Dipartimento di Matematica

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Dipartimento di Matematica

F. Conforto, A. Jannelli, R. Monaco, T. Ruggeri

ON THE RIEMANN PROBLEM FOR A SYSTEM OF BALANCE LAWS MODELLING A REACTIVE GAS MIXTURE

Rapporto interno N. 23, settembre 2005

Politecnico di Torino

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On the Riemann problem for a system of balance laws modelling a reactive gas mixture

F. Conforto^1∗, A. Jannelli¹, R. Monaco², T. Ruggeri³

1 Dept. of Mathematics, University of Messina, Italy

2 Dept. of Mathematics, Technical University of Turin, Italy

3 Dept. of Mathematics and C.I.R.A.M., University of Bologna, Italy

Abstract

This paper deals with the Riemann problem for a gas mixture undergoing reversible and irreversible bimolecular reactions governed by a suitable closure at Euler level of the Boltzmann equation, which results to be a set of balance laws. The aim of the paper is to investigate questions like compatibility with entropy principle and dissipative character of the system in order to better understand the effects of the source terms on the structure of the solution of the Riemann problem and its relation with the corresponding solution in absence of chemical reactions, which is here deduced explicitly. Numerical simulations show the space-time evolution of the solution profiles for the proposed reactive system.

PACS: 05.20.Dd, 47.70.Fw, 47.40.-x

Keywords: hyperbolic systems, balance laws, Riemann problem, reactive mixtures

1 Introduction

In a recent study [1], the propagation of steady detonation waves driven by a bimolecular reaction of type A1 + A2 A3 + A4 has been studied.

The authors started from the full Boltzmann equation, extended to a gas mixture undergoing the above reaction, deriving the reactive hydrodynamic

∗Corresponding author: Fiammetta Conforto; Dipartimento di Matematica, Universit`a di Messina, Contrada Papardo, Salita Sperone 31, 98166 Messina, Italy; phone: +39 090 6765063; fax: +39 090 393502; e-mail: [email protected]

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equations at the Euler level and determining an exact expression of the reaction rate. By using these equations the steady detonation problem was formulated and solved, by means of an appropriate numerical scheme. A natural continuation of investigation on detonation waves, in authors’ opin- ion, consists in the study of the onset of the detonation itself. This kind of study may be approached by formulating a Riemann problem for the afore- mentioned reactive Euler equations in the sense that, at the initial time, in the space domain of the gas mixture a jump discontinuity occurs and its consequent relevant motion is then driven by the chemical processes governing the hydrodynamic equations. Moreover, it seems natural to compare the time-space evolution of the Riemann problem for the reactive hydrodynamic equations, with the one obtained by using the Euler equations for the same gas mixture in the non-reactive case. From a mathematical point of view, this corresponds to compare the time-space evolution of the same hyperbolic system which, in the reactive case, presents balance laws for densities, while, in the chemically inert one, is a set of conservation equations. As well-known [2]-[5], the Riemann problem for a system of conservation laws, under the assumption of a not too large initial jump, admits an unique solution consisting in a superposition of shocks, contact discontinuities, rarefaction waves and constant states. In Sect.7 the Riemann problem for the inert mixture is faced and solved to conclude that in absence of chemical reactions the mixture, described at Euler level, behaves like a perfect gas since the characteristic velocities of the system are the classical ones of Eu- ler equations. Thus, for what concerns the non-reactive case, the solution is formed by four constant states connected by a contact discontinuity and different combinations of shocks and rarefactions.

Conversely, the same problem in presence of source terms cannot be analytically solved since rarefaction waves and constant states are not compatible solutions of the problem. What one should expect in this case is that the solution is, in some sense, a ‘deformation’ of the solution in the inert case. Therefore, after presenting the model in Sect.2 and sketching its thermodynamic properties in Sect.3, Sect.4-6 are devoted to the symmetrization of the system in order to investigate its dissipative character [6]-[17]. In fact, thanks to some recent results on existence of smooth solution for dissipative quasi-linear hyperbolic systems compatible with an entropy principle of hyperbolic-parabolic type [18]-[22], it is possible to get information on the effect of source terms on the solution structure and on its long time behavior. The aim of this paper consists in pointing out the relation between the solution of the Riemann problem for the reactive mixture with the ones for the inert case and for a perfect gas (Sect.7), and in perform-

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ing numerical simulations to show the main features due to the presence of chemical reactions. Moreover, in order to better appreciate the influence of chemical processes on the space-time evolution of the initial jump, the simulations have been carried on by considering both reversible and exothermic irreversible reactions.

On the numerical ground, discussed at the end of Sect.7, the simulations have been performed by using the Strang splitting method [23], with the hydrodynamic part solved by the Nessyahu and Tadmor scheme [24].

