Well-balanced DG scheme for Euler equations with gravity

Transcription

1 Well-balanced DG scheme for Euler equations with gravity Praveen Chandrashekar Center for Applicable Mathematics Tata Institute of Fundamental Research Bangalore Higher Order Numerical Methods for Evolutionary PDEs: Applied Mathematics Meets Astrophysical Applications (15w5134) BIRS, Banff Centre, May, 2015 Supported by Airbus Foundation Chair at TIFR-CAM, Bangalore 1 / 55

2 Euler equations with gravity q t + f x = where ρ ρu q = ρu, f = p + ρu 2, E (E + p)u Calorically ideal gas p = (γ 1) 0 ρ Φ x ρu Φ = gravitational potential [E 12 ρu2 ], γ = c p c v > 1 2 / 55

3 Hydrostatic solutions Fluid at rest u e = 0 Mass and energy equation satisfied t ρ + x (ρu) = 0, t E + x [(E + p)u] = ρu x Φ Momentum equation dp e dx = ρ dφ e dx Need additional assumptions to solve this equation (1) 3 / 55

4 Hydrostatic solutions: Isothermal case Thermally ideal gas model p = ρrt, R = gas constant Isothermal hydrostatic state, i.e., T e (x) = T e = const, then ( ) Φ(x) p e (x) exp = const (2) RT e Density ρ e (x) = p e(x) RT e 4 / 55

5 Hydrostatic solutions: Polytropic case Polytropic hydrostatic state p e ρ ν e = α = const, ν > 1 constant (3) From (1) and (3), we obtain ναρe ν 1 (x) + Φ(x) = β = const (4) ν 1 5 / 55

6 Scope of present work Nodal DG scheme Gauss-Lobatto-Legendre nodes Arbitrary quadrilateral cells in 2-D Well-balanced for hydrostatic solutions isothermal solutions under ideal gas polytropic solutions (with Markus Zenk) Any consistent numerical flux Discontinuous density: contact preserving flux 6 / 55

7 Finite element method Conservation law with source term stationary solution q e q t + f(q) x = s(q) f(q e ) x = s(q e ) Weak formulation: Find q(t) V such that d (q, ϕ) + a(q, ϕ) = (s(q), ϕ), dt ϕ V FEM with quadrature: Find q h (t) V h such that d dt (q h, ϕ h ) h + a h (q h, ϕ h ) = (s h (q h ), ϕ h ) h, ϕ h V h 7 / 55

8 Finite element and well-balanced Stationary solution is not a polynomial, q e / V h. Let q e,h = Π h (q e ), Π h : V V h, interpolation or projection FEM is well-balanced if a h (q e,h, ϕ h ) = (s h (q e,h ), ϕ h ) h, ϕ h V h because if q h (0) = q e,h = q h (t) = q e,h t 8 / 55

9 Mesh and basis functions Partition domain into disjoint cells C i = (x i 1, x 2 i+ 1 ), x i = x 2 i+ 1 x 2 i 1 2 Approximate solution inside each cell by a polynomial of degree N I i 1 I i I i 1 9 / 55

10 Mesh and basis functions Map C i to a reference cell, say [0, 1] x = ξ x i + x i 1 2 (5) On reference cell, ξ j, 0 j N are Gauss-Lobatto-Legendre nodes, roots of (1 ξ 2 )P N (ξ) in [ 1, +1] l j (ξ) = Lagrange polynomials using GLL points l j (ξ k ) = δ jk, 0 j, k N Basis functions in physical coordinates ϕ j (x) = l j (ξ), 0 j N Derivative of ϕ j : apply the chain rule of differentiation d dx ϕ j(x) = l j(ξ) dξ dx = 1 x i l j(ξ) 10 / 55

11 Mesh and basis functions x j C i denote the physical locations of the GLL points x j = ξ j x i + x i 1, 0 j N 2 11 / 55

12 Discontinuous Galerkin Scheme Consider the single conservation law with source term q t + f = s(q, x) x Solution inside cell C i is polynomial of degree N q h (x, t) = Approximate the flux N q j (t)ϕ j (x), q j (t) = q h (x j, t) j=0 f(q h ) f h (x, t) = N f(q h (x j, t))ϕ j (x) = j=0 N f j (t)ϕ j (x) j=0 12 / 55

