Traffic model examples
Problem 1. Consider a traffic model with umax = 1 and ρmax = 4. We
then
ρ
ρ
u = umax 1 −
=1− ,
ρmax
4
2ρ
ρ
0
and c(ρ) = q (ρ) = umax 1 −
=1−
q(ρ) = ρ u = ρ 1 −
4
ρmax
have
ρ
.
2
The density function ρ(x, t) satisfies
∂ρ ρ ∂ρ
+ 1−
= 0.
∂t
2 ∂x
Find the solution ρ(x, t) if
(a)


1, x < 0,
ρ(x, 0) = 2, 0 < x < 1,


3 x > 1.
(b)


1, x < 0,
ρ(x, 0) = 3, 0 < x < 1,


2 x > 1.
(c)


3, x < 0,
ρ(x, 0) = 2, 0 < x < 1,


1 x > 1.
Solution.
(a) We compute the characteristics x = x(t) starting at a point (x0 , 0) on the
x-axis. We have




1/2, x0 < 0,

x0 + t/2, x0 < 0,
ρ(x0 , 0)
x0 (t) = 1−
= 0, 0 < x0 < 1,
and thus x(t) = x0 , 0 < x0 < 1,


2


−1/2 x0 > 1.
x0 − t/2 x0 > 1.
1
2
If 0 < t < 1, each point in region 0 < x < t/2 contains belongs to 2
characteristics, one for which x0 < 0 and ρ(x0 , 0) = 1 = ρ− and one for
which 0 < x0 < 1 and ρ(x0 , 0) = 2 = ρ+ . This produced a curve of
discontinuity xs (t) satisfying
ρ+ + ρ−
q(ρ+ ) − q(ρ− )
1+2
=
u
1
−
x0s (t) =
=
1
−
= 1/4,
max
ρ+ − ρ−
ρmax
4
i.e. xs (t) = t/4. Also, If 0 < t < 1, each point in region 1 − t/2 <
x < 1 contains belongs to 2 characteristics, one for which 0 < x0 < 1 and
ρ(x0 , 0) = 2 = ρ− and one for which x0 > 1 and ρ(x0 , 0) = 3.
This produced a curve of discontinuity xp (t) satifying
ρ+ + ρ−
2+3
q(ρ+ ) − q(ρ− )
=
u
= −1/4,
1
−
=
1
−
x0p (t) =
max
ρ+ − ρ−
ρmax
4
and xp (t) = 1 − t/4. Since the curve of discontinuity xs (t) = t/4 and
xp (t) = 1 − t/4 intersect when t = 2 at x = 2, we can thus define the
solution for 0 < t < 2 as


1, x < t/4,
ρ(x, t) = 2, t/4 < x < 1 − t/4,


3 x > 1 − t/4.
At time t = 2, the solution is
(
1, x < 1/2,
ρ(x, 2) =
3 x > 1/2.
The characteristics originating at a point (x0 , 2) satisfy
ρ(x0 , 0)
x0 (t) = 1−
=
2
(
1/2, x0 < 1/2,
−1/2 x0 > 1/2.
(
and thus
x(t) =
x0 + (t − 2)/2, x0 < 1/2,
x0 − (t − 2)/2 x0 > 1/2.
and these two different types of characteristics intersect in the region t >
2 and 1/2 − (t − 2)/2 < x < 1/2 + (t − 2)/2 resulting in a shock-wave
discontinuity xs (t) satisfying
q(ρ+ ) − q(ρ− )
ρ+ + ρ−
1+3
x0s (t) =
=
u
1
−
=
1
−
= 0,
max
ρ+ − ρ−
ρmax
4
i.e. xs (t) = 1/2. Thus, for t > 2, the solution is
(
1, x < 1/2,
ρ(x, t) =
3 x > 1/2.
3
4
(b) We compute the characteristics x = x(t) starting at a point (x0 , 0) on the
x-axis. We have




1/2, x0 < 0,

x0 + t/2, x0 < 0,
ρ(x0 , 0)
0
x (t) = 1−
= −1/2, 0 < x0 < 1,
and thus x(t) = x0 − t/2, 0 < x0 < 1,


2


0, x0 > 1.
x0 , x0 > 1.
If 0 < t < 2, each point in region 1 − t/2 < x < t/2 contains belongs to
2 characteristics, one for which x0 < 0 and ρ(x0 , 0) = 1 = ρ− and one
for which 0 < x0 < 1 and ρ(x0 , 0) = 3 = ρ+ . This produced a curve of
discontinuity xs (t) satisfying
ρ+ + ρ−
1+3
q(ρ+ ) − q(ρ− )
=
u
1
−
=
1
−
= 0,
x0s (t) =
max
ρ+ − ρ−
ρmax
4
i.e. xs (t) = 0. Also, If 0 < t < 2, the region 1 − t/2 < x < 1 contains does
not contain any characteristic. We look for a solution of the form
ρ(x, t) = ψ((x − 1)/t),
for some unknown function ψ which is not constant. We have
−(x − 1)/t2 ψ 0 ((x − 1)/t) + (1 − ψ((x − 1)/t)/2) (1/t) ψ 0 ((x − 1)/t) = 0
or
(−(x − 1)/t + (1 − ψ((x − 1)/t)/2)) , ψ 0 ((x − 1)/t) = 0
and thus
ψ((x − 1)/t) = 2 (1 − (x − 1)/t)
For 0 < t < 2, we have thus

