The Riemann proble for fluid mechanics: the case of a perfect gaz
R. Abgrall
May 16th, 2016
Contents
1 The Euler equations in one dimension: equivalent forms of the equations
1
2 Particular solutions.
3
3 General case, Riemann invariants
3.1 Simple waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 Behavior of simple waves. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
4
6
4 Discontinuous solution, Rankine Hugoniot relations.
4.1 Practical evaluation of the states in a shock. . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2 Admissible shocks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
9
11
5 Parametrization of shock and expansion
5.1 Parametrization of expansion waves . .
5.2 Parametrization of shock waves . . . . .
5.3 Final parametrisation . . . . . . . . . .
waves
12
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
6 Solution of the Riemann problem
1
14
The Euler equations in one dimension: equivalent forms of the
equations
We recall the conservative and various non conservative forms of the Euler equations. If W = (ρ, ρu, E)
where ρ is the density, u the velocity and E the total energy, i.e
1
E = ρ(e + u2 )
2
and if the gas is a perfect gas p = ρRT , calorically perfect ρe = cv T , we easily get
p = (γ − 1)ρe
In what follows, we often denote κ = γ − 1.
The conservative form of the Euler equations is
∂ρ ∂(ρu)
+
=0
∂t
∂x
∂(ρu) ∂(ρu2 + p)
+
=0
∂t
∂x
∂E
∂(u(E + p))
+
=0
∂t
∂x
1
and we need to add the second principle of thermodynamics
∂(ρs) ∂(ρus)
+
=≥ 0.
∂t
∂x
The Lagrangian for of the equations is, with d/dt the material derivative
∂φ
∂φ
dφ
=
+u ,
dt
∂t
∂x
dρ
∂u
+ρ
=0
dt
∂x
du ∂p
ρ
+
=0
dt
∂x
∂u
∂e
+p
=0
ρ
∂t
∂x
We notice that the energy equation also writes, using mass conservation, as
ρ
de p dρ
−
=0
dt
ρ dt
i.e., introducing the specific volume v = 1/ρ and the Gibbs relation
T ds = de + pdv = de −
p
dρ,
ρ2
we see that the energy equation implies
ds
= 0.
dt
Remark 1.1. Lagrangian coordinates The one dimensional Lagrangian form can also be written in conservation form via a change of variable. We introduce the mass coordinate
dm = ρdx
. then we get
dv
∂u
−
=0
dt
∂m
∂p
du
+
=0
dt
∂m
∂ve ∂(up)
+
= 0.
∂t
∂m
We also have
∂(ρe)
∂(ρe)
∂u
+u
+ (p − ρe)
=0
∂t
∂x
∂x
and since the gas is caloricaly perfect,
p = κρe,
we get
∂p
∂u
∂p
+u
+ ρc2
=0
∂t
∂x
∂x
where c2 = γp/rho is the square of the sound speed. This relation is true for any divariant gaz, setting
c2 =
∂p
.
∂ρ |s
2
2
Particular solutions.
We look for solutions U = (ρ, u, p)T of the form U (x, t) = V (x − ϕ(t). We write the problem as


u ρ
0
∂U 
∂U
+ 0 u 1/ρ
=0
∂t
∂x
2
0 ρc
u
and set

u ρ
A = 0 u
0 ρc2

0
1/ρ .
u
We get
∂U
∂U
+A
=0
∂t
∂x
and see that
∂U
∂U
= V 0 (x − ϕ(t)) and
= ϕ0 (t)V 0 (x − ϕ(t)).
∂x
∂t
Hence, setting ξ = x − ϕ(t), we get
− ϕ0 (t)Id − A V 0 (ξ) = 0.
This relation shows that
• either V 0 = 0,
• or V 0 (ξ) is an eigenvector of A and ϕ0 (t) is one of its eigenvalues.
This leads to look at the eigenvalues of A, and if this matrix is diagonalisable.
The characteristic polynomial of A is
2
2
P (λ) = det(A − λId) = (u − λ) (u − λ) − c
thus λ = u where λ = u ± c. Since c 6= 0, the eigenvalues are different, and then the matrix is diagonalizable
in R3 .
The eigenvectors are
• for λ = u, we have
 
