1 Infinite horizon optimal control and stabilizability of linear descriptor systems Mihaly Petreczky‡ and Sergiy Zhuk† arXiv:1312.7547v2 [math.OC] 7 Jan 2014 ‡ Ecole des Mines de Douai, Douai, France, [email protected], † IBM Research, Dublin, Ireland, [email protected] Abstract In this paper we construct an infinite horizon LQG optimal controller for a linear stationary differential-algebraic equation (DAE) and prove that a stabilizability from the given initial state is necessary and sufficient for the controller existence. Our approach is based on ideas from linear geometric control theory that allows one to represent solutions of DAEs as outputs of LTI system. It is applicable for generic DAEs without imposing additional regularity assumptions. I. I NTRODUCTION Consider a linear Differential-Algebraic Equation (DAE) in the following form: d(Ex) = Ax(t) + Bu(t), dt Ex(t0 ) = Ex0 , (1) where E, A ∈ Rc×n and B ∈ Rn×m are given matrices. In this paper, we are interested in the following two problems. For the given positive definite matrices Q0 , Q, R find a control law u∗ (t) such that: 1) Finite horizon optimal control u∗ minimizes the cost functional Z t1 T T J(u, x0 , t1 ) = inf (x(t1 ) E Q0 Ex(t1 ) + xT (s)Qx(s)ds + u(s)T Ru(s)ds) x satisfies (1) 0 2) Infinite horizon optimal control u∗ minimizes the cost functional J(u, x0 ) = lim sup J(u, x0 , t1 ) t1 →∞ DRAFT 2 Contribution of the paper We show that the optimal control problem above has a solution for a given x0 if and only if (1) is stabilizable from x0 . We also establish a link between DAEs and classical LTI system, which allows us to re-use existing theory and understand better the notion of stabilizability for DAEs. More precisely, we solve the infinte-horizon LQ control problem by transforming it to a classical LTI optimal control problem. In order to do we show that the solutions of DAEs of the form (1) can be represented as outputs of a linear time-invariant system (Al , Bl , Cl , Dl ): (x, u) satisfies (1), if there exists an input v such that ṗ = Al p + Bl v (xT , uT )T = Cl p + Dl v The corresponding LTI (Al , Bl , Cl , Dl ) can be obtained by using geometric theory of LTI systems [1], [2], more specifically, by computing the largest output nulling controlled invariant subspace. The LTI (Al , Bl , Cl , Dl ) can be easily computed using basic matrix operations and the computational tools available from geometric control theory. In fact, we show the existence of a class of LTIs whose outputs are solutions of DAEs and we also show that all LTIs corresponding to the same DAE are feedback equivalent. The infinite horizon LQ control problem for (1) can then be reformulated as a classical LQ control problem. The latter has a solution, if (Al , Bl , Cl , Dl ) is stabilizable, which is equivalent to stabilizability of (1). Note that we do not need to use any assumptions in order to show the correspondence above: namely, the DAE (1) can be represented as an output of an associated LTI system for any F, A, B. In particular, we do not require uniqueness of a solution of the DAE at hand, i.e. the DAE (1) might not have solutions from any initial state or it might have several solutions starting from a certain initial state. The only condition we need for existence of an optimal controller is that the DAE is ℓ-stabilizable. Motivation DAEs have a wide range of applications: namely, without claiming completeness, we mention robotics [3], cyber-security [4] and modelling [5]. The motivation for considering the optimal control problem above is twofold. First, it is interesting on its own right. Second, from [6] it follows that the solution of the optimal control problem yields a minimax observer for the DAE with uncertain but bounded input and incomplete and noisy observations. Related work To the best of our knowledge, the main result of this paper is new. A preliminary version of this paper appeared in [7]. With respect to [7] the main difference is that we include DRAFT 3 detailed proofs, and we provide necessary and sufficient conditions for existence of a solution for the infinite horizon optimal control problem. The finite horizon optimal control problem was already presented in [8]. The main difference with respect to [8] is that we consider infinite horizon case, give detailed proofs for the finite horizon case and present a generic procedure of reducing the finite horizon optimal control problem to a classical finite horizon LQ problem for LTI systems. This procedure contains the reduction algorithm proposed in [8] as a particular case. The literature on optimal control for DAE is vast. For an overview we refer the reader to [9], [10] and the references therein. To the best of our knowledge, the most relevant references are [11]–[16]. In [11], [12] only regular DAEs were considered. In contrast, we allow any DAE which is stabilizable from the designated initial state. The infinite-horizon LQ control problem for non-regular DAE was also addressed in [13], however there it is assumed that the DAE has a solution from any initial state. In terms of results, [14]–[17] seem to be the closest to our paper. However, they neither imply nor they are implied by our resuts. The main differences are as follows. 1) In [14]–[17] the cost function does not have the final cost term x(t1 )T Q0 x(t1 ). Note that the presence of this terms is indispensable to make the duality work (see [6, Theorem 1] for the further details), when the results of this paper are applied to the minimax observer design problem [6]. 2) We are interested in existence of a solution from a particular initial condition, as opposed to [14]–[16]. 3) We present necessary and sufficient conditions for existence of an optimal solution. In contrast, [14]–[17] present only sufficient conditions for existence of a solution. Moreover, those sufficient conditions appear to be more restrictive than the ones of this paper. This is not suprising, since [14]–[17] consider a different cost function and impose different additional restrictions on the optimal solution. As a result, we can weaken some of the conditions imposed in [14]–[17]. 4) We require only Ex to be absolutely continous, not x. This leads to subtle but important technical differences in the solution. 