EMAFF: MAgnetic Equations with FreeFem++

Transcription

1 EMAFF: MAgnetic Equations with FreeFem++ The Grad-Shafranov equation & The Current Hole Erwan DERIAZ Bruno DESPRÉS Gloria FACCANONI š Kirill Pichon GOSTAF Lise-Marie IMBERT-GÉRARD Georges SADAKA Remy SART CEMRACS 10, August 25, 2010

2 Outline 1 Introduction 2 The Equilibrium The model Analytical Solutions and Computational Examples 3 The Current Hole 4 Conclusion Perspectives 2 / 21

3 Introduction Outline 1 Introduction 2 The Equilibrium The model Analytical Solutions and Computational Examples 3 The Current Hole 4 Conclusion Perspectives 3 / 21

4 Introduction FreeFem++ FreeFem ++1 is a software to solve partial differential equations numerically, based on finite element methods. For the moment this platform is restricted to the numerical simulations of problems which admit a variational formulation. Our goal will be to evaluate the FreeFem ++ tool on basic magnetic equations arising in Fusion Plasma / 21

5 Introduction Geometry configuration Z y λ b φ R 0 e a x ε = a R 0 R 5 / 21

6 Introduction Reduced Resistive MHD ψ t = (1 + εx)[ψ, ϕ] + η(j J c), ω ϕ 1 = 2εω + (1 + εx)[ω, ϕ] + t y 1 + εx [ψ, J] + ν ω, J = ψ, ω = ϕ. ψ is the magnetic flux, ϕ is the velocity potential, J is the toroidal current density, ω is the vorticity, J c is the non-ohmic driven current density that sets a constant profile likely to be perturbed with fluctuations, η stands for the resistivity, ν is the viscosity, [, ] is the Poisson brackets is the Grad-Shafranov operator is the laplacian restricted to the poloidal section. 6 / 21

7 Introduction EMAFF Project EMAFF-I: the Grad-Shafranov equation ψ = R 2 dp dψ + F df dψ ; EMAFF-II: the current hole in cylindrical case (i.e. ε = 0) ψ t ω t = [ω, ϕ] + [ψ, J] + ν ω, ω = ϕ. = [ψ, ϕ] + η(j Jc), J = ψ, 7 / 21

8 The Equilibrium Outline 1 Introduction 2 The Equilibrium The model Analytical Solutions and Computational Examples 3 The Current Hole 4 Conclusion Perspectives 8 / 21

9 The Equilibrium The Grad-Shafranov Equation In cylindrical coordinates (R, Z) R ( ) ( 1 ψ + 2 ψ R R R Z 2 = R 2 dp dψ + F df ). dψ Soloviev equilibrium dp dψ = α = cst, F df dψ = β = cst In Cartesian coordinates (x, y) ( (1 + εx) div 1 a 2 ( ψ 1 + εx )) = ( αr0 2 (1 + εx) 2 + β ). 9 / 21

10 The Equilibrium Example I - Soloviev equilibrium Compute the magnetic flux ψ solution of the PDE on the domain Ω defined by Ω = {(R, Z) R = R ψ R R 2 ψ R 2 2 ψ Z 2 = f 0(R 2 + R0 2 ) 2a cos α R 0, Z = ar 0 sin α, α = 0 : 2π}. 10 / 21

11 The Equilibrium Example I - Soloviev equilibrium Compute the magnetic flux ψ solution of the PDE on the domain Ω defined by Ω = {(R, Z) R = R Analytical solution ψ(r, Z) = f 0R 2 0 a2 2 1 ψ R R 2 ψ R 2 2 ψ Z 2 = f 0(R 2 + R0 2 ) ( 1 2a cos α R 0, Z = ar 0 sin α, α = 0 : 2π}. ( ) 2 ( Z R R0 + (R R 0) 2 ) 2 ). a a 2aR 0 10 / 21

12 The Equilibrium Example I - Soloviev equilibrium Computational parameters then a = 0.5, f 0 = 1, R 0 = 1; ψ(r, Z) = R2 4 R4 8 Z 2 2 solution of the Grad-Shafranov equation c ψ = R and ψ Ω = 0. ψ < 0 ψ > 0 ψ = 0 10 / 21

13 The Equilibrium Example I - Soloviev equilibrium Mesh with N Ω = 200 elements on the border. Magnetic flux ψ ex. Computed magnetic flux ψ with P 1 finites elements type. E(ψ): = ψ ψ ex L 2 ψ ex L 2 = / 21

14 The Equilibrium Example II - Soloviev equilibrium Compute the magnetic flux ψ solution of the PDE ( ) 1 ψ + 2 ψ x 1 + εx x y 2 = ( αr0 2 (1 + εx) 2 + β ) a 2 (1 + εx) on the domain Ω defined by Ω = {(x, y) R 2 ψ(x, y) = 0}. Analytical solution ( ψ(x, y) = 1 x ε 2 (1 x 2 ) ((1 ε2 4 ) 2 ) (1 + εx) 2 + λx (1 + ε ) ) ( a ) 2 2 x b y α = 4(a2 + b 2 )ε + a 2 (2λ ε 3 ) 2R 2 0 εa2 b 2, β = λ b 2 ε. 11 / 21

15 The Equilibrium Example II - Soloviev equilibrium IsoValue IsoValue ψex x ψex IsoValue Mesh for N Ω = 128. ψ N IsoValue ψ x 11 / 21

