# Viscous Regularization of Delta Shock Wave Solution for a Simplified Chromatography System

Authors: Xiumei Li; Chun Shen
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

The delta shock wave for a simplified chromatography system is obtained in the Riemann solution when-1<v-<0<v+. In fact, the result in this paper is the reasonable generalization of the result in Sun (2011) with v-=0<v+ which is exactly the critical situation for -1<v-<0<v+. The self-similar viscosity vanishing approach is also used to check the delta shock wave solution.


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## Body

## 1. Introduction

The theory of nonlinear chromatography system is the foundation of chromatography separation process and also plays an important role in the modern industry. Recently, the delta shock wave has been captured numerically and experimentally by Mazzotti et al. [1, 2] in the Riemann solutions for the local equilibrium model of two-component nonlinear chromatography. Since then, extensive attention has been drawn on the formation of delta shock wave solution to all kinds of chromatography systems [3–6].In [5], Sun considered the Riemann problem for a simplified chromatography system:
(1)vt+(v1+v)x=0,wt+(w1+v)x=0,
with the initial data
(2)(v,w)(x,0)=(v±,w±),±x>0,
where v± and w± are all given constants. She discovered that a delta shock wave solution also appears in the situation v-=0<v+ by employing the self-similar viscosity vanishing approach.In the present paper, we extend the result from the critical situationv-=0<v+ in [5] to the general situation -1<v-<0<v+ and discover that the delta shock wave also appears in the Riemann solution to (1) and (2). Actually, in the situation -1<v-<0<v+, the Riemann solution to (1) and (2) is a delta shock wave connecting two constant states (v±,w±) whose propagation speed and strength are, respectively,
(3)σ=1(1+v-)(1+v+),β(t)=(w-v+-w+v-)t(1+v-)(1+v+).
On the other hand, in the situation v-=0<v+ in [5], the propagation speed and strength of delta shock wave are, respectively,
(4)σ=11+v+,β(t)=w-v+t1+v+.
If we let v-→0+ in the situation -1<v-<0<v+, then the result is identical with that for the situation v-=0<v+ in [5]. Thus, one can see that the situation v-=0<v+ is the critical one, where the delta shock wave can be obtained, and the result in this paper is the reasonable generalization of the result in [5].Similar to [5], we also consider the following viscous regularization system:
(5)vt+(v1+v)x=εtvxx,wt+(w1+v)x=0,
which enables us to see that the delta shock wave is the reasonable choice of the Riemann solution to (1) and (2) when -1<v-<0<v+.In addition, we emphasize that our result here is also reasonable in the sense of chromatography engineer since a simplified chromatography system can also be derived from the other nonlinear chromatography system:(6)∂tu1+∂x(u11-u1+u2)=0,∂tu2+∂x(u21-u1+u2)=0,
by introducing the change of variables
(7)v=u2-u1,w=u1+u2,
which is different from the derivation in [5]. Here u1, u2 are the nonnegative functions of the variables (x,t)∈R×R+ to denote the Langmuir-like and the anti-Langmuir-like components, respectively, which express transformations of the concentrations of two solutes. It is clear that the delta measure appears in w for the system (1) is equivalent to saying that the delta measures appear in both the state variables u1 and u2 for the system (6). It is easily derived that the strengths of the delta measures in u1 and u2 are equal to each other. In other words, the phenomena of singular concentrations in the two components u1 and u2 appear simultaneously for the system (6).The system (1) is the simplest example for systems of Temple class [7]. In other words, the shock curves coincide with the rarefaction curves in the phase plane. Compared with general systems of conservation laws, well-posed results for systems of Temple class [8] are available for a much larger class of initial data due to the above feature. Furthermore, thanks to the particular structure, wave interactions for the systems of Temple class can also be dealt with completely [9–12], which enables us to