Iterative methods to compute center and center-stable manifolds with application to the optimal output regulation problem

Transcription

1 Iterative methods to compute center and center-stable manifolds with application to the optimal output regulation problem Noboru Sakamoto, Branislav Rehak N.S.: Nagoya University, Department of Aerospace Engineering, Graduate School of Engineering and B.R.: UTIA Academy of Sciences of the Czech Republic December 14, 2011 CDC /17

2 Outline of the talk To present a new method for computing center and center-stable manifolds - Iterative method - Suitable for computer implementation This method will be applied to the optimal output regulation - A constructive design method - No need to solve regulator equation - Center algorithm solves the regulator equation - Center-stable algorithm computes optimal controller 2/17

3 Setting of the problem Consider the following system of ordinary differential equations: ẋ = Ax + X(x, y, z) ẏ = By + Y(x, y, z) ż = Cz + Z(x, y, z) A R n x n x, Reλ(A) < 0 B R n y n y, Reλ(B) =0 C R n z n z, Reλ(C) > 0 The functions X, Y, Z are continuously differentiable X, Y, Z together with all of their first derivatives vanish at the origin. 3/17

4 Integral equations (Kelly 1966) The center manifold integral equation: t x(t) = e A(t s) X(x(s), y(s), z(s))ds t y(t) =e Bt y 0 + e B(t s) Y(x(s), y(s), z(s))ds z(t) = t 0 e C(t s) Z(x(s), y(s), z(s))ds These functions move on x = ϕ 1 (y), z = ϕ 2 (y) The center-stable manifold integral equation: x(t) =e At x 0 + y(t) =e Bt y 0 + z(t) = t t 0 t 0 e A(t s) X(x(s), y(s), z(s))ds e B(t s) Y(x(s), y(s), z(s))ds e C(t s) Z(x(s), y(s), z(s))ds These functions move on z = ψ(x, y) 4/17

5 Iterative algorithms In the center-stable manifold case: x k +1 y k +1 z k +1 (t, x 0, y 0 )= x 1 (t) =e At x 0, y 1 (t) =e Bt y 0, z 1 (t) =0 e At x 0 + t 0 ea(t s) X(x k (s), y k (s), z k (s)) ds e Bt y 0 + t 0 eb(t s) Y(x k (s), y k (s), z k (s)) ds t In the center manifold case x k +1 y k +1 z k +1 (t, y 0)= e C(t s) Z(x k (s), y k (s), z k (s)) ds x 1 (t) =0, y 1 (t) =e Bt y 0, z 1 (t) =0 t ea(t s) X(x k (s), y k (s), z k (s)) ds e Bt y 0 + t 0 eb(t s) Y(x k (s), y k (s), z k (s)) ds t e C(t s) Z(x k (s), y k (s), z k (s)) ds 5/17

6 Existence of limit Theorem i) There exists δ c > 0 such that for all y 0, y 0 <δ c the sequence converges pointwise to the solutions on x = ϕ 1 (y), z = ϕ 2 (y) (convergence to the center manifold). ii) There exists δ cs > 0 such that for all (x 0, y 0 ), (x 0, y 0 ) <δ cs the sequence converges pointwise to the solutions on z = ψ(x, y) (convergence to the center-stable manifold). 6/17

7 Optimal output regulation - equations Denote System: ẋ = f(x)+g(x)u, x(t) R n, f(0) =0 Exosystem: ẇ = s(w), w(t) R p, s(0) =0 Error (output) equation: e = h(x, w) Regulator equation: A = f h (0), B = g(0), C = (0, 0), x x S = s w (0), Q = h (0, 0). w π s(w) =f(π(w)) + g(π(w))σ(w), h(π(w), w) =0 w 7/17

8 Optimal output regulation - assumptions The exosystem is Lyapunov stable at w = 0, Poisson stable around w = 0 and all eigenvalues of S are purely imaginary. (A, B) is stabilizable, (C, A) is detectable The number of inputs the number of outputs (square) The system has well-defined relative degree (rel.deg. =1, det L g h(0, 0) 0) 8/17

