On Linear-Quadratic Control Theory of Implicit Difference Equations

Transcription

1 On Linear-Quadratic Control Theory of Implicit Difference Equations Daniel Bankmann Technische Universität Berlin 10. Elgersburg Workshop 2016 February 9, 2016 D. Bankmann (TU Berlin) Control of IDEs February 9, / 18

2 Linear IDEs Consider the IDE (implicit difference equation) Eσx j = Ax j + B, Ex 0 = Ex 0, (1) σ denotes the shift operator, i. e., σx j = x j+1, (x j ) (K n ) N0, ( ) (K m ) N0 are some sequences. x 0 is some consistent shift variable. The IDE is assumed to be regular, i. e., det(ze A) 0. Given x 0, minimize J σ (x, u) := j=0 [ xj ] [ Q ] [ ] S xj S R subject to (1) with Q = Q, R = R and lim j Ex j = 0. D. Bankmann (TU Berlin) Control of IDEs February 9, / 18

3 Motivation Optimal control problems have a variety of applications, e. g., Trajectory-following problems, e. g., in Robotics Reaching a goal with minimal energy consumption Discrete-time models are obtained by discretization methods from the continuous-time models. D. Bankmann (TU Berlin) Control of IDEs February 9, / 18

4 Discretized Stokes Problem Solve [ ] M 0 σ 0 0 [ ] v = p [ K + M D T D 0 ] [ ] v + p [ ] B u, Mv 0 0 = Mv 0. v describes the velocity p describes the pressure M is the mass matrix K is the stiffness matrix Minimize, j=0 i. e., Q = 0, S = 0, R = I m, such that lim j Mv j = 0. D. Bankmann (TU Berlin) Control of IDEs February 9, / 18

5 Discrete Time Algebraic Riccati Equations Solve A XA X + Q (A XB + S)(B XB + R) 1 (B XA + S ) = 0 for Hermitian X fulfilling B XB + R 0. Under some circumstances invertibility of B XB + R is not guaranteed, e. g., when R is indefinite (e. g., differential games) Can be rewritten as [ A XA X + Q A ] [ ] XB + S K [K ] B XA + S B = L. XB + R L D. Bankmann (TU Berlin) Control of IDEs February 9, / 18

6 Discrete Time Algebraic Riccati Equations Solve A XA X + Q (A XB + S)(B XB + R) 1 (B XA + S ) = 0 for Hermitian X fulfilling B XB + R 0. Under some circumstances invertibility of B XB + R is not guaranteed, e. g., when R is indefinite (e. g., differential games) In the IDE case [ A XA E X E + Q A XB + S ] [ ] K [K ] L, B XA + S B XB + R = V Σ L where V Σ denotes the system space, i. e., the space where all solutions (x, u) evolve. D. Bankmann (TU Berlin) Control of IDEs February 9, / 18

7 Discrete Time Algebraic Riccati Equations Solve A XA X + Q (A XB + S)(B XB + R) 1 (B XA + S ) = 0 for Hermitian X fulfilling B XB + R 0. Under some circumstances invertibility of B XB + R is not guaranteed, e. g., when R is indefinite (e. g., differential games) In the IDE case [ (V Σ ) A XA E XE + Q A ] [ ] XB + S B XA + S B V Σ K [K ] = L, XB + R L where V Σ is a basis matrix of V Σ. D. Bankmann (TU Berlin) Control of IDEs February 9, / 18

8 Discrete-Time KYP Lemma Solve (discrete-time) Kalman-Yakubovich-Popov (KYP) inequality [Kalman 1963; Popov 1961; Yakubovich 1962] [ A PA E PE + Q A ] PB + S B PA + S B PB + R V Σ 0, P = P Popov function [ (ze A) Φ(z) := 1 ] [ ] [ B Q S (ze A) 1 ] B S K m m (z), R I m where G(z) := G(z 1 ) I m D. Bankmann (TU Berlin) Control of IDEs February 9, / 18

9 Discrete-Time KYP Lemma Theorem (KYP Lemma for IDEs) Let (E, A, B, Q, S, R) Σ w m,n(k) and the system space V Σ be given with corresponding Popov function Φ(z) K m m (z). If there exists some P K n n that is a solution of the KYP inequality, then Φ(e iω ) 0 for all ω R with det(e iω E A) 0. If on the other hand (E, A, B) is R-controllable and Φ(e iω ) 0 for all ω R with det(e iω E A) 0, then there exists a solution P K n n of the KYP inequality. R-controllability R-controllability means controllability on the set of reachable states R. In the IDE case R K n. D. Bankmann (TU Berlin) Control of IDEs February 9, / 18

