classification of pdes, method of characteristics, traffic ... · classification of pdes, method...

Classification of PDEs,Method of Characteristics,

Traffic Flow ProblemGIAN 2016-17l

R. von Fay-Siebenbuurgen

SP2RC, School of Mathematics and Statistics

University of Sheffield

email: [email protected]

web: robertus.staff.shef.ac.uk

September 2016, Varanasi, India

With special thanx to Dr V. Fedun & C. Malins

Table of contents

• Background

– Linear PDEs

– Classification

• Quasi-linear first-order PDEs

– Some properties of characteristics

– Model of traffic flow

– Small amplitude disturbances from a uniform state

– The initial value problem (IVP)

– Shocks

– The Riemann problem

– Additional refinements

• 1-D linear convection-dominated problems

– 1-D linear convection equation

– Numerical dissipation and dispersion

• 1-D Burger’s equation

– Useful properties

– Explicit schemes

– Implicit schemes

ii

I Classification of PDEs 1

1 Background

1.1 Linear PDEs

Classification of linear PDE of 2nd order in two indepen-

dent variables

A∂2u

∂x2+B

∂2u

∂x∂y+C

∂2u

∂y2+D

∂u

∂x+E

∂u

∂y+Fu+G = 0,

(1)

where A,B, ..., G are constant coefficients.

Three categories of PDEs can be distinguished:

B2 − 4AC < 0 elliptic PDE

B2 = 4AC parabolic PDE

B2 − 4AC > 0 hyperbolic PDE

Classification depends only on the highest-order deriva-

tives in each independent variables.

If A,B, ..., G are functions of x, y, u, ux or uy the classi-

fication still can be used with local interpretation.


Well-posed mathematical problem

• the solution exist

• the solution is unique

• the solution depends continuously on the auxiliary

(IC/BC) data

Well-posed computational problem

• the computational solution exist

• the computational solution is unique

• the computational solution depends continuously on

the approximate auxiliary datra

BC & IC

Notation: If computational domain R, boundary ∂R,

normal to boundary n, tangetial to boundary s.

• Dirichlet condition, e.g. u = f on R

• Neumann (derivative) condition, e.g.∂u∂n

= f or ∂u∂s

= g on ∂R

• mixed or Robin condition, e.g.∂u∂n + ku = f, k > 0 on ∂R


1.2 Classification

Classification by characteristics

• For a single 1st-order hPDE w. two independent vari-

ables,

A∂u

∂t+B

∂u

∂x= C (2)

a single real characteristics exists through ∀ point, and

the characteristic direction is defined

dx

dt=

A

B(3)

Along the characteristics directions Eq. (2) reduces to

du

dt=

C

A&

du

dx=

C

B(4)

Eq. (4) can be integrated as ODE along the grid defined

by Eq. (3) provided the initial data are given on a non-

characteristics line


• Concept of characteristic directions for a 2nd-order PDEw. two independent variables can be established. Since

only highest derivatives determine the category of PDE,

Eq. (1) can be written as

A∂2u

∂x2+B

∂2u

∂x∂y+ C

∂2u

∂y2+H = 0, (5)

where H contains all other terms. It is possible to ob-

tain ∀ ∈ R two directions along which the integration

of Eq. (5) involves only two total differentials. The exis-

tence of these (characteristics) directions relates directly

to the category of PDE.

Introduce

P =∂u

∂x,Q =

∂u

∂y,R =

∂2u

∂x2, S =

∂2u

∂x∂y, T =

∂2u

∂y2.

Further, a curve K is introduced in R on which Eq. (5)

is satisfied. Along a tangent to K

dP = Rdx + Sdy & dQ = Sdx + Tdy,

where dy/dx defines the slope of tangent to K, and,

Eq (5) can be written as

AR + BS + CT +H = 0. (6)


Eq. (6) can be written as

S

[

A

(dy

dx

)2

− B

(dy

dx

)

+ C

]

−{[

A

(dP

dx

)

+H

]dy

dx+ C

dQ

dx

}

= 0 (7)

If dy/dx is chosen such that

A

(dy

dx

)2

−B

(dy

dx

)

+ C = 0 (8)

the solutions to Eq. (8) define the characteristic direc-

tions.

Conclusion:

• for a hPDE two real characteristics exist

• for a pPDE one real characteristic exists

• for an ePDE the characteristics are complex

The discriminantB2−4AC determines both type of PDE

and nature of characteristics.


• System of equations

Two-component system of 1st order PDE

A11∂u

∂x+B11

∂u

∂y+ A12

∂v

∂x+B12

∂v

∂y= E1 (9)

A21∂u

∂x+B21

∂u

∂y+ A22

∂v

∂x+B22

∂v

∂y= E2 (10)

After re-arranging Eqs (9)-(10)

[(A11dy −B11dx) (A21dy −B21dx)

(A12dy −B12dx) (A22dy −B22dx)

] [L1

L2

]

= 0

(11)

where L1, L2 are suitable multipliers. Since the system is

homogeneous in Li, it is necessary that

det[Ady −Bdx] = 0. (12)


For non-trivial solution, i.e. Eq. (12) gives

DIS = (A11B22 − A21B12 + A22B11 − A12B21)2

−4(A11A22 −A21A12)(B11B22 − B21B12) (13)

Classification:

DIS Roots Classification

positive 2 real hyperbolic

zero 1 real parabolic

negative 2 complex elliptic


• System of n 1st order PDEs

[

A

(∂y

∂x

)(k)

−B

]

L(k) = 0, k = 1, ..., n (14)

where Li are suitable multipliers.

