Particle in Cell method

Transcription

1 Particle in Cell method Birdsall and Langdon: Plasma Physics via Computer Simulation Dawson: Particle simulation of plasmas Hockney and Eastwood: Computer Simulations using Particles we start with an electrostatic 1d-1v collissionless plasma: Lecture by G. Lapenta Vlasov equation for distribution function f s : f s t + v f s x + q se m s f s v =0 q s, m s : charge and mass of the species Poisson equation: 2 x 2 = charge density follows from distribution function (x, t) = f s (x, v, t)dv s

2 moving finite elements or superparticles: f p f s (x, v, t) = p f p (x, v, t) Ansatz: tensor product of shape functions: f p (x, v, t) =N p S x (x x p (t))s v (v v p (t)) S x, S v : shape functions N p : number of real particles in the superparticle Assumptions: 1. compact support 2. normalisation: = x or v 3. symmetric shapes: S ( p)d =1 S ( p) =S ( p )

3 Selection of particle shape S v (v v p )= (v v p ) b-splines first b-spline: flat-top function b 0 ( ) b 0 ( )= 1 if < 1/2 0 otherwise subsequent b-splines: b l ( )= d b 0 ( )b l 1 ( )

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5 spatial shape function S x (x x p )=b l x x p p: size of superparticle p Remarks: few PIC codes use splines of order 1 but most use order 0 Vlasov equation looks linear, but is nonlinear due to E-field. Equation for superparticle: f p t + v f p x + qe m f p v =0 The arbitrary functional form chosen for the elements does not satisfy exactly the Vlasov equation. The usual procedure of the finite element method is to require that the moments of the equations be satisfied. Notation:... := dx dv

6 Moment 0 f p t + v f p x + qe m f p v =0 second and third term = 0 = N p = const Moment 1 x xf p t + xv f p x + qxe m f p v =0 last term = 0

7 consider first term: xf p = N p S v (v v p )dv xs x (x x p )dx = N p (x + x p )S x (x )dx = N p x p using symmetry of S x consider second term: Z Z vdv dx = Z vf p dxdv = hvf p i hvf p i = N p Z = N p Z Z vs v (v v p )dv S x (x (v 0 + v p )S v (v 0 )dv 0 x p )dx thus dx p = N p v p using symmetry of S v dt = v p

8 Moment 1 v vf p t + v 2 f p x + qev m f p v =0 second term = 0 first term has already been computed: vf p = N p v p remaining term: qs E dx v m s f p v dv = qs E m s dx f p dv = q s E f s m s This defines average electric field E p acting on computational particle: q s E q s f s = N p E p m s m s with E p = = S v (v S x (x v p )dv x p )E(x)dx S x (x x p )E(x)dx thus dv p dt = q s m s E p

9 Equations of motion Field equations: finite volume approach dn p dt dx p dt dv p dt = 0 = v p = q s m s E p cell centres: x i cell faces: x i+1/2 cell-averaged values i of potential Poisson equation: i+1 2 i + i 1 ( x) 2 = q i with cell-averaged density: i = 1 x x i+1/2 x i 1/2 (x)dx

10 recall definition of b-spline of order 0 xi+1/2 (x)dx = x i 1/2 b 0 x x i x (x)dx recall definition of the density xi+1/2 x i 1/2 (x)dx = p b 0 x x i x S x (x x p )dx define interpolation function W (x i x p )= S x (x x p )b 0 x x i x dx and get: q p x W (x i x p ) with q p = q s N p i = p

11 very simple expression, when we choose: p = x =) xi x p W (x i x p )=b l+1 x Electric field from central di erences E i = i+1 i 1 2 x Efieldisconstantineachcell: E(x) = X i E i b 0 xi x p x recall definition of E p : E p = X i E i Z b 0 x x xi S x (x x i ) and recall definition of interpolation function E p = X i E i W (x x i )

12 Simplest symplectic integrator: leap-frog xp n+1 = xp n + tvp n+1/2 v n+3/2 p = v n+1/2 p + te p (x n p ) second order accurate initial step: Euler forward v 1/2 = v 0 p + te p (x 0 p) Stability c t < x pe < 2 x < De

