Numerical Methods for Conservation Laws , WPI Spring 2006

Numerical Methods for ConservationLaws

WPI, January 2006

C. Ringhofer

[email protected],C2b2

2 h2∂xu http://math.la.asu.edu/∼chris

Last update: Jan 24, 2006

1

LITERATURE

1. Numerical Methods for Conservation Laws, by Randall J.

LeVeque, Birkhauser-Verlag, Basel, 1990. ISBN 3-7643-

2464-3

2. J. Strickwerda: Finite difference schemes and partial dif-

ferential equations, Wadsworth and Brooks/Cole, 1992.

2

OUTLINE

6 lectures

Week 1: Mo 11:00-12:30, Tue 12:30-14:00 Wed 11:00-12:30

Week 2: Mo 11:00-12:30, Tue 12:30-14:00 Wed 12:30-14:00

Materials: http://math.la.asu.edu/ chris → Courses

Topics:

L1 Conservation laws and conservative formulation, Differ-

ence methods, Linear waves, Diffusion, Dispersion, CFL

3

conditions. Examples: Wave equations, diffusion equa-tions, Schroedinger equations.

L2 Nonlinear equations, Characteristics, Shocks, Weak so-lutions, The Rankine - Hugoniot condition. Examples:Traffic Flow.

L3 Godunov methods, Local Riemann problems, Shock cap-turing, Systems, local linearization and Roe matrices. Ex-amples: Traffic flow, Fluids, Compressible Euler.

L4 Higher order methods, TVD schemes.

L5 Convergence analysis, Stability and compactness.

L6 Topics: Random methods, Entropy, Boundary conditions.Examples: Boltzmann equations.

LECTURE 1

Conservation Laws

UΩ(t): amount of ’mass’ in Ω at time t

F : flux through the boundary ∂Ω

UΩ(t+∆t) = UΩ(t)−∆t∫∂Ω

F ·n dσ = UΩ(t)−∆t∫Ω∇x ·F dx

Size of |Ω| and ∆t → 0 :

∂tu+∇x · F = 0, UΩ(t) =∫Ωu(x, t) dx

Examples: F = b(u)u (transport), F = −a(u)∇xu (diffusion)

Wave speeds

Linear problems and plane wave solutions

F (u) = bu− a∇xu, u(x, t) = exp[iξ(x− vt)]

4

v : velocity, vξ : frequency

u(x, t) = exp[iξ(x− bt)− a|ξ|2t]

wave speed = b (property of the medium). amplitude dampedwith exp[−a|ξ|2t].

Dispersion:

∂tu+ i∆xu = 0, v = v(ξ) = ξ

wave speed dependent on the frequency!

hyperbolicity ⇐⇒ v only medium dependentdispersivity ⇐⇒ v frequency dependentparabolicity ⇐⇒ exponential damping

Systems: ∂tu+ ∂xj[Aju]

Difference methods always exhibit all three features,

regardless of the type of equation!

Difference methods for advection diffusion equations

A linear example:

∂tu+ ∂x[−a∂xu+ bu] = 0, u(x,0) = uI(x)

Discretization: Grid: xj = jh, j ∈ Z, tn = nk, n ∈ N, ap-

proximate u(xj, tn) ≈ U(xj, tn)

Difference approximation:

U(xj, tn+1) = U(xj, tn)+

k

h2[(a+

h

2b)U(xj−1, tn)− 2aU(xj, tn) + (a−

h

2b)U(xj+1, tn)]

U(xj, t0) = uI(xj)

advance in time: U(∗, tn) → U(∗, tn+1), tn+1 − tn = k

Notation: The translation operator:

Tu(x) = u(x+ h), T−1u(x) = u(x− h)

U(x, t+ k) = 1 +k

h2[(a+

h

2b)T−1 − 2a+ (a−

h

2b)T ]U(x, t)

Linear wave ansatz: U(x, t) = exp[iξ(x− vt)]

exp[−iξvk] = 1− a4k

h2sin2(

ξh

2)− ib

k

hsin(ξh)

e−iξv = qe−iξvr, vr ∈ R

U(x, t) = qt exp[iξ(x− vrt)]

q2k = [1−a4k

h2sin2(

ξh

2)]2+[b

k

hsin(ξh)]2, tan(ξkvr) =

bkh sin(ξh)

1− a4kh2 sin2(ξh2 )

Wave speeds always dependent on frequencies.

