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Journal of Scientific Computing (2022) 93:83
https://doi.org/10.1007/s10915-022-02028-x
Exact and Numerical Solutions of the Riemann Problem for a
Conservative Model of Compressible Two-Phase Flows
Ferdinand Thein1,4 ·Evgeniy Romenski2,3 ·Michael Dumbser3
Received: 21 April 2022 / Revised: 16 July 2022 / Accepted: 28 September 2022 /
Published online: 9 November 2022
© The Author(s) 2022
Abstract
In this work we study the solution of the Riemann problem for the barotropic version of the
conservative symmetric hyperbolic and thermodynamically compatible (SHTC) two-phase
flow model introduced in Romenski et al. (J Sci Comput 42(1):68, 2009, Quart Appl Math
65(2):259–279, 2007). All characteristic fields are carefully studied and explicit expressions
are derived for the Riemann invariants and the Rankine–Hugoniot conditions. Due to the
presence of multiple characteristics in the system under consideration, non-standard wave
phenomena can occur. Therefore we briefly review admissibility conditions for discontinu-
ities and then discuss possible wave interactions. In particular we will show that overlapping
rarefaction waves are possible and moreover we may have shocks that lie inside a rarefaction
wave. In contrast to nonconservative two phase flow models, such as the Baer–Nunziato
system, we can use the advantage of the conservative form of the model under consideration.
Furthermore, we show the relation between the considered conservative SHTC system and
the corresponding barotropic version of the nonconservative Baer–Nunziato model. Addi-
tionally, we derive the reduced four equation Kapila system for the case of instantaneous
relaxation, which is the common limit system of both, the conservative SHTC model and
the non-conservative Baer–Nunziato model. Finally, we compare exact solutions of the Rie-
mann problem with numerical results obtained for the conservative two-phase flow model
under consideration, for the non-conservative Baer–Nunziato system and for the Kapila limit.
BFerdinand Thein
ferdinand.thein@ovgu.de; thein@igpm.rwth-aachen.de
Evgeniy Romenski
evrom@math.nsc.ru
Michael Dumbser
michael.dumbser@unitn.it
1Institute for Analysis and Numerics, Otto-von-Guericke University Magdeburg, PSF 4120, 39016
Magdeburg, Germany
2Sobolev Institute of Mathematics, Novosibirsk, Russia
3Department of Civil, Environmental and Mechanical Engineering, University of Trento, Via
Mesiano, 77, 38123 Trento, Italy
4Institut für Geometrie und Praktische Mathematik, RWTH Aachen, Templergraben 55, 52056 Aachen,
Germany
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83 Page 2 of 60 Journal of Scientific Computing (2022) 93 :83
The examples underline the previous analysis of the different wave phenomena, as well as
differences and similarities of the three systems.
Keywords Conservative model of compressible two fluid flow ·Thermodynamically
compatible hyperbolic systems ·Exact solution of the Riemann problem ·Resonance for
non-strictly hyperbolic systems ·Finite volume schemes ·Comparison with the
Baer-Nunziato model
Mathematics Subject Classification 35L03 ·35Q35 ·76Txx
1 Introduction
The aim of the paper is to construct exact solutions for the Riemann problem of the
one-dimensional barotropic version of the conservative Symmetric Hyperbolic and Ther-
modynamically Compatible (SHTC) model of compressible two-phase flows introduced in
[64,69]. The results obtained can be useful for a qualitative understanding of the physical
processes occurring in two-phase flows, for a comparative analysis of various models, and
also for testing numerical methods.
Note that the existence of a well-developed theory of exact solutions of the Riemann prob-
lem ensured the success in the development of modern shock-capturing numerical methods
[31] for solving the Euler equations of compressible gas dynamics, which are of fundamental
importance in science and engineering, in particular aerospace engineering and astrophysics.
See [78] for an exhaustive overview of shock capturing schemes based on the exact or approx-
imate solution of the Riemann problem. And even at present, the known exact solutions of the
Euler equations are successfully used to test new numerical methods for solving hyperbolic
systems of equations. Therefore, in our opinion, the construction of exact solutions for the
equations of two-phase flows will influence the formation of a common point of view in
the field of multiphase flow modeling and the development of new numerical methods for
solving problems related to this area.
In contrast to single-phase gas dynamics, there is still no universally accepted model of
compressible multiphase flows, even for the case of only two phases, see, for example [72].
The generally accepted approach to design a two-phase flow model is based on the assumption
that a mixture is a system of two interacting single phase continua, see, for example, [40].
The most widely used PDE system is the Baer–Nunziato model [2], various modifications of
which have been applied by many authors to study different types of flows, including flows
with phase transitions and chemical reactions, see, for example recent papers [11,27,73]
and references therein.
In this paper, we will study a model developedon the basis of the theory of SHTC systems,
which was developed in [32–34,57,60,62]. This theory connects the local well-posedness of
the governing partial differential equations of continuum physics (symmetric hyperbolicity
in the sense of Friedrichs) with the fulfillment of the laws of thermodynamics (the law of
conservation of energy and the law of increasing entropy). The general master system of
SHTC equations can be derived from an underlying variational principle [57], and the study
of numerous models of continuum mechanics has shown that their governing differential
equations belong to the SHTC class of PDEs. This theory can be used to formulate new, well-
posed models of processes in complex media, including unified models for the description
of viscous Newtonian and non-Newtonian fluids and nonlinear elasto-plastic solids at large
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Journal of Scientific Computing (2022) 93 :83 Page 3 of 60 83
deformations [17,18,56,58], its extension to general relativity [66], SHTC models of rupture
dynamics [28,76] and flows in deformable porous media [67,68].
The object of study in this paper is the SHTC model of compressible two-phase flow, the
general equations of which are formulated and discussed in [63,64,69,70]. An all Mach
number flow solver for the barotropic SHTC two-phase model was recently forwarded in
[49]. In the model under consideration, the two-phase medium is assumed to be a single
continuum, the properties of which take into account the features of the two-phase flow. This
means that the element of the medium is described by the average field of phase velocities
and pressures for the mixture, but the flow of phases through this element is allowed. We
consider the simplest model of a one-dimensional flow of a barotropic mixture. In this case,
the SHTC equations can be written in a fully conservative form and therefore allow a direct
formulation of discontinuous solutions. Nevertheless, the construction of exact solutions for
the Riemann problem turns out to be a rather difficult task.
It is necessary to note that the governing equations of the Baer–Nunziato model [1,2]in
general can not be completely written in a conservative form, even in the one-dimensional
case. This creates difficulties in the definition of discontinuous solutions of the shock-wave
type. But it turns out that the barotropic Baer–Nunziato system can be presented in a conser-
vative form for one particular choice of interfacial velocity and interfacial pressure. And this
conservative system is exactly the same as the barotropic case of the SHTC model studied
in this paper. We discuss the similarities and differences between the different models by
rewriting the equations of the SHTC model in the form of the Baer–Nunziato model for two
interacting single phase continua. We also consider the reduced one-dimensionallimit system
obtained in the stiff pressure and velocity relaxation limit i.e. the resulting single-velocity
single-pressure approximation of the SHTC and Baer–Nunziato models for the barotropic
flow and show that they are identical and reduce to the well-known Kapila model [41].
The conservative form of the SHTC model of compressible two-phase flows has advan-
tages not only for the construction of exact solutions of the Riemann problem, but also when
using advanced shock capturing numerical methods. For the numerical simulations shown in
this paper we will therefore rely on classical second order high resolution shock capturing
TVD finite volume schemes as presented in [78].
Degenerate behaviour in hyperbolic systems is a topic of constant interest. However, one
has to be careful identifying the source of degeneracy. Non-strictly hyperbolic systems, i.e.
systems with multiple eigenvalues, naturally arise in multi dimensions [15]. Specific cases
of non-strictly hyperbolic systems where studied in [3,42,48] and recently by Freistühler
and Pellhammer [25]. Subject of study in the aforementioned references are mostly systems
of two equations where the coincidence of eigenvalues often occurred in points where also
the character of the related field changes from genuinely nonlinear to linearly degenerated.
In particular Freistühler showed in [23] that coinciding eigenvalues in the presence of a
discontinuity must belong to linearly degenerated fields. If the Euler equations or the present
SHTC system are considered, it can be directly verified that such a situation is related to
the vanishing of the fundamental derivative G. This may happen for certain equations of
state, but in general does not necessarily imply multiple eigenvalues. A prominent example
is the system of Euler equations and we highly recommend [50,53,79]. Another reason for
degeneracies may be the loss of an eigenvector as described in [13,74]. Systems with missing
eigenvectors are often called weakly hyperbolic and systems with coinciding eigenvalues
are sometimes also called hyperbolic degenerate. For further reading and a more detailed
survey we recommend [10]. Coinciding eigenvalues may lead to difficult situations, but as
long as there is a full set of eigenvectors which span the complete space the system is still
diagonisable. For the construction of complete Riemann solvers a full set of eigenvectors is of
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83 Page 4 of 60 Journal of Scientific Computing (2022) 93 :83
great importance, see e.g. [20,54,59]. However, we want to emphasize that the consequences
depend crucially on the system under consideration. In some cases the numerical methods
break down when applied to weakly hyperbolic systems, see e.g. [12], whereas in other
situations a numerical treatment of weakly hyperbolic systems is still possible, see e.g. [35,
36].
The rest of the paper is organized as follows: in Sect.2we present the SHTC system under
consideration in this paper and study its eigenstructure. In Sect. 3we briefly summarize
degeneracies and admissibility conditions. The wave relations, i.e. the Riemann invariants
and the Rankine–Hugoniot conditions of the model are presented in Sect. 4, while the possible
wave configurations are shown in Sect.5. The relation of the conservative SHTC system
with the nonconservative Baer–Nunziato model and the common Kapila limit are shown in
Sect. 6. Some examples of exact solutions and corresponding numerical results are presented
in Sect. 7. The paper closes with some concluding remarks and an outlook to future work in
Sect. 8.
2 Conservative Barotropic SHTC Model for Compressible Two-Phase
Flows
2.1 Multi-dimensional Case
The PDE system for compressible barotropic two-phase flows was discussed in Romenski et
al. [64,69]. Written in terms of the specific energy E=E(α1,c1,ρ,w
k)it reads
∂ρα1
∂t+∂ρα1uk
∂xk
=ξ1,(2.1a)
∂ρc1
∂t+∂(ρc1uk+ρEwk)
∂xk
=ξ2,(2.1b)
∂ρ
∂t+∂ρuk
∂xk
=ξ3,(2.1c)
∂ρui
∂t+∂(ρuiuk+pδik +ρwiEwk)
∂xk
=ξ4,(2.1d)
∂wk
∂t+∂(wlul+Ec1)
∂xk
+ul∂wk
∂xl
−∂wl
∂xk=ξ5.(2.1e)
Here, α1is the volume fraction of the first phase, which is connected with the volume fraction
of the second phase α2by the saturation law α1+α2=1, ρis the mixture mass density,
which is connected with the phase mass densities ρ1,ρ
2by the relation ρ=α1ρ1+α2ρ2.
The phase mass fractions are defined as c1=α1ρ1/ρ, c2=α2ρ2/ρ and it is easy to see
that c1+c2=1. The mixture velocity is given by ui=c1ui
1+c2ui
2and wi=ui
1−ui
2is
the relative phase velocity. The equations describe the balance law for the volume fraction,
the balance law for the mass fraction, the conservation of total mass, the total momentum
balance law and the balance for the relative velocity. The phase interaction is present via
algebraic source terms in (2.1a)and(2.1e), which are proportional to thermodynamic forces.
These source terms are phase pressure relaxation to the common value
ξ1=−ρφ/θ1=−ρEα1/θ1(2.2)
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Journal of Scientific Computing (2022) 93 :83 Page 5 of 60 83
and interfacial friction
ξ5=−λk/θ2=−Ewk/θ2.(2.3)
The coefficients θ1,θ
2characterize the rate of pressure and velocity relaxation and can depend
on parameters of state. Due to mass and momentum conservation throughout this work we
assume ξ2=ξ3=ξ4=0.
2.2 Discussion of the Mixture Equation of State
We now want to further specify the derivatives of the generalized energy E. Due to the
relation for the mass fractions and the concentrations we write α≡α1,α2=1−αand
c≡c1,c2=1−c, when appropriate. The mixture equation of state (EOS) is defined as
the sum of the mass averaged phase equations of state and the kinematic energy of relative
motion
E(α, c,ρ,w
1,w
2,w
3)=e(α, c,ρ)+c1c2
wiwi
2=e(α, c,ρ)+c(1−c)wiwi
2,(2.4)
e(α, c,ρ) =c1e1(ρ1)+c2e2(ρ2)=ce1cρ
α+(1−c)e2(1−c)ρ
1−α.
(2.5)
where ei(ρi)is the specific internal energy of the i-th phase and is assumed to be known.
Special attention has to be paid when the derivatives of the internal energies are needed.
Remark 2.1 (isentropic vs. isothermal) When we want to calculate the derivative of the inter-
nal energy with respect to the density in the barotropic case, we have to specify whether we
are considering an isentropic or an isothermal process. If we keep the entropy constant we
have the well known derivative
∂e
∂ρ s
=p
ρ2.(2.6)
However, if the temperature is held constant the derivative is given by
∂e
∂ρ T
=p
ρ2+T∂s
∂ρ T
.(2.7)
Note that the relations for the barotropic case were not given in previous work and in particular
the isothermal case is not a straightforward simplification of the general case. The detailed
calculations are given in the Appendix Aand we will just summarize the results needed here.
We now introduce the mixture pressure
p=α1p1+α2p2,(2.8)
the specific enthalpy of phase i
hi(ρi)=ei(ρi)+pi(ρi)
ρi
(2.9)
and the specific Gibbs energy of phase i
gi(ρi)=ei(ρi)−Ts
i(ρi)+pi(ρi)
ρi
.(2.10)
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83 Page 6 of 60 Journal of Scientific Computing (2022) 93 :83
The Gibbs energy, or sometimes free enthalpy, may be generalized to the chemical potential
μwhen more substances are involved, see [44]. For the mixture EOS we have the following
derivatives
∂E
∂α =∂e
∂α,∂E
∂c=∂e
∂c+(1−2c)wiwi
2,∂E
∂ρ =∂e
∂ρ and ∂E
∂wi
=c(1−c)wi.