2 Governing equations

In [1] a mixture of four gases, undergoing a chemical reaction A1+ A2 A3+ A4, has been treated starting from the Boltzmann equation, and the hydrodynamic equations have been derived by performing a closure at Euler level. In 1-D, these equations read as

∂n_i

∂t + ∂

∂x(n_iv) = R_i, i = 1, 2, 3, 4 (1)

∂

∂t(ρv) + ∂

∂x

¡ρv²+ p¢

= 0 , (2)

∂

∂t

¡ρe^tot¢ + ∂

∂x

¡ρe^totv + pv¢

= 0 , (3)

where the total energy is given by the sum of kinetic, thermal and chemical ones:

ρe^tot= 1

2ρv²+3

2p + Ech= 1

2ρv²+3 2nkT +

X4 i=1

E_in_i, (4) and pressure p is defined by the state equation of perfect gases:

p = nkT . (5)

Here ni, i = 1, 2, 3, 4, are the number densities of the four species, v is the mean velocity, T the temperature. Moreover, n = P4

i=1n_i and ρ = P4

i=1m_in_i are the total number and mass densities, respectively, being m_i the molecular masses, Ei are the constant internal chemical bond energies and k is the Boltzmann constant. Because of mass conservation through the chemical reaction, the relation m1+ m2 = m3+ m4 must hold and the four species are ordered in such a way that the activation energy of the reaction E = E3 + E4− E¹− E² is a positive constant, meaning that the reaction A3+ A4 → A¹+ A2 is exothermic.

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In Eqs. (1), R_i = σ_iR, σ_i = 1 if i = 1, 2, σ_i = −1 if i = 3, 4, are the reaction rates and R is given by [1], [25]

R (n1, . . . , n4, T ) = 4πβµ34

kT

"

α1n3n4− α²n1n2

µm3m4

m1m2

¶³/2

exp µ

− E kT

¶#

× (χ

π

s2πkT µ34

exp

·

−µ34χ² 2kT

¸ +

µkT µ34 − χ²

¶ · 1 − erf

µrµ34

2kTχ

¶¸)

(6) Here, β is the Arrhenius parameter, χ the exothermic threshold velocity and µ34= m3m4/ (m3+ m4). By choosing α1 = α2 = 1, eq. (6) is the reaction rate of the reversible reaction A1 + A2 A3 + A4; the particular cases α1 = 1, α2 = 0 and α1 = 0, α2 = 1 correspond to exothermic irreversible (A1 + A2 ← A³+ A4) and endothermic irreversible (A1+ A2 → A³+ A4) reactions, respectively. Moreover, χ = 0 in endothermic reactions.

Eqs. (1) can be re-arranged in such a way to involve the conservation equation of total mass and, thus, system (1)-(3) is re-written in the following form:

∂ρ

∂t + ∂

∂x(ρv) = 0, (7)

∂

∂t(ρv) + ∂

∂x(ρv²+ p) = 0, (8)

∂

∂t

¡ρe^tot¢ + ∂

∂x

¡ρe^totv + pv¢

= 0, (9)

∂ρj

∂t + ∂

∂x(ρjv) = mjRj, j = 1, 2, 3. (10) where ρ_i = m_in_i are the mass densities of each species.

Let us note that eqs. (1)-(3) are the so-called reactive Euler equations.

At equilibrium, meaning both mechanical and chemical equilibrium, eqs.

(1)-(3), as well as (7)-(10), reduce to Euler equations for a inert gas mixture.

Finally, it’s worth noticing that the system is hyperbolic and admits the classical eigenvalues of Euler equations, with the contact one of multiplicity four.

3 Thermodynamic features

Thermodynamic properties of multi-component reactive mixtures of gases have been widely investigated by Giovangigli and Massot in [26]. By following the outline in [26], let’s observe that, from the state equation (5), the

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total pressure p can be written as:

p = X4

i=1

p_i= kT X4

i=1

n_i = T X4 i=1

k_iρ_i, (11)

where we set k_i = k/m_i and

p_i = n_ikT = k_iρ_iT

is the pressure of the i−th species. Moreover, the internal energy density e takes into account both thermal and chemical contributions:

ρe = X4

i=1

ρ_ie_i = X4 i=1

µ3

2kT n_i+ E_in_i

¶

= X4 i=1

ρ_i µ3

2k_iT + eE_i

¶

, (12)

ei being the internal energy density of the i−th species given by:

e_i = 3

2k_iT + eE_i, (13)

where eEi = Ei/mi. The enthalpy per unit mass of the i−th species, hⁱ, is h_i = e_i+p_i

ρ_i = 5

2k_iT + eE_i. (14)

Finally, it is useful to introduce the chemical potential µ_i of the i−th species µi= ei+pi

ρ_i − sⁱT = hi− sⁱT = 5

2kiT + eEi− sⁱT . (15) From the kinetic theory [27], the expression of the entropy density of each species, si, can be obtained (unless for an additive constant):

s_i = 3

2k_iln (kT ) − kiln n_i

m³_i^/2, (16)

and the entropy density of the reacting mixture, S, is given by:

ρS = X4 i=1

ρ_is_i = 3

2kn ln (kT ) − k X4

i=1

n_iln n_i

m³_i^/2. (17) In what follows, we choose as field vector

u= (ρ, v, T, ρ1, ρ2, ρ3)^T

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and in terms of the field variables we have:

ρ4 = ρ − ρ¹− ρ²− ρ³, n4 = 1

m4 (ρ − ρ¹− ρ²− ρ³) (18) n =

X4 i=1

ni= 1 m4

ρ + X3 j=1

µ 1 mj − 1

m4

¶

ρj (19)

p = nkT =



k4ρ + X3 j=1

bkjρj



 T (20)

Ech = X4 i=1

E_in_i = eE4ρ + X3 j=1

Eb_jρ_j (21)

where bk_j = k_j − k⁴ and bE_j = eE_j − eE4, for j = 1, 2, 3.

