On divergence-free reconstruction schemes for CED and MHD

Transcription

1 On divergence-free reconstruction schemes for CED and MHD Praveen Chandrashekar Center for Applicable Mathematics Tata Institute of Fundamental Research Bangalore , India Oberseminar, Dept. of Mathematics University of Würzburg, 26 April 2018 Supported by Airbus Foundation Chair at TIFR-CAM, Bangalore 1 / 59

2 Joint work with Dinshaw Balsara, Univ. of Notre Dame Arijit Hazra, TIFR-CAM, Bangalore Rakesh Kumar, TIFR-CAM, Bangalore 2 / 59

3 Maxwell Equations Linear hyperbolic system B t + E = 0, B = magnetic flux density E = electric field D t H = J D = electric flux density H = magnetic field J = electric current density B = µh, D = εe, J = σe µ, ε R 3 3 symmetric ε = permittivity tensor µ = magnetic permeability tensor σ = conductivity B = 0, D = ρ (electric charge density), ρ t + J = 0 3 / 59

4 Ideal MHD equations Nonlinear hyperbolic system Compressible Euler equations with Lorentz force ρ t + (ρv) = 0 ρv + (pi + ρv v B B) t = 0 E + ((E + p)v + (v B)B) t = 0 B (v B) t = 0 Magnetic monopoles do not exist: = B = 0 4 / 59

5 Objectives Divergence conforming schemes for Maxwell Exactly divergence-free schemes for MHD High order accurate, non-oscilatory schemes Upwind-type schemes, approximate Riemann solvers Efficient explicit time stepping 5 / 59

6 Some existing methods for MHD Exactly divergence-free methods Yee scheme ([1], (1966)) Constrained transport ([2] Evans & Hawley (1989)) Divergence-free reconstruction ([3] Balsara (2001)) Globally divergence-free scheme ([4] Li et al. (2011)) Approximate methods Locally divergence-free schemes ([5] Li & Shu (2005)) Using Godunov s symmetrized version of MHD [?] (Powell [?], [6], Gassner, C & K [7]) Divergence cleaning methods (Dedner et al. [?]) 6 / 59

7 Approximation of magnetic field If B = 0, it is natural to look for approximations in H(div, Ω) = {B L 2 (Ω) : div(b) L 2 (Ω)} To approximate functions in H(div, Ω) on a mesh T h with piecewise polynomials, we need the following compatibility condition. Theorem (See [8], Proposition 3.2.2) Let B h : Ω R d be such that 1 B h K H 1 (K) for all K T h 2 for each common face F = K 1 K 2, K 1, K 2 T h, the trace of normal component n B h K1 and n B h K2 is the same. Then B h H(div, Ω). Conversely, if B h H(div, Ω) and (1) holds, then (2) is also satisfied. 7 / 59

8 Approximation spaces B n on the faces must be continuous: directly approximate B n by polynomial functions P k on the faces. e + y 2 3 x, y [ 1 2, ] [ 1 2, ] B y + (ξ) ξ = x x c x, η = y y c y e x B x (η) C B + x (η) e + x k B x ± (η) = a ± j φ j(η) j=0 B y (ξ) 0 1 e y k B y ± (ξ) = b ± j φ j(ξ) j=0 8 / 59

9 Approximation spaces φ j are mutually orthogonal polynomials on the interval [ 1 2, 1 2 ] given by φ 0 (ξ) = 1, φ 1 (ξ) = ξ, φ 2 (ξ) = ξ , φ 3(ξ) = ξ ξ φ 4 (ξ) = ξ ξ Diagonal elements of the mass matrix: m i = 1 2 φ 2 1 i (ξ)dξ 2 m 0 = 1, m 1 = 1 12, m 2 = 1 180, m 3 = ,... The mutual orthogonality implies that φ j (ξ)dξ = 0, j = 1, 2,... 9 / 59

