Lecture 6) Multidimensional Schemes for Hyperbolic

Conservation Laws: Bringing it Together

1

ByProf. Dinshaw S. Balsara (dbalsara@nd.edu)

Les Houches Summer School in Computational Astrophysicshttp://www.nd.edu/~dbalsara/Numerical-PDE-Course

1

7.1) Introduction

This is the chapter where we bring together all our ideas on hyperbolic conservation laws.

Multi-d conservation law:

Whether it is hyperbolic depends on eigenstructure of :

Written in conservation form:

Question: What are the benefits of conservation form? Why do we have overbars?

( ) ( )U F U G U 0t x y+ + =

( ) ( ) F U G UA = Λ ; B

U UR L

∂ ∂≡ ≡

∂ ∂

( ) ( )1 1/2 1/2 1/2 1/21/2, 1/2,, , , 1/2 , 1/2U U F F G G

n n n n n ni j i ji j i j i j i j

t tx y

+ + + + ++ − + −

∆ ∆= − − − −

∆ ∆

2

Avoiding spurious oscillations: Non-linear hybridization; TVD Reconstruction

Hyperbolic conservation law becomes simple in characteristic variables:

Above equation still holds locally, though not globally when non-linearities are introduced.

Riemann problem is essential for dealing with non-linearities.

Various kinds of approximate Riemann solvers were studied.

We will always begin with examples of second order schemes. Then we will show how the methodology is extended to higher order schemes. I.e. go from the familiar to that which may be less familiar.

( ) ( ) + 0 with U 1,..,m m m m mt xw w w x l x m M= = ∀ =λ

3

7.2) Reconstructing the Solution for Conservation Laws – Part I, TVD & PPM Reconstruction7.2.1) TVD Reconstruction in Conserved, Primitive or Characteristic VariablesThe Problem: Let U be an M-component solution vector for a 2D conservation law. At a time tn in a zone (i,j), we start with zone-averaged values. We wish to reconstruct the solution:

The Solution: Three ways to do it:

A)Reconstruct on the conserved variables.B) Reconstruct on the primitive variables.C) Reconstruct on the characteristic variables.

( ) ( ) ( ){ }

, , , ,

, , ,Reconstruct

U , U + U + U where ;

means we use the mesh function U to obtain U and Uion

in each zone.

n ni j i j x i j y i j i j

ni j x i j y i j

x y x y x x x x y y y y= ∆ ∆ ≡ − ∆ ≡ − ∆

∆ ∆

4

A) Componentwise Reconstruction on the conserved variables:

( )

( )

th, ,

, , , ,

, 1, , , 1,

, , 1

Let be the component of U .

For each of the components, we want:, + +

with:

, ;

m ni j i j

m m m mi j i j x i j y i j

m m m m mx i j i j i j i j i j

m my i j i j

u m

u x y u u x u y

u Limiter u u u u

u Limiter u

+ −

+

= ∆ ∆

∆ = − −

∆ = −

( ), , , 1,m m mi j i j i ju u u −−

Question: What are the advantages and disadvantages of this form of limiting?

5

B) Componentwise Reconstruction on the primitive variables:

( )

, ,

th, ,

, , , ,

,

From each vector U , obtain a vector of primitive variables V

Let be the component of V .

For each of the components, we want:, + +

with:

n ni j i j

m ni j i j

m m m mi j i j x i j y i j

mx i j

v m

v x y v v x v y

v

= ∆ ∆

∆ =

( )( )

1, , , 1,

, , 1 , , , 1

, ;

,

m m m mi j i j i j i j

m m m m my i j i j i j i j i j

Limiter v v v v

v Limiter v v v v

+ −

+ −

− −

∆ = − −

Question: What are the advantages and disadvantages of this form of limiting? 6

C) Reconstruction on the characteristic variables:For a linear system, + 0 is valid all over. Not so for a non-linear system. Question: Why?

When the eigenvalues and eigenvectors become solution-dependent, we only have alocal representatio

m m mt xw wλ =

( ) ( ) ( )

( ) ( ){ }

, , ,

1,

within each zone.

I.e., U , U and U have become solution-dependent.

In simple words, the left eigenvectors in zone 1, , given by U : 1,..

n of the eigensys

., ,

will not be

e

t mm n m n m n

i j i j i j

m ni j

l r

i j l m M+

λ

+ =

( )( ){ },

orthonormal with the right eigenvectors in zone , , given by

U : 1,..., . Consequently, each z

Advantage of thi

one de

s Limi

fines its own .

-Difft erent wave families may hing a : e

v d

m ni j

i j

r local eigenspacem M=

ifferent properties. Some

may be linearly degenerate with the result that they don't self-steepen (think contact discontinuity in Euler flow). Others may be genuinely non-linear so that they do self-s

Different waveteepen

(think families havesound waves in Euler flow). to be treated differently.7

project into an eigenspace specific to the zonWe can only the solution from neighboring zones that is

as follows:

First, obtain the characteristic variables from the neighboring zones and

e ( , )

the cu

i j

( ) ( ) ( ), ; , 1, , ; , , , ; , 1,

,

rrent zone:-

U U ; U U ; U U

(pa

1,..,

Second, limit the local characteristic v

y attention

ariab

to the subscrip

le

ts)

:

m m n n m m n n m m n ni j L i j i j i j C i j i j i j R i j i j

mi j

w l w l w l m M

w Limi

− += = = ∀ =

∆ =

( )

( )

, ; , ; , ; , ;

, , ,1

, 1,..,

Third, project the variation in the characteristic variables into the local space of right eigenvectors:

U U

m m m mi j R i j C i j C i j L

Mm m n

x i j i j i jm

ter w w w w m M

w r=

− − ∀ =

∆ = ∆∑

Question: What do subscripts “L”, “C” and “R” stand for? What are the advantages and disadvantages of this form of limiting? 8

Results from piecewise linear limiting. Advect Gaussian, square wave, triangle and ellipse. See if shape is preserved. But what is that scheme to the right?

9

7.2.2) Going Beyond Piecewise Linear: Piecewise Parabolic ReconstructionIt is only logical to think that going beyond piecewise linear reconstruction would yield better dividends. The piecewise parabolic method, PPM, is predicated on that viewpoint.

It retains a TVD-like flavor in that it prohibits new extrema from developing.

It is acceptable to reconstruct the variation one dimension at a time. Therefore, we focus on a one-dimensional reconstruction.

