5 Modelling Traffic Flow
5
1
Modelling Traffic Flow
5.1
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
∂z
and zy =
.
∂x
∂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:
P zx + Qzy = R
(1)
where
P = P (x, y, z), Q = Q(x, y, z), R = R(x, y, z)
are given continuous functions.
(2)
5 Modelling Traffic Flow
2
• Consider the family of curves in the (x, y) plane satisfying
dy Q
dx P
dx dy
=
or
=
or
= .
dx P
dy Q
P
Q
Suppose z is known at a point A(x, y). There is one curve
ΓA of this family through A, and along ΓA
µ
dz = zxdx + zy dy =
using Eq. (1). Hence
dz
dx
=
¶
R
Q
zx + zy dx = dx
P
P
R
P
(3)
along ΓA, and so:
dx dy dz
=
= .
P
Q
R
(4)
Eqs. (4) are known as the associated equations for Eq. (1),
and are equivalent to Eq. (1). For let each term in Eq. (4)
be ds, so dx = P ds, dy = Qds, dz = Rds. Substitute
in Eq. (3) to get
Rds = P zxds + Qzy ds ⇒ P zx + Qzy = R,
i.e. Eq. (1).
5 Modelling Traffic Flow
3
y
G
+
C
A
GA
x
0
GFigure 1: 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
dy
is nowhere equal to Q
to C, i.e. provided that, on C, dx
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 Q
P
is independent of z the characteristics are independent of
the boundary conditions. In particular Q
P is independent
of z for a linear PDE. If P and Q are constants, then the
characteristics are parallel straight lines.
5 Modelling Traffic Flow
4
Example 1
Solve zx − zy = 1 with z = x2 on y = 0.
Solution
The associated equations are
dx
dy
dz
dy
dz
=
=
⇒
= −1,
= 1.
1
−1
1
dx
dx
Thus the characteristics are x + y = α and on the characteristics z − x = β.
y
x+y=a
C
0
Figure 2: Characteristics of Example 1
x
5 Modelling Traffic Flow
5
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.
¡
2
¢
Thus z = x + α − α on x + y = α.
¡
2
¢
Eliminate α to get z = x + (x + y) − (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
5 Modelling Traffic Flow
6
Example 2
Solve yzx + xzy = z with z = x3 on y = 0 and z = y 3
on x = 0.
Solution
The associated equations are
dx dy dz
=
=
⇒
y
x
z
dy x
= ⇒ x2 − y 2 = α
dx y
are characteristics, where α =const. Then
⇒
dz
dz
z
dx
⇒
=√ 2
=√ 2
(x − α)
(x − α)
dx
z
¡
¢¢
√¡ 2
ln z = ln x +
x − α + β0
¡
¢¢
√¡ 2
z = β x+
x −α
0
on a characteristic, where β = eβ =const.
2
2
¡
√¡
2
2
2
¢¢
α=x −y ⇒z = β x+
x −x +y
= β(x + y) or β(x − y).
5 Modelling Traffic Flow
7
Case 1: z = β(x + y)
In this case we find, that as in Ex 1 (B) above,
z
is constant when x2 − y 2 is constant.
x+y
¡ 2
¢
2
The GS is therefore z = (x + y)f x − y , and it remains to determine f .
y
a <0
B
C
x
A
0
a >0
a >0
a <0
a =0
a =0
Figure 3: 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.
5 Modelling Traffic Flow
8
Suppose α > 0.
1
2
Consider A(α1 , 0) at which
3
3
2
z = x = α1 .
So from the GS ⇒
3
2
1
2
α1 = α1 f (α1) ⇒ f (α1) = α1 ⇒ z = (x + y)(x2 − y 2)
for x2 > y 2.
Suppose α < 0.
1
Consider B(0, (−α2) 2 ) at which
3
z = y 3 = (−α2) 2 .
So from the GS ⇒
3
1
(−α2) 2 = (−α2) 2 )f (α2) ⇒ f (α2) = −α2
⇒ z = (x + y)(y 2 − x2) for x2 < y 2
In summary