1
r0 = 0 .
0
• for λ = u + εc with ε = ±1, we have


ρ
r± =  εc 
ρc2
We look for the eigenforms. Let us consider X = (A, B, C)T where A has the dimension of a density, B
of a velocity and C of a pressure. Writing
X = α0 (X)r0 + α1 (X)r1 + α−1 r−1 ,
we see that
α0 = A −
α1 =
C
c2
1
2ρc2 (ρcB
α−1 =
+ C)
1
2ρc2 (−ρcB
3
+ C).
The αi s are the linear forms that are dual of the ri .
Let us come back to computing the solutions. We write V 0 = (A0 , B 0 , C 0 ) and have 3 cases.
• If λ = u, V 0 is colinear to r0 so that α1 (X) = α−1 (X) = 0 and thus
u0 = 0 and p0 = 0.
This shows that p and u are constant. The density is arbitrary.
• For λ = u + c. V 0 is colinear to r1 so that
α0 (X) = 0 and α−1 (X) = 0.
We write these relations and thanks to the perfect gas EOS,
ρp0
= 0 i.e.
γp
p0
u0 +
=0
cρ
ρ0 −
γ
ρ0
p0
=
ρ
p
The first relation shows that pρ−γ = p0 ρ−γ
where p0 et ρ0 are arbitrary. This relation shows that p is
0
a function of ρ. Similarly, we have
γ−1
p0 ρ
p
,
c2 = γ = γ
ρ
ρ0 ρ0
This shows that the pressure and the accoustic impedance ρc only depend on the density. Hence, we
can parametrize this kind of solution via the density ρ: we can assume that ξ = ρ and hence we have
Z ρ 0
p
dρ = 0
u − u0 +
ρ0 cρ
This leads to
Z
p0
2
dρ =
c.
cρ
γ−1
For a wave associated to λ = u + c, we get
pρ−γ = p0 ρ−γ
and u +
0
2
2
c = u0 +
c0 .
γ−1
γ−1
• If λ = u − c, a similar calculation provides
pρ−γ = p0 ρ−γ
and u −
0
3
2
2
c = u0
c0 .
γ−1
γ−1
General case, Riemann invariants
3.1
Simple waves
We come back on
and we use the linear forms of A since
∂U
∂U
+A
=0
∂t
∂x
1
αi A = λi αi
1 In
the general
Pof A diagonalisable with the eigenvectors ri and linear forms αi , we have for any
P case, if λi is an eigenvalue
vector X X = i αi (X)ri and then AX = i λi αi (X)ri . This shows that αj AX = λj αj (X).
4
We have
α0
α1
∂U
∂t
∂U
∂t
α−1
+ uα0
∂U
∂x
+ (u + c)α1
∂U
∂t
=0
∂U
∂x
+ (u − c)α−1
=0
∂U
∂x
=0
If ξ = t or ξ = x, we get
α0
α1
∂U
∂ξ
∂U
∂t
α−1
This shows that
∂ρ
1 ∂p
− 2
∂t
c ∂t
∂u
1 ∂p
+
∂t
ρc ∂t
−
∂U
∂t
∂ρ
1 ∂p
− 2
∂ξ
c ∂ξ
=
1 ∂u
1 ∂p
+
2c ∂ξ
ρc ∂ξ
=
1
2c
−
∂u
1 ∂p
+
∂ξ
ρc ∂ξ
∂ρ
1 ∂p
+u
− 2
=0
∂x c ∂x
+ (u + c)
1 ∂p
∂u
+
∂t
ρc ∂t
The first relation writes
i.e.
=
1 ∂p
∂u
+
∂x ρc ∂x
=0
∂u
1 ∂p
+ (u − c) −
+
=0
∂x ρc ∂x
dρ
u dp
+ 2
=0
dt
c dt
ds
∂p
= 0 because c2 =
: The flow is isentropic.
dt
∂ρ |s
For the next two relations, we first notice that
Z
Z
1 ∂p
∂
1 ∂p
∂
dp
dp
=
et
=
ρc ∂t
∂t
ρc
ρc ∂x
∂x
ρc
i.e.
∂
∂
+ (u ± c)
∂t
∂x
u±
This shows that
RI± = u ±
Z
Z
dp
=0
ρc
dp
ρc
stay constant along the characteristics
dx
= u ± c.
dt
The flow is characterized by
s = cte
dp
dx
u+
= cte on
=u+c
ρc
dt
Z
dp
dx
u−
= cte on
=u−c
ρc
dt
Z
5
(1)
For a perfect gas with constant specific heats, we have seen that
