The global solutions of the perturbed Riemann problem for the Leroux system are constructed explicitly under the suitable assumptions when the initial data are taken to be three piecewise constant states. The wave interaction problems are widely investigated during the process of constructing global solutions with the help of the geometrical structures of the shock and rarefaction curves in the phase plane. In addition, it is shown that the Riemann solutions are stable with respect to the specific small perturbations of the Riemann initial data.
National Natural Science Foundation of China1144100211271176Natural Science Foundation of Shandong ProvinceZR2014AM0241. Introduction
We are concerned with the Leroux system [1] which is represented in the form (1)ut+u2+ρx=0,ρt+ρux=0,in which u and ρ stand for the velocity and density, respectively. It was shown in [2] that system (1) can be derived as a hydrodynamic limit under Eulerian scaling for a two-component lattice gas. The Leroux system (1) has been widely used in various fields; for example, it may be utilized to investigate the stochastic dynamics in the stochastic particle system [3] and a deposition growth in the biological chemotaxis-mechanism model [4].
It is easy to see that the Leroux system (1) is strictly hyperbolic in the upper-half plane ρ≥0 provided that (u,ρ)≠(0,0) for both the eigenvalues of system (1) are distinct. In addition, the Leroux system (1) is genuinely nonlinear for it is so in both of the characteristic fields. The main feature of system (1) lies in the fact that the k-shock curves (k=1,2) coincide with the k-rarefaction curves in the (u,ρ) phase plane attributed to the special structure of system (1). Thus, system (1) belongs to the so-called Temple class [5]. In comparison with general systems of hyperbolic conservation laws, the well-posed result for the Temple class can be obtained for a much larger class of initial data [6, 7]. For the related works about the Leroux system (1), the entropy solutions were obtained in [8] when the initial data were taken in the form of the sum of Dirac measures and bounded variation functions. The global existence of weak solutions to the Cauchy problem for system (1) was obtained in [9] by constructing four families of Lax-type entropies and entropy fluxes and in [10] by using a new technique from the div-curl lemma in the compensated compactness theorem. In addition, the global bounded entropy solution of system (1) was also achieved in [11] for the bounded measurable initial data by combining the kinetic formulation and the compensated compactness method.
In this paper, we are interested in constructing the global solutions in a fully explicit form to the particular Cauchy problem for system (1) when the initial data are taken to be three piecewise constant states as(2)u,ρx,0=u-,ρ-,-∞<x<0,um,ρm,0<x<x0,u+,ρ+,x0<x<+∞,in which x0>0 is arbitrarily small. System (1) is of interest for the reason that it is one of the simplest, nonstrictly hyperbolic systems of Temple class. Some useful information for Temple class can be achieved by studying the particular Cauchy problem (1) and (2). This type of initial data (2) has been widely used such as in [12–14] to study the wave interaction problem [15–19] for different hyperbolic systems of conservation laws. It is worthwhile to notice that the initial data (2) may be regarded as the specific small perturbations of the corresponding Riemann initial data:(3)u,ρx,0=u-,ρ-,-∞<x<0,u+,ρ+,0<x<+∞.Thus, the particular Cauchy problem (1) and (2) is called as the perturbed Riemann problem (or the double Riemann problems) below.
To deal with the perturbed Riemann problem (1) and (2) it is essential to study various possible interactions of elementary waves for the Leroux system (1). We find that there exist 16 different combinations of Riemann solutions at the initial points (0,0) and (x0,0). It is clear that the interaction between the 2-wave starting from (0,0) and the 1-wave starting from (x0,0) happens at first which will play a critical role in the construction of solution to the perturbed Riemann problem (1) and (2). It is well known in [20] that only the existence result can be obtained by studying the Goursat problem and the global solution cannot be constructed in a completely explicit form for the perturbed Riemann problem (1) and (2) when the 2-rarefaction wave from (0,0) and the 1-rarefaction wave from (x0,0) occur. Thus, some assumptions should be taken to avoid the above situation. If the interaction between the 2-wave from 0,0 and the 1-wave from (x0,0) is completed, then the remaining problem is just to study the interaction between the waves of the same family which is easy to be dealt with for the Temple class. For simplicity, we restrict ourselves to considering the above situation; that is, the solution of the Riemann problem at the origin (0,0) is two shock waves S1+S2. In order to meet the above requirements, let us make the following assumption.
Assumption 1.
Assume that um<u- together with (4)max0,u-2+4ρ--u-um-u-2+ρ-<ρm<-u--u-2+4ρ-um-u-2+ρ-;then the solution of the Riemann problem originating from the origin (0,0) is two shock waves for the perturbed Riemann problem (1) and (2).
In fact, we want to investigate the perturbed Riemann problem (1) and (2) by fixing the Riemann solutions from (0,0) and then changing the Riemann solutions from (x0,0). That is to say, for the given left state (u-,ρ-), we first fix the intermediate state (um,ρm) and then change the right state (u+,ρ+). With the method described above, we have the following theorem to describe the main result of this paper.
Theorem 2.
If the initial data (2) satisfy Assumption 1, then the global solutions of the perturbed Riemann problem (1) and (2) can be constructed in a completely explicit form. Furthermore, the limits of the global solutions to the perturbed Riemann problem (1) and (2) are identical with the corresponding ones of the Riemann problem (1) and (3) when the limit x0→0 is taken.
