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 Mahmed Bouzidi, Pierre Lallemand, Mahdi Tekitek Edition du 27 septembre 2005.

 On Direct and Adjoint Lattice Boltzmann Equations Scope of the lecture 1) On Taylor and Chapman Enskog expansions for discrete Boltzmann dynamics 2) Inverse methodology for a linear thermal problem 3) Adjoint Lattice Boltzmann Equation ICMMES Conference, Hong Kong, July 28, 2005

 General framework x : a node of the lattice ∆ t : the (small) time step components v α v j : discrete celerity in the lattice, j , α = 1 , 2 then x + v j ∆ t is an other node of the lattice f j ( x , t ) : density of particles having the velocity v j at node x and at discrete time t . � m 0 = ρ ≡ f j : density of matter j � m α = q α ≡ j f j : component number α of the momentum v α j W ≡ ( ρ , q 1 , q 2 ) : conserved variables ICMMES Conference, Hong Kong, July 28, 2005

 General framework 2 5 6 3 1 0 7 8 4 Neighbourhood of a vertex x for the D2Q9 LBE model ICMMES Conference, Hong Kong, July 28, 2005

d’Humi`  eres (1992) representation of the discrete dynamics � m k ≡ j f j ( k ≥ 3) : other components of the momentum M k j remark that we have M 0 j = 1 and M α j = v α j m k eq and f j eq : equilibrium momenta and velocity distribution • collision step ∗ ≡ m i ≡ m i m i eq : conserved momenta ( i = 0 , 1 , 2) during the collision ∗ = (1 − s k ) m k + s k m k m k k ≥ 3 : nonconserved momenta eq , Classical stability condition for the explicit Euler scheme : 0 < s k < 2 � ( M − 1 ) j f j k m k ∗ ≡ ∗ : particle distribution after the collision k • advection step f j ( x , t + ∆ t ) = f j ∗ ( x − v j ∆ t , t ) ICMMES Conference, Hong Kong, July 28, 2005

 Taylor expansion at the order zero f j ( x , t + ∆ t ) ≡ f j ∗ ( x − v j ∆ t , t ) f j ( x , t + ∆ t ) = f j ( x , t ) + O(∆ t ) f j ∗ ( x − v j ∆ t , t ) = f j ∗ ( x , t ) + O(∆ t ) � ∗ = m k + O(∆ t ) m k M k j f j then ∗ = j ∗ − m k = O(∆ t ) m k ∗ − m k ≡ − s k ( m k − m k m k but eq ) and s k > 0 if k ≥ 3 m k = m k thus eq + O(∆ t ) m k ∗ = m k 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

 Taylor expansion at the first order f j ( x , t + ∆ t ) ≡ f j ∗ ( x − v j ∆ t , t ) = f j ( x , t ) + ∆ t ∂ t f j + O(∆ t 2 ) f j ( x , t + ∆ t ) ∗ ( x , t ) − ∆ t v β ∗ + O(∆ t 2 ) f j ∗ ( x − v j ∆ t , t ) = f j j ∂ β f j we take the moment of order k of this identity � m k + ∆ t ∂ t m k + O(∆ t 2 ) = m k j v β M k j ∂ β f j ∗ + O(∆ t 2 ) ∗ − ∆ t j we use the previous Taylor expansion at the order zero � m k + ∆ t ∂ t m k j v β eq + O(∆ t 2 ) = m k M k j ∂ β f j eq + O(∆ t 2 ) ∗ − ∆ t j ∂ t ρ + ∂ β q β = O(∆ t ) • k = 0 : conservation of mass � F α β ≡ j v β v α j f j introduce the tensor eq j ∂ t q α + ∂ β F α β = O(∆ t ) • k = α : conservation of impulsion ICMMES Conference, Hong Kong, July 28, 2005

 Taylor expansion at the first order � m k + ∆ t ∂ t m k j v β eq + O(∆ t 2 ) = m k M k j ∂ β f j eq + O(∆ t 2 ) ∗ − ∆ t j m k − m k ∗ ≡ s k ( m k − m k but eq ) if k ≥ 3 � � introduce θ k = ∂ t m k j v β eq + v β M k j ∂ β f j M k j ( ∂ t f j j ∂ β f j eq + eq ≡ eq ) j j θ i = O(∆ t ) : for i = 0 , 1 , 2 , Euler equations of gas dynamics eq − ∆ t m k = m k θ k + O(∆ t 2 ) then for k ≥ 3 : s k � 1 � ∆ t θ k + O(∆ t 2 ) m k ∗ = m k eq − − 1 s k � 1 � � k ∂ β θ k + O(∆ t 2 ) ( M − 1 ) j ∂ β f j ∗ = ∂ β f j eq − ∆ t − 1 s k k ≥ 3 ICMMES Conference, Hong Kong, July 28, 2005

