Olivier C - ENSEEIHTcots.perso.enseeiht.fr/resources/2014-AIMS-cots-slides.pdfBocop TOOLBOX Direct...
Transcript of Olivier C - ENSEEIHTcots.perso.enseeiht.fr/resources/2014-AIMS-cots-slides.pdfBocop TOOLBOX Direct...
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Geometric and numerical methods in the saturation
problem of an ensemble of spin particles.
Olivier COTS
Post-Doc INRIA Sophia AntipolisMcTAO Team
AIMS 2014, Madrid.
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CONTEXT
• Saturation problem of an ensemble of spin particles in MRI.
– Mayer optimal control problem
– Dynamics: Bloch equation
• Geometric tool: Pontryagin Maximum Principle
• Numerical methods
– Indirect method (HamPath ): shooting, homotopy. . .
– Direct method (Bocop): state and control discretizations⇒ NLP problem
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COLLABORATORS
[1] with B. Bonnard (Bourgogne Univ.),S. Glaser, M. Lapert, D. Sugny & Y. Zhang, (Bourgogne & Munich Univ.)
IEEE Trans. Automat. Control (2011).
[2] with B. Bonnard,Math. Models Methods Appl. Sci. (2012).
[3] with B. Bonnard, M. Claeys (LAAS Toulouse) & P. Martinon (INRIA Saclay)Acta Appl. Math. (2013).
[4] . . .
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MAGNETIC MOMENT
Particles of spin-1/2 (proton, neutron, electron. . . ) have magnetic moment.
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ORIENTATION OF THE MAGNETIC MOMENT
There are two possible orientations for a spin-1/2 held in a stationary magnetic field ~B.
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MAGNETIZATION VECTOR
(c) The magnetization vector ~M = (Mx,My,Mz) is the sum of the magnetic moments.
~M is non zero and pointing in the same direction as ~B.
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BLOCH EQUATION
Bloch equation.dM(t)
dt= γ B(t)∧M(t),
with γ the gyromagnetic ratio.
z Precession around B, with frequency:
ω =γ
2π|B|.
z Do not change the norm or the direction of M.
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NUCLEAR MAGNETIC RESONANCE EXPERIMENTS
Two magnetic fields.
Intense stationary field: B0 = (0, 0, Bz)
Control RF-field: B1(t) = (Bx(t),By(t),0)
z M(t) lives on the sphere with radius M0 = M(0).
Two relaxation effects. longitudinal (T1) and transversal (T2)
M = R(M,T1,T2)+ γ B∧M
z M(t) lives in the Bloch ball: |M(t)| ≤M0.
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MEDICAL IMAGING
Medical imaging. The norm of the magnetization vector corresponds to gray scale.
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NORMALIZED DISSIPATIVE BLOCH EQUATION
Normalized dissipative Bloch equation.
x = −Γx + u2zy = −Γy − u1zz = γ (1− z) + u1y−u2x
• State q = (x,y,z) = M/M0 ∈ B(0,1) : normalized magnetization vector
• γ = 1/T1, Γ = 1/T2 : parameters of the molecule
• Control u = (u1,u2), |u| ≤ 1 : normalized RF-field
• N = (0,0,1) : equilibrium
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MULTISATURATION PROBLEM
Saturation of an ensemble of N spinsa.
c(q(t f )) =1N
N
∑i=1|qi(t f )|2 −→min
u(·), t f fixed
q = F0(q)+u1 F1(q)+u2 F2(q), |u| ≤ 1
q(0) = q0
State q = (q1, · · · ,qN),
qi = (xi,yi,zi) ∈ B(0,1), i = 1, . . . ,N
qi(0) = (0,0,1), i = 1, . . . ,N
|qi|a J. S Li & N. Khaneja, Control of inhomogeneous quantum ensembles, Phys. Rev. A., 2006.
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MULTISATURATION PROBLEM WITH B1 INHOMOGENEITY
Saturation of an ensemble of N spins, with B1 inhomogeneity.
c(q(t f )) =1N
N
∑i=1|qi(t f )|2 −→min
u(·), t f fixed
q = F0(q)+u1 F1(q)+u2 F2(q), |u| ≤ 1
q(0) = q0
Scaling factors: ai, i = 1, . . . ,NF0(q) = ∑
i=1,N−Γxi
∂
∂xi−Γyi
∂
∂yi+ γ(1− zi)
∂
∂ zi
F1(q) = ∑i=1,N
ai
(−zi
∂
∂yi+ yi
∂
∂ zi
)
F2(q) = ∑i=1,N
ai
(zi
∂
∂xi− xi
∂
∂ zi
)
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PONTRYAGIN MAXIMUM PRINCIPLE
Hamiltonian. H(q, p,u) = 〈p, q〉= H0+u1H1+u2H2, Hi = 〈p,Fi(q)〉.
