DSpace at VNU: Subadjoint Equations of Index-1 Linear Singular Difference Equations

DSpace at VNU: Subadjoint Equations of Index-1 Linear Singular Difference Equations tài liệu, giáo án, bài giảng , luận...

Vietnam J Math (2013) 41:81–96 DOI 10.1007/s10013-013-0003-9 Subadjoint Equations of Index-1 Linear Singular Difference Equations Le Cong Loi Received: 12 December 2011 / Published online: 30 January 2013 © Vietnam Academy of Science and Technology (VAST) and Springer Science+Business Media Singapore 2013 Abstract For index-1 linear singular difference equations, the concept of the subadjoint equation is analysed in full detail based on the Kronecker normal form and the inherent regular ordinary difference equation We show that this subadjoint equation is of index by solving backwards In addition, the discrete Lagrange identity and some relationships between the original equation and its subadjoint equation are established Keywords Linear singular difference equations · Differential algebraic equations · Adjoint equations · Subadjoint equations · Discrete Lagrange identity Mathematics Subject Classification (2010) 39A10 · 39A09 Introduction Linear singular difference equations (LSDEs) of the form Ak xk+1 = Bk xk + qk , k ∈ N, where N denotes either the set of natural numbers including zero or its subset, Ak , Bk ∈ Rm×m , qk ∈ Rm are given, and the matrix Ak is singular for all k ∈ N have been studied intensively in [2, 3, 12, 13, 15, 16, 19] In this paper, the data qk that occurs in the above general LSDEs will not be involved in the analysis; hence we will assume that qk = Further, we only consider the case where N is a finite set; i.e., N is a time interval [0, N − 1], where N is a fixed positive integer In other words, we are only interested in the following homogeneous equations: Ak xk+1 = Bk xk , B k = 0, , N − L.C Loi ( ) Department of Mathematics, Vietnam National University, Hanoi, Vietnam e-mail: loilc@vnu.edu.vn (1) 82 L.C Loi The continuous-time version of (1), namely, the differential algebraic equation (DAE) A(t)x + B(t)x = 0, t ∈ J := [t0 , T ], (2) where A, B ∈ C(J, Rm×m ), has been treated in many works (see [4–12] and the references therein) In the theory of homogeneous DAEs, the adjoint equations have been studied [4, 6, 8] and have already found various applications in many problems, such as boundary value problems [5], sensitivity analysis of DAEs [9, 10], and control theory of singular systems (see [7, 8, 11, 12]) In addition, the adjoint equations which are also called the dual equations of regular ordinary difference equations and their applications have been studied in [1, 14, 17, 18, 20] The adjoint equation of an LSDE was been introduced in [12, 16, 19]; however, very few duality results in dynamical system theory were obtained In this paper we give a twofold study First, based on the Kronecker normal form and the inherent regular ordinary difference equation of the index-1 LSDE, we define its adjoint equation In what follows, this adjoint equation will be called subadjoint, and we will explain it in more detail later This result coincides with the one introduced in [16, 19] Second, we investigate some properties of the subadjoint equation such as its index property and the discrete Lagrange identity The outline of this paper is as follows For the convenience of the reader, in Sect we recall the notions used in the following sections In Sect we derive the subadjoint equation of the index-1 LSDE from the Kronecker