# The Existence of Zero Waves for Nonsimplified Chromatography System

Authors: Maozhou Lin; Lihui Guo; Yinsong Bai
Journal: Mathematical Problems in Engineering
 (2020)

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## Abstract

In this paper, we mainly consider Riemann problem for the widely used nonsimplified chromatography system with initial data consisting of three pieces of constant states. Through phase plane analysis, the solutions of the nonsimplified chromatography system are established. When the different initial data tend to−1 from the right side, the existence of zero shock wave, zero delta shock wave, and zero rarefaction wave is obtained via analyzing its wave interaction. Finally, the correctness of the main conclusions is verified by numerical simulation, and the numerical results are in good agreement with the theoretical solutions of several experimental cases.


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## 1. Introduction

Chromatography plays an important role in modern industry. It is used to separate two chemical components especially by chemists and engineers. The research of the nonlinear chromatography system is, in particular, a significant part of the theory of chromatography. Mazzotti [1–3] captured the delta shock wave for local equilibrium model (1) by numerical and experimental methods:(1)ut+u+au1−u+vx=0,vt+v+bv1−u+vx=0,where b>a>0 are constants. u and v, which describe concentrations of two chemical species, are positive function with 1−u+v>0. The complexity of system (1) is due to hyperbolicity in the region a1+v+b1−u2−4ab1−u+v>0 and ellipticity in the rest of the phase plane.In order to make the deep research for (1), one could consider the following chromatography system:(2)ut+u1−u+vx=0,vt+v1−u+vx=0.In 2013, Wang [4] had found the Riemann solution of (2) contains delta shock wave with Dirac delta function support on both u and v. It is clear that (2) belongs to the Temple class [5, 6]; in other words, the shock curves and rarefaction curves are connected in the phase plane. One can refer to [7–13].Recently, Ambrosio et al. [14] introduced the change of variables:(3)p=v−u,q=v+u,and derived the following simplified chromatography system:(4)pt+p1+px=0,qt+q1+px=0.They obtain the well-posedness. In 2011, Sun [15] discovered that solutions of (4) and (6) contain delta shock wave in case of p−=0<p+ via self-similar vanishing viscosity approach. In 2015, Shen [16] considered the asymptotic behaviors of solutions of the perturbed Riemann problem (4) and (6) near the singular curve. For other related results, one can refer to [17–24].The zero waves describe the equilibrium of separation; under this state, there is no exchange of kinetic between two chemical species. Zero waves play an important role in analyzing the chromatography systems [1, 2, 23]. Moreover, it plays a key role in the study of traffic flow, production flow, and the supply chain with limited buffers [25–29].At the same time, solving partial differential equations is always an interesting and difficult task. In recent decades, many kinds of analytical methods have been proposed. There are also some numerical methods, which are powerful treatments in solving nonlinear equations. Such as numerical methods of two-phase flow equations are proposed and developed by Zeidan et al. [30–34], which provide perfect approaches for the study of wave structure and our future studies. In this paper, we construct a solution of (2) with Riemann initial data:(5)u,vx,t=u−,v−,−∞<x<0,um,vm,0<x<1,u+,v+,1<x<+∞,and analyse the interaction of various waves. We find interesting phenomena: when the differences v−−u− and vm−um in (5) tend to −1+, the zero waves occur. Now, we give our main results.Theorem 1.
According to different zero waves produced, we divide the solution of (2) and (5) into three types:(1)
When−1<v−−u−<vm−um<0, on the one hand, the solution performs as a zero shock wave if v+−u+<0 and v−−u−⟶−1+. On the other hand, its solution performs as a zero shock wave in the interval 0,1 and a zero delta shock wave in the interval 1,∞ if v+−u+>0 and v−−u−⟶−1+.(2)
When−1<v−−u−<0<vm−um and v−−u−<v+−u+, the solution performs as a zero delta shock wave when v−−u−⟶−1+.(3)
When−1<vm−um<0 and vm−um<minv−−u−,v+−u+, a zero rarefaction wave is alive in the interval 0,1 when vm−um⟶−1+.This paper is divided into seven sections. In Section2, we review the Riemann problem (2) and (6) and discover the zero waves appear when the differences v−−u− and v+−u+ in (5) tend to −1+. In Section 3, we construct solution of Riemann problem (2) and (5), and then we obtain zero shock wave by studying the interaction of waves. In Section 4 and Section 5, we obtain, by analogy, zero delta shock wave and zero rarefaction wave of system (2) and (5). In Section 6, we verify the presentation of zero waves by numerical simulations. Finally, conclusions are drawn in Section 7.