13 Elektromagnetic PIC FDTD and Yee mesh (1966) Particle push: Boris scheme (1970) conservative charge deposition (1991, 2001) Literature: inspired by lectures by A. Spitkovsky J. Villasenor, Bunemann T. Zh. Esirkepov T. Umeda

14 d dt "mv = q(e + v c # B) FDTD: second order in space and time Yee mesh: div B = 0

15 Equations of motion: leap-frog: looks implicit

16 Solution: Boris (1970) explicit with t t = q t 2 t m Bt =2arctan(t t )

17 Proof: We know: We want to proof: v + v = q t 2m (v+ + v ) B v + v = v s v = v + v t, t = q t 2m B, s = 2t 1+t 2 thus: v + v = v s +(v t) s v s = v B q t 2 2m 1+t 2 = v + B q t 2m = (v + v ) B q t 2m 2 1+t 2 +(v+ v ) 2 1+t t 2 +(v+ v ) 1 1+t 2

18 (v t) s = (v s) t 2 q t 1 = 2m 1+t 2 [(v+ v ) B] B + q t 2m = v + v = (v + v ) 1 1+t 2 + q t 2m t 2 B [(v+ v ) B] 1 1+t 2 (v+ v ) B (v + v ) B = q t 2m [(v+ + v ) B] B =: C B B (C B) =CB 2 BB C = q t 2m [(v+ + v ) B]B 2 = t 2 (v + v )= q t 2 B 2 [(v + + v ) B] q t 2m 2m t 2 = q t 2 B 2 = 2m

19 PIC Vlasov I) x n+1/2 x n 1/2 t = v n ˆf n+1 (x n+1/2, v n )= x ( t) f n (x n 1/2, v n ) II) j n = v n S(x ) x = x n+1/2 + x n 1/2 2 = x n + O( t 2 ) j n = v n f (x n, v n ) f (x n, v n )= f n (x n 1/2, v n )+ˆf n (x n+1/2, v n ) 2 = f n (x n, n v )+O( t 2 ) III) E n+1/2 E n 1/2 t = B n j n IV) B n+1 B n t = E n+1/2 V) Boris Boris v n+1 B t v n = Bn+1 + B n 2 = q m E n+1/2 (x n+1/2 )+ v n+1 + v n = B n+1/2 + O( t 2 ) 2 B (x n+1/2 ) f n+1 (x n+1/2, v n+1 )= v ( t)ˆf (x n+1/2, v n )

20 Charge and current deposition B/t = c( E) = B/t =0 Yee grid "# $ E "t = c# $ (# % B) & 4'# $ J "# "# "t = $% & J continuity equation is again an initial condition

21 ) =) projection method =) parabolic =) hyperbolic

22 If one uses standard weighting, charge is not conserved. Villasenor, Bunemann (1992): count current through faces Problem: lots of if -statements current deposition as expensive as particle push optimized deposit by Umeda (2003): zig-zag curve Esirkepov (2001) arbitrary form factor, easy and precise used in PSC (Ruhl) and EPOCH (UK code based on PSC)

23 Esirkepov: E n+1 E n = + B n+1/2 J n+1/2, dt B n+1/2 B n 1/2 = E n, dt + E n = ρ n, B n+1/2 = 0, u n+1/2 α x n+1 α u n 1/2 α dt x n α dt = un+1/2 α γα n+1/2, γ α = ( 1 + (u α ) 2) 1/2. = 2π q α m e m α e ( E n( x n α,t) + un α B n( ) x n α γ,t), α

24 E n = ( E 1 i,j+1/2,k+1/2,e2 i+1/2,j,k+1/2,e3 i+1/2,j+1/2,k) n, B n+1/2 = ( B 1 i+1/2,j,k,b2 i,j+1/2,k,b3 i,j,k+1/2) n+1/2, ρ n = ρ n i+1/2,j+1/2,k+1/2, J n+1/2 = ( Ji,j+1/2,k+1/2 1, J i+1/2,j,k+1/2 2, J 3 ( ) ( + fi+1,j,k f i,j,k f i,j,k = dx ( fi,j,k f i 1,j,k f i,j,k = ( dx, f i,j+1,k f i,j,k dy, f i,j,k f i,j 1,k dy i+1/2,j+1/2,k) n+1/2, ± ±, f i,j,k+1 f i,j,k dz, f i,j,k f i,j,k 1 dz ), ). = + + = 0, + = + =,

