THE VACUUM IN NONISENTROPIC GAS DYNAMICS

Availableonlineatwww.sciencedirect.comScienceDirectActa Mathematica Scientia 2012, 32B(1): 339-351数学物理学报http://actams.wipm.ac.cnTHE VACUUM IN NONISENTROPICGAS DYNAMICS*Dedicated to Professor Constantine M. Dafermos on the occasion of his 70th birthdayGeng ChenDepartment of Mathematics, Pennsylvania State University, University Park, PA, 16802, USAE-mail: chengmath. psu. eduRobin YoungDepartment of Mathematics, University of Massachusetts, Amherst, MA 01003, USAE-mailAbstract We investigate the vacuum in nonisentropic gas dynamics in one space vari-able, with the most general equation of states allowed by thermodynamics. We recallhysical constraints on the equations of state and give explicit and easily checkable condi-tions under which vacuums occur in the solution of the riemann problem. We then presenta class of models for which the Riemann problem admits unique global solutions withoutKey words nonisentropic gas dynamics; conservation laws; vacuum; large data; Rie-mann problem2010 MR Subject Classification 35L65; 76N15: 35L671 IntroductionWe consider the Euler equations of fluid dynamics in one space dimensionpt t(pu0(p)t+((1.1)E)t+(u(epu+pE +p))x=0,describing conservation of mass, momentum and energy, respectively. Here p is the density, p isthe pressure, u is the velocity, and E is the specific intcrnal energy of the fuid: the equationsare closed by specifying an equation of state or constitutive relation, describing how the thermodynamic variables are related. This equation of state中国煤化工 structure ofthe fuid and is subject to physical constraints such as tlC N M H DynamicsReceived November 27, 2011. Youngs research supported in part by NSF Applied Mathematics GrantNumber DMS-0908190340ACTA MATHEMATICA SCIENTIAVol 32 Ser. BThe Cauchy problem for (1.1 )is well understood when the initial data have small totalvariation [1, 7, 8, 16], but little is known for solutions with large data[18, 19]. One of the maindifficulties is the possible occurrence of vacuum. Several authors have studied the vacuum inisentropic gas dynamics(obtained by dropping the energy equation)[10, 12, 15, 20), and morerecently nonisentropic polytropic ideal (y-law) gases 3-5Our goal here is to describe in detail which gases admit vacuums in the solution. Itwell known that y-law gases require vacuum in order to solve the global Riemann problem (16whereas an isothermal gas (y= 1)does not. Also, a better existence theory is available forthe isothermal system, due to a degenerate wave curve structure(14, but there is no natural3 x 3 physically consistent analogue of the isothermal system. In [17], Temple considers aclass of 33 constitutive relations with simplified structure, but this system violates somethermodynamic constraints, given in 11, 13 Our intention is to present a class of physicallyconsistent constitutive laws which do not admit vacuum, thus removing the issue of vacuum forthese equations of state, in order to focus more fully on the effects of nonlinear wave interacWe begin by collecting all the physical conditions that restrict the equation of state. Werecall the solution of the Riemann problem and give a necessary and sufficient condition for theoccurrence of vacuums in the general solution of the Riemann problem. By exhibiting specifiexamples, we describe a class of constitutive laws which satisfy all our physical constraints andfor which the Riemann problem does not contain a vacuum state We then present the simplestclass of such equations of state, namelKoes/E2-1(ln(r+1)p=Koes/e,(In-+1)where 