 (x + y)(x2 − y 2) for x2 > y 2
z=
0
for x2 = y 2
¡
¢

(x + y) y 2 − x2 for x2 < y 2
(5)
5 Modelling Traffic Flow
9
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 − y 2).
However, this GS gives
¡
2
zx = g x − y
2
¢
0
¡
2
+ 2x(x − y)g x − y
2
¢
¡
¢
¡
¢
zy = −g x2 − y 2 − 2y(x − y)g 0 x2 − y 2
Therefore
yzx + xzy = −(x − y)g(x2 − y 2) = −z
which is not our original PDE, therefore we dismiss this
second case, z = β(x − y), as spurious solution.
5 Modelling Traffic Flow
5.2
10
Some properties of characteristics
• We begin by considering Eq. (5). It is clear that z is
everywhere continuous, and that zx, zy are everywhere
continuous except possibly on the lines
x = ±y
(x2 − y 2) = 0.
From Eq. (5) we find
x2 > y 2 : zx = (x2 − y 2) + 2x(x + y) = (x + y)(3x − y)
zy = (x2 − y 2) − 2y(x + y) = (x + y)(x − 3y)
x2 < y 2 : zx = (y 2 − x2) − 2x(x + y) = (x + y)(y − 3x)
zy = (y 2 − 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.
However, 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.
5 Modelling Traffic Flow
11
We investigate the possibility of discontinuities in zx and
zy for Eq.(1), but we shall suppose z is everywhere continuous. (The standard terminology is that we are looking
for weak discontinuities whereas discontinuities in z itself
are strong discontinuities).
Suppose C is approached from + and −. Then
y
+
dr =(dx,dy)
A
C
x
0
Figure 4: Jump across characteristics
δz
+
δz −
∂z +
∂z +
=
δx +
δy
∂x
∂y
∂z −
∂z −
=
δx +
δy
∂x
∂y
where (δx, δy) is along C. Subtract.
5 Modelling Traffic Flow
12
Because z is continuous, δz + = δz −. Hence
·
∂z
δx
∂x
¸+
·
∂z
+ δy
∂y
−
¸+
= 0.
(6)
−
where the square brackets denote the jump in the expression across C.
Since Eq. (1) is satisfied on both sides and since, by hypothesis, P , Q, R are continuous
·
∂z
P
∂x
¸+
·
∂z
+Q
∂y
−
¸+
= 0.
(7)
−
The necessary condition for
·
∂z
∂x
¸+
·
6= 0 and
−
∂z
∂y
¸+
6= 0 is
−
δx δy
= ,
P
Q
i.e.
dx dy
= .
P
Q
Thus C must be a characteristic ΓA.
(8)
5 Modelling Traffic Flow
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• 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.
D
A
BA
C
B
Figure 5: (i) z determined throughout rectangle; (ii) z not deter-
mined in BCD
We can also consider situations in which z itself is discontinuous 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:
5 Modelling Traffic Flow
14
• In (i) there appear to be two characteristics through
a point, whereas
• in (ii) there is a region containing no characteristics.
Figure 6: Characteristics leading to (i) shocks; or to (ii) centred
fans (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.
5 Modelling Traffic Flow
15
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) = 1 (x ≥ 1);
f (x) = x (0 ≤ x < 1),
Case 2 :
f (x) = 0 (x < 0), f (x) = −x (0 ≤ x < 1),
f (x) = −1 (x ≥ 1).
The associated equations Eq. (4) are
dt dx dρ
=
=
1
ρ
0
where the last is to be interpreted as dρ = 0.
Thus
ρ = α,
dx
= α ⇒ x − αt = β are characteristics.
dt
5 Modelling Traffic Flow
16
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)
(9)
t
slope
0
x
1
f(x)
x
Figure 7: Characteristics for Example
Alternatively, from Eq. (9)
x = ρt + ξ
so that we can write Eq. (9) implicitly as
ρ = f (x − ρt)
(10)
5 Modelling Traffic Flow
17
Case 1: From Eq. (10)
ρ = 0 (x < 0),
ρ = x − ρt (0 ≤ x − ρt < 1)
⇒
ρ=
x
(0 ≤ x < 1 + t),
1+t
ρ = 1 (x ≥ 1 + t),
i.e.