Classification:

The character of the system depends on the solution to

Eq. (12):

• If n real roots → system is hyperbolic

• If ν real roots, 1 ≤ ν ≤ n− 1, and ∄ complex roots

→ system is parabolic

• If ∄ real roots → system is elliptic.

Note: Most important whether ePDE or non-ePDE, since

latter preclude time-like behavior.


Classification by Fourier Analysis

Useful

• if higher-order derivatives appear

• since can indicate the expected behavior of solution

(oscillatory, exp. growth, etc.)

Example

A∂2u

∂x2+B

∂2u

∂x∂y+ C

∂2u

∂y2= 0, (15)

Introduce

u = Fu

where F is the Fourier transform defined by

u =

∫ ∞

−∞

∫ ∞

−∞u(x, y) exp(−iσxx) exp(−iσyy)dxdy

(16)

Eq. (15) transforms into

[A(iσx)2 +B(iσxiσy) + C(iσy)

2]u (17)

often called the characteristic polynomial or the symbol

of the PDE.


Wave equation

Simplest hPDE is wave equation

∂2u

∂t2− ∂2u

∂x2= 0 (18)

For

IC: u(x, 0) = sinπx, ∂u/∂t(x, 0) = 0

and

BC: u(0, t) = u(1, t) = 0 ⇒

u(x, t) = sinπx cosπt

Note: lack of attenuation is feature of linear hPDE.

hPDEs produce real characteristics. E.g. for wave equa-

tion Eq. (18) characteristics directions are given by

dx/dy = ±1.

see Fig. 1


Figure 1: Characteristics or the wave equation


Wave representation

Q: Whether the discretisation process represents waves

of short of long wavelength with the same accuracy?

• Significance of grid coarseness:

- FDM replaces a continuous fnc g(x) → with vector of

nodal values (gj).

- Choice of grid spacing ∆x depends on smoothness of

g(x) (Fig. 2a) poor choice, Fig. 2b reasonable choice)

Figure 2: Discrete representation of g(x). (a) coarse, (b) reasonable


A F-rep of g(x)

g(x) =∞∑

m=−∞gme

imx (19)

gm, amplitude of Fourier mode of wavelength λ = 2π/m,

gm =1

2π

∫ 2π

0

g(x)e−imxdx. (20)

A F-rep of vector of nodal (gj)

gj =J∑

m=1

gmeimj∆x (21)

where the modal amplitude gj is given

gm = ∆x

J∑

j=1

gje−imj∆x. (22)

Wavelength shorter than cut-off wavelength λ = 2∆x

cannot be represented. ⇒ (gj) should be interpreted as

a long-wave approx of g(x).


• Accuracy of representing waves

Accuracy may be assessed by comparing FD approximate

solutions to progressive waves

T (x, t) = R{e[im(x−qt)]} = cos[m(x− qt)]. (23)

At (j, n)-th node the exact values of 1st/2nd derivatives

of T :

∂T

∂x= −m sin[m(xj − rtj)] (24)

∂2T

∂x2= −m2 cos[m(xj − rtj)] (25)

- Substitution into three-point formula (3PT)

[∂T

∂x

]n

j

≈(T n

j+1 − T nj−1)

2∆x

and calculating the amp ratio of 1st derivatives gives

AR(1)3PT =[∂T ∂x]nj[∂T ∂x]

=sin(m∆x)

m∆x. (26)


- Using CD approx

(T nj−1 − 2T n

j + T nj+1)

∆x2

gives the amp ratio for 2nd derivatives

AR(2)3PT =

(sin(m∆x/2)

m∆x/2

)2

. (27)

Summary table:

Amplitude ratio Amplitude ratio

Derivative LW (λ = 20∆x) SW (λ = 4∆x)

dT

dx0.64 0.984

d2T

dx20.992 0.405

II Method of Characteristics 16

2 Quasi-linear first-order PDEs

• We consider only two independent variables (x, y) and

an unknown z = z(x, y) satisfying a first order PDE.

This PDE is quasi-linear if it is linear in its highest order

terms, i.e.

zx =∂z

∂xand zy =

∂z

∂y.

Thus

zzx + zy = 0 is quasi-linear (and non-linear)

(zx)2 + zy = 0 is not quasi-linear.

The most general first-order quasi-linear PDE is:

Pzx +Qzy = R (28)

where

P = P (x, y, z), Q = Q(x, y, z), R = R(x, y, z) (29)

are given continuous functions.


• Consider the family of curves in the (x, y) plane satis-

fying

dy

dx=

Q

Por

dx

dy=

P

Qor

dx

P=

dy

Q.

