Stable Numerical Scheme for the Magnetic Induction Equation with Hall Effect

Transcription

1 Stable Numerical Scheme for the Magnetic Induction Equation with Hall Effect Paolo Corti joint work with Siddhartha Mishra ETH Zurich, Seminar for Applied Mathematics 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey

2 Outline Formulation and motivation of the problem Theoretical analysis Numerical methods Time integration DG Formulation P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 2

3 Magnetic Reconnection Change in topology of the magnetic field U in U out U out U in Figure: Schematic of a reconnection. Magnetic energy kinetic and thermal energy Dissipation P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 3

4 MHD Equations The equations ρ t = (ρu) (ρu) t = E = t B = E t { ρuu + (p + B2 {(E + p B2 2 are coupled through the equation of state ) I 3 3 BB } 2 ) u+e B E = p γ 1 + ρu2 2 + B2 2 To complete the formulation of the problem we need to state some equation for E P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 4 }

5 Ideal MHD Standard model for E: Ohm s Law E = u B Problem: no dissipation frozen condition. P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 5

6 Ideal MHD Standard model for E: Ohm s Law E = u B Problem: no dissipation frozen condition. We need to add dissipation Resistive MHD: E = u B ηj not sufficient for fast reconnection. P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 5

7 Ideal MHD Standard model for E: Ohm s Law E = u B Problem: no dissipation frozen condition. We need to add dissipation Resistive MHD: E = u B ηj not sufficient for fast reconnection. We need another model... Numerical simulation and laboratory experiment Hall Effect P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 5

8 Generalized Ohm s Law E = u B+ηJ+ δ i J B δ i p L 0 ρ L 0 ρ + ( δe L 0 ) 2 [ ] 1 J ρ t +(u ) J P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 6

9 Generalized Ohm s Law E = u B+ηJ+ δ i J B δ i p L 0 ρ L 0 ρ + ( δe L 0 ) 2 [ ] 1 J ρ t +(u ) J Resistivity Hall effect Electron pressure Electron inertia P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 6

10 Generalized Ohm s Law E = u B+ηJ+ δ i J B δ i p L 0 ρ L 0 ρ + ( δe L 0 ) 2 [ ] 1 J ρ t +(u ) J Resistivity Hall effect Electron pressure Electron inertia J is the electric current given by Ampère s law J = B X.Qian, J.Bablás, A. Bhattacharjee, H.Yang (2009) P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 6

11 General Induction Equation P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 7

12 General Induction Equation Faraday s law Generalized Ohm law Ampère s law p isotropic. B t = E P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 7

13 General Induction Equation Faraday s law Generalized Ohm law Ampère s law p isotropic. Are combined to obtain t [ B+ ( δe ( δe L 0 L 0 B t ) 2 1 ρ ( B) ] = E = (u B) η ( B) ) 2 1 ρ ((u )( B)) δ i L 0 1 ρ (( B) B) P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 7

14 General Induction Equation Faraday s law Generalized Ohm law Ampère s law p isotropic. Are combined to obtain t [ B+ ( δe ( δe L 0 L 0 B t ) 2 1 ρ ( B) ] = E = (u B) η ( B) ) 2 1 ρ ((u )( B)) δ i L 0 1 ρ (( B) B) This equation preserve the divergence of the magnetic field d dt ( B) = 0 P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 7

15 Symmetrized Equation Using the identity (u B) = (B )u B( u)+u( B) (u )B P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 8

16 Symmetrized Equation Using the identity (u B) = (B )u B( u)+u( B) (u )B Since the magnetic field is solenoidal B = 0 we subtract u( B) to the right side of the equation [ ( ) 2 δe B+ ( B)] = t L 0 (B )u B( u) (u )B η ( B) ( ) 2 δe 1 ρ ((u )( B)) δ i 1 (( B) B) (1) L 0 ρ L 0 P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 8

