12. Non-Equilibrium Phase Transitions

12.A. Introduction

12.B. Non-Equilibrium Stability Criteria

12.C. The Schlogl Model

12.D. The Brusselator

12.E. The Rayleigh- Benard Instability

S12.A. Fluctuations Near A Non-Equilibrium Phase Transition

12.A. Introduction

This chapter deals with phase transitions in systems that are far from absolute

thermodynamic equilibrium. To keep the problem manageable, we shall consider

only chemical and hydrodynamic systems that can be described in terms of

thermodynamic densities that are sufficiently well-behaved functions of space and

time. Physically, this means the systems are in local equilibrium and the time scales

of interest are large compared to any local relaxation times.

If a system is moved far enough away from absolute equilibrium, nonlinear effects

become dominant. Now, nonlinear equations can admit multiple solutions that have

different regions of stability. This means, by changing the parameters of a nonlinear

system, we can induce transitions between phases represented by the stable solutions.

In the linear regime near absolute equilibrium, the stable state is a steady state

characterized by minimum entropy production if one of the thermodynamic forces is

held fixed. This state is said to be on the thermodynamic branch.

Nonequilibrium phase transitions occur when the thermodynamic branch becomes

unstable. If space-time symmetries are broken during the transition, the new state

may exhibit oscillatory pattern in space and/or time. These patterns are called

dissipative structures since energy, and sometimes matter, must be supplied to

maintain them.

Simple models of nonequilibrium phase transitions to be discussed are

1. Chemical systems

(a) Schlogl model with 1st order-like phase transition.

(b) Brusselator model with dissipative structures.

2. Hydrodynamic systems

(a) Rayleigh-Benard instability.

12.B. Non-Equilibrium Stability Criteria

12.B.0. Entropy Production

12.B.1. Stability Conditions Near Equilibrium

12.B.2. Stability Conditions Far From Equilibrium

12.B.0. Entropy Production

A nonequilibrium system is dissipative since, to satisfy the 2nd law, there must be

some mechanisms that increase its entropy more than that offerred by the spontaneous

decay to absolute equilibrium. Thus, stability conditions are related to the rate of

entropy production.

To begin, consider a closed, isolated, hydrodynamic system of constant V. The total

time rate of change of entropy of the system is given by integrating over V the entropy

balance equation

Ds s

ss

t

v J (10.24)

to get

V

dS sdV

dt t

Ds s

V V

dV s dV v J

Ds s

A V

d s dV A v J (12.1)

s

V

dV (12.2)

where A is the surface bounding V and the surface integral in (12.1) vanishes because

the system is closed and isolated. The 2nd law becomes

0s

V

dSdV

dt (12.3)

which can be satisfied if the local entropy production s is nonnegative.

According to (10.179), we can write

0s i ii

J (12.4)

where iJ is a local flux and i the corresponding local force.

12.B.1. Stability Conditions Near Equilibrium

Near equilibrium in the linear regime, we can write

i ij jj

J L (12.5)

where ij jiL L (Onsager's relation). Putting (12.5) into (12.4) gives

0s ij i ji j

L (12.6)

which means the symmetric matrix ijLL is non-negative. Furthermore, for

spontaneous processes, it is positive definite. The necessary condition for minimum

entropy production is

0 sij ik j jk i

i jk

L

kj j ik i

j i

L L

ki ik ii

L L (12.12)

2 ki ii

L 2 kJ

Thus, if there is no constraint on the system, all i are independent and we have

0iJ for all i

which is just the absolute equilibrium state. On the other hand, if k of the forces, say,

, 1i i k are held fixed, the necessary conditions for minimum entropy

production in a system subject to n forces are

0iJ for 1, ,i k n (12.13)

When the minimum entropy production state is reached, the system becomes

stationary so that the k fluxes , 1, ,iJ i k will become constant. Such a state is

called a kth order stationary state. Near this steady state, we can write

0

0 01 1

1

, , , ,n

ss k n s k n i

i k i

02

1 1

1

2

n ns

i ji k j k i j

where the subscript 0 denotes steady state values. Using (12.12), we have

0 01 1

1 1

1, , , ,

2

n n

s k n s k n ij i ji k j k

L

(12.14)

The quantity

, 1

10

2

n

ij i ji j k

P dV L

(12.15)

is called the excess entropy production. It is positive definite since L is. Since

the steady state is the minimum entropy production state, we have

, 1

nj

ij ii j k

dd PdV L

dt dt

1

nj

jj k

ddV J

dt

1

n

ij i jj k

dV g J J

0 (12.16)

where we've used [see section 10.D],

1 χ L M α gα

andd

dt

α

J

Thus, P is a Liapounov function [see Appendix C].

