# The Perturbed Riemann Problem for Special Keyfitz-Kranzer System with Three Piecewise Constant States

Authors: Yuhao Jiang; Chun Shen
Journal: Advances in Mathematical Physics
 (2017)

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## Abstract

The Riemann problem for a special Keyfitz-Kranzer system is investigated and then seven different Riemann solutions are constructed. When the initial data are chosen as three piecewise constant states under suitable assumptions, the global solutions to the perturbed Riemann problem are constructed explicitly by studying all occurring wave interactions in detail. Furthermore, the stabilities of solutions are obtained under the specific small perturbations of Riemann initial data.


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## Body

## 1. Introduction

The Keyfitz-Kranzer system is a pair of conservation laws in the following form [1]:(1)ut+ϕux=0,vt+ϕvx=0,in which ϕ=ϕ(u,v) is a given function. System (1) arises as a model for the stretched elastic string to describe the propagation of forward longitudinal and transverse waves [1, 2]. It is also used to illustrate certain features of the solar wind in magnetohydrodynamics [3, 4]. Furthermore, it is usually assumed that ϕ is a function of r, where r=u2+v2, which is called the symmetric Keyfitz-Kranzer system and thus widely investigated such as in [3, 5]. Thus, system (1) was also taken as an example for a nonstrictly hyperbolic system of conservation laws. In addition, it should be pointed out that the nonsymmetric Keyfitz-Kranzer system(2)ρt+ρϕρ,u1,u2,…,unx=0,ρuit+ρuiϕρ,u1,u2,…,unx=0,i=1,2,…,n,has also been proposed and widely investigated in [6–8]; also see [9, 10] about the Riemann problem for some special forms of system (2).Recently, it has been assumed in [11], where ϕ is a function of r=au+bv, that system (1) can be simplified into the form(3)rt+rϕrx=0,ut+uϕrx=0.One may also see [12] where ϕ is a function of r=u/v. It is easily shown that system (3) owns two eigenvalues λ1=ϕ and λ2=ϕ+rϕr; thus system (3) is also nonstrictly hyperbolic. More precisely, the characteristic field associated with λ1 is linearly degenerate and the characteristic field associated with λ2 is genuinely nonlinear when r(rϕ)rr≠0 or otherwise linearly degenerate when r(rϕ)rr=0. In order to ensure r(rϕ)rr≠0, the assumptions ϕr>0, (rϕ)rr>0, and ϕ(0)=0 were made and then the delta shock wave was captured in [11]. In the present paper, we give up the assumption ϕr>0 and want to discover some new interesting nonlinear phenomena. More precisely, we take the detailed example ϕ(r)=(c1r+c2/r)r, such that system (3) is simplified into(4)rt+c1r+c2rrx=0,ut+c1r+c2rux=0,in which c1 and c2 are two given positive constants, which enable us to deal with system (4) in completely explicit forms.It is easily shown that system (4) has two eigenvalues λ1=2c1r and λ2=c1r+c2/r; thus we can deduce that λ1>λ2 for r>c2/c1 and otherwise λ1<λ2 for 0<r<c2/c1. It is easy to see that the special form (4) of the Keyfitz-Kranzer system (1) is also nonstrictly hyperbolic in the quarter (r,u) phase plane. One of the main features of system (4) lies in that the shock curves have the same representation with the rarefaction ones for the λ2-characteristic family in the quarter (r,u) phase plane thanks to the special form of system (4), which belongs to the so-called Temple class for some hyperbolic systems of conservation laws [13, 14]. Compared with general hyperbolic systems of conservation laws, the well-posed result for Temple class may be achieved in a more general sense of initial data.In this paper, we want to construct the global solutions to the particular Cauchy problem for system (4) in fully explicit forms when the three piecewise