5) Finally, similarly to [14], [16], we solve the LQ problem by reducing it to an LTI LQ problem. Unlike in [14], [16], where Silverman’s algorithm was used, we obtain correDRAFT 4 sponding LTI through geometric arguments. The construction of [14], [16] then becomes a particular case of our construction, if we allow only absolutely continuos x. Optimal control of non-linear and time-varying DAEs was also addressed in the literature, see [18]–[20]. However, they do not seem to be directly applicable to the problem of the current paper, as they look at finite horizon optimal control problems. Furthermore, the cited papers focus more on the (very challenging) problem of existence of a solution rather than on algorithms The idea of presenting solutions of general DAEs as outputs of LTIs appeared in [8], [13], [14], [16]. In [8], [14], [16] an algorithm was presented to compute such an LTI. In this paper we show that a whole family of such LTIs can be obtained using classical results from geometric control theory, and thus the algorithms of [8], [14], [16] are specific instances of the classical algorithms of geometric control theory. The construction of [13] represents a special case of the results of this paper. Note that relating DAEs with ordinary differential equations via Weierstrass canonical forms is a classical trick. However, this trick requires either smoothness constraints on admissible inputs or usage of distributions. Despite the superficial difference, there is a connection between our approach and that of based on quasi-Weistrass canonical forms [21]. We postpone the detailed discussion until Remark 8. There is a long history of geometric ideas in DAEs, see for example [22], [23] and the references in the surveys [10], [24]. However, to the best of our knowledge, these ideas were not used to relate LTIs and DAEs. Outline of the paper This paper is organized as follows. Subsection I-A contains notations, section II describes the mathematical problem statement, section III presents the main results of the paper. A. Notation In denotes the n × n identity matrix; for an n × n matrix S, S > 0 means xT Sx > 0 for all x ∈ Rn ; kSk denotes the Frobenious norm. F + denotes the pseudoinverse of the matrix F . I := [0, t] for t ≤ +∞ and L2 (I, Rn ) denotes the space of all square-integrable functions f : I → Rn , L2loc (I, Rn ) = {f ∈ L2 (I 1 , Rn ), ∀I 1 ( I, I 1 is compact }. To simplify the notation we will often write L2 (0, t) and L2loc (0, t) referring to L2 (I, Rn ) and L2loc (I, Rn ) respectively. Note that L2 (0, t) = L2loc (0, t) if t < +∞. Finally, f |A stands for the restriction of a function f onto a set A. DRAFT 5 II. P ROBLEM STATEMENT Consider the DAE Σ: dEx = Ax(t) + Bu(t) and Ex(0) = Ex0 . dt (2) Here x0 ∈ Rn is a fixed initial state and A, E ∈ Rc×n , B ∈ Rc×m Notation 1 (Dx0 (t1 ) and Dx0 (∞)): . For any t1 ∈ [0, +∞] denote by I the interval [0, t1 ] ∩ [0, +∞) and denote by Dx0 (t1 ) the set of all pairs (x, u) ∈ L2loc (I, Rn ) × L2loc (I, Rm ) such that Ex is absolutely continuous and (x, u) satisfy (2) almost everywhere. A state x0 will be called differentiably consistent if Dx0 (t1 ) 6= ∅. The set of differentiably consistent initial states is denoted by Vc . Note that we did not assume that the DAE is regular, and hence there may exist initial states x0 such that Dx0 (t1 ) is empty for some t1 ∈ [0, +∞]. Note that our definition of differentiable consistency is slightly different from the one used in the literature, see [24]: for an initial state x0 to be consistent, we only require that there exists a locally integrable solution such that Ex(0) = Ex0 , we require neither that x(0) = x0 or continuity (differentiability) of x. Later on we will show that if x0 is differentiably consistent for some t1 > 0, then, in fact, there exists a trajectory (x, u) ∈ Dx0 (∞) defined on the whole time axis (see Lemma 2). This then implies that the set of consistent states of the DAE (2) is equal to the set of consistent initial differentiable variables in the terminology of [24]. From [24, Corollary 4.3] it also follows that every state is differentiably consistent if and only if the DAE is impulse controlable, i.e. if and only if for any matrix Z such that imZ = ker E, h i h i Rank E, AZ, B = Rank E, A, B . Next, we formulate the optimal control problem already mentioned in the introduction. Take R ∈ Rm×m , Q, ∈ Rn×n , Q0 ∈ Rc×c and assume that R > 0, Q > 0 Q0 ≥ 0. For any initial state x0 ∈ Rn , and any trajectory (x, u) ∈ Dx0 (t), t > t1 define the cost functional J(x, u, t1 ) = x(t1 )T E T Q0 Ex(t1 )+ Z t1 + (xT (s)Qx(s) + uT (s)Ru(s))ds. (3) 0 DRAFT 6 Problem 1 (Finite-horizon optimal control problem): The problem of finding (x∗ , u∗ ) ∈ Dx0 ([0, t1 ]) such that J(x∗ , u∗ , t1 ) = inf (x,u)∈Dx0 ([0,t1 ]) J(x, u, t1 ) < +∞ . is called the finite-horizon control problem and (x∗ , u∗) is called the solution of the finite-horizon control problem. Problem 2 (Infinite horizon optimal control problem): For every (x, u) ∈ D(∞), define J(x, u) = lim sup J(x, u, t1 ) . t1 →∞ Let x0 be a differentiably consistent initial state of the DAE (2). The infinite horizon optimal control problem for (2) and initial state x0 is the problem of finding (x∗ , u∗ ) such that (x∗ , u∗ ) ∈ Dx0 (∞) and J(x∗ , u∗) = lim sup inf t1 →∞ (x,u)∈Dx0 (t1 ) J(x, u, t1 ) < +∞ . (4) The pair (x∗ , u∗) will be called the solution of the infinite horizon optimal control problem for the initial state x0 . Remark 1: The proposed formulation of the infinite horizon control problem is not necessarily the most natural one. We could have also required the (x∗ , u∗ ) ∈ D(∞) to satisfy J(x∗ , u∗ ) = inf (x,u)∈D(∞) J(x, u). The latter means that the cost induced by (x∗ , u∗ ) is the smallest among all the trajectories (x, u) which are defined on the whole time axis. It is easy to see that formulation above implies that J(x∗ , u∗) = inf (x,u)∈D(∞) J(x, u). Another option is to use lim instead of lim sup in the definition of J(x∗ , u∗ ) and in (4). In fact, the solution we are going to present remains valid if we replace lim sup by lim. Remark 2 (Solution as feedback): In the sequel we will show that the solution of the optimal control problem (x∗ , u∗) is such that u∗ (t) = K(t)x∗ (t) for some state feedback matrix K(t) of suitable dimensions. Note, however, that for DAEs the feedback law does not determine the control input uniquely, since even autonomous DAEs may admit several solutions starting from the same initial state. If the DAE has at most one solution from any initial state, in particular, if the DAE is regular, then the feedback law above determines the optimal trajectory x∗ uniquely. In the general case, one could follow the behavioral interpretation adopted by [25]. There, the interconnection of a feedback controller u = Kx, where K is a constant matrix, with the DRAFT 7 original system interpreted as the following DAE d (Ê x̂) = Âx̂ dt E 0 A 0 , Â = , and x̂(t) = (x(t), u(t)). where Ê = 0 0 K I Remark 3: In the sequel we will show that the optimal solution (x∗ , u∗ ) of the infinite horizon optimal control problem is such that u∗ = Kx∗ and limt→∞ x∗ (t) = 0. Hence, if the DAE admits at most one solution from any initial state, then the closed-loop system is globally asymptotically stable, i.e. for any initial state the corresponding solution converges to zero. In order to state the conditions under which Problem 2 has a solution, we will need the notion of stabilizability. Definition 1 (Stabilizability): The DAE (2) is said to be stabilizable for a differentiably consistent initial state x0 ∈ Rn , if there exists (x, u) ∈ Dx0 (∞) such that limt→∞ Ex(t) = 0. The DAE (2) is said to be stabilizable, if it is stabilizable for all x0 ∈ Rn such that x0 is differentiably consistent. III. M AIN RESULTS In §III-A we present the construction of an LTI system whose outputs are solutions of the original DAE. In §III-B we reformulate the optimal control problem as a linear quadratic infinite horizon control problem for LTIs. The solution of the latter problem yields a solution to the original control problem. A. DAE systems as solutions to the output zeroing problem Consider the DAE system (2). In this section we will study solution set Dx0 (t1 ), t1 ∈ [0, +∞] of (2). It is well known that for any fixed x0 and u, (2) may have several solutions or no solution at all. In the sequel, we will use the tools of geometric control theory to find a subset X of Rc , such that Dx0 (t1 ) 6= ∅ for any x0 ∈ E −1 (X ) and all t1 ∈ [0, +∞]. Furthermore, we provide a complete characterization of all such solutions as outputs of an LTI system. Theorem 1: Consider the DAE system (2). There exists a linear system S = (Al , Bl , Cl , Dl ) defined on state-space X = Rn̂ , input space U = Rk and output space Rn+m , and a linear subspace X ⊆ Rc such that the following holds. DRAFT 8 1) RankDl = k. 2) Let Cs and Ds be the matrices formed by the first n rows of Cl and Dl respectively. Then EDs = 0, RankECs = n̂, X = imECs . 