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Non-linear Wave Propagation andNon-Equilibrium Thermodynamics - Part 3

Tommaso Ruggeri

Department of Mathematics and Research Center of Applied Mathematics

University of Bologna

Universita Cattolica del Sacro CuoreDipartimento di Matematica e Fisica ”Niccolo Tartaglia”

January 23-25, 2017

Tommaso Ruggeri Non-linear Wave Propagation and Non-Equilibrium Thermodynamics - Part 3

Overview

1 Constitutive Equations

2 Elasticity and Thermoelasticity

3 Fluids

4 Heat equation

5 Hyperbolic Systems

6 Wave equation and method of the characteristics

7 A non linear example: the Burgers equation

8 Weak solutions and Shock waves

9 Shock waves in Euler fluid


Balance Law of Energy

If we enlarge the framework from mechanics to thermodynamics we have also thebalance law of energy. In this case

= ⇢e +⇢v2

2, �

i

= �tij

vj

+ qi

, f = F · v + r = ⇢b · v + r .

Then

@

@t

✓1

2⇢v2 + ⇢e

◆+

@

@xi

⇢✓1

2⇢v2 + ⇢e

◆vi

� tij

vj

+ qi

�= ⇢b

i

vi

+ r . (1)

Therefore in thermo-mechanics we have the following system of Balance Laws:8>>>>>>>>><

>>>>>>>>>:

@⇢

@t+@⇢v

i

@xi

= 0

@⇢vj

@t+

@

@xi

(⇢vi

vj

� tij

) = ⇢bj

(j = 1, 2, 3)

@

@t

�1

2

⇢v2 + ⇢e�+

@

@xi

��1

2

⇢v2 + ⇢e�vi

� tij

vj

+ qi

= ⇢b

j

vj

+ r .

(2)


Generic System of Balance Laws

The previous system of balance laws is a particular case of

@F0

@t+@Fi

@x i= f, $ @F↵

@x↵= f (3)

F

0 =

0

BBBB@

⇢⇢v

1

⇢v2

⇢v3

1

2

⇢v2 + ⇢e

1

CCCCA; F

i =

0

BBBB@

⇢vi

⇢vi

v1

� ti1

⇢vi

v2

� ti2

⇢vi

v3

� ti3�

1

2

⇢v2 + ⇢e�vi

� tij

vj

+ qi

1

CCCCA; f =

0

BBBB@

0⇢b

1

⇢b2

⇢b3

⇢bj

vj

+ r

1

CCCCA.

(4)In general F

0, Fi e f Rn vectors.


Lagrangian form of Balance Laws

It is convenient in some case to use Lagrangian variables (X, t) referring to theinitial configuration. We know

dV = JdV ⇤, nd⌃ = F

C

n

⇤d⌃⇤. (5)

Therefored

dt

Z

V

dV = �Z

⌃

�i

ni

d⌃+

Z

V

fdV (6)

can be rewritten

d

dt

Z

V

⇤J dV ⇤ = �

Z

⌃

⇤�

i

FC

iA

n⇤A

d⌃⇤ +

Z

V

⇤fJdV ⇤. (7)

Using Cauchy-Green theoremZ

V

⇤

@

@t(J ) dV ⇤ +

Z

V

⇤

@�i

FC

iA

@XA

dV ⇤ =

Z

V

⇤fJdV ⇤

and assuming regularity conditions

@ ⇤

@t+@�⇤

A

@XA

= f ⇤ (8)


where

⇤ =⇢⇤

⇢ , �⇤

A

= �i

FC

iA

, f ⇤ =⇢⇤

⇢f . (9)

For example the Balance Laws of mass, momentum and energy becomes inLagrangian variables:

8>>>>>>><

>>>>>>>:

⇢ = ⇢⇤/J

@⇢⇤vj

@t� @T

jA

@XA

= ⇢⇤bj

@

@t

✓⇢⇤

v2

2+ ⇢⇤e

◆+

@

@XA

(QA

� TiA

vi

) = ⇢⇤bi

vi

+ r⇤.