(x < 0)
0
ρ = x/(1 + t) (0 ≤ x < 1 + t)

1
(x ≥ 1 + t)
(11a)
Case 2: Likewise,

(x < 0)
0
ρ = −x/(1 − t) (0 ≤ x < 1 − t)

−1
(x ≥ 1 − t)
(11b)
The solution Eq. (11b) breaks down at t = 1; as the
sketch on the hand-out shows, t he characteristics intersect at t = 1 in Case 2 and the profile of ρ against x
becomes triple-valued (but this cannot occur in reality).
5 Modelling Traffic Flow
5.3
18
Model of traffic flow
We assume:
1. One lane of traffic in direction of Ox with no overtaking.
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)
(12)
The meaning of Eq. (12) is that each driver adjusts his,
her or its speed to local conditions exclusively, whereas
most drivers look ahead and adjust speed where appropriate. These assumptions give a car flowrate q(ρ) with
q(ρ) = ρv(ρ)
(13)
5 Modelling Traffic Flow
q(x,t)
q(x+dx,t)
r (x,t)
x
t
19
x+dx
r (x,t+dt)
x
t+dt
x+dx
Figure 8: 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
Z x2
ρ(x, t)dx.
x1
The rate of change of this must be the net flowrate, viz.
½ Z x2
¾
∂
ρ(x, t)dx = [q(x, t)]xx12
(14)
∂t
x1
5 Modelling Traffic Flow
20
If x1 = x, x2 = x + δx, Eq. (14) becomes
⇒
∂
∂q
ρδx = − δx
∂t
∂x
∂ρ ∂q
+
= 0.
∂t ∂x
(15)
• 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. (12) is zero.
We also assume v(ρ) decreases as ρ increases, with a maximum of V when ρ = 0.
These assumptions are shown schematically...
5 Modelling Traffic Flow
21
v(r)
V
1
0
q(r)
VP
1
r
P
0
a
1
r
P
Figure 9: Modelling car flow
With Eq. (13), Eq. (15) becomes
d
∂ρ
∂ρ
+
(ρv(ρ))
=0
∂t dρ
∂x
or
∂ρ
∂ρ
+ c(ρ)
= 0,
∂t
∂x
c(ρ) =
d
(ρv(ρ)) = v(ρ) + ρv 0(ρ)
dρ
(16)
5 Modelling Traffic Flow
22
c(r)
V
1
0
a
1
r
P
Figure 10: 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.
5 Modelling Traffic Flow
5.4
23
Small amplitude disturbances from a uniform state
• Before studying the full non-linear problem, it is instructive to consider a simpler one. Suppose that there
is almost a uniform state with ρ = ρ0 and
ρ = ρ0 + ρ0 with |ρ0| ¿ ρ0.
(17)
Linearise Eq. (16) - as with sound waves earlier - to get
Either
∂ρ0
∂ρ0
+ c (ρ0)
= 0.
∂t
∂x
(18)
dt
dx
dρ0
=
=
⇒
1
c (ρ0)
0
ρ0 = const. on x − c(ρ0)t = const.
Or put ξ = x − c (ρ0) t ⇒
µ 0¶
µ 0¶
µ 0¶
∂ρ
∂ρ
∂ρ
∂ρ0 ∂ρ0
=
,
=
− c (ρ0)
⇒
∂x
∂ξ
∂t x
∂t ξ
∂ξ
µ 0¶
∂ρ
= 0.
∂t ξ