Note that (12.15,16) are stability criteria for steady states in the linear regime. They

need not be satisfied by steady states in the nonlinear regime.

12.B.2. Stability Conditions Far From Equilibrium

12.B.2.1. General Arguments

12.B.2.2. Stability of Nonlinear Chemical Reactions

12.B.2.3. Exercise 12.1

12.B.2.1. General Arguments

1. Stability Relations

2. Example: Chemical System

12.B.2.1.1. Stability Relation

Using (12.4), the total entropy production can always be written as

i ii

P dV J (12.17)

Taking the time derivative gives

i ii i

i

P JdV J

t t t

(12.18)

JP P

t t

(12.18a)

where

ii

i

PdV J

t t

(12.18b)

J ii

i

P JdV

t t

(12.18c)

In the linear regime where i ij jj

J L with ij jiL L , we have

jii ij i

i ij

JL

t t

iji j

ij

Lt

ii

i

Jt

so that (12.18) implies

10

2i

ii

P PdV J

t t t

(12.19)

where the inequality is obtained from (12.16).

In the nonlinear regime, we still have (proof shown later)

0ii

i

PdV J

t t

(12.20)

but there is no general constraint on J P

t

. Stability analysis of the system therefore

requires, besides (12.20), one more relation.

12.B.2.1.2. Example: Chemical System

Consider a chemically reacting system of N types of molecules held far from

equilibrium by its large affinity. We shall assume both pressure and temperature to

be uniform throughout the system so that all effects of diffusion, viscosity, and

thermal conduction can be neglected. The entropy production thus simplifies to

1

1 rcP dV J A

T

(12.22)

[see (10.179)], where r is the number of reactions. Therefore

1

1 rcP A

dV Jt T t

(12.23)

From (10.179), we can write the balance equation for the jth kind of molecules as

1

rj c

j

cJ

t

(12.24)

Using the definition of the affinity

1

N

j jj

A

we have

1

Nj

jj

A

t t

1 1 1

r N rjc c

jj

AJ J

t t

1

Nj j

j

c

t t

so that (12.23) becomes

1

1 Nj j

j

P cdV

t T t t

(12.25)

Since the pressure and temperature are held constant, the chemical potentials j can

only be functions of the concentrations ic . Therefore

1k

Nj j i

i i PT c i

c

t c t

(12.26)

and (12.25) becomes

1 1

1

k

N Nj ji

j i i PT c i

P ccdV

t T c t t

(12.27a)

Now, as shown in section 2.H.2 and in particular eq(2.179), the matrix k

j

i PT c ic

is positive semi-definite so that (12.27a) implies

0P

t

(12.27)

Thus, condition (12.20) is indeed valid for the present case.

12.B.2.2. Stability of Nonlinear Chemical Reactions

Consider the reaction

1

21 1 2 2 3 3 4 4

k

kA A A A (12.28)

with

1 2 3 431 1 2 2 3 4

Nk N N k N N

t

3 41 2

1 2

2 3 41 1 2

1 1 2

1k N N

k N Nk N N

(12.29)

Now,

i ic

t t

i J ii

NM

t

where i i iM and iN is the molar density, iM the molecular weight, and

i i iM N is the mass density.

For a system of ideal gases,

3 4

1 2

3 4

1 2

, lnN N

A RT F T PN N

3 4

1 2

3 4

1 2

ln ,N N

RT K T PN N

(12.30)

where , exp ,K T P F T P and 4

1

, lni

ii i

i

F T P P

is independent of

iN . At equilibrium,

0A and 0iN

t

3 4

1 2

3 4

1 2

, 1N N

K T PN N

and3 4

1 2

2 3 4

1 1 2

1 0k N N

k N N

which means

2

1

,k

K T Pk

Hence,

3 33

3

NJ J

t M

3 4

1 2

1 2

3 41 1 2

1 2

1 ,N N

k N N K T PN N

1 2

1 1 2 1 expB

Ak N N

k T

(12.31)

thus relating the reaction rate J to the affinity A. For small A, we have

1 211 2

3 B

k AJ N N

k T

Now, for constant T and P, we have

, 1

1

2k j

ni

i ji j j PTN

S N NT N

, 1

1

2k j

ni

i ji j j PTn

n nT n

Using

ii

i

Vn

M

i

i

cV

M [ iM is the molar weight ]

we have

, 1

1

2k j

ni

i ji j j iPTc

Sc c

V T c M

Stability of Steady State of Purely Chemical Reactions

Near equilibrium, we have

0 21

2local local local localS S S S (12.32)

with0

2

, 1

1 10

2 2k j

ni

local i ji j j iPTc

S c cV T c M

(12.33)

where the inequality results from the fact that the matrix

0

k j

i

j PTcc

is non-negative.

