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.
I Classification of PDEs 2
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
I Classification of PDEs 3
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
I Classification of PDEs 4
• 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)
I Classification of PDEs 5
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.
I Classification of PDEs 6
• 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)
I Classification of PDEs 7
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
I Classification of PDEs 8
• 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.
I Classification of PDEs 9
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.
I Classification of PDEs 10
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
I Classification of PDEs 11
Figure 1: Characteristics or the wave equation
I Classification of PDEs 12
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
I Classification of PDEs 13
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).
I Classification of PDEs 14
• 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)
I Classification of PDEs 15
- 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.
II Method of Characteristics 17
• 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).
II Method of Characteristics 18
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.
II Method of Characteristics 19
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
II Method of Characteristics 20
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
II Method of Characteristics 21
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).
II Method of Characteristics 22
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.
II Method of Characteristics 23
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)
II Method of Characteristics 24
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.
II Method of Characteristics 25
• 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.
II Method of Characteristics 26
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.
II Method of Characteristics 27
• 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.
II Method of Characteristics 28
• 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:
II Method of Characteristics 29
• 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.
II Method of Characteristics 30
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.
II Method of Characteristics 31
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)
II Method of Characteristics 32
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)
III Traffic Flow Problem 34
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)
III Traffic Flow Problem 35
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...
III Traffic Flow Problem 36
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)
III Traffic Flow Problem 37
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.
III Traffic Flow Problem 38
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)
III Traffic Flow Problem 39
• 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.
III Traffic Flow Problem 40
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)
III Traffic Flow Problem 41
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
III Traffic Flow Problem 42
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
)]
III Traffic Flow Problem 43
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
III Traffic Flow Problem 44
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.
III Traffic Flow Problem 45
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.]
III Traffic Flow Problem 46
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)
III Traffic Flow Problem 47
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.
III Traffic Flow Problem 48
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
III Traffic Flow Problem 49
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)
III Traffic Flow Problem 50
• 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.
III Traffic Flow Problem 51
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...
III Traffic Flow Problem 52
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:
III Traffic Flow Problem 53
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.
III Traffic Flow Problem 54
• 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)
III Traffic Flow Problem 55
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
III Traffic Flow Problem 56
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)
III Traffic Flow Problem 57
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.
III Traffic Flow Problem 58
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.
III Traffic Flow Problem 59
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.
III Traffic Flow Problem 60
• 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)
III Traffic Flow Problem 61
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.
III Traffic Flow Problem 62
• 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).
III Traffic Flow Problem 63
III Traffic Flow Problem 64
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)
III Traffic Flow Problem 65
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!
III Traffic Flow Problem 66
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.
III Traffic Flow Problem 67
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.
III Traffic Flow Problem 68
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.