Riemann problem for the zero-pressure flow in gas dynamics

\ol 11 \o. 5PROGRESS IN NATURAL SCIENCEMay 2001SCIENTIFIC PAPERSRiemann problem for the zero- pressure flow in gas dynamicsLI Jiequan(李杰权) and LI Wei(荔炜)I. Institule of Mathemalrs, Chinese Aeadeiny of Scenes, Beijing 100080.China2. Department of Mathematit's, Northwest I niersits, Xi an 710069, ChinaReceIved July 31, 2000: mv ised September 21, 2000Abstract The Ruemann problem for zer-pressure flow in gas dynames in une dimension and two dimensions isinvestigated. Through studying the generalized RankIme-Hueoniot condit ns of delta-shosk waves, the one-dimensionalH:mann solution is proposel which exhibits four different stru tures when the initial density insoles Dirac measure. Fcthe twodimensional cHwe. the Riemann solution wth two pieces of nitial constant states separated at a smooth curve IsKeywords: zero-pressure now in gas dynamics. Riemann problem, delta-shocks. generalized Rankine-Hugoniot condition, entropy conditionConsider a zero-pressure flow in gas dynamicsdiv(pu)=0,where p(x,t)≥0andu(x,1)=(u1,…,u)(x∈,t∈[0,)) are the density and the meanseboeis respectively, x represents the partial derivatives with respect to the spare component x, (k=I,.). This model can be viewed as a direct result of imposing the pressure p=0 on the Eulerj=1,…,d(2)Besides, model(1)is referred to as the adhesion particle dynamics system to explain the formation ofthe large-scale structures in the universe. In this paper, d is 1 or 2The distinct feature is that Eqs. (1)have repeated eigenvalues and incomplete associated eigenvectors.The C solution of (1)can be given by the explicit formulaProject partly suppored hy the \ational Natural Scienee Foundation of China( Grant \o. 6987108.5中国煤化工CNMHGPROGRESS IN NATURAL SCIENCEVoln(x,1)=ao((x,t)xwhere(Pe, uo)(x0 EC(R )are the initial density and veloctiy, (x, t)is the inverse of flow(xo)and aEquation(3)shows that C solutions of (1)may not exist after a finite time. More preciselye density itself and the gradient of velocity beeome a singular measure if the flow map is compres-sive. So we have to seek discontinuous solutions. For bounded BV solutions, the normal directionalcomponents of the velocity on both sides of a discontinuity must be identical, otherwise there is nobounded solution. Noticing that the solution must develop Dirac measure in the density for somerases, we introduce delta-shocks in the solution of (1). For the exact definition of the delta-shockwave. the reader may refer to Reference [2.The study of delta-shocks began in 1990 when Tan et al. found that no classical weak solutionexisted for some values of initial data in the investigation of the Reimann problem for a 2 x 2 simpfied model of the Euler equation(4)t),+(t2)and that delta- shock wavecessarily used as parts in their solutions. Since then some effortsave been made to search for the physical background of the delta-shock and to perfect itsal theory. Sheng et al. 4 set forth from the splitting method of the numerical scheme of the Eulerquation. and studied the Riemann problem for (1)in one dimension and in two dimensions with theI anishing viscosity method when the initial density is just a function of total bounded variation. Laterit was found that(1 )is just the consequence of neglecting the pressure effect on the Euler equationor an adhesion particle dynamics model in astrophysicsdelta-shock can be explained ag theconcentration of particles after collision. Besides, Bouchut also discussed this