# The Perturbed Riemann Problem for the Aw-Rascle Model with Modified Chaplygin Gas Pressure

Authors: Yujin Liu; Wenhua Sun
Journal: Advances in Mathematical Physics
 (2018)

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## Abstract

This paper is concerned with the perturbed Riemann problem for the Aw-Rascle model with the modified Chaplygin gas pressure. We obtain constructively the solutions when the initial values are three piecewise constant states. The global structure and the large-time asymptotic behaviors of the solutions are discussed case by case. Further, we obtain the stability of the corresponding Riemann solutions as the initial perturbed parameter tends to zero.


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## Body

## 1. Introduction

In the present paper we study the Aw-Rascle (AR) macroscopic model of traffic flow which is expressed by(1)ρt+ρux=0,ρu+pt+ρuu+px=0,where ρ≥0, u≥0 are the density and the velocity, respectively, and p is the velocity offset which is called the “pressure" inspired from gas dynamics. The AR model was established to remedy the deficiencies of second order models of car traffic presented by Daganzo [1] and was independently obtained by Zhang [2]. For the related work, we refer to [3–10] and the references cited therein.In [10], Wang obtained the Riemann solutions to (1) with(2)p=-B1ρα,0<α≤1,B1>0,which is called the generalized Chaplygin gas. Wang observed that the delta shock wave occurs in the solutions. Let B1=1 and α=1, we call it the Chaplygin gas which was introduced by Chaplygin [11]. Chaplygin gas (see [12–15]) is a good candidate for dark energy.In [16], we constructed the solutions of the elementary wave interactions for the generalized AR model (1) with (2) by resolving two initial value problems case by case and found that the Riemann solutions are globally stable.In [17], Wang, Liu, and Yang constructed the Riemann solutions for the AR model (1) with(3)p=B2ρ-B1ρα,B2>0,which is called the modified Chaplygin gas. The authors analyzed the limit behavior when the pressure tends to zero.In [18], Cheng and Yang studied the Riemann problem for (1) with(4)p=B2ρ-B1ρ,and they investigated the limits of the solutions when the pressure tends to the Chaplygin gas pressure.In our present paper, we study the wave interactions for (1) with the state equation (3) which can be regarded as a combination of the perfect fluid and Chaplygin gas. In order to construct the solutions of the all possible wave interactions, we consider (1) and (3) with the following three piecewise constants:(5)u,ρx,0=u-,ρ-,x<-ζ,um,ρm,-ζ<x<ζ,u+,ρ+,x>ζ,where the perturbation parameter ζ>0 is sufficiently small. The initial data (5) is a small perturbation on the corresponding Riemann initial values(6)u,ρx,0=u±,ρ±,±x>0,where u±,ρ±>0. We will face the interesting question that whether the solutions of the perturbed Riemann problem (1), (3), and (5) converge to the corresponding Riemann solutions of (1), (3), and (6) as ζ tends to zero.The paper is organized as follows. In Section2, for readers’ convenience we restate the results of the Riemann problem (1), (3), and (6). In Section 3, the elementary wave interactions are investigated case by case. And we obtain that the perturbed Riemann solutions of the modified AR model (1) and (3) converge to the corresponding Riemann solutions when ζ tends to zero which shows that the corresponding Riemann solutions of (1), (3), and (6) are globally stable.