16 The Equilibrium Example II - Soloviev equilibrium E(ξ): = ξ ξ ex L 2 ξ ex L 2 = Ω (ξ ξ ex) 2 dω Ω (ξ ex) 2 dω 10 0 Convergence Order of E(ψ) 10 0 Convergence Order of E( ψ/ x)) log 10 (E(ψ)) ~h 2 log 10 (E( ψ/ x) 10 2 ~h log 10 (Nx) N Ω E(ψ) log 10 (Nx) N Ω E ( x ψ ) P 1 elements provide 2 convergence rate for the magnetic flux while gradients are approximated by one order of magnitude less. 11 / 21

17 The Equilibrium Example III - polynomial (nonlinear) RHS The equation c ψ(r, Z) = G(R, Z; ψ) with RHS G(R, Z; ψ) polynomial in ψ (nonlinear). Iterative method idea: numerical linearization of the RHS. The general scheme { ψ 0 given c ψ k+1 = ψ k+1 G 1(R, Z; ψ k ) + G 2(R, Z; ψ k ) 12 / 21

18 The Equilibrium Example III - polynomial (nonlinear) RHS The test case solution of ψ ex = 6 9aR 2 + k 1 (z + c 1 ) 2 c ψ = ψ 2 ( k 1 + 2a(k 1 9a)R 2 ψ ) Schemes the start point ψ 0 = ψ ex cos(r), the discretization of the RHS RHS = k 1ψ β k+1 ψ k 2 β + 2a(k 1 9a)R 2 ψ δ k+1ψ k 3 δ. 12 / 21

19 The Equilibrium Example III - polynomial (nonlinear) RHS (1) (2) (3) (4) β δ (1) (2) (3) (4) 0.1 (1) (2) (3) (4) e N Ω ψ ψex H 1 (Ω) ψ ex H 1 (Ω) N Ω ψ ψex H 1 (Ω) ψ ex H 1 (Ω) with elements P 1 with elements P 2 12 / 21

20 The Current Hole Outline 1 Introduction 2 The Equilibrium The model Analytical Solutions and Computational Examples 3 The Current Hole 4 Conclusion Perspectives 13 / 21

21 The Current Hole The equations Reduced model with simplified geometry (ε = 0, x Ω, t [0, T ]) t ψ = [ψ, ϕ] + η(j J c ), t ω = [ω, ϕ] + [ψ, J] + ν ω, J = ψ, ω = ϕ with Poisson brackets initial conditions boundary conditions [a, b]: = a x 1 b a x 2 x 2 b x 1 ϕ(0, x) = ω(0, x) = 0, x Ω, J(0, x) = J c, x Ω, ψ(0, x) = 1 J c, x Ω; ϕ(t, x) = ω(t, x) = ψ(t, x) = J(t, x) = 0, x Ω, t [0, T ]. 14 / 21

22 The Current Hole The FreeFem++ space discretization Finite elements with P 1 (order 2) or P 2 (order 3). Variational formulation: solved by u V h (P 1 or P 2 ) and Au = f v, Au = v, f, v h 15 / 21

23 The Current Hole Discretization in time Finite elements favor implicit time discretization. Crank-Nicholson (order 2) with linearization tu = F (u) un+1 un δt = 1 (F (un) + F (un+1)) 2 with u n+1 = u n + δu, then we have to solve ( Id δt ) F (un) δu = δt F (u n). 2 }{{} M 16 / 21

24 The Current Hole Matricial Formulation - I First choice for our problem: ( ) ψ ϕ n+1 = ( ) ψ + ϕ n ( ) δψ δϕ which implies to solve ( ) ( ) δψ [ψ M = δt n, ϕ n ] + η ψ n ηj c δϕ [ ϕ n, ϕ n ] + [ψ n, ψ n ] + ν 2 ϕ n with ( Id + δt M: = 2 [ψ n, ] δt 2 [ψ n, ] + δt 2 [ ψ n, ] δt 2 [ ϕ n, ] + δt 2 [ ϕ n, ] δt ν [ϕ n, ] δt η 2 δt ) = we can not use P 1 elements we introduce J = ψ and ω = ϕ to obtain a new Matricial Formulation / 21

25 The Current Hole Matricial Formulation - II δψ [ψ n, ϕ n ] + η(j n J c ) M δϕ δj = δt [ω n, ϕ n ] + [ψ n, J n ] + ν ω n 0 δω 0 with Id + δt 2 [ϕ n, ] δt 2 [ψ n, ] δt η 2 Id 0 δt 2 M: = [J n, ] δt 2 [ω n, ] δt 2 [ψ n, ] Id + δt 2 [ϕ n, ] δt ν 0 0 Id 0 Id / 21

26 The Current Hole Computational Tests Tests I: P 1 : see the video P 2 : still in progress Tests II: P 1 (G. HUYSMANS paper): see the video P 2 : see the video Tests III: P 1 (J c initial Laplace): see the video 19 / 21

27 Conclusion Perspectives Outline 1 Introduction 2 The Equilibrium The model Analytical Solutions and Computational Examples 3 The Current Hole 4 Conclusion Perspectives 20 / 21

28 Conclusion Perspectives Conclusion Perspectives GS-equation FreeFem ++ is very easy to handle and adapted to solve this kind of problems in general geometry, good agreement with the (limited!) literature, Work in progress: polynomial nonlinear cases several solutions: how to select the physical one? initial condition and FE type: how to choose them? The current-hole effective simulation, no need of any initial perturbation (included in space approximation), different geometries, many test cases with FreeFem / 21