model in one dimensionfrom the point of view of cold plasma. low pressure and evanescent viscositySinee the delta-shocks appear in the solution, it is natural to consider that when the initial dataontains Dirac measures, the viscosity vanishing method is too complicated to solve this problemThus we gave generalized Rankine-Hugoniot conditions to define one-dimensional and two-dimensionaldelta-shocks2.Although the generalized Rankine-Hugoniot conditions are rather complicated inorm. we can diseuss them explicitly1)Ii. J. \ote on the compressible Euler equations with zero temperature, Appl. Math. Let, in press, 2000中国煤化工CNMHG丶o.5L I et al.: RTEMA\\ PROBLEM FOR ZERO-PRESSURE FLOW333In this paper we study the Riemann problem for Eos (1)respectively. The crucial arguments are how to solve the generalized Rankine-HugThe above discussion shows that the riemann solutions can be used as a building block to establishthe existence of the general Cauchy problem1 Generalized Rankine Hugoniot conditions of delta shocksIn this section we will give a precise definition of solutions of (1)in the sense of distributionsThe equivalent solution can be defined in the sense of measure. Then we will offer the generalizedRankine-Hugoniot conditions of delta-shocks and the so-called entropy conditions to ensure theDefinition 1.(p, u)is a solution of (1)in the sense of distributions if?2+(q)· V dadt=0,0,c)×Rj=l,d: d= I or 20xP9+(pn)… V dadt=0for every o(t, x)ECC(0, )R), where V is the gradient operatorDefinition 2. The weighted Dirac delta function wd is a distribution in D'(ic,d, x R)(i)u(t)6∈m([c,dXR) supported on a smooth curve L:x=x(t)(c≤l≤d)isdefined by(n(t)12,9)-w(1)g(t,x(t)dr,g(t,x)∈C:(0,。」×R).(6)(ii)w(t,s)8sED'(c,djx R2)supported on a smooth surface of「x=x(t,s)S(a≤l≤b,c≤s≤d)y= y(t,s),is define byu(t,s)p(t, x(t, s).y(t, s))dtdg(t,x,y)∈C(.0,∞)×R(7)onal case of (1)at the outset. For a bounded discontinuity xx(t)in the sense that the solution is BV, the Rankine. Hugoniot condition is中国煤化工CNMHGPROCRESS 1\ NATURAL SCIE\CEThroughout this paper Iu and I, u are the limit values of u on the left-hand side and righl-handside of x=x(t)Identity(&)shows that the bounded discontinuity is just a slip line denoted by J, and twostates,(p:, u,)(x, L)and(e, u, )(x, t), can be connected by J if and only if I (t)=I(L). When the initial data of Eqs. (1)are decreasing, the characteristie lines overlap and boundedsolutions do not exist. So we introduce della-shock solutions which are measure solutions in the form(P,u)(x,t)x ua(t)>Iu(t)(11)Here we should point out that the entropy inequality5(u)+2s(u)≤(12)for ever convex function S: R-R cannot ensure the uniqueness: 5For the two-dimensional case of (1), if we consider bounded weak solutions as similar to thoseof the one-dimensional case, then we obtain the Rankine- Hugoniot condition ofn;=-llnx-ln、=-lnfor a bounded discontinuity, where(n,, n,, n, )is the normal direction of the discontinuity. Identity(13)shows that the components of velocity in the normal direction of a discontinuity on both sides ofthe discontinuity must be consistent. Therefore it is necessary to consider della- shock solutions asimilar to those in the one-dimensional case中国煤化工CNMHGu ct al.: RIEMANN PROBJ FM FOR ZERO. PRESSLRE FLOWLet a delta-shock solution be(p2,u,t2)(x,y,t)(p,u,n)(x,y,t)={(u(t,s)8(x-x(t,s),y-y(t,s),u(t,s),t(t,s),(14)x =x\t,s),is a discontinuity surface with a suitable parameter s, Then the generalized Rankine-Hugoniot condition is a svstem of the first-order parial differential equations(t,s)P,pu,pan,, n3t(I ou],[pu2),L puv ]).