Z
2
dp
=
c + Cte.
ρc
γ−1
Definition 3.1 (Rieman invariants, simple waves). We call simple wave any solution such that one of the
Riemann invariants
Z
dp
u±
,
ρc
i.e.
2
u±
c
γ−1
for a caloricaly perfect gas, stay uniform in the flow.
We call wave (C + ) the waves defined by
dx
2
= u + c andu +
c = cte
dt
γ−1
and wave (C − ) the waves defined by
2
dx
= u − c and u −
c = cte
dt
γ−1
We see that the solutions studied in section 2 are particular cases of simple waves. Another particular
case are the simple centered waves, i.e the solution of the type U (x, t) = V (x/t).
3.2
Behavior of simple waves.
We assume (without loss of generality) that
u−
2
a = J−
γ−1
u+
2
a = JC+ .
γ−1
uniformly. On any (C + ), we also have
By the index C, we wish to indicate that the constant JC+ depends on the characteristic (C + ) which is chosen.
Note that pρ−γ is also uniform in the whole domain.
The first remark is that
J − + JC+
γ−1 +
u=
and c =
(J − JC− ).
2
2
Since u and c are constant on a (C + ). This shows that if we consider a (C + ) we also have a constant u + c
and thus the characteristics are straight lines.
They are X(t) = x0 + t(u + c)0 t where (u + c)0 is the value of u + c at x0 and t = 0.
Now we look at the behavior or u + c : We know that it is a constant on any characteristic (C + ), but what
happens when we change characteristics ? We can parametrize u + c only in term of the density. Indeed,
(γ−1)/2
because the flow is isentropic and
a = a0 ρρ0
u = J− +
2
2a0 ρ (γ−1)/2
a = J− +
= G(ρ).
γ−1
γ − 1 ρ0
Now let us look at the behavior of u + c in the wave. Since pρ−γ is constant, as well as u −
everything can be parametrized in term of the density. Let us look at the behavior of de
du
2 dc
=
,
dρ
γ − 1 dρ
6
d(u+c)
dρ .
2
γ−1 c,
We have
then
d(u + c)
du
dc
=
+
dρ
dρ dρ
2
dc
= 1+
γ − 1 dρ
γ + 1 dc
γ − 1 dρ
Using the entropy relation, we see that
2
c =
c20
ρ
ρ0
γ−1
where
γ+1
d(u + c)
=
c>0
dρ
2ρ
This shows that the slope u + c increases when ρ increases. Similarly, we would have got, in a simple wave
such that
2
u+
c
γ−1
is uniform,
d(u − c)
γ+1
=−
c<0
dρ
2ρ
and the slope u − c decreases when ρ increases.
Remark 3.1. In fact, we have shown that in a (C + ), we have
d(u − c)
γ+1
=−
c<0
dρ
2ρ
and in a C − ,
d(u + c)
γ+1
=
c > 0.
dρ
2ρ
4
Discontinuous solution, Rankine Hugoniot relations.
A discontinuity of the Euler equations satisfies the Rankine Hugoniot relations that, in that case, write
∆ρu = σ∆ρ,
(a)
∆ ρu2 + p = σ∆ρu,
(b)
∆ (ρe + p)u = σ∆ ρe .
(c)
(2)
We denote by (0) and (1) the two states on the left and right of the shock. We also introduce the relative
speeds
vi = ui − σ
We get (using for example the Galilean invariance of the Euler equations, or

