A High Order Conservative Semi-Lagrangian Discontinuous Galerkin Method for Two-Dimensional Transport Simulations

Transcription

1 Motivation Numerical methods Numerical tests Conclusions A High Order Conservative Semi-Lagrangian Discontinuous Galerkin Method for Two-Dimensional Transport Simulations Xiaofeng Cai Department of Mathematics University of Houston Joint work with Wei Guo (Michigan State University) and Jingmei Qiu (University of Houston) IMAGe 2017 Theme of the Year Workshop on Multiscale Geoscience Numerics 1 / 32

2 Motivation Numerical methods Numerical tests Conclusions Outline 1 Motivation The transport equation 2 Numerical methods A new SLDG formulation for 1D transport problems A nonsplitting SLDG formulation for 2D transport problems 3 Numerical tests 2D linear problem The rotational problem The swirling deformation problem 4 Conclusions 2 / 32

3 Motivation Numerical methods Numerical tests Conclusions The transport equation Transport equation We are concerned with the transport equation u + (au) = 0, t where a is the advection coefficient and could depend on space and time. 3 / 32

4 Motivation Numerical methods Numerical tests Conclusions The transport equation Existing numerical methods Lagrangian approach Tracking macro-particle trajectories, e.g., Particle-in-cell (PIC). No CFL restriction. inherent numerical noise O(1/ N). Eulerian approach Grid-based scheme. Arbitrary order of accuracy in both space and time. CFL restriction for stability with explicit time-stepping method. e.g. The RKDG scheme suffers the stringent CFL stability restriction 1, where k is the degree of polynomials. 2k+1 Semi-Lagrangian (SL) approach Combination of Eulerian approach and Lagrangian approach. Grid-based scheme. Numerical solution is updated by following the characteristics. Allowing for a very large CFL number. 4 / 32

5 Motivation Numerical methods Numerical tests Conclusions The transport equation High order SLDG methods SLDG method: low numerical dissipation; compactness; flexibility for boundary and parallel implementation; superconvergence. could take about 10 to 25+ times the time stepping size of RKDG with accuracy and stability, leading to gain in efficiency. mass conservation can be preserved. A history of SLDG: Qiu-Shu, Rossmanith-Seal (JCP, 2011): 1D SLDG schemes. Guo-Nair-Qiu (MWR, 2014): SLDG scheme on cubed-sphere. Dimensional splitting error of 1D SLDG schemes. 5 / 32

6 Motivation Numerical methods Numerical tests Conclusions The transport equation High order SLDG methods (CONT.) Desired properties of mutli-dimensional SLDG: 1 no dimensional splitting error 2 third order accuracy 3 mass conservation 4 large time stepping size 5 oscillations free 6 positivity preservation 6 / 32

7 Motivation Numerical methods Numerical tests Conclusions A new SLDG formulation for 1D transport problems A nonsplitti One-dimensional SLDG Consider a one-dimensional linear transport problem u t + (a(x, t)u) = 0 x with appropriate initial and boundary conditions. The solution and test function space of DG is {v h : v h Ij P k (I j ), the set of polynomials of degree at most k}, where j = 1,, N x. 7 / 32

8 Motivation Numerical methods Numerical tests Conclusions A new SLDG formulation for 1D transport problems A nonsplitti Weak formulation of characteristic Galerkin method We consider an adjoint problem for the test function ψ(x, t): ( ψt + a(x, t)ψ x = 0, ψ(t = t n+1 ) = Ψ(x). Let Ij e (t) is a dynamic interval bounded by characteristics emanating from cell boundaries of I j at t = t n+1, and then it can be shown that Z d u(x, t)ψ(x, t)dx = 0. dt ei j (t) Z I j u n+1 Ψdx = Z I j u(x, t n )ψ(x, t n )dx. Guo, Nair and Qiu, MWR, / 32

9 Motivation Numerical methods Numerical tests Conclusions A new SLDG formulation for 1D transport problems A nonsplitti Evaluation of R I j u(x, tn )ψ(x, t n )dx: Guo, Nair and Qiu. 1 collect the intervals and sub-intervals denoted dx(t) dt x j 1 2 = a(x(t), t) x j 1 2 x j+ 1 2 I j,1 I j,2 x j,1,ig x j,1,ig x j+ 1 2 t n+1 t n by I j,l. 2 Locate (k + 1) Gauss-Legendre-Lobatto (GLL) points (red circles) over I j,l denoted as x j,l,ig. 3 Trace characteristics forward in time from x j,l,ig to x j,l,ig (green circles). 4 R I j P l u(x, t n )ψ(x, t n )dx = Š Pig ω igu(x j,l,ig, tn )Ψ(x j,l,ig )l(i j,l ) Recently, there is a 2D extension of the this scheme by (D. Lee, R. Lowrie, M. Petersen, T. Ringler, M. Hecht, JCP, 2016) 9 / 32

