Ting ZHANGWancheng SHENG

Department of Mathematics,College of Science,Shanghai University,Shanghai 200444,China

E-mail:zhangting123@i.shu.edu.cn;mathwcsheng@shu.edu.cn

Abstract The Riemann problem for the chromatography equations in a conservative form is considered.The global solution is obtained under the assumptions that the initial data are taken to be three piecewise constant states.The wave interaction problems are discussed in detail during the process of constructing global solutions to the perturbed Riemann problem.In addition,it can be observed that the Riemann solutions are stable under small perturbations of the Riemann initial data.

Key words Riemann problem;chromatography equation;wave interactions;Temple class;hyperbolic conservation laws

1 Introduction

In this article,we consider the chromatography equations in a simple conservative form

where the variables u,v denote the positive functions of(x,t)∈R×R+.System(1.1)is derived from the mass balances of the two-component chromatography system,which we can refer to[1–3].

Chromatography was a procedure of separating chemical components in a fl uid phase and was extensively applied in chemistry and engineer.Multicomponent chromatography was investigated theoretically in lots of articles,such as[4,5].Especially,the chromatographic separation of two chemical species was studied rigorously by introducing the Langmuir isotherm.There were varieties of mathematical models in analyzing the behavior of chromatographic columns like[6–9].

Recently,the chromatography equations of di ff erent forms were developed widely.Most of them studied simple forms of the chromatographyequations for the local equilibrium theory.For example,Mazzotti[8–10]studied the delta shock wave numerically and experimentally in the Riemann solutions of two-component nonlinear chromatography model with special Langmuir isotherm

Shen[11]discussed the wave interactions and stability of the Riemann solutions for the chromatography equations

A change of variables v=u+v and w=u?v of equations(1.2)were introduced in[12],and Ambrosio et al.studied it as an example of continuity transport equations by using new well-posedness results.Also,in[13],Sun used the self-similar viscosity vanishing method to investigated the equations and proved the existence and uniqueness of solutions involving the delta shock.Another change of variables w=u+v and θ=v/u of equations(1.2)were taken by Bressan[14]and that article mainly focus on the ODEs with discontinuous vector fi elds.

Moreover,Tsikkou[15]recently transformed a chromatography model arised in[9]into a hyperbolic system in a simple conservative form and studied the singular shocks in this model.Singular shocks were discovered by Key fi tz and Kranzer fi rst in[16–18].They belong to a type of weak solutions of very low regularity.Singular shocks appear in lots of models except the chromatography model,such as in isothermal,isentropic gas dynamics[16];in a model of gravity-driven,particle-laden thin- fi lm fl ow[19,20];in the Brio system studied in plasma and the classical shallow-water system[21];and possibly in a model for chemotaxis[22].In[23],the interactions of a singular shock wave with another wave were investigated.In this aricle,we can fi nd that singular shocks appear in the study of the Riemann problem of a chromatography model(1.1)and the wave interactions are considered.

It is easy to see that system(1.1)is strictly hyperbolic and genuinely nonlinear in both characteristic fi elds.Furthermore,system(1.1)belongs to the type of equations called Temple class because the k-shock curves and the k-rarefaction curves(k=1,2)are coincide in the u?v plane.The equations of Temple class which have many interesting features arise from di ff erent areas such as oil reservoir simulation,nonlinear wave motion in elastic strings,multi-component chromatography and Leroux system.The well-posed results for these systems can be drawn for a much larger class of initial data compared with normal hyperbolic conservation laws.The Riemann problem for the Temple class can be resolved in an explicit form in the large,and wave interactions have simpli fi ed structures.There are lots of references related to the results about the Temple class such as[6,24–29].

In this article,our main purpose is to investigate interactions of elementary waves for the chromatography equations(1.1).It is noteworthy that the wave interaction problems are signi fi cant in the applications of the chromatography systems,for example,they compare with the numerical and experimental results,separate the two chemical components in the chemical fi elds and so on.Also,they are the basic problems for the general mathematical theory of the chromatography systems.In order to get various possible interactions completely,we usually need to consider the initial value problem with three pieces constant initial data

where ε>0 is arbitrarily small.Actually,the initial data(1.4)is a local perturbation of the corresponding Riemann initial data

Thus,the initial problem(1.1)and(1.4)is called as the perturbed Riemann problem below.