Now,

1

1 rki

i kk

cJ

t

1

1 rki

i kk

cJ

t

Also,

1k j

ni

i jj j PTc

cc

Hence, using

1

nk

k j jj

A

we have

1

nk

k j jj

A

, 1

k i

njk

j ii j i PTc

cc

Consider now,0

2

, 1

1 1

2 2k j

nji i

local j ii j j iPTc

ccS c c

V t T c t t M

Since,

j i

ii TPn

G

n

1

j ii TPc

G

n c

we have

21

k j k j i

i

j j iTPc TPc

G

c n c c

21

k i k j i

j

i i jTPc TPc

G

c n c c

k ik j

ji

j i TPcTPcc c

Therefore,0

2

, 1

1 1

2k j

ni i

local ji j j iPTc

cS c

V t T c t M

0

, 1 1

1

k j

n rki

j i ki j kj PTc

c JT c

1

1 r

k kk

A JT

so that

2 21 1

2 2total localS dV St t V

1

1 r

k kk

dV A JT

Now,

2 210total localS dV S

V

Therefore, if 2 0totalSt

, the steady state is stable [see Lyapounov function].

Consider the entropy production

1

1 r

k kk

P dV A JT

1

1 r

k kk

P dV J AT

where | means fixed J. In the steady state, 0kJ corresponding to the forces kA

that are not held fixed vanish [see Minimum Entropy Production]. Therefore,

21

2k k kJ J J

so that

1

1 r

k kk

P dV J AT

21

2 totalSt

which means for stable states,

0P

Example

Consider the unimolecular reaction

1

2

k

kX A

with 1A X and

1 2X

X A

dNk N k N

dt AdN

dt

In the steady state,

0XdN

dt

so that

0 2

1X A

kN N

k

i i

i

M NJ

t

1 i

i

N

t

1 X

X

N

t

1 A

A

N

t

1 2X Ak N k N

,i i

i X A

A

2

1

ln A

X

k NRT

k N

In the steady state, 0X XN N so that 0A . Hence, the steady state is also the

equilibrium state. Now, let AN be held fixed away from the equilibrium value so

that

1 XJ k N

X

X

NA RT

N

1

P A JT

2

1 0X

X

Nk R

N

Therefore, the steady state is stable.

12.B.2.3. Exercise 12.1

Consider the auto-catalytic reaction

1

2

2k

kA X X

with

21 2

XA X X

dNk N N k N J

dt

where the sign of the 1k ( 2k ) term is + (–) since there is one more (less) X after the

() reaction. Also, the net stoichiometric coefficient for X is 1X . Thus,

22

1

ln X

A X

k NA RT

k N N

2

1

ln X

A

k NRT

k N

For the steady state,

0XdN

dt

0

0 1

2

0X

X A

N

kN N

k

The case 0 1

2X A

kN N

k also implies 0A so that it is an equilibrium state.

Now, let AN be held fixed away from the equilibrium value so that

01 22A X XJ k N k N N

0X

X

NA RT

N

201 20

2A X XX

RP k N k N N

N

For 0 0XN , we have P

so that the state is unstable.

For 0 1

2X A

kN N

k , we have 2

2 0XP Rk N

so that the state is stable.

12.C. The Schlogl Model

Consider the reactions

1

2

2 3k

kA X X

3

4

k

kX B (12.34)

with rates

2 31 2 3 4

XA X X X B

dNk N N k N k N k N

dt (12.46)

2 31 2

AA X X

dNk N N k N

dt

3 4B

X B

dNk N k N

dt

X A BdN dN dN

dt dt dt

currents

2 31 1 2

1 AA X X

A

dNJ k N N k N

dt

2 3 4

1 BX B

B

dNJ k N k N

dt

and affinities

32

1 21

ln X

A X

k NA RT

k N N

2

1

ln X

A

k NRT

k N

42

3

ln B

X

k NA RT

k N

At equilibrium, we have 1 2 0A A so that using overbars to indicate equilibrium

values, we have

2 1X Ak N k N and 4 3B Xk N k N (12.44)

Let

2 4

1 3

A

B

N k kR

N k k (12.45)