Suppose z is known at a pointA(x, y). There is one curve

ΓA of this family through A, and along ΓA

dz = zxdx + zydy =

(

zx +Q

Pzy

)

dx =R

Pdx (30)

using Eq. (28). Hence dzdx

= RPalong ΓA, and so:

dx

P=

dy

Q=

dz

R. (31)

Eqs. (31) are known as the associated equations for Eq. (28),

and are equivalent to Eq. (28). For let each term in

Eq. (31) be ds, so dx = Pds, dy = Qds, dz = Rds.

Substitute in Eq. (30) to get

Rds = Pzxds +Qzyds ⇒ Pzx +Qzy = R,

i.e. Eq. (28).


y

x0

A

C

G

G

+

GA

-

Figure 3: Characteristic curves

• Suppose z is given along a curve C in the (x, y) plane.

Through each point A on C, we can continue the solution

along ΓA in both directions provided ΓA is not parallel

to C, i.e. provided that, on C, dydx is nowhere equal to Q

P .

The curves ΓA are known as the characteristics. Provided

the characteristics do not intersect, we obtain a region

bounded by Γ+ and Γ− within which z is known. If QP

is independent of z the characteristics are independent of

the boundary conditions. In particular QPis independent

of z for a linear PDE. If P and Q are constants, then the

characteristics are parallel straight lines.


Example 1

Solve zx − zy = 1 with z = x2 on y = 0.

Solution

The associated equations are

dx

1=

dy

−1=

dz

1⇒ dy

dx= −1,

dz

dx= 1.

Thus the characteristics are x + y = α and on the char-

acteristics z − x = β.

y

x0

Cx+y=a

Figure 4: Characteristics of Example 1


The curve C is y = 0 and each point on C is intercepted

by exactly one characteristic. We can proceed in two

ways.

(A) When y = 0, x + y = α ⇒ x = α and z = α2.

Hence from z = x + β ⇒ β = α2 − α.

Thus z = x +(α2 − α

)on x + y = α.

Eliminate α to get z = x +((x + y)2 − (x + y)

)⇒

z = (x + y)2 − y.

(B) Since z − x is constant when x + y is constant ⇒

z− x = f(x+ y) for some function f . But z = x2 when

y = 0 ⇒ x2 − x = f(x).

Thus z = x +((x + y)2 − (x + y)

)⇒

z = (x + y)2 − y


Example 2

Solve yzx + xzy = z with z = x3 on y = 0 and z = y3

on x = 0.

Solution

The associated equations are

dx

y=

dy

x=

dz

z⇒

dy

dx=

x

y⇒ x2 − y2 = α

are characteristics, where α =const. Then

dz

dx=

z√(x2 − α)

⇒ dz

z=

dx√(x2 − α)

⇒

ln z = ln(x +

√ (x2 − α

))+ β′

z = β(x +

√ (x2 − α

))

on a characteristic, where β = eβ′=const.

α = x2 − y2 ⇒ z = β(x +

√ (x2 − x2 + y2

))

= β(x + y) or β(x− y).


Case 1: z = β(x + y)

In this case we find, that as in Ex 1 (B) above,

z

x + yis constant when x2 − y2 is constant.

The GS is therefore z = (x + y)f(x2 − y2

), and it re-

mains to determine f .

y

x0

C

A

B

>0a

=0a

<0a

=0a

<0a

>0a

Figure 5: Characteristics of Example 2

We are given z on both axes. At the common point

O, z = 0 from both prescriptions. The characteristics

through O are x = ±y (α = 0) and on these z = 0. The

result is obviously symmetric about both axes.


Suppose α > 0.

Consider A(α121 , 0) at which

z = x3 = α321 .

So from the GS ⇒

α321 = α

121f(α1) ⇒ f(α1) = α1 ⇒ z = (x + y)(x2 − y2)

for x2 > y2.

Suppose α < 0.

Consider B(0, (−α2)12) at which

z = y3 = (−α2)32 .

So from the GS ⇒(−α2)

32 = (−α2)

12)f(α2) ⇒ f(α2) = −α2

⇒ z = (x + y)(y2 − x2) for x2 < y2

In summary

z =

(x + y)(x2 − y2) for x2 > y2

0 for x2 = y2

(x + y)(y2 − x2

)for x2 < y2

(32)


Case 2: z = β(x− y)

In this case it can be similarly deduced that the GS of

the PDE is

z = (x− y)g(x2 − y2).

However, this GS gives

zx = g(x2 − y2

)+ 2x(x− y)g′

(x2 − y2

)

zy = −g(x2 − y2

)− 2y(x− y)g′

(x2 − y2

)

Therefore

yzx + xzy = −(x− y)g(x2 − y2) = −z

which is not our original PDE, therefore we dismiss this

second case, z = β(x− y), as spurious solution.


• We begin by considering Eq. (32). It is clear that z is

everywhere continuous, and that zx, zy are everywhere

continuous except possibly on the lines

x = ±y (x2 − y2) = 0.

From Eq. (32) we find

x2 > y2 : zx = (x2 − y2) + 2x(x + y) = (x + y)(3x− y)

zy = (x2 − y2)− 2y(x + y) = (x + y)(x− 3y)

x2 < y2 : zx = (y2 − x2)− 2x(x + y) = (x + y)(y − 3x)

zy = (y2 − x2) + 2y(x + y) = (x + y)(3y − x).

Thus as x → −y from either side, zx → 0 and zy → 0.

