1. The Euler equations

The Euler equations are often used as a simplification of the Navier-Stokes equations as a modelof the flow of a gas. In one space dimension these represent the conservation of mass, momentumand energy, and read

(1)

ρρvE

t

+

ρvρv2 + pv(E + p)

x

= 0.

Here ρ denotes the density of the gas, v the velocity, p the pressure and E the energy. To closethis system, i.e., to reduce the number of unknowns to the number of equations, one can add aconstitutive “law” relating these. Such laws are often called equations of state and are deducedfrom thermodynamics. For a so called ideal polytropic gas the equation of state takes the form

E =p

γ − 1+

1

2ρv2,

where γ > 1 is a constant spesific to the gas. For air, γ ≈ 1.4. Solving for p, we get

(2) p = (γ − 1)E − γ − 1

2ρv2 = (γ − 1)E − γ − 1

2

q2

ρ,

where the momentum q = ρv. Inserting this in the Euler equations ρρvE

t

+

ρvγ−32 ρv2 + (γ − 1)E

v(γE − γ−1

2 ρv2)x

= 0.

In the conserved variables ρ, q and E, this system of conservation laws read

(3)

ρqE

t

+

q(3−γ2

)q2

ρ + (γ − 1)E

γEqρ −(γ−12

)q3

ρ2

x

= 0.

Set

u =

ρqE

and f(u) =

q(3−γ2

)q2

ρ + (γ − 1)E

γEqρ −(γ−12

)q3

ρ2

.

Then the Jacobian df(u) reads

df(u) =

0 1 0(γ−32

)q2

ρ2 (3− γ) qρ γ − 1

−γEqρ2 + (γ − 1) q3

ρ3 γEρ −3(γ−1)

2q2

ρ2 γ qρ

.

Introducing the enthalpy as

H =E + p

ρ

= γE

ρ−(γ − 1

2

)q2

ρ2

=γ

ρ

(p

γ − 1

)+

1

2v2,

the Jacobian can be rewritten as

df(u) =

0 1 0(

γ−32

)v2 (3− γ) v γ − 1(

γ−12

)v3 − vH H − (γ − 1) v2 γv

.

To find its eigenvalues, we compute the determinant

det (λI − df(u)) = λ[(λ− (3− γ)v) (λ− γv) + (γ − 1)

((γ − 1) v2 −H

)]1

2

+3− γ

2v2 (λ− γv) + (γ − 1)

(vH − γ − 1

2v3)

= λ[λ2 − 3vλ+ γ(3− γ)v2 + (γ − 1)2v2 + (γ − 1)H

]+

3− γ2

v2λ− 1

2(γ + 1)v3 + (γ − 1)Hv

= λ

[λ2 − 3vλ+ 2v2 +

1

2(γ + 1) v2 − (γ − 1)H

]− 1

2(γ + 1)v3 + (γ − 1)vH

= λ

[(λ− v)(λ− 2v) +

1

2(γ + 1)v2 − (γ − 1)H

]− 1

2(γ + 1)v3 + (γ − 1)vH

= (λ− v)

[λ(λ− 2v) +

1

2(γ + 1)v2 − (γ − 1)H

]= (λ− v)

[(λ− v)2 −

(v2 − 1

2(γ + 1)v2 + (γ − 1)H

)]= (λ− v)

[(λ− v)2 −

(γ − 1

2(2H − v2)

)].

This can be simplified further by introducing the sound speed c, by

c2 =γp

ρ.

We then calculate

2H − v2 = 2γE

ρ− (γ − 1)v2 − v2

= 2γE

ρ− γv2

= γ

(2E

ρ− v2

)=γ

ρ

(2E − ρv2

)=γ

ρ

2p

γ − 1.

Therefore

det(λI − df(u)) = (λ− v)[(λ− v)2 − c2

].

Thus the eigenvalues of the Jacobian are

(4) λ1(u) = v − c, λ2(u) = v, λ3(u) = v + c.

As for the corresponding eigenvectors, we write these as ri = (1, yi, zi)1, and we see that yi = λi,

and

zi =1

γ − 1

(λ2i −

1

2(γ − 3)v2 + λi(γ − 3)v

).

