39

Compressible Euler in 1d

Everyone knows that heat can produce motion. That it possesses vast mo-
tive power no one can doubt, in these days when the steam engine is every-
where so well known. The study of these engines is of great interest, their
importance is enormous, their use is continually increasing, and they seem
designed to produce a great revolution in the civilized world. (Sadi Carnot
in “Reflections of the motive power of fire and on machines fitted to develop
that power”, 1824).

By 2045 one thousand dollars’ worth of computation will be equal to 10

26

cps, so the intelligence created per year (at a total cost of about $10

12

) will

be about one billion times more powerful than all human intelligence today
(Ray Kurzweil in “The Singularity Is Near”, 2005)

39.1 The Compressible Euler Equations in 1d

We now proceed to formulate a deterministic form of the 2nd Law and then
verify that EG2 satisfies the 2nd Law. For definiteness we may assume the
gas to be perfect, but the 2nd Law will take the same form for a general state
equation.

To get some familiarity with the compressible Euler equations, we start

considering the case of one space dimension. We thus consider a compressible
inviscid perfect gas enclosed in a tube represented by the interval Ω = (0, 1)
in space over a time interval I = (0, ˆ

t ] with zero as initial time and final time

ˆ

t and we assume that the force f = 0. The compressible Euler equations take
the form: Find ˆ

u = (ρ, m, e) depending on (x, t)

∈ Q = [0, 1] × [0, 1] such that

˙

ρ + (ρu)



= 0,

in Q,

˙

m + (mu + p)



= 0,

in Q,

˙e + (eu + pu)



= 0,

in Q,

u(0, t) = u(1, t) = 0,

t

∈ I,

ˆ

u(

·, 0) = ˆu

0

,

in Ω,

(39.1)

358

39 Compressible Euler in 1d

where v



=

∂v
∂x

, ρ is density, u velocity and m = ρu momentum, the total

energy e = k + θ is the sum of the kinetic energy k = ρu

2

/2 and internal heat

energy θ = ρT , the pressure p = (γ

− 1)ρT and γ > 1. For a mono-atomic gas

γ = 5/3.

We may express the 1st Law of thermodynamics as conservation of mass,

momentum and total energy. We may then say that the Euler equations (39.1)
express the 1st Law. Strictly speaking, the 1st Law in classical form expresses
conservation of total energy, while conservation of momentum corresponds
to Newton’s 2nd law and conservation of mass reflects that mass cannot be
destroyed. However, it is convenient to let the 1st Law express conservation of
both mass, momentum and total energy, so that in short the Euler equations
express the 1st Law.

39.2 Euler is Formally Reversible

The Euler equations (39.1) are formally reversible: Changing the sign of the
velocity u and the direction of time (the sign of the time-derivative ˙u), the
Euler equations obviously remain unchanged. This means that if ˆ

u(t) is a

solution to the Euler equations in forward time on [0, ˆ

t ] taking the initial

value ˆ

u(0) to the final value ˆ

u(ˆ

t), we obtain a solution in backward time

taking ˆ

u(ˆ

t) back to ˆ

u(0) by reversing the velocity at time ˆ

t. Of course we can

view this solution as proceeding in forward time by just continuing counting
time forward after the velocity reversal.

We conclude that, formally, the Euler equations admit a perpetuum mo-

bile, which is a machine running forever without consuming any energy, for
ever bouncing back and forth by repeated reversal of the velocity at two given
time instances.

39.3 All Wrong

The problem with the above argument showing reversibility of Euler solu-
tions, is that it is all wrong. How come? We shall answer this question below
by showing that the trouble is that the Euler equations do not admit exact
solutions, which can be reversed as indicated. In contrast approximate EG2
solutions do exist, but these solutions turn out to be irreversible as we will
see.

We sum up: the exact solutions which would have been reversible if they

had existed, do not exist. The approximate EG2 solutions which do exist, are
not reversible. This is the main lesson of this part of the book, and resolves
the main open problem of classical thermodynamics: Loschmidt’s Mystery.

39.4 The 2nd Law in Local Form

359

39.4 The 2nd Law in Local Form

We now present a formal argument leading to the 2nd Law. We then start
from the following viscous version of the momentum equation

˙

m + (mu + p)



− νu



= 0,

in Q,

(39.2)

where ν is a positive viscosity, which we will let tend to zero. The basic oper-
ation is then to multiply the viscous momentum equation (39.2) by uφ, with
φ a non-negative smooth test function. We then obtain integrating over the
space-time domain Q, and performing an integration by parts in the viscous
term,



Q

(ρ ˙uu + ˙

ρuu + uρu



u + uρ



uu + u



ρuu + p



u)φ dxdt

(39.3)

=

−



Q

ν(u



)

2

φ dxdt

−



Q

νu



uφ



dxdt,

where ν(u



)

2

is the intensity of the viscous dissipation. Assuming now that



Q

ν(u



)

2

dxdt stays bounded as ν tends to zero, and further that the velocity

u stays bounded, we see using Cauchy’s inequality that the second term in
the right hand side of (39.3) tends to zero as ν tends to zero. We conclude
that for all non-negative smooth test functions φ,



Q

(ρ ˙uu + ˙

ρuu + uρu



u + uρ



uu + u



ρuu + p



u)φ dxdt =

−



Q

δφ dxdt,

where

δ = lim

ν

→0

ν(u



)

2

,

(39.4)

and since φ is variable,

ρ ˙uu + ˙

ρuu + uρu



u + uρ



uu + u



ρuu + p



u =

−δ, in Q.