10 Motivation Numerical methods Numerical tests Conclusions A new SLDG formulation for 1D transport problems A nonsplitti Evaluation of R I j u(x, tn )ψ(x, t n )dx: New formualtion 1 locate the boundary points of the upstream x j,p x j 1 2 x j+ 1 2 cell, x j± 1 ; and collect all the overlapping 2 intervals or sub-intervals as Ij,l t n+1 2 Choose (k + 1) interpolation points x j,q, such as the GLL points over I j. x j 1 2 Z I j 3 Locate the feet x x j,q by tracking the j+ 1 2 t n characteristics backward from t n+1 to t n, (x j,p, Ψ(x j,p)) ψ and we have ψ(x j,q (x), tn ) = Ψ(x j,q ). 4 (x j,p interpolate, Ψ(x j,p)) ψ (x). u(x, t n )ψ(x, t n )dx = X l Z I j,l u(x, t n )ψ (x, t n )dx 10 / 32

11 Motivation Numerical methods Numerical tests Conclusions A new SLDG formulation for 1D transport problems A nonsplitti Two-dimensional SLDG Consider a two-dimensional linear transport problem u t + x (a(x, y, t)u) + (b(x, y, t)u) = 0 y with appropriate initial and boundary conditions. The solution space and test function space of DG is {v h : v h Aj P k (A j ), the set of polynomials of degree at most k}, where j = 1,, N x N y is the Cartesian grid. 11 / 32

12 Motivation Numerical methods Numerical tests Conclusions A new SLDG formulation for 1D transport problems A nonsplitti Weak formulation of characteristic Galerkin method We consider an adjoint problem for the test function ψ(x, y, t): ( ψt + a(x, y, t)ψ x + b(x, y, t)ψ y = 0, ψ(t = t n+1 ) = Ψ(x, y). Let Aj Ü (t) be the dynamic element bounded by characteristics emanating from cell edges of A j, then it can be shown that ψ P k (A j ), Z d u(x, y, t)ψ(x, y, t)dxdy = 0. dt ea j (t) Z A j u(x, y, t n+1 )Ψ(x, y)dxdy = Guo, Nair and Qiu, MWR, Z A j u(x, y, t n )ψ(x, y, t n )dxdy. 12 / 32

13 Motivation Numerical methods Numerical tests Conclusions A new SLDG formulation for 1D transport problems A nonsplitti Evaluation of R A j u(x, y, tn )ψ(x, y, t n )dxdy v 4 A l A j A j A l A j,l v 4 A j Characteristics tracing: Locate four vertices of upstream cell A j : v 1, v 2, v 3 and v 4 by solving the characteristics equations, 8 >< >: dx(t) dt dy(t) dt = a(x(t), y(t), t), = b(x(t), y(t), t), x(t n+1 ) = x(v q ), y(t n+1 ) = y(v q ), starting from the four vertices of A j : v 1, v 2, v 3 and v / 32

14 Motivation Numerical methods Numerical tests Conclusions A new SLDG formulation for 1D transport problems A nonsplitti Evaluation of R A j u(x, y, tn )ψ(x, y, t n )dxdy v 4 A j Two observations: 1 ψ(x, y, t n ) may not be a polynomial. v 4 A l A j 2 u(x, y, t n ) is a piecewise polynomial function on background cells. Strategies: We can reconstruct ψ (x, y) approximating ψ(x, y, t n ) on A j by a least square strategy, based on A l A j A j,l ψ(x(v q ), y(v q ), t n ) = Ψ(x(v q ), y(v q )), q = 1, 2, 3, 4. Evaluation of the integrand over the upstream cell has to be done subregion-by-subregion. 14 / 32

15 Motivation Numerical methods Numerical tests Conclusions A new SLDG formulation for 1D transport problems A nonsplitti v 4 A l A j A l A j A j,l v 4 A j Subregions can be of irregular shape. It is difficult to obtain the integral on a subregion. = Z Z = X l = X l A j u(x, y, t n+1 )Ψ(x, y)dxdy A j u(x, y, t n )ψ(x, y, t n )dxdy Z I A j,l A j,l u(x, y, t n )ψ(x, y, t n )dxdy P dx + Qdy with P x = uψ, due to the Green theorem. a y + Q a CSLAM: Lauritzen, Nair, and Ullrich, JCP, / 32

16 Motivation Numerical methods Numerical tests Conclusions A new SLDG formulation for 1D transport problems A nonsplitti L q s 2 s 3 s 4 c 1 S q s1 Figure: Search algorithm for outer (left, L q) and inner (right, S q) segments. X I l A j,l P dx + Qdy = 2 X 4 No q=1 Z X Z N i + L q q=1 S q 3 5 [P dx + Qdy]. 16 / 32

