the example of rayleigh-benard convection. pattern-forming instabilities: the example of...

The example of Rayleigh-Benard convection

Pattern-forming instabilities:The example of Rayleigh-Benard convection

Rayleigh-Benard convection.Boussinesq approximation,

incompressible flow

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∇⋅ r

u = 0

∂t T +r u ⋅∇T = κ ∇ 2 T

∂t

r u +

r u ⋅∇

r u = −

1

ρ 0

∇p +r g

Δρ

ρ 0

+ ν ∇ 2 r u

r u = u,v,w( )r g = 0,0,−g( )

Rayleigh-Benard convection.Boussinesq approximation,

incompressible flow

€

∇⋅ r

u = 0

∂t T +r u ⋅∇T = κ ∇ 2 T

∂t

r u +

r u ⋅∇

r u = −

1

ρ 0

∇p −r g α T − T0( ) + ν ∇ 2 r

u

Δρ = ρ − ρ 0 = −α ρ 0 T − T0( )r u = u,v,w( ) ,

r g = 0,0,−g( )

Rayleigh-Benard convection.Boussinesq approximation,

incompressible flow

€

∇⋅ r

u = 0

∂t T +r u ⋅∇T = κ ∇ 2 T

∂t

r u +

r u ⋅∇

r u = −

1

ρ 0

∇p −r g α T + ν ∇ 2 r

u

Δρ = ρ − ρ 0 = −α ρ 0 T − T0( )r u = u,v,w( ) ,

r g = 0,0,−g( )

Rayleigh-Benard convection.Boussinesq approximation,

incompressible flow

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rω =∇×

ru

∇ ⋅r u = 0

∂t T +r u ⋅∇T = κ ∇ 2 T

∂t

r ω +

r u ⋅∇

r ω =

r ω ⋅∇( )

r u − α ∇ ×

r g T( ) + ν ∇ 2 r

ω

Δρ = ρ − ρ 0 = −α ρ 0 T − T0( )r u = u,v,w( ) ,

r g = 0,0,−g( )

Rayleigh-Benard convection: static state

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ru = 0,0,0( ) ,

r ω = 0,0,0( )

∂

∂t= 0

0 = κ ∇ 2 T = κ∂ 2T

∂z2

0 =−α ∇ ×r g T ( ) ⇒

∂ T

∂ x=

∂ T

∂ y= 0

Rayleigh-Benard convection

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Boundary conditions

u = v = w = 0 z = 0,D no slip

w = 0

∂u

∂z=

∂u

∂z= 0 z = 0,D free slip

T = T2 z = 0

T = T1 z = D fixed temperature

κ∂T

∂z= C z = 0,D fixed flux

Rayleigh-Benard convection: static state with pure conduction

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ru = 0,0,0( ) ,

r ω = 0,0,0( )

T = T (z) = T2 −T2 − T1

Dz = T2 − β z

Fixed temperature b.c.

Rayleigh-Benard convection.Boussinesq approximation,

incompressible flow

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rω =∇×

ru , T = T (z) + θ = T2 − β z + θ

∇ ⋅r u = 0

∂tθ +r u ⋅∇θ − β w = κ ∇ 2 θ

∂t

r ω +

r u ⋅∇

r ω =

r ω ⋅∇( )

r u − α ∇ ×

r g θ( ) + ν ∇ 2 r

ω

r u = u,v,w( ) ,

r g = 0,0,−g( )

Rayleigh-Benard convection:non dimensional formulation

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x,y,z( ) = D ˜ x , ˜ y , ˜ z ( ) , t =D2

κ˜ t , θ = T2 − T1( ) ˜ θ

u,v,w( ) =κ

D˜ u , ˜ v , ˜ w ( ) ,

r ω =

κ

D2

r ̃ ω

∇ ⋅r u = 0

∂tθ +r u ⋅∇θ −w =∇ 2 θ

1

σ∂t

r ω +

r u ⋅∇

r ω −

r ω ⋅∇( )

r u [ ] = R∇ × ˆ z θ( ) +∇ 2 r

ω

R =gα D3 T2 − T1( )

νκ, σ =

ν

κ

Rayleigh-Benard convection

Important parameters:

R = g D3T2-T1) / = a = L D

Rayleigh-Benard convection.

If R < Rcrit conductionT(x,y,z,t)=Tcond (z)=T2 - z

(u,v,w)=(0,0,0)

If R > Rcrit convectionT= Tcond +

(u,v,w) non zero

Rayleigh-Benard convection.

Linear stability analysis

2D Rayleigh-Benard convection(non dimensional formulation)

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u,v,w( ) = u,0,w( ) ,r ω = 0,ω,0( )

∇ ⋅r u = 0

∂tθ +r u ⋅∇θ −w =∇ 2 θ

1

σ∂t

r ω +

r u ⋅∇

r ω −

r ω ⋅∇( )

r u [ ] = R∇ × ˆ z θ( ) +∇ 2 r

ω

2D Rayleigh-Benard convection

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ru = u,0,w( ) ,

r ω =∇ ×

r u = 0,ω,0( )

u = −∂ψ

∂z, w =

∂ψ

∂ x, ω = −∇2ψ

∂θ

∂t+ ψ ,θ[ ] −

∂ψ

∂ x=∇ 2 θ

1

σ

∂∇ 2ψ

∂t+ ψ ,∇ 2ψ[ ]