The wave interaction problem for the Temple class has been widely investigated recently, such as for the pressureless gas dynamics equations [21], the isentropic Chaplygin gas dynamics equations [22, 23], the Aw-Rascle traffic flow model [14], and various types of chromatography systems [24–27]. It is worthwhile to notice that all the above systems are not genuinely nonlinear, in which at least one of the characteristic fields is linearly degenerate. Thus, the wave interaction problem for the above systems has relatively simple structures. Otherwise, both of the characteristic fields are genuinely nonlinear for the Leroux system (1), which is obviously different from the above systems, such that the wave interaction problem is more difficult to be dealt with. To our knowledge, the wave interaction problem for the system of Temple class with two genuinely nonlinear characteristic fields has not been paid attention to before.
We can investigate the wave interaction problem for system (1) in the explicit form by combining the method of characteristics together with the geometrical structures of the rarefaction and shock curves in the (u,ρ) phase plane when the initial data are taken to be three piecewise constant states (2). The global solutions of the perturbed Riemann problem (1) and (2) are constructed completely through obtaining the exact result of each interaction during the process of constructing the solutions under our assumption, which is able to reveal more properties of system (1) than those of the solutions to the Riemann problem (1) and (3). Furthermore, we can see that the solutions of the Riemann problem (1) and (3) are stable with respect to the small perturbations (2) of the Riemann initial data (3) if the limit x0→0 is taken.
The paper is organized in the following way. In Section 2, the Riemann solutions of (1) and (3) are restated for convenience. In Section 3, the wave interaction problems are investigated widely for system (1). Under the suitable assumptions, the global solutions of the perturbed Riemann problem (1) and (2) are constructed in explicit forms when the initial data are taken to be three piecewise constant states. In the end, the stability of Riemann solutions is analyzed with respect to the specific small perturbation of the Riemann initial data and the conclusion is drawn in Section 4.
2. The Riemann Problem for Leroux System (1) and (3)
In this section, we are dedicated to recalling some main results about the Riemann problem for the Leroux system (1) with the Riemann initial data (3), which has been extensively investigated, for example, in [1, 8]. We can also refer to such as [15, 28–31] for the general knowledge about the Riemann problem for hyperbolic systems of conservation laws.
By a simple calculation, it is shown that system (1) has two characteristic velocities (eigenvalues):(5)λ1=3u-u2+4ρ2,λ2=3u+u2+4ρ2.Thus, system (1) is strictly hyperbolic in the upper-half (u,ρ) phase plane (ρ≥0) provided that (u,ρ)≠(0,0). The corresponding right eigenvectors for system (1) are(6)r1→=1,-u-u2+4ρ2T,r2→=1,u2+4ρ-u2T.It is easy to get ∇λi·ri→=2≠0(i=1,2), in which ∇=(∂/∂u,∂/∂ρ). Thus, both of the characteristic fields are genuinely nonlinear provided that (u,ρ)≠(0,0), which implies that (1) is a genuinely nonlinear system except for the origin. The waves associated with the two characteristic fields will be either shock waves or rarefaction waves which are determined by the choice of initial data. The Riemann invariants along with the characteristic fields may be selected as(7)ω1=-u-u2+4ρ2,ω2=-u+u2+4ρ2.
For smooth solutions, by taking the self-similar transformation ξ=x/t, system (1) may be rewritten in the form(8)2u-ξ1ρu-ξuρ=00.
For the given left state (u-,ρ-), the two rarefaction wave curves can be expressed, respectively, by(9)R1u-,ρ-:ξ=λ1=3u-u2+4ρ2,ρ-ρ-=-u--u-2+4ρ-2u-u-,(10)R2u-,ρ-:ξ=λ2=3u+u2+4ρ2,ρ-ρ-=u-2+4ρ--u-2u-u-.It is clear to see that both of the rarefaction wave curves are straight lines in the (u,ρ) phase plane. By direct calculation, one finds that uξ=1/2>0 and ρu<0 for the 1-rarefaction wave and uξ=1/2>0 and ρu>0 for the 2-rarefaction wave. Thus, the 1-rarefaction wave is made up of the half-branch of R1(u-,ρ-) with u≥u- and ρ≤ρ-, while the 2-rarefaction wave is made up of the half-branch of R2(u-,ρ-) with u≥u- and ρ≥ρ-.
On the other hand, for discontinuous solutions, the Rankine-Hugoniot conditions at a discontinuous curve x=x(t) can be expressed as(11)-σu+u2+ρ=0,-σρ+ρu=0,where σ=dx/dt and u=ur-ul with ul=uxt-0,t and ur=uxt+0,t, and so forth, which implies that(12)ρr-ρlur-ul=-ul±ul2+4ρl2.Then, for the given left state u-,ρ-, the two shock wave curves can also be expressed, respectively, by(13)S1u-,ρ-:xt=σ1=u+u--u-2+4ρ-2,ρ-ρ-=-u--u-2+4ρ-2u-u-,(14)S2u-,ρ-:xt=σ2=u+u-+u-2+4ρ-2,ρ-ρ-=u-2+4ρ--u-2u-u-.It is remarkable that both of the shock wave curves are also straight lines in the (u,ρ) phase plane. One can see that the classical Lax entropy condition implies that ρ>ρ- and u<u- for the 1-shock wave curve and ρ<ρ- and u<u- for the 2-shock wave curve.