10 Approximation spaces and hence B ± x (η)dη = a ± 0, B ± y (ξ)dξ = b ± 0 Given the data B x ±, B y ± on the faces, we want to recontruct a divergence free vector field (B x (ξ, η), B y (ξ, η)) inside the cell. Necessary condition: B n = B = 0 (1) Writing this condition in the reference coordinates yields C C or 1 2 y (B 1 x + (η) Bx (η))dη + x 2 C (B + y (ξ) B y (ξ))dξ = 0 B n = (a + 0 a 0 ) y + (b+ 0 b 0 ) x = 0 (2) 10 / 59

11 Raviart-Thomas polynomials Space of tensor product polynomials Q r,s = span{ξ i η j : 0 i r, 0 j s} Approximate the vector field inside the cells by tensor product polynomials B x (ξ, η) = B y (ξ, η) = k+1 i=0 j=0 k a ij φ i (ξ)φ j (η) Q k+1,k k k+1 b ij φ i (ξ)φ j (η) Q k,k+1 i=0 j=0 RT(k) = Q k+1,k Q k,k+1, dim RT(k) = 2(k + 1)(k + 2) div RT(k) Q k,k 11 / 59

12 Brezzi-Douglas-Marini polynomials The BDM polynomials are of the form BDM(k) = (P k ) 2 + r (x k+1 y) + s (xy k+1 ) dim BDM(k) = (k + 1)(k + 2) + 2, div BDM(k) P k 1 In reference coordinates, we take the polynomials to be of the form B x (ξ, η) = k i,j=0 i+j k a ij φ i (ξ)φ j (η) + r (ξk+1 M k+1 ) y (k + 1)ξηk + s y B y (ξ, η) = k i,j=0 i+j k b ij φ i (ξ)φ j (η) r (k + 1)ξk η x s (ηk+1 M k+1 ) x 12 / 59

13 Brezzi-Douglas-Marini polynomials where M k = 1 2 ξ k dξ. This scaling ensures that the last two terms are 1 2 divergence-free, e.g., the terms with coefficient r satisfy [ 1 (ξ k+1 M k+1 ) x ξ y ] + 1 y [ (k + ] 1)ξk η = 0 η x 13 / 59

14 Brezzi-Douglas-Fortin-Marini polynomials The BDFM polynomials are of the form BDFM(k) = (P k+1 ) 2 \ (0, x k+1 ) \ (y k+1, 0) dim BDFM(k) = (k + 2)(k + 3) 2, div BDFM(k) P k In reference cell coordinates, the polynomials can be written as k+1 k i k+1 B x (ξ, η) = a ij φ i (ξ)φ j (η), B y (ξ, η) = b ij φ i (ξ)φ j (η) i=0 j=0 k j j=0 i=0 Remark: Note that the Raviart-Thomas space is the largest space among the three spaces considered here, and in fact we have BDM(k) BDFM(k) RT(k) 14 / 59

15 Conditions for the reconstruction For all the three polynomial spaces B x (± 1 2, η) P k and B y (ξ, ± 1 2 ) P k We can try to satisfy the conditions B x (± 1 2, η) = B± x (η) η [ 1 2, ], B y(ξ, ± 1 2 ) = B± y (ξ) ξ [ 1 2, ] In addition, we have to make the vector field divergence-free B = 1 B x x ξ + 1 B y y η = 0, ξ, η [ 1 2, +1 2 ] Do we have enough equations? Can we solve them? Note: All three polynomial spaces contain P k, which is enough to obtain (k + 1) th order accuracy. We are free to throw away any basis function outside P k without affecting the order of accuracy. 15 / 59