(10)

where T ⌘ (TiA

) is the Piola-Kirchhoff stress tensor

TjA

= tij

FC

iA

() T = tF

C . (11)

Q ⌘ (QA

) the Lagrangian heat flux

QA

= qi

FC

iA

() Q = F

CT

q (12)

and

r⇤ = r⇢⇤

⇢. (13)


Constitutive Equations and Universal Principles

To close the system we need the constitutive equations that to be physicalconsistent need to verifies the following two principles:

1 The Frame Indi↵erence Principle – Objectivity that require that theconstitutive equations are independent of the Observer

2 The Entropy Principle, that require that admissible constitutive equationsare such that every solution of the closed system is compatible with thesecond law of thermodynamics

Objectivity Principle

R

RT

R-1

=

e

e

e

e’

e’

e’

1

1

2

2

33

C

C’

x

x’ RT

= x

Figure: Material Indi↵erence

Therefore the stress, the heat flux, the internal energy, are the same in C and in C0

except for a similitude transformation:

t

0 = R

T

tR, q

0 = R

T

q, e0 = e. (14)


The Entropy Principle

i) There exists an additive and objective scalar, which we call entropy;

ii) The entropy density s the flux of the entropy �i

are constitutive functions tobe determined;

iii) The entropy production ⌃ is non-negative for all thermodynamic processes.

@⇢s

@t+

@

@xi

(⇢s vi

+ �i

) = ⌃ (15)

This formulations is due to Muller. In the classical vision (Coleman-Noll)was considered the so called Clausius-Duhem entropy principle in which

� =q

#. (16)

In this lectures we consider the simple case of Clausius-Duhem:

@⇢s

@t+

@

@xi

⇣⇢s v

i

+qi

#

⌘= ⌃ � 0. (17)


In the case of energy production the entropy principle is modified in

@⇢s

@t+

@

@xi

⇣⇢s v

i

+qi

#

⌘=

r

#+ ⌃, ⌃ � 0. (18)

In Lagrangian variables the balance law of entropy becomes:

@⇢⇤s

@t+

@

@XA

✓Q

A

#

◆=

r⇤

#+ ⌃⇤; ⌃⇤ � 0 (19)

with

r⇤ =⇢⇤

⇢r , ⌃⇤ =

⇢⇤

⇢⌃ .


Elastic body

A body is elastic if the stress tensor depend only on the gradient of deformation:

t ⌘ t (F) . (20)

In this case we have

@2⇢⇤ui

@t2�@T

iA

⇣@u

k

@XB

⌘

@XA

= ⇢⇤bi

, (21)

that can be rewritten as system of first order:

8>>><

>>>:

⇢⇤@v

i

@t� @T

iA

(FkB

)

@XA

= ⇢⇤bi

@FiA

@t� @v

i

@XA

= 0

(22)

in the unknown field v ⌘ v (X,t) and F ⌘ F (X,t) .


Consequences of the Objectivity Principle in elasticity

Let S the so called second tensor of Piola-Kirchhoff:

S = F

�1

T 2 Sym , t =1

JFSF

T . (23)

TheoremNecessary and su�cient condition such that the objectivity principle hold is thatthe second tensor of Piola-Kirchhoff depends on F only trough thedeformation matrix E:

S ⌘ S (E) or that is the same S ⌘ S (C) .

Proof: As the body is elastic we have:

t (F0) = R

T

t(F)R 8 R 2 Rot con F

0 = R

T

F

thent

�R

T

F

�= R

T

t (F)R 8 R 2 Rot . (24)


Substituting (23)2

in (24) we have

1

JR

T

FS

�R

T

F

�F

T

R =1

JR

T

FS (F)FT

R 8R 2 Rot

i.e.S

�R

T

F

�= S (F) 8R 2 Rot (25)

Recalling the polar theoremF = ˆ

RU (26)

and requiring that (25)is satisfied also for R = ˆ

R we obtain

S (U) = S (F) ,

and therefore S depends on F only trough the dilatation U or equivalently S

depends on C

�C = U

2

�or S depends on E

�E = 1

2

(C� I)�:

S ⌘ S (E) c .v .d .

From (23)2

we have that the dependence of the stress tensor t on F must be

t(F) =1

JFS(E)FT (27)

while the first Piola-Kirchhoff tensor

T(F) = FS(E).


Thermoelastic body

A body is thermoelastic if the constitutive equations depends on the gradient ofdeformation, the temperature and eventually of gradient of temperature:

8<

:

t ⌘ t (F,#)e ⌘ e (F,#)q = �� (F,#)r#.

(28)

The last equation of (28) is the Fourier law in which � denotes the heatconductibility.