It is clear that the shock curves coincide with the rarefaction curves in the phase plane. Thus, system (1) belongs to the so-called Temple class [5]. If the first- and second-family wave curves are denoted by W1(u-,ρ-)=S1(u-,ρ-)∪R1(u-,ρ-) and W2(u-,ρ-)=S2(u-,ρ-)∪R2(u-,ρ-), respectively, then the lines of W1(u-,ρ-) and W2(u-,ρ-) are just tangent to the parabola {Γ:u2+4ρ=0} in the (u,ρ) phase plane. Furthermore, it can be seen that the slopes of the two tangent lines originating from (u-,ρ-) to the parabola {Γ:u2+4ρ=0} are exactly calculated by the corresponding Riemann invariants in formula (7). Let us draw Figure 1 in the (u,ρ) phase plane to illustrate the situation. For the given left state (u-,ρ-), in view of the right state (u+,ρ+) in the different regions, the solutions of the Riemann problem (1) and (3) can be summarized as follows: S1+S2 when (u+,ρ+)∈ I(u-,ρ-), S1+R2 when (u+,ρ+)∈ II(u-,ρ-), R1+S2 when (u+,ρ+)∈ III(u-,ρ-), and R1+R2 when (u+,ρ+)∈ IV(u-,ρ-).
For the given left state u-,ρ-, the half-upper (u,ρ) phase plane is shown for the Leroux system (1).
For simplicity, let us introduce the notations,(15)k1-=-u--u-2+4ρ-2,k2-=-u-+u-2+4ρ-2,to stand for the slopes of the two tangent lines originating from (u,ρ) to the right-hand and left-hand sides of the parabola {Γ:u2+4ρ=0} in the (u,ρ) phase plane.
For the Riemann problem (1) and (3), the intermediate state (u∗,ρ∗) is the one which can be connected with u-,ρ- on the left-hand side by 1-wave and can be connected with (u+,ρ+) on the right-hand side by 2-wave. No matter what the Riemann initial data (3) are taken, we always have the following relations:(16)k1-=-u--u-2+4ρ-2=-u∗-u∗2+4ρ∗2=k1∗,k2∗=-u∗+u∗2+4ρ∗2=-u++u+2+4ρ+2=k2+.The similar notations have also been adopted here and below and we do not give the detailed explanation any more without confusion. Thus, it can be seen that the intermediate state (u∗,ρ∗) can be calculated by (17)u∗,ρ∗=k1-u--k2+u++ρ+-ρ-k1--k2+,ρ-+k1-k2+u--u++ρ+-ρ-k1--k2+,for all kinds of Riemann initial data (3) due to the special form of system (1).
3. Construction of Global Solutions to the Perturbed Riemann Problem (1) and (2)
In this section, the main purpose is to construct the global solutions of the perturbed Riemann problem (1) and (2) under Assumption 1. In other words, we assume that the Riemann solution starting from the origin (0,0) is always two shock waves. In this situation, we will construct the global solutions of the perturbed Riemann problem (1) and (2) case by case by investigating the wave interaction problems in detail. If we put all the elementary wave curves across the states (u-,ρ-) and (um,ρm) together, then the upper-half (u,ρ) phase plane is divided into nine different regions shown in Figure 2. According to the different Riemann solutions originating from the other initial point (x0,0), our discussion may be divided into the following four cases.
If (um,ρm)∈ I(u-,ρ-), then the nine different regions in the half-upper (u,ρ) phase plane are shown for the perturbed Riemann problem (1) and (2).
Case 1 (S+S and S+S).
We first consider the situation that there are also two shock waves emitting from the other initial point (x0,0). Obviously, the occurrence of this case depends on the conditions u+<um and (18)max0,ρm+um2+4ρm-umu+-um2<ρ+<ρm+-um-um2+4ρmu+-um2.
For convenience, we use S3 and S4 to denote the two shock waves, respectively. In this case, the state (u+,ρ+) should locate in the region V1 in Figure 2 only. When the time t is small enough, the solution of the perturbed Riemann problem (1) and (2) may be represented succinctly as (see Figure 3) (19)u-,ρ-+S1+u1,ρ1+S2+um,ρm+S3+u2,ρ2+S4+u+,ρ+.It follows from (17) that the intermediate states (u1,ρ1) and (u2,ρ2) may be given, respectively, by (20)u1,ρ1=k1-u--k2mum+ρm-ρ-k1--k2m,ρ-+k1-k2mu--um+ρm-ρ-k1--k2m,(21)u2,ρ2=k1mum-k2+u++ρ+-ρmk1m-k2+,ρm+k1mk2+um-u++ρ+-ρmk1m-k2+,in which k1- and k2+ are given by (16) and k1m and k2m are calculated, respectively, by(22)k1m=-um-um2+4ρm2,k2m=-um+um2+4ρm2.It can be seen from Figure 3 that the relations u+<u2<um<u1<u-, ρ1>max(ρm,ρ-), and ρ2>max(ρm,ρ+) can be established directly.
The interaction between S+S and S+S is displayed when (u+,ρ+)∈V1.
Lemma 3.
The shock wave S2 collides with the shock wave S3 in finite time.
Proof.
The propagation speeds of S2 and S3 can be computed, respectively, by (23)σ2=um+u1+u12+4ρ12=um-k11,(24)σ3=u2+um-um2+4ρm2=u2-k2m.