For a = 0 waves are always amplified (instability!)

Difference discretizations are always dispersive.

Explicit difference discretizations always need some amountof diffusion to be stable (artificial diffusion).

a = kα, c = kh, z = sin(ξh2 )

q2k = [1− 4αc2z1]2 + b2c2z2(1− z2) ≤ 1, ∀z ∈ [0,1]

⇒ b2c2 ≤ 2akc2 < 1

CFL condition: k|b|h < 1

Artificial diffusion: a = O(k)

Also true for nonlinear problems and problems with non

- constant coefficients!

LECTURE 2

CHARACTERISTICS AND SHOCKS

Characteristics for linear and scalar problems:

∂tu+ a(x, t)∂xu = 0, u(x,0) = uI(x)

Characteristic:

Try to find a solution of the form u(ξ(t), t) = const

d

dtξys(t) = a(ξys, t), ξys(s) = y,

⇒d

dtu(ξys(t), t) = 0, u(ξys(t), t) = u(y, s)

forward: u(ξy0(t), t) = uI(y),

backward: The solution u(x, t) is given by

u(y, s) = u(ξys(t), t) = uI(ξys(0)),

5

Conservative formulation:

∂tu+ ∂x[a(x, t)u] = 0, u(x,0) = uI(x)

Characteristic:

d

dtξys(t) = a(ξys, t), ξys(s) = y,

⇒d

dtu(ξys(t), t) = −u(ξys(t), t)∂xa(ξys(t), t),

d

dtξys(t) = a(ξys, t),

d

dtvys(t) = −vys∂xa(ξys, t),

vys(t) = u(ξys(t), t), ξys(s) = y, vys(s) = u(y, s),

forward:d

dtξy0 = a(ξy0, t),

d

dtvy0 = −vy0∂xa(ξy0, t),

ξy0(0) = y, vy0(0) = uI(y),

Characteristics can never intersect for linear problems!

Find the solution: Given (y, s)

backward solve:

ξ′ys(t) = a(ξys, t), ξys(s) = y → ξys(0) := ξ0ys

forward solve:

v′ys(t) = −∂xa(ξys, t)vys, vys(0) = uI(ξ0ys) → vys(s) = u(ξys(s), s) = u(y, s)

Example: ∂tu+ ∂x[xu] = 0

Nonlinear problems:

∂tu+ ∂xf(x, t, u) = 0, u(x,0) = uI(x)

d

dtξys = ∂uf(ξys, t, vys),

d

dtvys = −∂xf(ξys, t, vys),

For nonlinear problems characteristics can intersect. The

solution will develop a discontinuity (shock)!

Example: Burger’s equation ∂tu+∂xu2

2 = 0, u(x,0) = uI(x)

THE CONCEPT OF WEAK SOLUTIONS

∂tu+ ∂xf(u) = 0, u(x,0) = uI(x) (1)

u discontinuous ⇒ need to re- define derivatives

ψ sufficiently smooth test function, ψ ∈ C∞0

Definition: u is a weak solution of (1) iff∫ ∞0

dt∫dx [u∂tψ+ f(u)∂xψ] = −

∫dx [ψ(x,0)uI(x)] ∀ψ

Remark: u differentiable ⇒ u classical solution.

Example:

∂tu+ ∂x[H(x)u+1

2H(−x)u] = 0, u(x,0) = uI(x)

THE RANKINE - HUGONIOT CONDITION

Relates shockheight to shockspeed if there is a shock.

∂tu+ ∂xf(u) = 0,

Assume:

1. u differentiable away from the shock curve x = γ(t).

2. u is a weak solution.

Implies:

γ′(t)(u+ − u−)(γ(t), t) = (f+ − f−)(γ(t), t)

Example: Traffic flow

∂tu+ ∂x[c(1−u

u0)u] = 0, u(x,0) = uI(x) =

u0 for x > 012u0 for x < 0

LECTURE 3

GODUNOV’S METHOD

Scalar problems:

∂tu+ ∂xf(u) = 0

Equation for averages:

Unj =1

h

∫ xj+1

xju(x, tn)dx , F

nj =

1

k

∫ tn+1

tnf(u(xj, t))dt .