13 Discontinuous Galerkin Scheme Gauss-Lobatto-Legendre quadrature N φ(x)ψ(x)dx (φ, ψ) h = x i ω q φ(x q )ψ(x q ) C i q=0 Semi-discrete DG: For 0 j N d dt (q h, ϕ j ) h + ( x f h, ϕ j ) h +[ ˆf i+ 1 f h (x )]ϕ 2 i+ 1 j (x ) i [ ˆf (6) i 1 f h (x + )]ϕ 2 i 1 j (x + ) = (s i 1 h, ϕ j ) h 2 2 where ˆf i+ 1 = ˆf(q, q + ) is a numerical flux function. This is also 2 i+ 1 i called a DG Spectral Element Method. 13 / 55

14 Numerical flux Consistency of numerical flux ˆf(q, q) = f(q) Def: Contact property The numerical flux ˆf is said to satisfy contact property if for any two states w L = [ρ L, 0, p] and w R = [ρ R, 0, p] we have ˆf(q(w L ), q(w R )) = [0, p, 0] states w L, w R in the above definition correspond to a stationary contact discontinuity. Contact Property = exactly supports stationary contact Examples: Roe, HLLC, etc. 14 / 55

15 Approximation of source term: isothermal case Let T i = temperature corresponding to the cell average value in cell C i Rewrite the source term in the momentum equation as (Xing & Shu) s = ρ Φ ( ) ( x = ρr T Φ i exp R T i x exp Φ ) R T i Source term approximation: For x C i ( ) s h (x) = ρ h (x)r T Φ(x) N ( i exp R T exp Φ(x ) j) i x R T ϕ j (x) (7) i Source term in the energy equation 1 ρ h (ρu) h s h j=0 15 / 55

16 Approximation of source term: polytropic case Define H(x) inside each cell C i as H(x) = ν [ ] ν 1 ν 1 ln (β i Φ(x)), x C i να i α i, β i : constants to be chosen. Rewrite source term s(x) = ρ Φ x = ν 1 ρ(β i Φ(x)) exp( H(x)) ν x exp(h(x)), The source term is approximated as x C i s h (x) = ν 1 ρ h (x)(β i Φ(x)) exp( H(x)) ν x [ ν β i = max 0 j N ν 1 ] p j + Φ(x j ) ρ j N exp(h(x j ))ϕ j (x) j=0, α i = p j ρ ν j 16 / 55

17 Well-balanced property Well-balanced property Let the initial condition to the DG scheme (6), (7) be obtained by interpolating the isothermal/polytropic hydrostatic solution corresponding to a continuous gravitational potential Φ. Then the scheme (6), (7) preserves the initial condition under any time integration scheme. Proof: For continuous hydrostatic solution, q h (0) is continuous. By flux consistency ˆf i+ 1 f h (x ) = 0, ˆfi 2 i+ 1 1 f h (x + ) = i 1 2 Above is true even if density is discontinuous, provided flux satisfies contact property. = density and energy equations are well-balanced 17 / 55

18 Well-balanced property Momentum eqn: flux f h has the form N f h (x, t) = p j (t)ϕ j (x), j=0 p j = pressure at the GLL point x j Isothermal initial condition, T i = T e = const. The source term evaluated 18 / 55

19 Well-balanced property at any GLL node x k is given by s h (x k ) = ( ) Φ(xk ) N ρ h (x k )RT }{{} e exp RT e p j=0 k = ( ) Φ(xk ) N p k exp RT e = = = j=0 N ( Φ(xk ) p k exp RT e j=0 j=0 ( exp Φ(x ) j) RT e x ϕ j(x k ) ( exp Φ(x ) j) RT e x ϕ j(x k ) ) exp ( Φ(x ) j) RT e x ϕ j(x k ) N ( ) ( Φ(xj ) p j exp exp Φ(x ) j) RT e RT e x ϕ j(x k ) N j=0 p j x ϕ j(x k ) = x f h(x k ) 19 / 55

20 Well-balanced property Since x f h (x k ) = s h (x k ) at all the GLL nodes x k we can conclude that ( x f h, ϕ j ) h = (s h, ϕ j ) h, 0 j N = scheme is well-balanced for the momentum equation The proof for polytropic case is similar. 20 / 55

