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Hyperbolic Balance Laws for Multilane Traffic Flow Model
Hsin-Yi Lee∗
Department of Mathematics National Central University
The 27th Annual Meeting on Differential Equations and Related Topics
January 27, 2019
∗Jointly with Shih-Wei Chou and John Hong
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Content
History: LWR mode ⇒ PW model ⇒ AR model
Goals: Multilane model (applying AR model)
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LWR model Lightihll, Whitham (1955) & Richards (1956) ρt+ (ρve(ρ))x = 0, where qe(ρ) = ρve(ρ).
(conservation law of vehicles)
We have two important results if traffic is in equilibrium.
1. (Anisotropy) ve′(ρ)≤ 0 ⇒ qe′(ρ) = ve(ρ) + ρve′(ρ)≤ ve(ρ).
2. (Acceleration) a = dv /dt =−ρx(ve′(ρ))2ρ
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Drawbacks of LWR model: Traffic is not in equilibrium 1. Instability at the vacuum：slow drivers in the light traffic 2. drivers’ behavior：a = vx(...), not a =−ρx(...)
cf.
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PW moel Payne (1971) & Whitham (1974)
ρt+ (ρv )x = 0,
vt+ vvx+A′eρ(ρ)ρx = ve(ρ)τ−v.
NOTE. Different choices on Ae(ρ), "pressure" of traffic 1. Payne A′e(ρ) = 2τ1 |ve′(ρ)|
2. Whitham A′e(ρ) = D
3. Zhang (1998): A′e(ρ) = (ρve′(ρ))2 4. Others ...
But ...
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PW model
ρt+ (ρv )x = 0,
vt+ vvx +A′eρ(ρ)ρx = ve(ρ)τ−v.
Drawbacks of PW model (Daganzo, 1995)
Violate anisotropy : v +√A′e(ρ) (wave) > v (vehicle) Diffusion: negative speed
Ref. Daganzo, Requiem (安魂曲) for second-order fluid approximation of traffic flow, Transportation Res., Part B, 1995.
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Fortunately, AR model overcomes these drawbacks Daganzo mentioned.
{ ρt+ (ρv )x = 0, αt+ v αx = 0, where α = v + cργ ( γ > 0).
This model obeys anisotopy,
instability at the vacuum, non-diffusion effect
reasonable drivers’ behavior
Ref. Aw and Rascle, Resurrection (復活，Risurrezione) of
"second" order models of traffic flow, SIAM, J. Appl. Math, 2000.
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A Multilane Model (Greenberg, Klar and Rascle, 2003) For an unidirectional one-dimensional road with n lanes, the macroscopic variables are the density ρ of vehicles and the average speed v across all the lanes. Then
ρ = Σi =ni =1ρi and ρv = Σi =ni =1ρivi (total flux), where ρi and vi are respectively the density and the average speed of vehicles in the i -th lane.
When traffic is high, lane changing and passing is difficult.
⇒ the equilibrium speed for vehicles is low.
When traffic is lower, these actions become easier.
⇒ the equilibrium speed for vehicles is higher.
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With the aid of Kerner’s three-phase traffic theory, Greenberg, Klar, and Rascle give the following figure.
For simplicity, we apply Sopasakis’ argument (2002): ρ1= ρ2.
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Multilane model (Greenberg, Klar, and Rascle, 2003)
ρt+ (m− cρ2)x = 0 mt+ (mρ(m− cρ2))x =
{ ρw
1(ρ)−(m−cρ2)
τ , ρ < ρ∗,
ρw2(ρ)−(m−cρ2)
τ , ρ≥ ρ∗, where m≡ ρv + cρ2 and ρ̸= 0.
Multilane model (balance laws) Ut+ F (U)x = G (U)
= AR model (conservation laws) Ut+ F (U)x = 0 + Kerner’s theory, 1998 (source term) G (U) Our goals are:
(1) to solve Riemann problem for the AR model
(2) to get an approximate solution for the multilane model.
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.Definition (Riemann’s problem) ..
...
The Riemann problem is the initial-value problem for the conservation laws
Ut+ F (U)x = 0 in R × (0, ∞) with the piecewise-constant initial data
U(x , 0) =
{ UL, x < 0, UR, x > 0.
We call uL and uR the left and right initial data, respectively.
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Since the flux F is a smooth function, the system we consider is of the form Ut+ DF (U)Ux = 0. More precisely,
[ ρ m
]
t
+
[ −2cρ2 1
−(mρ)2− mc 2mρ − cρ ] [
ρ m
]
x
= [
0 0
] .
this gives the distinct real eigenvalues of DF (U) λ1(U) = mρ − 2cρ < λ2(U) = mρ − cρ (= v) so that the system is strictly hyperbolic since ρ̸= 0.