constant states are taken for the initial conditions as follows:(5)r,ux,0=r-,u-,-∞<x<0,rm,um,0<x<x0,r+,u+,x0<x<+∞,in which x0>0 is arbitrarily small. This type of initial data (5) has been widely used to study the wave interaction problem [15–17] for different hyperbolic systems of conservation laws. It is worthwhile to notice that the initial data (5) may be regarded as a special small perturbation of the corresponding Riemann initial data(6)r,ux,0=r-,u-,-∞<x<0,r+,u+,0<x<+∞.Thus, the particular Cauchy problem (4) and (5) is usually called the perturbed Riemann problem (or the double Riemann problems) in literature.The first task of this paper is to construct the solutions to the Riemann problem (4) and (6) when the Riemann initial data (6) lie in the quarter (r,u) phase plane. More precisely, we find seven different combinations for the solutions to the Riemann problem (4) and (6) according to the choices of Riemann initial data (6). In particular, some composite waves are needed be introduced in the constructions of Riemann solutions. The second task of this paper is to construct the global solutions to the perturbed Riemann problem (4) and (5), which is essential to study various possible interactions of elementary waves for system (4). For simplicity, we restrict ourselves to consider the situation that the condition r-,rm,r+>c2/c1 is satisfied for the initial data (5). Under this assumption, the global solutions to the perturbed Riemann problem (4) and (5) are constructed in fully explicit forms by studying all the wave interactions appearing in the construction processes of solutions. Furthermore, we can see that the solutions of the Riemann problem (4) and (6) are stable under the particular small perturbation (5) of Riemann initial data (6) when the limit x0→0 is taken in the solutions. In addition, it should be pointed out that the wave interaction problem for the Temple class has been widely investigated recently, such as for the Aw-Rascle model [18, 19], the Chaplygin gas model [20–23], and the chromatography model [24–28].The paper is organized in the following way. In Section2, the Riemann problem (4) and (6) is investigated and the Riemann solutions are constructed for seven different cases. In Section 3, under the suitable assumptions, the global solutions to the perturbed Riemann problem (4) and (5) are constructed in fully explicit forms by investigating all appearing wave interactions 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 perturbations (5) of the Riemann initial data (6).


## 2. The Riemann Problem for the Special Keyfitz-Kranzer System (4)

In this section, we are devoted to investigating the Riemann problem for system (4) associated with the Riemann initial data (6). Let us see, for example, [1, 3, 11, 12] about the related Riemann problem for system (4). We can also refer to [15, 29] for the general knowledge about the Riemann problem for hyperbolic systems of conservation laws.By a simple calculation, it is obtained that there are two eigenvalues for system (4) as follows:(7)λ1=2c1r,λ2=c1r+c2r.We have λ1>λ2 for r>c2/c1; otherwise we have λ1<λ2 for 0<r<c2/c1. Thus, system (4) is nonstrictly hyperbolic in the quarter (r,u) phase plane (r>0,u>0). The corresponding right eigenvectors for system (4) are given, respectively, by(8)r1→=r,uT,r2→=0,1T.It is easy to get ∇λ1·r1→=2c1r and ∇λ2·r2→=0, in which ∇=(∂/∂r,∂/∂u). The characteristic field is genuinely nonlinear for λ1 provided that r>0 and is always linear degeneracy for λ2. The waves associated with the first characteristic field will be either shock waves (denoted by S) or rarefaction waves (denoted by R), which are determined by the choice of initial data. The waves associated with the second characteristic