For i = 1 we find that

z1 =1

γ − 1

(v2 − 1

2(γ − 3)v2 + v(γ − 3)v

)+

1

γ − 1

(c2 − 2cv − (γ − 3)cv

)=

1

2v2 +

c2

γ − 1− cv

1Recall that this is in (ρ, q, E) coordinates.

3

=

(1

2v2 +

γp

ρ(γ − 1)

)− cv

= H − cv.For i = 3 we similarly calculate

z3 = H + cv,

and for i = 2 it is straightforward to see that z2 = v2/2. Summing up we have the followingeigenvalues and eigenvectors

(5)

λ1(u) = v − c, r1(u) =

1v − cH − cv

,

λ2(u) = v, r2(u) =

1v

12v

2

,

λ3(u) = v + c, r3(u) =

1v + cH + cv

.

At this point it is convenient to introduce the concept of an i–Riemann invariant. An i–Riemanninvariant is a function R = R(ρ, q, E) such that R is constant along the integral curves of ri. Inother words, an i–Riemann invariant satisfies

∇R(u) · ri(0) = 0.

The usefulness of this is that if we can find for each of the three eigenvectors, two Riemanninvariants R(u) and R̃(u), then we can possibly solve the equations

R(ρ, q, E) = R(ρl, ql, El), R̃(ρ, q, E) = R̃(ρl, ql, El)

to obtain a formula for the rarefaction waves. This is the equivalent to finding an implicit solutionof the ordinary differential equation, u̇ = r(u), defining the rarefaction curves.

It turns out that we have the following Riemann invariants

i = 1, Riemann invariants:

{S,

v + 2cγ−1 ,

i = 2, Riemann invariants:

{v,

p,

i = 3, Riemann invariants:

{S,

v − 2cγ−1 ,

where we have introduced the entropy S

S = − log

(p

ργ

).

Verify these statements!Now we can try to obtain solution formulas for the rarefaction curves, for i = 1, this curve is

given by

p = pl

(ρ

ρl

)γ, v = vl +

2clγ − 1

(1−

(ρ

ρl

)(γ−1)/2).

This curve is parameterized by ρ. We must check which half of the curve to use, this will be thepart where λ1 = v − c is increasing. On the curve we have

v(ρ)− c(ρ) = vl +2clγ − 1

(1−

(ρ

ρl

)(γ−1)/2)−(γp(ρ)

ρ

)1/2

4

= vl +2clγ − 1

(1−

(ρ

ρl

)(γ−1)/2)−(γplρl

)1/2(ρ

ρl

)(γ−1)/2

= vl +2clγ − 1

(1−

(ρ

ρl

)(γ−1)/2)− cl

(ρ

ρl

)(γ−1)/2

= vl +2clγ − 1

(1− γ + 1

2

(ρ

ρl

)(γ−1)/2).

Since γ > 1, we see that v(ρ)− c(ρ) is decreasing in ρ, and for the 1-rarefaction wave we must useρ < ρl. Since p(ρ) is increasing in ρ, this also means that we use the part where p < pl. Thereforewe can use p as a parameter in the curve for v and write the 1-rarefaction curve as

v1(p) = vl +2clγ − 1

(1−

(p

pl

)(γ−1)/(2γ)), p ≤ pl.

General theory tells us that (at least for p close to pl) this curve can be continued smoothly as a1-shock curve.

To find the rarefaction curve of the third family, we adopt the viewpoint that ur is fixed, andwe wish to find u as a function of ur. In the same way as for v1 this leads to the formula

v3(p) = vr +2crγ − 1

(1−

(p

pr

)(γ−1)/(2γ)), p ≤ pr.

To find how the density varies along the rarefaction curves we can use that the entropy s isconstant, leading to

ρ

ρl=

(p

pl

)1/γ

.

Now we turn to the computation of the Hugoniot loci. We view the left state ul as fixed, andtry to find the right state u, recall the notation JuK = u−ul. The Rankine–Hugoniot relations for(1) are

(6)

s JρK = JρvK

s JρvK =qρv2 + p

y

s JEK = Jv(E + p)K ,

where s no longer denotes the entropy, but the speed of the discontinuity. Now we introduce newvariables by

w = v − s and m = ρw.