4 Entropy principle and symmetrization

In [27], at kinetic level, an H-theorem has been proved:

∂_th⁰+ ∂_xh¹= Σ ≤ 0 (22) with

h⁰ = −ρS , h¹= −ρSv . (23)

At macroscopic level, (22) represents an entropy principle [10] and it is well-known that a hyperbolic system of balance laws

∂_tF⁰(u) + ∂_xF¹(u) = F(u) (24) compatible with (22), h⁰ being a convex function of the densities u ≡ F⁰, can be put into symmetric form [6]-[9]. In particular in [8], [9] it has been proven that the original system assumes the special symmetric form

∂²h⁰⁰

∂u⁰∂u⁰∂_tu⁰+ ∂²h⁰¹

∂u⁰∂u⁰∂_xu⁰ = F(u⁰) (25) once the so-called main field, u⁰,

dh⁰ = u⁰· dF⁰, dh¹ = u⁰· dF¹, Σ = u⁰· F ≤ 0 ,

is chosen. The potentials h⁰⁰ and h⁰¹ are related to h⁰ and h¹, respectively, by:

h⁰⁰= u⁰· F⁰− h⁰, h⁰¹= u⁰· F¹− h¹,

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h⁰⁰ is the Legendre transform of h⁰ and u⁰ is the dual field of u:

u⁰ = ∂h⁰

∂u , u= ∂h⁰⁰

∂u⁰ . In the present case (see also [26]),

h⁰⁰= nk = p

T , h⁰¹= nkv = pv T and

u⁰ =¡

Λ^ρ, Λ^v, Λ^T, Λ¹, Λ², Λ³¢

(26) being

Λ^ρ= 1 T

µ

µ4−v² 2

¶

(27) Λ^v = v

T (28)

Λ^T = −1

T (29)

Λ^j = µb_j

T j = 1, 2, 3 (30)

with b

µ_j = µ_j− µ⁴= 5

2bkjT + bE_j− bsjT , bsj = s_j − s⁴, j = 1, 2, 3 . Moreover

Σ = −R T

( kT ln

"

n3n4

n1n2

µm1m2

m3m4

¶³/2# + E

)

. (31)

The sign of the entropy production −Σ has been checked to conclude that it is positive in the cases of reversible and exothermic irreversible reactions, in both of which temperature grows and chemical energy decreases; conversely, entropy production is negative for an endothermic irreversible reaction. In a reversible reaction, in spite of the presence of endothermic reactions, the exothermic character is dominant.

The transformation u⁰ ≡ u⁰(u) can be easily inverted (by convexity arguments), yielding in our case:

ρ_i = m_i µ

−km_i Λ^T

¶³₂ exp

µ

−s_i k_i

¶

, i = 1, 2, 3, 4 , (32)

ρ = X4

i=1

ρi, v = −Λ^v

Λ^T , T = − 1

Λ^T , (33)

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where, in the main field variables, entropy densities s_i are given by s_j = −Λ^T

"

h_j −1 2

µΛ^v Λ^T

¶2

+ Λ^ρ Λ^T + Λ^j

Λ^T

#

, j = 1, 2, 3 , (34)

s4 = −Λ^T

"

h4−1 2

µΛ^v Λ^T

¶2

+ Λ^ρ Λ^T

#

, (35)

and enthalpies h_i= −5ki/¡ 2Λ^T¢

+ eE_i, i = 1, 2, 3, 4.

5 Equilibrium manifold and equilibrium subsys- tem

Within the context of the theory of hyperbolic principal subsystems [12]-[14]

the fact that the governing system exhibits both conservation laws, eqs.(7)- (9) for total mass, momentum and energy, and balance laws, eqs.(10) for three of the four mass densities, suggests to split the main field vector u⁰into two parts u⁰ ≡ (v⁰, w⁰), where v⁰ are the variables relevant to conservation equations and w⁰ are those relative to balance laws; the symmetric system (25), in the main field variables, reads as

∂²h⁰⁰

∂v⁰∂v⁰∂_tv⁰+ ∂²h⁰¹

∂v⁰∂v⁰∂_xv⁰ = 0 , (36)

∂²h⁰⁰

∂w⁰∂w⁰∂_tw⁰+ ∂²h⁰¹

∂w⁰∂w⁰∂_xw⁰ = g¡ v⁰, w⁰¢

. (37)

In [12], [14] a equilibrium principal subsystem is defined as the system

∂²h⁰⁰(v⁰, 0)

∂v⁰∂v⁰ ∂_tv⁰+∂²h⁰¹(v⁰, 0)

∂v⁰∂v⁰ ∂_xv⁰= 0 , (38) obtained by fixing w⁰≡ w⁰e= 0.

It was also proved [12] that a principal subsystem is still hyperbolic and symmetric and it is still compatible with an entropy principle (sub-entropy law) with convex sub-entropy. Moreover, the eigenvalues of a principal subsystem are always included between the minimum and the maximum eigenvalues of the full system (sub-characteristic conditions).

In [12] an equilibrium state has been identified as the state for which the entropy production −Σ vanishes and reaches its minimum value. In addition it has been proven that this is equivalent to the conditions:

g¡ v⁰, 0¢

= 0 , ∂g

∂v⁰

¡v⁰, 0¢

= 0 , D¡ v⁰, 0¢

is negative definite , (39)

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where

D= 1 2

"

∂g

∂w⁰ + µ ∂g

∂w⁰

¶T#

. (40)

The condition (39)3 determines the dissipative character of the system [12].