solve the Cauchy problem for arbitrary data of bounded variation.The self-similar viscosity vanishing approach was first proposed by Dafermos [13] to construct Riemann solutions for broad classes of 2×2 systems that include the equations of isentropic, Lagrangian, and gas dynamics. This method was widely used in [14] for the admissibility of weak solutions of the Riemann problem for hyperbolic systems of conservation laws, and the admissibility criterion was called the wave-fan criterion in [15]. Special attention was also paid in [6, 16–22] to the formation of the delta shock waves in the Riemann solutions for some systems of hyperbolic conservation laws.There exist numerous excellent papers for the related equations and results about the measure-valued solutions for hyperbolic systems of conservation laws. The well-known examples are the transport equations [21, 23–27] (or called the pressureless Euler equations) and the Chaplygin gas dynamics equations [28, 29] which have been extensively studied. For the other related equations and results, we can refer to [30–35] and the related references therein.The paper is organized as follows. In Section2, the Riemann problem of (1) and (2) is considered. In particular, the delta shock wave does occur in the Riemann solution when -1<v-<0<v+. In Section 3, we show that there exists a smooth solution for the regularized system (5) with the initial data (2). Then, by letting ε→0, we prove that the limit of the regularized solution is exactly the corresponding Riemann solution to (1) and (2) when -1<v-<0<v+.


## 2. The Riemann Problem for (1) and (2) in the Phase Space v>-1

In this section, we generalize the Riemann problem for (1) and (2) from the phase space (v,w)∈[0,∞)×(-∞,∞) in [5] to the phase space (v,w)∈(-1,∞)×(-∞,∞). Compared with the result in [5] that there are three kinds of the Riemann solutions to (1) and (2) in the phase space (v,w)∈[0,∞)×(-∞,∞), there exist six kinds of the Riemann solutions to (1) and (2) in the phase space (v,w)∈(-1,∞)×(-∞,∞)(Figure 1). In what follows, we first depict the properties of (1) and then list the Riemann solutions to (1) and (2) in the phase space (v,w)∈(-1,∞)×(-∞,∞) case by case. Finally, the generalized Rankine-Hugoniot relations of delta shock wave are derived in detail. We can also refer to [3, 4, 12] for the related results about the other nonlinear chromatography systems.The Riemann solutions to (1) and (2) in all the situations.
(a)
R
→
+
J when v->v+>0
(b)
S
→
+
J when 0<v-<v+
(c)
R
1
→
+
R
2
→ when -1<v+<0<v-
(d)
J
+
S
→ when -1<v-<v+<0
(e)
J
+
R
→ when -1<v+<v-<0
(f)
δ
S when -1<v-<0<v+The characteristic eigenvalues of (1) are λ1=1/(1+v)2 and λ2=1/(1+v), respectively. Thus, it is clear to see that (1) is strictly hyperbolic in the phase space v>0 and -1<v<0 and is nonstrictly hyperbolic on the line v=0. The corresponding right characteristic vectors are r1→=(v,w)T and r2→=(0,1)T, respectively. Thus λ1 is genuinely nonlinear for v≠0, and λ2 is always linearly degenerate. Therefore, the associated waves are rarefaction waves R or shock waves S for the characteristic field λ1 and contact discontinuities J for the characteristic field λ2. The Riemann invariants along the characteristic fields are w/v and v, respectively.Now we are in a position to deliver the Riemann solutions to (1) and (2) according to the value of v as follows.(1)
Ifv->v+>0, then the Riemann solution of (1) and (2) can be expressed as R→+J:(8)(v,w)(x,t)={(v-,w-),x<λ1(v-)t,(tx-1,w-v-(tx-1)),λ1(v-)t≤x≤λ1(v+)t,(v+,v+w-v-),λ1(v+)t<x<τt,(v+,w+),x>τt,in which the propagation speed of R is ξ=1/(1+v)2 for v from v- to v+ such that 0<ξ<1 and that of J is τ=1/(1+v+).(2)
If0<v-<v+, then the Riemann solution of (1) and (2) can be expressed as S→+J:
(9)(v,w)(x,t)={(v-,w-),x<σt,(v+,v+w-v-),σt<x<τt,(v+,w+),x>τt,
in which the propagation speed of S is σ=1/((1+v-)(1+v+)) and that of J is τ=1/(1+v+). It is clear to see that 0<σ<τ<1.(3)
If-1<v+<0<v-, then the Riemann solution of (1) and (2) can be expressed as R1→+R2→:
(10)(v,w)(x,t)={(v-,w-),x<λ1(v-)t,(tx-1,w-v-(tx-1)),λ1(v-)t≤x<t,(tx-1,w+v+(tx-1)),t<x≤λ1(v+)t,(v+,w+),x>λ1(v+)t.