Let us assume a piecewise parabolic profile that is valid over [-0.5,0.5], which spans the zone “i”:

Question: Why might it be beneficial to choose this form?

( ) 2 1ˆ ˆ 12i i x xxu x u u x u x = + + −

10

i−2 i−1 i i+1 i+2 i+3

ui;Rui;L

First, realize that .

Thus, a parabola in zone "i" would beleft and

is already specified and can't be

right edge values of "u"

within t

h

fully specified if the

were to beat sp

cha

ecizon fie

ng d

e

e

c

d.

Fo

u

; ;

; ; ;

; ;

;ˆ ˆ ; 3 6

Quest

us on zone "i". Given

ion: So, how do we obta

and , we

in and with third or better accura

o

c

+ 3 btain:

y?i

i L i R

x i R i L xx i R i i L

L i R

u uu u u u u u

u

u

u

= − = −

11

Shock-like profile shown by dotted curve.Mesh function (average of shock-like profile) shown by dashed slabs.Reconstructed profile shown by solid curve. Observe how it changes during the course of the reconstruction procedure!

i−2 i−1 i i+1 i+2 i+3

ui;Rui;L

; ;Question: So, how do we obtain and with third or better accuracy?

i L i Ru u

( )

( )0.5

1.

2 30 1 2 3

5

Answer: :Consider the cubic , which is to be fitted to zones from " 1"

. This can

Reconstruction

be done by imp

via primitive func

osing the four conditito " 2"

i

s

t on

onq x q q x q x q x

i i

q x dx−

−

= + + +

− +

∫ ( ) ( ) ( )0.5 1.5 2.5

1 1 20.5 0.5 1.5

; ; ; i i i iu q x dx u q x dx u q x dx u− + +−

= = = =∫ ∫ ∫ 12

( ) ( )

( ) ( ) ( ) ( )

; 1

; 1 1 1 2 1

1

1

;

2

1;

1 1 1 1 with

7 1 12 12

We also set .This parabolic profile is shown by the solid curve

and 2

in

6 2

2

i R

i R i i i i i

i i i i

i i i i i i

i L i R

u u u u u u u u u u

u u u u

u

u

u

u

u

+ + + + + −

+ −

+

+

= + − − ∆ − ∆ ∆ = −

+ −

∆

=

+

= −

=

the previous figure.

Question: What are the problems with the solid curve in the previous figure? 13

i−2 i−1 i i+1 i+2 i+3

ui;Rui;L

; ;Question: So, how do we obtain and with third or better accuracy?

i L i Ru u

i−2 i−1 i i+1 i+2 i+3

( ) ( ) ( ) ( )1 2 1 1; 1 1

Second, consider the previous formula:1 1 with and 2 2

But now enforce a limiter. I.e., unlimited slopes above are replaced wit

1 1

h the

6

2 i i ii R i i i i i i i iu u u u u u u u u u u u+ + + + + −∆ = −= + − − ∆ ∆ =− −∆

( ) ( )1 2 1 1

1; ;

1 1

limited slopes below:

We also set . The res

,

ult is

;

shown above. It is still unacceptable. Question: Why?

,

i L i R

i i i i i i i i i i

u

u MC u

u

u u u u MC u u u u+ + + + +

+

−∆ = − − ∆ =

=

− −

14

i−2 i−1 i i+1 i+2 i+3

( )( )

; ;

; ;

Third, in the spirit of not introducing new extrema, we first require that the mean value,

, should lie between min , and

max , . This gives us our first condition

for enforcing monoto

i i L i R

i L i R

u u u

u u

( )( ); ; ; ;

nicity:

This condition would .

But it leaves zones " " and " 1" unchanged!Question: How do we detect and cure those z

and if 0

flatten zone " 2

es?

"

on

i L i i R i i R i i i Lu u u u u u

i

i i

u u

+

→ → − − ≤

+

( ) ( )

( )

2 ˆ ˆ2

ˆ ˆintroduces a new ex

1ˆ ˆAnswer: Notice

tremum in zone " " centered at 0 if : 0.5 2 0.5

Without changing

that has extrema at 12

The method

Solution : t

e x xx

x xx

i i x xx x u u

i x

u x u u x

u

u x

u

= + + − = −

=

<

− < −

ˆ ˆhe sign of , reduce the value ofQuestion: Why will this w

.ork?

xx xxu u15

( ) ( ) ( )

( ) ( )

2; ;

; ; ; ; ; ;

2; ;

; ; ; ; ;

1 3 2 if > 2 6

1 3 2 if >

An automatic procedure for shifting the extremum out of the zo

2

e:

6

n

i R i Li L i i R i R i L i i R i L

i R i Li R i i L i R i L i i

u uu u u u u u u u

u uu u u u u u u

− → − − − +

−→ − − − − ( );

Compare the parabolic profile above to the profile from an MC limiter. Question: What do you see? Compare results of advection test with PPM & MC and discuss.

R i Lu +

i−2 i−1 i i+1 i+2 i+3 i−2 i−1 i i+1 i+2 i+3

PPM

MC

16

7.3) Reconstructing the Solution for Conservation Laws – Part II, Non-Oscillatory ReconstructionThe TVD and classical PPM reconstruction do clip local extrema, whether they are physical or not. Over thousands of timesteps, this can add up. (There are newer variants of PPM that avoid this.)

Clipping extrema scheme is second order.

To go to higher order, we must avoid clipping physically valid extrema.

In Essentially Non-Oscillatory (ENO) schemes, we permit the introduction of new extrema in the reconstruction as long as the original mesh function also has such extrema.

Weighted Essentially Non-Oscillatory (WENO) reconstruction schemes are a modern variant.

Comes in two flavors: Finite Difference and Finite Volume 17

7.3.1) Weighted Essentially Non-Oscillatory (WENO) Reconstrutionin One DimensionThink of the minmod limiter as choosing from a left slope (corresponding to a left-biased stencil) and a right slope (corresponding to a right-biased stencil).

Either reconstruction is second order. Minmod chooses the stencil with the smaller slope.

The choice is made using non-linear hybridization.

WENO takes these two ideas much further:

1) We do a much more detailed stencil analysis, often between two or more stencils.2) The non-linear hybridization between competing stencils is also more sophisticated.

Let us see how WENO reconstructs a smooth and a discontinuous profile.