Then the first equation in (6) reads

sρ− sρl = ρw + sρ− ρlwl − sρl,which implies that JmK = 0. Similarly the second equation reads

sρw + s2ρ− sρwl − s2ρl = ρ(w + s)2 − ρl(wl + s)2 + JpK

s JmK + s2 JρK = ρw2 + 2ρw + s2ρ− ρlw2l − 2ρlwl − s2ρl + JpK

s2 JρK =qρw2 + p

y+ s2 JρK .

Hence Jmw + pK = 0. Finally the third equation in (6) reads

sE − sEl = Ew + Es+ pw + ps− Elwl − Els− plwl − plswhich implies

0 =

(E

ρ− Elρl

)m+ pw − plwl + s JpK

=

(E

ρ− Elρl

+p

ρ− plρl

)m− sm JwK

5

= m

sE + p

ρ− sw

{

= m

sc2

γ − 1+

1

2(w + s)2 − sw

{

= m

sc2

γ − 1+

1

2w2

{

Hence the Rankine–Hugoniot conditions are equivalent to

(7)

JmK = 0

Jmw + pK = 0

m

sc2

γ − 1+

1

2w2

{= 0.

We immediately find one solution by setting m = 0, which implies JpK = 0 and JvK = 0. This isthe contact discontinuity. Hence we assume that m 6= 0 to find the other Hugoniot loci.

Now we introduce auxiliary parameters

π =p

pl, and z =

ρ

ρl.

Using these we have that

(8)c2

c2l=π

zand

w

wl=

1

z.

Then the third equation in (7) reads

c2lγ − 1

+1

2w2l =

c2lγ − 1

π

z+

1

2w2l

1

z2

which can be rearranged as

c2l2

γ − 1

(1− π

z

)= w2

l

(1

z2− 1

),

so that

(9)

(wlcl

)2

=2

γ − 1

z(z − π)

1− z2.

Next recall that p = ρc2/γ. Using this, the second equation in (7) reads

ρc2

γ+ ρw2 =

ρlc2l

γ+ ρlw

2l

or

z

(c2

γ+ w2

)=c2lγ

+ w2l

which again can be rearranged as

z

(c2l π

γz+ w2

l

1

z2

)=c2lγ

+ w2l .

Dividing by c2l , we can solve for (wl/cl)2, viz.

(10)

(wlcl

)2

=1

γ

z(π − 1)

z − 1.

Equating (10) and (9) and solving for z yields

(11) z =βπ + 1

π + β,

6

where

β =γ + 1

γ − 1.

Using this expression for z in (9) we get(wlcl

)2

=2

γ − 1

πβ+1π+β

(πβ+1π+β − π

)1− (πβ+1)2

(π+β)2

=2

γ − 1

(πβ + 1)(1− π2)

(π2 − 1)(1− β2)

=2

γ − 1

πβ + 1

β2 − 1.

Note that γ > 1 implies β > 1 so that this is always well defined. Since wl = vl − s, we can usethis to get an expression for the shock speed,

(12) s = vl ∓ cl

√2

γ − 1

βπ + 1

β2 − 1,

where we use − for the first family and + for the third.Next, using (8),

v − svl − s

=1

z

which can be used to express v as a function of π.

v = vl ∓ cl

√2

γ − 1

βπ + 1

(β2 − 1)± π + β

πβ + 1cl

√2

γ − 1

πβ + 1

(β2 − 1)

= vl ∓ cl

√2

γ − 1

1

(β2 − 1)(πβ + 1)

((β − 1)(π − 1)

πβ + 1

)= vl ∓ 2cl

1√2γ(γ − 1)

π − 1

(πβ + 1)1/2,

where we take the − sign for the first family and the + sign for the third. To see how the densityvaries along the Hugoniot loci, we use that ρ = ρlz, or

(13) ρ = ρlπβ + 1

π + β,

which holds for both the first and the third family.Recall that for the first family we obtained a rarefaction solution if p ≤ pl, hence we have a

shock solution if p ≥ pl. This means that the whole solution curve for waves of the first family isgiven by

(14) v1(p) = vl + 2cl

1

γ−1

(1−

(ppl

)(γ−1)/(2γ)), p ≤ pl,

1√2γ(γ−1)

(1− p

pl

)(1 + β p

pl

)−1/2, p ≥ pl.