17 Motivation Numerical methods Numerical tests Conclusions A new SLDG formulation for 1D transport problems A nonsplitti Implementation procedure 1 Characteristics tracing (High order Runge-Kutta method) for four vertices of upstream cell A j : v 1, v 2, v 3 and v 4 and connect the four vertices as a quadrilateral approximation to the upstream cell. 2 Least square approximation of test function ψ(x, y, t n ) on upstream cells. 3 Search algorithm of line segments as edges of sub-regions. 4 Line integral evaluations (Gaussian quadrature) along outer segments along upstream cell edges and inner segments along background cells edges. 17 / 32

18 Motivation Numerical methods Numerical tests Conclusions A new SLDG formulation for 1D transport problems A nonsplitti Third order P 2 SLDG scheme v 7 A j η v 7 A l A A j j,l v 1 v 2 v 3 ξ Figure: Schematic illustration of the P 2 SLDG formulation in two dimension. Right: An upstream edge approximated by constructing a quadratic through three points, v 1, v 2, v / 32

19 Motivation Numerical methods Numerical tests Conclusions A new SLDG formulation for 1D transport problems A nonsplitti Quadratic curved edge: Construction η Construct a coordinate transformation x y to ξ η such that the coordinates of v 1 and v 3 are ( 1, 0) and (1, 0) in ξ η space, respectively. x = x 3 x 1 2 ξ + y3 y1 2 + x1+x3 2, y = y3 y1 2 ξ x3 x1 2 + y1+y3 2 v 1 v 2 v 3 ξ Get the ξ η coordinate for the point v 2 as (ξ 2, η 2 ). Based on ( 1, 0), (ξ 2, η 2 ) and (1, 0), we construct the parabola: η = η 2 ξ 2 2 1(ξ2 1). 19 / 32

20 Motivation Numerical methods Numerical tests Conclusions A new SLDG formulation for 1D transport problems A nonsplitti Quadratic curved edge: Line integral evaluation P Ni R q=1 S q [P dx + Qdy] is evaluated in the same way as P 1 case. P No R q=1 L q [P dx + Qdy] is evaluated by the following parameterization on each edge. x(ξ) = x 3 x 1 ξ + y 3 y 1 η ξ (ξ2 1) + x 3+x 1, 2 y(ξ) = y 3 y 1 ξ x 3 x 1 η ξ (ξ2 1) + y 3+y 1. 2 Hence, Z L q [P dx+qdy] = Z ξ (q+1) ξ (q) [P (x(ξ, η), y(ξ, η))x (ξ)+q((ξ, η), y(ξ, η))y (ξ)]dξ, where (ξ (q), η (q) ) and (ξ (q+1), η (q+1) ) are the starting point and the end point of L q in ξ η coordinate, respectively. 20 / 32

21 Motivation Numerical methods Numerical tests Conclusions A new SLDG formulation for 1D transport problems A nonsplitti Implementation procedure 1 Characteristics tracing for nine vertices of upstream cell A j. 2 Least square approximation of test function ψ(x, y, t n ) on upstream cells. 3 Construction of quadratic curves in approximating edges of upstream cells. 4 Search algorithm of line segments as edges of sub-regions. 5 Line integral evaluations along outer segments along upstream cell edges and inner segments along background Cartesian grids. 21 / 32

22 Motivation Numerical methods Numerical tests Conclusions A new SLDG formulation for 1D transport problems A nonsplitti Properties of SLDG Mass conservation: mass conservation is guaranteed due to the weak formulation and the CSLAM idea. Accuracy: high order accurate in the DG framework. Efficiency: allowing for large time step evolution by characteristics tracing mechanism. Controlling oscillations: a WENO limiter by Zhong and Shu (2013). Positivity preserving: positivity limiter by Zhang and Shu (2011). The numerical solution u(x, y, t n ) in cell A j is modified by eu(x, y) eu(x, y) = θ(u(x, t, t n ) u) + u, θ = min u m u ª, 1 where u is the cell average of the numerical solution and m is the minimum value of u(x, y, t n ) over A j., 22 / 32

23 Motivation Numerical methods Numerical tests Conclusions 2D linear problem The rotational problem The swirling deforma 2D linear problem: u t + u x + u y = 0 Table: Errors of SLDG solutions. u(x, y, 0) = sin(x + y) at T = π. mesh L 2 error Order L 2 error Order P 1 SLDG CF L = 1 CF L = E E E E E E E E P 2 SLDG CF L = 1 CF L = E E E E E E E E / 32

