Did numerical methods for hyperbolic problems take a ...

Did numerical methods for hyperbolic problems take a wrong turning? Phil Roe University of Michigan

16th International Conference on Hyperbolic Problems, Aachen, Germany August 2016

A wrong turning?

Phil Roe

HYP 2016

1/41

Godunov’s question Sergei Godunov published the one-dimensional version of his method in 1959, but it was not well-known in the West until the later 1970s. At this time, a large number of extensions and alternative (Godunov-type) methods were discovered. It was, and still is, widely held that his conceptual breakthrough was to pose the following question; How would the data evolve if it was projected onto a set of piecewise constant states? Was this really such a good question?

A wrong turning?

Phil Roe

HYP 2016

2/41

“Reminiscences about difference schemes”, S.K. Godunov, JCP, 153, 1999 The first question we faced in our attempts to generalize our method for two dimensions was how to find a solution of the two-dimensional Riemann problem with arbitrary initial conditions. ......... In order for a two-dimensional method to be absolutely similar to the one-dimensional one we needed an analytical solution of the gas-dynamic equations with four initial discontinuities coming together in a single point. Naturally, we did not have these solutions at that time and they are still unknown (for general initial conditions). At this point, a roguish suggestion was made which was to use only the solutions of a classical Riemann problem involving only planar waves. Thus, the interaction of the four cells with a common vertex was neglected altogether. This removed a nice physical interpretation underlying the construction of the one-dimensional scheme. Quite naturally, there were many arguments during the discussion of this hardly justifiable suggestion. L. V. Brushlinkii........... carried out the analytical solution for an acoustic wave propagating in stationary media, which took into account the interaction of the cells sharing a common vertex. His solution was implemented in a scheme completely analogous to the one-dimensional one. To our surprise and pleasure there were no significant differences. Afterwards, only the simpler model was used. A wrong turning?

Phil Roe

HYP 2016

3/41

A selection of 2d Riemann problems

A wrong turning?

Phil Roe

HYP 2016

4/41

Dimensional Splitting Many of the early Godunov-type codes employed dimensional splitting. Instead of directly solving un+1 = un − ∆t(∂x Fn + ∂y Gn ) the solution was sought to a pair of one-dimensional problems u∗ = un − ∆t∂x Fn

un+1 = u∗ − ∆t∂y G∗ This splitting (always?) converges to a correct solution of the two-dimensional problem, but the convergence is very slow for discontinuous problems.

A wrong turning?

Phil Roe

HYP 2016

5/41

An oblique acoustic wave Solution to an oblique discontinuity in the acoustic equations. At left, exact solution, at right exact solution to the split system after one full timestep. y

y

x

x

The jump across an oblique discontinuity with orientation is an eigenvector of A cos θ + B sin θ but in a dimensionally split solution it must be represented by the eigenvectors of A in the x-step, and by the eigenvectors of B in the y -step. This gives the correct representation if and only if AB = BA . A wrong turning?

Phil Roe

HYP 2016

6/41

Dimensionally-split exact solution to a discontinuous acoustic wave If it were possible to reduce the dissipation in the one-dimensional solvers, the solution would look more and more like this. For example Glimms random choice method gives VERY poor results in two dimensions. We are only saved from our bad physics by the dissipation of our bad numerics.!

Roe, Num. Meth. PDEs, 7, 1991 A wrong turning?

Phil Roe

HYP 2016

7/41

Myth # 1; discontinuous representations are beneficial To solve ut + ux = 0, the DG1 method assigns two degrees of freedom to each cell and makes a piecewise linear reconstruction. Even so, the results are third- order accurate. Many papers prove this, but why?

1. A discontinuous basis is a good way to represent discontinuous solutions. 2. Continuous Galerkin (in the old sense) is global; discontinuous Galerkin is local. A wrong turning?

Phil Roe

HYP 2016

8/41

van Leer’s Scheme V

uL

ū

uR

Two degrees of freedom per cell permit a quadratic reconstruction if sharing is allowed. The semi-discrete version of Scheme V is DG1.

B. van Leer,J. Comput. Phys, 1977. A wrong turning?

Phil Roe

HYP 2016

9/41

The explanation of superconvergence 1. This result generalizes to arbitrary order (for linear advection). 2. A superconvergent solution can be obtained for any DG scheme (for this problem) by combining p + 1 degrees of freedom from the current element with p degrees of freedom from the upwind element to make the most continuous posible (2p+1) reconstruction 3. Thus, the “superiority” of discontinuous Galerkin methods lies in their having a continuous interpretation! 4. But, superconvergence is generally not observed with DG schemes for sufficiently complex problems. Could this explain why not?

Roe, Comm. in Comp. Phys, accepted A wrong turning?