The Riemann problem is one of the important basic theory in studying the chromatography equations.It is known that the solutions of the Riemann problem(1.1)and(1.5)consist three constant states separated by rarefaction waves or shock waves or singular shocks.So,we can see that there exist 36 di ff erent combinations of Riemann solutions at the initial points(?ε,0)and(ε,0).It is obvious that the fi rst interaction of the 2-wave starting from(?ε,0)and the 1-wave starting from(ε,0)collide at fi rst,which is important when we construct the solutions to the perturbed Riemann problem(1.1)and(1.4).

As is well known in[30],only the existence result can be obtained by studying the Goursat problem and it is impossible to construct the global solution for the perturbed Riemann problem(1.1)and(1.4)in a completely explicit form in the case that the 2-rarefaction wave from(?ε,0)meets the 1-rarefaction wave from(ε,0).Thus,we should make some assumptions to avoid this situation.If the interaction between the 2-wave from(?ε,0)and the 1-wave from(ε,0)is completed,then it is easy to study the interactions between the waves of the same family for the Temple class.For simplicity,we restrict ourselves to consider the above situation that the solution of the Riemann problem at the origin(?ε,0)is two shock waves S1+S2.Thus,we make the following assumption to meet the above requirements.

Proposition 1.1Providing that vm>vland

hold,then for the perturbed Riemann problem(1.1)and(1.4),the solutions initiating from the point(?ε,0)are two shock waves.

As a matter of fact,we expect to study the global solutions to the perturbed Riemann problem(1.1)and(1.4).The Riemann solutions starting from(?ε,0)is fi xed and the Riemann solutions at(ε,0)changes.Namely,for the given left state(ul,vl),we fi rst fi x the intermediate state(um,vm)and then change the right state(ur,vr).Using this method,we can present the main theorem of this note as follows.

Theorem 1.2If Proposition 1.1 is satis fi ed for the initial data(1.4),then the global solutions to the perturbed Riemann problem(1.1)and(1.4)can be constructed explicitly.Moreover,if ε→ 0 is taken,then the limits of this global solutions are identical with the corresponding ones of the Riemann problem(1.1)and(1.5).

The wave interaction problem for the Temple class was studied extensively.For instance,for the pressureless gas dynamics equations[31],the isentropic Chaplygin gas dynamics equations[32–34],the Aw-Rascle traffic fl ow model[35]and various types of chromatography systems[36–39],these systems are not genuinely nonlinear in every characteristic fi elds,which means that at least one of the characteristic fi elds is linear degenerate.Therefore,the wave interaction problem for these systems is relatively simple.On the other hand,for our chromatography equations(1.1),both characteristic fi elds are genuinely nonlinear.Obviously,it is much harder to solve this kind of wave interaction problems and it has rarely been concerned about.In this note,we mainly discuss the interactions of the shock waves and the rarefaction waves explicitly by using the method of characteristics and analyzing the geometrical structures of waves in the u?v plane.We investigate the wave interaction problem through making a small perturbation of the Riemann data(1.5).Namely,when the initial data are three piecewise constant states,we construct the global solutions of the perturbed Riemann problem(1.1)and(1.4)under the assumption of Proposition 1.1.This kind of method has also been applied to the delta shock problem[31],the traffic fl ow[35],and the combustion model[40].We can also refer to[41–46]about the general wave interaction problem for hyperbolic conservation laws.Moreover,the solutions of the Riemann problem(1.1)and(1.5)are stable when there is a small perturbation as ε→ 0.

The article is arranged as follows.In Section 2,we simply recall some preliminaries about the derivation of the chromatography model and the Riemann problem(1.1)and(1.5).In Section 3,we mainly discuss the wave interactions of all kinds for system(1.1).The global solutions to the perturbed Riemann problem(1.1)and(1.4)are constructed in explicit forms when the initial data are taken to be three piecewise constant states under the appropriate assumptions,and the stability of Riemann solutions are analyzed with respect to the speci fi c small perturbation of the Riemann initial data.The conclusion is drawn in Section 4.

