The Riemann Problem with Delta Data for Zero-Pressure Gas Dynamics∗

Li WANG

1 Introduction

In this paper,we are concerned with the one-dimensional system of conservation laws of mass,momentum and energy in zero-pressure gas dynamics characterized by

where ρ and u represent the density and velocity,respectively,H= ρτ is the internal energy,τ is the internal energy per unit mass.The zero-pressure gas dynamics system is a very important one to approach the full Euler equations,which can describe the motion of free particles which stick under collision and explain the formation of large-scale structures in the universe(see[1–2]).The zero-pressure gas dynamics system consisting of conservation laws of mass and momentum has been investigated extensively.For related results,see[3–8]and the papers cited therein.However,it is well known that for the media which can be considered as having no pressure,we must take into account energy transport.Therefore it is very necessary to consider the energy conservation law in zero-pressure gas dynamics(see[9–10]).

For the Riemann problem of system(1.1),it is not difficult to see that the delta shock and vacuum do occur(see[9]).In this paper,we will investigate the Riemann problem with delta initial data and possible interactions of delta shock waves and contact vacuum state for system(1.1).As for delta shock waves,we refer readers to[11–17]and the references cited therein for more details.

In this article,we first consider the Riemann problem for(1.1)with initial data

where?>0 is sufficiently small.We constructively obtain the solutions for the problem(1.1)–(1.2).Moreover,a new kind of nonclassical wave is obtained,namely,a delta contact discontinuity,which is a Dirac delta function supported on a contact discontinuity.Let?→0,under the stability theory of weak solutions,we obtain solutions for the system(1.1)with delta initial data

where δ is the standard Dirac delta function.With the help of generalized Rankine-Hugoniot conditions and the entropy condition,we obtain the global existence of generalized solutions for the problems(1.1)with(1.3).Moreover,if we let ω0=0,u0=0 and h0=0,the solutions of(1.1)with(1.3)correspond to the solutions of Riemann problem for(1.1).This method has been used in[18]to study the Riemann problem with delta initial data for the 1-D Chaplygin gas equations.

This paper is organized as follows.In Section 2,we recall and present some known results about the system(1.1)and its Riemann solution with constant initial data.In Section 3,we construct solutions of(1.1)–(1.2)case by case.Then letting? → 0,we obtain the generalized solutions of(1.1)with(1.3).

2 The Riemann Problem with Constant Initial Data

In this section,we study solutions of system(1.1)by considering the Riemann problem with initial data

The details can be found in[9].The system(1.1)has a triple eigenvalue λ=u and two right eigenvectors r1=(1,0,0)Tand r2=(0,0,1)T,which satisfy∇λ·ri=0,i=1,2.Thus the system(1.1)is linearly degenerate.Furthermore,we obtain the Riemann solutions of(1.1)with(2.1)containing contact discontinuities,vacuum or delta shock wave.

For the case u−≤u+,the solution can be expressed as

where u(ξ)satisfies that u(u−)=u−and u(u+)=u+(see Figure 2.1).

For the caseu−>u+,the overlap of the characteristic results in aδ-wave connecting the two states(ρ−,u−,H−)and(ρ+,u+,H+).The solution of(1.1)with(2.1)is

wherex(t),ω(t),h(t)anduδsatisfy the generalized R-H conditions

where[ρ]=ρ+− ρ−,with initial datax(0)=0,ω(0)=0,h(0)=0.By simple calculation,we obtain

and

Thus,we have obtained the solutions of system(1.1)with(2.1).

Figure 2.1 u− ≤u+

Figure 2.2 u−>u+

3 The Riemann Problem With Delta Initial Data

In this section,we first consider the solution to system(1.1)with(1.2).Then letting?→0,we get the solution of(1.1)with(1.3)under the stability theory of weak solutions.According to the relations amongu−,u+andu0,we discuss the Riemann problem case by case.

Case 1u−≤u0≤u+

In this case,whentis small enough,the solution of the initial value problem(1.1)–(1.2)can be expressed briefly as follows(see Figure 3.1(a)):

where“+” means“followed by”.The propagation speed of theandisu0.Thus we see that thecan not overtakeJ02at a finite time.So far,the solutions of(1.1)with(1.2)have been constructed completely.Letting?→0,we obtain a solution of(1.1)with(1.3)as follows(see Figure 3.1(b)):

where the propagation speed of theδSisu0.