25 dropping indices ρ n+1 ρ n + + J n+1/2 = 0, dt B n+1/2 B n 1/2 = 0. dt + E = ρ and B = 0 are initial conditions continuity equation ρ n+1 i+1/2,j+1/2,k+1/2 ρn i+1/2,j+1/2,k+1/2 dt + J 2 i+1/2,j+1,k+1/2 J 2 i+1/2,j,k+1/2 dy + J 1 i+1,j+1/2,k+1/2 J 1 i,j +1/2,k+1/2 dx + J 3 i+1/2,j+1/2,k+1 J 3 i+1/2,j+1/2,k dz ± = 0.

26 charge density ρ i,j,k = α Q α S i,j,k (x α,y α,z α ), S i,j,k (x α,y α,z α ) = S(X i x α,y j y α,z k z α ), = X i,y j,z k denote coordinates of the grid, (x α,y α,z α ) re the form-factor can be interpreted as a charge dens density decomposition: due to linearity of continuity equation, consider single superparticle vector W defined by J 1 i+1,j,k J 1 i,j,k = Qdx dt W 1 i,j,k, J 2 i,j+1,k J 2 i,j,k = Qdy dt W 2 i,j,k, J 3 i,j,k+1 J 3 i,j,k = Qdz dt W 3 i,j,k. continuity equation: W 1 + W 2 + W 3 = S(x + x,y + y,z+ z) S(x, y, z).

27 Here ( x, y, z) is 3-dimensional shift of the quasi-particle due to the motion. Shift of the quasi-particle generates eight functions S(x, y, z), S(x + x, y, z), S(x, y + y, z), S(x, y, z + z), S(x + x,y + y, z), S(x + x,y,z+ z), S(x, y + y,z+ z), S(x + x,y + y,z+ z). must increase proportionally. (2) The quasi-particle trajectory over time step is treated as a straight lin decompose any three-dimensional shift of the form-factor S(x, y, z) into three one-dimensional shifts: S(x + x,y + + y,z+ + z) + S(x, y, z) = S(x + x,y,z) S(x, y, z) We assume + S(x + x,y + y,z) S(x + x,y,z) + S(x + x,y + y,z+ z) S(x + x,y + y, z). (1) The vector (Wi,j,k 1,W2 i,j,k,w3 i,j,k ) is a linear combination of eight functions, Eq. (21). (2) W 1 + W 2 + W 3 = S(x + x,y + y,z+ z) S(x, y, z). (2) The sum of components of the vector is equal to the finite difference of form-factors, Eq. (20). (3) If one of the shifts x, y, z is zero, the corresponding component of W is also zero: x = 0 W 1 = 0, y = 0 W 2 = 0, z = 0 W 3 = 0. (4) If S(x, y, z) is symmetrical with respect to the permutation of (x, y), S(x, y, z) = S(y, x, z),and x = y, then W 1 = W 2. The same property is assumed for symmetries with respect to permutations of pairs (x, z) and (y, z). Only one linear combination of eight functions,obeys the conditions (1) (4):

28 Solution: W 1 = 1 3 S(x + x,y + y,z+ z) 1 3 S(x, y + y,z+ z) S(x + x,y,z + z) 1 6S(x, y, z + z) S(x + x,y + y,z) 1 6S(x, y + y,z) S(x + x,y,z) 1 3S(x, y, z), W 2 = 1 3 S(x + x,y + y,z+ z) 1 3 S(x + x,y,z + z) S(x, y + y,z+ z) 1 6 S(x, y, z + z) S(x + x,y + y,z) 1 6S(x + x,y,z) S(x, y + y,z) 1 3S(x, y, z), W 3 = 1 3 S(x + x,y + y,z+ z) 1 3 S(x + x,y + y,z) S(x, y + y,z+ z) 1 6 S(x, y + y,z) S(x + x,y,z + z) 1 6S(x + x,y,z) S(x, y, z + z) 1 3 S(x, y, z).