It is obvious to get um>u2 and k11<0<k2m from Figure 3. Thus, we have (25)σ2-σ3=um-u2-k11+k2m>0,which implies that σ2>σ3. That is to say, the shock wave S2 collides with S3 in finite time. The intersection (x1,t1) is determined by (26)x1=σ2t1,x1-x0=σ3t1,which yields (27)x1,t1=um+u1+k21x0u1-u2+um+2k21,x0u1-u2+um+2k21.
The collision between the two shock waves S2 and S3 occurs at the point (x1,t1), where we again have a local Riemann problem with the initial data (u1,ρ1) and (u2,ρ2). In order to solve this problem, we must determine the relative position of (u2,ρ2) based on (u1,ρ1) in the (u,ρ) phase plane which can be described below.
Lemma 4.
The state (u2,ρ2) lies in region I with respect to the state (u1,ρ1). In other words, the solution of the local Riemann problem at the point (x1,t1) is also two shock waves.
Proof.
With u2<um<u1 and ρm<min(ρ1,ρ2) in mind, in order to show that the state (u2,ρ2) lies in region I with respect to the state (u1,ρ1) in the (u,ρ) phase plane, it is sufficient to show that the line S1(um,ρm) always lies below the line S1(u1,ρ1). It can be obtained from (13) that (28)S1u1,ρ1:ρ-ρ1=-u1-u12+4ρ12u-u1=k11u-u1,S1um,ρm:ρ-ρm=-um-um2+4ρm2u-um=k1mu-um.Noticing that u1>um, ρ1>ρm and k11<k1m<0, one can easily obtain that the inequality (29)k11u-u1+ρ1-k1mu-um-ρm>0holds for u<um<u1.
Therefore, it can be concluded from Section 2 that the solution of the Riemann problem with the initial data (u1,ρ1) and (u2,ρ2) at the point (x1,t1) is also two shock waves. Let us use S5 and S6 to denote them, respectively (see Figure 3). Analogously, the intermediate state (u3,ρ3) between S5 and S6 can also be obtained by (30)u3,ρ3=k11u1-k22u2+ρ2-ρ1k11-k22,ρ1+k11k22u1-u2+ρ2-ρ1k11-k22,in which (u1,ρ1) and (u2,ρ2) are given by (20) and (21), respectively. In addition, the relations k11=k1- and k22=k2+ can be established easily. Thus, the conclusion may be drawn.
More precisely, let us first compare the propagation speeds of S3 and S5 and then compare those of S2 and S6, respectively. In a word, we use the following lemma to state the results.
Lemma 5.
The inequalities σ3<σ5 and σ2>σ6 can be established for the corresponding propagation speeds of shock waves.
Proof.
The propagation speeds of S3 and S5 are calculated, respectively, by (24) and (31)σ5=u3+u1-u12+4ρ12=u3-k21.Taking into account u2<u3 and k2m=k21, we have (32)σ3-σ5=u2-u3<0.Thus, the shock wave S3 decelerates backward when it passes through S2.
On the other hand, the propagation speeds of S2 and S6 are calculated, respectively, by (23) and (33)σ6=u2+u3+u32+4ρ32=u2-k13.Noticing that u2<um and k11=k13, we also have (34)σ2-σ6=um-u2>0.Thus, the shock wave S2 decelerates forward when it passes through S3.
Now, we are in the position to consider the coalescence of two shock waves belonging to the same family which is shown below.
Lemma 6.
The two shock waves S1 and S5 belonging to the first family coalesce into a new shock wave of the first family. Similarly, the two shock waves S6 and S4 belonging to the second family coalesce into a new shock wave of the second family.
Proof.
We start with the interaction between S1 and S5. The propagation speed of S1 is (35)σ1=u1+u--u-2+4ρ-2=u1-k2-,which, together with (31), yields (36)σ1-σ5=u1-u3+k21-k2->0,in which u1>u3 and k21>k2->0 have been used. Hence, S1 catches up with S5 in finite time and the intersection (x2,t2) is determined by (37)x2=σ1t2,x2-x1=σ5t2-t1,which yields (38)x2,t2=u1+u-+k1-um-u3+k21-k11x0u1-u2+um+2k21u--u3,um-u3+k21-k11x0u1-u2+um+2k21u--u3.
It can be seen from the (u,ρ) phase plane in Figure 3 that the two states (u-,ρ-) and (u3,ρ3) can be connected by a shock wave of the first family directly. Thus, after this time t2, the two shock waves S1 and S5 coalesce into a new shock wave which is denoted by S7, whose propagation speed is(39)σ7=u3+u--u-2+4ρ-2=u3-k2-.It is easy to get σ1>σ7>σ5 from u1>u3 and k21>k2->0.
Then, we turn our attention to the interaction between S6 and S4. The propagation speed of S4 is (40)σ4=u++u2+u22+4ρ22=u+-k12.Thus, it follows from (33) and (40) that (41)σ6-σ4=u2-k13-u++k12>0,in which u2>u+ and k13<k12<0 have been used. As before, S6 catches up with S4 in finite time and the intersection (x3,t3) can be calculated by (42)x3,t3=x0+u++u2+k22u3+k23+k21x0u1-u2+um+2k21u3-u+,u3+k23+k21x0u1-u2+um+2k21u3-u+.It can also be seen from the (u,ρ) phase plane in Figure 3 that the two states (u3,ρ3) and (u+,ρ+) can be connected by a shock wave of the second family directly. Thus, after this time t3, the two shock waves S6 and S4 coalesce into a new shock wave which is denoted by S8, whose propagation speed is (43)σ8=u++u3+u32+4ρ32=u+-k13.In addition, it can also turn out that σ6>σ8>σ4 holds from u2>u+ and k13<k12<0. In other words, the propagation speed of S8 is between those of S4 and S6. The proof is completed.
