The Riemann problem of relativistic Euler system with Synge energy
TTHE RIEMANN PROBLEM OF RELATIVISTICEULER SYSTEM WITH SYNGE ENERGY
TOMMASO RUGGERI, QINGHUA XIAO, AND HUIJIANG ZHAO
Abstract.
In this paper, we study the Riemann problem of relativistic Euler system forrarefied monatomic and diatomic gases when the constitutive equation for the energy isthe Synge equation that is the only one compatible with the relativistic kinetic theory.The Synge equation is involved with modified Bessel functions of the second kind andthis makes the relativistic Euler system quite complex. Based on delicate estimates of themodified Bessel functions of the second kind, we provide a detailed investigation of basichyperbolic properties and the structure of elementary waves, especially for the structureof shock waves and in this way, the mathematical theory of the Riemann problem for theserelativistic Euler system, which is analogous to the corresponding theory of the classicalones, is rigorously provided.
Contents
1. Introduction 22. Entropy principle, symmetric form and growth of entropy 72.1. Main field and symmetric form 72.2. Entropy growth across a shock wave 82.3. Consequences for the relativistic Euler system 93. Relativistic rarefied monatomic gas 103.1. Characteristic velocities 103.2. Strict hyperbolicity 113.3. Sub-luminal characteristic velocities 113.4. Genuine nonlinearity 133.5. Riemann invariants 143.6. Structure of the shock curves 143.7. Monotonicity of the velocity on rarefaction curves 224. Relativistic Euler system for Diatomic gas 225. Proof of the main theorem 1.1 275.1. Vacuum condition 275.2. Existence of solutions to the Riemann problem 296. Conclusions 29Appendix 1: Modified Bessel functions and properties 30Appendix 2: Estimates of the ratio K ( γ ) K ( γ ) Date : January 14, 2020. a r X i v : . [ m a t h - ph ] J a n RUGGERI, XIAO, AND ZHAO Introduction
One of the main problems in hyperbolic systems is the
Riemann problem . This problemwas proposed by Riemann considering a gas that is initially separated into two regions by athin diaphragm. The gases in the two regions are in different equilibrium thermodynamicstates, respectively. The question raised by Riemann is what happens when the diaphragmis put away. In literature, by extension of this problem, the Riemann problem deals withevery solution of a system of conservation laws in one-space dimension along the x axiswhen the initial data composed of two different constant states ( u L , u R ) are connected witha jump at x = 0.The Riemann problem for hyperbolic conservation systems was completely solved mainlyby P. Lax [1]. It was shown that the solution of the Riemann problem for hyperbolic systemsof conservation laws is a combination of the rarefaction waves, contact waves, and shockwaves (see e.g. [2] and references therein).A huge literature of the Riemann problem exists, in particular, many numerical resultshave been obtained by using the Riemann solvers (see e.g., [3]).For the classical Euler system, there have been enormous works (see [2,4–6] for instance).For brevity, we only list some of them: the global existence, as well as the sharp decay rate,was obtained for the entropy solutions with small amplitude in the celebrated work of Glimmand Lax [7]; the “large data” global existence theorem for weak solutions was initiated byNishida [8]. It is well known that the Boltzmann equation is related to the systems of fluiddynamics for rarefied gas. This fact is revealed in the works such as [9, 10] for the shockprofile solutions of the classical Boltzmann equation, [11–13] for the hydrodynamic limitsfrom classical Boltzmann equation to Euler system with waves and [14–17] for the nonlinearstability of waves and boundary layers of the classical Boltzmann equation.The aim of this paper is to consider the problematic of Riemann problem in the relativisticframework. Let V α and T αβ be the particle-particle flux and energy-momentum tensor,respectively [18–22]: V α := ρu α , T αβ := ph αβ + ec u α u β . (1.1)Then, the field equations for relativistic single fluid are the conservation of particle numbersand energy-momentum tensor in Minkowski space: ∂ α V α = 0 , ∂ α T αβ = 0 , (1.2)where ρ = nm is the density, n is the particle number, m is the mass in rest frame, u α ≡ ( u = Γ c, u i = Γ v i ) is the four-velocity vector, Γ = 1 / (cid:112) − v /c is the Lorentzfactor, v i is the velocity, h αβ = u α u β /c − g αβ is the projector tensor, g αβ is the metrictensor with signature (+ , − , − , − ), p is the pressure, e = ρ ( c + ε ) (1.3)is the energy, that is the sum of internal energy ( ε is the internal energy density) and theenergy in the rest frame, c is the light velocity; ∂ α = ∂/∂x α , x α ≡ ( x = ct, x i ) are thespace-time coordinates and the greek indices run from 0 to 4 while the Latin indices from1 to 3 and, as usual, contract indices indicate summation. HE RIEMANN PROBLEM OF RELATIVISTIC EULER SYSTEM WITH SYNGE ENERGY 3
For two dimensional space-time case, the system (1.2) with (1.1) is expressed as ∂ t (cid:18) ρc √ c − v (cid:19) + ∂ x (cid:18) ρcv √ c − v (cid:19) = 0 ,∂ t (cid:18) ( e + p ) vc − v (cid:19) + ∂ x (cid:18) ( e + p ) v c − v + p (cid:19) = 0 ,∂ t (cid:18) ( e + p ) v c − v + e (cid:19) + ∂ x (cid:18) ( e + p ) c vc − v (cid:19) = 0 . (1.4)We need the constitutive equation p ≡ p ( ρ, e ) (1.5)to close the system (1.4). This is usually obtained, in parametric form, through the thermaland caloric equation of state p ≡ p ( ρ, T ) , e ≡ e ( ρ, T ) , (1.6)where T is the temperature.To the system (1.4) with (1.5) we prescribe the Riemann initial data u ( x ) = (cid:26) u L = ( ρ L , v L , e L ) , x < , u R = ( ρ R , v R , e R ) , x > , (1.7)where u L and u R are two different constant states: u L (cid:54) = u R .In 1948, Taub [23] derived the equations (1.4) for a relativistic fluid and the Rankine-Hugoniot equations of the shock waves assuming as constitutive functions the pressure andthe internal energy of polyatomic polytropic classical case: p = k B m ρT, ε = D k B m T, → ε = pρ ( k − , → p = ( k − e − ρc ) , (1.8)where D = 2 / ( k −
1) is related to the degree of freedom and k = c p /c V > k B is the Boltzmann constant. Smoller and Temple [24] considered asconstitutive equation p = σ e ( σ is a constant) that substantially corresponds to the ultra-relativistic regime as we will see later. In this case the authors took into account only thesecond and third equation of the system (1.4) because the first equation is independent, andestablished the global existence of entropy solutions to the Cauchy problem with arbitraryinitial data of finite total variation (see also Wissman [25]). Chen [26] extended this result tothe case of a constitutive equation corresponding to an isentropic classical gas for which p = σ ρ k and discussed the Riemann problem of the relativistic Euler system (1.4). The sameauthor in [27] considered as constitutive equations (1.8) that corresponds to a polyatomicclassical gas. On the other hand, for smooth solutions to the ultra-relativistic Euler systemin (3 + 1)-dimensional space-time, Makino-Ukai [28, 29] established the local existence ofsolutions with data away from vacuum applying Friedrichs-Lax-Kato’s theory, and Lefloch-Ukai [30] further extended it to the case with vacuum; the singularity formation of smoothsolutions was studied by Pan-Smoller [31]. For more works about the relativistic Eulersystem, we refer the interested readers to [32–35] and the references therein.The previous constitutive equations in [24, 26–29] are too much simplified, either onlyverified in the ultra-relativistic limit or verified in the classical limit. To have more realisticequations in the relativistic regime, at least for rarefied gas, we need to justify this at the RUGGERI, XIAO, AND ZHAO mesoscopic scale using the kinetic theory. If we take into account the relativistic kineticframework, we have the Boltzmann-Chernikov equation: p α ∂ α f = Q, (1.9)where f ≡ f ( x α , p β ) is the distribution function, p α is the four-momentum with the property p α p α = m c , and Q is the collisional term. Taking the first 2-blocks of the tensorialmoments, we have: V α = mc (cid:90) (cid:60) f p α d (cid:126)P , T αβ = c (cid:90) (cid:60) f p α p β d (cid:126)P , (1.10)with d (cid:126)P = dp dp dp p . In the case of non-degenerate gases, the constitutive equations (1.6) can be calculated viakinetic theory with the J¨uttner equilibrium distribution function f J = nγK ( γ ) 14 πm c e − γmc u β p β , as follows: p = mnc γ = k B m ρT , (1.11) e = nmc K ( γ ) (cid:20) K ( γ ) − γ K ( γ ) (cid:21) , (1.12)where K j ( γ ) , ( j = 0 , , , . . . , ) are the modified second order Bessel functions, and γ is adimensionless variable defined as γ = mc k B T . (1.13)We recall that a fluid can be considered in a relativistic context if γ is very small. This meansthat the bodies are so hot that the mean kinetic energy of particles becomes comparable withtheir rest energy or even surpasses that energy or the mass is extremely small. Thereforeit is of considerable interest in several areas of astrophysics and nuclear physics. The twolimits γ → γ → ∞ correspond respectively to the ultra-relativistic limit and classicallimit.The expression of energy (1.12) is called the Synge energy [22]. In the classical limit( γ → ∞ ), by taking into account the expansion of the Bessel functions: K ( γ ) = (cid:114) π γ − / e − γ (cid:20) γ + o (cid:18) γ (cid:19)(cid:21) ,K ( γ ) = (cid:114) π γ − / e − γ (cid:20) γ + o (cid:18) γ (cid:19)(cid:21) , the Synge energy converges to eρ = c + ε, with ε = 32 k B m T. The expression of internal energy ε shows that both classical and relativistic kinetic theoriesare valid only for rarefied monatomic gases. In fact, the usual expression in classical theoryof internal energy (for polyatomic polytropic gas) is (1.8), where D = 3 + f i is related tothe degrees of freedom of a molecule given by the sum of the space dimension 3 for the HE RIEMANN PROBLEM OF RELATIVISTIC EULER SYSTEM WITH SYNGE ENERGY 5 translational motion and the contribution from the internal degrees of freedom f i ≥ D = 3. In theultra-relativistic limit γ →
0, the Synge energy equation (1.12) converges to e = 3 p .In this context, we mention the work of Speck and Strain [36] where the local existenceof smooth solutions to the Relativistic Euler system derived from relativistic Boltzmannequation was presented with the energy currents method introduced by Christodoulou [37].Recently, a big effort was made to construct a Rational Extended Thermodynamics(RET) theory, in the classical framework, that goes beyond the monatomic gas case. Infact, Ruggeri and Sugiyama with coworkers gave a series of papers in these years on thissubject and the results are summarized in their recent book [38]. Pennisi and Ruggeri gen-eralized this idea to the relativistic framework for a gas with internal structure both in thecase of dissipative gas [39] and the most simple case of Euler fluid [40]. They started fromthe classical ideas for polyatomic gases introduced first by Borgnakke and Larsen [41] andproposed a generalized Boltzmann-Chernikov equation that has the same form of (1.9) buthas the extended distribution function f ≡ f ( x α , p β , I ), depending on an extra variable I that takes into account the energy due to the internal degrees of freedom of a molecule.The authors considered instead of (1.10), the following moments: V α = mc (cid:90) (cid:60) (cid:90) + ∞ f p α φ ( I ) d (cid:126)P d I ,T αβ = 1 mc (cid:90) (cid:60) (cid:90) + ∞ f (cid:0) mc + I (cid:1) p α p β φ ( I ) d (cid:126)P d I . (1.14)The meaning of (1.14) is that the energy and the momentum in relativity are components ofthe same tensor and we expect that, besides the energy at rest, there is a contribution fromthe degrees of freedom of the gas due to the internal structure, as in the case of a classicalpolyatomic gas. φ ( I ) is the state density of the internal mode, that is, φ ( I ) d I representsthe number of the internal states of a molecule having the internal energy between I and I + d I .In [39], using the Maximum Entropy Principle (MEP), the authors found the equilibriumdistribution function that generalizes the J¨uttner one: f E = nγA ( γ ) K ( γ ) 14 πm c e − γmc (cid:104)(cid:16) I mc (cid:17) u β p β (cid:105) , (1.15)with A ( γ ) given by A ( γ ) = γK ( γ ) (cid:90) + ∞ K ( γ ∗ ) γ ∗ φ ( I ) d I , where γ ∗ = γ + I k B T .