(2.11)
The derivatives for the internal energy e(α, c,ρ)=c1e1(ρ1)+c2e2(ρ2)of the mixture are
∂e
∂α =p2−p1
ρ,∂e
∂c=h1(ρ1)−h2(ρ2), ∂e
∂ρ =p
ρ2(2.12)
for the isentropic case and
∂e
∂α =p2−p1
ρ+T(c1s1−c2s2), ∂e
∂c=g1(ρ1)−g2(ρ2),
∂e
∂ρ =p
ρ2−T(c1s1−c2s2)
ρ(2.13)
for the isothermal case.
2.3 One-Dimensional Model
In this paper we focus on the one dimensional case and thus the equations simplify to
∂αρ
∂t+∂αρu
∂x=ξ1,(2.14a)
∂ρc
∂t+∂(ρcu +ρEw)
∂x=ξ2,(2.14b)
∂ρ
∂t+∂ρu
∂x=ξ3,(2.14c)
∂ρu
∂t+∂ρu2+p+ρwEw
∂x=ξ4,(2.14d)
∂w
∂t+∂(wu+Ec)
∂x=ξ5.(2.14e)
Note, that the curl term in equation (2.1e) vanishes. Applying the obtained results and relations
and introducing the notation
i(ρi)=hi(ρi), isentropic
gi(ρi), isothermal
the system can be rewritten in the following form:
∂α1ρ
∂t+∂α1ρu
∂x=ξ1,(2.15a)
∂α1ρ1
∂t+∂α1ρ1u1
∂x=ξ2,(2.15b)
∂ρ
∂t+∂ρu
∂x=ξ3,(2.15c)
∂(α1ρ1u1+α2ρ2u2)
∂t+∂α1ρ1u2
1+α2ρ2u2
2+α1p1(ρ1)+α2p2(ρ2)
∂x=ξ4,(2.15d)
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Journal of Scientific Computing (2022) 93 :83 Page 7 of 60 83
∂w
∂t+∂
∂x1
2u2
1−1
2u2
2+1(ρ1)−2(ρ2)=ξ5,(2.15e)
For the isentropic case the total energy inequality, which serves as mathematical entropy
inequality, reads
2
i=1
∂αiρiei+1
2u2
i
∂t+∂αiρiuihi+1
2u2
i
∂x≤0,(2.16)
while for the isothermal case we have the inequality
2
i=1
∂αiρiei−Ts
i+1
2u2
i
∂t+∂αiρiuigi+1
2u2
i
∂x≤0.(2.17)
Inequality (2.17) is the mathematical formulation of the physical statement that the free
energy of a system under consideration is minimized in an isothermal process. Moreover, this
formulation is a straightforward generalization of the inequality obtained for the isothermal
Euler equations, see [15,71,77]. We want to note that it is beneficial to derive the isothermal
model from the more general model including the thermal impulse. More detailed information
are given in the Appendix.
2.4 Conservative Formulation
The system (2.15a)-(2.15e) can be written in the conservative form given by
∂
∂t
W+∂
∂x
F(W)=,
where the vector of conserved quantities reads
W=(w1,w
2,w
3,w
4,w
5)T≡(α1ρ,α1ρ1,ρ,α
1ρ1u1+α2ρ2u2,u1−u2)T.
Using the equations obtained so far we can write the conservative flux as follows
F(W)=
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
w1
w4
w3
w2
(w3−w2)w5+w4
w3
w4
w2(w3−w2)w5+w4
w32
+(w3−w2)w4−w2w5
w32
+w1
w3
p1(W)+w3−w1
w3
p2(W)
1
2w52(w3−w2)w5+w4
w3
−w5+1(W)−2(W)
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
.
(2.18)
2.5 Primitive Formulation
We want to reformulate the barotropic system (2.15) in terms of the primitive variables
α1,ρ
1,ρ
2,u1and u2. The other quantities are then obtained using the relations
α2=1−α1,ρ=α1ρ1+(1−α1)ρ2,ρu=α1ρ1u1+(1−α1)ρ2u2,w=u1−u2.
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83 Page 8 of 60 Journal of Scientific Computing (2022) 93 :83
Here we assume α1∈(0,1)and ρ1,ρ
2>0, i.e. we exclude vacuum states. We further
introduce the speed of sound aiof phase igiven by the following relation
a2
i=⎧
⎪
⎪
⎨
⎪
⎪
⎩
ρi∂hi
∂ρis
,isentropic,
ρi∂gi
∂ρiT
,isothermal.
(2.19)
For a thermodynamically consistent equation of state the speed of sound is well defined and
we do not have to differ between the two cases for the mathematical considerations. With
W≡(α1,ρ
1,ρ
2,u1,u2)the Jacobian of this system is given by
A(W)=
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
u0000
ρ1
α1
(u1−u)u10ρ10
ρ2
α2
(u−u2)0u20ρ2
p1−p2
ρ
a2
1
ρ1
0u10
p1−p2
ρ0a2
2
ρ2
0u2
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
(2.20)
and the source terms can be transformed using the following matrix
B(W)=
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
1
ρ0−α1
ρ00
−ρ1
α1ρ
1
α1
ρ1
ρ00
ρ2
α2ρ−1
α2
c1
α2
+c200
0u2−u1
ρ−u2
ρ
1
ρc2
0u2−u1
ρ−u2
ρ
1
ρ−c1
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
.(2.21)
Now we can write the system in the following compact form
∂tW+A(W)∂xW=B(W).
The eigenvalues can be computed as
λ1±=u1±a1,λ
C=u,λ
2±=u2±a2(2.22)
and we have (up to scaling) the following right eigenvectors
R1±=⎛
⎜
⎜
⎜
⎜
⎜
⎝
0
1
0
±a1
ρ1
0
⎞
⎟
⎟
⎟
⎟
⎟
⎠
,RC=⎛
⎜
⎜
⎜
⎜
⎝
ε1ε2
δ1ε2
δ2ε1
(u−u1)ε2γ1
−(u−u2)ε1γ2
⎞
⎟
⎟
⎟
⎟
⎠
,R2±=⎛
⎜
⎜
⎜
⎜
⎜
⎝
0
0
1
0
±a2
ρ2
⎞
⎟
⎟
⎟
⎟
⎟
⎠
.(2.23)
Here we introduced the following abbreviations
δ1=p1−p2
ρ−(u−u1)2
α1
,δ
2=p1−p2
ρ+(u−u2)2
α2
,
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Journal of Scientific Computing (2022) 93 :83 Page 9 of 60 83
ε1=(u−u1)2−a2
1
ρ1
,ε
2=(u−u2)2−a2
2
ρ2
,
γ1=α1(p1−p2)−ρa2
1
α1ρ1ρ,γ
2=−
α2(p1−p2)+ρa2
2
α2ρ2ρ.
The subscripts 1,2 refer to the corresponding phases and the pair (λC,RC)takes a special
role as we will see in a moment. We further want to investigate the fields and see whether
they are genuine nonlinear or linearly degenerated. The gradients of the eigenvalues with
respect to the given variables W=(α1,ρ
1,ρ
2,u1,u2)are given by
∇Wλ1±=⎛
⎜
⎜
⎜
⎜
⎜
⎝
0
±∂a1
∂ρ1
0
1
0
⎞
⎟
⎟
⎟
⎟
⎟
⎠
,∇WλC=
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
ρ1ρ2
ρ2(u1−u2)
α1c2(u1−u2)
ρ
α2c1(u2−u1)
ρ
c1
c2
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
,∇Wλ2±=⎛
⎜
⎜
⎜
⎜
⎜
⎝
0
0
±∂a2
∂ρ2
0
1
⎞
⎟
⎟
⎟
⎟
⎟
⎠
.(2.24)
We immediately obtain
∇Wλ1±·R1±=±
∂a1
∂ρ1
±a1
ρ1
=±1
ρ1
∂(ρ1a1)
∂ρ1
=±
a1
ρ1
G1,
∇Wλ2±·R2±=±
∂a2
∂ρ2
±a2
ρ2
=±1
ρ2
∂(ρ2a2)
∂ρ2
=±
a2
ρ2
G2.
Here we have introduced the fundamental derivative Gand the following relation holds
G=1+ρ
a
∂a
∂ρ ⇔1
ρ
∂(ρa)
∂ρ =a
ρG.
For more details on Gand its crucial influence on the flow we recommend Menikoff and
Plohr [50]andMüllerandVoss[53]. Throughout this work we assume G>0 and thus the
fields 1±and 2±are genuine nonlinear. Indeed for an ideal gas we have
G=γ+1
2,isentropic with γ>1
1,isothermal
We further have the remarkable property that
R1+,R1−⊥R2+,R2−and ∇Wλ1±∈R1+,R1−,∇Wλ2±∈R2+,R2−.
Thus the eigenvectors R1±and R2±span a four dimensional hyperplane in the state space
where αis constant. It remains to discuss the field C. Therefore we use
u−u1=(c1−1)u1+c2u2=−c2(u1−u2)=−c2w,
u−u2=c1u1+(c2−1)u2=c1(u1−u2)=c1w.
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83 Page 10 of 60 Journal of Scientific Computing (2022) 93 :83
We obtain
∇WλC·RC=
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
ρ1ρ2
ρ2(u1−u2)
α1c2(u1−u2)
ρ
α2c1(u2−u1)
ρ
c1
c2
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
·⎛
⎜
⎜
⎜
⎜
⎝
ε1ε2
δ1ε2
δ2ε1
(u−u1)ε2γ1
−(u−u2)ε1γ2
⎞
⎟
⎟
⎟
⎟
⎠
=w
ρ2(ρ1ρ2ε1ε2+α1α2ρ2δ1ε2−α1α2ρ1δ2ε1
−α1ρ1α2ρ2ε2γ1−α1ρ1α2ρ2ε1γ2)
=w
ρ2(c2w)2−a2
1(c1w)2−a2
2
+α1α2(c1w)2−a2
2p1−p2
ρ−(c2w)2
α1
−α1α2(c2w)2−a2
1p1−p2
ρ+(c1w)2
α2
−α2(c1w)2−a2
2α1(p1−p2)−ρa2
1
ρ
+α1(c2w)2−a2
1α2(p1−p2)+ρa2
2
ρ
=w
ρ2(c2w)2−a2
1(c1w)2−a2
2−α2(c1w)2−a2
2(c2w)2
−α1(c2w)2−a2
1(c1w)2+α2(c1w)2−a2
2a2
1+α1(c2w)2−a2
1a2
2
=0.
Thus this field is linearly degenerated and hence discontinuities associated to this field are
contact waves. In view of the above results it is clear that each character of the present fields is
independent of the flow. Note that up to now we cannot exclude situations where eigenvalues
coincide, i.e. have a multiplicity larger than one. Different possible phenomena related to
these special situations will be discussed in Sect. 5.
3 Degeneracies of the System and Admissibility Conditions
By construction, the system under consideration is symmetric hyperbolic in the sense of
Friedrichs [26]. The initial value problem (Cauchy problem) for such a system is well-posed
locally in time [4,15]. But the question of the solvability of the Cauchy problem in the large
for symmetric hyperbolic systems is still a challenging problem and, for example, nontrivial
interesting phenomena (such as the resonance effect) can arise. As mentioned in the previous
section we now will have to study under which conditions this system is hyperbolic in the
sense that we have real eigenvalues and a full set of eigenvectors, or, more precisely, under
which conditions certain degeneracies may occur. Furthermore another crucial point is, as
mentioned before, that up to now the order of the waves is not clear, i.e. the order of the
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Journal of Scientific Computing (2022) 93 :83 Page 11 of 60 83
eigenvalues. Thus we will briefly review results concerning the admissibility conditions for
discontinuities which will play an important role throughout this work.
3.1 Coincidence of Eigenvalues, Parabolic Degeneracy and Hyperbolic Resonance
We now want to investigate situations where the eigenvectors may become linearly dependent.
Considering the eigenvectors (2.23) it is obvious that the eigenvectors Ri±become linearly
dependent in each phase iff ai=0. Since we assume a strictly positive speed of sound in
each phase this is excluded and the four eigenvectors Ri±remain linearly independent. For
the field (λC,RC)we can state for a fixed field (λi±,Ri±)that
RC∈Ri±⇔λi±=λC.(3.1)
This can be seen by direct calculations which are given in the Appendix Afor convenience.
We want to discuss the consequences and interpretation of this case later on. We note that in
the extreme situation that three eigenvalues coincide, which is the maximum for reasonable
EOS, the eigenvector RCbecomes the null vector. This situation can only happen when λC
coincides with one eigenvalue of each phase and thus we have
(u−u1)2=a2
1and (u−u2)2=a2
2.
3.2 Admissibility Conditions for Discontinuities Revisited
In the previous Sect. 2we presented the eigenvalues (2.22) and eigenvectors (2.23)ofthe
system under consideration (2.15). As noted before different situations may occur due to
coinciding eigenvalues. Thus it is important to review suitable criteria to single out admissible
solutions. In particular we are interested in the case when discontinuities occur. For a given
and fixed state Wwe have for the eigenvalues
λ(1)(W)≤··· ≤ λ(p)(W)(3.2)
with p≤nand nbeing the total number of the possible eigenvalues. In our case we have
n=5 and further
λ(1)(W),...,λ
(p)(W)⊆{λ1±(W), λC(W), λ2±(W)}.