field are always contact discontinuities (denoted by J).For the Riemann problem (4) and (6), it is invariant under uniform stretching of coordinates: (x,t)→(αx,αt) with α>0, such that we may consider solutions in the self-similar form(9)r,ux,t=r,uξ,ξ=xt.By carrying out the self-similar transformation ξ=x/t, the Riemann problem (4) and (6) may be reformulated as the ordinary different equations(10)-ξrξ+2c1rrξ=0,c1r+c2r-ξuξ+c1u-c2ur2rξ=0,associated with the boundary values at infinity (r,u)(±∞)=(r±,u±).For smooth solutions, (10) is equivalent to(11)2c1r-ξ0c1u-c2ur2c1r+c2r-ξdrdu=00.It means either constant state (r,u)=const or singular solution, which is a wave of the first characteristic family,(12)ξ=c1r+c2r,dr=0,or a wave of the second characteristic family,(13)ξ=2c1r,c1u-c2ur2dr+c1r+c2r-2c1rdu=0.For the given left state (r-,u-), by integrating (12), the contact discontinuity curves in the quarter (r,u) phase plane, which are the state set that can be connected on the right by a contact discontinuity, are as follows:(14)J:ξ=c1r+c2r,r=r-.Similarly, by integrating (13), the rarefaction wave curves in the quarter (r,u) phase plane, which are the state set that can be connected on the right by a rarefaction wave, are as follows:(15)R:ξ=2c1r,ur=u-r-,r>r-.On the other hand, for a bounded discontinuity atξ=σ, the Rankine-Hugoniot conditions read(16)-σr+c1r+c2rr=0,-σu+c1r+c2ru=0,in which σ=dx/dt is the propagation speed of the discontinuity and [r]=r+-r- is the jump across the discontinuity with r-=r(x(t)-0,t) and r+=r(x(t)+0,t). When the two states (r±,u±) can be connected by only one discontinuous wave directly, by solving (16), one can see if it is a contact discontinuity corresponding to a wave of the first characteristic family; then they should satisfy(17)J:σ=c1r-+c2r-=c1r++c2r+,r-=r+.Otherwise, if it is a shock wave corresponding to a wave of the second characteristic family, then they should satisfy(18)S:σ=c1r-+r+,u-r-=u+r+,r+<r-,in which the entropy condition is taken into account.Using these elementary waves, we are now in a position to construct the solutions to the Riemann problem (4) and (6) by the analysis method in the quarter (r,u) phase plane. It can be seen from (14), (15), (17), and (18) that if the Riemann initial data (6) satisfy r-=r+ or u-/r-=u+/r+, then the Riemann solution contains only a single wave. For the other situations, we can construct the following seven different combinations for the solutions to the Riemann problem (4) and (6) according to the choices of Riemann initial data (6).(1) When0<r+<r-<c2/c1, the solution can be expressed by the symbol S+J (see Figure 1), which is given by(19)r,ux,t=r-,u-,x<σt,r+,r+u-r-,σt<x<λ2r+t,r+,u+,x>λ2r+t.The symbol S+J is used to stand for a shock wave S followed by a contact discontinuity J. In what follows, similar symbols are also used and not explained again without confusion. Here and below, the symbol σ=c1(r-+r+) is used to indicate the propagation speed of the shock wave S and λ2r+=c1r++c2/r+ is used to indicate the propagation speed of the contact discontinuity.Figure 1
The Riemann solution of (4) and (6) is shown for 0<r+<r-<c2/c1.(2) When0<r-<r+<c2/c1, the solution is R+J (see Figure 2) in the form(20)r,ux,t=r-,u-,x<λ1r-t,x2c1t,u-x2c1r-t,λ1r-t<x<λ1r+t,r+,r+u-r-,λ1r+t<x<λ2r+t,r+,u+,x>λ2r+t.Figure 2
The Riemann solution of (4) and (6) is shown for 0<r-<r+<c2/c1.(3) Whenr->r+>c2/c1, the solution is J+S (see Figure 3) given by(21)r,ux,t=r-,u-,x<λ2r-t,r-,r-u+r+,λ2r-t<x<σt,r+,u+,x>σt,in which the propagation speed of the contact discontinuity is λ2r-=c1r-+c2/r- and the propagation speed of the shock wave is also σ=c1(r-+r+).Figure 3
