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An asymptotic preserving multi-dimensional ALE

method for a system of two compressible flows coupled

with friction

Stéphane del Pino, Emmanuel Labourasse, Guillaume Morel

To cite this version:

Stéphane del Pino, Emmanuel Labourasse, Guillaume Morel.

An asymptotic preserving

multi-dimensional ALE method for a system of two compressible flows coupled with friction. 2017.

�hal-01505238�

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An asymptotic preserving multi-dimensional ALE method for a

system of two compressible flows coupled with friction

S. Del Pinoa,∗, E. Labourassea,∗, G. Morela

aCEA, DAM, DIF, F-91297 Arpajon, France.

Abstract

We present a multi-dimensional asymptotic preserving scheme for the approximation of a mix-ture of compressible flows. Fluids are modeled by two Euler systems of equations coupled with a friction term.

The asymptotic preserving property is mandatory for this kind of model, to derive a scheme that behaves well in all regimes (i.e. whatever the friction parameter value is). The method we propose is defined in ALE coordinates, using a Lagrange plus remap approach. This imposes a multi-dimensional definition and analysis the scheme.

Keywords: Compressible gas dynamics, multi-fluid, finite-volumes, unstructured meshes, asymptotic preserving, arbitrary-Lagrangian-Eulerian (ALE)

1. Introduction

1

A multifluid model is a model for a fluid mixture for which each fluid is described by is

2

own full set of variables (for instance density, velocity and energy). The model is generally

3

closed in a way that defines interactions between the constituents, depending on the envolved

4

physic. These models are widely used in different communities. One very popular model of this

5

kind is the Baer-Nunziato model [1] for deflagration-to-detonation transition of reactive flows.

6

Many numerical methods to approximate this model have been designed, let us just cite a few

7

of them [2, 3, 4, 5, 6]. Such kind of models is also used in plasma physics to account for

plas-8

mas collision or Non-Local-Thermodynamic-Equilibrium (NLTE) Ion-Electron interactions [7].

9

Scannapieco and Cheng [8] also derive similar kind of model for turbulent flows and apply it to

10

describe a mixing zone driven by Rayleigh-Taylor or Richtmyer-Meshkov instabilities [9].

11

In this paper, we present a multi-dimensional scheme to approximate solutions of two

com-12

pressible inviscid fluids coupled with friction, refer to equation (1). This model is a slightly

13

simplified version of the Scannapieco-Cheng [8] model where friction is considered uniform in

14

space. It can also be viewed as a simplification of the model proposed in [7] for which the

elec-15

tron effect is neglected or of the Baer-Nunziato model [1], neglecting the interfacial terms and in

16

the case where there is no phase transition.

17

Corresponding authors

Email addresses: stephane.delpino@cea.fr (S. Del Pino), emmanuel.labourasse@cea.fr (E. Labourasse), guillaume.morel@cea.fr (G. Morel)

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Our goal in this paper is to address two difficulties. First one is inherent to this kind of model

18

and rely to the asymptotic preserving (AP) property [10, 11, 12] in the high friction regime or

19

infinite friction regime. In the former regime, the fluids interpenetration follows a diffusion law.

20

In the latter one, the mixture evolves as a single fluid, see (4)–(5). If no attention is paid to

21

these regimes, the scheme will fail to capture it at a reasonnable calculation cost. The second

22

difficulty comes from the numerical framework we consider. We want our scheme to be able to

23

deal with Arbitrary-Lagrange-Euler (ALE) frame and unstructured meshes in order to properly

24

handle highly deformed calculation domains.

25

While authors [13, 14, 15] propose an asymptotic discretization for the system (1) in 1D in

26

the Eulerian frame, no asymptotic preserving scheme has been yet published for 2D unstructured

27

meshes for this model. Even for simpler model, only few unstructured asymptotic preserving

28

schemes have been developped (refer for instance to Berthon and Turpault [16] and Franck et

29

al. [17, 18]). The scheme we propose in section 4 has connections with [19, 20], where an

30

Euler with friction system is studied. However, it is not a direct extension of [19] to the

bi-31

fluid case. The scheme presented in this work is split into two steps. In the first step we solve

32

two Euler systems of equations coupled by friction. Since each fluid has its own velocity, the

33

Lagrangian mesh of each fluid will evolve separately during this step. Then, in the second step,

34

the conservative variables vector of each of the fluid will be projected onto a common mesh (not

35

necessarily identical to the initial mesh).

36

In the section 2 of this paper, we recall the properties of the model we consider, that are

37

conservation, hyperbolicity, and asymptotic limit model. In section 3, we recall the basis of

38

the solver (Glace [21] or Eucclhyd [22]) used to compute the Lagrangian step. The section 4

de-39

scribes the Lagrangian step of the proposed scheme. It is demonstrated that the scheme preserves

40

the properties of conservation, stability and consistency with respect to the continuous model for

41

all regimes (independantly of the value of the friction parameter). Then in section 5, our ALE

42

strategy is described. Finally, section 6 is devoted to numerical experiments on several problems

43

(Sod shock tube, triple point and Rayleigh-Taylor). Some comparisons with a non-AP scheme

44

are provided.

45

2. A two fluids model with friction

46

Let us consider a mixture of two fluids f1and f2. In the following, we will denote by

“multi-47

fluid model”, a model for which each fluid α ∈ { f1, f2} is represented by its own set of variables:

48

(ρα, uα, Eα). Conversely, we will refer as “mono-fluid model”, a model describing a mixture

49

where mean quantities are considered (ρ, u, E), each fluid position being precised by an

addi-50

tional equation on the concentration (e.g. cα= ραραβ). 51

In this part, we present a simplified version of Scannapieco-Cheng’s model where the

inter-52

action between the two constituents reduces to a friction term. In semi-Lagrangian coordinates,

53

for each fluid α ∈ { f1, f2} (β denoting the other fluid), the model writes

54 ρα Dαtτ α= ∇ · uα, ραDα tu α= −∇pανρδuα, ραDα tE α= −∇ · (pαuα) − νρδuα· u, (1)

where ρα, uα and Eα respectively denote the mass density, the velocity and the total energy

55

density of fluid α. Also, τα = ρ1α denotes the specific volume. The pressure pα satisfies the 56

equation of state pα := pα(ρα, eα), where eα, the internal energy density, is defined by eα :=

57

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Eα− 1

2ku

αk2. The total density ρ and the mean velocity u are defined as ρ := ρα+ ρβ and

58

ρu := ραuα+ ρβuβ. The term δuα is the velocity difference, the δ(·)αoperator being defined by

59

δφα = −δφβ = φαφβ. Finally, ν is the friction parameter. Also, remark that the Lagrangian

60

derivative Dαt := ∂t+ uα· ∇, is obviously not the same for each fluid.

61

The entropy ηαdefined by Gibbs formula Tαdηα = deα+pαdταsatisfies the following entropy

62 inequality 63 TαDαtηα≥ντ α τβδuα·δuα≥ 0. (2)

Prior to establishing a numerical scheme that discretizes this set of six equations, we recall

64

some properties of the model itself.

65

Property 1 (Conservation). The model (1) is conservative in volume and mass for each fluid.

66

Also, it is conservative in the sum of momenta and in the sum of the total energies of the two

67

fluids.

68

Proof. Conservation of mass and volume is obvious since the first equation of (1) is the

continu-69

ity equation written for each fluid.

70

Conservation of momenta sum and total energies sum requires more cautiousness, since

La-71

grangian derivative are not the same for each fluid. To establish them one rewrites (1) in an

72

Eulerian framework.

73

Developing Lagrangian derivatives and using the identity ∂t(ρατα)= 0 elementary

calcula-74

tions allow to rewrite (1) as

75

∂tρα+ ∇ · (ραuα)= 0,

∂t(ραuα)+ ∇ · (ραuα⊗ uα)+ ∇pα+ νρδuα= 0,

∂t(ραEα)+ ∇ · (ραEαuα)+ ∇ · (pαuα)+ νρδuα· u= 0.

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Summing the two later equations over α gives a system of the conservative form ∂tU+∇· F(U) =

0, where U= ρ αuα+ ρβuβ ραEα+ ρβEβ ! , and F(U)= ρ αuα⊗ uα+ ρβuβ⊗ uβ+ pα+ pβI ραEαuα+ ρβEβuβ+ pαuα+ pβuβ ! , where I is the identity matrix of R2×2.