normal form and the inherent regular ordinary difference equation The properties of the subadjoint equation are analysed in detail in Sect Finally, in Sect we discuss some open problems Basic Concepts Adjoint or dual equations have been introduced in the theory of DAEs In [8], the dual equation of (2) has the form AT (t)z = −B T (t)z, t ∈ J (3) This equation appeared in several papers; the reader can refer to [4–10] Note that in those papers (3) is called the adjoint equation and is considered as being solving backwards For linear singular discrete-time systems, in [16, 19] the authors have introduced a dual equation of (1) which is given by ATk−1 λk−1 = BkT λk , k = 1, , N − (4) In those papers, this equation is called a subdual equation of (1), and we will interpret this terminology later Further, (4) is solved in the backward direction In what follows, we will only use the term adjoint Moreover, we consider that the original equations are solved in the forward direction, whereas their adjoint equations are solved backwards In the theory of LSDEs, the index-1 concept for (1) has been introduced in [2, 3, 13, 15], and in this paper we also assume that rankAk ≡ r, ≤ r ≤ m − Definition [2, 3] Equation (1) is called index-1 tractable (index-1 for short) if the following conditions hold: Subadjoint Equations of Index-1 Linear Singular Difference Equations 83 (i) rank Ak = r, k = 0, , N − 1, (ii) Sk ∩ ker Ak−1 = {0}, k = 1, , N − 1, where Sk = {z ∈ Rm : Bk z ∈ im Ak } In the following, we always assume that dim S0 = r and, since rank AN−1 = r, it follows that dim(ker AN−1 ) = m − r Let Qk be a projector onto ker Ak and put Pk := I − Qk Since rank Ak = rank Ak−1 , there exists Tk ∈ GL(Rm ) such that Tk |ker Ak is an isomorphism from ker Ak onto ker Ak−1 Such an operator can be constructed as follows Since Qk ∈ Rm×m is a projector onto kerAk , there exists a nonsingular matrix Zk ∈ Rm×m such that ¯ k−1 , Qk = Zk QZ ¯ = diag(Or , Im−r ) and Or , Im−r stand for r × r zero and (m − r) × (m − r) idenwhere Q tity matrices, respectively To simplify the notation we will again denote by Tk the matrix induced by the operator Tk It is easy to verify that Tk = Zk−1 Zk−1 In practice, to find any projector onto kerAk one can consider a singular value decomposition (SVD) of the matrix Ak , Ak = Uk Σk VkT , (5) where Σk = diag(σk(1) , , σk(r) , 0, , 0) with σk(i) , i = 1, , r being singular values of Ak , and where Uk , Vk are orthogonal matrices Clearly, ¯ kT Qk = Vk QV (6) is an orthogonal projector onto kerAk (see [3, 15] for details) Moreover, Tk = Vk−1 VkT (7) and Tk |ker Ak is an isomorphism from ker Ak onto ker Ak−1 In the following sections we focus on a subadjoint equation of (1); hence, in what follows, for simplicity we will use orthogonal projectors Qk and operators Tk given by (6) and (7), respectively Together with (1), we introduce the following matrices: Gk := Ak + Bk Tk Qk or ¯ kT , Gk := Ak + Bk Vk−1 QV k = 1, , N − Since the SVD (5) of the matrix Ak is not unique, the matrix Gk depends on the chosen SVDs of Ak We need the following lemmas Lemma [2, 3] The following assertions are equivalent: ¯ kT is nonsingular; (i) the matrix Gk := Ak + Bk Vk−1 QV (ii) Sk ∩ ker Ak−1 = {0}; (iii) Rm = Sk ⊕ ker Ak−1 Due to Lemma 1, if (1) is of index 1, then the matrix Gk is nonsingular Moreover, its nonsingularity does not depend on the choice of SVDs of Ak 84 L.C Loi For the index-1 LSDE (1), we have the decomposition Rm = Sk ⊕ ker Ak−1 , k = 1, , N − (r+1) Let {sk(1) , , sk(r) } and {hk−1 , , h(m) k−1 } denote certain bases of Sk and kerAk−1 , respectively In other words, Sk = span sk(i) r i=1 , k = 0, , N − and ker Ak−1 = span h(i) k−1 m i=r+1 , k = 1, , N Lemma [2] Suppose (1) is of index and Ak = Uk Σk VkT is an arbitrary SVD of Ak Then, the following relations hold: ¯ kT G−1 ˜ k−1 := Vk−1 QV (i) Q k Bk is the canonical projector onto ker Ak−1 along Sk for every k = 1, , N − 1; −1 ¯ Z˜ k−1 ˜ k−1 = Z˜ k−1 Q , where Z˜ k−1 is a matrix, whose columns are vectors sk(1) , , sk(r) (ii) Q (r+1) and hk−1 , , h(m) k−1 for every k = 1, , N − Derivations of Subadjoint Equations 3.1 Kronecker Normal Form We begin this subsection by recalling the Kronecker normal form of the index-1 LSDEs First, via a scaling of the equations Ek , k = 0, , N − and a transformation of variables xk = Fk−1 x¯k , k = 0, , N , the index-1 LSDE (1) can be transformed into the following equation: A¯ k x¯k+1 = B¯ k x¯k , k = 0, , N − 1, where A¯ k = Ek Ak Fk and B¯ k = Ek Bk Fk−1 , k = 0, , N − Moreover, both matrices Ek and Fk are nonsingular for all k The above equation is said to be in Kronecker normal form if A¯ k = diag(Ir , Om−r ), B¯ k = diag(Wk , Im−r ), k = 0, , N − The following theorem was proved Theorem [2] Suppose (1) is of index Then there exist nonsingular scaling matrices Ek and transforming variable matrices Fk such that (1) is transformed into the Kronecker normal form diag(Ir , Om−r )x¯k+1 = diag(Wk , Im−r )x¯k , k = 0, , N − (8) Remark From the proof of Theorem (see [2, Theorem 3.3]), the scaling matrices Ek and the transforming variable matrices Fk for the index-1 equation (1) can be constructed as follows Due to Lemmas and 2, the matrix ¯ Z˜ k−1 ˜ k := Ak + Bk Z˜ k−1 Q G Subadjoint Equations of Index-1 Linear Singular Difference Equations 85 is nonsingular, and so is the matrix ¯ ˜ k Z˜ k = Ak Z˜ k + Bk Z˜ k−1 Q ¯ k := G G ¯ −1 ˜ Thus, we can use the scaling Ek = G k and the transformation of variables Fk = Zk As discussed in Remark 1, in order to perform the scaling matrices Ek , k = 0, , N − and the transforming variable matrices Fk , k = −1, , N − we must specify the matrices ¯ k = 0, , N − and Z˜ k for k = −1, , N − Assume that ¯ k := Ak Z˜ k + Bk Z˜ k−1 Q, G (i) r m m } families of vectors {h(i) −1 i=r+1 and {sN }i=1 are linearly independent in R , such that m i=r+1 S0 ∩ span h(i) −1 r i=1 span sN(i) = {0}, ∩ ker AN−1 = {0} With these assumptions, we obtain all necessary scaling equations and transforming variables, reducing the index-1 equation (1) to the Kronecker normal form (8) Remark For (1) and (8), it is easy to check that −1 (Sk ∩ ker Ak−1 ), S¯k ∩ ker A¯ k−1 = Fk−1 k = 1, , N − Therefore, the index-1 property of LSDEs is invariant under the above-mentioned scaling equations and linear transformations We now derive a subadjoint equation of the index-1 equation (1) by using its Kronecker normal form (8) Put x¯k(1) , x¯k(2) x¯k := where x¯k(1) ∈ Rr and x¯k(2) ∈ Rm−r Obviously, (8) is equivalent to the system (1) = Wk x¯k(1) , x¯k+1 = x¯k(2) , k = 0, , N − (9) The first equation of (9) is an ordinary difference equation; hence, according to [1, 14, 17, 18, 20] its adjoint equation is given by T (1) ¯ k+1 , u¯ (1) k = Wk u k = 0, , N − (10) In addition, the second equation of (9) can be considered as a special ordinary difference equation Hence, its adjoint equation has the following form: = u¯ (2) k+1 , k = 0, , N − (11) Equations (10) and (11) can be rewritten in a compact form: diag(Ir , Om−r )T