24 Motivation Numerical methods Numerical tests Conclusions 2D linear problem The rotational problem The swirling deforma 2D linear problem: SLDG vs. RKDG Table: 2D linear equation. u t + u x + u y = 0, x [ π, π], y [ π, π] with the initial condition u(x, y, 0) = sin(x + y) and the periodic boundary conditions in both x and y directions. T = π. scheme Mesh CF L L 2 error L error CPU time (s) P 1 SLDG E E P 1 SLDG E E P 1 SLDG E E P 1 SLDG E E P 1 RKDG E E P 2 SLDG E E P 2 SLDG E E P 2 SLDG E E P 2 SLDG E E P 2 RKDG E E For computing a time step, the cost of P 1 SLDG is around 2 times larger than that of P 1 RKDG. the cost of P 2 SLDG is around 2 times larger than that of P 2 RKDG. 24 / 32

25 Motivation Numerical methods Numerical tests Conclusions 2D linear problem The rotational problem The swirling deforma Rigid body rotation: u t (yu) x + (xu) y = 0 Table: Errors of SLDG solutions. u(x, y, 0) = exp( x 2 y 2 ) at T = 2π. mesh L 2 error Order L 2 error Order P 1 SLDG t = 0.5 x t = 2.5 x E E E E E E E E P 2 SLDG t = 0.5 x t = 2.5 x E E E E E E E E / 32

26 Motivation Numerical methods Numerical tests Conclusions 2D linear problem The rotational problem The swirling deforma Rigid body rotation: SLDG vs. RKDG Table: Rigid body rotation. u t (yu) x + (xu) y = 0, x [ 2π, 2π], y [ 2π, 2π] with the initial condition u(x, y, 0) = exp( x 2 y 2 ). T = 2π. scheme Mesh CF L L 2 error L error CPU time (s) P 1 SLDG E E P 1 SLDG E E P 1 SLDG E E P 1 SLDG E E P 1 RKDG E E P 2 SLDG E E P 2 SLDG E E P 2 SLDG E E P 2 SLDG E E P 2 RKDG E E For computing a time step, the cost of P 1 SLDG is around 4 times larger than that of P 1 RKDG. the cost of P 2 SLDG is around 3 times larger than that of P 2 RKDG. 26 / 32

27 Motivation Numerical methods Numerical tests Conclusions 2D linear problem The rotational problem The swirling deforma The swirling deformation problem Table: u t sin(y) cos 2πt 3 πu x + sin(x) cos2 y 2 cos 2 x 2 initial condition is a smooth cosine bell. T = 1.5. cos 2πt 3 πu = 0. The y mesh L 2 error Order L 2 error Order P 1 SLDG CFL = 1 CFL = E E E E E E E E P 2 SLDG CFL = 1 CFL = E E E E E E E E P 2 SLDG-QC CFL = 1 CFL = E E E E E E E E / 32

28 Motivation Numerical methods Numerical tests Conclusions 2D linear problem The rotational problem The swirling deforma The swirling deformation problem x u t cos 2 sin(y)g(t)u 2 x y + sin(x) cos 2 g(t)u = 0 2 y with Š πt g(t) = cos T π, x [ π, π], y [ π, π], The initial condition as shown on the right. 28 / 32

29 Motivation Numerical methods Numerical tests Conclusions 2D linear problem The rotational problem The swirling deforma Figure: Swirling deformation problem. Third order SLDG scheme: T = 0.75 (left) and T = 1.5 (right). The numerical mesh is with CFL = / 32

30 Motivation Numerical methods Numerical tests Conclusions 2D linear problem The rotational problem The swirling deforma Exact Exact P 1 SLDG+WENO+BP + P SLDG P 2 SLDG P 2 SLDG-QC P 2 SLDG+WENO+BP P 2 SLDG-QC+WENO+BP u u x x Figure: Plots of the 1D cuts of the numerical solution for the swirling deformation problem at y = π 2. The solid line depicts the exact solution. The numerical mesh has a resolution of / 32

31 Motivation Numerical methods Numerical tests Conclusions 2D linear problem The rotational problem The swirling deforma (a) P 2 SLDG (b) P 2 SLDG-QC Figure: The relative mass error for SLDG. The numerical mesh has a resolution of t = 2.5 x. 31 / 32

32 Motivation Numerical methods Numerical tests Conclusions Summary and future work We develop a robust truly multi-dimensional SLDG solver with no dimensional splitting error mass conservation large time stepping size positivity-preserving Ongoing work Cai-Guo-Qiu (2017): Couple the SLDG solver with high order characteristics tracing schemes for nonlinear Vlasov systems. In collaboration with R. Nair: Multi-D SLDG scheme on cubed-sphere. General boundary conditions. Adaptive mesh refinement for multi-d SLDG. SLDG scheme for transport equations with source terms and diffusive terms. Qiu and Russo, JSC, / 32