76

Property 2 (Hyperbolicity). The model (1) is hyperbolic.

77

Proof. The proof is straightforward but calculatory, see [15] for details.

78

Asymptotic model. Whenν → +∞, (1) behaves has the following five equations model ρDtu= −∇



pα+ pβ , (4)

while, for each fluidα ∈ { f1, f2},β denoting the other one, one has

ραD tτα= ∇ · u, ραD tEα= − ρα ρu · ∇  pα+ pβ− pα∇ · u, (5) where u is the same velocity for both fluids, and thus the Lagrangian derivative is also the same.

79

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Formal derivation (established in [15]). Let = ν−1so that (1) rewrites 80 ραDα tτ α= ∇ · uα, ρα Dαtu α= −∇pα1 ρδu α, ραDα tE α= −∇ · (pαuα) −1 ρδuα· u. (6)

We will now study its limit while  → 0+focusing first on the momentum equations since the

81

friction term’s goal is to impose that δu0−→ 0.→0

82

Developing the Lagrangian derivatives and dividing each momentum equation by ρα > 0, one has ∂tuα+ (∇uα) uα= − ∇pα ρα − 1  ρ ραδuα.

Since fluid β satisfies the same equation and recalling that δφα= −δφβ= φα−φβ, one gets ∂t(δuα)+ δ ((∇u) u)α= −δ ∇p ρ !α −1 λδuα, where λ= ρ2 ραρβ.

We now perform an Hilbert expansion for all variables in the equation, that is φ= φ0+ φ1+

83

O(2). One has

84 ∂t(δuα,0)+ δ ((∇u)u)α,0= −δ ∇p ρ !α,0 −λ0 1 δuα,0+ δuα,1 ! −λ1δuα,0+ O(). (7)

Multiplying this equation by  one has λ0δuα,0= O(), which gives δuα,0= 0 when  → 0 since

85

λ > 0.

86

So, when  → 0, formula (7) recasts

87 δuα,1= −1 λ0δ ∇p ρ !α,0 . (8)

Now, we perform an Hilbert expansion for the whole system (6), neglecting the non negative powers of . Choosing α ∈ { f1, f2}, β being the other one, it reads

ρα,0Dα tτ α,0=∇ · uα,0, ρα,0Dα tu α,0= − ∇pα,0ρ0 1 δuα,0+ δuα,1 ! −ρ1δuα,0, ρα,0Dα tE α,0= − ∇ · pα,0uα,0−ρ0 1 δuα,0· u α,0+ δuα,1· uα,0+ δuα,0· uα,1! −ρ1δuα,0· uα,0,

Since we just established δuα,0 = 0, one has u0 = u0 = uα,0 = uβ,0. Also, since Dα tφ =

∂tφ + uα,0· ∇φ + O(), Lagrangian derivatives are the same when  → 0, so that using (8) the

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system simplifies to ρα,0D tτα,0=∇ · u0, ρα,0 Dtu0= − ∇pα,0+ ρ0 1 λ0δ ∇p ρ !α,0 , ρα,0D tEα,0= − ∇ ·  pα,0u0 + ρ0 1 λ0δ ∇p ρ !α,0 · u0,

Recalling λ=ρρα2ρβ and developing δ

∇p

ρ

α,0

, momentum equation rewrites

ρα,0D tu0= − ∇pα,0+ ρα,0ρβ,0 ρ0 ∇pα,0 ρα,0 − ∇pβ,0 ρβ,0 ! , = −ρρα,00 ∇pα,0+ pβ,0 .

Proceeding the same way with total energy equation, one gets

ρα,0 DtEα,0= − ∇ ·  pα,0u0 + ρ α,0ρβ,0 ρ0 ∇pα,0 ρα,0 − ∇pβ,0 ρβ,0 ! · u0, = − ρρα,00  ∇pα,0+ ∇pβ,0· u0− pα,0∇ · u0, 88

Remark 1. Defining E := ραEα+ρρ βEβ andτ := ρ−1, it is easy to check that ifα, ρβ, u, Eα, Eβ) is

a solution of the asymptotic model (4)–(5), one has ρDtτ = ∇ · u, ρDtu= −∇  pα+ pβ , ρDtE= −∇ ·  pα+ pβu .

One recognizes Euler equations for the mixture. The mixing pressure follows Dalton’s law as

89

one could have expected since we consider here non-reactive gases.

90

However, notice that unless each fluid follows a barotropic equation of state (pα = pα(ρα)),

91

equation (5) must be solved to determine eα.

92

3. Cell-centered schemes

93

We recall briefly the muti-dimensional finite volume schemes [23, 24, 22], since it is the

94

basis of this work. For convinience, we use the notations defined in [21]. In the following, for

95

all cell j, and for any quantity φ, one defines its mean value φj:=V1j

R

jφ, where Vj:= Rj1 is the

96

cell volume. Also, let us denote the cell’s mass as mj:= Rjρ = ρjVj, which is constant in time in

97

semi-Lagrangian coordinates (dtmj= 0).

98

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x y j r+ 1 r r −1 Cjr N− jr N+jr

Figure 1: Illustration of Cjrand Nijrvectors at vertex r for a polygonal cell j.

We consider first-order schemes, so that one has the following relations d dt Z j 1= mjdtτj, d dt Z j ρj= 0, d dt Z j ρjuj= mjdtuj, d dt Z j ρjEj= mjdtEj.

Let Jrdenote the set of cells connected to node r and let Rjthe set of nodes of cell j. Also,

99

let us introduce Cjr:= ∇xrVj, the gradient of the volume of the polygonal cell j, according to the 100

position of one of its vertices r. Then, the cell-centered schemes we consider in this paper have

101

the following structure: for any cell j of the mesh one has

102 mjdtτj= X r∈Rj Cjr· ur, dtmj=0, mjdtuj= − X r∈Rj Fjr, mjdtEj= − X r∈Rj Fjr· ur, (9)

where the fluxes urand Fjrare defined for any node r

∀ j ∈ Jr, Fjr= Cjrpj− Ajr(ur− uj), (10)

and X

j∈Jr

Fjr= 0. (11)

In one hand, relation (10) is the matrix form of the acoustic Riemann solver (see for instance [25,

103

26]), while in the other hand (11) imposes conservation.

104

In the following to simplify notations, we omit sets Rjand Jrwhen there is no confusion.

105

• If Ajr:= ρjcj Cjr⊗Cjr

kCjrk , then (9)–(11) defines the Glace scheme [24, 21]. 106 • Let N+jr = −12(xr+1− xr)⊥and N−jr = − 1 2(xr− xr−1) ⊥. If A jr := ρjcj N+ jr⊗N+jr kN+jrk + N− jr⊗N − jr kN− jrk  , the 107

scheme (9)–(11) is Eucclhyd [22, 26]. One has N+jr+ N−

jr= Cjr, see figure 1.

108

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These schemes are conservative in volume, mass, momentum and total energy. One easily

109

shows that they are entropy stable. These results can be found in [24, 22, 21, 26], for instance.

110

Also, a consistency result has been established in [27].

111

4. Asymptotic Preserving scheme in semi-Lagrangian coordinates

112

We shall now present a multi-dimensional finite-volume scheme written in semi-Lagrangian

113

coordinates that preserves the asymptotic.

114

In this section, we present the Lagrangian step of our ALE method. In this step, each fluid is

115

associated to its own mesh. If the meshes may evolve differently, we assume that they coincide

116

at the begining of the Lagrangian step. The rezoning/remapping procedure that is detailed in

117

section 5 is used to ensure that the meshes will coincide for the next Lagrangian step.

118

We first focus on the semi-discrete continuous in time scheme. Most of the properties of the

119

scheme are proved using this simpler formulation without any lost of generality. In paragraph 4.2,

120

we describe the fully discrete scheme. It is analysed in the remaining of this section.

