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On Direct and Adjoint Lattice Boltzmann Equations Fran cois Dubois - - PowerPoint PPT Presentation
On Direct and Adjoint Lattice Boltzmann Equations Fran cois Dubois - - PowerPoint PPT Presentation
ICMMES Conference, Hong Kong, July 25-29, 2005 On Direct and Adjoint Lattice Boltzmann Equations Fran cois Dubois Numerical Analysis and Partial Differential Equations Department of Mathematics, Universit e Paris Sud conjoint work with
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General framework
x : a node of the lattice ∆t : the (small) time step vj : discrete celerity in the lattice, components vα
j , α = 1, 2
then x + vj ∆t is an other node of the lattice f j(x , t) : density of particles having the velocity vj at node x and at discrete time t . m0 = ρ ≡
- j
f j : density of matter mα = qα ≡
- j
vα
j f j : component number α of the momentum
W ≡ ( ρ , q1 , q2 ) : conserved variables
ICMMES Conference, Hong Kong, July 28, 2005
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General framework
1 3 6 5 2 4 8 7
Neighbourhood of a vertex x for the D2Q9 LBE model
ICMMES Conference, Hong Kong, July 28, 2005
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d’Humi` eres (1992) representation of the discrete dynamics
mk ≡
- j
M k
j f j (k ≥ 3) : other components of the momentum
remark that we have M 0
j = 1 and M α j = vα j
mk
eq and f j eq : equilibrium momenta and velocity distribution
- collision step
mi
∗ ≡ mi ≡ mi eq : conserved momenta (i = 0 , 1, 2) during the collision
mk
∗ = (1 − sk) mk + sk mk eq ,
k ≥ 3 : nonconserved momenta Classical stability condition for the explicit Euler scheme : 0 < sk < 2 f j
∗ ≡
- k
(M −1)j
k mk ∗ :
particle distribution after the collision
- advection step
f j(x , t + ∆t) = f j
∗(x − vj ∆t , t)
ICMMES Conference, Hong Kong, July 28, 2005
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Taylor expansion at the order zero
f j(x , t + ∆t) ≡ f j
∗(x − vj ∆t , t)
f j(x , t + ∆t) = f j(x , t) + O(∆t) f j
∗(x − vj ∆t , t) = f j ∗(x , t) + O(∆t)
then mk
∗ =
- j
M k
j f j ∗ = mk + O(∆t)
mk
∗ − mk = O(∆t)
but mk
∗ − mk ≡ −sk (mk − mk eq)
and sk > 0 if k ≥ 3 thus mk = mk
eq + O(∆t)
mk
∗ = mk eq + O(∆t)
coming back in the space of velocity distribution : f j = f j
eq + O(∆t)
f j
∗ = f j eq + O(∆t) .
ICMMES Conference, Hong Kong, July 28, 2005
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Taylor expansion at the first order
f j(x , t + ∆t) ≡ f j
∗(x − vj ∆t , t)
f j(x , t + ∆t) = f j(x , t) + ∆t ∂tf j + O(∆t2) f j
∗(x − vj ∆t , t) = f j ∗(x , t) − ∆t vβ j ∂βf j ∗ + O(∆t2)
we take the moment of order k of this identity mk + ∆t ∂tmk + O(∆t2) = mk
∗ − ∆t
- j
M k
j vβ j ∂βf j ∗ + O(∆t2)
we use the previous Taylor expansion at the order zero mk + ∆t ∂tmk
