München 2012

TUM 2012

The Entropy Principle

and its Mathematical Impact

Hans Wilhelm Alt (TU M¨ unchen, Germany)

Contains:

• The history of the Second Law of Thermodynamics

• The distributional Entropy Principle ∂ _t H + divΨ ≥ 0

• Example: Collision of two masses

• Example: Hyperbolic shock waves

• Example: Surface tension

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Recent history

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¸

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c 5 ¬' #$ 1 % Æ 5 \ 5 j ^ % 5 Ǟ\ ¸ \ #$5

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ç

è

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Û

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\ % #ÿ 5 5 #ž ' [ 5 ¡ 5 #ž #ž[ $

#8 1 1 3 5 \ 3 = [ ·J Ž 3 [ ^ " . 5 ›ÿ5 j ·

[S.R. de Groot & P. Mazur. Non-equilibrium Thermodynamics

North Holland 1962]

[I. M¨ uller. Thermodynamics. Pitman 1985]

[I. M¨ uller & T. Ruggeri. Rational Extended Thermodynamics

Springer Tracts in Natural Philosophy 37, 1998]

[D. Bedeaux. Nonequilibrium Thermodynamics and Statistical

Physics of Surfaces. Advance in Chemical Physics, Vol. LXIV, Wiley 1986]

Entropy principle

Standard definition for a set P of processes

For each solution in P there exists an entropy pair (H, Ψ) with

∂ _t H + div Ψ ≥ 0 in D ⁰ (Ω) , Ω ⊂ R ^× R ⁿ

Equation transforms as a scalar equation (i.e. ζ ◦ Y = ζ ^∗ ) (⇒ H objective scalar, Ψ objective vector )

Entropy principle means for test funcions ζ :

h −∂ _t ζ , H i + h −∇ζ , Ψ i ≥ 0 for ζ ≥ 0, ζ ∈ C ₀ ^∞ (Ω; R ⁾ Goes back to entropy principle for funcions in (t, x):

h ξ , H i =

Z

Ω

ξη d(t, x) for ξ ∈ C ₀ ^∞ (Ω) (η entropy) similar Ψ = ψL _n+1 x ^{Ω (ψ} entropy flux), such that

σ := ∂ _t η + div ψ ≥ 0 pointwise in Ω ⊂ R ^× R ⁿ

Of course, there are constitutive relations for (η, ψ) depending on P

Example 1 : System of hyperbolic conservation laws

∂ _t u _k + divq _k (u) = f _k (u) (k = 1, . . . , N )

Constitutive ansatz : η = η(u) , _b ψ = ψ(u) , ^b u = (u ₁ , . . . , u _N )

∂ _t η = ^X

k

η 0 k ∂ _t u _k = − ^X

k

η 0 k divq _k + ^X

k

η 0 k f _k

0 ≤ σ := ∂ _t η + divψ = divψ − ^X

k

η 0 k divq _k + ^X

k

η 0 k f _k

= ^X

l

ψ 0 l − ^X

k

η 0 k q _k 0 l

• ∇u _l + ^X

k

η 0 k f _k

Requirement: This holds for all solutions of the system

Result : Entropy principle is satisfied, if ψ 0 l (u) = ^X

k

η 0 k (u)q _k 0 l (u) for all l ⇒ D ² ηDq = (Dq) ^T D ² η

X k

η 0 k (u)f _k (u)≥0 ⇒ ∇η • f ≥0

Entropy principle has consequence for u 7→ f _k (u) and u 7→ q _k ⁰ _l (u) of the system

Example 2 : Compressible fluid

∂ _t % + div(%v) = 0

∂ _t (%v) + div(%v v ^T + Π) = f (Π pressure tensor)

∂ _t e + div(ev + Π ^T v + q) = v • f (e total energy) e = ε + %

2 | v | ² (ε inner energy)

Constitutive ansatz : η = η(%, ε) _b (η is objective scalar)

% ˙ + %divv = 0 ( ˙ = ∂ _t + v • ∇ )