21 2-D Euler equations with gravity q t + f x + g y = s q = vector of conserved variables, (f, g) = flux vector and s = source term, given by ρ q = ρu ρv, f = E s = ρu p + ρu 2 ρuv (E + p)u 0 ρ Φ x ρ Φ y ( u Φ, g = ) x + v Φ y ρv ρuv p + ρv 2 (E + p)v 21 / 55

22 2-D hydrostatic solution: Isothermal case Momentum equation p e = ρ e Φ Assuming the ideal gas equation of state p = ρrt and a constant temperature T = T e = const, we get ( ) Φ(x, y) p e (x, y) exp = const (8) RT e We will exploit the above property of the hydrostatic state to construct the well-balanced scheme. 22 / 55

23 Mesh and basis functions A quadrilateral cell K and reference cell ˆK = [0, 1] [0, 1] y 1 4 K 2 3 T K η 4 ˆK 3 x 1 2 ξ 1-D GLL points ξ r [0, 1], 0 r N 23 / 55

24 Mesh and basis functions Tensor product of GLL points N = 1 N = 2 N = 3 Basis functions in cell K: i (r, s) ϕ K i (x, y) = l r (ξ)l s (η), (x, y) = T K (ξ, η) Computation of x f h requires derivatives of the basis functions x ϕk i (x, y) = l r(ξ)l s (η) ξ x + l r(ξ)l s(η) η x 24 / 55

25 DG scheme in 2-D Consider a single conservation law with source term Solution inside cell K q h (x, y, t) = q t + f x + g x = s M qi K (t)ϕ K i (x, y), M = (N + 1) 2 i=1 Approximate fluxes inside the cell by interpolation at the same GLL nodes f h (x, y) = M f(q h (x i, y i ))ϕ K i (x, y) = i=1 M f i ϕ K i (x, y) i=1 g h (x, y) = M g(q h (x i, y i ))ϕ K i (x, y) = i=1 M g i ϕ K i (x, y) i=1 25 / 55

26 DG scheme in 2-D Quadrature on element K using GLL points (φ, ψ) K = N r=0 s=0 Quadrature on the faces of the cell K N φ(ξ r, ξ s )ψ(ξ r, ξ s )ω r ω s J K (ξ r, ξ s ) (φ, ψ) K = e K (φ, ψ) e where (, ) e are one dimensional quadrature rules using the subset of GLL nodes located on the boundary of the cell. 26 / 55

27 DG scheme in 2-D Semi-discrete DG scheme: For 1 i M d ( qh, ϕ K ) i dt K + ( x f h + y g h, ϕ K ) i K ( ) + ˆFh F h, ϕk i = ( s h, ϕ K i K ) (9) Numerical flux function ˆF h = ˆF (q h, q+ h, n), n = (n x, n y ), unit outward normal to K Trace of flux on K from cell K F h = f h n x + g h n y 27 / 55

28 Approximation of source term: Isothermal case Let T K = temperature corresponding to the cell average value in cell K Rewrite source term in the x momentum equation s = ρ Φ ( ) ( x = ρr T Φ K exp R T K x exp Φ ) R T K Approximation of source term for (x, y) K ( ) s h (x, y) = ρ h (x, y)r T Φ(x, y) K exp R T K x M j=1 ( exp Φ(x ) j, y j ) R T ϕ K j (x, y) K (10) 28 / 55

29 Well-balanced property Let the initial condition to the DG scheme (9), (10) be obtained by interpolating the isothermal/polytropic hydrostatic solution corresponding to a continuous gravitational potential Φ. Then the scheme preserves the initial condition under any time integration scheme. 29 / 55

30 Limiter TVB limiter of Cockburn-Shu How to preserve hydrostatic solution? If residual in cell K is zero, then dont apply limiter in that cell. 30 / 55

31 Time integration scheme: dq dt = R(t, q) Second order accurate SSP Runge-Kutta scheme q (1) = q n + tr(t n, q n ) q (2) = 1 2 qn [q(1) + tr(t n + t, q (1) )] q n+1 = q (2) Third order accurate SSP Runge-Kutta scheme q (1) = q n + tr(t n, q n ) q (2) = 3 4 qn [q(1) + tr(t n + t, q (1) )] q (3) = 1 3 qn [q(2) + tr(t n + t/2, q (2) )] q n+1 = q (3) 31 / 55