Hence zx and zy are continuous on x + y = 0.

Hovewer, as x → y, zx jumps from +4x2 (x > y) to

−4x2 (x < y), and zy jumps from −4x2 (x > y) to +4x2

(x < y). Thus zx and zy are discontinuous across the

characteristic x = y.


2.1 Some properties of characteristics

We investigate the possibility of discontinuities in zx and

zy for Eq.(28), but we shall suppose z is everywhere con-

tinuous. (The standard terminology is that we are look-

ing for weak discontinuities whereas discontinuities in z

itself are strong discontinuities).

Suppose C is approached from + and −. Then

y

x0

C

A-

+ =( x, y)d ddr

Figure 6: Jump across characteristics

δz+ =∂z+

∂xδx +

∂z+

∂yδy

δz− =∂z−

∂xδx +

∂z−

∂yδy

where (δx, δy) is along C. Subtract.


• Weak discontinuities

Because z is continuous, δz+ = δz−. Hence

δx

[∂z

∂x

]+

−+ δy

[∂z

∂y

]+

−= 0. (33)

where the square brackets denote the jump in the expres-

sion across C.

Since Eq. (28) is satisfied on both sides and since, by

hypothesis, P , Q, R are continuous

P

[∂z

∂x

]+

−+Q

[∂z

∂y

]+

−= 0. (34)

The necessary condition for

[∂z

∂x

]+

−6= 0 and

[∂z

∂y

]+

−6= 0 is

δx

P=

δy

Q,

i.e.dx

P=

dy

Q. (35)

Thus C must be a characteristic ΓA.


• When Q/P is independent of z, the characteristics are

independent of the boundary conditions.

When Q/P depends on z, different boundary conditions

produce different sets of characteristics.

A B A B

D C

Figure 7: (i) z determined throughout rectangle; (ii) z not deter-mined in BCD

• Strong discontinuities

We can also consider situations in which z itself is discon-

tinuous at a point on the boundary. Then the shape of

the characteristics (in the case when Q/P depends on z)

will change discontinuously at that point. Qualitatively,

there are two possibilities:


• In (i) there appear to be two characteristics through

a point, whereas

• in (ii) there is a region containing no characteristics.

Figure 8: Characteristics leading to (i) shocks; or to (ii) centredfans (also called rarefication shocks)

Again, qualitatively these two situations will be relevant

to our models of traffic flow [(i) leads to shocks, (ii) leads

to centred fans - see later].

• We can obtain similar situations to (i) and (ii), but

even without initial discontinuities in the slopes of the

characteristics. The following example will connect well

with our models of traffic flow.


Example

Solve

ρt + ρρx = 0

with ρ = f(x) on t = 0. Consider two special cases:

Case 1 : f(x) = 0 (x < 0), f(x) = x (0 ≤ x < 1),

f(x) = 1 (x ≥ 1);

Case 2 : f(x) = 0 (x < 0), f(x) = −x (0 ≤ x < 1),

f(x) = −1 (x ≥ 1).

The associated equations Eq. (31) are

dt

1=

dx

ρ=

dρ

0where the last is to be interpreted as dρ = 0.

Thus

ρ = α,dx

dt= α ⇒ x− αt = β are characteristics.


Now consider the characteristic through x = ξ on t = 0.

On this characteristic ρ = α = f(ξ). Thus

x = f(ξ)t + ξ, ρ = f(ξ) (36)

t

x0 x

1

f( )xslope

Figure 9: Characteristics for Example

Alternatively, from Eq. (36)

x = ρt + ξ

so that we can write Eq. (36) implicitly as

ρ = f(x− ρt) (37)


Case 1: From Eq. (37)

ρ = 0 (x < 0), ρ = x− ρt (0 ≤ x− ρt < 1)

⇒

ρ =x

1 + t(0 ≤ x < 1 + t), ρ = 1 (x ≥ 1 + t),

i.e.

ρ =

0 (x < 0)

x/(1 + t) (0 ≤ x < 1 + t)

1 (x ≥ 1 + t)

(38a)

Case 2: Likewise,

ρ =

0 (x < 0)

x/(1− t) (0 ≤ x < 1− t)

−1 (x ≥ 1− t)

(38b)

The solution Eq. (38b) breaks down at t = 1; as the

sketch on the hand-out shows, t he characteristics inter-

sect at t = 1 in Case 2 and the profile of ρ against x

becomes triple-valued (but this cannot occur in reality).

III Traffic Flow Problem 33

3 Model of traffic flow

We assume:

1. One lane of traffic in direction of Ox with no over-

taking.

2. We can define a local car density ρ = ρ(x, t) as the

number of cars per unit length of road.

3. The local car velocity v(x, t) is a function of ρ alone,

i.e.

v = v(ρ) (39)

The meaning of Eq. (39) is that each driver adjusts his,

her or its speed to local conditions exclusively, whereas

most drivers look ahead and adjust speed where appro-

priate. These assumptions give a car flowrate q(ρ) with

q(ρ) = ρv(ρ) (40)


x t x+ xd

q(x,t) q(x+ x,t)dr (x,t)

x t+ tdx+ xd

r (x,t+ t)d

Figure 10: Car flow

Consider two values of x, viz. x1, x2 with x1 ≤ x ≤ x2.