17 Boundary Condition Ω R 3 is a smooth domain Ω in = {x Ω n u < 0} is the inflow boundary. Natural BC Inflow BC B n = 0 on Ω\ Ω in (2) B = 0 on Ω in J = 0 on Ω in (3) P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 9

18 Estimates Theorem For u C 2 (Ω) and B solution of (??) satisfying (??) and (??), then this estimate holds ( ( ) ) 2 d B 2 L dt 2 (Ω) + δe 1 L 0 ρ B 2 L 2 (Ω) ( ( ) ) 2 C 1 B 2 L 2 (Ω) + δe 1 L 0 ρ B 2 L 2 (Ω) (4) with C 1 a constant that depend on u and its derivative only. P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 10

19 Additional Boundary condition B = 0 on Ω in (5) Theorem For u C 2 (Ω) and B solution of (??) satisfying (??),then this estimate holds d dt B L 2 (Ω) C 2 B L 2 (Ω) (6) with C 2 a constant that depend on u and its derivative only. Remark If the solution satisfy all the three boundary conditions, then B H 1 (Ω) P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 11

20 Numerical Formulation Semi-discrete formulation d ) (ˆB(t)+D D ˆB(t) = R(ˆB(t), û) dt where ˆB and û are grid functions. Discretize the space with D. Resulting system of ODE satisfies similar estimates as the one we have for the continuous case. P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 12

21 Numerical Formulation Semi-discrete formulation d ) (ˆB(t)+D D ˆB(t) = R(ˆB(t), û) dt where ˆB and û are grid functions. Discretize the space with D. Resulting system of ODE satisfies similar estimates as the one we have for the continuous case. Integrate on time with Runge Kutta Method We have to compute a matrix inversion! P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 12

22 Discrete Operator One dimensional D x with summation by parts D x = P 1 x Q Q + Q = diag( 1, 0,, 0, 1) P x diagonal positive definite matrix. (D xˆv, ŵ) Px = (ˆv, D x ŵ) Px + v N w N v 0 w 0. Multidimensional operators d x = D x I y I z P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 13

23 Summation by Parts Writing the differential operator as D = d x d y d z Lemma (ˆv, Dŵ) P = (Dˆv, ŵ) P + B.T (ˆv, D ŵ) P = (D ˆv, ŵ) P + B.T (ˆv,(û D)ˆv)) P C ˆv P + B.T P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 14

24 Numerical Scheme [ ˆB+( δe t L 0 ( δe ) ˆB)] 2 D (D = ADV(û, ˆB) ηd (D ˆB) L 0 ) 2 1 ρ D ((ˆv D)ˆB) δ i 1 ( ) L 0 ρ D (D ˆB) ˆB (7) ˆB decays to zero st infinity neglect boundary terms. Assuming that ADV satisfy Lemma (ˆB,ADV(û, ˆB)) P C ˆB 2 P (8) P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 15

25 Theorem Solution of (??) with condition (??), satisfy d ( ( ) 2 ˆB 2 P dt + δe 1 L 0 C 1 ( ˆB 2 P + ρ D ˆB 2 P ( δe L 0 ) ) ) 2 1 ρ D ˆB 2 P can be shown using the same technique used in the continuous case. P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 16

26 Approximated Chain Rule Lemma (Mishra, Svärd) Let ū be a restriction of u C 2 (Ω), then there is a special average operator Āx, such that D x (ūw) = ūd x (w)+ū x Ā x (w)+ w where w Px C x w Px Average operators Ā x = Āx I y I z P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 17

27 Symmetric Advection ADV(û, ˆB) = (Ā(ˆB) D)û Ā x (ˆB)d x û 1 Ā y (ˆB)d y û 2 Ā z (ˆB)d z û 3 (û D)ˆB (9) where Ā(ˆB) = (Ā x (ˆB 1 ),Ā y (ˆB 2 ),Ā z (ˆB 3 )) (??) is valid Theorem (Mishra, Svärd) For numerical scheme with ADV defined as (??) we have d dt D ˆB P C ( D ˆB ) P + ˆB P P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 18

