Development of a stable coupling of the Yee scheme with linear current

Transcription

1 Development of a stable coupling of the Yee scheme with linear current Martin Campos Pinto (LJLL), Bruno Després (LJLL) Stéphane Heuraux (IJL), Filipe Da Silva (IPFN+IST) 15 octobre 2013 Munich: Pereverzev legacy 13/10/2013 p. 1 / 30

2 Outline Munich: Pereverzev legacy 13/10/2013 p. 2 / 30

3 The equations No magnetisation (µ = µ 0 ). Maxwell equations with a linear current derive from the linearization µ 0 H B 0 of the Vlasov-Maxwell system (for electrons) around a strong magnetic field B 0 : ε 0 t E + H = q e N e (x)u e, µ 0 t H + E = 0, m e t u e = q e (E + B 0 (x) u e ) νm e u e. Or, writing J = q e N e (x)u e, ε 0 t E = H J, µ 0 t H = E, t J = ε 0 ωpe 2 + ω c b J with ω p (x) = q 2 e N e(x) mε 0, ω c (x) = qe B 0(x) m e and b(x) = B 0(x) B 0 (x). Munich: Pereverzev legacy 13/10/2013 p. 3 / 30

4 Direct simulation of reflectometry configuration The domain is a parallepiped ( 1500 cells in x direction) with an antenna on the side : pulsation ω Cut-off : in O mode (TM), waves propagate if ω ω p (x). Cyclotron resonance : ω = ω c Hybrid resonance : ω 2 = ω p (x) 2 + ω 2 c Munich: Pereverzev legacy 13/10/2013 p. 4 / 30

5 Standard scheme of Xu-Yuan (2006) Based on the Yee scheme for the (E, H) field : general form is ε 0 t (E n+1 E n ) = RH n+ 1 2 J n+ 1 2 µ 0 t (Hn+ 3 2 H n+ 1 2 ) = R t E n+1 1 t (Jn+ 3 2 J n+ 1 2 ) = ε 0 ωpe 2 n+1 + ω c b 1 2 (Jn J n+ 1 2 ). Need to specify the operator h on the Yee grid Munich: Pereverzev legacy 13/10/2013 p. 5 / 30

6 X-mode equations X-mode=Transverse electric (O-mode not discussed in this talk). ε 0 t E x + y H z = J x, J x = en e u x, ε 0 t E y x H z = J y, J y = en e u y, µ 0 t H z + x E y y E x = 0, m e t u x = ee x + eu y Bz 0, m e t u y = ee y eu x Bz 0. Call VLC external Munich: Pereverzev legacy 13/10/2013 p. 6 / 30

7 Main difficulty In fusion plasmas, N e (x) has huge fluctuations along the main axis n e [ m -3 ] n e w/o frm n e w/ frm Negative density (num. or measurement artifact) induces automatically an instability, as well as strong spatial gradient at the plasma edge (phys.) or inside the plasma (phys.). Munich: Pereverzev legacy 13/10/2013 p. 7 / 30

8 Example of unstability (for large times) Magnetic field 20 log 10 H z (where H z = H z L ) vs. time step and level of fluctuations Expertise from F. Da Silva and S. Heuraux. Munich: Pereverzev legacy 13/10/2013 p. 8 / 30

9 Some references Xu-Yuan, FDTD Formulations for Scattering From 3-D Anisotropic Magnetized Plasma Objects, IEEE-2006 Bohner, Simulation of Microwave Propagation in a Fusion Plasma, Diploma Thesis, MaxPlanck, 2011 An unconditionally stable (?) time-domain discretization on cartesian meshes for the simulation of nonuniform magnetized cold plasma, JCP-2012, Tierens-Zutter Yu-Simpson, An E-J Collocated 3-D FDTD Model of Electromagnetic Wave Propagation in Magnetized Cold Plasma, Ieee Transations on antennas and propagation, 58-2, 2010 Smithe, Finite-difference time-domain simulation of fusion plasmas at radiofrequency time scales, Physics of plasmas, 2007 da Silva-Heuraux-Ribeiro-Scott, (2013). Development of a 2D full-wave JE-FDTD Maxwell X-mode code for reflectometry simulation (pp. 16). Presented at the 9th International Reflectometry Workshop. Munich: Pereverzev legacy 13/10/2013 p. 9 / 30

10 Outline Munich: Pereverzev legacy 13/10/2013 p. 10 / 30

11 Energy conservation at the continuous level For simplicity : constant density profile N e (x, t) = N e (x). Inside the computational domain (no boundaries assumed), the total energy is conserved in time, ( d ε0 E 2 + H 2 + m en e (x) u e 2 ) dv = 0. dt Ω 2 2µ 0 2 Using normalized variables Ê := 1 c E, Ĥ := µ 0H and Ĵ := 1 ω pcε 0 J, we have ( ) d Ê 2 dt 2 + Ĥ Ĵ 2 dv = 0. 2 Munich: Pereverzev legacy 13/10/2013 p. 11 / 30