⇒ Un+1j − Unj +

k

h(Fnj+1 − Fnj ) = 0

Up to here exact!Numerical approximation: compute Fnj from Unj−1, U

nj

Piecewise constant approximation: Assume u(x, tn) = Unj forxj < x < xj+1.Solve problem with piecewise constant initial data (the Rie-mann problem) exactly.

6

THE RIEMANN PROBLEM

∂tu+ ∂xf(u) = 0, u(x,0) =

(UL for x < 0UR for x > 0

)

Compute F = 1k

∫ k0 f(u(0, t))dt .

Characteristics: x′ = f ′(u), u′ = 0

Shock curve according to Rankine - Hugoniot:

γ′(UR − UL) = f(UR)− f(UL)

Case 1: f ′(UR) > 0, f ′(UL) > 0 ⇒ F = f(UL)

Case 2: f ′(UR) < 0, f ′(UL) < 0 ⇒ F = f(UR)

Case 3: f ′(UR) < 0 < f ′(UL) and γ′ > 0 ⇒ F = f(UL)

Case 4: f ′(UR) < 0 < f ′(UL) and γ′ < 0 ⇒ F = f(UR)

Case 5: f ′(UL) < 0 < f ′(UR): rarefaction wave and the

Barenblatt solution.

THE BARENBLATT SOLUTION7

∂tu+ ∂xf(u) = 0, set u(x, t) = g(x

t) ⇒ f ′(g(z)) = z

u(x, t) = (f ′)−1(xt ) gives continuous solution in Case 5!

Case 5: f ′(UL) < 0 < f ′(UR) ⇒ F = f(us) with f ′(us) = 0.

us is called the sonic point and u(x, t) = (f ′)−1(xt ) is called a

rarefaction wave.

Godunov:

γ′ :=f(Unj )− f(Unj−1)

Unj − Unj−1

Case 1: f ′(Unj ) > 0, f ′(Unj−1) > 0 ⇒ Fnj = f(Unj−1)

Case 2: f ′(Unj ) < 0, f ′(Unj−1) < 0 ⇒ Fnj = f(Unj )

Case 3: f ′(Unj ) < 0 < f ′(Unj−1) and γ′ > 0 ⇒ Fnj = f(Unj−1)

Case 4: f ′(Unj ) < 0 < f ′(Unj−1) and γ′ < 0 ⇒ Fnj = f(Unj )

Case 5: f ′(Unj−1) < 0 < f ′(Unj ) ⇒ Fnj = f(us) with f ′(us) =

0.

SYSTEMS AND APPROXIMATE RIEMANN SOLVERS

∂tu+ ∂xf = 0, u ∈ RN

Un+1j − Unj +

k

h(Fnj+1 − Fnj ) = 0, Fnj = F (Unj−1, U

nj , v0),

8

Flux Approximation

Solution with approximate flux function fUL,UR

∂tv+ ∂xfUL,UR(v) = 0, v0 := v(xj, t)

Requirements on F and fUL,UR:

Conservation:

F (UL, UR, UL) = f(UL), F (UL, UR, UR) = f(UR),

Shock speeds:

fUL,UR(UR)− fUL,UR(UL) = f(UR)− f(UL) (2)

Set:

F (UL, UR, v) = fUL,UR(v) + f(UL)− fUL,UR(UL) (3)

= fUL,UR(v) + f(UR)− fUL,UR(UR)

Step 1: Find fUL,UR satisfying (2)

Step2: Solve approximate conservation law for v0.

Step3: Compute F according to (3).

Approximate Riemann solvers:

∂tv+∂xfUL,UR(v) = 0, v(x,0) = UL, x < 0, v(x,0) = UR, x > 0

Linear approximation: fUL,UR(v) = A(UL, UR)v

A : Roe matrix

Shock speeds:

A(UL, UR)(UR − UL) = f(UR)− f(UL)

diagonalize A and solve N scalar linear problems.