## 2. Zero Waves in the Limit of Riemann Solutions to (2)

In this section, we review the solution to system (2) with initial data:(6)u,vx,0=u−,v−,x<0,u+,v+,x>0,where u± and v± are positive constants. In light of [4], we know characteristics satisfy λ1=1/1−u+v2 and λ2=1/1−u+v. System (2) is strictly hyperbolic if −1<v−u<0 or v−u>0 and nonstrictly hyperbolic if v−u=0. The characteristics λ1 and λ2 are genuinely nonlinear and linearly degenerate, respectively. Therefore, the associated waves are rarefaction waves R or shock waves S for λ1 and contact discontinuities J for λ2:(7)R:uu−=vv−,v+u<v−−u−,λ=11−u+v2,S:uu−=vv−,v+u>v−−u−,ω=11−u+v1−u−+v−.J:v−u=v−−u−,τ=11−u+v=11−u−+v−.Using the phase plane analysis method (see Figure1), Riemann solutions of (2) are constructed as follows:(1)
If0<v−−u−<v+−u+, then the Riemann solution is S←+J:(8)u,vx,t=u−,v−,ξ<ω,v+−u+u−v−−u−,v+−u+v−v−−u−,ω<ξ<τ,u+,v+,ξ>τ,in which the propagation speeds ofS← and J are ω=1/1−u−+v−1−u++v+ and τ=1/1−u++v+, respectively. In particular, if v+−u+=0, then ω=τ=1, and the line of S← is identical with J. Therefore, there exists the composite wave S←J.(2)
If0<v+−u+<v−−u−, then solution is R←+J:(9)u,vx,t=u−,v−,ξ<λ1u−,v−,uu−=vv−,λ1u−,v−<ξ<λ1u+,v+,v+−u+u−v−−u−,v+−u+v−v−−u−,λ1u−,v−<ξ<τ,u+,v+,ξ>τ.Ifv+−u+=0, then a composite wave R←J forms.(3)
If−1<v−−u−<v+−u+<0, then the Riemann solution is J+S⟶:(10)u,vx,t=u−,v−,ξ<τ,v−−u−u+v+−u+,v−−u−v+v+−u+,τ<ξ<ω,u+,v+,ξ>ω.in which the propagation speeds ofJ and S are τ=1/1−u−+v− and ω=1/1−u−+v−1−u++v+, respectively.We deduce from (10) that(11)limv−−u−⟶−1+τ=limv−−u−⟶−1+11−u−+v−=+∞,limv−−u−⟶−1+ω=limv−−u−⟶−1+11−u−+v−1−u++v+=+∞.This fact in turn implies that both the propagation speeds ofJ and S⟶ tend to +∞. In other words, both J and S⟶ coincide with the positive x-axis and the middle state v−−u−u+/v+−u+,v−−u−v+/v+−u+ disappears. Thus, zero shock wave [28] appears in this situation.(4)
If−1<v+−u+<0<v−−u−, then the solution is J+R⟶:(12)u,vx,t=u−,v−,ξ<τ,v−−u−u+v+−u+,v−−u−v+v+−u+,τ<ξ<λ1u−,v−,uu+=vv+,λ1u−,v−<ξ<λ1u+,v+,u+,v+,ξ>λ1u+,v+.in whichτ=1/1−u−+v−. Since v+−u+⟶−1+, then(13)limv+−u+⟶−1+λ1u+,v+=limv+−u+⟶−1+11−u++v+2=+∞.Equality (13) implies the propagation speed of the wave front of R⟶ tends to +∞. Then, zero rarefaction wave forms.(5)
If−1<v+−u+<v−−u−<0, then the Riemann solution is R←+R⟶:(14)u,vx,t=u−,v−,ξ<λ1u−,v−,uu−=vv−,λ1u−,v−<ξ<1,uu+=vv+,1<ξ<λ1u+,v+,u+,v+,ξ>λ1u+,v+.The line isx=t both the wave front of R← and the wave back of R⟶. As v+−u+⟶−1+, the zero rarefaction wave also appears in this condition.(6)
If−1<v−−u−<0<v+−u+, Wang [4] constructed delta shock wave:(15)u,vx,t=u−,v−,x<σt,βt,βtδx−σt,x=σt,u+,v+,x>σt,where(16)σ=11−u−+v−1−u++v+,βt=u−v+−u+v−t1−u−+v−1−u++v+.Figure 1
Phase plane.
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(b)He checked that the measure solution (15) with (16) satisfies the generalized Rankine–Hugoniot condition:(17)dxdt=σ,dβtdt=−σu+u1−u+v,dβtdt=−σv+v1−u+v.To ensure the uniqueness of delta shock wave solution, entropy condition should be satisfied provided−1<v−−u−<0<v+−u+,(18)λ1u+,v+<λ2u+,v+<σ<λ2u−,v−<λ1u−,v−,namely,(19)11−u++v+2<11−u++v+<σ<11−u−+v−<11−u−+v−2.That is to say, all the characteristics are incoming from both sides.For the measure solution (15) with (16), we deduce(20)limv−−u−⟶−1+σ=limv−−u−⟶−1+11−u−+v−1−u++v+=+∞,(21)limv−−u−⟶−1+βt=limv−−u−⟶−1+u−v+−u+v−t1−u−+v−1−u++v+=+∞.We apply (20) and (21) to deduce that both the propagation speed and the strength of the delta shock wave tend to +∞. Therefore, there is a delta shock wave.