At time t, the number of cars in this interval is

∫ x2

x1

ρ(x, t)dx.

The rate of change of this must be the net flowrate, viz.

∂

∂t

{∫ x2

x1

ρ(x, t)dx

}

= [q(x, t)]x1x2 (41)


If x1 = x, x2 = x + δx, Eq. (41) becomes

∂

∂tρδx = −∂q

∂xδx

⇒

∂ρ

∂t+

∂q

∂x= 0. (42)

• We need to model v(ρ).

We assume there is a maximum possible density P with

essentially “bumper-to-bumper” traffic. When ρ = P ,

we assume v(ρ) in Eq. (39) is zero.

We also assume v(ρ) decreases as ρ increases, with a max-

imum of V when ρ = 0.

These assumptions are shown schematically...


0

1

1 r

P

v( )r

V

0 1 r

P

q( )r

VP

a

Figure 11: Modelling car flow

With Eq. (40), Eq. (42) becomes

∂ρ

∂t+

d

dρ(ρv(ρ))

∂ρ

∂x= 0

or

∂ρ

∂t+ c(ρ)

∂ρ

∂x= 0,

(43)

c(ρ) =d

dρ(ρv(ρ)) = v(ρ) + ρv′(ρ)


0

1

1 r

P

c( )r

V

a

Figure 12: Model of car flow

The assumptions made about v(ρ) give a c(ρ) which is

monotonic decreasing and negative for ρ/P > α, where

α is the value of ρ/P for which q(ρ) is a maximum.


3.1 Small amplitude disturbances from a uniform state

• Before studying the full non-linear problem, it is in-

tructive to consider a simpler one. Suppose that there is

almost a uniform state with ρ = ρ0 and

ρ = ρ0 + ρ′ with |ρ′| ≪ ρ0. (44)

Linearise Eq. (43) - as with sound waves earlier - to get

∂ρ′

∂t+ c (ρ0)

∂ρ′

∂x= 0. (45)

Either

dt

1=

dx

c (ρ0)=

dρ′

0⇒

ρ′ = const. on x− c(ρ0)t = const.

Or put ξ = x− c (ρ0) t ⇒∂ρ′

∂x=

∂ρ′

∂ξ,

(∂ρ′

∂t

)

x

=

(∂ρ′

∂t

)

ξ

− c (ρ0)

(∂ρ′

∂ξ

)

⇒(∂ρ′

∂t

)

ξ

= 0.

Thus the GS of Eq. (45) is

ρ′ = f {x− c (ρ0) t}. (46)


• The characteristics of Eq. (45) are the straight lines

t

x0 x

c( )r 0<0

Figure 13: Characteristics of car flow are straight lines

x = ξ + c (ρ0) t. (47)

Eq. (46) shows that ρ′ is constant on each characteristic.

Eq. (46) represents a wave travelling to the right with

speed c (ρ0). If ρ0/P > α ⇒ c(ρ0) < 0.

This is a kinematic wave; c (ρ0) is the speed of the dis-

turbance, not of the cars.

This explains a common phenomenon on a busy road

when a sudden increase in density reaches you from ahead

with no apparent reason.


3.2 The initial value problem for Eq. (43)

•We wish to solve Eq. (43) subject to the initial condition

ρ(x, 0) = f(x) (48)

By the earlier methods - see especially the Example in

§ (5.2) - ρ is constant on the characteristics

dt

1=

dx

ρ⇒ dx

dt= ρ.

Since ρ is constant on a characteristic, the characteristics

are straight.

⇒

Thus, if c {f(ξ)} = F (ξ), the solution can be written for

t ≥ 0

ρ = f(ξ) on the straight line x = ξ + F (ξ)t. (49)


Example

Suppose

v(ρ) =V

P(P − ρ) (50)

and that ρ(x, 0) = f(x) satisfies

ρ =1

2(ρL + ρR)−

1

2(ρL − ρR) tanh

x

L(51)

where ρL, ρR and L are constants. Discuss the solution

given by Eq. (49) when

(i) ρL > ρR, and (ii) ρL < ρR.

Solution

From flow model Eq. (50) it follows

0 1r

P

v

V1

Figure 14: (a) Car flow v(ρ) = V (P − ρ)/P


q(ρ) =V

P

(Pρ− ρ2

), c(ρ) =

V

P(P − 2ρ). (52)

Note also that

0 1r

P

q

PV

14

12

0 1r

P

c

V1

12

1

Figure 12: (b) Car flow flux, and (c) speed of disturbance

ρ → ρL asx

L→ −∞

and

ρ → ρR asx

L→ +∞.