Thus the GS of Eq. (18) is
ρ0 = f {x − c (ρ0) t}.
(19)
5 Modelling Traffic Flow
24
• The characteristics of Eq. (18) are the straight lines
t
x
0
c(r0 ) < 0
x
Figure 11: Characteristics of car flow are straight lines
x = ξ + c (ρ0) t.
(20)
Eq. (19) shows that ρ0 is constant on each characteristic.
Eq. (19) 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 disturbance, 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.
5 Modelling Traffic Flow
5.5
25
The initial value problem for Eq. (16)
• We wish to solve Eq. (16) subject to the initial condition
ρ(x, 0) = f (x)
(21)
By the earlier methods - see especially the Example in
§ (5.2) - ρ is constant on the characteristics
dt
dx
dx
=
⇒
= c(ρ).
1
c(ρ)
dt
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. (22)
5 Modelling Traffic Flow
26
Example
Suppose
V
(P − ρ)
P
and that ρ(x, 0) = f (x) satisfies
(23)
v(ρ) =
1
1
x
(ρL + ρR ) − (ρL − ρR ) tanh
(24)
2
2
L
where ρL, ρR and L are constants. Discuss the solution
given by Eq. (22) when
ρ=
(i) ρL > ρR ,
and (ii ) ρL < ρR .
Solution
From flow model Eq. (23) it follows
v
V
1
0
1
r
P
Figure 12: (a) Car flow v(ρ) = V (P − ρ)/P
5 Modelling Traffic Flow
27
¢
V ¡
2
q(ρ) =
Pρ − ρ ,
P
Note also that
c(ρ) =
V
(P − 2ρ).
P
(25)
c
V
1
q
PV
1
4
0
0
1
2
1
1
2
1
r
P
r
P 1
Figure 12: (b) Car flow flux, and (c) speed of disturbance
ρ → ρL as
x
→ −∞
L
ρ → ρR as
x
→ +∞.
L
and
Also
F (ξ) = c {f (ξ)}
(26)
·
µ ¶¸
ξ
V
P − ρL − ρR + (ρL − ρR ) tanh
=
P
L
5 Modelling Traffic Flow
28
Case (i): ρL > ρR ⇒ F 0(ξ) > 0 for ∀ξ
c
c( rR )
r
x0
rL
0
rR
c( rL )
x
0
x
rL+ rR > P
rL > 1 P > rR
2
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
5 Modelling Traffic Flow
29
Case (ii): ρL < ρR ⇒ F 0(ξ) < 0 for ∀ξ
c
c( rL )
r
rL+ rR > P
rR > 1 P > rL
2
rR
x0
x
0
rL
0
c( rR )
x
t
x0
x
Figure 14: (a) Car flow, (b) profile of speed of disturbance, and (c)
characteristics for Case (ii). Note, from (c) that characteristics
eventually intersect
Characteristics eventually intersect ⇒ problem becomes
ill-posed.
5 Modelling Traffic Flow
30
If two characteristics intersect, any enclosed characteristic must meet one of them at an earlier time ⇒ earliest
intersection must be between neighbouring characteristics.
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 0(ξ)t} ∂ξ