32 Numerical Results 32 / 55

33 1-D hydrostatic test: Well-balanced property Potential Initial state u = 0, Φ = x, Φ = sin(2πx) ρ = p = exp( Φ(x)) Mesh ρu ρ E 25x e e e-14 50x e e e x e e e x e e e-14 Table: Well-balanced test on Cartesian mesh using Q1 and potential Φ = y 33 / 55

34 1-D hydrostatic test: Well-balanced property Mesh ρu ρ E 25x e e e-14 50x e e e x e e e x e e e-14 Table: Well-balanced test on Cartesian mesh using Q2 and potential Φ = x Mesh ρu ρ E 25x e e e-13 50x e e e x e e e x e e e-13 Table: Well-balanced test on Cartesian mesh using Q1 and Φ = sin(2πx) 34 / 55

35 1-D hydrostatic test: Well-balanced property Mesh ρu ρ E 25x e e e-13 50x e e e x e e e x e e e-13 Table: Well-balanced test on Cartesian mesh using Q2 and Φ = sin(2πx) 35 / 55

36 1-D hydrostatic test: Evolution of perturbations Potential Φ(x) = x Initial condition u = 0, ρ = exp( x), p = exp( x) + η exp( 100(x 1/2) 2 ) 36 / 55

37 1-D hydrostatic test: Evolution of perturbations Initial 100 cells 1000 cells Initial 50 cells 500 cells Pressure perturbation Pressure perturbation x (a) x (b) Figure: Evolution of perturbations for η = 10 2 : (a) Q1 (b) Q2 37 / 55

38 1-D hydrostatic test: Evolution of perturbations Initial 100 cells 1000 cells Initial 50 cells 500 cells Pressure perturbation Pressure perturbation x (a) x (b) Figure: Evolution of perturbations for η = 10 4 : (a) Q1 (b) Q2 38 / 55

39 Shock tube problem Potential Φ(x) = x and initial condition { (1, 0, 1) x < 1 (ρ, u, p) = 2 (0.125, 0, 0.1) x > Q1, 100 cells Q2, 100 cells FVM, 2000 cells Q1, 200 cells Q2, 200 cells FVM, 2000 cells Density 0.6 Density x x 39 / 55

40 2-D hydrostatic test: Well-balanced test Potential Φ = x + y Hydrostatic solution is given by ( ρ = ρ 0 exp ρ ) ( 0g (x + y), p = p 0 exp ρ ) 0g (x + y), u = v = 0 p 0 p 0 ρu ρv ρ E Q1, e e e e-13 Q1, e e e e-13 Q1, e e e e-13 Q2, e e e e-13 Q2, e e e e-13 Q2, e e e e / 55

41 2-D hydrostatic test: Evolution of perturbation ( p = p 0 exp ρ ) ( 0g (x + y) + η exp 100 ρ ) 0g [(x 0.3) 2 + (y 0.3) 2 ] p 0 p 0 41 / 55

42 2-D hydrostatic test: Evolution of perturbation y y x (a) x Figure: Pressure perturbation on mesh at time t = 0.15 (a) Q 1 (b) Q 2. Showing 20 contours lines between to (b) 42 / 55

43 2-D hydrostatic test: Evolution of perturbation y y x (a) x Figure: Pressure perturbation on mesh at time t = 0.15 (a) Q 1 (b) Q 2. Showing 20 contours lines between to (b) 43 / 55

44 2-D hydrostatic test: Discontinuous density Potential Temperature is discontinuous T = Hydrostatic pressure and density p = Φ(x, y) = y { T l, y < 0 T u, y > 0 { p 0 e y/rt l, y 0 p 0 e y/rtu, y > 0, ρ = Density is discontinuous at y = 0. { p/rt l, y < 0 p/rt u, y > 0 44 / 55

