CIMPA Summer School on Current Research on Finite Element Method Lecture 1. IIT Bombay. Introduction to feedback stabilization

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1 1/51 CIMPA Summer School on Current Research on Finite Element Method Lecture 1 IIT Bombay July 6 July 17, 215 Introduction to feedback stabilization Jean-Pierre Raymond Institut de Mathématiques de Toulouse

2 2/51 Q1. Controllability. How to drive a nonlinear system from a given initial state to another prescribed state by the action of some control during the time interval (, T )? Q2. Stabilizability. How to maintain a system at an unstable position, in the presence of perturbations by using some measurements to estimate the state at each time t, and by using the estimated state in a control law?

3 3/51 Plan of Lecture 1 1. Introduction 1.1. Problems and models. The inverted pendulum A finite dimensional model Two infinite dimensional models. Fluid flows in a neighbourhood of unstable solutions. A Burgers type equation 1.2. Feedback stabilization with total and partial information 2. Finite dimensional linear systems 2.1. The Duhamel formula 2.2. Controllability and Reachability 2.3. Stabilizability and its characterization 2.4. Construction of a Feedback 3. Algorithms for solving Riccati equations

4 4/51 1. Introduction 1.1. Models 1.1.i. The inverted pendulum m l u M The nonlinear system (M + m)x + m l θ cos θ m l θ 2 sin θ = u, m l θ + m x cos θ = m g sin θ.

5 5/51 M is the mass of the car, m is the mass of the pendulum, x(t) is the position of the car at time t, θ(t) is the angle between the unstable position and the pendulum measured clockwise, u is a force applied to the car. The Lagrance equations read as ( ) d L dt x L x = u, ( ) d L dt θ L θ =, where the Lagrangian L = T V, T is the kinetic energy and V is the potential energy. L = 1 2 (M + m) x 2 + m l x θ cos θ m l2 θ 2 mgl cos θ.

6 6/51 x mg sin θ cos θ = M + m sin 2 θ + ml M + m sin 2 θ θ 2 1 sin θ + M + m sin 2 θ u, θ m = M + m sin 2 θ θ 2 M + m sin θ cos θ + g Ml + ml sin 2 θ sin θ cos θ Ml + ml sin 2 θ u. As a first order system, we write x d x = F(x, x, θ, θ )+ dt θ θ 1 M + m sin 2 θ cos θ Ml + ml sin 2 θ u = F (x, x, θ, θ )+G(θ) u,

7 7/51 with F(x, x, θ, θ ) = mg sin θ cos θ M + m sin 2 θ + ml M + m θ 2 sin θ θ m M + m sin 2 θ θ 2 M + m sin θ cos θ + g Ml + ml sin 2 θ sin θ x.

8 8/51 We linearize F about (,,, ), we write F (x, x, θ, θ ) = F (,,, ) + DF(,,, ) x x θ θ + Ñ(θ, θ ), with F (,,, ) = (,,, ) T and 1 mg M DF (,,, ) = 1 g(m + m) Ml.

9 9/51 Similarly with G(θ)u = G()u + R(θ)u, G() = 1 M. 1 Ml

10 1/51 Setting z = (x, x, θ, θ ) T, we have z = Az + Bu + N(z) + R(z)u, z() = z, where A = 1 mg M 1 g(m + m) Ml, B = 1 M 1 Ml, z = (x, x 1, θ, θ 1 ) T and the two nonlinear terms N and R obey N() = and R() =.

11 11/ ii. Fluid flows in a neighbourhood of unstable solutions We consider a fluid flow governed by the N.S.E. (w and q are the volocity and the pressure) Given an unstable stationary solution w s. Find a boundary control u able to stabilize w exponentially when w() = w s + z. We consider the control of the wake behind a circular cylinder, or behind a square.

12 12/51 Control of the wake behind an obstacle Re = u s D/ν 5 < Re < 5 A fixed pair of vortices 5 < Re < 15 Vortex street

13 3/51 Flow around a cylinder with an outflow boundary condition Γ o Γ i Γ c Γ n or Γ o ν z n pn = on Γ n ( ) σ(z, p)n = ν z + ( z) T n pn = on Γ n.

14 14/51 The unstable stationary solution w s of the N.S.E. ν w s + (w s )w s + q s =, in Ω, div w s = in Ω, w s = u s on Γ i, + Other B.C. on Γ \ Γ i. The stabilization problem Find u s.t. w(t) w s Z as t, w t ν w + (w )w + q =, div w = in Q, w = u s on Σ i = Γ i (, ), w = u on Σ c = Γ c (, ), We choose + Other B.C. on Σ \ (Σ i Σ c ), w() = w in Ω. supp u Γ c.

