Optimal Control of Two-Phase Flow Harald Garcke, Michael Hinze, Christian Kahle RICAM special semester on Optimization WS1: New trends in PDE constrained optimization 14.10. – 18.10.2019 Christian Kahle Optimal Control of Two-Phase Flow 10/2019 1/32
Optimal control of two-phase flow Figure: without control Christian Kahle Optimal Control of Two-Phase Flow 10/2019 2/32
Optimal control of two-phase flow Figure: without control Figure: with control Christian Kahle Optimal Control of Two-Phase Flow 10/2019 2/32
Outline Setting The time discrete setting The fully discrete setting Numerical examples Christian Kahle Optimal Control of Two-Phase Flow 10/2019 3/32
Outline Setting The time discrete setting The fully discrete setting Numerical examples Christian Kahle Optimal Control of Two-Phase Flow 10/2019 3/32
Diffuse interface approach Setting: Two subdomains Ω 1 and Ω 2 separated by unknown Γ ǫ . Assumption: Γ ǫ of small thickness O( ǫ ) > 0 and components are mixed inside. Representation: Continuous order parameter ϕ for Ω 1 and Ω 2 . Christian Kahle Optimal Control of Two-Phase Flow 10/2019 4/32
Diffuse interface approach Setting: Two subdomains Ω 1 and Ω 2 separated by unknown Γ ǫ . Assumption: Γ ǫ of small thickness O( ǫ ) > 0 and components are mixed inside. Representation: Continuous order parameter ϕ for Ω 1 and Ω 2 . ϕ Ω 2 1 − 1 Ω 1 Γ ǫ ϕ ( x ) = 1 ⇔ x ∈ Ω 1 ϕ ( x ) = − 1 ⇔ x ∈ Ω 2 − 1 < ϕ ( x ) < 1 ⇔ x ∈ Γ ǫ Christian Kahle Optimal Control of Two-Phase Flow 10/2019 4/32
Diffuse interface approach Setting: Two subdomains Ω 1 and Ω 2 separated by unknown Γ ǫ . Assumption: Γ ǫ of small thickness O( ǫ ) > 0 and components are mixed inside. Representation: Continuous order parameter ϕ for Ω 1 and Ω 2 . ρ 2 ( x ) ϕ 1 Ω 2 ρ 2 ρ 1 ( x ) 1 ˜ ˜ ρ 1 − 1 0 Ω 1 Γ ǫ ϕ − 1 Ω 2 Ω 1 Γ ǫ , O( ǫ ) ϕ ( x ) = 1 ⇔ x ∈ Ω 1 ϕ ( x ) = ρ 1 ( x ) − ρ 2 ( x ) ϕ ( x ) = − 1 ⇔ x ∈ Ω 2 ρ 1 ˜ ρ 2 ˜ − 1 < ϕ ( x ) < 1 ⇔ x ∈ Γ ǫ Christian Kahle Optimal Control of Two-Phase Flow 10/2019 4/32
