# Delta Shocks and Vacuums to the Isentropic Euler Equations with the Flux Perturbation for van der Waals Gas

Authors: Jinhuan Wang; Yongbin Nie
Journal: Mathematical Problems in Engineering
 (2020)

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## Abstract

In this paper, we study the isentropic Euler equations with the flux perturbation for van der Waals gas, in which the density has both lower and upper bounds due to the introduction of the flux approximation and the molecular excluded volume. First, we solve the Riemann problem of this system and construct the Riemann solutions. Second, the formation mechanisms of delta shocks and vacuums are analyzed for the Riemann solutions as the pressure, the flux approximation, and the molecular excluded volume all vanish. Finally, some numerical simulations are demonstrated to verify the theoretical analysis.


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## Body

## 1. Introduction

In this paper, we consider the isentropic Euler equations with the flux perturbation:(1)ρt+ρ−2ϵux=0,ρut+ρ−ϵu2+Px=0,where u denotes the velocity, P denotes the pressure, and ρ denotes the density satisfying ρ≥2ϵ. Here, ϵ is a small positive perturbation parameter.When the flux perturbation and pressure both vanish, system (1) turns to be the transport equations:(2)ρt+ρux=0,ρut+ρu2x=0,which can describe the formation of large-scale structures in the universe [1, 2] and the motion of free particles which stick under collision [3]. We also call this system the zero-pressure gas dynamics. For more details, readers can see [4–6].In recent years, a lot of research work and achievement for the formation mechanisms of delta shocks and vacuums have been done by many scholars. Li [7] started the research on the isentropic Euler equations for perfect fluids in 2001, and then Chen and Liu [8, 9] made an in-depth study on isentropic and nonisentropic Euler equations for polytropic gas in 2003 and 2004. In [8, 9], formation mechanisms of delta shocks and vacuums were discussed as the polytropic gas pressure vanishes. Following that, the results were extended to various systems for different gas state equations in [10–16]. Furthermore, readers can also see [17] for the perturbed system of generalized pressureless gas dynamics model and [18] for the case of triangular conservation law system arising from “generalized pressureless gas dynamics model.” In addition, Yang and Liu [19] considered the limit behaviors of Riemann solutions of system (1) for polytropic gas as the flux approximation and the polytropic gas pressure both vanish. Flux perturbation can be regarded as the external shear force that causes the deformation of fluid particles and can be used to control some dynamic behaviors of fluid. Further, the flux approximation including the pressure perturbation portion is a more general physical consideration.Recently, Wang et al. [20] considered the isentropic Euler equations for van der Waals gas:(3)P=kρ1−bργ,where b denotes the molecular excluded volume satisfying 0≤bρ≪1, k denotes a positive constant, and γ denotes the adiabatic exponent with 1<γ<2. It is easy for us to find that when b=0, the state equation (3) just corresponds to the ideal gas. As mentioned in, at a high pressure or a low temperature, the behavior of real gas is not consistent with the ideal polytropic gas model but accords with the van der Waals gas one. Therefore, it is natural for us to explore the limit behavior of Riemann solutions of system (1) with the van der Waals gas (3) as the corresponding pressure vanishes.Compared to the works in [8, 19], the flux approximation and the molecular excluded volume are considered simultaneously in this paper, which makes the density has both lower and upper bounds. Further, we find that if only the flux approximation and the pressure tend to zero at the same time and the molecular excluded volume does not tend to zero, then the density always has an upper bound, so only vacuums may be generated, and delta shocks may not occur. This is the motivation for us to study the case where all of the pressure, the flux approximation, and the molecular excluded volume tend to zero simultaneously; that is, the triple parameters ϵ,k,b⟶0.In this paper, we rigorously prove that asϵ,k,b⟶0, any Riemann solution to the perturbation isentropic Euler equations (1) for van der Waals gas (3) containing two shocks converges to the delta shock solution of system (2) and any Riemann solution of equations (1) and (3) containing two rarefaction waves converges to the vacuum solution of system (2). By theory analysis, it is found that the introduction of the flux perturbation and the van der Waal gas pressure does not affect the formation of delta shocks and vacuums when the triple parameters ϵ,k,b⟶0 simultaneously, which means that our work can also be regarded as the extension of that in [8, 19].The remainder of this paper is arranged as follows. In Section2, we solve the Riemann problem of equations (1) and (3). In Section 3, the limit behaviors of Riemann solution of equations (1) and (3) are considered as the flux perturbation, the van der Waals gas pressure, and the molecular excluded volume all vanish. In Section 4, some numerical results are shown to verify the theoretical analysis of the formation of delta shocks and vacuum states.For the details of delta shocks and vacuums for the transport equations (2), we will not repeat them here. Readers can refer to Section 2 in [8, 11].