18

i−2 i−1 i i+1 i+2 i−2 i−1 i i+1 i+2

19

Smooth Gaussian & Shock-like profiles shown by dotted curves.Mesh function (average of shock-like profile) shown by dashed slabs.Reconstructed profile shown by solid curve. Observe how it changes during the course of the reconstruction procedure!

i−2 i−1 i i+1 i+2

i−2 i−1 i i+1 i+2

( ) 2

In zone " ", we wish to reconstruct at third order

ˆ ˆBecause there are two moments, and , weneed a stencil with three zo

1ˆ ˆ 12

nes.

First st -encil is and co

x xx

i i x xxu x u u x u x

left biased

i

u u

= + + −

( ) ( )1.5 0.5

; 2 ; 12.5 1.5

; 1 2

; 2 1

vers zone " ".

Again, :

;

ˆ = 2 + 0.5 + 1.5 ; ˆ = 0.5

reconstruct via primitive func

+ 0.5

Question

tion

: C

i L i i L i

L x i i i

L xx i i i

i

u x dx u u x dx u

u u u uu u u u

− −

− −− −

− −

− −

= =

⇒ −

−

∫ ∫

an you comment on the smoothness of the two left-biased stencils? (I.e., solid curves)

20

( ) ( )

( )

0.5 1.5

; 1 ; 11.5 0.5

; 1 1

; 1

reconstruct via primitive function

Second stencil is and covers zone " ".

Again, :

;

ˆ = 0.5 ; ˆ = 0.5

-

i C i i C i

C x i i

C xx i

centri

u x dx u u x dx u

u u

ally bi

u

a

u

e

u

s d

−

− +−

+ −

−

= =

⇒ +

∫ ∫

1 + 0.5

Question: Can you comment on the smoothness of the two centrally-biased stencils? (I.e., solid curves)

Questions: Which stencil is best for stabilityof smooth flow? What happens when th

i iu u +−

e flowis not smooth?

i−2 i−1 i i+1 i+2

i−2 i−1 i i+1 i+2 21

( ) ( )1.5 2.5

; 1 ; 20.5 1.5

; 1 2

;

Third stencil is and covers zone " ".

Again, :

; ;

ˆ = 1.5 + 2 0.5 ; ˆ =

reconstruct via primitiv

0

-

e funct

.5

ion

i R i i R i

R x i i i

R xx

i

u x dx u u x dx u

right bias

u u u

e

u

d

u

+ +

+ +

= =

⇒ − −

∫ ∫

1 2 + 0.5

Question: Can you comment on the smoothness of the two right-biased stencils? (I.e., solid curves)

i i iu u u+ +−

i−2 i−1 i i+1 i+2

i−2 i−1 i i+1 i+2 22

( );

We need to quantify the smoothness of a stencil.Insight: the smoothness relates to the derivatives:

Define for sten cil:

i LL

smoothne e

larger derivatives less smoothness

ach

I

ss ind

d

ica

ux

Sd

tors

x=

⇒

( )0.52 2

; ;0.5

2 2 2 2

2

;

2

2

; ;

2;

;

13ˆ ˆ + 3

Similarly:13 13ˆ ˆ ˆ ˆ + and + 3 3

Smoothness indicator is a misnomer, bec

au

+

s

iL L x L xx

C C x C xx R R x R xx

L dx IS u u

IS

d u

u u IS u u

xdx−

⇒ =

= =

∫

e the above measures actually measurehow rough the solution is in each of the stencils.

Questions: What do we expect for the 3 smoothness indicators for the Gaussian?What do we expect for the discontinuous profile?What would happen if we keep switching back and forth too rapidly between stencils?23

( ) ( ) ( )

Note:

Solution: .

Weight the stencils in inverse proportion to their smoothness ind

; ; ;

; ;

icators

CL RL C Rp p p

L C R

CLL C

L C R L C R

RR

L C R

w w wIS IS IS

www ww w w w w w

www w w

γγ γ= = =

+ ε + ε + ε

= =+ + + +

=+ +

; ; ;

;

, and are called .

, and are called .

The final reconstructed profile is given byˆ ˆ ˆ ˆ + +

; ˆ ˆ

1

+

-L C R

L C R

x L L x C C x R R x

xx

L C

L L xx C

R

li

non linea

w w w

r weightsw

near weigh

w w

u w u w u ww

s

uu w

t

u

γ γ γ

=

=

+ + =

; ;ˆ ˆ + C xx R R xxu w u

i−2 i−1 i i+1 i+2

i−2 i−1 i i+1 i+2 24

Different strokes for different blokes:

There is a certain amount of flexibility in the choice of the linear weights, and also, indeed,in the choice of smoothness measures.

CWENO 1 with 50 : L R Cγ = γ = γ =Question: What does this do?

Dispersion minimization: Requires adding fur

and =4.Central stencil is emphasized becaus

ther stencils.

Optimize accuracy: For right-

e it is mor

going waves we pick

e stable.p

With this choice we can achieve fifth order of accuracy, but only for finite di

0.1, 0.6, 0.3

increasing order of afference sch

ccuracemes.

This advantage of does not extend to finite volume sy c

L C Rγ = γ = γ =

hemes.

Question: The smoothness indicators are extremely nonlinear in the solution. Why is thata good thing? What does that give you?

In general, WENO works better if the reconstruction is carried out i characteristic variabn les.25

Question: Can you compare and contrast these results with those for TVD and PPM schemes?

26

7.3.2) WENO Reconstruction in Multiple Dimensions

Say for example that we want third order finite volume reconstruction over a rectangular zone. We then have

Our job is to reconstruct all the moments. Can be done in one of two ways:

1) Choose large, multidimensional stencils, each of which yields all the moments.

2) Do all x-moments by looking in the x-direction; all y-moments by looking in the y-direction. Then do something special for the few cross terms that remain.

We pick strategy 2) thought the book gives citations where 1) is done.

( ) 2 2, ,

1 1ˆ ˆ ˆ ˆ ˆ, + 12 12i j i j x xx y yy xyu x y u u x u x u y u y u x y = + + − + + −

27

( ) ( ){ }

( )1.5 1.5

, 1, 10.5 0.5

1, 1 ,

Using , , 1, 1 along with the consistency condition:

,

givesˆ ˆ ˆ ˆ ˆ=

Three other stencils are possible. Question: Can yo

y x

i j i jy x

xy i j i j x y xx yy

i j i j

u x y dx dy u

u u u u u u u

= =

+ += =

+ +

+ +

=

− − − − −

∫ ∫

2 2 2

u list the other three stencils?