Case 2 (S+S and S+R).
In this case, we turn our attention to the situation that the Riemann solution is a shock wave followed by a rarefaction wave starting from the initial point (x0,0). We use S3 and R4 to denote them, respectively (see Figure 4). On this occasion, the state (u+,ρ+) should lie in the region V2∪V3 in Figure 2, which should satisfy the conditions ρ+>ρm and(44)um-2ρ+-ρmum+um2+4ρm<u+<um+2ρ+-ρmum2+4ρm-um.In this situation, the local solution of the perturbed Riemann problem (1) and (2) for the sufficiently small time t may be indicated by the symbols as (45)u-,ρ-+S1+u1,ρ1+S2+um,ρm+S3+u2,ρ2+R4+u+,ρ+.Hereafter the intermediate states (u1,ρ1), (u2,ρ2), and (u3,ρ3) have the same presentations as those in Case 1.
The interaction between S+S and S+R is shown when (u+,ρ+)∈V2.
As before, S2 collides with S3 at the point (x1,t1) which may be also calculated by the formula (27) in Case 1. After the time t1, the new local Riemann problem with the left state (u1,ρ1) and the right state (u2,ρ2) can also be solved by two shock waves which are denoted by S5 and S6, respectively. Similarly, the two shock waves S1 and S5 of the first family coalesce into a new shock wave of the first family which is denoted by S7.
On the other hand, we consider the situation that the shock wave S6 penetrates the rarefaction wave R4 which can be summarized below.
Lemma 7.
The shock wave S6 is able to catch up with the wave back of the rarefaction wave R4 in finite time and consequently begins to penetrate R4. More precisely, if u+<u3, then S6 has the ability to cancel the whole R4 thoroughly. Otherwise, if u+>u3, then S6 penetrates R4 incompletely and finally has the characteristic line (46)x=x0+3u3+u32+4ρ32tin R4 as its asymptotic line.
Proof.
It is obvious to see that the propagation speed of S6 is given by (33) and that of the wave back in the rarefaction wave R4 is calculated by (47)ξ4u2,ρ2=3u2+u22+4ρ22=u2-k12.In view of k13<k12<0, we have (48)σ6-ξ4u2,ρ2=k12-k13>0,which implies that S6 keeps up with the wave back of R4 in finite time and the intersection (x3,t3) is computed by(49)x3-x1=σ6t3-t1=u2-k13t3-t1,x3-x0=ξ4u2,ρ2·t3=u2-k12t3,in which t1 is given by (27). Thus, we have (50)x3,t3=x0+u2-k12u3+k23+k21x0u1-u2+um+2k21u3-u2,u3+k23+k21x0u1-u2+um+2k21u3-u2.
Consequently, S6 begins to penetrate R4 after the time t3. It is remarkable that both S6 and R4 are attributed to the waves of the second family. It can also be seen from the (u,ρ) phase plane in Figure 4 that the state (u3,ρ3) and the varying state (u,ρ) along with the corresponding characteristic line in R4 can also be connected by a shock wave of the second family directly. Thus, during the process of penetration, the shock wave is denoted by S8 whose propagation speed is given by (51)σ8=u+u3+u32+4ρ32,where u varies from u2 to u+. Thus, the shock curve of S8 in the (x,t) plane is determined by (52)dxdt=u+u3+u32+4ρ32,x-x0=3u+u2+4ρ2t,xt3=x3.From the (u,ρ) phase plane in Figure 4 and the notation of k2 in this paper, we always have k2=k23 when u varies from u2 to u+. Thus, we can arrive at (53)x-x0=2u3+k23t-2u3-u2u3+k21+k23x0tu1+um-u2+2k21.In addition, by differentiating (53) with respect to t twice, we have (54)d2xdt2=12u3-u2u3+k21+k23x0u1+um-u2+2k21t3>0,which means that the shock wave S8 speeds up during the process of penetration.
Furthermore, there exist two possible situations according to the region in which the state (u+,ρ+) lies.
(a) If (u+,ρ+)∈V2, then we have u+<u3. In this situation, S8 is able to cancel R4 completely and terminates at the point (x4,t4) which is given by (55)x4-x0=3u++u+2+4ρ+2t4,x4-x0=2u3+k23t4-2u3-u2u3+k21+k23x0t4u1+um-u2+2k21;namely, we have (56)x4,t4=x0+2u++k2+u3-u2u3+k21+k23x0u1+um-u2+2k21u3-u+2,u3-u2u3+k21+k23x0u1+um-u2+2k21u3-u+2.After the penetration, the shock wave is denoted by S9 whose propagation speed is (57)σ9=u++u3+u32+4ρ32=u+-k13.
(b) If (u+,ρ+)∈V3, then we have u+>u3. Thus S8 cannot cancel the entire rarefaction wave R4 thoroughly and ultimately has the characteristic line in R4 whose expression is shown in (46) as the asymptotic line.
Case 3 (S+S and R+S).