## 2. Preliminaries

In this section we first sketch some results on the Riemann solutions to the modified AR model (1) and (3) and the detailed study can be found in [17].The AR model (1) has two eigenvalues: λ1=u-B2ρ-B1α/ρα, λ2=u, and the corresponding right eigenvectors are given by(7)r→1=-B2-B1αρ1+α,1T,r→2=0,1T.By a direct calculation, we have(8)∇λ1·r→1=-2B2-1-αB1αρ1+α<0,∇λ2·r→2≡0.Therefore, λ1 is genuinely nonlinear, and λ2 is always linearly degenerate. And we know that the Riemann solutions of (1), (3), and (6) can be constructed by shock wave, rarefaction wave, or the contact discontinuity connecting the two constant states (u-,ρ-) and (u+,ρ+).For the reason that system (1) and the initial values (6) are invariant under stretching of coordinates, i.e., (x,t)→(βx,βt) (β is constant), we look for the self-similar solution (u,ρ)(x,t)=(u,ρ)(η), η=x/t. Thus we get the following boundary value problem of the ordinary differential equations:(9)-ηρη+ρuη=0,-ηρu+B2ρ-B1ραη+ρuu+B2ρ-B1ραη=0,and (u,ρ)(±∞)=(u±,ρ±). By solving the above problem, we obtain the constant state solution (u,ρ)=constant, and the rarefaction wave solution(10)R:η=λ1=u-B2ρ-B1αρα,u=u--B2ρ+B1ρα+B2ρ--B1ρ-α,ρ<ρ-,u>u-.For a bounded discontinuity atη=σ, the Rankine-Hugoniot conditions are as follows:(11)-σρ+ρu=0,-σρu+B2ρ-B1ρα+ρuu+B2ρ-B1ρα=0,where [ρ]=ρr-ρl, ρl=ρ(σ-0), ρr=ρ(σ+0), etc.By solving (11) and using the Lax entropy inequalities, we get the shock wave solution(12)S:σ=u-B1ρα-B2ρ--B1ρ-1-α-ρ1-αρ-ρ-,u=u--B2ρ+B1ρα+B2ρ--B1ρ-α,ρ>ρ-,u<u-.Becauseλ2 is always linearly degenerate, from (11) we get the contact discontinuity(13)J:η=u=u-,ρ<ρ-orρ>ρ-.R, S, and J are called the elementary waves of (1). Notice that system (1) can be written by Yt+uYx=0 [19]; i.e., it is the Temple type; we can see that the shock curves coincide with the rarefaction curves in the phase plane (u,ρ) (Figure 1). This property simplifies the discussion of the wave interactions.Figure 1
Wave curves in the phase plane (u, ρ).Sinceuρ=-B2-B1α/ρα+1<0 and uρρ=αα+1B1/ρα+2>0, the above two wave curves are monotonic decreasing and convex. And we know that the shock wave curve interacts with ρ-axis, and the rarefaction wave curve has the u-axis as its asymptote.Based on the above analysis, we have the following result.Theorem 1.
There exists a unique Riemann solution for (1) and (4) with the initial data (6). When (u+,ρ+)∈ I or IV, the unique Riemann solution is R+J; when (u+,ρ+)∈ II or III, the unique Riemann solution is S+J.