∴ 1 + F 0(ξ)t = 0.
⇒
(27)
We get solutions of Eq. (27) with t > 0 only if ∃ ξ with
F 0(ξ) < 0. [Thus for ρL > ρR there are no intersections
and the solution given by Eq. (22) applies for ∀ t ≥ 0.]
5 Modelling Traffic Flow
31
The first positive t satisfying Eq. (27) occurs when
t = Tmin =
1
.
Max {−F 0(ξ)}
(28)
−∞<ξ<∞
While Eqs. (27) and (28) are general, we can calculate
Tmin in our particular case when Eq. (26) holds.
We find
V
1
−F 0(ξ) = (ρR − ρL) sech2
P
L
µ ¶
ξ
L
⇒
max {−F 0(ξ)} =
V (ρR − ρL)
PL
when ξ = 0. Then Eq. (28) gives
Tmin =
PL
.
V (ρR − ρL)
(29)
5 Modelling Traffic Flow
5.6
32
Shocks
• We can understand in another way why there is trouble
when ρL < ρR . With c0(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 unacceptable since ρ must be a single valued quantity.
r
r
rR
t=t 1< Tmin
t=0
rL
x
0
r
r
t >Tmin
t= Tmin
0
x
0
x
0
x
Figure 16: Development of shock
• Instead the wave breaks, and the model must be extended. A consistent extension conserves cars but allows
discontinuities in ρ to occur across a shock.
5 Modelling Traffic Flow
33
We cannot have characteristics crossing one another. Instead the picture is as shown schematically.
t
shock
x
0
Figure 17: Shock front and characteristics
shock
r q
r q
1
2
1
2
x=s(t)
x1
x2
Figure 18: Quantities at a shock front
5 Modelling Traffic Flow
34
From Eq. (14) ⇒
∂
∂t
Z
x1
Z
LHS =
=
s(t)
∂
ρdx +
∂t
Z
x2
ρdx = q1 − q2
s(t)
(Z
s(t)
s(t+δt)
Z
s(t)
)
∂ρ
1
dx +
−
ρdx
∂t
∂t
x1
x1
| x1 {z }
→0 as x1 →s−
½ Z x2
Z x2 ¾
Z x2
1
∂ρ
dx +
−
ρdx
+
∂t
∂t
s(t+δt)
s(t)
| s(t){z }
→0 as x2 →s+
1
∂t
(Z
Z
s(t)+δt
ρ1dx −
s(t)
)
s(t)+δt
ρ2dx
s(t)
= ṡ (ρ1 − ρ2)
by the mean value theorem.
⇒
ṡ (ρ1 − ρ2) = (q1 − q2)
(30a)
q1 − q2
.
ρ1 − ρ2
(30b)
ṡ =
5 Modelling Traffic Flow
35
• The position of the shock is fixed by the need to conserve cars. Without a shock the curve of ρ against x
would become (unacceptably) triple-valued. We insert
the shockRso that the shaded areas are equal, thus ensuring that ρdx = number of cars is unchanged. This is
known as (Whitham’s) equal area rule.
r
r2
r1
0
s(t)
x
Figure 19: Shock fitting: Whitham’s equal area rule
• As an application consider what happens as cars approach 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.
5 Modelling Traffic Flow
5.7
36
The Riemann problem
• We wish to consider the case when we solve Eqs. (16)
and (21) where there is a discontinuity in f (x). It will be
sufficient to consider the simplest possible case, viz.