15 5/51 Set z = w w s, p = q q s. The linearized (resp. nonlinear) equation is z t ν z + (w s )z + (z )w s + (z )z + p =, div z = in Q, z = u on Σ d, + Other B.C. on Σ \ Σ d, z() = z in Ω. with supp u Γ c.

16 6/ iii. A simplified model A Burgers type equation w t w + i (w 2 ) = in Q, w = u + g on Σ, z() = z in Ω. Let w s be the stationnary solution to w s + i (w 2 s ) = in Ω, w s = g on Γ. The L. E. and the nonlinear equation satisfied by z = w w s are z t z + 2 iw s z + 2 i z w s + i (z 2 ) =, z = u on Σ, z() = z = w w s in Ω.

17 1.2. Feedback stabilization 1.2.i Representation of systems The model of the inverted pendulum can be written in the form z = Az + Bu + N(z) + R(z)u, z() = z, where z = (x, x, θ, θ ), the two nonlinear terms N and R obey N() =, N () =, and R() =. The Burgers eq. can be written in the form z = Az + Bu + F(z), z() = z, where z = w w s, w is the velocity of the instationary flow, while w s is the stationary solution. The Navier-Stokes equations can be written in the form Πz = AΠz + Bu + F(z), Πz() = Πz, (I Π)z = (I Π)Lu. 7/51 The Leray projector Π is used to eliminate the pressure p in the equations.

18 8/51 Thus we have to deal with linear systems either of the form or of the form z = Az + Bu, z() = z, Πz = AΠz + Bu, Πz() = Πz, (I Π)z = (I Π)Lu. This last system is called a descriptor system. The approximate system obtained by a F.E.M. of the L.N.S.E. will be of the same type.

19 9/ ii. The open loop stabilization problem For these models, the problem consists in finding a control u L 2 (, ; U) such that z z,u(t) Z Ce ωt z Z, C >, ω >, provided that z Z is small enough. Existence of a stabilizing control. Except for finite dimensional systems, there is no general result for the stabilizability of nonlinear systems. Stabilization of the linearized model. Similar problems can be studied for the linearized model z = Az + Bu, z() = z. If we find a control u L 2 (, ; U) able to stabilize the linearized system (e.g. by solving an optimal control problem), there is no reason that u also stabilizes the nonlinear system.

20 2/ iii. Stabilization by feedback with full information One way for finding a control able to stabilize the nonlinear system is to look for a control stabilizing the linear system in feedback form, that is such that u(t) = K z z,u(t), K L(Z, U). For that we consider the optimal control problem (P) Minimize J(z, u) = 1 2 z = Az + Bu, z() = z, Cz(t) 2 Y dt where C L(Z, Y ), R = R >, R L(U). The solution to (P) is obtained by z = Az + BKz, z() = z, K = B P, ( ) Ru(t), u(t) U dt P L(Z ), P = P, A P + PA PBR 1 B P + CC =.

21 21/51 The optimal cost is J(z uopt, u opt ) = 1 2 ( Pz, z ) Z. The issues What are the conditions on (A, B) so that the above control problem has a solution? How to determine the operator K? How to choose the control problem so that the closed loop nonlinear system z = Az + BKz + F (z), z() = z, admits a solution exponentially decreasing? Is this programme useful for pratical applications?

22 22/ iv. Stabilization by feedback with partial information If the feedback K also stabilizes the nonlinear system, by solving z = Az + BKz + F (z) + µ, z() = z + µ, we know a control u(t) = Kz cln,z (t) stabilizing the nonlinear system. But in pratice µ is not necessarily known, we can also have other types of uncertainties µ. This is why we have to look for an estimation z e of the state z in function of measurements, and use the control law u(t) = K z e (t). model error noise z NS or Burgers system z measures y obs u K z e estimator L

23 23/51 The complete problem: estimation + feedback control fluid flow suction/blowing sensor Von Karman vortex sensor suction/blowing

24 24/51 2. Finite Dimensional Linear Systems 2.1. The Duhamel formula When A R n n, B R n m, the solution to system is defined by z = Az + Bu, z() = z, t (E) z z,u(t) = z(t) = e ta z + e (t s)a Bu(s) ds. The same formula holds true in C n. If (λ i ) 1 i r are the complex eigenvalues of A, we can define E(λ j ) = Ker(A λ j I), dim E(λ j ) = l j = geometric multiplicity of λ j, G(λ j ) = Ker((λ j I A) m(λ j ) ), the generalized eigenspace ass. to λ j, dim G(λ j ) = d(λ j ) = algebraic multiplicity of λ j.

