numerical methods for hyperbolic conservation laws lecture...

Numerical Methods for Hyperbolic Conservation LawsLecture 1

Wen Shen

Department of Mathematics, Penn State UniversityEmail: [email protected]

Oxford, Spring, 2018

Wen Shen (Penn State) Numerical Methods for Hyperbolic Conservation Laws Lecture 1Oxford, Spring, 2018 1 / 41

Course outline; tentative plan

1 Linear Equations

Basic theory, method of characteristics,Weak solutions, Riemann problemsFinite volume method: conservative, BV stability.High resolution methods: shock capturing, flux limiters, slope limitersConvergence, accuracy and StabilityVariable coefficient linear equations

2 Nonlinear equations

Examples: Traffic flow, Euler equation, shallow water equationsWeak solutions, shock formation, vanishing viscosity, entropy conditionsConservative methods: Godunov’s, Lax-Friedrich, E-schemes etc.Nonlinear systems, approximate Riemann solvers,Finite volume methods for nonlinear systems.

3 Additional topics:(if time permits)

Multidimensional problems;Wave front tracking; Center difference scheme;Conservation laws with discontinuous flux and applications.


Texts:

1 R.J. LeVeque, Numerical Methods for Conservation Laws. Birkauser 1992.

2 R.J. LeVeque, Finite Volume Methods for Hyperbolic Problems, CambridgeUniversity Press, 2002.

3 K.W. Morton and D. F. Mayers, Numerical Solution of Partial DifferentialEquations.

4 H. Holden, and N.H. Risebro, Front Tracking for Hyperbolic ConservationLaws, Springer Verlag, New York 2002.

5 C.-W. Shu, High order ENO and WENO schemes for computational fluiddynamics, Lecture Notes, Springer, (1999), 439-582.

6 E. Tadmor, Approximate solutions of nonlinear conservation laws. Lecturenotes in Mathematics 1697, C.I.M.E. course in Cetraro, Italy. Springer 1998.

7 Class handout of research papers.

Books [2] will be heavily used. Parts of [1] and [3] will be used.


Derivation of a conservation law, fluid/gas dynamics

Consider fluid/gas flowing through a 1D pipe.u(x , t): velocityq(x , t): tracer density = mass per unit length in 1D

Given any points x1 < x2:

Q=

∫ x2

x1

q(x , t) dx = total mass on the section [x1, x2]

Conservation of mass: Q(t) changes only due to the flow (flux) at x1 and x2.

Fi (t) =: flux at xi . Rate of mass passing xi per unit time.• Fi > 0: flow to the right• Fi < 0: flow to the left

d

dt

∫ x2

x1

q(x , t) dx = F1(t)− F2(t) : integral form of a conservation law


Here, flux at (x , t) = u(x , t) · q(x , t).• Assume u(x , t) is known: flux f (q, x , t) = u(x , t)q. variable coefficient• Assume u(x , t) ≡ u constant: flux f (q) = u · q. linear flux• If u depends on q such that u = u(q): flux f (q) = u(q) · q. nonlinear flux,autonomous, with the most important applications.

Put together: for any x1 < x2

d

dt

∫ x2

x1

q(x , t) dx = f (q(x1, t))− f (q(x2, t)) = −∫ x2

x1

∂

∂xf (q(x , t)) dx

Assume q(x , t) differentiable:∫ x2

x1

[∂

∂tq(x , t) +

∂

∂xf (q(x , t))

]dx , for all x1, x2

Differential form of the conservation law

∂

∂tq(x , t) +

∂

∂xf (q(x , t)) = 0. i.e., qt + f (q)x = 0


Advection equation: characteristics

Cauchy problem:qt + uqx = 0, q(x , 0) = q(x)

Explicit solutionq(x , t) = q(x − ut)

Observation:• q(x , t) is constant along rays in space-time for which x − ut = constant• Along the ray X (t) = x0 + ut, we have q(X (t), t) = q(x0).• Values of q advect with constant velocity u.Such rays X (t) are called characteristics.

d

dtq(X (t), t) = qt(X (t), t) + qx(X (t), t) · X ′(t) = qt + uqx = 0.


Boundary conditions:• If u > 0, BC is needed at x = a, say q(a, t) = g0(t).

Solution:

q(x , t) =

{g0(t − (x − a)/u), a < x < a + u(t − t0)

q(x − u(t − t0)), a + u(t − t0) < x < b

• If u < 0, BC is needed at x = b. Similar treatment.