It should be emphasized that the state (v,w) in R1→ is determined by w/v=w-/v- which is on the left of the line x=t; otherwise, the state (v,w) in R2→ is determined by w/v=w+/v+ which is on the right of the line x=t.(4)
If-1<v-<v+<0, then the Riemann solution of (1) and (2) can be expressed as J+S→:
(11)(v,w)(x,t)={(v-,w-),x<τt,(v-,v-w+v+),τt<x<σt,(v+,w+),x>σt,
in which the propagation speed of J is τ=1/(1+v-) and that of S is σ=1/(1+v-)(1+v+). Here we notice that 1<τ<σ.(5)
If-1<v+<v-<0, then the Riemann solution of (1) and (2) can be expressed as J+R→:(12)(v,w)(x,t)={(v-,w-),x<τt,(v-,v-w+v+),τt<x<λ1(v-)t,(tx-1,w+v+(tx-1)),λ1(v-)t≤x≤λ1(v+)t,(v+,w+),x>λ1(v+)t,in which the propagation speed of J is τ=1/(1+v-)>1 and that of R is ξ=1/(1+v)2>1 for v from v- to v+.(6)
For the case-1<v-<0<v+, motivated by the result in [22], we consider that the Riemann solution to (1) and (2) is a delta shock wave δS connecting the two constant states (v-,w-) and (v+,w+). In order to deal with it, we need the following definition of a delta shock wave solution in the framework introduced in [36] and used in [12, 37].Let us suppose thatΓ={γi∣i∈I} is a graph in the closed upper half-plane {(x,t)∣(x,t)∈(-∞,∞)×[0,∞)}, containing Lipschitz continuous arcs γi with i∈I, in which I is a finite index set. Let us also suppose that I0 is the subset of I containing all indices of arcs linking to the x-axis and Γ0={xk0∣k∈I0} is the set of initial points of the arcs γk with k∈I0.Definition 1.
Let(v,w) be a pair of distributions, where w is represented in the form
(13)w(x,t)=w^(x,t)+α(x,t)δ(Γ),
in which v,w^∈L∞(R×R+) and the singular part should be defined by
(14)α(x,t)δ(Γ)=∑i∈Iαi(x,t)δ(γi).
Let us consider the initial data of the form
(15)(v,w)(x,0)=(v0(x),w^0(x)+∑k∈I0αk(xk0,0)δ(x-xk0)),
in which v0,w^0∈L∞(R); then the earlier pair of distributions (v,w) are called a generalized delta shock wave solution of the simplified chromatography system (1) with the initial data (15) if the following integral identities
(16)∫R+∫R(vψt+v1+vψx)dxdt+∫Rv0(x)ψ(x,0)dx=0,∫R+∫R(w^ψt+w^1+vψx)dxdt+∑i∈I∫γiαi(x,t)∂ψ(x,t)∂l+∫Rw^0(x)ψ(x,0)dx+∑k∈I0αk(xk0,0)ψ(xk0,0)=0
hold for all test functions ψ∈Cc∞(R×R+), in which ∂ψ(x,t)/∂l denotes the tangential derivative of a function ψ on the graph γi and ∫γi denotes the line integral along the arc γi.With the above definition, similar to that in [5], we can also construct the Riemann solution to (1) and (2) for the situation -1<v-<0<v+ in the following theorem.Theorem 2.
