On Edge-Elements Reproducing the Discrete Kernel of the Curl Operator

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1 On Edge-Elements Reproducing the Discrete Kernel of the Curl Operator Tomáš Vejchodský 1 (vejchod@math.cas.cz) Pavel Šoĺın, Vojtěch Rybář 1 Institute of Mathematics, Žitná 25, Prague 1 ESCO 08, June , Jetřichovice

2 Time-harmonic Maxwell s equations where Ω R 2 curl ( µ 1 r curl E ) κ 2 ɛ r E = F in Ω E τ = 0 on Ω curl = ( / x 2, / x 1 ) curl E = E 2 / x 1 E 1 / x 2 τ = ( ν 2, ν 1 ) positively oriented unit tangent vector µ r = µ r (x) R relative permeability ɛ r = ɛ r (x) C 2 2 relative permittivity E = E(x) C 2 phaser of the electric field intensity F = F(x) C 2 κ R the wave number

3 Weak formulation E W : a(e, Φ) = F(Φ) Φ W W = {E H(curl, Ω) : E τ = 0 on Ω} a(e, Φ) = (µ 1 r curl E, curl Φ) κ 2 (ɛ r E, Φ) F(Φ) = (F, Φ) = F Φ dx Ω

4 Lowest-order edge elements E h W h : a(e h, Φ h ) = F(Φ h ) Φ h W h W h = {E h W : E h K [P 1 (K)] 2 K T h and E h K τ e = E h K τ e = const. e E, e = K K } T h... triangulation τ e... fixed tangent to e E E... edges in T h E = E b E i V... vertices in T h V = V b V i N v = N bv + N iv N e = N be + N ie Whitney functions: ψ e W h for all e E i ψ e τ e = 1/ e on e E i ψ e τ e = 0 on e e, e E

5 Lowest-order edge elements E h W h : a(e h, Φ h ) = F(Φ h ) Φ h W h W h = {E h W : E h K [P 1 (K)] 2 K T h and E h K τ e = E h K τ e = const. e E, e = K K } T h... triangulation τ e... fixed tangent to e E E... edges in T h E = E b E i V... vertices in T h V = V b V i N v = N bv + N iv N e = N be + N ie Whitney functions: ψ e W h for all e E i ψ e τ e = 1/ e on e E i ψ e τ e = 0 on e e, e E

6 Courant FE V h = {ϕ h H 1 0 (Ω) : ϕ h K P 1 (K) K T h } Courant functions: ϕ i V h for all B i V i ϕ i (x) = 1 at x = B i, B i V i ϕ i (x) = 0 at x = B j, B j B i, B j V ϕ i W h ϕ i τ e = σ { ie +1 e, where σ τe aims towards B ie = i 1 otherwise ϕ i = σ ie ψ e... unique way e ω(b i ) e ω(b i ): ϕ i τ e = σ ie e = σ ie ψ e τ e = e ω(b i ): ϕ i τ e = 0 = e ω(b i ) e ω(b i ) σ ie ψ e τ e σ ie ψ e τ e on e on e

7 De Rham diagram R H id 1 curl H(curl) H(div) div L curl V h Wh R = ker( ) R( ) = ker(curl) R(curl) = ker(div) R(div) = L 2 curl ψ e 0 e E i

8 Discrete kernel of curl Theorem S ϕ = span{ ϕ i, B i V i } = {ψ W h : curl ψ = 0} = K h Proof. (a) S ϕ K h curl ϕ = 0

9 Discrete kernel of curl Theorem S ϕ = span{ ϕ i, B i V i } = {ψ W h : curl ψ = 0} = K h Proof. (b) S ϕ K h ψ K h, curl ψ = 0 e E, e = B i B j, α j = α(b j ) given by α j α i = B k B i B j e ψ τ ij B i B j B k = K T h α i α i = α i α k α j α i = ψ τ K = curl ψ = 0 Define Φ V h : Φ(B i ) = α i B i V Φ W h Bj Φ e τ e e = Φ τ e = Φ = Φ(B j ) Φ(B i ) = e B i = ψ τ e = ψ e τ e e ψ = Φ e K K

10 Discrete kernel of curl Theorem S ϕ = span{ ϕ i, B i V i } = {ψ W h : curl ψ = 0} = K h Proof. (b) S ϕ K h ψ K h, curl ψ = 0 e E, e = B i B j, α j = α(b j ) given by α j α i = B k B i B j e ψ τ ij B i B j B k = K T h α i α i = α i α k α j α i = ψ τ K = curl ψ = 0 Define Φ V h : Φ(B i ) = α i B i V Φ W h Bj Φ e τ e e = Φ τ e = Φ = Φ(B j ) Φ(B i ) = e B i = ψ τ e = ψ e τ e e ψ = Φ e K K