In the present case, in (36), (37), v⁰ =¡

Λ^ρ, Λ^v, Λ^T¢

, w⁰ = ¡

Λ¹, Λ², Λ³¢ and the source term g = (m1R, m2R, −m³R) is the vector of the productions of mass densities ρ1, ρ2, ρ3.

Let us examine now the equilibria of the system. As first, we want to point out that, in order to get the closure of the Boltzmann equations for the reacting mixture at Euler level the assumption of mechanical equilibrium is necessary (a direct consequence of this assumption is the unique mean velocity v and the unique temperature T ). The system is at equilibrium when the mixture is also in chemical equilibrium.

In the case of reversible reactions, the gas is in chemical equilibrium if the number densities ni and the temperature T satisfy the mass action law:

n3n4− n¹n2

µm3m4

m1m2

¶³/2

exp µ

− E kT

¶

= 0 . (41)

In the case of exothermic irreversible reactions, chemical equilibrium is reached when one of the species A3 and A4 is completely burnt (n3n4= 0).

It is easy to observe that, at equilibrium, the source terms R_i of the only three balance equations (10) in the governing system (7)-(10) vanish because the quantity between square brackets in the expression of the reaction rate (6) is equal to zero for both kinds of reaction (R = 0).

Moreover, entropy production −Σ vanishes since R = 0.

For the previous considerations the equilibrium state u_e corresponds to

Λ^j_e= 0 ∀ j = 1, 2, 3, (42)

i.e.

µ1e = µ2e= µ3e= µ4e (43) meaning that at equilibrium chemical potentials are all equal [10].

Let us underline that, for reversible reactions, the quantity between curly brackets in the expression (31) of Σ vanishes too, as the equation

kT ln

"

n3n4

n1n2

µm1m2

m3m4

¶³/2#

+ E = 0

is an equivalent expression of the mass action law (41). The case of exothermic irreversible reaction has to be seen as the limiting case n3 → 0 (or

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equivalently, n4 → 0). In this case, the quantity between curly brackets in (31) now tends to −∞, but still Σ → 0. Moreover, when one of nα → 0, α = 3, 4, both entropy sα and enthalpy hα of the species accountable for equilibrium have to diverge to +∞, in accordance with (42) and (34), (35).

Even if this case may be considered as a limit one, numerical simulations show that the model works well.

It is interesting to point out that, in case of reversible reactions, the matrix D (v⁰, 0) is only negative semi-definite (its rank is 1), meaning that the system is not genuinely dissipative [12] and this is reasonable as the productions of the four species are not independent in a bimolecular reaction;

in fact, the number of lost molecules of both species A3 and A4 is the same and equals the number of gained molecules of both species A1 and A2 (and vice versa, when the reaction is reversible). In case of irreversible reactions, D(v⁰, 0) is the null matrix; it is proportional, by a the factor containing the product n3n4, to a matrix of rank 1.

It is now possible to conclude that the equilibrium subsystem of (7)-(10) is constituted by Euler equations for a perfect gas:

∂ρ

∂t + ∂

∂x(ρv) = 0 (44)

∂

∂t(ρv) + ∂

∂x

¡ρv²+ p¢

= 0, (45)

∂

∂t µ1

2ρv²+3 2p

¶ + ∂

∂x µ1

2ρv³+5 2pv

¶

= 0, (46)

as number and mass densities, and then chemical energy, are now constant.

6 The Kawashima-Shizuta condition

The last problem to be investigated is whether the hyperbolic system is dissipative or of composite type ([15]-[20]), in the sense that it has to be clear if the role of dissipation is dominant with respect to hyperbolicity, or not. One way to identify a system of composite type is the fulfillment of the so-called Kawashima-Shizuta condition, or genuine coupling [18]-[20].

A hyperbolic system of balance law (24) with convex entropy satisfies K-condition if in the equilibrium manifold, any characteristic eigenvector is not in the null space of ∇F, or equivalently, if every right eigenvector r⁰ of the symmetric system (36), (37) satisfy the requirement

r⁰_e ≡ r⁰¡ v⁰, 0¢

6=

µ X 0

¶

, (47)

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with X 6= 0.

In the present case, we have

r⁰1,6e=







±^v_Tê^ceê −3T²ê

³Ee4−^v2^e²

´

∓_T^cêe −²3T^vêê 2 3Te

³k1

ρ1e + _ρ^k⁴

4e

´ρ1e+_ρ^k⁴

4eρ_e−²3^beT^1ee

³k²

ρ^2e + _ρ^k_4e⁴´

ρ2e+_ρ^k_4e⁴ ρ_e−²3^beT^2e^e

³k3

ρ3e + _ρ^k⁴

4e

´ρ3e+_ρ^k⁴

4eρ_e−²3^beT^3e^e





 ,

r⁰₂_e=







−_ρ^k4e⁴ +_n_e^bk_kT¹ _e ³ e4−^v2²

´

bk1vê nêkTê

−_ne^bkkT¹ ^e k¹

ρ1e +_ρ^k⁴

4e +_n^bk_e¹_kT^be^1e_e

bk1be^2e n^ekT^e bk1be^3e nekTe







, r⁰₃_e =







−_ρ^k4e⁴ +_n_e^bk_kT² _e³ e4−^v2²

´

bk2vê nêkTê

−_ne^bkkT² ê bk²be^1e nêkTê k2

ρ^2e +_ρ^k_4e⁴ +_n^bk²_e_kT^be^2e_e

bk2be^3e nekTe





 ,

r⁰4e=







−_ρ^k4e⁴ +_n_e^bk_kT³_e ³ e4−^v2²

´

bk3vê nêkTê

−_ne^bkkT³ê bk³be^1e nêkTê bk3be^2e nêkTê k3

ρ3e +_ρ^k⁴

4e + _n^bk_e³_kT^be^3e_e







, r⁰5e=







k⁴

ρ^4e +_n_e^k_kT⁴_e ³ e4−^v2²

´

k4vê nêkTê

−_ne^kkT⁴ ^e k⁴

ρ4e +_n^k_e⁴_kT^be^1e_e

k4

ρ^4e +_n^k_e⁴_kT^be^2e_e

k⁴

ρ^4e +_n^k_e⁴_kT^be^3e_e







and therefore K-condition (47) holds.