Thus this model is anisotropic.
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Moreover, the corresponding eigenvectors can be taken as
r1(U) = −12c [ 1
m ρ
]
and r2(U) = −12c [ 1
m ρ + cρ
] . .Definition
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The kth characteristic field is said to be genuinely nonlinear if
∇λk(z)· rk(z)̸= 0 for all z.
The kth characteristic field is said to be linearly degenerate if
∇λk(z)· rk(z) = 0 for all z.
It follows from our system that
The first characteristic field is genuinely nonlinear since
∇λ1(U)· r1(U) = 1̸= 0.
The second characteristic field is linearly degenerate since
∇λ2(U)· r2(U) = 0.
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After some computations we have the three essential curves R1+(UL), S1−(UL) andC (UL).
to help us to solve the Riemann problem of conservation laws.
R1+(UL)={(ρ, m)T : mρL
L = mρ, ρ < ρL, m < mL}, The slope = mL/ρL = v + cρ = v0+ c· 0, where v0 is the vehicle speed at the vacuum.
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The three curves can be expressed in the v− ρ plane. That is, R1+(v , ρ)={(v, ρ)T : v = vL+ c(ρL− ρ), 0 < ρ < ρL}, S1−(v , ρ)={(v, ρ)T : v = vL+ c(ρL− ρ), ρL ≤ ρ ≤ ρmax}, C (v , ρ) ={(v, ρ)T : v = vL, 0 < ρ≤ ρmax}.
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LetD = {(v, ρ) : 0 ≤ v ≤ w1(ρ)), 0 < ρ≤ ρmax, 0≤ v ≤ vmax}.
Then it is the triangle region.
.Theorem (Aw and Rascle 2000) ..
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Given the left initial state UL. If the right initial state UR is in the regionD, then there exists a unique integral solution U of
Riemann’s problem, which is constant on lines through the origin.
NOTE. ThisD, the triangle region, is called the invariant region.
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But the Riemann’s problem obviously has no solution for such initial data in the following figure.
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Fortunately, extending their work, we can define the trapezoid
D(vL, ρL) =D ∩ {v : 0 ≤ v ≤ vL+ cρL}
as a new invariant region such that the Riemann’s problem has a unique solution for any (vR, ρR)∈ D(vL, ρL).
.
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In this case, uL1−→ uS 1B −→ uC 1R denoted by (1, 1, 1)S. .Example
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For (1, 1, 1)S we have
u(x , t) =
uL, x < st, s = λ1(uL)+λ2 1(uB), uB, st < x < σt,
uR, σt < x , σ = λ2(uB)+λ2 2(uR).
uB =( 1c((cρL+ vL)(cρL+ vL− vR)), 1c(cρL+ vL− vR) )
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In general, the horizontal line ρ = ρ∗ divides the invariant region D(uL) into two subregions, called Ω1 if ρ < ρ∗, and Ω2 if ρ∗ ≤ ρ.
Thus there are four cases that need to be considered in terms of (uLi, ujR)∈ (Ωi, Ωj), where i , j = 1, 2.
(1) For the case (uL1, uR1)∈ (Ω1, Ω1), there are three possibilities:
u1L−→ uR B1
−→ uC R1, u1L−→ uS B1
−→ uC R1, and uL1 −→ uS B2
−→ uC 1R, which are denoted by (1, 1, 1)R, (1, 1, 1)S, and (1, 2, 1)S, resp..
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(1) (uL1, u1R)∈ (Ω1, Ω1): (1, 1, 1)R, (1, 1, 1)S, (1, 2, 1)S
Analogous arguments can be applied to the others. That is,
(2) (uL1, u2R)∈ (Ω1, Ω2): (1, 1, 2)R, (1, 1, 2)S, (1, 2, 2)S
(3) (uL2, u1R)∈ (Ω2, Ω1): (2, 2, 1)R, (2, 1, 1)R, (2, 2, 1)S
(4) (uL2, u2R)∈ (Ω2, Ω2): (2, 2, 2)R, (2, 1, 2)R, (2, 2, 2)S
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In fact, there are twelve cases as follows. (R←→ S and 1 ←→ 2)
NOTE. We can get the geometrical information from the algebraic notations (i , j, k)R,S, and the converse is also true.
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Back to the balance laws (multiland model),
ρt+ (m− cρ2)x = 0 mt+ (mρ(m− cρ2))x =
ρw1(ρ)−(m−cρ2)
τ , ρ < R(m−cρρ 2),
ρw2(ρ)−(m−cρ2)
τ , ρ≥ R(m−cρρ 2), we consider
Uet+ F (U)e x = 0, U(x , 0) =e
{
UL, x < 0, UR, x > 0,
&
Ut+ F (U)x = G (U), U(x , 0) =
{
UL, x < 0, UR, x > 0.