The Riemann solution of (4) and (6) is shown for r->r+>c2/c1.(4) Whenr+>r->c2/c1, the solution is J+R (see Figure 4) in the form(22)r,ux,t=r-,u-,x<λ2r-t,r-,r-u+r+,λ2r-t<x<λ1r-t,x2c1t,u+x2c1r+t,λ1r-t<x<λ1r+t,r+,u+,x>λ1r+t.Figure 4
The Riemann solution of (4) and (6) is shown for r+>r->c2/c1.(5) Whenr+>c2/c1>r->0, the solution can be expressed by a composite wave R1+J+R2 (see Figure 5), which is given by(23)r,ux,t=r-,u-,x<λ1r-t,x2c1t,u-x2c1r-t,λ1r-t<x<2c1c2t,x2c1t,u+x2c1r+t,2c1c2t<x<λ1r+t,r+,u+,x>λ1r+t.Figure 5
The Riemann solution of (4) and (6) is shown for r+>c2/c1>r->0.(6) When0<r+<c2/c1<r- and r-r+<c2/c1, we can also construct the solution S+J (see Figure 6) in the form(24)r,ux,t=r-,u-,x<σt,r+,r+u-r-,σt<x<λ2r+t,r+,u+,x>λ2r+t.It can be calculated from (7) that(25)λ1r-,u-=2c1r-,λ2r-,u-=c1r-+c2r-,λ1r+,r+u-r-=2c1r+,λ2r+,r+u-r-=c1r++c2r+.In view of r+<c2/c1<r- and r-r+<c2/c1, it can be concluded that the entropy condition should obey(26)λ1r-,u->λ2r-,u->σ,λ1r+,r+u-r-<σ<λ2r+,r+u-r-,which implies that the shock wave is a 1-shock wave.Figure 6
The Riemann solution of (4) and (6) is shown when 0<r+<c2/c1<r- and r-r+<c2/c1 are satisfied.(7) When0<r+<c2/c1<r- and r-r+>c2/c1, the solution is J+S (see Figure 7) which is given by(27)r,ux,t=r-,u-,x<λ2r-t,r-,r-u+r+,λ2r-t<x<σt,r+,u+,x>σt.It follows from (7) that(28)λ1r+,u+=2c1r+,λ2r+,u+=c1r++c2r+,λ1r-,r-u+r+=2c1r-,λ2r-,r-u+r+=c1r-+c2r-.Based on r+<c2/c1<r- and r-r+>c2/c1, it can be checked that the entropy condition should satisfy(29)λ1r+,u+<λ2r+,u+<σ,λ2r-,r-u+r+<σ<λ1r-,r-u+r+,which implies that the shock wave is a 2-shock wave.Figure 7
The Riemann solution of (4) and (6) is shown when 0<r+<c2/c1<r- and r-r+>c2/c1 are satisfied.It can be concluded from the above discussions that the solutions to the Riemann problem (4) and (6) can be constructed completely when the Riemann initial data (6) lie in the quarter (r,u) phase plane. It is obvious to see that if we give up the assumptions ϕr>0, (rϕ)rr>0, and ϕ(0)=0 in system (3), then the Riemann solutions are also changed and different from each other.


## 3. Construction of Global Solutions to the Perturbed Riemann Problem (4) and (5)

In this section, we are dedicated to the constructions of global solutions to the perturbed Riemann problem (4) and (5). In this paper, for simplicity, we restrict ourselves only to consider the situation that the condition r-,rm,r+>c2/c1 is satisfied for the initial data (5). Under the above assumption, the wave interaction problems are investigated by employing the method of characteristics and then the global solutions to the perturbed Riemann problem (4) and (5) are constructed in fully explicit forms.Case 1 (J+S and J+S).
We first consider the situation that each of the Riemann solutions originating from the initial points(0,0) and (x0,0) is a contact discontinuity followed by a shock wave, respectively. This case arises when the condition c2/c1<r+<rm<r- is satisfied. In this case, when the time t is sufficiently small, the solution to the perturbed Riemann problem (4) and (5) can be represented as (see Figure 8)(30)r-,u-+J1+r1,u1+S2+rm,um+J3+r2,u2+S4+r+,u+.The intermediate states (r1,u1) and (r2,u2) may be given, respectively, by(31)r1,u1=r-,r-umrm,r2,u2=rm,rmu+r+.
First of all, we have the following lemma to describe the interaction between the shock waveS2 and the contact discontinuity J3.Figure 8
The interaction betweenJ+S and J+S is shown when r->rm>r+>c2/c1.Lemma 1.
The shock waveS2 keeps up with the contact discontinuity J3 in finite time and then the interaction between S2 and J3 happens. After the interaction, on the one hand, the contact discontinuity J5 when across the shock wave S2 is parallel to the contact discontinuity J1. On the other hand, the shock wave S6 cannot change its direction when it passes through the contact discontinuity J3.Proof.