## 2. Riemann Solutions to the Perturbation Euler Equations for van der Waals Gas

In this section, we solve the Riemann problem of the perturbation isentropic Euler equations (1) for the van der Waals gas (3) with Riemann initial data:(4)u,ρ0,x=u±,ρ±,where u±,ρ± are arbitrary constants and then we construct Riemann solutions of this system.For equations (1) and (3), the eigenvalues and corresponding right eigenvectors are(5)λ1=u−ρ−2ϵkγργ−21−bργ+1,r⟶1=ρ−2ϵkγργ−21−bργ+1,2ϵ−ρT,λ2=u+ρ−2ϵkγργ−21−bργ+1,r⟶2=ρ−2ϵkγργ−21−bργ+1,ρ−2ϵT,respectively. By direct calculation, we can obtain that ∇λl⋅rl⟶≠0l=1,2, which implies that both eigenvalues λll=1,2 are genuinely nonlinear. Thus, the elementary waves of this system contain rarefaction waves and shock waves.We look for the self-similar solutionu,ρξξ=x/t of equations (1) and (3) with (4) and obtain the two-point boundary value problem as follows:(6)−ξρξ+ρ−2ϵuξ=0,−ξρuξ+ρ−ϵu2+Pξ=0,P=kρ1−bργ,(7)u,ρ±∞=u±,ρ±.Now, we consider the smooth solution of (6) and get either the constant state solution u,ρξ=constant, or the backward rarefaction wave(8)R1:ξ=λ1=u−kγργ−2ρ−2ϵ1−bργ+1,ρ−2ϵdu+kγργ−2ρ−2ϵ1−bργ+1dρ=0,or the forward rarefaction wave(9)R2:ξ=λ2=u+kγργ−2ρ−2ϵ1−bργ+1,ρ−2ϵdu−kγργ−2ρ−2ϵ1−bργ+1dρ=0.By (8) and (9), we calculate that when ϵ is small enough,(10)dλ1dρ=−kγργ−3γ+1ρ−2ϵγ+4ϵ−6ϵbρ21−bργ+2kγργ−2ρ−2ϵ/1−bργ+1<0,(11)dλ2dρ=kγργ−3γ+1ρ−2ϵγ+4ϵ−6ϵbρ21−bργ+2kγργ−2ρ−2ϵ/1−bργ+1>0.From these two conditions, we find that the velocity of the backward rarefaction waveλ1 is monotonically decreasing with respect to ρ, while the forward one λ2 is increasing. Furthermore, integrating the second equations of (8) and (9), respectively, by λlu,ρ>λlu−,ρ−l=1,2 yields(12)R1:ξ=λ1=u−kγργ−2ρ−2ϵ1−bργ+1,u=u−−∫ρ−ρkγsγ−21−bsγ+1s−2ϵds,ρ<ρ−,(13)R2:ξ=λ2=u+kγργ−2ρ−2ϵ1−bργ+1,u=u−+∫ρ−ρkγsγ−21−bsγ+1s−2ϵds,ρ>ρ−.In theu,ρ-plane, we call the curve of the second equation of (12) the backward rarefaction wave curve, which is monotonically decreasing with respect to ρ by a direct calculation from the second equation of (8). Further, for this curve, we obtain that limρ⟶2ϵu=u−+∫2ϵρ−kγsγ−2/1−bsγ+1s−2ϵds. In fact, lims⟶2ϵs−2ϵ1/2kγsγ−2/1−bsγ+1s−2ϵ=kγ2ϵγ−2/1−2ϵbγ+1; then, the integral ∫2ϵρ−kγsγ−2/1−bsγ+1s−2ϵds is convergent according to Cauchy criterion. Therefore, this curve and the line ρ=2ϵ have an intersection point u1,2ϵ=u−+∫2ϵρ−kγsγ−2/1−bsγ+1s−2ϵds,2ϵ.Similarly, we call the curve of the second equation of (13) the forward rarefaction wave curve, which is monotonically increasing with respect to ρ from the second equation of (9). Since we have proved in [20] that limρ⟶1/b−∫ρ−ρkγsγ−2/1−bsγ+1sds⟶+∞ and it is obvious to find ∫ρ−ρkγsγ−2/1−bsγ+1sds<∫ρ−ρkγsγ−2/1−bsγ+1s−2ϵds, we obtain that limρ⟶1/b−u=+∞ from the second equation of (13).Now, we turn to a bounded discontinuity atξ=σ. For equations (1) and (3), the Rankine–Hugoniot compatibility conditions(14)σρ=ρ−2ϵu,σρu=ρ−ϵu2+kρ1−bργhold, where F=F−F− denotes the jump of function F across the discontinuity. By solving (14), in terms of the stability conditions(15)λ2u−,ρ−>λ1u−,ρ−>σ1,λ2u,ρ>σ1>λ1u,ρ,for λ1, and(16)λ2u−,ρ−>σ2>λ1u−,ρ−,σ2>λ2u,ρ>λ1u,ρ,for λ2, we obtain the backward shock S1 and the forward shock S2 as follows:(17)S1:σ1=u−−ρ−2ϵkρ1−bργ−ρ−1−bρ−γ1ρ−ρ−ρρ−−ϵρ+ρ−,u=u−−kρ1−bργ−ρ−1−bρ−γρ−ρ−ρρ−−ϵρ+ρ−,ρ>ρ−,(18)S2:σ2=u−+ρ−2ϵkρ1−bργ−ρ−1−bρ−γ1ρ−ρ−ρρ−−ϵρ+ρ−,u=u−−kρ1−bργ−ρ−1−bρ−γρ−ρ−ρρ−−ϵρ+ρ−,ρ<ρ−.In theu,ρ-plane, we call the curve of the second equations of (17) and (18) the backward shock curve (forward shock curve). From the second equation of (17), we can obtain(19)uρ=ρ−ρ/1−bργ−ρ−/1−bρ−γρ−−2ϵ+γργ−1ρ−ρ−/1−bργ+1ρρ−−ϵρ+ρ−−2k−1ρρ−−ϵρ+ρ−2kρ−ρ−/ρρ−−ϵρ+ρ−ρ/1−bργ−ρ−/1−bρ−γ<0,which means that the backward shock curve is monotonically decreasing. Similarly, we have uρ>0 for the forward shock wave, which implies that the forward shock curve is monotonically increasing. Further, a direct calculation gives that limρ⟶1/b−u=−∞ for the backward shock curve and limρ⟶2ϵ+u=u−−kρ−/1−bρ−γ−2ϵ/1−2ϵbγϵ−1 for the forward shock curve, which implies that this forward shock wave curve and the straight line ρ=2ϵ have an intersection point u2,2ϵ=u−−kρ−/1−bρ−γ−2ϵ/1−2ϵbγϵ−1,2ϵ.Based on the aforementioned analysis, we conclude that the elementary waves of equations (1) and (3) contain two rarefaction waves (R1 and R2) and two shock waves (S1 and S2). Using the curves of these elementary waves, we can divide the u,ρ-plane ρ≥2ϵ,0<bρ≪1 into four regions S1S2u−,ρ−, S1R2u−,ρ−, R1S2u−,ρ−, and R1R2u−,ρ− for any fixed left state u−,ρ−. Furthermore, for any fixed right state u+,ρ+, we can construct a unique Riemann solution, no matter which of the four regions the right state belongs to. Specifically, when u+,ρ+∈S1S2u−,ρ−, the Riemann solution can be constructed with two shocks (S1 and S2) besides a nonvacuum constant state in between. When u+,ρ+∈R1R2u−,ρ−, the Riemann solution can be constructed with two rarefaction waves (R1 and R2) besides an intermediate constant state which may be a constant-density solution (ρ=2ϵ). The discussion for the other two cases u+,ρ+∈R1S2u−,ρ− and u+,ρ+∈S1R2u−,ρ− is trivial, so in this paper we only consider the limit process for the cases u+,ρ+∈S1S2u−,ρ− and u+,ρ+∈R1R2u−,ρ−.