Let U = U−U. Then U(x , 0) = 0 for any nonzero x .e
Our goal is to obtain an approximate solution of U and use it to get an approximate solution of U.
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It follows from Ut+ F (U)x = G (U) and U =U + U thate (U + U)e t+ F (U + U)e x = G (U + U).e
Doing linearization on F and G, and applying the operator-splitting method, the perturbation U is the solution of the initial-value problem of ODE.
{
Ut = G (U) + DG (e U)U,e U(x , 0) = 0.
To solve it, we need to introduce a new parameter µ(x ) which depends only on the location x .
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.Definition ..
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Fixed a location x , the parameter is defined to be the ratio µ(x ) = Tx/T ,
where Tx is the observed time when the solution U exists in Ωe 1, and T is the observed time when the solutionU exists in Ωe 1∪ Ω2.
NOTE. Obviously, the parameter is decreasing from 1 to 0, and increasing from 0 to 1.
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.Theorem (Approximate Solution) ..
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The approximate solution of the balance laws (multilane model) is {
ρ =ρ,e
v = θv + (1e − θ)w(µ,ρ),e
where θ = e−t/τ and w (µ,ρ)e ≡ µw1(ρ) + (1e − µ)w2(ρ).e
If τ → 0, then θ = e−t/τ → 0, and thus the above theorem shows that the approximate speed v approaches w (µ,ρ), i.e.,e
τlim→0v = lim
θ→0(1− θ)w(µ,ρ) + θe ev = w (µ,ρ).e
Clearly, this gives the asymptotic behavior as the relaxation time τ tends to zero.
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In particular, if we consider the AR model with a single source term
w1−v
τ , then our result on the asymptotic behavior is similar to the result of Liu.
w1(ρ)e ≥ev⇒ v ≥ve
limτ→0v = w1(ρ) , similar to the result of Liu. (1991)e
.
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For Solving the initial-value problem of the balance laws
Ut+ F (U)x = G (U), U(x , 0) = U0(x ).
,
we has three steps.
Step 1. Solving the Riemann problem for the conservation lawsUet+ F (U)e x = 0. (U =U + U)e
Step 2. Using the solutionU of Riemann problem to obtain ane approximate solution for the solution U of balance laws.
Step 3. Apply the approximate solutions to construct the building blocks for the generalized Glimm scheme to solve the initial-value problem for the balance laws.
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Thank you
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Thank you
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Thank you
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.Example (Shock Wave) ..
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Given UL, if UR lies in S1−, then U(x , t) =
{
UL, x < st, UR, x > st, where s = mρL
L − c(ρR + ρL).
NOTE. Shock speed s = λ1(UL)+λ2 1(UR).
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.Example (Rarefaction Wave) ..
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Given UL, if UR lies in R1+, then
U(x , t) =
UL, xt ≤ λ1(UL),
V (xt), λ1(UL)≤ xt ≤ λ2(UR), UR, xt ≥ λ2(UR).
NOTE. V (xt) = [ 1
2c(mρL
L − xt) 2c1 mρL
L(mρL
L −xt) ]T
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.Example (2-contact discontinuity) ..
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Given UL, if UR lies in S2(= R2), then U(x , t) =
{
UL, x < σt, UR, x > σt, where σ = λ2(UL) = λ2(UR).
NOTE. λ2(U) = mρ − cρ = v, the average speed of vehicles.
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Define ϵ1 =−2((mρLL − cρL)− (mρRR − ρR)) and ϵ2 = mρL
L − mρRR. Then the next result (UL→ U → UR) is an application of the previous theorem.
NOTE.
U = UL+ ϵ1r1(UL),
UR = U + ϵ2r1(U) +ϵ222Drk(U)· rk(U).
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For Riemann’s problem, we have in general .Theorem (local solution of Riemann problem) ..
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Assume each kth characteristic field is either genuinely nonlinear or linearly degenerate. Suppose further that the left initial state UL is given. Then for each right initial state UR sufficiently close to UL
there exists an integral solution U of Riemann’s problem, which is constant on lines through the origin.
To our system, [ ρ
m ]
t
+
[ −2cρ2 1
−(mρ)2− mc 2mρ − cρ ] [ ρ
m ]
x
= [ 0
0 ]
,
we can give a complete solution of Riemann’s problem for any two "reasonable" constant states (UL= (ρL, mL)T and
UR = (ρR, mR)T).
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To our system, we have the following theorem.
.Theorem ..