11 Basis reproducing the discrete kernel of curl B W = {ψ e : e E i }... basis in W h, dim W h = #E i = N ie B K = { ϕ i : B i V i }... basis in K h, dim K h = #V i = N iv N iv N ie

12 Basis reproducing the discrete kernel of curl B W = {ψ e : e E i }... basis in W h, dim W h = #E i = N ie B K = { ϕ i : B i V i }... basis in K h, dim K h = #V i = N iv N iv N ie Euler s formula: Nt N e + N v = 1 3Nt = 2N ie + N be Nbe = N bv Nbe 3 3Niv = 3 + N ie N be N ie 3N ie N t = 15 N e = 26 N v = 12

13 Basis reproducing the discrete kernel of curl B W = {ψ e : e E i }... basis in W h, B K = { ϕ i : B i V i }... basis in K h, N iv N ie dim W h = #E i = N ie dim K h = #V i = N iv E i = E rem E keep #E rem = N iv #E keep = N ie N iv Is B = B K {ψ e : e E keep } a basis of W h?

14 Basis reproducing the discrete kernel of curl B W = {ψ e : e E i }... basis in W h, B K = { ϕ i : B i V i }... basis in K h, N iv N ie dim W h = #E i = N ie dim K h = #V i = N iv E i = E rem E keep #E rem = N iv #E keep = N ie N iv Is B = B K {ψ e : e E keep } a basis of W h? Theorem B is a basis of W h the only cycle in (V, E b E rem ) is (V b, E b ) G = (V i {B }, Erem) is a spannig tree B G

15 Basis reproducing the discrete kernel of curl B W = {ψ e : e E i }... basis in W h, B K = { ϕ i : B i V i }... basis in K h, N iv N ie dim W h = #E i = N ie dim K h = #V i = N iv E i = E rem E keep #E rem = N iv #E keep = N ie N iv Is B = B K {ψ e : e E keep } a basis of W h? Theorem B is a basis of W h the only cycle in (V, E b E rem ) is (V b, E b ) G = (V i {B }, Erem) is a spannig tree B G

16 Proof : by contradiction: B a basis in W h and G not a spanning tree G has an isolated component (V isol, E isol ): V isol V i, E isol E rem ϕ = ϕ = B i V isol ϕ i B i V isol ϕ i = B i V isol e ω(b i ) σ ie ψ e = e ω 0 (V isol ) ω 0 (V isol ) = { e = B i B j : B i V isol, B j V isol } σ ie ψ e e = Bi B j E isol, B i V isol, B j V isol ϕ = + σie ψ e + σ je ψ }{{} e + =0 (V isol, E isol ) isolated e ω 0 (V isol ) e E isol e E rem ψ e B

17 Proof : by contradiction: G spanning tree and B not a basis in W h ϕ = c i ϕ i = d e ψ e = ψ B k V i : c k 0 B i V i e E keep e k E keep : d ek 0 e E rem, e = B i B j, B i, B j V i c i c j c j c i ϕ τ e e = σ ie e = σ je e ψe τ e = 0 e E keep c i = c j = c B i, B j V i = 0 G is a spanning tree ẽ E rem, ẽ = B m B : B m V i, B V b c m 0 = ψ τ e e = ϕ τ e e = σ me ẽ ẽ c m = 0 B m c i = c = 0 B i V i B

18 What choice of E rem is the best? µ r = ɛ r = I, κ = 1 a(e, Φ) = (curl E, curl Φ) (E, Φ) { ϕk B S ij = a(φ i, Φ j ) Φ i, Φ j B Φ i = k V i ψ e e E keep cond(s) min cond(s) max cond(s)

19 What choice of E rem is the best? µ r = ɛ r = I, κ = 1 a(e, Φ) = (curl E, curl Φ) (E, Φ) { ϕk B S ij = a(φ i, Φ j ) Φ i, Φ j B Φ i = k V i ψ e e E keep cond(s) min cond(s) max cond(s)

20 What choice of E rem is the best? µ r = ɛ r = I, κ = 1 a(e, Φ) = (curl E, curl Φ) (E, Φ) { ϕk B S ij = a(φ i, Φ j ) Φ i, Φ j B Φ i = k V i ψ e e E keep cond(s) min cond(s) max cond(s)

21 Linear algebraic solver S = ( ) ( ) A11 A 12 κ 2 M11 M 12 A 21 0 M 21 M 22

22 Thank you for your attention Tomáš Vejchodský 1 (vejchod@math.cas.cz) Pavel Šoĺın, Vojtěch Rybář 1 Institute of Mathematics, Žitná 25, Prague 1 ESCO 08, June , Jetřichovice

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