SOLVING THE CONSTANT COEFFICIENT RIEMANN

PROBLEM FOR SYSTEMS

A(UL, UR)rm = λmrm, Df(UL)rLm = λLmr

Lm, Df(UR)rRm = λRmr

Rm

expand UL, UR in eigenvectors

UL =∑mwLmrm, UR =

∑mwRmrm,

solve

∂twm + λm∂xwm = 0, wm(x,0) =

(wLm for x < 0wRm for x > 0

)

wm(0, t) =

(wLm for λm > 0wRm for λm < 0

), v(0, t) =

∑mwm(0, t)rm

Problem:

Only shock solutions, no rarefaction waves.

except: λLm < 0 < λRm (rarefaction wave)

replace by two shocks: wm(0, t) = ws

conservation:

wLm(kλLm+C)+kwsm(λRm−λLm)+wRm(C−kλRm) = wLm(kλm+C)+wRm(C−kλm)

wsm =wLm(λm − λLm) + wRm(λRm − λm)

λRm − λLm

This is sometimes called the ’sonic fix’.

SUMMARY

Roe matrix:

Anj = A(Unj−1, Unj ), Anj (U

nj − Un−1

j ) = f(Unj )− f(Unj−1)

9

Fnj = f(Unj−1) +Anj (v0 − Unj−1), v0 =

∑mw0mrm

w0m =

(wLm for λm > 0wRm for λm < 0

), Unj−1 =

∑mwLmrm, Unj =

∑mwRmrm

except for the sonic fix:

Df(Unj )rnjm = λnjmrnjm, m = 1, .., N

If λnj−1,m < 0 < λnj,m then

w0m =

wLm(λm − λnj−1,m) + wRm(λnj,m − λm)

λnj,m − λnj−1,m

ONE WAY TO FIND A ROE MATRIX

A(UL, UR) =∫ 1

0Df(UL + s(UR − UL))ds

Example 1: Burger’s equation, f(u) = 12u

2

Example 2: Isothermal flow

f(ρ, φ) =

φφ2

ρ + ρ

,

A(UL, UR) =

(0 1

1− v2 2v

), v =

√φ2LρL

+

√φ2RρR√

ρL +√ρR, vLR =

φLRρLR

LECTURE 4

ODE METHODS AND THEIR PDE COUNTERPARTS

ODE: dudt = f(u, t)

Euler’s method (explicit): U(t+ k) = U(t) + kf(U, t)

Higher order methods I: Runge Kutta

Idea: Write in integral form

u(t+ k) = u(t) + k∫ 1

0f(u, t+ sk) ds

Use integration rule for the integral

Euler:∫ 10 f(u, t+ sk) ds ≈ f(U, t)

Second order:∫ 10 f(u, t+ sk) ds ≈ f(U1/2, t+ k/2); compute

10

u1/2 by intermediate Euler step

U1/2 = U(t) +k

2f(U, t), U(t+ k) = U(t) + kf(U1/2, t+

k

2)

Equivalent for hyperbolic PDE’s: Lax-Wendroff in staggered

grid formulation

∂tu+∂xf(u) = 0, U(t+k) = U(t)−k

h(T1/2−T−1/2)f(U(t+

k

2))

U(t+k

2) =

1

2(T1/2 + T−1/2)U(t)−

k

2h(T1/2 − T−1/2)]f(U(t))

Higher order methods have less aritficial diffusion!

∂tu+b∂xu = 0: Godunov: C|b|2 h2∂xu Lax - Wendroff: C2b2

2 h2∂xu,

C = kh

Higher order methods II: Multi-step

du

dt= f(u, t), U(t+ k) =

S∑s=0

αsU(t− sk) + kβsf(U, t− sk)

Problem 1: needs start up!

Stability:

U(nk) = gnU(0), gn+1 =S∑s=0

αsgn−s + k...

g solution of polynomial of degree S + 1, S + 1 roots!

Example: Leapfrog for ∂tu+ ∂xf(u)

U(t+ k) = U(t− k)−bk

h(T − T−1)f(U(t)),

growth function g for ∂tf + b∂xf = 0:

g2 = 1 + 2igbC sin(ξh), C =k

h

For |b|C ≤ 1: |g12| = 1, no artificial diffusion!!! (weaklyunstable)

MONOTONICITY AND THE MAXIMUM PRINCIPLE

Difference scheme for general nonlinear problem

Uj(t+ k) = gj(U1(t), .., UN(t))

Consistency ⇒gj(u, ...., u) = u, ∀j (or gj(u, ...., u) = u(1 + O(k)), ∀j in the

presence of source terms).