## 3. Analysis of Zero Shock Wave

Based on the fact that all situations occurring in dynamic pictures cannot be depicted by Riemann problem (2), we consider the initial valuing problem (2) and (5).In this section, we are interested in the interactions of zero shock wave with other elementary waves. Initial data with (5) together with −1<v−−u−<−1+ε<vm−um<0 ensure the appearance of zero shock wave at the point 0,0.Case 1.
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For smallt>0, if −1<v−−u−<v+−u+<vm−um<0 (Figure 2), then the solution of (2) and (5) is as follows:(22)u−,v−+J1+u1,v1+S1⟶+um,vm+J2+u2,v2+R⟶+u+,v+,in which(23)u1,v1=v−−u−umvm−um,v−−u−vmvm−um,u2,v2=vm−umu+v+−u+,vm−umv+v+−u+.
The propagation speeds ofS1⟶ and J2 are ω1=1/1−u−+v−1−um+vm and τ2=1/1−um+vm, respectively. Since ω1>τ2 provided −1<u−−v−<0, then the shock wave S1⟶ overtakes contact discontinuity J2 in finite time. The waves S1⟶ and J2 interact at the point:(24)x1,t1=1u−−v−,1−um+vm1−u−+v−u−−v−.
Fort>t1, notice the condition −1<v1−u1<v2−u2<0, a new contact discontinuity J3 and a new shock S2⟶ generate. By applying (23), the following equation can be obtained:(25)u3,v3=v−−u−u+v+−u+,v−−u−v+v+−u+.
The propagation speeds ofS2⟶ and the wave back in R⟶, respectively, are ω2=1/1−u−+v−1−um+vm and λ2u2,v2=1/1−um+vm2. Since ω2>λ2u2,v2, S2⟶ can catch up with R⟶ in finite time and the intersection point x2,t2 is given by(26)x2,t2=1−um+vmvm−um−v−−u−,1−u−+v−1−um+vm2vm−um−v−−u−.
Aftert2, the shock wave S3⟶ crosses the rarefaction wave R⟶ with a varying speed during the penetration. The shock wave S3⟶ satisfies the following equation:(27)dxdt=11−u−+v−1−u+v,x−1=t1−u+v2,x2,t2=1−um+vmvm−um−v−−u−,1−u−+v−1−um+vm2vm−um−v−−u−.
In view of (27), we can obtain that(28)x−1=t1−u−+v−−vm−um−v−−u−1−u−+v−.
The waveS3 and the wave front of R⟶ interact at the point(29)x3,t3=1+vm−um−v−−u−1−u−+v−v+−u+−v−−u−2,vm−um−v−−u−1−u−+v−1−u++v+2v+−u+−v−−u−2.
After the timet3, a new shock wave S4⟶ forms and the propagation speed of S4⟶ is 1/1−u++v+1−u−+v−.
Combining (24), (26), and (29), we discover(30)limv−−u−⟶−1+x1,t1=limv−−u−⟶−1+x2,t2=limv−−u−⟶−1+x3,t3=1,0.
Based on the above discussion, asv−−u−⟶−1+, the solution of (2) and (5) has the zero shock wave starting from the point 0,0.Figure 2