2< o< 1. Although it is implied by quoted results, we explicitly solve the globalRiemann problem for this pressure law, without use of the vacuum state. It is expected thatthis will inform the study of wave interactions, which is the subject of the authors ongoingresearch2 Thermodynamic ConstraintsTo set notation, we describe the thermodynamic constraints. The thermodynamic prop.ies of a fuid are embodied in the constitutive relation E= E(T, S), where T= l/p isspecific volume and S is the specific entropy. See[13 for a physical discussion of theseconstraints, and [15 for a detailed mathematical analysisFirst, we make the smoothness assumption,E=E(T,S)∈C2,forr∈R+andS∈R,which is true for most fluidsWe require the fuid to satisfy the Second Law of Thermodynamics, which asserts thatde=Tds-pdH中国煤化工(22)CNMHGwhere T is the temperature. This in turn implies thatEs=T, and Er=-pG. Chen &e R Young: THE VACUUM IN NONISENTROPIC GAS DYNAMICSWe assume the standard thermodynamic constraints: specific volume T, pressure p andtemperature T' satisfyT>0,p>0andT≥0;so that by(2.3)E-<0 and Es≥0.We assume stability of matter, which asserts that the energy is finiteElim e0 and S, see 11Next, we assume the thermodynamic stability constraint that the energy E=E(T, S)bejointly convexEr=-pr>0,Ess=Ts≥while alsoE·Ess≥ E-s and Ers=-ps≤0.6According to [13, thermodynamic stability yields(2.5) and (2.6), but our discussion requiresonly (2.5). Equation(2.5) in turn implies that the system is strictly hyperbolic away fromvacuumThe condition ps >0 states that the material expands upon heating at constantWe assume nonstrict inequality for ps to include isentropic gas dynamics, for whichPs≡0Our final condition is an energy condition, which states that if the pressure p(t, s)isspecified, then the energy e is well defined: that isE(,S)=/p(r,S)dr,here we have used(2. 3)and(2. 4). That is, we require that, for all S,p(T, S)dTP(p, S)dp<+∞,(27)where p is defined byP(p, S)=p(l/p, S)=p(r, S)(28)The energy condition(2. 7) imposes growth conditions on P(T, S), or equivalently restrthe pressure p near vacuum, namely pl0+, S)=0, and by IHospital's rulelim Pe(p, S)=limP(p, S)0(29)Note that our conditions alone are not sufficientCOTniceness of solutions to theRiemann problem: uniqueness is assured if and only中国煤化CNMHG例(T,E)≤E(210)is satisfied, for pressure p= p(r, E)as a function of T and E, see(15342ACTA MATHEMATICA SCIENTIAVol 32 Ser. B3 Vacuum in the Solution of Riemann ProblemsWe wish to investigate the circumstances in which a vacuum appears in the solution to aRiemann problem. We briefly recall the solution of the Riemann problem; see [16]3.1 Riemann ProblemWe begin by calculating the simple (rarefaction) wave curves. For smooth solutions, wereplace the third (energy ) equation of (1.1) by the entropy equationSt+uS=0and use(p, u, S)as the state variables. It is routine to calculate the eigensystem after writing(1. 1)in quasilinear form. The eigenvalues of (1.1)areA1=u-c, A2=u, A3=u+cand these are the wavespeeds of the backward, middle and forward waves, respectively, andc(p, S):=vpp(31)is the speed of sound. As is well known, the forward and backward waves are genuinely nonlinearand the middle waves linearly degenerate. The corresponding eigenvectors arePS0It follows that the equation of a backward simple wave isul=R(PL, S)-R(P, S),S=St(32)where the subscript I refers to the left state of the wave, and we defineR(p, S)cr, SThe equation of a forward simple wave curve isWr-u=R(Pr, S)-R(P, S),S=Sr,(34)where the subscript r refers to the right