Theory and Applications of BangBang and Singular Control Problems - PowerPoint PPT Presentation

Theory and Applications of Bang–Bang and Singular Control Problems Helmut Maurer Institut f¨ ur Numerische und Angewandte Mathematik Universit¨ at M¨ unster CEA–EDF–INRIA School, May 30 – June 1, 2007

Overview Basic tasks for solving optimal control problems • Necessary conditions: Minimum Principle of Pontryagin et al. ( 50 years ! ) • Numerical methods for verifying necessary conditions: (1) boundary value methods, (2) discretize and optimize. • (Second-order) Sufficient conditions • Sensitivity analysis and real-time control techniques Optimal control problems with control appearing linearly • bang–bang and singular control • Additional numerical method: direct optimization of switching times • Verification of second order sufficient conditions (SSC) – bang–bang control: Agrachev/Stefani/Zezza, Osmolovskii/Maurer, Sch¨ attler – singular control: work in progress: Dmitruk, Stefani, Vossen – state constraints: Ledzewicz/Sch¨ attler, Maurer/Vossen – applications to sensitivity analysis

Overview Examples • Control of a Van der Pol oscillator: bang-bang and singular control • Time–optimal control for a semiconductor laser • GODDARD problem: bang-singular control • Time–optimal control of a two-link robot • Optimal control of a fedbatch fermentation process: bang–singular control Joint work with Nikolai Osmolovskii, Christof B¨ uskens, Jang–Ho Robert Kim, Georg Vossen

Van der Pol oscillator: time-optimal control ☛✟ ☛✟ ☛✟ L ☛✟ ☛✟ I ✲ ✂✁✂✁✂✁✂✁ x 1 ( t ) = V ( t ) voltage I C I D I R ★✥ ★✥ ❄ ❄ ❄ u ( t ) = V 0 ( t ) control ✻ V t ❆ ❆✁ V 0 ✁ C D R ✲ ✧✦ ✧✦ ❄ Minimize the final time t f subject to x 1 ( t ) = x 2 ( t ) , ˙ x 1 (0) = 1 , x 2 ( t ) = − x 1 ( t ) + x 2 ( t )( p − x 1 ( t ) 2 ) + u ( t ) , ˙ x 2 (0) = 1 , x 1 ( t f ) 2 + x 2 ( t f ) 2 = r 2 , ( r = 0 . 2 ) − 1 ≤ u ( t ) ≤ 1 , t ∈ [0 , t f ] . Perturbation p , nominal value p 0 = 1 . 0 : discretize and optimize � − 1 for � , 0 ≤ t < t 1 Optimal bang–bang control u ( t ) = 1 , for t 1 < t ≤ t f

Nominal optimal solution for p 0 = 1 1.4 1 x 1 ( t ) x 2 ( t ) 0.8 1.2 0.6 1 0.4 0.8 0.2 0.6 0 0.4 -0.2 0.2 -0.4 0 -0.6 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 State trajectories x 1 ( t ) , x 2 ( t ) 1.8 1 λ 1 ( t ) λ 2 ( t ) 1.6 0.5 1.4 0 1.2 -0.5 1 -1 0.8 -1.5 0.6 0.4 -2 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 Adjoint variables λ 1 ( t ) , λ 2 ( t ) = σ ( t ) (switching function), σ ( t 1 ) � = 0 ˙

SSC and sensitivity analysis Optimization variables : z := ( t 1 , t f ) switching time t 1 = 0 . 713935566 , final time t f = 2 . 86419188 Φ( z ) = x 1 ( t f ) 2 + x 2 ( t f ) 2 = r 2 Compute Jacobian of terminal conditions and Hessian of Lagrangian: � � 188 . 066 − 7 . 39855 Φ z = ( − 0 . 0000264 , 0 . 3049115) , L zz = − 7 . 39855 3 . 06454 SSC hold ! Sensitivity derivatives exist (code NUDOCCCS, C. B¨ uskens) dt 1 dt f dp = − 0 . 344220 , dp = 1 . 395480 dx 1 dx 2 0.4 0.8 dp dp 0.35 0.7 0.3 0.6 0.25 0.5 0.2 0.4 0.15 0.3 0.1 0.2 0.05 0.1 0 0 -0.05 -0.1 0 0.5 1 1.5 2 2.5 3 0 0.5 1 1.5 2 2.5 3 Parametric sensitivity derivatives of state variables (scaled)