ˆ ˆ ˆThe smoothness indicators are given by : 4 4

The resulting non-linear hybridization is obvious.

xx yy xyIS u u u= + +

28

29

7.4) Evolving Conservation Laws Accurately in Time – Part I, Runge-Kutta MethodsSpatial reconstruction at second and third order has been thoroughly discussed. Even higher orders of spatial reconstruction are possible.

To get a balanced scheme, the temporal accuracy must match the spatial accuracy.

Easiest way to do that is given by Runge-Kutta methods.

The time evolution is conceptualized as the solution of an ordinary differential equation. Recall the terminology, “method of lines”, “semi-discrete approximation” from Lecture 2.

Each stage in a Runge-Kutta method is simple and multiple internal stages make up the RK method to give us the overall time accuracy.

Only some classes of RK methods preserve the TVD property – SSP-RK.

30

7.4.1) Runge-Kutta Time Stepping

( ) ( ) ( )

( ) ( )1

1

Strong Stability Pr

UConceptualize this as

eserving (S

an evolution in time: U F U G U

approximation is and :

Question: What does S

I

S

mpro

P do

ved Euler second

for us?

U U

SP

U

1U

o

U

de

2

r r)

x y

n n

n n

Lt

t L

+

∂= ≡ − −

∂

= + ∆

= + ( ) ( )( )

( ) ( )( ) ( ) ( )( )

( ) ( )( )

1 1

1

2 1 1

2 21

1 1U U2 2

Can we extend this to ? Yes!

U U U

3 1 1U U U U4 4 41 2 2U U U U 3 3 3

Above two schemes support a max CFL o

t

f

hir

unity in 1D; Question: What about

d order R

mu

K-SSP

lti-d?

n n

n

n n

t L

t L

t L

t L+

+ ∆

= + ∆

= + + ∆

= + + ∆

31

( )1

Question: To what orders can we continue this? Answer: From onwards, schemes become 4th order RK-SSP

Butcher

Fourt

quite complica

h order RK-SSP

ted -- !

sch

bar

eme

U U 0.391752227003

riers

n= + ( )( ) ( ) ( )( )( ) ( ) ( )( )( ) ( )

2 1 1

3 2 2

4 3

92 U

U 0.44437049406734 U 0.55562950593266 U 0.36841059262959 U

U 0.62010185138540 U 0.37989814861460 U 0.25189177424738 U

U 0.17807995410773 U 0.82192004589227 U 0.54497475

n

n

n

n

t L

t L

t L

∆

= + + ∆

= + + ∆

= + + ( )( )( )

( ) ( )

( )( ) ( )( )

3

21

3 4

3 4

021237 U

U 0.00683325884039 U 0.51723167208978 U

0.12759831133288 U 0.34833675773694 U

0.08460416338212 U 0.22600748319395 U

n n

t L

t L t L

+

∆

= +

+ +

+ ∆ + ∆

32

7.4.2) Second Order Accurate Runge-Kutta Scheme

( ) ( ) ( )

( ) ( )

,1/2, 1/2, , 1/2 , 1/2

,

1/2

1/2, , 1/2

1/2

We can now discretize the spatial parts of the governing equation as

U 1 1U F F G G

with

F F 1/ 2, ; G G , 1/ 2

i ji j i j i j i j

i j

y

i j i j

y x

Lt x y

x y dy x y dx

+ − + −

=

+ +

=−

∂= = − − − −

∂ ∆ ∆

≡ = ≡ =∫

1/2

1/2

integrationfinite volume formulat

The of the fluxes along the faces reminds us that we are dealing with a . The integration of the fluxes shoul

ionsufficiently high orded be done with o r

x=

=−∫

At second order things are sim

in order to realize the full accuracy of the method.

. We can replace the spatially averaged fluxes with the fluxes from the Riem

f spatial acc

ann problem t

uracy

hatple

have been evaluated at the centers of the zone faces.

33

; 1/

, , ,

, , 1; 1/

; , 1

, 1,

,

2, 2

2 ,/

,

; ,

In each zone ( , ), we assume we have U and undivided differences U and U ,to get:

1 1= U U ; = U U2 21= U U

UU

; U2

U

i j x i j y i j

i j x i j i j x i j

i j y i

R i

T i

L i j

i j

j

jB j

i j

+

+

+ +

+

+

∆ ∆

+ ∆ − ∆

+ ∆

( ) ( );

1 , 1 , 1

1/2, ;

/

1 ,, 1/2; 1/2,

2

; , 1/2, 1/ /22

1= U U2

F F , ; G G UUU U ,R i B i

i j y i j

i j RS i j RS T ijj j jL i

+ +

+ ++ + ++

− ∆

= =

The neighboring zones have to interact using a Riemann solver so that we get the flux at the zone boundaries!

34

7.4.3) Runge-Kutta Schemes at Higher Orders; Using Third Order as an Example

( ) ( ) ( )

( ) ( )

,1/2, 1/2, , 1/2 , 1/2

,

1/2

1/2, , 1/2

1/2

We still discretize the spatial parts of the governing equation as

U 1 1U F F G G

with

F F 1/ 2, ; G G , 1/ 2

i ji j i j i j i j

i j

y

i j i j

y x

Lt x y

x y dy x y dx

+ − + −

=

+ +

=− =−

∂= = − − − −

∂ ∆ ∆

≡ = ≡ =∫

1/2

1/2

Problem: , evaluating the above integrals becomes difficult.Questions:What does the tell us about numerical quadrature?How many ev

At higher or

aluations of

derSimpson ru the numerical flu

le

x=

∫

multiple Riemann solver evaluationsGaussian

x would it require?One option would be to use at multiple

along each bo points expensiundary. This can become very .