u¯ k = diag(Wk , Im−r )T u¯ k+1 , , u¯ (2)T )T where u¯ k := (u¯ (1)T k k k = 0, , N − 1, (12) 86 L.C Loi In the following, to simplify the notation we will write H −T := (H −1 )T = (H T )−1 , where H ∈ Rm×m is an invertible matrix Next, performing a transformation of variables −T u¯ k := Ek−1 uk , k = 1, , N for a subequation of (12), we obtain −T diag(Ir , Om−r )T Ek−1 uk = diag(Wk , Im−r )T Ek−T uk+1 , k = 1, , N − −T Multiplying the above equation by Fk−1 gives −1 −1 Ek−1 diag(Ir , Om−r )Fk−1 T −1 uk = Ek−1 diag(Wk , Im−r )Fk−1 T uk+1 , or equivalently, ATk−1 uk = BkT uk+1 , k = 1, , N − (13) 3.2 Inherent Regular Ordinary Difference Equation We briefly describe the decomposition technique for index-1 LSDEs (see [2, 3, 13, 15] for −1 details) Multiplying (1) by Pk G−1 k and Qk Gk , respectively, and carrying out a few technical computations, we decouple this equation into the system Pk xk+1 = Pk G−1 k Bk Pk−1 xk , ¯ kT G−1 = Vk−1 QV k Bk Pk−1 xk + Qk−1 xk , k = 1, , N − Decomposing the solution {xk }N k=1 of (1) into two parts, xk = Pk−1 xk + Qk−1 xk =: yk + zk , k = 1, , N − 1, we can rewrite the above system as yk+1 = Pk G−1 k B k yk , k = 1, , N − (14) and ¯ kT G−1 zk = −Vk−1 QV k B k yk , k = 1, , N − Clearly, we only need to initialise the P−1 -component of x0 , i.e., P−1 x0 − x = 0, (15) where x is an arbitrary vector in Rm It yields that the initial value problem (IVP) for the index-1 LSDE (1) and the initial condition (15) has a unique solution xk = P˜k−1 yk , k = 1, , N − 1, (16) N−1 where {yk }k=1 solves the IVP for (14) and y0 = y := P−1 x Here we put P˜k−1 := I − Q˜ k−1 and note that (14) is called the inherent regular ordinary difference equation of the index-1 LSDE (1) Subadjoint Equations of Index-1 Linear Singular Difference Equations 87 Based on a similar idea as in [6], we will show how to obtain the subadjoint equation (13) of the original equation (1) by using the inherent regular ordinary difference equation (14) It is well known that the adjoint equation of (14) is given by wk = Pk G−1 k Bk T wk+1 , k = 1, , N − 1, or T wk = BkT G−T k Pk wk+1 , k = 1, , N − (17) Putting vk := A†k−1 T k = 1, , N, T Pk−1 wk , where A†k−1 denotes the Moore–Penrose inverse of Ak−1 , and noting that † T T A†k−1 Ak−1 = Vk−1 Σk−1 Uk−1 Uk−1 Σk−1 Vk−1 T = Vk−1 diag(Ir , Om−r )Vk−1 T ¯ k−1 = I − Vk−1 QV = Pk−1 , we find that ATk−1 vk = A†k−1 Ak−1 T T T Pk−1 wk = Pk−1 wk , k = 1, , N T Multiplying (17) by Pk−1 and using the above equalities, we get ATk−1 vk = Ak G−1 k Bk Pk−1 T vk+1 , k = 1, , N − (18) Observing that ¯ kT Pk = Ak (I − Qk ) + Bk Vk−1 VkT Qk Pk = Ak , Gk Pk = Ak + Bk Vk−1 QV one has G−1 k Ak = Pk , k = 1, , N − This leads to the relations −1 T ¯ T Ak G−1 k Bk Qk−1 = Ak Gk Bk Vk−1 QVk Qk Vk Vk−1 T = Ak G−1 k (Gk − Ak )Qk Vk Vk−1 = and −1 T ATk−1 G−T k−1 Ak−1 = Ak−1 Gk−1 Ak−1 T = (Ak−1 Pk−1 )T = ATk−1 It follows that (18) can be rewritten as −T T T T ATk−1 G−T k−1 Ak−1 vk = Bk Gk Ak vk+1 , k = 1, , N − Hence, the last equation is the same as (13) Thus, we have shown another way of deriving the subadjoint equation of the index-1 LSDEs Now we come to the following definition Definition Equation (13) is called the subadjoint equation of the index-1 LSDE (1) 88 L.C Loi Remark Note that the subadjoint equation (13) coincides with (4), which has been stated in [16, 19] On the other hand, according to those papers, the size of the subadjoint equation (13) is smaller than that of the original equation (1) For this reason, the term ‘subadjoint’ rather than ‘adjoint’ should be employed for (13) We will discuss some properties of the subadjoint equation (13) of the index-1 LSDE (1) in the next section Properties of Subadjoint Equations We know that the index concept plays an important role in the theory of LSDEs First, we will introduce an index-1 concept for the subadjoint equation (13), which is considered as being solved backwards Similar to the original equation (1), with its subadjoint (13) we the following subspace: Skb = z ∈ Rm : BkT z ∈ im ATk−1 , k = 1, , N − Definition Equation (13) is called index-1 tractable (index-1 for short) if the following conditions hold: (i) rank ATk−1 = r, k = 1, , N − 1, (ii) Skb ∩ ker ATk = {0}, k = 1, , N − We recall that A¯ k := diag(Ir , Om−r ) = Ek Ak Fk , B¯ k := diag(Wk , Im−r ) = Ek Bk Fk−1 , ¯ −1 ˜ where Ek = G k and Fk = Zk ; hence, it yields ATk = Fk−T A¯ Tk Ek−T (k = 0, , N − 2), −T ¯ T −T Bk Ek (k = 1, , N − 1) BkT = Fk−1 Further, note that ker A¯ Tk = z1T , z2T T ∈ Rm : z1 ∈ Rr and z1 = and S¯kb := z ∈ Rm : B¯ kT z ∈ im A¯ Tk−1 = z1T , z2T T ∈ Rm : z2 ∈ Rm−r and z2 = It is also easy to see that ker ATk = EkT ker A¯ Tk (k = 0, , N − 2), Skb = EkT S¯kb (k = 1, , N − 1) This means that Skb ∩ ker ATk = EkT ker A¯ Tk ∩ S¯kb = E T {0} = {0} for all k = 1, , N − On the other hand, clearly rank ATk−1 = rank Ak−1 = r, k = 1, , N − Thus, we have completed the proof of the following useful result Subadjoint Equations of Index-1 Linear Singular Difference Equations 89 Theorem Let the original equation (1) be of index Then, its subadjoint equation (13) is also of index Example Let us consider equation (1) with the data ⎛ ⎞ ⎛ 1 k + 1⎠, Bk = ⎝ k Ak = ⎝ k + k+1 k+2 1 ⎞ k + 2⎠ k+1 (19) Observing that rank Ak = ∀k = 0, , N − 1, Sk = span (1, 0, 1)T , 0, 1, (k + 1)2 + k T , k = 0, , N − and ker Ak−1 = span (1, 0, −1)T , k = 1, , N Hence, for any k = 1, , N − we have Sk ∩ ker Ak−1 = {0} On the other hand, by direct computations, we find that rank ATk−1 = for all k = 1, , N − 1, ker ATk = span (−k − 1, 1, 0)T , k = 0, , N − 2, and Skb = span (1, 0, 0)T , (0, −k + 1, 2)T , k = 1, , N − It follows that Skb ∩ ker ATk = {0}, k = 1, , N − Thus, we come to the conclusion that (1) with the data (19) and its subadjoint equation are of the same index Example This example taken from [13] shows that if the original equation is not of index 1, then its subadjoint is not an index-1 equation Let ⎞ ⎞ ⎛ ⎛ 0 −1 k ⎠ , k = 0, , N − (20) Bk = ⎝ −k Ak = ⎝ k ⎠ , −k − 1 0 0 Since ker Ak−1 = span{(0, 0, 1)T } for all k = 1, , N and Sk = span (0, 0, 1)T , (1, k + 1, 0)T , k = 0, , N − 1, it yields that Sk ∩ ker Ak−1 = span (0, 0, 1)T , k = 1, , N − Thus, (1) with the data (20) is not an index-1 equation We now consider its subadjoint equation Noting that ker ATk = span (0, 0, 1)T , k = 0, , N − 90 L.C Loi and Skb = span (0, 0, 1)T , (−k, 1, 0)T , k = 1, , N − 1, we have Skb ∩ ker ATk = span (0, 0, 1)T , k = 1, , N − Therefore, neither (1) with the data (20) nor its subadjoint equation are of index Since rank ATk = r for all k = 0, , N − 1, i.e., dim(ker ATk−1 ) = dim(ker ATk ) for all k = 1, , N − 1, we can define Tb,k−1 ∈ GL(Rm ) such that Tb,k−1 |ker AT is an isomorphism k−1 b := I − Qbk−1 from kerATk−1 onto ker ATk Let Qbk−1 be any projector onto ker ATk−1 and Pk−1 To characterise equation (13) we need the notation Gb,k−1 := ATk−1 + BkT Tb,k−1 Qbk−1 , k = 1, , N − Remark In the index-1 linear DAE case, the relation between Qk and Qbk has been shown in [6, Remark 1] Since