state of the wave.Next we calculate the shock curves: these are described by the Rankine-Hugoniot conditionslpl=pul￡|pul=o2+p(35)SlPu+pE]=u(pu2中国煤化工where 5 is the shock speed and the brackets denoteCNMHGCrosS the shockWe simplify these as follows: the first equation can be writtenpr36)No.1G. Chen R. Young: THE VACUUM IN NONISENTROPIC GAS DYNAMICS343and, recalling that T=1/p, using this in the second equation and simplifying yieldspl[]= soul[]-lou2l[]=-u]2Next, denoting the average of a quantity by 9= gfr, manipulating the first two equations of(3.5) yields5p]=muu+,sop-瓦=(37)The third equation of (3.5) gives, after simplifying15P22+B=22++wpUsing(3.7) and again simplifying, we finally obtain[E]+=0which is the hugoniot curve for shocks. We conclude that this describes the shock curve fullyfirst, solve (3.8 to find the relation between p and S, then useobtained from the entropy condition [6, 9, to resolve u, and finally use(. 7)to determine SRecall that an entropy condition is required to choose admissible shocks and thus obtainuniqueness of Riemann solutions [6, 9. This condition states that pressure(and thus alsodensity)is bigger behind the shock, and leads to the negative square root in(3.9)above. Itfollows similarly that the density behind a(forward or backward) rarefaction wave is smallerthan the density ahead of the wave. Also, this implies that the state behind a shock cannot bee vacuum staIt is routine to describe contact discontinuities using( 3.5 ) namely, substitute ul=0 indirectly, to obtain=0,=0u=.(310)If a contact discontinuity is adjacent to a vacuum, then we combine the contact discotinuity and the vacuum region into a new vacuum region, called a non-isentropic vacuum onwhich the entropy density pS vanishes. The left and right hand limits of s on the left and rightboundaries of such a vacuum region are different. It follows that, if the vacuum is involved inthe solution of the Riemann problem, it can only be generated between two outgoing rarefactionwaves, see 10, 12, 2013.2 Vacuum ConditionLemma 3.1 The vacuum state exists in the solution of Riemann problems of (1. 1)if andonly if, for some s,R(0+,S)>-∞1中国煤化工where R is defined in(3.3)CNMHGProof We first prove that if the vacuum state exisuo u ne suluuiuuo ui iucanann problemsthen(3. 11)is satisfied. Recall that vacuum state only appears between two rarefactionIf a Riemann solution consists of forward and backward rarefactions, it follows that the velocit344ACTA MATHEMATICA SCIENTIAVol 32 Ser. Bu is monotone increasing as a function of x[6]. Parameterizing the forward wave by e, it followsfrom(3.4)thatur-u=R(Pr, Sr)-R(P, Sr);now since u> u, we get the uniform boundR(p,S)≥-tur+R(p,S)and allowing p-0 implies(3. 11)Now suppose (3. 11) holds for some S. We claim that the Riemann problem with dataUt=(1, 0, S), and U,=(1, ur, S)has a vacuum in the solution wheneverur>-2R(0+,S)To see this, assume ur >0 and resolve the Riemann problem into backward and forwardrarefactions using (3.2)and (3.4), to getum=R(1, S)-R(pm, S)=-R(Pm, S)andur-um=R(1, S)-R(Pm, S)=-R(pm, S)with no contact as Sr= St=S. Adding, we must solve2 R(pm, S)and so if ur>-2 R(O+, S), no such pm can be found and a vacuum is required to solve theRiemann problem; see also[16, 20We now introduce an easily checkable condition which implies(3. 11), so is a sufficientcondition for existence of Riemann solutions with vacuum. This pressure near vacuum conditiondescribes the rate at which p-0: for some value of S, there exist pore numbers Eo, ao andMo, such thatp(p,S)≤Mop0 whenever p∈(0,60(3.12)Note that we require this condition at only one S: by continuity, we would generally expectthe condition to hold in an open set of S values. Note also that polytropic ideal gases satisfy(3. 