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 Harald Garcke, Michael Hinze, - - PowerPoint PPT Presentation
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
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2 ((1 + ϕ)log(1 + ϕ) + (1 − ϕ)log(1 − ϕ)) + θϕ 2 (1 − ϕ2),
4 (1 − ϕ2) 2,
2 (1 − ϕ2)iff ∣ϕ∣ ≤ 1,
2 (1 − (ξϕ)2) + s 2 (max(0,ξϕ − 1)2 + min(0,ξϕ + 1)2) + θ.
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ρ1+ρ2 2
2
η1+η2 2
2
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t2 t1
t2 t1
t2 t1
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i=1 fi(x)uV [i],
i=1 gi(x)uB[i],
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L2(0,T ;RsV ) + αB∥uB∥2 L2(0,T ;RsB ))
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⋆ ∶= 1 τ ∫ tk tk−1 u⋆(t) dt, v k∣∂Ω = Buk B, ϕ0 = uI
V w dx = 0∀w ∈ Hσ(Ω),
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2 dx+σ ∫Ω
2 dx+σ ∫Ω
V )v k dx
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V ∈ L2(Ω)n,
B ∈ H
1 2 (∂Ω), BuI ∈ H1(Ω) ∩ L∞(Ω) be given data.
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B, ϕ0 = uI
V w dx−∫Ω ρ0g ⋅ w = 0∀w ∈ Hσ(
τ)
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k=1 ∈ Hσ(Ω)K, (ϕk)K k=1 ∈ (H1(Ω) ∩ C(Ω)) K, (µk)K k=1 ∈ W 1,3(Ω)K
τ) for k = 1 and
k=1∥ℓ∞(H1(Ω)) ≤ C (v 0,uI,uV ,uB),
k=1∥ℓ∞(H1(Ω)∩C(Ω)) ≤ C (v 0,uI,uV ,uB),
k=1∥ℓ∞(W 1,3(Ω)) ≤ C (v 0,uI,uV ,uB).
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k=1 ∈ Hσ(Ω)K, (ϕk)K k=1 ∈ H2(Ω)K, (µk)K k=1 ∈ H2(Ω)K
τ) for k = 1 and
k=1∥ℓ∞(H1(Ω)) ≤ C (v 0,uI,uV ,uB),
k=1∥ℓ∞(H2(Ω)) ≤ C (v 0,uI,uV ,uB),
k=1∥ℓ∞(H2(Ω)) ≤ C (v 0,uI,uV ,uB).
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k=1) ∶=1
L2(0,T ;RsV ) + αB∥uB∥2 L2(0,T ;RsB ))
τ) and (CHNSτ)
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h triangulation of Ω at time instance tk,
1 ∶= {v ∈ C(Ω)∣v∣T ∈ P 1 ∀T ∈ T k h },
2 ∶= {v ∈ C(Ω)n ∣v∣T ∈ (P 2)n ∀T ∈ T k h , (div(v),q) = 0∀q ∈ V k 1 },
1 prolongation, e.g. H1-prolongation.
h,µk h ∈ V k 1 ,
h ∈ V k 2
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⋆ ∶= ⨏ tk tk−1 u⋆(t) dt, v k h ∣∂Ω = Πh(Buk B), ϕ0 h = uI
h
h
h − ρk−2 h
h
h
h
h
h ,w) + ∫Ω 2ηk−1 h
h ∶ Dw dx
h∇ϕk−1 h
h
V )w dx = 0∀w ∈ V k 2 ,
h − P kϕk−1 h
h ⋅ ∇ϕk−1 h
h ⋅ ∇Ψ dx = 0∀Ψ ∈ V k 1 ,
h ⋅ ∇Φ dx−∫Ω µk hΦ dx
h) + (W−)′(P kϕk−1 h
1 .
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h,µk h,v k h be a solution to (CHNSh), and uB ≡ 0.
h
h ∣ 2 dx+σ ∫Ω
h∣2 + 1
h) dx
h
h − v k−1 h
h − ∇P kϕk−1h∣2 dx
h
h ∣2 dx+τ ∫Ω m∣∇µk h∣2 dx
h
h
2 dx+σ ∫Ω
h
h
h
h dx+∫Ω(Buk V )v k h dx
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h )K k=1 ∈ (V k 2 )K k=1, (ϕk h)K k=1,(µk h)K k=1 ∈ (V k 1 )K k=1, such that (v k h ,ϕk h,µk h) is
h )K k=1∥ℓ∞(H1(Ω)) ≤ C (v 0,uI,uV ,uB),
h)K k=1∥ℓ∞(W 1,4(Ω)) ≤ C (v 0,uI,uV ,uB),
h)K k=1∥ℓ∞(W 1,3(Ω)) ≤ C (v 0,uI,uV ,uB).
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h)K k=1) ∶=1
h − ϕd∥2
L2(0,T ;RsV ) + αB∥uB∥2 L2(0,T ;RsB ))
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h,v ⋆ h ,ϕ⋆ h,µ⋆ h) denote a stationary point of (Ph). Then there
V,h ⇀ u⋆ V ∈ UV ,
B,h ⇀ u⋆ B ∈ UB,
h
I,h → u⋆ I ∈ H1(Ω),
h
h
h
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0.2 0.4 0.6 0.8 1 5 10 15 20 time ∥u(t)∥ Strength of control 0.2 0.4 0.6 0.8 1 0.5 0.6 0.7 0.8 time Center of mass
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1 2(1 − ϕ2)
uI∈H1(Ω)∩L∞(Ω),∣uI∣≤1J(uI)
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I = −0.8
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I
I
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