eq + O(∆t2) = mk ∗ − ∆t
- j
M k
j vβ j ∂βf j eq + O(∆t2)
- k = 0 :
∂tρ + ∂β qβ = O(∆t) conservation of mass introduce the tensor F α β ≡
- j
vα
j vβ j f j eq
- k = α :
∂tqα + ∂β F α β = O(∆t) conservation of impulsion
ICMMES Conference, Hong Kong, July 28, 2005
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Taylor expansion at the first order
mk + ∆t ∂tmk
eq + O(∆t2) = mk ∗ − ∆t
- j
M k
j vβ j ∂βf j eq + O(∆t2)
but mk − mk
∗ ≡ sk (mk − mk eq)
if k ≥ 3 introduce θk = ∂tmk
eq +
- j
M k
j vβ j ∂βf j eq ≡
- j
M k
j ( ∂tf j eq + vβ j ∂βf j eq )
for i = 0, 1, 2, θi = O(∆t) : Euler equations of gas dynamics then for k ≥ 3 : mk = mk
eq − ∆t
sk θk + O(∆t2) mk
∗ = mk eq −
1 sk − 1
- ∆t θk + O(∆t2)
∂βf j
∗ = ∂βf j eq − ∆t
- k≥3
1 sk − 1
- (M −1)j
k ∂βθk + O(∆t2)
ICMMES Conference, Hong Kong, July 28, 2005
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Taylor expansion at the second order
f j(x , t + ∆t) ≡ f j
∗(x − vj ∆t , t)
f j(x , t + ∆t) = f j(x , t) + ∆t ∂tf j + 1
2 ∆t2 ∂2 ttf j + O(∆t3)
f j
∗(x − vj ∆t , t) = f j ∗(x , t) − ∆t vβ j ∂βf j ∗ + 1 2 ∆t2 vβ j vγ j ∂2 βγf j ∗ + O(∆t3)
we take the moment of order i (0 ≤ i ≤ 2) of this identity mi + ∆t ∂tmi + 1 2 ∆t2 ∂2
ttmi + O(∆t3) =
= mi
∗ − ∆t
- j
M i
j vβ j ∂βf j ∗ + 1
2 ∆t2
j
M i
j vβ j vγ j ∂2 βγf j ∗ + O(∆t3)
we use the microscopic conservation mi
∗ = mi
and the previous Taylor expansion at the order one : ∂tmi + 1 2 ∆t ∂2
ttmi + O(∆t2) = −
- j
M i
j vβ j ∂βf j eq +
+ ∆t
- j, k≥3
M i
j vβ j
1 sk −1
- (M −1)j
k ∂βθk + 1
2 ∆t
- j
M i
j vβ j vγ j ∂2 βγf j eq + O(∆t2)
ICMMES Conference, Hong Kong, July 28, 2005
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Taylor expansion at the second order
∂tmi +
- j
M i
j vβ j ∂βf j eq = ∆t
- j, k≥3
M i
j vβ j
1 sk − 1
- (M −1)j
k ∂βθk +
+ ∆t 2
- − ∂2
ttmi +
- j
M i
j vβ j vγ j ∂2 βγf j eq
- + O(∆t2)
- i = 0 :
conservation of mass M 0
j ≡ 1 and the first sum is null
∂2
ttρ = −∂2 tβqβ = −∂β∂tqβ = ∂2 βγ F β γ =
- j
vβ
j vγ j ∂2 βγf j eq
and the second sum is null. Then ∂tρ + ∂β qβ = O(∆t2)
- i = α :
conservation of impulsion ∂tqα +
- j
vα
j vβ j ∂βf j eq = ∆t
- k≥3
1 sk − 1
j
vα
j vβ j (M −1)j k
- ∂βθk +
+ ∆t 2
- − ∂2
ttqα +
- j
vα
j vβ j vγ j ∂2 βγf j eq
- + O(∆t2)
ICMMES Conference, Hong Kong, July 28, 2005
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From Taylor to Chapman-Enskog
−∂2
ttqα +
- j
vα
j vβ j vγ j ∂2 βγf j eq = ∂t∂βF α β +
- j
vα
j vβ j vγ j ∂2 βγf j eq
= ∂β
- j
vα
j vβ j
- ∂tf j
eq + vγ j ∂γf j eq
- = ∂β
- j
vα
j vβ j
- k
(M −1)j
k θk
= ∂β
- k≥3
j
vα
j vβ j (M −1)j k
- θk + O(∆t)
introduce the tensor Λα β
k
≡
- j
vα
j vβ j (M −1)j k
∂tqα + ∂βF α β = ∆t
- k≥3
1 sk − 1
- Λα β
k
∂βθk + ∆t 2
- k≥3
Λα β
k
∂βθk ∂tqα + ∂βF α β − ∆t
- k≥3
1 sk − 1 2
- Λα β
k
∂βθk = O(∆t2) : Chapman-Enskog !