The two-phase flow model [Abels, Garcke, Grün, 2012] v velocity, p pressure, ϕ phase field variable, µ chemical potential ρ∂ t v + (( ρv + J ) ⋅ ∇ ) v − div ( 2 ηDv ) + ∇ p = − ϕ ∇ µ + ρg, div v = 0 , ∂ t ϕ + v ⋅ ∇ ϕ − div ( m ∇ µ ) = 0 , − σǫ∆ϕ + σǫ − 1 W ′ ( ϕ ) = µ, where 2 Dv = ∇ v + ( ∇ v ) t , J = − ρ ′ ( ϕ ) m ( ϕ ) ∇ µ . ρ ( ϕ ) density, g gravity, 1 η ( ϕ ) viscosity, W s ǫ interfacial width, m ( ϕ ) mobility. σ surface tension, − 1 σ = c W σ phys , ϕ 1 Christian Kahle Optimal Control of Two-Phase Flow 10/2019 5/32
The free energy density W logarithmic: W log ( ϕ ) = θ 2 (( 1 + ϕ ) log ( 1 + ϕ ) + ( 1 − ϕ ) log ( 1 − ϕ )) + θ ϕ 2 ( 1 − ϕ 2 ) , polynomial: W poly ( ϕ ) = 1 4 ( 1 − ϕ 2 ) 2 , double-obstacle: W ∞ ( ϕ ) = 1 2 ( 1 − ϕ 2 ) iff ∣ ϕ ∣ ≤ 1 , ∞ else, relaxed double-obstacle: 2 ( max ( 0 ,ξϕ − 1 ) 2 + min ( 0 ,ξϕ + 1 ) 2 ) + θ . W s ( ϕ ) = 1 2 ( 1 − ( ξϕ ) 2 ) + s ∞ 1 1 1 1 W ∞ W s W log W poly − 1 − 1 − 1 − 1 ϕ ϕ ϕ ϕ 1 1 1 1 Christian Kahle Optimal Control of Two-Phase Flow 10/2019 6/32
Functions depending on ϕ ρ ( ϕ ) = η ( ϕ ) = W ( ϕ ) ϕ + ρ 2 − ρ 1 + η 2 − η 1 ρ 1 + ρ 2 η 1 + η 2 ϕ ϕ 2 2 2 2 Christian Kahle Optimal Control of Two-Phase Flow 10/2019 7/32
The formal energy inequality Theorem Let v,ϕ,µ denote a sufficiently smooth solution (if exists) and let 2 ∣ ∇ ϕ ( t )∣ 2 + 1 E ( t ) = ∫ Ω 2 ρ ( t )∣ v ( t )∣ 2 dx + σ ∫ Ω ǫ W ( ϕ ( t )) dx 1 ǫ denote the energy of the system. Let v ∣ ∂Ω = 0 hold. Then it holds dt E ( t ) = −∫ Ω 2 η ( ϕ )∣ Dv ∣ 2 dx −∫ Ω m ( ϕ )∣ ∇ µ ∣ 2 dx +∫ Ω gv dx , d E ( t 2 ) + ∫ ∫ Ω m ( ϕ ( s ))∣ ∇ µ ( s )∣ 2 dxds +∫ ∫ Ω 2 η ( ϕ ( s ))∣ Dv ( s )∣ 2 dxds t 2 t 2 t 1 t 1 = E ( t 1 ) + ∫ ∫ Ω gv ( s ) dxds t 2 t 1 Christian Kahle Optimal Control of Two-Phase Flow 10/2019 8/32
Applied Controls B u V = B u B = ϕ 0 = B u I = u I ∑ s V i = 1 f i ( x ) u V [ i ] , ∑ s B i = 1 g i ( x ) u B [ i ] , f i ∈ L 2 ( Ω ) n g i ∈ H 1 / 2 ( ∂Ω ) n u V ∈ L 2 ( 0 ,T ; R s V ) = U V , u B ∈ L 2 ( 0 ,T ; R s B ) = U B , u I ∈ K ∶= { v ∈ H 1 ( Ω ) ∩ L ∞ ( Ω )∣∣ v ∣ ≤ 1 , ( v, 1 ) = const } = U I , u = ( u V ,u B ,u I ) ∈ U = U V × U B × U I . Christian Kahle Optimal Control of Two-Phase Flow 10/2019 9/32
