The Existence of Zero Waves for Nonsimplified Chromatography System

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. +rough 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.


Introduction
Chromatography plays an important role in modern industry. It is used to separate two chemical components especially by chemists and engineers. e research of the nonlinear chromatography system is, in particular, a significant part of the theory of chromatography. Mazzotti [1][2][3] captured the delta shock wave for local equilibrium model (1) by numerical and experimental methods: where b > a > 0 are constants. u and v, which describe concentrations of two chemical species, are positive function with 1 − u + v > 0. e complexity of system (1) is due to hyperbolicity in the region a(1 + v) + b(1 − u) 2 − 4ab(1 − 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: 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][8][9][10][11][12][13].
Recently, Ambrosio et al. [14] introduced the change of variables: and derived the following simplified chromatography system: ey 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][18][19][20][21][22][23][24]. e 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][26][27][28][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.
ere 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][31][32][33][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: and analyse the interaction of various waves. We find interesting phenomena: when the differences v − − u − and v m − u m 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 − < v m − u m < 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 is paper is divided into seven sections. In Section 2, 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.

Zero Waves in the Limit of Riemann
Solutions to (2) In this section, we review the solution to system (2) with initial data: where u ± and v ± are positive constants. In light of [4], we know characteristics satisfy e characteristics λ 1 and λ 2 are genuinely nonlinear and linearly degenerate, respectively. erefore, the associated waves are rarefaction waves R or shock waves S for λ 1 and contact discontinuities J for λ 2 : Using the phase plane analysis method (see Figure 1), Riemann solutions of (2) are constructed as follows: in which the propagation speeds of S ← 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. erefore, there exists the composite wave S ← J.
If v + − u + � 0, then a composite wave R ← J forms.
in which the propagation speeds of J and S are τ We deduce from (10) that is fact in turn implies that both the propagation speeds of J and S → tend to +∞. In other words, both J and S → coincide with the positive x-axis and the middle state (v + − u + )) disappears. us, zero shock wave [28] appears in this situation.
Equality (13) implies the propagation speed of the wave front of R → tends to +∞. en, zero rarefaction wave forms.
e line is x � 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. Wang [4] constructed delta shock wave: He checked that the measure solution (15) with (16) satisfies the generalized Rankine-Hugoniot condition: To ensure the uniqueness of delta shock wave solution, entropy condition should be satisfied provided namely, 1 at is to say, all the characteristics are incoming from both sides.
For the measure solution (15) with (16), we deduce We apply (20) and (21) to deduce that both the propagation speed and the strength of the delta shock wave tend to +∞. erefore, there is a delta shock wave.

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) Figure 2), then the solution of (2) and (5) is as follows: in which e propagation speeds of S 1 → and J 2 are overtakes contact discontinuity J 2 in finite time. e waves S 1 → and J 2 interact at the point: For t > t 1 , notice the condition − 1 < v 1 − u 1 < v 2 − u 2 < 0, a new contact discontinuity J 3 and a new shock S 2 → generate. By applying (23), the following equation can be obtained: e propagation speeds of S 2 → and the wave back in R → , respectively, are in finite time and the intersection point (x 2 , t 2 ) is given by After t 2 , the shock wave S 3 → crosses the rarefaction wave R → with a varying speed during the penetration. e shock wave S 3 → satisfies the following equation: In view of (27), we can obtain that e wave S 3 and the wave front of R → interact at the point After the time t 3 , a new shock wave S 4 → forms and the propagation speed of (26), and (29), we discover Mathematical Problems in Engineering Based on the above discussion, as (v − − u − ) ⟶ (− 1) + , the solution of (2) and (5) has the zero shock wave starting from the point (0, 0). Figure 3), from Case 1, it is clear to see that the interaction point (x 1 , t 1 ) of S 1 → and J 2 is given by (24), and the states (23) and (25), respectively. Since in finite time and the interaction point is as follows: In view of (24) and (31), we see When (v − − u − ) ⟶ (− 1) + , the propagation speeds of all shock waves and contact discontinuities tend to +∞.
us, there is a zero shock wave starting from (0, 0) in the Our plan is to show that a delta shock wave δS 1 emits from the point Figure 4). e state (u 1 , v 1 ) is given by (23).
Owing to ω 1 > σ 1 , then S 1 → overtakes δS 1 at the point It is easy to see that a new delta shock δS 2 forms, with Notice that propagation speed of contact discontinuity J is τ 1 and τ 1 > σ 2 , and J 1 and δS 2 interact at the point For t > t 2 , there is a new delta shock δS 3 with same propagation speed of δS 2 . It follows from (16) that the strengths of the all delta shock waves satisfy the following equation: Mathematical Problems in Engineering We then combine (33) and (34) to discover Based on the above analysis, all the waves overlap with the positive x-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. e above discussions verify the type (1) of eorem 1.