ICMMES Conference, Hong Kong, July 28, 2005
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Unidimensional heat equation with the D2Q9 LBE model
periodic boundary conditions for the y direction equilibrium velocity = 0 equilibrium energy = −2 density equilibrium (energy)2 = density relaxation of velocities : vx
∗ = vx + s1 (0 − vx)
κ = 1 3 1 s1 − 1 2
- senergy
= 1 s(energy)2 = 1 sfluxofenergy = 1, 2 sXX = sXY = 1, 4
ICMMES Conference, Hong Kong, July 28, 2005
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Inverse methodology for a linear thermal problem
∂u ∂t − ∂ ∂x
- κ∂u
∂x
- = 0 ,
0 ≤ x ≤ L , t > 0 ∂u ∂n(x = 0 , t) = Φ(t) , t > 0 ∂u ∂n(x = L , t) = 0 , t > 0 u(x , t = 0) = 0 , 0 ≤ x ≤ L unknown flux function Φ(t)
- bserved values ψ(xk, t) ≈ u(xk, t)
Inverse problem : does exists a function
- ψ(xk, •)
- k −
→ Φ(•) ? “cost” error functional : J(Φ) = 1 2
- kobserved
T
- t=0
| u(xk, t) − ψ(xk, t) |2
ICMMES Conference, Hong Kong, July 28, 2005
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Inverse methodology for a linear thermal problem
use the linearity of the problem ! Elementary discrete solution : θ(j ∆x , n ∆t) ≡ θj(n ∆t) ∂θ ∂t − ∂ ∂x
- κ ∂θ
∂x
- = 0 ,
0 ≤ x ≤ L , t > 0 ∂θ ∂n(x = 0 , t = 1) = 1 , 0 < t < ∆t ∂θ ∂n(x = 0 , t) = 0 , t > ∆t ∂θ ∂n(x = L , t) = 0 , t > 0 θ(x , t = 0) = 0 , 0 ≤ x ≤ L then Φ(t) =
N
- n=1
ϕn δ(t − n ∆t) and u(j ∆x , t) =
N
- n=1
ϕn θj(t − n ∆t) .
ICMMES Conference, Hong Kong, July 28, 2005
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Inverse methodology for a linear thermal problem
Data for Bouzidi experiment ∆x = 1 ∆t = 1 0 ≤ x ≤ L = 110 0 ≤ t ≤ T = 500 Φ(t) ≡ 0 if t ≥ N = 70
- bservation points : k = 10, 20, 30, 40, 50
ICMMES Conference, Hong Kong, July 28, 2005
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Inverse methodology for a linear thermal problem
0.1 0.2 0.3 0.4 0.5 0.6 0.7 20 40 60 80 100 grid number initial time time=50 time=100 time=150 time=200 time=250 time=300 time=350 time=400 time=450
Elementary temperature θj(n ∆t) for a unity flux
ICMMES Conference, Hong Kong, July 28, 2005
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Inverse methodology for a linear thermal problem
0.02 0.04 0.06 0.08 0.1 2 4 6 8 10 12 14 grid number (zoom near the left boundary) initial time time=50 time=100 time=150 time=200 time=250 time=300 time=350 time=400 time=450
Elementary temperature θj(n ∆t) for a unity flux
ICMMES Conference, Hong Kong, July 28, 2005
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Inverse methodology for a linear thermal problem
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 2 4 6 8 10 12 grid number (zoom near the left boundary) initial time time=2 time=3 time=4 time=5 time=6 time=7 time=8 time=9 time=10
Elementary temperature θj(n ∆t) for a unity flux ; first times
ICMMES Conference, Hong Kong, July 28, 2005
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Inverse methodology for a linear thermal problem
0.02 0.04 0.06 0.08 0.1 4 6 8 10 12 grid number (zoom near the left boundary) initial time time=2 time=3 time=4 time=5 time=6 time=7 time=8 time=9 time=10
the elementary temperature is always positive
ICMMES Conference, Hong Kong, July 28, 2005
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Inverse methodology for a linear thermal problem
- 0.002
0.002 0.004 0.006 0.008 0.01 0.012 0.014 4 6 8 10 12 14 grid number (zoom near the left boundary) initial time time=2 time=3 time=4 time=5 time=6 time=7 time=8 time=9 time=10
even after a big zoom !