9 The Hamiltonian H D : Optimal output regulation Cost function: J = e 2 + ė 2 dt H D = px T (f + gu)+pw T s(w)+1 h(x, w) L f h(x, w)+(l g h(x, w))u + L s h(x, w) 2, 2 The control vector ū minimizing H D : ū = (L g h) 1 { (L g h) T g(x) T p x + L f h + L s h }. The Hamilton-Jacobi equation is then { f g(lg h) 1 (L f h + L s h) } + pw T s(w) p T x 1 2 pt x g(l gh) 1 (L g h) T g T p x h(x, w) 2 = 0 9/17

10 Associated Hamiltonian System The Hamiltonian system is ẋ =(A B(CB) 1 CA)x B(CB) 1 QSw B(B T C T CB) 1 B T p x + N 1 (x, w, p x ) ẇ =Sw + N 2 (w) ṗ x =C T Cx C T Qw (A B(CB) 1 CA) T p x + N 3 (x, w, p x ) ṗ w = Q T Cx Q T Qw + S T Q T (B T C T ) 1 B T p x S T p w + N 4 (x, w, p x, p w ). Define the Hamiltonian matrix H as d dt [x, w, p x, p w ] T = H [x, w, p x, p w ] T +[N 1, N 2, N 3, N 4 ] T 10 / 17

11 Block diagonalization The linear regulator equation A Riccati equation Lyapunov equation ΠS = AΠ+BΣ, CΠ+Q = 0 PĀ + Ā T P PR B P + C T C = 0; Ā = A B(CB) 1 CA, R B = B(B T C T CB) 1 B T where VA c + Ac T V = B(BT C T CB) 1 B T A c = A B(CB) 1 CA B(B T C T CB) 1 B T P Linear symplectic coordinate transformations T 1 = I Π I I Π T I, T 2 = I 0 V 0 0 I 0 0 P 0 PV + I I 11 / 17

12 New Hamiltonian system [x T, w T, p T x, p T w ] T = T 1 2 T 1 1 [xt, w T, p T x, p T w ] T Hamiltonian system with block-diagonalized linear part: ẋ = A c x + N1 (x, w, p x ) ẇ = Sw + N2 (w ) ṗ x = A c T p x + N3 (x, w, p x ) ṗ w = ST p w + N4 (x, w, p x, p w ) stable center1 unstable center2 Center manifolds: x = π 1 (w ), p x = π 2 (w ) CM algorithm computes π 1, π 2 Regulator equation Center-stable manifold p x = π 3 (x, w ) C-SM algorithm computes π 3 Feedback controller u = u(w, x) 12 / 17

13 A numerical example Consider the example with unstable linearization ( ) ( )( ) ( ẋ x1 x 3 = + 1 ẋ x 2 2x x3 2 exosystem: ẇ = 0 ) ( ) u The goal is to design a feedback law u = u(x, w) that achieves x 1 = w as t in an optimal way: J = (x 1 w) 2 +(ẋ 1 ) 2 dt 13 / 17

14 A numerical example Hamiltonian system: ẋ 1 x 1 x 3 1 ẋ 2 x 2 2x x3 2 ẇ w 0 = H + ṗ 1 p 1 6x 2 1 p 2 ṗ 2 p 2 3x 2 2 p 2 ṗ w p w 0 ; H = Block-diagonalization center-stable manifold algorithm (Matlab) back to the original coordinates feedback controller u(x, w) 14 / 17

15 Outline Iterative methods Output regulation Numerical example Simulation results reference (w(0)) = 0.2 reference (w(0)) = 0.2 Linear controller response 15 / 17

16 Conclusions New methods for computing center and center-stable manifolds Iterative algorithms, Matlab codes Applications to the optimal output regulation problem Center-stable algorithm directly computes feedback law One does not need to solve the regulator equation 16 / 17

17 Controller expression u = (L g h) { 1 (L g h) T g(x) T p x + L f h + } L s h p x1 (x, w) =1.2289x e 5 x x x e 5 x 1 x x 2 1 x e 5 x x 1x x w x 1 w x 2 1 w x 2w x 1 x 2 w x 2 2 w e 5 w x 1 w x 2 w w 3 p x2 (x, w) = x x x x x 1 x x 2 1 x x x 1x x w x 1 w x 2 1 w x 2w x 1 x 2 w x 2 2 w w x 1 w x 2 w w 3 17 / 17