At a steady state with

0XdN

dt 0A BdN dN

dt dt

we have

0 2 0 3 01 2 3 4 0A X X X Bk N N k N k N k N

0 3 0 2 01 3 4

2 2 2

0AX X X B

k N k kN N N N

k k k

0 3 0 2 0 0X X XN aN kN b (12.47)

where

1

2

Ak Na

k 4

2B

kb N

k 3

2

kk

k

For fixed a and k, b is a cubic function of 0XN , i.e.,

0 3 0 2 0X X Xb N aN kN

The extrema of b are given by

0 2 00

3 2 0X XX

dbN aN k

dN (12.48)

with solutions

213

3XN a a k (12.49)

so that

3 2

X X X Xb b N N a N kN

Now,

20

0 2

0

6 2 0

0X

X

d bN a

dN

for 0

3X

aN

Also,

b as 0XN

0b as 0 0XN

Thus, assuming 2 3a k so that XN are real, the graph of 0Xb b N is

Therefore, there are 3 phases as indicated by the intersects of the line B with the curve.

Of these, the one with 0X X XN N N is unstable.

For 2 3a k , there is no extrema so that there is only an inflection point at 0

3X

aN

b

b –

b +

N--X N +

XN0

X

Stable,

steady

phase

Unstable

Stable,

steady

phase

B

Thus, the phase diagram in the b a plane is

b

N0Xa /3

b–(a) curve

b+(a) curve

Hysteresis Region

a

b–

3k

N0X N+

X

N0X < N–

X

Single

Phase

2 stable, 1 unstable phases

b+

12.D. The Brusselator

12.D.1. The Brusselator – A NonLinear Chemical Model

12.D.2. Boundary Conditions

12.D.3. Linear Stability Analysis

12.D.1. The Brusselator – A NonLinear Chemical Model

Belousov- Zhabotinski reaction

Ce ion catalyzed oxidation of Molonic acid by Bromate in Sulfuric acid medium:

1. Well-stirred:

Periodic changes of Br , 4 3/Ce Ce concentrations (spatially

homogeneous) color changes (chemical clock)

2. Unstirred, shallow dish:

Travelling waves of Br , 4 3/Ce Ce concentrations.

See R.J.Field, R.M.Noyes, J.Chem.Phys. 60, 1877 (74).

3 Variable Intermediates Model (Oregonator)

See R.J.Field, J.Chem.Phys. 63, 2289 (75).

Simplified Model (Brusselator)

See I.Prigogine, Lefever, J.Chem.Phys. 48, 1695 (68).

The Brusselator consists of 4 reactions involving 6-components:

1k

A X

2k

B X Y D

3

2 3k

X Y X

4k

X E (12.50)

Adding these reactions gives

4 4A B X Y X Y D E or A B D E (net reaction)

Thus, X and Y are catalysis. Futhermore, the k3 reaction shows that X is autocatalytic

(it appears on both sides of the reaction) and provides the non-linearity.

In a typical run, A, B are kept in excess while D, E are removed instantly so that

concentrations of A, B, D and E are fixed. Thus, the affinity A is large and the

system is far from chemical equilibrium. X, Y then exhibit variations in time / space.

Rate Equations

The rate equations are

2 21 2 3 4'

XX X A B X X Y X

ND N k N k N N k N N k N

t

(12.51)

2 22 3'

YY Y B X X Y

ND N k N N k N N

t

(12.52)

where XD and YD are diffusion co-efficients. Note that the 2k and 3k

reactions contribute equal amounts but with opposite signs to'XN

t

and'YN

t

.

In order to simplify the notations, we would like to absorb the k's into the N's and

write (1) as

2 2X

XD X A BX X Y X

t

(12.53)

2 2Y

YD Y BX X Y

t

(12.54)

To this end, we set

X XX N 'tt t etc

so that 7 's for X, Y, A, B, t, DX, and DY. Eq(12.53-4) then become

2 2 2

' X

X XD X X X A A B X B X X Y X Y X X

t

ND N N N N N N N

t

2 2 2

' Y

Y YD Y Y Y B X B X X Y X Y

t

ND N N N N N

t

Rearranging, we have

2 2

' X

X A tD t X X A B t B X X Y t X Y t X

X

ND N N N N N N N

t

2 2 2

' Y

Y B X tD t Y Y B X X t X Y

Y

ND N N N N N

t

Comparing with (12.51-2) gives

1XD t 1

YD t

1A t

X

k

2B X t

Y

k

2B t k 23X t k

3X Y t k 4t k

Thus, there are 8 equations for 7 variables. However, the 4 equations involving k2