Phil Roe

HYP 2016

10/41

Comparison of dissipation

1. DG1+RK3 is unstable for ν > 0.4 and needs three evaluations for each advance. Advantage = 7.5. 2. van Leer V (and FD) improve as ν approaches 1.0.

green pink blue brown

FD DG1 with perfect integration DG1 with RK3 van Leer V

3. van Leer V typically needs half the grid points. Advantage=22 ∗ 7.5 = 30.0

4. van Leer V approaches dG1 as ν → 0.

note θ ∈ [0, 2π] A wrong turning?

Phil Roe

HYP 2016

11/41

Myth #2 Riemann solvers are essential Riemann solvers are needed only to resolve the discontinuities between elements, but the discontinuities themselves are unnecessary and misleading.

A wrong turning?

Phil Roe

HYP 2016

12/41

Myth #2 Riemann solvers are essential Riemann solvers are needed only to resolve the discontinuities between elements, but the discontinuities themselves are unnecessary and misleading.

A wrong turning?

Phil Roe

HYP 2016

13/41

A fresh start - the war of words

1. Genuinely multidimensional 2. Truly multidimensional 3. Fully multidimensional 4. Really, really, really multidimensional, honestly!

A wrong turning?

Phil Roe

HYP 2016

14/41

A fresh start - the wish-list 1. Conservative 2. Third-order accuracy (MUCH better than second) 3. Fully discrete. 4. Not expensive 5. Low storage and computationally intensive (for exascale) 6. Works on poor-quality unstructured grids. 7. Not based on one-dimensional thinking but should recover regular upwinding in one dimension. 8. Does not employ discontinuous representation. 9. Does not use Riemann solvers 10. Applicable over the full range of Mach number. 11. Need not be applicable to all conservation laws, but may exploit specific properties of Euler and NS. A wrong turning?

Phil Roe

HYP 2016

15/41

A fresh start - the new ingredients We construct a finite-volume method in which the fluxes are independent degrees of freedom. Initially we consider only the Euler equations, and only third-order accuracy. We will make use of the following underestimated observations 1. Except in one dimension, acoustic waves do NOT resemble advection processes. 2. At the linear level, advection commutes with acoustics (and with everthing else) 3. Fluxes need not be predicted in conservative variables. 4. Fluxes can be one order less accurate than conserved variables, so second-order will be enough. 5. Nonlinear effects are required only to leading order. 6. Linear acoustics can locally be solved exactly in n dimensions. A wrong turning?

Phil Roe

HYP 2016

16/41

The data structure The degrees of freedom are

p p p

A wrong turning?

p

C

p

p

1. Point values of the primitive variables (or anything else, e.g. s, ln ρ, ln p) at points on the boundaries of the control volume. These apply on BOTH sides of the boundary. Because they are shared, storage is greatly reduced, especially in three dimensions. (e.g. 20 values per vertex replaced by one) 2. Cell averages of the conserved variables. We use piecewise parabolic reconstruction enriched with a cubic bubble.

Phil Roe

HYP 2016

17/41

A new approach to acoustics van Leers “new approach to numerical advection” can be generalised as follows. Let ∂t u = Au be any set of equations for which the linearized initial value problem can be solved “easily”. Project the data onto some set of basis function, and solve the locally linearized problems. Project the solution back onto the initial data and repeat. An example is the three-dimensional scalar wave equation ∂tt φ = c 2 ∇2 φ solved by φ(x, t) = Mct {φ(x, 0)} + ∂t [tMct {∂t φ(x, t)}] where MR {φ(x)} is the mean value over a sphere of radius R.

A wrong turning?

Phil Roe

HYP 2016

18/41

Poisson integral formulas for the acoustic system This can be directly applied to pressure in the acoustic equations ∂t p + c∇ · v = 0

∂t + c∇ · v = 0

but not to the velocity components unless the flow is irrotational. In the formula for pressure we can insert the time derivatives to obtain Z ct p(x, t) = Mct {p(x, 0)} − ctMct {∇ · v} + tMct {∇2 p(x, t)} dt 0

The corresponding expression for the velocity is Z ct v(x, t) = Mct {p(x, 0)} − ctMct {∇p} + tMct {∇(∇ · v(x, t))} dt 0

These formulas are valid at all Mach numbers, but IMEX may be required as M → 0. A wrong turning?

Phil Roe

HYP 2016

19/41

Reduction to one dimension In one dimension the spherical mean becomes MR {u(x)} =

1 2R

Z

R

u(x) dx −R

Inserting this into the Poisson integral gives the standard one-dimensional characteristic equations. For two-dimensional problems the integral is over a disc. Archimedes, ca. 250 BCE A wrong turning?

Phil Roe

HYP 2016

20/41

Reduction to two dimensions The two dimensional case is handled by considering data for which there is no dependence on one coordinate.