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Assume that each kth characteristic field is normalized, that is,
∇λk(U)· rk(U) = 1 and lk(U)· rk(U) = 1 for each k, and the left initial state UL is given. Then
(a) there exists a parametrization of R1+(UL)：
ϵ7→ UL+ ϵr1(UL), where ϵ = 2c(ρL− ρ) > 0, (b) there exists a parametrization of S1−(UL)：
ϵ7→ UL+ ϵr1(UL), where ϵ = 2c(ρL− ρ) < 0,
(c) there exists a parametrization of R2(UL) = S2(UL)：
ϵ7→ UL+ ϵr2(UL) +ϵ22Dr2(UL)· r2(UL), where ϵ = c(ρL− ρ).
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As a consequence of the previous result, we give a rigorous proof on the following theorem about instability.
.Theorem (Aw-Rascle, 2000) ..
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When one of Riemann data is near the vacuum, that is, the light traffic, if and only if the solution presents instabilities.
NOTE.
Near the vacuum (i.e., the density is very low), the solution of Riemann’s problem is very sensitive to the data.
For example, any driver in everyday life can observe that there is ahead very light traffic of very slow drivers.
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Ref. B.-C Huang, S.-W. Chou, J. M. Hong, and C.-C. Yen, Global transonic solutions of planetary atmospheres in a hydrodynamics region–hydrodynamic escape problem due to gravity and heat, SIAM, J. Math, Anal., Vol. 48. No. 6 (2016), pp. 4268–4310.
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.Theorem (Approximate Solution) ..
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The approximate solution of the balance laws (multilane model) is {
ρ =ρ,e
v = (1− θ)w(µ,ρ) + θe ev ,
where θ = e−t/τ and w (µ,ρ)e ≡ µw1(ρ) + (1e − µ)w2(ρ).e By the theorem, it is clear that v−v = (1e − θ)(w(µ,ρ)e −v ).e A necessary and sufficient condition for the inequality
v ≥ve to be true is that the following holds
ω(µ,ρ)e ≥ev . In addition, v =v if and only if ω(µ,e ρ) =e v .e
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Since we introduce the parameter µ, the convex combination speed w (µ,ρ) appears. So v is a convex combination ofe ev and w (µ,ρ).e In particular,
(1, 1, 1)R,S ⇒ µ = 1 ⇒ {
limτ→0v = w1(ρ),e w1(ρ)e ≥ve⇒ v ≥ev .
(2, 2, 2)R,S ⇒ µ = 0 ⇒ {
limτ→0v = w2(ρ),e see the following figure.
Hsin-Yi Lee. Hyperbolic Balance Laws for Multilane Traffic Flow Model January 27, 2019
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It follows from Ut+ F (U)x = G (U) and U =U + U thate (U + U)e t+ F (U + U)e x = G (U + U).e
Performing the Taylor series expansion on F and G atU yields thate F (U + U) = F (e U) + DF (e U)U + higher order terms,e and
G (U + U) = G (e U) + DG (e U)U + higher order termse respectively. Removing higher order terms from two expansions and putting the remaining terms into the balance laws yields
Uet+ Ut+ F (U)e x + (DF (U)U) = G (e U) + DG (e U)U.e This with the conservation lawsUet+ +F (U)e x = 0 gives
Ut+ (DF (U)U)e x = G (U) + DG (e U)U.e
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We now have
Ut+ (DF (U)U)e x = G (U) + DG (e U)U.e
Apply the operator-splitting method, i.e., the term ∂x(DF (U)U)e can be erased from the above equations:
Ut = G (U) + DG (e U)Ue
Hence we have the initial-value problem of ODE as follows.
{
Ut = G (U) + DG (e U)U,e U(x , 0) = 0.
We introduce a new parameter µ = µ(x ), a function depending only on the space x so that the initial-value problem of ODE can be solved easily.
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2. Diffusion
.Example (Traffic light on the red) ..
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Consider the stationary PW model with the initial condition ρ(x , 0) =
{
1, if− 1 ≤ x ≤ 0,
0, otherwise, and v (x , 0) = 0, and the boundary condition
ρ(0, t) = 1 and v (0, t) = 0 for all t ≥ 0.
The natural traffic evolution should be that nothing happens if the red light remains. However, there are some cars in the interval (−∞, −1). This shows that these cars travel with negative speed, which is completely unrealistic.
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Before introducing the multilane model, we mention that both of the PW model and AR model are derived from the car-following model.
On the other hand, recall the acceleration equation from the LWR model:
a = vx(−ρve′(ρ)) =−ρx(ve′(ρ))2ρ.
If we choose a =−ρx(ve′(ρ))2ρ, this gives the PW model, and if we choose a = vx(−ρve′(ρ)), this gives the AR model.