12), ao being given by the adiabatic exponent y>0.Theorem 3.2 The energy condition(2.7)and pressure near vacuum condition (3.12)together imply that vacuums exist in the solution of some Riemann problemsProof According to(2.9), the energy condition implies that pp+0 as p-0, so thecomparison (3. 12) makes sense. It then suffices by Lemma 3.1 to show that, for S given by(3.12), equation(3.11)is satisfied. From(3. 3), for p< Eo, we haveR(E0,S)-R(p,S)=E/vYH中国煤化工CNMHGMoand taking the limit p-0 gives the required lower bound for R(O+, S)No. 1G. Chen R. Young: THE VACUUM IN NONISENTROPIC GAS DYNAMICS3454 Gas Dynamics Without VacuumWe now write down a class of constitutive laws for gases which do not admit vacuums inthe solution of the Riemann problem. These gases satisfy all the constraints of Section 2, butdo not satisfy the vacuum condition (3. 11). By(3.12), there is a vacuum for any y-law gas(N p?)with y>1, but no vacuum for an isothermal gas (y= 1). We thus look for presssurelaws between these cases, which restricts our equation of stateseparable energies, that is, energieE(p,f(S)g(p). Specifically, we consider energies given by the expressionE(7,S)=f(S)k2(g)dy(4.1)where f and k are C2 functions satisfying certain conditions described belowWe assume that f(S)is C2, positive, increasing and convex∫(S)>0,f(S)>0,f"(S)≥0imf(S)=∞ and lim f(S)(4.3)We assume k(y) is a C2, convex, decreasing function defined on(0, oo)k(y)>0,k'(y)<0,k"(y)>0,(44)subject to the growth limitsk2dk(y)dy(4.5)describing growth near vacuum,T→∞, and limitsk(W'y)(4.6)describing infinite density.Theorem 4.1 The above conditions imply that for gases having energy(4.1), the Rie-mann problem has a unique global solution which does not admit the vacuum state.Moreoversuch gases satisfy all constraints of Section 2 with the exception of (2.6)Proof By(4. 1) and(2.3), we haveP(T, S)=-k(In z)f(S),where we have set z=T +1. It is easy to check that the standard thermodynamic constraintsstability of matter, and thermodynamic stability (2中国煤化工ergy conditionfollows from the first equation in(4.5)We now check that the vacuum condition(3. 11)faiCNMHGe. We calculatePr(T, S)k2(In z)-2*(In z)k'(In z)∫(S)<0,346ACTA MATHEMATICA SCIENTIAVol 32 Ser. Band, since p=1/T, we writei(p,S)=-p-(r,S)r2,and so(33/pR(e, S√-p(r,S)ln(1+1/p)-√f(S)√k2(u)-2k()k()d(4.8)In 2where w=In z= In(1+T). It now follows from(4.4),(4.5)thatln(1+1/p)R(p,S)<-Vf(S)k(w)dwas p-0 for all S, so that(3. 11) is never satisfiedExistence and uniqueness of Riemann solutions with arbitrary data now follows fromSmith's medium condition(2.10), see [15]. Smith has an extra assumption, namely, he requireslim E(p, 7)=0,(4.9)when E is described as E=E(p, T). Because our energy is separable, we haveE(p,T)fna k2(y)dyk2(In z)(4.10)Since→lasr→0, for small T, we writeE(p, T)k2(y)dy+Ji k2(y)dy:2(In z)By our assumptions on k ,(4.9)follows if we show that the limit门k2(y)dr0asg→0This in turn follows from(4.6)and I'Hospital's ruleTo check the medium condition(2. 