The speeds ofS2 and J3 can be calculated, respectively, by(32)σ2=c1r-+rm,τ3=c1rm+c2rm.We can easily get(33)σ2-τ3=c1r--c2rm>0,which means that σ2>τ3. That is to say, the shock wave S2 meets J3 in finite time. The intersection (x1,t1) is determined by(34)x1=σ2t1,x1-x0=τ3t1,which enables us to gain(35)x1,t1=c1r-+rmrmx0c1r-rm-c2,rmx0c1r-rm-c2.The interaction between S2 and J3 happens at the point (x1,t1), such that we can get a new Riemann problem with the initial date (r1,u1) and (r2,u2) again. It can be deduced from r1=r->rm=r2>c2/c1 that the Riemann solution is a new contact discontinuity denoted by J5 followed by a new shock wave denoted by S6 originating from the point (x1,t1). The intermediate state (r3,u3) between J5 and S6 is given by(36)r3,u3=r-,r-u+r+.The propagation speeds of J5 and S6 can be computed, respectively, by(37)τ5=c1r1+c2r1=c1r-+c2r-,(38)σ6=c1r3+r2=c1r-+rm.That is to say, the contact discontinuities J5 and J1 are parallel and the shock wave cannot change its direction when it passes through the contact discontinuity.Consequently, we consider the coalescence of two shock wavesS4 and S6 belonging to the same family.Lemma 2.
The two shock wavesS4 and S6 merge into a new shock wave denoted by S7.Proof.
The propagation speed ofS4 is(39)σ4=c1rm+r+,which together with (38) yields(40)σ6-σ4=c1r--r+>0,which means that S6 catches up with S4 in finite time. The intersection (x2,t2) is determined by(41)x2=σ6t2,x2-x0=σ4t2,which yields(42)x2,t2=r-+rmx0r--r+,x0c1r--r+.We can see from the (r,u) phase plane in Figure 8 that the two states (r3,u3) and (r+,u+) can be connected by a shock wave directly, for the reason that it follows from (36) that u3/r3=u+/r+. It means that the two shock waves S4 and S6 merge into a new shock wave which is denoted by S7 whose propagation speed is(43)σ7=c1r-+r+.It is easy to get σ4<σ7<σ6. In other words, the propagation speed of S7 is between those of S4 and S6. The proof is completed.Case 2 (J+S and J+R).
In this case, we investigate the situation that the Riemann solution is a contact discontinuity followed by a shock wave originating from the point(0,0) and a contact discontinuity followed by a rarefaction wave originating from the point (x0,0) when the initial data (5) need to satisfy both c2/c1<rm<r- and c2/c1<rm<r+. When t is small enough, the solution of the perturbed Riemann problem (4) and (5) may be expressed in the symbol (see Figures 9 and 10):(44)r-,u-+J1+r1,u1+S2+rm,um+J3+r2,u2+R4+r+,u+.The intermediate states (r1,u1), (r2,u2), and (r3,u3) have the same expressions as those in Case 1. As in Case 1, S2 collides with J3 at the point (x1,t1) which is given by the same formula (35). Similarly, a new local Riemann problem is generated at the point (x1,t1) with the left state (r1,u1) and the right state (r2,u2), which gives rise to a contact discontinuity denoted by J5 and a shock wave denoted by S6. The propagation speeds of J5 and S6 can also be calculated by the formulae (37) and (38), such that we have τ1=τ5 and σ6=σ2. Now, we need to deal with the situation that the shock wave S6 penetrates the rarefaction wave R4.Figure 9
The interaction betweenJ+S and J+R is shown when r->r+>rm>c2/c1.Figure 10
The interaction betweenJ+S and J+R is shown when r+>r->rm>c2/c1.Lemma 3.
The shock waveS6 overtakes the wave back of the rarefaction wave R4 in finite time. More precisely, if r+<r-, then S6 is able to cancel the whole R4 thoroughly (see Figure 9). Otherwise, if r+>r-, then S6 passes through R4 incompletely and has the line x-x0=2c1r-t in R4 as its asymptote (see Figure 10).Proof.