121

4.1. Continuous in time semi-discrete scheme

122

Let ω ∈ [0, 2], for each fluid α ∈ { f1, f2}, β denoting the other fluid, we define the scheme

123 mαjdtταj = X r Cjr· uαr, dtmαj =0, mαjdtuαj = − X r Fαjr−ωX r νρrBjrδuαj − (1 − ω) X r νρrBjrδuαr, mαjdtEαj = − X r Fαjr· uαr − X r νρruTrBjrδuαr + ω X r νρruTjrBjr(δuαr −δuαj), (12)

where the fluxes are given by

jr= Cjrpαj− A α jr(u α r − u α j) − νρrBjrδuαr, and (13) X j Fαjr= 0. (14)

In order to write (12), we introduced ραr := #J1

r P j∈Jrρ α j and ρr := ρ α r + ρ β r. Also, we set 124 ur := ρα ruαr+ρ β ruβr ρα r+ρ β r and ujr := ρα ruαj+ρ β ruβj ρα r+ρ β r

. Bjr are symmetric and positive definite matrices such that

125

P

r∈RjBjr= VjI. Matrices A

α

jrare the standard “hydro-matrices” as defined in section 3.

126

Remark 2. One can choose Bjr := VjrI, where Vjr is the volume of the subcell associated to

127

vertex r of cell j. Another obvious choice could be for instance Bjr:= #R1jVjI. 128

Observe that simple calculations allow to write

129 ρrur= ρruαr −ρ β rδuαr and ρrujr= ρruαj−ρ β rδuαj. (15) 7

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Injecting (13) in (12), and using (15), one gets the alternative form 130 mαjdtταj = X r Cjr· uαr, dtmαj =0, mαjdtuαj = X r Aαjr(uαr − uαj)+ ωνX r ρrBjrδuαr −δu α j , mαjdtEαj = − X r Cjrpαj · u α r + X r uαrTAαjr(uαr − uαj)+ νX r ρβr tδuαrBjrδuαr −ωνX r

ρβrtδuαjBjr(δuαr −δu α j)+ ων X r ρrtuαjBjr(δuαr −δu α j). (16)

This form enlightens the fact that knowing the fluxes (uαr, uβr) at any vertex r is enough to define

131

the scheme. We shall now show that these nodal velocities are well defined.

132

Injecting (13) in (14) allows to calculate (uαr, uβr). Obviously, as soon as ν , 0, both nodal

velocities are coupled at vertex r. Omitting boundary conditions in the sake of simplicity, for each vertex of the mesh (uαr, u

β

r) is the unique solution of the following linear system:

X j       Aαjr+ νρrBjr −νρrBjr −νρrBjr A β jr+ νρrBjr       | {z } Aνr:= uαr uβr ! =X j       Aαjrj + Cjrpαj Aβjrj+ Cjrp β j       | {z } br:= .

Proof. Since matrices Aαjr and Bjr are symmetric, Aνr is symmetric. To prove that (uαr, u β r) is

unique, it remains to show that it is positive definite. Elementary calculation gives, ∀(vα, vβ) ∈ R2× R2, (vα, vβ)TAνr(vα, vβ)=vαT         X j Aαjr         vα+ vβT         X j Aβjr         vβ + (vα− vβ)T         X j νρrBjr         (vα− vβ),

which is strictly positive if (vα, vβ) , (0, 0) since matricesP

jAαjr and P jνρrBjr are positive 133 definite. 134

The scheme being well-defined, we now establish its properties.

135

4.1.1. Nodal velocitiesa priori estimates

136

Here, we establish estimates for the nodal velocities with regard to the frictionless case. These

137

are actually some instantaneous stability results with regard to the mono-fluid schemes [21, 22],

138

i.e.velocity fluxes are controled by the frictionless ones.

139

Property 3 (A priori estimates). For each fluid α ∈ { f1, f2}, let uα,νr denote the nodal velocities

at vertex r. Let Aαr := PjAαjrand Br := PjBjr. Letβ denote the other fluid, then one has the

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following relations, ∀ν ≥ 0 uα,νr TAαruα,νr + uβ,νr T Aβru β,ν r ≤ uα,0r T Aαruα,0r + uβ,0r T Aβru β,0 r , (17)  uα,νr − uβ,νr T Br  uα,νr − uβ,νr  ≤ 1 2νρr  uα,0r TAαruα,0r + u β,0 r T Aβru β,0 r  , (18) and uα,νr − uβ,νr T Br  uα,νr − uβ,νr  ≤uα,0r − uβ,0r T Br  uα,0r − uβ,0r  . (19)

Let us first comment these estimates. The estimate (17) is a stability results. It shows that

140

the nodal velocity k(uα,νr , u β,ν r )kA0 r is bounded by k(u α,0 r , u β,0 r )kA0

r independently of ν. It shows that 141

friction nodal velocities are stable with regard to the classic frictionless case for a given state.

142

The second estimate (18) shows that the nodal velocity difference kδuα,νr kBris at most O(ν

−1/2) 143 according to k(uα,0r , u β,0 r )kA0r. 144

The last inequality (19) states that the nodal velocity difference is bounded by the frictionless

145

case independently of ν in the k · kBrnorm, which is purely geometric. 146

Proof of Property 3. ∀ν ≥ 0, (uα,ν r , u

β,ν

r ) is the unique solution of

r + νρrBr −νρrBr −νρrBr A β r+ νρrBr ! uα,νr uβ,νr ! = br, with br :=       P jCjrpαj P jCjrp β j      . So, since bris independent of ν, one has

147 ∀ν ≥ 0,  A0r + νρr∆r  uνr= A0ru0r, (20) where A0r := Aαr 0 0 Aβr ! , ∆r := Br −Br −Br Br ! and uνr := u α,ν r uβ,νr ! . Multiplying on the left by uνr yields uνrTAr0uνr + νρruνr

T

ruνr = uνr T

A0ru0r. Since Br is a positive

matrix∆ris also positive, and since νρr≥ 0, one gets

∀ν ≥ 0, uν r T A0ru ν r ≤ u ν r T A0ru 0 r. Finally, A0

r being symmetric and positive definite, the simple following Youngs inequality,

rTA0ru 0 r ≤ 1 2u ν r T A0ru ν r+ 1 2u 0 r T A0ru 0 r, allows to prove (17). 148

The proof of (18) follows the same way. Multiplying (20) on the left by uνr, one has ∀ν ≥ 0, νρruνr T ruνr+ u ν r T A0ru ν r = u ν r T A0ru 0 r.

Then, using the same Youngs inequality, one gets after a few arrangements

∀ν ≥ 0, νρruνr T ruνr+ 1 2u ν r T A0ruνr≤ 1 2u 0 r T A0ru 0 r,

which yeilds to (18) since A0r is positive.

149

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The third inequality is a bit more difficult to establish. Let us introduce the quadratic form Jνv:=12v

T

A0r + νρr∆r



v − br· v. So, since uνris the unique solution of the linear system, one has

∀ν ≥ 0, ∀v, Jν uνr ≤ J

ν v.

In the particular case v= u0

r, one gets Jνuνr ≤ J

ν u0

r

. It is then easy to check that

u0 r = 1 2u 0 r T A0r+ νρr∆r  u0r − br· u0r = J 0 u0 r + νρr 2 u 0 r T ∆ru0r.

So, one has established a first inequality

150 Jνuν r ≤ J 0 u0 r + νρr 2 u 0 r T ∆ru0r. (21)

Similarly, since u0ris the unique solution of the linear system in the case ν= 0, one has J0u0 r

≤ J0 uνr,

which can be written as

J0u0 r ≤ J ν uνr− νρr 2 u ν r T ruνr.

This actually gives a lower bound to Jνuν

rwhich combined with its upper bound (21) yields

J0u0 r+ νρr 2 u ν r T ruνr≤ J 0 u0 r + νρr 2 u 0 r T ∆ru0r.

Since νρris positive, elementary calculations allow to write (19).

151

4.1.2. Conservativity

152

Property 4 (Conservation). The scheme defined by (12)–(14) ensures conservation of mass and

153

volume for each fluidα or β. It also ensures that the sum of the fluids’ momenta and total energies

154

are conserved.

155

Proof. Conservations of mass and volume for each fluid are obvious since the associated balance

156

equations are unchanged with regard to the mono-fluid schemes (see for instance [24, 21, 22,

157

26]).

