Geometric Control and Homotopic Methods for solving a Bang-Singular - PowerPoint PPT Presentation

Geometric Control and Homotopic Methods for solving a Bang-Singular problem Applied and Numerical Optimal Control Spring School & Workshop, Ensta ParisTech O. Cots 23-27 April 2012 O. Cots (IMB Bourgogne) Results on contrast problem April 2012 1 / 40

Introduction - Collaborations Figure: (LHS) Hard pulse of 90 ◦ . (RHS) Optimal solution. – Bernard Bonnard (Univ. Bourgogne) – Jean-Baptiste Caillau (Univ. Bourgogne) – Joseph Gergaud (N7 - IRIT) – Steffen J. Glaser (Univ. Munich) – Dominique Sugny (Univ. Bourgogne) – . . . O. Cots (IMB Bourgogne) Results on contrast problem April 2012 2 / 40

Contrast problem | q 2 ( t f ) | 2 = � y 2 2 ( t f ) + z 2 �  2 ( t f ) → max   q i = ( y i , z i ) , | q i | ≤ 1 , i = 1 , 2     � ˙ = − Γ 1 y 1 − y 1 u z 1     ˙ = γ 1 ( 1 − z 1 ) + γ i = 1 / ( 32 . 3 T i 1 . 10 − 3 ) z 1 u y 1    � ( P ) ˙ = − Γ 2 y 2 − Γ i = 1 / ( 32 . 3 T i 2 . 10 − 3 ) y 2 u z 2 z 2 ˙ = γ 2 ( 1 − z 2 ) + u y 2      | u | ≤ 2 π    q 1 ( 0 ) = ( 0 , 1 ) = q 2 ( 0 )      q 1 ( t f ) = ( 0 , 0 ) Spin 1 ( y 1 , z 1 ) - Deoxygenated blood : T 11 = 1350 ms, T 12 = 50 ms Spin 2 ( y 2 , z 2 ) - Oxygenated blood : T 21 = 1350 ms, T 22 = 200 ms Spin 1 ( y 1 , z 1 ) - Cerebrospinal fluid : T 11 = 2000 ms, T 12 = 200 ms Spin 2 ( y 2 , z 2 ) - Water : T 21 = 2500 ms, T 22 = 2500 ms Blood: T min = 6 . 7980978 Fluid: T min = 20 . 2301921 O. Cots (IMB Bourgogne) Results on contrast problem April 2012 3 / 40

Pontryagin Maximum Principle We have a (smooth) Mayer problem of the form: q = ( q 1 , q 2 ) ∈ R 4 , u ∈U c ( q ( t f )) , q = F ( q ) + uG ( q ) , ˙ | u | ≤ 2 π, u ∈ R , min where y 2 2 ( t f ) + z 2 c ( q ( t f )) = − � 2 ( t f ) � i = 1 , 2 ( − Γ i y i ) ∂ ∂ y i + ( γ i ( 1 − z i )) ∂ F ( q ) = � ∂ z i i = 1 , 2 − z i ∂ ∂ y i + y i ∂ G ( q ) = � ∂ z i The Hamiltonian is: H ( q , p , u ) = � p , F ( q ) + uG ( q ) � � u = 2 π sign ( H G ) is bang if H G � = 0 And the Maximum Principle gives: u is singular if H G = 0 The boundary conditions are: q 1 ( t f ) = 0 (zero magnetization of the first spin) p 0 ≤ 0 . p 2 ( t f ) = − 2 p 0 q 2 ( t f ) , O. Cots (IMB Bourgogne) Results on contrast problem April 2012 4 / 40

Simplest possible solution The north pole N = (( 0 , 1 ) , ( 0 , 1 )) = x ( 0 ) , is an equilibrium point for the singular control system ( u s = 0 ): start with a bang. Due to symmetry of revolution: the first bang is either u = + 2 π or u = − 2 π . A bang solution amounts roughly to a rotation of each plane ( y i , z i ) around 0 . At t f , q 1 ( t f ) = 0 , and there is no improvement by rotation (because T 21 > T 22 ) of the contrast: no bang at the end. Lemma The simplest BC-extremal in the contrast problem is of the form B + S . 1 1 6 5 0.5 0.5 4 z 1 0 z 2 0 3 2 −0.5 −0.5 1 0 −1 −1 −1 −0.5 0 0.5 1 −1 −0.5 0 0.5 1 y 1 y 2 0 0.2 0.4 0.6 0.8 1 Figure: Trajectories of spin 1, 2 and the control associated for the Blood case. O. Cots (IMB Bourgogne) Results on contrast problem April 2012 5 / 40