ρ 0 v0 = ρ 1 v1 ,





ρ0 v02 + p0 = ρ1 v12 + p1 ,





(ρ0 ε0 + ρ0 v02 /2 + p0 )v0 = (ρ1 ε1 + ρ1 v12 + p1 )v1 .
7
a calculation),
(a)
(b)
(c)
(3)
We set M = ρ0 v0 = ρ1 v1 , the mass flux that crosses the discontinuity. We have two possible cases: Either
M = 0, and then
u0 = u1 = σ,
(4)
p0 = p1
and the only possibility for having a discontinuity is to have
ρ0 6= ρ1 .
This kind of discontinuity is a contact discontinuity.
When M 6= 0, all the variables are discontinuous accross the discontinuity: this is a shock wave. In that
case, we easily get

M = (u1 − u0 )/(τ1 − τ0 ),
(a)





M 2 = −(p1 − p0 )/(τ1 − τ0 ),
(b)
(5)





ε1 − ε0 + (p1 + p0 )(τ1 − τ0 )/2 = 0, (c)
where τ = 1/ρ is the specific heat
Proof. We set M = ρ0 v0 = ρ1 v1 the mass flow accross the shock. Since M 6= 0, we get
ρ0 v02 = p0 = M v0 + p = ρ1 v12 + p1 = M v1 + p1
i.e.
M =−
i.e.
−1 p1 − p0
p1 − p0
=
v1 − v0
M τ1 − τ0
M2 = −
The third relation is equivalent to
p1 − p0
.
τ1 − τ0
ε0 + v02 /2 + p0 τ0 = ε1 + v12 /2 + p1 τ1
Again, we have used that M 6= 0. This is rewritten as
1
∆ε + ∆v 2 + ∆(pτ ) = 0.
2
Note that if f represents p, v or τ , we denote
f=
f0 + f1
2
and then
1
∆ε + ∆v 2 + ∆(pτ ) = ∆ε + v∆v + p∆τ + τ ∆p
2
v v
= ∆ε + p∆
+
∆p + M ∆v
M
D
v = ∆ε + p∆
M
=0
where again we ave used τ M = v and the jump relation on the momentum.
8
Remark 4.1.
1. The relations (a) and (b) can be combined to get
M = −(p1 − p0 )/(u1 − u0 ).
(6)
2. The relation (b) shows that the pressure and the specific volume have an opposite behavior.
3. The relation (c) is a pure thermodynamic relation. Indeed, since the thermodynamical state is completely defined by two state variables, we can write ε = ε(p, τ ) and (c) is a relation between p and τ .
This relation is called the Hugoniot relation. It is valid without any assumption on the equation of
state.
Now we look more closely to the Hugoniot relation. The Hugoniot function is
H(p, τ ) = ε(p, τ ) − ε0 +
(p + p0 )
(τ − τ0 ).
2
The graph of this function in the (p, τ ) (denoted as H) is called the Hugoniot locus. It is the set of all
thermodynamic states that can be connected by a shock. to (τ0 , p0 ). For a perfect gas, we have
H(p, τ ) =
(p + p0 )
pτ − p0 τ0
+
(τ − τ0 ),
γ−1
2
i.e.
2µ2 H(p, τ ) = (τ − µ2 τ0 )p − (τ0 − µ2 τ )p0
with µ2 = (γ − 1)/(γ + 1) . This shows that the Hugoniot locus is an hyperbola with asymptots τ = µ2 τ0
and p = −µ2 p0 .
The pressure is negative for τ > µ−2 τ0 and only the part defined for τ ∈ [µ2 τ0 , µ−2 τ0 ] has a physical meaning.
4.1
Practical evaluation of the states in a shock.