## 3. Formation of Delta Shocks and Vacuums asε,k,b⟶0

### 3.1. Formation of Delta Shocks

In this subsection, the formation of delta shocks is considered in the Riemann solutions of equations (1) and (3) in the case u+,ρ+∈S1S2u−,ρ− with u−>u+ as ϵ,k,b⟶0.


#### 3.1.1. Limit Behavior of the Riemann Solutions asϵ,k,b⟶0

For the caseu+,ρ+∈S1S2u−,ρ−, we assume that u∗Π,ρ∗Π is the intermediate state. Then the left and right states of the backward shock S1 (the forward shock S2) are u−,ρ− and u∗Π,ρ∗Π (u∗Π,ρ∗Π and u+,ρ+), respectively. Thus, we have(20)σ1Π=u−−ρ∗Π−2ϵk1ρ∗Πρ−−ϵρ∗Π+ρ−ρ∗Π−ρ−ρ∗Π1−bρ∗Πγ−ρ−1−bρ−γ,u∗Π=u−−kρ∗Π−ρ−ρ∗Πρ−−ϵρ∗Π+ρ−ρ∗Π1−bρ∗Πγ−ρ−1−bρ−γ,ρ∗Π>ρ−,for S1, and(21)σ2Π=u∗Π+ρ+−2ϵk1ρ∗Πρ+−ϵρ++ρ∗Πρ+−ρ∗Πρ+1−bρ+γ−ρ∗1−bρ∗Πγ,u+=u∗Π−kρ+−ρ∗Πρ∗Πρ+−ϵρ++ρ∗Πρ+1−bρ+γ−ρ∗Π1−bρ∗Πγ,ρ+<ρ∗Π,for S2. Here σiΠ (i=1,2) denotes the speed of Si.Now, we are ready to present our main results for the formation of delta shocks.Theorem 1.
Letu−>u+ and u+,ρ+∈S1S2u−,ρ−. For any fixed ϵ,k,b>0, suppose that uΠ,ρΠ is a Riemann solution of equations (1) and (3) with (4) containing two shocks S1 and S2. Then as ϵ,k,b⟶0, ρΠ and ρΠuΠ converge, in the sense of distributions, to the sums of a step function and a δ-measure with weights(22)1+σ2−1/2σρ−ρut,1+σ2−1/2σρu−ρu2t,respectively, which just form the delta shock solution of system (2) with (4).Before we prove this theorem, some lemmas should be presented. The proof process of these lemmas is similar to that in [10], so we do not repeat it here.Lemma 1.
limϵ,k,b⟶0ρ∗Π=+∞.Lemma 2.
limϵ,k,b⟶0kρ∗Π/1−bρ∗Πγ=ρ+ρ−u−−u+2/ρ++ρ−2.Lemma 3.
Setlimϵ,k,b⟶0u∗Π=σ; then limϵ,k,b⟶0u∗Π=limϵ,k,b⟶0σ1Π=limϵ,k,b⟶0σ2Π=σ∈u+,u−.Lemma 4.
(23)limϵ,k,b⟶0∫σ1Πσ2Πρ∗Πdξ=σρ−ρu,limϵ,k,b⟶0∫σ1Πσ2Πρ∗Πu∗Πdξ=σρu−ρu2.Lemma 5.
For above quantityσ, we can obtain(24)σ=ρ−+ρ+−1ρ−u−+ρ+u+.On the basis of the above results, we find that asϵ,k,b⟶0, the velocity of backward shock σ1Π, the velocity of forward shock σ2Π, and the intermediate velocity u∗Π of equations (1) and (3) tend to the same quantity σ, which is proposed for the delta shock solution of system (2), and ρ∗Π becomes singular simultaneously.


#### 3.1.2. Proof of Theorem 1

Now, we give the rigorous proof of Theorem 1.Proof of Theorem 1.Step 1.
Setξ=x/t. By the two-shock Riemann solution(25)uΠ,ρΠξ=u−,ρ−,ξ<σ1Π,u∗Πξ,ρ∗Πξ,σ1Π<ξ<σ2Π,u+,ρ+,ξ>σ2Π,
we can obtain the weak formulations(26)−∫−∞+∞ρΠuΠ−ξ−2ϵuΠϕ′dξ+∫−∞+∞ρΠϕdξ=0,(27)−∫−∞+∞ρΠuΠuΠ−ξ−ϵuΠ2ϕ′dξ−∫−∞+∞kρΠ1−bρΠγϕ′dξ+∫−∞+∞ρΠuΠϕdξ=0,
for any test functionϕ∈C01−∞,+∞.Step 2.
Here, we prove that the limit functions ofρΠuΠ and ρΠ are the sums of a step function and a δ-measure. We rewrite the first term on the left of (27) and obtain(28)∫−∞+∞ρΠuΠuΠ−ξ−ϵuΠ2ϕ′dξ=∫−∞σ1Π+∫σ1Πσ2Π+∫σ2Π+∞ρΠuΠuΠ−ξ−ϵuΠ2ϕ′dξ.
With the analysis of integration by part, noticing Lemmas3 and 4, we have(29)limϵ,k,b⟶0∫−∞σ1Π+∫σ2Π+∞ρΠuΠuΠ−ξ−ϵuΠ2ϕ′dξ=σρu−ρu2ϕσ+∫−∞+∞Hξ−σϕdξ,(30)limϵ,k,b⟶0∫σ1Πσ2ΠρΠuΠuΠ−ξ−ϵuΠ2ϕ′dξ=limϵ,k,b⟶0∫σ1Πσ2Πρ∗Πu∗Πu∗Π−ξ−ϵu∗Π2ϕ′dξ=limϵ,k,b⟶0ρ∗Πu∗Πσ2Π−σ1Πϕσ2Π−ϕσ1Πσ2Π−σ1Πu∗Π−σ2Πϕσ2Π−σ1Πϕσ1Πσ2Π−σ1Π+1σ2Π−σ1Π∫σ1Πσ2Πϕdξ−limϵ,k,b⟶0ϵu∗Π2ϕσ2Π−ϕσ1Π=σρu−ρu2σϕ′σ−σϕ′σ−ϕσ+ϕσ=0.
A combination of (29) and (30) gives that(31)limϵ,k,b⟶0∫−∞+∞ρΠuΠuΠ−ξ−ϵuΠ2ϕ′dξ=ϕσσρu−ρu2+∫−∞+∞Hξ−σϕdξ,where(32)Hξ−σ=ρ−u−,ξ<σ,ρ+u+,ξ>σ.
Similarly, we calculate the second integral on the left side of (27), noticing Lemmas 1–3, and obtain that(33)limϵ,k,b⟶0∫−∞+∞kρΠ1−bρΠγϕ′dξ=limϵ,k,b⟶0∫−∞σ1Π+∫σ1Πσ2Π+∫σ2Π+∞kρΠ1−bρΠγϕ′dξ=limϵ,k,b⟶0kρ−Π1−bρ−Πγϕσ1Π−ρ+Π1−bρ+Πγϕσ2Π+ρ∗Π1−bρ∗Πγϕσ2Π−ϕσ1Π=0.
Substituting (31) and (33) into (27), we have(34)limϵ,k,b⟶0∫−∞+∞ρΠuΠ−Hξ−σϕdξ=ϕσσρu−ρu2,for any ϕ∈C01−∞,+∞.
In the same way, from (26), we can get that(35)limϵ,k,b⟶0∫−∞+∞ρΠ−H˜ξ−σϕdξ=ϕσσρ−ρu,for any ϕ∈C01−∞,+∞, where(36)H˜ξ−σ=ρ−,ξ<σ,ρ+,ξ>σ.Step 3.
We turn to prove the weights of theδ-measures. For any test function ψ∈C0∞R+×R, by (34), we obtain(37)limϵ,k,b⟶0∫0+∞∫−∞+∞ψt,xρΠxtuΠxtdxdt=limϵ,k,b⟶0∫0+∞∫−∞+∞ψt,ξtρΠξuΠξdξtdt=∫0+∞∫−∞+∞ψt,ξtHξ−σdξtdt+∫0+∞ψt,σtσρu−ρu2tdt=∫0+∞∫−∞+∞ψt,xHx−σtdxdt+∫0+∞ψt,σtσρu−ρu2tdt,which together with definition (2.3) in [11] yields that(38)∫0+∞ψt,σtσρu−ρu2tdt=ψ⋅,⋅,w1⋅δS,where(39)w1t=1+σ2−1/2σρu−ρu2t.
Similarly, from (35), we obtain(40)limϵ,k,b⟶0∫0+∞∫−∞+∞ψt,xρΠxtdxdt=∫0+∞∫−∞+∞ψt,xH˜x−σtdxdt+∫0+∞ψt,σtσρ−ρutdt,and then we get that(41)∫0+∞ψt,σtσρ−ρutdt=ψ⋅,⋅,w2⋅δS,where(42)w2t=1+σ2−1/2σρ−ρut.
This completes the proof of Theorem 1.