The pressure and the energy for polyatomic gases, compatible with the distribution function(1.15) are [39]: p = nmc γ = k B m ρT ,e = nmc A ( γ ) K ( γ ) (cid:90) + ∞ (cid:20) K ( γ ∗ ) − γ ∗ K ( γ ∗ ) (cid:21) φ ( I ) d I . (1.16) RUGGERI, XIAO, AND ZHAO
We remark that the pressure has the same expression for a monatomic and for a polyatomicgas, while (1.16) is the generalization of the Synge energy to the case of polyatomic gases.The macroscopic internal energy in the classical limit, when γ → ∞ , converges to the oneof a classical polyatomic gas (1.8), provided that the measure φ ( I ) = I a , where the constant a = D − . (1.17)In the ultra-relativistic limit it was proved in [40] that the generalized Synge equation(1.16) for a gas with internal structure coincides with the one postulated by Smoller andTemple and other authors but with a precise value of the constant that is related to thedegree of freedom p = σ e, with σ = , ∀ − < a ≤ , a +1 , ∀ a > , with a given by (1.17).For a → −
1, the polyatomic equations converge to the monatomic ones [42]. The poly-atomic gas theory is very complex, but when a = 0 we will prove that the integral in (1.16)can be written in an analytical way and this case corresponds to the diatomic gas.As also noted in [36], due to complexity of constitutive equations (1.11), (1.12) (monatomicgas) or (1.16) with a = 0 (diatomic gas) , basic issues such as the maps’ invertibility betweenfluid dynamic variables which are expressed as functions of any two of them, and hyperbol-icity of the relativistic Euler system are difficult to verify. We will show that despite therelativistic Euler system (1.4) with state relations (1.11), (1.12) (monatomic gas) or (1.16)with a = 0 (diatomic gas) is very complicated, similar results as the Riemann problem ofclassical Euler system can be obtained.In order to formulate the main result of the Riemann problem, as in [6], we define S i ( u L ) = { ( p, v, S ) : ( p, v, S ) on i-shock waves from u L } , R i ( u L ) = { ( p, v, S ) : ( p, v, S ) on i-rarefaction waves from u L } , i = 1 , , where u L is the left point on the i − th wave S i ( u L ) or R i ( u L ). And we further define S pi ( u L ) = { ( p, v ) : ( p, v ) ∈ Projection of S i ( u L ) } , R pi ( u L ) = { ( p, v ) : ( p, v ) ∈ Projection of R i ( u L ) } , T pi ( u L ) = S pi ( u L ) ∪ R pi ( u L ) , i = 1 , . The principal aim of this paper is to prove the following theorem:
Theorem 1.1.
For the relativistic Euler system (1.4), let its constitutive equations be givenas in (1.11), (1.12) (monatomic gas) or (1.16) with a = 0 (diatomic gas) and its Riemanninitial data be given in (1.7). Then a vacuum occurs in the solution of the Riemann problemif ¯ r L < ¯ s R , where ¯ r L and ¯ s R are the 1-Riemann invariant and 3-Riemann invariant corresponding tothe left state u L and right state u R , respectively. In the opposite case, namely ¯ r L ≥ ¯ s R , the HE RIEMANN PROBLEM OF RELATIVISTIC EULER SYSTEM WITH SYNGE ENERGY 7
Riemann problem admit a unique solution. As in Figure 1, the p − v plane is divided intofour parts by the curves T p ( u L ) and T p ( u L ) . Figure 1.
Projected wave curves on the p − v plane.Before presenting the problematic of the Riemann problem in these cases, it is convenientto recall first the concept of entropy principle , main field , symmetrization and entropy growthacross the shock for a general hyperbolic system of balance laws which are essential to thefollowing analysis.2. Entropy principle, symmetric form and growth of entropy
The relativistic system (1.2) belongs to a general system of N balance laws for the field u ( x β ) ∈ R N : ∂ α F α ( u ) = f ( u ) , (2.1)where F α , ( α = 0 , , ,
3) and f are column vectors in R N representing densities-fluxes andproduction terms, respectively. Now, any theory of continuum needs to be compatible withthe entropy principle which requires that system (2.1) has a natural entropy-entropy fluxpair h α satisfying a supplementary balance law: ∂ α h α ( u ) = Σ( u ) , (2.2)where Σ is the entropy production term, which is nonnegative according to the second law ofthermodynamics. We also assume that h = h α ξ α is a convex function of the field u = F α ξ α ,where ξ α is a constant time-like congruence.2.1. Main field and symmetric form.
In [43], Ruggeri and Strumia observed that (2.1)and (2.2) form a overdetermined quasilinear hyperbolic system. Thus, in order for anysmooth solution of (2.1) to satisfy the entropy law (2.2), the equation (2.2) must be obtainedas a linear combination of the equations of system (2.1): there exists a vector u (cid:48) ( u ) ∈ R N such that ∂ α h α ( u ) − Σ( u ) ≡ u (cid:48) ( u ) · ( ∂ α F α ( u ) − f ( u )) . (2.3)Since (2.3) is an identity, by comparing the differential terms and production terms, onehas: d h α = u (cid:48) · d F α , Σ = u (cid:48) · f ≥ . (2.4) RUGGERI, XIAO, AND ZHAO
Next, introduce potentials h (cid:48) α defined as follows: h (cid:48) α = u (cid:48) · F α − h α . Then, it follows from (2.4) that dh (cid:48) α = d u (cid:48) · F α . Now, if one chooses u (cid:48) as a new field, one has F α = ∂h (cid:48) α ∂ u (cid:48) . (2.5)Then ∂ α F α = ∂ α (cid:18) ∂h (cid:48) α ∂ u (cid:48) (cid:19) = (cid:18) ∂ h (cid:48) α ∂ u (cid:48) ∂ u (cid:48) (cid:19) ∂ α u (cid:48) . (2.6)Combine (2.1) and (2.6) to rewrite the original system (2.1) in the form (cid:18) ∂ h (cid:48) α ∂ u (cid:48) ∂ u (cid:48) (cid:19) ∂ α u (cid:48) = f ( u (cid:48) ) . (2.7)Since h (cid:48) = h (cid:48) α ξ α is the Legendre transform of h = h α ξ α , it follows from (2.4) that dh = d ( h α ξ α ) = u (cid:48) · d u , ↔ u (cid:48) = ∂h∂ u , i.e. u (cid:48) is the dual field of (multiplying (2.5) by ξ α ) u = ∂h (cid:48) ∂ u (cid:48) . We observe that the map u (cid:48) ( u ) is globall invertible (see [43]). Then one concludes that theoriginal system (2.1) is expressed as the form (2.7) if we choose the field u (cid:48) . This is a veryspecial symmetric system according with the Friedrichs definition. In fact all matrices aresymmetric and (cid:18) ∂ h (cid:48) α ∂ u (cid:48) ∂ u (cid:48) (cid:19) ξ α = ∂ h (cid:48) ∂ u (cid:48) ∂ u (cid:48) is positive definite. This result given in [43] generalizes Boillat’s symmetrization result [44]in covariant formalism. This symmetrization holds only for the new field u (cid:48) . This is whythis field u (cid:48) was called the main field by Ruggeri and Strumia [43]. The system (2.7) is alsofrequently called as Godunov system , since Godunov was the first one who symmetrizes theEuler system for fluids and physical systems arising from a variational principle [45]. Theinterested reader can read a brief history on the symmetrization procedure for a systemcompatible with an entropy principle in Chapter 2 of [38].2.2.
Entropy growth across a shock wave.
Let Ω be a connected open set of V and S an hyper-surface cutting Ω into two open subsets Ω , Ω . Let φ ( x α ) = 0 be an equationof Ω referred to any coordinate frame: we shall identify Ω with a shock hyper-surface forthe field u . Then it is known that the Rankine-Hugoniot conditions must hold : φ α [[ F α ( u )]] = 0 , where the square bracket indicates the jump in Ω:[[ F α ( u )]] = F α ( u L ) − F α ( u R ) . HE RIEMANN PROBLEM OF RELATIVISTIC EULER SYSTEM WITH SYNGE ENERGY 9
Formally the Rankine-Hugoniot equations are obtained from the field equations (2.1) throughthe correspondence rule ∂ α → φ α [[ ]] , f → . But the previous rule does not hold when it is applied on the supplementary equation (2.2)since η = φ α [[ h α ]] , (2.8)is generally non vanishing. We can decompose φ α = − s ξ α + ζ α with ξ α and ζ α respectivelyconstant time-like and space-like congruences and s is a shock velocity. Therefore (2.8)becomes η = − s [[ h ]] + ζ α [[ h α ]] . Ruggeri and Strumia proved that if h is convex, then η is an increasing function of s andthe positive branch (admissible shocks) requires that the shock velocity s is greater thanthe corresponding characteristic velocity evaluated in the unperturbed equilibrium state: η > s > λ, with λ = λ ( u R ) . Consequences for the relativistic Euler system.
In the case of relativistic Eulerfluid (1.2), (1.1), the entropy law is (2.2) with h α = − ρSu α , Σ = 0 , (2.9)where S is the entropy density satisfying the Gibbs equation: T dS = dε − pρ dρ. (2.10)The system is symmetric hyperbolic in the main field [43] : u (cid:48) ≡ T (cid:0)(cid:0) g + c (cid:1) , − u β (cid:1) , (2.11)where g is the chemical potential g = ε + pρ − T S.
In the same paper [43] (see also [46]), it was proved that the convexity of entropy isequivalent that the maximum characteristic velocity in the rest frame satisfies the sub-luminal condition and the specific heat at constant pressure c p is positive: p e = ∂p∂e (cid:12)(cid:12)(cid:12)(cid:12) S < , c p = k B m + c V > , (2.12)where c V = dε/dT is the specific heat at constant volume. The two conditions in (2.12) areequivalent to the hyperbolicity and sub-luminal conditions:0 < p e < . (2.13)We will prove in the following that for relativistic Euler system with Synge energy, theinequalities (2.13) can be verified. We can conclude that the system of relativistic Euler issymmetric hyperbolic in the main field given by (2.11), the entropy is convex and it growsacross a shock wave. Relativistic rarefied monatomic gas
In this section, we analyze basic properties of the relativistic Euler system (1.4) withconstitutive equations (1.11), (1.12) and (3.2). Due to the complexity of the relativisticEuler system and the modified Bessel functions, the analysis of basic properties of (1.4)such as the strict hyperbolicity and genuine nonlinearity is far from trivial.First of all, from the property of the Bessel functions Appendix 1 (6.2), we can rewritethe constitutive equation (1.12) as the following form e = c ρ K ( γ ) K ( γ ) + 3 p = p (cid:18) γ K ( γ ) K ( γ ) + 3 (cid:19) . (3.1)And it is also convenient to write the expression of n as a function of γ and entropy density S (see e.g. [20, 36]): n = 4 πe m c h − e − SkB K ( γ ) γ e γ K γ ) K γ ) . (3.2)3.1. Characteristic velocities.
The relativistic Euler system (1.2) or (1.4) is a particularcase of a general system of conservation laws: ∂ t u + ∂ x F ( u ) = 0 , (3.3)with u ≡ c (cid:0) V , T , c T (cid:1) = (cid:18) ρc √ c − v , ( e + p ) vc − v , ec + pv c − v (cid:19) , F ( u ) ≡ (cid:0) V , T , c T (cid:1) = (cid:18) ρcv √ c − v , ( e + p ) v c − v + p, ( e + p ) c vc − v (cid:19) . (3.4)Eigenvalues of (3.3) are λ = ( e p ( p, S ) − c v − (cid:112) e p ( p, S ) c ( c − v ) e p ( p, S ) c − v ,λ = v,λ = ( e p ( p, S ) − c v + (cid:112) e p ( p, S ) c ( c − v ) e p ( p, S ) c − v , (3.5)with e p = 1 /p e . Eigenvectors r i , ( i = 1 , ,
3) corresponding to λ i are r = (cid:101) r (cid:32) , − √ e p ( c − v )( e + p ) c , (cid:33) ,r = (0 , , ,r = (cid:101) r (cid:32) , √ e p ( c − v )( e + p ) c , (cid:33) , (3.6)where (cid:101) r is an arbitrary scalar function to be determined later. Denote ¯ λ = λ/c as thecharacteristic velocities (3.5) in the unity of light velocity and let ˆ λ = ¯ λ | v =0 , i.e. thecharacteristic velocity in the unity of light velocity evaluated in the rest frame. Then, from HE RIEMANN PROBLEM OF RELATIVISTIC EULER SYSTEM WITH SYNGE ENERGY 11 (3.5) and reference [40], we haveˆ λ = 0 , ˆ λ , = ∓ √ e p = ∓√ p e = ∓ (cid:115) r + 1 − r (cid:48) γ ( r + 1)( r − r (cid:48) γ ) = ∓ (cid:114) c p c V pp + e , with r = ep and r (cid:48) = dr ( γ ) dγ . (3.7)3.2. Strict hyperbolicity.
This part is devoted to the proof of the strict hyperbolicity forthe system (3.3). In fact, we have the following proposition:
Proposition 3.1.
The system (3.3) with constitutive equations (1.11) , (3.1) and (3.2) isstrictly hyperbolic.Proof. First we prove that c V = dεdT > . (3.8)In fact, from (1.3) and (1.13), we have dγdT = − γT , ε = ρc (cid:20) K ( γ ) K ( γ ) + 3 γ − (cid:21) . (3.9)Then we use (3.9) and Appendix 3 (6.24) to have c V = ∂ε∂T = ρc ∂∂γ (cid:20) K ( γ ) K ( γ ) + 3 γ − (cid:21) ∂γ∂T , = − k B ρm (cid:34) γ (cid:18) K ( γ ) K ( γ ) (cid:19) + 3 γ K ( γ ) K ( γ ) − γ − (cid:35) > . Then the strict hyperbolicity of the system (3.3) with constitutive equations (1.11), (3.1)and (3.2) follow from (2.12) and (3.7). (cid:3)
Sub-luminal characteristic velocities.
We want to prove in this subsection thatcharacteristic velocities of the system (3.3) are sub-luminal.
Theorem 3.1.
For the system (3.3) with relations (1.11), (3.1) and (3.2), we have ¯ λ max = ¯ λ ( γ, v ) < , ˆ λ max = ¯ λ ( γ, < √ . (3.10) Proof.