Because Riemann problem evaluation

ve

is expensive, we want to make as fewevaluations of the Riemann problem as possible -- A - approach .quadrature free

35

36

( ) 2 2, , , ; , ; , ; , ; , ;

Start with a reconstructed profile in each zone ( , ):1 1ˆ ˆ ˆ ˆ ˆU , U U U U U + U

12 12

Let us take the HLL Riemann solver as an example.

i j i j i j x i j y i j xx i j yy i j xy

i j

x y x y x y x y = + + + − + −

( ) ( ) ( )( ) ( ) ( )

, ; 1/2, ,

1, ; 1/2, 1,

U , is used to obtain the left state : U U 1/ 2, 0

U , is used to obtain the right state : U U 1/ 2, 0These can be used to obtain and , the extr

ci j L i j i j

ci j R i j i j

L R

x y x y

x y x yS S

+

+ + +

= = =

= = − =

( ) ( ) ( )1/2, ; 1/2, ; 1/2,

emal wave speeds for the HLL RS.If these wave speeds are frozen, we obtain the HLL flux as:

F F F

S SS S

S S

US SS S

R L

R L Ri j L i

Lj i j

L

R

R L

R

y y y+ + +

= −

+

−

− −

( ) ( )( ); 1/2, ; 1/2,U

Question: What does freezing the wave speeds, and , give us?

R i j L i j

L R

y y

S S

+ +−

37

( ) ( ) ( ) ( ); 1/2, , ; 1/2, 1,

The reconstruction within each zone allows us to write:U U 1/ 2, ; U U 1/ 2,

The above expressions are available analytically and can be integrated once

L i j i j R i j i jy x y y x y+ + += = = = −

( ) ( )( )

( ) ( )

1/2

; 1/2, ; 1/2,1/2

1, 1, ; 1, ; , , ; , ;

1; 1/2, ,

and for all:

U U

1 1 1 1ˆ ˆ ˆ ˆ U U U U U U2 6 2 6

Fluxes treated similarly:

U U 1/ 2, 1/ 2

y

R i j L i jy

i j i j x i j xx i j i j x i j xx

L i j i j

y y dy

x y

=

+ +=−

+ + +

+

−

= − + − + +

= = = −

∫

( ) ( )( ) ( )

( ) ( )( ) ( )( ) ( )( )

2; 1/2, ,

3; 1/2, ,

1/21 2 3

; 1/2, ; 1/2, ; 1/2, ; 1/2,1/2

; U U 1/ 2, 0 ;

U U 1/ 2, 1/ 2

1 2 1F F U + F U + F U6 3 6

L i j i j

L i j i j

y

L i j L i j L i j L i jy

x y

x y

y dy

+

+

=

+ + + +=−

= = =

= = =

=∫

38

7.5) Evolving Conservation Laws Accurately in Time – Part II, Predictor-Corrector Schemes7.5.1) Second Order Accurate Predictor-Corrector Schemes

( )

( ) ( ) ( )

, , , ,

, ,

Say we have used the reconstruction in zone ( , ) at time to obtain:U , U + U + U

We refer to U and U as the .

Conside 1r : ,1/ 2 / 2, , / 2, , 1

n

n ni j i j x i j y i j

x i j y i j

i j tx y x y

nodal po

m

i j

odes

ii s j jnt i −

= ∆ ∆

∆ ∆

++

( )( )

and .

We evaluate from immediately inside the zone , being considered.

, 1/ 2

nodal valu i jes

i j −

(i,j) ( )1/ 2,i j+( )1/ 2,i j−

( ), 1/ 2i j +

( ), 1/ 2i j −

39

( ) ( ) ( ) ( )

, , 1/

, 1/

1/2, , ,2

,2

,

Consider : , , and .

1 1We now have : U + U ; U U ;2 2

1/ 2, , 1/ 21/

U

2 ,

U

2, 1

U U

/

n ni j x i j i j x i j

ni

n

nj

i j

j

i

i

nodal points

nodal valu

i j

es

i jji j i

+ −

+

+ −+

= ∆

−

= − ∆

=

( )

( ) ( )( ) ( ) ( )( ),

, , ,

1/2, 1/2, , 1/2 , 1/

, 1/2

2

1 1 + U ; U U2 2

then gives us the time rate of update within each zone , as:

1 1F U F U G U G

U

UU

n

i

n nj y i j i j y

ni j i j

n n n ni j i j i j i j

j

Predict i j

x

or Step

t y+ +

−

− −

∆ = − ∆

=∂ −

− − −∆ ∆∂

(i,j)1/2,Un

i j+1/2,Uni j−

, 1/2Uni j+

, 1/2Uni j−

The Predictor Step now gives us an “evolution-in-the-small” for the PDE as follows:-

40

1/2 1/2; 1/2, , ; 1/2, 1, , 1,

,

then assembles the on either side of

each zone boundary:

1 1 1 1 UU U

space- and time-centered value

+ ; U U U2 2

U2

s

U2

n n n n nL i j i j R i j i

nn

x jj

j x ii ji

t t

Corrector St p

t

e

+ ++ + + +

∂ ∂≡ + ∆ ≡ − ∆ + ∆

∆ ∂ ∂

( )

1,

1/2 1/2; , 1/2 , , ; , 1/2 , 1 , 1

, , 1

1 1/2 1/2, , ; 1/2, ; 1/2, ; 1/

1 1 U 1 1 UU U + U ; U U U2 2 2 2

: U U F U , U F U

n

i j

n nn n n n n nB i j i j y i j T i j i j y i j

i j i j

n n n ni j i j RS L i j R i j RS L i

t

t tt t

tFinal updatex

+

+ ++ + + +

+

+ + ++ + −

∂ ∂ ≡ ∆ + ∆ ≡ − ∆ + ∆ ∂ ∂ ∆

= − −∆

( )( )( ) ( )( )

1/2 1/22, ; 1/2,

1/2 1/2 1/2 1/2; , 1/2 ; , 1/2 ; , 1/2 ; , 1/2

, U

G U , U G U , U

n nj R i j

n n n nRS B i j T i j RS B i j T i j

ty

+ +−

+ + + ++ + − −

∆− −∆

The Corrector Step is the step where the neighboring zones interact with each other through the Riemann solver.

41

7.5.2) ADER Time Stepping for Second, and Higher Order Time Accuracy

ADER == Arbitrary DERivatives in space and time.It is a method for starting with a high order spatial reconstruction and rapidly doing the time update with higher order of accuracy.

I.e., the Predictor-Corrector methods at second order go on to become full-fledged ADER methods at all orders in space and time.

The important part is the ADER-Predictor step. It starts with the spatially reconstucted modes with any order of spatial accuracy within a zone. It gives us the temporal “evolution-in-the-small” of the PDE up to the same order of space-time accuracy. Here, “evolution-in-the-small” means locally within the zone of interest.