## 3. Wave Interactions and Large-Time Behaviors

In this section, we consider the perturbed problem (1) and (3) with initial data (5). The data (5) is a small perturbation to the corresponding Riemann initial values (6). We want to determine whether the solutions of the perturbed Riemann problem (1), (3), and (5) converge to the corresponding Riemann solutions of (1), (3), and (6) as ζ tends to zero, where (uζ,ρζ)(x,t) are the solutions of (1), (3), and (5).In order to cover all the cases completely, we divide our discussions into the following four cases:u+<um<u-, u+>um>u-, um>u±, and um<u±.Case 1 (u+<um<u-).
In this case, we consider the interaction ofS+J emitted from (-ζ,0) and R+J emitted from (ζ,0) (Figure 2), where “+” means “followed by”. The occurrence of this case depends on the condition u+<um<u-.
The propagation speed ofJ1 and S2 is τ1=um=u1 and σ2=u2-B1/ρ2α--B2ρm-B1((ρm1-α-ρ21-α)/(ρ2-ρm)), respectively. Since σ2<u2=u+<um=τ1, we know that J1 will overtake S2 at a point (x0,t0) which is determined by(14)x0+ζ=umt0,x0-ζ=σ2t0.It follows that(15)x0,t0=um+σ2ζum-σ2,2ζum-σ2.At the interaction point (x0,t0), the new Riemann problem with the initial data (u1,ρ1) and (u2,ρ2) is formed, which is resolved by a new S3 and a new J. For the reason that S3 propagates with the same speed as that of S2, we obtain that the propagation direction of S2 is unchanged. From τ=τ2 we get that J has the same propagation speed as that of J2 which implies that J is parallel to J2.
Whent>t0, due to (16)σ1=u1-B1ρ1α-B2ρ--B1ρ-1-α-ρ11-αρ1-ρ-,σ3=u3-B1ρ3α-B2ρ1-B1ρ11-α-ρ31-αρ3-ρ1,and ρ-<ρ1<ρ3, by a direct computation it follows that σ1-σ3>0. Then we know S1 will overtake S3 at (x1,t1) which satisfies(17)x1+ζ=σ1t1,x1-x0=σ3t1-t0.From (17) we get (x1,t1)=(σ1x0-σ1σ3t0+σ3ζ)/(σ1-σ3),(x0+ζ-σ3t0)/(σ1-σ3). When t>t1, a new shock wave S will occur and its propagating speed is u3-B1/ρ3α-B2ρ--B1((ρ-1-α-ρ1-α)/(ρ3-ρ-))<u3, which yields that S cannot overtake J3 forever.
Whenζ tends to zero, we get that (x0,t0)→(0,0) and (x1,t1)→(0,0); J and J2 will coincide with each other. Thus, as the time t is large enough, the solution of the perturbed Riemann problem is S+J. For this case, the limit of the perturbed Riemann solution to (1), (3), and (5) is no other than the corresponding Riemann solution to the initial problem (1), (3), and (6) which implies the global stability of the Riemann solution to (1), (3), and (6).Figure 2
Wave interactions asu+<um<u-.Case 2 (u+>um>u-).
In this case, we investigate the interaction ofR+J emitted from (-ζ,0) and R+J emitted from (ζ,0) (Figure 3).
Due toτ1=um>λ2=um-B2ρm-B1α/ρmα, we know that J1 will overtake R2 at the point (x0,t0) which satisfies(18)x0+ζ=umt0,x0-ζ=λ2t0.A direct computation yields that(19)x0=um+λ2um-λ2ζ,t0=2ζum-λ2.
Whent>t0, J1 begins to penetrate R2, and during the penetration J1 satisfies(20)dxdt=u,x-ζ=u-B2ρ-B1αραt,u-um=-B2ρ+B1ρα+B2ρm-B1ρmα,xt0=x0,ρ2≤ρ≤ρm.
Differentiating with respect tot in the second equation of (20), we know that(21)dxdt=u-B2ρ-B1αρα+dudt-B2dρdt+B1α2ρ-α-1dρdtt.Differentiating with respect to t in the third equation of (20), we obtain that(22)dudt=-B2-B1αρ-α-1dρdt.Substituting the first equation of (20) into the above representations yields that(23)dρdt=B2ρ+B1α-2B2-B1αρ-α-1+B1α2ρ-α-1t<0,and(24)d2xdt2=dudt=B2ρ+B1αρα1t-B2-B1αρ-α-1-2B2-B1αρ-α-1+B1α2ρ-α-1>0.From the above discussions, we conclude that J1 accelerates during the process of penetration.
From (23) and the condition ρ2<ρ<ρm, we conclude that the contact discontinuity J1 can penetrate completely the rarefaction wave R2 in a finite time.Figure 3