From (1.11), (3.1) and (3.2), we obtain p = 4 πe m c h − e − SkB K ( γ ) γ e γ K γ ) K γ ) ,e = p (cid:18) γ K ( γ ) K ( γ ) + 3 (cid:19) . (3.11)We use Appendix 1 (6.2) and (6.3) to have ddγ (cid:18) K ( γ ) γ (cid:19) = − K ( γ ) + γK ( γ ) γ ,ddγ (cid:18) γ K ( γ ) K ( γ ) (cid:19) = γ (cid:18) K ( γ ) K ( γ ) (cid:19) + 4 K ( γ ) K ( γ ) − γ. (3.12)Applying (3.12) in (3.11) yields ∂ γ pp = γ (cid:18) K ( γ ) K ( γ ) (cid:19) + 3 K ( γ ) K ( γ ) − γ − γ , and e p = ep + p∂ γ p ddγ (cid:18) γ K ( γ ) K ( γ ) (cid:19) = 3 + γ K ( γ ) K ( γ ) + γ (cid:16) K ( γ ) K ( γ ) (cid:17) + 4 K ( γ ) K ( γ ) − γγ (cid:16) K ( γ ) K ( γ ) (cid:17) + 3 K ( γ ) K ( γ ) − γ − γ = 3 + γ (cid:16) K ( γ ) K ( γ ) (cid:17) + 4 γ (cid:16) K ( γ ) K ( γ ) (cid:17) − γ K ( γ ) K ( γ ) − γγ (cid:16) K ( γ ) K ( γ ) (cid:17) + 3 K ( γ ) K ( γ ) − γ − γ > , (3.13)by Appendix 3 (6.24) and (6.25). Then, from (3.5) and (3.13), we have¯ λ max − √ e p − v − √ e p c )( c − v ) e p c − v < . Then we obtain the first inequality in (3.10). Moreover, we use (3.13) to haveˆ λ = 1 √ e p = √ p e < √ . (cid:3) Figure 2.
The behavior of ˆ λ versus γ for a = − a = 0 (Diatomic gas). HE RIEMANN PROBLEM OF RELATIVISTIC EULER SYSTEM WITH SYNGE ENERGY 13
From (3.7), we can plot ˆ λ versus γ and we see also from Figure 2 (with a = −
1) that inthe ultra-relativistic limit ( γ → / √ γ .Therefore we proved that the inequality (2.13) can be verified and from the general resultsof [43] addressed in Section 2, the following theorem holds: Theorem 3.2.
The system (3.3) satisfies the entropy law (2.2) with (2.9) in space-timeform: ∂ t ( ρS Γ) + ∂ x ( ρS Γ v ) = 0 . (3.14) The system is symmetric hyperbolic with respect to the main field (2.11) : u (cid:48) ≡ T (cid:18) e + pρ − T S, Γ v, Γ c (cid:19) . The entropy density h = − ρS Γ is a convex function with respect to the field u given by (3.4) , and the entropy grows acrossthe shock. Genuine nonlinearity.
In this subsection, we show that λ = v is linearly degenerate,while λ and λ are genuinely nonlinear.Denote ∇ λ as the gradient of λ with respect to ( p, v, S ). Then λ = v is linearlydegenerate since ∇ λ · r = (0 , , · (0 , , T = 0 . (3.15)We now turn to show that λ is genuinely nonlinear. Since the genuine nonlinearity of λ can be done similarly, we omit the details. The eigenvalue λ satisfies ∇ λ = (cid:18) ∂λ ∂p , ∂λ ∂v , ∂λ ∂S (cid:19) = (cid:32) e pp c ( c − v )2 √ e p ( e p c − v ) ( v + √ e p c ) , ( e p − c ( v + √ e p c ) ( e p c − v ) ,e pS c ( c − v )2 √ e p ( e p c − v ) ( v + √ e p c ) (cid:19) . Then taking into account of (3.6) , we have ∇ λ · r = (cid:101) re pp c ( c − v )2 √ e p ( e p c − v ) ( √ e p c + v ) − (cid:101) r √ e p ( e p − c ( c − v )( e + p )( e p c − v ) ( √ e p c + v ) = (cid:101) r ( √ e p c + v ) c ( c − v )2( e + p ) √ e p ( e p c − v ) [( e + p ) e pp − e p ( e p − . (3.16) Proposition 3.2.
For any γ ∈ (0 , ∞ ) , it holds that ( e + p ) e pp − e p ( e p − < . (3.17) Then the eigenvalue λ is genuinely nonlinear and we can choose the function (cid:101) r properlysuch that: ∇ λ · r = 1 . (3.18)Since the proof of Proposition 3.2 is quite long, we put it in Appendix 4. Riemann invariants.
In this part, we solve Riemann invariants for each eigenvalue λ i , ( i = 1 , , λ is (0 , ,
1) in (3.15), it is trivial to find the two Riemanninvariants of λ are v and p. We now solve the Riemann invariants for λ . According to the definition, the correspondingRiemann invariant w satisfies0 =( w p , w v , w S ) · r =˜ r ( w p , w v , w S ) · (cid:32) , − √ e p ( c − v )( e + p ) c , (cid:33) =˜ r (cid:104) w p − √ e p ( c − v )( e + p ) c w v (cid:105) . (3.19)It is straightforward to see that S is one of the Riemann invariants satisfying (3.19). Wefurther solve (3.19) to find that the other Riemann invariant should be constant along thecurve determined by √ e p dp ( e + p ) c = − dvc − v . (3.20)Solving this differential equation to get the other Riemann invariant ln (cid:16) c + vc − v (cid:17) + (cid:82) √ e p dp ( e + p ) c .Thus the two Riemann invariants of λ are S and ¯ r = 12 ln (cid:16) c + vc − v (cid:17) + (cid:90) p √ e p dp ( e + p ) c . Similarly, the two Riemann invariants of λ are S and ¯ s = 12 ln (cid:16) c + vc − v (cid:17) − (cid:90) p √ e p dp ( e + p ) c . Structure of the shock curves.
In this part, we study the structure of the shockcurves. It is divided into four parts: the rigorous derivation of the Hugoniot curve; verifi-cation of the Lax entropy conditions; monotonicity of the entropy along the shock curves;monotonicity of the velocity along the shock curves.3.6.1.
Hugoniot curve.
We first rigorously derive the Hugoniot curve of the shock curves.
Proposition 3.3.
Denote ( n L , e L , p L , v L ) as the proper number density, proper energy den-sity, pressure and the velocity of the fluid in the left of the shock curve and ( n, e, p, v ) as thecorresponding variables at right. Then it holds that e + pn ( e + p L ) = e L + p L n L ( e L + p ) . (3.21) Remark 3.1.
It should be pointed out that (3.21) has been proved in [21] and [27], and wegive a different proof here.
HE RIEMANN PROBLEM OF RELATIVISTIC EULER SYSTEM WITH SYNGE ENERGY 15
Proof.
Let s be the shock speed. According to the Rankine-Hugoniot conditions, it holdsthat s (cid:16) cρ √ c − v − cρ L (cid:113) c − v L (cid:17) = cρv √ c − v − cρ L v L (cid:113) c − v L ,s (cid:104) ( e + p ) vc − v − ( e L + p L ) v L c − v L (cid:105) = ( e + p ) v c − v + p − ( e L + p L ) v L c − v L − p L ,s (cid:104) ( e + p ) v c − v + e − ( e L + p L ) v L c − v − e L (cid:105) =( e + p ) c vc − v − ( e L + p L ) c v L c − v L . (3.22)From (3.22), it is straightforward to get n (cid:104) p − p L √ c − v − ( e L + p L ) v L ( v L − v ) √ c − v ( c − v L ) (cid:105) = n L (cid:104) p − p L (cid:113) c − v L − ( e + p )( v L − v ) v (cid:113) c − v L ( c − v ) (cid:105) ,n (cid:104) ( p + e L ) v √ c − v − ( e L + p L ) v L ( c − vv L ) √ c − v ( c − v L ) (cid:105) = n L (cid:104) − ( e + p L ) v L (cid:113) c − v L + v ( e + p )( c − v L v ) (cid:113) c − v L ( c − v ) (cid:105) . Namely, n [( p − p L )( c − v L ) + ( e L + p L ) v L ( v − v L )] (cid:112) c − v = n L [( p − p L )( c − v ) + ( e + p ) v ( v − v L )] (cid:113) c − v L , (3.23) n [( p + e L ) v ( c − v L ) − ( e L + p L ) v L ( c − vv L )] (cid:112) c − v = n L [ − ( e + p L ) v L ( c − v ) + ( e + p ) v ( c − vv L )] (cid:113) c − v L . (3.24)Applying (3 . − v (3 .
23) and (3 . − v L (3 . n ( e L + p L )( v − v L )( c − vv L ) (cid:112) c − v = n L ( e + p L )( v − v L )( c − v ) (cid:113) c − v L ,n ( p + e L )( v − v L )( c − v L ) (cid:112) c − v = n L ( e + p )( v − v L )( c − vv L ) (cid:113) c − v L . (3.25)Note that ( v − v L )( c − vv L ) (cid:113) c − v L (cid:112) c − v < . (3.25) can be further simplified as n ( e L + p L )( c − vv L ) = n L ( e + p L ) (cid:112) c − v (cid:113) c − v L ,n ( p + e L ) (cid:113) c − v L (cid:112) c − v = n L ( e + p )( c − vv L ) . (3.26)We multiply (3 . by (3 . and divide the resulting equation by ( c − vv L ) (cid:113) c − v L √ c − v to have n ( e L + p L )( p + e L ) = n L ( e + p L )( e + p ) . Then (3.21) follows. (cid:3)
Lax entropy conditions.
In this subsection, we show that similar to the non-relativisticEuler system [6, 47], the Lax entropy conditions are satisfied globally for the shock curves.
Proposition 3.4.
The Lax entropy conditions hold wholly along the shock curves for therelativistic Euler system (1.4) with constitutive equations (1.11), (3.1) and (3.2). Namely,for a shock curve u = u ( (cid:15) ) , (cid:15) ≤ with shock speed s = s ( (cid:15) ) , one has λ ( (cid:15) ) < s ( (cid:15) ) < λ (0) , (cid:15) < . Proof.
For brevity, we only prove the inequality λ ( (cid:15) ) < s ( (cid:15) ) for the 1-shock curves since theremaining parts can be proven in a similar way. Our proof will be done by contradiction.Assume (cid:15) be the first point such that λ ( (cid:15) ) = s ( (cid:15) ) , (cid:15) <
0. Corresponding to our systemgiven in the form (3.3), the jump condition is s [[ u ]] = [[ F ( u )]]. We differentiate it withrespect to (cid:15) , and multiply the resulted system by the left eigenvector (cid:96) ( (cid:15) ) at (cid:15) = (cid:15) tohave s (cid:48) [[ u ]] + s u (cid:48) = dF u (cid:48) ,s (cid:48) (cid:96) · [[ u ]] = ( λ − s ) (cid:96) · u (cid:48) = 0 . (3.27)(3 . implies s (cid:48) ( (cid:15) ) = 0 or (cid:96) · [[ u ]] = 0. Now suppose s (cid:48) ( (cid:15) ) = 0. Then it follows from(3 . that u (cid:48) ( (cid:15) ) = r ( (cid:15) ), and( s − λ ) (cid:48) ( (cid:15) ) = −∇ λ ( (cid:15) ) · r ( (cid:15) ) = − . (3.28)Since ( s − λ ) (cid:48) (0) <
0, (3.28) implies that there exists some point (cid:15) ∈ ( − (cid:15) ,
0) such that λ ( (cid:15) ) = s ( (cid:15) ). This contradicts with our choice of (cid:15) . Therefore, our proof can be completedif we can show that (cid:96) · [[ u ]] (cid:54) = 0 . (3.29)Now we turn to proof of (3.29). For the system (3.3) with respect to ( n, v, p ), we can choosethe left eigenvector (cid:96) = ( (cid:96) , (cid:96) , (cid:96) ) corresponding to λ in (3.6): (cid:96) = ( e + p )( e − pe p ( p, S )) √ c − v ρ ,(cid:96) = (cid:20) ( e + p − pe p ) v + p (cid:113) e p ( p, S ) c (cid:21) c,(cid:96) = − (cid:20) ( e + p − pe p ) c + p (cid:113) e p ( p, S ) v (cid:21) HE RIEMANN PROBLEM OF RELATIVISTIC EULER SYSTEM WITH SYNGE ENERGY 17
We further choose a coordinate system in which v L = 0. Note that (cid:96) · [[ u ]] = ( e + p )( e − pe p ( p, S )) √ c − v ρ (cid:16) ρc √ c − v − ρ L (cid:17) + (cid:104) ( e + p − pe p ( p, S )) v + p (cid:113) e p ( p, S ) c (cid:105) ( e + p ) cvc − v − (cid:104) ( e + p − pe p ( p, S )) c + p (cid:113) e p ( p, S ) v (cid:105)(cid:104) ( e + p ) v c − v + e − e L (cid:105) =( e + p )( e − pe p ( p, S )) (cid:16) c − ρ L ρ (cid:112) c − v (cid:17) − ( e + p − pe p ( p, S ))( e − e L ) c + p (cid:113) e p ( p, S )[( e + p ) − ( e − e L )] v. (3.30)On the other hand, we let v L = 0 in (3 . and (3 . to have( e + p ) svc − v = ( e + p ) v c − v + p − p L ,s (cid:104) ( e + p ) v c − v + e − e L (cid:105) = ( e + p ) c vc − v . Furthermore, we can get s ( e − e L ) = ( e + p ) v ( c − v ) c − v − ( p − p L ) v = ( e + p L ) v. (3.31)And we let v L = 0 in (3 .