and k3 are not independent. They reduce to 2 equations by setting X Y . Thus,

we have

4t k

4

1 1XD

t k

4

1 1YD

t k

2 2

4B

t

k k

k

2 3 3

4X

t

k k

k

3

4X Y

k

k

21 3 1 3 1

34 4 4

XA

t

k k k k k

k k k

i.e.,

4 't k t

4

XX

DD

k

4

YY

DD

k

2

4B

kB N

k

23 1

34

A

k kA N

k

3

4X

kX N

k 3

4Y

kY N

k

Steady Uniform State

For a steady state

0X Y

t t

For a uniform state2 2 0X Y

Hence, for a steady, uniform state, eq(2) becomes

21 0A B X X Y

2 0BX X Y The 2nd equation gives

0X orB

XY

Now, the 0X solution can be discarded since it means nothing happens.

For theB

XY

solution, the 1st equation becomes

210

B B BA

Y Y

00

0

BY

A and 0 0X A (12.55)

where we've used the subscript 0 to indicate the steady, uniform state. In terms of

the usual parameters, we have

0 0X A 2

0 03 3 13

4 4X A

k k kN N

k k 0 01

4X A

kN N

k

00

0

BY

A

3 003 2 4

2 04 4 3 1

BY

A

k k k NN

k k k k N

00 2 4

03 1

BY

A

k k NN

k k N

Reverse Reactions

Now, if the reverse reactions are allowed with reaction constants ik , we have

2 21 2 3 4'

XX X A B X X Y X

ND N k N k N N k N N k N

t

31 2 3 4X Y D X Ek N k N N k N k N

and similarly for YN , AN , …, etc. In chemical equilibrium, the ki and ki' reactions

cancel pairwise, e.g., 4 4X Ek N k N , …, etc. Hence, the 0XN and 0

YN solutions

are also solutions of the chemical equilibrium (with appropriate 0EN and 0

DN ). In

other words, the steady, uniform states are on the thermodynamics branch.

Stability

To study the stability of the steady, uniform state, we set

0X X X A (12.56)

0

BY Y Y

A (12.57)

Eq(2) then becomes

22X

BD A B A A A

t A

22Y

BD B A A

t A

Keeping only terms linear in and , we have

2 22XD A B BA B A AB At

2 21XD B A

2 22YD B BA B A ABt

2 2YD B A

Rearranging, we have

2 21XD B At

(12.58)

2 2YD A B

t

(12.59)

Note that these equations govern the kinetics of the system near 0 , the

steady, homogeneous, state. Since these equations are linear,

1. Different components of the Fourier transform of and don't mix.

2. Study of 1 component ,k is enough.

12.D.2. Boundary Conditions

Particular solutions require the specification of boundary conditions. We'll assume

the system to be contained in a box of sides L and set

, ,

ˆii

i x y z

n

L

k e

Case I: Dirichlet Boundary Conditions

0X X and 0Y Y on boundary.

0 on boundary.

, , sin sin sin tx y zt k x k y k z e r k (12.60)

, , sin sin sin tx y zt k x k y k z e r k (12.61)

Case II: Neumann Boundary Conditions

ˆ Xn ˆ 0Y n on boundary. ( ˆ surfacen )

ˆ n ˆ 0 n on boundary.

, , cos cos cos tx y zt k x k y k z e r k (12.62)

, , cos cos cos tx y zt k x k y k z e r k (12.63)

In both cases, we have2 2k

t

and similarly for . Here,

2 2 2 2x y zk k k k

22 2 2x y zn n n

L

2

mL

where m is an integer. Note that

0m 0x y zn n n

Hence, 0m is not allowed for case I since it would imply 0 identically.

However, 0m is not allowed for case II.

Since the time dependence of and are te , we have

1. real

(a) 0 0 for t large, i.e., the system return to the spatially homogeneous

steady state.

(b) 0

grows as t grows. The linearized equations will become invalid so that

the state is unstable. It bifurcates to another state, thus inducing a phase

transition. The new state obeys a set of non-linear equations and hence

cannot be discussed using our linearized analysis. However, the condition

0 does indicate the on-set of the bifurcation. Finally, the new state is

expected to exhibit spatial oscillations.

2. complexLet R Ii . Time oscillation is then denoted by I .

(a) 0R .

The system gradually returns to the steady state.

(b) 0R .

Bifurcate to new state, possibly oscillatory in both time and space.

( chemical clock, travelling waves, etc.)