25 25/51 We can decompose R n into real generalized eigenspaces R n = r j=1 G R(λ j ), G R (λ j ) = G R ( λ j ) = vec{reg(λ j ), ImG(λ j )}, AG R (λ j ) G R (λ j ), A = QΛQ 1, Λ 1 Λ 2 Λ =... Λ r, Λ j = Jj 1 J 2 j... J l j j, and e ta = Qe tλ Q 1 = Q e tλ 1 e tλ 2... e tλr Q 1.

26 6/51 In particular We can rewrite the equation in the form that is e ta G R (λ j ) G R (λ j ). z = Az + Bu, z() = z, Q 1 z = Q 1 AQQ 1 z + Q 1 Bu, Q 1 z() = Q 1 z, ζ = Λ ζ + Bu, ζ() = ζ, where ζ = Q 1 z, B = Q 1 B, ζ = Q 1 z. The above system is equivalent to ζ i = Λ i ζ i + B i u, ζ i () = ζ,i, 1 i r.

27 27/ Controllability and Reachability of Finite Dimensional Systems In this part, Z = R n or Z = C n and U = R m or U = C m. We make the identifications Z = Z and U = U. The operator L T : L 2 (, T ; U) Z L T u = T The reachable set from z at time T e (T s)a Bu(s) ds. R T (z ) = e TA z + Im L T. Exact controllability. The pair (A, B) is exactly controllable at time T if R T (z ) = Z for all z Z. Reachability. A state z T is reachable from z = at time T < if there exists u L 2 (, T ; U) such that L T u = z,u (T ) = T e (T s)a Bu(s) ds = z T.

28 8/51 The system (A, B) is reachable at time T < iff Im L T = Z. A finite dimensional system is reachable at time T iff it is controllable at time T. The system is reachable (or exactly controllable) at time T < iff the matrix (called controllability Gramian ) is invertible. T WA,B T = e ta BB e ta dt Idea of the proof. L T is surjective iff its adjoint operator L T L(Z, L2 (, T ; U)), (L T φ) ( ) = B e (T )A φ, is injective. This last condition is equivalent to the existence of α > such that T B e sa φ 2 U ds = L T φ 2 L 2 (,T ;U) α φ 2 Z, φ Z.

29 9/51 Assume that the system (E) is reachable at time T. For a given z T Z, the control is such that Moreover and for all u such that ū(t) = B e (T t)a (W T A,B ) 1 e TA z T z,ū (T ) = T e (T s)a Bū(s) ds = z T. ū 2 L 2 (,T ;U) = ((W T A,B ) 1 z T, z T ) Z z,u (T ) = ū L2 (,T ;U) u L2 (,T ;U), T e (T s)a Bu(s) ds = z T.

30 /51 Finite Dimensional System. The system (A, B) is controllable at time T if and only if one of the following conditions is satisfied. T WA,B T = e ta BB e ta dt >, T >, rank [ B AB... A n 1 B ] = n, λ C, rank [A λi B] = n, λ C, A φ = λ φ and B φ = φ =, K L(Z, U), σ(a + BK ) can be freely assign (with the cond. that complex eigenvalues are in conjugate pairs).

31 31/51 Applications. Let us study the controllability of the linearized inverted pendulum. 1 mg 1 M M A =, B =, 1 g(m + m) 1 Ml Ml The controllability matrix is [B AB A 2 B A 3 B] = 1 M 1 M mg M 2 l mg M 2 l 1 g(m+m) M M 2 l 2 1 g(m + m) M M 2 l 2.

32 32/51 Its determinent is det[b AB A 2 B A 3 B] g(m + m) = M 3 l 2 + mg M 3 l 2 ( g(m + m) M 3 l 2 + mg M 3 l 2 ( g(m + m) M 3 l 2 + mg M 3 l 2 ) ) = g2 M 4 l 4. The linearized inverted pendulum is controllable at any time T >.