both Qk and Qbk are arbitrary, the equality QTk = Qbk may not hold For example, consider the orthogonal projector Pk := A†k Ak and Pkb := (ATk )† ATk or, equiv¯ Therefore, the necessary and alently, Pk = Vk P¯ VkT and Pkb = Uk P¯ UkT with P¯ := I − Q sufficient condition for QTk = Qbk is that Vk = Uk holds Obviously, this is not valid for an arbitrary matrix Ak ; however, it is true for symmetric matrices Ak T ¯ k−1 is an orthogonal projector onto ker ATk−1 According to the SVD (5), Qbk−1 := Uk−1 QU T T and Tb,k−1 |ker AT is an isomorphism from ker Ak−1 onto ker ATk , where Tb,k−1 := Uk Uk−1 k−1 Similarly as in Sect 2, in the following we will use the orthogonal projectors Qbk−1 and the operators Tb,k−1 Therefore, T ¯ k−1 Gb,k−1 = ATk−1 + BkT Uk QU , k = 1, , N − Now we come to the following lemmas from the well-known results with A = ATk−1 , A¯ = ATk and B = BkT (see [13, Lemmas A.1 and A.2]) and Theorem Lemma The following conditions are equivalent: T ¯ k−1 is nonsingular; (i) the matrix Gb,k−1 := ATk−1 + BkT Uk QU b T m (ii) R = Sk ⊕ ker Ak ; (iii) Skb ∩ ker ATk = {0} Lemma Suppose (1) is of index Then, the following relations hold: b T = G−1 Pk−1 b,k−1 Ak−1 , (21) −1 T T b ¯ T G−1 b,k−1 Bk = Gb,k−1 Bk Pk + Uk−1 QUk (22) b T T T ˜ bk := Uk QU ¯ k−1 Further, Q G−1 b,k−1 Bk is the canonical projector onto ker Ak along Sk −1 ˜ −1 T ¯ Z˜ b,k ˜ bk The canonical Z˜ b,k+1 Q¯ Z˜ b,k , Gb,k := ATk + Bk+1 and P˜kb := I − Q Let Q˜ bk = Z˜ b,k Q projectors mentioned in Lemmas and corresponding to (1) and (13) are connected by the following proposition This result can be considered as a discrete analogue of that in [4, Lemma 1] Subadjoint Equations of Index-1 Linear Singular Difference Equations 91 ˜ bk be the canonical projectors associProposition Let (1) be of index and let Q˜ k−1 , Q ated with the original equation (1) and its subadjoint equation (13), respectively Then, the following relations hold: −1 ¯ kT G−1 ¯ T Vk−1 QV k = Uk QUk−1 Gb,k−1 P˜k−1 = ATk−1 G−1 b,k−1 T , T (23) , P˜kb = Ak G−1 k T (24) , ˜ k ˜ Tb,k = G G (25) Proof We have ¯ kT = Ak−1 + Uk−1 QU ¯ kT Bk Vk−1 QV ¯ kT GTb,k−1 Vk−1 QV ¯ kT Bk Vk−1 QV ¯ kT = Uk−1 QU ¯ kT Ak + Bk Vk−1 QV ¯ kT = Uk−1 QU ¯ kT Gk = Uk−1 QU It implies that −T ¯ kT G−1 ¯ T Vk−1 QV k = Gb,k−1 Uk−1 QUk , which yields relation (23) Using formula (23) gives −1 ¯ kT G−1 ¯ T ˜ k−1 = Vk−1 QV Q k Bk = Uk QUk−1 Gb,k−1 T T ¯ T Bk = G−T b,k−1 Bk Uk QUk−1 T T T ¯ k−1 ¯ k−1 Qbk−1 = BkT Uk QU , we have Further, observing that ATk−1 Qbk−1 = and BkT Uk QU b Q˜ k−1 = G−T b,k−1 Qk−1 T GTb,k−1 Therefore, b P˜k−1 = G−T b,k−1 I − Qk−1 T GTb,k−1 , or equivalently, b P˜k−1 = G−T b,k−1 Gb,k−1 Pk−1 T Applying equality (21), we come to the conclusion that the first formula of (24) is valid The second relation of (24) can be checked in a similar way Applying Theorem 1, we get Ak = Ek−1 Ir 0 Fk−1 and Bk = Ek−1 Wk 0 Im−r −1 , Fk−1 ˜ −1 where Fk = Z˜ k and Ek = Z˜ k−1 G k It implies that ˜ k = Ek−1 Fk−1 G Furthermore, it is clear that Z˜ b,k = EkT , and hence, (26) 92 L.C Loi ˜ b,k = Fk−T G Ir T Wk+1 0 Ek−T + Fk−T ¯ k−T = Ek−1 Fk−1 = Fk−T (P¯ + Q)E T Im−r −T T ¯ k−T Ek+1 QE Ek+1 Taking into account the relation (26), we arrive at (25) Thus, Proposition is proved Similarly as at the beginning of Sect 3.2, we decouple the subadjoint equation (13) b into an ordinary difference equation and algebraic relations Multiplying Pk−1 G−1 b,k−1 and −1 b Qk−1 Gb,k−1 on both sides of (13) and applying the formulae (21) and (22), we decouple equation (13) into the system b b T b uk = Pk−1 G−1 Pk−1 b,k−1 Bk Pk uk+1 , b T T b ¯ k−1 = Uk QU G−1 b,k−1 Bk Pk uk+1 + Qk uk+1 k = 1, , N − With the notation αk+1 := Pkb uk+1 and βk+1 := Qbk uk+1 the above system becomes b T αk = Pk−1 G−1 b,k−1 Bk αk+1 , k = 1, , N − (27) and T T ¯ k−1 βk+1 = −Uk QU G−1 b,k−1 Bk αk+1 , k = 1, , N − Therefore, associated with the subadjoint equation (13), a final condition on uN should be given Moreover, this condition must be stated as follows: b PN−1 uN − uN = 0, (28) where uN is an arbitrary vector in Rm Once again, (27) is called the inherent regular ordinary difference equation of the subadjoint equation (13) Recalling that the subadjoint equation is solved backwards, we now arrive at the following theorem Theorem Suppose that the original equation (1) is of index Then the final value problem (FVP) for its subadjoint equation (13) and (28) has a unique solution: b αk , uk = P˜k−1 k = 1, , N, (29) where {αk }N k=1 is a solution of the FVP for the inherent regular ordinary difference equation (27) and the final condition b uN αN = α N := PN−1 At the end of this section, we will discuss other properties of the subadjoint equation (13) that have been considered in the regular ordinary difference equations [1, 14, 18] Now let {xk } and {uk } be solutions of (1) and (13), respectively Then, from these equations, it is easy to see that uTk+1 Ak xk+1 = uTk Ak−1 xk ∀k = 1, , N − In other words, there exists a constant c such that uTk Ak−1 xk = c ∀k = 1, , N (30) Subadjoint Equations of Index-1 Linear Singular Difference Equations 93 Note that when Ak = I , formula (30) is the well-known discrete Lagrange identity for the regular ordinary difference equations Let k0 be a positive integer satisfying ≤ k0 < N From the formulae (14) and (16), we see that every solution {xk } of (1) satisfies the following relation: xk = Φ(k, k0 )xk0 for all k : k0 < k ≤ N, where Φ(k, k0 ) := P˜k−1 k−k0 Pk−i G−1 k−i Bk−i Pk0 −1 i=1 Now we prove that Φ(k, k0 ) does not depend on the choice of SVDs of the matrices Ai (i = k0 , , k − 1) Indeed, we have T ˜ ˜ l = Al + Bl Z˜ l−1 Q¯ Z˜ l−1 = Gl Pl + Bl Vl−1 VlT Vl Vl−1 G Zl−1 Q¯ Z˜ l−1 T ˜ Obviously, since Vl Vl−1 Zl−1 Q¯ Z˜ l−1 x ∈ ker Al for all x ∈ Rm , T ˜ T ˜ ¯ Z˜ l−1 Ql Vl Vl−1 Zl−1 Q¯ Z˜ l−1 = Vl Vl−1 Zl−1 Q It follows that T ˜ ˜ l = Gl Pl + Bl Vl−1 QV ¯ lT Ql Vl Vl−1 ¯ Z˜ l−1 G Zl−1 Q T ˜ Zl−1 Q¯ Z˜ l−1 = Gl Pl + Gl Ql Vl Vl−1 T ˜ Zl−1 Q¯ Z˜ l−1 = Gl Pl + Vl Vl−1 Further, observing that −1 T ˜ ¯ lT Vl−1 QV Zl−1 Q¯ Z˜ l−1 P˜l + Z˜ l Z˜ l−1 Pl + Vl Vl−1 −1 T ˜ ¯ lT + Vl Vl−1 = Pl P˜l + Pl Q˜ l Z˜ l Z˜ l−1 Zl−1 Z˜ l−1 Q˜ l P˜l Vl−1 QV T ˜ Ql−1 Ql−1 Vl−1 VlT + Vl Vl−1 T = Pl + Vl Vl−1 Ql−1 Vl−1 VlT = I, we have T ˜ Zl−1 Q¯ Z˜ l−1 Pl + Vl Vl−1 −1 −1 ¯ lT = P˜l + Z˜ l Z˜ l−1 Vl−1 QV Therefore, we get T ˜ ¯ ˜ −1 G ˜ −1 G−1 l = Pl + Vl Vl−1 Zl−1 QZl l Applying the above equality and Lemma A.2 in [13] yields −1 T ˜ ¯ ˜ −1 G ˜ −1 ˜ −1 Pl G−1 l Bl Pl−1 = Pl Gl Bl = Pl Pl + Ql Vl Vl−1 Zl−1 QZl l B l = Pl Gl B l This leads to the relation Φ(k, k0 ) = P˜k−1 Pk−1 k−k0 i=1 ˜ −1 G k−i Bk−i , 94 L.C Loi or equivalently, Φ(k, k0 ) = P˜k−1 k−k0 ˜ −1 G k−i Bk−i for all k : k0 < k ≤ N, (31) i=1 i.e., Φ(k, k0 ) does not depend on the choice of SVDs of Ai Similar arguments can be used for the subadjoint equation (13) Letting k1 be a positive integer such that ≤ k0 < k1 ≤ N and using the formulae (27) and (29), we obtain uk = Ψ (k1 , k)uk1 , ≤ k < k1 , where {uk } is a solution of (13) and b Ψ (k1 , k) := P˜k−1 k1 −1−k