The two-phase flow model with controls v velocity, p pressure, ϕ phase field variable, µ chemical potential ρ∂ t v + (( ρv + J ) ⋅ ∇ ) v − div ( 2 ηDv ) + ∇ p = − ϕ ∇ µ + ρg + B u V , div v = 0 , ∂ t ϕ + v ⋅ ∇ ϕ − div ( m ∇ µ ) = 0 , − σǫ∆ϕ + σǫ − 1 W ′ ( ϕ ) = µ, where 2 Dv = ∇ v + ( ∇ v ) t , J = − ρ ′ ( ϕ ) m ( ϕ ) ∇ µ , v ∣ ∂Ω = B u B , ϕ ( 0 ) = u I . ρ ( ϕ ) density, g gravity, 1 η ( ϕ ) viscosity, W s ǫ interfacial width, m ( ϕ ) mobility. σ surface tension, − 1 σ = c W σ phys , ϕ 1 Christian Kahle Optimal Control of Two-Phase Flow 10/2019 10/32
The optimal control problem The optimal control problem ϕ d : desired distribution, α V + α B + α I = 1 min J ( u I ,u V ,u B ,ϕ ) ∶ = 1 2 ∥ ϕ ( T ) − ϕ d ∥ 2 2 ∣ ∇ u I ∣ 2 + ǫ − 1 W u ( u I ) dx 2 ( α I ∫ Ω + α ǫ ( P ) α V ∥ u V ∥ 2 L 2 ( 0 ,T ; R sV ) + α B ∥ u B ∥ 2 L 2 ( 0 ,T ; R sB ) ) s.t. two-phase fluid dynamics, i.e. ϕ ≡ ϕ ( u V ,u B ,u I ) Christian Kahle Optimal Control of Two-Phase Flow 10/2019 11/32
Outline Setting The time discrete setting The fully discrete setting Numerical examples Christian Kahle Optimal Control of Two-Phase Flow 10/2019 11/32
A weak formulation Abbreviate a ( u,v,w ) ∶= 1 2 (( u ⋅ ∇ ) v,w ) − 1 2 (( u ⋅ ∇ ) w,v ) The model satisfies ∂ t ρ ( ϕ ) + div ( ρ ( ϕ ) v + J ) = −∇ µ ⋅ ∇ ρ ′ ( ϕ ) If ρ ( ϕ ) is linear (mass conservation) ρ∂ t v + (( ρv + J ) ⋅ ∇ ) v − div ( 2 ηDv ) = µ ∇ ϕ, ∂ t ( ρv ) + div ( ρv ⊗ v ) + div ( v ⊗ J ) − div ( 2 ηDv ) = µ ∇ ϕ. Then a weak formulation is 2 ( ρ∂ t v + ∂ t ( ρv ) ,w ) + a ( ρv + J,v,w ) + 2 ( ηDv,Dw ) = ( µ ∇ ϕ,w ) ∀ w ∈ H σ 1 Christian Kahle Optimal Control of Two-Phase Flow 10/2019 12/32