Fourier Components

In terms of and , we have

2 21 xB D k A

2 2yB A D k

or

2 2

2 2

10x

y

B D k A

B A D k

(12.64)

i.e.,

21

2

0c A

B c

where

21 1 xc B D k and 2 2

2 yc A D k

2 21 2 1 2 0c c c c A B (12.65)

with solutions

2 21 2 1 2 1 2

14

2c c c c c c A B

2 21 2 1 2

14

2c c c c A B

(12.67)

12.D.3. Linear Stability Analysis

12.D.3.0. General Discussion

12.D.3.1. Real Frequency Dispersion

12.D.3.2. Complex Frequency Dispersion

12.D.3.0. General Discussion

12.D.3.1. Real Frequency Dispersion

2 21 2 4c c A B (12.68)

Now, consider the case 2 1c c so that 0 and 0 iff

2 2 21 2 1 2 4c c c c A B

or 21 2 0c c A B (12.69)

2 2 2 21 0x yB D k A D k A B

2 2 2 2 2 2 4 2 0y y x x yBA A BD k D k A D k D D k A B

22 2

21 x

xy y

D AB A D k

D D k

2 2 22

2 21 x

xy y

D A L mA D

D D m L

mB (12.70)

where 2 2 2x y zm n n n . Writing

mB mm

2mdB

dm m

0Set

extremum at

0m

2 2 2

2 2y x

A L L

D D

2

2x y

AL

D D

Since B for 0 orm , 0m is a minimum.

Consider the case of a 1-dim system. We have 2m n so that

0 0n mx y

A L

D D is not an integer.

Note that only n integer are solutions to the kinetic equation. Thus, for CB B

(the 2n line), bifurcation occurs.

B

BC

0 1 2 3 4n0

Bm

Unstable

12.D.3.2. Complex Frequency Dispersion

2 21 2 4c c A B (12.71)

Since everything is positive, we can write

1 2 2c c A B

or

2 21 2y xB A D D k A B

221 0y xB A D D k

22 1y xB A D D k

Since

20B A

we have

2 1y xD D k or2

2y x

LD D

m (12.72)

The real part of is 1 2

1

2c c . Therefore, 0 if 1 2c c , i.e.,

2 2 21 x yB D k A D k

or 2 21 x yB A D D k

2

22

1 x y

mA D D

L

mB (12.73)

Note that for case II of the Neumenn B.C., 0n is allowable, which gives rise to a

homogeneous time clock. Else, we have travelling waves.

Unstable

n1 2 3

BBm

12.E. The Rayleigh- Benard Instability

12.E.1. Hydrodynamic Equations And Boundary Conditions

12.E.2. Linear Stability Analysis

12.E.1. Hydrodynamic Equations And Boundary Conditions

Fluid flow in pipe:

1. low velocity, high viscosity smooth, steady.

2. high velocity, low viscosity turbulence.

Benard's problem: (sometimes called Rayleigh-Benard problem)

For small 0 dT T T , fluid is at rest.

For large enough T, Bernard's instability set in and convective flow cells appear.

Td

T0

Glass

plates

Fluid with viscosity and thermal conductivity

Gravity

Td < T0

Side View

Hexagonal

cells

Covered

container

Top View

Consider the Navier-Stokes equations

0t

v (conservation of mass) (12.75)

2 1

3P

t

v

vv F v v

(conservation of momentum) (12.76)

:u

u K Tt

v

Π v(conservation of energy) (12.77)

For a thin fluid layer occupying the entire x–y plane,

0

0 0P

T TT

0 0 0P T T (12.78)

where the thermal expansivity

1P

P

V

V T

1

PT

is evaluated at 0 . Rearranging, we have

0 01 P T T z

Open

container

Rings

Steady State, No Flow ( v 0 )

The Navier-Stokes equations become

0 P F ˆP g z (12.80)2 0K T (12.81)

where ˆg F z and g is the gravitational acceleration. Since the only variations

are in the z–direction, we have

dPg

dz 0 01 P T T g

2

20

d T

dz

Using x to denote the steady state value of x, the 2nd equation gives

T z c c z ( c, c' are constants )

With the boundary conditions

00T T and dT d T

we have

00

dT TT z T z

d

(12.82)

Hence,

00 1 d

P

dP T Tz g

dz d

with solution

2000

2d

P

T TP z P z z g

d

Setting 0 dT Ta

d

, we have,

0T z T a z

20 0

1

2 PP z P z a z g

(12.84)

where 0 0P P . Note that

0 01 P T T and 0 1 Pa z (12.83)

Stability

Let the deviation of X from its steady state value be

, ,X t X t X z r r

Since 0v , we have , ,t t v r v r . For small v, the energy u is mostly

thermal so that we can write

vu c T and vu c T

Keeping only terms linear in X, the Navier-Stokes equations become

0t

v (12.85)

2 1ˆ3

P gt

v

z v v (12.86)

0u u u K Tt

v (12.87)

where the term :Π v does not contribute since Π is proportional to partials of v.