Also

F (ξ) = c {f(ξ)} (53)

=V

P

[

P − ρL − ρR + (ρL − ρR) tanh

(ξ

L

)]


Case (i): ρL > ρR ⇒ F ′(ξ) > 0 for ∀ξ

0

r

r

x

R

Lr

0 x

rR

Lrc( )

c( )

x0

rRL

r + > P12L

r > > rRP

c

t

x0

x

Figure 13: (a) Car flow, (b) profile of speed of disturbance, and(c) characteristics for Case (i). Note, that ρ is constant on each

characteristic


Case (ii): ρL < ρR ⇒ F ′(ξ) < 0 for ∀ξ

0

r

x

rR

Lr

0 x

Lrc( )

rR

c( )

x0

rRL

r + > P

Rr > > r

L12

P

c

t

x0x

Figure 14: (a) Car flow, (b) profile of speed of disturbance, and (c)characteristics for Case (ii). Note, from (c) that characteristicseventually intersect

Characteristics eventually intersect ⇒ problem becomes

ill-posed.


If two characteristics intersect, any enclosed characteris-

tic must meet one of them at an earlier time ⇒ earliest

intersection must be between neighbouring characteris-

tics.

t

x

Figure 15: Intersecting characteristics

Suppose these are

x = {ξ + F (ξ)t}

x = {(ξ + ∂ξ) + F (ξ + ∂ξ)t}= {ξ + F (ξ)t} + {1 + F ′(ξ)t} ∂ξ

⇒

∴ 1 + F ′(ξ)t = 0. (54)

We get solutions of Eq. (54) with t > 0 only if ∃ ξ with

F ′(ξ) < 0. [Thus for ρL > ρR there are no intersections

and the solution given by Eq. (49) applies for ∀ t ≥ 0.]


The first positive t satisfying Eq. (54) occurs when

t = Tmin =1

Max−∞<ξ<∞

{−F ′(ξ)}. (55)

While Eqs. (54) and (55) are general, we can calculate

Tmin in our particular case when Eq. (53) holds.

We find

−F ′(ξ) =V

P(ρR − ρL)

1

Lsech2

(ξ

L

)

⇒

max {−F ′(ξ)} =V (ρR − ρL)

PL

when ξ = 0. Then Eq. (55) gives

Tmin =PL

V (ρR − ρL). (56)


3.3 Shocks

•We can understand in another way why there is trouble

when ρL < ρR. With c′(ρ) < 0, low densities propagate

forward relative to high densities. The profile of ρ against

x inevitably steepens as t increases and has a vertical

section at t = Tmin. Were we to continue, the profile

would develop the triple-valued shape - clearly unaccept-

able since ρ must be a single valued quantity.

0

r

x

rR

Lr t=0

0

r

x

t=t1< Tmin

0

r

x

t=Tmin

0

r

x

t >Tmin

Figure 16: Development of shock

• Instead the wave breaks, and the model must be ex-

tended. A consistent extension conserves cars but allows

discontinuities in ρ to occur across a shock.


We cannot have characteristics crossing one another. In-

stead the picture is as shown schematically.

t

x

shock

0

Figure 17: Shock front and characteristics

x

shock

1 2x

x=s(t)

r1

q1

r2

q2

Figure 18: Quantities at a shock front


From Eq. (41) ⇒

∂

∂t

∫ s(t)

x1

ρdx +∂

∂t

∫ x2

s(t)

ρdx = q1 − q2

LHS =

∫ s(t)

x1

∂ρ

∂tdx

︸︷︷︸→0asx1→s−

+1

∂t

{∫ s(t+δt)

x1

−∫ s(t)

x1

}

ρdx

+

∫ x2

s(t)

∂ρ

∂tdx

︸︷︷︸→0asx2→s+

+1

∂t

{∫ x2

s(t+δt)

−∫ x2

s(t)

}

ρdx

=1

∂t

{∫ s(t)+δt

s(t)

ρ1dx−∫ s(t)+δt

s(t)

ρ2dx

}

= s (ρ1 − ρ2)

by the mean value theorem.

⇒s (ρ1 − ρ2) = (q1 − q2) (57a)

s =q1 − q2ρ1 − ρ2

. (57b)


• The position of the shock is fixed by the need to con-

serve cars. Without a shock the curve of ρ against x

would become (unacceptably) triple-valued. We insert

the shock so that the shaded areas are equal, thus ensur-

ing that∫ρdx = number of cars is unchanged. This is

known as (Whitham’s) equal area rule.

0

r

x

r1

r2

s(t)

Figure 19: Shock fitting: Whitham’s equal area rule

• As an application consider what happens as cars ap-

proach a stationary queue behind a red traffic light so

that ρR = P , ρL < P . On meeting the queue cars stop

and the lengthening of the queue is achieved by a shock

wave propagating backwards.


3.4 The Riemann problem

• We wish to consider the case when we solve Eqs. (43)

and (48) where there is a discontinuity in f(x). It will be

sufficient to consider the simplest possible case, viz.

ρ(x, 0) = f(x) =

ρL (x < 0)

ρR (x > 0)

(58)

• Then Eq. (49) gives the solution as

ρ = ρL on x = ξ + c (ρL) t (ξ < 0)

ρ = ρR on x = ξ + c (ρR) t (ξ > 0)

(59)

Consider first the case ρL > ρR. The characteristic dia-

gram is easy to draw...


t

x0

rR

r=Lrr=

AB

Figure 20: Fan of characteristics

They are either parallel to OA with slope c (ρL); to the

left ofOA, ρ = ρL. Or they are parallel toOB with slope

c(ρR); to the right of OB, ρ = ρL. But what happens in

OAB?