ADER-CG studied here. “CG” stands for Continuous Galerkin. Question: What does a stiff source term do?

42

1D ADER Predictor In Brief ; Toro & Titarev (2002), Titarev & Toro (2002), Gassner et al (2007), Dumbser, Enaux & Toro (2008), Dumbser et al. (2008) Toro & Castro (2008), Gassner et al (2011), Balsara et al. (2009, 2013)

Mathematical Statement of our goal:-

Within a 1D zone start with spatial reconstruction, i.e. spatial modes:-

20 x xx

1ˆ ˆ ˆw + w x + w u ( xx, t=0 )2

= 1

−

Use governing equation to obtain space-time reconstruction, i.e. space-time modes:-

t xGoverning Equation : u + f(u) = 0∂ ∂

2t t

20 x x t tx xˆ ˆ ˆ+ u t + u (x, t) u t1ˆ ˆ ˆw + + w x + w u x t = x Dynamics!

2!!

1← −

The process is iterative, but the iterations converge very fast!

43

STEP I) Start with the reconstructed initial polynomial within each zoneexpressed in a higher order modal basis set (i.e. orthogonal Hermitepolynomials) (1d here for simplicity; modes shown with caret):-

20 x xx

1ˆ ˆ ˆw + w x + w u ( xx, t=0 )2

= 1

−

We can use the above to obtain the spatial representation of the fluxes at t=0; i.e. evaluate only once:

t xGoverning Equation : u + f(u) = 0∂ ∂

20 x xx

1ˆ ˆ ˆf + f x + f x f (x, t =0) = 12

−

How? Answer: a) Evaluate u1 = u(x=0,t=0), u2 = u(x=.5,t=0), u3 = u(x=-.5,t=0). b) Find f1 = f(u1), f2 = f(u2), f3 = f(u3) . c) Obtain modal representation of flux. 3rd order in 1D is first non-trivial case.

1 23

45

6

x

t

x=0.5x= −0.5

t=0

t=1

Space-time element

44

STEP I, cont’d)We wish to obtain the space-time representation of the conserved vars. within a zone:-

2t tt x0 t

2x xx

1ˆ ˆ ˆw + w x + w x 12

ˆ ˆ ˆ+ u t +u (x, u t +t) = u x t −

Similarly, we wish to obtain a space-time representation of the fluxes:(the fluxes don’t need to be saved, so they don’t take up storage)

2t tt x0 t

2x xx

1ˆ ˆ ˆf + f x + f x 12

ˆ ˆ ˆ+ f t +f (x, f t +t) = f x t −

t tt xt t tt xtˆ ˆ ˆˆ ˆ ˆWe start the with : u u = u = 0 and f f = f = 0iteration = =

t xGoverning Equation : u + f(u) = 0∂ ∂

V. Imp. Questions: 1) Given approx. u(x,t), how to obtain improved f(x,t)?

2) How to obtain an improved space-time representation? t tt xtˆ ˆ ˆi.e. u , u and u

45

STEP II) Pick a space-time element. Pick a set of nodal points in a space-time element. 1d shown here; has been extended to multi-d.

u (xi , ti ) can be evaluated at each such nodal point “i = 4, 5, 6”.

STEP III) Use nodal values to find nodal fluxes fi = f ( u (xi , ti ) ) .

STEP IV) Obtain modal representation of fluxes:Notice: Finite Difference-like forms

( )( )( )

( )

0 1 2 3

x 2 3

xx 2 1 3

t 1 2 3 4 5 6

tt 1 2 3 4 5 6

xt 2

f̂ = 4f + f + f 6 ;

f̂ = f f ;

f̂ = 2 f 2 f + f ;

f̂ = f 2 f 2 f + 2 f + 2 f f ;

f̂ = 2 f f f + f + f f ;

f̂ = 2 f

−

−

− − −

− − − −

− −( )3 4 5f f + f− Notice: t=0 variables only evaluated once; saves float pts.

1 23

45

6

x

t

Nodal points atdifferent time levels

Space-time element in (x,t)Spans [-.5,.5]x[0,1]

x=0.5x= −0.5

t=0

t=1

46

STEP V) Use space-time basis functions as test functions to derive update equations:

( ) ( ) ( )21 2 3x,t = t ; x,t = t ; x,t = x tφ φ φ

Make Galerkin, i.e. weak-form integration over the space-time element :-

( ) ( )t x u x,t + f x,t 0∂ ∂ =

2 20 x xx t tt xt

1ˆ ˆ ˆ ˆ ˆ ˆu (x, t) = w + w x + w x + u t + u t + u x t12

−

2 20 x xx t tt xt

1ˆ ˆ ˆ ˆ ˆ ˆf (x, t) = f + f x + f x + f t + f t + f x t12

−

( ) ( ) ( )[ ] [ ]

t x1/2,1/2 , 0,1

Galerkin formulation:

x, t u x,t + f x,t dx dt = 0 for = 1, 2, 3ix t

i∈ − ∈

∂ ∂ ∫∫ φ

47

For an Nth order scheme, even in multi-dimensions, this process can be shown to converge in (N−1) iterations. – Picard Iteration

ADER can now be used in Predictor-Corrector fashion to obtain a one-step update scheme for conservation laws.

We’ve described the Predictor Step. Next, we describe efficient evaluation of the Corrector Step.

This gives us the update equations:-(Notice that this is the part where we do dynamics, because we relate gradients of flux to the time-rate of change of the conserved variables!!!)

t x tt xt xt xxˆ ˆ ˆˆ ˆ ˆu = f ; u = f 2 ; u = 2f− − −

Go back to STEP II and iterate!