An energy stable time discretization [Garcke, Hinze, K. 2016] B , ϕ 0 = u I ⋆ ∶= 1 t k − 1 u ⋆ ( t ) dt , v k ∣ ∂Ω = B u k τ ∫ t k u k τ ∫ Ω ( ρ k − 1 + ρ k − 2 v k − ρ k − 2 v k − 1 ) w dx 1 2 + a ( ρ k − 1 v k − 1 + J k − 1 ,v k ,w ) + ∫ Ω 2 η k − 1 Dv k ∶ Dw dx +∫ Ω ϕ k − 1 ∇ µ k ⋅ w dx −∫ Ω ρ k − 1 g ⋅ w dx −∫ Ω B u k V w dx = 0 ∀ w ∈ H σ ( Ω ) , τ ∫ Ω ( ϕ k − ϕ k − 1 ) Ψ dx −∫ Ω ϕ k − 1 v k ⋅ ∇ Ψ dx 1 +∫ Ω m ∇ µ k ⋅ ∇ Ψ dx = 0 ∀ Ψ ∈ H 1 ( Ω ) , σǫ ∫ Ω ∇ ϕ k ⋅ ∇ Φ dx −∫ Ω µ k Φ dx ǫ ∫ Ω (( W + ) ′ ( ϕ k ) + ( W − ) ′ ( ϕ k − 1 )) Φ dx = 0 ∀ Φ ∈ H 1 ( Ω ) . + σ (CHNS τ ) Christian Kahle Optimal Control of Two-Phase Flow 10/2019 13/32
Energy inequality Theorem Let k ≥ 2 , ϕ k ,µ k ,v k be a solution to (CHNS τ ) , and u B ≡ 0 . Then the following energy inequality holds 2 ∫ Ω ρ k − 1 ∣ v k ∣ 2 ∣ ∇ ϕ k ∣ 2 + 1 ǫ W ( ϕ k ) dx 1 2 dx + σ ∫ Ω ǫ 2 ∫ Ω ρ k − 2 ∣ v k − v k − 1 ∣ 2 dx + σǫ 2 ∫ Ω ∣ ∇ ϕ k − ∇ ϕ k − 1 ∣ 2 dx + 1 + τ ∫ Ω 2 η k − 1 ∣ Dv k ∣ 2 dx + τ ∫ Ω m ∣ ∇ µ k ∣ 2 dx 2 ∫ Ω ρ k − 2 ∣ v k − 1 ∣ ≤ 1 2 ∣ ∇ ϕ k − 1 ∣ 2 + 1 ǫ W ( ϕ k − 1 ) dx 2 dx + σ ∫ Ω ǫ + ∫ Ω ρ k − 1 gv k dx + ∫ Ω ( B u k V ) v k dx Christian Kahle Optimal Control of Two-Phase Flow 10/2019 14/32
Existence of a unique solution Theorem Let Ω denote a polygonally / polyhedrally bounded Lipschitz domain. Let v k − 1 ∈ H σ ( Ω ) , ϕ k − 2 ∈ H 1 ( Ω ) ∩ L ∞ ( Ω ) , ϕ k − 1 ∈ H 1 ( Ω ) ∩ L ∞ ( Ω ) , and µ k − 1 ∈ W 1 , 3 ( Ω ) be given data. Further let B u k V ∈ L 2 ( Ω ) n , 2 ( ∂Ω ) , B u I ∈ H 1 ( Ω ) ∩ L ∞ ( Ω ) be given data. B ∈ H 1 B u k Then there exists a weak solution ϕ k ∈ H 1 ( Ω ) ∩ C ( Ω ) , µ k ∈ W 1 , 3 ( Ω ) , v k ∈ H σ ( Ω ) to (CHNS τ ) . Furthermore, it can be found by Newton’s method. Christian Kahle Optimal Control of Two-Phase Flow 10/2019 15/32
Initialization step B , ϕ 0 = u I For k = 1 we solve: v 1 ∣ ∂Ω = B u 1 τ ∫ Ω ( ρ 1 + ρ 0 v 1 − ρ 0 v 0 ) w dx + a ( ρ 1 v 0 + J 1 ,v 1 ,w ) 1 2 V w dx − ∫ Ω ρ 0 g ⋅ w = 0 ∀ w ∈ H σ ( + ∫ Ω 2 η 1 Dv 1 ∶ Dw dx − ∫ Ω µ 1 ∇ ϕ 0 w dx − ∫ Ω B u 1 τ ∫ Ω ( ϕ 1 − ϕ 0 ) Ψ dx −∫ Ω ϕ 0 v 0 ⋅ ∇ Ψ dx 1 +∫ Ω m ∇ µ 1 ⋅ ∇ Ψ dx = 0 ∀ Ψ ∈ H 1 ( σǫ ∫ Ω ∇ ϕ 1 ⋅ ∇ Φ dx −∫ Ω µ 1 Φ dx ǫ ∫ Ω (( W + ) ′ ( ϕ 1 ) + ( W − ) ′ ( ϕ 0 )) Φ dx = 0 ∀ Φ ∈ H 1 ( + σ (CHNS I τ ) Christian Kahle Optimal Control of Two-Phase Flow 10/2019 16/32
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