According to the general theory of Hanouzet and Natalini [18], Ruggeri and Serre [19], Yong [21], if K-condition holds, together with the convexity of h⁰ and the dissipative character of the production, the existence of global smooth solution for small initial data is assured and for large t, the solution converges to an equilibrium state.

For what concerns the Riemann problem, there exists a conjecture [22]

that, by rescaling space and time, for large t the solution converges to the solution of the Riemann problem for the equilibrium subsystem. This is one of the questions that have been investigated numerically and discussed in Subsect.7.4.

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7 The Riemann problem

7.1 Solution in the non-reactive case

In this section, the Riemann problem for the hydrodynamic equations (7)- (10), in absence of chemical reactions (R = 0 in (10)), will be dealt with.

Let u_` and ur be the Riemann data at t = 0. We recall that in the case of systems of conservation laws, there exists a neighborhood N of u_` such that, if u_r ∈ N, the Riemann problem has a unique solution. This solution consists of at most (n + 1) constant states separated by shocks, centered simple waves or contact discontinuities, n being the dimension of u.

In the case of systems of balance laws, an analogous theorem does not exist; in fact, centered wave solutions are not admitted in presence of source terms.

In the present case, the solution is substantially equivalent to the Euler one. Nevertheless, we briefly consider all possible solutions admitted for the inert mixture.

As already said, the eigenvalues of (7)-(10) are

λ1 = v − c , λ2,3,4,5= v , λ6= v + c ; c²= 5 3

nkT ρ = 5

3 p ρ (48) meaning that the initial jump can be thought as the composition of three waves: two related to the characteristic velocities λ1 and λ6, respectively, which can be either shocks or rarefactions, and one, related to λ2,3,4,5, being a contact shock propagating with velocity s = v. The gas is then divided into four regions: besides the state ur of the gas ahead the fastest wave and the state u_` of the gas behind the slowest one, there are the unknown intermediate states denoted by ue and bu.

As well-known [2]-[5], the two intermediate states, connected by the contact shock, can be found once the shock curves and the rarefaction curves through u_` and u_r have been determined.

Let us start with the shocks compatible with the states ur and u_`, de- noting by s+ the velocity of the shock propagating into u_r and by s⁻ the velocity of the shock propagating from u_`. Observe that Rankine-Hugoniot conditions (R-H) for the two systems (1)-(3) and (7)-(10) are equivalent. By

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applying R-H conditions between u_r and eu, we get e

ρ_j = ρ_jr 4M+²

M+² + 3, j = 1, 2, 3, (49) e

ρ = ρr

4M+²

M₊² + 3, (50)

ev = vr+ 3c_rM+² − 1 4M+

, (51)

Te = T_r

¡5M+² − 1¢ ¡

M+² + 3¢

16M₊² , (52)

e

p = p_r5M₊² − 1

4 , (53)

where

M+= s+− vr

cr

, c²_r = 5 3

p_r ρr

= 5 3

n_rkT_r ρr

; (54)

M+is the Mach number with respect to the state u_r ahead the shock front and M+> 0 as s+ > v_r. We are discussing only the cases in which the initial jump is such that propagation occurs from the left to the right; this means that the only two possible solutions exhibit a 3-shock [29] and a 1-shock both propagating to the right, or a 3-shock propagating to the right and a 1-rarefaction propagating to the left. The other two cases are somehow specular.

Analogously, by applying R-H conditions betweenub and u_`, we get b

ρ_j = ρ_j`M−² + 3 4M−²

, j = 1, 2, 3, (55)

b

ρ = ρ_`M−² + 3 4M−²

, (56)

bv = v`+ 3c_` M−² − 1 q¡5M−² − 1¢ ¡

M−² + 3¢ , (57) Tb = T_` 16M−²

¡5M−² − 1¢ ¡

M−² + 3¢ , (58)

b

p = p_` 4

5M−² − 1, (59)

where

M−= s−− bv

bc , bc² = 5 3

b p b

ρ; (60)

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here M⁻ is the Mach number with respect to the state u, with Mb ⁻< 0 as s− < bv.

Let us underline that, from the requirement T > 0, the following restric- tions follow:

M− < − 1

√5, M+> 1

√5.