• The problem arises because of the discontinuity and can

be solved by considering a limit process in which ρ takes

all the values from ρR to ρL, and all the characteristics

go through the origin. Thus

ρ = k (ρR < ρ < ρL)

on x = c(k)t.

The solution is therefore:


ρ =

ρL : x < c (ρL) t

k on x = c(k)t : c (ρL) < c(k) < c (ρR)

ρR : x > c (ρR) t

(60)

x

Figure 21: Centered fan or expansion fan corresponding or rarefi-cation wave

The characteristic diagram is augmented by a centred

fan or an expansion fan or expansion wave or rarefication

wave.


• Conversely, when ρL < ρR the characteristic diagram

shows immediately trouble whose only resolution is a

shock starting from t = 0 with speed, given by Eq. (57b)

as U , where

t

x

shock

0

Figure 22: Schematic shock with speed U

U =q (ρL)− q (ρR)

(ρL − ρR)(61)


3.5 Additional refinements

• The model assumptions leading to Eq. (43) are too

simple. One extension is to suppose that q is a function

of the density gradient ∂ρ/∂x as well as ρ, thus allowing

drivers to reduce their speed to account for an increasing

density ahead. A simple assumption is to take

q = Q(ρ)− νρx (62)

where ν is a positive constant. Thus q decreases if ρx is

positive, i.e. it there is an increasing density ahead. Use

of Eq. (62) in Eq. (42) gives

ρt + c(ρ)ρx = νρxx, c(ρ) = q′(ρ) (63)

• Seek solutions of Eq. (63) of the form

ρ = ρ(X) X = x− Ut (64)

where U is a constant still to be determined. Substitution

in Eq. (63) ⇒

−Uρ′(X) + c(ρ)ρ′(X) = νρ′′(X).

Since c(ρ) = Q′(ρ) we have


Q(ρ)− Uρ + C = νρ′(X) (65)

where C is a constant. Suppose ρ → ρL as X → −∞and ρ → ρR as X → +∞ ⇒

Q (ρL)− UρL + C = Q (ρR)− UρR + C = 0,

⇒

U =Q (ρR)−Q (ρL)

ρR − ρL(66)

This is exactly Eq. (57b) but (for the moment) in a dif-

ferent context.

• Since ρ → ρL as X → −∞ and ρ → ρR as X → +∞,

ρ′(x) = 0 at ρ = ρL and ρ = ρR. We suppose ρL and ρRare simple zero’s of

Q(ρ)− Uρ + C,

and more precisely we shall suppose ρL < ρR and

Q(ρ)−Uρ+C = α (ρ− ρL) (ρR − ρ) (α > 0) (67)


With α > 0,

c(ρ) = Q′(ρ) = α (ρR − ρ)− α (ρ− ρL)

and

c′(ρ) = α (ρL − ρR) < 0.

We can always approximate Q(ρ) by a quadratic. Then

Eq. (65) becomes

νdρ

dX= α (ρ− ρL) (ρR − ρ)

with solution

(ρR − ρ

ρ− ρL

)

=

(ρR − ρ0ρ0 − ρL

)

e−X

L , L =ν

α (ρR − ρL)(68)

where ρ = ρ0 at X = 0. We note that ρ → ρL as

X → −∞ and that ρ → ρR as X → +∞ as required.


The transition between ρ ∼ ρL and ρ ∼ ρR occupies a

thickness of order L.

As L diminishes, i.e. as ν diminishes for fixed α and

(ρR − ρL) the transition takes place more sharply ⇒ a

shock is approached.

r

xL

rR

Lr

r0

Figure 23: Development of shock front

• The model in this section can be taken further in the

case when Eq. (67) holds.


Multiply Eq. (63) by c′(ρ) ⇒c′(ρ)ρt + c(ρ)c′(ρ)ρx − νc′(ρ)ρxx

∴ ct + ccx = ν ∂2c∂x2

− νc′′(ρ)(∂ρ∂x

)2

because∂c

∂x= c′(ρ)

∂ρ

∂xand therefore

∂2c

∂x2= c′′(ρ)

(∂ρ

∂x

)2

+ c′(ρ)∂2ρ

∂x2.

In the case when Q(ρ) is quadratic, i.e. Eq. (67) holds,

c′′(ρ) = 0 since c(ρ) = Q′(ρ). Thus

ct + ccx = νcxx. (69)

This is known as Burger’s equation and, remarkably, it

can be solved explicitly by means of the transformation

c = −2νφx

φ(70)

discovered independently by E. Hopf (1950) and J.D.Cole

(1951). Use of Eq. (70) transforms Eq. (69) into the

standard linear equation (after one integration w.r.t. x):

φt = νφxx. (71)

It can be shown that this is also consistent with the shock

structure.


• A second refinement is that there is a time lag in driver

response. One way of handling this is to take Eq. (62)

and deduce from it that v = q/ρ satisfies

v = V (ρ)− ν

ρρx, V (ρ) =

Q(ρ)

ρ. (72)

Then regard this as a velocity which the driver tries to

achieve. The acceleration of the car is

Dv

Dt=

∂v

∂t+ v

∂v

∂x[see Notes after § (4.3)] and the model is

vt + vvx = −1

τ

{

v − V (ρ) +ν

ρρx

}

, (73)

where τ is a measure of the response time. Eq. (73) is to

be solved together with Eq. (42), i.e.

