MHD Simulations for Fusion Applications Lecture 4 Implicit Methods and the M3D- C 1 Approach Stephen C. Jardin Princeton Plasma Physics Laboratory CEMRACS ‘10 Marseille, France July 22, 2010 1
Outline of Remainder of Lecture Yesterday Today • Galerkin method • Split implicit time C 1 finite elements • differencing for MHD • Examples • Accuracy, spectral pollution, and • Split implicit time representation of the differencing vector fields • Projections of the split implicit equations • Energy conserving subsets • 3D solver strategy • Examples
2-Fluid MHD Equations: ∂ + ∇ • n = ( V ) 0 continuity n ∂ t ∂ B = −∇× ∇ = μ = ∇× E i B 0 J B Maxwell 0 ∂ t ∂ V + •∇ + ∇ = × − ∇ + μ ∇ 2 ( V V ) J B i Π V momentum nM p ∂ i G V t 1 ( ) + × = η + × − ∇ E V B J J B Ohm's law p e ne ∂ ⎛ ⎞ 3 3 p + ∇ = − ∇ + η −∇ + + 2 i V i V i q electron energy e ⎜ ⎟ p p J Q S Δ ∂ 2 ⎝ 2 e ⎠ e e F e t ∂ ⎛ ⎞ 3 3 p 2 + ∇ = − ∇ + μ ∇ −∇ − + i V i V i q ion energy i ⎜ ⎟ p p V Q S Δ ∂ i i i Fi 2 ⎝ 2 ⎠ t μ viscosity V fluid velocity number density n Idea l MHD η resistivity electron pressure Β magnetic field p e Resistive MH D q ,q heat fluxes ion pressure J current density p i e i 2- flui d MHD equipartition ≡ + Q E electric field p p p Δ e i μ permeability ≡ ρ 3 mass density electron charge e nM 0 i
The split-implicit time advance 4
Consider a simple 1-D Hyperbolic System of Equations (Wave Equation) ⎡ ⎤ ⎛ ⎞ ⎛ ⎞ + − + − + − 1 1 1 n n n n n n u u v v v v + − + − = θ 1/2 1/2 + − θ 1/2 1/2 j j ⎢ j j j j ⎥ ⎜ ⎟ (1 ) ⎜ ⎟ c ⎜ ⎟ ⎜ ⎟ δ δ δ ⎢ ⎥ t x x ⎝ ⎠ ⎝ ⎠ ⎣ ⎦ ⎡ ⎤ ⎛ ⎞ ⎛ ⎞ + − + − + − 1 1 1 n n n n n n v v u u u u + + + + 1/2 1/2 = θ 1 + − θ 1 ⎢ ⎥ j j ⎜ j j ⎟ (1 ) ⎜ j j ⎟ c ⎜ ⎟ ⎜ ⎟ δ δ δ ⎢ ⎥ t ⎝ x ⎠ ⎝ x ⎠ ⎣ ⎦ Substitute from second equation into first: ⎡ ⎤ ⎛ + + + ⎞ ⎛ ⎞ ⎛ ⎞ 1 − 1 + 1 − + − 2 2 n n n n n n n n u u u u u u v v + − + − + − + 1 = + δ 2 θ 2 1 1 + θ − θ 1 1 + δ 1/2 1/2 ⎢ ⎥ ( ) ⎜ j j j ⎟ (1 ) ⎜ j j j ⎟ ⎜ j j ⎟ n n u u tc tc ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ δ δ δ 2 2 j j ⎢ ⎥ x x x ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎣ ⎦ δ ( ) ( ) tc ⎡ ⎤ + + + = + θ − + − θ − 1 1 1 (1 ) n n n n n n v v u u u u ⎣ ⎦ + + + + 1/2 1/2 δ 1 1 j j j j j j x These two equations are completely equivalent to those on the previous page, but can be solved sequentially! Only first involves Matrix Inversion … Diagonally Dominant 5