158

Summing momenta equations in (12) for both fluids gives

jdtuαj + m β jdtu β j = − X r Fαjr−X r Fβjr −ωX r νρrBjr(δuαj + δu β j) − (1 − ω) X r νρrBjr(δuαr + δu β r).

Recalling that by definition, δuαj + δuβj = 0, one has mαjdtuαj + m β jdtu β j= − X r Fαjr−X r Fβjr.

The conservativity proof is ended in a standard way. One now sums these equations over the cells which gives

X j mαjdtuαj+ X j mβjdtuβj = − X j X r∈Rj Fαjr−X j X r∈Rj Fβjr, 10

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which rewrites, X j mαjdtuαj+ X j mβjdtu β j = − X r X j∈Jr Fαjr−X r X j∈Jr Fβjr.

This proves that momenta sum is conserved using (14) and recalling that cell masses are

La-159

grangian.

160

Conservation of total energies sum is obtained in the exact same way.

161

4.1.3. Stability

162

Before proving this result, let us recall that the fully discrete scheme’s stability is presented

163

bellow (see paragraph 4.2).

164

Property 5 (Entropy). The first-order continuous in time scheme defined by (12)–(14) satisfies, ∀ω ∈ [0, 2], the following entropy inequality ∀α ∈ { f1, f2}

jjdtηαj ≥  1 − ω 2  X r νρβrtδuαrBjrδu β r+ ω 2 X r νρβr tδuαjBjrδuαj ≥ 0.

This inequality is consistent with (2).

165

Let us establish a simple technical Lemma that will be useful in the following and to

demon-166

strate Property 5.

167

Lemma 1. Let M denote a symmetric matrix of Rd×d. Letω ∈ R, then

∀v, w ∈ Rd, vTMv −ωwTM(v − w)=  1 − ω 2  vTMv+ω 2w TMw+ω 2(w − v) TM(w − v).

Proof. Let ξ := vTMv −ωwTM(v − w). Obviously, one has

ξ = vTMv+ ωwTMw −ωwTMv.

Since M is symmetric, one has −2wTMv= (v − w)TM(v − w) − vTMv − wTMw. Injecting this

168

equality in the expression of ξ ends the demonstration.

169

Corollary 1. Let M denote a symmetric and positive matrix of Rd×d. Letω ≥ 0, then

∀v, w ∈ Rd, vTMv −ωwTM(v − w) ≥  1 − ω 2  vTMv+ω 2w TMw.

Proof. This is a direct consequence of Lemma 1, since ωM is a positive matrix.

170

We can now give the proof of Property 5.

171

Proof of Property 5. Gibbs formula reads T dη= de + pdτ, so that one has Tαjdtηαj = dteαj + p

α jdtταj,

which rewrites also

jjdtηαj = m α jdtEαj − u α j · m α jdtuαj + p α jm α jdtταj. 11

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Using (16), one gets mαjjdtηαj = − X r Cjrpαj · u α r + X r uαrTAαjr(uαr − uαj)+ νX r ρβr tδuαrBjrδuαr −ωνX r

ρβr tδuαjBjr(δuαr −δu α j)+ ων X r ρrtuαjBjr(δuαr −δu α j) + uα j·        X r Aαjr(uαr − uαj)+ ωνX r ρrBjrδuαr −δu α j        + X r Cjr· uαrp α j, which simplifies as mαjjdtηαj = X r (uαr−uαj)TAαjr(uαr−uαj)+νX r ρβrδuαr TB jrδuαr−ων X r ρβrδuαj TB jr(δuαr−δu α j).

Since Bjr matrices are symmetric and positive and since ω ≥ 0, one can apply Corollary 1 to

obtain mαjjdtηαj ≥ X r (uαr − uαj)TAαjr(uαr − uαj) + 1 −1 2ω ! νX r ρβrδuαr T Bjrδuαr + 1 2ων X r ρβrδuαj T Bjrδuαj,

Matrix Aαjrbeing positive, one finally has

jjdtηαj ≥ 1 − 1 2ω ! νX r ρβrδuαr TB jrδuαr + 1 2ων X r ρβrδuαj TB jrδuαj,

which is positive as soon as ω ∈ [0, 2].

172

4.1.4. Asymptotic preserving

173

We now establish the main result of this paper. It consists in stating that when the friction

174

parameter ν tends to infinity, the scheme (12)–(14) behaves asymptotically as a scheme that is

175

consistent with the asymptotic model (4)–(5).

176

To this end, we first compute the asymptotic scheme by means of Hilbert expansions, then we

177

show its consistency with the asymptotic model. This later result relies strongly on B. Despr´es’s

178

work [27].

179

Asymptotic scheme. Letω , 0. If ∀α ∈ { f1, f2}, ∀ j, (ραj, uαj, Eαj) are constant cell data, then the

scheme (12)–(14), behaves asymptotically as (mαj + mβj)dtuj= − X r Fαjr−X r Fβjr, (22) dtVj= mαjdtταj = X r Cjr· ur, (23) dtmαj = 0, mαjdtEαj = − X r Cjrpαj · ur+ X r uTrjr(ur− uj) − ρα jρ β j ρj X r uTjδ Ajr ρj !α (ur− uj), (24) 12

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where uj= uαj = u β

j, and where nodal velocities ur= uαr = u β r satisfy Fαjr+ Fβjr= Cjr  pαj+ pβj−Aαjr+ Aβjr(ur− uj), and X j Fαjr= 0. (25)

Formal derivation. Let α ∈ { f1, f2}, β denoting the other fluid. Let us introduce  := ν−1. One

rewrites (16) as mαjdtταj = X r Cjr· uαr, (26) dtmαj =0, mαjdtuαj = X r Aαjr(uαr − uαj) −1 ω X r ρrBjr(δuαj −δu α r), (27) mαjdtEαj = − X r Cjrpαj · u α r + X r tuα r A α jr(u α r − u α j)+ 1  X r ρβr(δuαr) TB jrδuαr +1ωX r (ρruαj T ρβ rδuαj T)B jr(δuαr −δu α j), (28) and 180 X j Aαjrr +X j 1 ρrBjrδuαr = X j Aαjrj +X j Cjrpαj. (29)

Following the analysis of the asymptotic model, we perform an Hilbert expansion.

181

The first information one gets is from equation (29) that writes

X j Aα,0jr uα,0r +X j 1 ρ0rBjrδuα,0r + X j ρ0 rBjrδuα,1r + X j ρ1 rBjrδuα,0r =X j Aα,0jr uα,0j +X j Cjrpα,0j + O(),

so that multiplying this equation by  leads to ρ0 r( P jBjr)δuα,0r = 0 that is 182 δuα,0 r = 0, (30) sinceP

jBjris symmetric positive definite and ρr = ρ0r + O() > 0 so that ρ0r > 0 when  → 0.

183

One gets volume conservation equation (23).

184

Now, the momentum equation (27) is considered, using (30), one has

jdtuα,0j = X r Aα,0jr (uα,0r − uα,0j ) −1 ω X r ρ0 rBjrδuα,0j −ωX r ρ1 rBjrδuα,0j −ω X r ρ0 rBjr(δuα,1j −δu α,1 r )+ O(), which gives 185 δuα,0 j = 0. (31) 13

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Using, (31) and (30), one defines u0 j := u α,0 j = u β,0 j and u 0 r := u α,0 r = u β,0 r . 186

So, Hilbert expansions of equations (26), (27) and (28) simplify as

jdtτα,0j = X r Cjr· u0r, mαjdtu0j= X r Aα,0jr (u0r − u0j) − ωX r ρ0 rBjr(δuα,1j −δuα,1r ), (32) mαjdtEα,0j = − X r (Cjrpα,0j − A α,0 jr (u 0 r− u 0 j)) · u 0 r + ωX r ρ0 ru α,0 j T Bjr(δuα,1r −δu α,1 j ), (33)

Our aim is now to evaluate the term ωP

rρ0ru α,0

j T

Bjr(δuα,1r −δuα,1j ). To do so, we divide

momentum equation (32) by ραj(> 0), which gives

Vjdtu0j = 1 ρα j X r Aα,0jr (u0r− u0j) − ωX r ρ0 r ρα j Bjr(δuα,1j −δuα,1r ).