In this case, we consider that the Riemann solution emanating from the initial point (x0,0) is a rarefaction wave followed by a shock wave. This case happens if and only if the conditions 0≤ρ+<ρm and (58)um+2ρ+-ρmum2+4ρm-um<u+<um-2ρ+-ρmum+um2+4ρmare satisfied. In other words, the state (u+,ρ+) should locate in the region V4∪V5 in Figure 2. We use R3 and S4 to denote the rarefaction wave and the shock wave, respectively. When t is small enough, the solution of the perturbed Riemann problem (1) and (2) may be displayed in the symbol form (59)u-,ρ-+S1+u1,ρ1+S2+um,ρm+R3+u2,ρ2+S4+u+,ρ+.Let us first consider the situation that the shock wave S2 penetrates the rarefaction wave R3 and use the following lemma to depict it.
Lemma 8.
The shock wave S2 is able to penetrate the whole rarefaction wave R3 thoroughly in finite time and then a transmitted rarefaction wave is generated during the process of penetration.
Proof.
It can be seen that the propagation speed of S2 is given by (23) and that of the wave back in the rarefaction wave R3 is given by (60)ξ3um,ρm=3um-um2+4ρm2=um-k2m.
Noticing that k2m>0>k11, we can obtain (61)σ2-ξ3um,ρm=k2m-k11>0,which means that S2 penetrates the wave back of R3 at the point which may be computed by (62)x1=σ2t1,x1-x0=ξ3um,ρm·t1;namely, (63)x1,t1=um+u1+k21x0u1+2k21,x0u1+2k21.
After the time t1, the shock wave S2 begins to penetrate the rarefaction wave R3 and is denoted by S6 during the process of penetration. This penetration gives rise to a transmitted rarefaction wave which is denoted by R5. In order to compare the propagation speeds of rarefaction waves before and after penetration when across S6, we have the following description. The propagation speeds of the matched characteristic lines in R3 and R5 can be calculated, respectively, by (64)ξ3u-,ρ-=3u--u-2+4ρ-2,(65)ξ5u+,ρ+=3u+-u+2+4ρ+2,in which the state u-,ρ- in R3 becomes the matched one u+,ρ+ in R5 when across S2. They should obey the relation (66)ρ+-ρ-u+-u-=u-2+4ρ--u-2=u+2+4ρ+-u+2,in which um<u-<u2 and u1<u+<u3. For the match characteristic lines in R3 and R5, taking into account u-<u+, it follows from (64), (65), and (66) that ξ3(u-,ρ-)<ξ5(u+,ρ+). That is to say, the rarefaction wave accelerates when it passes through the shock wave S2.
On the other hand, during the process of penetration, the curve of the shock wave S6 is determined by (67)dxdt=u-+u++u+2+4ρ+2,x-x0=3u--u-2+4ρ-2t,xt1=x1.An easy computation leads to(68)x-x0=-k1m+2k11t+um-k21+k1m+2k11x0tu1+2k21.In addition, the following follows from (68): (69)d2xdt2=-14um-k21+k1m+2k11x0u1+2k21t3>0,which means that the shock wave S6 also speeds up during the process of penetration. It is obvious to see that the shock wave S2 is able to penetrate the whole rarefaction wave R3 thoroughly in finite time for the reason that the states (u3,ρ3) and (u2,ρ2) can be connected directly by a shock wave of the second family which is denoted by S7 after penetration. The penetration terminates at the point (x2,t2) which is given by (70)x2-x0=3u2-u22+4ρ22t2,x2-x0=-k1m+2k11t2+um-k21+k1m+2k11x0t2u1+2k21;namely, (71)x2,t2=x0+2u2+k12um-k21+k1m+2k112x04u1+2k21u2+k1m+k112,um-k21+k1m+2k112x04u1+2k21u2+k1m+k112.Thus, the conclusion can be drawn and the proof is completed.
The shock wave S7 continues to move forward and consequently overtakes the shock wave S4 at the point which is given by (72)x3,t3=x0+u++u2+k22x2-x0-u2+u3+k23t2u+-u3,x2-x0-u2+u3+k23t2u+-u3,in which (x2,t2) is given by (71). As before, they also coalesce into a new shock wave denoted by S8 due to the fact that the states (u3,ρ3) and (u+,ρ+) can also be connected directly by a shock wave of the second family. On the other hand, let us consider that the shock wave S1 penetrates the rarefaction wave R5 which is described in the following lemma.
Lemma 9.
If u->u3, then the shock wave S1 is able to penetrate the whole rarefaction wave R5 thoroughly. Otherwise, if u-<u3, then S1 penetrates R5 incompletely and finally has the characteristic line associated with the state (u-,ρ-) in R5 as its asymptotic line.
Proof.
The propagation speed of S1 and that of the wave back in R5 are computed, respectively, by (73)σ1=u1+u--u-2+4ρ-2=u1-k2-,(74)ξ5u1,ρ1=3u1-u12+4ρ12=u1-k21.Thus, one can arrive at σ1>ξ5(u1,ρ1) by taking into account k21>k2->0. Equivalently, the shock wave S1 catches up with the wave back of R5 in finite time. In fact, the intersection is determined by (75)σ1t4=ξ5u1,ρ1·t4-t1+x1,in which (x1,t1) is given by (63), which enables us to have (76)x4,t4=u1+u-+k1-um-u1+k21-k11x0u1+2k21u--u1,um-u1+k21-k11x0u1+2k21u--u1.