3) The state x0 is differentiably consistent if and only if Ex0 ∈ X . 4) Fix t1 ∈ [0, +∞] and set I = [0, t1 ] ∩ [0, +∞). For any g ∈ L2loc (I, Rk ), let v = vS (v0 , g) : I → Rn̂ , (5) ν = νS (v0 , g) : I → Rn+m be the state and output trajectory of S which corresponds to the initial state v0 and input g, i.e. v̇ = Al v + Bl g and v(0) = v0 (6) ν = Cl v + Dl g . Define the map M = (ECs )+ : Rc → Rn̂ . Assume that Ex0 ∈ X . Then the following holds. a) If (x, u) ∈ Dx0 (t1 ), then there exists an input g ∈ L2loc (I, Rk ), such that x(t) for t ∈ I a.e. νS (M(Ex0 ), g)(t) = u(t) (7) b) Conversely, for any g ∈ L2loc (I, Rk ), there exists (x, u) ∈ Dx0 (t1 ) such that νS (M(Ex0 ), g)(t) = (xT (t), uT (t))T for all t ∈ I. c) If (7) holds for some (x, u) ∈ Dx0 (t1 ),then M(Ex(t)) = vS (M(Ex(0), g)(t) =: x(t) − Cl v(t)) for t ∈ I a.e. v(t) for all t ∈ I, and g(t) = Dl+ ( u(t) Proof of Theorem 1: There exist suitable nonsingular matrices S and T such that Ir 0 , SET = (8) 0 0 where r = RankE. Let e A SAT = A21 A12 B , SB = 1 A22 B2 e ∈ Rr×r and B1 ∈ Rr×m . Define be the decomposition of E, A, B such that A i h i h e = A21 . e = A , B and C G = A12 , B1 , D 22 2 DRAFT 9 Consider the following linear system S e + Gq ṗ = Ap e + Dq e z = Cp . (9) Now, it is easy to see that if (x, u) solves (2) then (p, q) defined by T −1 x = (pT , q1T )T , q = (q1T , uT )T , q1 ∈ Rn−r solve the linear system (9) and the output z is zero almost everywhere. Recall from [1, Section 7.3] the problem of making the output zero by choosing a suitable input. Recall from [1, Definition 7.8] the concept of a weakly unobservable subspace of a linear system. If we apply this concept to S, then an initial state p(0) ∈ Rr of S is weakly unobservable, if there exists an input function q ∈ L2 ([0, +∞), Rk ) such that the resulting output function z of S(Σ) equals zero, i.e. z(t) = 0 for almost all t ∈ [0, +∞). Following the convention of [1], let us denote the set of all weakly unobservable initial states by V(S). As it was remarked in [1, Section 7.3], V(S) is a vector space and in fact it can be computed. Moreover, if p(0) ∈ V(S) and for the particular choice of q, z = 0, then p(t) ∈ V(S) for all t ≥ 0. Let I = [0, t] or I = [0, +∞). Let q ∈ L2 (I, Rn−r+m ) and let p0 ∈ Rr . Denote by p(p0 , q) and z(p0 , q) the state and output trajectory of (9) which corresponds to the initial state p0 and input q. For technical purposes we will need the following easy extension of [1, Theorems 7.10–7.11]. Theorem 2: 1) V = V(S) is the largest subspace of Rr for which there exists a linear map Fe : Rr → Rm+n−r such that e + GFe)V ⊆ V and (C e+D e Fe)V = 0 (A (10) 2) Let Fe be a map such that (10) holds for V = V(S). Let L ∈ R(m+n−r)×k for some k be e ∩ G−1 (V(S)) and RankL = k. a matrix such that imL = ker D For any interval I = [0, t] or I = [0, +∞), and for any p0 ∈ Rr , q ∈ L2loc (I, Rk ), z(p0 , q)(t) = 0 for t ∈ I a.e. if and only if p0 ∈ V and there exists g ∈ L2loc (I, Rk ) such that: q(t) = Fep(p0 , q)(t) + Lg(t) for t ∈ I a.e. Proof of Theorem 2: Part 1 is a reformulation of [1, Theorem 7.10]. For I = [0, +∞), Part 2 is a restatement of [1, Theorem 7.11]. For I = [0, t1 ], the proof is simillar to [1, Theorem 7.11]. DRAFT 10 We are ready now to finalize the proof of Theorem 1. The desired linear system S = (Al , Bl , Cl , Dl ) may be obtained as follows. Consider the linear system below. e + GFe)p + GLg ṗ = (A (xT , uT )T = C̄p + D̄g T 0 I T 0 0 r and D̄ = . C̄ = 0 Im Fe 0 Im L h i−1 Consider a basis b1 , . . . , br of R such that b1 , . . . , bn̂ span V. Let R = b1 , . . . , br be In̂ 0 . Set Al = the corresponding basis transformation. It then follows that R(V) = im 0 0 h i h i I I −1 n̂ −1 n̂ e e R( A + G F )R , C = C̄R , B = In̂ 0 In̂ 0 RGL, and Dl = D̄. That l l 0 0 is, Al , Bl , Cl are the matrix representations in the basis b1 , . . . , bn̂ of the linear maps Al = e + GFe)|V : V → V, Cl = C̄|V : V → Rn×m , Bl = GL : Rk → V. Define (A p X = {ET | p ∈ V}. 0 r It is easy to see that this choice of (Al , Bl , Cl , Dl ) and X satisfies the conditions of the theorem. The proof of Theorem 1 is constructive and yields an algorithm for computing (Al , Bl , Cl , Dl ) from (E, A, B). This prompts us to introduce the following terminology. Definition 2: A linear system S = (Al , Bl , Cl , Dl ) described in the proof of Theorem 1 is called the linear system associated with the DAE (2), and the map M is called the corresponding state map. Remark 4: Notice that the dimension n̂ of the associated linear system satsifies n̂ ≤ RankE ≤ max{c, n}. h i Remark 5: It is easy to see from the construction that Cl Dl is full column rank, i.e. it represents an injective linear map. Indeed, if Cl p + Dl g = 0, then Cs p + Ds g = 0 and hence, by using EDs = 0, E(Cs p + Ds g) = ECs p = 0. Since ECs is full column rank, it follows that p = 0. Hence, Cl p + Dl g = Dl g = 0, but Dl is full column rank and hence g = 0. That is, Cl p + Dl g = 0 implies p = 0, g = 0. DRAFT 11 Remark 6 (Comparison with [8]): The system (As1 , As2 ) described in [8, Proposition 3] is e = A0 , G = A0 , C e = A0 , related to a linear associated system as follows. If we identify A 1 e= D A04 , imP0s , As1 |V(S) 2 3 e + GFe)|V(S), As L (A 2 and define F̃ as in (10) then V(S) = and = = GL. e + GFe)p + GLw such that p(0) ∈ V(S), then p̂ = p Moreover, if (p, w) is a solution to ṗ = (A and q̂ = Lw is a solution to [8, eq. (22)]. The proof is by induction on k. More precisely, recall e + Gq ∈ Xk , Cx e + Dq e = 0}, then that if we set X0 = Rr and Xk+1 = {x ∈ Xk | ∃q ∈ Rn−r , Ax for some k ∗ ≤ r, Xk∗ +1 = Xk∗ = V(S). If W is a subspace, denote by PW the projection onto W . We claim that T 1) Xk = ki=1 ker(I − PimAi−1 Ai−1 3 ), 4 P k k 0 + 0 e k + k e e e 2) Aki = (A+G e , F1 = −(A4 ) A3 , Fk+1 = −(A4 ) A3 , j=1 Fj )PXk , A2 = GPG−1 (Xk−1 ∩ker D) e+D e Pk Fei )|X = 0, 3) (C k i=1 4) ImAki ⊆ Xk−1 and in particular, Aki (Xk−1) ⊆ Xk−1 . e ⊕ (G−1 (Xk−1) ∩ ker D) e ⊥ and ImAk ⊆ Xk . 