⎧ ⎨ ⎩

⎫ ⎬ ⎭= R

∂θ

∂ x+∇ 2∇ 2ψ

2D Rayleigh-Benard convection:Linearization around the static state

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ru = u,0,w( ) ,

r ω =∇ ×

r u = 0,ω,0( )

u = −∂ψ

∂z, w =

∂ψ

∂ x, ω = −∇2ψ

∂θ

∂t+ ψ ,θ[ ] −

∂ψ

∂ x=∇ 2 θ

1

σ

∂∇ 2ψ

∂t+ ψ ,∇ 2ψ[ ]

⎧ ⎨ ⎩

⎫ ⎬ ⎭= R

∂θ

∂ x+∇ 2∇ 2ψ

2D Rayleigh-Benard convection:Linearization around the static state

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∂∂t

−∂ψ

∂ x=∇ 2 θ

1

σ

∂∇ 2ψ

∂t= R

∂θ

∂ x+∇ 2∇ 2ψ

u = −∂ψ

∂z= U(z)exp ikx + λ t( )

w =∂ψ

∂ x= W (z)exp ikx + λ t( )

θ = Θ(z)exp ikx + λ t( )

2D Rayleigh-Benard convection:Linearization around the static state

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∂∂t

−∂ψ

∂ x=∇ 2 θ

1

σ

∂∇ 2ψ

∂t= R

∂θ

∂ x+∇ 2∇ 2ψ

∂ψ

∂ x= sin π z( )exp ikx + λ t( )

θ = sin π z( )exp ikx + λ t( )

2D Rayleigh-Benard convection:Linearization around the static state

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Mψ x

θ

⎛

⎝ ⎜

⎞

⎠ ⎟= λ

ψ x

θ

⎛

⎝ ⎜

⎞

⎠ ⎟

M =−σ p2 σ k 2R

p2

1 −p2

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟

p2 = π 2 + k 2

ψ x = sin π z( )exp ikx + λ t( ) , θ = sin π z( )exp ikx + λ t( )

2D Rayleigh-Benard convection:Linearization around the static state

€

λ2 + σ +1( )p2λ + σ p4 1−k 2R

p6

⎛

⎝ ⎜

⎞

⎠ ⎟= 0

1−k 2R

p6< 0 ⇒ λ > 0

R > R0 =p6

k 2=

π 2 + k 2( )

3

k 2convection

Rayleigh-Benard convection.

Linear stability analysis

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R0 =π 2 + k 2

( )3

k 2

Rcrit (min) =27π 4

4

kcrit (min) =π

2

2D Rayleigh-Benard convection:Threshold to convection

Rayleigh-Benard convection:above Rcrit, convective motion occurs.

This takes the form of parallel rolls

The convective rolls saturate the instability

Amplitude expansion

€

∂∂t

+ ψ ,θ[ ] −∂ψ

∂ x=∇ 2 θ

1

σ

∂∇ 2ψ

∂t+ ψ ,∇ 2ψ[ ]

⎧ ⎨ ⎩

⎫ ⎬ ⎭= R

∂θ

∂ x+∇ 2∇ 2ψ

ψ = εψ1 + ε 2ψ 2 + ... , θ = εθ1 + ε 2θ2 + ...

T = ε 2 t , R = R0 + ε 2R2

for simplicity : σ >>1

Amplitude expansion

€

∂∂t

+ ψ ,θ[ ] −∂ψ

∂ x=∇ 2 θ

R∂θ

∂ x+∇ 2∇ 2ψ = 0

ψ = εψ1 + ε 2ψ 2 + ... , θ = εθ1 + ε 2θ2 + ...

T = ε 2 t , R = R0 + ε 2R2

for simplicity : σ >>1

Amplitude expansion: first order

€

ψ1,x

θ1

⎛

⎝ ⎜

⎞

⎠ ⎟= A(T)exp(ikx) + c.c.[ ]

1

p−2

⎛

⎝ ⎜

⎞

⎠ ⎟ sinπ z

Amplitude expansion: second order

Amplitude expansion: third order

€

∇4ψ 3 + R0θ3,xx = −R2θ1,xx

ψ 3,x +∇ 2θ3 = θ1,T +ψ1,xθ2,z

Amplitude expansion: third order

€

∇6θ3 − R0θ3,xx = p2 AT −k 2R2

p4A +

1

2A

2A

⎛

⎝ ⎜

⎞

⎠ ⎟exp(ikx)sin π z( )

−k 2 + 9π 2

( )2

2 k 2 + π 2( )

A2Asin 3π z( ) + c.c.

Amplitude expansion: third order

The Fredhom alternative:eliminate the secular term

and get a solvability condition:

The Landau equation €

AT =k 2R2

p4A −

1

2A

2A

Amplitude expansion: third order The Landau equation with real coefficients

€

aT =k 2R2

p4a −

1

2a3

Amplitude expansion: third order The Landau equation with real coefficients

Pitchfork bifurcation at R2=0€

ak 2R2

p4−

1

2a2

⎛

⎝ ⎜

⎞

⎠ ⎟= 0

a0 = 0

a1,2 = ±2k 2R2

k 2 + π 2( )

Stability of the rolls: Busse balloon

Stability of the rolls

Stability of the rolls