ε ˙ + εdivv = −divq − Dv • Π (here no force f ⁾ 0 ≤ σ := ∂ _t η + divψ = ˙ η + ηdivv + div(ψ − ηv)

= η 0 % % ˙ + η 0 ε ε ˙ + ηdivv + div(ψ − ηv)

= −η 0 ε divq + Dv •

(η − %η 0 % − εη 0 ε )Id − η 0 ε Π + div(ψ − ηv)

= ∇η 0 ε • q + Dv •

(η − %η 0 % − εη 0 ε )Id − η 0 ε Π +div(ψ − ηv − η 0 ε q)

Temperature : θ with ¹

θ = η 0 ε (%, ε) > 0

1

θ = η 0 ε Use temperature and obtain

0 ≤ σ := ∂ _t η + divψ

= ∇η 0 ε • q + Dv •

(η − %η 0 % − εη 0 ε )Id − η 0 ε Π +div(ψ − ηv − η 0 ε q)

= ∇ 1 θ

• q + Dv • 1

θ

pId − Π + div(ψ − ηv − 1 θ q )

where p = ¹

η _{0 ε} (η − %η 0 % − εη 0 ε ) = −ε + ^{η−%η 0 %}

η _{0 ε}

Requirement : σ ≥ 0 for all solutions of the system Result : Entropy principle is satisfied, if

Π = pId − S (Momentum tensor)

ψ = ηv + 1

θ q (Clausius-Duhem flux) η = %η 0 % + (ε + p)η 0 ε (Gibbs relation)

σ = 1

θ Dv • S + ∇ 1 θ

• q ≥ 0 (Dissipative terms)

Entropy principle has consequences for the system

Gibbs relation

Let (%, ε) 7→ η(%, ε) be the entropy and define

θ = 1

η ⁰ _ε > 0 temperature p pressure f = ε − θη internal free energy v _s = 1

% specific volume η _s = η

% etc. specific quantities The following is equivalent :

• η = %η 0 % + (ε + p)η 0 ε (Gibbs Relation)

• η _s 0 % + (ε _s + p _s )η _s 0 ε = 0

• dη _s = 1

θ dε _s + p

θ dv _s (Second Law: “dS = ¹

T dQ”)

• df _s = −η _s dθ − p dv _s

• d(ε _s + p _s ) = θ dη _s + v _s dp (ε _s + p _s is the enthalpy)

This follows by computing differential forms with (%, ε) as unknowns

Hence the classical formulas hold e.g. for homogeneous systems

Classical Thermodynamics

Zeroth Law There exists for every thermodynamic system in equilibrium a property called temperature. Equality of temperature is a necessary and sufficient condition for thermal equilibrium.

First Law There exists for every thermodynamic system a property called the energy. The change of energy of a system is equal to the mechanical work done on the system in an adiabatic process. In a non-adiabatic process, the change in energy is equal to the heat added to the system minus the mechanical work done by the system.

Second Law There exists for every thermodynamic system in equilibrium an extensive scalar property called the entropy, S , such that in an infinitesimal reversible change of state of the system, dS = dQ/T , where T is the absolute temperature and dQ is the amount of heat received by the system. The entropy of a thermally insulated system cannot decrease and is constant if and only if all processes are reversible.

[MIT, Lecture on Thermodynamics (Spakovszky, Fall 2008)]

(This is the so-called “axiomatic formulation”)

Classical Thermodynamics

Zeroth Law There exists for every thermodynamic system in equilibrium a property called temperature. Equality of temperature is a necessary and sufficient condition for thermal equilibrium.

θ absolute temperature

First Law There exists for every thermodynamic system a property called the energy

e (total) energy

∂ _t e + divϕ = ...