The propagation speeds ofS6 and the wave back of R4 are calculated, respectively, by(45)σ6=c1r-+rm,ξ4r2,u2=2c1r2=2c1rm,such that we have(46)σ6-ξ4r2,u2=c1r--rm>0.It means that S6 overtakes the wave back of R4 in finite time. The intersection (x2,t2) is computed by(47)x2=σ6t2=c1r-+rmt2,x2-x0=ξ4r2,u2t2=2c1rmt2,which enables us to have(48)x2,t2=r-+rmx0r--rm,x0c1r--rm.After the time t2, S6 begins to penetrate R4 with a varying propagation speed since it has a rarefaction wave on its right. During the process of penetration, the shock wave is denoted by S7 which can be calculated by(49)dxdt=c1r+r3,x-x0t=2c1r,xt2=x2,in which we have r3=r- and r varies from rm to r+. We can get(50)x=2c1r-t-2c1x0r--rmt+x0,and by differentiating (50) with respect to t twice, we have(51)d2xdt2=12c1x0r--rmt3>0.It means that S7 begins to accelerate during the process of penetration. Now, there exist two possible situations to happen which depend on the relation between r- and r+.
(1) Ifr+<r-, then S7 is able to cancel R4 completely and ends at the point (x3,t3) (see Figure 9) which is given by(52)x3-x0=2c1r+t3,x3-x0=2c1r-t3-2c1x0r--rmt3,which enables us to get(53)x3,t3=x0r-2+r+2-2r+rmr--r+2,x0r--rmc1r--r+2.The shock wave is denoted by S8 when it penetrates R4 completely, whose propagation speed is(54)σ8=c1r++r-.
(2) Otherwise, ifr+>r-, then S7 cannot cancel R4 completely and has the line x-x0=2c1r-t in R4 as its asymptote (see Figure 10).Case 3 (J+R and J+S).
In this case, we consider that the Riemann solution originating from the point(0,0) is a contact discontinuity followed by a rarefaction wave and the Riemann solution originating from the point (x0,0) is a contact discontinuity followed by a shock wave when the initial data (5) need to satisfy both c2/c1<r-<rm and c2/c1<r+<rm (see Figures 11 and 12). When t is small enough, the solution of the Riemann problem (4) and (5) may be displayed in the symbol form (see Figure 11):(55)r-,u-+J1+r1,u1+R2+rm,um+J3+r2,u2+S4+r+,u+.The intermediate states (r1,u1) and (r2,u2) are given, respectively, by(56)r1,u1=r-,r-umrm,r2,u2=rm,rmu+r+.Figure 11
The interaction betweenJ+R and J+S is shown when c2/c1<r+<r-<rm.Figure 12
The interaction betweenJ+R and J+S is shown when c2/c1<r-<r+<rm.Lemma 4.
The wave front ofR2 collides with the contact discontinuity J3 in finite time. Consequently, on the one hand, the contact discontinuity penetrates the rarefaction wave completely. On the other hand, the rarefaction wave also passes through the contact discontinuity without changing its direction.Proof.
The propagation speeds of the wave frontR2 and J3 can be computed, respectively, by(57)ξ2rm,um=2c1rm,τ3=c1rm+c2rm.We can get(58)ξ2rm,um-τ3=c1rm-c2rm>0.It means that the wave front of R2 overtakes J3 at a point which can be calculated by(59)x1=ξ2rm,umt1,x1-x0=τ3t1,which yields(60)x1,t1=2c1rm2x0c1rm2-c2,x0rmc1rm2-c2.
After the timet1, the contact discontinuity J3 begins to penetrate the rarefaction wave R2. During the process of penetration, it is denoted by J5 which can be calculated by(61)dxdt=c1r+c2r,xt=2c1r,xt1=x1,in which r varies from rm to r-. We can get(62)x=2c1c2t2+c1rmx0t.By differentiating (62) with respect to t twice, we have(63)d2xdt2=-12c1c2t2+c1rmx0t-3/2c1rmx02<0.It means that J5 begins to decelerate during the process of penetration. It is clear that J5 is able to cancel R2 completely and ends at the point (x2,t2) which is given by(64)x2=2c1r-t2,x2=2c1c2t22+c1rmx0t2,such that we have(65)x2,t2=2c1r-rmx0c1r-2-c2,x0rmc1r-2-c2.At the same time, the rarefaction wave R2 also passes through the contact discontinuity J5 without changing its direction and is denoted by R7 after penetration, for the reason that the state variable r cannot change and only the state variable u changes during the process of penetration.