TheoremNecessary and su�cient condition such that the objectivity principle hold is thatthe second tensor of Piola-Kirchhoff, internal energy and heat conductivitydepends on F only trough the deformation matrix E and on the temperature:

S ⌘ S (E,#) , e ⌘ e(E,#), � ⌘ �(E,#). (29)


The field equations

8>>>>>>>>><

>>>>>>>>>:

⇢⇤@v

i

@t� @T

iA

@XA

= ⇢⇤bi

@FiA

@t� @v

i

@XA

= 0

⇢⇤@

@t

✓v2

2+ e

◆� @

@XA

(TiA

vi

� QA

) = ⇢⇤bi

vi

+ r⇤.

(30)

can be rewritten taking into account that e ⌘ e(F,#) in the form

8>>>>>>>>>>><

>>>>>>>>>>>:

⇢⇤@v

i

@t=@T

iA

@XA

+ ⇢⇤bi

@FiA

@t=

@vi

@XA

@#

@t= 1

@e

@#

⇢✓TiA

⇢⇤� @e

@FiA

◆@v

i

@XA

� 1

⇢⇤@Q

A

@XA

+r⇤

⇢⇤

�(31)


The Fourier law become in Lagrangian variables:

QA

= ��FC

iA

@#

@xi

= ��FC

iA

@#

@XB

@XB

@xi

= ��JF�1

Ai

F�1

Bi

@#

@XB

i.e.Q = ��JF�1

�F

�1

�T

Grad # Q = ��(E,#) JC�1Grad #

where

Grad # ⌘✓@#

@X1

,@#

@X2

,@#

@X3

◆.


Entropy principle in thermoelasticity

We now require the compatibility with the entropy principle, i.e. every solution of(31) must be solution of (19) assuming that the entropy principle is also aconstitutive equation:

s ⌘ s (F,#) . (32)

we have the following

Theorem (Entropy Principle)

Necessary an su�cient condition such that the entropy principle is satisfied is thatthere exists a scalar function the free energy, ⌘ (E,#), such that

s = �@ @#

, S =⇢⇤@

@E, e = � #

@

@#. (33)

Moreover the heat conductivity must be non negative:

� (E,#) � 0.


Proof : From (19) and (32) we have

⇢⇤✓@s

@FiA

@FiA

@t+@s

@#

@#

@t

◆+

1

#

@QA

@XA

� 1

#2Q

A

@#

@XA

� r⇤

#= ⌃⇤ � 0.

then

⇢⇤

8><

>:@s

@FiA

+

@s

@#@e

@#

✓TiA

⇢⇤� @e

@FiA

◆9>=

>;@v

i

@XA

+

8><

>:1

#�

@s

@#@e

@#

9>=

>;

✓@Q

A

@XA

� r⇤◆+

(34)

+1

#2�JC�1Grad # ·Grad # = ⌃⇤ � 0.

then 8>>>>>>><

>>>>>>>:

#@s

@#=@e

@#

#@s

@FiA

=@e

@FiA

� TiA

⇢⇤

� � 0.

(35)


The first two of (35) are equivalent to the so called Gibbs equation (localequilibrium)

#ds = de � 1

⇢⇤T · dF (36)

or equivalently

#ds = de � 1

⇢⇤S·dE. (37)

Let = e � #s, (38)

the free energy we obtain

d = �s d#+1

⇢⇤S·dE. (39)

and we proved the conditions (33). Taking into account that

⌃⇤ =�J

#2C

�1Grad# · Grad# � 0. (40)

then � � 0 and theorem is proved, In particular the first Piola-Kirchhofftensor must be in the form

T = ⇢⇤F@

@E. (41)


Ideal Fluids and Euler system

Definition

A fluid is ideal if the specific stress is normal and have pressure character:

t

n

= �pn

n pn

� 0 8 n. (42)

For the Cauchy theorem we have

pn

n = p1

n1

e

1

+ p2

n2

e

2

+ p3

n3

e

3

and therefore we obtains soon the Pascal result

pn

= p1

= p2

= p3

= p 8 n.

Thent

n

= �p n p � 0 8 n (43)

that implies that the stress tensor is isotropic

t = �pI, t ⌘

��

�p 0 00 �p 00 0 �p

��, t

ij

= �p�ij

. (44)


In an ideal fluid is supposed also negligible the heat conductivity and then:

q = 0. (45)

Therefore the balance laws assume the form8>>>>>>>>><

>>>>>>>>>:

@⇢

@t+@⇢v

i

@xi

= 0

@⇢vj

@t+

@

@xi

(⇢vi

vj

+ p�ij

) = ⇢bj

(j = 1, 2, 3)

@

@t

✓⇢v2

2+ ⇢e

◆+

@

@xi

⇢✓⇢v2

2+ ⇢e + p

◆vi

�= ⇢b

j

vj

+ r .