Second Law There exists for every thermodynamic system in equilibrium an extensive scalar property called the entropy, S , such that in an infinitesimal ... change ... of the system, dS = dQ/T

η ^entropy , η 0 ε = 1

θ , ε internal energy

The entropy of a thermally insulated system cannot decrease

∂ _t η + divψ ≥ 0

Example 3 : Collision of particles

Two particles with mass m ^α which move with speed v ^α at space points x ^α Collision : At (t _∗ , x _∗ ), i.e. x _∗ = x ¹ (t _∗ ) = x ² (t _∗ ). Let

h ζ , µ µ µ _x ^α i = Z

R

ζ(t, x ^α (t)) dt for ζ ∈ C ₀ ^∞ ( R ^× R ^{n ;} R ⁾ Distributional mass-momentum-energy balance

∂ t

X

α

m ^α µ µ µ _x ^α

+ div X

α

m ^α v ^α µ µ µ _x ^α

= 0

∂ t

X

α

m ^α v ^α µ µ µ _x ^α

+ div X

α

m ^α v ^α v ^{α T} µ µ µ _x ^α

= X

α

f ^α µ µ µ _x ^α

∂ _t X

α

e ^α µ µ µ _x ^α

+ div X

α

e ^α v ^α µ µ µ _x ^α

= X

α

v ^α • f ^α µ µ µ _x ^α , e ^α = ε ^α + m ^α

2 |v ^α | ² is equivalent to v ^α (t, x(t)) = ˙ x ^α (t) and m ^α , ε ^α locally constant in t 6= t _∗ and

m ^α ¨ x ^α = f ^{α for} α = 1, 2 and t 6= t _∗ m ¹ ₋ + m ² ₋ = m ¹ ₊ + m ² ₊ (mass conservation in t _∗ )

m ¹ ₋ v ₋ ¹ + m ² ₋ v ₋ ² = m ¹ ₊ v ₊ ¹ + m ² ₊ v ² ₊ (momentum conservation in t _∗ ) X

α

ε ^α ₋ + m ^α ₋

2 |v ₋ ^α | ²

= X

α

ε ^α ₊ + m ^α ₊

2 |v ₊ ^α | ²

(energy conservation in t ∗ )

What is the entropy principle?

Use an entropy η ^α and the claim is: With an entropy production h h δ δ δ _{(t ∗ ,x ∗ )} = ∂ _t X

α

η ^α µ µ µ _x ^α

+ div X

α

η ^α v ^α µ µ µ _x ^α

≥ 0 This identity is equivalent to

−(ζh)(t _∗ , x _∗ ) = −

ζ , h δ δ δ _{(t ∗ ,x ∗ )}

=

*

∂ _t ζ , X

α

η ^α µ µ µ _x ^α +

+

*

∇ζ , X

α

η ^α v ^α µ µ µ _x ^α +

= X

α

Z

R

(∂ _t ζ )(t, x ^α (t))η ^α (t, x(t)) dt + Z

R

(∇ζ )(t, x ^α (t))• (η ^α v ^α )(t, x ^α (t))

| {z } η ^α (t, x ^α (t)) ˙ x ^α (t)

dt

= X

α

Z

R \{t ∗ }

d dt

ζ (t, x ^α (t))

η ^α (t, x ^α (t)) dt = X

α

Z

R \{t ∗ }

d dt

ζ(t, x ^α (t))η ^α (t, x ^α (t))

dt (if η ^α = ^η b

α (m ^α , ε ^α ) and since (m ^α , ε ^α ) is locally constant for t 6= t _∗ )