Note that the steady state satisfies

0u K Tt

0t

ˆ 0P g z

2 0T Now, (12.87) can be written as

2 0u

u u u K Tt t

v v

which, with the help of (12.85), gives

2 0u

u K Tt

v

2 0v

Tc T K T

t

v

2 0v z

Tc av K T

t

Next, we can rewrite (12.85) as

0t

v v

Since 0 1 Pa z z , we have

z

dz d dv

t dt dz dz

v

Thus,

0t

v where

Since 0 01 P T T where T T T , we have

0 P

T

t t

Hence,

0 P T

t

v

Now, 3 410 ~ 10 1P while

T

t

is expected to be small for small T.

Therefore,

0 v ( Boussineq approximation )

and

0 1 Pa z 0

The linearized N-S eqs thus reduce to

0 v (12.89)

20 ˆP g

t

v

z v (12.90)

2

0

0zv

T Kav T

t c

(12.91)

with

0 P T (12.88)

Now, 12.90 gives

20 ˆg

t

ω

z

ω

where ω v is the vorticity. Setting the kinematic viscosity as0

and

using (12.88), we have

2ˆPg Tt

ω

z

ω(12.92)

Since ˆf z is in the x–y plane, we have

2zzt

(9)

Using

vω

2 v v 2 v

where the last equality is due to 0 v . Similarly,

ˆT z ˆT z 2 ˆT z

2ˆTT

z

z

Thus, 12.92 gives

2 2 4ˆP

Tg T

t z

vz v

so that

2 2 2 22 4

2 2 2 2z

P z

v Tg T v

t z x y z

2 22 4

2 2z

P z

vg T v

t x y

(12.94)

Thus, the z–components of and v are given by eqs(9) and (12.94), respectively.

The situation of interest is the transition from the steady state to the Benard instability

state (see figure). The actual patterns of vz in the x–y plane are determined by the

boundary conditions.

Boundary Conditions

Since the upper and lower plates are held at fixed temperatures 0T and dT , we have

, ,0; , , ; 0T x y t T x y d t for all t

Also, the fluid is bounded between the plates so that

, ,0; , , ; 0z zv x y t v x y d t for all t

These 2 conditions are general boundary conditions valid for all types of surfaces.

In addition, there will be further constraints dependent on the properties of the plates.

I. Rigid Surfaces (No Tangential Flow)

, ,0; , , ; 0x y t x y d t v v for all t

Thus, on the boundaries,

0v

0x y

v v

0yxz

vv

y x

0zv

z

[since 0 v everywhere]

II. Smooth Surfaces (e.g. Free Surfaces)

This means flows in the x–y plane do not exert force in the z direction. Hence,

// 0z boundary where // denotes components in the x–y plane.

Thus,

, ,0; , ,0;xz yzx y t x y t , , ; , , ; 0xz yzx y d t x y d t

Since v is proportional to , we have

0yx z zvv v v

z x z y

on boundaries

Since 0zv for all x and y on boundaries, we have

0yx z zvv v v

z x z y

for all x and y on boundaries

Thus,

22yz x

vv

z z y z x

0yx

vv

y z x z

on boundaries

Also,

0yx zvv v

x y z

v

implies2

20yz x

vv v

z x z y z

2

2zv

z

on boundaries

12.E.2. Linear Stability Analysis

In summary, the linear stability equations are

2

0

0zv

T Kav T

t c

(6)

2zzt

(9)

2 22 4

2 2z P zv g T vt x y

(10)

Looking for solutions periodic in the x–y plane, we write

// //, , , expX t X z i t r k k r

where

// ˆxkk x ˆyk y and // x yk x k y k r

Hence,

XX

t

and2

2 2//2

X k Xz

After cancelling the common exponential factor, the stability equations become

22//2

0

0zv

KT av k T

c z

(12.102)

22//2z zk

z

(12)

22 22 2 2// // //2 2z P zk v g k T k v

z z

(12.103)

