A generalized Holland model for wave diffraction by thin wires Xavier Claeys Joint work with Francis Collino Marc Duruflé POEMS, Inria Rocquencourt, France Generalized Holland model p.1/26
Motivation Incident wave Hypothesis: λ = wavelength Diffracted wave perfectly conducting wire. ε λ. ɛ = thickness of the wire Context: simulation of EM wave propagation in media including perfectly Context: conducting thin wires (ex: antennas). Wish: propose a volumic method with no mesh refinement. Wish: Different from integral equation approach (Rogier, Fedoryuk, Mazari.) Generalized Holland model p.2/26
Motivation Holland method Heuristic method used for straight wires in FDTD schemes based on: Assumption: the current is constant across any section of the wire. Assumption: the field is electrostatic close to the wire. Technical ingredient: averaging operator in a region as large as a cell close to the wire (parameter). R.Holland and L.Simpson Finite difference analysis of emp coupling to thin struts and wires, IEEE Trans. Electromagn. Compat. 1981. Generalized Holland model p.3/26
Model problem Hypothesis 2D acoustic diffraction. Dirichlet condition on the boundary of obstacles. Small obstacle (section of a wire). Radius: ε 0. u 0 Γ ε Ω R = D(0, R), Γ R = Ω R. Ω ε = D(0, ε), Γ ε = Ω ε. u 0 (x) = e ik x (incident wave). Γ R Generalized Holland model p.4/26
Model problem Hypothesis 2D acoustic diffraction. Dirichlet condition on the boundary of obstacles. Small obstacle (section of a wire). Radius: ε 0. u 0 Γ ε Ω R = D(0, R), Γ R = Ω R. Ω ε = D(0, ε), Γ ε = Ω ε. u 0 (x) = e ik x (incident wave). Γ R Generalized Holland model p.5/26
Model problem Hypothesis 2D acoustic diffraction. Dirichlet condition on the boundary of obstacles. Small obstacle (section of a wire). Radius: ε 0. u 0 Γ ε 8 >< >: u ε H 1 (Ω R \ Ω ε ), u ε + k 2 u ε = 0 in Ω R \ Ω ε, u ε = 0 on Γ ε, u ε u 0 outgoing. Γ R Generalized Holland model p.6/26
eu ε = (1 χ ε ) u 0 + Asymptotic analysis Approximate field π u 2 0(0) 2π u 0 (0) v 0 + χ ε V0 ε. ln(1/ε) + λ 0 ln(1/ε) + λ 0 X.Claeys, H.Haddar and P.Joly Etude d un problème modèle pour la diffraction par des fils minces par développements asymptotiques raccordés. cas 2-d. Research Report 5839, INRIA,2006. Generalized Holland model p.7/26
eu ε = (1 χ ε ) u 0 + Asymptotic analysis Approximate field π u 2 0(0) 2π u 0 (0) v 0 + χ ε V0 ε. ln(1/ε) + λ 0 ln(1/ε) + λ 0 1.5χ ε Cut-off function χ ε Gauge function λ 0 = π γ, γ: Euler constant. 2 More complicated in general. 1 0.5 Incident field u 0 0 x ε 2 ε 3 3 0.5 0 0.2 0.4 0.6 0.8 1 1.2 1.4 u 0 (x) = e ik x. Generalized Holland model p.8/26
eu ε = (1 χ ε ) u 0 + Asymptotic analysis Approximate field π u 2 0(0) 2π u 0 (0) v 0 + χ ε V0 ε. ln(1/ε) + λ 0 ln(1/ε) + λ 0 Far field v 0 Close Field V ε 0 v 0 (x) = 1 i H(1) 0 (k x ). V ε 0 (x) = 1 2π ln x ε 1 Ω R \Ω ε. Close Field V ε 0 Incident field u 0 u 0 (x) = e ik x. V ε 0 (x) = 1 2π ln x ε 1 Ω R \Ω ε. Generalized Holland model p.9/26