For the case-1<v-<0<v+, the Riemann problem (1) and (2) has the piecewise smooth Riemann solution in the form
(17)(v,w)(x,t)={(v-,w-),x<σt,(vδ,β(t)δ(x-σt)),x=σt,(v+,w+),x>σt,
where
(18)vδ=v-+v++v-v+,σ=11+vδ=1(1+v-)(1+v+),β(t)=(w-v+-w+v-)t(1+v-)(1+v+).
The measure solution (17) with (18) also satisfies the generalized Rankine-Hugoniot condition as follows:
(19)dxdt=σ,dβ(t)dt=σ[w]-[w1+v],σ[v]=[v1+v].Proof.
Let us check that theδ-measure solution (17) with (18) obeys (1) in the sense of distributions. In other words, we only need to check that (17) with (18) satisfies
(20)∫0∞∫-∞∞(vψt+v1+vψx)dxdt=0,∫0∞∫-∞∞(wψt+w1+vψx)dxdt=0.
Actually, we can verify the first equation in (20) as follows:
(21)∫0∞∫-∞∞(vψt+v1+vψx)dxdt=∫0∞∫-∞σt(v-ψt+v-1+v-ψx)dxdt+∫0∞∫σt∞(v+ψt+v+1+v+ψx)dxdt=∫0∞∫-∞0v-ψtdxdt+∫0∞∫0σtv-ψtdxdt+∫0∞∫-∞σtv-ψx1+v-dxdt+∫0∞∫σt∞(v+ψt+v+1+v+ψx)dxdt=∫0∞∫x/σ+∞v-ψtdtdx+∫0∞∫-∞σtv-ψx1+v-dxdt+∫0∞∫0x/σv+ψtdtdx+∫0∞∫σt∞v+ψx1+v+dxdt=-∫0∞v-ψ(x,xσ)dx+∫0∞v-1+v-ψ(σt,t)dt+∫0∞v+ψ(x,xσ)dt-∫0∞v+1+v+ψ(σt,t)dt=0,
in which the fact that ψ(x,t) is compactly supported in the region R×R+ has been used.
Consequently, for the second equation of (20), we have
(22)∫0∞∫-∞∞(wψt+w1+vψx)dxdt=∫0∞∫-∞σt(w-ψt+w-ψx1+v-)dxdt+∫0∞∫σt∞(w+ψt+w+ψx1+v+)dxdt+∫0∞β(t)(ψt(σt,t)+σψx(σt,t))dt=∫0∞∫0σtw-ψtdxdt+∫0∞∫-∞σtw-ψx1+v-dxdt+∫0∞∫σt∞w+ψtdxdt+∫0∞∫σt∞w+ψx1+v+dxdt+∫0∞β(t)(ψt(σt,t)+σψx(σt,t))dt=∫0∞(w+-w-)ψ(x,xσ)dx+∫0∞(w-1+v--w+1+v+)ψ(σt,t)dt+∫0∞β(t)dψ(σt,t)=[w+-w-(1+v-)(1+v+)+w-1+v--w+1+v+-β′(t)]×∫0∞ψ(σt,t)dt=0,
in which the changes of variables and integration by parts have been used.
Through the earlier verification, we can conclude that (17) with (18) is indeed the piecewise smooth Riemann solution to (1) and (2) in the sense of distributions.Remark 3.
If we take the limitv-→0 in the Riemann solution (17) with (18), then we can obtain the result which is identical with that in [5]. In other words, the Riemann solution to (1) and (2) for 0=v-<v+ can be seen as the critical situation for -1<v-<0<v+. Thus, the result in [5] can be seen as the special situation in this paper.Remark 4.