Variable coefficients

Assume u = u(x).qt + (u(x)q)x = 0

Characteristic curves X (t): X ′(t) = u(X (t))→ tracks the motion of particles carried along by the fluid

Along X (t):

d

dtq(X (t), t) = qt + X ′(t)qx = qt + u(X (t)) · qx

= qt + (u(X (t)) · q)x − u′(X (t))q = − u′(X (t))q

→ Solution q is NOT constant along characteristics!

NB! The PDE can be solved by solving two ODEs.


Non-conservative advection equation:

qt + u(x)qx = 0

Characteristics: X ′(t) = u(X (t)) , and q is constant along X (t).


Diffusion

At a molecular level, particles tend to spread out, in −∇q direction.

Fick’s law of diffusion in 1D: net flux = −βqx

If u = 0, and β =constant, then

qt = βqxx , (heat equation)

If β = β(x), thenqt = (β(x)qx)x

If it’s in a flow: flux = uq − βqx , and

qt + uqx = βqxx , (advection-diffusion equation)


Example in gas dynamics

ρ: gas density. u: velocity

Conservation of mass:

ρt + (ρu)x = 0 (continuity equation)

Conservation of momentum:ρ(x , t)u(x , t) : density of momentumMeaning:

∫ x2

x1ρ(x , t)u(x , t) dx = total momentum on [x1, x2] at t

Momentum flux at (x , t):• advective flux: = (ρu)u = ρu2. (i.e., the momentum carried past (x , t)) –macroscopic• micro-scopic: caused by pressure p

Add up, momentum flux = ρu2 + p


Need a third equation for p: equation of state

.Example: isentropic gas where entropy of gas is constant:

p = κργ=P(ρ), (γ ≈ 1.4 for air)

2× 2 system: {ρt + (ρu)x = 0

(ρu)t + (ρu2 + P(ρ))x = 0

orqt + f (q)x = 0

where

q =

(ρρu

)=

(q1

q2

), f (q) =

(ρu

ρu2 + P(ρ)

)=

(q2

(q2)2/q1 + P(q1)

)


Quasilinear form

Rewrite qt + f (q)x = 0 asqt + f ′(q)qx = 0

where f ′(q): Jacobian matrix with Ji,j = ∂fi∂qj

.

If f ′(q) = A = constant matrix,

qt + Aqx = 0, (linear system )

Let q ∈ Rm,A ∈ Rm×m.

Definition of hyperbolicity:

The system qt + Aqx = 0 is called hyperbolic if A is diagonalizable with realeigenvalues.


This means: A has a set of eigenvalues λ1 ≤ λ2 ≤ · · · ≤ λm,and a complete set of linearly independent right eigenvectors r1, r2, · · · , rmsuch that

A = RΛR−1, R = [r1, r2, · · · , rm], Λ = diag{λ1, λ2, · · · , λm}

One has:

R−1qt + R−1Aqx = 0, [R−1q]t + [R−1AR][R−1q]x = 0

wt + Λwx = 0 where w = R−1q

A decoupled system:

(wp)t + λp(wp)x = 0, p = 1, 2, · · · ,m


Special cases of hyperbolic systems:

• A is symmetric: A = AT . A is diagonalizable with real eigenvalues.Symmetric hyperbolic

• A has distinct real eigenvalues λ1 < λ2 < · · · < λm.→ all rp are linearly independent.Strictly hyperbolic

• A has real eigenvalues but not diagonalizable.Weakly hyperbolic


Connection to second order wave equation

ptt = c20pxx (2nd order hyperbolic PDE)

Equivalence to first order equation: let

q1 = pt , q2 = −px .

Then(q1)t + c2

0 (q2)x = 0, (q2)t + (q1)x = 0

qt + Aqx = 0, A =

(0 c2

0

1 0

)

λ1 = −c0, λ2 = c0, r1 =

(c0

−1

), r2 =

(c0

1

)⇒ strictly hyperbolic


Quasi-linear systems

qt + A(q, x , t)qx = 0 (I )

Definition. (I) is hyperbolic at (q, x , t) if A(q, x , t) satisfies the hyperbolicitycondition at this point.


Cauchy Problem

qt + Aqx = 0, A = RΛR−1, w = R−1q, wt + Λwx = 0

Initial condition given q(x , 0) = q(x).

Let w(x) = R−1q(x).pth equation:

(wp)t + λp(wp)x = 0,

wp(x , t) = wp(x − λpt, 0) = wp(x − λpt)

q(x , t) = R · w(x , t)


Superposition of waves

Then

q(x , t) = R · w(x , t) = [r1, r2, · · · , rm][w1,w2, · · · ,wp]T =m∑

p=1

wp(x , t)rp

→ superposition of waves of different familiesdecompose q in the space spanned by {rp}wp(x , t)=strength of pth wave. (eigen-coefficient)Advected with speed λp, along line X (t) = x0 + λpt

Definition. X (t) with X ′(t) = λp is called the characteristic of the pth family, orp-characteristics.