48

Flowchart for Initializing ADER-CG Iteration

20 x xx

1ˆ ˆ ˆStart with : u (x, t) = w + w x + w x 12

−

u (xi , ti =0) evaluated at each spatial nodal point “i = 1, .., 3”.x1=−0.5 ; x2 = 0.0 ; x3=0.5

Use nodal values, u (xi , ti ) to find nodal fluxes fi = f (u (xi , ti )), i=1,..,3

( ) ( ) ( )i i

0 1 2 3 x 2 3 xx 2 1 3

ˆUse nodal fluxes f to obtain modal fluxes fˆ ˆ ˆf = 4f + f + f 6 ; f = f f ; f = 2 f 2 f + f− −

t tt xt t tt xtˆ ˆ ˆˆ ˆ ˆWe start the with : u u = u = 0 and f f = f = 0iteration = =

49

Flowchart for ADER-CG Iteration2 2

0 x xx t tt xt1ˆ ˆ ˆ ˆ ˆ ˆStart with : u (x, t) = w + w x + w x + u t + u t + u x t

12 −

u (xi , ti ) evaluated at each space-time nodal point “i = 4, 5, 6”. (x4, t4) = (-.5,.5) ; (x5, t5) = (0.5,0.5) ; (x6,t6) = (0,1)

Use nodal values, u (xi , ti ) to find space-time nodal fluxes fi = f (u (xi , ti ))

( ) ( )i i t 1 2 3 4 5 6

tt 1 2 3 4 5 6 xt 2 3 4 5

ˆ ˆUse nodal fluxes f to obtain modal fluxes f : f = f 2 f 2 f + 2 f + 2 f f ;ˆ ˆf = 2 f f f + f + f f ; f = 2 f f f + f

− − −

− − − − − − −

t x tt xt xt xxˆ ˆ ˆˆ ˆ ˆUpdate equations : u = f ; u = f 2 ; u = 2f− − −

N-1 iterations for an Nth order scheme.

50

1D ADER Corrector: Rapid strategy for obtaining numerical fluxes

Dumbser, Kaser & Toro (2008), Balsara et al. (2009), Balsara et al. (2013)

Say for the governing equation :

We have obtained the space-time representation of “u” and “f”,i.e. within each zone we have:

t x u + f(u) = 0∂ ∂

2 20 x xx t tt xt

1ˆ ˆ ˆ ˆ ˆ ˆu (x, t) = w + w x + w x + u t + u t + u x t12

−

2 20 x xx t tt xt

1ˆ ˆ ˆ ˆ ˆ ˆf (x, t) = f + f x + f x + f t + f t + f x t12

−

Question: How do we rapidly obtain the numerical flux after wehave obtained the space-time representation?

51

( ) ( ) ( ) ( ) ( )( )

( )

1/2 ; 1/2 ; 1/2 ; 1/2 ; 1/2

1/21/2

Consider the v. simple case of the HLL flux:

The numerical flux is then given by :

HLLi L i R i R i L i

HL

R

L HLLii

L R L

R L R L R L

S S Sf t f t f t u t u t

f f

SS S S S S

t

S+ + + + +

++

= − + −

≡

− − −

( )

1

0

1/2 ; 1/2 ; 1/2 ; 1/2 ; 1/2

If we freeze and in some intelligent way (described later) then we can write:

t

L R

HLL

iR L R

L i R i R i LL

R L R R Li

L

dt

f f fS S S SS S S S S

u uS

λ λ=

+ + + + +

≡ − + − − − −

∫

( )

( )

( )

( )

1

; 1/2; 1/20

1

; 1/2; 1/20

1

; 1/2; 1/20

1

; 1/2; 1/20

Define:

;

;

;

;

L iL it

R iR it

L iL it

R iR it

f f t dt

f f t dt

u u t dt

u u t dt

++=

++=

++=

++=

≡

≡

≡

≡

∫

∫

∫

∫x

ti i+1

; 1/2 ; 1/2;

L i L iu f

+ + ; 1/2 ; 1/2;

R i R iu f

+ +

( ) ( )( ) ( )

; 1/2

; 1/2

1/ 2, &

1/ 2, L i i

L i i

u t u x t

f t f x t+

+

≡ =

≡ =

( ) ( )( ) ( )

; 1/2 1

; 1/2 1

1/ 2, &

1/ 2, R i i

R i i

u t u x t

f t f x t+ +

+ +

≡ = −

≡ = −

RSLS

t=0

t=1

52

Above example showed how time-averaging can be done. For multidimensional problems, the averaging can be done in space & time.Can also be done for other Riemann solvers.

Question: How does one obtain SL and SR ?

x

ti i+1i-1

; 1/2bL iu + ; 1/2bR iu +

LS RS

( ) ( ); 1/2 ; 1/2 11/ 2, 1/ 2 and 1/ 2, 1/ 2are honest to goodness physical states at the space-time barycenters of the faces.They can be used to obtain and

bL i i bR i i

L R

u u x t u u x t

λ λ

+ + +≡ = = ≡ = − =

These ideas extend to multidimensions; they can also be used to yield the electric fields for MHD.

53

7.5.2.1) Multidimensional ADER-CG Predictor Step

( ) 2 2, , ; , ; , ; , ;, , ;

Within a zone we start with the third order spatial reconstruction:

U , ,

To endow time-dependence wi

1 1ˆ ˆ ˆ ˆ ˆU U U U U + U 1

th

1

t

2

i

2

n

i j i j x i j y i j xx i j yy i jj xyi x y x yt xy yx + + + − + −=

( ) 2 2, , ; , ; , ; , ; ,

2, ; ,

, ;

; ,

1 1ˆ ˆ ˆ ˆ ˆU U U

ˆ ˆ ˆ+ U + U

he zone, we want (Zero new terms to begin with):

U , ,

U U + U

12 12

+ Ui j t i j t

i j i j x i j y i j xx i j yy i j x

t

yi j

i j

x y x

t

x y t x

t

y y + + + − + −

=

; , ;

Question: Why do we add just these terms? (Hint:

ˆ +

Think Taylor serie )

U

s

xt i j ytx t y t

54

Recall that we needed the gradients of the x- and y- fluxes to obtain the predictor step. Same holds true here. So let us write the space-time moments of the x- and y-fluxes:

( )

( )

2, ; ,

2 2, , ; , ; , ; , ;

; , ; ,

,

,,

, ;

,

;

1 1ˆ ˆ ˆ ˆ ˆ ˆF F F F F + F 12

ˆ ˆ ˆ ˆ+ F + F + F +

12

ˆ ˆG G

F , ,

F

G , ,

i j t i j tt i j xt i

i j i j x i j y i j xx i j yyi j

i

i j x

j y

y

ij i

t

j

x y x y x yx y t

x y t

t t x t y t

+ + + − + −

=

+=

2, ;

2 2; , ; , ; ,

, ; , ; ,

;

;

; ,

Question: How would you obtain the spat

ˆ ˆ ˆ ˆ+ G + Gial mo

1 1ˆ ˆ ˆ ˆG

men

G G +G

+

G + G

12 1

ts

2

i j t

j x i j

i j tt

y i j x

i j x

x i j yy i y

i yt

j

j

x

tt t x

x y x y

y

x

t t

y + + − + −

( )

( ) ( ) ( )

,

, , ,

of the fluxes? What about temporal moments? How do we relate the moments of the fluxes to the moments of U , , ?

