Full-wave simulation of tokamak plasma heating

Full-wave simulation of tokamak plasma heating Institut Elie Cartan de Nancy, UMR 7502, Université de Lorraine, CNRS & INRIA (project-team CALVI). B.P. 70239, 54506 Vandœuvre les Nancy Cedex, France. Université Pierre et Marie Curie, 21 juin 2012

Joint work with: Jean R. Roche (IECN) Takashi Hattori (IECN) Christoph Kirsch (ex-iecn) Pierre Bertrand (Institut Jean Lamour, U. Lorraine) Jean-Hugues Chatenet (Institut Jean Lamour, U. Lorraine) Yves Peysson (CEA Cadarache)

Outline

Outline 1 Physical and mathematical modelling 1 Physical assumptions 2 Mixed and augmented formulations 3 Spectral properties of the dielectric tensor 4 Well-posedness of the variational formulations 5 Discretisation (Fourier Finite element) 6 A first numerical test 2 The issue of accessibility 1 Electromagnetic dispersion relation 2 Accessibility criteria 3 More physical test cases

Part One: Physical and mathematical modelling

The name of the game

The name of the game (simplified)

Physical assumptions Strong external, time-invariant magnetic field B ext. Harmonic time dependence of self-consistent electromagnetic fields (of the form e iωt, with ω > 0).

Physical assumptions Strong external, time-invariant magnetic field B ext. Harmonic time dependence of self-consistent electromagnetic fields (of the form e iωt, with ω > 0). Plasma response described by the matrix: S id 0 K = id S 0 + i γ 0 0 0 e 0 0 0 ε 0 0 P 0 ω 0 0 1 }{{}}{{} ε r Landau term (1) Stix frame: third coordinate B ext. Cold plasma approximation of ε r is used, valid in the lower hybrid (LH) regime ω ci ω ω ce. (More on this later.)

Electromagnetic waves: model problem The domain Ω is a torus (tokamak plasma volume). curl curl E (ω/c) 2 KE = 0 in Ω, (2) div(ke) = 0 in Ω. (3)

Electromagnetic waves: model problem The domain Ω is a torus (tokamak plasma volume). curl curl E (ω/c) 2 KE = 0 in Ω, (2) div(ke) = 0 in Ω. (3) Boundary condition: Ω = Γ A Γ C = antenna + the rest E n = 0 on Γ C. (Phoney?) (4) On Γ A several possible models... Neumann: curl E n = iωµ 0 J s (5) Dirichlet: E n = E A n (6) Silver Müller: (E + i (c/ω) curl E n) n = E n (7) where J s = J s (0, 0, 1) is a current source equivalent to the electric field E A excited at the antenna.

Variational Formulations (1) Neumann BC on Γ A (fss) variational space for (2): H C 0 (curl; Ω) = {v H(curl; Ω) : v n = 0 on Γ C} Eq. (3): constraint mixed (unaugmented) formulation: Find (E, p) H C 0 (curl; Ω) H1 0 (Ω) or H1 0,C (Ω) such that a pl (E, F ) + b pl (F, p) = L(F ), F H C 0 (curl; Ω); (8) b pl (E, q) = 0, q H0,(C) 1 (Ω), (9) where H0,C 1 (Ω) = {w H1 (Ω) : w = 0 on Γ C } and a pl (E, F ) := (curl E curl F ) (ω/c) 2 (KE F ), b pl (F, p) := (KF grad p), L(F ) := iωµ 0 J s, F H 1/2 (Γ A ),H 1/2 (Γ A ). Above, ( ) = scalar product in L 2 (Ω).

Variational Formulations (2) Other approaches: consider the space X C N (K; Ω) := HC 0 (curl; Ω) H(div K; Ω) Mixed augmented formulation: Find (E, p) X C N (K; Ω) L2 (Ω) such that a s (E, F ) + b(f, p) = L(F ), F X C N (K; Ω); (10) b(e, q) = 0, q L 2 (Ω), (11) where a s (E, F ) := (curl E curl F ) + s (div KE div KF ) (ω/c) 2 (KE F ), b(f, p) := (div KF p).