16 Reconstruction at degree k = 0 RT(0) = BDM(0) = BDFM(0) Constant approximation on the faces B x ± (η) = a ± 0, B± y (ξ) = b ± 0 Vector field inside the cell B x (ξ, η) = a 00 + a 10 ξ, B y (ξ, η) = b 00 + b 01 η Matching this to face solution B x (± 1 2, η) = B± x (η), B y (ξ, ± 1 2 ) = B± y (ξ) we obtain a 00 ± 1 2 a 10 = a ± 0, b 00 ± 1 2 b 01 = b ± 0 whose solution is given by a 00 = 1 2 (a 0 + a+ 0 ), b 00 = 1 2 (b 0 + b+ 0 ), a 10 = a + 0 a 0, b 01 = b + 0 b 0 16 / 59

17 Reconstruction at degree k = 0 The divergence of the reconstructed vector field is B = 1 B x x ξ + 1 B y y η = a 10 x + b 01 y = 1 x y [(a+ 0 a 0 ) y + (b+ 0 b 0 ) x] = 0 since we have assumed that (2) holds. Thus the reconstructed field is automatically divergence-free if the face solution satisfies the necessary compatibility condition. 17 / 59

18 Reconstruction at degree k = 1 The face solution is a polynomial of degree one and is of the form B x ± (η) = a ± 0 + a± 1 η B x ± (η) = b ± 0 + b± 1 ξ Matching the cell solution to the face solution yields 8 equations. 18 / 59

19 Reconstruction using RT(1) The vector field inside the cells is assumed to be of the form B x (ξ, η) = 2 1 a ij φ i (ξ)φ j (η), B y (ξ, η) = i=0 j=0 1 2 b ij φ i (ξ)φ j (η) i=0 j=0 and there are 12 coefficients to be determined. Matching the cell solution with the face solution a 00 a 10 /2 + a 20 /6 = a 0 (3) a 01 a 11 /2 + a 21 /6 = a 1 (4) a 00 + a 10 /2 + a 20 /6 = a + 0 (5) a 01 + a 11 /2 + a 21 /6 = a + 1 (6) b 00 b 01 /2 + b 02 /6 = b 0 (7) b 10 b 11 /2 + b 12 /6 = b 1 (8) b 00 + b 01 /2 + b 02 /6 = b + 0 (9) b 10 + b 11 /2 + b 12 /6 = b + 1 (10) 19 / 59

20 Reconstruction using RT(1) The divergence of the vector field is a polynomial in Q 1 div(b) x y = (a 10 y + b 01 y) + (2a 20 y + b 11 x)ξ + (a 11 y + 2b 02 x)η + (2a 21 y + 2b 12 x)ξη and the divergence-free condition yields four equations a 10 y + b 01 y = 0 (constant) (11) 2a 20 y + b 11 x = 0 (ξ) (12) a 11 y + 2b 02 x = 0 (η) (13) 2a 21 y + 2b 12 x = 0 (ξη) (14) We have 12 coefficients and 12 equations. However these equations are not linearly independent. To show this, combine the equations in the form [(5) (3)] y + [(9) (7)] x which yields a 10 y + b 01 x = (a + 0 a 0 ) y + (b+ 0 b 0 ) x = 0 (15) 20 / 59

21 Reconstruction using RT(1) by (2). Hence we see that equation (11) is contained in the remaining equations. Another way to see this result is by noticing that since the cell solution matches the face solution, then B n = a 10 y + b 01 x = 0 C and hence equation (11) is already contained in the face matching equations (3)-(10). We thus have only 11 equations (i.e., (3)-(14) but excluding (11)) for the 12 unknown coefficients. 21 / 59

22 Reconstruction using RT(1) We can solve for some of the variables from the above equations a 00 = 1 2 (a 0 + a + 0 ) (b+ 1 b 1 ) x y a 10 = a + 0 a 0 b 00 = 1 2 (b 0 + b + 0 ) (a+ 1 a 1 ) y x b 01 = b + 0 b 0 a 11 = a + 1 a 1 a 20 = 1 2 (b+ 1 b 1 ) x y b 11 = b + 1 b 1 b 02 = 1 2 (a+ 1 a 1 ) y x The remaining unknowns are a 01, a 21, b 10, b 21 which satisfy the equations a a 21 = 1 2 (a 1 + a+ 1 ) (16) b b 12 = 1 2 (b 1 + b+ 1 ) (17) a 21 y + b 12 x = 0 (18) 22 / 59