(46)

For prescribed thermal and caloric equation of state

p ⌘ p(⇢,#), e ⌘ e(⇢,#). (47)

the system is closed and is called Euler system.


Fourier-Navier-Stokes Dissipative fluids

Definition

A real fluid have character of pressure only in equilibrium.

Then in nonequilibrium

t = �pI+ �, tij

= �p�ij

+ �ij

, (48)

where � ⌘k�ij

k 2 Sym is the viscous stress tensor and is assumed that depends inthe symmetric part of velocity gradient D:

� ⌘ �(D), �(0) = 0, D =1

2

�rv + (rv)T

�. (49)

For a real fluid the heat flux is not zero and depends on gradient of thetemperature

q ⌘ q(r#), q(0) = 0. (50)

The most simple case is to suppose linear constitutive equations. For the heat fluxwe have seen the Fourier law:

q = �� r#, � ⌘ �(⇢,#). (51)


While for the viscous stress tensor there is the assumptions of Navier Stokes) :

� = � divv I+ 2µ D, (52)

i.e.

�ij

= �@v

k

@xk

�ij

+ µ

✓@v

i

@xj

+@v

j

@xi

◆.

It is more convenient to use orthogonal tensors and for this reason is convenient todecompose the matrix D in the deviatoric part (traceless) and the isotropic part:

D = D

D +1

3divv I

and the (52) becomes� = ⌫ divv I+ 2µ D

D (53)

con

⌫ =1

3(3�+ 2µ).

The scalars ⌫ e and µ are the so called bulk viscosity and shear viscosityrespectively and they depends on the density and temperature:

⌫ ⌘ ⌫(⇢,#), µ ⌘ µ(⇢,#). (54)


The stress tensor becomes

t = �(p + ⇡) I+ 2µD

D (55)

where we have put⇡ = �⌫ divv.

For this reason ⇡ is called dynamic pressure, while

t

D = �D = 2µDD

is the deviatoric viscous stress tensor. A fluid is called Stokesian if ⌫ = 0 (exampleare the Monatomic gases). The balance laws for Fourier-Navier-Stokes fluidsbecomes

8>>>>>>>>><

>>>>>>>>>:

@⇢

@t+@⇢v

i

@xi

= 0

@⇢vj

@t+

@

@xi

(⇢vi

vj

+ p�ij

� �ij

) = ⇢bj

(j = 1, 2, 3)

@

@t

✓⇢v2

2+ ⇢e

◆+

@

@xi

⇢✓⇢v2

2+ ⇢e + p

◆vi

� �ij

vj

+ qi

�= ⇢b

j

vj

+ r

(56)Tommaso Ruggeri Non-linear Wave Propagation and Non-Equilibrium Thermodynamics - Part 3

To close the system we need the constitutive equations:

p ⌘ p(⇢,#) (pressure), e ⌘ e(⇢,#) (internal energy),

� ⌘ �(⇢,#) (heat conductivity), (57)

⌫ ⌘ ⌫(⇢,#) e µ ⌘ µ(⇢,#) (bulk and shear viscosity).

It is easy to prove that the previous system can be rewritten fro classical solutionsin the form: 8

>>>>>>>><

>>>>>>>>:

d⇢

dt+ ⇢ div v = 0

⇢dv

j

dt+

@

@xi

(p�ij

� �ij

) = ⇢bj

⇢de

dt+ p div v � � ·D+ div q = r

(58)

Taking into account that e ⌘ e(⇢,#) the last equation can be write

d#

dt=

1

⇢ @e@#

⇢r +

✓⇢2@e

@⇢� p

◆div v + � ·D� div q

�. (59)


Entropy principle for a fluid

The entropy principle require that every solutions of the fluid system satisfy also

⇢ds

dt+

@

@xi

⇣qi

#

⌘� r

#= ⌃ � 0. (60)

As the entropy density is a constitutive function s ⌘ s(⇢,#) we have(1

#�

@s@#@e@#

)(div q� r) + div v

(�⇢2 @s

@⇢+

✓⇢2@e

@⇢� p

◆ @s@#@e@#

)

+@s@#@e@#

� ·D� 1

#2q ·r# = ⌃ � 0. (61)

Then 8>>><

>>>:

#@S

@#=@e

@#

#@S

@⇢=@e

@⇢� p

⇢2.