With this we allow for situations where the order of the eigenvalues changes and that eigen-
values may coincide. We start with the original work by Lax [45] and assume p=n.In
particular we consider a strictly hyperbolic system. Further we consider the states left and
right of a discontinuity denoted by WLand WR. According to Lax a k-shock with the speed
Ssatisfies the condition
λ(k)(WL)>S>λ
(k−1)(WL), (3.3a)
λ(k+1)(WR)>S>λ
(k)(WR). (3.3b)
From these inequalities we deduce that n−(k−1)characteristics impinge from the left of
the discontinuity and kfrom the right. Following the presentation given in Dafermos [15]
the situation can be generalized as follows. We consider the eigenvalues for the left and right
state
λ(1)(WL)≤ ··· ≤ λ(i−1)(WL)<S<λ
(i)(WL)≤ ··· ≤ λ(n)(WL),
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83 Page 12 of 60 Journal of Scientific Computing (2022) 93 :83
λ(1)(WR)≤ ··· ≤ λ(j)(WR)<S<λ
(j+1)(WR)≤··· ≤ λ(n)(WR),
with the agreement of λ(0)(WL)=−∞and λ(n+1)(WR)=∞. If the inequalities are
satisfied with i=jwe obtain the Lax condition (3.3) given above and the shock is called
compressive.Inthecaseofi<jthe shock is called overcompressive and for i>jthe shock
is called undercompressive. Now we want to deal with the non strict situation, i.e. there is
no well-defined ordering of the eigenvalues and they may coincide. In Keyfitz and Kranzer
[42] a generalization to non-strictly hyperbolic systems is given as follows
1) n+1 characteristics enter the shock and n−1leaveitor
2) n−1 characteristics enter and leave the shock whereas the remaining two are tangent to
the shock and belong to linearly degenerated fields.
When an i-shock has n−1 leaving characteristics and the corresponding eigenvectors fulfill
det R(1)(WL),..., R(i−1)(WL), WR−WL,R(i+1)(WR),...,R(n)(WR)= 0(3.4)
then the shock is called evolutionary, see again [15] or the book of Kulikovskii et al. [43]. We
want to end this brief review of admissibility conditions with the results presented in [1]. In
[1] the results obtained in the references given above are basically collected and generalized
to incorporate different situations that are of interest for non-strictly hyperbolic systems. Let
us consider the situation (3.2). Having peigenvalues we conclude that we have punknowns
on each side of the discontinuity and additionally the shock speed S, i.e. N=2p+1
unknowns. These Nunknowns can be determined as follows. Let us assume we have m
relations across the discontinuity, e.g. the jump conditions. Second, the unknowns should be
determined by the flow using the characteristics. Let the number of incoming characteristics
be i, the number of outgoing characteristics oand the number of coinciding characteristics be
c. The incoming and coinciding characteristics are determined by the past and thus provide
further information. Hence in order to determine the unknowns we demand
N=i+c+m.(3.5)
In [1] this is called evolutionarity condition and a discontinuity is called evolutionary if the
condition (3.5) holds (in agreement with the results above). This implies that a discontinuity is
evolutionary iff o=m−1, see [1]. A closely related concept was introduced by Freistühler in
[24]. The idea is quite similar as again a linearised problem is investigated. The eigenvalues
are distinguished, using their relation to the speed of the discontinuity, into slow and fast
characteristics. In this sense outgoing characteristics are slow characteristics on the left
side of the discontinuity and fast ones on the right side. For the incoming characteristics
the situation is reversed. Coinciding characteristics move at the characteristic speed of the
discontinuity.
4 Wave Relations
We now want to obtain the relations that are valid across the different types of waves. In
particular we derive the Riemann invariants which are constant across rarefaction waves and
the Rankine Hugoniot jump conditions across discontinuities.
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4.1 Rarefaction Waves
Given a smooth solution we may apply a nonlinear transformation of the variables aiming
to simplify the system using another appropriate choice of variables. Such a special set of
variables is given by the Riemann invariants,cf.[15,22,75]. The Riemann invariants for
R1±and R2±can be calculated using
dα1
ds=0,dρμ
ds=±aμ
ρμ
Gμ−1
,dρν
ds=0,duμ
ds=±±aμ
ρμ
Gμ−1aμ
ρμ
and duν
ds=0.
Here μ, ν ∈{1,2},μ = νwhere μ=1for R1±and μ=2for R2±. The relations for ρμ
and uμcan be combined to
duμ
dρμ
=±
aμ
ρμ
.
Thus we obtain the following invariants
α1=const ., R±=uμ±aμ
ρμ
dρμ,ρ
ν=const .and uν=const .(4.1)
Furthermore, the slope inside a left rarefaction wave is given by
dx
dt=x
t=λμ−=uμ−aμ(4.2)
and hence we obtain that the solution inside the rarefaction fan is given by
uμ=x
t+aμand F(ρ) =uμ−uμ, L+ρ
ρμ,L
aμ
ρμ
dρμ=0.(4.3)
Here, ρis obtained as the root of F(ρ ). Similar we obtain the results for right rarefaction
waves corresponding to λμ+
dx
dt=x
t=λμ+=uμ+aμ,uμ=x
t−aμand F(ρ) =uμ, R−uμ−ρμ,R
ρ
aμ
ρμ
dρμ=0.
(4.4)
Although the Riemann invariants (4.1) state that ρνand uνremain constant, we prefer the
notion that these quantities are not affected by the rarefaction wave. As we see later on they
might change due to other waves.
4.2 Shock Waves
In the presence of discontinuities, such as shock waves, the situation is different. However,
since the present system is conservative, corresponding Rankine Hugoniot jump conditions
F(W)=SW.
have to hold at a shock with speed S, see for example [15,46]. In the present case we obtain
the following conditions across discontinuities
α1ρu=Sα1ρ,(4.5a)
α1ρ1u1=Sα1ρ1,(4.5b)
ρu=Sρ,(4.5c)
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α1ρ1u2
1+α2ρ2u2
2+α1p1+α2p2=Sα1ρ1u1+α2ρ2u2,(4.5d)
1
2u2
1−u2
2+1−2=Su1−u2.(4.5e)
The jump conditions (4.5a)–(4.5c) can be reformulated to
α1ρ(u−S)=0,(4.6)
α1ρ1(u1−S)=0,(4.7)
ρ(u−S)=0.(4.8)
Using the third jump condition (4.8) we obtain for the first equation (4.6)
ρ(u−S)α1=0.
We introduce the abbreviations
Q=−ρ(u−S), Q1=−ρ1(u1−S)and Q2=−ρ2(u2−S)
for the mixture mass flux and the mass fluxes of the phases, respectively. In particular we
have
Q=−ρ(u−S)=−(ρu−ρS)=−(α1ρ1u1+α2ρ2u2−(α1ρ1+α2ρ2)S)
=−(α1ρ1(u1−S)+α2ρ2(u2−S))=α1Q1+α2Q2.(4.9)
Thus we can derive a jump condition for the partial mass flux of the second phase, using the
continuity of the mixture mass flux (4.8) and the continuity mass flux of the first phase (4.7),
i.e.
0=Q(4.9)
=α1Q1+α2Q2(4.7)
=α2Q2.(4.10)
Using the jump conditions for the partial mass fluxes (4.7)and(4.10) we obtain for the fourth
jump condition (4.5d)
α1ρ1u2
1+α2ρ2u2
2+α1p1+α2p2=Sα1ρ1u1+α2ρ2u2
⇔0=α1ρ1u1(u1−S)+α2ρ2u2(u2−S)+α1p1+α2p2
=α1ρ1(u1−S)u1+α1p1+α2ρ2(u2−S)u2+α2p2
=−α1Q1u1+α1p1−α2Q2u2+α2p2.(4.11)
The fifth jump condition (4.5e) can be reformulated as follows
1
2u2
1−u2
2+1−2=Su1−u2
⇔0=1
2u2
1−u2
2−S(u1−u2)+1−2
=1
2(u1−u2)(u1+u2−2S)+1−2
=1
2(u1−u2)(u1−S+u2−S)+1−2
=1
2(u1−u2)(u1−S)+1+1
2(u1−u2)(u2−S)−2
=1
2(u1−S)2+1+1
2(S−u2)(u1−S)
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+−1
2(u2−S)2−2+1
2(u1−S)(u2−S)
=1
2(u1−S)2+1−1
2(u2−S)2+2.(4.12)
Summarizing we have the following system of jump conditions
ρ(u−S)α1=0,(4.13a)
α1ρ1(u1−S)=0,(4.13b)
ρ(u−S)=0,(4.13c)
−α1Q1u1+α1p1−α2Q2u2+α2p2=0,(4.13d)
1
2(u1−S)2+1−1
2(u2−S)2+2=0.(4.13e)
Note that (4.13c) may be replaced by (4.10). Further, equations (4.13d)and(4.13e) may also
be reformulated to
−Qu+ρc1c2w2+p=0,(4.14)
1
2Q1
ρ12−1
2Q2
ρ22+1−2=0.(4.15)
To give the complete picture we briefly summarize the jump conditions according to the
system (2.14) in terms of the mixture EOS
ρ(u−S)α1=0,(4.16a)
ρc1(u−S)+ρEw=0,(4.16b)
ρ(u−S)=0,(4.16c)
ρ(u−S)u+p+ρwEw=0,(4.16d)
w(u−S)+Ec=0.(4.16e)
We also want to emphasize the analogy to the jump conditions of the Euler equations, cf. [15,
78]. Equation (4.13d) is basically a volume fraction weighted combination of an individual
momentum jump condition for each phase as it appears in the Euler equations. Moreover, a
single jump term in (4.13e) agrees with the jump bracket of the energy equation for the Euler
equations.
4.3 Exploiting the Jump Conditions for a Lax Shock
In the following we assume that we have a Lax-shock, i.e. with no tangential eigenvalues.
In particular this implies u= Swith Sbeing the shock speed. Indeed, as we will see this
cannot happen for shock waves. From equation (4.13a) we thus have that α1is continuous
across a shock corresponding to the eigenvalues λ1±and λ2±. Up to now we have made no
further assumption on α1. With the aim to further exploit and simplify the jump conditions
(4.13a)-(4.13e) we assume α1∈(0,1)from now on. First we have, due to the continuity
of α1(4.13a), that also the mass fluxes Q1and Q2are continuous, see (4.13b)and(4.10).
Hence the partial mass fluxes may be written outside the jump brackets in (4.15), i.e.
Q2
1
21
ρ2
1−Q2
2
21
ρ2
2+1−2=0.(4.17)
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Further we can use the continuity of the partial mass fluxes and write with μ∈{1,2}
uμ,R−S=− Qμ
ρμ,R
,uμ,L−S=− Qμ
ρμ,L
⇒uμ=−Qμ1
ρμ.(4.18)
Using (4.18) together with the momentum jump condition (4.11)weget
0=α1Q2
11
ρ1+p1+α2Q2
21
ρ2+p2.(4.19)
Thus equations (4.17)and(4.19) form a linear system for the partial mass fluxes
⎛
⎜
⎜
⎜
⎝
α11
ρ1α21
ρ2
1
21
ρ2
1−1
21
ρ2
2⎞
⎟
⎟
⎟
⎠
=:M
·Q2
1
Q2
2=−⎛
⎝
α1p1+α2p2
1−2⎞
⎠.(4.20)
The inverse is given by
M−1=1
det(M)
⎛
⎜
⎜
⎜
⎜
⎝
−1
21
ρ2
2−α21
ρ2
−1
21
ρ2
1α11
ρ1⎞
⎟
⎟
⎟
⎟
⎠
,(4.21)
det(M)=−
1
2α11
ρ11
ρ2
2+α21
ρ2
11
ρ2.(4.22)
Note that across a shock the phase densities necessarily jump and thus the determinant is not
equal to zero, see Appendix A. Hence we obtain for the partial mass fluxes
Q2
1=1
det(M)1
21
ρ2
2(α1p1+α2p2)+α21
ρ21−2,(4.23)
Q2
2=1
det(M)1
21
ρ2
1(α1p1+α2p2)+α11
ρ12−1.(4.24)
Furthermore, these partial mass fluxes are also related to each other using
Q1=−ρ1(u1−S)∧Q2=−ρ2(u2−S)⇒u1−u2=−Q1
ρ1
+Q2
ρ2
.
Thus one may express one partial mass flux by the other. Up to now we only have the squares
of the mass flux. The correct sign of the corresponding root is given by the Lax condition for
the present shock. The Lax criterion states for a shock in phase μthat
λμ±(WL)>S>λ
μ±(WR)
where WLand WRare the left and right states adjacent to the shock. For a left shock in
phase μ∈{1,2}we have
uμ,L−aμ,L>S>uμ, R−aμ,R
⇔uμ,L−S>aμ,L∧aμ, R>uμ,R−S
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⇔−ρμ,L(uμ,L−S)<−ρμ,Laμ,L∧−ρμ, R(uμ,R−S)>−ρμ, Raμ,R
⇔−ρμ,Raμ,R<Qμ<−ρμ,Laμ, L<0.
Similar we obtain for a right shock
uμ,L+aμ,L>S>uμ, R+aμ,R
⇔uμ,L−S>−aμ,L∧−aμ, R>uμ,R−S
⇔−ρμ,L(uμ,L−S)<ρ
μ,Laμ,L∧−ρμ, R(uμ,R−S)>ρ
μ,Raμ, R
⇔ρμ,Laμ,L>Qμ>ρ
μ,Raμ, R>0.
Thus the partial mass flux has a strict sign. Hence across a λ1±-shock the sign of Q1is
determined and for a λ2±-shock the sign of Q2is given, respectively. Therefore we choose
the following
⎧
⎪
⎨
⎪
⎩
Q2=ρ2Q1
ρ1+w,λ
1±−Shock,
Q1=ρ1Q2
ρ2−w,λ
2±−Shock.
(4.25)
The values for the densities and wmay assumed to be given, i.e. the values on one side of
the shock. Once we have obtained the partial mass flux we can eliminate a further unknown
using the velocity jump condition (4.18)
uμ=−Qμ1
ρμ,μ∈{1,2}.