26) to have n L √ c − v nc = e L + p L e + p L , and v = ( p − p L )( e − e L )( p + e L )( p L + e ) c . (3.32)Note that s < v < v L = 0. We can use (3.31) and (3 .
32) to have p > p L , e > e L on the 1-shock curves . Then we combine (3.30) and (3.32) to obtain (cid:96) · [[ u ]] =( e − e L ) ce + p L (cid:104) ( e − pe p ( p, S ))( e + p ) − ( e + p − pe p ( p, S ))( e + p L ) (cid:105) + p ( p + e L ) (cid:113) e p ( p, S ) v < ( e − e L ) ce + p L (cid:104) − ( p − p L ) pe p ( p, S ) − ( e + p ) p L (cid:105) < . (cid:3) Entropy Growth across the shock waves.
Taking into account the results of Section2, and the proof that the characteristic velocities are sub-luminal, we conclude that:
Corollary 3.1.
For the system of relativistic Euler fluid with Synge energy, the entropygrows across the shock waves.
Taking into account the Gibbs equation (2.10), the constitutive equation (1.11) and theexpression of e given in (1.3), we have mk B dS = (cid:18) r (cid:48) ( γ ) − r ( γ ) γ (cid:19) dγ − dρρ , with r = ep = γ K ( γ ) K ( γ ) − . Then mk B S = γ K ( γ ) K ( γ ) − ln γ + ln K ( γ ) − ln ρ + const. (3.33)Every classical solution of the differential system (1.2), thanks to the Gibbs equation (2.10),also satisfies the supplementary entropy law (2.2) with (2.9), which is expressed as the form(3.14) in two dimensional space-time. Without loss of generality, we let v L = 0. Then theentropy production along the shock (2.8) becomes now η = ( − s + v ) ρ Γ S + sρ L S L > . And taking into account the first RH condition of equations (1.4), we have η = − sρ L ( S − S L ) > ∀ s > λ. (3.34)This is exact what we want to prove. Notice that for the 1-shock curves, s < S > S L ;while for the 3-shock curves, s > S < S L . Taking into account (3.33), we have a = -1 a = 0 η ^ s λλ a=-1a=0 Figure 3.
Growth of entropy across the shock ( a = −
1: Monatomic gas, a = 0: Diatomic gas). HE RIEMANN PROBLEM OF RELATIVISTIC EULER SYSTEM WITH SYNGE ENERGY 19 definitively ˆ η = mck B ρ L η = − ¯ s (cid:26) r ( γ ) − r ( γ L ) + ln (cid:18) K ( γ ) K ( γ L ) γ L γ ρ L ρ (cid:19)(cid:27) > . By numerical solution of the RH equations, we can plot ˆ η as a function of ¯ s = s/c . Thefigure is in perfect agreement with the theoretical results and we can see that ˆ η grows andis positive when s > λ . In this case, as in the classical Euler, the entropy growth conditionis equivalent to the Lax conditions.3.6.4. Monotonicity of the velocity.
In this subsection, we discuss the monotonicity of thevelocity along the shock curves.
Proposition 3.5.
For the relativistic Euler system (1.4) with constitutive equations (1.11),(3.1) and (3.2), dvdp < for the 1-shock curves, and dvdp > for the 3-shock curves.Proof. We only prove dvdp < dvdp > v L = 0 to have v = ( p − p L )( e − e L )( p + e L )( p L + e ) c . Differentiate the above equation with respect to p to have dv dp = 2 v dvdp = c ( e L + p L )( p + e L ) ( p L + e ) (cid:104) ( e − e L )( e + p L ) + ( p − p L )( p + e L ) dedp (cid:105) . Note the fact v < v L = 0. To show dvdp <
0, it is equivalent to derive( e − e L )( e + p L ) + ( p − p L )( p + e L ) dedp > . (3.35)On the other hand, we use (3.1) and (3.21) to have dedp = γ K ( γ ) K ( γ ) + 3 + (cid:104) γ (cid:16) K ( γ ) K ( γ ) (cid:17) + 4 K ( γ ) K ( γ ) − γ (cid:105) p dγdp ,e L + p L n L dpdγ = m c ddγ (cid:104)(cid:16) γ K ( γ ) K ( γ ) + 4 γ (cid:17)(cid:16) γ K ( γ ) K ( γ ) + 3 + p L p (cid:17)(cid:105) = m c (cid:110)(cid:104) γ (cid:16) K ( γ ) K ( γ ) (cid:17) + 2 γ K ( γ ) K ( γ ) − γ − γ (cid:105)(cid:16) γ K ( γ ) K ( γ ) + 3 + p L p (cid:17) + (cid:16) γ K ( γ ) K ( γ ) + 4 γ (cid:17)(cid:104) γ (cid:16) K ( γ ) K ( γ ) (cid:17) + 4 K ( γ ) K ( γ ) − γ − p L p dpdγ (cid:105)(cid:111) , (3.36)and e L + p L n L = m c p (cid:16) K ( γ ) γK ( γ ) + 4 γ (cid:17) e + p L e L + p (3.37) We combine (3.36) and (3.37) to have1 p dpdγ = (cid:104) γ (cid:16) K ( γ ) K ( γ ) (cid:17) + 13 γ (cid:16) K ( γ ) K ( γ ) (cid:17) + (14 − γ ) K ( γ ) K ( γ ) − γ − γ (cid:105) p (cid:16) γ K ( γ ) K ( γ ) + 4 (cid:17)(cid:16) e + p L e L + p p + p L (cid:17) + (cid:104) γ (cid:16) K ( γ ) K ( γ ) (cid:17) + 2 K ( γ ) K ( γ ) − γ − γ (cid:105) p L (cid:16) γ K ( γ ) K ( γ ) + 4 (cid:17)(cid:16) e + p L e L + p p + p L (cid:17) ,dedp = γ K ( γ ) K ( γ ) + 3 + B ( γ ) (cid:16) γ K ( γ ) K ( γ ) + 4 (cid:17)(cid:16) e + p L e L + p p + p L (cid:17)(cid:104) B ( γ ) − B ( γ ) (cid:105) p + B ( γ ) p L . (3.38)Here and below, we use the following notations: B ( γ ) =: γ (cid:16) K ( γ ) K ( γ ) (cid:17) + 7 γ (cid:16) K ( γ ) K ( γ ) (cid:17) + (8 − γ ) K ( γ ) K ( γ ) − γ − γ ,B ( γ ) =: γ (cid:18) K ( γ ) K ( γ ) (cid:19) + 2 K ( γ ) K ( γ ) − γ − γ ,B ( γ ) =: γ (cid:16) K ( γ ) K ( γ ) (cid:17) + 4 K ( γ ) K ( γ ) − γ. Note that from Appendix 3 Proposition 6.3, one has B ( γ ) = (cid:104) γ (cid:18) K ( γ ) K ( γ ) (cid:19) + 3 K ( γ ) K ( γ ) − γ − γ (cid:105)(cid:16) γ K ( γ ) K ( γ ) + 4 (cid:17) < ,B ( γ ) = (cid:104) γ (cid:18) K ( γ ) K ( γ ) (cid:19) + 3 K ( γ ) K ( γ ) − γ − γ (cid:105) − K ( γ ) K ( γ ) − γ < ,B ( γ ) = 1( K ( γ )) (cid:104) γK ( γ ) + 4 K ( γ ) (cid:16) K ( γ ) γ + K ( γ ) (cid:17) − γ (cid:16) K ( γ ) γ + K ( γ ) (cid:17) (cid:105) =1( K ( γ )) (cid:104)(cid:16) γ + 4 γ (cid:17) ( K ( γ )) − γ ( K ( γ )) (cid:105) > , (3.39)and B ( γ ) − B ( γ ) = γ (cid:16) K ( γ ) K ( γ ) (cid:17) + 6 γ (cid:16) K ( γ ) K ( γ ) (cid:17) + (6 − γ ) K ( γ ) K ( γ ) − γ − γ = γ (cid:18) K ( γ ) K ( γ ) (cid:19) + 4 γ (cid:18) K ( γ ) K ( γ ) (cid:19) − γ K ( γ ) K ( γ ) − γ +2 (cid:104) γ (cid:18) K ( γ ) K ( γ ) (cid:19) + 3 K ( γ ) K ( γ ) − γ − γ (cid:105) < ,B ( γ ) − B ( γ ) − B ( γ ) = (cid:104) γ (cid:18) K ( γ ) K ( γ ) (cid:19) + 3 K ( γ ) K ( γ ) − γ − γ (cid:105)(cid:16) γ K ( γ ) K ( γ ) + 2 (cid:17) < . (3.40) HE RIEMANN PROBLEM OF RELATIVISTIC EULER SYSTEM WITH SYNGE ENERGY 21
Now we use (3.35) and (3.38) to have( e − e L )( e + p L ) + ( p − p L )( p + e L ) × (cid:20) γ K ( γ ) K ( γ ) + 3 + ( γ K ( γ ) K ( γ ) + 4) B ( γ )( e + p L p + e L p + p L )[2 B ( γ ) − B ( γ )] p + B ( γ ) p L (cid:21) > . By (3.39) and (3.40), the equation above can be simplified as( p − p L )( p + e L ) (cid:110) e − e L p − p L e + p L e L + p B ( γ ) + B ( γ )+ (cid:16) γ K ( γ ) K ( γ ) + 3 (cid:17)(cid:16) B ( γ ) + B ( γ ) (cid:17)(cid:111) p L +( p − p L )( e + p L ) (cid:110)(cid:16) e − e L p − p L + e L + pe + p L ep (cid:17) [2 B ( γ ) − B ( γ )]+ (cid:16) γ K ( γ ) K ( γ ) + 4 (cid:17) B ( γ ) (cid:111) p < . (3.41)Before verifying (3.41), we first show that e − e L p − p L > . (3.42)In fact, we differentiate (3.33) with respect to p and use (3.34) to have mk B dSdp = − p − γ dγdp + (cid:104) K (cid:48) ( γ ) K ( γ ) + ddγ (cid:16) γ K ( γ ) K ( γ ) (cid:17)(cid:105) dγdp = − p + (cid:104) γ (cid:16) K ( γ ) K ( γ ) (cid:17) + 3 K ( γ ) K ( γ ) − γ − γ (cid:105) dγdp > . (3.43)Then we combine (3.38), (3.39), (3.40) and (3.43) to have B ( γ ) (cid:16) e + p L p + e L − (cid:17) p < [ B ( γ ) − B ( γ )] ( p − p L ) . That is, e − e L p − p L − > B ( γ ) − B ( γ ) B ( γ ) p + e L p > . This inequality implies (3.42). Now we use (3.40) and (3.42) to have e − e L p − p L e + p L e L + p B ( γ ) + B ( γ ) < B ( γ ) + B ( γ ) < ,e − e L p − p L + e L + pe + p L ep = ( ep − e L p L )( e + p )( p − p L )( e + p L ) p = e + pp (cid:16) e − e L ) − ( p − p L )] p L ( p − p L )( e + p L ) (cid:17) > e + pp = γ K ( γ ) K ( γ ) + 4 . Then (3.41) holds since 2 B ( γ ) − B ( γ ) + B ( γ ) = 2[ B ( γ ) − B ( γ )] + [ B ( γ ) + B ( γ )] < (cid:3) Monotonicity of the velocity on rarefaction curves.
In this subsection, we con-sider the monotonicity of the velocity on rarefaction curves.
Proposition 3.6.
For the relativistic Euler system (1.4) with constitutive equations (1.11),(3.1) and (3.2), dvdp < on the 1-rarefaction curves, and dvdp > on the 3-rarefaction curves.Proof. Here we also only prove the case for 1-rarefaction curves, the other case for 3-rarefaction curves can be proved similarly. From (3.20), we have dvdp = − (cid:112) ( c − v ) e p ( e + p ) c < . (cid:3) Relativistic Euler system for Diatomic gas
In this section, we analyze basic properties of the relativistic Euler system (1.4) in thecase of gas with internal structure (polyatomic gas). Contents in this section are almostalong the same line of Section 3. Therefore, we will present the corresponding results in abrief way.For the relativistic Euler system for polyatomic gas, the corresponding constitutive equa-tions are given in (1.16) and (3.1). We can rewrite them also in this equivalent form: p = k B nT = k B m ρT, (4.1) e = nmc A ( γ ) K ( γ ) (cid:90) ∞ (cid:104) K ( γ ∗ ) γ ∗ + K ( γ ∗ ) (cid:105) φ ( I ) d I , (4.2) n = 4 πm c h − A ( γ ) K ( γ ) γ e − SkB exp (cid:110) (cid:90) ∞ (cid:104) K ( γ ∗ ) γ ∗ + K ( γ ∗ ) (cid:105) φ ( I ) d I (cid:111) , (4.3)where A ( γ ) = γK ( γ ) (cid:90) ∞ K ( γ ∗ ) γ ∗ I a d I . Then for a = 0, we use Appendix 1 (6.3) to have A ( γ ) = γK ( γ ) (cid:90) ∞ K ( γ ∗ ) γ ∗ d I = γK ( γ ) mc γ K ( γ ) γ = mc K ( γ ) γK ( γ ) . In the rest part of this paper, we focus on the case a = 0 (diatomic gases). Namely, φ ( γ ) = 1and A ( γ ) = mc K ( γ ) γK ( γ ) (cid:90) ∞ (cid:104) K ( γ ∗ ) γ ∗ + K ( γ ∗ ) (cid:105) φ ( I ) d I = mc γ (cid:104) K ( γ ) γ + K ( γ ) (cid:105) . Then, in our case discussed below, the constitutive equations given in (4.1), (4.2) and (4.3)take the special form: p = k B nT, (4.4) e = nk B T (cid:18) γK ( γ ) K ( γ ) + 3 (cid:19) = p (cid:18) γK ( γ ) K ( γ ) + 3 (cid:19) , (4.5) n = 4 πe m c h − K ( γ ) γ e γK γ ) K γ ) e − SkB . (4.6) HE RIEMANN PROBLEM OF RELATIVISTIC EULER SYSTEM WITH SYNGE ENERGY 23
Proposition 4.1.