 Taylor expansion at the second order f j ( x , t + ∆ t ) ≡ f j ∗ ( x − v j ∆ t , t ) = f j ( x , t ) + ∆ t ∂ t f j + 1 2 ∆ t 2 ∂ 2 tt f j + O(∆ t 3 ) f j ( x , t + ∆ t ) 2 ∆ t 2 v β f j ∗ ( x − v j ∆ t , t ) = f j ∗ ( x , t ) − ∆ t v β j ∂ β f j j v γ βγ f j ∗ + 1 j ∂ 2 ∗ + O(∆ t 3 ) we take the moment of order i (0 ≤ i ≤ 2) of this identity m i + ∆ t ∂ t m i + 1 tt m i + O(∆ t 3 ) = 2 ∆ t 2 ∂ 2 � 2 ∆ t 2 � ∗ + 1 j v β j v β j v γ j ∂ 2 ∗ + O(∆ t 3 ) = m i M i j ∂ β f j M i βγ f j ∗ − ∆ t j j we use the microscopic conservation m i ∗ = m i and the previous Taylor expansion at the order one : � ∂ t m i + 1 tt m i + O(∆ t 2 ) = − 2 ∆ t ∂ 2 j v β M i j ∂ β f j eq + j � 1 � � � k ∂ β θ k + 1 j v β ( M − 1 ) j j v β j v γ M i M i j ∂ 2 βγ f j eq + O(∆ t 2 ) + ∆ t − 1 2 ∆ t j s k j, k ≥ 3 j ICMMES Conference, Hong Kong, July 28, 2005

 Taylor expansion at the second order � 1 � � � ∂ t m i + k ∂ β θ k + j v β j v β ( M − 1 ) j M i j ∂ β f j M i eq = ∆ t − 1 j s k j j, k ≥ 3 � � � + ∆ t tt m i + j v β j v γ − ∂ 2 j ∂ 2 + O(∆ t 2 ) M i βγ f j eq 2 j • i = 0 : conservation of mass M 0 j ≡ 1 and the first sum is null � tβ q β = − ∂ β ∂ t q β = ∂ 2 βγ F β γ = ∂ 2 tt ρ = − ∂ 2 v β j v γ j ∂ 2 βγ f j eq j ∂ t ρ + ∂ β q β = O(∆ t 2 ) and the second sum is null. Then • i = α : conservation of impulsion � 1 � � � � � � ∂ t q α + ∂ β θ k + j v β j v β j ( M − 1 ) j v α j ∂ β f j v α eq = ∆ t − 1 k s k j k ≥ 3 j � � � + ∆ t tt q α + j v β j v γ − ∂ 2 j ∂ 2 + O(∆ t 2 ) v α βγ f j eq 2 j ICMMES Conference, Hong Kong, July 28, 2005

 From Taylor to Chapman-Enskog � � tt q α + eq = ∂ t ∂ β F α β + j v β j v γ j v β j v γ − ∂ 2 j ∂ 2 j ∂ 2 v α βγ f j v α βγ f j eq j j � � � j v β eq + v γ v α ∂ t f j j ∂ γ f j = ∂ β eq j j � � j v β ( M − 1 ) j v α k θ k = ∂ β j j k � � � � θ k + O(∆ t ) j v β j ( M − 1 ) j v α = ∂ β k k ≥ 3 j � Λ α β j v β j ( M − 1 ) j v α introduce the tensor ≡ k k j � 1 � � � ∂ β θ k + ∆ t ∂ t q α + ∂ β F α β = ∆ t Λ α β Λ α β ∂ β θ k − 1 k k s k 2 k ≥ 3 k ≥ 3 � 1 � � − 1 ∂ t q α + ∂ β F α β − ∆ t ∂ β θ k = O(∆ t 2 ) : Chapman-Enskog ! Λ α β k 2 s k k ≥ 3 ICMMES Conference, Hong Kong, July 28, 2005

 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 ∗ = v x + s 1 (0 − v x ) v x relaxation of velocities : � 1 � κ = 1 − 1 3 s 1 2 s energy = 1 s (energy) 2 = 1 s fluxofenergy = 1 , 2 s XX = s XY = 1 , 4 ICMMES Conference, Hong Kong, July 28, 2005

 Inverse methodology for a linear thermal problem � � ∂u ∂t − ∂ κ∂u = 0 , 0 ≤ x ≤ L , t > 0 ∂x ∂x ∂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 ) observed values ψ ( x k , t ) ≈ u ( x k , t ) � � Inverse problem : does exists a function ψ ( x k , • ) k �− → Φ( • ) ? T � � J (Φ) = 1 | u ( x k , t ) − ψ ( x k , t ) | 2 “cost” error functional : 2 t =0 k observed ICMMES Conference, Hong Kong, July 28, 2005

 Inverse methodology for a linear thermal problem use the linearity of the problem ! Elementary discrete solution : θ ( j ∆ x , n ∆ t ) ≡ θ j ( n ∆ t ) � � ∂θ ∂t − ∂ κ ∂θ = 0 , 0 ≤ x ≤ L , t > 0 ∂x ∂x ∂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 N N � � Φ( t ) = ϕ n δ ( t − n ∆ t ) and u ( j ∆ x , t ) = ϕ n θ j ( t − n ∆ t ) . n =1 n =1 ICMMES Conference, Hong Kong, July 28, 2005

 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 observation points : k = 10 , 20 , 30 , 40 , 50 ICMMES Conference, Hong Kong, July 28, 2005

 Inverse methodology for a linear thermal problem initial time time=50 0.7 time=100 time=150 time=200 0.6 time=250 time=300 time=350 time=400 0.5 time=450 0.4 0.3 0.2 0.1 0 0 20 40 60 80 100 grid number Elementary temperature θ j ( n ∆ t ) for a unity flux ICMMES Conference, Hong Kong, July 28, 2005

 Inverse methodology for a linear thermal problem initial time time=50 time=100 0.1 time=150 time=200 time=250 time=300 time=350 0.08 time=400 time=450 0.06 0.04 0.02 0 2 4 6 8 10 12 14 grid number (zoom near the left boundary) Elementary temperature θ j ( n ∆ t ) for a unity flux ICMMES Conference, Hong Kong, July 28, 2005