The same relation can be written for fluid β. The difference of these two equations writes, recalling that δφα= −δφβ, 0=X r δ        A0 jr ρj        α (u0r− u0j) − ρj ρα jρ β j ωX r ρ0 rBjr(δuα,1j −δuα,1r ).

Injecting this relation in (33) gives the limit scheme total energy balance equation (24). The

187

momentum equation (22) is obtained the same way or by simply summing equations (32) for

188

both fluids α and β.

189

In order to establish that the scheme is asymptotic preserving, it remains to show that the

190

limit scheme (22)–(25) is consistent with the asymptotic model (4)–(5).

191

Before establishing this result, we recall the fundamental result by B. Despr´es [27], that we

192

adapt to the present context.

193

Property 6 (B. Despr´es). Let mj:= mαj+ m β j,ρj:= ραj+ ρ β j,τj= ρ−1j and Ej:= ρα jE α j+ρ β jE β j ρj . Then, the scheme dtmj= 0, mjdtτj= X r Cjr· ur, mjdtuj= − X r Fjr, mjdtEj= − X r Fjr· ur, where Fjr= Cjr(pαj + p β j) − (A α jr+ A β jr)(ur− uj), and X j (Aαjr+ Aβjr)ur= X j (Aαjr+ Aβjr)uj+ X j Cjr(pαj + p β j), 14

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is weakly consistent with the following system of equations ρDtτ = ∇ · u,

ρDtu= −∇(pα+ pβ),

ρDtE= −∇ · (pα+ pβ)u.

Proof. The proof can be found in [27].

194

Property 7. The limit scheme (22)–(25) is weakly consistent with the asymptotic model (4)–(5).

195

Proof. Consistency for volume, mass and momentum is a direct consequence of Property 6, it

196

remains to show the consistency for total energy.

197

We rewrite equation (5) using a more convenient form

ρα

DtEα= −∇ · (pα+ pβ)u+ pβ∇ · u+

ρβ

ρ∇(p

α+ pβ) · u.

As a starting point we recall (25) for fluid α

jdtEαj = − X r Cjpαj· ur+ X r uTrjr(ur− uj) − ρα jρ β j ρj X r uTjδ Ajr ρj !α (ur− uj), that we rewrite mαjdtEαj = − X r Cj(pαj + p β j) · ur+ X r uTr(Aαjr+ Aβjr)(ur− uj) +X r Cjpβj· ur− X r uTrjr(ur− uj) − ρα jρ β j ρj X r uTj         Aαjr ρα j − Aβjr ρβj         (ur− uj).

Simple algebraic manipulations on the later term allow to write

jdtEαj = − X r Cj(pαj + p β j) · ur+ X r uTr(Aαjr+ Aβjr)(ur− uj) +X r Cjp β j· ur− X r (ur− uj)TA β jr(ur− uj) − ρβj ρj uTj X r  Aαjr+ Aβjr(ur− uj).

• According to Property 6 the term 1 Vj        −X r Cj(pαj+ p β j) · ur+ X r uTr(Aαjr+ Aβjr)(ur− uj)       , is weakly consistent with−∇ · (pα+ pβ)u

x j . 198 • Also sinceV1 j( P

rCj· ur) is weakly consistent with ∇ · u,

1 Vj       p β j X r Cj· ur        ≈pβ∇ · u xj . 15

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• Now, sinceP rCjr = 0, one has −X r Fjr = X r (Aαjr+ Aβjr)(ur− uj),

so that Property 6 implies that

1 Vj         − ρβj ρj uTjX r  Aαjr+ Aβjr(ur− uj)         ≈ ρ β ρ∇(pα+ pβ) · u ! xj . To conclude, it remains to prove for the remaining term

1 Vj        −X r (ur− uj)TA β jr(ur− uj)       ≈ 0. Let ζαdenote its limit:

1 Vj        −X r (ur− uj)TAβjr(ur− uj)        −→ Vj→0 ζα. We have shown ρα jdtEαj ≈ −∇ · (p α+ pβ)u+ pβ∇ · u+ρβ ρ∇(pα+ pβ) · u ! x j + ζα.

Since the same result holds for fluid β, simple calculations lead to ρα jdtEαj + ρ β jdtE β j = ρjdtEj≈  −∇ · (pα+ pβ)u x j + ζα+ ζβ. According to Property 6 ρjdtEj≈  −∇ · (pα+ pβ)u x j , so that ζα+ ζβ≈ 0. 199

Actually, one has

1 Vj        X r (ur− uj)TAβjr(ur− uj)       + 1 Vj        X r (ur− uj)TAαjr(ur− uj)       → 0,

since Aαjrand Aβjrare positive matrices, one has finally 1 Vj        X r (ur− uj)TAαjr(ur− uj)       → 0 and 1 Vj        X r (ur− uj)TAβjr(ur− uj)       → 0, which ends the proof.

200

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4.2. Discrete scheme

201

We now describe the fully discrete scheme. According to the previously established results, let ω ∈]0, 2]. One defines the following scheme for each fluid α ∈ { f1, f2}, β denoting the other

one. mαj τα j n+1τα j n ∆t = X r Cnjr· uαrn, (34) mαjjn+1− uαjn ∆t = − X r Fα,njr −ωX r νρn rB n jrδu α j n+1 − (1 − ω)X r νρn rB n jrδu α r n, (35) mαjjn+1− Eαjn ∆t = − X r Fα,njr · uαrn−X r νρn r tun rB n jrδu α r n+ ωX r νρn r tun+1 jr B n jrδu α r n −δuαjn+1 , (36) where the fluxes are computed explicitly as

Fα,njr = Cnjrjn− Aα,njr (uαr n − uαjn) − νρnrBnjrδu α r n, (37) and X j Aα,njrrn+X j νρn rB n jrδu α r n =X j Aα,njrjn+X j Cnjrjn. (38)

To complete the scheme definition, observe that we introduced the following mean velocities

202 unjr+1=ρ α rnuαjn+1+ρ β r n uβjn+1 ρα rn+ρβr n and u n r = ρα rnuαrn+ρ β r n uβr n ρα rn+ρβr n , which rewrite 203 ρn ru n+1 jr = ρ n ru α j n+1

−ρβrnδuαjn+1 and ρnrunr = ρnrrn−ρβrnδuαrn. (39) Similarly to the semi-discrete case, for convinience, we inject the fluxes expression into

momen-204

tum and total energy balance equation and use (39)

205 mαjjn+1− uαjn ∆t = X r Aα,njr (uαrn− uαjn)+ ωνX r ρn rB n jr(δu α r n −δuαjn+1), (40) mαjjn+1− Eαjn ∆t = − X r Cnjrjn· uαr n+X r tuα r n Aα,njr (uαr n − uαjn) + νX r ρβrn tδuαr n Bnjrδuαrn+ ωνX r ρn r tuα j n+1 Bnjrδuαrn−δuαjn+1 −ωνX r ρβrn tδuαj n+1Bn jrδu α r nδuα j n+1 . (41)

4.3. Stability of the discrete scheme

206

In this section we establish that the scheme is stable for arbitrary equation of state: there

207

exists∆t > 0 such that for each fluid α ∈ { f1, f2}, ταj

n+1 > 0, eα j n+1> e(T = 0) and ηα j n+1ηα j n. 208

For the sake of simplicity, and without loss of generality, we will consider in the following the

209

case eαjn+1> 0.

210

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Actually, we will provide explicit timesteps for the positivity of density and internal energy,

211

but we will only show that the increasing physical entropy timestep will be greater that the

212

one of the mono-fluid case for given velocity fluxes, for which we established Property 3. The

213

main reason is that there only exists existence results for entropy stability for cell-centered

semi-214

Lagrangian schemes (even in 1D), see [28, 29].

215

4.3.1. Positivity of density

216

Since p= p(ρ, e) one has to ensure that density cannot be made negative.

217

Property 8 (Positivity of density). Assuming that ∀α ∈ { f1, f2}, ∀ j ∈ M, ραj

n > 0. Denoting

Cαnthe set of compressive cells for each fluidα, Cαn :=nj ∈ M/ PrCn jr· u α rn< 0 o , there exists ∆tρ> 0 such that, ∀α ∈ { f1, f2}, ∀ j ∈ Cαn, ∆tρ< Vαjn −P rCnjr· u α rn . Then, the scheme (34)–(38) defined by∆t ∈ ]0, ∆tρ] ensures that

∀ω ∈ [0, 2], ∀α ∈ { f1, f2}, ∀ j ∈ M, ραj n+1> 0.