5) ker Ak4 = (G−1 (Xk ) ∩ ker D) 4 6) Xk = {x ∈ Xk−1 | ∃v : Ak3 x + Ak4 v = 0} = {x ∈ Xk−1 | ∃q : Ak1 x + Ak2 q ∈ Xk−1 }. e + Gq ∈ Xk−1 , Cx e + Dq e = 0. If q Moreover, for any x ∈ Xk , there exists q such that Ax Pk e is as above, then v = q − i=1 Fi x has the property that Ak3 x + Ak4 v = 0. Conversely, for every x ∈ Xk , assume that Ak3 x + Ak4 v = 0. It then follows that q = PG−1 (Xk−1 )∩ker De (v) + Pk e e e e i=1 Fi x is such that Ax + Gq ∈ Xk−1 and Cx + Dq = 0. The claims above can be shown by induction on k. As it was noted above, there exists a smallest k ∗ such that Xk∗ = Xk∗ +1 . Let s = k ∗ + 1. Notice that As3 = (I − Pker(I−P I − Pker(I−P ∗ ∗ Ak 3 ) im(Ak 4 ) k ∗ +1 A2 k ∗ +1 . Since ImA1 k ∗ +1 , ImA2 ∗ ∗ Ak ) im(Ak ) 3 4 Ak1 ∗ +1 , and ⊆ Xk∗ = Xk∗ +1 and Xk∗ +1 ⊆ ker(I − Pim(Ak4 ∗ ) A3 , it follows that As3 = As4 = 0. Note moreover, that (A3s−1 , A4s−1 ) 6= 0. Indeed, if this k∗ was the case, it would then follow that Xs−1 = {x ∈ Xs−2 | ∃v : A3s−1 x + As4 v = 0} = Xs−2 , and this would contradict to the minimality of k ∗ . Hence, it then follows that the procedure described in [8, Proposition 3] stops for some s if and only if Xs−1 = Xs . Moreover, from the e + GFe)PV , Fe = Ps Fei , As V ⊆ V, claims above it follows that V = Xs = ImP s , As = (A 0 1 i=1 1 e +D e Fe), and that if L is any matrix such that ImL = G−1 (V)∩ker D, e then As L = GL. V ⊆ ker(C 2 Remark 7 (Impulse controllability): Recall from [24] that the DAE (2) is impulse controllable, h i h i if for any matrix Z such that imZ = ker E, Rank E, AZ, B = Rank E, A, B . Let S, T DRAFT 12 be non-singular matrices such that SET = Choose Z = T 0 0 0 In−r from right we get that e A SAT = A21 Ir 0 , r = RankE. Consider the decomposition 0 0 A12 B , SB = 1 . A22 B2 h i , By multiplying E, AZ, B by S from the left and by diag(T, In , Im ) h i h i Ir , 0, A12 , B1 = r + Rank A22 , B2 . Rank E, AZ, B = Rank 0, 0, A22 B2 h i By multiplying E, A, B by S from the left and by diag(T, T, Im ) from the right, h i h i e Ir , 0, A, A12 , B1 = r + Rank A12 , A22 , B2 . Rank E, A, B = Rank 0, 0, A12 , A22 B2 Hence, h i h i Rank A22 , B2 = Rank A12 , A22 , B2 . h i e i.e. imC e ⊆ imD. e e e Using the notation of Theorem 1, it then follows that Rank C, D = RankD, e and Fe = −D e + C. e In this case , V(S) = Rr , L is any matrix such that imL = ker D Note that the linear system associated with (E, A, B) is not unique. There are two sources of non-uniqueness: 1) The choice of the matrices S and T in (8). 2) The choice of Fe and L and R in Theorem 2. However, we can show that all associated linear systems are feedback equivalent. Definition 3 (Feedback equivalence): Two linear systems Si = (Ai , Bi , Ci , Di ), i = 1, 2 are said to be feedback equivalent, if there exist a linear state feedback matrix K and a non-singular square matrices U, T such that (T (A1 + B1 K)T −1 , T B1 U, (C1 + D1 K)T −1 , D1 U) = S2 . We will call (T, K, U) feedback equivalence from S1 to S2 . It is not difficult to see that the state feedback equivalence is an equivalence relation. Lemma 1: Let Si , i = 1, 2 be two linear systems which are obtained from the proof of Theorem 1 and let Mi , i = 1, 2 be the corresponding state maps. Then S1 and S2 are feedback equivalent. Moreover, if (T, K, U) is the corresponding feedback equivalence, then T M1 = M2 . DRAFT 13 The proof of Lemma 1 can be found in the appendix. Lemma 2: If there exists t1 > 0 such that Dx0 (t1 ) 6= ∅, then Dx0 (∞) 6= ∅. In other words, if x0 is differentiably consistent, then there exists a solution (x, u) ∈ Dx0 (∞). Proof: Let S = (Al , Bl , Cl , Dl ) be the linear system associated with the DAE. From Theorem 1 it then follows that if x0 is differentiably consistent for some t1 > 0, then any solution (x, u) ∈ Dx0 (t1 ) arises as the output of S corresponding to some input g and the initial state M(Ex0 ). This solution is defined for all t > 0 and so belongs to (x, u) ∈ Dx0 (∞). In particular, if g = 0, then the output (x, u) of S from the initial state M(Ex0 ) is smooth. In the sequel we will need stabilizability of the linear systems associated with a DAE. In the rest of the section we will show that stabilizability of the former is equivalent to stabilizability of the DAE. Recall from [1] the notion of the stabilizability subspace Vg of a linear system S = (A, B, C, D): Vg is the set of all inital states x0 of S , for which there exists an input u such that the corresponding state trajectory x starting from x0 has the property that limt→∞ x(t) = 0. Recall from [1] that there exists a feedback matrix F such that (A + BF )Vg ⊆ Vg , and the restriction of (A + BF ) to Vg can be represented by a stable matrix. In fact, if L is a matrix such that imL = B −1 (Vg ), then the following holds. Let x0 be an initial state and let u be an input and let x be the corresponding state trajectory starting from x0 . Then limt→∞ x(t) = 0 if and only if x0 ∈ Vg and there exists a signal w such that ẋ(t) = (A + BF )x(t) + BLw(t) and u(t) = F x(t) + Lw(t). Lemma 3: The DAE (2) is stabilizable from x0 if and only if M(Ex0 ) belongs to the stabilizability subspace Vg of S for every associated linear system S = (Al , Bl , Cl , Dl ) and the corresponding map M. In particular, the DAE is stabilizable if and only if S is stabilizable. Proof Lemma 3: Since feedback equivalence preserves stabilizability subspaces and all associated linear systems are feedback equivalent, existence of an associated linear systems which is stabilizable from M(Ex0 ) is equivalent to saying that all associated linear systems are stabilizable from M(Ex0 ). Assume that one of the associated linear systems S = (Al , Bl , Cl , Dl ) is stabilizable from M(Ex0 ). Let Vg be the stabilizability subspace of S . It then follows that M(Ex0 ) ∈ Vg , and there exists a feedback Fl such that the restriction of Al + Bl Fl to Vg is stable and hence for any v0 = M(Ex0 ), there exists a state v̇ = (Al + Bl Fl )v with v(0) = v0 , such that limt→∞ v(t) = 0. Consider the output (xT , uT )T = (Cl + Dl Fl )v. It then follows that Ex(0) = Ex0 and (x, u) is a solution of the DAE. Moreover, limt→∞ x(t) = DRAFT 14 limt→∞ (Cs + Ds Fl )v(t) = 0. That is, for this particular u, limt→∞ x(t) = 0 and thus the DAE is stabilizable from x0 . Conversely, assume that the DAE is stabilizable from x0 , and choose any associated linear system (Al , Bl , Cl , Dl ). Let (x, u) ∈ Dx0 (∞) be a solution of the DAE such that limt→∞ Ex(t) = 0. It then follows that there exist an input g such that v̇ = Al v + Bl g, (xT , uT )T = Cl v + Dl g, v(0) = v0 = M(Ex0 ) and M(Ex(t)) = v(t). In particular, limt→∞ v(t) = ME limt→∞ x(t) = 0. That is, there exists an input g, such that the corresponding state trajectory v of S starting from M(Ex0 ) converges to zero. But this is precisely the definition of stabilizability of (Al , Bl , Cl , Dl ) from M(Ex0 ). Remark 8 (Relationship to Wang sequence and quasi-Weierstrass canonical forms): Recall from [24] that the augmented Wong sequence is defined as follows V0 = Rn , Vi+1 = A−1 (EVi +imB), T ∗ and that the limit V ∗ = ∞ i=0 Vi is achieved in finite number of steps: V = Vk for some k ∈ N. It is not difficult to see that V(S) from the proof of Theorem 1 correspond to the limit V∗ of the augmented Wong sequence Vi for the DAE (2): V ∗ = {(p, F p + Lq)T | p ∈ V(S)}. Hence, h i ∗ if (Al , Bl , Cl , Dl ) is as in Theorem 1, then