If we take the limitv-→-1 in the Riemann solution (17) with (18), then both the propagation speed and strength of the delta shock wave will tend to +∞. In fact, both λ1=1/(1+v)2 and λ2=1/(1+v) tend to +∞ if we let v→-1 on the right. Thus, the line v=-1 in the phase space (v,w) can be seen as a singular curve.Remark 5.
The Riemann solution to (1) and (2) can be constructed simply if we take the Riemann initial data (2) in the phase space (v,w)∈(-∞,-1)×(-∞,∞). However, it is difficult to construct the global Riemann solution to (1) and (2) if we require the Riemann initial data (2) in the phase space (v,w)∈(-∞,∞)×(-∞,∞). In this situation, we at least need to make some criterions on the line v=-1.


## 3. Viscous Regularization of Delta Shock Wave Solution

In this section, we are only interested in the viscous regularization of delta shock wave solution to the Riemann problem (1) and (2). Thus, we confine our attention only to the situation -1<v-<0<v+.Performing the self-similar transformationξ=x/t, we get the boundary value problem:
(23)-ξvξ+(v1+v)ξ=εvξξ,-ξwξ+(w1+v)ξ=0,
with the boundary conditions
(24)(v,w)(±∞)=(v±,w±).By observing (23), we know that the solution of the first equation in (23) does not rely on the second one in (23). So, we can first consider the first equation in (23), namely,
(25)-ξvξ+(v1+v)ξ=εvξξ,v(±∞)=v±.With the similar deduction and calculation in Theorem 4.1 in [5], we can also obtain the same conclusion as follows.Theorem 6.
For each fixedε>0, there exists a unique and monotonic smooth solution vε(ξ) of the boundary value problem (25) for -1<v-<0<v+.Now, we turn our attention to the second equation in (23) and consider the existence of the solution w(ξ). Based on Theorem 6, also with the similar deduction and calculation in Theorem 4.2 in [5], we can obtain the following theorem.Theorem 7.
For each fixedε>0, assume that w(ξ) is a weak solution of the following boundary value problem:
(26)-ξwξ+(w1+vε)ξ=0,w(±∞)=w±,
where vε(ξ) is the solution of (25).
Then, one haswε(ξ)∈L1(-∞,+∞) which can be expressed as(27)wε(ξ)={w-exp(∫-∞ξv˙ε(s)1+vε(s)-s(1+vε(s))2)dsforξ<ξαε,w+exp(∫ξ+∞-v˙ε(s)1+vε(s)-s(1+vε(s))2)dsforξ>ξαε,where ξαε is the unique fixed point of ξ=1/(1+vε(ξ)).It follows from Theorem6 that vε(ξ) is strictly monotonically increasing with respect to ξ when -1<v-<0<v+, which enables us to see that both 1/(1+vε(ξ)) and 1/(1+vε(ξ))2 are strictly monotonically decreasing with respect to ξ when -1<v-<0<v+. Thus, there exist unique fixed points of the following equations, and we use ξαε and ξβε to express as
(28)ξαε=11+vε(ξαε),(29)ξβε=1(1+vε(ξβε))2,
respectively. Consequently, we take the limits ξα=limε→0+ξαε and ξβ=limε→0+ξβε, which will play very critical roles in the following discussion.Lemma 8.
For anyη>0, draws the conclusion that
(30)limε→0+vε(ξ)={v-,forξ<ξβ-η,v+,forξ>ξβ+η
is in uniform in the earlier intervals.Proof.
It is easy to takeε>0 sufficiently small such that ξβε<ξβ+η/4 for ξβ=limε→0+ξβε. If we take ξ1=ξβ+η/2, then it is clear that ξβε<ξ1-η/4. From Theorem 6, we know that v˙ε(ξ)>0 if -1<v-<0<v+. Then (25) can be written as
(31)1ε(-ξ+1(1+vε(ξ))2)=v¨ε(ξ)v˙ε(ξ).