For a perfect gas, the Hugoniot function can easily give an explicit relation between the two states accross
a shock. Indeed, since
pτ
,
ε=
γ−1
we have
and combining with
1
∆ε =
τ̄ ∆p + p̄∆τ
γ−1
∆ε + p̄∆τ = 0,
we get
τ̄ ∆p +
i.e.
γ+1
p̄∆τ = 0
2
τ0 − τ1
1 p0 − p1
=−
.
τ0 + τ1
γ p0 + p1
We solve and get
We also have
ρ0
τ1
v1
(γ + 1)p0 + (γ − 1)p1
=
=
=
.
ρ1
τ0
v0
(γ − 1)p0 + (γ + 1)p1
(γ + 1) − (γ − 1) ρρ10
p1
=
.
p0
(γ + 1) ρρ10 − (γ − 1)
9
p
A0
p
0
µ
-
-2
τ0
τ
2
µ τ0 τ0
2
µ p0
Figure 1: Hugoniot locus for a perfect gas.
We can determine the ratio ρ0 /ρ1 in term of M0 = v0 /c0 the Mach number before the shock. We have
c20 +
γ−1 2
γ−1 2
v0 = c21 +
v1 ,
2
2
introducing x = ρ0 /ρ1 and knowing that v1 = xv0 ,
1+
Then
γ−1 2
c2
γ−1 2 2
M0 = 02 +
M0 x .
2
c1
2
c20
p1 ρ0
(γ + 1) − (γ − 1)x
=
=x
2
c1
p0 ρ1
(γ + 1)x − (γ − 1)
which is introduced in (7):
1+
so that
i.e.
γ−1 2
(γ + 1) − (γ − 1)x γ − 1 2 2
M0 = x
+
M0 x
2
(γ + 1)x − (γ − 1)
2
1 − x2
M2
= 0 (1 − x2 )
(γ + 1)x − (γ − 1)
2
1
M2
= 0
(γ + 1)x − (γ − 1)
2
10
(7)
and then
γ−1
2
ρ0
.
=
+
ρ1
γ + 1 (γ + 1)M02
(8)
p1
2γM02
γ−1
=
−
p0
γ+1
γ+1
(9)
From this we get
We also have a relation between the Mach numbers before and after the shock M1 = v1 /c1 ,
M12 =
2 + (γ − 1)M02
2γM02 − (γ − 1)
which can be rewritten as
2γM12 M02 − (γ − 1)(M12 + M22 ) = 2.
We can also see that
M02 ≤ 1 if and only if M12 ≥ 1.
4.2
Admissible shocks
The problem is the following: Given a state (p1 , τ1 ) that we have just computed, is this state physically
admissible, or not ? Is this state compatible with the second principle of thermodynamics?
The second principle requests that the entropy, taken here as p/ργ , of a fluid particle that one follows in
its movement must increase with time:
Ds
≥ 0.
Dt
For C1 solutions, it is easy to see that l
Ds
= 0,
Dt
i..e the smooth solutions are compatible with the second principle of thermodynamics. What about discontinuous solutions ?
In order to look at this problem, we first consider the case v0 > 0, i.e. v1 > 0. In that case we need,
because of the second principle, s1 > s0 . We have
s1 − s0 = log(p1 /p0 ) + γ log(ρ0 /ρ1 )
(γ + 1) − (γ − 1) ρρ10
= log
+ γ log g(ρ0 /ρ1 )
(γ + 1) ρρ10 − (γ − 1)
We compute the first derivative with respect to x = ρ0 /ρ1 : the derivative est
(γ 2 − 1)
(x − 1)2
x (γ + 1)x − (γ − 1) (γ − 1)x − (γ + 1)
is negative for ρ0 ≤ ρ1 : we need ρ0 ≤ ρ1 in such a shock.
If v0 < 0, i.e. v1 < 0, we need ρ0 ≥ ρ1 . These results can generalised for a very large class of EOS.
Proposition 4.1. A shock is admissible if and only if it satisfies any of the equivalent conditions
• If M > 0 then