(n,, r,n),(16b)a(wig)where u and r are the respective components of velocity in x and y directions,(n,, n,, n,uz ds3as'as'as is the normal direction of S oriented from(1)to(r)((1)is the ab-breviation of the side where the state is(pr, up, Dp)),p]=p-r. For the definiteness, we alwaysdenote by(r)the side of(x-xo, y-yo) (n,, n, )>0 for all t >0, where(xo, yo, t )is a pointFor the two-dimensional delta-shock S, the geometrical entropy condition is that the characteristie lines on both sides of S are incoming. In other words,r,)·(n1,n,)<(aa,n2)·(nx,n1)<(a,E2)·(n2,n,)(17)As we have pointed out above, system(16)has exactly a quintupledent of all variables. In this sense, it is somewhat similar to a degeneratelystem of hyperbolicpartial differential equations中国煤化工CNMHGPROGRESS IN \ATL RAL SCIENCERiemann solution in one dimensionNow we start to discuss the Riemann problem for(1 )in one dimension with the standard charac-teristic method using the generalized Rankine- Hugoniot condition(10). Due to obvious reasons, weonsider the Riemann initial data involving singular measures0where pr, mn and p, are not all zero. Otherwise the solution is trivialAt first, we solve svstem (10) with the initial conditionx(0)=0,m(0)= ma and u2(0)=u0(19)satisfying u:>uo>by iously we have「pultuue-ou]x= mao -]tIt follows thatIp] -pu. x mou0-pu]t-(mo- pu ]t )ug(2la)d(Lolx'+(Lpu'Jt(21b)t".pu: /2mo-lpu t(mo -Lou n)+u(t)leldu(t)=(mo+ 2mo(Ipu])t+Pp,(u1-u,)2)2(23)Furthermore, from(20)中国煤化工CNMHGlI et al. HIEMANN PROBLEM FOR ZERO-PRESSURE FLOTou Ix mouo-L puE(x(1),x(t),(1)=|yea+vpaPrLemma I. The solutions of (10)and(19)possess the following properties0, then u≡(iii)x'(t) is a monotone function of tThe proof of this Lemma is a simple matter, so the details are omittedNow we start to construct the Riemann solutions of(1)and(18). For simplicidenote a particular particle by its Lagrange coordinate xo, and the vacuum state by Vac. According tothe relation of ur, uo and u, we solve the riemann problem case by caseCase 1. u, uo u. This is a typical case, a delta-shock emits from the origin. Solving(10)and (19), we obtain a fusion solution(p,un)(x,t1)={(no(t).8,u2(t),x=x(t),(28)x>x(twhere x(t), w(t)and ua (t)are defined in Equations(22)-(24)Case2.u1≥u,≥uo( for the case where un≥l≥l,, the structure of solution is similar)It is easy to see that the particles at xo <0 collide with the particle at the origin at first, and thetrajectory, mass and velocity of the fused particles are respectivelyx=()=m+42-(mn+m(-0))=)m2+2mop1(u1-L0)中国煤化工CNMHGPROGRESS EN NATURAL SCTEIol. IIu-uAt the time twhile x (t)=u.Io(ut-urthe particles at xo<0 start to collide with those with xo >0. At this moment(31)After that,x(t), wo(t)and u,(t)are defined by the generalized Rankine-Hugoniot condition(10)and the initial condition(31). The details are omitted. The solution can be expressed asP,叫(1(t)8,u2(t),x1(t)(p,u)(x,t)=x1x>utx(t)(p,u)(x,t)≥Case 3. uo u0. The solution for this case can be expressed中国煤化工CNMHG[I et al. RIEMANN PROBLEM FOR ZERO-PRFSSURE FLOWt(x(t)8(x-x)(t)),u2(t))(t)Case 4. u suos u,. For this case, the solution is simple and can be written as≤4,L,dt0(38)then the characteristic lines from state 1 will intersect those from state 2, where [v]=cr-ti,etcOtherwise they will not intersectProof. Inequality(38)is equivalent to中国煤化工CNMHGPROGRESS IN NATURAL SCIE.CEVol. IlThe characleristic directions of the left state(pI, uI, u,) and the right