By removing M+from (49)-(53), the parametric equations of the 3-shock curve, s3(u_r), through u_r, can be easily derived









 e ρ_j = ρ_jr

ρ_r ρe

ev = v^r+ 3cr(eρ − ρ^r) p3eρ (4ρ_r− eρ) T = Te _rρ_r

e ρ

4eρ − ρr

4ρ_r− eρ

(61)

for ρ_r < eρ < 4ρ_r and ρ_jr < eρ_j < 4ρ_jr, ev > vr, eT > T_r. In (v, p)-plane, they

read as 









ev = vr+ 3c_r(eρ − ρr) p3eρ (4ρr− eρ) e

p = pr4eρ − ρ^r 4ρ_r− eρ

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for ρ_r< eρ < 4ρ_r, ev > vr, ep > p_r Analogously, the 1-shock curve through u_`, denoted by s1(u_`) is, in parametric form









 b ρ_j = ρ_j`

ρ_` ρb

bv = v`− 3c_`(bρ − ρ`) p3bρ (4ρ_`− bρ) T = Tb `

ρ_` b ρ

4bρ − ρ`

4ρ_`− bρ

(63)

for ρ` < bρ < 4ρ` and ρj`< bρj < 4ρj`, bv < v^`, bT > T`. In (v, p)-plane









bv = v^`− 3c_`(bρ − ρ`) p3bρ (4ρ_`− bρ) b

p = p_`4bρ − ρ`

4ρ_`− bρ

(64)

for ρ_` < bρ < 4ρ_`, bv < v`, bp > p_`.

It is also possible to derive rarefaction waves compatible with system (7)-(10) in absence of source terms; in fact the system of ODE coming from

du

dz = αr (65)

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with u = u (z), z = x/t, r being the generic right eigenvector and α a constant, can be easily integrated.

In particular, by integrating (65), with r = r6 = (ρ, c, 2T /3, ρ1, ρ2, ρ3)^T, between u_r and the state eu satisfying the constraint eρ_j < ρ_jr, eρ < ρ_r, ev < vr, eT < T_r, the parametric equations of the rarefaction curve r3(u_r) can be obtained:









 e ρ_j = ρ_jr

ρ_r ρe ev = vr+ 3c_r

"µ e ρ ρ_r

¶¹/3

− 1

#

T = Te _r µ ρe

ρr

¶2/3

(66)

In the (v, p)-plane, they read as:











ev = vr+ 3c_r

"µ e ρ ρr

¶1/3

− 1

#

e p = p_r

µ ρe ρ_r

¶⁵/3 (67)

with eρ < ρ_r, ev < vr, ep < p_r.

Analogously, by integrating (65), with r = r1= (ρ, −c, 2T/3, ρ¹, ρ2, ρ3)^T, between u_` and a state ub such that bρ_j < ρ_j`, bρ < ρ_`, bv > v`, bT < T_`, the parametric equations of the rarefaction curve r1(u_`) is obtained









 b ρ_j = ρ_j`

ρ_` ρb bv = v`− 3c`

"µ b ρ ρ_`

¶¹/3

− 1

#

T = Tb _` µρb

ρ_`

¶²/3

(68)

In the (v, p)-plane, they are:











bv = v`− 3c`

"µ b ρ ρ_`

¶¹/3

− 1

#

b p = p_`

µρb ρ_`

¶⁵/3 (69)

with bρ < ρ_`, bv > v`, bp < p_`.

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Once the shocks and the rarefactions have been stated, the further step in order to find the solution to the Riemann problem is to impose that the two unknown intermediate states ue and bu are connected by the contact shock propagating with velocity s = v = λ2,3,4,5. As well known [28], this leads to the conditions ½

ev = bv = s e

p = bp (70)

meaning that the projections of eu andub on the (v, p)-plane coincides.

The solution of the Riemann problem now consists in the determination of the unknowns eρ_j, bρ_j, j = 1, 2, 3, eρ, bρ, v^∗ = ev = bv, p^∗ = ep = bp and, eventually, the shock velocities M+and M−in the four admissible structures of the profile. Let us examine each case.

1. eu∈ s³(u_r) ∧ bu∈ r¹(u_`)

The relations to be taken into account are (49)-(53) relating ue to u_r and M+, (68) and (69) relating ub to u_` and bρ, together with (70), in the unknowns eρj, bρj, j = 1, 2, 3, eρ, bρ, v^∗, eT , bT and M+.

From (53) and (69)2, by imposing (70)2, we obtain the explicit expression of M+ as function of bρ

M+= s

1 5+4

5π µρb

ρ_`

¶5/3

, (71)

satisfying the constraint M+ > p

1/5. From (51) and (69)1, by imposing (70)1, and using (71), one gets an equation for bρ

√5 5 c_r

π³

b ρ ρ`

´⁵/3

r − 1 1 + 4π³

b ρ ρ`

´⁵/3 + c_`

"µ b ρ ρ_`

¶¹/3

− 1

#

= v_`− vr

3 , (72)

where π = p_`/p_r. Note that, from the conditions p^∗ = ep > p_r and p^∗= bp < p_`, it follows that

p_`> p_r ⇔ π > 1 . (73) Lax conditions for a 3-shock must hold [29]; in terms of the Mach number M+, they read as

M+> 1 , v^∗− vr

c_r < M+< v^∗+ ec− vr

c_r . (74)

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By imposing (74)1to the expression (71) for M+, we get the constraint ρ_`

π³^/5 < bρ < ρ_`. (75) The knowledge of bρ by (72), and then of M+ from (71), at least numerically, allows to recover all the unknowns, by using (49)-(52) and (68).

2. eu∈ s³(ur) ∧ bu∈ s¹(u_`)

In this case, the conditions to be taken into account are (49)-(53) relatingue to u_r and M+, (55)-(59) relatingub to u_` and M⁻, together with (70), in the unknowns eρ_j, bρ_j, j = 1, 2, 3, eρ, bρ, v^∗, eT , bT , M+ and M⁻.