The reconstructed data is integrated over the disc of radius ∆t centered on each boundary point. For stability the disc must not escape from the union of elements to which the point belongs. Published for irrotational flow by Eymann and Roe, AIAA 2011-3840, and by Hagstrom, App. Num. Meth. 93, 2015 A wrong turning?

Phil Roe

HYP 2016

21/41

The required integrals

Z

θQ

Z

R

Ip = θP

0

(r cos θ)p √ r dr dθ R2 − r 2

I0 =R(θQ − θP ) πR I1 = (yQ − yP ) 2 R R3 I2 = (xQ yP − xP yQ ) + (θQ − θP ) 3 3

For a third-order scheme, I0 and I1 are required, and perhaps I2 .

A wrong turning?

Phil Roe

HYP 2016

22/41

Verification in one dimension

The one-dimensional and two-dimensional results are of equal quality A wrong turning?

Phil Roe

HYP 2016

23/41

Results for linear acoustics (1)

A wrong turning?

Phil Roe

HYP 2016

24/41

Results for linear acoustics (2)

A wrong turning?

Phil Roe

HYP 2016

25/41

Results for non-linear acoustics This is not an overshoot!

The outgoing wave steepens into a shock.

A wrong turning?

Phil Roe

HYP 2016

26/41

Nonlinear advection with d > 1 ∂t v + v · ∇v = 0 is the multidimensional Burgers’ equation, describing particles that coast under inertia, as in dust clouds. It is not in conservation form. Adding conservation gives ∂t ρ + ∇ · (ρv) =0

∂t (ρv) + ∇ · (ρv ⊗ v) =0

(1) (2)

known as the pressureless Euler equations. To second order x∗ =x − ∆t v · ∇v

(3)

vn+1 =v(x∗ )

(4)

∗

ρ =ρ(x )/det(I + ∆t∇v)

(5)

The final term represents convergence and twisting of streamtubes. This is an inertial effect and belongs with the advection. A wrong turning?

Phil Roe

HYP 2016

27/41

Accuracy for pressureless Euler 4 0.9 0.85 0.8 0.75 0.7 0.65 0.6 0.55 0.5 0.45 0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05

y

2

0

-2

-4

-4

-2

0

2

4

x Figure 6: Density at tf = 0.2, ∆t = 0.05, N = 800

−1

−1

10

10

−2

−2

10

10

−3

−3

10 ρ u v

−4

10

−5

|ε|2

|ε|1

10

−5

10

10 3

−6

3

−6

10

10

−7

10 −3 10

ρ u v

−4

10

−7

−2

10 DOF−1/2

(a) L1 norm error

−1

10

10 −3 10

−2

−1

10 DOF−1/2

10

(b) L2 norm error

Figure 7: Local truncation error convergence

A wrong turning?

Phil HYP The exact same parameters areRoe used to find the error convergence results.

2016

28/41

Accuracy for pressureless Euler Table 7: ρ error Level 1 2 3 4 5

√ −1 DOF 3.949763e-02 2.007644e-02 1.012169e-02 5.081907e-03 2.546244e-03

Level 1 2 3 4 5

√ −1 DOF 3.949763e-02 2.007644e-02 1.012169e-02 5.081907e-03 2.546244e-03

Level 1 2 3 4 5

√ −1 DOF 3.949763e-02 2.007644e-02 1.012169e-02 5.081907e-03 2.546244e-03

ε1 2.701100e-03 4.064032e-04 5.550211e-05 7.321861e-06 9.236982e-07

O(ε1 ) 2.7990 2.9070 2.9399 2.9957

ε2 4.742808e-03 7.104399e-04 9.546539e-05 1.252790e-05 1.580521e-06

O(ε2 ) 2.8056 2.9307 2.9475 2.9957

Table 8: u error ε1 2.798819e-03 2.216802e-04 2.135123e-05 1.825973e-06 2.083604e-07

O(ε1 ) 3.7472 3.4169 3.5690 3.1409

O(ε2 )

ε2 5.307710e-03 3.802758e-04 3.584019e-05 2.989206e-06 3.346169e-07

3.8954 3.4486 3.6053 3.1687

ε2 2.878431e-03 2.603886e-04 2.329468e-05 1.834389e-06 2.058630e-07

3.5508 3.5247 3.6887 3.1650

Table 9: v error

A wrong turning?