Van der Pol oscillator: singular control t f J ( x, u ) = 1 ( x 2 1 + x 2 � Minimize 2 ) dt , t f = 4 , 2 0 subject to x 1 = x 2 , ˙ x 1 (0) = 0 , x 2 = − x 1 + x 2 (1 − x 2 ˙ 1 ) + u, x 2 (0) = 1 , − 1 ≤ u ( t ) ≤ 1 . Hamiltonian H ( x, λ, u ) = 1 x 2 1 + x 2 − x 1 + x 2 (1 − x 2 � � � � + λ 1 x 2 + λ 2 1 ) + u 2 2 ˙ Adjoint ODEs : λ 1 = − H x 1 = − x 1 + λ 2 (1 + 2 x 1 x 2 ) , λ 1 ( t f ) = 0 , ˙ λ 2 = − H x 2 = − x 2 − λ 1 − λ 2 (1 − x 2 1 ) , λ 2 ( t f ) = 0 . Switching function: σ = H u = λ 2 Singular feedback control of order 1 : σ = ˙ σ = ¨ σ ≡ 0 u = u sing ( x ) = 2 x 1 − x 2 (1 − x 2 ⇒ 1 )

Van der Pol oscillator: singular control Optimal control is bang–bang–singular   − 1 for 0 ≤ t < t 1 = 1 . 3667     u = 1 for t 1 < t < t 2 = 2 . 4601 2 x 1 − x 2 (1 − x 2 for  1 ) t 2 < t ≤ t f = 4    x 1 u x 2 σ = λ 2

SSC for switching times and sensitivity analysis Optimize with respect to z = ( t 1 , t 2 ) Hessian of the Lagrangian � � 215 . 4 − 10 . 54 L zz = is positive definite. − 10 . 54 0 . 5665 But: No sufficient conditions for the control problem ars available ! Sensitivity analysis: perturbation p in the dynamics x 2 = − x 1 + x 2 ( p − x 2 x 1 = x 2 , ˙ ˙ 1 ) + u Nominal parameter p 0 = 1 : dz � � 0 . 2831 dp ( p 0 ) = − ( L zz ) − 1 L zp = . 2 . 2555

Time–optimal control of a semiconductor laser Dokhane, Lippi: “Minimizing the transition time for a semiconductor laser with homogeneous transverse profile,” IEE Proc.-Optoelectron. 149 , 1 (2002). Kim, Lippi, Maurer: “Minimizing the transition time in lasers by optimal control methods. Single mode semiconductor laser with homogeneous transverse profile”, Physica D 191 , 238–260 (2004). S ( t ) : photon density; N ( t ) : carrier density; I ( t ) : current (control) S = dS dt = − S ˙ + Γ G ( N, S ) S + βBN ( N + P 0 ) τ p N = dN dt = I ( t ) ˙ − R ( N ) − Γ G ( N, S ) S q G ( N, S ) = G p ( N − N tr )(1 − ǫS ) (optical gain) R ( N ) = AN + BN ( N + P 0 ) + CN ( N + P 0 ) 2 (recombination) Initial and terminal conditions (stationary points): S (0) = S 0 , N (0) = N 0 ( for I ( t ) ≡ 20 . 5 mA ) ( for I ( t ) ≡ 42 . 5 mA ) S ( t f ) = S f , N ( t f ) = N f

Laser: time-optimal bang-bang control Minimize the final time t f subject to the control bounds I min ≤ I ( t ) ≤ I max for 0 ≤ t ≤ t f I ( t ) 70 S ( t ) 6 / 10 5 60 / mA 5 50 4 40 3 30 2 20 1 10 0 0 600 t/ ps t/ ps -40 -20 0 20 40 60 80 0 100 200 300 400 500 time–optimal bang-bang control uncontrolled versus controlled