23 Reconstruction using RT(1) We have four remaining unknowns but only three equations. Hence we have to make additional assumptions or simplificiations in order to solve the problem. For second order accuracy, it is enough to include P 1 in our approximation space and hence we can set a 21 = b 12 = 0. Then the remaining two coefficients are given by a 01 = 1 2 (a 1 + a+ 1 ), b 10 = 1 2 (b 1 + b+ 1 ) 23 / 59

24 Reconstruction using BDM(1) The vector field inside the cell is of the form B x (ξ, η) = a 00 + a 10 φ 1 (ξ) + a 01 φ 1 (η) + r y (ξ ) + 2 s y ξη B y (ξ, η) = b 00 + b 10 φ 1 (ξ) + b 01 φ 1 (η) 2 r x ξη s x (η ) This has 8 coefficients and matching these polynomials to the corresponding solution on the faces, we get 8 equations whose solution yields a 00 = 1 2 (a 0 + a + 0 ) (b+ 1 b 1 ) x y a 10 = a + 0 a 0 a 01 = 1 2 (a 1 + a + 1 ) r = 1 2 (b 1 b + 1 ) x b 00 = 1 2 (b 0 + b + 0 ) (a+ 1 a 1 ) y x b 10 = 1 2 (b 1 + b + 1 ) b 01 = b + 0 b 0 s = 1 2 (a+ 1 a 1 ) y 24 / 59

25 Reconstruction using BDM(1) The divergence of the reconstructed polynomial is given by div(b) x y = a 10 y + b 01 x = (a + 0 a 0 ) y + (b+ 0 b 0 ) x = 0 since the face solution satisfies equation (2). Hence in this case, the divergence-free property of the reconstructed field follows automatically if the face data is consistent. A careful check of this solution shows that the reconstructed vector field inside the cell is identical to the solution obtained with RT(1). 25 / 59

26 Reconstruction using BDFM(1) The vector field inside the cell is given by B x (ξ, η) = a 00 + a 10 ξ + a 01 η + a 20 (ξ ) + a 11ξη B y (ξ, η) = b 00 + b 10 ξ + b 01 η + b 11 ξη + b 02 (η ) which is same as the polynomial used by Balsara, and the divergence is a polynomial in P 1 div(b) x y = (a 10 y+b 01 y)+(2a 20 y+b 11 x)ξ+(a 11 y+2b 02 x)η If the above cell solution matches the face solution, then B n = a 10 y + b 01 x = 0 C Note that this is same as the constant term in the divergence and hence the constant term is already zero if the face solution satisfies (2). We have 26 / 59

27 Reconstruction using BDFM(1) 10 coefficients to determine B. Matching with the face solution gives us 8 equations and we get 2 equations from the divergence-free condition, so that we have a total of 10 equations whose solution is given by a 00 = 1 2 (a 0 + a + 0 ) (b+ 1 b 1 ) x y a 10 = a + 0 a 0 a 01 = 1 2 (a 1 + a + 1 ) a 20 = 1 2 (b 1 b + 1 ) x y b 00 = 1 2 (b 0 + b + 0 ) (a+ 1 a 1 ) y x b 10 = 1 2 (b 1 + b + 1 ) b 01 = b + 0 b 0 b 02 = 1 2 (a 1 a + 1 ) y x b 11 = b + 1 b 1 a 11 = a + 1 a 1 We verify that with the above solution, the constant term in the divergence is indeed zero since a 10 y + b 01 y = (a + 0 a 0 ) y + (b+ 0 b 0 ) x = 0 27 / 59

28 Reconstruction using BDFM(1) and hence the reconstructed polynomial is indeed divergence-free if the face solution is compatible with (2). It is easy to see that the BDFM(1) polynomial is identical to the RT(1) and BDM(1) polynomial, which is not a suprising result that the space BDM(1) is a subset of BDFM(1). 28 / 59