(62)

i.e. the Gibbs equation hold

#ds = de � p

⇢2d⇢. (63)


The residual inequality becomes for Fourier-Navier-Stokes

⌃ =1

#

�� div v I+ 2µDD

�·D+

�

#2|r#|2 =

=1

#

⇣� (div v)2 + 2µkDDk2

⌘+

�

#2|r#|2 � 0,

that implies�(⇢,#) � 0, µ(⇢,#) � 0, �(⇢,#) � 0. (64)

Introducing the free energy = e � #s (65)

we obtaind =

p

⇢2d⇢� sd#. (66)

and then

p = ⇢2@

@⇢, s = �@

@#, e = � #

@

@#. (67)


Heat equation

In the case of a rigid heat conductor we have only the energy balance

⇢⇤@e

@t+ div q = r (68)

with the constitutive Fourier equation:

q = ��r# (69)

ande ⌘ e(#) e � ⌘ �(#). (70)

Substituting the (69) and (70) in (68) we have

⇢⇤ cV

(#)@#

@t� div (�r#) = r (71)

with

cV

(#) = e0(#) =de

d#(specific heat).

Then (71) becomes:@#

@t� µ�#� ⌫ (grad#)2 = r (72)


where

µ(#) =�(#)

⇢⇤cV

(#); ⌫(#) =

�0(#)

⇢⇤cV

(#); �0(#) =

d�(#)

d#.

in the simple case in which � and cV

are constants and we have no supply r = 0,the equation assume the usual form of heat equation:

@#

@t� µ�# = 0. (73)

that is the typical parabolic equation If we prescribe the initial data

#(x, 0) = #0

(x), (74)

the solution is

#(x, t) =1

(4⇡µt)3/2

Z+1

�1#0

(y) exp

✓� (y � x)2

4µt

◆dy, (75)

and for an initial data having support compact we have the so called heat paradoxof infinite propagation.


Cattaneo Equation

Carlo Cattaneo propose to modify the Fourier law:

q = ��r#+ �⌧r# (# = @#/@t), (76)

where ⌧ is a relaxation time. The (76) can be rewritten as :

q = ��✓1� ⌧

@

@t

◆r#. (77)

If ⌧ is small enough the inverse operator is✓1� ⌧

@

@t

◆�1

' 1 + ⌧@

@t(78)

then from (78) and (77) we obtain the Cattaneo equation

⌧@q

@t+ q = ��r#. (79)

Combining this equation with the energy equation we obtain the hyperbolictelegraphist equation

⌧@2#

@t2+@#

@t� µ�# = 0. (80)


Hyperbolic Systems

The hyperbolic systems of continuum mechanic are balance laws

@F↵(u)

@x↵= f(u) (81)

with F

↵ e f local function of u. The (81) can be rewritten

A

↵(u)@u

@x↵= f(u), A

↵ =@F↵

@u. (82)

Definition (Hyperbolic System)

A system (82) is hyperbolic in the time direction ifa) detA0 6= 0;b) The following eigenvalue problem

(An

� �A0)d = 0, (An

= A

ini

) (83)

8n 2 R3 : knk = 1, admits real eigenvalues � and the eigenvectors d are linearlyindependent


Example of Euler system

Choosing as fieldu ⌘ (⇢, v

1

, v2

, v3

, s)T

the Euler system becomes in the form (82) with:

A

0 ⌘

0

BBBB@

1 0 0 0 00 1 0 0 00 0 1 0 00 0 0 1 00 0 0 0 1

1

CCCCA, A

1 ⌘

0

BBBB@

v1

⇢ 0 0 0p⇢/⇢ v

1

0 0 pS

/⇢0 0 v

1

0 00 0 0 v

1

00 0 0 0 v

1

1

CCCCA,

A

2 ⌘

0

BBBB@

v2

0 ⇢ 0 00 v

2

0 0 0p⇢/⇢ 0 v

2

0 pS

/⇢0 0 0 v

2

00 0 0 0 v

2

1

CCCCA, A

3 ⌘

0

BBBB@

v3

0 0 ⇢ 00 v

3

0 0 00 0 v

3

0 0p⇢/⇢ 0 0 v

3

pS

/⇢0 0 0 0 v

3

1

CCCCA,


then ( vn

= v · n)