2 Preliminaries

In this section,we give some basic knowledge of what we are about to use.We fi rst introduce the derivation of the chromatography model(1.1).Then,we recall some main results about the Riemann problem for the chromatography equations(1.1)with the initial data(1.5).The general knowledge about the Riemann problem for the hyperbolic conservation laws can be referred in[41–43,46,47].

2.1 Derivation of the Chromatography Equations(1.1)

In this article,we consider the equilibrium chromatographyof two interacting solutes Ai,i=1,2.For an ideal chromatographiccolumn with void fraction ε and Z as the characteristic length of the column,we use ciand nito represent the concentration of the solute Aiin the fl uid phase and solid phase.Then,the mass balance equations are

where z ∈ (?∞,∞)and t∈ [0,∞)are the space and time coordinates,and V is the interstitial fl uid velocity.

If we de fi ne the dimensionless parameter ν=and the Langmuir isotherm

where N,Kiare experimentally determined constants,and we assume that K1

where fi=ci+νni,i=1,2.

Let u1=K1c1,u2=K2c2,and κi= νNKi,i=1,2,then,κ1< κ2,and equations(2.3)can be changed to

Then,rescale the coordinates by x′= κ1x and y′= τ?x so that equations(2.4)yields to

Next,we let

then,we obtain

At last,we introduce the variables u,v again by

and rescale the space and time coordinates x= γx′,t=y′.Finally,we obtain the system we are going to study

Through the derivation,we know that u>0,v>0,and γ∈(0,1).

2.2 The Riemann Problem(1.1)and(1.5)

Let U(x,t)=(u(x,t),v(x,t)),then system(1.1)can be written in this form

where

From|λE ? dF|=0,we can get the characteristic values:

and the corresponding right eigenvectors are

We consider the region

it is clear to see that λ2> λ1>0,which means that system(1.1)is strictly hyperbolic in this region.On the curve v2? 4u=0,system(1.1)is non-strictly hyperbolic because λ1= λ2and

By a direct calculation,we can see

such that both characteristic fi elds are genuinely nonlinear.Therefore,the waves are either rarefaction waves or shock waves for the fi rst and second families.The Riemann invariants along the characteristic fi elds are

Given Ul=(ul,vl),we consider what kind of state U=(u,v)can be connected to Ulon the right by a shock wave or rarefaction wave.

First,we suppose that the solutions are smooth.Let ξ=,then,system(1.1)can be rewritten to

namely,

When Uξ=0,U is constant.When Uξ6=0,U is not constant,and we can get two rarefaction waves.From

we can obtain ξ= λ1,2,consequently,we have=u2λ1,2from(2.17).

When ξ= λ1,2,from(2.17),we have

then,we can obtain that for both characteristic fi elds,ul>u,vl>v.So,the rarefaction waves in the u?v plane are

Next,when the solutions are discontinuous,we have already known that the Rankine-Hugoniot jump condition σ[U]=[F(U)]is satis fi ed on the discontinuous curve,where σ =and[U]=Ur?Ul.So,we have

then we obtain

In view of the entropy conditions,the i-shock waves(i=1,2)satisfy the following inequalities

then we have u>ul,v>vl.Through a calculation,we can obtain the fi rst and second derivatives at the point(ul,vl),

So,the shock waves in the u?v plane are

It is clear to see that the shock waves and the rarefaction waves are coincide in the u?v phase plane,which means that system(1.1)is called Temple class.For the given left state Ul=(ul,vl),if we denote the two families of curves by Wi(ul,vl)=Si(ul,vl)∪Ri(ul,vl)(i=1,2),then the lines of W1(ul,vl)and W2(ul,vl)are tangent to the parabola?Γ:v2?4u=0?at the points A and B.So,the region ?,mentioned as(2.12),is divided into six parts and can be depicted in Figure 1.When the right state Ur=(ur,vr)is in the region I(ul,vl),it is connected with Ulon the right by a back shock wave S1and a front rarefaction wave S2.We simply write the solution to the Riemann problem(1.1)and(1.5)as S1+S2.Similarly,when Ur∈II(ul,vl),III(ul,vl),or IV(ul,vl),the Riemann solution is S1+R2,R1+R2or R1+S2,respectively.