29 Reconstruction at degree k = 2 The face solution is a polynomial of degree two and is of the form B ± x (η) = a ± 0 + a± 1 η + a± 2 (η ) B ± y (ξ) = b ± 0 + b± 1 ξ + b± 2 (ξ ) Since we have three coefficients on each face, matching the cell solution to the faces yields a total of 12 equations. 29 / 59

30 Reconstruction using RT(2) The vector field inside the cell of the form B x (ξ, η) = 3 2 a ij φ i (ξ)φ j (η), B y (ξ, η) = i=0 j=0 2 3 b ij φ i (ξ)φ j (η) i=0 j=0 and there are 24 coefficients need to be determined. We thus have 12 equations from matching with the face solution and 8 equations from the divergence-free condition for a total of 20 equations, but 24 unknown coefficients. We set the highest degree terms a 32 = b 23 = 0, a 22 = b 22 = a 31 = b 13 = 0 and a 21 = b 12 = 0, which allows us to solve the reconstruction problem. 30 / 59

31 Reconstruction using BDM(2) The vector field has the form B x (ξ, η) = a 00 +a 10 ξ+a 01 η+a 20 (ξ )+a 11ξη+a 02 (η )+ r y ξ3 + 3s y ξη2 B y (ξ, η) = b 00 +b 10 ξ +b 01 η+b 20 (ξ )+b 11ξη+b 02 (η 2 1 3r 12 ) x ξ2 η s x η3 We need to find 14 coefficients to determine B inside the cell. By matching the solution to the faces, we obtain 12 equations and we get two equations from the divergence-free condition (excluding the constant term), for a total of 14 equations, which can be solved. 31 / 59

32 Reconstruction using BDFM(2) The field inside the cell is approximated by B x (ξ, η) = a 00 + a 10 ξ + a 01 η + a 20 (ξ ) + a 11ξη + a 02 (η )+ a 30 (ξ ξ) + a 21(ξ )η + a 12ξ(η ) B y (ξ, η) = b 00 + b 10 ξ + b 01 η + b 20 (ξ ) + b 11ξη + b 02 (η )+ b 21 (ξ )η + b 12ξ(η ) + b 03(η η) which has 18 coefficients. We get 12 equations from matching the face solution and 5 equations from divergence-free condition, for a total of 17 equations. We can set a 21 = b 12 = 0 and complete the solution. 32 / 59

33 Reconstruction at degree k = 3 The face solution is a polynomial of degree three and is of the form B ± x (η) = a ± 0 φ 0(η) + a ± 1 φ 1(η) + a ± 2 φ 2(η) + a ± 3 φ 3(η) B ± x (η) = b ± 0 φ 0(ξ) + b ± 1 φ 1(ξ) + b ± 2 φ 2(ξ) + b ± 3 φ 3(ξ) Since we have four coefficients on each face, matching the cell solution to the faces yields a total of 16 equations. 33 / 59

34 Reconstruction using BDFM(3) The vector field has the form B x(ξ, η) = a 00 + a 10ξ + a 01η + a 20(ξ ) + a11ξη + a02(η )+ a 30(ξ 3 3 ξ) + 20 a21(ξ2 1 )η + 12 a12ξ(η2 1 ) + 12 a03(η3 3 η)+ 20 a 40(ξ ξ2 + 3 ) a13ξ(η3 3 η) + 20 a31(ξ3 3 ξ)η+ 20 a 22(ξ )(η2 1 ) 12 P4 \ {y4 } B y(ξ, η) = b 00 + b 10ξ + b 01η + b 20(ξ ) + b11ξη + b02(η )+ b 30(ξ ξ) + b21(ξ )η + b12ξ(η ) + b03(η η)+ b 22(ξ )(η ) + b13ξ(η η) + b31(ξ ξ)η+ b 04(η η ) P4 \ {x4 } which has a total of 28 coefficients. The divergence-free condition gives 9 equations and matching the face solution gives 16 equations for a total of 26 equations. 34 / 59