28 Divergence Preserving The operator d x, d y and d z commute D (D ŵ) = 0 Lemma Scheme (??) with (??) satisfy ADV(û, ˆB) = D (û ˆB) (10) d dt D ˆB = 0 Can we obtain (??) for (??)? P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 19

29 Divergence Preserving The operator d x, d y and d z commute D (D ŵ) = 0 Lemma Scheme (??) with (??) satisfy ADV(û, ˆB) = D (û ˆB) (10) d dt D ˆB = 0 Can we obtain (??) for (??)? If D ˆB 0 = 0 yes, symmetrizing the problem. P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 19

30 Time Integration Semi-discrete formulation d dt [(I+A)ˆB(t)] = R(ˆB(t), û(t)) Large dimension dim(ˆb) = dim(r) = 3 N x N y N z A matrix: discrete curl curl operator Time evolution has to invert this matrix, e.g. Euler method ˆB n+1 = ˆB n + t(i+a) 1 R(ˆB(t), û(t)) P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 20

31 Fast Approximated Solution: Auxilary Space joint work with Ralf Hiptmair Known problem for edge element formulation: Main idea: P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 21

32 Fast Approximated Solution: Auxilary Space joint work with Ralf Hiptmair Known problem for edge element formulation: Main idea: Restriction cell centered edge element P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 21

33 Fast Approximated Solution: Auxilary Space joint work with Ralf Hiptmair Known problem for edge element formulation: Main idea: Restriction cell centered edge element Multigrid solution P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 21

34 Fast Approximated Solution: Auxilary Space joint work with Ralf Hiptmair Known problem for edge element formulation: Main idea: Restriction cell centered edge element Multigrid solution Prolong back edge element cell centered P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 21

35 Fast Approximated Solution: Auxilary Space joint work with Ralf Hiptmair Known problem for edge element formulation: Main idea: Restriction cell centered edge element Multigrid solution Prolong back edge element cell centered Prolongation P should satisfy P(ker(A FE )) ker(a FD ) P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 21

36 Fast Approximated Solution: Auxilary Space joint work with Ralf Hiptmair Known problem for edge element formulation: Main idea: Restriction cell centered edge element Multigrid solution Prolong back edge element cell centered Prolongation P should satisfy P(ker(A FE )) ker(a FD ) Problem: A FE has a local kernel, A FD for d x = D x I y has a global kernel! P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 21

37 Solution: Averages! For Outline MHD Theoretical Analysis Numerical Scheme Time Integration Discontinous Galerkin Conclusion Tests d x = D x A y with a D a second order central difference and Aw i = w i+1 + 2w i + w i 1 4 in this case we have a compact kernel Figure: Local Star Shaped Kernel of A FD in 2D. P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 22

38 Prolongation operator in 2D: P : FD space FE conform space P T 1/2 1/2 1/2 1/2 Figure: Schematic for P. We have P(ker(A FE )) ker(a FD ) P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 23

39 Prolongation operator in 2D: P : FD space FE conform space P T 1/2 1/2 1/2 1/2 Figure: Schematic for P. We have P(ker(A FE )) ker(a FD ) New Problem: dim(ker(a FD )) dim(ker(a FE )) P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 23

40 Additonal modes prevent fast convergence. Cleaning Comparison between scheme with and without correction 10 2 N=64 N=64 with correction 10 3 N=128 N=128 with correction N= N=256 with correction Figure: Algorithm with and without cancellation of bad modes. P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 24

41 Still not effective... P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 25

42 Still not effective... Soultion: Same framework (FD): new ˆD and ˆP. OR New formulation DG more suitable space. P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 25

43 System of equations B +U 1 B+(u )B = Ẽ t J = B Ẽ = ηj+αj B+β( J t +(u )J) U 1 depends on u i x j, α = δ i L 0 ρ and β = δ2 e L 2 0 ρ P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 26