An alternate derivation: ∂ ∂ u v ∂ ∂ ∂ = ⎡ ⎤ u v c = + θδ n ∂ ∂ c v t ⎢ ⎥ t x ∂ ∂ ∂ ⎣ ⎦ t x t Expand RHS in Taylor ∂ ∂ v u = ∂ ∂ ∂ ⎡ ⎤ series in time to time- v u c = + θδ n ∂ ∂ c u t ⎢ ⎥ t x ∂ ∂ ∂ center ⎣ ⎦ t x t 6
An alternate derivation: ∂ ∂ u v ∂ ∂ ∂ = ⎡ ⎤ u v c = + θδ n ∂ ∂ c v t ⎢ ⎥ t x ∂ ∂ ∂ ⎣ ⎦ t x t Expand RHS in Taylor ∂ ∂ v u = ∂ ∂ ∂ ⎡ ⎤ series in time to time- v u c = + θδ n ∂ ∂ c u t ⎢ ⎥ t x ∂ ∂ ∂ center ⎣ ⎦ t x t Substitute from second ⎡ ⎤ ∂ ∂ ⎛ ∂ ∂ ⎞ ⎡ ⎤ u u = + θδ + θδ n n ⎢ ⎥ c v t c ⎜ u t ⎟ ⎢ ⎥ equation into first ∂ ∂ ∂ ∂ ⎣ ⎦ ⎝ ⎠ ⎣ ⎦ t x x t ∂ ∂ ∂ ⎡ ⎤ v u = + θδ n c u t ⎢ ⎥ ∂ ∂ ∂ ⎣ ⎦ t x t 7
An alternate derivation: ∂ ∂ u v ∂ ∂ ∂ = ⎡ ⎤ u v c = + θδ n ∂ ∂ c v t ⎢ ⎥ t x ∂ ∂ ∂ ⎣ ⎦ t x t Expand RHS in Taylor ∂ ∂ v u = ∂ ∂ ∂ ⎡ ⎤ series in time to time- v u c = + θδ n ∂ ∂ c u t ⎢ ⎥ t x ∂ ∂ ∂ center ⎣ ⎦ t x t Substitute from second ⎡ ⎤ ∂ ∂ ⎛ ∂ ∂ ⎞ ⎡ ⎤ u u = + θδ + θδ n n ⎢ ⎥ c v t c ⎜ u t ⎟ ⎢ ⎥ equation into first ∂ ∂ ∂ ∂ ⎣ ⎦ ⎝ ⎠ ⎣ ⎦ t x x t ∂ ∂ ∂ ⎡ ⎤ v u = + θδ n c u t ⎢ ⎥ ∂ ∂ ∂ ⎣ ⎦ t x t Use standard centered difference in time: ⎡ ⎤ ⎡ ⎤ ∂ ∂ ∂ 2 2 − θ δ + = + θ − θ δ + δ 2 2 2 1 2 2 1 ( ) 1 (1 )( ) n n n ⎢ ⎥ ⎢ ⎥ t c u t c u tc v ∂ 2 ∂ 2 ∂ ⎣ ⎦ ⎣ ⎦ x x x ∂ ∂ ⎡ ⎤ + = + δ θ + + − θ 1 1 (1 ) n n n n v v tc u u ⎢ ⎥ ∂ ∂ ⎣ ⎦ x x 8
An alternate derivation: ∂ ∂ u v ∂ ∂ ∂ = ⎡ ⎤ u v c = + θδ n ∂ ∂ c v t ⎢ ⎥ t x ∂ ∂ ∂ ⎣ ⎦ t x t Expand RHS in Taylor ∂ ∂ v u = ∂ ∂ ∂ ⎡ ⎤ series in time to time- v u c = + θδ n ∂ ∂ c u t ⎢ ⎥ t x ∂ ∂ ∂ center ⎣ ⎦ t x t Substitute from second ⎡ ⎤ ∂ ∂ ⎛ ∂ ∂ ⎞ ⎡ ⎤ u u = + θδ + θδ n n ⎢ ⎥ c v t c ⎜ u t ⎟ ⎢ ⎥ equation into first ∂ ∂ ∂ ∂ ⎣ ⎦ ⎝ ⎠ ⎣ ⎦ t x x t ∂ ∂ ∂ ⎡ ⎤ v u = + θδ n c u t ⎢ ⎥ ∂ ∂ ∂ ⎣ ⎦ t x t Use standard centered difference in time: ⎡ ⎤ ⎡ ⎤ ∂ ∂ ∂ 2 2 − θ δ + = + θ − θ δ + δ 2 2 2 1 2 2 1 ( ) 1 (1 )( ) n n n ⎢ ⎥ ⎢ ⎥ t c u t c u tc v ∂ 2 ∂ 2 ∂ ⎣ ⎦ ⎣ ⎦ x x x ∂ ∂ ⎡ ⎤ + = + δ θ + + − θ 1 1 (1 ) n n n n v v tc u u ⎢ ⎥ ∂ ∂ ⎣ ⎦ x x This is the same operator as before when centered spatial differences used 9
W (next 3 vgs) δ