ε1 1.855756e-03 1.802475e-04 1.602904e-05 1.374465e-06 1.533301e-07

Phil Roe

O(ε1 ) 3.4457 3.5334 3.5651 3.1737

O(ε2 )

HYP 2016

29/41

Putting it together (1) The primitive variables are updated using a Lax-Wendroff procedure. Write the Euler equations with u = (ρ, u, v , p)T ut + Aux + Buy = 0 utt + At ux + Autx + Bt uy + Buty = 0

so that un+1 =un + tut + 21 t 2 utt =un − t(Aux + Buy ) − 21 t 2 (At ux + Autx + Bt uy + Buty ) =un − t(A + 12 tAt )ux − t(B + 21 tBt )uy + 21 t 2 A(Aux + Buy )x + 21 t 2 B(Aux + Buy )y ...................and then some algebra...................... =un − t(A∗ ux + B∗ uy ) + 12 t 2 (A2 uxx + (BA + AB)uxy + B2 uyy )

(6)

where

A wrong turning?

A∗ = A + 21 t(At − AAx − BAy )

(7)

B∗ = B + 21 t(At − AAx − BBt )

(8)

Phil Roe

HYP 2016

30/41

Putting it together (2)

The effect is to give a local linearization with some modified coefficients (NOT modified derivatives). This is discretized in such a way that if the coefficients were constant the linear solution with streamline tracing and Poisson integrals would be recovered. This simple procedure is one of the attractions of p = 3

A wrong turning?

Phil Roe

HYP 2016

31/41

The travelling vortex problem-full Euler equations The figures show contours of density after one revolution around a box about three times larger than the part displayed. In the top figure, the exact solution (lines) is superposed on the numerical solution (colors). In the bottom figure the medium mesh is displayed two of our grids are coarser than this; two are finer.

A wrong turning?

Phil Roe

HYP 2016

32/41

Density errors Fast Vortex after 1 period

0

10

−1

10

−1

−2

10

10

Velocity Error

Velocity Error

10

−3

10

FVM2 AF DG1 DG2 2nd−order 3rd−order

−4

10

−5

10

Fast Vortex after 1 period

0

10

−3

10

−2

−3

10

FVM2 AF DG1 DG2

−4

10

−5

−1

10 h

−2

10

10

2

10

4

10

6

work units

10

8

10

1. No superconvergence- DG1 is second-order, DG2 is third-order. 2. “Work units” allow for number of residual evaluations, but make no assumptions about the cost of each residual. A wrong turning?

Phil Roe

HYP 2016

33/41

Velocity errors Fast Vortex after 1 period

0

10

−1

10

−1

−2

10

10

Velocity Error

Velocity Error

10

−3

10

FVM2 AF DG1 DG2 2nd−order 3rd−order

−4

10

−5

10

Fast Vortex after 1 period

0

10

−3

10

A wrong turning?

−2

FVM2 AF DG1 DG2

−4

−5

10

Phil Roe

−3

10

10

−1

10 h

−2

10

2

10

4

10

HYP 2016

6

work units

10

8

10

34/41

A transonic vortex

A much more nonlinear version of the same problem. −0.3 > M < 1.5, 0.4 < ρ < 1.0 A wrong turning?

Phil Roe

HYP 2016

35/41

Dissipative terms

The plan is to use some hyperbolic version of Navier-Stokes (Nishikawa, Peshkov, moment models). The viscous terms become isomorphic to the elastodynamic equations, whose initial value problem also has an exact solution in spherical means. Initial work on the advection diffusion equation indicates that third-order accuracy is achievable in the derivatives. Previous work on the hyperbolic formulation indicates a great tolerance for distorted grids.

A wrong turning?

Phil Roe

HYP 2016

36/41

Limiting-the final frontier? The outcome of pure advection is always bounded and boundedness-preserving schemes are well developed (Shu). Acoustics is NOT usefully bounded. Entropy stability is too weak. We are exploring the idea of limiting in time, in order to preserve the symmetries of space discretization.

Roe, Lung, Maeng AIAA 2015-2193 A wrong turning?

Phil Roe

HYP 2016

37/41

What’s in? Whats out?

IN Finite differences, Lax-Wendroff, operator splitting OUT Discontinuous reconstructions, Riemann solvers, semi-discretizations.

A wrong turning?

Phil Roe

HYP 2016

38/41

So, was it a good question?

Yes, it was an excellent question. It has been extraordinarily fruitful, both in theory and practice. But it may not have been the best question. If I had to try to pose a better question, it might be this. What does the data teach us about what is going on? But maybe I should wait a bit before committing myself.......

A wrong turning?

Phil Roe

HYP 2016

39/41

Are we there yet?

A wrong turning?

Phil Roe

HYP 2016

40/41

Acknowledgements Timothy B. Eymann, Doreen Fan, Brad Maeng, Ju Wang, Kyle Ding, Praveen Chakravarthy, Prof. Krzystof Fidkowski.

NASA Cooperative Agreement NNX12AJ70A

A wrong turning?

Phil Roe

HYP 2016

41/41