When−1<v−−u−<v+−u+<vm−um<0.Case 2.
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In this case, for−1<v−−u−<vm−um<v+−u+<0 (see Figure 3), from Case 1, it is clear to see that the interaction point x1,t1 of S1⟶ and J2 is given by (24), and the states u1,v1, u2,v2, and u3,v3 satisfy (23) and (25), respectively.
Sinceω3=1/1−u−+v−1−um+vm>ω2=1/1−um+vm1−u++v+, for −1<v−−u−<v+−u+<0, then S3⟶ overtakes the S2⟶ in finite time and the interaction point is as follows:(31)x2,t2=1−u++v+v+−u+−v−−u−,1−u−+v−1−um+vm1−u++v+v+−u+−v−−u−.
In view of (24) and (31), we see(32)limv−−u−⟶−1+x1,t1=limv−−u−⟶−1+x2,t2=1,0.
Whenv−−u−⟶−1+, the propagation speeds of all shock waves and contact discontinuities tend to +∞. Thus, there is a zero shock wave starting from 0,0 in the limit v−−u−⟶−1+.Figure 3
When−1<v−−u−<vm−um<v+−u+<0.Case 3.
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Our plan is to show that a delta shock waveδS1 emits from the point 1,0 as −1<v−−u−<vm−um<0<v+−u+ (see Figure 4). The state u1,v1 is given by (23).
Owing toω1>σ1, then S1⟶ overtakes δS1 at the point(33)x1,t1=1−u++v+v+−u+−v−−u−,1−u−+v−1−um+vm1−u++v+v+−u+−v−−u−.
It is easy to see that a new delta shockδS2 forms, with t>t1, −1<v−−u−<vm−um<0<v+−u+. The propagation speed of δS2 is σ2=1/1−u−+v−1−u++v+. Notice that propagation speed of contact discontinuity J is τ1 and τ1>σ2, and J1 and δS2 interact at the point(34)x2,t2=v+−u+−vm−um1−u++v+v+−u+−v−−u−v+−u+,v+−u+−vm−um1−u−+v−1−u++v+v+−u+−v−−u−v+−u+.
Fort>t2, there is a new delta shock δS3 with same propagation speed of δS2. It follows from (16) that the strengths of the all delta shock waves satisfy the following equation:(35)β1t=umv+−u+vm1−um+vm1−u++v+t,0≤t≤t1,β2t=β1t1+u1v+−u+v11−u1+v11−u++v+t−t1,t1≤t≤t2,β3t=β2t2+u−v+−u+v−1−u−+v−1−u++v+t−t2,t≥t2.
We then combine (33) and (34) to discover(36)limv−−u−⟶−1+x1,t1=limv−−u−⟶−1+x2,t2=1,0,limv−−u−⟶−1+β1t=limv−−u−⟶−1+β2t=limv−−u−⟶−1+β3t=∞.
Based on the above analysis, all the waves overlap with the positivex-axis when v−−u−⟶−1+. Hence, at the interval 0,1 and the point 1,0, zero shock wave and zero delta shock wave occur, respectively.
The above discussions verify the type1 of Theorem 1.Figure 4
When−1<v−−u−<vm−um<0<v+−u+.