4.4 Entropy Inequality
Given the (mathematical) entropy inequalities (2.16)and(2.17) we have in the presence of
a discontinuity
0≥
2
i=1
−Sαiρiei+1
2u2
i+αiρiuihi+1
2u2
i(4.26)
for the isentropic case and
0≥
2
i=1
−Sαiρiei−Ts
i+1
2u2
i+αiρiuigi+1
2u2
i(4.27)
for the isothermal case, respectively. With algebraic manipulations using the continuity of
the partial mass fluxes (4.7), (4.10) and the jump condition for the momentum (4.13d)we
obtain for both cases
0≥−
2
i=1
αiQii+1
2(ui−S)2.(4.28)
Note that the bracket terms correspond to the terms in the jump condition for the relative
velocity (4.13e). Thus we can replace the term corresponding to one phase by the other if
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83 Page 18 of 60 Journal of Scientific Computing (2022) 93 :83
(4.13e) holds and obtain
0≥−Q1+1
2(u1−S)2or 0 ≥−Q2+1
2(u2−S)2.(4.29)
In particular the inequality for the isothermal case is perfectly analogous to the one used in
[37].
4.5 Contact Wave
The wave corresponding to the linearly degenerated field (λC,RC)is a contact wave and
because of
0=∇
WλC·RC=∇
Wu·RC
it is immediately clear that we have the Riemann invariant
u=const .(4.30)
The characteristic condition for the contact wave is λC(WL)=S=λC(WR)which further
gives for the velocity of the contact S=u. Thus we have Q=−ρ(u−S)=0 and hence
α1may be non-zero. Indeed, since α1is continuous across the other waves the jump of α1
is given by the initial data. From the continuity of the partial mass fluxes we obtain
0=Q=α1Q1+α2Q2
0=α1Q1=−α1ρ1(u1−u)=−α1ρ1c2w=−ρc1c2w.(4.31)
For the mixture momentum at the contact we obtain using (4.14)and(4.31)
0=−Qu+ρc1c2w2+p=ρc1c2w2+p=ρc1c2ww+p.(4.32)
From the jump condition for the relative velocity (4.13e)weget
0=1
2(u1−S)2+1−1
2(u2−S)2+2
=1
2(c2w)2−(c1w)2+1−2
=1
2c2
2−c2
1w2+1−2
=1
2(c2−c1)w2+1−2.(4.33)
Thus we have the following equations at the contact
u=0,(4.34a)
ρc1c2w=0,(4.34b)
ρc1c2ww+p=0,(4.34c)
1
2(c2−c1)w2+1−2=0.(4.34d)
Note that ρc1c2w2is a dynamic mixture pressure related to the relative velocity. By intro-
ducing the generalized total mixture pressure ¯p=ρc1c2w2+pas the sum of the dynamic
pressure and the static mixture pressure pwe can rewrite (4.34c)as
0=ρc1c2ww+p=ρc1c2w2+p=¯p.(4.35)
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We again give the jump conditions in terms of the mixture EOS for completeness:
u=0,(4.36a)
ρEw=0,(4.36b)
p+ρwEw=0,(4.36c)
Ec=0.(4.36d)
Finally, it is easy to see from (4.29)andu=Sthat the entropy inequality is fulfilled with
the right side being identically zero.
5 Wave Configurations and Relations
In this section we want to study particular wave configurations. The crucial point is, as
mentioned before, that up to now the order of the waves is not clear, i.e. the order of the
eigenvalues. In particular eigenvalues may coincide in certain points. Thus the reviewed
results concerning the admissibility conditions for discontinuities will play a crucial role
throughout this section.
5.1 Contact
We first want to discuss phenomena related to the linearly degenerates field (λC,RC), i.e.
the contact wave. Obviously we can determine the position of the contact wave due to the
fact that λC=uis the convex combination of the individual phase velocities, i.e.
min{u1,u2}≤u=c1u1+c2u2≤max{u1,u2}
⇔min{λ1−,λ
2−}<λ
C<max{λ1+,λ
2+}.
Let us now consider the situation that we have an isolated shock (w.l.o.g.) corresponding to
λμ−moving with speed S. A priori it is not obvious at all whether we may encounter the
situation that u=S. In the following we will show that this is not possible. In this situation
we have a discontinuity with the tangential eigenvalues λ−
C=λ+
C=S. Here a superscript −
refers to the state W−left of the discontinuity and a superscript +refers to the state W+right
of the discontinuity, respectively. According to the results obtained by Keyfitz et al. [42] cited
above this corresponds to the second situation. Thus we would further need n−1 incoming
and n−1 outgoing characteristics, with n=5 in our case. If we consider the results given
in [1] we obtain the same results with m=5 jump conditions, c=2 tangential eigenvalues,
N=2n+1=2·5+1=11 unknowns and thus i=4 incoming and o=m−1=4
outgoing characteristics. Hence it is clear that for an evolutionary discontinuity we must not
have further coinciding eigenvalues. Let us picture the situation more precisely and assume
w.l.o.g. that we have a shock corresponding to λ1−. Thus we would have
λ−
1−>S>λ
+
1−and λ−
C=λ+
C=S.(5.1)
We can conclude immediately that λ−
1+>Sand S>λ
−
2−. Using the continuity of α1Q1
it follows that λ+
1+>Sand hence S>λ
+
2−. Since we now already have four ingoing
characteristics it follows that λ−
2+and λ+
2+must be outgoing characteristics. The situation
I={λ−
1−,λ
+
1−,λ
−
1+,λ
+
2−},
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Fig. 1 Shock with u=S(example)
C={λ−
C,λ
+
C},
O={λ+
1+,λ
−
2−,λ
−
2+,λ
+
2+}
is depicted as an example in Fig.1.
So clearly this situation is excluded, since the information related to the eigenvalue λ2+
is coming out of the discontinuity. Another short argument would be to say that we are in
a strictly hyperbolic situation on both sides of the discontinuity and hence only a classical
wave is allowed. Note that this is not by far that obvious for systems with multiple linearly
degenerated fields. Thus we can state that in our system an evolutionary discontinuity with
u=Sis a contact and vice versa. Even more important is the statement that due to u= S
for a shock we always have the continuity of αacross the shock.
This also gives another view on the results obtained above in Sect. 3and in particular the
situation of coinciding eigenvectors described by (3.1). In this particular situation αwill not
even jump across RCand thus remains constant in the complete fan. Hence we can exclude
this situation by simply prescribing different values for αinitially. Or in other words this
situation may only occur when αL=αRholds for the initial states of the Riemann problem.
We therefore could interpret the case (3.1) as a consequence of a redundant αequation. With
αconstant everywhere we can reformulate the system as
∂
∂tα=0,
∂
∂t
˜
W+∂
∂x
˜
F(α, ˜
W)=0
with ˜
W=(w2,w
3,w
4,w
5)T
and ˜
F(α, ˜
W)=(F2(α, ˜
W), F3(α, ˜
W), F4(α, ˜
W), F5(α, ˜
W)))T.
The situation λi±=λCthen corresponds to the case of hyperbolic resonance discussed by
Isaacson and Temple [39]. In the literature you also find the phrasing parabolic degeneracy
or weak hyperbolicity for missing eigenvectors, but we think resonance is the term best suited
here.
123
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Fig. 2 Contact inside a rarefaction (example)
5.1.1 Contact Inside Rarefaction
Let us assume that a contact lies inside a rarefaction wave, see e.g. the sketch shown in Fig. 2.
It cannot be attached to one side (or both) of the rarefaction, because then we would have
the situation of hyperbolic resonance discussed above (i.e. coincidence of eigenvalues and
eigenvectors). Thus the contact will tear the rarefaction wave into two parts. Assume we have
a rarefaction wave corresponding to λμ−and thus we consider the following situation
λ−
μ−<λ
−
C=S=λ+
C<λ
+
μ−.
According to the results obtained above concerning the admissibility of discontinuities we
have c=2 coinciding characteristics and thus need i=o=4 characteristics going in
and out. From the given relation for the eigenvalues λμ−and λCwe can directly conclude
λ+
μ+>S. We can then conclude that that λ+
ν−<S. Using the inequalities for λμ−we see
that Qμ<0 across the contact. Since 0 =Q=αμQμ+ανQνwe thus have Qν>0. Using
Qμ<0 we yield λ−
μ+>S.FromQν>0 we yield λ−
ν−<Sand hence finally λ+
ν+<S.
Altogether we therefore obtain
I={λ−
μ+,λ
+
ν−,λ
−
ν+,λ
+
ν+},C={λ−
C,λ
+
C}and O={λ−
μ−,λ
+
μ−,λ
+
μ+,λ
−
ν−}.
Such a discontinuity violates the admissibility criteria given above due to the fact, that the
characteristics λ−
μ−and λ+
μ−do not contribute any information to the discontinuity. Note that
as mentioned before the phrasing outgoing characteristic seems to be not quite suited here
for these two eigenvalues. However, the results remain valid and one could use the notation
of slow and fast characteristics with respect to the side of the discontinuity, cf. [24]. In this
sense the situation is analogue to that of a contact coinciding with a shock.
5.2 Overlapping Rarefaction Waves
A possible wave configuration might be two rarefaction waves that overlap, see Fig.3.
In this situation one rarefaction wave belongs to phase one and the other to phase two,
respectively. The rarefaction fans are cones given by
C1=(t,x)|0<t,S1,Lt≤x≤S1,Rt
C2=(t,x)|0<t,S2,Lt≤x≤S2,Rt 123
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83 Page 22 of 60 Journal of Scientific Computing (2022) 93 :83
Fig. 3 Overlapping rarefaction waves (example)
C∗=C1∩C2
In the case of an empty intersection we have the classical situation and the solution is obtained
using the corresponding eigenvector. If however, C∗=∅we have
C∗=(t,x)|0<t,max{S1,L,S2,L}t≤x≤min{S1,R,S2,R}t
Due to the special structure of the eigenvectors the obtained invariants remain unchanged and
the formulas (4.3)and(4.4) stay valid. Indeed if we write down the (homogeneous) system
using the primitive variables W=(α1,ρ
1,ρ
2,u1,u2)we have the Jacobian (2.20). For a
rarefaction wave α1remains constant and thus the system simplifies to
∂ρ1
∂t+∂ρ1u1
∂x=0,(5.2a)
∂ρ1u1
∂t+∂ρ1u2
1+p1
∂x=0,(5.2b)
∂ρ2
∂t+∂ρ2u2
∂x=0,(5.2c)
∂ρ2u2
∂t+∂ρ2u2
2+p2
∂x=0.(5.2d)
Hence the system decouples into two barotropic Euler systems for each phase and the rarefac-
tion waves can be obtained individually. The solution is then obtained as the superposition
of the individual solutions.
5.3 Shock Interacting with a Rarefaction Wave
Another situation that can occur is a shock which lies inside a rarefaction fan, see Fig.4.
Let us assume for the moment that we have a left shock corresponding to phase μ∈{1,2}
and a left rarefaction corresponding to phase ν∈{1,2},μ = ν. Clearly, the phase μis
only affected by the shock wave due to the structure of the eigenvector corresponding to
the rarefaction wave. For the phase νthe situation is more complicated. We have for the
eigenvalue λν−=ξ:= x/tcorresponding to the rarefaction wave
λν−(WL)≤ξ≤λν−(WR)
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Fig. 4 Shock inside a rarefaction wave - Case (i)
where WLand WRdenote the states left and right of the rarefaction wave. Let us denote the
shock speed with Sand quantities left of the shock are denoted with a superscript −and a
+when they are on the right, respectively. There are four cases which are possible in this
situation
(i) λ−
ν−=S=λ+
ν−,
(ii) λ−
ν−<S<λ
+
ν−,
(iii) λ−
ν−=S<λ
+
ν−,
(iv) λ−
ν−<S=λ+
ν−.
Further we demand the Lax condition λ−
μ−>S>λ
+
μ−.
Case (i): The first case can be excluded since it implies linear degeneracy of the field
(λν−,Rν−),see[23,42]. This is obviously not the case as long as we have Gν= 0, which
we may assume for our EOS. Further discussion of the fundamental derivative can be found
in [50,53].
Case (ii): Considering the characteristics in the second case we obviously have
λ−
ν−<S<λ
−
μ−and λ+
μ−<S<λ
+
ν−.
Thus we directly obtain
S<λ
+
ν−<u+
ν<λ
+
ν+and S<λ
−
μ−<u−
ν<λ
−
μ+.
Due to the continuity of the mass fluxes we further yield
u+
ν>S⇔0>Q+
ν=Q−
ν⇔u−
ν>S⇒λ−
ν+>S,
u−
μ>S⇔0>Q−
μ=Q+
μ⇔u+
μ>S⇒λ+
μ+>S.
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Fig. 5 Shock inside a rarefaction wave - Case (ii)
Since uis a convex combination of the phase velocities we also conclude λ−
C>Sand
λ+
C>S. Summarizing we have the following situation
I={λ−
μ−,λ
+
μ−,λ
−
μ+,λ
−
ν+,λ
−
C},C=∅,O={λ−
ν−,λ
+
ν−,λ
+
ν+,λ
+
μ+,λ
+
C}.
Therefore this situation is not admissible. According to the admissibility criteria we have
N=11 unknowns m=5 equations and thus we would need i=6 incoming and o=4
outgoing characteristics.
Case (iii): Now, for the third case we have
λ−
ν−<S<λ
−
μ−<λ
−
μ+
and λ+
μ−<S=λ+
ν−<λ
+
ν+.
We have for the mass flux of phase μthat Qμ<0. Thus we have u+
μ>Sand hence
λ+
μ+>S. For the mass flux of phase νwe obtain
λ−
ν−=u−
ν−a−
ν<S=λ+
ν−=u+
ν−a+
ν
⇔u−
ν−S<a−
ν∧a+
ν=u+
ν−S
⇔−ρ−
ν(u−
ν−S)>−ρ−
νa−
ν∧−ρ+
ν(u+
ν−S)=−ρ+
νa+
ν
⇔−ρ−
νa−
ν<Qν=−ρ+
νa+
ν<0.
Thus we also have Q<0 and hence
0>Q=−ρ−(u−−S)⇔λ−
C=u−>S
0>Q=−ρ+(u+−S)⇔λ+
C=u+>S.
We therefore obtain that λ−
Cis an ingoing characteristic and λ+
Can outgoing characteristic. In
the situation under consideration we have as before N=11 unknowns and m=5 equations.