Asymptotic analysis eu ε = (1 χ ε ) u 0 + Approximate field π u 2 0(0) 2π u 0 (0) v 0 + χ ε V0 ε. ln(1/ε) + λ 0 ln(1/ε) + λ 0 Matched asymptotic expansions H = H 1 (Ω R ) H 1 (Ω R \ Ω R ) u ε eu ε H ε 1/2 ε u ε u 0 H ln ε 1/2 ln ε 1 Γ ε Ω R \ Ω R Generalized Holland model p.10/26
Usual formulation Usual formulation without any obstacle a(u, v) = R Ω R u v k 2 uv + R Γ R v T R u u, v H 1 (Ω R ) 8 >< >: Find (u ε, p ε ) H 1 (Ω R ) H 1/2 (Γ ε ) such that a(u ε, v)+b ε (p ε, v) = R Γ R (T R u 0 + u 0 n )v v H1 (Ω R ), b ε (q, u ε ) = 0, q H 1/2 (Γ ε ). Generalized Holland model p.11/26
Usual formulation Fictitious domain formulation b ε (q, v) = R Γ ε q v q H 1/2 (Γ ε ), v H 1 (Ω R ). 8 >< >: Find (u ε, p ε ) H 1 (Ω R ) H 1/2 (Γ ε ) such that a(u ε, v)+b ε (p ε, v) = R Γ R (T R u 0 + u 0 n )v v H1 (Ω R ), b ε (q, u ε ) = 0, q H 1/2 (Γ ε ). Remarks u ε Γ ε = 0 and pε = [ uε n ] Γ ε on Γε. u ε is extended by 0 in Ω ε. Generalized Holland model p.12/26
Asymptotic analysis Approximate jump of the normal derivative p ε = [ uε n ] Γ ε for uε = 0 in Ω ε. ep ε = [ euε n ] Γ ε = 2πu 0(0) V0 ε ln 1/ε + λ 0 n = 2πu 0(0) 1 ln 1/ε + λ 0 2πε P 0. v 2 H 1/2 (Γ ε ) = ε2 X n Z (1 + n 2 ) 1/2 Z 2π 0 v(θ)e inθ dθ 2. Matched asymptotic expansions p ε 0 H 1/2 (Γ ε ) κ ln ε 1. p ε ep ε H 1/2 (Γ ε ) κ ε. Generalized Holland model p.13/26
Simplified problem Reduced space of Lagrange multipliers Remind ep ε = [ euε n ] Γ ε P 0. Idea: replace H 1/2 (Γ ε ) by P 0. Initial fictitious domain formulation 8 >< >: Find (u ε ap, p ε ap) H 1 (Ω R ) H 1/2 (Γ ε ) such that a(u ε ap, v) + b ε (p ε ap, v) = R Γ R (T R u 0 + u 0 n )v v H1 (Ω R ), b ε (q, u ε ap) = 0, q H 1/2 (Γ ε ). Generalized Holland model p.14/26
Simplified problem Reduced space of Lagrange multipliers Remind ep ε = [ euε n ] Γ ε P 0. Idea: replace H 1/2 (Γ ε ) by P 0. Simplified fictitious domain formulation 8 >< >: Find (u ε ap, p ε ap) H 1 (Ω R ) P 0 such that a(u ε ap, v) + b ε (p ε ap, v) = R Γ R (T R u 0 + u 0 n )v v H1 (Ω R ), b ε (q, u ε ap) = 0, q P 0. Generalized Holland model p.15/26
Simplified problem Reduced space of Lagrange multipliers Remind ep ε = [ euε n ] Γ ε P 0. Idea: replace H 1/2 (Γ ε ) by P 0. Simplified fictitious domain formulation 8 >< >: Find (u ε ap, p ε ap) H 1 (Ω R ) P 0 such that a(u ε ap, v) + b ε (p ε ap, v) = R Γ R (T R u 0 + u 0 n )v v H1 (Ω R ), b ε (q, u ε ap) = 0, q P 0. Remarks: Uniform Inf-Sup condition on a and b ε with respect to ε. Continuity of b ε : b ε κ p ln 1/ε. Well posed and nearly stable problem for ε sufficiently small. Generalized Holland model p.15/26
Simplified problem Asymptotic analysis H = H 1 (Ω R ) H 1 (Ω R \ Ω R ) u ε ap u 0 H ln ε 1/2 ln ε 1 eu ε u ε H ε 1/2 ε u ε ap u ε H ε 1/2 ε u ε ap provides an approximation of u ε of the same quality as eu ε. p ε ap 0 H 1/2 (Γ ε ) = ε pε ap κ ln ε 1. p ε ap p ε H 1/2 (Γ ε ) = ε pε ap p ε κ ε. Generalized Holland model p.16/26