s1 ≥ s0



p1 ≥ p0
;
ρ
 1 ≥ ρ0


u0 ≥ u1
11
(10)
• if M < 0, then

s0 ≥ s1



p1 ≥ p0
ρ
 0 ≥ ρ1


u1 ≥ u0
(11)
Proof. We deal only with the case M > 0, the case M < 0 is similar. Since the entropy decreases with τ ,
we have s1 ≥ s0 if and only if tau0 ≥ τ1 and since p is decreasing in τ , these two conditions are equivalent
to p1 ≥ p0 . We now look at the fourth condition. We have
ρ0 (u0 − σ) = ρ1 (u1 − σ)
and in an admissible shock, ρ1 ≥ ρ0 for M > 0: this leads to
u1 − σ ≤ u0 − σ
i.e. u1 ≤ u0 . Reciprocicaly, if u1 ≤ u0 , then ρ1 ≥ ρ0 : this show that the last condition is equivalent to any
of the first three ones.
In term of the Mach number, we can also characterize the admissible shocks
Proposition 4.2. Let us define M0 =
|u0 − σ|
|u1 − σ|
et M1 =
A shock is admissible if and only if
a0
a1
• If M > 0 then M0 ≥ 1 and M1 ≤ 1,
• If M < 0 then M0 ≤ 1 and M1 ≥ 1.
Proof. If M > 0, we know that ρ1 ≥ ρ0 , and since:
ρ0
γ−1
2
=
+
,
ρ1
γ + 1 (γ + 1)M02
we see that
ρ0
ρ1
≤ 1 implies M02 ≥ 1. Since M0 =
v0
a0
≥ 0, we have M0 ≥ 1, and then M1 ≤ 1
If we consider the case M > 0 (i.e. u1 > σ), an admissible shock is defined by the two conditions
u0 − σ > a0 and u1 − σ < a1
i.e.
u0 − a0 > σ > u1 − a1
If the case M < 0 (i.e. u0 < σ) the condition to have an admissible discontinuity are
⇔ u0 + a0 > σ > u1 + a1
T hese are the Lax conditions.
5
Parametrization of shock and expansion waves
.
12
5.1
Parametrization of expansion waves
Recall that in an expansion wave connection the states (ρ0 , u0 , p0 ) to (ρ, u, p), the entropy stays constant as
well as on Riemann invariant. Since
2
a
R± = u ±
γ−1
and
pρ−gamma = p0 ρ−γ
0 .
Hence, we have
γ+1
2
p 2γ
2
R± = u ±
= u0 ±
γ − 1 p0
γ−1
i.e.
2
u − u0 ±
γ−1
p
p0
γ+1
2γ
−1 =0
(12)
is an equation of the expansion wave in the coordinates (u, p).
5.2
Parametrization of shock waves
Remember we have
M (τ − τ0 ) = −(u − u0 )
M (u − u0 ) = p − p0
, so
p
u − u0 = ± −(p − p0 )(τ − τ0 ).
The sign is related to the sign of M , since we must have p − p0 ≥ 0 from the admissibility condition.
Remember that
τ1
(γ + 1)p0 + (γ − 1)p1
=
,
τ0
(γ − 1)p0 + (γ + 1)p1
so that
r
p
τ
1
−( − 1)( − 1)
p0 τ0
p0
τ0
s
1
(γ + 1)p0 + (γ − 1)p1
p
±√
−( − 1)
p0 τ0
p0
(γ − 1)p0 + (γ + 1)p1
u − u0 = ± √
i.e.
1
u − u0 = √
p 0 τ0
s
(γ + 1) pp10 + (γ − 1)
p
− 1)
p0
(γ − 1) pp10 + (γ + 1)
(13a)
(γ + 1) pp01 + (γ − 1)
p
−( − 1)
p0
(γ − 1) pp01 + (γ + 1)
(13b)
−(
when M > 0 and
1
u − u0 = − √
p0 τ0
s
when M < 0. In these relations, we need to have pp0 ≥ 1.
It is easy to show that the shock curves (13) and the expansion curves (12) have a second order contact
(same slope, same second derivatives)
13
5.3
Final parametrisation
We define the shock/expansion curves:
• D+ :
ϕ+ (u, p; u0 , p0 ) =
• D− :
ϕ− (u, p; u0 , p0 ) =