Singular extremals The Hamiltonian is H ( q , p , u ) = H F ( q , p ) + u H G ( q , p ) , H F := � p , F � , H G := � p , G � . Let z ( · ) = ( q ( · ) , p ( · )) be a singular extremal, then H G ( z ( . )) = 0 identically. Differentiating with respect to time,  H G ( z ( t ))= 0   (2 constraints) dH G dt ( z ( t )) = { H , H F } ( z ( t )) = { H G , H F } ( z ( t ))= 0 d 2 H G dt 2 ( z ( t )) = {{ H G , H F } , H F } ( z ( t )) + u s ( t ) {{ H G , H F } , H G } ( z ( t )) = 0 Besides, + we can restrict p to H s = h because of the homogeneity : u s ( q , α p ) = u s ( q , p ) . + generalized Legendre-Clebsch condition has to be satisfied that is {{ H G , H F } , H G } ≥ 0 . ⇒ for each q ( 0 ) ∈ R 4 , p ( 0 ) ∈ R 4 lives in half a straight line and the singular flow starting from q ( 0 ) is a surface of dimension 2. O. Cots (IMB Bourgogne) Results on contrast problem April 2012 6 / 40

Singular extremals in the Fluid case 1 1 0.5 0.5 z1 z2 0 0 −0.5 −0.5 −1 −1 −1 −0.5 0 0.5 1 −1 −0.5 0 0.5 1 y1 y2 Figure: Singular surface with blowing up control in red ( {{ H G , H F } , H G } tends to 0 ). O. Cots (IMB Bourgogne) Results on contrast problem April 2012 7 / 40

Singular extremals in the Fluid case 1 0.06 0.04 0.02 0.5 0 −0.02 z1 z2 0 −0.04 −0.06 −0.5 −0.08 −0.1 −0.12 −1 −0.1 −0.05 0 0.05 −1 −0.5 0 0.5 1 y1 y2 Figure: Zoom of singular surface with blowing up control in red ( {{ H G , H F } , H G } tends to 0 ). O. Cots (IMB Bourgogne) Results on contrast problem April 2012 8 / 40

Local optimality of singular arcs Définition Let z ( · ) be a reference singular solution of # — H s on [ 0 , t f ] . The variational equation on the tangent space to Σ s := { H G = { H F , H G } = 0 } δ z = d # ˙ — H s ( z ( t )) δ z dH G = d { H F , H G } = 0 is called Jacobi equation. A Jacobi field J ( t ) = ( δ q , δ p ) is a non-zero solution of Jacobi equation. It is said semi-vertical at time t if δ q ( t ) ∈ R G ( q ( t )) . A time t ∈ ( 0 , t f ] is said to be conjugate if there exists a Jacobi field J ( t ) semi-vertical at 0 and t . Theorem Under strict Legendre-Clebsch condition (i.e. {{ H G , H F } , H G } > 0 ) and additional generic assumptions, the absence of conjugate times on ( 0 , t f ) is necessary for local optimality of a singular arc. O. Cots (IMB Bourgogne) Results on contrast problem April 2012 9 / 40

Conjugate points fot singular extremals in the Fluid case −0.6 0.02 −0.65 0 −0.7 −0.02 z1 z2 −0.04 −0.75 −0.06 −0.8 −0.08 −0.85 −0.1 −0.1 −0.05 0 0.05 −0.3 −0.2 −0.1 0 y2 y1 Figure: Zoom of singular surface with conjugate points in red. ⇒ We can expect more intricate structures than BS in the fluid case. O. Cots (IMB Bourgogne) Results on contrast problem April 2012 10 / 40