Figure 1 Characteristic curves of shock and rarefaction waves in u?v plane

In addition,there are cases that no classical Riemann solutions exist.We can observe that v2=4u is an invariant curve for(1.1),and if U(x,t)is a smooth solution on this curve then u satis fi es the equation

Therefore,if Ur∈V(ul,vl)in Figure 1,then the solution consists a back rarefaction wave from Ulto UA,a rarefaction solution to equation(2.30)from UAto a point UC,and a front rarefaction wave from UCto Ur,where UCis the point where R2(Ur)is tangent to v2=4u.

When the state Urlies in V I(ul,vl),there exists no classical Riemann solution.Using a similar method in[15],we know that a special weak solution called singular shock appears in this situation.The singular shock satis fi es

To make it simple,we use the notationsandto denote the slopes of the two lines through(ui,vi)tangent to the parabola?Γ:v2?4u=0?in the u?v plane.For the Riemann problem(1.1)and(1.5),we assume that the state(um,vm)is connected with(ul,vl)on the left by 1-wave and connected with(ur,vr)on the right by 2-wave.Then,we have

Thus,with a calculation as[48],for the Riemann initial data(1.5),the intermediate state(um,vm)of the solution is denoted by

Without confusion,we use similar notations here and below.

3 Wave Interactions

For the perturbed Riemann problem(1.1)and(1.4),our purpose is to fi nd the global solutions under Proposition 1.1,in which the main task is to deal with the wave interaction problem.As we discussed in Section 2,when(ul,vl)is given,there are six cases of the location of(um,vm).Here we suppose that the state(um,vm)lies in the fi rst quadrant of(ul,vl),that is the Riemann solution originating from the point(?ε,0)in the x?t plane is always two shock waves,denoted by S1and S2in the following discussion.For this case,we will give a detailed discussion about the wave interactions case by case.We can separate region ? as(2.13)into 15 parts if we put together all the wave lines of the states(ul,vl)and(um,vm)in the u?v plane,as is shown in Figure 2.When the right state(ur,vr)lies in each part,we can get di ff erent kinds of wave interactions.The discussion can be divided into the following six cases due to the Riemann solutions starting from the point(ε,0).

Case 1S+S and S+S

In this situation,there are two subcases.

Subcase(a)At fi rst,we are concerned about the situation when the state(ur,vr)lies in the fi rst part in Figure 2,that is,(ur,vr)∈V1.In this case,the Riemann solutions initiating from the point(ε,0)are also two shock waves,denoted by S3and S4,respectively.It is clear that the conditions vm

hold.Then,we can see that when the time t is sufficiently small,the solution of the perturbed Riemann problem(1.1)and(1.4)is written by(see Figure 2)

The intermediate states(u1,v1)and(u2,v2)can be obtained as(2.32)by

Figure 2 If(um,vm)∈ I(ur,vr),then the region ? is divided into 15 parts

Lemma 3.1The shock waves S2and S3collide in fi nite time.

ProofWith a calculation,we can obtain the following propagation speeds of the shock waves S2and S3,respectively,

It is clear from Figure 3 that>and u1 σ3,which means the shock wave S2overtakes S3in fi nite time.

The interaction point(x1,t1)is determined by

which leads to

When the shock waves S2and S3collide at the point(x1,t1),there is a new local Riemann problem with the initial data(u1,v1)and(u2,v2).To solve this problem,we must determine which region(u2,v2)belongs to with respect to(u1,v1)in the u?v plane.

Lemma 3.2The state(u2,v2)lies in the fi rst quadrant of(u1,v1),that is the local Riemann problem at the point(x1,t1)is resolved by two shock waves,denoted by S5and S6.Furthermore,the propagation speeds of the shock waves satisfy σ3< σ5and σ2> σ6.

ProofFirst,if we can show that the line S1(um,vm)always lies above the line S1(u1,v1)in the u?v plane,then we will have proved that state(u2,v2)lies in the fi rst quadrant of(u1,v1).From(2.28),we have

As is shown in Section 2,the Riemann problem at(x1,t1)is resolved by two shock waves,denoted by S5and S6,respectively(see Figure 3).Similarly,it can be obtained that the intermediate state(u3,v3)between S5and S6is

Now we consider the interaction between two shock waves of the same family.