Left eigenvalues

LetL = R−1

`1, `2, · · · , `m: rows of L`pA = λp`p

Thenw = R−1q = Lq, wp = `pq

Solution

q(x , t) =m∑

p=1

[`p q(x − λpt)] rp

Solution of linear equations: Superposition of m waves, each wave travelsindependently with no change in shape.


Simple waves

pth wave:shape: wp(x)rpspeed: λp

Given an index i . If the initial data is given such that

wp(x) ≡ wp(constant) for p 6= i

thenq(x , t) = wi (x − λi t)ri +

∑p 6=i

wprp = q(x − λi t)

→ Initial data propagates with speed λi .

This is called a simple wave.


Domain of dependence and range of influence

Take m = 3 for example, and assume λ1 < 0 < λ2 < λ3.

NB! Finite propagation speed given bounded domain of dependence and range ofinfluence. (leads to stability conditions in numerical methods)

Different from diffusion equation.


Riemann Problems for linear hyperbolic systems

• If q(x) is smooth, so is the solution q(x , t)• If q(x) has a jump, we allow it to propagate along various characteristics

Riemann initial data:

q(x) =

{ql , x < 0

qr , x > 0

For m = 1, scalar equation qt + uqx = 0, we have

q(x , t) = q(x − ut).

For m ≥ 2: decompose ql , qr as

ql =m∑

p=1

wl,prp, qr =m∑

p=1

wr ,prp


pth advection equation

(qp)t + λp(qp)x = 0, I.C. wp(x) =

{wl,p, x < 0

wr ,p, x > 0

which gives

wp(x , t) = wp(x − λpt) =

{wl,p, x − λpt < 0

wr ,p, x − λpt > 0

The jumps are traveling with 3 velocities.


Across the pth characteristics, the jump in q is

(wr ,p − wl,p)rp=αprp

NB! The jump is an eigenvector of A.Idea can be generalized to nonlinear systems.

Summarize:• Decompose the initial jump qr − ql into eigenvectors of A

qr − ql = α1r1 + · · ·+ αmrm, where αp = `p(qr − ql)

• Jump in the pth wave is Wp = αprp.• Solution

q(x , t) = ql +∑

p:λp<x/t

Wp = qr −∑

p:λp>x/t

Wp = ql +∑p

H(x − λpt)Wp

H(x): Heaviside function


Phase plane for m = 2, q = (q1, q2), and λ1 < λ2

Observations:• A discontinuity with ql , qr as left and right states can propagate as a singlejump (simple wave) only if qr − ql is an eigenvector of A.

• Self-similar, i.e., solution is constant along any rays x/t = ξ.

• ql is connected in r1 direction, while qr is connected in r2 direction


How to locate qm in the phase plane:

For m > 2: similar ideas, but hard to draw.


Finite Volume Methods

qt + f (q)x = 0, f (q) = Aq

Uniform mesh: (∆x ,∆t) :

xi− 12

= ∆x(i − 1

2), tn = n∆t

Grid cell: Ci = (xi− 12, xi+ 1

2)

Cell average:

Qni ≈

1

∆x

∫ xi+ 1

2

xi− 1

2

q(x , tn)dx =1

∆x

∫Ci

q(x , tn)dx

Apply conservation law on Ci :

d

dt

∫Ci

q(x , t)dx = f (q(xi− 12, t))− f (q(xi+ 1

2, t))


Integrate over [tn, tn+1] in t:∫Ci

q(x , tn+1)dx −∫Ci

q(x , tn)dx =

∫ tn+1

tn

f (q(xi− 12, t))dt −

∫ tn+1

tn

f (q(xi+ 12, t))dt

Divided by ∆x :

Qn+1i = Qn

i −1

∆x

[∫ tn+1

tn

f (q(xi+ 12, t))dt −

∫ tn+1

tn

f (q(xi− 12, t))dt

]= Qn

i −1

∆x

[∆t · Fi+ 1

2−∆t · Fi− 1

2

]where

Fi+ 12=

1

∆t

∫ tn+1

tn

f (q(xi+ 12, t)) dt (average flux)

=⇒ Qn+1i = Qn

i −∆t

∆x

[Fi+ 1

2− Fi− 1

2

]Fi+ 1

2: numerical flux


Numerical flux: It is natural to assume that Fi− 12

= F(Qni−1,Q

ni )

⇒ Qn+1i = Qn

i −∆t

∆x

[F(Qn

i ,Qni+1)−F(Qn

i−1,Qni )]

⇒ explicit, conservative, 3-point computation stencil

(i ,n)(i−1,n) (i+1,n)

(i ,n+1)


Conservative method

Summing ∆xQn+1i over any set of consective cells

∆xJ∑

i=I

Qn+1i = ∆x

J∑i=I

Qni −∆t

[FJ+ 1

2− FI− 1

2

](intermediate terms all cancel out!)

