(h k n K k ) n K k = E j n K j η(h j n K j ) n K j where η > 0 is a parameter (function) on Σ j,k (e.g. η = R( µ r /ɛ r )).

40 12 13 Maxwell UWVF Let X k = E k n K k η(h k ) T Ωk and Y k = ξ k n K k η(ψ k ) T Ωk and let F k (Y k ) = ξ k n K k η(ψ k ) T Ωk then, for a tetrahedron surrounded by four other tetrahedra 1 Ω k η X ky k ds = 1 j Σ k,j η X jf k (Y k ) ds. Boundary faces are handled using the boundary condition in place of X j. 12 O. Cessenat. Application d une nouvelle formulation variationnelle aux équations d ondes harmoniques. Problèmes de Helmholtz 2D et de Maxwell 3D. PhD thesis, Université Paris IX Dauphine, T. Huttunen, M. Malinen, and P.B. Monk. Solving Maxwell s equations using the Ultra Weak Variational Formulation. J. Comput. Phys., 223:731 58, 2007.

41 The Discrete UWVF: plane waves For each element Ω k we choose p k directions d j, j = 1,..., p k on the unit sphere [Sloan et al.] and for each direction two polarizations q (1,2) j then define (choosing κ = ɛ r µ r ) X h k = 2 p k s=1 j=1 ( x j,s q s j n K k ηκ ) (d j q s j kµ ) T r exp(iκd j x) The test function is, for 1 r p k, 1 t 2 ( Yk h = q t r nk k ηκ ) (d r q t r kµ ) T exp(iκd r x) r Ωk In this case F k (Yk h ) is easy to compute: ( F k (Yk h ) = q t r nk k ηκ ) (d r q t r kµ ) T exp(iκd r x) r Ωk Ωk

42 Results for wood at 5GHz Using a Perfectly Matched Layer, and keeping a fixed mesh but varying the number of directions with the local wavelength. E x E y E x E y z (m) z (m) z (m) z (m) y (m) 70 db y (m) 70 db y (m) 70 db y (m) 70 db E z 0 E 0 E z 0 E 0 z (m) z (m) z (m) z (m) y (m) 70 db y (m) 70 db y (m) 70 db y (m) 70 db ɛ r = 3 ɛ r = 8

43 Scattering from a sphere Sphere of radius 0.25 inside cube [.5,.5] 3, κ = 100, ɛ = µ = 1 (λ = 0.06). PML width 0.1 (uses 3,474,770 degrees of freedom). Real E 1, UWVF Real E 2, UWVF Real E 2, UWVF

44 Iterative solution The UWVF linear system can be solved by simple iterative scheme. We use BiCGStab Bi CGStab Relative residual Iteration BiCGStab convergence for a problem having 3,474,770 degrees of freedom using a 24 processor cluster (2.8GHz P4, 48Gb memory total, 1000BaseT). Solution time is 451s using 25.3 GB memory (109 iterations).

45 Elastic UWVF: Navier s equation The elasticity problem: Navier s equation with boundary conditions µ u + (λ + µ) ( u) + ω 2 ρu = 0 T (n) (u) iσu = Q( T (n) (u) iσu) + g in Ω on Γ where the traction operator T (n) (u) is defined by T (n) (u) = 2µ u + λn u + µn u. n where n is an outward unit normal, g is the source term, ω is the angular frequency of the field, u is the time-harmonic displacement vector, λ and µ are the Lamé constants and ρ is the density of the medium.

46 The Elastic UWVF 14 [T. Luostari] We can derive the following UWVF of finding X k such that k k σ 1 X k ( T (n k ) (e k ) iσe k ) Ω k k Γ k Qσ 1 X k (T (n k ) (e k ) iσe k ) = k j k,j σ 1 X j (T (n k ) (e k ) iσe k ) Γ k σ 1 g (T (n k ) (e k ) iσe k ) (1) for all plane wave (S- or P-wave) test functions e k. Here X k = T (n k ) (u k ) iσu k on Ω k. So far, in 3D numerical simulations we use an ad hoc choice for coupling parameter (flux parameter) where I is the unit matrix. σ = ωρr{c P }I (2) 14 2D: T. Huttunen, P. Monk, F. Collino, and J.P. Kaipio. The Ultra Weak Variational Formulation for elastic wave problems. SIAM J. Sci. Comput., 25: , 2004.

47 Plane wave propagation in a unit cube The exact solution is of the form u = A 1 d exp(iκ P x d) + A 2 d SH exp(iκ S x d) + A 3 d SV exp(iκ S x d) where the wave numbers are κ P = ω/c P, κ S = ω/c S, the direction d [ ], d SH = d, d SV = d d and the amplitudes A 1 = A 2 = A 3 = 1. In addition, u P = 0 and u SH = u SV = 0. As a boundary condition we choose Q = 0.

48 Results for p-convergence 10 1 κ P = and κ S = p P =0.25p S p P =(1/3)p S p P =0.5p S p P =(2/3)p S p P =p S Relative error (%) max(cond(d k )) Number of basis functions Number of basis functions Results when κ P = , κ SH = κ SV = with different ratios between p P /p S and mesh size is fixed (24 tetrahedra in the unit cube).

49 Key questions We are starting to investigate: 1 Convergence to evanescent modes (e.g. Rayleigh waves) 2 Choice of number of directions 3 Choice of the coupling operator σ

50 Other developments Hybridization with finite element methods (acoustics) 15 PML 16 and coupling to integral equations 17 Fluid-solid problem in 2D P. Monk, J. Schöberl, and A. Sinwel. Hybridizing Raviart-Thomas elements for the Helmholtz equation. Electromagnetics, 30:149 76, T. Huttunen, J.P. Kaipio, and P. Monk. The perfectly matched layer for the ultra weak variational formulation of the 3D Helmholtz equation. Int. J. Numer. Meth. Eng., 61: , E. Darrigrand and P. Monk. Coupling of the Ultra-Weak Variational Formulation and an integral representation using a Fast Multipole Method in electromagnetism. J. Comput. Appl. Math., 204:400 7, T. Huttunen, J. Kaipio, and P. Monk. An ultra-weak method for acoustic fluid-solid interaction. J. Comput. Appl. Math., 213:166 85, 2008.

51 Conclusions The method is applicable to a variety of wave propagation problems using a common framework Excellent accuracy can be obtained if the complete family is well matched to the problem Robustness is an issue (particularly ill-conditioning)