33 2.3. Stabilizability of F.D.S. 2.3.i. Open loop stabilizability. System (A, B) is open loop stabilizable in Z when for any initial condition z Z, there exists a control u L 2 (, ; U) s.t. z z,u(t) 2 Z dt <. Stabilizability by feedback. System (A, B) is stabilizable by feedback when there exists an operator K L(Z, U) s. t. A + BK is exponentially stable in Z. Open loop stabilizability is equivalent to stabilizability by feedback for finite dimensional systems or for parabolic systems with Dirichlet boundary conditions like the heat equation, the linearized Burgers equation, the Stokes equation, the linearized Navier-Stokes equations. 3/51 The precise assumptions are the semigroup generated by (A, D(A)) on Z is analytic, B L(U, (D(A )) )

34 34/ ii. Characterization of the stabilizability of F.D.S. The finite dimensional case. System (E) is stabilizable if and only if one of the following conditions is satisfied. (i) λ, Reλ, rank [A λi B] = n, (ii) λ, Reλ, A φ = λ φ and B φ = φ =, (iii) λ, Reλ, Ker(λI A ) Ker(B ) = {}, (iv) K L(Z, U), σ(a + BK ) is stable. Conditions (ii), (iii), (iv) are equivalent to the stabilizability of (E) for Infinite Dimensional Systems under the previous conditions on (A, B) (analyticity, compactness, degree of unboundness of B).

35 5/51 FDS Example of a stabilizing feedback. The system (A, B) is controllable iff T W A,B T = e ta BB e ta dt, is invertible for all T >. Indeed e TA W T A,B eta = T e (T t)a BB e (T t)a dt = Assume that (A, B) is controllable, then is a stabilizing feedback. Idea of the proof. The mapping K = B (W T A,B ) 1 z ((W T A,B ) 1 z, z) Z T is a Lyapunov function of the closed loop linear system. e τa BB e τa dτ = W T A,B.

36 36/51 If W A,B T is invertible a fortiori W A,B is also invertible (W A,B W A,B T ). Link between this feedback and the Riccati approach where (W A,B ) 1 = P, P L(Z ), P = P >, A P + PA PBB P =. The control is that of minimal norm.

37 37/51 FDS Characterization of the stabilizability in terms of Gramians. We assume that Z = Z s Z u, Z u = Nu j=1 G R(λ j ), Z s = r j=n u+1 G R(λ j ), Reλ j > if 1 j N u, Reλ j < if N u + 1 j r. Recall that We also have e ta Z u Z u and e ta Z s Z s. Z = Z = Z s Z u, Z u = Nu j=1 G R (λ j), Z s = r j=n u+1 G R (λ j), e ta Z u Z u and e ta Z s Z s.

38 38/51 Let π u the projection onto Z u along Z s and set π s = I π u. The system π u z = z u = A u z u + π u Bu, π s z = z s = A s z s + π s Bu, can be written as a system in Z Z of the form (using matrix notation) ( ) ( ) ( ) ( ) ( ) ( ) z u Au zu πu Bu zu () πu z = +, =. A s z s π s Bu z s () π s z z s If the system (A, B) is stabilizable then the system (A u, B u ) = (π u A, π u B) is also stabilizable. The converse is true. Assume that (A u, B u ) = (π u A, π u B) is stabilizable and let K L(Z u, U) be a stabilizing feedback. Then the system ( ) ( ) ( ) z u Au + B u K zu = B s K is also stable. z s A s z s

39 39/ Characterization of the stabilizability of F.D.S. When σ(a), the following conditions are equivalent (a) (A, B) is stabilizable, (b) (A u, B u ) = (π u A, π u B) is stabilizable, (c) The Gramian W A u,b u = is invertible. e tau B u B ue ta u (d) There exists α > such that for all φ Zu, (O.I.) (W A u,b u φ, φ) Z = Bue ta u 2 φ dt U α φ 2 Z. The operator P u = (W A u,b u ) 1 L(Z u, Zu ), P u = Pu, provides a stabilizing feedback A u B u BuP u is exponentially stable. dt

40 4/51 Recall that B u = π u B and B u = B π u, where π u is the projection onto Z u along Z s. The operator P u satisfies the following Algebraic Bernoulli equation (a degenerate Algebraic Riccati equation) If we set P u A u + A up u P u B u B up u =. P = π up u π u. Then P L(Z ) is such that P = P and solves the following (A.B.E.) A.B.E. P L(Z ), P = P, A P + PA PBB P =, A BB P generates an exponentially stable semigroup.