⇒: Over the full domain, we have exact conservation except for fluxes at theboundaries!

View the method as FDM:

Qn+1i − Qn

i

∆t+

Fi+ 12− Fi− 1

2

∆x= 0


The diffusion equation

qt = (β(x)qx)x

flux: f (qx , x) = −β(x)qx

numerical flux:

F(Qi−1,Qi ) = −βi− 12· Qi − Qi−1

∆x, βi− 1

2≈ β(xi− 1

2)

⇒ FDM:

Qn+1i = Qn

i +∆t

∆x2

[βi+ 1

2(Qn

i+1 − Qni )− βi− 1

2(Qn

i − Qni−1)

]If β ≡constant:

Qn+1i = Qn

i +∆t

∆x2β[Qn

i+1 − 2Qni + Qn

i−1

]


Crank-Nicolson method

Numerical flux:

F ni− 1

2= F(Qi−1,Qi ) = − 1

∆xβi− 1

2· 1

2

[(Qn

i − Qni−1) + ((Qn+1

i − Qn+1i−1 )

]This gives

Qn+1i = Qn

i +∆t

2∆x2

[βi+ 1

2(Qn

i+1 − Qni )− βi− 1

2(Qn

i − Qni−1)

+βi+ 12(Qn+1

i+1 − Qn+1i )− βi− 1

2(Qn+1

i − Qn+1i−1 )

]


Conditions on numerical flux

Consistency:

If Qni−1 = Qn

i = q, then F = f (q), i.e., F(q, q) = f (q)

Continuity:

There exists a constant L so that

|F(Qi−1,Qi )− f (q)| ≤ L ·max{|Qi − q|, |Qi−1 − q|}

Consistency and Continuity are necessary for convergence.


Numerical domain of dependence, stability condition

Consider explicit 3-point stencil FDM:

Domain of dependence determined by ∆t/∆x .

A necessary stability condition:

A numerical method can be converged only if its numerical domain of dependencecontains the true domain of dependence of the PDE.


Example: qt + uqx = 0 with a 3-point stencil method.∣∣∣∣∆t

∆x

∣∣∣∣ ≤ ∣∣∣∣1u∣∣∣∣ , ν=

∣∣∣∣ u∆t

∆x

∣∣∣∣ ≤ 1

ν: the Currant number

⇒ The CFL condition

For system: ν = ∆t∆x ·maxp |λp|

3-point stencil: CFL condition is ν ≤ 1


The Lax-Friedrichs method

qt + f (q)x = 0, f (q) = Aq

Qn+1i =

1

2(Qn

i−1 + Qni+1)− ∆t

2∆x(f (Qn

i+1)− f (Qni−1))

(This looks more like a FDM. Actually also FVM with central scheme)

numerical flux:

F(Qni−1,Q

ni ) =

1

2[f (Qn

i−1) + f (Qni )]− ∆x

2∆t[Qn

i − Qni−1]

CFL condition: ν ≤ 1

consistency: F(q) = 12 (f (q) + f (q))− 0 = f (q)

Accuracy (formally): first order in t, second order in x .


An unstable flux

A naive attempt for numerical flux

F(Qni−1,Q

ni ) =

1

2[f (Qn

i−1) + f (Qni )]

Scheme:

Qn+1i = Qn

i −∆t

2∆x[f (Qn

i+1)− f (Qni−1)]

Unfortunately, the method is generally unstable, even with CFL condition.


REA (Reconstructed-evolve-average) algorithm

Example: qt + uqx = 0, u > 0

∆xQn+1i = ∆tuQn

i−1 + (∆x −∆tu)Qni

= ∆xQni −∆tu(Qn

i − Qni−1)

Denote the wave (i.e., the jump)

Wni− 1

2= Qn

i − Qni−1

If u > 0:

Qn+1i = Qn

i −∆tu

∆xWn

i− 12

If u < 0:

Qn+1i = Qn

i −∆tu

∆xWn

i+ 12


Combine both cases into one formula: let

u+= max(u, 0), u−= min(u, 0)

F ni− 1

2= u−Qn

i + u+Qni−1

Qn+1i = Qn

i −∆t

∆x

[u+Wi− 1

2+ u−Wi+ 1

2

]Special case: if ν = u ∆t

∆x = 1, it gives the exact solution!


numerical methods for hyperbolic conservation laws lecture...

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