Analysis of Zero Delta Shock Wave
In this section, we mainly consider zero delta shock wave interacts with other waves. Taking sufficiently ε together with the Riemann solution of (2) has a zero delta shock wave at the point (0, 0).
In this case, we are concentrated on the interaction of δS with R ← + J (see Figure 5). e state between R ← and J is e propagation speeds of δS 1 and the wave back of R e strength of δS 1 is given by By substituting (37) into (38), we can obtain the following equation: After the time t 1 , there is δS 2 which has the same expression of (28).
Next, we compute the strength of δS 2 , by virtue of splitting the delta function method introduced by Nedeljkov and Oberguggeenberger in [35]. is method has also been used in [36,37]. Now, we construct the delta shock wave x � x(t) which satisfies the following equation: is a split delta function and β 2 (t) � β − 2 (t) + β + 2 (t) denotes the strength of δS 2 . Furthermore, we see from (40) and (41) where H, δ, and δ ′ are the functions of x − x(t); besides, x(t) 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 According to (28), we have Replacing σ 2 by (46) in (44), we conclude that Integrating (47) with respect to t, recall the initial condition (39), thus 1 , then the delta shock wave δS 2 will cross the whole R ← at the point After t 2 , there is a new delta shock with δS 3 whose propagation velocity and strength, respectively, are given by where β 2 (t 2 ) can be calculated by (48).
e δS 3 interacts with J at point (x 3 , t 3 ) which is resolved by the following equation: Recalling (49), the following is clear: After t 3 , there is a new delta shock wave δS 4 which has the same speed with δS 3 . e strength of δS 4 is given by Passing to limits as (26), (49), and (53), we derive Since (v − − u − ) ⟶ (− 1) + , then all the propagation speeds and strength of each delta shock wave tend to infinite.
us, the solution of this case has a zero delta shock wave for Based on the analysis of Case 4, δS 1 overtakes R ← (see Figure 6). e intersection point (x 1 , t 1 ) and the strength of delta shock wave at t 1 are given by (26) and (39), respectively. e δS 2 satisfies (28) and its strength is given by (48). e interaction point of the curve of δS 2 and forward wave of R ← (or back wave of R → ) can be calculated by the following equation: It follows from (56) that

Mathematical Problems in Engineering
In view of (48) and (57), we conclude It is easy to see that the state (u, v) of R → satisfies u/u + � v/v + . At the point (x 2 , t 2 ), a new local Riemann problem containing a Dirac delta measure is formulated as here we assume that θ ⟶ 0 and satisfies At the time t 2 , the initial data become (59) and (60), we construct the solution with the following form: where given nearby (x 2 , t 2 ) for sufficiently small θ denotes shock wave curve of S 1 .
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 where δ and δ ′ are the functions of In light of (63) and (64), we thus deduce that Similarly, we claim that with a Dirac delta function, then there is a delta contact discontinuity δJ in [35]. After t 2 , δS 2 splits into δJ and S 1 which are connected by intermediate state e δJ continues to move forwards with the constant speed τ � 1/(1 − u − + v − ) and the strength β 2 (t 2 ) is given by (58). Besides, S 1 continues to cross the R → with the initial speed During the process, the state of the left-hand side is (u 1 , v 1 ) and the right-hand side is In the physics plane, the curve of S 1 satisfies the following equation: After t 3 , the shock wave is denoted by S 2 which has the propagation speed ω 2 � 1/((1 − u − + v − )(1 − u + + v + )).
As (v − , u − ) ⟶ (− 1) + , it is clear to see that all intersection points tend to (1, 0) and all speeds of waves tend to infinite. us, 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.
In this case, we consider the interaction of δS with S ← + J, and the middle state of S ← and J is ( It is clear to see that δS 1 catches up S ← in finite time, and the intersection point (x 1 , t 1 ) can be seen in (33) (see Figure 7). Like as before, due to − 1 < v − − u − < 0 < v + − u + , after the interaction of δS 1 and S ← , there is a new delta shock wave δS 2 whose propagation speed is For , δS 2 interacts with J 1 at the point (x 2 , t 2 ) which satisfies the following equation: Together with (33), we deduce that When the delta shock wave passes through J 1 , it continues to move forward without change of direction. e strengths of delta shock waves are given by in which t 1 and t 2 can be obtained by (33) and (71). If (v − − u − ) ⟶ (− 1) + , then the propagation speeds and strengths of all delta shock tend to infinite. erefore, there is a zero delta shock in the positive x-axis as In this section, we mainly testify the type (2) of eorem 1.