35 Reconstruction using BDFM(3) We can first solve for some of the coefficients completely in terms of the face solution a 00 = 1 2 (a 0 + a+ 0 ) (b+ 1 b 1 ) x y a 10 = a + 0 a (b+ 2 b 2 ) x y a 20 = 1 2 (b+ 1 b 1 ) x y (b+ 3 b 3 ) x y a 30 = 1 3 (b+ 2 b 2 ) x a 03 = 1 2 (a 3 + a+ 3 ) y b 00 = 1 2 (b 0 + b+ 0 ) (a+ 1 a 1 ) y x b 01 = b + 0 b (a+ 2 a 2 ) y x b 02 = 1 2 (a+ 1 a 1 ) y x (a+ 3 a 3 ) y x b 30 = 1 2 (b 3 + b+ 3 ) b 03 = 1 3 (a+ 2 a 2 ) y x a 12 = a + 2 a 2 a 40 = 1 4 (b+ 3 b 3 ) x y a 13 = a + 3 a 3 b 21 = b + 2 b 2 b 31 = b + 3 b 3 b 04 = 1 4 (a+ 3 a 3 ) y x 35 / 59

36 Reconstruction using BDFM(3) The remaining coefficients satisfy the following equations a a21 = 1 2 (a 1 + a + 1 ) a a22 = 1 2 (a 2 + a + 2 ) b b12 = 1 2 (b 1 + b + 1 ) 15a 11 y b 22 x = 15(a + 1 a 1 ) y 15b 11 x a 22 y = 15(b + 1 b 1 ) x b 12 x + a 21 y = 0 3a 31 y + 2b 22 x = 0 b b22 = 1 2 (b 2 + b + 2 ) 3b 13 x + 2a 22 y = 0 Let us find a minimal solution by setting the fourth order terms to zero since they are not required to get fourth order accuracy, Then we get a 31 = a 22 = b 13 = b 22 = 0 a 02 = 1 2 (a 2 + a+ 2 ), b 20 = 1 2 (b 2 + b+ 2 ), a 11 = a + 1 a 1, b 11 = b + 1 b 1 36 / 59

37 Reconstruction using BDFM(3) and the remaining unknowns satisfy a a 21 = 1 2 (a 1 + a+ 1 ) =: r 1 (19) b b 12 = 1 2 (b 1 + b+ 1 ) =: r 2 (20) b 12 x + a 21 y = 0 (21) We have four unknowns but only three equations. We cannot throw away any of these coefficients since they are all at or below third degree, and we must retain all of them in order to get fourth order accuracy. The only way to complete the reconstruction is to provide additional information. Let us assume that we know the value of ω such that b 10 a 01 = ω (22) 37 / 59

38 Reconstruction using BDFM(3) Note that ω provides information about the mean value of the curl of the vector field in the cell ω = B y ξ (0, 0) B x (0, 0) η Then we can solve the equations (19)-(22) to obtain a 01 = [ ] 1 y 1 + y r 1 x + r 2 ω x b 10 = ω + a 01 a 21 = 6(r 1 a 01 ) b 12 = 6(r 2 b 10 ) We will know the value of ω from the initial condition and we have to device a scheme to evolve it forward in time using the induction equation. 38 / 59

39 4 th order scheme of Balsara B inside cells B x (P 4 \ {y 4 }) + span{x 4 y, x 2 y 3 } B y (P 4 \ {x 4 }) + span{xy 4, x 3 y 2 } a 21 and b 12 determined using WENO a 41 = b 14 and a 23 = b 32 (energy minimzation) 39 / 59