 u − u0 −
s
−( pp0
√1
p0 τ 0



 u − u0 −
2
γ−1




 u − u0 +
√1
p0 τ 0



 u − u0 +
2
γ−1
p
p0
− 1)
γ+1
2γ
(γ − 1) pp10 + (γ + 1)
−1
s
−( pp0 − 1)
p
p0
γ+1
2γ
(γ + 1) pp01 + (γ − 1)
(γ + 1) pp01 + (γ − 1)
(γ − 1) pp01 + (γ + 1)
−1
if
p0
p1
≥1
else.
if
p0
p1
≥1
else.
Both functions ϕ± are C 2 , and a shock/expansion curve is ϕ± (u, p; u0 , p0 ) = 0.
Now we are in a position to solve the Riemann problem.
6
Solution of the Riemann problem
Consider the states (ρL , uL , pL ) and (ρR , uR , pR ) A general results states :
Théorème 6.1. Consider a strickly hyperbolic (1D) system where all the fields are either genuinely non
linear or linearly degenerate. Then for any UL ∈ Ω (Ω is the set where the fux and entropy are defined),
there exists a neighborhood of uL in Ω such that for any uR in that neighborhood, the Riemann problem has
a unique weak solution that consists of p + 1 constant states separated by rarefaction waves, admissible shock
waves or contact discontinuities. Moreover, a solution of this kind is unique.
In the coordinates (p, u) a contact discontinuity is simply a point. So the problem amount to looking at
the interesection of the curves ϕ− (u, p; uL , pL ) = 0 and ϕ+ (u, p; uR , pR ) = 0.
We give the example of the Sod case, where uL = uR = 0 (cf. figure 2). Cette situation correspond
état droit
état gauhe
ρd , ud = 0, Pd , γd
ρg , ug = 0, Pg , γg
X0
X
Figure 2: Initial conditions
à une expérience de type tube à choc. Initialement deux fluides au repos sont séparés par une membrane
positionnée en x0 . On suppose que la pression du fluide situé à gauche de la membrane est plus élevée que la
pression du fluide à droite (Pg > Pd ). Les deux fluides obéissent à une équation d’état du type gaz parfait de
coefficients polytropiques γg et γd . A t = 0 la membrane est enlevée et les fluides se mettent en mouvement.
On suppose que les parois latérales du tubes sont suffisamment éloignées de l’interface de sorte qu’on ne
prendra pas en compte de phénomènes de réflexion d’ondes.
14
P
(Id )
(Ig )
Pg
P⋆
Pd
u⋆
0
u
Figure 3: Tube à choc : courbes d’interaction dans le plan (u, P )
En vertu de l’hypothèse Pg > Pd , l’examen des courbes (Ig ) et (Id ) (cf. figure 3) montre que nous avons
à faire une interaction du type détente-choc avec Pd < P ? < Pg . Sur le diagramme (x, t) (cf. figure 4) nous
observons que l’onde de choc se déplace vers la droite avec une vitesse Dd et que l’onde de détente se propage
vers la gauche. Le faisceau de détente se compose de caractéristiques (C − ). Une discontinuité de contact est
présente et elle se déplace à la vitesse u? .
t
X = u⋆ t
C−
X = Dd t
ρ⋆g , u⋆, P⋆
ρ⋆d , u⋆, P⋆
ρd , ud , Pd
ρg , ug , Pg
X0
X
Figure 4: Tube à choc : diagramme (x, t)
Après cette description nous passons maintenant au calcul des variables de l’écoulement.
L’onde de choc est définie par la polaire de choc de pôle (ud , Pd ) et nous avons :
P ? − Pd = Md (u? − ud ) ,
avec Md > 0 qui désigne la masse balayée par unité de temps par l’onde de choc définie par :
r
Md =
ρd
((γd + 1)P ? + (γd − 1)Pd ).
2
15
(14)
L’onde de détente est définie par la courbe de détente de pôle (ug , Pg ) et il vient :


g −1
? γ2γ
g
P
2
− 1 .
u? − ug = −
cg 
γg − 1
Pg
Puisque ug = ud = 0 on obtient P ? solution de l’équation non linéaire :