ICMMES Conference, Hong Kong, July 28, 2005
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Inverse methodology for a linear thermal problem
5 10 15 20 25 30 35 40 20 40 60 80 100 grid number initial time time=50 time=100 time=150 time=200 time=250 time=300 time=350 time=400 time=450
“measured” temperature for a complex flux Φ(t)
ICMMES Conference, Hong Kong, July 28, 2005
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Inverse methodology for a linear thermal problem
5 10 15 20 1 2 3 4 5 6 7 8 grid number (zoom near the left boundary) initial time time=2 time=3 time=4 time=5 time=6 time=7 time=8 time=9 time=10
“measured” temperature for a complex flux Φ(t)
ICMMES Conference, Hong Kong, July 28, 2005
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Inverse methodology for a linear thermal problem
u(j ∆x , t) =
N
- n=1
ϕn θj(t − n ∆t) J(Φ) = 1 2
- k
T
- t=0
| u(xk, t) − ψ(xk, t) |2 J(Φ) = 1 2
- k
T
- t=0
|
N
- n=1
ϕn θj(t − n ∆t) − ψ(xk, t) |2 minimize the error functional J(•) relatively to the function Φ(•) δJ =
- k
T
- t=0
- N
- n=1
ϕn θj(t − n ∆t) − ψ(xk, t)
N
- m=1
δϕm θj(t − m ∆t)
- ∂J
∂ϕm =
N
- n=1
k T
- t=0
θj(t−n ∆t) θj(t−m ∆t)
- ϕn −
- k,t
ψ(xk, t) θj(t−m ∆t)
ICMMES Conference, Hong Kong, July 28, 2005
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Inverse methodology for a linear thermal problem
matrix Am n =
- k
T
- t=0
θj(t − n ∆t) θj(t − m ∆t) right hand side bn =
- k
T
- t=0
ψ(xk, t) θj(t − m ∆t) solve the linear system of order N = 70 : A • ϕ = b . What sensibility relatively to the errors of measure ?
ICMMES Conference, Hong Kong, July 28, 2005
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Inverse methodology for a linear thermal problem
5 10 15 20 100 200 300 400 500 time heat flux at the left boundary
Computed heat flux at the left boundary
ICMMES Conference, Hong Kong, July 28, 2005
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Inverse methodology for a linear thermal problem
5 10 15 20 10 20 30 40 50 60 70 time heat flux at the left boundary
Computed heat flux at the left boundary
ICMMES Conference, Hong Kong, July 28, 2005
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Inverse methodology for a linear thermal problem
Lallemand experiment : periodic boundary conditions ∂u ∂t − ∂ ∂x
- κ∂u
∂x
- = 0 ,
0 ≤ x ≤ L , t > 0 u(x = 0 , t) = u(x = L , t) t > 0 u(x = 0 , t) = 1 , 0 < t < ∆t u(x = 0 , t) = 0 , t > ∆t Two methods : LBE / Heun finite difference method ∂2u ∂x2 = five points fourth order accurate scheme Heun scheme for the temporal evolution :
- u = u(n ∆t) + ∆t ∂u
∂t
n
u((n + 1) ∆t) = u(n ∆t) + 1 2 ∂u ∂t
n
+
- ∂u
∂t
- ICMMES Conference, Hong Kong, July 28, 2005
SLIDE 28
Inverse methodology for a linear thermal problem
0.01 0.02 0.03 0.04 0.05 50 100 150 200 250 300 350 400 abscissa time=10 time=50 time=100 time=500
Evolution of the “elementary” temperature
ICMMES Conference, Hong Kong, July 28, 2005
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Inverse methodology for a linear thermal problem
0.01 0.02 0.03 0.04 0.05 10 20 30 40 50 abscissa (zoom near the left boundary) time=10 time=50 time=100 time=500 Gaussian reference
Evolution of the “elementary” temperature
ICMMES Conference, Hong Kong, July 28, 2005