## 4. Analysis of Zero Delta Shock Wave

In this section, we mainly consider zero delta shock wave interacts with other waves. Taking sufficientlyε together with −1<v−−u−<−1+ε<0<vm−um and v−−u−⟶−1+, the Riemann solution of (2) has a zero delta shock wave at the point 0,0.Case 4.
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In this case, we are concentrated on the interaction ofδS with R←+J (see Figure 5). The state between R← and J is u1,v1=v+−u+um/vm−um,v+−u+vm/vm−um. The propagation speeds of δS1 and the wave back of R⟵ are σ1=1/1−u−+v−1−um+vm and λ1um,vm=1/1−um+vm2. If −1<v−−u−<0<vm−um, then the δS will meet R⟵ at point(37)x1,t1=1−um+vmvm−um−v−−u−,1−u−+v−1−um+vm2vm−um−v−−u−.
The strength ofδS1 is given by(38)β1t=u−vm−umv−1−u−+v−1−um+vmt.
By substituting (37) into (38), we can obtain the following equation:(39)β1t1=u−vm−umv−1−um+vmvm−um−v−−u−.
After the timet1, there is δS2 which has the same expression of (28).
Next, we compute the strength ofδS2, by virtue of splitting the delta function method introduced by Nedeljkov and Oberguggeenberger in [35]. This method has also been used in [36, 37]. Now, we construct the delta shock wave x=xt which satisfies the following equation:(40)vx,t=v−,x<xt,tx−1−1vmvm−um,x>xt+β2−tδ−x−xt+β2+tδ+x−xt,(41)ux,t=u−,x<xt,tx−1−1umvm−um,x>xt,+β2−tδ−x−xt+β2+tδ+x−xt,in which β2tδx−xt=β2−tδ−x−xt+β2+tδ+x−xt is a split delta function and β2t=β2−t+β2+t denotes the strength of δS2. Furthermore, we see from (40) and (41) that(42)vt=vmvm−um⋅H2x−1t−tx−1−1vmvm−um−v−δσ2+β2−′t+β2+′tδ−σ2β2+t+β2−tδ′,(43)v1−u+vx=−vmvm−um⋅H2x−1t+vmvm−um−vmvm−um⋅x−1t−v−1−u−+v−δ+β2+t+β2−tδ′,where H, δ, and δ′ are the functions of x−xt; besides, xt is given by (28). We substitute (42) and (43) into the first equation of (2) and compare the coefficients of δ and δ′, we obtain(44)−σ2tx−1−1vmvm−um−v−+β2−′t+β2+′t+vmvm−um1−x−1t−v−1−u−+v−=0,(45)−σ2β2+t+β2−t+β2+t+β2−t=0.
According to (28), we have(46)σ2=11−u−+v−1−u+v=11−u−+v−⋅x−1t.
Replacingσ2 by (46) in (44), we conclude that(47)β2−′t+β2+′t=v−−u−1−u−+v−+vm−um−v−−u−1−u−+v−t⋅u−vm−v−um1−u−+v−vm−um.
Integrating (47) with respect to t, recall the initial condition (39), thus(48)β2t=u−vm−v−um1−u−+v−vm−umv−−u−1−u−+v−t+2vm−um−v−−u−1−u−+v−t−1+um−vm.
Since−1<v−−u−<0<v1−u1, then the delta shock wave δS2 will cross the whole R← at the point(49)x2,t2=1+vm−um−v−+u−1−u−+v−v+−u+−v−+u−2,vm−um−v−+u−1−u−+v−1−u++v+v+−u+−v−+u−2.
Aftert2, there is a new delta shock with δS3 whose propagation velocity and strength, respectively, are given by(50)σ3=11−u−+v−1−u++v+,(51)β3t=β2t2+v+−u+vm−um⋅u−vm−umv−1−u−+v−1−u++v+t−t2,where β2t2 can be calculated by (48).
TheδS3 interacts with J at point x3,t3 which is resolved by the following equation:(52)x3−1=t31−u++v+,x3−x2=t3−t21−u−+v−1−u++v+.
Recalling (49), the following is clear:(53)x3,t3=1+vm−um−v−+u−1−u−+v−v+−u+−v−+u−u−−v−,vm−um−v−+u−1−u++v+1−u−+v−v+−u+−v−+u−u−−v−.
Aftert3, there is a new delta shock wave δS4 which has the same speed with δS3. The strength of δS4 is given by(54)β4t=β3t3+u−v+−u+v−1−u−+v−1−u++v+t−t3,t>t3.
Passing to limits asv−−u−⟶−1+ in (26), (49), and (53), we derive(55)limv−−u−⟶−1+x1,t1=limv−−u−⟶−1+x2,t2=limv−−u−⟶−1+x2,t2=1,0.
Sincev−−u−⟶−1+, then all the propagation speeds and strength of each delta shock wave tend to infinite. Thus, the solution of this case has a zero delta shock wave for v−−u−⟶−1+.Figure 5
When−1<v−−u−<0<v+−u+<vm−um.Case 5.
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Based on the analysis of Case4, δS1 overtakes R← (see Figure 6). The intersection point x1,t1 and the strength of delta shock wave at t1 are given by (26) and (39), respectively. The δS2 satisfies (28) and its strength is given by (48). The interaction point of the curve of δS2 and forward wave of R← (or back wave of R⟶) can be calculated by the following equation:(56)x2−1=t21−u−+v−−vm−um−v−−u−1−u−+v−,x2−1=t2,