= X

α

ζ (t _∗ , x ^α (t _∗ ))(η ₋ ^α − η ₊ ^α ) = ζ(t _∗ , x _∗ ) X

α

(η ₋ ^α − η ₊ ^α ) that is, if η ^α = ^η b

α (m ^α , ε ^α ) for t 6= t _∗ ,

then the entropy principle is equivalent to h(t ∗ , x ∗ ) + X

α

η ^α ₋ = X

α

η ₊ ^α h(t _∗ , x _∗ ) ≥ 0

X

α

m ^α ₋ = X

α

m ^α ₊

X

α

m ^α ₋ v ₋ ^α = X

α

m ^α ₊ v ₊ ^α

X

α

ε ^α ₋ + m ^α ₋

2 |v ₋ ^α | ²

= X

α

ε ^α ₊ + m ^α ₊

2 |v ₊ ^α | ²

Example 4 : Shock solution of gas equations

The equations for a fluid are

∂ _t % + div(%v) = 0

∂ _t (%v ) + div(%v v ^T + Π) = f , Π = pId − S

∂ _t e + div(ev + Π ^T v + q) = v• f , e = ε + %

2 | v | ²

∂ _t η + div ηv + 1 θ q

= 1

θ Dv•S + ∇ 1 θ

•q ≥ 0 , η = ^η b ^{(%, ε) = %η 0 % + (ε + p)η 0 ε} Neglecting S ≈ 0 and q ≈ 0 one considers

weak solutions (that is, L ^∞ -solutions) of the gas equations

∂ _t % + div(%v) = 0

∂ _t (%v ) + div(%v v ^T + pId) = f , η = %η _% + (ε + p)η _ε

∂ t e + div((e + p)v) = v• f , e = ε + %

2 | v | ²

These are distributional solutions. One considers only solutions satisfying

∂ _t η + div(ηv) ≥ 0 , η = ^{η(%, ε)} b

Remark: For smooth solutions (%, v, ε) one has ∂ _t η + div(ηv) = 0.

What is the meaning of this inequality? It defines the correct shocks!

Case: An interface Γ with Ω = Ω ¹ ∪ Γ ∪ Ω ² ⊂ R ^× R ⁿ

If the L ^∞ -solution has the form

%L _n+1 = X

m

% ^m µ µ µ _Ω ^m , similar v ^m , p ^m , ε ^m , f ^m , h ξ , µ µ µ _Ω ^m i = Z

R

Z

Ω ^m _t

ξ(t, x) dx dt the above distributional differential equations can be written as

∂ t

X

m

% ^m µ µ µ _Ω ^m

+ div X

m

% ^m v ^m µ µ µ _Ω ^m

= 0

∂ _t X

m

% ^m v ^m µ µ µ _Ω ^m

+ div X

m

(% ^m v ^m v ^{m T} + p ^m Id)µ µ µ _Ω ^m

= X

m

f ^m µ µ µ _Ω ^m

∂ _t X

m

e ^m µ µ µ _Ω ^m

+ div X

m

(e ^m + p ^m )v ^m µ µ µ _Ω ^m

= X

m

v ^m • f ^m µ µ µ _Ω ^m

where p ^m = ^p(% b

m , ε ^m ) and e ^m = ε ^m + ^% ₂ ^m | v ^m | ² .

This is equivalent to the differential equations for (% ^m , v ^m , ε ^m ) in Ω ^m , and on Γ X

m

% ^m (v ^m − v Γ )•ν _Ω ^m = 0 X

m

(% ^m (v ^m − v _Γ )•ν _Ω ^m v ^m + p ^m ν _Ω ^m ) = 0 X

m

(e ^m (v ^m − v Γ )•ν _Ω ^m + p ^m v ^m •ν _Ω ^m ) = 0



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 

 



 

 

 





 

 

 

 

 

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 

M ^{is defined} M · v _tan ¹ = M · v _tan ² p ¹ + M · λ ¹ = p ² + M · λ ² M · ε ¹ _s + p ¹ _s + |λ ¹ | ²

2

= M · ε ² _s + p ² _s + |λ ² | ² 2

Definition: This is called a “shock”, if in the mass conservation

M ^:= % ¹ λ ¹ = % ² λ ² 6= 0 , λ ^m := (v ^m − v _Γ )•ν _Ω ¹

For shocks, that is M 6= 0, the momentum and energy balance on Γ say:

v _tan ¹ = v _tan ²

p ¹ + % ¹ |λ ¹ | ² = p ² + % ² |λ ² | ² ε ¹ _s + p ¹ _s + |λ ¹ | ²

2

= ε ² _s + p ² _s + |λ ² | ² 2 Here λ ^m := (v ^m − v _Γ )•ν _Ω ¹

[L.D. Landau, E.M. Lifschitz. Lehrbuch der Theoretischen Physik VI.