Necessary condition (PMP). If u is optimal, then ∃ p : [0, t f ]→ Rn such as a.e.:
q(t) =∂H∂ p
(q(t), p(t),u(t)), p(t) =−∂H∂q
(q(t), p(t),u(t))
andH(q(t), p(t),u(t)) = max
|v|≤2π
H(q(t), p(t),v).
We have: u(t) =
(H1,H2)√H2
1 +H22
is bang if√
H21 +H2
2 6= 0
u(t) is singular if H1 = H2 = 0
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KEY ROLES OF SINGULAR EXTREMALS AND SINGLE-INPUT CASE (u2 = 0)
Hamiltonian. H(q, p,u) = 〈p, q〉= H0+u1H1+u2H2, Hi = 〈p,Fi(q)〉.
Singular set. Σ := H1 = H2 = 0.
Differentiating H1, H2:{
H1 = {H,H1}= H01−u2 H12 = 0H2 = {H,H2}= H02+u1 H12 = 0
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KEY ROLES OF SINGULAR EXTREMALS AND SINGLE-INPUT CASE (u2 = 0)
Hamiltonian. H(q, p,u) = 〈p, q〉= H0+u1H1+u2H2, Hi = 〈p,Fi(q)〉.
Singular set. Σ := H1 = H2 = 0.
Differentiating H1, H2:{
H1 = {H,H1}= H01−u2 H12 = 0H2 = {H,H2}= H02+u1 H12 = 0
Definitions. Let F1, F2 be two vector fields and z = (q, p).
Lie bracket: F12 = [F1,F2](q) =∂F1
∂q(q)F2(q)−
∂F2
∂q(q)F1(q)
Poisson bracket: H12 = {H1,H2}(z) = dH1 (#—H 2)(z) = H[F1,F2](z),
#—H 2 =
(∇pH2−∇qH2
).
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KEY ROLES OF SINGULAR EXTREMALS AND SINGLE-INPUT CASE (u2 = 0)
Stratification of Σ := H1 = H2 = 0.
The singular extremals must satisfy:{H01−u2 H12 = 0H02+u1 H12 = 0
Σ1 := Σ\H12 = 0
z us =(−H02,H01)
H12
z Singular extremals in Σ1 are either not admissible (|u|> 1),
or saturating (|u|= 1)
or not optimal (Goh conditiona: H12 = 0).
a B. Bonnard & M. Chyba, Singular trajectories and their role in control theory, Springer-Verlag, Berlin, 2003.
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KEY ROLES OF SINGULAR EXTREMALS AND SINGLE-INPUT CASE (u2 = 0)
Stratification of Σ := H1 = H2 = 0.
The singular extremals must satisfy:{H01−u2 H12 = 0H02+u1 H12 = 0
If H12 = 0 then H01 = H02 = 0. Differentiating H01, H02, we geta:
Aus+b = 0, with A =
[H011 H012H021 H022
]and b =
[H010H020
].
Σ2 := H1 = H2 = H12 = H01 = H02 = 0\det A = 0
a Y. Chitour, F. Jean & E. Trélat, Genericity results for singular curves, J. Differential Geom., 2006.
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KEY ROLES OF SINGULAR EXTREMALS AND SINGLE-INPUT CASE (u2 = 0)
Stratification of Σ := H1 = H2 = 0.
Σ3 := H1 = H2 = H12 = H01 = H02 = det A = 0, with A =
[H011 H012H021 H022
]
Imposing u2 = 0, each substate qi = (xi,yi,zi), i = 1, . . . ,N, is restricted to xi = 0.
z The corresponding extremals force the surface det A = 0 to be invariant.
z Remaining relationsa : H1 = H01 = H010+u1,s H011 = 0.
z Legendre-Clebsch condition: − ∂
∂u∂ 2
∂ t2∂H∂u
= H011 ≤ 0.
a I. Kupka, Geometric theory of extremals in optimal control problems., Trans. Amer. Math. Soc., 1987.
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KEY ROLES OF SINGULAR EXTREMALS AND SINGLE-INPUT CASE (u2 = 0)
Stratification of Σ := H1 = H2 = 0.
Σ = Σ1∪Σ2∪Σ3
• Σ1 of codimension 2: only saturating singular extremals
• Σ2 of codimension 5: feedback control for N = 2
• Σ3 of codimension 6: contains single-input case (u2 = 0)
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SINGLE-INPUT CASE (u2 = 0): PRELIMINARIES
Single-input case.
The singular extremals are defined by:
H1 = H01 = H010+u1,s H011 = 0.
Defining Hs = H0+u1,s H1, the singular extremals z = (q, p) are solutions of:
dzdt
=#—H s(z), z ∈ Σ
′1 := {z, H1(z) = H01(z) = 0}, u1,s =−
H010
H011.
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SINGLE-INPUT CASE (u2 = 0): PRELIMINARIES
Single-input case with two spins (N = 2).
• The dynamics on q ∈ R4 can be reduced to:
dqdt
= F0(q)−H010(q,λ )H011(q,λ )
F1(q), with λ a one-dimensional parameter.
