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