By integrating (31) over (ξ1,ξ), we arrive at
(32)v˙ε(ξ)=v˙ε(ξ1)exp∫ξ1ξ1ε(-s+1(1+vε(s))2)ds.
Integrating (32) over (ξ1,+∞), we get
(33)v+-vε(ξ1)=v˙ε(ξ1)∫ξ1+∞exp(∫ξ1ξ‍1ε(-s+1(1+vε(s))2)ds)dξ.
Becausevε(ξ) is monotonically increasing in the real axis, we have v˙ε(ξ1)≥0 and v-≤vε(s)≤v+. Thus, one can conclude that(34)v+-v-≥v+-vε(ξ1)=v˙ε(ξ1)∫ξ1+∞exp(∫ξ1ξ‍1ε(-s+1(1+vε(s))2)ds)dξ≥v˙ε(ξ1)∫ξ1+∞exp1ε(-12(ξ-ξ1)(ξ+ξ1)+ξ-ξ1(1+v+)2)dξ=v˙ε(ξ1)∫0+∞exp1ε(-12ζ2+ζ(1(1+v+)2-ξ1))dζ≥v˙ε(ξ1)·εC1for 0<ε<1 and C1 is a constant independent of ε, in which the change of variables ζ=ξ-ξ1 has been used. So, we can get 0≤v˙ε(ξ1)≤(v+-v-)/(εC1). Together with (32), we know that
(35)v˙ε(ξ)≤v+-v-εC1exp(∫ξ1ξ1ε(-s+1(1+vε(s))2)ds).
In view of (29), we have
(36)-s+1(1+vε(s))2=1(1+vε(s))2-1(1+vε(ξβε))2-(s-ξβε),
noticing thatξβε+η/4<ξ1<s, by applying the mean value theorem, and then there exists ξ2∈(ξβε,s) such that
(37)-s+1(1+vε(s))2=(-2v˙ε(ξ2)(1+vε(ξ2))3-1)(s-ξβε)≤-η4.
So, we get the following inequality:
(38)v˙ε(ξ)≤v+-v-εC1exp(-η4ε(ξ-ξ1)),
which means that v˙ε(ξ)→0 uniformly on ξ>ξβ+η.
For anyξ>ξβ+η, we have
(39)v+-vε(ξ)=∫ξ+∞v˙ε(s)ds.
Obviously, we know that limε→0+vε(ξ)=v+ uniformly on ξ>ξβ+η in view of (38).
Now, we try to prove thatlimε→0+vε(ξ)=v- uniformly on ξ<ξβ-η. Similarly, it is easy to take ε>0 sufficiently small such that ξβε>ξβ-η/4. If we take ξ3=ξβ-η/2, then we also have ξβε>ξ3+η/4. By integrating (25) over (ξ,ξ3), we have
(40)v˙ε(ξ)=v˙ε(ξ3)exp∫ξ3ξ1ε(-s+1(1+vε(s))2)ds.
Integrating (25) over (-∞,ξ3), we have
(41)vε(ξ3)-v-=v˙ε(ξ3)∫-∞ξ3exp(∫ξ3ξ1ε(-s+1(1+vε(s))2)ds)dξ.
Thus, we get the following result:(42)v+-v-≥vε(ξ3)-v-=v˙ε(ξ3)∫-∞ξ3exp(∫ξ3ξ1ε(-s+1(1+vε(s))2)ds)dξ≥v˙ε(ξ3)∫-∞ξ3exp1ε(-12(ξ-ξ3)(ξ+ξ3)+ξ-ξ3(1+v-)2)dξ=v˙ε(ξ3)∫-∞0exp1ε(-12ζ2+ζ(1(1+v-)2-ξ3))dζ≥v˙ε(ξ3)·εC2for 0<ε<1 and C2 is a constant independent of ε, in which the change of variables ζ=ξ-ξ3 has also been used. Thus we have v˙ε(ξ3)≤(v+-v-)/(εC2), together with (40), and we also know that
(43)v˙ε(ξ)≤v+-v-εC2exp(∫ξ3ξ1ε(-s+1(1+vε(s))2)ds).