A

n

= A

i ni

⌘

0

BBBBBBBBBBBBBBBB@

vn

⇢ n1

⇢ n2

⇢ n3

0

p⇢⇢

n1

vn

0 0pS

⇢n1

p⇢⇢

n2

0 vn

0pS

⇢n2

p⇢⇢

n3

0 0 vn

pS

⇢n3

0 0 0 0 vn

1

CCCCCCCCCCCCCCCCA

that have eigenvalues

�(1) = vn

� c ; �(2) = �(3) = �(4) = vn

; �(5) = vn

+ c , (84)

and eigenvectors

d

(1) ⌘ (⇢,�cn1

,�cn2

,�cn3

, 0)T , d

(5) ⌘ (⇢, cn1

, cn2

, cn3

, 0)T ,

(85)

d

(2) ⌘ (�ps

, 0, 0, 0, c2)T , d

(3) ⌘ (0,�n3

, 0, n1

, 0)T , d

(4) ⌘ (0,�n2

, n1

, 0, 0)T ,


Example of Cattaneo system

Instead to construct the matrix is more convenient to apply the following rulefrom the system

@

@t! ��, @

@xi

! ni

�, f ! 0 (86)

obtaining immediately(A

n

� �A0)�u = 0 (87)

from which we deduce that �u coincide qith the right eigenvector d.For example from Cattaneo system

8>>><

>>>:

⇢⇤cV

(#)@#

@t+@q

i

@xi

= r

⌧(#)@q

i

@t+ �(#)

@#

@xi

= �qi

,


we have 8<

:

�⇢⇤cV

(#)� �#+ �qn

= 0

�� ⌧(#) �qi

+ �(#) ni

�# = 0(88)

con �qn

= �qi

ni

. Da (88) and we have

�(1) = �r

�

⇢⇤ cV

⌧; �(2) = �(3) = 0; �(4) =

r�

⇢⇤ cV

⌧;

d

(1) ⌘✓1,

�

�1

⌧n

◆; d

(2) ⌘ (0,w1

); d

(3) ⌘ (0,w2

); d

(4) ⌘✓1,

�

�4

⌧n

◆;

con w

1

e w

2

: w1

· n = w

2

· n = w

1

·w2

= 0 and the system is hyperbolic provided⌧ > 0:


Wave equation and method of the characteristics

Let consider the wave equation in one space dimension

Utt

� c2Uxx

= 0. (89)

LetUt

= v , Ux

= w . (90)

that the equation can be rewritten as system of first order⇢

vt

� c2wx

= 0wt

� vx

= 0, (91)

that belong on the formu

t

+ Au

x

= 0 (92)

with

u = (v ,w)T , A =

✓0 �c2

�1 0

◆. (93)

The eigenvalues and the right eigenvectors are

�(1) = �c ; �(2) = c ; (94)

d

(1)⌘✓

c1

◆; d

(2)⌘✓

�c1

◆. (95)


Let l the left eigenvector of A, i.e.:

l A = �l (96)

thenl

(1)⌘ (1, c) ; l

(2)⌘ (1,�c). (97)

We assign the initial data

U(x , 0) = '(x), Ut

(x , 0) = (x). (98)

i.e. for the first order system

v(x , 0) = (x), w(x , 0) = '0(x); con '0(x) =d'

dx(99)

Multiplying the (92) for the left eigenvector we have

l {ut

+ Au

x

} = 0 , l {ut

+ �ux

} = 0 (100)

and we define the characteristic line in the space-time as:

dx

dt= �. (101)


x

C 1 2

x2

C

t

x1

Pt

x

Figure: Characteristic lines

In the present case we have two characteristic lines

C1 ) dx

dt= �(1) ) x = �ct + x

1

(102)

C2 ) dx

dt= �(2) ) x = ct + x

2

(103)


We consider the directional derivative

d

dt= @

t

+ � @x

(104)

obtaining

l · dudt

= 0. (105)

Thend

dt(l(1) · u) =0 (su C1),

d

dt(l(2) · u) =0 (su C2). (106)

Thereforel

(1) · u(x , t) = l

(1) · u(x1

, 0) (107)

l

(2) · u(x , t) = l

(2) · u(x2

, 0) (108)

Let u0

(x) the initial data of u:

u(x , 0) = u

0

(x).

Therefore 8<

:

l

(1) · u(x , t) = l

(1) · u0

(x + ct)

l

(2) · u(x , t) = l

(2) · u0

(x � ct).(109)


In the present case we have the algebraic system:8<

:

v(x , t) + c w(x , t) = (x + ct) + c'0(x + ct)

v(x , t)� c w(x , t) = (x � ct)� c'0(x � ct).