Linear ideal MHD response can be analyzed with δ W approach ∂ ξ let = V introduce a displacement vector ξ ∂ t linearize equations about =0 V ∂ 2 ξ 1 1 ( ) ( ) ρ + ∇ = ∇× × + ∇× × B B B B p 1 1 1 ∂ μ μ 2 t 0 0 ( ) = ∇× × B ξ B 1 2 ( ) + ∇ = − ∇ i ξ i ξ p p p 1 3 or, ∂ 2 ξ ρ = F ( ) ξ Linearized equation of motion ∂ 2 t 1 ( ) ( ) ( ) Ideal MHD = ⎡ ∇× × + ∇× × ⎤ + ∇ ∇ + ∇ i i F ξ ( ) B Q Q B ξ 5 ξ p p ⎣ ⎦ μ 3 operator 0 ( ) ≡ = ∇× × Q B ξ B 1
δ W is the potential energy for a given displacement field ∂ 2 ξ ρ = F ξ ( ) ∂ 2 t 1 ( ) ( ) ( ) μ ⎡ ⎤ = ∇× × + ∇× × + ∇ ∇ + ∇ F ξ ( ) B Q Q B ξ i i ξ 5 ⎣ ⎦ p p 3 0 ( ) ≡ = ∇ × × Q B ξ B 1 x e ω = † Assume ( , ) ξ ξ ( ) and take dot product with - ξ , i t 1 x t 2 and integrate over volume to get perturbed energy ∫ ∫ ρω τ = − τ ≡ δ 2 2 † † 1 ξ 1 ξ F ξ i ( ) ( ξ ξ , ) d d W 2 2 δ < → if 0 for any displacement field ξ instabilit y W
The plasma motion will be such as to minimize δ W This is the term associated with the fast wave…it ∇ ≅ − i ξ 2 ξ i κ ~ 0 is positive definite with a large multiplier. Any ⊥ ⊥ unstable plasma motion will make this term small [ ] ⎡ ⎤ 2 2 + ∇ + + ∇ 2 2 1 Q 1 i ξ 2 ξ i κ 5 i ξ B p ∫ ⊥ ⊥ ⊥ ⎢ μ μ ⎥ 3 δ = τ 1 0 0 W d ( )( ) 2 ⎢ ⎥ − ∇ − σ × 2 ξ i κ ξ i ξ B Q i p ⎣ ⎦ ⊥ ⊥ ⊥ ⊥ = b B / unit vector in direction of field B = ∇ κ b i b curvature of magnetic field σ = 2 J B i / parallel current density B ( ) = ∇× × Q ξ B perturbed magnetic field = Q perpendicular component of Q ⊥ = ξ perpendicular component of ξ ⊥
Now apply the split implicit advance to the basic 3D (ideal) MHD equations: 1 [ ] Ideal MHD Equations for velocity, � ρ = ∇× × −∇ V B B p 0 μ magnetic field, and pressure: 0 [ ] � = ∇× × B V B Symmetric Hyperbolic System = − ∇ − γ ∇ � V i i V p p p 7-waves 14
Now apply the split implicit advance to the basic 3D (ideal) MHD equations: 1 [ ] Ideal MHD Equations for velocity, � ρ = ∇× × −∇ V B B p 0 μ magnetic field, and pressure: 0 [ ] � = ∇× × B V B Symmetric Hyperbolic System = − ∇ − γ ∇ � V i i V p p p 7-waves 1 ( ) ( ) ( ) � � � ⎡ ⎤ ρ = ∇× + θδ × + θδ −∇ + θδ � V B B B B t t p tp ⎣ ⎦ 0 μ 0 Taylor Expand in ( ) � � ⎡ ⎤ = ∇× + θδ × B V V B t ⎣ ⎦ Time as before ( ) ( ) � � = − + θδ ∇ − γ ∇ + θδ � V V i i V V p t p p t 15
Recommend
More recommend