More on plasma response (1) Experimentalists / coupling with other codes maps of external magnetic field B ext (x), density n s (x) and temperature T s (x) of each species s (electrons e, various ions). Compute maps of plasma and cyclotron frequencies: n s qs 2 ω ps =, ω cs = q s B ext, ε 0 m s m s then plasma parameter and collision frequency Λ = 4π 3 n e ( ) ε0 k B T 3/2 e n e qe 2 ; ν c = ω pe ln Λ. Λ

More on plasma response (2) Setting α(x) = ω + iν c (x), the entries of ε r are: S = 1 α ωps 2 ω α s 2 ωcs 2 (13) D = 1 ω cs ωps 2 ω α s 2 ωcs 2 (14) P = 1 1 ωps 2 (15) ωα s

More on plasma response (2) Setting α(x) = ω + iν c (x), the entries of ε r are: S = 1 α ωps 2 ω α s 2 ωcs 2 (13) D = 1 ω cs ωps 2 ω α s 2 ωcs 2 (14) P = 1 1 ωps 2 (15) ωα Landau damping rate (phenomenological): ( ) π ω 2 n e qe 2 m 1/2 e γ e = 4(k B T e ) 3/2 k 3 exp ω2 m e 4k B T e k 2. (16) Relies on plane wave analysis: k = k B ext / B ext. s

Spectral properties of K (1) Eigenvalues and uniform (in x), unitary eigenvector basis λ 1 = S + D, λ 2 = S D, λ 3 = P + i γ e ε 0 ω := P L (17) i 1 0 2 u 1 = 1 2, u 2 = i, u 3 = 0. (18) 2 2 0 0 1

Spectral properties of K (1) Eigenvalues and uniform (in x), unitary eigenvector basis λ 1 = S + D, λ 2 = S D, λ 3 = P + i γ e ε 0 ω := P L (17) i 1 0 2 u 1 = 1 2, u 2 = i, u 3 = 0. (18) 2 2 0 0 1 There always holds: Im λ i (x) > 0, hence: Im(v K(x)v) ζ(x) v v, v C 3. Suitable bounds on n s, T s, B ext : v C 3, x Ω, v K(x)v Im(v K(x)v) ζ v 2. (19)

Spectral properties of K (2) On the contrary, in the LH frequency range: Re λ 1 (x) > 0, Re λ 2,3 (x) < 0. Without the imaginary parts (damping), the well-posedness proof breaks down!!! (Still OK if ν c = 0 or γ e = 0, but not both.)

Well-posedness of the variational formulations (1) Lemma (Sébelin, Bénilan 97) Let V H two Hilbert spaces, with continuous embedding. Let a(, ) be a sesquilinear form on V satisfying the inequalities: Re a(v, v) α v 2 V λ v 2 H, Im a(v, v) γ v 2 H, (20) for some strictly positive constants α, γ, λ and all v V. Then a(, ) is V -elliptic, i.e., a(v, v) α v 2 V. With the bound (19): Im(Kv v) ζ v L 2, v L 2 (Ω). a pl (u, v) = (curl u curl v) (ω/c) 2 (Ku v) satisfies (20) on V = H(curl; Ω), with H = L 2 (Ω);

Well-posedness of the variational formulations (1) Lemma (Sébelin, Bénilan 97) Let V H two Hilbert spaces, with continuous embedding. Let a(, ) be a sesquilinear form on V satisfying the inequalities: Re a(v, v) α v 2 V λ v 2 H, Im a(v, v) γ v 2 H, (20) for some strictly positive constants α, γ, λ and all v V. Then a(, ) is V -elliptic, i.e., a(v, v) α v 2 V. With the bound (19): Notice: Im(Kv v) ζ v L 2, v L 2 (Ω). a pl (u, v) = (curl u curl v) (ω/c) 2 (Ku v) satisfies (20) on V = H(curl; Ω), with H = L 2 (Ω); a s (u, v) = a pl (u, v) + s (div Ku div Kv) satisfies (20) on V = H(curl, div K; Ω) if Re s > 0 and Im s 0. coercivity ( Fredholm alternative); no BC required.

Well-posedness of the variational formulations (2) Augmented formulation (12) well-posed on the closed subspace X C N (K; Ω) by Lax Milgram.