V = im Cs Ds . In [21] a relationship between the quasi-Weierstrass form of regular DAEs and and space V ∗ for B = 0 was established. This indicates that there might be a deeper connection between quasi-Weierstrass forms and associated linear systems. The precise relationship remains a topic of future research. B. Solution of the optimal control problem for DAE We apply Theorem 1 in order to solve the Problem 2. Let S = (Al , Bl , Cl , Dl ) be a linear system associated with Σ and let M and Cs be defined as in Theorem 1. Consider the following linear quadratic control problem. For every initial state v0 , for every interval I containing [0, t1 ] and for every g ∈ L2loc (I, U) define a cost functional J(v0 , g, t): J (v0 , g, t1) = v T (t1 )E T CsT Q0 ECs v(t1 )+ Z t1 Q 0 ν(t)dt + ν T (t) 0 0 R v̇ = Al v + Bl g and v(0) = v0 ν = Cl v + Dl g. DRAFT 15 For any g ∈ L2loc ([0, +∞)) and v0 ∈ Rn̂ , define J (v0 , g) = lim sup J (v0 , g, t1). t1 →∞ In our next theorem we prove that the problem of minimizing the cost function J for (2) is equivalent to minimizing the cost function J for the associated LTI S . Theorem 3: Consider the associated LTI S = (Al , Bl , Cl , Dl ) and for any g ∈ L2loc (I), I = [0, t1 ] or I = [0, +∞), denote by yS (v0 , g) : I → Rn+m the output trajectory of S which corresponds to the initial state v0 and input g. • Assume that x0 is a differentiably consistent initial state of (1). For any g ∈ L2loc (I), I = [0, t1 ] or I = [0, +∞) and any (xT , uT )T = yS (M(Ex0 ), g) we have: J(x, u, t1 ) = J (M(Ex0 ), g, t1) and if I = [0, +∞) • J(x, u) = J (M(Ex0 ), g) (11) For all t1 > 0, (x∗ , u∗ ) ∈ Dx0 (t1 ) is a solution of the finite horizon optimal control problem if and only if there exists g ∗ ∈ L2 (0, t1 ) such that (x∗ T , u∗ T )T = yS (M(Ex0 ), g ∗ ) and J (v0 , g ∗, t1 ) = • inf g∈L2 ([0,t1 ]) J (v0 , g, t1) < +∞ . (12) The tuple (x∗ , u∗ ) ∈ Dx0 (∞) is a solution of the infinite horizon optimal control problem if and only if there exists an input g ∗ ∈ L2loc (0, +∞) such that (x∗T , u∗T )T = yS (M(Ex0 ), g ∗), and J (v0 , g ∗ ) = lim sup inf t1 →∞ g∈L2 (0,t1 ) J (v0 , g, t1) < +∞. (13) Proof of Theorem 3: Equation (11) follows by routine manipulations and by noticing that the first n rows of Cl v(t) + Dl g(t) equals Cs v(t) + Ds g(t) and as EDs = 0 (see Theorem 1), E(Cs v(t) + Ds g(t)) = ECs v(t). The rest of theorem follows by noticing that any element of Dx0 (t1 ) or Dx0 (∞) arises as an output of S for some g ∈ L2loc (I, Rk ). Next theorem is based on classical results (see [26]) and provides the solution of the finite horizon optimal control problem (Problem 1.). Theorem 4: Assume that x0 is a differentially consistent initial state of (1). The finite horizon optimal control problem has a unique solution of the form (x∗T (t), u∗T (t))T = (Cl − Dl K(t1 − t))v(t) v̇(t) = (Al − Bl K(t1 − t))v(t) and (14) v(0) = M(Ex0 ) . DRAFT 16 where P (t) and K(t) satisfy the following Riccati differential equation Ṗ (t) = ATl P (t) + P (t)Al − K T (t)(DlT SDl )K(t) + ClT SCl and P (0) = (ECs )T Q0 ECs K(t) = (DlT SDl )−1 (BlT P (t) + DlT SCl ) (15) and S = diag(Q, R) where Cs is defined as in Theorem 1. Moreover, the optimal value of the cost function is J(x∗ , u∗ , t1 ) = (M(Ex0 ))T P (t1 )(M(Ex0 )). (16) Proof: Let us first apply the feedback transformation g = F̂ v + Uw to S = (Al , Bl , Cl , Dl ) with U = (DlT SDl )−1/2 and F̂ = −(DlT SDl )−1 DlT SCl , as described in [1, Section 10.5, eq. (10.32)]. Note that Dl is injective and hence U is well defined. Consider the linear system v̇ = (Al + Bl F̂ )v + Bl Uw and v(0) = v0 (17) For any w ∈ L2loc (I), I = [0, t1 ] the state trajectory v of (17) equals the state trajectory of S for the input g = F̂ v + Uw and initial state v0 . Moreover, all inputs g of S can be represented in such a way as it follows from Theorem 2. Define now Ic(v0 , w, t) = v T (t)CsT E T Q0 ECs v(t)+ Z t + (v T (t)(Cl + Dl F̂ )T S(Cl + Dl F̂ )v(t) + w T (t)w(t))dt, 0 where v is a solution of (17), and Cs is defined as in the theorem’s statement (so that Cs has n rows). It is easy to see that J (v0 , g, t) = Ic(v0 , w, t) for g = F̂ v + Uw and any initial state v0 of S . Consider now the problem of minimizing Ic(v0 , w, t). The solution of this problem can be found using [26, Theorem 3.7]. Notice that (15) is equivalent to the Riccati differential equation described in [26, Theorem 3.7] for the problem of minimizing Ic(v̂0 , w, t). Hence, by [26, Theorem 3.7], (15) has a unique positive solution P , and for the optimal input w ∗ , g ∗ = F̂ v ∗ + Uw ∗ = −K(t1 − t)v(t) satisfies (12) and v̇(t) = (Al − Bl K(t1 − t))v(t) and v(0) = v0 . From Theorem 3 and Part 4b of Theorem 1 it then follows that (x∗ T , g ∗T )T = Cl v ∗ + Dl g ∗ is the solution of the Problem 1 and that (16) holds. In our last theorem we present a solution to the infinite horizon optimal control problem (Problem 2). Let S = (Al , Bl , Cl , Dl ) be the LTI associated with (1). Let Vg denote the stabilizability DRAFT 17 subspace of S . From [27] it then follows that Vg is Al -invariant and imBl ⊆ Vg . Hence, there exists a basis transformation T such that h i Âg ⋆ B̂ M , T Bl = g , Cl T −1 = Ĉg ⋆ , T M = g T Al T −1 = 0 ⋆ 0 ⋆ Denote by Sg = (Âg , B̂g , Ĉg , D̂g ), where Dg = Dl . The LTI Sg represents the restriction of S to the subspace Vg . It follows that Sg is stabilizable. Definition 4 (Stabilizable associated LTI): We call the LTI Sg defined above the stabilizable LTI associated with (1) and we call Mg the associated map. Using Theorem 3 and the classical results on infinite horizon LQ control, we obtain the following. Theorem 5: The following are equivalent: • (i) The infinite horizon optimal control problem is solvable for x0 • (ii) M(Ex0 ) belongs to the stabilizability subspace Vg of the associated LTI S • (iii) lim supt1 →∞ inf (x,u)∈Dx0 (t1 ) J(x, u, t1 ) < +∞ Consider the stabilizable LTI Sg = (Ag , Bg , Cg , Dg ) and the map Mg . If M(F T ℓ) ∈ Vg , then there exists a unique positive definite matrix P such that: 0 = P Ag + ATg P − K T (DgT SDg )K + CgT SCg , K= (DgT SDg )−1 (BgT P + DgT SCg ) (18) and S = diag(Q, R) It then follows that Ag − Bg K is a stable matrix and (x∗ , u∗ ) defined by v˙∗ = (Ag − Bg K)v ∗ ∗T ∗T T v ∗ (0) = Mg (Ex0 ) (x , u ) = (Cg − Dg K)v (19) ∗ is a solution of the infinite horizon optimal control problem (Problem 2) and J(x∗ , u∗ ) = (Mg Ex0 )T P Mg Ex0 . (20) Proof of Theorem 5: (i) =⇒ (ii) If (x∗ , u∗ ) is a solution of the infinite horizon optimal control problem, then by Theorem 3, there exists an input g ∗ ∈ L2loc (0, +∞) such that J (M(Ex0 ), g ∗) < +∞. Let v0 = M(Ex0 ). We claim that if J (v0 , g ∗) < +∞, then limt→∞ v ∗ (t) = 0 for the state trajectory v ∗ of S which corresponds to the input g ∗ and starts from v0 . The latter is equivalent DRAFT 18 to v ∗ (0) = v0 = M(Ex) ∈ Vg . Let us prove that limt→∞ v ∗ (t) = 0. To thisend, notice R∞ Q 0 Q 0 ν(t)dt < +∞, and as is positive that J (v0 , g ∗ ) < +∞ implies 0 ν(t)T 0 R 0 R R∞ R∞ Q 0 ν(t)dt < +∞ for some µ > 0 definite, it follows that µ 0 ν T (t)ν(t)dt < 0 ν(t)T 0 R and so ν ∈ L2 (0, +∞). Consider the decomposition ν(t) = (xT (t), uT (t))T , where