•Moreover in the exceptional case (H0 = 0), we have:
ue1,s =−D′(q)/D(q)
withD = det(F0,F1,F10,F101), D′ = det(F0,F1,F10,F100)
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SINGLE-INPUT CASE (u2 = 0): NUMERICAL RESULTS
Saturation of an ensemble of 2 spins in the single-input case.
c(q(t f )) =12
2
∑i=1|qi(t f )|2 −→min
u(·), t f fixed
q = F0(q)+u1 F1(q), |u1| ≤ 1
q(0) = q0
State q = (q1,q2), qi = (yi,zi), |qi| ≤ 1 and qi(0) = (0,1).
F0(q) =2
∑i=1
(−Γyi)∂
∂yi+(γ(1− zi))
∂
∂ zi, F1(q) =
2
∑i=1
ai
(−zi
∂
∂yi+ yi
∂
∂ zi
),
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SINGLE-INPUT CASE (u2 = 0): NUMERICAL RESULTS
Saturation of an ensemble of 2 spins in the single-input case.
F0(q) =2
∑i=1
(−Γyi)∂
∂yi+(γ(1− zi))
∂
∂ zi, F1(q) =
2
∑i=1
ai
(−zi
∂
∂yi+ yi
∂
∂ zi
),
• Blood case: (γ,Γ) =(
11.35
,1
0.05
)• t f = λTmin, with Tmin = 6.7981 and λ ∈ [1,2].
• (a1,a2) = (1,1− ε), with ε = 0.3.
• Numerical methods: Direct (Bocopa) and Indirect (HamPath b).
ahttp://bocop.orgbhttp://cots.perso.math.cnrs.fr/hampath
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Bocop TOOLBOX
Direct approach. Optimal control problem (OCP)⇒ finite–dimensional optimization prob-lem (NLP), obtained by a discretization in time applied to the state and control variables, aswell as the dynamics equation.
(OCP)
t ∈ [0, t f ] → {t0 = 0, . . . , tN = t f}q(·),u(·) → X = {q0, . . . ,qN,u0, . . . ,uN−1, t f}Criterion → min c(qN)Dynamics → (ex : Euler) qi+i = qi+h f (qi,ui)Adm. Cont. → −1≤ ui ≤ 1Bnd. Cond. → Φ(q0,qN) = 0
⇒ (NLP){
min F(X) = c(qN)LB≤C(X)≤UB
Bocop package. IPOPT solver with MUMPS and ADOL-C for Automatic Differentiation.
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HamPath PACKAGE
efun
E(z0,z f ,λ )
hfun
H(z,λ )
dE #—
H d#—
H
h(z0,λ ) h′(z0,λ ) T (c(s))
ssolve :Simple–Multiple Shooting
hampath :Homotopy
expdhvfun :Var. Eqs.
tapenade
dopri5 dopri5
dopri5
hybrj
QR
dopri5
Fortran Routines
Coeur Fortran: Minpack, Lapack, Blas . . .
Matlab Functions
tapenade
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SINGLE-INPUT CASE (u2 = 0): NUMERICAL RESULTS
Blood case with N = 2, λ = 1.1, ε = 0.3.
Method c(q(t f )) initial costate p(0) t1, t2, t3 CPUBocop 2.56×10−4 (2.57,−3.46,−2.96,−4.92)×10−4 (0.043,0.338,0.382) 0.66sHamPath 2.56×10−4 (1.82,−3.85,−3.47,−4.48)×10−4 (0.040,0.347,0.385) 0.43s
Solution from direct (black dots) and indirect (blue line) methods. Control structure is 2BS.
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SINGLE-INPUT CASE (u2 = 0): NUMERICAL RESULTS
Homotopy on transfert time (parameter λ = t f/Tmin).
(Left) The mean distance to the origin(1
2 ∑2i=1 |qi(t f )|
)decreases linearly in log scale.
(Right) The switching times indicate a decreasing duration t2− t1 for the first singular arc.
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SINGLE-INPUT CASE (u2 = 0): NUMERICAL RESULTS
Homotopy on transfert time (parameter λ = t f/Tmin): controls.
For λ = 1,1.5 and 2, the control structure remains 2BS, with the duration of the first singulararc decreasing for larger λ .
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SINGLE-INPUT CASE (u2 = 0): NUMERICAL RESULTS
Multisaturation with N = 11 spins, with ai ∈ [0.7,1].
(Left) The trajectory of each spin for λ = 1 i.e. t f = Tmin.
(Right) A closeup on the final positions of all spins for λ = 1,1.125 and 1.25.
We observe that the spins tend to spread regularly around the origin, and get closer for largertransfer times.
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MULTICONTRAST PROBLEM IN MRI
We have two different samples (de-oxygenated and oxygenated bloods).
Equilibrium state⇒ both samples are white. Optimal control applied⇒ maximized contrast.
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REFERENCES
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