Additionally we have c=1 coinciding wave and thus we need i=5 incoming and o=4
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Fig. 6 Shock inside a rarefaction wave - Case (iii)
Fig. 7 Shock inside a rarefaction wave - Case (iv)
outgoing characteristics. This implies that λ−
ν+is also an ingoing characteristic. Summing
up we have the following situation
I={λ−
μ−,λ
+
μ−,λ
−
μ+,λ
−
ν+,λ
−
C},C={λ+
ν−}and O={λ+
μ+,λ
−
ν−,λ
+
ν+,λ
+
C}.
Case (iv): Finally,forthefourthcasewehave
λ−
ν−=S<λ
−
μ−and λ+
μ−<S<λ
+
ν−.
Using similar arguments as for the third case we obtain the following situation
I={λ−
μ−,λ
+
μ−,λ
−
μ+,λ
−
ν+,λ
−
C},C={λ−
ν−}and O={λ+
μ+,λ
+
ν−,λ
+
ν+,λ
+
C}.
Now we have two possible configurations which seem to be allowed. A priori it is not
obvious whether one of these cases can be ruled out or if both may occur. Thus we will make
use of the energy inequality (4.28) to investigate both cases.
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Energy Inequality Case (iii): As in (4.29) we use the entropy inequality for the phase ν
which defines the rarefaction, i.e.
0≥−Qν+1
2(uν−S)2.
From the characteristics λ−
C,λ
+
Cwe directly conclude Q<0 and hence we expect
0≥ν+1
2(uν−S)2(5.3)
for this configuration in order to be admissible. We rewrite the kinetic energy in terms of the
mass flux and assume the right state to be given. Thus it is possible to write the jump bracket
as a function in the density ρ−
νof the left side of the discontinuity, i.e.
0≥ν+1
2(uν−S)2=+
ν−ν(ρ−
ν)+Q2
ν
21
(ρ+
ν)2−1
(ρ−
ν)2,
with Qν=−ρ+
νaν(ρ+
ν).
Investigating this function gives the following
f(ρ) := +
ν−ν(ρ) +Q2
ν
21
(ρ+
ν)2−1
ρ2,f(ρ+
ν)=0,
f(ρ) =−
aν(ρ)2
ρ+Q2
ν
ρ3,f(ρ+
ν)=0,
f(ρ ) =−2aν(ρ)2
ρ2Gν+3aν(ρ)2
ρ2−3Q2
ν
ρ4,f(ρ +
ν)<0.
In order to get further insight we need to discuss the function g(ρ) =ρaν(ρ ) which is
sometimes called Lagrangian wave speed [50]. More precisely, we already discussed it
investigating the characteristic fields and we have for phase ν
g(ρ) =ρaν(ρ ) > 0andg(ρ) =aνGν>0,∀ρ>0.(5.4)
Thus we can conclude for all 0 <ρ<ρ
+
ν
f(ρ) =−
aν(ρ)2
ρ+Q2
ν
ρ3=1
ρ3Q2
ν−ρ2aν(ρ)2>0,
and f(ρ ) =−2aν(ρ)2
ρ2Gν−3
ρ4Q2
ν−ρ2aν(ρ)2<0.
Due to this monotonicity behaviour, ρ−
ν<ρ
+
νand since f(ρ+
ν)=0 we can conclude
ν+1
2(uν−S)2=f(ρ−
ν)<0.
Hence configuration (iii) respects the mathematical entropy inequality.
Energy Inequality Case (iv): We use a similar argumentation as before. Now we assume
the left state to be given and write the jump bracket as a function in the density ρ+
νof the
right side of the discontinuity, i.e.
0≥ν+1
2(uν−S)2=ν(ρ+
ν)−−
ν+Q2
ν
21
(ρ+
ν)2−1
(ρ−
ν)2,
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Journal of Scientific Computing (2022) 93 :83 Page 27 of 60 83
with Qν=−ρ−
νaν(ρ−
ν).
Investigating this function gives the following
f(ρ) := ν(ρ +
ν)−−
ν+Q2
ν
21
ρ2−1
(ρ−
ν)2,f(ρ−
ν)=0,
f(ρ) =aν(ρ )2
ρ−Q2
ν
ρ3,f(ρ−
ν)=0,
f(ρ ) =2aν(ρ)2
ρ2Gν−3aν(ρ)2
ρ2+3Q2
ν
ρ4,f(ρ −
ν)>0.
Again we use the monotonicity of the Lagrangian wave speed (5.4) to conclude for all
0<ρ
−
ν<ρ
f(ρ) =aν(ρ )2
ρ−Q2
ν
ρ3=1
ρ3ρ2aν(ρ)2−Q2
ν>0,
Due to this monotonicity behaviour, ρ−
ν<ρ
+
νand since f(ρ−
ν)=0 we can conclude
ν+1
2(uν−S)2=f(ρ+
ν)>0.
Hence configuration (iv) violates the energy inequality and finally we can say that only case
(iii) is possible. The treatment for the case considering λμ+and λν+is completely analogous
showing that the shock and the rarefaction characteristics coincide now on the left side.
If we reconsider the arguments above it becomes clear that the side on which the rarefaction
characteristic coincides with the discontinuity is basically defined by the mixture mass flux
Q. Thus we can extend the above results to the cases of a λμ+shock interacting with a λν−
rarefaction or the analogous case with λμ−and λν+.For Q>0inthefirstcaseweexemplary
have the situation
I={λ+
μ−,λ
−
μ+,λ
+
μ+,λ
−
ν+,λ
+
C},C={λ−
ν−}and O={λ+
ν−,λ
+
ν+,λ
−
μ−,λ
−
C}.
and for Q<0
I={λ+
μ−,λ
−
μ+,λ
+
μ+,λ
−
ν+,λ
+
C},C={λ+
ν−}and O={λ−
ν−,λ
+
ν+,λ
−
μ−,λ
−
C}.
The other cases can be discussed as before. In particular the case Q=0 is not admissible
since it would contradict the genuine non-linearity of the field λν.
5.4 Shock Resonance
Due to the already mentioned result by Freistühler [23] and the results above we can exclude
multiple eigenvalues near a shock discontinuity. However, we have to discuss the situation
that we have a shock in each phase and both move at the same speed S. Hence we first
consider the situation
λ−
μ−>S>λ
+
μ−and λ−
ν−>S>λ
+
ν−.
We directly verify u−
μ>S,u−
ν>Sand hence u−>S. Due to the continuity of Qwe also
have u+>S. Furthermore we have λ−
μ+>Sand λ−
ν+>Sand this gives in total seven
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83 Page 28 of 60 Journal of Scientific Computing (2022) 93 :83
incoming characteristics, i.e.
I={λ−
μ−,λ
+
μ−,λ
−
ν−,λ
+
ν−,λ
−
μ+,λ
−
ν+,λ
−
C}.
There is no coinciding characteristic and we have three outgoing characteristics
O={λ+
μ+,λ
+
ν+,λ
+
C}.
Thus such a configuration violates the admissibility conditions. Let us close with the case
λ−
μ−>S>λ
+
μ−and λ−
ν+>S>λ
+
ν+.
We directly verify u−
μ>S,S>u+
νand hence we have
λ−
μ+>S,0>Qμand S>λ
+
ν−,Qν>0.
From the continuity of the partial mass fluxes we conclude u+
μ>Sand S>u−
νand hence
we already have the following six incoming and two outgoing characteristics
I={λ−
μ−,λ
+
μ−,λ
−
ν+,λ
+
ν+,λ
−
μ+,λ
+
ν−},O={λ+
μ+,λ
−
ν−}.
Thus the situation of λ−
C=λ+
C=Scan be excluded. But due to the continuity of the mass
flux Qthis implies that for Q<0or Q>0 we have seven incoming and three outgoing
characteristics and thus this situation is also not admissible. With this we have discussed
most wave patterns that can appear in a Riemann problem for the model under consideration,
including the interaction of two characteristic families.
6 Related Models
It is further interesting to discuss how other well established models are related to the studied
system at hand. In particular we discuss Kapila’s limit of the barotropic SHTC model and
the relation to the Baer–Nunziato model in the following.
6.1 Instantaneous Relaxation Limit of the Barotropic SHTC Model
As it is noted above there are two relaxation terms in the system (2.1). One for the pressure
relaxation (2.2) and one for the velocity relaxation (2.3), respectively. Since in real processes
these relaxation processes can be quite fast, it is useful to obtain asymptotic limits of the
solution for small values of the relaxation times. In the paper [51] the simplified equations
for instantaneous pressure and relative velocity relaxations are accurately derived by the
asymptotic analysis as a relaxation limit of the general Baer–Nunziato two-pressure two-
velocity model for two-phase compressible fluid flows. In this section, based on the results
obtained in [51] we derive a reduced SHTC system of PDEs for the instantaneous velocity
and pressure relaxation. Note that it seems intuitive that the time scale of pressure relaxation
is less than the time scale of velocity relaxation, because the physical mechanism of pressure
relaxation is the pressure waves propagation. Thus, one can consider a possibility to study
the instantaneous relaxation limit separately for the pressure and then for the velocity. If to
consider only the pressure relaxation limit then we arrive at a single pressure two-velocity
system of governing equations. It turns out that this PDE system is not hyperbolic. If we
want to deal with hyperbolic equations we should consider instantaneous velocity relaxation
together with the instantaneous pressure relaxation.
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Journal of Scientific Computing (2022) 93 :83 Page 29 of 60 83
We will not repeat the rigorous asymptotic analysis as done in the [51], but will rely on
the assumptions that follow from the rigorous theory. So, for the sake of simplicity we start
with the instantaneous velocity relaxation. Let us consider equation (2.1e)
∂wk
∂t+∂(wlul+Ec1)
∂xk
+ul∂wk
∂xl
−∂wl
∂xk=−Ewk
θ2
=−
c1c2wk
θ2
,(6.1)
An asymptotic expansion of the relative velocity wk=wk
0+θ2wk
1+... for small θ2gives us
wk
0=0,w
k
1=− 1
c1c2
∂ec1
∂xk
.(6.2)
For our purpose we only need to account for the zeroth order term of the expansion wk
0=0
which gives us a single velocity approximation of the model. The second term of the expansion
wk
1=− 1
c1c2
∂ec1
∂xkcan be interpreted as a phase diffusion Fick’s law. Indeed this can be seen
when the expansion is inserted into the equation for the mass fraction (2.1b) which results in
0=∂ρc1
∂t+∂(ρc1uk+ρEwk)
∂xk
=∂ρc1
∂t+∂(ρc1uk)
∂xk
+∂(ρc1c2wk)
∂xk
=∂ρc1
∂t+∂(ρc1uk)
∂xk
−∂
∂xkρθ2
∂ec1
∂xk.
Now we substitute wk=0 into the system (2.1) and remove the equation for the relative
velocity. This gives a single velocity model for two-phase flows
∂ρui
∂t+∂(ρuiuk+pδik)
∂xk
=0,(6.3a)
∂ρ
∂t+∂ρuk
∂xk
=0,(6.3b)
∂ρc1
∂t+∂ρc1uk
∂xk
=0,(6.3c)
∂ρα1
∂t+∂ρα1uk
∂xk
=−
ρφ
θ1
,(6.3d)
where uiis the single velocity of the flow.
The above system (6.3) is equivalent to
∂ρui
∂t+∂(ρuiuk+(α1p1+α2p2)δik)
∂xk
=0,(6.4a)
∂α1ρ1
∂t+∂α1ρ1uk
∂xk
=0,(6.4b)
∂α2ρ2
∂t+∂α2ρ2uk
∂xk
=0,(6.4c)
∂α1
∂t+uk∂α1
∂xk
=p1−p2
θ1
.(6.4d)
The latter is obtained with the use of the definitions c1=α1ρ1/ρ, ρ =α1ρ1+α2ρ2.
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Let us now consider the instantaneous pressure relaxation limit for (6.4) assuming θ1→0.
Note that θ1→0givesus p1=p2as a zeroth order approximation, but it is not correct to
simply put p1=p2=Pinto the equations, because this would give us
∂α1
∂t+uk∂α1
∂xk
=0.(6.5)
This equation for α1means that the volume fraction does not change along the trajectory,
although the pressure can change. But if the phase pressures change, then α1must change,
too, due to the different phase compressibility coefficients.
The correct way to derive equations for the instantaneous pressure relaxation limit is to
account for the following consequences of p1=p2=P:
dp1=K1
ρ1
dρ1=dp2=K2
ρ2
dρ2,K1
ρ1
∂ρ1
∂t=K2
ρ2
∂ρ2
∂t,K1
ρ1
∂ρ1
∂xk
=K2
ρ2
∂ρ2
∂xk
,(6.6)
where Ki=ρia2
iis the phase bulk modulus, aiis the phase speed of sound. Now we use the
above equation (6.6), the phase mass conservation equations (6.4b)and(6.4c)toderivethe
following equation for the volume fraction
∂α1
∂t+uk∂α1
∂xk
+α1α2(K1−K2)
α1K2+α2K1
∂uk
∂xk
=0.(6.7)
Thus, in case of instantaneous velocity and pressure relaxation we arrive at the following
system
∂ρui
∂t+∂(ρuiuk+Pδik)
∂xk
=0,(6.8a)
∂α1ρ1
∂t+∂α1ρ1uk
∂xk
=0,(6.8b)
∂α2ρ2
∂t+∂α2ρ2uk
∂xk
=0,(6.8c)
∂α1
∂t+uk∂α1
∂xk
+α1α2(K1−K2)
α1K2+α2K1
∂uk
∂xk
=0.(6.8d)
The above equations are exactly the same as is in the Kapila model.