U , , F , , G , ,Governing equation: +

i j

i j i j i j

x y t

x y t x y t x y tt tt x x y

∂ ∂ ∂∆ ∆+

∂ ∆ ∂ ∆ ∂

( )

( ) ( ) ( ) ( )1/21 1/2, , ,

0 1/2 1/2

= 0

with a , , :

U , , F , , G , ,, , + = 0

yt xi j i j i j

t y x

yGalerkin Projection test function x y t

x y t x y t x y tt tx y t dx dy dtt x x y y

φ

φ== =

= =− =−

∂ ∂ ∂∆ ∆+ ∂ ∆ ∂ ∆ ∂

∫ ∫ ∫

55

( )

( )

, ; , ; , ; , ; , ; , ;

2

, ;

To understand how it goes, let us explicitly try out , , . Show that we get:4 2 2ˆ ˆˆ ˆ ˆ ˆU + U = F G F G3 3 3

Similarly, with , , :3ˆ ˆU + U2

i j t i j tt i j x i j y i j xt i j yt

i j t

x y t t

x y t t

φ =

− − − −

φ =

( )

( )

, ; , ; , ; , ; , ;

, ; , ; , ;

, ; , ;

3 3ˆ ˆˆ ˆ = F G F G4 4

ˆˆ ˆSimilarly, with , , : U = 2 F G

ˆ ˆSimilarly, with , , : U = F

i j tt i j x i j y i j xt i j yt

i j xt i j xx i j xy

i j yt i j xy

x y t x t

x y t y t

− − − −

φ = − −

φ = −

, ;

, ; , ; , ; , ;

ˆ 2 G

Recall that this is a finite element-like procedure. We can resort to iterative improvement of the

ˆ ˆ ˆ ˆmoments : U , U , U and U .Quest

i j yy

i j t i j tt i j xt i j yt

−

ion: How would such an iteration work? Answer: Break it into two parts & see.It turns out that, at third order, the process converges in two iterations!

56

57

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )

2 1, ; , ,

From the spatial nodal points (Only evaluated at the start of the iteration.):0,0,0 , 1/ 2,0,0 , 1/ 2,0,0 , 0,1/ 2,0 , 0, 1/ 2,0 ,

1/ 2,1/ 2,0 , 1/ 2,1/ 2,0 , 1/ 2, 1/ 2,0 , 1/ 2, 1/ 2,0We get:

F̂ 2 F 2 F F

{}

i j xx i j i j

once− −

− − − −

= − +( ) ( )

( )

3 4 1 5, , ; , , ,

6 7 8 9, ; , , , ,

2 3 4 5 1, ; , , , ; , , , , , ; , ;

ˆ ; F 2 F 2 F F ;

F̂ F F F F ;

ˆ ˆ ˆ ˆ ˆF F F ; F F F ; F F F F 12

Notice the fi

i j i j yy i j i j i j

i j xy i j i j i j i j

i j x i j i j i j y i j i j i j i j i j xx i j yy

= − +

= − − +

= − = − = + +

( ) ( ) ( ) ( ) ( ) ( )

( )11 12 2 3, ; , , , , ,

nite difference-like forms!

From the space-time nodal points (Evaluated for iteration):0,0,1/ 2 , 1/ 2,0,1/ 2 , 1/ 2,0,1/ 2 , 0,1/ 2,1/ 2 , 0, 1/ 2,1/ 2 , 0,0,1

We get:ˆ ˆF 2 F F F F ; F

{ }

i j xt i j i j i j i j i

each− −

= − − + ( )( )

13 14 4 5; , , , ,

15 10 1 15 1, ; , , , , ; , , , ;

2 F F F F ;

ˆ ˆ ˆF 2 F 2 F F ; F F F F

j yt i j i j i j i j

i j tt i j i j i j i j t i j i j i j tt

= − − +

= − + = − −

58

7.5.2.2) Multidimensional ADER-CG Corrector Step

The construction from the previous sub-section has given us the space and time representation of the conserved variables and the fluxes.

We wish to obtain the numerical flux in a quadrature-free fashion. In doing so, we wish to use all the space-time information that we have generated.

( ) ( ) ( )

( ) ( )( )

1/2, ; 1/2, ; 1/2,

; 1/2, ; 1/2,

S SF , F , F ,S S S S

S S U , U ,S S

The extremal speeds, S and S , are obtained from:

U

R Li j L i j R i j

R L R L

R LR i j L i j

R L

L R

y t y t y t

y t y t

+ + +

+ +

= − − −

+ − −

( ) ( ) ( ) ( ); 1/2, , ; 1/2, 1,U 1/ 2, 0, 1/ 2 ; U U 1/ 2, 0, 1/ 2As before, we freeze these speeds. Question: What does this buy us?

c cL i j i j R i j i jx y t x y t+ + += = = = = = − = =

59

60

( )1/21

; 1/2, , , ; , ;0 1/2

, ; , ; , ;

; 1/2,

1 1ˆ ˆU , = U U U2 6

1 1 1ˆ ˆ ˆ U U U2 2 3

Do analogously for F

yt

L i j i j i j x i j xxt y

i j t i j xt i j tt

L i j

y t dy dt

y

==

+= =−

+

+ +

+ + +

∫ ∫

( )

( )1/21

; 1/2, 1, 1, ; 1, ;0 1/2

1, ; 1, ; 1, ;

, .

1 1ˆ ˆU , = U U U2 6

1 1 1ˆ ˆ ˆ U U U2 2 3

Do analogously

yt

R i j i j i j x i j xxt y

i j t i j xt i j tt

t

y t dy dt==

+ + + += =−

+ + +

− +

+ − +

∫ ∫

( ); 1/2, for F , .R i j y t+

61

7.8) Numerical Tests for Euler Flow7.8.1) Demonstrating Accuracy on Multidimensional Test ProblemsWe study scheme performance for vortex flow with increasing resolution.

Questions: On a coarse mesh, with just a few zones across, which scheme performs better? As the mesh is refined, which scheme performs better? How do you measure order of accuracy?

62

5.8.2) Forward Facing Step Test Problem

63

7.8.3) Double Mach Reflection Problem