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Inverse methodology for a linear thermal problem
correlation matrix for observation at the position k ∆x
- A(k)
- m n =
T
- t=0
θj(t − n ∆t) θj(t − m ∆t)
- rthogonal decomposition :
A = U • S • V t U and V orthogonal matrices S diagonal matrices condition number = min
j
Sj max
j
Sj
ICMMES Conference, Hong Kong, July 28, 2005
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Inverse methodology for a linear thermal problem
- 18
- 16
- 14
- 12
- 10
- 8
- 6
- 4
- 2
5 10 15 20 abscissa of the observation point kappa = 1 kappa = 2 kappa = 3 kappa = 4 kappa = 5
Logarithm of the condition number ; Boltzmann scheme
ICMMES Conference, Hong Kong, July 28, 2005
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Inverse methodology for a linear thermal problem
- 18
- 16
- 14
- 12
- 10
- 8
- 6
- 4
- 2
5 10 15 20 abscissa of the observation point kappa = 1 kappa = 2 kappa = 3 kappa = 4 kappa = 5
Logarithm of the condition number ; Heun scheme
ICMMES Conference, Hong Kong, July 28, 2005
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Adjoint Lattice Boltzmann Equation
- LBE discrete dynamics
f j(x, (n + 1) ∆t) = f j
∗(x − vj ∆t , n ∆t) ≡ (A • C(f, σ))(x , n ∆t)
the observation : ψ(xk, t) is supposed to be a simple function ϕ(•) of the state f(xk, t) : ϕ(f(xk, t)) ≈ ψ(xk, t)
- introduce the cost function
J(σ) ≡ 1 2
- k
t=T −∆t
- t=0
| ϕ(f(xk, t)) − ψ(xk, t) |2
- treat the LBE dynamics as a constraint (Pontryaguine, J.L. Lions) :
introduce the associated Lagrangian L ≡ J(σ) +
- x
t=T −∆t
- t=0
- p(x, t+∆t) , f(x, t+∆t) − (A • C(f, σ))(x , t)
- ICMMES Conference, Hong Kong, July 28, 2005
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Adjoint Lattice Boltzmann Equation
δL = δ 1 2
- k
t=T −∆t
- t=0
| ϕ(f(xk, t)) − ψ(xk, t) |2
- +
+
- x
t=T −∆t
- t=0
δ
- p(x, t + ∆t) , f(x, t + ∆t) − (A • C(f, σ))(x , t)
=
- x
k t=T −∆t
- t=0
δx = xk
- ϕ(f(xk, t)) − ψ(xk, t)
∂ϕ ∂f δf(x, t) +
t=T −∆t
- t=0
- δp(x, t + ∆t) , f(x, t + ∆t) − (A • C(f, σ))(x , t)
- +
t=T −∆t
- t=0
- p(x, t + ∆t) , δf(x, t + ∆t) − (A • dCt(f, σ))• δf(x , t)
- −
t=T −∆t
- t=0
- p(x, t + ∆t) , A • ∂C
∂σ (f, σ)
- δσ
- ICMMES Conference, Hong Kong, July 28, 2005
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Abel transform : integrate by parts the discrete sum
t=T −∆t
- t=0
- p(x, t + ∆t) , δf(x, t + ∆t) − (A • dCt(f, σ))• δf(x , t)
- =
=
t=T
- t=∆t
- p(x, t) , δf(x, t)
- −
t=T −∆t
- t=0
- p(x, t+∆t) , (A • dCt(f, σ))• δf(x , t)
- =
- p(x, T) , δf(x, T)
- +
t=T −∆t
- t=∆t
- p(x, t) , δf(x, t)
- −
t=T −∆t
- t=∆t
- dCt• At • p(x, t+∆t) , δf(x , t)
- −
- dCt• At • p(x, ∆t) , δf(x , 0)
- =
- p(x, T) , δf(x, T)
- −
- dCt• At • p(x, ∆t) , δf(x , 0)
- +
t=T −∆t
- t=∆t
- p(x, t) − dCt• At • p(x, t + ∆t) , δf(x, t)
- ICMMES Conference, Hong Kong, July 28, 2005
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Variation of the Lagrangian
δL =
- x
- −
- dCt• At • p(x, ∆t) , δf(x , 0)
- +
- p(x, T) , δf(x, T)
- +
t=T −∆t
- t=∆t
- p(x, t) − dCt• At • p(x, t + ∆t) +
+
- k
δx = xk
- ϕ(f(xk, t)) − ψ(xk, t)
- ,
δf(x, t)
- +
t=T −∆t
- t=0
- δp(x, t + ∆t) , f(x, t + ∆t) − (A • C(f, σ))(x , t)
- −
t=T −∆t
- t=0
- p(x, t + ∆t) , A • ∂C
∂σ (f, σ)
- δσ
- and if the discrete dynamics LBE is satisfied,
δL ≡ δJ !!