6.2 Comparison with the Baer–Nunziato Model
It is of further interest to compare the system (2.15) with the one dimensional barotropic
Baer–Nunziato model given by the following equations
∂α1
∂t+uI
∂α1
∂x=ζ1,(6.9a)
∂α1ρ1
∂t+∂α1ρ1u1
∂x=ζ2,(6.9b)
∂α2ρ2
∂t+∂α2ρ2u2
∂x=ζ3,(6.9c)
∂α1ρ1u1
∂t+∂α1ρ1u2
1+α1p1(ρ1)
∂x−pI
∂α1
∂x=ζ4,(6.9d)
∂α2ρ2u2
∂t+∂α2ρ2u2
2+α2p2(ρ2)
∂x−pI
∂α2
∂x=ζ5,(6.9e)
123
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It was already mentioned in [70]and[69] that for smooth solutions the systems can be
reformulated into each other in one space dimension. However, we want to recall this trans-
formation with a slightly different purpose. In the following we want to show the equivalence
of these two systems (2.15)and(6.9) for smooth solutions and in particular that there is no
freedom of choice for the interface quantities uI,pIand the sources in this case. This is
of special interest since there are several possible choices for the interface quantities in the
context of these Baer–Nunziato type models, see e.g. [1].
(i) For the choice ζ2=ξ2, equation (6.9b) and equation (2.15b) are equal.
(ii) The sum of (6.9b)and(6.9c) gives the mixture mass balance (2.15c) with ξ3=ζ2+ζ3.
(iii) Subtracting (2.15b) from (2.15c) we obtain (6.9c) with ζ3=ξ3−ξ2.
(iv) The balance equations (2.15a)and(6.9a) for the volume fraction are equivalent for the
choices uI=uand ζ1=(ξ1−α1ξ3)/ρ.
(v) With α2=1−α1and ξ4=ζ4+ζ5the sum of (6.9d)and(6.9e) gives the mixture
momentum balance (2.15d).
To show how the barotropic SHTC model (2.15) can be derived from the Baer–Nunziato
model (6.9), we need to derive the balance for the relative velocity (2.15e). Therefore we
want to reformulate the partial momentum balances (6.9d)
ζ4=∂α1ρ1u1
∂t+∂α1ρ1u2
1+α1p1(ρ1)
∂x−pI
∂α1
∂x
=u1∂α1ρ1
∂t+∂α1ρ1u1
∂x+α1ρ1∂u1
∂t+1
2
∂u2
1
∂x+∂α1p1(ρ1)
∂x−pI
∂α1
∂x
⇔ζ4−u1ζ2
α1ρ1
=∂u1
∂t+1
2
∂u2
1
∂x+1
α1ρ1
∂α1p1(ρ1)
∂x−pI
α1ρ1
∂α1
∂x(6.10)
and (6.9e)
ζ5−u2ζ3
α2ρ2
=∂u2
∂t+1
2
∂u2
2
∂x+1
α2ρ2
∂α2p2(ρ2)
∂x−pI
α2ρ2
∂α2
∂x.(6.11)
Subtracting equation (6.11) from equation (6.10)gives
∂(u1−u2)
∂t+1
2
∂u2
1−u2
2
∂x+1
α1ρ1
∂α1p1(ρ1)
∂x
−1
α2ρ2
∂α2p2(ρ2)
∂x−pI
α1ρ1
∂α1
∂x+pI
α2ρ2
∂α2
∂x
=∂(u1−u2)
∂t+1
2
∂u2
1−u2
2
∂x+p1
α1ρ1
+p2
α2ρ2∂α1
∂x
+a2
1
ρ1
∂ρ1
∂x−a2
2
ρ2
∂ρ2
∂x−pI1
α1ρ1
+1
α2ρ2∂α1
∂x=ζ4−u1ζ2
α1ρ1
−ζ5−u2ζ3
α2ρ2
.
Using relation (2.19)and pI=(α2ρ2p1+α1ρ1p2)/ρ we obtain
∂(u1−u2)
∂t+
∂1
2u2
1−1
2u2
2+1(ρ1)−2(ρ2)
∂x=ζ4−u1ζ2
α1ρ1
−ζ5−u2ζ3
α2ρ2
=ξ5
123
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Thus we have shown that the SHTC model can be derived from the Baer–Nunziato model
with the following choices
uI=u,pI=α2ρ2p1+α1ρ1p2
ρ,=Bζ
with B=⎛
⎜
⎜
⎜
⎜
⎜
⎝
ρα
1α100
01 0 0 0
01 1 0 0
00 0 1 1
0−u1
α1ρ1
u2
α2ρ2
1
α1ρ1
−1
α2ρ2
⎞
⎟
⎟
⎟
⎟
⎟
⎠
It remains to show how the partial momentum balances can be derived using the SHTC
model. We start with the balance for the relative velocity (2.15e)
ξ5=∂(u1−u2)
∂t+
∂1
2u2
1−1
2u2
2+1(ρ1)−2(ρ2)
∂x
=α1ρ1
α1ρ1⎛
⎜
⎜
⎝
∂u1
∂t+
∂1
2u2
1+1(ρ1)
∂x⎞
⎟
⎟
⎠−α2ρ2
α2ρ2⎛
⎜
⎜
⎝
∂u2
∂t+
∂1
2u2
2+2(ρ2)
∂x⎞
⎟
⎟
⎠
⇔α1ρ1α2ρ2ξ5+α2ρ2u1ξ2−α1ρ1u2ζ3
=α2ρ2∂α1ρ1u1
∂t+∂α1ρ1u2
1+α1p1(ρ1)
∂x−p1
∂α1
∂x
−α1ρ1∂α2ρ2u2
∂t+∂α2ρ2u2
2+α2p2(ρ2)
∂x−p2
∂α2
∂x(6.12)
Now we multiply the mixture momentum balance (2.15d) with α1ρ1and add it to the previous
equation (6.12) to obtain
∂α1ρ1u1
∂t+∂α1ρ1u2
1+α1p1(ρ1)
∂x−α2ρ2p1+α1ρ1p2
ρ
∂α1
∂x
=1
ρ(α1ρ1α2ρ2ξ5+α2ρ2u1ξ2−α1ρ1u2ζ3+α1ρ1ξ4)
With the same choice for the interface pressure as before and the corresponding source term
we have obtained the partial momentum balance for the first phase. The second balance can
be obtained in a similar way (multiply with α2ρ2and subtract)
∂α2ρ2u2
∂t+∂α2ρ2u2
2+α2p2(ρ2)
∂x−α2ρ2p1+α1ρ1p2
ρ
∂α2
∂x
=−1
ρ(α1ρ1α2ρ2ξ5+α2ρ2u1ξ2−α1ρ1u2ζ3−α2ρ2ξ4)
Summarizing the relations we have
uI=u,pI=α2ρ2p1+α1ρ1p2
ρ,ζ=C
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with C=⎛
⎜
⎜
⎜
⎜
⎜
⎝
1
ρ0−α1
ρ00
01 000
0−1100
0c2u1+c1u2−c1u2c1c1c2ρ
0−(c2u1+c1u2)c1u2c2−c1c2ρ
⎞
⎟
⎟
⎟
⎟
⎟
⎠
One immediately verifies BC =Iand thus the equivalence in the smooth case is proven.
Following from this equivalence, we verify that the systems share the same Jacobian (2.20)
and thus the same eigenvalues (2.22) and eigenvectors (2.23). However, there are crucial
differences between the Baer–Nunziato system (6.9) and the SHTC system (2.15). Apart
from the obvious distinction that system (2.15) can be written in conservative form, the
most remarkable difference are the jump conditions for discontinuities. For system (6.9)the
equation for αand the phase momentum equations are not in conservative form and hence it
is in general not possible to write down jump conditions, cf. [51]. However, by the argument
that αstays constant across the shock the equations are decouple into Euler systems for each
phase with corresponding jump conditions, independent of the particular choice made for the
interface velocity and pressure. In contrast to system (2.15) where the phases remain coupled
across discontinuities and a shock in one phase also affects the other. Note that the particular
choice for uIand pImade above is also mentioned as a consistent choice in a work by
Hérard [38]. Moreover similar choices have been discussed before by Glimm et al. in [30],
for the isentropic single temperature case in Coquel et al. [14] and in Gallouët et al. [29].
In particular in [29] a general form of the interface velocity is discussed which includes the
mixture velocity presented here. However, the pressure is slightly different since it includes
the sound speeds of the present phases, but nevertheless the structure is similar.
For a detailed hyperbolicity analysis of the Baer–Nunziato model the reader is referred to
the work of Embid and Baer, see [21].
7 Numerical Results
For the SHTC model, all numerical results shown in this section have been obtained
with a classical second order MUSCL-Hancock scheme, see [78] for details. For the non-
conservative Baer–Nunziato system, which we solve for comparison at the end of this section,
we employ a second-order path-conservative version of the MUSCL Hancock scheme, see
e.g. [9,16,19,47,55]. In all cases we use the simple Rusanov-flux (local Lax-Friedrichs
flux) as approximate Riemann solver.
7.1 Exact Versus Numerical Solution of the Homogeneous System
The exact solution for the examples shown in this section is obtained as follows:
1. Assume that the eigenvalues λi−are located left of λCand the eigenvalues λi+on the
right, respectively.
2. Prescribe the state left of the contact, α1on the right and then solve for remaining
quantities on the right.
3. Choose eigenvalue next to the contact on the left side. Prescribe wave type (shock/
rarefaction) and choose wave speed (head speed for a rarefaction).
4. Repeat with remaining eigenvalue on the left side while the admissibility is constantly
checked and thus certain waves might be excluded already.
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Table 1 Primitive states of Riemann problem RP1
ULU∗
LU∗∗
LU∗∗
R¯
UU
∗
RUR
α10.70.70.70.30.30.30.3
ρ11.2449 0.47883 0.47883 0.30577 0.40186 0.41275 0.60312
ρ21.2969 1.2969 1.1064 0.894 0.894 0.73436 0.73436
u1−1.2638 −0.18865 −0.18865 −0.24825 0.01399 0.040001 0.43059
u2−0.38947 −0.38947 −0.14351 −0.15416 −0.15416 −0.40507 −0.40507
5. Repeat the afore mentioned steps for the right side.
6. Sample the complete solution.
7.1.1 Shock in Rarefaction
The exact solution is obtained by using a contact centered inverse construction of the solution
as described above. The states of this first Riemann problem (RP1 )are given in Table 1.Due
to the shock inside the rarefaction we have an additional state behind the shock given by ¯
U.
The following computation was performed using the ideal gas EOS
pi(ρi)=ργi
i,i∈{1,2}with γ1=1.4,γ
2=2
and the parameters
x=0.5·10−4m,CCFL =0.25,tend =0.25 s and x∈[−1,1]m.
In Figs. 8and 9the numerical results together with the exact solution are shown. The overall
wave structure of the solution is depicted in Fig.9. It consists of a λ2−- rarefaction (green)
which is completely contained inside the λ1−- rarefaction (blue). The contact (red) is (by
construction) the middle wave. On the right we have a λ1+-rarefaction (blue) which contains a
λ2+-shock (green) inside. The dashed blue line marks the intermediate tail of the rarefaction,
i.e. the position when the rarefaction starts again after the shock. Looking at the densities and
velocities in Fig. 8we verify that a rarefaction only affects the related phase. However, the
interaction of the two rarefactions manifests itself in the mixture quantities ρ,u,w.Forthe
volume fraction we observe that it only jumps at the contact as shown before and takes the
initial values on each side. Special attention has to be paid to the density of the first phase.
Right of the contact the rarefaction starts (not affecting phase two) until the shock occurs.
There the density jumps according to the jump conditions. Continuing on the right we have
a plateau of the right state of the shock and then the rarefaction continues again.
The eigenvalues are shown in the upper right panel of Fig. 9. One clearly sees the overlap-
ping eigenvalues in the region where the rarefaction waves coincide. Moreover, one can see
that in most of the states the system is strictly hyperbolic, but the ordering of the eigenvalues
changes throughout the whole domain.
7.1.2 Solution Without Contact
The states of this second Riemann problem RP2 are given in Table 2and we have α1=0.5
for all states.
123
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(a)
(b)
Fig. 8 Exact solution (black) and numerical solution (red) of Riemann problem RP1 (Color figure online)
123
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Fig. 9 Wave structure of Riemann problem RP1 (top): phase one (blue), contact (red) and phase 2 (green).
Eigenvalues of RP1 (bottom) (Color figure online)
123
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Table 2 Primitive states of
Riemann problem RP2 ULU∗
LU∗∗
LU∗∗
RU∗
RUR
ρ12.9194 2.9194 2 2 0.43057 0.42256
ρ21.5773 1 1 1 1.2486 0.58056
u1−0.53404 −0.53404 0 0 −1.8225 −1.876
u2−0.72386 0 0 0 0.09954 −0.93653
The following computation was performed using the ideal gas EOS
pi(ρi)=ργi
i,i∈{1,2}with γ1=1.4,γ
2=2
and the parameters
x=0.5·10−4m,CCFL =0.25,tend =0.25 s and x∈[−1,1]m.
In Figs. 10 and 11 the numerical results together with the exact solution are shown. It consists
of a λ2−- rarefaction (green) which overlaps with the λ1−- rarefaction (blue). The contact is
(by construction) not visible since all quantities stay constant across it. On the right we have
an isolated λ1+-shock (blue) which is followed by an isolated λ2+-shock (green). Looking
at the densities and velocities in Fig. 10 we again verify that a rarefaction only affects the
related phase. The interaction of the two rarefactions where they overlap can be observed by
the changing slope in the mixture quantities ρ,u,w.
The eigenvalues are shown at the right in Fig. 11. One clearly sees the overlapping eigen-
values in the region where the rarefaction waves coincide. Moreover one can see that in most
of the states the system is strictly hyperbolic, but the ordering of the eigenvalues changes
throughout the whole domain three times.
7.1.3 Symmetric Double Rarefaction
The following example (RP3) is taken from a paper of Romenski and Toro [70]. The exact
solution is obtained by using a contact centered inverse construction of the solution as
described previously. The states are given in Table 3and we have α1=0.9 for all states.
The following computation was performed using the following EOS
pi(ρi)=Aiρi
ρ(i)
ref γi
+Bi,i∈{1,2},
with A1=105Pa,γ
1=1.4,ρ
(1)
ref =1 kgm−3,B1=0Pa,
and A2=8.5·108Pa,γ
2=2.8,ρ
(2)
ref =103kgm−3,B2=8.4999 ·108Pa.
and the parameters
x=0.01
5·103m,CCFL =0.25,tend =0.11 ·10−5sandx∈[0,0.01]m.