Then the solution

v(x , t) =1

2{ (x + ct) + (x � ct) + c ['0(x + ct)� '0(x � ct)]} (110)

w(x , t) =1

2c{ (x + ct)� (x � ct) + c ['0(x + ct) + '0(x � ct)]} . (111)

or

U(x , t) =1

2c{�(x + ct)� �(x � ct) + c ['(x + ct) + '(x � ct)]} (112)

with

�(⇠) =

Z ⇠

0

(⌧)d⌧.


Linear systems

In the case of a generic first order system of N equations if we represent the initialdata in the basis of right eigenvectors

u(x , t) =NX

j=1

⇧j(x , t)d(j) (113)

proceeding in the same way with the method of characteristic it is possible toprove that the solution is a combination of N waves:

u(x , t) =NX

j=1

⇧j

0

(x � �j t)d(j). (114)


A non linear example: the Burgers equation

Let consider the non linear Burgers equation

ut

+ uux

= 0. (115)

The characteristc isdx

dt= u (x , t) (116)

Butdu

dt= u

t

+ �ux

= ut

+ uux

= 0. (117)

i.e.u(x , t) = u

0

(x0

). (118)

Therefore also in this case the charcateristc is a line and

x = x0

+ u0

(x0

) t (119)


x

t

x

Pt

0

x

a)

x

t

x1x

2

Pt

cr

b)

Figure: Caratteristica e tempo critico dell’equazione di Burgers

but the slope of the line depends on x0

(see figure).


The determination of the critical time is simple In fact the (118) and (119) aresolution in paramteric form: for a fixed time t the depenmdence of u from x istyrouth the paarmeter x

0

. Then invertibility is lost when dx/dx0

= 0 and then:

tc

(x0

) = � 1

u0(x0

).

the critical time istcr

= infx

0

{tc

(x0

) > 0}. (120)

−0.5 0 0.5 1 1.50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

u(x,0)

x−0.5 0 0.5 1 1.50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

u (x, t )c

/a x/a

Figure: Critical time for Burtgers equation


Weak solutions

t

C

Σ

να

Figure: weak solutions

Let a system of balance laws

@↵F↵(u) = f(u). (121)

Let C a domain in the space-time and multiply (121) for a test function �, withsupport in C and we have

Z

C� (@↵F

↵ � f) dC = 0 (122)


or Z

C@↵ (�F↵) dC �

Z

C(F↵@↵�+ f�) dC = 0. (123)

Using the Gauss-Green theoremZ

⌃

⌫↵ (�F↵) d⌃�Z

C(F↵@↵�+ f�)dC = 0. (124)

then (assuming zero initial data)

Z

CF

↵@↵�dC +

Z

Cf�dC = 0. (125)

A solution of (125) for any test function � is called a weak solution of (121).


Shock waves

If exists a regular surface � with unit normal n fronte d' onda

campo perturbato

ϕ = ϕ(x, t)

Γ

n!

1u

0u

campo imperturbato

Figure: Onda d’urto

moving with normal velocity s separating the space in twosub-spaces in which there are classical smooth solutionsu

0

and u

1

such that their limit values in the surfaceare di↵erent we call this kind of solution a shock wave.Let denoting the jump with a square bracket:

[u] = u

1

��'� � u

0

| '+ (su �).

We want to prove that a shock wave is aweak solution if and only if across the surface there existssome compatibility conditions called Rankine-Hugoniotconditions. Let condider a surface in the spacetime � of normal '↵ (see Figura 7). We have from (124)

Z

⌃

+[�+

'+

↵�F↵+

d⌃⇤ �Z

C+

�F

↵+

@↵�+ f

+

��dC+ = 0

Z

⌃

�[��'�↵�F

↵�d⌃

⇤ �Z

C�

�F

↵�@↵�+ f��

�dC� = 0.


t

σ

ϕα ν α

+

-

σ

σ

C+

C-

Σ

Σ+

-

Figure: Soluzioni deboli tipo onde d’urto.