Time–optimal bang-bang control Time–optimal control is bang-bang: � � I max , 0 ≤ t < t 1 , , t 1 = 29 . 523 ps , t f = 56 . 894 ps I ( t ) = I min , t 1 ≤ t ≤ t f . Switching function and strict bang–bang property: t ( ps ) σ ( t ) = λ N ( t ) , σ ( t 1 ) = 0 , σ ( t 1 ) � = 0 ˙

Sufficient conditions, Sensitivity analysis, Switch–on–off Jacobian of terminal conditions is regular: � � 0 . 199855 0 Φ z = − 1 . 5556 · 10 − 4 − 2 . 52779 · 10 − 3 First order sufficient conditions hold ! Sensitivity derivatives p = I max : dt 1 /dp = − 0 . 55486 , dt 2 /dp = 0 . 22419 , p = I min : dt 1 /dp = − 0 . 24017 , dt 2 /dp = 0 . 57532 . Laser: simultaneous switch–on and switch–off: S ( t 2 ) = S f , N ( t 2 ) = N f   for I max , 0 ≤ t < t 1     I min , for t 1 ≤ t ≤ t 2       I ( t ) = − − − − − , − − − − − − −− I min , for t 2 < t < t 3         I min , for t 3 ≤ t ≤ t f   PROBLEM: arclengths are not synchronized: t 1 � = t 3 − t 2 and t 2 − t 1 � = t f − t 4 OPTIMIZE with respect to I max I max and I 0 , I ∞

GODDARD problem: maximize altitude of a rocket A.E. Bryson, Y.C. Ho: Applied Optimal Control , Ginn and Company, 1969. H. Maurer: Numerical solution of singular control problems using multiple shooting techniques , J. of Optimization Theory 18 , pp. 235–257 (1976). Maximize h ( t f ) ( final time t f is free ) ˙ subject to h = v, � � v = 1 ˙ cu − D ( v, h ) − γ ( h ) , m m = − u, ˙ h (0) = v (0) = 0 , m (0) = m 0 , m ( t f ) = m f , 0 ≤ u ( t ) ≤ u max = 9 . 52551 . r 2 D ( v, h ) = αv 2 exp( − βh ) , Drag Gravitation γ ( h ) = g 0 ( r 0 + h ) 2 . 0 α = 0 . 01227 , β = 0 . 145 × 10 − 3 , c = 2060 , Data : m 0 = 214 . 839 , m f = 67 . 9833 , g 0 = 9 . 81 , r 0 = 6 . 371 × 10 6 [ m ]

GODDARD problem: singular feedback control � � 1 � � Hamiltonian: H ( h, v, m, λ, u ) = λ h v + λ v c · u − D ( v, h ) − γ ( h ) − λ m u. m Switching function: σ = H u = c λ v m − λ m . Singular feedback control of order 1: σ ≡ ˙ σ ≡ ¨ σ ≡ 0 and H [ t ] ≡ 0 u sing ( h, v, m ) = D c + m ( c − v ) D h + D v γ + cD vv γ − cD vh v + cmγ h D + 2 cD v + c 2 D vv u ( t ) = u max | singular | 0 Control structure: u σ 10 2000 8 1500 6 1000 4 500 2 0 0 −500 0 50 100 150 200 0 50 100 150 200 250 Switching times and final time: t 1 = 4 . 11525 , t 2 = 46 . 04063 , t f = 212 . 90299 .

GODDARD problem: SSC w.r.t. t 1 , t 2 , t f Optimization : arc lengths ξ 1 = t 1 , ξ 2 = t 2 − t 1 , ξ 3 = t f − t 2   − 244422 . 57 − 19472 . 17 − 1611 . 55 ˜  , L ξξ = − 19472 . 17 − 1488 . 13 − 119 . 50  − 1611 . 55 − 119 . 50 9 . 36 � − 55 . 76 ˜ 0 . 00 � Φ ξ = − 4 . 20 . L ξξ is positive definite on ker( ˜ ˜ Φ ξ ) . Reduced Hessian � � 58 . 04 1 . 77 N ∗ ˜ ˜ L ξξ ˜ N = , 1 . 77 9 . 36 N span ker( ˜ ˜ where the columns of Φ ξ ) . Georg Vossen: Dissertation, Universit¨ at M¨ unster, 2005.