The intermediate state(r3,u3) between J6 and R2 can also be calculated by (36). The states (r1,u1) and (r3,u3) can be connected directly by a contact discontinuity denoted with J6 for the reason that r1=r3=r-. It is clear to see that the two contact discontinuities J6 and J1 are parallel to each other.Next we consider the situation that the shock waveS4 penetrates the rarefaction wave R5 and have the following lemma to describe the result.Lemma 5.
The shock waveS4 can overtake the wave front of the rarefaction wave R7 in finite time. More precisely, if r+<r3, then S4 can cancel the whole R7 thoroughly (see Figure 11). Otherwise, if r+>r3, then S4 penetrates R7 incompletely and finally has the line x=2c1r+t in R7 as its asymptote (see Figure 12).Proof.
The propagation speeds ofS4 and those of the wave front in the rarefaction wave R7 can be computed, respectively, by(66)σ4=c1rm+r+,ξ7rm,um=2c1rm.It is easy to calculate(67)ξ7rm,um-σ4=2c1rm-c1rm+r+>0,which implies that the wave front of R7 overtakes S4 in finite time. The intersection (x3,t3) is computed by(68)x3=ξ7rm,umt3=2c1rmt3,x3-x0=σ4r2,u2t3=c1rm+r+t3,such that we have(69)x3,t3=2rmx0rm-r+,x0c1rm-r+.
After the timet3, S4 begins to penetrate R7 with a varying speed since it has a rarefaction wave on its right. During the process of penetration, the shock wave is denoted by S8 which can be calculated by(70)dxdt=c1r++r,xt=2c1r,xt3=x3,in which r varies from r2(=rm) to r3(=r-). With the similar calculation as before, we can get(71)x=2c1r+t+2c1x0rm-r+t.By differentiating (71) with respect to t twice, we have(72)d2xdt2=-12c1x0rm-r+2c1x0rm-r+t-3/2<0.It means that S8 begins to reduce the speed during the process of penetration.
As before, there also exist two possible situations to happen which depend on the relation betweenr- and r+.
(1) Ifr+<r-, then S8 can cancel R7 completely and end at the point (x4,t4) which is computed by(73)x4=2c1rmt4,x4=2c1r+t4+2c1x0rm-r+t4,such that we get(74)x4,t4=2rmx0rm-r+,x0c1rm-r+.The shock wave is denoted by S9 when it penetrates R7, whose propagation speed is σ9=c1(r-+r+).
(2) Ifr+>r3, then S8 cannot cancel R7 completely and finally has the line x=2c1r+t in R7 as its asymptote.Case 4 (J+R and J+R).
Finally, we consider the situation that both the Riemann solutions originating from the points(0,0) and (x0,0) are a contact discontinuity followed by a rarefaction wave when the initial data (5) meet c2/c1<r+<rm<r+. When the time t is small enough, the solution to the perturbed Riemann problem (4) and (5) may be represented succinctly as (see Figure 13)(75)r-,u-+J1+r1,u1+R2+rm,um+J3+r2,u2+R4+r+,u+.The calculation is similar to that in Case 3. Thus, we omit the detail and only draw Figure 13 to illustrate this case.
Up to now, the discussion for all kinds of interactions has been accomplished when the conditionr-,rm,r+>c2/c1 is required. The global solutions to the perturbed Riemann problem (4) and (5) are constructed completely in explicit forms when r-,rm,r+>c2/c1. Thus, we can summarize the main results in this section by the theorem below.Figure 13
The interaction betweenJ+R and J+R is shown when c2/c1<r+<rm<r+.Theorem 6.
Under the assumption that the initial data (5) need to satisfy r-,rm,r+>c2/c1, the large-time asymptotic states of the global solutions to the perturbed Riemann problem (4) and (5) are identical with the corresponding ones to the Riemann problem (4) and (6). In other words, the solutions to the Riemann problem (4) and (6) can be obtained if the limits x0→0 of the global solutions to the perturbed Riemann problem (4) and (5) are taken when r-,rm,r+>c2/c1. Thus, it turns out that the Riemann solutions are stable with respect to the specific small perturbation (5) of Riemann initial data (6) when r-,rm,r+>c2/c1.In fact, if the initial data (5) satisfy 0<r-,rm,r+<c2/c1, the situation can be dealt with similarly. For the other situations, the wave interactions are very complicated for the perturbed Riemann problem (4) and (5) and thus left for our future work.



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