### 3.2. Formation of Vacuums

In this subsection, the limit of Riemann solutions of equations (1) and (3) is considered in the case u+,ρ+∈R1R2u−,ρ− with u−<u+ as ϵ,k,b⟶0. For this case, we assume that u∗Π,ρ∗Π is the intermediate state. Then the left and right states of the backward rarefaction wave R1 (the forward rarefaction wave R2) are u−,ρ− and u∗Π,ρ∗Π (u∗Π,ρ∗Π and u+,ρ+), respectively. Then, for the backward rarefaction wave R1, we have(43)ξ=uΠ−kγρΠγ−2ρΠ−2ϵ1−bρΠγ+1,u−−kγρ−γ−2ρ−−2ϵ1−bρ−γ+1<ξ<u∗Π−kγρ∗Πγ−2ρ∗Π−2ϵ1−bρ∗Πγ+1,ρ−>ρ∗Π.For the forward rarefaction waveR2, we have(44)ξ=uΠ+kγρΠγ−2ρΠ−2ϵ1−bρΠγ+1,u∗Π+kγρ∗Πγ−2ρ∗Π−2ϵ1−bρ∗Πγ+1<ξ<u++kγρ+γ−2ρ+−2ϵ1−bρ+γ+1,ρ+>ρ∗Π.Now, we can get the following theorem.Theorem 2.
Letu−<u+ and u+,ρ+∈R1R2u−,ρ−. For any fixed ϵ,k,b>0, suppose that uΠ,ρΠ is a Riemann solution of equations (1) and (3) with (4) containing two rarefaction waves R1 and R2. Then, there exists a positive constant a, and the constant-density solution ρ=2ϵ occurs in the solution when 0<ϵ,k,b<a. As ϵ,k,b⟶0, R1 and R2 become two contact discontinuities, which connect the vacuum ρ=0 and constant states u±,ρ±.Proof.
Setk=b=ϵ=a. Since u∗Π,ρ∗Π is on the curve R1u−,ρ−, we obtain(45)u∗Π=u−−∫ρ−ρ∗Πaγsγ−21−asγ+1s−2ads≤u−+∫2aρ−aγsγ−21−asγ+1s−2ads≡Aa.
Whenu−<u+<Aa, the constant-density state does not appear in the solution, which means that there exists a positive constant a1 and when u−<u+<Aa1 the Riemann solution only contains two rarefaction waves (R1 and R2) and the two given constant states u±,ρ±. Meanwhile, when Aa<u+, the constant-density state occurs, which means that there exists a positive constant a2, and when Aa2<u+ the intermediate state between the backward rarefaction wave R1 and the forward rarefaction wave R2 is just a constant-density state.
We setga=∫2aρ−aγsγ−2/1−asγ+1s−2ads+u−−u+, a∈a2,a1. Since(46)∫2aρ−aγsγ−21−asγ+1s−2ads<aγ1−aρ−γ+1∫2aρ−sγ−2s−2ads=aγ1−aρ−γ+1∫0ρ−s+2aγ−2sds<a1γ1−a1ρ−γ+1∫0ρ−s+2a2γ−2sds,with ∫0ρ−s+2a2γ−2/sds being convergent according to Cauchy criterion, we can prove that the integral ∫2aρ−aγsγ−2/1−asγ+1s−2ads is uniformly convergent in a, and then the function ga is continuous with respect to a. In addition, it is easy to get that ga1ga2<0. Thus, there exists a0∈a2,a1, such that ga0=0.
Therefore, when0<ϵ,k,b<a0, the intermediate state u∗Π,ρ∗Π becomes a constant state with(47)u∗Π,ρ∗Πξ=ξ,2ϵ,u1Π≤ξ≤u2Π.
Here,(48)u1Π=u−+∫2ϵρ−kγsγ−21−bsγ+1s−2ϵds,u2Π=u+−∫2ϵρ+kγsγ−21−bsγ+1s−2ϵds.
Then, it is obvious to getlimϵ,k,b⟶0ρ∗Π=0. Furthermore, we get that(49)limϵ,k,b⟶0u1Π=u−,limϵ,k,b⟶0u2Π=u+,limϵ,k,b⟶0uΠξ=ξ,u−<ξ<u+.
Based on above analysis, we conclude that the limit solution for this case is a solution to transport equations (2) with (4), containing a vacuum state (ρ=0) and two contact discontinuities (ξ=x/t=u±) on either side. This completes the proof of Theorem 2.