41 1/51 Moreover, the feedback B P provides the control of minimal norm in L 2 (, ; U) u(t) = B Pe t(a BB P) z, and the spectrum of A BB P is σ(a BB P) = { Reλ + i Imλ λ σ(a), Reλ > } {λ λ σ(a), Reλ < }. To obtain a better exponential decay we can replace A by A + ωi and determine the corresponding feedback B P ω. In that case the spectrum of A BB P ω is as follows

42 42/51 Spectrum of A and of A BB P ω ω ω

43 43/51 4. An algorithm for solving Riccati equations We consider the A.R.E. A.R.E. P L(Z ), P = P, A P + PA PBR 1 B P + C C =, A BR 1 B P generates an exponentially stable semigroup, where the unknown P belongs to L(R n ), A L(R n ), B L(R m ; R n ), where R m is the discrete control space, C L(R n ; R l ), where R l is the observation space. We make the following assumption. (H 1 ) The pair (A, B) is stabilizable (H 2 ) (The pair (A, C) is detectable or) A has no eigenvalue on the imaginary axis. Equation (A.R.E) admits a unique solution. The matrix [ H = A CC ] BR 1 B A is called the Hamiltonian matrix associated with equation (A.R.E.).

44 44/51 It satisfies [ I I ] 1 [ I H I ] = H. Therefore the matrix H and H are similar and they have the same eigenvalues. On the other hand H and H have also the same set of eigenvalues. Thus if λ is an eigenvalue of H, then λ is also an eigenvalue of H with the same multiplicity. Let us denote by λ 1, λ 2,..., λ n, λ 1, λ 2,..., λ n, the eigenvalues of H, where Re(λ i ) for i = 1,..., n. Under assumptions (H 1 ) and (H 2 ), it can be shown that the matrix H has no pure imaginary eigenvalues, that is Re(λ i ) > for i = 1,..., n. There exists a matrix V whose columns are eigenvectors, or generalized eigenvectors of H, such that [ ] V 1 J HV =, J

45 45/51 where J is composed of Jordan blocks corresponding to eigenvalues with negative real part, and [ ] V11 V V = 12, V 21 V 22 is such that V = [ V11 V 21 ], is a matrix whose columns are eigenvectors corresponding to eigenvalues with negative real parts. It may be proved that V 11 is nonsingular and the unique positive semidefinite solution of equation (A.R.E.) is given by P = V 21 V 1 11.

46 46/51 A very simple example Set ( ) ( ) λ 1 A =, B =, λ 1 with λ >, R 1 = I, and C =. Then λ 1 λ 1 H =. λ 1 λ In that case (C = ), we replace the condition (A, C) is detectable by the condition A BB P is stable.

47 7/51 We have HV 1 = λv 1 with V 1 = ( 1, 2λ,, 4λ 2 ) T. The solution to (H + λi)v 2 = V 1, is V 2 = ( 1/λ, 1, 4λ 2, ) T. We obtain the solution to the (A.B.E.) (the degenerate (A.R.E.)) ( ) ( ) 4λ /λ P =, 4λ 2 2λ 1 ( ) ( ) ( ) 4λ 2 1 1/λ 8λ 3 4λ 2 P = =. 4λ 2 2λ 1 4λ 2 4λ

48 48/51 Moreover ( λ 1 A BB P = 4λ 2 3λ ).

49 9/51 Another simple example A reduced inverted pendulum Instead of studying the equations of the inverted pendulum, we can consider the simple model θ sin θ = u, where θ is the angular displacement from the unstable vertical equilibrium, and u is taken as a control. The linearized system about is z = Az + Bu, z() = z, where z = (θ, ζ) T, A = ( 1 1 ), B = ( We choose R 1 = I and C =. Then H = ), ζ = θ. λ = 1 and λ = 1 are the two eigenvalues of H of multiplicity 2.

50 5/51 Rather than computing the matrix Riccati equation, we can equivalently determine directly the control of minimal norm stabilizing the system. The eigenvalues of A are λ 1 = 1 and λ 2 = 1. They are both of multiplicity 1. We have E(λ 1 ) = {(θ, ζ) R 2 θ = ζ} and E(λ 2 ) = {(θ, ζ) R 2 θ = ζ}. We can rewrite the system as follows ( ) ( θ + ζ θ + ζ = 2 2 ( ) ( ζ θ ζ θ = 2 2 ) u, ) u. The first equation corresponds to the projected system onto the unstable subspace. The feedback of minimal norm stabilizing the unstable system is obtained by solving the one dimensional Riccati equation p >, 2p p 2 /4 =.

51 1/51 Thus p = 8 and the feedback law is u(t) = ( θ + ζ 2 ) = 2(θ + ζ). Thus the closed loop linear system is ( ) ( θ = 1 ζ 1 2 ) ( θ ζ ). We notice that the two eigenvalues of the generator of this system are λ = 1.