b T b Pk−1+i G−1 b,k−1+i Bk+i Pk1 −1 i=0 We now prove that Ψ (k1 , k) does not depend on the choice of SVDs of the matrices Ai (k − ≤ i ≤ k1 − 2) Using equality (21) and similar techniques for the original equation, we find b T ˜ ¯ ˜ −1 ˜ −1 G−1 b,l−1 = Pl−1 + Ul−1 Ul Zb,l QZb,l−1 Gb,l−1 Combining the above equality with relation (22), we get −1 b T b b T Pl−1 G−1 b,l−1 Bl Pl = Pl−1 Gb,l−1 Bl −1 b b T ˜ −1 = Pl−1 + Qbl−1 Ul−1 UlT Z˜ b,l Q¯ Z˜ b,l−1 Pl−1 G b,l−1 Bl b ˜ −1 Gb,l−1 BlT = Pl−1 Thus, b Ψ (k1 , k) = P˜k−1 k1 −1−k T ˜ −1 G b,k−1+i Bk+i , (32) i=0 which implies that Ψ (k1 , k) does not depend on the choice of SVDs of Ai Using equalities (24) and (25) and the fact that T ˜ ¯ ˜ −1 G ˜ −1 Ak G−1 k = Ak Pk + Vk Vk−1 Zk−1 QZk k T ˜ ˜ −1 Zk−1 Q¯ Z˜ k−1 G = Ak Pk + Qk Vk Vk−1 k ˜ −1 = Ak G k , we can rewrite expression (32) as follows: T ˜ −T Ψ (k1 , k) = G k−1 Ak−1 k1 −1−k T ˜ −T G k−1+i Bk+i for all k : ≤ k < k1 (33) i=0 Remark The solution of the subadjoint equation (13) can be directly obtained from the data by using formula (33) Subadjoint Equations of Index-1 Linear Singular Difference Equations 95 From equations (31) and (33), we see that the property Ψ (k1 , k0 ) = Φ T (k1 , k0 ), ≤ k0 < k1 ≤ N, being valid for the regular ordinary difference equations (see [18]) is not true for index-1 LSDEs On the other hand, using (31) and (33) we can conclude that the following pair of IVP and FVP: Ak xk+1 = Bk xk , Pk0 −1 xk0 − x k0 = 0, k = k0 , , N ATk−1 uk = BkT uk+1 , Pkb1 −1 uk1 − uk1 = 0, k = 1, , k1 − 1, and has solutions xk = Φ(k, k0 )x k0 , k > k0 , uk = Ψ (k1 , k)u , k < k1 , k1 respectively In particular, the IVP (1) and (15), and the FVP (13) and (28) have the corresponding solutions xk = Φ(k, 0)x ∀k: < k ≤ N, uk = Ψ (N, k)uN ∀k: ≤ k < N and It implies that uTk Ak−1 xk = uN T L(k)x0 for all k = 1, , N − 1, where L(k) := Ψ T (N, k)Ak−1 Φ(k, 0) ˜ −1 ˜ Using the fact that G k−1 Ak−1 = Pk−1 and some elementary computations, we get −1 L(k) = EN−1 W 0 −1 , F−1 k = 1, , N − 1, where W := N i=1 WN−i This means that L(k) is not dependent on the index k; i.e., we obtain the discrete Lagrange identity (30) again Remark For index-1 linear DAEs, a necessary and sufficient condition for an adjoint equation of an inherent regular ordinary differential equation of (2) to coincide with an inherent regular ordinary differential equation of the adjoint equation (3), has been established in [4] Unfortunately, it is difficult to establish a similar condition for index-1 LSDEs 96 L.C Loi Conclusion and Future Work In this paper we have thoroughly analysed the derivations of the 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(1996) 19 Sreedhar, J., Dooren, P.V.: Periodic descriptor systems: solvability and conditionability IEEE Trans Autom Control 44, 310–313 (1999) 20 Willigenburg, L.G.V., Koning, W.L.D.: Linear systems theory revisited Automatica 44, 1686–1692 (2008) ... solution of the subadjoint equation (13) can be directly obtained from the data by using formula (33) Subadjoint Equations of Index-1 Linear Singular Difference Equations 95 From equations (31)... useful result Subadjoint Equations of Index-1 Linear Singular Difference Equations 89 Theorem Let the original equation (1) be of index Then, its subadjoint equation (13) is also of index Example... Q˜ k−1 and note that (14) is called the inherent regular ordinary difference equation of the index-1 LSDE (1) Subadjoint Equations of Index-1 Linear Singular Difference Equations 87 Based on