Lemma:

∂gj

∂uk≥ 0 ∀j, k ⇒ ‖U(t+ k)‖∞ ≤ ‖U(t)‖∞,

Implicit schemes:

g+j (U(t+ k)) = g−j (U(t))

g+j (.., Us(t), ..) = g+j (.., Us(t), ..), ∀j

Define:

g+j (v) = G+j (.., v, v, ..), g−j (v) = G−j (.., v, v, ..)

Assume:

(A1)∂G+

j

∂Us≤ 0 for j 6= s,

∂G−j∂Us

≥ 0 ∀j, s

(A2) g+j (v) ≥ g−j (w) ⇒ v ≥ w ∀j

Lemma: Assumptions (A1),(A2) imply ‖U(t+k)‖∞ ≤ ‖U(t)‖∞

Monotone schemes build (nonlinear) averages over the solu-

tion with nonnegative coefficients!

Example: Lax - Wendroff

Methods of order higher than order 1 cannot be monotone!

THE TVD PROPERTY

The discrete W∞1 norm:

∫|∂xu| dx

Non-oscillatory schemes:

Definition: A scheme is TVD ⇐⇒∑j

|(T − 1)Un+1j | ≤

∑j

|(T − 1)Unj |

The linear case:

Write scheme solely in terms of derivatives.

Un+1 = Un − T−1(AnV n) +BnV n, V n := (T − 1)Un

Theorem:

A,B ≥ 0, A+B ≤ 1 ⇒ TV D

LECTURE 5

FLUX LIMITER METHODS

Un+1 = Un − c(T − 1)Fn

Idea:

Lower order method with numerical flux FL. Higher order

method with numerical flux FH. Lower order method non-

oscillatory.

Combine:

F = FL + Φ(FH − FL)

Smooth part: φ ≈ 1; Non-oscillatory otherwise, i.e. total

method is TVD.

11

Total method formally only first order but higher order in

smooth regions.

THE FLUX LIMITER

Condition 1: (Smooth order)

Φ = φ((1− T−1)U

(T − 1)U), φ(1) = 1

Condition 2: Choose φ such that total method is TVD.

Derive φ for linear case and use in general.

∂tu+ a∂xu = 0, a > 0

Example: Upwind and Lax Wendroff

FL = aT−1U, FH =a

2(1 + T−1)U −

ca2

2(1− T−1)U

FH − FL =a

2(1− ac)(1− T−1)U

Un+1 = 1− ν(1− T−1)[1 +φ

2(1− ν)(T − 1)]Un, ν := ac

Un+1 = Un−ν[1−(1− ν)

2T−1φ]T−1V n−ν

φ

2(1−ν)V n, V = (T−1)U

A = ν[1−(1− ν)

2φ] + ν

Tφ

2(1− ν)

TV

V

A = ν1 +(1− ν)

2(−φ+ Tφ

TV

V)

12

A ≥ 0 ⇒ |φ(θj)

θj− φ(θj−1)| ≤ 2, θj =

T−1V

V

Two standard choices:

1. Superbee:

φ(θ) = max0,min1,2θ,minθ,2

2. Van Leer:

φ(θ) =θ+ |θ|1 + |θ|

13

LECTURE 6

BOUNDARY CONDITIONS AND GHOSTPOINTS FOR

HYPERBOLIC SYSTEMS

∂tu+ ∂x[A(x, t)u] = 0, x ≥ 0, Bu(0, t) = ub,

u ∈ RN , B ∈ RK×N

Relation of B to A: Influx given in terms of outflux !

Diagonalization:

A(0, t) = EDE−1, D =

(D+ 00 D−

), D+ > 0, D− ≤ 0

Important!! dim(D) = K ×K

Projections: partition E

E = (E+, E−), E+ : N ×K, E−1 =

(R+R−

), R+ : K ×N

14

Necessary: ∃(BE+)−1

Ghostpoint:

R−U(−h, t) = R−(2U(0, t)− U(h, t))

R+U(−h, t) = (BE+)−1[ub −BE−R−(2U(0, t)− U(h, t))]

U(−h, t) = E

((BE+)−1

0

)ub+E

(−(BE+)−1BE−R−

R−

)(2U(0, t)−U(h, t))