It follows from (56) that(57)x2,t2=1+vm−um−v−+u−1−u−+v−u−−v−2,vm−um−v−+u−1−u−+v−u−−v−2.
In view of (48) and (57), we conclude(58)β2t2=u−vm−v−umu−−v−.
It is easy to see that the stateu,v of R⟶ satisfies u/u+=v/v+. At the point x2,t2, a new local Riemann problem containing a Dirac delta measure is formulated as(59)vt=t2=v−,x<x2θ,x>x2+β2t2δx−x2−t−t21−u−+v−,(60)ut=t2=u−,x<x2θu+v+,x>x2+β2t2δx−x2−t−t21−u−+v−,here we assume that θ⟶0 and satisfies x−1/t=1/1+θ2.
At the timet2, the initial data become (59) and (60), we construct the solution with the following form:(61)v=v−,x<x2,v−−u−v+−u+v+,x2<x<xt,θ,x>xt,+β2t2δx−x2−t−t21−u−+v−,(62)u=u−,x<x2,v−−u−v+−u+u+,x2<x<xt,θu+v+,x>xt,+β2t2δx−x2−t−t21−u−+v−,where xt=x2+t−t2/1−u−+v−1−θu+/v++θ given nearby x2,t2 for sufficiently small θ denotes shock wave curve of S1.
Next, we clarify equalities (61) and (62) satisfy equations (2) with initial data (59) and (60) in the weak sense. Combining (61) and (62), we deduce(63)vt=v−−u−v+−u+v+δ⋅−11−u−+v−+β2t2δ′−11−u−+v−,(64)v1−u+vx=v−−u−v+1−u−+v−v+−u+−v−1−u−+v−δ+β2t21−u−+v−δ′,where δ and δ′ are the functions of x−x2−t−t2/1−u−+v−. In light of (63) and (64), we thus deduce that(65)vt+v1−u+vx=0.
Similarly, we claim that(66)ut+u1−u+vx=0.
Since contact discontinuity linex=x2+t−t2/1−u−+v− with a Dirac delta function, then there is a delta contact discontinuity δJ in [35]. After t2, δS2 splits into δJ and S1 which are connected by intermediate state u1,v1=v−−u−/v+−u+v+,v−−u−/v+−u+u+. The δJ continues to move forwards with the constant speed τ=1/1−u−+v− and the strength β2t2 is given by (58). Besides, S1 continues to cross the R⟶ with the initial speed ω1=1/1−u−+v−. During the process, the state of the left-hand side is u1,v1 and the right-hand side is u,v=t/x−1−1u+/v+−u+,t/x−1−1v+/v+−u+. In the physics plane, the curve of S1 satisfies the following equation:(67)x−1=t1−u−+v−−vm−um−v−+u−1−u−+v−.
According to−1<v−−u−<v+−u+<0, the S1 can penetrate the whole R⟶ completely at the point x3,t3, which is(68)x3,t3=1+vm−um−v−+u−1−u−+v−v+−u+−v−+u−2,vm−um−v−+u−1−u−+v−1−u++v+2v+−u+−v−+u−2.
Aftert3, the shock wave is denoted by S2 which has the propagation speed ω2=1/1−u−+v−1−u++v+.
Asv−,u−⟶−1+, it is clear to see that all intersection points tend to 1,0 and all speeds of waves tend to infinite. Thus, a zero delta shock will appear in the interval 0,1. For x>1, the zero delta shock wave also emerges since the superposition of a zero delta contact discontinuity and a zero shock wave.Figure 6
When−1<v−−u−<v+−u+<0<vm−um.Case 6.
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In this case, we consider the interaction ofδS with S←+J, and the middle state of S← and J is u1,v1=v+−u+um/vm−um,v+−u+vm/vm−um. It is clear to see that δS1 catches up S← in finite time, and the intersection point x1,t1 can be seen in (33) (see Figure 7). Like as before, due to −1<v−−u−<0<v+−u+, after the interaction of δS1 and S←, there is a new delta shock wave δS2 whose propagation speed is σ2=1/1−u−+v−1−u++v+.
Forσ2>τ1=1/1−u++v+, δS2 interacts with J1 at the point x2,t2 which satisfies the following equation:(69)x2−x1=t2−t11−u−+v−1−u++v+,x2−1=t21−u++v+.
Together with (33), we deduce that(70)x2,t2=v−−u−−vm+um1−u−+v−v+−u+−v−+u−v−−u−+1,v−−u−−vm+um1−u−+v−1−u++v+v+−u+−v−+u−v−−u−.
When the delta shock wave passes throughJ1, it continues to move forward without change of direction. The strengths of delta shock waves are given by(71)β1t=u−vm−umv−1−u−+v−1−um+vmt,0≤t≤t1,(72)β2t=β1t1+u−v1−u1v−1−u−+v−1−u1+v1t−t1,t1≤t≤t2,(73)β3t=β2t2+u−v+−u+v−1−u−+v−1−u++v+t−t2,t≥t2,in which t1 and t2 can be obtained by (33) and (71).
Ifv−−u−⟶−1+, then the propagation speeds and strengths of all delta shock tend to infinite. Therefore, there is a zero delta shock in the positive x-axis as v−−u−⟶−1+.
In this section, we mainly testify the type2 of Theorem 1.Figure 7
When−1<v−−u−<0<vm−um<v+−u+.