In Figs. 12 and 13 the numerical results together with the exact solution are shown. For the
numerical solution we used the Force Flux together with Godunov’s method as exemplary
shownin[78]. This example again shows the behaviour of rarefaction waves quite nicely.
Each rarefaction is seen individually in the corresponding phase but the interaction of the
overlapping rarefaction waves is observed in the mixture quantities. In Fig.13 on can see that
123
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(a)
(b)
Fig. 10 Exact solution (black) and numerical solution (red) of Riemann problem RP2 (Color figure online)
123
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Fig. 11 Wave structure of RP2 (top): phase one (blue), contact (red) and phase 2 (green). Eigenvalues of RP2
(bottom) (Color figure online)
123
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Table 3 Primitive states for Riemann problem RP3
ULU∗
LU∗∗
LU∗∗
RU∗
RUR
ρ1789.79932 160 160 160 160 789.79932
ρ21270.0579 1270.0579 200 200 1270.0579 1270.0579
u1−1942.0873 0 0 0 0 1942.0873
u2−1722.9353 −1722.9354 0 0 1722.9354 1722.9354
the rarefaction waves of phase one are contained in the rarefaction waves of phase two and
that the corresponding eigenvalues coincide where they overlap.
7.1.4 Symmetric Double Shock
The following example is taken from a paper of Romenski and Toro [70]. The exact solution is
obtained by using a contact centered inverse construction of the solution as described above.
The states are given in Table 4and we have α1=0.9 for all states.
The following computation was performed using the following EOS
pi(ρi)=Aiρi
ρ(i)
ref γi
+Bi,i∈{1,2},
with A1=105Pa,γ
1=1.4,ρ
(1)
ref =1 kgm−3,B1=0Pa,
and A2=8.5·108Pa,γ
2=2.8,ρ
(2)
ref =103kgm−3,B2=8.4999 ·108Pa.
and the parameters
x=0.01 ·10−4m,CCFL =0.25,tend =0.22 ·10−5sandx∈[0,0.01]m.
In Figs. 14 and 15 the numerical results together with the exact solution are shown. For the
numerical solution we used the Force Flux together with Godunov’s method as exemplary
shown in [78]. This is another good test as it shows four separated shock waves one in each
phase. As noted before, in contrast to the Baer–Nunziato type systems, the shocks affect
every phase. In Fig.15 we can see that the shock waves related to phase one are slower than
the shocks related to phase two. Moreover the order of the eigenvalues changes across every
shock.
7.2 Comparison of the SHTC Model with the Baer–Nunziato Model
In this last section we present a comparison of the numerical solutions obtained for the
conservative SHTC model discussed in this paper, and the well-known Baer–Nunziato model.
We show the solution of two Riemann problems, without and with stiff relaxation source
terms.
The test problem RP5 has the following initial data: ρ1,L=ρ1,R=2, ρ2,L=ρ2,R=1,
u1,L=−2, u1,R=+2, u2,L=−1, u2,R=+1, αL=0.7, αR=0.3. The computational
domain is the interval [−1,1], which is discretized with 10000 uniform grid cells and the
final time of the simulation is t=0.1. In Fig. 16 we compare the exact solution of the
Riemann problem for the SHTC system with the numerical solutions obtained for the SHTC
system and the Baer–Nunziato model without any relaxation source terms. Since the wave
123
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(a)
(b)
Fig. 12 Exact solution (black) and numerical solution (red) of Riemann problem RP3 (Color figure online)
123
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Fig. 13 Wave structure of RP3 (top): phase one (blue), contact (red) and phase 2 (green). Eigenvalues of RP3
(bottom) (Color figure online)
123
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Table 4 Primitive states for Riemann problem RP4
ULU∗
LU∗∗
LU∗∗
RU∗
RUR
ρ1131.01705 142.98406 1079 1079 142.98406 131.01705
ρ21040.1358 2983.4101 2706 2706 2983.4101 1040.1358
u13075.6226 2677.4348 0 0 −2677.4348 −3075.6226
u23033.3793 −38.030561 0 0 38.030561 −3033.3793
structure consists of two rarefaction waves in both phases, the numerical results of the SHTC
model and the numerical results obtained for the Baer–Nunziato system agree perfectly
well with each other and with the exact solution of the Riemann problem, as expected. In
Fig. 17 we show the results obtained for the same initial data, but with stiff pressure and
velocity relaxation, choosing the relaxation parameters as θ1=10−3and θ2=10−8,which
corresponds to the Kapila limit of both systems. We find that the numerical solutions obtained
for the conservative SHTC model and the Baer–Nunziato model agree perfectly well with
each other.
The last test problem RP6 has the following initial data: ρ1,L=2, ρ1,R=1, ρ2,L=1,
ρ2,R=2, u1,L=u1,R=0, u2,L=u2,R=0, αL=0.7, αR=0.3. The computational
domain is again the interval [−1,1], which is discretized with 10000 uniform grid cells and
the final time of the simulation is t=0.25. In Fig. 18 we compare the exact solution of the
Riemann problem for the SHTC system with the numerical solutions obtained for the SHTC
system and the Baer–Nunziato model without any relaxation source terms. Since the wave
structure consists of a shock and a rarefaction wave in both phases, the numerical results of the
SHTC model and the numerical results obtained for the Baer–Nunziato system do not agree
with each other any more, since the jump conditions of the non-conservative Baer–Nunziato
system and the conservative SHTC model do not coincide. In particular, one can observe how
the shock waves in one phase leave the other phase unchanged in the Baer–Nunziato model,
while in the conservative SHTC system a shock in one phase always affects the other phase,
which is a consequence of the jump conditions. In Fig. 19 we show the results obtained for
the same initial data, but with stiff pressure and velocity relaxation, choosing the relaxation
parameters as θ1=10−3and θ2=10−8, which corresponds again to the Kapila limit of both
systems. Despite the visible discrepancies observed in the homogeneous case, we find that
the numerical solutions obtained in the stiff relaxation limit of the conservative SHTC model
and the Baer–Nunziato model agree perfectly well with each other. This indicates that for
numerical purposes it may be more beneficial to solve the SHTC system in the stiff relaxation
limit, since standard numerical schemes for conservation laws can be applied, while in the
Baer–Nunziato model special numerical techniques for the treatment of nonconservative
terms are needed.
8 Conclusion
In this paper we have presented the exact solution of the Riemann problem for the barotropic
conservative two-phase model proposed in [64,69,70], which belongs to the SHTC class
of symmetric hyperbolic and thermodynamically compatible systems. We have discussed
the characteristic fields of the system and the admissibility criteria of shock waves. The
Riemann invariants as well as the Rankine–Hugoniot jump conditions have been provided.
123
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(a)
(b)
Fig. 14 Exact solution (black) and numerical solution (red) of Riemann problem RP4 (Color figure online)
123
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Fig. 15 Wave structure of Riemann problem RP4 (top): phase one (blue), contact (red) and phase 2 (green).
Eigenvalues of RP4 (bottom) (Color figure online)
123
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(a)
(b)
Fig. 16 Exact solution of the SHTC system (black), numerical solution of the SHTC system (red) and numerical
solution of the Baer–Nunziato model (blue) of Riemann problem RP5a without relaxation source terms (Color
figure online)
123
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Fig. 17 Numerical solution of the SHTC system (black) and of the Baer–Nunziato model (red) of Riemann
problem RP5b with stiff relaxation source terms. From top left to bottom right: mixture density ρ, mixture
velocity u, mixture pressure pand volume fraction α1(Color figure online)
A particular feature of the model under consideration is that some eigenvalues of the two
phases may coincide, creating particular wave phenomena for the case where a shock in one
phase interacts with a rarefaction wave in the other phase. We have discussed possible wave
patterns and based on the mathematical entropy inequality associated with the system and the
admissibility criteria of shock waves we have ruled out non-admissible wave configurations.
Independent of the potential physical applications of the model discussed in this paper, its
mathematical structure and properties as well as the possible wave configurations that may
appear in the solution of the Riemann problem make it a very interesting object of study from
a mathematical point of view. Concerning the properties of the class of SHTC systems it
will also be interesting to investigate the Kawashima-Shizuta condition, see e.g. [71], when
source terms are considered. This was, however, beyond the scope of this work.
At the end of the paper we show exact solutions of some example Riemann problems,
providing also a detailed comparison with numerical results obtained with a second order
TVD scheme. For two problems, we also provide a comparison with numerical solutions
obtained for the barotropic Baer–Nunziato model, both, for the homogeneous case without
source terms and for the case of stiff pressure and velocity relaxation. In the case of smooth
solutions, the results of the SHTC model and the Baer–Nunziato model agree perfectly well
with each other, as expected, while in the presence of shock waves the numerical solutions
123
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83 Page 48 of 60 Journal of Scientific Computing (2022) 93 :83
(a)
(b)
Fig. 18 Exact solution of the SHTC system (black), numerical solution of the SHTC system (red) and numerical
solution of the Baer–Nunziato model (blue) of Riemann problem RP6a without relaxation source terms (Color
figure online)
123
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Journal of Scientific Computing (2022) 93 :83 Page 49 of 60 83
Fig. 19 Numerical solution of the SHTC system (black) and of the Baer–Nunziato model (red) of Riemann
problem RP6b with stiff relaxation source terms. From top left to bottom right: mixture density ρ, mixture
velocity u, mixture pressure pand volume fraction α1(Color figure online)
disagree in the homogeneous case. On the contrary, in the presence of stiff velocity and
pressure relaxation source terms, the numerical results obtained in the relaxation limit of the
SHTC system and of the Baer–Nunziato model (Kapila limit) agree perfectly well with each
other.
Future work will consider the extension of the Riemann solver presented in this paper to
the non-barotropic case, solving the full model [64,69] and as also shown in the appendix. We
furthermore plan to develop provably thermodynamically compatible finite volume schemes
for the SHTC system, following the ideas outlined in [6–8].
Acknowledgements M.D. is member of the INdAM GNCS group and acknowledges the financial support
received from the Italian Ministry of Education, Universityand Research (MIUR) in the frame of the PRIN 2017
project Innovative numerical methods for evolutionary partial differential equations and applications.The
work of E.R. is supported by Mathematical Center in Akademgorodok, the agreement with Ministry of Science
and High Education of the Russian Federation Number 075-15-2022-281. F.T. gratefully acknowledges the
support by the research training group Energy, Entropy and Dissipative Dynamics (EDDy) of the Deutsche
Forschungsgemeinschaft (DFG) - Project No. 320021702/GRK2326.
123
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83 Page 50 of 60 Journal of Scientific Computing (2022) 93 :83
Funding Open Access funding enabled and organized by Projekt DEAL.
Declarations
Conflict of interest The authors declare that they have no conflict of interest
Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which
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appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence,
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regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder.
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A Appendix
A.1 Derivation of the Barotropic Submodel
A PDE system for compressible two-phase, two-temperature flow using a hyperbolic heat
conduction model, was previously discussed in Romenski et al. [64,69]. Note, that there is
another version of the SHTC model for hyperbolic heat transfer originally proposed in [61]
and used in [5,17,18,65]. The mentioned version seems to be more suitable from the point
of view of the general theory of SHTC models, since its PDE system can be derived from
the variational principle [57]. However, since this heat transfer model is not implemented in
the SHTC two-phase flow model, we use the equations formulated in [69]. Moreover, the
conclusions drawn in this section should be the same for both models, since the difference
in the equations for the thermal impulse does not affect the final result. Written in terms of
the generalized energy the system under consideration reads as
∂ρα1
∂t+∂ρα1uk
∂xk
=−φ, (A.1a)
∂ρc1
∂t+∂(ρc1uk+ρEwk)
∂xk
=−ψ, (A.1b)
∂ρ
∂t+∂ρuk
∂xk
=0,(A.1c)
∂ρui
∂t+∂(ρuiuk+pδik +ρwiEwk)
∂xk
=0,(A.1d)
∂wk
∂t+∂(wlul+Ec1)
∂xk
+ul∂wk
∂xl
−∂wl
∂xk=−1
ρλ0.(A.1e)
∂ρ ji
1
∂t+∂(ρ ji
1uk+ES1δik)
∂xk
=−λi
1,(A.1f)
∂ρ ji
2
∂t+∂(ρ ji
2uk+ES2δik)
∂xk
=−λi
2,(A.1g)
∂ρ S1
∂t+
∂(ρS1uk+Ejk
1)
∂xk
=1−π1,(A.1h)
123
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Journal of Scientific Computing (2022) 93 :83 Page 51 of 60 83
∂ρ S2
∂t+
∂(ρS2uk+Ejk
2)
∂xk
=2−π2,(A.1i)
Here, α1is the volume fraction of the first phase which is connected with the volume fraction
of the second phase α2by the saturation law α1+α2=1, ρis the mixture mass density
which is connected with the phase mass densities ρ1,ρ
2by the relation ρ=α1ρ1+α2ρ2.
The phase mass fractions are defined as c1=α1ρ1/ρ, c2=α2ρ2/ρ and it is easy to see that
c1+c2=1. Eventually, ui=c1ui
1+c2ui
2is the mixture velocity, wi=ui
1−ui
2is the phase
relative velocity. The quantities jk
iis the thermal impulse of phase iand Si=cisiis the
partial entropy of phase i. The equations describe the balance law for the volume fraction,
the balance law for the mass fraction, the conservation of total mass, the total momentum
conservation law, the balance for the relative velocity, the two equations for the partial thermal
impulses and the tow equations for the partial entropies. Moreover by construction the system
is equipped with a conservation law for the total energy
∂ρ E+1
2ulul
∂t+
∂ρukE+1
2ulul+p
ρ+wlEwl+ρEcEwk+Ejk
iESi
∂xk
=0.