Observe that, as expected, only compressive cells ( j ∈ Cαn) can lead to negative densities, so

218

in the case of non-compressive flows,∆tρmay be arbitrarly large. Also, in the case of trianglular

219

meshes, this constrain implies that no cell will tangle during the timestep.

220

Proof. Obviously, this is equivalent to show that ταjn+1= ρα1 jn+1

> 0. According to (34), one has τα j n+1= τα j n+ ∆t mαj X r Cnjr· uαr n.

So, one has the following alternative:

221

• if j < Cαnthat isP

rCnjr· u α r

n≥ 0, then ∀∆t > 0 one has τα j

n+1> 0,

222

• else if j ∈ Cαn, one hasP

rCnjr· u α r n < 0, then ∀∆t < τα j n mαj −P

rCnjr·uαrn, one has τ

α j n+1 > 0. 223 Since m α j −P rCnjr·u α

rn > 0, the existence of such a ∆t > 0 is obvious. 224

225

4.3.2. Positivity of internal energy

226

First, as a primary result, we give internal energy variation for fluid α ∈ { f1, f2}, β denoting

the other one. Internal energy is updated as

jn+1= eαjn+ ∆t mαj        X r t(uα j n− uα r n)Aα jr n(uα j n− uα r n) −X r pαjnCnjr· uαrn        + ν∆t mαj        X r ρβrn tδuαr n Bnjrδuαr n+ ωX r ρβrn tδuαj n+1 Bnjr(δuαjn+1−δuαrn)        − ∆t 2 2mαj2        X r Aαjrn(uαjn− uαrn)        2 + ∆t2 2mαj2       ων X r ρn rB n jr(δu α j n+1 −δuαrn)        2 . (42) 18

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Proof. Rewriting eαjn+1= −12kuαjn+1k2+ Eα j

n+1and using (41), one gets after a few arrangements

jn+1= 1 2 u α j n 2 −1 2 u α j n+1 2 + eα j n − ∆t mαj        X r pαjnCnjr· uαrn+X r tuα r n Aαjrn(uαjn− uαrn)        + ν∆t mαj        X r ρβrn tδuαr n Bnjrδuαr nωX r ρβrn tδuαr n+1 Bnjrδuαr nδuα j n+1 +ωX r ρn r tun+1 j B n jrδu α r n −δuαjn+1        . (43) As a first step one estimates kinetic energy variation

−∆Kαj = 1 2ku α j n k2−1 2ku α j n+1 k2 = uαjn+ uαjn+1 2 ·  uαjn− uαjn+1 , which rewrites using (40)

−∆Kαj =       u α j n − ∆t 2mαj        X r Aαjrn(uαjn− uαrn)+ ωνX r ρn rB n jr(δu α j n+1 −δuαrn)               · ∆t mαj        X r Aαjrn(uαjn− uαrn)+ ωνX r ρn rB n jr(δu α j n+1δuα r n)       , that is −∆Kαj = ∆t mαj        X r tuα j n Aαjrn(uαjn− uαrn)+ ωνX r tuα j nρn rB n jr(δu α j n+1 −δuαrn)        − ∆t 2 2mαj2        X r Aαjrn(uαjn− uαrn)+ ωνX r ρn rB n jr(δu α j n+1δuα r n)        2 . So, one has

jn+1= eαjn+ ∆t mαj        X r t(uα j n − uαrn)Aαjrn(uαjn− uαrn) −X r pαjnCnjr· uαrn − ∆t 2mαj        X r Aαjrn(uαjn− uαrn)+ ωνX r ρn rB n jr(δu α j n+1δuα r n )        2        + ν∆t mαj        X r ρβrn tδuαr n Bnjrδuαr n+ ωX r ρβrn tδuαj n+1 Bnjr(δuαjn+1−δuαrn)        + ων∆t mαj X r ρn r t(uα j n − uαjn+1)Bnjr(δuαjn+1−δuαrn),

which using (38) is nothing but (42).

227

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Actually, (42) can be rewritten as eαjn+1= ehαj n+1+ ν∆t mαj        X r ρβrn tδuαr n Bnjrδuαrn+ ωX r ρβrn tδuαj n+1 Bnjr(δuαjn+1−δuαrn)        + ∆t2 2mαj2        ωνX r ρn rB n jr(δu α j n+1δuα r n)        2 , (44) where ehαjn+1denotes the obtained internal energy without friction: i.e. injecting nodal velocities

228

rninto the classic mono-fluid scheme. The remaining terms can be viewed as the heating do to

229

friction.

230

Since ω ∈]0, 2], using Corollary 1 allows to minorate eαjn+1

jn+1≥ ehαj n+1+ ν∆t mαj         1 −ω 2  X r ρβrn tδuαr nBn jrδu α r n+ω 2 X r ρβrn tδuαj n+1Bn jrδu α j n+1        + ∆t2 2mαj2       ων X r ρn rBnjr(δu α j n+1 −δuαrn)        2 , (45) which implies eαjn+1≥ ehαj

n+1, since friction terms are positive.

231

Property 9 (Positivity of internal energy). Assuming that ∀α ∈ { f1, f2}, ∀ j ∈ M, eαjn> 0, there

exists∆te> 0 such that the scheme (34)–(38) ensures that

∀ω ∈ [0, 2], ∀∆t ∈]0, ∆te[, ∀α ∈ { f1, f2}, ∀ j ∈ M, eαj n+1> 0.

Proof. The proof is obvious since eαjn+1≥ ehαjn+1and since ehαjn+1(∆t) is a polynomial of degree 2

232

satisfying ehαj

n+1(0)= eα j

n > 0. ∆teis nothing but the smallest root of these polynomial for each

233

cells of each fluid.

234

4.3.3. Entropy stability for general equations of state

235

In the previous paragraph, we provided explicitly a choice of∆t > 0 that ensures positivity

236

of internal energy and density for the proposed scheme, but this is not sufficient for stability. In

237

this section, we give an existence result of a strictly positive timestep∆t that ensures production

238

of physical entropy for arbitrary physical equation of state.

239

Let U=τ, uT, ETand let η be the entropy of the fluid. Gibbs formula reads T dη= de+pdτ. Following [23, 30], we estimate the entropy change, by means of a third-order Taylor expansion, due to the proposed scheme:

η(Uα j n+1 ) − η(Uαjn)= (Uαjn+1− Uαjn)T ∂η ∂U Uα j n +1 2(U α j n+1 − Uαjn)T ∂ 2η ∂U2 Uαjn (Uαjn+1− Uαjn)+ O(Uαjn+1− Uαjn)3 . 20

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One has ∂U∂η Uα j n = 1

jn(pαjn, −uαjn, 1)Tand the variable change V= (p, −u, η)T reads

(Uαjn+1− Uαjn)T ∂ 2η ∂U2 Uα jn (Uαjn+1− Uαjn)= (Vαjn+1− Vαjn)T ∂ 2η ∂V2 Vα jn (Vαjn+1− Vαjn) + O (Uαjn+1− Uαjn)3 , where, see [28, 23] for instance,

∂2η ∂V2 Vα jn = − 1 Tαjn             (ρc)αjn−2 0 0 0 1 0 0 0 0            . Let O1 := (Uαj n+1− Uα j n)T ∂η ∂U Uα j

n, using (34), (40) and (41), one gets

O1= 1 Tαjn ∆t mαj        pαjnX r Cnjr· uαrn −tuαjn        X r Aα,njr (uαr n− uα j n )+ ωνX r ρn rB n jr(δu α r nδuα j n+1 )        −X r Cnjrjn· uαrn+X r tuα r n Aα,njr (uαrn− uαjn) + νX r ρβrn tδuαr nBn jrδu α r nωνX r ρβrn tδuαr n+1Bn jrδu α r nδuα j n+1 +ωνX r ρn r tuα j n+1 Bnjrδuαrn−δuαjn+1        . which simplifies as O1= 1 Tαjn ∆t mαj        X r t(uα r n− uα j n)Aα,n jr (u α r n− uα j n) + νX r ρβrn tδuαr n Bnjrδuαrn−ωνX r ρβrn tδuαr n+1 Bnjrδuαrn−δuαjn+1        + 1 Tαjn ∆t mαj        ωνX r ρn r t (uαjn+1− uαjn)Bnjrδuαr nδuα j n+1        . Now using Lemma 1, one gets

O1= 1 Tαjn ∆t mαj        1 − 1 2ω ! νX r ρβr tδuαr n Bnjrδuαrn+1 2ων X r ρβrtδuαj n+1 Bnjrδuαjn+1        + 1 Tαjn ∆t mαj        X r t (uαr n− uα j n )Aα,njr (uαr n− uα j n )+ ωνX r ρβrn tδuαr nδuα j n+1 Bnjrδuαr nδuα j n+1        + 1 Tαjn ∆t mαj        ωνX r ρn r t(uα j n+1− uα j n)Bn jrδu α r nδuα j n+1        . 21

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Observe that later term is second-order in time, so that one retrieves as expected the entropy

240

production of the continuous in time scheme established in Property 5 page 11.