Figure 3 The interaction between S+S and S+S when(ur,vr)∈V1

Lemma 3.3The shock waves S1and S5interact in fi nite time and form a new back shock wave which is denoted by S7in Figure 3.Similarly,the shock waves S6and S4also interact in fi nite time,but form a new front shock wave which is denoted by S8in Figure 3.

ProofAt fi rst,we are concerned with the coalescence of the two shock waves S1and S5belonging to the fi rst family.The propagation speeds of S1and S5are calculated by

It is also easy to see from Figure 3 that the states(u3,v3)and(ur,vr)can be connected by a front shock wave,which is denoted by S8.The speed of S8is

Noticing that u3 σ8> σ4.It can be calculated that σ7< σ8.The proof is completed. ?

Subcase(b)Then,we consider another situation of Case 1 when(ur,vr)∈V14.The former discussion is the same as Lemmas 3.1 to 3.3,but we see that σ7> σ8in the end.So,the shocks S7and S8interact at the point(x4,t4),which is determined by

According to Figure 4,at the point(x4,t4),there is a new Riemann problem with the initial data(ul,vl)and(ur,vr).Then,the solution of the Riemann problem is obtained in the following lemma.

Figure 4 The interaction between S+S and S+S when(ur,vr)∈V14

Lemma 3.4When the state(ur,vr)∈V14,we can see from Figure 2 that(ur,vr)also lies in the V I part of the state(ul,vl).Thus,the solution after the interactions is a singular shock wave,denoted by Ssingular(see Figure 4).Furthermore,the speed of Ssingularis between S7and S8.

ProofFrom the position of(ul,vl),(u3,v3),and(ur,vr),we know that

Then,we add them to get

as vl

With a tedious calculation,we can obtain that σ8< σSsingular< σ7. ?

Case 2S+S and S+R

In this case,we consider the situation when the state(ur,vr)belongs to V2∪V3in Figure 2,that is,the Riemann problem initiating from(ε,0)is resolved by a back shock wave and a front rarefaction wave.Then,the local solution of the perturbed Riemann problem(1.1)and(1.4)is denoted by(see Figure 5 and Figure 6)

Figure 6 The interaction between S+S and S+R when(ur,vr)∈V3

The intermediate states(u1,v1),(u2,v2)and(u3,v3)are presented as those in Case 1 here and below.And,(ur,vr)satis fi es the inequalities

Like in Case 1,the same conclusions of Lemmas 3.1 and 3.2 can also be established here,so is the former part of Lemma 3.3.In addition,the interaction of the shock wave S6and the rarefaction wave R4of the same family can be depicted in the following lemma.

Lemma 3.5The shock wave S6overtakes the rarefaction wave R4in fi nite time,and then starts to penetrate R4.To be more speci fi cally,(a)if ur>u3,then S6is able to penetrate the rarefaction wave R4completely;(b)if ur

ProofWith a direct calculation,we obtain the propagation speeds of S6and the wave back of R4are(3.16)and

After the time t3,S6starts to penetrate R4.As we can see in Figure 5 and Figure 6,there is a shock wave S8of the second family connecting the state(u3,v3),and the varying state(u,v)along the characteristic lines in R4.During the penetration,the propagation speed of S8is

in which u varies from u2to ur.Therefore,the shock wave S8in the x?t plane is determined by

It means that the shock curve S8is convex in the x?t plane,that is the speed of the shock wave is faster during the penetration.

More precisely,there are two possible subcases depending on the region where the state(ur,vr)belongs to.

(a)If(ur,vr)∈V2,that is ur>u3,then S8has the ability to penetrate the whole R4thoroughly,and the ending point is denoted by(x4,t4).From(3.32)and(3.35),we have

Thus,the point(x4,t4)is determined by

After the time t4,the shock wave is denoted by S9whose propagation speed is

(b)If(ur,vr)∈V3,that is ur

Case 3S+S and R+S

In this case,there are also two situations,and we will discuss them respectively.

Subcase(a)First,we turn our attention to the situation when the state(ur,vr)lies in V4∪V5in Figure 2,which means the Riemann problem starting from the point(ε,0)is resolved by a back rarefaction wave and a front shock wave.Then the local solution of the perturbed Riemann problem(1.1)and(1.4)is denoted by(see Figure 7 and Figure 8)

The intermediate states(u1,v1),(u2,v2)and(u3,v3)are also presented as those in Case 1 hereafter.The state(ur,vr)satis fi es the following inequalities

We fi rst consider the interaction of the shock wave S2and the rarefaction wave R3and discuss it in the following lemma.