## 3.1. Formation of Delta Shocks

In this subsection, the formation of delta shocks is considered in the Riemann solutions of equations (1) and (3) in the case u+,ρ+∈S1S2u−,ρ− with u−>u+ as ϵ,k,b⟶0.


### 3.1.1. Limit Behavior of the Riemann Solutions asϵ,k,b⟶0

For the caseu+,ρ+∈S1S2u−,ρ−, we assume that u∗Π,ρ∗Π is the intermediate state. Then the left and right states of the backward shock S1 (the forward shock S2) are u−,ρ− and u∗Π,ρ∗Π (u∗Π,ρ∗Π and u+,ρ+), respectively. Thus, we have(20)σ1Π=u−−ρ∗Π−2ϵk1ρ∗Πρ−−ϵρ∗Π+ρ−ρ∗Π−ρ−ρ∗Π1−bρ∗Πγ−ρ−1−bρ−γ,u∗Π=u−−kρ∗Π−ρ−ρ∗Πρ−−ϵρ∗Π+ρ−ρ∗Π1−bρ∗Πγ−ρ−1−bρ−γ,ρ∗Π>ρ−,for S1, and(21)σ2Π=u∗Π+ρ+−2ϵk1ρ∗Πρ+−ϵρ++ρ∗Πρ+−ρ∗Πρ+1−bρ+γ−ρ∗1−bρ∗Πγ,u+=u∗Π−kρ+−ρ∗Πρ∗Πρ+−ϵρ++ρ∗Πρ+1−bρ+γ−ρ∗Π1−bρ∗Πγ,ρ+<ρ∗Π,for S2. Here σiΠ (i=1,2) denotes the speed of Si.Now, we are ready to present our main results for the formation of delta shocks.Theorem 1.
Letu−>u+ and u+,ρ+∈S1S2u−,ρ−. For any fixed ϵ,k,b>0, suppose that uΠ,ρΠ is a Riemann solution of equations (1) and (3) with (4) containing two shocks S1 and S2. Then as ϵ,k,b⟶0, ρΠ and ρΠuΠ converge, in the sense of distributions, to the sums of a step function and a δ-measure with weights(22)1+σ2−1/2σρ−ρut,1+σ2−1/2σρu−ρu2t,respectively, which just form the delta shock solution of system (2) with (4).Before we prove this theorem, some lemmas should be presented. The proof process of these lemmas is similar to that in [10], so we do not repeat it here.Lemma 1.
limϵ,k,b⟶0ρ∗Π=+∞.Lemma 2.
limϵ,k,b⟶0kρ∗Π/1−bρ∗Πγ=ρ+ρ−u−−u+2/ρ++ρ−2.Lemma 3.
Setlimϵ,k,b⟶0u∗Π=σ; then limϵ,k,b⟶0u∗Π=limϵ,k,b⟶0σ1Π=limϵ,k,b⟶0σ2Π=σ∈u+,u−.Lemma 4.
(23)limϵ,k,b⟶0∫σ1Πσ2Πρ∗Πdξ=σρ−ρu,limϵ,k,b⟶0∫σ1Πσ2Πρ∗Πu∗Πdξ=σρu−ρu2.Lemma 5.
For above quantityσ, we can obtain(24)σ=ρ−+ρ+−1ρ−u−+ρ+u+.On the basis of the above results, we find that asϵ,k,b⟶0, the velocity of backward shock σ1Π, the velocity of forward shock σ2Π, and the intermediate velocity u∗Π of equations (1) and (3) tend to the same quantity σ, which is proposed for the delta shock solution of system (2), and ρ∗Π becomes singular simultaneously.