ICMMES Conference, Hong Kong, July 28, 2005
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Adjoint Lattice Boltzmann Equation
Lagrange multiplyer p(x, t) “adjoint state” integrate by parts the evaluation of δL eliminate the terms in δf in order to impose δL ≡ δJ then obtain a “backward linearized equation” for multiplyer p(x, t) also called “Adjoint Lattice Boltzmann Equation” : p(x, t) = dCt• At • p(x, t + ∆t) −
- k
δx = xk (ϕ(f(xk, t)) − ψ(xk, t)) ∂ϕ ∂f final condition p(x, T) = 0 gradient of the cost function relatively to to the parameter σ : ∂J ∂σ = −
t=T −∆t
- t=0
- p(x, t + ∆t) , A • ∂C
∂σ (f, σ)
- ICMMES Conference, Hong Kong, July 28, 2005
SLIDE 38
Gradient algorithm for optimization
minimize J(•) relatively to the parameter σ initialization : fix σ = σ0 (1) compute ∂J ∂σ :
- simulate the LBE process with a given σ and obtain f(x, t ; σ)
- simulate the ALBE process with a given f(•, • ; σ) :
final condition p(x, T) = 0 p(x, t) = dCt• At • p(x, t+∆t) −
- k
δx = xk (ϕ(f(xk, t)) − ψ(xk, t)) ∂ϕ ∂f and obtain p(x, t ; σ)
- ∂J
∂σ = −
t=T −∆t
- t=0
- p(x, t + ∆t ; σ) , A • ∂C
∂σ (f(x, t ; σ) , σ)
- evaluate a new value of σ :
σnew = σ − ρ ∂J ∂σ , ρ > 0 . go to (1)
ICMMES Conference, Hong Kong, July 28, 2005
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ALBE : first results proposed by Tekitek
−14 −10 −7 −4 −1 −4.87299 −4.8729 −4.8728 −4.8727 −4.8726 −4.87256 −4.87299 −4.8729 −4.8728 −4.8727 −4.8726 −4.87256
Numerical evaluation of the cost function by finite differences
ICMMES Conference, Hong Kong, July 28, 2005
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ALBE : first results proposed by Tekitek
0.1686 1 1.429 −1.9.10−6 −1.10−6 0.0 1.10−6 2.10−6 3.10−6 4.10−6 5.10−6 6.10−6 6.7.10−6 0.1686 1 1.429 −1.9.10−6 −1.10−6 1.10−6 2.10−6 3.10−6 4.10−6 5.10−6 6.10−6 6.7.10−6
Gradient of the cost function computed with finite differences (×) and with the adjoint method (+)
ICMMES Conference, Hong Kong, July 28, 2005
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ALBE : first results proposed by Tekitek
10 20 30 40 −5.0 −4.0 −3.0 −2.0 −1.0 0.0 −5.0 −4.0 −3.0 −2.0 −1.0 0.0
Logarithm of the error vs iteration of the gradient method
ICMMES Conference, Hong Kong, July 28, 2005
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ALBE : first results proposed by Tekitek
10 20 30 −15 −14 −13 −12 −11 −10 −9 −8 −7 −6 −5 −4 −3 −2 −1 −15 −14 −13 −12 −11 −10 −9 −8 −7 −6 −5 −4 −3 −2 −1
Logarithm of the s5 error (top), of the s8 error (middle) and of the cost function (bottom) vs gradient iteration
ICMMES Conference, Hong Kong, July 28, 2005
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ALBE : first results proposed by Tekitek
1 100 200 300 −10 −9 −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 1 100 200 300 −10 −9 −8 −7 −6 −5 −4 −3 −2 −1 1 2 3
Nonlinear problem : logarithm of the gradient and of the cost function vs gradient iteration
ICMMES Conference, Hong Kong, July 28, 2005
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