Finite element approximation 8 >< >: Discrete simplified fictitious domain formulation Find (u ε,h ap, p ε,h ap ) V h P 0 such that a(u ε,h ap, v h ) + b ε (p ε,h ap, v h ) = R (T Γ R R u 0 + u 0 n )vh v h V h, b ε (q h, u ε,h ap ) = 0, q h P 0. Hypothesis: Uniformly regular triangulation and ε < κ h Error estimate: u ε ap u ε,h ap H 1 κ p ln 1/ε inf v h V h uε ap v h H 1 Generalized Holland model p.17/26
Finite element approximation 8 >< >: Discrete simplified fictitious domain formulation Find (u ε,h ap, p ε,h ap ) V h P 0 such that a(u ε,h ap, v h ) + b ε (p ε,h ap, v h ) = R (T Γ R R u 0 + u 0 n )vh v h V h, b ε (q h, u ε,h ap ) = 0, q h P 0. Hypothesis: Uniformly regular triangulation and ε < κ h Error estimate: u ε ap u ε,h ap H 1 κ p ln 1/ε inf v h V h uε ap v h H 1 Problem: inf v h V h uε ap v h H 1 is not small: Problem: inf uε v h V h ap v h H 1 κ h 1 2 δ u ε ap 3 δ and u ε H 2 ap 3 δ κ H 2 ε. 1 2 δ Generalized Holland model p.17/26
Numerical locking We use Q 3 finite elements. u ε ap u ε,h ap H u ε ap H 10 1 10 0 10 1 10 2 10 3 10 4 10 5 10 6 10 1 10 0 h/λ H = H 1 (Ω R \ Ω R ) Ω R = D(0, 3) Ω R = D(0, 1.5) eps = 1e 2 eps = 1e 3 eps = 1e 4 eps = 1e 5 ε/λ slope 10 2 0.9051 10 3 0.8697 10 4 0.6163 10 5 0.5074 Generalized Holland model p.18/26
Additional shape function ϕ ε ad(x) = χ 0 (x)v ε 0 = χ 0 (x) 1 2π V h e = V h span{ϕ ε ad(x)}. ln x ε 1 Ω R \Ω ε. χ 0 ϕ ε ad 1.4 1 1.2 0.8 1 0.6 0.8 0.6 0.4 0.4 0.2 0 1 2 0 3 3 0 0.2 0.4 0.6 0.8 1 1.2 1.4 x 0.2 0 ε 1 2 3 3 0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 x Generalized Holland model p.19/26
Additional shape function ϕ ε ad(x) = χ 0 (x)v ε 0 = χ 0 (x) 1 2π V h e = V h span{ϕ ε ad(x)}. ln x ε 1 Ω R \Ω ε. Augmented discrete simplified fictitious domain formulation 8 >< >: Find (u ε,h ap, p ε,h ap ) Ve h P 0 such that a(u ε,h ap, v h ) + b ε (p ε,h ap, v h ) = R (T Γ R R u 0 + u 0 n )vh b ε (q h, u ε,h ap ) = 0, q h P 0. v h V h e, P.Ciarlet, B.Jung, S.Kaddouri, S.Labrunie, J.Zou The Fourier Singular Complement Method. Part I&II. Numer.Math. 2005 M.Bourlard, M.Dauge, M-S.Lubuma, S.Nicaise Coefficients of the singularities for elliptic boundary value problems on domains with conical points. III: Finite element methods on polygonal domains. SIAM J.Numer.Anal. 1992 Generalized Holland model p.19/26
Additional shape function ϕ ε ad(x) = χ 0 (x)v ε 0 = χ 0 (x) 1 2π V h e = V h span{ϕ ε ad(x)}. ln x ε 1 Ω R \Ω ε. Augmented discrete simplified fictitious domain formulation 8 >< >: Find (u ε,h ap, p ε,h ap ) Ve h P 0 such that a(u ε,h ap, v h ) + b ε (p ε,h ap, v h ) = R (T Γ R R u 0 + u 0 n )vh b ε (q h, u ε,h ap ) = 0, q h P 0. v h V h e, Lemma: u ε u ε,h ap H 1 (Ω R ) κ p ln 1/ε (ε 1/2 + h). Generalized Holland model p.20/26
Locking free We use Q 3 finite elements. 10 1 10 0 H = H 1 (Ω R \ Ω R ) Ω R = D(0, 3) Ω R = D(0, 1.5) u ε ap u ε,h ap H u ε ap H 10 1 10 2 10 3 10 4 10 5 10 6 10 2 10 1 10 0 h/λ eps = 1e 2 eps = 1e 3 eps = 1e 4 eps = 1e 5 ε/λ slope 10 2 3.8883 10 3 4.1136 10 4 4.2392 10 5 4.3103 Generalized Holland model p.21/26