From (53) and (59), by imposing (70)2, we obtain the explicit expression of M− as function of M+

M⁻= − s1

5 µ

1 + 16π 5M₊² − 1

¶

, (76)

satisfying the constraint M⁻ < −p

1/5 provided that M+ > p 1/5.

From (51) and (57), by imposing (70)1, and using (76), one gets an equation for the quantity M+

c_rM+² − 1 M+

+ c_` 5M+² − 1 − 4π q

5π¡

5M+² − 1 + π¢ =

4 (v_`− vr)

3 . (77)

Let us note that the knowledge of M+ from eq. (77), and then of M− from (76), allows to recover all the unknowns from (49)-(52) and (55)-(58).

Note that, from the conditions v^∗= ev > vr and v^∗= bv < v`, it follows that

v_` > v_r. (78)

Finally, the admissibility of the two shocks must be checked. M+ still have to verify Lax conditions for a 3-shock (74). Conversely, M− must satisfy Lax conditions for a 1-shock:

−1 < M⁻< 0 , M−< v_`− c`− v^∗

bc . (79)

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By imposing condition (79)1 to the expression (76) for M⁻ and taking into account condition (74)1 on M+, we get the constraint

M+ > max (

1, r1

5+ 4 5π

)

; (80)

condition (80) reads M+>

r1 5 +4

5π > 1 ⇔ π > 1 ⇔ p_` > p_r, (81) M+> 1 >

r1 5 +4

5π ⇔ 0 < π < 1 ⇔ p_` < p_r. (82) 3. eu∈ r³(u_r) ∧ bu∈ s¹(u_`)

The system to be solved, in the unknowns eρ_j, bρ_j, j = 1, 2, 3, eρ, bρ, v^∗, T , be T , and M−, is determined by conditions (66) and (67) relating ue to u_r and eρ, (55)-(59) relating bu to u_` and M−, together with (70).

From (69)2 and (59), by imposing (70)2, we obtain the explicit expression of M− as function of eρ

M⁻= − s

1 5+4

5π µρe

ρ_r

¶−5/3

, (83)

satisfying the constraint M⁻ < −p

1/5. From (67)1 and (57), by imposing (70)1, and using (83), one gets an equation for eρ

cr

"µ ρe ρ_r

¶¹/3

− 1

# +

√5 5 c_`

1 π

³ρe ρr

´5/3

r − 1 1 +⁴_π³

e ρ ρ^r

´⁵/3 = v_`− vr

3 . (84)

Note that, from the conditions p^∗= ep < pr and p^∗= bp > p`, it follows that

p_` < p_r, (85)

and, by imposing Lax condition (79)1 to the expression (83) for M⁻, we get the constraint

ρ_rπ³^/5< eρ < ρ_r; (86) Moreover, if we observe that the left hand side of the equation for eρ is negative, it follows that

v_` < vr. (87)

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The knowledge of eρ by (84), and then of M⁻ from (83), allows to recover all the unknowns from (66) and (55)-(58).

4. eu∈ r³(u_r) ∧ bu∈ r¹(u_`)

In this case the conditions to be taken into account are (66) and (67) relatingeu to ur and eρ, (68) and (69) relating buto u_` and bρ, together with (70), and the problem consists in finding the unknowns eρ_j, bρ_j, j = 1, 2, 3, eρ, bρ, v^∗, eT , bT .

From (67)2 and (69)2, by imposing (70)2, we obtain the expression of e

ρ as function of bρ

e

ρ = π³^/5

r ρ ;b (88)

from (67)1 and (69)1, by imposing (70)1, and using (88), bρ is derived explicitly

b ρ = ρ_`

"₁

3(v_`− vr) + c_r+ c_` c_rπ¹^/5+ c_`

#³

. (89)

Note that, from the conditions v^∗= ev < vr and v^∗= bv > v`, it follows that

v_` < v_r. (90)

The knowledge of bρ from (89), and then of eρ from (88), allows to recover all the unknowns from (66) and (68).

Let us underline that the solution in the case of an inert mixture is absolutely analogous to the case of the perfect gas [2]-[5].

7.2 Riemann problem in the reactive case

As already said, centered wave type solutions, as well as constant ones, are not compatible with systems of balance laws. Therefore, the exact solution of the Riemann problem can not be deduced. The only way to investigate the problem is by using numerical procedures. In Subsect.7.4, some numerical simulations are shown in order to point out the influence of chemical reactions, i.e. of source terms, on the solution to eqs. (7)-(10), with R 6= 0, with respect to the non-reactive case.

Nevertheless, it is possible to get an estimate of the intermediate states (at least for very short times) by using a simplified solution procedure proposed in a recent paper [30]. This procedure may be useful also for systems of conservation laws as, in general, the ODEs to be solved in order to deter- mine the centered waves solutions can be integrated only numerically. This

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method consists in solving the Riemann problem by using only R-H conditions, which are always algebraic equations, like case 2 in Subsect.7.1, where the solution exhibits two shocks. This means that the resulting intermediate states lie, in any case, on the shock curves through u_` and u_r. Once all the unknown have been determined, the validity of the Lax conditions has to be checked. If one of the Lax conditions (74), (79) (or both) is not satisfied, the corresponding intermediate state, ue or bu, (or both) are not exact, as they lie on the unstable part of the shock curve, rather than on the suitable rarefaction curve; nevertheless, for sufficiently small initial jump, they are in good agreement with the exact ones. Moreover, for systems of balance laws, again under the assumption of small initial jump, the proposed procedure yields a good approximation of the intermediate states for times close to t = 0. It is clear that the space-time evolution of the intermediate states is perturbed by the presence of source terms.