Figure 7 The interaction between S+S and R+S when(ur,vr)∈V4

Figure 8 The interaction between S+S and R+S when(ur,vr)∈V5

Lemma 3.6The shock wave S2is able to cross the whole rarefaction wave R3completely in fi nite time.Then,during the process of penetration,a transmitted rarefaction wave is generated and the shock wave speeds up.

ProofIt is easy to obtain that the propagation speeds of S2and the wave back of R3are given by(3.15)and

namely,

After the time t1,the shock wave S2starts to penetrate the rarefaction wave R3,and it is denoted by S6during the penetration.When R3crosses S6,a transmitted rarefaction wave which is denoted by R5is generated.We use(u?,v?)and(u+,v+)to denote the corresponding states in R3and R5,then the propagation speeds of the corresponding characteristic lines are

They also should satisfy the following relation

where um>u?>u2and u1>u+>u3.With u?>u+in mind,we can see from(3.45),(3.46),and(3.47)that ξ3(u?,v?)< ξ5(u+,v+).Namely,the rarefaction wave speeds up after the penetration.

Meanwhile the shock wave S6is determined by

Through a tedious calculation,we have

After the time t2,the shock wave S7keeps on moving forwards,and then catches up with the shock wave S4since σ7> σ4,in which σ7and σ4are propagation speed of S7and S4,respectively.The intersection(x3,t3)is computed like before,then we have

Similarly as we described in Case 1,S7and S4coalesce into a new front shock wave S8.On the other hand,the interaction of the shock wave S1and the rarefaction wave R5is discussed in the following lemma.

Lemma 3.7The shock wave S1is able to overtake the rarefaction wave R5in fi nite time.To be more precisely,(a)if ulu3,then S1cannot penetrate the whole R5and fi nally is asymptotic to the characteristic line in R5whose speed is

ProofAs before,the propagation speeds of S1and the wave back of R5are(3.17)and

in which the point(ˉx,ˉt)lies on the shock curve S6and the state(u+,v+)is on the corresponding characteristic lines of R5.It is impossible to obtain the explicit form of S9for now as R5is not a centered rarefaction wave.According to the region where the state(ur,vr)is located,there are two possible subcases as follows.

(a) If(ur,vr)∈V4,that is,ul

(b)If(ur,vr)∈V5,that is,ul>u3,then S9cannot cancel the whole R5,and eventually is asymptotic to the characteristic line in R5whose speed is(see Figure 8). ?

Subcase(b)When the state(ur,vr)∈V13,it can be seen that the former part of the interactions is the same as Subcase(a)in Case 3.The di ff erence is that the speed of S10is faster than S8here.Therefore,in Figure 7,the shocks S8and S10intersect at fi nite time and form a singular shock(see Figure 4).

Lemma 3.8It can be seen from Figure 2 that if the state(ur,vr)∈V13,then(ur,vr)also lies in the VI part of the state(ul,vl).Therefore,the solution after the interactions is also a singular shock wave,whose speed is also between S10and S8.

The proof of this lemma is similar to Lemma 3.4.

Case 4S+S and R+R

In this part,we consider the situation when the state(ur,vr)lies in the region V6∪V7∪V8∪ V9.In this case,the Riemann problem initiated from the point(ε,0)is resolved by two rarefaction waves,denoted by R3and R4,respectively.Thus,for sufficiently small t,the solution of the perturbed Riemann problem(1.1)and(1.4)can be presented by

The following conditions

are satis fi ed.

The interaction of the shock wave S2and the rarefaction wave R3is the same as we described in Case 3.Namely,S2is able to penetrate the whole R3completely in fi nite time,and a transmitted rarefaction wave R5is generated.Then,the interaction of S1and R5is like that in Lemma 3.6.If ulu3,that is,(ur,vr)∈V8∪V9,then S1cannot cancel R5completely.On the other hand,the interaction of the shock wave S7and the rarefaction wave R4is also like that in Lemma 3.4.It is easy to obtain that if ur>u3,that is,(ur,vr)∈V6∪V8,then S7is capable of canceling the entire R4.Otherwise,if ur

Therefore,we just need to combine the conclusions above and we can obtain the following main results of this case.Here we omit the details.