10 Lur e Equations For q := rk K(z) Φ(z) find X = X K n n, K K q n, and L K q m such that ] [ ] K [K ] L. [ A XA E XE + Q A XB + S B XA + S B XB + R = V Σ L If X is a solution of the KYP inequality, we can always find K K p n and L K p m solving the Lur e equation, p q. From now on E = I n, i. e., solve [ A XA X + Q A ] [ ] XB + S K [K ] B XA + S B = L. XB + R L D. Bankmann (TU Berlin) Control of IDEs February 9, / 18

11 Construction of Deflating Subspace Consider the pencil 0 I n 0 0 A B ze A = z A 0 0 I n Q S K 2n+m 2n+m [z]. B S R Find deflating subspace, i. e., Y K 2n+m n+m, Z K 2n+m n+q, Ẽ, Ã K n+q n+m with rk K(z) (zẽ Ã) = n + q such that (ze A)Y = Z(zẼ Ã). D. Bankmann (TU Berlin) Control of IDEs February 9, / 18

12 Construction of Deflating Subspace Consider the pencil 0 I n 0 0 A B ze A = z A 0 0 I n Q S K 2n+m 2n+m [z]. B S R Find deflating subspace, i. e., Y K 2n+m n+m, Z K 2n+m n+q, Ẽ, Ã K n+q n+m with rk K(z) (zẽ Ã) = n + q such that (ze A)Y = Z(zẼ Ã). Ansatz: Y µ X 0 Z µ I n 0 Y = Y x = I n 0, Z = Z x = A X K, Y u 0 I m Z u B X L and [ ] zin A B zẽ Ã =. K L D. Bankmann (TU Berlin) Control of IDEs February 9, / 18

13 Application to Infinite Horizon Optimal Control Suppose, we have a solution (X, K, L) of the Lur e equation, j 2 j 1 and optimal (x, u) with lim j x j = 0. Then x j 2 Xx j2 x j 1 Xx j1 j 2 1 = σ(xk Xx k) xk Xx k, σx k = x k+1 k=j 1 j 2 1 [ ] [ xj A = XA X A ] [ ] XB xj B XA B XB k=j 1 D. Bankmann (TU Berlin) Control of IDEs February 9, / 18

14 Application to Infinite Horizon Optimal Control Suppose, we have a solution (X, K, L) of the Lur e equation, j 2 j 1 and optimal (x, u) with lim j x j = 0. Then x0 Xx 0 = σ(xk Xx k) xk Xx k, σx k = x k+1 = k=0 [ xj k=0 ] [ A XA X A ] [ ] XB xj B XA B XB D. Bankmann (TU Berlin) Control of IDEs February 9, / 18

15 Application to Infinite Horizon Optimal Control Suppose, we have a solution (X, K, L) of the Lur e equation, j 2 j 1 and optimal (x, u) with lim j x j = 0. Then x0 Xx 0 = σ(xk Xx k) xk Xx k, σx k = x k+1 = = = k=0 [ xj k=0 [ xj k=0 [ xj k=0 ] [ A XA X A ] [ ] XB xj B XA B XB ] ([ A XA X + Q A XB + S B XA + S B XB + R ] [ ] K [K ] [ x L j L ] [ ]) [ ] Q S xj S R ] J σ (x, u) = Kx + Lu 2 l J σ (x, u) 2 D. Bankmann (TU Berlin) Control of IDEs February 9, / 18

16 Application to Infinite Horizon Optimal Control Suppose, optimal (x, u) fulfills Kx + Lu 2 l 2 = 0. Using the deflating subspace it holds that 0 σi n A B X 0 [ ] σa I n Q S I n 0 xj σb S u R 0 I j m [ ] [ ] σin A B xj = Z, Z K 2n+m n+q. K L D. Bankmann (TU Berlin) Control of IDEs February 9, / 18