Under the relations (4.4)-(4.6), the system (3.3) is strictly hyperbolicand the corresponding characteristic velocities are sub-luminal.Proof.
Corresponding to (3.8) and (3 . , we use (4.5) to have ε = c (cid:20) K ( γ ) K ( γ ) + 3 γ − (cid:21) , and c V = dεdT = ρc ∂∂γ (cid:20) K ( γ ) K ( γ ) + 3 γ − (cid:21) dγdT = − k B ρm (cid:34) γ (cid:18) K ( γ ) K ( γ ) (cid:19) + γ K ( γ ) K ( γ ) − γ − (cid:35) > , since for γ ≤ γ (cid:18) K ( γ ) K ( γ ) (cid:19) + γ K ( γ ) K ( γ ) − γ − < γ − ≤ , (4.7)and for γ > γ (cid:18) K ( γ ) K ( γ ) (cid:19) + γ K ( γ ) K ( γ ) − γ − ≤ γ (cid:18) − γ + 38 γ (cid:19) + γ −
12 + 38 γ − γ − −
52 + 964 γ < , (4.8)by Appendix 2 (6.14). Then c p > e p > e p = γ K ( γ ) K ( γ ) + 3 + γ (cid:16) K ( γ ) K ( γ ) (cid:17) + 2 K ( γ ) K ( γ ) − γγ (cid:16) K ( γ ) K ( γ ) (cid:17) + K ( γ ) K ( γ ) − γ − γ =3 + γ (cid:16) K ( γ ) K ( γ ) (cid:17) + 2 γ (cid:16) K ( γ ) K ( γ ) (cid:17) − ( γ + 2) K ( γ ) K ( γ ) − γγ (cid:16) K ( γ ) K ( γ ) (cid:17) + K ( γ ) K ( γ ) − γ − γ > , by (4.7) (4.8) and Appendix 3 (6.28). (cid:3) The behavior of ˆ λ is plotted in Figure 1 ( a = 0). By the same arguments in Subsection3.4, we can also show that the second eigenvalue of the system (3.3) is linear degenerateunder the relations (4.4)-(4.6). Moreover, we have the following proposition. Proposition 4.2.
Under the relations (4.4)-(4.6), the inequality corresponding to (3.17)holds. The first and third eigenvalues of the system (3.3) are genuinely nonlinear.
Proof.
As in the proof of Proposition 3.2 in Appendix 4, we only need to show( e + p ) e pp − e p ( e p −
1) = e p ( − e p + 3) + ep (cid:16) e p − ep − (cid:17) + e + p∂ γ p ∂ γ (cid:16) p∂ γ p ddγ (cid:16) γ K ( γ ) K ( γ ) (cid:17)(cid:17) < − (cid:16) γ K ( γ ) K ( γ ) + 3 (cid:17) K ( γ ) K ( γ ) + γ γ (cid:16) K ( γ ) K ( γ ) (cid:17) + K ( γ ) K ( γ ) − γ − γ + (cid:16) γ K ( γ ) K ( γ ) + 4 (cid:17)(cid:34) γ γ (cid:16) K ( γ ) K ( γ ) (cid:17) + K ( γ ) K ( γ ) − γ − γ − (cid:16) K ( γ ) K ( γ ) + 4 γ (cid:17) × γ (cid:16) K ( γ ) K ( γ ) (cid:17) + 4 (cid:16) K ( γ ) K ( γ ) (cid:17) + (cid:16) γ − γ (cid:17) K ( γ ) K ( γ ) − γ (cid:16) γ (cid:16) K ( γ ) K ( γ ) (cid:17) + 3 K ( γ ) K ( γ ) − γ − γ (cid:17) (cid:35) = − (cid:16) γ K ( γ ) K ( γ ) + 4 (cid:17)(cid:16) K ( γ ) K ( γ ) + γ (cid:17)(cid:16) γ (cid:16) K ( γ ) K ( γ ) (cid:17) + K ( γ ) K ( γ ) − γ − γ (cid:17) × I ( γ ) (cid:105) < , (4.9)where I ( γ ) = γ (cid:104) − (cid:16) K ( γ ) K ( γ ) (cid:17) (cid:105) − (cid:16) K ( γ ) K ( γ ) (cid:17) − γ K ( γ ) K ( γ ) + 10 + 12 γ . We first use (4.7) and (4.8) to have γ (cid:18) K ( γ ) K ( γ ) (cid:19) + K ( γ ) K ( γ ) − γ − γ < γ (cid:18) K ( γ ) K ( γ ) (cid:19) + K ( γ ) K ( γ ) − γ − γ < . (4.10)Then we combine (4.9) and (4.10) to see that in order to prove (3.17), one only need toshow I ( γ ) > . (4.11)Now we come to prove (4.11). We first note the fact I ( γ ) ≥ − − γ + 12 γ . This inequality implies that (4.11) holds for γ ≤ γ , where γ = − √ > . > γ satisfies γ + 9 γ −
12 = 0 . HE RIEMANN PROBLEM OF RELATIVISTIC EULER SYSTEM WITH SYNGE ENERGY 25
Next we show that (3.17) holds for γ ∈ ( γ , √ I ( γ ) > ( γ − − (cid:16) − γ − γ (cid:17) − γ (cid:16) − γ − γ (cid:17) + 10 + 12 γ > − (cid:104) − γ − γ + ( γ − γ (cid:105) − γ + 9( γ − γ + 10 + 12 γ > − γ − γ + 3 + 9 γ − γ − γ = − − γ + 252 γ > , γ ∈ ( γ , √ . Finally, we prove (3.17) holds for γ ∈ ( √ , ∞ ). We use Appendix 2 (6.13) to obtain I ( γ ) ≥ γ (cid:16) γ − γ − γ (cid:17) (cid:16) − γ (cid:17) − (cid:16) − γ + 38 γ + 316 γ (cid:17) − γ (cid:16) − γ + 38 γ + 316 γ (cid:17) + 10 + 12 γ =14 (cid:16) − γ − γ (cid:17) (cid:16) − γ + 14 γ (cid:17) − (cid:16) − γ + 1 γ − γ + 964 γ + 9256 γ (cid:17) + 10 − γ + 332 γ − γ − γ = (cid:16) − γ + 116 γ (cid:17)(cid:16) − γ − γ + 916 γ + 964 γ (cid:17) − γ + 112 γ − γ − γ − γ − γ = (cid:16)
34 + − (cid:17) γ + (cid:16) − (cid:17) γ + (cid:16) − − − (cid:17) γ + (cid:16) − − (cid:17) γ + (cid:16) × − (cid:17) γ =498 γ − γ − γ − γ − γ > , for γ ∈ ( √ , ∞ ). (cid:3) Corresponding to the monotonicity of the velocity in Proposition 3.5, we have the fol-lowing proposition.
Proposition 4.3.
For the relativistic Euler system (1.4) with constitutive equations (4.4)-(4.6), dvdp < for the 1-shock curves, and dvdp > for the 3-shock curves. Proof.
As in Proposition 3.5, we choose proper coordinate system such that v L = 0. Similarto the derivation in Proposition 3.5, we can show that dvdp < e − e L )( e + p L ) + ( p − p L )( p + e L ) dedp > . and (4.12) dedp = γ K ( γ ) K ( γ ) + 3 + ¯ B ( γ ) (cid:16) γ K ( γ ) K ( γ ) + 4 (cid:17) (cid:16) e + p L e L + p p + p L (cid:17)(cid:2) B ( γ ) − ¯ B ( γ ) (cid:3) p + ¯ B ( γ ) p L . (4.13)Here and below, we use the following notations:¯ B ( γ ) =: γ (cid:16) K ( γ ) K ( γ ) (cid:17) + 5 γ (cid:16) K ( γ ) K ( γ ) (cid:17) − γ K ( γ ) K ( γ ) − γ − γ , ¯ B ( γ ) =: γ (cid:18) K ( γ ) K ( γ ) (cid:19) − γ − γ , ¯ B ( γ ) =: γ (cid:16) K ( γ ) K ( γ ) (cid:17) + 2 K ( γ ) K ( γ ) − γ. We use Appendix Proposition 6.1, 6.2, 6.4 to have¯ B ( γ ) = (cid:104) γ (cid:18) K ( γ ) K ( γ ) (cid:19) + K ( γ ) K ( γ ) − γ − γ (cid:105)(cid:16) γ K ( γ ) K ( γ ) + 4 (cid:17) < , ¯ B ( γ ) = γ (cid:16) K ( γ ) K ( γ ) (cid:17) + 2 K ( γ ) K ( γ ) − γ > , (4.14)and ¯ B ( γ ) − ¯ B ( γ ) = γ (cid:16) K ( γ ) K ( γ ) (cid:17) + 4 γ (cid:16) K ( γ ) K ( γ ) (cid:17) − γ K ( γ ) K ( γ ) − γ − γ = (cid:104) γ (cid:18) K ( γ ) K ( γ ) (cid:19) + 2 γ (cid:18) K ( γ ) K ( γ ) (cid:19) − ( γ + 2) K ( γ ) K ( γ ) − γ (cid:105) +2 (cid:104) γ (cid:18) K ( γ ) K ( γ ) (cid:19) + K ( γ ) K ( γ ) − γ − γ (cid:105) < , ¯ B ( γ ) − ¯ B ( γ ) − ¯ B ( γ ) = (cid:104) γ (cid:18) K ( γ ) K ( γ ) (cid:19) + K ( γ ) K ( γ ) − γ − γ (cid:105)(cid:16) γ K ( γ ) K ( γ ) + 2 (cid:17) < . (4.15)Now we use (4.12) and (4.13) to have( e − e L )( e + p L ) + ( p − p L )( p + e L ) × (cid:20) γ K ( γ ) K ( γ ) + 3 + ( γ K ( γ ) K ( γ ) + 4) ¯ B ( γ )( e + p L p + e L p + p L )[2 ¯ B ( γ ) − ¯ B ( γ )] p + ¯ B ( γ ) p L (cid:21) > . HE RIEMANN PROBLEM OF RELATIVISTIC EULER SYSTEM WITH SYNGE ENERGY 27
By (4.14) and (4.15), the equation above can be further simplified as( p − p L )( p + e L ) (cid:110) e − e L p − p L e + p L e L + p ¯ B ( γ ) + ¯ B ( γ )+ (cid:16) γ K ( γ ) K ( γ ) + 3 (cid:17)(cid:16) ¯ B ( γ ) + ¯ B ( γ ) (cid:17)(cid:111) p L +( p − p L )( e + p L ) (cid:110)(cid:16) e − e L p − p L + e L + pe + p L ep (cid:17) [2 ¯ B ( γ ) − ¯ B ( γ )]+ (cid:16) γ K ( γ ) K ( γ ) + 4 (cid:17) ¯ B ( γ ) (cid:111) p < . (4.16)By almost the same derivation of (3.35) e − e L p − p L > . Then we further use (4.14), (4.15) to have e − e L p − p L e + p L e L + p ¯ B ( γ ) + ¯ B ( γ ) < ¯ B ( γ ) + ¯ B ( γ ) < ,e − e L p − p L + e L + pe + p L ep = ( ep − e L p L )( e + p )( p − p L )( e + p L ) p = e + pp (cid:16) e − e L ) − ( p − p L )] p L ( p − p L )( e + p L ) (cid:17) >e + pp = γ K ( γ ) K ( γ ) + 4 . Then (4.16) holds since 2 ¯ B ( γ ) − ¯ B ( γ ) + ¯ B ( γ ) = 2[ ¯ B ( γ ) − ¯ B ( γ )] + [ ¯ B ( γ ) + ¯ B ( γ )] < (cid:3) The behavior of the production of entropy across the shock is given in Figure 3 with a = 0. By the same arguments as in Proposition 3.6 for the monatomic gas, we also havethe monotonicity of the velocity on rarefaction curves. Proposition 4.4.
For the relativistic Euler system (1.4) with constitutive equations (4.4)-(4.6), dvdp < on the 1-rarefaction curves, and dvdp > on the 3-rarefaction curves. Proof of the main theorem 1.1
In this section, we are devoted to the proof of the main Theorem 1.1. We first discussthe condition under which vacuum occurs. Then for the case that no vacuum occurs, basedon the analysis about the structure of shock curves and the monotonicity of the velocityon rarefaction curves in Section 3 and 4, we solve the Riemann problem of the relativisticEuler system (1.4) with constitutive equations given in (1.11), (3.1) and (3.2) or in (4.4),(4.5) and (4.6).5.1.
Vacuum condition.