Well-posedness of the variational formulations (2) Augmented formulation (12) well-posed on the closed subspace X C N (K; Ω) by Lax Milgram. (K grad u grad v) elliptic on H0 1 (Ω) well-posedness of K u := div(k grad u) = f H 1 (Ω). "Warped" Hodge decomposition for all u L 2 (Ω): u = grad ϕ + u T, with: ϕ H 1 0 (Ω), div Ku T = 0. inf-sup condition for b pl (v, p) = (Kv grad p) on H 0 (curl; Ω) H 1 0 (Ω) and HC 0 (curl; Ω) H1 0,(C) (Ω) inf-sup condition for b(v, p) = (div Kv p) on X N (K; Ω) L 2 (Ω) and X C N (K; Ω) L2 (Ω)

Dimension reduction (1)

Dimension reduction (2) The domain Ω is described by cylindrical coordinates (R, Z, φ), which are related to the cartesian ones (x, y, z) by x = R cos φ, y = R sin φ, z = Z. The 3D problem is reduced to a series of 2D ones by expanding all fields u(r, Z, φ) as Fourier series in φ: u(r, Z, φ) = ν Z u ν (R, Z ) e iνφ, (21) where the coefficients u ν (R, Z ) are defined on a meridian section Ω := {(R, Z ) R 2 (R, Z, π/2) Ω}. Furthermore: curl u = ν Z (curl ν u ν ) e iνφ, div u = ν Z(div ν u ν ) e iνφ, (22) where curl ν, div ν are obtained by setting φ iν in the expressions of curl ν, div ν in cylindrical coordinates.

Dimension reduction (3) Simplifying assumptions: External magnetic field B ext (x) = B ext (R, Z ) e φ. Particles constrained to move and diffuse freely along field lines of B ext

Dimension reduction (3) Simplifying assumptions: External magnetic field B ext (x) = B ext (R, Z ) e φ. Particles constrained to move and diffuse freely along field lines of B ext plasma properties averaged in φ n s, T s, hence S, D, P, are independent of φ. The r.h.s. J s has a single Fourier mode: J s = J ν 0 s assign a definite value to k (x) = ν 0 /R in (16) the whole matrix K is independent of φ. As a consequence: e iν 0φ div Ku = ν Z(div ν Ku ν ) e iνφ, and, by orthogonality: a s (u, v) = ν Z a s,(ν) (u ν, v ν ), b(v, p) = ν Z b (ν) (v ν, p ν ).

Discretisation: Fourier Finite Element Method A 2D problem is obtained for each pair of Fourier coefficients (E ν, p ν ), ν Z, set in the spaces X C N (K; Ω) L 2 ( Ω), traces on Ω of the 3D spaces: a s,(ν) (E ν, F ) + b (ν) (F, p ν ) = L ν (F ), F X C N (K; Ω); (23) b (ν) (E ν, q) = 0, q L 2 ( Ω). (24) The r.h.s. and the solution are zero except when ν = ν 0. We use a Taylor-Hood P 2 -iso-p 1 finite element (continuous fields and multipliers) to discretize the MAVF (23) (24). ( As,(ν) B ) ( ) ( ) (ν) E h Lν ν B (ν) 0 p ν 2h =, (25) 0 where E ν h and p ν 2h nodal values of E ν and p ν. Notice that the matrix of the system (25) is complex-valued and non-hermitian.

Essential boundary conditions In a local basis {n, τ, e φ = n τ }, we have: E 1 B 1 0 E n = EB 2 0 = E 3 EB 3 B. 0 EB 2 The boundary condition E n = 0 on Γ C becomes: (EB 2, E B 3 ) = (0, 0). (26) Denoting P = transformation matrix between {e R, e Z, e φ } and {n, τ, n τ } on Γ C nodes, identity elsewhere: v M u = v f v B (P MP) u B = v B (P f ). Thus, one sets (E 2 B, E 3 B ) = (0, 0) in P MP and P f. Generalisation to E n = E A n on Γ A obvious.