17 Application to Infinite Horizon Optimal Control Suppose, optimal (x, u) fulfills Kx + Lu 2 l 2 = 0. Using the deflating subspace it holds that 0 σi n A B X 0 [ ] σa I n Q S I n 0 xj σb S u R 0 I j m [ ] [ ] σin A B xj = Z = 0, Z K 2n+m n+q. K L D. Bankmann (TU Berlin) Control of IDEs February 9, / 18

18 Application to Infinite Horizon Optimal Control Suppose, optimal (x, u) fulfills Kx + Lu 2 l 2 = 0. Using the deflating subspace it holds that 0 σi n A B X 0 [ ] σa I n Q S I n 0 xj σb S u R 0 I j m [ ] [ ] σin A B xj = Z = 0, Z K 2n+m n+q. K L Thus, µ j := Xx j solves the boundary value problem 0 I n 0 µ 0 A B µ A 0 0 σ x = I n Q S x, B 0 0 u 0 S R u x 0 = x 0, lim µ j = 0. j D. Bankmann (TU Berlin) Control of IDEs February 9, / 18

19 Palindromic Matrix Pencils Reformulation of the BVP [Schröder 2008] with yields m j := µ j µ j+1 0 I n 0 m 0 A B m A Q S σ x = I n Q S x, B S R u 0 S R u x 0 = x 0, m j = µ j with corresponding palindromic matrix pencil 0 I n 0 0 A B za A = z A Q S I n Q S K 2n+m 2n+m [z]. B S R 0 S R j=0 D. Bankmann (TU Berlin) Control of IDEs February 9, / 18

20 Deflating Subspace for Palindromic Matrix Pencils Find deflating subspace Y of 0 I n 0 0 A B za A = z A Q S I n Q S B S R 0 S R given a solution of [ A XA X + Q A ] [ ] XB + S K [K ] B XA + S B = L, XB + R L i. e., (za A)Y = Z(zẼ Ã) for Ẽ, Ã K n+q n+m. D. Bankmann (TU Berlin) Control of IDEs February 9, / 18

21 Deflating Subspace for Palindromic Matrix Pencils Find deflating subspace Y of 0 I n 0 0 A B za A = z A Q S I n Q S B S R 0 S R given a solution of [ A XA X + Q A ] [ ] XB + S K [K ] B XA + S B = L, XB + R L i. e., (za A)Y = Z(zẼ Ã) for Ẽ, Ã K n+q n+m. ] Y = does not work due to the variable transformation. [ X 0 I n 0 0 I m D. Bankmann (TU Berlin) Control of IDEs February 9, / 18

22 Deflating Subspace for Palindromic Matrix Pencils From µ j = Xx j = X [ I n 0 ] [ ] x j it follows that m j = µ j µ j+1 = X (x j+1 x j ) = X [ A I n B ] [ ] x j. D. Bankmann (TU Berlin) Control of IDEs February 9, / 18

23 Deflating Subspace for Palindromic Matrix Pencils From µ j = Xx j = X [ I n 0 ] [ ] x j it follows that m j = µ j µ j+1 = X (x j+1 x j ) = X [ A I n B ] [ ] x j. D. Bankmann (TU Berlin) Control of IDEs February 9, / 18

24 Deflating Subspace for Palindromic Matrix Pencils From µ j = Xx j = X [ I n 0 ] [ ] x j it follows that m j = µ j µ j+1 = X (x j+1 x j ) = X [ A I n B ] [ ] x j. Leads to the deflating subspace Y µ X (A I n ) XB Y = Y x = Y u, Z = I n 0 0 I m Z µ Z x Z u = I n 0 (I n A )X K, B X L and [ ] zẽ Ã = zin A B. (z 1)K (z 1)L D. Bankmann (TU Berlin) Control of IDEs February 9, / 18

25 Deflating Subspace for Palindromic Matrix Pencils Theorem Let (I n, A, B, Q, S, R) Σ w m,n(k) and consider the palindromic pencil za A. Further, let q = rk K(z) Φ(z) and assume that rk [ I n A B ] = n. Then the following are equivalent: There exists a solution of the (discrete time) Lur e equation. The KYP [ ] inequality [ has a solution and there exist Yµ Zµ Y =, Z = Y x Y u Z x Z u the matrix [ I n A ],Ẽ, Ã, such that (ze A)Y = Z(zẼ Ã) and B ] [ ] Y x has full row rank n; Y u the space Y = im Y is the largest space such that Y (A A)Y = 0. D. Bankmann (TU Berlin) Control of IDEs February 9, / 18