ρt + (ρv)x = 0. (74)


4 1-D Linear convection-dominated problems

4.1 1-D linear convection equation

∂T

∂s+ u

∂T

∂x= 0 (75)

• Schemes: FTCD, Upwind Differencing, Leapfrog, lax-

Wendroff, Crank-Nicolson ⇒ Handout pp.278-279

Observe: stencil of schemes, algebraic form, leading term

of truncation error, and stability.


• Example: linear convection of a truncated since wave

Solution of Eq. (75) subject to IC/BC:T (x, 0) = sin(10πx) for 0 ≤ x ≤ 0.1,

= 0 for 0.1 < x ≤ 1.0,

T (0, t) = 0 and T (1, t) = 1.

Note: Observe dissipation (Figs. 9.2-9.4), dispersion (slow-

down) and oscillatory wake (Figs. 9.3-9.4).


4.2 Numerical dissipation and dispersion

-Most (M)HD phenomena are governed by hPDEs, they

contain no or little dissipation ⇒

-Solutions are characterised by wave-trains that propa-

gate with little or no loss of amplitude ⇒

-Numerical schemes ‘should not’ introduce non-physical

dissipation, and non-physical dispersion (see Figs. 9.2-4).

Solution for propagating plane wave subject to dissipa-

tion & dispersion

T = RTampe−p(m)teim[x−q(m)t] (76)

Here p(m): amplitude attenuation; q(m): propagation

speed. It’s instructive to consider two related eqs:

∂T

∂t+ u

∂T

∂x− α

∂2T

∂x2= 0 (77)

∂T

∂t+ u

∂T

∂x+ β

∂3T

x3= 0 (78)


Eq. (77) is the transport equation, for plane waves:

p(m) = αm2, q(m) = 0 (79)

amplitude attenuated by diff term but propagation speed

unaffected.

Eq. (78) is the linear Korteweg-de Vries eq. For plane

waves:

p(m) = 0, q(m) = u− βm2 (80)

amplitude unaltered but propagation speed depends on

wavelength.

⇒:

-dissipation→ attenuation↔ even-ordered spatial deriva-

tives;

-dispersion → propagation of waves with different wave

number m at different speeds q(m) ↔ odd-ordered spa-

tial derivatives.

⇛ Through discretisation the truncation error consist,

typically, of higher even- and odd-ordered derivatives!


5 1-D Burger’s Equation

5.1 Useful properties

Burger’s (1948)

∂u

∂t+ u

∂u

∂x− ν

∂2u

∂x2= 0 (81)

- Similar to transport equation, except the convective

term is nonlinear

- If ν = 0: inviscid Burger’s equation

- Effect of viscosity: (a) reduces amplitude; (b) prevents

multi-valued solutions ⇒ Burger’s eq is very suitable for

testing comp algorithms

- Cole-Hopf transformation allows exact solution for a

wide range of IC/BCs

- Alternative way to handling nonlinear convective form

by using conservation form:

∂u

∂t+

∂F

∂x− ν

∂2u

∂x2= 0, F =

1

2u2 (82)

Note: Aliasing can be a problem in astro/geophysical

problems where dissipation is generally small and cannot

reduce errors caused by energy reappearance from scales

≤ 2∆x to longer wavelength.


5.2 Explicit schemes

• FTCS finite difference rep

un+1j − unj∆t

+unj (u

nj+1 − unj−1)

2∆x−ν(unj−1 − 2unj + unj+1)

∆x2= 0

(83)

The contribution unj to convective term is the local so-

lution at node j, n. Since this is available directly the

various schemes mentioned before can be applied to BE.

• FTCS of the conservative form (Eq. 82)

un+1j − unj∆t

+F nj+1 − F n

j−1

2∆x−

ν(unj−1 − 2unj + unj+1)

∆x2= 0

(84)

A potentially more accurate treatment of the convective

term is provided by a four-point upwind discretisation,

e.g. the truncation error is only O(∆x2).

• Lax-Wendroff scheme

Less economic because evaluation at half-steps (j − 12,

j + 12) is involved.


5.3 Implicit schemes

Note: It is not so straightforward as for linear equations.

• Crank-Nicolson formulation

∆un+1j

∆t= −1

2Lx(F

nj +F n+1

j ) +1

2νLxx(u

nj + un+1

j ) (85)

where

∆un+1j = un+1

j − unj ;Lx =(−1, 0, 1)

2∆x;Lxx =

(1,−2, 1)

∆x2.

Problem: Difficult to reduce to a sys of lin tridiagonal eqs

for the solution un+1j and to use the very efficient Thomas

algorithm because of the nonlin implicit term F n+1j .

Solution: introduce a correction term with split scheme!

(I.e., Taylor series expansion of F n+1j is made about the

nth time level resulting in a tridiagonal logarithm)

∆un+1j

∆t= −1

2Lx(2F

nj + unj∆un+1

j ) +1

2νLxx(u

nj + un+1

j )

(86)

Here the truncation error: O(∆t2,∆x2) and uncondition-

ally stable in the von Neumann sense.

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