Stable Numerical Scheme for the Magnetic Induction Equation with - PowerPoint PPT Presentation

Stable Numerical Scheme for the Magnetic Induction Equation with Hall Effect Paolo Corti joint work with Siddhartha Mishra ETH Zurich, Seminar for Applied Mathematics 24-29th June 2012, Hyp 2012, Padova

Outline MHD Theoretical Analysis Discontinuous Galerkin One Dimensional Model Numerical Examples Conclusion Outline Formulation and motivation of the problem Theoretical analysis DG Formulation 1D Model Numerical Tests P. Corti 24-29th June 2012, Hyp 2012, Padova p. 2

Outline MHD Theoretical Analysis Discontinuous Galerkin One Dimensional Model Numerical Examples Conclusion Magnetic Reconnection Change in topology of the magnetic field U in U out U out U in Figure: Schematic of a reconnection. Magnetic energy ⇒ kinetic and thermal energy Dissipation P. Corti 24-29th June 2012, Hyp 2012, Padova p. 3

Outline MHD Theoretical Analysis Discontinuous Galerkin One Dimensional Model Numerical Examples Conclusion MHD Equations The equations ∂ρ ∂ t = −∇ · ( ρ u ) � � � � ∂ ( ρ u ) p + B 2 ρ uu ⊤ + I 3 × 3 − BB ⊤ = −∇ ∂ t 2 �� � � E + p − B 2 ∂ E u + E × B ∂ t = −∇ 2 ∂ B ∂ t = −∇ × E are coupled through the equation of state γ − 1 + ρ u 2 p + B 2 E = 2 2 To complete the formulation of the problem we need to state some equation for E P. Corti 24-29th June 2012, Hyp 2012, Padova p. 4

Outline MHD Theoretical Analysis Discontinuous Galerkin One Dimensional Model Numerical Examples Conclusion Ideal MHD Standard model for E : Ohm’s Law E = − u × B Problem: no dissipation ⇒ “frozen” condition. P. Corti 24-29th June 2012, Hyp 2012, Padova p. 5

Outline MHD Theoretical Analysis Discontinuous Galerkin One Dimensional Model Numerical Examples Conclusion Ideal MHD Standard model for E : Ohm’s Law E = − u × B Problem: no dissipation ⇒ “frozen” condition. We need to add dissipation Resistive MHD: E = − u × B + η J not sufficient for fast reconnection. P. Corti 24-29th June 2012, Hyp 2012, Padova p. 5

Outline MHD Theoretical Analysis Discontinuous Galerkin One Dimensional Model Numerical Examples Conclusion Ideal MHD Standard model for E : Ohm’s Law E = − u × B Problem: no dissipation ⇒ “frozen” condition. We need to add dissipation Resistive MHD: E = − u × B + η J not sufficient for fast reconnection. We need another model... Numerical simulation and laboratory experiment ⇒ Hall Effect P. Corti 24-29th June 2012, Hyp 2012, Padova p. 5

Outline MHD Theoretical Analysis Discontinuous Galerkin One Dimensional Model Numerical Examples Conclusion Generalized Ohm’s Law � δ e � 2 1 � ∂ J � ↔ J × B p E = − u × B + η J + δ i − δ i ∇ ∂ t + ( u · ∇ ) J + L 0 L 0 L 0 ρ ρ ρ P. Corti 24-29th June 2012, Hyp 2012, Padova p. 6

Outline MHD Theoretical Analysis Discontinuous Galerkin One Dimensional Model Numerical Examples Conclusion Generalized Ohm’s Law � δ e � 2 1 � ∂ J � ↔ J × B p E = − u × B + η J + δ i − δ i ∇ ∂ t + ( u · ∇ ) J + L 0 L 0 L 0 ρ ρ ρ Resistivity Hall effect Electron pressure Electron inertia P. Corti 24-29th June 2012, Hyp 2012, Padova p. 6

Outline MHD Theoretical Analysis Discontinuous Galerkin One Dimensional Model Numerical Examples Conclusion Generalized Ohm’s Law � δ e � 2 1 � ∂ J � ↔ J × B p E = − u × B + η J + δ i − δ i ∇ ∂ t + ( u · ∇ ) J + L 0 L 0 L 0 ρ ρ ρ Resistivity Hall effect Electron pressure Electron inertia J is the electric current given by Ampère’s law J = ∇ × B X.Qian, J.Bablás, A. Bhattacharjee, H.Yang (2009) P. Corti 24-29th June 2012, Hyp 2012, Padova p. 6

Outline MHD Theoretical Analysis Discontinuous Galerkin One Dimensional Model Numerical Examples Conclusion General Induction Equation P. Corti 24-29th June 2012, Hyp 2012, Padova p. 7

Outline MHD Theoretical Analysis Discontinuous Galerkin One Dimensional Model Numerical Examples Conclusion General Induction Equation Faraday’s law ∂ B ∂ t = −∇ × E Generalized Ohm law Ampère’s law ↔ p isotropic. P. Corti 24-29th June 2012, Hyp 2012, Padova p. 7

Outline MHD Theoretical Analysis Discontinuous Galerkin One Dimensional Model Numerical Examples Conclusion General Induction Equation Faraday’s law ∂ B ∂ t = −∇ × E Generalized Ohm law Ampère’s law ↔ p isotropic. Are combined to obtain � � � δ e � 2 1 ∂ B + ρ ∇ × ( ∇ × B ) = ∇ × ( u × B ) − η ∇ × ( ∇ × B ) ∂ t L 0 � δ e � 2 1 ρ ∇ × (( u · ∇ )( ∇ × B )) − δ i 1 ρ ∇ × (( ∇ × B ) × B ) − L 0 L 0 P. Corti 24-29th June 2012, Hyp 2012, Padova p. 7