After the time t4, the shock wave begins to penetrate R5 with a varying propagation speed and is denoted by S9 during the process of penetration. The curve of S9 is determined by(77)dxdt=u++u--u-2+4ρ-2,x-x¯=3u+-u+2+4ρ+2t-t¯,xt4=x4,in which the point (x¯,t¯) is located on the curve of S6 and the characteristic line with the state (u+,ρ+) is departing from it. For our knowledge, it is impossible to calculate the explicit form for the curve of S9 due to the fact that R5 is a noncentered rarefaction wave. Depending on the region in which the state (u+,ρ+) lies, there are two possible situations as follows:
If (u+,ρ+)∈V4, then we have u3<u- (see Figure 5). For this subcase, S9 is able to cancel the entire R5 thoroughly. The shock wave is denoted by S10 after penetration whose propagation speed is given by (39).
If (u+,ρ+)∈V5, then we have u3>u- (see Figure 6). In this situation, S9 cannot penetrate the whole R4 completely and at last has the characteristic line associated with the state (u-,ρ-) in R5 as its asymptotic line.
The interaction between S+S and R+S is shown when (u+,ρ+)∈V4.
The interaction between S+S and R+S is shown when (u+,ρ+)∈V5.
Case 4 (S+S and R+R).
In the end, we consider the situation that there are two rarefaction waves originating from the point (x0,0). This case arises when the conditions u+>um and(78)max0,ρm+-um2+4ρm-umu+-um2<ρ+<ρm+um2+4ρm-umu+-um2are satisfied. The two rarefaction waves are denoted by R3 and R4, respectively. In this case, the state (u+,ρ+) should locate in the region V6∪V7∪V8∪V9 in Figure 2. The solution of the perturbed Riemann problem (1) and (2) for sufficiently small t may be symbolized as (79)u-,ρ-+S1+u1,ρ1+S2+um,ρm+R3+u2,ρ2+R4+u+,ρ+.
As in Case 3, the shock wave S2 is able to penetrate the entire R3 thoroughly in finite time and a transmitted rarefaction wave R5 is generated. It can be seen from Lemma 9 that if u->u3, then S1 is able to penetrate the whole R5 thoroughly. The condition (u+,ρ+)∈V6∪V7 should be satisfied in this situation. Otherwise, if u-<u3, then S1 cannot penetrate the whole R5 thoroughly. In this situation, (u+,ρ+)∈V8∪V9 should be satisfied. On the other hand, we turn our attention back to the situation that the shock wave S7 penetrates the rarefaction wave R4. With the similar process as before, we can see that if u+<u3, then S7 is able to penetrate the whole R4 thoroughly. In this situation, (u+,ρ+)∈V6∪V9 should be satisfied. Otherwise, if u+>u3, then S7 cannot penetrate the whole R4 completely. In this situation, we need to require (u+,ρ+)∈V7∪V8.
Thus, this case is just the combination of the results ahead and we omit the details. In fact, we can summarize the main results for the case below.
If (u+,ρ+)∈V6, then we have u+<u3<u- (see Figure 7). For this subcase, S1 is able to penetrate the whole R5 and S7 is also able to penetrate the whole R4, such that the large-time behavior of solution is S+S.
If (u+,ρ+)∈V7, then we have u3<min(u-,u+). For this subcase, S1 is able to penetrate the whole R5, but S7 cannot penetrate the whole R4 completely, such that the large-time behavior of solution is S+R.
If (u+,ρ+)∈V8, then we have u-<u3<u+ (see Figure 8). For this subcase, S1 cannot penetrate the whole R5 and S7 also cannot penetrate the whole R4, such that the large-time behavior of solution is R+R.
If (u+,ρ+)∈V9, then we have max(u-,u+)<u3. For this subcase, S1 cannot penetrate the whole R5 completely, but S7 is able to penetrate the whole R4 thoroughly, such that the large-time behavior of solution is R+S.
The interaction between S+S and R+R is shown when (u+,ρ+)∈V6.
The interaction between S+S and R+R is shown when (u+,ρ+)∈V8.
4. Conclusion
So far, the discussion for all kinds of interactions has been accomplished. The global solutions of the perturbed Riemann problem (1) and (2) are constructed completely in explicit forms under Assumption 1. It is clear to see that the large-time asymptotic states of the global solutions to the perturbed Riemann problem (1) and (2) are identical with the corresponding ones to the Riemann problem (1) and (3). In other words, the solutions to Riemann problem (1) and (3) can be obtained if the limits x0→0 of the global solutions to the perturbed Riemann problem (1) and (2) are taken. Thus, it turns out that the Riemann solutions are stable with respect to the specific small perturbation (2) of the Riemann initial data (3). Based on the above analysis, the proof of Theorem 2 has been finished.
In fact, we can give suitable assumptions such that the Riemann solution originating from the origin (0,0) is R+S. By adopting the method developed in this paper, similar results can also be obtained. On the other hand, the reasonable assumptions can also be made such that the Riemann solution originating from the initial point (x0,0) is S+S or S+R. In this situation, we need to fix the Riemann solution at (x0,0) and then change the Riemann solution at (0,0). This is to say, for the given right state (u+,ρ+), we first fix the intermediate state (um,ρm) and then change the left state (u-,ρ-). Then, the global solutions to the perturbed Riemann problem (1) and (2) can also be constructed as above. In fact, the above assumptions are made in order to avoid the situation that the forward rarefaction wave collides with the backward one, whose solution cannot be constructed in an explicit form.