### 3.1.2. Proof of Theorem 1

Now, we give the rigorous proof of Theorem 1.Proof of Theorem 1.Step 1.
Setξ=x/t. By the two-shock Riemann solution(25)uΠ,ρΠξ=u−,ρ−,ξ<σ1Π,u∗Πξ,ρ∗Πξ,σ1Π<ξ<σ2Π,u+,ρ+,ξ>σ2Π,
we can obtain the weak formulations(26)−∫−∞+∞ρΠuΠ−ξ−2ϵuΠϕ′dξ+∫−∞+∞ρΠϕdξ=0,(27)−∫−∞+∞ρΠuΠuΠ−ξ−ϵuΠ2ϕ′dξ−∫−∞+∞kρΠ1−bρΠγϕ′dξ+∫−∞+∞ρΠuΠϕdξ=0,
for any test functionϕ∈C01−∞,+∞.Step 2.
Here, we prove that the limit functions ofρΠuΠ and ρΠ are the sums of a step function and a δ-measure. We rewrite the first term on the left of (27) and obtain(28)∫−∞+∞ρΠuΠuΠ−ξ−ϵuΠ2ϕ′dξ=∫−∞σ1Π+∫σ1Πσ2Π+∫σ2Π+∞ρΠuΠuΠ−ξ−ϵuΠ2ϕ′dξ.
With the analysis of integration by part, noticing Lemmas3 and 4, we have(29)limϵ,k,b⟶0∫−∞σ1Π+∫σ2Π+∞ρΠuΠuΠ−ξ−ϵuΠ2ϕ′dξ=σρu−ρu2ϕσ+∫−∞+∞Hξ−σϕdξ,(30)limϵ,k,b⟶0∫σ1Πσ2ΠρΠuΠuΠ−ξ−ϵuΠ2ϕ′dξ=limϵ,k,b⟶0∫σ1Πσ2Πρ∗Πu∗Πu∗Π−ξ−ϵu∗Π2ϕ′dξ=limϵ,k,b⟶0ρ∗Πu∗Πσ2Π−σ1Πϕσ2Π−ϕσ1Πσ2Π−σ1Πu∗Π−σ2Πϕσ2Π−σ1Πϕσ1Πσ2Π−σ1Π+1σ2Π−σ1Π∫σ1Πσ2Πϕdξ−limϵ,k,b⟶0ϵu∗Π2ϕσ2Π−ϕσ1Π=σρu−ρu2σϕ′σ−σϕ′σ−ϕσ+ϕσ=0.
A combination of (29) and (30) gives that(31)limϵ,k,b⟶0∫−∞+∞ρΠuΠuΠ−ξ−ϵuΠ2ϕ′dξ=ϕσσρu−ρu2+∫−∞+∞Hξ−σϕdξ,where(32)Hξ−σ=ρ−u−,ξ<σ,ρ+u+,ξ>σ.
Similarly, we calculate the second integral on the left side of (27), noticing Lemmas 1–3, and obtain that(33)limϵ,k,b⟶0∫−∞+∞kρΠ1−bρΠγϕ′dξ=limϵ,k,b⟶0∫−∞σ1Π+∫σ1Πσ2Π+∫σ2Π+∞kρΠ1−bρΠγϕ′dξ=limϵ,k,b⟶0kρ−Π1−bρ−Πγϕσ1Π−ρ+Π1−bρ+Πγϕσ2Π+ρ∗Π1−bρ∗Πγϕσ2Π−ϕσ1Π=0.
Substituting (31) and (33) into (27), we have(34)limϵ,k,b⟶0∫−∞+∞ρΠuΠ−Hξ−σϕdξ=ϕσσρu−ρu2,for any ϕ∈C01−∞,+∞.
In the same way, from (26), we can get that(35)limϵ,k,b⟶0∫−∞+∞ρΠ−H˜ξ−σϕdξ=ϕσσρ−ρu,for any ϕ∈C01−∞,+∞, where(36)H˜ξ−σ=ρ−,ξ<σ,ρ+,ξ>σ.Step 3.
We turn to prove the weights of theδ-measures. For any test function ψ∈C0∞R+×R, by (34), we obtain(37)limϵ,k,b⟶0∫0+∞∫−∞+∞ψt,xρΠxtuΠxtdxdt=limϵ,k,b⟶0∫0+∞∫−∞+∞ψt,ξtρΠξuΠξdξtdt=∫0+∞∫−∞+∞ψt,ξtHξ−σdξtdt+∫0+∞ψt,σtσρu−ρu2tdt=∫0+∞∫−∞+∞ψt,xHx−σtdxdt+∫0+∞ψt,σtσρu−ρu2tdt,which together with definition (2.3) in [11] yields that(38)∫0+∞ψt,σtσρu−ρu2tdt=ψ⋅,⋅,w1⋅δS,where(39)w1t=1+σ2−1/2σρu−ρu2t.
Similarly, from (35), we obtain(40)limϵ,k,b⟶0∫0+∞∫−∞+∞ψt,xρΠxtdxdt=∫0+∞∫−∞+∞ψt,xH˜x−σtdxdt+∫0+∞ψt,σtσρ−ρutdt,and then we get that(41)∫0+∞ψt,σtσρ−ρutdt=ψ⋅,⋅,w2⋅δS,where(42)w2t=1+σ2−1/2σρ−ρut.
This completes the proof of Theorem 1.


## 3.1.1. Limit Behavior of the Riemann Solutions asϵ,k,b⟶0

For the caseu+,ρ+∈S1S2u−,ρ−, we assume that u∗Π,ρ∗Π is the intermediate state. Then the left and right states of the backward shock S1 (the forward shock S2) are u−,ρ− and u∗Π,ρ∗Π (u∗Π,ρ∗Π and u+,ρ+), respectively. Thus, we have(20)σ1Π=u−−ρ∗Π−2ϵk1ρ∗Πρ−−ϵρ∗Π+ρ−ρ∗Π−ρ−ρ∗Π1−bρ∗Πγ−ρ−1−bρ−γ,u∗Π=u−−kρ∗Π−ρ−ρ∗Πρ−−ϵρ∗Π+ρ−ρ∗Π1−bρ∗Πγ−ρ−1−bρ−γ,ρ∗Π>ρ−,for S1, and(21)σ2Π=u∗Π+ρ+−2ϵk1ρ∗Πρ+−ϵρ++ρ∗Πρ+−ρ∗Πρ+1−bρ+γ−ρ∗1−bρ∗Πγ,u+=u∗Π−kρ+−ρ∗Πρ∗Πρ+−ϵρ++ρ∗Πρ+1−bρ+γ−ρ∗Π1−bρ∗Πγ,ρ+<ρ∗Π,for S2. Here σiΠ (i=1,2) denotes the speed of Si.Now, we are ready to present our main results for the formation of delta shocks.Theorem 1.
Letu−>u+ and u+,ρ+∈S1S2u−,ρ−. For any fixed ϵ,k,b>0, suppose that uΠ,ρΠ is a Riemann solution of equations (1) and (3) with (4) containing two shocks S1 and S2. Then as ϵ,k,b⟶0, ρΠ and ρΠuΠ converge, in the sense of distributions, to the sums of a step function and a δ-measure with weights(22)1+σ2−1/2σρ−ρut,1+σ2−1/2σρu−ρu2t,respectively, which just form the delta shock solution of system (2) with (4).Before we prove this theorem, some lemmas should be presented. The proof process of these lemmas is similar to that in [10], so we do not repeat it here.Lemma 1.
limϵ,k,b⟶0ρ∗Π=+∞.Lemma 2.
limϵ,k,b⟶0kρ∗Π/1−bρ∗Πγ=ρ+ρ−u−−u+2/ρ++ρ−2.Lemma 3.
Setlimϵ,k,b⟶0u∗Π=σ; then limϵ,k,b⟶0u∗Π=limϵ,k,b⟶0σ1Π=limϵ,k,b⟶0σ2Π=σ∈u+,u−.Lemma 4.
(23)limϵ,k,b⟶0∫σ1Πσ2Πρ∗Πdξ=σρ−ρu,limϵ,k,b⟶0∫σ1Πσ2Πρ∗Πu∗Πdξ=σρu−ρu2.Lemma 5.
For above quantityσ, we can obtain(24)σ=ρ−+ρ+−1ρ−u−+ρ+u+.On the basis of the above results, we find that asϵ,k,b⟶0, the velocity of backward shock σ1Π, the velocity of forward shock σ2Π, and the intermediate velocity u∗Π of equations (1) and (3) tend to the same quantity σ, which is proposed for the delta shock solution of system (2), and ρ∗Π becomes singular simultaneously.