1, x < 0,



3, 0 < x < 1 − t/2,
ρ(x, t) =

2 (1 − (x − 1)/t), 1 − t/2 < x < 1,



2, x > 1.
For t > 2 no too large, the characteristics where ρ = 1 and those where
ρ(x, t) = 2 (1 − (x − 1)/t) intersect producing a curve of discontinuity xs (t)
satisfying
q(ρ+ ) − q(ρ− )
ρ+ + ρ−
1 + 2 (1 − (xs (t) − 1)/t)
0
xs (t) =
= umax 1 −
= 1−
.
ρ+ − ρ−
ρmax
4
We have
x0s (t) − xs (t)/(2t) = 1/4 − 1/(2t)
or
t−1/2 x0s (t) − xs (t) t−3/2 /2 = (t−1/2 xs (t))0 = (1/4) t−1/2 − t−3/2 /2,
5
i.e.
t−1/2 xs (t) = (1/2) t1/2 + t−1/2 + C
or
xs (t) = (1/2) t + 1 + C t1/2 .
√
Since xs (2) = 0, xs (t) = t/2 + 1 − 2 t1/2 . Note that xs (t) is increasing for
t > 2 and xs (8) = 1. For 2 < t < 8, we have thus

√ 1/2

1, x < t/2 + 1 − 2 t , √
ρ(x, t) = 2 (1 − (x − 1)/t), t/2 + 1 − 2 t1/2 < x < 1,


2, x > 1.
At time t = 8, the solution is
(
1, x < 1,
ρ(x, 2) =
2 x > 1.
For t > 8, the curve of discontinuity xs (t) satisfies
q(ρ+ ) − q(ρ− )
ρ+ + ρ−
1+2
1
x0s (t) =
=
u
1
−
=
1
−
= .
max
+
−
ρ −ρ
ρmax
4
4
and thus xs (t) = 1/4 (t − 8) + 1 for t > 8. For t > 8, we have thus
(
1, x < 1/4 (t − 8) + 1,
ρ(x, t) =
2, x > 1/4 (t − 8) + 1.
6
7
(c) We compute the characteristics x = x(t) starting at a point (x0 , 0) on the
x-axis. We have




−1/2, x0 < 0,

x0 − t/2, x0 < 0,
ρ(x0 , 0)
0
= 0, 0 < x0 < 1,
and thus x(t) = x0 , 0 < x0 < 1,
x (t) = 1−


2


1/2 x0 > 1.
x0 + t/2 x0 > 1.
Each point (x, t) in the region x < −t/2 belongs to exactly one characteristics originating from a point (x0 , 0) with x0 < 0, so ρ(x, t) = 3.
Each point (x, t) in the region 0 < x < 1 belongs to exactly one characteristics originating from a point (x0 , 0) with 0 < x0 < 1, so ρ(x, t) = 2.
Each point (x, t) in the region x > 1 + t/2 belongs to exactly one characteristics originating from a point (x0 , 0) with x0 > 1, so ρ(x, t) = 1.
There are 2 regions not covered by characteristics: the region where −t/2 <
x < 0 and the region where 1 < x < 1 + t/2.
For the region −t/2 < x < 0, we look for a solution which is constant on
lines passing through (0, 0), i.e. of the form ρ(x, t) = ψ(x/t). We have using
the PDE:
−x/t2 ψ 0 (x/t) + (1 − ψ(x/t)/2) (1/t) ψ 0 (x/t) = 0
or
(−x/t + (1 − ψ(x/t)/2)) ψ 0 ((x − 1)/t) = 0
and thus
ψ(x/t) = 2 (1 − x/t).
For the region 1 < x < 1 + t/2, we look for a solution which is constant on
lines passing through (1, 0), i.e. of the form ρ(x, t) = ψ((x − 1)/t). Exactly
as in part (b), we find that
ψ((x − 1)/t) = 2 (1 − (x − 1)/t).
The solution is thus


3, x < −t/2,





2 (1 − x/t), −t/2 < x < 0,
ρ(x, t) = 2, 0 < x < 1,



2 (1 − (x − 1)/t), 1 < x < 1 + t/2,



1, x > 1 + t/2.
8