## 5. Analysis of Zero Rarefaction Wave

In this section, we are interested in the zero rarefaction wave interacting with the other waves. As−1<vm−um<−1+ε, for sufficiently small ε and vm−um⟶−1+, there is a zero rarefaction wave at the point 0,0.According tov−−u−<0 or not, we separate our discussion into two parts. To be more precise, if v−−u−<0 and −1<vm−um<−1+ε, for sufficiently small ε, then the Riemann solution of (2) at the point 0,0 is J+R⟶. Otherwise, the Riemann solution is R←+R⟶ at the point 0,0. Firstly, we study the v−−u−>0 and then consider the situation v−−u−<0.Case 7.
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When−1<vm−um<v−−u−<v+−u+<0 (see Figure 8), the middle state satisfies u1,v1=v−−u−/vm−umum,v−−u−/vm−umvm and u2,v2=vm−um/v+−u+u+,vm−um/v+−u+v+, respectively. The propagation speeds of J2 and the wave front of R1⟶ are τ2=1/1−um+vm and λ21um,vm=1/1−um+vm2, respectively.
It is easy to see thatR⟶ overtakes J2 in finite time, and the interaction point is(74)x1,t1=1um−vm,1−um+vm2um−vm.
Based on the above discussions, we derive thatJ2 begins to penetrate R1⟶ at point x1,t1. During the process of penetration, there is a new contract discontinuity J3 satisfying the following equation:(75)x=t+um−vm.
WhenJ3 crosses the whole R1⟶, there is a new wave J4 whose propagation speed is same with J1.
Besides, we denoteR2⟶ after R1⟶ penetrates J3. The R2⟶ has the same propagation speed with R1⟶. The shock S2⟶ will penetrate R2⟶, and the new shock S2⟶ appears after t2. The wave S2⟶ penetrates whole R2⟶ at the point x3,t3:(76)x3,t3=v+−u+−vm+um1−u++v+v+−u+−v−+u−2,v+−u+−vm+um1−u++v+1−u−+v−2v+−u+−v−+u−2.
Aftert3, shock wave S2 turns into S3 with a speed ω3=1/1−u−+v−1−u++v+.
In the same way, forvm−um⟶−1+, it is easy to deduce that there is a zero rarefaction wave in the interval 0,1.Figure 8
When−1<vm−um<v−−u−<v+−u+<0.Case 8.
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In this case, it is similar to the analysis of Case7. The difference is that the shock wave cannot penetrate R2⟶ completely.Case 9.
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Similarly, Case5, for vm−um⟶−1+, δS coincides with positive x-axis at the point 1,0 and δS penetrates R⟶ completely. Furthermore, if vm−um⟶−1+, the zero rarefaction wave occurs in the interval 0,1.Case 10.
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From Figure9, the delta shock wave interaction R⟶ at the point is as follows:(77)x1,t1=1−u++v+v+−u+−vm+um,1−um+vm21−u++v+v+−u+−vm+um.
The strength ofδS1 at the t1 is given by(78)β1t1=umv+−u+vm1−um+vmv+−u+−vm+um.
Whent>t1, a new delta shock wave δS2 is generated satisfying(79)x=t1−u++v++v+−u+−vm+um1−u++v+.
TheδS2 penetrates R⟶ completely at the point(80)x2,t2=v+−u+−vm+um1−u++v+v+−u+,v+−u+−vm+um1−u++v+v+−u+.
As in Case5, we construct the delta shock wave as follows:(81)vx,t=vmvm−umtx−1,x<xt,v+,x>xt,+β2−tδ−x−xt+β2+tδ+x−xt,(82)ux,t=umvm−umtx−1,x<xt,u+,x>xt,+β2−tδ−x−xt+β2+tδ+x−xt,where xt is given by (79). Applying (81), (82) implies(83)vt=vmvm−um12xt−vmvm−umH2xt−v+−vmvm−umtx−1δσ2+β2−′t+β2+′tδ−β2−t+β2+tδ′σ2,(84)v1−u+vx=−vmvm−um12xt+vmvm−umH2xt+v+1−u++v+−vmvm−um1−xtδ+β2−t+β2+tδ′,where σ2 is the propagation speed of δS2. Substituting (83) and (84) into first equation of (2) and comparing the coefficients of δ and δ′, by virtue of (79), we obtain(85)β2′t=β2−′t+β2+′t=v+um−vmu+1−u++v+vm−umv+−u+1−u++v+−v+−u+−vm+um1−u++v+t.
Recalling (78), we deduce that(86)β2t=v+um−vmu+1−u++v+vm−umv+−u+1−u++v+t−2v+−u+−vm+um1−u++v+t−1+um−vm.
After the timet2, δS2 is decomposed into a shock wave S1 and a delta contact discontinuity δJ with the middle state u1,v1=v+−u+/v−−u−u+,v+−u+/v−−u−v+ between them. Consequently, δJ continues to move forwards with the constant speed 1/1−u++v+ and the invariant strength β2t2. On the other hand, S1 continues to penetrate R← with a varying propagation speed, which satisfies the following equation:(87)dxdt=11−u++v+xt,x2,t2=v+−u+−vm+um1−u++v+v+−u+,v+−u+−vm+um1−u++v+v+−u+,thus(88)x=t1−u++v++v+−u+−vm+umv+−u+1−u++v+.
For−1<0<v+−u+<v−−u−, S1 is able to penetrate whole R←.
Forvm−um⟶−1+, it is clear to see that zero rarefaction wave only exists in the interval 0,1 since the delta shock wave begins to penetrate the rarefaction wave from the point 1,0.Figure 9
When−1<vm−um<0<v−−u−<v+−u+.Case 11.
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In this case, it is easy to find similar to Case10, and S1 cannot penetrate R← completely and has the asymptote line x=t/1−u++v+2.Case 12.
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As in this case, we are concerned with the interaction ofR←+R⟶ and J+S. In the beginning, the situation is similar to Case 7. More precisely, J has asymptote x=t and do not able to enter the region of R←. On the other hand, R⟶ intersects with J denoted R1⟶, and the shock wave S cannot penetrate whole R1⟶ and finally has the asymptote x=t/1−u++v+2.
Therefore, we have proved the type3 of Theorem 1.