## 3.1.2. Proof of Theorem 1

Now, we give the rigorous proof of Theorem 1.Proof of Theorem 1.Step 1.
Setξ=x/t. By the two-shock Riemann solution(25)uΠ,ρΠξ=u−,ρ−,ξ<σ1Π,u∗Πξ,ρ∗Πξ,σ1Π<ξ<σ2Π,u+,ρ+,ξ>σ2Π,
we can obtain the weak formulations(26)−∫−∞+∞ρΠuΠ−ξ−2ϵuΠϕ′dξ+∫−∞+∞ρΠϕdξ=0,(27)−∫−∞+∞ρΠuΠuΠ−ξ−ϵuΠ2ϕ′dξ−∫−∞+∞kρΠ1−bρΠγϕ′dξ+∫−∞+∞ρΠuΠϕdξ=0,
for any test functionϕ∈C01−∞,+∞.Step 2.
Here, we prove that the limit functions ofρΠuΠ and ρΠ are the sums of a step function and a δ-measure. We rewrite the first term on the left of (27) and obtain(28)∫−∞+∞ρΠuΠuΠ−ξ−ϵuΠ2ϕ′dξ=∫−∞σ1Π+∫σ1Πσ2Π+∫σ2Π+∞ρΠuΠuΠ−ξ−ϵuΠ2ϕ′dξ.
With the analysis of integration by part, noticing Lemmas3 and 4, we have(29)limϵ,k,b⟶0∫−∞σ1Π+∫σ2Π+∞ρΠuΠuΠ−ξ−ϵuΠ2ϕ′dξ=σρu−ρu2ϕσ+∫−∞+∞Hξ−σϕdξ,(30)limϵ,k,b⟶0∫σ1Πσ2ΠρΠuΠuΠ−ξ−ϵuΠ2ϕ′dξ=limϵ,k,b⟶0∫σ1Πσ2Πρ∗Πu∗Πu∗Π−ξ−ϵu∗Π2ϕ′dξ=limϵ,k,b⟶0ρ∗Πu∗Πσ2Π−σ1Πϕσ2Π−ϕσ1Πσ2Π−σ1Πu∗Π−σ2Πϕσ2Π−σ1Πϕσ1Πσ2Π−σ1Π+1σ2Π−σ1Π∫σ1Πσ2Πϕdξ−limϵ,k,b⟶0ϵu∗Π2ϕσ2Π−ϕσ1Π=σρu−ρu2σϕ′σ−σϕ′σ−ϕσ+ϕσ=0.
A combination of (29) and (30) gives that(31)limϵ,k,b⟶0∫−∞+∞ρΠuΠuΠ−ξ−ϵuΠ2ϕ′dξ=ϕσσρu−ρu2+∫−∞+∞Hξ−σϕdξ,where(32)Hξ−σ=ρ−u−,ξ<σ,ρ+u+,ξ>σ.
Similarly, we calculate the second integral on the left side of (27), noticing Lemmas 1–3, and obtain that(33)limϵ,k,b⟶0∫−∞+∞kρΠ1−bρΠγϕ′dξ=limϵ,k,b⟶0∫−∞σ1Π+∫σ1Πσ2Π+∫σ2Π+∞kρΠ1−bρΠγϕ′dξ=limϵ,k,b⟶0kρ−Π1−bρ−Πγϕσ1Π−ρ+Π1−bρ+Πγϕσ2Π+ρ∗Π1−bρ∗Πγϕσ2Π−ϕσ1Π=0.
Substituting (31) and (33) into (27), we have(34)limϵ,k,b⟶0∫−∞+∞ρΠuΠ−Hξ−σϕdξ=ϕσσρu−ρu2,for any ϕ∈C01−∞,+∞.
In the same way, from (26), we can get that(35)limϵ,k,b⟶0∫−∞+∞ρΠ−H˜ξ−σϕdξ=ϕσσρ−ρu,for any ϕ∈C01−∞,+∞, where(36)H˜ξ−σ=ρ−,ξ<σ,ρ+,ξ>σ.Step 3.
We turn to prove the weights of theδ-measures. For any test function ψ∈C0∞R+×R, by (34), we obtain(37)limϵ,k,b⟶0∫0+∞∫−∞+∞ψt,xρΠxtuΠxtdxdt=limϵ,k,b⟶0∫0+∞∫−∞+∞ψt,ξtρΠξuΠξdξtdt=∫0+∞∫−∞+∞ψt,ξtHξ−σdξtdt+∫0+∞ψt,σtσρu−ρu2tdt=∫0+∞∫−∞+∞ψt,xHx−σtdxdt+∫0+∞ψt,σtσρu−ρu2tdt,which together with definition (2.3) in [11] yields that(38)∫0+∞ψt,σtσρu−ρu2tdt=ψ⋅,⋅,w1⋅δS,where(39)w1t=1+σ2−1/2σρu−ρu2t.
Similarly, from (35), we obtain(40)limϵ,k,b⟶0∫0+∞∫−∞+∞ψt,xρΠxtdxdt=∫0+∞∫−∞+∞ψt,xH˜x−σtdxdt+∫0+∞ψt,σtσρ−ρutdt,and then we get that(41)∫0+∞ψt,σtσρ−ρutdt=ψ⋅,⋅,w2⋅δS,where(42)w2t=1+σ2−1/2σρ−ρut.
This completes the proof of Theorem 1.