Wave interactions asu+>um>u-.After the penetration, a newR and a new J will occur. Due to u1-B2ρ1-B1/ρ1=um-B2ρm-B1/ρm and u3-B2ρ3-B1/ρ3=u2-B2ρ2-B1/ρ2, the propagation direction of R keeps unchanged during the whole penetration. Notice that J is parallel to J2 since τ=τ2.Whenζ tends to zero, the points (x0,t0) and (x1,t1) tend to (0,0) and J coincides with J2. Since the propagation speed of the wave back in R equals that of the wave front in R1, it follows that R coincides with R1 when ζ tends to zero. Thus, we know for this case the result is R+J for large enough time. And the perturbed Riemann solution converges to the corresponding Riemann solution.Case 3 (um>u±).
In this case, we discuss the interaction ofR+J emitted from (-ζ,0) and S+J emitted from (ζ,0) (Figure 4).
By similar discussions to that in Case1, a new shock S and a new J will occur after the interaction of J1 with S1. Notice that S propagates with S1 at the same speed and J is parallel to J2 due to τ=τ2. J1 intersects with S1 at (x0,t0) which is determined by (14).
Because the propagation speed of the wave front inR1 is larger than the propagation speed in S, it follows that S will penetrate R1 at (x1,t1) which satisfies(25)x1+ζ=λ1t1,x1-x0=σt1-t0,where λ1=u1-B2ρ1-B1α/ρ1α and σ=u3-B1/ρ3α--B2ρ1-B1((ρ11-α-ρ31-α)/(ρ3-ρ1)) are, respectively, the propagation speed of the wave front in R1 and the propagation speed of S.
By a direct computation, we get(x1,t1)=(λ1x0-λ1σt0+σζ)/(λ1-σ),(x0+ζ-σt0)/(σ1-σ). When t>t1, S begins to penetrate R1 and we have(26)dxdt=u-B1ρα-B2ρ--B1ρ-1-α-ρ1-αρ-ρ-,x+ζ=u-B2ρ-B1αραt,u-u-=-B2ρ+B1ρα+B2ρ--B1ρ-α,xt1=x1,u-≤u≤u1,ρ1≤ρ≤ρ-.From the first and the second expression in (26) we know that(27)dudt+-B2+B1α2ρ-α-1dρdt=B2ρ-ρ-+B1α-1/ραt,and from the third expression we obtain that(28)dudt=-B2-B1αρ-α-1dρdt.Substituting (28) into (27), due to ρ<ρ- we have(29)dρdt=B2ρ-ρ-+B1α-1/ρα-2B2-B1αρ-α-1+B1α2ρ-α-1t>0.From (28)(30)dudt=-B2-B1αρ-α-1dρdt<0.From (26), (28), and (29), we have(31)d2xdt2=-B2-B11-αρ-αρ--ρρ-ρ-2-B1ρ1-α-ρ-1-αρ-ρ-2dρdt<0,telling us that S decelerates during the above penetration. From (29) we conclude that for the case u->u+, S will penetrate R1 completely due to ρ3>ρ- at the finite time. While for the other case u-≤u+, S cannot penetrate R1 forever because t→∞ when ρ→ρ3 and S has x+ζ=(u3-B2ρ3-B1α/ρ3α)t as its asymptote.Figure 4 (a)
Wave interactions asum>u->u+ (b)
Wave interactions asu-≤u+<umFor the large time, the solution isS+J (Figure 4(a)) as u+<u-, and the solution is R+J (Figure 4(b)) as u+≥u-. For this case, we obtain the stability of the Riemann solution to the initial problem (1), (3), and (6).Case 4 (um<u±).
In this case, we study the interaction ofS+J emitted from (-ζ,0) and R+J emitted from (ζ,0) (Figure 5).
By similar discussions to that of Case2, a new R and a new J will occur after the intersection of J1 and R1. Notice that R keeps the same propagation speed and J is parallel to J2 because τ=τ2.
Similarly to Case3, as u+<u-, we obtain that S1 will penetrate R completely in a finite time and a new shock will occur whose propagation speed is less than that of J. It implies that for the large time the solution is S+J (Figure 5(a)). As u+≥u-, S1 cannot penetrate R forever and it has x-ζ=(u--B2ρ--B1α/ρ-α)t as its asymptote, which shows that for the large time the solution is R+J (Figure 5(b)).
For this case, we get the global stability of the corresponding Riemann solution.Figure 5 (a)
Wave interactions asum<u+<u- (b)
Wave interactions asu+≥u->umNow we have discussed all possible wave interactions in the phase plane(u,ρ). Based on the above discussions, we obtain our main result.Theorem 2.
The limits of the perturbed Riemann solutions of (1), (3), and (5) are exactly the corresponding Riemann solutions of (1), (3), and (6). The asymptotic behavior of the perturbed Riemann solutions is governed completely by the states (u±,ρ±). The Riemann solutions of the initial value problem (1), (3), and (6) are stable under such small perturbation to the initial values.



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