To give a vacuum condition, we first define vacuum. We saythat vacuum occurs if e = 0 . In fact, we have obtained that the pressure p is monotonic along 1-curves and 3-curves,and is constant along 2-contact discontinuity waves. Note that e = p (cid:16) γ K ( γ ) K ( γ ) + 3 (cid:17) for monatomic gas and e = p (cid:16) γ K ( γ ) K ( γ ) + 3 (cid:17) for diatomic gas. It is easy to see that the vacuummay occur only when a 1-rarefaction wave interacts a 3-rarefaction wave.Denote (¯ r L , ¯ s L ) and (¯ r R , ¯ s R ) as the Riemann invariants at x < x >
0, respectively.¯ r L , ¯ s L , ¯ r R and ¯ s R can be written as¯ r L = 12 ln (cid:16) c + v L c − v L (cid:17) + (cid:90) p L √ e p dp ( e + p ) c =12 ln (cid:16) c + v L c − v L (cid:17) + (cid:90) e L de ( e + p ) √ e p c , ¯ s L = 12 ln (cid:16) c + v L c − v L (cid:17) − (cid:90) p L √ e p dp ( e + p ) c =12 ln (cid:16) c + v L c − v L (cid:17) − (cid:90) e L de ( e + p ) √ e p c , ¯ r R = 12 ln (cid:16) c + v R c − v R (cid:17) + (cid:90) p R √ e p dp ( e + p ) c =12 ln (cid:16) c + v R c − v R (cid:17) + (cid:90) e R de ( e + p ) √ e p c , ¯ s R = 12 ln (cid:16) c + v R c − v R (cid:17) − (cid:90) p R √ e p dp ( e + p ) c =12 ln (cid:16) c + v R c − v R (cid:17) − (cid:90) e R de ( e + p ) √ e p c . Note that ¯ r is a constant along a 1-rarefaction curve, velocity v and pressure p are con-stant along a 2-contact discontinuity wave, and ¯ s is a constant along a 3-rarefaction curve.Therefore, there exist states ( e , v , S ) and ( e , v , S ) with v = v , S = S L , S = S R suchthat ¯ r L − ¯ s R = 12 ln (cid:16) c + v L c − v L (cid:17) + (cid:90) e L de ( e + p ) √ e p c − (cid:104)
12 ln (cid:16) c + v R c − v R (cid:17) − (cid:90) e R de ( e + p ) √ e p c (cid:105) =12 ln (cid:16) c + v c − v (cid:17) + (cid:90) e de ( e + p ) √ e p c − (cid:104)
12 ln (cid:16) c + v c − v (cid:17) − (cid:90) e de ( e + p ) √ e p c (cid:105) = (cid:90) e de ( e + p ) √ e p c + (cid:90) e de ( e + p ) √ e p c . Then ¯ r L ≤ ¯ s R implies that (cid:90) e de ( e + p ) √ e p c + (cid:90) e de ( e + p ) √ e p c ≤ . Namely, e , e ≤ r L ≤ ¯ s R . HE RIEMANN PROBLEM OF RELATIVISTIC EULER SYSTEM WITH SYNGE ENERGY 29
Existence of solutions to the Riemann problem.
Finally, we discuss the solutionsto the Riemann problem for the case ¯ r L > ¯ s R . In Section 3 and Section 4, we have provedthat dvdp < T p ( u L ) and dvdp > T p ( u L ) for the monatomic gas case and diatomic gascase, respectively. Denote v = f ( p ; p L , v L , S L ) and v = f ( p ; p R , v R , S R )as the curves T p ( u L ) and the backward 3-curve, respectively. Then the p − v plane is dividedinto four parts by the curves T p ( u L ) and T p ( u L ). Moreover, if ( p L , v L ) is above the curve f ( p ; p R , v R , S R ), i.e., v L > f ( p L ; p R , v R , S R ) , the 1-curve of the Riemann problem should be a rarefaction curve; while if ( p L , v L ) is belowthe curve f ( p ; p R , v R , S R ), i.e., v L < f ( p L ; p R , v R , S R ) , the 1-curve of the Riemann problem should be a shock curve.Correspondingly, the 3-curve of the Riemann problem should be a rarefaction curve if( p R , v R ) is above the curve T p ( u L ), i.e., v R > f ( p R ; p L , v L , S L ) , and the 3-curve of the Riemann problem should be a rarefaction curve if ( p R , v R ) is belowthe curve T p ( u L ), i.e., v R < f ( p R ; p L , v L , S L ) . Since dvdp < T p ( u L ) and dvdp > p M , v M ) can beuniquely solved by v = f ( p ; p L , v L , S L ) , v = f ( p ; p R , v R , S R ) . Having obtained the 1-curve and 3-curve, we need to determine the 2-contact disconti-nuity wave to fully solve the Riemann problem. In fact, since the velocity v and pressure p are constants on the 2-contact discontinuity wave, we only need to obtain the entropy. Notethat the entropy is a constant along a rarefaction curve and is monotonic along a shockcurve. Then we can uniquely determine the entropy for the left state and right state of the2-contact discontinuity wave by the values of p on the 1-curve and 3-curve, respectively.6. Conclusions
In this paper, we consider the relativistic Euler equation in one space dimension. Theconstitutive equations for the closure of the differential system come from the relativisticBoltzmann-Chernikov equation that involves the Synge energy in the case of a monatomicgas and the generalized Synge energy in the case of a polyatomic gas. These constitutiveequations are more appropriate than the ones present in literatures to study the Riemannproblem. In fact, using the Synge equation we discover the constitutive equations of previouspapers listed in the introduction are valid only in the classical limit ( γ → ∞ ) or in the ultra-relativistic limit ( γ → γ small). This more physical case is mathematically difficult because the modified Besselfunctions of the second kind appear in the constitutive equations. Nevertheless, we are ableto prove rigorously the well-posedness of the Riemann problem at least for monatomic anddiatomic gases. For these kinds of gases, our results reduce to one of the previous studiesas limit cases of the classical regime or ultra-relativistic framework. Appendix 1: Modified Bessel functions and properties
In this part, we recall expressions of the modified Bessel functions and their basic prop-erties. Moreover, with our observation, a simple corollary is also presented. Now we givethe modified Bessel functions and collect their basic properties:
Lemma 6.1. [20, 48, 49] Let K j ( γ ) be the Bessel functions defined by K j ( γ ) = (2 j ) j !(2 j )! 1 γ j (cid:90) λ = ∞ λ = γ e − λ ( λ − γ ) j − / dλ, ( j ≥ . (6.1) Then the following identities hold: K j ( γ ) = j − ( j − j − γ j (cid:82) λ = ∞ λ = γ e − λ ( λ − γ ) j − / dλ, ( j > ,K j +1 ( γ ) = 2 j K j ( γ ) γ + K j − ( γ ) , ( j ≥ , (6.2) K j ( γ ) < K j +1 ( γ ) , ( j ≥ , and ddγ (cid:16) K j ( γ ) γ j (cid:17) = − K j +1 ( γ ) γ j , ( j ≥ , (6.3) K j ( γ ) = (cid:114) π γ e − γ (cid:32) γ j,n ( γ ) γ − n + n − (cid:88) m =0 A j,m γ − m (cid:33) , ( j ≥ , n ≥ , (6.4) where expressions of the coefficients in (6.4) are A j, = 1 A j,m = (4 j − j − ) · · · (4 j − (2 m − ) m !8 m , ( j ≥ , m ≥ , | γ j,n ( γ ) | ≤ e [ j − / γ − | A j,n | , ( j ≥ , n ≥ . (6.5) On the other hand, according to [49] (in Page 80), the Bessel functions defined in (6.1)can also be written in the following form: K ( γ ) = − ∞ (cid:88) m =0 ( γ ) m m ! m ! (cid:104) ln (cid:16) γ (cid:17) − ψ ( m + 1) (cid:105) ,K n ( γ ) = 12 n − (cid:88) m =0 ( − m ( n − m − m ! (cid:16) γ (cid:17) − n +2 m +( − n +1 ∞ (cid:88) m =0 ( γ ) n +2 m m !( m + n )! × (cid:104) ln (cid:16) γ (cid:17) − ψ ( n + m ) − ψ ( n + m + 1) (cid:105) , (6.6) ψ (1) = − C E , ψ ( m + 1) = − C E + m (cid:88) k =1 k , m ≥ ,K ( γ ) = 1 γ + ∞ (cid:88) m =0 ( γ ) m +1 m !( m + 1)! (cid:104) ln (cid:16) γ (cid:17) − ψ ( m + 1) − ψ ( m + 2) (cid:105) , where C E = 0 . . . . is the Euler’s constant. HE RIEMANN PROBLEM OF RELATIVISTIC EULER SYSTEM WITH SYNGE ENERGY 31
From Lemma 6.1, we immediately have the following corollary:
Corollary 6.1.
For K j ( j ≥ defined in Lemma 6.1, it holds that K ≤ K K , (6.7)3 (cid:18) K ( γ ) K ( γ ) (cid:19) + 6 γ K ( γ ) K ( γ ) − ≥ . (6.8) Proof.
From (6.1), we have K ( γ ) = (cid:82) λ = ∞ λ = γ e − λ ( λ − γ ) − / dλ,K ( γ ) = γ (cid:82) λ = ∞ λ = γ e − λ ( λ − γ ) / dλ,K ( γ ) = γ (cid:82) λ = ∞ λ = γ e − λ ( λ − γ ) / dλ. These equations imply (6.7) by H¨ o lder’s inequality.Taking j = 1 in (6.2) and inserting it to (6.7) yield K ( γ ) ≤ K ( γ ) (cid:18) γ K ( γ ) + K ( γ ) (cid:19) . Then (6.8) follows. (cid:3)
Appendix 2: Estimates of the ratio K ( γ ) K ( γ ) In this subsection, we concentrate on estimates of K ( γ ) K ( γ ) . Our estimates are divided intotwo cases according to different expressions of K m ( m ≥
1) given in Lemma 6.1: the case γ ∈ (0 , √
2] by (6.6) and the case γ ∈ [1 . , ∞ ) by (6.1). Here γ = 1 . . . . is a constantsatisfying ln (cid:16) γ (cid:17) + C E = 0 . We first estimate K ( γ ) K ( γ ) for the first case γ ∈ (0 , √
2] by using the expressions (6.6).
Proposition 6.1.
For γ ∈ [ γ , √ , it holds that K ( γ ) K ( γ ) ≤ − γ − γ . (6.9) And for γ ∈ (0 , γ ] , we have (cid:16) K ( γ ) K ( γ ) (cid:17) + γ K ( γ ) K ( γ ) − > . Moreover, we have γ (cid:112) γ + 1 + 1 ≤ K ( γ ) K ( γ ) ≤ γ (cid:20) − (cid:16) ln( γ C E (cid:17)(cid:21) . (6.10) Proof.