## 3.2. Formation of Vacuums

In this subsection, the limit of Riemann solutions of equations (1) and (3) is considered in the case u+,ρ+∈R1R2u−,ρ− with u−<u+ as ϵ,k,b⟶0. For this case, we assume that u∗Π,ρ∗Π is the intermediate state. Then the left and right states of the backward rarefaction wave R1 (the forward rarefaction wave R2) are u−,ρ− and u∗Π,ρ∗Π (u∗Π,ρ∗Π and u+,ρ+), respectively. Then, for the backward rarefaction wave R1, we have(43)ξ=uΠ−kγρΠγ−2ρΠ−2ϵ1−bρΠγ+1,u−−kγρ−γ−2ρ−−2ϵ1−bρ−γ+1<ξ<u∗Π−kγρ∗Πγ−2ρ∗Π−2ϵ1−bρ∗Πγ+1,ρ−>ρ∗Π.For the forward rarefaction waveR2, we have(44)ξ=uΠ+kγρΠγ−2ρΠ−2ϵ1−bρΠγ+1,u∗Π+kγρ∗Πγ−2ρ∗Π−2ϵ1−bρ∗Πγ+1<ξ<u++kγρ+γ−2ρ+−2ϵ1−bρ+γ+1,ρ+>ρ∗Π.Now, we can get the following theorem.Theorem 2.
Letu−<u+ and u+,ρ+∈R1R2u−,ρ−. For any fixed ϵ,k,b>0, suppose that uΠ,ρΠ is a Riemann solution of equations (1) and (3) with (4) containing two rarefaction waves R1 and R2. Then, there exists a positive constant a, and the constant-density solution ρ=2ϵ occurs in the solution when 0<ϵ,k,b<a. As ϵ,k,b⟶0, R1 and R2 become two contact discontinuities, which connect the vacuum ρ=0 and constant states u±,ρ±.Proof.
Setk=b=ϵ=a. Since u∗Π,ρ∗Π is on the curve R1u−,ρ−, we obtain(45)u∗Π=u−−∫ρ−ρ∗Πaγsγ−21−asγ+1s−2ads≤u−+∫2aρ−aγsγ−21−asγ+1s−2ads≡Aa.
Whenu−<u+<Aa, the constant-density state does not appear in the solution, which means that there exists a positive constant a1 and when u−<u+<Aa1 the Riemann solution only contains two rarefaction waves (R1 and R2) and the two given constant states u±,ρ±. Meanwhile, when Aa<u+, the constant-density state occurs, which means that there exists a positive constant a2, and when Aa2<u+ the intermediate state between the backward rarefaction wave R1 and the forward rarefaction wave R2 is just a constant-density state.
We setga=∫2aρ−aγsγ−2/1−asγ+1s−2ads+u−−u+, a∈a2,a1. Since(46)∫2aρ−aγsγ−21−asγ+1s−2ads<aγ1−aρ−γ+1∫2aρ−sγ−2s−2ads=aγ1−aρ−γ+1∫0ρ−s+2aγ−2sds<a1γ1−a1ρ−γ+1∫0ρ−s+2a2γ−2sds,with ∫0ρ−s+2a2γ−2/sds being convergent according to Cauchy criterion, we can prove that the integral ∫2aρ−aγsγ−2/1−asγ+1s−2ads is uniformly convergent in a, and then the function ga is continuous with respect to a. In addition, it is easy to get that ga1ga2<0. Thus, there exists a0∈a2,a1, such that ga0=0.
Therefore, when0<ϵ,k,b<a0, the intermediate state u∗Π,ρ∗Π becomes a constant state with(47)u∗Π,ρ∗Πξ=ξ,2ϵ,u1Π≤ξ≤u2Π.
Here,(48)u1Π=u−+∫2ϵρ−kγsγ−21−bsγ+1s−2ϵds,u2Π=u+−∫2ϵρ+kγsγ−21−bsγ+1s−2ϵds.
Then, it is obvious to getlimϵ,k,b⟶0ρ∗Π=0. Furthermore, we get that(49)limϵ,k,b⟶0u1Π=u−,limϵ,k,b⟶0u2Π=u+,limϵ,k,b⟶0uΠξ=ξ,u−<ξ<u+.
Based on above analysis, we conclude that the limit solution for this case is a solution to transport equations (2) with (4), containing a vacuum state (ρ=0) and two contact discontinuities (ξ=x/t=u±) on either side. This completes the proof of Theorem 2.


## 4. Numerical Results

In this section, we simulate the formation process of delta shocks and vacuums. In order to discretize equations (1) and (3), we use the fifth-order weighted essentially nonoscillatory scheme and third-order Runge-Kutta method with 150×150 cells. For the sake of convenience, we take γ=1.4.To illustrate the formation of the delta shock, we take the following initial data:(50)ρ−,u−=3.0,2.0,ρ+,u+=2.0,0.5.Here, we begin withϵ=0.1, k=1, and b=0.05 and then choose ϵ=0.05, k=0.09, and b=0.001, and finally we choose ϵ=0.00001, k=0.00001, and b=0.00005. The corresponding numerical results at t=2.0 are listed in Figures 1–3, which present the formation process of a delta shock in the two-shock solution of equations (1) and (3) as the van der Waals gas pressure, the molecular excluded volume, and the flux approximation all vanish.Figure 1
(a) Density and (b) velocity of the delta shock forϵ=0.1, k=1, and b=0.05.
(a)(b)Figure 2
(a) Density and (b) velocity of the delta shock forϵ=0.05, k=0.09, and b=0.001.
(a)(b)Figure 3
(a) Density and (b) velocity of the delta shock forϵ=0.00001, k=0.00001, and b=0.00005.
(a)(b)From Figures1–3, it is easy to find that when the values of ϵ, k, and b become smaller and smaller, the locations of S1 and S2 get closer and closer, and the intermediate density ρ∗Π increases dramatically, and at the same time the velocity turns to be a step function. When ϵ, k, and b vanish, two shocks S1 and S2 coincide with each other and a delta shock wave develops, while the velocity remains a step function.For the formation of vacuum states, we take the following initial data:(51)ρ−,u−=1.0,0.5,ρ+,u+=2.0,2.0.In this case, we start withϵ=0.1, k=1, and b=0.05 and then choose ϵ=0.05, k=0.09, and b=0.001, and finally we choose ϵ=0.00001, k=0.00001, and b=0.00005. The corresponding numerical results at t=2.0 are shown in Figures 4–6, which describe the formation of a vacuum state in the Riemann solution containing two rarefaction waves (R1 and R2) and a nonvacuum intermediate state for equations (1) and (3) as the van der Waals gas pressure, the molecular excluded volume, and the flux approximation vanish.Figure 4
(a) Density and (b) velocity of the vacuum forϵ=0.1, k=1, and b=0.05.
(a)(b)Figure 5
(a) Density and (b) velocity of the vacuum forϵ=0.05, k=0.09, and b=0.001.
(a)(b)Figure 6
(a) Density and (b) velocity of the vacuum forϵ=0.00001, k=0.00001, and b=0.00005.
(a)(b)Figures4–6 show that as ϵ, k, and b decrease, the locations of R1 and R2 get closer and closer, and ρ∗Π tends to zero leading to an inside vacuum state as ϵ,k,b⟶0, and at the same time the velocity tends to a linear function. These numerical results illustrate that Riemann solution of equations (1) and (3) consisting of R1 and R2 converges to the vacuum solution of system (2) including a vacuum state ρ=0 as ϵ,k,b⟶0.To sum up, all of above numerical simulations completely support the theoretical analysis.



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