The limit case t f = T min . Single spin problem (see [BCG + 11])  t f → min  q = ( y , z ) , | q | ≤ 1   � ˙    = − Γ y − y u z   3 γ  < Γ ( P 1 ) ˙ = γ ( 1 − z ) + z u y 2    | u | ≤ 2 π  q ( 0 ) = ( 0 , 1 )     q ( t f ) = ( 0 , 0 )  The dynamics: q = F ( q ) ˙ + u G ( q ) The Hamiltonian: H ( q , u , p ) = H F ( q , p ) + u H G ( q , p ) , H i = < p , F i > , i = F , G ⇒ The singular extremals are those contained in H G = 0 . [BCG + 11] B. Bonnard, O. Cots, S. Glaser, M. Lapert, and D. Sugny. Geometric optimal control of the contrast imaging problem in nuclear magnetic resonance. math.u-bourgogne.fr , 2011. O. Cots (IMB Bourgogne) Results on contrast problem April 2012 11 / 40

Single spin: singular extremals ⇒ The singular extremals are those contained in H G = 0 . � = < p , G > = 0 H G ⇒ det ( G , [ G , F ]) = y ( − 2 δ z + γ ) = 0 ˙ = < p , [ G , F ] > = H G 0 with δ = γ − Γ . γ The singular lines are y = 0 and z 0 = 2 δ . The singular control u s ( q , p ) is computed from ¨ H G = {{ H G , H F } , H F } ( z ) + u s {{ H G , H F } , H G } ( z ) = 0 . γ Horizontal line ( z 0 = 2 δ ): optimal, according to Generalized Legendre-Clebsch condition (see [BC03]) and u s ( q ) = γ ( 2 Γ − γ ) / ( 2 δ y ) ⇒ u s ∈ L 1 , u s / ∈ L 2 Vertical line ( y = 0 ): optimal for z 0 < z < 1 (GLC) and u s ( q ) = 0 . [BC03] B. Bonnard and M. Chyba. Singular trajectories and their role in control theory, mathématiques & applications, vol. 40. lavoisier.fr , Jan 2003. O. Cots (IMB Bourgogne) Results on contrast problem April 2012 12 / 40

Single spin: global synthesis Red: bang ( u = 2 π ) Blue: bang ( u = − 2 π ) Green: singular ( u = u s ) B: saturation O. Cots (IMB Bourgogne) Results on contrast problem April 2012 13 / 40

Summary: single spin => contrast problem Solutions of the contrast problem are made of Bang-Singular sequences. We get T min . The solution for the limit case t f = T min is B + SB + S . To solve the contrast problem, we use an indirect method (multiple shooting) ⇒ we need to know the structure a priori. ⇒ we use an homotopic approach to capture the structure and initialize the multiple shooting method. O. Cots (IMB Bourgogne) Results on contrast problem April 2012 14 / 40

Homotopy � t f 0 | u | 2 − λ ( t ) dt → min  � y 2 2 ( t f ) + z 2 � − 2 ( t f ) + ( 1 − λ )   q i = ( y i , z i ) , | q i | ≤ 1 , i = 1 , 2     � y 1 ˙ = − Γ 1 y 1 − u z 1     z 1 ˙ = γ 1 ( 1 − z 1 ) + u y 1  γ i = 1 / ( 32 . 3 T i 1 )   ( P λ ) � ˙ y 2 = − Γ 2 y 2 − u z 2 Γ i = 1 / ( 32 . 3 T i 2 ) ˙ = γ 2 ( 1 − z 2 ) + z 2 u y 2      | u | ≤ 2 π    q 1 ( 0 ) = ( 0 , 1 ) = q 2 ( 0 )     q 1 ( t f ) = ( 0 , 0 )  � t f � y 2 2 ( t f ) + z 2 � 0 | u | 2 − λ ( t ) dt → min Homotopy ( P λ ) : − 2 ( t f ) + ( 1 − λ ) The Hamiltonian: H ( q , u , p , λ ) = H F ( q , p ) + u H G ( q , p ) + ( 1 − λ ) | u | 2 − λ ( P λ ) -extremals are admissible for ( P ) . O. Cots (IMB Bourgogne) Results on contrast problem April 2012 15 / 40