Boundary Conditions

General

0B

T 0 0T T d

0z Bv 0 0z zv v d for all // ,k

Rigid Surface

0z B 0 0z z d

0z

B

v

z

0

0z z

z z d

v v

z z

for all // ,k

Smooth Surface

0z

Bz

0

0z z

z z dz z

2

20z

B

v

z

2 2

2 2

0

0z z

z z d

v v

z z

for all // ,k

Solutions

(11) gives

22//2

0

1z

v

Kv k T

a c z

Substituting into (13)

2 22 2 2// // //2 2

0z Pk k v g k Tz z

gives

2 2 22 2 2 2// // // //2 2 2

0

10P

v

Kk k k T g k T

a z z c z

2 2 22 2 2 20// // // //2 2 2

0

0vP

v

K ck k k T g k T

a c z z z K

2 2 2 22 2 2 0 0 //// // //2 2 2

v v Pc a c g kk k k T T

z z z K K

Introducing the dimensionless variablez

Zd

, we have

2 2

2 2 2

1

z d Z

and

2 22

2 2 2

1A Ad

z d Z

Now, set

2 2 2//k d 2d s

20 0v vc cd s s

K K

where 0 vc

K

26 4 2 20 // 0v P v Pa c g k a c g

d d RK K

where 40 v Pa c gR d

K

is the Rayleigh number. Thus,

2 2 22 2 2 2

2 2 2s s T R T

Z Z Z

Doing the same to zv gives in the same equation for zv . Since the B.C. are more

specific on zv , we shall work on it henceforth. Following the standard technique,

we set

22

2 zG vZ

2 2 22 2 2

2 2 2 zF s G s vZ Z Z

so that we have

22 2

2 zs F R vZ

1

0

dZ F both sides gives

1 122 2

20 0

zdZ F s F R dZ F vZ

Consider

1 2

20

I dZ F FZ

1 1

0 0

F F FF dZ

Z Z Z

Now, F is a sum ofn

zn

v

Z

with 0,2,4,n . Hence,F

Z

s a sum ofn

zn

v

Z

with

1,3,5,n . Note that

For a rigid surface, 0n

zn

v

Z

for all n.

For a smooth surface, 0n

zn

v

Z

for even n.

Therefore,

1

0

0F

FZ

for both cases.

Hence,

21

0

FI dZ

Z

and

21 1

22 2

0 0

z

FdZ s F R dZ F v

Z

Since

22

2 zG vZ

22

2F s G

Z

Thus,

1

0

zJ dZ F v 1 2

22

0

zdZ v s GZ

1 2

22

0

zz

vdZ G s v G

Z

[integration by parts twice]

1

2

0

zdZ G s v G [2

22z

z

vG v

Z

]

1 22 2

20

z zdZ G s v vZ

21 22 22

20

zz

vdZ G s v

Z

[integration by parts once]

Hence,

221 1 2

2 2 22 2 22

0 0

zz

F vdZ s F R dZ G s v

Z Z

Since only s is complex, the imaginary part gives

21 1 2

2 22 22

0 0

Im Im zz

vdZ s F R dZ s v

Z

or

21 2

2 22 22

0

Im 0zz

vs dZ F R v

Z

Since , 0R , the integrand is always positive unless 0zv . Thus, other than the

steady state, we must have Im 0s , i.e., s is real and likewise2

s

d

. Hence,

0s system is unstable.

0s system is stable.

so that 0s is the transition point. Setting 0s , we have

322 2

2 z zv R vZ

[eigen-equation]

Note that //k and R a T . Now, the imposition of boundary conditions

will in general restrict the eigenvalues 2R to discrete values. Thus, k// and T are

related. Note that the reciprocal of k// is proportional to the cell size.

As we increase R from 0, the system will become unstable as 2R reaches the 1st

eigenvalue.

If both plates are smooth boundaries,2

20

mz

m

v

Z

at 0,1Z for all m.

sinzv A n Z 1,2,n

2

2

2z

z

vn v

Z

so that the eigen-equation becomes

32 2 2n R

i.e.,

32 2

2

nR

The smallest R is therefore

32 2

2R

This may be further reduced by adjusting so that2

0dR

d . Thus,

2 32 2 2 2

2 2 4

3dR

d

22 2

2 2 24

3

22 2

2 24

2 0

which gives only one real solution as2

2

2

2.222

C

and

22 2C

C

dd

32 2

2

C

CC

R

32

42

3272 657.514

2

Ref:

1. E.N.Lorenz, J.Atmos.Sci. 20,130 (63). [Chaos]

2. S.Chandrasekar, "Hydrodynamics and Hydromagnetic Stability" (61)

3. E.Ott, "Chaos in Dynamical System" (93)