10), we use(4.10) to write( T,e)=ek2(InJina k2(y)dywhere z=T +1. The medium condition(2.10)is2k(In z)k'(Inz)Jin:k2(y)dy-k2(In a) Sm k2(y)dy +k4(ln z)n p2(2Jm2829y)dy)2which simplifies to中国煤化工CNMHGN(=24()+2()20)umwhich must hold for all z >1G. Chen R. Young: THE VACUUM IN NONISENTROPIC GAS DYNAMICSy(4.5),k(x)→08z→∞, which in turn implies that N(z)→0asz→∞. Furthermore, by(4.4)N(z)=-2k'k2+2k"/k2-k/k2+k3>0so N is increasing with limit 0, and thus N(a)<0 for all zWe remark that our conditions do not suffice to prove the convexity of E(, S), namelythe first condition of(2.6)E·EsS≥EsThis condition is easily seen to be implied by the dual assumptions that f is log-convexf(S)f"(S)≥P2(S),(4.11)and k satisfies the condition(k(2)-2k(2)/k2(y)dy>k3(2)for all z5 Concrete ExampleWe now present the simplest equation of state which does not allow for a vacuum in thesolution of the Riemann problem. Ideal polytropic (y-law) gases with adiabatic constant y>1admit vacuums, while an isothermal gas(y= 1)does not, but the isothermal gas does notsatisfy the finite energy stability of matter condition. We thus consider equations of statethat fall between these two cases. Since(3. 12)implies existence of vacuums, this restricts theequation of state to those for which(3. 12 )failsThis class of gases we present here are polytropic, satisfyingbut do not satisfy the ideal gas law pT= RT. For a polytropic gas of the form(4.1),(2.3)implies that we must havef(S)=Koe/crand this trivally satisfies conditions(4.2),(4.3)and(411)We choose k=k(z) as simple as possible so that properties(4.4)and(4.5) hold, namely2)=zfor someo∈(,1It is then easy to check that(4.4),(4.5),(4.6)and(4. 12 )holdWith these choices,(4. 1)becomesKoeS/crIn(T+中国煤化工(5.2)CNMHGand(4.7)becomesKS/crr+1(n(x+1)348ACTA MATHEMATICA SCIENTIAVol 32 Ser. BIt follows from Theorem 4.1 that the Riemann problem has a unique solution withoutvacuum. In fact, Smith's strong condition holds, namelyE(T,p)>0.This is easily seen by eliminating es/cr from(5.2)and (5.3), to getE(T, P)=p(T+1)In(T2a-1differentiatingAlthough we have an abstract proof of existence and uniqueness of Riemann solutions, wefind it instructive to prove this directlyLemma 5.1 There is a unique solution to the Riemann problem for(1.1)with equationof state(5.2), which does not include the vacuum, for arbitrary Riemann dataProof Using(4.8)and simplifying, we haveln(+1)R(7,S)Koes20(54)Following [2, we now make convenient changes of variables. First, setp: In(T +1)>0, so that T+1and defineh: =e-S/2CT R(TIt is clear that p, T, and h are equivalent coordinates, with o and r decreasing and h increasingas functions of p, and since o∈(号,1], we havelim hand lim h=∞.(5Next, definer=(ln(T+1)so that(o, m)can be used in place of (T, S)or(p, S)It follows from(5. 2)and(5.3)thatE(56)while alsoeS/2r=mφ°andR=hmφ°The simple wave curves(3. 2),(3.4)are described byuxr-=(ha-hb)maφa,a(5.7)where the subscripts denote the behind, ahead, right arFor rarefaction waves, the sound speed c decreases fromH中国煤化工andp=o(h)by ha>hb. Similarly, by(3.10), a contact discontinuity is aescrnbea byCNMHGcharacterized(58)G. Chen R Young: THE VACUUM IN NONISENTROPIC GAS DYNAMICS349It remains to calculate the shock curves. Using(5.6)in(3.8), we get20-7(P6m2-pamla)+hoe-ppm2中b-e)=0which becomesφbHere, %a>b, or equivalently ha< hb, since the sound speed c is greater behind the shock[6,9]. It is clear that the function f in(5. 9)makes sense only if the function inside the squareroot is nonnegative. To check that this holds, consider the functionyfor a>y>0. Denote the numerator and denominator of g bygn(a, 9)11and2qd(a,y):11respectively. It is immediate that for a >0qn(x,x)>0,qn(x,0)>0,gd(x,x)>0,qd(x,0)<0.Since qd(a, y) is strictly increasing with respect to y, for each a, ad(a, y)=0 has a uniquesolution 0 < y= p(a)0 for s>y>0. Hence, for fixed xq(a, y)>0 as long as p(a)bim、9(a,φb)=-∞(513)To express the composite wave curves, we defineG(hh). 9(a, pb), ha hb