Outline MHD Theoretical Analysis Discontinuous Galerkin One Dimensional Model Numerical Examples Conclusion General Induction Equation Faraday’s law ∂ B ∂ t = −∇ × E Generalized Ohm law Ampère’s law ↔ p isotropic. Are combined to obtain � � � δ e � 2 1 ∂ B + ρ ∇ × ( ∇ × B ) = ∇ × ( u × B ) − η ∇ × ( ∇ × B ) ∂ t L 0 � δ e � 2 1 ρ ∇ × (( u · ∇ )( ∇ × B )) − δ i 1 ρ ∇ × (( ∇ × B ) × B ) − L 0 L 0 This equation preserve the divergence of the magnetic field d dt ( ∇ · B ) = 0 P. Corti 24-29th June 2012, Hyp 2012, Padova p. 7

Outline MHD Theoretical Analysis Discontinuous Galerkin One Dimensional Model Numerical Examples Conclusion Symmetrized Equation Using the identity ∇ × ( u × B ) = ( B · ∇ ) u − B ( ∇ · u ) + u ( ∇ · B ) − ( u · ∇ ) B P. Corti 24-29th June 2012, Hyp 2012, Padova p. 8

Outline MHD Theoretical Analysis Discontinuous Galerkin One Dimensional Model Numerical Examples Conclusion Symmetrized Equation Using the identity ∇ × ( u × B ) = ( B · ∇ ) u − B ( ∇ · u ) + u ( ∇ · B ) − ( u · ∇ ) B Since the magnetic field is solenoidal ∇ · B = 0 we subtract u ( ∇ · B ) to the right side of the equation � � � δ e � 2 ∂ B + ∇ × ( ∇ × B ) = ∂ t L 0 ( B · ∇ ) u − B ( ∇ · u ) − ( u · ∇ ) B − η ∇ × ( ∇ × B ) � δ e � 2 1 ρ ∇ × (( u · ∇ )( ∇ × B )) − δ i 1 ρ ∇ × (( ∇ × B ) × B ) − (1) L 0 L 0 P. Corti 24-29th June 2012, Hyp 2012, Padova p. 8

Outline MHD Theoretical Analysis Discontinuous Galerkin One Dimensional Model Numerical Examples Conclusion Boundary Condition Ω ⊂ R 3 is a smooth domain ∂ Ω in = { x ∈ ∂ Ω | n · u < 0 } is the inflow boundary. Natural BC η ( B × n ) = 0 on ∂ Ω \ ∂ Ω in (2) Inflow BC B = 0 on ∂ Ω in δ i J = 0 on ∂ Ω in (3) P. Corti 24-29th June 2012, Hyp 2012, Padova p. 9

Outline MHD Theoretical Analysis Discontinuous Galerkin One Dimensional Model Numerical Examples Conclusion Estimate Theorem For u ∈ C 2 (Ω) and B solution of (1) satisfying (2) and (3), then this estimate holds � � � δ e � 2 1 d � B � 2 ρ �∇ × B � 2 L 2 (Ω) + dt L 0 L 2 (Ω) � � � δ e � 2 1 ≤ C 1 � B � 2 ρ �∇ × B � 2 L 2 (Ω) + (4) L 2 (Ω) L 0 with C 1 a constant that depend on u and its derivative only. P. Corti 24-29th June 2012, Hyp 2012, Padova p. 10

Outline MHD Theoretical Analysis Discontinuous Galerkin One Dimensional Model Numerical Examples Conclusion System of Equations with Auxiliary Variables ∂ B ∂ t + U 1 B + ( u ∇ ) B = − η ∇ × J − α ∇ × ˜ E 1 − β ∇ × ˜ E 2 J = ∇ × B E 1 = J × B ˜ E 2 = ( ∂ J ∂ t + ( u ∇ ) J ) ˜ Boundary Condition: η ( B × n ) = 0 on ∂ Ω B = G 1 on ∂ Ω in β J = G 2 on ∂ Ω in . U 1 depends on ∂ u i δ 2 δ i e ∂ x j , α = L 0 ρ and β = L 2 0 ρ P. Corti 24-29th June 2012, Hyp 2012, Padova p. 11

Outline MHD Theoretical Analysis Discontinuous Galerkin One Dimensional Model Numerical Examples Conclusion Define � � ( v , w ) T h = v ( x ) w ( x ) dx K K ∈T h � � � v , w � Faces = v ( x ) w ( x ) ds e f ∈ Faces where T h a triangulation of Ω . Faces can be F h set of faces in T h . F I h set of inner faces in T h . Γ h set of boundary faces in T h . Γ + h set of outflow boundary faces in T h . Γ − h set of inflow faces boundary in T h . P. Corti 24-29th June 2012, Hyp 2012, Padova p. 12

Outline MHD Theoretical Analysis Discontinuous Galerkin One Dimensional Model Numerical Examples Conclusion To have unique valued on faces we define: averages { . } normal jumps [ [ . ] ] N tangential jumps [ [ . ] ] T P. Corti 24-29th June 2012, Hyp 2012, Padova p. 13