(a)If(ur,vr)∈V6,that is,ul

(b)If(ur,vr)∈V7,that is,u3>max{ul,ur},then S1is able to cancel the entire R5,while S7cannot cross R5completely(see Figure 10).

(c)If(ur,vr)∈V8,that is,u3

(d)If(ur,vr)∈V9,that is,ul>u3>ur,then S1is not able to cancel the entire R5,and S7also cannot cancel R5completely(see Figure 12).

Figure 9 The interaction between S+S and R+R when(ur,vr)∈V6

Figure 10 The interaction between S+S and R+R when(ur,vr)∈V7

Case 5S+S and?R

Then,we turn to the case that the state(ur,vr)lies in the region V10∪V11∪V12,in which(ur,vr)satis fi es the conditions

When the time t is small,the Riemann solution of(1.1)and(1.4)is denoted by

where?R represents a rarefaction wave with the propagation speed from λ1(um)to λ2(ur).It follows from a similar discussion that after the wave interactions,we obtain three di ff erent kinds of solutions:(The pictures are similar to Figure 11 and Figure 12)

Figure 11 The interaction between S+S and R+R when(ur,vr)∈V8

Figure 12 The interaction between S+S and R+R when(ur,vr)∈V9

(a)If(ur,vr)∈V10,then we see from Figure 2 that(ur,vr)∈III(ul,vl).So the global solution of the Riemann problem(1.1)and(1.4)is←?R+?→S.

(b)If(ur,vr)∈V11,then we see from Figure 2 that(ur,vr)∈IV(ul,vl).So the global solution of the Riemann problem(1.1)and(1.4)is←?R+?→R.

(c)If(ur,vr)∈V12,then we see from Figure 2 that(ur,vr)∈V(ul,vl).So the global solution of the Riemann problem(1.1)and(1.4)is?R.

Case 6S+S and Ssingular

At last,we consider the case that no classical Riemann solutions exist at the initial point(ε,0),that is,the state(ur,vr)lies in V15.Here,the following relations

hold.Then,we can write the local Riemann solution of(1.1)and(1.4)when the time t is sufficiently small as

Figure 13 The interaction between S+S and R+R when(ur,vr)∈V15

The proofs of Lemma 3.10 and Lemma 3.12 are similar to Lemma 3.4.

Above all,we know that in this case,the global solution of the Riemann problem(1.1)and(1.4)is singular shocks.

4 Conclusions

So far,we discuss the interactions for all cases.It can be seen from the above results that we fi nish the construction of the global solutions of the perturbed Riemann problem(1.1)and(1.4)in explicit forms provided that Proposition 1.1 holds.It is clear that the limit global solutions to the perturbed Riemann problem(1.1)and(1.4)when ε→ 0 are identical with the corresponding ones to the Riemann problem(1.1)and(1.5).That is to say,the Riemann solutions are stable for the small perturbation(1.4)of the Riemann initial data(1.5).According to our discussion above,the proof of Theorem 1.2 is accomplished.

Actually,if the Riemann solution originating from the point(?ε,0)is resolved by R+S,then,we can also obtain the similar results as those in this article under appropriate assumptions.We made the assumptions in order to avoid the interaction of the front rarefaction wave and the back rarefaction wave,as the global solutions cannot be constructed explicitly in this situation.Furthermore,we may make suitable assumptions to fi x the Riemann solution initiating from the point(ε,0)is S+S or S+R and change the Riemann solution at the point(?ε,0).Namely,for the given right state(ur,vr),we fi rst fi x the intermediate state(um,vm),and change the left state(ul,vl).Then,the global solutions to the perturbed Riemann problem(1.1)and(1.5)can also be constructed as we discuss in this note.

It is worth noticing that the wave interactions for system(1.1)has important applications in the chromatography equations.Although system(1.1)is a simpler transformation of the chromatography system,the results we obtain here can be applied to the more general chromatography system of the Temple class and for more general initial data.Moreover,the method we use to discuss the wave interactions is a very important tool to study the behavior of the chromatographic column.

Acknowledgements The frist author would like to express her gratitude to all those who helped her during the writing of this thesis.