We first prove (6.9). From (6.6), we get for γ ∈ [ γ , √
2] that γ K ( γ ) K ( γ ) ≤ − [ln( γ ) + C E ] (cid:104) γ + γ + γ e γ (cid:105) + γ + − γ e γ [ln( γ ) + C E ] (cid:104) γ + γ + γ e γ × (cid:105) + 1 − γ − γ + − γ e γ × . Then, (6.9) holds if we can show[ln( γ C E ] (cid:104) γ + γ γ e γ (cid:105) + γ − γ e γ ≤ (cid:110) [ln( γ C E ] (cid:104) γ γ
16 + γ e γ × (cid:105) + 1 − γ − γ
64 + − γ e γ × (cid:111) × (1 + γ − γ ) . Namely, f ( γ ) :=1 + γ − γ −
14 (1 + γ − γ ) γ − γ
64 [5( γ − γ ) + 21] − (cid:104) γ − γ ) e γ + 27 e γ (cid:105) γ ×
36 + (cid:104) ln( γ C E (cid:105)(cid:110) γ γ − γ ) + γ
16 (5 + γ − γ )+ γ × (cid:104) (1 + γ − γ ) e γ + 6 e γ (cid:105)(cid:111) ≥ . (6.11)Note that for γ ∈ [ γ , √ f (cid:48) ( γ ) =1 − γ − γ γ − γ ) − γ
16 [5( γ − γ ) + 21] − γ − γ × (cid:104) γ − γ ) e γ + 27 e γ (cid:105) + γ γ − γ ) + γ
16 (5 + γ − γ )+ γ × (cid:104) (1 + γ − γ ) e γ + 6 e γ (cid:105) − γ × (cid:110) (cid:104) γ (1 + γ − γ )20 (cid:105) e γ + 27 γ e γ (cid:105)(cid:111) + (cid:104) ln( γ C E (cid:105)(cid:110) (3 + γ − γ ) γ + γ γ γ − γ ) + γ
16 + γ (cid:104) (1 + γ − γ ) e γ + 6 e γ (cid:105) + γ × (cid:104)(cid:16) γ (1 + γ − γ )20 (cid:17) e γ + 3 γ e γ (cid:105)(cid:111) . HE RIEMANN PROBLEM OF RELATIVISTIC EULER SYSTEM WITH SYNGE ENERGY 33
We can further obtain that for γ ∈ [ γ , √ f (cid:48)(cid:48) ( γ ) =4 − γ + γ + (2 γ − γ − γ − γ (cid:104) γ − γ ) e γ + 68 e γ (cid:105) − γ (cid:110)(cid:104) γ (1 + γ − γ )20 (cid:105) e γ + 27 γ e γ (cid:111) − γ × (cid:110)(cid:104) − γ + 3 γ γ (1 + γ − γ )80 (cid:105) e γ + (cid:16) γ (cid:17) e γ (cid:111) + (cid:104) ln( γ C E (cid:105)(cid:110) − γ + 3 γ ++ (15 − γ ) γ γ γ (cid:104) (1 + γ − γ ) e γ + 6 e γ (cid:105) + γ (cid:104)(cid:16) γ (1 + γ − γ )20 (cid:17) e γ + 3 γ e γ (cid:105) + γ × × (cid:110)(cid:104) − γ + 3 γ
20 + γ (1 + γ − γ )400 (cid:105) e γ + (cid:16)
37 + 3 γ (cid:17) e γ (cid:111)(cid:111) < . Then we have f ( γ ) ≥ min { f ( γ ) , f ( √ } for γ ∈ [ γ , √ f ( γ ) =1 − γ − γ − γ × (cid:104) e γ + 27 e γ (cid:105) > , and f ( √
2) =1 + √ − γ − √ − γ − √ − γ ) + 2116 − (cid:104) √ − γ )144 e + 27144 e (cid:105) + (cid:104) ln( √
22 ) + C E (cid:105) × (cid:110) √ − γ + 5 + √ − γ (cid:104) (1 + √ − γ ) e + 6 e (cid:105)(cid:111) > . Then (6.11) holds.Now we turn to the proof of (6.10). We first verify the left inequality of (6.10). For γ ∈ (0 , γ ] , we use (6.6) to have K ( γ ) K ( γ ) ≥ − (cid:2) ln( γ ) + C E (cid:3) γ − (cid:2) ln( γ ) + C E − (cid:3) γ (cid:2) ln( γ ) + C E − (cid:3) γ + (cid:2) ln( γ ) + C E − (cid:3) γ . (6.10) holds if we have f ( γ ) = − (cid:104) ln( γ C E (cid:105)(cid:104) (cid:112) γ + 1 + γ (3 + (cid:112) γ + 1)4 + γ (cid:105) + γ (2 + (cid:112) γ + 1)4 + 5 γ − > . In fact, for γ ∈ (0 , γ ), we have f (cid:48) ( γ ) = − γ (cid:104) (cid:112) γ + 1 + γ (3 + (cid:112) γ + 1)4 + γ (cid:105) + γ (cid:112) γ + 1 + γ (2 + (cid:112) γ + 1)2 + 5 γ − (cid:104) ln( γ C E (cid:105) × (cid:104) γ (cid:112) γ + 1 + γ (3 + (cid:112) γ + 1)2 + γ (cid:112) γ + 1 + γ (cid:105) < (1 + (cid:112) γ + 1) (cid:104) − γ + γ γ γ + 1) (cid:105) − γ (cid:104) ln( γ C E (cid:105) (3 + (cid:112) γ + 1) < − (cid:112) γ + 1)5 γ − γ (cid:104) ln( γ C E (cid:105) (3 + (cid:112) γ + 1) < . Then f ( γ ) ≥ f ( γ ) >
0. The left inequality of (6.10) holds.We finally treat the right inequality of (6.10). For γ ∈ (0 , γ ], we use (6.6) to have γ K ( γ ) K ( γ ) ≤ − (cid:2) ln( γ ) + C E (cid:3) γ − (cid:2) ln( γ ) + C E − (cid:3) γ e γ (cid:2) ln( γ ) + C E − (cid:3) γ e γ . To prove the right inequality of (6.10), we only need to derive the following inequality − (cid:104) ln( γ C E (cid:105) γ − (cid:104) ln( γ C E − (cid:105) γ e γ ≤ (cid:104) (cid:16) ln( γ C E − (cid:17) γ e γ (cid:105) γ (cid:20) − (cid:16) ln( γ C E (cid:17)(cid:21) . That is, ˜ f ( γ ) = γ (cid:110) − (cid:104)(cid:16) ln( γ C E (cid:17) − (cid:105) e γ + (cid:104) (cid:16) ln( γ C E (cid:17) − (cid:16) ln( γ C E (cid:17) + 1116 (cid:105) e γ (cid:111) − ≤ γ ∈ (0 , γ ]. Note the fact˜ f (cid:48) ( γ ) =2 γ (cid:110) − (cid:104)(cid:16) ln( γ C E (cid:17) − (cid:105) e γ + (cid:104) (cid:16) ln( γ C E (cid:17) − (cid:16) ln( γ C E (cid:17) + 1116 (cid:105) e γ (cid:111) − γe γ − γ e γ +4 (cid:16) ln( γ C E (cid:17) γe γ + γ (cid:104) − (cid:16) ln( γ C E (cid:17) + 1 (cid:105) e γ + 5 γ (cid:104) (cid:16) ln( γ C E (cid:17) − (cid:16) ln( γ C E (cid:17) + 1116 (cid:105) e γ ≥ γ (cid:110)(cid:16) γ (cid:17) e γ + (cid:104) − (cid:16) ln( γ C E (cid:17) + 55 γ − (cid:105) e γ (cid:111) > . HE RIEMANN PROBLEM OF RELATIVISTIC EULER SYSTEM WITH SYNGE ENERGY 35
Here we used the estimate − (cid:16) ln( γ ) + C E (cid:17) + γ ≥ for γ ∈ (0 , γ ]. Then we have˜ f ( γ ) ≤ γ (cid:16) e γ + 1116 e γ (cid:17) − < γ ∈ (0 , γ ]. (6.12) is verified. (cid:3) For later use, we also need two different estimates:
Proposition 6.2.
Let γ ∈ ( √ , ∞ ) . Then K ( γ ) K ( γ ) satisfies: − γ ≤ K ( γ ) K ( γ ) ≤ − γ + 38 γ + 316 γ . (6.13) Moreover, for γ ∈ (2 , ∞ ) , it holds that K ( γ ) K ( γ ) ≥ − γ + 38 γ − γ + 63128 γ − γ ,K ( γ ) K ( γ ) ≤ − γ + 38 γ − γ + 63128 γ + 78 γ . (6.14) Proof.
Compared to the proof of (6.13), the proof of (6.14) is more tedious but simpler. Forbrevity, we only prove (6.13). From (6.5), one has A , = − , A , = 92 × , A , = − × ,A , = 3 × × , A , = − × × , and (6.15) A , = 38 , A , = − × , A , = 1052 × ,A , = − × , A , = 21 × × . (6.16)Moreover, for γ > r , ≤ e − γ | A , | = 75 e − γ , r , ≤ e γ | A , | = 105 e γ , (6.17) r , ≤ e − γ | A , | = 75 × e − γ × ,r , ≤ e γ | A , | = 105 × e γ × , (6.18) r , = 2 e − γ | A , | ≤ × × × e − γ ,r , = 2 e γ | A , | ≤ × × × e γ . Firstly, we show that the inequality on the left side of (6.13) is true. We use (6.5), (6.15),(6.16) and (6.17) to have K ( γ ) K ( γ ) ≥ − γ + γ − γ e − γ γ − γ + γ e γ . Then, it suffice to show that1 − γ + 9128 γ − γ e − γ ≥ (cid:18) γ − γ + 1058 γ e γ (cid:19) (cid:18) − γ (cid:19) =1 − γ − (cid:18)
316 + 15128 (cid:19) γ + (cid:18) e γ + 15256 (cid:19) γ − × γ e γ . Namely, 38 γ + 1052 × γ e γ ≥ (cid:18) e − γ + 1058 e γ + 15256 (cid:19) γ ,γ − (cid:18) e − γ + 3564 e γ + 532 (cid:19) γ + 35128 e γ ≥ . (6.19)Denote f ( γ ) =: γ − (cid:16) e − γ + e γ + (cid:17) γ + e γ . It is easy to check that f (1 . > , f (cid:48) ( γ ) > γ ≥ . Then (6.19) holds for γ ∈ [1 . , ∞ ) ⊂ ( √ , ∞ ).We now continue to verify the inequality on the right side of (6.13). Similarly, from (6.5),(6.15), (6.16) and (6.18), we get K ( γ ) K ( γ ) ≤ − γ + γ − × γ + × e − γ × γ γ − γ + × γ − × e γ × γ . The proof can be completed if we can show the following inequality:1 − γ + 9128 γ − × γ + 75 × e − γ × γ ≤ (cid:16) γ − γ + 1052 × γ − × e γ × γ (cid:17) × (cid:16) − γ + 38 γ + 316 γ (cid:17) =1 − γ + 9128 γ + (cid:18)
316 + 964 + 15256 + 1051024 (cid:19) γ + (cid:18) − − × − × × e γ (cid:19) γ + (cid:18) − × × + 105 × e γ (cid:19) γ + (cid:18) × − × × e γ (cid:19) γ − × e γ γ . HE RIEMANN PROBLEM OF RELATIVISTIC EULER SYSTEM WITH SYNGE ENERGY 37
This inequality can be simplified as9 γ + (cid:32) − − × e γ + 75 × e − γ × (cid:33) γ + (cid:16) − + 105 × e γ × (cid:17) γ + (cid:16) × − × e γ × (cid:17) γ − × e γ × ≥ . (6.20)Denote the function on the left side of (6.20) as f ( γ ). It can be verified that f ( √ > , f (cid:48) ( γ ) > γ ∈ [1 , ∞ ) . Therefore, (6.20) holds for γ ∈ [ , ∞ ) (cid:3) Appendix 3: Essential estimates and solution of the conjectures in [36]
In this part, we present estimates essential to the analysis in the rest of our paper.The estimates are also closely related to the two conjectures in [36]. For convenience ofdiscussion, we first list these conjectures.
Conjecture 6.1.
The first conjecture of [36] reads: The map ( n, γ ) → ( H ( n, γ ) , P ( n, γ )) is auto-diffeomorphism of the region (0 , ∞ ) × (0 , ∞ ) , where the maps H and P are definedas follows: S = H ( n, γ ) = k B ln (cid:18) πe m c h − K ( γ ) nγ e γ K γ ) K γ ) (cid:19) ,p = P ( n, γ ) = nmc γ . With relations in (1.11)-(3.2), one can deduce the local resolvability of any one of thevariables n, T, S and p in terms of any two of the others whenever one knows that thenecessary derivatives are non-zero. In fact, as in the analysis of Lemma 3.5 in [36], thenegativity of ∂p∂γ (cid:12)(cid:12)(cid:12) S : ∂ γ | S pp = γ (cid:18) K ( γ ) K ( γ ) (cid:19) + 3 K ( γ ) K ( γ ) − γ − γ < , γ > ∂ X | Y to denote partial differentiation with respect to the variable X while Y is held constant.Authors in [36] also made another conjecture which is about the speed of sound in therelativistic setting, a stronger statement than Conjecture 6.1: Conjecture 6.2.
The second conjecture of [36] reads: Under the relations (1.11)-(3.2), pcan be written as a smooth, positive function of
E, S on the domain (0 , ∞ ) × (0 , ∞ ) , i.e.,the kinetic equation of state: p = p ( e, S ) is well-defined for all ( e, S ) ∈ (0 , ∞ ) × (0 , ∞ ) . Furthermore, on (0 , ∞ ) × (0 , ∞ ) , we havethat < ∂p∂e (cid:12)(cid:12)(cid:12) S ( e, S ) = ∂p ( e, S ) ∂e (cid:12)(cid:12)(cid:12) S < . (6.22)As is noted in [36], proving (6.22) is equivalent to proving the following inequality3 < ∂e∂p (cid:12)(cid:12)(cid:12) S ( p, S ) = 3 + γ K ( γ ) K ( γ ) + γ (cid:16) K ( γ ) K ( γ ) (cid:17) + 4 K ( γ ) K ( γ ) − γγ (cid:16) K ( γ ) K ( γ ) (cid:17) + 3 K ( γ ) K ( γ ) − γ − γ < ∞ , (6.23)since ∂p∂e (cid:12)(cid:12)(cid:12) S ( e, S ) = (cid:16) ∂e∂p (cid:12)(cid:12)(cid:12) S ( p, S ) (cid:17) − . The speed of sound, the square of which is defined to be ∂p∂e (cid:12)(cid:12)(cid:12) S ( e, S ), is a fundamental quantity in the relativistic Euler system. 0 < ∂p∂e (cid:12)(cid:12)(cid:12) S ( e, S ) < ∂p∂e (cid:12)(cid:12)(cid:12) S ( e, S ) plays a fundamental role in the well-posedness theory of the relativistic Euler system, see Remark 2.1 there. Remark 6.1.
For γ ∈ (0 , ] ∪ [70 , ∞ ) , the authors in [36] verified (6.21) and made Con-jectures 6.1 and 6.2 based on numerical observations in the remained region γ ∈ ( , .Later, Juan [50] gave a proof of the two conjectures for any range of γ . We will present twoestimates which implies the two conjectures since these estimates are basic in our analysis. In the following proposition, based on Lemma 6.1 and Corollary 6.1, we present estimatesmore accurate than (6.21) and (6.23).
Proposition 6.3.
Let γ ∈ (0 , ∞ ) and K j ( γ )( j ≥ be the functions defined in Lemma 6.1.Then it holds that γ (cid:18) K ( γ ) K ( γ ) (cid:19) + 3 γ K ( γ ) K ( γ ) − γ − < , (6.24) γ (cid:18) K ( γ ) K ( γ ) (cid:19) + 4 (cid:18) K ( γ ) K ( γ ) (cid:19) − γ K ( γ ) K ( γ ) − < . (6.25) Proof.