To find the density along this solution curve we have the formula

(15) ρ1(p) = ρl

(ppl

)1/γ, p ≤ pl,

1+β ppl

β+ ppl

, p ≥ pl.

Since the second family is linearly degenerate, we can use the whole integral curve of r2. Usingthe Riemann invariants, this is given simply as

(16) v = vl, p = pl.

7

For the third family, we take the same point of view as for the shallow water equations; we keepur fixed and look for states u such that the Riemann problem

u(x, 0) =

{u x < 0,

ur x > 0,

is solved by a wave (shock or rarefaction) of the third family. By much the same calculations asfor the first family we end up with

(17) v3(p) = vr − 2cr

1

γ−1

(1−

(ppr

)(γ−1)/(2γ)), p ≤ pr (rarefaction part),

1√2γ(γ−1)

(1− p

pr

)(1 + β p

pr

)−1/2, p ≥ pr (shock part).

Regarding the density along this curve, it will change according to

(18) ρ3(p) = ρr

(ppr

)1/γ, p ≤ pr,(

1+β ppr

β+ ppr

), p ≥ pr.

Now for any ρl, the curve v1(p) is a strictly decreasing function of p for p in the range [0,∞) takingvalues in the set (−∞, vl + 2cl/(γ− 1)]. Similarly, for any ρr v2(p) is a strictly increasing functionof p, taking values in the set [vr − 2cr/(γ − 1),∞). It follows that these curves will intersect inone point (pm, vm) if

vr −2crγ − 1

≤ vl +2clγ − 1

.

If this does not hold, then v1 does not intersect v3 and we have a solution with vacuum. Howeverif they intersect, then we can find the solution of the Riemann problem by finding (pm, vm).

1.0.1. Sod’s shock tube problem. We consider an initial value problem similar to the dam breakproblem for shallow water. The initial velocity is everywhere zero, but the pressure to the left ishigher than the pressure on the right. Specifically we set

(19) p(x, 0) =

{12 x < 0,

1 x ≥ 0,v(x, 0) = 0, ρ(x, 0) = 2.

We have used γ = 1.4.In Figure 1 we show the solution to this Riemann problem in the (p, v) plane and in the (x, t)

plane. We see that the solution consists of a leftward moving rarefaction wave of the first family,followed by a contact discontinuity and a shock wave of the third family. In Figure 2 we show the

0 2 4 6 8 10 123

2

1

0

1

2

3

4

5

6

7

8

p

v

Solution in the (p, v) plane

(pm, vm)

(pr, vr) (pl, vl)

v1(p)

v2(p)

3 2 1 0 1 20

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x

t

Solution in the (x, t) plane

Figure 1. The solution of the Riemann problem (19).

8

pressure, velocity, density and the Mach number as functions of x/t. The Mach number is definedto be |v|/c, so that if this is larger than 1, the flow is called supersonic. The solution found hereis actually supersonic between the contact discontinuity and the shock wave.

3 2 1 0 1 20

2

4

6

8

10

12

x

p

Pressure

4 3 2 1 0 1 2 30

0.2

0.4

0.6

0.8

1

1.2

1.4

x

v

Velocity

4 3 2 1 0 1 2 31

2

3

4

5

6

7

x

!

Density

3 2 1 0 1 20

0.2

0.4

0.6

0.8

1

1.2

x

Mach

num

ber

Figure 2. Pressure, velocity, density and the Mach number for the solution of (19).

1.1. The Euler equations and entropy. We shall show that the physical entropy is in fact alsoa mathematical entropy for the Euler equations, in the sense that

(20) (ρS)t + (vρS)x ≤ 0,

weakly for any weak solution u = (ρ, q, E) which is the limit of solutions to the viscous approxi-mation.