7.3 The numerical method

In this section, we present the numerical method used in order to solve the hydrodynamic equations, considered in section 2, with discontinuous initial data. By using a fractional step approach, one alternates between solving the homogeneous hyperbolic equations, ignoring the source terms, and solving the time-dependent ordinary differential equation with sources, ignoring the space derivative terms. In this context, for the homogeneous nonlinear hyperbolic equations, we use the non oscillatory central scheme of Nessyahu and Tadmor (NT) [24] that is second-order accurate in regions of smooth flow and first-order accurate near discontinuities. The equations with source terms are handled by a second order Runge-Kutta scheme.

In the last years, high-resolution non-oscillatory central schemes have become popular in solving nonlinear hyperbolic conservation law because of the simplicity in their construction. In fact, by the NT scheme, we obtain second order accuracy in smooth regions without knowledge of the characteristic structure of equations. No Riemann solvers and no field-by-field decomposition need to be solved in the computation of the smooth numerical fluxes along the mid-cells. The results obtained by analytical study of the Riemann problem on the homogeneous hyperbolic equations are in accordance with the numerical results obtained implementing this method without requiring complicate Riemann solvers.

Regarding the fractional step method we use the Strang splitting approach which consists in advancing first by a half time step on PDE equation, then by a full time step on the ODE equation, and ends with a half

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time step on the first equation again. As far as the accuracy question is concerned, the Strang splitting [23] yields second order accuracy in time if each sub-problem is solved by a second order accurate method.

For what concerns the NT method, let us observe that the central scheme is implemented as a predictor-corrector method

wⁿ⁺

1 2

j = w¯ⁿ_j −1 2λf_j⁰

¯ wⁿ⁺¹

j+¹₂ = 1 2

£w¯ⁿ_j + ¯wⁿ_j+1¤ + 1

8

£w⁰_j− w⁰j+1

¤− λ

· f(wⁿ⁺

1 2

j+1 ) − f(wⁿ⁺

1 2

j )

¸ , where ¯w_jⁿrepresents the approximate cell-average at the discrete time level tⁿ and λ = ∆t/∆x. f (wⁿ⁺

1 2

j ) denotes the numerical flux evaluated at the point-wise values at the half-time steps. w_j⁰/∆x and f_j⁰/∆x are approxima- tions of the space derivatives of the field and of the flux, such that

w⁰_j ' ∆x · wx(x_j, tⁿ) + O (∆x)² , f_j⁰ ' ∆x · fx(x_j, tⁿ) + O (∆x)² . For the stability of this scheme it is necessary to require the following CFL (Courant-Friedrichs-Lewy) condition

λ max

j ρ¡ A¡

¯ wⁿ_j¢¢

< 1 2,

where ρ(·) represents the spectral radius of the Jacobian matrix A (·) =

∂f_i/∂w_k.

In order to ensure that this scheme is non-oscillatory, the numerical derivatives should satisfy

0 ≤ w⁰j· sgn³

∆w_j±¹

2

´≤ Cost¯¯¯MinMod³

∆w_j+¹

2, ∆w_j−¹

2

´¯¯¯ , 0 ≤ fj⁰· sgn³

∆f_j±¹

2

´≤ Cost¯¯¯MinMod³

∆f_j+¹

2, ∆f_j−¹

2

´¯¯¯ , where ∆w_j+¹

2 = wj+1− w^j and M inM od stands for the slope limiter M inM od (a, b) = 1

2[ sgn (a) + sgn (b)] · min(|a| , |b|) . 7.4 Numerical results

In this section we present the simulations concerning the integration of the reactive hydrodynamic equation with the aim of showing the influence of chemical source terms on the shape of the profiles.

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Simulations are performed for the reaction H2O + H OH + H2

by setting the following parameters

m1= 0.018 m2 = 0.001 m3 = 0.017 m4= 0.002 E = 63300 χ = 3927 β = 0.5.

Fig.1 shows the density profiles at different times obtained by solving the Riemann problem in absence of chemical reactions in the classical case v_`= v_r = 0, ρ_`> ρ_rand T_`> T_r(i.e. case 1 in Subsect. 7.1). The numerical results are obtained with final time t_max = 100, on a computational domain of dimension [−100, 100] with ∆x = 0.1 and time step ∆t, chosen in order to ensure a Courant number equal to 0.4. We always apply the M inM od slope limiter function for each wave; moreover, Neumann boundary conditions are used. In addition to the choice v_` = vr = 0, we set the following arbitrary values:

n1`= 0.06, n2`= 0.05, n3` = 0.15, n4` = 0.25, T_` = 10³ n1r= 0.03, n2r= 0.02, n3r = 0.1, n4r= 0.2, T_r= 300 ;

in this case, as shown in Fig.1, the profile exhibits a shock propagating to the right, a rarefaction propagating to the left and a contact discontinuity propagating with velocity v.

−1000 −80 −60 −40 −20 0 20 40 60 80 100

2 4 6 8x 10⁻³

x

ρ

t=0 t=100

Figure 1: Mass density profiles at different times for the inert mixture.

If initial states u_` and u_r are both equilibrium states, the solution of the Riemann problem with non zero source terms coincides with the one