241

One now focuses on the second-order term of the entropy variation

O2:= 1 2(V α j n+1− Vα j n)T ∂2η ∂V2 Vα j n (Vαjn+1− Vαjn), which rewrites O2= 1 2(∆Ψ) T         (ρc)αjn−2 0 0 1       ∆Ψ, with ∆Ψ = pαjn+1− pαjn −uαjn+1+ uα j n ! .

One has to estimate pαjn+1− pαjn. Assuming that the equation of state p : (τ, e) → p(τ, e) is regular enough, one has

jn+1− pαjn= (ταjn+1−ταjn) ∂p ∂τ jn+ (e α j n+1− eα j n) ∂p ∂e jn+ O(∆t 2).

Using (34) and (42) and keeping only first-order terms, one has

jn+1− pαjn= ∆t mαj        X r Cnjr· uαrn        ∂p ∂τ jn + ∆t mαj               X r t(uα j n − uαrn)Aαjrn(uαjn− uαrn) −X r pαjnCnjr· uαrn        + ν        X r ρβrn tδuαr nBn jrδu α r n+ ωX r ρβrn tδuαj n+1Bn jr(δu α j n+1δuα r n)        +ων        X r ρn r t(uα j n − uαjn+1)Bnjr(δuαjn+1−δuαrn)               ∂p ∂e jn+ O(∆t 2).

Then, using (40), one gets

O2= − 1 Tαjn ∆t2 2mαj2 "(        X r Cnjr· uαrn        ∂p ∂τ jn +        X r t(uα j n − uαrn)Aαjrn(uαjn− uαrn) −X r pαjnCnjr· uαrn + νX r ρβrn tδuαr nBn jrδu α r n+ ωνX r ρβrn tδuαj n+1Bn jr(δu α j n+1δuα r n) +ωνX r ρn r t(uα j n − uαjn+1)Bnjr(δuαjn+1−δuαrn)        ∂p ∂e jn )2  (ρc)αjn−2 +        X r Aα,njr (uαr n− uα j n )+ ωνX r ρn rBnjr(δu α r nδuα j n+1 )        2 # + O(∆t3 ). 22

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Finally, putting all the pieces together, one has η(Uα j n+1 ) − η(Uαjn) = 1 Tαjn ∆t mαj        1 −1 2ω ! νX r ρβr tδuαr n Bnjrδuαr n+1 2ων X r ρβr tδuαj n+1 Bnjrδuαjn+1        + 1 Tαjn ∆t mαj       a − ∆t mαj(b+ c) + O(∆t 2)       , with a ≥ 0 and b ≥ 0. 242

Thus it remains to study the positiveness of a −m∆tα j

(b+c)+O(∆t2). There are two possibilities.

243

Case a> 0. In that case, there obviously exists ∆t > 0 such that Tαjnmαj η(Uα j n+1) − η(Uα j n) ∆t ≥ 1 − 1 2ω ! νX r ρβr tδuαr nBn jrδu α r n +1 2ων X r ρβr tδuαj n+1 Bnjrδuαjn+1.

Case a= 0. If a = 0, one has X r t(uα r n− uα j n)Aα,n jr (u α r n− uα j n)+ ωνX r ρβrn tδuαr nδuα j n+1 Bnjrδuαrn−δuαjn+1 = 0.

Since ω ≥ 0 and since Aα,njr and Bn

jrare positive matrices, all the terms in the sum are zeros. Let

us first focus ont(uα r n− uα j n)Aα,n jr (u α r n− uα j

n) = 0 terms. Two cases occur. In case of Eucclhyd

scheme, Aα,njr is positive definite so that one has uαrn= uαjn. For Glace scheme

t(uα r n− uα j n)Aα,n jr (u α r n− uα j n)= (ρc) α j n kCn jrk C n jr· (u α r n− uα j n) 2 = 0. So, for both scheme, one has Cn

jr· u α r n = Cn jr · u α j n and Aα,n jr (u α r n− uα j n) = 0. Recalling that 244 P

rCnjr= 0, one also has Prpαj nCn jr· u α r n= 0. 245

One now analyzes ωδuαrn−δuαjn+1TBn jrδu

α r

nδuα j

n+1 = 0. Here again two cases occur

246

ω = 0 or ω > 0. In that second case, since Bn

jrare positive definite, this implies δu α r n−δuα j n+1= 0. 247

Finally, if a= 0, one has Tαjnmαj η(Uα j n+1) − η(Uα j n) ∆t = ν X r ρβr tδuαr n Bnjrδuαrn − ∆t 2mαj        νX r ρβrn tδuαr nBn jrδu α r n        2 ∂p ∂e 2 jn  (ρc)αjn−2+ O(∆t2).

Before enunciating the result, one should remark that in the general case, one has

η(Uα j n+1) − η(Uα j n)= 1 Tαjn ∆t mαj        (a+ aν) − ∆t mαj(b+ c) + O(∆t 2)        , 23

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with a ≥ 0, aν ≥ 0 and b ≥ 0. Again, one has two alternatives a+ aν > 0 or a + aν = 0. In the

248

first case, there exists∆t such that η(Uαjn+1) − η(Uαjn) > 0. In the second case, one has a= aν= 0

249 so as previously, a= 0 =⇒ Cn jr· u α r n = Cn jr· u α j n, Aα,n jr (u α r n− uα j n)= 0 and δuα r nδuα j n+1= 0. 250

Also, since aν = 0 and since Bnjris positive definite one has δuαrn = 0, so the scheme (34)–(36)

251

gives Uαjn+1= Uαjnand finally one has ∀∆t > 0, η(Uαjn+1)= η(Uαjn).

252

The obtained results are summarized in the following Property.

253

Property 10 (Entropy). Let U :=τ, uT, ET and letη the entropy. There exists ∆tη > 0 , such

254

that ∀α, β ∈ { f1, f2}, such thatα , β, if the pressure law pα: (ρ, e) → pα(ρ, e) is a differentiable

255

function, then the scheme (34)–(38) defined by∆t = ∆tηand ∀ω ∈ [0, 2], ensures that,

256

1. the scheme is entropy stable:

∀ j ∈ M, η

jn+1≥ηUαjn ,

2. and ∀ j ∈ M, one has the following alternative. If ∀r ∈ Rj, Cnjr· uαrn = Cnjr· u α j n and δuα r nδuα j n+1= 0, thenjnmαj η(Uα j n+1) − η(Uα j n) ∆t ≥ν X r ρβr tδuαr n Bnjrδuαrn+ O(∆t), else Tαjnmαj η(Uα j n+1) − η(Uα j n) ∆t ≥ν X r ρβrtδuαr n Bnjrδuαrn. 257

Remark 3. Let us comment point 2 of property 10. Actually, this is a consistency result with

258

regard to (2). In the first case (if ∀r ∈ Rj, Cnjr· u α rn= Cnjr· u α j nandδuα rn−δuαj n+1= 0), the scheme 259

gives following valuesραjn+1= ραjn, uαjn+1= uαjnand eαjn+1= eαjn+m∆tα j ν Prρ β r nt δuα r nBn jrδu α r n. In 260

this case, the scheme acts simply as a first-order ODE solver. Since then dη = de and since η is

261

strictly convex, a time integration error is to be expected.