12 Classical stability analysis for the Yee scheme With normalized variables, the Yee scheme (J = 0) reads 1 t (Ê n+1 Ê n ) = crĥ n t (Ĥn+ 1 2 Ĥ n 1 2 ) = cr t Ê n where {Ê := 1 c E Ĥ := µ 0 H. In particular, the energy Ê n := Ê n 2 h + Ĥn h satisfies Ê n+1 Ê n = c t ( RĤ n+ 1 2, Ê n+1 + Ê n R t Ê n, Ĥ n Ĥ n 1 2 ) hence E n := Ê n c t Ê n, RĤn 1 2 is constant. Moreover, Ê n, RĤn R Ên = Ê n (1 c t 2 R ) En = Stability in the energy norm : for c t < 2/ R = h/ 3 Munich: Pereverzev legacy 13/10/2013 p. 12 / 30

13 Stability analysis for an abstract Yee+J scheme With Ê, Ĥ and Ĵ := 1 ω pcε 0 J, the abstract Yee+J scheme is 1 t (Ê n+1 Ê n ) = crĥ n+ 1 2 ω p Ĵ n t (Ĥn+ 1 2 Ĥn 1 2 ) = cr t Ê n 1 t (Ĵn+ 1 2 Ĵ n 1 2 ) = ω p Ê n Ĵ + ω c b n Ĵ n 1 2 h 2 Here the energy Ê n := Ê n 2 + Ĥ n Ĵ n satisfies t ( ω p Ĵ n+ 1 2, Ê n+1 + Ê n ω p Ê n, Ĵ n Ĵ n 1 2 ) provided V, b h V = 0 for all V. Stability in the energy norm : for t 2 (12c2 h 2 + ω p 2 L ) 1 2 < 1. Munich: Pereverzev legacy 13/10/2013 p. 13 / 30

14 Remark on average cross products One can use local averages to define a 2nd order cross product, (b h V ) x := b y {V z } b z {V y } (b h V ) y := (b h V ) z := Then if b(x) = B 0(x) B 0 is uniform, V, b h V = 0 holds for all V previous analysis applies. If b(x) is not uniform this is not so clear... Munich: Pereverzev legacy 13/10/2013 p. 14 / 30

15 Improved stability for a new Yee+J scheme Discretizing the current on t n, t n+1,... yields a new scheme 1 t (Ê n+1 Ê n ) = crĥ n+ 1 Ĵ n ω Ĵn p 2 1 t (Ĥn+ 1 2 Ĥn 1 2 ) = cr t Ê n 1 t (Ĵn+1 Ĵn 1 ) = ω p {Ê} n+ 1 Ĵ n ω c b Ĵn h The energy Ê n satisfies Ê n+1 Ê n =c t ( RĤ n+ 1 2, Ê n+1 + Ê n R t Ê n, Ĥ n Ĥ n ) 1 2 t ( ω p {Ĵ} n+ 1 2, 2{Ê} n+ 1 2 ω p {Ê} n+ 1 2, 2{Ĵ} n+ ) 1 2 once again provided V, b h V = 0 for all V. Stability in the energy norm : for c t < 2/ R = h/ Munich: Pereverzev legacy 13/10/2013 p. 15 / 30

16 Outline Munich: Pereverzev legacy 13/10/2013 p. 16 / 30

17 Example of the Xu-Yuan scheme Based on the Yee scheme for the (E, H) field : general form is ε 0 t (E n+1 E n ) = RH n+ 1 2 J n+ 1 2 µ 0 t (Hn+ 3 2 H n+ 1 2 ) = R t E n+1 1 t (Jn+ 3 2 J n+ 1 2 ) = ε 0 ωpe 2 n+1 + ω c b 1 2 (Jn J n+ 1 2 ). Munich: Pereverzev legacy 13/10/2013 p. 17 / 30

18 Example of the Xu-Yuan scheme Based on the Yee scheme for the (E, H) field : general form is ε 0 t (E n+1 E n ) = RH n+ 1 2 J n+ 1 2 µ 0 t (Hn+ 3 2 H n+ 1 2 ) = R t E n+1 1 t (Jn+ 3 2 J n+ 1 2 ) = ε 0 ωpe 2 n ω c b h 2 (Jn J n+ 1 2 ). Need an explicit solver with the b h operator. Munich: Pereverzev legacy 13/10/2013 p. 17 / 30

19 Problem with the X-Y approach Consider once again the cross product by local averages (b h V ) x := b y {V z } b z {V y } (b h V ) y := (b h V ) z := The result ε 0 t (E n+1 E n ) = RH n+ 1 2 J n+ 1 2 µ 0 t (Hn+ 3 2 H n+ 1 2 ) = R t E n+1 1 t (Jn+ 3 2 J n+ 1 2 ) = ε 0 ωpe 2 n ω c b h 2 (Jn J n+ 1 2 ). is a global scheme which needs a linear solver to invert the matrice. Munich: Pereverzev legacy 13/10/2013 p. 18 / 30

