Optimal Control Problems in Eye Movement. Bijoy K. Ghosh Department of Mathematics and Statistics Texas Tech University, Lubbock, TX

Transcription

1 Optimal Control Problems in Eye Movement Bijoy K. Ghosh Department of Mathematics and Statistics Texas Tech University, Lubbock, TX

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3 We study the human oculomotor system as a simple mechanical control system. It is a well known physiological fact that all eye movements obey Listing s Law which states that eye rotations are always restricted to a given fixed plane called the Listing s plane.

4 In fact, Listing s constraint is sufficient to ensure that corresponding to any specific gaze direction, the torsion is predetermined. In other words, orientation of the eye is completely specified by its gaze. Thus objects in space are not viewed with a rotational ambiguity.

5 All these ideas are very interesting and biomechanists are interested in the study of how the brain is able to satisfy the Listing s constraint through muscle based actuation of the eye ball. The subject matter of my talk is the following: Assume that the Listing s Law is satisfied, how can we rotate the eye ball optimally from one gaze direction to

6 It is natural to start with SO (3), the space of 3 by 3 rotation matrices, as the configuration space of the eye ball. We denote by List the subset of which is described as follows: -- SO (3) List = R SO(3) v =[v 1,v 2 ] R 2 \{0} 3R v 1 v 2 0 = v 1 v 2 0

7 It is natural to start with SO (3), the space of 3 by 3 rotation matrices, as the configuration space of the eye ball. We denote by List the subset of which is described as follows: -- SO (3) The space List is the set of all 3 by 3 rotation matrices that has axis of rotation perpendicular to the vector

8 It turns out that elements of List are rotation matrices of the form : q2 0 + q1 2 q2 2 q3 2 2(q 1 q 2 q 0 q 3 ) 2(q 1 q 3 + q 0 q 2 ) 2(q 1 q 2 + q 0 q 3 ) q0 2 + q2 2 q1 2 q3 2 2(q 2 q 3 q 0 q 1 ) 2(q 1 q 3 q 0 q 2 ) 2(q 2 q 3 + q 0 q 1 ) q0 2 + q3 2 q1 2 q2 2 where q 3 =0, and (q 0,q 1,q 2 ) is a unit vector. It is not at all clear, apriori, why this would be the case.

9 It turns out that there is a correspondence between rotation matrices and unit quaternions. A quaternion Q is a 4 tuple of real numbers written as a 0 ~1 + a 1 ~ i + a2 ~ j + a3 ~ k and we have two obvious maps: vec : Q R 3, a 7 (a 1,a 2,a 3 ) called the vector part of the quaternion.

10 It turns out that there is a correspondence between rotation matrices and unit quaternions. A quaternion Q is a 4 tuple of real numbers written as a 0 ~1 + a 1 ~ i + a2 ~ j + a3 ~ k and we have two obvious maps: scal : Q R, a 7 a 0 called the scalar part of the quaternion.

11 The space of unit quaternions will be identified with the unit sphere in R 4, and is denoted by S 3. Each q S 3 can be written as q =cos(α/2) ~1 +sin(α/2) n 1 ~ i +sin(α/2) n2 ~ j +sin(α/2) n3 ~ k where α [0, π] and n =(n 1,n 2,n 3 ) is a unit vector in R 3. We denote by rot the standard map from S 3 into SO 3, which maps q to a rotation around the axis n by a clockwise angle α. There are two explicit ways to describe this map. First it is easy to verify that, rot (q)(v 1,v 2,v 3 )=vec ³ q ³v 1 ~ i + v2 ~ j + v2 ~ j 1 q

12 Alternatively, we can think of rot(q) as follows: rot(q) = q0 2 + q1 2 q2 2 q 2 3 2(q 1 q 2 q 0 q 3 ) 2(q 1 q 3 + q 0 q 2 ) 2(q 1 q 2 + q 0 q 3 ) q0 2 + q2 2 q1 2 q3 2 2(q 2 q 3 q 0 q 1 ) 2(q 1 q 3 q 0 q 2 ) 2(q 2 q 3 + q 0 q 1 ) q0 2 + q3 2 q1 2 q2 2

13 The main point is that counterclockwise rotation by an angle α with respect to an axis n =(n 1,n 2,n 3 ), an unit vector, is given precisely by the unit quaternion cos (α/2) ~1+sin (α/2) n 1 ~ i+sin (α/2) n2 ~ j+sin (α/2) n3 ~ k where Listing s constraint is imposed by restricting n 3 =0

14 Alternative we can say that unit quaternions with the fourth coordinate 0 precisely corresponds to rotation matrices that satisfy the Listing s constraint. Thus we have a map: ρ :[0, π] [0, 2π] S 3 cos (φ/2) ρ (θ, φ) = cos (θ)sin(φ/2) sin (θ)sin(φ/2) 0

15 Riemannian metric on List We assume that eye is a perfect sphere and its inertia tensor is equal to. I 3 3 This is associated with the left invariant Riemannian metric on given by SO 3 hω (e i ), Ω (e j )i I = δ i,j Where Ω (e k )= 0 δ 3,k δ 2,k δ 3,k 0 δ 1,k δ 2,k δ 1,k 0 and {δ i,j } denotes the Kronecker delta function