Numerical test We consider a 2D test problem analogous to (23) (24), with K = I (and s = 1 in the augmented formulation): curl ν curl ν E η 2 E = f in Ω, (27) div ν E = g in Ω, (28) E n = 0 on Γ C, (29) curl ν E n = h on Γ A. (30) The r.h.s. f, g, h are chosen so that the exact solution is E(R, Z ) = (R R 0, Z, 0) H 0 (curl, Ω) H(div, Ω) and ν = 220. This gives rise to six independent tests, for the real and imaginary parts of the three coordinates of E.

Figure: The real part of the test solution E.

Figure: The imaginary part of the test solution E.

Part Two: The issue of accessibility

Dispersion relation (1) We look for plane wave solutions of the form Ee i(k x ωt) to Maxwell s equations in the plasma. we set E = Ee ik x in the propagation equation (2): curl curl E (ω/c) 2 KE = 0. No BC is assumed (propagation in an infinite medium). Therefore: ik (ik E) (ω/c) 2 K E = 0. Multiplying by (c/ω) 2, we get the dispersion relation n (n E) + KE = 0. (31) where n = c ω k is the vector index of refraction.

Dispersion relation (2) Take a (local) Stix frame (e 1, e 2, e 3 ), with e 3 in the direction of B ext and k e 2 = 0. Thus: n = (n, 0, n ).

Dispersion relation (2) Take a (local) Stix frame (e 1, e 2, e 3 ), with e 3 in the direction of B ext and k e 2 = 0. Thus: n = (n, 0, n ). The dispersion relation (31) is rewritten as: ME = 0, (32) where: M = S n 2 id n n id S n 2 n2 0 n n 0 P L n 2. To have a non-trivial solution, there must hold det M = 0: Sn 4 [(S + P L)(S n 2 ) D2 ]n 2 + P L[(S n 2 )2 D 2 ] = 0. (33)

Accessibility (1) To simplify the discussion, we assume there is no damping (ν c = γ e = 0). Effect of actual damping on this analysis: A propagating wave will still propagate (with a little attenuation). An evanescent wave will not become propagating. Therefore, the entries S, D, P L = P of K are real. Under this assumption, we say there is strict accessibility if n R 3 solution to (33); vague accessibility if n C 3 solution to (33), with Im n Re n.

Accessibility (1) To simplify the discussion, we assume there is no damping (ν c = γ e = 0). Effect of actual damping on this analysis: A propagating wave will still propagate (with a little attenuation). An evanescent wave will not become propagating. Therefore, the entries S, D, P L = P of K are real. Under this assumption, we say there is Recall: strict accessibility if n R 3 solution to (33); vague accessibility if n C 3 solution to (33), with Im n Re n. in the LH regime, there holds: P < 0 < S < D and S, D P. (34)

Accessibility (2) We fix n R (Phenomenology...) and write r = S n 2. Then, det M is a second degree polynomial in n 2 : det M = Q(n 2 ) = a r n 4 b r n 2 + c r, where: a r = S, b r = (S + P)r D 2, c r = P(r 2 D 2 ). Let r := b 2 r 4 a r c r be the discriminant of Q. 1 If r 0 and Q has a non-negative root, there is strict accessibility. 2 If r 0 and both roots are negative, there is strict inaccessibility (all waves are evanescent). 3 If r < 0, there cannot be strict accessibility, but there may be vague accessibility.

Strict Accessibility One calculates: r = (S P) 2 r 2 2D 2 (S + P) r + D 2 (D 2 + 4SP). 1 r 0 r / (r, r + ), with: r ± := D2 (S + P) ± δ (S P) 2, δ := 4D 2 SP(D 2 (S P) 2 ). Actually, in the LH regime (34), δ > 0.

Strict Accessibility One calculates: r = (S P) 2 r 2 2D 2 (S + P) r + D 2 (D 2 + 4SP). 1 r 0 r / (r, r + ), with: r ± := D2 (S + P) ± δ (S P) 2, δ := 4D 2 SP(D 2 (S P) 2 ). Actually, in the LH regime (34), δ > 0. 2 Both roots negative (strict inaccessibility) iff b r a r < 0 and c r a r > 0 r > r := D2 S + P and r D.

The case r < 0 We have vague accessibility iff n C s.t. det M = 0 and Re(n ) Im(n ).