x(t) ∈ Rn . By Theorem 1 it follows that M(Ex(t)) = v ∗ (t) and hence v ∗ ∈ L2 (0, +∞). As g ∗ is a linear function of x, u and v (see Theorem 1) it follows that g ∗ ∈ L2 (0, +∞). Recalling that Rt Rt v̇ ∗ (t) = Al v ∗ (t) + Bl g ∗ (t) we write: v ∗ (t) = v ∗ (τ ) + Al τ v ∗ (s)ds + Bl τ g ∗(s)ds for τ < t. R∞ Rt 1 Since τ kv ∗ (s)kds ≤ (t − τ ) 2 kv ∗ kL2 (0,+∞) where kv ∗ k2L2 (0,+∞) := 0 kv ∗ (s)k2Rr ds, it follows 1 that: kv ∗ (t) − v ∗ (τ )kRr ≤ (t − τ ) 2 (kAl kkvkL2(0,+∞) + kBl kkg ∗kL2 (0,+∞) ). Hence v ∗ is uniformly continous. This and v ∗ ∈ L2 (0, +∞) together with Barbalat’s lemma imply that limt→∞ v ∗ (t) = 0. Hence the optimal state trajectory v ∗ converges to zero, i.e. g ∗ is a stabilizing control for v0 , and thus v0 = M(Ex0 ) ∈ Vg . (ii) =⇒ (i) Assume now that v0 = M(Ex0 ) ∈ Vg . Let us recall the construction of the LTI Sg and the basis transformation T used in the definition of Sg . It then follows that h i Il 0 T (Vg ) = im , where l = dim Vg . Define the map Π = Il 0 T . It then follows that for 0 0 any v0 ∈ Vg , and for any g ∈ L2loc (I), (xT , uT )T is the output of S and t 7→ v(t) is the state trajectory of S starting from the initial state v0 and driven by input g, if and only if (xT , uT )T is the output of Sg and t 7→ Π(v)(t) is the state trajectory of Sg starting from Π(v0 ) and driven by the input g. In other words, the state trajectories of Sg correspond to the state trajectories of S which start in Vg , and for these trajectories Sg and S produce the same output. For any initial state v̂0 of Sg define now the cost function I (v̂0 , g, t1) as I (v̂0 , g, t1 ) = v T (t1 )(E Ĉs )T Q0 E Ĉs v(t1 )+ Z t1 Q 0 ν(t)dt + ν T (t) 0 0 R v̇ = Ag v + Bg g and v(0) = v̂0 ν = Cg v + Dg g. DRAFT 19 h iT T T where Ĉs is such that Cg = Ĉs , Ĉinp , such that Ĉs has n rows. Notice that from the Il definition of Cg it follows that Cg = Cl Π+ (notice that Π+ = T −1 ) and hence Ĉs = Cs Π+ . 0 Define now I (v̂0 , g) = lim supt1 →∞ I (v̂0 , g, t1). It is not hard to see that: I (Π(v0 ), g, t1) = J (v0 , g, t1 ) and I (Π(v0 ), g) = J (v0 , g) (21) for any initial state v0 of S such that v0 ∈ Vg . Consider now the problem of minimizing limt1 →+∞ I (v̂0 , g, t1 ). To this end, let us first apply the feedback transformation g = F̂ v + Uw to Sg = (Ag , Bg , Cg , Dg ) with U = (DgT SDg )−1/2 and F̂ = −(DgT SDg )−1 DgT SCg , as described in [1, Section 10.5, eq. (10.32)]. Consider the linear system v̇ = (Ag + Bg F̂ )v + Bg Uw and v(0) = v̂0 (22) For any w ∈ L2loc (I), where I = [0, t1 ] of I = [0, +∞), the state trajectory v of (22) equals the state trajectory of Sg for the input g = F̂ v + Uw and initial state v̂0 . Moreover, all inputs g of Sg can be represented in such a way. Define now Ic(v̂0 , w, t) = v T (t)(E Ĉs )T Q0 E Ĉs v(t)+ Z t + (v T (t)(Cg + Dg F̂ )T S(Cg + Dg F̂ )v(t) + w T (t)w(t))dt, 0 where v is a solution of (17). Due to the construction of F̂ and U, kS 1/2 (Cg + Dg F̂ )v(t) + Dg Uw(t)k2 = v T (t)(Cg + Dg F̂ )T S(Cg + Dg F̂ )v(t) + w T (t)w(t). Using these remarks, it is then easy to see that I (v̂0 , g, t) = Ic(v̂0 , w, t) for g = F̂ v + Uw. Consider now the problem of minimizing limt→∞ Ic(v̂0 , w, t). The solution of this problem can be found using [26, Theorem 3.7] To this end, notice that (Ag + Bg F̂ , Bg U) is stabilizable and (S 1/2 (Cg + Dg F̂ ), Ag + Bg F̂ ) is observable. Indeed, it is easy to see that stabilizability of (Ag , Bg ) implies that of (Ag + Bg F̂ , Bg U). Observability of (S 1/2 (Cg + Dg F̂ ), Ag + Bg F̂ ) can be derived as follows. Recall from Theorem 1 that ECs is of full column rank and EDs = 0. Let D̂s be the sub-matrix of Dg formed by its first n rows. Hence, E D̂s = EDs = 0. Moreover, since E Ĉs equals the matrix representation of the restriction of the map E(Cs + Ds F̂ ) = ECs to Vg , it follows that E Ĉs is of full column rank if ECs is injective which is the case (by Theorem 1). Hence, E(Ĉs + D̂s F̂ ) = E Ĉs is of full column rank, and thus the pair (Ĉs + D̂s F̂ , Ag + Bg F̂ ) DRAFT 20 is observable. Let us now return to the minimization problem. Notice that (18) is equivalent to the algebraic Riccati equation described in [26, Theorem 3.7] for the problem of minimizing limt→∞ Ic(v̂0 , w, t). Hence, by [26, Theorem 3.7], (18) has a unique positive definite solution P , and Ag + Bg F̂ − Bg UU T BgT P = Ag − Bg K is a stable matrix. From [26, Theorem 3.7], there exists w ∗ such that limt→∞ Ic(v̂0 , w ∗ , t) is minimal and v̂ T P v̂0 = limt→∞ Ic(v̂0 , w ∗, t). On 0 the other hand, [26, Theorem 3.7] also implies that Hence, g ∗ = F̂ v ∗ + Uw ∗ satisfies I (v̂0 , g ∗) = inf g∈L2loc ([0,+∞)) v̂0T P v̂0 = limt1 →∞ inf w∈L2 (0,t1 ) Ic(v̂0 , w, t1). I (v0 , g) = v̂0T P v̂0 = lim inf t→∞ g∈L2 (0,t1 ) I (v̂0 , g, t) , where v̇ ∗ = (Ag + Bg F̂ )v ∗ + Bg Uw ∗ , v ∗ (0) = v̂0 . A routine computation reveals that (v ∗ , g ∗) satisfies v̇ ∗ = Ag v ∗ + Bg g ∗ and g ∗ = −Kv ∗ and v ∗ (0) = v̂0 . From Theorem 3 and Part 4b of Theorem 1 it then follows that (x∗T , u∗T )T = Cg v ∗ + Dg g ∗ is the solution of the infinite horizon optimal control problem and that (x∗ , u∗ ) satisfies (19) and (20). (i) =⇒ (iii) If (x∗ , u∗ ) ∈ Dx0 (∞) is a solution of the infinite horizon optimal control problem, then, by definition, +∞ > J(x∗ , u∗ ) = lim supt1 →∞ inf (x,u)∈Dx0 ([0,t1 ]) J(x, u, t1 ). (iii) =⇒ (ii) From (11) it follows that the condition of (iii) implies that there exists M > 0 such that for all t1 > 0, inf g∈L2 (0,t1 ) J (M(Ex0 ), g, t1) ≤ M. (23) Recall that S = diag(Q, R) and for each g ∈ L2 (I), [0, t1 ] ⊆ I, and initial state v0 , define Z t1 I (v0 , g, t1) = (Cl v(t) + Dl g(t))T S(Cl v(t) + Dl g(t))dt 0 v̇ = Al v + Bl g, v(0) = v0 It then follows that I (v0 , g, t1) ≤ J (v0 , g, t1 ) for any t1 > 0 and I (v0 , g, t1 ) is non-decreasing in t1 . Hence, (23) implies that ∀t1 ∈ (0, +∞) : inf g∈L2 (0,t1 ) I (v0 , g, t1 ) < M. From classical LTI theory [1], [26] it follows that if H is the solution of the Riccati equation Ḣ(t) = ATl H(t) + H(t)Al − K T (t)(DlT SDl )K(t) + ClT SCl and H(0) = 0 (24) K(t) = (DlT SDl )−1 (BlT H(t) + DlT SCl ) DRAFT 21 then v0T H(t1 )v0 = inf g∈L2 (0,t1 ) I (v0 , g, t1). The latter may be easily seen applying the state feedback transformation g = F̂ v + Uw with F, U defined as in the proof of Theorem 4 and solve the resulting standard LQ control problem for the transformed system. It then follows that (24) is the differential Riccati equation which is asscociated with this problem. Note that the matrix Ḣ(t) is symmetric and positive semi-definite. Indeed, symmetry follows immediately from (24) and semi-definiteness may be proved as follows: for every state v0 , v0T H(t)v0 is a non-decreasing differentiable function of t and hence its derivative v0T Ḣ(t)v0 is non-negative. Define the set V = {v0 | sup t1 ∈(0,+∞) v0T H(t1 )v0 < +∞}. (25) By assumption (iii) it follows that M(Ex0 ) ∈ V . From [1, Theorem 10.13] it follows that1 Vg ⊆ V . It is also easy to see that V is a linear space. We will show that V = Vg , from which M(Ex0 ) ∈ Vg follows. First, we will argue that V is invariant with respect to Al . To this end, consider v0 ∈ V and set v1 = e−Al t v0 . For any t1 and any g ∈ L2 (0, t1 ), define ĝ ∈ L2loc (0, t1 + t) as ĝ(s) = 0, s ≤ t and ĝ(s) = g(s − t) if s > t. Consider v̇ = Al v + Bl ĝ, v(0) = v1 . It then follows that v(s) = eAs t v1 for s ≤ t, v(t) = v0 and if ṙ = Al r + Bl g, r(0) = v0 , then v(t + s) = r(s) for all s ∈ [0, +∞). 