45 2-D hydrostatic test: Discontinuous density Case 1, T l = 1, T u = 2 Q1, 25x e e e e-13 Q1, 50x e e e e-13 Q2, 25x e e e e-13 Q2, 50x e e e e-13 Table: Well-balanced test for Rayleigh-Taylor problem Case 2, T l = 2, T u = 1 Q1, 25x e e e e-13 Q1, 50x e e e e-13 Q2, 25x e e e e-13 Q2, 50x e e e e-13 Table: Well-balanced test for Rayleigh-Taylor problem 45 / 55

46 Order of accuracy Exact solution of Euler equations with gravity given by (Xing & Shu) ρ = sin(π(x + y t(u 0 + v 0 ))), u = u 0, v = v 0 p = p 0 + t(u 0 + v 0 ) x y cos(π(x + y t(u 0 + v 0 )))/π Compute solution error in L 2 norm at time t = 0.1 1/h ρu ρv ρ E Error Rate Error Rate Error Rate Error Rate e e e e e e e Table: Convergence of error for degree N = 1 1/h ρu ρv ρ E Error Rate Error Rate Error Rate Error Rate e e e e e e e e e e e e e e e e Table: Convergence of error for degree N = 2 46 / 55

47 Radial Rayleigh-Taylor problem cells 956 cells 47 / 55

48 Radial Rayleigh-Taylor problem: Well-balanced test Radial potential Hydrostatic solution Φ(x, y) = r = x 2 + y 2 ρ = p = exp( r) ρu ρv ρ E Q1, 30x e e e e-14 Q1, 50x e e e e-14 Q1, 100x e e e e-14 Q2, 30x e e e e-14 Q2, 50x e e e e-14 Q2, 100x e e e e-14 Table: Well balanced test for radial Rayleigh-Taylor problem on Cartesian mesh 48 / 55

49 Radial Rayleigh-Taylor problem: Well-balanced test ρu ρv ρ E Q1, e e e e-16 Q1, e e e e-16 Q1, e e e e-15 Q2, e e e e-13 Q2, e e e e-13 Q2, e e e e-13 Table: Well balanced test for radial Rayleigh-Taylor problem on unstructured mesh 49 / 55

50 Radial Rayleigh-Taylor problem: Perturbations Initial pressure and density where { { e r r r 0 e r r r i p = e r α +r 0 (1 α), ρ = α r > r 0 r > r i 1 α e r α +r (1 α) 0 α r i = r 0 (1 + η cos(kθ)), α = exp( r 0 )/(exp( r 0 ) + ρ ) The density jumps by an amount ρ > 0 at the interface defined by r = r i whereas the pressure is continuous. Following [2], we take ρ = 0.1, η = 0.02, k = / 55

51 Radial Rayleigh-Taylor problem: Perturbations Cartesian mesh of and Q1 basis 51 / 55

52 Extensions: Gauss-Legendre nodes Define variables (isothermal case) Interpolate onto GLL nodes DG scheme in 1-D v = [ρ exp(φ/r T i ), u, p exp(φ/r T i )] v h = Π h v(q h ) d dt (q h, ϕ j ) h + ( x f h (q h ), ϕ j ) h +[{ ˆf(q(v h ), q(v+ h )) f(q(v h ))}ϕ j ] i+ 1 2 [{ ˆf(q(v h ), q(v+ h )) f(q(v+ h ))}ϕ+ j ] i 1 2 = (s h (q h ), ϕ j ) h Well-balanced on adaptive meshes with hanging nodes 52 / 55

53 Summary Well-balanced DG scheme for Euler with gravity Preserves isothermal hydrostatic solutions for ideal gas model Preserves polytropic hydrostatic solution Any numerical flux can be used Cartesian, unstructured meshes 53 / 55

54 Thank You 54 / 55

55 References [1] R. Käppeli and S. Mishra, Well-balanced schemes for the Euler equations with gravitation, J. Comput. Phys., 259 (2014), pp [2] Randall. J. LeVeque and Derek. S. Bale, Wave propagation methods for conservation laws with source terms, in Hyperbolic Problems: Theory, Numerics, Applications, Rolf Jeltsch and Michael Fey, eds., vol. 130 of International Series of Numerical Mathematics, Birkhäuser Basel, 1999, pp [3] Yulong Xing and Chi-Wang Shu, High order well-balanced WENO scheme for the gas dynamics equations under gravitational fields, J. Sci. Comput., 54 (2013), pp / 55