To this end, it is convenient to introduce the internal specific energy, defined by

e =1

ρ

(E − 1

2ρv2).

Then the Euler equations read

(21)

ρt + (ρv)x = 0

(ρv)t +(ρv2 + p

)x

= 0(ρ

(e+

1

2v2))

t

+

(1

2ρv2 + ρev + pv

)x

= 0.

9

For classical solutions, this is equivalent to the nonconservative form

(22)

ρt + vρx + ρvx = 0

vt + vvx +1

ρpx = 0

et + vex +p

ρvx = 0.

Verify this!We have that

S = − log

(p

ργ

)= − log

((γ − 1)e

ργ−1

)= (γ − 1) log(ρ)− log(e)− log(γ − 1).

Thus we see that

Sρ =γ − 1

ρ> 0 and Se = −1

e< 0.

These inequalities are general, and thermodynamical mumbo-jumbo implies that they hold for anyequation of state, not only for polytropic gases.

For classical solutions we can compute

St = Sρρt + Seet

= −γ − 1

ρ(vρx + ρvx) +

1

e

(vex +

p

ρvx

)= −

((γ − 1)− p

eρ

)vx −

((γ − 1)

ρxρ− ex

e

)v

= −vSx.Therefore

St + vSx = 0

for smooth solutions to the Euler equations. This states that the entropy of a “particle” of thegas remains constant as the particle is transported with the velocity v. Furthermore,

(ρS)t = ρtS + ρSt

= −(ρv)xS − ρvSx= − (vρS)x .

Thus for smooth solutions the specific entropy η(u) = ρS(u) is conserved

(23) (ρS)t + (ρvS)x = 0.

Of course, combining this with (21) and viewing the entropy as an independent unknown, wehave four equations for three unknowns, so we cannot automatically expect to have a solution.Sometimes one considers models where the energy is not conserved but the entropy is, so calledisentropic flow. In models of isentropic flow the third equation in (21) is replaced by the conser-vation of entropy (23).

To show that (20) holds for viscous limits, we must show that the map

u 7→ η(u) = ρ(u)S(ρ(u), e(u)) is convex.

We have that η is convex if its Hessian d2η is a positive definite matrix. For the moment we usethe convention that all vectors are column vectors, and for a vector a, aT denotes its transpose.Then we compute

d2η = d2 (ρ(u)S(u))

= ∇ρ (∇S)T

+∇S (∇ρ)T

+ ρd2S.

10

Now we view S(u) as S(ρ(u), e(u)) and compute

d2S(ρ(u), e(u)) = Sρρ∇ρ (∇ρ)T

+ Sρe

(∇e (∇ρ)

T+∇ρ (∇e)T

)+ See∇e (∇e)T + Sed

2e.

If we use this in the previous equation, we end up with

d2η(u) = (ρSρρ + 2Sρ)∇ρ (∇ρ)T

+ ρSρe

(∇ρ (∇e)T +∇e (∇ρ)

T)

+ ρSee∇e (∇e)T − SeC,

where C is given by

C = −(ρd2e+∇ρ (∇e)T +∇e (∇ρ)

T).

Trivially we have that ∇ρ = (1, 0, 0)T . Furthermore

e(u) =E

ρ− 1

2

q2

ρ2,

so we have

∇e =

(−Eρ2

+q2

ρ3,− q

ρ2,

1

ρ

)T=

1

ρ

(−e+

1

2v2,−v, 1

)T.

The Hessian of e is given by

d2e =

2 Eρ3 − 3 q2

ρ4 2 qρ3 − 1

ρ2

2 qρ3 − 1

ρ2 0

− 1ρ2 0 0

=1

ρ2

2e− 2v2 2v −12v 1 0−1 0 0

.

Next,

∇ρ (∇e)T +∇e (∇ρ)T

=1

ρ

100

(−e+ 12v

2 −v 1)

+1

ρ

−e+ 12v

2

−v1

(1 0 0)

=1

ρ

−2e+ v2 −v 1−v 0 01 0 0

.

