Well-balanced DG scheme for Euler equations with gravity

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1 Well-balanced DG scheme for Euler equations with gravity Praveen Chandrashekar Center for Applicable Mathematics Tata Institute of Fundamental Research Bangalore Dept. of Mathematics, IIT Delhi 26 October, 2016 Supported by Airbus Foundation Chair at TIFR-CAM, Bangalore 1 / 59

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 / 59

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 / 59

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 / 59

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 / 59

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 / 59

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 / 59

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 How to construct a h, s h to achieve well-balancing? 8 / 59

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 / 59

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 / 59

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 / 59

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 / 59

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 / 59

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 / 59

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 / 59

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 / 59

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 / 59

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 / 59

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 / 59

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 / 59

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 / 59

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 / 59

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 / 59

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 / 59

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 / 59

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 / 59

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 / 59

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 / 59

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 / 59

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 / 59

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 / 59

32 Numerical Results 32 / 59

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 / 59

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 / 59

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 / 59

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 / 59

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 / 59

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 / 59

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 / 59

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 / 59

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 / 59

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 / 59

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 / 59

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 / 59

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 / 59

46 2-D hydrostatic test: Discontinuous density Ligther fluid on top 46 / 59

47 2-D hydrostatic test: Discontinuous density Heavier fluid on top 47 / 59

48 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 48 / 59

49 Radial Rayleigh-Taylor problem cells 956 cells 49 / 59

50 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 50 / 59

51 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 51 / 59

52 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 [?], we take ρ = 0.1, η = 0.02, k = / 59

53 Radial Rayleigh-Taylor problem: Perturbations Cartesian mesh of and Q1 basis 53 / 59

54 Well-balanced versus non well-balanced (a) (b) (c) Well balanced test for radial Rayleigh-Taylor problem on mesh. (a) Initial density, (b) well-balanced scheme, density at t = 1.5, (c) non well-balanced scheme, density at t = / 59

55 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 55 / 59

56 Extensions Define variables v = [ρ/α, u, p/β] Interpolate v onto V h v h = Π h v(q h ) Source DG scheme in 1-D s h (x, t) = ρ h(x, t) p 0 α h (x) ρ 0 x d dt (q h, ϕ j ) h + ( x f h (v h ), ϕ j ) h N β j ϕ j (x) j=0 +[{ 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 56 / 59

57 Extensions GL nodes can be used Well-balanced for any equation of state Well-balanced on adaptive meshes with hanging nodes 57 / 59

58 Thank You 58 / 59

59 References 59 / 59

Well-balanced DG scheme for Euler equations with gravity

Well-balanced DG scheme for Euler equations with gravity Praveen Chandrashekar praveen@tifrbng.res.in Center for Applicable Mathematics Tata Institute of Fundamental Research Bangalore 560065 Higher Order