We first prove (6.24). Its proof is divided into two cases: γ ∈ (0 , √
2] and γ ∈ ( √ , ∞ ). Firstly, by (6.2), we can rewrite (6.24) as( γ + 3) (cid:18) K ( γ ) K ( γ ) (cid:19) + (cid:18) γ + 12 γ (cid:19) K ( γ ) K ( γ ) + 12 γ − γ − > . (6.26)Noting K ( γ ) , K ( γ ) > γ ∈ (0 , ∞ ) and12 γ − γ − > , γ ∈ (0 , √ , HE RIEMANN PROBLEM OF RELATIVISTIC EULER SYSTEM WITH SYNGE ENERGY 39 (6.24) holds when γ ∈ (0 , √ γ ∈ ( √ , ∞ ), we use (6.13) to have( γ + 3) (cid:18) K ( γ ) K ( γ ) (cid:19) + (cid:18) γ + 12 γ (cid:19) K ( γ ) K ( γ ) + 12 γ − γ − ≥ ( γ + 3) (cid:18) − γ (cid:19) + (cid:18) γ + 12 γ (cid:19) (cid:18) − γ (cid:19) + 12 γ − γ − γ − γ + 134 − γ + 34 γ + γ −
12 + 12 γ − γ − γ − γ − γ > . This yields (6.26). Then (6.24) follows.Now we turn to prove (6.25). The proof is also done in two cases, γ ∈ (0 , √
2] and γ ∈ ( √ , ∞ ), separately. We use (6.2) again to rewrite (6.25) as (cid:18) K ( γ ) K ( γ ) (cid:19) + (cid:18) γ + 6 γ (cid:19) (cid:18) K ( γ ) K ( γ ) (cid:19) + 12 γ K ( γ ) K ( γ ) − γ − γ + 8 γ > . (6.27)We first show that (6.27) is true when γ ∈ (0 , √ (cid:18) K ( γ ) K ( γ ) (cid:19) + (cid:18) γ + 6 γ (cid:19) (cid:18) K ( γ ) K ( γ ) (cid:19) + 12 γ K ( γ ) K ( γ ) − γ − γ + 8 γ = K ( γ ) K ( γ ) (cid:34)(cid:18) K ( γ ) K ( γ ) (cid:19) + 2 γ K ( γ ) K ( γ ) − (cid:35) + (cid:18) γ + 4 γ (cid:19) (cid:34)(cid:18) K ( γ ) K ( γ ) (cid:19) + 2 γ K ( γ ) K ( γ ) − (cid:35) +13 (cid:18) γ + 4 γ (cid:19) + (cid:16) − γ (cid:17) K ( γ ) K ( γ ) − γ − γ + 8 γ > (cid:18) γ − (cid:19) K ( γ ) K ( γ ) − γ − γ + 8 γ > , when γ ∈ (0 , √ γ ∈ (0 , √ γ − > , − γ − γ + 8 γ ≥ . When γ ∈ ( √ , ∞ ), similar to proof of (6.24), we use (6.13) to obtain (cid:16) K ( γ ) K ( γ ) (cid:17) + (cid:18) γ + 6 γ (cid:19) (cid:18) K ( γ ) K ( γ ) (cid:19) + 12 γ K ( γ ) K ( γ ) − γ − γ + 8 γ ≥ (cid:18) γ + 1 + 112 γ (cid:19) (cid:18) − γ + 14 γ (cid:19) + 12 γ + 2 γ − γ − γ = γ + 194 γ − γ + 118 γ + 12 γ + 2 γ − γ − γ =34 γ + 274 γ + 278 γ > . (cid:3) Remark 6.2.
In the proof of Proposition 6.3, instead of working on the ratio K ( γ ) K ( γ ) directly,we transformed the ratio K ( γ ) K ( γ ) into the ratio K ( γ ) K ( γ ) and divided our proof of (6.24) and (6.25)into two cases, γ ∈ (0 , √ and γ ∈ ( √ , ∞ ) . The motivations for this are as follows: fromthe expansion of K j ( γ ) in (6.4) and (6.5), we can see that it works well for γ which isa little larger than , and vice versa; the estimate of remained term | r j,n ( γ ) | seems moreaccurate when j is smaller due to the coefficient e [ j − / γ − in the estimate, which increasemore rapidly than the normal exponential function; when γ is small, we can make use ofthe simple inequality (6.8) from the observation (6.7). Remark 6.3.
Estimates (6.24) and (6.25) are more accurate than (6.21) and (6.22). ThenConjecture 6.1 and 6.2 are correct and the main results in [36] can be extended to includingthe whole case γ ∈ (0 , ∞ ) . The range for speed of sound (cid:114) ∂p∂e (cid:12)(cid:12)(cid:12) S ( E, S ) is (0 , √ ) for γ ∈ (0 , ∞ ) . Moreover, estimates (6.24) and (6.25) are of essential importance in this paper. Proposition 6.4.
Let γ ∈ (0 , ∞ ) and K j ( γ )( j ≥ be the functions defined in Lemma 6.1.Then it holds that γ (cid:18) K ( γ ) K ( γ ) (cid:19) + 2 γ (cid:18) K ( γ ) K ( γ ) (cid:19) − ( γ + 2) K ( γ ) K ( γ ) − γ < . (6.28) Proof.
For γ ≤ , it is straightforward to get (6.28) by the fact K ( γ ) K ( γ ) <
1. For the case γ >
2, we use (6.13) to have K ( γ ) K ( γ ) ≤ − γ + 12 γ , and γ (cid:18) K ( γ ) K ( γ ) (cid:19) + 2 γ (cid:18) K ( γ ) K ( γ ) (cid:19) − ( γ + 2) K ( γ ) K ( γ ) − γ ≤ γ K ( γ ) K ( γ ) (cid:104) γ (cid:16) − γ + 12 γ (cid:17) + 2 (cid:16) − γ + 12 γ (cid:17)(cid:105) − ( γ + 2) K ( γ ) K ( γ ) − γ ≤ γ K ( γ ) K ( γ ) (cid:104) γ (cid:16) − γ + 54 γ − γ + 14 γ (cid:17) + 2 − γ + 1 γ (cid:17)(cid:105) − ( γ + 2) K ( γ ) K ( γ ) − γ ≤ K ( γ ) K ( γ ) (cid:16) γ −
34 + 12 γ + 14 γ (cid:17) − γ < . (cid:3) Appendix 4: Proof of Proposition 3.2 for the genuine nonlinearity
Proof of Proposition 3.2:
We only need to prove (3.17). If this is done, (3.18) follows
HE RIEMANN PROBLEM OF RELATIVISTIC EULER SYSTEM WITH SYNGE ENERGY 41 immediately from (3.16) and (3.17). From (3.13), we can further obtain e pp = e p p − ep + 1 ∂ γ p ∂ γ (cid:18) p∂ γ p ddγ (cid:18) γ K ( γ ) K ( γ ) (cid:19)(cid:19) =1 p γ (cid:16) K ( γ ) K ( γ ) (cid:17) + 4 K ( γ ) K ( γ ) − γγ (cid:16) K ( γ ) K ( γ ) (cid:17) + 3 K ( γ ) K ( γ ) − γ − γ +1 p γ (cid:16) K ( γ ) K ( γ ) (cid:17) + 3 K ( γ ) K ( γ ) − γ − γ × (6.29) ddγ (cid:16) γ (cid:16) K ( γ ) K ( γ ) (cid:17) + 4 K ( γ ) K ( γ ) − γγ (cid:16) K ( γ ) K ( γ ) (cid:17) + 3 K ( γ ) K ( γ ) − γ − γ (cid:17) . Then we use (3.13) and (6.29) to have( e + p ) e pp − e p ( e p −
1) = e p ( − e p + 3) + ep (cid:16) e p − ep − (cid:17) + e + p∂ γ p ∂ γ (cid:16) p∂ γ p ddγ (cid:16) γ K ( γ ) K ( γ ) (cid:17)(cid:17) < − (cid:16) γ K ( γ ) K ( γ ) + 3 (cid:17) K ( γ ) K ( γ ) + γ γ (cid:16) K ( γ ) K ( γ ) (cid:17) + 3 K ( γ ) K ( γ ) − γ − γ + (cid:16) γ K ( γ ) K ( γ ) + 4 (cid:17)(cid:34) γ γ (cid:16) K ( γ ) K ( γ ) (cid:17) + 3 K ( γ ) K ( γ ) − γ − γ − (cid:16) K ( γ ) K ( γ ) + 4 γ (cid:17) × (cid:104) γ (cid:16) K ( γ ) K ( γ ) (cid:17) + 10 (cid:16) K ( γ ) K ( γ ) (cid:17) + (cid:16) γ − γ (cid:17) K ( γ ) K ( γ ) − γ (cid:105)(cid:16) γ (cid:16) K ( γ ) K ( γ ) (cid:17) + 3 K ( γ ) K ( γ ) − γ − γ (cid:17) (cid:35) = − (cid:16) γ K ( γ ) K ( γ ) + 4 (cid:17) (cid:16) K ( γ ) K ( γ ) + γ (cid:17)(cid:18) γ (cid:16) K ( γ ) K ( γ ) (cid:17) + 3 K ( γ ) K ( γ ) − γ − γ (cid:19) × I ( γ ) (cid:105) , (6.30)where I ( γ ) = γ (cid:16) K ( γ ) K ( γ ) (cid:17) + 4 γ (cid:16) K ( γ ) K ( γ ) (cid:17) − (2 γ + 9) (cid:18) K ( γ ) K ( γ ) (cid:19) − (cid:16) γ + 33 γ (cid:17) K ( γ ) K ( γ ) + γ + 12 + 12 γ . Noting γ (cid:18) K ( γ ) K ( γ ) (cid:19) + 3 K ( γ ) K ( γ ) − γ − γ < , in order to show (3.17), one suffices to prove I ( γ ) > , (6.31) in (6.30). By using K ( γ ) = γ K ( γ ) + K ( γ ), we can rewrite (6.31) as I ( γ ) = (cid:16) γ + 12 + 12 γ (cid:17)(cid:16) K ( γ ) K ( γ ) (cid:17) + (cid:16) γ + 63 γ + 96 γ (cid:17)(cid:16) K ( γ ) K ( γ ) (cid:17) + (cid:16) − γ − γ + 288 γ (cid:17)(cid:16) K ( γ ) K ( γ ) (cid:17) ++ (cid:16) − γ − γ − γ + 384 γ (cid:17) K ( γ ) K ( γ ) + γ − γ − γ + 192 γ > . (6.32)Now we come to prove (6.32). It is easy to find that (6.32) holds for γ ∈ (0 , r ].Now we turn to show that (6.32) holds for γ ∈ ( r , ∞ ). Rewrite I ( γ ) as I ( γ ) = (cid:16) K ( γ ) K ( γ ) − γ (cid:17)(cid:104)(cid:16) γ + 12 + 12 γ (cid:17)(cid:16) K ( γ ) K ( γ ) (cid:17) + (cid:16) γ + 7 γ γ + 12 γ + 90 γ (cid:17)(cid:16) K ( γ ) K ( γ ) (cid:17) + (cid:16) − γ + 3 γ + 54 + 51 γ + 1472 γ + 84 γ + 243 γ (cid:17) K ( γ ) K ( γ ) − γ − γ − − γ + 48 γ + 1414 γ + 201 γ + 5252 γ (cid:105) − γ − γ + 454 γ + 8918 γ + 162 γ + 2434 γ = (cid:16) K ( γ ) K ( γ ) − γ (cid:17)(cid:110)(cid:16) K ( γ ) K ( γ ) − γ (cid:17) × (cid:104)(cid:16) γ + 12 + 12 γ (cid:17)(cid:16) K ( γ ) K ( γ ) (cid:17) + (cid:16) γ + 3 γ + 24 + 51 γ + 24 γ + 84 γ (cid:17) K ( γ ) K ( γ ) + γ + 5 γ + 954 + 90 γ + 72 γ + 156 γ + 201 γ (cid:105) +4 γ + 21 + 1532 γ + 75 γ + 6214 γ + 324 γ + 162 γ (cid:111) − γ − γ + 454 γ + 8918 γ + 162 γ + 2434 γ Note that − γ − γ + 454 γ + 8818 γ + 162 γ + 2434 γ > , for γ ≤ . Then (6.32) holds for γ ∈ (0 , γ >
4. For the case γ >
4, we use (6.14) to have K ( γ ) K ( γ ) ≥ − γ + 38 γ − γ . HE RIEMANN PROBLEM OF RELATIVISTIC EULER SYSTEM WITH SYNGE ENERGY 43
Moreover, we have (cid:16) γ + 21 + 1532 γ (cid:17)(cid:16) γ − γ (cid:17) − γ − γ +454 γ + 8918 γ + 162 γ + 2434 γ = (cid:16) (cid:17) γ + 132316 γ + 162 γ + 2434 γ > γ > Acknowledgments:
The work of T. Ruggeri was supported by GNFM (INdAM), the work of Q.H. Xiao was supported by grants from Youth Innovation Promotion Association and the NationalNatural Science Foundation of China under contract 11871469 and the work of H. J. Zhao wassupported in part by the grants from National Natural Science Foundation of China under contracts11671309 and 11731008.
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Commun. Pure Appl. Anal. (3), 1341-1347, 2013.(T. Ruggeri) Department of Mathematics and Alma Mater Research Center on Applied Mathematics AM ,University of Bologna, Bologna, Italy E-mail address : [email protected] (Q.H. Xiao) Wuhan Institute of Physics and Mathematics, Chinese Academy of Sciences, Wuhan 430071,China
E-mail address : [email protected] (H.-J. Zhao) School of Mathematics and Statistics, Wuhan University, Wuhan 430072, China
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