(A.2)
In this work the source terms are of no special interest. The details can be found in [64,
69]. We want to present a derivation of the barotropic submodel in order to emphasize the
differences in the conservation law for the total energy since this will affect the mathematical
entropy for the system given above. By barotropic we understand that the internal energy of
each phase solely depends on the corresponding phase density. More precisely we consider
either an isentropic (constant entropy) or an isothermal (constant temperature) process. Let
us first consider an isentropic process. This implies that the derivatives ESivanish which
leads to the following conservation law for the total energy
∂ρ E+1
2ulul
∂t+
∂ρukE+1
2ulul+p
ρ+wlEwl+ρEcEwk
∂xk
=0.(A.3)
The isothermal case is a bit more involved. The reason is, that for an isothermal process we
necessarily need the thermal impulse. We first rewrite the equation for the total energy (A.2)
∂ρ E+1
2ulul
∂t+
∂ρukE+1
2ulul+p
ρ+wlEwl+ρEcEwk
∂xk
+
∂Ejk
iESi
∂xk
=∂ρ E+1
2ulul
∂t+
∂ρukE+1
2ulul+p
ρ+wlEwl+ρEcEwk
∂xk
+ESi
∂Ejk
i
∂xk
+Ejk
i
∂ESi
∂xk
=0
Since we have an isothermal process the partial space derivative of ESi=Tivanishes
identically. We further reformulate the partial entropy balances (A.1h)and(A.1i), i.e.
∂Ejk
1
∂xk
=1−π1−∂ρ S1
∂t−∂(ρS1uk)
∂xk
,
∂Ejk
2
∂xk
=2−π2−∂ρ S2
∂t−∂(ρS2uk)
∂xk
123
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83 Page 52 of 60 Journal of Scientific Computing (2022) 93 :83
and insert these two equations into the equation for the total energy. From now on we also
assume that the flow has a single temperature T=T1=T2and thus the sources πivanish.
This gives
∂ρ E+1
2ulul
∂t+
∂ρukE+1
2ulul+p
ρ+wlEwl+ρEcEwk
∂xk
+ESi
∂Ejk
i
∂xk
=∂ρ E+1
2ulul
∂t+
∂ρukE+1
2ulul+p
ρ+wlEwl+ρEcEwk
∂xk
+T
2
i=1!i−∂ρ Si
∂t−∂(ρSiuk)
∂xk"=0.
Before we further simplify the energy equations for the isentropic and the isothermal case
we need to precisely discuss the mixture equation of state in each case. In the following we
will discuss the isentropic and isothermal case separately.
A.1.1 Isentropic Case
Using relation (2.6) we obtain the following derivatives for the mixture internal energy (2.5)
∂e
∂α =c∂e1
∂ρ1s−cρ
α2+(1−c)∂e2
∂ρ2s
(1−c)ρ
(1−α)2
=(1−c)2ρ
(1−α)2
p2
ρ2
2
−c2ρ
α2
p1
ρ2
1
=p2−p1
ρ,(A.4)
∂e
∂c=e1+c∂e1
∂ρ1s
ρ
α−e2−(1−c)∂e2
∂ρ2s
ρ
1−α
=e1+cρ
α
p1
ρ2
1
−e2+(1−c)ρ
1−α
p2
ρ2
2=h1(ρ1)−h2(ρ2), (A.5)
∂e
∂ρ =c∂e1
∂ρ1s
c
α+(1−c)∂e2
∂ρ2s
1−c
1−α
=c2
α
p1
ρ2
1
+(1−c)2
1−α
p2
ρ2
2
=αp1+(1−α)p2
ρ2=p
ρ2.(A.6)
Here we have introduced the specific enthalpy of phase i
hi(ρi)=ei(ρi)+pi
ρi
and the mixture pressure
p=α1p1+α2p2.
Note that the derivatives with respect to αand ρare equal to the corresponding derivatives
in the general case, see [69]. However, the derivative with respect to cis different (enthalpy
vs. chemical potential). Using these results we obtain the following for the mixture EOS
∂E
∂α =∂e
∂α =p2−p1
ρ,(A.7)
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Journal of Scientific Computing (2022) 93 :83 Page 53 of 60 83
∂E
∂c=∂e
∂c+(1−2c)wiwi
2=h1(ρ1)−h2(ρ2)+(1−2c)wiwi
2,(A.8)
∂E
∂ρ =∂e
∂ρ =p
ρ2,(A.9)
∂E
∂wi
=c(1−c)wi.(A.10)
A.1.2 Review Isothermal Thermodynamics
Due to the extra term with the derivative of the entropy in (2.7) we need to revisit some
thermodynamic relations before we calculate the derivatives. First, we have the following
well known Maxwell relation (omitting the phase index for the following considerations)
∂p
∂Tρ
=−ρ2∂s
∂ρ T
(A.11)
which bases on the equality of second order derivatives of the internal energy depending
on the canonical variables ρand s, exemplary see [44]. This relation then can be used to
derive a more complicated Maxwell relation which is also known but not that obvious. More
precisely we want to show the following relation
∂(ρe)
∂ρ T
=−T2⎛
⎜
⎝
∂g
T
∂T⎞
⎟
⎠ρ
.(A.12)
This is also a well established relation, see [44,52], but for the convenience of the reader
we show how it can be derived. Here gdenotes the specific Gibbs energy which may be
generalized to the chemical potential μwhen more substances are involved, see [44]. The
Gibbs energy (or sometimes free enthalpy) is given by
g=e−Ts +p
ρ.
The differentials for the internal energy and the Gibbs energy depending on the variables ρ
and Tare given by
de=Tds+p
ρ2dρ=T∂s
∂ρ T
+p
ρ2dρ+T∂s
∂ρ T
dT,(A.13)
dg=1
ρdp−sdT=1
ρ∂p
∂ρ T
dρ+1
ρ∂p
∂Tρ
−sdT.(A.14)
Now we can show the stated relation (A.12)
∂(ρe)
∂ρ T
=e+ρ∂e
∂ρ T
(2.7)
=e+ρp
ρ2+T∂s
∂ρ T
(A.11)
=e+p
ρ−T
ρ∂p
∂Tρ
(2.10)
=g+Ts−T
ρ∂p
∂Tρ
(A.14)
=g−T∂g
∂Tρ
=−T2⎛
⎜
⎝
∂g
T
∂T⎞
⎟
⎠ρ
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83 Page 54 of 60 Journal of Scientific Computing (2022) 93 :83
Since we are considering the isothermal case we obtain for this specific case
−T2⎛
⎜
⎝
∂g
T
∂T⎞
⎟
⎠ρ
=g.
A.1.3 Isothermal Case
Now we can calculate the derivatives of the mixture internal energy (2.5)using(A.12)
∂e
∂α =c∂e1
∂ρ1T−cρ
α2+(1−c)∂e2
∂ρ2T
(1−c)ρ
(1−α)2
=−c
αρ1∂e1
∂ρ1T
+1−c
1−αρ2∂e2
∂ρ2T
=1−c
1−α∂(ρ2e2)
∂ρ2T
−c
α∂(ρ1e1)
∂ρ1T
+c
αe1−1−c
1−αe2
=1−c
1−α(g2−e2)−c
α(g1−e1)=p2−p1
ρ+T(c1s1−c2s2), (A.15)
∂e
∂c=e1+c∂e1
∂ρ1T
ρ
α−e2−(1−c)∂e2
∂ρ2T
ρ
1−α
=e1+ρ1∂e1
∂ρ1T
−e2+ρ2∂e2
∂ρ2T
=g1(ρ1)−g2(ρ2), (A.16)
∂e
∂ρ =c∂e1
∂ρ1T
c
α+(1−c)∂e2
∂ρ2T
1−c
1−α
=c
ρρ1∂e1
∂ρ1T
+1−c
ρρ2∂e2
∂ρ2T
=c
ρ∂(ρ1e1)
∂ρ1T
+1−c
ρ∂(ρ2e2)
∂ρ2T
−ce1+(1−c)e2
ρ
=cg1+(1−c)g2
ρ−ce1+(1−c)e2
ρ=
cp1
ρ1
−Ts
1+(1−c)p2
ρ2
−Ts
2
ρ
=α1p1+α2p2
ρ2−T(c1s1−c2s2)
ρ=p
ρ2−T(c1s1−c2s2)
ρ.(A.17)
Note that in the general case we also obtain the difference of the Gibbs energies (or chemical
potentials) for the derivative with respect to c,see[69]. Using these results we obtain the
following for the mixture EOS
∂E
∂α =∂e
∂α =p2−p1
ρ+T(c1s1−c2s2), (A.18)
∂E
∂c=∂e
∂c+(1−2c)wiwi
2=g1(ρ1)−g2(ρ2)+(1−2c)wiwi
2,(A.19)
∂E
∂ρ =∂e
∂ρ =p
ρ2−T(c1s1−c2s2)
ρ,(A.20)
123
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Journal of Scientific Computing (2022) 93 :83 Page 55 of 60 83
∂E
∂wi
=c(1−c)wi.(A.21)
A.1.4 Simplifying the Energy Equations
Using the results obtained above we can further simplify the energy equations. For the
isentropic case we the relations (A.7)–(A.10) can be used to yield
0=∂ρ E+1
2ulul
∂t+
∂ρukE+1
2ulul+p
ρ+wlEwl+ρEcEwk
∂xk
=
2
i=1
∂αiρiei+1
2u2
i
∂t+∂αiρiuk
ihi+1
2u2
i
∂xk
For the isothermal case we first rewrite the energy equality again using Si=cisiand
S=S1+S2, i.e.
∂ρ E+1
2ulul
∂t+
∂ρukE+1
2ulul+p
ρ+wlEwl+ρEcEwk
∂xk
+T
2
i=1!i−∂ρ Si
∂t−∂(ρSiuk)
∂xk"
⇔−T
2
i=1
i=∂ρ E+1
2ulul
∂t+
∂ρukE+1
2ulul+p
ρ+wlEwl+ρEcEwk
∂xk
−T
2
i=1!∂ρ Si
∂t+∂(ρSiuk)
∂xk"
=∂ρ E−TS+1
2ulul
∂t+
∂ρukE−TS+1
2ulul+p
ρ+wlEwl+ρEcEwk
∂xk
.
Here F=E−TSis generalized mixture free energy and we can further write =1+2
for the non-negative total entropy production. Now we use relations (A.18)–(A.21) to obtain
−T=∂ρ E−TS+1
2ulul
∂t+
∂ρukE−TS+1
2ulul+p
ρ+wlEwl+ρEcEwk
∂xk
=
2
i=1
∂αiρiei−Ts
i+1
2u2
i
∂t+∂αiρiuk
igi+1
2u2
i
∂xk
.
In [64] a single temperature model is presented. The governing equations are the same as in
the isothermal case, the energy equation reads
2
i=1#∂αiρiei+1
2u2
i
∂t+∂αiρiuk
ihi+1
2u2
i
∂xk$−∂ρc1c2(uk
1−uk
2)(s1−s2)T
∂xk
=0.
However, after some manipulations this equation may also be written in the same form as
it is used in this work. Thus the thermal impulses are again basically hidden in the energy
equality.
123
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83 Page 56 of 60 Journal of Scientific Computing (2022) 93 :83
A.2 Calculation for Coinciding Eigenvectors
We want to give the detailed calculation for the statement (3.1). Let us first assume that the
vector RCis a multiple of a fixed vector Ri±,see(2.23). We can directly conclude that
εi=0⇔0=(u−ui)2−a2
i⇔u=ui±ai⇔λC=λi±.
Note that the specific correct sign ±is determined by the non-zero components of the eigen-
vector. To show the reverse direction we assume that λCcoincides with a fixed λi±,see
(2.22). From this we get that the corresponding εivanishes. To simplify the notation let us
assume w.l.o.g. that i=1 and thus ε1=0. We further obtain
δ1=p1−p2
ρ−(u−u1)2
α1
=α1(p1−p2)−ρa2
1
α1ρ=ρ1γ1.
Thus we get
δ1ε2=ρ1γ1ε2=ρ1
a1
a1γ1ε2=±
ρ1
a1
(u−u1)γ1ε2.
This shows the desired relation.
A.3 Regularity of the Mass Flux System Matrix
In Sect. 4.2 we derived a linear system for the squares of the partial mass fluxes (4.20). It
remains to verify that the matrix is regular, i.e. the determinant
det(M)=−
1
2α11
ρ11
ρ2
2+α21
ρ2
11
ρ2
does not vanish. We can immediately verify that
sgn 1
ρ11
ρ2
2=sgn 1
ρ2
11
ρ2
and thus the determinant will only vanish iff at least one of densities does not jump across the
shock. If the partial density would not jump across the shock we obtain from the continuity of
the corresponding partial mass flux that the associated partial velocity also would not jump.
From the Lax condition of the eigenvalues we hence can conclude that for a shock in phase
μ∈{1,2}the partial density of phase μmust jump. Let us now consider the other phase
ν∈{1,2},ν= μ.Ifweassumeρν=0 we obtain for (4.18)
uν=−Qν1
ρν=0.
Thus we have for the momentum jump condition (4.13d) and for the relative velocity jump
condition (4.13e)
−Qμuμ+pμ=0,(A.22)
1
2(uμ−S)2+μ=0.(A.23)
123
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Journal of Scientific Computing (2022) 93 :83 Page 57 of 60 83
Equation (A.23) can be rewritten using
1
2(uμ−S)2=−
1
2pμ1
ρ+
μ
+1
ρ−
μ
and thus
0=1
2(uμ−S)2+μ=μ−1
2pμ1
ρ+
μ
+1
ρ−
μ.(A.24)
Using the reformulated jump conditions (4.19)and(4.17)wehaveinthecaseρν=0
Q2
μ
21
ρ2
μ−μ=0,Q2
μ1
ρμ+pμ=0.
Since we know that ρμjumps we can combine these two equations and obtain
0=2μ
1
ρ2
μ+pμ
1
ρμ
⇔0=2μ1
ρμ+pμ1
ρ2
μ
=2!μ+1
2pμ1
ρ+
μ
+1
ρ−
μ"1
ρμ
⇔0=μ+1
2pμ1
ρ+
μ
+1
ρ−
μ.(A.25)
Summing up (A.24)and(A.25)gives
0=2μ⇔0=ρμ.
This is a contradiction since the partial density ρμmust jump as argued above. Hence both
partial densities jump across a shock and thus the determinant does not vanish.
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