Hydrodynamik 3. Auflage Akademie-Verlag 1974]

The entropy principle is

∂ _t H + divΨ ≥ 0 , H = X

m

η ^m µ µ µ _Ω ^m , Ψ = X

m

η ^m v ^m µ µ µ _Ω ^m , η ^m = ^η(% b

m , ε ^m ) That is, the entropy inequality is satisfied in distributional sense.

It is equivalent to

∂ _t η ^m + div(η ^m v ^m ) = 0 in Ω ^m (since η ^m = ^η(% b

m , ε ^m )) X

m

η ^m (v ^m − v Γ )•ν _Ω ^m ≥ 0 on Γ

and this is (in the shock case) equivalent to η _s ^m = ^{η m}

% ^m

M · (η _s ¹ − η _s ² ) ≥ 0 or

( η _s ¹ ≥ η _s ² if M > 0 η _s ¹ ≤ η _s ² if M < 0

[L.D. Landau, E.M. Lifschitz. Lehrbuch der Theoretischen Physik VI.

Hydrodynamik 3. Auflage Akademie-Verlag 1974]

Weak and strong equations

A single balance law is an equality (resp. inequality) of the form

∂ _t E + divQ = (resp. ≤) F in D ⁰ (Ω) Theorem It is equivalent

∂ _t

X ²

m=1

e ^m µ µ µ _Ω ^m + e ^s µ µ µ _Γ

| {z }

= E

+ div

X ²

m=1

q ^m µ µ µ _Ω ^m + q ^s µ µ µ _Γ

| {z }

= Q

= (resp. ≤)

2

X

m=1

f ^m µ µ µ _Ω ^m + f ^s µ µ µ _Γ

| {z }

= F

and

1. ∂ _t e ^m + divq ^m = (resp. ≤) f ^m for m = 1, 2 in Ω ^m 2. (q ^s − e ^s v _Γ )(t, x) ∈ T _x (Γ _t ) for all (t, x) ∈ Γ

3. ∂ _t ^Γ e ^s + div ^Γ q ^s = (resp. ≤) f ^s +

2 X m=1

(q ^m − e ^m v _Γ ) • ν _Ω ^m on Γ

This includes “Rankine-Hugoniot” conditions and “Kotchine” conditions We have that ∂ _t ^Γ e ^s + div ^Γ q ^s = ∂ _t ^Γ e ^s − e ^s κ•v _Γ + div ^Γ (q ^s − e ^s v _Γ )

[H.W. Alt. The Entropy Principle for Interfaces. Fluids and Solids.

AMSA 19, pp. 585-663, 2009]

[T. Alts, K. Hutter. Continuum Description of the Dynamics and Thermodyna-

mics of Phase Boundaries Between Ice and Water. J.Non-Equilib.Thermodyn.]

[D. Bedeaux. Nonequilibrium Thermodynamics and Statistical

Physics of Surfaces. Advance in Chemical Physics Vol. LXIV, Wiley 1986]

[D. Bedeaux. Nonequilibrium Thermodynamics and Statistical

Physics of Surfaces. Advance in Chemical Physics Vol. LXIV, Wiley 1986]

Example 5 : Surface tension

We consider two fluids (e.g. water and oil) with surface tension

∂ _t % ^m µ µ µ _Ω ^m

+ div % ^m vµ µ µ _Ω ^m

= 0 for m = 1, 2 v continuous at Γ

∂ _t X

m

% ^m vµ µ µ _Ω ^m ) + div X

m

(% ^m v ⊗ v + Π ^m )µ µ µ _Ω ^m + Π ^s µ µ µ _Γ

= X

m

f ^m µ µ µ _Ω ^m

∂ _t X

m

e ^m µ µ µ _Ω ^m + ε ^s µ µ µ _Γ

+ div X

m

(e ^m v + (Π ^m ) ^T v + q ^m )µ µ µ _Ω ^m + (Π ^s ) ^T vµ µ µ _Γ

= X

m

v• f ^m µ µ µ _Ω ^m

θ continuous at Γ, i.e. η ¹ 0 ε (% ¹ , ε ¹ ) = η ² 0 ε (% ² , ε ² )

It is the purpose to determine ε ^s and Π ^s . Here we assume for the two fluids that the standard entropy principle holds for m = 1, 2

σ ^m := ∂ _t η ^m + div η ^m v + 1 θ q ^m

= 1

θ Dv•S ^m + ∇ 1 θ

•q ^m ≥ 0 in Ω ^m and quantities for the two fluids are given as above.