Ifω0=0,u0=0 andh0=0,it is easy to see that the solution of(1.1)with(1.3)is consistent with the Riemann solutions of(1.1)with(2.1),which implies that the solution is stable with respect to the perturbation in this case.

Figure 3.1(a)

Figure 3.1(b)

Case 2u0

Similarly to the analysis in Case 1,we seek the solution of the following form(see Figure 3.2):

andδSsatisfies the following generalized Rankine-Hugoniot conditions:

where[ρ]=0−ρ−,with initial data

For more details,we refer readers to[18–19]and the references cited therein.

Figure 3.2

Next,we only need to solve the initial value problem(3.2)with(3.3).From(3.2),we have

and

By virtue of(3.4)–(3.5),we have

Solving(3.7)withω(0)=ω0,we have

From(3.4)and(3.7),we have

Using(3.5)–(3.6)and(3.8)–(3.9),we obtain

Remark 3.1 From(3.9),we have

This implies thatδSdoes not penetrate vacuum completely and it converts toδJwhent→∞.

Remark 3.2 From(3.4),(3.6)and(3.8),ifω0=0,u0=0,h0=0,then(ω,u,h)=(0,u−,0).This is consistent with the results of the Riemann problem(1.1)with(2.1).It implies that the solution constructed here is stable under some perturbations.

Case 3u+

Similarly to the analysis in Case 1,we seek the solution of the following form(see Figure 3.3):

andδSsatisfies the following generalized Rankine-Hugoniot conditions:

where[ρ]=ρ+−ρ−,with initial data

Figure 3.3

Now,we are going to solve the initial value problem(3.13)with(3.14).

Integrating(3.13)from 0 totwith initial data(3.14),we have

Canceling ω in the first and second equation of(3.5),we have

Integrating(3.16)from 0 to t,we have

Solving(3.17),we get

and

From(3.15),we have

and

Remark 3.3 It is seen that

So from(3.22),we have

Remark 3.4 If ω0=0,u0=0,h0=0,then

This is consistent with the results of the Riemann problems(1.1)with(2.1).It implies that the solution constructed here is stable under some perturbations.

Case 4u0

Similarly to the analysis in Case 2,we know that,in this case,whentis small enough,the solution is the same as that in Case 2.More precisely,according to(3.9),there exists a uniquet1,such thatuδ(t1)=u+.

When 0≤t≤t1,the solution is the same as that in Case 2.Whent>t1,theδ-shock wave will overtakeJ+in finite timet=t2,and we seek the solution of the following form(see Figure 3.4):

Figure 3.4

For 0≤t≤t1,the solution is the same as that in Case 2,and here we omit the details.

Fort1≤t≤t2,

andδS1satisfies the following generalized Rankine-Hugoniot conditions:

where[ρ]=0−ρ−,with initial data

wherex1=x(t1),ω1=ω(t1),u1=u(t1)andh1=h(t1).The solution of(3.26)–(3.27)(x1,ω1,u1δ,h1)(t)can be obtained similarly as in Case 2.

Fort>t2,wheret2is determined byx1(t2)=u+t2,

and δS2satisfies the following generalized Rankine-Hugoniot conditions

where[ρ]= ρ+− ρ−,with initial data

where x2=x1(t2), ω2= ω1(t2),u2=u1(t2)and h2=h1(t2).The solution of(3.29)–(3.30)(x2,ω2,u2δ,h2)(t)can be obtained similarly as in Case 3.

So far,we have finished the discussion for all kinds of interactions.We summarize our results in the following.

Theorem 3.1 The solution of(1.1)with(1.3)can be obtained by letting?→0 for the solutions of(1.1)with(1.2).If we let ω0=0,u0=0,h0=0,the solution of(1.1)with(1.3)corresponds exactly to the solution of Riemann problem(1.1)with(2.1).It implies that the solution constructed here is stable under some perturbation.

AcknowledgementThe author would like to thank the referees for their comments which helped to improve and clarify this paper.

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