The case r < 0 We have vague accessibility iff n C s.t. det M = 0 and Re(n ) Im(n ). We calculate: n 2 = (b r ± i r )/(2a r ) = Re(n ) 2 = b r + br 2 r, Im(n ) 2 = b r + b 2 r r. 4a r 4a r Necessary condition: b r 0. In this case: Re(n ) Im(n ) = κ + 1 + κ, where: κ := b2 r r. Thus: vague accessibility b r 0 and κ large enough

The case r < 0 (continued) We investigate the condition ɛ 2 b 2 r + r 0, for ɛ > 0 given. The l.h.s. is a second-degree polynomial in r, it is positive outside its roots: ρ ɛ ± = (ɛ 2 + 1)(S + P)D 2 ± δ ɛ (ɛ 2 + 1)(S 2 + P 2 ) 2(ɛ 2 1)SP, with: δ ɛ = 4D 2 SP[(ɛ 2 + 1)(D 2 S 2 P 2 ) 2(ɛ 2 1)SP]. In the LH regime (34), δ ɛ > 0 and ρ ɛ r ρ ɛ +.

Accessibility: conclusions Proposition 1 Assume ω ci ω ω ce and ν c = γ e = 0. Then for ɛ small enough. r < ρ ɛ < r < ρ ɛ + < r + < S, 2 Let λ = D2 P. One has strict accessibility iff n > S r S + 2 Sλ + λ. One has vague accessibility ( Re(n ) / Im(n ) > K ) iff S ρ ɛ < n < S r, with ɛ = 2K /(K 2 1).

"Physical" test case We consider an electron-proton plasma with densities n e = n i = 10 19 m 3 Plasma and cyclotron frequencies (rad/s): (constant). ω pe 1.7827 10 11, ω pi 4.1638 10 9, ω ce 5.6202 10 11, ω ci 3.0659 10 8, for an external magnetic field of intensity B ext = 3.2 T. Lower hybrid frequency ω LH = ω ci ω ce 1.3127 10 10 rad/s. Electron temperature: k B T e = 1 kev ν c, γ e. We present a series of tests on the physical problem (2) (5) and its 2D VF (23) (24), for the values ω = ω LH, 2ω LH or ω LH /2 of the frequency, and several values of the Fourier index ν.

Influence of parameters Parallel wave vector: k (x) = ν R or n (x) = νc Rω.

Influence of parameters Parallel wave vector: k (x) = ν R or n (x) = νc Rω. ω = const., ν = { more accessible γ e 0 for ν large enough,

Influence of parameters Parallel wave vector: k (x) = ν R or n (x) = νc Rω. ω = const., ν = ν = const., ω = { more accessible γ e 0 for ν large enough, { less accessible γ e 0 for ω large enough. Note: k ν/r 0 in γ e = K = constant. On the other hand we should have an a priori estimate E ν C ν α rhs (in the suitable spaces). So, for a fixed ω, there is an optimal ν where the wave emitted by the antenna penetrates most the plasma.

Figure: The real part of E for ω = ω LH and ν = 200

Figure: The real part of E for ω = ω LH and ν = 300

Figure: The real part of E for ω = ω LH and ν = 400

Figure: The real part of E for ω = ω LH /2 and ν = 100

Figure: The real part of E for ω = ω LH /2 and ν = 200

Figure: The real part of E for ω = ω LH /2 and ν = 300

Figure: The real part of E for ω = 2 ω LH and ν = 400

Figure: The real part of E for ω = 2 ω LH and ν = 600

Figure: The real part of E for ω = 2 ω LH and ν = 700

Figure: The real part of E for ω = 2 ω LH and ν = 800

Provisional conclusions

In progress and to do Feasibility of simulations incorporating most of the physics

In progress and to do Feasibility of simulations incorporating most of the physics (to do: variable plasma properties) Simulations confirm accessibility analysis

In progress and to do Feasibility of simulations incorporating most of the physics (to do: variable plasma properties) Simulations confirm accessibility analysis Non-hermitianness crucial for well-posedness, but precludes the use of many numerical methods To do: comparison of antenna boundary conditions More realistic test-cases: transversal wave length domain size Domain decomposition