26 Deflating Subspace for IDEs In the IDE case find deflating subspace Y of 0 E 0 0 A B za A = z A Q S E Q S B S R 0 S R given a solution of [ A XA E XE + Q A XB + S ] [ ] K [K ] L, B XA + S B XB + R = V Σ L i. e., (za A)Y = Z(zẼ Ã) for Ẽ, Ã K n+q n+m. D. Bankmann (TU Berlin) Control of IDEs February 9, / 18

27 Deflating Subspace for IDEs In the IDE case find deflating subspace Y of 0 E 0 0 A B za A = z A Q S E Q S B S R 0 S R given a solution of [ A XA E XE + Q A XB + S ] [ ] K [K ] L, B XA + S B XB + R = V Σ L i. e., (za A)Y = Z(zẼ Ã) for Ẽ, Ã K n+q n+m. Part of the Solution: [ ] X (A E) + G1 XB + G Y = 2 V1 Σ V2 Σ, where im [ ] G 1 G 2 ker E, and [ V1 Σ V2 Σ ] [ ] x = u [ ] x, u [ ] x V u Σ. D. Bankmann (TU Berlin) Control of IDEs February 9, / 18

28 Concluding Remarks Conclusions We have seen the discrete time counterpart of the KYP Lemma for DAEs. We have seen the discrete time counterpart of Lur e equations and corresponding deflating subspaces for (explicit) difference equations. The results on Lur e equations for DAEs can also be adapted to Lur e equations on IDEs. D. Bankmann (TU Berlin) Control of IDEs February 9, / 18

29 Concluding Remarks Conclusions We have seen the discrete time counterpart of the KYP Lemma for DAEs. We have seen the discrete time counterpart of Lur e equations and corresponding deflating subspaces for (explicit) difference equations. The results on Lur e equations for DAEs can also be adapted to Lur e equations on IDEs. What was not shown The spectral structure of palindromic matrix pencils can be characterized w.r.t. the solvability of the KYP inequality. Certain discretizations display relations between continuous time results and their discrete time counterpart. D. Bankmann (TU Berlin) Control of IDEs February 9, / 18

30 Concluding Remarks Conclusions We have seen the discrete time counterpart of the KYP Lemma for DAEs. We have seen the discrete time counterpart of Lur e equations and corresponding deflating subspaces for (explicit) difference equations. The results on Lur e equations for DAEs can also be adapted to Lur e equations on IDEs. What was not shown The spectral structure of palindromic matrix pencils can be characterized w.r.t. the solvability of the KYP inequality. Certain discretizations display relations between continuous time results and their discrete time counterpart. Thank you for your attention! D. Bankmann (TU Berlin) Control of IDEs February 9, / 18

31 References Byers, R., D. S. Mackey, V. Mehrmann, and H. Xu (2008). Symplectic, BVD, and palindromic approaches to discrete-time control problems. Preprint. Kalman, R. E. (1963). Lyapunov functions for the problem of Lur e in automatic control. Proceedings of the National Academy of Sciences of the United States of America 49.2, p Mehrmann, V. (1991). The Autonomous Linear Quadratic Control Problem. Ed. by M. Thoma and A. Wyner. Vol Lecture Notes in Control and Information Sciences. Heidelberg: Springer. Popov, V. M. (1961). On absolute stability of non-linear automatic control systems. Automatika i Telemekhanika 12, pp Reis, T. (2011). Lur e equations and even matrix pencils. Linear Algebra and its Applications 434.1, pp Reis, T., O. Rendel, and M. Voigt (2015). The Kalman Yakubovich Popov inequality for differential-algebraic systems. Linear Algebra and its Applications 485, pp Schröder, C. (2008). Palindromic and even eigenvalue problems-analysis and numerical methods. Dissertation. Institut für Mathematik, Technische Universität Berlin. Voigt, M. (2015). On Linear-Quadratic Optimal Control and Robustness of Differential-Algebraic Systems. Doctoral Dissertation. Fakultät für Mathematik, Otto-von-Guericke-Universität Magdeburg. Berlin: Logos. Yakubovich, V. (1962). Solution of certain matrix inequalities in the stability theory of nonlinear control systems. 143, pp D. Bankmann (TU Berlin) Control of IDEs February 9, / 18