1 Indeed, recall from the system Sg = (Ag , Bg , Cg , Dg ) and the map Π. Recall that Sg is stabilizable and hence by [1, R∞ Theorem 10.19] for every v0 ∈ Vg there exists an input g such that 0 ((Cg v(s) + Dg g(s))T S(Cg v(s) + Dg g(s))ds < +∞ for v̇ = Ag v + Bg g, v(0) = Π(v0 ). Consider the state trajectory r(t) where ṙ = Al r + Bl g, r(0) = v0 . It then Rt follows that Π(r) = v and hence Cl r + Dl g = Cg v + Dg g. Therefore, I (v0 , g, t1 ) = 0 1 (Cl r(t) + Dl g(t))T S(Cl r(t) + Rt R∞ Dl g(t)) = 0 1 ((Cg v(s) + Dg g(s))T S(Cg v(s) + Dg g(s))ds ≤ 0 ((Cg v(s) + Dg g(s))T S(Cg v(s) + Dg g(s))ds < +∞. R∞ Hence, v0H (t1 )v0 ≤ I (v0 , g, t1 ) < 0 ((Cg v(s) + Dg g(s))T S(Cg v(s) + Dg g(s))ds. DRAFT 22 It then follows that J (v1 , ĝ, t + t1 ) = T Z T t1 +t v (t1 + t)(ECs ) Q0 ECs v(t1 + t) + (Cl v(s) + Dl ĝ(s))T S(Cl v(s) + Dl ĝ(s))ds = 0 Z t1 +t r(t1 )(ECs )T Q0 ECs r(t1 ) + (Cl v(s) + Dl ĝ(s))T S(Cl v(s) + Dl ĝ(s))ds+ t Z t (Cl v(s))T S(Cl v(s))ds = (26) 0 Z t1 r(t1 )(ECs )T Q0 ECs r(t1 ) + (Cl r(s) + Dl g(s))T S(Cl r(s) + Dl g(s))ds+ 0 Z t (Cl eAl s v1 )T S(Cl eAl s v1 )ds = 0 Z t J (v0 , g, t1) + (Cl eAl s v1 )T S(Cl eAl s v1 )ds 0 Since v0 ∈ V , it then follows that there exists Γ > 0 such that for any t1 there exists g ∈ L2 (0, t1 ) such that v0T H(t1 )v0 = J (v0 , g, t1) ≤ Γ. Hence, from (26) it follows that Z t T v1 H(t1 + t)v1 = inf J (v1 , ĝ, t + t1 ) ≤ Γ + (Cl eAl s v1 )T S(Cl eAl s v1 )ds 2 ĝ∈L (0,t1 +t) 0 and hence sup t1 ∈(0,+∞) v1T H(t1 )v1 ≤ sup t1 ∈(0,+∞) v1T H(t1 + t)v1 ≤ Γ + Z t (Cl eAl s v1 )T S(Cl eAl s v1 )ds < +∞. 0 In the last step we used that v1T H(t + t1 )v1 ≥ v T H(t1 )v for all t, t1 ∈ [0, +∞). Hence, v1 = e−Al t v0 ∈ V . Since V is a linear space, and t is arbitrary, it then follows that Al v0 = − dtd e−Al t v0 |t=0 = limδ→0 − (e −Al δ v −v ) 0 0 δ ∈ V . That is, V is Al invariant. Notice that the controllability subspace of S is contained in Vg ⊆ V and that imBl is contained in the controllability subspace S . Hence, imBl ⊆ V . Notice that for any b ∈ V , the function bT H(t)b is monotonically non-decreasing in t and it is bounded, hence limt→∞ bT H(t)b exists and it is finite. Notice for any b1 , b2 ∈ V , (b1 + b2 )T H(t)(b1 + b2 ) = bT1 H(t)b1 + 2bT1 H(t)b2 + bT2 H(t)b2 and as b1 + b2 ∈ V , the limit on both sides exists and so the limit limt→∞ bT1 H(t)b2 exists. Consider now a basis b1 , . . . , br of V and for any t define Ĥi,j (t) = bTi H(t)bj . It then follows that the matrix Ĥ(t) = (Hi,j (t))i,j=1,...,r is positive semi-definite, symmetric and there exists a positive semi-definite matrix Ĥ+ such that Ĥ+ = limt→∞ Ĥ(t). From (24) and Al V ⊆ V it follows that Al bi ∈ V for all i = 1, . . . , r DRAFT 23 and hence limt→t bTi H(t)Al bj , limt→t bTi ATl H(t)bj exist for all i, j = 1, . . . , r. Notice that the term bTi K(t)T (DlT SDl )K(t)bj is the sum of the terms bTi (BlT H(t))T (DlT SDl )−1 (BlT H(t))bj (DlT SCl bi )T (DlT SDl )−1 (BlT H(t)bj ), (DlT SCl bj )T (DlT SDl )−1 (BlT H(t)bi ) and the term (DlT SCl bi )T (DlT SDl )−1 (DlT SCl bj ). From imBl ⊆ V and the fact that for any x, z ∈ V , the limit limt→∞ xT H(t)z exists, it follows that the limits limt→∞ BlT H(t)x = limt→∞ xT H(t)Bl for all x ∈ V . Applying this remark to x = bi and x = bj , it follows that limt→∞ bTi K(t)T (DlT SDl )K(t)bj ˙ exists. Hence, for any i, j = 1, . . . , r, the limit of Ĥi,j (t) = bTi Ḣ(t)bj exists as t → ∞ and hence ˙ the limit limt→∞ Ĥ(t) =: Z exists. Moreover, since Ḣ(t) is symmetric and positive semi-definite, it follows that Z is symmetric and positive semi-definite. We claim that Z is zero. To this end it is sufficient to show that limt→∞ bT Ḣ(t)b = 0 for any b ∈ V . Indeed, from this it follows that Zi,j = limt→∞ 0.5((bi + bj )T Ḣ(t)(bi + bj ) − bTi Ḣ(t)bi − bTj H(t)bj ) = 0 for any i, j = 1, . . . , r. Now, fix b ∈ V and assume that c = limt→∞ bT Ḣ(t)b 6= 0. Set now h(t) = bT H(t)b. It then follows that ḣ(t) = bT Ḣ(t)b and thus there exists T > 0 such that for all t > T , ḣ(t) > 2c > 0 (note that Ḣ(t) is positive semi-definite). Hence, h(t) = Rt Rt h(0) + 0 ḣ(s)ds > T ḣ(s)ds > (T − t) 21 . Hence, h(t) is not bounded, which contradicts to the assumption that b ∈ V . Hence, Z = 0 and thus, it follows that Ĥ+ satisfies the algebraic Riccati equation 0 = Âl Ĥ+ + Ĥ+ Âl − K T (Dl SDl )K + ĈlT S Ĉl K= (DlT SDl )−1 (B̂lT Ĥ+ + (27) DlT S Ĉl ) where Âl , B̂l , Ĉl are defined as follows: Âl and Ĉl are the matrix representations of the linear maps Al and Cl restricted to V , B̂l is the matrix representation of the map Rk ∋ g 7→ Bl g ∈ V in the basis b1 , . . . , br of V chosen as above. Note that for Âl and B̂l to be well defined, we had to use the facts imBl ⊆ V and Al V ⊆ V , which allow us to view Bl as a linear map Bl : Rk → V and Al |V as Al |V : V → V . Notice that Dl is injective as a linear map and recall from Remark 4 that Cl p + Dl g = 0 implies that p = 0, g = 0 and hence the largest output nulling subspace V(Σ̂) of the linear system Σ̂ = (Âl , B̂l , Ĉl , Dl ) is zero. Then from [1, Theorem 10.19], V(Σ̂) = 0 and (27) it follows that Σ̂ is stabilizable. Since Σ is just the restriction of S to V , it then follows that every state from V is stabilizable and hence V = Vg . Remark 9 (Infinite-horizon case): The proof of Theorem 5 implies that in the formulation of DRAFT 24 associated linear control problem, we can replace lim sup by lim. Remark 10 (Computation and existence of a solution): The existence of solution for Problem 2 and its computation depend only on the matrices (E, A, B, Q, R, Q0 ). Indeed, a linear system S associated with (E, A, B) can be computed from (E, A, B), and the solution of the associated LQ problem can be computed using S and the matrices Q, Q0 , R. Notice that the only condition for the existence of a solution is that is stabilizability from x0 . IV. C ONCLUSIONS We have presented a solution to infinite horizon linear quadratic control problem for linear DAEs and we provided sufficient and necessary conditions for existence of a solution in terms of stabilizability. Note that the DAEs were not assumed to be regular, or to have a unique solution from any initial state. Furthermore, the conditions we formulated depend only on the matrices of the DAE and on the initial state. R EFERENCES [1] H. Trentelman, A. A. Stoorvogel, and M. Hautus, Control theory of linear systems. Springer, 2005. [2] G. Basile and G. Maro, Controlled and conditioned invariants in linear systems theory, 1991. [3] W. Schiehlen, “Force coupling versus differential algebraic description of constrained multibody systems,” Multibody System Dynamics, vol. 4, pp. 317–340, 2000. [4] F. Pasqualetti, F. Dorfler, and F. Bullo, “Attack detection and identification in cyber-physical systems - part i: Models and fundamental limitations,” Tech. Rep., arXiv:1202.6144v2. 