Then

C = −1

ρ

2e− 2v2 2v −12v 1 0−1 0 0

− 1

ρ

−2e+ v2 −v 1−v 0 01 0 0

=1

ρ

v2 −v 0−v 1 00 0 0

.

Now introduce the matrix D by

D =

1 v 12v

2 + e0 ρ ρv0 0 ρ

.

We have that D is invertible, and thus d2η is positive definite if and only if Dd2ηDT is positivedefinite. Now

Dd2η(u)DT = (ρSρρ + 2Sρ)D∇ρ (D∇ρ)T

+ ρSρe

(D∇ρ (D∇e)T +D∇e (D∇ρ)

T)

+ ρSeeD∇e (D∇e)T − SeDCDT .

We compute

D∇ρ =

100

,

11

D∇e =

001

,

DCDT =

0 0 00 1 00 0 0

,

and using this

Dd2η(u)DT = (ρSρρ + 2Sρ)

1 0 00 0 00 0 0

+ ρSρe

0 0 10 0 01 0 0

+ ρSee

0 0 00 0 00 0 1

− Se0 0 0

0 1 00 0 0

=

ρSρρ + 2Sρ 0 Sρe0 −Se 0Sρe 0 See

=

γ−1ρ 0 0

0 1e 0

0 0 1e2

.

Hence A has three positive eigenvalues, and is positive definite, therefore also d2η is positivedefinite, and η is convex.

3 2 1 0 1 22

1.5

1

0.5

0

0.5

1

1.5Entropy

S

x3 2 1 0 1 2

4

3

2

1

0

1

2

3

4

5Specific entropy

x

!

Figure 3. The entropy and specific entropy for the solution of the Riemannproblem (19).

In Figure 3 we show the entropy and the specific entropy for the solution of Riemann problem(19). The entropy decreases as the shock and the contact discontinuity passes, while it is constantacross the rarefaction wave.

Analogously to the shallow water equations, we can also check whether (20) holds for thesolution of the Riemann problem. We know that this will hold if and only if

−s JρSK + JρvSK ≤ 0.

Using the expression giving the shock speed, (12), we calculate

−s JρSK + JρvSK = S (−s JρK + JρvK) + ρl (−s JSK + vl JSK)

= ±ρlcl

√2

γ − 1

βπ + 1

β2 − 1JSK ,

12

where we use the + sign for the first family and the − sign for the second. Hence the entropy willdecrease if and only if JSK < 0 for the first family, and JSK > 0 for the third family.

Note in passing that for the contact discontinuity, s = v, and thus

−s JρSK + JρvSK = −v JρSK + v JρSK = 0.

Therefore, as expected, entropy is conserved across a contact discontinuity.We consider shocks of the first family, and view JSK as a function of π = p/pl. Recall that for

these shocks π > 1.

JSK = S − Sl

= log

(p

ρl

)− log

(p

ρ

)= γ log(z)− log(π)

= γ log

(βπ + 1

π + β

)− log(π)

=: h(π).

To check if h(π) < 0 = h(1) we differentiate

h′(π) = γβ2 − 1

(π + β)(βπ + 1)− 1

π

=1

π(π + β)(βπ + 1)

(γ(β2 − 1)π − (π + β)(βπ + 1)

)=

β

π(π + β)(βπ + 1)

((γ − 1)π(β − 1)− π2 − 1

)=

β

π(π + β)(βπ + 1)(2π − π2 − 1)

= − β

π(π + β)(βπ + 1)(π + 1)2 < 0.

Thus S is monotonously decreasing along the Hugoniot locus of the first family. We also see that(20) holds only if p ≥ pl for waves of the first family.

For shocks of the third family, an identical computation shows that (20) holds only if p ≤ pl.Exercise. Show that the Lax inequalities

vl − cl > s > v1(p)− c1(p),

if and only if p ≥ pl, where v1 is given by (14), c1 by (15) and s is the speed of the shock, givenby (12) using the − sign. Similarly we have

v3(p) + c3(p) > s > vr + cr,

where v3, c3 and s are given by (17), (18) and (12) (with the + sign) respectively.