 ρL
ρ(x, 0) = f (x) =

(x < 0)
(31)
ρR
(x > 0)
• Then Eq. (22) gives the solution as
ρ = ρL on x = ξ + c (ρL) t (ξ < 0)
(32)
ρ = ρR on x = ξ + c (ρR ) t (ξ > 0)
Consider first the case ρL > ρR . The characteristic diagram is easy to draw...
5 Modelling Traffic Flow
37
t
B
A
r = rL
r = rR
x
0
Figure 20: Fan of characteristics
They are either parallel to OA with slope c (ρL); to the
left of OA, ρ = ρL. Or they are parallel to OB 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
on x = c(k)t.
The solution is therefore:
(ρR < ρ < ρL)
5 Modelling Traffic Flow







ρ=






ρL
:
38
x < c (ρL) t
k on x = c(k)t : c (ρL) < c(k) < c (ρR )
ρR
:
(33)
x > c (ρR ) t
x
Figure 21: Centered fan or expansion fan corresponding to rarefi-
cation wave
The characteristic diagram is augmented by a centred
fan or an expansion fan or expansion wave or rarefication
wave.
5 Modelling Traffic Flow
39
• 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. (30b)
as U , where
t
shock
0
x
Figure 22: Schematic shock with speed U
U=
q (ρL) − q (ρR )
(ρL − ρR )
(34)
5 Modelling Traffic Flow
5.8
40
Additional refinements
• The model assumptions leading to Eq. (16) 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
(35)
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. (35) in Eq. (15) gives
ρL + c(ρ)ρx = νρxx, c(ρ) = q 0(ρ)
(36)
• Seek solutions of Eq. (36) of the form
ρ = ρ(X)
X = x − Ut
(37)
where U is a constant still to be determined. Substitution
in Eq. (36) ⇒
−U ρ0(X) + c(ρ)ρ0(X) = νρ00(X).
Since c(ρ) = Q0(ρ) we have
5 Modelling Traffic Flow
41
Q(ρ) − U ρ + C = νρ0(X)
(38)
where C is a constant. Suppose ρ → ρL as X → −∞
and ρ → ρR as X → +∞ ⇒
Q (ρL) − U ρL + C = Q (ρR ) − U ρR + C = 0,
⇒
Q (ρR ) − Q (ρL)
(39)
ρR − ρL
This is exactly Eq. (30b) but (for the moment) in a different context.
U=
• Since ρ → ρL as X → −∞ and ρ → ρR as X → +∞,
ρ0(x) = 0 at ρ = ρL and ρ = ρR . We suppose ρL and ρR
are simple zero’s of
Q(ρ) − U ρ + C,
and more precisely we shall suppose ρL < ρR and
Q(ρ)−U ρ+C = α (ρ − ρL) (ρR − ρ)
(α > 0) (40)
5 Modelling Traffic Flow
42
With α > 0,
c(ρ) = Q0(ρ) = α (ρR − ρ) − α (ρ − ρL)
and
c0(ρ) = α (ρL − ρR ) < 0.
We can always approximate Q(ρ) by a quadratic. Then
Eq. (38) becomes
ν
dρ
= α (ρ − ρL) (ρR − ρ)
dX
with solution
µ
ρR − ρ
ρ − ρL
¶
µ
=
ρR − ρ0
ρ0 − ρL
¶
X
e− L ,
L=
ν
α (ρR − ρL)
(41)
where ρ = ρ0 at X = 0. We note that ρ → ρL as
X → −∞ and that ρ → ρR as X → +∞ as required.
5 Modelling Traffic Flow
43
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
rR
r0
rL
L
x
Figure 23: Development of shock front
• The model in this section can be taken further in the
case when Eq. (40) holds.
5 Modelling Traffic Flow
44
Multiply Eq. (36) by c0(ρ) ⇒
c0(ρ)ρt + c(ρ)c0(ρ)ρx = νc0(ρ)ρxx
∴
because
ct + ccx =
∂ 2c
ν ∂x
2
00
− νc (ρ)
³ ´2
∂ρ
∂x
∂c
∂ρ
= c0(ρ)
∂x
∂x
and therefore
µ ¶2
∂ 2c
∂ρ
∂ 2ρ
00
0
= c (ρ)
+ c (ρ) 2 .
∂x2
∂x
∂x
In the case when Q(ρ) is quadratic, i.e. Eq. (40) holds,
c00(ρ) = 0 since c(ρ) = Q0(ρ). Thus
ct + ccx = νcxx.
(42)
This is known as Burger’s equation and, remarkably, it
can be solved explicitly by means of the transformation
φx
c = −2ν
(43)
φ
discovered independently by E. Hopf (1950) and J.D.Cole
(1951). Use of Eq. (43) transforms Eq. (42) into the
standard linear equation (after one integration w.r.t. x):
φt = νφxx.
(44)
It can be shown that this is also consistent with the shock
structure.
5 Modelling Traffic Flow
45
• A second refinement is that there is a time lag in driver
response. One way of handling this is to take Eq. (35)
and deduce from it that v = q/ρ satisfies
ν
v = V (ρ) − ρx,
ρ
V (ρ) =
Q(ρ)
.
ρ
(45)
Then regard this as a velocity which the driver tries to
achieve. The acceleration of the car is
Dv ∂v
∂v
=
+v
Dt
∂t
∂x
[see Notes after § (4.3)] and the model is
¾
½
ν
1
v − V (ρ) + ρx ,
vt + vvx = −
τ
ρ
(46)
where τ is a measure of the response time. Eq. (46) is to
be solved together with Eq. (15), i.e.
ρt + (ρv)x = 0.
THE END
(47)