g −1
r
? γ2γ
g
2
ρ
P
d
P ? = Pd −
cg
((γd + 1)P ? + (γd − 1)Pd ) 
− 1 .
γg − 1
2
Pg
(15)
(16)
On peut résoudre cette équation non linéaire à l’aide d’une méthode de point fixe. Pour initialiser le processus
itératif on pourra utiliser la solution approchée (ũ, P̃ ) définie précédemment. Une fois connu P ? on calcule
u? au moyen de (14). On évalue ensuite Md pour en déduire la vitesse du choc Dd :
Dd =
Md
.
ρd
En utilisant la première relation de Rankine-Hugoniot nous obtenons la densité ρ?d :
ρ?d = ρd
Dd
.
Dd − u?
(17)
La zone en détente est une onde simple dans laquelle l’invariant de Riemann J + est constant, par
conséquent les caractéristiques (C − ) sont des droites de pentes u − c. Le long des (C + ) on a J + = cte. Nous
en déduisons :
0
u − c = x−x
t ,
2
u + γg −1 c = γg2−1 cg .
Tous calculs faits nous obtenons :
(
γ −1
0
c(x, t) = γg2+1 cg − γgg +1 x−x
t ,
2
2 x−x0
u(x, t) = γg +1 cg − γg +1 t .
(18)
La pression est la densité sont obtenues en considérant l’isentrope de pôle (τg , Pg ) :

h
i 2γg

 P (x, t) = Pg c(x,t) γg −1 ,
cg
h
i 2

 ρ(x, t) = ρ c(x,t) γg −1 .
g
cg
(19)
Pour calculer ρ?g nous utilisons la constance de J + et il vient :
c?g = cg −
γg − 1 ?
u ,
2
et nous en déduisons :
ρ?g
= ρg
c?g
cg
γ 2−1
g
.
Nous récapitulons les résultats obtenus dans la section suivante.
Pour t > 0 nous obtenons la solution suivante :
• x ≤ x0 − cg t
16
(20)

 ρ(x, t) = ρg ,
u(x, t) = 0,

P (x, t) = Pg ,
• x0 − cg t ≤ x ≤ x0 + (u? − c?g )t

i 2
h
γg −1 x−x0 γg −1
2


c
−
,
ρ(x,
t)
=
ρ
g
g

γg +1
γg +1
t

2 x−x0
2
u(x, t) = γg +1 cg − γg +1 t ,

h
i 2γg


γg −1 x−x0 γg −1
 P (x, t) = P
2
c
−
,
g γg +1 g
γg +1
t
• x0 + (u? − c?g )t ≤ x ≤ x0 + u? t

 ρ(x, t) = ρ?g ,
u(x, t) = u? ,

P (x, t) = P ? ,
• x0 + u? t ≤ x ≤ x0 + Dd t

 ρ(x, t) = ρ?d ,
u(x, t) = u? ,

P (x, t) = P ? ,
• x ≥ x0 + Dd t

 ρ(x, t) = ρd ,
u(x, t) = 0,

P (x, t) = Pd .
A titre d’illustration nous donnons en figure 5 les profils de densité, pression, vitesse et énergie interne
que l’on obtient pour l’application numérique suivante : x0 = 0.5, ρg = 1, Pg = 1, ρd = 0.125 et Pd =
0.1 avec γd = γg = 1.4. Cette configuration est connue dans la littérature sous le nom de tube à choc
de Sod. La solution correspondante est fréquemment utilisée comme référence pour valider des schémas
numériques. L’écoulement est caractérisé par les valeurs suivantes : P ? = 0.303130178, u? = 0.92745262,
Dd = 1.75215573, ρ?d = 0.265573712 et ρ?g = 0.426319428.
17
1
1
t=0.2
0.9
0.8
0.8
0.7
0.7
0.6
0.6
t=0.2
ρ
P
0.9
0.5
0.5
0.4
0.4
0.3
0.3
0.2
0.2
0.1
0
0.1
0.2
0.3
0.4
0.5
X
0.6
0.7
0.8
0.9
0.1
1
0
0.1
0.2
0.3
densité
0.4
0.5
X
0.6
0.7
0.8
0.9
1
pression
1
3
t=0.2
t=0.2
0.9
2.8
0.8
2.6
0.7
2.4
0.5
ε
u
0.6
2.2
0.4
0.3
2
0.2
1.8
0.1
0
1.6
0
0.1
0.2
0.3
0.4
0.5
X
0.6
0.7
0.8
0.9
1
0
0.1
0.2
0.3
0.4
0.5
X
0.6
18
vitesse
énergie interne
0.7
0.8
0.9
1