Then Z

�+

'+

↵�F↵+

d�+ �Z

C+

�F

↵+

@↵�+ f

+

��dC+ = 0 (126)

Z

��'�↵�F

↵�d�

� �Z

C�

�F

↵�@↵�+ f��

�dC� = 0. (127)


Rankine-Hugoniot equations

Summing (126) e (127) we haveZ

�+

'+

↵�F↵+

d�+ +

Z

��'�↵�F

↵�d�

� �Z

C(F↵@↵�+ f�) dC = 0. (128)

The last integral vanish and therefore we mush have in the surface:Z

�'↵

�F

↵+

� F

↵�� d� = 0. (129)

and then �F

↵+

� F

↵��'↵ = 0 (130)

i.e.[F↵]'↵ = 0. (131)

This means that the normal components in the space time of F↵ must becontinuous across the surface. Dividing space and time

'0

= �s, 'i

= ni

(132)

and assuming u ⌘ F

0 we can rewrite the R-H conditions in the usual form:

�s [u] +⇥F

i

⇤ni

= 0 (133)


or explicitly�su

1

+ F

i (u1

) ni

= �su0

+ F

i (u0

) ni

, (134)

The R-H conditions formally can be write from the di↵erential system with theoperator rule

@t

! �s [·] , @i

! ni

[·] , f ! 0. (135)

Let

s

(u) = �su+ F

i (u) ni

. (136)

then the R-H implies

s

1

0

s 1

u

u

Ψ

Ψ

(u )

0(u )

s

(u1

) = s

(u0

) . (137)

This require the non invertibility of the function s

(u).


We have@

s

@u= �sI+ A

ini

(138)

then bifurcation point are the one when s meet an unperturbed eigenvalue �0

(k-shocks)det (A

n

� �I)u

0

= 0. (139)

bif

orc

azio

ne

ort

ogonal

e

λ(1)

0

λ(2)

0

λ(3)

0

λ(r)

0

λ(j)

0

u =u

u =

u (u

0 , s

)

1

1 0

urto caratteristico

1

u

1

⌘ u

1

(u0

, s,n) . (140)


Shock waves in Euler fluid

The R-H for Euler system becomes:

�s [⇢] + [⇢vn

] = 0 (141)

�s [⇢ v] + [⇢ vn

v + p n] = 0 (142)

�s

⇢v2

2+ ⇢e

�+

✓⇢v2

2+ ⇢e + p

◆vn

�= 0 (143)

where vn

= v · n. Let introduce the Mach number and the specific volume

M0

=s � v

0n

c0

, V =1

⇢, (144)

then the solution of the R-H are

p = p0

+2�

� + 1p0

(M2

0

� 1). (145)

V = V0

� 2

� + 1V0

M2

0

� 1

M2

0

(146)

v = v

0

+2c

0

� + 1

M2

0

� 1

M0

n. (147)


From (145) and (146) we have

[p]

[V ]= �c2

0

M2

0

V 2

0

0. (148)

Then we have two possibilities

i) [p] > 0 e [V ] < 0 : corresponding to M2

0

> 1,

ii) [p] < 0 e [V ] > 0 : corresponding to M2

0

< 1.

Mathematically both are acceptable but which of the two is physical consistent?For this reason we calculate the R-H relative to the entropy law

⌘ = s[⇢S ]� [⇢Svn

] = [⇢(s � vn

)S ] (149)

If the weak solution of teh system is also weak solution of the entropy law ⌘ mustbe zero, while ⌘ is not null. In fact we have

⌘ = ⇢0

c0

cV

M0

log

⇢✓2 +M2

0

(� � 1)

M2

0

(� + 1)

◆�2M2

0

� + 1� �

1 + �

�(150)


Entropy growth across the shock

0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

M1

η

0

c0ρ

0c

V

_____

Figure: Entropy growth across the shock

As ⌘ have the meaning of the production of entopy across the shock we need torequire

⌘ � 0

and thenM2

0

> 1.


1 3 5 7 90

20

40

60

80

100

M0

p / p

0(a)

1 3 5 7 90.2

0.4

0.6

0.8

1

M0

V / V

0

(b)

1 3 5 7 90

2

4

6

8

M0

[vn] /

c0

(c)

1 3 5 7 915

10

15

20

25

30

M0

θ / θ

0

(d)

1 3 5 7 91

2

3

4

5

M0

c / c

0

(e)

1 3 5 7 90.4

0.5

0.6

0.7

0.8

0.9

1

M0

M

(f)


We note that

limM

0

!±1

V

V0

=� � 1

� + 1.

In reality we have another solution of the R-H equations: the characteristic shock

vn

= v0n

= s, p = p0

, (151)

with[v

T

] arbitrario, [⇢] arbitrario, v

T

= v � vn

n (152)

where v

T

is the tangential component of the fluid velocity. In this case ⌘ = 0.


non-linear wave propagation and non-equilibrium ...musesti/merla/nonequilibrium/... · overview 1...

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