Analysis of Zero Rarefaction Wave
In this section, we are interested in the zero rarefaction wave interacting with the other waves. As − 1 < v m − u m < − 1 + ε, for sufficiently small ε and (v m − u m ) ⟶ (− 1) + , there is a zero rarefaction wave at the point (0, 0).
According to v − − u − < 0 or not, we separate our discussion into two parts. To be more precise, if v − − u − < 0 and − 1 < v m − u m < − 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 sit- e propagation speeds of J 2 and the wave front of R 1 �→ and It is easy to see that R → overtakes J 2 in finite time, and the interaction point is Based on the above discussions, we derive that J 2 begins to penetrate R 1 �→ at point (x 1 , t 1 ). During the process of penetration, there is a new contract discontinuity J 3 satisfying the following equation: When J 3 crosses the whole R 1 �→ , there is a new wave J 4 whose propagation speed is same with J 1 .
Besides, we denote R 2 �→ after R 1 �→ penetrates J 3 . e R 2 �→ has the same propagation speed with R 1 �→ . e shock S 2 → will penetrate R 2 �→ , and the new shock S 2 → appears after t 2 . e wave S 2 → penetrates whole R 2 �→ at the point (x 3 , t 3 ): After t 3 , shock wave S 2 turns into S 3 with a speed In the same way, for (v m − u m ) ⟶ (− 1) + , it is easy to deduce that there is a zero rarefaction wave in the interval [0, 1].
In this case, it is similar to the analysis of Case 7. e difference is that the shock wave cannot penetrate R 2 �→ completely.  Figure 9, the delta shock wave interaction R → at the point is as follows: e strength of δS 1 at the t 1 is given by When t > t 1 , a new delta shock wave δS 2 is generated satisfying e δS 2 penetrates R → completely at the point As in Case 5, we construct the delta shock wave as follows: where x(t) is given by (79). Applying (81), (82) implies where σ 2 is the propagation speed of δS 2 . Substituting (83) and (84) into first equation of (2) and comparing the coefficients of δ and δ ′ , by virtue of (79), we obtain Recalling (78), we deduce that After the time t 2 , δS 2 is decomposed into a shock wave S 1 and a delta contact discontinuity δJ with the middle state between them. Consequently, δJ continues to move forwards with the constant speed 1/(1 − u + + v + ) and the invariant strength β 2 (t 2 ). On the other hand, S 1 continues to penetrate R ← with a varying propagation speed, which satisfies the following equation: 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).
In this case, it is easy to find similar to Case 10, and S 1 cannot penetrate R ← completely and has the asymptote line As in this case, we are concerned with the interaction of R ← + 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 R 1 �→ , and the shock wave S cannot penetrate whole R 1 �→ and finally has the asymptote x � t/(1 − u + + v + ) 2 . erefore, we have proved the type (3) of eorem 1.

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. e following numerical simulations correspond to the zero shock wave of Case 1 (Figure 2). We take the initial data as follows: is implies that the v − − 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. at is to say, if then the velocity tends to ∞. ere is a zero shock wave. e numerical simulations (Figures 12 and 13) correspond to the zero delta shock wave of Case 4 ( Figure 5) with initial data as follows: at 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 there is a zero delta shock wave. e numerical simulations (Figures 14 and 15) correspond to the zero rarefaction waves of Case 7 (Figure 8) with the initial data as follows: is implies that the v m − u m 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 (u m , v m ) � (1.9, 1) disappears. at is to say, the wave velocity of rarefaction wave R 1 �→ increases. is is very consistent with our conclusion when − 1 < v m − u m < 0 and v m − u m < min v − − u − , v + − u + , a zero rarefaction wave is alive in the interval [0, 1] when (v m − u m ) ⟶ (− 1) + .

Conclusions
In this paper, we mainly study the zero waves of the non-

Data Availability
No data were used to support this study.

Conflicts of Interest
e authors declare that they have no conflicts of interest.