40 Discontinuous Galerkin method Let us first consider B x which is stored on the vertical faces e x. y B x t + E z η = 0 Multiply by test function φ i (η), i = 0, 1,..., k, and integrate over the vertical face 1 2 y y 1 2 B x + 1 t φ 2 idη Ê x+ φ i 1 η dη + (Ẽφ i) 3 (Ẽφ i) 1 = 0 (23) 2 Bx 1 t φ 2 idη Ê x φ i 1 η dη + (Ẽφ i) 2 (Ẽφ i) 0 = 0 (24) 2 Ê x± = Numerical flux on face from 1-D Riemann solver Ẽ = Numerical flux at vertex from 2-D Riemann solver 40 / 59

41 Discontinuous Galerkin method Similarly, for the B y component stored on the horizontal faces, we first transform the induction equation x B y t E z ξ = 0 multiply this by a test function φ i (ξ), i = 0, 1,..., k, and integrate over the horizontal edge 1 2 x x 1 2 B y + 1 t φ 2 idξ + Ê y+ φ i 1 ξ dξ (Ẽφ i) 3 + (Ẽφ i) 2 = 0 (25) 2 By 1 t φ 2 idξ + Ê y φ i 1 ξ dξ (Ẽφ i) 1 + (Ẽφ i) 0 = 0 (26) 2 41 / 59

42 Fourth order scheme ω = b 10 a 01 = 12 (B y ξ B x η)dξdη Derive an evolution equation for ω using the induction equation ( 1 dω 12 dt = By t ξ B ) ( x 1 t η E dξdη = x ξ ξ + 1 ) E y η η dξdη Performing an integration by parts and using a numerical flux on the faces which is based on a 1-D Riemann solver, we obtain a semi-discrete DG scheme 1 dω 12 dt [ = 1 1 x y [ Ê x dη Ê y dξ Ê x+ 2 dη 1 2 Ê y+ dξ Edξdη ] Edξdη ] 42 / 59

43 Divergence-free property DG equations for i = 0 for which φ 0 = 1 and combine them in the form (23)-(24)+(25)-(26) which leads to 1 2 y t (B+ x Bx 2 )dη + x 1 2 t (B+ y Ẽ3 + Ẽ2 + Ẽ1 Ẽ0 = 0 B y )dξ + Ẽ3 Ẽ1 Ẽ2 + Ẽ0 which simplifies to d B n = d dt dt [(a+ 0 a 0 ) y + (b+ 0 b 0 ) x] = 0 C This shows that if the face solution C B n = 0 at t = 0, then the DG scheme preserves this property at future times also. 43 / 59

44 Conservation property Is the 2-D conservation law satisfied for B? 1 B x dxdy = a 00 = 1 x y 2 (a 0 + a+ 0 ) (b+ 1 b 1 ) x y C The mean value a 00 in the cell evolves as d dt a Ẽ 2 00 = Ẽ0 Ẽ 3 Ẽ1 1 2 y 2 y x y e + Ê y+ Ẽ 3 + dx + Ẽ2 y 2 y = 1 x y e + Ê y+ dx + 1 y x y e Ê y dx y which is just the 2-D conservation law for B x. + 1 x y e Ê y Ẽ 1 + dx Ẽ0 y 2 y Thus the divergence-free reconstruction + DG scheme is conservative in the usual sense, even though we do not explicitly enforce this in the scheme. 44 / 59

45 Numerical results: Induction equation B x + E t y = 0, B y E t x = 0 E = (v B) z = v y B x v x B y Algorithm 1: Divergence-free reconstruction scheme Allocate memory for all variables; Set initial condition on the faces; Set time counter to zero; while final time is not reached do Copy current solution into old solution; for each RK stage do Loop over cells and do divergence-free reconstruction; Loop over vertices and compute vertex flux; Loop over faces and update the solution; If degree==3, loop over cells and update ω; end Update time counter; end 45 / 59