44 Define (v, w) Th = v(x)w(x) dx K T K h v, w Faces = v(x)w(x) ds e f Faces where T h a triangulation of Ω. Faces can be F h set of faces in T h. F I h set of inner faces in T h. Γ h set of boundary faces in T h. Γ + h set of outflow boundary faces in T h. Γ h set of inflow faces boundary in T h. P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 27

45 To have unique valued on faces we define: averages {.} normal jumps [[.]] N tangential jumps [[.]] T P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 28

46 Find B h, E h, J h V h ( ) Bh +U 2 B h, t B h ( ) B h,(u ) B h + ( E T h, B h + h )T h T h i ûbi h,[[ B i h]] N Γ + h FI h Êh, B h Fh = (u n)g 1, B h Γ h ( Jh, Ēh) ( ) B T h, Ēh + B h T h,[[ēh]] T h F I h = 0 B h V h Ēh V h ( E h (η β( u))j h α(j h B h ) β J ) h t, J h + ( B h,(u ) J ) h T h T h β i ûj i h,[[ J i h]] N Γ + h FI h = (u n)g 2, J h Γ h J h V h P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 29

47 DG Fluxes Upwind with c = nu /2. LDG ub h i = {ubi h}+c[[b h]] i N uj h i = {uj h}+c[[j i h]] i N B h = {B h }+b[[b h ]] T { {Eh } b[[e Ê h = h ]] T +a[[b h ]] T internal edges {E h }+a[[b h ]] T boundary edges With this Fluxes Energy estimate. P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 30

48 Knowing B n h, En h, Jn h find Bn+1 h, E n+1 h, J n+1 h V h 1 ( B n+1 h B n t h, B ) h + ( U 2 B n h, B ) h ( B n T T h h,(u ) B h + )T h h ( E n+1 h, B h ) T h + i = (u n)g 1 (t n ), B h Γ h û(bi ) n h,[[ B i h]] N Γ + h FI h Ên+1 h, B h Fh ( J n h, Ēh) ( B n ) T h h, Ēh + B n T h h,[[ēn h ]] T F I h = 0 Ēh V h ( E n+1 h ηj n+1 h +β( u))j n h α(jn+1 β ( ) J n+1 h J n h t, J h + ( B n h,(u ) J ) h β T T h h i h B n h ), J h ) T h B h V h û(j i ) n h,[[ J i h ]] N Γ + h FI h = (u n)g 2 (t n ), J h Γ h J h V h P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 31

49 Matrix formulation Algorithm B n+1 B n for i N iter do r R n C n DG Bn+1 c L n DG c = r B n+1 B n+1 + c r R n C n DG Bn+1 ρ P r γ C n FE γ = ρ c Pγ B n+1 B n+1 + ĉ i i + 1 end for (γ1m+γ n 2A n DG ) ˆB n+1 = R n }{{} C n DG P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 32

50 1D Model t b + v x b η xx b β( xxt b + x (u xx b)) = 0 Auxiliary variables t b + u x b x e = 0 j = x b e = ηj +β( t j + u x j) P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 33

51 Preconditioner P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 34

52 Conclusion The solution of the induction equation with Hall term is in H 1 (Ω). We can build symmetric and divergence preserving methods. The spatially discretized systems possess similar estimates as continuous one. stability granted for exact-time evolution of the discrete system. Preconditioner for time evolution. DG Formulation. P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 35

53 Conclusion The solution of the induction equation with Hall term is in H 1 (Ω). We can build symmetric and divergence preserving methods. The spatially discretized systems possess similar estimates as continuous one. stability granted for exact-time evolution of the discrete system. Preconditioner for time evolution. DG Formulation. Thank You! P. Corti 17-19th August 2011, Pro*Doc Retreat, Disentis Abbey p. 35