Similarly, by applying the mean value theorem, then there exists ξ4∈(s,ξβε) such that
(44)-s+1(1+vε(s))2=(-2v˙ε(ξ4)(1+vε(ξ4))3-1)(s-ξβε)≥η4.
Combining (43) together with (44), we get
(45)v˙ε(ξ)≤v+-v-εC2exp(-η4ε(ξ3-ξ)),
which implies that v˙ε(ξ)→0 uniformly on ξ<ξβ-η.
Thus, for anyξ<ξβ-η, we have
(46)vε(ξ)-v-=∫-∞ξv˙ε(s)ds.
Obviously, we can see that limε→0+vε(ξ)=v- uniformly on ξ<ξβ-η in view of (45).Lemma 9.
Letξα and ξβ be defined as above. Then, one has
(47)ξα=ξβ=1(1+v-)(1+v+).Proof.
Take anyϕ(ξ)∈C0∞(ξ1,ξ2) where ξ1<ξβ<ξ2. It follows from (25) that
(48)∫ξ1ξ2εvξξ·ϕ(ξ)dξ=∫ξ1ξ2(-ξvξ+(v1+v)ξ)ϕ(ξ)dξ.
By employing integration by parts and applying the fact thatϕ is compactly supported in (ξ1,ξ2), we have
(49)∫ξ1ξ2εvϕ′′(ξ)dξ=∫ξ1ξ2(v(ϕ(ξ)+ξϕ′(ξ))-v1+vϕ′(ξ))dξ.
Taking the limit ε→0 in (49), it can be obtained from Lemma 8 that
(50)∫ξ1ξβ‍(v-(ϕ(ξ)+ξϕ′(ξ))-v-1+v-ϕ′(ξ))dξ+∫ξβξ2‍(v+(ϕ(ξ)+ξϕ′(ξ))-v+1+v+ϕ′(ξ))dξ=0.
Through simplifying (50), we get
(51)ϕ(ξβ)·(v-ξβ-v-1+v--v+ξβ+v+1+v+)=0,
which leads to ξβ=1/(1+v-)(1+v+) for the arbitrary of ϕ(ξ).
Let us turn back to calculateξα; we have
(52)(1+v+)ξβ=11+v-≥ξα=limε→0+ξαε=limε→0+11+vε(ξαε)≥11+v+=(1+v-)ξβ.
It is clear that1/(1+v-)>ξβ>1/(1+v+) for -1<v-<0<v+. Furthermore, if we assume that 1/(1+v-)>ξα>ξβ, through taking η=(ξα-ξβ)/2>0 here, then we can take ε>0 sufficiently small such that ξαε>ξα-η=ξβ+η for ξα=limε→0+ξαε. It can be derived from Lemma 8 that limε→0+vε(ξαε)=v+. Thus, we arrive at ξα=1/(1+v+) by applying (52), which contradicts the assumption 1/(1+v-)>ξα>ξβ.
On the other hand, if we assume thatξβ>ξα>1/(1+v+), by taking η=(ξβ-ξα)/2>0 now, then we can also take ε>0 sufficiently small such that ξαε<ξα+η=ξβ-η. It can be derived from Lemma 8 that limε→0+vε(ξαε)=v-. Thus, we arrive at ξα=1/(1+v-) by applying (52), which also contradicts the assumption ξβ>ξα>1/(1+v+).
From the above discussion, we can get the relationξα=ξβ=1/(1+v-)(1+v+). The proof is completed.Lemma 10.
For anyη>0, one can obtain
(53)limε→0+wε(ξ)={w-,forξ<ξβ-η,w+,forξ>ξβ+η,
uniformly in the above intervals.The process of this proof is similar to that of Lemma 5.3 in [5], and thus we omit it.Theorem 11.