20 Solution : use clustered cross-products Instead, choose a pattern (α, β, γ) { 1, +1} 3 and define the first order cross product with local clusters : (b h V ) x i+ α 2,j,k := b y V z i,j,k+ γ b z{v y } 2 i,j+ β 2,k (b h V ) y i,j+ β 2,k := b zv x i+ α 2,j,k b x {V z } i,j,k+ γ 2 (b h V ) z i,j,k+ γ := b xv y 2 i,j+ β 2,k b y {V x } i+ α 2,j,k The resulting scheme ε 0 t (E n+1 E n ) = RH n+ 1 2 J n+ 1 2 µ 0 t (Hn+ 3 2 H n+ 1 2 ) = R t E n+1 1 t (Jn+ 3 2 J n+ 1 2 ) = ε 0 ωpe 2 n ω c b h 2 (Jn J n+ 1 2 ). can be solved with a local procedure (i.e. solution is explicit and local). Munich: Pereverzev legacy 13/10/2013 p. 19 / 30

21 Abstract criterion The criterion for explicit scheme writes : (b h ) 4 = (b h ) 2. Indeed one has the implications J αb h J = Z, J α 2 (b h ) 2 J = (I + αb h )Z, (1 + α 2 )(b h ) 2 J = (b h ) 2 (I + αb h )Z, J = (I + αb h )Z + α2 1 + α 2 (b h) 2 (I + αb h )Z. This algebra is enough to compute the solution by means of explicit and local formulas (for MXYK and new Kernel). Munich: Pereverzev legacy 13/10/2013 p. 20 / 30

22 Short summary The coupling of the Yee scheme and a linear current is for : (V, b h V ) = 0 for : (b h ) 4 = (b h ) 2 Solution (so far) is clustered first order product Additional and natural condition is that ω p and ω c are the same within a cluster. Munich: Pereverzev legacy 13/10/2013 p. 21 / 30

23 Outline Munich: Pereverzev legacy 13/10/2013 p. 22 / 30

24 Data N e Zoom around the foot of the ramp Cut of the electronic density in the horizontal direction. An additional kink (in red) is sometimes added at x = 500 to evaluate the effect of an extremely strong gradient. Munich: Pereverzev legacy 13/10/2013 p. 23 / 30

25 H z plot With the kick and 30% noise Without the kick but 40% noise An instability shows up near x = 500 cells on the left, near x = 1000 Munich: Pereverzev legacy 13/10/2013 p. 24 / 30

26 20 log 10 H z L With respect to the time and to the level of noise. With the kick on the left, without the kick on the right. Munich: Pereverzev legacy 13/10/2013 p. 25 / 30

27 With the first order vectorial product 20 log 10 H z L, with respect to the time and to the level of noise. The computation is done We observe unconditional stability, with however more amplitude for a higher level of noise. The number of time steps is much greater than in previous figure to illustrate the long time stability of the method. Munich: Pereverzev legacy 13/10/2013 p. 26 / 30

28 Energy dissipation Initial data is a Dirac mass, at the exact foot of the electronic density ramp. The external magnetic field used in this set of runs was B 0 = 0.95T. The plasma density N e (x) is linear, with its edge at x = 500 grid point. The number of iterations considered is N = 700 (far from PML layer). (slide courtesy of F. Da Silva) Munich: Pereverzev legacy 13/10/2013 p. 27 / 30

29 Additional remarks and Need to use clustered multiplications by scalar fields, consistent with clustered cross products. Counter-intuitive : the stable and explicit scheme is globally first order (and not second order like the standard Yee scheme). Possibility to average in time by alternating the cluster patterns (α, β, γ) in { 1, +1} 3 Work in progress for direct simulation of time-dependent densities N e = N e (x, t) (Doppler reflectometry) A paper is being written Munich: Pereverzev legacy 13/10/2013 p. 28 / 30

30 An open question (is it really?) Look at ε 0 t E + H = q e N e (x)u e, µ 0 t H + E = 0, m e t u e = q e (E + B 0 (x) u e ) νm e u e plus harmonic forcing on the boundary, plus initial condition, plus friction ν > 0. Assume resonance configuration (cyclotron, hybrid,...) : do we have lim lim = lim lim T T ν 0 + ν 0 +? In other words, do we have Limit absorption=limit amplitude? If not, which one is the correct physical solution? Munich: Pereverzev legacy 13/10/2013 p. 29 / 30

31 lim ν 0 + lim T (L.M. Imbert-Grard) Munich: Pereverzev legacy 13/10/2013 p. 30 / 30