46 Periodic plane wave solution The exact solution is given by B = ( y A z, x A z ) where A z (x, y, t) = cos(2π(x + y 2t)), v = (1, 1) k = 1 k = 2 k = k = 1 k = 2 k = L 1 -error 10-6 L 2 -error N N Convergence of error at time t = 1 units 46 / 59

47 Magnetized vortex solution Initial condition is given by B 0 = ( y A z, x A z ) where A z = exp(α(1 r 2 )), r 2 = x 2 + y 2, (x, y) [ 5, +5] [ 5, +5] with α = 2 and the velocity field is constant and given by v = (1, 1) k =1 k =2 k = k =1 k =2 k = L 1 -error L 2 -error N N Convergence of error at time t = 10 units 47 / 59

48 Rotating magnetic vortex The initial condition is given by B 0 = ( y A z, x A z ) where A z (x, y) = 1 10 exp[ 20((x 1/2)2 + y 2 )] and the velocity field is v = (y, x). The exact solution is a pure rotation of the initial condition and is given by [ ] cos t sin t B(r, t) = R(t)B 0 (R( t)r), R(t) = sin t cos t 48 / 59

49 Rotating magnetic vortex k =1 k =2 k = k =1 k =2 k = L 1 -error L 2 -error N N Domain [ 1, +1] [ 1, +1]: error at t = 2π 49 / 59

50 Rotating magnetic vortex k =1 k =2 k = k =1 k =2 k = L 1 -error L 2 -error N N Domain [0, 1] [0, 1]: error at t = π/2 50 / 59

51 Numerical results: 2-D Maxwell equation B z t D x t D y t H z y + H z x + E y x E x y = J x (27) = J y (28) = 0 (29) where (E x, E y ) = 1 ε (D x, D y ), H z = 1 µ B z, (J x, J y ) = σ ε (D x, D y ) Flux reconstruction scheme on faces for D x, D y, and DG scheme on cells for B z All the tests currently have: σ = 0 and D = 0 51 / 59

52 Plane wave propagation 10 2 Error in B z k=1 k=2 k= Error in D k=1 k=2 k= L 1 error norm L 1 error norm Grid size Grid size / 59

53 Refraction of gaussian pulse by disc p r = 5 4 tanh(( x2 + y )/0.08) mesh, k = 1, solution at t = 0 and t = s 53 / 59

54 Refraction of plane wave: ε0 ε 2.25ε0 r = tanh(x/10 8 ) mesh, k = 1, solution at t = 0 and t = s 54 / 59

55 Total internal reflection of plane wave: 4ε0 ε ε mesh, k = 1, solution at t = 0 and t = s 55 / 59

56 Summary Divergence-free reconstruction using RT(k), BDM(k), BDFM(k) k = 1, 2: face solution determines reconstruction k = 0: all are same RT(1): set a 21 = b 12 = 0 BDM(1), BDFM(1): can be solved RT(1) = BDM(1) = BDFM(1) = Balsara RT(2): set a32 = b 23 = a 22 = b 22 = a 31 = b 13 = a 21 = b 12 = 0 BDM(2): can be solved BDFM(2): set a 21 = b 12 = 0 RT(2) = BDM(2) = BDFM(2) = Balsara BDFM(3): Need additional information ω DG scheme: preserves divergence, is conservative Ongoing work MHD with multi-d Riemann solvers Maxwell: stiff source terms, IMEX-type scheme WENO-type limiters AMR with hanging nodes: flux reconstruction scheme on faces 3-D 56 / 59

57 References [1] K. Yee, Numerical solution of initial boundary value problems involving maxwell s equations in isotropic media, IEEE Transactions on Antennas and Propagation 14 (3) (1966) doi: /tap [2] C. R. Evans, J. F. Hawley, Simulation of magnetohydrodynamic flows - A constrained transport method, The Astrophysical Journal 332 (1988) 659. doi: / [3] D. S. Balsara, Divergence-Free Adaptive Mesh Refinement for Magnetohydrodynamics, Journal of Computational Physics 174 (2) (2001) doi: /jcph / 59

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