The more general entropy principle for the system is Σ := ∂ _t X

m

η ^m µ µ µ _Ω ^m + η ^s µ µ µ _Γ ) + div X

m

η ^m v + 1 θ q ^m

µ µ µ _Ω ^m + (η ^s v + q ^s )µ µ µ _Γ

≥ 0

Exploitation of the entropy principle:

For the distributional entropy principle Σ := ∂ _t X

m

η ^m µ µ µ _Ω ^m + η ^s µ µ µ _Γ ) + div X

m

η ^m v + 1 θ q ^m

µ µ µ _Ω ^m + (η ^s v + q ^s )µ µ µ _Γ

≥ 0 we compute

Σ = X

m

σ ^m µ µ µ _Ω ^m + σ ^s µ µ µ _Γ ≥ 0 (σ ^m as above)

which is equivalent to σ ^m ≥ 0 in Ω ^m and σ ^s ≥ 0 in Γ. It is (since (v − v _Γ )•ν = 0) σ ^s = ∂ _t ^Γ η ^s + div ^Γ (η ^s v + q ^s ) − X

m

1

θ q ^m •ν _Ω ^m , X

m

1

θ q ^m •ν _Ω ^m = 1 θ

X

m

q ^m •ν _Ω ^m X

m

q ^m •ν _Ω ^m = ∂ _t ^Γ ε ^s + div ^Γ (ε ^s v + Π ^s v) − v• X

m

Π ^m ν _Ω ^m (energy) X

m

Π ^m ν _Ω ^m = div ^Γ Π ^s (momentum) X

m

q ^m •ν Ω ^m = ∂ _t ^Γ ε ^s + div ^Γ (ε ^s v) + D ^Γ v•Π ^s and therefore

σ ^s = ∂ _t ^Γ η ^s + div ^Γ (η ^s v + q ^s ) − 1

θ ∂ _t ^Γ ε ^s + div ^Γ (ε ^s v)

− 1

θ D ^Γ v•Π ^s v = v _Γ + v _tan

= ˙ η ^s − 1

θ ε ˙ ^s + div ^Γ q ^s ˙ = ∂ _t ^Γ + v•∇ ^Γ = ∂ t + v Γ •∇ + v•∇ ^Γ = ∂ t + v•∇

+D ^Γ v• (η ^s − 1

θ ε ^s )(Id − ν ⊗ ν) − 1

θ Π ^s

For the entropy principle Σ ≥ 0 it remains σ ^s ≥ 0 with σ ^s = ˙ η ^s − 1

θ ε ˙ ^s + div ^Γ q ^s + D ^Γ v• (η ^s − 1

θ ε ^s )(Id − ν ⊗ ν ) − 1 θ Π ^s

Result : The entropy principle is satisfied, if q ^s = 0 and Π ^s = −γ (Id − ν ⊗ν ) − S ^s , γ = ε ^s − θη ^s

η ^s = η _b ^s (ε ^s ) , η ^s _{0 ε s} = 1

θ , ε ^s = ε _b ^s (θ)

and if the above properties for the fluids are satisfied.

The remaining inequality on the surface is σ ^s = 1

θ D ^Γ v • S ^s ≥ 0 (for example, if S ^s = 0)

Mathematical literature for the isothermal case:

[I.V. Denisova. Solvability in weighted H¨ older spaces for a problem governing the

evolution of two compressible fluids. Zap. Nauchn. Sem. 295, pp. 57-89 (2003)]