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Malabre, “Generalized linear systems: geometric and structural approaches,” Linear Algebra Appl., vol. 122-124, pp. 591 – 621, 1989. [23] V. Armentano, “The pencil (se-a) and controllability-observability for generalized linear systems,” SIAM J. Control Optimization, vol. 24, pp. 616–638, 1986. [24] T. Berger and T. Reis, “Controllability of linear differential-algebraic systems: a survey,” in Surveys in Differential-Algebraic Equations I, ser. Differential-Algebraic Equations Forum, A. Ilchmann and T. Reis, Eds. Springer Berlin Heidelberg, 2013, pp. 1–61. [25] J. Willems and J. Polderman, An Introduction to Mathematical Systems Theory: A Behavioral Approach. Springer Verlag, New York, 1998. [26] H. Kwakernaak and R. Sivan, Linear Optimal Control Systems. [27] F. M. Callier and C. A. Desoer, Linear System Theory. Wiley-Interscience, 1972. Springer-Verlag, 1991. A PPENDIX Proof of Lemma 1: We will use the following terminology in the sequel. Consider two linear systems (A1 , B1 , C1 , D1 ) and (A2 , B2 , C2 , D2 ) with n states, p outputs and m inputs. A tuple (X, F, G, V, U) of matrices, X ∈ Rn×n , U ∈ Rm×m , V ∈ Rp×p , F ∈ Rm×n , G ∈ Rn×p such that X, U and V are non-singular, is said to be a feedback equivalence with output injection DRAFT 26 from (A1 , B1 , C1 , D1 ) to (A2 , B2 , C2 , D2 ), if X(A1 + B1 F + GC1 + GD1 F )X −1 = A2 V (C1 + D1 F )X −1 = C2 X(B1 + GD1 )U = B2 and V D1 U = D2 If G = 0, V = Ip , then (X, F, G, U) is just a feedback equivalence and (A1 , B1 , C1 , D1 ) and (A2 , B2 , C2 , D2 ) are feedback equivalent. In this case (i.e. when G = 0, V = Ip ), we denote this transformation by (T, F, U). Consider two ssociated linear systems Si = (Ai,l , Bi,l , Ci,l , Di,l ), i = 1, 2 each of which correspond to the choosing T = Ti , S = Si , Fe = Fi , L = Li , R = Ri , i = 1, 2 in the proof of n×n Theorem 1. Consider the maps Mi ,i = 1, 2 which correspond to Si , i = 1, 2. Let Si , Ti ∈ R Ir 0 , i = 1, 2. Let be invertable, such that Si ETi = 0 0 Ai A12,i Si ÂTi = A21,i A22,i B1,i , Si B̂ = B2,i h i Gi = A12,i , B1,i h i ei = A21,i and D ei = A , B C 22,i 2,i and consider the linear systems Si ṗi = Ai pi + Gi qi ei p1 + D e i qi zi = C for i = 1, 2. Denote by Vi = V(Si ) the set of weakly unobservable states of Si ,i = 1, 2. Denote by F (Vi ), i = 1, 2, the set of all state feedback matrices F ∈ R(n+m)×r such that ei + D e i F )Vi = 0. Pick Fi ∈ F (Vi ), i = 1, 2 and pick full column rank (Ai + Gi F )Vi ⊆ Vi , (C e i . In order to prove the lemma, it is matrices Li , i = 1, 2 such that imLi = G−1 (Vi ) ∩ ker D i enough to show that RankL1 = RankL2 = k, and there exist invertable linear maps X ∈ Rr×r , DRAFT 27 U ∈ Rk×k , and a matrix F ∈ Rk×r such that X(V1 ) = V2 (28a) (A1 + G1 F1 + G1 L1 F )V1 ⊆ V1 (28b) ∀x ∈ V1 : X(A1 + G1 F1 + G1 L1 F )x = (A2 + G2 F2 )Xx (28c) XG1 L1 U = G2 L2 T 0 0 T 0 0 1 r×k = 2 r×k 0 Im L1 U 0 Im L2 (28d) ∀x ∈ V1 : T1 0 Ir T2 0 I x = r Xx, 0 Im (F1 + L1 F ) 0 Im F2 (28e) (28f) where 0r×k denotes the r × k matrix with all zero entries. Indeed, n̂ = note that (28)implies that h i In̂ Ti 0 I I r Ri−1 n̂ , dim V1 = dim V2 . Since Al,i = In̂ 0 Ri (Ai +GFi )Ri−1 , Cl,i = 0 0 Im Fi 0 h i Ti 0 0 I 0 , and imGi Li ⊆ Vi , Ri (Vi ) = im n̂ , Bl,i = In̂ 0 Ri Gi Li , Dl,i = 0 Im Li 0 0 h i In̂ In̂ if (X, F, U) satisfy (28), it then follows that ( In̂ 0 R2 XR1−1 , F R1−1 , U) is a 0 0 feedback equivalence between S1 and S2 . Finally, (T̂ , F̂ , Û) is a feedback equivalence from S1 to S2 , and Cs,1, Cs,2, Ds,1 , Ds,2 denote the first n rows of Cl,1 , Cl,2, Dl,1 , Dl,2 respectively, then ECs,2 = E(Cs,1 + Ds,1 F̂ )T̂ −1 = ECs,1 T̂ −1 , since EDs,1 = 0. Hence, M2 = (ECs,2)+ = (ECs,1T̂ −1 )+ = T̂ (ECs,1 )+ = T̂ M1 . In order to find the matrices F, U, X, notice that R 0 11 T2−1 T1 = R21 R22 for R11 ∈ Rr×r , R22 ∈ R(n−r)×(n−r) , R21 0 p̄ ∈ R(n−r)×r . Indeed, assume T2−1 T1 = for q q̄ DRAFT 28 some q, q̄ ∈ Rn−r , p̄ ∈ Rr . Then p̄ 0 = S2 ET2 T2−1 T1 = 0 q 0 S2 S1−1 S1 ET1 = S2 S1−1 0 = 0. q 0 0 Hence, T2−1 T1 = from which the statement follows. In a simillar fashion q q̄ H11 H21 , S2 S1−1 = 0 H22 where H11 ∈ Rr×r , H22 ∈ R(c−r)×(c−r) , H12 ∈ Rr×(c−r) , moreover, H11 = R11 . Indeed, p p S2 S1−1 = S2 S1−1 S1 ET1 = 0 0 p p̄ S2 ET2 T2−1 T1 = 0 0 for some p̄ ∈ Rr . Finally, H 0 R 0 11 = S2 S1−1 (S1 ET1 ) = (S2 ET2 )T2−1 T1 = 11 . 0 0 0 0 Hence, R11 = H11 . From S2 ÂT2 = S2 S1−1 S1 ÂT1 (T2−1 T1 )−1 it and S2 B̂ = S2 S1−1 S1 B̂ follows that e1 + ĜD e 1 F̂ )R−1 A2 = R11 (A1 + G1 F̂ + ĜC 11 e 1 )Û G2 = R11 (G1 + ĜD (29) e 2 = V̂ D e 1 Û and C e2 = V̂ (C e1 + D e 1 F̂ )R−1 D 11 −1 −1 −R22 R12 R22 0 −1 , Ĝ = R11 , and V̂ = H22 . where F̂ = H12 , Û = 0 0 Im DRAFT 29 We then claim that the following choice of matrices X = R11 and U = L+ 1 ÛL2 F = L+ 1 (F̂ (30) + ÛF2 R11 − F1 ) satisfies (28). We prove (28a) – (28f) one by one. Proof of (28a): Indeed, from (29) it then follows that S1 and S2 are related by a feedback equivalence with output injection (R11 , F̂ , Ĝ, Û, V̂ ). From [1, page 169, Exercise 7.1] it follows that V2 = R11 (V1 ) = XV1 . Proof of (28b): e2 + D e 2 F2 )V2 = {0}, and (A2 + G2 F2 )V2 ⊆ V2 . From the definition of F2 it follows (C e2 , D e 2 , A2 , G2 from (29) and using that V2 = R11 V1 and that Substituting the expressions for C R11 , V̂ are invertable, it follows that for all x ∈ V1 , e1 + D e 1 (F̂ + ÛF2 R11 ))x = 0 (C e1 + ĜD e 1 (F̂ + ÛF2 R11 ))x = (A1 + G1 (F̂ + Û F2 R11 ) + ĜC (31) (A1 + G1 (F̂ + Û F2 R11 ))x ∈ V1 Hence, (A1 + G1 F1 + G1 F )V1 ⊆ V1 . e1 x + Proof of (28c): Since from the definition of F1 it follows that (A1 + G1 F1 )x ∈ V1 , (C e 1 F1 x) = 0, for all x ∈ V1 , from (31), it then follows that for all x ∈ V1 , G1 (F̂ + ÛF2 R11 − D e 1 (F̂ x + Û F2 R11 − F1 )x = 0. Hence, (F̂ + ÛF2 R11 − F1 )x ∈ imL1 and hence F1 )x ∈ V1 and D L1 F x = L1 L+ 1 (F̂ + ÛF2 R11 − F1 )x = (F̂ + ÛF2 R11 − F1 )x for all x ∈ V1 . From this it follows that x ∈ V1 : F1 x + L1 F x = (F̂ + ÛF2 R11 )x. (32) From (32) it then follows that (A1 + G1 F1 + G1 L1 F )x = A1 x + G1 (F̂ + Û F2 R11 )x for all x ∈ V1 . From this and (29), (28c) follows. e 1 Û)∩(R11 G1 Û )−1 (V2 ) = Û −1 (ker D e 1 ∩G1 (V1 )) = Proof of (28d): Recall that imL2 = ker(V̂ D Û −1 imL1 . Since Û is invertable, it follows that RankL1 = RankL2 = k and that L1 U = L1 L+ 1 ÛL2 = ÛL2 . (33) DRAFT 30 e 2 ÛL2 = 0, it follows that XG1 L1 U = T G1 Û L2 = XG1 Û L2 + Hence, using (29) and D e 2 Û L2 = G2 L2 . X ĜD Proof of (28e): It is easy to see that (28e) is equivalent to −1 T2 0 T1 0 0r×k 0r×k = . 0 Im 0 Im L1 U L2 We will show (34) To this end, T 2 0 R11 R21 0 notice that −1 0 T1 0 T2−1 T1 0 = = Im 0 Im 0 Im 0 0 0 R11 = R R22 0 21 Û −1 0 0 Im Hence, −1 T2 0 0 Im T1 0 0 0r×k 0r×k = . −1 Im L1 U Û L1 U (34) (35) (36) Using L1 U = Û L2 proven above in (33), it follows that Û −1 L1 U = L2 and hence (36) implies (34). Proof of (28f): Again, it is enough to show that ∀x ∈ V1 : −1 T 0 T 0 Ir I 2 1 x = r R11 x. 0 Im 0 Im (F1 + L1 F ) F2 (37) From (35) it follows that −1 T 0 T 0 Ir 2 1 = 0 Im 0 Im (F1 + L1 F ) R11 R21 . −1 + Û (F1 + L1 F ) 0 (38) DRAFT 31 Notice that Û −1 F̂ = T −R21 0 and hence, using (32), R 21 x + Û −1 (F1 + L1 F )x = 0 R 21 x + F̂ x + F2 R11 x = F2 R11 x 0 for all x ∈ V1 . Combining this with (38), (37) follows easily. DRAFT