262

To sum up, we proved that the proposed scheme is stable, meaning that there exists 0 <∆t ≤

263

min(∆tρ, ∆te, ∆tη) such that the scheme is entropy stable and preserves positivity of density and

264

internal energy. Moreover, it is consistent with (2).

265

4.3.4. Minoration of∆tη

266

As stated before, to prove that the scheme is asymptotic preserving, it remains to show that

267

limν→+∞∆tη, 0. Even if we will not provide here an explicit value, we will give a lower bound

268

independent of ν.

269

Property 11. Letω ∈]0, 2]. ∀ j ∈ M, letτn j, u n j T, En j T

denote the initial state of fluidα ∈ { f1, f2}. Let {ur}r∈Rj, be an arbitrary set of nodal velocities (or velocity fluxes). Then, if ∀ν ≥ 0,

τν,n+1 j , e

ν,n+1 j



denotes the thermodynamic state obtained by scheme (34)–(36), one has ητν,n+1 j , e ν,n+1 j  ≥ητ0,nj +1, e0,nj +1 ,

whereη := η(τ, e) is the physical entropy expressed according to the independent variables τ

270

and e.

271

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Proof. Gibbs formula reads ∇τ,eη = T1p1, where T := T(τ, e) is a positive function. So, for any

272

τ, η(τ, ·) is an increasing function.

273

Since (34) is independent of ν and according to (45), one has ∀{ur}r∈Rj, ∀ν ≥ 0, ∀∆t τ ν,n+1 j = τ 0,n+1 j and eν,n+1j ≥ e 0,n+1 j , so ∀{ur}r∈Rj, ∀ν ≥ 0, ∀∆t η  eν,n+1j , τν,n+1j ≥ηe0,nj +1, τ0,nj +1 . 274

Remark 4. Property 11 establishes that, for a given set of velocities {ur}r∈Rj, the maximum 275

timestep required for the scheme to be entropy stable is greater than the mono-fluid timestep

276

independently ofν.

277

However, one emphisises that velocities {ur}r∈Rj are actually functions of ν as expressed 278

in (38). It turns out that entropy stability timestep depends on ν though the velocity fluxes and

279

can be either bigger or smaller than the mono-fluid timestep.

280

Example 1 (∆tη0> ∆tην). Let us consider two fluides in the monodimensional ]0, 1[ domain. Let

281

the first fluidα be a very light fluid at rest ρα =  with 0 <   1, uα= 0 and eαbeing set such

282

that the sound speed cα= 1. Let the second fluid β be the initial state of a Sod shock tube.

283

If ν = 0, one has obviously ∀r, uαr = 0. So, (34)–(36) implies Uαj

n+1 = Uα j n, which is 284 unconditionally stable. 285

Choosingν  1 and solving (38) implies that δuαr is arbitrary small, and we write ur= uαr =

286

uβr, and the sum equations (38) can be rewritten has

287 X j (Aαjr+ Aβjr)ur= X j Cjrp β j, (46)

since uα = uβ = 0 and pα is constant (recalling thatP

jCjr = 0). Since ραcα = , Aαjr is

288

neglectable with regard to Aβjr. So, one hasP

jA β

jrur= PjCjrpβj, that is the timestep for fluidα

289

is the same as one imposed by the fluidβ which is much smaller than the arbitrary one ontained

290

in the mono-fluid caseν = 0.

291

Example 2 (∆tη0 < ∆tην). Let us now consider a similar example. The two fluids in ]0, 1[ are

292

now described as follows. Let fluidα be in the initial state of a Sod shock tube. Fluid β is this

293

time a very heavy fluid at rest: ρβ = 1 with0 <   1, uβ = 0 and eβ being set such that the

294

sound speed cβ = 1.

295

Ifν = 0, stability for the fluid α is the one of the mono-fluid Sod shock tube.

296

Choosingν  1, (46) holds again. Since, ρβ = 1 is very large, one gets in the limit ur = 0,

297

which provides unconditionally stability for both fluids.

298

4.3.5. On the importance of the implicit velocities in (34)–(36)

299

Using the notations defined in section 4.2, let us consider the fully explicit scheme that con-sists in replacing momentum and total energy updates in (34)–(38) by their explicit counterparts

jjn+1− uα j n ∆t = − X r Fα,njr −ωX r νρn rB n jrδu α j n− (1 − ω)X r νρn rB n jrδu α r n,jjn+1− Eαjn ∆t = − X r Fα,njr · uαrn−X r νρn r tun rB n jrδu α r n+ ωX r νρn r tun jrB n jrδu α r nδuα j n . 25

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Using this scheme, one easily checks that internal energy variation reads eαjn+1= eαjn+ ∆t mαj        X r t(uα j n− uα r n)Aα jr n(uα j n− uα r n) −X r pαjnCnjr· uαrn        + ν∆t mαj        X r ρβrn tδuαr n Bnjrδuαr n+ ωX r ρβrn tδuαj n Bnjr(δuαjn−δuαrn)        − ∆t 2 2mαj2        X r Aαjrn(uαjn− uαrn)+ ωνX r ρn rB n jr(δu α j n −δuαrn)        2 . that is eαjn+1= ehαj n+1+ ν∆t mαj        X r ρβrn tδuαr n Bnjrδuαrn+ ωX r ρβrn tδuαj n Bnjr(δuαjn−δuαrn)        − ∆t 2 mαj2        ωνX r ρn rB n jr(δu α j nδuα r n)        ·        X r Aαjrn(uαjn− uαrn)        − ∆t 2 2mαj2       ων X r ρn rB n jr(δu α j n −δuαrn)        2 , where ehαjn+1still denotes the obtained internal energy without friction. The later term being a

300

negative factor of ν2, in the explicit case, ∀∆t > 0 for large values of ν, one can have eα j

n+1 <

301

ehαjn+1. So even if a similar result to Property 10 can be established (existence of an entropy

302

stable timestep), one cannot prove an equivalent of Property 11. If cell velocities are explicit,

303

one eventually gets lim

ν→+∞∆t e= lim

ν→+∞∆t

η= 0 for a given set of nodal velocities {u r}r∈Rj. 304

5. ALE scheme

305

The semi-Lagrangian scheme presented in this paper is defined assuming that both fluid

306

meshes are identical at the begining of the timestep. One understands easily that this is of huge

307

help in the construction of an asymptotic preserving scheme. One could imagine a purely

La-308

grangian approach, but even dealing with a non-AP approach seems very difficult since one

309

would have to consider meshes intersections and complex geometrical calculations.

310

Thus, the algorithm, we propose in this paper, consists in ensuring that for each timestep both

311

fluids meshes coincide. To do so an ALE formulation is mandatory.

312

Figure 2 depicts the general ALE case. Our ALE method is a Lagrange-rezoning-advection

313

procedure which ensures that the solution is defined at time tn+1on a unique mesh.

314

• At time tnsolutions are discretized on the meshes Mn α= Mnβ

315

• In a first step (Lagrangian phase), each mesh evolves in a different way ˜Mn+1

α , M˜nβ+1.

316

Each mesh being defined by ˜xα,n+1r = xnr+ ∆tu α,n r .

317

• Then the meshes are smoothed in a way to obtain new meshes such that Mn+1

α ≡ Mnβ+1. For

318

each fluid α, it allows to define an arbitrary velocity vα,n+1r such that xnr+1= ˜xα,n+1r +∆tvα,n+1r .

319

Figure

Figure 1: Illustration of C jr and N i jr vectors at vertex r for a polygonal cell j.
Figure 2 depicts the general ALE case. Our ALE method is a Lagrange-rezoning-advection
Figure 2: Left: at time t = t n , both fluid share the same mesh. Middle: at the end of the Lagrangian phase, one gets two di ff erent meshes, one for each fluid
Figure 3: ν = 100. ρ α + ρ β profile. AP-scheme (left) gives a much better solution than the non-AP scheme (right).
+6

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Pour l’algorithme A∗ il faut appliquer la proc´edure Recherche-Avec-Graphe RAG en triant les sommets dans OU V ERT selon la fonction suivante : f x = gx + hx En prenant gx coˆ