16 To compute the Riemannian metric on List induced from, we define SO 3 g 11 = h θ, θ i g 12 = h θ, φ i g 22 = h φ, φ i

17 Using the map ρ :[0, π] [0, 2π] S 3 cos (φ/2) ρ (θ, φ) = cos (θ)sin(φ/2) sin (θ)sin(φ/2) 0 it follows that we have the following diagram: List T (θ,φ) List ρ S 3 ρ Tρ(θ,φ) S 3

18 We can compute the following: µ µ ρ θ µ ρ φ = sin (φ/2) 1 sin (θ)sin(φ/2) 2 cos (θ)cos(φ/2) 1 cos (θ)sin(φ/2) 2 sin (θ)cos(φ/2) 0 0

19 The Riemannian metric on List is easily computed by taking inner products and we obtain the following: g 11 =sin 2 (φ/2) g 12 =0 g 22 = 1 4 Finally we obtain g =sin 2 (φ/2) dθ dφ2

20 Using a Riemannian metric one is able to measure distance between points on a manifold. It is a standard calculation in differential geometry to treat the Riemannian Metric as the kinetic energy of the eye in motion and compute the path taken by the eye that would minimize this energy, while disregarding the potential energy. This path would be called the Geodesic

21 Equations of geodesics are given in the form of a differential equation. For our problem it is given by θ + 1 tan (φ/2) θ φ =0 φ sin (φ) φ 2 =0 θ Recall that is the angle for the axis of rotation with respect to the horizontal line on the Listing s plane. φ on the other hand is the angle of rotation with respect to the given axis.

22 If we define state variables as x 1 = θ ; x 3 = φ ; x 2 = θ x 4 = φ then the geodesic is given by: ẋ 1 = x 2 ẋ 2 = x 2 x 4 cot (x 3 /2) ẋ 3 = x 4 ẋ 4 = x 2 2 sin (x 3 ) This dynamical system has a singularity when is an even multiple of φ and equilibrium when is an odd multiple of and is zero. φ π φ π

23 The space of (θ, φ) isactuallythe surface of a doughnut in R 3. The geodesic is defined as a dynamical system on this doughnut. This dynamical system has a singularity when is an even multiple of φ and equilibrium when is an odd multiple of and is zero. φ π φ π We would like to ask for example How to deal with this singularity? Is the equilibrium stable, unstable or saddle?

24 It turns out that to deal with the singularity we can define state variables as x 1 = θ ; x 3 = φ ; and the geodesic is given by: ẋ 1 = x 2 tan (x 3 /2) ẋ 2 = x 2 x cot 2 (x 3 /2) ẋ 3 = x 4 tan (x 3 /2) x 2 = θ/ tan (φ/2) x 4 = φ/ tan (φ/2) ẋ 4 = x 2 2 sin (x 3 )tan(x 3 /2).5 x 4 2 sec 2 (x 3 /2) This dynamical system has a singularity when is an odd multiple of φ and equilibrium when is an even multiple of and is zero. φ π x 4 π

25 Thus there is no way to get out of singularity but one can define the geodesic on open charts that cover the whole space and switch between charts.

26 So far we have not considered any potential energy and also did not consider any external torque as a control input. If we consider the potential energy given by V (θ, φ) =A cos 2 (φ/2) The geodesic equations are modified as follows: θ + θ φ cot (φ/2) = csc 2 (φ/2) τ θ φ φ 2 sin (φ)+ 1 2 sin (φ) 2A =4τ φ

27 We can now wish to control the state from (θ 0, 0, φ 0, 0) to (θ 1, 0, φ 1, 0) in T units of time, while minimizing the control energy R T 0 h (τ θ (t)) 2 +(τ φ (t)) 2i dt We use the Hamiltonian H (z,λ) =λ ż 1 2 τ 2 θ + τ 2 φ and the Hamilton s equations: ż = H λ, λ = H z

28 We obtain a state space system given by: ż 1 ż 2 ż 3 ż 4 λ 1 λ 2 λ 3 λ 4 = z 2 z 2 z 4 cot (z 3 /2) + λ 2 csc 4 (z 3 /2) z 4 z 2 sin (z 3 )+2Acos (z 3 )+16λ 4 0 λ 1 + λ 2 z 4 cot (z 3 /2) 2λ 4 z 2 sin (z 3 ) λ 2 z 2 cot (z 3 /2) + λ 3 where : = 1 2 λ 2z 2 z 4 csc 2 (z 3 /2) + λ 2 2 cot (z 3/2) csc 4 (z 3 /2) λ 4 z λ 4A cos (z 3 )

29 The optimal controls are given by: τ θ = λ 2 sin 2 (z 3 /2) τ φ =4λ 4

30 An example of an optimal and a geodesic path

31 θ and φ as a function of time for the optimal path

32 The trajectories of the corresponding optimal torques

33 x0 = [1.5,.1e-3,3,.0001] V=0,-1,-5 e-5

34 x0 = [1.5,.1e-3,3,.001] V=0, -.1, -1, -10, -100 e-5

35 V=0

36 V=.1

37 V=1

38 V=10

39 V=100

40 V=1000

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