Let-1<v-<0<v+ and let (vε(ξ),wε(ξ)) be the solution of (23) and (24). Then, the limit of (vε(ξ),wε(ξ)) as ε→0 is (17) with (18), namely, the corresponding Riemann solution of (1) and (2).Proof.
To prove it, we should consider the limit behavior ofwε(ξ) in the neighborhood of ξ=σ=1/(1+v-)(1+v+). We take ϕ(ξ)∈C0∞(ξ1,ξ2) with ξ1<σ<ξ2 and satisfy ϕ(ξ)=ϕ(σ) on the neighborhood of ξ=σ such as N(μ)=[σ-μ,σ+μ] for sufficiently small μ>0, which is called the sloping test functions in [22].
Then, by (26), we have
(54)∫ξ1ξ2(-ξwξε+(wε1+vε)ξ)ϕ(ξ)dξ=∫ξ1ξ2(wεϕ+wε(ξ-11+vε)ϕ′)dξ=0.
It follows fromϕ(ξ)=ϕ(σ) for ξ∈N(μ) that
(55)limε→0+∫ξ1ξ2(wε(ξ-11+vε)ϕ′)dξ=∫ξ1σ-μ(w-(ξ-11+v-)ϕ′)dξ+∫σ+μξ2(w+(ξ-11+v+)ϕ′)dξ=w-(σ-μ-11+v-)ϕ(σ)-w-∫ξ1σ-μϕdξ-w+(σ+μ-11+v+)ϕ(σ)-w+∫σ+μξ2ϕdξ.
By taking the limitμ→0 above, we have
(56)limε→0+∫ξ1ξ2(wε(ξ-11+vε)ϕ′)dξ=(-σ[w]+[w1+v])ϕ(σ)-∫ξ1ξ2(w-H(σ-ξ)+w+H(ξ-σ))ϕ(ξ)dξ,
in which H denotes the Heaviside function.
Combining (54) together with (56), we have
(57)limε→0+∫ξ1ξ2(wε-w-H(σ-ξ)-w+H(ξ-σ))ϕ(ξ)dξ=(σ[w]-[w1+v])ϕ(σ)
for all sloping test functions ϕ(ξ)∈C0∞[ξ1,ξ2]. With the same technique in [5, 22], one can see that the conclusion can also be drawn for any test function ϕ(ξ)∈C0[ξ1,ξ2].
By taking the limitsξ1→σ- and ξ2→σ+ in (57), we can obtain
(58)w(ξ)=limε→0+wε(ξ)=w-H(σ-ξ)+w+H(ξ-σ)+(σ[w]-[w1+v])δ(ξ-σ).
Thus the limit of wε(ξ) is exactly the corresponding Riemann solution (17) with (18) for w when -1<v-<0<v+.
Finally, the value ofv(ξ) at the discontinuity point ξ=σ should be calculated. Then, (26) can be changed into the following form:
(59)∫ξ1ξ2w·(ϕ+ξϕ′-ϕ′1+v)dξ=0.
By substituting (58) into (59), through a simple calculation, one can arrive at
(60)∫ξ1σ-(ϕ+ξϕ′-ϕ′1+v)dξ+∫σ+ξ2(ϕ+ξϕ′-ϕ′1+v)dξ+(σ[w]-[w1+v])(ϕ(σ)+σϕ′(σ)-ϕ′(σ)1+vδ)=0;
namely,
(61)w-(σ-11+v-)ϕ(σ)-w+(σ-11+v+)ϕ(σ)+(σ[w]-[w1+v])(ϕ(σ)+σϕ′(σ)-ϕ′(σ)1+vδ)=0.
Since ϕ is arbitrary, one can obtain σ=1/(1+vδ)=1/(1+v-)(1+v+), which implies that vδ=v-+v++v+v-.



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*Source: 893465-2013-10-09.xml*