state(p2,u2,t2)are(u1,1,1)andc2=(u2,t2,2)and the tangential direction at any point of r in(x, y)plane isT=(1,f'(x),0)(41)we obtaic1XT=(-f'(x),1,u1f'(x)-t1)c2xT=(-f"(x),1,u2f'(x)-n2)he planes passing any point(s, (s))of r and with the normal direction n, and n2 are respect(t(43)2equality(39)implies that)f(x)+(y1-f(s5)<-(x2-s)f(s)+(y2-f(s)in which(x1, yI, t)and(x2, y2, t) are points of n, and 12, respectively. Thus we have)·(f'(s),-1)<0,(45)which shows that Lemma 2 holdsLemma 2 means that a two-dimensional delta-shock wave must emit from the initial discontinuityr: y=/(x)when(38)is satisfied. In what follows, we will discuss the Riemann problem in twoCase5.v]-[ulf'(x)>0In order to use the generalized Rankine-Hogoniot condition(16), we parametrize Ias∫(s),and we just consider the case of w =0 initially and [o*0. As the case in which [p]=0 is ratherimple, we omit it. First, we notice that中国煤化工CNMHGLI et al.: RIEMANN PRORIEM FOR ZERO- PRESSURE FLOWIplou']] 2=-P1P2[u]2By the generalized Rankine-Hogoniot condition(16), we havea(8=(LpL u2].pu12)ay+(Lpu l[ pu]-Le1. ut /)do1pP21-1(49)A comparison of(48)with(49)leads to the identity[o,a(uua)\-fuller(50a)rewritten a(lpx]-[u].p]-[pl([lu3-[ul2)i:=0(50bIt follows that(Lul. puj-[ul p]-[](leJug-[ulva))w=0inee u>0 when (>0 and I]#0, we obtainut+Wjla=(51)n2L[vlx+luly =-(u1t2-u,)t-v]s+.ulf(s).中国煤化工CNMHGPROGRESS IN NATURAL SCIENCEwhich is differentiated with respect to s, and ist]+[u]f'(s)Hence we getf'(s)We calculateasu1t2-u2E1 d+Gulf(s)(55)Therefore, we havea=((s)-1[p+(m1(2:(r()-)f(s)-1)(ple2-m])Always regarding "g as at, we getIt follows that(:()-1)y-(=r()-m-(:(-)/,:r()-(ah+)havem2-(2()-1(m1-:mP中国煤化工CNMHGL et al.: RIEMANN PROBLEM FOR ZERO-PRESSURE FLOW343ot-1f"(s)-1)(m2lt+[pcJf"(s).(From(57)and(58), we obtain()-1)(pl5))t8([2]+[prl/(s)「pitsy-[ply=(plt+[pf(s)t;-(pc2t+[pmf(s),(59)ince∫'(s)-1≠0. Note that(59) is equivalent to7(2y2-(m+|l1(),)=-(11+m(,)2+[pmlf(s)t=0.(60)Solving(60), we getp]±√p1pLu])t+f(s)(61)And from(53)we havep±√p1P;u)t(un, va)√1u1±√P2"2yP1±√P1±√P2By the entropy condition(17), we haveLaf'(x)It is easy to check thatP1a=√2l2yP-√p22/2中国煤化工CNMHGPROGRESS IN NATLRAL SCIENCEdoes not satisfy(63). Thus we chooseM,)=P1a+√Pu2yfe2as our(x,y)=Y“+y2:1+,P+P21+f()(66)Finally, we obtaini/(s)-1)(pt-[m])t√p1P2(-uJ/'(s)+[r)(67)Case6.[:]-Lulf'(x)<0. For this case, the structure of solution is simple and the soluion can be written asx≥xp2,42,U2)t x?.for any fixed t=T>0 and y= Y, where x, and x2 are the intersection points of the straight linet= T, y= Y and the surfaces y=f(x-tu1)+Itr and y=f(x-tu2)+ti2 respectivelyRemark. For the general case, the structure of solution is just the combination of solution ofCases 5 and 6ReferencesI Shandarin, S. F, et al. The large-scale structure of the universe: turbulence, intermittency. structures in a self-gravitstingmedium, Rev. Mod. Phvs., 1989, 61: 18.2 L, J. et al. Generalized Rankine-Hugomot conditions of weighted Dirac delta waves of transportation equations. Nonlinear PDEand Related Areas(ed. Chen Guiqiang et al. ) Singapor: World Scientifie, 1998, 219-2323 Tan, D. et al. Two-dimensional Riemann problem for a hyperpolic system of nonlinear conservation laws(1)Four-J cases. JDifferential Equations. 1994, 111: 203al.The Riemann problem for transportation equations in gas dynamics, Memoirs of AMs, 1999, 137(634): I5 Bouchut. F. On zero-pressure gas dynamics, Advances in Kinetic Theory and Computing, Series on Advances in Mathenaticsfor Appl:ed Sciences, Singapore: World Scientific, 1944, 22: I7-10中国煤化工CNMHG