Competing Interests
The authors declare that there is no conflict of interests regarding the publication of this article.
Acknowledgments
This work is partially supported by National Natural Science Foundation of China (nos. 11441002 and 11271176) and Shandong Provincial Natural Science Foundation (no. ZR2014AM024).
LerouxA. Y.Approximation des systems hyperboliques1981RocquencourtFritzJ.TóthB.Derivation of the Leroux system as the hydrodynamic limit of a two-component lattice gas2004249112710.1007/s00220-004-1103-xMR20772512-s2.0-3543029206PopkovV.SchützG. M.Why spontaneous symmetry breaking disappears in a bridge system with PDE-friendly boundaries200412P1200410.1088/1742-5468/2004/12/p120042-s2.0-84901008115TóthB.ValkóB.Perturbation of singular equilibria of hyperbolic two-component systems: a universal hydrodynamic limit2005256111115710.1007/s00220-005-1314-9MR21343382-s2.0-17444377883TempleB.Systems of conservation laws with invariant submanifolds1983280278179510.2307/1999646MR716850ZBL0559.35046CanonE.On some hyperbolic systems of temple class201275114241425010.1016/j.na.2012.03.013MR29219862-s2.0-84860381594SerreD.Solutions à variations bornées pour certains systèmes hyperboliques de lois de conservation198768213716810.1016/0022-0396(87)90189-6MR8920212-s2.0-45949114786GramchevT.Entropy solutions to conservation laws with singular initial data199524572173310.1016/0362-546x(94)e0055-lMR13190812-s2.0-0013519492LuY. G.MantillaI.RendonL.Convergence of approximated solutions to a nonstrictly hyperbolic system200116579Zbl0988.35114LuY.-G.Global entropy solutions of Cauchy problem for the Le Roux system201660616610.1016/j.aml.2016.04.003MR3505854ChengZ.On application of kinetic formulation of the Le Roux system200952126327210.1017/S0013091507000697MR24758932-s2.0-68349131223ChenG.EndresE. E.JenssenH. K.Pairwise wave interactions in ideal polytropic gases2012204378783610.1007/s00205-012-0497-4MR29171222-s2.0-84862796924GreenbergJ. M.On the elementary interactions for the quasilinear wave equation On the elementary interactions for the quasilinear wave equation ∂γ/∂t-∂v/∂x=0 and ∂v/∂t-∂σ(γ)/∂x=0197143325349SunM.Interactions of elementary waves for the Aw-Rascle model20096961542155810.1137/0807314022-s2.0-67649252412ChangT.HsiaoL.The Riemann problem and interaction of waves in gas dynamics198941Longman Scientific and TechnicalGlimmJ.The interaction of nonlinear hyperbolic waves198841556959010.1002/cpa.3160410505MR948072LaiG.Interaction of jump-fan composite waves in a two-dimensional jet for van der Waals gases201556606150410.1063/1.4922443MR33698712-s2.0-84935875134MentrelliA.RuggeriT.SugiyamaM.ZhaoN.Interaction between a shock and an acceleration wave in a perfect gas for increasing shock strength200845449851710.1016/j.wavemoti.2007.09.005MR24068452-s2.0-40649109102Raja SekharT.SharmaV. D.Riemann problem and elementary wave interactions in isentropic magnetogasdynamics201011261963610.1016/j.nonrwa.2008.10.036MR2571237ZBL1183.352302-s2.0-70449679122LiT. T.199432New York, NY, USAJohn Wiley & SonsMR1291392ShenC.SunM.Interactions of delta shock waves for the transport equations with split delta functions2009351274775510.1016/j.jmaa.2008.11.005MR2473979ZBL1159.350422-s2.0-56649088130GuoL.ZhangY.YinG.Interactions of delta shock waves for the Chaplygin gas equations with split delta functions2014410119020110.1016/j.jmaa.2013.07.0822-s2.0-84884983096QuA.WangZ.Stability of the Riemann solutions for a Chaplygin gas2014409134736110.1016/j.jmaa.2013.07.018MR3095045ZBL1306.350732-s2.0-84883458792GuoL.PanL.YinG.The perturbed Riemann problem and delta contact discontinuity in chromatography equations201410611012310.1016/j.na.2014.04.016MR32096882-s2.0-84899789664ShenC.Wave interactions and stability of the Riemann solutions for the chromatography equations2010365260961810.1016/j.jmaa.2009.11.037MR2587063ZBL1184.352072-s2.0-74849089113ShenC.The asymptotic behaviors of solutions to the perturbed Riemann problem near the singular curve for the chromatography system20152217610110.1080/14029251.2015.996442MR32867352-s2.0-84918513358SunM.Interactions of delta shock waves for the chromatography equations201326663163710.1016/j.aml.2013.01.002MR30280672-s2.0-84875239435BressanA.200020Oxford, UKOxford University PressOxford Lecture Series in Mathematics and Its ApplicationsDafermosC. M.Hyperbolic conversation laws in continuum physics2000Berlin, GermanySpringerSerreD.1999/2000Cambridge, UKCambridge University PressSmollerJ.1994New York, NY, USASpringer10.1007/978-1-4612-0873-0MR1301779