Using next mass conservation in the form

u

2

2

( ˙

ρ + (ρu)



) = 0, this equality can

be written

∂

∂t

(ρ

u

2

2

) + u(ρ

u

2

2

)



+ u



ρ

u

2

2

+ p



u =

−δ,

or

˙k + (ku)



+ up



=

−δ.

(39.5)

Combined with the energy equation written in the form

˙k + ˙θ + (ku)



+ (θu)



+ p



u + pu



= 0,

this gives

˙

θ + (θu)



+ pu



= δ,

(39.6)

or, using again mass conservation,

360

39 Compressible Euler in 1d

˙

T + uT



+

pu



ρ

=

δ
ρ

.

(39.7)

We have now formally derived the 2nd Law in local form in any of the equiv-
alent expressions (39.5)-(39.7), where δ

≥ 0. For a shock with a discontinuous

velocity, the viscous dissipation δ > 0 is of the size of the discontinuity squared
and thus is not small. We understand that for a shock there is a significant
transfer of kinetic energy to internal heat energy corresponding to the δ-term
reflecting viscous dissipation in the shock.

We note that the basic ingredient in the proof of the 2nd Law in any of

the forms (39.5)-(39.7) is multiplication of the momentum equation by the
velocity u, or more precisely uφ, and using the sign of the viscous dissipation
in the momentum equation. The key observation is that the intensity of the
viscous dissipation has a non-zero limit as ν tends to zero, and that δ > 0 for
a shock.

39.5 The 2nd Law in Global Form

By partial integration in space we have



Ω

pu



dx =

−



Ω

p



u dx,

using that u(0, t) = u(1, t) = 0. Integrating (39.5) and (39.6) in space, we thus
obtain the following global form of the 2nd Law:

˙

K

− W = −∆,

˙

Θ + W = ∆,

(39.8)

where a capital letter indicates integration in space so that

K =



Ω

k dx,

Θ =



Ω

θ dx,

∆ =



Ω

δ dx,

and

w = pu



,

W =



Ω

w dx,

represents local and global work performed by the pressure p on the velocity
u, and where ∆ > 0 for solutions with shocks. We note that w > 0 under
expansion with u



> 0.

In particular, we have by summation the global form of the 1st Law

˙

E

≡

d

dt



Ω

e dx = ˙

K + ˙

Θ = 0,

(39.9)

stating that the integral (or totality) in space of the total energy e is constant
in time.

39.7 Compression and Expansion

361

39.6 Irreversibility by the 2nd Law

We understand that the global 2nd Law (39.8) for solutions with shocks with
∆ > 0, states an irreversible transfer of kinetic energy to heat energy. On
the other hand, the sign of W is variable and thus the corresponding energy
transfer may go in either direction. In a cyclic process, ∆ represents heat lost
in cooling.

To couple to the above argument with velocity and time reversal at final

time, suppose we formally perform these operations on a given forward so-
lution and denote the time reversed solution by u(t). The global 2nd Laws
(39.8) would then transform to

˙

K

− W = ∆,

˙

Θ + W =

−∆,

(39.10)

with a change of sign of the ∆-term resulting from reversing the sign of velocity
and time. This is a false 2nd Law as we will see.

We now compare with the true 2nd Law for a process starting at final time

with the initial value u(ˆ

t ) and going backward in time (thus forgetting the

forward solution producing the value u(ˆ

t )). The true 2nd Law for this process

takes the form:

˙

K

− W = −∆,

˙

Θ + W = ∆.

(39.11)

We understand that the difference between the two processes from ˆ

t to

0 is the sign of the ∆-terms. In a real process satisfying the true 2nd Law
(39.11), there is transfer from kinetic to internal heat energy, while in the
fictitious process satisfying the false 2nd Law (39.10), the transfer would be
from heat to kinetic energy, thus violating the true 2nd Law. We conclude
that the formally reversed process and the true reversed process both starting
with the value u(ˆ

t ) and going backward in time, would be different for t < ˆ

t

since the true solution would satisfy the true 2nd Law (39.11), while the false
formally reversed process would satisfy the false 2nd Law (39.10).

39.7 Compression and Expansion

We see from the 2nd Law (39.8) that there is a transfer of kinetic energy to
heat energy if W < 0, that is under compression with u



< 0, and a transfer

from heat to kinetic energy if W > 0, that is under expansion with u



> 0. In

addition, there is always a transfer from kinetic to heat energy since always
∆

≥ 0 with ∆ > 0 for shocks.

Returning to Joule’s experiment, we see by the 2nd Law that contraction

back to the original volume from the final rest state in the double volume
is impossible, because the only way the gas can be set into motion is by
expansion.