## 6. Numerical Simulations

In this section, we present some representative numerical simulations for the formation of zeros waves, mentioned in this paper, when the differences of initial data tend to−1 from the right side. In this paper, we employ the upwind scheme to validate our main results.The following numerical simulations correspond to the zero shock wave of Case1 (Figure 2). We take the initial data as follows:(89)u−=1.9,v−=1,um=3.5,vm=3,u+=2.7,v+=2,(90)u−=1.99,v−=1,um=3.5,vm=3,u+=2.7,v+=2.This implies that thev−−u− from −0.9 (Figure 10) tends to −0.99 (Figure 11). It can be clearly seen from the numerical results that when v−−u− approaches −0.99 from −0.9, u, v and the region of negative states all increase. Namely, the velocity of shock wave increases. That is to say, if −1<v−−u−<vm−um<0, v+−u+<0 and if v−−u−⟶−1+, then the velocity tends to ∞. There is a zero shock wave.Figure 10
(a)u and (b)v for v−−u−=0.9 with v−=1, u−=1.9, vm=3.5, um=3, v+=2, and u+=2.7.
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(b)Figure 11
(a)u and (b)v for v−−u−=0.99 with v−=1, u−=1.99, vm=3.5, um=3, v+=2, and u+=2.7.
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(b)The numerical simulations (Figures12 and 13) correspond to the zero delta shock wave of Case 4 (Figure 5) with initial data as follows:(91)u−=1.9,v−=1,um=1,vm=3,u+=50,v+=51,(92)u−=1.99,v−=1,um=1,vm=3,u+=50,v+=51.That is to say, v−−u− from −0.9 (Figure 12) tends to −0.99 (Figure 13). As can be seen from the following numerical results, the negative state region and the values of u and v are getting more larger. As v−−u− tends to −1, the shock velocity and u, v approach ∞. In other words, when −1<v−−u−<0<vm−um, v−−u−<v+−u+, and v−−u−⟶−1+, there is a zero delta shock wave.Figure 12
(a)u and (b)v for v−−u−=0.9 with u−=1.9, v−=1, um=1, vm=3, u+=50, and v+=51.
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(b)Figure 13
(a)u and (b)v for v−−u−=0.99 with u−=1.99, v−=1, um=1, vm=3, u+=50, and v+=51.
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(b)The numerical simulations (Figures14 and 15) correspond to the zero rarefaction waves of Case 7 (Figure 8) with the initial data as follows:(93)u−=1.5,v−=1,um=1.9,vm=1,u+=2.3,v+=2,(94)u−=1.5,v−=1,um=1.99,vm=1,u+=2.3,v+=2.Figure 14
(a)u and (b)v for vm−um=0.9 with u−=1.5, v−=1, um=1.9, vm=1, u+=2.3, and v+=2.
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(b)Figure 15
(a)u and (b)v for vm−um=0.99 with u−=1.5, v−=1, um=1.99, vm=1, u+=2.3, and v+=2.
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(b)This implies that thevm−um from −0.9 (Figure 14) tends to −0.99 (Figure 15). In order to better explain the generation of zero rarefaction wave on the interval 0,1, we subdivide −1,3 on the space. It can be seen from the following numerical simulations that in a very short period of time, when v−−u− approaches −0.99 from −0.9, the intermediate state um,vm=1.9,1 disappears. That is to say, the wave velocity of rarefaction wave R1⟶ increases. This is very consistent with our conclusion when −1<vm−um<0 and vm−um<minv−−u−,v+−u+, a zero rarefaction wave is alive in the interval 0,1 when vm−um⟶−1+.


## 7. Conclusions

In this paper, we mainly study the zero waves of the nonsimplified chromatographic equation. From the analysis of Section3, one can know when −1<v−−u−<vm−um<0, on the one hand, the solution performs as a zero shock wave if v+−u+<0 and v−−u−⟶−1+. On the other hand, its solution performs as a zero shock wave in the interval 0,1 and a zero delta shock wave in the interval 1,∞ if v+−u+>0 and v−−u−⟶−1+. From the analysis of Section 4, when −1<v−−u−<0<vm−um and v−−u−<v+−u+, the solution performs as a zero delta shock wave when v−−u−⟶−1+. From the analysis of Section 5, when −1<vm−um<0 and vm−um<minv−−u−,v+−u+, a zero rarefaction wave is alive in the interval 0,1 when vm−um⟶−1+.



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