Generalized Holland model Projection on the approximation space For ϕ H 1 (Ω R ),! P h (ϕ) V h such that a(ϕ P h (ϕ), v h ) = 0, v h V h. δ h = Id Ph. Generalized Holland model p.22/26
Generalized Holland model Projection on the approximation space For ϕ H 1 (Ω R ),! P h (ϕ) V h such that a(ϕ P h (ϕ), v h ) = 0, v h V h. δ h = Id Ph. Decomposition in singular and regular part Or u ε,h ap = eu ε,h reg + p ε,h s ϕ ε ad, u ε,h reg V h. Or u ε,h ap = u ε,h reg + p ε,h s δ h (ϕ ε ad), u ε,h reg = eu ε,h reg + P h (ϕ ε ad) V h. Generalized Holland model p.22/26
Generalized Holland model Projection on the approximation space For ϕ H 1 (Ω R ),! P h (ϕ) V h such that a(ϕ P h (ϕ), v h ) = 0, v h V h. δ h = Id Ph. Decomposition in singular and regular part Or u ε,h ap = eu ε,h reg + p ε,h s ϕ ε ad, u ε,h reg V h. Or u ε,h ap = u ε,h reg + p ε,h s δ h (ϕ ε ad), u ε,h reg = eu ε,h reg + P h (ϕ ε ad) V h. Q 3 ε = 0.01 h = 0.02597 Re{ϕ ε ad } Re{δ h(ϕ ε ad )} Generalized Holland model p.22/26
Generalized Holland model Augmented discrete simplified fictitious domain formulation 8 >< >: Find (u ε,h reg, p ε,h ap ) V h P 0 such that a(u ε,h reg, v h ) + b ε (p ε,h ap, v h ) = R Γ R (T R u 0 + u 0 n )vh b ε (q h, u ε,h reg ) L ε,h c(p ε,h ap, q h ) = G ε,h (q h ), q h P 0 v h V h With L ε,h = bε`1, δ h (ϕ ε ad) 2 a(δ h (ϕ ε ad), δ h (ϕ ε ad)) and c(p ε,h ap, q h ) = p ε,h ap q h. Generalized Holland model p.23/26
Generalized Holland model Augmented discrete simplified fictitious domain formulation 8 >< >: Find (u ε,h reg, p ε,h ap ) V h P 0 such that a(u ε,h reg, v h ) + b ε (p ε,h ap, v h ) = R Γ R (T R u 0 + u 0 n )vh b ε (q h, u ε,h reg ) L ε,h c(p ε,h ap, q h ) = G ε,h (q h ), q h P 0 v h V h F.Collino and F.Millot Fils et méthodes d éléments finis pour les équations de Maxwell. Le modèle de Holland revisité. Research Report 3472, INRIA, 1998. 8 >< >: Find (u ε,h hol Holland formulation, p ε,h hol ) V h P 0 such that a(u ε,h hol, v h ) + b ε (p ε,h hol, v h ) = R (T Γ R R u 0 + u 0 n )vh b ε (q h, u ε,h hol ) L hol c(p ε,h hol, q h ) = 0, q h P 0 v h V h Generalized Holland model p.24/26
Generalized Holland model 8 >< >: Find (u ε,h hol Holland formulation, p ε,h hol ) V h P 0 such that a(u ε,h hol, v h ) + b ε (p ε,h hol, v h ) = R (T Γ R R u 0 + u 0 n )vh b ε (q h, u ε,h hol ) L ε,h hol c(p ε,h hol, q h ) = 0, q h P 0 v h V h H3: There exists κ > and ν > 0 such that κh ν < ε. Lemma: Under H3 there exists κ > 0 (idpdt of ε and h) such that G ε,h < κ h q. Lemma: Under H3, for L hol = L ε,h there exists κ > 0 (idpdt of ε and h) such that u ε,h reg u ε,h hol H 1 (Ω R ) + p ε,h ap p ε,h hol H 1/2 (Γ ε ) κ hq. Generalized Holland model p.25/26
Conclusions Method compatible with high order FEM. Generalization of the Holland method for arbitrary meshes. Analytical expression of L hol. Weakly locking method. Not local with respect to the mesh. Generalized Holland model p.26/26